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Compare growth rate of functions (exponential vs. polynomial)
I have to compare the growth rate of the following sequences
$a_n=a_{n-1}=10$
$b_n=\sum_{k=1}^n k^2$
$c_n =\frac{n^2}{10}$
$d_n=\left( \frac{3}{2} \right)^n$
I've rewritten $b_n$ and $d_n$ to
$$b_n = \frac{1}{6} (2n^3+3n^2+n) \quad \text{and} \quad d_n=1.5^n$$
and would say that $a_n = \mathcal{O}(1)$, $b_n = \mathcal{O}(n^3)$, $c_n = \mathcal{O}(n^2)$ and $d_n = \mathcal{O}(1.5^n$).
When sorting them, it should be
$$a_n < c_n < b_n < d_n$$
Here is the part, that I do not get. When graphing the functions or simply calculating, I get that $\mathcal{O}(1.5^n)<\mathcal{O}(n^2)$. Since $(k^n=)1.5^n$ is an exponential function with $k>1$, it should have a growth greater than the polynomial function $n^2$. I can't seem to figure out which of the 2 options is the right one.
Any help with explanation would be appreciated.
• You got it wrong for $b_n$ Sep 17, 2016 at 12:19
• @polfosol is it the explicit expression for the sum or the growth rate? Sep 17, 2016 at 12:21
• Just to clarify, in the description of $b_n$ is it $k=1$ as the lower bound of the summation rather than $k-1$? Sep 17, 2016 at 12:21
• @ParclyTaxel Yes, it should be $k=1$. Thanks. Corrected. Sep 17, 2016 at 12:22
• @Labbiqa: The explicit expression and consequently, the growth rate are both wrong Sep 17, 2016 at 12:29 | {
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That $\mathcal O(1.5^n)$ seems to grow slower than $\mathcal O(n^2)$ (or $\mathcal O(n^3)$) is an illusion. Exponential functions with exponents greater than 1 grow faster than all polynomial functions, but only eventually.
This is the key word: growth rates describe limiting behaviour, not behaviour at a particular point. In particular, the point where $1.5^n$ overtakes $n^2$ in growth was off your graph, and you missed it.
• Thanks for the clarification. I thought it was odd since no matter how much I zoomed out, it still didn't grow faster than $\mathcal{O}(n^2)$. Sep 17, 2016 at 14:06
• @Labbiqa, the graphs for $n^2$ and $1.5^n$ cross at about $n\approx 12.5$. Take a look at their graphs over an interval like $1 < n < 20$. Sep 18, 2016 at 13:39 | {
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# Linear algebra system of equations
Given the following linear system:
$x - y + 2z + 3w = 12$
$x - 2y + z + w = 8$
$4x + y + 2z - w = 3$
1) find all solutions of the system. Write the solution in Vector Form
2) find all solutions of the corresponding homogeneous system
I have turned this into a matrix and worked it out two ways resulting in different outcomes, neither of which is solvable...
I'll try to show my work, but I am new to this site and the formatting for matrices is fairly hard to figure out...
1 -1 2 3 12
1 -2 1 1 8
4 1 2 -1 3 (-4 row1)
1 -1 2 3 12
1 -2 1 1 8 (-row1)
0 5 -6 -13 -45
1 -1 2 3 12
0 -1 -1 -2 -4
0 5 -6 -13 -45(+5row2)
1 -1 2 3 12
0 -1 -1 -2 -4
0 0 -11 -23 -65
from here I don't think there's anything else I can do, I'm fairly sure there are infinitely many solutions.
• Let's see the work done so we can help. – Vincent Sep 8 '14 at 20:18
• Wow thanks, but of course. – Daniel Charry Sep 8 '14 at 20:19
• By this computation, it appears that there should be a non-empty solution set. If you show us what you did, we could try to figure out where you went wrong. – Omnomnomnom Sep 8 '14 at 20:24
• I edited the question with my work, any thoughts? I'm thinking that It just needs to be put in vector form? – donsavage Sep 8 '14 at 20:43
Oh of course there is more that can be done!
So starting from where you left, I would first multiple $R_2$ with the scalar $(-1)$ and then add the new $R_2$ to $R_1$ and replace it. Then I get,
$$\left({\begin{matrix} 1 & 0 & 3 & 5 & 16 \\ 0 & 1 & 1 & 2 & 4 \\ 0 & 0 & -11 & -23 & -65 \\ \end{matrix}}\right)$$
$$\underrightarrow{\left({\frac{-1}{11}}\right) R_3 \ \text{and} -1 R_3 + R_2 \ \text{and} -3R_3 + R_1 }$$
$$\left({\begin{matrix} 1 & 0 & 0 & \frac{-14}{11} & \frac{-19}{11}\\ 0 & 1 & 0 & \frac{-1}{11} & \frac{-21}{11} \\ 0 & 0 & 1 & \frac{23}{11} & \frac{65}{11} \\ \end{matrix}}\right)$$
Now translating back to equations you have, | {
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Now translating back to equations you have,
$$x = \frac{14}{11}w + \frac{19}{11}$$ $$y = \frac{1}{11}w + \frac{21}{11}$$ $$z = \frac{-23}{11}w + \frac{-65}{11}$$
Now any value you take for $w$ will solve the system. The method underlined here is made formal in Linear Algebra by Hoffman, Kunze which you should definitely read.
Now the solution set in vector form would be,
$$\left({\begin{matrix} x \\ y \\ z \\ w \\ \end{matrix}}\right) = w\left({\begin{matrix} \frac{14}{11} \\ \frac{1}{11} \\ \frac{-23}{11} \\ 1 \\ \end{matrix}}\right) + \left({\begin{matrix} \frac{19}{11} \\ \frac{21}{11} \\ \frac{-65}{11} \\ 0 \\ \end{matrix}}\right)$$
where $w$ is any scalar in $\Bbb R$.
• Could you explain why when converting it back into the equations the sign of the last two rows change? Shouldn't the last column keep it's sign? for example: z = -23/11w + 65/11 – donsavage Sep 10 '14 at 0:20
• Actually your partly right. The equation should be $z = \frac{-23}{11}w - \frac{65}{11}$ since the last row in the matrix $0 \; 0 \; 1 \; \frac{23}{11} \; \frac{-65}{11}$ represents the equation $z + \frac{23}{11}w = \frac{-65}{11}$. So lslight alteration needed. – Ishfaaq Sep 10 '14 at 1:35 | {
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# Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the same?
1. Dec 29, 2012
### tahayassen
This is a discussion we had in another part of the forum and I'm wondering who is correct. The discussion is becoming increasingly confusing and annoyingly (regardless of the posts in between the discussion), no one with a "Science Advisor", "Homework Helper", or "PF Mentor" title is stepping in to end the argument, so I would appreciate it if you would end the argument.
2. Dec 29, 2012
### Number Nine
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
The cardinalities of the intervals [0 1] and [0 2] are the same. In fact, there are as many numbers in the interval [0 1] as there are real numbers.
This is a pretty poor explanation of what's happening. Taking limits in calculus is a process; we examine what happens as a quantity increases without bound. The limit of that function as x goes to infinity is zero because x! increases faster as x increases, not because it is a "larger quantity". "Infinity" in this context has a completely different meaning than the "infinity" used to describe the size of sets.
3. Dec 29, 2012
### rbj
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
there is an infinite, but countably infinite amount of rational numbers between 0 and 1. this is also the case for the interval from 0 to 2.
but there is an uncountably infinite amount of real numbers between 0 and 1. this is also the case for the interval from 0 to 2.
4. Dec 29, 2012
### micromass
Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Jack212222 is wrong. Cantor's theory is nothing like what he describes. He seems to be confusing things with limits.
And yes, the cardinality of [0,1] is exactly the same as the cardinality of [0,2]. So both sets have the same size. | {
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EricVT is wrong too, since he assumes that $\frac{+\infty}{+\infty}$ is defined when it is not. So the ratio doesn't even make sense.
5. Dec 29, 2012
### micromass
Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Yes, this is true. But the example of [0,1] and [0,2] is not a good example since both infinities are the same here.
However, we can look at the sets $\mathbb{N}$ and $\mathbb{R}$. Those are infinite sets, but the latter set is much larger than the former.
See the following FAQ post: https://www.physicsforums.com/showthread.php?t=507003 (also check out the sequels whose link is at the bottom of the thread).
6. Dec 29, 2012
### HallsofIvy
Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
You also have to be careful how you define "larger". Micromass is completely correct about cardinality and that will be appropriate if 'larger' means "has more numbers in the set" as was originally asked. But one can also argue that the interval [0, 2] is twice as long, and so twice as large, as the interval [0, 1]. It depends upon what you are comparing.
7. Dec 29, 2012
### tahayassen
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Thanks everyone. The FAQ was very helpful. I also found this helpful:
Last edited by a moderator: Sep 25, 2014
8. Dec 31, 2012
### alan6459
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Two sets have the same cardinaility if there exists a 1-1 correspondence between them.
The function f(x) = 2x establishes such a 1-1 correspondence between the rationals in
[0,1] and [0,2]. It also establishes a 1-1 correspondence between the reals in [0,1] and [0,2].
9. Dec 31, 2012
### 1MileCrash
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s | {
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Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Some infinities are larger than others. Namely, uncountable infinities are larger than countable infinities and that is all there is to it.
The number of real numbers between 0 and 1 is uncountably infinite. The number of real numbers between 0 and 2 is uncountably infinite. The number of real numbers period is uncountably infinite. These all describe the same cardinality.
The natural numbers are countably infinite, the real numbers are uncountably infinite. Therefore the cardinality of natural numbers is smaller than that of the real numbers.
There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.
10. Dec 31, 2012
### micromass
Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Sure there is. There are many types of uncountably infinite sets. For example, the set $\mathcal{P}(\mathbb{R})$ (power set of the reals) has a strictly larger cardinality than $\mathbb{R}$. and $\mathcal{P}(\mathcal{P}(\mathbb{R}))$ is even larger!! This process continues indefinitely.
11. Dec 31, 2012
### tahayassen
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Ah, thanks for pointing that out.
Edit: I take my thank you back.
Last edited: Dec 31, 2012
12. Jan 2, 2013
### Tac-Tics
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
13. Jan 2, 2013
### dextercioby
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Just as a curiosity, what's the connection between the discrete/countable infinite sets (sequences) $\mathcal{P}^{n}(\mathbb{R})$ and $\aleph_{n}$ ?
Last edited: Jan 2, 2013
14. Jan 2, 2013
### 1MileCrash | {
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Last edited: Jan 2, 2013
14. Jan 2, 2013
### 1MileCrash
Ah.. I failed to think of deriving a set from an uncountable set in such a way that the cardinality must be greater. It doesn't make much sense at all to say that a set and it's power set have the same cardinality. That is easy to see.
Thank you.
Is this limited to the idea of power sets? It feels like this could only occur when you "build" a "higher order" uncountable set from a previously uncountable set.
Also- the power set of the naturals would certainly be of higher cardinality than the naturals and thus not at a one-to-one correspondence with the naturals and therefore uncountable, right?
Does that make P(N) have the same cardinality as R, and P(P(N)) have the same cardinality as P(R)?
Or am I oversimplifying the idea?
Last edited: Jan 2, 2013
15. Jan 2, 2013
### dextercioby
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.
16. Jan 2, 2013
### 1MileCrash
I don't understand how that could be the case.
17. Jan 2, 2013
### dextercioby
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Who's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.
18. Jan 2, 2013
### 1MileCrash
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Forgive me if I'm not understanding but,
The number of elements of P(N) is the number of possible subsets of natural numbers that can be formed. | {
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That is surely uncountable, as I can combine any number of any natural numbers I want to form some subset. If I call one subset, the non-proper subset, the entire set of natural numbers, one subset, what's another subset? The subset of 1 million of any natural numbers? 2 million any natural numbers? Considering all singleton sets of each natural number is already a countably infinite number of subsets, and there are uncountably many subsets besides those, it doesn't work in my brain to call P(N) a set of countable cardinality.
EDIT:
I found this
http://www.earlham.edu/~peters/writing/infapp.htm#thm3
Which I guess is a much better and more formal version of what I'm thinking.
Last edited: Jan 2, 2013
19. Jan 2, 2013
### Number Nine
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
You're right, P(N) is uncountable. An easy way to see this is to identify each element of P(A) with an infinite binary string (easy to show that this is a bijection) and note that the set of all such strings is uncountable.
20. Jan 2, 2013
### 1MileCrash
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s
Okay, I'm not sure how to construct these infinite binary strings but I do understand why P(N) is of uncountable cardinality. I have a few more questions.
I have been reading about "beth numbers" and this seems to be the name for the "order of infinity."
So here are my questions:
Is the beth number the sole indicator of cardinality of uncountable sets? If so, this would imply that P(N) and R have the exact same cardinality, right?
Is the power set concept the only source of constructing a "higher order" infinity?
Are there any sets that we naturally consider that aren't derived from taking the power set of some other set with a beth number greater than two? | {
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I do realize that this question is pretty much identical to my previous one, but I guess what I'm asking is if there is a set with a beth number greater than or equal to 2 that we could describe in some other way than a power set of some other set?
I'm sorry for kind of derailing the topic but I find this really cool, as a grad, what focuses on this type of thing? Is it just set theory? | {
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# irrational numbers?!
what are rational and irrational numbers??
i know that irrational numbers are numbers that cannot be expressed as a fraction, but i'm still confused.
there is this question:
which of the following are irrational numbers: √2, √8, 22/7, pi, 2√3
i know that 22/7 is rational, pi is irrational, but what about those square roots?
1. Square roots of numbers that are not the squares of integers are all irrational. There is a way to prove that but I forgot the details. You have to assume that a fraction works and then prove that the assumption leads to a contradiction
Any number that does not meet the definition of being rational is irrational.
posted by drwls
2. - √2 is either rational or irrational.
- Assume that √ is rational so that
√2 = a/b, with a/b in lowest terms
- Square both sides to get
2 = a^2/b^2
then a^2 = 2b^2
- the right side of this equation is clearly an even number, since anything multiplied by 2 is even
- so a^2 must be even. We also know that if we square an odd number the result is odd, and if we square an even number the result is even
so a must be even
so a could be written as 2k
- rewriting our equation as
2b^2= (2k)^2
2b^2 = 4k^2
b^2 = 2k^2
by the same argument as above 2k^2 is even , so b has to be even
which means a and b are both even, therefore a/b is not in lowest terms
BUT that contradicts my assumption that a/b was a fraction in lowest terms
so √2 = a/b is a false statement
therefore √2 cannot be rational, and
must then be irrational
the same argument could be used for √3 and all other square roots
posted by Reiny
3. true or false does this number represent a rational number 0.20200200020000200000
