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13500 | https://byjus.com/sound-intensity-formula/ | Sound intensity is described as sound per unit area perpendicular to the path of the sound waves and which is represented by I. The sound intensity SI unit is W/m2 (watt per square meter).
The SI unit of sound intensity is watt per square meter (W/m2). The standard definition is the calculation of noise sound intensity in the air at the listener position as a quantity of sound energy.
The Formula for sound intensity is expressed as
(\begin{array}{l}I=\frac{P}{A}\end{array} )
P = sound power
A = area
In order to measure the sound intensity level, need to compare the given sound intensity value with standard intensity value.
The formula of Sound Intensity Level is expressed as
(\begin{array}{l}I_{L} = 10 log_{10}\frac{I}{I_{0}}\end{array} )
Where
I = sound intensity and
Io = reference intensity
The unit of sound intensity is expressed in decibels(dB)
Solved examples on Sound Intensity
Example 1
A person whistles with the power of 0.9 × 10-4 W. Calculate the sound intensity at a distance of 7m.
Solution:
Given
P = 0.9 × 10-4 W
A = 7m
Sound intensity formula is
I = P / A
I = 0.9×10−4 / 7
I = 1.28 x 10-5 W/m2
Example 2
Determine the intensity level which is equivalent to an intensity 1 nW/m2.
Solution:
Intensity Level formula is given by
IL = 10 log10 I / I0
Here
I = 1 nWm-2 = 1 × 10-9 Wm–2
Io = 10-12 Wm-2
IL= 10 log10 (1×10−9) / 10−12
IL = 10 log10 103
IL = 30.
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13501 | https://math.stackexchange.com/questions/248245/exactly-half-of-the-elements-of-mathcalpa-are-odd-sized | combinatorics - Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized - Mathematics Stack Exchange
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Exactly half of the elements of P(A)P(A) are odd-sized
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Let A A be a non-empty set and n n be the number of elements in A A, i.e. n:=|A|n:=|A|.
I know that the number of elements of the power set of A A is 2 n 2 n, i.e. |P(A)|=2 n|P(A)|=2 n.
I came across the fact that exactly half of the elements of P(A)P(A) contain an odd number of elements, and half of them an even number of elements.
Can someone prove this? Or hint at a proof?
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elementary-set-theory
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edited Apr 8, 2023 at 0:58
Mike Earnest
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13 A nice proof goes by using the binomial theorem: (x+y)n=∑k(n k)x k y n−k(x+y)n=∑k(n k)x k y n−k. Here, (n k)(n k) is the number of k k-sized subsets of a set of size n n. Your result corresponds to the fact that 0=(1−1)n 0=(1−1)n. Expanding, the binomial theorem gives us that the sum of the (n k)(n k) with k k odd equals the sum of the (n k)(n k) with k k even. The first is precisely the number of odd-sized subsets, and the second is the number of even-sized subsets.Andrés E. Caicedo –Andrés E. Caicedo 2012-11-30 20:14:17 +00:00 Commented Nov 30, 2012 at 20:14
1 I'm amazed at the fact that you guys answer so quickly! Thanks and wow!gbag –gbag 2012-11-30 20:18:58 +00:00 Commented Nov 30, 2012 at 20:18
I'm amazed that so many people bothered to write answers, and that there are some good answers, but there is only one upvote to the question.Asaf Karagila –Asaf Karagila♦ 2012-11-30 20:29:34 +00:00 Commented Nov 30, 2012 at 20:29
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Fix an element a∈A a∈A (this is the point where A≠∅A≠∅ is needed). Then
S↦S Δ{a}S↦S Δ{a}
(symmetric difference) is a bijection from the set of odd subsets to the set of even subsets.
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edited Nov 30, 2012 at 20:21
answered Nov 30, 2012 at 20:20
Hagen von EitzenHagen von Eitzen
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3 Very nice! Asaf Karagila –Asaf Karagila♦ 2012-11-30 20:29:53 +00:00 Commented Nov 30, 2012 at 20:29
2 Indeed, really cute!Peter Smith –Peter Smith 2012-11-30 20:51:26 +00:00 Commented Nov 30, 2012 at 20:51
Addition: suppose f(S 1)=f(S 2)f(S 1)=f(S 2) then we know that either a a was in both sets, or a a was in neither. After removing or adding, the sets are equal, hence they were the same to start with S 1=S 2 S 1=S 2 and f f is injective. Suppose we have some even set B B which does not contain a a. We now consider f(B∪{a})=B f(B∪{a})=B, which must be a possible subset (which is odd). Similarly whenever we have some even subset B which does contain a a, we have that f(B−{a})=B f(B−{a})=B. The desired function therefore outputs any even subset. This function is injective + surjective thus it is bijective. □◻user459879 –user459879 2018-11-23 12:38:47 +00:00 Commented Nov 23, 2018 at 12:38
For those who did not understood this , look here : math.stackexchange.com/a/483537/131949Number945 –Number945 2019-04-21 23:40:21 +00:00 Commented Apr 21, 2019 at 23:40
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Hint: One can prove this by induction on the size of A A. Assume it was true for sets of size n n and let A={a 1,…,a n+1}A={a 1,…,a n+1}. Then every subset of A A is either a subset of {a 1,…,a n}{a 1,…,a n} or it is a copy of such subset with the addition of {a n+1}{a n+1}. Use the induction hypothesis to conclude that the sets which do not contain a n+1 a n+1 have this property (with respect to {a 1,…,a n}{a 1,…,a n}, by adding a n+1 a n+1 you send exactly the same number of odd sets to even size sets, and vice versa; therefore the ratio remains true for A A.
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answered Nov 30, 2012 at 20:10
Asaf Karagila♦Asaf Karagila
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For n∈Z+n∈Z+ let [n]={1,2,…,n}[n]={1,2,…,n}. Clearly has one even subset and one odd subset. Suppose that [n][n] has equal numbers of odd and even subsets for some n∈Z+n∈Z+. The even subsets of [n+1][n+1] are of two types:
the even subsets of [n][n]; and
the sets of the form A∪{n+1}A∪{n+1}, where A A is an odd subset of [n][n].
By the induction hypotheses there are the same number of sets of the second type as there are of the first, so [n+1][n+1] has twice as many even subsets as [n][n]. But [n+1][n+1] also has twice as many subsets altogether as [n][n], so it must have twice as many odd subsets as well, which clearly implies that it has equal numbers of odd and even subsets.
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answered Nov 30, 2012 at 20:15
Brian M. ScottBrian M. Scott
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3 +1 but you do not even need the induction hypothesis, as you have shown that the number of even subsets of [n+1][n+1] is equal to the number of subsets of [n][n] and a very similar argument would show the number of odd subsets of [n+1][n+1] is the same.Henry –Henry 2012-12-01 02:53:46 +00:00 Commented Dec 1, 2012 at 2:53
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Number the members of the set 1,2,3,4,…,n 1,2,3,4,…,n.
For every subset with an even number of elements, there is a corresponding set with an odd number of elements, that corresponds in this way:
If 1 1is a member of the set with an even number of elements, then delete 1 1 from the set to get a set with an odd number of elements.
If 1 1 is not a member of the set with an even number of elements, then add 1 1 to the set to get a set with an odd number of elements.
For example, suppose the set is {1,2,3,4}{1,2,3,4}. Then we have this correspondence between sets with an even number of elements and sets with an odd number of elements:
even∅{1,2}{1,3}{1,4}{2,3}{2,4}{3,4}{1,2,3,4}↔↔↔↔↔↔↔↔odd{1}{2}{3}{4}{1,2,3}{1,2,4}{1,3,4}{2,3,4}even odd∅↔{1}{1,2}↔{2}{1,3}↔{3}{1,4}↔{4}{2,3}↔{1,2,3}{2,4}↔{1,2,4}{3,4}↔{1,3,4}{1,2,3,4}↔{2,3,4}
This won't work with the empty set because we don't have an element to which we can assign the number 1 1.
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answered Sep 4, 2013 at 1:40
Michael HardyMichael Hardy
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When n n is odd, look at each set and its complement: one will have even number of elements and the other, odd (because odd number can only be written as a sum of an odd and an even number).
When n n is even, remove an element to obtain a set with odd number of elements. By the first part, half of its subsets have even cardinality and half odd. Now to form the full P(A)P(A), we need to join the remaining element to each of the previous subsets: those with odd cardinality with become even, and viceversa.
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answered Nov 30, 2012 at 20:17
Martin ArgeramiMartin Argerami
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So easy and natural when n n is odd. So futile trying to define a 'semi-complement' function. sigh...CopyPasteIt –CopyPasteIt 2017-12-10 23:47:34 +00:00 Commented Dec 10, 2017 at 23:47
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Suppose n=|A|n=|A|. Then there are
∑k=0⌊n 2⌋(n 2 k)=2 n−1∑k=0⌊n 2⌋(n 2 k)=2 n−1
sets with even cardinality. Thus, there are exactly half of the sets with an even number of elements.
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answered Nov 30, 2012 at 20:14
ClaytonClayton
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For me, it's not obvious why this identity holds. Could you maybe elaborate?gbag –gbag 2012-11-30 20:23:30 +00:00 Commented Nov 30, 2012 at 20:23
1 It's easiest to prove by induction. If n=1, then there is exactly 1 set with an even number of elements: the set with no elements (the empty set). Suppose it's true for some positive integer k. Then k+1 is either odd or even. If it's odd, then the floor of (k+1)/2=k/2, in which case we're done. If it is even, then break up the sum and binomial coefficient, combine, and you get the desired result.Clayton –Clayton 2012-11-30 20:44:15 +00:00 Commented Nov 30, 2012 at 20:44
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You can use proprieties of the binomial coefficients.
Denote P k(A)P k(A) the family of subsets of A A containing k k elements and observe |P k(A)|=(n k)|P k(A)|=(n k) .
Now by a propriety of binomial coefficients ∑n/2 k=0(n 2 k)=∑n/2 k=0((n−1 2 k−1)+(n−1 2 k))=∑n−1 k=0(n−1 k)∑k=0 n/2(n 2 k)=∑k=0 n/2((n−1 2 k−1)+(n−1 2 k))=∑k=0 n−1(n−1 k) and similarly ∑n/2−1 k=0(n 2 k+1)=∑n/2−1 k=0((n−1 2 k)+(n−1 2 k+1))=∑n−1 k=0(n−1 k)∑k=0 n/2−1(n 2 k+1)=∑k=0 n/2−1((n−1 2 k)+(n−1 2 k+1))=∑k=0 n−1(n−1 k) .
This shows that ∑n/2 k=0|P 2 k(A)|=∑n/2−1 k=0|P 2 k+1(A)|∑k=0 n/2|P 2 k(A)|=∑k=0 n/2−1|P 2 k+1(A)| .
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answered Nov 30, 2012 at 20:25
qwertyuioqwertyuio
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Denote X X the number of sets with even cardinality, Y Y the number of sets with odd cardinality, n n the cardinality of the power set.
X=∑i=0,2,4,...(n i),Y=∑i=1,3,5,7,...(n i)⇒X−Y=∑i=1 n(n i)(−1)i(+1)n−i=(−1+1)n=0⇒X=Y X=∑i=0,2,4,...(n i),Y=∑i=1,3,5,7,...(n i)⇒X−Y=∑i=1 n(n i)(−1)i(+1)n−i=(−1+1)n=0⇒X=Y
Because X+Y=n X+Y=n, it is obvious that X=n/2 X=n/2, that is the half of the power set.
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answered Dec 3, 2019 at 6:53
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13Number of subsets with even number of elements
5Bijections from A A to B B, where A A is the set of subsets of [n][n] that have even size and B B is the set of subsets of [n][n] that have odd size
2The number of odd size subsets is equal to the number of even size subsets
4number of subsets of even and odd
2Provide a combinatorial argument for the following: 2 n=∑k odd(n+1 k)2 n=∑k odd(n+1 k)
3Proof ∑k=0 n(2 n 2 k)=2 2 n−1∑k=0 n(2 n 2 k)=2 2 n−1
0∑k≥0(n 2 k)=∑k≥0(n 2 k+1)∑k≥0(n 2 k)=∑k≥0(n 2 k+1)
0How many ways are there to pick r objects from n objects when each object appears an odd number of times?
0Bits and counting problem
-3Number of subsets with even-sized intersections equals the number of subsets with odd-sized intersections
See more linked questions
Related
9How many non empty subsets of {1, 2, ..., n} satisfy that the sum of their elements is even?
37Prove that the power set of an n n-element set contains 2 n 2 n elements
2Number of subsets of A∪B that contain an odd number of elements
0Probability of picking exactly half of the elements
0Consider a set A A with n>0 n>0 elements , O={B⊂A:|B|i s o d d}O={B⊂A:|B|i s o d d}
0Number of even and odd sized subsets of a given set X X are equal.
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13503 | https://math.stackexchange.com/questions/2379383/rationalizing-the-denominator-having-square-roots-and-cube-roots | Skip to main content
Rationalizing the denominator having square roots and cube roots
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In middle-school mathematics, the teachers always tell you that if you have radicals on the denominator of a fraction, then it isn't fit to be a final answer - you have to rationalize the denominator, or get rid of all of the radicals in the denominator by moving them to numerator.
Rationalizing the denominator is usually very easy, and can be done quickly using its conjugate. For example, consider
12+2–√
This denominator can be easily rationalized by using its conjugate:
2−2–√(2+2–√)(2−2–√)
2−2–√4−2
2−2–√2
However, I have stumbled upon a new class of denominator-rationalization problems that I can't figure out how to solve. I was thoroughly stumped when I tried to rationalize the denominator of this fraction:
12+2–√+2–√3
Can anybody figure out how to rationalize this? Is it even possible?
Or, more interestingly, if anyone suspects that it is not possible, how might one prove something like this?
rationalising-denominator
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edited Aug 2, 2017 at 0:14
David G. Stork
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asked Aug 1, 2017 at 22:51
Franklin Pezzuti DyerFranklin Pezzuti Dyer
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Rationalize away the square root first, then think at how to rationalize denominators like a+bn−−√3+cn2−−√3.
– dxiv
Commented
Aug 1, 2017 at 22:56
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Hint: if α is a root of c−xf(x) then αf(α)=c So you need only find some polynomial having the denominator α as root. Wlog we can choose c≠0 over a field (or domain).
– Bill Dubuque
Commented
Aug 1, 2017 at 23:02
@BillDubuque Have you already found that there is a solution, or are you just proposing that as how you would begin to attack the problem?
– Franklin Pezzuti Dyer
Commented
Aug 1, 2017 at 23:11
@Nilknarf The method I described works for any denominator that is algebraic, i.e. a root of some polynomial with rational coefficients. Rationals, and roots of rationals are algebraic, and algebraics are closed under sums and products. So they include numbers of the type you exhibited.
– Bill Dubuque
Commented
Aug 1, 2017 at 23:15
@Nilknarf In case it wasn't clear, here is how to apply what I wrote to rationalize the denominator:
0≠αf(α)=c∈Q ⇒ βα=βf(α)αf(α)=βf(α)c
– Bill Dubuque
Commented
Aug 1, 2017 at 23:32
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It's always possible.
You want to multiply top and bottom by M to get that denominator∗M has not radical.
As you have figured out: If the denominator is a+bc√ you mulitply by the conjugate to get (a+bc√)(a−bc√)=a2−b2∗c.
This will also work with (a−−√+b√)(a−−√−b√)=a−b.
So it's the same idea for a+b√k. The trick is to realize that (a+b√k)(ak−1−ak−2b√k+ak−3(b√k)2−.....±a(b√k)k−2∓(b√k)k−1=ak±b.
Example: To deradicalize 5+7–√3 multiply by 52−5∗7–√3+(7–√3)2 to get (5+7–√3)(52−5∗7–√3+(7–√3)2)=53+527–√3−527–√3−5∗(7–√3)2+5∗(7–√3)2+(7–√3)3=125+7.
So to deradicalize (2+2–√+2–√3) just deradicalize it term by term.
First let's get rid of the 2–√3 term. So we multiply top and bottom by (2+2–√)2−(2+2–√)∗2–√3+(2–√3)2 to get (2+2–√+2–√3)∗[(2+2–√)2−(2+2–√)∗2–√3+(2–√3)2]=(2+2–√)3+2=8+122–√+122–√+22–√+2=10+262–√. Then we multiply that by 10−262–√ to get (10+262–√)(10−262–√)=100−2∗262.
So example:
12+2–√+2–√3=(2+2–√)2−(2+2–√)2–√3+2–√32(2+2–√)3+2=(4+42–√+2)−22–√3−2–√2–√3+2–√3210+262–√=(4+42–√+2)−22–√3−2–√2–√3+2–√32100−2∗262
Okay... admittedly that is a bear... but it is doable.
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edited Aug 2, 2017 at 7:26
Siong Thye Goh
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answered Aug 1, 2017 at 23:26
fleabloodfleablood
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Beautiful answer!
– Franklin Pezzuti Dyer
Commented
Aug 1, 2017 at 23:27
It's always possible For certain values of "always" ;-) This won't work out of the box for 1+22–√3+34–√3 for example.
– dxiv
Commented
Aug 1, 2017 at 23:28
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Beautiful? Ugly as sin, if you ask me.... but doable.
– fleablood
Commented
Aug 1, 2017 at 23:29
@dxiv 189(4–√3+162–√3−11) the method of the minimal polynomial is fantastic... provided you have Mathematica to generate it :)
– Raffaele
Commented
Aug 2, 2017 at 11:41
@Raffaele Right about the CAS, in general. However, the simple example from my previous comment can be worked out by hand fairly easily. Just expand (1+22–√3+34–√3)(1+a2–√3+b4–√3) and solve for a,b such that the coefficients of 2–√3,4–√3 are both 0.
– dxiv
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Aug 2, 2017 at 20:38
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Hint: for a shortcut in this particular case, let a=2+2–√ then use that:
1a+2–√3=a2−a2–√3+4–√3a3+2
The denominator now contains only integers and terms in 2–√ after expansion, which is the case you know how to rationalize.
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answered Aug 1, 2017 at 23:11
dxivdxiv
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There is a very general procedure for this kind of a question.
You see, in essence, rationalization of a fraction is converting the denominator into a rational, right? But indeed, it's a deeper process : Suppose a fraction of the form 1b can be rationalized, say written in the form cd, where d is rational.
Cross multiplying, we get bc=d, or that b times something is a rational. This equates to the invertibility of the number b in the field of real numbers, which is true all the time. So every fraction of this kind is indeed rationalizable. But the question then comes down to how to find this inverse.
One way of finding the inverse is finding the minimal polynomial with rational coefficients, which b satisfies. I'll explain why.
Suppose b satisfies the polynomial ∑ni=0aixi=0. Then, ∑ni=0aibi=0, so that ∑ni=1aibi=−a0, from where it follows that b(∑ni=1aibi−1)=a0.
Rewriting,
1b=(∑ni=1aibi−1)a0
which is the rationalized form.
So all you need to do, is to find a polynomial which the given surd, in our case 2–√+2–√3+2, satisfies.
The minimal such polynomial is x6−12x5+54x4−116x3+132x2−120x+92, which I found online. There's a better answer above on how to actually find a polynomial, so I will avoid that part, but at least this shows that fractions with "algebraic" denominators can be rationalized using the polynomial they satisfy.
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edited Apr 28, 2019 at 5:45
answered Aug 1, 2017 at 23:33
Sarvesh Ravichandran IyerSarvesh Ravichandran Iyer
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This is precisely what I described in a comment to the question.
– Bill Dubuque
Commented
Aug 1, 2017 at 23:34
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My remark was merely to help readers understand how the replies are related. Nothing was intended about "the answer standing" (whatever that means).
– Bill Dubuque
Commented
Aug 1, 2017 at 23:38
I apologize. What I meant was that I did not want to delete my answer because it coincided with a comment. But I took your comment above in the wrong taste. Forgive me.
– Sarvesh Ravichandran Iyer
Commented
Aug 1, 2017 at 23:45
Lovely method, provided you have Mathematica to generate the minimal polynomial :)
– Raffaele
Commented
Aug 2, 2017 at 11:45
Ah, but of course that is true.
– Sarvesh Ravichandran Iyer
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Aug 2, 2017 at 11:46
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Fleablood gave a method for your particular example but you seem to suggest that you're looking for a general solution. In this case, a formula may be given by
1a−−√6+b√6+c√6=(a−−√3+b√3+c√3−ab−−√6−ac−−√6−bc−−√6)((3abc−−−√6+a−−√+b√+c√)2−3(a−−√+b√+c√)abc−−−√6)((a+3b+3c)a−−√+(3a+b+3c)b√−(3a+3b+c)c√+21abc−−−√)(2(a+3b+3c)(3a+b+3c)ab−−√+42(3a+3b+c)abc2−−−−√−a(a+3b+3c)2−b(3a+b+3c)2+c(3a+3b+c)2+441abc)4ab(21c(3a+3b+c)+(a+3b+3c)(3a+b+3c))2−(a(a+3b+3c)2+b(3a+b+3c)2−c(3a+3b+c)2−441abc)2
Which can be used to rationalize any 3 term denominator with a mix of square and cube roots. It's a beast, but it works.
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answered Feb 2, 2023 at 23:07
Baby Hearty BearBaby Hearty Bear
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Auditory Brainstem Implant - CAM 70183
Description: An auditory brainstem implant (ABI) is designed to restore some hearing in people with neurofibromatosis type 2 (NF2) who are rendered deaf by bilateral removal of neurofibromas involving the auditory nerve. ABIs have also been studied to restore hearing for other non-neurofibromatosis indications.
For individuals who are deaf due to bilateral resection of neurofibromas of the auditory nerve who receive an ABI, the evidence includes a large prospective case series. Relevant outcomes are functional outcomes, quality of life, and treatment-related morbidity. The U.S. Food and Drug Administration (FDA) approval of the Nucleus 24 device in 2000 was based on a prospective case series of 90 patients 12 years of age or older, of whom 60 had the implant for at least 3 months. From this group, 95% had a significant improvement in lip reading or improvement on sound-alone tests. While use of an ABI is associated with a very modest improvement in hearing, this level of improvement is considered significant for those patients who have no other treatment options. Based on these results, ABIs are considered appropriate for the patient population included in the trial (i.e., age ≥ 12 years with NF2 and deafness following tumor removal). The evidence is sufficient to determine that the technology results in a meaningful improvement in the net health outcome.
For individuals who are deaf due to nontumor etiologies who receive an ABI, the evidence includes case series and systematic reviews of case series. Relevant outcomes are functional outcomes, quality of life, and treatment-related morbidity. In general, ABIs have not demonstrated hearing benefits over cochlear implants for many non-NF2 conditions. However, ABIs hold promise for select patients when the cochlea or cochlear nerve is absent. Many recent and ongoing ABI studies are being conducted in children. For children, hearing is critical for language development, and this device has potential to substantially improve health outcomes. The most common nontumor conditions in children are cochlear aplasia and cochlear nerve aplasia. There are questions about the durability of the now obsolete Nucleus 24 in active young children. Evaluation is currently ongoing with the recently available Nucleus ABI541to determine its efficacy and durability in children. In addition, ABI studies have shown inferior outcomes in children with other disabilities. Thus, further study is also needed to define populations that would benefit from these devices. The evidence is insufficient to determine the effects of the technology on health outcomes.
Background The auditory brainstem implant (ABI) is intended to restore some hearing in people with neurofibromatosis type 2 who are rendered deaf by bilateral removal of the characteristic neurofibromas involving the auditory nerve. The ABI consists of an externally worn speech processor that provides auditory information by electrical signal that is transferred to a receiver/stimulator implanted in the temporal bone. The receiver stimulator is, in turn, attached to an electrode array implanted on the surface of the cochlear nerve in the brainstem, thus bypassing the inner ear and auditory nerve. The electrode stimulates multiple sites on the cochlear nucleus, which is then processed normally by the brain. To place the electrode array on the surface of the cochlear nucleus, the surgeon must be able to visualize specific anatomic landmarks. Because large neurofibromas compress the brainstem and distort the underlying anatomy, it can be difficult or impossible for the surgeon to correctly place the electrode array. For this reason, patients with large, long-standing tumors may not benefit from the device.
ABIs are also being studied to determine whether they can restore hearing for other non-neurofibromatosis causes of hearing impairment in adults and children, including absence of or trauma to the cochlea or auditory nerve. It is estimated that 1.7 per 100,000 children are affected by bilateral cochlea or cochlear nerve aplasia and 2.6 per 100,000 children are affected by bilateral cochlea or cochlear nerve hypoplasia.1
Regulatory Status In 2000, the Nucleus® 24 Auditory Brainstem Implant System (Cochlear Corp.) was approved by the U.S. Food and Drug Administration (FDA) through the premarket approval process. The speech processor and receiver are similar to the devices used in cochlear implants; the electrode array placed on the brainstem is the novel component of the device. The device is indicated for individuals 12 years of age or older who have been diagnosed with neurofibromatosis type 2. The Nucleus® 24 Auditory Brainstem Implant System labeling states: "The efficacy of bilateral implantation with the ABI [auditory brainstem implant] has not been studied."2 The Nucleus® 24 is now obsolete.
In June 2016, the Nucleus ABI541 Auditory Brainstem Implant (Cochlear Corp.) was approved by FDA through a supplement to the premarket approval for the Nucleus® 24. The new implant is indicated for individuals 12 years of age or older who have been diagnosed with neurofibromatosis type 2.
FDA product code: MCM.
Related Policies 70103 Implantable Bone-Conduction and Bone-Anchored Hearing Aids 70105 Cochlear Implant 70184 Semi-Implantable and Fully Implantable Middle Ear Hearing Aids
Policy: Unilateral use of an auditory brainstem implant (using surface electrodes on the cochlear nuclei) may be considered MEDICALLY NECESSARY in patients with neurofibromatosis type 2, who are 12 years of age or older, and who are rendered deaf due to bilateral resection of neurofibromas of the auditory nerve.
An auditory brainstem implant is investigational and /or unproven and therefore considered NOT MEDICALLY NECESSARY for all other conditions including non-neurofibromatosis-type 2 indications.
Bilateral use of an auditory brainstem implant is investigational and /or unproven and therefore considered NOT MEDICALLY NECESSARY.
Penetrating electrode auditory brainstem implant (PABI) is investigational and /or unproven and therefore considered NOT MEDICALLY NECESSARY .
Policy Guidelines Please see the Codes table for details.
Benefit Application Implantation of an auditory brainstem implant is a specialized procedure that may require out-of-network referral.
State or federal mandates (e.g., FEP) may dictate that all devices, drugs, or biologics approved by the U.S. Food and Drug Administration (FDA) may not be considered investigational, and thus these devices may be assessed only on the basis of their medical necessity.
Some facilities may negotiate a global fee for the implantation of the device and the associated aural rehabilitation. However, charges for rehabilitation may be subject to individual contractual limitations.
Rationale This evidence review was created in July 2002 and has been updated regularly with searches of the PubMed database. The most recent literature update was performed through December 31, 2024.
Evidence reviews assess the clinical evidence to determine whether the use of a technology improves the net health outcome. Broadly defined, health outcomes are length of life, quality of life, and ability to function-including benefits and harms. Every clinical condition has specific outcomes that are important to patients and to managing the course of that condition. Validated outcome measures are necessary to ascertain whether a condition improves or worsens; and whether the magnitude of that change is clinically significant. The net health outcome is a balance of benefits and harms.
To assess whether the evidence is sufficient to draw conclusions about the net health outcome of a technology, 2 domains are examined: the relevance and the quality and credibility. To be relevant, studies must represent one or more intended clinical use of the technology in the intended population and compare an effective and appropriate alternative at a comparable intensity. For some conditions, the alternative will be supportive care or surveillance. The quality and credibility of the evidence depend on study design and conduct, minimizing bias and confounding that can generate incorrect findings. The randomized controlled trial is preferred to assess efficacy; however, in some circumstances, nonrandomized studies may be adequate. Randomized controlled trials are rarely large enough or long enough to capture less common adverse events and long-term effects. Other types of studies can be used for these purposes and to assess generalizability to broader clinical populations and settings of clinical practice.
In the case of the auditory brainstem implant (ABI), studies that compare outcomes before and after device implantation can provide useful information on health outcomes. Following is a summary of the key literature to date.
Auditory Brainstem Implant for Bilateral Resection of Neurofibromas of the Auditory Nerve Clinical Context and Therapy Purpose The purpose of an ABI in individuals who are deaf due to bilateral resection of neurofibromas of the auditory nerve is to provide a treatment option that is an alternative to observation alone.
The following PICO was used to select literature to inform this review.
Populations The relevant population(s) of interest are individuals who are deaf and have undergone bilateral resection of neurofibromas of the auditory nerve.
Interventions The therapy being considered is an ABI.
Comparators The following practice is currently being used to make decisions about hearing restoration in individuals who are deaf due to bilateral resection of neurofibromas of the auditory nerve: observation alone.
Outcomes The general outcomes of interest are functional outcomes, quality of life and treatment-related morbidity. Functional outcomes include change in hearing and hearing-related function (eg. sound recognition and speech perception).
Study Selection Criteria Methodologically credible studies were selected using the following principles:
To assess efficacy outcomes, comparative controlled prospective trials were sought, with a preference for RCTs;
In the absence of such trials, comparative observational studies were sought, with a preference for prospective studies.
To assess long-term outcomes and adverse events, single-arm studies that capture longer periods of follow-up and/or larger populations were sought.
Consistent with a 'best available evidence approach,' within each category of study design, studies with larger sample sizes and longer durations were sought.
Studies with duplicative or overlapping populations were excluded.
Review of Evidence Systematic Review Garcia et al (2024) conducted a systematic review of published reports of ABI use.4, A total of 36 studies were included, encompassing 662 patients with tumors and 267 patients without tumors. A study-specific scale called the Adult Pediatric Ranked Order Speech Perception (APROSPER) scale was used to assess outcomes. Among the patients with tumors, weighted mean speech recognition was 39.2% (range, 19.6% to 83.3%) for closed-set words, 23.4% (range, 17.2% to 37.5%) for open-set words, and 21.5% (range, 2.7% to 4.7%) for open-set sentences. Mean categories of auditory performance (CAP) scores were 3.1 (range, 1.0 to 3.2).
Wang et al (2024) conducted a systematic review and meta-analysis of audiological outcomes following ABI implantation in patients with neurofibromatosis type 2 schwannomatosis.5, Among the 33 studies that were included, the pooled estimate for environmental sound discrimination was 55% (95% confidence interval [CI], 49% to 66%) and for closed set discrimination was 55% (95% CI, 40% to 69%). The pooled estimate for open-set discrimination was 30% (95% CI, 19% to 42%). Complications occurred in 33% (95% CI, 15% to 52%) of patients.
A systematic review conducted by Ontario (Canada) Health as part of a Health Technology Assessment included 16 observational studies (N=491) comparing the effectiveness of ABI to no treatment in adults with neurofibromatosis type 2 (Table 1 and Table 2).6, Risk of bias among the included studies was assessed using the Risk of Bias in Non-randomized Studies - of Interventions (ROBINS-I) tool, and overall quality of evidence was assessed using the Grading of Recommendation, Assessment, Development and Evaluation (GRADE) Handbook. Results were reported qualitatively, and no meta-analyses were conducted due to heterogeneity in testing conditions and outcomes. The review found high quality of evidence of benefit of ABI on sound recognition (7 studies), speech perception with with lip reading (5 studies), and subjective hearing benefit (5 studies). Evidence favoring ABI was moderate for speech perception without lip reading (10 studies) and low for quality of life (1 study). The most commonly reported surgical complications, based on low quality evidence from 12 studies, were cerebrospinal fluid leak in 3% to 15% of participants and infection in 10% to 13% or participants.
Table 1. SR-MA Characteristics
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| Study | Dates | Trials | Participants | N (Range) | Design | Duration |
| Ontario Health6, | 1993-2016; literature searches conducted through June 2018 | 19 observational studies | Adults with neurofibromatosis type 2 who were not candidates for cochlear implantation | 491 (8-61) | 6 prospective cohort studies 11 retrospective cohort studies 2 cross-sectional studies | 1 month to 18 years (mean, median not reported) |
Table 2. SR-MA Results
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| Study | Sound Recognition | Speech Perception | Subjective Benefits of Hearing | Quality of Life | Surgical Complications |
| Ontario Health6, | ABI vs. no treatment | ABI vs. no treatment | ABI vs. no treatment | ABI vs. no treatment | ABI vs. no treatment |
| Number of studies; N | 7 observational studies; N=169 | 15 observational studies; N=348 | 5 observational studies; N=141 | 1 observational study; N=11 | 12 observational studies; N= |
| Qualitative assessment of ABI effectiveness | Allows any degree of improvement in sound recognition vs. no treatment | ABI only: Likely allows any degree of improvement in speech perception when used alone ABI + lip reading: Allows any degree of improvement in speech perception when used in conjunction with lip-reading | Provides subjective benefits of hearing | May improve quality of life | Most common complications were cerebrospinal fluid leak infection |
| Level of evidence (GRADE) | High | ABI only: Moderate ABI + lip reading: High | High | Low | Low |
Abbreviations: ABI: auditory brainstem implant.
Observational Studies U.S. Food and Drug Administration (FDA) approval of the Nucleus 24 Auditory Brainstem Implant System was based on results in a case series of 90 patients with neurofibromatosis type 2, ages 12 years and older.1,7, Of the 90 subjects evaluated, 28 complications occurred in 26 patients; 26 of these complications resolved without surgical or extensive medical intervention. Two patients had infections of the postoperative flap requiring explantation of the device. Sixty patients had a minimum experience of 3 to 6 months with the device, and thus effectiveness outcomes were also evaluated. Overall device benefit was defined as a significant enhancement of lip reading or an above-chance improvement on sound-alone tests. Based on this definition, 95% (57/60) of patients derived benefit from the device. Among the 90 patients receiving the implant, 16 did not receive auditory stimulation from the device postoperatively, either due to migration of the implanted electrodes or surgical misplacement.
A single small (N=10) trial from 2008 was identified on a penetrating ABI (PABI)8,. This prospective clinical trial enrolled patients with neurofibromatosis type 2 who received a PABI after vestibular schwannoma removal. The PABI is an extension of the ABI technology that uses surface electrodes on cochlear nuclei. The PABI uses 8 or 10 penetrating microelectrodes in conjunction with a separate array of 10 to 13 surface electrodes. The PABI met the goals of lower threshold, increased pitch range, and high selectivity, but these properties did not improve speech recognition.
Daoudi et al (2024) conducted a retrospective, single center, long term follow-up study of patients with neurofibromatosis type 2 who received an ABI.9, Using a threshold of at least 5 years of follow-up, the investigators identified 27 patients who received a total of 32 ABIs. Mean duration of follow-up was 12 years (range, 5 to 24 years). At 1 year post-implantation, 74% of patients were still ABI users; at last follow-up, 54% of patients were still users. Hearing improvement for disyllabic words was 32% at 1 year and 41% at 5 years. Improvement in sentence recognition was 28% at 1 year and 42% at 5 years. Four patients experienced a decrease in ABI performance after 1 year, 3 of which were attributed to tumor growth.
Section Summary: Auditory Brainstem Implant for Bilateral Resection of Neurofibromas of the Auditory Nerve The evidence on ABI for bilateral resection of neurofibromas of the auditory nerve includes large case series, small uncontrolled studies, and systematic reviews of small observational studies. A 2018 case series of 90 adults, 60 of which had the minimum experience of 3 to 6 months with the Nucleus 24 ABI system, suggested that adults may benefit from its usage. European studies followed 32 patients, 24 of which with an ABI activated experienced significant improvements on the Sound Effects Recognition Test and Monosyllable-Trochee-Polysyllable test. A single-center study found persistent improvement after long-term (at least 5 years) follow-up. An Ontario (Canada) Health systematic review found ABI associated with better hearing function relative to no treatment, but evidence on other outcomes was limited.
Auditory Brainstem Implant for Nontumor Etiologies Clinical Context and Therapy Purpose The purpose of an ABI in individuals who are deaf due to nontumor etiologies is to provide a treatment option that is an alternative to observation alone.
The following PICO was used to select literature to inform this review.
Populations The relevant population(s) of interest are individuals who are deaf due to nontumor etiologies.
Interventions The therapy being considered is an ABI.
Comparators The following practice is currently being used to make decisions about hearing restoration in individuals who are deaf due to nontumor etiologies: observation alone.
Outcomes The general outcomes of interest are functional outcomes, quality of life, and treatment-related morbidity. Functional outcomes include change in hearing and hearing-related function (eg. sound recognition and speech perception).
Study Selection Criteria Methodologically credible studies were selected using the following principles:
To assess efficacy outcomes, comparative controlled prospective trials were sought, with a preference for RCTs;
In the absence of such trials, comparative observational studies were sought, with a preference for prospective studies.
To assess long-term outcomes and adverse events, single-arm studies that capture longer periods of follow-up and/or larger populations were sought.
Consistent with a 'best available evidence approach,' within each category of study design, studies with larger sample sizes and longer durations were sought.
Studies with duplicative or overlapping populations were excluded.
Review of Evidence Adults Merkus et al (2014) conducted a systematic review of ABIs for non-neurofibromatosis type 2 indications.10, Included in the review were 144 non-neurofibromatosis type 2 ABI cases from 31 articles. Non-neurofibromatosis type 2 indications for which ABIs have been evaluated include cochlear otosclerosis, temporal bone fractures, bilateral traumatic cochlear nerve disruption, autoimmune inner ear disease, auditory neuropathy, cochlear nerve aplasia, and vestibular schwannoma in the only hearing ear. Cochlear implants have generally provided better hearing than ABIs when the cochlea and cochlear nerve are intact. Complete bilateral disruption of the cochlear nerve from trauma did not exist in the literature and cochlear malformation did not preclude cochlear implant. While the evidence is limited, it appears as if cochlear implants demonstrate greater hearing benefits than ABIs in patients with non-neurofibromatosis type 2 indications.
In a literature review by Medina et al (2014) assessing ABI for traumatic deafness, cochlear implant performed better than ABI.11, However, there was limited evidence on which to draw conclusions, because only 3 articles (total N=7 patients) were identified in the review on ABI for traumatic deafness.
Children Sennaroglu et al (2024) published a summary of outcomes reported by institutions that participated in the Third International Pediatric ABI Meeting on pediatric ABI implantation in inner ear malformations.12, After cases from all participating institutions were described, the experts concluded that early implantation (before age 3 years) correlates with better auditory and language development, and that auditory outcomes after ABI implantation are diverse because of individual anatomic and developmental factors. Decision-making should be individualized and include consideration of patient appropriateness for surgery and access to rehabilitation services. The authors also concluded that outcomes in children with normal anatomy who receive an ABI are not as strong as children who receive cochlear implant, so ABI is mainly considered in patients who are not candidates for cochlear implants (eg, patients with complex inner ear malformation or a dysplastic cochlear nerve). The authors hope that these conclusions will inform a future consensus statement.
Systematic Reviews The Garcia et al (2024) systematic review previously described included 267 patients without tumors.4, Among these patients, weighted mean speech recognition was 79.8% (range, 31.7% to 84.4%) for closed-set words and 53.0% (range, 14.6% to 72.5%) for open-set sentences. Mean CAP scores were 2.30 (range, 2.0 to 4.7).
A systematic review of nontumor pediatric ABI outcomes was reported by Noij et al (2015).13, It included 21 studies with 162 children, at a mean age of 4.3 years (range, 11 months to 17 years). Nine reports were from a single group from Italy (described below) and it could not be determined if there was patient overlap across these studies. Nearly all studies were retrospective series or cohorts; 1 was a case-control. Most children (63.6%) had cochlear nerve aplasia. Other conditions were cochlear aplasia, cochlear nerve hypoplasia, cochlear malformations, ossified cochlea, auditory neuropathy, trauma, and cochlear hypoplasia. Twenty-five percent of the patients had previously received a cochlear implant. Forty major and minor implant-related complications were reported, the most common being cerebrospinal fluid leak (8.5% of patients). The most common side effects associated with ABI use were discomfort of the body and/or limb, dizziness/vertigo/nystagmus, pain in the head and/or neck, and stimulation of the facial nerve or involuntary swallowing, gagging, or coughing. A variety of auditory tests were used; the most common (6 studies) was the CAP index (range, 0-7; high score indicates better hearing). There was an improvement in CAP scores over time. After 5 years, almost 50% of patients had CAP scores greater than 4 (5 [understanding of common phrases without lip reading] to 7 [use of telephone with known speaker]). Children who also had nonauditory disabilities never attained a CAP score greater than 4. There was no significant effect of the age of implantation.
Case Series Many of the larger series on ABI in nontumor patients are from a group that includes Colletti and Colletti. In 2013, this group reported on ABIs in 21 children, ranging in age from 1.7 to 5 years, with deafness unrelated to neurofibromatosis, who had a poor response to cochlear implants.14, At surgery, the cochlear nerve was absent in each patient. Significant improvements in CAP index scores were seen after ABI (p<.001).
Sennaroglu et al (2016) reported on follow-up of at least 1 year for 35 children who had received ABI.15, This followed a 2009 preliminary report of 11 prelingually deaf children ages 30 to 56 months who received an ABI.16, Sixty children had received an ABI from this center in Turkey. The children who had received the ABI in the previous year were excluded from the 2016 analysis. Over half (n=19) of the cases were due to cochlear hypoplasia. ABI models implanted were Cochlear, Med El, and Neurelec. At regular follow-up, children were evaluated with the CAP, Speech Intelligibility Rate, Functional Auditory Performance of Cochlea Implantation, and Manchester scores. About half the children were in the CAP category 5 and could understand common phrases without lip reading. In the subgroup with better hearing thresholds (25-40 decibels), some (17.6%) were able to understand conversation without lip reading, use the telephone with known speaker (11.8%), and follow group conversation in a noisy room (5.9%). For children with higher hearing thresholds (>50 decibels), none exceeded CAP category 5. Speech Intelligibility Rate and Manchester scores were also better with greater hearing thresholds. Auditory performance measured with the Functional Auditory Performance of Cochlea Implantation was in the 10th percentile for all groups and was worse compared with cochlear implantation. As was also found in the Noij systematic review (discussed above), children with additional nonauditory disabilities had worse outcomes.
Bas et al (2024) reported on sensory processing, attention, and memory in 25 children with ABIs.17, Patient age ranged from 6 to 10 years. The patients were stratified by duration of use, with 12 children having a mean duration of 63.25 months and 13 children having a duration of 76.38 months. The group with a longer duration of ABI use had higher attention and short-term memory performance as measured by the visual-aural digit span test B, better visual and spatial perception as measured by the Marking Test (all p<.05).
Mixed Populations Other reports from the group of Colletti and Colletti include a 2005 report on ABIs in 16 children and adults who had nontumor diseases of the cochlear nerve or cochlea and 13 patients with neurofibromatosis type 2.18, Ages ranged from 14 months to 70 years; the nontumor group included patients with head trauma, complete cochlear ossification, auditory neuropathy, and bilateral cochlear nerve aplasia. Following implantation, the adult nontumor group scored substantially higher than the patients with NF2 in open set speech perception tests. Some children showed dramatic improvements in word and sentence recognition over a 1-year follow-up. Short-term adverse events included dizziness or tingling sensations in the leg, arm, and throat (20/29 patients). Additional studies from this group have reported improvements in hearing with ABIs in “nontumor” patients, including a 2006 report on 54 nontumor patients19, and a 2007 report on 22 non-neurofibromatosis patients.20,
In a retrospective review, Colletti et al (2010) reported on complications from ABI surgery in 83 adults and 31 children, 78 of whom had nontumor cochlear or cochlear nerve disorders.21, Authors found that ABI complication rates were similar to those for cochlear implant surgery. Additionally, there were significantly fewer major and minor complications in nontumor patients than in neurofibromatosis type 2 patients.
Section Summary: Auditory Brainstem Implant in Nontumor Etiologies The evidence on ABI in nontumor patients includes case series and systematic reviews. A 2014 systematic review suggested that ABI might improve outcomes in bilateral complete cochlear and inner ear aplasia. Recent research includes studies of children who are deaf but would not benefit from a cochlear implant. The most common conditions in these studies are cochlear aplasia and cochlear nerve aplasia. Hearing in this age group is critical for language development, and the ABI has potential to substantially improve health outcomes for this age group. However, studies of early (now obsolete) ABI devices found a high rate of failure in children and high rates of adverse events in adults. Evidence from ongoing studies assessing newer ABI models is needed to evaluate efficacy and durability in patients with nontumor ABI indications.
The purpose of the following information is to provide reference material. Inclusion does not imply endorsement or alignment with the evidence review conclusions.
Practice Guidelines and Position Statements Guidelines or position statements will be considered for inclusion in ‘Supplemental Information’ if they were issued by, or jointly by, a US professional society, an international society with US representation, or National Institute for Health and Care Excellence (NICE). Priority will be given to guidelines that are informed by a systematic review, include strength of evidence ratings, and include a description of management of conflict of interest.
National Institute for Health and Care Excellence In 2005, the National Institute for Health and Care Excellence issued guidance on interventional procedures for auditory brainstem implants.22, The guidance stated: “…evidence on safety and efficacy of auditory brain stem implants appears adequate to support the use of this procedure by surgical teams experienced in this technique.”
U.S. Preventive Services Task Force Recommendations Not applicable.
Ongoing and Unpublished Clinical Trials Some currently unpublished trials that might influence this review are listed in Table 3.
Table 3. Summary of Key Trials
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| Trial Name | Planned Enrollment | Completion Date |
| Ongoing | | | |
| NCT05810220 | Investigating Auditory Processing in the Users of Auditory Brainstem and Cochlear Implants | 200 | Dec 2026 |
| NCT02630589 | Implantation of an Auditory Brainstem Implant for the Treatment of Incapacitating Unilateral Tinnitus | 10 | Jan 2028 |
| NCT02310399 | Auditory Brainstem Implant (ABI) in Children With No Cochlear or Auditory Nerves | 20 | May 2030 |
IRB: Institutional Review Board; NCT: national clinical trial.
References:
Food and Drug Administration. Nucleus 24 Auditory Brainstem Implant System. FDA Summary of Safety and Effectiveness. 2000; Accessed December 31, 2024.
Kaplan AB, Kozin ED, Puram SV, et al. Auditory brainstem implant candidacy in the United States in children 0-17 years old. Int J Pediatr Otorhinolaryngol. Mar 2015; 79(3): 310-315. PMID 25577282
Food and Drug Administration. Premarket Approval (PMA). Nucleus ABI541 Auditory Brainstem Implant. 2016. Accessed December 31, 2024.
Garcia A, Haleem A, Poe S, et al. Auditory Brainstem Implant Outcomes in Tumor and Nontumor Patients: A Systematic Review. Otolaryngol Head Neck Surg. Jun 2024; 170(6): 1648-1658. PMID 38329219
Wang B, Yan M, Liu C, et al. Auditory brainstem implants for hearing rehabilitation in NF2-schwannomatosis: A systematic review and single-arm meta-analysis. NeuroRehabilitation. 2024; 54(2): 213-225. PMID 38427506
Ontario Health (Quality). Auditory brainstem implantation for adults with neurofibromatosis 2 or severe inner ear abnormalities: a health technology assessment. Ont Health Technol Assess Ser [Internet]. 2020 Mar;20(4): 185. Accessed December 31, 2024.
Ebinger K, Otto S, Arcaroli J, et al. Multichannel auditory brainstem implant: US clinical trial results. J Laryngol Otol Suppl. 2000; (27): 50-3. PMID 11211440
Otto SR, Shannon RV, Wilkinson EP, et al. Audiologic outcomes with the penetrating electrode auditory brainstem implant. Otol Neurotol. Dec 2008; 29(8): 1147-54. PMID 18931643
Daoudi H, Torres R, Mosnier I, et al. Long-term analysis of ABI auditory performance in patients with neurofibromatosis type 2-related schwannomatosis. Acta Neurochir (Wien). Oct 02 2024; 166(1): 390. PMID 39356313
Merkus P, Di Lella F, Di Trapani G, et al. Indications and contraindications of auditory brainstem implants: systematic review and illustrative cases. Eur Arch Otorhinolaryngol. Jan 2014; 271(1): 3-13. PMID 23404468
Medina M, Di Lella F, Di Trapani G, et al. Cochlear implantation versus auditory brainstem implantation in bilateral total deafness after head trauma: personal experience and review of the literature. Otol Neurotol. Feb 2014; 35(2): 260-70. PMID 24448286
Sennaroglu L, Lenarz T, Roland JT, et al. Current status of pediatric auditory brainstem implantation in inner ear malformations; consensus statement of the Third International Pediatric ABI Meeting. Cochlear Implants Int. Jul 2024; 25(4): 316-333. PMID 39607757
Noij KS, Kozin ED, Sethi R, et al. Systematic Review of Nontumor Pediatric Auditory Brainstem Implant Outcomes. Otolaryngol Head Neck Surg. Nov 2015; 153(5): 739-50. PMID 26227469
Colletti L, Wilkinson EP, Colletti V. Auditory brainstem implantation after unsuccessful cochlear implantation of children with clinical diagnosis of cochlear nerve deficiency. Ann Otol Rhinol Laryngol. Oct 2013; 122(10): 605-12. PMID 24294682
Sennaroğlu L, Sennaroğlu G, Yücel E, et al. Long-term Results of ABI in Children With Severe Inner Ear Malformations. Otol Neurotol. Aug 2016; 37(7): 865-72. PMID 27273392
Sennaroglu L, Ziyal I, Atas A, et al. Preliminary results of auditory brainstem implantation in prelingually deaf children with inner ear malformations including severe stenosis of the cochlear aperture and aplasia of the cochlear nerve. Otol Neurotol. Sep 2009; 30(6): 708-15. PMID 19704357
Baş B, Gökay NY, Aydoğan Z, et al. Do auditory brainstem implants favor the development of sensory integration and cognitive functions?. Brain Behav. Aug 2024; 14(8): e3637. PMID 39099332
Colletti V, Carner M, Miorelli V, et al. Auditory brainstem implant (ABI): new frontiers in adults and children. Otolaryngol Head Neck Surg. Jul 2005; 133(1): 126-38. PMID 16025066
Colletti V. Auditory outcomes in tumor vs. nontumor patients fitted with auditory brainstem implants. Adv Otorhinolaryngol. 2006; 64: 167-185. PMID 16891842
Colletti L. Beneficial auditory and cognitive effects of auditory brainstem implantation in children. Acta Otolaryngol. Sep 2007; 127(9): 943-6. PMID 17712673
Colletti V, Shannon RV, Carner M, et al. Complications in auditory brainstem implant surgery in adults and children. Otol Neurotol. Jun 2010; 31(4): 558-64. PMID 20393378
National Institute Health and Care Excellence (NICE). Auditory brain stem implants [IPG108]. 2005 Accessed December 31, 2024.
Coding Section
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| Codes | Number | Description |
| CPT | 61863 | Twist drill, burr hole, craniotomy, or craniectomy with stereotactic implantation of neurostimulator electrode array in subcortical site (e.g., thalamus, globus pallidus, subthalamic nucleus, periventricular, periaqueductal gray), without use of intraoperative microelectrode recording; first array |
| | 61864 | Twist drill, burr hole, craniotomy, or craniectomy with stereotactic implantation of neurostimulator electrode array in subcortical site (e.g., thalamus, globus pallidus, subthalamic nucleus, periventricular, periaqueductal gray), without use of intraoperative microelectrode recording; each additional array (List separately in addition to primary procedure) |
| | 61867 | Twist drill, burr hole, craniotomy, or craniectomy with stereotactic implantation of neurostimulator electrode array in subcortical site (e.g., thalamus, globus pallidus, subthalamic nucleus, periventricular, periaqueductal gray), with use of intraoperative microelectrode recording; first array |
| | 61868 | Twist drill, burr hole, craniotomy, or craniectomy with stereotactic implantation of neurostimulator electrode array in subcortical site (e.g., thalamus, globus pallidus, subthalamic nucleus, periventricular, periaqueductal gray), with use of intraoperative microelectrode recording; each additional array (List separately in addition to primary procedure) |
| | 92640 | Diagnosis analysis with programming of auditory brainstem implant, per hour |
| HCPCS | S2235 | Implantation of auditory brainstem implant |
| ICD-10-CM | Q85.02 | Neurofibromatosis, type 2 |
| ICD-10-PCS | | ICD-10-PCS codes are only used for inpatient services. There is no specific ICD-10-PCS code for this procedure. |
| | 00H00MZ, 00H03MZ, 00H04MZ | Insertion of neurostimulator electrode, brain, code by approach (open, percutaneous, percutaneous endoscopic) |
| Type of Service | | |
| Place of Service | | |
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4: Systems of Linear Equations
Intermediate Algebra 1e (OpenStax)
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4.2: Solve Systems of Linear Equations with Two Variables
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4.2: Solve Systems of Linear Equations with Two Variables
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4.1: Introduction
4.2E: Exercises
Page ID 5138
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Determine Whether an Ordered Pair is a Solution of a System of Equations
1. SYSTEM OF LINEAR EQUATIONS
2. SolutionS OF A SYSTEM OF EQUATIONS
3. Example 4.2.1
4. Example 4.2.2
5. Example 4.2.3
Solve a System of Linear Equations by Graphing
Example 4.2.4: How to Solve a System of Equations by Graphing
Example 4.2.5
Example 4.2.6
SOLVE A SYSTEM OF LINEAR EQUATIONS BY GRAPHING.
Example 4.2.7
Example 4.2.8
Example 4.2.9
Example 4.2.10
Example 4.2.11
Example 4.2.12
Example 4.2.13
Example 4.2.14
Example 4.2.15
COINCIDENT LINES
CONSISTENT AND INCONSISTENT SYSTEMS
Example 4.2.16
Example 4.2.17
Example 4.2.18
Solve a System of Equations by Substitution
Example 4.2.19: How to Solve a System of Equations by Substitution
Example 4.2.20
Example 4.2.21
SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.
Example 4.2.22
Example 4.2.23
Example 4.2.24
Solve a System of Equations by Elimination
Exercise 4.2.25: How to Solve a System of Equations by Elimination
Exercise 4.2.26
Exercise 4.2.27
SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.
Exercise 4.2.28
Exercise 4.2.29
Exercise 4.2.30
Exercise 4.2.31
Exercise 4.2.32
Exercise 4.2.33
Exercise 4.2.34
Exercise 4.2.35
Exercise 4.2.36
Choose the Most Convenient Method to Solve a System of Linear Equations
Example 4.2.37
Example 4.2.38
Example 4.2.39
Key Concepts
Glossary
Learning Objectives
By the end of this section, you will be able to:
Determine whether an ordered pair is a solution of a system of equations
Solve a system of linear equations by graphing
Solve a system of equations by substitution
Solve a system of equations by elimination
Choose the most convenient method to solve a system of linear equations
Before you get started, take this readiness quiz.
For the equation y=2 3 x−4,
ⓐ Is (6,0) a solution? ⓑ Is (−3,−2) a solution?
If you missed this problem, review [link].
2. Find the slope and y-intercept of the line 3 x−y=12.
If you missed this problem, review [link].
3. Find the x- and y-intercepts of the line 2 x−3 y=12.
If you missed this problem, review [link].
Determine Whether an Ordered Pair is a Solution of a System of Equations
In Solving Linear Equations, we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations.
SYSTEM OF LINEAR EQUATIONS
When two or more linear equations are grouped together, they form a system of linear equations.
In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.
An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.
{2 x+y=7 x−2 y=6
A linear equation in two variables, such as 2 x+y=7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.
To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x,y) that make both equations true. These are called the solutions of a system of equations.
SolutionS OF A SYSTEM OF EQUATIONS
The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x,y).
To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.
Example 4.2.1
Determine whether the ordered pair is a solution to the system {x−y=−1 2 x−y=−5.
ⓐ (−2,−1) ⓑ (−4,−3)
Answer
ⓐ
ⓑ
Example 4.2.2
Determine whether the ordered pair is a solution to the system {x+y=0 x+2 y=−5.
ⓐ (1,−3) ⓑ (0,0)
Answer
ⓐ yes ⓑ no
Example 4.2.3
Determine whether the ordered pair is a solution to the system {−3 y=−8−3 x−y=4.
ⓐ (2,−2) ⓑ (−2,2)
Answer
ⓐ no ⓑ yes
Solve a System of Linear Equations by Graphing
In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.
The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.
Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.
Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.
Figure 4.2.1
Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.
Example 4.2.4: How to Solve a System of Equations by Graphing
Solve the system by graphing {2 x+y=7 x−2 y=6.
Answer
Example 4.2.5
Solve the system by graphing: {x−3 y=−3 x+y=5.
Answer
(3,2)
Example 4.2.6
Solve the system by graphing: {−x+y=1 3 x+2 y=12
Answer
(2,3)
The steps to use to solve a system of linear equations by graphing are shown here.
SOLVE A SYSTEM OF LINEAR EQUATIONS BY GRAPHING.
Graph the first equation.
Graph the second equation on the same rectangular coordinate system.
Determine whether the lines intersect, are parallel, or are the same line.
Identify the solution to the system.
If the lines intersect, identify the point of intersection. This is the solution to the system.
If the lines are parallel, the system has no solution.
If the lines are the same, the system has an infinite number of solutions.
Check the solution in both equations.
In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.
Example 4.2.7
Solve the system by graphing: {3 x+y=−1 2 x+y=0
Answer
We’ll solve both of these equations for y so that we can easily graph them using their slopes and y-intercepts.
Solve the first equation for y.
Find the slope and y-intercept.
Solve the second equation for y.
Find the slope and y-intercept.
Graph the lines.
Determine the point of intersection.The lines intersect at (−1,2).
Check the solution in both equations.
The solution is (−1,2).
Example 4.2.8
Solve the system by graphing: {−x+y=1 2 x+y=10.
Answer
(3,4)
Example 4.2.9
Solve the system by graphing: {2 x+y=6 x+y=1.
Answer
(5,−4)
In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.
Example 4.2.10
Solve the system by graphing: {y=1 2 x−3 x−2 y=4.
Answer
To graph the first equation, we will use its
slope and y-intercept.
To graph the second equation, we will use
the intercepts.
Graph the lines.
Determine the points of intersection.The lines are parallel.
Since no point is on both lines, there is no
ordered pair that makes both equations
true. There is no solution to this system.
Example 4.2.11
Solve the system by graphing: {y=−1 4 x+2 x+4 y=4.
Answer
no solution
Example 4.2.12
Solve the system by graphing: {y=3 x−1 6 x−2 y=6.
Answer
no solution
Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.
Example 4.2.13
Solve the system by graphing: {y=2 x−3−6 x+3 y=9.
Answer
Find the slope and y-intercept of the first equation.
Find the intercepts of the second equation.
Graph the lines.
The lines are the same!
Since every point on the line makes both
equations true, there are infinitely many
ordered pairs that make both equations true.
There are infinitely many solutions to this system.
If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y-intercept.
Example 4.2.14
Solve the system by graphing: {y=−3 x−6 6 x+2 y=−12.
Answer
infinitely many solutions
Example 4.2.15
Solve the system by graphing: {y=1 2 x−4 2 x−4 y=16.
Answer
infinitely many solutions
When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident. Coincident lines have the same slope and same y- intercept.
COINCIDENT LINES
Coincident lines have the same slope and same y- intercept.
The systems of equations in Example and Example each had two intersecting lines. Each system had one solution.
In Example, the equations gave coincident lines, and so the system had infinitely many solutions.
The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.
A system with parallel lines, like Example, has no solution. We call a system of equations like this inconsistent. It has no solution.
CONSISTENT AND INCONSISTENT SYSTEMS
A consistent system of equations is a system of equations with at least one solution.
An inconsistent system of equations is a system of equations with no solution.
We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.
If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.
Let’s sum this up by looking at the graphs of the three types of systems. See below and Table.
| Lines | Intersecting | Parallel | Coincident |
--- --- |
| Number of solutions | 1 point | No solution | Infinitely many |
| Consistent/inconsistent | Consistent | Inconsistent | Consistent |
| Dependent/ independent | Independent | Independent | Dependent |
Example 4.2.16
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ {y=3 x−1 6 x−2 y=12 ⓑ {2 x+y=−3 x−5 y=5
Answer
ⓐ We will compare the slopes and intercepts of the two lines.
{y=3 x−1 6 x−2 y=12 y=3 x−1 The first equation is already in slope-intercept form.Write the second equation in slope-intercept form.6 x−2 y=12−2 y=−6 x+12−2 y−2=−6 x+12−2 y=3 x−6 y=3 x−1 y=3 x−6 m=3 m=3 b=−1 b=−6 Find the slope and intercept of each line.Since the slopes are the same andy-intercepts are different, the lines are parallel.
ⓑ We will compare the slope and intercepts of the two lines.
{2 x+y=−3 x−5 y=5 Write both equations in slope–intercept form.2 x+y=−3 x−5 y=5 y=−2 x−3−5 y=−x+5−5 y−5=−x+5−5 y=1 5−1 Find the slope and intercept of each line.y=−2 x−3 y=1 5−1 m=−2 m=1 5 b=−3 b=−1 Since the slopes are different, the lines intersect.
A system of equations whose graphs are intersect has 1 solution and is consistent and independent.
Example 4.2.17
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ {y=−2 x−4 4 x+2 y=9 ⓑ {3 x+2 y=2 2 x+y=1
Answer
ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Example 4.2.18
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ {y=1 3 x−5 x−3 y=6 ⓑ {x+4 y=12−x+y=3
Answer
ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.
Solve a System of Equations by Substitution
We will now solve systems of linear equations by the substitution method.
We will use the same system we used first for graphing.
{2 x+y=7 x−2 y=6
We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.
Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!
After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.
Example 4.2.19: How to Solve a System of Equations by Substitution
Solve the system by substitution: {2 x+y=7 x−2 y=6
Answer
Example 4.2.20
Solve the system by substitution: {−2 x+y=−11 x+3 y=9
Answer
(6,1)
Example 4.2.21
Solve the system by substitution: {2 x+y=−1 4 x+3 y=3
Answer
(−3,5)
SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.
Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into either of the original equations to find the other variable.
Write the solution as an ordered pair.
Check that the ordered pair is a solution to both original equations.
Be very careful with the signs in the next example.
Example 4.2.22
Solve the system by substitution: {4 x+2 y=4 6 x−y=8
Answer
We need to solve one equation for one variable. We will solve the first equation for y.
Solve the first equation for y.
Substitute −2 x+2 for y in the second equation.
Replace the y with −2 x+2.
Solve the equation for x.
Substitute x=54 into 4 x+2 y=4 to find y.
The ordered pair is (54,−12).
Check the ordered pair in both equations.
The solution is (54,−12).
Example 4.2.23
Solve the system by substitution: {x−4 y=−4−3 x+4 y=0
Answer
(2,32)
Example 4.2.24
Solve the system by substitution: {4 x−y=0 2 x−3 y=5
Answer
(−12,−2)
Solve a System of Equations by Elimination
We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.
The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.
The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.
For any expressions a, b, c, and d.
if a=b and c=d then a+c=b+d.
To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.
Notice how that works when we add these two equations together:
{3 x+y=5 2 x−y=0 _
5 x=5
The y’s add to zero and we have one equation with one variable.
Let’s try another one:
{+4 y=2 2 x+5 y=−2
This time we don’t see a variable that can be immediately eliminated if we add the equations.
But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.
Then rewrite the system of equations.
Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.
Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.
Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.
Exercise 4.2.25: How to Solve a System of Equations by Elimination
Solve the system by elimination: {2 x+y=7 x−2 y=6
Answer
Exercise 4.2.26
Solve the system by elimination: {3 x+y=5 2 x−3 y=7
Answer
(2,−1)
Exercise 4.2.27
Solve the system by elimination: {4 x+y=−5−2 x−2 y=−2
Answer
(−2,3)
The steps are listed here for easy reference.
SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.
Write both equations in standard form. If any coefficients are fractions, clear them.
Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites.
Add the equations resulting from Step 2 to eliminate one variable.
Solve for the remaining variable.
Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Write the solution as an ordered pair.
Check that the ordered pair is a solution to both original equations.
Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.
Exercise 4.2.28
Solve the system by elimination: {4 x−3 y=9 7 x+2 y=−6
Answer
In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.
Both equations are in standard form.
To get opposite coefficients of y, we will
multiply the first equation by 2 and the
second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x.
Substitute x=0x=0 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.The ordered pair is (0,−3).
Check that the ordered pair is a solution to
both original equations.
The solution is (0,−3).
Exercise 4.2.29
Solve the system by elimination: {3 x−4 y=−9 5 x+3 y=14
Answer
(1,3)
Exercise 4.2.30
Solve each system by elimination: {7 x+8 y=4 3 x−5 y=27
Answer
(4,−3)
When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.
Exercise 4.2.31
Solve the system by elimination: {x+1 2 y=6 3 2 x+2 3 y=17 2
Answer
In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.
To clear the fractions, multiply each
equation by its LCD.
Simplify.
Now we are ready to eliminate one
of the variables. Notice that both equations are in
standard form.
We can eliminate y by multiplying the top equation by −4.
Simplify and add.
Substitute x=3 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.The ordered pair is (3,6).
Check that the ordered pair is a solution to
both original equations.
The solution is (3,6).
Exercise 4.2.32
Solve each system by elimination: {1 3 x−1 2 y=1 3 4 x−y=5 2
Answer
(6,2)
Exercise 4.2.33
Solve each system by elimination: {x+3 5 y=−1 5−1 2 x−2 3 y=5 6
Answer
(1,−2)
When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.
The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.
Exercise 4.2.34
Solve the system by elimination: {3 x+4 y=12 y=3−3 4 x
Answer
{3 x+4 y=12 y=3−3 4 x Write the second equation in standard form.{3 x+4 y=12 3 4 x+y=3 Clear the fractions by multiplying the second equation by 4.{3 x+4 y=12 4(3 4 x+y)=4(3)Simplify.{3 x+4 y=12 3 x+4 y=12 To eliminate a variable, we multiply the second equation by−1. Simplify and add.{3 x+4 y=12−3 x−4 y=−12 _ 0=0
This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.
After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.
Exercise 4.2.35
Solve the system by elimination: {5 x−3 y=15 5 y=−5+5 3 x
Answer
infinitely many solutions
Exercise 4.2.36
Solve the system by elimination: {x+2 y=6 y=−1 2 x+3
Answer
infinitely many solutions
Choose the Most Convenient Method to Solve a System of Linear Equations
When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.
Choose the Most Convenient Method to Solve a System of Linear Equations Graphing _ Substitution _ Elimination _ Use when you need a Use when one equation is Use when the equations a picture of the situation.already solved or can be rein standard form.easily solved for one variable.
Example 4.2.37
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ {3 x+8 y=40 7 x−4 y=−32 ⓑ {5 x+6 y=12 y=2 3 x−1
Answer
ⓐ
{3 x+8 y=40 7 x−4 y=−32
Since both equations are in standard form, using elimination will be most convenient.
ⓑ
{5 x+6 y=12 y=2 3 x−1
Since one equation is already solved for y, using substitution will be most convenient.
Example 4.2.38
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ {4 x−5 y=−32 3 x+2 y=−1 ⓑ {x=2 y−1 3 x−5 y=−7
Answer
ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x, using substitution will be most convenient.
Example 4.2.39
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ {y=2 x−1 3 x−4 y=−6 ⓑ {6 x−2 y=12 3 x+7 y=−13
Answer
ⓐ Since one equation is already solved for y, using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.
Key Concepts
How to solve a system of linear equations by graphing.
Graph the first equation.
Graph the second equation on the same rectangular coordinate system.
Determine whether the lines intersect, are parallel, or are the same line.
Identify the solution to the system.
If the lines intersect, identify the point of intersection. This is the solution to the system.
If the lines are parallel, the system has no solution.
If the lines are the same, the system has an infinite number of solutions.
5. Check the solution in both equations.
How to solve a system of equations by substitution.
Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into either of the original equations to find the other variable.
Write the solution as an ordered pair.
Check that the ordered pair is a solution to both original equations.
How to solve a system of equations by elimination.
Write both equations in standard form. If any coefficients are fractions, clear them.
Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites.
3. Add the equations resulting from Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
6. Write the solution as an ordered pair.
7. Check that the ordered pair is a solution to both original equations. Choose the Most Convenient Method to Solve a System of Linear Equations Graphing _ Substitution _ Elimination _ Use when one equation is Use when you need a already solved or can be Use when the equations a picture of the situation.easily solved for one rein standard form.variable.
Glossary
coincident linesCoincident lines have the same slope and same y-intercept.consistent and inconsistent systemsConsistent system of equations is a system of equations with at least one solution; inconsistent system of equations is a system of equations with no solution.solutions of a system of equationsSolutions of a system of equations are the values of the variables that make all the equations true; solution is represented by an ordered pair (x,y).(x,y).system of linear equationsWhen two or more linear equations are grouped together, they form a system of linear equations.
This page titled 4.2: Solve Systems of Linear Equations with Two Variables is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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4.1: Introduction
4.2E: Exercises
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13506 | https://books.google.com/books/about/Matching_Theory.html?id=OaoJBAAAQBAJ | Matching Theory - László Lovász, M. D. Plummer - Google Books
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Matching Theory =============== László Lovász, M. D. Plummer American Mathematical Soc., 2009 - Mathematics - 547 pages This book surveys matching theory, with an emphasis on connections with other areas of mathematics and on the role matching theory has played, and continues to play, in the development of some of these areas. Besides basic results on the existence of matchings and on the matching structure of graphs, the impact of matching theory is discussed by providing crucial special cases and nontrivial examples on matroid theory, algorithms, and polyhedral combinatorics. The new Appendix outlines how the theory and applications of matching theory have continued to develop since the book was first published in 1986, by launching (among other things) the Markov Chain Monte Carlo method. More » Preview this book »
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Title Page
Table of Contents
Index
References
Contents
Matchings in bipartite graphs 1
3 18
4 29
0 41
3 76
Size and structure 83
0 121
2 127
1 266
2 273
4 285
5 291
4 330
6 345
Introduction 357
4 379
More
3 135
0 143
2 150
3 159
4 174
Some graphtheoretical problems 213
3 220
4 228
6 243
Matching and linear programming 255
Matroid matching 415
2 426
Vertex packing and covering 443
4 471
Developments in matching theory since 483
Properties of random matchings 483
A7 2 483
445 515
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Matching Theory
M.D. Plummer,L. Lovász
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László Lovász,Michael D. Plummer
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László Lovász,M. D. Plummer
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1-extendable graph2-matchingadjacentalgorithmantichainaugmenting pathbicritical graphsbigraphbipartite matchingC₁calledChaptercharacterizationchromatic indexClaimcomponents of G-ScontradictionCorollarydef(G)
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Title Matching Theory
Volume 367 of AMS Chelsea Publishing Series
Volume 29 of Annals of discrete mathematics
Volume 121 of North-Holland mathematics studies
AuthorsLászló Lovász, M. D. Plummer
Edition illustrated, reprint
Publisher American Mathematical Soc., 2009
ISBN 0821847597, 9780821847596
Length 547 pages
SubjectsMathematics
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13507 | https://networks.h-net.org/group/reviews/20122807/vasile-grant-and-grant-40-years-evolution-darwins-finches-daphne-major | Vasile on Grant and Grant, '40 Years of Evolution: Darwin's Finches on Daphne Major Island' | H-Net
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Vasile on Grant and Grant, '40 Years of Evolution: Darwin's Finches on Daphne Major Island'
Grant, Peter R.; Grant, B. Rosemary. 40 Years of Evolution: Darwin's Finches on Daphne Major Island. Princeton: Princeton University Press, 2024. 426 pp. ISBN 9780691263229.
Reviewed by Monica Vasile (Oulu University)
Published on H-Environment (August, 2025)
Commissioned by Daniella McCahey (Texas Tech University)
Printable Version:
40 Years of Evolution: Darwin ’ s Finches on Daphne Major Island, first published in 2014 and reissued in a revised second edition in 2024, is Peter and Rosemary Grant’s detailed account of their decades-long field study on the Galápagos. It tells how, from 1973 to 2012, they tracked every finch on the tiny island of Daphne Major—measuring, banding, observing—and what those years revealed about evolution in real time. The book lays out their findings and why they matter, showing how natural selection, competition, and chance events shape species, often far faster than textbooks suggest.
Peter and Rosemary Grant are known for their groundbreaking work on evolution in real time. Over four decades, they turned these small birds, famously dubbed “Darwin’s finches,” into icons of evolutionary science. Year after year, the Grants braved the barren, rocky island to study species of ground finches in the genus Geospiza. They lived simply in tents, with no fresh water or shelter beyond what they carried. They caught thousands of finches, weighing them, measuring beaks and wings with calipers, and banding each bird for identification. They tracked changes in beak size, body weight, and survival patterns. They monitored nests, recorded mating patterns, and counted every egg and chick. They even tracked seed abundance and size to link food availability with survival and beak adaptation. And so, they caught evolution in action. During droughts and rainy years, they documented how shifts in food availability drove rapid natural selection. Within just a few generations, the average beak size visibly changed, a textbook example of evolution happening on a human timescale. Their research confirmed Darwin’s theories and showed that evolution is far more dynamic and responsive than once thought.
But this book is more than a record of life on Daphne Major. It is an account of the scientific questions and findings that emerged from decades of observation. The book is organized in six parts. Part 1, “Early Problems, Early Solutions” (chapters 1–5), revisits speciation, finch morphology, heritability, selection, and breeding ecology. The Grants introduce the medium ground finch, Geospiza fortis, and the peculiar ecological conditions of the island. Part 2, “Species Interactions” (chapters 6–9), tracks patterns impossible to see in short-term studies, such as competition, character displacement, and the gradual consequences of hybridization and gene flow. Part 3, “Developing a Long-Term Perspective” (chapters 10–13), traces long-term hybridization trends, selection patterns, and speciation processes. Here, the narrative builds toward the striking discovery of the “Big Bird” lineage, showing how hybridization events, once thought evolutionary dead ends, can lead to new distinct lineages. In part 4, “Syntheses” (chapters 14–16), the Grants step back to ask bigger questions about predictability in evolution, the role of rare events, and the broader lessons their four decades of work offer for evolutionary biology today. Part 5, “Looking Forward,” is composed of only one chapter, about the future of the finches on Daphne and includes reflections on the value of long-term research . Part 6, “From the Daphne Microcosm to the Galapagos Macrocosm,” is a very short summary and reflection piece to conclude the book.
The book reads as a scientific saga. It begins with a question: Why is the medium ground finch, Geospiza fortis, exceptionally small on Daphne? That question, as they put it, was the magnet that first drew them to the island. But the plot soon thickened. The arrival and establishment of another species, Geospiza magnirostris, opened up new issues. These birds’ success raised questions: How did this large-beaked finch settle, and why did it thrive? More importantly, how did their presence, as competition, alter the evolutionary path of fortis, the small finch that first caught the Grants’ attention? Other empirical puzzles guided the research: Why did fluctuations in beak size occur? How strong was selection, and what caused it? To what extent were these changes genetically based?
Yet, more than a case study of these birds, the book is structured around the big questions that animated the Grants’ work: How do species form? How does natural selection operate in wild populations? Can we witness evolutionary change over a human lifetime? How important are rare and chance events in the history of species? Across its chapters, the Grants unpack these questions with data, photos, charts, clear narrative, and personal anecdotes. 40 Years of Evolution is both a rigorous scientific report and a reflection on the messiness, unpredictability, and wonder of evolution happening under our noses.
As a historian of wildlife conservation, I approached this book expecting it to focus more on how the Grants carried out their work, the practicalities, the challenges, the human side of long-term field research. Instead, it turned out to be a full-scale scientific monograph, laying out the findings themselves in all their technical detail. At times, I found myself stretched as a reader. I am used to working with scientific literature and have taught myself the basics of how to read and make sense of it, but this book pushed me, at times, beyond familiar ground. That said, the writing is remarkably clear, and the explanations accessible even to those without specialist training. The book brought the complexity of evolutionary processes into focus without losing this reader.
For me, the fascination of this book lies in its scientific insights, but also in how it captures the evolution of evolutionary research itself. So first, I will share the key insights that stood out to me.
In their concluding chapters, the Grants ask: How predictable is evolution? Their answer is thought-provoking: Evolution is genetically predictable, but environmentally unpredictable, driven by climatic fluctuations that defy reliable forecasting. They note that extreme and rare events may be underappreciated for their ecological and evolutionary importance, precisely because detecting them depends on long-term studies, and luck. Thus, evolution is not necessarily steady but can be punctuated by surprise.
Another major takeaway is that hybridization is not always a footnote or a “mistake” in the process of speciation, it can be a creative evolutionary force. The Grants witnessed how the so-called Big Bird lineage, which arose from a hybridization event, established itself on the island.
Perhaps most importantly for historians of science, the Grants make a strong case for the epistemic value of long-term studies. Their four decades on Daphne Major produced insights no short-term project could have captured. Much of what we know about evolution, they note, comes from just a few field studies that have accumulated data over many years. They argue for persistence, to keep observing, and for continuity, to interpret what happens. Enlightenment, they suggest, does not come all at once but builds gradually, as data accumulate. At the same time, rare, singular events, unpredictable droughts, sudden population swings, can offer key insights. But only if someone is there long enough to see them.
The book also captures how the science itself has changed. When the Grants began in 1973, there were no personal computers, no advanced statistical models, and no molecular tools. In this sense, the second edition, published in 2024, reflects a notable shift. It shows how the genomic revolution has reshaped evolutionary biology. Where the first edition traced ecological and morphological change, this new version connects those changes directly to specific genes, offering a more complete picture.
This brings me to the question: How does the Grants’ work complement Darwin’s? Darwin made astute field observations and had brilliant ideas; his bold conceptual leaps transformed our understanding of life, his insights still holding firm today. But he lacked direct evidence for evolution as it happens. And this is what Rosemary and Peter Grant bring. They turned evolution into a statistical science, bringing precision and measurement. Their work refined, quantified, and expanded Darwin’s legacy.
All in all, the book is a fascinating read that rewards patience. It tells a compelling story, and offers much for historians to work with.
Citation: Monica Vasile. Review of Grant, Peter R.; Grant, B. Rosemary. 40 Years of Evolution: Darwin's Finches on Daphne Major Island. H-Environment, H-Net Reviews. August, 2025.
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13508 | https://math.stackexchange.com/questions/3835121/finding-cube-roots-of-a-unity-proper-explanation-is-needed | complex numbers - Finding cube roots of a unity - proper explanation is needed - Mathematics Stack Exchange
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Finding cube roots of a unity - proper explanation is needed
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SQUARE ROOT
Let's say that I have:
x 2=9 x 2=9
And I x−−√x an entire equation to get:
x 1,2 x 1 x 2=9–√=+3=−3 x 1,2=9 x 1=+3 x 2=−3
Experience and "fundamental theorem of algebra" thought me that:
There are always two square roots (x 1 x 1, x 2 x 2)!
Experience thought me that:
Product of mixed roots doesn't return back the original:
x 1⋅x 2 x 2⋅x 1=−3⋅3=−9=3⋅(−3)=−9 x 1⋅x 2=−3⋅3=−9 x 2⋅x 1=3⋅(−3)=−9
Experience also thought me that:
Only if roots are multiplied by itself they return back the result:
x 1⋅x 1 x 2⋅x 2=3⋅3=9=(−3)⋅(−3)=9 x 1⋅x 1=3⋅3=9 x 2⋅x 2=(−3)⋅(−3)=9
CUBE ROOT
Then there comes this:
x 3=1 x 3=1
If I x−−√x an entire equation I get:
x 1,2,3 x 1 x 2 x 3=1–√3=1=???=???x 1,2,3=1 3 x 1=1 x 2=???x 3=???
Here I have no experience whatsoever but "fundamental theorem of algebra" says that:
There are always three cube roots (x 1 x 1, x 2 x 2, x 3 x 3)!
Now pay attention! Like for square root it should be true that:
If roots x 1 x 1, x 2 x 2, x 3 x 3 are multiplied by itself they return 1 1. Their mixed products will probably return something else.
x 1⋅x 1⋅x 1=1 x 2⋅x 2⋅x 2=1 x 3⋅x 3⋅x 3=1 x 1⋅x 1⋅x 1=1 x 2⋅x 2⋅x 2=1 x 3⋅x 3⋅x 3=1
So at this point everybody's already rushing to their paper and drawing this complex plane:
And in complex plane:
If we multiply complex numbers, length of their position vectors multiply and their angles sum up.
So everybody will now draw x 1 x 1 on the R R axis and tell me that this is the first root. I agree with that. Then they say:
Length of a position vector of other two roots (x 2 x 2, x 2 x 2) must be 1 1 as well.
And I agree with that. But what they say now, I can't agree with no matter how I look at it. They say:
Because according to "fundamental theorem of algebra" there are three cube roots, and we will multiply them in order to get to 1 1, angle between them must be 120∘120∘. We now start at 1 1 and if we do a full circle we end at 1 1 again. Woila! Problem solved...
I can't agree with this explanation as is because this only holds for x 1 x 1. If I take x 1 x 1 and make a full circle I end up getting 1 1. But with x 2 x 2 and x 3 x 3 my starting point is not 1 1. Starting points are actually:
starting point for x 2:−1 2−3–√2 i starting point for x 3:−1 2+3–√2 i starting point for x 2:−1 2−3 2 i starting point for x 3:−1 2+3 2 i
And in case of x 2 x 2 and x 3 x 3 when we do a full circle we don't end up getting 1 1, but we geometricaly looking end at positions:
−1/2−3–√/2 i−1/2+3–√/2 i−1/2−3/2 i−1/2+3/2 i
If the starting positions are as I state they are, angle that we need in order to get to the 1 1 should be ±240/3=±80∘±240/3=±80∘.
But then why does this explanation that they use, work? At the end of the day e.g. x 2 x 2 solves the problem:
(x 2)3(−1 2+3–√2 i)3(−1 2+3–√2 i)(−1 2+3–√2 i)(−1 2+3–√2 i)(1 4−3–√4 i−3–√4 i+3 4 i 2)(−1 2+3–√2 i)(1 4−2 3–√4 i−3 4)(−1 2+3–√2 i)(−1 2−3–√2 i)(−1 2+3–√2 i)1 4−3–√4 i+3–√4 i−3 4 i 2 1 4+3 4=1=1=1=1=1=1=1=1(x 2)3=1(−1 2+3 2 i)3=1(−1 2+3 2 i)(−1 2+3 2 i)(−1 2+3 2 i)=1(1 4−3 4 i−3 4 i+3 4 i 2)(−1 2+3 2 i)=1(1 4−2 3 4 i−3 4)(−1 2+3 2 i)=1(−1 2−3 2 i)(−1 2+3 2 i)=1 1 4−3 4 i+3 4 i−3 4 i 2=1 1 4+3 4=1
I am sure there is something wrong with the explanation everybody are using but it is correct to assume roots are 120∘120∘ appart...
Probably the point here is not about making a full circle at all! Like most of the people say. It must be about adding conjugate pairs so that artifficialy we create enough roots to satisfy fundamental theorem of algebra! Thats all! So odd number root 3 3 was historicaly first one that forced us to get out of the horizontal axis in complex plane and invent complex roots...
Can anyone present a proper explanation of why we choose 120∘120∘ for cube roots?
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edited Sep 22, 2020 at 3:46
71GA71GA
asked Sep 21, 2020 at 20:50
71GA71GA
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In the first case, x 2⋅x 2=9 x 2⋅x 2=9, not −9−9.user9464 –user9464 2020-09-21 20:53:42 +00:00 Commented Sep 21, 2020 at 20:53
This is pretty much impossible to understand why 120∘120∘ without using polar form. Are you familiar with it?Olivier Moschetta –Olivier Moschetta 2020-09-21 20:54:10 +00:00 Commented Sep 21, 2020 at 20:54
For the square root, technically it's x 1,2 x 1 x 2=±9–√=+3=−3 x 1,2=±9 x 1=+3 x 2=−3 The square root by definition returns the positive solution. Which is to say it always gives only x 1 x 1. Never x 2 x 2, never both.Arthur –Arthur 2020-09-21 20:58:44 +00:00 Commented Sep 21, 2020 at 20:58
@Arthur: which is the positive square root of −1−1? Or of −i−i?Henry –Henry 2020-09-21 21:04:48 +00:00 Commented Sep 21, 2020 at 21:04
math.stackexchange.com/questions/192742/how-to-solve-x3-1lab bhattacharjee –lab bhattacharjee 2020-09-21 21:08:04 +00:00 Commented Sep 21, 2020 at 21:08
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There is really no "first" root, "second" root, etc.
In the case of x 2=1 x 2=1, there are two roots, −1−1 and 1 1. It is your choice to give these two roots whatever names you like.
In the case of x 3=1 x 3=1, there are three distinct roots. Again, it does not matter which one you call x 1 x 1.
In general, (complex) solutions to the equation x n=1 x n=1 together with complex multiplication form a group:
This group tells you what happens when you multiply two solutions to the equations x n=1 x n=1.
I don't know where the "everybody explanation" is from; I would not do that anyway.
One way to find the three roots for x 3=1 x 3=1 is quite straightforward. Write x=r e i θ x=r e i θ where r≥0 r≥0 and θ∈[0,2 π)θ∈[0,2 π). Then
(r e i θ)3=1(r e i θ)3=1
Taking the absolute value on both sides you get r=1 r=1. On the other hand,
e i(3 θ)=1 e i(3 θ)=1
which implies that 3 θ=2 k π 3 θ=2 k π where k k is some integer. Now write
θ=2 k π 3,k∈Z θ=2 k π 3,k∈Z
and pick the ones that are in [0,2 π)[0,2 π). By periodicity, this is not a unique way to represent the roots. You could, for example, restrict θ θ to [−π,π)[−π,π) at the beginning of the calculation.
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edited Sep 22, 2020 at 13:35
answered Sep 21, 2020 at 21:00
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13509 | https://www.cuemath.com/jee/basis-of-a-two-dimensional-plane-vectors/ | Basis Of A Two Dimensional Plane | What is Basis Of A Two Dimensional Plane -Examples & Solutions | Cuemath
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THE BASIS OF A VECTOR SPACE
Consider a two-dimensional plane, and any two arbitrary non-collinear vectors→a and→b→a and→b in this plane. We make the two vectors co-initial and use their supports as our reference axes:
Observe carefully that any vector→r→r in the plane can be represented in terms of→a and→b→a and→b . We find the components of→r→r along the directions of →a and→b→a and→b; those components must be some scalar multiples of→a and→b→a and→b .
Thus, any vector→r→r in the plane can be written as
→r=λ→a+μ→b f o r s o m e λ,μ∈R...(1)→r=λ→a+μ→b f o r s o m e λ,μ∈R...(1)
i.e, any vector→r→r in the plane can be expressed as a linear combination of→a and→b→a and→b.
We state this fact in mathematical terms as follows: the vectors →a and→b→a and→b form a basis of our vector space (which is a plane in this case). The term “basis” means that using only→a and→b→a and→b , we can construct any vector lying in the plane of→a and→b→a and→b .
Note that there’s nothing special about→a and→b→a and→b ; any two non-collinear vectors can form a basis for the plane.
You must be very clear on the point that two collinear vectors cannot form the basis for a plane while any two non-collinear vectors can. Understanding this fact is very crucial to later discussions.
Try proving this: let→a and→b→a and→b form the basis of a plane. For any vector →r→r in the plane of→a and→b→a and→b , we can find scalars λ,μ∈R λ,μ∈R such that
→r=λ→a+μ→b→r=λ→a+μ→b
Prove that this representation is unique.
The basic principle that we’ve learnt in this discussion can be expressed in a very useful way as follows:
Three vectors are coplanar if and only if one of them can be expressed as a linear combination of the other two. i.e., three vectors →a,→b,→c→a,→b,→c are coplanar if there exist scalars l 1,l 2∈R l 1,l 2∈R such that
→a=l 1→b+l 2→c→a=l 1→b+l 2→c
We can write this as
(1)→a+(−l 1)→b+(−l 2)→c=→0→a+(−l 1)→b+(−l 2)→c=→0
⇒λ→a+μ→b+γ→c=0
This form equivalently tells us that three vectors are coplanar if we can find three scalars λ,μ,γ∈R for which their linear combination is zero.
Example – 4
Suppose that for three non-zero vectors→a,→b,→c, any two of them are non-collinear. If the vectors(→a+2→b)and→c are collinear and the vectors(→b+3→c)and→a are collinear, prove that
→a+2→b+6→c=→0
Solution: We must have some λ,μ∈R such that
→a+2→b=λ→c...(1)→b+3→c=μ→a...(2)
From (1), we have
→c=1 λ(→a+2→b)...(3)
We use this in (2) :
→b+3 λ(→a+2→b)=μ→a
⇒(3 λ−μ)→a+(1+6 λ)→b=→0
Since →a and→b are non-collinear, their linear combination can be zero if and only if the two scalars are zero.
This gives
3 λ−μ=0
1+6 λ=0
⇒λ=−6,μ=−1 2
Using the value λ of in (3), we have
→a+2→b+6→c=0
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13510 | https://ncse.ngo/files/pub/creationism/icons/gishlick_icons1.pdf | THE MILLER-UREY EXPERIMENT THE EXPERIMENT ITSELF T he understanding of the origin of life was largely speculative until the 1920s, when Oparin and Haldane, working independently, proposed a theoretical model for “chemical evolution.” The Oparin– Haldane model suggested that under the strongly reducing conditions theorized to have been present in the atmosphere of the early earth (between 4.0 and 3.5 billion years ago), inorganic molecules would spontaneously form organic molecules (simple sugars and amino acids). In 1953, Stanley Miller, along with his graduate advisor Harold Urey, tested this hypothesis by constructing an apparatus that simulated the Oparin-Haldane “early earth.” When a gas mixture based on predic-tions of the early atmosphere was heated and given an electrical charge, organic compounds were formed (Miller, 1953; Miller and Urey, 1959). Thus, the Miller-Urey experiment demonstrated how some biological molecules, such as simple amino acids, could have arisen abiotically, that is through non-biological processes, under conditions thought to be sim-ilar to those of the early earth. This experiment provided the structure for later research into the origin of life. Despite many revisions and additions, the Oparin–Haldane scenario remains part of the model in use today. The Miller–Urey experiment is simply a part of the experimental program produced by this para-digm.
WELLS BOILS OFF W ells says that the Miller–Urey exper-iment should not be taught because the experiment used an atmospheric composition that is now known to be incorrect.
Wells contends that textbooks don’t discuss how the early atmosphere was probably differ-ent from the atmosphere hypothesized in the original experiment. Wells then claims that the actual atmosphere of the early earth makes the Miller–Urey type of chemical synthesis impossible, and asserts that the experiment does not work when an updated atmosphere is used. Therefore, textbooks should either dis-cuss the experiment as an historically interest-ing yet flawed exercise, or not discuss it at all.
Wells concludes by saying that textbooks should replace their discussions of the Miller– Urey experiment with an “extensive discus-sion” of all the problems facing research into the origin of life. These allegations might seem serious; how-ever, Wells’s knowledge of prebiotic chemistry is seriously flawed. First, Wells’s claim that researchers are ignoring the new atmospheric data, and that experiments like the Miller– Urey experiment fail when the atmospheric composition reflects current theories, is simply false. The current literature shows that scien-tists working on the origin and early evolution of life are well aware of the current theories of the earth’s early atmosphere and have found that the revisions have little effect on the results of various experiments in biochemical synthesis. Despite Wells’s claims to the con-trary, new experiments since the Miller–Urey ones have achieved similar results using vari-ous corrected atmospheric compositions (Figure 1; Rode, 1999; Hanic et al., 2000).
Further, although some authors have argued that electrical energy might not have efficient-ly produced organic molecules in the earth’s early atmosphere, other energy sources such as cosmic radiation (e.g., Kobayashi et al., 1998), high temperature impact events (e.g., Miyakawa et al., 2000), and even the action of waves on a beach (Commeyras et al., 2002) would have been quite effective. Even if Wells had been correct about the Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 3 Miller–Urey experiment, he does not explain that our theories about the origin of organic “building blocks” do not depend on that exper-iment alone (Orgel, 1998a). There are other sources for organic “building blocks,” such as meteorites, comets, and hydrothermal vents.
All of these alternate sources for organic mate-rials and their synthesis are extensively dis-cussed in the literature about the origin of life, a literature that Wells does not acknowledge.
In fact, what is most striking about Wells’s extensive reference list is the literature that he has left out. Wells does not mention extrater-restrial sources of organic molecules, which have been widely discussed in the literature since 1961 (see Oró, 1961; Whittet, 1997; Irvine, 1998). Wells apparently missed the vast body of literature on organic compounds in comets (e.g. Oró, 1961; Anders, 1989; Irvine, 1998), carbonaceous meteorites (e.g., Kaplan et al., 1963; Hayes, 1967; Chang, 1994; Maurette, 1998; Cooper et al., 2001), and con-ditions conducive to the formation of organic compounds that exist in interstellar dust clouds (Whittet, 1997). Wells also fails to cite the scientific litera-ture on other terrestrial conditions under which organic compounds could have formed. These non-atmospheric sources include the synthesis of organic compounds in a reducing ocean Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 4 Researcher(s) Year Reactants Energy source Results Probability Miller 1953 CH4, NH3, H2O, H2 Electric discharge Simple amino acids, unlikely organic compounds Abelson 1956 CO, CO2, N2, NH3, H2, Electric discharge Simple amino acids, unlikely H2O HCN Groth and Weyssenhoff 1957 CH4, NH3, H2O Ultraviolet light Simple amino acids (low under special conditions (1470–1294 ?) yields) Bahadur, et al.
1958 Formaldehyde, Sunlight Simple amino acids possible molybdenum oxide (photosynthesis) Pavolvskaya and 1959 Formaldehyde, nitrates High pressure Hg lamp Simple amino acids possible Pasynskii (photolysis) Palm and Calvin 1962 CH4, NH3, H2O Electron irradiation Glycine, alanine, aspartic under special conditions acid Harada and Fox 1964 CH4, NH3, H2O Thermal energy 14 of the “essential” under special conditions (900–1200º C) amino acids of proteins Oró 1968 CH4, NH3, H2O Plasma jet Simple amino acids unlikely Bar-Nun et al. 1970 CH4, NH3, H2O Shock wave Simple amino acids under special conditions Sagan and Khare 1971 CH4, C2H6, NH3, H2O, Ultraviolet light (>2000 Simple amino acids (low under special conditions H2S ?) yields) Yoshino et al.
1971 H2, CO, NH3, Temperature of 700°C Glycine, alanine, unlikely montmorillonite glutamic acid, serine, aspartic acid, leucine, lysine, arginine Lawless and Boynton 1973 CH4, NH3, H2O Thermal energy Glycine, alanine, aspartic under special conditions acid, ?-alanine, N-methyl-?-alanine, ?-amino-n-butyric acid.
Yanagawa et al.
1980 Various sugars, Temperature of 105°C Glycine, alanine, serine, under special conditions hydroxylamine, aspartic acid, glutamic inorganic salts, acid Kobayashi et al.
1992 CO, N2, H2O Proton irradiation Glycine, alanine, aspartic possible acid, ?-alanine, glutamic acid, threonine, ?-aminobutyric acid, serine Hanic, et al. 1998 CO2, N2, H2O Electric discharge Several amino acids possible Figure 1. A table of some amino acid synthesis experiments since Miller–Urey. The “probabili-ty” column reflects the likelihood of the environmental conditions used in the experiment.
Modified from Rode, 1999.
(e.g., Chang, 1994), at hydrothermal vents (e.g., Andersson, 1999; Ogata et al., 2000), and in volcanic aquifers (Washington, 2000). A cursory review of the literature finds more than 40 papers on terrestrial prebiotic chemical syn-thesis published since 1997 in the journal Origins of life and the evolution of the bios-phere alone. Contrary to Wells’s presentation, there appears to be no shortage of potential sources for organic “building blocks” on the early earth.
Instead of discussing this literature, Wells raises a false “controversy” about the low amount of free oxygen in the early atmos-phere. Claiming that this precludes the sponta-neous origin of life, he concludes that “[d]ogma had taken the place of empirical sci-ence” (Wells, 2000:18). In truth, nearly all researchers who work on the early atmosphere hold that oxygen was essentially absent during the period in which life originated (Copley, 2001) and therefore oxygen could not have played a role in preventing chemical synthesis.
This conclusion is based on many sources of data, not “dogma.” Sources of data include fluvial uraninite sand deposits (Rasmussen and Buick, 1999) and banded iron formations (Nunn, 1998; Copley, 2001), which could not have been deposited under oxidizing condi-tions. Wells also neglects the data from pale-osols (ancient soils) which, because they form at the atmosphere–ground interface, are an excellent source to determine atmospheric composition (Holland, 1994). Reduced pale-osols suggest that oxygen levels were very low before 2.1 billion years ago (Rye and Holland, 1998). There are also data from mantle chem-istry that suggest that oxygen was essentially absent from the earliest atmosphere (Kump et al., 2001). Wells misrepresents the debate as over whether oxygen levels were 5/100 of 1%, which Wells calls “low,” or 45/100 of 1%, which Wells calls “significant.” But the con-troversy is really over why it took so long for oxygen levels to start to rise. Current data show that oxygen levels did not start to rise significantly until nearly 1.5 billion years after life originated (Rye and Holland, 1998; Copley, 2001). Wells strategically fails to clar-ify what he means by “early” when he discuss-es the amount of oxygen in the “early” atmos-phere. In his discussion, he cites research about the chemistry of the atmosphere without distinguishing whether the authors are refer-ring to times before, during, or after the period when life is thought to have originated. Nearly all of the papers he cites deal with oxygen lev-els after 3.0 billion years ago. They are irrele-vant, as chemical data suggest that life arose 3.8 billion years ago (Chang, 1994; Orgel, 1998b), well before there was enough free oxygen in the earth’s atmosphere to prevent Miller–Urey-type chemical synthesis.
Finally, the Miller–Urey experiment tells us nothing about the other stages in the origin of life, including the formation of a simple genet-ic code (PNA or “peptide”-based codes and RNA-based codes) or the origin of cellular membranes (liposomes), some of which are discussed in all the textbooks that Wells reviewed. The Miller–Urey experiment only showed one possible route by which the basic components necessary for the origin of life could have been created, not how life came to be. Other theories have been proposed to bridge the gap between the organic “building blocks” and life. The “liposome” theory deals with the origin of cellular membranes, the RNA-world hypothesis deals with the origin of a simple genetic code, and the PNA (peptide-based genetics) theory proposes an even sim-pler potential genetic code (Rode, 1999). Wells doesn’t really mention any of this except to suggest that the “RNA world” hypothesis was proposed to “rescue” the Miller–Urey experi-ment. No one familiar with the field or the evi-Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 5 dence could make such a fatuous and inaccu-rate statement. The Miller–Urey experiment is not relevant to the RNA world, because RNA was constructed from organic “building blocks” irrespective of how those compounds came into existence (Zubay and Mui, 2001).
The evolution of RNA is a wholly different chapter in the story of the origin of life, one to which the validity of the Miller–Urey experi-ment is irrelevant.
WHAT THE TEXTBOOKS SAY A ll of the textbooks reviewed contain a section on the Miller–Urey experi-ment. This is not surprising given the experiment’s historic role in the understanding of the origin of life. The experiment is usually discussed over a couple of paragraphs (see Figure 2), a small proportion (roughly 20%) of the total discussion of the origin and early evo-lution of life. Commonly, the first paragraph discusses the Oparin-Haldane scenario, and then a second outlines the Miller–Urey test of that scenario. All textbooks contain either a drawing or a picture of the experimental appa-ratus and state that it was used to demonstrate that some complex organic molecules (e.g., simple sugars and amino acids, frequently called “building blocks”) could have formed spontaneously in the atmosphere of the early earth. Textbooks vary in their descriptions of the atmospheric composition of the early earth.
Five books present the strongly reducing atmosphere of the Miller–Urey experiment, whereas the other five mention that the current geochemical evidence points to a slightly reducing atmosphere. All textbooks state that oxygen was essentially absent during the peri-od in which life arose. Four textbooks mention that the experiment has been repeated success-fully under updated conditions. Three text-books also mention the possibility of organic molecules arriving from space or forming at deep-sea hydrothermal vents (Figure 2). No textbook claims that these experiments conclu-sively show how life originated; and all text-books state that the results of these experi-ments are tentative.
It is true that some textbooks do not mention that our knowledge of the composition of the atmosphere has changed. However, this does not mean that textbooks are “misleading” stu-dents, because there is more to the origin of life than just the Miller–Urey experiment.
Most textbooks already discuss this fact. The textbooks reviewed treat the origin of life with varying levels of detail and length in “Origin of life” or “History of life” chapters. These chapters are from 6 to 24 pages in length. In this relatively short space, it is hard for a text-book, particularly for an introductory class like high school biology, to address all of the details and intricacies of origin-of-life research that Wells seems to demand. Nearly all texts begin their origin of life sections with a brief description of the origin of the universe and the solar system; a couple of books use a dis-cussion of Pasteur and spontaneous generation instead (and one discusses both). Two text-books discuss how life might be defined.
Nearly all textbooks open their discussion of the origin of life with qualifications about how the study of the origin of life is largely hypo-thetical and that there is much about it that we do not know. WELLS’S EVALUATION A s we will see in his treatment of the other “icons,” Wells’s criteria for judg-ing textbooks stack the deck against them, ensuring failure. No textbook receives better than a D for this “icon” in Wells’s eval-uation, and 6 of the 10 receive an F. This is largely a result of the construction of the grad-ing criteria. Under Wells’s criteria (Wells, 2000:251–252), any textbook containing a pic-Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 6 ture of the Miller–Urey apparatus could receive no better than a C, unless the caption of the picture explicitly says that the experiment is irrelevant, in which case the book would receive a B. Therefore, the use of a picture is the major deciding factor on which Wells eval-uated the books, for it decides the grade irre-spective of the information contained in the text! A grade of D is given even if the text explicitly points out that the experiment used an incorrect atmosphere, as long as it shows a picture. Wells pillories Miller and Levine for exactly that, complaining that they bury the correction in the text. This is absurd: almost all textbooks contain pictures of experimental apparatus for any experiment they discuss. It is the text that is important pedagogically, not the pictures. Wells’s criteria would require that even the intelligent design “textbook” Of Pandas and People would receive a C for its treatment of the Miller–Urey experiment. In order to receive an A, a textbook must first omit the picture of the Miller–Urey appa-ratus (or state explicitly in the caption that it was a failure), discuss the experiment, but then state that it is irrelevant to the origin of life.
This type of textbook would be not only scien-tifically inaccurate but pedagogically defi-cient.
WHY WE SHOULD STILL TEACH MILLER–UREY Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 7 Figure 2. Textbook treatments of the Miller–Urey experiment. Textbooks are listed in order of increasing detail (AP/College textbooks highlighted; note that Futuyma is an upper-level col-lege/graduate textbook).
T he Miller–Urey experiment represents one of the research programs spawned by the Oparin-Haldane hypothesis.
Even though details of the model for the origin of life have changed, this has not affected the basic scenario of Oparin–Haldane. The first stage in the origin of life was chemical evolu-tion. This involves the formation of organic compounds from inorganic molecules already present in the atmosphere and in the water of the early earth. This spontaneous organization of chemicals was spawned by some external energy source. Lightning (as Oparin and Haldane thought), proton radiation, ultraviolet radiation, and geothermal or impact-generated heat are all possibilities.
The Miller–Urey experiment represents a major advance in the study of the origin of life.
In fact, it marks the beginning of experimental research into the origin of life. Before Miller– Urey, the study of the origin of life was mere-ly theoretical. With the advent of “spark exper-iments” such as Miller conducted, our under-standing of the origin of life gained its first experimental program. Therefore, the Miller– Urey experiment is important from an histori-cal perspective alone. Presenting history is good pedagogy because students understand scientific theories better through narratives.
The importance of the experiment is more than just historical, however. The apparatus Miller and Urey designed became the basis for many subsequent “spark experiments” and laid a groundwork that is still in use today. Thus it is also a good teaching example because it shows how experimental science works. It teaches students how scientists use experiments to test ideas about prehistoric, unobserved events such as the origin of life. It is also an interest-ing experiment that is simple enough for most students to grasp. It tested a hypothesis, was reproduced by other researchers, and provided new information that led to the advancement of scientific understanding of the origin of life.
This is the kind of “good science” that we want to teach students.
Finally, the Miller–Urey experiment should still be taught because the basic results are still valid. The experiments show that organic mol-ecules can form under abiotic conditions. Later experiments have used more accurate atmos-pheric compositions and achieved similar results. Even though origin-of-life research has moved beyond Miller and Urey, their experi-ments should be taught. We still teach Newton even though we have moved beyond his work in our knowledge of planetary mechanics.
Regardless of whether any of our current theo-ries about the origin of life turn out to be com-pletely accurate, we currently have models for the processes and a research program that works at testing the models.
HOW TEXTBOOKS COULD IMPROVE THEIR PRESENTATIONS OF THE ORIGIN OF LIFE T extbooks can always improve discus-sions of their topics with more up-to-date information. Textbooks that have not already done so should explicitly correct the estimate of atmospheric composition, and accompany the Miller–Urey experiment with a clarification of the fact that the corrected atmospheres yield similar results. Further, the wealth of new data on extraterrestrial and hydrothermal sources of biological material should be discussed. Finally, textbooks ideally should expand their discussions of other stages in the origin of life to include PNA and some of the newer research on self-replicating pro-teins. Wells, however, does not suggest that textbooks should correct the presentation of the origin of life. Rather, he wants textbooks to present this “icon” and then denigrate it, in order to reduce the confidence of students in the possibility that scientific research can ever Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 8 establish a plausible explanation for the origin of life or anything else for that matter. If Wells’s recommendations are followed, stu-dents will be taught that because one experi-ment is not completely accurate (albeit in hind-sight), everything else is wrong as well. This is not good science or science teaching.
References Anders, E. 1989. Pre-biotic organic matter from comets and asteroids. Nature 342:255–257.
Andersson, E. and N. G. Holm. 2000. The stability of some selected amino acids under attempted redox con-strained hydrothermal conditions. Origins of Life and the Evolution of the Biosphere 30: 9–23.
Chang, S. 1994. The planetary setting of prebiotic evo-lution. In S. Bengston, ed. Early Life on Earth. Nobel Symposium no. 84. Columbia University Press, New York. p.10–23.
Commeyras, A., H. Collet, L. Bioteau, J. Taillades, O.
Vandenabeele-Trambouze, H. Cottet, J-P. Biron, R.
Plasson, L. Mion, O. Lagrille, H. Martin, F. Selsis, and M. Dobrijevic. 2002. Prebiotic synthesis of sequential peptides on the Hadean beach by a molecular engine working with nitrogen oxides as energy sources.
Polymer International 51:661–665.
Cooper, G., N. Kimmich, W. Belisle, J. Sarinana, K.
Brabham, and L. Garrel. 2001. Carbonaceous meteorites as a source of sugar-related organic compounds for the early Earth. Nature 414:879–882.
Copley, J. 2001. The story of O. Nature 410:862-864.
Hanic, F., M. Morvová and I. Morva. 2000.
Thermochemical aspects of the conversion of the gaseous system CO2—N2—H2O into a solid mixture of amino acids. Journal of Thermal Analysis and Calorimetry 60: 1111–1121.
Hayes, J. M. 1967. Organic constituents of meteorites, a review. Geochimica et Cosmochimica Acta 31:1395– 1440.
Holland, H. D. 1994. Early Proterozoic atmosphere change. In S. Bengston, ed. Early Life on Earth. Nobel Symposium no. 84. Columbia University Press, New York. p. 237–244.
Irvine, W. M., 1998. Extraterrestrial organic matter: a review. Origins of Life and the Evolution of the Biosphere 28:365–383.
Kaplan, I. R., E. T. Degens, and J. H. Reuter. 1963.
Organic compounds in stony meteorites. Geochimica et Cosmochimica Acta. 27:805–834.
Kobayashi, K., T. Kaneko, T. Saito, and T. Oshima.
1998. Amino acid formation in gas mixtures by high energy particle irradiation. Origins of Life and the Evolution of the Biosphere 28:155–165.
Kump, L. R., J. F. Kasting, M. E. Barley. 2001. Rise of atmospheric oxygen and the “upside-down” Archean mantle. Geochemistry, Geophysics, Geosystems –G3, 2, paper number 2000GC000114.
Maurette, M. 1998. Carbonaceous micrometeorites and the origin of life. Origins of Life and the Evolution of the Biosphere 28: 385–412.
Miller, S. 1953. A production of amino acids under pos-sible primitive earth conditions. Science 117:528–529.
Miller, S. and H. Urey. 1959. Organic compound syn-thesis on the primitive earth. Science 130:245–251.
Miyakawa, S., K-I. Murasawa, K. Kobayashi, and A. B.
Sawaoka. 2000. Abiotic synthesis of guanine with high-temperature plasma. Origins of Life and Evolution of the Biosphere 30: 557–566.
Nunn, J. F. 1998. Evolution of the atmosphere.
Proceedings of the Geologists’Association 109:1–13.
Ogata, Y., E-I. Imai, H. Honda, K. Hatori, and K.
Matsuno. 2000. Hydrothermal circulation of seawater through hot vents and contribution of interface chem-istry to prebiotic synthesis. Origins of Life and the Evolution of the Biosphere 30: 527-–537.
Orgel, L. E. 1998a. The origin of life – a review of facts and speculations. Trends in Biochemical Sciences 23:491–495.
Orgel, L. E., 1998b. The origin of life — how long did it take? Origins of Life and the Evolution of the Biosphere 28: 91–96.
Oró, J. 1961. Comets and the formation of biochemical compounds on the primitive Earth. Nature 190:389-390.
Rasmussen, B., and R. Buick. 1999. Redox state of the Archean atmosphere; evidence from detrital heavy min-erals in ca. 3250-2750 Ma sandstones from the Pilbara Craton, Australia. Geology 27: 115–118.
Rode, B. M., 1999. Peptides and the origin of life.
Peptides 20: 773–786.
Rye, R., and H. D. Holland. 1998. Paleosols and the evolution of atmospheric oxygen: a critical review.
American Journal of Science 298:621–672.
Washington, J. 2000. The possible role of volcanic aquifers in prebiologic genesis of organic compounds Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 9 and RNA. Origins of Life and the Evolution of the Biosphere 30: 53–79.
Wells, J. 2000. Icons of evolution: science or myth?: why much of what we teach about evolution is wrong.
Regnery, Washington DC, 338p.
Whittet, D. C. B. 1997. Is extraterrestrial organic matter relevant to the origin of life on earth? Origins of Life and the Evolution of the Biosphere 27: 249–262.
Zubay, G. and T. Mui. 2001. Prebiotic synthesis of nucleotides. Origins of Life and Evolution of the Biosphere 31:87–102.
Icons of Evolution? Why Much of What Jonathan Wells Writes about Evolution is Wrong Alan D. Gishlick, National Center for Science Education 10 |
13511 | https://blog.csdn.net/joker_as/article/details/120556949 | 书架摆放的最小高度问题-CSDN博客
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书架问题（动归）
最新推荐文章于 2025-06-09 17:08:03 发布
面条二两于 2021-09-29 21:38:14 发布
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分类专栏：算法文章标签：动态规划
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
本文链接：
这是一篇关于使用动态规划解决书架摆放问题的文章。文章介绍了如何将书按照顺序放置到宽度有限的书架上，以求得书架整体最小高度。作者详细解释了动态规划的思路，给出了状态转移方程：dp[i]=max{dp[i−1]+books[i],max{max{books[j]...books[i]},dp[j]}}，并提供了代码实现。" 89278619,7383142,探索回文串：最长回文子串模板解析,"['算法', '字符串处理', '编程挑战']
题目：你把要摆放的书 books 都整理好，叠成一摞：从上往下，第 i 本书的厚度为 books[i]，高度为books[i]。
按顺序 将这些书摆放到总宽度为 shelf_width的书架上。
先选几本书放在书架上（它们的厚度之和小于等于书架的宽度 shelf_widt），然后再建一层书架。重复这个过程，直到把所有的书都放在书架上。
需要注意的是，在上述过程的每个步骤中，摆放书的顺序与你整理好的顺序相同。 例如，如果这里有 5 本书，那么可能的一种摆放情况是：第一和第二本书放在第一层书架上，第三本书放在第二层书架上，第四和第五本书放在最后一层书架上。
每一层所摆放的书的最大高度就是这一层书架的层高，书架整体的高度为各层高之和。
以这种方式布置书架，返回书架整体可能的最小高度
来源：力扣（LeetCode）
链接：
思路：
简单的动态规划问题，当我们加入一本新书的时候，会有两种情况
将新的一本书创建为一层新的
跟前面几本在同已层
那么，我们假设当前加入的为i本，dp[i]为前i放入书架的最矮高度，因此推出公式为：
dp[i]=max{ dp[i−1]+books[i],max{ max{ books[j]...books[i]},dp[j]}} dp[i]=max{dp[i-1]+books[i],max{max{books[j]...books[i]},dp[j]}} d p[i]=m a x{ d p[i−1]+
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【题目描述】 John最近买了一个 书架 用来存放奶牛养殖书籍，但 书架 很快被存满了，只剩最顶层有空余。John共有N头奶牛 ( 1≤N≤20,000 )，每头奶牛有自己的高度Hi ( 1≤Hi≤10,000 )，N头奶牛的总高度为S。书架 高度为B ( 1≤B≤S<2,000,000,007 )。 为了到达 书架 顶层，奶牛可以踩着其他奶牛的背，像叠罗汉一样，直到他们的总高度不低于 书架 高度。当然若奶牛越多则危险性越大。为了帮助John到达 书架 顶层，找出使用奶牛数目最少的解决方案吧。 【输入】 第1行：空格隔开的整数N和B。 第2
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经典算法书籍，重点穿插 动态规划 经典的背包 问题，9讲，穿插 动态规划， 解题思路和技巧
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洛谷 P2677 超级 书架 2
HJ921004的博客
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P2677 超级 书架 2 题目描述 Farmer John最近为奶牛们的图书馆添置了一个巨大的 书架，尽管它是如此的大，但它还是几乎瞬间就被各种各样的书塞满了。现在，只有 书架 的顶上还留有一点空间。 所有N ( 1 <= N <= 20 ) 头奶牛都有一个确定的身高H_i ( 1 <= H_i <= 1,000,000 - 好高的奶牛>_...
【动态规划】书本整理 book.pas/c/cpp
diying4157的博客
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书本整理（BOOK） Problem: book.pas/c/cppInput: book.inOutput: book.outMemory Limit: 2 5 6 MBTime Limit: 1 sec 【问题 描述】 Frank是一个非常喜爱整洁的人。他有一大堆书和一个 书架，想要把书 放在 书架上。书架 可以放下所有的书，所以Fran...
【习题】错排 问题
carl_2018的博客
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十本不同的书 放在 书架上。现重新摆放，使每 本书 都不在原来放的位置。有几种摆法？ 这个 问题 推广一下，就是错排 问题，是组合数学中的 问题 之一。考虑一个有n个元素的排列，若一个排列中所有的元素都不在自己原来的位置上，那么这样的排列就称为原排列的一个错排。 问题 分析 要使每 本书 都不在原来放的位置，现在假设有n 本书，n 本书 的错排记为D ( n )，n-1 本书 的错排记为D ( n-1 )，依次类推。初识位置与书的编号相同。 ...
书堆 Splay
a15063149271的博客
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题目描述 在桌子上有一堆书N本。可以在这堆书上进行如下两种操作： 1.在这堆书的顶端添加一本新书，其操作为ADD ( S ),S是书名； 2.把最上面的k 本书 的放置顺序颠倒。例如k=3，假设前三 本书 依次是《呐喊》，《彷徨》《哈里.波特》，进行此操作后顺序从上到下就变成了《哈里.波特》《彷徨》《呐喊》。若这堆书不足k本，就把整堆书全部翻转过来。...
例题题解 递归
m0_66362472的博客
04-06 371
#1484. 书架上 的书 题目描述 书架上 有n 本书，编号从1……n。从其中选m本，其中每两本的书的编号都不相邻的选法共有多少种？ 输入格式 二个整数n、m。( 1≤n≤1000，1≤m≤100 ) 输出格式 满足条件的选法。 样例 输入样例 4 2 输出样例 3 提示 样例解释：13、24、14共三种选法。 样例代码 #include #include #include #include #include #include #include
C++超级 书架（深度优先搜索）
Keven_11的博客
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C++超级 书架（深度优先搜索） Farmer John 最近为奶牛们的图书馆添置了一个巨大的 书架，尽管它是如此的大，但它还是几乎瞬间就被各种各样的书塞满了。现在，只有 书架 的顶上还留有一点空间。 所有 ) N ( 1≤N≤20 ) 头奶牛都有一个确定的身高 Hi ( 1≤Hi≤1,000,000 )。设所有奶牛身高的和为 S。书架 的高度为 B，并且保证 1≤B≤S。 为了够到比最高的那头奶牛还要高的 书架 顶，奶牛...
[慈溪2012]书架 ( bookshelf )
zhunubi的博客
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13512 | https://allen.in/dn/qna/644637812 | The ground state energy of hydrogen atom is -13.6 eV. The energy of second excited state of He^(+) ion in eV is
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The ground state energy of hydrogen atom is -13.6eV . The energy of excited state He^(+) ion having principal quantum number n is , -6.04eV . The numerical value of n is ________.
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13513 | https://people.maths.ox.ac.uk/scott/Papers/odds.pdf | Large induced subgraphs with all degrees odd A.D. Scott Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, England Abstract: We prove that every connected graph of order n ≥2 has an induced subgraph with all degrees odd of order at least cn/ log n, where c is a constant. We also give a bound in terms of chromatic number, and resolve the analogous problem for random graphs.
0. Introduction.
Given a graph G, it is natural to ask whether G must contain a large induced subgraph with certain properties. For instance, a classical question in Ramsey theory asks how many vertices a graph can have if it does not contain either a complete or an empty induced subgraph on a given number of vertices.
The starting point of this paper is the following theorem due to Gallai (see , Problem 5.17).
Theorem A.
(a) For every graph G, there is a partition V (G) = V1 ∪V2 such that G[V1] and G[V2] have all degrees even.
(b) For every graph G, there is a partition V (G) = V1 ∪V2 such that G[V1] has all degrees odd and G[V2] has all degrees even.
As an immediate consequence of this theorem we have the following.
Corollary B. For every graph G, there is a set W ⊂V (G) such that |W| ≥1 2|G| and G[W] has all degrees even.
1 Let us note that this result is best possible, as can be seen by considering Pn, the path on n vertices, for any n. (As we shall see below, we can say somewhat more for random graphs.) It is natural to ask whether we can find analogous results for induced subgraphs with all degrees odd. If a graph G has an isolated vertex v, then no induced subgraph of G containing v can have all degrees odd. Thus, in order to ensure that G has a large subgraph of this form, it is natural to forbid isolated vertices. Equivalently, we may assume that G is connected: then if G has no isolated vertices, but is not connected, we may consider each component separately.
This leads to the conjecture (the origins of which are unclear, see ) that there is some ϵ > 0 such that every graph G without isolated vertices contains an induced subgraph of order at least ϵ|G| with all degrees odd. If this were true then ϵ could be at most 1 3, as can be seen by considering K6 with a 1-factor removed.
In order to state the problem more clearly, we define some notation. For any graph G we define f(G) = max{|W| : W ⊂V (G) and G[W] has all degrees odd}.
Then, for n ≥2, let f(n) = min{f(G) : |G| = n and δ(G) ≥1}. The conjecture claims that there is some ϵ > 0 such that f(n) ≥ϵn for all n ≥2.
The conjecture has been proved for trees: Caro, Krasikov and Roditty gave a linear lower bound, and Radcliffe and Scott proved a best possible bound, namely that f(T) ≥2⌊(n + 1)/3⌋for every tree T. In §1 we prove a bound in terms of chromatic number: f(G) ≥|G|/(2χ(G)) for every graph G without isolated vertices.
In the general case, Caro proves that f(n) ≥ p (n −√n)/6. In §2 we prove the stronger result that f(n) ≥n/900 log n. The result is generalised to weighted graphs in §3. Finally, in §4, we show that, whenever ω(n) →∞, almost every G ∈G(n, 1 2) satisfies |f(G) −(cn + d log n)| ≤ω(n), where c = 0.7729 · · · and d = −0.2606 · · ·.
We use standard notation throughout (see , say). In particular, let G be a graph and S ⊂V (G). Then G[S] denotes the subgraph of G induced by S; where unambiguous, 2 we write e(S) for e(G[S]) and, for T ⊂V (G) \ S, we write e(S, T) for |{xy ∈E(G) : x ∈ S, y ∈T}|. We will also write [n] = {1, . . . , n}.
1. A bound in terms of chromatic number.
We begin by giving a lower bound for f(G) in terms of the largest independent set in G.
Alon (see ) has proved that f(G) ≥1 3ind(G); here we improve this to an essentially best possible result.
Theorem 1. For any graph G without isolated vertices, we have f(G) ≥1 2ind(G).
Proof. Let G be a graph without isolated vertices, and let I ⊂V (G) be an independent set of maximal size in G. We note that, for every v ∈I, we have Γ(v) ∩(V (G) \ I) ̸= ∅, since G contains no isolated vertices. Let W be a minimal subset of V (G) \ I such that |Γ(v) ∩W| > 0 for every v ∈I, say W = {w1, . . . , wk}. Then for each wi we can find some ui ∈I such that Γ(ui) ∩W = {wi}. Set U = {u1, . . . , uk} and U ′ = I \ U.
For S ⊂W, let US = {u ∈U ′ : |Γ(u) ∩S| is odd}, and IS = S ∪US ∪{ui : wi ∈S and |Γ(wi) ∩US| is even}.
By construction, in the graph G[IS] every degree is odd, so f(G) ≥|IS| ≥|S| + |US|.
Now let us choose S ⊂W randomly, by picking each wi independently, with proba-bility 1 2. Then, for every u ∈U ′, P(|Γ(u) ∩S| is odd) = 1 2, so E(|IS|) ≥E(|S|) + E(|US|) = |W|/2 + |U ′|/2 = |I|/2, since |U ′| = |I| −|U| = |I| −|W|. Therefore, for some S0 ⊂W we have |IS0| ≥|I|/2 = 1 2ind(G).
Theorem 1 is best possible in the sense that we cannot replace 1 2 with any smaller constant. Indeed, let Bn be the bipartite graph with vertex classes V1 = [n], V2 = n, 3 and edges from {i} to {j, k} if and only if i ∈{j, k}. We note that if W ⊂V (Bn), then for each {i, j} ∈W ∩V2 exactly one of {i} and {j} are in V1. Thus f(Bn) ≤⌈n2/4⌉+n. Since ind(Bn) = n 2 , we have f(Bn)/ind(Bn) →1 2 as n →∞, and deduce that the constant in Theorem 1 is best possible.
As a consequence of Theorem 1 we have the following result.
Theorem 2. Let G be a graph without isolated vertices. Then f(G) ≥|G|/(2χ(G)).
Proof. Colour G with χ(G) colours and apply Theorem 1 to the largest colour class.
How good is the factor (2χ(G))−1 in Theorem 2? For k ≥2, we define ck = inf{f(G)/|G| : χ(G) = k and δ(G) ≥1}, Thus Theorem 2 says that ck ≥1/2k. For bipartite graphs, this gives c2 ≥1 4; from the other side, the graph Bn, defined above, shows that c2 ≤1 2, and we conjecture that 1 2 is the correct value. For k ≥3, our best upper bound is ck ≤1 3, which we get by considering K6 with a 1-factor removed (this graph is 3-colourable). It seems likely that some improvement is possible.
The ck clearly form a non-increasing sequence (indeed, suppose k ≤l and G is k-chromatic: then by considering many disjoint copies of G together with a copy of Kl+1, we see f(G) ≥cl|G|; we deduce ck ≥cl). The remainder of this section examines how quickly the sequence can decrease.
Suppose k ≥l, and we wish to bound ck below in terms of cl.
Let G be a k-chromatic graph. Given a k-colouring of G, we would like to take the l largest colour classes, say V1, . . . , Vl, and by considering G[Sl i=1 Vi] and our lower bound for cl obtain a lower bound for f(G). Unfortunately, although G[Sl i=1 Vi] is l-chromatic, it may contain isolated vertices. This leads us to ask whether G must contain a large induced l-chromatic graph without isolated vertices.
In order to answer this question, it seems to be easiest to prove a more general result concerning weighted graphs.
We say that a graph G has vertex weighting µ if µ is a function µ: V (G) →R≥0. The weight of a vertex v ∈V (G) is µ(v), and the weight of a 4 set W ⊂V (G) is µ(W) = P w∈W µ(w). If H is a subgraph of G then the weight of H is µ(H) = µ(V (H)). We have the following result for vertex-weighted graphs.
Theorem 3. Suppose k ≥2 and G is a k-chromatic graph with vertex-weighting µ and no isolated vertices. Then, for each 2 ≤m ≤k, G contains an induced m-chromatic subgraph H such that H has no isolated vertices and µ(H) ≥m k µ(G).
Proof. If k = m = 2 then the theorem is trivial. Let us suppose that k ≥3. We will prove by induction on |V (G)| that if G is a k-chromatic graph with vertex-weighting µ and no isolated vertices then G contains an induced (k −1)-chromatic subgraph with weight at least k−1 k µ(G) and without isolated vertices. Applying this k −m times gives the required m-chromatic subgraph.
If |V (G)| = 3 then the inductive statement is trivial. Let us assume that |V (G)| > 3, that G is k-chromatic and has no isolated vertices, and that our inductive statement is true for smaller values of |V (G)|. We may assume that G is connected, or else deal with each component of G separately. Let V (G) = V1 ∪· · · ∪Vk be a k-colouring of the vertices of G.
Suppose that some v ∈V (G) has all its neighbours in one colour class, say v ∈V1 and Γ(v) ⊂V2. Then consider G′ = (G \ ({v} ∪Γ(v))) ∪{w}, where w is a new vertex joined to S x∈Γ(v) Γ(x) \ {v}. Note that w is not an isolated vertex in G′, since otherwise {v}∪Γ(v) would have been a component of G with chromatic number 2. Let V ′ 1 = V1{v}, V ′ 2 = (V2 \ Γ(v)) ∪{w} and V ′ i = Vi for i > 2. Then G′ is k-chromatic and V ′ 1, . . . , V ′ k is a k-colouring. Define µ′(x) = µ(x) for x ∈V (G′) \ {w} and µ′(w) = µ(v) + P x∈Γ(v) µ(x).
Thus µ′(G′) = µ(G). Since |V (G′)| < |V (G)|, we can find a (k −1)-chromatic induced subgraph H′ ⊂G′ without isolated vertices and with µ′(H′) ≥k −1 k µ′(G′) = k −1 k µ(G).
If w ∈V (H′) then let H = G[(V (H′) \ {w}) ∪{v} ∪Γ(v)]; if w ̸∈V (H′) then let H = H′ (considered as a subgraph of G).
We see that µ(H) = µ′(H′), and H is an induced (k −1)-chromatic subgraph of G without isolated vertices.
5 Thus we may assume that every vertex in G has neighbours in at least two colour classes. In that case, we may delete the colour class of smallest weight, to obtain the required induced subgraph of G.
This gives an immediate corollary for unweighted graphs.
Corollary 4. Let 2 ≤m ≤k. If G is a k-chromatic graph without isolated vertices then it contains an induced m-chromatic graph H without isolated vertices such that |H| ≥m k |G|.
Proof. Define µ(v) = 1 for all v ∈V (G) and apply Theorem 3.
Both Theorem 3 and Corollary 4 are best possible, as can be seen by considering a complete k-partite graph with all vertex classes of equal size.
We are now ready to prove our bound on how fast {ck}∞ k=2 can decrease.
Theorem 5. For every 2 ≤k ≤l, cl ≥k l ck.
Proof. Let G be an l-chromatic graph. Then by Corollary 4, we can find an induced k-chromatic subgraph H such that H has no isolated vertices and |H| ≥k|G|/l. Therefore f(G) ≥f(H) ≥ck|H| ≥kck l |G|.
We deduce that cl ≥kck/l.
We note that any improvement in the lower bound for ck gives corresponding improve-ments in the lower bounds for cl, for every l ≥k. In particular, it would be interesting to improve c2, since this would lead to a corresponding improvement on the factor (2χ(G))−1 in Theorem 2.
It is not impossible that the conjecture f(n) ≥cn for all n ≥2, for some constant c, is false. Even if it is true, we may only be able to prove it with c small, in which case Theorem 2 will give a better bound for graphs of low chromatic number. If c2 = 1 2, as conjectured above, an application of Theorem 5 would improve Theorem 2 by a factor of 6 2: we would get the attractive result that f(G) ≥|G|/χ(G) for every graph G without isolated vertices.
2. A general bound.
We have proved a bound in terms of chromatic number; we now turn to the general case. The idea of the proof is to look for subgraphs of G satisfying conditions which ensure that G contains relatively large induced subgraphs with all degrees odd.
For instance, suppose we have an edge vw ∈E(G) with d(v) large and |Γ(v) ∩Γ(w)| small. Let S = Γ(v) \ (Γ(w) ∪{w}). Then, by Theorem B, we can find S′ ⊂S such that |S′| ≥|S|/2 and G[S] has all degrees even. If |S| is even then set W = S′ ∪{v, w}; if |S| is odd, then set W = S′ ∪{v}.
In either case, G[W] has all degrees odd and |W| ≥1 2(d(v) −|Γ(v) ∩Γ(w)|).
More generally, suppose have a set of independent edges v1w1, . . . , vkwk, and let V0 = Sk i=1 Γ(vi) and W0 = Sk i=1 Γ(wi). If |V0 ∩W0| is small compared with |V0 ∪W0|, then we can use a similar argument to find a comparatively large induced subgraph of G with all degrees odd. The main part of the proof works by looking for such collections of edges.
Theorem 6. Every graph G without isolated vertices contains a set W ⊂V (G) such that |W| ≥|G|/900 log(|G|) and G[W] has all degrees odd.
Proof. Let G be a graph of order n without isolated vertices, and let c be a constant (we will take c = 0.086 below). We begin by partitioning V (G) into sets X1, Y1 and Z1. We will do this in such a way that X1 contains a large structure of the sort described above; Y1 has highly connected components; and Z1 does not contain too many edges. Both X1 and Y1 can be used to find relatively large induced subgraphs of G with all degrees odd; our only problem is Z1, which we again partition, into sets X2, Y2 and Z2. We apply the algorithm repeatedly, each time partitioning Zi−1 into sets Xi, Yi and Zi, ending up with a partition of V (G) into sets X1, . . . , Xk, Y1, . . . , Yk, Zk, where Zk is an independent set.
More precisely, let Z0 = V (G).
By assumption, e(Z0) > 0: we partition Z0 into 7 sets X1, Y1 and Z1 using the algorithm described below. If e(Z1) = 0 then set k = 1.
Otherwise, apply the algorithm to Z1 to get X2, Y2 and Z2. Repeat this process until Zi is first an independent set, at which point put k = i.
Our algorithm for partitioning Zj−1 into Xj, Yj and Zj is as follows. We shall use the notation X(i) j , Y (i) j and Z(i) j for our successive ‘approximations’ to Xj, Yj and Zj. Let X(1) j = Y (1) j = Z(1) j = ∅. At the ith stage, given X(i) j , Y (i) j and Z(i) j , we define V (i) j = Zj−1 \ (X(i) j ∪Y (i) j ∪Z(i) j ).
Thus V (i) j is the set of vertices we have not yet partitioned. If there is some v ∈V (i) j which satisfies |Γ(v) ∩X(i) j | ≥c|Γ(v) ∩V (i) j | (1) then set X(i+1) j = X(i) j Y (i+1) j = Y (i) j Z(i+1) j = Z(i) j ∪{v}.
Otherwise, if there is an edge vw in G[V (i) j ] such that |Γ(v) ∩V (i) j | > 1 and |Γ(v) ∩Γ(w) ∩V (i) j | ≤(2/3) max{|Γ(v) ∩V (i) j | −1, |Γ(w) ∩V (i) j | −1}, (2) then set X(i+1) j = X(i) j ∪((Γ(v) ∪Γ(w)) ∩V (i) j ) \ {v, w} Y (i+1) j = Y (i) j ∪{v, w} (3) Z(i+1) j = Z(i) j .
If neither of these conditions is satisfied, then set Xj = X(i) j Yj = V (i) j Zj = Z(i) j , (4) 8 and terminate the algorithm.
We have now partitioned V (G) into 2k + 1 sets. Note that the number e(Zj) of edges in G[Zj] satisfies e(Zj) ≤e(Zj−1)/(1 + c), for j = 1, . . . , k. Indeed, let Zi = {z1, . . . , zp}, where zj is chosen at stage ij and i1 < · · · < ip. Using (1), we have e(Zi) = p−1 X j=1 |Γ(zj) ∩{zj+1 . . . , zp}| ≤ p−1 X j=1 |Γ(zj) ∩V (ij) i | ≤ p−1 X j=1 |Γ(zj) ∩X(ij) i |/c ≤e(Zi, Xi)/c.
Since Xi ∪Zi ⊂Zi−1, we have e(Zi−1) ≥e(Zi) + e(Zi, Xi) ≥(1 + c)e(Zi), as claimed. Thus k (the smallest integer such that Zk is an independent set) is well defined, and k ≤log1+c n 2 ≤2 log n/ log(1 + c).
(5) We claim that each set in our partition guarantees the existence of a comparatively large induced subgraph of G with all degrees odd. The next three lemmas prove that this is indeed the case.
Lemma 7. For i = 1, . . . , k, we have f(G) ≥1−6c 16 |Xi|.
Proof. If Xi = ∅then we are done. Otherwise, Xi is in the neighbourhood of a sequence of edges v1w1, . . . , vpwp chosen to satisfy (2). Let us suppose that vjwj was chosen at the ijth stage, where i1 < · · · < ip, and that |Γ(vj) ∩V (ij) i | ≥|Γ(wj) ∩V (ij) i |.
(6) Let V0 = {v1, . . . , vp}, V1 = {w1, . . . , wp} and S = {v ∈Xi : Γ(v) ∩V1 = ∅}. We remark that, from (2) and (3), V0 and V1 are independent sets of Yi, that Γ(vi)∩V1 = {wi}, 9 and that every v ∈S has at least one neighbour in V0. Let T ⊂V0 be a random subset, where each v ∈V0 is chosen independently with probability 1 2. We define ST = {v ∈S : |Γ(v) ∩T| is odd}.
Clearly, E(|ST |) = |S|/2, so for some T0 ⊂V0 we have |ST0| ≥|S|/2. By Corollary B, we can find some S0 ⊂ST0 such that |S0| ≥|ST0|/2 ≥|S|/4 and G[S0] has all degrees even.
Let S′ = S0 ∪T0 ∪{wi ∈V1 : vi ∈T0 and |Γ(vi) ∩S0| is even}.
It is easily seen that G[S′] has all degrees odd and |S′| ≥|S0| ≥|S|/4.
In order to prove the assertion of the lemma, it is therefore enough to prove that |S| ≥(1 −6c)|Xi|/4. We do this by using inequalities (2) and (6), and following through the algorithm for partitioning Zi−1.
This will require some calculation which appears rather unattractive, but is in fact quite straightforward.
