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# Count the number of trend changes in a vector.
23 views (last 30 days)
Sebastian Holmqvist on 12 Jul 2012
I need to count the number of times a vector increases/decreases. E.g
vec = [0 0 1 2 4 4 2 0 1 2 3]
diff(vec)
ans =
[0 1 1 2 0 -2 -2 1 1 1]
would evidently yield 4 trend changes. I don't wish to count the zero as a trend.
Update:
If you were to plot(vec) it's easier to see what I mean. I wish to simply count the number of row changes in the plot. A flat line (derivative of 0) isn't considered a row change in my application.
Take this as an example:
vec = [3 3 3 5 5 4 3 2 4 5 5 5];
plot(vec)
As you can see from the plot, it's 4 row changes (even though it's 8 in pure diff).
Update:
Ok, I'll try to do better.
Imagine a matrix that describes position over time, where row is position and column is time. When going in a straight line (along a single row) I wish to do nothing. But when changing course over to another row-line I wish to count each new "course" as an occurrence.
E.g
• Travelling at row 5 and then moving up to row 6 and then going straight, that's one course change.
• [5 5 5 6 6 6] => 1 "course change"
• Travelling at row 5 and then moving up to row 7 is similarly one change.
• [5 5 5 7 7 7] => 1 "course change"
• Travelling at row 5, moving up to 7 and then directly back to row 6 is then two changes.
• [5 5 5 7 6 6 6] => 2 "course changes"
• Travelling at row 5, moving up to 7 and then to 12 and then going straight is two changes as the row increment is 2 and then 5 (not a straight course).
• [5 5 5 7 12 12 12] => 2 "course changes"
So basically I wish to count each time you change position (or row) in a straight course.
Sebastian Holmqvist on 30 Jul 2012
I see your point there. I edited accordingly.
Andrei Bobrov on 16 Jul 2012
Edited: Andrei Bobrov on 16 Jul 2012
EDIT
a = diff(vec(:));
out = nnz(unique(cumsum([true;diff(a)~=0]).*(a~=0)));
Sebastian Holmqvist on 17 Jul 2012
Great explanation! Thanks!
Doug Hull on 12 Jul 2012
You would need to remove the zeros from the first diff vector and diff it again, then count number of non-zeros. The only complication is the case where the first diff vector looks like:
[1 1 0 1 1]
Once the zero is removed, it looks constant. That can be dealt with, but would be messy.
Sebastian Holmqvist on 12 Jul 2012
Good idea. But it doesn't work with this example:
>> vec = [3 3 3 5 5 4 3 2 4 5 5 5];
>> d_vec = diff(vec)
d_vec =
0 0 2 0 -1 -1 -1 2 1 0 0
>> d_vec(d_vec==0) = []
d_vec =
2 -1 -1 -1 2 1
>> diff(d_vec)
ans =
-3 0 0 3 -1
>> sum(diff(d_vec)~=0)
ans =
3
Which would be 3 changes with your argumentation. But with my (somewhat fuzzy) specification there should be 4. plot(vec) should make it more clear.
Sebastian Holmqvist on 16 Jul 2012
I think I've cracked it. My logic follows from;
1. I wish to count the number of first derivative changes.
2. I wish to exclude derivatives == 0.
3. Derivative element trends are to be treated as single elements.
First off, I wrote a small function to remove all local dupes.
function vec = rmDupes(vec)
% Remove all duplicate subsequential elements in a vector
% Iterate backwards to avoid changing indexing
for i = length(vec):-1:2
if vec(i) == vec(i-1)
vec(i) = [];
end
end
By not removing the zero elements before removing the dupes I don't get the case where [1 2 3 3 4 5] counts as one.
vec = [1 1 2 2 3 4 1 1]
d_vec = diff(vec)
d_vec = rmDupes(d_vec)
sum(d_vec ~= 0)
ans =
3
If anyone can detect any flaws in my solution, or has a better suggestion, I would be more than grateful if you would give me some feedback! | HuggingFaceTB/finemath | |
# Inverse Trigonometric Functions
## Trigonometry Inverse
We already know about inverse operations. As we know addition and subtraction are inverse operations, and multiplication and division are inverse operations. Each operation has the opposite of its inverse. Similarly, inverse functions of the basic trigonometric functions are said to be inverse trigonometric functions. It also termed as arcus functions, anti trigonometric functions or cyclometric functions.
The inverse of g is denoted by ‘g -1’.
Let y = f(y) = sin x, then its inverse is y = sin-1x.
In this article let us study the inverse of trigonometric functions like sine, cosine, tangent, cotangent, secant, and cosecant functions.
### Introduction to Inverse Trig Function
The inverse trigonometric functions actually perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We know that trig functions are especially applicable to the right angle triangle. These six important functions are used to find the angle measure in a right triangle when two sides of the triangle measures are known.
The convention symbol to represent the inverse trigonometric function using arc-prefix like arcsin(x), arccos(x), arctan(x), arccsc(x), arcsec(x), arccot(x).
• sin-1x, cos-1x, tan-1x etc. denote angles or real numbers whose sine is x, cosine is x and tangent is x, provided that the answers given are numerically smallest available. These are also termed as arcsin x, arccosine x etc.
• If there are two angles, one positive and the other negative having the same numerical value, then a positive angle should be taken.
• Principal values, domains of inverse circular functions and range of inverse trig functions:
S. No. Function Domain Range 1. y = sin-1x -1 ≤ x ≤ 1 -π/2 ≤ y ≤ π/2 2. y = cos-1x -1 ≤ x ≤ 1 0 ≤ y ≤ π 3. y = tan-1x x ∈ R -π/2 < x < π/2 4. y = cot-1x x ∈ R 0 < y < π 5. y = cosec-1x x ≤ -1 or x ≥ 1 -π/2 ≤ y ≤ π/2, y ≠ 0 6. y = sec-1x x ≤ -1 or x ≥ 1 0 ≤ y ≤ π, y ≠ π/2
Inverse Trigonometric Functions Graphs
There are particularly six inverse trig functions for each trigonometric ratio. The inverse of six important trigonometric functions are:
• Arcsine
• Arccosine
• Arctangent
• Arccotangent
• Arcsecant
• Arccosecant
Graphs of all Inverse Circular Functions
1.Arcsine
y = sin-1x, |x| ≤ 1, y ∈ [-π/2, π/2]
• sin-1x is bounded in [-π/2, π/2].
• sin-1x is an increasing function.
• In its domain, sin-1x attains its maximum value π/2 at x = 1 while its minimum value is -π/2 which occurs at x = -1.
2.Arccosine
y = cos-1x, |x| ≤ 1, y ∈ [0, π]
• cos-1x is bounded in [0, π].
• cos-1x is a decreasing function.
• In its domain, cos-1x attains its maximum value π at x = -1 while its minimum value is 0 which occurs at x = 1.
3.Arctangent
y = tan-1x,where x ∈ R, y ∈ (- $\frac{π}{2}$, $\frac{π}{2}$)
• tan-1x is bounded in (-π/2, π/2).
• tan-1x is an increasing function.
4.Arccotangent
y = cot-1x where x ∈ R, y ∈ (0, π)
• cot-1x is bounded in (0, π).
• cot-1x is a decreasing function.
5.Arccosececant
y = cosec-1x, |x| ≥ 1, y ∈ [-π/2, 0) ∪ (0, π/2].
• cosec-1x is bounded in [-π/2, π/2].
• cosec-1x is a decreasing function.
• In its domain, cosec-1x attains its maximum value π/2 at x = 1 while its minimum value is -π/2 which occurs at x = -1.
6. Arcsecant
y = sec-1x, |x| ≥ 1, y ∈ [0, π/2) ∪ (π/2, π].
1. sec-1x is bounded in [0, π].
2. sec-1x is an increasing function.
3. In its domain,sec-1x attains its maximum value π at x = -1 while its minimum value is 0 which occurs at x = 1.
Here is the list of all the inverse trig functions with their notation, definition, domain and range of inverse trig functions
## Inverse Trigonometric Functions Table
Function Name Notation Definition Domain of x Range Arcsine or inverse sine y = sin-1(x) x=sin y −1 ≤ x ≤ 1 − π/2 ≤ y ≤ π/2 -90°≤ y ≤ 90° Arccosine or inverse cosine y=cos-1(x) x=cos y −1 ≤ x ≤ 1 0 ≤ y ≤ π 0° ≤ y ≤ 180° Arctangent or Inverse tangent y=tan-1(x) x=tan y For all real numbers − π/2 < y < π/2 -90°< y < 90° Arccotangent or Inverse Cot y=cot-1(x) x=cot y For all real numbers 0 < y < π 0° < y < 180° Arcsecant or Inverse Secant y = sec-1(x) x=sec y x ≤ −1 or 1 ≤ x 0≤y<π/2 or π/2
The derivatives of inverse trig functions are first-order derivatives. Let us check here the derivatives of all the six inverse functions.
## Inverse Trigonometric Functions Derivatives
Inverse Trig Function dy/dx sin-1(x) $\frac{1}{\sqrt{(1 – x^{2})}}$ cos-1(x) $\frac{-1}{\sqrt{(1 – x^{2})}}$ tan-1(x) $\frac{1}{\sqrt{(1 + x^{2})}}$ cot-1(x) $\frac{-1}{\sqrt{(1 + x^{2})}}$ sec-1(x) $\frac{1}{[|x| \sqrt{(x^{2} – 1)]}}$ cosec-1(x) $\frac{-1}{[|x| \sqrt{(x^{2} – 1)]}}$
### Solved Examples
Example 1: Find the value of tan-1(tan 9π/ 8 )
Solution:
tan-1(tan9π/8)
= tan-1tan ( π + π/8)
= tan-1 (tan(π/ 8))
=π/ 8
Example 2: Find sin (cos-13/5).
Solution:
Suppose that, cos-1 3/5 = x
So, cos x = 3/5
We know, sin x = $\sqrt{1 – cos2x}$
So, sin x = $\sqrt{1 – 9/25}$ = 4/5
This implies, sin x = sin (cos-1 3/5) = 4/5FAQs (Frequently Asked Questions)
1. What are the six trigonometric functions?
Trigonometry means the science of measuring triangles.Trigonometric functions can be simply defined as the functions of an angle of a triangle i.e. the relationship between the angles and sides of a triangle are given by these trig functions.
The six main trigonometric functions are as follows:
• Sine (sin)
• Cosine (cos)
• Tangent (tan)
• Secant (sec)
• Cosecant (csc)
• Cotangent (cot)
These functions are used to relate the angles of a triangle with the sides of that triangle where the triangle is the right angled triangle.. Trigonometric functions are important when studying triangles. To define these functions for the angle theta, begin with a right triangle. Each function relates the angle to two sides of a right triangle. | HuggingFaceTB/finemath | |
Published March 12, 2018 by
# Nth Even Fibonacci Number
Nth Even Fibonacci Number
I this post, we'll learn about how to calculate Nth Even Fibonacci Number.
The Fibonacci numbers are the numbers in the following integer sequence, called the Fibonacci sequence, and characterized by the fact that every number after the first two is the sum of the two preceding ones. In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation,
`````` Fn = Fn-1 + Fn-2 with seed values
F0 = 0 and F1 = 1.``````
The even number Fibonacci sequence is : 0, 2, 8, 34, 144, 610, 2584…. We have to find nth number in this sequence.
``The formula of even Fibonacci number = ((4*evenFib(n-1)) + evenFib(n-2));``
The Below Code shows how to calculate Nth Even Fibonacci Number in Java.
p.p1 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco} p.p2 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco; min-height: 15.0px} p.p3 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco; color: #4e9072} p.p4 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco; color: #7e504f} span.s1 {color: #931a68} span.s2 {color: #7e504f} span.s3 {color: #000000} span.s4 {color: #0326cc} span.s5 {text-decoration: underline} span.Apple-tab-span {white-space:pre} public class NthEvenFibonacciNumber { p.p1 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco} p.p2 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco; min-height: 15.0px} p.p3 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco; color: #4e9072} p.p4 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Monaco; color: #7e504f} span.s1 {color: #931a68} span.s2 {color: #7e504f} span.s3 {color: #000000} span.s4 {text-decoration: underline} span.s5 {color: #0326cc} span.Apple-tab-span {white-space:pre} public static void main(String[] args) { // Here we are passing nth Number int nthNumber = 3; //Calling the getEvenNumber Method And Printing the Nth Even Fibonacci Number number System.out.println(getEvenNumber(nthNumber)); } public static int getEvenNumber(int nthNumber) { if(nthNumber < 1) { return nthNumber; } else if(nthNumber == 1) { return 2; } else { //Formula to calculate Nth Even Fibonacci Number Fn = 4*(Fn-1) + Fn-2 return ((4 * getEvenNumber(nthNumber - 1)) + getEvenNumber(nthNumber - 2)); } } }
Output :- 34
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# How power produced by a solar cell is affected by its distance from a light source.
How power produced by a solar cell is affected by its distance from a light source.
Theory
I expect that as the solar cells are moved closer the light source the higher the power they will produce.
As light moves away from the source it spreads out at all angles the further you are from the source the less intense the light will be.
P = I V
Power = current x volts
P = FA
Power = flux x area
Flux = P/A Area of 4 solar cells that I will use is 0.0077m2
This means that the higher the flux the higher the power as we are not changing the area of the solar cells as this remains is constant. So as flux increases so will the power and this will happen as the cells are moved closer to the light.
Diagram
Using what I know about space and the way light intensity changes the relation I expect is. Power = 1/distance2
Plan
To measure power by a solar cell and distance light source is away form it, then to plot the distance against power to get a strong inverse relationship.
Then to plot a graph of power against 1/distance this should then give a straight line and hopefully proportionality that is what I wish to prove. To prove that power produce by a solar cell is reciprocal of distance.
I placed a light on a table pointed at an array of solar cells mounted on a stand fixed with a clamp. I will move the solar cell closer to the light source and take measurements of the volts and amps that the array produces so I can work out how much power is produced at different distances.
• Input variable is the distance the solar cells are from the light source.
• Output variable is the power given out by the solar cells.
Diagram of apparatus
The circuit I used for measuring the power was set out like this. The voltmeter in parallel measuring the potential difference across the cells and the a ammeter in series measuring the current flowing through the circuit.
As you can see I have used a resistor in the circuit this is to minimise wasted energy lost by the power supply or solar cell. If the circuit did not have a resistor in the current would flow round ... | HuggingFaceTB/finemath | |
Download Download Presentation # HEAPSORT COUNTING SORT RADIX SORT
Download Presentation ## HEAPSORT COUNTING SORT RADIX SORT
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
2. HEAPSORT • O(n lg n) worstcaselikemergesort. • Likeinsertionsort, but unlike merge sort, heapsort sorts in place: • Combines the best of both algorithms. To understand heapsort, we’llcover heaps and heap operations.
3. Heap data structure a binary tree is a tree data structure in which each node has at most two child nodes • The (binary) heap data structure is an array object that we can view as anearlycompletebinarytree • A.length, number of elements in the array, • A.heap-size,how many elements in the heap are stored withinarrayA. • 0 <=A.heap-size <=A.length • Height of node = # of edges on a longest simple path from the nodedown toa leaf. • Height of heap = height of root = (lg n). • A heap can be stored as an array A. • Root of tree is A. • Parent of A[i ] = A[i/2]. • Left child of A[i ] = A[2i ]. • Right child of A[i ] = A[2i + 1].
4. Heap data structure • There are two kinds of binary heaps: max-heaps and min-heaps. In both kinds,thevalues in the nodes satisfy a heap property, the specifics of which depend onthekind of heap. • In a max-heap, the max-heap property is that for every node iotherthantheroot, • A[PARENT(i)] >=A[İ] • In a min-heap, the min-heap property is that for every node iotherthantheroot, • A[PARENT(i)] <=A[İ] • Fortheheapsortalgorithmweusemax-heaps.
5. Maintainingtheheapproperty • In order to maintain the max-heap property, we call the procedure MAX-HEAPIFY. • Its inputs are an array A and an index i into the array.
6. Building a heap • We can use the procedure MAX-HEAPIFY in a bottom-up manner to convert anarray A[1…n], where n = A.length, into a max-heap.
7. ççç
8. Heapsortalgorithm • The HEAPSORT procedure takes time O(nlgn), since the call to BUILD-MAXHEAPtakes time O(n)and each of the n -1 calls to MAX-HEAPIFY takes time O(lgn) The complexity of BuildHeap appears to be Θ(n lg n) – n calls to Heapify at a cost of Θ(lg n)per call, but this result can be improved to Θ(n). The analysis is in the book. The intuition is thatmost of the calls to heapify are on very short heaps.
9. CountingSort • Counting sort assumes that each of the n input elements is an integer in the range0 to k • Counting sort determines, for each input element x, the number of elements lessthan x. • It uses this information to place element x directly into its position in theoutputarray.
10. CountingSort
11. CountingSort
12. RadixSort • Radix sort is the algorithm used by the card-sorting machines you now find only incomputermuseums. • Radix sort solves the problem of card sorting by sorting onthe least significant digit first. • The algorithm then combines the cards into a singledeck, with the cards in the 0 bin preceding the cards in the 1 bin preceding thecards in the 2 bin, and so on. Then it sorts the entire deck again on the second-leastsignificant digit and recombines the deck in a like manner. • The process continuesuntil the cards have been sorted on all d digits. Remarkably, at that point the cardsare fully sorted on the d-digit number.
13. RadixSort • In a typical computer, which is a sequential random-access machine, we sometimesuse radix sort to sort records of information that are keyed by multiple fields. • For example, we might wish to sort dates by three keys: year, month, and day. Wecould run a sorting algorithm with a comparison function that, given two dates,compares years, and if there is a tie, compares months, and if another tie occurs,compares days. • Alternatively, we could sort the information three times with astable sort: first on day, next on month, and finally on year.
14. RadixSort • The code for radix sort is straightforward. The following procedure assumes thateach element in the n-element array A has d digits, where digit 1 is the lowest-orderdigit and digit d is the highest-order digit. | HuggingFaceTB/finemath | |
# Disable one angle of rotation
I'd like to disable one angle of rotation of an object rotating in 3D space. Imagine a camera rotating around and displaying objects as they are in space. I'd like this object to be fixed on the horizontal axis (always in the center of the camera view) and follow the camera rotation on other two angles ( yaw and pitch).
Before multiplying the position matrix with the view matrix, I tried to anull the roll rotation by extracting Euler angles from the view matrix and then recreating it with roll value of zero. Something along these lines:
1. Take view matrix and extract Euler angles
2. Create the same matrix by replacing roll with zero
It's giving somewhat strange results, and there must be a better way to do this.
• So in other words, if there was a horizontal line across the camera viewer, it would always appear to be parallel to the horizon, correct? Is it critical you do this with Euler angles or would you be willing to do it with quaternions? Commented Mar 27, 2013 at 17:23
A quaternion solution which can compute a rotation quaternion as a composition of first a yaw turn then a pitching turn:
I'm thinking of the usual right-handed $i,j,k$ axes in three space. We suppose that the camera begins looking along the $i$ axis, and that the $i,j$ plane is horizontal.
To accomplish a yaw turn through an angle of $\psi$ radians, we can apply the transformation $x\mapsto qxq^{-1}$ where $q=\cos(\psi/2)+\sin(\psi/2)k$.
At that point, we could rotate around $qjq^{-1}$ to perform a pitch turn. Set $h=qjq^{-1}$. A pitch up by $\theta$ radians is accomplished by the transformation $x\mapsto pxp^{-1}$ where $p=\cos(\theta/2)+\sin(\theta/2)h$.
The composition would just be given by $R=pq$, mapping $x\mapsto RxR^{-1}$. One would have to multiply out what $pq$ is in terms of $\psi$ and $\theta$, but that isn't too hard.
Unfortunately I am not adept at converting this solution to Euler angles or rotation matrices, but fortunately there is a wiki devoted to that subject.
• Thanks, but my quaternion knowledge is rather limited, unfortunately. Commented Mar 28, 2013 at 10:26
• @user1304844 Yeah, sorry if it doesn't directly help. It's the least complicated theoretical solution that I'm handy with. What texts do you have to help you solve the problem? Commented Mar 28, 2013 at 12:06
• Actually you did help. I ended up reading your solution over and over again and studying quaternions for a day. What I did was: 1.Take rotation matrix and convert to quaternion; 2. Set Y component to zero (since the y component get multiplied with the Y axis of the vector. If there is no rotation, y is 0); 3. Normalize the quaternion; 4. OPTIONAL: Convert it back to a matrix4f (since openGL works very well with matrices) and set translation values from the original matrix ( the last column of the original matrix gets copied) Commented Mar 29, 2013 at 10:31
• @user1304844 Awesome! Good luck with your studies! Commented Mar 29, 2013 at 13:14
• For those of you who need to read up on quaternions like me, better explained has a fabulous tutorial: betterexplained.com/articles/… Commented Aug 14, 2017 at 0:32 | HuggingFaceTB/finemath | |
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## In this why are we not converting speed to feet?
This topic has 2 expert replies and 2 member replies
shibsriz@gmail.com Master | Next Rank: 500 Posts
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#### In this why are we not converting speed to feet?
Sun Jul 14, 2013 10:42 pm
70. At a certain instant in time, the number of cars, N,
traveling on a portion of a certain highway can be
estimated by the formula
N= 20Ld
600+ s2
where L is the number of lanes in the same direction,
d is the length of the portion of the highway, in feet,
and s is the average speed of the cars, in miles per
hour. Based on the formula, what is the estimated
number of cars traveling on aA-mile portion of the
highway if the highway has 2 lanes in the same
direction and the average speed of the cars is
40 miles per hour? (5,280 feet = 1mile)
ganeshrkamath Master | Next Rank: 500 Posts
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Test Date:
August 12, 2013
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Tue Aug 20, 2013 8:16 pm
shibsriz@gmail.com wrote:
70. At a certain instant in time, the number of cars, N,
traveling on a portion of a certain highway can be
estimated by the formula
N= 20Ld
600+ s2
where L is the number of lanes in the same direction,
d is the length of the portion of the highway, in feet,
and s is the average speed of the cars, in miles per
hour. Based on the formula, what is the estimated
number of cars traveling on aA-mile portion of the
highway if the highway has 2 lanes in the same
direction and the average speed of the cars is
40 miles per hour? (5,280 feet = 1mile)
This is a question from OG12 and is pretty straightforward. You have not copied it correctly.