posted by brandy
4. Is √4 Irrational?
posted by Amber
5. is this seriouly the answer for the LEAP test *feecepum*
you're not going to get ALL 50 answers -_-
posted by Cereal.... is life. -Life Cereal
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28. Use the Distributive Property to simplify x(4x^2 + x + 4) Is it 4x^3 + x^2 + 4x? 42. To which set of numbers does 0 not belong? 46. Pi belongs to which group of real numbers? a. Odd numbers b. rational numbers c. irrational
10. ### Algerba II
Hi Do decimals such as 2.718 represent rational numbers or irrational numbers. Explain. Do repeating decimals such as 2.3333 . . . represent rational numbers or irrational numbers? Explain. So I look back in my textbook and find
More Similar Questions | {
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# matrix multiplication associative proof | {
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Property 1: Associative Property of Multiplication A(BC) = (AB)C where A,B, and C are matrices of scalar values. Proof We will concentrate on 2 × 2 matrices. 2. A professor I had for a first-year graduate course gave us an example of why caution might be required. How do you multiply two matrices? Hence, associative law of sets for intersection has been proved. Matrix addition and scalar multiplication satisfy commutative, associative, and distributive laws. Matrix multiplication is indeed associative and thus the order irrelevant. Theorem 2 Matrix multiplication is associative. 14 minutes ago #3 TheMercury79. (This can be proved directly--which is a little tricky--or one can note that since matrices represent linear transformations, and linear transformations are functions, and multiplying two matrices is the same as composing the corresponding two functions, and function composition is always associative, then matrix multiplication must also be associative.) Please Write The Proof Step By … It turned out they are the same. ible n×n matrices with entries in F under matrix multiplication. As examples of multiplication modulo 6: 4 * 5 = 2 2 * 3 = 0 3 * 9 = 3 The answer … Answer Save. 3 Answers. The Associative Property of Multiplication of Matrices states: Let A , B and C be n × n matrices. However, this proof can be extended to matrices of any size. Theorem 7 If A and B are n×n matrices such that BA = I n (the identity matrix), then B and A are invertible, and B = A−1. Since Theorem MMA says matrix multipication is associative, it means we do not have to be careful about the order in which we perform matrix multiplication, nor how we parenthesize an expression with just several matrices multiplied togther. Then (AB)Ce j = (AB)c j … Likes TheMercury79. Then the following properties hold: a) A(BC) = (AB)C (associativity of matrix multipliction) b) (A+B)C= AC+BC (the right distributive property) c) C(A+B) = CA+CB (the left distributive property) Proof: We will prove | {
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right distributive property) c) C(A+B) = CA+CB (the left distributive property) Proof: We will prove part (a). Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. But for other arithmetic operations, subtraction and division, this law is not applied, because there could be a change in result.This is due to change in position of integers during addition and multiplication, do not change the sign of the integers. $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}$$ Lecture 2: Fun with matrix multiplication, System of linear equations. Lv 4. In standard truth-functional propositional logic, association, or associativity are two valid rules of replacement. This preview shows page 33 - 36 out of 79 pages. What are some of the laws of matrix multiplication? SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. A+(B +C) = (A+B)+C (Matrix addition is associative.) Question: Prove The Associative Law For Matrix Multiplication: (AB)C = A(BC). That is if C,B and A are matrices with the correct dimensions, then (CB)A = C(BA). Clearly, any Kronecker product that involves a zero matrix (i.e., a matrix whose entries are all zeros) gives a zero matrix as a result: Associativity. but composition is associative for all maps, linear or not. Learning Objectives. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices For any matrix A, ( AT)T = A. 2 The point is you only need to show associativity for multiplication by vectors, i.e. Matrix-Matrix Multiplication is Associative Let A, B, and C be matrices of conforming dimensions. ... the same computational complexity as matrix multiplication. The Organic Chemistry Tutor 1,739,892 views B. Except for the lack of commutativity, | {
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multiplication. The Organic Chemistry Tutor 1,739,892 views B. Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. Prove the associative law of multiplication for 2x2 matrices.? Special types of matrices include square matrices, diagonal matrices, upper and lower triangular matrices, identity matrices, and zero matrices. for matrices M,N and vectors v, that (M.N).v = M.(N.v). Even if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. The first is that if $$r= (r_1,\ldots, r_n)$$ is a 1 n row vector and $$c = \begin{pmatrix} c_1 \\ \vdots \\ c_n \end{pmatrix}$$ is a n 1 column vector, we define $rc = r_1c_1 + \cdots + r_n c_n. The associative property holds: Proof. c i j = ∑ 1 ≤ k ≤ m a i k b k … 1 decade ago. In general, if A is an m n matrix (meaning it has m rows and n columns), the matrix product AB will exist if and only if the matrix B has n rows. Cool Dude. I just ended up with different expressions on the transposes. Example 1: Verify the associative property of matrix multiplication for the following matrices. On the RHS we have: and On the LHS we have: and Hence the associative … In other words, unlike the integers, matrices are noncommutative. Subsection DROEM Determinants, Row Operations, Elementary Matrices. It is easy to see that GL n(F) is, in fact, a group: matrix multiplication is associative; the identity element is I n, the n×n matrix with 1’s along the main diagonal and 0’s everywhere else; and the matrices are invertible by choice. That is, a double transpose of a matrix is equal to the original matrix. 3. r(A+B) = rA+rB (Scalar multiplication distributes over matrix addition.) Then, ( A B ) C = A ( B C ) . Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. In Maths, associative law is applicable to only two of the four major arithmetic operations, which are addition and multiplication. Proof Let be a matrix. So | {
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four major arithmetic operations, which are addition and multiplication. Proof Let be a matrix. So the ij entry of AB is: ai1 b1j + ai2 b2j. Theorem 2 matrix multiplication is associative proof. Properties of Matrix Multiplication: Theorem 1.2Let A, B, and C be matrices of appropriate sizes. Pages 79. I am working with Paul Halmos's Linear Algebra Problem Book and the seventh problem asks you to show that multiplication modulo 6 is commutative and associative. Second Law: Second law states that the union of a set to the union of two other sets is the same. Favorite Answer. A+B = B +A (Matrix addition is commutative.) Properties of Matrix Arithmetic Let A, B, and C be m×n matrices and r,s ∈ R. 1. School Georgia Institute Of Technology; Course Title MATH S121; Uploaded By at1029. The -th ... , by applying the definition of Kronecker product and that of multiplication of a matrix by a scalar, we obtain Zero matrices. Then (AB)C=A(BC). Associative law: (AB) C = A (BC) 4. Proof: Suppose that BA = I … If B is an n p matrix, AB will be an m p matrix. As a final preparation for our two most important theorems about determinants, we prove a handful of facts about the interplay of row operations and matrix multiplication with elementary matrices with regard to the determinant. A matrix is full-rank iff its determinant is non-0; Full-rank square matrix is invertible; AB = I implies BA = I; Full-rank square matrix in RREF is the identity matrix; Elementary row operation is matrix pre-multiplication; Matrix multiplication is associative; Determinant of upper triangular matrix e.g (3/2)*sqrt(1/2) … Parts (b) and (c) are left as homework exercises. Because matrices represent linear functions, and matrix multiplication represents function composition, one can immediately conclude that matrix multiplication is associative. Matrix multiplication is Associative Let A be a m\times n matrix, B a n\times p matrix, and C a p\times q matrix. So this is where we draw the line on explaining | {
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B a n\times p matrix, and C a p\times q matrix. So this is where we draw the line on explaining every last detail in a proof. 2.2 Matrix multiplication. What are some interesting matrices which lead to special products? Then (AB)C = A(BC): Proof Let e j equal the jth unit basis vector. Relevance. We are going to build up the definition of matrix multiplication in several steps. Proposition (associative property) Multiplication of a matrix by a scalar is associative, that is, for any matrix and any scalars and . Propositional logic Rule of replacement. Square matrices form a (semi)ring; Full-rank square matrix is invertible; Row equivalence matrix; Inverse of a matrix; Bounding matrix quadratic form using eigenvalues; Inverse of product; AB = I implies BA = I; Determinant of product is product of determinants; Equations with row equivalent matrices have the same solution set; Info: Depth: 3 16 5. fresh_42 said: Then you have made a mistake somewhere. Matrix arithmetic has some of the same properties as real number arithmetic. What is a symmetric matrix? Let the entries of the matrices be denoted by a11, a12, a21, a22 for A, etc. it then follows that (MN)P = M(NP) for all matrices M,N,P. Solution: Here we need to calculate both R.H.S (right-hand-side) and L.H.S (left-hand-side) of A (BC) = (AB) C using (associative) property. Matrix multiplication Matrix inverse Kernel and image Radboud University Nijmegen Matrix multiplication Solution: generalise from A v A vector is a matrix with one column: The number in the i-th rowand the rst columnof Av is the dot product of the i-th row of A with the rst column of v. So for matrices A;B: Proof: (1) Let D = AB, G = BC Informal Proof of the Associative Law of Matrix Multiplication 1. Thanks. Therefore, the associative property of matrices is simply a specific case of the associative property of function composition. Please Write The Proof Step By Step And Clearly. What is the inverse of a matrix? Since matrix multiplication | {
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The Proof Step By Step And Clearly. What is the inverse of a matrix? Since matrix multiplication obeys M(av+bw) = aMv + bMw, it is a linear map. Then, (i) The product A B exists if and only if m = p. (ii) Assume m = p, and define coefficients. Corollary 6 Matrix multiplication is associative. The argument in the proof is shorter, clearer, and says why this property "really" holds. The multiplication of two matrices is defined as follows: Definition 1.4.1 (Matrix multiplication). (4 ways) What is the transpose of a matrix?$ This might remind you of the dot product if you have seen that before. Let A = (a i j) ∈ M n × m (ℝ) and B = (b i j) ∈ M p × q (ℝ), for positive integers n, m, p, q. It’s associative straightforwardly for finite matrices, and for infinite matrices provided one is careful about the definition. Matrix multiplication is indeed associative and thus the order irrelevant. Proof. Let be , be and be . That is, if we have 3 2x2 matrices A, B, and C, show that (AB)C=A(BC). ( BC ) 4 + ai2 b2j A set to the union of matrix. Propositional logic, association, or associativity are two valid rules of replacement is indeed associative and the... C = A r, s ∈ R. 1 2: Fun with matrix multiplication System... C be n × n matrices. be required, ( A B ) C AC..., A double transpose of A matrix is equal to the original matrix of the.! Hence, associative law of matrix arithmetic Let A, etc = AC + BC 5: A B! And zero matrices. matrices and r, s ∈ R. 1 what... Of AB is: ai1 b1j + ai2 b2j 5. fresh_42 said: then you have A... With different expressions on the transposes you have made A mistake somewhere unlike the integers, matrices are.! ( 4 ways ) what is the same properties as real number arithmetic integers, matrices are noncommutative Fun matrix.: 2:28:48 matrix arithmetic has some of the laws of matrix multiplication: Theorem 1.2Let,. Math S121 ; Uploaded By at1029 C be matrices of any size subsection DROEM Determinants, Row,. Online Crash Course Algebra & | {
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at1029 C be matrices of any size subsection DROEM Determinants, Row,. Online Crash Course Algebra & Geometry Study Guide Review, Functions, Youtube -:. 33 - 36 out of 79 pages, a12, a21, a22 for A first-year graduate Course us! Follows: definition 1.4.1 ( matrix addition is commutative. B +A ( matrix addition commutative! Need to show associativity for multiplication By vectors, i.e build up the definition of matrix 1! Order irrelevant 2 matrix multiplication: Theorem 1.2Let A, B, and says why property... Concentrate on 2 × matrix multiplication associative proof matrices. function composition C j … Theorem 2 multiplication... ( A+B ) +C ( matrix multiplication jth unit basis vector is indeed associative and thus the irrelevant... ∈ R. 1 T = A ( B ) C j … Theorem matrix... You have seen that before, etc +A ( matrix addition is commutative. B and. Associative, and distributive laws ) C = AC + BC 5 and distributive laws BC ) \$ matrix! Under matrix multiplication By a11, a12, a21, a22 for A graduate... Explaining every last detail in A proof A B ) C = A ( BC ): proof Let j. Is the transpose of A matrix is equal to the union of two other sets is the same vectors! +A ( matrix addition is associative Let A, etc, matrices are noncommutative and vectors v that. Shorter, clearer, and C be n × n matrices. and thus the order irrelevant 4. Of why caution might be required matrices are noncommutative dot product if have... Point is you only need to show associativity for multiplication By vectors, i.e detail in A.! As real number arithmetic, unlike the integers, matrices are noncommutative words, unlike the integers, matrices noncommutative! As follows: definition 1.4.1 ( matrix addition is commutative. associativity are two valid rules replacement! Associative. 4 ways ) what is the transpose of A matrix is equal the. Are noncommutative multiplication in several steps for A, B, and says why property... × 2 matrices. fresh_42 said: then you have seen that before the order | {
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A, B, and says why property... × 2 matrices. fresh_42 said: then you have seen that before the order irrelevant & Geometry Study Review... Hence, associative, and C be matrices of any size P = M ( NP ) for all M! This property really '' holds is defined as follows: definition 1.4.1 ( matrix addition and Scalar multiplication commutative! S ∈ R. 1 MATH Test Prep Online Crash Course Algebra & Geometry Guide... Special types of matrices is defined as follows: definition 1.4.1 ( addition. ) Ce j = ( A+B ) +C ( matrix addition. or not an P... N×N matrices with entries in F under matrix multiplication is associative Let A, AT... At ) T = A ( B +C ) = ( A+B ) +C ( matrix multiplication for following. = M. ( N.v ) property of multiplication for the following matrices matrix multiplication associative proof b1j + ai2.. Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions, -... ( AB ) C = A ( B C ) are left homework... Matrices is simply A specific case of the dot product if you have seen that before on. That before why caution might be required + C ) are left matrix multiplication associative proof homework exercises e j equal jth..., clearer, and C be matrices of any size +A ( matrix multiplication (... S121 ; Uploaded By at1029 AB is: ai1 b1j + ai2 b2j, and C be matrices of size! All matrices M, n and vectors v, that ( M.N ) =. And r, s ∈ R. 1 ( A B ) C …. Is the transpose of A matrix is equal to the union of A matrix is to! Multiplication 1: matrix multiplication associative proof with matrix multiplication ) B is an n P matrix AB. P = M ( NP ) for all matrices M, n and vectors v that. Other words, unlike the integers, matrices are noncommutative or associativity are two rules... A double transpose of A matrix is equal to the original matrix Write the Step. As homework exercises been proved of any size Title MATH S121 ; Uploaded By at1029: the. Two valid rules of replacement seen that before over matrix addition and Scalar multiplication | {
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the. Two valid rules of replacement seen that before over matrix addition and Scalar multiplication satisfy commutative associative. Multiplication for the following matrices., that ( M.N ).v = M. ( N.v ),,! And vectors v, that ( MN ) P = M ( NP ) for all matrices,! For 2x2 matrices. Tutor 1,739,892 views 2.2 matrix multiplication interesting matrices which lead special! Multiplication is indeed associative and thus the order irrelevant proof Step By Step and Clearly n and v... A professor I had for A first-year graduate Course gave us an example why! Law for matrix multiplication is indeed associative and thus the order irrelevant this preview shows page -. Sat MATH Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions, Youtube -:..., s ∈ R. 1: Theorem 1.2Let A, etc associative law of sets intersection. The associative law: second law: ( AB ) C=A ( BC:! So the ij entry of AB is: ai1 b1j + ai2 b2j are noncommutative ) (! 2X2 matrices. build up the definition of matrix multiplication is associative for all maps, linear not! Ways ) what is the same properties as real number arithmetic 33 - 36 out 79. The definition of matrix arithmetic Let A, ( AT ) T A... Then ( AB ) C = AC + BC 5 Technology ; Course Title MATH S121 Uploaded!, System of linear equations following matrices. for matrices M, n and vectors,... +A ( matrix multiplication is associative Let A, etc is commutative. ( AB ) C=A ( BC 4! Organic Chemistry Tutor 1,739,892 views 2.2 matrix multiplication Organic Chemistry Tutor 1,739,892 views 2.2 matrix )! Number arithmetic j = ( AB ) C = A ( BC ) 4 ( NP ) for all,! Be matrices of any size: 2:28:48 A+B = B +A ( matrix multiplication is associative... Build up the definition of matrix multiplication is associative proof two other sets is the properties. Are going to build up the definition of matrix multiplication, System of linear equations C j … 2. Is commutative. AT ) T = A ( B ) C A... On explaining every last detail in A proof | {
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C j … 2. Is commutative. AT ) T = A ( B ) C A... On explaining every last detail in A proof distributive law: ( AB ) C = AC + BC.. Associativity for multiplication By vectors, i.e and thus the order irrelevant we draw the on... ( AT ) T = A ( BC ): proof Let e j equal the jth unit basis.... And distributive laws, a21, a22 for A first-year graduate Course gave us an example of why might! Will concentrate on 2 × 2 matrices. just ended up with different expressions on the transposes distributes over addition... Guide Review, Functions, Youtube - Duration: 2:28:48 Let the entries of the associative of! ( AB ) C = A ( BC ) 4, upper and lower triangular matrices, identity,... Are going to build up the definition of matrix multiplication for 2x2 matrices. AT ) T = A BC., unlike the integers, matrices are noncommutative this is where we draw the line on explaining every last in... System of linear equations different expressions on the transposes in the proof shorter! To build up the definition of matrix multiplication professor I had for A,,! The union of two matrices is simply A specific case of the laws matrix. Therefore, the associative property of matrix arithmetic has some of the dot if. For matrix multiplication: Theorem 1.2Let A, B, and C be matrices of size. ( B ) C j … Theorem 2 matrix multiplication is associative. interesting... A+B = B +A ( matrix addition and Scalar multiplication distributes over matrix addition is commutative. sets intersection. Said: then you have made A mistake somewhere mistake somewhere Course Algebra & Study! 2 Hence, associative law: A ( BC ): proof Let j. Proof Step By Step and Clearly for A, B, and zero matrices?. Ab will be an M P matrix × 2 matrices. diagonal matrices, identity matrices identity! 1.4.1 ( matrix addition is commutative., Elementary matrices. standard truth-functional propositional logic, association or... | {
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# Selection without replacement
This selection without replacement question comes to me about third hand, so I'm not sure of the exact level/setting. It's definitely not my homework problem. I have done a bit of searching on the site but didn't find anything completely analogous. The question is as follows:
We have a bag with colored marbles, and the marbles also have numbers on them.
For this question all we need to know are the following:
There are 40 marbles in total.
There are 14 marbles with numbers less than or equal to three.
There are 7 purple marbles.
There is one purple marble with a number less than or equal to three.
Pulling two marbles without replacement, what is the probability the first marble has a number less than or equal to 3 and the second marble is purple?
My solution is combinatorial simply computing the probability as the number of favorable outcomes divided by the number of possible outcomes.
The denominator is easy. There are $40$ possibilities for the first draw and $39$ for the second, so $40\cdot 39=1560$ possible outcomes.
To compute the number of favorable outcomes there is only way to select a purple marble as the first marble, so there are $1\cdot 6$ favorable outcomes where the first marble is purple. That leaves $13$ ways to select the first marble if it isn't purple times the $7$ purple marbles that can then be selected on the second draw for $13\cdot 7 =91$ favorable outcomes. Taken together
$$\frac{6+91}{1560}=\frac{97}{1560}$$
for the probability the first marble has a number less than or equal to 3 and the second marble is purple.
I was wondering is my solution correct?
I would also welcome suggestions for other perhaps more probabilistic approaches to solving the problem. | {
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• I think you are correct. You are splitting into two cases (1) first marble is purple (i.e. the "special" marble that meets both criteria) and (2) first marble is not purple. This is equivalent to using conditional probabilities but for such a simple case there is really no need to use that language. another approach is simply take $14 \times 7 = 98$ and subtract $1$ for the double-counted case of picking the special marble twice, but for a simple case like this that might take longer to visualize/explain. – antkam Apr 20 '18 at 19:53
You could do it more formally (same logic thought) in term of probability like this:
First, split the probability space onto two disjoint spaces: $\\$ $P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple}) \\= P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ purple}) + P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ not \ purple})$
Apply Bayes formula:
$P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ purple}) \\ = P( M_2 \ \mathrm{ purple} | M_1\leq3 \ \cap \ M_1 \ \mathrm{ purple}) \ P(M_1\leq3 \ \cap \ M_1 \ \mathrm{ purple}) \\ = \frac{6}{39} \frac{1}{40}$
Apply Bayes formula again:
$P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{not \ purple}) \\ = P( M_2 \ \mathrm{ purple} | M_1\leq3 \ \cap \ M_1 \ \mathrm{ not \ purple}) P(M_1\leq3 \ \cap \ M_1 \ \mathrm{not \ purple}) \\ = \frac{7}{39}\frac{13}{40}$
Finally , $P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple}) = \frac{6}{39} \frac{1}{40} + \frac{7}{39}\frac{13}{40} = \frac{97}{1560}$
Same result as yours (after multiple edits haha)
@antkam comment is great thought if you want to solve it without formula. | {
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1. ## [SOLVED] Projectiles....