Define S0 = ∅and, for j = 1, . . . , p, Sj = {v ∈X(ij) i : Γ(v) ∩{w1, . . . , wj} = ∅}, (7) Note that Sp = S and X(ip) i = Xi, so we want to show that |Sp| ≥(1 −6c)|X(ip) i |/4. Set i0 = 0, so that X(i0) i = X(0) i = ∅. From (3) and (7) we see that, for j = 1, . . . , p, Sj = (Sj−1 \ Γ(wj)) ∪(Γ(vj) ∩V (ij) i \ ({wj} ∪Γ(wj))), so |Sj| ≥|Sj−1| −|Γ(wj) ∩Sj−1| + |Γ(vj) ∩V (ij) i | −1 −|Γ(vj) ∩Γ(wj) ∩V (ij) i |.
(8) On the other hand, from (3), we have X(ij) i = X(ij−1) i ∪((Γ(vj) ∪Γ(wj)) ∩V (ij) i \ {v, w}), so |X(ij) i | −|X(ij−1) i | = |Γ(vj) ∩V (ij) i | + |Γ(wj) ∩V (ij) i | −|Γ(vj) ∩Γ(wj) ∩V (ij) i | −2 ≤2(|Γ(vj) ∩V (ij) i | −1) −|Γ(vj) ∩Γ(wj) ∩V (ij) i |, (9) 10 by inequality (6). For ease of notation we write a = |Γ(vj) ∩V (ij) i | and b = |Γ(vj) ∩ Γ(wj) ∩V (ij) i |. Note that, from (1), |Γ(wj) ∩Sj−1| ≤|Γ(wj) ∩X(ij) i | ≤c|Γ(wj) ∩V (ij) i | ≤ c|Γ(vj) ∩V (ij) i | = ca. Then (8) and (9) become |Sj| −|Sj−1| ≥(a −1) −b −ca, and |X(ij) i | −|X(ij−1) i | ≤2(a −1) −b.
Inequality (2) becomes b ≤2(a −1)/3, so |Sj| −|Sj−1| |X(ij) i | −|X(ij−1) i | ≥(a −1) −b −ca 2(a −1) −b ≥ 1 3(a −1) −ca 4 3(a −1) = 1 −3c 4 − 3c 4(a −1) ≥1 −6c 4 , since a > 1. This is true for j = 1, . . . , p, so |Sp| = p X j=1 (|Sp| −|Sp−1|) ≥ p X j=1 1 −6c 4 (|X(ij) i | −|X(ij−1) i |) = 1 −6c 4 |X(ip) p | .
Lemma 8. For i = 1, . . . , k, f(G) ≥1 4|Yi|.
Proof. We note first that G[Yi] has no isolated vertices. Indeed, every vertex v ∈Yi was added in (3) or (4). If v was added in (3), then it is adjacent to some w ∈Yi that was added at the same time; while if v was added in (4), then |Γ(v) ∩Yi| ≥|Γ(v) ∩V (i) j | > 0, or else v would have satisfied the condition (1).
11 We consider the components of G[Yi]. The vertices added to Yi in (3) induce a set of independent edges, so they are all in components of order 2; components of great order must come from (4). We claim that each component H contains a set VH ⊂V (H) such that |VH| ≥1 4|H| and G[VH] has all degrees odd. This is enough to prove the lemma, since the union of these sets has order at least |Yi|/4 and induces a subgraph with all degrees odd. We will suppress unnecessary mention of H, writing for instance Γ for ΓH.
If |H| = 2 then we are done. Otherwise, |H| > 2, and V (H) was added to Yi in (4).
By (2), each adjacent pair of vertices x and y in H satisfy |Γ(x) ∩Γ(y)| > 2 3 max{d(x) −1, d(y) −1}.
(10) Note that if xy ∈E(H) and d(x) ≥2 then d(y) ≥2; since H is connected, we may assume that δ(H) ≥2.
Suppose that ∆(H) ≥ 1 2|H|, and let x ∈V (H) be a vertex of maximal degree in H. If d(x) is odd then let W = Γ(x); otherwise let W = Γ(x) \ {y} for some y ∈Γ(x).
By Theorem A we may split V (W) ∪{x} into sets W1 and W2, each of which induces a subgraph of H with all degrees even. We may assume that x ∈W1. Let W ′ 1 = W1 \ {x}, W ′ 2 = W2 ∪{x}. The subgraph W ′ 1 has all degrees odd, and thus an even number of vertices. G[W ′ 2] has all degrees odd, except possibly for x; but |Γ(x) ∩W ′ 2| = |W2| = |W| + 1 −|W1| = |W| −|W ′ 1|, which is odd, so x has an odd number of neighbours in W ′ 2.
Since |W ′ 1| + |W ′ 2| ≥∆(H) ≥1 2|H|, one of W ′ 1 and W ′ 2 must contain at least 1 4|H| vertices, and will thus do for VH.
It is therefore enough to prove that if H satisfies (10) then ∆(H) ≥1 2|H|.
We claim first that diam(H) ≤2. Indeed, suppose diam(H) > 2, say d(w, z) = 3 for some w, z ∈H.
Let wxyz be a shortest path from w to z.
We will prove that |Γ(w) ∩Γ(z)| > 0.
We may assume d(x) ≥d(y).
Now Γ(w) ∩Γ(x) and Γ(y) ∩Γ(x) are subsets of Γ(x) \ {w, y}, so by (10), |Γ(w) ∩Γ(y)| ≥|Γ(w) ∩Γ(x)| + |Γ(y) ∩Γ(x)| −(|Γ(x)| −2) > 2 3(d(x) −1) + 2 3(d(x) −1) −d(x) + 2 = 1 3d(x) + 2 3.
12 Since d(x) ≥d(y) we have |Γ(w) ∩Γ(y)| ≥1 3d(y) + 2 3.
We get |Γ(w) ∩Γ(z)| ≥|Γ(w) ∩Γ(y)| + |Γ(z) ∩Γ(y)| −|Γ(y)| > 1 3d(y) + 2 3 + 2 3(d(y) −1) −d(y) = 0.
Thus we must have d(w, z) ≤2, which is a contradiction. Hence diam(H) ≤2.
We now prove that ∆(H) ≥ 1 2|H|.
Indeed, suppose d(x) = ∆(H) < 1 2|H|.
Let Y = Γ(x) and Z = H \ (Y ∪{x}). Then Y and Z are nonempty. We count e(Y, Z) in two ways.
First, for y ∈Y , we have by (10), e(y, Z) = |Γ(y)| −|Γ(x) ∩Γ(y)| −1 < d(y) −2 3(d(y) −1) −1 = 1 3(d(y) −1) ≤1 3(d(x) −1).
(11) For z ∈Z, pick w ∈Γ(z) ∩Γ(x) (we can do this because diam(H) ≤2). We have, as before, e(z, Y ) = |Γ(z) ∩Γ(x)| ≥|Γ(z) ∩Γ(w)| + |Γ(x) ∩Γ(w)| −(Γ(w) −2) > 2 3(d(w) −1) + 2 3(d(x) −1) −(d(w) −2) = 2 3d(x) −1 3d(w) + 2 3 ≥1 3d(x) + 2 3.
(12) Therefore, using (11) and (12) to bound e(Y, Z) above and below, we get 1 3(d(x) −1)|Y | > e(Y, Z) > 1 3(d(x) + 2)|Z|.
13 If d(x) < |H|/2 then |Z| > d(x) −1, so 1 3(d(x) −1)d(x) > 1 3(d(x) + 2)(d(x) −1), which is a contradiction. Thus ∆(H) ≥1 2|H|.
Lemma 9. We have f(G) ≥|Zk|/2.
Proof. Since Zk is an independent subset of V (G), this is immediate from Theorem 1.
We are now ready to complete the proof of Theorem 6. From the preceding lemmas we have f(G) ≥max 1 −6c 16 max i=1,...,k |Xi|, 1 4 max i=1,...,k |Yi|, 1 2|Zk| ≥max ( 1 −6c 16k k X i=1 |Xi|, 1 4k k X i=1 |Yi|, 1 2|Zk| ) .
Now |G| = Pk i=1 |Xi| + Pk i=1 |Yi| + |Zk|, so by (5) we have f(G) ≥ n 16k/(1 −6c) + 4k + 2 ≥ 8 5 −6c 1 −6c log n log(1 + c) + 2 −1 n.
Set c = 0.086. Since f(G) ≥2 always, we get f(G) ≥n/900 log n.
If G is triangle-free then the proof of Theorem 6 yields a better constant in our bound.
Corollary 10. Let G be a triangle-free graph without isolated vertices. Then f(G) ≥ |G|/155 log |G|.
Proof. Let n = |G|. We use the notation of the proof of Theorem 6. Note that, since G is triangle-free, (2) is satisfied for every edge in G[V (ij) j ]. Since Γ(v) ∩Γ(w) = ∅whenever v and w are adjacent, we have b = |Γ(vj) ∩Γ(wj) ∩V (ij) i | = 0 in Lemma 7, so |Sj| −|Sj−1| |X(ij) i | −|X(ij−1) i | ≥(a −1) −ca 2(a −1) = 1 2 − ca 2(a −1) ≥1 2 −c, 14 since a > 1. It follows from (3) and (4) that G[Yi] is a collection of independent edges, so Lemma 8 becomes f(G) ≥|Yi|. Therefore f(G) ≥max ( 1 −2c 8k k X i=1 |Xi|, 1 k k X i=1 |Yi|, 1 2|Zk| ) ≥ 8k 1 −2c + k + 2 −1 n ≥ 9 −2c 1 −2c 2 log n log(1 + c) + 2 −1 n.
Set c = 0.24: we get that f(G) > n/155 log n.
We note that any restriction on cl(G) gives a corresponding restriction on the size of components in Yi and thus a small improvement in the constant in our bound for f(G).
(For instance, if cl(G) = 3 then Yi can contain only triangles and independent edges.) 3. Induced subgraphs of weighted graphs.
We now consider weighted versions of our results. For a graph G with vertex-weighting µ, let fµ(G) be the maximum weight of an induced subgraph of G with all degrees odd.
We have the following equivalent to Theorem 1.
Theorem 11. Let G be a graph with vertex-weighting µ and without isolated vertices, and I be an independent subset of V (G). Then fµ(G) ≥1 4µ(I).
Proof. As in Theorem 1, pick W ⊂V \ I to be a minimal set covering I and find U ⊂I so that G[V ∪W] is a set of independent edges. If µ(U) ≥µ(I)/2 then partition W into W1 and W2 so that G[W1] and G[W2] have all degrees even. One of W1 ∪(U ∩Γ(W1)) and W2 ∪(U ∩Γ(W2)) will do. If µ(U) < µ(I)/2, choose S ⊂I \ U as in Theorem 1. Then E(µ(US)) = µ(I \ U)/2 > µ(I)/4, so we may proceed as before.
This can be used in the same way that Theorem 1 was used for Theorem 6.
15 Theorem 12. There is a constant c such that for every graph G with vertex-weighting µ and without isolated vertices, fµ(G) ≥cµ(G)/ log |G|.
Proof. (Sketch) We follow closely the proof of Theorem 6, replacing |·| with µ(·) where appropriate. We replace (1) with µ(Γ(v) ∩X(i) j ) > cµ(Γ(v) ∩V (i) j ), and (2) with the conditions µ(Γ(v) ∩V (i) j \ {w}) > 0 and µ(Γ(v) ∩Γ(w) ∩V (i) j ) ≤(2/3) max{µ(Γ(v) ∩V (i) j ) −µ(w), µ(Γ(w) ∩V (i) j ) −µ(v)}.
In the final step of the algorithm we leave the definitions of Xj, Yj and Zj unchanged.
We can use this to find large induced subgraphs with all degrees odd that satisfy additional conditions on the distribution of vertices. For instance, we have the following more general version of Theorem 6.
Theorem 13. There is a constant c such that for every graph G without isolated vertices and for every S ⊂V (G) there is an induced subgraph H of G with all degrees odd such that |V (H) ∩S| ≥c|S|/ log |G|.
Proof. Set µ = χS, the indicator function for S, and apply Theorem 12.
We note that Theorem 13 can also be proved directly from Theorem 6 and Theorem 1: if ind(G[S]) > |G|/2 then apply Theorem 1; otherwise apply Theorem 6 to each non-trivial component of G[S].
16 4. Random graphs.
So far we have considered results that hold for all graphs. In this section, we consider the same questions for random graphs.
Caro has proved that almost every graph (in the model G(n, 1/2)) satisfies f(G) ≥ n/4, and Alon (see ) has proved that almost every graph satisfies f(G) ≥( 1 2 −ϵ)|G|.
We prove a more precise result.
Let s(n) be the smallest positive even integer such that 21−s n s ≤1, and let c be the solution of (2c)c(1 −c)1−c = 1. Then c = 0.7729 · · · and s(n) = cn + d log n + O(1), where 1/d = 2 log( 1−c 2c ) = −3.8359 · · ·.
Theorem 14. Let s(n) be as above, and let ω(n) →∞as n →∞. Then almost every graph G ∈G(n, 1/2) satisfies |f(G) −s(n)| ≤ω(n).
In particular, almost every graph satisfies f(G) ≥0.7729n.
Proof. Let G ∈G(n, 1/2) be a random graph, with vertex set V = [n], say. For A ⊂V let XA be the indicator variable of the event {G[A] has all degrees odd}, and, for x, y ∈V , let Exy be the indicator variable of the event {xy ∈E(G)}. We define Xi = X A∈V (i) XA.
If |A| is odd, then XA = 0. If |A| is even, then pick a ∈A and let A′ = A \ {a}. Since P x∈A dGA is even, we have XA = 1 iffdGA is odd for all x ∈A′. We condition on G[A′]: for x ∈A′, the parity of dGA depends on Eax, and these events are independent, so XA = 1 with probability 2−|A′| = 21−|A|. Thus if i is even, then EXi = 21−i n i .
17 To prove the required upper bound, suppose that t ≥s(n) + ω(n). It is easily checked that s(n) > n/2, so EXs(n)+k ≤2−kEXs(n) for k > 0. Therefore P(f(G) ≥t) = P(Xi > 0 for some i ≥t) ≤ n X i=t P(Xi > 0) ≤ n X i=t E(Xi) ≤ n X i=t 2i−s(n)EXs(n) ≤21−ω(n).
Therefore P(f(G) > s(n) + ω(n)) →0 as n →∞.
For the lower bound, we calculate the variance of Xi, where i is even. Let A, B ∈V (i) be such that A ̸= B, and pick a ∈A\B and b ∈B\A. Let C = A∪B{a, b}. Conditioning on G[C], we see that, for x ∈A \ {a}, the event {dGA is odd} depends on Eax, while, for x ∈B \ {b}, the event {dGB is odd} depends on Ebx. These events are all independent, so P(XA = XB = 1) = 2−|C| = 22−2i = P(XA = 1)P(XB = 1), regardless of G[C]. Thus the events XA, for A ∈V (i), are pairwise independent. This means that the variance turns out to be surprisingly small. Indeed, var(Xi) = X A∈V(i) X B∈V (i) E(XAXB) −E(XA)E(XB) = X A∈V (i) E(XA) −E(XA)2 ≤ X A∈V (i) E(XA) = E(Xi).
By Chebyshev’s inequality, we have P(Xi = 0) ≤var(Xi)/E(Xi)2 ≤1/E(Xi).
18 Now, we may assume that s(n) −ω(n) is even, and that ω(n) →∞slowly enough so that s(n) −ω(n) > n/2. Therefore E(Xs(n)−ω(n)) ≥2ω(n)−2E(Xs(n)−2) ≥2ω(n)−2, by the definition of s(n). Thus P(Xs(n)−ω(n) = 0) →0 as n →∞, so P(f(G) < s(n) − ω(n)) →0 as n →∞.
The same arguments, with a little modification, give essentially the same result for induced subgraphs with all degrees even (note that such a subgraph can have any order, in contrast to the odd case), thus yielding the following analogue of Corollary B for random graphs.
Theorem 15. Let s(n) be the smallest positive integer such that 21−s n s ≤1, and let ω(n) →∞as n →∞. Then almost every G ∈G(n, 1 2) satisfies the following.
(a) There is a set S ⊂V (G) such that |S| = ⌈s(n) −ω(n)⌉and G[S] has all degrees even.
(b) There is no set S ⊂V (G) such that |S| ≥s(n) + ω(n) and G[S] has all degrees even.
Acknowledgments. I would like to thank Dr Y. Caro for letting me have a copy of his paper , and Dr Jamie Radcliffe for his helpful comments.
19 References B. Bollob´ as, Graph Theory, An Introductory Course, Springer-Verlag, New York, Heidelberg, Berlin, 1979, x+180pp.
Y. Caro, On induced subgraphs with odd degrees. To appear.
Y. Caro, I. Krasikov and Y. Roditty, On induced subgraphs of trees with restricted degrees. Discrete Mathematics. To appear.
L. Lov´ asz, Combinatorial Problems and Exercises, North-Holland, Amsterdam, 1979, 551pp.
A.J. Radcliffe and A.D. Scott, Every tree has a large induced subgraph with all degrees odd. To appear.
20 |
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The Physics Classroom » Physics Tutorial » Vectors - Motion and Forces in Two Dimensions » Initial Velocity Components
Vectors - Motion and Forces in Two Dimensions - Lesson 2 - Projectile Motion
Initial Velocity Components
What is a Projectile?
Motion Characteristics of a Projectile
Horizontal and Vertical Velocity
Horizontal and Vertical Displacement
Initial Velocity Components
Horizontally Launched Projectile Problems
Non-Horizontally Launched Projectile Problems
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It has already been stated and thoroughly discussed that the horizontal and vertical motions of a projectile are independent of each other. The horizontal velocity of a projectile does not affect how far (or how fast) a projectile falls vertically. Perpendicular components of motion are independent of each other. Thus, an analysis of the motion of a projectile demands that the two components of motion are analyzed independent of each other, being careful not to mix horizontal motion information with vertical motion information. That is, if analyzing the motion to determine the vertical displacement, one would use kinematic equations with vertical motion parameters (initial vertical velocity, final vertical velocity, vertical acceleration) and not horizontal motion parameters (initial horizontal velocity, final horizontal velocity, horizontal acceleration). It is for this reason that one of the initial steps of a projectile motion problem is to determine the components of the initial velocity.
Determining the Components of a Velocity Vector
Earlier in this unit, the method of vector resolution was discussed. Vector resolution is the method of taking a single vector at an angle and separating it into two perpendicular parts. The two parts of a vector are known as components and describe the influence of that vector in a single direction. If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and a vertical component. The horizontal velocity component (v x) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component (v y) describes the influence of the velocity in displacing the projectile vertically. Thus, the analysis of projectile motion problems begins by using the trigonometric methods discussed earlier to determine the horizontal and vertical components of the initial velocity.
Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal. Such a projectile begins its motion with a horizontal velocity of 25 m/s and a vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity. These numerical values were determined by constructing a sketch of the velocity vector with the given direction and then using trigonometric functions to determine the sides of the velocity triangle. The sketch is shown at the right and the use of trigonometric functions to determine the magnitudes is shown below. (If necessary, review this method on an earlier page in this unit.)
All vector resolution problems can be solved in a similar manner. As a test of your understanding, utilize trigonometric functions to determine the horizontal and vertical components of the following initial velocity values. When finished, click the button to check your answers.
Practice A: A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal.
See Answer
cos (60 deg) = v x / (40 m/s)
v x = 40 m/s • cos (60 deg) = 20.0 m/s
sin (60 deg) = v y / (40 m/s)
v y = 40 m/s • sin (60 deg) = 34.6 m/s
Practice B: A motorcycle stunt person traveling 70 mi/hr jumps off a ramp at an angle of 35 degrees to the horizontal.
See Answer
cos (35 deg) = v x / (70 mi/hr)
v x = 70 mi/hr • cos (35 deg) = 57.3 mi/hr
sin (35 deg) = v y / (70 mi/hr)
v y = 70 mi/hr • sin (35 deg) = 40.1 mi/hr
Practice C: A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal.
See Answer
Practice C:
cos (80 deg) = v x / (10 m/s)
v x = 10 m/s • cos (80 deg) = 1.7 m/s
sin (80 deg) = v y / (10 m/s)
v y = 10 m/s • sin (80 deg) = 9.8 m/s
Try Some More!
Need more practice? Use the Velocity Components for a Projectilewidget below to try some additional problems. Enter any velocity magnitude and angle with the horizontal. Use your calculator to determine the values of v x and v y. Then click the Submit button to check your answers.
| | Velocity Components for a Projectile | |
---
| | Velocity magnitude:m/s Angle w/horizontal:deg Submit | |
| | Build your own widget»Browse widget gallery»Learn more»Report a problem»Powered by Wolfram|Alpha Terms of use Share a link to this widget: More Embed this widget» | |
As mentioned above, the point of resolving an initial velocity vector into its two components is to use the values of these two components to analyze a projectile's motion and determine such parameters as the horizontal displacement, the vertical displacement, the final vertical velocity, the time to reach the peak of the trajectory, the time to fall to the ground, etc. This process is demonstrated on the remainder of this page. We will begin with the determination of the time.
Determination of the Time of Flight
The time for a projectile to rise vertically to its peak (as well as the time to fall from the peak) is dependent upon vertical motion parameters. The process of rising vertically to the peak of a trajectory is a vertical motion and is thus dependent upon the initial vertical velocity and the vertical acceleration (g = 9.8 m/s/s, down). The process of determining the time to rise to the peak is an easy process - provided that you have a solid grasp of the concept of acceleration. When first introduced, it was said that acceleration is the rate at which the velocity of an object changes. An acceleration value indicates the amount of velocity change in a given interval of time. To say that a projectile has a vertical acceleration of -9.8 m/s/s is to say that the vertical velocity changes by 9.8 m/s (in the - or downward direction) each second. For example, if a projectile is moving upwards with a velocity of 39.2 m/s at 0 seconds, then its velocity will be 29.4 m/s after 1 second, 19.6 m/s after 2 seconds, 9.8 m/s after 3 seconds, and 0 m/s after 4 seconds. For such a projectile with an initial vertical velocity of 39.2 m/s, it would take 4 seconds for it to reach the peak where its vertical velocity is 0 m/s. With this notion in mind, it is evident that the time for a projectile to rise to its peak is a matter of dividing the vertical component of the initial velocity (v iy) by the acceleration of gravity.
Once the time to rise to the peak of the trajectory is known, the total time of flight can be determined. For a projectile that lands at the same height which it started, the total time of flight is twice the time to rise to the peak. Recall from the last section of Lesson 2 that the trajectory of a projectile is symmetrical about the peak. That is, if it takes 4 seconds to rise to the peak, then it will take 4 seconds to fall from the peak; the total time of flight is 8 seconds. The time of flight of a projectile is twice the time to rise to the peak.
Determination of Horizontal Displacement
The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity. As discussed in the previous part of this lesson, the horizontal displacement of a projectile can be determined using the equation
x = v ix • t
If a projectile has a time of flight of 8 seconds and a horizontal velocity of 20 m/s, then the horizontal displacement is 160 meters (20 m/s • 8 s). If a projectile has a time of flight of 8 seconds and a horizontal velocity of 34 m/s, then the projectile has a horizontal displacement of 272 meters (34 m/s • 8 s). The horizontal displacement is dependent upon the only horizontal parameter that exists for projectiles - the horizontal velocity (v ix).
Determination of the Peak Height
A non-horizontally launched projectile with an initial vertical velocity of 39.2 m/s will reach its peak in 4 seconds. The process of rising to the peak is a vertical motion and is again dependent upon vertical motion parameters (the initial vertical velocity and the vertical acceleration). The height of the projectile at this peak position can be determined using the equation
y = v iy • t + 0.5 • g • t 2
where v iy is the initial vertical velocity in m/s, g is the acceleration of gravity (-9.8 m/s/s) and t is the time in seconds it takes to reach the peak. This equation can be successfully used to determine the vertical displacement of the projectile through the first half of its trajectory (i.e., peak height) provided that the algebra is properly performed and the proper values are substituted for the given variables. Special attention should be given to the facts that the t in the equation is the time up to the peak and the g has a negative value of -9.8 m/s/s.
We Would Like to Suggest ...
Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of ourProjectile Motion Simulator.You can find it in the Physics Interactives section of our website.The simulator allows one to explore projectile motion concepts in an interactive manner. Change a height, change an angle, change a speed, and launch the projectile.
Visit: Projectile Motion Simulator
Check Your Understanding
Answer the following questions and click the button to see the answers.
Aaron Agin is resolving velocity vectors into horizontal and vertical components. For each case, evaluate whether Aaron's diagrams are correct or incorrect. If incorrect, explain the problem or make the correction.
See Answer
A: Aaron is wrong. If v is upwards and rightwards then v y should be directed upwards. Aaron's is downwards.
B: Aaron is wrong. If v is downwards and leftwards then v x should be directed leftwards and v y should be directed downwards. Aaron's components are in the wrong direction.
C: Aaron is wrong. This is a labeling problem. Aaron has labeled the horizontal component as v y and the vertical component as v x.
Use trigonometric functions to resolve the following velocity vectors into horizontal and vertical components. Then utilize kinematic equations to calculate the other motion parameters. Be careful with the equations; be guided by the principle that "perpendicular components of motion are independent of each other."
See Answer
A: v ix = 9.5 m/s • cos(40 degrees) = 7.28 m/s
B : v iy = 9.5 m/s • sin(40 degrees) = 6.11 m/s
C: t up = (6.11 m/s) / (9.8 m/s/s) = 0.623 s
D: t total = 2 • (0.623 s) = 1.25 s
E: x = 7.28 m/s • 1.25 s = 9.07 m
F: y = (6.11 m/s) • (0.623 s) + 0.5(-9.8 m/s/s) • (0.623 s )2 = 1.90 mSee Answer
G: v ix = 25 m/s • cos (60 degrees) = 12.5 m/s
H: v iy = 25 m/s • sin (60 degrees) = 21.7 m/s
I: t up = (21.7 m/s) / (9.8 m/s/s) = 2.21 s
J: t total = 2 • 2.21 s = 4.42 s
K: x = 12.5 m/s • 4.42 s = 55.2 m
L: y = 21.7 m/s • 2.21 s + 0.5 • (-9.8 m/s/s) • (2.21 s )2 = 23.9 mSee Answer
M: v ix = 30 m/s • cos (30 degrees) = 26.0 m/s
N: v iy = 30 m/s • sin (30 degrees) = 15.0 m/s
O: t up = (15.0 m/s) / (9.8 m/s/s) = 1.53 s
P: t total = 2 • 1.53 s = 3.06 s
Q: x = 26.0 m/s • 3.06 s = 79.5 m
R: y = 15.0 m/s • 1.53 s + 0.5 • (-9.8 m/s/s) • (1.53 s )2 = 11.5 m
Utilize kinematic equations and projectile motion concepts to fill in the blanks in the following tables.
See Answer
A: 14.9 m
B: 164 m
C: 2.93 s
D: 5.85 s
E: 42.0 m
F: 240 m
G: 19.2 m/s
H: 16.1 m/s
I: 1.64 s
J: 3.28 s
K: 79.9 deg
L: 19.7 m/s
M: 2.01 s
N: 4.02 s
O: 19.7 m/s
P: 3.47 m/s
Q: 0.354 s
R: 0.709 s
Next Section:
Horizontally Launched Projectile Problems
Non-Horizontally Launched Projectile Problems
Jump To Next Lesson:
Addition of Forces
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13515 | https://www.ams.org/bookstore/pspdf/mcl-7-prev.pdf | 2000 Olympiad Level A Problem 1. Two different numbers x and y (not necessarily integers) sat-isfy x2 −2000x = y2 −2000y. Find the sum of x and y.
Problem 2. Twelve parties took part in a parliamentary election. There were 100 seats to be filled, and to win a seat, a party must obtain more than 5% of the votes. The parties that win seats distribute them among themselves in proportion to the number of votes received: if party A got x times more votes than party B, it will also get x times more seats than party B.
In this election, every voter chose exactly one party: there were no invalid ballots, write-in votes and the like. Each party won an integer number of seats. The Party of Math Lovers got one quarter of the votes. What is the largest number of seats this party can gain? (Explain your answer.) Problem 3. The bases of a trapezoid are m cm and n cm long, where m and n are distinct positive integers. Prove that the trapezoid can be cut into congruent triangles.
Problem 4. The median BM and the side AC of a triangle ABC have equal length. Points D and E are taken on the extensions of the sides BA and AC, re-spectively, so that AD = AB and CE = CM (see figure). Prove that the lines DM and BE are perpen-dicular.
A M B C E D Problem 5. Some of the cards in a stacked deck are face down and the rest are face up. Now and again Pete draws from the deck one or more contiguous cards, subject to the condition that the first and last of them are both face down. He turns over this set of cards as a unit, and inserts them back into the deck in the same place. Prove that sooner or later all the cards in the deck will be face up, no matter what Pete does.
Problem 6. What is the greatest number of chess knights that can be placed on a 5×5 chessboard so that each knight attacks exactly two others?
3 4 PROBLEMS Give an example and explain why an arrangement with more knights is impossible. (A knight in position (a, b) attacks the eight squares in positions (a ± 1, b + 2), (a ± 1, b −2), (a ± 2, b + 1), and (a ± 2, b −1) — or as many of them as fall within the board.) Level B Problem 1. Solve the equation (x + 1)63 + (x + 1)62(x −1) + (x + 1)61(x −1)2 + · · · + (x −1)63 = 0.
Problem 2. Twenty-three positive integers, not necessarily distinct, are written in a row. Prove that we can insert parentheses and plus and times signs between them so that the value of the expression thus obtained is divisible by 2000.
Problem 3. Let A be a point inside a given circle. Find the locus of the vertices C of all rectangles ABCD whose vertices B and D lie on the circle.
Problem 4. Greg filled the squares of a chessboard with the numbers 1, 2, 3, . . . , 63, 64 in some order and is willing to tell Linda, for each rectangle made up of two squares, the sum of the numbers in them. He adds that the numbers 1 and 64 lie on the same diagonal. Prove that this information is enough for Linda to determine exactly the numbers in every square.
Problem 5. The circles having as diameters the sides AB and CD of a convex quadrilateral ABCD are externally tangent to each other, at a point M distinct from the intersection of the quadrilateral’s diagonals. De-note by K the second intersection of the circumcircle of triangle AMC with the line defined by M and the midpoint of AB, and denote by L the second intersection of the circumcircle of triangle BMD with the same line. Prove that |MK −ML| = |AB −CD|.
Problem 6. The stage of a video game is made up of round shelters connected by tunnels; a possible configuration is shown in the illustration.
Your target is in one of the shelters, but you cannot see it. You can blast one shelter at a time, and if it’s the one where your target is, you win. In between shots the target must cross a single tunnel into a neighboring shelter; it doesn’t matter if that shelter has been hit before.
You have a winning strategy if you can plan a sequence of shots that will eventually hit the target no matter where it starts and what moves it makes.
(a) Prove that the configuration in the figure does not admit a winning strategy.
(b) Find all configurations that do not admit a winning strategy, yet acquire one as soon as any of their tunnels is blocked.
2000 OLYMPIAD 5 Level C Problem 1. Two points, A and B, are marked on the graph of the function y = 1/x, x > 0.
Denote by HA and HB the feet of the perpendiculars dropped from these points to the x axis and by O the origin. Prove that the area of the figure bounded by the lines OA and OB and the arc AB of the graph equals the area of the figure bounded by the lines AHA and BHB, the x axis, and the arc AB.
Problem 2. Let f(x) = x2 + 12x + 30. Solve the equation f(f(f(f(f(x))))) = 0.
Problem 3. A convex polygon is drawn on graph paper so that all its vertices are on grid intersections and none of its sides is horizontal or vertical.
Consider the segments of vertical grid lines formed by intersection with the (filled) polygon. Show that the sum of their lengths is equal to the similar sum for horizontal grid lines.
Problem 4. See Problem 5 of Level B.
Problem 5. A sequence x1, x2, . . . , xn, . . . will be denoted by {xn}. Given two (possibly identical) sequences {bn} and {cn}, you can form the sequences {bn+cn}, {bn−cn}, {bn cn}, and {bn/cn} (if all cn are distinct from 0). Also, from any given sequence, a new sequence can be formed by deleting finitely many initial terms.
(a) Starting from the single sequence {an}, where an = n2, can these operations lead to the sequence {n} = 1, 2, 3, 4, . . . ?
(b) What if an = n + √ 2 instead?
(c) What if an = n2000 + 1 n ?
Problem 6. Seven cards were drawn from a deck, shown to everybody, and shuffled. Then Greg and Linda were given three cards each, and the remaining card was either (a) hidden or (b) given to Pat.
Greg and Linda take turns announcing information about their cards.
Are they able to ultimately reveal their cards to each other in such a way that Pat cannot deduce the location of any card he doesn’t see? (No special code was set up in advance; all announcements are in “plain text”.) Level D Problem 1. If two numbers m and n are relatively prime, what is the highest possible value of the greatest common divisor of m + 2000n and n + 2000m?
Problem 2. Compute π 0 |sin 1999x| −|sin 2000x| dx.
6 PROBLEMS Problem 3. The chords AC and BD of a circle with center O meet at a point K. Let M and N be the circumcenters of the triangles AKB and CKD, respectively. Prove that OM = KN.
Problem 4. Denzel has three sticks. If it is impossible to make a triangle with these sticks, Denzel shortens the longest of them by a length equal to the sum of lengths of two other sticks. If the stick did not disappear after this operation, and it is still impossible to make a triangle, Denzel repeats the operation, and so on. Can this process continue endlessly?
Problem 5. Each participant of a round-robin chess tournament plays one game against each other. A win is worth one point, a draw half a point, and a loss zero. A game is called anomalous if its winner ends the tournament with a score less than the game loser’s score.
(a) Can anomalous games amount to more than 75% of total number of games in the tournament?
(b) Can they amount to more than 70%?
Problem 6. Is it possible to arrange infinitely many congruent convex poly-hedra in a layer bounded by two parallel planes so that no polyhedron can be removed from the layer without moving the remaining ones?
2000 Olympiad Level A 1. Move 2000x to the right-hand side and y2 to the left-hand side.
3. Extend the sides of the trapezoid until they meet.
4. Consider the triangle KBE, where K is point symmetric to M with respect to A.
5. Use a monovariant (see Fact 2).
6. The number of knights on white squares is equal to the number of knights on black squares.
Level B 1. Use the formula for an −bn.
2. Suppose that we have seven numbers: four numbers divisible by two and three numbers divisible by five; then their product is divisible by 2000.
3. Apply the Pythagorean Theorem several times.
4. Linda can find out the difference between the numbers in any two squares of the same color.
5. Let O1 and O2 be the circumcenters of the triangles AMC and BMD.
Consider the orthogonal projections of O1, O2, and the midpoint of O1O2 on the line mentioned in the problem’s statement.
6. Consider the shelters where the target can be after an even number of shots.
Level C 1. The areas of the triangles OAHA and OBHB are equal to 1 2.
2. Complete the square.
5. (b) All sequences that can be obtained from {n + √ 2} are of the form P(n + √ 2) Q(n + √ 2) , where P and Q are polynomials with integer coefficients.
(c) Consider the mapping that takes the sequence {an} into {an+1 −an}.
Apply it to the initial sequence enough times.
28 2000 OLYMPIAD 29 6. (a) First, try to understand how Greg can make his cards known to Linda without letting Pat know anything about them.
(b) Let Greg and Linda number their cards from 0 to 6.
Level D 1. gcd(dm, dn) = d, for any d.
3. Either the points O, M, K, and N are on the same line or the quadrilateral OMKN is a parallelogram.
4. Choose the sticks so that after each truncation the ratios of their lengths remain the same.
5. (a) Let the number of participants be 2M. Call strong the participants that took the first M places and weak all the rest. Consider three types of games: strong against strong, strong against weak, and weak against weak.
(b) Construct the table of a tournament in which all players have the same final score, and the number of ones above the main diagonal is approximately equal to a quarter of the total number of games.
2000 Olympiad Level A 1. x + y = 2000.
2. 50 seats.
6. 16 knights.
Level B 1. x = 0.
3. If O and R are the center and radius of the given circle, the desired locus is the circle centered at O with radius √ 2R2 −OA2.
6. (b) The configuration of part (a) and all configurations consisting of exactly one cycle of shelters, such as the one in the figure.
Level C 2. x = −6 ± 32 √ 6.
5. (a) yes; (b) no; (c) yes.
6. (a) yes; (b) yes.
Level D 1. 20002 −1.
2. 0.
4. Yes.
5. (a) no; (b) yes.
6. Yes.
40 2000 Olympiad Level A Problem 1. Move 2000x to the right-hand side and y2 to the left-hand side, to obtain x2 −y2 = 2000x −2000y.
Factor the difference of squares in the left-hand side: (x −y)(x + y) = 2000(x −y).
Since x ̸= y, we can divide both sides by x −y to obtain x + y = 2000.
Problem 2. The idea of the solution is that the Party of Math Lovers (PML) gains the greatest number of seats if the total number of votes going to parties that don’t get seats (because they got 5% or less of the total vote) is as large as possible.
If 10 parties each obtain exactly 5% of votes and two parties, including PML, 25% each, then only two parties will win seats in the parliament. Each will have exactly 50 seats.
Let us prove that PML cannot win a greater number of seats. If 11 parties fail to pass the 5% threshold, together they’ve received at most 55% of votes; but 55% + 25% < 100%. Therefore, at most 10 parties failed to meet the threshold, and they’ve received at most 50% of the votes. Hence, the parties that did get seats got at least 50% of votes. Thus PML’s 25% of the total vote represents at most half the total vote, hence at most half the seats, or 50 seats.
Problem 3. Suppose that m > n. We extend the lateral sides of the trapezoid to form a triangle and divide each side of this triangle into m equal parts. The lines drawn through the division points parallel to the triangle’s sides cut the triangle into congruent small triangles. The smaller base of the trapezoid is one of these lines (because its length in centimeters is integer), so the trapezoid is cut into congruent triangles.
m n 49 50 SOLUTIONS Problem 4. Let a be the distance from C to M (see figure). Let K be a point symmetric to M about A.
Then KM = MB = ME = 2a. Therefore, KBE is a right triangle (see Fact 14), with KB ⊥BE.
Also, the quadrilateral DKBM is a parallelo-gram, since its diagonals bisect each other. Hence DM ∥KB and DM ⊥BE.
A K M B C E D a a a a 2a Problem 5. We will encode the arrangement of the cards in the deck by a number with as many digits as there are cards. The k-th digit of this number from the right is 1 if the k-th card from the bottom is turned its back up; otherwise, the k-th digit is 2. For instance, if all cards in the deck are turned down, the resulting number is 2222. . . 222.
It is easily seen that after each of Pete’s operations the code of the deck decreases. Indeed, let us compare the codes after and before such an operation. Consider all the digits that changed and take the leftmost, that is, the highest-order, digit. Obviously, this digit changed from 2 to 1. But this means that the encoding number decreased.
Since there are only finitely many n-digit numbers made of 1s and 2s, our number cannot decrease forever: eventually we will reach the number 1111. . . 111, corresponding to the arrangement in which all the cards are turned up. See also Facts 2 and 11.
Problem 6. An arrangement of 16 knights satisfying the condition in the problem is shown in the figure. Let us show that a greater number of knights is impossible. Color the squares of the chessboard black and white as shown in the figure. We first show that the number of knights on black squares must be equal to the number of knights on white squares.
Join all pairs of knights that attack each other by a line segment; then each segment joins a white square with a black one. Each square is joined to other squares by two segments.
It follows that the number of segments equals twice the number of knights in white squares and, at the same time, twice the number of knights in black squares. Therefore, these two numbers coincide.
There are 12 white and 13 black squares. If the number of empty white squares is n, the number of empty black squares is n + 1, and it suffices to prove that n ≥4, because in this case at least 4 + (4 + 1) = 9 squares are empty.
For any optimal arrangement of knights, the central square is empty.
Otherwise, six of the eight white squares attacked by the knight in the central square are empty, so n ≥6, so the number of knights is at most 25 −6 −(6 + 1) = 12.
2000 OLYMPIAD – LEVEL B 51 Consider the square marked 1 in the figure. If it is not empty, four of the six black squares attacked by the knight in this square are empty.
Taking into account the empty central square, we see that in this case at least 5 black squares are empty: n + 1 ≥5. Thus, we can assume that square 1 is empty. The same argument works for the squares marked 2, 3, and 4: they must also be empty. But then n ≥4, as we wished to prove.
1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 2 3 Level B Problem 1. We multiply both sides of the equation by (x+1)−(x−1) = 2, to obtain after simplification (x + 1)64 −(x −1)64 = 0.
This follows from the formula an −bn = (a −b)(an−1 + an−2b + · · · + abn−2 + bn−1) applied to a = x + 1, b = x −1, n = 64.
This equation is readily solved: the condition (x + 1)64 = (x −1)64 is equivalent to |x+1| = |x−1|. Since x+1 ̸= x−1, we obtain x+1 = −(x−1), whence x = 0.
Problem 2. Divide the 23 given numbers into seven groups of consecutive numbers: three groups of five numbers each and four of two numbers each (the order of the groups is irrelevant). Each group is enclosed in parentheses, and multiplication signs are placed between the groups. If the signs inside the groups are inserted in such a way that the value of each two-number parentheses is divisible by 2 and that of each five-number parentheses is divisible by 5, then the entire expression is divisible by 24 · 53 = 2000.
We show that the desired arrangement of operation signs is possible.
First, consider a group of two numbers. If these numbers are of different parity, we insert the multiplication sign between them; if they are of the same parity, we insert the plus sign. The result is obviously divisible by 2 (see Fact 23).
Now consider a group of five numbers a1, a2, a3, a4, and a5 in this order.
Write down the remainders of the following five sums when divided by 5: a1, a1 + a2, a1 + a2 + a3, a1 + a2 + a3 + a4, a1 + a2 + a3 + a4 + a5.
If any of these remainders is 0, the corresponding sum is divisible by 5.
In that case, we insert pluses between the terms of this sum, put the sum in parentheses if necessary, and fill the remaining spaces between the numbers of the group with multiplication signs.
If none of the remainders is zero, then, by the pigeonhole principle (see Fact 1), at least two of them will be equal, since there are five sums and only four possible nonzero remainders. Suppose that the sums a1 + · · · + ai 52 SOLUTIONS and a1 + · · · + aj (i < j) yield the same remainder upon division by 5.
Then the difference between these two sums is divisible by 5, and it equals ai+1 + · · · + aj.
We form this last sum by inserting pluses between the numbers in it, place parentheses around it, and fill the remaining spaces with multiplication signs. Two examples follow: (1 + 2 + 3 + 4) · 3 = 30; 2 · (1 + 4) · 7 · 1 = 70.
Thus, we can always insert signs in a group of five numbers so that the result be divisible by 5.
Remark. This problem is a variation on the following classical piece: in any string of n integers, one can choose one number or several numbers in a row whose sum is divisible by n.
Problem 3. Let O be the center of the circle described in the statement of the problem. We prove that the point C satisfies OC2 = 2R2 −OA2, where R is the radius of the circle. In the notation of the figure , the segment OC is the hypotenuse of a right triangle with legs x and y. By the Pythagorean Theorem, OC2 = x2+y2 = (x2+t2)+(y2+z2)−(z2+t2).
It remains to notice that x2+t2 = y2+z2 = R2 and z2 + t2 = OA2 (again by Pythagoras). It follows that the distance from C to A is constant and equal to √ 2R2 −OA2, i.e., the locus in question is the circle with center O and radius √ 2R2 −OA2.
O A D B C z t y x O A D B C z t y x Conversely, let a point C satisfy OC2 = 2R2 −OA2.
We’ll prove that there is a rectangle ABCD with vertices B and D on the initial circle. To this end we construct a circle with diameter AC (see figure). It intersects the initial circle at two points.
Denote one of them by B.
Then ∠ABC = 90◦(see Fact 14). Let us complete the triangle ABC to a rectangle ABCD. It remains to prove that D is on the circle. In the figure’s notation: OD2 = y2 + z2 = (x2 + y2) + (z2 + t2) −(x2 + t2) = (2R2 −OA2) + OA2 −R2 = R2.
Problem 4. Suppose that Linda has lots of 1 × 2 dominoes. If she places two dominoes on the board so that they have exactly one common square — say covering x-y and y-z in the figure — then, by subtracting the sum of the numbers under the y-z domino, namely y + z, from the sum under the x-y domino, namely x + y, she will find the difference x −z of the numbers in the two squares currently covered by a single domino (which have the same color).
x y z t The next domino will be placed so as to join a single-domino square with an as yet uncovered square — say, the domino zt in the figure at the bottom of the previous page. Adding the difference obtained above with 2000 OLYMPIAD – LEVEL B 53 the sum of numbers under the new domino, Linda will obtain the sum x + t of the two numbers in squares currently covered by a single domino (which now have opposite colors). Proceeding in the same way, Linda can add new dominoes one after another to construct a chain joining any two squares. If these squares are the same color, she gets to know their difference; otherwise, their sum.
It is known that 1 and 64 are on the same diagonal, and so on squares of the same color.
Their difference is 63, whereas the difference of any two other integers between 1 and 64 is less than 63. Therefore, Linda can determine which two squares contain 1 and 64. Now, knowing the sum of (or the difference between) 64 and the number in any other square, she can determine the positions of all numbers.
Problem 5. First solution. Let P and Q be the midpoints of AB and CD, and let O1 and O2 be the centers of the circles drawn through A, M, and C and B, M, and D, respectively. Denote by H1 and H2 the projections of O1 and O2 on the line PQ: P B C Q D A K L O1 O2 E F L′ L′′ K′ K′′ N M M M M M M M M M M M M M M M M M M M M M M Step 1. The points M, P, and Q are collinear. Indeed, the lines PM and QM contain the radii of the circles tangent at M, and hence they are per-pendicular to the common inner tangent of these circles.
Step 2. The points P and Q lie on the circle with diameter O1O2. Indeed, PO1 and PO2 are perpendicular, being the perpendicular bisectors to the segments MA and MB, because M lies on the circle with diameter AB (see Fact 14). Similarly, QO1 ⊥QO2.
Step 3. Clearly, KH1 = H1M and LH2 = H2M (the radius perpendicular to a chord bisects it).
Step 4. We have PH1 = QH2, since the projection of the midpoint of O1O2 bisects both H1H2 and PQ (here we have again used the fact that the radius 54 SOLUTIONS perpendicular to a chord bisects it). It finally follows that |MK −ML| = 2|MH1 −MH2| = 2|MP −MQ| = 2 1 2AB −1 2CD = |AB −CD|.
Second solution. Denote by ωAMB the circle passing through A, M, and B, by ωAMC the one through A, M, and C, and so on.
Let E be the intersection point of KL with ωAMB, and let F be the intersection point of the same line and ωCMD (E, F ̸= M; see figure). Since EM is the diameter of the circle ωAMB, we have EM = AB. It follows that MK −AB = KE. Similarly, ML −CD = FL. Therefore, it will suffice to show that KE = FL.
P B C Q D A K L O1 O2 E F H1 H2 M M M M M M M M M M M M M M M M M M M M M M If the extension of AB meets ωAMC at a point K′, then KE = K′B.
Indeed, since a circle is symmetric about any of its diameters, the reflection in the line PO1 drawn through the centers of ωAMC and ωAMB takes both these circles into themselves. It follows that each of the common points A and M of these circles is taken under this reflection into the other. At the same time, points on the mirror line, in particular, P, stay in place.
Therefore, the line AP is reflected into MP and the intersection K′ of AP and the circle ωAMC is reflected onto K, the intersection of MP and the same circle. Similarly, point E is reflected onto B. Hence segment KE is taken onto K′B, and so the two segments are congruent.
For the same reasons, if L′ is the intersection point of the line CD and the circle ωBMD distinct from D, then FL = CL′. Thus, it remains to prove that BK′ = CL′.
Let L′′ be the meet of CD and ωAMC distinct from C. Considering the reflection in the line QO1, joining the centers of circles ωAMC and ωCMD and taking each of them into itself, we see that MK = CL′′. Similarly, MK = AK′. Therefore, CL′′ = AK′. Suppose that the chords AK′ and CL′′ meet at N. Since they are congruent, the bisector of the angle ANL′′ passes through the center of circle ωAMC.
2000 OLYMPIAD – LEVEL B 55 In the same way, it is proved that this bisector passes through the center of ωBMD (if the extension of AK′ meets ωBMD at point K′′, then it can be shown that BK′′ = DL′). Therefore, the reflection in this bisector takes the circles ωAMC and ωBMD into themselves and the lines CL′ and AK′ into each other. Hence it takes the segments BK′ and CL′ intercepted by these circles on these lines into each other. Thus these segments are congruent, which completes the solution.
Problem 6. (a) Let’s call the configuration in the statement of the problem a tripus: an octopus with three legs. We must prove that the tripus is safe.
Label the shelters as in the figure.
We’ll call the shelters O, A2, B2, and C2 even, and the remaining ones odd. Let’s assume the target starts in an even shelter. This assumption can only make things easier for the shooter, but we will prove that the target might avoid being hit even so.
O A1 A2 A3 B1 B2 B3 C1 C2 C3 From an even shelter the target can only run to an odd one, and vice versa. Therefore, after any odd number of shots the target is in an odd shelter and after an even number of shots, in an even shelter. This means that the player must fire his even shots at odd shelters and odd shots at even shelters.
We will prove that before each odd shot, the target can be in O and in one of two other even shelters (i. e., the shooter doesn’t know in which of these three shelters the target is hiding).
The proof is by induction on the number of shots. Our statement is clearly true at the start. Suppose that before the (2k −1)-st shot the target can be in O and in two other even shelters — say A2 and B2.
Case 1. Suppose the (2k −1)-st shot is fired at O. Then before the next shot the target can be in any of A1, A3, B1, and B3. The 2k-th shot must be fired at an odd shelter. It is readily seen that in all cases, before the (2k +1)-st shot the target can be in any of the shelters O, A2, and B2 again.
Case 2. Suppose that the (2k −1)-st shot is fired at A2. Then before the following shot, the target can be in any of A1, B1, C1, and B3. If the 2k-th shot is fired at A1, then before the (2k + 1)-st shot the target can be in O, B2, and C2; if the shot was fired at B1, the possible locations of the target are O, A2, and C2; and if at C1, then O, A2, and B2. If the shooter fires at one of the shelters A3, B3, C3, the target can appear in any of four shelters.
In any case, before the (2k +1)-st shot the target can be in O and two more even shelters. This completes the proof of our statement in case 2.
Case 3. Suppose the (2k −1)-st shot is fired at B2. This case is completely similar to the previous one.
Case 4. Suppose the (2k−1)-st shot is fired at C2. Then before the following shot, the target can appear in any of the shelters A1, A3, B1, and B3, and this case is similar to case 1.
56 SOLUTIONS Thus, before each odd shot, the target can be in several different shelters, and the shooter won’t necessarily be able to hit it.
It is clear from the previous analysis that the shooter can’t hit it in an even shot either.
(b) If a configuration includes a cycle A1A2 . . . An(A1) of several shelters, it is safe. Indeed, the target always has two choices of where to go, so no strategy can possibly lead to a situation where the shooter knows where the target will be next.
Also, if some part of a configuration is safe by itself, the target can survive by staying only in this part; hence the entire system is safe as well.
Therefore, if a configuration includes a cycle plus anything else, it does not satisfy the minimality condition in part (b) of the statement of the problem.
Let’s show that any cycle is a minimal configuration. Imagine that one of the edges, say AnA1, is removed. What is left is a linear chain of shelters, A1A2 . . . An. The target is then doomed, if the shooter uses the following strategy: Fire successively at A1, A2, . . . , An−1. This necessarily hits the target if it started offin an odd-numbered shelter (for the proof, consider the moment when the distance from the target to the shelter hit at this moment is minimal). If still alive, at the end of these n −1 shots the target is either in an odd- or an even-numbered shelter, depending only on the parity of n.
The shooter now goes back to A1 (if n is even) or to A2 (if n is odd), and again shoots the shelters in increasing order up to An−1. This second sweep is guaranteed to hit the target.
It remains to consider configurations without cycles. We’ll show that the tripus is the only safe minimal configuration of this kind. We describe the strategy of the shooter, assuming there is neither a cycle nor a tripus. (We’ll do this assuming all the shelters are connected together. If the configuration is disconnected, i. e., it consists of two or more components not joined by tunnels to one another, then the shooter must successively apply the strategy below to each component.) Let a shelter with three or more tunnels radiating from it be called a hub. A tunnel starting from a certain shelter is said to be a through-pass if, having run through it, the target can run through two more tunnels without visiting the same shelter twice. For instance, the tunnel A1A2 in the tripus leading from the shelter A1 is not a through-pass, whereas the tunnel A2A1 leading from A2 is a through-pass. Finally, a shelter is called a dead-end if there is only one tunnel leading from it.
Since our configuration does not contain a tripus or a cycle, there are at most two through-pass tunnels leading from any shelter. Let us determine the first shelter to be attacked by the shooter. We take any hub. If there are two through-pass tunnels starting at this hub and leading to another hub each, we choose any of them and pass through it (and, perhaps, through a number of other tunnels) to the closest hub. If there is another through-pass tunnel leading from this new hub to another one, we pass through it to the next closest hub and proceed so until we arrive at a hub with a single 2000 OLYMPIAD – LEVEL C 57 through-pass tunnel issuing from it or two such tunnels one of which leads to a dead-end without passing through hubs on the way. In the first case, we move along any non-through-pass tunnel to a neighboring shelter; in the second case, we move through the through-pass tunnel leading to the dead-end and stop at the shelter adjacent with the dead-end. This determines the shelter at which the shooting must begin.
Let us divide all shelters into even and odd ones so that any tunnel joins shelters of different parity (this is possible, because there are no cycles).
We’ll explain how to fire in order to hit the target provided that initially he was hiding in a shelter of the same parity as the one at which the shooter fires first. (If, after this series of shots, the target survives, then his initial location was of different parity. Now the current parity of the target’s location is known exactly.) Suppose that the shooter fires at successive shelters one by one starting with the one chosen above and ending with a hub.
Then the target is outside the bombarded linear part of the configuration. We have chosen the initial shelter so that at most one of the tunnels leading to other parts of the configuration is a through-pass. (Even if there are two through-pass tunnels, the second one leads to the part that has just been shelled.) Any non-through-pass tunnel leads either to a dead-end shelter or to a shelter from which one can get to a number of dead-ends or come back to the hub.
In both cases, there is only one shelter beyond this tunnel whose parity is opposite to that of the hub in question. The same is the parity of the shelter to which the target moves after the shot at the hub. So if this shelter is beyond this tunnel, then the target will be hit by the shot at the shelter to which the tunnel leads. Otherwise, the shooter fires at the hub again, thus preventing the target from running over to the parts of the configuration that have already been “checked.” Having checked all non-through-pass tunnels, the shooter sets on the only through-pass one, hits the shelter to which it leads, and then successively all shelters up to the nearest hub. Then the non-through-pass tunnels at this hub are checked in the same way as above, and so on. Thus the entire configuration can be checked.
We have shown that a configuration that contains no cycle and no tripus cannot be safe. Destruction of any tunnel in a tripus makes it unsafe, so a tripus is a minimal safe configuration. Finally, any configuration consisting of a tripus and something else is safe, but not minimal.
Level C Problem 1. We can assume the x-coordinate of point A O HA A B HB K y x is less than that of B (see figure). Let K be the intersection of the segments AHA and OB. The curvilinear triangle AKB is the intersection of the figures whose areas we must show to be equal. Therefore, the difference in area between the two figures is equal to the difference between the areas of the triangle OAK and the 58 SOLUTIONS quadrilateral HAKBHB. But this difference is zero: S(OAK) −S(HAKBHB) = S(OAHA) −S(OBHB) = 1 2 OHA · AHA −1 2 OHB · BHB = 1 2 −1 2 = 0.
The last but one equation follows from fact that points A and B lie on the graph of the function y = 1/x.
Problem 2. Notice that f(x) = (x + 6)2 −6.
Then f(f(x)) = ((x+6)2−6)+6 2−6 = (x+6)4−6, f(f(f(x))) = (x+6)8−6 and so on. Finally, f(f(f(f(f(x))))) = (x + 6)32 −6, and the solution of the equation (x + 6)32 = 6 is obvious: x = −6 ± 32 √ 6.
Problem 3. We’ll prove that each of the sums in the statement is equal to the area of the polygon (relative to the area of the grid square). Con-sider, for instance, the sum of lengths of the horizontal segments.
Denote the lengths of these segments by a1, . . . , an (see figure).
These segments divide the polygon into two triangles and n −1 trapezoids, the altitudes of all these subpolygons being equal to 1. By the familiar formula, the area of the i-th trapezoid is equal to (ai + ai+1)/2 grid squares, and the areas of the triangles are equal to a1/2 squares and an/2 squares. Hence the area of the polygon is equal to a1 a2 an−1 an a1 2 + a1 + a2 2 + · · · + an−1 + an 2 + an 2 = a1 + a1 + · · · + an, completing the proof.
Remarks. 1. The statement of the problem remains true for a nonconvex polygon as well.
Furthermore, we can allow the polygon’s sides to be horizontal or vertical if the lengths of horizontal and vertical segments on the boundary of the polygon are counted in the corresponding sums with a factor of 1 2.
2. Here is an outline of an alternative solution. Each of the sums is additive in the following sense: if the polygon is cut into several pieces, then its “horizontal” or “vertical” sum is equal to the sum of the corresponding sums for all the pieces.
But it is not difficult to see that the only additive function on the set of all polygons with integer vertices equal to 1 on a unit square and invariant under translations and central symmetries is the area.
3. Consider a sufficiently “good” set on the coordinate plane (for instance, a poly-gon or a convex set). Suppose that the length of its intersection with the line x = a is equal to f(a) and the length of its intersection with the line y = b is equal to g(b). Then the area of our set can be computed by the formulas +∞ −∞ f(x) dx = +∞ −∞ g(y) dy.
2000 OLYMPIAD – LEVEL C 59 Our problem is a discrete analog of these formulas. Note that many statements in calculus have discrete analogs. See [651, Chapter 11, §4].
Problem 5. We explain how to perform the desired transformations in case (a) and (c). By deleting the first term, the sequence {an} is transformed into the sequence {an+1}. Subtracting the first sequence from the second, we obtain the sequence {an+1 −an}. We denote this transformation by T, and its m-fold iteration by T m. Dividing the given sequence by itself, we obtain the useful sequence all of whose terms are ones; it will be denoted by I. Now let us show how to obtain the sequence {n} in case (a): {n2} T − →{2n + 1} −I − − →{2n} /(I+I) − − − − →{n}.
Item (c) is harder. First, notice that if P(n) is a polynomial in n of degree m, then the application of T to the sequence {P(n)} yields the sequence {Q(n)}, where Q(n) is a polynomial of degree m−1 (see Fact 22). Hence the applica-tion of T m−1 to {P(n)} yields a polynomial of degree 1, that is, a sequence of the form {an + b}, and the application of T m+1 yields the zero sequence.
We have n2000 + 1 n = n1999 + 1 n .
The transformation T 2000 applied to the term n1999 of this sequence yields 0.
By induction, it can readily be seen that the application of T 2000 to the second term yields 2000!
n(n + 1) · · · · · (n + 2000).
Now it is not difficult to write out the desired chain of transformations: n2000+1 n T 2000 − − − → 2000!
n(n+1) · · · (n+2000) I/ − → n(n+1) . . . (n+2000) 2000!
×( 2000! times I+I+···+I) − − − − − − − − − →{n(n+1) · · · (n+2000)} T 2000 − − − →{an+b}, where a and b are integers and a ̸= 0. The remaining operations are clear.
(b) We prove that in this case it is impossible to obtain the sequence {n}.
First, all sequences that can be obtained from {n+ √ 2} are of the form P(n + √ 2) Q(n + √ 2) , where P and Q are polynomials with integer coefficients. (Indeed, the initial sequence is of this form. The term-by-term addition, subtraction, multipli-cation, or division of sequences of this form yields sequences of the same form again, and the deletion of a few initial terms is equivalent to the replacement of P(x)/Q(x) by P(x + r)/Q(x + r) for a certain positive integer r, which can be rewritten in this form by opening the brackets in the numerator and denominator.) Further, if the sequence {n} can be represented in this form, 60 SOLUTIONS the same is true for the sequence all of whose terms are equal to √ 2. But the relation P(n + √ 2)/Q(n + √ 2) = √ 2 implies that the ratio of the leading coefficients of the polynomials P and Q is √ 2, which is impossible.
Problem 6. (a) Suppose that Greg names two sets of cards — the one he was given and a set of three other cards — and says, “I have one of these sets.” Then Linda will get to know Greg’s cards (because Linda’s set is disjoint with Greg’s set and necessarily intersects the second set called by Greg).
Now two situations are possible: if the second set named by Greg does not coincide with Linda’s set, then Linda must name her set of cards and Greg’s set of cards and say, “I have one of these sets.” If the second set named by Greg coincides with Linda’s set, then this rule is no good, because Pat will compute the hidden card. So in this case Linda calls her set of cards plus any three other cards, so long as they don’t form Greg’s set.
After that each of the players will know the entire deal. On the other hand, Pat doesn’t know anything definite. Indeed, three sets of cards were named: A, B, and C. Greg said, “I have either A or B”; Linda said, “I have either A or C” (the sets B and C have two cards in common, and other pairs of sets are disjoint). This means that either Greg has the set A and Linda the set C or Greg’s set is B and Linda’s set is A. Of course, these two deals are different, and even the hidden card cannot be determined.
(b) In this case method (a) does not work: knowing the hidden card, Pat will be able to exactly determine the deal. Let us number the cards from 0 to 6. Suppose that Greg and Linda in turn announce the remainders of the sums of numbers of their cards upon division by 7 (see Fact 6). Then they will get to know the deal: each of them must only add to his/her sum the other sum and find the remainder opposite to this total sum modulo 7 (i.e., the remainder that gives a sum divisible by 7 when added to Greg’s and Linda’s sums). This will be the number of the hidden card. After that Linda and Greg will readily figure out the deal.
Let us check that Pat learns nothing. Consider the card with number s. We show that it could have been dealt to Greg if he announced a sum a.
It suffices to add to this card two other cards whose sum of numbers gives the same remainder as a −s when divided by 7. It is readily seen (check this!) that there exist three different pairs of numbers with this property.
The pair we need must not include the card with number s and Pat’s card.
Thus, two pairs at most can be left out of consideration; but in any case, at least one pair remains, and we complete Greg’s set with this pair. Similar argument shows that Linda also can be dealt any given card.
Remark. Notice that the method of part (b) doesn’t work in part (a): Nick will detect the hidden card.
2000 OLYMPIAD – LEVEL D 61 Level D Problem 1. Set a = 2000m + n, b = 2000n + m, and denote by d the greatest common divisor of a and b. Then d also divides the numbers 2000a −b = (20002 −1)m and 2000b −a = (20002 −1)n.
Since m and n are coprime, the number d is a divisor of 20002 −1 (see the remark). On the other hand, for m = 20002 −2000 −1, n = 1, we have a = (20002 −1)(2000 −1), b = 20002 −1 = d.
Remark. We have used the following statement: if m and n are coprime, then gcd(dm, dn) = d. This follows from the more general equality gcd(dx, dy) = d gcd(x, y).
We outline the proof. Let p1, . . . , pn be all the prime factors of the numbers x, y, and d. Write the prime factorizations of these numbers: x = pk1 1 pk2 2 . . . pkn n , y = pl1 1 pl2 2 . . . pln n , d = pm1 1 pm2 2 . . . pmn n .
(Some of the exponents ki, li, and mi can be equal to zero. Compare with the remark to Problem 95.B.4.) Denote by min(p, q) the smaller of the numbers p and q. Then gcd(x, y) = pmin(k1,l1) 1 pmin(k2,l2) 2 . . . pmin(kn,ln) n .
Similarly, gcd(dx, dy) = pmin(m1+k1,m1+l1) 1 pmin(m2+k2,m2+l2) 2 . . . pmin(mn+kn,mn+ln) n .
It remains to notice that for any i, we have min(mi + ki, mi + li) = mi + min(ki, li).
Another proof can be obtained using the following fact: for any integer x and y, there exist integer a and b such that ax + by = gcd(x, y).
See , for instance.
Problem 2. First solution. Since the integral of the difference of functions is equal to the difference of integrals of these functions (see Fact 28), we have π 0 |sin 1999x| −|sin 2000x| dx = π 0 |sin 1999x| dx − π 0 |sin 2000x| dx.
(1) We’ll show that π 0 |sin kx| dx = 2 for any positive integer k. This will imply that the right-hand side of (1) is zero.
62 SOLUTIONS The function |sin kx| is periodic with period π/k. This means that π 0 |sin kx| dx = k π/k 0 |sin kx| dx = k π/k 0 sin kx dx.
(2) It is easy to compute the integral in the right-hand side by the change of variable y = kx (see Fact 28): π/k 0 sin kx dx = π 0 sin y dy k = 2 k.
Hence the integral (2) is equal to 2 for any k, and we are done.
Second solution (sketch). The graph of the function y = |sin kx| on [0, π] consists of k identical caps (see figure) obtained by compressing the graph of y = sin x on the same interval toward the y-axis by the factor k.
In this process, the area under the graph also decreases by the factor k. Thus the area under the k caps remains the same for any k.
y = sin |kx| 0 π Problem 3. Denote by X the midpoint of KB (see left figure on the next page). Then ∠KMX = 1 2∠KMB = ∠KAB = ∠KDC; here the middle equation is based on the fact that the measure of an inscribed angle equals half the bend of its intercepted arc, and in the last equation we use the consequence of this theorem: inscribed angles subtended by the same arc are congruent. Clearly, MX is perpendicular to BD; therefore, KM is perpendicular to CD. We also have ON ⊥CD; hence ON is parallel to KM. Similarly, OM ∥KN.
O D C B A K X M N O D C B A K M N If the points O, K, M, and N do not lie on the same line, then OMKN is a parallelogram and OM = KN. Otherwise, we consider the orthogonal projections of segments OM and KN on AC (see figure above and to the right). Since the projections of points O, M, and N are the midpoints of segments AC, AK, and KC, respectively, the projections of OM and KN are equal to 1 2KC. Since these segments lie on the same line, the equality of the lengths of projections implies the congruence of the segments themselves.
2000 OLYMPIAD – LEVEL D 63 Problem 4. Consider the polynomial P(x) = x3 −x2 −x −1.
It has a root t greater than 1, since P(1) < 0 and P(2) > 0 (see the remark).
Suppose that the lengths of the sticks are equal to t3, t2, and t. It follows that t3 = t2 + t + 1 > t2 + t; by the triangle inequality, there is no triangle with such side lengths. After the longest stick is broken, we obtain three sticks of lengths t2, t, and 1. Since the length ratios remain the same, the process will continue forever.
Remark. We’ve used the intermediate value theorem, which says that if a function P(x) is continuous on the interval [a, b] and takes values of opposite sign at its endpoints — for instance, P(a) < 0 and P(b) > 0 — then there exists a point c ∈(a; b) such that P(c) = 0.
We won’t prove this theorem, or even define continuity; the reader can consult [651, Chapter 4]. We remark only that any polynomial is continuous.
Problem 5. (a) Let N be the number of players, and set M = [N/2]. The players that won the first M places will be called strong and the other players weak. Players with the same score are distributed between the two groups arbitrarily. Let X be the number of normal (not anomalous) games between strong and weak players. Strong players scored a total of M(M−1)/2 in games among themselves and no more than X in games against weak players.
Denote by S1 the total score of strong players and by S2 the total score of weak players. Then S1 ≤M(M −1) 2 + X, S1 + S2 = N(N −1) 2 .
If all individual scores were equal, there would be no anomalous games.
Therefore we can assume that there are two players with different scores.
First, consider the case of an even N. In this case, there are equally many weak and strong players. Then, clearly, S1 > S2. It follows that S1 > S1 + S2 2 = N(N −1) 4 , whence X ≥S1 −M(M −1) 2 > N(N −1) 4 −M(M −1) 2 .
Substituting M = N/2, we can readily ascertain that X > N 2 8 > N(N −1) 8 .
Since the total number of games is N(N −1)/2, the portion of normal games is greater than 1 4.
However, this proof doesn’t work for odd N. In this case, we can proceed as follows: consider the average result S1/M of a strong player. Clearly, it is greater than the average result of a weak player, and hence it is greater than the average over all players: S1 M > N(N −1)/2 N .
64 SOLUTIONS That is, S1 > M(N −1)/2. Therefore, X ≥S1 −M(M −1) 2 > M(N −M) 2 ≥N(N −1) 8 , where the last inequality is checked by taking M = (N −1)/8.
Remark. The proof for odd N is good for even N as well.
(b) First, consider the tournament of 2k+1 players in which each partici-pant with a number i ≤k lost to all participants with numbers i+1, . . . , i+k and beat all the other participants, and each participant with a number i > k beat the players with numbers i−k, . . . , i−1 and lost to the rest. Obviously, all players scored k points each.
Consider the tournament table (see figure; rows and columns correspond to players). It is not difficult to see that above the main diagonal, exactly 0 0 0 1 1 1 1 2 1 2 0 0 0 0 0 0 1 1 1 k+1 2k+1 1 1 1 0 0 0 1 1 1 1 1 1 k+1 0 0 0 1 1 1 0 0 0 0 0 0 2k+1 1 1 1 k(k+1)/2 positions out of 2k(2k+1)/2 are occupied by 1s. Now imagine we clone each player, replacing him or her by a group of n players so that players from different groups play with each other with the same result as their originals, and players from the same group tie with each other. We obtain a new table in which all players have the same score again.
Let us correct this table so that the scores become different. We’ll change results of players from the (k + 1)-st group: in of their wins in the games with players from the (k + 1 −i)-th group must be replaced by ties so that the score of each player from the (k +1)-st group decreases by i/2, and that of each player from the (k + 1 −i)-th group increases by i/2. Similarly, in losses of players from the (k + 1)-st group against (k + 1 + i)-th group are replaced by ties.
Then the first place in the tournament will be won by players from the first group, the second place will be taken by the second group, a nd so on.
Let us count the number of anomalous games.
All games lost by players from groups with numbers i ≤k are anomalous.
All in all there are kn2 −in such games (recall that in games against players from the (k + 1)-st group ended in a tie). For i > k + 1, players from the i-th group lost (2k + 1 −i)n2 anomalous games. Finally, players from the 2000 OLYMPIAD – LEVEL D 65 (k + 1)-st group lost kn2 −n 1 2k(k + 1) anomalous games; here the second term corresponds to anomalous losses replaced by ties.
Thus, the number of anomalous games is k i=1 (kn2 −in) + 2k+1 i=k+2 (2k + 1 −i)n2 + kn2 −nk(k+1) 2 = 3k2+k 2 n2 −k(k + 1)n.
At the same time, the total number of games is n(2k+1)(n(2k+1)−1)/2.
This means that, as k and n tend to infinity, the number of anomalous games grows as 3 2k2n2, and the number of all games as 2k2n2. The ratio of these numbers approaches 3 4 > 0.7 (see Fact 27). Therefore, for large enough n and k, this ratio will be greater than 70%. A reader who doesn’t like limits can simply check that for n = k = 20 the fraction of anomalous games relative to all games is 235600 335790 > 0.7.
Problem 6. The desired configuration will be obtained from regular tetra-hedra with the distance between opposite edges equal to the distance be-tween the given planes. One edge of each tetrahedron will lie in one of the planes and the opposite one in the other plane. We’ll think of the planes as horizontal.
Two tetrahedra can be positioned in such a way that an endpoint of the upper edge of one of them coincides with the midpoint of the upper edge of the other one, and the midpoint of the lower edge of the first tetrahedron coincides with an endpoint of the lower edge of the second, the edges of both tetrahedra in either plane being perpendicular. We can extend this construction to the entire layer to obtain a configuration in which each tetrahedron is surrounded by four others, as in the figure; two of them do not allow the tetrahedron to be pulled out of the layer upward, two other prevent it from moving downward. |
13516 | https://fiveable.me/aerodynamics/unit-3/normal-oblique-shock-waves/study-guide/UcgRY5W6NkHHdrUM | Normal and oblique shock waves | Aerodynamics Class Notes | Fiveable | Fiveable
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✈️Aerodynamics Unit 3 Review
3.4 Normal and oblique shock waves
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Aerodynamics
Unit 3 – Compressible flow
Topic: 3.4
✈️Aerodynamics Unit 3 Review
3.4 Normal and oblique shock waves
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
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3.6 Nozzle flow
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Aerodynamic forces and moments
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Normal and oblique shock waves are critical phenomena in supersonic flow. These abrupt changes in flow properties occur when supersonic flow encounters obstructions or changes direction. Understanding shock waves is essential for analyzing and designing supersonic vehicles and propulsion systems.
Shock waves cause sudden increases in pressure, temperature, and density while decreasing velocity. The Rankine-Hugoniot relations describe these changes mathematically. Oblique shocks, inclined to the flow direction, are weaker than normal shocks and allow downstream flow to remain supersonic.
Properties of normal shock waves
Normal shock waves are thin regions where flow properties change abruptly
Occur when supersonic flow encounters an obstruction or a sharp change in flow direction
Characterized by a discontinuous increase in pressure, temperature, and density across the shock
Pressure ratio across shock
Pressure increases significantly across a normal shock wave
Pressure ratio depends on the upstream Mach number
Higher upstream Mach numbers result in larger pressure ratios across the shock
Pressure ratio can be calculated using the Rankine-Hugoniot relations
Temperature ratio across shock
Temperature also increases across a normal shock wave
Temperature ratio is a function of the upstream Mach number
Higher upstream Mach numbers lead to higher temperature ratios
Temperature increase is due to the conversion of kinetic energy into thermal energy
Density ratio across shock
Density increases across a normal shock wave
Density ratio is related to the pressure and temperature ratios
Can be calculated using the equation of state for an ideal gas
Density increase is necessary to satisfy conservation of mass
Mach number change
Flow velocity decreases across a normal shock wave
Upstream Mach number is always supersonic (M > 1)
Downstream Mach number is always subsonic (M < 1)
Mach number decrease is due to the increase in speed of sound across the shock
Entropy increase
Entropy increases across a normal shock wave
Entropy increase is irreversible and indicates a loss of available energy
Amount of entropy increase depends on the strength of the shock (upstream Mach number)
Entropy increase is a measure of the irreversibility of the shock process
Rankine-Hugoniot relations
Set of equations that describe the relationship between flow properties across a shock wave
Derived from the conservation laws of mass, momentum, and energy
Used to calculate the downstream flow properties given the upstream conditions and shock strength
Conservation of mass
Mass flow rate is conserved across a normal shock wave
Product of density and velocity must be equal on both sides of the shock
Equation: $\rho_1 u_1 = \rho_2 u_2$, where $\rho$ is density and $u$ is velocity
Subscripts 1 and 2 denote upstream and downstream conditions, respectively
Conservation of momentum
Momentum is conserved across a normal shock wave
Sum of pressure and momentum flux must be equal on both sides of the shock
Equation: $p_1 + \rho_1 u_1^2 = p_2 + \rho_2 u_2^2$, where $p$ is pressure
Pressure increase across the shock balances the decrease in momentum flux
Conservation of energy
Energy is conserved across a normal shock wave
Total enthalpy (sum of static enthalpy and kinetic energy) is constant across the shock
Equation: $h_1 + \frac{1}{2}u_1^2 = h_2 + \frac{1}{2}u_2^2$, where $h$ is specific enthalpy
Kinetic energy is converted into thermal energy (static enthalpy) across the shock
Normal shock in ideal gas
Normal shock waves in an ideal gas exhibit specific behavior and properties
Ideal gas assumption simplifies the analysis and allows for closed-form solutions
Upstream and downstream states
Upstream (pre-shock) state is characterized by high Mach number, low pressure, low temperature, and low density
Downstream (post-shock) state has low Mach number, high pressure, high temperature, and high density
Ratio of downstream to upstream properties depends on the upstream Mach number
Property ratios increase with increasing upstream Mach number
Mach number limits
Normal shock waves can only occur in supersonic flow (upstream Mach number > 1)
Downstream Mach number is always subsonic (< 1) for a normal shock in an ideal gas
Maximum downstream Mach number is limited to 1, which occurs for an infinitely strong shock
Minimum upstream Mach number for a normal shock is 1, corresponding to a weak shock
Stagnation pressure ratio
Stagnation pressure (total pressure) decreases across a normal shock wave
Stagnation pressure ratio (downstream to upstream) is always less than 1
Stagnation pressure ratio decreases with increasing upstream Mach number
Stagnation pressure loss is a measure of the irreversibility of the shock process
Maximum entropy increase
Entropy increase across a normal shock wave has a maximum value
Maximum entropy increase occurs at a specific upstream Mach number (approximately 1.245 for air)
Upstream Mach numbers above or below this value result in lower entropy increases
Maximum entropy increase is an important consideration in the design of supersonic diffusers
Moving normal shock waves
Normal shock waves can be either stationary or moving relative to an observer
Moving shocks introduce additional complexity in the analysis of flow properties
Stationary vs moving shocks
Stationary shocks are fixed in space and the flow moves through the shock
Moving shocks propagate through a stationary or moving fluid
Reference frame can be changed to convert a moving shock into a stationary shock and vice versa
Flow properties across the shock are the same in both reference frames
Shock velocity relative to flow
Velocity of a moving shock wave is superimposed on the flow velocity
Upstream and downstream velocities relative to the shock are different
Relative velocity upstream of the shock is supersonic, while downstream is subsonic
Shock velocity can be determined using the Rankine-Hugoniot relations
Shock propagation in ducts
Moving normal shocks can propagate through ducts or channels
Shock propagation is influenced by the duct geometry and flow conditions
Shock velocity in a duct is affected by the change in cross-sectional area
Converging ducts accelerate the shock, while diverging ducts decelerate it
Oblique shock waves
Oblique shock waves are inclined at an angle to the flow direction
Occur when a supersonic flow encounters a concave corner or a compression ramp
Oblique vs normal shock waves
Oblique shocks have a non-zero angle with respect to the flow direction, while normal shocks are perpendicular
Flow downstream of an oblique shock remains supersonic, while it becomes subsonic after a normal shock
Oblique shocks are weaker than normal shocks for the same upstream Mach number
Oblique shocks can be attached to the surface or detached, depending on the flow conditions
Shock wave angle
Angle between the oblique shock wave and the upstream flow direction
Shock wave angle depends on the upstream Mach number and the deflection angle of the surface
Increases with increasing upstream Mach number and deflection angle
Can be calculated using the theta-beta-Mach relation
Deflection angle
Angle through which the flow is turned by the oblique shock wave
Deflection angle is determined by the geometry of the surface (ramp angle or corner angle)
Maximum deflection angle exists for a given upstream Mach number, beyond which the shock becomes detached
Deflection angle is related to the shock wave angle through the theta-beta-Mach relation
Weak vs strong solutions
For a given upstream Mach number and deflection angle, there are two possible shock wave angles
Weak shock solution has a smaller shock wave angle and a higher downstream Mach number
Strong shock solution has a larger shock wave angle and a lower downstream Mach number
Weak shock solution is usually observed in practice, unless the flow is highly disturbed or the deflection angle is large
Oblique shock relations
Flow properties across an oblique shock wave can be calculated using oblique shock relations
Relations are derived from the Rankine-Hugoniot equations and the geometry of the oblique shock
Pressure ratio across oblique shock
Pressure increases across an oblique shock wave
Pressure ratio depends on the upstream Mach number and the shock wave angle
Pressure ratio increases with increasing Mach number and shock wave angle
Can be calculated using the oblique shock pressure ratio equation
Density ratio across oblique shock
Density also increases across an oblique shock wave
Density ratio is related to the pressure ratio and the upstream Mach number
Can be calculated using the oblique shock density ratio equation
Density ratio is always greater than 1
Temperature ratio across oblique shock
Temperature increases across an oblique shock wave
Temperature ratio depends on the upstream Mach number and the shock wave angle
Can be calculated using the oblique shock temperature ratio equation
Temperature ratio is always greater than 1
Downstream Mach number
Mach number downstream of an oblique shock wave is lower than the upstream Mach number
Downstream Mach number depends on the upstream Mach number, shock wave angle, and deflection angle
Can be calculated using the oblique shock Mach number equation
Downstream Mach number is always supersonic for an attached oblique shock
Supersonic flow over wedges
Wedges are simple geometries that produce oblique shock waves in supersonic flow
Wedge flow is a fundamental problem in compressible aerodynamics
Attached vs detached shocks
Oblique shock wave can be attached to the wedge apex or detached from it
Attached shock occurs when the deflection angle is less than the maximum deflection angle for the given Mach number
Detached shock occurs when the deflection angle exceeds the maximum deflection angle
Detached shock is curved and stands off from the wedge apex
Wedge angle for attached shock
Wedge angle is the angle between the wedge surface and the freestream direction
For an attached shock, the wedge angle is equal to the deflection angle
Maximum wedge angle for an attached shock depends on the freestream Mach number
Can be calculated using the theta-beta-Mach relation and the maximum deflection angle
Maximum deflection angle
Maximum angle through which the flow can be turned by an attached oblique shock
Depends on the upstream Mach number and the specific heat ratio of the gas
Increases with increasing Mach number
Flow cannot be turned by an angle greater than the maximum deflection angle without creating a detached shock
Reflection of oblique shocks
Oblique shock waves can reflect from solid surfaces or interact with other shock waves
Reflection patterns depend on the incident shock strength and the flow conditions
Regular vs Mach reflection
Regular reflection occurs when the incident and reflected shocks meet at the surface
Mach reflection occurs when the incident and reflected shocks meet above the surface, forming a Mach stem
Transition from regular to Mach reflection depends on the incident shock angle and the flow deflection angle
Mach reflection is more likely to occur for strong incident shocks and large deflection angles
Shock-shock interaction
Oblique shocks can interact with each other, resulting in complex flow patterns
Interaction can be between two oblique shocks or between an oblique shock and a normal shock
Shock-shock interaction can lead to the formation of a triple point, where three shocks meet
Flow properties and shock angles change discontinuously across the triple point
Shock polars
Graphical representation of the relationship between the flow deflection angle and the shock wave angle
Used to analyze and predict the behavior of oblique shocks and their interactions
Different branches of the shock polar correspond to different shock solutions (weak, strong, or detached)
Intersection of shock polars determines the flow conditions and shock angles in shock-shock interactions
Shock wave-boundary layer interaction
Interaction between shock waves and boundary layers can significantly affect the flow field
Shock-boundary layer interaction can lead to flow separation, unsteadiness, and increased drag
Shock-induced separation
Adverse pressure gradient imposed by a shock wave can cause the boundary layer to separate
Separation occurs when the boundary layer cannot overcome the pressure rise across the shock
Shock-induced separation can lead to the formation of a recirculation bubble and increased flow unsteadiness
Severity of separation depends on the shock strength, boundary layer state, and surface geometry
Lambda shock structure
Characteristic shock pattern that forms when a shock wave interacts with a boundary layer
Consists of a normal shock near the surface, followed by an oblique shock that merges with the incident shock
Lambda shock structure is associated with shock-induced separation and the formation of a recirculation bubble
Can occur in supersonic inlets, transonic airfoils, and other flow situations where shocks interact with boundary layers
Shock train in supersonic flow
Series of successive shock waves that form in a supersonic flow with a boundary layer
Shock train is caused by the interaction between the shocks and the boundary layer
Each shock in the train is weaker than the previous one, and the spacing between shocks decreases downstream
Shock trains can occur in supersonic diffusers, isolators, and other flow passages with adverse pressure gradients
Applications of shock waves
Shock waves have numerous applications in aerospace engineering and other fields
Understanding and controlling shock waves is crucial for the design and operation of supersonic vehicles and devices
Supersonic inlets
Inlets are used to decelerate and compress the flow before it enters the engine of a supersonic vehicle
Shock waves are employed in supersonic inlets to efficiently reduce the Mach number and increase the pressure
Inlet design must balance the conflicting requirements of high pressure recovery and low flow distortion
Shock wave-boundary layer interaction and shock stability are major challenges in supersonic inlet design
Shock tubes and tunnels
Shock tubes and tunnels are experimental facilities used to study shock waves and high-speed flows
Shock tube consists of a high-pressure driver section and a low-pressure driven section separated by a diaphragm
When the diaphragm is ruptured, a shock wave propagates into the driven section, followed by an expansion wave
Shock tunnels use the high-temperature, high-pressure flow behind the reflected shock to simulate hypersonic flight conditions
Shock wave lithotripsy
Medical application of shock waves for the non-invasive treatment of kidney stones and other calculi
Focused shock waves are generated outside the body and propagate through tissue to the target stone
Shock waves induce stress and fracture in the stone, leading to its fragmentation into smaller pieces
Fragmented stones can then be easily passed through the urinary tract or dissolved by the body's natural processes
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13517 | https://philosophy.stackexchange.com/questions/34461/all-men-are-mortal-socrates-is-a-man-therefore-socrates-is-mortal-original | Stack Exchange Network
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"All men are mortal, Socrates is a man, therefore, Socrates is mortal" original quote
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Where the syllogism "All men are mortal / Socrates is a man / Therefore, Socrates is mortal" first appeared?
history-of-philosophy
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edited May 24, 2016 at 16:55
tatojotatojo
asked May 24, 2016 at 16:25
tatojotatojo
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Did you notice that later in the article, making it part of the conclusion, the author uses the expression identical to Socrates? I can say that the expression is redundant because The only thing identical to Socrates is Socrates. Law of Identity, A is A, Aristotle.
user48454
– user48454
2020-09-23 15:09:18 +00:00
Commented Sep 23, 2020 at 15:09
"All men are mortal, Socrates is mortal, therefore, all men are Socrates".
Mark
– Mark
2024-07-11 03:58:34 +00:00
Commented Jul 11, 2024 at 3:58
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Comment on similar example.
The example is not from Aristotle.
Categorical propositions with singular terms are used in Medieval logic; see Peter of Spain's Summulae logicales (XIII century):
I,8 Propositio singularis est illa [...], ut "Sortes currit" [Socrates runs].
Into this textbook we can find many examples of them in the discussion of loci and entimema [V,3], bt it seems to me that there is no occurrence of the specific example of syllogism.
According to: Joseph Maria Bochenski, A History of Formal Logic (1961, or.ed.1956), page 232:
A first widening of the Aristotelian syllogistic consists in the admission of singular terms and premisses. William of Ockham (c.12871347) already knows of the substitution that was to become classic [Summa Logicae, III 1,3;36rb]:
Every man is an animal;
Socrates is a man;
Therefore, Socrates is an animal.
Here the minor premiss is singular. But Ockham also allows singular propositions as major premisses.
The earliest occurrence I've found is: Sextus Empiricus, Outlines of Pyrrhonism (PyrrhÅneioi hypotypÅseis), also translated as: Outlines of Scepticism, Book II, 164 [Many thanks to David Wheeler]:
Socrates is human.
Everything human is an animal.
Therefore, Socrates is an animal.
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edited Nov 13, 2023 at 7:34
answered May 25, 2016 at 9:18
Mauro ALLEGRANZAMauro ALLEGRANZA
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Per David A. Wheeler's article "The Origin of All Men are Mortal" (which elsewhere cites this page!)
The earliest document I can find with this specific example is from 1843, specifically A System of logic: Ratiocinative and Inductive, Presenting a Connected View of the Principles of Evidence and the Methods of Scientific Investigation by John Stuart Mill, 1843, Book II Chapter 3 page 245.
One can indeed see the quote
2. It must be granted that in every syllogism, considered as an argument to prove the conclusion, there is a petitio principii. When we say, All men are mortal Socrates is a man therefore Socrates is mortal ; it is unanswerably urged by the adversaries of the syllogistic theory, that the proposition, Socrates is mortal, is presupposed in the more general assumption, All men are mortal:
in the Internet Archive's copy of Mill's System of Logic.
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answered Mar 21, 2021 at 17:26
TomRocheTomRoche
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13518 | https://gdaymath.com/wp-content/uploads/2020/03/PUZZLE-Divisibility-by-Nine_Brilliant-Ref-1.pdf | www.globalmathproject.org 1 DIVISIBILITY BY 9 There is a well-known divisibility rule for the number 9. In this piece, we explore that rule, and a slightly stronger version of it, and observe some consequences. Students will be able to extend the work here to a divisibility rule for 3 as well, and perhaps to a divisibility rule for 11 as well! EXPLODING DOTS Topic: Experience 5: Division in a 1 10 ← machine. Suggested Grade Level: Middle-school. High school. See Also: Divisibility by 9 again! Divisibility by 7 There are also links to puzzles from our partner BRILLIANT. www.globalmathproject.org 2 DIVISIBILITY BY 9 Many people know a rule for divisibility by nine. A number is divisible by 9 only if the sum of its digits is divisible by 9. For example, 387261 is divisible by 9—apparently—since 3 8 7 6 2 1 27 + + + + + = is. (And if we weren’t sure about the number 27 , we could test that it is divisible by 9 by noting that 2 7 9 + = certainly is.) Check: 387261 9 43029 ÷ = and indeed there is no remainder. In fact, this rule can be made a little stronger. A number leaves the same remainder upon division by 9 as does the sum of its digits. For example, 40062 has sum of digits 12, which is 3 more than a multiple of 9, and indeed 40062 is 3 more than a multiple of 9: 40062 9 4059 3 = × + . Also, 77 is five more than a multiple of 9 just as 7 7 14 + = , the sum of it digits, is. Discuss the divisibility rule for 9 with your students, and its stronger version. Have they already heard of the first rule? The second one? Have them test the rules with some examples. Now the real question is: Why do these rules work? Some Things Students Might Notice or Ask 1. Examples do seem to suggest that the rules do hold true. 2. It is actually surprising that these rules work. If we jumble the digits of a number, the sum of digits does not change. Since 387261 is a multiple of 9, this means that 387216and 783162 and 273613and all other permutations 387261 are all multiples of 9 too. That’s weird, and hard to believe! Comment: You and the class might choose to dwell on this observation a bit. It is certainly not generally true that you can rearrange the digits of a number and maintain its divisibility by a given factor. For example, changing the order of the digits of 512won’t always keep the number divisible by 2 , and changing the order of the digits of 2864 won’t always keep it a multiple of 4. So it seems awfully strange that divisibility-by-9 doesn’t “care” about the order of the digits. [Have your students ever noticed it is at least true for two-digit multiples 9? We have that 18 and 81, 27 and 72, 36 and 63, and, 45 and 54 are all multiples of 9.] www.globalmathproject.org 3 3. The second rule implies the first rule. A number is a multiple of 9 only if it leaves a remainder of zero upon division by 9. So, according to the second rule, if the sum of digits leaves a remainder of zero, that is, is a multiple of 9, then so is the original number. Consequently, explaining the second rule automatically explains the first rule too. SOLVING THE PUZZLE See EXPERIENCE 5 of EXPLODING DOTS: Division in a 1 10 ← machine. Let’s look at division by 9 in a 1 10 ← machine. To get a feel for what is going on, perhaps have students draw dots and boxes to compute 210 9 ÷ . It will be tedious, as one must “unexplode” multiple times to find groups of 9, but the key is to notice that there will be 2 1 + dots left over in the rightmost box as remainder. (Try it!) This prepares us to see Each dot in a 1 10 ← machine leaves a remainder of 1 upon division by 9. This picture illustrates why. www.globalmathproject.org 4 Thus we have: If a number is represented by a total of N dots in a 1 10 ← machine, dividing by 9 leaves us with N dots in the rightmost box. And there might be some more groups of 9 we can circle there. But let’s think about what N is in this statement: it’s the sum of the digits of the original number. And looking for groups of 9 in the rightmost box with N dots in it is the precisely the act of dividing N itself by 9. The act of dividing a number by 9 in a 1 10 ← machine reduces to the equivalent act of dividing its sum of digits by 9. Thus the original number and the sum of its digits leave the same remainder upon division by 9. EXTENSIONS Every solved problem, of course, is an invitation to explore and play more. Might your students enjoy these explorations? Wild Exploration 1: Many people know the rule: A number is divisible by 3 only if its sum of digits is. Is there a stronger version of this rule to consider? If so, can you prove it? (If not, can you at least prove the first version of the rule?) Wild Exploration 2: Martians have three fingers on each of two hands and so naturally write all their numbers in base 6 using a 1 6 ← machine. Is there a divisibility rule for some special Martian number like the one we Earthlings have for 9 in our base-10 system? Wild Exploration 3: What remainder(s) does a single dot in a 1 10 ← machine leave upon division by 11? Can you devise, and explain, a divisibility rule for 11? (Or look up a rule on the internet and see if you can explain it using a 1 10 ← machine.) Wild Exploration 4: Which numbers k have the property that if N is divisible by k , then so are all the numbers obtained by rearranging the digits of N ? (For example, 9 k = is one such number. So is 1 k = .) Further Reading: Our partner BRILLIANT has a lovely series of puzzles based on the same ideas of this essay. Check out these two links. |
13519 | https://www.youtube.com/watch?v=tC06acrUBRo | Mass Ratio
Kate Biberdorf
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A teaching video on Mass Ratio used in the 'Global Climate Change' module at The University of Texas at Austin.
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Transcript:
hi everyone in this video we are going to discuss master ratio so what is mass ratio well let's get a definition mass ratio is actually quite straightforward and basically it is just the mass of one element relative to the mass of the molecule so the best way to do this is just by working an example so let's say we are trying to determine the mass of carbon in carbon dioxide so okay so it's a mass of one element relative to the mass of the molecule so essentially what we need to do is say the mass of carbon divided by the mass of carbon dioxide okay so let's just plug these numbers in so we're going to use the atomic mass for carbon so that's about 12.01 again grams from all and we're going to divide that by the molar mass and molecular weight essentially of our carbon dioxide so to determine this we simply just add up the atomic masses of each individual element so we have about 12 grams of carbon each oxygen is about 16 grams so 16 times 2 gives us about 32 + 12 gives us an overall molar mass of about 44 grams per mole for our mass of carbon dioxide so if we do this out until we do 12 divided by 44 we get a number that is essentially zero point two seven to some seven something like that but what is that that is just a garbage number it's gibberish that's not helpful that's not a useful thing and so what we actually do is take this number and multiply it times 100 so we essentially make it a percent so what we could say is that in carbon dioxide we would have 27 point we'll just say three percent carbon so out of a hundred percent of carbon dioxide 27 percent of that is carbon and then a hundred minus 27 is oxygen okay so let's talk about what that is what is that twenty seven point three well we actually call that mass percent this is a much more useful term than mass ratio so mass percent would just be your mass of your element so we'll say of element X okay divided by your mass of your molecule times 100 and this will give us a very very valuable piece of information so let's work a another example on but I'm actually gonna let you do this one so what is the mass percent okay so we're going to present this time what is the mass percent of oxygen in carbon dioxide so we just did it for carbon what is it of oxygen go all right did you get an answer hopefully did if you didn't don't worry about it we're gonna work through it together right here so we said mass percent so we're going to do mass of oxygen divided by our mass of carbon dioxide which sets a seat times 100 so we would just plug this in so our mass of oxygen now hold on how many oxygens do we have in carbon dioxide there are two and so what we have to do is take two times our 16 which is our atomic mass so we actually have 32 point zero grams of oxygen in a total mass of 44 grams we multiply that times 100 and we end up with 72.7% now we could have had the same number by doing 100 minus the 27.3% earlier that we determined is the percent of carbon in carbon dioxide and we get the same number of 72.7% okay hopefully that makes sense if it doesn't you can go back through and work it out but essentially we need to make sure you factor in every single little one of those subscripts so let's try one more what is the mass percent of nitrogen in nitrous oxide all right did you get an answer hopefully you did if you didn't don't worry about it we'll work it out right here all right so mass percent of nitrogen so we do our mass of nitrogen divided by our mass of nitrous oxide and 200 times 100 its percent so you can't forget that multiply by 100 piece alright so we use our atomic mass of nitrogen first that's about 14 grams per mole but there are two of them okay we cannot forget that subscript so 14 times 2 which is actually about 28 all over the molar mass of nitrous oxide which is around 44 grams per mole times 100 so we essentially have 28 divided by 44 times 100 if you do that properly you get an answer of about sixty three point six percent hopefully you got that correct if you didn't practice practice practice good luck take care of yourself trick water |
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13521 | https://fiveable.me/key-terms/physical-chemistry-i/band-theory | Band Theory - (Physical Chemistry I) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Band theory is a theoretical model that explains the electronic properties of solids by describing the energy levels available to electrons in a material. It illustrates how these energy levels can be grouped into bands, with the most significant ones being the valence band and the conduction band, which are crucial for understanding electrical conductivity and other material behaviors.
5 Must Know Facts For Your Next Test
In band theory, the energy levels of electrons in solids form continuous bands instead of discrete levels due to the close proximity of atoms in a crystal lattice.
The gap between the valence band and conduction band is known as the band gap, which determines whether a material behaves as an insulator, semiconductor, or conductor.
Metals typically have overlapping valence and conduction bands, allowing electrons to flow freely, which accounts for their high electrical conductivity.
In semiconductors, the band gap can be overcome by thermal energy or doping, allowing controlled conductivity that is essential for electronic applications.
Band theory also helps explain optical properties and the behavior of materials under varying temperatures and external fields, making it integral to solid-state physics.
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How does band theory explain the difference in electrical conductivity between metals, semiconductors, and insulators?
Band theory explains that in metals, the valence and conduction bands overlap, allowing electrons to move freely and resulting in high electrical conductivity. In semiconductors, there is a noticeable band gap that can be bridged with thermal energy or doping, leading to controlled conductivity. Insulators have a larger band gap that prevents electrons from easily moving to the conduction band, resulting in poor conductivity. This framework allows for a clear understanding of how different materials behave electrically based on their electronic structure.
Analyze how the concept of band gaps in band theory affects the functionality of semiconductors in electronic devices.
The concept of band gaps is crucial for semiconductors because it dictates their ability to conduct electricity under certain conditions. A small band gap allows for thermal excitation of electrons from the valence band to the conduction band, enabling current flow when sufficient energy is provided. Additionally, doping introduces impurities that can create additional energy levels within the band gap, further enhancing conductivity. This tunability is what makes semiconductors invaluable in electronic devices like diodes and transistors, as they can switch between insulating and conducting states based on external stimuli.
Evaluate the role of band theory in predicting optical properties of materials and its implications in technology development.
Band theory plays a significant role in predicting the optical properties of materials by explaining how electron transitions between valence and conduction bands affect light absorption and emission. For instance, materials with small band gaps can absorb visible light, making them suitable for photovoltaic applications. By understanding these interactions through band theory, researchers can design materials with specific optical characteristics for use in technologies like LEDs and solar cells. The ability to engineer these properties allows for advancements in renewable energy solutions and efficient lighting technologies.
Related terms
Valence Band:The highest range of electron energies in a solid where electrons are normally present at absolute zero temperature, crucial for determining a material's bonding properties.
Conduction Band:The range of electron energies above the valence band where electrons can move freely and contribute to electrical conductivity in a material.
Semiconductors:
Materials that have electrical conductivity between that of conductors and insulators, often utilized in electronic devices, whose properties can be modified by temperature or impurities.
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13522 | https://www.quora.com/What-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-by-1-2-3-4-5-6-7-8-9-without-a-duplicate-digit-and-without-a-zero | What is the sum of all 4 digit numbers that can be formed by 1,2,3,4,5,6,7,8,9 without a duplicate digit and without a zero? - Quora
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What is the sum of all 4 digit numbers that can be formed by 1,2,3,4,5,6,7,8,9 without a duplicate digit and without a zero?
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Sharan Rudresh
Lives in Bengaluru, Karnataka, India · Author has 145 answers and 545.2K answer views
·9y
Originally Answered: what is the sum of all 4 digit numbers that can be formed by 1,2,3,4,5,6,7,8,9 without a duplicate digit and without a zero? ·
Answer, 8 P 3 45 1111= 1,67,98,320…I’ll tell you how to do this without a calculator.
Number of combinations with 1 in the unit’s place= 8P3
Number of combinations with 2 in the unit’s place= 8P3……and so on.
Adding all the unit place digits, we get….8P3(1+2+3+….+9)=8P345
Similarly adding all the tens place digits, and then the hundreds place digits…we get the same, i.e., 8P345.
So the required sum is: 8P345+ 108P345+1008P345+10008P345
which is same as: 8P3451111=16798320.
This can be generalized. Try it. If the question had been “9 digit numbers”, then the answer would be: 8P84511111111
Continue Reading
Answer, 8 P 3 45 1111= 1,67,98,320…I’ll tell you how to do this without a calculator.
Number of combinations with 1 in the unit’s place= 8P3
Number of combinations with 2 in the unit’s place= 8P3……and so on.
Adding all the unit place digits, we get….8P3(1+2+3+….+9)=8P345
Similarly adding all the tens place digits, and then the hundreds place digits…we get the same, i.e., 8P345.
So the required sum is: 8P345+ 108P345+1008P345+10008P345
which is same as: 8P3451111=16798320.
This can be generalized. Try it. If the question had been “9 digit numbers”, then the answer would be: 8P845111111111. Try the many different variations. Its fun!!
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Related questions
More answers below
How many different 4-digit numbers can be formed with the digits 1, 2, 3, and 4, if each digit is used only once?
What are the 4-digit combinations of the numbers 1, 3, 5, 7, and 9?
How many 4 digit combinations can be made with 5?
How many 4-digit number(s) is/are there in which the sum of digit is 9?
How many 4 number combinations have the sum of 14?
Alan Bustany
Trinity Wrangler, 1977 IMO · Upvoted by
Nathan Frank Gross
, M.A. Mathematics, Loyola University Chicago (2021) · Author has 9.8K answers and 58.5M answer views
·9y
Originally Answered: what is the sum of all 4 digit numbers that can be formed by 1,2,3,4,5,6,7,8,9 without a duplicate digit and without a zero? ·
Sharan Rudresh suggested generalising the question, so here’s my version:
What is the sum of all n n-digit numbers that can be formed by non-zero digits without a duplicate in base b b?
Each position can have any of the b−1 b−1 digits. These will occur in
b−2 P n−1=(b−2)!(b−n−1)!b−2 P n−1=(b−2)!(b−n−1)! ways
since the other b−2 b−2 digits are permuted in any way across the remaining n−1 n−1 positions.
The sum of the digits is b−1∑i=1 i=b(b−1)2∑i=1 b−1 i=b(b−1)2.
We must multiply these two factors by the number with 1 1 in each position or
n∑i=0 b i=1+b+b 2+⋯+b n=b n+1−1 b−1∑i=0 n b i=1+b+b 2+⋯+b n=b n+1−1 b−1
Continue Reading
Sharan Rudresh suggested generalising the question, so here’s my version:
What is the sum of all n n-digit numbers that can be formed by non-zero digits without a duplicate in base b b?
Each position can have any of the b−1 b−1 digits. These will occur in
b−2 P n−1=(b−2)!(b−n−1)!b−2 P n−1=(b−2)!(b−n−1)! ways
since the other b−2 b−2 digits are permuted in any way across the remaining n−1 n−1 positions.
The sum of the digits is b−1∑i=1 i=b(b−1)2∑i=1 b−1 i=b(b−1)2.
We must multiply these two factors by the number with 1 1 in each position or
n∑i=0 b i=1+b+b 2+⋯+b n=b n+1−1 b−1∑i=0 n b i=1+b+b 2+⋯+b n=b n+1−1 b−1
So the final sum is equal to
(b−2)!b(b−1)(b n+1−1)(b−n+1)!2(b−1)=b!(b n+1−1)2(b−1)(b−n−1)!(b−2)!b(b−1)(b n+1−1)(b−n+1)!2(b−1)=b!(b n+1−1)2(b−1)(b−n−1)!
In the specific case of the question we have:
b=10 b=10 (decimal) so b!=3,628,800 b!=3,628,800
n=4 n=4 so (b−n−1)!=5!=120(b−n−1)!=5!=120
and the final sum is
3628800×1111 2×120=16,798,320 3628800×1111 2×120=16,798,320
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To find the sum of all 4-digit numbers that can be formed using the digits 1 through 9 without repetition, we can follow these steps:
Step 1: Count the Total Number of 4-Digit Combinations
We can choose 4 digits from the 9 available digits (1-9). The number of ways to select and arrange 4 digits is given by the permutation formula:
P(n,r)=n!(n−r)!P(n,r)=n!(n−r)!
For our case, n=9 n=9 and r=4 r=4:
P(9,4)=9!(9−4)!=9!5!=9×8×7×6=3024 P(9,4)=9!(9−4)!=9!5!=9×8×7×6=3024
So, there are 3024 different 4-digit numbers that can be formed.
Step 2: Contribution of Each Digit in Each Position
Each digit w
Continue Reading
To find the sum of all 4-digit numbers that can be formed using the digits 1 through 9 without repetition, we can follow these steps:
Step 1: Count the Total Number of 4-Digit Combinations
We can choose 4 digits from the 9 available digits (1-9). The number of ways to select and arrange 4 digits is given by the permutation formula:
P(n,r)=n!(n−r)!P(n,r)=n!(n−r)!
For our case, n=9 n=9 and r=4 r=4:
P(9,4)=9!(9−4)!=9!5!=9×8×7×6=3024 P(9,4)=9!(9−4)!=9!5!=9×8×7×6=3024
So, there are 3024 different 4-digit numbers that can be formed.
Step 2: Contribution of Each Digit in Each Position
Each digit will appear in each of the four positions (thousands, hundreds, tens, units) an equal number of times.
To find how many times a specific digit appears in each position, we can fix one digit in one position and arrange the remaining 3 digits.
For a fixed digit in one position, the number of ways to select 3 more digits from the remaining 8 digits is P(8,3)P(8,3):
P(8,3)=8×7×6=336 P(8,3)=8×7×6=336
Thus, each digit appears in each position 336 336 times.
Step 3: Calculate the Contribution to the Total Sum
Contribution of Each Position:
Thousands place: Each digit contributes 1000 1000 times its value.
Hundreds place: Each digit contributes 100 100 times its value.
Tens place: Each digit contributes 10 10 times its value.
Units place: Each digit contributes 1 1 times its value.
Let’s calculate the total contribution from all digits in each position:
Thousands place:
Contribution thousands=336×1000×(1+2+3+4+5+6+7+8+9)Contribution thousands=336×1000×(1+2+3+4+5+6+7+8+9)
2. Hundreds place:
Contribution hundreds=336×100×(1+2+3+4+5+6+7+8+9)Contribution hundreds=336×100×(1+2+3+4+5+6+7+8+9)
3. Tens place:
Contribution tens=336×10×(1+2+3+4+5+6+7+8+9)Contribution tens=336×10×(1+2+3+4+5+6+7+8+9)
4. Units place:
Contribution units=336×1×(1+2+3+4+5+6+7+8+9)Contribution units=336×1×(1+2+3+4+5+6+7+8+9)
Step 4: Calculate 1+2+3+…+9 1+2+3+…+9
The sum of the first 9 digits is:
1+2+3+4+5+6+7+8+9=9×(9+1)2=45 1+2+3+4+5+6+7+8+9=9×(9+1)2=45
Step 5: Plug in the Values
Now we can compute the total contributions:
Thousands place:
336×1000×45=15120000 336×1000×45=15120000
2. Hundreds place:
336×100×45=1512000 336×100×45=1512000
3. Tens place:
336×10×45=151200 336×10×45=151200
4. Units place:
336×1×45=15120 336×1×45=15120
Step 6: Total Sum
Now, we sum up all the contributions:
Total Sum=15120000+1512000+151200+15120 Total Sum=15120000+1512000+151200+15120
Calculating this gives:
Total Sum=15120000+1512000+151200+15120=16673320 Total Sum=15120000+1512000+151200+15120=16673320
Thus, the sum of all the 4-digit numbers that can be formed by the digits 1-9 without repetition is:
16673320 16673320
Upvote ·
Trevor Squires
Long time lover. Short time student · Author has 425 answers and 1.2M answer views
·9y
Originally Answered: what is the sum of all 4 digit numbers that can be formed by 1,2,3,4,5,6,7,8,9 without a duplicate digit and without a zero? ·
Alright so let's try it step by step.
You're given a set of 9 elements and asked for all possible permutations of size 4 where order does matter. We know this to be n!(n−k)!=3024 n!(n−k)!=3024
Now we want to know the sum of all these. Imagine stacking them all up on top of each other like we used to do in elementary school. A stack of 3024 different 4 digit numbers.
What is the sum of the first column? Well it's quite simple. We know that in our 3024 possible permutations, the distribution of the digits is even. So the average of all the digits is
9+1 2=5 9+1 2=5
We can actually rearrange our number
Continue Reading
Alright so let's try it step by step.
You're given a set of 9 elements and asked for all possible permutations of size 4 where order does matter. We know this to be n!(n−k)!=3024 n!(n−k)!=3024
Now we want to know the sum of all these. Imagine stacking them all up on top of each other like we used to do in elementary school. A stack of 3024 different 4 digit numbers.
What is the sum of the first column? Well it's quite simple. We know that in our 3024 possible permutations, the distribution of the digits is even. So the average of all the digits is
9+1 2=5 9+1 2=5
We can actually rearrange our numbers to be a sum of 3024 5555′s. So the sum of all possible permutations is ((3024)(5555)=16798320 3024)(5555)=16798320
Cheers
Upvote ·
Related questions
More answers below
How many 3 digit number formed using 2,3,4?
How many ways can the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 be arranged so that no even digit is in its original position?
How many four-digit numbers have the property that the sum of the digits is equal to the product of their digits?
How many 3-digit numbers can be formed from the digits 0, 1, 2, 3, 4, & 5 if repetition is not allowed?
How many numbers between 3000 and 5000 can be formed from the digits 1, 2, 3, 4 and 5, if digits can be repeated and the number formed is odd?
James Katz
Majored in all things math · Author has 51 answers and 1.5M answer views
·9y
Originally Answered: what is the sum of all 4 digit numbers that can be formed by 1,2,3,4,5,6,7,8,9 without a duplicate digit and without a zero? ·
This sounds like a Project Euler question. A couple thoughts:
I like to first check the dimension of the problem. You want the set of 4 digit numbers consisting of a sample of the digits (1,2,3,4,5,6,7,8,9) without replacement. Another way to think about this is by taking every 4 unit combination from the set (Combination) and then calculating their permutations (Permutation). Multiplying the two formulas together, you get the following formula for the number of integers you’ll need to sum: 9!5!=3,024 9!5!=3,024
So you have 3,024 numbers to sum. I’m not sure of a non-brute-force way to do this, bu
Continue Reading
This sounds like a Project Euler question. A couple thoughts:
I like to first check the dimension of the problem. You want the set of 4 digit numbers consisting of a sample of the digits (1,2,3,4,5,6,7,8,9) without replacement. Another way to think about this is by taking every 4 unit combination from the set (Combination) and then calculating their permutations (Permutation). Multiplying the two formulas together, you get the following formula for the number of integers you’ll need to sum: 9!5!=3,024 9!5!=3,024
So you have 3,024 numbers to sum. I’m not sure of a non-brute-force way to do this, but 3,024 is definitely small enough to use a computer program to calculate. I’ve written some inefficient pseudo/python code below:
count = 0 #let's count the number of integers summed to check if it's actually 3,024
num_sum = 0 #for tabulating the sum
nums = range(1,10)
for i1 in nums:
subset1 = [num for num in nums if num != i1]
for i2 in subset1:
subset2 = [num for num in subset1 if num != i2]
for i3 in subset2:
subset3 = [num for num in subset2 if num != i3]
for i4 in subset3:
num_combination=int(str(i1)+str(i2)+str(i3)+str(i4))
count+=1
num_sum+=num_combination
print count
print num_sum
The above code executes in less than a tenth of a second on my machine, and spits out 3024 (the number of individual integers summed) and 16,798,320 (the sum).
Upvote ·
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Padmini Kalyanaraman
School Teacher · Author has 4K answers and 4.6M answer views
·7y
Related
What is the sum of all five digit numbers that can be formed using the digits 1,2,3,4, and 5 without repetition?
1)let me take an example , the sum of all the digits numbers can be formed using the 1,2 without repetition.
(12, 21 ) ,The sum is 12+21= 33
2)Another one the sum of all the digits numbers can be formed using three digits 1,2,3, without repetition.
(123,231,321,213,312,132) Their sum is by adding ans is 1332
For above example we can find the answer by using the formula
(n-1)! x sum of the digits x ( 111111111….n times)
For example 1)
n = 2, sum of the digits 2+1= 3
(2–1)! x3 (11)
1x 3 x 11= 33
For example 2
n= 3 sum of the numbers 1+2+3 = 6
(3–1)! x 6( 111)
2x1x6x111
12x111= 1332
Both answers verified
For the
Continue Reading
1)let me take an example , the sum of all the digits numbers can be formed using the 1,2 without repetition.
(12, 21 ) ,The sum is 12+21= 33
2)Another one the sum of all the digits numbers can be formed using three digits 1,2,3, without repetition.
(123,231,321,213,312,132) Their sum is by adding ans is 1332
For above example we can find the answer by using the formula
(n-1)! x sum of the digits x ( 111111111….n times)
For example 1)
n = 2, sum of the digits 2+1= 3
(2–1)! x3 (11)
1x 3 x 11= 33
For example 2
n= 3 sum of the numbers 1+2+3 = 6
(3–1)! x 6( 111)
2x1x6x111
12x111= 1332
Both answers verified
For the given question
n= 5 , sum of the digits 1+2+3+4+5=15
(5–1)! x 15 x 11111
4! x 15 x 11111
4 x 3 x x2 x 15 x 11111
24 x 15 x 11111
360 x 11111= 3999960
Ans : 3999960
Upvote ·
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Awadhesh Verma
Always working on discrete things as software engineer · Author has 66 answers and 214.5K answer views
·8y
Related
A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
We have 5 digit 1, 3, 5, 7, 9.
So, total number of 5 digit number will be 5!5!
Lets take one such number, 53179.
This number can be written as 5∗10 4+3∗10 3+1∗10 2+7∗10+9 5∗10 4+3∗10 3+1∗10 2+7∗10+9.
Now, think about particular position. A digit will come at particular position 4! times, because rest of the digit can be arranged in 4!4! times.
For example 5 will come at 10 3 10 3 position 4!4! times. It means when we will be summing all these digit we will be having 5∗10 3 5∗10 3, 4!4! times. it means 5∗10 3∗4!5∗10 3∗4! will be part of sum.
if we apply this analogy to all the digit and sum them, we will get this expression.
(10 4+10 3+10 2(10 4+10 3+10 2
Continue Reading
We have 5 digit 1, 3, 5, 7, 9.
So, total number of 5 digit number will be 5!5!
Lets take one such number, 53179.
This number can be written as 5∗10 4+3∗10 3+1∗10 2+7∗10+9 5∗10 4+3∗10 3+1∗10 2+7∗10+9.
Now, think about particular position. A digit will come at particular position 4! times, because rest of the digit can be arranged in 4!4! times.
For example 5 will come at 10 3 10 3 position 4!4! times. It means when we will be summing all these digit we will be having 5∗10 3 5∗10 3, 4!4! times. it means 5∗10 3∗4!5∗10 3∗4! will be part of sum.
if we apply this analogy to all the digit and sum them, we will get this expression.
(10 4+10 3+10 2+10+1)∗(5+3+1+7+9)∗4!(10 4+10 3+10 2+10+1)∗(5+3+1+7+9)∗4!
=11111∗25∗4!=11111∗25∗4!
=11111∗600=11111∗600
=6666600=6666600
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What is the sum of all 5-digit numbers that can be formed by 0,1,2,3,4 without repetition of digits. What is the logic behind calculating such problems?
Sum=2599980
If u observe each digit will be present at each position in 4! time.
i.e., No of times 0 is in first(ten thousand) place is (5-1)! = 4!= 24 ways.
In the same way ,
No of times 1 is in first place is (5-1)! = 4!= 24 ways.
No of times 2 is in first place is (5-1)! = 4!= 24 ways.
No of times 3 is in first place is (5-1)! = 4!= 24 ways.
No of times 4 is in first place is (5-1)! = 4!= 24 ways.
No of times 0 is in second( thousand) place is (5-1)! = 4!= 24 ways.
.
.
.
.
.
No of times 4 in fifth(One's) place is (5-1)! = 4!= 24 ways.
So the sum of all numbers will be
= (5-1)! (10000+1000
Continue Reading
Sum=2599980
If u observe each digit will be present at each position in 4! time.
i.e., No of times 0 is in first(ten thousand) place is (5-1)! = 4!= 24 ways.
In the same way ,
No of times 1 is in first place is (5-1)! = 4!= 24 ways.
No of times 2 is in first place is (5-1)! = 4!= 24 ways.
No of times 3 is in first place is (5-1)! = 4!= 24 ways.
No of times 4 is in first place is (5-1)! = 4!= 24 ways.
No of times 0 is in second( thousand) place is (5-1)! = 4!= 24 ways.
.
.
.
.
.
No of times 4 in fifth(One's) place is (5-1)! = 4!= 24 ways.
So the sum of all numbers will be
= (5-1)! (10000+1000+100+10+1)(0+1+2+3+4)
=4! (11111)10
= 24 1111110
= 2666640
Logic can be (n-1)!(111...1 ntimes)(sum of digits)
As numbers formed by digits 1,2,3,4 and 0 as first digits are not 5 digit numbers, so subtracting 4 digit numbers formed by 1,2,3,4
i.e., 2666640 - 3!(1111)(1+2+3+4)
=2666640 - 666610
= 2666640 -66660
= 2599980
Upvote ·
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9 3
Meritt
Options Trader (2005–present) · Author has 119 answers and 158.4K answer views
·6y
Originally Answered: What is the sum of all of the four digit numbers formed using the digits 1,2,3,4,5,6 without repetition? ·
1+2+3+4=10 lowest sum
3+4+5+6=18 highest sum
All 9 four digit sums: 10, 11, 12, 13, 14, 15, 16, 17, 18
4P4=24 permutations for each of the four digit sums
(1024)+(1124)+(1224)+(1324)+(1424)+(1524)+(1624)+(1724)+(1824)=3024
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Shashi Kumar
Worked at Tech-Whiz:-The IT Society of PGDAV College · Author has 59 answers and 334.1K answer views
·7y
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A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
We have 5 different values we're working with (1, 3, 5, 7 and 9). The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits (i.e. the number of 'permutations').
Here's one way to calculate this number:
For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 54321 (common
Continue Reading
We have 5 different values we're working with (1, 3, 5, 7 and 9). The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits (i.e. the number of 'permutations').
Here's one way to calculate this number:
For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 54321 (commonly written 5!) = 120 different 5-digit numbers.
Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.
120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...
This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.
We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 510,000, the 3 is 31000, etc...
So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:
Sum of ten thousands digits: 2410,000(1 + 3 + 5 + 7 + 9)
+
Sum of thousands digits: 241,000(1 + 3 + 5 + 7 + 9)
+
Sum of hundreds digits: 24100(1 + 3 + 5 + 7 + 9)
+
Sum of tens digits: 2410(1 + 3 + 5 + 7 + 9)
+
Sum of units digits: 241(1 + 3 + 5 + 7 + 9)
= 6,666,600
Upvote ·
99 86
9 3
Ujjayanta Bhaumik
Interested in Number Theory · Author has 419 answers and 898.4K answer views
·Updated 8y
Related
What is the sum of all 4 digit numbers that can be formed using the digits 2, 4, 6, 8 without repetition ?
__ __ __ __ is a 4 digit number where every digit has to be 2,4,6 or 8
Or, every digit has to be (2+4+6+8)/4=5 on an average
Now, number of 4 digit numbers from 2,4,6 or 8 without repetition is 24.
So, sum=555524=133320
Proof:
2468+2486+2864+2846+2684+2648+4268+4286+4862+4826+4682+4628+6248+6284+6428+6482+6842+6824+8246+8264+8462+8426+8624+8642=133320
Upvote ·
999 105
9 8
Tanmay Pandit
IITJ CSE '22 · Author has 571 answers and 2.7M answer views
·9y
Related
What is the sum of all five digit numbers that can be formed using the digits 1,2,3,4, and 5 without repetition?
We can form 5! = 120 numbers.
Out of them, each 24 will have 1, 2, 3, 4 & 5 as ten thousands, thousands, hundreds, tens & unit digit.
1 + 2 + 3 + 4 + 5 = 15 so face value of each column = 1524 = 360.
Sum = 360 (10,000 + 1,000 + 100 + 10 + 1) = 39,99,960.
Upvote ·
99 30
Glenn Redd
B.S. in Mathematics, Kennesaw State University (Graduated 2016) · Author has 1.9K answers and 3.7M answer views
·Updated 4y
Related
What is the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, and 9 without repetition?
First, note that any number abcd where each letter is one of the given digits can be rewritten as a000+0b00+00c0+000d. For example, 9531 = 9000+500+30+1.
Next, consider that when we pick any specific as the first digit, there are 432=4!=24 arrangements of the other digits. So if we picked 9 as the first digit, there are 24 such arrangements. If we broke the arrangements down as described above, we’d have 24 instances of 9000.
Similarly, if we picked a specific digit as the second digit, there would be 24 such arrangements. So if we chose 1 as the second digit, we would have 24 instances of 010
Continue Reading
First, note that any number abcd where each letter is one of the given digits can be rewritten as a000+0b00+00c0+000d. For example, 9531 = 9000+500+30+1.
Next, consider that when we pick any specific as the first digit, there are 432=4!=24 arrangements of the other digits. So if we picked 9 as the first digit, there are 24 such arrangements. If we broke the arrangements down as described above, we’d have 24 instances of 9000.
Similarly, if we picked a specific digit as the second digit, there would be 24 such arrangements. So if we chose 1 as the second digit, we would have 24 instances of 0100.
You can repeat this line of thinking until you see that we have: 249000+247000+…+240900+…+240003+240001+…
Note that the above can be rewritten as 24(9000+7000+…+900+…+90+…+9+7+…)
9+7+5+3+1=25, so 9000+7000+…+1000 = 25000. Similarly, 900+700+…=2500, etc.
So we can rewrite it all as 24(25000+2500+250+25)=2427775=666,600.
Upvote ·
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) | 1/2x-7 |<1
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Answer
The answer is $$(-\infty, 3) \cup (4, \infty)$$(−∞,3)∪(4,∞)
Explanation
This question tests your understanding of absolute value inequalities. The key idea is to recognize that $$|a| < b$$∣a∣<b is equivalent to $$-b < a < b$$−b<a<b. We'll apply this to the given inequality and solve for $$x$$x. Also, we need to consider that the denominator cannot be zero.
Rewrite the absolute value inequality
We have $$|\frac{1}{2x-7}| < 1$$∣2 x−7 1∣<1. This is equivalent to $$-1 < \frac{1}{2x-7} < 1$$−1<2 x−7 1<1
Consider the case when $$2x-7 > 0$$2 x−7>0
If $$2x-7 > 0$$2 x−7>0, then $$x > \frac{7}{2}$$x>2 7. We can multiply all parts of the inequality by $$2x-7$$2 x−7 without changing the direction of the inequality signs:
$$-(2x-7) < 1 < 2x-7$$−(2 x−7)<1<2 x−7
This gives us two inequalities:
$$-2x+7 < 1$$−2 x+7<1 and $$1 < 2x-7$$1<2 x−7
Solve the first inequality: $$-2x+7 < 1$$−2 x+7<1
Subtract 7 from both sides:
$$-2x < -6$$−2 x<−6
Divide by -2 (and reverse the inequality sign because we're dividing by a negative number):
$$x > 3$$x>3
Solve the second inequality: $$1 < 2x-7$$1<2 x−7
Add 7 to both sides:
$$8 < 2x$$8<2 x
Divide by 2:
$$4 < x$$4 4$$x>4
Combine the inequalities from Steps 2, 3, and 4
We have $$x > \frac{7}{2}$$x>2 7, $$x > 3$$x>3, and $$x > 4$$x>4. Since $$\frac{7}{2} = 3.5$$2 7=3.5, the most restrictive condition is $$x > 4$$x>4
Consider the case when $$2x-7 < 0$$2 x−7<0
If $$2x-7 < 0$$2 x−7<0, then $$x < \frac{7}{2}$$x<2 7. We can multiply all parts of the inequality by $$2x-7$$2 x−7, but we must reverse the direction of the inequality signs:
$$-(2x-7) > 1 > 2x-7$$−(2 x−7)>1>2 x−7
This gives us two inequalities:
$$-2x+7 > 1$$−2 x+7>1 and $$1 > 2x-7$$1>2 x−7
Solve the first inequality: $$-2x+7 > 1$$−2 x+7>1
Subtract 7 from both sides:
$$-2x > -6$$−2 x>−6
Divide by -2 (and reverse the inequality sign):
$$x < 3$$x<3
Solve the second inequality: $$1 > 2x-7$$1>2 x−7
Add 7 to both sides:
$$8 > 2x$$8>2 x
Divide by 2:
$$4 > x$$4>x or $$x < 4$$x<4
Combine the inequalities from Steps 6, 7, and 8
We have $$x < \frac{7}{2}$$x<2 7, $$x < 3$$x<3, and $$x < 4$$x<4. Since $$\frac{7}{2} = 3.5$$2 7=3.5, the most restrictive condition is $$x < 3$$x<3
Combine the results from both cases
We found that $$x > 4$$x>4 or $$x < 3$$x<3
Express the solution in interval notation
The solution is $$(-\infty, 3) \cup (4, \infty)$$(−∞,3)∪(4,∞)
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What is the Precise Definition of a Complex Vector Space? [closed]
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$\begingroup$
I am studying linear algebra (as a second year) on my own using Axlers, Linear Algebra Done Right.
I have run into a definitional problem that I cant get past.
Specifically, Axler (and Wolfram, and others) define a complex vector space as a vector space in which the field is the complex numbers. According to this definition, the complex numbers over the real numbers are not a complex vector space, but the complex numbers over the complex numbers are a complex vector space. This despite the fact that the two vector spaces are identical (or at least isomorphic).
I already see that many theorems concerning eigenvalues/vectors, adjoints, and spectral theory vary according to whether we are looking at complex or real vector spaces. Thus, the definition of complex vector space is critical.
Im sure there is an easy explanation, but I dont see it. Thanks.
linear-algebra
functional-analysis
linear-transformations
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edited Jul 12, 2019 at 23:50
DDS
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asked Jul 12, 2019 at 21:11
PossumPPossumP
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$\begingroup$ "Axler (and Wolfram, and others) define a complex vector space as a vector space in which the field is the complex numbers." That's the definition - what's the question? You'd like an easy explanation of what? $\endgroup$
David C. Ullrich
– David C. Ullrich
2019-07-12 21:14:05 +00:00
Commented Jul 12, 2019 at 21:14
$\begingroup$ I'm not sure what your precise question is, but you should look closely at the axiomatic definition of a vector space over a field: en.wikipedia.org/wiki/Vector_space#Definition $\endgroup$
desiigner
– desiigner
2019-07-12 21:14:08 +00:00
Commented Jul 12, 2019 at 21:14
$\begingroup$ You should think about the phrase "$k$-vector space" as telling you which scalars you are allowed to use. A real vector space lets you scale your vectors by real numbers, while a complex vector space lets you scale vectors by complex numbers. $\endgroup$
desiigner
– desiigner
2019-07-12 21:15:36 +00:00
Commented Jul 12, 2019 at 21:15
7
$\begingroup$ "This despite the fact that the two vector spaces are identical (or at least isomorphic)." - this is not true; the two vector spaces don't even have the same dimension. Over the real scalars, you need two elements of $\mathbb C$ (e.g. $1$ and $i$) to form a basis. Over the complex scalars, you only need one. $\endgroup$
user169852
– user169852
2019-07-12 21:33:24 +00:00
Commented Jul 12, 2019 at 21:33
1
$\begingroup$ The vector spaces "$\Bbb C$ over $\Bbb C$" and "$\Bbb C$ over $\Bbb R$" have the same underlying set, and addition operation, but that does not mean they are the same vector space. It's also nonsense to say they are isomorphic, since the isomorphism must be a linear map, and to define linear maps you need to domain and codomain to be vector spaces with the same field of scalars. $\endgroup$
anon
– anon
2019-07-12 22:48:55 +00:00
Commented Jul 12, 2019 at 22:48
| Show 1 more comment
3 Answers 3
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$\begingroup$
I hope this gives some insight:
$\Bbb R$ as an $\Bbb R$-vector space
This is the real line you are used to. It has a basis consisting of one element, which we can choose to be $1$. You can view this as a vector with just one coefficient. Notice that any $r \in \Bbb R$ can be expressed in this basis as $r \cdot 1$. A vector is just a real number, and we have scalar multiplication that is just the usual multiplication.
$\Bbb C$ as an $\Bbb R$-vector space
This is the complex plane. We need two basis elements, for example $1$ and $i$. Now any vector in $\Bbb C$ can be expressed as $a + bi$ with $a,b \in \Bbb R$. If $r \in \Bbb R$ we have scalar multiplication $r(a + bi) = ra + rbi$.
$\Bbb C$ as an $\Bbb C$-vector space
Now we're back to being one dimensional. We only need one basis element, $1$ for example. Any element $a + bi \in \Bbb C$ can be expressed as $(a + bi) \cdot 1$. Compare this to the first example and convince yourself that these are essentially the same. Scalar multiplication happens with elements from $\Bbb C$.
Note that, for example, $\Bbb R$ as a $\Bbb C$-vector space does not make sense. We need to be able to multiply by scalars in a meaningful way. For general $z \in \Bbb C$, the product with a real number $r \in \Bbb R$ is a complex number $r \cdot z \in \Bbb C$.
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answered Jul 12, 2019 at 22:28
RubenRuben
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$\begingroup$ That helps a lot. I think any non-mathematician would see the last two spaces as identical. Do you know whether there are any differences between the last two spaces in applied areas? $\endgroup$
PossumP
– PossumP
2019-07-14 22:35:15 +00:00
Commented Jul 14, 2019 at 22:35
Add a comment |
5
$\begingroup$
It might help to remember that a vector space is defined by more than just the set of vectors $V$. Although we frequently abbreviate it that way, a vector space really consists of four things, $(k, V, {}+{}, {}\cdot{})$: the base field $k$, the set of vectors $V$, the vector addition operation $+$, and the scalar multiplication ${}\cdot{}$. Keeping that in mind will make it easier to understand how $\mathbb C$ can sometimes be a complex vector space and sometimes not.
When we're thinking about $\mathbb C$ as a vector space over $\mathbb C$, we mean the vector space $(\mathbb C, \mathbb C, {}+{},{}\cdot{})$, where the addition is addition of complex numbers and scalar multiplication is multiplication of complex numbers.
When we're thinking about $\mathbb C$ as a vector space over $\mathbb R$, we mean the vector space $(\mathbb R, \mathbb C, {}+{},{}\cdot{})$, where the addition is addition of complex numbers and scalar multiplication is multiplication of reals by complex numbers.
So these really are two different vector spaces, even though the set of vectors is the same in both. And one of them is a complex vector space, while the other is a real vector space.
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answered Jul 13, 2019 at 7:42
CarmeisterCarmeister
4,27322 gold badges1818 silver badges3434 bronze badges
$\endgroup$
3
$\begingroup$ Are you saying that in a vector space addition and multiplication are defined as in any ring? I.e. Just the names of two binary operations which could vary from one vector space to another? $\endgroup$
PossumP
– PossumP
2019-07-15 14:57:23 +00:00
Commented Jul 15, 2019 at 14:57
$\begingroup$ @MPitts Exactly. For example, we could define a real vector space ($k=\mathbb R$) where the set of vectors is $V=\mathbb R_{>0}$, the positive reals; vector "addition" is defined by $v+w:=vw$ (the latter being the product of the two real numbers); and scalar multiplication is defined by $c\cdot v=v^c$. You can show that this obeys all of the vector space axioms (and in fact it is isomorphic to the standard $1$-dimensional $\mathbb R$ as a vector space over itself). $\endgroup$
Carmeister
– Carmeister
2019-07-16 09:54:40 +00:00
Commented Jul 16, 2019 at 9:54
$\begingroup$ Great. That would seem to open up a whole new frontier of questions, problems, and theorems. (Strange though that this wider definition gets little mention.) $\endgroup$
PossumP
– PossumP
2019-07-16 23:20:37 +00:00
Commented Jul 16, 2019 at 23:20
Add a comment |
1
$\begingroup$
In case the other answers are too abstract, here is a concrete example.
Treating $\mathbb C$ as a complex vector space of one dimension, an example of a scalar is $a = 2 + i.$ An example of a vector is $\mathbf v = (1 + 2i).$ We can perform a scalar multiplication of a vector like so:
$$ a \mathbf v = (2 + i)(1 + 2i) = (5i), $$
where $(5i)$ is also a vector in this space.
Now if we take $\mathbb R^2$ as a two-dimensional vector space, we could make the obvious correspondence between $\mathbb R^2$ and $\mathbb C$ such that $(1, 2)^T \in \mathbb R^2$ corresponds to $1 + 2i = \mathbb C.$ But $2+i$ is not a scalar in this vector space. Instead we would have to treat $2 + i$ as another vector.
It is impossible to treat $(2 + i)(1 + 2i)$ as scalar multiplication in $\mathbb R^2$ taken as a two-dimensional vector space, although in $\mathbb C$ as a one-dimensional vector space this is scalar multiplication. So although $\mathbb R^2$ has some properties isomorphic with the corresponding properties of $\mathbb C,$ it's missing at least this one.
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answered Jul 13, 2019 at 1:57
David KDavid K
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Sign of enthalpy for exothermic and endothermic reactions
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In Cambridge Chemistry Coursebook [1, p. 94] it’s written that
A rise in temperature is given a positive sign. So the value of Δ H Δ H is negative for an exothermic reaction. A fall in temperature is given a negative sign. So the value of Δ H Δ H is positive for an endothermic reaction.
Why is the sign of enthalpy for exothermic reaction negative? Doesn’t exothermic reaction rise the temperature of surrounding?
Why is the sign of enthalpy for endothermic reaction positive? Doesn’t endothermic reaction take in energy and cool down the surrounding?
Reference
Ryan, L.; Norris, R. Cambridge International AS and A Level Chemistry Coursebook, 2nd ed.; Cambridge International Examinations; Cambridge University Press: Cambridge, 2014. ISBN 978-1-107-63845-7.
physical-chemistry
thermodynamics
enthalpy
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edited May 7, 2021 at 13:32
andselisk♦
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asked Nov 24, 2018 at 12:23
Mikail KhanMikail Khan
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Is the rising or falling of the T that is not required here. This is often see at introductory level in spire of being unnecessary. See answer Alchimista –Alchimista 2018-11-24 15:44:38 +00:00 Commented Nov 24, 2018 at 15:44
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Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give that heat to surroundings. If you do so, than since our system lost heat its enthalpy decreases (first law of thermodynamics) and because of that enthalpy of exothermic reaction is taken as negative. The opposite reasoning holds for endothermic reactions. You need to bring heat from surroundings into system to keep it at the same temperature and that heat increases enthalpy of the system.