The correct question states "At a certain instant in time, the number of cars, N,
traveling on a portion of a certain highway can be
estimated by the formula
N=(20*L*d)/(600+s^2)
where L is the number of lanes in the same direction,
d is the length of the portion of the highway, in feet,
and s is the average speed of the cars, in miles per
hour. Based on the formula, what is the estimated
number of cars traveling on a 1/2-mile portion of the
highway if the highway has 2 lanes in the same
direction and the average speed of the cars is 40
miles per hour? (5,280 feet = 1 mile)
"
Solution:
N=(20*L*d)/(600+s^2)
L = 2
d = 1/2 miles = 1/2 * 5280 feet
s = 40 miles/hour
N = 20*2*1/2*5280/(600 + 40^2)
= 20*2*1/2*5280/(600 + 1600)
= 20*5280/2200
= 5280/110
= 528/11
= 48
The question explicitly says that d is in feet and s is in miles per hour.
Cheers
_________________
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### GMAT/MBA Expert
Rich.C@EMPOWERgmat.com Elite Legendary Member
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Tue Aug 20, 2013 4:14 pm
Hi ProGMAT,
We don't convert speed (from miles/hr to feet/hr) because the question doesn't ask us to. Notice the wording:
L is the number of lanes
d is the distance IN FEET
s is the speed IN MILES PER HOUR
Just follow the instructions, plug in the numbers and you'll have the answer to the question.
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at Rich.C@empowergmat.com
### GMAT/MBA Expert
Rich.C@EMPOWERgmat.com Elite Legendary Member
Joined
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Posted:
9195 messages
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GMAT Score:
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Sun Jul 14, 2013 10:56 pm
Hi shibsriz,
While this question is certainly wordy, you're given a formula, told what the variables are and given the numbers to plug in...
L = number of lanes = 2
d = length of highway in feet = 1 mile = 5280 feet
s = average speed of cars = 40
Now, plug these values into the formula and solve.
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at Rich.C@empowergmat.com
ProGMAT Senior | Next Rank: 100 Posts
Joined
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Posted:
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Target GMAT Score:
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Tue Aug 20, 2013 9:36 am
Rich.C@EMPOWERgmat.com wrote:
plug these values into the formula and solve.
Rich
In this question I am confused that why are we not converting speed to feet(its in mile only)?
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#### Solution for 817 is what percent of 2800:
817:2800*100 =
(817*100):2800 =
81700:2800 = 29.18
Now we have: 817 is what percent of 2800 = 29.18
Question: 817 is what percent of 2800?
Percentage solution with steps:
Step 1: We make the assumption that 2800 is 100% since it is our output value.
Step 2: We next represent the value we seek with {x}.
Step 3: From step 1, it follows that {100\%}={2800}.
Step 4: In the same vein, {x\%}={817}.
Step 5: This gives us a pair of simple equations:
{100\%}={2800}(1).
{x\%}={817}(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
\frac{100\%}{x\%}=\frac{2800}{817}
Step 7: Taking the inverse (or reciprocal) of both sides yields
\frac{x\%}{100\%}=\frac{817}{2800}
\Rightarrow{x} = {29.18\%}
Therefore, {817} is {29.18\%} of {2800}.
#### Solution for 2800 is what percent of 817:
2800:817*100 =
(2800*100):817 =
280000:817 = 342.72
Now we have: 2800 is what percent of 817 = 342.72
Question: 2800 is what percent of 817?
Percentage solution with steps:
Step 1: We make the assumption that 817 is 100% since it is our output value.
Step 2: We next represent the value we seek with {x}.
Step 3: From step 1, it follows that {100\%}={817}.
Step 4: In the same vein, {x\%}={2800}.
Step 5: This gives us a pair of simple equations:
{100\%}={817}(1).
{x\%}={2800}(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
\frac{100\%}{x\%}=\frac{817}{2800}
Step 7: Taking the inverse (or reciprocal) of both sides yields
\frac{x\%}{100\%}=\frac{2800}{817}
\Rightarrow{x} = {342.72\%}
Therefore, {2800} is {342.72\%} of {817}.
Calculation Samples | HuggingFaceTB/finemath | |
# Quick Puzzles - Second Set
1. Draw a 3 by 3 grid like that shown and colour the squares either Red , White ,Blue so that each Red touches a White eachWhite touches a Blue each Blue touches a Red 2. Make a copy of a triangle like that shown with four circles along each edge. Use the numbers 1 to 9. putting one in each circle so that the four numbers along each edge add up to the same total. Find two different ways of doing this. 3. How many acute angles can be found in this drawing? 4. On this diagram you may start at any square and move up or down or across (but NOT diagonally) into the next square. No square may be used twice. The digits in each square are written down in the order they are used. What is the largest number that can be made? 5. Copy the diagram on the right but re-arrange the numbers so that any pair which are side by side, add up to the same total as the pair opposite. At the moment it is only true for 1+10 = 5+6. 6. The French Flag (known as the tricolour) is coloured Red , White ,and Blue in the way shown. Using the same three colours, how many different flags would it be possible to make? 7. Copy out the diagram like that on the right. Use the numbers 1 to 8, putting one in each circle, in such a way that no two circles which are connected by a single straight line, contain two consecutive numbers. 8. Copy this grid and then colour it with 4 blue squares, 3 red squares , 3 green squares, 3 yellow squares , and 3 white squares .No two squares of the same colour must be in line, either vertically, horizontally or diagonally. 9. Draw out a 3 by 3 grid like that shown. Place the numbers - 2 2 2 3 3 3 4 4 4 in it so that, when any line of three numbers is added up in any direction (including diagonally), the total is always 9 10.Arrange the numbers 1 to 9, using each one only once, placing only one number in each cell so that the totals in both directions ( up and down and across) are the same. How may different ways are there of doing this?
Go to top PreviousPuzzles Page NextPuzzles Page Puzzles Index Page CIMT Home Page | HuggingFaceTB/finemath | |
# Chapter 9 - Section 9.3 - Exponential Functions - Exercise Set - Page 560: 61
No
#### Work Step by Step
We are given the function $f(x)=1.5x^{2}$. In general, exponential functions are written in the form $f(x)=b^{x}$ (where $b\gt0$, $b\ne1$, and $x$ is a real number). In other words, exponential functions consist of a constant that is raised to the power of a variable. Therefore, the given function is not an exponential function, because it consists of a variable raised to the power of a constant.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | HuggingFaceTB/finemath | |
# trig integrals
• Sep 16th 2008, 02:28 PM
skabani
trig integrals
i have a calc test coming up and im not sure how to do the following, if you could show me how, it would be greatly appreciated...thanks!
1. integral (where b= pi/2, a = 0) sin^2x cos^2 x dx
2. integral (where b= pi/4, a = 0) sec^4 x tan^4 x dx
3. integral tan^6(ay) dy
• Sep 16th 2008, 03:36 PM
Soroban
Hello, skabani!
Here are the first two.
. . I'll let someone else explain the third one.
Quote:
$\displaystyle 1)\;\;\int^{\frac{\pi}{2}}_0 \sin^2\!x\cos^2\!x\,dx$
We need two identities:
. . $\displaystyle 2\sin\theta\cos\theta \:=\:\sin2\theta\qquad\qquad \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}$
We have: .$\displaystyle \sin^2\!x\cos^2\!x \;=\;\frac{1}{4}\left(4\sin^2\!x\cos^2\!x\right) \;=\;\frac{1}{4}(2\sin x\cos x)^2 \;=\;\frac{1}{4}\sin^22x$
The integral becomes: .$\displaystyle \frac{1}{4}\int^{\frac{\pi}{2}}_0\sin^22x\,dx \;=\; \frac{1}{4}\int^{\frac{\pi}{2}}_0 \frac{1-\cos4x}{2}\,dx$
And we have: . $\displaystyle \frac{1}{8}\int^{\frac{\pi}{2}}_0(1 - \cos4x)\,dx\quad\hdots$ . Got it?
Quote:
$\displaystyle 2)\;\;\int^{\frac{\pi}{4}}_0 \sec^4\!x\tan^4\!x\,dx$
We have: .$\displaystyle \sec^2\!x\cdot\sec^2\!x\cdot\tan^4\!x \;=\;\sec^2\!x\cdot(\tan^2\!x + 1)\cdot\tan^4\!x \;=\;\sec^2\!x\cdot(\tan^6\!x+\tan^4\!x)$
The integral becomes: .$\displaystyle \int^{\frac{\pi}{4}}_0 (\tan^6\!x + \tan^4\!x)\,\sec^2\!x\,dx$
Let $\displaystyle u \:=\:\tan x \quad\Rightarrow\quad du \:=\:\sec^2\!x\,dx$
Substitute: .$\displaystyle \int^b_a(u^6 + u^4)\,du \quad\hdots$ . Okay?
• Sep 16th 2008, 03:44 PM
skabani
great thanks so much!
i hope someone can help me out with the third one, before tonight, for my exam tomorrow...
• Sep 16th 2008, 04:06 PM
Krizalid
For the last one, first put $\displaystyle z=\alpha y,$ then just worry about $\displaystyle \int{\tan ^{6}z\,dz}.$ Now, note that,
$\displaystyle \tan ^{6}z=\tan ^{4}z\tan ^{2}z=\tan ^{4}z\left( \sec ^{2}z-1 \right)=\tan ^{4}z\sec ^{2}z-\tan ^{4}z.$
Make it in the same fashion for $\displaystyle \tan ^{4}z.$ | HuggingFaceTB/finemath | |
# Thread: differentiating through the integral
1. ## differentiating through the integral
The problem is as follows:
Solve for f(x), given
$f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}$
What I did:
1. Differentiate both sides wrt x, and partially differentiate through the integral.
$f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}$
$f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}$
$f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}$
$f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}$
2. Differentiate both sides wrt x again.
$f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}$
$f''(x) = - xf'(x) - f(x) + f(x)$
$f''(x) = - xf'(x)$
This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)
2. Originally Posted by compliant
The problem is as follows:
Solve for f(x), given
$f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}$
What I did:
1. Differentiate both sides wrt x, and partially differentiate through the integral.
$f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}$
$f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}$
$f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}$
$f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}$
2. Differentiate both sides wrt x again.
$f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}$
$f''(x) = - xf'(x) - f(x) + f(x)$
$f''(x) = - xf'(x)$
This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)
Let y = f(x). Then you have
$\frac{d^2 y}{dx^2} = - x \frac{dy}{dx}$.
Substitute $\frac{dy}{dx} = p$ .... (1):
$\frac{dp}{dx} = - x p$ .... (2)
DE (2) is seperable. Solve it for p as a function of x. Then substitute your expression for p into DE (1) and solve for y as a function of x. | HuggingFaceTB/finemath | |
## Difference of Square Binomial Expansion Question
To simplify
$(3-x)^4(3+x)^4$
notice that
$9-x^2=(3-x)(3+x)$
so
\begin{aligned} (3-x)^4(3+x)^4 &= (9-x^2)^4 \\ &= {}^4C_4 9^4(-x^2)^0+ {}^4C_3 9^3(-x^2)^1+{}^4C_2 9^2(-x^2)^2 \\ &+ {}^4_1 9^1(-x^2)^3+{}^4C_0 9^0 (-x^2)^4 \\ &= 6561-2916x^2+486x^4-9x^6+x^8 \end{aligned}
Refresh | HuggingFaceTB/finemath | |
# Eggs and Boxes (Age 6)
Age 6
## The Activities
1. Topics: Numbers, Codes, Algebra: The Cat in Numberland, Chapter 3, by I. Ekeland. In this chapter the letters come to visit the numbers, and we learn about letter/number ciphers and letters standing in for numbers.
2. Topic: Algebra: I made problems of the form “X + 3 = 5” using unit cubes from Base Ten Blocks and a small cardboard box. I.e., I would secretly put 2 blocks into the box and close it, put 3 blocks next to it, and then say “There are 5 blocks total, how many are in the box?”
3. Topic: Primes: I introduced the idea of primes using Base Ten Blocks: a number N is a prime if the only rectangle you can make using N blocks is 1 X N. I gave different numbers to each kid and had them figure out whether it was prime or not.
4. Topics: Combinations, Combinatorics: I printed a bunch of “Easter eggs” with a top and bottom section. Using five different colors of crayons, I asked the kids to make as many different eggs as they could, coloring each section in solid colors (not stripes/dots/etc.). I taped each one to the wall (stacking repeats).
## How Did It Go?
We had four kids this week.
#### Cat In Numberland
The algebra in this chapter is tricky because it includes addition, subtraction, multiplication, and division; most of the kids don’t know multiplication or division yet.
#### Box Algebra
This worked pretty well. The kids understood what was going on right away, and they were always excited when I opened the box and dumped out the blocks inside to see if their guess was correct. At the end they made a problem for me, which was something like “X + 3 = 39” (of course, they used as many blocks as they could).
#### Rectangle Primes
We did up to about 14. I kept track of each result. The only odd composite number <= 14 is 9, so for the most part they just needed to check a 2 row rectangle. Proving something is prime is tricky, of course, and whenever a kid said that something was prime, I always asked them “did you check 3-wide”? Whoever had 9 didn’t initially check 3×3.
#### Easter Eggs
The kids were really into this activity and worked very hard to get all the combinations. They got all 10 two-color combinations pretty quickly and without help (first two rows in picture above) — but there was no pattern to which color was on top vs. bottom. Then one of the kids realized that you could flip the colors. They quickly got 6 more, but the next 3 took them a lot longer to find, and I had to help them find the last one. This got them to 20, but they didn’t think of having the same color on top and bottom. I suggested it to them and they quickly made the last 5. Then I rearranged them so that there were same color tops along the rows and same color bottoms along the columns. I realized afterwards that I should have made this chart before I gave them the hint about same color top/bottom, because then there would have been gaps and I could ask them what went in the gaps. | HuggingFaceTB/finemath | |
If you find any mistakes, please make a comment! Thank you.
## Stabilizer commutes with conjugation
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.1
Solution: First we prove that $\mathsf{stab}_G(b) = g \mathsf{stab}_G(a) g^{-1}$.
($\subseteq$) If $x \in \mathsf{stab}_G(b)$, then $x \cdot b = b$. Now $$g^{-1}xg \cdot a = g^{-1}x \cdot b = g^{-1} \cdot b = a,$$ so that $g^{-1}xg \in \mathsf{stab}_G(a)$. Hence $x \in g \mathsf{stab}_G(a) g^{-1}$.
($\supseteq$) Let $x \in g \mathsf{stab}_G(a) g^{-1}$. Then $$gxg^{-1} \cdot b = gx \cdot a = g \cdot a = b,$$ so that $gxg^{-1} \in \mathsf{stab}_G(b)$. Hence $g \mathsf{stab}_G(a) g^{-1} \subseteq \mathsf{stab}_G(b)$.
Suppose now that the action of $G$ on $A$ is transitive; that is, for all $a,b \in A$, there exists $g \in G$ such that $b = g \cdot a$. We show that, if $K$ is the kernel of the action, $K = \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}$.
($\subseteq$) Let $x \in K$. Then $x \cdot a = a$ for all $a \in A$. Let $g \in G$. Then $$g^{-1}xg \cdot a = g^{-1} \cdot (x \cdot (g \cdot a)) = g^{-1} \cdot (g \cdot a) = g^{-1}g \cdot a = 1 \cdot a = a,$$ so that $g^{-1}xg \in \mathsf{stab}_G(a)$ for all $g \in G$. Thus $x \in g \mathsf{stab}_G(a) g^{-1}$ for all $g \in G$, hence $x \in \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}$.
($\supseteq$) Let $x \in \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}$, and let $b \in A$. Because the action of $G$ on $A$ is transitive, we have $b = h \cdot a$ for some $h \in G$. Now $x = hyh^{-1}$ for some $y \in \mathsf{stab}_G(a)$, thus $$x \cdot b = hyh^{-1} \cdot b = hy \cdot a = h \cdot a = b.$$ Hence $x$ stabilizes $b$; since $b \in A$ is arbitrary, $x \in K$. | HuggingFaceTB/finemath | |
It’s time to bring the maths lesson outdoors. But don’t worry, it’s merely your local supermarket.
The supermarket is one of the best examples of a place where maths is real. It’s a great place for practising measurement, estimation, and quantity. Since trips to the supermarket usually affect everyone in the family, the following activities include various levels of difficulty within the activity.
Allowing your children to participate in weighing, counting, and figuring price per unit versus price per kilogramme will help improve their ability to estimate and predict amounts with accuracy.
## Weighing In
What You’ll Need
A weighing scale, either from the supermarket or your home
What to Do
1. Help your child examine the scale in the supermarket or the one you have at home. Explain that kilogrammes are divided into smaller parts called grams and 1,000 grams equal a kilogramme.
2. Gather the produce you are purchasing, and estimate the weight of each item before weighing it. If you need 500 grams of grapes, ask your child to place the first bunch of grapes on the weighing scale, and then estimate how many more or fewer grapes are needed to make exactly 500 grams.
3. Let your child hold an item in each hand and guess which item weighs more. Then use the scale to check.
4. Ask questions to encourage thinking about measurement and estimation. You might want to ask your child: How much do you think 6 apples will weigh? More than 500 grams, less than 500 grams, or equal to 500 grams? How much do the apples really weigh? Do they weigh more or less than you estimated? Will 6 potatoes weigh more or less than the apples? How much do potatoes cost per 100 grams? If they cost 10 cents per 100 grams, what is the total cost?
Let your child experiment with the store scale by weighing different products.
Parent Pointer: There are many opportunities to increase estimation and measurement skills by weighing objects in the produce section of the supermarket.
Reference: http://math.com/ | HuggingFaceTB/finemath | |
# What is 1% of 792?
## Solution and how to solve it
1% of 792 is 7.92
Calculating the percent of a number is simple, but can be a bit tricky if you aren't careful. Luckily, it only requires a few basic operations to get to the solution: multiplication and division. If you haven't learned what percentage is yet, or would like a little refresher, feel free to check out our introduction to percentage page. To solve percentage problems, it may be useful to use a calculator. But, they can also be solved by hand or in your head (if you practice enough). So, how did we get to the solution that 1 percent of 792 = 7.92?
• Step 1: As we know, a whole of something is equal to 100%. In this case, we want to find what 1% of 792 is. We know that 100% of 792 is, well, just 792.
• Step 2: If 100% of 792 is 792, then we can get 1% of 792 by dividing it by 100. Let's do 792 / 100. This is equal to 7.92. Now we know that 1% is 7.92.
• Step 3: Now that we know what 1% of 792 is, we just need to multiply it by 1 to get our solution! 7.92 times 1 = 7.92. That's all there is to it!
### When is this useful?
Percentage is one of the most commonly used math concepts in day-to-day life. You can use it to calculate a gratuity on a restaurant bill, or to grade your score on an exam. It is useful to know your percentages well! | HuggingFaceTB/finemath | |
# For her phone service, Martina pays a monthly fee of \$28 , and she pays an additional \$0.06 per minute of use. The least she has been charged in a month is \$88.72 . What are the possible numbers of minutes she has used her phone in a month? Use m for the number of minutes, and solve your inequality for m .
1
by laneishalanaee
2015-01-23T19:56:51-05:00
If you make an equation first, 88.72 ≤28+.06m, you can figure it out.
First you subtract 28 from 88.72 getting 60.72≤.06m. Then you divide both sides by .06. Then you get your final answer, which is an inequality, 1012≤m.
That means that Martina uses at least 1012 minutes of service. | HuggingFaceTB/finemath | |
## Skewed Distribution Mean Median
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# If k is a positive integer such that 5,880k is a perfect square, what
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If k is a positive integer such that 5,880k is a perfect square, what is the least possible value of k?
(A) 2
(B) 6
(C) 15
(D) 30
(E) 140
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Re: If k is a positive integer such that 5,880k is a perfect square, what [#permalink]
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20 Mar 2018, 23:49
1
IMO D
5880 = 2^3 * 7^2 * 3 * 5
Therefore, to make it a perfect square, the minimum value of k should be = 2 * 3 * 5 = 30
Thus, D.
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Re: If k is a positive integer such that 5,880k is a perfect square, what [#permalink]
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21 Mar 2018, 01:32
Bunuel wrote:
If k is a positive integer such that 5,880k is a perfect square, what is the least possible value of k?
(A) 2
(B) 6
(C) 15
(D) 30
(E) 140
Bunuel there seems something wrong with the question. Is the integer at hand 5,880*k or 5,880k which is a concatenation of 5,880 & k. Is this an ambiguity or am I making a mistake in reading it properly.
Best,
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Re: If k is a positive integer such that 5,880k is a perfect square, what [#permalink]
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21 Mar 2018, 01:46
Bunuel wrote:
If k is a positive integer such that 5,880k is a perfect square, what is the least possible value of k?
(A) 2
(B) 6
(C) 15
(D) 30
(E) 140
Bunuel there seems something wrong with the question. Is the integer at hand 5,880*k or 5,880k which is a concatenation of 5,880 & k. Is this an ambiguity or am I making a mistake in reading it properly.
Best,
5,880k is 5,880*k, so 5,880 multiplied by k (also recall that multiplication sign is often omitted). It cannot be a five-digit integer 5880k because if it were it would have been explicitly mentioned.
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Re: If k is a positive integer such that 5,880k is a perfect square, what [#permalink]
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22 Mar 2018, 14:47
Bunuel wrote:
If k is a positive integer such that 5,880k is a perfect square, what is the least possible value of k?
(A) 2
(B) 6
(C) 15
(D) 30
(E) 140
We are given that 5,880k is a perfect square. We must remember that all perfect squares break down to unique prime factors, each of which has an exponent that is a positive multiple of 2. So, let’s break down 5,880 into its prime factors to determine the minimum value of k.