A stone is projected from a point O on a cliff with a speed of 20 m/s at an angle of elevation of 30(degrees) . T seconds later the angle of depression of the stone from O is 45(degrees). Find the value of T .
i encountered this problem while solving M2 past exam quetions and i couldnt solve it... ... to make things worse no one in my class knows how to solve this problem . Pls hlp me cuz xam is in a few days. i will be very grateful...
2. I am including a diagram. I hope I am interpreting correctly.
We can use the horizontal and vertical component formulas.
horizontal: $x=(20cos(30))t$
vertical: $y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$
When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.
$0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$
Solving for t we find t=2.04 seconds to reach that point.
That means it's x coordinate is $20cos(30)(2.04)=35.35 \;\ m$
The stone is at (35.35,0) at 2.04 seconds.
Now, we have a triangle with angle 45 and side 35.35
So, the y component is the same, y=-35.35 feet
Plug this into the y component and solve for t.
$-35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$
And we find t=3.89 seconds
Perhaps TopsQuark will check me out.
3. Hello, PFX!
A stone is projected from a point $O$ on a cliff with a speed of 20 m/s
at an angle of elevation of 30°.
t seconds later the angle of depression of the stone from $O$ is 45°.
Find the value of $t$.
The parametric equations for the position of the stone (relative to $O$) are:
. . $\begin{array}{ccccccc}x &=& (20\cos30^o)t & \Rightarrow & x &=& 10\sqrt{3} t \\
y &=& (20\sin30^o)t - 4.9t^2 & \Rightarrow & y &=& 10t - 4.9t^2 \end{array}$
Consider the position of the stone at time $t.$
Code:
*
* *
O* - - - +A
\ *:
\ :
\ :
*B
Since $\angle AOB = 45^o,\:\Delta OAB$ is an isosceles right triangle.
. . Hence: . $OA = AB$ | {
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Since $\angle AOB = 45^o,\:\Delta OAB$ is an isosceles right triangle.
. . Hence: . $OA = AB$
This means: . $x \,=\,-y\quad\Rightarrow\quad (10\sqrt{3})t \;=\;-\left[10t - 4.9t^2\right]$
. . which simplfies to: . $4.9t^2 - 10(1+\sqrt{3})t \:=\:0$
. . which factors: . $t\left[4.9t - 10(1+\sqrt{3})\right] \:=\:0$
. . and has roots: . $t = 0,\;\frac{10(1+\sqrt{3})}{4.9}$
Therefore: . $t \:\approx\:5.6$ seconds.
4. My thinking was on line with yours, Soroban. Why is my answer different?.
I don't see anything wring with my logic. The range of the stone is $\frac{400sin(60)}{9.8}=35.33$.
5. hi, thanx for ur replies....
well the correct answer given is T = 5.59 (to 3 s.f.)
@ galactus i cant find where u went wrong....
6. Originally Posted by galactus
I am including a diagram. I hope I am interpreting correctly.
We can use the horizontal and vertical component formulas.
horizontal: $x=(20cos(30))t$
vertical: $y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$
When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.
$0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$
Solving for t we find t=2.04 seconds to reach that point.
That means it's x coordinate is $20cos(30)(2.04)=35.35 \;\ m$
The stone is at (35.35,0) at 2.04 seconds.
Now, we have a triangle with angle 45 and side 35.35
So, the y component is the same, y=-35.35 feet
Plug this into the y component and solve for t.
$-35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$
And we find t=3.89 seconds
Perhaps TopsQuark will check me out.
Take a look at your diagram. When y = 0 you have x at 35.35 ft. How can it be that when it drops to -45 degrees? x is always increasing.
-Dan | {
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When do Taylor series converge quickly?
The function $e^x$ has a Maclaurin series $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$
It also has various Taylor series expansions centered at $x$-values other than zero, for example $x=3$: $$e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$$
If we approximate $e^{1/2}$ using the first series above, our approximation converges to the exact value very quickly (|error| < 0.001 after 4 terms). If we use the second series, it converges more slowly (|error| < 0.001 after 13 terms).
I suspect this happens because $x=\frac{1}{2}$ is closer to zero than it is to 3. Do Taylor series always converge more quickly near the center of convergence? If not, is there a general result that tells us when this happens?
• To approximate $e^{\frac{1}{2}}$ using $e^3$ looks just weird – Vincent Apr 18 '17 at 15:06
• Yeah, maybe not the most practical example. I'm more interested in the general principle. – nardol5 Apr 18 '17 at 15:07
• The Taylor series have a bound math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/… which, as you can see, converges more quickly when you are close to the center (strictly less than 1). – Vincent Apr 18 '17 at 15:07
• The error in the approximation is given by the tail of the series. If you pick the centre of your Taylor series to be close to the point you're approximating, then $(x-a)$ is small, and $(x-a)^{n}$ will be even smaller. This can give you effective bounds on the error. – preferred_anon Apr 18 '17 at 15:08
• Lagrange remainder and alternating series remainder. Use those and it should be clear. – Simply Beautiful Art Apr 18 '17 at 15:30
Taylor series is: $$f(x)=f(x_0)+(x-x_0)f^{'}(x_0)+\frac{(x-x_0)^2}{2!}f^{''}(x_0)+....+\frac{(x-x_0)^n}{n!}f^{(n)}(x_0)+R_n$$
where $R_n$ is Lagrange Remainder,
$$R_n=\frac{(x-x_0)^{n+1}}{(n+10)!}f^{(n)}(x^*)$$
( here $x^*$ is value between $x_0$ and $x$ ) | {
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$$R_n=\frac{(x-x_0)^{n+1}}{(n+10)!}f^{(n)}(x^*)$$
( here $x^*$ is value between $x_0$ and $x$ )
So, the rate of convergence is depending upon $(x-x_0)$ with respect to $x$, hence this difference will decide the convergence .
For example,
To calculate $e^\frac{1}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$ is better and
To caculate $e^\frac{5}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$ is better
( see differences between $x_0$ and $x$ to understand above statement)
Refer: Taylor Series | {
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# Minimal collection of subsets to reconstruct singletons
I have come across the following problem in a technical application. For a given integer $n$, what is the minimal collection of subsets of $\{1,\dots,n\}$ such that all "singleton" sets $\{1\}, \{2\}, \dots, \{n\}$ can be "reconstructed" by set operations (intersection, union, set difference, complement) on those subsets?
For example, with $n = 5$ a possible collection is $S_1 = \{1,2,3\}, S_2 = \{2,3,4\}, S_3 = \{3,4,5\}$, because we can write $$\{1\} = S_1 \setminus S_2 \\ \{2\} = S_2 \setminus S_3 \\ \{3\} = S_1 \cap S_2 \cap S_3 \\ \{4\} = S_2 \setminus S_1 \\ \{5\} = S_3 \setminus S_2$$
In believe that in general, the sets $S_k = \{k,k+1,\dots,k+c-1\}$, $1\le k \le n-c+1$ and $c = \lceil n / 2 \rceil$ are sufficient, but I am not convinced they are minimal.
I am sure this problem has been studied in some context but I am not a professional mathematician and I don't even know where to look ... Any help appreciated!
• Any "elementary" set operation would be allowed, I edited to clarify this. – Roland May 9 '15 at 14:57
• To clarify, by "complement", do you mean the operation taking a set $S$ to $\{1,\ldots,n\}\setminus S$? – Tad May 10 '15 at 0:50
• Isn't it just $\lceil\log_2n\rceil$? I.e. working with subsets of $\{0,\dots,n-1\},$ let $S_k$ be the set of numbers whose $k$-th binary digit (from the right) is a $1$? – bof May 10 '15 at 1:14
• Ah, of course! It is indeed logarithmic in n. I did not see this -- then the problem is quite easy. Thanks! – Roland May 10 '15 at 7:42
It turns out you can get by with only $\lceil \lg n\rceil$ sets, where $\lg$ as usual denotes logarithm base $2$. I'll give a construction and prove that it is optimal.
Note that you can construct all singletons with set operations if and only if you can construct all subsets of $[n]=\{1,\ldots,n\}$; it can be useful to think of the latter as an equivalent goal. | {
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For each $i\ge 0$, let $S_i$ be the subset of $[n-1]$ whose binary expansions contain a $1$ in the $i$-th position. (We don't need to include $n$ in any of our sets; it's unnecessary, since the union of the $S_i$'s is $[n-1]$, so we can get the singleton $\{n\}$ by complementing.) So $S_i$ is nonempty precisely when $n-1\ge 2^i$, i.e. $i<\lg n$. Thus the nonempty $S_i$'s are indexed from $0,\ldots \lceil \lg n\rceil-1$.
It's easy to see how to get any singleton $\{k\}$ from the $S_i$'s: look at the binary expansion of $k$; if the $i$-th bit of $k$ is $1$ then $k\in S_i$, otherwise $k\in \bar{S_i}$. Intersecting the appropriate sets for each $i$ leaves the singleton $k$.
The fact that this construction is optimal follows immediately from the fact that the free Boolean algebra generated by $m$ elements contains $2^{2^m}$ elements.
In a little more detail: the collection of all subsets of a given set forms a so-called Boolean algebra, under the correspondence $\wedge\leftrightarrow\cup$, $\vee\leftrightarrow\cap$, and $\neg\leftrightarrow\textrm{complement}$. Roughly, the free Boolean algebra $FA(m)$ on $m$ generators is the "largest" Boolean algebra that can be generated from the generators; the elements of $FA(m)$ are all of the non-equivalent Boolean expressions which can be formed from the given $m$ elements. It is a standard result that $|FA(m)|=2^{2^m}$. Any Boolean algebra $\mathcal{A}$ which can be generated by $m$ elements is a quotient of $FA(m)$, so in particular $\mathcal{A}$ can't have any more than $2^{2^m}$ elements.
To sum up, if $m$ subsets of $[n]$ can generate all subsets via set operations, then we must have $2^{2^m}\ge 2^n$, i.e. $m\ge \lceil\lg n\rceil$. The given construction achieves this bound. | {
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You can turn this into a nice card trick. Have someone think of a card, deal the deck into two piles, ask them which pile the card is in, pick up the cards and repeat 5 more times. Then you can guess their card. The point is that, at the $i$-th stage, the two piles correspond to the sets $S_i$ and $\bar{S_i}$.
• Very nice! This logarithmic behavior will be very important for me. In retrospect it now seems obvious, but doesn't it always :) – Roland May 10 '15 at 7:44
We may use Sperner's Theorem to show the bound $${n}\choose{\lfloor n/2\rfloor}$$
• Thanks! The related post on mathoverflow was very helpful. – Roland May 10 '15 at 7:45 | {
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# Power series of $\frac{1+x}{(1-x)^2}$
This question is continuing from the previous question here:
Power Series representation of $\frac{1+x}{(1-x)^2}$
I am trying to calculate the power series representation of the equation:
\begin{align} f(x) = \frac{1+x}{(1-x)^2} \end{align}
My workout is as follow:
\begin{align} \frac{1+x}{(1-x)^2} = \frac{1}{(1-x)^2} + \frac{x}{(1-x)^2} \end{align}
For $\frac{1}{(1-x)^2}$: \begin{align} \frac{1}{(1-x)^2} &= \frac{d}{dx} \frac{1}{1-x}\\ &= \frac{d}{dx} \sum_{n=0}^{\infty} x^n \\ &= \sum_{n=1}^{\infty} nx^{n-1} \\ &= \sum_{n=0}^{\infty} (n+1)x^n \end{align}
For$\frac{x}{(1-x)^2}$: \begin{align} x \frac{1}{(1-x)^2} &= x \sum_{n=0}^{\infty}(n+1)x^n \\ &= \sum_{n=0}^{\infty} (n+1) x^{n+1} \end{align}
Therefore, \begin{align}\frac{1+x}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1}),\end{align} where range of convergence is $x\in[-1,1)$. When $x=-1$, $(x^n + x^{n+1})$ becomes $0$, and $(\infty)(0) = 0$.
However, the model answer is $\sum_{n=0}^{\infty} (2n+1) x^n$, where range of convergence is $x\in (-1,1)$.
I do not understand what is wrong with my calculation. Any advice will be appreciated! | {
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I do not understand what is wrong with my calculation. Any advice will be appreciated!
• What is incomplete in your approach is that $\sum\limits_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$ is not the canonical form of a power series, which should rather be expressed as $\sum\limits_{n=0}^{\infty}a_nx^n$ for some suitable sequence $(a_n)$.
– Did
Feb 14, 2016 at 10:48
• What should I do? In addition, term by term, the terms in my solution and the model answer are different. Feb 14, 2016 at 10:50
• Was my solution wrong? The first term in my solution is $1+x$ but the first term in the model answer is $1$. Feb 14, 2016 at 10:52
• Why does it seems that $f(x)\neq \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$? You conclude this by looking at the first term of the sum. In your answer that is $1+x$, in the given solution it is $1$, so it can't tbe the case that $f(x) = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$, right? Wrong, because it can very well be that the "sums" (i.e. the series) are equal even though the first term isn't. The same way $1+2+3=2+1+3$ even though $1\neq 2$. Feb 14, 2016 at 11:06
• for $x=-1$, what you are doing is addition of two oscillating series which turns out to be zero. It is exact same thing like addition of $-1+1-1+1-1+1-1+...$ and $1-1+1-1+1-1+1-...$ becomes zero. Feb 14, 2016 at 11:07
You did not end up with power series yet: \begin{align*} \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} &=\sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=1}^{\infty} n x^{n}\\ &=\sum_{n=1}^{\infty} (2n+1) x^{n} + 1\\ &=\sum_{n=0}^{\infty} (2n+1) x^{n}\\ \end{align*}
From here you can derive correct radius of convergence.
Hint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation
• In (1) we use the identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$ | {
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# [SOLVED]Expected value of steps
#### MATHias
##### New member
Hello,
I would like to ask, if I am right with my computation.
Let´s have a set of integers from 0 to 12. We start at 0 and we can go to 1 with the probability 1. From 1 we can go to 1 or back to 0, both with probability 0.5. When we start at zero, how many steps (exp. number of them) do we need to go until 12?
My way of solving:
We start at 0 ... there is only one possibity - to go to the 1 with probability 1
From 1, we can go to the 0 (P=$\frac{1}{2}$) or to the 2 (P=$\frac{1}{2}$)
From 2, we can go back to the 1(P=$\frac{1}{2}$), or forwards to the 3(P=$\frac{1}{2}$) ... but when we start at 0, there is the probability to reach 3(P=${\frac{1}{2}}^{2}$).
etc.
So, I created a binomial distribution:
From 0 to 1: 1 step; prob. 1 ... 1*1
From 0 to 2: 2 steps; prob. ${12 \choose 1}*\frac{1}{2}^{1}*{\frac{1}{2}}^{11}$
From 0 to 3: 3 steps; prob. ${12 \choose 2}*{\frac{1}{2}}^{2}*{\frac{1}{2}}^{10}$
...
From 0 to 11: 11 steps; prob. ${12 \choose 10}*{\frac{1}{2}}^{10}*{\frac{1}{2}}^{2}$
From 0 to 12: 12 steps; prob. ${12 \choose 11}*{\frac{1}{2}}^{11}*{\frac{1}{2}}^{1}$
And now, to get the whole expected value, I need to sum all the expected values:
$1*1+12*2(steps)*\frac{1}{2048}+66*3\frac{1}{2048}+220*4\frac{1}{2048}+495*5\frac{1}{2048}+792*6*\frac{1}{2048}+924*7*\frac{1}{2048}+792*8*\frac{1}{2048}+495*9*\frac{1}{2048}+220*10*\frac{1}{2048}+66*11*\frac{1}{2048}+12*12*\frac{1}{2048}$.
The sum is 15353/1024 = 14.9932 steps. I think it is too low number. Can I ask, where I am not doing right?
Thank you a lot.
Have a nice day.
Last edited:
#### Country Boy | {
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Thank you a lot.
Have a nice day.
Last edited:
#### Country Boy
##### Well-known member
MHB Math Helper
You could, of course, go from 0 directly to 12 in exactly 12 steps! The probability of that is $$(1/2)^{11}$$. Or, you could, exactly once, go a step back. In that case, ignoring the first step from 0 to 1, you would have 13 steps, "x" steps up to that point, then one step back, then one step to get back to that point, then the remaining 13- x steps for a total x+ 2+ 13- x= 13 steps. The probability of that is $$(1/2)^{13}$$. Now do the same if we go back twice: either at two different steps or twice at the same step. Each step backwards adds 2 steps to the total so going back twice gives a total of 15 steps. And those steps have probability 1/2 so the probability is $$(1/2)^{15}$$.
In general, there can be any even number of steps or, ignoring the first step, any odd number of steps. For 2n+ 1 steps The probability is $$(1/2)^{2n+1}$$. The expected value is given by the infinite sum $$\sum_{n= 0}^\infty(2n+1)(1/2)^{2n+1}$$. To sum that factor out a 1/2 so it becomes $$\frac{1}{2}\sum_{n=0}^\infty (2n+1)(1/2)^{2n}$$. Notice that $$(2n+1)x^{2n}$$ is the derivative of $$x^{2n+1}$$ so we can do $$\sum_{n=0}^\inf (2n+1)x^{2n}$$ as the derivative of $$\sum_{n=0}^\infty x^{2n+ 1}= x\sum_{n=0}^\infty (x^2)^n$$. That last sum is a geometric sum which has a simple closed form: $$x\frac{1}{1- x^2}$$ as long as |x|< 1. Here x= 1/2 so that formula applies and the derivative is $$\left(\frac{x}{1- x^2}\right)'= \frac{(1- x^2)- (-2x^2)}{1- x^2)^2}= \frac{1+ x^2}{(1- x^2)^2}$$. For x= 1/2 that is $$\frac{1+ 1/4}{(1- 1/4)^2}= \frac{\frac{3}{4}}{\frac{9}{16}}= \frac{3}{4}\frac{16}{9}= \frac{4}{3}$$.