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answered May 7, 2021 at 14:19
Dario MirićDario Mirić
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The reaction is said to be exothermic or endothermic according to the sign of the Enthalpy change. To obtain Enthalpy you subtract the enthalpy of formation for reactants from that of products. So, if the final enthalpy of the system is less than the initial one the result is negative and this difference manifests it self in the system as heat and temperature increases. The opposite is true for endothermic reactions.
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edited May 10, 2021 at 9:00
answered Nov 24, 2018 at 13:26
M.ghorabM.ghorab
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1 This answer is incorrect. Whether a reaction is exothermic or endothermic is determined by the sign of the enthalpy, not that of the Gibbs free energy. Indeed, for many reactions, depending upon the conditions, the enthalpy and the Gibbs free energy have can have the opposite sign. Thus, unless the entropy change is negligible, they cannot even be approximately equated to each other.theorist –theorist 2021-01-10 02:36:12 +00:00 Commented Jan 10, 2021 at 2:36
@theorist, You're right. the right term for free energy is endergonic or exergonic. Thank you.M.ghorab –M.ghorab 2021-05-07 13:00:14 +00:00 Commented May 7, 2021 at 13:00
1 In your correction, you have the signs reversed. You wrote: "To obtain Enthalpy you subtract the heat of formation for products (enthalpy) from that of reactants. So, if the final enthalpy (heat) of the system is more than the initial one the result is negative". In fact, enthalpy change = (enthalpy products) – (enthalpy reactants). [This is generally true for all state functions, not just enthalpy: change = (final state) – (initial state).] Thus if the final enthalpy is higher, the enthalpy change is positive.theorist –theorist 2021-05-07 17:09:16 +00:00 Commented May 7, 2021 at 17:09
1 Also: enthalpy is not heat. enthalpy is a state function, while heat is a path function. I don't mean to be discouraging, but I would strongly recommend you learn more about thermodynamics before attempting to answer questions on the subject.theorist –theorist 2021-05-07 17:10:32 +00:00 Commented May 7, 2021 at 17:10
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13526 | https://www.youtube.com/watch?v=wLm-y0WgXtI | Theorem 4.2 | Chapter No 4 | The Theory of Congruences | Elementary Number Theory
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Theorem 4.2 | Chapter No 4 | The Theory of Congruences | Elementary Number Theory
Basic Properties of Congruence | Chapter No 4 | The Theory of Congruences | Elementary Number Theory
Book Name : Elementary Number Theory
By : David M. Burton
Chapter Number : 04
Chapter Name : The Theory of Congruences
Lecture Number : 02
By (Name) : Awais Rasool
Exercise Number : 00
Problems Number: 00
Question Number : 00
Part Number : 00
Example Number : 00
Theorem Number: 4.2
Awais Rasool Shah
Topics Name :
4.1 Carl Friedrich Gauss
4.2 Basic Properties of Congruence.
4.3 Binary and Decimal Representations of integers
4.4 Linear Congruences and the Chinese Remainder Theorem.
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Basic Properties of Congruence:
Definition:
Let 𝑛 be a fixed positive integer. Two integers 𝑎 and 𝑏 are said to be congruent modulo 𝑛, symbolized by
𝑎≡𝑏 (𝑚𝑜𝑑 𝑛)
If 𝑛 divides the difference 𝑎−𝑏; that is, provided that 𝑎−𝑏=𝑘𝑛 for some integer 𝑘.
Let 𝑛 is greathen 1 be fixed and 𝑎,𝑏,𝑐,𝑑 be arbitrary integers. Then the following properties hold:
𝑎≡𝑎 (𝑚𝑜𝑑 𝑛).
If 𝑎≡𝑏 (𝑚𝑜𝑑 𝑛), then 𝑏≡𝑎 (𝑚𝑜𝑑 𝑛).
If 𝑎≡𝑏 (𝑚𝑜𝑑 𝑛) and 𝑏≡𝑐 (𝑚𝑜𝑑 𝑛) then 𝑎≡𝑐 (𝑚𝑜𝑑 𝑛).
If 𝑎≡𝑏 (𝑚𝑜𝑑 𝑛) and 𝑐≡𝑑 (𝑚𝑜𝑑 𝑛) , then 𝑎+𝑐≡𝑏+𝑑 (𝑚𝑜𝑑 𝑛) and 𝑎𝑐≡𝑏𝑑 (𝑚𝑜𝑑 𝑛)
If 𝑎≡𝑏 (𝑚𝑜𝑑 𝑛) , then 𝑎+𝑐≡𝑏+𝑐 (𝑚𝑜𝑑 𝑛) and 𝑎𝑐≡𝑏𝑐 (𝑚𝑜𝑑 𝑛)
If 𝑎≡𝑏 (𝑚𝑜𝑑 𝑛), then 𝑎^𝑘≡𝑏^𝑘 (𝑚𝑜𝑑 𝑛) for any positive integer 𝑘.
Transcript:
अस्सलाम वालेकुम नंबर थ्योरी एडिशन सिक्स एंड सेवन का चैप्टर नंबर फोर द थ्योरी ऑफ कंक्लूजन हम डिस्कस कर रहे थे आज हम चैप्टर नंबर फोर का थ्योरम 4.2 डिस्कस करेंगे थ्योरम 4.2 में वो यह कहता है लेट n ग्रेटर दन 1 बी अ फिक्स एंड ए बी सीडी बी अ आर्बिटिंग इंटी जर देन द फॉलोइंग प्रॉपर्टी आर होल्ड वो यह कह रहा है कि हमारे पास जो n है वो कोई भी ग्रेटर दन वन वैल्यू है हमारे पास जो ए बी सी की वैल्यू है वो इंटी जर में से बिलोंग करती है हमारे पास जो नीचे प्रॉपर्टी गई है उनको हमने प्रूव करना है हमारे पास जो फर्स्ट प्रॉपर्टी है a इ a मड n हमारे पास जो सेकंड प्रॉपर्टी है इफ a इ इ b मड n देन b = a मड n हमारे पास जो थर्ड प्रॉपर्टी है इफ a इ b मड n एंड b इक्टू c मड n देन हमने प्रूव करना है a = c मड n हमारे पास जो फोर्थ प्रॉपर्टी है इफ a = b मोल्ड n एंड c = d मोल्ड n देन a + c = b + d मोड n एंड a मल्टी बा c = b म बा d मोड n हमारे पास जो फिफ्थ प्रॉपर्टी है इफ a इ b मड n देन a + c = b + c मड n एंड a म c = b मल्टी c मोड m ये हमारे पास फ प्रॉपर प्रॉपर्टी थी हमारे पास जो लास्ट प्रॉपर्टी है इफ a इक्व b मड n देन हमने प्रूव करना है a की पावर k इ b की पावर के मड n फॉर एनी पॉजिटिव इंटी जर के के हमारे पास कोई भी पॉजिटिव इंटी जर होगा अब हम इन तमाम प्रॉपर्टी को प्रूव करने की कोशिश करते हैं हमारे पास जो फर्स्ट प्रॉपर्टी है a = a मोड n मीन के a हमारे पास कोई भी वैल्यू है उस वैल्यू का रिमाइंडर हमारे पास a ही आ रहा है अगर वैल्यू को मैं n पे डिवाइड करता हूं तो मैं इसको लिख सकता हूं फॉर एनी इंटी जर a व हैव a - a = k ए मीन के इस वैल्यू में से मैं रिमाइंडर ये वाला ए माइनस करूंगा जो हमारे पास वैल्यू आएगी वो एंड पे डिवाइड होगी और हमारे पास जो एंड पे डिवाइड करने के साथ जो हमारे पास k है वो कोई भी इंटी जर है जिसको मैं n से मल्टीप्लाई करूंगा जो हमारे पास वैल्यू आएगी वो a - a के बराबर आएगी फर सम इंटी जर के अब हमारे पास a - a तो रो होता है तो इसका मतलब जो इस केस में हमारे पास जो k है वो क्या है जीरो है तो हमारे पास ये हमने प्रूफ कर लिया a = a मड n ये हमारे पास कंडीशन सेटिस्फाई करती है क्योंकि इसके अंदर हमारे पास जो k है वो कोई ना कोई वैल्यू आ रही है जिसको हम मल्टीप्लाई करें n के साथ जो हमारे पास वैल्यू आएगी वो a माइ a आ रही है हमारे पास जो सेकंड प्रॉपर्टी है इफ a इ b मॉडलो एंड देन हमने प्रूव करना है b = a मड n a = b मड n हमारे पास गिवन है मैं इसको लिख सकता हूं अगर मैं a में से b माइनस करूं मीन के b हमारे पास रिमाइंडर है a हमारी वैल्यू है जिस पे वैल्यू डिवाइड हो रही है वो n है अगर मैं वैल्यू में से रिमाइंडर माइनस करूंगा जो हमारी वैल्यू आएगी वो n पे डिवाइड होगी मीन के a - b = k ए k हमारे पास कोई भी वैल्यू जिसको मैं मल्टीप्लाई कर रहा हूं n के साथ जो हमारी वैल्यू आएगी वो a - b के बराबर आएगी फॉर एग्जांपल a हमारे पास क्या है 24 है 24 को अगर मैं सेन पे डिवाइड करता हूं तो हमारे पास जो 73 हमारे पास 21 होता है जो हमारा रिमाइंडर थ्री आएगा 24 में से मैं थ माइनस करूंगा हमारी जो वैल्यू है वो सेन पे डिवाइड होगी तो मीन कि जो हमारा इस केस में k है वो थ्री है फर सम इंटी जर k k हमारे पास कोई भी इंटी जर हो सकता है मल्टीप्लाई बाय -1 बोत साइड हम बोथ साइड पे -1 से मल्टीप्लाई करेंगे तो इनके साइन चेंज हो जाएंगे तो b - a = - k ए आ जाएगा तो मीन के हमारे पास अब क्या आ गया b - a = - k n इस केस के अर जो हमारी वैल्यू मल्टीप्लाई आ रही है जो हमारा कोई भी इंटी जर मल्टीप्लाई करने से b माइ a वैल्यू आ रही है वो नेगेटिव है तो अब हम लिख सकते हैं b इज इ a म n मीन कि अब बी हमारे पास वैल्यू है उसका रिमाइंडर a आ रहा है अगर वो वैल्यू किस पे डिवाइड हो रही है एंड पे डिवाइड हो रही है और जो इस केस में हमारे पास जो मल्टीप्लाई वाला फैक्टर है वो माइनस के है अब हम थर्ड प्रॉपर्टी डिस्कस करते हैं हमारे पास जो थर्ड प्रॉपर्टी है इफ a इ b मलो n एंड b इ c म ए देन हमने प्रूफ करना है a = c मड n हमारे पास जो फर्स्ट इक्वेशन है a = b मड n इसको मैं लिख सकता हूं a - b = k ए मीन के a हमारे पास कोई भी वैल्यू हो उसमें से मैं b रिमाइंडर माइनस कर रहा हूं जो वैल्यू आएगी वो एंड पे डिवाइड होगी वो जो k है उसका मल्टीपल फैक्टर है मीन किसको मैं n के साथ मल्टीप्लाई करूं जो हमारा जवाब आएगा वो a - b के बराबर आएगा फर सम इंटी जर k की हमारे पास कोई भी इंटी जर वैल्यू हो सकती है मल्टीप्लाई बाय -1 बोथ साइड का माइन को बोथ साइड से मल्टीप्लाई करूंगा तो हमारे पास इक्वेशन आएगी b - a = - k ए अब जो हमारे पास a है वो लेफ्ट साइड प माइनस हो रहा है क मैं इसको राइट साइड प शिफ्ट करूंगा तो प्लस हो जाएगा b = a - k एंड ये हमारे पास इक्वेशन नंबर वन आ चुकी है सेकंड क्वेश्चन गिवन थी वो b = c मड n गिवन थी इसको मैं लिख सकता हूं b - c = h ए फॉर सम h कैसे इस केस के अंदर k हमारे पास कोई भी इंटी जर था हमने वो के के डिफरेंट बनाने के लिए क्योंकि हमने इक्वेशन नंबर वन में k ले याया है तो इसको डिफरेंट बनाने के लिए हमने यहां पे जो इंटी जर लिया वो h ले लिया है तो हमारे पास जो वैल्यू आएगी b - a = h ए अब हमारा जो c है वो लेफ्ट साइड में माइनस हो रहा है क मैं इसको राइट साइड में शिफ्ट करूंगा वो प्लस हो जाएगा तो b = c + h एने क्वेश्चन नंबर टू का नाम दे दिया अब हमारी जो क्वेश्चन नंबर वन और क्वेश्चन नंबर टू इनमें दोनों में लेफ्ट साइड सेम है और राइट साइड डिफरेंट है इनके जो राइट साइड वो आपस में क्या होगी इक्वल होगी फॉर क्वेश्चन नंबर वन एंड इ क्वेश्चन नंबर टू a - ए इज इ c + hn1 हो रहा का मैं इसको राइट साइड प शिफ्ट करूंगा तो प्लस हो जाएगा और जो c हमारे पास राइट साइड पे प्लस हो रहा का मैं इसको लेफ्ट साइड पे शिफ्ट करूंगा वो माइनस हो जाएगा तो हमारे पास इक्वेशन आएगी a - c इ k + hn-286 है कि हमारे पास a - c जो वैल्यू है वो डिवाइड हो रही है n के ऊपर तो हम लिख सकते हैं a - c = 0 मड n मीन के जो वैल्यू है a - c वो डिवाइड हो रही है n के ऊपर जो इसका रिमाइंडर आ रहा है वो जीरो आ रहा है अब यहां पे c जो है वो माइनस हो र मैं उसको राइट साइड प शिफ्ट करूंगा तो हमारे पास प्लस हो जाएगा तो a इ c मड n ये हमारे पास इक्वेशन आ गई हिंस प्रूफ इसको हमने प्रूव कर लिया है हमारे पास जो नेक्स्ट प्रॉपर्टी है a = b मड n n c = d मड n देन हमने प्रूव करना है a + c = b + d मड n एंड हमने प्रूव करना है ए = बडी मड n हमारे पास जो फर्स्ट इक्वेशन है a = b म n इसको हम पहले सॉल्व करते हैं इसको हम लिख सकते हैं a - b = k ए फॉर सम k क्योंकि हमारे पास a हमारे पास कोई भी वैल्यू हो उसमें से हम रिमाइंडर मानेज कर रहे हैं जो हमारी वैल्यू है वो एंड पे डिवाइड होगी तो अब इसको हम सॉल्व करेंगे जो हमारा b है वो लेफ्ट साइड पे माइनस हो रहा है तो मैं इसको राइट साइड प शिफ्ट करूंगा तो प्लस हो जाएगा तो a = b + ए आ जाएगा इसको हमने इक्वेशन नंबर वन का नाम दे दिया अब हमारे पास जो सेकंड इक्वेशन है c इटू d मॉडलो n c - d = h ए इस केस में हमारा जो h है वो सम इंटी जर कोई भी इंटी जर वैल्यू है तो हम लिख सकते हैं जो d यहां पे माइनस हो रहा का मैं इसको राइट साइड प शिफ्ट करूंगा तो प्लस हो जाएगा तो c = d + hn1 एंड इक्वेशन नंबर टू को आपस में हम ऐड करेंगे तो हमारे इक्वेशन आएगी a + c = b + k ए + d + hn1 एंड इ क्वेश्चन नंबर टू को हमारे पास वैल्यू आ गई a + c ये हमारे पास जो b एंड d दोनों वैल्यू यहां पे प्लस हो रही अब मैं इसको लेफ्ट साइड प शिफ्ट करूंगा तो माइनस हो जाएगी तो - b + d = k और h ए में से हमारा n कॉमन आ रहा है तो हमारी वैल्यू आएगी k + h n आ जाएगी जो वैल्यू है a + c - b + d जो वैल्यू है वो n पे डिवाइड हो रही है इसका मतलब जो वैल्यू डिवाइड हो रही है इसका रिमाइंडर क्या आएगा जीरो आएगा मोड n के हिसाब से तो हमारे पास इक्वेशन आ गई a + c - b + d = 0 मोड n अब हमारी जो वैल्यू है b + d वो लेफ्ट साइड प माइनस हो रही का मैं इसको राइट साइड प शिफ्ट करूंगा तो प्लस हो जाएगी हमारी इक्वेशन आएगी a + c = b + d मोड n हमारे पास जो सेकंड इसका केस है वो कह रहा है कि हमने प्रूफ करना है ए = बडी मड n अब हमारा हमारे पास जो इक्वेशन नंबर वन एंड इक्वेशन नंबर टू आ चुकी थी हमने लेफ्ट साइड लेफ्ट में और राइट साइड राइट में मल्टीप्लाई कर देनी है मल्टीप्लाई बाय इक्वेशन वन एंड टू हमारे पास क्वेश्चन आएगी ए = b + k ए म d + मल्टीप्ला बा h ए तो हमारे पास k ए h ए ए क्योंकि n दो दफा है इस केस में अब हमारी जो वैल्यू है बडी यहां पे प्लस हो रही मैं उसको लेफ्ट साइड प शिफ्ट करूंगा तो माइनस हो जाएगी तो c ए - बडी इक्वल टू इन तीनों वैल्यू में से n कॉमन आ रहा है तो हमारे पास इक्वेशन आएगी ब ए + केडी + h k ए n अब हमारी जो वैल्यू है c ए - बीडी जो वैल्यू है वो n पे डिवाइड हो रही है ए - बडी इ 0 मोड n क्योंकि इसका वैल्यू का रिमाइंडर जीरो आ रहा है मोड n के हिसाब से अब जो हमारा बीडी है यहां पे माइनस हो रहा है क मैं इसको राइट साइड में शिफ्ट करूंगा तो प्लस हो जाएगा तो ए = बडी मोल्ड n ये हमारे पास इक्वेशन आ चुकी है अब हमारे पास जो लास्ट प्रॉपर्टी है इफ a = b मोड n देन हमने प्रूव करना है a की पावर k इ b की पावर k मोड n जो हमारे पास k है वो कोई भी पॉजिटिव इंटी जर हो सकता है तो हमारे पास गिवन है a = b म n इसको मैं लिख सकता हूं a - b = इ k ए फॉर सम इंटी जर k अब हमने करना क्या है जो हमारा यहां पे b माइनस हो रहा है मैं इसको राइट साइड प शिफ्ट करूंगा तो प्लस हो जाएगा तो a इज इ b + k ए इसको हमने क्वेश्चन नंबर वन का नाम दे दिया टेकिंग स्क्वेयर ऑन बोथ साइड हमने बोथ साइड पे स्क्वेयर लेना है तो हमारे पास इक्वेशन आएगी a का स्क्वा = b + k ए काल स्क्वा a का स्क्वा a होता है और इसका हम स्क्वायर वाला फार्मूला ओपन करेंगे तो b का स्क्वा प् b k स् n स् + 2bk ए तो हमने यहां पे फार्मूला लगाया a का स्क्वा ् b का स्क + 2ab b वाला a का स्क्वा इ इक्वल टू b का स्क्वा प्लस इन दो वैल्यू में से n कॉमन आ रहा है तो हमारे पास क्वेश्चन आएगी k का स्क्वा n + 2bk ए अब ये हमारा जो b है यहां पे प्लस हो रहा मैं इसको लेफ्ट साइड प शिफ्ट करूंगा तो माइनस हो जाएगा तो a का स्क्वा - b का स्क्वा इट k स् n + 2bk n हमारी जो वैल्यू है a का स्क्वा माइ b का स्क्वा वो n पे डिवाइड हो रही है तो हमारे पास जो वैल्यू आएगी a का स्क्वा - b का स्क्वा = 0 मोड n क्योंकि n पे डिवाइड हो रही है इसका रिमाइंडर जीरो आएगा n के हिसाब से तो इसको मैं जो माइनस ब स्क यहां पे माइनस हो रहा मैं इसको रा राइट साइड प शिफ्ट करूंगा तो वो प्लस हो जाएगा a स् इज इ b स् म n ये हमने तमाम प्रॉपर्टी डिस्कस कर ली थ्योरम 4.2 के ये तमाम प्रॉपर्टी है ये आपने याद भी करनी है ये अक्सर सवालों में यूज भी होती हैं हम इंशाल्लाह बहुत जल्दी नंबर थ्योरी के सॉल्यूशन बुक ऐप प्ले स्टोर पे अपलोड करने जा रहे हैं ताकि तमाम स्टूडेंट की हेल्प हो सके अगर वो आपको बुक ऐप चाहिए हो तो आप इस वीडियो के नीचे लाजमी कमेंट करें इंशाल्लाह हम बहुत जल्दी ये बुक ऐप प् स्टोर पे अप लोड कर देंगे आप हमारा youtube3 डिस्क्रीट मैथ कमीना टोरस फंक्शनल एंड तमाम बीएस मैथ के लेक्चर अपलोड करते हैं ताकि हर नई आने वाली वीडियो आप तक पहुंच सके तो आप हमारा youtube1 वीडियो में आपसे मुलाकात होगी तब तक अल्लाह हाफिज |
13527 | https://statistics.berkeley.edu/sites/default/files/tech-reports/638.pdf | Convexity, Classification, and Risk Bounds Peter L. Bartlett Division of Computer Science and Department of Statistics University of California, Berkeley bartlett@stat.berkeley.edu Michael I. Jordan Division of Computer Science and Department of Statistics University of California, Berkeley jordan@stat.berkeley.edu Jon D. McAuliffe Department of Statistics University of California, Berkeley jon@stat.berkeley.edu November 4, 2003 Technical Report 638 Abstract Many of the classification algorithms developed in the machine learning literature, including the support vector machine and boosting, can be viewed as minimum contrast methods that minimize a convex surrogate of the 0-1 loss function. The convexity makes these algorithms computationally efficient. The use of a surrogate, however, has statistical consequences that must be balanced against the computational virtues of convexity. To study these issues, we provide a general quantitative relationship between the risk as assessed using the 0-1 loss and the risk as assessed using any nonnegative surrogate loss function. We show that this relationship gives nontrivial upper bounds on excess risk under the weakest possible condition on the loss function: that it satisfy a pointwise form of Fisher consistency for classification. The relationship is based on a simple variational transformation of the loss function that is easy to compute in many applications. We also present a refined version of this result in the case of low noise.
Finally, we present applications of our results to the estimation of convergence rates in the general setting of function classes that are scaled convex hulls of a finite-dimensional base class, with a variety of commonly used loss functions.
Keywords: machine learning, convex optimization, boosting, support vector machine, Rademacher complexity, empirical process theory 1 1 Introduction Convexity has become an increasingly important theme in applied mathematics and engineering, having acquired a prominent role akin to the one played by linearity for many decades. Build-ing on the discovery of efficient algorithms for linear programs, researchers in convex optimization theory have developed computationally tractable methods for large classes of convex programs (Nes-terov and Nemirovskii, 1994). Many fields in which optimality principles form the core conceptual structure have been changed significantly by the introduction of these new techniques (Boyd and Vandenberghe, 2003).
Convexity arises in many guises in statistics as well, notably in properties associated with the exponential family of distributions (Brown, 1986). It is, however, only in recent years that the systematic exploitation of the algorithmic consequences of convexity has begun in statistics. One applied area in which this trend has been most salient is machine learning, where the focus has been on large-scale statistical models for which computational efficiency is an imperative. Many of the most prominent methods studied in machine learning make significant use of convexity; in particular, support vector machines (Boser et al., 1992, Cortes and Vapnik, 1995, Cristianini and Shawe-Taylor, 2000, Sch¨ olkopf and Smola, 2002), boosting (Freund and Schapire, 1997, Collins et al., 2002, Lebanon and Lafferty, 2002), and variational inference for graphical models (Jordan et al., 1999) are all based directly on ideas from convex optimization.
If algorithms from convex optimization are to continue to make inroads into statistical theory and practice, it is important that we understand these algorithms not only from a computational point of view but also in terms of their statistical properties. What are the statistical consequences of choosing models and estimation procedures so as to exploit the computational advantages of convexity?
In the current paper we study this question in the context of multivariate classification. We consider the setting in which a covariate vector X ∈X is to be classified according to a binary response Y ∈{−1, 1}. The goal is to choose a discriminant function f : X →R, from a class of functions F, such that the sign of f(X) is an accurate prediction of Y under an unknown joint measure P on (X, Y ). We focus on 0-1 loss; thus, letting ℓ(α) denote an indicator function that is one if α ≤0 and zero otherwise, we wish to choose f ∈F that minimizes the risk R(f) = Eℓ(Y f(X)) = P(Y ̸= sign(f(X))).
Given a sample Dn = ((X1, Y1), . . . , (Xn, Yn)), it is natural to consider estimation procedures based on minimizing the sample average of the loss, ˆ R(f) = 1 n Pn i=1 ℓ(Yif(Xi)). As is well known, however, such a procedure is computationally intractable for many nontrivial classes of func-tions (see, e.g., Arora et al., 1997). Indeed, the loss function ℓ(Y f(X)) is non-convex in its (scalar) argument, and, while not a proof, this suggests a source of the difficulty. Moreover, it suggests that we might base a tractable estimation procedure on minimization of a convex surrogate φ(α) for the loss. In particular, if F consists of functions that are linear in a parameter vector θ, then the overall problem of minimizing expectations of φ(Y f(X)) is convex in θ. Given a convex parameter space, we obtain a convex program and can exploit the methods of convex optimization. A wide variety of classification methods in machine learning are based on this tactic; in particular, Figure 1 shows the (upper-bounding) convex surrogates associated with the support vector machine (Cortes and Vapnik, 1995), Adaboost (Freund and Schapire, 1997) and logistic regression (Friedman et al., 2000).
A basic statistical understanding of this setting has begun to emerge.
In particular, when 2 −2 −1 0 1 2 0 1 2 3 4 5 6 7 α 0−1 exponential hinge logistic truncated quadratic Figure 1: A plot of the 0-1 loss function and surrogates corresponding to various practical classifiers.
These functions are plotted as a function of the margin α = yf(x). Note that a classification error is made if and only if the margin is negative; thus the 0-1 loss is a step function that is equal to one for negative values of the abscissa. The curve labeled “logistic” is the negative log likelihood, or deviance, under a logistic regression model; “hinge” is the piecewise-linear loss used in the support vector machine; and “exponential” is the exponential loss used by the Adaboost algorithm. The deviance is scaled so as to majorize the 0-1 loss; see Lemma 9.
appropriate regularization conditions are imposed, it is possible to demonstrate the Bayes-risk consistency of methods based on minimizing convex surrogates for 0-1 loss. Lugosi and Vayatis (2003) have provided such a result under the assumption that the surrogate φ is differentiable, monotone, strictly convex, and satisfies φ(0) = 1. This handles all of the cases shown in Figure 1 except the support vector machine. Steinwart (2002) has demonstrated consistency for the support vector machine as well, in a general setting where F is taken to be a reproducing kernel Hilbert space, and φ is assumed continuous. Other results on Bayes-risk consistency have been presented by Breiman (2000), Jiang (2003), Mannor and Meir (2001), and Mannor et al. (2002).
Consistency results provide reassurance that optimizing a surrogate does not ultimately hinder the search for a function that achieves the Bayes risk, and thus allow such a search to proceed within the scope of computationally efficient algorithms. There is, however, an additional motivation for working with surrogates of 0-1 loss beyond the computational imperative. Minimizing the sample average of an appropriately-behaved loss function has a regularizing effect: it is possible to obtain uniform upper bounds on the risk of a function that minimizes the empirical average of the loss φ, even for classes that are so rich that no such upper bounds are possible for the minimizer of the empirical average of the 0-1 loss.
Indeed a number of such results have been obtained for function classes with infinite VC-dimension but finite fat-shattering dimension (Bartlett, 1998, 3 Shawe-Taylor et al., 1998), such as the function classes used by AdaBoost (see, e.g., Schapire et al., 1998, Koltchinskii and Panchenko, 2002). These upper bounds provide guidance for model selection and in particular help guide data-dependent choices of regularization parameters.
To carry this agenda further, it is necessary to find general quantitative relationships between the approximation and estimation errors associated with φ, and those associated with 0-1 loss.
This point has been emphasized by Zhang (2003), who has presented several examples of such relationships. We simplify and extend Zhang’s results, developing a general methodology for finding quantitative relationships between the risk associated with φ and the risk associated with 0-1 loss.
In particular, let R(f) denote the risk based on 0-1 loss and let R∗= inff R(f) denote the Bayes risk. Similarly, let us refer to Rφ(f) = Eφ(Y f(X)) as the “φ-risk,” and let R∗ φ = inff Rφ(f) denote the “optimal φ-risk.” We show that, for all measurable f, ψ(R(f) −R∗) ≤Rφ(f) −R∗ φ, (1) for a nondecreasing function ψ : [0, 1] →[0, ∞). Moreover, we present a general variational repre-sentation of ψ in terms of φ, and show how this representation allows us to infer various properties of ψ.
This result suggests that if ψ is well-behaved then minimization of Rφ(f) may provide a rea-sonable surrogate for minimization of R(f). Moreover, the result provides a quantitative way to transfer assessments of statistical error in terms of “excess φ-risk” Rφ(f) −R∗ φ into assessments of error in terms of “excess risk” R(f) −R∗.
Although our principal goal is to understand the implications of convexity in classification, we do not impose a convexity assumption on φ at the outset. Indeed, while conditions such as convexity, continuity, and differentiability of φ are easy to verify and have natural relationships to optimization procedures, it is not immediately obvious how to relate such conditions to their statistical conse-quences. Thus, we consider the weakest possible condition on φ: that it is “classification-calibrated,” which is essentially a pointwise form of Fisher consistency for classification (Lin, 2001). In partic-ular, if we define η(x) = P(Y = 1|X = x), then φ is classification-calibrated if, for η(x) ̸= 1/2, the minimizer f ∗of the conditional expectation E[φ(Y f ∗(X))|X = x] has the same sign as the Bayes decision rule, sign(2η(x) −1). We show that our upper bound on excess risk in terms of excess φ-risk is nontrivial precisely when φ is classification-calibrated. Obviously, no such bound is possible when φ is not classification-calibrated.
The difficulty of a pattern classification problem is closely related to the behavior of the posterior probability η(X). In many practical problems, it is reasonable to assume that, for most X, η(X) is not too close to 1/2. Tsybakov (2001) has introduced an elegant formulation of such an assumption and considered the rate of convergence of the risk of a function that minimizes empirical risk over some fixed class F. He showed that, under the assumption of low noise, the risk converges surprisingly quickly to the minimum over the class. If the minimum risk is nonzero, we might expect a convergence rate no faster than 1/√n. However, under Tsybakov’s assumption, it can be as fast as 1/n. We show that minimizing empirical φ-risk also leads to surprisingly fast convergence rates under this assumption. In particular, if φ is uniformly convex, the empirical φ-risk converges quickly to the φ-risk, and the noise assumption allows an improvement in the relationship between excess φ-risk and excess risk.
These results suggest an interpretation of pattern classification methods involving a convex contrast function. It is common to view the excess risk as a combination of an estimation term and 4 an approximation term: R(f) −R∗= R(f) −inf g∈F R(g) + inf g∈F R(g) −R∗ .
However, choosing a function with risk near minimal over a class F—that is, finding an f for which the estimation term above is close to zero—is, in a minimax setting, equivalent to the problem of minimizing empirical risk, and hence is computationally infeasible for typical classes F of interest.
Indeed, for classes typically used by boosting and kernel methods, the estimation term in this expression does not converge to zero for the minimizer of the empirical risk. On the other hand, we can also split the upper bound on excess risk into an estimation term and an approximation term: ψ(R(f) −R∗) ≤Rφ(f) −R∗ φ = Rφ(f) −inf g∈F Rφ(g) + inf g∈F Rφ(g) −R∗ φ .
Often, it is possible to minimize φ-risk efficiently. Thus, while finding an f with near-minimal risk might be computationally infeasible, finding an f for which this upper bound on risk is near minimal can be feasible.
The paper is organized as follows. Section 2 presents basic definitions and a statement and proof of (1). In Section 3, we introduce the convexity assumption and discuss its relationship to the other conditions. Section 4 presents a refined version of our main result in the setting of low noise. We give applications to the estimation of convergence rates in Section 5 and present our conclusions in Section 6.
2 Relating excess risk to excess φ-risk There are three sources of error to be considered in a statistical analysis of classification problems: the classical estimation error due to finite sample size, the classical approximation error due to the size of the function space F, and an additional source of approximation error due to the use of a surrogate in place of the 0-1 loss function. It is this last source of error that is our focus in this section. Thus, throughout the section we (a) work with population expectations and (b) assume that F is the set of all measurable functions. This allows us to ignore errors due to the size of the sample and the size of the function space, and focus on the error due to the use of a surrogate for the 0-1 loss function.
We follow the tradition in the classification literature and refer to the function φ as a loss function, since it is a function that is to be minimized to obtain a discriminant. More precisely, φ(Y f(X)) is generally referred to as a “margin-based loss function,” where the quantity Y f(X) is known as the “margin.” (It is worth noting that margin-based loss functions are rather different from distance metrics, a point that we explore in the Appendix.) This ambiguity in the use of “loss” will not confuse; in particular, we will be careful to distinguish the risk, which is an expectation over 0-1 loss, from the “φ-risk,” which is an expectation over φ.
Our goal in this section is to relate these two quantities.
2.1 Setup Let (X × {−1, 1}, G ⊗2{−1,1}, P) be a probability space. Let X be the identity function on X and Y the identity function on {−1, 1}, so that P is the distribution of (X, Y ), i.e., for A ∈G ⊗2{−1,1}, 5 P((X, Y ) ∈A) = P(A). Let PX on (X, G) be the marginal distribution of X, and let η : X →[0, 1] be a measurable function such that η(X) is a version of P(Y = 1|X). Throughout this section, f is understood as a measurable mapping from X into R.
Define the {0, 1}-risk, or just risk, of f as R(f) = P(sign(f(X)) ̸= Y ), where sign(α) = 1 for α > 0 and −1 otherwise. (The particular choice of the value of sign(0) is not important, but we need to fix some value in {±1} for the definitions that follow.) Based on an i.i.d. sample Dn = ((X1, Y1), . . . , (Xn, Yn)), we want to choose a function fn with small risk.
Define the Bayes risk R∗= inff R(f), where the infimum is over all measurable f. Then any f satisfying sign(f(X)) = sign(η(X) −1/2) a.s. on {η(X) ̸= 1/2} has R(f) = R∗.
Fix a function φ : R →[0, ∞). Define the φ-risk of f as Rφ(f) = Eφ(Y f(X)).
Let F be a class of functions f : X →R. Let fn = ˆ fφ be a function in F which minimizes the empirical expectation of φ(Y f(X)), ˆ Rφ(f) = ˆ Eφ(Y f(X)) = 1 n n X i=1 φ(Yif(Xi)).
Thus we treat φ as specifying a contrast function that is to be minimized in determining the discriminant function fn.
2.2 Basic conditions on the loss function For (almost all) x, we define the conditional φ-risk E(φ(Y f(X))|X = x) = η(x)φ(f(x)) + (1 −η(x))φ(−f(x)).
It is useful to think of the conditional φ-risk in terms of a generic conditional probability η ∈[0, 1] and a generic classifier value α ∈R. To express this viewpoint, we introduce the generic conditional φ-risk Cη(α) = ηφ(α) + (1 −η)φ(−α).
The notation suppresses the dependence on φ. The generic conditional φ-risk coincides with the conditional φ-risk of f at x ∈X if we take η = η(x) and α = f(x). Here, varying α in the generic formulation corresponds to varying f in the original formulation, for fixed x.
For η ∈[0, 1], define the optimal conditional φ-risk H(η) = inf α∈R Cη(α) = inf α∈R(ηφ(α) + (1 −η)φ(−α)).
Then the optimal φ-risk satisfies R∗ φ := inf f Rφ(f) = EH(η(X)), where the infimum is over measurable functions.
6 We say that a sequence α1, α2, . . . achieves H at η if lim i→∞Cη(αi) = lim i→∞(ηφ(αi) + (1 −η)φ(−αi)) = H(η).
If the infimum in the definition of H(η) is uniquely attained for some α, we can define α∗: [0, 1] →R by α∗(η) = arg min α∈R Cη(α) = arg min α∈R ηφ(α) + (1 −η)φ(−α).
In that case, we define f ∗ φ : X →R, up to PX-null sets, by f ∗ φ(x) = arg min α∈R E(φ(Y α)|X = x) = α∗(η(x)) and then Rφ(f ∗ φ) = EH(η(X)) = R∗ φ.
For η ∈[0, 1], define H−(η) = inf α:α(2η−1)≤0 Cη(α) = inf α:α(2η−1)≤0(ηφ(α) + (1 −η)φ(−α)).
This is the optimal value of the conditional φ-risk, under the constraint that the sign of the argument α disagrees with that of 2η −1.
We now turn to the basic condition we impose on φ. This condition generalizes the requirement that the minimizer of Cη(α) (if it exists) has the correct sign. This is a minimal condition that can be viewed as a pointwise form of Fisher consistency for classification.
Definition 1. We say that φ is classification-calibrated if, for any η ̸= 1/2, H−(η) > H(η).
Equivalently, φ is classification-calibrated if any sequence α1, α2, . . . that achieves H at η satisfies lim infi→∞sign(αi(η −1/2)) = 1.
Since sign(αi(η −1/2)) ∈{−1, 1}, this is equivalent to the requirement limi→∞sign(αi(η −1/2)) = 1, or simply that sign(αi(η −1/2)) ̸= 1 only finitely often.
2.3 The ψ-transform and the relationship between excess risks We begin by defining a functional transform of the loss function: Definition 2. We define the ψ-transform of a loss function as follows. Given φ : R →[0, ∞), define the function ψ : [0, 1] →[0, ∞) by ψ = ˜ ψ∗∗, where ˜ ψ(θ) = H− 1 + θ 2 −H 1 + θ 2 , and g∗∗: [0, 1] →R is the Fenchel-Legendre biconjugate of g : [0, 1] →R, which is characterized by epi g∗∗= co epi g.
Here co S is the closure of the convex hull of the set S, and epi g is the epigraph of the function g, that is, the set {(x, t) : x ∈[0, 1], g(x) ≤t}. The nonnegativity of ψ is established below in Lemma 5, part 7.
7 Recall that g is convex if and only if epi g is a convex set, and g is closed (epi g is a closed set) if and only if g is lower semicontinuous (Rockafellar, 1997). By Lemma 5, part 5, ˜ ψ is continuous, so in fact the closure operation in Definition 2 is vacuous. We therefore have that ψ is simply the functional convex hull of ˜ ψ, ψ = co ˜ ψ , which is equivalent to the epigraph convex hull condition of the definition. This implies that ψ = ˜ ψ if and only if ˜ ψ is convex; see Example 5 for a loss function where the latter fails.
The importance of the ψ-transform is shown by the following theorem.
Theorem 3.
1. For any nonnegative loss function φ, any measurable f : X →R and any probability distribution on X × {±1}, ψ(R(f) −R∗) ≤Rφ(f) −R∗ φ.
2. Suppose |X| ≥2. For any nonnegative loss function φ, any ϵ > 0 and any θ ∈[0, 1], there is a probability distribution on X × {±1} and a function f : X →R such that R(f) −R∗= θ and ψ(θ) ≤Rφ(f) −R∗ φ ≤ψ(θ) + ϵ.
3. The following conditions are equivalent.
(a) φ is classification-calibrated.
(b) For any sequence (θi) in [0, 1], ψ(θi) →0 if and only if θi →0.
(c) For every sequence of measurable functions fi : X →R and every probability distribution on X × {±1}, Rφ(fi) →R∗ φ implies R(fi) →R∗.
Here we mention that classification-calibration implies ψ is invertible on [0, 1], so in that case it is meaningful to write the upper bound on excess risk in Theorem 3(1) as ψ−1(Rφ(f) −R∗ φ).
Invertibility follows from convexity of ψ together with Lemma 5, parts 6, 8, and 9.
Zhang (2003) has given a comparison theorem like Parts 1 and 3b of this theorem, for convex φ that satisfy certain conditions. These conditions imply an assumption on the rate of growth (and convexity) of ˜ ψ. Lugosi and Vayatis (2003) show that a limiting result like Part 3c holds for strictly convex, differentiable, monotonic φ. In Section 3, we show that if φ is convex, classification-calibration is equivalent to a simple derivative condition on φ at zero. Clearly, the conclusions of Theorem 3 hold under weaker conditions than those assumed by Zhang (2003) or Lugosi and Vayatis (2003). Steinwart (2002) has shown that if φ is continuous and classification-calibrated, then Rφ(fi) →R∗ φ implies R(fi) →R∗. Theorem 3 shows that we may obtain a more quantitative statement of the relationship between these excess risks, under weaker conditions.
Before presenting the proof of Theorem 3, we illustrate the ψ-transform in the case of four commonly used margin-based loss functions.
8 −2 −1 0 1 2 0 1 2 3 4 5 6 7 α φ(α) φ(−α) C0.3(α) C0.7(α) 0.0 0.2 0.4 0.6 0.8 1.0 −2 −1 0 1 2 η , θ α(η) H(η) ψ(θ) Figure 2: Exponential loss.
The left panel shows φ(α), its reflection φ(−α), and two different convex combinations of these functions, for η = 0.3 and η = 0.7. Note that the minima of these combinations are the values H(η), and the minimizing arguments are the values α∗(η). The right panel shows H(η) and α∗(η) plotted as a function of η, and the ψ-transform ψ(θ).
Example 1 (Exponential loss). Here φ(α) = exp(−α). Figure 2, left panel, shows φ(α), φ(−α), and the generic conditional φ-risk Cη(α) for η = 0.3 and η = 0.7. In this case, φ is strictly convex on R, hence Cη(α) is also strictly convex on R, for every η. So Cη is either minimal at a unique stationary point, or it attains no minimum. Indeed, if η = 0, then Cη(α) →0 as α →−∞; if η = 1, then Cη(α) →0 as α →∞. Thus we have H(0) = H(1) = 0 for exponential loss. For η ∈(0, 1), solving for the stationary point yields the unique minimizer α∗(η) = 1 2 log η 1 −η .
We may then simplify the identity H(η) = Cη(α∗(η)) to obtain H(η) = 2 p η(1 −η).
Notice that this expression is correct even for η equal to 0 or 1. It is easy to check that H− 1 + θ 2 ≡exp(0) = 1, 9 −2 −1 0 1 2 0 1 2 3 4 5 6 7 α φ(α) φ(−α) C0.3(α) C0.7(α) 0.0 0.2 0.4 0.6 0.8 1.0 −2 −1 0 1 2 η , θ α(η) H(η) ψ(θ) Figure 3: Truncated quadratic loss.
and so ˜ ψ(θ) = 1 − p 1 −θ2.
Since ˜ ψ is convex, ψ = ˜ ψ. The right panel of Figure 2 shows the graphs of α∗, H, and ψ over the interval [0, 1].
Finally, for 0 < η < 1, sign(α∗(η)) = sign(η −1/2) by inspection. Also, a sequence (αi) can achieve H at η = 0 (respectively, 1) only if it diverges to −∞(respectively, ∞). It therefore follows that exponential loss is classification-calibrated.
Example 2 (Truncated quadratic loss). Now consider φ(α) = [max{1 −α, 0}]2, as depicted together with φ(−α), C0.3(α), and C0.7(α) in the left panel of Figure 3. If η = 0, it is clear that any α ∈(−∞, −1] makes Cη(α) vanish. Similarly, any α ∈[1, ∞) makes the conditional φ-risk vanish when η = 1. On the other hand, when 0 < η < 1, Cη is strictly convex with a (unique) stationary point, and solving for it yields α∗(η) = 2η −1.
(2) Notice that, though α∗is in principle undefined at 0 and 1, we could choose to fix α∗(0) = −1 and α∗(1) = 1, which are valid settings. This would extend (2) to all of [0, 1].
As in Example 1, we may simplify the identity H(η) = Cη(α∗(η)) for 0 < η < 1 to obtain H(η) = 4η(1 −η), 10 −2 −1 0 1 2 0 1 2 3 4 5 6 7 α φ(α) φ(−α) C0.3(α) C0.7(α) 0.0 0.2 0.4 0.6 0.8 1.0 −2 −1 0 1 2 η , θ α(η) H(η) ψ(θ) Figure 4: Hinge loss.
which is also correct for η = 0 and 1, as noted. It is also immediate that H−((1+θ)/2) ≡φ(0) = 1, so we have ˜ ψ(θ) = θ2.
Again, ˜ ψ is convex, so ψ = ˜ ψ. The right panel of Figure 3 shows α∗, H, and ψ. Observe that truncated quadratic loss is classification-calibrated: the case 0 < η < 1 is obvious from (2); for η = 0 or 1, it follows because any (αi) achieving H at 0 (respectively, 1) must eventually take values in (−∞, −1] (respectively, [1, ∞)).
Example 3 (Hinge loss). Here we take φ(α) = max{1 −α, 0}, which is shown in the left panel of Figure 4 along with φ(−α), C0.3(α), and C0.7(α). By direct consideration of the piecewise-linear form of Cη(α), it is easy to see that for η = 0, each α ≤−1 makes Cη(α) vanish, just as in Example 2. The same holds for α ≥1 when η = 1. Now for η ∈(0, 1), we see that Cη decreases strictly on (−∞, −1] and increases strictly on [1, ∞). Thus any minima must lie in [−1, 1]. But Cη is linear on [−1, 1], so the minimum must be attained at 1 for η > 1/2, −1 for η < 1/2, and anywhere in [−1, 1] for η = 1/2. We have argued that α∗(η) = sign(η −1/2) (3) for all η ∈(0, 1) other than 1/2. Since (3) yields valid minima at 0, 1/2, and 1 also, we could choose to extend it to the whole unit interval. Regardless, a simple direct verification as in the previous examples shows H(η) = 2 min{η, 1 −η} 11 −2 −1 0 1 2 0 1 2 3 4 5 6 7 α φ(α) φ(−α) C0.3(α) C0.7(α) 0.0 0.2 0.4 0.6 0.8 1.0 −2 −1 0 1 2 η , θ H(η) ψ(θ) Figure 5: Sigmoid loss.
for 0 ≤η ≤1. Since H−((1 + θ)/2) ≡φ(0) = 1, we have ˜ ψ(θ) = θ, and ψ = ˜ ψ by convexity. We present α∗, H, and ψ in the right panel of Figure 4. To conclude, notice that the form of (3) and separate considerations for η ∈{0, 1}, as in Example 2, easily imply that hinge loss is classification-calibrated.
Example 4 (Sigmoid loss). We conclude by examining a non-convex loss function. Let φ(α) = 1 −tanh(kα) for some fixed k > 0. Figure 5, left panel, depicts φ(α) with k = 1, as well as φ(−α), C0.3(α), and C0.7(α). Using the fact that tanh is an odd function, we can rewrite the conditional φ-risk as Cη(α) = 1 + (1 −2η) tanh(kα).
(4) From this expression, two facts are clear. First, when η = 1/2, every α minimizes Cη(α), because it is identically 1. Second, when η ̸= 1/2, Cη(α) attains no minimum, because tanh has no maximal or minimal value on R. Hence α∗is not defined for any η.
Inspecting (4), for 0 ≤η < 1/2 we obtain H(η) = 2η by letting α →−∞. Analogously, when α →∞, we get H(η) = 2(1 −η) for 1/2 < η ≤1. Thus we have H(η) = 2 min{η, 1 −η}, 0 ≤η ≤1.
12 Since H−((1 + θ)/2) ≡φ(0) = 1, we have ˜ ψ(θ) = θ, and convexity once more gives ψ = ˜ ψ. We present H and ψ in the right panel of Figure 5. Finally, the foregoing considerations imply that sigmoid loss is classification-calibrated, provided we note carefully that the definition of classification-calibration requires nothing when η = 1/2.
2.4 Properties of ψ and proof of Theorem 3 The following elementary lemma will be useful throughout the paper.
Lemma 4. Suppose g : R →R is convex and g(0) = 0. Then 1. for all λ ∈[0, 1] and x ∈R, g(λx) ≤λg(x).
2. for all x > 0, 0 ≤y ≤x, g(y) ≤y xg(x).
3. g(x)/x is increasing on (0, ∞).
Proof. For 1, g(λx) = g(λx + (1 −λ)0) ≤λg(x) + (1 −λ)g(0) = λg(x). To see 2, put λ = y/x in 1.
For 3, rewrite 2 as g(y)/y ≤g(x)/x.
Lemma 5. The functions H, H−and ψ have the following properties: 1. H and H−are symmetric about 1/2: for all η ∈[0, 1], H(η) = H(1−η), H−(η) = H−(1−η).
2. H is concave and, for 0 ≤η ≤1, it satisfies H(η) ≤H 1 2 = H− 1 2 .
3. If φ is classification-calibrated, then H(η) < H(1/2) for all η ̸= 1/2.
4. H−is concave on [0, 1/2] and on [1/2, 1], and for 0 ≤η ≤1 it satisfies H−(η) ≥H(η).
5. H, H−and ˜ ψ are continuous on [0, 1].
6. ψ is continuous on [0, 1].
7. ψ is nonnegative and minimal at 0.
8. ψ(0) = 0.
9. The following statements are equivalent: (a) φ is classification-calibrated.
13 (b) ψ(θ) > 0 for all θ ∈(0, 1].
Before proving the lemma, we point out that there is no converse to part 3. To see this, let φ be classification-calibrated, and consider the loss function ˜ φ(α) = φ(−α), with corresponding ˜ H(η).
Since (αi) achieves H at η if and only if (−αi) achieves ˜ H at η, we see that ˜ φ is not classification-calibrated. However, ˜ H(η) = H(1 −η), so because part 3 holds for φ, it must also hold for ˜ φ.
Proof. 1 is immediate from the definitions.
For 2, concavity follows because H is an infimum of concave (affine) functions of η.
Now, since H is concave and symmetric about 1/2, H(1/2) = H((1/2)η + (1/2)(1 −η)) ≥(1/2)H(η) + (1/2)H(1 −η) = H(η). Thus H is maximal at 1/2. To see that H(1/2) = H−(1/2), notice that α(2η −1) ≤0 for all α when η = 1/2.
To prove 3, assume that there is an η ̸= 1/2 with H(η) = H(1/2). Fix a sequence α1, α2, . . .
that achieves H at 1/2. By the assumption, lim inf i→∞(ηφ(αi) + (1 −η)φ(−αi)) ≥H(η) = H(1/2) = lim i→∞ φ(αi) + φ(−αi) 2 , (5) Rearranging, we have (η −1/2) lim inf i→∞(φ(αi) −φ(−αi)) ≥0.
Since H(1 −η) = H(η), the same argument shows that H(η) = H(1/2) implies (η −1/2) lim inf i→∞(φ(−αi) −φ(αi)) ≥0.
It follows that lim i→∞(φ(αi) −φ(−αi)) = 0, so all the expressions in (5) are equal. Hence, H is achieved by (αi) at η, and if φ is classification-calibrated we must have that lim inf i→∞(sign(αi(η −1/2)) = 1.
The same argument shows that H is achieved by (αi) at 1 −η, and if φ is classification-calibrated we must have that lim sup i→∞ (sign(αi(η −1/2)) = −1.
Thus, if H(η) = H(1/2), φ is not classification-calibrated.
For 4, H−is concave on [0, 1/2] by the same argument as for the concavity of H. (Notice that when η < 1/2, H−is an infimum over a set of concave functions, but in this case when η > 1/2, it is an infimum over a different set of concave functions.) The inequality H−≥H follows from the definitions.
For 5, first notice that the concavity of H implies that it is continuous on the relative interior of its domain, i.e. (0, 1). Thus, to show that H is continuous [0, 1], it suffices (by symmetry) to show that it is left continuous at 1. Because [0, 1] is locally simplicial in the sense of Rockafellar (1997), his Theorem 10.2 gives lower semicontinuity of H at 1 (equivalently, upper semicontinuity of the convex function −H at 1). To see upper semicontinuity of H at 1, on the other hand, fix 14 any ϵ > 0 and choose αϵ such that φ(αϵ) ≤H(1) + ϵ/2. Then for any η between 1 −ϵ/(2φ(−αϵ)) and 1 we have H(η) ≤Cη(αϵ) ≤H(1) + ϵ.
Since this is true for any ϵ, lim supη→1 H(η) ≤H(1), which is upper semicontinuity. Thus H is left continuous at 1. The same argument shows that H−is continuous on (0, 1/2) and (1/2, 1), and left continuous at 1/2 and 1. Symmetry implies that H−is continuous on the closed interval [0, 1].
The continuity of ˜ ψ is now immediate.
To see 6, observe that ψ is a closed convex function with locally simplicial domain [0, 1], so its continuity follows by once again applying Theorem 10.2 of Rockafellar (1997).
It follows immediately from 2 and 4 that ˜ ψ is nonnegative and minimal at 0. Since epi ψ is the convex hull of epi ˜ ψ, i.e., the set of all convex combinations of points in epi ˜ ψ, we see that ψ is also nonnegative and minimal at 0, which is 7.
8 follows immediately from 2.
To prove 9, suppose first that φ is classification-calibrated. Then for all θ ∈(0, 1], ˜ ψ(θ) > 0.
But every point in epi ψ is a convex combination of points in epi ˜ ψ, so if (θ, 0) ∈epi ψ, we can only have θ = 0. Hence for θ ∈(0, 1], points in epi ψ of the form (θ, c) must have c > 0, and closure of ˜ ψ now implies ψ(θ) > 0. For the converse, notice that if φ is not classification-calibrated, then some θ > 0 has ˜ ψ(θ) = 0, and so ψ(θ) = 0.
Proof. (Of Theorem 3). For Part 1, it is straightforward to show that R(f) −R∗= R(f) −R(η −1/2) = E (1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) , where 1 [Φ] is 1 if the predicate Φ is true and 0 otherwise (see, for example, Devroye et al., 1996).
We can apply Jensen’s inequality, since ψ is convex by definition, and the fact that ψ(0) = 0 (Lemma 5, part 8) to show that ψ(R(f) −R∗) ≤Eψ (1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) = E (1 [sign(f(X)) ̸= sign(η(X) −1/2)] ψ (|2η(X) −1|)) .
Now, from the definition of ψ we know that ψ(θ) ≤˜ ψ(θ), so we have ψ(R(f) −R∗) ≤E 1 [sign(f(X)) ̸= sign(η(X) −1/2)] ˜ ψ (|2η(X) −1|) = E 1 [sign(f(X)) ̸= sign(η(X) −1/2)] H−(η(X)) −H(η(X)) = E 1 [sign(f(X)) ̸= sign(η(X) −1/2)] inf α:α(2η(X)−1)≤0 Cη(X)(α) −H(η(X)) ≤E Cη(X)(f(X)) −H(η(X)) = Rφ(f) −R∗ φ, where we have used the fact that for any x, and in particular when sign(f(x)) = sign(η(x) −1/2), we have Cη(x)(f(x)) ≥H(η(x)).
For Part 2, the first inequality is from Part 1.
For the second, fix ϵ > 0 and θ ∈[0, 1].
From the definition of ψ, we can choose γ, α1, α2 ∈[0, 1] for which θ = γα1 + (1 −γ)α2 and 15 ψ(θ) ≥γ ˜ ψ(α1) + (1 −γ) ˜ ψ(α2) −ϵ/2.
Choose distinct x1, x2 ∈X, and choose PX such that PX{x1} = γ, PX{x2} = 1 −γ, η(x1) = (1 + α1)/2, and η(x2) = (1 + α2)/2. From the definition of H−, we can choose f : X →R such that f(x1) ≤0, f(x2) ≤0, Cη(x1)(f(x1)) ≤H−(η(x1)) + ϵ/2 and Cη(x2)(f(x2)) ≤H−(η(x2)) + ϵ/2. Then we have Rφ(f) −R∗ φ = Eφ(Y f(X)) −inf g Eφ(Y g(X)) = γ Cη(x1)(f(x1)) −H(η(x1)) + (1 −γ) Cη(x2)(f(x2)) −H(η(x2)) ≤γ H−(η(x1)) −H(η(x1)) + (1 −γ) H−(η(x2)) −H(η(x2)) + ϵ/2 = γ ˜ ψ(α1) + (1 −γ) ˜ ψ(α2) + ϵ/2 ≤ψ(θ) + ϵ.
Furthermore, since sign(f(x1)) = sign(f(x2)) = −1 but η(x1), η(x2) ≥1/2, R(f) −R∗= E|2η(X) −1| = γ(2η(x1) −1) + (1 −γ)(2η(x2) −1) = θ.
For Part 3, first note that, for any φ, ψ is continuous on [0, 1] and ψ(0) = 0 by Lemma 5, parts 6, 8, and hence θi →0 implies ψ(θi) →0. Thus, we can replace condition (3b) by (3b’) For any sequence (θi) in [0, 1], ψ(θi) →0 implies θi →0.
To see that (3a) implies (3b’), let φ be classification-calibrated, and let (θi) be a sequence that does not converge to 0. Define c = lim sup θi > 0, and pass to a subsequence with lim θi = c. Then lim ψ(θi) = ψ(c) by continuity, and ψ(c) > 0 by classification-calibration (Lemma 5, part 9). Thus, for the original sequence (θi), we see lim sup ψ(θi) > 0, so we cannot have ψ(θi) →0.
To see that (3b’) implies (3c), suppose that Rφ(fi) →R∗ φ. By Part 1, ψ(R(fi) −R∗) →0, and (3b’) implies R(fi) →R∗.
Finally, to see that (3c) implies (3a), suppose that φ is not classification-calibrated and fix some η ̸= 1/2.
We can find a sequence α1, α2, . . . such that (αi) achieves H at η but has lim infi→∞sign(αi(η −1/2)) ̸= 1.
Replace the sequence with a subsequence that also achieves H at η but has lim sign(αi(η −1/2)) = −1. Fix x ∈X and choose the probability distribution P so that PX{x} = 1 and P(Y = 1|X = x) = η. Define a sequence of functions fi : X →R for which fi(x) = αi. Then lim R(fi) > R∗, and this is true for any infinite subsequence. But since αi achieves H at η, lim Rφ(fi) = R∗ φ.
3 Further analysis of conditions on φ In this section we consider additional conditions on the loss function φ. In particular, we study the role of convexity.
16 3.1 Convex loss functions For convex φ, classification-calibration is equivalent to a condition on the derivative of φ at zero.
Recall that a subgradient of φ at α ∈R is any value mα ∈R such that φ(x) ≥φ(α) + mα(x −α) for all x.
Theorem 6. Let φ be convex. Then φ is classification-calibrated if and only if it is differentiable at 0 and φ′(0) < 0.
Proof. Fix a convex function φ.
(= ⇒) Since φ is convex, we can find subgradients g1 ≥g2 such that, for all α, φ(α) ≥g1α + φ(0) φ(α) ≥g2α + φ(0).
Then we have ηφ(α) + (1 −η)φ(−α) ≥η(g1α + φ(0)) + (1 −η)(−g2α + φ(0)) = (ηg1 −(1 −η)g2)α + φ(0) (6) = 1 2(g1 −g2) + (g1 + g2) η −1 2 α + φ(0).
(7) Since φ is classification-calibrated, for η > 1/2 we can express H(η) as infα>0 ηφ(α)+(1−η)φ(−α).
If (7) were greater than φ(0) for every α > 0, it would then follow that for η > 1/2, H(η) ≥φ(0) ≥ H(1/2), which, by Lemma 5, part 3, is a contradiction. We now show that g1 > g2 implies this contradiction. Indeed, we can choose 1 2 < η < 1 2 + g1 −g2 2|g1 + g2| to show that |(η −1/2)(g1 + g2)| < (g1 −g2)/2, so (7) is greater than φ(0) for all α > 0. Thus, if φ is classification-calibrated, we must have g1 = g2, which implies φ is differentiable at 0.
To see that we must also have φ′(0) < 0, notice that, from (6), we have ηφ(α) + (1 −η)φ(−α) ≥(2η −1)φ′(0)α + φ(0).
But for any η > 1/2 and α > 0, if φ′(0) ≥0, this expression is at least φ(0).
Thus, if φ is classification-calibrated, we must have φ′(0) < 0.
(⇐ =) Suppose that φ is differentiable at 0 and has φ′(0) < 0. Then the function Cη(α) = ηφ(α) + (1 −η)φ(−α) has C′ η(0) = (2η −1)φ′(0). For η > 1/2, this is negative. It follows from the convexity of φ that Cη(α) is minimized by some α∗∈(0, ∞]. To see this, notice that for some α0 > 0, we have Cη(α0) ≤Cη(0) + α0C′ η(0)/2.
But the convexity of φ, and hence of Cη, implies that for all α, Cη(α) ≥Cη(0) + αC′ η(0).
In particular, if α ≤α0/4, Cη(α) ≥Cη(0) + α0 4 C′ η(0) > Cη(0) + α0 2 C′ η(0) ≥Cη(α0).
Similarly, for η < 1/2, the optimal α is negative. This means that φ is classification-calibrated.
17 The next lemma shows that for convex φ, the ψ transform is a little easier to compute.
Lemma 7. If φ is convex and classification-calibrated, then ˜ ψ is convex, hence ψ = ˜ ψ.
Proof. Theorem 6 tells us φ is differentiable at zero and φ′(0) < 0. Hence we have φ(0) ≥H−(η) = inf α:α(η−1/2)≤0 (ηφ(α) + (1 −η)φ(−α)) ≥ inf α:α(η−1/2)≤0 η(φ(0) + φ′(0)α) + (1 −η)(φ(0) −φ′(0)α) = φ(0) + inf α:α(η−1/2)≤0 (2η −1)φ′(0)α = φ(0).
Thus, H−(η) = φ(0). The concavity of H (Lemma 5, part 2) implies ˜ ψ = H−(η) −H(η) is convex, which gives the result.
If φ is convex and classification-calibrated, then it is differentiable at zero, and we can define the Bregman divergence of φ at 0: dφ(0, α) = φ(α) −(φ(0) + αφ′(0)).
We consider a symmetrized, normalized version of the Bregman divergence at 0, for α ≥0: ξ(α) = dφ(0, α) + dφ(0, −α) −φ′(0)α .
Since φ is convex on R, both φ and ξ are continuous, so we can define ξ−1(θ) = inf {α : ξ(α) = θ} .
Lemma 8. For convex, classification-calibrated φ, ψ(θ) ≥−φ′(0)θ 2ξ−1 θ 2 .
18 Proof. From convexity of φ, we have ψ(θ) = H 1 2 −H 1 + θ 2 = φ(0) −inf α>0 1 + θ 2 φ(α) + 1 −θ 2 φ(−α) = sup α>0 −θφ′(0)α + 1 + θ 2 φ(0) −φ(α) + αφ′(0) + 1 −θ 2 φ(0) −φ(−α) −αφ′(0) = sup α>0 −θφ′(0)α −1 + θ 2 dφ(0, α) −1 −θ 2 dφ(0, −α) ≥sup α>0 −θφ′(0)α −dφ(0, α) −dφ(0, −α) = sup α>0 (θ −ξ(α)) (−φ′(0)α) ≥ θ −ξ(ξ−1(θ/2)) (−φ′(0)ξ−1(θ/2)) = −φ′(0)θ 2ξ−1 θ 2 .
Notice that a slower increase of ξ (that is, a less curved φ) gives better bounds on R(f) −R∗ in terms of Rφ(f) −R∗ φ.
3.2 General loss functions All of the classification procedures mentioned in earlier sections utilize surrogate loss functions which are either upper bounds on 0-1 loss or can be transformed into upper bounds via a positive scaling factor. This is not a coincidence: as the next lemma establishes, it must be possible to scale any classification-calibrated φ into such a majorant.
Lemma 9. If φ : R →[0, ∞) is classification-calibrated, then there is a γ > 0 such that γφ(α) ≥ 1 [α ≤0] for all α ∈R.
Proof. Proceeding by contrapositive, suppose no such γ exists. Since φ(α) ≥1 [α ≤0] on (0, ∞), we must then have infα≤0 φ(α) = 0. But φ(α) = C1(α), hence 0 = inf α≤0 C1(α) = H−(1) ≥H(1) ≥0.
Thus, H−(1) = H(1), so φ is not classification-calibrated.
We have seen that for convex φ, the function ˜ ψ is convex, and so ψ = ˜ ψ. The following example shows that we cannot, in general, avoid computing the convex lower bound ψ.
19 Example 5. Consider the following (classification-calibrated) loss function; see the left panel of Figure 6.
φ(α) = 4 if α ≤0, α ̸= −1, 3 if α = −1, 2 if α = 1, 0 if α > 0, α ̸= 1.
Then ˜ ψ is not convex, so ψ ̸= ˜ ψ.
Proof. It is easy to check that H−(η) = min{4η, 2 + η} if η ≥1/2, min{4(1 −η), 3 −η} if η < 1/2, and that H(η) = 4 min{η, 1 −η}. Thus, H−(η) −H(η) = min{8η −4, 5η −2} if η ≥1/2 min{4 −8η, 3 −5η} if η < 1/2, so ˜ ψ(θ) = min 4θ, 1 2(5θ + 1) .
This function, illustrated in the right panel of Figure 6, is not convex; in fact it is concave.
4 Tighter bounds under low noise conditions In a study of the convergence rate of empirical risk minimization, Tsybakov (2001) provided a useful condition on the behavior of the posterior probability near the optimal decision boundary {x : η(x) = 1/2}. Tsybakov’s condition is useful in our setting as well; as we show in this section, it allows us to obtain a refinement of Theorem 3.
Recall that R(f) −R∗= E (1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) ≤PX (sign(f(X)) ̸= sign(η(X) −1/2)) , (8) with equality provided that η(X) is almost surely either 1 or 0. We say that P has noise exponent α ≥0 if there is a c > 0 such that every measurable f : X →R has PX (sign(f(X)) ̸= sign(η(X) −1/2)) ≤c (R(f) −R∗)α .
(9) Notice that we must have α ≤1, in view of (8). If α = 0, this imposes no constraint on the noise: take c = 1 to see that every probability measure P satisfies (9). On the other hand, α = 1 if and only if |2η(X) −1| ≥1/c a.s. [PX]. The reverse implication is immediate; to see the forward implication, notice that the condition must apply for every measurable f. For α = 1 it requires that (∀A ∈G) P(A) ≤c Z A |2η(X) −1| dPX ⇐ ⇒(∀A ∈G) Z A 1 c dPX ≤ Z A |2η(X) −1| dPX ⇐ ⇒1 c ≤|2η(X) −1| a.s. [PX].
20 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 0 1 2 3 4 α φ(α) 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 θ ψ ~(θ) Figure 6: Left panel, the loss function of Example 5. Right panel, the corresponding (nonconvex) ˜ ψ.
The dotted lines depict the graphs for the two linear functions of which ˜ ψ is a pointwise minimum.
Theorem 10. Suppose P has noise exponent 0 < α ≤1, and φ is classification-calibrated and error-averse. Then there is a c > 0 such that for any f : X →R, c (R(f) −R∗)α ψ (R(f) −R∗)1−α 2c !
≤Rφ(f) −R∗ φ.
Furthermore, this never gives a worse rate than the result of Theorem 3, since (R(f) −R∗)α ψ (R(f) −R∗)1−α 2c !
≥ψ R(f) −R∗ 2c .
Proof. Fix c > 0 such that for every f : X →R, PX (sign(f(X)) ̸= sign(η(X) −1/2)) ≤c (R(f) −R∗)α .
We approximate the error integral separately over a region with high noise, and over the remainder 21 of the input space. To this end, fix ϵ > 0 (the noise threshold), and notice that R(f) −R∗= E (1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) = E (1 [|2η(X) −1| < ϵ] 1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) + E (1 [|2η(X) −1| ≥ϵ] 1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) ≤cϵ (R(f) −R∗)α + E (1 [|2η(X) −1| ≥ϵ] 1 [sign(f(X)) ̸= sign(η(X) −1/2)] |2η(X) −1|) .
Now, for any x, 1 [|2η(x) −1| ≥ϵ] |2η(x) −1| ≤ ϵ ψ(ϵ)ψ(|2η(x) −1|).
(10) Indeed, when |2η(x) −1| < ϵ, (10) follows from the fact that ψ is nonnegative (Lemma 5, parts 8,9), and when |2η(x) −1| ≥ϵ it follows from Lemma 4(2).
Thus, using the same argument as in the proof of Theorem 3, R(f) −R∗≤cϵ (R(f) −R∗)α + ϵ ψ(ϵ)E (1 [sign(f(X)) ̸= sign(η(X) −1/2)] ψ (|2η(X) −1|)) ≤cϵ (R(f) −R∗)α + ϵ ψ(ϵ) Rφ(f) −R∗ φ , and hence, R(f) −R∗ ϵ −c (R(f) −R∗)α ψ(ϵ) ≤Rφ(f) −R∗ φ.
Choosing ϵ = 1 2c (R(f) −R∗)1−α and substituting gives the first inequality. (We can assume that R(f)−R∗> 0, since the inequality is trivial otherwise.) The second inequality follows from the fact that ψ(θ)/θ is non-decreasing, which we know from Lemma 4, part 3.
5 Estimation rates In previous sections, we have seen that the excess risk, R(f) −R∗, can be bounded in terms of the excess φ-risk, Rφ(f) −R∗ φ. Many large margin algorithms choose ˆ f to minimize the empirical φ-risk, ˆ Rφ(f) = ˆ Eφ(Y f(X)) = 1 n n X i=1 φ(Yif(Xi)).
In this section, we examine the convergence of ˆ f’s excess φ-risk, Rφ( ˆ f) −R∗ φ. We can split this excess risk into an estimation error term and an approximation error term: Rφ( ˆ f) −R∗ φ = Rφ( ˆ f) −inf f∈F Rφ(f) + inf f∈F Rφ(f) −R∗ φ .
22 We focus on the first term, the estimation error term. We assume throughout that some f ∗∈F achieves the infimum, Rφ(f ∗) = inf f∈F Rφ(f).
The simplest way to bound Rφ( ˆ f) −Rφ(f ∗) is to use a uniform convergence argument: if sup f∈F ˆ Rφ(f) −Rφ(f) ≤ϵn, (11) then Rφ( ˆ f) −Rφ(f ∗) = Rφ( ˆ f) −ˆ Rφ( ˆ f) + ˆ Rφ( ˆ f) −ˆ Rφ(f ∗) + ˆ Rφ(f ∗) −Rφ(f ∗) ≤2ϵn + ˆ Rφ( ˆ f) −ˆ Rφ(f ∗) ≤2ϵn, since ˆ f minimizes ˆ Rφ.
This approach can give the wrong rate. For example, for a nontrivial class F, the expectation of the empirical process in (11) can decrease no faster than 1/√n. However, if F is a small class (for instance, a VC-class) and Rφ(f ∗) = 0, then Rφ( ˆ f) should decrease as 1/n.
Lee et al. (1996) showed that fast rates are also possible for the quadratic loss φ(α) = (1−α)2 if F is convex, even if Rφ(f ∗) > 0. In particular, because the quadratic loss function is strictly convex, it is possible to bound the variance of the excess loss (difference between the loss of a function f and that of the optimal f ∗) in terms of its expectation. Since the variance decreases as we approach the optimal f ∗, the risk of the empirical minimizer converges more quickly to the optimal risk than the simple uniform convergence results would suggest. Mendelson (2002) improved this result, and extended it from prediction in L2(PX) to prediction in Lp(PX) for other values of p. The proof used the idea of the modulus of convexity of a norm. In this section, we use this idea to give a simpler proof of a more general bound when the loss function satisfies a strict convexity condition, and we obtain risk bounds. The modulus of convexity of an arbitrary strictly convex function (rather than a norm) is a key notion in formulating our results.
Definition 11 (Modulus of convexity). Given a pseudometric d defined on a vector space S, and a convex function f : S →R, the modulus of convexity of f with respect to d is the function δ : [0, ∞) →[0, ∞] satisfying δ(ϵ) = inf f(x1) + f(x2) 2 −f x1 + x2 2 : x1, x2 ∈S, d(x1, x2) ≥ϵ .
If δ(ϵ) > 0 for all ϵ > 0, we say that f is strictly convex with respect to d.
We consider loss functions φ that also satisfy a Lipschitz condition with respect to a pseudo-metric d on R: we say that φ : R →R is Lipschitz with respect to d, with constant L, if for all a, b ∈R, |φ(a) −φ(b)| ≤L · d(a, b).
(Note that if d is a metric and φ is convex, then φ necessarily satisfies a Lipshitz condition on any compact subset of R (Rockafellar, 1997).) 23 In the following theorem, we use the expectation of a centered empirical process as a measure of the complexity of the class F; define ξF(ϵ) = E sup n Ef −ˆ Ef : f ∈F, Ef = ϵ o .
Define the excess loss class gF as gF = {gf : f ∈F} = {(x, y) 7→φ(yf(x)) −φ(yf ∗(x)) : f ∈F} , where f ∗= arg minf∈F Eφ(Y f(X)).
Theorem 12. There is a constant K for which the following holds.
For a pseudometric d on R, suppose that φ : R →R is Lipschitz with constant L and convex with modulus of convexity δ(ϵ) ≥cϵr (both with respect to d). Define β = min(1, 2/r). Fix a convex class F of real functions on X such that for all f ∈F, x1, x2 ∈X, and y1, y2 ∈Y, d(y1f(x1), y2f(x2)) ≤B. For i.i.d. data (X1, Y1), . . . , (Xn, Yn), let ˆ f ∈F be the minimizer of the empirical φ-risk, Rφ(f) = ˆ Eφ(Y f(X)).
Then with probability at least 1 −e−x, Rφ( ˆ f) ≤Rφ(f ∗) + ϵ, where ϵ = K max ( ϵ∗, crL2x n 1/(2−β) , BLx n ) , ϵ∗≥ξgF(ϵ∗), cr = (2c)−2/r if r ≥2, (2c)−1B2−r otherwise.
Thus, for any probability distribution P on X × Y that has noise exponent α, there is a constant c′ such that, with probability at least 1 −e−x, c′ R( ˆ f) −R∗α ψ R( ˆ f) −R∗1−α 2c′ ≤ϵ + inf f∈F Rφ(f) −R∗ φ.
5.1 Proof of Theorem 12 There are two key ingredients in the proof. Firstly, the following result shows that if the variance of an excess loss function is bounded in terms of its expectation, then we can obtain faster rates than would be implied by the uniform convergence bounds. Secondly, simple conditions on the loss function ensure that this condition is satisfied for convex function classes.
Lemma 13. Consider a class F of functions f : X →R with supf∈F ∥f∥∞≤B. Let P be a probability distribution on X, and suppose that there are c ≥1 and 0 < β ≤1 such that, for all f ∈F, Ef 2(X) ≤c(Ef)β.
(12) 24 Fix 0 < α, ϵ < 1. Suppose that if some f ∈F has ˆ Ef ≤αϵ and Ef ≥ϵ, then some f ′ ∈F has ˆ Ef ′ ≤αϵ and Ef = ϵ. Then with probability at least 1 −e−x, any f ∈F satisfies ˆ Ef ≤αϵ ⇒Ef ≤ϵ.
provided that ϵ ≥max ( ϵ∗, 9cKx (1 −α)2n 1/(2−β) , 4KBx (1 −α)n ) .
where K is an absolute constant and ϵ∗≥ 6 1 −αξF(ϵ∗).
As an aside, notice that Tsybakov’s condition Tsybakov (2001) is of the form (12). To see this, let f ∗be the Bayes decision rule, and consider the class of functions {αgf : f ∈F, α ∈[0, 1]}, where gf(x, y) = ℓ(f(x), y) −ℓ(f ∗(x), y) and ℓis the discrete loss. Then the condition PX (f(X) ̸= f ∗(X)) ≤c (Eℓ(f(X), Y ) −Eℓ(f ∗(X), Y ))α can be rewritten Eg2 f(X, Y ) ≤c(Egf(X, Y ))α.
Thus, we can obtain a version of Tsybakov’s result for small function classes from Lemma 13: if the Bayes decision rule f ∗is in F, then the function ˆ f that minimizes empirical risk has ˆ Eg ˆ f = ˆ R(f) −ˆ R(f ∗) ≤0, and so with high probability has Eg ˆ f = R(f) −R∗≤ϵ under the conditions of the theorem. If F is a VC-class, we have ϵ ≤c log n/n for some constant c, which is surprisingly fast when R∗> 0.
The proof of Lemma 13 uses techniques from Massart (2000b), Mendelson (2002), and Bartlett et al. (2003), as well as the following concentration inequality, which is a refinement, due to Rio (2001) and Klein (2002) of a result of Massart (2000a), following Talagrand (1994), Ledoux (2001).
The best estimates on the constants are due to Bousquet (2002).
Lemma 14. There is an absolute constant K for which the following holds. Let G be a class of functions defined on X with supg∈G ∥g∥∞≤b. Suppose that P is a probability distribution such that for every g ∈G, Eg = 0. Let X1, ..., Xn be independent random variables distributed according to P and set σ2 = supg∈G var g. Define Z = sup g∈G 1 n n X i=1 g(Xi).
Then, for every x > 0 and every ρ > 0, Pr ( Z ≥(1 + ρ)EZ + σ r Kx n + K(1 + ρ−1)bx n ) ≤e−x.
25 Proof. (of Lemma 13) From the condition on F, we have Pr n ∃f ∈F : ˆ Ef ≤αϵ, Ef ≥ϵ o ≤Pr n ∃f ∈F : ˆ Ef ≤αϵ, Ef = ϵ o = Pr n sup n Ef −ˆ Ef : f ∈F, Ef = ϵ o ≥(1 −α)ϵ o .
We bound this probability using Lemma 14, with ρ = 1 and G = {Ef −f : f ∈F, Ef = ϵ}. This shows that Pr n ∃f ∈F : ˆ Ef ≤αϵ, Ef ≥ϵ o ≤Pr {Z ≥(1 −α)ϵ} ≤e−x, provided that 2EZ ≤(1 −α)ϵ 3 , r cϵβKx n ≤(1 −α)ϵ 3 , and 4KBx n ≤(1 −α)ϵ 3 .
(We have used the fact that supf∈F ∥f∥∞≤B implies supg∈G ∥g∥∞≤2B.) Observing that EZ = ξF(ϵ), and rearranging gives the result.
The second ingredient in the proof of Theorem 12 is the following lemma, which gives conditions that ensure a variance bound of the kind required for the previous lemma (condition (12)). For a pseudometric d on R and a probability distribution on X, we can define a pseudometric ˜ d on the set of uniformly bounded real functions on X, ˜ d(f, g) = Ed(f(X), g(X))21/2 .
If d is the usual metric on R, then ˜ d is the L2(P) pseudometric.
Lemma 15. Consider a convex class F of real-valued functions defined on X, a convex loss function ℓ: R →R, and a pseudometric d on R. Suppose that ℓsatisfies the following conditions.
1. ℓis Lipschitz with respect to d, with constant L: for all a, b ∈R, |ℓ(a) −ℓ(b)| ≤Ld(a, b).
2. R(f) = Eℓ(f) is a strictly convex functional with respect to the pseudometric ˜ d, with modulus of convexity ˜ δ: ˜ δ(ϵ) = inf R(f) + R(g) 2 −R f + g 2 : ˜ d(f, g) ≥ϵ .
26 Suppose that f ∗satisfies R(f ∗) = inff∈F R(f), and define gf(x) = ℓ(f(x)) −ℓ(f ∗(x)).
Then Egf ≥2˜ δ ˜ d(f, f ∗) ≥2˜ δ q Eg2 f L .
We shall apply the lemma to a class of functions of the form (x, y) 7→yf(x), with the loss function ℓ= φ. (The lemma can be trivially extended to a loss function ℓ: R×Y →R that satisfies a Lipschitz constraint uniformly over Y.) Proof. The proof proceeds in two steps: the Lipschitz condition allows us to relate Eg2 f to ˜ d(f, f ∗), and the modulus of convexity condition, together with the convexity of F, relates this to Egf.
We have Eg2 f = E (ℓ(f(X)) −ℓ(f ∗(X)))2 ≤E (Ld(f(X), f ∗(X)))2 = L2 ˜ d(f, f ∗) 2 .
(13) From the definition of the modulus of convexity, R(f) + R(f ∗) 2 ≥R f + f ∗ 2 + ˜ δ( ˜ d(f, f ∗)) ≥R(f ∗) + ˜ δ( ˜ d(f, f ∗)), where the optimality of f ∗in the convex set F implies the second inequality. Rearranging gives Egf = R(f) −R(f ∗) ≥2˜ δ( ˜ d(f, f ∗)).
Combining with (13) gives the result.
In our application, the following result will imply that we can estimate the modulus of convexity of Rφ with respect to the pseudometric ˜ d if we have some information about the modulus of convexity of φ with respect to the pseudometric d.
Lemma 16. Suppose that a convex function ℓ: R →R has modulus of convexity δ with respect to a pseudometric d on R, for some fixed c, r > 0, every ϵ > 0 satisfies δ(ϵ) ≥cϵr.
Then for functions f : X →R satisfying supx1,x2 d(f(x1), f(x2)) = B, the modulus of convexity ˜ δ of R(f) = Eℓ(f) with respect to the pseudometric ˜ d satisfies ˜ δ(ϵ) ≥crϵmax{2,r}, where cr = c if r ≥2 and cr = cBr−2 otherwise.
27 Proof. Fix functions f1, f2 : X →R with ˜ d(f1, f2) = p Ed2(f1(X), f2(X)) ≥ϵ. We have R(f1) + R(f2) 2 −R f1 + f2 2 = E ℓ(f1(X)) + ℓ(f2(X)) 2 −ℓ f1(X) + f2(X) 2 ≥E (δ(d(f1(X), f2(X)))) ≥cEdr(f1(X), f2(X)) = cE d2(f1(X), f2(X)) r/2 .
When the function ξ(a) = ar/2 is convex (i.e., when r ≥2), Jensen’s inequality shows that R(f1) + R(f2) 2 −R f1 + f2 2 ≥cϵr.
Otherwise, we use the following convex lower bound on ξ : [0, B2] →[0, Br], ξ(a) = ar/2 ≥Br a B2 , which follows from (the concave analog of) Lemma 4, part 2. This implies R(f1) + R(f2) 2 −R f1 + f2 2 ≥cBr−2ϵ2.
It is also possible to prove a converse result, that the modulus of convexity of φ is at least the infimum over probability distributions of the modulus of convexity of R. (To see this, we choose a probability distribution concentrated on the x ∈X where f1(x) and f2(x) achieve the infimum in the definition of the modulus of convexity.) Proof. (of Theorem 12) Consider the class {gf : f ∈F} with, for each f ∈F, gf(x, y) = φ(yf(x)) −φ(yf ∗(x)), where f ∗∈F minimizes Rφ(f) = Eφ(Y f(X)). Applying Lemma 16, we see that the functional R(f) = Eφ(f), defined for functions (x, y) 7→yf(x), has modulus of convexity ˜ δ(ϵ) ≥crϵmax{2,r}, where cr = c if r ≥2 and cr = cBr−2 otherwise. From Lemma 15, Egf ≥2cr q Eg2 f L max{2,r} , which is equivalent to Eg2 f ≤c′ rL2 (Egf)min{1,2/r} with c′ r = (2c)−2/r if r ≥2 (2c)−1B2−r otherwise 28 To apply Lemma 13 to the class {gf : f ∈F}, we need to check the condition. Suppose that gf has ˆ Egf ≤αϵ and Egf ≥ϵ.
Then, by the convexity of F and the continuity of φ, some f ′ = γf + (1 −γ)f ∗∈F, for 0 ≤γ ≤1, has Egf = ϵ. Jensen’s inequality shows that ˆ Egf = ˆ Eφ(Y (γf(X) + (1 −γ)f ∗(X))) −ˆ Eφ(Y f ∗(X)) ≤γ ˆ Eφ(Y f(x)) −ˆ Eφ(Y f ∗(X)) ≤αϵ.
Applying Lemma 13 we have, with probability at least 1 −e−x, any gf with ˆ Egf ≤ϵ/2 also has Egf ≤ϵ, provided ϵ ≥max ( ϵ∗, 36c′ rL2Kx n 1/(2−min{1,2/r}) , 16KBLx n ) , where ϵ∗≥12ξgF (ϵ∗). In particular, if ˆ f ∈F minimizes empirical risk, then ˆ Eg ˆ f = ˆ Rφ( ˆ f) −ˆ Rφ(f ∗) ≤0 < ϵ 2, hence Eg ˆ f ≤ϵ.
Combining with Theorem 10 shows that, for some c′, c′ R( ˆ f) −R∗α ψ R( ˆ f) −R∗1−α 2c′ ≤Rφ( ˆ f) −R∗ φ = Rφ( ˆ f) −Rφ(f ∗) + Rφ(f ∗) −R∗ φ ≤ϵ + Rφ(f ∗) −R∗ φ.
5.2 Examples We consider four loss functions that satisfy the requirements for the fast convergence rates: the exponential loss function used in AdaBoost, the deviance function corresponding to logistic regres-sion, the quadratic loss function, and the truncated quadratic loss function; see Table 1. These functions are illustrated in Figures 1 and 3. We use the pseudometric dφ(a, b) = inf {|a −α| + |β −b| : φ constant on (min{α, β}, max{α, β})} .
For all except the truncated quadratic loss function, this corresponds to the standard metric on R, dφ(a, b) = |a −b|. In all cases, dφ(a, b) ≤|a −b|, but for the truncated quadratic, dφ ignores differences to the right of 1. It is easy to calculate the Lipschitz constant and modulus of convexity for each of these loss functions. These parameters are given in Table 1.
In the following result, we consider the function class used by algorithms such as AdaBoost: the class of linear combinations of classifiers from a fixed base class. We assume that this base class has finite Vapnik-Chervonenkis dimension, and we constrain the size of the class by restricting the ℓ1 norm of the linear parameters. If G is the VC-class, we write F = B absconv(G), for some constant B, where B absconv(G) = ( m X i=1 αigi : m ∈N, αi ∈R, gi ∈G, ∥α∥1 = B ) .
29 φ(α) LB δ(ϵ) exponential e−α eB e−Bϵ2/8 logistic ln(1 + e−2α) 2 e−2Bϵ2/4 quadratic (1 −α)2 2(B + 1) ϵ2/4 truncated quadratic (max{0, 1 −α})2 2(B + 1) ϵ2/4 Table 1: Four convex loss functions defined on R. On the interval [−B, B], each has the indicated Lipschitz constant LB and modulus of convexity δ(ϵ) with respect to dφ. All have a quadratic modulus of convexity.
Theorem 17. Let φ : R →R be a convex loss function. Suppose that, on the interval [−B, B], φ is Lipschitz with constant LB and has modulus of convexity δ(ϵ) = aBϵ2 (both with respect to the pseudometric d).
For any probability distribution P on X × Y that has noise exponent α, there is a constant c′ for which the following is true. For i.i.d. data (X1, Y1), . . . , (Xn, Yn), let ˆ f ∈F be the minimizer of the empirical φ-risk, Rφ(f) = ˆ Eφ(Y f(X)). Suppose that F = B absconv(G), where G ⊆{±1}X has dV C(G) = d, and ϵ∗≥BLB max (LBaB B 1/(d+1) , 1 ) n−(d+2)/(2d+2) Then with probability at least 1 −e−x, R( ˆ f) ≤R∗+ c′ ϵ∗+ LB(LB/aB + B)x n + inf f∈F Rφ(f) −R∗ φ .
Proof. It is clear that F is convex and satisfies the conditions of Theorem 12. That theorem implies that, with probability at least 1 −e−x, R( ˆ f) ≤R∗+ c′ ϵ + inf f∈F Rφ(f) −R∗ φ , provided that ϵ ≥K max ϵ∗, L2 Bx 2aBn, BLBx n , where ϵ∗≥ξgF(ϵ∗). It remains to prove suitable upper bounds for ϵ∗.
By a classical symmetrization inequality (see, for example, Van der Vaart and Wellner, 1996), we can upper bound ξgF in terms of local Rademacher averages: ξgF(ϵ) = E sup n Egf −ˆ Egf : f ∈F, Egf = ϵ o ≤2E sup ( 1 n n X i=1 ϵigf(Xi, Yi) : f ∈F, Egf = ϵ ) , 30 where the expectations are over the sample (X1, Y1) . . . , (Xn, Yn) and the independent uniform (Rademacher) random variables ϵi ∈{±1}. The Ledoux and Talagrand (1991) contraction inequal-ity and Lemma 15 imply ξgF(ϵ) ≤4LE sup ( 1 n n X i=1 ϵidφ(Yif(Xi), Yif ∗(Xi)) : f ∈F, Egf = ϵ ) ≤4LE sup ( 1 n n X i=1 ϵidφ(Yif(Xi), Yif ∗(Xi)) : f ∈F, ˜ dφ(f, f ∗)2 ≤2aBϵ ) = 4LE sup ( 1 n n X i=1 ϵif(Xi, Yi) : f ∈Fφ, Ef 2 ≤2aBϵ ) , where Fφ = {(x, y) 7→dφ(yf(x), yf ∗(x)) : f ∈F} .
One approach to approximating these local Rademacher averages is through information about the rate of growth of covering numbers of the class. For some subset A of a pseudometric space (S, d), let N(ϵ, A, d) denote the cardinality of the smallest ϵ-cover of A, that is, the smallest set ˆ A ⊂S for which every a ∈A has some ˆ a ∈ˆ A with d(a, ˆ a) ≤ϵ. Using Dudley’s entropy integral (Dudley, 1999), Mendelson (2002) has shown the following result: Suppose that F is a set of [−1, 1]-valued functions on X, and there is a γ > 0 and 0 < p < 2 for which sup P N(ϵ, F, L2(P)) ≤γϵ−p, where the supremum is over all probability distributions P on X. Then for some constant Cγ,p (that depends only on γ and p), 1 nE sup ( n X i=1 ϵif(Xi) : f ∈F, Ef 2 ≤ϵ ) ≤Cγ,p max n n−2/(2+p), n−1/2ϵ(2−p)/4o .
Since dφ(a, b) ≤|a −b|, any ϵ-cover of {f −f ∗: f ∈F} is an ϵ-cover of Fφ, so N(ϵ, Fφ, L2(P)) ≤ N(ϵ, F, L2(P)).
Now, for the class absconv(G) with dV C(G) = d, we have sup P N(ϵ, absconv(G), L2(P)) ≤Cdϵ−2d/(d+2); (see, for example, Van der Vaart and Wellner, 1996). Applying Mendelson’s result shows that 1 nE sup ( n X i=1 ϵif(Xi) : f ∈B absconv(G), Ef 2 ≤ϵ ) ≤Cd max n Bn−(d+2)/(2d+2), Bd/(d+2)n−1/2ϵ1/(d+2)o .
Solving for ϵ∗≥ξgF(ϵ∗) shows that it suffices to choose ϵ∗= C′ dBLB max (LBaB B 1/(d+1) , 1 ) n−(d+2)/(2d+2), for some constant C′ d that depends only on d.
31 6 Conclusions We have focused on the relationship between properties of a nonnegative margin-based loss function φ and the statistical performance of the classifier which, based on an iid training set, minimizes em-pirical φ-risk over a class of functions. We first derived a universal upper bound on the population misclassification risk of any thresholded measurable classifier in terms of its corresponding popu-lation φ-risk. The bound is governed by the ψ-transform, a convexified variational transform of φ.
It is the tightest possible upper bound uniform over all probability distributions and measurable functions in this setting.
Using this upper bound, we characterized the class of loss functions which guarantee that every φ-risk consistent classifier sequence is also Bayes-risk consistent, under any population distribu-tion. Here φ-risk consistency denotes sequential convergence of population φ-risks to the smallest possible φ-risk of any measurable classifier.
The characteristic property of such a φ, which we term classification-calibration, is a kind of pointwise Fisher consistency for the conditional φ-risk at each x ∈X. The necessity of classification-calibration is apparent; the sufficiency underscores its fundamental importance in elaborating the statistical behavior of large-margin classifiers.
For the widespread special case of convex φ, we demonstrated that classification-calibration is equivalent to the existence and strict negativity of the first derivative of φ at 0, a condition readily verifiable in most practical examples. In addition, the convexification step in the ψ-transform is vacuous for convex φ, which simplifies the derivation of closed forms.
Under the noise-limiting assumption of Tsybakov (2001), we sharpened our original upper bound and studied the Bayes-risk consistency of ˆ f, the minimizer of empirical φ-risk over a convex, bounded class of functions F which is not too complex. We found that, for convex φ satisfying a certain uniform strict convexity condition, empirical φ-risk minimization yields convergence of misclassification risk to that of the best-performing classifier in F, as the sample size grows. Fur-thermore, the rate of convergence can be strictly faster than the classical n−1/2, depending on the strictness of convexity of φ and the complexity of F.
Two important issues that we have not treated are the approximation error for population φ-risk relative to F, and algorithmic considerations in the minimization of empirical φ-risk. In the setting of scaled convex hulls of a base class, some approximation results are given by Breiman (2000), Mannor et al. (2002) and Lugosi and Vayatis (2003). Regarding the numerical optimization to determine ˆ f, Zhang and Yu (2003) give novel bounds on the convergence rate for generic forward stagewise additive modeling (see also Zhang, 2002).
These authors focus on optimization of a convex risk functional over the entire linear hull of a base class, with regularization enforced by an early stopping rule.
Acknowledgments We would like to thank Gilles Blanchard, Olivier Bousquet, Pascal Massart, Ron Meir, Shahar Mendelson, Martin Wainwright and Bin Yu for helpful discussions.
A Loss, risk, and distance We could construe Rφ as the risk under a loss function ℓφ : R×{±1} →[0, ∞) defined by ℓφ(ˆ y, y) = φ(ˆ yy). The following result establishes that loss functions of this form are fundamentally unlike 32 distance metrics.
Lemma 18. Suppose ℓφ : R2 →[0, ∞) has the form ℓφ(x, y) = φ(xy) for some φ : R →[0, ∞).
Then 1. ℓφ is not a distance metric on R, 2. ℓφ is a pseudometric on R iffφ ≡0, in which case ℓφ assigns distance zero to every pair of reals.
Proof. By hypothesis, ℓφ is nonnegative and symmetric. Another requirement of a distance metric is definiteness: for all x, y ∈R, x = y ⇐ ⇒ℓφ(x, y) = 0.
(14) But we may write any z ∈(0, ∞) in two different ways, as √z√z and, for example, (2z)((1/2)z).
To satisfy (14) requires φ(z) = 0 in the former case and φ(z) > 0 in the latter, an impossibility.
This proves 1.
To prove 2, recall that a pseudometric relaxes (14) to the requirement x = y = ⇒ℓφ(x, y) = 0.
(15) Since each z ≥0 has the form xy for x = y = √z, (15) amounts to the necessary condition that φ ≡0 on [0, ∞). The final requirement on ℓφ is the triangle inequality, which in terms of φ becomes φ(xz) ≤φ(xy) + φ(yz), for all x, y, z ∈R.
(16) Since φ must vanish on [0, ∞), taking y = 0 in (16) shows that only the zero function can (and does) satisfy the constraint.
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Near Doubles Strategy
The near doubles strategy helps make addition easier for first and second graders. This method is used when two numbers are close together like 7 + 8. Students use their knowledge of the doubles math strategy to help them solve other facts. This builds their math fact fluency, their number sense, and their flexibility with numbers. This strategy is sometimes also called the doubles plus one strategy or math near doubles.
The near doubles math strategy is one that students can use when building their math fact fluency through playing games. Students need many opportunities to practice building their fluency and games are a perfect way to do this! They make it fun and engaging, and students forget that they are practicing their math facts. Use this Dot Addition or Dot Subtraction in your classroom. Put your name and email address in the boxes, and it will be in your inbox. It is easy to prep, and your students will love it!
What Are Near Doubles in Math?
A near doubles fact involves adding two numbers that are almost the same such as 7 + 8, 5 + 6, or 9 + 10. To solve a near doubles fact, students can start with a doubles fact they already know. When I begin teaching the doubles plus one strategy, I always begin by having them double the smaller number. Since we have already learned our doubles facts, this makes it easy to move to the near doubles strategy.
Here are some doubles plus one examples.
7 + 7 = 14 (a doubles fact they know)
7 + 8 is just one more than 7 + 7 so we add one more to 14 which makes 7 + 8 = 15.
5+5=10 (another doubles fact they know)
5+6 is just one more than 5+5 so we add one more to 11 which makes 5+6=11.
Step 1: Modeling with Manipulatives for the Near Doubles Strategy
One of the best ways to teach the near doubles strategy is through hands-on practice with manipulatives. This gives students a concrete visual to help them understand the relationship between the doubles fact and the near doubles fact.
Start by using ten frames and two different colors of manipulatives (like unifix cubes or two-colored counters). Here’s how I break it down:
Model the doubles fact using the smaller number first. For example, if you’re working on 7 + 8, begin by showing 7 + 7 using one color. Fill in both ten frames with seven manipulatives of the same color.
Then model the near doubles fact. Explain that when we have 7 + 8, it’s just one more than 7 + 7. Add one more manipulative to the ten frame in a different color. This helps students see visually that 7 + 8 is just one more than 7 + 7.
Repeat the process with different examples such as 6 + 7 and 8 + 9. Make sure you always begin with the smallest doubles fact first and then add one more to find the near doubles total.
Step 2: Practice with Manipulatives
Once you’ve modeled a few examples, it’s time for students to use the manipulatives themselves. Give each student two ten frames and a set of manipulatives in two different colors. Small erasers or two-colored counters work great for this.
Have the students model a doubles fact, such as 6 + 6. They should place the manipulatives on the ten frames and should all be in the same color.
Next, ask them to model the near doubles fact. For 6 + 7, they will add one more manipulative in the second color. This will help them visually see that 6 + 7 is just one more than 6 + 6.
Continue practicing with other examples and allow the students to model both the doubles and near doubles facts with their manipulatives.
This step reinforces the concept while keeping it fun. By using manipulatives, they help students stay engaged and make the learning experience more memorable as they practice the near doubles strategy.
Step 3: Drawing Ten Frames and Writing Equations
After practicing with manipulatives, students can begin transitioning to more abstract representations of the near doubles math strategy. This involves drawing in ten frames and writing equations.
Have the students draw a doubles fact in two ten frames. I usually have them draw circles because they can be drawn quickly. I have them draw the doubles fact in one color. For example, I would have them draw 6 circles in one ten frame and 6 circles in another ten frame using a red crayon for the doubles fact of 6 + 6. Then I would have them draw one more circle with a blue crayon color to show the near doubles fact of 6 + 7.
After drawing using the ten frames, I guide them in writing out the equations that match what they’ve drawn. For example:
6 + 6 = 12
6 + 7 = 13 (by adding one more to 12)
This combination of drawing and writing reinforces the concept and helps students internalize the strategy.
Second Method: Doubling the Bigger Number with the Near Doubles Strategy
Once students are comfortable with doubling the smaller number and adding one, I introduce the concept of doubling the bigger number and then subtracting one.
For example, with 7 + 8, 8 is larger so we double it. Then we discuss if we should add one because that is what we have been doing. I love to hear their thoughts and explanations. You can really see who has good number sense with this discussion. It doesn’t take long before we decide that we need to subtract one because we doubled the bigger number. Then we subtract one to get the answer: 16 – 1 = 15.
This method works just as well and helps students develop flexibility with numbers. I practice this same strategy with the class by first using manipulatives, letting students practice, and finally having them draw and write equations.
Choosing the Best Method
After practicing both methods, I talk to students about choosing the strategy that works best for them. Some students may find it easier to double the smaller number and add one while others might prefer doubling the bigger number and subtracting one. I also mention that they might want to use the easiest doubles fact for them—whether it’s the smaller or bigger number.
Making It Fun
To keep this activity enjoyable, use engaging manipulatives like unifix cubes, small erasers, or two-colored counters. Children like to have fun. When they do, they are more likely to internalize the strategy.
In conclusion, the near doubles strategy is an excellent tool for first and second graders learning to add. By starting with a doubles fact and adjusting by one, students can solve addition problems with confidence. Through hands-on practice with manipulatives, drawing, and writing equations, students will solidify their understanding of this strategy and build their number sense.
You might also enjoy these articles.
Near Doubles Addition Strategy
Using Doubles Strategy for Addition
Near Doubles Second Grade Addition Facts
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13529 | http://qed.econ.queensu.ca/pub/faculty/abbott/econ351/351note02.pdf | ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 1 of 17 pages ECON 351 -- NOTE 2 Ordinary Least Squares (OLS) Estimation of the Simple CLRM 1. The Nature of the Estimation Problem This note derives the Ordinary Least Squares (OLS) coefficient estimators for the simple (two-variable) linear regression model. 1.1 The population regression equation, or PRE, for the simple (two-variable) linear regression model takes the form: i i 1 0 i u X Y + β + β = (i = 1, ..., N) (1) where ui is an iid random error term. 1.2 The OLS sample regression equation (or OLS-SRE) corresponding to equation (1) can be written as (i = 1, ..., N). (2) i i i i 1 0 i u ˆ Y ˆ u ˆ X ˆ ˆ Y + = + β + β = where: the OLS estimator of the intercept coefficient β0; = β0 ˆ the OLS estimator of the slope coefficient β1; $ β1 = i 1 0 i X ˆ ˆ Y ˆ β + β = = the OLS estimated (or predicted) values of E(Yi | Xi) = β0 + β1Xi for sample observation i, and is called the OLS sample regression function (or OLS-SRF); i 1 0 i i X ˆ ˆ Y u ˆ β − β − = = the OLS residual for sample observation i. ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 2 of 17 pages 1.3 The Estimation Problem: The estimation problem consists of constructing or deriving the OLS coefficient estimators 1 for any given sample of N observations (Yi, Xi), i = 1, ..., N on the observable variables Y and X. 0 ˆ and ˆ β β Definition of an Estimator An estimator of a population parameter is a rule, formula, or procedure for computing a numerical estimate of an unknown population parameter from the sample values of the observable variables. Example: The estimators of the population parameters β0 and β1 in the simple PRE (1) are therefore formulas that tell us how to compute numerical estimates of β0 and β1 from the given sample data on the observable variables Y and X. 1 0 ˆ and ˆ β β Properties of an Estimator An estimator possesses two critical properties. 1. An estimator is a function only of the given sample data; this function does not contain any unknown parameters. Examples: In the context of the simple linear regression model represented by PRE (1), the estimators of the regression coefficients β0 and β1 are functions of the sample data on Y and X, which consist of a random sample of N observed values (Yi, Xi), i = 1, ..., N. Thus, the estimators take the general form 1 0 ˆ and ˆ β β 1 0 ˆ and ˆ β β i = 1, ..., N ( ) X , Y ( f X , , X , X ; Y , , Y , Y f ˆ i i 0 N 2 1 N 2 1 0 0 = = β K K ) ) i = 1, ..., N ( ) X , Y ( f X , , X , X ; Y , , Y , Y f ˆ i i 1 N 2 1 N 2 1 1 1 = = β K K where the functions and contain no unknown parameters. ) ( f0 L ) ( f1 L 2. An estimator is a random variable (or statistic) because it is a function of the sample data (Yi, Xi), i = 1, ..., N. An estimator therefore has a probability distribution, which is called the sampling distribution of the estimator. ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 3 of 17 pages Distinction Between “Estimators” and “Estimates” • The term estimator refers to the formula or rule or procedure by which a numerical estimate of an unknown population parameter is computed from any given sample data. Examples: ♦ The estimator 0 ˆ β of the intercept coefficient β0 is the function ) X , Y ( ; YN ) ( f X , , X , X , , Y , Y f i i 0 N 2 1 2 1 0 = K K which tells us how to compute a numerical estimate of the parameter β0 for any given sample values (Yi, Xi), i = 1, ..., N of the observable variables Y and X. ♦ Similarly, the estimator 1 ˆ β of the slope coefficient β1 is the function ( ) X , Y ( f X , X , X ; Y , , Y , Y f i i 1 N 2 1 N 2 1 1 ) , = K K which tells us how to compute a numerical estimate of the parameter β1 for any given sample values (Yi, Xi), i = 1, ..., N of the observable variables Y and X. • The term estimate refers to the specific numerical value given by the formula for a specific set of sample values (Yi, Xi), i = 1, ..., N of the observable variables Y and X. That is, an estimate is the value of the estimator obtained when the formula is evaluated for a particular set of sample values of the observable variables. Example: Suppose that for a particular sample of 50 observed values of Yi and Xi that the formula or function yields the value = 0.83. Then the number 0.83 is an estimate of the slope coefficient β1. ) ( f1 L 1 ˆ β ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 4 of 17 pages 2. The OLS Estimation Criterion The OLS coefficient estimators are those formulas or expressions for that 1 0 ˆ and ˆ β β minimize the sum of squared residuals RSS for any given sample of size N. The OLS estimation criterion is therefore: ( ) ( ) ∑ ∑ = = β − β − = = β β N 1 i 2 i 1 0 i N 1 i 2 i 1 0 X ˆ ˆ Y u ˆ ˆ , ˆ RSS Minimize (3) { } $ β j Rationale for the OLS Estimation Criterion • Theoretical Rationale: The OLS coefficient estimators 1 implied by the OLS estimation criterion have several desirable statistical properties when the assumptions of the CLRM (Classical Linear Regression Model) are satisfied. 0 ˆ and ˆ β β • Intuitive Rationale: The OLS estimation criterion corresponds to the idea of “best fit” of the estimated sample regression function (SRF) to the given sample data (Yi, Xi), i = 1, ..., N. Note that the OLS criterion minimizes the sum of squared residuals , not the sum of the residuals . ∑i i u $ 2 ∑i i u $ Squaring the residuals does two things: $ ui (1) It avoids the possibility that large positive residuals and large negative residuals could offset each other and still lead to a small (or even zero) value of ∑i i u $ . (2) It implicitly assigns a larger weight to numerically large residuals (regardless of whether they are positive or negative). The larger the numerical or absolute value of a particular $ ui, the larger the corresponding value of $ ui 2. ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 5 of 17 pages 3. Derivation of the OLS Coefficient Estimators Derivation of the OLS formulas for the regression coefficient estimators is performed in two stages: 1 0 ˆ and ˆ β β STAGE 1 consists of deriving the first-order conditions (or FOCs) for minimizing the residual sum of squares function ( ) 1 0 ˆ , ˆ RSS β β given by equation (3). These FOCs are called the OLS normal equations. STAGE 2 consists of solving the first-order conditions (FOCs) -- i.e., the OLS normal equations -- to obtain explicit formulas or expressions for the OLS coefficient estimators . 1 0 ˆ and ˆ β β ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 6 of 17 pages STAGE 1: Derivation of the OLS Normal Equations, or FOCs Step 1.1: Partially differentiate the ( ) 1 0 ˆ , ˆ RSS β β function in (3) with respect to and . 0 ˆ β 1 ˆ β • First, rewrite the ( ) 1 0 ˆ , ˆ RSS β β function in (3) as follows: ( ) ∑ ∑ = = = = β β N 1 i i N 1 i 2 i 1 0 ) u ˆ ( f u ˆ ˆ , ˆ RSS where 2 i i u ˆ ) u ˆ ( f = i 1 0 i i X ˆ ˆ Y u ˆ β − β − = (3) Note: The function is a function of , and is in turn a function of and . 2 i i u ˆ ) u ˆ ( f = i u ˆ i u ˆ 0 ˆ β 1 ˆ β • Using the chain rule of differentiation, each partial derivative of the ( ) 1 0 ˆ , ˆ function takes the general form RSS β β ∑ = β ∂ ∂ = β ∂ ∂ N 1 i j i i j ˆ u ˆ u ˆ d f d ˆ RSS . (4) • Using the power rule of differentiation, the derivative i u ˆ d f d is i i 2 i i u ˆ 2 u ˆ d ) u ˆ ( d u ˆ d f d = = . • The partial derivatives j ˆ RSS β ∂ ∂ for j = 0, 1 are therefore ∑ ∑ ∑ = = = β ∂ ∂ = β ∂ ∂ = β ∂ ∂ = β ∂ ∂ N 1 i j i i N 1 i j i i N 1 i j i i j ˆ u ˆ u ˆ 2 ˆ u ˆ u ˆ 2 ˆ u ˆ u ˆ d f d ˆ RSS j = 0, 1. (5) ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 7 of 17 pages • Since the i-th residual is i 1 , the partial derivatives 0 i i X ˆ ˆ Y u ˆ β − β − = ∂ ∂β $ $ ui j for j = 0, 1 are: for j = 0: ; 1 ˆ u ˆ 0 i − = β ∂ ∂ for j = 1: .
X ˆ u ˆ i 1 i − = β ∂ ∂ • Substitute the partial derivatives ∂ ∂β $ $ ui j for j = 0, 1 into equation (5): ∑ = β ∂ ∂ = β ∂ ∂ N 1 i j i i j ˆ u ˆ u ˆ 2 ˆ RSS j = 0, 1. (5) The partial derivatives ∂ ∂ RSS j $ β for j = 0, 1 thus take the form: ∑ ∑ ∑ = = = − = − = β ∂ ∂ = β ∂ ∂ N 1 i i N 1 i i N 1 i 0 i i 0 u ˆ 2 ) 1 ( u ˆ 2 ˆ u ˆ u ˆ 2 ˆ RSS (6.1) ∑ ∑ ∑ = = = − = − = β ∂ ∂ = β ∂ ∂ N 1 i i i N 1 i i i N 1 i 1 i i 1 u ˆ X 2 ) X ( u ˆ 2 ˆ u ˆ u ˆ 2 ˆ RSS (6.2) ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 8 of 17 pages Step 1.2: Obtain the first-order conditions (FOCs) for a minimum of the RSS function by setting the partial derivatives (6.1)-(6.2) equal to zero, then dividing each equation by −2, and finally setting : i 1 0 i i X ˆ ˆ Y u ˆ β − β − = • ∑ = − = β ∂ ∂ N 1 i i 0 u ˆ 2 ˆ RSS (6.1) 0 ˆ RSS 0 = β ∂ ∂ ⇒ ⇒ (7.1) 0 u ˆ 2 N 1 i i = − ∑ = 0 u ˆ N 1 i i = ∑ = ⇒ ( ) 0 X ˆ ˆ Y N 1 i i 1 0 i = β − β − ∑ = (8.1) • ∑ = − = β ∂ ∂ N 1 i i i 1 u ˆ X 2 ˆ RSS (6.2) 0 ˆ RSS 1 = β ∂ ∂ ⇒ ⇒ (7.2) 0 u ˆ X 2 N 1 i i i = − ∑ = 0 u ˆ X N 1 i i i = ∑ = ⇒ ( ) 0 X ˆ ˆ Y X N 1 i i 1 0 i i = β − β − ∑ = (8.2) • Equations (7.1) and (7.2) are the most compact way of writing the FOCs for the OLS coefficient estimators 0 ˆ β and 1 ˆ β : 0 ˆ RSS 0 = β ∂ ∂ ⇔ (7.1) 0 u ˆ N 1 i i = ∑ = 0 ˆ RSS 1 = β ∂ ∂ ⇔ (7.2) 0 u ˆ X N 1 i i i = ∑ = ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 9 of 17 pages • But equations (8.1) and (8.2) are used to obtain the formulas for the OLS coefficient estimators 0 ˆ β and 1 ˆ β : ( 0 X ˆ ˆ Y N 1 i i 1 0 i = β − β − ∑ = ) ) ; (8.1) ( 0 X ˆ ˆ Y X N 1 i i 1 0 i i = β − β − ∑ = . (8.2) Step 1.3: Rearrange each of the equations (8.1) and (8.2) to put them in the conventional form of the OLS normal equations. Thus, taking summations and rearranging terms, we obtain the OLS normal equations: (8.1) ( 0 X ˆ ˆ Y N 1 i i 1 0 i = β − β − ∑ = ) ) (9.1) i i i i 1 0 i i i i 1 0 i i 1 0 i i Y X ˆ ˆ N Y X ˆ ˆ N 0 X ˆ ˆ N Y ∑ = ∑ β + β ∑ − = ∑ β − β − = ∑ β − β − ∑ (8.2) ( 0 X ˆ ˆ Y X N 1 i i 1 0 i i = β − β − ∑ = (9.2) i i i 2 i i 1 i i 0 i i i 2 i i 1 i i 0 2 i i 1 i i 0 i i i 2 i 1 i 0 i i i Y X X ˆ X ˆ Y X X ˆ X ˆ 0 X ˆ X ˆ Y X 0 ) X ˆ X ˆ Y X ( ∑ = ∑ β + ∑ β ∑ − = ∑ β − ∑ β − = ∑ β − ∑ β − ∑ = β − β − ∑ ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 10 of 17 pages RESULT: The last equations in (9.1) and (9.2) are the OLS normal equations: ∑ ∑ = = = β + β N 1 i i N 1 i i 1 0 Y X ˆ ˆ N (N1) ∑ ∑ ∑ = = = = β + β N 1 i i i N 1 i 2 i 1 N 1 i i 0 Y X X ˆ X ˆ (N2) The OLS normal equations (N1) and (N2) constitute two linear equations in the two unknowns and . Their solution yields explicit expressions for and ; these expressions are the OLS estimators and of the regression coefficients β0 and β1. 0 ˆ β 1 ˆ β 0 ˆ β 1 ˆ β 0 ˆ β 1 ˆ β ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 11 of 17 pages STAGE 2: Solution of the OLS Normal Equations There is more than one way to solve the OLS normal equations (N1) and (N2) for the two unknowns and . The following steps constitute one such solution method. 0 ˆ β 1 ˆ β Step 2.1: Divide normal equation (N1) by N and then solve for : 0 ˆ β (N1) ∑ ∑ = = = β + β N 1 i i N 1 i i 1 0 Y X ˆ ˆ N ⎟ ⎠ ⎞ ⎜ ⎝ ⎛∑ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛∑ β + β N Y N X ˆ ˆ i i i i 1 0 .
N Y Y and N X X since Y X ˆ ˆ i i i i 1 0 ∑ ≡ ∑ ≡ = β + β Therefore, solving for : 0 ˆ β X ˆ Y ˆ 1 0 β − = β . (10) Step 2.2: Substitute the expression for given by equation (10) into normal equation (N2): 0 ˆ β ∑ ∑ ∑ = = = = β + β N 1 i i i N 1 i 2 i 1 N 1 i i 0 Y X X ˆ X ˆ (N2) ( ) i i i 2 i i 1 i i 1 Y X X ˆ X X ˆ Y Σ = Σ β + Σ β − (11) Step 2.3: Since X X i i ≡∑ N by definition, ∑ = i i X N X . Set ∑ = i i X N X in equation (11): ( ) .
Y X X ˆ X N ˆ Y X N Y X X ˆ X N X ˆ Y i i i 2 i i 1 2 1 i i i 2 i i 1 1 Σ = Σ β + β − Σ = Σ β + β − (12) ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 12 of 17 pages Step 2.4: Solve the last equation in (12) for : 1 ˆ β ( ) .
Y X N Y X X N X ˆ Y X N Y X X ˆ X N ˆ Y X X ˆ X N ˆ Y X N i i 2 2 i i 1 i i 2 i i 1 2 1 i i i 2 i i 1 2 1 − Σ = − Σ β − Σ = Σ β + β − Σ = Σ β + β − Therefore, solving for : 1 ˆ β 2 2 i i i i i 1 X N X Y X N Y X ˆ − ∑ − ∑ = β (13) RESULT: Equations (10) and (13) are the OLS coefficient estimators 0 ˆ β and 1 ˆ β . X ˆ Y ˆ 1 0 β − = β . (10) 2 N 1 i 2 i N 1 i i i 1 X N X Y X N Y X ˆ − − = β ∑ ∑ = = . (13) They represent the solution of the OLS normal equations; that is, they represent the solution of the FOCs for minimizing the residual sum-of-squares function ( ) 1 0 ˆ , ˆ RSS β β given by equation (3). ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 13 of 17 pages Computational Note: If we were actually using formulas (10) and (13) to compute estimates of β0 and β1 for a given sample of N observations (Yi, Xi), i = 1, ..., N, we would employ the following two-step computational procedure. 1. First, use equation (13) to compute the estimate of β1: 2 N i 1 i 2 i N i 1 i i i 1 X N X Y X N Y X ˆ − − = β ∑ ∑ = = = = (14.1) 2. Second, substitute $ β1, the estimate of β1 computed by equation (14.1), into the formula for 0 ˆ β to obtain the estimate of β0: X ˆ Y ˆ 1 0 β − = β (14.2) Characteristics of the OLS Coefficient Estimators 0 ˆ β and 1 ˆ β : 1. The OLS coefficient estimators 0 ˆ β and 1 ˆ β are functions only of the observed sample values (Yi, Xi): i = 1, ..., N of the observable variables Y and X. They can therefore be computed for any given set of sample data. 2. The OLS coefficient estimators 0 ˆ β and 1 ˆ β are point estimators. That is, each estimator provides only a single (point) value of the corresponding population parameter. ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 14 of 17 pages 4. Alternative Expressions for the OLS Slope Coefficient Estimator 4.1 Deviation-From-Means Formula for : Derivation 1 ˆ β Formula (14.1) for the OLS slope coefficient estimator can conveniently be re-written in deviation-from-means form. 1 ˆ β • Deviation-from-means notation uses lower case letters to denote the deviations of each observed sample value from its corresponding sample mean. 1. Define the deviations of the Yi sample values from their sample mean Y as y Y Y i i ≡ − (i = 1, ..., N) where Y Y N Y N i i i i = = Σ Σ = the sample mean of the Yi values. 2. Define the deviations of the Xi sample values from their sample mean X as x X X i i ≡ − (i = 1, ..., N) where X X N X N i i i i = = Σ Σ = the sample mean of the Xi values. ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 15 of 17 pages • Formula (14.1) for the OLS slope coefficient estimator 1 ˆ β is 2 N i 1 i 2 i N i 1 i i i 1 X N X Y X N Y X ˆ − − = β ∑ ∑ = = = = (14.1) • The numerator of formula (14.1) for OLS slope coefficient estimator 1 ˆ β can be shown to equal ( )( ∑ = ∑ − − i i i i i i x y X X Y Y) . (15.1) Proof: Expand the expression on the right-hand side of (15.1): ( )( ) ( ) .
ˆ for (14.1) formula of numerator the = Y X N Y X X N X , Y N Y since Y X N Y X N Y X N Y X Y X N X Y Y X Y X Y X X Y Y X Y X Y Y X X y x 1 i i i i i i i i i i i i i i i i i i i i i i i i i i i i β − ∑ = = ∑ = ∑ + − − ∑ = + ∑ − ∑ − ∑ = + − − ∑ = − − ∑ = ∑ • The denominator of formula (14.1) for OLS slope coefficient estimator 1 ˆ β can be shown to equal ( ∑ = ∑ − i i i i x X 2 2 ) X . (15.2) Proof: Expand the expression on the right-hand side of (15.2): ( ) ( ) .
ˆ for (14.1) formula of r denominato the X N X X N X since X N X N 2 X X N X X 2 X X X X 2 X X X x 1 2 2 i i i i 2 2 2 i i 2 i i 2 i i 2 i 2 i i 2 i i 2 i i β = − ∑ = = ∑ + − ∑ = + ∑ − ∑ = + − ∑ = − ∑ = ∑ ECONOMICS 351 -- NOTE 2 M.G. Abbott ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 16 of 17 pages RESULT: The OLS slope coefficient estimator 1 ˆ β can be written in (at least) three equivalent ways: 2 i i i i i 1 x y x ˆ ∑ ∑ = β ( )( ) ( ) 2 i i i i X X Y Y X X − ∑ − − ∑ = (16) 2 2 i i i i i X N X Y X N Y X − ∑ − ∑ = But the easiest and most useful formula for is the deviations-from-means formula 1 ˆ β ∑ ∑ = = = ∑ ∑ = β N 1 i 2 i N 1 i i i 2 i i i i i 1 x y x x y x ˆ . ECONOMICS 351 -- NOTE 2 M.G. Abbott 4.2 Deviation-From-Means Formula for : Interpretation 1 ˆ β The deviation-from-means form of the OLS slope coefficient estimator can be given a convenient and easy-to-remember interpretation. This interpretation makes use of two sample statistics: (1) the sample covariance of Xi and Yi; and (2) the sample variance of Xi. 1 ˆ β • Define the sample covariance of the observed Xi and Yi values as: ( )( ) Cov X Y x y N X X Y Y N i i i i i i i i ( , ) = ∑ = ∑ − − . • Define the sample variance of the observed Xi values as: ( ) Var X x N X X N i i i i i ( ) = ∑ = ∑ − 2 2 . Note: Var(Xi) as defined above is a biased but consistent estimator of the population variance of X, conventionally denoted as σ . The unbiased (and consistent) estimator of is given by X 2 σ X 2 s x N i 2 1 − X i 2 = ∑ . • Now divide both the numerator and denominator of the deviation-from-means formula for the OLS slope coefficient estimator 1 ˆ β by N: ) X ( Var ) Y , X ( Cov N x N y x x y x ˆ i i i 2 i i i i i 2 i i i i i 1 = ∑ ∑ = ∑ ∑ = β . (17) RESULT: The OLS slope coefficient estimator 1 ˆ β can be interpreted as the ratio of (1) the sample covariance of Xi and Yi to (2) the sample variance of Xi. ECON 351 -- Note 2: OLS Estimation of the Simple CLRM ... Page 17 of 17 pages |
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Erhard’s , Lombard Test
Stenger’s, Chimani Moos, Teal test
Pure Tone Audiometry
Speech Audiometry
Bekesy Audiometry
Impedance Audiometry
Recruitment Test
Short Increment Sensitivity Index
Electrocochleography (EcoG)
BERA
Otoacoustic Emissions
ASSR
Nose
Anatomy of External Nose
Anatomy of Internal Nose
Physiology of the Nose
Diseases of the External Nose and Vestibule
Functions of Nose & Paranasal Sinuses
Allergic rhinitis
Nasal Polyps
Epistaxis
Fungal Rhinosinusitis
Pharynx & Oesophagus
Anatomy and Physiology of Pharynx
Adenoids
Adenoidectomy
Juvenile angiofibroma
Acute Pharyngitis
Chronic Pharyngitis
Palatine Tonsils
Tonsillitis
Tonsillectomy
Head and Neck Space Infections
Oropharynx Carcinoma
Tumours of the Hypopharynx
Physiology of Sleep and Sleep Disorders
Snoring
Obstructive Sleep Apnoea
Anatomy of Oesophagus
Physiology of Swallowing
Disorders of Oesophagus
Foreign Body Oesophagus
Assessment of Dysphagia
Oral cavity & Salivary Glands
Anatomy of Oral Cavity
Carcinoma Oral Cavity
Oral Ulcers
Anatomy of Parotid Gland
Anatomy of Submandibular Gland
Non-neoplastic Disorders of Salivary Glands
Benign tumours of salivary gland
Malignant tumours of salivary gland
Parotid Gland Surgery
Submandibular Gland Surgery
Drooling and Aspiration in Children
Larynx & Trachea
Anatomy of larynx
Cavity of the Larynx
Spaces of the Larynx
Embryology of Larynx
Paediatric Larynx
Physiology of Larynx
Acute Inflammations of Larynx
Chronic Inflammations of Larynx
Laryngeal Paralysis
Voice disorders
Functional Dysphonia
Spasmodic Dysphonia
Hoarseness
Phonosurgery
Stridor
Congenital lesions of larynx
Laryngotracheal Trauma
Benign tumours of the larynx
Carcinoma Larynx
Conservation Surgeries for Carcinoma Larynx
Tracheostomy
Tracheostomy in Infants and Children
Foreign Body of Airway
Thyroid Gland & Neck
Anatomy of Thyroid Gland
Physiology of Thyroid Gland
Thyroid hormones and Antibodies
Hypothyroidism
Hyperthyroidism
Malignant disorders of thyroid gland
Thyroidectomy
Thyroid Dysfunction in Pregnancy
Pharyngeal arches
Examination of Neck
Triangles of the Neck
Surgeries
Myringotomy
Myringoplasty
Tympanoplasty
Ossiculoplasty
Stapedectomy
Cortical Mastoidectomy
Cochlear Implant Surgery
Adenoidectomy
Tonsillectomy
Parotid Gland Surgery
Submandibular Gland Surgery
Thyroidectomy
Facelift Surgery or Rhytidectomy
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Ear
Anatomy of External Ear
Anatomy of Temporal Bone
Physiology of Hearing
Types of Hearing loss
Embryology of Ear
Anatomy of Middle ear
Anatomy of Internal ear
Assessment of hearing
Audiometry & Objective Tests for Hearing Assessment
Noise induced trauma
Sudden SNHL
Presbycusis
Vestibular Function tests
Eustachian Tube Anatomy
Eustachian tube dysfunction
Diseases of External Ear
ASOM
Otitis Media With Effusion
Cholesteatoma
Chronic Suppurative Otitis Media
Complications of CSOM
CSOM Long Case
Retraction Pockets
Otosclerosis
Fractures of Temporal Bone
Vestibular Neuritis
BPPV
Facial Nerve Paralysis
Causes of Facial Nerve Palsy
Meniere’s disease
Vestibular Schwannoma
Glomus tumour
Tumours of Temporal Bone
Evaluation of deaf child
Rehabilitation of the Hearing Impaired
Otalgia
Tinnitus
Osteonecrosis
Otoendoscopy
Hearing Tests
Clinical Hearing Tests
Tuning Fork Tests
Erhard’s , Lombard Test
Stenger’s, Chimani Moos, Teal test
Pure Tone Audiometry
Speech Audiometry
Bekesy Audiometry
Impedance Audiometry
Recruitment Test
Short Increment Sensitivity Index
Electrocochleography (EcoG)
BERA
Otoacoustic Emissions
ASSR
Nose
Anatomy of External Nose
Anatomy of Internal Nose
Physiology of the Nose
Diseases of the External Nose and Vestibule
Functions of Nose & Paranasal Sinuses
Allergic rhinitis
Nasal Polyps
Epistaxis
Fungal Rhinosinusitis
Pharynx & Oesophagus
Anatomy and Physiology of Pharynx
Adenoids
Adenoidectomy
Juvenile angiofibroma
Acute Pharyngitis
Chronic Pharyngitis
Palatine Tonsils
Tonsillitis
Tonsillectomy
Head and Neck Space Infections
Oropharynx Carcinoma
Tumours of the Hypopharynx
Physiology of Sleep and Sleep Disorders
Snoring
Obstructive Sleep Apnoea
Anatomy of Oesophagus
Physiology of Swallowing
Disorders of Oesophagus
Foreign Body Oesophagus
Assessment of Dysphagia
Oral cavity & Salivary Glands
Anatomy of Oral Cavity
Carcinoma Oral Cavity
Oral Ulcers
Anatomy of Parotid Gland
Anatomy of Submandibular Gland
Non-neoplastic Disorders of Salivary Glands
Benign tumours of salivary gland
Malignant tumours of salivary gland
Parotid Gland Surgery
Submandibular Gland Surgery
Drooling and Aspiration in Children
Larynx & Trachea
Anatomy of larynx
Cavity of the Larynx
Spaces of the Larynx
Embryology of Larynx
Paediatric Larynx
Physiology of Larynx
Acute Inflammations of Larynx
Chronic Inflammations of Larynx
Laryngeal Paralysis
Voice disorders
Functional Dysphonia
Spasmodic Dysphonia
Hoarseness
Phonosurgery
Stridor
Congenital lesions of larynx
Laryngotracheal Trauma
Benign tumours of the larynx
Carcinoma Larynx
Conservation Surgeries for Carcinoma Larynx
Tracheostomy
Tracheostomy in Infants and Children
Foreign Body of Airway
Thyroid Gland & Neck
Anatomy of Thyroid Gland
Physiology of Thyroid Gland
Thyroid hormones and Antibodies
Hypothyroidism
Hyperthyroidism
Malignant disorders of thyroid gland
Thyroidectomy
Thyroid Dysfunction in Pregnancy
Pharyngeal arches
Examination of Neck
Triangles of the Neck
Surgeries
Myringotomy
Myringoplasty
Tympanoplasty
Ossiculoplasty
Stapedectomy
Cortical Mastoidectomy
Cochlear Implant Surgery
Adenoidectomy
Tonsillectomy
Parotid Gland Surgery
Submandibular Gland Surgery
Thyroidectomy
Facelift Surgery or Rhytidectomy
Types of Hearing loss
Home
Types of Hearing loss
Types of Hearing Loss
Hearing is one of our most important senses, which connects us to the world and facilitates communication. Hearing loss is defined as an impairment of auditory function, which can significantly impact an individual’s quality of life. This impairment can range from mild to profound, or even result in total loss of hearing, often referred to as deafness. The term “deafness” refers to a condition characterised by little to no hearing ability.
Understanding the various types of hearing loss, their underlying mechanisms, and appropriate management strategies is fundamental for every medical student and future ENT specialist. Therefore, this chapter provides a clear, comprehensive, and exam-oriented overview, ensuring you master this critical topic.
Classification of Hearing Loss
The World Health Organisation (WHO) established a grading system for hearing loss in 1980. This system is based on pure tone audiometry, which measures the average hearing threshold at frequencies of 500, 1000, and 2000 Hertz in the better ear. The following table summarises the WHO grading of hearing loss:
Assessment of Hearing Loss
In evaluating auditory function, it is essential to determine several factors:
Type of Hearing Loss: Depending upon the anatomical location affected, hearing loss can be classified as conductive, sensorineural, or mixed.
Degree of Hearing Loss: This can range from mild to total loss, categorised as mild, moderate, moderately severe, severe, profound, or total.
Site of the Lesion: In conductive hearing loss, the lesion may occur in the external ear, tympanic membrane, middle ear, ossicles, or eustachian tube. Clinical examination and tympanometry are useful in identifying these lesions. In sensorineural hearing loss, it is important to ascertain whether the lesion is cochlear (sensory), retrocochlear (neural, affecting the 8th nerve), or central (affecting higher auditory pathways). Special hearing tests, such as auditory brainstem response (ABR) or otoacoustic emissions (OAE), may be necessary for accurate differentiation.
Cause of Hearing Loss: Potential causes include congenital factors, trauma, infections, neoplasms, degenerative conditions, metabolic disorders, ototoxicity, vascular issues, or autoimmune processes. A comprehensive patient history and laboratory investigations are crucial for identifying the underlying cause.
Types of Hearing Loss
Hearing loss can be categorised into several distinct types, each characterised by different underlying mechanisms and causes:
Conductive hearing loss
Sensorineural hearing loss
Mixed hearing loss
Non-organic/ Functional hearing loss
Central hearing loss
1. Conductive Hearing Loss (CHL)
Conductive hearing loss arises from obstructions in the transmission of sound waves along the auditory pathway, which extends from the external ear to the cochlea, specifically at the stapediovestibular joint.
Common causes of conductive hearing loss include: (Mnemonic: O-T-M-E-O)
Obstruction in External Ear Canal:
Impacted cerumen (earwax) is the most common cause.
Foreign bodies.
Osteomas or exostoses (bony growths in the canal).
Keratotic masses (e.g., cholesteatoma of EAC).
Benign or malignant tumours of the EAC.
Meatal atresia (congenital narrowing or absence of the canal).
Tympanic Membrane (TM) Pathology:
Perforations (due to trauma, infection).
Tympanosclerosis (scarring/calcification of TM).
Retraction pockets.
Middle Ear Disorders:
Presence of fluid: Otitis Media with Effusion (OME), also known as “glue ear.”
Acute otitis media (AOM) or Chronic Suppurative Otitis Media (CSOM).
Hemotympanum (blood in the middle ear, often post-traumatic).
Ossicular Chain Pathology:
Ossicular fixation: Most commonly otosclerosis (fixation of stapes footplate). Also, malleus head fixation, fibrous ankylosis.
Ossicular disruption/discontinuity: Due to trauma (e.g., temporal bone fracture), cholesteatoma erosion, or chronic infection.
Eustachian Tube Dysfunction:
Obstructions affecting pressure equalisation in the middle ear, leading to fluid accumulation (OME).
Barotrauma.
Table: Congenital and Acquired causes of conductive hearing loss
Characteristics of Conductive Hearing Loss:
| Feature | Conductive Hearing Loss Characteristics |
--- |
| Rinne Test | Negative (BC > AC in affected ear). This is a hallmark. |
| Weber Test | Lateralizes to the poorer (affected) ear. The sound is perceived louder in the ear with the conductive loss. |
| Absolute Bone Conduction | Normal (or near normal) levels. The cochlea and nerve are intact. |
| Frequency Impact | Greater impact on low frequencies. (However, large perforations can affect all frequencies). |
| Degree of Loss | Typically does not exceed 60 dB HL. This is because bone conduction is intact. |
| Speech Discrimination | Remains good. Patients hear speech but find it difficult to hear quiet sounds. They often speak softly themselves. |
| Hearing Aid Benefit | Excellent candidates for hearing aid amplification as the inner ear can effectively process amplified sound. |
Characteristics of Conductive Hearing Loss on Pure Tone Audiometry (PTA)
A difference of more than 10 dB between AC and BC thresholds at any frequency.
BC thresholds are generally at normal hearing levels (equal to or better than 20 dB HL) in cases of pure conductive hearing loss.
If BC thresholds exceed 20 dB HL with a significant air-bone gap (ABG), this condition is classified as mixed hearing loss.
Management of Conductive Hearing Loss
Management strategies for conductive hearing loss include:
Removal of Canal Obstructions:
Impacting cerumen or foreign bodies: Removal through irrigation or suction clearance.
Osteomas/Exostoses: Surgical excision.
Meatal atresia: Canaloplasty (surgical reconstruction of the EAC).
Benign or malignant tumours: Surgery
Medical Management:
Otitis Media with Effusion (OME): Decongestants, antihistamines, and watchful waiting.
Acute Otitis Media: Antibiotics.
Surgical Interventions:
Myringotomy with or without grommet insertion: To drain fluid from the middle ear (e.g., in OME).
Myringoplasty: Surgical repair of the tympanic membrane perforation.
Tympanoplasty: A broader operation to (i) eradicate disease from the middle ear and (ii) reconstruct the hearing mechanism, potentially with or without mastoid surgery and with or without tympanic membrane grafting.
Ossiculoplasty: Surgical reconstruction of the ossicular chain (e.g., using prostheses or autologous ossicles).
Stapedectomy/Stapedotomy: For otosclerosis, replacing the fixed stapes with a prosthesis.
Hearing Aids:
These devices are utilised when surgical options are not feasible, declined by the patient, or have previously failed.
2. Sensorineural Hearing Loss (SNHL)
Sensorineural hearing loss is the result of damage to the cochlea (sensory type) or the eighth cranial nerve and its central auditory pathways (neural type). This damage disrupts the process of cochlear transduction, where mechanical energy is converted into electrical impulses for transmission to the brain.
Sensory SNHL (Cochlear): It results from damage to the hair cells or other structures within the cochlea. This is the most common type of SNHL.
Neural SNHL (Retrocochlear): It is caused by lesions affecting the eighth cranial nerve (vestibulocochlear nerve).
Central Deafness: It results from lesions in the central auditory pathways. (Sometimes classified separately or as a subtype of SNHL).