5,880 = 588 x 10 = 12 x 49 x 10 = 2^3 x 3^1 x 5^1 x 7^2
So, in order for 5,880k to be a perfect square, k must contain the factors 2 x 3 x 5 = 30 (so that 5,880k = 2^4 x 3^2 x 5^2 x 7^2), so 30 is the smallest possible value of k.
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If k is a positive integer such that 5,880k is a perfect square, what [#permalink]
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25 Mar 2018, 12:31
1
Solution
• $$5880* k$$ can be written in prime factorization form as $$2^3 * 3 *5 * 7^2 * k$$.
• Since $$5880 * k$$ is a perfect square, hence $$5880 * k=n^2$$, where $$n$$ is any integer.
o $$n^2 = 2^3 * 3 * 5 * 7^2 * k$$
o After taking square root on both the sides, we can write:
o $$n= 2*7* \sqrt{(2*3*5*k)}$$
oFor $$n$$ to be an integer,$$\sqrt{(2 * 3 * 5 * k)}$$ must be a perfect square.
oHence,$$k = 2 * 3 * 5 *$$ (square of any positive integer)
.
• The least possible value of the square of a positive integer is 1.
o Hence, k=2 * 3 * 5 = 30
.
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Re: If k is a positive integer such that 5,880k is a perfect square, what [#permalink]
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25 Mar 2018, 23:22
Bunuel wrote:
If k is a positive integer such that 5,880k is a perfect square, what is the least possible value of k?
(A) 2
(B) 6
(C) 15
(D) 30
(E) 140
The rule says, " A perfect square always has even number of powers of prime factors".
Prime factorisation of 5880 = 2^3*3^1*5^1*7^2
Therefore, to make 5880 a perfect square we need to multiply it by a set of 2*3*5 = 30
Hence (D)
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Re: If k is a positive integer such that 5,880k is a perfect square, what [#permalink] 25 Mar 2018, 23:22
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# Puzzle No. 030 - Cookie Conundrum
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• Southern Street - Minnie
• 30 Picarats
### EditHints
1
Visualizing exactly what the puzzle is describing as a scene in your mind might be all you need to figure this one out
2
Try and break the problem down into numbers. If one out of 15 cookies is eaten, that means 14 are left. If you divide those 14 in half and pass them to two people, they each get seven. That's taken one minute so far.
If you think about the problem in concrete mathematical terms like this, you should be able to figure it out.
3
The two people who get seven cookies each eat one cookie, which takes one minute. They both have six cookies left and each divide those in half, then pass three cookies to two people each. That's two minutes so far.
S
The people who receive three cookies each eat one of them, so they each have two remaining. That's three minutes elapsed to this point.
If you divide these two cookies in half and pass one cookie to each of the next people in line, how many more minutes will it take to finish the cookies?
4
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1. Chapter 7 Class 12 Integrals (Term 2)
2. Serial order wise
3. Ex 7.5
Transcript
Ex 7.5, 1 𝑥/((𝑥 + 1) (𝑥 + 2) ) Solving integrand 𝑥/((𝑥 + 1) (𝑥 + 2) ) We can write it as 𝑥/((𝑥 + 1) (𝑥 + 2) ) " " = 𝐴/((𝑥 + 1) ) + 𝐵/((𝑥 + 2) ) 𝑥/((𝑥 + 1) (𝑥 + 2) ) " " = (𝐴(𝑥 + 2) + 𝐵 (𝑥 + 1))/((𝑥 + 1) (𝑥 + 2) ) Cancelling denominator 𝑥 = 𝐴(𝑥+2) + 𝐵(𝑥+1) Putting x=−1 in (1) −1=𝐴(−1+2) + 𝐵(−1+1) −1=𝐴×1+ 𝐵×0 −1=𝐴 𝐴=−1 Similarly Putting y=−2 in (1) −2 = 𝐴(−2+2) + 𝐵(−2+1) −2 = 𝐴×0+ 𝐵×(−1) −2 = − 𝐵 𝐵 = 2 Hence we can write it as 𝑥/((𝑥 + 1) (𝑥 + 2) ) = (−1)/((𝑥 + 1) ) + 2/((𝑥 + 2) ) Therefore ∫1▒𝑥/((𝑥 + 1) (𝑥 + 2) ) 𝑑𝑥 = ∫1▒(−1)/((𝑥 + 1) ) 𝑑𝑥 + ∫1▒2/((𝑥 + 2) ) 𝑑𝑥 = −1∫1▒1/((𝑥 + 1) ) 𝑑𝑥 + 2∫1▒1/((𝑥 + 2) ) 𝑑𝑥 = −〖log 〗|𝑥+1|+2 〖log 〗|𝑥+2|+𝐶 = −〖log 〗|𝑥+1|+〖log 〗〖|𝑥+2|^2 〗+𝐶 = 〖log 〗|(𝑥 + 2)^2/(𝑥 + 1)|+𝐶 = 〖𝐥𝐨𝐠 〗〖(𝒙 + 𝟐)^𝟐/|𝒙 + 𝟏| 〗 +𝑪
Ex 7.5 | HuggingFaceTB/finemath | |
# The Logic of Chance/Chapter I
THE LOGIC OF CHANCE.
CHAPTER I.
ON CERTAIN KINDS OF GROUPS OR SERIES AS THE FOUNDATION OF PROBABILITY.
§ 1. It is sometimes not easy to give a clear definition of a science at the outset, so as to set its scope and province before the reader in a few words. In the case of those sciences which are more immediately and directly concerned with what are termed objects, rather than with what are termed processes, this difficulty is not indeed so serious. If the reader is already familiar with the objects, a simple reference to them will give him a tolerably accurate idea of the direction and nature of his studies. Even if he be not familiar with them, they will still be often to some extent connected and associated in his mind by a name, and the mere utterance of the name may thus convey a fair amount of preliminary information. This is more or less the case with many of the natural sciences; we can often tell the reader beforehand exactly what he is going to study. But when a science is concerned, not so much with objects directly, as with processes and laws, or when it takes for the subject of its enquiry some comparatively obscure feature drawn from phenomena which have little or nothing else in common, the difficulty of giving preliminary information becomes greater. Recognized classes of objects have then to be disregarded and even broken up, and an entirely novel arrangement of the objects to be made. In such cases it is the study of the science that first gives the science its unity, for till it is studied the objects with which it is concerned were probably never thought of together. Here a definition cannot be given at the outset, and the process of obtaining it may become by comparison somewhat laborious.
The science of Probability, at least on the view taken of it in the following pages, is of this latter description. The reader who is at present unacquainted with the science cannot be at once informed of its scope by a reference to objects with which he is already familiar. He will have to be taken in hand, as it were, and some little time and trouble will have to be expended in directing his attention to our subject-matter before he can be expected to know it. To do this will be our first task.
§ 2. In studying Nature, in any form, we are continually coming into possession of information which we sum up in general propositions. Now in very many cases these general propositions are neither more nor less certain and accurate than the details which they embrace and of which they are composed. We are assuming at present that the truth of these generalizations is not disputed; as a matter of fact they may rest on weak evidence, or they may be uncertain from their being widely extended by induction; what is meant is, that when we resolve them into their component parts we have precisely the same assurance of the truth of the details as we have of that of the whole. When I know, for instance, that all cows ruminate, I feel just as certain that any particular cow or cows ruminate as that the whole class does. I may be right or wrong in my original statement, and I may have obtained it by any conceivable mode in which truths can be obtained; but whatever the value of the general proposition may be, that of the particulars is neither greater nor less. The process of inferring the particular from the general is not accompanied by the slightest diminution of certainty. If one of these ‘immediate inferences’ is justified at all, it will be equally right in every case.
But it is by no means necessary that this characteristic should exist in all cases. There is a class of immediate inferences, almost unrecognized indeed in logic, but constantly drawn in practice, of which the characteristic is, that as they increase in particularity they diminish in certainty. Let me assume that I am told that some cows ruminate; I cannot infer logically from this that any particular cow does so, though I should feel some way removed from absolute disbelief, or even indifference to assent, upon the subject; but if I saw a herd of cows I should feel more sure that some of them were ruminant than I did of the single cow, and my assurance would increase with the numbers of the herd about which I had to form an opinion. Here then we have a class of things as to the individuals of which we feel quite in uncertainty, whilst as we embrace larger numbers in our assertions we attach greater weight to our inferences. It is with such classes of things and such inferences that the science of Probability is concerned.
§ 3. In the foregoing remarks, which are intended to be purely preliminary, we have not been able altogether to avoid some reference to a subjective element, viz. the degree of our certainty or belief about the things which we are supposed to contemplate. The reader may be aware that by some writers this element is regarded as the subject-matter of the science. Hence it will have to be discussed in a future chapter. As however I do not agree with the opinion of the writers just mentioned, at least as regards treating this element as one of primary importance, no further allusion will be made to it here, but we will pass on at once to a more minute investigation of that distinctive characteristic of certain classes of things which was introduced to notice in the last section.
In these classes of things, which are those with which Probability is concerned, the fundamental conception which the reader has to fix in his mind as clearly as possible, is, I take it, that of a series. But it is a series of a peculiar kind, one of which no better compendious description can be given than that which is contained in the statement that it combines individual irregularity with aggregate regularity. This is a statement which will probably need some explanation. Let us recur to an example of the kind already alluded to, selecting one which shall be in accordance with experience. Some children will not live to thirty. Now if this proposition is to be regarded as a purely indefinite or, as it would be termed in logic, ‘particular’ proposition, no doubt the notion of a series does not obviously present itself in connection with it. It contains a statement about a certain unknown proportion of the whole, and that is all. But it is not with these purely indefinite propositions that we shall be concerned. Let us suppose the statement, on the contrary, to be of a numerical character, and to refer to a given proportion of the whole, and we shall then find it difficult to exclude the notion of a series. We shall find it, I think, impossible to do so as soon as we set before us the aim of obtaining accurate, or even moderately correct inferences. What, for instance, is the meaning of the statement that two new-born children in three fail to attain the age of sixty-three? It certainly does not declare that in any given batch of, say, thirty, we shall find just twenty that fail: whatever might be the strict meaning of the words, this is not the import of the statement. It rather contemplates our examination of a large number, of a long succession of instances, and states that in such a succession we shall find a numerical proportion, not indeed fixed and accurate at first, but which tends in the long run to become so. In every kind of example with which we shall be concerned we shall find this reference to a large number or succession of objects, or, as we shall term it, series of them.
A few additional examples may serve to make this plain.
Let us suppose that we toss up a penny a great many times; the results of the successive throws may be conceived to form a series. The separate throws of this series seem to occur in utter disorder; it is this disorder which causes our uncertainty about them. Sometimes head comes, sometimes tail comes; sometimes there is a repetition of the same face, sometimes not. So long as we confine our observation to a few throws at a time, the series seems to be simply chaotic. But when we consider the result of a long succession we find a marked distinction; a kind of order begins gradually to emerge, and at last assumes a distinct and striking aspect. We find in this case that the heads and tails occur in about equal numbers, that similar repetitions of different faces do so also, and so on. In a word, notwithstanding the individual disorder, an aggregate order begins to prevail. So again if we are examining the length of human life, the different lives which fall under our notice compose a series presenting the same features. The length of a single life is familiarly uncertain, but the average duration of a batch of lives is becoming in an almost equal degree familiarly certain. The larger the number we take out of any mixed crowd, the clearer become the symptoms of order, the more nearly will the average length of each selected class be the same. These few cases will serve as simple examples of a property of things which can be traced almost everywhere, to a greater or less extent, throughout the whole field of our experience. Fires, shipwrecks, yields of harvest, births, marriages, suicides; it scarcely seems to matter what feature we single out for observation[1]. The irregularity of the single instances diminishes when we take a large number, and at last seems for all practical purposes to disappear.
In speaking of the effect of the average in thus diminishing the irregularities which present themselves in the details, the attention of the student must be prominently directed to the point, that it is not the absolute but the relative irregularities which thus tend to diminish without limit. This idea will be familiar enough to the mathematician, but to others it may require some reflection in order to grasp it clearly. The absolute divergences and irregularities, so far from diminishing, show a disposition to increase, and this (it may be) without limit, though their relative importance shows a corresponding disposition to diminish without limit. Thus in the case of tossing a penny, if we take a few throws, say ten, it is decidedly unlikely that there should be a difference of six between the numbers of heads and tails; that is, that there should be as many as eight heads and therefore as few as two tails, or vice versâ. But take a thousand throws, and it becomes in turn exceedingly likely that there should be as much as, or more than, a difference of six between the respective numbers. On the other hand the proportion of heads to tails in the case of the thousand throws will be very much nearer to unity, in most cases, than when we only took ten. In other words, the longer a game of chance continues the larger are the spells and runs of luck in themselves, but the less their relative proportions to the whole amounts involved.
§ 4. In speaking as above of events or things as to the details of which we know little or nothing, it is not of course implied that our ignorance about them is complete and universal, or, what comes to the same thing, that irregularity may be observed in all their qualities. All that is meant is that there are some qualities or marks in them, the existence of which we are not able to predicate with certainty in the individuals. With regard to all their other qualities there may be the utmost uniformity, and consequently the most complete certainty. The irregularity in the length of human life is notorious, but no one doubts the existence of such organs as a heart and brains in any person whom he happens to meet. And even in the qualities in which the irregularity is observed, there are often, indeed generally, positive limits within which it will be found to be confined. No person, for instance, can calculate what may be the length of any particular life, but we feel perfectly certain that it will not stretch out to 150 years. The irregularity of the individual instances is only shown in certain respects, as e.g. the length of the life, and even in these respects it has its limits. The same remark will apply to most of the other examples with which we shall be concerned. The disorder in fact is not universal and unlimited, it only prevails in certain directions and up to certain points.
§ 5. In speaking as above of a series, it will hardly be necessary to point out that we do not imply that the objects themselves which compose the series must occur successively in time; the series may be formed simply by their coming in succession under our notice, which as a matter of fact they may do in any order whatever. A register of mortality, for instance, may be made up of deaths which took place simultaneously or successively; or, we might if we pleased arrange the deaths in an order quite distinct from either of these. This is entirely a matter of indifference; in all these cases the series, for any purposes which we need take into account, may be regarded as being of precisely the same description. The objects, be it remembered, are given to us in nature; the order under which we view them is our own private arrangement. This is mentioned here simply by way of caution, the meaning of this assertion will become more plain in the sequel.
I am aware that the word ‘series’ in the application with which it is used here is liable to some misconstruction, but I cannot find any better word, or indeed any as suitable in all respects. As remarked above, the events need not necessarily have occurred in a regular sequence of time, though they often will have done so. In many cases (for instance, the throws of a penny or a die) they really do occur in succession; in other cases (for instance, the heights of men, or the duration of their lives), whatever may have been the order of their actual occurrence, they are commonly brought under our notice in succession by being arranged in statistical tables. In all cases alike our processes of inference involve the necessity of examining one after another of the members which compose the group, or at least of being prepared to do this, if we are to be in a position to justify our inferences. The force of these considerations will come out in the course of the investigation in Chapter VI.
The late Leslie Ellis[2] has expressed what seems to me a substantially similar view in terms of genus and species, instead of speaking of a series. He says, “When individual cases are considered, we have no conviction that the ratios of frequency of occurrence depend on the circumstances common to all the trials. On the contrary, we recognize in the determining circumstances of their occurrence an extraneous element, an element, that is, extraneous to the idea of the genus and species. Contingency and limitation come in (so to speak) together; and both alike disappear when we consider the genus in its entirety, or (which is the same thing) in what may be called an ideal and practically impossible realization of all which it potentially contains. If this be granted, it seems to follow that the fundamental principle of the Theory of Probabilities may be regarded as included in the following statement,—The conception of a genus implies that of numerical relations among the species subordinated to it.” As remarked above, this appears a substantially similar doctrine to that explained in this chapter, but I do not think that the terms genus and species are by any means so well fitted to bring out the conception of a tendency or limit as when we speak of a series, and I therefore much prefer the latter expression.
§ 6. The reader will now have in his mind the conception of a series or group of things or events, about the individuals of which we know but little, at least in certain respects, whilst we find a continually increasing uniformity as we take larger numbers under our notice. This is definite enough to point out tolerably clearly the kind of things with which we have to deal, but it is not sufficiently definite for purposes of accurate thought. We must therefore attempt a somewhat closer analysis.
There are certain phrases so commonly adopted as to have become part of the technical vocabulary of the subject, such as an ‘event’ and the ‘way in which it can happen.’ Thus the act of throwing a penny would be called an event, and the fact of its giving head or tail would be called the way in which the event happened. If we were discussing tables of mortality, the former term would denote the mere fact of death, the latter the age at which it occurred, or the way in which it was brought about, or whatever else in it might be the particular circumstance under discussion. This phraseology is very convenient, and will often be made use of in this work, but without explanation it may lead to confusion. For in many cases the way in which the event happens is of such great relative importance, that according as it happens in one way or another the event would have a different name; in other words, it would not in the two cases be nominally the same event. The phrase therefore will have to be considerably stretched before it will conveniently cover all the cases to which we may have to apply it. If for instance we were contemplating a series of human beings, male and female, it would sound odd to call their humanity an event, and their sex the way in which the event happened.
If we recur however to any of the classes of objects already referred to, we may see our path towards obtaining a more accurate conception of what we want. It will easily be seen that in every one of them there is a mixture of similarity and dissimilarity; there is a series of events which have a certain number of features or attributes in common,—without this they would not be classed together. But there is also a distinction existing amongst them; a certain number of other attributes are to be found in some and are not to be found in others. In other words, the individuals which form the series are compound, each being made up of a collection of things or attributes; some of these things exist in all the members of the series, others are found in some only. So far there is nothing peculiar to the science of Probability; that in which the distinctive characteristic consists is this;—that the occasional attributes, as distinguished from the permanent, are found on an extended examination to tend to exist in a certain definite proportion of the whole number of cases. We cannot tell in any given instance whether they will be found or not, but as we go on examining more cases we find a growing uniformity. We find that the proportion of instances in which they are found to instances in which they are wanting, is gradually subject to less and less comparative variation, and approaches continually towards some apparently fixed value.
The above is the most comprehensive form of description; as a matter of fact the groups will in many cases take a far simpler form; they may appear, e.g. simply as a succession of things of the same kind, say human beings, with or without an occasional attribute, say that of being left-handed. We are using the word attribute, of course, in its widest sense, intending it to include every distinctive feature that can be observed in a thing, from essential qualities down to the merest accidents of time and place.
§ 7. On examining our series, therefore, we shall find that it may best be conceived, not necessarily as a succession of events happening in different ways, but as a succession of groups of things. These groups, on being analysed, are found in every case to be resolvable into collections of substances and attributes. That which gives its unity to the succession of groups is the fact of some of these substances or attributes being common to the whole succession; that which gives their distinction to the groups in the succession is the fact of some of them containing only a portion of these substances and attributes, the other portion or portions being occasionally absent. So understood, our phraseology may be made to embrace every class of things of which Probability can take account.
§ 8. It will be easily seen that the ordinary expression (viz. the ‘event,’ and the ‘way in which it happens’) may be included in the above. When the occasional attributes are unimportant the permanent ones are sufficient to fix and appropriate the name, the presence or absence of the others being simply denoted by some modification of the name or the addition of some predicate. We may therefore in all such cases speak of the collection of attributes as ‘the event,’—the same event essentially, that is—only saying that it (so as to preserve its nominal identity) happens in different ways in the different cases. When the occasional attributes however are important, or compose the majority, this way of speaking becomes less appropriate; language is somewhat strained by our implying that two extremely different assemblages are in reality the same event, with a difference only in its mode of happening. The phrase is however a very convenient one, and with this caution against its being misunderstood, it will frequently be made use of here.
§ 9. A series of the above-mentioned kind is, I apprehend, the ultimate basis upon which all the rules of Probability must be based. It is essential to a clear comprehension of the subject to have carried our analysis up to this point, but any attempt at further analysis into the intimate nature of the events composing the series, is not required. It is altogether unnecessary, for instance, to form any opinion upon the questions discussed in metaphysics as to the independent existence of substances. We have discovered, on examination, a series composed of groups of substances and attributes, or of attributes alone. At such a series we stop, and thence investigate our rules of inference; into what these substances or attributes would themselves be ultimately analysed, if taken in hand by the psychologist or metaphysician, it is no business of ours to enquire here.
§ 10. The stage then which we have now reached is that of having discovered a quantity of things (they prove on analysis to be groups of things) which are capable of being classified together, and are best regarded as constituting a series. The distinctive peculiarity of this series is our finding in it an order, gradually emerging out of disorder, and showing in time a marked and unmistakeable uniformity.
The impression which may possibly be derived from the description of such a series, and which the reader will probably already entertain if he have studied Probability before, is that the gradual evolution of this order is indefinite, and its approach therefore to perfection unlimited. And many of the examples commonly selected certainly tend to confirm such an impression. But in reference to the theory of the subject it is, I am convinced, an error, and one liable to lead to much confusion.
The lines which have been prefixed as a motto to this work, “So careful of the type she seems, so careless of the single life,” are soon after corrected by the assertion that the type itself, if we regard it for a long time, changes, and then vanishes and is succeeded by others. So in Probability; that uniformity which is found in the long run, and which presents so great a contrast to the individual disorder, though durable is not everlasting. Keep on watching it long enough, and it will be found almost invariably to fluctuate, and in time may prove as utterly irreducible to rule, and therefore as incapable of prediction, as the individual cases themselves. The full bearing of this fact upon the theory of the subject, and upon certain common modes of calculation connected with it, will appear more fully in some of the following chapters; at present we will confine ourselves to very briefly establishing and illustrating it.