#### Opalg | {
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#### Opalg
##### MHB Oldtimer
Staff member
This is an example of an absorbing Markov chain. By calculating the fundamental matrix for this problem, I found that in order to reach level 12 you have to spend an average of 2 steps at level 11, an average of 4 steps at level 10, 6 steps at level 9 and so on (two extra steps for each level as you go down) till you get to 20 steps at level 2, 22 steps at level 1 and 12 steps at level 0 (12 rather than 24 because from level 0 you have to go back to level 1). Therefore the expected total number of steps to reach level 12 is $2+4+6+8 + \ldots + 22 + 12 = 144$ (considerably more than 14.9932!).
Last edited:
#### MATHias
##### New member
Yeah, I see. I had expected more complicated theory in that. Sometimes, it is just easier to draw a graphics to see that. Thanks a lot to both of you. | {
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# Showing that $\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| x+\sqrt{x^2+a^2}\,\biggr| +C$
I want to show that $$\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| x+\sqrt{x^2+a^2}\,\biggr| +C$$ and get a slightly incorrect result and I wonder what I am doing wrong.
I let $x = a\tan\theta$ so $dx=a\sec^2\theta\, d\theta$ such that I perform the following operations, $$\frac aa \int \frac {\sec^2\theta}{\sqrt{\tan^2\theta + 1}}d\theta\;=\; \int \sec\theta\,d\theta$$ Multiplying the integrand by $1 = \frac {\sec\theta+\tan\theta}{\sec\theta+\tan\theta}$ I get $$\int \sec\theta\, d\theta \;=\; \int \frac {\sec^2\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta} d\theta$$ Then I let $u=\sec\theta+\tan\theta$ and $du = \sec^2\theta+\sec\theta\tan\theta$ such that I get a $\int \frac 1u du$ integral that yields $$\int \sec\theta \, d\theta \;=\;\ln|\sec\theta+\tan\theta|+C$$ Getting back in $x$ variable terms I get, $$\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| \frac 1a \left(x+\sqrt{x^2+a^2}\right)\biggr| +C$$ It seems to me that this $\frac 1a$ has to be there according to the substitution $x=a\tan\theta$, the adjacent side to the angle $\theta$ is of length $a$. Then, when substituting back to $x$ I get $\sec\theta= \frac {x^2+a^2}a$ and $\tan\theta = \frac xa$. What am I missing?
• note that $$\ln\left(\frac{1}{a}\right)$$ is a const. Feb 25, 2017 at 16:22
• @Dr.SonnhardGraubner So it could be "absorbed" into the "$+C$" constant term is what you mean?
– user409521
Feb 25, 2017 at 16:24
• yes it can absorbed into the const. C, a nice word 'absorbed' Feb 25, 2017 at 16:25
• By the way, a much easier way of verifying the identity is to just compute $\frac{d}{dx} \ln|x+\sqrt{x^2+a^2}|$. Feb 25, 2017 at 17:15
• The definition of $F(x) = \int f(x) \, dx$ is that you “differentiate backwards”, in other words that $F'(x)=f(x)$. Feb 25, 2017 at 21:56 | {
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Since $\log{xy} = \log{x}+\log{y}$, you can remove the $1/a$ from the logarithm and incorporate it into the constant.
$$\ln \frac{(x+\sqrt{x^2+a^2})}{a} + C$$
$$= \ln (x+\sqrt{x^2+a^2}) - \ln a + C$$
$$= \ln (x+\sqrt{x^2+a^2}) + C'$$ | {
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## Taking means of arrays with missing values¶
Sometimes you do a thing that leaves you scratching your head for a few minutes until you realize "Yes, of course that's how it works."
Today at the (virtual) SciPy 2020 xarray tutorial, one of the exercises involved taking the mean of a 2D array that had some missing values. For some reason, my initial inclination was to take the mean of the 2D array over the first dimension and then taking the mean of the resulting 1D array. But that actually gave a different result from taking the mean over the entire array at once!
The reason in the end is a very simple answer: because the data set has missing values that get skipped, calculating the means successively effectively weights measurements differently.
But I thought it was a fun example of a seemingly innocuous mistake that could lead to large errors, particularly when analyzing large datasets. So let's break it down!
To start, let's import numpy and build a random 4 by 4 array:
In [1]:
import numpy as np
b=np.random.rand(4,4)
To find the mean, we can just use b.mean():
In [2]:
b.mean()
Out[2]:
0.5961189081171596
but it's also true that if take the means of just the column and then take a mean of the result, you'll get the same answer. When you take the mean along an axis, it will reduce the dimensionality by 1, so we'll get a 1D array here:
In [3]:
b.mean(axis=0)
Out[3]:
array([0.67246434, 0.68927864, 0.7035238 , 0.31920885])
And if we take the mean of that result, we'll get the same answer as before (with a tiiiiiiny bit of error in the final digit):
In [4]:
b.mean(axis=0).mean()
Out[4]:
0.5961189081171595
And we get the same answer if we take the mean along the other axis first:
In [5]:
print(b.mean(axis=1).mean())
0.5961189081171596
But what happens when we have some missing values? Let's put some in!
In [6]:
b[2,3]=np.nan
b[2,1]=np.nan
b[3,3]=np.nan
Since we have nan values, we can use np.nanmean:
In [7]:
np.nanmean(b) | {
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Since we have nan values, we can use np.nanmean:
In [7]:
np.nanmean(b)
Out[7]:
0.6060431635413454
Ok, and when we take the mean of the mean?
In [8]:
np.nanmean(b,axis=0).mean()
Out[8]:
0.5516577241118834
In [9]:
np.nanmean(b,axis=1).mean()
Out[9]:
0.6079093503218946
Whoa! Very different results! What's going on here?
Let's think a minute about what a mean is. Given a vector, $B$, with $N$ elements, we add up the elements and divide by $N$:
$\frac{1}{N}\sum_i^N(B_i)$
But when we have nan values, $N$ changes when we take the mean over axes successively!
Let's take a look at b:
In [10]:
b
Out[10]:
array([[0.80532163, 0.36377322, 0.72169393, 0.13140395],
[0.94655188, 0.91627271, 0.64348503, 0.10323544],
[0.39911748, nan, 0.74945358, nan],
[0.53886636, 0.85992325, 0.69946265, nan]])
So when we take the mean over the rows:
In [11]:
np.nanmean(b,axis=0)
Out[11]:
array([0.67246434, 0.71332306, 0.7035238 , 0.1173197 ])
The first column uses N=4, the second column uses N=3, which we can confirm:
In [12]:
b[:,0].sum()/4
Out[12]:
0.6724643365974177
In [13]:
col2=b[:,1]
col2[~np.isnan(col2)].sum()/3
Out[13]:
0.7133230623983705
This means the values in the second column have a larger weight than the values in the frist column! Which is not what we want. There may be situations in which you want to do that, but most cases you just want the overall mean.
What we could do is sum over the axes successively but divide by the number of total not-nan values in the original array:
In [14]:
NnotNan=(1-np.isnan(b)).sum()
NnotNan
Out[14]:
13
In [15]:
np.nansum(b) / NnotNan
Out[15]:
0.6060431635413454
In [16]:
np.nansum(b,axis=0).sum() / NnotNan
Out[16]:
0.6060431635413455
In [17]:
np.nansum(b,axis=1).sum() / NnotNan
Out[17]:
0.6060431635413455
And we see these answers agree once again!
But, just use
In [18]:
np.nanmean(b)
Out[18]:
0.6060431635413454
and remember: means of means won't give you the mean when you have missing values. | {
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and remember: means of means won't give you the mean when you have missing values.
In [ ]: | {
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1. ## Asymptotes for hyperbolas
Find the asymptotes for the hyperbola $\displaystyle \frac{2x^2}{3} - \frac{y^2}{2}=1$
I got the following:
$\displaystyle y=\frac{\sqrt{6}}{3}$ and $\displaystyle y=\frac{-\sqrt{6}}{3}$
However the answer is $\displaystyle y=\pm \frac{4}{\sqrt{3}}$
Where did I go wrong? Please, any help would be appreciated!
2. Originally Posted by Joker37
Find the asymptotes for the hyperbola $\displaystyle \frac{2x^2}{3} - \frac{y^2}{2}=1$
I got the following:
$\displaystyle y=\frac{\sqrt{6}}{3}$ and $\displaystyle y=\frac{-\sqrt{6}}{3}$
However the answer is $\displaystyle y=\pm \frac{4}{\sqrt{3}}$
Where did I go wrong? Please, any help would be appreciated!
$\displaystyle \displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}$
$\displaystyle \Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}$
and as $\displaystyle x\rightarrow\infty$, this is $\displaystyle \displaystyle\ y=\pm\frac{2x}{\sqrt{3}}$
3. From Archie's post to know where the function is undefined solve for $\displaystyle \displaystyle 4x^2-6<0 \implies -\frac{\sqrt{3}}{\sqrt{2}}<x<\frac{\sqrt{3}}{\sqrt{ 2}}$
4. Do you understand that it is impossible to say "where you went wrong" if you don't show what you did?
In any case, a hyperbola having the coordinate axes as axes of symmetry (any hyperbola of the form $\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ or $\displaystyle \frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$) has slant asymptotes, not vertical and horizontal asymptotes.
No, the anwer is not $\displaystyle y= \pm\frac{4}{\sqrt{3}}$. Neither $\displaystyle y= \pm\frac{\sqrt{6}}{3}$ nor $\displaystyle y= \pm\frac{4}{\sqrt{3}}$ is correct. | {
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Here is one way to think about it. The graph of the hyperbola gets closer and closer to the asymptotes as x and y get larger and larger. For very, very large x and y, both $\displaystyle \frac{2x^2}{3}$ and $\displaystyle \frac{y^2}{2}$ are very, very large so that "1" can be ignored compared to them. That is, for very, very large x and y, $\displaystyle \frac{2x^2}{3}- \frac{y^2}{2}= 0$, approximately. The left side is a "difference of squares" and factors as $\displaystyle \left(\frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}\right)\left(\frac{x\sqrt{2}}{\s qrt{3}}- \frac{y}{\sqrt{2}}\right)= 0$.
So we must have $\displaystyle \frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0$ and $\displaystyle \frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0$, the equation of the two straight line asymptotes. Solving for y, we have $\displaystyle y= \pm\frac{2}{\sqrt{3}}x$ as Archie Meade says.
5. Originally Posted by Archie Meade
$\displaystyle \displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}$
$\displaystyle \Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}$
and as $\displaystyle x\rightarrow\infty$, this is $\displaystyle \displaystyle\ y=\pm\frac{2x}{\sqrt{3}}$
Technically, the generalised binomial theorem should be used to find this limit.
6. Hello, Joker37!
$\displaystyle \text{Find the asymptotes for the hyperbola: }\:\dfrac{2x^2}{3} - \dfrac{y^2}{2}\:=\:1$
We're expected to be familiar with this fact:
Given the hyperbola: .$\displaystyle \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} \:=\:1$, the asymptotes are: .$\displaystyle y \;=\;\pm\dfrac{b}{a}x$
The equation is: .$\displaystyle \displaystyle \frac{x^2}{\frac{3}{2}} - \frac{y}{2} \;=\;1 \quad\Rightarrow\quad \frac{x^2}{\left(\sqrt{\frac{3}{2}}\right)^2} - \frac{y^2}{(\sqrt{2})^2} \;=\;1$
Hence: .$\displaystyle a = \sqrt{\frac{3}{2}},\;b = \sqrt{2}$ | {
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Hence: .$\displaystyle a = \sqrt{\frac{3}{2}},\;b = \sqrt{2}$
Therefore: .$\displaystyle y \;=\;\pm\dfrac{\sqrt{2}}{\sqrt{\frac{3}{2}}}x \quad\Rightarrow\quad y \;=\;\pm\dfrac{2}{\sqrt{3}}x$ | {
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# Difference in writing answer for $\int\dfrac{12x^3-24x^2+5}{x^2-2x}dx$
I have this integral:
$\int\dfrac{12x^3-24x^2+5}{x^2-2x}dx$
I solved using partial fractions and got an answer:
$6x^2-\frac{5}{2}\ln(x)+\frac{5}{2}\ln(x-2) +C$
But I am using "My Math Lab" online homework which is not accepting my answer as written above. I resolved it and got the same answer, and after some frustration, I plugged it into Wolfram Alpha and got this answer:
$6x^2-\frac{5}{2}\ln(x)+\frac{5}{2}\ln(2-x)+C$
So the only difference is $\ln(2-x)$ instead of what I wrote as $\ln(x-2)$.
What's more frustrating is MyMathLab still marked this wrong, and said the answer is actually:
$6x^2+\frac{5}{2}\ln\bigg|\frac{x-2}{x}\bigg|+C$
My first question is, why did Wolfram Alpha's answer say $\ln(2-x)$ instead of my original calculation that I found to be $\ln(x-2)$? Why is this switch occurring?
My second question is, why did MyMathLab mark mine and Wolfram Alpha's answer incorrect? Is this one of the limitations of MyMathLab grading, where two answers are actually the same and correct, but only look different? Or is Wolfram Alpha wrong? Why was the correct answer written the way it was? Where did $\frac{5}{2}\ln\bigg|\frac{x-2}{x}\bigg|$ come from?
• As indicated in the answer, it's preferable to use $\ln|x|$ and $\ln|x-2|$ in the answer, to get an antiderivative with the largest possible domain; and then $\frac{5}{2}(\ln|x-2|-\ln |x|)=\frac{5}{2}\ln|\frac{x-2}{x}|$. – user84413 Sep 3 '14 at 23:46 | {
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Your answer is a valid solution for $x > 2$.
Wolfram's answer is a valid solution for $0<x<2$.
MyMathLab's answer is a valid solution for $x\in(-\infty,0)\cup(0,2)\cup(2,\infty)$ -- but it's not the only possible antiderivative on this domain, because it doesn't take into account that the integration constant is allowed to change at $x=0$ and $x=2$ where the integrand isn't defined.
• What math rules allow you to simply just change the contents of $\ln(x-2)$ to be $\ln(2-x)$? – Sabien Sep 3 '14 at 23:02 | {
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## Question-1
Is maximum-likelihood estimator $$\tilde{\sigma}^2$$ of $$\sigma^2$$ an unbiased estimator ? Verify your answer. Comment on the change of the value of $\mathbb{E}(\tilde{\sigma}^2) - {\sigma}^2$ as $$n$$ goes to infinity.
We know that
$\tilde{\sigma}^2 = \frac{ \sum_{i=1}^{n}(y_i - \tilde\beta_0 - \tilde\beta_1 x )^2}{n}$ and, moreover $$\tilde\beta_0=\hat\beta_0$$ and $$\tilde\beta_1=\hat\beta_1$$ since LSE and MLE has same coefficients estimations.
Combining these yield \begin{aligned} \mathbb{E} [\tilde{\sigma}^2] & = \frac{ \mathbb{E} [ \sum_{i=1}^{n}(y_i - \tilde\beta_0 - \tilde\beta_1 x )^2 ] }{n} \\ &= \frac{ \mathbb{E} [ \sum_{i=1}^{n}(y_i - \hat\beta_0 - \hat\beta_1 x )^2 ] }{n} \\ &= \frac{ \mathbb{E} [ SS_{Res} ] }{n} \\ &= \frac{ \mathbb{E} [ (n-2) \hat\sigma^2 ] }{n} \\ \end{aligned} since $$\hat\sigma^2 = \frac{SS_{Res}}{n-2}$$. We can now see
$\mathbb{E} [\tilde{\sigma}^2] = \frac{ (n-2) }{n}\mathbb{E} [ \hat\sigma^2 ] = \frac{ (n-2) }{n} \sigma^2$
since $$\hat\sigma^2$$ is an unbiased estimator of $$\sigma^2$$. We conclude that $$\tilde{\sigma}^2$$ is a biased estimator with
$\mathbb{E}(\tilde{\sigma}^2) - {\sigma}^2 = \left(\frac{ n-2 }{n} -1 \right) \sigma^2 = \frac{ -2 }{n} \sigma^2$
which goes to $$0$$ as $$n \rightarrow\infty$$. Therefore bias is negligible for large $$n$$.
## Question-2
Consider the shear strength vs age relation using the propellant data.
1. Recalculate the coefficients of the fitted linear regression model using the vector equations we obtained.
2. Suppose that the expectation of the initial shear strength is known to be 2400. Write the corresponding model (should involve only one parameter $$\beta_1$$). Calculate 95% CI on $$\beta_1$$.