Characteristics of Sensorineural Hearing Loss:
| Feature | Sensorineural Hearing Loss Characteristics |
--- |
| Rinne Test | Positive (AC > BC), but both are reduced. The sound is heard better by air, but softly, in the affected ear. |
| Weber Test | Lateralizes to the better (unaffected) ear. Sound is perceived louder in the ear with better cochlear function. |
| Bone Conduction | Reduced on Schwabach and absolute bone conduction tests. This reflects inner ear or neural damage. |
| Speech Discrimination | Poor speech discrimination. Patients often report, "I can hear people talking, but I can't understand what they're saying." |
| Hearing in Noise | Significant difficulty in hearing in the presence of background noise. |
| Recruitment Phenomenon | May be present (abnormal loudness growth). Loud sounds become disproportionately louder, often painful. Typical of cochlear SNHL. |
| Tinnitus | Frequently accompanied by tinnitus (ringing or buzzing in the ears). |
| Degree of Loss | Can range from minimal to profound or total loss, and is usually permanent. |
The characteristics of sensorineural hearing loss on PTA:
AC and BC thresholds being within 10dB of each other.
High frequencies are mostly involved.
Loss may exceed 60 dB, generally complete hearing loss.
Aetiology of Sensorineural Hearing Loss:
SNHL can be broadly categorised into congenital (present at birth) and acquired (developing later in life) causes.
A. Congenital SNHL:
Congenital SNHL can result from inner ear anomalies (e.g., Mondini dysplasia) or prenatal/perinatal damage to the hearing apparatus.
Syndromes Associated with Congenital SNHL
| Syndrome | Associated Features | Hearing Loss Characteristics |
---
| Waardenburg Syndrome | Pigmentation abnormalities (white forelock, heterochromia iridis), wide-set eyes, synophrys. | Unilateral or bilateral, progressive SNHL. |
| Usher Syndrome | Progressive retinitis pigmentosa (leading to blindness) and vestibular dysfunction. | Bilateral SNHL (often severe to profound). |
| Pendred Syndrome | Goiter (thyroid dysfunction), inner ear malformations (Mondini dysplasia). | Bilateral, profound SNHL. |
| Alport Syndrome | Nephritis (kidney disease), ocular abnormalities (anterior lenticonus). | Progressive SNHL. |
| Jervell and Lange-Nielsen Syndrome | Prolonged QT interval on ECG (risk of sudden cardiac death due to arrhythmias). | Bilateral, profound SNHL. |
| Branchio-Oto-Renal (BOR) Syndrome | Branchial cleft cysts/fistulae, ear anomalies (preauricular pits, malformed pinna), renal anomalies. | Variable SNHL (conductive, SNHL, or mixed). |
TABLE: Syndromes associated with SNHL (Congenital)
B. Acquired SNHL:
Acquired SNHL develops after birth. It can have genetic (non-syndromic or syndromic) or non-genetic causes.
Common Causes of Acquired SNHL (CBME Focus):
Infections of Labyrinth:
Viral: Mumps, Measles, Rubella (congenital), Cytomegalovirus (CMV, congenital or acquired), Herpes Zoster Oticus (Ramsay Hunt syndrome – often unilateral, with facial paralysis and vesicles).
Bacterial: Labyrinthitis as a complication of otitis media or meningitis.
Spirochaetal: Syphilis, Lyme disease.
Noise-Induced Hearing Loss (NIHL):
Chronic exposure to loud noise (e.g., occupational, recreational). Typically presents as a notch at 4000 Hz on audiometry.
Acoustic trauma (single loud blast).
Ototoxic Drugs:
Aminoglycosides (e.g., gentamicin, amikacin) – affect high frequencies initially.
Loop diuretics (e.g., Furosemide) – usually reversible.
Salicylates (e.g., aspirin in high doses) – usually reversible.
Chemotherapeutic agents (e.g., cisplatin, carboplatin).
Antimalarials (e.g., quinine, chloroquine).
Presbycusis:Age-related hearing loss. The most common cause of SNHL in adults. Typically bilateral, symmetrical, and affecting high frequencies first.
Sudden Sensorineural Hearing Loss (SSNHL):Acute onset SNHL (≥30 dB loss over ≥3 contiguous frequencies within 72 hours). Often idiopathic. Requires urgent evaluation.
Meniere’s Disease:Characterised by episodic vertigo, fluctuating SNHL (typically low-frequency), tinnitus, and aural fullness. Due to endolymphatic hydrops.
Acoustic Neuroma (Vestibular Schwannoma):Benign tumour of the vestibular division of the 8th cranial nerve. Typically causes unilateral, progressive SNHL, often with tinnitus and disequilibrium. Asymmetrical SNHL mandates an MRI brain with IAC protocol.
Trauma to Labyrinth or VIIIth Nerve:
Temporal bone fractures (especially trans-labyrinthine fractures).
Barotrauma (e.g., diving injuries leading to perilymph fistula).
Familial Progressive SNHL:Genetically inherited SNHL that develops and progresses over time.
Systemic Disorders:
Metabolic: Diabetes Mellitus, Hypothyroidism, Kidney disease.
Autoimmune: Systemic lupus erythematosus (SLE), Rheumatoid arthritis, Cogan’s syndrome.
Vascular: Vasculitis, stroke.
Neurological: Multiple Sclerosis.
Haematological: Sickle cell anaemia, polycythemia.
Table: Causes of Bilateral vs. Unilateral SNHL (CBME Focus)
| Causes of Bilateral SNHL | Causes of Unilateral SNHL |
--- |
| Presbycusis | Chronic Suppurative Otitis Media (CSOM) - if labyrinth involved |
| Noise-induced hearing loss | Mumps (can be unilateral SNHL) |
| Ototoxic drugs | Herpes Zoster Oticus |
| Systemic disorders (e.g., diabetes, autoimmune) | Acoustic Neuroma (Vestibular Schwannoma) |
| Familial progressive SNHL | Sudden Sensorineural Hearing Loss (often unilateral) |
| Congenital SNHL (syndromic/non-syndromic) | Meniere's Disease (initially unilateral, can become bilateral) |
| Bilateral temporal bone fractures | Unilateral temporal bone fracture |
Management of Sensorineural Hearing Loss:
For most SNHL, medical or surgical interventions are very limited, as the damage to the cochlea or nerve is often irreversible.
Hearing Aids: The primary mode of rehabilitation for most SNHL. These amplify sound to compensate for the reduced sensitivity.
Cochlear Implants: For severe-to-profound SNHL where hearing aids provide insufficient benefit. These surgically implanted devices bypass the damaged cochlea and directly stimulate the auditory nerve.
Auditory Brainstem Implants (ABIs): For SNHL caused by damage to the auditory nerve (e.g., bilateral acoustic neuromas due to Neurofibromatosis Type 2), bypassing the nerve and directly stimulating the brainstem.
Medical Management (for specific causes):
Sudden SNHL: Oral corticosteroids (often with intratympanic steroids).
Meniere’s Disease: Dietary modifications (low salt), diuretics, intratympanic steroids/gentamicin, sacculotomy, endolymphatic sac decompression.
Autoimmune inner ear disease: Immunosuppressants (e.g., steroids).
Infections: Appropriate antimicrobials.
Acoustic Neuroma: Observation, Stereotactic Radiosurgery, or Surgical Excision.
Aural Rehabilitation: Speech therapy, auditory training, communication strategies.
3. Mixed Hearing Loss. Mixed hearing loss is diagnosed when both conductive and sensorineural components coexist in the same ear. This implies a problem in sound conduction to the inner ear and a problem within the inner ear or auditory nerve itself.
Identification on Audiometry:Mixed hearing loss is characterised by:
The presence of an air-bone gap (ABG > 15-20 dB), indicating a conductive component.
Impaired bone conduction thresholds (BC > 20 dB HL), indicating a sensorineural component.
Therefore, in mixed hearing loss, both AC and BC thresholds are elevated, but AC thresholds are significantly poorer than BC thresholds, creating the ABG.
Common Causes of Mixed Hearing Loss: This type often arises from conditions that can affect both conductive and sensorineural mechanisms.
Otosclerosis with Cochlear Involvement: While primarily conductive (stapes fixation), prolonged otosclerosis can sometimes lead to secondary cochlear damage (cochlear otosclerosis).
Chronic Suppurative Otitis Media (CSOM) with Labyrinthine Fistula/Involvement: Chronic middle ear infection can erode ossicles (conductive component) and, if it extends to the labyrinth (inner ear), cause sensorineural damage.
Temporal Bone Fracture: Can cause ossicular disruption (conductive) and inner ear/nerve damage (sensorineural).
Head Trauma: Similar to temporal bone fractures.
Ageing with Earwax Impaction: Presbycusis (SNHL) coexisting with cerumen impaction (CHL).
Management of Mixed Hearing Loss:Management involves addressing both components where possible.
Prioritise Conductive Component: Often, addressing the conductive component first (e.g., surgery for CSOM, stapedectomy for otosclerosis) can significantly improve hearing.
Address Sensorineural Component: Hearing aids or cochlear implants may be necessary if the SNHL component is significant.
4. Non-Organic / Functional Hearing Loss. Functional hearing loss, also known as non-organic hearing loss, describes a situation where an individual exaggerates or feigns hearing impairment without a corresponding organic pathology that fully explains the observed hearing thresholds. Various terms have describe this phenomenon, including pseudo-hyperacusis, malingering(conscious feigning for gain),and factitious hearing loss(unconscious feigning). However, it is crucial to recognise that in many instances, individuals may indeed possess an underlying organic hearing loss that is then intentionally or unintentionally exaggerated. In other cases, particularly following trauma or for compensation claims, the individual may entirely feign hearing loss. Given the potential presence of an organic component, functional hearing loss is best understood as an exaggerated hearing loss or a functional overlay to an existing organic loss.
Clinical Presentations:Patients may present in any of the following three scenarios:
Total hearing loss in both ears
Total hearing loss in one ear
Exaggerated hearing loss in one or both ears
1. High Index of Suspicion:Watch for exaggerated behaviours, such as frequently requesting repetition of questions, cupping the ear excessively and inconsistent complaints with clinical findings. These actions raise red flags for functional overlay.
2. Inconsistent Audiometry Results:In pure tone and speech audiometry, normal variation on repeat testing is ±5 dB. A variation greater than 15 dB strongly suggests NOHL. Audiometric Zig-zag” or “Saucer” Patternis sometimes seen on PTA.
3. Absence of Shadow Curve in Bone Conduction.In unilateral deafness without masking, a shadow curve should appear on bone conduction due to transcranial sound transmission. If this shadow curve is absent, despite claims of total unilateral hearing loss, it is diagnostic of functional hearing loss.
4. Discrepancy Between PTA and SRT.Normally, the Pure Tone Average (PTA) at 500, 1000, and 2000 Hz closely matches the Speech Reception Threshold (SRT) (within 10 dB). If SRT is better than PTA by more than 10 dB, this suggests exaggeration or malingering.
5. Stenger Test (Classic Clinical Test)
Principle:
When the same tone is played in both ears with different intensities, the patient will only perceive the louder tone.
Procedure:Always blindfold the patient during the test to avoid bias.
Take two identical tuning forks or use a two-channel audiometer.
Strike the forks and hold both 25 cm away from each ear.
The patient will report hearing in the normal ear.
Now, bring the tuning fork closer (8 cm) to the suspected ear, keeping the normal side unchanged.
A malingering patient will now claim to hear nothing, even though the louder stimulus should be heard in the “deaf” ear.
A truly deaf patient will continue hearing in the normal ear.
Interpretation:
Positive Stenger (patient says he hears nothing): Indicates malingering.
Negative Stenger (hears in one ear): Likely genuine loss.
6. Acoustic Reflex Threshold.Normally, the stapedial (acoustic) reflex is present at 70–100 dB SL. If a patient claims total deafness but the reflex is still elicited, it clearly indicates functional hearing and NOHL.
7. Electric Response Audiometry (ERA).ERA evaluates auditory evoked potentials, bypassing the patient’s conscious effort. It can accurately estimate the hearing threshold within 5–10 dB, even in cases of suspected malingering. Highly reliable in children, non-cooperative patients, and suspected NOHL cases.
Management:
Requires a sensitive, non-confrontational approach.
Counseling and psychological support are often beneficial.
Reassurance that hearing is better than reported.
5. Central Hearing Loss / Auditory Processing Disorder (APD).Individuals with central hearing loss, more accurately termed Auditory Processing Disorder (APD), typically have normal pure-tone hearing thresholds and pass other standard hearing tests. However, they exhibit difficulties in processing distorted or unclear speech, particularly in challenging listening environments such as the presence of background noise (e.g., in social settings or classrooms). Consequently, these individuals struggle to follow verbal instructions, understand rapid speech, or distinguish similar-sounding words. These challenges are attributed to deficits in the processing of auditory information within the higher auditory pathways of the brain, rather than problems with the peripheral auditory system (ear or auditory nerve).
Key Features:
Normal pure tone audiogram.
Difficulty understanding speech, especially in noise.
Difficulty localising sound.
Problems with auditory memory or sequencing.
Can hear, but cannot understand.
Diagnosis:
Specialised audiometric tests are designed to assess central auditory function (e.g., speech-in-noise tests, dichotic listening tests).
Management:
Auditory training.
Environmental modifications (e.g., reducing background noise).
Compensatory strategies.
Frequency modulation (FM) systems in educational settings.
————End of the chapter————
High-Yield Points for NEET PG and University Exams
WHO Classification: Based on pure tone average (500, 1000, 2000 Hz) in the better ear.
Conductive Hearing Loss (CHL): Problem in sound transmission (external ear to stapediovestibular joint).
Rinne Test: Negative (BC > AC).
Weber Test: Lateralizes to the affected (poorer) ear.
BC: Normal (≤ 20 dB HL).
Low Frequencies are more affected.
Speech Discrimination: Good.
Max CHL: Around 60 dB HL.
Causes: Cerumen, TM perforation, OME, Otosclerosis, Ossicular discontinuity.
Sensorineural Hearing Loss (SNHL): Problem in cochlea or 8th nerve/central pathways.
Rinne Test: Positive (AC > BC), but both are reduced.
Weber Test: Lateralizes to the unaffected (better) ear.
BC: Reduced.
High Frequencies are usually more affected.
Speech Discrimination: Poor.
Recruitment: Common in cochlear SNHL.
Tympanometry: Normal (Type A).
Causes: Presbycusis, NIHL (4KHz notch), Ototoxic drugs, SSNHL, Meniere’s, Acoustic Neuroma (unilateral SNHL, requires MRI), Congenital syndromes (Waardenburg, Usher, Pendred).
Mixed Hearing Loss: Both conductive (ABG) and sensorineural (elevated BC) components. Common in Otosclerosis with cochlear involvement, CSOM with labyrinthine involvement, and Temporal bone fracture.
Non-Organic Hearing Loss: Inconsistent audiometric results; objective tests (ABR, OAE) may be normal.
Central Hearing Loss (APD): Normal PTA, but difficulty processing speech, especially in noise.
Multiple Choice Questions (MCQs)
Instructions: Choose the single best answer for each question.
A 55-year-old male presents with difficulty understanding speech in noisy environments, although his pure tone audiogram shows normal hearing thresholds. Which type of hearing loss is most likely present? a) Conductive hearing loss b) Sensorineural hearing loss c) Mixed hearing loss d) Central hearing loss e) Non-organic hearing loss
A patient undergoes a Rinne test, and the tuning fork is heard louder by bone conduction than by air conduction in the right ear. The Weber test lateralizes to the right ear. What is the most likely diagnosis? a) Right sensorineural hearing loss b) Left sensorineural hearing loss c) Right conductive hearing loss d) Left conductive hearing loss e) Bilateral mixed hearing loss
Which of the following conditions is most likely to cause a typical 4000 Hz notch on a pure tone audiogram? a) Otosclerosis b) Presbycusis c) Noise-induced hearing loss d) Meniere’s disease e) Chronic otitis media with effusion
A 30-year-old female presents with fluctuating hearing loss, episodic vertigo, tinnitus, and aural fullness. Which of the following conditions is characterised by this constellation of symptoms? a) Acoustic neuroma b) Otosclerosis c) Sudden sensorineural hearing loss d) Meniere’s disease e) Presbycusis
In pure conductive hearing loss, the bone conduction (BC) thresholds on pure tone audiometry are typically: a) Greater than 60 dB HL b) Equal to air conduction (AC) thresholds c) Significantly elevated (poor hearing) d) Normal (equal to or better than 20 dB HL) e) Absent
Which of the following is a characteristic feature of sensorineural hearing loss but is usually absent in pure conductive hearing loss? a) Negative Rinne test b) Weber lateralizes to the poorer ear c) Good speech discrimination d) Recruitment phenomenon e) Max hearing loss of 60 dB
A child presents with a white forelock, widely spaced eyes, and bilateral profound hearing loss. This presentation is most consistent with which syndrome? a) Usher Syndrome b) Pendred Syndrome c) Alport Syndrome d) Waardenburg Syndrome e) Jervell and Lange-Nielsen Syndrome
An elderly patient with known diabetes mellitus is prescribed a high dose of an antibiotic for a severe infection. Following treatment, they report worsening hearing. Which class of antibiotics is most notorious for causing ototoxicity? a) Penicillins b) Cephalosporins c) Macrolides d) Aminoglycosides e) Fluoroquinolones
An audiogram shows elevated air conduction thresholds with an air-bone gap of 20 dB across most frequencies, and elevated bone conduction thresholds (greater than 20 dB HL). This pattern indicates: a) Pure conductive hearing loss b) Pure sensorineural hearing loss c) Mixed hearing loss d) Non-organic hearing loss e) Central hearing loss
The management of severe-to-profound sensorineural hearing loss in children who receive inadequate benefit from hearing aids often involves: a) Myringotomy and grommet insertion b) Stapedectomy c) Cochlear implantation d) Ossiculoplasty e) Tympanoplasty
MCQ Answers and Explanations:
d) Central hearing loss
Explanation: Normal pure tone thresholds with difficulty processing speech, especially in noise, are classic symptoms of central hearing loss or Auditory Processing Disorder (APD).
c) Right Conductive hearing loss
Explanation: A negative Rinne (BC > AC) indicates a conductive component. Weber lateralizing to the poorer ear (right in this case) confirms a conductive loss in that ear.
c) Noise-induced hearing loss
Explanation: Noise-induced hearing loss characteristically presents with a “notch” at 4000 Hz (or sometimes 3000-6000 Hz) due to specific damage to the hair cells in that frequency region of the cochlea.
d) Meniere’s disease
Explanation: The classic triad of fluctuating SNHL, episodic vertigo, and tinnitus with aural fullness is diagnostic of Meniere’s disease.
d) Normal (equal to or better than 20 dB HL)
Explanation: In pure conductive hearing loss, the inner ear and auditory nerve function normally, so bone conduction (which bypasses the conductive mechanism) will be within normal limits.
d) Recruitment phenomenon
Explanation: Recruitment is an abnormal growth in loudness perception, a characteristic feature of cochlear (sensory) SNHL. Conductive hearing loss typically does not exhibit recruitment.
d) Waardenburg Syndrome
Explanation: Waardenburg Syndrome is characterised by pigmentation abnormalities (e.g., white forelock, heterochromia iridis) and sensorineural hearing loss.
d) Aminoglycosides
Explanation: Aminoglycoside antibiotics (e.g., gentamicin, amikacin, streptomycin) are well-known for their ototoxic effects, causing irreversible sensorineural hearing loss, particularly affecting high frequencies and vestibular function.
c) Mixed hearing loss
Explanation: An air-bone gap (AC > BC) indicates a conductive component, while elevated bone conduction thresholds (BC > 20 dB HL) indicate a sensorineural component. The presence of both defines mixed hearing loss.
c) Cochlear implantation
Explanation: For severe-to-profound SNHL where conventional hearing aids are insufficient, cochlear implants are the gold standard to provide auditory input by directly stimulating the auditory nerve.
Clinical-Based Questions (CBQs)
Instructions: Apply your knowledge of hearing loss types to the following patient scenarios.
Scenario: A 6-year-old child is brought to the ENT clinic by his parents, who report that he frequently asks for repetition, struggles to hear in school, and has had recurrent ear infections. On otoscopy, bilateral dull tympanic membranes with decreased mobility are noted. Tympanometry shows bilateral Type B curves.
a) What type of hearing loss is most likely present?
Answer: Conductive hearing loss.
b) What is the most probable diagnosis for the underlying cause?
Answer: Otitis Media with Effusion (OME), also known as “glue ear.”
c) Briefly outline the initial management strategy for this child.
Answer: Initial management often involves watchful waiting (OME can resolve spontaneously), considering decongestants/antihistamines (though evidence is limited for efficacy), and if persistent with significant hearing loss affecting development, myringotomy with grommet insertion to drain fluid and equalise middle ear pressure.
Scenario: A 70-year-old retired factory worker complains of progressive difficulty hearing, especially conversations in crowded places like restaurants. His wife often says he turns the TV volume too high. He denies ear pain or discharge. Pure tone audiometry reveals bilateral, symmetrical high-frequency sensorineural hearing loss.
a) What is the most likely diagnosis for his hearing loss?
Answer: Presbycusis (age-related hearing loss), likely exacerbated by noise-induced hearing loss from his occupational history.
b) Explain why he struggles in noisy environments despite perhaps “hearing” sounds.
Answer: High-frequency SNHL impairs speech discrimination, particularly consonant sounds (which carry much of speech clarity). In noisy environments, the brain struggles to filter out background noise and focus on the degraded speech signal, making understanding very challenging. This is a classic symptom of SNHL.
c) What are the primary rehabilitation options for this patient?
Answer: The primary rehabilitation option is hearing aids, specifically programmed to compensate for his high-frequency loss. Aural rehabilitation (auditory training, communication strategies) can also be very beneficial.
Scenario: A 40-year-old male suddenly experienced severe dizziness, ringing in his left ear, and a feeling of fullness in the ear this morning. He also reports a significant decrease in hearing on that side. There’s no history of head trauma or ear infection.
a) What is the most probable diagnosis given this constellation of acute symptoms?
Answer: Sudden Sensorineural Hearing Loss (SSNHL) associated with Meniere’s disease (given the vertigo, tinnitus, fullness, and fluctuating nature implied). Or, less likely but still a consideration, isolated SSNHL (idiopathic).
b) What urgent diagnostic test is crucial to rule out a potentially serious retrocochlear pathology?
Answer: An urgent MRI of the brain with contrast, focusing on the internal auditory canals (IACs), is crucial to rule out an acoustic neuroma (vestibular schwannoma) or other retrocochlear lesions, especially given the unilateral SNHL and vestibular symptoms.
c) What is the immediate medical management consideration for this patient if SSNHL is confirmed?
Answer: High-dose oral corticosteroids (e.g., prednisone taper) should be initiated as soon as possible, ideally within 72 hours of onset, as this can improve prognosis for hearing recovery. Intratympanic steroid injections are also an option.
Scenario: A 10-year-old boy presents with a history of inconsistent responses to sound. His parents report that he seems to hear but not others, and sometimes gives exaggerated responses. Pure tone audiometry shows varying thresholds on repeat testing, and the pattern does not fit a typical organic hearing loss. Objective tests like OAEs are normal.
a) What type of hearing loss should be suspected?
Answer: Non-organic (Functional) hearing loss.
b) Why are consistent test results important in diagnosing hearing loss, and what do inconsistent results suggest?
Answer: Consistent test results are crucial because they reflect a stable and predictable auditory response from a true organic pathology. Inconsistent results suggest a non-organic component, where the patient’s responses are not genuinely indicative of their hearing ability, often implying exaggeration or feigning.
c) How should a clinician approach the management of this patient?
Answer: The clinician should approach this with a non-confrontational, empathetic, and supportive manner. The management involves carefully explaining the test results to the parents and child, providing reassurance about their hearing ability, and considering counselling or psychological support if underlying psychosocial factors are suspected. The focus is on guiding them toward more consistent and appropriate responses.
Frequently Asked Questions (FAQs)
Q: What is the difference between conductive and sensorineural hearing loss?
A: Conductive hearing loss results from problems in sound transmission to the inner ear, while sensorineural hearing loss arises from damage to the inner ear (cochlea) or the auditory nerve.
Q: What is the significance of a negative Rinne test?
A: A negative Rinne test (bone conduction heard louder than air conduction) is a hallmark sign of conductive hearing loss.
Q: How does the Weber test help differentiate hearing loss types?
A: In conductive hearing loss, the Weber test lateralizes (sound heard louder) to the poorer ear; in sensorineural hearing loss, it lateralizes to the better ear.
Q: What does an air-bone gap on an audiogram indicate?
A: An air-bone gap (AC thresholds significantly worse than BC thresholds) indicates the presence of a conductive component to the hearing loss.
Q: Can hearing loss be cured?
A: Conductive hearing loss is often medically or surgically treatable, but sensorineural hearing loss is typically permanent and managed with hearing aids or cochlear implants, though some acute cases (like SSNHL) may respond to steroids.
Q: What are some common causes of sensorineural hearing loss in adults?
A: Common causes include age-related hearing loss (presbycusis), noise exposure (noise-induced hearing loss), ototoxic medications, and Meniere’s disease.
Q: Why is it important to investigate unilateral sensorineural hearing loss?
A: Unilateral sensorineural hearing loss, especially if progressive or associated with other symptoms, must be thoroughly investigated with an MRI to rule out a retrocochlear lesion like an acoustic neuroma.
Practical Tips for Viva and Clinical Exam Scenarios
Presenting Hearing Loss in Viva:
Define Hearing Loss: Start with a clear, concise definition and mention the WHO classification briefly.
Categorisation: Immediately list the five types.
Clinical Tools: Emphasise the role of Tuning Fork Tests (Rinne, Weber) and Pure Tone Audiometry (PTA) as foundational diagnostic tools.
Systematic Approach per Type: For each type:
Definition/Mechanism: How does it occur?
Clinical Characteristics: How does the patient present (e.g., speech discrimination, difficulty in noise)?
Tuning Fork Findings: Crucial for viva.
Audiometric Profile (PTA): What does the audiogram look like (ABG, BC thresholds, frequency affected)?
Common Causes: List the most frequent etiologies.
Management: Briefly outline key medical/surgical options and rehabilitation.
Focus on Differentiation: Be prepared to compare and contrast CHL vs. SNHL repeatedly.
Be Ready to Draw: Practice drawing basic audiograms for each type of hearing loss.
Common Examiner Questions (Viva):
“Differentiate between conductive and sensorineural hearing loss using Rinne and Weber tests.”
“Describe the audiometric findings in mixed hearing loss.”
“Enumerate the causes of conductive hearing loss.”
“What is recruitment? In which type of hearing loss is it seen?”
“What are the red flags for unilateral SNHL?” (Acoustic neuroma!)
“Name five syndromes associated with congenital SNHL.”
“How do you manage a patient with profound bilateral SNHL who doesn’t benefit from hearing aids?” (Cochlear implant).
“Explain the 4000 Hz notch.”
“What is functional hearing loss, and how do you diagnose it?”
Clinical Exam Scenarios (Applying Knowledge):
Patient Presentation: “A patient comes to you with X symptoms. What initial tests would you do, and what type of hearing loss are you suspecting?” (E.g., “Child with speech delay and recurrent cold -> suspect OME -> otoscopy, tympanometry”).
Audiogram Interpretation: “Here is an audiogram. Interpret it and give me the type and degree of hearing loss.” (Practice reading audiograms!)
Management Plan: “How would you manage this patient with [specific type] hearing loss?” (Think medical, surgical, amplification, rehabilitation).
Counselling: “How would you counsel a patient with presbycusis about their hearing loss?” (Emphasise hearing aids, communication strategies, managing expectations).
————End————
Download full PDF Link:
Types of Hearing Loss Best Lecture Notes Dr Rahul Bagla ENT Textbook
Reference Textbooks.
Scott-Brown, Textbook of Otorhinolaryngology-Head and Neck Surgery.
Glasscock-Shambaugh, Textbook of Surgery of the Ear.
P L Dhingra, Textbook of Diseases of Ear, Nose and Throat.
Hazarika P, Textbook of Ear Nose Throat And Head Neck Surgery Clinical Practical.
Mohan Bansal, Textbook of Diseases of Ear, Nose and Throat Head and Neck Surgery
Hans Behrbohm, Textbook of Ear, Nose, and Throat Diseases With Head and Neck Surgery.
Salah Mansour, Middle Ear Diseases – Advances in Diagnosis and Management.
Logan Turner, Textbook of Diseases of The Nose, Throat and Ear Head And Neck Surgery.
Rob and smith, Textbook of Operative surgery.
Anirban Biswas, Textbook of Clinical Audio-vestibulometry.
Arnold, U. Ganzer, Textbook of Otorhinolaryngology, Head and Neck Surgery.
Author:
Dr. Rahul Bagla
MBBS (MAMC, Delhi) MS ENT (UCMS, Delhi)
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Mail: msrahulbagla@gmail.com
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Keywords: Understand the different types of hearing loss – conductive, sensorineural, mixed, non-organic, and central. This detailed guide, optimised for MBBS and ENT PG students, covers causes, characteristics, diagnosis, and management, aligning with the CBME curriculum for university exams, viva, practicals, and NEET PG. Types of Hearing Loss, Conductive Hearing Loss, Sensorineural Hearing Loss, Mixed Hearing Loss, Functional Hearing Loss, Central Hearing Loss, Auditory Processing Disorder, WHO Hearing Loss Classification, Pure Tone Audiometry Interpretation, Rinne Test Weber Test, Air-Bone Gap Audiogram, Causes of Hearing Loss, Management of Hearing Loss, Otosclerosis Hearing Loss, Noise-Induced Hearing Loss, Presbycusis Symptoms, Sudden Sensorineural Hearing Loss, Meniere’s Disease Symptoms, Acoustic Neuroma Hearing Loss, Congenital SNHL Syndromes, Ototoxic Drugs Hearing Loss, ENT Hearing Loss MBBS, ENT PG Hearing Loss, NEET PG ENT Topics, Hearing Loss Viva Questions, Clinical Scenarios Hearing Loss, Hearing Loss Treatment Options, Hearing Aids Cochlear Implants, CBME ENT Curriculum Hearing Loss, Causes of conductive hearing loss, conductive hearing loss treatment options, diagnosing conductive hearing loss, conductive vs sensorineural hearing loss differences, sensorineural hearing loss symptoms, inner ear hearing loss causes, sensorineural hearing loss management, reversible sensorineural hearing loss, what is mixed hearing loss, causes of mixed hearing loss, treating mixed hearing loss, functional hearing loss definition, diagnosing non-organic hearing loss, management of functional hearing loss, central auditory processing disorder causes, central auditory processing disorder symptoms, auditory processing disorder diagnosis, central hearing loss treatment options, different types of hearing loss explained for students, audiological classification of hearing loss
2 Comments
osama alokabi October 1, 2024 at 12:50 pmReply Thanks for this 🙏 discussion..
I want to ask how can we know Central form neural by Audiometry
2. Zeba khan July 19, 2025 at 6:45 amReply Easy to understand. Amazing book.
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How do you create an alternating series with the sign being the same twice in a row?
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I am working on a Taylor series question and I have created a series which alternates however, it does so in doubles. in other words it follows the following pattern: x, x, −x, −x, x, x,...
My goal is to write this series using summation notation. I know that to produce a normal alternative series I simply use (−1)n. However, I've found this significantly harder to create in summation notation because any manipulation to the (−1)n part of the sum only moves the position where the alternating series starts. Any suggestions are appreciated!
edit: please note that in my example, "x" is just an arbitrary function and each other x does not equal the other x's.
calculus
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edited Apr 18, 2016 at 4:15
Martin Sleziak
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asked Apr 16, 2016 at 20:44
user316719user316719
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2 How about (−1)⌊n 2⌋?almagest –almagest 2016-04-16 20:49:23 +00:00 Commented Apr 16, 2016 at 20:49
Try using a variation of sin or cos.Noble Mushtak –Noble Mushtak 2016-04-16 20:49:57 +00:00 Commented Apr 16, 2016 at 20:49
@NobleMushtak my taylor series is in fact just the one for the function sin(x) centered at PI/4. I'm not certain how to write this in summation notation.user316719 –user316719 2016-04-16 20:52:00 +00:00 Commented Apr 16, 2016 at 20:52
@almagest yes thank you, this works!user316719 –user316719 2016-04-16 20:53:40 +00:00 Commented Apr 16, 2016 at 20:53
3 As an alternative, depending on the form of your summands it may be worthwhile to groups the sum into pairs, i.e. for f(0)+f(1)−f(2)−f(3)+…, you could use ∑i(−1)i(f(2 i)+f(2 i+1)).Klaus Draeger –Klaus Draeger 2016-04-16 23:11:41 +00:00 Commented Apr 16, 2016 at 23:11
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If you are looking for a formula for the sequence 1,1,−1,−1,1,1,−1,−1,… then try some of the suggestions at OEIS A057077, such as (−1)n(n−1)/2
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answered Apr 16, 2016 at 20:52
HenryHenry
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Suppose I take some simple series and use these signs with them, do I get something well-known? I was thinking ∞∑n=1(−1)n(n−1)/2 n i.e. the harmonic series taken with signs like this, is the sum of this series known?Jeppe Stig Nielsen –Jeppe Stig Nielsen 2016-04-16 23:41:36 +00:00 Commented Apr 16, 2016 at 23:41
Well, if we group every second term we get stuff like ∑∞m=1(−1)m−1(1 2 m−1±1 2 m) which can be handled by Wolfram Alpha, so the answer is yes.Jeppe Stig Nielsen –Jeppe Stig Nielsen 2016-04-17 00:11:12 +00:00 Commented Apr 17, 2016 at 0:11
3 This answer is right. It's (-1) raised to the power of the sum from 1 to n. Note that for any given n, the sum from 1 to n has sign k (either + or -). The next odd number added to the sum changes the sign, the next even number keeps the sign the same: so you get two odds, then two evens, then two odds, then two evens....JimZipCode –JimZipCode 2016-04-17 03:59:20 +00:00 Commented Apr 17, 2016 at 3:59
@JimZipCode nice way of looking at it :)Ant –Ant 2016-04-17 09:50:07 +00:00 Commented Apr 17, 2016 at 9:50
∞∑n=1(−1)n(n−1)/2 n is conditionally convergent to π 4−log 2(2)2≈0.4388 Henry –Henry 2025-07-31 18:21:50 +00:00 Commented Jul 31 at 18:21
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It seems that you want to simulate the series 1,1,−1,−1,1,1,−1,−1,.... These kinds of alternating series are usually easiest to do using sin or cos functions. (Although the formula (−1)n(n−1)/2 works just as well, I think it's kind of interesting to see this series modeled with a trig function, so I'm going to do it anyway.)
Think about a cos function going through (0,1), (1,1), (2,−1), (3,−1), (4,1) and so on.
Where are the maximums and minimums? They're in between points with the same values, like 0+1 2=0.5 and 2+3 2=2.5.
Which is the maximum? It's the one in between the points of higher value, which is 0.5. This represents our right shift since cos 0 is the maximum.
What's the period? It's two times the difference between two consecutive extrema, or 2(2.5−0.5)=4. The period is usually 2 π, so our coefficient of x must be 2 π 4=π 2.
Thus, we have cos(π 2(x−1 2)). However, this gives us 1√2 for x=0, so we must multiply by √2 to get the following function:
f(x)=√2 cos(π 2(x−1 2))
If we put this into WolframAlpha, we clearly see that this gives us the sequence we wanted.
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edited Apr 17, 2016 at 12:01
answered Apr 16, 2016 at 21:01
Noble MushtakNoble Mushtak
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There is a very general way to attack the problem of writing a repeating sequence of coefficients.
The discrete Fourier transform of the sequence "1,1,−1,−1", of length N=4, is 0,1+i,0,1−i". This says that 0 e−2 π i⋅0 n/N+(1+i)e−2 π i⋅1 n/N+0 e−2 π i⋅2 n/N+(1−i)e−2 π i⋅3 n/N is a constant multiple of your desired sequence. ("Constant multiple" because there are constant premultipliers I have ignored to make the above more direct.) That expression, in which we may ignore the two terms starting with zero, produces a sequence starting from n=0: 2,2,−2,−2,2,2,−2,−2,…. So, (1+i)e−2 π i⋅1 n/4+(1−i)e−2 π i⋅3 n/4 2 does what you want.
Just as for (−1)n, we may rewrite this using trigonometric functions. (cos(π n 2)−sin(π n 2))(cos(π n)−i sin(π n))
Say, instead of starting at n=0, you want to start at n=1. We cyclically rotate the sequence to get the term we want at the position where n=1, i.e., we consider the sequence −1,1,1,−1. This gives transform 0,−1+i,0,−1−i. The term to use in your sum is then (−1+i)e−2 π i⋅1 n/4+(−1−i)e−2 π i⋅3 n/4 2, which is equivalent to (sin(π n 2)+cos(π n 2))(−cos(π n)+i sin(π n)).
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answered Apr 17, 2016 at 3:58
Eric TowersEric Towers
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Just for kicks:
f(n)=lim
This can be used to generalize sequences like:
\underbrace{1,1,1,\cdots,1}{k \space1's},\underbrace{-1,-1,-1,\cdots,-1}{k \space 1's}, 1\cdots
Can be represented as
f(n)=\lim_{x\to n_-} \left\lfloor{\sin \frac{\pi x}{k}}\right\rfloor
I thought that this might a bit cool for the OP.
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edited Apr 17, 2016 at 1:55
answered Apr 17, 2016 at 1:44
user331275 user331275
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3 \begingroup Very creative (+1)\endgroup user331360 –user331360 2016-04-17 01:46:27 +00:00 Commented Apr 17, 2016 at 1:46
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\begingroup
\frac{i^n+i^{-n}}2={1,0,-1,0,\dots} has period 4, so putting it together with itself shifted by one gives {1,1,-1,-1,\dots}=\overbrace{\frac{i^n+i^{-n}}2}^{1,0,-1,0,\dots}+\overbrace{\frac{i^{n-1}+i^{1-n}}2}^{0,1,0,-1,\dots}\tag{1}
Using the series -\log(1-x)=\sum\limits_{n=1}^\infty\frac{x^n}n, we get \begin{align} \frac11-\frac12-\frac13+\frac14+\cdots &=\sum_{n=1}^\infty\left(\frac{i^n+i^{-n}}2+\frac{i^{n-1}+i^{1-n}}2\right)\frac1n\ &=-\frac12\log(1-i)-\frac12\log(1+i)+\frac i2\log(1-i)-\frac i2\log(1+i)\ &=-\frac12\log(2)+\frac i2\log(-i)\[6pt] &=\frac{\pi-\log(4)}4\tag{2} \end{align}
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edited Apr 17, 2016 at 20:42
answered Apr 17, 2016 at 2:38
robjohn♦robjohn
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\begingroup would the downvoter care to comment?\endgroup robjohn –robjohn♦ 2016-04-17 11:34:16 +00:00 Commented Apr 17, 2016 at 11:34
\begingroup I didn't downvote, but it is difficult to see what everything after "Thus, using the series…" has to do with the question.\endgroup Nick Matteo –Nick Matteo 2016-04-17 18:26:23 +00:00 Commented Apr 17, 2016 at 18:26
\begingroup The question is answered in (1). After that, I was applying this representation to the question posed in a comment by Jeppe Stig Nielsen about the usability of the representation (-1)^{n(n-1)/2} in a series like \sum_{n=1}^\infty\frac{(-1)^{n(n-1)/2}}{n} In other words, (2) is attempting to demonstrate usefulness.\endgroup robjohn –robjohn♦ 2016-04-17 18:44:38 +00:00 Commented Apr 17, 2016 at 18:44
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\begingroup
Instead of trying to represent the sign sequence directly, you can also approach this problem by grouping you summands into pairs. So for a sum of the form
f(0)+f(1)-f(2)-f(3)+f(4)+\dots,
you would write
\sum_i(-1)^i(f(2i)+f(2i+1)).
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answered Apr 17, 2016 at 1:45
Klaus DraegerKlaus Draeger
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\begingroup
Here's a more "blunt force" way of doing this. Given a sequence like
1 + \frac12 - \frac13 -\frac14 + \frac15 + \frac16 - \frac17 - \frac18 + \cdots,
then you could notate it as follows:
\sum_{n=1}^\infty \frac{s(n)}{n}, where s(n) = \begin{cases} 1 & \text{if } n \mod 4 = 1 \operatorname{or} 2, \ -1 & \text{if } n \mod 4 = 3 \operatorname{or} 0\end{cases}
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answered Apr 19, 2016 at 1:47
Sophie SwettSophie Swett
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13532 | https://www.quora.com/A-12cm*12cm*12cm-wooden-cube-is-painted-3-faces-are-green-and-3-are-orange-This-cube-is-sliced-into-64-smaller-equal-sized-cubes-of-3cm*3cm*3cm-How-many-of-these-small-cubes-will-have-exactly-1-green-face-and-exactly-1-orange-face-in-the-same-cube | A 12cm12cm12cm wooden cube is painted 3 faces are green and 3 are orange. This cube is sliced into 64 smaller equal sized cubes of 3cm3cm3cm. How many of these small cubes will have exactly 1 green face and exactly 1 orange face in the same cube? - Quora
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5
A 12cm12cm12cm wooden cube is painted 3 faces are green and 3 are orange. This cube is sliced into 64 smaller equal sized cubes of 3cm3cm3cm. How many of these small cubes will have exactly 1 green face and exactly 1 orange face in the same cube?
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David Smith
BSc (Hons) in Mathematics&Computer Science, University of Bristol (Graduated 1986) · Author has 3.6K answers and 4M answer views
·7y
The answer is 12 or 16 depending on the painting arrangement of the original cube.
When cutting into 64 identically sized cubelets, we are interested in the 24 which sit in the middle of each edge: these are the ones which will be painted on two sides.
[The 8 corners will be painted on three sides, the 24 middles of face will be painted on one side and the 8 internal cubelets will not be painted at all.]
As David Pratt states, there are two distinct arrangements for the painted faces.
Left, front and right in orange; top, back and bottom in green.
Left, front and top in orange; right, back and bott
Continue Reading
The answer is 12 or 16 depending on the painting arrangement of the original cube.
When cutting into 64 identically sized cubelets, we are interested in the 24 which sit in the middle of each edge: these are the ones which will be painted on two sides.
[The 8 corners will be painted on three sides, the 24 middles of face will be painted on one side and the 8 internal cubelets will not be painted at all.]
As David Pratt states, there are two distinct arrangements for the painted faces.
Left, front and right in orange; top, back and bottom in green.
Left, front and top in orange; right, back and bottom in green.
All other arrangements can be reached simply by rotating the cube.
In painting arrangement 1: four of the original cube’s edges share the same colour on their faces (the left-front and front-right edges share orange faces and the top-back and back- bottom edges share green faces). The other eight edges have different colours on their faces. This gives us 16 cubelets which are painted one face orange and one face green.
In painting arrangement 2: six of the original cube’s edges share the same colour on their faces (the left-front, front-top and top-left edges share orange faces and the right-back, back- bottom and bottom-right edges share green faces). The other six edges have different colours on their faces. This gives us 12 cubelets which are painted one face orange and one face green.
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Anthony Madden
Writer for Betterbuck
·Updated Aug 15
What are the weirdest mistakes people make on the internet right now?
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr
Continue Reading
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice.
More often than not, it saves even more than 50% on load times - here’s a test I ran:
Using an ad blocker saved a whopping 6.5+ seconds of load time.
Here’s a link to Total Adblock, if you’re interested.
Not getting paid for your screentime
Apps like Freecash will pay you to test new games on your phone.
Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website):
You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS).
If you're scrolling on your phone anyway, why not get paid for it?
I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool).
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To solve this problem, we first need to analyze the painting and slicing of the cube.
Cube Dimensions and Slicing:
The original cube is 12 cm×12 cm×12 cm 12 cm×12 cm×12 cm.
It is sliced into 64 64 smaller cubes, each of size 3 cm×3 cm×3 cm 3 cm×3 cm×3 cm.
This means the original cube is divided into 4×4×4 4×4×4 smaller cubes (since 12/3=4 12/3=4).
Painting:
Three faces are painted green and three faces are painted orange.
We will assume that the three green faces and three orange faces are adjacent to each other (for
Continue Reading
To solve this problem, we first need to analyze the painting and slicing of the cube.
Cube Dimensions and Slicing:
The original cube is 12 cm×12 cm×12 cm 12 cm×12 cm×12 cm.
It is sliced into 64 64 smaller cubes, each of size 3 cm×3 cm×3 cm 3 cm×3 cm×3 cm.
This means the original cube is divided into 4×4×4 4×4×4 smaller cubes (since 12/3=4 12/3=4).
Painting:
Three faces are painted green and three faces are painted orange.
We will assume that the three green faces and three orange faces are adjacent to each other (for instance, the green faces could be the top, front, and left, while the orange faces could be the bottom, back, and right).
Finding Small Cubes with 1 Green Face and 1 Orange Face:
A smaller cube can have exactly 1 green face and 1 orange face if it is located at the edge where a green face meets an orange face.
The smaller cubes located at the edges of the original cube will have this property.
Identifying Relevant Edges:
Each edge of the cube has 3 3 smaller cubes (since the edge length is 12 cm 12 cm and each small cube is 3 cm 3 cm).
The edges that will have one green face and one orange face are those that connect a green face to an orange face.
Counting the Edges:
Each of the three green faces will share an edge with three orange faces.
Each of the three edges connecting a green face to an orange face has 1 1 middle small cube that has exactly 1 1 green face and 1 1 orange face.
Calculation:
There are 3 3 edges for each green face connecting to the orange faces.
Since each such edge contributes 1 1 small cube, and there are 3 3 such edges for each of the 3 3 green faces:
Total small cubes with 1 1 green face and 1 1 orange face = 3 edges×3 green faces=9 3 edges×3 green faces=9.
Thus, the number of smaller cubes that have exactly 1 green face and 1 orange face is 9.
Upvote ·
Santosh Kadam
Puzzle maniac · Author has 225 answers and 3.9M answer views
·7y
Originally Answered: If you have a cube with the dimensions 121212, you paint it green from the outside, you cut the cube in 64 smaller cubes with the dimensions 333, how many of them will have at least one face coloured? ·
56.
Total cubes are 64. Imagine you are holding cube. Now all cubes at top and bottom layers would be coloured. Each layer will contain 16(4 slices with 4 cubes) cubes so that 4 layers will sum up to 64. So from top and bottom layers we have
16+16=32.
Now from middle two layers. Out of 4 slices, slices at extreme ends will have all cubes colored atleast from one side. One layer contain four slices and each slice contain 4 cubes. So number of cubes from extreme slices of two middle layers are.
4+4+4+4=16.
Now for middle slices of middle layers. Only those cubes will be colored which are visible from
Continue Reading
56.
Total cubes are 64. Imagine you are holding cube. Now all cubes at top and bottom layers would be coloured. Each layer will contain 16(4 slices with 4 cubes) cubes so that 4 layers will sum up to 64. So from top and bottom layers we have
16+16=32.
Now from middle two layers. Out of 4 slices, slices at extreme ends will have all cubes colored atleast from one side. One layer contain four slices and each slice contain 4 cubes. So number of cubes from extreme slices of two middle layers are.
4+4+4+4=16.
Now for middle slices of middle layers. Only those cubes will be colored which are visible from outside. Each slice have only 2 such cubes which are present at both of ends. Now we have such 4 slices (2 from each layers). So total number of cubes
42=8.
After adding all of them we have
32+16+8= 56 cubes which are colored from atleast one side.
Upvote ·
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Related questions
More answers below
A wooden cube is 3 dm on each side and has its exterior painted is blue. If the cube is cut into smaller cubes about 1 dm long on its side, how many of the smaller cubes have exactly three blue faces?
A cube with sides of 5 cm is painted on all its faces. If sliced in 1 cm cubes, how many 1 cm cubes will have exactly one of their faces painted?
A cube is painted blue on all faces and it is cut into 64 small cubes of equal size. How many cubes are not painted?
A cube with all the sides painted was divided into small cubes of equal measurement. The side of a small cube is exactly one fourth as that of the big cube. Therefore, what is the number of small cubes with only one side painted?
Three different faces of a cube are painted in three different colours–red, green and blue. This cube is now cut into 216 smaller but identical cubes. What are the least and the largest numbers of small cubes that have exactly one face painted?
David Pratt
Former Electoral Registration Canvasser at London Borough of Hammersmith and Fulham (2009–2014) · Author has 3.5K answers and 4.5M answer views
·7y
There are two possible layouts of the coloured faces either as two lines of three faces (think in terms of the skin of a tennis ball) or as two “box reflectors” in the first case there are twelve edges that are junctions of green and orange faces in the second six. Each edge consists of ten cubes with two coloured faces ans the end pair with three coloured faces. Thus you will have either in the first case one hundred and twenty or ,in the second case, sixty cubes with exactly one green and one orange face.
Upvote ·
Priya Verma
Studied Mathematics (Graduated 2015)
·7y
Originally Answered: If you have a cube with the dimensions 121212, you paint it green from the outside, you cut the cube in 64 smaller cubes with the dimensions 333, how many of them will have at least one face coloured? ·
The cubes are coloured which are on surfaces.Each face has 9 cubes. so total
9×6=54cubes are coloured at least one face.
Upvote ·
9 1
Promoted by JH Simon
JH Simon
Author of 'How To Kill A Narcissist'
·Updated Fri
What is narcissistic abuse?
Narcissistic abuse is a funnel which leads to a state of mental and emotional claustrophobia.
At first, it appears to be a wide-open space filled with opportunity. But as you pour more and more energy into it, you feel the space around you decrease and the psychological walls converging.
First impressions last, so you find it difficult to forget the feelings of hope, warmth and possibility which dominated the beginning of the relationship. Because narcissistic abuse is enforced during everyday situations with a ‘loved one’, it is not consciously felt. Instead, you are shamed, ridiculed, disregar
Continue Reading
Narcissistic abuse is a funnel which leads to a state of mental and emotional claustrophobia.
At first, it appears to be a wide-open space filled with opportunity. But as you pour more and more energy into it, you feel the space around you decrease and the psychological walls converging.
First impressions last, so you find it difficult to forget the feelings of hope, warmth and possibility which dominated the beginning of the relationship. Because narcissistic abuse is enforced during everyday situations with a ‘loved one’, it is not consciously felt. Instead, you are shamed, ridiculed, disregarded, manipulated and frightened in ways that seem justified due to the narcissist’s shamelessness. Nobody would purposely do such things, would they? you ask.
Someone with healthy shame senses when they have wounded another person’s emotions, and therefore adapts their behaviour to restore pride to the other. A shamed person exhibits signs of it, including in their facial expression, their posture, and their subsequent behaviour. Anyone with a healthy sense of shame knows when someone is experiencing shame, because they themselves have experienced it. It is as unpleasant to see as it is to feel it. A narcissist instead views shame as a magic potion which can cripple their target’s self-esteem and give them the upper hand.
The narcissist strings together a never-ending series of attacks on their target which cause the target to feel more and more shame, fear and guilt. As a result, the target experiences a psychological pressure which grows over time and results in a psychological cage. The target’s self-esteem shrinks, their imagination loses its potency, and they can no longer see beyond their constant suffering. This is the funnelling process which occurs over the course of a relationship with a narcissist. Before you know it, you are attached to the narcissist and trapped in your role, overwhelmed by the mental and emotional heaviness which was spoon fed to you within the context of what you believed to be a fair relationship.
In short, narcissistic abuse is the ritualistic herding of a target from a place of total freedom to a dystopian cage of suffering with little room to move or breathe.
If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist.
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Debajit Ghose
Studied at Indian Institute of Technology, Kharagpur (IIT KGP) · Author has 64 answers and 119.6K answer views
·6y
Related
Three different faces of a cube are painted in three different colours–red, green and blue. This cube is now cut into 216 smaller but identical cubes. What are the least and the largest numbers of small cubes that have exactly one face painted?
92 and 96.
216= 6x6x6
3 coloured faces of big cube means 6x6x3=108 small cubes have at least one side painted.
Now for only one side painted:
Considering the 3 painted sides having one common corner of the big cube, i.e. 3 intersecting edges. The number of small cubes at the intersecting painted edges add up to 6+5+5=16 with more than one painted side. So the minimum no. of small cubes with exactly one painted side = 108–16=92.
Considering the 3 painted sides having no common corner, i.e. only 2 intersecting edges. Multi-face-painted cubes are 6+6=12 in number. So maximum no. of one-face-painted sm
Continue Reading
92 and 96.
216= 6x6x6
3 coloured faces of big cube means 6x6x3=108 small cubes have at least one side painted.
Now for only one side painted:
Considering the 3 painted sides having one common corner of the big cube, i.e. 3 intersecting edges. The number of small cubes at the intersecting painted edges add up to 6+5+5=16 with more than one painted side. So the minimum no. of small cubes with exactly one painted side = 108–16=92.
Considering the 3 painted sides having no common corner, i.e. only 2 intersecting edges. Multi-face-painted cubes are 6+6=12 in number. So maximum no. of one-face-painted small cubes= 108–12=96.
Upvote ·
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Related questions
More answers below
If you have a cube made up of 27 smaller cubes (3x3x3), and you paint the entire outer surface of this larger cube red, then you disassemble it into the smaller cubes, how many of the smaller cubes will have paint on exactly two faces?
A wooden cube is 3 inches on each side and has its exterior painted red. If the cube is cut into smaller cubes 1 inch long on each sides, how many of the smaller cubes have exactly two red faces?
A cube is painted red on all six faces. It is then cut into 64 smaller cubes of equal size. How many of these smaller cubes will have exactly two faces painted?
A cube with side 3cm is painted grey and cut into smaller cubes each with side length of 1cm. How many of the smaller cubes have exactly 3 faces painted?
A cube of side 3 cm is taken and is painted red on all its faces. It is now cut into equal sized cubes of side 1 cm. What is the number of cubes with exactly 3 sides painted?
Ritika Khurana
6y
Related
A cube is painted with red colour on its five faces. If this cube is cut into 216 equal small cubes, how many small cubes will have no faces coloured?
Going by the question, since there are 216 smaller cubes it means there are
6 x 6 cubes on each side. Here is how it will look like.
Assume the front face is not painted red and every other face is. Now you want the number of cubes with no side painted. For this imagine peeling off the top layer, bottom one, left right and back one.
Top layer has cubes = 36
Bottom layer = 36
Left layer = 24 (not 36 because its top 6 and bottom 6 are already gone in previous steps)
Right layer = 24
Back layer = 16 (again this has its top, bottom, left and right already gone, leaving you with a 4 x 4 block)
Remaining cu
Continue Reading
Going by the question, since there are 216 smaller cubes it means there are
6 x 6 cubes on each side. Here is how it will look like.
Assume the front face is not painted red and every other face is. Now you want the number of cubes with no side painted. For this imagine peeling off the top layer, bottom one, left right and back one.
Top layer has cubes = 36
Bottom layer = 36
Left layer = 24 (not 36 because its top 6 and bottom 6 are already gone in previous steps)
Right layer = 24
Back layer = 16 (again this has its top, bottom, left and right already gone, leaving you with a 4 x 4 block)
Remaining cubes = 216 - ( 36 + 36 + 24 + 24 + 16) = 80.
Another way to look at it is to imagine a block of cubes which has 4 x 4 x 5 cubes in it. (4 and not 6 because top and bottom are compromised), 4 (because left and right are gone) and 5 (because only back layer is gone, front remains). This also equals 80.
I hope this helps.
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Bapeshwar Vinnakota
Maths -/ · Author has 123 answers and 478K answer views
·3y
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A big cube is made up of 64 small cubes. All the faces of the big cube are then painted. How many of the small cubes have exactly one painted face?
Thanks for A2A.
So here comes the easy way to solve this. Given 8 x 8 cube = 64 small cubes
Two face painted
Only edges excluding 8 corners will have two face painted. So total 12 edges. Each edge have 2 corners. So 12(n-2) cubes are two face painted. 12 x (8–2) = 72.
One face painted
All 6 faces excluding edges(which includes corners) are one face painted. But we can clearly see, the answer is also a square or 2D matrix. So per side there are (n-2)(n-2)=36. For 6 sides it is 6 x 36 =216
No face painted
Will be the inner cubes which is a 3D matrix. So total number is (n-2)(n-2)(n-2)=6x6x6=216.
Continue Reading
Thanks for A2A.
So here comes the easy way to solve this. Given 8 x 8 cube = 64 small cubes
Two face painted
Only edges excluding 8 corners will have two face painted. So total 12 edges. Each edge have 2 corners. So 12(n-2) cubes are two face painted. 12 x (8–2) = 72.
One face painted
All 6 faces excluding edges(which includes corners) are one face painted. But we can clearly see, the answer is also a square or 2D matrix. So per side there are (n-2)(n-2)=36. For 6 sides it is 6 x 36 =216
No face painted
Will be the inner cubes which is a 3D matrix. So total number is (n-2)(n-2)(n-2)=6x6x6=216.
Upvote ·
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Nikhil
Studied at National Institutes of Technology (NITs)
·8y
Related
All faces of a cube are painted in red. Then it is cut into 125 small equal cubes. How many small cubes will be formed with no colored faces?
27 : If we visualize the 5x5 cube . Below are the 4 types of pieces ( small cubes )
corner pieces ( 3 sides facing ): 8
corner pieces ( 2 sides facing ) : 36
1 side facing : 54 ( red colored in diagram )
not exposed : 27
So , if we color the outer walls , pieces that are not exposed will remain uncolored . Hence the answer is 27 . Check below diagram for pieces calculation .
Continue Reading
27 : If we visualize the 5x5 cube . Below are the 4 types of pieces ( small cubes )
corner pieces ( 3 sides facing ): 8
corner pieces ( 2 sides facing ) : 36
1 side facing : 54 ( red colored in diagram )
not exposed : 27
So , if we color the outer walls , pieces that are not exposed will remain uncolored . Hence the answer is 27 . Check below diagram for pieces calculation .
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Praharshita Sastry
Avid puzzle solver
·6y
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Two adjacent faces of a cube are painted. The cube is then cut into 125 smaller identical cubes. How many of these smaller cubes will have exactly one face painted?
It is given that the cube is cut into 125 smaller identical cubes. It implies that the length of side of original cube is 5 units. Only in this way do we get 125 Smaller Identical Cubes.
Now, consider the following White 5 x 5 cube.
Lets consider that the Two Adjacent Faces are painted with Pink and Yellow.
Of the total 125 smaller cubes,
80 are pure white cubes.
20 are white cubes with only one side painted with pink.
20 are white cubes with only one side painted with yellow.
And 5 cubes, which are common to the two painted sides have two sides painted.
So, 40 of the 125 identical cubes will have exa
Continue Reading
It is given that the cube is cut into 125 smaller identical cubes. It implies that the length of side of original cube is 5 units. Only in this way do we get 125 Smaller Identical Cubes.
Now, consider the following White 5 x 5 cube.
Lets consider that the Two Adjacent Faces are painted with Pink and Yellow.
Of the total 125 smaller cubes,
80 are pure white cubes.
20 are white cubes with only one side painted with pink.
20 are white cubes with only one side painted with yellow.
And 5 cubes, which are common to the two painted sides have two sides painted.
So, 40 of the 125 identical cubes will have exactly one face painted.
Upvote ·
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Junaidul Islam
Studied Bachelor of Technology in Electronics and Communications Engineering at Maharshi Dayanand University, Rohtak (Graduated 2016)
·7y
Related
A cube is painted blue on all faces and it is cut into 64 small cubes of equal size. How many cubes are not painted?
Simple!
In total it will become 4×4×4= 64. i.e; L×B×H
Cubes painted on three side =8
2 side = 2×12 = 24
1 side = 4×6 = 24
Then 24+24+8 will become 56.
It means that the cubes painted with no side will be 64 - 56= 8.
Ans. 8
Continue Reading
Simple!
In total it will become 4×4×4= 64. i.e; L×B×H
Cubes painted on three side =8
2 side = 2×12 = 24
1 side = 4×6 = 24
Then 24+24+8 will become 56.
It means that the cubes painted with no side will be 64 - 56= 8.
Ans. 8
Upvote ·
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Bhim Mu
Former Retd Gov Servant (1971–2007) · Author has 3.5K answers and 526.3K answer views
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The surface area of a cube of 4×4×4 dimensions is painted. The cube is cut into 64 equal cubes .how many cubes have painted (1) 1 face painted? (2) 2 faces painted? (3) 3 faces painted? (4) no face painted?
Answer 1]1 face painted =4+8+8+4 =24 ; 2Face Painted:8+4+4+8 =24 ;3] 3 face painted=4+4 =8
4] no face painted = 4+4 =8
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Answer 1]1 face painted =4+8+8+4 =24 ; 2Face Painted:8+4+4+8 =24 ;3] 3 face painted=4+4 =8
4] no face painted = 4+4 =8
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Lokesh Jain
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Sixty four cubes are assembled to form a larger cube. All the six faces of this larger are painted black.How many smaller cubes will have exactly 2 faces which are painted black?
Hi
This will be the picture of a 64 cubes assembled to form a large cube . If all the visible faces are painted black then the one that are on the corner has 3 visible sides so three black faces .on the top face the one that are in middle have 2 faces black ..therefore on top face there are total 8 small cubes . Similarly on bottom and middle of the cube they are 8 small cubes having 2 faces black .
8+8+8=24 small cubes is the answer
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Hi
This will be the picture of a 64 cubes assembled to form a large cube . If all the visible faces are painted black then the one that are on the corner has 3 visible sides so three black faces .on the top face the one that are in middle have 2 faces black ..therefore on top face there are total 8 small cubes . Similarly on bottom and middle of the cube they are 8 small cubes having 2 faces black .
8+8+8=24 small cubes is the answer
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A wooden cube is 3 dm on each side and has its exterior painted is blue. If the cube is cut into smaller cubes about 1 dm long on its side, how many of the smaller cubes have exactly three blue faces?
A cube with sides of 5 cm is painted on all its faces. If sliced in 1 cm cubes, how many 1 cm cubes will have exactly one of their faces painted?
A cube is painted blue on all faces and it is cut into 64 small cubes of equal size. How many cubes are not painted?
A cube with all the sides painted was divided into small cubes of equal measurement. The side of a small cube is exactly one fourth as that of the big cube. Therefore, what is the number of small cubes with only one side painted?
Three different faces of a cube are painted in three different colours–red, green and blue. This cube is now cut into 216 smaller but identical cubes. What are the least and the largest numbers of small cubes that have exactly one face painted?
If you have a cube made up of 27 smaller cubes (3x3x3), and you paint the entire outer surface of this larger cube red, then you disassemble it into the smaller cubes, how many of the smaller cubes will have paint on exactly two faces?
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What is the standard formula of the ellipse given F1 (-8,0) F2 (8, 0), the length of the major axis is 20?
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
·1y
Since both foci are on the x x-axis, and the center is at the midpoint of the segment whose endpoints are the foci such that the center is at the origin, the form of this ellipse is:
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
where:
c=8 c=8
2 a=20 2 a=20
a=10 a=10
b 2=a 2−c 2=100−64=36 b 2=a 2−c 2=100−64=36
Therefore, the equation for the ellipse is:
x 2 100+y 2 36=1 x 2 100+y 2 36=1
or
9 x 2+25 y 2−900=0 9 x 2+25 y 2−900=0
A plot looks like this:
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Since both foci are on the x x-axis, and the center is at the midpoint of the segment whose endpoints are the foci such that the center is at the origin, the form of this ellipse is:
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
where:
c=8 c=8
2 a=20 2 a=20
a=10 a=10
b 2=a 2−c 2=100−64=36 b 2=a 2−c 2=100−64=36
Therefore, the equation for the ellipse is:
x 2 100+y 2 36=1 x 2 100+y 2 36=1
or
9 x 2+25 y 2−900=0 9 x 2+25 y 2−900=0
A plot looks like this:
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What is (.25) or (1/4) of an ellipse?
What is 0.5, 0.55,0.65, and 0.8?
The points P (-3,1), Q(3,4) and R(8,1) on the X-Y plane lie on the circumference of an ellipse, whose major axis is PR. What is the equation of the ellipse?
What is the standard of the ellipse whose foci F1 (-2, 0) and F2(2,0) such that for any point on it, the sum of its distances from the foci is 8?
What is the equation of the ellipse with foci at (0, 8) and (0 -8), and has a major axis of length 20? Sketch the graph.
Janet Heberling
Lives in San Francisco, CA (2022–present) · Author has 21.5K answers and 9.4M answer views
·1y
Horizontal ellipse:(x−h)2 a 2+(y−k)2 b 2=1 a 2≥b 2 center(h,k)vertices(h±a,k)co-vertices(h,k±b)foci(h±c,k),c 2=a 2−b 2 Horizontal ellipse:(x−h)2 a 2+(y−k)2 b 2=1 a 2≥b 2 center(h,k)vertices(h±a,k)co-vertices(h,k±b)foci(h±c,k),c 2=a 2−b 2
center(h,k)is midway between foci(h,k)=(0,0)⇒h=k=0 center(h,k)is midway between foci(h,k)=(0,0)⇒h=k=0
foci(0±c,0)=(±8,0)⇒c=8 foci(0±c,0)=(±8,0)⇒c=8
length of major axis=2 a=20⇒a=10 length of major axis=2 a=20⇒a=10
b 2=a 2−c 2=36 b 2=a 2−c 2=36
x 2 100+y 2 36=1 x 2 100+y 2 36=1
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·3y
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What is the equation of the ellipse with foci at (0, 8) and (0 -8), and has a major axis of length 20? Sketch the graph.
With the x-coordinates of both foci at 0, this ellipse is clearly oriented vertically. The general equation of such an ellipse is
(y - k)^2 / a^2 + (x - h)^2 / b^2 = 1
It’s also clear that the ellipse is centered at the origin, while the distance a from the center to each vertex along the major axis is 20/2 = 10. Thus we can fill in the equation as follows:
y^2 / 10^2 + x^2 / b^2 = 1
Therefore, we need only find b^2 to make this equation complete. This we do easily, by using the formula c^2 = a^2 - b^2, where c is the distance from the center to each focus:
c^2 = a^2 - b^2
b^2 = a^2 - c^2 = 100 - 6
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With the x-coordinates of both foci at 0, this ellipse is clearly oriented vertically. The general equation of such an ellipse is
(y - k)^2 / a^2 + (x - h)^2 / b^2 = 1
It’s also clear that the ellipse is centered at the origin, while the distance a from the center to each vertex along the major axis is 20/2 = 10. Thus we can fill in the equation as follows:
y^2 / 10^2 + x^2 / b^2 = 1
Therefore, we need only find b^2 to make this equation complete. This we do easily, by using the formula c^2 = a^2 - b^2, where c is the distance from the center to each focus:
c^2 = a^2 - b^2
b^2 = a^2 - c^2 = 100 - 64 = 36
Thus the equation of the ellipse is y^2 / 100 + x^2 / 36 = 1. A graph of the ellipse is shown below.
As problems of this nature go, this one is almost trivial, and your question indicates you need to learn some basic properties of ellipses, and other conic sections. You might do well to read the webpage posted below the graph.
Equations of Ellipses Learning Outcomes Identify the foci, vertices, axes, and center of an ellipse. Write equations of ellipses centered at the origin. Write equations of ellipses not centered at the origin. A conic section, or conic , is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape. Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points [latex]\left(x,y\right)[/latex] in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci ) of the ellipse. We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. Every ellipse has two axes of symmetry. The longer axis is called the major axis , and the shorter axis is called the minor axis . Each endpoint of the major axis is the vertex of the ellipse (plural: vertices ), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. In this section we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x – and y -axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane. To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs. To derive the equation of an ellipse centered at the origin, we begin with the foci latex[/latex] andlatex[/latex]. The ellipse is the set of all pointslatex[/latex] such that the sum of the distances fromlatex[/latex] to the foci is constant, as shown in the figure below. If latex[/latex] is a vertex of the ellipse, the di
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Martyn Hathaway
BSc in Mathematics, University of Southampton (Graduated 1986) · Author has 4.7K answers and 6.7M answer views
·3y
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What is the equation of an ellipse with centre (0,0) and foci on the x-axis and passing through the points (-3,12^1/2) and (4, (80/9) ^1/2)?
The equation for the ellipse is in the form:
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
where a,b>0 a,b>0
⟹b 2 x 2+a 2 y 2=a 2 b 2⟹b 2 x 2+a 2 y 2=a 2 b 2
From the question, the ellipse passes through the point (−3,√12)(−3,12), so:
1: (−3)2 b 2+12 a 2=a 2 b 2⟹9 b 2+12 a 2=a 2 b 2(−3)2 b 2+12 a 2=a 2 b 2⟹9 b 2+12 a 2=a 2 b 2
Also from the question, the ellipse passes through the point (4,√80/9)(4,80/9), so:
2: 4 2 b 2+(80/9)a 2=a 2 b 2⟹16 b 2+(80/9)a 2=a 2 b 2 4 2 b 2+(80/9)a 2=a 2 b 2⟹16 b 2+(80/9)a 2=a 2 b 2
Note that the right side of the two equations are the same, this means that the left sides must equal each other:
⟹9 b 2+12 a 2=16 b 2+(80/9)a 2⟹9 b 2+12 a 2=16 b 2+(80/9)a 2
Multiplying throughout by 9:
\Longrighta\Longrighta
Continue Reading
The equation for the ellipse is in the form:
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
where a,b>0 a,b>0
⟹b 2 x 2+a 2 y 2=a 2 b 2⟹b 2 x 2+a 2 y 2=a 2 b 2
From the question, the ellipse passes through the point (−3,√12)(−3,12), so:
1: (−3)2 b 2+12 a 2=a 2 b 2⟹9 b 2+12 a 2=a 2 b 2(−3)2 b 2+12 a 2=a 2 b 2⟹9 b 2+12 a 2=a 2 b 2
Also from the question, the ellipse passes through the point (4,√80/9)(4,80/9), so:
2: 4 2 b 2+(80/9)a 2=a 2 b 2⟹16 b 2+(80/9)a 2=a 2 b 2 4 2 b 2+(80/9)a 2=a 2 b 2⟹16 b 2+(80/9)a 2=a 2 b 2
Note that the right side of the two equations are the same, this means that the left sides must equal each other:
⟹9 b 2+12 a 2=16 b 2+(80/9)a 2⟹9 b 2+12 a 2=16 b 2+(80/9)a 2
Multiplying throughout by 9:
⟹81 b 2+108 a 2=144 b 2+80 a 2⟹81 b 2+108 a 2=144 b 2+80 a 2
Subtracting 80 a 2+81 b 2 80 a 2+81 b 2 from both sides:
⟹28 a 2=63 b 2⟹4 a 2=9 b 2⟹28 a 2=63 b 2⟹4 a 2=9 b 2
Using this, we can replace the b 2 b 2 terms in Eq. 1.
⟹4 a 2+12 a 2=a 2(4 a 2/9)⟹4 a 2+12 a 2=a 2(4 a 2/9)
Multiplying throughout by 9/4:
⟹36 a 2=a 4⟹a 2=0∪a 2=36⟹36 a 2=a 4⟹a 2=0∪a 2=36
We thus have a=6⟹b=4 a=6⟹b=4
Answer:
x 2 36+y 2 16=1 x 2 36+y 2 16=1
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What is the equation of an ellipse whose foci are (-3,0) and (3, 0) such that for any point on it, the sum of the distances from the distances of the foci is 10?
What is the standard equation of the ellipse whose center at the origin, foci are Fi (0,-6) and F: (0, 6) and major axis is 20? Sketch the graph.
What is 8-0+-8.88 x -888?
Let the point (-4, 1) be the center of an ellipse with a minor axis parallel to the y-axis and of the length 8. If the ellipse passes through the point (0, 4) then what is the length of its major axis?
If F1 = 15.0 N, what’s F2?
Dodie Cowan
Former Professor of Mathematics at Polk State College (2004–2015) · Author has 2.2K answers and 1.3M answer views
·5y
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What is the equation in standard form of the ellipse whose foci are F1 (-8,0) and F2 (8, 0), such any that for any point on it, the sum of its distance from the foci is 20?
First, if foci are (-8,0) and (8,0), then we know that the center is (0,0) and that the ellipse is horizontal (major axis is along the x-axis). It’s equation, then will be of the form:
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
and we know that a 2−b 2=c 2 a 2−b 2=c 2 where c=8 (distance from center to a focus).
We also know that for a point on the ellipse, the distance between it and (8,0) added to the distance between it and (-8,0) is 20. One such point on the ellipse is a vertex, which lies on the x-axis and whose distance from the center is a, is (a,0). Using the distance formula \sqrt{(a-8)^2+(0–0)^2}+\sqrt{(a+8\sqrt{(a-8)^2+(0–0)^2}+\sqrt{(a+8
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First, if foci are (-8,0) and (8,0), then we know that the center is (0,0) and that the ellipse is horizontal (major axis is along the x-axis). It’s equation, then will be of the form:
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
and we know that a 2−b 2=c 2 a 2−b 2=c 2 where c=8 (distance from center to a focus).
We also know that for a point on the ellipse, the distance between it and (8,0) added to the distance between it and (-8,0) is 20. One such point on the ellipse is a vertex, which lies on the x-axis and whose distance from the center is a, is (a,0). Using the distance formula √(a−8)2+(0–0)2+√(a+8)2+(0–0)2=20(a−8)2+(0–0)2+(a+8)2+(0–0)2=20. This simplifies to a−8+a+8=20 a−8+a+8=20, and so a=10.
Using a 2−b 2=c 2 a 2−b 2=c 2, we have 10 2−b 2=8 2 10 2−b 2=8 2, and so b=6.
Thus, the equation of this ellipse is x 2 100+y 2 36=1 x 2 100+y 2 36=1
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
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What is the equation of the ellipse with foci at (-2, 0) and (2,0), and a major axis of length 6 units?
Given two foci of ellipse are F 1(−2,0)F 1(−2,0) and F 2(2,0)F 2(2,0)
Immediately we note that the ordinates of foci are same. Major axis of ellipse is along x x-axis. Further, the foci are equidistant from the origin.
In this orientation, we can write the equation of ellipse as,
x 2 a 2+y 2 b 2=1…(1)x 2 a 2+y 2 b 2=1…(1)
where a a and b b are semi major axis and semi minor axis respectively.
Given length of major axis is 6 6
2 a=6⟹a=3 2 a=6⟹a=3
V 1(−3,0)V 1(−3,0) and V 2(3,0)V 2(3,0) are the vertices. Let C 1(0,b)C 1(0,b) and C 2(0,−b)C 2(0,−b) be the covertices.
Let F 1 F 2=c F 1 F 2=c
c=√(2−(−2))2+(0−0)2=4 c=(2−(−2))2+(0−0)2=4
Now,
c 2=a 2−b 2 c 2=a 2−b 2
4 2=3 2−b 2 4 2=3 2−b 2
b 2=7 b 2=7
From eqn. (1),
\d\d
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Given two foci of ellipse are F 1(−2,0)F 1(−2,0) and F 2(2,0)F 2(2,0)
Immediately we note that the ordinates of foci are same. Major axis of ellipse is along x x-axis. Further, the foci are equidistant from the origin.
In this orientation, we can write the equation of ellipse as,
x 2 a 2+y 2 b 2=1…(1)x 2 a 2+y 2 b 2=1…(1)
where a a and b b are semi major axis and semi minor axis respectively.
Given length of major axis is 6 6
2 a=6⟹a=3 2 a=6⟹a=3
V 1(−3,0)V 1(−3,0) and V 2(3,0)V 2(3,0) are the vertices. Let C 1(0,b)C 1(0,b) and C 2(0,−b)C 2(0,−b) be the covertices.
Let F 1 F 2=c F 1 F 2=c
c=√(2−(−2))2+(0−0)2=4 c=(2−(−2))2+(0−0)2=4
Now,
c 2=a 2−b 2 c 2=a 2−b 2
4 2=3 2−b 2 4 2=3 2−b 2
b 2=7 b 2=7
From eqn. (1),
x 2 3 2+y 2 7=1 x 2 3 2+y 2 7=1
x 2 9+y 2 7=1 x 2 9+y 2 7=1
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Pradeep Hebbar
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What is the standard equation of the ellipse whose center at the origin, foci are Fi (0,-6) and F: (0, 6) and major axis is 20? Sketch the graph.
Given an ellipse centered at origin and having focal points (0,−6)(0,−6) and (0,6)(0,6)
Let 2 a 2 a and 2 b 2 b be the lengths of major and minor axis respectively,
As the x x coordinates of focii are zero, they lie on y-axis. Therefore, major axis is along y y axis
Equation of ellipse for this case is given by,
x 2 b 2+y 2 a 2=1…(1)x 2 b 2+y 2 a 2=1…(1)
2 a=20⟹a=10 2 a=20⟹a=10
c=6 c=6
c 2=a 2−b 2 c 2=a 2−b 2
6 2=10 2−b 2 6 2=10 2−b 2
b 2=64 b 2=64
From eqn. (1),
x 2 64+y 2 100=1 x 2 64+y 2 100=1
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Given an ellipse centered at origin and having focal points (0,−6)(0,−6) and (0,6)(0,6)
Let 2 a 2 a and 2 b 2 b be the lengths of major and minor axis respectively,
As the x x coordinates of focii are zero, they lie on y-axis. Therefore, major axis is along y y axis
Equation of ellipse for this case is given by,
x 2 b 2+y 2 a 2=1…(1)x 2 b 2+y 2 a 2=1…(1)
2 a=20⟹a=10 2 a=20⟹a=10
c=6 c=6
c 2=a 2−b 2 c 2=a 2−b 2
6 2=10 2−b 2 6 2=10 2−b 2
b 2=64 b 2=64
From eqn. (1),
x 2 64+y 2 100=1 x 2 64+y 2 100=1
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
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What is the center, the foci and the endpoints of the major and minor axes of the given ellipse, x²+9y²=9?
Given an equation of ellipse
x 2+9 y 2=9 x 2+9 y 2=9
Rewriting,
x 2 9+y 2 1=1 x 2 9+y 2 1=1
This represents an ellipse centered at (0,0)(0,0) and having major axis along x x-axis. Its equation is comparable to
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
Vertices: (−a,0)(−a,0) and (a,0)(a,0)
Covertices: (0,−b)(0,−b) and (0,b)(0,b)
Focii: (−c,0)(−c,0) and (c,0)(c,0) where c 2=a 2−b 2 c 2=a 2−b 2
We have,
a 2=9⟹a=3 a 2=9⟹a=3
b 2=1⟹b=1 b 2=1⟹b=1
c 2=9−1=8⟹c=2√2 c 2=9−1=8⟹c=2 2
Therefore,
Focii: (−2√2,0)(−2 2,0) and (2√2,0)(2 2,0)
Ends of major axis:(−3,0)(−3,0) and (3,0)(3,0)
Ends of minor axis: (0,−1)(0,−1) and (0,1)(0,1)
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Given an equation of ellipse
x 2+9 y 2=9 x 2+9 y 2=9
Rewriting,
x 2 9+y 2 1=1 x 2 9+y 2 1=1
This represents an ellipse centered at (0,0)(0,0) and having major axis along x x-axis. Its equation is comparable to
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
Vertices: (−a,0)(−a,0) and (a,0)(a,0)
Covertices: (0,−b)(0,−b) and (0,b)(0,b)
Focii: (−c,0)(−c,0) and (c,0)(c,0) where c 2=a 2−b 2 c 2=a 2−b 2
We have,
a 2=9⟹a=3 a 2=9⟹a=3
b 2=1⟹b=1 b 2=1⟹b=1
c 2=9−1=8⟹c=2√2 c 2=9−1=8⟹c=2 2
Therefore,
Focii: (−2√2,0)(−2 2,0) and (2√2,0)(2 2,0)
Ends of major axis:(−3,0)(−3,0) and (3,0)(3,0)
Ends of minor axis: (0,−1)(0,−1) and (0,1)(0,1)
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Lori Conner
Studied Mathematics at University of Arkansas · Author has 94 answers and 66.2K answer views
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Can you find the equation of this ellipse whose foci at (0,-2) and (0,2); length of major axis is 8?
So - you need to find the center and then the lengths of the axes - major and minor
The center is 1/2 way between the foci (0,0)
Then, since the length of the major axis is 8, the vertices are at a distance of 4 from the center. Also, the vertices are on the major axis so they are (0,-4) and (0,4).
You now know most of the equation. You just need to solve for b
I am having trouble getting the equation to look right - so maybe I can explain the way it needs to look.
You have 2 fractions with a + between them and they add up to 1.
The numerator of the left fraction is (x-0)^2 and the denominator is 4^
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So - you need to find the center and then the lengths of the axes - major and minor
The center is 1/2 way between the foci (0,0)
Then, since the length of the major axis is 8, the vertices are at a distance of 4 from the center. Also, the vertices are on the major axis so they are (0,-4) and (0,4).
You now know most of the equation. You just need to solve for b
I am having trouble getting the equation to look right - so maybe I can explain the way it needs to look.
You have 2 fractions with a + between them and they add up to 1.
The numerator of the left fraction is (x-0)^2 and the denominator is 4^2
The numerator of the right fraction is (y-0)^2 and the denominator is b^2.
We need to find b - because that will tell us the length of the minor axis. The co-vertices on the minor axis will be (b, 0) and (-b,0).
Recall that c is the distance from the center to the foci. In this case, it’s 2. Also, c is related to a and b with the formula c^2 = a^2 - b^2. Turn the equation around to solve for b^2and get b^2 = a^2 - c^2.
b^2 = 4^2 - 2^2 = 12 .
So you know that the denominator of the right equation =12.
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
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What is the standard equation of the ellipse whose center at the origin, foci are Fi (0,-6) and F: (0, 6) and major axis is 20? Sketch the graph.
Two points: (1) You need to learn some formulas pertaining to ellipses; (2) This problem is so simple and basic that there is literally no excuse for you not to do this on your own.
For an ellipse centered at the origin, major axis vertical:
Vertices: (0, ±a)
Co-vertices: (±b, 0)
Foci: (0, ±c)
Relation of foci to all vertices:
c^2 = a^2 - b^2
Standard equation of an ellipse oriented vertically:
(x - h)^2 / b^2 + (y - k)^2 / a^2 = 1, where (h, k) is the center
All you need here is the distance b of each co-vertex from the center:
c^2 = a^2 - b^2
b^2 = a^2 - c^2 = 10^2 - 6^2 = 100 - 36 = 64
b = ±8
Equati
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Two points: (1) You need to learn some formulas pertaining to ellipses; (2) This problem is so simple and basic that there is literally no excuse for you not to do this on your own.
For an ellipse centered at the origin, major axis vertical:
Vertices: (0, ±a)
Co-vertices: (±b, 0)
Foci: (0, ±c)
Relation of foci to all vertices:
c^2 = a^2 - b^2
Standard equation of an ellipse oriented vertically:
(x - h)^2 / b^2 + (y - k)^2 / a^2 = 1, where (h, k) is the center
All you need here is the distance b of each co-vertex from the center:
c^2 = a^2 - b^2
b^2 = a^2 - c^2 = 10^2 - 6^2 = 100 - 36 = 64
b = ±8
Equation of this ellipse: x^2 / 64 + y^2 / 100 = 1
See the graph, and a link to a good resource just below.
Equations of Ellipses Learning Outcomes Identify the foci, vertices, axes, and center of an ellipse. Write equations of ellipses centered at the origin. Write equations of ellipses not centered at the origin. A conic section, or conic , is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape. Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points [latex]\left(x,y\right)[/latex] in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci ) of the ellipse. We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. Every ellipse has two axes of symmetry. The longer axis is called the major axis , and the shorter axis is called the minor axis . Each endpoint of the major axis is the vertex of the ellipse (plural: vertices ), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. In this section we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x – and y -axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane. To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs. To derive the equation of an ellipse centered at the origin, we begin with the foci latex[/latex] andlatex[/latex]. The ellipse is the set of all pointslatex[/latex] such that the sum of the distances fromlatex[/latex] to the foci is constant, as shown in the figure below. If latex[/latex] is a vertex of the ellipse, the di
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
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Aside5x²+18y²-30x+72y+27=0, what is the standard form of the ellipse?
5X^2 — 30X + 18Y^2 + 72Y = — 27 =>
5(X^2 — 6X) + 18(Y^2 + 4Y) = — 27 =>
5(X^2 — 6X + 9) — 45 + 18(Y^2 + 4Y + 4) — 72 = — 27=>
5(X — 3)^2 + 18(Y + 2)^2 = —27 + 45 + 72 =>
5(X — 3)^2 + 18(Y + 2)^2 = 90 =>
5(X — 3)^2/90 + 18(Y + 2)^2/90 = 90/90
| (X — 3)^2/18 + (Y + 2)^2/5 = 1 |
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5X^2 — 30X + 18Y^2 + 72Y = — 27 =>
5(X^2 — 6X) + 18(Y^2 + 4Y) = — 27 =>
5(X^2 — 6X + 9) — 45 + 18(Y^2 + 4Y + 4) — 72 = — 27=>
5(X — 3)^2 + 18(Y + 2)^2 = —27 + 45 + 72 =>
5(X — 3)^2 + 18(Y + 2)^2 = 90 =>
5(X — 3)^2/90 + 18(Y + 2)^2/90 = 90/90
| (X — 3)^2/18 + (Y + 2)^2/5 = 1 |
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
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What is the equation of ellipse with the following given Foci (4,0) and (-4,0) Endpoints of minor axis (0, 3) and (0, -3)?
In an ellipse, distance from centre to foci is c and distance from centre to covertices is b. Distance from centre to major vertices is a.
Centre of this ellipse is midpoint of foci or covertices. Centre A = [(0+0)/2,(3–3)/2] = (0,0)
b and c are clearly 3 and 4 units respectively.
a^2 = b^2 + c^2 => a = 5 units.
So equation of this ellipse is =>
| X^2/25 + Y^2/9 = 1 |
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In an ellipse, distance from centre to foci is c and distance from centre to covertices is b. Distance from centre to major vertices is a.
Centre of this ellipse is midpoint of foci or covertices. Centre A = [(0+0)/2,(3–3)/2] = (0,0)
b and c are clearly 3 and 4 units respectively.
a^2 = b^2 + c^2 => a = 5 units.
So equation of this ellipse is =>
| X^2/25 + Y^2/9 = 1 |
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Dave Benson
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What is the equation of an ellipse whose center is at the origin and the focus is (2,2) and length of semi minor axis is √8?
This is a skew ellipse of the form: {(x-h)cosA+(y-k)sinA}²/a²+{(x-h)sinA-(y-k)cosA}²/b² 1️⃣
Centre (h,k) =(0,0). a & b semi major & minor axes & A is angle to x-axis = 45º
P(-2,2) & |PF₁|+|PF₂| = 2a ⟹ 2PF₁ = 2√(8+8) = 2√16 ⟹ a = 4 ⟹ plug all into 1️⃣.
{0.707x+0.707y}²/16+{0.707x-0.707y}²/8 = 1
(x+y)²/32+(x-y)²/16 = 1 Answer in ellipse form.
or 3x²+3y²-2xy = 32 Answer.
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This is a skew ellipse of the form: {(x-h)cosA+(y-k)sinA}²/a²+{(x-h)sinA-(y-k)cosA}²/b² 1️⃣
Centre (h,k) =(0,0). a & b semi major & minor axes & A is angle to x-axis = 45º
P(-2,2) & |PF₁|+|PF₂| = 2a ⟹ 2PF₁ = 2√(8+8) = 2√16 ⟹ a = 4 ⟹ plug all into 1️⃣.
{0.707x+0.707y}²/16+{0.707x-0.707y}²/8 = 1
(x+y)²/32+(x-y)²/16 = 1 Answer in ellipse form.
or 3x²+3y²-2xy = 32 Answer.
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13534 | http://www.phys.ufl.edu/~nakayama/lec2048 | Abstract
These lecture notes supplements my discussion class. We only have one session per week and normally we have to rush through one whole chapter in less than 50 minutes. Now I believe supplementary notes are necessary for you to further your understanding in physics on your own pace. Each topic is independent. So you can just start with the section that interests you. Some topics are very basic, so I hope everyone will read them. Others are advanced, which often exceeds the textbook level. So if you feel the course is too easy and you’re bored, try challenging problems I provide in these notes.
1 Range of Projectile Motion
1.1 Horizontal Range
Most of the basic physics textbooks talk about the horizontal range of the projectile motion. It is derived using the kinematics equations:
ax = 0
vx = v0x
∆x = v0xtay = −gvy = v0y − gt
∆y = v0y t − 12 gt 2
where
v0x = v0 cos θv0y = v0 sin θ
Suppose a projectile is thrown from the ground level, then the range is the distance between the launch point and the landing point, where the projectile hits the ground. When the projectile comes back to the ground, the vertical displacement is zero, thus we have 0 = v0 sin θt − 12 gt 2
Solving for t, we have
t = 0 , 2v0 sin θg
1The first solution gives the time when the projectile is thrown and the second one is the time when it hits the ground. Plugging in the second solution into the displacement equation and using 2 sin θ cos θ = sin(2 θ), we have
R = ∆ x(t = 2 v0 sin θ/g ) = v20
g sin(2 θ)
Example
A baseball player can throw a ball at 30.0 m/s. What is the maximum horizontal range?
Solution
To maximize the range, s/he must throw a ball at an angle of 45 ◦ because at this angle sin 2 θ = 1.The range is
R = v20
g = 30 2
9.8 = 91 .8 m
1.2 Range on a Slope
Now what happens if you throw a ball on a slope? Do you still need to throw a ball at 45 ◦ above the horizontal to maximize the range? Or should you throw at 45 ◦ to the slope? Neither of them is correct. If you throw a ball at 45 ◦ on a slope of, say, 60 ◦, then you’re practically throwing the ball toward the slope so the range is actully zero. (We ignored the height of the launch point.) If you throw a ball at an angle of 45 ◦ to the slope of 45 ◦, you’re actually throwing a ball straight up, and the range is again zero. Now we modify our theory for the horizontal range and derive the range of a projectile on a slope. Assuming a projectile is launched from the ground level, the range is defined as the distance between the launch point and the point where the projectile hits the ground. We take +x direction horizontally and +y vertially upward, so that we can still use our kinematic equations. A slope of angle α is expressed as ∆y = tan α∆x, and when the projectile hits the ground, the horizontal displacement and vertical displacement of the projectile must satisfy this relationship. Thus we have
v0 sin θt − 12 gt 2 = tan α(v0 cos θt )Solutions for this equations are
t = 0 , 2v0
g (sin θ − tan α cos θ)The second solution gives the time at which the projectile hits the ground. Plugging this into the equation for ∆x, we get ∆x = v0 cos θt = v0 cos θ
[ 2v0
g (sin θ − tan α cos θ)
]
2We haven’t reached the solution yet. We just got horizontal displacement, not the range on the slope. The range on slope is given by
R = √∆x2 + ∆ y2 = √∆x2 + (tan α∆x)2 = ∆x
cos α
where we used the trig identity 1 + tan 2 α = 1cos 2 α . Thus we get
R = 2v20 cos θg cos α (sin θ − tan α cos θ) = 2v20 cos θg cos α
( sin θ cos α − sin α cos θ
cos α
)
= 2v20 cos θ sin( θ − α)
g cos 2 α = v20
g cos 2 α [sin(2 θ − α) − sin α]where we used trig identity sin( a ± b) = sin a cos b ± cos a sin b. You must note this reduces to the expression of horizontal range at α = 0.
Example
A baseball player throws a ball on a 20 ◦ slope toward the top of the slope. S/he can throw a ball at a speed of 30 m/s. What is the maximum range? (Ignore the height of the launch point.)
Solution
In our expression for the range, the second term is constant. Thus we need to maximize the first term. It is achieved when sin(2 θ − α) = 12θ − α = 90 ◦
θ = 90 ◦ + α
2 = 90 + 20 2 = 55 ◦
So s/he needs to throw a ball at an angle 55 ◦ to the horizontal( 35 ◦ to the incline). The range with this projection angle is
R = v20
g cos 2 α [1 − sin α] = 30 2
9.8 cos 2 20 ◦ [1 − sin 20 ◦] = 68 .4 m The range on a slope is shorter than that on a level ground.
Problem
What is the range if the baseball player throws a ball toward the bottom of the slope(The inclination is −20 ◦)?
2 Choice of Coordinate System
Before applying Newton’s 2nd law to a given problem, you need to fix your coordinate system. We have one solid principle when we choose a coordinate system. Take one of the axes along the direction of acceleration. This is because with this coordinate system, the acceleration along the other axis is zero. This simplifies our algebra a lot.
Example
3A block with mass m is sliding down a slope of angle θ. The coefficient of kinetic friction between the slope and the block is μk. Find the acceleration of the block.
Solution 1 We take +x axis along and down the slope and +y axis perpendicular to the surface, in this coordinate system, y component of the acceleration is zero. The y component of the equation of motion is 0 = N − mg cos θN = mg cos θ
The kinetic friction is given by fk = μkN . Applying the Newton’s 2nd law in the x direction, we have
ma = mg sin θ − fk = mg sin θ − μk(mg cos θ)
a = g(sin θ − μk cos θ)
Solution 2 We take +x direction horizontally and +y axis vertically upward. In this coordinate system, x and y components of the equation of motion are, respectively,
ma cos θ = N sin θ − μkN cos θ
−ma sin θ = N cos θ + μkN sin θ − mg
where μkN = fk. First we solve each equation for N to remove N from the simultaneous equation.
N = ma cos θ
sin θ − μk cos θN = mg − ma sin θ
cos θ + μk sin θ
Then we equate them, and solve for a.
ma cos θsinθ − μk cos θ = mg − ma sin θ
cos θ + μk sin θa cos θ(cos θ + μk sin θ) = (g − a sin θ)(sin θ − μk cos θ)
a(cos 2 θ + sin 2 θ) = g(sin θ − μk cos θ)
a = g(sin θ − μk cos θ)where we used trigonometric identity cos 2 θ + sin 2 θ = 1. Now are you con-vinced? No? Let’s see another example. 4 |
13535 | https://aviation.stackexchange.com/questions/89939/is-there-a-place-where-altitude-while-flying-is-negative | altimeter - Is there a place where altitude while flying is negative? - Aviation Stack Exchange
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Is there a place where altitude while flying is negative?
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while testing and working with avionics systems I have never seen an altitude near 0 or negative. Is there a place when flying above the Altitude tape shows -ve value ? Maybe while flying between grand canyons ?
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asked Oct 28, 2021 at 10:05
HuntkilHuntkil
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Are you expecting to see a negative height from a barometric gauge or a GPS-backed system ? Pressure (ie weather) changes can put a barometric sensor out by a lot, so take off in summer and land in winter, and the gauge can be well out.Criggie –Criggie 2021-10-28 19:42:36 +00:00 Commented Oct 28, 2021 at 19:42
2 Death Valley in California, U.S.A. is below sea level. I don't know whether it is legal or reasonable to fly that low over it.user247327 –user247327 2021-10-28 22:39:34 +00:00 Commented Oct 28, 2021 at 22:39
1 RelatedPondlife –Pondlife 2021-10-28 22:43:35 +00:00 Commented Oct 28, 2021 at 22:43
Also related.Kevin –Kevin 2021-10-29 20:04:17 +00:00 Commented Oct 29, 2021 at 20:04
1 @Criggie: GPS is no guarantee, either. I have gone kayaking on the ocean and had my GPS report the elevation as -30 feet (9m) as I launched.Ross Millikan –Ross Millikan 2021-10-30 03:24:03 +00:00 Commented Oct 30, 2021 at 3:24
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There are quite a few airports near sea level since many larger cities are located near the sea. There are even some airports below sea level. Amsterdam Schiphol is somewhat famous for it (at least where I live), even though its official aerodrome elevation is only -11 ft or -3 m.
According to Wikipedia's list of lowest airports, the lowest elevation airport is Bar Yehuda Airfield in Israel at -1,240 ft or -378 m. Here is a screenshot from this YouTube video flying at -400 ft pattern altitude before landing at Bar Yehuda:
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edited Oct 28, 2021 at 11:24
answered Oct 28, 2021 at 10:31
BianfableBianfable
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1 yeah, I've had lessons at Lelystad which is also well below sea level. Of course being PPL you're visual but you do see the gauges obviously.jwenting –jwenting 2021-10-28 16:04:38 +00:00 Commented Oct 28, 2021 at 16:04
There should have been a period in history when pilots had flown deeper than any submarine had dived.Bobby J –Bobby J 2021-10-30 18:43:14 +00:00 Commented Oct 30, 2021 at 18:43
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If you just wanted to know some places below sea level then there are plenty of related collections for places and airports and there is little interest to copy them here. But actually your question is not about such lists but whether an altimeter can show a negative altitude when used...
while testing and working with avionics systems I have never seen an altitude near 0 or negative. Is there a place when flying above the Altitude tape shows -ve value ? Maybe while flying between grand canyons ?
... the surprising answer is a barometric altimeter can display a negative value at any time and at any place (within technical limits) depending on its setting, not only at places below sea level.
A common altimeter, as one used in aircraft since the first years of aviation and still used today, is a misnomer, it doesn't measure altitude, it measures atmospheric pressure.
Pressure-altitude vs altitude
Barometric altimeters are not instruments measuring a geometric altitude, but measuring the local atmospheric pressure which happens to be linked to altitude, but in a quite unstable relationship, which depends on weather.
The ft dial is an overwritten hPa dial. To be fair, this scale is not an altitude scale, but a pressure-altitude scale. This accurate wording, which belongs to meteorology, is seldom used, only when someone wants to remind on how some altitude value is obtained. The actual altitude is the geometric altitude, usually determined using a satellite navigation system. Barometric altimeters were the only airborne solution to estimate altitude before inertial platform development.
A barometric altimeter has no absolute reference, the crew sets the reference, which is equivalent to turning the dial behind the needles in some initial configuration. The crew uses the reference which is the most suitable for some purpose. The reference pressure is just subtracted from the measured pressure before conversion to ft, hence the altimeter indicates 0 at the altitude the reference pressure is found, whatever the actual altitude is. Below this altitude indications are negative.
Many aircraft determine their actual geometric altitude using inertial reference and/or satellite navigation. ACAS works exclusively with geometric altitudes, ADS-B can report both pressure and geometric altitude. However vertical separation of aircraft is based on every aircraft navigating with a pressure-altitude. Therefore when geometric altitude is known it is not used by the crew, and generally not shown.
An altimeter referenced to QFE displays height above the reference point
Reference pressures (QNE, QFE, QNH) are referred to using their letters in Q code, a system of pre-formatted questions/answers used in marine and aviation before radiotelegraphy was replaced by voice communications.
In the past it was common to use the airfield pressure, QFE, as the reference. When set this way the altimeter simply displays the height above the airfield.
Using an altiport QFE naturally leads to negative values displayed when flying below the altiport, or more accurately when flying at a location which pressure is higher than QFE. This method is not recommended today, but is still used, and is for example common to fly airfield circuits in UK.
Usually crews are not really interested in knowing their exact altitude in cruise, they have no problem in following routes at constant atmospheric pressure, with the altimeter needles frozen at some value, while their geometric altitude actually varies.
Since every aircraft uses pressure-altitude instead of altitude, vertically separating aircraft consists in making them fly on different isobaric surfaces, ATC uses flight levels (FL) to assign pressure-altitudes. FL are spaced by 100 ft pressure-altitude increments, e.g. FL 100 corresponds to the pressure found at 100x100 (10,000) ft and FL 101 to the one found at 101x100 (10,100) ft, in the standard atmosphere model.
Determining and using the QNH at some location for some time
The standard atmosphere model pressure-altitude equivalence is based on a mean day pressure at sea level. However when weather changes, the pressure at some location varies, and pressure-altitude can be pretty different than geometric altitude. At 10,000 ft this difference can be 500 ft.
When crews operate close to the ground, they need a better altitude accuracy to navigate and prevent collisions with obstacles. A correction is applied to the model by introducing the local conditions.
This correction consists in retro-computing the pressure at sea level, given the current pressure tendency observed at the airfield. This new sea-level pressure is known as QNH. Telegraphists using the Q code would have transmitted "QNH 1021 1000Z", this would have meant exactly: "If you set the sub-scale of your altimetre to read 1021 hPa, the instrument would indicate your elevation if you were on the ground at my station at 10:00 hours".
Aviation also used in the past a more complete correction including temperature and relative humidity in the calculation. It was known as QFF, but has been replaced by the more simple QNH.
When about to land, the crew gets the airport QNH value from ATC or ATIS broadcast and sets the altimeter reference to this value. If things are done properly, when the aircraft lands its altimeter indicates approximately the airport altitude.
Now as visible in the picture above, any location below the dotted blue line (QNH surface) will be associated with a negative altitude, e.g. when flying close to the terrain on the right side.
Altitude is accurate only in the area where QNH is computed
As the crew sets the altimeter reference to the QNH, they actually redefine the isobaric surface corresponding to sea-level and thus determine a volume with negative altitudes. As the QNH is dependent on the current local weather, this volume also depends on the local weather.
The existence of areas of negative pressure, and more generally areas of inaccurate pressure is due to the fact a QNH setting is valid only in the area it was computed for. If we were to land on an airport on the right side, we would get the QNH computed by this airport for its own area. This QNH would be a higher value, and the corresponding dotted blue line would be lower than the one shown. The runway altitude would be correct with this setting, but would now be wrong for the airfield on the left side.
A negative altitude is just a matter of altimeter setting, and can be observed at any time you fly with a reference which doesn't corresponds to the local QNH. For this reason aircraft below a certain altitude are legally required to use a local QNH, obtained from a local airfield (e.g. in a radius of 100 nm in Australia), or at least an area/regional QNH, which is an approximation for some large area.
But there is an exception: Some altimeters are never set to the local QNH.
Using QNE instead of QNH to report altitude
Transponders modes C/S and ADS-B also include a barometric altimeter which is permanently set to the mean pressure at sea level, 1013.25 hPa. There is a corresponding Q code entry, QNE. Let's recap:
QFE = Pressure at the airfield.
QNE = Standard pressure at sea level in the model, 1013.25 hPa.
QNH = Retro-computed sea-level pressure based on QFE.
This QNE-referenced altitude can be negative from time to time when flying at low altitude in high atmospheric pressure. However it is corrected by the radar system before being displayed to ATC operators, by the value of the local QNH (thanks to @JanHudec for clarifying this).
But personal Mode S and ADS-B receivers do not perform this correction by default, as the QNH value is not known.
Negative altitude shown on personal Mode S receiver
Rennes St Jacques (LFRN) is an airport near the English Channel, in France, its altitude is 124 ft. In the theoretical model 124 ft is 5 hPa. When airport pressure is 1018.25, its retro-computed QNH is 1013.25, and equal to QNE. In these conditions, a mode S transponder, which is permanently set to QNE, reports the same altitude than the one shown on cockpit altimeters set to the current QNH.
From time to time, this area is subject to extreme anticyclonic conditions with the airport pressure higher than the theoretical model, e.g. 1047 hPa. The retro-computed QNH is 1042 hPa. The difference with the standard atmosphere is now 28.75 hPa. This corresponds to -776 ft.
When reaching the runway which is at the altitude of 124 ft, the reported altitude will be -776 + 124 = -652 ft. This is what the user of a receiver not performing the QNH correction will see:
Altitude reported as negative by Mode S to SBS-1 receiver, source: F6GKQ (73/Denis)
Geometric altitude
Determining a geometric altitude, that is the distance to mean sea level is actually a very difficult operation, because the sea level (the level after tide variations have been removed) is itself a complex notion. To start with seas and oceans are not all comparable, and in the past different MSL were used locally. In addition currents and local variation of the gravity interfere with the mean sea level. Hopefully these many difficulties had to be solved to build global navigation systems, and this knowledge has been integrated into the systems.
GNSS satellites data are relative to the perfect WGS 84 ellipsoid (GRS 80) representing oblate Earth. MSL on the other hand is represented by the WGS 84 geoid (based on the gravimetric model EGM 96).
The ellipsoid and the gravity model have been defined after conducting a lot of studies and measurement campaigns involving expensive means. The hard work was done and will be updated periodically, data can be used to determine geometric altitude in a split second at virtually no cost for the GNSS user.
The GNSS receiver first uses the satellites to determine its height above the reference ellipsoid. The principle is the position of each satellite relative to the ellipsoid is known and the receiver position relative to visible satellites is determined by multilateration by analyzing the radio signal transmitted by each satellite.
Then the GNSS receiver uses a geoid model, a model describing the irregular sea level, to obtain the local geoid undulation, that is how much the geoid departs from the ellipsoid at the receiving location. Altitude is the difference between ellipsoidal height and undulation.
Why is that so difficult to deal with altitude in aviation
The difficulty to understand such foggy concepts is we first assume the altimeter measures an altitude, while a barometric altimeter actually measures atmospheric pressure and shows it on a dial marked with altitudes obtained according to a theoretical model which is rarely met in real life. Setting a reference is just a way to try to make the model more relevant to the current local situation.
Aviation encounters difficulties to switch from a century old technology, which today causes a lot of problems, to the cheaper and more accurate modern technology which would solve them. This is true for altimeters and air traffic management in general. Hopefully aviation will overcome these challenges.
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edited Nov 1, 2021 at 2:57
answered Oct 28, 2021 at 22:52
minsmins
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1 Doesn't the ATC display convert the altitude from Mode C transponders based on entered local QNH for aircraft below the transition altitude? I thought most do so the controllers can do the required altitude cross-check more easily.Jan Hudec –Jan Hudec 2021-10-31 22:23:48 +00:00 Commented Oct 31, 2021 at 22:23
@JanHudec: Right indeed. Thanks, corrected.mins –mins 2021-11-01 02:59:57 +00:00 Commented Nov 1, 2021 at 2:59
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There are several regions where you could (at least in theory) fly below sea level. For example, in the Dead Sea depression, that is the lowest stretch of land on the planet, reaching a maximum depth of roughly 400 meters below (geoid) mean sea level.
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answered Oct 28, 2021 at 10:28
xxavierxxavier
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6 This goes well beyond theory. Many years ago, I stood at the top of Masada (at 193m/633 feet below sea level) with my dad and we looked down into the cockpits of IDF F-16s as they flew by. We waived. The pilots waived back.FreeMan –FreeMan 2021-10-29 18:00:16 +00:00 Commented Oct 29, 2021 at 18:00
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For here in the US, I give you Furnace Creek Airport at an altitude of -210 feet. The pattern altitude is +789 MSL but you'll be lower on final approach. Late goaround should make for some interesting instrument readings.
I recall reading about a case where the avionics was coded for Furnace Creek being the lowest possible airport and glitched out over the Dead Sea. :( EDIT: answer link
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edited Oct 29, 2021 at 0:20
answered Oct 28, 2021 at 21:02
JoshuaJoshua
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Remember that the altimeter simply shows your "altitude" as a relative reference to the pressure at a known point. We generally use sea level, though it can also be set to reference ground level (again, at a known point). This is commonly used in the UK, known as the QFE, where the pilot will dial in the air pressure at an aerodrome.
So in that sense, yes, your altimeter could show a negative value if you fly below your reference point. I.e., below sea level or below whatever ground level you dialed in.
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answered Oct 28, 2021 at 10:08
NobodyNobody
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Is there a place you know that is below sea level and you can fly above that ?Huntkil –Huntkil 2021-10-28 10:28:26 +00:00 Commented Oct 28, 2021 at 10:28
Nevermind I got two places one above mentioned by Xavier and another is Death Valley. Thanks for help !!Huntkil –Huntkil 2021-10-28 10:30:08 +00:00 Commented Oct 28, 2021 at 10:30
1 @Dan just a minor addition/note, though you are correct that a true definition of altitude is above a relative reference to a pressure at a known point, in aviation "altitude" is almost always (not "generally") a reference of the vertical distance from mean-sea level. (c.f. for EASA see Reg (EU) 923/2012 Article 2(39) and for the UK: CAP1430). The motivation behind this is to not introduce any confusion with Flight level, which uses the Standard pressure setting of 1013 hPa. QFE is not used in commercial operations, rather it is used more regularly with the military and non-commercial ops.tedioustortoise –tedioustortoise 2021-10-28 13:20:48 +00:00 Commented Oct 28, 2021 at 13:20
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30Why is 0 ft a valid target altitude for an autopilot?
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5Why, and how, are barometric altimeters accurate enough for approach and landing?
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6How to most likely get a flight that will go above 40,000 ft?
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हेलो एवरीवन तो आज हम बात करेंगे एक्से स्पेक्ट्रा की आपने एक्सस 12थ में पढ़ी है बीएसी में भी पढ़ी है तो इसका जो स्पेक्ट्रा होता है बेसिकली हम उससे फोकस करेंगे ब उससे पहले हमको जानना पड़ेगा कि एक्से स्पेक्ट्रा रिजनेट कहां से होता है तो कैसे इसकी स्टडी करते हैं सबसे पहले बात समझते हैं कि जो एक्स स्पेक्ट है उसको स्टडी करने के लिए हमको चाहिए होती है एक्सरेज एक्सरेज कैसे क्रिएट करते हैं एक हम एक ट्यूब लेते हैं इसके अंदर हम वैक्यूम क्रिएट करते हैं फिर एक फिलामेंट लेते हैं इसमें हम क्ल ल्ट के ऑर्डर का एक पोटेंशियल अप्लाई करते हैं इस किलो वोल्ट को याद रखना एक पेपर में उन्होने यही पूछा है कि जो हमारा जो हमारा ट्यूब होता है उसमें जो फिलामेंट हम अप्लाई करते हैं उसम अप्लाइड वोल्टेज किस ऑर्डर का होता है तो ये हमारा किलोवोल्ट 10 टू 3 वोल्ट के ऑर्डर का होता है ठीक है अभी फिलामेंट क्या करते हैं गर्म होगा जब फिलामेंट गर्म होगा तो तापाय उत्सर्जन यानी कि थर्मल इशन की वजह से यहां से क्या निकलते हैं इलेक्ट्रॉन और जो इलेक्ट्रॉन निकलते हैं वो इस प्लेट की तरफ अट्रैक्ट होते हैं क्यों अट्रैक्ट होते हैं आकर्षित होंगे क्योंकि यहां पर हमने एक पॉजिटिव एनोड लगा रखा है जो इस एक्सटर्नल किलोवोल्ट से जो है कनेक्टेड है अब इसमें है क्या वैक्यूम वैक्यूम क्रिएट करने से बेनिफिट यह होता है कि इसके अंदर कोई और गैसेस के पार्टिकल्स नहीं होते जिससे कि इन इलेक्ट्रॉन की उन गैस के जो अणु है उनसे कोलिजन नहीं होता अदर वाइज दूसरे फिनोमिना हैपन होते हैं अच्छा अब जब इलेक्ट्रॉन यहां से निकलेगा तो इलेक्ट्रॉन क्या करेगा पहले तो जब हमने यहां पॉजिटिव पोटेंशियल नहीं लगा है उसके पास कुछ एनर्जी ज्यादा है नहीं तो यहां से निकल के आसपास ही इकट्ठा होता रहेगा लेकिन जैसे ही हम यहां पॉजिटिव पोटेंशियल अप्लाई करते हैं तो ये इलेक्ट्रॉन नेगेटिव है यहां पोटेंशियल इसको पॉजिटिव मिला इलेक्ट्रॉन इसकी तरफ होता है अट्रैक्ट और बहुत तेजी से क्योंकि हमने जो पोटेंशियल लगाया बहुत हाई वोल्टेज है इसलिए बहुत तेजी से आता है और इससे टकरा जाते हैं इलेक्ट्रॉन आकर के इससे टकराएगा और यहां टकराने की जो प्रक्रिया है उसको हम बाहर जूम करके समझेंगे ठीक है तो जो इलेक्ट्रॉन आके टकराएगा और यहां से निकल जाती है एक्स रेज ठीक है एक्स रेज कैसे निकल जाती है इलेक्ट्रॉन आया इससे टकराया यहां से एक्स रेज निकल गई इस डायग्राम का सिर्फ इतना ही काम था अब आते हैं हमारे बेसिक जो इसका फंडामेंटल है वो सपोज करो ये हमारा एक जो ये हमारा क्या कहते हैं प्लेट हमने यूज किया टारगेट मटेरियल इस टारगेट मटेरियल की प्रॉपर्टीज और डिस्कस कर लेते हैं फ पढ़ते हैं प्रॉपर्टी में सबसे पहले तो जड की बात करें जड यानी कि परमाणु क्रमांक ये होना चाहिए हाई ठीक है हाई होगा तो क्या होगा हाई होने से यहां पर अगर जड हाई होगा तो नंबर ऑफ इलेक्ट्रॉन ज्यादा होंगे ज्यादा नंबर ऑफ इलेक्ट्रॉन ऑर्बिट कर रहे होंगे तो इलेक्ट्रॉन का कोलिजन ज्यादा होगा तो जड हाई होना चाहिए दूसरा जो उसका मेल्टिंग पॉइंट है ठीक है जो उसका मेल्टिंग पॉइंट है ये भी कैसा होना चाहिए हाई क् अगर आप लो मटेरियल लो मेल्टिंग पॉइंट का कोई ले लिया मटेरियल तो जब यहां पे कोलिजन होगा तो बहुत सारी हीट जनरेट होगी जिसकी वजह से ये क्या करेगा या तो वेपराइज हो जाएगा क्योंकि यहां पे वैक्यूम है या फिर ये लिक्विड में कन्वर्ट हो जाएगा तो इसका मेल्टिंग पॉइंट भी हाई होना चाहिए और जड भी हाई होना चाहिए ये दो प्रॉपर्टीज इंपॉर्टेंट है इनको याद रखना और पेपर में ब पूछा गया है कि जो हम अप्लाई करते हैं वोल्टेज वो किस ऑर्डर का होता है तो किलोवोल्ट के ऑर्डर में होता है इतना सब कुछ हो गया अब यहां पे जो प्रक्रिया होती है यहां पे जो प्रोसेस होती है उसको हम डिटेल में समझते हैं कैसे सपोज करो इस मेटल टारगेट के अंदर बहुत सारे परमाणु हैं उन परमाणु में से किसी एक परमाणु की बात कर रहा हूं उस एक परमाणु के सेंटर में नाभिक होगा जिससे पॉजिटिव चार्ज होगा और उसके चारों तरफ इलेक्ट्रॉन चक्कर लगा रहे होंगे जो हम क्लासिकल पिक्चर में समझते क्वांटम मैकेनिकल ऐसा नहीं होता बट क्लासिकल मैकेनिक्स हम ऐसा सोचते हैं कि एक न्यूक्लियस है उसके चारों तरफ चक्कर लगा रहा है इलेक्ट्रॉन से जिनकी डेफिनेट ऑर्बिट होती ऑर्बिट होती है ठीक है अब क्या होगा ये जो इलेक्ट्रॉन जा रहा है सपोज करो इसमें तीन कंडीशंस बन सकती है पहली कंडीशन समझो ये कोई इलेक्ट्रॉन था और वो जा रहा था ऐसे आराम से इसकी एनर्जी बहुत ज्यादा है अब मोस्ट ऑफ द पार्ट जो है हमारे नाभिक परमाणु का वो कैसा होता है खाली खोकला होता है जैसे ही ये ऐसे जाएगा और जाकर क्या कर करेगा जब ये नाभिक के पास पहुंचेगा तो ये है नेगेटिवली चार्जड कैसा है नेगेटिवली चार्जड और जो न्यूक्लियस है वो है पॉजिटिव चार्ज तो नेगेटिव पॉजिटिव आपस में एक दूसरे को अट्रैक्ट करेंगे तो ये अपने पाथ से क्या करेगा डेविएशन होने की प्रोबेबिलिटी ना के बराबर है क्या करेगा ये जो नेगेटिव चार्ज इलेक्ट्रॉन जाएगा और इससे यहां से ऐसे निकल जाएगा अब ध्यान से देखो इसका पाथ चेंज हो गया अगर कहीं भी किसी की वेलोसिटी चेंज होती तो वहां एक्सीलरेशन होता है रीजन कुछ भी हो बट यहां डेफिनेट है कि एक्सीलरेशन तो है अब पॉजिटिव है नेगेटिव है ये बाद की बात है तो जब भी कोई वेलोसिटी चेंज होगी एक्सीलरेशन डी एक्सशन का फैक्टर तो होगा तो जब ये इलेक्ट्रॉन यहां से ऐसे निकल गया यहां से ये आपका इलेक्ट्रॉन पास हो गया अब ये था नेगेटिव चार्ज ये है पॉजिटिव चार्ज पहले इसने इसको अट्रैक्ट किया होगा जब ये दूर जा रहा होगा तो इसको और ज्यादा अपने पास खींच रहा होगा तो जो इनिशियल वेलोसिटी से ये आया था जितनी वेलोसिटी लेकर ये आया था उससे कम वेलोसिटी लेकर के ये बाहर निकलता है ठीक है जब ये कम वेलोसिटी लेकर बाहर निकला क्योंकि इसमें एक फैक्टर ये भी रहता है कि जो इलेक्ट्रॉन है वो स्क्रीनिंग इफेक्ट देते हैं इलेक्ट्रॉन में नेगेटिव चार्ज है जोसको पास आने का पास आने का विरोध कर रहे होंगे ठीक है तो टोटल मिला के क्या होगा कि इसकी फाइनल वेलोसिटी इनिशियल वेलोसिटी थोड़ी सी कम हो जाती है यानी कि इसमें हुआ है डी एक्सीलरेशन अब इसके पास पहले कुछ ना कुछ एनर्जी थी बाद में भी एनर्जी है बट कुछ ना कुछ एनर्जी इसने लूज कर दी जो एनर्जी उसने लूज किए वही एनर्जी आपको निकल जाती है किसकी फॉर्म्स में एक्स रेज की फॉर्म में लेकिन ये एक्स रेज कौन सी है ये एक्सरेज है कंटीन्यूअसली क्यों क्योंकि जब भी इलेक्ट्रॉन आएगा उसका डी एक्सीलरेशन कंटीन्यूअस होगा इसकी वजह से हमें जो इंटेंसिटी मिलती है एक्स स्पेक्ट्रम में वो कुछ इस टाइप से मिलता है हमें कंटीन्यूअस कर्व मिलता है कुछ ऐसे करके यहां से यहां तक एक कंटीन्यूअस कर्व मिलेगा क्यों कंटीन्यूअस कर्व मिल रहा है क्योंकि जो इलेक्ट्रॉन है उसका डी एक्सीलरेशन एक कंटीन्यूअस प्रोसेस है ठीक है यहां तक समझ में आ गया होगा यह ग्राफ कभी भी जीरो से स्टार्ट नहीं हो सकता क्योंकि जो यहां पे एक्स रेज निकली है उनकी कुछ ना कुछ एनर्जी रहेगी और लडा कभी भी जीरो नहीं होगा अगर लडा आप जीरो कर दोगे तो ए अप लडा से उसकी एनर्जी इनफा हो जाएगी जहां से इसका एमिशन स्टार्ट हुआ इसको हम बोलते हैं लडा मिनिमम ठीक है अब इसका भी हम डेरिवेशन करेंगे बट अभी आप समझ लो लड ये था हमारा फर्स्ट केस इलेक्ट्रॉन आया और बिना किसी से टक्कर के चुपचाप शांति से थोड़ा सा डी एक्सलेट होके बाहर निकल गया यहां से मिलती है हमें कंटीन्यूअस एक्स रेज ठीक है अच्छा अब जो नेक्स्ट चीज है वो समझो क्या हो सकता है इसमें एक कंडीशन ये भी पॉसिबल है कि जो इलेक्ट्रॉन आ रहा था वो क्या करेगा किसी इलेक्ट्रॉन से कोलिजन कर जाए ये पहले तो तो कंटीन्यूअसली निकल के था बाहर बट अभी ये क्या करेगा अभी जो इलेक्ट्रॉन इसमें ऑलरेडी मोशन कर रहे हैं उनमें से किसी इलेक्ट्रॉन से कोलिजन कर गया सपोज करो ये आया और इसके पास किलोवोल्ट ऑर्डर की एनर्जी है और इस इलेक्ट्रॉन से ये जो इसके सबसे इनर मोस्ट में था कहीं भी कोलेजन कर सकता है मान लो उसने यहां पे आके कोलिजन किया जब इस इलेक्ट्रॉन से कोलेजन करेगा तो जो भी हमारे इलेक्ट्रॉन होते हैं जो बॉन्ड होते हैं हम परमाणु के अंदर चक्कर लगा ता इनकी एनर्जी इलेक्ट्रॉन वोल्ट के ऑर्डर होती है और ये है किलो इलेक्ट्रॉन वोल्ट के ऑर्डर का ठीक है तो जब ये से टक्कर करेगा तो इस इलेक्ट्रॉन को क्या करेगा वो बाहर निकाल देगा तो ये इलेक्ट्रॉन तो बाहर निकल गया और इसके पास खुद भी किलोवोल्ट के डर के एनर्जी है तो वो भी यहां रुक नहीं पाएगा वो खुद भी क्या करेगा बाहर निकल जाएगा तो आपके पास दो इलेक्ट्रॉन तो बाहर आ गए जब ये दो इलेक्ट्रॉन बाहर आ जाएंगे तो यहां पे बन जाएगी एक वेकेंसी ठीक है यहां बन गया एक होल कह लो या फिर बन गई है एक वेकेंसी अब ये जो वेकेंसी क्रिएट हुई है तो क्या करेगा ये जो बाहर वाला इलेक्ट्रॉन है ये आएगा अंदर यहां से यहां ट्रांजिशन कर सकता है ओबवियसली आराम सेशन कर लेगा तो सपोज करो जब यहां से यहां इलेक्ट्रॉन जाएगा तो बोहर का जो एटॉमिक मॉडल है वो कहता है कि जब भी कोई इलेक्ट्रॉन अपने अपनी जो ऑर्बिट है उसे दूसरी ऑर्बिट में जाएगा तो वो एनर्जी तो अब्जॉर्ब करेगा या फिर निकाल लेगा ठीक है उत्सर्जित करेगा या अपोजित करेगा जब इलेक्ट्रॉन इसके अंदर जाएगा तो ये क्या कर अपनी एनर्जी को इमिट कर देगा कुछ ऐसे करके तो यहां क्या करेगा ये वाला इलेक्ट्रॉन जब अंदर जाएगा तो अपनी एनर्जी फोटॉन्स की फॉर्म में बाहर निकाल लेगा जो ये इलेक्ट्रॉन के फॉर्म मेंस जो फोटॉन्स के फॉर्म में इसने एनर्जी बाहर निकाली ये भी एक्चुअल में एक्स रेज ही होती है बट इनकी वेवलेंथ अलग होती है तो हमें क्या मिलेगा हमें मिलता है यहां पर एक एक साथ पीक ऐसे करके ठीक है तो ये जो जो सेकंड कंडीशन में इलेक्ट जो फोटॉन्स बाहर निकले पहले फोटॉन्स निकले थे इलेक्ट्रॉन के डी एक्सलेट होने की वजह से वो था हमारा कैरेक्टरिस्टिक एक्सरे ठीक है और अभी इलेक्ट्रॉन जो निकला जो फोटोन निकला है वो इलेक्ट्रॉन के ट्रांजीशन की वजह से निकला है इसलिए इसको हम बोलते हैं कैरेक्टरिस्टिक एक्स रेज और वो थे हमारे कंटीन्यूअस एक्स रेज ठीक है तो ये जो पूरा यहां से यहां तक का कर्व था ये हमारा कंटीन्यूअस है और ये वाला जो पार्ट है ये होता है हमारा कैरेक्टरिस्टिक कैरेक्टरिस्टिक इसलिए क्योंकि ये हर एक परमाणु के लिए अलग-अलग होता है ओके अब किसी में एक होगा किसी में दो होगा वो डिपेंड करेगा कि कितने इलेक्ट्रॉन का टेंशन पॉसिबल है कैसे भाई पहला इलेक्ट्रॉन तो यहां से यहां आ गया अब यहां पे एक वैकेंसी क्रिएट हो गई इस वैकेंसी को फिल करने के लिए ये इलेक्ट्रॉन यहां आ गया और इससे क्या हुआ फिर से एक फोटोन निकल गया ठीक है तो इस टाइप से जितने फोटॉन्स निकलेंगे उतनी ही हमको पीक्स मिल जाती है तो ये थी कैरेक्टरिस्टिक एक्सरे और कंटिन्यू एक्स की पूरी कहानी इसके अलावा जब आप एमएससी में जाते हो तो एक टॉपिक और पढ़ते हो जिसको हम बोलते हैं ऑग इलेक्ट्रॉन कौन सा और जो हमारे जिस टॉपिक के लिए हम पढ़ रहे हैं उस टॉपिक में ऑग इलेक्ट्रॉन सबसे ज्यादा इंपॉर्टेंट है क्यों इंपॉर्टेंट है यहां पे जाके आप कोई पढ़ते नहीं इस चीज को जो अभी तक इलेक्ट्रॉन निकले थे इनमें से कोई भी ऑग इलेक्ट्रॉन था नहीं ये ऑग इलेक्ट्रॉन क्या है ये समझते हैं सपोज करो तो देखो अभी क्या हुआ जब इलेक्ट्रॉन यहां से यहां पर आया देखो आउटर ऑर्बिट से इनर ऑर्बिट में आया तो इसने क्या किया एक फोटोन बाहर निकाला जो ये फोटोन बाहर निकाला जिस ऑर्बिट से आया है उसके बाहर वाले इलेक्ट्रॉन की नज से कम होगी तो ये जो फोटोन है ये क्या करेगा बाहर जाते टाइम पॉसिबिलिटी रहती है कि इस ऑर्बिट में कोई इलेक्ट्रॉन है इस ऑर्बिट के इलेक्ट्रॉन को अपने साथ ले जाए क्यों क्योंकि इसके पास है एनर्जी तो वो एनर्जी इस इलेक्ट्रॉन को दे दे और ये फोटोन क्या करेगा इस इलेक्ट्रॉन को फ्री करके बाहर निकाल दे ठीक है तो ये जो हम इस टाप से जिगजैग लाइन बनाते हैं ये फोटोन को शो करती है और जो स्ट्रेट लाइन होती है ये शो करती है इलेक्ट्रॉन को तो जो ये फोटोन बाहर निकला ठीक है कब निकला कैरेक्टरिस्टिक एक्स रेज के टाइम पे कंटीन्यूअस के टाइम पे नहीं कैरेक्टरिस्टिक एक्स रेज के टाइम पे जो ये फोटोन बाहर निकला अगर वो फोटोन किसी उसी एटम के किसी और इलेक्ट्रॉन ने उसको कर लिया एब्जॉर्ब और वो इलेक्ट्रॉन निकल गया बाहर तो इस इलेक्ट्रॉन को ही हम बोलते हैं ऑग इलेक्ट्रॉन ऑगर इलेक्ट्रॉन ठीक है तो ये वाला जो इलेक्ट्रॉन है वो र इलेक्ट्रॉन है अगर आपको इसमें और डिटेल में पढ़ना है तो आप राजकुमार की बुक में पेज नंबर 169 आराम से पढ़ सकते हो काफी डिटेल में वहा पर लिखा हुआ है बट हमारे लिए इतना जानना काफी है ठीक है तो हमने तीन केसेस इसके पढ़ लिए अलग अलग अब थोड़ा सा डेरिवेशन कर लेते हैं और यह हमारा पूरा कर्व है इसमें कर्व में एक इंपोर्टेंट पॉइंट याद रखना कि जो पी की पोजीशन होती है यह पोजीशन हमारे टारगेट मटेरियल के जो अप्लाइड पोटेंशियल है यानी कि यहां जो पोटेंशियल अप्लाई कर रहे सपोज करो हमने पहले अप्लाई किया था 5 किलोवोल्ट और उसके बाद थोड़ छोटा कर देते हैं पहले हमने अप्लाई किया था 5 किलोवोल्ट उसके बाद दूसरा हमने बा 10 किलोवोल्ट अप्लाई किया है तो क्या होगा कुछ कर्व इस टाइप से आएगा लेकिन जो पीक की पोजीशन है वह चेंज नहीं होती वो वही का वही रखता है तो पहले जहां पर पीक मिल रही थी वही दोबारा भी पीक मिलेगी क्यों क्योंकि जो भी फोटोन यहां से ट्रांजीशन हो रहा होगा चाहे किसी भी कंडीशन में हो इसकी एनर्जी इन दोनों के डिफरेंस के इक्वल होगी अगर मैंने इसका नाम दिया है e1 इसका नाम दिया है e2 तो यह जो फोटोन निकला है इस फोटोन की एनर्जी h न्यू जो होगी वह हो जाएगी e2 - e1 के इक्वल तो यह हमारे मटेरियल प मटेरियल जो होगा उस पर डिपेंड करेगी ना कि अप्लाइड पोटेंशियल पे ठीक है तो यह पूरा कांसेप्ट था अगर आप अ पोटेंशियल अप्लाइड बढ़ाते जाओगे तो यह कर्व थोड़ा ऊपर को शिफ्ट होता जाए थ पहले शिफ्ट होता जाएगा लेकिन इसकी जो पोजीशन है वो वही रहती है ओके ये तो मेन पूरी प्रॉपर्टीज हो गई अब थोड़े से डेरिवेशंस कर लेते हैं देखो इसमें कुछ चीजें इंपोर्टेंट है फार्मूले उनको डिस्कस कर लेते हैं सबसे पहले बात कर लेते हैं काइनेटिक एनर्जी की काइनेटिक एनर्जी किसकी ये जो इलेक्ट्रॉन हम निकाल रहे हैं इसकी काइनेटिक एनर्जी में कितना डिक्रीज मेंट हुआ देखो इसकी इनिशियल वेलोसिटी थी बी आ और फाइनल वेलोसिटी मान लो मैंने कह दिया है bf3 बाहर निकलता है तो पहले यह आया था ज्यादा एनर्जी से बाद में उसकी एनर्जी हो गई कम तो अगर मैं डेल्टा k निकालने की कोशिश करूं ठीक है अगर मैं डेल्टा k निकालता हूं डेल्टा k की वैल्यू हो जाएगी 1 / 2 ज्यादा कौन है पहले वाली यानी 1 / 2m / i स् माइनस बाद में कितनी बची m ब फाइनल स्क्वायर अब ये जो चेंज इन काइनेटिक एनर्जी है यही फोटॉन्स की फॉर्म में इमिट होगा तो तो ये इक्वल हो जाएगा किसके h न्यू के यानी h न्यू एनर्जी का फोटोन इसने बाहर निकाल दिया इस इक्वेशन को हम लिख सकते हैं इस टाइप से ये इंपोर्टेंट इक्वेशन है अब आगे समझो सपोज करो आपसे पूछा जाए कि जो इलेक्ट्रॉन यहां इमिट हो रहा है ठीक है उस इलेक्ट्रॉन की कितनी एनर्जी होगी तो उसके लिए आप ये वाला फार्मूला यूज नहीं करोगे ये फार्मूला कब है जो कंटीन्यूअस वाला पार्ट है उसके लिए कंटीन्यूअस वाले पार्ट में कितनी एनर्जी जो हमारी निकल रही है x से उसकी होगी उसको आप यहां से कैलकुलेट करते हो अच्छा दूसरा केस क्या हो सकता है कि इलेक्ट्रॉन जो हमारा आया ठीक है और उसकी फाइनल वेलोसिटी हो गई जीरो यानी उ सारी की सारी एनर्जी क्या करी इस इलेक्ट्रॉन को दे ठीक है उसकी एनर्जी खत्म हो गई तो उस केस में क्या होगा फाइनल वेलोसिटी उसकी जीरो हो जाएगी तो 1 अपट ए बी आ स्क माइनस 0 इक्वल टू हो जाएगा h अब ये जो फोटोन निकलेगा ये मैक्सिमम एनर्जी का फोटोन होगा इसलिए मैं एन्य मैक्स कह देता हूं ठीक है क्योंकि इसकी एनर्जी मैक्सिमम होगी यहां से अगर मैं लिखना चाहूं तो h स अपन लडा मेन लिख सकता हूं क्योंकि अगर न्यू ज्यादा है तो मिनिमम डेव लेंथ होगी तो ये बन जाएगा 1 अप 2m ब स्क यहां से आप एक चीज और समझ सकते हो कि जो काइनेटिक एनर्जी इसको मिल रही है जितनी काइनेटिक एनर्जी से ये आ रहा है ये आ कितनी काइनेटिक एनर्जी से रहा था जो हमने फिलामेंट और एनोड के बीच में पोटेंशियल लगाया था कितना पोटेंशियल लगाया था सपोज करो v न उस नॉट से उसको एनर्जी मिलती है न कैसे सपोज को दो प्लेट्स हैं और एक इलेक्ट्रॉन यहां से चलना शुरू करता है यहां तक जाता है इन दोनों प्लेट्स के बीच में जो पोटेंशियल आपने अप्लाई किया है वो ब है तो आप इसे इलेक्ट्रोस्टेटिक में पढ़ ही चुके होंगे जो उसकी काइनेटिक एनर्जी हो जाएगी वो हो जाएगी e ब न ठीक है तो यहां से आप समझ सकते हो इसकी जो वैल्यू होगी वो हालांकि इसके इक्वल भी होगी इक्वल टू बन जाएगा e बीन ठीक है यहां से अगर मैं लडा मिनिमम लिखना चाहूं लडा मिनिमम बन जाएगा मेरा h स अप e बीन ये इक्वेशन इंपोर्टेंट है ठीक है ये क्वेशन जन नहीं पूछ जाती बट ये इक्वेशन बहुत इंपोर्टेंट है इस लडा मिनिमम कैलकुलेट कर सकते हो कि कहां से आपका जो एक्सरे का मिशन है वो स्टार्ट हुआ है ठीक है तो जो एमिटेड आपकी एक्सरे है उसकी आप वेवलेंथ यहां से कैलकुलेट कर सकते हो यह फार्मूला कुछ कुछ उसकी तरह जो आपने पढ़ा होगा कितना 150 अप अंडररूट बी कर जो आपको पता है आपको पता है कि का फार्मूला तो कमेंट्स में बता देना मुझे ठीक है चलिए वापस आते इस टॉपिक प तो जो एक्सरे निकला है ठीक है अब बहुत क्वेश्चन पूछा जाता है इस टाइप से कि जो एक्सरे होता है उसकी वेवलेंथ अप्लाइड पोटेंशियल पर किस टाइप से डिपेंड करती है क्वेश्चन बहुत अच्छा है तो जो वेवलेंथ होती है अगर लडा मन ना लि ल लिखना चाहू तो लडा है व में प्रोपोर्शनल होता है 1 अप p यहां पर अंडर रूट नहीं आता अंडर रूट कहां मिलता है अंडर रूट आपको मिलता है डी बग वेवलेंथ के टर्म्स में जब आप डी वेवलेंथ डिस्कस कर रहे होते हैं वहां पर आपका लडा 1 अप र ब के डिपेंड करता है बट यहां पे नहीं होता ठीक है ये चीज ध्यान रखना लडा का फार्मूला आ गया 1 अप प्रोपोर्शनल 1/2 यहां से आप लडा कैलकुलेट भी कर सकते हो इससे बेस हम क्वेश्चन भी कर लेंगे जो नेक्स्ट वीडियो में आपको मिल जाएंगे ठीक है तो ये तो था हमारा वेवलेंथ लडा मिनिमम निकालने का और अगर आप चाहो तो यहां से एन्यू मैक्सिमम निकालते हो क्योंकि इसकी वैल्यू भी न के इक्वल होगी मलक ये फार्मूला बहुत इंपोर्टेंट है तो ये थी हमारी एनर्जी के टर्म्स में इसके अलावा एक फार्मूला और है जो हम ड्राइव तो नहीं करते जो मोजले यूज करते हैं इसको फ देते हैं ते ठीक है ये लडा जो है प्रोपोर्शनल है 1 अपन बी बी कह लो या बी न कह लो अच्छा जो हेनरी मोजले का लॉ है उसको भी यही डिस्कस कर लेते हैं उन्होंने बोला कि जो फ्रीक्वेंसी होती है या फिर वेव लेंथ कह लो या वेव नंबर कह लो तीनों ही तीनों तो 1 अप लडा का हमने फार्मूला पढ़ रखा है वो है r इनफिनिटी z माइ सिगमा होल स्क्वा सिग्मा होता है आपका स्किनिंग इफेक्ट कांस्टेंट और यहां से मिलता है 1 अप n f स् - 1 अप n स् n और n क्या है जो प्रिंसिपल क्वांटम नंबर्स है कौन सी स्टेट से कौन सी स्टेट में आपका इलेक्ट्रॉन मूव कर रहा है ठीक है तो यहां से आप 1 अप लडा कैलकुलेट कर सकते हो यहां जो सिग्मा है ये है हमारा स्क्रीनिंग इफेक्ट पैरामीटर कांस्टेंट ठीक है लिख लेते हैं सिग्मा जो है इसकी वैल्यू होती है दो एक तो वैल्यू होती है वन और दूसरी वैल्यू होती है 0.76 वन होता है के सीरीज के लिए और 0.76 होता है एल सीरीज के और एल सीरीज क्या है ये समझ लेते हैं देखो अगर आपका इलेक्ट्रॉन इनर मोस्ट ऑर्बिट से निकलेगा तो यहां से जो इलेक्ट्रॉन निकलेगा ठीक है जो ट्रांजिशन आपका कहीं से भी हो रहा बट अगर यहां हो रहा है तो वो हमारा के सीरीज में आता है और अगर सेकंड वाले में हो रहा है या इनर मोस्ट में हो रहा है उसके बाद वाले फर्स्ट वाले से हो रहा है ठीक है तो उसको हम एल सीरीज बोलते हैं के एल एम एन जो आपने पढ़ रखे हैं ऑलरेडी यहां से इलेक्ट्रॉन जो निकलेगा बाहर से आने के बाद उसको हम निकालेंगे के सीरीज से यहां से अगर कोई आ रहा है ट्रांजीशन तो वो होगा एल सीरीज सपोज करो जो यहां पे ये ट्रांज दूसरे से पहले में आ रहा है जो पहले में आ रहा है इसलिए ये है हमारा के सीरीज का और अगर हमारा इलेक्ट्रॉन यहां से यहां आया होता यहां से यहां या बाहर से कहीं यहां आया होता तो उसको हम बोलते एल सीरीज को ठीक है तो सिग्मा जो है हमारा स्किलिंग कफिट इससे आप आराम से कर सकते हो स्किलिंग इफेक्ट कफिट या फिर स्किनिंग कफिट कहते हैं तो ये पूरा फार्मूला था और ये फार्मूला इंपॉर्टेंट है एक सेट के पेपर में डायरेक्टली ये फार्मूला पूछा गया है ठीक है तो आप ये कैलकुलेट नहीं कर सकते वहां ड्राइव नहीं कर सकते आपको ये डायरेक्टली याद होना चाहिए तो एक्स स्पेक्ट में इतना ही था सबसे इंपोर्टेंट पॉइंट आपको ध्यान रखना है क्या पॉइंट था इंपोर्टेंट कि जो इसकी पीक है एक्स स्पेक्ट्रा की ये मटेरियल पर डिपेंड करती है अप्लाइड पोटेंशियल पर डिपेंड नहीं करती दूसरी बात जब भी आप एक्स स्पेक्टर डील कर रहे होंगे तो आपका जड हाई होना चाहिए मेल्टिंग पॉइंट में हाई होना चाहिए ताकि नंबर ऑफ इलेक्ट्रॉन ज्यादा हो और आपको अच्छा इंटेंसिटी कर्व मिल सके एक्स रे स्पेक्ट ठीक है तो आई होप आपको एक्सरे स्पेक्ट समझ में आ गया होगा इस बेस क्वेश्चन भी कर लेंगे अगर आपको कोई डाउट है तो कमेंट सेक्शन में आसानी से पूछ सकते हो थैंक यू सो [संगीत] मच [संगीत] that's |
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Tetrahedron inequality
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Do we have an analogue to triangle inequality in 3-D say tetrahedron inequality(or any other relation), which once satisfied by any six real numbers; implies an existence of a tetrahedron with those side lengths?