Let us take, for example, the average duration of life. This, provided our data are sufficiently extensive, is known to be tolerably regular and uniform. This fact has been already indicated in the preceding sections, and is a truth indeed of which the popular mind has a tolerably clear grasp at the present day. But a very little consideration will show that there may be a superior as well as an inferior limit to the extent within which this uniformity can be observed; in other words whilst we may fall into error by taking too few instances we may also fail in our aim, though in a very different way and from quite different reasons, by taking too many. At the present time the average duration of life in England may be, say, forty years; but a century ago it was decidedly less; several centuries ago it was presumably very much less; whilst if we possessed statistics referring to a still earlier population of the country we should probably find that there has been since that time a still more marked improvement. What may be the future tendency no man can say for certain. It may be, and we hope that it will be the case, that owing to sanitary and other improvements, the duration of life will go on increasing steadily; it is at least conceivable, though doubtless incredible, that it should do so without limit. On the other hand, and with much more likelihood, this duration might gradually tend towards some fixed length. Or, again, it is perfectly possible that future generations might prefer a short and a merry life, and therefore reduce their average longevity. The duration of life cannot but depend to some extent upon the general tastes, habits and employments of the people, that is upon the ideal which they consciously or unconsciously set before them, and he would be a rash man who should undertake to predict what this ideal will be some centuries hence. All that it is here necessary however to indicate is, that this particular uniformity (as we have hitherto called it, in order to mark its relative character) has varied, and, under the influence of future eddies in opinion and practice, may vary still; and this to any extent, and with any degree of irregularity. To borrow a term from Astronomy, we find our uniformity subject to what might be called an irregular secular variation.
§ 11. The above is a fair typical instance. If we had taken a less simple feature than the length of life, or one less closely connected with what may be called by comparison the great permanent uniformities of nature, we should have found the peculiarity under notice exhibited in a far more striking degree. The deaths from small-pox, for example, or the instances of duelling or accusations of witchcraft, if examined during a few successive decades, might have shown a very tolerable degree of uniformity. But these uniformities have risen possibly from zero; after various and very great fluctuations seem tending towards zero again, at least in this century; and may, for anything we know, undergo still more rapid fluctuations in future. Now these examples must be regarded as being only extreme ones, and not such very extreme ones, of what is the almost universal rule in nature. I shall endeavour to show that even the few apparent exceptions, such as the proportions between male and female births, &c., may not be, and probably in reality are not, strictly speaking, exceptions. A type, that is, which shall be in the fullest sense of the words, persistent and invariable is scarcely to be found in nature. The full import of this conclusion will be seen in future chapters. Attention is only directed here to the important inference that, although statistics are notoriously of no value unless they are in sufficient numbers, yet it does not follow but that in certain cases we may have too many of them. If they are made too extensive, they may again fall short, at least for any particular time or place, of their greatest attainable accuracy.
§ 12. These natural uniformities then are found at length to be subject to fluctuation. Now contrast with them any of the uniformities afforded by games of chance; these latter seem to show no trace of secular fluctuation, however long we may continue our examination of them. Criticisms will be offered, in the course of the following chapters, upon some of the common attempts to prove a priori that there must be this fixity in the uniformity in question, but of its existence there can scarcely be much doubt. Pence give heads and tails about equally often now, as they did when they were first tossed, and as we believe they will continue to do, so long as the present order of things continues. The fixity of these uniformities may not be as absolute as is commonly supposed, but no amount of experience which we need take into account is likely in any appreciable degree to interfere with them. Hence the obvious contrast, that, whereas natural uniformities at length fluctuate, those afforded by games of chance seem fixed for ever.
§ 13. Here then are series apparently of two different kinds. They are alike in their initial irregularity, alike in their subsequent regularity; it is in what we may term their ultimate form that they begin to diverge from each other. The one tends without any irregular variation towards a fixed numerical proportion in its uniformity; in the other the uniformity is found at last to fluctuate, and to fluctuate, it may be, in a manner utterly irreducible to rule.
As this chapter is intended to be little more than explanatory and illustrative of the foundations of the science, the remark may be made here (for which subsequent justification will be offered) that it is in the case of series of the former kind only that we are able to make anything which can be interpreted into strict scientific inferences. We shall be able however in a general way to see the kind and extent of error that would be committed if, in any example, we were to substitute an imaginary series of the former kind for any actual series of the latter kind which experience may present to us. The two series are of course to be as alike as possible in all respects, except that the variable uniformity has been replaced by a fixed one. The difference then between them would not appear in the initial stage, for in that stage the distinctive characteristics of the series of Probability are not apparent; all is there irregularity, and it would be as impossible to show that they were alike as that they were different; we can only say generally that each shows the same kind of irregularity. Nor would it appear in the next subsequent stage, for the real variability of the uniformity has not for some time scope to make itself perceived. It would only be in what we have called the ultimate stage, when we suppose the series to extend for a very long time, that the difference would begin to make itself felt[3]. The proportion of persons, for example, who die each year at the age of six months is, when the numbers examined are on a small scale, utterly irregular; it becomes however regular when the numbers examined are on a larger scale; but if we continued our observation for a very great length of time, or over a very great extent of country, we should find this regularity itself changing in an irregular way. The substitution just mentioned is really equivalent to saying, Let us assume that the regularity is fixed and permanent. It is making a hypothesis which may not be altogether consistent with fact, but which is forced upon us for the purpose of securing precision of statement and definition.
§ 14. The full meaning and bearing of such a substitution will only become apparent in some of the subsequent chapters, but it may be pointed out at once that it is in this way only that we can with perfect strictness introduce the notion of a ‘limit’ into our account of the matter, at any rate in reference to many of the applications of the subject to purely statistical enquiries. We say that a certain proportion begins to prevail among the events in the long run; but then on looking closer at the facts we find that we have to express ourselves hypothetically, and to say that if present circumstances remain as they are, the long run will show its characteristics without disturbance. When, as is often the case, we know nothing accurately of the circumstances by which the succession of events is brought about, but have strong reasons to suspect that these circumstances are likely to undergo some change, there is really nothing else to be done. We can only introduce the conception of a limit, towards which the numbers are tending, by assuming that these circumstances do not change; in other words, by substituting a series with a fixed uniformity for the actual one with the varying uniformity[4].
§ 15. If the reader will study the following example, one well known to mathematicians under the name of the Petersburg[5] problem, he will find that it serves to illustrate several of the considerations mentioned in this chapter. It serves especially to bring out the facts that the series with which we are concerned must be regarded as indefinitely extensive in point of number or duration; and that when so regarded certain series, but certain series only (the one in question being a case in point), take advantage of the indefinite range to keep on producing individuals in it whose deviation from the previous average has no finite limit whatever. When rightly viewed it is a very simple problem, but it has given rise, at one time or another, to a good deal of confusion and perplexity.
The common form of objection is given in the reply, that so far from paying an infinite sum, no sensible man would give anything approaching to £50 for such a chance. Probably not, because no man would see enough of the series to make it worth his while. What most persons form their practical opinion upon, is such small portions of the series as they have actually seen or can reasonably expect. Now in any such portion, say one which embraces 100 turns, the longest succession of heads would not amount on the average to more than seven or eight. This is observed, but it is forgotten that the formula which produced these, would, if it had greater scope, keep on producing better and better ones without any limit. Hence it arises that some persons are perplexed, because the conduct they would adopt, in reference to the curtailed portion of the series which they are practically likely to meet with, does not find its justification in inferences which are necessarily based upon the series in the completeness of its infinitude.
§ 16. This will be more clearly seen by considering the various possibilities, and the scope required in order to exhaust them, when we confine ourselves to a limited number of throws. Begin with three. This yields eight equally likely possibilities. In four of these cases the thrower starts with tail and therefore loses: in two he gains a single point (i.e. £1); in one he gains two points, and in one he gains four points. Hence his total gain being eight pounds achieved in four different contingencies, his average gain would be two pounds.
Now suppose he be allowed to go as far as ${\displaystyle n}$ throws, so that we have to contemplate ${\displaystyle 2^{n}}$ possibilities. All of these have to be taken into account if we wish to consider what happens on the average. It will readily be seen that, when all the possible cases have been reckoned once, his total gain will be (reckoned in pounds),
${\displaystyle 2^{n-2}+2^{n-3}\,.\,2+2^{n-4}\,.\,2^{2}+....+2\,.\,2^{n-3}+2^{n-2}+2^{n-1},}$
viz.${\displaystyle (n+1)2^{n-2}}$.
This being spread over ${\displaystyle 2^{n-1}}$ different occasions of gain his average gain will be ${\displaystyle {\tfrac {1}{2}}(n+1)}$.
Now when we are referring to averages it must be remembered that the minimum number of different occurrences necessary in order to justify the average is that which enables each of them to present itself once. A man proposes to stop short at a succession of ten heads. Well and good. We tell him that his average gain will be £5. 10s. 0d.: but we also impress upon him that in order to justify this statement he must commence to toss at least 1024 times, for in no less number can all the contingencies of gain and loss be exhibited and balanced. If he proposes to reach an average gain of £20, he will require to be prepared to go up to 39 throws. To justify this payment he must commence to throw ${\displaystyle 2^{39}}$ times, i.e. about a million million times. Not before he has accomplished this will he be in a position to prove to any sceptic that this is the true average value of a ‘turn’ extending to 39 successive tosses.
Of course if he elects to toss to all eternity we must adopt the line of explanation which alone is possible where questions of infinity in respect of number and magnitude are involved. We cannot tell him to pay down ‘an infinite sum,’ for this has no strict meaning. But we tell him that, however much he may consent to pay each time runs of heads occur, he will attain at last a stage in which he will have won back his total payments by his total receipts. However large ${\displaystyle n}$ may be, if he perseveres in trying ${\displaystyle 2^{n}}$ times he may have a true average receipt of ${\displaystyle {\tfrac {1}{2}}(n+1)}$ pounds, and if he continues long enough onwards he will have it.
The problem will recur for consideration in a future chapter.
1. The following statistics will give a fair idea of the wide range of experience over which such regularity is found to exist: “As illustrations of equal amounts of fluctuation from totally dissimilar causes, take the deaths in the West district of London in seven years (fluctuation 13•66), and offences against the person (fluctuation 13•61); or deaths from apoplexy (fluctuation 5•54), and offences against property, without violence (fluctuation 5•48); or students registered at the College of Surgeons (fluctuation 1•85), and the number of pounds of manufactured tobacco taken for home consumption (fluctuation 1•89); or out-door paupers (fluctuation 3•45) and tonnage of British vessels entered in ballast (fluctuation 3•43), &c.” [Extracted from a paper in the Journal of the Statistical Society, by Mr Guy, March, 1858; the ‘fluctuation’ here given is a measure of the amount of irregularity, that is of departure from the average, estimated in a way which will be described hereafter.]
2. Transactions of the Cambridge Philosophical Society, Vol. ix. p. 605. Reprinted in the collected edition of his writings, p. 50.
3. We might express it thus:—a few instances are not sufficient to display a law at all; a considerable number will suffice to display it; but it takes a very great number to establish that a change is taking place in the law.
4. The mathematician may illustrate the nature of this substitution by the analogies of the ‘circle of curvature’ in geometry, and the ‘instantaneous ellipse’ in astronomy. In the cases in which these conceptions are made use of we have a phenomenon which is continuously varying and also changing its rate of variation. We take it at some given moment, suppose its rate at that moment to be fixed, and then complete its career on that supposition.
5. So called from its first mathematical treatment appearing in the Commentarii of the Petersburg Academy; a variety of notices upon it will be found in Mr Todhunter’s History of the Theory of Probability. | open-web-math/open-web-math | |
# TRANSFORMATIONS OF FUNCTIONS
Transformations of Functions :
In this section, we will learn, how to do different types of transformations of functions like translation, stretch, compression and reflection.
To make the students to understand the different types of transformations, we have explained each kind of transformation with step by step explanation along with the corresponding figures.
## Transformations - Definition
A very simple definition for transformations is, whenever a figure is moved from one location to another location, a Transformation occurs.
If a figure is moved from one location another location, we say, it is transformation.
Our next question is, how will the transformation be
To know that, we have to be knowing the different types of transformations.
Now, let us come to know the different types of transformations.
## Different types of Transformations
The different types of transformations which we can do in the functions are
1. Horizontal Translation
2. Vertical Translation
3. Reflection through the x-axis
4. Reflection through the y-axis
5. Horizontal Expansions and Compressions
6. Vertical Expansions and Compressions
How different types of transformations occur in terms of x-coordinate and y-coordinate have been summarized below.
Summary of Transformation
Transformation
Appearance in Function
Transformation of Point
Vertical Translation
f(x) ---> f(x) + d
(x, y) --> (x, y + d)
Horizontal Translation
f(x) ---> f(x - c)
(x, y) --> (x + c. y)
Vertical Stretch or Compression
f(x) ---> af(x)
(x, y) --> (x, ay)
Horizontal Stretch or Compression
f(x) ---> f(kx)
(x, y) --> (x/k, y)
Reflection in x-axis
f(x) ---> -f(x)
(x, y) --> (x, -y)
Reflection in y-axis
f(x) ---> f(-x)
(x, y) --> (-x, y)
## Order of Transformations
In transformations of functions, if we have more than one transformations, we have to do the transformations one by one in the following order.
1. Stretches / Compressions
2. Reflections
3. Translations
## What is image and pre-image?
When a transformation occurs, the original figure is known as the pre-image and the new figure is known as the image.
It has been clearly shown in the below picture. When a figure is moved from one location to another location, we say that it is a transformation.
In this point, always students have the following questions.
How is this transformation made ?
More clearly, on what grounds is the transformation made ?
Is there any pre-decided rule to make transformation ?
Yes, there is a pre-decided rule to make each and every transformation.
The rule we apply to make transformation is depending upon the kind of transformation we make.
We have already seen the different types of transformations in functions.
For example, if we are going to make transformation of a function using reflection through the x-axis, there is a pre-decided rule for that. According to the rule, we have to make transformation.
The rule that we apply to make transformation using reflection and the rule we apply to make transformation using rotation are not same.
Hence, for each type of transformation, we may have to apply different rule. After having gone through the stuff given above, we hope that the students would have understood, how to do transformations of functions.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | HuggingFaceTB/finemath | |
# Centripetal Force
A Centripetal Force ( from Latin centrum , “centre” and peter , “to seek” ) is a force that causes a body to follow a curved path . Its direction is always orthogonal to the motion of the body and of the instantaneous center of curvature of the path towards the fixed point . Isaac Newton described it as “a force by which bodies are pulled or propelled, or in any way directed towards a point as the centre”. In Newtonian mechanics , the centripetal force of gravity provides the astronomicalclasses .
A common example of a centripetal force is the case in which a body moves along a circular path with a uniform speed. The centripetal force is directed at right angles to the motion and along the radius towards the center of the circular path. The mathematical description was obtained in 1659 by the Dutch physicist Christiaan Huygens.
## sources
The magnitude of the centripetal force on an object of mass m moving along a path with tangential speed v is radius r :
{\displaystyle F_{c}=ma_{c}={\frac {mv^{2}}{r}}}
{\displaystyle a_{c}=\lim _{\Delta {t}\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta {t}}}}
where is is the difference between centripetal acceleration and velocity vectors . Since the velocity vectors in the diagram above have a constant magnitude and since each one stands to their respective position vector, simple vector subtraction implies two similar isosceles triangles with favorable angles of the base and one in which a – The length of one leg , and the other of a base (position vector difference ) and the length of a leg :
a_c{\displaystyle \Delta {\textbf {v}}}{\displaystyle \Delta {\textbf {v}}}v{\displaystyle \Delta {\textbf {r}}}r
{\displaystyle {\frac {|\Delta {\textbf {v}}|}{v}}={\frac {|\Delta {\textbf {r}}|}{r}}}
{\displaystyle |\Delta {\textbf {v}}|={\frac {v}{r}}|\Delta {\textbf {r}}|}
Therefore, can be replaced with :
{\displaystyle |\Delta {\textbf {v}}|}{\displaystyle {\frac {v}{r}}|\Delta {\textbf {r}}|}
{\displaystyle a_{c}=\lim _{{\Delta }t\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta {t}}}={\frac {v}{r}}\lim _{{\Delta }t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta {t}}}=\omega \lim _{{\Delta }t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta {t}}}=v\omega ={\frac {v^{2}}{r}}}
The direction of the force is toward the center of the circle in which the object is moving, or the circle of oscillation (the circle best suited for the object’s local path, if the path is not circular). In the formula, speed is squared, so twice the speed requires four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity of the object about the center of the circle, related to the tangential velocity by the formula
v = \omega r
So That
{\displaystyle F_{c}=mr\omega ^{2}\,.}
Expressed using the orbital period T for one revolution of the circle ,
{\displaystyle \omega ={\frac {2\pi }{T}}\,}
the equation becomes
{\displaystyle F_{c}=mr\left({\frac {2\pi }{T}}\right)^{2}.}
In particle accelerators, the velocity can be much higher (closer to the speed of light in vacuum) so the same rest mass now exerts more inertia (relative mass) requiring more force for the same centripetal acceleration, so the equation becomes :
{\displaystyle F_{c}={\frac {\gamma mv^{2}}{r}}}
Where from
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
is the Lorentz factor .
Thus the centripetal force is given by:
{\displaystyle F_{c}=\gamma mv\omega }
which is the rate of change of relative momentum
{\displaystyle \gamma mv}.
## sources say
In the case of an object moving at the end of a rope in a horizontal plane, the centripetal force on the object is supplied by the tension of the rope. The rope example is an example involving a ‘pull’ force. Centripetal force can also be supplied as a ‘push’ force, as in the case where the normal reaction of the wall supplies the centripetal force to the wall of the die plunger .
This corresponds to Newton ‘s idea of a centripetal force, which is now called the centripetal force . When a satellite is in orbit around a planet , the instantaneous center of gravity is considered to be a centripetal force even though in the case of eccentric orbits, the gravitational force is directed toward the focus, and not toward the instantaneous center of curvature.
Another example of centripetal force arises in the helix which is detected when a charged particle moves in a uniform magnetic field in the absence of other external forces. In this case, the magnetic force is the centripetal force acting towards the helix axis.
## Analysis of multiple cases
Below are three examples of increasing complexity, along with the derivation of the formulas governing velocity and acceleration.
### Uniform circular motion
Uniform circular motion refers to the case of a constant rate of rotation. Here are two approaches to describe the matter.
#### calculus derivation
In two dimensions, the position vector , which has magnitude (length) and is directed at an angle above the x-axis, can be expressed in Cartesian coordinates using the unit vectors and :
\textbf{r}r\theta {\hat {x}}{\hat{y}}
\textbf{r} = r \cos(\theta) \hat{x} + r \sin(\theta) \hat{y}.
Assume uniform circular motion , which requires three things.
• The object moves only in a circle.
• The radius of the circle does not change in time.r
• The object moves with constant angular velocity around the circle. Therefore, where is the time.
\omega\theta = \omega tt
Now find the velocity and the acceleration of the motion by deriving the position with respect to time. va
\textbf{r} = r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y}
\dot{\textbf{r}} = \textbf{v} = - r \omega \sin(\omega t) \hat{x} + r \omega \cos(\omega t) \hat{y}
\ddot{\textbf{r}} = \textbf{a} = - r \omega^2 \cos(\omega t) \hat{x} - r \omega^2 \sin(\omega t) \hat{y}
\textbf{a} = - \omega^2 (r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y})
Note that the root expression of the word in parentheses is in Cartesian coordinates. As a result, r
\textbf{a} = - \omega^2 \textbf{r}.
The negative signifies that the acceleration is pointed toward the center of the circle (as opposed to the radius), so it is called “centripetal” (i.e. “centre-seeking”). While objects naturally follow a straight path (due to inertia), this centripetal acceleration describes a circular motion path due to a centripetal force.
#### Derivation using vectors
The image on the right shows the vector relation for uniform circular motion. The rotation itself is represented by the angular velocity vector , which is normal to the plane of the orbit (using the right-hand rule) and the magnitude is given by:
|\mathbf{\Omega}| = \frac {\mathrm{d} \theta } {\mathrm{d}t} = \omega \ ,
with the angular position in time t . In this subsection , d / d t is assumed constant, independent of time. The distance d of the particle in time d t along the circular path is
\mathrm{d}\boldsymbol{\ell} = \mathbf {\Omega} \times \mathbf{r}(t) \mathrm{d}t \ ,
which, by the properties of the vector cross product , has magnitude r d and is tangent to the direction for the circular path.
As a result,
\frac {\mathrm{d} \mathbf{r}}{\mathrm{d}t} = \lim_{{\Delta}t \to 0} \frac {\mathbf{r}(t + {\Delta}t)-\mathbf{r}(t)}{{\Delta}t} = \frac{\mathrm{d} \boldsymbol{\ell}}{\mathrm{d}t} \ .
In other words,
\mathbf{v}\ \stackrel{\mathrm{def}}{ = }\ \frac {\mathrm{d} \mathbf{r}}{\mathrm{d}t} = \frac {\mathrm{d}\mathbf{\boldsymbol{\ell}}}{\mathrm{d}t} = \mathbf {\Omega} \times \mathbf{r}(t)\ .
differentiate with respect to time,
{\displaystyle \mathbf {a} \ {\stackrel {\mathrm {def} }{=}}\ {\frac {\mathrm {d} \mathbf {v} }{d\mathrm {t} }}=\mathbf {\Omega } \times {\frac {\mathrm {d} \mathbf {r} (t)}{\mathrm {d} t}}=\mathbf {\Omega } \times \left[\mathbf {\Omega } \times \mathbf {r} (t)\right]\ .}
Lagrange’s formula states:
{\displaystyle \mathbf {a} \times \left(\mathbf {b} \times \mathbf {c} \right)=\mathbf {b} \left(\mathbf {a} \cdot \mathbf {c} \right)-\mathbf {c} \left(\mathbf {a} \cdot \mathbf {b} \right)\ .}
Applying Lagrange’s formula with the observation that • r ( t ) = 0 at all times,
\mathbf{a} = - {|\mathbf{\Omega|}}^2 \mathbf{r}(t) \ .
In words, the acceleration is always pointing directly opposite to the radial displacement r , and its magnitude is:
|\mathbf{a}| = |\mathbf{r}(t)| \left ( \frac {\mathrm{d} \theta}{\mathrm{d}t} \right) ^2 = r {\omega}^2
where vertical bars |…| denote the vector magnitude, which in the case of r ( t ) is just the radius r of the path . This result agrees with the previous section, although the notation is slightly different.