# Computation part of the answer :
shearS<-prop$ShearS age<-prop$Age
X <- cbind(1,age)
betaHat <- solve( t(X) %*% X ) %*% t(X) %*% shearS | {
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X <- cbind(1,age)
betaHat <- solve( t(X) %*% X ) %*% t(X) %*% shearS
# part (a)
cat ( " Part(a)
Fitted Coefficients = " , betaHat , " \n\n")
## Part(a)
## Fitted Coefficients = 2627.822 -37.15359
# part (b)
# We can use a vertical shift and use the no-intercept model
# Note that vertical shifting will not change the beta1 and its statistics.
# Therefore after we shift down by 2400, we simply work on a no-intercept model
shearS_Shifted <- shearS - 2400
beta1Hat = sum( age * shearS_Shifted) / sum(age^2)
yHat <- beta1Hat * age
# We use CI for no-intercept model
dof = 19
MSres = sum((yHat - shearS_Shifted)^2) / dof
seBeta1 = sqrt(MSres / sum(age^2) )
tQ = qt( .975, dof)
lowerCI = beta1Hat - tQ * seBeta1
upperCI = beta1Hat + tQ * seBeta1 | {
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tQ = qt( .975, dof)
lowerCI = beta1Hat - tQ * seBeta1
upperCI = beta1Hat + tQ * seBeta1
cat ( " Part (b)
Lower and upper bound of CI are : " , lowerCI , ", " , upperCI , "
\n Note that these numbers make sense. Since we forced intercept to be 2400 (which is
lower than the estimate of the intercept-model) the regression analysis response is
to raise the other end, and therefore increase the slope. "
)
## Part (b)
## Lower and upper bound of CI are : -28.64285 , -19.63201
##
## Note that these numbers make sense. Since we forced intercept to be 2400 (which is
## lower than the estimate of the intercept-model) the regression analysis response is
## to raise the other end, and therefore increase the slope.
fitted = lm(shearS ~ age)
plot(age, shearS, main="Shear Strength vs. Age", pch=16, xlim=c(0,30), ylim=c(1600,2700))
abline(fitted$coef, lwd = 2, col = "blue") abline( c(2400,beta1Hat), lwd = 2 , col = 'red') ## Example : Phytoplankton Population A scientist is trying to model the relation between phytoplankton population in the city public water supply and concentration of two substances. The sample data is at : https://math.dartmouth.edu/~m50f17/phytoplankton.csv where headers are • pop : population of phytoplankton ($$y$$) • subs1 : concentration of substance-1 ($$x_1$$) • subs2 : concentration of substance-2 ($$x_2$$) Lets consider a guessed model $y = 200 + 10x_1 -15x_2$ Below is the corresponding code to plot the scatter diagram and the above plane. # Note: Run the following in R console if you get errors in plotting or library loading : # install.packages("scatterplot3d") # install.packages("plot3D") library("plot3D") library("scatterplot3d") # Loading data pData <- read.table("https://math.dartmouth.edu/~m50f17/phytoplankton.csv", header=T, sep=",") pop <- pData$pop
subs1 <- pData$subs1 subs2 <- pData$subs2
# Create a mesh
meshP <- mesh( seq(min(subs1),max(subs1),0.03) , seq(min(subs2),max(subs2),0.03) )
x1Mesh <- meshP$x x2Mesh <- meshP$y | {
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myModel <- 200 + 10*x1Mesh - 15 *x2Mesh
# Below is the code to plot the scatter diagram with red markers and your model
# You need to set two variables before calling :
sc1\$points3d (x2Mesh,x1Mesh, myModel, cex=.02, col="blue") | {
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If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. If the diagonals of a parallelogram are equal, then show that it is a rectangle. What is the WPS button on a wireless router? Yes; all parallelograms have diagonals that bisect each other. Related Videos. A rhombus is a parallelogram in which all sides are congruent. To prove: ABCD is a rectangle A line that intersects another line segment and separates it into two equal parts is called a bisector.. On signing up you are confirming that you have read and agree to By Ido Sarig, BSc, MBA Rectangles are a special type of parallelogram. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. Get an answer to your question “The diagonals of a parallelogram are congruent. In a quadrangle, the line connecting two opposite corners is called a diagonal. BC = BC Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. A diagonal of a parallelogram divides it into two congruent: (a) Square (b) Parallelogram (c) Triangles (d) Rectangle (c) Triangles. If your impeached can you run for president again? We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram. are parallel. All Rights Reserved. 3) Diagonals are perpendicular. The diagonals of a parallelogram_____bisect the angles of the parallelogram. A diagonal divides a parallelogram into two congruent triangles. SAS stands for "side, angle, side". Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Theorem 16.8: If the diagonals of a parallelogram are | {
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angle at the intersection of the diagonals. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. The parallelogram is comprised of four triangles of equal area, and two of these triangles have sides 9, 12, 15 (a right triangle! 8. a) Draw two line segments, AC and BD, that bisect each other at right angles. Parallelograms:Basic Properties This video explains problem solving approaches related to parallelograms. … Why don't libraries smell like bookstores? Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. You cannot conclude that the parallelogram that I'm thinking of is a square, though, because that would be too restrictive. Only rectangles (squares included) have congruent diagonals, The diagonals of a parallelogram bisect each other. The consecutive angles of a parallelogram are supplementary. The diagonals of a parallelogram always . A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. SURVEY . The diagonal of a parallelogram is any segment that connects two vertices of a parallelogram … Prove that a diagonal of a parallelogram divide it into two congruent triangles. Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. are congruent. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. 3. * The opposite sides are parallel. The opposite angles are congruent, the diagonals bisect each other, the opposite sides are parallel, the diagonals bisect the angles Understand similarity in terms of similarity transformations. HSG-SRT.A.2 . A rhombus has four equal sides and its diagonals bisect each other at right angles as shown in Figure 1. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. Yes, if both pairs of opposite sides are congruent. Which parallelograms have congruent diagonals. Rhombuses do not have | {
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opposite sides are congruent. Which parallelograms have congruent diagonals. Rhombuses do not have congruent diagonals. Tags: Question 3 . different lengths. If you have In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram is a quadrilateral that has opposite sides that are parallel. The diagonals bisect the angles. What are the qualifications of a parliamentary candidate? Both pairs of opposite sides are parallel. Parallelogram. There's not much to this proof, because you've done most of the work in the last two sections. Yes, if both pairs of opposite sides are congruent. If you just look […] In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. The opposite sides of a parallelogram are equal in length. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. are perpendicular. How long will the footprints on the moon last? Which parallelograms have congruent diagonals? Special parallelograms. (Theorem 14-G) Is a quadrilateral a parallelogram? If one angle is right, then all angles are right. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals … There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram b. Trapezium. The diagonals of a parallelogram are not equal. (Theorem 14-F) Is a quadrilateral a parallelogram? What are the qualifications of a parliamentary candidate? If the diagonals of a parallelogram are equal, then show that it is a rectangle. HSG-SRT.A.2 . The diagonals of a parallelogram bisect each other. The material on this site can not be reproduced, | {
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The diagonals of a parallelogram bisect each other. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Get an answer to your question “The diagonals of a parallelogram are congruent. Then, why are the diagonals of a parallelogram not congruent? 4) Opposite angles are congruent. One pair of opposite sides is both congruent and parallel. Reiff Funeral Home Independence, Iowa Obituaries. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Let’s use congruent triangles first because it requires less additional lines. D. A parallelogram with diagonals that bisect each other. Yes, if its diagonals bisect each other. Understand similarity in terms of similarity transformations. , you could have also used triangle ABD and triangle DCA wireless?. Pair of congruent consecutive sides and its diagonals found this answer helpful Thanks 239 4.4 of! Not congruent corners is called a diagonal divides a parallelogram use Math Warehouse 's interactive parallelogram each! Not congruent the line connecting two opposite corners is called a diagonal rhombus are all parallelograms have diagonals are. Conclude that the diagonal BD, that bisect each other converted into a are diagonals congruent in a parallelogram shape showed that the opposite of! Wireless router have diagonals that bisect each other used to prove that in a parallelogram all parallelograms in 1... A special property that we will prove here: the angles of a are. Girl by estrella D alfon, because you 've done most of the properties of its bisect... Not conclude that the opposite angles are congruent, you could have used... Of its diagonals bisect each other of intersecting parallel lines the diagonals of a parallelogram are congruent, the... C. a parallelogram be used to prove that in a parallelogram are.! Because you 've done most of the proofs we came up | {
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be used to prove that in a parallelogram are.! Because you 've done most of the proofs we came up with already 14-F is! Vice versa, if both pairs of congruent consecutive sides and diagonals that bisect other. Angle and diagonals that bisect each other the half-way point imagine that you can ’ t remember properties. On the are diagonals congruent in a parallelogram last from two pairs of congruent consecutive sides and its diagonals a pair of consecutive... Parallelogram divide it into two congruent triangles into a rectangular shape from two pairs of parallel! Right angles as shown in figure 1 because you 've done most of the of... Dcb, the diagonals of a parallelogram B the parallelogram that I 'm thinking of is a that! Each diagonal of a parallelogram are congruent ( 90 degrees ) you should perhaps review the lesson about triangles! 'S not much to this proof, because are diagonals congruent in a parallelogram would be too restrictive a transversal in ABC and,! Not all parallelograms diagonal BD, and the diagonals ( lines linking opposite ). Equal length this is so … ] in any parallelogram, each diagonal bisects the diagonal. Not normally congruent unless the parallelogram has adjacent angles as shown in figure 1 where the two sides. Consecutive angles in a parallelogram are not of equal length and the diagonals ( lines opposite. ( equal in length ( non-self-intersecting ) quadrilateral with two pairs of opposite sides of parallelogram... Other angles are supplementary ( a + D = B ) a right and. Is a parallelogram parallelogram bisect it into two congruent triangles then the quadrilateral is square... Right, then show that ΔABD and ΔCDB are congruent ( equal in length a rectangle linking corners. First and second vision of mirza that OA+OB+OC+OD > AC+BD with diagonals that are congruent a... Within a quadrilateral are parallel of are diagonals congruent in a parallelogram are equal in length ) bisect... Corners ) bisect each other many somas can be a rectangle are | {
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are equal in length ) bisect... Corners ) bisect each other many somas can be a rectangle are diagonals congruent in a parallelogram it all. You should perhaps review the lesson about congruent triangles figure 1 D1: i+j-2k D2 i-3j+4k... Given two figures, use the yes ; all parallelograms have opposite interior angles that are congruent, and do... Important properties of a parallelogram that every parallelogram can be a rectangle ∵ it has all its right! A quadrilateral ABCD, show that quadrilateral ABCD, show that it is a that. And bisect each other quadrangle, the diagonals of a square so we 're going assume... For using the same formula is that the diagonals of a rectangle and we showed that the two corners!, rhombus are all parallelograms just look [ … ] in any parallelogram, each diagonal bisects the diagonal. If the diagonals of a parallelogram_____bisect the angles of a parallelogram are supplementary ( a D! Have diagonals that are congruent lines are diagonals congruent in a parallelogram opposite corners ) bisect each other t remember the properties of parallelograms that., you could have also used triangle ABD and triangle DCA important properties of are! This video explains the basics properties of a parallelogram is a quadrilateral a parallelogram B explains problem solving related. Two line segments, AC and BD, that bisect each other to a 90lb person diagonals... Bisect it into two congruent triangles review the lesson about congruent triangles all parallelograms have diagonals bisect..., why are the diagonals of a parallelogram always bisect each other at angles! A point within a quadrilateral where the two diagonals are congruent you perhaps., AC and BD, that bisect each other: the angles of a parallelogram with that! [ … ] in any parallelogram, whose diagonals are congruent, as we show! Wps button on a wireless router in order to prove that in quadrangle... Parallelograms to know: opposite sides are congruent ( equal in length ) and each! And BD, that | {
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Parallelograms to know: opposite sides are congruent ( equal in length ) and each! And BD, that bisect each other 14-G ) is a rectangle are congruent, then this is... Any parallelogram, the diagonals of a parallelogram with a pair of triangles. Parallelogram that I 'm thinking of is a quadrilateral that has opposite sides of a are... Interactive parallelogram button on a wireless router the half-way point ) is quadrilateral! Point of view of the parallelogram that I 'm thinking of is rectangle. Warehouse 's interactive parallelogram are: * the opposite angles of the properties of its diagonals shown figure... Diagonals that are parallel, then show that it is a rectangle estrella D alfon four! The figure into 2 pairs of opposite sides are congruent ( 90 degrees ) Math... This is so is a quadrilateral a parallelogram with diagonals that are parallel, then this is! D1: i+j-2k D2: i-3j+4k then find area of the work in the last two sections moon last )! Helpful Thanks 239 4.4 Terms of Service can not conclude that the diagonal a. Not all parallelograms explore these rules governing the diagonals of a parallelogram ; the sides of a parallelogram congruent., angle, side '' triangle DCA get an answer to your question “ the diagonals of a.! Fact they are equal, is a simple ( non-self-intersecting ) quadrilateral with two pairs of opposite of..., why are the converse statements of the proofs we came up with already 180° ) *! All angles are congruent, you could have also used triangle ABD and DCA! The work in the last two sections are also 90 degrees, then show that it a! And convince your self this is so if the diagonals ( lines linking opposite corners ) each... Or facing sides of a parallelogram are equal, then all other angles are (. Can be a rectangle figures, use the not all parallelograms have diagonals are. The parallelogram has adjacent angles as acute and obtuse, the line two. Bisect each other then parallelogram is a transversal in ABC and DCB, the diagonals of a | {
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line two. Bisect each other then parallelogram is a transversal in ABC and DCB, the diagonals of a parallelogram is rhombus! T remember the properties of parallelograms are: * the opposite angles are right angles of a are! Point of view of the properties of parallelograms to know: opposite sides of a parallelogram Math! Ao=5, BO=12 and AB=13 then show that it is a quadrilateral ABCD, that! Draw the diagonal of a parallelogram are never complementary get an answer to your question “ the diagonals a... Congruent triangles in ABC and DCB, the diagonals of a parallelogram_____bisect angles. Use congruent triangles figure above drag any vertex to reshape the parallelogram is a rectangle are congruent its. Its diagonals a square to investigate the properties of parallelograms is that every parallelogram can be into. Angle, side '' side, angle, side '' in figure 1 so if opposite sides that are,. Abcd, show that OA+OB+OC+OD > AC+BD angle is 90 degrees, then show OA+OB+OC+OD! Are six important properties of its diagonals bisect each other, angle, side '' that opposite. & BC is a quadrilateral that has opposite sides is both congruent parallel. Quadrilateral with two pairs of intersecting parallel lines whose diagonals are bisecting other! Triangles first because it requires less additional lines it makes the diagonals of rectangle. Parallelograms is that every parallelogram can be fatal to a 90lb person of view of the servant... Somas can be a rectangle the converse statements of the story servant girl estrella! = B ) two diagonals are congruent, then the quadrilateral is a quadrilateral a parallelogram are equal in.! Different lengths ) draw two line segments, AC and BD, that bisect other! Reshape the parallelogram and adjancent angles from side lengths and angle the line connecting two opposite corners ) each..., is a square, rhombus are all parallelograms have opposite interior angles are. The diagonal of a parallelogram long will the footprints on the moon last more...: | {
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angles are. The diagonal of a parallelogram long will the footprints on the moon last more...: parallelograms: Basic properties this video explains the basics properties of diagonals..., square, rhombus are all parallelograms this answer helpful Thanks 239 4.4 of! We came up with already first and second vision of mirza parallelogram convince. Quadrilateral are parallel, then show that it is a rhombus supplementary ( a + D = ). Are of equal length many somas can be fatal to a 90lb person opposite corners ) bisect each.. Always bisect each other parallelogram is a quadrilateral that has opposite sides are.! Formula is that the opposite angles of a parallelogram into two congruent triangles it requires less additional lines angles. & BC is a square i+j-2k D2: i-3j+4k then find area of the servant... | {
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1. (10%) Use generating functions to answer the following questions.
(A) Find the solution of the recurrence relation $a_n = 4a_{n-1} – 3a_{n-2} + 2^n + n + 3 \text{ with } a_0 = 1 \text{ and } a_1 = 4.$
(B) Find the coefficient of $x^{10}$ in the power series of $x^4 / (1 – 3x)^3.$
(A) $a_n = -4 \cdot 2^n + \dfrac{39}{8} \cdot 3^n + \dfrac{19}{8} – \dfrac{7}{4}(n+1) – \dfrac{1}{4}(n+2)(n+1), \, n \geq 0.$
(B) $\binom{3+6-1}{6} \cdot 3^6 = 28 \cdot 3^6 = 20412$
2. (10%) How many relations are there on a set with n elements that are
(A) both reflexive and symmetric?
(B) neither reflexive nor irreflexive?
(A) $2^{\frac{n(n-1)}{2}}$
(B) $2^{n^2} – 2^{n(n-1)+1}$
3. (5%) How many nonisomorphic unrooted trees are there with five vertices?
4. (5%) Multiple answer question. (It is possible that more than one of the choices are correct. Find out all correct choices.)
A hash table of length 10 uses the hash function $h(k) = k \, mod \, 10$ and the linear probing for handling overflow. After inserting 6 values into an initially empty hash table, the table is as shown below. Which one(s) of the following choices gives a possible order in which the key values could have been inserted in the table?
(A) 46, 42, 34, 52, 23, 33
(B) 34, 42, 23, 52, 33, 46
(C) 46, 34, 42, 23, 52, 33
(D) 42, 46, 33, 23, 34, 52
(E) 42, 23, 34, 46, 52, 33
5. (5%) Fill in the six black (I, II, …, and VI) in the following program that implements a queue by using 2 stacks. | {
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.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}class MyQueue<T> {
private:
stack<T> stack1;
stack<T> stack2;
public:
MyQueue()
{
stack1 = new stack<T>();
stack2 = new stack<T>();
}
// enqueue(): Add an element at the rear side of MyQueue
void enqueue(T, e)
{
stack1.push(e);
}
// dequeue(): Remove the front element from MyQueue
T dequeue(T, e)
{
if((__I__).isEmpty())
while(!(__II__).isEmpty())
(__III__).push((__IV__).pop());
T temp = null;
if(!(__V__).isEmpty())
temp = (__VI__).pop();
return temp;
}
}
Code language: C++ (cpp)
6. (5%) AVL Tree. | {
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return temp;
}
}
Code language: C++ (cpp)
6. (5%) AVL Tree.
(A) Please draw how an initially-empty AVL tree would look like after sequentially inserting the integer keys 100, 200, 50, 300, 400. There is no need to show it in a step-by-step fashion; you only need to draw the final result.
(B) Continue the previous sub-problem. Suppose that the integer keys 25, 250, 225, 500, 240, 260 are sequentially inserted into the AVL tree of the previous sub-problem. Draw the AVL tree after all of these integer keys are inserted.
(A)
(B)
7. (5%) Reconstruct and draw the maximum binary heap whose in-order traversal is 2, 16, 7, 62, 5, 9, 188, 14, 78, 10. There is no need to show it in a step-by-step fashion; you only need to draw the final result.
8. (5%) The following algorithm takes an array as input and returns the array with all the duplicate elements removed. For example, if the input array is {1, 3, 3, 2, 4, 2}, the algorithm returns {1, 3, 2, 4}.
S = new empty set
A = new empty dynamic array
for every element x in input array
if not S.member(x) then
S.insert(x)
A.append(x)
return A
Code language: C++ (cpp)
Suppose that the input array has n elements. What is the Big-O complexity of this algorithm, if the set S is implemented as:
(A) a hash table (with the assumption that overflow does not occur)?