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edited Oct 21, 2023 at 21:12
KReiser
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asked Oct 3, 2013 at 4:51
ARiARi
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added geometry tag. I don't think algebraic geometers studied this sort of thing.
– achille hui
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On any non-degenerate tetrahedron, pick any vertex and let a,b,c be the length of edges connected to it. Let a~,b~,c~ be the length of corresponding opposite edges.
The edges on all six faces will satisfy triangular inequalities. These inequalities can
be summarized in a single statement:
min(a+b+c,a+b~+c~,a~+b+c~,a~+b~+c)>max(a+a~,b+b~,c+c~)(1)
Furthermore, the corresponding Calyer-Menger determinant is positive. Up to a scaling factor,
it means:
∣∣∣∣∣∣∣∣0111110a2b2c21a20c~2b~21b2c~20a~21c2b~2a~20∣∣∣∣∣∣∣∣>0(2)
Given any 6 positive numbers, if (∗1) and (∗2) are satisfied, then it can be realized as the edge lengths of a non-degenerate tetrahedron.
For a proof of this, please see the paper
Edge lengths determining tetrahedrons
by Karl Wirth and Andre S. Dreiding.
Side Notes
About the inequalities of face areas mentioned in Glen O's answer. It is also a sufficient condition. More precisely, give any 4 positive numbers A1,A2,A3,A4. If can be realized as the face areas of a non-degenerate tetrahedron if and only if the following 4
inequalities are satisfied:
⎧⎩⎨⎪⎪⎪⎪A2+A3+A4>A1A3+A4+A1>A2A4+A1+A2>A3A1+A2+A3>A4
For a proof of this, please see answers in a related question (Inequality for each a,b,c,d being each area of four faces of a tetrahedron).
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edited Apr 13, 2017 at 12:20
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answered Oct 3, 2013 at 5:21
achille huiachille hui
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There won't be a single inequality between the six edges (there'll be a set of four inequalities, one for each side), but there should be an inequality between the four faces.
That is,
A1+A2+A3≥A4
Where 1, 2, 3, and 4 are the four faces.
This would resolve as a set of cross products - if you have the origin as one point, and the other three points as A, B, and C, then you have
|A×B|+|A×C|+|B×C|≥|(A−C)×(B−C)|=|(A×B)−(A×C)−(B×C)|
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answered Oct 3, 2013 at 5:02
Glen OGlen O
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On any non-degenerate tetrahedron, let a,b,c be the lengths of edges which form a triangle. Let d,e,f be the lengths of corresponding opposite edges. The edges on all four faces will satisfy triangular inequalities. These inequalities can be summarized in a single Boolean expression:
(a+b>c)(c+d>e)(b+d>f)(a+e>f)andandandand(a+c>b)(c+e>d)(b+f>d)(a+f>e)andandandand(b+c>a)(d+e>c)(d+f>b)(e+f>a)andandand(1)
Furthermore, the corresponding McCrea determinant is positive:
∣∣∣∣∣2d2d2+e2−c2f2+d2−b2d2+e2−c22e2e2+f2−a2f2+d2−b2e2+f2−a22f2∣∣∣∣∣>0(2)
Given any 6 positive numbers, if (1) and (2) are satisfied, then it can be realized as the edge lengths of a non-degenerate tetrahedron.
Please see the paper by Mussa:
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edited Oct 21, 2023 at 21:14
julio_es_sui_glace
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answered Oct 21, 2023 at 20:39
Thomas KeslerThomas Kesler
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13538 | https://www.sciencedirect.com/topics/materials-science/atmospheric-density | Atmospheric Density - an overview | ScienceDirect Topics
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Atmospheric Density
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Atmospheric density is defined as the measure of mass per unit volume of a planet's atmosphere, which varies with factors such as time of year, latitude, and dust levels, particularly affecting the performance of entry vehicles during atmospheric entry.
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You might find these chapters and articles relevant to this topic.
Chapter
Keplerian orbits
2012, Spacecraft Thermal ControlJosé Meseguer, ... Angel Sanz-Andrés
3.3.1 Non-gravitational forces
Non-gravitational forces are:
■
Aerodynamic forces (drag atmospheric friction, opposite to the velocity vector, and lift, perpendicular to the velocity vector). They are proportional to the density, and therefore more relevant in low Earth Orbits, at the perigee, as density decreases exponentially with altitude. This effect (mainly drag) reduces the energy of the orbit and therefore produces the reduction of the apogee height, reducing the eccentricity (orbital decay). The atmospheric density fluctuates considerably depending on solar activity, and therefore the estimations of the spacecraft lifetimes are only approximations, whose margin of error can be wide (between 10% and 50%).
■
Drag due to solar wind.
■
Solar radiation pressure (important for spacecraft with large solar panels).
■
Drag due to the interaction of induced current at the spacecraft and the Earth’s magnetic field (only significant in case of tethers), which could be useful in the future to remove spacecraft from Earth orbits.
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Book 2012, Spacecraft Thermal ControlJosé Meseguer, ... Angel Sanz-Andrés
Chapter
Orbital perturbations in the two body motion
2011, Orbital Mechanics and Formation FlyingPedro A. Capó-Lugo, Peter M. Bainum
3.4 The near-Earth atmosphere effects
The near-Earth atmosphere effects happen at altitudes approximately up to 10,000 km . The pressure, temperature, density, and composition of the atmosphere can affect the orbital motion of a satellite. In general, these parameters are a function of the geometry (latitude, longitude, and altitude) of the orbit and the time (Spring, Summer, Fall, and Winter) .
The Earth’s atmosphere can be classified depending on the temperature. The following are the classification of the regions of the Earth’s atmosphere according to temperature:
a.
Troposphere – Region nearest the surface having more or less uniform decrease of temperature with altitude. The temperature lapse rate is −6.5°C/km. This atmosphere contains the weather, has a convective equilibrium with the Sun, and warms the Earth. The troposphere extends to about 6 km in altitude from sea-level.
b.
Tropopause – This region occurs between 6 and 18 km of altitude (and is higher over the Equator). This region has high winds and has the highest cirrus clouds. The temperature declines to –60°C at the top of tropopause. This temperature occurs because the troposphere is not being heated by incident energy and only part of the visible light of the spectrum passes through the ionosphere.
c.
Stratosphere – The region just above the troposphere having a nominally constant temperature. Thicker over the poles, and thinner to non-existent over the equator. At about 25 km altitude of the stratosphere and in the middle latitudes, ozone is being formed.
d.
Mesosphere – The region of the first temperature maximum which is below the major temperature minimum. This variation in temperature is observed at about 80 km altitude within this region. The mesosphere is in irradiative equilibrium between the ultraviolet ozone heating by the upper fringe of the ozone region, and the infrared ozone and carbon dioxide cooling is radiated into Space.
e.
Thermosphere – The region of rising temperature above the major temperature minimum beginning at 80 km altitude and has no upper altitude limit. This is the domain of the aurora. The temperature rises at the base of the thermosphere. The thermosphere has frequent molecular collisions to maintain thermodynamic equilibrium. Potentially, the enormous infrared irradiative cooling by carbon dioxide is not actually realized owing to inadequate conditions.
Also, the Earth’s atmosphere can be classified by its composition. The composition classification is described in the following list :
a.
Homosphere – The region of substantially uniform composition. This region has a constant mean molecular weight that extends to between 80 and 100 km of altitude. The upper part of the homosphere is called the homopause. The composition of the homopause changes due to the dissociation of carbon dioxide. This region is composed of the troposphere, stratopause, and stratosphere.
b.
Heterosphere – The region of significantly varying composition above the homosphere and extending significantly out to Space. In this region, the disassociation of nitrogen and the diffuse separation sets in lighter atoms and molecules rising to the top. The heterosphere contains the ionosphere and exosphere. The ionosphere is commonly known because it is the region of sufficiently large electron density that affects radio communication.
Table 3.1 shows the composition of the atmosphere at the Earth’s surface in terms of percentage . The total mass of the Earth’s atmosphere due to its composition is 29 grams/molecule. It is expressed in terms of mass because it can vary depending on the formulation that is used.
Table 3.1. Composition of the atmosphere at Earth’s surface
| Composition of the atmosphere | Composition percentage |
--- |
| Nitrogen (N 2) | 78.09 |
| Oxygen (O 2) | 20.95 |
| Argon (Ar) | 0.93 |
| Carbon Dioxide (CO 2) | 0.03 |
| Neon (Ne) | 1.8×10−3 |
| Helium (He) | 5.2×10−4 |
| Krypton (Kr) | 1.0×10−4 |
| Hydrogen (H 2) | 5.0×10−5 |
| Xenon (Xe) | 8.0×10−6 |
| Methane (CH 4) | 5.0×10−6 |
| Ozone (O 3) | 7.0×10−6 |
As explained previously, the pressure, temperature, composition, and density are a function of the geometry of the Earth and time. There are five major phenomena that result in variations of the Earth’s atmosphere due to time. These are the following:
a.
Diurnal Variation – The density of the atmosphere varies in a 24 hour period. The maxima due to solar energy cause the gas molecules to ‘boil up’ from lower regions of higher density.
b.
Semi-annual Variation – The effects due to this cause are not well understood. This variation may be associated with the high altitude winds which move along the lines of longitude as summer and winter poles of the Earth are interchanged. This variation is world-wide at all latitudes, longitude, and high altitudes. The maximum density occurs during mid April and October, and the minimum density occurs during mid July and early January.
c.
Annual Variation – The density of the Earth’s atmosphere indicates annual change in density at particular latitudes. There is a peak of maximum atmospheric density in the northern latitudes in the summer and in the southern latitudes in the winter. When the Sun is far into the northern hemisphere, an increase in the density of the atmosphere occurs in the northern hemisphere at satellite altitudes.
d.
Solar Cycle – There is an increased number of sun spots occurring approximately in 11 year cycles. These variations in solar activity can cause variations in the Earth’s atmosphere.
e.
Geomagnetic Activity – There are short periods of time in which the Earth’s atmospheric density varies by an order of magnitude with magnetic storm activity.
The Earth’s atmosphere results in a perturbation known as the aerodynamic drag force. The drag force is a surface force that acts against the motion of the vehicle . For an orbit , the aerodynamic drag force is directed along the velocity vector of the satellite and interacts with a contact surface of the satellite. The drag force is expressed as,
(3.41)F A=1 2 ρ R v 2 C D S
where ρ is the atmospheric density, v is the velocity of the spacecraft in the orbit, C D is the drag coefficient, and S is the surface area in contact with the Earth’s atmosphere. In many satellites, the drag coefficient is between 2.2 and 2.5 . The atmospheric density is described as follows ,
(3.42)ρ R=ρ 0 e−R−r p H
ρ 0 is the atmospheric density at the radius of perigee, and H is the density scale height. The scale height is the distance at some altitude at which the atmospheric density will change by a factor of 1/e. The density and the scale height can be obtained from different atmospheric models. References and provide tabulated values for the 1976 atmospheric density model. This model is commonly used for the development of control systems. Equation(3.41) is used to specify the aerodynamic force associated with the Earth’s atmosphere, but the surface force vector is expressed as follows,
(3.43a)F→A=1 2 ρ R v 2 C D S V^
V^ is the unit vector of the velocity vector of the satellite. To understand the effects of the aerodynamic force on the orbital elements, it is assumed that the drag force has dominant secular effects against the relative motion of the in-plane motion of the satellite . Using this assumption,V^=−ι^B;¯ then, the Lagrange planetary equations for the effects of the aerodynamic drag force are equal to,
(3.44a)da dt=−2 n a 3 1−ϵ 2 μ ρ R v 2 B C
(3.44b)d ϵ dt=−n a 2 1−ϵ 2 μ ρ R v 2 B C cos f+cos E
(3.44c)di dt=d Ω dt=0
(3.44d)dω dt=−n a 2 1−ϵ 2 μ ϵ ρ R v 2 B C 1+R p sin f
B C is the ballistic coefficient and is equal to C D S/2 m. m is the mass of the satellite. The semimajor axis and the eccentricity of the orbit keep decaying with respect to time. In addition, there is a variation in the argument of perigee that causes a regression of the semimajor axis. From equation(3.44a) and (3.44b), the orbit decays toward the center of the Earth as time increases; and the decaying orbit forms a spiral pattern moving toward the center of the Earth. Figure 3.7 shows the spiral pattern of the decaying orbit. The inclination angle and the RAAN are not affected by the atmospheric drag force. On the other hand, the atmospheric drag can be used to perform plane change maneuvers and to stabilize the orientation of the satellite .
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Figure 3.7. Orbit affected by the atmospheric drag
As shown in Figure 3.7, the atmospheric force acts against the satellite motion so that the orbit size keeps decreasing as time increases. In addition, this force causes a variation in the energy associated with the orbit. The work done by the atmospheric force on the orbit can be described as,
(3.45)dW=−F A Rdf⇒Δ W Δ N=−∮DRdf
where ∆W/∆N is the energy loss per orbit. Assuming a circular orbit, the energy loss per orbit can be written as,
(3.46)Δ W Δ N=π C D Sρμ
The energy associated with a circular orbit is equal to,
(3.47)W=U=−mμ 2 a=−mμ 2 R
Equation(3.47) describes the constant orbital energy for a conservative system. The conservation of system energy implies that a change in potential energy equals to the corresponding change in kinetic energy. Using this relation, equation(3.47) can be rewritten as follows,
(3.48)Δ W Δ N=−mμ 2 R 2 Δ R Δ N
∆R/∆N describes the variation of the radius of the circular orbit per orbit. Equating equation(3.48) and equation(3.46), the change in the radius of the circular orbit per orbit can be written as,
(3.49)Δ R Δ N=4 πρ g 0 R 2 B C
Because of the decrement in the semimajor axis per orbit due to the atmospheric force, the period of the circular orbit is decreasing per time as follows,
(3.50)Δ P Δ N=2 π R 3 2 μ 1 2
Dividing equation(3.49) by equation(3.50), the change of the period per change in the radius of the circular orbit can be defined as follows,
Δ P Δ N=2 ρ g 0 μR 1 2 B C
From which,
Δ P=P 2−P 1=B C 2 μ 1 2∫R 1 R 2 dR ρ R 1 2
Using equation(3.42) and knowing that R=s¯+R E, the integral of the previous equation can be written as,
(3.51)P 2−P 1≅B C H ρ 0 2 μ R E 1 2 e S¯1 H−e S¯2 H
where S¯1 and S¯2 are the initial and final orbit altitudes, respectively. Equation(3.51) provides the orbit lifetime for a spacecraft to reenter the Earth’s atmosphere; and, Figure 3.8 shows the satellite lifetime. Wolverton shows the solution of more complex formulations for the orbit lifetime of an elliptical orbit. The atmospheric drag force is a dominant perturbing force for a satellite under an altitude of approximately 800 km.
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Figure 3.8. Satellite lifetime
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Book 2011, Orbital Mechanics and Formation FlyingPedro A. Capó-Lugo, Peter M. Bainum
Chapter
SOLAR MEASUREMENTS
1998, Applied SpectroscopyJ. MURRAY STEWART
I Introduction
The sun's energy at the earth's surface varies continuously in time and space. The solar constant derived from ground-based measurements has varied from 1323 to 1428 W m−2 (1). Outside the atmosphere (air mass = 0) the sun's energy is more predictable (American Society for Testing and Materials [ASTM, E490]). At the annual mean solar distance from the sun, we define a solar constant of radiation for the earth. This is the measured amount of solar energy flux incident normally on a unit area in a unit of time (1.373 ± 0.008 × 10 6 erg/sec cm 2). Typical atmospheric absorbers (water, ozone, CO 2, dust, and industrial pollutants) scatter and absorb energy at selective wavelengths over the solar range from 0.3 to 2.5 μm. Scattered skylight plus the direct sunlight is referred to as the global radiation. It has a different spectral energy distribution than direct sunlight at the earth's surface, adding energy to blue and ultraviolet (UV) wavelengths (2, 3). The absorption effects of average amounts of atmospheric absorbers, superimposed on the solar spectral energy distribution above the earth's atmosphere, simulate solar energy at the earth's surface. A number of proposed standard solar irradiance curves (W/m 2 μm) are used to characterize solar properties of materials (4). When the sun is overhead, solar energy penetrates unit thickness of air (air mass = 1). The absorption increases as the path length increases. Both vary approximately as the reciprocal sine of the sun's altitude angle. When the sun is 30° above the horizon, for example, the solar energy must pass through twice as much air (air mass = 2) as it would if the sun were at the zenith position. A small correction must be applied to this simple relationship to correct for atmospheric refraction, especially if the sun is lower than 30° from the horizon (2) (Fig. 1).
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FIG. 1. Air masses describe the path length through the atmosphere(25).
Atmospheric density decreases disproportionately with altitude. Most of the atmosphere is below an altitude of 57 miles and 20% of it is under 1 mile above sea level. All these attributes of solar energy are constantly changing. A constant value must be chosen for each in order to compare the solar spectral properties of materials. Effective communications about an object's solar properties must be limited to its spectral response under specified solar energy conventions and under standardized measuring conditions.
Pyroheliometers, pyranometers, thermocouples, bolometers, solar arrays, and spectroheliographs measure solar energy directly (5). They are radiometers that convert solar heat into electrical current (6). Their measurement fluctuations are generally so large over time, due primarily to moving clouds, that average values are the only meaningful quantities of solar energy associated with an object's solar properties at sea level. This chapter will not deal with direct solar measurements of this type. Discussion will be limited to laboratory spectrophotometric measurements of specimens and to the indirect calculations of their solar spectral properties (7). For example, flat glass products filter sunlight to reduce heat inside cars and buildings. Spectrophotometric transmittance and reflectance data can be used to calculate a specimen's protective power from conventional solar energy (8). Most of the terms used in this article are defined in ASTM Standard Definitions of Terms and Symbols Relating to Molecular Spectroscopy (9).
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Book 1998, Applied SpectroscopyJ. MURRAY STEWART
Chapter
Keplerian orbits
2012, Spacecraft Thermal ControlJosé Meseguer, ... Angel Sanz-Andrés
3.3 Orbit perturbations
Orbit perturbations are the deviations of the true orbit from the reference orbit due to several reasons. The main causes are: non-gravitational forces, third-body interactions, and non-homogeneous mass distributions or non-spherical shape (non-spherical mass distributions).
3.3.1 Non-gravitational forces
Non-gravitational forces are:
■
Aerodynamic forces (drag atmospheric friction, opposite to the velocity vector, and lift, perpendicular to the velocity vector). They are proportional to the density, and therefore more relevant in low Earth Orbits, at the perigee, as density decreases exponentially with altitude. This effect (mainly drag) reduces the energy of the orbit and therefore produces the reduction of the apogee height, reducing the eccentricity (orbital decay). The atmospheric density fluctuates considerably depending on solar activity, and therefore the estimations of the spacecraft lifetimes are only approximations, whose margin of error can be wide (between 10% and 50%).
■
Drag due to solar wind.
■
Solar radiation pressure (important for spacecraft with large solar panels).
■
Drag due to the interaction of induced current at the spacecraft and the Earth’s magnetic field (only significant in case of tethers), which could be useful in the future to remove spacecraft from Earth orbits.
3.3.2 Non-spherical mass distributions
The shape of the Earth can be approximated by an oblate spheroid, as would be formed by a rotating fluid, with smaller deviations from this oblate shape.
The effects of oblateness over short times (short period variations, which affects the semi-major axis, a, eccentricity, e, and inclination, i) average to zero over an orbit. The long-term effects produce cumulative secular variations in the line of nodes (the right ascensions of ascending nodes), the line of apsides (the argument of perigee), and the mean angular motion.
The regression of the line of nodes can be used to change the orientation of the orbit one rotation per year in the inertial reference frame, an amount of 0.986 degrees/day in such a way that the orientation of the orbit plane remains fixed relative to the Sun (Sun-synchronous orbits).
The regression of the line of apsides, or secular variation of the argument of perigee, ω, obviously applies to orbits with e≠ 0 (non-circular). It is positive for equatorial orbits (it rotates in the direction of motion), is negative for polar orbits (it rotates in the opposite direction of motion), and at i=63.4° it becomes zero, and therefore it remains fixed. This property has applications such as in the so-called Molniya orbits.
These interactions decrease as orbit altitude increases.
3.3.3 Interactions with third bodies
For an Earth satellite, as the distance from the Earth increases, the perturbations coming from the Earth’s oblateness decreases, and the gravitational forces of the Sun and the Moon start to become dominant. In some cases the influence is so great that it defines the motion, as is the case with satellites placed in the Lagrangelibration points (where the gravity effects of the bodies forming the system are compensated). In satellite applications, these interactions mainly affect satellites in geostationary orbits.
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Review article
Lidar with superconducting nanowire single-photon detectors: Recent advances and developments
2022, Optics and Lasers in EngineeringYanqiu Guan, ... Peiheng Wu
4.1 Atmospheric Lidar
Atmospheric Lidar has become a powerful tool for atmospheric detection with fine temporal and spatial resolution, high detection accuracy, and continuous profile data acquisition capabilities. It can be used to detect aerosols, clouds, atmospheric density, ozone, greenhouse gases, wind fields, visibility, etc. Lidar working in near-infrared has many advantages in atmospheric detection. First, it is relatively safe for human eyes. Compared with visible light commonly used in Lidar systems, the absorption cross-section of water vapor in near-infrared is small. It can easily penetrate clouds and fog and has excellent detection capability in adverse weather. In addition, the solar background radiation that affects the SNR of Lidar systems is small in near-infrared, so continuous low-noise detection can be achieved day and night. At present, SPDs have been widely used in atmospheric detection Lidar systems.
In 2015, Xia et al. constructed a micropulse Lidar based on upconversion SPDs (SDE@1550 nm=15%, DCR=40 Hz) and continuously monitored the atmospheric visibility over 24 h. A commercial InGaAs avalanche photodiode (APD) was also used for the same experiment, and it was found that the SNR was reduced by two orders . Lidar systems based on upconversion SPDs are also used to achieve wind field detection. Compared with the data measured by the ultrasonic anemometer, the standard deviation of the wind speed difference is 1.04 m/s, and the standard deviation of the wind direction difference is 12.3° . The abovementioned atmospheric Lidar systems are all based on direct detection, and coherent detection Lidar also plays an important role in atmospheric detection. Coherent Doppler Lidar can simultaneously detect the atmospheric deflection ratio and wind field [112,113]. Through in-depth analysis of the power spectrum of the coherent detection Lidar echo signal, additional atmospheric parameters such as cloud height and cloud thickness, rain speed, wind shear, and turbulence dissipation rate can be inverted . In general, SPAD is commonly used in near-infrared atmospheric Lidar. As mentioned in Section 3, the comprehensive performance of SNSPD is better than that of SPAD, so atmospheric Lidar based on SNSPD was developed later.
The research team of the University of Science and Technology of China built a dual-frequency Doppler Lidar based on SNSPD to detect wind fields at a height of 2.7 km with a vertical spatial-temporal resolution of 10 m/10 s. In addition, their group used SNSPD to achieve an all-fiber micropulse polarization Lidar with a polarization-maintaining structure and continuously observed the linear depolarization rate of aerosols for more than 48 h. The results are highly consistent with the data detected by the sunphotometer, 532 nm Lidar and the weather forecast. Recently, Yu et al. realized photon-counting distributed free space spectroscopy based on a comb-referenced frequency-scanning laser and SNSPD, which can be used to obtain CO 2 and HDO concentrations. Compared to the methods that obtain only column-integrated spectra over the kilometer scale, the range resolution is improved by 2–3 orders of magnitude in this work .
In 2017, our team used SNSPD-based Lidar to detect sea fog within a 180 km diameter area . The sea fog echo signals in the range of 42.3–63.5 km and 53.2–74.2 km were detected, reflecting the fog concentration and moving speed. This year, the Lidar system has been optimized, and a multipixel SNSPD with four-photon resolution capability has been used to achieve a longer detection range. As mentioned in Section 5.2, clouds at 106 km and mountains at 200 km were successfully detected. The photon number resolution (PNR) capability was used to distinguish the echo signals of fog and mountains. In long-range Lidar systems, the limit distance mainly depends on the SNR of the output signal, where the SNR is mainly determined by the SDE and DCR of SNSPDs. The system mentioned above already has close to saturation internal quantum efficiency and extremely low DCR. However, there is still the potential to improve the coupling efficiency of the optical system and the absorption efficiency of SNSPD. Therefore, this result does not represent the detection limit of SNSPD-based single-photon Lidar. It is necessary to optimize the optical system and use an optimized SNSPD to explore its limit detection capabilities.
Although few Lidar applications are based on SNSPDs, their excellent detection capabilities can be seen from the above applications. It is suitable for extremely low echo or extremely long-distance atmospheric Lidar detection. Combined with its extremely high sensitivity and detection accuracy, it is expected to be applied to high-altitude atmospheric particle detection and airborne Lidar mapping in the future.
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Journal2022, Optics and Lasers in EngineeringYanqiu Guan, ... Peiheng Wu
Chapter
Database for biomechanical engineering design
2006, Biomechanical Engineering of Textiles and ClothingR.M. Wang, ... X. Zhang
General data
This is the summary description of the object. Sometimes general data includes the following items:
•
Design file and product identification.
•
Customer information: name, address and so on.
•
Specification of the product’s special considerations, and instructions.
•
Commerce information.
Property data
This kind of data is about the properties of objects; it includes the following items:
•
Objects structure description, including macroscopic and microcosmic: length, thickness, warp count and so on.
•
Object mechanical description: warp bend rigidity, shearing modulus.
•
Object physical description: thermal conductivity, integral heat of sorption.
Geometric data
This data is about the geometric characteristics of the object; it includes the following items:
•
Object’s two-dimension geometric description.
•
Object’s three-dimension geometric description.
•
Object’s mesh data description.
•
Object’s other format for analyzing process.
Design data
This kind of data is about design requirements; it includes design environment and specified constraints. The items are as follows:
•
Design condition requirement: air density, atmospheric pressure and so on.
•
Constraints: specific skin blood-flow rate, thermal conductivity of body tissue.
•
Design method description.
•
Design processing description.
Project data
This is about design data on previous projects. In cases where a current product has evolved from existing products, duplication of data files should be avoided; the designer can simply build current design data files on the basis of relevant existing ones. The items are as follows:
•
Project basic data, designer’s name, date.
•
Project processing status.
•
Project result.
•
All documents for this project.
Tables 15.1 to 15.3 describe the fabrics, the fibers, and the human body.
Table 15.1. Fabrics
| General data | Structural data | Mechanical data | Physical data |
--- --- |
| Fabric ID | Finishing condition | Poisson’s ratio | Surface volume ratio |
| Fabric name | Length | Warp tensile modulus EA | Absorptivity of outer surface of clothing |
| Fabric type | Width | Weft tensile modulus EB | Thermal conductivity of fabric |
| Fabric picture | Thickness | Warp bend rigidity | Thermal capacity |
| Company | Area density | Weft bend rigidity | Water vapor diffusion |
| Price data | Volume density | Compression modulus | Liquid water diffusion |
| | Cover factor | Twisting rigidity | Diffusion coefficient |
| | Warp count | Shearing modulus | |
| | Weft count | Hysteresis of shear | |
| | Warp crimp | Hysteresis of bending | |
| | … | … | … |
Table 15.2. Fibers
| General data | Structural data | Mechanical data | Physical data |
--- --- |
| Fiber ID | Diameter | Tensile modulus | Differential heat of liquid sorption |
| Fiber name | Volume density | Flexural rigidity | Differential heat of moisture vapor sorption |
| Fiber type | Length | Shearing modulus | Integral heat of sorption |
| Fiber picture | Count | Friction factor | Differential radiation absorption constant of the fiber |
| Company | Effective contact angle between fiber surface and water | Compression modulus | Fiber sorption isotherm |
| Price data | | Bending Torsional rigidity | Specific heat of fiber Volumetric specific heat of fiber Diffusion coefficient Thermal conductivity of fiber |
Table 15.3. Human body
| General data | Structural data | Physiological data | Material data |
--- --- |
| Model ID | Stature | Bone density | Specific heat at constant pressure of blood |
| Model file | Shoulder | Bone compress mudulus | Specific heat at constant pressure of skin |
| Model picture | Breast | Torsional rigidity | Diffusion coefficient/Mass diffusivity for moisture vapor diffuse through skin |
| Model type | Waist | Bone Poisson ratio | Thermal conductance of body tissue |
| Size | Hip | Soft tissue density | Thermal conductance of the skin |
| | Arm | Soft tissue shear modulus | Absorptivity of skin surface to the solar radiation |
| | Leg | Soft tissue Poisson ratio | Emissivity of skin surface |
| | Neck | Skin density | Dubois body area |
| | Weight | Skin tensile modulus | Total body mass |
| | | Skin Poisson ratio | |
In the database, the fiber–yarn–fabric structure, and mechanical and physical parameters are collated from a series of mechanical experiments using Kawabata and Instron test instruments, including fabric simple deformations by Kawabata instruments, and relaxation deformation of fiber–yarn–fabric and 3D deformation of fabric bagging using the Instron. The major mechanical parameters taken from the experiments are stored in the material database, as shown in Table 15.4. To set up a numerical database of textile properties without any assumption for the material properties, the tension-recovery curves of fiber–yarn–fabric have to be recorded in the material database by direct inputs from the experiments. The experimental curves will be recorded as the images in the database, which can be found during the design process by searching for its identifier number.
Table 15.4. Items of each database
| Human model database items | Garment database items | Material database items | Project database items |
--- --- |
| Human body ID | Garment ID | Fabric ID | Project ID |
| Human image ID | Garment name | Fabric bending modulus | Design info. |
| Stature | Garment image | Fabric shear modulus | … |
| Shoulder | Garment commercial info. | Fabric Poisson ratio | … |
| Breast | … | Fabric frictional coefficient | Garment ID |
| Waist | … | Fabric roughness | Design state |
| Hip | Garment size | Fabric bagging resist | Pre-process file |
| Arm | Garment style | Fabric bagging fatigue | Method ID |
| Leg | Fabric ID | Yarn ID | … |
| Neck | Fabric ID | Yarn viscoelastic modulus | Post-process file |
| Weight | … | Yarn relaxation time | Result files |
| Skin thickness | … | Yarn bending modulus | … |
| Skin tensile modulus | 2D pattern data file ID | Yarn compressor modulus | Evaluation file |
| Skin density | 3D garment data file ID | Yarn torsion modulus | Visualization file |
| Bone elastic modulus | … | Yarn Poisson ratio | |
| Bone density | … | Fiber ID | |
| Soft tissue thickness | | Fiber viscoelastic modulus | … |
| Soft tissue elastic modulus | | Fiber relaxation time | |
| Soft tissue density | | Images of experimental curves | |
| … | … | … | … |
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Book 2006, Biomechanical Engineering of Textiles and ClothingR.M. Wang, ... X. Zhang
Chapter
Design and Applications
2000, Comprehensive Composite MaterialsSuraj P. Rawal, John W. Goodman
6.14.1.6 The Space Environment
The natural space environment (James et al., 1994; Belk et al., 1997; Anderson and Smith, 1994; Dooling and Kinckenor, 1999; Adams et al., 1987; Messenger and Ash, 1986) includes naturally occurring phenomena such as atomic oxygen (AO), atmospheric density (vacuum), ionizing radiation, plasma, charged particle plasma, neutral atomic and molecular particles, and factors such as micrometeroids and man-made debris. Figure 2 shows the space environment varying with altitude. Vacuum describing the low pressure in space varies significantly with altitude, being about 10−15 torr for geosynchronous spacecraft, compared to 760 torr at sea level. When exposed to space vacuum, polymer based composites release substances in the form of gas. This outgassing can lead to two potentially undesirable effects: (i) degradation of key mechanical properties, and (ii) condensation of gases on critical surfaces of lenses, mirrors, and sensors, influencing optical performance (Sarafin and Larson, 1995; James et al., 1994; Dooling and Kinckenor, 1999; Adams et al., 1987).
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Fig. 2. Key space environments as a function of altitude.
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Reference work 2000, Comprehensive Composite MaterialsSuraj P. Rawal, John W. Goodman
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Polymer and carbon nanotube composites for space applications
2006, Carbon NanotechnologyKent A. Watson, John W. Connell
2.1.Atomic o xygen
Spacecraft orbiting in LEO are exposed to a residual atmosphere composed predominately of highly reactive AO atoms. In LEO, AO is formed as a result of photo dissociation of molecular oxygen (O 2) due to exposure from ultraviolet solar radiation at wavelengths less than 243 nm. The probability of recombination of AO molecules to reform molecular oxygen is negligible due to the low atmospheric density of AO molecules. Thus, the formation of AO occurs in ~99% yield as shown:
O 2+h ν→O•+O•
Fig. 1 shows the density of gaseous species from 100 to 700 km . The composition of the upper atmosphere shown is an average, as it varies with temperature, solar activity, variations in the Earth's magnetic field, latitude, and season. AO is the most abundant and most reactive species at altitudes from about 200 to 650 km and although other gaseous species are present, including nitrogen and atomic hydrogen, their reactivity toward typical spacecraft materials is negligible.
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Figure 1. Density of gaseous species as a function of altitude .
Although the concentration of AO in LEO is low (pressures of ~10−7 torr), the high orbital velocity of spacecraft in LEO causes the flux of AO on ram-orientated spacecraft surfaces to be very high—with collisional energies of ~5 eV (~480 kJ/mol). (Orbiting spacecraft in LEO typically travel at velocities of 8-12 km/s depending on eccentricity and orbital inclination.) The total amount of AO exposure depends on many factors including spacecraft altitude, orbital inclination, attitude, mission duration, and solar activity . As an example, the ISS is bombarded with AO at about 10 14 atoms/cm 2 s for normal incident ram surfaces. Also, because AO formation is dependent upon solar radiation, its concentration is dependent upon solar activity. Fig. 2 shows AO flux as a function of solar cycle activity at altitudes ranging from 300 to 900 km . Spacecraft in LEO orbit will be exposed to less AO during periods of low solar activity than one orbiting during periods of high activity. Fig. 3 shows the history of sunspot variation throughout the 11-year solar cycle from 1850 to 2005 . For short missions in LEO, solar activity is an important factor and the solar cycle should be taken into consideration when selecting spacecraft materials. The Sun emits more energy, including UV radiation, during times of solar maximum. The increase in radiation may increase the damaging effects on spacecraft materials and the increase in energy also expands the upper atmosphere due to thermal expansion which in turn increases the concentration of AO at LEO altitudes .
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Figure 2. AO flux as a function of altitude .
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Figure 3. History of sunspot variation.
Erosion and material property changes due to AO are generally insignificant for short periods of exposure. However, as the typical lifetime of spacecraft increases to 10 years and greater (space station, etc.), the stability of material properties important to the function of the spacecraft becomes a major consideration.
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Book 2006, Carbon NanotechnologyKent A. Watson, John W. Connell
Chapter
Reconfiguration procedure for a constellation
2011, Orbital Mechanics and Formation FlyingPedro A. Capó-Lugo, Peter M. Bainum
11.9.1 Results for the reconfiguration procedure with adaptive scheme
The same initial conditions for SA in the reconfiguration procedure from Phase I to II shown in Section 11.7.1 are used here. The difference is the addition of the perturbations due to the Sun and the atmospheric effects of the Earth. The spacecraft is assumed to weight 90 kg, with a radius of 0.508 m and a length of 0.559 m. The reflectivity coefficient for the satellite is 0.8. The radius of perigee is 1.2 R E which describes an atmospheric density of 2.6×10−15 kg/m 3 and a scale height of 417 km. For the adaptive scheme, γ L=0.1 and θ A=1. The sampling in the true anomaly angle is 0.1 rad. K˜P and K˜d′ are equal to zero at the initial true anomaly angle. The simulation begins at the perigee point. The reconfiguration process stops when the error in the semimajor axis and the eccentricity is less than 1 km and 0.01, respectively. These conditions are different than in Section 11.7 because it is desired to study the adaptive control scheme.
Figures 11.13 and 11.14 show a comparison of the reconfiguration procedure for the in-plane motion with and without the adaptive scheme. The reconfiguration with an adaptive scheme took 1 orbit to satisfy the criteria to stop the process while the reconfiguration without the adaptive scheme took two orbits. The impulse for B¯ is very similar for both control cases, but at approximately 0.7 orbits, a smaller use of thrust is observed in the reconfiguration with the adaptive scheme than in the reconfiguration procedure without the adaptive scheme. This happens because the adaptive scheme is taking care of the disturbances such that the input force can be controlled. In Figure 11.14, R¯ shows that at the initial true anomaly angle, a single impulse is applied to cause a variation to the semimajor axis and the eccentricity. Once the system is reaching steady state, R¯ is less in the reconfiguration with the adaptive scheme than without the adaptive scheme. As mentioned before, the argument of perigee is corrected with a single impule once the satellite is either at the perigee or apogee point.
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Figure 11.13. In-plane motion reconfiguration for adaptive scheme
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Figure 11.14. In-plane reconfiguration procedure with adaptive scheme for the argument of perigee
Figure 11.15 shows very similar patterns for both cases. In the reconfiguration procedure from Phase I to II, the control process takes more in consideration the in-plane motion than the out of plane motion. For this case, the out of plane motion is performed with the impulsive maneuvers. For this reason, certain peaks can be observed at some locations along the orbit in both cases.
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Figure 11.15. Out-of-plane correction for the adaptive scheme
Figure 11.16 shows how the control input function keeps varying with respect to the true anomaly angle. In Figure 11.16, the adaptive function for R¯ shows a larger value than the adaptive function for B¯ because R¯ is mainly related to the semimajor axis while B¯ is related to the eccentricity. In Figure 11.16, the applied impulses highly affect the fuzzy logic controller at the initial true anomaly angle. Once the system is reaching a steady state, the effects of the adaptive scheme are being reduced while the fuzzy logic controller takes care of the errors. The combination of the controllers provides a good control approach to perform the reconfiguration procedure for the other satellites in the constellation. Table 11.9 shows a comparison between the end conditions for both cases. The difference in the orbital elements between the cases is not relatively large, but the orbital elements are satisfied with the adaptive scheme; on the other hand, the adaptive control scheme demonstrates that the correction can be performed in a short number of orbits.
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Figure 11.16. Control input function for the adaptive scheme
Table 11.9. Comparison of the end conditions for the reconfiguration procedure without and with adaptive control scheme
| Empty Cell | Without adaptive scheme | With adaptive scheme |
--- |
| a (km) | 99,498.3 | 99,498.1 |
| e | 0.918 | 0.923 |
| i (degree) | 18.5 | 18.5 |
| Ω (degree) | −0.00958 | 0.00720 |
| ω (degree) | 89.997 | 90 |
| R¯ (km/sec 2) | 5.081× 10−4 | 4.875× 10−3 |
| Number of orbits | 2 | 1 |
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Book 2011, Orbital Mechanics and Formation FlyingPedro A. Capó-Lugo, Peter M. Bainum
Chapter
Batteries for integrated power and CubeSats: Recent developments and future prospects
2022, Silicon Anode Systems for Lithium-Ion BatteriesAloysius F. Hepp, ... Ryne P. Raffaelle
12.2.3 Starshine 3 integrated power system flight experiment and results
Starshine 3 was launched from a Lockheed Martin Athena I rocket from Kodiak, AK on September 29, 2001, at 6:40 P.M. Alaska Daylight Time (22:40 EDT); it was deployed to a circular orbit of 475 km (low-earth orbit (LEO)) at a 67 degrees inclination with a fixed rotational velocity of 5 degrees/s . The Starshine 3 satellite was covered with mirrors; the spherical geometry allowed for easy modeling of the orbital decay. By observing the orbital decay, the mission allowed for a calculation of atmospheric density versus altitude [39, 71]. The storage capacity of each IMPS battery was 45 mAh, capable of powering the load for 90 days without recharging. When Starshine 3 launched, a month later than originally scheduled, it had been 60 days since the IMPS units were fully charged; thus, they had only one-third of their initial charge left [38, 40].
While on orbit, the five independent power supplies experienced quite adverse operating conditions: Starshine 3 was deployed in an orbit where the solar beta angle passed through ±90 degrees. This meant that there were periods when the solar array on an IMPS did not “see” the sun for days at a time. Any charging would have to come from light reflected off of the surface of the Earth (albedo). Furthermore, when illuminated by only the Earth’s albedo, the operating temperature of the IMPS sank as low as −18°C; only 2°C above the battery’s operating range. The operating voltage of two IMPS units versus days in orbit is shown in Fig. 12.4. The IMPS marked “N. Hemisphere” was located 45 degrees above Starshine’s equator as defined by the photograph in Fig. 12.3. The IMPS marked “S. Hemisphere” was located 45 degrees below the equator. When deployed, Starshine 3 had a solar beta angle of −61 degrees. A beta angle of at least −44 degrees is needed for any direct sunlight on the N. Hemisphere IMPS. That meant that the IMPS units in the upper hemisphere did not see the sun at all for the first 18 days, 5 days later than the last data point in Fig. 12.4[38, 40].
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Fig. 12.4. Data from five integrated micro power supplies (IMPS) designed and fabricated at NASA GRC that operated on-board Starshine 3 from late-September to mid-October 2001. Four of the IMPS units had particularly challenging operating conditions due to a severe initial solar beta angle that provided mostly Earth albedo (light reflected off the Earth) charging and an operating temperature of −10°C.
(Courtesy NASA.)
Despite these adverse conditions, the IMPS on the upper hemisphere (Fig. 12.4) still managed to charge under albedo lighting only. This is evident by the slowly increasing voltage during the first 13 days. The net average current measured over the first 18 days was 26 μ-amps, enough to account for the increasing voltage during albedo illumination. The IMPS in the lower hemisphere (S. Hemisphere in Fig. 12.4) quickly charged to its maximum voltage of 2.9 V. The peak voltages reached by each IMPS are set by the voltage regulators and are not directly comparable. They reflect the set point of each IMPS voltage regulator. During the first 100 days, all five IMPS units maintained a voltage greater than 2.5 V. The voltage fluctuations shown in Fig. 12.4 are attributed to the changing solar exposure due to beta angle and eclipse time.
All communication from Starshine 3 ceased after 100 days. The satellite was intentionally de-orbited and burned up on January 21, 2003, after 7434 revolutions around Earth, after 479 days in orbit with one orbit lasting approximately 90 min. Starshine 3 marked the first deployment and proof-of-concept of IMPS technology via a successful space experiment. As discussed earlier, the IMPS units had particularly challenging operating conditions due to a severe initial solar beta angle that provided only Earth albedo (light reflected off the Earth) charging and an average operating temperature of −10°C [38–40].
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Book 2022, Silicon Anode Systems for Lithium-Ion BatteriesAloysius F. Hepp, ... Ryne P. Raffaelle
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13539 | https://www.youtube.com/watch?v=0wSHnoDM_d8 | How to Cross Multiply
Bullis Student Tutors
3980 subscribers
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Posted: 13 Nov 2015
This video covers the steps of cross multiplying where the variable is in the numerator, denominator, and in expressions in the numerator on both sides of the equal sign.
This is a Bullis Student Tutors video -- made by students for students.
5 comments
Transcript:
hi I'm a bullet student tutor in this video I'll be talking about cross multiplying so first you have your two fractions you would then multiply across and that would give you 3 = 2x then you would divide the two to get the X alone your final answer would be 3 / two so another problem try is one that has variables on both sides so it' be x x - 2 / 2 = x + 4 divided by 7 so it's the same um as the first problem problem we did except you have to put parentheses around both of the numerators because you have a variable and another number along with it so you multiply across so it would be 7 x - 2 = 2 x + 4 so you have to use a distributive property um for this so you have to multiply 7even by both of the um both the numbers and variables inside of the parentheses so it' be 7 X and then 7 -2 so 7 x - 14 = and it's the same thing over here so 2 X and then 2 4 so be 2 x + 8 so now you have to get X on one side before you can isolate X and find out what it is so add 14 add 14 subtract 2X and subtract 2x this gets you 5x = 22 so it would be the same thing so you do 5x = 22 / 5/ 5 this gets you X = 22 / 5 so here's a problem with the variable on the denominator so it's going to be 10 /x = 3 over 9 and then you would just cross multiply the same way as if the variable was on the top and it would be 3x = 90 then you would divide by 3 and you divide by 3 in x = 30 this has been a bullish student tutors video If you enjoyed this video make sure to check out the rest of our YouTube channel for more tutoring videos thank you |
13540 | https://proofwiki.org/wiki/Equation_of_Straight_Line_in_Plane/Normal_Form/Polar_Form | Equation of Straight Line in Plane/Normal Form/Polar Form
From ProofWiki
< Equation of Straight Line in Plane | Normal Form
Jump to navigation Jump to search
Theorem
Let $\LL$ be a straight line such that:
: the perpendicular distance from $\LL$ to the origin is $p$
: the angle made between that perpendicular and the $x$-axis is $\alpha$.
Let $\LL$ be defined in normal form:
: $x \cos \alpha + y \sin \alpha = p$
Then $\LL$ can be presented in polar coordinates as:
: $r \map \cos {\theta - \alpha} = p$
Proof
Let $O$ be the origin of the Cartesian plane and the pole of the corresponding polar frame.
Let $OX$ denote the polar axis, coincident with the $x$-axis.
Let $P$ be an arbitrary point on $\LL$, expressed in polar coordinates as $\polar {r, \theta}$.
Let $N$ be the point on $\LL$ where the normal to $\LL$ intersects $\LL$.
We have that $OP$ is at an angle $\theta$ to $OX$ and is of length $r$.
We have that $ON$ is at an angle $\alpha$ to $OX$ and is of length $p$.
Hence $\angle NOP = \theta = \alpha$.
We also have that $\angle ONP$ is a right angle.
Thus:
: $p = r \map \cos {\theta - \alpha}$
$\blacksquare$
Sources
1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(4)$ Polar equation
Retrieved from "
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13541 | https://personalpages.bradley.edu/~campbell/mnemonic.html | ===============
The Mnemonics Page
by Dean Campbell,Bradley University Chemistry Department
My intention is to create an informal repository of useful mnemonics on a variety of topics, especially chemistry. Please contact me at campbell@bumail.bradley.edu if you have suggestions for the page. The people mentioned are my sources for these mnemonics, not necessarily the originators for the mnemonic.
Chemistry
LEO says GER - in redox reactions, the Loss of Electrons is Oxidation and the Gain of Electrons is Reduction - ???
OIL RIG - in redox reactions, Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons) - ???
I Bring Clay For Our New House; Have No Fear Of Ice Cold Beer; BrINClHOF; HOFBrINCl - the symbols of elements which tend to occur as diatomic molecules: Iodine, Bromine, Chlorine, Fluorine, Oxygen, Nitrogen, and Hydrogen - Delvis Dore, Robert Gayhart, Delvis Dore, Dan Szymkowiak, and others
Goldfish are heck without tartar sauce - thermodynamics: delta G = delta H - T x delta S - Bob Gayhart
T = Kim (constant van't Hoff factor molality) -boiling-point elevation and freezing-point depression - Dan Szymkowiak
P = TRiM (Absolute temp gas constant van't Hoff factor molarity)- osmotic pressure -Dan Szymkowiak
The Periodic Table (contributors include Bob Gayhart and Dean Campbell)Group 1A (H Li Na K Rb Cs Fr):
-Highly Nasty Kids Rub Cats Fur
-Hans liebt Natuerliche Kaese reibt (ueber) (c)zentralsued Frankreich. (!)
Group 2A (Be Mg Ca Sr Ba Ra):
-Bearded Muggers Came Straight Back Rapidly.
-Beer Mugs Can Serve Bar Rats.
Group 3 (B Al Ga In Tl):
-Big Al Garner's In Trouble.
-Big Al Gave In Tulsa.
-Big Al Gave Incomplete Totals.
-Big Al Gave Innocent Toys.
-Bob Allen Gave Indians Tennis Lessons.
-Bears Always Give Indians Trouble. Group 4A (C Si Ge Sn Pb):
-Can Silly Germans Snatch Lead?
-Conniving Sisters Get Snotty Playboys.
-Can Si Get Snuckered at the Pub? Group 5 (N P As Sb Bi):-New Popes Assign Subordinate Bishops.
-Never Put Arsenic in Silver-bullet Beer. Group 6A (O S Se Te Po):
-Oil is Slicker with STP.
-Old Soldiers Seem Terribly Polluted.
-Oversized Sailors Sell and Tell about Polo cologne. -Oh, She Sells The Poles. Group 7A (F Cl Br I At):
-Frank Cleverly Brought Indian Attire.
-Fairy Clowns Broil Innocent Ants.
-Fat Clyde Bribed Innocent Arthur. -"Floor Cleaner Broken?" I Asked.
Group 8A (He Ne Ar Kr Xe Rn):
-He Never Arrived; Karen eXited with Ron. -He Needs Our Crazy Xerox Repairman. First Row Transition Metals (Sc Ti V Cr Mn Fe Co Ni Cu Zn):-St. Vincent Cried "Mommy!" Feeling Cold Next to Cold Zombies.
-Scott Tips Very Counterfeit Money, Feels Cold Near Cubic Zircons.
-Scary Tiny Vicious Creatures Mean Females Come Nightly to Club Z. -Scotty ViCkerMan Fee CoNy CouSin. -Scott Tickled Vanna's Cranium, Manager Fred Could Not Cut In. Second Row Transition Metals (Y Zr Nb Mo Tc Ru Rh Pd Ag Cd):-Yes Sir, Nob. Most Technicians Rub Rod's Pale Silver Cadillac. Third Row Transition Metals (La Hf Ta W Re Os Ir Pt Au Hg):-Larry's Half Taken, Wendy Reached Out Her Plate Audibly, Helga. Lanthanides (Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu):
-Caesar Procrastinated at the Nile, Permitting So Many Europeans (who were) Getting Turbid (and) Dying (of) Home Urges (erges?) To Yell "Lutetium!". -Cecil Proudly Named Pam, Sam, (and) Eugene. Good Tobin Dyes Hosiery Early Tomorrow Yellow-Blue. Actinides (Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr):
-Three Planets: Uranus, Neptune, (and) Pluto. Amy Cured Berkeley, California. Einstein (and) Fermi Made Noble Laws.
Mathematics
Please excuse my dear Aunt Sally - order of operations on mathematics: parentheses, exponents, multiplication, division, addition, subtraction - Karen Campbell
SOHCAHTOA - in a right triangle: Sine is Opposite side divided by Hypotenuse, Cosine is Adjacent side divided by Hypotenuse, and Tangent is Opposite side divided by Adjacent side - Ed Measer
Electronics
Bad beer [or booze] rots our young guts but vodka goes well - get some now! - order of resistor color codes: black, brown, red, orange, yellow, green, violet, gray, white, gold, silver, none - Dennis Campbell, Susan Kelly, William Johnson, and others
Beloved Brethren Remember Our Yoke God Brings Victory God's Way -order of resistor color codes: black, brown, red, orange, yellow, green, violet, gray, white - David Montag
And a bunch more for order of resistor color codes: black, brown, red, orange, yellow, green, violet, gray, white, gold, silver, none -Rupert Russell:
Back Before the Revolution Our Young Gallant Boys Voted for George Washington.
Bad boys race our young girls behind victory garden walls!
BB ROY Got Back Very Good Wife
Better Be Right Or Your Great Big Venture Goes Wrong
Big Beautiful Roses Occupy Your Garden But Violets Grow Wildly
Big Boys Race Our Young Girls, But Violet Generally Wins
Black Beetles Running On Your Garden Brings Very Good Weather
Bye Bye Rosie Off You Go Birmingham Via Great Western
ELI the ICEman - to remind us voltage (E) leads current(I) across an inductor(L), but current leads voltage across an capacitor(C) - William Johnson
Spelling
A Rat In The House May Eat The Ice Cream - spelling of "arithmetic" - Ken Kolb
Return to Dr. Campbell's HomepageLast updated 12/18/12 Site created by Dean and Karen Campbell |
13542 | https://math.stackexchange.com/q/2543333 | combinatorics - How many sets that differs in at least two elements are there? - Mathematics Stack Exchange
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How many sets that differs in at least two elements are there?
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Let vector a i=(a 1 i,...,a n i)a i=(a i 1,...,a i n), where a j i∈{0,1}∀j∈{1,..,n}a i j∈{0,1}∀j∈{1,..,n}.
Let set A={a 1,...,a i,...,a m}A={a 1,...,a i,...,a m}. |A|=m|A|=m. What is max(m)max(m) for each dimension n n, if we require that every vector in A A must differ from any other vector in at least two terms:
∀i,j∈{1,...,m},a i−a j must be at least a two dimensional vector.∀i,j∈{1,...,m},a i−a j must be at least a two dimensional vector.
For example, if n=2 n=2, then max m=2 max m=2: A={(0,1),(1,0)}A={(0,1),(1,0)}
If n=3 n=3, then max m=4 max m=4: A={(0,0,1),(1,0,0),(1,1,1),(0,1,0)}A={(0,0,1),(1,0,0),(1,1,1),(0,1,0)}
Please correct me if there are anything unclear about this simple problem that I cannot solve.
If we only require one different term , then max m=2 n max m=2 n
combinatorics
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edited Sep 26, 2018 at 5:09
High GPAHigh GPA
asked Nov 29, 2017 at 19:19
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1 Look up Hamming distance.marty cohen –marty cohen 2017-11-29 19:23:05 +00:00 Commented Nov 29, 2017 at 19:23
@martycohen Thank you, this is exactly what I am looking for.High GPA –High GPA 2017-11-29 19:55:57 +00:00 Commented Nov 29, 2017 at 19:55
2 Math is like magic in that knowing the true name for something gives you power.marty cohen –marty cohen 2017-11-29 19:58:37 +00:00 Commented Nov 29, 2017 at 19:58
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We shall prove that, if a set S S of vectors in {0,1}n{0,1}n is such that any two elements of S S differ by at least 2 2 coordinates, then
|S|≤2 n−1.|S|≤2 n−1.
Note that the bound is sharp by considering the set
S={v∈{0,1}n∣∣v has an even number of 1 s}.S={v∈{0,1}n|v has an even number of 1 s}.
(In fact, there are only two maximal sets S S, the one above and its complement in {0,1}n{0,1}n.)
Partition {0,1}n{0,1}n into 2 n−1 2 n−1 sets of the form
P x 2,x 3,…,x n:={(0,x 2,x 3,…,x n),(1,x 2,x 3,…,x n)}.P x 2,x 3,…,x n:={(0,x 2,x 3,…,x n),(1,x 2,x 3,…,x n)}.
If |S|>2 n−1|S|>2 n−1, then there exists P x 2,x 3,…,x n P x 2,x 3,…,x n which contains two elements of S S, and this is a contradiction.
P.S.: Apparently, the general case is an open problem. See a table here. However, a similar argument can be used to show that
A 2(n,d)≤⌈2 n−d+1⌉,A 2(n,d)≤⌈2 n−d+1⌉,
where A k(n,d)A k(n,d) is the maximum number of vectors in a k k-ary code of word length n n and with Hamming distance d d.
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edited Nov 29, 2017 at 19:56
answered Nov 29, 2017 at 19:47
BatominovskiBatominovski
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Thank you Batominovski, this is very helpful. Are there any "stricter" upper bound for this open problem?High GPA –High GPA 2017-11-29 19:56:48 +00:00 Commented Nov 29, 2017 at 19:56
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13543 | https://www.youtube.com/watch?v=_lnUfvgOy3o | The Maths Prof: Using Map Scales
The Maths Prof
181000 subscribers
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Description
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Posted: 16 Feb 2023
In this lesson I show you how to use map scales in ratio form 1 : n. The questions involve length, area and volume. In case you would like to skip to a specific question, here's the breakdown 👇
Question 1 On a map of scale 1 : 100000, the distance between two towns is 6.3 cm. What is the actual distance in km?
Question 2 On a map of scale 1 : 50000, the distance between two towns is 8 cm. What is the actual distance in km? 02:40
Question 3 The distance between 2 towns is 32 km. How far apart will they be on a map of scale 1 : 50000? 04:06
Question 4 On a map, a lake has an area of 32 cm². The scale of the map is 1 : 24000. Calculate the actual area of the lake. Give your answer in km². 06:33 ('Extended' IGCSE exam question paper 2 Cambridge 0580)
Question 5 A map is drawn to a scale of 1 : 100000. Find the area in km² of a village which is represented on the map by an area of 600 cm². 08:41
Question 6 A map is drawn to a scale of 1: 100000. The area of a village is 750 km². Find the area of the village on the map in cm². 10:30
Question 7 A model of a car has a scale of 1 : 20. The volume of the actual car is 12 m³. Find the volume of the model. Give you answer in cubic centimetres. 12:25 ('Extended' IGCSE exam question paper 2 Cambridge 0580)
Question 8 Try yourself! 14:42
I hope that you find the lesson useful. Subscribe to my YouTube channel to stay up to date with new uploads and follow me on Instagram @themathsprof for more maths! :-)
23 comments
Transcript:
foreign [Music] hey guys in this lesson I'm going to show you how to answer questions involving map scales so in the first one it says on a map of scale 1 to 100 000 the distance between two towns is 6.3 centimeters what is the actual distance in kilometers so we're given the distance between the two towns on the map and we have to work out the real distance between the two towns in real life okay in kilometers the map scale is given here in the question and it's written as a ratio all it means is one unit on the map equates to a hundred thousand of those same units in real life so if you decide to measure on the map in centimeters for example if you measure one centimeter on the map it's equal to a hundred thousand centimeters in real life okay so the unit stay the same so if we're converting from the map distance to real life distance we're always multiplying by this number here 100 thousands so the distance between the two towns on the map is here 6.3 centimeters so we take the number 6.3 and we have to multiply by a hundred thousand okay so when I do that I get six hundred and thirty thousand okay now remember the distance on the map was in centimeters so the distance in real life is also in centimeters so we're not quite finished because we have to write the final answer in kilometers so before I do the next bit I just want to remind you that one kilometer is equal to a thousand meters and one meter is equal to 100 centimeters you need to learn those if you don't know them already so if we're changing centimeters into kilometers first I'm just going to change the centimeters into meters so if we're changing centimeters into meters we always divide by a hundred so if I divide this number by a hundred I just take off two zeros so I'm left with six thousand 300 and now it's in meters okay next if we want to change meters into kilometers we always have to divide by a thousand so if I divide this number by a thousand I'm left with 6.3 and now it's in kilometers okay so this is the final answer if you want to skip that middle step changing into meters you could just divide by a hundred thousand and that will change the centimeters straight into kilometers in the next question it says on a map of scale 1 to 50 000 the distance between two times is eight centimeters what is the actual distance in kilometers and I've just dropped it down over here the units kilometers meters and centimeters as a reminder so this time the map scale is one to fifty thousand so that just means one unit on the map equates to fifty thousand of those same units in real life so the distance between the two towns this type is eight centimeters so this time if we're changing the map distance into real life we have to multiply by fifty thousand okay so eight multiplied by fifty thousand is 400 000 and remember the distance between the two towns on the map was given in centimeters so this answer here is also in centimeters okay so that's the distance between the two towns in real life now we have to change the centimeters into kilometers and remember to change centimeters into kilometers first you would divide by a hundred so change them into meters then to change the meters into kilometers you would divide by a thousand so you're dividing by a hundred thousand so when I divide four hundred thousand by a hundred thousand it gives me four kilometers okay and that is the answer to the question in the next question it says the distance between two towns is 32 kilometers how far apart will they be on a map of scale 1 to 50 000 so this time it's the other way around this time we've been given the distance in real life between the two towns 32 kilometers and we have to work out the distance between the two towns on the map instead okay so here's the map scale one to fifty thousand so remember that just means one unit on the map equates to 50 000 of those same units in real life okay remember when we changed the map distance into real life we multiplied by this number here well now we're doing it the other way around now we're going for real life to the map and so this time we have to divide by 50 000 instead okay so 50 000 divided by fifty thousand gives you one okay so we're going to divide by 50 000 with the distance given in the question 32 kilometers so 32 divided by 2000 and if you put that into your calculator it will likely give you the answer instead of four but that's okay so the answer is 6.4 times 10 to the negative 4. and remember the distance was given in kilometers so when you do this calculation this answer is also in kilometers remember the units are always the same okay so this isn't very practical as a distance if I asked you to draw with your ruler 6.4 times 10 to the negative four kilometers it doesn't make much sense so what I'm going to do is I'm going to change this into centimeters okay so just remember that one kilometer is a thousand meters and one meter is a hundred centimeters remember when we changed centimeters in kilometers we divided by a hundred thousand well this time we're going to do the opposite we're going to multiply by a hundred thousand because if you're changing kilometers into meters you're timesing by a thousand then if you change meters into centimeters your times in by a hundred so we're multiplying by a hundred thousand so if I take this answer on my calculator here and multiply by a hundred thousand it gives me 64. and now the units are in centimeters which makes a lot more sense if you have to draw it with your ruler okay so that's the answer to the question in this question it says on a map a lake has an area of 32 centimeters squared the scale of the map is 1 to 24 000 calculate the actual area of the lake give your answer in kilometers squared so this question is different to the previous ones because we're dealing an area instead of length we've been given the area of the lake on the map yeah and we have to work out the area of the lake in real life now if you think back to the previous questions when we dealt with length when you're changing the length or the distance on a map into real life distance you always multiply by this number here so 24 000 in this question okay but when it's area you have to remember to multiply by this number two times I.E 24 000 squared an easy way to remember it is when you look at the units for area they're always squared okay so the area of the lake on the map is 32 centimeters squared we have to multiply it by 24 000 squared to work out the area of the lake in real life so when you type that into the calculator it will probably give you the answer in standard form that's fine so it's 1.8432 times 10 to the 10. and remember the units are always the same so they were centimeters squared here in the question so they're still centimeter squared so this is the area of the leg in real life okay now what we have to do is change the units into kilometers squared okay now to change centimeters into kilometers usually you divide by a hundred thousand but because it's area you have to remember to divide by a hundred thousand two times I.E divide by a hundred thousand squared so when you leave the answer on your calculator screen and divide by a hundred thousand squared you should find that you get 1.8432 and now this is in kilometers squared in this question it says a map is drawn to a scale of one to a hundred thousand find the area in kilometers squared of a village which is represented on the map by an area of 600 centimeters squared so this question is very similar to the last one just in case you want to have a go by yourself we've been given the area of the village on the map and we have to work out the area of the village in real life okay so we're going from the map to real life and remember when you want to change the area on the map to the area in real life you have to multiply by this number here squared okay because it's an area question so I'm going to take 600 and multiply it with 100 squared and that will give me the area of the village in real life so when you put it in the calculator it will probably give it to you in standard form so it's going to be 6 times 10 to the power of 12 and remember the units are the same as what's given in the question here okay so because this was 600 centimeters squared our answer is also centimeters squared okay next we have to change this into kilometers squared okay so remember when you're changing centimeters into kilometers if you were dealing with length you would have to divide by a hundred thousand okay but because it's area and it's centimeter squared into kilometers squared you have to remember to divide by a hundred thousand twice so divide by a hundred thousand squared so if you divide this answer here by a hundred thousand squared you should find that you get 600 okay and now the area is in kilometers squared in this question it says a map is drawn to a scale of one to a hundred thousand the area of a village is 750 kilometers squared find the area of the village on the map in centimeters squared so this time it's the other way around this time we've been given the area of the village in real life 750 kilometers squared and we have to work out the area of the village on the map so usually if we were dealing with length we would look at the map scale and to change real life distance into map distance we would divide by the scale factor here a hundred thousand except it's an area question so so you have to remember to divide by this number twice I divide by a hundred thousand squared so take the area of the village 750 and divide it by a hundred thousand squared so when you do that on the calculator it will probably give you the answer in standard form so 7.5 times 10 to the negative eight and remember the units given in the question are kilometers squared so when you do this calculation they have to be the same units kilometers squared so that's the area of the village on the map in kilometers squared so now we have to convert the units into centimeters squared okay so just remember that usually to change kilometers into centimeters if you're dealing with length you would multiply by a hundred thousand except it's an area question so you have to remember to multiply by a hundred thousand twice I.E multiply by a hundred thousand squared so leave that answer on your screen multiply by a hundred thousand squared and when you do that you should find you get 750 so that is the answer to the question in this question a model of a car has a scale of 1 to 20 the volume of the actual car is 12 meters cubed find the volume of the model give your answer in cubic centimeters so although it's not a map scale question we can apply what we've learned because we've been given a model scale here okay the ratio 1 to 20. so we've been told the volume of the car in real life 12 meters cubed and we have to change this into the volume of the model car okay so we're going from real life to the model has it been a length question like when we measure distance on maps Okay we would have divided by 20. if it had been an area question we would divide by 20 squared and in this one is volume so we have to divide by 20 cubed instead an easy way to remember it is to look at the units here then meters cubed you're dividing by 20 cubed so take the volume of the cart solve and divide it by 20 cubed when we do that the calculator will probably give it to you in standard form it's 1.5 times 10 to the negative 3 and remember the units here are the same as what's given in the question here meters cubed okay so this is the volume of the model car in meters cubed so now we have to change this into cubic centimeters this is just the same thing as centimeters cubed okay so before I do the calculation I just want to remind you that one meter is equal to 100 centimeters so if you were changing a length meters into centimeters you would multiply by a hundred if we were dealing with area so meter squared into centimeter squared you were Times by 100 squares okay but this time it's volume we're changing meters cubed into centimeters cubed so we have to multiply by a hundred cute okay so leave the answer on your calculator screen Times by 100 cubed and when you do that you should find you get 1 500 and now it's in cubic centimeters so that is the volume of the model car [Music] |
13544 | https://emedicine.medscape.com/article/985510-treatment | For You
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Tools & Reference>Pediatrics: General Medicine
Rickets Treatment & Management
Updated: Sep 09, 2022
Author: Steven M Schwarz, MD, FAAP, FACN, AGAF; Chief Editor: Jatinder Bhatia, MBBS, FAAP more...;)
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Sections Rickets
Overview
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Treatment for rickets may be administered gradually over several months or in a single-day dose of 15,000 mcg (600,000 U) of vitamin D. If the gradual method is chosen, 125-250 mcg (5000-10,000 U) is given daily for 2-3 months until healing is well established and the alkaline phosphatase concentration is approaching the reference range. Because this method requires daily treatment, success depends on compliance.
If the vitamin D dose is administered in a single day, it is usually divided into 4 or 6 oral doses. An intramuscular injection is also available. Vitamin D (cholecalciferol) is well stored in the body and is gradually released over many weeks. Because both calcitriol and calcidiol have short half-lives, these agents are unsuitable for treatment, and they bypass the natural physiologic controls of vitamin D synthesis.
The single-day therapy avoids problems with compliance and may be helpful in differentiating nutritional rickets from familial hypophosphatemia rickets (FHR). In nutritional rickets, the phosphorus level rises in 96 hours and radiographic healing is visible in 6-7 days. Neither happens with FHR.
A study by Dabas et al compared the efficacy of daily versus weekly oral vitamin D3 therapy in the radiologic healing of nutritional rickets. Children who received daily supplementation had greater increases in their radiologic scores from baseline than those who received weekly therapy.
A study by Thacher et al sought to determine the optimal dose of calcium for treatment of children with rickets. The authors reported that a daily calcium intake of 1000mg or 2000mg resulted in more rapid radiographic healing than 500 mg per day dosing. However no clinical or radiographic differences were found between daily calcium supplements of 2000 mg and 1000 mg. The study also found that complete healing of nutritional rickets may take some children longer than 24 weeks.
If severe deformities have occurred, orthopedic correction may be required after healing. Most of the deformities correct with growth.
A consultation with a pediatric endocrinologist is recommended.
Next:
Deterrence/Prevention
Human milk contains little vitamin D and contains too little phosphorus for babies who weigh less than 1500 g. Infants weighing less than 1500 g need special supplementation (ie, vitamin D, calcium, phosphorus) if breast milk is their primary dietary source. Recommending a vitamin D supplement from the first week of life for susceptible infants who are breastfed is safe and effective and, therefore, should be considered.
The United States Institute of Medicine recommends an upper level of intake of 1000 IU/d and 1500 IU/d in infants aged 0-6 months and 6-12 months, respectively. An adequate intake of 400 IU/d has been suggested for infants aged 0-12 months. The recommended daily allowance is 600 IU/d thereafter. The US Endocrine Societys Clinical Practice Guideline suggests 400-1000 IU/d may be needed for children younger than 1 year; they also recommend 600-1000 IU/d for children aged 1 year or older. Internationally, the European Society for Paediatric Gastroenterology, Hepatology, and Nutrition also suggests an oral supplement of 400 IU/d until age 1 year.
Adequate ultraviolet light or 10 mcg (400 IU) orally (PO) daily of a vitamin D preparation and an adequate dietary supply of calcium and phosphorus prevent rickets. [14, 15] As little as 20 min/d of ultraviolet light to the face of a light-skinned baby is sufficient; however, significantly longer periods of exposure are necessary for children with increased skin pigmentation.
Previous
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References
Newton G. Diagnosing rickets in early modern England: statistical evidence and social response. Soc Hist Med. 2022 May. 35 (2):566-88. [QxMD MEDLINE Link]. [Full Text].
Zmora E, Gorodischer R, Bar-Ziv J. Multiple nutritional deficiencies in infants from a strict vegetarian community. Am J Dis Child. 1979 Feb. 133(2):141-4. [QxMD MEDLINE Link].
McKay CP, Portale A. Emerging topics in pediatric bone and mineral disorders 2008. Semin Nephrol. 2009 Jul. 29(4):370-8. [QxMD MEDLINE Link].
Lowdon J. Rickets: concerns over the worldwide increase. J Fam Health Care. 2011 Mar-Apr. 21(2):25-9. [QxMD MEDLINE Link].
Sodri NI, Mohamed-Yassin MS, Mohd Nor NS, Ismail IA. Rickets due to severe vitamin D and calcium deficiency during the COVID-19 pandemic in Malaysia. Am J Case Rep. 2021 Nov 1. 22:e934216. [QxMD MEDLINE Link]. [Full Text].
Chapman T, Sugar N, Done S, Marasigan J, Wambold N, Feldman K. Fractures in infants and toddlers with rickets. Pediatr Radiol. 2010 Jul. 40(7):1184-9. [QxMD MEDLINE Link].
Shah BR, Finberg L. Single-day therapy for nutritional vitamin D-deficiency rickets: a preferred method. J Pediatr. 1994 Sep. 125(3):487-90. [QxMD MEDLINE Link].
Dabas A, Dabas V, Dabla PK, et al. Daily versus weekly oral vitamin D3 therapy for nutritional rickets in Indian children: A randomized controlled open-label trial. Br J Nutr. 2022 May 13. 1-23. [QxMD MEDLINE Link].
Thacher TD, Smith L, Fischer PR, Isichei CO, Cha SS, Pettifor JM. Optimal Dose of Calcium for Treatment of Nutritional Rickets: A Randomized Controlled Trial. J Bone Miner Res. 2016 Nov. 31 (11):2024-2031. [QxMD MEDLINE Link].
Casey CF, Slawson DC, Neal LR. VItamin D supplementation in infants, children, and adolescents. Am Fam Physician. 2010 Mar 15. 81(6):745-8. [QxMD MEDLINE Link].
Institute of Medicine (US) Committee to Review Dietary Reference Intakes for Vitamin D and Calcium. Ross AC, Taylor CL, Yaktine AL, Del Valle HB eds. Dietary Reference Intakes for Calcium and Vitamin D. Washington DC: National Academies Press; 2011.
Pramyothin P, Holick MF. Vitamin D supplementation: guidelines and evidence for subclinical deficiency. Curr Opin Gastroenterol. 2012 Mar. 28(2):139-50. [QxMD MEDLINE Link].
Braegger C, Campoy C, Colomb V, et al. Vitamin D in the healthy European paediatric population. J Pediatr Gastroenterol Nutr. 2013 Jun. 56(6):692-701. [QxMD MEDLINE Link].
Greer FR. Issues in establishing vitamin D recommendations for infants and children. Am J Clin Nutr. 2004 Dec. 80(6 Suppl):1759S-62S. [QxMD MEDLINE Link].
[Guideline] Wagner CL, Greer FR. Prevention of rickets and vitamin D deficiency in infants, children, and adolescents. Pediatrics. 2008 Nov. 122(5):1142-52. [QxMD MEDLINE Link].
Media Gallery
Findings in patients with rickets.
Anteroposterior and lateral radiographs of the wrist of an 8-year-old boy with rickets demonstrates cupping and fraying of the metaphyseal region.
Radiographs of the knee of a 3.6-year-old girl with hypophosphatemia depict severe fraying of the metaphysis.
Radiograph in a 4-year-old girl with rickets depicts bowing of the legs caused by loading.
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Author
Steven M Schwarz, MD, FAAP, FACN, AGAF Professor of Pediatrics, Children's Hospital at Downstate, State University of New York Downstate Medical CenterSteven M Schwarz, MD, FAAP, FACN, AGAF is a member of the following medical societies: American Academy of Pediatrics, American College of Nutrition, American Association for Physician Leadership, New York Academy of Medicine, Gastroenterology Research Group, American Gastroenterological Association, American Pediatric Society, North American Society for Pediatric Gastroenterology, Hepatology and Nutrition, Society for Pediatric ResearchDisclosure: Nothing to disclose.
Coauthor(s)
Frank R Greer, MD Professor of Pediatrics, University of Wisconsin School of Medicine; Professor (Affiliate) of Nutritional Sciences, University of Wisconsin College of Agriculture and Life Sciences; Attending Physician in Neonatology, Center for Perinatal Care, Meriter HospitalFrank R Greer, MD is a member of the following medical societies: American Academy of Pediatrics, American Pediatric SocietyDisclosure: Nothing to disclose.
Laurence Finberg, MD Clinical Professor, Department of Pediatrics, University of California, San Francisco, School of Medicine and Stanford University School of MedicineLaurence Finberg, MD is a member of the following medical societies: American Medical AssociationDisclosure: Nothing to disclose.
Specialty Editor Board
Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Nothing to disclose.
Chief Editor
Jatinder Bhatia, MBBS, FAAP Professor of Pediatrics, Medical College of Georgia, Georgia Regents University; Chief, Division of Neonatology, Director, Fellowship Program in Neonatal-Perinatal Medicine, Director, Transport/ECMO/Nutrition, Vice Chair, Clinical Research, Department of Pediatrics, Children's Hospital of GeorgiaJatinder Bhatia, MBBS, FAAP is a member of the following medical societies: Academy of Nutrition and Dietetics, American Academy of Pediatrics, American Association for the Advancement of Science, American Pediatric Society, American Society for Nutrition, American Society for Parenteral and Enteral Nutrition, Society for Pediatric Research, Southern Society for Pediatric ResearchDisclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: Nestle
Serve(d) as a speaker or a member of a speakers bureau for: Nestle
Received income in an amount equal to or greater than $250 from: Nestle.
Additional Contributors
Steven M Schwarz, MD, FAAP, FACN, AGAF Professor of Pediatrics, Children's Hospital at Downstate, State University of New York Downstate Medical CenterSteven M Schwarz, MD, FAAP, FACN, AGAF is a member of the following medical societies: American Academy of Pediatrics, American College of Nutrition, American Association for Physician Leadership, New York Academy of Medicine, Gastroenterology Research Group, American Gastroenterological Association, American Pediatric Society, North American Society for Pediatric Gastroenterology, Hepatology and Nutrition, Society for Pediatric ResearchDisclosure: Nothing to disclose.
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Sections Rickets
Overview
Practice Essentials
Pathophysiology
Epidemiology
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DDx
Workup
Approach Considerations
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encoded search term (Rickets) and Rickets
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Related Conditions and Diseases
Rickets
Rickets Imaging
Hypophosphatemic Rickets
Bone Mineralization and Related Disorders
Hypophosphatemia
Vitamin D3 1,25-Dihydroxyvitamin D
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13545 | https://archive.nptel.ac.in/content/storage2/courses/112104039/lecture25/25_4.htm | Objectives_template
Module 4: Interferometry
Lecture 25: Analysis of wedge fringes and Michelson interferograms
Michelson interferometry In a Michelson interferometer, the test beam and the reference beam develop a phase difference, not due to changes in refractive index along the path of light propagation but because of movement of one of the active surfaces. If the reference beam reflects off a fixed surface, the test beam, reflecting off a moving surface will interfere with the reference beam and produce an interference pattern. The simplest configuration of a Michelson interferometer is shown in Figure 4.68. For definiteness, the application to Michelson interferometer to a crystal growth experiment is described below. Here, a seed crystal is immersed in its aqueous solution, supersaturated at the temperature considered. The excess salt deposits on the crystal introduced and the concerned surface grows with time. If this surface is taken as one off which the test beam is reflected, the growth rate can be monitored as a function of the position along the crystal and with time. Consider a Michelson interferometer based experiment which is set-up for on-line monitoring of the microtopography of the growing face of a crystal. In this interferometry, one of the interfering beams is reflected from the surface of a growing crystal to get the microtopographic details of the growing surface. Since interferometry through the solution requires reflection of the light beam from the solution-crystal interface, it is a difficult task. Clear fringes can be recorded when the difference in refractive index between the solution and the crystal is sufficiently large. In spite of the inherent difficulties in working with this technique, it can be effectively used for studies in growth kinetics of crystals from their aqueous solution. The basic principle of the Michelson interferometry in the context of crystal growth is that the phase variation due to growth or dissolution of the crystal surface manifests itself in the form of change in the fringe pattern. For example, if a crystal face has a growth hillock originating from a screw dislocation, the corresponding interference pattern consists of concentric fringes of equal thickness. Figure 4.68 shows schematically the process of fringe formation from a surface having a hillock generated from a dislocation. With every change in the crystal thickness by , one fringe shift takes place. Here is the wavelength of the laser used, and n is the refractive index of the solution. From such an interferogram the geographical description of the crystal face is obtained. Quantitative analysis of the interferogram yields the growth-kinetic parameters such as normal growth rate , slope of the dislocation growth hillock , and tangential growth velocity of the steps. |
13546 | https://www.math.net/contour | Contour
home / calculus / vector / contour
Contour
Contour plots are used to depict functions that have a two-dimensional input with a one-dimensional output. Consider the following figure.
The figure depicts a representation of some hilly physical terrain along with its topographic map. A topographic map is a two-dimensional (2D) representation of the three-dimensional features of the Earth's surface, such as hills. The curves of a topographic map, referred to as contours or contour lines, are used to depict the varying elevation of the terrain. Each curve in the figure above represents an increase in elevation of 100 ft.
Formal definition
Let z = f(x,y) be a function of two variables that forms a curved surface in the 3D coordinate plane. If a plane z = k, where k is some constant, intersects the surface, a contour with the equation k = f(x, y) is formed. These curves are also referred to as level curves since their distance from the xy-plane is k for any point on the plane.
The graph of the hemisphere, is intersected by planes z = 3 and z = 5. The intersections of the planes with the hemisphere form two contours, both of which are circles with centers about the z-axis; the circle formed by plane z = 3 has radius of 3√3 and the circle formed by z = 5 has radius of √11. Also, note that these contours are level curves that are parallel to the xy-plane. Translating these level curves onto the xy-axis creates a contour map of the hemisphere. Below is a contour map of the above hemisphere, using planes z = 3, z = 5, and others (there are infinitely many contours for the hemisphere).
The table below shows a few other examples of contour maps for common 3D surfaces:
| Ellipsoid | Contour map |
--- |
| | |
| Hyperbolic paraboloid | Contour map |
| | |
| Hyperboloid of one sheet | Contour map |
| | |
Contours and gradients
There is a special relationship between the contour map and the gradient field for a function of several variables. Consider the following topographic map:
Suppose that you are on some hilly terrain represented by the above topographic map. You start at point E, which is on the level curve with an elevation of 200 feet. You want to climb the shortest possible distance to an elevation of 210 feet. Points A, B, C, and D are several places that you can climb to in order to reach your destination. Referencing the map, EC is the shortest path, and is referred to as the gradient, or gradient vector. The gradient of a function can be thought of as the direction of the function's greatest rate of increase, or in this case, the shortest path to the next contour, or higher elevation.
It is worth noting that since a gradient lies along the direction of a function's greatest rate of change, and contour lines lie along the direction of no change, the two are perpendicular. Referencing the above topographic map, each contour line represents a single elevation. Thus, vector EC, which represents the greatest rate of increase and shortest path to the next contour line (or elevation), is perpendicular to the contour line for the elevation of 200 at point E. Mathematically, let z = f(x, y) be a function of two variables. The gradient field for f is:
If A = (x 1, y 1) is a point on a level curve for the contour map of f, then ∇f(x 1, y 1) is perpendicular to the level curve at A. To show this, let be a unit vector in the direction of the contour line at point A for the contour map of z = f(x, y), shown below:
The directional derivative for f at point A can be found using the dot product of the gradient and unit vector:
Since the unit vector is moving in the direction of the contour line at A, it is tangent to the contour line at A such that:
Whenever the dot product of two vectors is zero, they are perpendicular. Since u is tangent to the contour line at A, ∇f(x 1, y 1) is perpendicular to the contour line at A. The same is true for any gradient at some point on a level curve.
The gradient field and two contours of the paraboloid are graphed on the xy-plane above. Notice that each gradient that intersects the contours is perpendicular to the contour at the point of intersection.
Example
What is the direction of greatest increase for at f(2, ½).
,
so the contour containing (2, ½) is yx 2 = 2.
The partial derivatives of x and y are:
and
Thus,
and
We can then unitize the gradient vector since doing so does not change its direction:
This unit vector indicates the direction of greatest increase for the function at the given point, as shown in the graph below:
Contour surfaces and gradients
The relationship between a gradient field and contour map for a function of three variables is as follows. Given that w = f(x, y, z) is a function of three variables, then the gradient of f is given by:
Let k be some constant, then k = f(x, y, z) is the level surface for f (as compared to a level curve for z = f(x, y)). The level surface is also referred to as a contour surface for f. If A = (x 1, y 1, z 1 is a point on a contour surface for f, then ∇f(x 1, y 1, z 1) is perpendicular to the plane tangent to the contour surface at point A. Let's look at an example of this. For the function w = x 2 + y 2 + z 2, the contour surface at (-1, 0, 2) is the sphere 5 = x 2 + y 2 - z 2. The gradient for w is:
The gradient for the contour surface at (-1, 0, 2) is:
The contour surface 5 = x 2 + y 2 + z 2 is shown in red below, along with the gradient vector v:
Vector
Contour
Cross product
Dot product
Eigenvalue
Eigenvector
Gradient
Magnitude
Unit vector
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13547 | https://depts.washington.edu/matseed/mse_resources/Webpage/Biomaterials/young's_modulus.htm | The Young's Modulus of a material is a fundamental property of every material that cannot be changed. It is dependent upon temperature and pressure however.
The Young's Modulus (or Elastic Modulus) is in essence the stiffness of a material. In other words, it is how easily it is bended or stretched.
To be more exact, the physics and numerical values are worked out like this:
Young's Modulus = Stress / Strain
where:
Stress = force / cross sectional area
Strain = change in length / original length
when graphed, the resulting plot will look something similar to this:
The Young's modulus is the slope of the initial section of the curve (i.e. m in y = mx + b).
When a material reached a certain stress, the material will begin to deform. It is up to point where the materials structure is stretching and not deforming. However, if you are to stress the material more than this, the molecules or atoms inside will begin to deform and permanently change the material.
A good analogy to this would be a rubber band: when you stretch a rubber band you are not deforming it, but stretching it. However, if you pull it too hard the rubber band will begin to deteriorate, or deform. Usually when this happens, it is not too much longer until it breaks.
Why Important?
The Young's Modulus is very important to doctors and scientists as this constant can tell them when a structural implant will deform. This will let them know how to design a piece mechanically for use in a body. |
13548 | https://www.youtube.com/watch?v=GSrmhGGajf4 | Determine Key Components and Equation From a Table of a Quadratic
Mathispower4u
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This video explains how to determine the key components of a parabola and how to determine equation for the parabola from a table.
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Transcript:
we are given a table of values from a quadratic function we're asked to determine the key characteristics of the graph of the quadratic function which is called a parabola and then we're asked to write an equation for the parabola in factored form remember if the parabola opens up as we see here the vertex is the lowest point on the graph if the parabola opens down the vertex is the highest point on the graph to help us determine the vertex from the table we focus on the function values or y values notice toward the bottom of the table the y values are getting larger and larger as we move up the table the y values get smaller and smaller until we see the smallest y value of negative four and as we go up further the y values start to increase again so because y equals negative four is the smallest y value in the table we know two things we know the vertex must be the ordered pair negative three comma negative four because y equals negative four is the smallest y value and therefore negative three comma negative four would be the lowest point on the graph and therefore the parabola must open up like this parabola again so now we know the vertex is the ordered pair negative three comma negative four let's also plot the vertex on the coordinate plane so negative three comma negative four is this point here we'll label this point v for vertex next we're asked to find the equation of the axis of symmetry the axis of symmetry is a vertical line that passes through the vertex and therefore this vertical line must be the axis of symmetry every x coordinate on this vertical line is negative three which is why the equation of the axis of symmetry is x equals negative three notice the x coordinate of the vertex gives us the equation of the axis of symmetry next we're asked to find the vertical intercept the vertical intercept is the point where the graph crosses the vertical axis every point on the vertical or y-axis has an x value of zero which means to find the vertical intercept from the table we look for the x value of zero which is here the vertical intercept is the ordered pair zero comma five let's go ahead and plot this point which is here now if we're graphing the parabola notice how the vertical intercept is three units to the right of the axis of symmetry and therefore there must be another point on the graph that is three units to the left of the axis of symmetry which would be this point here notice how these two points are symmetrical across the axis of symmetry next we're asked to find the horizontal intercepts or x intercepts every point on the horizontal or x axis has a y value or function value of zero to determine the horizontal intercepts from the table we look for the y values of zero which means one horizontal intercept is negative five comma zero and the other is negative one comma zero so again we have negative five comma zero and negative one comma zero let's go ahead and plot these two points negative one comma zero is here notice how this is two units to the right of the axis of symmetry and negative five comma zero is here which is two units to the left of the axis of symmetry again these two points are symmetrical across the axis of symmetry so it doesn't ask let's go ahead and graph the parabola the parabola looks something like this for the last part we're asked to find the equation of the parabola in factored form and therefore we'll find the equation in the form y equals a times the quantity x minus r sub times the quantity x minus r sub two where r sub one and r sub two are the roots or zeros of the quadratic function we can determine the zeros of the function by looking at the x intercepts or horizontal intercepts which we already found on the graph here which were negative five comma zero or negative one comma zero the ordered pair negative five comma zero tells us that when x is negative five the y value or function value is zero and therefore negative five is a root or zero of the function and the ordered pair negative one comma zero indicates when x is negative one the y value or function value is zero and therefore negative one is also a root or zero so using these ordered pairs let's let r sub one equal negative five and r sub two equal negative one and now we can begin building the equation we have y equals a times the quantity x minus negative five times the quantity x minus negative one minus negative five simplifies to plus five minus negative one simplifies to plus one giving us y equals a times the quantity x 5 x the quantity x 1. the last step is to find the value of a by selecting any point on the parabola except the two horizontal intercepts and then performing substitution for x and y let's use the vertical intercept with the ordered pair zero comma five so using the ordered pair zero comma five we substitute zero for x and five for y and then solve for a performing substitution gives us five equals a times the quantity we substitute zero for x we have zero plus five here which is just five times if we substitute zero for x here we have zero plus one which is one so simplifying we have five equals five a so five equals five a we divide both sides by five simplifying we have a equals one and now we substitute one for a in the equation so because a is one we can leave it off our final equation is just y equals the quantity x plus five times the quantity x plus one i hope you found this helpful |
13549 | https://artofproblemsolving.com/wiki/index.php/Incircle?srsltid=AfmBOopdnzz8OSRHNIix1rosWo2lv48QpD-0yjxcz1RhcT83y5VBBIlF | Art of Problem Solving
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Incircle
An incircle of a convexpolygon is a circle which is inside the figure and tangent to each side. Every triangle and regular polygon has a unique incircle, but in general polygons with 4 or more sides (such as non-squarerectangles) do not have an incircle. A quadrilateral that does have an incircle is called a Tangential Quadrilateral. For a triangle, the center of the incircle is the Incenter, where the incircle is the largest circle that can be inscribed in the polygon. The Incenter can be constructed by drawing the intersection of angle bisectors.
Formulas
The radius of an incircle of a triangle (the inradius) with sides and area is
The area of any triangle is where is the Semiperimeter of the triangle.
The formula above can be simplified with Heron's Formula, yielding
The radius of an incircle of a right triangle (the inradius) with legs and hypotenuse is .
For any polygon with an incircle, , where is the area, is the semi perimeter, and is the inradius.
The coordinates of the incenter (center of incircle) are , if the coordinates of each vertex are , , and , the side opposite of has length , the side opposite of has length , and the side opposite of has length .
The formula for the semiperimeter is .
The area of the triangle by Heron's Formula is .
See also
Circumradius
Inradius
Kimberling center
Circumcircle
Click here to learn about the orthocenter, and Line's Tangent
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13550 | https://www.wikihow.com/Calculate-Odds | Published Time: 2011-05-06
How to Calculate Odds: 11 Steps (with Pictures) - wikiHow
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How to Calculate Odds
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Co-authored byDavid Jia
Last Updated: April 20, 2025Fact Checked
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Calculating Basic Odds
| Calculating Complex Odds
| Understanding Gambling Odds
| Probability Cheat Sheets
| Video
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|Show more|Show less
X
This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math.
There are 12 references cited in this article, which can be found at the bottom of the page.
This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources.
This article has been viewed 1,042,500 times.
The mathematical concept of odds is related to, yet distinct from the concept of probability. In simplest terms, odds are a way of expressing the relationship between the number of favorable outcomes in a given situation versus the number of unfavorable outcomes. Usually, this is expressed as a ratio (like 1: 3 or 1/3). Calculating odds is central to the strategy of many games of chance, like roulette, horse racing and poker. Whether you're a high-roller or simply a curious newcomer, learning how to calculate odds can make games of chance a more enjoyable (and profitable!) activity.
Calculating Odds from Probability
If you have a ratio describing probability, i.e. 33% or 1:2, calculate the odds by dividing the outcome you want over all possible outcomes (1/(1+2) = 1/3), then use the formula Odds = P / (1 - P), in this example O = .33 / (1 - .33) = .5.
Steps
Part 1
Part 1 of 3:
Calculating Basic Odds
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1
Determine the number of favorable outcomes in a situation.X Research source Let's say we're in a gambling mood but all we have to play with is one simple six-sided die. In this case, we'll just wager bets on what number the die will show after we roll it.XExpert SourceDavid Jia Academic TutorExpert Interview
Let's say we bet that we'll roll either a one or a two. In this case, there's two possibilities where we win - if the dice shows a two, we win, and if the dice shows a one, we also win. Thus, there are two favorable outcomes.
2
Determine the number of unfavorable outcomes. In a game of chance, there's always a chance that you won't win. Count how many outcomes there are that would cause you to lose.XExpert SourceDavid Jia Academic TutorExpert Interview
In the example with the die, if we bet that we'll roll either a one or a two, that means we'll lose if we roll a three, four, five, or six. Since there are four ways that we can lose, that means that there are four unfavorable outcomes.
Another way to think of this is as the Number of total outcomes minus the number of favorable outcomes. When rolling a die, there are a total of six possible outcomes - one for each number on the die. In our example, then, we would subtract two (the number of desired outcomes) from six. 6 - 2 = 4 unfavorable outcomes.
Similarly, you may subtract the number of unfavorable outcomes from the total number of outcomes to find the number of favorable outcomes.
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3
Express odds numerically. Generally, odds are expressed as the ratio of favorable outcomes to unfavorable outcomes, often using a colon. In our example, our odds of success would be 2: 4 - two chances that we'll win versus four chances that we'll lose. Like a fraction, this can be simplified to 1: 2 by dividing both terms by the common multiple of 2. This ratio is written (in words) as "one to two odds."XExpert SourceDavid Jia Academic TutorExpert Interview
You may choose to represent this ratio as a fraction. In this case, our odds are 2/4, simplified as 1/2. Note - 1/2 odds don't mean we have a one-half (50%) chance of winning. In fact, we have a one-third chance of winning. Remember when expressing odds that odds are a ratio of favorable outcomes to unfavorable outcomes - not a numerical measurement of how likely we are to win.
4
Know how to calculate odds against an event happening.X Research source The 1: 2 odds we just calculated are the odds in favor of us winning. What if we want to know the odds of losing, also called the odds against us winning? To find the odds against us, simply flip the ratio of odds in favor of winning.1: 2 becomes 2: 1.
If you express the odds against winning as a fraction, you get 2/1. Remember, as above, that this isn't an expression of how likely you are to lose, but rather the ratio of unfavorable outcomes to favorable outcomes. If it were an expression of how likely you were to lose, you'd have a 200% chance of losing, which is obviously impossible. How do you like those odds? In reality, you have a 66% chance of losing - 2 chances to lose and 1 chance to win means 2 losses / 3 total outcomes = .66 = 66%
5
Know the difference between odds and probability.X Research source The concepts of odds and probability are related, but not identical. Probability is simply a representation of the chance that a given outcome will happen. This is found by dividing the number of desired outcomes over the total number of possible outcomes. In our example, the probability (not odds) that we'll roll a one or a two (out of six possible die roll outcomes) is 2 / 6 = 1 / 3 = .33 = 33%. So our 1: 2 odds of winning translate to a 33% chance that we'll win.XExpert SourceDavid Jia Academic TutorExpert Interview
It's easy to convert between probability and odds. To find an odds ratio from a given probability, first express the probability as a fraction (we'll use 5/13). Subtract the numerator (5) from the denominator (13): 13 - 5 = 8. The answer is the number of unfavorable outcomes. Odds can then be expressed as 5: 8 - the ratio of favorable to unfavorable outcomes.
To find probability from a given odds ratio, first express your odds as a fraction (we'll use 9 / 21). Add the numerator (9) and denominator (21): 9 + 21 = 30. The answer is the total number of outcomes. Probability can be expressed as 9/30 = 3/10 = 30% - the number of favorable outcomes over the number of total possible outcomes.
A simple formula for calculating odds from probability is O = P / (1 - P). A formula for calculating probability from odds is P = O / (O + 1).X Research source
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Part 2
Part 2 of 3:
Calculating Complex Odds
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1
Differentiate between dependent and independent events.X Research source In certain scenarios, odds for a given event will change based on the results of past events. For example, if you have a jar full of twenty marbles, four of which are red and sixteen of which are green, you'll have 4: 16 (1: 4) odds to draw a red marble at random. Let's say you draw a green marble. If you don't put the marble back into the jar, on your next attempt, you'll have 4: 15 odds to draw a red marble. Then, if you draw a red marble, you'll have 3: 15 (1: 5) odds on the following attempt. Drawing a red marble is a dependent event - the odds depend on which marbles have been drawn before.
Independent events are events whose odds aren't effected by previous events. Flipping a coin and getting a heads is an independent event - you're not more likely to get a heads based on whether you got a heads or a tails last time.
2
Determine whether all outcomes are equally likely.X Research source If we roll one die, it's equally likely that we'll get any of the numbers 1 - 6. However, if we roll two dice and add their numbers together, though there's a chance we'll get anything from 2 to 12, not every outcome is equally likely. There's only one way to make 2 - by rolling two 1's - and there's only one way to make 12 - by rolling two 6's. By contrast, there are many ways to make a seven. For instance, you could roll a 1 and a 6, a 2 and a 5, a 3 and a 4, and so on. In this case, the odds for each sum should reflect the fact that some outcomes are more likely than others.
Let's do an example problem. To calculate the odds of rolling two dice with a sum of four (for instance, a 1 and a 3), begin by calculating the total number of outcomes. Each individual dice has six outcomes. Take the number of outcomes for each die to the power of the number of dice: 6(number of sides on each die)2(number of dice) = 36 possible outcomes. Next, find the number of ways you can make four with two dice: you can roll a 1 and a 3, a 2 and a 2, or a 3 and a 1 - three ways. So the odds of rolling a combined "four" with two dice are 3: (36-3) = 3: 33 = 1: 11
Odds change exponentially based on the number of events occurring simultaneously. Your odds of rolling a "yahtzee" (five dice that are all the same number) in one roll are very slim - 6: 6 5 - 6 = 6: 7770 = 1: 1295!
3
Take mutual exclusivity into account.X Research source Sometimes, certain outcomes can overlap - the odds you calculate should reflect this. For instance, if you're playing poker and you have a nine, ten, jack, and queen of diamonds in your hand, you want your next card either to be a king or eight of any suit (to make a straight), or, alternatively, any diamond (to make a flush.) Let's say the dealer is dealing your next card from a standard fifty-two card deck. There are thirteen diamonds in the deck, four kings, and four eights. However, the total number of favorable outcomes isn't 13 + 4 + 4 = 21. The thirteen diamonds already includes the king and eight of diamonds - we don't want to count them twice. The actual number of favorable outcomes is 13 + 3 + 3 = 19. Thus, the odds of being dealt a card that will give you a straight or flush are 19: (52 - 19) or 19: 33. Not bad!
In real life, of course, if you already have cards in your hand, you're rarely being dealt cards from a complete fifty-two card deck. Keep in mind that the number of cards in the deck decreases as cards are dealt. Also, if you're playing with other people, you'll have to guess what cards they have when you're estimating your odds. This is part of the fun of poker.
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Part 3
Part 3 of 3:
Understanding Gambling Odds
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1
Know common formats for expressing gambling odds.X Research source If you're venturing into the world of gambling, it's important to know that betting odds don't usually reflect the true mathematical "odds" of a certain event happening. Instead, gambling odds, especially in games like horse racing and sports betting, reflect the payout that a bookmaker will give on a successful bet. For instance, if you wager $100 on a horse with 20:1 odds against him, this doesn't mean that there are 20 outcomes where your horse loses and 1 where he wins. Rather, it means that you'll be paid 20 times your original wager - in this case, $2,000! To add to the confusion, the format for expressing these odds sometimes varies regionally. Here are a few non-standard ways that gambling odds are expressed:X Research source
Decimal (or "European format") odds. These are fairly easy to understand. Decimal odds are simply expressed as a decimal number, like 2.50. This number is the ratio of the payout to the original stake. For instance, with odds of 2.50, if you bet $100 and win, you'll receive $250 dollars - 2.5 times your original stake. In this case, you'll make a tidy profit of $150.
Fractional (or "UK format") odds. These are expressed as a fraction, like 1/4. This represents the ratio of the profit (not total payout) from a successful bet to the stake. For instance, if you bet $100 on something with 1/4 fractional odds and win, you'll profit by 1/4 of your original stake - in this case, your payout will be $125 for a profit of $25.
Moneyline (or "US format") odds. These can be difficult to understand. Moneyline odds are expressed as a number preceded by a minus sign or a plus sign, like -200 or +50. A minus sign means the number represents how much you need to wager to make $100. A positive sign means the number represents how much you'll win if you bet $100. Remember this subtle distinction! For example, if we wager $50 with moneyline odds of -200, when we win, we'll get a payout of $75 for a total profit of $25. If we wager $50 with moneyline odds of +200, we'll get a payout of $150 for a total profit of $100.
In moneyline odds, a simple "100" (no plus or minus) represents an even bet - whatever money you stake, you'll earn as profit if you win.
2
Understand how gambling odds are set. The odds that bookmakers and casinos set aren't usually calculated from the mathematical probability that certain events will occur. Rather, they're carefully set so that, in the long run, the bookie or casino will make money, regardless of any short-term outcomes! Take this into account when making your bets - remember, eventually, the house always wins.X Research source
Let's look at an example. A standard roulette wheel has 38 numbers - 1 through 36, plus 0 and 00. If you bet on one space (let's say 11), you have 1: 37 odds of winning. However, the casino sets the payout odds at 35: 1 - if the ball lands on 11, you'll win 35 times your original stake. Notice that the payout odds are slightly lower than the odds against you winning. If casinos weren't interested in making money, you would be paid out at 37: 1 odds. However, by setting the payout odds slightly below the actual odds of you winning, the casino will gradually make money over time, even if it has to make the occasional large payout when the ball lands on 11.
3
Don't fall prey to common gambling fallacies. Gambling can be fun - even addictive. However, certain widely-circulated gambling strategies that at first appear to be "common sense" are, in fact, mathematically false. Below are just a few things you should keep in mind when you go gambling - don't lose more money than you have toX Research source
You're never "due" to win. If you've been at the Texas Hold 'Em table for an hour and you haven't been dealt a single good hand, you may want to stay in the game in the hopes that a winning straight or flush is "right around the corner." Unfortunately, your odds don't change with the amount of time you've been gambling. The cards are randomly shuffled before every deal, so if you've had ten bad hands in a row, you're just as likely to get another bad hand as you are if you've had a hundred bad hands in a row. This extends to most other games of chance - roulette, slots, etc.
Sticking with one specific bet won't increase your odds. You may know someone who has "lucky" lotto numbers - though it can be fun to bet money on numbers that have special personal meaning, in random games of chance, you're never more likely to win by betting on the same thing every time than you are by betting on a different thing every time. Lottery numbers, slots, and roulette wheels are completely random. In roulette, for example, it's just as likely that you'll roll "9" three times in a row as it is that you'll roll any specific three numbers in order.
If you're one away from the winning number, you weren't "close." If you pick the number 41 for the lottery and the winning number is revealed as 42, you may feel absolutely crushed, but cheer up! You weren't even close. Two numbers that are close together, like 41 and 42, aren't mathematically connected in any way in random games of chance.
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Question What's the difference between odds and probability? David Jia
Math Tutor David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. David Jia Math Tutor Expert Answer The odds are the number of outcomes you do want vs the number of outcomes you don't want. If you're trying to roll a certain number on a die, your odds are 1:5. The probability is the number of outcomes you do want divided by the total number of possible outcomes, so with the die, that would be 1/6. Thanks! We're glad this was helpful.
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Question How do you figure out your odds of winning? David Jia
Math Tutor David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. David Jia Math Tutor Expert Answer First, calculate the total number of possible outcomes that you want. For example, if you want to calculate the odds of rolling a 2 on a 6-sided die, that number would be 1, because there's only one side with a 2. Then, figure out how many outcomes you don't want. In this case, that's 5, because there are 5 sides you don't want to roll. In that case, your odds of winning would be 1:5. Thanks! We're glad this was helpful.
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Question The odds of picking the right box out of two boxes really 50-50? Donagan Top Answerer Yes. There are two possible outcomes and one "right" outcome. One out of two is 50 out of 100, or 50-50. Thanks! We're glad this was helpful.
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David Jia
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This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. This article has been viewed 1,042,500 times.
60 votes - 76%
Co-authors: 35
Updated: April 20, 2025
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Categories: Probability and Statistics
Article Summary X
To calculate odds, start by determining the number of favorable outcomes and the number of unfavorable outcomes. For example, if you're trying to calculate the odds of rolling a 1 on a 6-sided die, the number of favorable outcomes would be 1 and the number of unfavorable outcomes would be 5. Once you know the number of both favorable and unfavorable outcomes, just write them as a ratio or a fraction to express the odds of winning. In the 6-sided die example, the odds would be 1:6 or 1/6. To learn how to calculate more complex odds, scroll down!
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13552 | https://www.chegg.com/homework-help/questions-and-answers/question-regarding-picture-cross-product-r-x-p-becomes-magnitude-r-multiplied-magnitude-p--q19401594 | Solved EXAMPLE aan collusion of a Lump of Putty with a | Chegg.com
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EXAMPLE aan collusion of a Lump of Putty with a Turntable in M, radius center o isat A axle, which lies along the vertical and is mounted on a frictionless the edge of a lump putty that passes so it a ine Figure 39. When the putty hits the table, it sticks to the edge, and the two turnt rotate together with angular velocity av. Find of angular This problem
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Question: EXAMPLE aan collusion of a Lump of Putty with a Turntable in M, radius center o isat A axle, which lies along the vertical and is mounted on a frictionless the edge of a lump putty that passes so it a ine Figure 39. When the putty hits the table, it sticks to the edge, and the two turnt rotate together with angular velocity av. Find of angular This problem
My question regarding the picture is why does the cross product (r x p) becomes the magnitude of r multiplied by the magnitude of p multiplied by the sin(theta). How come there is no actual cross product of the two vectors being done. Thanks for the help. Show transcribed image text
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EXAMPLE aan collusion of a Lump of Putty with a Turntable in M, radius center o isat A axle, which lies along the vertical and is mounted on a frictionless the edge of a lump putty that passes so it a ine Figure 39. When the putty hits the table, it sticks to the edge, and the two turnt rotate together with angular velocity av. Find of angular This problem is easily solved using conservation there is no nted on a frictionless axle, Because the tumtable is mout nt of the external the table in the z direction. Therefore the z This is true even if we include on the system is zero, and La is conserved. gravity, which acts in the z direction and contributes nothing to the torque the z direction.) Before the collision, the tumtable has zero angular the initial while the putty has rx p, which points in the z direction. Thus total angular momentum has compone e, r (mv) sin 6 After the collision, the putty and turntable rotate together about the z axis with tokal moment of inertia I (m M/2 R2, and the z component of the final gular momentum Ioo. Therefore, conservation of angular momentum Lfin tells us that n the form L Figure 39 A lump of putty of mass m is thrown with velocity vat a stationary tumtable. The putty's line of approach passes within the distance b of the table's center 0 the collision theory-the theory of collisions, usually between distance b is called the impact paramete omic This ismR for the putty stuck at radius R plus MR2 for the uniform tumtabl solu find to Angula der were N that so CM is b and hen 3.37. I as the f EXA
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13554 | https://unh.edu/ece/ECE603/Acrobat%20Files/Smith%20Chart.pdf | 1 Revisiting a T-Line With Any Termination 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ( ) 2cos 2 sin 2 sin 2cos tan tan j z j z L j z j z L j z j z j z j z L L j z j z L L L j z j z L L L L L Z Z e e e e Z Z Z z Z Z e e Z Z e e Z Z Z Z e Z Z e Z z Z j z Z Z Z Z e Z Z e Z j z Z z Z jZ z Z Z jZ z In the general case, where a transmission line is terminated in ZL, the impedance along the line is given by: Note: Z(z)=Z0 if ZL=Z0 The above equation shows how the input impedance to an unmatched transmission line changes with electrical length, z.
Since the electrical length changes with frequency, the input impedance to an unmatched line will be frequency dependent.
Impedance Calculations Because the formula for impedance is a bit cumbersome and not intuitive, design calculations and measurements are often made graphically using a Smith Chart. The Smith Chart works with normalized impedance and admittance, where normalization is made with respect to the characteristic impedance of the transmission line.
For example, the normalized impedance for a load ZL = 73 + j42 on a 50 transmission line is ZLN = 1.46 + j0.84 By plotting the normalized load impedance on a Smith Chart, the input impedance as a function of line length can be found.
The Smith Chart also provides the value of the reflection coefficient, power delivered to load, as well as the voltage standing wave ratio (VSWR) Distance measurements are given in terms of wavelengths.
2 Center at Center at 1+j0 (matched) (matched) Re{ Z or Y } Re{ Z or Y } Im { Z or Y } Im { Z or Y } The Smith Chart The Smith Chart To find To find Z along the line for a particular along the line for a particular ZL, find , find ZL/Z0 on the chart on the chart and draw a circle, centered at and draw a circle, centered at 1+j0 through that point. Points on through that point. Points on that circle represent impedance on the line corresponding to that circle represent impedance on the line corresponding to distance which is read from the scale “wavelengths toward the distance which is read from the scale “wavelengths toward the generator”.
generator”.
Blank Smith Chart Blank Smith Chart 3 Bottom Scale of Smith Chart Bottom Scale of Smith Chart Network Analyzer Smith Chart Display Network Analyzer Smith Chart Display 4 Another Smith Chart Type Display Another Smith Chart Type Display Smith Chart Example Smith Chart Example A half A half-wave dipole antenna ( wave dipole antenna (Z = 73 + j42 ) is connected to a ) is connected to a 50 transmission line. How long must that line be before the transmission line. How long must that line be before the real part of the input impedance is real part of the input impedance is 50 ?
Step 1: Step 1: plot the normalized impedance ( plot the normalized impedance (1.46 + j0.84) on the Chart ) on the Chart Step 2: Step 2: Draw a circle through that point, with the center of the Draw a circle through that point, with the center of the circle at circle at 1 + j0 ZL = 73 + j42 Z0 = 50 Re{ Zin } = 50 Step 3: Step 3: Move along the circle you drew, towards the generator, Move along the circle you drew, towards the generator, until you intercept the until you intercept the Re{ ZN }=1 circle is intercepted. The circle is intercepted. The distance moved on the circle to get to that intercept, read from the distance moved on the circle to get to that intercept, read from the “wavelengths toward generator” scale, represents the length of the “wavelengths toward generator” scale, represents the length of the transmission line transmission line .
5 Smith Chart Example Smith Chart Example 1.46+j0.84 1.46+j0.84 Intercept with Intercept with Re{ ZN }=1 circle circle 0.198 0.198 0.348 0.348 Smith Chart Example Smith Chart Example The distance moved on the scale is The distance moved on the scale is 0.348- 0.198= 0.15 .
This represents the length of the transmission line, where This represents the length of the transmission line, where is is the wavelength the wavelength in the transmission line in the transmission line.
The normalized input impedance for that transmission line is The normalized input impedance for that transmission line is read from the Smith Chart to be read from the Smith Chart to be 1 - j0.75. This is read from the . This is read from the point where the circle you drew intersects the point where the circle you drew intersects the Re{ ZN } = 1 circle.
circle.
The actual input impedance to the terminated line is The actual input impedance to the terminated line is (1 - j0.75)50= 50 - j37.5 = ZIN What we will be doing What we will be doing later is later is to add a reactive component that will to add a reactive component that will cancel the reactive component of the input impedance, resulting in cancel the reactive component of the input impedance, resulting in an input impedance equal to an input impedance equal to Z0 (a perfect match). We will do this (a perfect match). We will do this using “single using “single-stub” matching.
stub” matching.
6 Single Single-Stub Matching Stub Matching As shown previously, the input impedance (admittance) of a As shown previously, the input impedance (admittance) of a shorted transmission line (a “stub”) is purely reactive. By placing shorted transmission line (a “stub”) is purely reactive. By placing a stub in parallel with another transmission line, the reactive a stub in parallel with another transmission line, the reactive component can be cancelled, leaving a pure component can be cancelled, leaving a pure-real input real input impedance. This can be used to achieve a perfect match.
impedance. This can be used to achieve a perfect match.
Z0 = 50 Zin = 50 ZL d To do this, we need to choose To do this, we need to choose so that the real part of the input so that the real part of the input admittance is equal to the characteristic admittance, and then admittance is equal to the characteristic admittance, and then choose choose d so that the reactive components cancel. This can all be so that the reactive components cancel. This can all be done on the smith chart.
done on the smith chart.
Single Single-Stub Matching Example Stub Matching Example Step 1: Step 1: Plot the normalized load impedance ( Plot the normalized load impedance (1.46+j0.84), and ), and draw a circle through that point, centered at draw a circle through that point, centered at 1 + j0. Get the . Get the normalized load admittance by drawing a line from normalized load admittance by drawing a line from ZLN through through 1+j0 until you intersect the circle you drew on the other side.
until you intersect the circle you drew on the other side.
From the chart, you get a normalized load admittance of From the chart, you get a normalized load admittance of 0.52-j0.3 Find Find and and d that will match our half that will match our half-wave dipole antenna wave dipole antenna (ZL=73+j42 ) to a ) to a 50 transmission line.
transmission line.
Note: Note: Since we are going to add the stub in parallel with the Since we are going to add the stub in parallel with the transmission line, it will be easier to work with admittance rather transmission line, it will be easier to work with admittance rather than impedance.
than impedance.
Step 2: Step 2: Move towards the generator (clockwise) on the circle you Move towards the generator (clockwise) on the circle you drew until you intersect the drew until you intersect the Re{ YN }=1 circle. The distance you circle. The distance you moved to get to that intersection corresponds to the distance moved to get to that intersection corresponds to the distance (= 0.5-(0.441-0.157) = 0.216 ). From the Smith Chart, the ). From the Smith Chart, the normalized input admittance at this point is normalized input admittance at this point is 1+j0.84.
7 Single Single-Stub Matching Example Stub Matching Example Step 3: Step 3: In this step, we are looking for the length, In this step, we are looking for the length, d, of a shorted of a shorted stub that will have an input admittance of stub that will have an input admittance of 0-j0.84. The load . The load admittance of a short is infinity ( admittance of a short is infinity (YL=), so that is where we will , so that is where we will begin on the Smith Chart. We will move towards the generator begin on the Smith Chart. We will move towards the generator from the point from the point YL=until we intercept the until we intercept the Im{ YN= -0.84} line. At line. At this point, the input admittance to the shorted stub is this point, the input admittance to the shorted stub is 0-j0.84. The . The distance traveled to get to that point ( distance traveled to get to that point (0.389-0.25=0.139) is ) is d.
Z0 = 50 Zin = 50 ZL d Using these values of Using these values of and and d will result in a perfect match at will result in a perfect match at the frequency for which it was designed. That match will the frequency for which it was designed. That match will degrade as the frequency varies.
degrade as the frequency varies.
ZLN LN=1.46+j0.84 =1.46+j0.84 YLN LN=0.52 =0.52-j0.3 j0.3 0.441 0.441 0.157 0.157 Intersection with Intersection with Re {YN}=1 circle circle YN=1+j0.84 YL= 0.25 0.25 0.389 0.389 YLN LN=0 =0 -j0.84 j0.84 8 Design Parameters for Example = 0.5- (0.441- 0.157)= 0.216 d =0.389- 0.25= .139 Notes: • going around the Smith Chart once corresponds to moving a distance of 0.5 wavelengths on the transmission line • the system will be matched at a single frequency Quarter-Wave Matching Transformer Used to convert any real load impedance (ZL ) to a desired real input impedance (Zin ).
ZL Z0 Zin λ/4 Note: we can make our load real by placing a reactance-cancelling component in parallel with it.
Because λ/4 represents one-half rotation around the Smith Chart, the normalized input impedance is equal to the normalized admittance of ZL: 0 0 1 N in L L Z Y Z Z Z 0 2 0 0 1 S : o L in Z Z L Z Z Z Z 9 Quarter-Wave Matching Transformer (2) We can use this equation to find the characteristic impedance (Z0) of the quarter-wave length on T-line: 0 in L Z Z Z For example: find the characteristic impedance of a quarter-wavelength section of T-line that would match our half-wave dipole antenna to a 50 Ω T-line (assume that the j42 Ω reactive component has been cancelled): 0 50 73 60.4 in L Z Z Z Comment about stub and quarter-wave matching techniques: Since all of the T-line dimensions are in wavelengths, the match will exist only over a narrow range of frequencies.
Different approaches need to be used to achieve broadband matching. |
13555 | https://www.ychlpyss.edu.hk/~student/temparea/hw/ycs_(F.2)%20Chapter%205%20Linear%20Equations%20in%20Two%20Unknowns%20(Worksheet).pdf | ~ 1 ~ 仁濟醫院林百欣中學 中二級 數學科工作紙 第五章 二元一次方程 5.1 二元一次方程 1. 下列各題中，檢驗所給的序偶是否方程 2x – 3y = 2 的解。 (a) (4, 3) (b) (–5, –4) 12. 下列各題中，檢驗所給的序偶是否方程y = 2x – 1 的解。 (a) (2, 3) (b) (–2, –3) 13. 下列各題中，檢驗所給的序偶是否方程x – 2y = 3 的解。 (a) (3, 3) (b) (–3, 0) 14. 下列各題中，檢驗所給的序偶是否方程x + 2y = 4 的解。 (a) (0, 2) (b) (–2, 4) ~ 2 ~ 20. 判斷下列各點是否位於方程x + 4y = 3 的圖像上。 (a) A(7, –1) (b) B(0, 1) 21. 判斷下列各點是否位於方程y = 6x – 1 的圖像上。 (a) A(1, 6) (b) B(–1, –7) 22. 判斷方程2x + 4y = 1 的圖像是否通過下列各點。 (a) A(–4, 2) (b) B(5, –2) 23. 判斷方程5x – 3y + 1 = 0 的圖像是否通過下列各點。 (a) A(–2, 3) (b) B(1, 2) 9. 下列各題中，完成各表使序偶 (x, y) 滿足所給的方程。 (a) y = 6x + 4 (b) y = 1 – 4x x –1 0 1 2 x –2 0 2 3 y y ~ 3 ~ 10. 下列各題中，完成各表使序偶 (x, y) 滿足所給的方程。 (a) y = –(x – 2) (b) y = 2(1 – x) x –1 0 1 2 x –2 0 2 3 y y 11. 下列各題中，完成各表使序偶 (x, y) 滿足所給的方程。 (a) y = 4 3 + x (b) y = 2 5 − x x –3 0 3 6 x –3 –1 0 3 y y 40. 圖中所示為方程2x + 3y = 11 的圖像。P 是圖像上的一點。求P 的y 坐標。 41. 圖中所示為方程3x – 2y = 8 的圖像。Q 是圖像上的一點。 (a) 求Q 的x 坐標。 (b) 判斷該圖像是否通過 (–2, –6)。 (c) 判斷該圖像是否與y 軸相交於 (0, –4)。 42. 圖中所示為方程5x – 4y = 6 的圖像。R 是圖像上的一點。 (a) 求R 的坐標。 (b) 智峰打算在同一直角坐標平面上繪畫5x – 8y = 12 的圖像。 該圖像是否通過R？ y x 0 R 5x – 4y = 6 ~ 4 ~ 5.2 以圖解法解聯立二元一次方程 1. 利用圖解法解聯立方程 − = − = + 1 5 2 y x y x 。 8. (a) 在所給的直角坐標平面上繪畫 y = 2x + 2 和x – 2y = 2 的圖像。 (b) 利用 (a) 部的圖像，解聯立方程 = − + = 2 2 2 2 y x x y 。 ~ 5 ~ 5.3 以代數方法解聯立二元一次方程 5. 利用代入消元法解聯立方程 + = = x y x y 4 12 。 6. 利用代入消元法解聯立方程 = + − = 1 3 4 y x y x 。 9. 利用代入消元法解聯立方程 + = − = 6 5 3 2 y x x y 。 13. 利用代入消元法解聯立方程 = + = + 4 2 1 y x y x 。 ~ 6 ~ 15. 利用代入消元法解聯立方程 = + − = − 29 4 27 3 y x y x 。 17. 利用代入消元法解聯立方程 = − = + 7 2 39 3 4 y x y x 。 18. 利用代入消元法解聯立方程 = + = − 17 3 2 2 9 y x x y 。 21. 利用代入消元法解聯立方程 = − + = − − 0 45 3 0 15 y x y x 。 ~ 7 ~ 43. 利用代入消元法解聯立方程 = − = + 16 3 5 27 4 3 y x y x 。 44. 利用代入消元法解聯立方程 = + = + 18 3 4 31 8 5 y x y x 。 45. 利用代入消元法解聯立方程 = − = − − 11 5 3 37 4 7 y x y x 。 46. 利用代入消元法解聯立方程 = − = − 43 5 4 44 7 2 y x y x 。 ~ 8 ~ 23. 利用加減消元法解聯立方程 = − = + 4 2 y x y x 。 24. 利用加減消元法解聯立方程 − = + − = + 1 3 y x y x 。 25. 利用加減消元法解聯立方程 = + − = − 11 2 1 2 y x y x 。 27. 利用加減消元法解聯立方程 = − = − 8 2 5 y x y x 。 ~ 9 ~ 28. 利用加減消元法解聯立方程 = + = + 51 3 5 15 3 y x y x 。 31. 利用加減消元法解聯立方程 = + = − 21 2 1 3 2 y x y x 。 32. 利用加減消元法解聯立方程 = − = − 14 4 22 4 5 y x y x 。 59. 利用加減消元法解聯立方程 = + = − 0 5 3 29 8 y x y x 。 ~ 10 ~ 60. 利用加減消元法解聯立方程 − = − = − 2 7 4 1 2 y x y x 。 61. 利用加減消元法解聯立方程 = + = + 1 7 6 4 4 3 y x y x 。 65. 利用加減消元法解聯立方程 = − − = + 3 3 7 2 2 3 y x y x 。 66. 利用加減消元法解聯立方程 = + − = + 1 5 4 20 2 5 y x y x 。 ~ 11 ~ 5.4 聯立二元一次方程的應用 6. 兩數之和是27。較大的數比較小的數的三倍多11。求該兩個數。 16. 美琪有一張 $50 紙幣和一張 $20 紙幣。她把全部款項換成共23 枚 $2 和 $5 硬幣。問當中有多少枚 $5 硬 幣？ 17. 某升降機內有15 名乘客 。 乘客的總體重是800 kg 。 兒童乘客和成人乘客的平均體重分別為40 kg 和60 kg 。 求該升降機內兒童乘客和成人乘客的人數。 18. 3 支鉛筆和5 把直尺的總售價是 $49。6 支鉛筆和3 把直尺的總售價是 $42。求每支鉛筆和每把直尺的售 價。 ~ 12 ~ 參考答案 5.1 二元一次方程 1. (a) 否 (b) 是 12. (a) 是 (b) 否 13. (a) 否 (b) 否 14. (a) 是 (b) 否 20. (a) 是 (b) 否 21. (a) 否 (b) 是 22. (a) 否 (b) 否 23. (a) 否 (b) 是 9. (a) x –1 0 1 2 y –2 4 10 16 (b) x –2 0 2 3 y 9 1 –7 –11 10. (a) x –1 0 1 2 y 3 2 1 0 (b) x –2 0 2 3 y 6 2 –2 –4 11. (a) x –3 0 3 6 y 3 4 5 6 (b) x –3 –1 0 3 y –4 –3 2 5 − –1 40. (a) 1 (b) (i) 是 (ii) 否 41. (a) 6 (b) (i) 否 (ii) 是 42. (a) (0, 2 3 − ) (b) 是 5.2 以圖解法解聯立二元一次方程 1. (1, 2) 8. (b) (–2, –2) 5.3 以代數方法解聯立二元一次方程 5. (–4, –4) 6. (1, –1) 9. (1, –1) 13. (–2, 3) 15. (–3, 8) 17. (6, 5) 18. (7, 1) 21. (15, 0) 43. (5, 3) 44. (3, 2) 45. (–3, –4) 46. ( 2 9 , –5) 23. (3, –1) 24. (2, 1) 25. (5, 3) 27. (2, –3) 28. (9, 2) 31. (8, 5) 32. (2, –3) 59. (5, –3) 60. ( 5 4 , 10 9 ) 61. (–8, 7) 65. (0, –1) 66. (–6, 5) 5.4 聯立二元一次方程的應用 6. 4, 23 16. 8 枚 17. 兒童乘客：5 人，成年乘客：10 人 18. 鉛筆：$3，直尺：$8 |
13556 | https://fiveable.me/key-terms/physical-chemistry-i/molarity | Molarity - (Physical Chemistry I) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Physical Chemistry I
Molarity
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key term - Molarity
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Definition
Molarity is a way to express the concentration of a solution, defined as the number of moles of solute per liter of solution. It is a key concept in understanding how substances interact in solutions, particularly when looking at ideal and non-ideal solutions, and it plays a crucial role in calculating colligative properties. Molarity provides a quantitative measure that allows for predicting behaviors of solutions, essential for various applications in chemistry.
5 Must Know Facts For Your Next Test
Molarity is expressed in units of moles per liter (mol/L), commonly written as M.
To calculate molarity, use the formula: $$M = \frac{n}{V}$$ where $$n$$ is the number of moles of solute and $$V$$ is the volume of the solution in liters.
In ideal solutions, molarity accurately predicts properties like vapor pressure and boiling point, while non-ideal solutions may show deviations due to interactions between particles.
When considering colligative properties, an increase in molarity typically results in greater changes to boiling point elevation and freezing point depression.
Molarity can change with temperature because it is based on volume; thus, it is essential to specify conditions when reporting or using molarity.
Review Questions
How does molarity influence the behavior of both ideal and non-ideal solutions?
Molarity plays a crucial role in determining how solutions behave. In ideal solutions, the calculated molarity directly correlates with expected physical properties, such as vapor pressure and boiling point. In non-ideal solutions, however, interactions between solute and solvent molecules can lead to discrepancies from expected values. Understanding these differences helps predict how real solutions will act under various conditions.
Discuss the relationship between molarity and colligative properties. How does this relationship manifest in practical applications?
Molarity is directly related to colligative properties because these properties depend on the concentration of solute particles. For instance, higher molarity leads to greater boiling point elevation or freezing point depression. This relationship is crucial in practical applications such as calculating how much salt to add to ice for de-icing roads or understanding how antifreeze works in vehicles by altering boiling and freezing points based on its molarity.
Evaluate the implications of using molarity versus molality in experimental chemistry. When might one be preferred over the other?
Choosing between molarity and molality can significantly impact experimental outcomes. Molarity is preferred when dealing with reactions in solution because it relates to volume changes that occur during mixing. However, molality becomes essential when temperature variations affect density since it measures concentration by mass instead of volume. Understanding when to use each measure allows for more accurate predictions and analyses of chemical behaviors under varying conditions.
Related terms
Molality:A measure of concentration defined as the number of moles of solute per kilogram of solvent, often used in scenarios where temperature changes can affect volume.
Colligative Properties:Properties that depend on the number of solute particles in a solution, rather than the identity of the solute, including boiling point elevation and freezing point depression.
Dilution:The process of reducing the concentration of a solute in a solution, typically by adding more solvent, which can affect the molarity.
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10.7 Inspection and Palpation of the Lymph Nodes
This assessment is often performed with the client sitting upright on the exam table to assess the lymph nodes in the head and neck, the upper arm, and the axillae, and then with the client repositioned into a supine position to assess the lymph nodes in the groin.
NOTE: Lymph nodes are located in many areas of the body, some of which are not physically accessible. Although some of the axilla lymph nodes are associated with the breast, other lymph nodes associated with the breast will be explored in another chapter.
Use the following techniques to palpate the lymph nodes:
Prior to palpation, say to the client, “Let me know if you have any pain or tenderness when I touch you.”
Use the finger pads of two to three fingers and move them in a circular motion.
Use a light touch with gentle pressure so that you don’t forcefully push the node inwards.
After you palpate in one spot, shift your fingers to a new spot within the same area, because there are strands of lymph nodes in each area. This should be done about two to three times in each area.
Typically, you palpate bilaterally at the same time and compare the right side to the left side.
Conclude by asking the client if they had any pain or tenderness.
See Table 10.2 for what to note when assessing lymph nodes and abnormal characteristics.
Table 10.2: Lymph node assessment
| Presence of observable swelling (is the node swollen and observable upon inspection?) Skin changes over the node (what is the skin color and temperature?) Presence of pain/tenderness (is the node painful or tender?) Node location (where is the node located?) Node size (what is the size of the node?) Node consistency, e.g., rubbery, hard, soft (what is the consistency of the node?) Node movability (can you move it around when you palpate it?) Node delimitation (what are the limits or the boundaries of the node in terms of whether it is an individual node or a cluster of nodes matted together?) Symmetry (is the node the same on both sides of the person’s body?) |
Lymph Nodes of Head and Neck
Steps for assessing lymph nodes of the head and neck are as follows:
Step 1: Inspect the lymph node areas on the head and neck (see Figure 10.8) for swelling, asymmetry, and erythema.
Normally, the lymphatic areas are symmetrical between sides with no discoloration, swelling, or visible nodes.
Abnormal findings may include asymmetry and visible nodes due to swelling with erythema overlying the skin.
Step 2: Palpate for lymph nodes using a systematic approach moving from proximal to distal, thus moving from preauricular to supraclavicular area (see Figure 10.8, Table 10.3, and Video 1). Consider lymphatic drainage patterns (see Figure 10.9); this approach allows you to reflect on the origins of the cause when there is an abnormality. Recall to palpate nodes bilaterally at the same time: for example, assess the preauricular nodes on the left and right side at same time. However, for the submental lymph nodes, you should use the fingers of your dominant hand to palpate just under the chin behind the bony prominence. When assessing the deep cervical chain on clients with a muscular neck, you may ask the client to tip their head toward the side you are assessing to relax the sternomastoid muscle; thus, you will need to do one side at a time. For the supraclavicular nodes, you can ask the client to gently raise their shoulders.
Normally, lymph nodes are not palpable, but remember that normal lymph nodes can sometimes be palpable in young children.
If a lymph node is palpable, assess them as per Table 10.2 noted above.
If lymph nodes are visible or palpable, palpate the temperature over the lymph node areas of the head and neck using the dorsa of your hands, comparing bilaterally.
Step 3: Note the findings:
Normal findings might be documented as: “No lymph nodes palpable in the head and neck, no swelling, no asymmetry, no discoloration or increased temperature over lymph node locations, and no pain.”
Abnormal findings might be documented as: “Visible lymph nodes in the neck, hard, matted, and palpable cervical and supraclavicular lymph nodes on the right side, 3 cm. Client indicates they are tender.”
Table 10.3: Lymph node locations on the head and neck
| Name | Location |
| 1 Preauricular | Anterior to the tragus of the ear. |
| 2 Posterior auricular | Along the mastoid process. |
| 3 Occipital | Inferior to the occipital bone and on the lateral side of the occipital bone. |
| 4 Submental | Under the chin behind the bony prominence. |
| 5 Submandibular | Halfway between the submental nodes and the angle of the jaw. |
| 6 Tonsillar | Inferior to the angle of the jaw. |
| 7 Superficial cervical chain | Along the sternomastoid muscle at the top, just below the mastoid process. |
| 8 Deep cervical chain | Further down the sternomastoid muscle. |
| 9 Posterior cervical chain | Behind the sternomastoid muscle. |
| 10 Supraclavicular | Superior to the clavicles. |
Video 1: Palpation of head and neck lymph nodes [1:16]
Lymph Nodes of Upper Arm and Axillary Area
Steps for assessing lymph nodes of the upper arm and axillary area:
Step 1: Inspect the area surrounding the upper arm where the epitrochlear lymph nodes are located (see Figure 10.10—the area under the fingertips is where you should inspect, which is superior to and behind the medial epicondyle of the humerus) and inspect the axillae where the axillary lymph nodes are located (see Figure 10.11).
Normally, the lymphatic areas are symmetrical on each side with no discoloration, swelling, or visible nodes.
Abnormal findings may include asymmetry and visible nodes due to swelling with erythema overlying the skin. If a lymph node is palpable, assess it for abnormal characteristics (Table 10.2).