When the rate of rotation is kept constant in the analysis of non-uniform circular motion, that analysis agrees.
A beauty of the vector approach is that it is clearly independent of any coordinate system.
#### Example: banked turn
The upper panel in the image to the right shows a ball in circular motion on an edge curve. The curve is bent at an angle to the horizontal , and the road surface is considered to be slippery. The objective is to find out what the angle of the bank should be so that the ball does not slip off the road. Intuition tells us that, on a flat curve with no banking, the ball will simply slide off the road; Whereas with very fast banking, the ball will slide towards the center unless it travels a sharp curve.
The lower panel of the image above indicates the forces on the ball, in addition to any acceleration occurring in the direction of the path. There are two forces; The force of gravity of a ball vertically down through the center is g , where m is the mass of the ball and g is the gravitational acceleration; The second is the upward normal force exerted by the road on the road surface nBut is placed at right angles. The centripetal force demanded by the rotational motion is also shown above. This centripetal force is not the third force exerted on the ball, but must be imparted by the net force on the ball as a result of the vector addition of the normal force and the gravitational force. The resultant or net force found on the ball by the vector addition must be equal to the normal force exerted by the road and the vertical force due to the centripetal force determined by the centripetal force needed to travel along a circular path. The rotational motion is maintained as long as this net force provides the centripetal force necessary for the motion.
The horizontal mesh force on the ball is the horizontal component of the force on the road, whose magnitude is H | = M | n | Sin . The vertical component of the force from the road must counteract the force of gravity: | V | = M | n | Because = m | G |, which means | n | , g | / because , , Substituting in the above formula for h | Produces a horizontal force:
|\mathbf{F}_\mathrm{h}| = m |\mathbf{g}| \frac { \mathrm{sin}\ \theta}{ \mathrm {cos}\ \theta} = m|\mathbf{g}| \mathrm{tan}\ \theta \ .
On the other hand, with velocity. V | On a circular path of radius r , the kinematics says that the force required to make the ball make continuous turns is an inward centripetal force of radius c of magnitude:
|\mathbf{F}_\mathrm{c}| = m |\mathbf{a}_\mathrm{c}| = \frac{m|\mathbf{v}|^2}{r} \ .
As a result, the ball is in a constant path when the angle of the road is set to satisfy the condition:
m |\mathbf{g}| \mathrm{tan}\ \theta = \frac{m|\mathbf{v}|^2}{r} \ ,
OR
\mathrm{tan}\ \theta = \frac {|\mathbf{v}|^2} {|\mathbf{g}|r} \ .
As the angle of bank approaches 90°, the tangent function approaches infinity, so that | , Larger values are obtained for V | 2 / r . In words, this equation states that for maximal speed (large | v |) the road is more steep (for a larger value ) , and the faster turns (for small r ) road too Should be trusted more rapidly, which accords with intuition. when angledoes not satisfy the above condition, then the horizontal component of the force exerted by the road does not provide the true centripetal force, and an additional frictional force called tangent is applied to the road surface to provide the difference. If friction cannot do this (that is, the coefficient of friction is exceeded), the ball slides to a different radius where equilibrium can be felt.
These considerations also apply to air flight. See FAA Pilot’s Manual.
### non-uniform circular motion
As a generalization of the uniform circular motion case, suppose that the angular rate of rotation is not constant. The acceleration now has a tangent component, as shown in the image to the right. This case is used to demonstrate the derivation strategy based on the polar coordinate system.
Let r ( t ) be a vector that describes the position of the point mass as a function of time. Since we are assuming circular motion, let r ( t ) = R u r , where R is a constant (radius of the circle) and r is the unit vector representing the point mass from the origin The direction of r is described by , the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, stands for RAnd points in the direction of increasing . These polar unit vectors can be expressed as Cartesian unit vectors in the x and y directions, denoted i and j , respectively:
r = cos + sin j
AndU = -sin i + cos j . _
To find the velocity one can differentiate:
\mathbf{v} = r \frac {\mathrm{d} \mathbf{u}_\mathrm{r}}{\mathrm{d}t} = r \frac {\mathrm{d}}{\mathrm{d}t} \left( \mathrm{cos}\ \theta \ \mathbf{i} + \mathrm{sin}\ \theta \ \mathbf{j}\right)
= r \frac {d \theta} {dt} \left( -\mathrm{sin}\ \theta \ \mathbf{i} + \mathrm{cos}\ \theta \ \mathbf{j}\right)\,
= r \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{u}_\mathrm{\theta} \,
= \omega r \mathbf{u}_\mathrm{\theta} \,
where is the angular velocity d / d t .
This result for velocity corresponds to the requirements that the velocity should be directed tangentially to the circle, and that the magnitude of the velocity should be rω . differentiate again, and pay attention to
{\frac {\mathrm{d}\mathbf{u}_\mathrm{\theta}}{\mathrm{d}t} = -\frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{u}_\mathrm{r} = - \omega \mathbf{u}_\mathrm{r}} \ ,
We find that the acceleration, a is:
\mathbf{a} = r \left( \frac {\mathrm{d}\omega}{\mathrm{d}t} \mathbf{u}_\mathrm{\theta} - \omega^2 \mathbf{u}_\mathrm{r} \right) \ .
Thus, the radial and tangential components of acceleration are:
\mathbf{a}_{\mathrm{r}} = - \omega^{2} r \ \mathbf{u}_\mathrm{r} = - \frac{|\mathbf{v}|^{2}}{r} \ \mathbf{u}_\mathrm{r} \ ~~And~~ \ \mathbf{a}_{\mathrm{\theta}} = r \ \frac {\mathrm{d}\omega}{\mathrm{d}t} \ \mathbf{u}_\mathrm{\theta} = \frac {\mathrm{d} | \mathbf{v} | }{\mathrm{d}t} \ \mathbf{u}_\mathrm{\theta} \ ,
where | V | = r is the magnitude of velocity.
These equations express mathematically that, in the case of an object that moves along a circular path with a varying speed, the acceleration of the body can be decomposed into a perpendicular component that changes the direction of motion (central acceleration). , and a parallel, or tangent component, which varies speed.
### normal planar motion
Kinetic vector in plane polar coordinates. Note that the setup is not restricted to 2d space, but a plane in any higher dimension.
#### polar coordinates
The above results can probably be more easily obtained in polar coordinates, and also extended to normal speeds within a plane, as shown next. The aircraft crew has polar coordinates consisting of a radial unit vector and an angular unit vector , as shown above. A particle at position r is described by:
\mathbf{r} = \rho \mathbf{u}_{\rho} \ ,
where the notation is used instead of r to describe the distance of the path from the origin to emphasize that this distance is not fixed, but varies with time. The particle in the same direction as the unit vector U and always travels along the point r ( t ). The unit vector U also travels with the particle and remains orthogonal to U . Thus, U and form a local Cartesian coordinate system attached to the particle and bound by the path the particle travels. If its tail coincides, as seen in the circle on the left of the image above, it is observed that the unit vectors and U have a right angled pair with the tips of the unit circle that trace back and trace back to this circle. forward on the circumference with the same angle ( t ) as r ( t ).
When the particle is moving, its velocity is
\mathbf{v} = \frac {\mathrm{d} \rho }{\mathrm{d}t} \mathbf{u}_{\rho} + \rho \frac {\mathrm{d} \mathbf{u}_{\rho}}{\mathrm{d}t} \ .
To evaluate the derivative of velocity, the unit vector is needed. Because u is a unit vector , its magnitude is fixed, and it can only change in direction, that is, its change d u has a component only perpendicular to u . The trajectory when r ( t ) rotates by a sum d , u , which points in the same direction as r ( t ), also rotates by d . See picture above. Therefore, the change in is
\mathrm{d} \mathbf{u}_{\rho} = \mathbf{u}_{\theta} \mathrm{d}\theta \ ,
Or
\frac {\mathrm{d} \mathbf{u}_{\rho}}{\mathrm{d}t} = \mathbf{u}_{\theta} \frac {\mathrm{d}\theta}{\mathrm{d}t} \ .
Similarly, the rate of change of is found. As with U , U is a unit vector and can only rotate without resizing To remain orthogonal to u while the trajectory ( t ) rotates a sum d , u , which is orthogonal to r ( t ), also rotates by d . See picture above. Therefore, the change d is orthogonal to u u and proportional to d(see above image):
\frac{\mathrm{d} \mathbf{u}_{\theta}}{\mathrm{d}t} = -\frac {\mathrm{d} \theta} {\mathrm{d}t} \mathbf{u}_{\rho} \ .
The image above shows the sign to be negative: to maintain orthogonality, if dis positive with d , then dmust decrease.
In the expression for the velocity of the derivative substitute :
\mathbf{v} = \frac {\mathrm{d} \rho }{\mathrm{d}t} \mathbf{u}_{\rho} + \rho \mathbf{u}_{\theta} \frac {\mathrm{d} \theta} {\mathrm{d}t} = v_{\rho} \mathbf{u}_{\rho} + v_{\theta} \mathbf{u}_{\theta} = \mathbf{v}_{\rho} + \mathbf{v}_{\theta} \ .
In order to get the acceleration, the differentiation is done for the second time:
\mathbf{a} = \frac {\mathrm{d}^2 \rho }{\mathrm{d}t^2} \mathbf{u}_{\rho} + \frac {\mathrm{d} \rho }{\mathrm{d}t} \frac{\mathrm{d} \mathbf{u}_{\rho}}{\mathrm{d}t} + \frac {\mathrm{d} \rho}{\mathrm{d}t} \mathbf{u}_{\theta} \frac {\mathrm{d} \theta} {\mathrm{d}t} + \rho \frac{\mathrm{d} \mathbf{u}_{\theta}}{\mathrm{d}t} \frac {\mathrm{d} \theta} {\mathrm{d}t} + \rho \mathbf{u}_{\theta} \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2} \ .
Substituting derivatives u and u , the speed of the particle is:
\mathbf{a} = \frac {\mathrm{d}^2 \rho }{\mathrm{d}t^2} \mathbf{u}_{\rho} + 2\frac {\mathrm{d} \rho}{\mathrm{d}t} \mathbf{u}_{\theta} \frac {\mathrm{d} \theta} {\mathrm{d}t}-\rho \mathbf{u}_{\rho}\left( \frac {\mathrm{d} \theta} {\mathrm{d}t}\right)^2 + \rho \mathbf{u}_{\theta} \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2} \ ,
= \mathbf{u}_{\rho} \left[ \frac {\mathrm{d}^2 \rho }{\mathrm{d}t^2}-\rho\left( \frac {\mathrm{d} \theta} {\mathrm{d}t}\right)^2 \right] + \mathbf{u}_{\theta}\left[ 2\frac {\mathrm{d} \rho}{\mathrm{d}t} \frac {\mathrm{d} \theta} {\mathrm{d}t} + \rho \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2}\right]
= \mathbf{u}_{\rho} \left[ \frac {\mathrm{d}v_{\rho}}{\mathrm{d}t}-\frac{v_{\theta}^2}{\rho}\right] + \mathbf{u}_{\theta}\left[ \frac{2}{\rho}v_{\rho} v_{\theta} + \rho\frac{\mathrm{d}}{\mathrm{d}t}\frac{v_{\theta}}{\rho}\right] \ .
As a special example, if the particle moves in a circle of constant radius r , then d / d = 0, v = v , and:
\mathbf{a} = \mathbf{u}_{\rho} \left[ -\rho\left( \frac {\mathrm{d} \theta} {\mathrm{d}t}\right)^2 \right] + \mathbf{u}_{\theta}\left[ \rho \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2}\right]
{\displaystyle =\mathbf {u} _{\rho }\left[-{\frac {v^{2}}{r}}\right]+\mathbf {u} _{\theta }\left[{\frac {\mathrm {d} v}{\mathrm {d} t}}\right]\ }
Where from
v = v_{\theta}.
These results agree with the above results for non-uniform circular motion. See also article on non-uniform circular motion. If this acceleration is multiplied by the particle mass, the leading term is the centripetal force and the negative of the second term related to the angular acceleration is sometimes called the Euler force.
For trajectories other than circular motion, for example, above the more general trajectory, the instantaneous center of rotation and the radius of curvature of the trajectory to the coordinate system defined in Fig . and length to | r ( t )| = . Consequently, in the general case, it is not straightforward to separate the centripetal and Euler terms from the above general acceleration equation. To deal with this issue directly, local coordinates are preferable, as discussed further.
#### Local coordinates
Local coordinates mean a set of coordinates that the particle travels along, and has an orientation determined by the particle’s path. Unit vectors are formed as shown in the image on the right, both tangent and normal to the path. This coordinate system is sometimes referred to as intrinsic or path coordinates or nt-coordinates , for common-tangents, referring to these unit vectors. These coordinates are a very special example of the more general concept of local coordinates from the theory of differential forms.
The distance along the path of the particle is the length of the arc s , which is assumed to be a known function of time.
s = s(t) \ .
A center of curvature is defined at each position s located a distance (on a line with radius of curvature normal) from the position n ( s ). The required distance ( arc length at s) s is defined as the rate of rotation of the tangent to the curve, which in turn is determined by the path itself. If the orientation of the tangent relative to some initial position is ( s ) , then ( s ) is defined by the derivative d / ds :
\frac{1} {\rho (s)} = \kappa (s) = \frac {\mathrm{d}\theta}{\mathrm{d}s}\ .
The radius of curvature is usually taken (that is, as an absolute value), while the curvature is a signed quantity , taken as positive.
A geometric approach to find the center of curvature and radius of curvature uses a finite process that leads to the oscillating circle. See image above.
Using these coordinates, motion along the path is viewed as a succession of circular paths of ever-changing center, and each position constitutes non-uniform circular motion with radius at that position . Then the place value of the angular rate of rotation is given by:
\omega(s) = \frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\mathrm{d}\theta}{\mathrm{d}s} \frac {\mathrm{d}s}{\mathrm{d}t} = \frac{1}{\rho(s)}\ \frac {\mathrm{d}s}{\mathrm{d}t} = \frac{v(s)}{\rho(s)}\ ,
With local speed v given by :
v(s) = \frac {\mathrm{d}s}{\mathrm{d}t}\ .
As for the other examples above, because unit vectors cannot change magnitude, their rate of change is always perpendicular to their direction (see left-hand insert in the image above):
\frac{d\mathbf{u}_\mathrm{n}(s)}{ds} = \mathbf{u}_\mathrm{t}(s)\frac{d\theta}{ds} = \mathbf{u}_\mathrm{t}(s)\frac{1}{\rho} \ ;\frac{d\mathbf{u}_\mathrm{t}(s)}{\mathrm{d}s} = -\mathbf{u}_\mathrm{n}(s)\frac{\mathrm{d}\theta}{\mathrm{d}s} = - \mathbf{u}_\mathrm{n}(s)\frac{1}{\rho} \ .
Consequently, velocity and acceleration are:
\mathbf{v}(t) = v \mathbf{u}_\mathrm{t}(s)\ ;
and using the chain-law of differentiation:
\mathbf{a}(t) = \frac{\mathrm{d}v}{\mathrm{d}t} \mathbf{u}_\mathrm{t}(s) - \frac{v^2}{\rho}\mathbf{u}_\mathrm{n}(s) \ ;
with tangential acceleration
\frac{\mathrm{\mathrm{d}}v}{\mathrm{\mathrm{d}}t} = \frac{\mathrm{d}v}{\mathrm{d}s}\ \frac{\mathrm{d}s}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}s}\ v \ .
In this local coordinate system, the acceleration resembles the expression for non-uniform circular motion with a local radius (s), and the centripetal acceleration is recognized as the second term
Extending this approach to three-dimensional space curves leads to the Frenet–Serat formulas.
##### Alternative approach
, given the image above, one might wonder whether sufficient account has been taken of the difference in curvature between ( s ) and ( s + ds as computing d in arc length) s = s ) d Assurance at this point can be found using the more formal approach outlined below. This approach also ties in with the article on curvature.
To introduce the unit vectors of a local coordinate system, one approach is to start in Cartesian coordinates and describe the local coordinates in terms of these Cartesian coordinates. In terms of arc length s , the path may be described as:
\mathbf{r}(s) = \left[ x(s),\ y(s) \right] \ .
Then an incremental displacement along the path d s is described by:
\mathrm{d}\mathbf{r}(s) = \left[ \mathrm{d}x(s),\ \mathrm{d}y(s) \right] = \left[ x'(s),\ y'(s) \right] \mathrm{d}s \ ,
where primes are introduced to represent the derivatives with respect to s . The magnitude of this displacement is d s , indicating that:
\left[ x'(s)^2 + y'(s)^2 \right] = 1 \ .(Eq. 1)
This displacement is essentially the tangent to the curve at s , indicating that the unit vector tangent to the curve is:
\mathbf{u}_\mathrm{t}(s) = \left[ x'(s), \ y'(s) \right] \ ,
Whereas the external unit vector common to the curve is
\mathbf{u}_\mathrm{n}(s) = \left[ y'(s),\ -x'(s) \right] \ ,
The orthogonality can be verified by showing that the vector dot product is zero. The unit magnitude of these vectors is the result of Eq. 1. Using the tangent vector, the angle tangent to the curve is given by:
\sin \theta = \frac{y'(s)}{\sqrt{x'(s)^2 + y'(s)^2}} = y'(s) \ ;~~And~~\cos \theta = \frac{x'(s)}{\sqrt{x'(s)^2 + y'(s)^2}} = x'(s) \ .
The radius of curvature is introduced purely formally (without the need for geometric interpretation):
\frac{1}{\rho} = \frac{\mathrm{d}\theta}{\mathrm{d}s}\ .
The derivative of for sin sin can be found from:
\frac{\mathrm{d} \sin\theta}{\mathrm{d}s} = \cos \theta \frac {\mathrm{d}\theta}{\mathrm{d}s} = \frac{1}{\rho} \cos \theta \ = \frac{1}{\rho} x'(s)\ .
Now:
\frac{\mathrm{d} \sin \theta }{\mathrm{d}s} = \frac{\mathrm{d}}{\mathrm{d}s} \frac{y'(s)}{\sqrt{x'(s)^2 + y'(s)^2}}= \frac{y''(s)x'(s)^2-y'(s)x'(s)x''(s)} {\left(x'(s)^2 + y'(s)^2\right)^{3/2}}\ ,
in which the denominator is unity. With this formula for the derivative of the sine, the radius of curvature becomes:
\frac {\mathrm{d}\theta}{\mathrm{d}s} = \frac{1}{\rho} = y''(s)x'(s) - y'(s)x''(s)\ = \frac{y''(s)}{x'(s)} = -\frac{x''(s)}{y'(s)} \ ,
where the equivalence of forms arises from the differentiation of Eq. 1 :
x'(s)x''(s) + y'(s)y''(s) = 0 \ .
With these results, the acceleration can be found:
\mathbf{a}(s) = \frac{\mathrm{d}}{\mathrm{d}t}\mathbf{v}(s)= \frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\mathrm{d}s}{\mathrm{d}t} \left( x'(s), \ y'(s) \right) \right]
= \left(\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\right)\mathbf{u}_\mathrm{t}(s) + \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right) ^2 \left(x''(s),\ y''(s) \right)
= \left(\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\right)\mathbf{u}_\mathrm{t}(s) - \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right) ^2 \frac{1}{\rho} \mathbf{u}_\mathrm{n}(s) \ ,
As can be verified by taking the dot product with the unit vectors t ( s ) and n ( s ). This result for acceleration is the same as for circular motion based on radius . Using this coordinate system in the inertial frame, it is easy to identify the force normal to the trajectory as the centripetal force and the tangential force parallel to the trajectory. From a qualitative point of view, the path can be approximated by the arc of a circle for a finite time, and a particular radius of curvature applied for a finite time, analysis of centrifugal and Euler forces based on circular motion along that radius. can be done. ,
This result for acceleration agrees with a previously found result. However, in this method, the question of the change in radius of curvature along s is completely formally handled in conformity with a geometric interpretation, but it is not relied upon, thereby neglecting the variation in the image above. Is to avoid any question .
##### Example: circular motion
To illustrate the above formulas, let x , y be given as:
x = \alpha \cos \frac{s}{\alpha} \ ; \ y = \alpha \sin\frac{s}{\alpha} \ .
Then,
x^2 + y^2 = \alpha^2 \ ,
which can be recognized as a circular path around the origin with radius α . The position corresponds to s = 0 [ α , 0], or 3 o’clock. To use the above formalism, the derivative needs to be:
y^{\prime}(s) = \cos \frac{s}{\alpha} \ ; \ x^{\prime}(s) = -\sin \frac{s}{\alpha} \ ,
y^{\prime\prime}(s) = -\frac{1}{\alpha}\sin\frac{s}{\alpha} \ ; \ x^{\prime\prime}(s) = -\frac{1}{\alpha}\cos \frac{s}{\alpha} \ .
With these results, one can verify that:
x^{\prime}(s)^2 + y^{\prime}(s)^2 = 1 \ ; \ \frac{1}{\rho} = y^{\prime\prime}(s)x^{\prime}(s)-y^{\prime}(s)x^{\prime\prime}( s) = \frac{1}{\alpha} \ .
Unit vectors can also be found:
\mathbf{u}_\mathrm{t}(s) = \left[-\sin\frac{s}{\alpha} \ , \ \cos\frac{s}{\alpha} \right] \ ; \ \mathbf{u}_\mathrm{n}(s) = \left[\cos\frac{s}{\alpha} \ , \ \sin\frac{s}{\alpha} \right] \ ,
which show that s = 0 is located at the position [ , 0] and s = / 2 at [0, ] , for which the motif agrees with x and y . In other words, s is measured counterclockwise around the circle at 3 o’clock. In addition, the derivatives of these vectors can be found:
\frac{\mathrm{d}}{\mathrm{d}s}\mathbf{u}_\mathrm{t}(s) = -\frac{1}{\alpha} \left[\cos\frac{s}{\alpha} \ , \ \sin\frac{s}{\alpha} \right] = -\frac{1}{\alpha}\mathbf{u}_\mathrm{n}(s) \ ;
\ \frac{\mathrm{d}}{\mathrm{d}s}\mathbf{u}_\mathrm{n}(s) = \frac{1}{\alpha} \left[-\sin\frac{s}{\alpha} \ , \ \cos\frac{s}{\alpha} \right] = \frac{1}{\alpha}\mathbf{u}_\mathrm{t}(s) \ .