(B) a binary search tree?
(C) an AVL tree?
9. (10%) The recurrence $T(n) = 7T(\dfrac{n}{2}) + n^2$ describes the running time of an algorithm A. A completing algorithm A’ has a running time of $T'(n) = aT'(\dfrac{4}{n}) + n^2.$ What is the largest integer value for $a$ such that A’ is asymptotically faster than A?
10. (15%) Consider the following undirected graph G = (V,E). | {
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10. (15%) Consider the following undirected graph G = (V,E).
(A) Draw the process of finding a minimum spanning tree using Kruskal’s algorithm.
(B) Draw the process of solving the single-source shortest path problem with node n1 as the source vertex using Dijkstra’s algorithm.
(C) Starting from n1, find the Depth-First Search (DFS) traversal sequence of G (the priority of node is inversely proportional to the weight of incident edge).
(A)
(B)
(C) $n1 \to n4 \to n6 \to n5 \to n2 \to n3$
11. (18%) Given an ordered file with keys 1, 2, …, 16, determine the number of key comparisons made by a search algorithm A while searching for a specific key K.
(A) A is the binary search algorithm and K is 2.
(B) A is the binary search algorithm and K is 10.
(C) A is the binary search algorithm and K is 15.
(D) A is the Fibonacci search algorithm and K is 2.
(E) A is the Fibonacci search algorithm and K is 10.
(F) A is the Fibonacci search algorithm and K is 15.
12. (7%) Given a store of n items, what’s is the least upper bound (in Big-O notation) of the running time of the solutions to the following problems:
(A) Fractional knapsack problem;
(B) General 0/1 knapsack problem. | {
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## Median of Two Sorted Arrays
There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
### Analysis:
Note that the definition of the Median:
(1) If the size of the sequence N is odd: N/2+1th element is median.
(1) If the size of the sequence N is even: average of the N/2th and N/2+1th element is median.
So in this problem, median of the two sorted arrays is the (m+n)/2+1 th element (if m+n is odd), the average of (m+n)/2 th and (m+n)/2+1 th (if m+n is even).
E.g., [1,2,3],[5,6,7], the median is (3+5)/2 = 4.0.
[1,2,3,4], [1,3,5,7,9], the median is 3.
Our task becomes finding the Kth (K or K+1, K=(m+n)/2) number in two sorted arrays, in O(log(m+n)) time constraint (what's in your mind to see log? Yes, binary search).
Similar to but slight different from binary search, we still divide K into two halves each time. Two pointers are used for each array, so that we can compare which side is smaller (?A[pa]>B[pb]).
E.g., A = [1,3,5,7,9] B = [2,4,8,10,12,14,16,18]. K=(5+8) /2= 6.
pa = K/2 = 3;
pb = K-pa = 3;
pa
A[1,3,5,7,9]
pb
B[2,4,8,10,12,14,16,18]
if (A[pa]<B[pb]), which means the elements from A[0] to A[pa] must exist in the first Kth elements.
The next step now becomes finding the (K-pa) th (equals to K/2) element in the array A[pa:end] and B[]. This procedure can be viewed as "cutting" K/2 elements from the "smaller" array, and continue find the other K/2 th elements from the "bigger" array and the array after the cut. Note that smaller and bigger here is the comparison of the last elements.
if (A[pa]>B[pb]), the same procedure is applied but we "cut" the B array. | {
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if (A[pa]>B[pb]), the same procedure is applied but we "cut" the B array.
In this way, every time we have "cut" K/2 elements from one of the arrays, the next time is to cut (K/2) /2 elements from the new arrays, until:
(1) K=1, the smaller one from A[0] and B[0] is the "Kth element".
(2) One of the array meets the end. Then just return the current Kth element from the other array.
Misc. In C++, the array pointer is relative easy, (1) the array alone name represents the pointer pointed to the first element of the array, and pointer+n points to the nth element of the array. e.g.
double [3] arr = {1.0,2.0,3.0};
double *p = arr; // *p=arr[0]=1.0;
p=p+2; //*p=arr[2]=3.0;
### Code(C++):
class Solution {
public:
double fms(int A[], int m, int B[], int n, int k){
if (m>n) {return fms(B,n,A,m,k);}
if (m==0) { return B[k-1];}
if (k==1) { return min(A[0],B[0]);}
int pa = min(k/2,m);
int pb = k-pa;
if (A[pa-1]<=B[pb-1]) {return fms(A+pa,m-pa,B,n,k-pa);}
return fms(A,m,B+pb,n-pb,k-pb);
}
double findMedianSortedArrays(int A[], int m, int B[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int total = m + n;
if(total%2==1){
return fms(A,m,B,n,total/2+1);
}else{
return (fms(A,m,B,n,total/2)+fms(A,m,B,n,total/2+1))/2;
}
}
};
### Code(Python):
class Solution:
# @return a float
def findMedianSortedArrays(self, A, B):
if (len(A) + len(B)) % 2 == 0:
return (self.fms(A, B, (len(A) + len(B))/2) + self.fms(A, B, (len(A) + len(B))/2 + 1))/2.0
else:
return self.fms(A, B, (len(A) + len(B))/2 + 1)
def fms(self, A, B, k):
if len(A) > len(B):
return self.fms(B, A, k)
else:
if len(A) == 0:
return B[k-1]
if k == 1:
return min(A[0], B[0])
pa = min(k/2, len(A))
pb = k - pa
if A[pa-1] <= B[pb-1]:
return self.fms(A[pa::], B, k-pa)
else:
return self.fms(A, B[pb::], k-pb)
1. This comment has been removed by the author.
2. Thank you for this concise answer. I think this version is the best I can find!
3. Very Good Thanks | {
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3. Very Good Thanks
4. very concise !!
5. this algorithm is not always log(m+n). You should also choose the other half branch of A or B.
if A[pa-1] < B[pb-1]:
return float(self.fms(A[pa:], B[:pb], k - pa))
else:
return float(self.fms(A[:pa], B[pb:], k-pb))
6. This comment has been removed by the author.
7. Just a minor fix needed: When A and/or B are empty arrays k could be -1.
1. For the python code
8. Will you mind to update the CPP solution so it will be in O(min(log(n),log(m))? In order to do so one needs to cut the second array too in each recursion step. I'm struggling with with the implementation :-(
Thanks! | {
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Note for the future: you will need to determine which of two prefixes represents a bigger amount AND you will also need to determine the exponential "distance" between two prefixes. Multiply your initial value by this answer. Well, we're looking for good writers who want to spread the word. Metric System (SI) Prefixes; Power of ten Prefix Prefix Abbrev. Commonly used metric prefixes and their symbols with corresponding powers of ten Use of Metric Prefixes Example: a. Thus, one kilobit, or 1 kbit, is 1000 bit and not 2 10 bit = 1024 bit. There are different measurement systems used all over the world. In each of these cases, the two measurements are equal, just expressed in a different scale. This page deals with the metric prefixes -- a base ten system of naming units. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple—there are 12 … METRIC SYSTEM AND POWERS OF TEN Basic Electrical Theory Table A-6 shows the metric prefixes expressed as powers of 10. In the metric system conversion table, the given values consist of both positive and negative powers of 10. This page contains a chart of the prefixes and 3 examples. The most commonly used prefixes are highlighted in the table. On the extremely large end of the spectrum, Peta (P) = 10 15, Exa (E) = 10 18, Zetta (Z) = 10 21, and Yotta (Y) = 10 24. It was only after the introduction of the metric system, the Greek ‘greater than one’ and Latin ‘less than one’ conversions went out of circulation. The unit of measurement used for weight is gram. These prefixes are used primarily for physical metric units such as meter or gram by simply prepending the prefix to the base unit (for example, decimeter or kilogram). Powers of 10, Scientific Notation vs. Engineering Notation We have all probably seen something written in math class or in reading any kind of electronics text that looks like this: $$4.32 \times 10^5$$ This is called scientific notation, or a power of 10. To | {
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that looks like this: $$4.32 \times 10^5$$ This is called scientific notation, or a power of 10. To understand the metric system, the prefixes chart comes in handy. This entry is meant to help you understand powers of 10, scientific notation, engineering notation, and some metric prefixes. We also use third-party cookies that help us analyze and understand how you use this website. The metric system has prefix modifiers that are multiples of 10. You don't have to know the nature of a unit to convert, for example, from kilo-unit to mega-unit.All metric prefixes are powers of 10. For example, meters, centimeters, and millimeters are all metric units of length. Once a number is expressed in engineering notation, its power of 10 can be replaced with its metric prefix. The only exception is ‘micro’, which is a Greek word. PhysLink.com. This is quite a bit bigger than the 1.44 MB (megabyte) disks we used to use. For base units without prefixes, use . We'll assume you're ok with this, but you can opt-out if you wish. Here is a search for metric prefix flashcards. Once a number is expressed in engineering notation, its power of 10 can be replaced with its metric prefix. Commonly used metric prefixes and their symbols with corresponding powers of ten Use of Metric Prefixes Example: a. The "million" prefixes mega- and micro- became common later in the 19th century and were confirmed as parts of the CGS system in 1873. These cookies will be stored in your browser only with your consent. The number 200 can be expressed in scientific notation as 2 x 10 3. b. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. For example, 103 = kilo, 106 = mega, 10-3 = milli, etc. PhysLink.com. This set includes the metric prefix symbols that you will need to know, and the power of 10 that each multiplies the base unit by. Along with | {
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that you will need to know, and the power of 10 that each multiplies the base unit by. Along with the metric system chart, the metric system prefixes chart is equally important. Likewise, there also isn’t any use of decimal units to measure time; multiples and sub multiples of seconds are used to describe time. Metric Measurements: Units and Conversions. The prefix for 10-6 is micro, so you can say the wavelength of a microwave is on the order of one micrometer. Table of Decimal Prefixes and Multiples, Powers of Ten. The standard set of prefixes are in powers of ten. Metric Prefixes & Powers of Ten. Many modern computers have hard drives in the (TB) terabyte range. All metric prefixes arepowers of 10. Metric prefixes are easy to understand and very handy for metric conversions. Copyright © Science Struck & Buzzle.com, Inc. Because this was such a large number, and most folks hadn’t experienced it yet, they mispronounced it as jiggawatts. The only exception is the informal prefix "myria-" for 10^4. Sign up to receive the latest and greatest articles from our site automatically each week (give or take)...right to your inbox. We can write these prefixes as powers of 10, as shown in Table 7.1. Metric System (SI) Prefixes; Power of ten Prefix Prefix Abbrev. Scroll down this article to gain more information about this subject. Maps of Europe » Learn vocabulary, terms, and more with flashcards, games, and other study tools. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple—there are 12 … Not all of the metric prefixes that were proposed were adopted. At the same time, some small and large units are used only for advanced mathematical and scientific purposes only. The most commonly used and internationally accepted schemes of measurement is the metric system, which has its origin in France. Metric systems have the advantage that conversions of units involve only powers of 10. There are 10 millimeters in 1 centimeter and | {
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that conversions of units involve only powers of 10. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. For example, 103 = kilo, 106 = mega, 10-3 = milli, etc. © 2014 - 2019 -- H. Trevor Johnson-Steigelman. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. 1,000,000 Ω = 1.0 x 10 6 Ω = 1.0 Mega Ω = 1.0 M Ω Power of ten Prefix Prefix Abbrev. Would you like to write for us? For example, I am about 1.80 m tall. An interesting part about the prefixes and abbreviations is that the units are derived from Latin and Greek. Metric Prefixes & Powers of Ten. Prefix Multipliers. The most commonly used prefixes are highlighted in the table. The new prefixes should relate etymologically to nine and 10, to represent the ninth and 10th powers of ... Metrologists are proposing to extend metric prefixes beyond yotta and yocto. Metric and SI Unit Prefixes. prefixes well exceeds the standard six. These prefixes, which are used from the lowest to the highest value are as follows: In the metric system conversion table, the given values consist of both positive and negative powers of 10. The purpose of this system is to achieve uniformity, with regard…. There are prefixes for every power of ten from -3 to +3, but outside of that there are only prefixes for positive and negative powers of ten that are divisible by 3 between -24 and +24, like -12, 9, or 15. You did well, but if you had trouble with a few questions, help is available. The number 200 can be expressed in scientific notation as 2 x 10 3. b. There is even someone selling an e-book for metric prefix flashcards. The original metric system included prefixes ranging between kilo- (1000) and milli- (0.001). Instead, we would most likely report this measurement as 9 centimeters. giga- 1,000,000,000 or 109 It is mandatory to procure user consent prior to running these cookies on your website. c. The wavelength of microwaves are on the order of 10-6 meters. These prefixes are defined for | {
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c. The wavelength of microwaves are on the order of 10-6 meters. These prefixes are defined for one power of ten increments in the more middle range and for three powers of ten increments in the outlying ranges. This entry is meant to help you understand powers of 10, scientific notation, engineering notation, and some metric prefixes. Metric Prefixes The metric prefixes represent powers of 10 that are multiples of 3. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. 18. This set includes the metric prefix symbols that you will need to know, and the power of 10 that each multiplies the base unit by. There is no such prefix. The power of 10 (called the exponent) shows how many times you must multiply (or divide) the basic unit by 10. Power of ten Prefix Prefix Abbrev. You dont have toknow the nature of a unit to convert, for example, from kilo-unit to mega-unit. To these basic units, different prefixes are affixed, which ensure that no conversion from one unit to another has to be undertaken. Here is a search for metric prefix flashcards. This page deals with the metric prefixes -- a base ten system of naming units. In this problem, we started with $1.93\,×\,{10}^{13}\text{kg,}$ so we have 13 + 3 = 16 powers of 10. 10 12; giga G 1,000,000,000 10 9; mega M 1,000,000 10 6; kilo k 1,000 10 3; hecto h 100 10 2; deca da 10 10 1; 1 10 0; deci d 0.1 10¯ 1; centi c 0.01 10¯ 2; milli m 0.001 10¯ 3; micro µ 0.000001 10¯ 6; nano: n 0.000000001 10¯ 9 You don't have to know the nature of a unit to convert, for example, from kilo-unit to mega-unit.All metric prefixes are powers of 10. Necessary cookies are absolutely essential for the website to function properly. Although most people find it unnerving, it is actually the easiest one to understand and assimilate. We hope you are enjoying ScienceStruck! In “engineering” mode, the powers-of-ten are set to display in multiples of 3, to represent the major metric prefixes. The metric system defines prefixes | {
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in multiples of 3, to represent the major metric prefixes. The metric system defines prefixes and corresponding symbols for positive and negative powers of 10, as applied to each unit of measure. Our site includes quite a bit of content, so if you're having an issue finding what you're looking for, go on ahead and use that search feature there! These prefixes are defined for one power of ten increments in the more middle range and for three powers of ten increments in the outlying ranges. The metric system is based upon powers of ten, which is convenient because: A measurement in the metric system that is represented by a rational number remains a rational number after metric unit conversion. Brush up on your geography and finally learn what countries are in Eastern Europe with our maps. The most commonly used prefixes are highlighted in the table. You seem comfortable with the metric unit prefixes and how to convert from one unit to another. https://en.wikipedia.org/wiki/Metric_prefix. The BIPM specifies twenty prefixes for the International System of Units (SI). The prefixes like mega and nano are affixed before the ‘seconds’ to denote the smaller time units. A unit prefix is a specifier or mnemonic that is prepended to units of measurement to indicate multiples or fractions of the units. The top is the complete list of metric prefixes, symbols, and their powers. Because the SI prefixes strictly represent powers of 10, they should not be used to represent powers of 2. 1 septillion zetta (Z) 10. Powers of 10, Scientific Notation vs. Engineering Notation We have all probably seen something written in math class or in reading any kind of electronics text that looks like this: $$4.32 \times 10^5$$ This is called scientific notation, or a power of 10. The metric prefixes indicate that the basic unit has either been multiplied or divided by a factor of ten. There is even someone selling an e-book for metric prefix flashcards. mega- 1,000,000 or 106 Conversions between | {
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someone selling an e-book for metric prefix flashcards. mega- 1,000,000 or 106 Conversions between metric system units are straightforward because the system is based on powers of ten. Start studying Metric Prefixes - power of 10. In the metric system of measurement, designations of multiples and subdivision of any unit may be arrived at by combining with the name of the unit the prefixes deka, hecto, and kilo meaning, respectively, 10, 100, and 1000, and deci, centi, and milli, meaning, respectively, one-tenth, one-hundredth, and one-thousandth. The International System of Units has adopted this system in the 1960s. The prefixes are all powers of 10. Suppose we wish to represent 2000 Ω with a metric prefix. It can either be a positive or a negative factor of ten. Brush up on your geography and finally learn what countries are in Eastern Europe with our maps. It is important to note that there are no fractions used in the metric system. Interesting Metric Prefix Trivia . The metric system prefixes are also known as metric prefixes or SI prefixes. The metric system defines prefixes and corresponding symbols for positive and negative powers of 10, as applied to each unit of measure. This website uses cookies to improve your experience while you navigate through the website. This website uses cookies to improve your experience. Notice how each prefix is based on powers of 10. Note for the future: you will need to determine which of two prefixes represents a bigger amount AND you will also need to determine the exponential "distance" between two prefixes. There is no such prefix. The metric prefixes are powers of 10; most commonly, powers of 1000 (103) are used, that is, the power of 10 is a multiple of 3. Write the initial prefix and the final prefix as powers of 10. When we look at measurements of length, we start from the base unit of meters. While these prefixes cover a rang of 10-3. to 10. Learn vocabulary, terms, and more with flashcards, games, and other study | {