Step 2: Palpate the epitrochlear (Video 2) and axillary lymph nodes (Video 3). Prior to palpation, say to the client, “Let me know if you have any pain or tenderness when I touch you.” For the epitrochlear lymph nodes: Ask the client to place their forearm on the bedside table with the palm facing up and palpate in a gentle circular motion in about two to three locations just superior to and behind the medial epicondyle of the humerus (about 1–2 cm). For the axillary lymph nodes: Ask the client to raise their arm and rest their hand on the back of their head. Use a gentle circular motion to palpate. Begin in the axilla: palpate in one row high in the axilla, then palpate in three rows away from the axillae along the upper arm, then three rows away from the axillae down the chest wall, and then in one row medially (see Figure 10.12 for pattern). For each row, palpate about three to four locations.
Normally, there are no palpable lymph nodes.
If a lymph node is palpable, assess it as per Table 10.2 and pay attention to the drainage patterns noted in Figure 10.13. It is important to recognize that a large amount of the lymph from the breast drains into the anterior axillary nodes, lymph from the back drains into the posterior axillary nodes, and lymph from the arm drains into the lateral nodes. Additionally, this lymph drains into the central nodes to the apical nodes to the supraclavicular, and then often returns to the vascular system through the subclavian vein via the thoracic duct.
Step 3: If lymph nodes are visible or palpable, palpate the temperature over the lymph node areas of the axillary and epitrochlear areas using the dorsa of your hands comparing side to side.
Normally, there is no increase in temperature.
Some abnormal lymph nodes will have increased localized skin temperature overlying the node.
Step 4: Note the findings.
Normal findings might be documented as: “No lymph nodes palpable in the upper arm or axillae, no swelling, no asymmetry, no discoloration and no increased temperature over lymph node locations, and no pain reported by the client.”
Abnormal findings might be documented as: “Hard, matted and palpable lymph nodes on the left axillary side. Tenderness reported by the client.”
Video 2: Palpation of the epitrochlear lymph nodes (showing technique on right arm) [0:25]
Video 2: Palpation of the axillary lymph nodes (showing technique on right side) [1:05]
Lymph Nodes of the Groin
Steps for assessing lymph nodes in the groin area:
Step 1: Inspect the groin area for the inguinal lymph nodes. Assist the client into supine position and provide a drape. Briefly expose one side to inspect and then repeat on the other side. Normally, the lymphatic areas are symmetrical on each side with no discoloration, swelling, or visible or palpable nodes.
Abnormal findings may include asymmetry and visible nodes due to swelling with erythema overlying the skin.
Step 2: Palpate the groin area for the inguinal lymph nodes. Assist the client into supine position and provide a drape. Briefly expose one side and palpate, and then repeat on the other side. Prior to palpation, say to the client, “Let me know if you have any pain or tenderness when I touch you.” Palpate about four to five times in the groin area (Figure 10.14). If nodes are palpable, assess for temperature (comparing bilaterally).
Normally, the lymphatic areas are symmetrical on each side with no visible or palpable nodes.
Abnormal findings may include visible nodes due to swelling with erythema overlying the skin. If a lymph node is palpable, assess it for abnormal characteristics (Table 10.2).
Step 3: Note the findings.
Normal findings might be documented as: “No lymph nodes palpable in the groin, no swelling, asymmetry, discoloration or increased temperature over lymph node locations, and no pain reported.”
Abnormal findings might be documented as: “Palpable, soft right-sided inguinal lymph node. Tenderness reported by the client.”
Priorities of Care
Lymph nodes with abnormal characteristics require further investigation and should be reported to the physician or nurse practitioner. Depending on the client, you may ask about whether they have been recently unwell or have noted any other concerns in the affected area (pain, itching, rashes). You might also ask whether they shave or have had a recent cat scratch or bite. Lymph node swelling can be associated with certain infections such as skin and yeast infections; sexually transmitted infections; and eye, ear, and throat infections. In rare situations, it can also be associated with cancer, particularly when accompanied by other symptoms such as fatigue, unexplained weight loss, persistent fever, fainting, or breathing issues, and if the lymph nodes have continued to enlarge with no obvious signs of infection.
Contextualizing Inclusivity
The lymphatic system undergoes changes as a person ages; the lymphatic vessels become thinner and the lymph nodes atrophy. As a result, older people may have more difficulty fighting infection and are more prone to lymphedema.
Always use a trauma-informed approach, particularly when assessing the inguinal lymph nodes due to their location. Explain what you are doing and why, ask permission to touch, provide a drape, and ask if the client would like someone present (a friend, family member, or another health care provider).
Activity: Check Your Understanding
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Nursing Physical Assessment Copyright © 2024 by Barbara Gawron and Meenu James is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.
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TABLE OF CONTENTS
Definition
How to find area
Area formula
Squares and rectangles
Parallelogram, Triangles, and Pentagons
Circle
Ellipse
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AREA
What is Area in Math? — Definition & Formula
Written by
Malcolm McKinsey
January 11, 2023
Fact-checked by
Paul Mazzola
Definition
How to find area
Area formula for various shapes
Area definition in math
In geometry, area is the amount of space a flat shape, like a polygon, circle or ellipse, takes up on a plane. The area of a shape is always measured in square units.
Once you know how square units relate to area, you can find the area of just about any two-dimensional shape.
Get free estimates from geometry tutors near you.
Search
How to find the area of a shape
Flat shapes have two dimensions:
Width
Length
A square, for instance, has a width equal to its length because all side lengths are the same. An ellipse has width and length, too.
We can easily see how the square could be divided up into small, square units like on a coordinate plane. You cannot easily see how an ellipse could be made up of little squares, but it can be.
Since it has width and length, it covers a space, and that space, even with the curving sides of the ellipse, can be divided up into square units:
Counting square area of square and rectangle
Counting the square units in the square is easy:one, two, three, etc..
But, how can you count all the square units in the ellipse? How do you decide what part of a square is under the top curve? What about the curves at the left and right ends?
Fortunately, mathematics has a fast way to add up all the square units without actually counting them.
Square units are the measurement unit for area because plane figures or flat shapes can always be divided into squares of known dimensions, like these:
m m 2{mm}^{2}mm 2
c m 2{cm}^{2}c m 2
f t 2{ft}^{2}f t 2
y d 2{yd}^{2}y d 2
k m 2{km}^{2}km 2
m i 2{mi}^{2}mi 2
Whether you are finding the of area of a quadrilateral like a trapezoid and a rhombus, or any other closed figure, the area will always be squared.
Area formula
The area formula you use depends on which shape you are trying to find the area for.
Area of squares and rectangles
To find the area of simple shapes like a square or the area of a rectangle, you only need its width,w, and length,l(or base,b). The area is length times width:
A =lxw
The area is always squared. You will always express area as square units, derived from the linear units.
Here is a rectangle90 meterswide and120 meterslong (the largest size of a FIFA soccer field). What is its area of this rectangle?
Area of a rectangle example
A=l×w A=l\times w A=l×w
A=120 m×90 m A=120m\times 90m A=120 m×90 m
A=10,800 m 2 A=\mathrm{10,800}{m}^{2}A=10,800 m 2
Because the soccer field is measured in linear meters, its area is square meters. The area of the rectangle is10,800 meterssquared.
The area of a square formula is actually even easier than writing outlength×widthbecause all sides are equal:
A=s 2 A={s}^{2}A=s 2
Here is a square with sides15 inches long, the same size as the bases on an MLB baseball field. Calculating area for this square looks like this:
Area of a square example
A=s 2 A={s}^{2}A=s 2
A=15 2 A={15}^{2}A=15 2
A=225 i n 2 A=225{in}^{2}A=225 in 2
Area of other shapes
All the other polygons do not easily divide into square units. Take a look at aparallelogram.
The two sides cut right across many square units. Of course, a parallelogram is just a knocked-over rectangle.
So, mathematically, if we could cut off one end and attach it to the other, we would have the area in square units. We can do exactly that, since the area of a parallelogram with a base,b, and width or height,h, is found using this formula:
A=b×h A=b\times h A=b×h
That is the same formula as for a square or rectangle!
If you divide a parallelogram along a diagonal, what do you have? Two triangles. That means the area of any triangle is half the area of a parallelogram with the same base length and height. A parallelogram, remember, uses the same formula as a rectangle.
Finding area of a parallelogram
Thearea of a triangleis, then, half the base,b, multiplied times the height,h:
A=1 2 b h A=\frac{1}{2}bh A=2 1bh
Here is a right triangle, a sail from a 45-foot Morgan sailboat with a base 20 1 4 20\frac{1}{4}20 4 1 feet and a height 44 1 2 44\frac{1}{2}44 2 1 feet. What is its area?
Area of a triangle example
For convenience in multiplying, you can change the fractions to decimals:
A=1 2 b h A=\frac{1}{2}bh A=2 1bh
A=1 2(20.25 f t×44.5 f t)A=\frac{1}{2}(20.25ft\times 44.5ft)A=2 1(20.25 f t×44.5 f t)
A=1 2(901.125 f t 2)A=\frac{1}{2}(901.125f{t}^{2})A=2 1(901.125 f t 2)
A=450.5625 f t 2 A=450.5625{ft}^{2}A=450.5625 f t 2
The area of the triangle sail is approximately450.6 square feet.
How about the home plate of an MLB baseball field? We can calculate the area of the home-platepentagonby considering it as two shapes:
A rectangle 17 inches×8.5 inches
An isosceles right triangle with legs 12 inches
First, we'll use the formula to find the area of the rectangle, which comes out to 144.5 i n 2 144.5{in}^{2}144.5 in 2.
Next, we'll use the formula to find the area of the triangle, which comes out to 72 i n 2 72{in}^{2}72 in 2.
Then, we add these two areas to find the total area, which 216.5 i n 2 216.5{in}^{2}216.5 in 2
Area of a pentagon example - home plate
Find the area of a circle
Some two-dimensional shapes are not even polygons, like our ellipse, or a circle. Thearea of a circlewith radius(r)is found using this formula:
A=π r 2 A=\pi {r}^{2}A=π r 2
If you have a circle with a radius of 4 cm, you can calculate the area of the circle easily with the formula above:
A=π(4)2 A=\pi {\left(4\right)}^{2}A=π(4)2
A=π(16)A=\pi (16)A=π(16)
A=3.14(16)A=3.14(16)A=3.14(16)
A≈50.24 A\approx 50.24 A≈50.24
The area of the circle is approximately50.24 square centimeters.
Find the area of an ellipse
Anellipse's areais found using its two axes, the major axis (length from the center) usually designated as aa, and the minor axis (width from the center), usually designated as bb, with this formula:
A=π a b A=\pi ab A=πab
Area of a circle and ellipse
Whether you are dealing with a regular polygon or an irregular plane figure, you can find the area!
Fact-checked and reviewed for accuracy.We adhere to stricteditorial integrity.
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Table of contents
Definition
How to find area
Area formula
Squares and rectangles
Parallelogram, Triangles, and Pentagons
Circle
Ellipse
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Supplements and Featured Publications
Understanding the Mechanisms to Maintain Glucose Homeostasis: A Review for Managed Care [CPE]Volume18 Issue 1 Suppl
Understanding the Kidneys' Role in Blood Glucose Regulation
Author(s):
Curtis L. Triplitt, PharmD, CDE
D
While not traditionally discussed, the kidneys’ contributions to maintaining glucose homeostasis are significant and include such functions as release of glucose into the circulation via gluconeogenesis, uptake of glucose from the circulation to satisfy their energy needs, and reabsorption of glucose at the level of the proximal tubule. Renal release of glucose into the circulation is the result of glycogenolysis and gluconeogenesis, respectively involving the breaking down and formation of glucose-6-phosphate from precursors (eg, lactate, glycerol, amino acids). With regard to renal reabsorption of glucose, the kidneys normally retrieve as much glucose as possible, rendering the urine virtually glucose free. The glomeruli filter from plasma approximately 180 grams of -glucose per day, all of which is reabsorbed through glucose transporter proteins that are present in cell membranes within the proximal tubules. If the capacity of these transporters is exceeded, glucose appears in the urine. The process of renal glucose reabsorption is mediated by active (sodium-coupled glucose cotransporters) and passive (glucose transporters) transporters. In hyperglycemia, the kidneys may play an exacerbating role by reabsorbing excess glucose, ultimately contributing to chronic hyperglycemia, which in turn contributes to chronic glycemic burden and the risk of microvascular consequences. This article provides an extensive review of the kidneys’ role in normal human physiology, the mechanisms by which they contribute to glucose regulation, and the potential impact of glucose imbalance on the kidneys.
(Am J Manag Care. 2012;18:S11-S16)Published studies over the last 60 years have provided considerable evidence regarding the ability of the kidneys to make and release glucose under various physiologic conditions. Yet traditionally, the kidneys have not been considered an important source of glucose (except during acidosis or after prolonged fasting), with most clinical discussions on glucose dysregulation centering on the intestine, pancreas, liver, adipose tissue, and muscle.1-3 More recently, however, the full significance of the kidneys’ contribution to glucose homeostasis, under both physiologic and pathologic conditions, has become well recognized, and is thought to involve functions well beyond glucose uptake and release. Besides the liver, the kidney is the only organ capable of generating sufficient glucose (gluconeogenesis) to release into the circulation, and it is also responsible for filtration and subsequent reabsorption or excretion of glucose.2-4 These findings have provided considerable insight into the myriad of pathophysiologic mechanisms involved in the development of hyperglycemia and type 2 diabetes mellitus (T2DM).5,6 This article provides a review of the kidneys’ role in normal human physiology, the mechanisms by which they contribute to glucose regulation, and the potential impact of glucose imbalance on the kidneys.
Overview of Renal Physiology
The kidneys are essentially designed to filter large quantities of plasma, reabsorb substances that the body must conserve, and secrete substances that must be eliminated. These basic functions are critical to regulation of fluid and electrolyte balance, body fluid osmolality, acid-based balance, excretion of metabolic waste and foreign chemicals, arterial pressure, hormone secretion, and, most relevant to this discussion, glucose balance.7,8 The 2 kidneys produce a total of approximately 120 mL/min of ultrafiltrate, yet only 1 mL/min of urine is produced. The basic urine-forming unit of the kidney is the nephron, which serves to filter water and small solutes from plasma and reabsorb electrolytes, amino acids, glucose, and protein. The nephron, of which there are approximately 1 million in each kidney, consists of a filtering apparatus (the glomerulus) that is connected to a long tubular portion that reabsorbs and conditions the glomerular ultrafiltrate. Fluid filtered from the glomerular capillaries flows into the tubular portion, which is made up of a proximal tubule, the Loop of Henle, and the distal tubule, all of which assist in reabsorbing essential substances and converting filtered fluid into urine.7
Evaluation of renal function is an important part of care, and with that, creatinine clearance (CrCl) or glomerular filtration rate (GFR), most frequently estimated (eGFR), are considered most useful in determining the degree of renal insufficiency and the stage of chronic kidney disease in accordance with the National Kidney Foundation classification system. Since alterations in all renal functions (ie, filtration, secretion, reabsorption, endocrine and metabolic function) have been associated primarily with GFR, this quantitative index may be used to measure any functional changes that result from kidney-related disease progression, therapeutic intervention, or toxic insult.9
Mechanisms of Glucose Homeostasis in the Kidneys
As described in greater detail in the first article in this supplement, maintenance of glucose homeostasis is crucial in preventing pathological consequences that may result from hyperglycemia or hypoglycemia. Chronically uncontrolled hyperglycemia leads to a higher risk of macrovascular and microvascular complications, such as cardiovascular disease, nephropathy, neuropathy, and retinopathy.10 Hypoglycemia, on the other hand, may lead to a myriad of central nervous system complications (eg, confusion, behavioral changes, seizures, loss of consciousness, and even death), since the brain is the body’s largest consumer of glucose in the fasting or “postabsorptive” state.10,11 Maintenance of glucose homeostasis involves several complementary physiologic processes, including glucose absorption (in the gastrointestinal tract), glycogenolysis (in the liver), glucose reabsorption (in the kidneys), gluconeogenesis (in the liver and kidneys), and glucose excretion (in the kidneys).10,12
As alluded to previously, the kidneys are capable of synthesizing and secreting many important hormones (eg, renin, prostaglandins, kinins, erythropoietin) and are involved in a wide variety of metabolic processes such as activation of vitamin D3, gluconeogenesis, and metabolism of numerous endogenous compounds (eg, insulin, steroids).9 With respect to renal involvement in glucose homeostasis, the primary mechanisms include release of glucose into the circulation via gluconeogenesis, uptake of glucose from the circulation to satisfy the kidneys’ energy needs, and reabsorption of glucose at the level of the proximal tubule.13
Glycogenolysis and Gluconeogenesis
Renal release of glucose into the circulation is the result of glycogenolysis and gluconeogenesis. Glycogenolysis involves the breakdown of glycogen to glucose-6-phosphate from precursors (eg, lactate, glycerol, amino acids) and its subsequent hydrolysis (via glucose-6-phosphatase) to free glucose. Conversely, gluconeogenesis involves formation of glucose-6-phosphate from those same precursors and subsequent conversion to free glucose. Interestingly, the liver and skeletal muscles contain most of the body’s glycogen stores, but only the liver contains glucose-6-phosphatase. As such, the breakdown of hepatic glycogen leads to release of glucose, whereas the breakdown of muscle glycogen leads to release of lactate. Lactate (generated via glycolysis of glucose by blood cells, the renal medulla, and other tissues) may be absorbed by organs and reformed into glucose.2
With regard to glucose utilization, the kidney may be perceived as 2 separate organs, with glucose utilization occurring predominantly in the renal medulla and glucose release limited to the renal cortex. These activities are separated as a result of differences in the distribution of various enzymes along the nephron. To this point, cells in the renal medulla (which, like the brain, are obligate users of glucose) have significant glucose-phosphorylating and glycolytic enzyme activity, and can therefore phosphorylate and accumulate glycogen. However, since these cells lack glucose-6-phosphatase and other gluconeogenic enzymes, they cannot release free glucose into the circulation. On the other hand, renal cortex cells do possess gluconeogenic enzymes (including glucose-6-phosphatase), and therefore can make and release glucose into the circulation. But because these cells have little phosphorylating capacity, they cannot synthesize glycogen.2
The magnitude of renal glucose release in humans is somewhat unclear, with inconclusive evidence regarding the contribution of the kidneys to total body gluconeogenesis.4 One analysis of 10 published studies concluded that the renal contribution to total body glucose release in the postabsorptive state is approximately 20%. Based on the assumption that gluconeogenesis accounts for approximately half of all circulatory glucose release during the fasting state, renal gluconeogenesis is projected, although not conclusively proven, to potentially be responsible for approximately 40% of all gluconeogenesis.2 Taking into consideration the potential contribution of renal gluconeogenesis, the kidneys appear to play a substantial role in overall glucose release in normal as well as pathophysiologic states (eg, hepatic insufficiency, counterregulation of hypoglycemia). To this point, evidence suggests that in patients with T2DM, renal glucose release is increased in both the postprandial and postabsorptive states, implicating the kidneys’ contribution to the hyperglycemia that characterizes this condition.4 In one study, a 3-fold increase in renal glucose release was observed in patients with diabetes versus those without.14 In contrast, hepatic glucose release increased by only 30% in the diabetic state. Potential mechanisms involved in excessive renal glucose release in T2DM include fasting gluconeogenesis, decreased postprandial insulin release, insulin resistance (known to suppress renal/hepatic insulin release), increased free fatty acid (FFA) concentrations (FFAs stimulate gluconeogenesis), greater availability of gluconeogenic precursors, and increased glycogenolysis.3 Again, it is clear that there is a renal contribution to glucose output in the body, but the actual contribution in individual patients with T2DM is still controversial.
Glucose Reabsorption
D
In addition to their important role in gluconeogenesis, the kidneys contribute to glucose homeostasis by filtering and reabsorbing glucose. Under normal conditions, the kidneys retrieve as much glucose as possible, rendering the urine virtually glucose free. The glomeruli filter from plasma approximately 180 grams of -glucose per day, all of which is reabsorbed through glucose transporter proteins that are present in cell membranes within the proximal tubules.4 If the capacity of these transporters is exceeded, glucose appears in the urine. This maximum capacity, known as the tubular maximum for glucose (TmG), ranges from 260 to 350 mg/min/1.73 m2 in healthy adults and children, and corresponds to a plasma glucose level of approximately 200 mg/dL.4 Once the TmG (the threshold) is reached and transporters are unable to reabsorb all the glucose (as in T2DM), glucosuria ocurrs.7,15 The correlation between the degree of hyperglycemia and degree of glucosuria becomes linear when blood glucose concentrations have increased beyond a threshold.4 It should be noted that slight differences between individual nephrons and the imprecise nature of biological systems may alter this linear concentration/reabsorption curve, as indicated by a splay from the theoretical as the TmG is approached.4 As such, glucosuria may potentially develop before the expected TmG is reached. Glucosuria may also occur at lower plasma glucose concentrations in certain conditions of hyperfiltration (eg, pregnancy), but as a consequence of hyperfiltration rather than significant hyperglycemia.12
Renal Glucose Transporters
The transport of glucose (a polar compound with positive and negative charged areas, making it soluble in water) into and across cells is dependent on specialized carrier proteins in 2 gene families: the facilitated glucose transporters (GLUTs) and the sodium-coupled glucose cotransporters (SGLTs). These transporters control glucose transport and reabsorption in several tissue types, including the proximal renal tubule, small intestine, blood-brain barrier, and peripheral tissues (Table).13,16 GLUTs are involved in the passive transport of glucose across cell membranes, facilitating its downhill movement as it equilibrates across a membrane. SGLTs, on the other hand, mediate active transport of glucose against a concentration gradient by means of cotransport with sodium. Of the various SGLT proteins expressed in the kidneys, SGLT2 is considered most important; based on animal studies, it is responsible for reabsorbing 90% of the glucose filtered at the glomerulus.4 SGLT1 contributes to the other 10% of glucose reabsorbed in the proximal tubule. This predominant role of SGLT2 in renal reabsorption of glucose raises the prospect of therapeutically blocking this protein in patients with diabetes. Of the various GLUT proteins expressed in the kidneys, GLUT2 is the major transporter, releasing into circulation the glucose reabsorbed by SGLTs in the proximal tubular cells (Figure).4,17
In examining disorders involving renal glucose transport, gene mutations within SGLTs lead to inherited disorders of renal glucosuria, including familial (primary) renal glucosuria (FRG) and glucose-galactose malabsorption (GGM). FRG, an autosomal recessive or autosomal dominant disorder resulting from several different SGLT2 mutations, is characterized by persistent glucosuria in the absence of hyperglycemia or general renal tubular dysfunction. Because the majority of patients with FRG have no clinical manifestations, FRG is commonly described as a “nondisease” and is synonymous with the condition known as benign glucosuria. Even the most severe form of FRG (type O), where nonfunctioning mutations within the SGLT2 gene result in a complete absence of renal tubular glucose reabsorption, is associated with a favorable prognosis. Because FRG is generally asymptomatic, affected individuals are identified through routine urinalysis.4
GGM, a more serious autosomal recessive disease caused by mutation of the SGLT1 transporter, is characterized by intestinal symptoms that manifest within the first few days of life and result from failure to absorb glucose and galactose from the intestinal tract. The resultant severe diarrhea and dehydration may be fatal if a glucose- and galactose-free diet is not initiated. In some patients with GGM, glucosuria is present but typically mild, while in others, no evidence of abnormal urinary glucose excretion exists, affirming the minor role of SGLT1 in renal glucose reabsorption of glucose.4
Gene mutations involving GLUTs are associated with more severe consequences, as these transporters are more widespread throughout the major organ systems. Compared with SGLT2 and SGLT1, which are present mostly in the renal system, GLUT2 is a widely distributed facilitative glucose transporter that has a key role in glucose homeostasis through its involvement in intestinal glucose uptake, renal reabsorption of glucose, glucosensing in the pancreas, and hepatic uptake and release of glucose.4 Mutations of the gene encoding this protein result in Fanconi-Bickel syndrome, a rare autosomal recessive glycogen storage disease that encompasses a multitude of complications (glucose and galactose intolerance, postprandial hyperglycemia, fasting hypoglycemia, tubular nephropathy, hepatomegaly, renomegaly, rickets, and stunted growth). Because GLUT2 is involved in the tubular reabsorption of glucose, glucosuria is a feature of the nephropathy.4
Impact of Hyperglycemia on the Kidneys
While renal glucose reabsorption is a glucose-conserving mechanism in normal physiologic states, it is known to contribute to hyperglycemia in conditions such as T2DM.15 Renal glucose reabsorption tends to increase with plasma glucose levels, up to plasma concentrations of 180 mg/dL to 200 mg/dL.7 Among patients with diabetes, an excess of approximately 13 grams of glucose is taken up from the systemic circulation, of which 85% is attributed to increased renal glucose uptake.3 Evidence suggesting a higher TmG in patients with diabetes compared with healthy controls attests to the increased state of renal glucose reabsorption seen in chronic hyperglycemia, which in turn can increase the risk of microvascular complications.13,18 Over time, the glomeruli become damaged and are unable to filter blood efficiently and glomerular membranes leak protein (more than 50% of the protein is albumin) into the urine.19 In patients with diabetes, the kidneys may be particularly susceptible to the effects of hyperglycemia, as many kidney cells are unable to sufficiently decrease glucose transport rates to prevent intracellular hyperglycemia in states of increased glucose concentration.19
Diabetic Nephropathy
Diabetes has become the most common single cause of endstage renal disease (ESRD) in the United States and Europe; this is most likely due to several evolving factors, including an increased prevalence of T2DM, longer life spans among patients with diabetes, and better formal recognition of renal insufficiency.20 Based on the most current (2008) US statistics from the American Diabetes Association, diabetes accounted for more than 40% of new cases of kidney failure, with 48,374 patients with diabetes beginning treatment for ESRD, and 202,290 people with diabetes-related ESRD on chronic dialysis or undergoing a kidney transplant.20 Compared with patients with type 1 diabetes mellitus, a considerably smaller fraction of those with T2DM progress to ESRD, but due to the much higher prevalence of T2DM, these individuals constitute over half of those with diabetes on dialysis. Considerable racial/ethnic variability exists in this regard, with Native Americans, Hispanics (especially Mexican Americans), and African Americans at much greater risk of developing ESRD than non-Hispanic whites with T2DM.20
Dialysis is a very expensive therapy, costing more than $50,000 per patient per year. Total medical spending for the approximately 400,000 patients with ESRD (representing those with and without diabetes) was $22.8 billion in 2001, an almost 3-fold increase over the 1991 to 2001 decade. ESRD spending represents 6.4% of the total Medicare budget, a 33% increase from 4.8% in 1991. The epidemic growth in ESRD cases has led to skyrocketing utilization of healthcare resources.21
The earliest clinical evidence of nephropathy is the appearance of low, but abnormal, levels (≥30 mg/day or 20 μg/min) of albumin in the urine (referred to as microalbuminuria).20 Although the course for each patient with T2DM is different, once albumin is detected in the urine, the chance of progression to more persistent albuminuria, progressive decline in GFR, raised arterial blood pressure, and increased cardiovascular morbidity and mortality is increased. Since undetected T2DM may be present for many years, a higher proportion of individuals with T2DM (vs type 1 diabetes mellitus) have microalbuminuria and overt nephropathy shortly after diagnosis. Without specific interventions, 20% to 40% of patients with T2DM and microalbuminuria progress to overt nephropathy; however, within 20 years of onset of overt nephropathy, only 20% will have progressed to ESRD.20 This may be attributable to the greater risk of dying from associated coronary artery disease than progressing to ESRD among the older diabetic population. As interventions for coronary artery disease continue to improve, however, more patients with T2DM may survive long enough to develop renal failure.20
Increasing evidence demonstrates that the onset and course of diabetic nephropathy may be significantly altered by several interventions (eg, tight glucose control, use of angiotensin-converting enzyme inhibitors and angiotensin receptor blockers), all of which have their greatest impact if instituted early. As such, annual screening for microalbuminuria is critical since it leads to early identification of nephropathy. Well-known data from the Diabetes Control and Complications Trial and the United Kingdom Prospective Diabetes Study established that intensive glycemic control may significantly reduce the risk of developing microalbuminuria and overt nephropathy.20 Recent research (eg, the Action in Diabetes and Vascular Disease: Preterax and Diamicron MR Controlled Evaluation [ADVANCE] trial) offers more perspective on the effects of tight glucose control and reduction of nephropathy.22 ADVANCE evaluated progression to major macrovascular events (death from cardiovascular causes, nonfatal myocardial infarction, or nonfatal stroke) and major microvascular events (new or worsening nephropathy or retinopathy) in 11,140 patients with T2DM randomly assigned to undergo standard or intensive glucose control (glycated hemoglobin level ≤6.5%). After a median of 5 years, intensive glucose control produced a 10% relative reduction in the combined outcome of major macrovascular and microvascular events, primarily as a result of a 21% relative reduction in the risk of developing new or worsening nephropathy. The intensive glucose control group was also associated with a 9% reduction in new onset microalbuminuria, but a higher incidence of severe hypoglycemia (2.7% vs 1.5% in the standard control group). The observed reduction in nephropathy is important, since indices of renal impairment are strongly associated with future risk of major vascular events, ESRD, and death in patients with diabetes.
Conclusion
The regulation of glucose production, uptake, reabsorption, and elimination is handled by several organs, most notably (historically) the pancreas and liver. While not traditionally discussed, the kidneys’ contributions to maintaining glucose homeostasis are multifaceted and include such functions as gluconeogenesis and glucose reabsorption, the latter being mediated by active (SGLT) and passive (GLUT) transporters. Under normal circumstances, glucose filtered by glomeruli is completely reabsorbed, but in conditions of hyperglycemia or reduced resorptive capacity, glucosuria may occur. In hyperglycemia, the kidneys may play an exacerbating role by reabsorbing excess glucose, ultimately contributing to chronic hyperglycemia, and subsequently to pancreatic β-cell failure, insulin resistance, and decreased glucose uptake. Hyperglycemia in turn detrimentally affects the kidneys by damaging glomeruli, ultimately causing microalbuminuria and nephropathy. Knowledge of the kidneys’ role in glucose homeostasis and the effect of glucose dysregulation on the kidneys is critical to optimal management of T2DM and prevention of associated renal complications.Author affiliations: Department of Medicine, Division of Diabetes, University of Texas Health Science Center at San Antonio; and Texas Diabetes Institute, San Antonio, TX.
Funding source: This activity is supported by an educational grant from Bristol-Myers Squibb and AstraZeneca LP.
Author disclosure: Dr Triplitt reports being a consultant or a member of the advisory board for Roche and Takeda Pharmaceuticals. He also reports being a member of the speakers’ bureau for Amylin, Eli Lilly, and Pfizer.
Authorship information: Concept and design; drafting of the manuscript; and critical revision of the manuscript for important intellectual content.
Address correspondence to: E-mail: Curtis.Triplitt@uhs-sa.com.
Meyer C, Dostou JM, Welle SL, Gerich JE. Role of human liver, kidney, and skeletal muscle in postprandial glucose homeostasis. Am J Physiol Endocrinol Metab. 2002;282(2):E419-E427.
Gerich JE, Meyer C, Woerle HJ, Stumvoll M. Renal gluconeogenesis: its importance in human glucose homeostasis. Diabetes Care. 2001;24(2):382-391.
Meyer C, Woerle HJ, Dostou JM, Welle SL, Gerich JE. Abnormal renal, hepatic, and muscle glucose metabolism following glucose ingestion in type 2 diabetes. Am J Physiol Endocrinol Metab. 2004;287(6):E1049-E1056.
Marsenic O. Glucose control by the kidney: an emerging target in diabetes. Am J Kidney Disease. 2009;53(5):875-883.
Shaefer CF. The ever-expanding universe. Physician’s Corner. 2008;3(4):204-207.
Defronzo R. Banting Lecture. From the triumvirate to the ominous octet: a new paradigm for the treatment of type 2 diabetes mellitus. Diabetes. 2009;58(4):773-795.
Guyton AC, Hall JE. Urine formation in the kidneys: I: glomerular filtration, renal blood flow, and their control. In: Textbook of Medical Physiology. 9th ed. Philadelphia, PA: W. B. Saunders Company; 1996:315-330.
Reilly RF, Jackson EK. Regulation of renal function and vascular volume. In: Goodman & Gilman’s The Pharmacological Basis of Therapeutics. 12th ed. New York, NY: McGraw-Hill; 2011:671-720.
DiPiro J, Talbert RL, Yee GC, et al, eds. Pharmacotherapy: A Pathophysiologic Approach. 6th ed. New York, NY: McGraw-Hill; 2002.
Gerich JE. Physiology of glucose homeostasis. Diabetes Obes Metab. 2000;2(6):345-350.
Cryer PE, Davis SN, Shamoon H. Hypoglycemia in diabetes. Diabetes Care. 2003;26(6):1902-1912.
Moe OW, Wright SH, Palacín M. Renal handling of organic solutes. In: Brenner BM, Rector FC, eds. Brenner & Rector’s The Kidney. Vol. 1. 8th ed. Philadelphia, PA: Saunders Elsevier; 2008:214-247.
Wright EM, Hirayama BA, Loo DF. Active sugar transport in health and disease. J Int Med. 2007;261(1):32-43.
Meyer C, Stumvoll M, Nadkarni J, Dostou J, Mitrakou A, Gerich J. Abnormal renal and hepatic glucose metabolism in type 2 diabetes mellitus. J Clin Invest. 1998;102(3):619-624.
Ganong WF. Renal function and micturition. In: Review of Medical Physiology. 21st ed. New York, NY: Lange Medical Publishing; 2003:702-732.
Farber SJ, Berger EY, Earle DP. Effect of diabetes and insulin on the maximum capacity of the renal tubules to reabsorb glucose. J Clin Invest. 1951;30(2):125-129.
Komoroski B, Vachharajani N, Boulton D, et al. Dapagliflozin, a novel SGLT2 inhibitor, induces dose-dependent glucosuria in healthy subjects. Clin Pharmacol Ther. 2009;85(5):520-526.
Mogensen CE. Maximum tubular reabsorption capacity for glucose and renal hemodynamics during rapid hypertonic glucose infusion in normal and diabetic subjects. Scand J Clin Lab Invest. 1971;28(1):101-109.
Forbes JM, Coughlan MT, Cooper ME. Oxidative stress as a major culprit in kidney disease in diabetes. Diabetes. 2008;57(6):1446-1454.
Molitch ME, DeFronzo RA, Franz MJ, et al; American Diabetes Association. Nephropathy in diabetes. Diabetes Care. 2004;27(1):S79-S83.
Rodby RA. Pharmacoeconomic challenges in the management of diabetic nephropathy. J Manag Care Pharm. 2004;10(5)(suppl A):S6-S11.
ADVANCE Collaborative Group; Patel A, MacMahon S, Chalmers J, et al. Intensive blood glucose control and vascular outcomes in patients with type 2 diabetes. N Engl J Med. 2008;358(24):2560-2572.
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13560 | https://en.wikipedia.org/wiki/Similarity_transformation | Jump to content
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From Wikipedia, the free encyclopedia
Similarity transformation may refer to:
Similarity (geometry), for shape-preserving transformations
Matrix similarity, for matrix transformations of the form A → P−1AP
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13561 | https://brainly.com/question/35255732 | [FREE] a. Determine the equation of the plane parallel to the xy-plane passing through the point (2, 6, 7). b. - brainly.com
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a. Determine the equation of the plane parallel to the x y-plane passing through the point (2,6,7).
b. Find an equation of the plane passing through the points (−10,−8,−1), (6,−8,−9), and (−7,−8,2).
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a. The equation of the plane parallel to the xy-plane passing through the point (2, 6, 7) is z = 7.
b. The equation of the plane passing through the points (-10, -8, -1), (6, -8, -9), and (-7, -8, 2) is 0(x + 10) + 48(y + 8) + 0(z + 1) = 0.
Explanation
Equation of a Plane Parallel to the xy-plane
To find the equation of a plane parallel to the xy-plane passing through the point (2, 6, 7), we can use the point-normal form of the equation. In this case, the normal vector will have a z-component of 1 since the plane is parallel to the xy-plane.
The equation of the plane is given by:
z = 7
Equation of a Plane Passing Through Three Points
To find the equation of a plane passing through the points (-10, -8, -1), (6, -8, -9), and (-7, -8, 2), we can use the cross product of two vectors method.
Step 1: Find two vectors in the plane using the given points.
Vector 1: (-10, -8, -1) - (6, -8, -9) = (-16, 0, 8)
Vector 2: (-10, -8, -1) - (-7, -8, 2) = (-3, 0, -3)
Step 2: Take the cross product of the two vectors to find the normal vector of the plane.
Normal Vector = (-16, 0, 8) x (-3, 0, -3) = (0, 48, 0)
Step 3: Use the point-normal form of the equation to find the equation of the plane.
The equation of the plane passing through the three points is:
0(x + 10) + 48(y + 8) + 0(z + 1) = 0
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The equation of the plane parallel to the xy-plane passing through the point (2, 6, 7) is z = 7. The equation of the plane passing through the points (-10, -8, -1), (6, -8, -9), and (-7, -8, 2) is y = -8.
Explanation
a. The equation of the plane parallel to the xy-plane passing through the point (2, 6, 7) can be determined by recognizing that planes parallel to the xy-plane have a constant z-value. Since the plane passes through the point (2, 6, 7), its z-coordinate is 7. Therefore, the equation of the plane is:
z=7
This means that for any point (x, y) on this plane, the z-coordinate will always be 7.
b. To find the equation of the plane passing through the points (−10,−8,−1), (6,−8,−9), and (−7,−8,2), we first need to find two vectors that lie on the plane. We can calculate these vectors using the given points:
Vector 1 from point 1 to point 2:
v 1=(6,−8,−9)−(−10,−8,−1)=(16,0,−8)
Vector 2 from point 1 to point 3:
v 2=(−7,−8,2)−(−10,−8,−1)=(3,0,3)
Next, we take the cross product of these two vectors to find the normal vector of the plane:
n=v 1×v 2=i^16 3j^0 0k^−8 3
Calculating the determinant gives:
n=(0⋅3−0⋅−8,−(16⋅3−−8⋅3),16⋅0−0⋅0)=(0,−48,0)
This normal vector tells us that the y-component is significant while the x and z-components are zero, indicating that the plane is horizontal. We write the equation of the plane using the normal vector and point-normal form:
0(x+10)+48(y+8)+0(z+1)=0
Simplifying this yields:
48(y+8)=0 or y=−8
Thus, the equation of the plane is:
y=−8
Examples & Evidence
For example, if you have a plane parallel to the xy-plane at z = 5, this plane will contain points like (1, 1, 5), (3, 4, 5), and (-2, -3, 5). Similarly, for the plane y = -4, points such as (0, -4, 0) and (2, -4, -1) will lie in this plane.
The method for determining equations of planes is based on foundational concepts in analytical geometry, which is well established in mathematical literature. The properties of planes parallel to the coordinate planes can be found in standard textbooks.
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Community Answer Determine the equation of the plane parallel to the yzyz-plane passing through the point (−7,−4,−6)(-7,-4,-6).
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13562 | https://www.colocationamerica.com/mbps-to-mb-converter | The standard measurement for bandwidth is in megabits per second, not megabytes.
This can cause some users confusion as they try to determine their download and upload speeds in relation to the file sizes they’re trying to access or send.
Use the converter below to calculate megabits per second (Mbps or Mb/s) to megabytes per second (MB/s) and vice versa.
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In short, transfer speeds. For most, an important factor in choosing a colocation or dedicated server provider is how quickly the client can access their data.
The faster the speed, the faster the user can upload their files and get to work more efficiently.
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When looking for a data center provider, you may notice a difference between how speeds are displayed on different products like dedicated servers vs colocation.
For example, the “speed” you see when looking at a provider’s dedicated server page is usually listed in TBs. This is the amount of data you’re allotted to download or upload typically per month.
Colocation, on the other hand, usually lists Mbps for speed and that number refers to the speed of the data being transferred and not your monthly allotment of bandwidth.
The difference is usually due to dedicated servers being rented servers and colocation being rented space (owned server).
Be sure to look at this distinction when you’re shopping for a hosting provider.
What’s the Difference Between Megabyte (MB) and Megabit (Mb)
1 megabyte (MB) is roughly equal to 8 megabits (mb).
Megabytes per second is not considered standard for network equipment but is much more common for labeling file sizes.
Additionally, you might be familiar with your Internet connection speeds at home which are described in megabits per second.
So when one sees a file downloading at 3 MB/s and they’re paying for what they think is 25 MB/s download speeds, they’ve simply confused the matter. What they’re really paying for is 25 mbps or megabits.
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13563 | https://www.effortlessmath.com/math-topics/how-to-find-unit-rates-and-rates/?srsltid=AfmBOori-fQJaPfcq5U80-BReQI1A-tStqFuWtjSXaAm5D1p4ypoa4I- | How to Find Unit Rates and Rates? - Effortless Math: We Help Students Learn to LOVE Mathematics
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How to Find Unit Rates and Rates?
When two quantities from different units are compared and expressed as a ratio, we refer to it as "rate". In this step-by-step guide, you get familiarized with the concept of unit rates and rates.
Rate is the ratio of two different quantities with different units, while the unit rate expresses the number of units of the first quantity for one unit of the second quantity. At the unit rate, the denominator is always one unit.
A step-by-step guide to unit rates and rates
In math, a rate is a ratio that compares two different quantities that have different units. For example, if we say Jim types 60 60 words in a minute, then his rate of typing is 60 60 words per minute. The word “per” indicates that we are dealing with a rate. The word “per” can be replaced with the “/” symbol in problems.
Unit rate
The unit rate is different from the rate at which a certain number of units of the first quantity are compared to one unit of the second quantity. In other words, we can say that the second quantity in the comparison is always 1 1. For example, there are 60 60 seconds in one minute. So as a unit rate, we can express it as 60 60 seconds per minute. Here, the word ‘per minute’ refers to one minute.
Ratio definition
A Ratio is used to compare two or more similar quantities or numbers with similar units. It is often written with a colon, and when used in words, we call the ratio of one quantity “to” the second quantity. For example, the ratio of girls and boys in a class is 4 4 to 3 3, or 4:3 4:3.Some problems may give you only two numbers, such as comparing two quantities, while some may have more than two quantities. For example, the ratio of different ingredients used in a recipe.
How is the rate calculated?
In general, we can write the formula for rate as the ratio between two quantities with various units. By putting this in the ratio format, we get,
𝑅 𝑎 𝑡 𝑒=𝑄 𝑢 𝑎 𝑛 𝑡 𝑖 𝑡 𝑦 1 𝑄 𝑢 𝑎 𝑛 𝑡 𝑖 𝑡 𝑦 2 R a t e=Q u a n t i t y 1 Q u a n t i t y 2
The steps to calculate the rate are given below:
Step 1: Write two quantities that are in different units.
Step 2: Find the ratio of quantity 1 1 to quantity 2 2.
Step 3: Express the ratio in the simplest way to find the rate.
Step 4:The answer will proceed with 𝑢 𝑛 𝑖 𝑡 1 𝑢 𝑛 𝑖 𝑡 2 u n i t 1 u n i t 2 as the unit for the value of the rate found.
To better understand this issue, we give an example. John cycled for 2 2 hours and traveled 24 24 miles. To calculate the speed at which he rode, let us use the rate formula, which is 𝑅 𝑎 𝑡 𝑒=𝑄 𝑢 𝑎 𝑛 𝑡 𝑖 𝑡 𝑦 1 𝑄 𝑢 𝑎 𝑛 𝑡 𝑖 𝑡 𝑦 2 R a t e=Q u a n t i t y 1 Q u a n t i t y 2. Given, quantity 1=24 1=24 miles, quantity 2=2 2=2 hours. Substituting the values in the formula, we get, 𝑅 𝑎 𝑡 𝑒=24 𝑚 𝑖 𝑙 𝑒 𝑠 2 ℎ 𝑜 𝑢 𝑟 𝑠 R a t e=24 m i l e s 2 h o u r s. Here, the speed is the rate.
So, 𝑟 𝑎 𝑡 𝑒=12 𝑚 𝑖 𝑙 𝑒 𝑠 ℎ 𝑜 𝑢 𝑟 r a t e=12 m i l e s h o u r or 12 12 miles per hour.
Unit rate is also a comparison between two quantities that have different units except that the quantity at the denominator is always one. Putting this in the ratio format, we get, 𝑈 𝑛 𝑖 𝑡 𝑅 𝑎 𝑡 𝑒=𝑄 𝑢 𝑎 𝑛 𝑡 𝑖 𝑡 𝑦 1 𝑂 𝑛 𝑒 𝑈 𝑛 𝑖 𝑡 𝑜 𝑓 𝑄 𝑢 𝑎 𝑛 𝑡 𝑖 𝑡 𝑦 2 U n i t R a t e=Q u a n t i t y 1 O n e U n i t o f Q u a n t i t y 2.
The rate “miles per minute” gives the distance traveled per unit of time. To calculate the unit rate, we divide the denominator with the numerator in such a way that the denominator becomes 1 1. For example, if 250 250 miles in 2 2 hours, the unit rate would be 250 𝑚 𝑖 𝑙 𝑒 𝑠 2 ℎ 𝑜 𝑢 𝑟 𝑠 250 m i l e s 2 h o u r s, which is equal to 125 𝑚 𝑖 𝑙 𝑒 𝑠 ℎ 𝑜 𝑢 𝑟 125 m i l e s h o u r. In other words, the denominator is always 1 1 at the unit rate.
Unit Rates and Rates – Example 1:
A printer prints 80 80 pages of an e-book in 40 40 seconds. Find the unit rate of the number of pages printed per second.
Solution:
The number of copies created in 40 40 seconds =80=80. To find the unit rate, divide the total number of pages printed by the total number of seconds. We get,
𝑈 𝑛 𝑖 𝑡 𝑟 𝑎 𝑡 𝑒=80 40 U n i t r a t e=80 40
=2=2
Therefore, the printer prints 2 2 pages per second or 2 2 𝑝 𝑎 𝑔 𝑒 𝑠 𝑠 𝑒 𝑐 𝑜 𝑛 𝑑 p a g e s s e c o n d.
Exercises for Unit Rates and Rates
Jane loves to cook and bakes wonderful cakes. He bakes 45 45 cakes in 9 9 hours. Can you figure out how many cakes he bakes per hour?
William and his family traveled 1,500 1,500 miles on vacation and consumed 60 60 gallons of gas. Can you help him calculate the average number of 𝑚 𝑖 𝑙 𝑒 𝑠 𝑔 𝑎 𝑙 𝑙 𝑜 𝑛 m i l e s g a l l o n?
1. 5 𝑐 𝑎 𝑘 𝑒 𝑠 𝑝 𝑒 𝑟 ℎ 𝑜 𝑢 𝑟 5 c a k e s p e r h o u r
2. 25 𝑚 𝑖 𝑙 𝑒 𝑠 𝑔 𝑎 𝑙 𝑙 𝑜 𝑛 25 m i l e s g a l l o n
by: Effortless Math Team about 3 years ago (category: Articles)
Effortless Math Team
3 weeks ago
Effortless Math Team
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13564 | https://simple.wikipedia.org/wiki/Relative_atomic_mass | Relative atomic mass - Simple English Wikipedia, the free encyclopedia
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A relative atomic mass (also called atomic weight; symbol: A r) is a measure of how heavy atoms are. It is the ratio of the average mass per atom of an element from a given sample to 1/12 the mass of a carbon-12 atom. In other words, a relative atomic mass tells you the number of times an average atom of an element from a given sample is heavier than one-twelfth of an atom of carbon-12. The word relative in relative atomic mass refers to this scaling relative to carbon-12. Relative atomic mass values are ratios;: 1 relative atomic mass is a dimensionless quantity. Relative atomic mass is the same as atomic weight, which is the older term.
The number of protons an atom has determines what element it is. However, most elements in nature consist of atoms with different numbers of neutrons.: 17 An atom of an element with a certain number of neutrons is called an isotope. For example, the element thallium has two common isotopes: thallium-203 and thallium-205. Both isotopes of thallium have 81 protons, but thallium-205 has 124 neutrons, 2 more than thallium-203, which has 122. Each isotope has its own mass, called its isotopic mass. A relative isotopic mass is the mass of an isotope relative to 1/12 the mass of a carbon-12 atom. The relative isotopic mass of an isotope is roughly the same as its mass number, which is the number of protons and neutrons in the nucleus. Like relative atomic mass values, relative isotopic mass values are ratios with no units.
We can find the relative atomic mass of a sample of an element by working out the abundance-weighted mean of the relative isotopic masses.: 17 For example, if a sample of thallium is made up of 30% thallium-203 and 70% thallium-205,
A r=(203×30)+(205×70)100=(6090)+(14350)100=20440 100=204.4{\displaystyle A_{r}={\frac {(203\times 30)+(205\times 70)}{100}}={\frac {(6090)+(14350)}{100}}={\frac {20440}{100}}=204.4}
Two samples of an element that consists of more than one isotope, collected from two widely spaced sources on Earth, are expected to have slightly different relative atomic masses. This is because the proportions of each isotope are slightly different at different locations.
A standard atomic weight is the mean value of relative atomic masses of a number of normal samples of the element. Standard atomic weight values are published at regular intervals by the Commission on Isotopic Abundances and Atomic Weights of the International Union of Pure and Applied Chemistry (IUPAC). The standard atomic weight for each element is on the periodic table.
Often, the term relative atomic mass is used to mean standard atomic weight. This is not quite correct, because relative atomic mass is a less specific term that refers to individual samples. Individual samples of an element could have a relative atomic mass different to the standard atomic weight for the element. For example, a sample from another planet could have a relative atomic mass very different to the standard Earth-based value.
Relative atomic mass is not the same as:
atomic mass (symbol: m a), which is the mass of a single atom, commonly expressed in unified atomic mass units
mass number (symbol: A), which is the sum of the number of protons and the number of neutrons in the nucleus of an atom
atomic number (symbol: Z), which is the number of protons in the nucleus of an atom.
References
[change | change source]
↑ Jump up to: 1.01.1"Atomic weight: The Name, its History, Definition, and Units". Commission on Isotopic Abundances and Atomic Weights of the International Union of Pure and Applied Chemistry. Archived from the original on 2013-12-15. Retrieved 2016-01-07.{{cite web}}: CS1 maint: bot: original URL status unknown (link)
↑Daintith, John, ed. (2008). A Dictionary of Chemistry (Sixth ed.). Oxford University Press. p.457. ISBN978-0-19-920463-2.
↑ Jump up to: 3.03.13.2Otter, Chris; Stephenson, Kay, eds. (2008). Salters Advanced Chemistry: Chemical Ideas (Third ed.). Heinemann. ISBN978-0-435631-49-9.
↑Moore, John T. (2010). Chemistry Essentials For Dummies. Wiley. p.44. ISBN978-0-470-61836-3.
Other websites
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"Standard Atomic Weights". Commission on Isotopic Abundances and Atomic Weights of the International Union of Pure and Applied Chemistry. Retrieved 2016-01-07.
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13565 | https://brainly.com/question/38714581 | [FREE] A boy weighs 200 lb on Earth's surface. a) What is his weight 500 miles above Earth's surface? - brainly.com
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A boy weighs 200 lb on Earth's surface.
a) What is his weight 500 miles above Earth's surface?
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The weight of a person decreases as they move away from the Earth's surface. This can be calculated by converting the weight to a mass in kg, calculating the new distance from the center of the Earth, and using the formula for weight in the field of an object of given mass at a given distance.
Explanation
The weight of a person or object decreases as the distance from the center of the Earth increases. If a boy weighs 200 lb on the Earth's surface, his weight would be different 500 miles above the Earth's surface due to the decrease in gravitational pull. The force of gravity decreases with the square of the distance from the center of the Earth (inverse square law). This means that if the Earth's radius is increased (by going 500 miles above the surface), the boy's weight would also decrease.
Let's calculate:
First, convert the 200 lb to kg, since weight in physics is usually dealt with in Newtons and the conversions require SI units. Using that 1 lb is roughly 0.4536 kg, we get a mass of about 90.72 kg. The weight on Earth is then mg = 90.72kg 9.81m/s² = 889.866 N.
The radius of Earth is about 6371 km, and 500 miles is approximately 804.672 km. So, the boy is 6371+805 = 7176km = 7.176 10^6 m from the center of the Earth when 500 miles above the surface. So, the weight 500 miles above the Earth's surface would be (G Mass of Earthmass of the boy)/(distance from the center of the Earth)^2. This would provide the weight in Newtons, which could again be converted to lbs if desired.
Note that this is an approximation; depending on the accuracy of the constants and the conversion factors used, your exact numerical result might vary slightly.
Learn more about Weight Variation with Altitude here:
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The boy's weight decreases as he moves 500 miles above Earth's surface due to the reduced gravitational pull. After calculating, his weight at that altitude would be approximately 400 lb. Understanding this requires taking into account the mass of the Earth and the distance from its center.
Explanation
To find the boy's weight 500 miles above the Earth's surface, we need to understand how weight changes with distance from the center of the Earth due to gravity. Weight is the force exerted by gravity on an object and decreases as you move farther from the Earth's center.
Let's go through the steps to calculate his new weight at that altitude:
Convert Weight to Mass:
The boy weighs 200 lb. First, we convert pounds to kilograms:
1 lb ≈ 0.4536 kg,
Mass = 200 lb × 0.4536 kg/lb ≈ 90.72 kg
Calculate the Radius of Earth plus Altitude:
The average radius of the Earth is about 6371 km. Since 500 miles is approximately 804.67 km, we add this to the radius of the Earth:
Total Distance from the center of the Earth = 6371 km + 804.67 km ≈ 7175.67 km
(which is 7175670 meters)
Calculate the Weight at that Altitude:
The weight (W) is calculated using the formula:
W = (G × M × m) / r²
where:
G is the gravitational constant ≈ 6.674 × 10⁻¹¹ N(m²/kg²)
M is the mass of Earth ≈ 5.972 × 10²⁴ kg
m is the mass of the boy (90.72 kg)
r is the distance from Earth’s center (7175670 m)
Plugging in the values:
W = (6.674 × 10⁻¹¹ N(m²/kg²) × 5.972 × 10²⁴ kg × 90.72 kg) / (7175670 m)²
W ≈ 1786.5 N
1. Convert Weight from Newtons to Pounds:
To convert from Newtons back to pounds, we use:
1 N ≈ 0.2248 lb,
Weight in pounds ≈ 1786.5 N × 0.2248 lb/N ≈ 400 lb
Thus, the boy's weight at 500 miles above Earth's surface would be approximately 400 lb.
Examples & Evidence
For example, an astronaut on the International Space Station, which is about 400 km above the Earth, experiences about 90% of their weight due to the gravity at that height, resulting in a feeling of weightlessness while still being pulled by Earth's gravity.
The calculations rely on the gravitational force formula and Earth's mass, which are established in physics; distance from the center of the Earth directly impacts weight as described by the inverse square law.
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Why is it a bad idea to use prime factorisation to find the GCD of really large numbers? What should we use instead?
Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968)
·
Upvoted by
Vance Faber
, Ph. D. Mathematics and
Daniel McLaury
, Ph.D. Student in Mathematics at University of Illinois at Chicago · Author has 16.4K answers and 29.1M answer views
·
6y
Why is it a bad idea to use prime factorisation to find the GCD of really large numbers? What should we use instead?
Finding prime divisors of large numbers is not easy—well, it’s easy enough, but it’s very time consuming. Potentially, you have to try dividing by all the primes up to the square root of the number. And you have to know which those primes are.
Instead you can use Euclid’s algorithm. Divide the greater of the two numbers by the smaller. Replace the greater by the remainder and repeat until the remainder is zero. The GCD is the last divisor.
The simplest way to see why this works is
Why is it a bad idea to use prime factorisation to find the GCD of really large numbers? What should we use instead?
Finding prime divisors of large numbers is not easy—well, it’s easy enough, but it’s very time consuming. Potentially, you have to try dividing by all the primes up to the square root of the number. And you have to know which those primes are.
Instead you can use Euclid’s algorithm. Divide the greater of the two numbers by the smaller. Replace the greater by the remainder and repeat until the remainder is zero. The GCD is the last divisor.
The simplest way to see why this works is that division is repeated subtraction, and the sum and difference of two numbers have exactly the same common divisors.
Example: the GCD of 17925 and 3456 is the GCD of 3456 and 645, which is also the GCD of 645 and 231, or 183and 231, or 231 and 48, or 48and 39, or 9 and 39, or 3 and 9, or 3 and 0. So the GCD is 3.
Even with these small numbers, I think this is quicker than trying to find the prime divisors. Well, it depends if there are no very small common prime divisors. The first number is 3×52×239 and the second is 27×33.
Jeff Suzuki
Mathematician and historian of mathematics.
·
Author has 4.4K answers and 9.9M answer views
·
6y
Factoring is the hardest easy problem in mathematics. It’s easy to explain what we want to do: write a number as a product of two smaller numbers.
It’s so hard to do in practice that it’s the basis for computer security.
How hard is it? Here are two numbers, both under a thousand: 851 and 899.
Now find their GCD. First, try to factor them. Even if I hand you a calculator and let you use it, I’m guessing it will take you several minutes on a calculator to find the factorizations of both numbers.
So how do we actually find the GCD of large numbers? (Note that “large” is “anything over 100”). The bas
Factoring is the hardest easy problem in mathematics. It’s easy to explain what we want to do: write a number as a product of two smaller numbers.
It’s so hard to do in practice that it’s the basis for computer security.
How hard is it? Here are two numbers, both under a thousand: 851 and 899.
Now find their GCD. First, try to factor them. Even if I hand you a calculator and let you use it, I’m guessing it will take you several minutes on a calculator to find the factorizations of both numbers.
So how do we actually find the GCD of large numbers? (Note that “large” is “anything over 100”). The basic principle is this:
The GCD of two numbers will also divide their sum or difference.
So the GCD of 851 and 899 will also divide 899±851. Now, 899+ 851 is a bigger number, and if I don’t want to try to factor 899 or 851 by hand because they’re too large, then I don’t really want to try to factor an even larger number. But notice that 899−851=48.
What can we do with that? First, the GCD of 851 and 899 must also divide 48. There are lots of things that divide 48…but most of them are even. In fact, the only odd number that divides 48 is 3…which does not divide 851 or 899.
That means the GCD of 899 and 851 is 1 (which always divides an integer): the two numbers are relatively prime.
This idea generalizes into what’s called the Euclidean algorithm:
Robby Goetschalckx
PhD in Computer Science & Artificial Intelligence, KU Leuven (Graduated 2009)
·
Upvoted by
Luís Sequeira
, PhD Mathematics, University of Lisbon (2001) and
Mpendulo Cele
, MSc Mathematics & Algebra, Stellenbosch University (2020) · Author has 6.4K answers and 9.2M answer views
·
6y
Originally Answered: Why is prime factorization not useful when finding the GCD of large numbers?
·
For small numbers that you can factorize on sight, it is. For example, if you ask me for the GCD of 240 and 180, I’ll see — on sight — that the answer is 60.
For big numbers — as in, tens or even hundreds of digits, or more, this is not feasible. Finding the prime factorization of large numbers is not very efficient. At least, it is not nearly as efficient as Euclid’s method. Using Euclid’s method, or even better: the version that uses modulo instead of the difference, is highly efficient.
The worst-case inputs for Euclid’s algorithm are two consecutive terms of the Fibonacci sequence, and even
For small numbers that you can factorize on sight, it is. For example, if you ask me for the GCD of 240 and 180, I’ll see — on sight — that the answer is 60.
For big numbers — as in, tens or even hundreds of digits, or more, this is not feasible. Finding the prime factorization of large numbers is not very efficient. At least, it is not nearly as efficient as Euclid’s method. Using Euclid’s method, or even better: the version that uses modulo instead of the difference, is highly efficient.
The worst-case inputs for Euclid’s algorithm are two consecutive terms of the Fibonacci sequence, and even for those, the number of steps needed is logarithmic in terms of the original numbers — whereas prime factorization, as far as we know, needs O(√n) time.
Related questions
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Assistant
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·
1y
Using prime factorization to find the greatest common divisor (GCD) of very large numbers can be impractical for several reasons:
Complexity of Factorization
: Prime factorization becomes computationally expensive as the numbers grow larger. The time required to factor large numbers increases significantly, making it infeasible for very large integers.
Storage Requirements
: Storing the prime factors of large numbers can require substantial memory, especially if the numbers have large prime factors.
Error-Prone
: The process of factorization can be error-prone, and mistakes in identifying prime fact
Using prime factorization to find the greatest common divisor (GCD) of very large numbers can be impractical for several reasons:
Complexity of Factorization: Prime factorization becomes computationally expensive as the numbers grow larger. The time required to factor large numbers increases significantly, making it infeasible for very large integers.
Storage Requirements: Storing the prime factors of large numbers can require substantial memory, especially if the numbers have large prime factors.
Error-Prone: The process of factorization can be error-prone, and mistakes in identifying prime factors can lead to incorrect results.
Alternative Method: Euclidean Algorithm
Instead of using prime factorization, the Euclidean algorithm is a much more efficient method to compute the GCD of two integers. The Euclidean algorithm is based on the principle that the GCD of two numbers also divides their difference. Here’s how it works:
Given two numbers a and b (where a>b):
Replace a with b and b with amodb (the remainder of a divided by b).
Repeat this process until b becomes 0. The non-zero remainder at this point will be the GCD of the original two numbers.
Example of the Euclidean Algorithm
To find the GCD of 48 and 18:
48mod18=12 (so replace 48 with 18 and 18 with 12)
18mod12=6 (replace 18 with 12 and 12 with 6)
12mod6=0 (replace 12 with 6 and 6 with 0)
The GCD is 6.
Benefits of the Euclidean Algorithm
Efficiency: The Euclidean algorithm runs in logarithmic time relative to the size of the smaller number, making it very efficient even for large integers.
Simplicity: It requires only basic arithmetic operations, which are straightforward to implement programmatically.
In summary, for very large numbers, it is better to use the Euclidean algorithm for finding the GCD rather than relying on prime factorization.
Narayan Dalai
Studied Mathematics & Science &Electrical Engg at Diploma in Electrical Engineering (IAF)
·
Author has 2.5K answers and 2.6M answer views
·
6y
When we want to find the GCD of Two Large Numbers we normally test where any of the Numbers is divisible by 2,3,5,7,11,13 .Here I will show you a short cut method to test the divisibility by 7,11,&13 . If it a 3 digit Number, try to divide it physically. It will be easily apparent whether the number is divisible by 7,11 &13. If the numbers are of appreciable length, you group 3 digit at a time from right (ie from units digit) and number it one. Do like that till end. The last group may not be 3 digits. Group it 2 or 1 digit as the case be. Sum odd number groups and even number groups separatel
When we want to find the GCD of Two Large Numbers we normally test where any of the Numbers is divisible by 2,3,5,7,11,13 .Here I will show you a short cut method to test the divisibility by 7,11,&13 . If it a 3 digit Number, try to divide it physically. It will be easily apparent whether the number is divisible by 7,11 &13. If the numbers are of appreciable length, you group 3 digit at a time from right (ie from units digit) and number it one. Do like that till end. The last group may not be 3 digits. Group it 2 or 1 digit as the case be. Sum odd number groups and even number groups separately. Find out if the difference is divisible by 7 or 11 or 13 to test the divisibility by the respective numbers.
If not divisible by 2,3,5,7,11,13 we will take recourse of Successive Divisions methods to arrive at the GCD. In the process of Successive Divisions, at one stage one number divides without any Remainder, & the very number is the required GCD. I suppose we are all acquainted with the process .
This may suffice for the purpose the question is asked for. □ANSWER.
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Anil Bapat
Have studied Mathematics up to pre-Degree Level
·
Author has 2.8K answers and 3.7M answer views
·
6y
Originally Answered: Why is prime factorization not useful when finding the GCD of large numbers?
·
Why is prime factorization not useful when finding the GCD of large numbers?
When we talk of large Numbers, we are talking about 500–600 digit numbers or even beyond that. Till now, it has been a challenge to factorize such big numbers and in view of that Prime factorization method is not realy useful when you want to find the GCD of such big numbers. Even the best computers and the best know algorithms might take substantially large amount of time as you try to find the factors and the whole exercise turns out to be futile as too much precious time gets lost i the process.
On the other hand, th
Why is prime factorization not useful when finding the GCD of large numbers?
When we talk of large Numbers, we are talking about 500–600 digit numbers or even beyond that. Till now, it has been a challenge to factorize such big numbers and in view of that Prime factorization method is not realy useful when you want to find the GCD of such big numbers. Even the best computers and the best know algorithms might take substantially large amount of time as you try to find the factors and the whole exercise turns out to be futile as too much precious time gets lost i the process.
On the other hand, this property (that such big numbrs are difficult to factorize) is being used to develope Encryption Algorithms that are difficult to crack, even after many many years.
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David Vanderschel
PhD in Mathematics & Physics, Rice (Houston neighborhood) (Graduated 1970)
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Author has 37.4K answers and 49.8M answer views
·
6y
Originally Answered: Why is prime factorization not useful when finding the GCD of large numbers?
·
Because, for large numbers, prime factorization can itself be very hard, no matter how you intend to use those factors. The Euclidean algorithm will yield a result in a straightforward way. It will be a factor of both, but not necessarily prime.
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Pooja Chauhan
Upvoted by
Jeff Nelson
, Invented Chromebook. Former Math PhD student. and
Kumar Utkarsh
, M.Sc Mathematics, Indian Institute of Technology, Madras (2019) · Author has 88 answers and 515.1K answer views
·
Updated 11y
Related
Why are prime numbers important in real life? What practical use are prime numbers?
I love this question. My views:
First there was zero. Then there was one. Philosophically, first there was nothing and then there was something. Lets assume this emergence of something from nothing happened. Even if it happened in infinitely small, potentially infinite number of steps, well, it did happen.
Now, what if the universe had 0 and 1. What all can it do? Lets not forget it also has the "knowledge" of how to go from 0 to 1. Not intelligent knowledge, nor deliberate knowledge, but existential knowledge. If I had to bet on the next actions of such a universe, I would say, it will eventu
I love this question. My views:
First there was zero. Then there was one. Philosophically, first there was nothing and then there was something. Lets assume this emergence of something from nothing happened. Even if it happened in infinitely small, potentially infinite number of steps, well, it did happen.
Now, what if the universe had 0 and 1. What all can it do? Lets not forget it also has the "knowledge" of how to go from 0 to 1. Not intelligent knowledge, nor deliberate knowledge, but existential knowledge. If I had to bet on the next actions of such a universe, I would say, it will eventually make 2.
Once the universe evolved addition, there was no stopping it. It made 3, 4, 5 and so on. But there was a caveat. To reach the idea of every number, a lot of steps had to be taken first. Eventually, those steps were internalized and 4 existed as naturally as 1+1+1+1. But that step was still a step, however fast and however internalized.
With infinite copies of infinite number of numbers existing for an infinite amount of time, it is highly probable that some would come together, (may be keep looking at each other waiting for the other to make eye contact first) and eventually evolve multiplication.
My O my!
Now addition looks such an old-old-way of doing things. You don't need a community of workers to get things done slowly. Two heroes could come together, take out their wands, chant multiplicus and "reach" a new number in one step.
Once the universe evolved multiplication, it discovered that its heartbeat increased, its breath became shallow, it sweated and found to its delight, that it enjoyed creation. Addition was "happening". It was passive. Multiplication was "doing". It was adrenaline.
Once the universe evolved multiplication, everything had to be a multiple. The design was so elegant, and so next generation, that everything had to be garbed in it. Everything had to be defined as a multiple.
Now, here is the thing: The elegant design had a flaw. There were dissenters. Some numbers, loners since the beginning, could only be reached by addition, the old fashioned way.
Although the speed of reaching those numbers had become so fast that it was immaterial how many 1s you needed to step on, it was still a point that, the only way to reach them was to wander through those 1s.
These loners, these old fashioned addition-reachable supremacist numbers are Primes. They are the real PRIMALS. They are old, Old.
They "know" how the world was created. From 0 and 1. They "remember".
(Fig: Primes Till 1000)
I am so glad that this thought of mine, however naive, is finally out in the world.
It feels (thinks) even better when you see the universe as human brain and every new number just floating around till something says Aha!
When we are born it takes some time to get used to the idea of "2". Notice babies and toddlers: They understand zero and one immediately. Even before they start to understand, what is called Object Permanence (It means they know that when Mom is hiding behind the blanket, she still exists), even before Object Permanence, they know nothing and something.
Then, they learn the concept of zero and many before they learn numbers, another manifestation of nothing and something. Then they learn the concept of zero, one, and many - most of the time on their own. Once we teach them, 2, 3 and so on, there is little work involved in teaching them more numbers.
By the time we are adults, numbers are so internalized that we do not notice a difference in thinking of 1 or 100. But there is that very very fast speed of thought to account for.
I would say that we "perceive" 0 and 1, and we "think" all other numbers. And Primes are the proof - we take an infinitely small bit longer thinking of 67 than thinking of 98.
I wish there was a way to check this hypothesis.
Mark Gritter
recreational mathematician
·
Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 5.7K answers and 11.7M answer views
·
9y
Related
What would you do if you detect a fast way for prime number factorisation of any given number?
Let's be clear: the difficulty of the factoring problem is the basis of of one public-key cryptographic technique, but others exist. We have backups implemented in software, right now, like elliptic curves.
What does public-key cryptography protect? It doesn't protect your bank account. Your password does that. In order to make use of a weakness in public-key cryptography to intercept your password, an attacker must be positioned to intercept traffic. This is not common. Feasible, yes, but not automatic.
(I'm going to assume you're not using public-key encryption for email since so few peo
Let's be clear: the difficulty of the factoring problem is the basis of of one public-key cryptographic technique, but others exist. We have backups implemented in software, right now, like elliptic curves.
What does public-key cryptography protect? It doesn't protect your bank account. Your password does that. In order to make use of a weakness in public-key cryptography to intercept your password, an attacker must be positioned to intercept traffic. This is not common. Feasible, yes, but not automatic.
(I'm going to assume you're not using public-key encryption for email since so few people do.)
Finally, has TLS-based security ever been broken? Yes, lots of times. Many of the attacks are due to downgrading encryption to a scheme which is feasible to break. Heck, many users click through warnings anyway.
So, the stakes are not very high. We're not talking about a collapse of the world banking system. Airplanes will not fall from the sky. I love Sneakers, but computer security doesn't work like that. We might see more password leakage until people adjust by deploying other protocols. (Guess what: we've seen massive password leaks too!)
Therefore, the best thing to do is to publish. Odds are, if a good factoring algorithm exists, then a three-letter-acronym agency has it or will soon. So you'd be doing us a favor. And if the weakness exists, then it's better to know today rather than continuing to depend on a system that will be broken tomorrow when the negative consequences probably grow with time rather than shrink.
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Steve Verrill
Studied at University of Southampton
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Alex Eustis
, Ph.D. Mathematics, University of California, San Diego (2013) and
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987)
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11y
Related
Why are prime numbers important in real life? What practical use are prime numbers?
Gas turbine and steam turbine shafts vibrate at low harmonics, such as 2,3,4,5 times the rotational frequency. As such, 60 blades, for example would be a very bad number, because the vibration from the blades would only enhance these low order vibrations. Much better to have 59 or 61 blades (both prime numbers) as this will give a blade-passing vibration at 59 or 61 times the rotational speed, and there will be no interactions with the low-order shaft vibrations.
Kip Fisher
Former Risk Analyst and Sometime Math Tutor
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Author has 682 answers and 480K answer views
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Updated Sat
Related
Why is it unnecessary to test divisibility by large numbers when checking if 1,009 is prime?
To establish that a given number is prime, it is sufficient to show that no smaller prime divides it.
It is almost universal practice to test divisibility by primes in increasing order, first testing 2, then 3, then 5, then 7, then 11, and so on.
Testing in increasing order saves labor. Let me illustrate why.
We can see at a glance that 2 does not go evenly into 1009.
Now ask yourself, is there any possibility that 3 goes into 1009 2 times evenly?
The answer is no, because we have already established that 2 does not go evenly into 1009.
If 3 goes evenly into 1009, it must go in at least 3 times.
I
To establish that a given number is prime, it is sufficient to show that no smaller prime divides it.
It is almost universal practice to test divisibility by primes in increasing order, first testing 2, then 3, then 5, then 7, then 11, and so on.
Testing in increasing order saves labor. Let me illustrate why.
We can see at a glance that 2 does not go evenly into 1009.
Now ask yourself, is there any possibility that 3 goes into 1009 2 times evenly?
The answer is no, because we have already established that 2 does not go evenly into 1009.
If 3 goes evenly into 1009, it must go in at least 3 times.
It turns out that 3 does not go evenly into 1009.
Next, ask yourself if there is any possibility that 5 goes into 1009 either 2 times or 3 times evenly.
The answer is no because neither 2 nor 3 goes evenly into 1009.
If 5 goes evenly into 1009, it must go in at least 5 times.
It turns out that 5 does not go evenly into 1009.
The pattern continues. If 7 goes evenly into 1009, it must go in at least 7 times.
But 7 does not go evenly into 1009.
If 11 goes evenly into 1009, it must go in at lease 11 times.
But 11 does not go evenly into 1009.
How long must we keep this up? We must keep it up until we encounter either a prime that goes evenly into 1009 or, importantly, a prime whose square is larger than 1009.
We end up trying 13, 17, 19, 23, 29, and 31. none of them goes evenly into 1009.
We do not have to try the next prime, which is 37, because 37^2 is more than 1009. 37^2=1369.
BEWARE! The stopping rule works only if we have meticulously established that no smaller prime divides 1009. That is why we (almost) always test primes in increasing order.
As long as we test primes in increasing order, we can stop when we first encounter a prime whose square is too large.
In the case of 1009, we can stop after testing all of the primes up to and including 31. 31^2 is less than 1009. But 37^2 is more than 1009.
Because no prime whose square is less than 1009 is a factor of 1009, we conclude that 1009 is prime. We do not need to test divisibility by any prime whose square is larger than 1009. And that is why we do not need to test divisibility by “large numbers.”
Alberto Cid
M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008)
·
Author has 2K answers and 3.8M answer views
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3y
Related
What is an efficient way to find the least common multiple of two extremely large numbers without prime factorization?
You don’t need prime factorization.
You can use GCD.
GCD can be computed without prime factorization.
GCD(a, b) = g
then g is the greatest integer such that
a = g·x
b = g·y
And, since g is the greatest, then x and y are coprimes, relative primes.
That is,
g includes the common primes,
x includes the extra prime factors needed for “a”
y includes the extra prime factors needed for “b”
So, LCM(a, b) = g·x·y = L
That’s a common multiple: L = y·a = x·b
And it’s the least… because no multiple of “b” can be also a multiple of “a” unless it includes the extra prime factors… so, you need b and als
You don’t need prime factorization.
You can use GCD.
GCD can be computed without prime factorization.
GCD(a, b) = g
then g is the greatest integer such that
a = g·x
b = g·y
And, since g is the greatest, then x and y are coprimes, relative primes.
That is,
g includes the common primes,
x includes the extra prime factors needed for “a”
y includes the extra prime factors needed for “b”
So, LCM(a, b) = g·x·y = L
That’s a common multiple: L = y·a = x·b
And it’s the least… because no multiple of “b” can be also a multiple of “a” unless it includes the extra prime factors… so, you need b and also x…
L = b·x = g·y·x = a·y = g·x·y
a·b = g·x · g·y = g^2 · x·y
So: a·b = g·L
Then :
LCM(a, b) = a·b / GCD(a, b)
To find the GCD you don’t need prime factorization, because Euclid’s Algorithm doesn’t need it. This algorithm is more than 2200 years old!!
Example:
I’ve seen you asked about LCM(p-1, q-1) where p and q are primes.
p = 1973
q = 4099
GCD(4098, 1972) =
= GCD(4098–1972, 1972) =
= GCD(2126, 1972) =
= GCD(1972, 154) =
= GCD(154, 1972–154·12) =
= GCD(154, 124) =
= GCD(124, 30) =
= GCD(30, 4) =
= GCD(4, 2)
= 2
Then: LCD(4098, 1972) = 4098·1972/2 = 4040628
I didn’t do the factorization of 1972 or 4098
But:
1972 = 2·2·17·29
4098 = 2·3·683
Senia Sheydvasser
PhD in Mathematics
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Author has 2.5K answers and 39.7M answer views
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Updated 7y
Related
How is it useful to have/use prime numbers and to apply prime factorization to composite numbers? What can you obtain?
As always, I’m glad I checked the comments, which included the following addition: “I’m not referring to how can you use them in cryptography… just what can you do directly in math and in abstract concepts.”
A dangerous sentiment to express to me, James. But I am happy to oblige.
Prime factorization is actually very helpful, as is decomposition in principle. So, before I get into the particular types of things that factorization helps with, let’s talk about decomposing things in general.
Mathematical objects can be very, very complicated. It can be difficult to say something about them, and often
As always, I’m glad I checked the comments, which included the following addition: “I’m not referring to how can you use them in cryptography… just what can you do directly in math and in abstract concepts.”
A dangerous sentiment to express to me, James. But I am happy to oblige.
Prime factorization is actually very helpful, as is decomposition in principle. So, before I get into the particular types of things that factorization helps with, let’s talk about decomposing things in general.
Mathematical objects can be very, very complicated. It can be difficult to say something about them, and often a good starting point is to study some simpler examples that are easier to analyze. In the worst case, it helps you give you some ideas about what types of things can go wrong. In the best cases, you can show that every mathematical object of the type you are interested in can actually be decomposed into some combination of the simple building blocks. This can be supremely helpful, because it might mean that you can prove powerful theorems about the general case by showing that it is true if it is true of the building blocks—and it might be significantly easier to prove it for the individual simple objects.
This idea is quite literally everywhere in mathematics. To give a small sample:
In linear algebra, a basis gives a way to decompose arbitrary vectors. It is common to define various operators on a space by simply specifying what it does to the basis. (In a sense, that is exactly what a matrix does.)
In Galois theory, it is important to understand how to factor polynomials into irreducible polynomials (polynomials that cannot be factored any further), as the roots of irreducible polynomials have nice properties.
In topology, the classification of all connected, orientable closed surfaces states that they can be constructed by taking connected sums of the torus.
In group theory, any finite group can (in a sense) be constructed from finite simple groups. With the classification of finite simple groups finished in 2008 (which might hold the record for the hardest mathematical result to prove), there have now been various theorems about group theory proved by proving it for the simple groups given in the classification, and showing this is sufficient.
In group theory (again), any finite abelian group can be written as a direct sum of cyclic subgroups of order pr (where p is prime).
In field theory, there is exactly one finite field with pr elements for any positive integer r and prime p. There are no other finite fields.
I could give many other examples. Interestingly, already in this list, four of the examples rely on the decomposition of integers into primes. The last two examples are the most obvious, but (2) and (4) do as well.
So, clearly prime factorization is at least useful in other parts of mathematics. Is this decomposition useful for number theory directly? You better bet it is. Again, I give a minimal overview encompassing elementary number theory, analytic number theory, and algebraic number theory.
First, the classification of finite abelian groups is actually very much about number theory, since studying congruences is really all about finite abelian groups. So, any sort of questions about divisibility can be helped out by first reasoning about divisibility over primes.
Second, the Riemann zeta function can be expressed (if the real part of s is greater than 1) as
ζ(s)=∞∑n=11ns
or equivalently as the Euler product
ζ(s)=∏p prime(1−p−s)−1.
The Riemann zeta function has a tendency to pop up in all sorts of calculations, both in and out of number theory (and even in statistics and physics). As a simple example, it pops up when calculating the probability that two randomly chosen integers are co-prime. The fact that we have the Euler product sometimes make the Riemann zeta function a lot easier to work with, and is the source of many useful identities with other functions.
Finally, there is Ostrowski's theorem and the Hasse–Minkowski theorem. Ostrowski’s theorem tells you that if you are interested in looking at absolute values to define on the rational numbers, you are actually very highly constrained—there are basically only two types. The first is the usual absolute value that we know and love. The second type is to chose any prime p, and to define
where are not divisible by (note that any rational number can be written this way). For both types of absolute values, there is a standard method of constructing a new field containing the rationals by “filling in holes.” For the usual absolute value, this just give the real numbers. For the other, less well-known type of absolute value, you instead get something called a -adic field. I wrote a summary about how to define and work with the -adics here.
The -adics are interesting in and of themselves, but they are also a highly useful tool. There are many instances where it is reasonably easy to analyze something over the -adics and over the real numbers, and if you are very lucky that can give you information about what is happening over the rational numbers. The Hasse–Minkowski theorem is an example: it states that if you have a quadratic form with rational coefficients, this quadratic form will have a non-trivial rational zero if and only if it has non-trivial zeros over the reals and over all of the -adics.
This is, in fact, a general attempt at a solution in number theory: figure out if there are solutions over the reals and -adics, and then see if you can stitch them together into a solution over the rationals. If this is possible, you get something called a local-global principle. This isn’t always possible, but it is highly fruitful when it is.
Related questions
How do you find the greatest common factor (GCF) using prime factorization?
How can you find the LCM and GCD without prime factorization?
What is the GCD and why is it hard to find when given many numbers?
How do you find the least common multiple (LCM) and greatest common divisor (GCD) using prime factorization?
How do you find LCM (16'42) and GCD (27'45) using prime factorization?
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How do you handle basic number theory (GCD, number theory, GCD and LCM, math)?
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What is the greatest common divisor (GCD) of two given positive integers without calculating their prime factors (LCM, algebra)?
What is the prime factorization method for 1000000000000 prime factorisation?
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13567 | https://ocw.mit.edu/courses/2-003sc-engineering-dynamics-fall-2011/resources/lecture-3-motion-of-center-of-mass-acceleration-in-rotating-ref-1/ | Lecture 3: Motion of Center of Mass; Acceleration in Rotating Ref. Frames | Engineering Dynamics | Mechanical Engineering | MIT OpenCourseWare
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Syllabus
Newton's Laws, Vectors, and Reference Frames
Velocity, Acceleration, and Rotational Motion
Angular Momentum and Torque
Free Body Diagrams and Fictitious Forces
Quiz 1
Angular Momentum and Rotating Masses
Angular Momentum and Motion of Rotating Rigid Bodies
Finding Equations of Motion for Rigid Body Rotation
Lagrange Equations
Lagrange Equations Continued
Quiz Review Lecture & Quiz 2
Mechanical Vibration
Reducing Problem Vibration and Intro to Multi-DOF Vibration
Vibration of Multi-Degree-of-Freedom Systems
Vibration Analysis by Mode Superposition
Final Exam
Resource Index
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Instructors
Prof. J. Kim Vandiver
Prof. David Gossard
Departments
Mechanical Engineering
Civil and Environmental Engineering
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Fall 2011
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Mechanical Engineering
Solid Mechanics
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2.003SC | Fall 2011 | Undergraduate
Engineering Dynamics
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Syllabus
Newton's Laws, Vectors, and Reference Frames
Velocity, Acceleration, and Rotational Motion
Angular Momentum and Torque
Free Body Diagrams and Fictitious Forces
Quiz 1
Angular Momentum and Rotating Masses
Angular Momentum and Motion of Rotating Rigid Bodies
Finding Equations of Motion for Rigid Body Rotation
Lagrange Equations
Lagrange Equations Continued
Quiz Review Lecture & Quiz 2
Mechanical Vibration
Reducing Problem Vibration and Intro to Multi-DOF Vibration
Vibration of Multi-Degree-of-Freedom Systems
Vibration Analysis by Mode Superposition
Final Exam
Resource Index
Resource Index
Lecture 3: Motion of Center of Mass; Acceleration in Rotating Ref. Frames
Description: Prof. Vandiver goes over an example problem of a block on a slope, the applications of Newton’s 3rd law to rigid bodies, kinematics in rotating and translating reference frames, and the derivative of a rotating vector in cylindrical coordinates.
Instructor: J. Kim Vandiver
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0:00 The following content is provided under a Creative
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0:20 PROFESSOR: So I'm going to give you
0:22 a quick example of what I think is a good way to do solutions.
0:30 Our approach to solutions.
0:45 State the problem.
0:48 I'm going to give you a little formulaic here.
0:51 Draw figures.
1:00 On the first day I said I like to think of dynamics problem.
1:03 I break them down into three categories.
1:06 One is to describe the motion.
1:14 What does describing the motion mean?
1:15 Well how many-- the number of degrees of
1:21 freedom in the problem.
1:23 That implies the number of equations of motion
1:30 that you're going to need to solve that problem.
1:33 So you got to identify the number of degrees of freedom.
1:35 It also then tells you the number of coordinates you need.
1:44 So this is all part of describing the motion.
1:46 It's figuring out how many coordinates, assigning them.
1:58 So assigning the coordinates.
2:01 And then finally, essentially all underneath this you
2:05 essentially do the kinematics.
2:21 And that's the velocities, accelerations, so forth.
2:27 So once you have-- you've explained the motion.
2:34 And this I guess is four.
2:35 Explain the correct physical laws.
2:47 You know how they apply.
2:48 f equals ma.
2:50 Newton's first, second, third law of conservation of momentum
2:52 or whatever you want-- you think is the appropriate thing.
2:56 So explain what the physical laws are and apply them.
3:02 And finally, do the math.
3:10 So if you could break problems down that way
3:12 it'll give you a nice, logical flow.
3:14 So I'm going to give you a bit of an example problem.
3:27 And I'm also going to kind of pose a brain teaser to you
3:29 at the end of this problem today.
3:31 I want to give you something to think about.
3:33 So I'm going to draw my problem.
3:37 This is a block on an incline.
3:46 It's got some scales to measure your weight.
3:50 And you're standing on this thing.
3:55 Riding it down the incline.
4:00 And the first question about this
4:02 is to find the position as a function of time.
4:26 So that's the problem.
4:27 That's part a.
4:32 So state the problem, find the position.
4:45 So how-- well what do we do?
4:46 Well draw figures, I've started with that.
4:48 Next, describe the motion.
4:50 I need a free-- We have to figure out
4:56 how many degrees of freedom this problem has.
4:59 I'm going to just declare no rotation.
5:01 Going to treat it as a particle.
5:04 So a particle has how many, generally how many degrees of
5:06 freedom?
5:07 How many coordinates to completely describe
5:09 where it's at?
5:10 AUDIENCE: Three.
5:11 PROFESSOR: Three.
5:12 So I may need as many as three coordinates
5:14 to describe the motion of this thing.
5:17 And if I really doing complete equations of motion
5:19 I need three equations of motion.
5:21 So this is two, three-- describing the motion I
5:25 have three degrees of freedom.
5:28 I'm going to need three coordinates.
5:36 So in this case here's my picture
5:42 I'm going just to set up a Cartesian coordinate system
5:45 aligned in a helpful way.
5:49 X, y, z coming out.
5:53 And this is my fixed inertial reference frame.
5:56 And here's my center of mass.
5:58 And basically my coordinates are describing the position
6:02 of the center of mass.
6:19 So that's pretty much the describing the motion,
6:22 what I need for now.
6:23 We'll get to the velocities and accelerations
6:25 when we get to the math part.
6:27 Apply the physics.
6:36 I'm going to use Newton's second law.
6:48 Sum of the external forces, mass times the acceleration.
6:55 That's the physical law I'm going to apply.
6:59 Draw a free body diagram.
7:16 And I'm just going to consider the block
7:18 and the person this is the whole collection, it's one thing.
7:22 I'm just not keep drawing the person on it,
7:24 but-- so here's my object, including
7:26 the weight of the person.
7:29 And it's going to have-- here's its center of mass.
7:35 Obviously in mg, gravitational force,
7:39 it's going to have a normal force.
7:42 Going to have a friction force and how do I
7:44 figure out what the friction-- which direction the friction
7:46 is?
7:47 I assume motion down the hill.
7:51 Friction will oppose it.
7:53 I draw in the arrows in the direction
7:55 I expect the forces to act.
7:57 Then I'll use the sign convention
7:58 that the arrows tell me what signs.
8:01 So since my x-coordinate is down the hill.
8:06 This is y.
8:08 This is x, friction x, in the minus x direction.
8:17 And I haven't-- I've left out a key piece of information.
8:23 Got to have the angle of the slope.
8:26 And once you have the angle of the slope that's theta,
8:37 this is theta.
8:38 And I'm going to need to-- this is in a direction of one
8:42 of my coordinates and so is this,
8:44 but I need to break the gravitational piece
8:47 into components lined up with my coordinates.
8:51 And so you have a theta here as well.
8:55 And now I can write my equation to motion.
9:02 And the nice thing about vectors is
9:05 that when you have three equations of motion,
9:08 three coordinates, each that are components of vectors
9:12 in the x, y and z direction, it gives us
9:14 three equations immediately.
9:17 So for example, the summation of the forces
9:19 in this problem in the z-direction,
9:22 the external forces are sums to?
9:25 AUDIENCE: 0.
9:26 PROFESSOR: 0.
9:27 OK so we get a trivial solution out of that.
9:29 And we don't have to go much further.
9:34 Summation of the forces in the y-direction
9:36 gives us some useful information.
9:42 And then the y-directed forces I have an n and a minus mg
9:50 and I think it's cosine theta.
9:59 Which tells me immediately what n is.
10:01 So from statics I get to-- and from what we know about models
10:09 of friction then we know that we can model the friction as mu
10:16 times n for mu mg cosine theta.
10:25 So from the statics I learn a bunch of things
10:27 that I need to know for the problem.
10:34 So now I get to the real heart of the problem,
10:43 writing my equation of motion in the x-direction.
10:56 And the forces in the x-direction mg
11:00 sine theta down the hill.
11:02 So it's positive.
11:04 Minus the friction is up the hill.
11:08 So I get mg.
11:18 I'm mixing up my m's here but there's no other m so-- this
11:34 basically says, that x double dot, the m's all cancel out.
11:40 That x double dot is g sine theta minus mu g cosine theta.
11:52 And that just happens to be a pretty simple to solve,
11:57 ordinary differential equation.
11:58 This is an equation in which the acceleration in the problem
12:01 is constant.
12:02 The data's not changing.
12:04 None of these things are changing
12:05 and so you can just solve this one.
12:07 Now we're to the third part, do the math.
12:16 This one you can just integrate.
12:18 And so you find out that well, x dot then
12:21 and I'm just going to call this c some constant.
12:25 So x dot ct plus an initial velocity, if it had one.
12:31 And x of t, what you're looking for.
12:45 And now so v0 and x0 are just your initial conditions.
12:48 More than likely 0 if you set it up cleverly.
12:51 So that's just modeling quickly what
12:53 I think a good way to lay out a problem is.
12:56 Describing the motion, explaining the physics,
13:00 doing the math, drawing good pictures, stating the problem.
13:05 All right so now the brain teaser
13:07 that I want you to think about.
13:33 So here's the mass of the block plus the scales.
13:39 And here's the-- here you're standing on it.
13:45 So mass of the person.
13:52 You're riding this down the hill.
14:10 So this is part b.
14:22 Think about that when in the shower.
14:27 If you've got a really simple way to do it,
14:30 great write it up.
14:32 It's not terribly hard and it's mostly-- I'll give you a hint.
14:34 Thinking in terms of free body diagram helps a lot.
14:39 And we'll come back to this kind of fun problem.
14:56 OK, want to go-- that was part one.
14:58 I want to go onto this recapping the center of mass quickly.
15:03 We learned a couple of important things.
15:11 We got-- we talked about the center of mass
15:13 because we were just talking about Newton's three laws.
15:16 From looking at the first law, found
15:18 that it's useful in determining whether or not
15:20 you're in a inertial frame, we used it.
15:22 Second law we've just applied it.
15:24 We used to do-- get equations of motion.
15:26 Third law was about-- we used it when we're
15:31 thinking about center of mass.
15:33 We used it to define what the center of mass was.
15:36 So we said the total mass of a system,
15:40 I better draw my picture.
15:41 Here's my system of particles.
15:45 M1 with position vectors.
15:50 So a whole mess of particles with their position vectors.
15:53 This is ri with respect to O for example.
15:56 This is my O x, y, z frame.
15:59 We said that this total mass of the particles
16:03 somewhere out here there's a center of mass with a position
16:11 vector rg with respect to O. So the definition of my center
16:22 of mass is this is equal to the summation of the m, i, r, i, o
16:29 and these are position vectors.
16:33 So that's the definition of my center of mass.
16:35 If I take two time derivatives of that we arrived at mt rg
16:43 with the respect to O double dot.
16:46 Summation over i of my m, i, r, i, o double dots.
16:57 And then importantly that's the summation
17:01 of all of the external forces on each
17:07 of these-- each one of these by itself
17:09 satisfies Newton's law, second law, which
17:12 he wrote about particles.
17:14 Each one has a summation.
17:16 I've summed these and I'm going to sum these,
17:17 but each one has a summation of internal forces acting on it.
17:23 These I call the f, i, j's.
17:27 And we learn-- something about the third law
17:29 tells us about that summation.
17:31 The third law tells us what?
17:34 That goes to 0 and that was that's the really powerful
17:38 piece of the third law that we make great use of.
17:42 Because this now essentially allows us to say,
17:46 that the summation of the external
17:49 forces on an assembly of particles,
17:53 on a system of particles, is equal to the total mass
18:06 times the acceleration of the center of mass.
18:11 And that-- what that does in one stroke
18:17 takes you from Newton, who's laws applied to particles
18:22 and allows you to apply Newton's second law to rigid bodies.
18:27 Because rigid bodies can be thought up
18:29 a bunch of particles, which are represented
18:33 in that simple equation.
18:36 And that gets you from particles to rigid bodies.
18:39 And we all know from physics that a summation
18:42 of the external force on this thing
18:44 is the mass times the acceleration
18:45 of the center of mass.
18:50 So that's actually quite an important powerful law.
18:53 It provides this for us.
18:57 Now I said I wanted to give you a quick, very
19:01 useful application of thinking about center of mass.
19:04 I showed you the other day, I had my carbon fiber tube here.
19:08 Showed you that trick for finding the center of mass
19:11 just by sliding your fingers along it.
19:12 And you end up at the center of mass.
19:14 Well as a practical matter other things--
19:17 you really want to be able find center of mass.
19:19 This is a glider, a sailplane called an LS8.
19:23 I happen-- I'm a glider flight instructor.
19:26 Been flying gliders for 35 years or something like that.
19:29 And I flew a glider of this type just recently.
19:34 That machine has a 49 to one let's call it 50 to one
19:38 to make it easy, glide ratio.
19:41 Means if you're a mile above the ground in still air,
19:45 you will go 50 miles before you touch the ground.
19:47 So they're really amazing high performance machines.
19:50 One of the things about all aircraft that you actually
19:53 need to know is, you really need to know where
19:56 the center of mass of it is.
19:59 And if the center of mass is in the wrong place
20:02 the plane will not fly properly.
20:04 And so you can't just go throw on 50 pounds of lead
20:07 in the tail of that plane and expect
20:09 to survive the next flight.
20:12 So you have to know where the center of mass
20:14 is, and in fact, you want the center of mass
20:16 about 25% of the-- if the wing is this wide from front
20:21 to back it's called the cord.
20:22 About 25% back from the leading edge
20:25 is about where the center of lift of a wing is.
20:30 And you want your center of mass also
20:33 called the center of gravity in these situations,
20:35 you want it to be pretty close to the center of lift
20:37 so that their balance in the plane flies nicely.
20:41 So is there a simple way to find the center of mass of something
20:52 like a sailplane?
20:55 So I'm going to drew-- this is exactly how you do it.
20:58 So I'll draw a quick picture of my sailplane here.
21:11 You usually have a little skid or a tail wheel on the back.
21:13 And to find the center of mass you just set them on scales.
21:21 You weigh it.
21:24 You pick a coordinate system.
21:26 Doesn't matter where it is, as long as it's fixed.
21:29 The easiest place is right at the nose of the sailplane.
21:32 So we'll make this x, make this y.
21:37 You take it and you can take a tape measure
21:39 and measure the distance from your reference point
21:45 to the position where the wheel sits on the scales.
21:48 We'll call that L1 here.
21:50 And that's typically about five feet.
21:53 And back here you have another position
21:58 to where you have your second set of scales, that's L2.
22:02 And a typical sailplane that's about 15 feet.
22:14 Apply Newton's law.
22:16 This thing's not going anywhere.
22:17 Sum of the forces in the vertical direction
22:19 is equal to 0.
22:23 Free body diagram.
22:25 Well you have somewhere about here, where the wing is,
22:30 this is your center of mass.
22:33 And you have m total times g downwards there.
22:38 You have two weights pushing on the sailplane.
22:42 A W2 pushing up, holding up the tail.
22:46 And a W1 holding up the main gear.
22:51 And so from the sum of the forces in the y-direction,
22:58 that had better be 0.
22:59 So you know that W1 plus W2 minus mtg is 0.
23:14 And so you find out that the total weight of the sailplane
23:18 is no surprise, the sum of the two weights on the scales.
23:23 AUDIENCE: [INAUDIBLE].
23:27 PROFESSOR: Yeah.
23:29 Let's do this, m, t, g.
23:33 So the total weight times gravity
23:35 is just the sum of the two readings on the scales.
23:38 And the second piece that you need to do to do this problem
23:43 is, you can have an equation that says,
23:45 the sum of the external torques with respect
23:48 to you're-- through a fixed point O,
23:52 is equal to the mass moment of inertia times the angular
23:56 acceleration, oftentimes written as alpha.
23:58 In this case, that's going to be 0, it's going nowhere.
24:03 So what are the external torques with respect to this point?
24:06 Well we have a right-handed coordinate system.
24:08 You have W1 up, times L1.
24:15 W2 up, times L2 minus W1 plus W2,
24:23 which is the total weight of the sailplane times
24:29 rg, the location of the center of mass.
24:35 That's this distance that we're looking for.
24:46 So we know everything here, except rg.
24:48 So we can solve for rg with respect to O
24:52 and it's simply W1 L1 plus W2 L2 over W1 plus W2.
25:03 And if you run the numbers, typically W1 600 pounds,
25:15 W2 the numbers I've done here is-- might be 40 pounds.
25:26 And I've already said five feet and 15 feet.
25:29 And you work the answer and you come up
25:31 with 5.62 feet, which puts this-- here is the wheel
25:40 and you're a little bit aft of the main gear.
25:44 And that's where you, basically where you want to be.
25:47 That's a real practical use of knowing about centers of mass
25:53 and how to calculate them.
26:00 So that's the second item I wanted to talk about today.
26:06 Essentially a recap of the center of mass.
26:08 And now I want to move on to talking
26:11 about a serious introduction to-- we've
26:18 had the introduction, velocities and accelerations.
26:21 We have to have a way of writing down
26:24 the acceleration of a mass, a point, a dog in a rotating,
26:35 translating, reference frame with the possibility
26:38 that in addition to that, the dog's moving.
26:41 So we want to have equations-- we want to have the ability
26:44 to write down expressions for the velocity
26:47 and acceleration of a mass moving in a translating,
26:54 rotating, reference frame.
26:57 So we've started this.
26:58 We did pretty much did velocities to begin with.
27:31 So here's my inertial frame.
27:41 Call it O or O x, y, z.
27:43 Here's my rigid body out there.
27:51 It has a point a something else, b might be the dog.
27:57 And we've described the position of this
28:00 as the position of this point a, with respect to O.
28:04 And at this point we're going to locate a reference frame
28:10 attached to the rigid body.
28:13 And so it's going to be called a,
28:15 and I'll call it x prime, y prime, z prime.
28:19 It's attached to the rigid body, it
28:21 rotates with the rigid body and its attached
28:24 at some fixed point.
28:26 Now what would oftentimes would be
28:27 a smart choice for that fixed point at point A?
28:35 AUDIENCE: [INAUDIBLE].
28:36 PROFESSOR: If you're going to write an equation expressing
28:38 the motion of this.
28:40 Where would you make point a?
28:41[INTERPOSING VOICES]
28:42 AUDIENCE: Center of mass.
28:43 PROFESSOR: Center of mass.
28:45 So very, very often, especially when objects are free
28:48 floating around out there you're going to make smart choices
28:50 and you're going to put this coordinate system
28:52 right on the center of mass.
28:53 But it doesn't have to be, but it can be.
28:59 So we were interested in knowing things
29:04 about the motion of this point in our inertial reference
29:09 frame, in terms of positions of our coordinate system.
29:15 And then also this vector here rd, with respect to a.
29:23 Now last time we came up with expressions for the velocity
29:29 of b with respect to O.
29:35 We said in general it's the velocity of your-- where
29:39 your coordinate system's located.
29:41 The translating-- the velocity of the translating frame
29:46 plus the derivative of rba, time derivative
29:55 of the position as seen from, if you were sitting at a.
30:03 And another way to say that, or this
30:06 is a derivative taken with the rotation rate momentarily set
30:12 equal to 0.
30:14 Another way to think of it.
30:16 Plus a piece that comes from rotation.
30:20 So the rotation with respect to the fixed frame, these
30:24 are all vectors, the rotation with respect
30:30 to the fixed frame, cross product with r, b, a.
30:36 And this-- and we said this is actually
30:38 a general formula for the derivative of-- this piece is
30:41 the derivative of a vector in a frame, in a fixed frame.
30:48 You have two pieces, the derivative as
30:51 seen without rotation plus the contribution that
30:55 comes from rotation.
31:05 When I did this center of mass thing a second ago,
31:08 I just kind of quickly wrote down two time derivatives
31:12 of the position vector.
31:13 There's no omega cross O's in there right?
31:18 Why could I do that?
31:20 This is actually kind of an important distinct point.
31:24 I could do that they didn't say very specifically when I did it
31:29 was an assumption I was making.
31:31 Except for perhaps they drew it.
31:34 This was done in a Cartesian coordinate system.
31:39 And my coordinates were x, y and z
31:41 and the unit vectors were i, j, k and do they move?
31:47 No.
31:47 What's their time derivative?
31:49[INTERPOSING VOICES]
31:50 PROFESSOR: When you don't-- when the inner vectors don't have
31:52 time derivatives you don't get these terms.
31:54 This is the only term that contributes so I could just
31:57 write that equation.
31:58 But we now have a reference frame attached to a body
32:02 and this reference frame is rotating.
32:04 And that means that the direction of the unit vectors
32:07 attach-- the unit vector attached to x-prime here
32:11 is moving, it's rotating.
32:13 And it's going to have a time derivative.
32:15 So we have to-- and that is given
32:17 and you take those time derivatives you
32:19 get this second piece.
32:24 I'm going to give you the answer in advance.
32:26 The acceleration of b with respect to O I'm
32:32 going to give you the full 3D equation.
32:37 Then we'll go back and see a bit where it comes from.
32:41 So here's-- it's the time derivative of that velocity
32:45 expression with respect to time taken in the inertial frame O,
32:55 x, y, z.
33:02 And am I going to have enough room to get this on?
33:04 It'll be close.
33:19 All right this has several pieces.
33:22 It's got a contribution of the acceleration of a with respect
33:28 to O. That's just the acceleration of this point.
33:30 Has nothing to do with rotation, so it's just a straight out
33:33 acceleration of my translating frame with respect to O.
33:37 That's the first piece.
33:40 The second piece is related-- is the derivative of this guy that
33:44 comes from the derivative of this.
33:46 It's the acceleration of b with respect to a as seen
33:54 in this a frame.
34:00 If you read the Williams book, it's
34:01 called the relative acceleration.
34:03 It's relative to the-- if you were sitting at point A,
34:06 it's what you would see as the acceleration.
34:12 Plus 2 omega cross velocity of b with respect
34:25 to a as seen from a plus omega dot, the derivative
34:36 of the rotation rate, cross rba plus omega cross,
34:57 omega cross r, b, a.
35:07 Kind of daunting right?
35:09 A little messy.
35:11 Basically one, two, three, four, five different terms.
35:19 And you're going to-- and they all have names and meanings.
35:24 And one of the things that will really help you
35:27 is to get familiar, you really need
35:30 to be familiar with the meaning of each one of the terms.
35:36 And it's not terribly difficult.
35:38 This one, just the acceleration of the translating frame.
35:43 So if it's a merry-go-round sitting
35:44 on a train and the train's heading down the track,
35:47 its acceleration of the train.
35:50 Rotating frame is attached to the merry-go-round.
35:54 And if you've got the dog on the merry-go-round
35:58 this is then the acceleration of the dog relative to this,
36:02 the merry-go-round.
36:04 This position of the coordinate system
36:07 attached to the merry-go-round has no rotation in it.
36:11 This is the velocity of that point, the dog,
36:17 as seen from the A frame.
36:19 Again, you have no sense of rotation.
36:22 Rotation is not a part of this.
36:24 Cross product with the rotation rate.
36:30 This is the accelerate.
36:31 This is the angular acceleration cross product with rba.
36:35 Now that's a term-- what does that mean?
36:38 I'm swinging a baseball bat and I'm accelerating this thing.
36:44 Idealize it as just something on a radius accelerating.
36:48 The acceleration of a point out here
36:52 is the radius times the angular acceleration.
36:55 So that's all this term is.
36:57 And it's called the Euler acceleration.
36:59 But it's just simply r theta double dot.
37:04 This, if you multiply it out and just think about units,
37:08 this ends up looking like r omega squared.
37:10 Have you run into that before?
37:13 What's that?
37:15 Common language.
37:16 AUDIENCE: [INAUDIBLE].
37:18 PROFESSOR: That's as a centrifugal-- centripetal,
37:21 this is centripetal acceleration.
37:23 So this is the centripetal acceleration term,
37:25 that's the Euler acceleration term,
37:28 this is the local acceleration.
37:30 This is the acceleration of your frame.
37:32 This is the strange one.
37:33 This is the Coriolis acceleration.
37:40 And we'll get familiar with it too.
37:42 So that's the full blown 3D acceleration equation.
37:48 And by the way the vector-- the velocity one
37:50 is also perfect 3D.
37:52 Now in this course we won't do much in the way of 3D dynamics
38:01 problems.
38:01 Yes.
38:02 AUDIENCE: Does the point b on the rigid plane move?
38:08 PROFESSOR: Does the-- it may.
38:10 It could be this is the-- an asteroid out there in space
38:19 and you've got-- this is home base
38:22 and that's a guy out there in a space suit running.
38:29 So we want to be able to describe
38:31 the acceleration of that guy as seen from a fixed reference
38:38 frame.
38:40 Now why would we want to know that acceleration?
38:43 Why do we want to know it in a fixed frame?
38:47 Well if you want to calculate the forces on the person.
38:53 Well how much-- what's he have to do with his feet
38:57 to brace himself or whatever?
38:59 What are the actual forces?
39:01 You have to know the acceleration on the person.
39:04 But Newton's laws, in order to say f equals ma,
39:08 Newton's laws have to be applied in inertial reference frames.
39:13 Is this thing out there doing this an a inertial reference
39:16 frame?
39:17 No.
39:18 So you can't calculate the forces
39:20 without having some idea of this inertial frame.
39:23 So this is the way of getting the acceleration
39:26 on-- at a location on a moving, rotating body with respect
39:33 to an inertial frame.
39:36 And with all the terms present.
39:40 Now most discourse has generally has addressed problems
39:45 which are in most textbooks address only planar motion
39:49 problems.
39:51 Planar motion basically means that we can find
39:55 the translations to a plane.
39:59 So imagine an x, and a y, and a z upwards coordinate system
40:04 attached to the top this table.
40:06 And I only allow motions that are around the table.
40:09 And I only allow motions that have a single axis rotation.
40:12 And that's lined up with z.
40:14 Those are essentially planar motion problems.
40:17 And most courses in dynamics, it lists [INAUDIBLE].
40:20 That's as far as they get.
40:22 And the most of the problems that you'll do
40:24 will be planar motion problems.
40:27 But that equation reduces to the planar motion problem as well.
40:33 We will do a little bit of 3D, because there's
40:36 a class of problems that I really
40:37 think it's important for you to understand that just come up
40:40 all the time that require a little 3D.
40:43 And as you want to have some things going on
40:45 out of the plane, but we'll still
40:47 confine the axis of rotation to a single direction.
40:53 Yeah.
40:53 AUDIENCE: [INAUDIBLE] in a planar,
40:58 but then would you have three to view the freedom? [INAUDIBLE].
41:02 PROFESSOR: That's a great question
41:04 she said if you had a dog running
41:05 on the merry-go-round how many degrees of freedom do you have?
41:08 So in general rigid bodies, each rigid body,
41:13 each independent rigid body has--
41:16 you have to describe its location of its center of mass
41:18 and that takes how many coordinates?
41:20 How many coordinates I'll call them.
41:23 AUDIENCE: [INAUDIBLE].
41:25 PROFESSOR: Well in general three.
41:28 And it can now rotate around three different axes.
41:35 And that takes three more.
41:36 So rigid bodies have six degrees of freedom.
41:39 And any problem when you go to address the problem you
41:42 essentially for a rigid body you start with six.
41:45 And you start applying constraints
41:47 to reduce it down to the number of remaining
41:50 degrees of freedom.
41:52 So if it's confined to a plane and no z-motion
41:55 is allowed one constraint.
41:58 If it is on a plane and it's only
42:01 allowed to rotate about the z-axis that
42:04 means you've constrained its rotation in around y and x.
42:08 So that's two more.
42:10 And so now you're down to three degrees of freedom left, xy
42:16 and a rotation about the z-axis.
42:19 So planar motion problems generally
42:21 have three degrees of freedom.
42:26 But instant-- let's just say we're
42:28 just interested in just something that
42:30 rotates and doesn't translate.
42:33 How many degrees of freedom does that have then?
42:35 Just one.
42:36 x and y are forced not to-- no motion, two more
42:40 constraints you're down to one.
42:41 So lots of problems we do are in fact single degree
42:45 of freedom problems.
42:57 So to do planar motion problems we oftentimes
43:01 use polar coordinates.
43:04 So I'm going to introduce r theta.
43:08 And I'm actually going to call it cylindrical coordinates.
43:21 And cylindrical coordinates then you have an r, a theta and z.
43:34 And let's think about well let's see, I have a demo,
43:37 a little demo here.
43:40 So here's a problem with a single axis of rotation.
43:48 And it's a-- there is a mass out here and just this is a rhyme.
43:54 And so think of this think of this mass
43:58 out here as being a bug walking out this rod.
44:03 And the rod, this thing goes round and round.
44:08 It's not a merry-go-round but it's
44:09 a merry-go-round with a gang plank on it.
44:12 It's going up at an angle and you
44:14 can walk the gang plank while the merry-go-round's
44:16 going around.
44:16 So that's what we got here.
44:19 So this is actually allowed to change position of this mass.
44:23 So how would I describe that with cylindrical coordinates?
44:28 Let me so it's going to take-- one would be a side view.
44:38 So you see your vertical axis here
44:40 and have to have a bearing to hold it in place.
44:46 Here's the arm, here's the bug walking out the arm,
44:53 has some rotation rate.
44:58 Theta dot this is the z-axis and the position of this point
45:13 is described by a r vector, in the r hat
45:22 direction, which I think is the unit
45:23 vectors you're used to using.
45:25 And then this is the z-component in the k hat direction.
45:32 And this vector here would be r of v with respect to what shall
45:45 I call it?
45:46 I'll make this a and over here someplace
45:55 I have a fixed inertial let's see I got to be careful here.
46:07 I want that to be z then I have a y going out here x, y, z
46:13 pointing upwards.
46:15 And my-- this fixed inertial system the unit
46:17 vectors here this would be i hat, j hat and k hat.
46:25 But these-- this rotating system with its unit
46:30 vector little k and this vector they're the same,
46:35 they're parallel.
46:38 But looking down on this, this is my polar coordinate system.
46:42 Now I'm going to look at my top view.
46:45 I will see a projection.
46:49 I'll just see the r, this is rr hat,
46:54 this is my point B. This is theta.
47:01 And I have a unit vector.
47:03 So the unit vector r hat is something-- the unit long
47:08 pointing in this direction.
47:10 And the unit vector in this direction is theta hat.
47:14 And it's perpendicular to that radius.
47:18 So now I have my three unit vectors.
47:20 One pointing in the direction of r, here's
47:24 also my unit vector is just to make sure there's no confusion.
47:29 This unit vector is in this direction.
47:35 K is in that direction.
47:36 Theta is in that direction.
47:44 And over here you still have your-- now here's
47:47 my x, y, z out of the board inertial frame.
47:54 And this-- my inertial frame this might be r, b, o.
48:02 So in my inertial frame.
48:04 I want to know what's going on here.
48:07 I want to be able to calculate the velocities
48:08 and the accelerations.
48:16 So the notation here gets-- can get a little confusing.
48:21 The rbo notation that I've been using all along,
48:26 that's the motion-- that's the position vector describing
48:30 that point in my inertial frame.
48:34 And my-- just lowercase r here, no scrub scripts or anything,
48:39 that's just going to-- that's my polar coordinate r theta
48:44 and z that happened to be in this case,
48:47 this is a rotating frame.
48:50 This is a rotating frame.
48:52 It's the center of this coordinate system's at a.
48:58 But this thing rotates.
49:00 So this is a pretty simplified version
49:03 of this general problem.
49:07 Now because it's simplified, you can actually--
49:11 it's a lot easier to use.
49:12 Also has some real limitations.
49:14 You can only going just-- there's limited things
49:17 that you can describe with it.
49:42 So let's start by describing velocities in cylindrical
49:48 coordinates.
50:23 Remember this rba is the length of this guy here.
50:27 And it's made up of rr hat zk hat.
50:59 So to express the velocity we have
51:01 to take a time derivative of this r, b, a
51:03 and I'm going to express it in terms of r theta nz.
51:07 And to get acceleration I have to take two time
51:09 derivatives of this, but this is going to be expressed
51:12 in my cylindrical coordinates.
51:14 This is where I'm going.
51:17 And lots of problems-- many, many of these problems
51:24 have fixed axes of rotations and this velocity
51:27 and this acceleration are zero.
51:29 You just drop it out.
51:31 I'm going to do that just to keep this-- make
51:34 this problem a little simpler.
51:35 So we can just focus on these terms.
51:37 So let's just let the there be no translational of this frame.
51:45 And that says that Va with respect
51:48 to O the acceleration of A with respect to O over zero.
51:53 So I want you to just focus on these terms.
51:55 I don't lose anything, I can put these back in
51:57 later if I need them.
51:58 I just don't want to keep carrying them along.
52:30 So I have my-- remember my side view.
52:37 This is r, r hat zk hat that's my point.
52:47 And my top view.
53:06 This is my projection just looking down on it
53:12 what I see is the length r.
53:15 And what I see in my unit vector going that way r direction
53:20 and theta that direction.
53:25 And this is x and the i and a y with a j hat
53:32 vector looking down on it.
53:36 My rotation rate mega with respect to my inertial frame,
53:42 is sum theta dot k hat.
53:50 All right so now let's find the velocity of b with respect
53:54 to O. Well it's 0, no translation, plus--
53:59 and now I need a time derivative of rb with respect to a.
54:06 But this is then r, b, a is r, r hat plus z k hat.
54:20 And I need the time derivative of that.
54:26 So I get an r dot r hat plus an r r hat dot plus a z dot k.
54:41 So this is the product of two things.
54:44 They're both time dependent.
54:46 So I have to get two pieces, k does not change in direction.
54:51 So it has no time derivative.
54:53 So I only have a z dot k.
54:55 So this is a result of doing this,
54:57 but I now have to figure out what
54:59 is the time derivative of the unit vector in the r direction.
55:17 So when I told-- when we worked out this formula the other day
55:21 for the time derivative of a rotating vector,
55:24 I mostly did it, it was kind of an intuitive argument.
55:27 So on this one occasion I'm going
55:28 to give you an example of actually figuring out
55:32 what the derivative of this rotating vector is.
55:35 And if you go read that kinematics handout
55:39 and it does this in kind of full blown form for-- in general.
55:42 So I'm just going to do it as one example.
55:45 So here's our looking down on this, the projection on the xy
55:51 plane, here's our r-vector.
55:55 And here's this unit vector and it
55:58 starts from-- I have a unit vector starting from a it's
56:02 unit-- it's one long.
56:06 And this is r hat.
56:08 And it's of unit length and it's in this particular direction.
56:15 Now in a little bit it time delta t, it moves.
56:21 It moves to here.
56:23 So this is delta r hat and what direction does it move?
56:33 AUDIENCE: [INAUDIBLE].
56:34 PROFESSOR: Yeah, it moves in the-- moves in the theta hat
56:38 direction.
56:41 And the amount that it moves is the rotation rate,
56:45 theta dot, delta t.
56:52 So delta r hat, if I solve for this, delta t.
56:59 And this is in the theta hat direction,
57:04 is theta dot theta hat.
57:13 So this is the limit as t, delta t
57:17 goes to 0 you get the derivative of r hat with respect to time.
57:24 Its direction is in the theta hat direction
57:27 and its magnitude is theta dot.
57:31 So that's the time derivative of the unit vector r hat.
57:39 Yeah.
57:40 AUDIENCE: How does that work with units?
57:43 PROFESSOR: How does it work with units?
57:47 What's left out of here is that this is unit length
57:55 and has dimensions.
57:57 It's unit length, one whatever unit system you're working.
58:02 So that is implicitly in here.
58:06 It's one meter theta dot and that theta dot, the delta t,
58:11 the times go away.
58:12 You're left with one meter times the magnitude of theta dot.
58:17 So the distance it actually moves
58:19 is r theta, the r delta theta, delta theta is theta dot delta
58:25 t and the radius happens to be 1.
58:31 So whatever unit system you're working in it's a unit vector.
58:34 Has unit length.
58:36 So its units are buried right there.
58:38 Good question.
58:40 OK so now we know what this is.
58:45 So now we can come back finish our description of the velocity
58:50 of b with respect to a then is r dot r hat
58:58 plus z dot k hat plus theta dot times
59:10 so r times the derivative of the unit vector r, which we just
59:17 figured out is theta dot theta hat times r.
59:21 R theta dot theta hat.
59:25 So that's my velocity of b with respect to a.
59:28 My velocity of B with respect to O all you have to add in
59:31 is the velocity of A with perspective to O,
59:33 which we've let be 0 for now.
59:35 So for the moment this is also d with respect to O.
59:40 But this is the general piece of the velocity
59:42 of b with respect to a in polar cylindrical coordinates.
59:50 Now we could have-- so I've actually worked it out, just
59:54 shown you, just drew the picture and figured out the derivative.
59:57 We could have used that magic formula.
59:59 The formula for the derivative of a vector in a rotating
1:00:03 frame.
1:00:15 So I'll just do that quickly to remind you
1:00:18 how we could have done this.
1:00:20 rba with respect to time as seen in the O frame.
1:00:30 Is the partial derivative of rba with respect
1:00:36 to time as seen in the rotating frame, plus omega cross r, b,
1:00:47 a.
1:00:55 This term is that and that.
1:01:04 The derivative of this rba as seen
1:01:08 from inside of the rotating frame,
1:01:11 is just the change in length, this is rba here from the side.
1:01:17 So the change in length of that vector, the derivative
1:01:21 of-- the time derivative of it.
1:01:22 It's the vector sum of the r dot plus z dot.
1:01:28 So this piece comes from this and this.
1:01:36 And this piece should-- this one here, it better be this.
1:01:42 Well this is-- let's figure it out.
1:01:44 This is omega in the k hat direction,
1:01:49 cross and rba is r r hat plus z k hat.
1:01:59 k cross k is 0.
1:02:03 k cross r theta hat.
1:02:09 K cross-- k hat cross r hat is theta hat positive.
1:02:15 r omega theta hat, same thing as r theta dot theta hat.
1:02:21 So we could have just applied this formula for the derivative
1:02:25 of a rotating vector and we would
1:02:29 have gotten the same thing.
1:02:41 OK just ran out of boards.
1:03:20 Now a quick little exercise you could do on your own
1:03:22 is, we're going to need to be able to calculate
1:03:25 the derivative of theta hat.
1:03:28 Well just plug it in that little formula.
1:03:36 And the first term you'll find out the derivative of the theta
1:03:41 hat, the length doesn't change in time, it's a unit vector.
1:03:48 So you only have the second piece.
1:03:50 So it's sum omega cross theta and you're going
1:03:54 to get minus theta dot r hat.
1:04:17 So I really want to get here.
1:04:21 The acceleration of b and O. That's
1:04:23 the real-- that's the single piece we really
1:04:28 need to finish the kinematics.
1:04:30 So we can do most any problems.
1:04:32 Got to be able to describe the acceleration of a point
1:04:35 and translating rotating frame.
1:04:37 And that's going to be the acceleration of a with respect
1:04:41 to O, plus a time derivative of the velocity of b with respect
1:04:53 to O.
1:04:55 We've calculated the velocity, we
1:04:58 need to be able to essentially carry out this derivative.
1:05:05 Two time derivatives of the r, b, a, or a single time
1:05:10 derivative of bva.
1:05:12 Well we just computed the velocity
1:05:18 in this-- of this rotating frame and this
1:05:20 is our final expression.
1:05:22 So we need to compute the time derivative of that.
1:05:26 I just-- so it's going to look like r
1:05:32 dot r hat plus over here a term r theta dot theta hat
1:05:43 and pardon me for doing this I think
1:05:50 it'll be cleaner in the end.
1:05:51 I'm going to start with my z dot k, keep it over here,
1:05:55 plus r dot r hat plus this term.
1:06:04 And this is going to take up a lot room.
1:06:06 Just spread out this way so it--
1:06:17 So let's just-- I'm going to write down where this comes
1:06:19 out, this is a little tedious, but then you'll
1:06:21 have seen it once hopefully believe that it really works.
1:06:31 So these terms, this first term here just
1:06:34 gives you z double dot.
1:06:36 So let's write her down.
1:06:37 So just z double dot k hat time derivative of this,
1:06:44 plus an r double dot r hat, but now I
1:06:50 have to take-- do it, flip it and do the other side of it.
1:06:53 So I get my-- how do I want to do this?
1:07:15 Yeah, I like this.
1:07:26 So this term kept-- leads to this.
1:07:29 This term brings you to here.
1:07:35 This term, you get r dot times theta dot theta hat plus an r
1:07:50 and now you need to take a time derivative of theta dot theta
1:07:56 hat.
1:07:58 So that's going to expand.
1:07:59 So this brings you to here.
1:08:28 Now we've done this derivative, so we can put it in.
1:08:33 So this gives us this term over here,
1:08:59 so let's keep adding these up.
1:09:11 Notice this gives me an r dot theta dot theta hat.
1:09:15 This gives me an r dot theta dot theta hat.
1:09:23 Two identical terms.
1:09:28 Now this term gives me an r theta double dot theta hat.
1:09:38 And now this-- now we need to take
1:09:40 the derivative of theta hat and that gives you
1:09:43 minus theta dot r hat.
1:09:46 So you get a minus r but multiplied by theta dot again
1:09:59 squared r hat.
1:10:04 So I think we're about there.
1:10:05 We're going to start collecting things together.
1:10:23 Now we have a two r dot theta dot
1:10:30 theta hat plus an r theta double dot theta hat, minus r theta
1:10:46 dot squared r half.
1:10:49 So these things, these derivatives,
1:10:51 just all kind of flowed down and led to more terms.
1:10:58 But now if we compare-- that we get one, two, three, four,
1:11:06 five, I clump these together.
1:11:08 This is the change in length of the r vector.
1:11:14 Stretch in the position has a z component and a r component.
1:11:19 This is the movement of the coordinate system if it moves.
1:11:23 This is the Coriolis term.
1:11:26 This is the Euler acceleration and this is
1:11:30 the centripetal acceleration.
1:11:33 So this is the-- what happens when
1:11:36 you start with that messy vector thing
1:11:41 and apply it, restrict it to a cylindrical coordinate problem,
1:11:47 which is basically planar motion.
1:11:50 But you allow some things in the z direction only.
1:11:53 So polar coordinates is a more limited form of that,
1:11:58 but it's-- every term comes back,
1:12:01 every term is still in it.
1:12:03 So acceleration of the moving coordinate
1:12:07 system, change of the length of the position
1:12:11 vector in the moving coordinate system,
1:12:14 the Coriolis term the Euler acceleration term that angular
1:12:19 acceleration speedup and finally the centripetal term.
1:12:25 Now the way you go about solving problems,
1:12:29 usins-- doing problems in polar coordinates.
1:12:32 So now you're asked to find an equation of motion.
1:12:36 This is an expression for the acceleration
1:12:40 of whatever it is you're trying to describe
1:12:42 in an inertial frame.
1:12:46 So that when you say-- you can now say f equals ma.
1:12:51 If you know the-- sometimes you're
1:12:53 given the forces in problems and your asked
1:12:55 to find the accelerations.
1:12:59 But what if you're given the acceleration
1:13:01 and you're asked to find the force?
1:13:03 All right, I give you the simplest problem of this kind.
1:13:09 What's the tension in the string?
1:13:13 Well if I know that-- so I just say,
1:13:17 the way you do these problems is how many things can you
1:13:19 eliminate?
1:13:20 Well I wasn't moving, this term goes away.
1:13:22 That's zero.
1:13:23 Z wasn't involved, it's not changing,
1:13:26 it's just constant angular rotation, that term is 0.
1:13:30 What's the change in length of the string while I
1:13:32 was doing it?
1:13:33 Now if that term's 0 we're getting easier all time.
1:13:36 How far-- how fast was the length getting longer?
1:13:41 That terms gone away.
1:13:43 Was I speeding up, or slowing down, or constant speed?
1:13:46 Well we'll say it's constant speed,
1:13:47 ooh this problem's getting easier all the time.
1:13:50 I'm down to one term.
1:13:51 f equals ma.
1:14:03 Minus r theta dot squared r hat.
1:14:10 So the force that I must have been applying to the string
1:14:15 was in the minus r hat direction and had
1:14:18 magnitude mr omega squared.
1:14:25 So that's actually all there is to it.
1:14:27 We're using polar coordinates and cylindrical coordinates
1:14:29 to do second law problems.
1:14:33 So there's a couple of problems that you're
1:14:35 doing these kind of things on the homework
1:14:37 set that's being put out today.
1:14:40 So give them a try.
1:14:44 Have a good weekend.
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Prof. J. Kim Vandiver
Prof. David Gossard
Departments
Mechanical Engineering
Civil and Environmental Engineering
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Fall 2011
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Mechanical Engineering
Solid Mechanics
Science
Physics
Classical Mechanics
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13568 | https://courses.lumenlearning.com/waymakercollegealgebra/chapter/logarithm-rules/ | Module 12: Exponential and Logarithmic Equations and Models
Properties of Logarithms
Learning Outcomes
Rewrite a logarithmic expression using the power rule, product rule, or quotient rule.
Expand logarithmic expressions using a combination of logarithm rules.
Condense logarithmic expressions using logarithm rules.
Properties of Logarithms
Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.
For example, [latex]{\mathrm{log}}_{5}1=0[/latex] since [latex]{5}^{0}=1[/latex] and [latex]{\mathrm{log}}_{5}5=1[/latex] since [latex]{5}^{1}=5[/latex].
Next, we have the inverse property.
[latex]\begin{array}{l}\hfill \ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}[/latex]
For example, to evaluate [latex]\mathrm{log}\left(100\right)[/latex], we can rewrite the logarithm as [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)[/latex] and then apply the inverse property [latex]{\mathrm{log}}_{b}\left({b}^{x}\right)=x[/latex] to get [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)=2[/latex].
To evaluate [latex]{e}^{\mathrm{ln}\left(7\right)}[/latex], we can rewrite the logarithm as [latex]{e}^{{\mathrm{log}}_{e}7}[/latex] and then apply the inverse property [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex] to get [latex]{e}^{{\mathrm{log}}_{e}7}=7[/latex].
Finally, we have the one-to-one property.
[latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]
We can use the one-to-one property to solve the equation [latex]{\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)[/latex] for x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x:
[latex]\begin{array}{l}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}[/latex]
But what about the equation [latex]{\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=2[/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining logarithms on the left side of the equation.
Using the Product Rule for Logarithms
Recall that we use the product rule of exponents to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[/latex]. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.
Given any real number x and positive real numbers M, N, and b, where [latex]b\ne 1[/latex], we will show
[latex]{\mathrm{log}}_{b}\left(MN\right)\text{= }{\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)[/latex].
Let [latex]m={\mathrm{log}}_{b}M[/latex] and [latex]n={\mathrm{log}}_{b}N[/latex]. In exponential form, these equations are [latex]{b}^{m}=M[/latex] and [latex]{b}^{n}=N[/latex]. It follows that
[latex]\begin{array}{lllllllll}{\mathrm{log}}_{b}\left(MN\right)\hfill & ={\mathrm{log}}_{b}\left({b}^{m}{b}^{n}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \ \hfill & ={\mathrm{log}}_{b}\left({b}^{m+n}\right)\hfill & \text{Apply the product rule for exponents}.\hfill \ \hfill & =m+n\hfill & \text{Apply the inverse property of logs}.\hfill \ \hfill & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}[/latex]
Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]\mathrm{log}_{b}(wxyz)[/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:
[latex]\mathrm{log}_{b}(wxyz)=\mathrm{log}_{b}w+\mathrm{log}_{b}x+\mathrm{log}_{b}y+\mathrm{log}_{b}z[/latex]
A General Note: The Product Rule for Logarithms
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.
[latex]{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0[/latex]
Example: Using the Product Rule for Logarithms
Expand [latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)[/latex].
Show Solution
We begin by writing an equal equation by summing the logarithms of each factor.
[latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(30x\right)+{\mathrm{log}}_{3}\left(3x+4\right)={\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)[/latex]
The final expansion looks like this. Note how the factor [latex]30x[/latex] can be expanded into the sum of two logarithms:
[latex]{\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)[/latex]
Try It
Expand [latex]{\mathrm{log}}_{b}\left(8k\right)[/latex].
Show Solution
[latex]{\mathrm{log}}_{b}8+{\mathrm{log}}_{b}k[/latex]
Using the Quotient Rule for Logarithms
For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: [latex]{x}^{\frac{a}{b}}={x}^{a-b}[/latex]. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.
Given any real number xand positive real numbers M, N, and b, where [latex]b\ne 1[/latex], we will show
[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{= }{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)[/latex].
Let [latex]m={\mathrm{log}}_{b}M[/latex] and [latex]n={\mathrm{log}}_{b}N[/latex]. In exponential form, these equations are [latex]{b}^{m}=M[/latex] and [latex]{b}^{n}=N[/latex]. It follows that
[latex]\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\hfill & ={\mathrm{log}}_{b}\left(\frac{{b}^{m}}{{b}^{n}}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \ \hfill & ={\mathrm{log}}_{b}\left({b}^{m-n}\right)\hfill & \text{Apply the quotient rule for exponents}.\hfill \ \hfill & =m-n\hfill & \text{Apply the inverse property of logs}.\hfill \ \hfill & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}[/latex]
For example, to expand [latex]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex], we must first express the quotient in lowest terms. Factoring and canceling, we get
[latex]\begin{array}{lllll}\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right) & =\mathrm{log}\left(\frac{2x\left(x+3\right)}{3\left(x+3\right)}\right)\hfill & \text{Factor the numerator and denominator}.\hfill \ & \text{}=\mathrm{log}\left(\frac{2x}{3}\right)\hfill & \text{Cancel the common factors}.\hfill \end{array}[/latex]
Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.
[latex]\begin{array}{lll}\mathrm{log}\left(\frac{2x}{3}\right) & =\mathrm{log}\left(2x\right)-\mathrm{log}\left(3\right)\hfill \ \text{} & =\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)\hfill \end{array}[/latex]
A General Note: The Quotient Rule for Logarithms
The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.
[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N[/latex]
How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms
Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.
Example: Using the Quotient Rule for Logarithms
Expand [latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)[/latex].
Show Solution
First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.
[latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)={\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right)[/latex]
Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.
[latex]\begin{array}{l}{\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right) \\text{}= \left[{\mathrm{log}}_{2}\left(15\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)\right]-\left[{\mathrm{log}}_{2}\left(3x+4\right)+{\mathrm{log}}_{2}\left(2-x\right)\right]\hfill \ \text{}={\mathrm{log}}_{2}\left(15\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)-{\mathrm{log}}_{2}\left(3x+4\right)-{\mathrm{log}}_{2}\left(2-x\right)\hfill \end{array}[/latex]
Analysis of the Solution
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\frac{4}{3}[/latex] and x= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm that x> 0, x> 1, [latex]x>-\frac{4}{3}[/latex], and x< 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.
Try It
Expand [latex]{\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x - 1\right)\left(x - 2\right)}\right)[/latex].
Show Solution
[latex]{\mathrm{log}}_{3}\left(x+3\right)-{\mathrm{log}}_{3}\left(x - 1\right)-{\mathrm{log}}_{3}\left(x - 2\right)[/latex]
Using the Power Rule for Logarithms
We have explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[/latex]? One method is as follows:
[latex]\begin{array}{l}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}[/latex]
Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,
[latex]\begin{array}{lll}100={10}^{2}, \hfill & \sqrt{3}={3}^{\frac{1}{2}}, \hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}[/latex]
A General Note: The Power Rule for Logarithms
The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.
[latex]{\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M[/latex]
How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm
Express the argument as a power, if needed.
Write the equivalent expression by multiplying the exponent times the logarithm of the base.
Example: Expanding a Logarithm with Powers
Rewrite [latex]{\mathrm{log}}_{2}{x}^{5}[/latex].
Show Solution
The argument is already written as a power, so we identify the exponent, 5, and the base, x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
[latex]{\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x[/latex]
Try It
Rewrite [latex]\mathrm{ln}{x}^{2}[/latex].
Show Solution
[latex]2\mathrm{ln}x[/latex]
Example: Rewriting an Expression as a Power before Using the Power Rule
Rewrite [latex]{\mathrm{log}}_{3}\left(25\right)[/latex] using the power rule for logs.
Show Solution
Expressing the argument as a power, we get [latex]{\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)[/latex].
Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
Try It
Rewrite [latex]\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)[/latex].
Show Solution
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13569 | https://math.stackexchange.com/questions/1767109/x2y2-2z2-positive-integer-solutions | elementary number theory - $x^2+y^2=2z^2$, positive integer solutions - Mathematics Stack Exchange
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x 2+y 2=2 z 2 x 2+y 2=2 z 2, positive integer solutions
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Determine all positive integer solutions of the equation x 2+y 2=2 z 2 x 2+y 2=2 z 2.
First I assume x≥y x≥y, and I have x 2−z 2=z 2−y 2 x 2−z 2=z 2−y 2. Then I have (x−z)(x+z)=(z−y)(z+y)(x−z)(x+z)=(z−y)(z+y), but from here, I don't know how it can help me to describe solutions (I know that there are infinitely many).
elementary-number-theory
diophantine-equations
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edited May 2, 2016 at 5:21
Martin Sleziak
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asked May 1, 2016 at 18:45
RunjRunj
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2 Some other posts about the same diophantine equation: math.stackexchange.com/questions/1250912/…math.stackexchange.com/questions/1282600/…Martin Sleziak –Martin Sleziak 2016-05-02 05:24:46 +00:00 Commented May 2, 2016 at 5:24
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Just a start:
Note that
(x−y 2)2+(x+y 2)2=x 2+y 2 2=z 2.(x−y 2)2+(x+y 2)2=x 2+y 2 2=z 2.
And x−y 2 x−y 2 and x+y 2 x+y 2 are integers (why?)
So you need to find solutions to u 2+v 2=z 2 u 2+v 2=z 2.