To obtain velocity and acceleration, there is a time-dependence necessary for s . For counterclockwise motion at variable speed v ( t ):
s(t) = \int_0^t \ dt^{\prime} \ v(t^{\prime}) \ ,
where v ( t ) is speed and t is time, and s ( t =0) = 0. then:
\mathbf{v} = v(t)\mathbf{u}_\mathrm{t}(s) \ ,
\mathbf{a} = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s) + v\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s) = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s)-v\frac{1}{\alpha}\mathbf{u}_\mathrm{n}(s)\frac{\mathrm{d}s}{\mathrm{d}t}
\mathbf{a} = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s)-\frac{v^2}{\alpha}\mathbf{u}_\mathrm{n}(s) \ ,
where it is already established that α = . This acceleration is the standard result of non-uniform circular motion. | HuggingFaceTB/finemath | |
# Rle Iterator Problem
## Description
LeetCode Problem 900.
We can use run-length encoding (i.e., RLE) to encode a sequence of integers. In a run-length encoded array of even length encoding (0-indexed), for all even i, encoding[i] tells us the number of times that the non-negative integer value encoding[i + 1] is repeated in the sequence.
• For example, the sequence arr = [8,8,8,5,5] can be encoded to be encoding = [3,8,2,5]. encoding = [3,8,0,9,2,5] and encoding = [2,8,1,8,2,5] are also valid RLE of arr.
Given a run-length encoded array, design an iterator that iterates through it.
Implement the RLEIterator class:
• RLEIterator(int[] encoded) Initializes the object with the encoded array encoded.
• int next(int n) Exhausts the next n elements and returns the last element exhausted in this way. If there is no element left to exhaust, return -1 instead.
Example 1:
``````1
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Input
["RLEIterator", "next", "next", "next", "next"]
[[[3, 8, 0, 9, 2, 5]], , , , ]
Output
[null, 8, 8, 5, -1]
Explanation
RLEIterator rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5].
rLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now .
rLEIterator.next(2); // exhausts 2 terms, returning -1. This is because the first term exhausted was 5,
but the second term did not exist. Since the last term exhausted does not exist, we return -1.
``````
Constraints:
• 2 <= encoding.length <= 1000
• encoding.length is even.
• 0 <= encoding[i] <= 10^9
• 1 <= n <= 10^9
• At most 1000 calls will be made to next.
## Sample C++ Code
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class RLEIterator {
public:
int curInd;
vector<int> seq;
RLEIterator(vector<int>& A) {
seq = A;
curInd = 0;
}
int next(int n) {
while (curInd < seq.size()) {
if (seq[curInd] >= n) {
seq[curInd] -= n;
return seq[curInd + 1];
} else {
n -= seq[curInd];
curInd += 2;
}
}
return -1;
}
};
/**
* Your RLEIterator object will be instantiated and called as such:
* RLEIterator* obj = new RLEIterator(encoding);
* int param_1 = obj->next(n);
*/
`````` | HuggingFaceTB/finemath | |
Mathematics 5
Conics
Applications
# Mathematics Conics Ellipse
### 1. Ellipse: Construction:
An ellipse is defined as:
dist (P,F) + dist(P,F') = 2a
P is any point on the curve. F and F' are the two focii of the ellipse; and "dist" is the distance between the two related points.
Let's express the above equation:
sqrt [(c - x)2 + (0 - y)2 ] + [(- c - x)2 + (0 - y)2]1/2 = 2a
[(c - x)2 + y2 ]1/2 + [(- c - x)2 + y2 ]1/2 = 2a
[c2 - 2cx + x2 + y2]1/2 = 2a - [c2 + 2cx + x2 + y2]1/2
c2 - 2cx + x2 + y2 = 4a2 - 4a [c2 + 2cx + x2 + y2]1/2 + c2 + 2cx + x2 + y2
a [c2 + 2cx + x2 + y2]1/2 = a2 + cx
a2 [c2 + 2cx + x2 + y2] = a4 + 2cx a2 + c2x2
a2 [c2 + x2 + y2] = a4 + c2x2
a2 [ x2 + y2] - c2x2 = a2[a2 - c2]
x2[a2 - c2] + a2 y2] = a2[a2 - c2]
x2 b2 + a2 y2 = a2b2
Then:
x2/ a2 + y2/ b2 = 1
x2/ a2 + y2/ b2 = 1
That is the equation of an ellipse.
## 2.1. Eccentricity of an ellipse
The eccentricity "e" of a conic is defined as the ratio :
e = dist(P,F)/dist(P,L)
Applied to the point P = A, and then to the point P = A', yields:
e = dist(A,F)/dist(A,L)
e = (a - c)/(d - a) (1)
e = dist(A',F)/dist(A',L)
e = (a + c)/(d + a) (2)
Solving for "e":
ed - ea = a - c
ed + ea = a + c
2ed = 2a, then: e = a/d, or
Subtracting (2) from (1) gives: -2 ea = - 2c, then: e = c/a
e = c/a , with: 0 < e < 1
## 2.2. Polar equation of en ellipse
Cosine law gives:
PF2 = r2 + c2 - 2rc cosθ
We have :
PL = d - rcosθ
dist(P,F) = e dist(P,L), yields:
sqrt[r2 + c2 - 2r c cosθ] = e(d - r cosθ)
[r2 + c2 - 2r c cosθ]1/2 = e(d - r cosθ)
r2 + c2 - 2r c cosθ = e2d2 + e2 r2 cos2 θ - 2rde2 cosθ
r2[1 - e2 cos2 θ] + 2r cosθ [ de2 - c] + c2 - e2d2 = 0
Since ed = a, ae = c, and b2 = a2 - c2 then:
r2[1 - e2 cos2 θ] = b2
r = b/[1 - e2 cos2 θ]1/2
r = b/[1 - e2 cos2 θ]1/2 =
[(a2 - c2)/(1 - e2 cos2 θ)]1/2
To simplify, lest' consider F = O, that is the focus is at the origin c = 0, therefore:
ro = dist(F,P) = e dist(P,L)
d - c = do
dist(P,L) = do - rocos θo; then:
ro = e[do - rocos θo]
ro[1 + e cos θo] = edo
ro = edo/[1 + e cos θo]
ro = edo/(1 + e cos θo)
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Helga and Rob like corn on cob. Each prefers to eat the gene
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Helga and Rob like corn on the cob. Each prefers to eat the genetically modified type corn, which has perfect cylindrical shape.
Helga likes to eat her corn by chewing circular strips of equal width. Rob prefers to go along the height of the cylinder, munching straight strips of the same width as Helga's strips. If Helga eats half as many circular strips as Rob eats straight strips, what is the ratio between the height and the radius of the corn on the cob?
$$(A)\, 1:2$$
$$(B)\, 2:1$$
$$(C)\, 1:\pi$$
$$(D)\, \pi:1$$
$$(E)\, 1:\sqrt{\pi}$$
[Reveal] Spoiler: OA
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EvaJager wrote:
Helga and Rob like corn on cob. Each prefers to eat the genetically modified type corn, which has perfect cylindrical shape.
Helga likes to eat her corn by chewing circular strips of equal width. Rob prefers to go along the height of the cylinder, munching straight strips of the same width as Helga's strips. If Helga eats half as many circular strips as Rob eats straight strips, what is the ratio between the height and the radius of the corn on cob?
$$(A)\, 1:2$$
$$(B)\, 2:1$$
$$(C)\, 1:\pi$$
$$(D)\, \pi:1$$
$$(E)\, 1:\sqrt{\pi}$$
Say the the width of the strip each eats is $$x$$.
Since Helga eats the corn circularly, then the number of strips she eats is $$\frac{h}{x}$$.
Since Rob eats the corn along the height, then the number of strips he eats is $$\frac{2\pi{r}}{x}$$.
We are told that $$\frac{h}{x}=\frac{1}{2}*\frac{2\pi{r}}{x}$$ --> $$\frac{h}{r}=\pi$$.
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Re: Helga and Rob like corn on cob. Each prefers to eat the gene [#permalink]
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17 Aug 2012, 19:55
Bunuel wrote:
EvaJager wrote:
Helga and Rob like corn on cob. Each prefers to eat the genetically modified type corn, which has perfect cylindrical shape.
Helga likes to eat her corn by chewing circular strips of equal width. Rob prefers to go along the height of the cylinder, munching straight strips of the same width as Helga's strips. If Helga eats half as many circular strips as Rob eats straight strips, what is the ratio between the height and the radius of the corn on cob?
$$(A)\, 1:2$$
$$(B)\, 2:1$$
$$(C)\, 1:\pi$$
$$(D)\, \pi:1$$
$$(E)\, 1:\sqrt{\pi}$$
Say the the width of the strip each eats is $$x$$.
Since Helga eats the corn circularly, then the number of strips she eats is $$\frac{h}{x}$$.
Since Rob eats the corn along the height, then the number of strips he eats is $$\frac{2\pi{r}}{x}$$.
We are told that $$\frac{h}{x}=\frac{1}{2}*\frac{2\pi{r}}{x}$$ --> $$\frac{h}{r}=\pi$$.
Please explain how you obtained the number of strips eaten by each of them; how was the equality obtained? How can we use the information about the way of their eating corn to determine the number of strips eaten by each of them.
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Re: Helga and Rob like corn on cob. Each prefers to eat the gene [#permalink]
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20 Aug 2012, 09:51
The circumference of the cylinder is 2*pi*r.
Dividing by x gives the number of vertical strips ,because each strip is of width x.
The height of the cylinder is h.Dividing by x will give you the number of horizontal strips.
Now apply the data given in the question and equate.
Hope this helps .
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# CFA Level I: Hypothesis Testing
Good evening,
As I keep practicing towards the Level I exam, I want to finish my review of the Quantitative Finance material. This post will hence be the follow up of my previous post. I will here discuss how you test a hypothesis on some statistical measure.
## General Concept
The main concept is as follows, you make an initial hypothesis which is called the null hypothesis, $H_0$, and which is the statement you want to reject. If $H_0$ is rejected, we hence can accept the alternative hypothesis $H_a$. Of course in statistics, you can never be sure of anything. Hence, you can only reject the null hypothesis with a certain confidence level $\alpha$. It is important to understand that if you can’t reject the null hypothesis, it does not mean that you can accept it! Hence, rejecting the null hypothesis is more powerful than failing to reject it. So, if you want to prove some statement, you should test for its opposite as the null hypothesis; if you can reject it, then you can accept the alternative hypothesis which is your original statement. A test can either be one-tailed or two-tailed, depending on what you want to test.
## First example
Let’s take a simple example: you measure a sample of the returns of the S&P that you assume to be normal. You measure a sample volatility $\mu_s$ and a sample standard deviation $s$. By the central limit theorem, we now that the estimate of the mean follows a law $\mathcal{N}(\mu_0, \frac{\sigma^2}{n})$ where $\mu_0$ is the population mean and $\sigma$ is the population standard deviation. Now, you cannot use your measure $\mu_s$ to say that you found the true population mean, because it’s just a sample statistic. However, what you can say is that, given the fact that you found the sample statistic $\mu_s$ and a given confidence interval $\alpha$, it is highly improbable that the population mean $\mu_0$ was equal to some value $x$. Hence the null hypothesis is $\mu_0=x$.
Let’s assume that you found that $\mu_s=0.1$, $s=0.25\%$ and $n=250$. You want to show that $\mu_0 \neq 0$, so you null hypothesis is $\mu_0 = 0$ and with a significance level $\alpha=5\%$. You can compute the z-statistic as follows:
$$z= \frac{\mu_s-\mu_0}{\frac{s}{\sqrt{n}}} \sim \mathcal{N}(0,1)$$
Now if $\mu_0$ was equal to 0, you know that there is 5% chance that the estimate $\mu_s$ would lie outside the range $\mu_0 \pm z_{2.5\%} = 0 \pm 1.96 = \pm 1.96$ (because the sample statistic can lie on both side of the distribution). The z-statistic was computed at 6.33, which is more than 1.96, so you can reject the null hypothesis $\mu_0=0$ and accept the alternative hypothesis $\mu \neq 0$.
This was a two-tailed hypothesis. One-tailed hypothesis would be looking at only one side of the distribution: $H_0 = \mu_0 \geq 0$ or $H_0 = \mu_0 \leq 0$. As a rule of thumb, you can remember the the null hypothesis always contains the “equal” sign.
## P-values
The p-value is the probability (assuming the null hypothesis is true) to have a sample statistic at least as extreme as the one being measure. You compute it by look at the test statistic (6.33 in the previous example) and you find the probability (using the Z-table) that a test statistic can be above that value (and you multiply it by 2 if it is two-tailed). In this case, the p-value is very close to 0. Now, if the p-value of the test statistic is below the significance level of the test, you can reject the null hypothesis. This is useful if you want to discuss statistics without having to impose a certain significance level $\alpha$ to the reader; you can just display the p-value and let him decide whether it’s good enough or not.
There are some other hypothesis tests presented in the curriculum, but this is the main framework to remember. It’s pretty easy, and it allows you to score a lot of points in the quantitative finance part of the exam.
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# Homework Help: Force on a dielectric slab
1. Sep 3, 2014
### albega
...partially inserted between the plates of a parallel plate capacitor.
Imagine a parallel plate capacitor, of plate area l*d, separation d, filled with a linear dielectric of susceptibility X. Now imagine sliding the dielectric out of the plates in the direction that the plates have length l by a distance x. The idea is to find the force on the dielectric.
You can do this for two cases - constant charge and constant voltage. Both give the force as
F=-ε0XwV2/2d along the x direction where V is the p.d between the plates. I understand both of these derivations.
However understanding the results physically is proving to be an issue.
First of all if we let x=0 so the dielectric slab is perfectly between the plates, why is the force non-zero - apparently there is still a force in the negative x direction which does not make sense at all, from the symmetry of the scenario for starters.
Secondly, at constant charge, F=-ε0XwV2/2d but V is not a constant and has x dependence - this makes me happy as I would expect the force to vary with x. However at constant voltage, F=-ε0XwV2/2d but now V is a constant, and so is this telling me the force is the same on the slab wherever it is in the universe (provided it remains in the correct 'slot' between the plates). This just wouldn't make sense surely because as x gets very large, the fringe field causing the force should be vanishingly small.
If anyone can answer these I would be grateful! I have a few more issues about it but not understanding the above probably means I'm not in a position to worry about them just yet.
2. Sep 3, 2014
### TSny
The analysis is based on Griffith's equation (4.62) for the capacitance when the dielectric is displaced $x$ from being fully inserted: $$C = \frac{\varepsilon_0 a}{d}(\varepsilon_r l - \chi_e x)$$
If you think about how this expression is derived, for what range of $x$ does this expression hold?
In particular, is it valid for $x > l$ ?
Is it valid for $x << l \,$?
Consider the assumptions made about a parallel plate capacitor when deriving $C = \frac{\varepsilon A}{d}$.
EDIT: Watch a qualitative demo video here .
Last edited: Sep 3, 2014
3. Sep 4, 2014
### albega
I can see it's not valid for x>l, as the capacitance should then be C=ε0wl/d. So if the capacitance is constant for x>l, does this mean the force on the slab is zero when the slab is out of the capacitor 'slot'? I think the video would sort of agree with that.
For x<<l, I'm not too sure - I'm guessing it's because we require that d is much less than the two dimensions of the capacitor (why exactly do we need this though?), and if x<<l, this isn't the case for the part of the capacitor in the vacuum, invalidating the capacitance expression...
Onto another problem - Griffiths states 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound'.
However although the force expressions are the same in each case, F=-ε0XwV2/2d if you consider them in their separate contexts, they appear to be different. For constant voltage, if you pick some voltage V, F is the same for all x. In the constant charge case, if you pick some V, you get the same F as in the constant voltage case but because V is not being held constant, you can only get this value of V at a certain value of x such that V=Q/C (as C depends on x). So then the variation of force with x appears to be different, despite the expressions being the same - so doesn't that contradict the above statement, or am I misunderstanding it?
Last edited: Sep 4, 2014
4. Sep 4, 2014
### TSny
There will still be some force on the dielectric when it it outside the capacitor. But the force decreases rapidly with distance once its outside. There is a "bowing out" or "fringing" of the field of the plates at the edges of the plates. This field that exists outside of the region between the plates would create some force on the dielectric when it is outside.
Yes, that's right.
I agree with what you are saying. How the force varies with x is different for keeping Q constant and keeping V constant. But, as you noted, in both cases you can express the force at any position with the same equation F=-ε0XwV2/2d. Griffiths is right when he says that the force at some position of the dielectric ultimately depends on just the distribution of free and bound charges at that position. For a particular x, the distribution of the charges will generally be different when keeping V constant as compared to keeping Q constant. So, the force at some x will not be the same for the two cases, in general. But in either case, the force can be expressed by the same equation.
I agree with you that Griffiths statement [ 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound' ] could be interpreted as implying that the force should vary with x in the same way for the two cases. As you noted, that would be incorrect. | HuggingFaceTB/finemath | |
# Is $n$ a multiple of 5 if $n^2$ is?
Say $n$ is a natural number and $n^2$ is a multiple of 5. Does that mean $n$ is a multiple of 5?
This seems to be true. If we look at squares:
\begin{align*} 2^2 &= 4 & 3^2 &= 9 & 4^2 &= 16 & \mathbf{5^2} &\mathbf= \mathbf{25} \\ 6^2 &= 36 & 7^2 &= 49 & 8^2 &= 64 & 9^2 &= 81 \\ \mathbf{ {10}^2 }&\mathbf= \mathbf{100} & {11}^2 &= 121 & {12}^2 &= 144 & {13}^2 &= 169 \\ {14}^2 &= 196 & \mathbf{ {15}^2 }&\mathbf= \mathbf{225} & {16}^2 &= 256 & {17}^2 &= 289 \\ {18}^2 &= 324 & {19}^2 &= 361 & \mathbf{ {20}^2 }&\mathbf= \mathbf{400} & {21}^2 &= 441\\ \end{align*}
Seems to be that $n^2$ is a multiple of 5 only when $n$ is. Can we prove it?
If $n^2$ is a multiple of 5, then $n^2 = 5k$ for some $k \in \mathbb{N}$. Then $n = 5k/n = 5(k/n)$, which means $n$ is a multiple of 5 as long as $k/n$ is an integer… which is true if $k$ is a multiple of $n$. Is $k$ a multiple of $n$?
Well, $k = n^2/5 = n(n/5)$, which means $k$ is a multiple of $n$ if $n$ is a multiple of 5… ah, we’re in a loop here. Let’s try something else.
Maybe we need to use something specific about 5. Is this property true for all numbers? What about 4: is $n$ a multiple of 4 whenever $n^2$ is?
\begin{align*} \mathbf{2^2} &\mathbf= \mathbf{4} & 3^2 &= 9 & \mathbf{4^2} &\mathbf= \mathbf{16} & 5^2 &= 25 \\ \mathbf{6^2} &\mathbf= \mathbf{36} & 7^2 &= 49 & \mathbf{8^2} &\mathbf= \mathbf{64} & 9^2 &= 81 \\ \mathbf{ {10}^2} &\mathbf= \mathbf{100} & {11}^2 &= 121 & \mathbf{ {12}^2} &\mathbf= \mathbf{144} & {13}^2 &= 169 \\ \end{align*}
No! 4, 36, and 100 are multiples of 4, but 2, 6, and 10 are not multiples of 4.
There must be something different about 5 that makes this work for it. 5 is prime… can that help?
The fundamental theorem of arithmetic is often handy when dealing with primes. Let’s represent $n$ as its prime factorization:
$n = p_1p_2\cdots p_n$
Then:
$n^2 = p_1^2p_2^2\cdots p_n^2$
If $n^2$ is a multiple of 5, then 5 must be in $n^2$’s prime factorization. Which means 5 must be one of the primes $p_i$, which in turn means that $n$ itself must also have $5$ as a factor. Woohoo!
So we’ve proved this is true for 5. And, clearly, it’s true for any other prime number. We know it’s not true for 4. Is this true only for primes? Let’s look at 6:
\begin{align*} 2^2 &= 4 & 3^2 &= 9 & 4^2 &= 16 & {5^2} &= {25} \\ \mathbf{6^2} &\mathbf= \mathbf{36} & 7^2 &= 49 & 8^2 &= 64 & 9^2 &= 81 \\ {10}^2 &= {100} & {11}^2 &= 121 & \mathbf{12}^2 &\mathbf= \mathbf{144} & {13}^2 &= 169 \\ {14}^2 &= 196 & {15}^2 &= {225} & {16}^2 &= 256 & {17}^2 &= 289 \\ \mathbf{18}^2 &\mathbf= \mathbf{324} & {19}^2 &= 361 & { {20}^2 }&= {400} & {21}^2 &= 441\\ \end{align*}
Hm… seems like this actually might be true for 6, even though 6 isn’t prime. Can we prove it? It turns out we can make a pretty similar argument:
If $n^2 = p_1^2p_2^2\cdots p_n^2$ is a multiple of 6, 6 won’t be any of the $p_i$ (because 6 isn’t a prime). But 6 = 2*3, which means one of the $p_i$ must be 2 and another of the $p_i$ must be 3.
In case it isn’t clear why: if $m$ is a multiple of 6, then $m = 6k = 2\cdot3\cdot k$. We can prime-factorize $k = q_1q_2\cdots q_n$, which means $m = 2\cdot3\cdot q_1\cdot\cdots\cdot q_n$.
This means $n$’s prime factorization contains 2 and 3, which means $n$ is a multiple of 6. QED
So hang on… since every number has a prime factorization, can’t we make this argument for any number? What goes wrong with 4?
If $n^2 = p_1^2p_2^2\cdots p_n^2$ is a multiple of 4, since 4 = 2*2, one of the $p_i$ must be 2 and another of the $p_i$ must be… ah, 2 again. So we can’t actually prove that $n$ is a multiple of 4, but we can prove that it’s a multiple of 2.