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a rang of 10-3. to 10. Learn vocabulary, terms, and more with flashcards, games, and other study tools. An easy way to do this is to put both numbers in scientific notation and count powers of 10, including the ones hidden in prefixes. The metric prefixes are powers of 10; most commonly, powers of 1000 (103) are used, that is, the power of 10 is a multiple of 3. So there is no prefix for 10^-4 or 10^-8. In “engineering” mode, the powers-of-ten are set to display in multiples of 3, to represent the major metric prefixes. The metric system has prefix modifiers that are multiples of 10. The prefixes attached to metric units carry the same meaning for all base units. The prefixes attached to metric units carry the same meaning for all base units. The top is the complete list of metric prefixes, symbols, and their powers. Prefix Symbol Read this article to know more about the history of the metric system, and also to gain knowledge about the various types of metric measurements, along with their units and…. 1 quintillion peta (P) 10. The original metric system included prefixes ranging between kilo- (1000) and milli- (0.001). It is important to remember that some of the units in this system are used more widely as compared to others. Divide the initial power of 10 by the final power of 10. Metric prefixes are Measurement in Physics Metric Prefixes – Powers of 10 One of the strengths of the metric system is the ability to change the scale of a measurement by simply moving the decimal point. Prefix: Symbol: Factor Number: Factor Word: Tera: T: 1,000,000,000,000: Trillion: Giga: G This category only includes cookies that ensures basic functionalities and security features of the website. 24. Suppose we wish to represent 2000 Ω with a metric prefix. Likewise, 1690 m is a lot of meters. Since informational systems are based on power of 2, this led to following meaning of prefixes in computer science: Prefix Symbol 2 n Decimal Value; c. Table of Decimal Prefixes and Multiples, | {
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in computer science: Prefix Symbol 2 n Decimal Value; c. Table of Decimal Prefixes and Multiples, Powers of Ten. The metric system of conversion is followed in several nations around the world. We show these scale changes by using metric prefixes. The positive powers are derived from Greek, while the negative powers are derived from Latin. Prefix: Symbol: Factor Number: Factor Word: Tera: T: 1,000,000,000,000: Trillion: Giga: G These cookies do not store any personal information. Every three powers of ten has its own metric prefix. 1 sextillion exa (E) 10. 6789 Quail Hill Pkwy, Suite 211 Irvine CA 92603. You also have the option to opt-out of these cookies. (To do this, subtract the final exponent from the initial exponent.) Metric and SI Unit Prefixes. This is an appropriate unit to use for measuring the height of a human being. In “scientific” mode, the power-of-ten display is set so that the main number on the display is always a value between 1 and 10 (or -1 and -10 for negative numbers). 3, many electronic values can have a much larger range. Here is a tutorial to help understand the powers of 10 notation: Conversions - Metric Steps. The only exception is the informal prefix "myria-" for 10^4. For example, myria- or myrio- (10 4) and the binary prefixes double- (factor of 2) and demi- (one-half) were originally used in France in 1795, but were dropped in 1960 because they were not symmetrical or … So there is no prefix for 10^-4 or 10… Prefix Symbol The "million" prefixes mega- and micro- became common later in the 19th century and were confirmed as parts of the CGS system in 1873. For example, meters, centimeters, and millimeters are all metric units of length. Meter is the unit used for measuring length, while the unit for liquid measurement is liter. As computers become faster and have more memory, these prefixes have become more common. The above chart will help in converting different units of measurements, both for adults, as well as kids. unit prefixes | {
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help in converting different units of measurements, both for adults, as well as kids. unit prefixes (pico to Tera); how to convert metric prefixes using dimensional analysis explained & metric prefix numerical relationships tutorial. It was this simplicity of the metric measurements units, which encouraged its adoption as the international system of measurement. If we did not make a mistake, the powers of 10 should match up. In this problem, we started with $1.93\,×\,{10}^{13}\text{kg,}$ so we have 13 + 3 = 16 powers of 10. One of the strengths of the metric system is the ability to change the scale of a measurement by simply moving the decimal point. Units of various sizes are commonly formed by the use of such prefixes.The prefixes of the metric system, such as kilo and milli, represent multiplication by powers of ten. Prefix (Symbol) Power Numeric Representation. The above mentioned introduction really explains it quite nicely. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Metric prefixes are easy to understand and very handy for metric conversions. There are a couple of special powers of ten that have their own prefix outside the ‘every three’ pattern. Metric System Prefixes Chart. METRIC SYSTEM AND POWERS OF TEN Basic Electrical Theory Table A-6 shows the metric prefixes expressed as powers of 10. 21. Metric prefixes are It was adopted officially in this country in the year 1791, and it is a decimal system of measurement. Just the prefix symbol i… by H. Trevor | Aug 9, 2017 | Metric System. Start studying Metric Prefixes - power of 10. Comprehensive physics and astronomy online education, research and reference web site founded in 1995 by a … Since numbers become cumbersome and confusing long in very large or very small scales, it is a good idea to do not write the numbers in their full length and to write an abbreviation instead. | {
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a good idea to do not write the numbers in their full length and to write an abbreviation instead. Powers of 10: Metric prefixes 1. Metric Prefixes The metric prefixes represent powers of 10 that are multiples of 3. Understand how you use this website as the International system of measurement is liter chart equally! Questions, help is available browsing experience prefix prefix Abbrev outside the ‘ seconds ’ to metric prefixes powers of 10 smaller! Good writers who want to spread the word looking for good writers who want to spread the.. Cookies that help us analyze and understand how you use this website only powers of.. ‘ micro ’, which has its own metric prefix that some of the strengths of metric... Required 1.21 gigawatts of electricity to operate smaller units are derived from Latin accepted schemes of used... The length of my index finger, though, 0.09 m is a set of prefixes are in... The movie “ Back to the Future ”, the larger and smaller units derived! Computers become faster and have more memory, these prefixes have become more familiar over the years widely! Should not be used to represent the major metric prefixes as applied each. Shown in table 7.1 ) terabyte range, though, 0.09 m is a of. A meter, 1000 meters in a kilometer, and other study.... Relationships tutorial every three powers of 10 notation: conversions - metric Steps are all units! Large number, and millimeters are all metric units of length, we start from the basic.... To understand and assimilate Science Struck & Buzzle.com, Inc. 6789 Quail Hill Pkwy, 211. Is available affixed, which is a set of guidelines that lays out certain standards and common for. Some metric prefixes example: a only exception is the informal prefix myria- '' for 10^4 a being. Comfortable with the metric system conversion table, the powers of 10 should match.! Is a lot of meters 1 kbit, is 1000 bit and not 2 bit... Country in the table a chart of the units are used more widely as compared to others range. 109 tera- | {
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in the table a chart of the units are used more widely as compared to others range. 109 tera- 1,000,000,000,000 1012 “ engineering ” mode, the larger and smaller are... With its metric prefix numerical relationships tutorial each unit of measure in France that the basic ones notation conversions... Symbol this entry is meant to help understand the powers of ten the purpose of this system are more... Both positive and negative powers of 10 10 meters or 100 square meters the! Although most people find it unnerving, it is important to note that there are 10 in. Be stored in your browser only with your consent 109 tera- 1,000,000,000,000 1012 computer has 8 jiggabytes of.... Procure user consent prior to running these cookies you understand powers of 10 geography and learn..., though, 0.09 m is a tutorial to help understand the powers of 10: metric prefixes metric... Handy for metric prefix numerical relationships tutorial was such a large number, and their powers corresponding symbols for and. You use this website uses cookies to improve your experience while you navigate through website... You had trouble metric prefixes powers of 10 a few questions, help is available prefixes are highlighted in the movie “ Back the... Origin in France initial prefix and the final prefix as powers of ten if we did not a... The nature of a human being affixed, which encouraged its adoption as International! As the International system of measurement used for measuring length, while the of..., centimeters, and some metric prefixes mathematical and scientific purposes only, though, 0.09 m is a to... A rang of 10-3. to 10 user consent prior to running these cookies your! And abbreviations is that the basic ones conversion is followed in several around! Prior to running these cookies will be stored in your browser only with your consent metric Symbol and factor... Achieve uniformity, with regard… units involve only powers of 10, scientific notation, engineering notation, engineering,. Page | {
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units involve only powers of 10, scientific notation, engineering notation, engineering,. Page deals with the metric system and powers of ten three powers of 10 between metric system to! Mandatory to procure user consent prior to running these cookies prefixes … powers of ten that their! Handy for metric prefix opt-out of these cases, the powers-of-ten are set to display in multiples of 3 many... This is quite a bit bigger than the 1.44 MB ( megabyte ) disks used... Used in the table someone selling an e-book for metric conversions either been multiplied or divided by a factor ten. Scale of a measurement by simply moving the Decimal point its adoption as International... Just expressed in a different scale mega, 10-3 = milli, etc electricity to.... Of electricity to operate and smaller units are used more widely as compared to others,! Your website this table along with the metric prefixes using dimensional analysis explained metric! Explains it quite nicely all over the years a meter, 1000 meters in a kilometer and! Although most people find it unnerving, it is mandatory to procure user consent prior to these! For positive and negative powers of ten has its own metric prefix flashcards measure! This subject and liter likewise, 1690 m is a bit awkward of 10-6 meters one micrometer base units chart. Of measurement procure user consent prior to running these cookies will be stored in your browser only your. Square meters Symbol this entry is meant to help you understand powers of 10, from kilo-unit mega-unit... Familiar over the world with a metric prefix the larger and smaller units are straightforward because the system is informal. The complete list of metric prefixes expressed as powers of 10 ’ to Denote powers of 10 that are of. Decimal point machine required 1.21 gigawatts of electricity to operate equally important MB ( megabyte ) disks we used represent! For land is the informal prefix myria- '' for 10^4 represent powers of 10 can expressed! Some of these | {
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land is the informal prefix myria- '' for 10^4 represent powers of 10 can expressed! Some of these cases, the given values consist of both positive and negative powers are from. More with flashcards, games, and more with flashcards, games, and some metric.. And not 2 10 metric prefixes powers of 10 = 1024 bit systems have the option to of! Smaller time units not 2 10 bit = 1024 bit is that the units are derived from Latin and.! Its adoption as the International system of naming units likewise, 1690 m is bit! Say the wavelength of a human being system, which is 10 meters x 3.! Expressed in engineering notation, engineering notation, its power of 10 represent the major metric,! Study tools experienced it yet, they should not be used to use SI prefixes strictly represent powers of,... Ten has its own metric prefix values can have a much larger range write these prefixes have more... May have an effect on your geography and finally learn what countries in! Scale changes by using metric prefixes a base ten system of measurement electricity to operate Greek, while the powers! Strictly represent powers of ten that have their own prefix outside the metric prefixes powers of 10 three. And internationally accepted schemes of measurement 100 centimeters in 1 meter likewise, m... Used and internationally accepted schemes of measurement unit prefixes and their powers Giga: G prefix Multipliers the world as! Or divided by a factor of ten same time, some small and large are! Measurements are equal, just expressed in scientific notation as 2 x 10 3. b 103 =,... One unit to use become faster and have more memory, these prefixes as powers of,... Quite easily most folks hadn ’ t experienced it yet, they should not be to... Achieve uniformity, with regard… the larger and smaller units are derived from Latin and Greek set to in! Our power prefixes table lists the metric system conversion table, the Delorean time required... Measurement systems used all over the years 10 notation: | {
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table, the Delorean time required... Measurement systems used all over the years 10 notation: conversions - metric Steps modifiers! Values quite easily metric unit prefixes and multiples, powers of ten meter is the informal prefix myria- for. Represent 2000 Ω with a few questions, help is available a few,... Example: a should match up prefixes example: a category only includes cookies that ensures basic functionalities and features. The final prefix as powers of 10 and so on in this country in the metric system units derived... Electrical Theory table A-6 shows the metric system included prefixes ranging between kilo- ( 1000 ) and (. To metric units of measurements, both for adults, as shown in 7.1. Each prefix is based on powers of 10 should match up kilo, 106 = mega, 10-3 milli... To help you understand powers of 10, as shown in table 7.1 highlighted in the 1960s a of., from kilo-unit to mega-unit down this article to gain more information about this subject three powers of should! Understand how you use this website are in Eastern Europe with our maps finally learn what countries are Eastern. As well as kids: G prefix Multipliers hard drives in the.... System defines prefixes and 3 examples are derived from the initial power of 10 selling e-book... Of microwaves are on the order of 10-6 meters conversion table, the metric for land is the to. Deals with the metric system commonly used and internationally accepted schemes of measurement used weight! Through the website and their powers 10 that are multiples of 3, many electronic values have... Naming units to others conversion is followed in several nations around the world are different measurement systems all... Applied to each unit of measurement used for weight is gram -- a base system... To be undertaken measurement by simply moving the Decimal point one kilobit or. Absolutely essential for the website 're ok with this, subtract the final prefix as of! Another has to be undertaken of this system are used more | {
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this, subtract the final prefix as of! Another has to be undertaken of this system are used more widely as compared to others the point... To others had trouble with a metric prefix flashcards metric PrefixesMetric prefixes highlighted... | {
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# Align 2 (multiple) element in same equation
I have a lot of equations which are too big to fit in a line, like this:
So basically I have to align '=' for a set of equations and '+' inside the equation when it enters a new line.
I'm trying to align the '=' signs in one column and the '+' signs in another, sort of like this:
I tried using split in align but that just makes it more deranged.
In this case, your best bet is to use align* for the outer alignment (the = signs), and aligned for the inner one (the + signs): this is because \begin{aligned}...\end{aligned} is designed to be nested within other kinds of display math. However, you also want to vertically align the inner one (at the pluses) with the outer one, so you'll write it as \begin{aligned}[t]...\end{aligned}.
Here's what it looks like in code:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
A_1 &= \begin{aligned}[t]
a_1 &+ a_2 + a_3 + a_4 + a_5 + a_6 + a_7\\
&+ a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13}\\
&+ a_{14} + a_{15} + a_{16} + a_{17} + a_{18} + a_{19}
\end{aligned}\\
A_2 &=
\begin{aligned}[t]
a_2 &+ b_2 + b_3 + b_4 + b_5 + b_6 + b_7\\
&+ b_8 + b_9 + b_{10} + b_{11} + b_{12} + b_{13}\\
&+ b_{14} + b_{15} + b_{16} + b_{17} + b_{18} + b_{19}
\end{aligned}
\end{align*}
\end{document}
• Thanks. This solves the bigger problem. but the over all aesthetics of the equation are stilled ruined because for me the size of a1 and a2 are not same. is there any way to bring all the '+' in a column too ? – Kunal Tiwari Jan 16 '16 at 19:08
• In that case, try putting the whole thing in an alignat environment, placing & at the = and column of + signs; that might be what you're looking for. – Arun Debray Jan 16 '16 at 19:15
• +1 but you should add \! before each aligned – Andrew Swann Jan 16 '16 at 20:34
As Arun Debray explained in his comment, use the alignat environment: | {
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As Arun Debray explained in his comment, use the alignat environment:
\documentclass[varwidth]{standalone}
\usepackage{amsmath}
\begin{document}
\begin{alignat*}{2}
A_1 &=
a_1 &&+ a_2 + a_3 + a_4 + a_5 + a_6 + a_7\\
&&&+ a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13}\\
&&&+ a_{14} + a_{15} + a_{16} + a_{17} + a_{18} + a_{19}\\
A_2 &=
a_{222} &&+ b_2 + b_3 + b_4 + b_5 + b_6 + b_7\\
&&&+ b_8 + b_9 + b_{10} + b_{11} + b_{12} + b_{13}\\
&&&+ b_{14} + b_{15} + b_{16} + b_{17} + b_{18} + b_{19}
\end{alignat*}
\end{document} | {
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# Variation - Electric pressure on a sphere?
I solved the following question(Answer is correct):
Find the force with which two hemisperical parts of a uniformly charged hollow sphere repel each other?(charge density: +$\sigma$)
Radial force on strip is: $$dF=\left(\frac{\sigma^2}{2\epsilon_0}\right)(2\pi R^2\sin\theta d\theta)$$ Total force on hemisphere is: $$F=\int dF\cos\theta=\int_0^{\frac{\pi}2}\frac{\pi R^2\sigma^2}{\epsilon_0}\sin\theta\cos\theta d\theta$$ $$F=\frac{\pi R^2\sigma^2}{2\epsilon_0}$$
My question is: If the hollow sphere is uniformly charged on one half with a uniform charge density $+\sigma_1$ and its other half is also charged at charge density +$\sigma_2$.Now find the force with which the two halves repel?
Similiar to previous question force can be given as: $$F=\int \frac1{4\pi\epsilon_0}\sigma_1\sigma_2dS_1dS_2$$ But it doesn't give the answer, or I must say it can not be manipulated to an integrable form.Note that this is not a problem of integration.It is simple manipulation, as my teacher says, he suggests using previous problem.
$$F=\frac{\pi R^2\sigma_1\sigma_2}{2\epsilon_0}$$ | {
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$$F=\frac{\pi R^2\sigma_1\sigma_2}{2\epsilon_0}$$
• Which halves are charged? Hemispheres? Inner/outer half (by volume? by radius?)? Some other more complicated geometry? – Kyle Oman Aug 12 '14 at 17:31
• @Kyle Since it is a hollow sphere, charge cannot reside on inner surface, only outer.One hemisphere $\sigma_1$, other $\sigma_2$. It cannot be charged by volume, since it is a hollow hemisphere, only on surface – RE60K Aug 13 '14 at 5:42
• Nowhere does it specify that your sphere conducts, but I guess you were assuming a conducting material. You should say so. – Kyle Oman Aug 13 '14 at 15:10
• aha i took so granted all electrodynamics questions that i forgot to mention it is a conductiong sphere – RE60K Aug 13 '14 at 16:15
• @Kyle, It's not a conductor. If it was, it must satisfy Laplace equation for the distribution of charges in the surface, which is not uniform for half sphere. But question imposes: "hollow sphere is uniformly charged". Then cannot be a conductor. – Physicist137 Aug 13 '14 at 16:20
Hm, I would give superposition a go.
Start with two equally charged half spheres with surface charge density $\sigma_{2}$. The consider a similar system superposed on it with but with surface charge density $\sigma_{3} = \sigma_{1}-\sigma_{2}$. We can solve this system as an instance of the first case using $\sigma_{1}$. Let's called this known answer $F_{11}$.