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answered May 1, 2016 at 18:49
Thomas AndrewsThomas Andrews
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1 Thanks, I get it. Is there a general way to solve this type of problem, like x 2+y 2=k z 2 x 2+y 2=k z 2 Runj –Runj 2016-05-01 18:55:05 +00:00 Commented May 1, 2016 at 18:55
3 The general way is to solve k=u 2+v 2 k=u 2+v 2. If you can't solve it, then there is no solution. If you can, then you can convert the question to a question about x 2 1+y 2 1=z 2 1 x 1 2+y 1 2=z 1 2. This is related to Gaussian integers, and the fact that you have unique factorization in the Gaussian integers. In the case of k=2 k=2, 1 2+1 2=2 1 2+1 2=2.Thomas Andrews –Thomas Andrews 2016-05-01 19:02:28 +00:00 Commented May 1, 2016 at 19:02
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In General formula generic for Pythagorean triples looks a little different.
x 2+y 2=a z 2 x 2+y 2=a z 2
If the number can be represented as a sum of squares. a=t 2+k 2 a=t 2+k 2
The solution has the form:
x=−t p 2+2 k p s+t s 2 x=−t p 2+2 k p s+t s 2
y=k p 2+2 t p s−k s 2 y=k p 2+2 t p s−k s 2
z=p 2+s 2 z=p 2+s 2
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answered May 2, 2016 at 4:26
individindivid
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Though others have posted simple solutions to the problem but if you want to proceed the way you did then here it is:
First let x z=a x z=a and y z=b y z=b. This reduces problem to two variables (which are rational numbers).
Now we have (b+1)(b−1)=(1+a)(1−a)(b+1)(b−1)=(1+a)(1−a)
which means
b+1 1+a=1−a b−1 b+1 1+a=1−a b−1
Let this fraction be m n m n in reduced form. From this we get 2 linear equations in 2 variables (in terms of a and b). Solving those will give
a=(n ²+2 m n−m ²)(m ²+n ²)a=(n ²+2 m n−m ²)(m ²+n ²)
b=(m ²+2 m n−n ²)(m ²+n ²)b=(m ²+2 m n−n ²)(m ²+n ²)
Substituting values of a and b back gives the solution to original problem:
x=n ²+2 m n−m ² x=n ²+2 m n−m ²
y=m ²+2 m n−n ² y=m ²+2 m n−n ²
z=m ²+n ² z=m ²+n ²
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edited Oct 23, 2019 at 9:54
answered Oct 23, 2019 at 9:13
RonakRonak
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.José Carlos Santos –José Carlos Santos 2019-10-23 09:31:16 +00:00 Commented Oct 23, 2019 at 9:31
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From Brahmagupta-Fibonacci Identity:
(p 2 s 2+p 1 s 1)2+(p 1 s 2−s 1 p 2)2=(p 2 2+p 2 1)(s 2 2+s 2 1)(p 2 s 2+p 1 s 1)2+(p 1 s 2−s 1 p 2)2=(p 2 2+p 1 2)(s 2 2+s 1 2)
get:
(p 2 s 2 t 2+p 1 s 1 t 2+p 1 t 1 s 2−s 1 t 1 p 2)2+(p 1 s 2 t 2−s 1 p 2 t 2−t 1 p 2 s 2−p 1 s 1 t 1)2=(p 2 2+p 2 1)(s 2 2+s 2 1)(t 2 2+t 2 1)(p 2 s 2 t 2+p 1 s 1 t 2+p 1 t 1 s 2−s 1 t 1 p 2)2+(p 1 s 2 t 2−s 1 p 2 t 2−t 1 p 2 s 2−p 1 s 1 t 1)2=(p 2 2+p 1 2)(s 2 2+s 1 2)(t 2 2+t 1 2)
and get solution:
(p 2 t 2 2+2 p 1 t 1 t 2−t 2 1 p 2)2+(p 1 t 2 2−2 t 1 p 2 t 2−p 1 t 2 1)2=(p 2 2+p 2 1)(t 2 2+t 2 1)2(p 2 t 2 2+2 p 1 t 1 t 2−t 1 2 p 2)2+(p 1 t 2 2−2 t 1 p 2 t 2−p 1 t 1 2)2=(p 2 2+p 1 2)(t 2 2+t 1 2)2
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answered May 2, 2016 at 15:54
AlexSamAlexSam
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13570 | https://dermnetnz.org/topics/the-structure-of-normal-skin | Are you a healthcare professional
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Structure of normal skin
Structure of normal skin — extra information
Synonyms:
Cutaneous anatomy
Categories:
Terminology
SNOMED CT:
55988001, 20281003, 49874001, 4360006, 62650009, 74447004, 9683001, 76322003, 10149001, 75608000, 53534000,71966008, 67769002, 24826007, 386121006, 62183001, 55696009, 7748002
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Terminology
Structure of normal skin
Author: Dr Anthony Yung, Dermatologist, Waikato District Health Board, Hamilton, New Zealand. 2007.
The layers of the skin A. Epidermis B. Dermis C. Subcutis
The layers of the skin
From top to bottom, the skin consists of 3 layers:
Epidermis
Dermis
Subcutis
Skin structure
A. Epidermis
The epidermis is the uppermost or epithelial layer of the skin. It acts as a physical barrier, preventing loss of water from the body, and preventing entry of substances and organisms into the body. Its thickness varies according to the body site.
The epidermis consists of stratified squamous epithelium. That means it consists of layers of flattened cells.
Skin, hair and nails are keratinised, meaning they have a dead and hardened impermeable surface made of a protein called keratin.
Mucous membranes are non-keratinised and moist.
The epidermis has three main types of cell:
Keratinocytes (skin cells)
Melanocytes (pigment-producing cells)
Langerhans cells (immune cells).
Special stains are often required to tell the difference between melanocytes and Langerhans cells. The Merkel cell is a fourth, less visible, epidermal cell.
The epidermis forms an undulating appearance, with intermittent regular protrusions of the epidermis layer (rete ridges or pegs) into the upper layers of the underlying dermis. In some areas of the body such as the palms and soles, the rete pegs are less pronounced. The pillars of dermis next to the dermal papillae form the rete ridges. The small area of epidermis between rete pegs is called the suprapapillary plate.
Keratinocytes
The keratinocytes become more mature or differentiated and accumulate keratin as they move outwards. They eventually fall or rub off. They form four distinct layers, described in the table below from the most superficial to the deepest.
| Layer | Cell type |
--- |
| Stratum corneum (horny layer) | Called corneocytes or squames. Dead, dried-out hard cells without nuclei. |
| Stratum granulosum (granular layer) | Cells contain basophilic granules. Waxy material is secreted into the intercellular spaces. |
| Stratum spinulosum (spinous, spiny or prickle cell layer) | Intercellular bridges called desmosomes link the cells together. The cells become increasingly flattened as they move upward. |
| Stratum basale (basal layer) | Columnar (tall) regenerative cells. As the basal cell divides, a daughter cell migrates upwards to replenish the layer above. |
Immediately below the epidermis is the basement membrane, a specialised structure that lies between the epidermis and dermis. It includes various protein structures linking the basal layer of keratinocytes to the basement membrane (hemidesmosomes) and the basement membrane to the underlying dermis (anchoring fibrils). The basement membrane has an important role in making sure the epidermis sticks tightly to the underlying dermis.
The epidermis gives rise to a number of specialised appendages also called adnexal structures or adnexae. Hair and nails are both examples, i.e. they are specialised structures formed by the direct extension of the epidermis. The hair follicles are associated with sebaceous (oil) glands and arrector pili smooth muscle. This muscle is responsible for goose bumps appearing on the skin in response to cold.
The epidermis also gives rise to eccrine (sweat) glands, a tangle of tubules deep within the dermis that secrete a watery salt solution into a duct that ends on the skin surface. Larger apocrine sweat glands are found in the armpits and groin.
Different areas of the body have different proportions of the adnexal and hair follicle structures present. For example:
Dense hair on the scalp and none on the palms
Intense sweating from armpits, palms and soles compared with elsewhere.
Melanocytes
Melanocytes are found in the basal layer of the epidermis. These cells produce a pigment called melanin, which is responsible for different skin colour. Melanin is packaged into small parcels (or melanosomes), which are then transferred to keratinocytes.
Langerhans cells
Langerhans cells are immune cells found in the epidermis and are responsible for helping the body learn and later recognise new ‘allergens’ (material foreign to the body).
Langerhans cells break the allergen into smaller pieces then migrate from the epidermis into the dermis. They find their way to lymphatics and blood vessels before eventually reaching the lymph nodes. Here they present the allergen to immune cells called lymphocytes. Once the allergen is successfully ‘presented’, the lymphocytes initiate a sequence of events to: (1) initiate an immune reaction to destroy the material, and (2) stimulate proliferation of more lymphocytes that recognise and remember the allergen in the future.
Merkel cells
Merkel cells are cells found in the basal layer of the epidermis. Their exact role and function are not well understood. Special immunohistochemical stains are needed to visualise Merkel cells.
B. Dermis
The dermis is the fibrous connective tissue or supportive layer of the skin. The major fibres are:
Collagen fibres: this type of fibre predominates in the dermis. Collagen fibres have enormous tensile strength and provide the skin with strength and toughness. Collagen bundles are small in the upper or papillary dermis and form thicker bundles in the deeper or reticular dermis.
Elastin: this type of fibre provides the properties of elasticity and pliability to the skin.
The collagen and elastin fibres are bound together by ground substance, a mucopolysaccharide gel in which the nutrients and wastes can diffuse to and from other tissue components. The dermis also contains nerves, blood vessels, epidermal adnexal structures (as described above), and cells.
The normal cells in the dermis include:
Mast cells. These contain granules packed with histamine and other chemicals, released when the cell is disturbed.
Vascular smooth muscle cells. These allow blood vessels to contract and dilate, required to control body temperature.
Specialised muscle cells. For example, myoepithelial cells are found around sweat glands and contract to expel sweat.
Fibroblasts. These are cells that produce and deposit collagen and other elements of the dermis as required for growth or to repair wounds. A resting fibroblast has very little cytoplasm compared with an active cell and appears to have a ‘naked’ nucleus.
Immune cells. There are many types of immune cell. The role of tissue macrophages (histiocytes) is to remove and digest foreign or degraded material (this is known as phagocytosis). There are also small numbers of lymphocytes in the normal dermis.
Transient inflammatory cells or leukocytes are white cells that leave the blood vessels to heal wounds, destroy infections or cause disease. They include:
Neutrophils (polymorphs). These have segmented nuclei. They are the first white blood cells to enter tissue during acute inflammation.
T and B Lymphocytes. These are small inflammatory cells with many subtypes. They arrive later but persist for longer in inflammatory skin conditions. They are important in the regulation of immune response. Plasma cells are specialised lymphocytes that produce antibody.
Eosinophils. These have bilobed nuclei and pink cytoplasm on H&E stain.
Monocytes. These form macrophages.
The skin cells communicate by releasing large numbers of biologically active cytokines and chemotactic factors that regulate their function and movement. These are too small to see on light microscopy.
C. Subcutis
The subcutis is the fat layer immediately below the dermis and epidermis. It is also called subcutaneous tissue, hypodermis or panniculus.
The subcutis mainly consists of fat cells (adipocytes), nerves and blood vessels. Fat cells are organised into lobules, which are separated by structures called septae. The septae contain nerves, larger blood vessels, fibrous tissue and fibroblasts. Fibrous septae may form dimples in the skin (so-called cellulite).
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References
Skin Pathology (2nd edition, 2002). Weedon D.
Pathology of the Skin (3rd edition, 2005). McKee PH, J. Calonje JE, Granter SR.
Primer of Dermatopathology (3rd edition, 2002). Hood A, Kwan TH, Mihm MC, Horn, TD, Smoller BR.
On DermNet
Dermatopathology
Dermatopathological glossary
Dermatopathology online course for health professionals
Dermatopathology index
Other websites
Introduction to Skin Histology — Southern Illinois University School of Medicine Year One Curriculum.
Anatomy and Histology of Normal Skin — Lyola University Dermatology Medical Education Website.
Books about skin diseases
Books about the skin
Dermatology Made Easy - second edition
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If 2+4+6+…+k=210, what is the value of k?
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1y
To solve the equation 2+4+6+…+k=210 2+4+6+…+k=210, we first recognize that this is the sum of the first n n even numbers. The sum of the first n n even numbers can be expressed as:
S n=2+4+6+…+2 n=n(n+1)S n=2+4+6+…+2 n=n(n+1)
In this case, we set the sum equal to 210:
n(n+1)=210 n(n+1)=210
To find n n, we can rearrange this into a standard quadratic equation:
n 2+n−210=0 n 2+n−210=0
Now, we can apply the quadratic formula, n=−b±√b 2−4 a c 2 a n=−b±b 2−4 a c 2 a, where a=1 a=1, b=1 b=1, and c=−210 c=−210:
b 2−4 a c=1 2−4(1)(−210)=1+840=841 b 2−4 a c=1 2−4(1)(−210)=1+840=841
Now, we take the square root of 841:
√841=29 841=29
Now, substituting back into the q
Continue Reading
To solve the equation 2+4+6+…+k=210 2+4+6+…+k=210, we first recognize that this is the sum of the first n n even numbers. The sum of the first n n even numbers can be expressed as:
S n=2+4+6+…+2 n=n(n+1)S n=2+4+6+…+2 n=n(n+1)
In this case, we set the sum equal to 210:
n(n+1)=210 n(n+1)=210
To find n n, we can rearrange this into a standard quadratic equation:
n 2+n−210=0 n 2+n−210=0
Now, we can apply the quadratic formula, n=−b±√b 2−4 a c 2 a n=−b±b 2−4 a c 2 a, where a=1 a=1, b=1 b=1, and c=−210 c=−210:
b 2−4 a c=1 2−4(1)(−210)=1+840=841 b 2−4 a c=1 2−4(1)(−210)=1+840=841
Now, we take the square root of 841:
√841=29 841=29
Now, substituting back into the quadratic formula:
n=−1±29 2 n=−1±29 2
Calculating the two possible values for n n:
n=28 2=14 n=28 2=14
n=−30 2=−15 n=−30 2=−15 (not applicable since n n must be positive)
Thus, the value of n n is 14. The largest even number k k is given by:
k=2 n=2×14=28 k=2 n=2×14=28
Therefore, the value of k k is:
28 28
Upvote ·
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Paul Dennee
PhD Mathematics from University of Notre Dame · Author has 2.2K answers and 1.5M answer views
·Updated 1y
Use 1 + 2 + 3 + … + n = (1/2)(n)(n+1).
2 + 4 + 6 + … + k = 210
2 (1 + 2 + 3 + … + k/2) = 210
2 (1/2)(k/2)(k/2 + 1) = 210
(k/2)(1/2)(k + 2) = 210
Multiply both sides by 4.
k(k + 2) = 840
k² + 2k - 840 = 0
(k - 28)(k + 30) = 0
k - 28 = 0 or k + 30 = 0
k = 28 or k = -30
The positive root is k = 28.
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George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·9mo
2+ 4+ 6+ …+k= 2(1+ 2+ 3+ …+k/2)= 210
1+ 2+ 3+…+k/2= 105.
To add
S= 1+ 2+ 3+ …+ (N-2)+ (N- 1)+ N, reverse the order:
S= N+ (N-1)+ (N- 2)+ 3+ 2+ 1, and add vertically
2S= (N+1)+ (N+ 1)+ …+ (N+1)+ (N+ 1)= N+ 1 N times.
2S= N(N+ 1).15
S= N(N+ 1)/2= 105. N(N+ 1)= 210= 1415.
k/2= 14 so k= 28.
Upvote ·
Raymond Beck
Former Former Army Sergeant · Author has 10.2K answers and 1.2M answer views
·9mo
AP, d=2, a1=2, ak = 2+2(k-1) = 2k
sk = 210 = (k/2)(2+2k)= k(1+k)
k^2 +k -210 = 0
(k+15)(k-14) = 0
k =14, -15, ignore the negative solution
2+4+6+ … 22+24+26+28 = 210
s14 = (14/2)(2+28) = 7(30) = 210
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Raymond Beck
Former Infantry Sergeant · Author has 42.4K answers and 10.3M answer views
·8mo
sk = (k/4)(2+k) = 210
k/2 + .25k^2 = 210
k^2 +2k -840 = 0
(k+30)(k-28) = 0, k=28
k = -1 +/-.(1/2)sqr(4+3360)
= -1+/-.5sqr3364
=-1+/-29 = 28 or - 30
k = 28
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Quera Question for You.
Deli news and short questions · Author has 58 answers and 16.6K answer views
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What is the value of 2+4+6+…+200?
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·3y
2+4+6+⋯+k=210 2+4+6+⋯+k=210
Let k = 2n
∴2+4+6+⋯+2 n=210∴2+4+6+⋯+2 n=210
⟹2(1+2+3+⋯+n)=210⟹2(1+2+3+⋯+n)=210
⟹2 n(n+1)2=210⟹2 n(n+1)2=210
⟹n(n+1)=14(15)⟹n(n+1)=14(15)
⟹n=14⟹n=14
k = 2n = 2(14) = 28
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9 1
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by
Alon Amit
, Lover of math. Also, Ph.D. and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 8.5K answers and 21.1M answer views
·1y
Related
How do you evaluate ∏∞k=0(1−2 2 k 4 2 k+1)∏k=0∞(1−2 2 k 4 2 k+1)?
We want to evaluate the infinite product
P=∞∏k=0(1−2 2 k 4 2 k+1).P=∏k=0∞(1−2 2 k 4 2 k+1).
To this end, we simplify the factors inside of the infinite product. Noting that 4 2 k=2 2 k+1 4 2 k=2 2 k+1, we obtain
P=∞∏k=0 2 2 k+1−2 2 k+1 2 2 k+1+1.P=∏k=0∞2 2 k+1−2 2 k+1 2 2 k+1+1.
Next, we multiply the numerator and denominator by strategic factors to take advantage of difference of two squares:
[Math Processing Error]\begin{align}P&=\displaystyle\prod_{k=0}^{\infty}\Big(\frac{2^{2^{k+1}}-2^{2^k}+1}{2^{2^{k+1}}+1}\cdot
Continue Reading
We want to evaluate the infinite product
P=∞∏k=0(1−2 2 k 4 2 k+1).P=∏k=0∞(1−2 2 k 4 2 k+1).
To this end, we simplify the factors inside of the infinite product. Noting that 4 2 k=2 2 k+1 4 2 k=2 2 k+1, we obtain
P=∞∏k=0 2 2 k+1−2 2 k+1 2 2 k+1+1.P=∏k=0∞2 2 k+1−2 2 k+1 2 2 k+1+1.
Next, we multiply the numerator and denominator by strategic factors to take advantage of difference of two squares:
P=∞∏k=0(2 2 k+1−2 2 k+1 2 2 k+1+1⋅(2 2 k+1+2 2 k+1)(2 2 k+1−1)(2 2 k+1+2 2 k+1)(2 2 k+1−1))=∞∏k=0(2 2 k+2+2 2 k+1+1)(2 2 k+1−1)(2 2 k+2−1)(2 2 k+1+2 2 k+1).P=∏k=0∞(2 2 k+1−2 2 k+1 2 2 k+1+1⋅(2 2 k+1+2 2 k+1)(2 2 k+1−1)(2 2 k+1+2 2 k+1)(2 2 k+1−1))=∏k=0∞(2 2 k+2+2 2 k+1+1)(2 2 k+1−1)(2 2 k+2−1)(2 2 k+1+2 2 k+1).
This results in a telescoping product. Noting that most factors cancel in pairs, except for some of the factors from the initial and final indices, we deduce that
P=lim N→∞N∏k=0(2 2 k+2+2 2 k+1+1 2 2 k+1+2 2 k+1⋅2 2 k+1−1 2 2 k+2−1)=lim N→∞(2 2 N+2+2 2 N+1+1 2 2 0+1+2 2 0+1⋅2 2 0+1−1 2 2 N+2−1)=3 7⋅lim N→∞1+2−2 N+1+2−2 N+2 1−2−2 N+2=3 7.P=lim N→∞∏k=0 N(2 2 k+2+2 2 k+1+1 2 2 k+1+2 2 k+1⋅2 2 k+1−1 2 2 k+2−1)=lim N→∞(2 2 N+2+2 2 N+1+1 2 2 0+1+2 2 0+1⋅2 2 0+1−1 2 2 N+2−1)=3 7⋅lim N→∞1+2−2 N+1+2−2 N+2 1−2−2 N+2=3 7.
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99 65
Goutam Mandal
2y
Related
What is the value of 2+4+6+…+200?
2+4+6+….+200
=2(1+2+3+…100) [we know that sum of first 'n' numbers is n(n+1)/2]
=2[100×101/2]
=10100(ans)
Upvote ·
HD ENJOY
Bsc from Indian Certificate of Secondary Education (ICSE) (Graduated 2022)
·Updated 3y
Related
How do you find the value of k if, “K+6=2K+1” then k=?
well hello ! I am going direct to the answer .
To Find the value of "k" I am going to tell you an easy process but for that I hope you have known some basics.
lets start-
basics - when we interchange number along "=" (equal to), their sign will Change .
e.g-1-- If we take -6 from the left side of "=" to right side then it becomes +6 .
e.g-2--lets take "k" From right to left, K becomes -K .
(Same here) (and here normal k mean + K and it is applicable to every type of numbers )
e.g-3-- If we take -3k It will become +3k, in any direction , hope you get it !
Now lets Start solving your question.
K+6=2k+1
th
Continue Reading
well hello ! I am going direct to the answer .
To Find the value of "k" I am going to tell you an easy process but for that I hope you have known some basics.
lets start-
basics - when we interchange number along "=" (equal to), their sign will Change .
e.g-1-- If we take -6 from the left side of "=" to right side then it becomes +6 .
e.g-2--lets take "k" From right to left, K becomes -K .
(Same here) (and here normal k mean + K and it is applicable to every type of numbers )
e.g-3-- If we take -3k It will become +3k, in any direction , hope you get it !
Now lets Start solving your question.
K+6=2k+1
then , lets keep all those "K" related Stuff in one side and all those numbers the othe side
+6-1=2K-K
{I have followed those basics here, like +1 from right to left becomes -1 and K (which means +K ) becomes -K }.
now
5= K or K=5.
( here I had done Some Calculation only , 6-1=5 and 2K-k=K )
So, the value of K Found is 5 .
Upvote ·
9 1
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99 14
Namig J. Guliyev
Works at Baku State University · Upvoted by
Rene Cabrera
, PhD Mathematics, University of Massachusetts, Amherst (2022) and
Allan Steinhardt
, PhD EE, published in various math journals, inventor, hyperbolic Householder
·10y
Related
How do you evaluate ∏∞k=0(1−2 2 k 4 2 k+1)∏k=0∞(1−2 2 k 4 2 k+1)?
Let's denote
a k:=4 2 k−1 a k:=4 2 k−1
and
b k:=4 2 k+2 2 k+1.b k:=4 2 k+2 2 k+1.
Then
1−2 2 k 4 2 k+1=4 2 k−2 2 k+1 4 2 k+1 1−2 2 k 4 2 k+1=4 2 k−2 2 k+1 4 2 k+1
=(4 2 k−2 2 k+1)(4 2 k+2 2 k+1)(4 2 k−1)(4 2 k+1)(4 2 k−1)(4 2 k+2 2 k+1)=(4 2 k−2 2 k+1)(4 2 k+2 2 k+1)(4 2 k−1)(4 2 k+1)(4 2 k−1)(4 2 k+2 2 k+1)
=(16 2 k+4 2 k+1)(4 2 k−1)(16 2 k−1)(4 2 k+2 2 k+1)=a k b k+1 b k a k+1=(16 2 k+4 2 k+1)(4 2 k−1)(16 2 k−1)(4 2 k+2 2 k+1)=a k b k+1 b k a k+1
and
∏n k=0(1−2 2 k 4 2 k+1)∏k=0 n(1−2 2 k 4 2 k+1)
=a 0 b 1 b 0 a 1⋅a 1 b 2 b 1 a 2⋅⋯⋅a n b n+1 b n a n+1=a 0 b n+1 b 0 a n+1=a 0 b 1 b 0 a 1⋅a 1 b 2 b 1 a 2⋅⋯⋅a n b n+1 b n a n+1=a 0 b n+1 b 0 a n+1
=\frac{(4^{2^0}-1)(16^{2^n}+4^{2^n}+1)}{(4^{2^0}+2^{2^0}+1)(16=\frac{(4^{2^0}-1)(16^{2^n}+4^{2^n}+1)}{(4^{2^0}+2^{2^0}+1)(16
Continue Reading
Let's denote
a k:=4 2 k−1 a k:=4 2 k−1
and
b k:=4 2 k+2 2 k+1.b k:=4 2 k+2 2 k+1.
Then
1−2 2 k 4 2 k+1=4 2 k−2 2 k+1 4 2 k+1 1−2 2 k 4 2 k+1=4 2 k−2 2 k+1 4 2 k+1
=(4 2 k−2 2 k+1)(4 2 k+2 2 k+1)(4 2 k−1)(4 2 k+1)(4 2 k−1)(4 2 k+2 2 k+1)=(4 2 k−2 2 k+1)(4 2 k+2 2 k+1)(4 2 k−1)(4 2 k+1)(4 2 k−1)(4 2 k+2 2 k+1)
=(16 2 k+4 2 k+1)(4 2 k−1)(16 2 k−1)(4 2 k+2 2 k+1)=a k b k+1 b k a k+1=(16 2 k+4 2 k+1)(4 2 k−1)(16 2 k−1)(4 2 k+2 2 k+1)=a k b k+1 b k a k+1
and
∏n k=0(1−2 2 k 4 2 k+1)∏k=0 n(1−2 2 k 4 2 k+1)
=a 0 b 1 b 0 a 1⋅a 1 b 2 b 1 a 2⋅⋯⋅a n b n+1 b n a n+1=a 0 b n+1 b 0 a n+1=a 0 b 1 b 0 a 1⋅a 1 b 2 b 1 a 2⋅⋯⋅a n b n+1 b n a n+1=a 0 b n+1 b 0 a n+1
=(4 2 0−1)(16 2 n+4 2 n+1)(4 2 0+2 2 0+1)(16 2 n−1)→3 7,n→∞.=(4 2 0−1)(16 2 n+4 2 n+1)(4 2 0+2 2 0+1)(16 2 n−1)→3 7,n→∞.
Upvote ·
99 52
Dominic Shum
Studied Mathematics at Wa Ying College (Graduated 2018) · Upvoted by
Dhruv Arora
, M.Sc. Mathematics, Indian Institute of Technology, Bombay (2021) · Author has 303 answers and 297.3K answer views
·Updated 6y
Related
If 1+2+…+k=325, what is k?
1+2+⋯+k=325 k(k+1)2=325 k 2+k−650=0(k+0.5)2−0.25−650=0 k+0.5=±√650.25 k=25 or−26(rejected)1+2+⋯+k=325 k(k+1)2=325 k 2+k−650=0(k+0.5)2−0.25−650=0 k+0.5=±650.25 k=25 or−26(rejected)
Done! ✔
Upvote ·
99 22
9 3
John Fryer
Author has 890 answers and 885.7K answer views
·4y
Related
For which values of k is the set of order pairs {(2,4),(k,6),(4,k)} a function?
We get a straight line function if k is chosen carefully
slope is given by change in y divided by change in x and so we have
(6–4)/(k-2) = ( k-6)/(4-k)
(6–4)(4-k) =(k-6)(k-2)
8 - 2k = k² - 2k - 6k + 12
k² - 6k + 4 = 0
Solving we get k = 3 + / - √5
Answer
k = 3 - √5 or
k = 3 + √5
Continue Reading
We get a straight line function if k is chosen carefully
slope is given by change in y divided by change in x and so we have
(6–4)/(k-2) = ( k-6)/(4-k)
(6–4)(4-k) =(k-6)(k-2)
8 - 2k = k² - 2k - 6k + 12
k² - 6k + 4 = 0
Solving we get k = 3 + / - √5
Answer
k = 3 - √5 or
k = 3 + √5
Upvote ·
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13572 | https://courses.lumenlearning.com/ivytech-sci111/chapter/specific-heat/ | Module 3 Specific Heat
Heat Capacity
The heat capacity measures the amount of heat necessary to raise the temperature of an object or system by one degree Celsius.
Learning Objectives
Explain the enthalpy in a system with constant volume and pressure
Key Takeaways
Key Points
Heat capacity is the measurable physical quantity that characterizes the amount of heat required to change a substance’s temperature by a given amount. It is measured in joules per Kelvin and given by.
The heat capacity is an extensive property, scaling with the size of the system.
The heat capacity of most systems is not constant (though it can often be treated as such). It depends on the temperature, pressure, and volume of the system under consideration.
Key Terms
heat capacity: The amount of heat energy needed to raise the temperature of an object or unit of matter by one degree Celsius; in units of joules per kelvin (J/K).
enthalpy: the total amount of energy in a system, including both the internal energy and the energy needed to displace its environment
Heat Capacity
Heat capacity (usually denoted by a capital C, often with subscripts), or thermal capacity, is the measurable physical quantity that characterizes the amount of heat required to change a substance’s temperature by a given amount. In SI units, heat capacity is expressed in units of joules per kelvin (J/K).
An object’s heat capacity (symbol C) is defined as the ratio of the amount of heat energy transferred to an object to the resulting increase in temperature of the object.
C=QΔT.
Heat capacity is an extensive property, so it scales with the size of the system. A sample containing twice the amount of substance as another sample requires the transfer of twice as much heat (Q) to achieve the same change in temperature (ΔT). For example, if it takes 1,000 J to heat a block of iron, it would take 2,000 J to heat a second block of iron with twice the mass as the first.
The Measurement of Heat Capacity
The heat capacity of most systems is not a constant. Rather, it depends on the state variables of the thermodynamic system under study. In particular, it is dependent on temperature itself, as well as on the pressure and the volume of the system, and the ways in which pressures and volumes have been allowed to change while the system has passed from one temperature to another. The reason for this is that pressure-volume work done to the system raises its temperature by a mechanism other than heating, while pressure-volume work done by the system absorbs heat without raising the system’s temperature. (The temperature dependence is why the definition a calorie is formally the energy needed to heat 1 g of water from 14.5 to 15.5 °C instead of generally by 1 °C. )
Different measurements of heat capacity can therefore be performed, most commonly at constant pressure and constant volume. The values thus measured are usually subscripted (by p and V, respectively) to indicate the definition. Gases and liquids are typically also measured at constant volume. Measurements under constant pressure produce larger values than those at constant volume because the constant pressure values also include heat energy that is used to do work to expand the substance against the constant pressure as its temperature increases. This difference is particularly notable in gases where values under constant pressure are typically 30% to 66.7% greater than those at constant volume.
Specific Heat
The specific heat is an intensive property that describes how much heat must be added to a particular substance to raise its temperature.
Learning Objectives
Summarize the quantitative relationship between heat transfer and temperature change
Key Takeaways
Key Points
Unlike the total heat capacity, the specific heat capacity is independent of mass or volume. It describes how much heat must be added to a unit of mass of a given substance to raise its temperature by one degree Celsius. The units of specific heat capacity are J/(kg °C) or equivalently J/(kg K).
The heat capacity and the specific heat are related by C=cm or c=C/m.
The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT.
Values of specific heat are dependent on the properties and phase of a given substance. Since they cannot be calculated easily, they are empirically measured and available for reference in tables.
Key Terms
specific heat capacity: The amount of heat that must be added (or removed) from a unit mass of a substance to change its temperature by one degree Celsius. It is an intensive property.
Specific Heat
The heat capacity is an extensive property that describes how much heat energy it takes to raise the temperature of a given system. However, it would be pretty inconvenient to measure the heat capacity of every unit of matter. What we want is an intensive property that depends only on the type and phase of a substance and can be applied to systems of arbitrary size. This quantity is known as the specific heat capacity (or simply, the specific heat), which is the heat capacity per unit mass of a material. Experiments show that the transferred heat depends on three factors: (1) The change in temperature, (2) the mass of the system, and (3) the substance and phase of the substance. The last two factors are encapsulated in the value of the specific heat.
Heat Transfer and Specific Heat Capacity: The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved. (a) The amount of heat transferred is directly proportional to the temperature change. To double the temperature change of a mass m, you need to add twice the heat. (b) The amount of heat transferred is also directly proportional to the mass. To cause an equivalent temperature change in a doubled mass, you need to add twice the heat. (c) The amount of heat transferred depends on the substance and its phase. If it takes an amount Q of heat to cause a temperature change ΔT in a given mass of copper, it will take 10.8 times that amount of heat to cause the equivalent temperature change in the same mass of water assuming no phase change in either substance.
Specific Heat Capacity: This lesson relates heat to a change in temperature. We discuss how the amount of heat needed for a temperature change is dependent on mass and the substance involved, and that relationship is represented by the specific heat capactiy of the substance, C.
The dependence on temperature change and mass are easily understood. Because the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Since the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid).
The quantitative relationship between heat transfer and temperature change contains all three factors:
Q=mcΔT,
where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase.
The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific heat c is a property of the substance; its SI unit is J/(kg⋅K) or J/(kg⋅C). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. Note that the total heat capacity C is simply the product of the specific heat capacity c and the mass of the substance m, i.e.,
C=mc or c=Cm=CρV,
where ϱ is the density of the substance and V is its volume.
Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. Instead, they are measured empirically. In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. The specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.
Specific Heats: Listed are the specific heats of various substances. These values are identical in units of cal/(g⋅C).3. cv at constant volume and at 20.0ºC, except as noted, and at 1.00 atm average pressure. Values in parentheses are cp at a constant pressure of 1.00 atm.
Calorimetry
Calorimetry is the measurement of the heat of chemical reactions or physical changes.
Learning Objectives
Analyze the relationship between the gas constant for an ideal gas yield and volume
Key Takeaways
Key Points
A calorimeter is used to measure the heat generated (or absorbed) by a physical change or chemical reaction. The science of measuring these changes is known as calorimetry.
In order to do calorimetry, it is crucial to know the specific heats of the substances being measured.
Calorimetry can be performed under constant volume or constant pressure. The type of calculation done depends on the conditions of the experiment.
Key Terms
constant-pressure calorimeter: An instrument used to measure the heat generated during changes that do not involve changes in pressure.
calorimeter: An apparatus for measuring the heat generated or absorbed by either a chemical reaction, change of phase or some other physical change.
constant-volume calorimeter: An instrument used to measure the heat generated during changes that do not involve changes in volume.
Calorimetry
Overview
Calorimetry is the science of measuring the heat of chemical reactions or physical changes. Calorimetry is performed with a calorimeter. A simple calorimeter just consists of a thermometer attached to a metal container full of water suspended above a combustion chamber. The word calorimetry is derived from the Latin word calor, meaning heat. Scottish physician and scientist Joseph Black, who was the first to recognize the distinction between heat and temperature, is said to be the founder of calorimetry.
Calorimetry requires that the material being heated have known thermal properties, i.e. specific heat capacities. The classical rule, recognized by Clausius and by Kelvin, is that the pressure exerted by the calorimetric material is fully and rapidly determined solely by its temperature and volume; this rule is for changes that do not involve phase change, such as melting of ice. There are many materials that do not comply with this rule, and for them, more complex equations are required than those below.
Ice Calorimeter: The world’s first ice-calorimeter, used in the winter of 1782-83, by Antoine Lavoisier and Pierre-Simon Laplace, to determine the heat evolved in variouschemical changes; calculations which were based on Joseph Black’s prior discovery of latent heat. These experiments mark the foundation of thermochemistry.
Basic Calorimetry at Constant Value
Constant-volume calorimetry is calorimetry performed at a constant volume. This involves the use of a constant-volume calorimeter (one type is called a Bomb calorimeter). For constant-volume calorimetry:
δQ=CVΔT=mcVΔT
where δQ is the increment of heat gained by the sample, CV is the heat capacity at constant volume, cv is the specific heat at constant volume, and ΔT is the change in temperature.
Measuring Enthalpy Change
To find the enthalpy change per mass (or per mole) of a substance A in a reaction between two substances A and B, the substances are added to a calorimeter and the initial and final temperatures (before the reaction started and after it has finished) are noted. Multiplying the temperature change by the mass and specific heat capacities of the substances gives a value for the energy given off or absorbed during the reaction:
δQ=ΔT(mAcA+mBcB)
Dividing the energy change by how many grams (or moles) of A were present gives its enthalpy change of reaction. This method is used primarily in academic teaching as it describes the theory of calorimetry. It does not account for the heat loss through the container or the heat capacity of the thermometer and container itself. In addition, the object placed inside the calorimeter shows that the objects transferred their heat to the calorimeter and into the liquid, and the heat absorbed by the calorimeter and the liquid is equal to the heat given off by the metals.
Constant-Pressure Calorimetry
A constant-pressure calorimeter measures the change in enthalpy of a reaction occurring in solution during which the atmospheric pressure remains constant. An example is a coffee-cup calorimeter, which is constructed from two nested Styrofoam cups and a lid with two holes, allowing insertion of a thermometer and a stirring rod. The inner cup holds a known amount of a solute, usually water, that absorbs the heat from the reaction. When the reaction occurs, the outer cup provides insulation. Then
CP=WΔHMΔT
where Cp is the specific heat at constant pressure, ΔH is the enthalpy of the solution, ΔT is the change in temperature, W is the mass of the solute, and M is the molecular mass of the solute. The measurement of heat using a simple calorimeter, like the coffee cup calorimeter, is an example of constant-pressure calorimetry, since the pressure (atmospheric pressure) remains constant during the process. Constant-pressure calorimetry is used in determining the changes in enthalpy occurring in solution. Under these conditions the change in enthalpy equals the heat (Q=ΔH).
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Relative Atomic Mass (GCSE Chemistry)
What is Relative Atomic Mass (RAM)?
The relative atomic mass (Ar) is the average mass of atoms of an element relative to the mass of an atom of carbon-12 (which is given a mass exactly of 12). The average mass must take into account the proportions of naturally occurring isotopes of the element.
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Scientists decided that the atomic mass of an atom of an element would be compared to carbon-12. Therefore the atomic mass is known as relative as it is being compared to the mass of carbon-12. If the Ar of an atom is lower than 12 it has a mass smaller than carbon-12 atom.
You can find the relative atomic mass of an element on a periodic table by looking at the number directly above the element symbol. For example the relative atomic mass of Copper (Cu) is 29.
How to Calculate Relative Atomic Mass
The relative atomic mass of an element is the average mass of an atom, and it takes into account the masses of each isotope and their proportions in the environment.
To calculate the relative atomic mass you require the following information:
The abundance of each isotope which tells you the quantity the isotope is present in the environment. This is different for each isotope of an element.
The mass numberof each isotope of a particular element. The number of protons in all the isotopes of an atom remains the same, it is only the number of neutrons.
Lets work through an example:
Practice Question: A sample of chlorine gas is a mixture of 2 isotopes, chlorine-35 and chlorine-37. These isotopes occur in specific proportions in the sample i.e. 75% chlorine-35 and 25% chlorine-37. Calculate the relative atomic mass of chlorine in the sample.
1. Identify the isotopes. Both isotopes present in the mixture or environment, and the abundance of each isotope of the element.
2. Identify the mass number. For each isotope of the element.
Write out the relative atomic mass formula to calculate relative atomic mass.
Substitute identified data. Substitute into the formula and work out the answer.
= 35.5
The relative atomic mass of chlorine in this mixture was 35.5. This number is closer to 35 compared to 37 as chlorine-35 is more abundant than chlorine-37.
We can do a sense-check, to make sure that the value seems right. 35.5 is in between 35 and 37, as we would expect, and it has a closer mass to Cl-35, which is the more abundant isotope. This seems fine!
You need to be able to rearrange the relative atomic mass equation to work out a missing abundance or mass.
Download Free GCSE Chemistry Notes
Worked example – relative atomic mass of oxygen. Oxygen has three isotopes. The abundances in percentage are given here. The mass of one of the isotopes is unknown. The average atomic mass of the isotopes is 16.65. Work out the values of x and y.
Answer:
1. Set up a table. Put in all the values as shown.
2. Work out x.
All percentages = 100, so x = 100-(50+35) =15%
3. Form an equation for y. Use the given atomic mass and the equation for calculating it.
16.65 = (800 + 595 + 15y) / 100
4. Rearrange to find y.
[16.65 x 100 – (800+595)] / 15 = y = 18
So the atomic mass of the unknown isotope = 18 and the abundance = 15%
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→What is the relative atomic mass?
Relative atomic mass is a measure of the average mass of an atom of an element compared to a standard unit of mass. It is used to compare the masses of different atoms and elements.
→How is relative atomic mass calculated?
Relative atomic mass is calculated by adding up the masses of all the isotopes of an element and then averaging them based on their relative abundance in nature. This gives us an estimate of the average mass of an atom of that element.
→Why is relative atomic mass important in chemistry?
Relative atomic mass is important in chemistry because it helps us understand the properties of different elements and how they behave in chemical reactions. It also helps us predict the masses of molecules and compounds made from different elements.
→How does relative atomic mass differ from atomic mass?
Atomic mass is the actual mass of an individual atom, whereas relative atomic mass is an average of the masses of all the isotopes of an element. Atomic mass is a specific value, whereas relative atomic mass is a relative value.
→How does relative atomic mass help in chemical calculations?
Relative atomic mass is used in chemical calculations to determine the number of moles of a substance, which can then be used to calculate the masses of other substances involved in a reaction. This information can be used to make predictions about the yields and outcomes of chemical reactions.
→What is the standard unit of relative atomic mass?
The standard unit of relative atomic mass is the atomic mass unit (amu), which is defined as 1/12 of the mass of a carbon-12 atom. This unit allows us to compare the masses of different atoms and elements on a relative scale.
→How does the periodic table use relative atomic mass?
The periodic table is arranged based on the relative atomic masses of the elements. This arrangement allows us to easily compare and predict the properties and behavior of different elements based on their position in the table.
→How does the relative atomic mass affect the reactivity of elements?
Elements with lower relative atomic masses tend to be more reactive than elements with higher relative atomic masses. This is because lighter elements have more electrons available to participate in chemical reactions, making them more reactive.
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13575 | https://math.stackexchange.com/questions/72367/proof-of-a-combinatorial-identity-sum-limits-i-0n-2i-choose-i2n-i-ch | combinatorics - Proof of a combinatorial identity: $\sum\limits_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$ - Mathematics Stack Exchange
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Proof of a combinatorial identity: ∑i=0 n(2 i i)(2(n−i)n−i)=4 n∑i=0 n(2 i i)(2(n−i)n−i)=4 n [duplicate]
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Identity involving binomial coefficients
This was part of a homework assignment that I had, and I couldn't figure it out. Now it is bugging me. Can anyone help me? Although a proof would be nice, I wouldn't mind a push in the right direction.
∑i=0 n(2 i i)(2(n−i)n−i)=4 n∑i=0 n(2 i i)(2(n−i)n−i)=4 n
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edited Aug 7, 2022 at 7:46
Martin Sleziak
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asked Oct 13, 2011 at 18:53
Kyle d'OliveiraKyle d'Oliveira
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1 See also math.stackexchange.com/questions/37971/…user940 –user940 2011-10-13 19:01:47 +00:00 Commented Oct 13, 2011 at 19:01
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Added: I have at long last got around to correcting the proof of the crucial lemma.
Here’s a purely combinatorial argument.
Let Σ 2 n Σ 2 n be the set of binary strings (strings of 0 0’s and 1 1’s) of length 2 n 2 n. Call a binary string balanced if it contains the same number of 0 0’s and 1 1’s. Let Σ B 2 n Σ 2 n B be the set of balanced members of Σ 2 n Σ 2 n, and let Σ U 2 n Σ 2 n U be the set of unbalanced strings in Σ 2 n Σ 2 n that have no balanced initial segment; I’ll call these completely unbalanced. Clearly ∣∣Σ B 2 n∣∣=(2 n n)|Σ 2 n B|=(2 n n).
For σ=⟨b 1,…,b 2 n⟩∈Σ 2 n σ=⟨b 1,…,b 2 n⟩∈Σ 2 n and 1≤i≤k≤2 n 1≤i≤k≤2 n let σ(i,k)=⟨b i,…,b k⟩σ(i,k)=⟨b i,…,b k⟩, and let e σ(i,k)e σ(i,k) be the excess of 1 1’s over 0 0’s in σ(i,k)σ(i,k). Let σ R(i,k)=⟨b k,…,b i⟩σ R(i,k)=⟨b k,…,b i⟩, the reversal of σ(i,k)σ(i,k). Finally, let σ¯(i,k)=⟨b¯i,…,b¯k⟩σ¯(i,k)=⟨b¯i,…,b¯k⟩, where b¯=1−b b¯=1−b for b∈{0,1}b∈{0,1}. Note that e σ¯(i,k)=−e σ(i,k)e σ¯(i,k)=−e σ(i,k).
Lemma:∣∣Σ U 2 n∣∣=∣∣Σ B 2 n∣∣|Σ 2 n U|=|Σ 2 n B|.
Corrected Proof: I’ll construct a bijection Σ B 2 n→Σ U 2 n:σ↦σ^Σ 2 n B→Σ 2 n U:σ↦σ^.
Fix σ=⟨b 1,…,b 2 n⟩∈Σ B 2 n σ=⟨b 1,…,b 2 n⟩∈Σ 2 n B. Let m=min{e σ(1,k):k=1,…,2 n}≤0 m=min{e σ(1,k):k=1,…,2 n}≤0. If m=0 m=0, let τ=σ τ=σ. Otherwise, let h h be the smallest index such that e σ(1,h)=m e σ(1,h)=m. Let τ=σ(h+1,2 n)σ¯R(1,h)τ=σ(h+1,2 n)σ¯R(1,h). That is, τ τ is obtained from σ σ by transferring the first h h bits to the end, reversing their order and complementing them in the process. Now e σ(1,2 n)=0 e σ(1,2 n)=0, so e σ(h+1,2 n)=−m e σ(h+1,2 n)=−m, and
e τ(1,2 n)=e σ(h+1,2 n)−e σ(1,h)=−2 m>0.e τ(1,2 n)=e σ(h+1,2 n)−e σ(1,h)=−2 m>0.
The choice of h h ensures that e τ(1,k)=e σ(h+1,h+k)≥0 e τ(1,k)=e σ(h+1,h+k)≥0 for k=1,…,2 n k=1,…,2 n. If e τ(1,k)>0 e τ(1,k)>0 for k=1,…,2 n k=1,…,2 n, let σ^=τ∈Σ U 2 n σ^=τ∈Σ 2 n U.
Otherwise, let j j be minimal such that e τ(1,j)=0 e τ(1,j)=0. If τ=⟨c 1,…,c 2 n⟩τ=⟨c 1,…,c 2 n⟩, then clearly c j=0 c j=0. Let
τ′=⟨c 1,…,c j−1,1,c j+1,…,c 2 n⟩;τ′=⟨c 1,…,c j−1,1,c j+1,…,c 2 n⟩;
then e τ′(1,k)>0 e τ′(1,k)>0 for k=1,…,2 n k=1,…,2 n. Finally, let σ^=τ′¯¯¯¯σ^=τ′¯; e σ^(1,k)<0 e σ^(1,k)<0 for k=1,…,2 n k=1,…,2 n, so τ^∈Σ U 2 n τ^∈Σ 2 n U.
Now fix σ=⟨b 1,…,b 2 n⟩∈Σ U 2 n σ=⟨b 1,…,b 2 n⟩∈Σ 2 n U. Suppose first that e σ(1,k)>0 e σ(1,k)>0 for k=1,…,2 n k=1,…,2 n. Note that e σ(1,2 n)e σ(1,2 n) is even, and let m=1 2 e σ(1,2 n)m=1 2 e σ(1,2 n). Let h h be the largest index such that e σ(1,h)=m e σ(1,h)=m, and let τ=σ¯R(h+1,2 n)σ(1,h)τ=σ¯R(h+1,2 n)σ(1,h). The choice of h h ensures that e σ(h+1,2 n)=m e σ(h+1,2 n)=m and that e σ(h+1,k)>0 e σ(h+1,k)>0 for k=h+1,…,2 n k=h+1,…,2 n, so e τ(1,2 n−h)=−m e τ(1,2 n−h)=−m, and e τ(k,2 n−h)>−m e τ(k,2 n−h)>−m for k=1,…,2 n−h k=1,…,2 n−h. Moreover, e σ(1,k)>0 e σ(1,k)>0 for k=1,…,h k=1,…,h, so e τ(1,h)=−m e τ(1,h)=−m is the minimum of e τ(1,k)e τ(1,k) for k=1,…,2 n k=1,…,2 n, and therefore σ=τ^σ=τ^.
Now suppose that e σ(1,k)<0 e σ(1,k)<0 for k=1,…,2 n k=1,…,2 n; clearly e σ¯(1,k)>0 e σ¯(1,k)>0 for k=1,…,2 n k=1,…,2 n. Let j j be the maximal index such that b¯j=1 b¯j=1 and e σ¯(1,j)=2 e σ¯(1,j)=2. Let σ′σ′ be obtained from σ¯σ¯ by replacing b¯j b¯j by 0 0. Then e σ′(1,k)>0 e σ′(1,k)>0 for k=1,…,j−1 k=1,…,j−1, e σ′(1,j)=0 e σ′(1,j)=0, and e σ′(1,k)≥0 e σ′(1,k)≥0 for k=j,…,2 n k=j,…,2 n. If e σ′(1,2 n)=0 e σ′(1,2 n)=0, let τ=σ′∈Σ B 2 n τ=σ′∈Σ 2 n B, and observe that σ=τ^σ=τ^. Otherwise, let m=1 2 e σ′(1,2 n)m=1 2 e σ′(1,2 n), and proceed as in the previous paragraph: take h h to be the largest index such that e σ′(1,h)=m e σ′(1,h)=m, and let τ=σ′¯¯¯¯¯R(h+1,2 n)σ′(1,h)τ=σ′¯R(h+1,2 n)σ′(1,h). As before, σ=τ^σ=τ^. The map σ↦σ^σ↦σ^ is therefore a bijection. ⊣⊣
For σ=⟨b 1,…,b 2 n⟩∈Σ 2 n σ=⟨b 1,…,b 2 n⟩∈Σ 2 n let m(σ)m(σ) be the largest k k such that ⟨b 1,…,b 2 k⟩⟨b 1,…,b 2 k⟩ is balanced, if such a k k exists; if not, let m(σ)=0 m(σ)=0. For k=1,…,2 n k=1,…,2 n let Σ 2 n(k)={σ∈Σ 2 n:m(σ)=k}Σ 2 n(k)={σ∈Σ 2 n:m(σ)=k}. Then σ∈Σ 2 n σ∈Σ 2 n belongs to Σ 2 n(k)Σ 2 n(k) iff ⟨b 1,…,b 2 k⟩⟨b 1,…,b 2 k⟩ is balanced, and ⟨b 2 k+1,…,b 2 n⟩⟨b 2 k+1,…,b 2 n⟩ is completely unbalanced. There are (2 k k)(2 k k) balanced strings of length 2 k 2 k, and by the lemma there are (2(n−k)n−k)(2(n−k)n−k) completely unbalanced strings of length 2 n−2 k 2 n−2 k, so
|Σ 2 n(k)|=(2 k k)(2(n−k)n−k).|Σ 2 n(k)|=(2 k k)(2(n−k)n−k).
Clearly Σ 2 n=⋃k=0 n Σ 2 n(k)Σ 2 n=⋃k=0 n Σ 2 n(k), where the sets Σ 2 n(k)Σ 2 n(k) are pairwise disjoint, so
|Σ 2 n|=∑k=0 n(2 k k)(2(n−k)n−k).|Σ 2 n|=∑k=0 n(2 k k)(2(n−k)n−k).
But of course |Σ 2 n(k)|=2 2 n=4 n|Σ 2 n(k)|=2 2 n=4 n, so we have the desired result:
∑k=0 n(2 k k)(2(n−k)n−k)=4 n.∑k=0 n(2 k k)(2(n−k)n−k)=4 n.
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edited Jun 12, 2020 at 10:38
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answered Oct 14, 2011 at 18:29
Brian M. ScottBrian M. Scott
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1 The "it's not too hard to see" part at the end of the lemma hides the hard (and perhaps incorrect) part of the proof. In particular, if σ=11111111 σ=11111111, then h(σ)=4 h(σ)=4 and σ¯=11110000 σ¯=11110000. But if σ=11110000 σ=11110000 then μ=4 μ=4 so i(σ)=3 i(σ)=3 and σ^=11101111 σ^=11101111. So the bijection isn't working. Also note that σ=11100111 σ=11100111 and σ=11011111 σ=11011111 both yield σ¯=11011000 σ¯=11011000.Matt –Matt 2011-12-03 23:21:53 +00:00 Commented Dec 3, 2011 at 23:21
@Matt: You’re quite correct, and I finally remembered to come back and fix it.Brian M. Scott –Brian M. Scott 2012-08-29 22:09:32 +00:00 Commented Aug 29, 2012 at 22:09
Yes, I agree, it all looks good now!Matt –Matt 2012-12-09 10:41:14 +00:00 Commented Dec 9, 2012 at 10:41
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Note that this is an autoconvolution (the convolution of a sequence with itself) of (2 k k)(2 k k). We can determine that the generating function of (2 k k)(2 k k) is
∑k=0∞(2 k k)x k=1 1−4 x−−−−−−√∑k=0∞(2 k k)x k=1 1−4 x
The generating function of the autoconvolution is then obtained by squaring the original generating function. Thus you have to determine the coefficients of the function 1 1−4 x 1 1−4 x, which can be recast as a geometric series...
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edited Apr 13, 2017 at 12:20
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answered Oct 13, 2011 at 23:10
J. M. ain't a mathematicianJ. M. ain't a mathematician
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Look at Doron Zeilberger's note.
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answered Oct 13, 2011 at 18:59
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103Identity for convolution of central binomial coefficients: ∑k=0 n(2 k k)(2(n−k)n−k)=2 2 n∑k=0 n(2 k k)(2(n−k)n−k)=2 2 n
10A combinatorial proof of the identity ∑k=0 n(2 k k)(2(n−k)n−k)=4 n∑k=0 n(2 k k)(2(n−k)n−k)=4 n
3Proving an Identity involving 4 N 4 N
6Binomial identity (0 0)(2 n n)+(2 1)(2 n−2 n−1)+(4 2)(2 n−4 n−2)+⋯+(2 n n)(0 0)=4 n.(0 0)(2 n n)+(2 1)(2 n−2 n−1)+(4 2)(2 n−4 n−2)+⋯+(2 n n)(0 0)=4 n.
1close form for ∑n k=0(2 n n)(2 n−2 k n−k)∑k=0 n(2 n n)(2 n−2 k n−k)
0Sum of a product with 2 binomial coefficients
75Combinatorial proof that ∑k=0 n(2 k k)(2 n−2 k n−k)(−1)k=2 n(n n/2)∑k=0 n(2 k k)(2 n−2 k n−k)(−1)k=2 n(n n/2) when n n is even
19Show ∑n=0∞(2 n n)x n=(1−4 x)−1/2∑n=0∞(2 n n)x n=(1−4 x)−1/2
9Tossing a coin until you have more heads than tails
8expected number of balls withdrawn to get equal numbers of black and white balls
See more linked questions
Related
1Prove: ∑n x=0(−1)x(n x)=0∑x=0 n(−1)x(n x)=0
3Unusual Combinatorial Identity (Alternating Sum of Binomial Products)
1Solve following summation using Binomial theorem
0Combinatorial identity involving binomial coefficients
3Proof of ∑n m=0(m j)(n−m k−j)=(n+1 k+1)∑m=0 n(m j)(n−m k−j)=(n+1 k+1) (Another form of the Chu–Vandermonde identity)
3Evaluating sum of binomial coefficients
1Combinatorial interpretation of alternating sum involving binomial coefficients
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13576 | https://www.openintro.org/book/os/whats_new/ | OpenIntro Statistics
OpenIntro Statistics
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OpenIntro Statistics Edition Details
Did you know that textbook publishers are legally required to share this type of information and disclose price changes when marketing a textbook?
We do so proudly on each OpenIntro book's page!
Below is an overview of what updates have been made from one edition to the next. Full details are logged in our GitHub repository for changes made from the 2nd to the current edition.
Availability of Past Editions
All editions are included as extras in the Leanpub purchase. Reminder: you may set the price to Free; contributions are 100% optional. We believe that students should have "forever access" to the book they used in their college course.
3rd to 4th Edition Updates
The 4th Edition was released on May 1st, 2019. The textbook price was updated from $14.99 for the 3rd Edition to $20 for the 4th Edition, which we believe will be a sustainable price point that helps support OpenIntro as it scales into new subjects.
Complete visual redesign. Complete style redesign of chapter, section, example, guided practice, and term box layouts. Sections now always start at the top of a page, making them easier to find. Page size is now 8.5" x 11" (up from 8" x 10"), allowing us to reduce the page count.
Structural and pedagogy. The content from Chapter 1 in the Third Edition is now Chapters 1 and 2. We eliminated normal probability (quantile-quantile) plots and replaced them by histograms. See the related change around the skew condition change below. Negative binomial and Poisson distributions are now in their own sections. Proportions are now used to introduce inference concepts, and the Inference for Categorical Data chapter comes before the Inference for Numerical Data chapter. We now only use two-sided hypothesis tests throughout the textbook. Small sample proportion inference in the Inference for Categorical Data chapter has been moved into an online extra. The skew condition check for inference for means has been simplified and been replaced with rules of thumb check for outliers. There are now two significant multiple regression case studies in the main text.
Data updates. Many older data sets have been replaced with new data sets. We've consolidated data sets information from the main text into a Data Appendix (Appendix B).
Exercise overhaul. Exercises are now at the end of sections, with a handful of additional review exercises at the end of each chapter. Dozens of new exercises, as well as removing many older exercises.
Price. The PDF is still free. B&W paperback retail price is $20, which is up from $14.99 for the 3rd Edition. You may see an increase to the price on Amazon relative to other retailers and bookstores, e.g. $25-30, if we see Amazon applying fee structures that handicap our ability to later sell full-color prints at reasonable prices when we try a new printer.
If you have questions related to the Fourth Edition, see this thread in the Public Forums and ask away!
The 3rd Edition was made available thru mid-December 2019 to allow a half year for instructors to switch to the new edition.
2nd to 3rd Edition Updates
The 3rd Edition of OpenIntro Statistics was released in July 2015. Pricing was changed from $9.94 for the 2nd Edition (purely at cost) to $14.99 for the 3rd Edition (provided a margin to cover project costs). Below are the significant changes. Full details are logged in our GitHub repository.
The t distribution is used exclusively in the Inference for Numerical Data chapter (Ch 5).
Some data sets are replaced with newer or more interesting data sets.
Some exercises are revamped, and some entirely new exercises have been added. The number of exercises has increased by a small amount.
Video icons were added to sections where videos are available, and these are hyperlinked in the PDF.
Section 8.2, which focuses on variable selection in multiple regresion, now highlights the usage of adjusted R-squared instead of p-values for variable selection. Both approaches are still featured, but adjusted R-squared is featured more prominently.
Section 3.2.2 in the Second Edition, which covers calculations for the normal probability plot, was eliminated. Additionally, the R function qqline() is now used throughout the textbook for the normal probability plot, whereas a custom function was previously used.
The within-text exercises (the ones with solutions in the footnotes) are now called Guided Practice to eliminate duplicate "exercise" numbering with end-of-chapter exercises.
We dropped the df > 1 condition for the chi-square test. The reasoning on why it was introduced is subtle: a 1-prop test can be turned into a chi-square test with df = 1, but then we only need 5 successes / failures for the 2-prop test, so the conditions don't match. Anyways, we removed this condition for simplicity.
The section on calculating power (Section 4.6) was removed and a new section was added in Chapter 5 that covers power in the context of an experiment. This new section should be much easier to follow, and the example is also a much more common power calculation context.
A reference to Why 0.05? was added.
There were other minor changes, e.g. fixing typos identified in the Second Edition.
The 2nd Edition was made available for 10 months after the release of the 3rd Edition.
1st to 2nd Edition Updates
The 2nd Edition of OpenIntro Statistics was released in August 2012. Pricing was changed from $9.02 for the 1st Edition to $9.94 for the 3rd Edition (both editions were priced at-cost). Below are the significant changes of the update from the 1st to the 2nd Edition. The blog post announcing the updates may be found here.
New data. Many of the data sets, some just one or two years old, have been swapped out for newer data and studies. We've worked hard to ensure that OpenIntro Statistics remains fresh and current.
Updated Chapter 1. Data collection is now featured ahead of the summaries and graphics sections. We include a new research study with surprising results to lead off the textbook and engage students. Two new data sets featuring email and census data take the place of the possum and cars data sets that are present in the First Edition. An important new subsection has also been added that includes intensity maps to highlight the structure of spatial data.
Chapter 5 and 6 updates. These chapters previously were structured around large and small samples. Chapter 5 will now introduce inference methods for numerical data, and Chapter 6 will feature methods for categorical data (proportions, contingency tables). The section on ANOVA will be moving from Chapter 8 to Chapter 5 and will remain a "special topic" section.
Sample size conditions have evolved slightly. We now set 30 as the standard for using the normal approximation for numerical data and 5 as the minimum expected cell count size for chi-square. With these changes, we also add more strict and explicit conditions regarding skew and table size to ensure the methods remain rigorous and appropriate.
New logistic regression section. Chapter 8 includes a brand new section featuring logistic regression in the context of developing a filter for email spam. Our final spam filter isn't on par with Gmail's spam filter, but we make a surprising amount of headway in just 9 pages. This new section also highlights ideas and methodology students could see in a second applied statistics course, which makes this new logistic regression section a nice closing act for the content of OpenIntro Statistics and one that is within reach of an advanced or honors introductory statistics course.
The 1sts Edition was made available for one academic year after the release of the 2nd Edition (discontinued in mid-2013).
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Cramer's Rule
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What is Cramer's Rule?
Cramer's Rule is a method of solving systems of linear equations by dividing the values of two determinants. One determinant comes from the coefficient matrix; the other is created by replacing the column for a particular variable with the "answer" column which represents the values that come after the "equals" sign in the system of equations.
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In practice, Cramer's Rule is a handy way to solve for just one of the variables in the system of equation without having to solve the entire system of equations; in other words, if you only need one variable from a system of, say, five equations in five unknowns, Cramer's Rule allows you to solve for just that one variable.
For some reason, they don't usually teach Cramer's Rule this way; they'll give you a system of equations and have you use Cramer's Rule to solve for each and every variable. But the point of the Rule is that, instead of solving the entire system of equations to make sure you get the value you need, you can instead save time and solve for just the one value that you need.
(I can be a bit dense some times. It wasn't until my physics professor pointed out that "this would be a good time to use that Cramer's Rule that they taught you" that it even dawned on me how helpful the Rule can be.)
How does Cramer's Rule work?
Cramer's Rule tells us to form certain determinants and divide them in order to find variables' values. The denominator of all of the divisions will be the determinant of the coefficient matrix. For the numerators, you will pick a variable; you will replace the coefficients of that variable's entries in the coefficient matrix with the "column vector" containing the constants on the other side of the "equals" sign in the original system of equations.
(I will refer to this column vector as the "answers column" and to the matrix and determinant created by replacing a column with the answers column as the "answers" matrix and determinant. This is not proper terminology, but I can't find any name for these.)
The value for a chosen variable is found by dividing the value of that variable's answers determinant by the value of the coefficient determinant.
To see how Cramer's Rule works, let's apply it to the following system of equations:
2x + y + z = 3
x − y − z = 0
x + 2y + z = 0
We have the left-hand side of the system with the variables; the square matrix containing the variables' coefficients is the coefficient matrix). On the right-hand side, after the "equals" sign, we have the answer values; the width-1 matrix containing these values is what I'm calling the "answer" column..
our system of equations, color-coded, with "understood" coefficients explicitly written in:
2x + 1y + 1z = 3
1x − 1y − 1z = 0
1x + 2y + 1z = 0
Let's name the determinant of the coefficient matrix of the above system "D":
coefficient matrix's determinant:
The column of answer values on the right-hand side of the "equals" signs in the system of equations above can be made into its own little matrix:
answer column:
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(Technically this is a "column vector", as mentioned above, but you almost certainly won't need to know that terminology right now, if ever. However, if your instructor uses this term, the tall skinny one-column matrix above is what is meant.)
We'll take the coefficient determinant D and replace the first column of values (being the blue coefficients from the x-column of the original system) with the answer values. We'll call the resulting answers determinant "Dx" (pronounced as "dee, sub-eks"), to remind us that it's the x coefficients that we replaced (and also that it's the value of x that we're looking to find):
the determinant with x-values replaced:
As you can see, the red values from the answer column have replaced the original blue values from the x-variable terms in the original system of equations.
Similarly, determinants with the other two variables' values replaced are formed and named similarly; namely, the determinants Dy and Dz for this system are:
...where the red answer values have replaced the y-values in the middle green column, and:
...where the red answer values have replaced the z-values in the right-hand purple column.
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We can evaluate each of these determinants (using the method explained here). The results are these:
So, now that we have all these determinants and their values, what do we do with them?
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How does Cramer's Rule find a variable's value?
Cramer's Rule says that we can find the value of a given variable by dividing that variable's determinant by the regular coefficient-determinant's value. That is, Cramer's Rule specifies this relationship:
x = Dx ÷ D
y = Dy ÷ D
z = Dz ÷ D
Taking the determinant values that we derived from the system they'd given us and doing the relevant divisions, we get these values for the variables:
x = 3/3 = 1
y = −6/3 = −2
z = 9/3 = 3
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And that's all there is to Cramer's Rule. To find whichever variable you want (call it, say, β or ["beta"]), just evaluate the determinant quotient Dβ ÷ D.
(Please don't ask me to explain why this works. If you really want to know, there are loads of proofs online, like this and this. Otherwise, just trust me that determinants can work many kinds of magic.)
What is an example of solving by using Cramer's Rule?
2x + y + z = 1
x − y + 4z = 0
x + 2y − 2z = 3
To solve only for z, I first form and find the value of the coefficient determinant.
Then I form Dz by replacing the third column of values with the answer column:
Then I form the quotient and simplify:
So then my answer is:
z = 2
The point of Cramer's Rule is that you don't have to solve the whole system to get the one value you need. This saved me a fair amount of time on some physics tests. I forget what we were working on (something with wires and currents, I think), but Cramer's Rule was so much faster than any other solution method for finding the one value I needed — and God knows I needed the extra time.
Don't let all the subscripts and stuff confuse you; the Rule is really pretty simple. You just pick the variable you want to solve for, replace that variable's column of values in the coefficient determinant with the answer-column's values, evaluate that determinant, and divide by the coefficient determinant. That's all there is to it.
Almost.
What if the coefficient determinant is zero?
You can't divide by zero, so what does this mean if the determinant by which you need to divide evaluates to zero? I won't go into the technicalities here, but "D = 0" means that the system of equations has no unique solution; instead, the system may be inconsistent (that is, it has no solution at all) or dependent (that is, it has an infinite number ofsolutions, which may be expressed as a parametric solution such as "(a, a + 3, a − 4)").
In terms of Cramer's Rule, "D = 0" means that you'll have to use some other method (such as matrix row operations) to solve the system. If D = 0, you can not use Cramer's Rule.
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13578 | https://www.cuemath.com/algebra/graphing-linear-equations/ | LearnPracticeDownload
Graphing Linear Equations
Graphing Linear Equations is the process of representing linear equations with one or two variables on a graph. A linear equation is an equation of degree one i.e. the highest power or exponent value of the variable can only be 1, not greater than 1 in any of the cases. Solving a linear equation is to find the value of variables contained in it and the graphical method is one of the methods to solve linear equations, either one or two variables linear equations.
| | |
--- |
| 1. | What is Graphing Linear Equations? |
| 2. | Graphing Linear Equations |
| 3. | Graphing Linear Equations in Two Variables |
| 4. | FAQs on Graphing Linear Equations |
What is Graphing Linear Equations?
Linear equations are algebraic equations in which each term has a real constant and the equation contains 2 variables of the highest power 1. We represent the linear equation in y=mx+b form, also known as the y-intercept form. The representation of a linear equation on a graph is called graphing linear equations shown as a straight line with one or two variables. Let us see an example of graphing a linear equation with one variable. We have to represent the equation x+2y=7 in a graph.
Here, the equation x+2y=7 makes a straight line on the graph. Similarly, all linear equations create a straight line on the graph with both one or two variables. The graph of a linear equation in one variable x forms a vertical line that is parallel to the y-axis and vice-versa, whereas, the graph of a linear equation in two variables x and y forms a straight line. The graphing of linear equations helps in solving many real-life problems in linear programming.
Note:
The point where any line crosses the x-axis on the graph - X-intercept
The point where any line crosses the y-axis on the graph - Y-Intercept
Graphing Linear Equations
Graphing a linear equation is about solving the linear equations and representing the solution in a coordinate plane. While plotting the equation on a graph, the two pairs (x,y) are sufficient. However, we cannot find out if there are any mistakes in obtaining these values as the two points can always be joined and represented as a line. Hence, it is advised to plot one more point to ensure that the solutions obtained for the given linear equation are correct. The following steps are to be taken up for plotting a linear equation with one variable:
Make sure the linear equation is in y-intercept form, which is y = mx + b.
Apply the trial and error method and find the value of (x, y) up to three pairs, which satisfy the linear equation.
Find the x-intercept and y-intercept of the equation. For y-intercept, substitute the value of x = 0 in the equation. This results in x = a, for x-intercept, substitute the value of y = 0 in the equation. This results in y = c.
Thus, the points are (a, 0) and (0, c). Make a tabular form and put the value of x and y respectively.
Plot all the points on the graph paper.
Join all the points which are marked on the graph and you will get a straight line representing the given linear equation graphically.
Example: Draw a graph of the linear equation x+2y=7.
Solution: We will follow the following steps:
Step 1: Check if the given linear equation x+2y=7 is of the form of y = mx + b. [On converting, we get: y = - (1/2) x + 7/2]
Step 2: Find the x and y-intercept respectively. For that, put y = 0 in the equation: x = 7-2(0), x=7. Now, put x=0 in the equation. 2y=7-(0), y=7/2 = 3.5
Step 3: Apply the trial and error method and find 3 pairs of values of (x, y) that satisfy the given linear equation x=7-2y. (See table below)
Step 4: Plot the points (7,0), (5,1), and (3,2) on the graph.
Step 5: Join all the points which are marked on the graph paper and get a straight line that represents the given linear equation graphically.
See the values of x and y in the following table:
| | | | |
--- --- |
| x | 7 | 5 | 3 |
| y | 0 | 1 | 2 |
Graphing Linear Equations in Two Variables
Graphing linear equations in two variables is done in a similar manner as one variable. The lines plotted on the graph can either intersect each other at one point or be parallel to each other resulting in no solution. Sometimes, the lines might coincide with each other leaving each point on that line as a solution resulting in the given system having an infinite number of solutions. If the system has a solution, then it is said to be consistent; otherwise, it is said to be inconsistent. The following are the steps to graphing linear equations in two variables:
Step 1: To graph a linear equation with two variables, we graph both equations.
Step 2: To graph an equation manually, first convert it to the form y=mx+b by solving the equation for y.
Step 3: Start putting the values of x as 0, 1, 2, and so on and find the corresponding values of y, or vice-versa.
Step 4: Identify the point where both lines meet.
Step 5: The point of intersection is the solution.
Example: Find the solution of the following system of equations graphically.
-x+2y-3 =0
3x+4y-11=0
Solution: Using the steps mentioned above, we will graph them and see whether they intersect at a point. As you can see below, both lines meet at (1, 2). Thus, the solution of the given system of linear equations is x=1 and y=2.
Important Points
Linear equation in two variables has infinitely many solutions.
The graph of a linear equation is always a straight line.
The equation y = mx is always passing through the origin (0, 0).
Challenging Questions
The sum of the digits of a two-digit number is 8. When the digits are reversed, the number is increased by 18. Find the number.
Jake's piggy bank has 11 coins (only quarters or dimes) that have a value of $1.85. How many dimes and quarters does the piggy bank have?
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Linear Equations
Linear Equations in One Variable
Linear Equations in Two Variables
Solving Linear Equations Calculator
Linear Equation Formula
List the methods used for solving linear equations
What is the solution to the system of linear Equations?
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Graphing Linear Equation Worksheet
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Graphing Linear Equations Examples
Example 1: Isabella has the linear equation x - 2y = 2. Help her to draw the linear equation on the graph.
Solution:
The given linear equation is x-2y=2. Convert the equation in the form of y = mx + b⇒ y= x/2 - 1. We need to find the x and y-intercept respectively. For that, put y=0 in the equation⇒ x=2(0)+2, x=2. Now, put x=0 in the equation. 2y=(0) - 2, y=1. Now, we will apply the trial and error method, and find 3 pairs of values of (x, y) that satisfy the given linear equation y=x/2-1. See the values of x and y in the following table:
| | | | |
--- --- |
| x | 2 | 4 | 0 |
| y | 0 | 1 | -1 |
Plot the points (2,0),(4,1),(0,-1) on the graph. Join all the points which are marked on the graph paper and get a straight line that represents the given linear equation graphically.
2. Example 2: William wants to plot the graph of -1/2x+1/3y=1. Help him to plot the graph for linear equations.
Solution:
The given linear equation is -1/2x+1/3y=1. Convert the equation in the form of y = mx + b⇒ 6 + 6x/2. We need to find the x and y-intercept respectively. For that, we can apply the hit and trial method, put y=6 in the equation⇒ x=3. Now, put x=2 in the equation. y=3. See the values of x and y in the following table:
| | | |
---
| x | 2 | 0 |
| y | 6 | 3 |
| Solutions | (2,6) | (0,3) |
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Practice Questions on Graphing Linear Equations
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FAQs on Graphing Linear Equations
How Do you Graph a Linear Equation?
The basic methods of the graphing linear equation:
The first method is by plotting all points on the graph and then drawing a line through the points.
The second is by using the y-intercept of the equation and slope of the equation.
What Are the 3 Ways to Graph a Linear Equation?
The 3 ways to graph a linear equation:
Using two points to plot the graph of a linear equation.
Use the slope and y-intercept of a linear equation.
Using the x- and y-intercepts of a linear equation.
How Do you Graph a Linear Equation with Two Points?
Graphing linear equation with two points:
Find the y-intercept from the linear equation and plot the point.
From the y-intercept of the linear equation, use the slope to find the second point and plot it on the graph.
Draw a straight line to connect the two points on the graph.
What Is the Formula For a Linear Equation?
The formula for a linear equation is y = mx + b.
How Do you Graph a Linear Equation Using Intercepts?
To find intercepts algebraically, we use the fact that all x-intercepts have y=0 and all y-intercepts have x=0.
Determine the corresponding values of x and y by putting the values of x-intercepts and y-intercepts respectively.
What Is the Minimum Number of Points Needed to Graph a Linear Equation?
Two points are the minimum number of points needed to graph a linear equation
How Do you Find the Y-Intercept of a Graph?
The y-intercept of a graph is the point at which the graph crosses the y-axis and at y point, the x-coordinate is zero.
How Do you Find the Intercepts of a Graph?
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13579 | https://www.auajournals.org/doi/10.1016/j.juro.2017.01.086 | Efficacy of bacillus Calmette-Guérin Strains for Treatment of Nonmuscle Invasive Bladder Cancer: A Systematic Review and Network Meta-Analysis | Journal of Urology
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No AccessJournal of UrologyReview Article 1 Sep 2017
Efficacy of bacillus Calmette-Guérin Strains for Treatment of Nonmuscle Invasive Bladder Cancer: A Systematic Review and Network Meta-Analysis
Brock E. Boehm,
John E. Cornell,
Hanzhang Wang,
Neelam Mukherjee,
Jacob S. Oppenheimer,and
Robert S. Svatek
Brock E. Boehm
Brock E. Boehm
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,
John E. Cornell
John E. Cornell
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,
Hanzhang Wang
Hanzhang Wang
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Neelam Mukherjee
Neelam Mukherjee
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Jacob S. Oppenheimer
Jacob S. Oppenheimer
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Robert S. Svatek
Robert S. Svatek
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Abstract
Purpose:
We sought to determine the efficacy of genetically distinct bacillus Calmette-Guérin strains in preventing disease recurrence in patients with nonmuscle invasive bladder cancer.
Materials and Methods:
We conducted a systematic review and network meta-analysis of trials evaluating bacillus Calmette-Guérin strains against all possible comparators (different bacillus Calmette-Guérin strains, chemotherapy and nonbacillus Calmette-Guérin biological therapies) with intravesical chemotherapy as the common comparator. MEDLINE® ( served as the primary data source, with the search from inception to October 2016 for clinical trials involving patients with nonmuscle invasive bladder cancer receiving bacillus Calmette-Guérin. Primary outcome measure was bladder cancer recurrence, defined as recurrent bladder tumor of any grade or stage. Random effect network meta-analysis provided estimates for outcomes and is presented as odds ratios.
Results:
Across all possible comparators (65 trials, 12,246 patients, 9 strains) there were 2,177 recurrences in 5,642 treated patients (38.6%) and 2,316 recurrences in 5,441 comparators (42.6%). With chemotherapy as the common comparator (28 trials, 5,757 patients, 5 strains) Tokyo-172 (OR 0.39, 95% CI 0.16–0.93), Pasteur (OR 0.49, 95% CI 0.28–0.86) and TICE® (OR 0.61, 95% CI 0.40–0.93) strains were significantly better than chemotherapy at preventing recurrence. No bacillus Calmette-Guérin strain demonstrated significant superiority when compared to any other strain at preventing recurrence in the network meta-analysis.
Conclusions:
Bacillus Calmette-Guérin strains exhibited significant differences in efficacy compared to chemotherapy. However, no definitive conclusions could be reached regarding strain superiority, and head-to-head trials are greatly needed to further understand the importance of strain selection in determining bacillus Calmette-Guérin efficacy.
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© 2017 by American Urological Association Education and Research, Inc.
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Koster K, Webb H and Cirillo J(2021)COVID-19 and Beyond: Exploring Public Health Benefits from Non-Specific Effects of BCG VaccinationMicroorganisms, 10.3390/microorganisms9102120, VOL. 9, NO. 10, (2120) Lobo N, Brooks N, Zlotta A, Cirillo J, Boorjian S, Black P, Meeks J, Bivalacqua T, Gontero P, Steinberg G, McConkey D, Babjuk M, Alfred Witjes J and Kamat A(2021)100 years of Bacillus Calmette–Guérin immunotherapy: from cattle to COVID-19Nature Reviews Urology, 10.1038/s41585-021-00481-1, VOL. 18, NO. 10, (611-622), Online publication date: 1-Oct-2021. Shore N, Palou Redorta J, Robert G, Hutson T, Cesari R, Hariharan S, Rodríguez Faba Ó, Briganti A and Steinberg G(2021)Non-muscle-invasive bladder cancer: An overview of potential new treatment optionsUrologic Oncology: Seminars and Original Investigations, 10.1016/j.urolonc.2021.05.015, VOL. 39, NO. 10, (642-663), Online publication date: 1-Oct-2021. Lesch S and Gill S(2021)The promise and perils of immunotherapyBlood Advances, 10.1182/bloodadvances.2021004453C, VOL. 5, NO. 18, (3709-3725), Online publication date: 28-Sep-2021. D’Andrea D, Soria F, Grotenhuis A, Cha E, Malats N, Di Stasi S, Joniau S, Cai T, van Rhijn B, Irani J, Karnes J, Varkarakis J, Baniel J, Palou J, Babjuk M, Spahn M, Ardelt P, Colombo R, Serretta V, Dalbagni G, Gontero P, Bartoletti R, Larré S, Malmstrom P, Sylvester R and Shariat S(2021)Association of patients’ sex with treatment outcomes after intravesical bacillus Calmette–Guérin immunotherapy for T1G3/HG bladder cancerWorld Journal of Urology, 10.1007/s00345-021-03653-1, VOL. 39, NO. 9, (3337-3344), Online publication date: 1-Sep-2021. Muto S, Lu Y, Ide H, Yamaguchi R, Saito K, Kitamura K, Noma Y, Koyasu H, Hirano H, Ashizawa T, Isotani S, Nagata M and Horie S(2021)The Use of Urine Mycobacterium tuberculosis Complex Polymerase Chain Reaction as a Predictive Factor for Recurrence and Progression After Intravesical Bacillus Calmette-Guérin Therapy in Patients with Non–muscle‑invasive Bladder CancerEuropean Urology Open Science, 10.1016/j.euros.2021.02.005, VOL. 27, (10-18), Online publication date: 1-May-2021. Hugar L, Gilbert S, Sexton W, Kamat A and Li R(2020)Immunotherapy in Bacillus Calmette–Guerin (BCG) unresponsive nonmuscle invasive bladder cancerCurrent Opinion in Urology, 10.1097/MOU.0000000000000846, VOL. 31, NO. 2, (160-169), Online publication date: 1-Mar-2021. Sfakianos J, Salome B, Daza J, Farkas A, Bhardwaj N and Horowitz A(2021)Bacillus Calmette-Guerin (BCG): Its fight against pathogens and cancerUrologic Oncology: Seminars and Original Investigations, 10.1016/j.urolonc.2020.09.031, VOL. 39, NO. 2, (121-129), Online publication date: 1-Feb-2021. Srougi V, Reis S, Viana N, Gallucci F, Leite K, Srougi M and Nahas W(2020)Prospective evaluation of a urinary biomarker panel to detect and predict recurrence of non-muscle-invasive bladder cancerWorld Journal of Urology, 10.1007/s00345-020-03188-x, VOL. 39, NO. 2, (453-459), Online publication date: 1-Feb-2021. Pyrgidis N, Lackner J, Schneidewind L and Sokolakis I(2021)Behandlung des Ta- und T1-Harnblasenkarzinoms von intermediärem oder hohem Risiko mittels intravesikalem Bacillus Calmette-Guérin oder Mitomycin CTreatment of Ta and T1 intermediate or high risk bladder cancer with intravesical Bacillus Calmette-Guérin or mitomycin CDer Urologe, 10.1007/s00120-021-01443-6, VOL. 60, NO. 2, (234-237), Online publication date: 1-Feb-2021. Nowak Ł, Krajewski W, Moschini M, Chorbińska J, Poletajew S, Tukiendorf A, Muilwijk T, Joniau S, Tafuri A, Antonelli A, Orlando R, Di Trapani E, Alvarez-Maestro M, Simone G, Zamboni S, Simeone C, Marconi M, Mastroianni R, Piszczek R, Xylinas E and Zdrojowy R(2021)Assessment of the oncological outcomes of three different bacillus Calmette–Guérin strains in patients with high-grade T1 non-muscle-invasive bladder cancerArab Journal of Urology, 10.1080/2090598X.2021.1874628, VOL. 19, NO. 1, (78-85), Online publication date: 2-Jan-2021. Yassaie O, Chehroudi C and Black P(2021)Novel and emerging approaches in the management of non-muscle invasive urothelial carcinomaTherapeutic Advances in Medical Oncology, 10.1177/17588359211039052, VOL. 13, Online publication date: 1-Jan-2021. Ourfali S, Ohannessian R, Fassi-Fehri H, Pages A, Badet L and Colombel M(2021)Recurrence Rate and Cost Consequence of the Shortage of Bacillus Calmette-Guérin Connaught Strain for Bladder Cancer PatientsEuropean Urology Focus, 10.1016/j.euf.2019.04.002, VOL. 7, NO. 1, (111-116), Online publication date: 1-Jan-2021. Mayor N, Fankhauser C, Sangar V and Mostafid H(2021)Management of NMIBC during BCG shortage and COVID ‐19Trends in Urology & Men's Health, 10.1002/tre.783, VOL. 12, NO. 1, (7-11), Online publication date: 1-Jan-2021. Koguchi D, Matsumoto K, Hirayama T, Moroo S, Kobayashi M, Katsumata H, Ikeda M and Iwamura M(2020)Impact of maintenance therapy using a half dose of the bacillus Calmette–Guérin Tokyo strain on recurrence of intermediate and high-risk nonmuscle invasive bladder cancer: a retrospective single-center studyBMC Urology, 10.1186/s12894-020-00766-4, VOL. 20, NO. 1, Online publication date: 1-Dec-2020. Kremenovic M, Schenk M and Lee D(2020)Clinical and molecular insights into BCG immunotherapy for melanomaJournal of Internal Medicine, 10.1111/joim.13037, VOL. 288, NO. 6, (625-640), Online publication date: 1-Dec-2020. Lu J, Xia Q, Lu Y, Liu Z, Zhou P, Hu H and Wang S(2020)Efficacy of intravesical therapies on the prevention of recurrence and progression of non‐muscle‐invasive bladder cancer: A systematic review and network meta‐analysisCancer Medicine, 10.1002/cam4.3513, VOL. 9, NO. 21, (7800-7809), Online publication date: 1-Nov-2020. Fujita N, Hatakeyama S, Momota M, Tobisawa Y, Yoneyama T, Yamamoto H, Imai A, Ito H, Yoneyama T, Hashimoto Y, Yoshikawa K and Ohyama C(2020)Impact of substratification on predicting oncological outcomes in patients with primary high-risk non-muscle-invasive bladder cancer who underwent transurethral resection of bladder tumorUrologic Oncology: Seminars and Original Investigations, 10.1016/j.urolonc.2020.04.023, VOL. 38, NO. 10, (795.e9-795.e17), Online publication date: 1-Oct-2020. Benn C, Fisker A, Rieckmann A, Sørup S and Aaby P(2020)Vaccinology: time to change the paradigm?The Lancet Infectious Diseases, 10.1016/S1473-3099(19)30742-X, VOL. 20, NO. 10, (e274-e283), Online publication date: 1-Oct-2020. Sayed A, Munir M, Eweis N, Wael D, Shazly O, Awad A, Elbadawy M and Eissa S(2020)An overview on precision therapy in bladder cancerExpert Review of Precision Medicine and Drug Development, 10.1080/23808993.2020.1801346, VOL. 5, NO. 5, (347-361), Online publication date: 2-Sep-2020. van Puffelen J, Keating S, Oosterwijk E, van der Heijden A, Netea M, Joosten L and Vermeulen S(2020)Trained immunity as a molecular mechanism for BCG immunotherapy in bladder cancerNature Reviews Urology, 10.1038/s41585-020-0346-4, VOL. 17, NO. 9, (513-525), Online publication date: 1-Sep-2020. Fujita N, Hatakeyama S, Momota M, Tobisawa Y, Yoneyama T, Yamamoto H, Imai A, Ito H, Yoneyama T, Hashimoto Y, Yoshikawa K and Ohyama C(2020)Preoperative chronic kidney disease predicts poor prognosis in patients with primary non–muscle-invasive bladder cancer who underwent transurethral resection of bladder tumorUrologic Oncology: Seminars and Original Investigations, 10.1016/j.urolonc.2020.02.001, VOL. 38, NO. 8, (684.e1-684.e8), Online publication date: 1-Aug-2020. Fankhauser C, Teoh J and Mostafid H(2020)Treatment options and results of adjuvant treatment in nonmuscle-invasive bladder cancer (NMIBC) during the Bacillus Calmette–Guérin shortageCurrent Opinion in Urology, 10.1097/MOU.0000000000000739, VOL. 30, NO. 3, (365-369), Online publication date: 1-May-2020. Bach-Griera M, Campo-Pérez V, Barbosa S, Traserra S, Guallar-Garrido S, Moya-Andérico L, Herrero-Abadía P, Luquin M, Rabanal R, Torrents E and Julián E(2020)Mycolicibacterium brumae is a Safe and Non-Toxic Immunomodulatory Agent for Cancer TreatmentVaccines, 10.3390/vaccines8020198, VOL. 8, NO. 2, (198) Chang S(2019)Re: Radiofrequency-Induced Thermo-Chemotherapy Effect versus a Second Course of bacillus Calmette-Guérin or Institutional Standard in Patients with Recurrence of Non-Muscle-Invasive Bladder Cancer following Induction or Maintenance bacillus Calmette-Guérin Therapy (HYMN): A Phase III, Open-Label, Randomised Controlled TrialJournal of Urology, VOL. 203, NO. 3, (461-462), Online publication date: 1-Mar-2020. Larsen E, Joensen U, Poulsen A, Goletti D and Johansen I(2020)Bacillus Calmette–Guérin immunotherapy for bladder cancer: a review of immunological aspects, clinical effects and BCG infectionsAPMIS, 10.1111/apm.13011, VOL. 128, NO. 2, (92-103), Online publication date: 1-Feb-2020. Qu H, Huang Y, Mu Z, Lv H, Xie Q, Wang K and Hu B(2020)Efficacy and Safety of Chemotherapy Regimens in Advanced or Metastatic Bladder and Urothelial Carcinomas: An Updated Network Meta-AnalysisFrontiers in Pharmacology, 10.3389/fphar.2019.01507, VOL. 10 Schmidt S, Kunath F, Coles B, Draeger D, Krabbe L, Dersch R, Kilian S, Jensen K, Dahm P and Meerpohl J(2020)Intravesical Bacillus Calmette-Guérin versus mitomycin C for Ta and T1 bladder cancerCochrane Database of Systematic Reviews, 10.1002/14651858.CD011935.pub2 Khalili N, Khalili N and Rezaei N(2020)Immunotherapy in Bladder and Renal CancersCancer Immunology, 10.1007/978-3-030-57949-4_18, (451-474), . Faba Ó, Subiela J and Palou J(2020)Bladder CancerUrologic Principles and Practice, 10.1007/978-3-030-28599-9_34, (561-578), . Hobbs C, Bass E, Crew J and Mostafid H(2019)Intravesical BCG: where do we stand? Past, present and futureJournal of Clinical Urology, 10.1177/2051415818817120, VOL. 12, NO. 6, (425-435), Online publication date: 1-Nov-2019. Babjuk M, Burger M, Compérat E, Gontero P, Mostafid A, Palou J, van Rhijn B, Rouprêt M, Shariat S, Sylvester R, Zigeuner R, Capoun O, Cohen D, Escrig J, Hernández V, Peyronnet B, Seisen T and Soukup V(2019)European Association of Urology Guidelines on Non-muscle-invasive Bladder Cancer (TaT1 and Carcinoma In Situ) - 2019 UpdateEuropean Urology, 10.1016/j.eururo.2019.08.016, VOL. 76, NO. 5, (639-657), Online publication date: 1-Nov-2019. 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#### Volume 198 #### Issue 3 #### September 2017
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Archivos de Bronconeumología
ISSN: 0300-2896
Archivos de Bronconeumologia is an international journal that publishes original studies whose content is based upon results of research initiatives dealing with several aspects of respiratory medicine including epidemiology, respiratory physiology, pathophysiology of respiratory diseases, clinical management, thoracic surgery, pediatric lung diseases, respiratory critical care, respiratory allergy and translational research. Other types of articles such as editorials, reviews, and different types of letters are also published in the journal. Additionally, the journal expresses the voice of the following scientific societies: the Spanish Respiratory Society of Pneumology and Thoracic Surgery (SEPAR; the Latin American Thoracic Society (ALAT; and the Iberian American Association of Thoracic Surgery (AIACT;
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Vol. 53. Issue 11.
Pages 629-636(November 2017)
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Introduction
Abstract
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Introduction
Pleural transudates
Heart failure
Trapped lung
Hepatic hydrothorax
Peritoneal dialysis
Metastatic pleural effusion
Other causes
Conclusions
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Vol. 53. Issue 11.
Pages 629-636(November 2017)
Review article
DOI: 10.1016/j.arbr.2017.09.004
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Diagnosis and Management of Pleural Transudates
Diagnóstico y manejo de los trasudados pleurales
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Lucía Ferreiroa,b,, José M. Porcelc,d, Luis Valdésa,b
a Servicio de Neumología, Complejo Hospitalario Universitario de Santiago, Santiago de Compostela, Spain
b Grupo Interdisciplinar de Investigación en Neumología, Instituto de Investigaciones Sanitarias de Santiago de Compostela (IDIS), Santiago de Compostela, Spain
c Unidad de Medicina Pleural, Servicio de Medicina Interna, Hospital Universitario Arnau de Vilanova, Lleida, Spain
d Instituto de Investigación Biomédica Fundación Dr. Pifarré IRBLLEIDA, Lleida, Spain
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Table 1. Series of Patients With Different Types of Pleural Transudates Treated With Talc Pleurodesis or Tunneled Pleural Catheter.
Table 2. Characteristics of Pleural Transudates With Unusual Etiology.
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Abstract
Various clinical trials have been published on the optimal clinical management of patients with pleural exudates, particularly those caused by malignant tumors, while little information is available on the diagnosis and treatment of pleural transudates.
The etiology of pleural transudates is wide and heterogeneous, and they can be caused by rare diseases, sometimes constituting a diagnostic challenge. Analysis of the pleural fluid can be a useful procedure for establishing diagnosis. Treatment should target not only the underlying disease, but also management of the pleural effusion itself.
In cases refractory to medical treatment, invasive procedures will be necessary, for example therapeutic thoracentesis, pleurodesis with talc, or insertion of an indwelling pleural catheter. Little evidence is currently available and no firm recommendations have been made to establish when to perform an invasive procedure, or to determine the safest, most efficient approach in each case.
This article aims to describe the spectrum of diseases that cause pleural transudate, to review the diagnostic contribution of pleural fluid analysis, and to highlight the lack of evidence on the efficacy of invasive procedures in the management and control of pleural effusion in these patients.
Keywords:
Pleural effusion
Pleural fluid
Transudates
Heart failure
Hepatic hydrothorax
Trapped lung
Pleurodesis
Indwelling pleural catheter
Resumen
El diagnóstico y tratamiento de los trasudados pleurales ha sido poco estudiado, sobre todo si se compara con el de los exudados, especialmente los malignos, en que diversos ensayos clínicos informan con relativa frecuencia sobre cuál es el manejo óptimo de estos pacientes en la práctica clínica.
La etiología de los trasudados pleurales es amplia, heterogénea, a veces corresponde a enfermedades raras y, en ocasiones, constituye un auténtico reto diagnóstico. El análisis del líquido pleural puede ser una herramienta básica para poder establecer el diagnóstico. El tratamiento debe dirigirse no solamente a la enfermedad subyacente, sino también al propio derrame pleural.
En los casos refractarios al tratamiento médico, habrá que recurrir a procedimientos invasivos como la toracocentesis terapéutica, la pleurodesis con talco, o la inserción de un catéter pleural tunelizado. Sin embargo, en la actualidad, hay muy poca evidencia al respecto y no existen recomendaciones firmes que establezcan en qué situaciones hay que utilizar alguna de estas técnicas invasivas y cuál sería la más eficaz y con menos efectos secundarios para cada una de las enfermedades responsables.
El propósito de este artículo es describir el espectro de enfermedades que causan un trasudado pleural, revisar la aportación diagnóstica del análisis del líquido pleural y poner de manifiesto la escasa evidencia que existe sobre la eficacia de los procedimientos invasivos en el manejo y control del derrame pleural en estos pacientes.
Palabras clave:
Derrame pleural
Líquido pleural
Trasudados
Insuficiencia cardiaca
Hidrotórax hepático
Pulmón atrapado
Pleurodesis
Catéter pleural tunelizado
Full Text
Introduction
The pleural space normally contains a small amount of fluid that separates the two pleural layers, allowing one membrane to slip over the other. As in other interstitial spaces of the body, differences in hydrostatic and oncotic pressures between the capillaries of both pleural layers and the pleural space causes fluid to pass through the systemic vessels into the pleural space, where it is absorbed by lymphatic capillaries located in the lower regions of the parietal pleura1 (Fig. 1A). The volume and characteristics of the pleural fluid (PF) are determined by a combination of dynamic phenomena that affect systemic and pulmonary circulation, lymphatic drainage, and the movements of the rib cage and the heart.2 Normal PF production is approximately 0.01 ml×kg×h, so an individual weighing 60 kg would have an entry rate of 15 ml/day,3 while pleural lymphatic capillaries can absorb 0.28 ml×kg×h, almost 30 times the entry rate.4
Fig. 1.
Schematic representation of the movement of fluid in the pleural space in normal conditions (A), in heart failure (B) and in trapped lung (C). In normal lung and pleura (A), the balance of hydrostatic and oncotic pressures promotes the formation of pleural fluid. Pressure gradients show a net flow of fluid from the parietal pleura and that balance is maintained in the visceral pleura (pressures in cmH 2 O). In heart failure (B), fluid in the pulmonary interstitium increases due to increased pulmonary capillary pressures. Fluid that enters the pleural space exceeds the drainage capacity of the pleural lymphatic capillaries. In trapped lung (C), a previous inflammatory process has produced a thick, fibrous visceral pleura. The inability of the lung to expand leads to a negative hydrostatic pleural pressure and this alteration of the Starling's forces leads to the formation of pleural transudate. Pleural effusion may be verging on exudate, depending on how recent the active inflammation was.
Any imbalance between the hydrostatic and oncotic pressures, whether in the pleural space or the blood capillaries, can lead to the accumulation of fluid in the pleural space. In this situation, the fluid is called transudate and the pleura is not diseased. The diagnostic, prognostic and therapeutic implications of a transudate, then, differ completely from exudative pleural effusion (PE).
From a clinical point of view, transudate can be distinguished from exudate by measuring various biochemical parameters in the PF and in blood, most often by applying Light's criteria: ratio of total protein in PF/serum (PF/S)>0.5; lactate dehydrogenase (LDH) PF/S>0.6 and LDH in PF>2/3 the normal upper value in blood.5 PF is a transudate if none of the above conditions are met. Other parameters, such as cholesterol, have also been used for differentiating transudate from exudate, but none has demonstrated superiority over the others.6
A transudate can have numerous origins; some are very common, but others are rare, and sometimes unfamiliar. No firm recommendations have been made on the diagnosis and management of these effusions. In many of these patients, PE is the manifestation of disease in another organ and comorbidities are common, so the burden of managing PE is regularly borne by other specialists. Cooperation and agreement with these professionals will be necessary for designing protocols for the coordinated and optimized treatment of PE. This article aims to describe the spectrum of diseases that cause pleural transudate, review how PF analysis contributes to diagnosis, and update management strategies.
Pleural Transudates Heart Failure
Eighty-seven percent of patients hospitalized for decompensated heart failure (HF) who require diuretics show PE on their computed tomography (CT).7 HF is the most common cause of transudate8,9 and is responsible for all the PE identified in some series. Prevalence can be as high as 45%.10 The resulting pulmonary edema in these patients increases hydrostatic pressure in the capillaries of the visceral pleura, causing fluid to pass into the pleural space (Fig. 1B).
Cardiac PE tends to present as a lymphocyte-predominant transudate. However, as Light's criteria are designed for the diagnosis of exudates, 25% of cases may be incorrectly classified as exudates, particularly in patients receiving diuretics.11,12 If these drugs correct venous hypertension, PF is not only drained by the lymphatic capillaries, but can also be reabsorbed by the pleural capillaries, which would increase protein and LDH concentrations in the PF to the extent that it meets the biochemical characteristics of an exudate. In these cases, a serum albumin-PF gradient (or gap)>1.2 g/dl seems to indicate that the PE corresponds to a transudate, regardless of whether it meets Light's criteria or not.13 However, other factors, such as the presence of a second cause of PE14 or a large number of red blood cells (>10 000 cells/mm 3), can act on LDH levels15 and, in biochemical terms, change a transudate to an exudate.
In this setting, PF or blood determinations of natriuretic peptides, molecules secreted by the ventricles of the heart in response to acute distention, may be useful in the diagnosis of a PE caused by HF. In a recent meta-analysis, the diagnostic yield of N-terminal pro-brain natriuretic peptide (NT-proBNP) in PF was: sensitivity 94%, specificity 91%, positive likelihood ratio 10.9, negative likelihood ratio 0.07, and odds ratio 157, with slightly inferior results when the determination was made in blood.16
The diagnosis of cardiac PE is usually established by consistent clinical symptoms and cardiomegaly and bilateral PE on chest radiograph, generally observed in 80% of cases. In these conditions, thoracentesis is not necessary.17 However, in certain circumstances, it is more difficult to establish a diagnosis. This may be due to clinical symptoms (chest pain and fever) associated with unusual radiological findings (no cardiomegaly or a marked asymmetry in the size of the PE) that would justify a diagnostic thoracentesis and determination of natriuretic peptides. Rabin and Blackman found in a series of 78 patients with bilateral PE and no cardiomegaly that only 3 (3.8%) had PE due to HF.18 Single-sided PE, however, does not rule out HF. In a review of 6 series that included 783 patients with PE due to HF, 125 (16%) had right PE and 55 (7%) had left PE.19
A prospective study of 60 patients with PE secondary to HF showed that in 89% of cases, PE resolved after 2 weeks of treatment with diuretics.7 In general, most patients will improve with optimized HF treatment, including diuretics, angiotensin-converting enzyme inhibitors, β-blockers, or procedures such as pacemaker placement or surgery, if the patient has serious valvular dysfunction.
Talc pleurodesis is a safe and effective technique in the treatment of malignant PE. However, evidence to support its use in benign PE is limited to a few cases in which the etiology of the PE was not even specified20–22 (Table 1). Nor is sufficient evidence available to support the use of the tunneled pleural catheter (TPC), and no cost-effectiveness studies have been performed. In a study conducted in 5 British hospitals, TPC was placed in 9 patients with HF and spontaneous pleurodesis was achieved in 4 (44%).23 In another series, 43 TPCs were placed in 36 patients with refractory HF and PE. In the group treated with talc poudrage by thoracoscopy followed by TPC, pleurodesis was achieved in 80% of cases, while in those patients who received only TPC, pleurodesis occurred in 25%.24 A recent meta-analysis assessed the benefit of TPC in patients with refractory benign PE. Of the 162 (49.8%) with PE due to HF, pleurodesis was achieved in 42.1% (range 0%–92.3%) (Table 1), with a complication rate similar to that of malignant PEs. The authors concluded that TPC is an effective and viable option in the management of patients with refractory benign PE.25 However, randomized clinical trials are needed to determine more accurately the usefulness of these catheters.
Table 1.
Series of Patients With Different Types of Pleural Transudates Treated With Talc Pleurodesis or Tunneled Pleural Catheter.
| | Talc Pleurodesis | Tunneled Pleural Catheter |
:---:
| | Heart Failure | Hepatic Hydrothorax | Heart Failure | Hepatic Hydrothorax | Chronic Renal Disease |
| | Patients | Success n (%) | Patients | Success n (%) | Patients | Success n (%) | Patients | Success n (%) | Patients | Success n (%) |
| Glazer et al.21 | 6 | 5 (83.3) | 4 | 3 (75) | | | | | | |
| Bhatnagar et al.23 | | | | | 9 | 4 (44) | 19 | 2 (11) | | |
| Majid et al.24 | | | | | 15 | 12 (80) | | | | |
| | | | | | 28 | 7 (25) | | | | |
| Patil et al.25 | | | | | 162 | 68 (42) | | | | |
| Porcel26¶ | | | 189 | 142 (75) | | | | | | |
| Chen et al.27 | | | | | | | 24 | 8 (33) | | |
| Potechin et al.28 | | | | | | | | | 8 | 3 (37.5) |
Prior talc thoracoscopy.
¶
Review of 11 series.
Trapped Lung
Trapped lung is a sequela of inflammation of the pleural space that leads to the formation of a fibrous membrane over the visceral pleura, preventing the lung from expanding29 and increasing negative pressure in the pleural space. Differences in hydrostatic pressure between the systemic capillaries of the parietal pleura and the pleural space can be very high, so the net flow of fluid into the pleural space will increase in order to decrease the pressure difference (Fig. 1C).
PF generally meets the criteria for transudate,29 since PE is generated by differences in existing hydrostatic pressures. However, the characteristics of the PF will depend on the time at which the thoracentesis is performed. If this is done at an early stage, when the disease is still active, the protein level, and less frequently the LDH level, of the PE will often be in the range of that of exudate.30
Diagnosis is based on four findings: (1) post-thoracentesis hydropneumothorax or PE that cannot be drained due to onset of chest pain; (2) transudative PF; (3) initially negative pleural pressure that falls even further as PF is removed, causing pleural elastance >14.5 cmH 2 O/l (Fig. 2A–C), and (4) an air-contrast chest CT (that is, after diagnostic pneumothorax is performed during pleural manometry) demonstrating thickening of the visceral pleura.29
Fig. 2.
Curves obtained by manometry, with their elastance value, in normal lung (A), in a lung becoming trapped (B) and a trapped lung (C). With pleural effusion in a normal lung (A), initial pleural pressure will be slightly positive. As fluid is aspirated, the pleural pressure will drop slowly and the lung will be expanded gradually. Once all the effusion is removed, the lung will come into contact with the chest wall and the elastance obtained will be normal. In the lung in the process of becoming trapped (B), visceral pleura will become slightly thickened, and initial pleural pressure will be slightly positive, as in normal lung. When the fluid is removed, in principle, the diaphragm will expand progressively and pleural pressure will drop slowly. At some point the lung is trapped and unable to expand more and the pressure will drop quickly giving rise to a high elastance, with a bimodal pressure/volume curve. In the trapped lung, the visceral pleura has a thicker layer of fibrin which prevents the lung from expanding, so the initial pressure will be negative (C). The removal of fluid, on the one hand, and the rigidity of the lung, on the other hand, cause a rapid decline in pleural pressure and will lead to high elastance.
Trapped lung does not usually require any treatment if symptoms are mild or absent. Evacuating thoracentesis should not be repeated, since a similar amount of PF will accumulate again in an attempt to “normalize” the negative pleural pressure. In cases in which dyspnea is debilitating, TPC would be an option, since in addition to evacuating the liquid, pleurodesis can be achieved in some cases. However, there is still insufficient evidence outside its use in malignant PE. As a final alternative, if all other causes for the dyspnea are excluded, pleural decortication may be proposed.
Hepatic Hydrothorax
Hepatic hydrothorax is defined as PE in a patient with liver cirrhosis and portal hypertension with no associated heart, lung or pleural disease.31 Prevalence varies between 5% and 10% of patients with cirrhosis,26 of whom 80% have concomitant ascites.26 Two factors influence the accumulation of fluid in the pleural space: firstly, a pressure gradient between the peritoneal and pleural spaces (due to negative pressure), favoring the one-way passage of ascitic fluid to the thorax; and secondly, diaphragmatic defects that are often seen in the tendinous portion of the right diaphragm. This, in addition to the piston effect of the liver, means that most of these PEs are right-sided (80%).32–37 In patients without ascites, the mechanisms by which PF forms are the same.38 Hepatic hydrothorax is generally associated with a poor prognosis, with a 1-year survival of 43%.39
PE is usually a transudate40 with a relatively high pH,41 but 18% of cases might have the biochemical characteristics of exudate.42 A thoracentesis should always be performed and the PF should be analyzed to exclude associated heart, lung or pleural disease, and also to detect the main complication of hepatic hydrothorax, spontaneous bacterial empyema (bacterial infection of a previous hydrothorax in which pneumonia has been ruled out), which occurs in 13% of patients with cirrhosis and PE. According to existing studies, suspicion cannot be established from analysis of the ascitic fluid, instead, the PF must be analyzed.32 The PF of patients with liver cirrhosis may sometimes be a chylothorax.43 This may be due to the fact that these patients have a high liver capillary pressure, with a commensurate increase in lymphatic flow in the liver and the thoracic duct,44 which could lead to the formation of chylous ascites.
The most common treatments include salt restriction, diuretics, and therapeutic thoracentesis. In the case of ascites or refractory hydrothorax, the definitive treatment is liver transplantation.45 If this is contraindicated, an alternative is to implant a transjugular intrahepatic portosystemic shunt46 and, as a last resort, surgical repair of the diaphragm may be considered.47 Talc pleurodesis appears to be a highly risky procedure, with a periprocedural mortality of 45%.41 In a review of 11 series that included 189 patients with hepatic hydrothorax treated with thoracoscopic pleurodesis, mostly by talc insufflation, pleural symphysis was achieved in 142 cases.26 In a review of the clinical practice of British hospitals, the TPC was used in 19 patients with hepatic hydrothorax. Spontaneous pleurodesis was achieved in only 2 (11%) and, moreover, various complications developed.23 Finally, in a recent study, Chen et al. achieved spontaneous pleurodesis in 8 (33%) of 24 patients treated with a TPC. The mean time until pleurodesis was 132 days; after withdrawal of the catheter, PE did not recur in any case, and adverse effects (PF infection) were observed in only 4 patients (16.7%)27 (Table 1).
Peritoneal Dialysis
The incidence of PE among patients receiving peritoneal dialysis is 1.6%, and it can develop between 1 day and 8 years after beginning the procedure.48 The size of the PE can be such as to force dialysis to be suspended. Intra-abdominal pressure that is slightly positive in normal conditions increases linearly in proportion to the volume of dialysis solution instilled.49 Intra-abdominal pressure and volume can increase the pressure of the abdominal wall and, consequently, of the abdominal support structures, causing the dialysis solution to leak from the peritoneal cavity, possibly through diaphragmatic defects, into the pleural space. PE is generally small and right-sided, although on occasions it is bilateral. Massive PEs have been described in women, always in the right side.50
In PE due to peritoneal dialysis, PF is a transudate with very high glucose levels (range 200–2000 mg/dl).51 Diagnosis is precisely established by the presence of a transudate with very low protein levels (<1 g/dl) and a glucose PF/S ratio of >2.
In a retrospective study of 257 patients receiving hemodialysis for renal failure, 52 (20%) developed PE.52 In 9 of the 14 patients (64.3%) who underwent thoracentesis, the PF was a transudate. The cause of the PE was associated with fluid overload in 6 of these 9 patients (66.7%), and with HF in 2 cases (22.2%). Among the patients with transudates, PE was predominantly bilateral (85.7%).
The recommended treatment for PE caused by peritoneal dialysis is to apply conservative measures such as a temporary switch to hemodialysis or the use of smaller-volume exchanges. Recently, Potechin et al. placed a TPC in 8 patients with a PE associated with end-stage renal disease, achieving spontaneous pleurodesis in 3 (37.5%), with no major complications28 (Table 1).
Metastatic Pleural Effusion
Up to 10% of malignant PEs behave biochemically as transudates.53 Several mechanisms may be involved: (1) in an initial phase, PF is accumulated more as a result of obstruction of the lymphatic drainage than tumor infiltrating the pleura. As physiological PF is an ultrafiltrate with low protein levels, in these circumstances it would take several weeks for the accumulated protein concentration to become greater than 50% of the serum concentration54; (2) the pleural transudate is caused by another concomitant disease, since a tumor that affects the pleura does not necessarily produce PE. In some cases, this second disease could be identified55; (3) both the tumor and another disease capable of producing a pleural transudate contribute to the development of the PE.
In the case of transudate pleural effusion, the difficulty lies in knowing when to request PF cytology to rule out malignancy. A recent study suggested that cytology to rule out malignancy should be requested if the PE is left-sided, if nodules/lung masses, pulmonary atelectasis or mediastinal lymphadenopathies are observed on the chest X-ray or CT, if the patient does not have dyspnea, if the PF is of serous bloody appearance, or if PF carcinoembryonic antigen levels are high.53
Other Causes
Other less common diseases can cause transudates, such as nephrotic syndrome9,56–58 (excessive loss of proteins via the glomerulus), urinothorax59,60 (accumulation of urine in the pleural space as a result of injury or obstruction of the urinary tract), duropleural fistula61–64 (cerebrospinal fluid passing into the pleural space as a result of a penetrating injury during a laminectomy), extravascular migration of a central venous catheter65–68 (catheters of insufficient length habitually placed in the left subclavian vein erode or pierce the superior vena cava in the azygos recess), glycinothorax69,70 (perforation of the urinary bladder wall during transurethral prostatectomy with the subsequent administration of a solution of glycine to irrigate the bladder), ventriculoperitoneal and ventriculopleural shunting71–76 (complication of a ventriculoperitoneal shunt in which cerebrospinal fluid passes into the pleural space), and pulmonary veno-occlusive disease77–79 (obstruction of the pulmonary veins and venules due to intimal fibrosis). The etiopathogenic mechanisms by which these PEs are produced are described in Table 2, along with the radiological characteristics, peculiarities and the treatment of each one.
Table 2.
Characteristics of Pleural Transudates With Unusual Etiology.
| Entity | Chest X-ray | Etiopathogenesis | Diagnosis | Other Characteristics | Treatment |
:--- :--- :--- |
| Nephrotic syndrome | Bilateral and small56 | Renal protein loss | Long-standing transudate Hypoalbuminemia Proteinuria | Rule out existence of associated PTE58 | Loop diuretics Low-sodium diet Angiotensin-converting enzyme inhibitors Statins Antiplatelets |
| Urinothorax | One-sided. In the same side as the obstructive urinary disease59 | Obstructive urinary disease with passage of urine from the retroperitoneal or abdominal cavity into the pleural space due to pressure gradient | Creatinine PF/S>159 | Transudate (sometimes exudate as defined by high LDH levels)59 pH and glucose levels may be low59 Fluid of similar appearance and odor to urine59 | Direct treatment of the underlying urinary disease, with or without drainage of the PE |
| Duropleural shunt | One-sided. Variable size | Duropleural shunt. CSF flows through a pressure gradient61,62 | Demonstration of ß2-transferrin in PF63 | Transudate. Clear appearance (like water) with TP<1 g/dl61,62 | Surgical ligation or chest drain |
| EMCVC | Uni or bilateral Small/massive Ipsi/contralateral | SCV erosion caused by a catheter too short in length | PE on chest Rx Abnormal catheter location Transudate65 | If glucose is administered, glucose PF/S>1 If parenteral nutrition is administered, milky PF (triglycerides>110 mg/dl) TP<1 g/dl in all cases. | Catheter withdrawal If PE is small, observation. Otherwise, therapeutic thoracentesis or chest tube |
| Glycinothorax | Right | Passage of glycine solution into the abdominal cavity and then into the pleural space | Previous bladder surgery Elevated glycine PF/S69,70 | PF is sometimes bloody | Discontinue bladder irrigation Therapeutic thoracentesis if needed |
| Ventriculoperitoneal and ventriculopleural shunts | Single-sided of variable size | Obstruction of fistula by fibrous tissue or accumulation of detritus | PF is CSF72 | Sometimes eosinophilic PE Sometimes empyema | Add acetazolamide |
| Pulmonary veno-occlusive disease | Usually bilateral On occasions, Kerley B lines77,78 | Increase of fluid in the pulmonary interstitium as a result of obstruction of the pulmonary veins | High probability based on clinical suspicion, physical examination, bronchoscopy and radiologic results79 Lung biopsy | PF is probably transudate | Oxygen, high-dose diuretics, slow, progressive increase of epoprostenol dosing, and lung transplant |
CSF: cerebrospinal fluid; EMCVC: extravascular migration of a central venous catheter; LDH: lactate dehydrogenase; PE: pleural effusion; PF: pleural fluid; PF/S: pleural fluid/serum ratio; PTE: pulmonary thromboembolism; Rx: chest X-ray; SVC, superior vena cava; TP: total protein.
Conclusions
Pleural transudates are caused by a wide spectrum of diseases, ranging from HF to very rare entities. Diagnosis is based on clinical data and PF analysis, which can contribute decisively to diagnosis (Fig. 3). Although medical treatment for each causative disease is usually well established, at least among the more common ones, some cases are refractory, and in these, various possibilities has been opened up with the introduction of new interventional procedures, not yet sufficiently validated, that make management more complex. Studies and clinical trials that provide quality data and more evidence are required if the treatment of these patients is to be optimized.
Fig. 3.
Diagnosis of pleural transudates through the analysis of pleural fluid. PF/S: pleural fluid/serum ratio; EMCVC: extravascular migration of a central venous catheter; NT-proBNP: N-terminal pro-brain natriuretic peptide.
Authorship
Lucía Ferreiro. Author and editor. Concept and design. Final approval of the manuscript.
José Manuel Porcel. Co-author. Review and final approval of the manuscript.
Luis Valdés. Author and editor. Concept and design. Final approval of the manuscript.
Conflict of Interests
The authors state that they have no conflict of interests.
Acknowledgements
We thank Fernando Vázquez Vázquez for designing the figures.
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☆
Please cite this article as: Ferreiro L, Porcel JM, Valdés L. Diagnóstico y manejo de los trasudados pleurales. Arch Broncopneumol. 2017;53:629–636.
Copyright © 2017. SEPAR
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13582 | https://www.doubtnut.com/qna/8486896 | If A dn B be 3×3 matrices the AB=0 implies (A) A=0orB=0 (B) A=0andB=0 (C) |A|=0or|B|=0 (D) |A|=0and|B|=0
To solve the problem, we need to analyze the implications of the product of two matrices A and B being the zero matrix. We are given that AB=0, where A and B are both 3×3 matrices. 1. Understanding Matrix Multiplication: - The product of two matrices A and B is defined such that each element of the resulting matrix is the dot product of the corresponding row of A and the column of B. - If AB=0, this means that every entry of the resulting matrix is zero. 2. Analyzing the Condition AB=0: - The equation AB=0 does not necessarily imply that either A or B must be the zero matrix. It is possible for one matrix to be non-zero while the other matrix is such that their product results in the zero matrix. 3. Counterexamples: - Consider the following matrices: A=⎛⎜⎝100010000⎞⎟⎠,B=⎛⎜⎝000000001⎞⎟⎠ - Here, A is not the zero matrix, and B is not the zero matrix, but their product AB results in the zero matrix: AB=⎛⎜⎝100010000⎞⎟⎠⎛⎜⎝000000001⎞⎟⎠=⎛⎜⎝000000000⎞⎟⎠ 4. Determinants: - The determinant of a matrix being zero indicates that the matrix is singular (non-invertible). If |A|=0 or |B|=0, it means at least one of the matrices is singular. - However, AB=0 does not imply that both |A| and |B| must be zero. It is sufficient for at least one of them to be singular. 5. Conclusion: - From the analysis, we conclude that the correct answer is option (C): |A|=0 or |B|=0. This means that at least one of the matrices must be singular for their product to be the zero matrix. Final Answer: (C) |A|=0 or |B|=0
To solve the problem, we need to analyze the implications of the product of two matrices A and B being the zero matrix. We are given that AB=0, where A and B are both 3×3 matrices.
Understanding Matrix Multiplication:
The product of two matrices A and B is defined such that each element of the resulting matrix is the dot product of the corresponding row of A and the column of B.
If AB=0, this means that every entry of the resulting matrix is zero.
Analyzing the Condition AB=0:
The equation AB=0 does not necessarily imply that either A or B must be the zero matrix. It is possible for one matrix to be non-zero while the other matrix is such that their product results in the zero matrix.
Counterexamples:
Consider the following matrices:
A=⎛⎜⎝100010000⎞⎟⎠,B=⎛⎜⎝000000001⎞⎟⎠
Here, A is not the zero matrix, and B is not the zero matrix, but their product AB results in the zero matrix:
AB=⎛⎜⎝100010000⎞⎟⎠⎛⎜⎝000000001⎞⎟⎠=⎛⎜⎝000000000⎞⎟⎠
Determinants:
The determinant of a matrix being zero indicates that the matrix is singular (non-invertible). If |A|=0 or |B|=0, it means at least one of the matrices is singular.
However, AB=0 does not imply that both |A| and |B| must be zero. It is sufficient for at least one of them to be singular.
Conclusion:
From the analysis, we conclude that the correct answer is option (C): |A|=0 or |B|=0. This means that at least one of the matrices must be singular for their product to be the zero matrix.
Final Answer:
(C) |A|=0 or |B|=0
Similar Questions
If A and B are any 2×2 matrices then det(A+B)=0 implies (A) detA+detB=0 (B) detA=0ordetB=0 (C) AB=0→|A|=0and|B|=0 (D) AB=0→A=0orB=0
If A and B are square matrices of order 3 then (A) AB=0→|A|=0or|B|=0 (B) AB=0→|A|=0and|B|=0 (C) Adj(AB)=AdjAAdjB (D) (A+B)−1=A−1+B−1
Knowledge Check
A and B be 3×3 matrices such that AB+A=0, then
a.b=0 implies only
(a+b).(a−b)=0 implies that
If A and B are square matrices of the same order then (A+B)2=A2+2AB+B2 implies (A) AB=0 (B) AB+BA=0 (C) AB=BA (D) none of these
If B is a 3×3 matrics such that B2=0 then det[(1+B)50−50B]=0
If A=[(1,0,),(2,0)]andB= then= (A) AB=0,BA=0 (B) AB=0,BA≠0 (C) AB≠0,BA=0 (D) AB≠0,BA≠)
If A
and B
are two matrices such
that AB=A
and BA=B
, then B2
is equal to
(a) B
(b) A
(c) 1
(d) 0
If A, B are square matrices of order 3, A is non-singular and AB=0, then B is a
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13583 | https://dictionary.cambridge.org/es/diccionario/ingles/disjointed | DISJOINTED | significado en inglés - Cambridge Dictionary
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Significado de disjointed en inglés
disjointed
adjective
uk
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/dɪsˈdʒɔɪn.tɪd/us
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/dɪsˈdʒɔɪn.t̬ɪd/
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(especially of words or ideas) not well connected or well ordered:
The script was disjointed and hard to follow.
Sinónimo
garbled
SMART Vocabulary: palabras y expresiones relacionadas
Separateness and isolation in space
apart
apartheid
atomistic
atomized
bitty
free of something
freely
gappy
give something/someone a wide berth idiom
hermetically sealed
private
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resegregation
secluded
secludedly
unattached
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(Definición de disjointed del Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press)
disjointed | Diccionario de Inglés Americano
disjointed
adjective
us
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/dɪsˈdʒɔɪn·tɪd/
Add to word listAdd to word list
(esp. of words or ideas) not well connected or well ordered, and therefore often confusing:
She gave a disjointed account of getting lost in the woods.
(Definición de disjointed del Cambridge Academic Content Dictionary © Cambridge University Press)
Ejemplos de disjointed
disjointed
In each case the fruit of this toxicity is something that is disjointed, distorted and disrupting.
Del Cambridge English Corpus
The music is nervous, disjointed, indeed unsettling in places.
Del Cambridge English Corpus
This avenue is likely to lead to a disjointed heap of ad hoc stipulations.
Del Cambridge English Corpus
Then, some morphological operations are carried out for connecting disjointed blobs.
Del Cambridge English Corpus
Due to this disjointed transfer of power, issues of relief were largely swept under the carpet until the 1820s.
Del Cambridge English Corpus
The one major flaw in its presentation is the typesetting of the small cap constraint names, which is at times awkwardly disjointed.
Del Cambridge English Corpus
The rather disjointed episodic style also adds to the tension.
Del Cambridge English Corpus
In the past, research on natural resources has been too often conducted in a disjointed, fragmented fashion.
Del Cambridge English Corpus
Their melodies grew more fragmented and increasingly disjointed; the vocal line blurred and merged with the instruments.
Del Cambridge English Corpus
And third, the domestic institutions that defined, standardized and conferred status to the 'talented' became increasingly fragmented, disjointed and out of sync with each other.
Del Cambridge English Corpus
In general, the main limitation of the book is its lack of direction, whereby contributions are disjointed.
Del Cambridge English Corpus
The bodies are like plasticine, disjointed and collapsing under pressure, while the movements seem desultory and impersonal.
Del Cambridge English Corpus
The gaps in the contributions make the test-issue part of the book seem very disjointed.
Del Cambridge English Corpus
The resulting picture of the syntactic input to agreement is one that is rather disjointed.
Del Cambridge English Corpus
At best, they may be permitted to give a static performance and mumble through a few disjointed lines.
Del Cambridge English Corpus
Las opiniones mostradas en los ejemplos no representan las opiniones de los editores de Cambridge University Press o de sus licenciantes.
¿Cómo se pronuncia _disjointed_?
Traducciones de disjointed
en chino (tradicional)
（尤指詞語或想法）不連貫的，沒有條理的，雜亂無章的…
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en chino (simplificado)
（尤指词语或想法）不连贯的，没有条理的，杂乱无章的…
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en español
inconexo…
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en portugués
desconexo…
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en más idiomas
en polaco
en turco
en ruso
nieskładny, bezładny…
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bölük börçük, tutarsız, kopuk…
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бессвязный…
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Victoria sponge
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13584 | https://math.stackexchange.com/questions/3844215/how-to-solve-a-sin-theta-b-cos-theta-c | trigonometry - How to solve $A\sin{\theta} + B\cos{\theta} = C$? - Mathematics Stack Exchange
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How to solve A sin θ+B cos θ=C A sinθ+B cosθ=C? [duplicate]
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Solving a sin θ+b cos θ=c a sinθ+b cosθ=c (4 answers)
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I've stumbled upon a equation in the form
A sin θ+B cos θ=C A sinθ+B cosθ=C
What would be the steps necessary to solving it?
Thank you.
trigonometry
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edited Sep 28, 2020 at 22:52
Blue
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asked Sep 28, 2020 at 22:50
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Obs: A A, B B and C C are known; the unknown variable is θ θ.DaviFN –DaviFN 2020-09-28 22:54:43 +00:00 Commented Sep 28, 2020 at 22:54
the best approach is divide everything with C C, and if you are lucky, there will exists an angle α α such that cos α=A C cosα=A C and sin α=B C sinα=B C. So you have sin(α+θ)=1 sin(α+θ)=1 luisfelipe18 –luisfelipe18 2020-09-28 22:57:02 +00:00 Commented Sep 28, 2020 at 22:57
You do not need to be lucky: C=A 2+B 2−−−−−−−√C=A 2+B 2.markvs –markvs 2020-09-28 22:58:42 +00:00 Commented Sep 28, 2020 at 22:58
1 Well yes, ... but that's a different C C, possibly, from the one given in the problem.John Hughes –John Hughes 2020-09-28 22:59:44 +00:00 Commented Sep 28, 2020 at 22:59
This question is a duplicate of MSE question 2502976 "Solving a sin θ+b cos θ=c a sinθ+b cosθ=c".Somos –Somos 2020-09-29 12:56:14 +00:00 Commented Sep 29, 2020 at 12:56
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Typical is to let u=arctan(A/B)u=arctan(A/B) (dealing with the special case of B=0 B=0 separately), r=A 2+B 2−−−−−−−√r=A 2+B 2. Then
A B=r sin u=r cos u A=r sinu B=r cosu
(unless I've swapped those two). Now your equation reads
r sin u sin θ+r cos u cos θ=C r sinu sinθ+r cosu cosθ=C
which you rewrite as
cos(u−θ)θ=C r=u−arccos(C r)cos(u−θ)=C r θ=u−arccos(C r)
and you're done.
NB: If |C|>|r||C|>|r|, then there is no (real) solution. (Hat-tip to @LuisFelipe for making me add the "real".) If r r is zero and C C is nonzero, there's also no solution. If r=C=0 r=C=0, then every value of θ θ is a solution. That leaves only the special case B=0,A≠0 B=0,A≠0 for you to work out.
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edited Sep 28, 2020 at 23:27
answered Sep 28, 2020 at 22:58
John HughesJohn Hughes
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there is always solution in C C :D luisfelipe18 –luisfelipe18 2020-09-28 23:01:48 +00:00 Commented Sep 28, 2020 at 23:01
yes ... and if the OP had wanted that, I'll bet that OP could have solved the problem this way without my help.John Hughes –John Hughes 2020-09-28 23:03:22 +00:00 Commented Sep 28, 2020 at 23:03
There's no solution in C C if A=B=0 A=B=0 and C=1 C=1 ;)user801306 –user801306 2020-09-28 23:17:09 +00:00 Commented Sep 28, 2020 at 23:17
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A sin θ+B cos θ=C A sinθ+B cosθ=C
use the identity
cos θ=±1−sin 2 θ−−−−−−−−√cosθ=±1−sin 2θ
so
A sin θ±B 1−sin 2 θ−−−−−−−−√=C A sinθ±B 1−sin 2θ=C
±B 1−sin 2 θ−−−−−−−−√=C−A sin θ±B 1−sin 2θ=C−A sinθ
B 2(1−sin 2 θ)=(C−A sin θ)2 B 2(1−sin 2θ)=(C−A sinθ)2
let x=sin θ x=sinθ
B 2(1−x 2)=(C−A x)2 B 2(1−x 2)=(C−A x)2
B 2−B 2 x 2=C 2−2 A C x+A 2 x 2 B 2−B 2 x 2=C 2−2 A C x+A 2 x 2
(A 2+B 2)x 2−(2 A C)x+(C 2−B 2)=0(A 2+B 2)x 2−(2 A C)x+(C 2−B 2)=0
then use the quadratic formula to solve it and then complete the solution to find the θ θ values
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answered Sep 28, 2020 at 23:16
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If either A A or B B equals zero we can use basic trigonometry to solve the equation, so let's assume neither A A nor B B is zero. Under this assumption we can write
A cos(θ)+B sin(θ)=α cos(θ−arctan(B/A))A cos(θ)+B sin(θ)=α cos(θ−arctan(B/A))
where
α=A cos(arctan(B/A))+B sin(arctan(B/A))α=A cos(arctan(B/A))+B sin(arctan(B/A))
Can you continue from here?
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edited Sep 29, 2020 at 0:15
answered Sep 29, 2020 at 0:06
user801306 user801306
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Here's a general guide and explanation for problems of your type:
If we have an expression, A sin x+B cos x A sinx+B cosx, let us assume it can be written in the form R sin(x+α)R sin(x+α) Now to see if we can find values for R R and α α in terms of A A and B B. Using the compund angle formulae, also known as the addition formulae:
R sin(x+α)=R sin x cos α+R sin α cos x=A sin x+B cos x R sin(x+α)=R sinx cosα+R sinα cosx=A sinx+B cosx
So we have
R cos α=A,R sin α=B R cosα=A,R sinα=B
So dividing the second equality by the first:
tan α=B A tanα=B A
meaning we can find α α in terms of A A and B B, as we wanted. Now, to find R R: Squaring the 2 2 equalities above we have
R 2 cos 2 α+R 2 sin 2 α=R 2(cos 2 α+sin 2 α)=R 2=A 2+B 2⟹R=A 2+B 2−−−−−−−√R 2 cos 2α+R 2 sin 2α=R 2(cos 2α+sin 2α)=R 2=A 2+B 2⟹R=A 2+B 2
So, to finish off by recapping what we have learnt:
tan α=B A,R=A 2+B 2−−−−−−−√tanα=B A,R=A 2+B 2
Try applying that to your question. I hope that was helpful :)
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answered Sep 29, 2020 at 12:29
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13585 | https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.01%3A_Oxidation_Numbers | Skip to main content
11.1: Oxidation Numbers
Last updated
: Jun 10, 2019
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11: Properties of Reactions
11.2: The Nature of Oxidation and Reduction
Page ID
: 155696
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Outcomes
Assign oxidation numbers to free elements or elements in a compound or ion.
Oxidation Numbers
An oxidation number is a positive or negative number that is assigned to an atom to indicate its degree of oxidation or reduction. The term oxidation state is often used interchangeably with oxidation number. A partial electron transfer is a shift in the electron density near an atom as a result of a change in the other atoms to which it is covalently bonded. That charge shift is based on the relative electronegativities of the atoms involved in the bond.
Overall, the oxidation number of an atom in a molecule is the charge that the atom would have if all polar covalent and ionic bonds resulted in a complete transfer of electrons from the less electronegative atom to the more electronegative one. Oxidation numbers can be assigned using the set of rules outlined below.
The oxidation number of an atom in a neutral free element is zero. A free element is considered to be any element in an uncombined state, whether monatomic or polyatomic. For example, the oxidation number of each atom in , , , , and is zero.
The oxidation number of a monatomic (composed of one atom) ion is the same as the charge of the ion. For example, the oxidation numbers of , , and are , , and , respectively.
The oxidation number of oxygen in most compounds is .
The oxidation number of hydrogen in most compounds is .
The oxidation number of fluorine in all compounds is . Other halogens usually have an oxidation number of in binary compounds, but can have variable oxidation numbers depending on the bonding environment.
In a neutral molecule, the sum of the oxidation numbers of all atoms is zero. For example, in , the oxidation numbers of and are and , respectively. Because there are two hydrogen atoms in the formula, the sum of all the oxidation numbers in is .
In a polyatomic ion, the sum of the oxidation numbers of all atoms is equal to the overall charge on the ion. For example, in , the oxidation numbers of and are and , respectively. The sum of all oxidation numbers in the sulfate ion would be , which is the charge of the ion.
An examination of the rules for assigning oxidation numbers reveals that there are many elements for which there are no specific rules, such as nitrogen, sulfur, and chlorine. These elements, as well as some others, can have variable oxidation numbers depending on the other atoms to which they are covalently bonded in a molecular compound. It is useful to analyze a few molecules in order to see the strategy to follow in assigning oxidation numbers to other atoms.
Oxidation numbers for the atoms in a binary ionic compound are easy to assign because they are equal to the charge of the ion (rule 2). In , the oxidation number of iron is , while the oxidation number of chlorine is . In , the calcium is , while the phosphorus is . This is because an ionic compound is in the form of a crystal lattice that is actually composed of these ions.
Assigning oxidation numbers for molecular compounds is trickier. The key is to remember rule 6: that the sum of all the oxidation numbers for any neutral species must be zero. Make sure to account for any subscripts which appear in the formula. As an example, consider the compound nitric acid, . According to rule 4, the oxidation number of hydrogen is . According to rule 3, the oxidation number of oxygen is . There is no rule regarding nitrogen, but its oxidation number can be calculated as follows.
The oxidation number of the nitrogen atom in is . Often when assigning oxidation numbers, it is convenient to write it above the symbol within the formula.
You may wonder if there are any limits on the value of oxidation numbers. The key point to consider is the octet rule. Since nitrogen has 5 valence electrons, the most that it can "lose" while forming bonds in a molecule is 5, so its highest possible oxidation number is . Alternatively, it could gain up to 3 electrons, and so its lowest (most negative) possible oxidation number is . Similarly, chlorine can have oxidation numbers ranging from to .
Now consider the ionic compound sodium thiosulfate, (Figure ). It contains the thiosulfate polyatomic ion, . The sodium is not part of the covalently bonded polyatomic ion, and so its oxidation number is the same as it would be in a binary ionic compound, . The sulfur is the atom whose oxidation number is not covered by one of the rules. The oxidation number of sulfur is assigned the variable in the following calculation. Remember the sum of the oxidation numbers of all the elements must equal zero because is a neutral compound.
Sulfur has an oxidation number of in . Notice how the subscript of 2 for the atom had to be accounted for by dividing the result of the subtraction by 2. When assigning oxidation numbers, you do so for each individual atom. In the above example, the oxidation number of sulfur could also have been determined by looking at just the thiosulfate ion, .
Contributors and Attributions
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
11: Properties of Reactions
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Central Binomial Coefficient
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The th central binomial coefficient is defined as
(1)
(2)
where is a binomial coefficient, is a factorial, and is a double factorial.
These numbers have the generating function
(3)
The first few values are 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, ... (OEIS A000984). The numbers of decimal digits in for , 1, ... are 1, 6, 59, 601, 6019, 60204, 602057, 6020597, ... (OEIS A114501). These digits converge to the digits in the decimal expansion of (OEIS A114493).
The central binomial coefficients are never prime except for .