4 happens to be a perfect square, but we’ll actually run into this problem for other numbers too. 12 = 2*2*3. If $n^2 = p_1^2p_2^2\cdots p_n^2$ is a multiple of 12, one of the $p_i$ is 2, one of the $p_i$ is 2 (again), and one of the $p_i$ is 3. So all we know for sure is that $n$ is a multiple of 2*3 = 6.
So it’s clear that this property holds for any $k$ that doesn’t have any repeats in its prime-factorization. These numbers are called square-free.
But even if our number isn’t square-free (like 4, or 12), we can still say something, which is that $n$ is a multiple of the number that you get when you get rid of all the duplicates in $k$’s prime factorization. The term for this is an integer’s radical:
\begin{align*} \operatorname{rad}({4}) &= \operatorname{rad}({2^2}) = 2 \\ \operatorname{rad}({5}) &= 5 \\ \operatorname{rad}({6}) &= \operatorname{rad}({2\cdot3}) = 2\cdot3 = 6 \\ \operatorname{rad}({12}) &= \operatorname{rad}({2^23}) = 2\cdot3 = 6 \\ \end{align*}
So the most general thing we can say is: if $n^2$ is a multiple of $k$, $n$ is a multiple of $\operatorname{rad}(k)$.
Inspired by Putnam 2017 A1.
May 23, 2020 | HuggingFaceTB/finemath | |
# LCM
Common multiple of three numbers is 3276. One number is in this number 63 times, second 7 times, third 9 times. What are the numbers?
Result
a = 52
b = 468
c = 364
#### Solution:
$a=3276/63=52$
$b=3276/7=468$
$c=3276/9=364$
Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!
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# Linear Inequality (One Variable)
How to solve a linear inequality (one variable): 2 examples and their solutions.
## Example 1
### Solution
Solve the inequality just like solving an equation.
First move +5 to the right side.
Then the left side is 7x.
And the right side is 19, change the sign of +5, -5.
19 - 5 = 14
So 7x ≥ 14.
To remove the coefficient 7,
divide both sides by 7.
When multiplying or dividing a minus number on both sides, the order of the inequality sign changes.
In this case, 7 is not a minus number.
So the order of the inequality sign doesn't change.
Then the left side is x.
The order of the inequality sign doesn't change.
And the right side is 14/7.
14/7 = 2
So x ≥ 2.
x ≥ 2
## Example 2
### Solution
First move 2 to the right side.
Then the left side is -3x.
And the right side is 8, change the sign of 2, -2.
8 - 2 = 6
So -3x < 6.
To remove the coefficient -3,
divide both sides by -3.
When multiplying or dividing a minus number on both sides, the order of the inequality sign changes.
In this case, -3 is a minus number.
So the order of the inequality sign does change.
Then the left side is x.
The order of the inequality sign does change: < → >.
And the right side is 6/(-3).
6/(-3) = -2
So x > -2.
x > -2 | HuggingFaceTB/finemath | |
# Difference between revisions of "2021 AIME I Problems/Problem 8"
## Problem
Find the number of integers $c$ such that the equation $$\left||20|x|-x^2|-c\right|=21$$has $12$ distinct real solutions.
## Solution 1 (Piecewise Function: Analysis and Graph)
We take cases for the outermost absolute value, then rearrange: $$\left|20|x|-x^2\right|=c\pm21.$$ Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: $$f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0 \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 020 \end{cases} \end{cases}.$$ We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ($f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$-axis) should facilitate the process of graphing. $[asy] /* Made by MRENTHUSIASM */ size(1200,300); real xMin = -65; real xMax = 65; real yMin = -50; real yMax = 125; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); real f(real x) { return abs(20*abs(x)-x^2); } real g(real x) { return 21; } real h(real x) { return -21; } draw(graph(f,-25,25),red,"y=\left|20|x|-x^2\right|"); draw(graph(g,-65,65),blue,"y=\pm21"); draw(graph(h,-65,65),blue); pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label("(-20,0)",A[0],(-1.5,-1.5),red,UnFill); label("(-10,100)",A[1],(-1.5,1.5),red); label("(0,0)",A[2],(0,-1.5),red,UnFill); label("(10,100)",A[3],(1.5,1.5),red); label("(20,0)",A[4],(1.5,-1.5),red,UnFill); add(legend(),point(E),40E,UnFill); [/asy]$ Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$
• For $f(x)=c+21$ to have $6$ distinct real solutions, we need $0\leq c<79.$
• For $f(x)=c-21$ to have $6$ distinct real solutions, we need $21
Taking the intersection of these two cases gives $21 from which there are $79-21-1=\boxed{057}$ such integers $c.$
~MRENTHUSIASM
## Solution 2 (Graphing)
Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$.
We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$, then rebounds up to $100$ at $x=-10$, then falls back down to $0$ at $x=0$.
The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$, falls back to $0$ at $x=10$, and rises to arbitrarily large values afterwards.
Now we analyze the $y$ (varied by $c$) values. At $y=k<0$, we will have no solutions, as the line $y=k$ will have no intersections with our graph.
At $y=0$, we will have exactly $3$ solutions for the three zeroes.
At $y=n$ for any $n$ strictly between $0$ and $100$, we will have exactly $6$ solutions.
At $y=100$, we will have $4$ solutions, because local maxima are reached at $x= \pm 10$.
At $y=m>100$, we will have exactly $2$ solutions.
To get $12$ distinct solutions for $y=|20|x|-x^2|=c \pm 21$, both $c +21$ and $c-21$ must produce $6$ solutions.
Thus $0 and $c+21<100$, so $c \in \{ 22, 23, \dots , 77, 78 \}$ is required.
It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $\boxed{057}$.
## Solution 3 (Graphing)
Let $y = |x|.$ Then the equation becomes $\left|\left|20y-y^2\right|-c\right| = 21$, or $\left|y^2-20y\right| = c \pm 21$. Note that since $y = |x|$, $y$ is nonnegative, so we only care about nonnegative solutions in $y$. Notice that each positive solution in $y$ gives two solutions in $x$ ($x = \pm y$), whereas if $y = 0$ is a solution, this only gives one solution in $x$, $x = 0$. Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\left|y^2-20y\right| = c \pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$
If $c < 21$, then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$. This means the equation's only solutions are in $\left|y^2-20y\right| = c + 21$. There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \ge 21$. $c$ also can't equal $21$, since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$
At this point, the equation $y^2-20y = c \pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$. Hence we have \begin{align*} -(c + 21) &> -100 \\ c + 21 &< 100 \\ c &< 79. \end{align*} Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $\boxed{057}$ such integers.
Note: Be careful of counting at the end, you may mess up and get $59$.
## Solution 4 (Algebra)
Removing the absolute value bars from the equation successively, we get \begin{align*} \left|\left|20|x|-x^2\right|-c\right|&=21 \\ \left|20|x|-x^2\right|&= c \pm21 \\ 20|x|-x^2 &= \pm c \pm 21 \\ x^2 \pm 20x \pm c \pm21 &= 0. \end{align*} The discriminant of this equation is $$\sqrt{400-4(\pm c \pm 21)}.$$ Equating the discriminant to $0$, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$. However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$. When $k<0$, $a>0$, $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$, $a>0$, then there will be $4$ real solutions, if they exist at all. In order to have $12$ solutions here, we thus need to ensure $-c+21<0$, so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$, so our final bounds are $21, giving us $\boxed{057}$ valid values of $c$.
## Remark
The graphs of $F(x)=\left||20|x|-x^2|-c\right|$ and $G(x)=21$ are shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp
Move the slider around for $21 to observe how they intersect for $12$ times.
~MRENTHUSIASM
## Video Solution
https://youtu.be/6k-uR71_jg0 ~mathproblemsolvingskills.com | open-web-math/open-web-math | |
When we multiply two or more polynomials together, we use our associative, commutative, and distributive properties. Basically, the simplest operation occurs when we multiply a monomial by another monomial. To perform this operation, we multiply the number parts together and the variable parts together. As we move into more difficult problems, such as a binomial times another binomial, we need to use our distributive property to ensure each term of the first polynomial (leftmost) is multiplied by each term of the second polynomial (rightmost). When we have something such as the product of two binomials, we can use a common shortcut known as FOIL (first terms, outer terms, inside terms, and last terms).
Test Objectives
• Demonstrate an understanding of the rules of exponents
• Demonstrate the ability to multiply a monomial by a polynomial
• Demonstrate the ability to use FOIL to multiply two binomials
• Demonstrate the ability to multiply more than two polynomials
Multiplying Polynomials Practice Test:
#1:
Instructions: Find each product.
$$a)\hspace{.2em}2(4x + 2)$$
$$b)\hspace{.2em}5x^3(3x + 2)$$
#2:
Instructions: Find each product.
$$a)\hspace{.2em}(3x - 2)(3x - 2)$$
$$b)\hspace{.2em}(x + 5)(4x - 1))$$
#3:
Instructions: Find each product.
$$a)\hspace{.2em}(4x + 4)(5x + 3)$$
$$b)\hspace{.2em}(5x + 2)(2x + 3)$$
#4:
Instructions: Find each product.
$$a)\hspace{.2em}(2x^2 - 3x - 4)(5x - 1)$$
$$b)\hspace{.2em}(5x^2 - 3x - 1)(4x - 1)$$
#5:
Instructions: Find each product.
$$a)\hspace{.2em}(6x^2 + 6xy - 8y^2)(7x + y)$$
$$b)\hspace{.2em}(x^2 - xy - 5y^2)(2x^2 - 5xy + 2y^2)$$
Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}8x + 4$$
$$b)\hspace{.2em}15x^4 + 10x^3$$
#2:
Solutions:
$$a)\hspace{.2em}9x^2 - 12x + 4$$
$$b)\hspace{.2em}4x^2 + 19x - 5$$
#3:
Solutions:
$$a)\hspace{.2em}20x^2 + 32x + 12$$
$$b)\hspace{.2em}10x^2 + 19x + 6$$
#4:
Solutions:
$$a)\hspace{.2em}10x^3 - 17x^2 - 17x + 4$$
$$b)\hspace{.2em}20x^3 - 17x^2 - x + 1$$
#5:
Solutions:
$$a)\hspace{.2em}42x^3 + 48x^2y - 50xy^2 - 8y^3$$
$$b)\hspace{.2em}2x^4 - 7x^3y - 3x^2y^2 + 23xy^3 - 10y^4$$ | HuggingFaceTB/finemath | |
Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10 1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise
3. Examples
Transcript
Example 19 A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water. Let the speed of boat in still water be x km/hr & let the speed of stream be y km/hr Now, Speed downstream = x + y Speed upstream = x – y Given that A boat goes 30 km upstream and 44 km downstream in 10 hours Time taken to go 30 km upstream + Time taken to go 44 km downstream We know that, Speed = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑇𝑖𝑚𝑒 Time = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑆𝑝𝑒𝑒𝑑 (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 30 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚) + (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 44 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚) = 10 𝟑𝟎/(𝒙 − 𝒚) + 𝟒𝟒/(𝒙 + 𝒚) = 10 Similarly, A boat goes 40 km upstream and 55 km downstream in 13 hours Time taken to go 40 km upstream + Time taken to go 55 km downstream We know that, Speed = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑇𝑖𝑚𝑒 Time = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑆𝑝𝑒𝑒𝑑 (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 40 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚) + (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 55 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚) = 13 𝟒𝟎/(𝒙 − 𝒚) + 𝟓𝟓/(𝒙 + 𝒚) = 13 Our equations are 30 (1/(𝑥 − 𝑦))+44(1/(𝑥 + 𝑦))=10 …(1) 40 (1/(𝑥 − 𝑦))+55(1/(𝑥 + 𝑦))=13 …(2) Let 1/(𝑥 − 𝑦) = u 1/(𝑥 + 𝑦) = v So, our equations become 30u + 44v = 10 40u + 55v = 13 Solving 30u + 44v = 10 …(3) 40u + 55v = 13 …(4) From (3) 30u + 44v = 10 30u = 10 – 44v u = (10 − 44𝑣)/30 Putting value of u in (4) 40u + 55v = 13 40((10 − 44𝑣)/30)+55𝑣=13 4((10 − 44𝑣)/3)+55𝑣=13 Multiplying both sides by 3 3 × 4((10 − 44𝑣)/3)+"3 ×" 55𝑣="3 ×" 13 4(10 – 44v) + 165𝑣= 39 40 – 176v + 165v = 39 – 176v + 165v = 39 – 40 – 11v = –1 v = 𝟏/𝟏𝟏 Putting v = 1/11 in equation (3) 30u + 44v = 10 30u + 44(1/11) = 10 30u + 4 = 10 30u = 10 – 4 30u = 6 u = 6/30 u = 𝟏/𝟓 So, u = 1/5 & v = 1/11 But we need to find x & y We know that u = 𝟏/(𝒙 − 𝒚) 1/5 = 1/(𝑥 − 𝑦) x – y = 5 v = 𝟏/(𝒙 + 𝒚) 1/11 = 1/(𝑥 + 𝑦) x + y = 11 So, our equations become x – y = 5 …(6) x + y = 11 …(7) Adding (6) and (7) (x – y) + (x + y) = 5 + 11 2x = 16 x = 16/2 x = 8 Putting x = 8 in (7) x + y = 11 8 + y = 11 y = 11 – 8 y = 3 So, x = 8, y = 3 is the solution of the given equation Hence Speed of boat in still water = x = 8 km/hr Speed of stream = y = 3 km/hr
Examples | HuggingFaceTB/finemath | |
Multiples of 280 Welcome to the Multiples of 280 page. Here we will first teach you everything you will ever need to know about the multiples of 280, and then give you a study guide summary of everything we taught you to make sure you remember it all. Use this page to look up facts and learn information about the multiples of 280. This page will make you a multiples of two hundred eighty expert!
Definition of Multiples of 280
Multiples of 280 are all the numbers that when divided by 280 equal an integer. Each of the multiples of 280 are called a multiple. A multiple of 280 is created by multiplying 280 by an integer.
Therefore, to create a list of multiples of 280, you start with 1 multiplied by 280, then 2 multiplied by 280, then 3 multiplied by 280, and so on for as long as you want. Thus, the list of the first five multiples of 280 is 280, 560, 840, 1120, and 1400. To see a larger list of multiples of 280, see the printable image of Multiples of 280 further down on this page. We also have a category where you can choose any nth multiple of 280.
Multiples of 280 Checker
The Multiples of 280 Checker below checks to see if any number of your choice is a multiple of 280. In other words, it checks to see if there is any number (integer) that when multiplied by 280 will equal your number. To do that, we divide your number by 280. If the the quotient is an integer, then your number is a multiple of 280.
Is a multiple of 280?
Least Common Multiple of 280 and ...
A Least Common Multiple (LCM) is the lowest multiple that two or more numbers have in common. This is also called the smallest common multiple or lowest common multiple and is useful to know when you are adding our subtracting fractions. Enter one or more numbers below (280 is already entered) to find the LCM.
Check out our LCM Calculator if you need more details about the Least Common Multiple or if you need the LCM for different numbers for adding and subtraction fractions.
nth Multiple of 280
As we stated above, 280 is the first multiple of 280, 560 is the second multiple of 280, 840 is the third multiple of 280, and so on. Enter a number below to find the nth multiple of 280.
th multiple of 280
Multiples of 280 vs Factors of 280
280 is a multiple of 280 and a factor of 280, but that is where the similarities end. All postive multiples of 280 are 280 or greater than 280. All positive factors of 280 are 280 or less than 280.
Below is the beginning list of multiples of 280 and the factors of 280 so you can compare:
Multiples of 280: 280, 560, 840, 1120, 1400, etc.
Factors of 280: 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280
As you can see, the multiples of 280 are all the numbers that you can divide by 280 to get a whole number. The factors of 280, on the other hand, are all the whole numbers that you can multiply by another whole number to get 280.
It's also interesting to note that if a number (x) is a factor of 280, then 280 will also be a multiple of that number (x).
Multiples of 280 vs Divisors of 280
The divisors of 280 are all the integers that 280 can be divided by evenly. Below is a list of the divisors of 280.
Divisors of 280: 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280
The interesting thing to note here is that if you take any multiple of 280 and divide it by a divisor of 280, you will see that the quotient is an integer.
Multiples of 280 Table
Below is an image of the first 100 multiples of 280 in a table. The table is in chronological order, column by column. The first column has the first ten multiples of 280, the second column has the next ten multiples of 280, and so on. The Multiples of 280 Table is also referred to as the 280 Times Table or Times Table of 280. You are welcome to print out our table for your studies.
Negative Multiples of 280
Although not often discussed or needed in math, it is worth mentioning that you can make a list of negative multiples of 280 by multiplying 280 by -1, then by -2, then by -3, and so on, to get the following list of negative multiples of 280:
-280, -560, -840, -1120, -1400, etc.
Multiples of 280 Summary
Below is a summary of important Multiples of 280 facts that we have discussed on this page. To retain the knowledge on this page, we recommend that you read through the summary and explain to yourself or a study partner why they hold true.
There are an infinite number of multiples of 280.
A multiple of 280 divided by 280 will equal a whole number.
280 divided by a factor of 280 equals a divisor of 280.
The nth multiple of 280 is n times 280.
The largest factor of 280 is equal to the first positive multiple of 280.
280 is a multiple of every factor of 280.
280 is a multiple of 280.
A multiple of 280 divided by a divisor of 280 equals an integer.
280 divided by a divisor of 280 equals a factor of 280.
Any integer times 280 will equal a multiple of 280.
Multiples of a Number
Here you can get the multiples of another number, all with the same attention to detail as we did for multiples of 280 on this page.
Multiples of
Multiples of 281
Did you find our page about multiples of two hundred eighty educational? Do you want more knowledge? Check out the multiples of the next number on our list! | HuggingFaceTB/finemath | |
Working Capital Turnover Ratio What is Working Capital Turnover Ratio?
Working Capital Turnover Ratio helps in determining that how efficiently the company is using its working capital (current assets – current liabilities) in the business and is calculated by diving the net sales of the company during the period with the average working capital during the same period.
Working Capital Turnover Ratio Formula
It signifies that how well a company is generating its sales with respect to the working capital of the company. The working capital of a company is the difference between the current assets and current liabilities of a company.
The formula for calculating this ratio is by dividing the sales of the company by the working capital of the company.
Working Capital Turnover Ratio Formula = Sales/ Working Capital
For eg:
Source: Working Capital Turnover Ratio (wallstreetmojo.com)
Interpretation
It signifies that how well a company is generating its sales with respect to the working capital of the company. The two variables to calculate this ratio is sales or turnover and the working capital of a company. The working capital of a company is the difference between the current assets and current liabilities of a company.
Examples
Example#1
Let us try to understand how to calculate the working capital of an arbitrary company by assuming the variables used to calculate working capital turnover.
Suppose for a company A, the sales for a particular company are \$4000.
For the calculation of working capital, the denominator is the working capital. Working capital, which is minus that is why it is important to take the average of working capital. Let’s assume that the working capital for the two respective periods is 305 and 295.
So, the following is the calculation of the Working Capital Turnover Ratio.
The result will be –
Example#2
Let us try to understand how to calculate this ratio of Tata Steel.
For the calculation of working capital, the denominator is the working capital. Working capital, which is current assets minus current liabilities, is a that is why it is important to take the average of working capital. The working capital for Tata steel for the two respective periods is -2946 and 9036
So, the following is the calculation of the Working Capital Turnover Ratio of Tata Steel.
The result will be –
The working capital ratio for Tata steel is 19.83
A higher ratio generally signals that the company is generating more revenue with its working capital. When the current assets are higher than the current liabilities, than the working capital will be a positive number. If the inventory level is lesser in comparison to the payables, than the working capital is low, which is in this case. That makes the working capital ratio very high. It is important to look at working capital ratio across ratio and also in comparison to the industry to make a good analysis of the working capital.
Example#3
Let us try to understand how to calculate the working capital turnover of Hindalco.
Working capital, which is current assets minus current liabilities, is a that is why it is important to take the average of working capital. The working capital for Hindalco for the two respective periods is 9634 and 9006. The below snapshot depicts the variables used to calculate this ratio.
The result will be –
The working capital ratio for Hindalco is 1.28.
A lower ratio generally signals that the company is not generating more revenue with its working capital. When the current assets are higher than the current liabilities, than the working capital will be a positive number. If the inventory level is lower in comparison to the payables, than the working capital is high, which is in this case. That makes the working capital very low. It is important to look at working capital ratio across ratio and also in comparison to the industry to make a good analysis of the formula.