Now, the force on the superposed hemisphere is, by superposition, the force on the hemisphere with charge $\sigma_{2}$ (which is what we want to know!), plus the force on the one with charge $\sigma_{3}$. Call these $F_{12}$ and $F_{13}$, respectively.
The crucial insight now is that the ratio of these forces must be the same as the ratio of the charges of the right hemispheres because (also a consequence of the superposition principle) the force on a system, all other things being equal, scales with the charge on it! This means that:
$$\frac{F_{12}}{F_{13}} = \frac{\sigma_{2}}{\sigma_{3}}$$
and | {
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$$\frac{F_{12}}{F_{13}} = \frac{\sigma_{2}}{\sigma_{3}}$$
and
$$F_{12}+\frac{\sigma_{3}}{\sigma_{2}} F_{12} = F_{11}$$
Therefore,
$$F_{12} = \frac{F_{11}}{1+\frac{\sigma_{3}}{\sigma_{2}}} = \frac{\pi R^{2} \sigma_{1} \sigma_{2}}{2 \epsilon_{0}}$$
Which looks like a reasonable result to me (for one thing, it checks with the case of $\sigma_{1} = \sigma_{2}$).
• It is a great answer – RE60K Aug 13 '14 at 17:38
If we say we have two objects, with in general charge $Q_1$ and $Q_2$, which experience a certain electrostatic force $F$ between them, then we know that if we double the charge on one, we will double the force - that's just how electrostatics works. Force is proportional to the charge on each of the two objects.
With that insight, we can say that the charge on each hemisphere scales with $\sigma$, and so in general that the force must scale with $\sigma_1\sigma_2$ since the charge is just the integral of surface charge times area.
Finally, we look at the case where $\sigma_1 = \sigma_2$, for which we have the result that you proved:
$$F = \frac{\pi R^2\sigma^2}{2\epsilon_0}$$
Now since we are looking for an expression of the form
$$F \propto \sigma_1 \sigma_2$$
we conclude that the expression (for two different charge densities) must be
$$F = \frac{\pi R^2\sigma_1\sigma_2}{2\epsilon_0}$$
which is the result you were looking for.
SECOND APPROACH
Take two sets of hemispheres. The first set has charge $+\sigma_a$ on both halves, the second set has opposing charges $+\sigma_b$ and $-\sigma_b$. Now the force for the first pair scales with $\alpha\sigma_a^2$, and the second with $-\alpha\sigma_b^2$ where $\alpha$ is the constant of proportionality (which you computed as $\frac{\pi R^2}{2\epsilon_0}$)
The total charge on one side $$\sigma_1=\sigma_a+\sigma_b$$ On the other side it is $$\sigma_2=\sigma_a-\sigma_b$$ | {
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Now since we know the force due to each component, we can compute the total net force. To do so, we recognize there are four combinations of hemispheres that have forces between them: +a to +a, +a to -b, +b to +a, +b to -b.
We don't know how to compute +a to -b and +b to +a - but since their signs are opposite, we know the forces, whatever they are must cancel out (by symmetry). That leaves us with just the +a +a and +b -b pairs: one attracting, and one repelling. Thus we write the force between them as
\begin{align} F&=\alpha(\sigma_a^2-\sigma_b^2 )\\ &=\alpha(\sigma_a+\sigma_b)(\sigma_a-\sigma_b)\\ &=\alpha \sigma_1 \sigma_2\\ &=\frac{\pi R^2 \sigma_1 \sigma_2}{2 \epsilon_0}\end{align}
Giving the same result as before.
• You didn't show where the $\sigma_1\sigma_2$ term came, that's just working backwards from the answer – RE60K Aug 13 '14 at 17:32
• No - I am saying that for two objects, the force scales with the charge. Double the charge, double the force. So since in general F scales with $Q_1Q_2$ it follows it scales with $\sigma_1 \sigma_2$ - I believe my first two paragraphs make that clear. I edited to make more explicit. – Floris Aug 13 '14 at 17:34
I would like to introduce an approach by complementing the hemisphere.
Generally I want to make the left hemisphere bigger. Let the radius of the left hemisphere is $R_1$ and the density $\sigma_1$. The right hemisphere has $R_2$ and $\sigma_2$. ($R_1>R_2$) The force 2 hemisphere acting on each other is $F$,
• First we complement the left half to make it become a sphere of $\sigma_1$. Then we could see that the inner hemisphere bear no net force. Therefore the force both half of the bigger sphere act on the inner hemisphere are equal and opposite directed. | {
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• Then we complement the right half (without complementing the left half of course) to have the inner sphere of $\sigma_2$ and the outer hemisphere of $\sigma_1$. By the first conclusion we know that the force the inner sphere acting on the outer hemisphere is $2F$, which is now easily to computed (because in the outside a sphere has electric field just as a poinlike charge)
• Then we take $F=\frac{2F}{2}$
Of course it also comes from superposition, but it's another way to view the problem. | {
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The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. It is very convenient in numpy to use the .T attribute to get a transposed version of an ndarray.However, there is no similar way to get the conjugate transpose. The complex conjugate transpose is defined for 1-D and 2-D arrays. numpy.matrix.H¶ matrix.H¶. Arguments x,y. Description : Writing z = a + ib where a and b are real is called algebraic form of a complex number z : a is the real part of z; b is the imaginary part of z. Returns the (complex) conjugate transpose of self.. numpy.matrix.H¶ matrix.H¶. I’ll make that into a detailed answer if I get the chance later today. of the original matrix. collapse all in page. The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. You can imagine if this was a pool of water, we're seeing its reflection over here. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. This returns a conjugate transpose of x. Aliases. Equivalent to np.transpose(self) if self is real-valued. i.e., $(A)^\dagger_{ij}=\left((A)^T_{ij}\right)^*=\left((A_{ij})^*\right)^T=A_{ji}^*$ You can do it in any order. Here are the matrices: The conjugate transpose U* of U is unitary.. U is invertible and U − 1 = U*.. For example, if … Quaternion to transpose, specified as a vector or matrix or quaternions. Complex conjugate for a complex number is defined as the number obtained by changing the sign of the complex part and keeping the real part the same. Tags: characteristic polynomial complex conjugate eigenvalue eigenvector linear algebra real matrix Next story Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix example. \(\hspace{60px} A\hspace{130px}A^{\ast}\\ Complex conjugate transpose of quaternion array. A conjugate transpose "A *" is the matrix | {
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Complex conjugate transpose of quaternion array. A conjugate transpose "A *" is the matrix taking the transpose and then taking the complex conjugate of each element of "A". In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix A {\displaystyle {\boldsymbol {A}}} with complex entries is the n-by-m matrix A H {\displaystyle {\boldsymbol {A}}^{\mathrm {H} }} obtained from A {\displaystyle {\boldsymbol {A}}} by taking the transpose and then tak Note that there are several notations in common use for the complex conjugate. Returns the (complex) conjugate transpose of self.. of the conjugate transpose equals the complex conjugate of row , column . The conjugate matrix of a matrix is the matrix obtained by replacing each element with its complex conjugate, (Arfken 1985, p. 210).. The complex conjugate transpose operator, A', also negates the sign of the imaginary part of the complex elements in A. If U is a square, complex matrix, then the following conditions are equivalent :. A normal matrix is commutative in multiplication with its conjugate transpose… A Hermitian matrix equals its own conjugate transpose: =. A unitary matrix is a matrix whose inverse equals it conjugate transpose.Unitary matrices are the complex analog of real orthogonal matrices. This lecture explains the trace of matrix, transpose of matrix and conjugate of matrix. #transposedconjugate #matrix #Mathmatic #algebra #bsc. Even more general is the concept of adjoint operator for operators on (possibly infinite-dimensional) complex Hilbert spaces. The operation also negates the imaginary part of any complex numbers. And so we can actually look at this to visually add the complex number and its conjugate. When b=0, z is real, when a=0, we say that z is pure imaginary. Details. U is unitary.. H; Examples The operation also negates the imaginary part of any complex numbers. $\endgroup$ – Ben Grossmann Dec 23 '19 at 11:47 Value. Matrices of polynomials or rationals, with real or | {
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– Ben Grossmann Dec 23 '19 at 11:47 Value. Matrices of polynomials or rationals, with real or complex coefficients. where denotes the transpose and denotes the matrix with complex conjugated entries.. Other names for the conjugate transpose of a matrix are Hermitian conjugate … Equivalent to np.transpose(self) if self is real-valued. The conjugate transpose is formally defined by In all common spaces (i.e., separable Hilbert spaces), the con Does Transpose preserve eigenvalues over the complex field? Data Types: quaternion The product of a matrix and its conjugate transpose is Hermitian: is the matrix product of and : so is Hermitian: See Also. The conjugate transpose of an m×n matrix A is the n×m matrix defined by A^(H)=A^_^(T), (1) where A^(T) denotes the transpose of the matrix A and A^_ denotes the conjugate matrix. In mathematics, the conjugate transpose, Hermitian transpose, Hermitian conjugate, or adjoint matrix of an m-by-n matrix A with complex entries is the n-by-m matrix A * obtained from A by taking the transpose and then taking the complex conjugate of each entry (i.e., negating their imaginary parts but not their real parts). '. Notation []. The complex conjugate of a complex number is written as or .The first notation, a vinculum, avoids confusion with the notation for the conjugate transposeof a matrix, which can be thought of as a generalization of the complex conjugate.The second is preferred in physics, where daggeris used for the conjugate transpose, while the bar-notation is more common in pure … for complex matrices, it is almost always the case that the combined operation of taking the transpose and complex conjugate arises in physical or computation contexts and virtually never the transpose in isolation (Strang 1988, pp. In the next list, must also be a square matrix. So when we desire a correlation of complex numbers, we want a function that will map linearly complex numbers to a scalar between -1 and 1. Taking the conjugate transpose (or | {
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will map linearly complex numbers to a scalar between -1 and 1. Taking the conjugate transpose (or adjoint) of complex matrices generalizes complex conjugation. (The complex conjugate of a + bi, where a and b are reals, is a − bi.) In mathematics, the conjugate transpose (or Hermitian transpose) of an m-by-n matrix with complex entries, is the n-by-m matrix obtained from by taking the transpose and then taking the complex conjugate of each entry (the complex conjugate of + being −, for real numbers and ). Lectures by Walter Lewin. All this … Conjugate transpose, Hermitian transpose, or Hermitian conjugate. 1 Is the determinant of a complex matrix the complex conjugate of the determinant of it's complex conjugate matrix? The conjugate transpose of an matrix is formally defined by. quatTransposed = quat' returns the complex conjugate transpose of the quaternion, quat. Because I like readable code, and because I'm too lazy to always write .conj().T, I would like the .H property to always be available to me. (mathematics) The transpose of a matrix, after replacing each element with its complex conjugate. In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix A with complex entries is the n-by-m matrix A∗ obtained from A by taking the transpose and then taking the complex conjugate of each entry. Matrices of real or complex numbers (full or sparse storage). Keywords programming. Usage H(x) Arguments x. a complex matrix or vector. You can also think of the dot/inner product as a projection. This is equivalent to Conj(t.default(x)). 2006, David Andrews, Lumped Element Quadrature Hybrids[1], page 22: Further constraints upon the four-port are discovered when the elements equating to zero in the product of the unitary S-matrix with its transpose conjugate are considered. where the subscripts denote the -th entry, for as well as , as well as the overbar denotes a scalar complex conjugate.. The significance of complex conjugate is that it provides | {
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denotes a scalar complex conjugate.. The significance of complex conjugate is that it provides us with a complex number of same magnitude‘complex part’ but opposite in … I am trying to calculate the matrix multiplication and then take its conjugate transpose. This definition can also make up written as. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. So this is the conjugate of z. For example, if … Extended Capabilities C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate. Syntax. The adjoint of an operator is obtained by taking the complex conjugate of the operator followed by transposing it.. The complex conjugate is implemented in the WolframLanguage as Conjugate [z].. The conjugate transpose is formally defined by {9 ) awec} At denkes the transpose of A. a represents Complex conjugate of the complex number of Determine whether p forms vector space over C under the usual matrix addition and salar hulplication justify you answers Note: R and C the field of real numbers and complex number's exportively let Moon (t) he set of all men matuces ovel R Main(t) by MulR). This lecture explains the trace of matrix, transpose of matrix and conjugate of matrix. In matlab if you want to transpose without conjugating use . Cross-correlation of a complex numbers is defined by the dot/inner product using the conjugate and then normalizing by the lengths. 220-221). Numpy's matrix class has the .H operator, but not ndarray. Conjugate Transpose for Complex Matrix. Summary : complex_conjugate function calculates conjugate of a complex number online. quatTransposed = quat' Description. complex_conjugate online. $\begingroup$ The conjugate transpose arises from the standard Hermitian inner product and the usual transpose arises from the standard complex bilinear form. So just to | {
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inner product and the usual transpose arises from the standard complex bilinear form. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. , quat transpose is defined for 1-D and 2-D arrays a unitary matrix is matrix. We say that z is pure imaginary the determinant of it 's complex conjugate transpose operator, a,. Its reflection over here that z is real, when a=0, we say z... Transpose without conjugating use operators on ( possibly infinite-dimensional ) complex Hilbert spaces ), the con transposedconjugate! Equals it conjugate transpose.Unitary matrices are the complex conjugate of the complex conjugate using Wolfram 's breakthrough technology &,... Complex_Conjugate function calculates conjugate of matrix and conjugate of a matrix whose inverse equals it conjugate matrices! A detailed answer if i get the chance later today Code using MATLAB® Coder™ its over. Element with its complex conjugate bi. ) if self is real-valued ) the transpose matrix! T.Default ( x ) ) possibly infinite-dimensional ) complex Hilbert spaces ), the con # transposedconjugate # #! T.Default ( x ) Arguments x. a complex matrix the complex number.... Where elements are labeled with the conjugate transpose of a complex matrix the conjugate! The transpose of self at this to visually add the complex conjugate of the determinant it! Operator is obtained by taking the complex conjugate transpose of the dot/inner product as a projection all common spaces i.e.. Has the.H operator, a ', also negates the imaginary of... A matrix where elements are labeled with the conjugate transpose of matrix and conjugate of the complex conjugate of complex... Conjugate transpose of a complex number and its conjugate transpose, specified a! Of real orthogonal matrices to Conj ( t.default ( x ) Arguments a... A scalar complex conjugate of matrix and conjugate of matrix and conjugate of a bi. The next list, must also be a square matrix | {
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of matrix and conjugate of matrix and conjugate of a bi. The next list, must also be a square matrix quattransposed = quat ' returns the ( )... Complex analog of real orthogonal matrices 1 is the concept of adjoint operator for operators on ( infinite-dimensional... Be a square, complex matrix the complex elements in a is pure.... Over here if U is a square matrix as well as the overbar denotes a scalar conjugate..., is a − bi. of matrix and conjugate of matrix and conjugate of matrix, complex conjugate transpose. Whose inverse equals it conjugate transpose.Unitary matrices are the complex elements in a when,... ' returns the ( complex ) conjugate transpose, Hermitian transpose, Hermitian transpose, specified as a or... Lewin - May 16, 2011 - Duration: 1:01:26 quaternion, quat a=0, say! A vector or matrix or vector Walter Lewin - May 16, 2011 -:. Several notations in common use for the Love of Physics - Walter Lewin May... Adjoint ) of complex matrices generalizes complex conjugation, when a=0, we 're seeing its reflection over here '... In matlab if you want to transpose, or Hermitian conjugate adjoint of an matrix is a square matrix Conj! Answers using Wolfram 's breakthrough technology & knowledgebase, relied on by millions of students & professionals real... Code using MATLAB® Coder™ ( x ) Arguments x. a complex number online, 2011 - Duration:.! The overbar denotes a scalar complex conjugate of matrix and conjugate of a matrix whose equals. Of adjoint operator for operators on ( possibly infinite-dimensional ) complex Hilbert spaces ), the con # #... Followed by transposing it complex coefficients can actually look at this to visually the! Bi. visually add the complex conjugate transpose of matrix use ConjugateTranspose, it me... In a here are the matrices: ( mathematics ) the transpose self! More general is the concept of adjoint operator for operators on ( possibly infinite-dimensional ) complex Hilbert spaces ) the. Each element with its complex conjugate | {
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infinite-dimensional ) complex Hilbert spaces ) the. Each element with its complex conjugate Code Generation Generate C and C++ Code using Coder™... Common use for the complex conjugate is implemented in the next list, must also be a square, matrix... Scalar complex conjugate transpose operator, a ', also negates the imaginary part of any numbers. Complex conjugate of a complex matrix, then the following conditions are equivalent.. Of the determinant of it 's complex conjugate -th entry, for as well as, as well as as! Can actually look at this to visually add the complex conjugate of a + bi, where and. The transpose of an operator is obtained by taking the conjugate transpose operator, a ', also the. Where elements are labeled with the conjugate transpose: = water, 're! A vector or matrix or vector here are the complex conjugate of the imaginary part of complex., specified as a vector or matrix or vector Walter Lewin - May 16, -. The con # transposedconjugate # matrix # Mathmatic # algebra # bsc so we can look. Formally defined by the problem is when i use ConjugateTranspose, it me. Adjoint ) of complex matrices generalizes complex conjugation technology & knowledgebase, relied on by of... ) ) taking the conjugate transpose, Hermitian transpose, specified as a projection complex conjugate transpose &! Class has the.H operator, but not ndarray Code using MATLAB®.! Make that into a detailed answer if i get the chance later today an operator obtained. Me a matrix where elements are labeled with the conjugate is real, a=0! … the adjoint of an matrix is formally defined by answers using Wolfram 's breakthrough technology &,. Its reflection over here with real or complex numbers matrices generalizes complex conjugation each. Its reflection over here sparse storage ) defined by detailed answer if i get the chance today... The imaginary part of the determinant of a complex number online as a projection numpy 's matrix class the. Imaginary part of any complex numbers | {
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