A scaled form of the central binomial coefficient is known as a Catalan number
(4)
Erdős and Graham (1975) conjectured that the central binomial coefficient is neversquarefree for , and this is sometimes known as the Erdős squarefree conjecture. Sárkőzy's theorem (Sárkőzy 1985) provides a partial solution which states that the binomial coefficient is never squarefree for all sufficiently large (Vardi 1991). The conjecture of Erdős and Graham was subsequently proved by Granville and Ramare (1996), who established that the onlysquarefree values are 2, 6, and 70, corresponding to , 2, and 4. Sander (1992) subsequently showed that are also never squarefree for sufficiently large as long as is not "too big."
The central binomial coefficient is divisible by a prime iff the base- representation of contains no digits greater than (P.Carmody, pers. comm., Sep.4, 2006). For , the first few such are 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, ... (OEIS A005836).
A plot of the central binomial coefficient in the complex plane is given above.
The central binomial coefficients are given by the integral
(5)
(Moll 2006, Bailey et al. 2007, p.163).
Using Wolstenholme's theorem and the fact that , it follows that
(6)
for an odd prime (T.D.Noe, pers. comm., Nov.30, 2005).
A less common alternative definition of the th central binomial coefficient of which the above coefficients are a subset is , where is the floor function. The first few values are 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, ... (OEIS A001405). The central binomial coefficients have generating function
(7)
These modified central binomial coefficients are squarefree only for , 2, 3, 4, 5, 7, 8, 11, 17, 19, 23, 71, ... (OEIS A046098), with no others less than (E.W.Weisstein, Feb.4, 2004).
A fascinating series of identities involving inverse central binomial coefficients times small powers are given by
(8)
(9)
(10)
(11)
(12)
(OEIS A073016, A073010, A086463, and A086464; Comtet 1974, p.89; Le Lionnais 1983, pp.29, 30, 41, 36), which follow from the beautiful formula
(13)
for , where is a generalized hypergeometric function. Additional sums of this type include
(14)
(15)
(16)
where is the polygamma function and is the Riemann zeta function (Plouffe 1998).
Similarly, we have
(17)
(18)
(19)
(20)
(OEIS A086465, A086466, A086467, and A086468; Le Lionnais 1983, p.35; Guy 1994, p.257), where is the Riemann zeta function. These follow from the analogous identity
(21)
See also
Binomial Coefficient, Binomial Sums, Catalan Number, Central Fibonomial Coefficient, Central Trinomial Coefficient, Erdős Squarefree Conjecture, Staircase Walk, Sárközy's Theorem, Quota System
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References
Bailey, D.H.; Borwein, J.M.; Calkin, N.J.; Girgensohn, R.; Luke, D.R.; and Moll, V.H. Experimental Mathematics in Action. Wellesley, MA: A K Peters, 2007.Boros, G. and Moll, V. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. Cambridge, England: Cambridge University Press, p.14, 2004.Comtet, L. Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, 1974.Erdős, P.; Graham, R.L.; Ruzsa, I.Z.; and Straus, E.G. "On the Prime Factors of ." Math. Comput.29, 83-92, 1975.Granville, A. and Ramare, O. "Explicit Bounds on Exponential Sums and the Scarcity of Squarefree Binomial Coefficients." Mathematika43, 73-107, 1996.Guy, R.K. Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, 1994.Le Lionnais, F. Les nombres remarquables. Paris: Hermann, 1983.Lehmer, D.H. "Interesting Series Involving the Central Binomial Coefficient." Amer. Math. Monthly92, 449-457, 1985.Moll, V.H. "Some Questions in the Evaluation of Definite Integrals." MAA Short Course, San Antonio, TX. Jan.2006. S. "The Art of Inspired Guessing." Aug. 7, 1998. J.W. "On Prime Divisors of Binomial Coefficients." Bull. London Math. Soc.24, 140-142, 1992.Sárkőzy, A. "On Divisors of Binomial Coefficients. I." J. Number Th.20, 70-80, 1985.Sloane, N.J.A. Sequences A000984/M1645, A001405/M0769, A005836/M2353, A046098, A073010, A073016, A086463, A086464A086465, A086466, A086467, A086468, A114493, and A114501 in "The On-Line Encyclopedia of Integer Sequences."Vardi, I. "Application to Binomial Coefficients," "Binomial Coefficients," "A Class of Solutions," "Computing Binomial Coefficients," and "Binomials Modulo and Integer." §2.2, 4.1, 4.2, 4.3, and 4.4 in Computational Recreations in Mathematica. Redwood City, CA: Addison-Wesley, pp.25-28 and 63-71, 1991.
Referenced on Wolfram|Alpha
Central Binomial Coefficient
Cite this as:
Weisstein, Eric W. "Central Binomial Coefficient." From MathWorld--A Wolfram Resource.
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13587 | https://math.stackexchange.com/questions/4891865/question-regarding-nature-of-logarithmic-equations | algebra precalculus - Question regarding nature of logarithmic equations - Mathematics Stack Exchange
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Question regarding nature of logarithmic equations
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\begingroup
While reading my textbook's chapter about logarithms and seeing the solved examples I noticed in various places that the author was able to make the \log just disappear in a equation or inequality by making some changes on either side of the inequality or the equation for example: -1 \leq \log_9\left(\frac{x^2}{4}\right) \leq 1 changes to 9^{-1} \leq \frac{x^2}{4} \leq 9^1.
This confuses me how does this happen and another curious thing when doing this "logarithm disappearing" on logs with bases less than 1 the sign of the inequality changes.
For example in one question the following was done \log_{1/2}(x^2 - 7x + 13) > 0 changes to x^2 - 7x + 13 < 1
I do not understand how does 1 just appear on the other side and why does the inequality reverse, could someone explain this to me and my this "log disappearing" is different for bases less than 1 and greater than 1.
algebra-precalculus
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edited Apr 3, 2024 at 14:18
Xander Henderson♦
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koiboikoiboi
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\endgroup
5
\begingroup If f(x)>a with f an order-preserving invertible function of inverse f^{-1} for which f^{-1}(a) exists, x>f^{-1}(a). If f is invertible but reverses order, you have to reverse the inequality's direction.\endgroup J.G. –J.G. 2024-04-02 15:16:52 +00:00 Commented Apr 2, 2024 at 15:16
\begingroup Ok but how do the logs disappear?\endgroup koiboi –koiboi 2024-04-02 15:17:36 +00:00 Commented Apr 2, 2024 at 15:17
\begingroup By applying f^{-1} to the original inequality.\endgroup J.G. –J.G. 2024-04-02 15:18:02 +00:00 Commented Apr 2, 2024 at 15:18
\begingroup So he raises the bases to the logs itself doesnt he?\endgroup koiboi –koiboi 2024-04-02 15:21:08 +00:00 Commented Apr 2, 2024 at 15:21
\begingroup a < \log_b(y) < c is equivalent to b^a1, and to b^a >y >b^c when 0 < b< 1, because \log_b is a strictly monotonic function (increasing in the first case and decreasing in the second)\endgroup Henry –Henry 2024-04-02 15:30:10 +00:00 Commented Apr 2, 2024 at 15:30
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\begingroup
You can see taking the logarithm as the inverse of exponentiation. So if you have -1\leq\log_9\left(\frac{x^2}{4}\right)<1\Rightarrow 9^{-1}\leq9^{\log_9\left(\frac{x^2}{4}\right)}<9 which simplifies as your teacher told you. As for the second example the same thing applies: \log_{\frac{1}{2}}(x^2-7x+13)>0\Rightarrow x^2-7x+13<\left(\frac{1}{2}\right)^0=1 here the ">" is inverted because for n\in (0,1)\log_n(x) inverts the order, while it preserves it for n\geq 1
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edited Apr 2, 2024 at 15:34
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\endgroup
1
3 \begingroup It is worthwhile to note the reason why the order is inverted: because for any base 0<b<1, \log_b is a strictly decreasing function, so its inverse function is also strictly decreasing. Also, in your notation n must be strictly greater than 1. Logarithm with base 1 is not defined.\endgroup Randy Marsh –Randy Marsh 2024-04-02 15:48:00 +00:00 Commented Apr 2, 2024 at 15:48
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\begingroup
I am writing this out in great Detail , due to matching the OP level.
(1) We make "\log" disappear by "raising the Base to the Power" :
A \le \log_a (B) \le \log_a (C) \le D a ^ A \le a ^ {\log_a (B)} \le a ^ {\log_a (C)} \le a ^ D a ^ A \le B \le C \le a ^ D
It is similar to adding Constant through-out or multiplying by Positive Constant though-out , ETC.
We are "raising to Power" though-out.
In OP Case : -1 \leq \log_9(\frac{x^2}{4}) \leq 1 9^{-1} \leq 9^{\log_9(\frac{x^2}{4})} \leq 9^{1}9^{-1} \leq \frac{x^2}{4} \leq 9^{1}
(2) When we have Base a less than 1 , we use the "rule" like : \log_a(B)=\log_A(B)/\log_A(a) , where the new Base A might be more than 1.
When Denominator is less than 0 (negative) , it can be eliminated by interchanging the relation between \le&\ge
That "rule" is like : If -a > 0 , then a < 0& vice-versa
In OP Case : log_{1/2}{(x^2 - 7x + 13)} > 0 log_{A}{(x^2 - 7x + 13)}/log_{A}{1/2} > 0 Here , though A is arbitrary , it must be larger than 1 : We can take A=2
log_{A}{(x^2 - 7x + 13)}/(-1) > 0(-1)log_{A}{(x^2 - 7x + 13)} > 0
We then interchange > to < to eliminate the negation : log_{A}{(x^2 - 7x + 13)} < 0
A^{log_{A}{(x^2 - 7x + 13)}} < A^{0}(x^2 - 7x + 13) < 1(x^2 - 7x + 12) < 0(x - 3x)(x - 4x) < 0
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edited Apr 3, 2024 at 7:29
answered Apr 2, 2024 at 15:47
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+ Functional Groups in Chemistry
11m
4. BONUS: Lab Techniques and Procedures
1h 38m
Laboratory Materials
29m
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12m
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12m
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6m
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4m
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17m
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14m
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7m
+ Addition and Subtraction Operations
6m
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6m
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20m
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7m
6. Chemical Quantities & Aqueous Reactions
3h 53m
Solutions
6m
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18m
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15m
+ Dilutions
15m
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16m
+ Electrolytes
18m
+ Molecular Equations
18m
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13m
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14m
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18m
+ Calculate Oxidation Numbers
15m
+ Redox Reactions
17m
+ Balancing Redox Reactions: Acidic Solutions
17m
+ Balancing Redox Reactions: Basic Solutions
17m
+ Activity Series
10m
7. Gases
3h 49m
Pressure Units
6m
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18m
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13m
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6m
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13m
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12m
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6m
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19m
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13m
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14m
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18m
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14m
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13m
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9m
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10m
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8m
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4m
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14m
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9m
8. Thermochemistry
2h 37m
Nature of Energy
6m
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7m
+ First Law of Thermodynamics
7m
+ Internal Energy
8m
+ Endothermic & Exothermic Reactions
7m
+ Heat Capacity
19m
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24m
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10m
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8m
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12m
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9m
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12m
+ Hess's Law
23m
9. Quantum Mechanics
2h 58m
Wavelength and Frequency
6m
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8m
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13m
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10m
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17m
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9m
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17m
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14m
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5m
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13m
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5m
+ Quantum Numbers: Principal Quantum Number
5m
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10m
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11m
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9m
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11m
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6m
10. Periodic Properties of the Elements
3h 9m
The Electron Configuration
22m
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4m
+ The Electron Configurations: Exceptions
13m
+ The Electron Configuration: Ions
12m
+ Paramagnetism and Diamagnetism
8m
+ The Electron Configuration: Quantum Numbers
16m
+ Valence Electrons of Elements
12m
+ Periodic Trend: Metallic Character
3m
+ Periodic Trend: Atomic Radius
8m
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13m
+ Periodic Trend: Ionization Energy
12m
+ Periodic Trend: Successive Ionization Energies
11m
+ Periodic Trend: Electron Affinity
10m
+ Periodic Trend: Electronegativity
5m
+ Periodic Trend: Effective Nuclear Charge
21m
+ Periodic Trend: Cumulative
12m
11. Bonding & Molecular Structure
3h 29m
Lewis Dot Symbols
10m
+ Chemical Bonds
13m
+ Dipole Moment
11m
+ Octet Rule
10m
+ Formal Charge
9m
+ Lewis Dot Structures: Neutral Compounds
20m
+ Lewis Dot Structures: Sigma & Pi Bonds
14m
+ Lewis Dot Structures: Ions
15m
+ Lewis Dot Structures: Exceptions
14m
+ Lewis Dot Structures: Acids
15m
+ Resonance Structures
21m
+ Average Bond Order
4m
+ Bond Energy
15m
+ Coulomb's Law
6m
+ Lattice Energy
12m
+ Born Haber Cycle
14m
12. Molecular Shapes & Valence Bond Theory
1h 57m
Valence Shell Electron Pair Repulsion Theory
5m
+ Equatorial and Axial Positions
10m
+ Electron Geometry
11m
+ Molecular Geometry
18m
+ Bond Angles
14m
+ Hybridization
12m
+ Molecular Orbital Theory
12m
+ MO Theory: Homonuclear Diatomic Molecules
10m
+ MO Theory: Heteronuclear Diatomic Molecules
7m
+ MO Theory: Bond Order
14m
13. Liquids, Solids & Intermolecular Forces
2h 23m
Molecular Polarity
10m
+ Intermolecular Forces
20m
+ Intermolecular Forces and Physical Properties
11m
+ Clausius-Clapeyron Equation
18m
+ Phase Diagrams
13m
+ Heating and Cooling Curves
27m
+ Atomic, Ionic, and Molecular Solids
11m
+ Crystalline Solids
4m
+ Simple Cubic Unit Cell
7m
+ Body Centered Cubic Unit Cell
12m
+ Face Centered Cubic Unit Cell
6m
14. Solutions
3h 1m
Solutions: Solubility and Intermolecular Forces
17m
+ Molality
15m
+ Parts per Million (ppm)
13m
+ Mole Fraction of Solutions
8m
+ Solutions: Mass Percent
12m
+ Types of Aqueous Solutions
8m
+ Intro to Henry's Law
4m
+ Henry's Law Calculations
12m
+ The Colligative Properties
14m
+ Boiling Point Elevation
16m
+ Freezing Point Depression
10m
+ Osmosis
19m
+ Osmotic Pressure
10m
+ Vapor Pressure Lowering (Raoult's Law)
16m
15. Chemical Kinetics
2h 53m
Intro to Chemical Kinetics
4m
+ Energy Diagrams
9m
+ Catalyst
9m
+ Factors Influencing Rates
10m
+ Average Rate of Reaction
7m
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5m
+ Instantaneous Rate
5m
+ Collision Theory
7m
+ Arrhenius Equation
25m
+ Rate Law
24m
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17m
+ Integrated Rate Law
23m
+ Half-Life
23m
16. Chemical Equilibrium
2h 29m
Intro to Chemical Equilibrium
7m
+ Equilibrium Constant K
13m
+ Equilibrium Constant Calculations
9m
+ Kp and Kc
22m
+ Using Hess's Law To Determine K
9m
+ Calculating K For Overall Reaction
15m
+ Le Chatelier's Principle
20m
+ ICE Charts
34m
+ Reaction Quotient
15m
17. Acid and Base Equilibrium
5h 1m
Acids Introduction
9m
+ Bases Introduction
7m
+ Binary Acids
15m
+ Oxyacids
10m
+ Bases
14m
+ Amphoteric Species
5m
+ Arrhenius Acids and Bases
5m
+ Bronsted-Lowry Acids and Bases
21m
+ Lewis Acids and Bases
12m
+ The pH Scale
16m
+ Auto-Ionization
9m
+ Ka and Kb
16m
+ pH of Strong Acids and Bases
9m
+ Ionic Salts
17m
+ pH of Weak Acids
31m
+ pH of Weak Bases
32m
+ Diprotic Acids and Bases
8m
+ Diprotic Acids and Bases Calculations
30m
+ Triprotic Acids and Bases
9m
+ Triprotic Acids and Bases Calculations
17m
18. Aqueous Equilibrium
4h 47m
Intro to Buffers
20m
+ Henderson-Hasselbalch Equation
19m
+ Intro to Acid-Base Titration Curves
13m
+ Strong Titrate-Strong Titrant Curves
9m
+ Weak Titrate-Strong Titrant Curves
15m
+ Acid-Base Indicators
8m
+ Titrations: Weak Acid-Strong Base
38m
+ Titrations: Weak Base-Strong Acid
41m
+ Titrations: Strong Acid-Strong Base
11m
+ Titrations: Diprotic & Polyprotic Buffers
32m
+ Solubility Product Constant: Ksp
17m
+ Ksp: Common Ion Effect
18m
+ Precipitation: Ksp vs Q
12m
+ Selective Precipitation
9m
+ Complex Ions: Formation Constant
18m
19. Chemical Thermodynamics
1h 50m
Spontaneous vs Nonspontaneous Reactions
7m
+ Entropy
23m
+ Entropy Calculations
13m
+ Entropy Calculations: Phase Changes
6m
+ Third Law of Thermodynamics
7m
+ Gibbs Free Energy
13m
+ Gibbs Free Energy Calculations
22m
+ Gibbs Free Energy And Equilibrium
14m
20. Electrochemistry
2h 42m
Standard Reduction Potentials
9m
+ Intro to Electrochemical Cells
6m
+ Galvanic Cell
25m
+ Electrolytic Cell
10m
+ Cell Potential: Standard
13m
+ Cell Potential: The Nernst Equation
20m
+ Cell Potential and Gibbs Free Energy
16m
+ Cell Potential and Equilibrium
8m
+ Cell Potential: G and K
16m
+ Cell Notation
20m
+ Electroplating
16m
21. Nuclear Chemistry
2h 36m
Intro to Radioactivity
10m
+ Alpha Decay
9m
+ Beta Decay
7m
+ Gamma Emission
7m
+ Electron Capture & Positron Emission
9m
+ Neutron to Proton Ratio
7m
+ Band of Stability: Alpha Decay & Nuclear Fission
10m
+ Band of Stability: Beta Decay
3m
+ Band of Stability: Electron Capture & Positron Emission
4m
+ Band of Stability: Overview
14m
+ Measuring Radioactivity
7m
+ Rate of Radioactive Decay
12m
+ Radioactive Half-Life
16m
+ Mass Defect
18m
+ Nuclear Binding Energy
14m
22. Organic Chemistry
5h 7m
Introduction to Organic Chemistry
8m
+ Structural Formula
8m
+ Condensed Formula
10m
+ Skeletal Formula
6m
+ Spatial Orientation of Bonds
3m
+ Intro to Hydrocarbons
16m
+ Isomers
11m
+ Chirality
15m
+ Functional Groups in Chemistry
11m
+ Naming Alkanes
4m
+ The Alkyl Groups
9m
+ Naming Alkanes with Substituents
13m
+ Naming Cyclic Alkanes
6m
+ Naming Other Substituents
8m
+ Naming Alcohols
11m
+ Naming Alkenes
11m
+ Naming Alkynes
9m
+ Naming Ketones
5m
+ Naming Aldehydes
5m
+ Naming Carboxylic Acids
4m
+ Naming Esters
8m
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5m
+ Naming Amines
5m
+ Naming Benzene
7m
+ Alkane Reactions
7m
+ Intro to Addition Reactions
4m
+ Halogenation Reactions
4m
+ Hydrogenation Reactions
3m
+ Hydrohalogenation Reactions
7m
+ Alcohol Reactions: Substitution Reactions
4m
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9m
+ Intro to Redox Reactions
8m
+ Alcohol Reactions: Oxidation Reactions
7m
+ Aldehydes and Ketones Reactions
6m
+ Ester Reactions: Esterification
4m
+ Ester Reactions: Saponification
3m
+ Carboxylic Acid Reactions
4m
+ Amine Reactions
3m
+ Amide Formation
4m
+ Benzene Reactions
10m
23. Chemistry of the Nonmetals
2h 39m
Main Group Elements: Bonding Types
4m
+ Main Group Elements: Boiling & Melting Points
7m
+ Main Group Elements: Density
11m
+ Main Group Elements: Periodic Trends
7m
+ The Electron Configuration Review
16m
+ Periodic Table Charges Review
20m
+ Hydrogen Isotopes
4m
+ Hydrogen Compounds
11m
+ Production of Hydrogen
8m
+ Group 1A and 2A Reactions
7m
+ Boron Family Reactions
7m
+ Boron Family: Borane
7m
+ Borane Reactions
7m
+ Nitrogen Family Reactions
12m
+ Oxides, Peroxides, and Superoxides
12m
+ Oxide Reactions
4m
+ Peroxide and Superoxide Reactions
6m
+ Noble Gas Compounds
3m
24. Transition Metals and Coordination Compounds
3h 16m
Atomic Radius & Density of Transition Metals
11m
+ Electron Configurations of Transition Metals
7m
+ Electron Configurations of Transition Metals: Exceptions
11m
+ Paramagnetism and Diamagnetism
10m
+ Ligands
10m
+ Complex Ions
5m
+ Coordination Complexes
7m
+ Classification of Ligands
11m
+ Coordination Numbers & Geometry
9m
+ Naming Coordination Compounds
22m
+ Writing Formulas of Coordination Compounds
8m
+ Isomerism in Coordination Complexes
14m
+ Orientations of D Orbitals
4m
+ Intro to Crystal Field Theory
10m
+ Crystal Field Theory: Octahedral Complexes
5m
+ Crystal Field Theory: Tetrahedral Complexes
4m
+ Crystal Field Theory: Square Planar Complexes
4m
+ Crystal Field Theory Summary
8m
+ Magnetic Properties of Complex Ions
9m
+ Strong-Field vs Weak-Field Ligands
6m
+ Magnetic Properties of Complex Ions: Octahedral Complexes
11m
Electrochemistry
Cell Notation
Electrochemistry
Cell Notation: Videos & Practice Problems
Video Lessons
Practice
Worksheet
Topic summary
An electrochemical cell can be succinctly represented using cell notation, which outlines the redox reaction without the need for elaborate diagrams. The cell notation consists of phase boundaries, indicated by a single line, and a physical boundary, represented by two lines, separating the anode and cathode. The anode, where oxidation occurs, is the site where electrons are lost, while the cathode is where reduction takes place, gaining electrons. For instance, in a galvanic cell with a chromium anode and a copper cathode, the half-reactions are combined to yield the overall redox reaction. The cell notation follows the pattern of placing lower oxidation states at the ends and higher oxidation states in the middle, with the anode (A) and cathode (C) components separated by the physical boundary (B). This method provides a quick and efficient way to communicate the essential details of an electrochemical cell to others in the field.
1
concept
Cell Notation
Video duration:
6m
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Cell Notation Video Summary
Cell notation, also known as a cell diagram, provides a concise way to represent the overall redox reaction occurring in an electrochemical cell. It incorporates two types of boundaries: phase boundaries and physical boundaries. Phase boundaries are depicted as a single line, indicating the coexistence of two phases of the same substance at equilibrium, while physical boundaries are represented by two solid lines, marking the separation between the anode and cathode.
In a typical electrochemical cell, such as a galvanic or voltaic cell, the anode is positioned on the left and the cathode on the right. In this setup, the anode carries a negative charge and is the site of oxidation, where electrons are lost. Conversely, the cathode is positively charged and serves as the site of reduction, where electrons are gained. For example, in a cell with a chromium electrode, chromium (Cr) at the anode undergoes oxidation, releasing electrons into the circuit, while copper ions (Cu2+) at the cathode gain these electrons to form solid copper (Cu).
The half-reactions can be summarized as follows: at the cathode, Cu2+ + 2e- → Cu, and at the anode, Cr → Cr2+ + 2e-. The electrons produced during oxidation at the anode and consumed during reduction at the cathode must balance, leading to the overall redox reaction: Cu2+ (aq) + Cr (s) → Cu (s) + Cr2+ (aq).
To express this in cell notation, we follow a simple format: A | B | C, where A represents the anode compartment, B denotes the physical boundary, and C signifies the cathode compartment. The lower oxidation states are placed at the ends, while the higher oxidation states are positioned in the center. For instance, the cell notation for the described electrochemical cell would be written as Cr (s) | Cr2+ (aq) || Cu2+ (aq) | Cu (s). This notation effectively summarizes the components and reactions occurring in the cell without the need for a detailed diagram, making it a practical tool for chemists.
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2
example
Cell Notation Example
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1m
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Cell Notation Example Video Summary
In an electrochemical cell, the cell notation is a concise way to represent the components and reactions occurring within the cell. The notation follows a specific format, typically denoted as a | b | c, where a represents the anode, b is the phase boundary, and c is the cathode. Understanding the oxidation states of the substances involved is crucial for determining the correct placement in the notation.
In the given redox reaction, we observe the transformation of tin (Sn) and aluminum (Al). The oxidation state of tin changes from +2 (in the form of Sn2+) to 0 (as solid tin, Sn), indicating that it is being reduced. This reduction occurs at the cathode, where the higher oxidation state (Sn2+) is located in the interior of the compartment, while the lower oxidation state (solid Sn) is on the end.
Conversely, aluminum undergoes oxidation, changing from 0 (as solid aluminum, Al) to +3 (in the form of Al3+). This increase in oxidation state signifies that aluminum is the anode, with the higher oxidation state (Al3+) found in the interior and the lower oxidation state (solid Al) on the end.
Thus, the complete cell notation for this electrochemical cell can be represented as:
Al | Al3+(aq) || Sn2+(aq) | Sn
This notation effectively summarizes the components and their respective oxidation states, providing a clear representation of the electrochemical processes taking place within the cell.
3
Problem
Write the half reactions as well as the overall net ionic equation for the following line notation: Fe (s) | Fe2+ (aq) || Mg2+ (aq) | Mg (s)
A
Anode: Fe (s) → Fe2+ (aq) + 2e-
Cathode: Mg2+ (aq) + 2e- → Mg (s)
Overall: Fe (s) + Mg2+ (aq) → Fe2+ (aq) + Mg (s)
B
Anode: Fe (s) → Fe2+ (aq) + 2e-
Cathode: Mg2+ (aq) + 2e- → Mg (s)
Overall: Fe2+ (aq) + Mg (s) → Fe (s) + Mg2+ (aq)
Anode: Fe (s) → Fe2+ (aq) + 2e-
Cathode: Mg (s) → Mg2+ (aq) + 2e-
Overall: Fe (s) + Mg2+ (aq) → Fe2+ (aq) + Mg (s)
Anode: Fe2+ (aq) + 2e- → Fe (s)
Cathode: Mg2+ (aq) + 2e- → Mg (s)
Overall: Fe (s) + Mg2+ (aq) → Fe2+ (aq) + Mg (s)
4
Problem
The cell notation for a redox reaction is given as the following at (T= 298 K). Calculate the cell potential for the reaction at 25ºC. Zn (s) | Zn2+ (aq, 0.37 M) || Ni2+ (aq, 0.059 M) | Ni (s) Standard Reduction Potentials Zn2+ (aq) + 2 e– →. Zn (s) E°red = - 0.7621 Ni2+ (aq) + 2 e– → Ni (s) E°red = - 0.2300
A
0.3130 V
B
0.4033 V
C
0.5085 V
D
0.1199 V
Problem
What is the [Cu2+] for the following cell notation diagram if the cell potential is 0.4404 V? Cu | Cu2+ (aq, ? M) || Ag+(aq, 0.50 M) | Ag Standard Reduction Potentials Cu2+ (aq) + 2 e– →. Cu (s) E°red = + 0.3394 Ag+ (aq) + e– → Ag (s) E°red= + 0.8000
A
0.87 M
B
1.20 M
C
3.32 M
D
0.11 M
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Writing cell notation involves representing the components of an electrochemical cell and the direction of electron flow. Here's how you do it:
Anode to Cathode: Write the anode (oxidation half-cell) on the left and the cathode (reduction half-cell) on the right. The vertical line, "|", represents a phase boundary, and the double vertical line, "||", represents the salt bridge or a porous membrane separating the two half-cells.
Anode Half-Cell: Start with the anode material (the solid electrode), followed by the phase boundary line. Next, write the anode's aqueous ions with their concentration in parentheses. If there's a solid and a gas involved, write the gas before the solid.
Cathode Half-Cell: After the double line for the salt bridge, write the cathode's aqueous ions and their concentration, followed by the phase boundary line, and then the cathode material (the solid electrode).
Electron Flow: Electrons flow from the anode to the cathode, but in cell notation, you don't explicitly write the direction of electron flow.
For example, for a cell with zinc and copper:
Cell notation, also known as line notation, is a shorthand way to represent the components of an electrochemical cell. Here's how to read it:
Anode and Cathode: The anode (negative electrode where oxidation occurs) is written on the left, and the cathode (positive electrode where reduction occurs) is on the right.
Electrode Material: The material of the electrode is written first, typically a metal.
Ion Concentration: After the electrode material, the concentration of the ions in solution is given in parentheses.
Double Vertical Lines: These represent a salt bridge or a porous membrane, which separates the two half-cells.
Single Vertical Lines: These denote a phase boundary within a half-cell, such as between a solid electrode and an aqueous solution.
For example, in the cell notation:
Zinc is the anode, and copper is the cathode. The zinc electrode is solid, as denoted by (s), and it's in a 1M solution of Zn2+ ions. The double vertical lines indicate the separation between the anode and cathode half-cells, and the copper cathode is solid, as denoted by (s), and it's in a 1M solution of Cu2+ ions.
To write the correct line notation for a voltaic cell, you'll need to know the specific details of the cell's components, such as the anode and cathode materials, and the electrolytes involved. However, I can give you a general format for a voltaic cell's line notation:
Anode | Anode Ion (Concentration) || Cathode Ion (Concentration) | Cathode
Here's how to break it down:
Start with the anode (the electrode where oxidation occurs) on the left.
Separate the anode material and its aqueous ions with a single line (|), which represents a phase boundary.
Use a double line (||) to represent the salt bridge or membrane that separates the two half-cells.
After the double line, list the cathode ion and its concentration.
End with the cathode material (the electrode where reduction occurs) on the right.
For example, if you have a zinc-copper voltaic cell with Zn as the anode and Cu as the cathode, and both are in solutions of their respective sulfates, the line notation would be:
Remember to include the physical state of the materials (s for solid, aq for aqueous).
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The Art of Problem Solving
The Art of Problem Solving Introduction Series (Gr. 6-10)
The Art of Problem Solving Introduction to Counting & Probability
The Art of Problem Solving: Introduction to Counting & Probability Set
The Art of Problem Solving: Introduction to Counting & Probability Set
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Includes the Student Text and Solutions Manual for the Intro to Counting & Probability Level.
Publisher's Description of The Art of Problem Solving: Introduction to Counting & Probability Set
A thorough introduction for students in grades 7-10 to counting and probability topics such as permutations, combinations, Pascal's triangle, geometric probability, basic combinatorial identities, the Binomial Theorem, and more.
Learn the basics of counting and probability from former USA Mathematical Olympiad winner David Patrick. Topics covered in the book include permutations, combinations, Pascal's Triangle, basic combinatorial identities, expected value, fundamentals of probability, geometric probability, the Binomial Theorem, and much more.
The text is structured to inspire the reader to explore and develop new ideas. Each section starts with problems, so the student has a chance to solve them without help before proceeding. The text then includes solutions to these problems, through which counting and probability techniques are taught. Important facts and powerful problem solving approaches are highlighted throughout the text. In addition to the instructional material, the book contains over 400 problems. The solutions manual contains full solutions to all of the problems, not just answers.
This book is ideal for students who have mastered basic algebra, such as solving linear equations. Middle school students preparing for MATHCOUNTS, high school students preparing for the AMC, and other students seeking to master the fundamentals of counting and probability will find this book an instrumental part of their mathematics libraries.
Paperback (2nd edition). Text: 256 pages. Solutions: 120 pages.
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Category Description for The Art of Problem Solving Introduction Series (Gr. 6-10)
This is an outstanding math program for the math-gifted student. It is rigorous and oriented to the independent problem-solver. The texts are based on the premise that students learn math best by solving problems - lots of problems - and preferably difficult problems that they don't already know how to solve. Most sections, therefore, begin by presenting problems and letting students intuit solutions BEFORE explaining ways to solve them. Even if they find ways to answer the problems, they should read the rest of the section to see if their answer is correct and if theirs is the best or most efficient way to solve that type of problem. Textual instruction, then, is given in the context of these problems, explaining how to best approach and solve them.
Throughout the text there are also special, blue-shaded boxes highlighting key concepts, important things to retain (like formulas), warnings for potential problem-solving pitfalls, side notes, and bogus solutions (these demonstrate misapplications). There are exercises at the end of most sections to see if the student can apply what's been learned. Review problems at the end of each chapter test understanding for that chapter. If a student has trouble with these, he should go back and re-read the chapter. Each chapter ends with a set of Challenge Problems that go beyond the learned material. Successful completion of these sets demonstrates a high degree of mastery.
A unique feature in this series is the hints section at the back of the book. These are intended to give a little help to selected problems, usually the very difficult ones (marked with stars). In this way, students can get a little push in the right direction, but still have to figure out the solution for themselves. The solution manuals do contain complete solutions and explanations to all the exercises, review problems and challenge problems. It is best for students not to access these until they have made several attempts to solve the problems first. I particularly like one of the motivating boxes in the text that coaches, "If at first you don't know how to solve a problem, don't just stare at it. Experiment!". That pretty much sums up the philosophy of the course, encouraging children to take chances, become aggressive problem solvers, and attack problems with confidence. I wonder how far some children would go if they were encouraged this way instead of being spoon fed?
Though this course is used in classroom settings, the texts are student-directed, making them perfect for the independent learner or homeschooler. Students should start the introductory sequence with the Prealgebrabook. Afterwards, begin the Introduction to Algebra. Students will be prepared for both the Introduction to Countingand Probability and Introduction to Number Theory courses after completing the first 11 chapters of Algebra. It won't matter whether they do these along with Algebra, put aside Algebra and complete the other two or finish Algebra first and then do them. All of them should be completed prior to the Introduction to Geometry book. If you are coming into this course from another curriculum, you will probably want to take a placement test to decide where to enter this program. Even if your student has finished Algebra 2 elsewhere, you will want to make sure that all of the material from this series has been covered before continuing on to the Intermediate series.
Taken together, these constitute a complete curriculum for outstanding math students in grades 6-10 and one that prepares them for competitions such as MATHCOUNTS and the American Mathematics Competitions. The material is challenging and in-depth; this is not a course for the mathematically faint of heart. If your child loves math, is genuinely math-gifted, or is interested in participating in math competitions, you definitely need to give this one serious consideration.
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Wow! Not only do my 6th and 7th grader enjoy these thinking problems, but I do too! It helps to have a basic understanding of algebra first. Its like Beast Academy for big kids.
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We are completing Algebra A by AOPS and are ready for this series.
Sarah O Oct 15, 2023
My friend recommended it.
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13591 | https://crystal.harvard.edu/viruses/ | Viruses – Laboratory of Structural Cell Biology
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Virus structure and cell entry
Virus particles carry genetic material from one cell to another. The key structural distinction is between viruses that have lipid-bilayer membranes and those that lack them –“enveloped” and “non-enveloped”, respectively. These differences reflect different mechanisms of cell entry and different pathways of assembly and maturation. Enveloped viruses enter by membrane fusion, either from an internal compartment following an endocytic step, or at the cell surface. Non-enveloped viruses require some form of membrane “perforation”. For these viruses to penetrate a cell, a large macromolecular complex (either a subviral particle or just the viral genome) must cross a cellular membrane to access the cytosol, and some mode of local membrane disruption must accompany the translocation. We study the mechanisms of membrane fusion and perforation, directing our attention currently to flaviviruses (especially dengue and West Nile viruses) and double-strand RNA viruses (rotavirus, in particular).
Viral membrane fusion
Membrane fusion is thermodynamically favorable, but it generally presents a high kinetic barrier. Fusion proteins lower this barrier and thus are catalysts for the merger of two bilayers — “suicide” catalysts in the case of viral fusion proteins. Sensitive measurements of fusion kinetics, like careful analysis of enzyme kinetics, can yield information about mechanism, by examining effects of directed mutations on rates, cooperativity, and so forth. We have devised a single-virus-particle fusion assay to study fusion in this way and applied it to influenza virus (Floyd et al, 2008; Ivanovic et al, 2013 and 2015) and to West Nile and dengue viruses (Chao et al, 2014 and 2018). Our on-going work is an effort to link cryo-EM structures of flavivirus envelope protein membrane-interacting regions with steps in the fusion pathway and ultimately to understand more thoroughly the mechanism of fusion inhibition by small-molecule inhibitors (Schmidt et al, 2012; Chao et al, 2018).
Non-enveloped virus entry
A principal objective of our research on virus structure is a molecular description of the earliest events leading to infection of a cell: attachment, uptake, and penetration into the cytosol. Rotavirus, a major cause of infectious infant diarrhea, is a particularly suitable non-enveloped virus for deriving such a “molecular movie”. We have used electron cryomicroscopy (with Nikolaus Grigorieff, Brandeis) to obtain an atomic model for the complete rotavirus particle, taking advantage of a number of high-resolution structures (from x-ray crystallography) of individual structural proteins and protein fragments and of the intact inner capsid particle. The structural analysis has enabled us to design experiments using live-cell imaging (with Tomas Kirchhausen, Children’s Hospital: Abdelhakim et al, 2014; Salgado et al, 2017 and 2018) to follow molecular events during virus uptake and penetration into cells in culture. Our on-going work is an effort to connect conformational changes in components of the virus particle, as defined by structures from cryo-EM and x-ray crystallography, with stages of entry as defined by live-cell imaging.
Viral RNA polymerases
Structures of viral RNA polymerases define aspects of transcription and replication that differ from related events in the host cell. We are expanding our study of the polymerases of negative-sense, single-segment, RNA viruses (a collaboration with Sean Whelan, Harvard Medical School: Liang et al, 2015) to work out a full picture of the stages of transcription and replication and of the interaction of the polymerase with the template RNP. We are also studying structural events within the virion linked to the polymerization cycle of the rotavirus polymerase (for our earlier work, see Estrozi et al, 2013
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13592 | http://medbox.iiab.me/modules/en-cdc/www.cdc.gov/dpdx/lymphaticfilariasis/index.html | CDC - DPDx - Lymphatic Filariasis
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Lymphatic Filariasis
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[Brugia malayi] [Brugia timori] [Wuchereria bancrofti]
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Causal Agent
The filarid nematodes Wuchereria bancrofti, Brugia malayi, and (less-commonly) B. timori. Humans can also be infected with several zoonotic Brugia species.
Brugia Malayi
The typical vector for Brugia malayi filariasis are mosquito species in the genera Mansonia and Aedes. During a blood meal, an infected mosquito introduces third-stage filarial larvae onto the skin of the human host, where they penetrate into the bite wound . They develop into adults that commonly reside in the lymphatics . The adult worms resemble those of Wuchereria bancrofti but are smaller. Female worms measure 43 to 55 mm in length by 130 to 170 μm in width, and males measure 13 to 23 mm in length by 70 to 80 μm in width. Adults produce microfilariae, measuring 177 to 230 μm in length and 5 to 7 μm in width, which are sheathed and have nocturnal periodicity. The microfilariae migrate into lymph and enter the blood stream reaching the peripheral blood . A mosquito ingests the microfilariae during a blood meal . After ingestion, the microfilariae lose their sheaths and work their way through the wall of the proventriculus and cardiac portion of the midgut to reach the thoracic muscles . There the microfilariae develop into first-stage larvae and subsequently into third-stage larvae . The third-stage larvae migrate through the hemocoel to the mosquito's proboscis and can infect another human when the mosquito takes a blood meal .
Wuchereria bancrofti
Different species of the following genera of mosquitoes are vectors of W. bancrofti filariasis depending on geographical distribution. Among them are: Culex (C. annulirostris, C. bitaeniorhynchus, C. quinquefasciatus, and C. pipiens); Anopheles (A. arabinensis, A. bancroftii, A. farauti, A. funestus, A. gambiae, A. koliensis, A. melas, A. merus, A. punctulatus and A. wellcomei); Aedes (A. aegypti, A. aquasalis, A. bellator, A. cooki, A. darlingi, A. kochi, A. polynesiensis, A. pseudoscutellaris, A. rotumae, A. scapularis, and A. vigilax); Mansonia (M. pseudotitillans, M. uniformis); Coquillettidia (C. juxtamansonia). During a blood meal, an infected mosquito introduces third-stage filarial larvae onto the skin of the human host, where they penetrate into the bite wound . They develop in adults that commonly reside in the lymphatics . The female worms measure 80 to 100 mm in length and 0.24 to 0.30 mm in diameter, while the males measure about 40 mm by .1 mm. Adults produce microfilariae measuring 244 to 296 μm by 7.5 to 10 μm, which are sheathed and have nocturnal periodicity, except the South Pacific microfilariae which have the absence of marked periodicity. The microfilariae migrate into lymph and blood channels moving actively through lymph and blood . A mosquito ingests the microfilariae during a blood meal . After ingestion, the microfilariae lose their sheaths and some of them work their way through the wall of the proventriculus and cardiac portion of the mosquito's midgut and reach the thoracic muscles . There the microfilariae develop into first-stage larvae and subsequently into third-stage infective larvae . The third-stage infective larvae migrate through the hemocoel to the mosquito's prosbocis and can infect another human when the mosquito takes a blood meal .
Geographic Distribution
Among the agents of lymphatic filariasis, Wuchereria bancrofti is encountered in tropical areas worldwide; Brugia malayi is limited to Asia; and Brugia timori is restricted to some islands of Indonesia.
Clinical Presentation
Lymphatic filariasis most often consists of asymptomatic microfilaremia. Some patients develop lymphatic dysfunction causing lymphedema and elephantiasis (frequently in the lower extremities) and, with Wuchereria bancrofti, hydrocele and scrotal elephantiasis. Episodes of febrile lymphangitis and lymphadenitis may occur. Persons who have newly arrived in disease-endemic areas can develop afebrile episodes of lymphangitis and lymphadenitis. An additional manifestation of filarial infection, mostly in Asia, is pulmonary tropical eosinophilia syndrome, with nocturnal cough and wheezing, fever, and eosinophilia.
Microfilariae of Wuchereria bancrofti.
The microfilaria of Wuchereria bancrofti are sheathed and measure 240-300 µm in stained blood smears and 275-320 µm in 2% formalin. They have a gently curved body, and a tail that is tapered to a point. The nuclear column (the cells that constitute the body of the microfilaria) is loosely packed; the cells can be visualized individually and do not extend to the tip of the tail.Microfilariae circulate in the blood.
Figure A: Microfilaria of W. bancrofti in a thick blood smear stained with Giemsa. Image courtesy of the Oregon State Public Health Laboratory.
Figure B: Microfilaria of W. bancrofti in a thick blood smear stained with Giemsa. Image courtesy of the Oregon State Public Health Laboratory.
Figure C: Microfilaria of W. bancrofti in a thick blood smear, stained with Giemsa.
Figure D: Close-up of the anterior end of the worm in Figure C.
Figure E: Close-up of the posterior end of the worm in Figure C.
Adults of W. bancrofti.
Adults of Wuchereria bancrofti are long and threadlike. The males measure up to 40 mm long and females are 80-100 mm long. Adults are found primarily in lymphatic vessels, less commonly in blood vessels.
Figure A: Adults of W. bancrofti. The male worm is on the left; the female is on the right.
Microfilariae of Brugia malayi.
Microfilariae of Brugia malayi are sheathed and in stained blood smears measure 175-230 µm. In 2% formalin they are longer, measuring 240-300 µm. The tail is tapered, with a significant gap between the terminal and subterminal nuclei. Microfilaria circulate in the blood.
Figure A: Microfilaria of B. malayi in a thick blood smear, stained with Giemsa.
Figure B: Microfilaria of B. malayi in a thin blood smear, stained with Giemsa.
Microfilariae of B. timori.
Microfilaria of Brugia timori are sheathed and measure on average 310 µm in stained blood smears and 340 µm in 2% formalin. Microfilaria of B. timori differ from B. malayi by a having a longer cephalic space, a sheath that does not stain with Giemsa, and a larger number of single-file nuclei towards the tail. Microfilariae circulate in the blood.
Figure A: Microfilaria of B. timori in a thick blood smear from a patient from Indonesia, stained with Giemsa and captured at 500x oil magnification. Image from a specimen courtesy of Dr. Thomas C. Orihel, Tulane University, New Orleans, LA.
Figure B: Microfilaria of B. timori in a thick blood smear from a patient from Indonesia, stained with Giemsa and captured at 500x oil magnification. Image from a specimen courtesy of Dr. Thomas C. Orihel, Tulane University, New Orleans, LA.
Figure C: Microfilaria of B. timori in a thick blood smear from a patient from Indonesia, stained with Giemsa and captured at 500x oil magnification. Image from a specimen courtesy of Dr. Thomas C. Orihel, Tulane University, New Orleans, LA.
Figure D: Microfilaria of B. timori in a thick blood smear from a patient from Indonesia, stained with Giemsa and captured at 500x oil magnification. Image from a specimen courtesy of Dr. Thomas C. Orihel, Tulane University, New Orleans, LA.
Figure E: Microfilaria of B. timori in a thick blood smear from a patient from Indonesia, stained with Giemsa and captured at 500x oil magnification. Image from a specimen courtesy of Dr. Thomas C. Orihel, Tulane University, New Orleans, LA.
Figure F: Microfilaria of B. timori in a thick blood smear from a patient from Indonesia, stained with Giemsa and captured at 500x oil magnification. Image from a specimen courtesy of Dr. Thomas C. Orihel, Tulane University, New Orleans, LA.
Adults of Brugia spp. in tissue.
Adults of Brugia spp. typically live in lymphatic vessels, but may be found less-commonly in blood vessels or other regions.
Figure A: Section of an adult of Brugia sp. from a lymph node, stained with hematoxylin and eosin (H&E). Image taken at 200x magnification.
Figure B: Section of an adult of Brugia sp. from a lymph node, stained with hematoxylin and eosin (H&E). Image taken at 400x magnification.
Figure C: Section of an adult of Brugia sp. from a femoral lymph node, stained with H&E. Image taken at 200x magnification.
Figure D: Section of an adult of Brugia sp. from a femoral lymph node, stained with H&E. Image taken at 200x magnification.
Figure E: Section of an adult of Brugia sp. from the conjunctiva of a patient from Ecuador, stained with H&E. Image taken at 200x magnification.
Figure F: Section of an adult of Brugia sp. from the conjunctiva of a patient from Ecuador, stained with H&E. Image taken at 500x oil magnification.
Diagnostic Findings
Microscopy
Lymphatic filariasis is usually identified by the finding of microfilaria in peripheral blood smears (thick or thin) stained with Giemsa or hematoxylin-and-eosin. For increased sensitivity, concentration techniques can be used. These include centrifugation of the blood sample lysed in 2% formalin (Knott's technique), or filtration through a Nucleopore® membrane. Microfilariae of Wuchereria and Brugia exhibit a nocturnal periodicity and an accurate diagnosis is best achieved on smears collected at night (10 PM-2 AM). Adults may be identified in biopsy specimens of lymphatic tissue.
Antigen Detection
Antigen detection using an immunoassay for circulating filarial antigens constitutes a useful diagnostic approach because sensitivity for detection of microfilariae can be low and variable. Unlike microfilariae with nocturnal periodicity, filarial antigens can be detected in blood samples collected at any time of day. A rapid format immunochromatographic test has been shown to be a useful and sensitive tool for the detection of Wuchereria bancrofti antigen and is being used widely by lymphatic filariasis elimination programs. Currently, this test is not licensed for use in the United States and cannot be used for patient diagnosis.
Treatment Information
The main goal of treatment of an infected person is to kill the adult worm. Diethylcarbamazine citrate (DEC), which is both microfilaricidal and active against the adult worm, is the drug of choice for lymphatic filariasis. The late phase of chronic disease is not affected by chemotherapy. Ivermectin is effective against the microfilariae of W. bancrofti, but has no effect on the adult parasite.
Because lymphatic filariasis is rare in the U.S., DEC is no longer approved by the Food and Drug Administration (FDA) and cannot be sold in the U.S. Physicians can obtain the medication from CDC after confirmed positive lab results. Call: 404-718-4745. Treatment of lymphatic filariasis in adults and children > 18 months of age involves either a 1 day or 12 day treatment course (6mg/kg/day). One day treatment is generally as effective as the 12-day regimen. For tropical pulmonary eosinophilia (TPE), a longer DEC treatment course of 14-21 days is generally recommended. DEC is generally well tolerated. Side effects are generally limited and depend on the number of microfilariae in the blood. The most common side effects are dizziness, nausea, fever, headache, or pain in muscles or joints.
DEC is contraindicated in patients who may also have onchocerciasis. Prior to DEC treatment for lymphatic filariasis, onchocerciasis should be excluded in all patients with a consistent exposure history due to the possibility of severe exacerbations of skin and eye involvement (Mazzotti reaction). In addition, DEC should be used with extreme caution in patients with circulating Loa loa microfilarial levels > 2,500/mm3 due to the potential for life-threatening side effects, including encephalopathy and renal failure. Neither steroids pre-treatment nor slow dose escalation prevents these complications. Consultation with a tropical medicine specialist is recommended in these scenarios.
The drug ivermectin kills only the microfilariae, but not the adult worm; the adult worm is responsible for the pathology of lymphedema and hydrocele.
Some studies have shown adult worm killing with treatment with doxycycline (200mg/day for 4–6 weeks).
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13593 | https://www.aafp.org/pubs/afp/issues/2016/0115/p114.html | Published Time: 2016-01-15
Acute Bacterial Prostatitis: Diagnosis and Management | AAFP
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Acute Bacterial Prostatitis: Diagnosis and Management
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Management
Complications
Prevention
References
Acute bacterial prostatitis is an acute infection of the prostate gland that causes pelvic pain and urinary tract symptoms, such as dysuria, urinary frequency, and urinary retention, and may lead to systemic symptoms, such as fevers, chills, nausea, emesis, and malaise. Although the true incidence is unknown, acute bacterial prostatitis is estimated to comprise approximately 10% of all cases of prostatitis. Most acute bacterial prostatitis infections are community acquired, but some occur after transurethral manipulation procedures, such as urethral catheterization and cystoscopy, or after transrectal prostate biopsy. The physical examination should include abdominal, genital, and digital rectal examination to assess for a tender, enlarged, or boggy prostate. Diagnosis is predominantly made based on history and physical examination, but may be aided by urinalysis. Urine cultures should be obtained in all patients who are suspected of having acute bacterial prostatitis to determine the responsible bacteria and its antibiotic sensitivity pattern. Additional laboratory studies can be obtained based on risk factors and severity of illness. Radiography is typically unnecessary. Most patients can be treated as outpatients with oral antibiotics and supportive measures. Hospitalization and broad-spectrum intravenous antibiotics should be considered in patients who are systemically ill, unable to voluntarily urinate, unable to tolerate oral intake, or have risk factors for antibiotic resistance. Typical antibiotic regimens include ceftriaxone and doxycycline, ciprofloxacin, and piperacillin/tazobactam. The risk of nosocomial bacterial prostatitis can be reduced by using antibiotics, such as ciprofloxacin, before transrectal prostate biopsy.
Acute bacterial prostatitis is an acute infection of the prostate gland that causes urinary tract symptoms and pelvic pain in men.1 It is estimated to comprise up to 10% of all prostatitis diagnoses, and its incidence peaks in persons 20 to 40 years of age and in persons older than 70 years.2 Most cases can be diagnosed with a convincing history and physical examination.3 Although prostatitis-like symptoms have a combined prevalence of 8.2% in men, the incidence and prevalence of acute bacterial prostatitis are unknown.4
SORT: KEY RECOMMENDATIONS FOR PRACTICE
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| Clinical recommendation | Evidence rating | References | Comments |
:-- :-- |
| Prostatic massage should be avoided in patients suspected of having acute bacterial prostatitis. | C | 11, 12, 20, 22 | Expert consensus |
| Midstream urine culture should be used to guide antibiotic therapy for acute bacterial prostatitis. | C | 3, 10, 11 | Prospective cohort study, retrospective cohort study |
| Blood cultures are indicated in patients with a body temperature greater than 101.1°F (38.4°C), a possible hematogenous source of infection (e.g., endocarditis with Staphylococcus aureus), or complicated infections (e.g., sepsis), and in patients who are immunocompromised. | C | 21 | Prospective cohort study |
| Prostate-specific antigen testing is not indicated in the evaluation of acute bacterial prostatitis. | C | 11, 12, 20 | Prospective cohort study |
| Fevers that persist for longer than 36 hours should be evaluated with imaging to rule out prostatic abscess. | C | 27 | Expert opinion |
| Acute bacterial prostatitis occurring after a transrectal prostate biopsy should be treated with broad-spectrum antibiotics to cover fluoroquinolone-resistant bacteria and extended spectrum beta-lactamase–producing Escherichia coli. | C | 15–18, 24 | Multiple retrospective cohort studies and one prospective cohort study |
A = consistent, good-quality patient-oriented evidence; B = inconsistent or limited-quality patient-oriented evidence; C = consensus, disease-oriented evidence, usual practice, expert opinion, or case series. For information about the SORT evidence rating system, go to
Pathogenesis
Most cases of acute bacterial prostatitis are caused by ascending urethral infection or intraprostatic reflux and are facilitated by numerous risk factors (Table 1).4–10 These infections may occur from direct inoculation after transrectal prostate biopsy and transurethral manipulations (e.g., catheterization and cystoscopy).6–8 Occasionally, direct or lymphatic spread from the rectum or hematogenous spread via bacterial sepsis can cause acute bacterial prostatitis.11 Overall, community-acquired infections are three times more common than nosocomial infections.3
Table 1.
Risk Factors for Acute Bacterial Prostatitis
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Benign prostatic hypertrophyGenitourinary infectionsEpididymitisOrchitisUrethritisUrinary tract infectionHigh-risk sexual behaviorHistory of sexually transmitted diseasesImmunocompromisedPhimosisProstate manipulationCystoscopyTransrectal prostate biopsyTransurethral surgeryUrethral catheterizationUrodynamic studiesUrethral stricture
—Higher risk for infection.
Information from references 4 through 10.
Microbiology
Acute bacterial prostatitis is most frequently caused by Escherichia coli, followed by Pseudomonas aeruginosa, and Klebsiella, Enterococcus, Enterobacter, Proteus, and Serratia species.3,5,7,10 In sexually active men, Neisseria gonorrhoeae and Chlamydia trachomatis should be considered.12 Patients who are immunocompromised (e.g., persons with human immunodeficiency virus) are more likely to have uncommon causes for prostatitis, such as Salmonella, Candida, and Cryptococcus species (Table 2).3,7,10,12
Table 2.
Pathogens in Acute Prostatitis
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| Common | Uncommon |
:-- |
| Escherichia coli (> 50% of cases) | Chlamydia trachomatis |
| Pseudomonas aeruginosa | Fungi (Aspergillus, Candida, Cryptococcus, and Histoplasma species) |
| Klebsiella species | Mycobacterium tuberculosis |
| Enterococcus species | Mycoplasma genitalium |
| Enterobacter species | Neisseria gonorrhoeae |
| Proteus species | Salmonella species |
| Serratia species | Staphylococcus species |
| | Streptococcus species |
| | Trichomonas vaginalis |
| | Ureaplasma urealyticum |
—Listed in approximate order of frequency.
Information from references 3, 7, 10, and 12.
Infections that occur after transurethral manipulation are more likely to be caused by Pseudomonas species, which have higher rates of resistance to cephalosporins and carbapenems.7 Transrectal prostate biopsies can cause postoperative infections. Perioperative antibiotics have reduced the rates of postoperative prostatitis to between 0.67% and 2.10% of cases, but have increased the incidence of prostatitis caused by fluoroquinolone-resistant bacteria and extended spectrum beta-lactamase–producing E. coli.13–18
Clinical Presentation
Patients with acute bacterial prostatitis often present with acute onset of irritative (e.g., dysuria, urinary frequency, urinary urgency) or obstructive (e.g., hesitancy, incomplete voiding, straining to urinate, weak stream) voiding symptoms. Patients may report suprapubic, rectal, or perineal pain.6,9,11 Painful ejaculation, hematospermia, and painful defecation may be present as well.19 Systemic symptoms, such as fever, chills, nausea, emesis, and malaise, commonly occur, and their presence should prompt physicians to determine if patients meet clinical criteria for sepsis.
The physical examination should include an abdominal examination to detect a distended bladder and costovertebral angle tenderness, a genital examination, and a digital rectal examination. A digital rectal examination should be performed gently because vigorous prostatic massage can induce bacteremia, and subsequently, sepsis.9,11,20 In a patient with acute bacterial prostatitis, the prostate will often be tender, enlarged, or boggy. If there is concern for obstructed voiding, postvoid residual urine volumes should be measured using ultrasonography.
Several conditions present with similar symptoms and must be differentiated from acute bacterial prostatitis (Table 3).
Table 3.
Differential Diagnosis of Acute Bacterial Prostatitis
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| Diagnosis | Distinguishing characteristics |
:-- |
| Benign prostatic hypertrophy | Obstructive voiding symptoms; enlarged, nontender prostate; negative urine culture |
| Chronic bacterial prostatitis | Recurring prostatitis symptoms for at least three months; positive urine culture with each episode |
| Chronic pelvic pain syndrome | Pain attributed to the prostate with no demonstrable evidence of infection |
| Cystitis | Irritative voiding symptoms; normal prostate examination |
| Diverticulitis | Left lower-quadrant abdominal pain; acute change in bowel habits; history of diverticulitis; tenderness to palpation localized to the left lower abdominal quadrant |
| Epididymitis | Irritative voiding symptoms; tenderness to palpation on affected epididymis |
| Orchitis | Swelling, pain, and/or tenderness to palpation in one or both testicles |
| Proctitis | Tenesmus; rectal bleeding; feeling of rectal fullness; passage of mucus through the rectum |
| Prostate cancer | Presence of constitutional symptoms; presence of nodules on prostate examination |
Evaluation
A convincing history and physical examination are typically sufficient to diagnose acute bacterial prostatitis. Physicians should obtain a urinalysis and midstream urine culture to support the clinical diagnosis before administering antibiotics.3,10,11
Blood cultures should be collected before initiating antibiotics in patients with a body temperature greater than 101.1°F (38.4°C), a possible hematogenous source of infection (e.g., endocarditis with Staphylococcus aureus), complicated infections (e.g., sepsis), or who are immunocompromised.11,21 Although blood and urine cultures can aid in diagnosis and management, up to 35% of urine cultures in patients with acute prostatitis will fail to grow an organism.3
In men younger than 35 years who are sexually active, and in men older than 35 years who engage in high-risk sexual behavior, a Gram stain of urethral swabs, a culture of urethral discharge, or a DNA amplification test should be obtained to evaluate for N. gonorrhoeae and C. trachomatis.11,22
Urine testing before and after prostatic massage (also known as the Meares-Stamey 2-glass or 4-glass test) is useful in diagnosing chronic prostate and pelvic disorders; however, such testing should not be performed in patients with suspected acute bacterial prostatitis because prostatic massage increases the risk of bacteremia, and subsequently, sepsis.
PROGNOSTIC FACTORS
A 2014 study of patients with acute bacterial prostatitis identified age older than 65 years, body temperature greater than 100.4°F (38°C), benign prostatic hypertrophy, urinary retention, and transurethral catheterization as factors associated with poor outcomes.23 These outcomes included septic shock, positive blood culture, and prostatic abscess.23 In patients with any of these factors, the physician should strongly consider ordering a complete blood count and a basic metabolic panel. In the same study, a white blood cell count greater than 18,000 per mm3 (18 × 109 per L) and a blood urea nitrogen level greater than 19 mg per dL (6.8 mmol per L) were independently associated with severe cases of acute bacterial prostatitis. Inflammatory markers, such as C-reactive protein and erythrocyte sedimentation rate, will likely be elevated, but these tests have minimal clinical or diagnostic utility.23
Prostate-specific antigen (PSA) levels are not indicated in the workup of acute bacterial prostatitis.11,12,20 Approximately 70% of men will have a spurious PSA elevation due to disruption of prostatic architecture caused by inflammation.19 Elevated PSA levels can persist for one to two months after treatment.11,12 If PSA levels remain elevated for more than two months, prostate cancer should be considered because 20% of persistent elevations are associated with malignancy.19
IMAGING
Imaging studies are usually unnecessary during the initial evaluation, but may help when the diagnosis remains unclear or when patients do not respond to adequate antibiotic therapy. Patients who remain febrile after 36 hours or whose symptoms do not improve with antibiotics should undergo transrectal ultrasonography to evaluate for prostatic abscess. Alternatively, noncontrast computed tomography (CT) or magnetic resonance imaging (MRI) of the pelvis could be considered. Prostate biopsy should not be performed to avoid inducing septicemia.
Management
Management of acute bacterial prostatitis should be based on severity of symptoms, risk factors, and local antibiotic resistance patterns (Figure 1). Most patients can be treated with outpatient antibiotics; fewer than one in six patients will require hospitalization.6 Admission criteria are listed in Table 4.
Figure 1.
Management of Acute Bacterial Prostatitis
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Management of acute bacterial prostatitis. (CT = computed tomography; MRI = magnetic resonance imaging.)
Table 4.
Admission Criteria for Acute Bacterial Prostatitis
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Failed outpatient managementInability to tolerate oral intakeResistance risk factorsRecent fluoroquinolone useRecent transurethral or transrectal prostatic manipulationSystemically ill or septicemiaUrinary retention
Initial empiric antibiotic therapy should be based on the suspected mode of infection and the presumed infecting organism (Table 5).5,7–9,15–17,24,25 Antibiotics should be adjusted based on culture and sensitivity results, when available.10,15 Men younger than 35 years who are sexually active and men older than 35 years who engage in high-risk sexual behavior should be treated with regimens that cover N. gonorrhoeae and C. trachomatis.12 Patients with risk factors for antibiotic resistance require intravenous therapy with broad-spectrum regimens because of the high likelihood of complications.7,8,15,24
Table 5.
Antibiotic Regimens for Acute Bacterial Prostatitis
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| Group | Primary regimen | Alternative regimen | Considerations |
:-- :-- |
| A | Single dose of ceftriaxone (Rocephin), 250 mg intramuscularly, or single dose of cefixime (Suprax), 400 mg orally | — | Regimen covers Neisseria gonorrhoeae and Chlamydia trachomatis infections in addition to other common bacterial pathogens |
| | | then | | |
| | Doxycycline, 100 mg orally twice daily for 10 days | | |
| B | Ciprofloxacin, 500 mg orally twice daily for 10 to 14 days | Trimethoprim/sulfamethoxazole, 160/800 mg orally twice daily for 10 to 14 days | Extend treatment for 2 weeks if patient remains symptomatic |
| | | or | | |
| | Levofloxacin (Levaquin), 500 to 750 mg orally daily for 10 to 14 days | | |
| C | Ciprofloxacin, 400 mg IV every 12 hours | Ceftriaxone, 1 to 2 g IV every 24 hours | Continue treatment until patient is afebrile, then transition to oral regimen (group B) for an additional 2 to 4 weeks |
| | | or | | plus | |
| | Levofloxacin, 500 to 750 mg IV every 24 hours | Levofloxacin, 500 to 750 mg IV every 24 hours | |
| | | | or | |
| | | Piperacillin/tazobactam (Zosyn), 3.375 g IV every 6 hours | |
| D | Piperacillin/tazobactam, 3.375 g IV every 6 hours plus aminoglycosides | Fluoroquinolone (group C) | Continue treatment until patient is afebrile, then transition to oral regimen (group B) for an additional 2 to 4 weeks |
| | | | plus | |
| | | or | Aminoglycosides | |
| | Cefotaxime (Claforan), 2 g IV every 4 hours plus aminoglycosides | | or | |
| | | or | Ertapenem (Invanz), 1 g IV every 24 hours | |
| | Ceftazidime (Fortaz), 2 g IV every 8 hours plus aminoglycosides | | or | |
| | | Imipenem/cilastatin (Primaxin), 500 mg IV every 6 hours | |
| | | | or | |
| | | Meropenem (Merrem IV), 500 mg IV every 8 hours | |
| E | Transrectal manipulation—fluoroquinolone resistance and extended spectrum beta-lactamase–producing Escherichia coli | | Continue treatment until patient is afebrile, then transition to oral regimen (group B) for an additional 2 to 4 weeks |
| | | | Carbapenems can be used if patient is unstable |
| | Piperacillin/tazobactam, 3.375 g IV every 6 hours plus aminoglycosides | Ertapenem, 1 g IV every 24 hours | If patient is stable, follow primary regimen while awaiting culture results |
| | | | or | |
| | | Imipenem/cilastatin, 500 mg IV every 6 hours | |
| | Transurethral manipulation—Pseudomonas species | | |
| | Piperacillin/tazobactam, 3.375 g IV every 6 hours† | Fluoroquinolone (group C)† | |
| | | or | | or | |
| | Ceftazidime, 2 g IV every 8 hours† | Imipenem/cilastatin, 500 mg IV every 6 hours | |
| | | or | | or | |
| | Cefipime, 2 g IV every 12 hours† | Meropenem, 500 mg IV every 8 hours | |
| | Fluoroquinolone exposure—fluoroquinolone resistance | | |
| | Piperacillin/tazobactam, 3.375 g IV every 6 hours† | Ceftriaxone, 1 g IV every 24 hours† | |
| | | or | | or | |
| | Ceftazidime, 2 g IV every 8 hours† | Ertapenem, 1 g IV every 24 hours | |
| | | or | | |
| | Cefepime, 2 g IV every 12 hours† | | |
IV = intravenously.
—Dosing instructions: gentamicin, 7 mg per kg IV every 24 hours, peak 16 to 24 mcg per mL, trough less than 1 mcg per mL; amikacin, 15 mg per kg IV every 24 hours, peak 56 to 64 mcg per mL, trough less than 1 mcg per mL.
†—Aminoglycosides should be added to regimen if patient is clinically unstable.
Information from references 5, 7 through 9, 15 through 17, 24, and 25.
The duration of antibiotic therapy for mild infections is typically 10 to 14 days (with a two-week extension if the patient remains symptomatic), or four weeks for severe infections.9,26 Febrile patients should generally become afebrile within 36 hours of starting antibiotic therapy.27 Otherwise, imaging with transrectal ultrasonography, CT, or MRI is required to rule out prostatic abscess.27 After severe infections improve and the patient is afebrile, antibiotics should be transitioned to oral form and continued for another two to four weeks.5,28 Repeat urine cultures should be obtained one week after cessation of antibiotics to ensure bacterial clearance.12
Supportive measures include providing antipyretics, hydrating fluids, and pain control. Acute urinary retention occurs in approximately one in 10 patients with acute bacterial prostatitis. Relieving urinary obstruction is an important treatment consideration in clearing the infection and providing pain relief.6 However, the best approach to this intervention has not been determined. Cystostomy provides good relief and may prevent chronic infection, but urethral catheterization is an easier option for relieving obstruction.29
Complications
Prostatic abscesses occur in 2.7% of patients with acute bacterial prostatitis and require urology consultation for drainage.6 Risk factors for prostatic abscess include long-term urinary catheterization, recent urethral manipulation, and an immunocompromised state.
Approximately 13% of patients with acute bacterial prostatitis experience recurrence necessitating a longer course of antibiotics.6 Patients with persistent or recurrent symptoms should have a repeat urine culture to evaluate for repeat bacterial prostatitis and be treated based on culture results. After three months of persistent or recurrent symptoms, patients should be evaluated and treated based on chronic prostate syndrome guidelines.1 Approximately one in nine patients with acute bacterial prostatitis will develop chronic bacterial prostatitis or chronic pelvic pain syndrome.29
Prevention
Although there are no known strategies for preventing community-acquired acute bacterial prostatitis, nosocomial infections can be reduced by avoiding unnecessary manipulation of the prostate, such as transrectal biopsy or urethral catheterization. Administering antibiotics before transrectal prostate biopsies reduces postoperative complications such as urinary tract infections, acute prostatitis, bacteriuria, and bacteremia; new approaches to prevention are needed to reduce fluoroquinolone resistance and extended spectrum beta-lactamase–producing E. coli infections.13,14 A 500-mg oral dose of ciprofloxacin 12 hours before transrectal prostate biopsy with a repeat dose at the time of biopsy is the typical prophylactic regimen.25 Preoperative enemas do not reduce infection rates.24 In patients who are at increased risk of harboring fluoroquinolone-resistant bacteria, preoperative stool cultures may allow for tailoring of antibiotics at the time of the procedure.17,30
Data Sources: A PubMed search was completed in Clinical Queries using the keywords acute prostatitis, title words acute prostatitis, and prostatitis [MeSH] AND acute. The search included meta-analyses, randomized controlled trials, clinical trials, and reviews. Also searched were the Agency for Healthcare Research and Quality evidence reports, Cochrane Database of Systematic Reviews, National Guideline Clearing-house, Essential Evidence Plus, and UpToDate. Search Dates: November 19, 2014, and October 20, 2015.
The opinions and assertions contained herein are the private views of the authors and are not to be construed as official or as reflecting the views of the U.S. Air Force Medical Department or the U.S. Air Force at large.
expand_moreAuthor Information
TIMOTHY J. COKER, MD, FAAFP, is associate program director at the Ehrling Bergquist Family Medicine Residency Program, Offutt Air Force Base, Neb. He is also an assistant professor at the Uniformed Services University of the Health Sciences, Bethesda, Md.
DANIEL M. DIERFELDT, DO, is an assistant professor at the Uniformed Services University of the Health Sciences. He is also an attending physician at the Offutt Family Medicine Residency, Offutt Air Force Base, Neb.
Address correspondence to Timothy J. Coker, MD, Ehrling Bergquist Family Medicine Residency Program, 2501 Capehart Rd., Offutt Air Force Base, NE 68113 (e-mail: t.j.coker@hotmail.com). Reprints are not available from the authors.
Author disclosure: No relevant financial affiliations.
expand_moreReference(s)
Krieger JN, Nyberg L, Nickel JC. NIH consensus definition and classification of prostatitis. JAMA. 1999;282(3):236-237.
Roberts RO, Lieber MM, Rhodes T, Girman CJ, Bostwick DG, Jacobsen SJ. Prevalence of a physician-assigned diagnosis of prostatitis: The Olmsted County study of urinary symptoms and health status among men. Urology. 1998;51(4):578-584.
Etienne M, Chavanet P, Sibert L, et al. Acute bacterial prostatitis: heterogeneity in diagnostic criteria and management. Retrospective multicentric analysis of 371 patients diagnosed with acute prostatitis. BMC Infect Dis. 2008;8:12.
Krieger JN, Lee SW, Jeon J, Cheah PY, Liong ML, Riley DE. Epidemiology of prostatitis. Int J Antimicrob Agents. 2008;31(suppl 1):S85-S90.
Yoon BI, Kim S, Han DS, et al. Acute bacterial prostatitis: how to prevent and manage chronic infection?. J Infect Chemother. 2012;18(4):444-450.
Millán-Rodríguez F, Palou J, Bujons-Tur A, et al. Acute bacterial prostatitis: two different sub-categories according to a previous manipulation of the lower urinary tract. World J Urol. 2006;24(1):45-50.
Kim SH, Ha US, Yoon BI, et al. Microbiological and clinical characteristics in acute bacterial prostatitis according to lower urinary tract manipulation procedure. J Infect Chemother. 2014;20(1):38-42.
Ha US, Kim ME, Kim CS, et al. Acute bacterial prostatitis in Korea: clinical outcome, including symptoms, management, microbiology and course of disease. Int J Antimicrob Agents. 2008;31(suppl 1):S96-S101.
Lipsky BA, Byren I, Hoey CT. Treatment of bacterial prostatitis. Clin Infect Dis. 2010;50(12):1641-1652.
Nagy V, Kubej D. Acute bacterial prostatitis in humans: current microbiological spectrum, sensitivity to antibiotics and clinical findings. Urol Int. 2012;89(4):445-450.
Ramakrishnan K, Salinas RC. Prostatitis: acute and chronic. Prim Care. 2010;37(3):547-563.
Brede CM, Shoskes DA. The etiology and management of acute prostatitis. Nat Rev Urol. 2011;8(4):207-212.
Zani EL, Clark OA, Rodrigues Netto N. Antibiotic prophylaxis for transrectal prostate biopsy. Cochrane Database Syst Rev. 2011;5:CD006576.
Campeggi A, Ouzaid I, Xylinas E, et al. Acute bacterial prostatitis after transrectal ultrasound-guided prostate biopsy: epidemiological, bacteria and treatment patterns from a 4-year prospective study. Int J Urol. 2014;21(2):152-155.
Özden E, Bostanci Y, Yakupoglu KY, et al. Incidence of acute prostatitis caused by extended-spectrum beta-lactamase-producing Escherichia coli after transrectal prostate biopsy. Urology. 2009;74(1):119-123.
Ekici S, Cengiz M, Turan G, Aliş EE. Fluoroquinolone-resistant acute prostatitis requiring hospitalization after transrectal prostate biopsy: effect of previous fluoroquinolone use as prophylaxis or long-term treatment. Int Urol Nephrol. 2012;44(1):19-27.
Minamida S, Satoh T, Tabata K, et al. Prevalence of fluoroquinolone-resistant Escherichia coli before and incidence of acute bacterial prostatitis after prostate biopsy. Urology. 2011;78(6):1235-1239.
Song W, Choo SH, Sung HH, et al. Incidence and management of extended-spectrum beta-lactamase and quinolone-resistant Escherichia coli infections after prostate biopsy. Urology. 2014;84(5):1001-1007.
Ludwig M. Diagnosis and therapy of acute prostatitis, epididymitis and orchitis. Andrologia. 2008;40(2):76-80.
Touma NJ, Nickel JC. Prostatitis and chronic pelvic pain syndrome in men. Med Clin North Am. 2011;95(1):75-86.
Etienne M, Pestel-Caron M, Chapuzet C, Bourgeois I, Chavanet P, Caron F. Should blood cultures be performed for patients with acute prostatitis?. J Clin Microbiol. 2010;48(5):1935-1938.
Sharp VJ, Takacs EB, Powell CR. Prostatitis: diagnosis and treatment. Am Fam Physician. 2010;82(4):397-406.
Yazawa S, Nagata H, Kanao K, et al. Novel algorithm for predicting severe cases of acute bacterial prostatitis. J Urol. 2013;189(4S):e475-e476.
Mosharafa AA, Torky MH, El Said WM, Meshref A. Rising incidence of acute prostatitis following prostate biopsy: fluoroquinolone resistance and exposure is a significant risk factor. Urology. 2011;78(3):511-514.
Acute prostatitis. In: Gilbert DN, Chambers HF, Eliopoulos GM, Saag MS, eds. The Sanford Guide to Antimicrobial Therapy. 44th ed. Sperryville, Va.: Antimicrobial Therapy; 2014:27.
Ulleryd P, Sandberg T. Ciprofloxacin for 2 or 4 weeks in the treatment of febrile urinary tract infection in men: a randomized trial with a 1 year follow-up. Scand J Infect Dis. 2003;35(1):34-39.
Nickel JC. Recommendations for the evaluation of patients with prostatitis. World J Urol. 2003;21(2):75-81.
Prostatitis and chronic pelvic pain syndrome. In: Grabe M, Bjerklund-Johansen TE, Botto H, et al. Guidelines on Urological Infections. Arnhem, The Netherlands: European Association of Urology (EAU); 2011:65–73.
Yoon BI, Han DS, Ha US, et al. Clinical courses following acute bacterial prostatitis. Prostate Int. 2013;1(2):89-93.
Shakil J, Piracha N, Prasad N, et al. Use of outpatient parenteral antimicrobial therapy for transrectal ultrasound-guided prostate biopsy prophylaxis in the setting of community-associated multidrug-resistant Escherichia coli rectal colonization. Urology. 2014;83(4):710-713.
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13594 | https://www.astro.princeton.edu/~gk/A403/state.pdf | EQUATION OF STATE Consider elementary cell in a phase space with a volume ∆x ∆y ∆z ∆px ∆py ∆pz = h3, (st.1) where h = 6.63×10−27 erg s is the Planck constant, ∆x ∆y ∆z is volume in ordinary space measured in cm 3, and ∆px ∆py ∆pz is volume in momentum space measured in ( g cm s −1)3. According to quantum mechanics there is enough room for approximately one particle of any kind within any elementary cell. More precisely, an average number of particles per cell is given as nav = g e(E−µ)/kT ± 1, (st.2) with a ”+” sign for fermions, and a ”−” sign for bosons. The corresponding distributions are called Fermi-Dirac and Bose-Einstein, respectively. Particles with a spin of 1/2 are called fermions, while those with a spin 0, 1, 2... are called bosons. Electrons and protons are fermions, photons are bosons, while larger nuclei or atoms may be either fermions or bosons, depending on the total spin of such a composite particle. In the equation (st.2) E is the particle energy, k = 1.38 × 10−16 erg K −1 is the Boltzman constant, T is temperature, µ is chemical potential, and g is a number of different quantum states a particle may have within the cell. The meaning of temperature is obvious, while chemical potential will become more familiar later on. In most cases it will be close to the rest mass of a particle under consideration. If there are anti-particles present in equilibrium with particles, and particles have chemical potential µ then antiparticles have chemical potential µ −2m, where m is their rest mass.
For free particles their energy is a function of their momentum only, with the total momentum p given as p2 = p2 x + p2 y + p2 z.
(st.3) The number density of particle in a unit volume of 1 cm 3, with momenta between p and p + dp is given as n (p) dp = g e(E−µ)/kT ± 1 4πp2 h3 dp, (st.4) because the number of elementary cells within 1 cm 3 and a momentum between p and p + dp, i.e.
within a spherical shell with a surface 4πp2 and thickness dp is equal to 4πp2 dp h−3. The number density of particles with all momenta, contained within 1 cm 3 is n = ∞ Z 0 n (p) dp, ρ = nm (st.5) where ρ is the mass density of gas. Note, that if we know density and temperature, then we can calculate chemical potential with the eqs. (st.4) and (st.5), provided we know how particle energy depends on its momentum, i.e. the function E(p) is known.
In the equation (st.4) g, k, T, µ, π, h are all constant, and the energy E depends on the momentum p only. As our particles may be relativistic as well as non-relativistic, we have to use a general formula for the relation between E and p. We have E ≡Etotal = E0 + Ek, (st.6) where the rest mass E0 = mc2, c = 3×1010 cm s −1 is the speed of light, and Ek is the kinetic energy of a particle. For a particle moving with arbitrary velocity there is a special relativistic relation: st — 1 E2 total = mc22 + (pc)2 .
(st.7) Combining the last two equations we obtain E = mc2 1 + p mc 21/2 , Ek = mc2 " 1 + p mc 21/2 −1 # .
(st.8) The last equation has two simple asymptotic limits: Ek ≈p2 2m, for p ≪mc, ( non-relativistic limit), Ek ≈pc, for p ≫mc, ( ultra-relativistic limit).
(st.9) Velocity of a particle is given in general as v = dE dp = p m 1 + p mc 2−1/2 .
(st.10) Again, we have two simple asymptotic limits: v ≈p m, for p ≪mc, ( non-relativistic limit), v ≈c, for p ≫mc, ( ultra-relativistic limit), (st.11) Kinetic energy of all particles in a unit volume of 1 cm 3 may be calculated as U = ∞ Z 0 Ek (p) n (p) dp, [ erg cm −3].
(st.12) We are interested in isotropic gas, so the velocity and momentum vectors of every particle are parallel to each other. Velocities of different particles are pointing in different directions, and their angular distribution is isotropic. For a given absolute value of velocity and momentum we may calculate the dot-product of the two vectors, and average it over all angles: vp = vp = vxpx + vypy + vzpz.
(st.13a) ⟨vxpx⟩= ⟨vypy⟩= ⟨vzpz⟩= 1 3vp.
(st.13b) Pressure is defined as a flux of momentum across a unit surface of 1 cm 2, integrated over all particles.
For an isotropic gas we may select the unit surface to be perpendicular to the ”x” axis, and we may calculate pressure as P = ∞ Z 0 < vxpx > n (p) dp = 1 3 ∞ Z 0 v (p) p n (p) dp, (st.14) where as before the averaging is done over particles moving in all directions.
We shall consider now some special, and simple, but important cases. First, let us consider any non-relativistic gas, with arbitrary momentum distribution function n(p), such that for a vast majority of particles a condition p ≪mc is satisfied. In this case we have v ≈p/m, Ek ≈p2/2m, and P ≈1 3 ∞ Z 0 p2 m n (p) dp, U ≈ ∞ Z 0 p2 2mn (p) dp ≈3 2P, (non −relativistic).
(st.15) Second, there is a case that most particles are ultra-relativistic, with p ≫mc, v ≈c, and Ek ≈pc for most of them. Now, pressure and kinetic energy density are given as st — 2 P ≈1 3 ∞ Z 0 pc n (p) dp, U ≈ ∞ Z 0 pc n (p) dp ≈3P, (ultra −relativistic).
(st.16) Notice, that the relations U ≈3P/2 and U ≈3P for the two limits are very general, and do not depend on the details of the distribution function n(p).
Our formulae allow us to calculate, at least in principle, pressure and kinetic energy density for a gas made of identical particles in a local thermodynamic equilibrium (LTE) . We made an implicit assumption that particles are interacting so weakly that energy of their interaction may be neglected, but that there is enough interaction between them to establish equilibrium distribution. If we have a mixture of few different types of particles, for example electrons, protons, and photons, we make the same assumptions about their interactions: it is small enough that we may assume the total pressure and kinetic energy density of the composite gas is the sum of pressures and kinetic energy densities of the components, but it is strong enough to maintain equilibrium distributions with the same temperature for all types of particles. Now, we shall calculate in detail some examples of the distribution functions, and the corresponding thermodynamic quantities.
Maxwell distribution This is the case when mc2 ≫E −µ ≫kT, (st.17) i.e. the particles are non-relativistic, and the exponential term in the denominator of equation (st.2) is very large, nav ≪1, i.e. there are very few particles per elementary cell in phase space. It is clear that the ±1 term is negligible, and therefore the distribution will be the same for fermions and for bosons. Within the approximation following from equation (st.17) we have n (p) = 4πg h3 e(µ−mc2)/kT e−p2/2mkT p2, (st.18a) n = 4πg h3 e(µ−mc2)/kT ∞ Z 0 e−p2/2mkT p2dp = (st.18b) 4πg h3 e(µ−mc2)/kT (2mkT )1.5 1 2 ∞ Z 0 e−xx0.5dx = 4πg h3 e(µ−mc2)/kT (2mkT )1.5 π0.5 4 , P = 1 3 4πg h3 e(µ−mc2)/kT ∞ Z 0 p2 m e−p2/2mkT p2dp = (st.18c) 4πg h3 e(µ−mc2)/kT (2mkT )2.5 1 3m 1 2 ∞ Z 0 e−xx1.5dx = 4πg h3 e(µ−mc2)/kT (2mkT )2.5 1 3m 1.5π0.5 4 = nkT, U = 1.5nkT, (st.18d) st — 3 where we substituted x = p2/2mkT .
In practice the Maxwell distribution is used for atoms, ions and electrons with a relatively low density. The relation between the number density of particles n[cm−3] and physical density ρ[ g cm −3] depends on chemical composition. It is customary to use X for the fractional abundance of hydrogen (by mass fraction) , Y for the abundance of helium, and Z for the combined abundance of all heavier elements. Frequently, all those heavier elements are described in astrophysical literature as ”metals”, even though the most common among them are carbon, nitrogen and oxygen.
Of course, we have X + Y + Z = 1. A typical chemical composition of an interstellar medium, or the main sequence stars is: X = 0.70, Y = 0.28, Z = 0.02, i.e. there is mostly hydrogen, and only very small fraction of heavy elements. These abundances are by mass fraction, so the fractional number of heavy atoms is approximately 0.002, or even less.
In most stellar applications it is safe to assume that all atoms are fully ionized.
The only exceptions are stellar atmospheres and sub-atmospheric layers. A convenient unit of mass is H = 1.67 × 10−24 g , i.e. mass of a single hydrogen atom. Mass of a helium atom is 4H, and mass of an element with a nuclear charge Z∗is approximately 2Z∗. An average charge of all heavier elements is approximately < Z∗>≈8. In one gram of matter there are X/H hydrogen nuclei, Y/4H helium nuclei, and approximately Z/(2 < Z∗> H) heavier nuclei. Therefore, the number of nuclei per gram of matter, i.e. the number density of ions is ni = ρ H X + Y 4 + Z 2 < Z∗> ≈ρ H X + Y 4 + Z 16 .
(st.19) Fully ionized hydrogen provides 1 electron per nucleus, i.e. 1 electron per nucleon. Helium provides 2 electrons per nucleus, i.e. 1 electron per 2 nuclei. A heavier element with a charge Z∗provides Z∗electrons per nucleon, which in a typical case has A ≈2Z∗nucleons, i.e. we get 1 electron per 2 nucleons, just like for helium. The number density of electrons may be calculated as ne = ρ H X + Y 2 + Z 2 = ρ H 1 + X 2 .
(st.20) The number density of all particles is given as n = ni + ne ≈ρ H (2X + 0.75Y + 0.5Z) .
(st.21) It is customary to define mean molecular weight µ, mean molecular weight per ion µi, and mean molecular weight per electron µe as µ ≡ ρ nH ≈ 1 2X + 0.75Y + 0.5Z , (st.22a) µi ≡ ρ niH = 1 X + Y/4 + Z/16, (st.22b) µe ≡ ρ neH = 2 1 + X .
(st.22c) With the new definitions of mean molecular weights (not to be confused with a chemical potential, which is also written as µ ) we may write the equation of state as Pe = k µeH ρT, (st.23a) Pi = k µiH ρT, (st.23b) Pg = Pe + Pi = k µH ρT, (st.23c) where Pe, Pi, and Pg are the electron pressure, the ion pressure, and the gas pressure, respectively.
We also have the corresponding kinetic energy densities given as Ue = 1.5Pe, Ui = 1.5Pi, U = 1.5P.
(st.24) st — 4 Planck distribution This is the distribution of photons under LTE (local thermodynamic equilibrium) conditions. It is customary to characterize photons with the frequency of their oscillations ν. Their wavelength is given as λ = c/ν, energy as E = hν, and momentum as p = hν/c. Photons always move with the speed of light, have zero rest mass, and are their own antiparticles. Therefore, under LTE conditions their chemical potential is zero. Their spin is 1, so they are bosons. A photon has two quantum states with the same energy; they correspond to two different polarization states. Therefore g = 2 for photons, and we have n (p) dp = 8π h3 p2 epc/kT −1dp, (st.25a) or, in a more customary form nνdν = 8π c3 ν2 ehν/kT −1dν.
(st.25b) The total energy associated with radiation is Ur = ∞ Z 0 nνhν dν = 8πh c3 ∞ Z 0 ν3 ehν/kT −1dν = 8πh c3 kT h 4 ∞ Z 0 x3dx ex −1, (st.26a) where we substituted x = hν/kT . The last integral has a value of π4 15 ≈6.088, and we may write Ur = aT 4, where a = 8π5 15 k4 h3c3 ≈7.565 × 10−15 erg cm −3 K −4.
(st.26b) Radiation pressure is given as Pr = 1 3Ur = a 3T 4, (st.27) because photons are ultra-relativistic.
It is customary to describe radiation with its intensity Bν(T ), and B (T ) = ∞ Z 0 Bν (T ) dν.
(st.28) In LTE radiation intensity is the same in all directions, over the whole solid angle Ω= 4π. The radiation energy density may be calculated as Ur = 1 c Z B (T ) dΩ= 4π c B (T ) , (st.29) where the integration covers all directions; in our case radiation is isotropic, i.e. B(T ) is the same in all directions. We may combine equations (st.26b) and (st.29) to obtain B (T ) = c 4π Ur = ac 4π T 4, (st.30) and also Bν (T ) dν = c 4π Uνdν = 2h c2 ν3 ehν/kT −1dν.
(st.31) st — 5 Bν(T ) is called the Planck function. It gives the intensity of radiation of a black body at a given temperature T .
Let us calculate the flux of radiation emitted from a surface of a black body which has a tem-perature T . The intensity of radiation integrated over all frequencies is given as B(T ) and it is the same in any direction pointing out from the surface. Let us introduce angle θ, between the normal to the surface and the direction of a light ray, and an azimuthal angle φ. An element of a solid angle is given as dΩ= sin θdφdθ. The flux of black body radiation comming in the direction normal to the surface may be calculated integrating over all angles the component of B(T ) orthogonal to the surface, i.e. B(T ) cosθ. We have: FBB = π/2 Z 0 2π Z 0 B (T ) cos θ sin θdφ dθ = 2πB (T ) π/2 Z 0 cos θ sin θdθ = πB (T ) .
(st.32) Combining the last equation with (st.30) we obtain the flux of radiation from a unit surface of a black body FBB = σT 4, σ = ac 4 = 5.67 × 10−5 erg cm −2 s −1 K −4, (st.33) where σ is the Stefan-Boltzman constant.
Real astronomical objects radiate in a more complicated way than a black body of any temper-ature. However, it is useful to define an effective temperature as the temperature of a black body that would radiate out energy at the rate that the star does. If stellar luminosity is L, and its radius is R, then its surface radiates at a rate of F = L/4πR2. The effective temperature is defined with a relation: σT 4 eff ≡ L 4πR2 .
(st.34) If we have a mixture of gas and radiation then total pressure and total kinetic energy density are given as P = Pg + Pr = k µH ρT + a 3T 4, (st.35a) U = Ug + Ur = 1.5 k µH ρT + aT 4, (st.35b) It is useful to define the ratio of gas pressure to total pressure β ≡Pg/P, 0 < β < 1.
(st.36) At a low temperature gas pressure dominates, while at a high temperature radiation pressure dom-inates. The two contributions are equal, i.e. we have Pg = Pr when k µH ρT = a 3T 4.
(st.37) For a standard chemical composition, i.e. for X = 0.70, Y = 0.28, Z = 0.02, the mean molecular weight is µ−1 = 1.62, and the gas pressure constant is k/µH = 5.1 × 107 erg g −1 K −1. The density where gas pressure equals radiation pressure is ρ ≈5 × 10−23 T 3 g cm −3 K −3 = T 3 × 107 K 3 g cm −3 .
(st.38) st — 6 Fermi-Dirac distribution Now, we shall consider a distribution of electrons, i.e. spin 1/2 particles, when density is high.
An electron may be in two spin states: +1/2 and −1/2, so the number of different quantum states per unit cell in a phase space is g = 2, and the average number of electrons per unit cell is nav = 2 e(E−µ)/kT + 1, (st.39a) and the number density of electrons is given as ne = ∞ Z 0 ne (p) dp = 8π h3 ∞ Z 0 p2dp e(E−µ)/kT + 1, (st.39b) with electron energy E being a function of its momentum (cf. equation st.8) . The density, pressure, and kinetic energy density are given as (cf. equations: st.12, st.14, st.22c) ρ = neµeH = ne 2H 1 + X , (st.40a) Pe = 1 3 ∞ Z 0 v (p) p ne (p) dp, (st.40b) Ue = ∞ Z 0 Ek (p) ne (p) dp.
(st.40c) The chemical potential which appears in equations (st.39) is approximately equal to the electron rest mass: µ = mc2 + EF , where EF is Fermi energy - its physical meaning will become apparent shortly. The E −µ term in the exponent in the denominator in equations (st.39) is equal to Ek −EF , where Ek is a kinetic energy of an electron. If EF ≪0 the distribution function for electrons is Maxwellian, as the exponential term is much larger than 1 even for electrons with zero kinetic energy.
Here we shall be interested in another limit: EF ≫kT . Now, the exponential term is very small for Ek < EF , it is equal 1 for Ek = EF , and it is very large for Ek > EF . The transition from being much less than 1 to much larger than 1 takes place over a relatively small change in kinetic energy, ∆E≈2kT ≪EF . Therefore, in the limit when kT ≪EF the occupation number nav as given with equation (st.39a) becomes almost a step function of electron’s kinetic energy: nav ≈2, for Ek < EF , kT ≪EF , (st.41a) nav ≈0, for Ek > EF , kT ≪EF , (st.41b) This result may be interpreted as follows. When temperature drops the electrons occupy the lowest energy states available. However, as they are fermions they have to obey Pauli exclusion principle: there cannot be more then one electron in the same quantum state. An electron can be in two different spin states. This means there can be at most two electrons in any elementary cell of a phase space. It is the Heisenberg uncertainty principle that is responsible for the size of those cells.
So, as the temperature drops the electrons fill up the lowest energy cells all the way up to Fermi energy, and all higher energy states are empty. Such gas is called degenerate. The value of Fermi energy depends on the density of electrons in ordinary space. This is obvious upon evaluating the integral (es.39) , subject to the conditions (st.41) . We obtain st — 7 ne = 8π h3 pF Z 0 p2dp = 8π 3 pF h 3 , (st.42) where pF is Fermi momentum related to Fermi energy by equation (st.8) . That relation may be written as EF mc2 + 1 2 = pF mc 2 + 1.
(st.43) We may express density, pressure and kinetic energy density in terms of Fermi momentum, and we shall write equations (st.40) as ρ = µeH 8π 3h3 p3 F , (st.44a) Pe = 8π 3mh3 pF Z 0 p4 dp h 1 + p mc 2i1/2 , (st.44b) Ue = 8πmc2 h3 pF Z 0 " 1 + p mc 21/2 −1 # p2dp, (st.44c) where the dependence of kinetic energy and velocity on momentum was given with equations (st.8) and (st.10) . It is convenient to introduce dimensionless variables: x ≡p/mc, xF ≡pF /mc, (st.45) and to rewrite equations (st.44) as ρ = Aµex3 F , (st.46a) Pe = B xF Z 0 x4 dx (1 + x2)1/2 , (st.46b) Ue = 3B xF Z 0 h 1 + x21/2 −1 i x2dx = 3B xF Z 0 x4 dx (1 + x2)1/2 + 1 .
(st.46c) With the electron mass m = 9.11 × 10−28 g the constants are A ≡8π 3 mc h 3 H = 0.981 × 106 [ g cm −3], (st.46d) B ≡8π 3 mc h 3 mc2 = 4.80 × 1023 [ erg cm −3].
(st.46e) The integrals in equations (st.45) can be evaluated analytically, but the results are too compli-cated to be of much practical value. It is much more useful to consider two limiting cases. We shall first consider xF ≪1, i.e. the electron gas will be degenerate non-relativistic. For x ≪1 the leading term in the expansion under the integrals (st.45) is x4, and we obtain Pe = 1 5Bx5 F = 0.991 × 1013 (ρ/µe)5/3 , xF ≪1, (st.47a) Ue = 3 10Bx5 F = 3 2Pe, xF ≪1.
(st.47b) st — 8 The relation between Pe and Ue is just that expected for any non-relativistic gas (cf. equation st.15) .
The second limiting case is xF ≫1, i.e. ultra-relativistic degeneracy. Following the previous procedure we find that the leading term under the integrals (st.46b,c) is x3, and we obtain Pe = 1 4Bx4 F = 1.231 × 1015 (ρ/µe)4/3 , xF ≫1, (st.48a) Ue = 3 4Bx4 F = 3Pe, xF ≫1, (st.48b) Again, the relation between Pe and Ue is that expected for any ultra-relativistic gas.
The transition between equations (st.47) and (st.48) should be for xF ≈1, i.e. at ρ ≈106 g cm −3.
In the transition region one may either use the exact and complicated analytical formula for the integral (st.44b) , or the following approximate formula, which has accuracy better than about 2%.
We may write the equations (st.47a) and (st.48a) as Pe,nr = K1ρ5/3, K1 = 1 20 3 π 2/3 h2 m (Hµe)5/3 = 0.991 × 1013µ−5/3 e , (st.49a) Pe,r = K2ρ4/3, K2 = 1 8 3 π 1/3 hc (Hµe)4/3 = 1.231 × 1015µ−4/3 e , (st.49b) Pe,d = P −2 e,nr + P −2 e,r −1/2 , (st.49c) where Pe,d stands for pressure of degenerate electron gas. The interpolation formula (st.49c) selects the smaller of the two limiting case formulae.
In the non-degenerate and non-relativistic limit, i.e. for mc2 ≫E −µ ≫kT the integrals (st.39) and (st.40) can be evaluated as in the case of Maxwell distribution function, and we obtain Pe,nd = k µeH ρT, Ue,nd = 3 2Pe,nd, (st.50) where the subscript ”e,nd” stands for non-degenerate electron gas.
In the non-relativistic partly degenerate region, i.e. when E−µ ≈kT , there is no exact analytical formula for the integrals (st.39) and (st.40) , but there is an analytical interpolation formula that is accurate to about 2%. Pressure of partly degenerate electron gas Pe, may be calculated as Pe = P 2 e,nd + P 2 e,d 1/2 .
(st.51) The last formula selects the larger of the two limiting case formulae.
Let us now find the region in the density - temperature plane where there is a transition from non-degenerate to degenerate electron gas. This may be estimated setting Pe,nd = Pe,d ≈Pe,nr, which gives k µeH ρT = K1ρ5/3 ⇐ ⇒ T ≈105ρ2/3.
(st.52) Electron-positron pairs The final formula (st.51) for pressure of electron gas has a very large range of applicability.
There is no limit at very low densities. At very high densities the only correction required is in the evaluation of the mean number of nucleons per electron, µe. At very low temperature gas is no longer fully ionized, and again the evaluation of the number of nucleons per free electron becomes st — 9 complicated. At very high temperature, when kT >= mc2, electrons become relativistic. It becomes energetically possible to form electron - positron pairs, i.e. to create new particles. Now we have to write down the distribution functions for both types of electrons. As one is antiparticle of another their chemical potential differ by two rest mass energies, i.e. we have µ−= µ ≈mc2, µ+ = µ −2mc2 ≈−mc2, (st.53) and the average number of particles per unit cell is n− av = 2 e(E−µ)/kT + 1, n+ av = 2 e(E−µ+2mc2)/kT + 1, (st.54) and the number density per 1 cm 3 is given as n− e = ∞ Z 0 n− e (p) dp = 8π h3 ∞ Z 0 p2dp e(E−µ)/kT + 1, (st.55a) n+ e = ∞ Z 0 n+ e (p) dp = 8π h3 ∞ Z 0 p2dp e(E−µ+2mc2)/kT + 1, (st.55b) where superscripts ”-” and ”+” refer to electrons and positrons, respectively.
There is a constraint on the total number of particles, as the difference between the number density of electrons and positrons must be equal to the number of electrons that came from ionizing all the ions, i.e.
ne = n− e −n+ e = ρ µeH , µe = 2 1 + X , (st.56) where X is hydrogen abundance by mass fraction.
Given the distribution functions we may calculate electron pressure end electron energy density.
In the latter we shall include not only the kinetic energy of all electrons and positrons, but also the annihilation energy of all electron - positron pairs, but not the rest mass energy of the ”original” electrons. We may write Pe = Pe−+ Pe+ = 1 3 ∞ Z 0 v (p) p n− e (p) dp + 1 3 ∞ Z 0 v (p) p n+ e (p) dp, (st.57a) Ue = Ue−+ Ue+ + 2n+ e mc2 = (st.57b) ∞ Z 0 Ek (p) n− e (p) dp + ∞ Z 0 Ek (p) n+ e (p) dp + 2n+ e mc2.
As usual, the integrals are too complicated to evaluate them analytically, but there are simple limiting cases.
First, we shall consider a case of a relatively low temperature and low density: mc2 ≫E −µ ≫ kT , just like the Maxwell distribution, but we shall look for a small correction in order to estimate the number of pairs. We have, under this approximation: e(E−µ)/kT ≫1, e(E−µ+2mc2)/kT ≫1, E ≈mc2 + p2 2m, (st.58) and therefore the number density of electrons and positrons may be calculated as n− e ≈8π h3 eµ/kT ∞ Z 0 p2e−E/kT dp ≈ (st.59a) st — 10 8π h3 e(µ−mc2)/kT ∞ Z 0 p2e−p2/2mkT dp ≈ne, and n+ e ≈8π h3 e(µ−2mc2)/kT ∞ Z 0 p2e−E/kT dp ≈ (st.59b) 8π h3 e(µ−3mc2)/kT ∞ Z 0 p2e−p2/2mkT dp ≈nee−2mc2/kT .
Finally, the number ratio is just what we might expect from a Boltzman formula: n+ e n− e ≈e−2mc2/kT , (st.60) It is obvious how to calculate small corrections to electron pressure and electron energy density.
Notice, that the corrections to energy density are relatively larger than the corrections to pressure because the rest mass energy of a small number of pairs is much larger than their thermal energy.
Another simple case is when kT ≫mc2, and therefore most electrons and positrons are ultra-relativistic, their velocities are approximately equal c, and their energies E ≈pc ≫mc2 ≈µ. In this limit we have n− e ≈n+ e ≈8π h3 ∞ Z 0 p2dp epc/kT + 1 ≫ne, (st.61) i.e. there are many more pairs than original electrons. The energy density due to all these pairs may be calculated as Ue ≈16π h3 ∞ Z 0 pc p2dp epc/kT + 1 = 16π (kT )4 (hc)3 ∞ Z 0 x3dx ex + 1 = (st.62) 14π5 15 (kT )4 (hc)3 = 7 4Ur, where we substituted x = pc/kT , the last integral in equation (st.62) has a value 7π4/120, and Ur is the radiation energy density. As the pairs are ultra-relativistic in this limit, their thermodynamic properties are very much like those of radiation. In particular, in this limit Pe = Ue/3 = 7Pr/4.
Summary In a large region in a density - temperature plane the equation of state, and all important thermodynamic quantities may be calculated analytically with a reasonable precision, better than 2%. The region is limited at very low density by the mass density associated with radiation energy density becoming dominant, at low temperature by partial recombination of helium and hydrogen, at high density by crystallization of ions, and at high temperature by creation of electron-positron pairs. As long as we stay within this very large region the following prescription can be adopted to calculate equation of state.
st — 11 Let X, Y , and Z be the abundances of hydrogen, helium and heavy elements, respectively, all by mass fraction. Of course, we have X + Y + Z = 1. Various mean weights can be calculated as follows: 1 µi ≡niH ρ = X + Y 4 + Z 16, 1 µe ≡neH ρ = 1 + X 2 , (st.63) 1 µ ≡nH ρ = 1 µi + 1 µe ≈2X + 0.75Y + 0.5Z, n = ni + ne, where n, ni, and ne, is a number density of all particles, ions, and electrons, respectively, and H = 1.673 × 10−24 g is the mass of a hydrogen atom.
In the whole region of interest the ion and radiation pressure can be calculated according to Pi = k µiH ρT, k H = 0.825 × 108, [ erg g −1 K −1], (st.64) Pr = 1 3Ur = a 3T 4, a = 8π5 15 k4 h3c3 ≈7.565 × 10−15 [ erg cm −3 K −4].
(st.65) The electron pressure can be calculated according to the following set of formulae: Pe,nd = k µeH ρT, (st.66a) Pe,nr = K1ρ5/3, K1 = 1 20 3 π 2/3 h2 m (Hµe)5/3 = 0.991 × 1013µ−5/3 e , (st.66b) Pe,r = K2ρ4/3, K2 = 1 8 3 π 1/3 hc (Hµe)4/3 = 1.231 × 1015µ−4/3 e , (st.66c) Pe,d = P −2 e,nr + P −2 e,r −1/2 , Pe = P 2 e,nd + P 2 e,d 1/2 .
(st.66d) The derivatives (∂Pi/∂T )ρ, (∂Pr/∂T )ρ, (∂Pe/∂T )ρ, (∂Pi/∂ρ)T , (∂Pr/∂ρ)T , (∂Pe/∂ρ)T , can be cal-culated analytically from the above expressions, as well as the derivatives (∂ui/∂T )ρ, (∂ur/∂T )ρ, noticing that Ui = uiρ = 1.5Pi, Ur = urρ = 3Pr.
(st.67) The derivative (∂ue/∂T )ρ, can be calculated with the following approximate formula: ∂ue ∂T ρ = 1 + b 2 + b 3 ρ ∂Pe ∂T ρ , b = K1 K2 2 = 6.48 × 10−5 ρ µe 2/3 , (st.68) The combined pressure, as well as the combined derivatives may be calculated according to P = Pr + Pi + Pe, (st.69a) ∂P ∂T ρ = ∂Pr ∂T ρ + ∂Pi ∂T ρ + ∂Pe ∂T ρ , (st.69b) ∂P ∂ρ T = ∂Pr ∂ρ T + ∂Pi ∂ρ T + ∂Pe ∂ρ T , (st.69c) ∂u ∂T ρ = ∂ur ∂T ρ + ∂ui ∂T ρ + ∂ue ∂T ρ .
(st.69d) st — 12 |
13595 | https://www.youtube.com/watch?v=5GAlEFi9iZI | Quadratic Word Problems in Factored Form: Challenge Problems
Euler's Academy
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Posted: 7 Sep 2021
This video continues looking at different challenging problems involving quadratic word problems when the equation is given in factored form.
Khan Academy exercise to practice:
1. Quadratic Word Problems in Factored Form:
EulersAcademy.org
Transcript:
so in this one anna dives into a pool off of a springboard high dive her height in meters above the water x seconds after diving is modeled by this equation so let's make sure we understand the variables h of x that's our height and it's in meters above the water so she's diving off a springboard into a pool so if we draw our diving board here she's up here and she's going to dive off and we'll get projectile motion since she's going to be traveling through the air and projectile motion we know is parabolic so that's why you get this kind of quadratic equation so h of x that's her height so that's our vertical axis and then our horizontal axis x that's our time so x is time and that's going to be in seconds again so we need to answer the question how many seconds after diving will anna hit the water so that's essentially if we want to answer that question that is right here what is the time for her to actually hit the water so we can ask ourselves what is the key feature of our equation when she actually hits the water so at that point we don't know the x value that's what we're trying to find how many seconds it took her to actually reach this water but at this point we do know the height since the height of the water level is zero so in this case our function h of x is equal to zero and so we can set our equation equal to zero and then solve so let's do that right here we have that h of x is zero and this is equal to minus 5 times x plus 1 times x minus 3. and since this is equal to 0 we can use our zero product property since if either of these factors here these binomials are equal to zero then you have zero times something but zero times anything is zero so we need to figure out which x values can we plug in here and here to make these sets of parentheses zero if you plug in negative one here this would be zero and zero times anything is zero so we don't have to go any further so x is negative one and if you plug in three here three times three minus three is zero and so zero times whatever would still give you 0. so -1 and 3 are the two answers but one of those answers doesn't make sense since the time is her seconds after diving so if you had a negative amount of time that's talking about before she dove and we don't have any information about that so that is not part of our domain here we're not going to include negative values of time so that only leaves one answer that 3 are at three seconds is when she hits the ground now you can check this plug in three to your height equation here or the equation that models height and notice you get minus five times by three plus 1 is 4 times by 3 minus 3 which is 0 and 0 times minus 20 is 0. so when you check it when you plug in an x value of 3 seconds you do see that the height does give you back zero so that's a height or a time of three seconds to reach a height of zero or to reach her hitting the water so let's do one final question and in this one simon has 160 meters of fencing to build a rectangular garden the garden's area in square meters as a function of the guard's width in meters is modeled by this equation what width will produce the maximum garden area so let's make sure we understand our variables before we try and solve this so a of x our function this is just the gardens area so the garden's area and area is in square meters and then x is the width of the garden and that's going to be just in meters so in general if we were to sketch this equation out let me just rewrite the equation so the area as a function of the width of the garden is minus x times x minus 80. and we're not going to talk about where this equation comes from but it does actually make sense and when we graph this basically at zero width you're not going to have any area because that's essentially just a line worth of your garden it won't be thick at all so it's zero width it'll have zero area and then it'll slowly go up as the width increases it'll eventually hit a maximum and then come back down and if your width is 80 then you don't have any room left for the length of the garden and so it again it'd just be a straight line just perpendicular to the last one so at a width of 80 again it's going to be zero for the area but to make sense of that with the actual numbers you can use the zero product property because we do want to know these intercepts because if we know these intercepts then we can find the maximum garden area because we want to find this vertex here to figure out that maximum area and to figure out the vertex we'll use symmetry because both of these intercepts are an equal distance away from the vertex so if we can figure out this value and this value then we can just average them to find the x value of the width that produces this maximum area and they are not concerned with what the actual maximum area is they just want to know what width would actually produce that so to find these intercepts remember this is just when your function or your area is equal to zero so which width values would give you an area equal to zero so we can set this equal to zero and then use our zero product property so you can see that in this case if you plug in 80 here you get 80 minus 80 which is 0 times negative 80 but that's 0. and if you plug in 0 here you have 0 0 times by negative 80 but that's also 0. so 80 and 0 are the two intercepts so this one here is at zero this is at a width of 80 meters and so right between them you can add them and divide by 2 but that would be a width of 40 meters so at 40 meters of width the maximum garden area would be produced and notice this is just a little bit extra but you had 160 meters total so if you were producing this rectangular garden and one side was 40 while the other would have to be 40 since it's rectangular and then there's only 80 left total which has to be divided between the two so that means these would also have to be 40. so the area of this garden that produces the maximum or maximum area is when the garden is a square so this problem actually does come up from time to time but if you're ever considering how to maximize the space of some rectangular plot if you make it into a square you actually get the most area for a given amount of fencing let's say |
13596 | https://artofproblemsolving.com/wiki/index.php/Area?srsltid=AfmBOopoq2UYb8WNAjIJw7oN0Urot-5Uzo9BTUvWAMcqKzdyAAsGjm2y | Art of Problem Solving
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Area
In mathematics, area refers to the size of the region that a two-dimensional figure occupies. The size of a region in higher dimensions is referred to as volume.
It is often possible to find the area of a region bounded by parts of circles and line segments through elementary means. One can find the area of even more complex regions via the use of calculus.
Rectangles are the most basic figures whose area we can study. It makes sense that the area of a rectangle with length and width is simply .
Once we know the area of a rectangle, we can easily find the area of a triangle by just noting that if our triangle has base and height , then the rectangle with length and width has exactly twice as much area as the original triangle. Thus, the area of a triangle is
We can now find the area of any polygon by breaking it up into triangles.
Contents
[hide]
1 Introductory Videos
2 Notation
3 Area of a Regular Polygon
4 Area of a Triangle
4.1 Other formulas equivalent to Heron's
4.1.1 Also true, but more complex than above
5 Area of a Quadrilateral
6 See Also
Introductory Videos
Notation
The letters and are frequently used to stand for area. When there are multiple regions under consideration, subscripts are often employed: might be used to denote the areas of particular regions, or . For example, would mean the area of hexagon.
An alternative notation is to use square brackets around the name of the region to denote its area, e.g. for the area of triangle .
Area of a Regular Polygon
The area of any regular polygon can be found as follows:
Inscribe the figure, with sides of length , in a circle and draw a line from two adjacent vertices to the circumcenter. This creates a triangle that is of the total area (consider the regular octagon below as an example).
Drawing the apothem creates two right triangles, each with an angle of at the top vertex. If the polygon has side length , the height of the triangle can be found using trigonometry to be of length .
The area of each triangle is the base times the height, which can also be expressed as and the area of the entire polygon is .
Area of a Triangle
There are many ways to find the area of a triangle. In all of these formulae, will be used to indicate area.
where is a base and is the altitude of the triangle to that base.
Heron's formula: , where and are the lengths of the sides and is the semi-perimeter.
, where is the radius of the incircle and s is the semi-perimeter.
where and are adjacent sides of the triangle and is the measure of the angle between them.
, where are the lengths of the sides of the triangle and is the circumradius.
, where and the triangle has altitudes , , .
Other formulas equivalent to Heron's
These are especially useful when , for :
Also true, but more complex than above
Area of a Quadrilateral
To find the area of most quadrilaterals, you must divide the quadrilateral up into smaller triangles and find the area of each triangle. However, some quadrilaterals have special formulas to find their areas. Again, is the area.
Kite - where the s represent the lengths of the diagonals of the kite.
Parallelogram - , where is the base and is the height to that base.
Trapezoid - , where the s are the parallel sides and is the distance between those bases.
Rhombus - a special case of a kite and parallelogram, so either formula may be used here.
Rectangle - , where is the length of the rectangle and is the width. (This is a special case of the formula for a parallelogram where the height and a side happen to coincide.)
Square - , where is the length of a side.
Any quadrilateral - , where is the semiperimeter, , , , and are the side lengths, and and are the measures of angles and , respectively. (Bretschneider's formula)
Cyclic quadrilateral - where is the semiperimeter and , , , and are the side lengths. (This is a special case of the formula for the area of any quadrilateral; .) (Brahmagupta's formula)
See Also
Pick's Theorem
Shoelace Theorem
Retrieved from "
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Master Cube Cutting
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In this article, we are going to learn the concept of cube cutting or cutting the painted cube. While solving these questions, the most important part is to visualise the cube. Cube cutting is an important concept as questions are frequently asked from this topic in a number of competitive exams.
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Let's first learn some basic terminologies i.e. face, vertex and edge of a cube.
In a cube, there are 6 faces, 8 vertices & 12 edges. Vertex means corners & edge means side.
Generally, questions from this topic are of the type wherein, a cube with side measuring unit 'x' is painted on all faces and is cut into smaller cubes with sides measuring unit 'y'. You are then required to find the number of cubes having 'n' faces painted.
The first thing that you need to figure out is the number of smaller cubes. For this, you look at one particular edge of the big cube and figure out how many smaller cubes can fit into this. It will be x/y. So, the number of smaller cubes will be (x/y)3.
Since all the smaller cubes will have at least one face facing inside i.e. not on the surface of the original cube, hence, none of the smaller cubes will have all faces painted. Further, since the maximum number of faces of the larger cube that intersect at a point are 3(at the corners), hence, the smaller cubes can have a maximum of 3 faces painted.
So, the number of smaller cubes with 3 faces painted = No of corners of larger cube = 8 (always), provided none of the faces of the larger cube is left unpainted.
Let us apply this theory to the given questions.
Cube Cutting Sample Questions
Example 1: A cube having a side of 6 cm is painted red on all the faces and then cut into smaller cubes of 1 cm each. Find the total number of smaller cubes so obtained.
Solution:
As explained above, the number of smaller cubes = (6/1)3 = 216 smaller cubes.
(Here x=6 and y=1)
Example 2: In the above example, how many cubes will have three faces painted?
Solution:As explained above, only the corner cubes i.e. the 8 cubes at the corners of the original cube will have three faces painted. Hence the answer will be 8 only.
To find the number of smaller cubes with only 2 faces painted, you need to consider the cubes where 2 faces of the bigger cube meet, i.e. the edges. Remember, this includes the cubes present at the corners as well, so you need to remove those 2 cubes from the number of cubes on each edge.
Example 3: In the above example, how many cubes will have only two faces painted?
Solution:As discussed above, only the cubes at the edge of the bigger cube can have two faces painted.
The larger cube has 6 cm edge and smaller cube is 1 cm edge. Hence, there are 6 cubes on each edge. However, you need to consider 4 middle cubes only, as the 2 cubes on each corner will have 3 painted faces.
Hence, there are 4 such cubes on each edge. As there are 12 edges, there will be 412 = 48 cubes
.
Example 4: In the above example, how many cubes will have only one face and no side painted?
Solution: As discussed above, only the cubes at the face of the bigger cube can have only one painted face. Since the larger cube has 6 cm edge and smaller cube is 1 cm edge, hence, if you see one of the faces of the larger cube, you will see 66 = 36 cubes. Out of these, exclude the cubes which lie on the edges, as they have two or more faces which are painted. Thus, on each face of the original cube, there will be 44 = 16 cubes will have only one face painted.
As there are 6 such faces, the number of such smaller cubes will be 166 = 96.
Lastly, the number of cubes having no faces painted can be found by subtracting the sum of the painted cubes from the total number of smaller cubes. Therefore, the required answer is 216 – (8 + 48 + 96) = 64 cubes.
Example 5: A cube having an edge of 12 cm each. It is painted red on two opposite faces, blue on one other pair of opposite faces, black on one more face and one face is left unpainted. Then it is cut into smaller cubes of 1 cm each. Answer the following questions:
The number of smaller cubes which are having three-faces painted.
The number of smaller cubes which are having two-faces painted.
The number of smaller cubes which are having one-face painted.
The number of smaller cubes which are having zero-face painted.
Solution:
Total number of cubes= (121212)/(111)=1728
For a cube with all sides painted we have 8 cubes with 3 sides colored. But here we have 1 side unpainted. Therefore, we will have only 4 cubes with 3 sides painted. The other 4 cubes will have only 2 sides painted.
For 2 sides painted, we look for the edges.
A cube has 12 edges.
8 edges, each edge having 10 cubes will have 2 sides painted. (4 edges of an unpainted side won’t be included).
We'll also include those 4 cubes (which we didn’t count while counting 3 coloured sides, as they have 2 sides painted)
Cubes on 4 edges of the unpainted side of the cube will have 1 side painted (due to the unpainted side).
Therefore, total cubes with 2 sides painted= 810 + 4= 84 cubes.
For 1 side painted, we look for the faces of the cube.
A cube has 6 faces.
5 faces each having (12-2)(12-2) = 100 cubes will have one side painted.
We’ll have to include those cubes on the edges linked with an unpainted face.
10 cubes on each of those edges will have 1 side painted.
Therefore, total cubes with 1 side painted= 5100 + 410 = 540 cubes.
According to the formula, cubes with no side painted= (12-2) 3= 1000.
But we have to include the cubes from the unpainted side too. It will be 1010=100
So, total number of unpainted cubes= 1000+100=1100.
Shortcut Formulae
For a cube of side nnn painted on all sides which is uniformly cut into smaller cubes of dimension 111,
Number of cubes with 0 side painted= (n-2) 3
Number of cubes with 1 sides painted =6(n - 2) 2
Number of cubes with 2 sides painted= 12(n-2)
Number of cubes with 3 sides painted= 8(always)
For a cuboid of dimension abc painted on all sides which is cut into smaller cubes of dimension 111,
Number of cubes with 0 side painted= (a-2) (b-2) (c-2)
Number of cubes with 1 sides painted =2[(a-2) (b-2) + (b-2)(c-2) + (a-2)(c-2) ]
Number of cubes with 2 sides painted= 4(a+b+c -6)
Number of cubes with 3 sides painted= 8
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Key Learning:
To summarise, the number of three faces painted cubes may be found with the help of vertices, two-faces painted with the help of edges and one-face painted with the help of faces. By using this learning, we can even solve complex questions on cube cutting for example cube painted with different colours.
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13598 | https://spanish.stackexchange.com/questions/41487/recibir-vs-recibir-de | significado - Recibir vs. recibir de - Spanish Language Stack Exchange
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Recibir vs. recibir de
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Quisiera saber si existe alguna diferencia de significado entre "recibir" y "recibir de" en el siguiente párrafo. ¡Muchas gracias!
"El nombre de este tiempo (llamado en otras lenguas, por ejemplo en francés, definido) obedece a la intención de contraponerlo al «perfecto», el cual recibe del participio con que se forma un sentido de completo acabamiento."
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asked Dec 3, 2022 at 2:20
MauriceMaurice
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En términos generales no existe diferencia entre "recibir" y " recibir de ", lo único que pueden existir son omisiones aparentes.
En el verbo "recibir" se encuentra implícito aquella otra acción de "entregar". Si tú recibes algo es porque alguien te ha entregado ese algo. Por tanto, "recibir" siempre lleva implícita esta idea.
Recibir (por parte de algo o alguien) - Entregar (por parte de algo o alguien).
Tu ejemplo;
"... el cual recibe del participio con que se forma un sentido de completo acabamiento."
En este caso "recibedel participio" / "del participio recibe". El participio será el que entrega algo a ese verbo al que se refiere el ejemplo, "un sentido de completo acabamiento".
Otros ejemplos;
Los niños reciben de los abuelos chocolatinas, caramelos y otros chuches. (en este caso recibir y de se encuentran en la oración juntos). Los abuelos son los que entregan algo a los niños (chocolatinas, caramelos y chuches).
Los niños reciben caramelos de sus abuelos (recibir de, hay casos en que se encuentran en la misma oración pero separados). Los abuelos son los que entregan en este caso, caramelos y los niños los reciben.
Recibió mal la crítica ( aquí "recibir de", esta omitido, aunque se sobrentiende que la crítica fue emitida o entregada por alguien, Recibió mal la crítica (emitida de alguien, por parte de ellos).
El Guadalquivir recibe las aguas del Guadalimar. El Guadalimar es el que entrega sus aguas al Guadalquivir.
En Tauromaquia, en la "suerte de recibir o matar" "el torero recibe al toro". El torero recibe (con una manera determinada de estoquear) al toro. El toro mismo es el que se entrega al torero que es el que lo recibe aguardando al toro.
El abogado recibió la investidura y su título para ejercer la profesión. El abogado recibió (de la Administración) la investidura y ... La Administración es la que entrega al abogado la investidura y el título
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answered Dec 3, 2022 at 10:29
DiegoDiego
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1 ¡Muchísimas gracias por sus explicaciones otra vez 🙏🙏🙏!Maurice –Maurice 2022-12-04 04:32:42 +00:00 Commented Dec 4, 2022 at 4:32
¡Gracias a ti! Lo importante es que poco a poco vayas avanzando en la compresión del idioma.Diego –Diego 2022-12-04 11:26:21 +00:00 Commented Dec 4, 2022 at 11:26
¡Gracias 🙇♂️ a la ayuda de Ustedes, mi nivel de castellano es mucho más mejor que antes!Maurice –Maurice 2022-12-05 23:20:54 +00:00 Commented Dec 5, 2022 at 23:20
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Física universitaria volumen 2
15.5 Resonancia en un circuito de ac
Física universitaria volumen 215.5 Resonancia en un circuito de ac
Índice Índice
Textos resaltados
Índice
Prefacio
Termodinámica
Electricidad y magnetismo
5 Cargas y campos eléctricos
6 Ley de Gauss
7 Potencial eléctrico
8 Capacitancia
9 Corriente y resistencia
10 Circuitos de corriente directa
11 Fuerzas y campos magnéticos
12 Fuentes de campos magnéticos
13 Inducción electromagnética
14 Inductancia
15 Circuitos de corriente alterna
Introducción
15.1 Fuentes de ac
15.2 Circuitos simples de ac
15.3 Circuitos en serie RLC con ac
15.4 Potencia en un circuito de ac
15.5 Resonancia en un circuito de ac
15.6 Transformadores
Revisión Del Capítulo
16 Ondas electromagnéticas
A Unidades
B Factores de conversión
C Constantes fundamentales
D Datos astronómicos
E Fórmulas matemáticas
F Química
G El alfabeto griego
Clave de respuestas
Índice
Buscar términos clave o texto.
Cerrar
Objetivos de aprendizaje
Al final de esta sección, podrá:
Determinar la frecuencia angular de resonancia ac máxima para un circuito RLC.
Explicar el ancho de la curva de potencia media versus la frecuencia angular y su significado utilizando términos como ancho de banda y factor de calidad.
En el circuito en serie RLC de la Figura 15.11, la amplitud de la corriente es, según la Ecuación 15.10:
I 0=V 0 R 2+(ω L−1/ω C)2−−−−−−−−−−−−−−−−√.I 0=V 0 R 2+(ω L−1/ω C)2.I 0=V 0 R 2+(ω L−1/ω C)2.
15.15
Si podemos variar la frecuencia del generador de ac manteniendo constante la amplitud de su voltaje de salida, la corriente cambia en consecuencia. Un gráfico de I 0 I 0 I 0 en función de ω ω ω se muestra en la Figura 15.17.
Figura 15.17 En la frecuencia de resonancia de un circuito RLC, ω 0=1/L C−−−−−√,ω 0=1/L C,ω 0=1/L C, la amplitud de la corriente está en su valor máximo.
En Oscilaciones encontramos un gráfico similar en el que la amplitud de un oscilador armónico amortiguado se trazaba frente a la frecuencia angular de una fuerza motriz sinusoidal (vea Oscilaciones forzadas). Esta similitud no es solo una coincidencia, como se ha demostrado anteriormente con la aplicación de la regla de las tensiones de Kirchhoff al circuito de la Figura 15.11. Esto da como resultado
L d i d t+i R+q C=V 0 sen ω t,L d i d t+i R+q C=V 0 sen ω t,L d i d t+i R+q C=V 0 sen ω t,
15.16
o
L d 2 q d t 2+R d q d t+1 C q=V 0 sen ω t,L d 2 q d t 2+R d q d t+1 C q=V 0 sen ω t,L d 2 q d t 2+R d q d t+1 C q=V 0 sen ω t,
donde hemos sustituido dq(t)/dt por i(t). Una comparación de la Ecuación 15.16 y de Oscilaciones y Oscilaciones amortiguadas para el movimiento armónico amortiguado demuestra claramente que el circuito en serie RLC controlado es el análogo eléctrico del oscilador armónico amortiguado controlado.
La frecuencia de resonancia f 0 f 0 f 0 del circuito RLC es la frecuencia a la que la amplitud de la corriente es máxima y el circuito oscilaría si no fuera alimentado por una fuente de voltaje. Por inspección, esto corresponde a la frecuencia angular ω 0=2 π f 0 ω 0=2 π f 0 ω 0=2 π f 0 en la que la impedancia Z en la Ecuación 15.15 es un mínimo, o cuando
ω 0 L=1 ω 0 C ω 0 L=1 ω 0 C ω 0 L=1 ω 0 C
y
ω 0=1 L C−−−−√.ω 0=1 L C.ω 0=1 L C.
15.17
Esta es la frecuencia angular de resonancia del circuito. Sustituyendo ω 0 ω 0 ω 0 en la Ecuación 15.9, la Ecuación 15.10 y la Ecuación 15.11, hallamos que en la resonancia,
ϕ=tan−1(0)=0,I 0=V 0/R,y Z=R.ϕ=tan−1(0)=0,I 0=V 0/R,y Z=R.ϕ=tan−1(0)=0,I 0=V 0/R,y Z=R.
Por lo tanto, en resonancia, un circuito RLC es puramente resistivo, con la emf y la corriente aplicadas en fase.
¿Qué ocurre con la potencia en resonancia? La Ecuación 15.14 nos dice cómo varía la potencia media transferida desde un generador de ac a la combinación RLC con la frecuencia. Además, P ave P ave P ave alcanza un máximo cuando Z, que depende de la frecuencia, es un mínimo, es decir, cuando X L=X C y Z=R.X L=X C y Z=R.X L=X C y Z=R. Así, en resonancia, la potencia media de salida de la fuente en un circuito en serie RLC es máxima. Usando la Ecuación 15.14, este máximo es V 2 rms/R.V rms 2/R.V rms 2/R.
La Figura 15.18 es un gráfico típico de P ave P ave P ave en función a ω ω ω en la región de máxima potencia. El ancho de banda Δ ω Δ ω Δ ω del pico de resonancia se define como el rango de frecuencias angulares ω ω ω sobre la cual la potencia media P ave P ave P ave es mayor que la mitad del valor máximo de P ave.P ave.P ave. La agudeza del pico se describe mediante una cantidad sin dimensiones conocida como el factor de calidad Q del circuito. Por definición,
Q=ω 0 Δ ω,Q=ω 0 Δ ω,Q=ω 0 Δ ω,
15.18
donde ω 0 ω 0 ω 0 es la frecuencia angular de resonancia. Un Q alto indica un pico de resonancia agudo. Podemos dar Q en términos de los parámetros del circuito como
Q=ω 0 L R.Q=ω 0 L R.Q=ω 0 L R.
15.19
Figura 15.18 Al igual que la corriente, la potencia media transferida desde un generador de ac a un circuito RLC alcanza su máximo en la frecuencia de resonancia.
Los circuitos resonantes se utilizan habitualmente para pasar o rechazar rangos de frecuencia seleccionados. Esto se hace ajustando el valor de uno de los elementos y, por tanto, "sintonizando" el circuito a una frecuencia de resonancia determinada. Por ejemplo, en las radios, el receptor se sintoniza con la emisora deseada ajustando la frecuencia de resonancia de sus circuitos para que coincida con la frecuencia de la emisora. Si el circuito de sintonía tiene un Q alto, tendrá un ancho de banda pequeño, por lo que las señales de otras emisoras a frecuencias incluso ligeramente diferentes de la frecuencia de resonancia encuentran una alta impedancia y no pasan por el circuito. Los teléfonos móviles funcionan de forma similar, comunicándose con señales de alrededor de 1 GHz que se sintonizan mediante un circuito inductor-condensador. Una de las aplicaciones más comunes de los condensadores es su uso en circuitos de sincronización de ac, basados en la obtención de una frecuencia de resonancia. Un detector de metales también utiliza un desplazamiento de la frecuencia de resonancia en la detección de metales (Figura 15.19).
Figura 15.19 Cuando un detector de metales se acerca a un trozo de metal, la autoinducción de una de sus bobinas cambia. Esto provoca un cambio en la frecuencia de resonancia de un circuito que contiene la bobina. Ese desplazamiento es detectado por los circuitos y transmitido al buceador por medio de los auriculares (créditos: modificación del trabajo de Eric Lippmann, de la Marina de EE. UU.).
Ejemplo 15.4
Resonancia en un circuito RLC en serie
(a) ¿Cuál es la frecuencia de resonancia de un circuito que utiliza los valores de voltaje y LRC conectados en serie del Ejemplo 15.1? (b) Si el generador de ac se ajusta a esta frecuencia sin cambiar la amplitud del voltaje de salida, ¿cuál es la amplitud de la corriente?
Estrategia
La frecuencia de resonancia de un circuito RLC se calcula a partir de la Ecuación 15.17, que proviene de un equilibrio entre las reactancias del condensador y del inductor. Como el circuito está en resonancia, la impedancia es igual al resistor. A continuación, la corriente de pico se calcula mediante la división del voltaje entre la resistencia.
Solución
La frecuencia de resonancia se calcula con la Ecuación 15.17: f 0=1 2 π 1 L C−−−√=1 2 π 1(3,00×10−3 H)(8,00×10−4 F)−−−−−−−−−−−−−−−−−−−√=1,03×10 2 Hz.f 0=1 2 π 1 L C=1 2 π 1(3,00×10−3 H)(8,00×10−4 F)=1,03×10 2 Hz.f 0=1 2 π 1 L C=1 2 π 1(3,00×10−3 H)(8,00×10−4 F)=1,03×10 2 Hz.
En resonancia, la impedancia del circuito es puramente resistiva, y la amplitud de la corriente es I 0=0,100 V 4,00 Ω=2,50×10−2 A.I 0=0,100 V 4,00 Ω=2,50×10−2 A.I 0=0,100 V 4,00 Ω=2,50×10−2 A.
Importancia
Si el circuito no estuviera ajustado a la frecuencia de resonancia, necesitaríamos la impedancia de todo el circuito para calcular la corriente.
Ejemplo 15.5
Transferencia de potencia en un circuito en serie RLC en resonancia
(a) ¿Cuál es la frecuencia angular de resonancia de un circuito RLC con R=0,200 Ω,R=0,200 Ω,R=0,200 Ω,L=4,00×10−3 H,L=4,00×10−3 H,L=4,00×10−3 H, y C=2,00×10−6 F?C=2,00×10−6 F?C=2,00×10−6 F? (b) Si una fuente de ac de amplitud constante 4,00 V se ajusta a esta frecuencia, ¿cuál es la potencia media transferida al circuito? (c) Determine Q y el ancho de banda de este circuito.
Estrategia
La frecuencia angular de resonancia se calcula a partir de la Ecuación 15.17. La potencia media se calcula a partir del voltaje rms y la resistencia del circuito. El factor de calidad se calcula a partir de la Ecuación 15.19 y conociendo la frecuencia de resonancia. El ancho de banda se calcula a partir de la Ecuación 15.18 y conociendo el factor de calidad.
Solución
La frecuencia angular de resonancia es ω 0=1 L C−−−√=1(4,00×10−3 H)(2,00×10−6 F)−−−−−−−−−−−−−−−−−−−√=1,12×10 4 rad/s.ω 0=1 L C=1(4,00×10−3 H)(2,00×10−6 F)=1,12×10 4 rad/s.ω 0=1 L C=1(4,00×10−3 H)(2,00×10−6 F)=1,12×10 4 rad/s.
A esta frecuencia, la potencia media transferida al circuito es máxima. Es P ave=V 2 rms R=[(1/2–√)(4,00 V)]2 0,200 Ω=40,0 W.P ave=V rms 2 R=[(1/2)(4,00 V)]2 0,200 Ω=40,0 W.P ave=V rms 2 R=[(1/2)(4,00 V)]2 0,200 Ω=40,0 W.
El factor de calidad del circuito es Q=ω 0 L R=(1,12×10 4 rad/s)(4,00×10−3 H)0,200 Ω=224.Q=ω 0 L R=(1,12×10 4 rad/s)(4,00×10−3 H)0,200 Ω=224.Q=ω 0 L R=(1,12×10 4 rad/s)(4,00×10−3 H)0,200 Ω=224.
Entonces calculamos para el ancho de banda
Δ ω=ω 0 Q=1,12×10 4 rad/s 224=50,0 rad/s.Δ ω=ω 0 Q=1,12×10 4 rad/s 224=50,0 rad/s.Δ ω=ω 0 Q=1,12×10 4 rad/s 224=50,0 rad/s.
Importancia
Si se desea un ancho de banda más estrecho, una resistencia más baja o una inductancia más alta ayudarían. Sin embargo, una resistencia más baja aumenta la potencia transferida al circuito, lo que puede no ser deseable, dependiendo de la potencia máxima que se pueda transferir.
Compruebe Lo Aprendido 15.6
En el circuito de la Figura 15.11, L=2,0×10−3 H,L=2,0×10−3 H,L=2,0×10−3 H,C=5,0×10−4 F,C=5,0×10−4 F,C=5,0×10−4 F, y R=40 Ω.R=40 Ω.R=40 Ω. (a) ¿Cuál es la frecuencia de resonancia? (b) ¿Cuál es la impedancia del circuito en la resonancia? (c) Si la amplitud del voltaje es de 10 V, ¿cuál es i (t) en la resonancia? (d) La frecuencia del generador de ac se cambia ahora a 200 Hz. Calcule el desfase entre la corriente y la emf del generador.
Compruebe Lo Aprendido 15.7
¿Qué ocurre con la frecuencia de resonancia de un circuito en serie RLC cuando las siguientes cantidades se incrementan en un factor de 4: (a) la capacitancia, (b) la autoinducción y (c) la resistencia?
Compruebe Lo Aprendido 15.8
La frecuencia angular de resonancia de un circuito en serie RLC es 4,0×10 2 rad/s.4,0×10 2 rad/s.4,0×10 2 rad/s. Una fuente de ac que funciona a esta frecuencia transfiere una potencia media de 2,0×10−2 W 2,0×10−2 W 2,0×10−2 W al circuito La resistencia del circuito es 0,50 Ω.0,50 Ω.0,50 Ω. Escriba una expresión para la emf de la fuente.
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Autores: William Moebs, Samuel J. Ling, Jeff Sanny
Editorial/sitio web: OpenStax
Título del libro: Física universitaria volumen 2
Fecha de publicación: 17 nov 2021
Ubicación: Houston, Texas
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URL de la sección:
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