Working Capital Turnover Ratio Calculator
You can use this calculator
Sales Working Capital Working Capital Turnover Ratio Formula
Working Capital Turnover Ratio Formula =
Sales = Working Capital
0 = 0 0
Relevance and Use
A higher working capital generally signals that the company is generating more revenue with its working capital. When the current assets are higher than the current liabilities, than the working capital will be a positive number. It is important to look at all the part that goes into the formula. It’s important to analyze whether the ratio is higher or lower due to a high level of inventory or the management of debtors or credits from whom the company buys raw materials or to whom they sell their finished goods. It is important to look at working capital ratio across ratio and also in comparison to the industry to make a good
You can download this Excel Template here – Working Capital Turnover Ratio Formula Excel Template
Recommended Articles
This article has been a guide to what is Working Capital Turnover Ratio? Here we discuss how to calculate Working Capital Turnover Ratio and its formula along with practical examples and downloadable excel sheets. You can learn more about accounting from the following articles – | HuggingFaceTB/finemath | |
#### What is 0.7 percent of 817?
How much is 0.7 percent of 817? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 0.7% of 817 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 0.7% of 817 = 5.719
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating zero point seven of eight hundred and seventeen How to calculate 0.7% of 817? Simply divide the percent by 100 and multiply by the number. For example, 0.7 /100 x 817 = 5.719 or 0.007 x 817 = 5.719
#### How much is 0.7 percent of the following numbers?
0.7 percent of 817.01 = 571.907 0.7 percent of 817.02 = 571.914 0.7 percent of 817.03 = 571.921 0.7 percent of 817.04 = 571.928 0.7 percent of 817.05 = 571.935 0.7 percent of 817.06 = 571.942 0.7 percent of 817.07 = 571.949 0.7 percent of 817.08 = 571.956 0.7 percent of 817.09 = 571.963 0.7 percent of 817.1 = 571.97 0.7 percent of 817.11 = 571.977 0.7 percent of 817.12 = 571.984 0.7 percent of 817.13 = 571.991 0.7 percent of 817.14 = 571.998 0.7 percent of 817.15 = 572.005 0.7 percent of 817.16 = 572.012 0.7 percent of 817.17 = 572.019 0.7 percent of 817.18 = 572.026 0.7 percent of 817.19 = 572.033 0.7 percent of 817.2 = 572.04 0.7 percent of 817.21 = 572.047 0.7 percent of 817.22 = 572.054 0.7 percent of 817.23 = 572.061 0.7 percent of 817.24 = 572.068 0.7 percent of 817.25 = 572.075
0.7 percent of 817.26 = 572.082 0.7 percent of 817.27 = 572.089 0.7 percent of 817.28 = 572.096 0.7 percent of 817.29 = 572.103 0.7 percent of 817.3 = 572.11 0.7 percent of 817.31 = 572.117 0.7 percent of 817.32 = 572.124 0.7 percent of 817.33 = 572.131 0.7 percent of 817.34 = 572.138 0.7 percent of 817.35 = 572.145 0.7 percent of 817.36 = 572.152 0.7 percent of 817.37 = 572.159 0.7 percent of 817.38 = 572.166 0.7 percent of 817.39 = 572.173 0.7 percent of 817.4 = 572.18 0.7 percent of 817.41 = 572.187 0.7 percent of 817.42 = 572.194 0.7 percent of 817.43 = 572.201 0.7 percent of 817.44 = 572.208 0.7 percent of 817.45 = 572.215 0.7 percent of 817.46 = 572.222 0.7 percent of 817.47 = 572.229 0.7 percent of 817.48 = 572.236 0.7 percent of 817.49 = 572.243 0.7 percent of 817.5 = 572.25
0.7 percent of 817.51 = 572.257 0.7 percent of 817.52 = 572.264 0.7 percent of 817.53 = 572.271 0.7 percent of 817.54 = 572.278 0.7 percent of 817.55 = 572.285 0.7 percent of 817.56 = 572.292 0.7 percent of 817.57 = 572.299 0.7 percent of 817.58 = 572.306 0.7 percent of 817.59 = 572.313 0.7 percent of 817.6 = 572.32 0.7 percent of 817.61 = 572.327 0.7 percent of 817.62 = 572.334 0.7 percent of 817.63 = 572.341 0.7 percent of 817.64 = 572.348 0.7 percent of 817.65 = 572.355 0.7 percent of 817.66 = 572.362 0.7 percent of 817.67 = 572.369 0.7 percent of 817.68 = 572.376 0.7 percent of 817.69 = 572.383 0.7 percent of 817.7 = 572.39 0.7 percent of 817.71 = 572.397 0.7 percent of 817.72 = 572.404 0.7 percent of 817.73 = 572.411 0.7 percent of 817.74 = 572.418 0.7 percent of 817.75 = 572.425
0.7 percent of 817.76 = 572.432 0.7 percent of 817.77 = 572.439 0.7 percent of 817.78 = 572.446 0.7 percent of 817.79 = 572.453 0.7 percent of 817.8 = 572.46 0.7 percent of 817.81 = 572.467 0.7 percent of 817.82 = 572.474 0.7 percent of 817.83 = 572.481 0.7 percent of 817.84 = 572.488 0.7 percent of 817.85 = 572.495 0.7 percent of 817.86 = 572.502 0.7 percent of 817.87 = 572.509 0.7 percent of 817.88 = 572.516 0.7 percent of 817.89 = 572.523 0.7 percent of 817.9 = 572.53 0.7 percent of 817.91 = 572.537 0.7 percent of 817.92 = 572.544 0.7 percent of 817.93 = 572.551 0.7 percent of 817.94 = 572.558 0.7 percent of 817.95 = 572.565 0.7 percent of 817.96 = 572.572 0.7 percent of 817.97 = 572.579 0.7 percent of 817.98 = 572.586 0.7 percent of 817.99 = 572.593 0.7 percent of 818 = 572.6 | HuggingFaceTB/finemath | |
# Construct
Construct a rhombus ABCD, if the size of the diagonal AC is 6 cm and diagonal BD 8 cm long.
a = 5 cm
### Step-by-step explanation:
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## Related math problems and questions:
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In the diamond ABCD, the diagonal e = 24 cm and the size of angle SAB is 28 degrees, where S is the intersection of the diagonals. Calculate the circumference of the diamond.
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Construct a square ABCD with cente S [3,2] and the side a = 4 cm. Point A lies on the x-axis. Construct square image in the displacement given by oriented segment SS'; S` [-1 - 4].
• Sides od triangle
Sides of the triangle ABC has length 4 cm, 5 cm and 7 cm. Construct triangle A'B'C' that are similar to triangle ABC which has a circumference of 12 cm. | HuggingFaceTB/finemath | |
# What is the R squared of a regression where none of the variables are collinear?
If a regression has 100 observations and 100 variables and none of them are collinear, what is the r squared? I know that it means that the full rank assumption is satisfied, so does this mean the r squared equals 1? Also, what would the t-statistic for the 5th coefficient be?
(Most of this a linear algebra question is disguise!)
If the $$100\times100$$ matrix $$X$$ is full-rank, that means the columns form a basis for $$\mathbb R^{100}$$. Since $$y\in\mathbb R^{100}$$, $$y$$ can be written as some linear combination of any basis for $$\mathbb R^{100}$$, such as the set of columns of $$X$$.
That is, the columns of $$X$$ perfectly predict $$y$$, and there is no prediction error (at least not in-sample). Consequently, $$y=\hat y$$, and $$R^2=1$$.
$$R^2=1-\dfrac{ \overset{n}{\underset{i=1}{\sum}}\left(y_i-\hat y_i\right)^2 }{ \overset{n}{\underset{i=1}{\sum}}\left(y_i-\bar y\right)^2 }\\ =1-\dfrac{ \left. \overset{n}{\underset{i=1}{\sum}}\left(y_i-\hat y_i\right)^2 \middle/ n \right. }{ \left. \overset{n}{\underset{i=1}{\sum}}\left(y_i-\bar y\right)^2 \middle/ n \right. } \\ 1-\dfrac{ \text{var}(y-\hat y) }{ \text{var}(y) }\\ =1-\dfrac{0}{\text{var}(y)}=1$$
(This assumes not all values of $$y$$ are equal, but if they are, that is not an interesting regression problem.)
With zero residual variance, it does not make much sense to talk about the t-stats for any of the coefficients, since t-stats divide by residual variance and dividing by zero is frowned upon.
• $\text{Var}$ (or $\text{var}$) could be a good alternative to $var$. Dec 2, 2022 at 16:17
If none of the explanatory variables is colinear (i.e., each pair has zero correlation) then the coefficient-of-determination for the regresion is equal to the sum of the coefficients-of-determination for individual simple linear regessions of each explanatory variable against the response variable. If you would like to learn more about the relationships between collinearity and the coefficient-of-determination in linear regression, as wel as a broader geometric view of regression, you can find some discussion of this topic in my paper O'Neill (2019).
• This is certainly true, yet I’m having a very difficult time reconciling it with my post (which also is correct…I hope). I think an implication of our two posts is that, if $99$ of the features are uncorrelated with the outcome, then that $100th$ feature must be perfectly correlated with the outcome, which I’m not sure I’d ever known until right now.
– Dave
Dec 2, 2022 at 5:20
• This is true but not useful, because in the present case, as @Dave's answer shows, we can determine $R^2$ without having to consider any of the partial $R^2$ values.
– whuber
Dec 2, 2022 at 14:25
• @Dave Perhaps the piece that will make you feel at home is to think about how strong of a constraint the pairwise non-colinearity is on explanatory variables. Dec 2, 2022 at 15:25
• @whuber: Yes, sorry all, in my haste to answer I didn't notice the piece of information that number of observations is equal to number of variables, so apply the other answer in that case. I'll leave this answer up anyway, in case you're interested in the more general case where number of observations is greater than number of variables.
– Ben
Dec 2, 2022 at 20:16 | HuggingFaceTB/finemath | |
## prime factoraization of 3600 by cube roots
Question
prime factoraization of 3600 by cube roots
in progress 0
4 weeks 2021-08-19T00:58:50+00:00 1 Answer 0 views 0
=2|3600
2 |1800
2|900
2 |450
5 |225
5 |45
3|9
3|3
|1
therefore,3600=(2×2)×(2×2)×(3×3)×(5×5)
=2×2×3×5=60 | HuggingFaceTB/finemath | |
1. ## Cones
I need help solving this question... i dont even know where to start... so confused:
Sand is being poured onto a conical pile at the rate of 9 m^3/h. Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3.
a. How fast is the altitude increasing when the radius of the base of the pile is 6m?
b. How fast is the radius of the base increasing when the height of the pile is 10 m?
2. Hello, Tascja!
Sand is being poured onto a conical pile at the rate of 9 m³/hr.
Friction forces in the sand are such that the slope of the sides of the conical pile is always 2/3.
Code:
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3k
The volume of a cone is: . $V \:=\:\frac{\pi}{3}r^2 h$
We are told that the ratio: . $\frac{h}{r} \,=\,\frac{2}{3}\quad\Rightarrow\quad\begin{Bmatri x}h = \frac{2}{3}r \\ r = \frac{3}{2}h\end{Bmatrix}$
a. How fast is the altitude increasing when the radius of the base of the pile is 6 m?
We have: . $V \:=\:\frac{\pi}{3}r^2 h$
Since $r = \frac{3}{2}h\!:\;\;V \:=\:\frac{\pi}{3}\left(\frac{3}{2}h\right)\!^2h \:=\:\frac{3\pi}{4}h^3$
Differentiate with respect to time: . $\frac{dV}{dt}\:=\:\frac{9\pi}{4}h^2\left(\frac{dh} {dt}\right)$ [1]
When $r = 6\!:\;h = \frac{2}{3}(6) = 4$ . . . and we are given: . $\frac{dV}{dt} = 9$
Substitute into [1]: . $9 \:=\:\frac{9\pi}{4}(4^2)\left(\frac{dh}{dt}\right)$
Therefore: . $\frac{dh}{dt} \:=\:\frac{1}{4\pi}$ m/hr.
b. How fast is the radius of the base increasing when the height of the pile is 10 m?
We have: . $V \:=\:\frac{\pi}{3}r^2 h$
Since $h = \frac{2}{3}r\!:\;\;V \:=\:\frac{\pi}{3}r^2\left(\frac{2}{3}r\right)\:=\ :\frac{2\pi}{9}r^3$
Differentiate with respect to time: . $\frac{dV}{dt}\:=\:\frac{2\pi}{3}r^2\left(\frac{dr} {dt}\right)$ [2]
When $h = 10\!:\;r = \frac{3}{2}(10) = 15$ . . . and we are given: . $\frac{dV}{dt} = 9$
Substitute into [2]: . $9 \:=\:\frac{2\pi}{3}(15^2)\left(\frac{dr}{dt}\right )$
Therefore: . $\frac{dr}{dt} \:=\:\frac{3}{50\pi}$ m/hr. | HuggingFaceTB/finemath | |
## Precalculus (6th Edition)
$x=\pm 3$
$\log _{5}\left( x+2\right) +\log _{5}\left( x-2\right) =1\Rightarrow \log _{5}\left( \left( x+2\right) \left( x-2\right) \right) =\log _{5}5$ $\Rightarrow \left( x+2\right) \left( x-2\right) =5\Rightarrow x^{2}-4=5\Rightarrow x^{2}=9\Rightarrow x=\pm 3$ | HuggingFaceTB/finemath | |
# Formula help
Hi, need help with a formula. I’m new to formulas, and trying to set up an excel sheet with formulas into Monday.
I have set up three number colums, and one formula colum at the end. In the formula colum I need to use both sum and minus, and collect the date from the three colums.
In excel it looks like this: =(N16+R16)-P16. For example =(2000+40)-60=1980. But when I try the same set up in the formula by choosing the correct colums from the list, and use sum and minus, it doesn’t work.
You may want to post your actual formula so that we can troubleshoot it. It should look like this:
``````MINUS(SUM({N16},{R16}),{P16})
``````
although you don’t need to use SUM and MINUS. This will work as well:
``````{N16}+{R16}-{P16}
``````
Hi thanks. Doesn’t look like is calculates it correctly.
But maybe the problem is furture down the spreadsheet.
In excel I have this formel: =L16-(L16M16). Looks like this =50000-(5000020%)=40000
L16=numbers
M16=percentage
The closes I have got to set this up is like this: MINUS(MULTIPLY({L16},{M16}),{L16}) but then the answer is -40 000,- not +40 000.
Anyway to solve this?
You’ve got your arguments of the minus function reversed. Try:
``````MINUS({L16},MULTIPLY({L16},{M16}))
`````` | HuggingFaceTB/finemath | |
The Infinity Workshop – SCOPES Digital Fabrication
### Lesson Details
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### Author Desmond Hearne Morrey
Informal educator
Desmond is a student at Oberlin College with a major in philosophy and planned minors in creative writing and mathematics. He is interested in alternative educational forms and particularly finding alternatives for test and memorization based math education. Desmond is… Read More
### Summary
This is a lesson to help students with a background in basic adding of fractions find some wonder in mathematical exploration. We start with a review of what fractions are and how we can represent them and their variants, manipulations, etc. Then we go into repeated addition and tending toward infinity. Themes in this lesson include fractions, asymptotes, and infinity. The lesson is structured as to give students a chance to struggle on difficult problems, work together, and explore math with a focus on aesthetics rather than practicality.
### What You'll Need
Teaching materials such as a whiteboard or chalkboard
Laser Cutter or Printer
## The Instructions
### Introduction and Warm Up
Have a class discussion about what fractions are and how to use them.
First we start with a discussion about fractions, what they are and how to use them. Students should already know everything we discuss in this first step, but it’s a good warm up for what comes next. Make sure students know the basics on adding fractions with different denominators and such. Ask the class as a whole to discuss different ways we can represent numbers and therefore fractions, i.e. as shapes such as the unit square, as numerals, or as points on a line. From here move the discussion toward how operations on numbers and fractions look in these different representations.
Some leading questions might be:
-What are fractions?
-What is the relationship between fractions and whole numbers? (What about when the number is the same as the denominator of the fraction?)
-How do we add fractions together? What about if they have different denominators?
-What are some ways we can represent fractions visually? (i.e. pie charts)
This is a class wide discussion, and it should be more relaxed. There don’t have to be right and wrong answers, but we can discuss which ideas are helpful/interesting and which are less so.
### An Impossible Task
Ask students to find a fraction that can be added to itself infinitely many times, without reaching infinity.
As a class or in groups of 3-4, ask students to try and find a fraction that, when added to itself infinitely many times, adds to a finite number. This is an impossible task, but let students struggle for a little while. If you’re soliciting answers from the whole class maybe call on students and write ideas up on the whiteboard, and then when there are no more ideas discuss if any of them work with the whole class. If you’re having students break out into small groups, have them work on the question for a few minutes and then ask each group to put a response on the board and discuss if any of them work with the class.
Now that students have had a chance to try to find such a number, student who think there exists a fraction that could be added to itself infinitely many times and add to a finite number stand on one side of the classroom, and have students who think otherwise stand on the other side. Have each side state some reasons or arguments for their position. Let students figure this one out themselves if you can.
If you want to show them a demonstration that no single fraction can be added to itself infinitely and add to a finite number, show the following proof:
Say we want to add 1/x to itself infinitely many times, such that x is any number. We can represent such an operation like so: 1/x + 1/x + 1/x + 1/x + …
Group the numbers so that they are in groups of x fractions, each group adding to 1, since 1/x times x = 1.
Therefore, this is equivalent to adding 1 to itself infinitely many times, which tends towards infinity.
In other words, summing 1/x an infinite number of times, is the same as summing 1 an infinite number of times, and then multiplying that sum by 1/x, which is still clearly infinite. ### A More Possible Task
Ask students to find a series of different fractions that sum to a finite number.
Now that students are (hopefully) agreed on the fact that there is no fraction that adds to a finite number after being added to itself an infinite number of times, ask students if they can find an infinite series of different fractions that add to themselves to form a finite number. Mention that, since the students obviously can’t list an infinite number of numbers to you, they’re going to need to find a pattern. An example (that does reach infinity) might be the harmonic series: 1 + 1/2 + 1/3 + 1/4 + 1/5 +… such that the denominator of the next term is always the next number on the number line. Hints could include:
-A reminder of your discussion of visual representations for fractions.
-The fact that the series of fractions is going to have to be getting smaller and smaller, since getting bigger and bigger clearly won’t work and you already tried staying the same.
-A demonstration of why the harmonic series tends toward infinity.
Have students work in groups again, (or if you had the previous discussion as a whole class, work as a whole class again). After 15 minutes (or longer if you want), have students put up their ideas on the board, and discuss each of them as a class. Discuss ways to prove that one of them works or doesn’t work. If no students have found the series 1/2^n, or 1/2+1/4+1/8+1/16+1/32+… demonstrate how it tends toward one by drawing a square, cutting it in half, then cutting one of the halves in half, and repeat so that you have something like this: If student’s weren’t able to find many series before, give them another chance. Each group )or each student, depending on how you’re structuring the lesson), should come up with a series and also a way to visually represent it within a shape. Here’s another example of the 1/2^n series, but on a circle: ### Fabrication of the Infinite
Have students put together a 2d representation of their series on the program Inkscape, and then laser cut the designs.
Students should now have a visual representation of their series drawn out with pen and paper. Now we’ll put them on Inkscape, and make cool sculptures with them! If student’s don’t know how to use Inkscape, give them a quick overview on the line tools, as well as the tools to create and edit basic shapes. If you’re looking for a refresher yourself, here’s the first video of a series of tutorials on youtube: https://www.youtube.com/watch?v=8f011wdiW7g you shouldn’t need more than the first 3 of these for this project.
Once each student has a design on Inkscape, either laser cut the designs such that there are cuts and folds where students want them, or print them on paper with a normal printer, and have students make cuts and folds themselves. This works better if you have a laser cutter, but paper, tape and scissors works okay as well. Here’s an example of one I made with the 1/2^n series: Here’s where I made cuts, colored over in red and folds in blue: ## Standards
• (Fab-Safety.1): I can safely conduct myself in a Fab Lab and observe operations under instructor guidance.
• (Fab-Modeling.1): I can arrange and manipulate simple geometric elements, 2D shapes, and 3D solids using a variety of technologies.
• (Fab-Fabrication.1): I can follow instructor guided steps that link a software to a machine to produce a simple physical artifact. | HuggingFaceTB/finemath | |
# JEE Main & Advanced Mathematics Geometric Progression Properties of G.P.
## Properties of G.P.
Category : JEE Main & Advanced
(1) If all the terms of a G.P. be multiplied or divided by the same non-zero constant, then it remains a G.P., with the same common ratio.
(2) The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common ratio of the original G.P.
(3) If each term of a G.P. with common ratio r be raised to the same power k, the resulting sequence also forms a G.P. with common ratio ${{r}^{k}}$.
(4) In a finite G.P., the product of terms equidistant from the beginning and the end is always the same and is equal to the product of the first and last term. i.e., if ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,......\,{{a}_{n}}$ be in G.P.
Then ${{a}_{1}}\,{{a}_{n}}={{a}_{2}}\,{{a}_{n-1}}={{a}_{3}}\,{{a}_{n-2}}={{a}_{4}}\,{{a}_{n-3}}=..........={{a}_{r}}\,.\,{{a}_{n-r+1}}$
(5) If the terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a G.P.
(6) If ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,.....,\,{{a}_{n}}......$ is a G.P. of non-zero, non-negative terms, then $\log {{a}_{1}},\,\log {{a}_{2}},\,\log {{a}_{3}},\,.....\log {{a}_{n}},\,......$ is an A.P. and vice-versa.
(7) Three non-zero numbers a, b, c are in G.P., iff ${{b}^{2}}=ac$.
(8) If first term of a G.P. of $n$ terms is $a$ and last term is $l,$ then the product of all terms of the G.P. is ${{(al)}^{n/2}}$.
(9) If there be $n$ quantities in G.P. whose common ratio is $r$ and ${{S}_{m}}$ denotes the sum of the first m terms, then the sum of their product taken two by two is $\frac{r}{r+1}\,{{S}_{n}}\,{{S}_{n-1}}$.
(10) If ${{a}^{{{x}_{1}}}},{{a}^{{{x}_{2}}}},{{a}^{{{x}_{3}}}},....,{{a}^{{{x}_{n}}}}$ are in G.P., then ${{x}_{1}},{{x}_{2}},{{x}_{3}},....,{{x}_{n}}$ will be are in A.P. ,
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# In how many ways can 4 women draw water from 4 taps if no tap remains unused?
Question:
In how many ways can 4 women draw water from 4 taps if no tap remains unused?
Solution:
Given: We have 4 women and 4 taps
To Find: Number of ways of drawing water
Condition: No tap remains unused
Let us represent the arrangement
The first woman can use any of the four taps $=4$ Ways
The second woman can use the remaining three taps $=3$ ways
The third woman can use the remaining two taps $=2$ ways
The fourth woman can use the remaining one tap $=1$ way
Total number of ways $=4 \times 3 \times 2 \times 1=24$
There is 24 number of ways in which 4 women can draw water from 4 taps such that no tap remains unused. | HuggingFaceTB/finemath | |
You'll have a hard time inverting a matrix if the determinant of the matrix … : If one of the pivoting elements is zero, then first interchange it's row with a lower row. If the determinant of the matrix is zero, then the inverse does not exist and the matrix is singular. If we calculate the determinants of A and B, we find that, x = 0 is the only solution to Ax = 0, where 0 is the n-dimensional 0-vector. with adj(A)ij=Cij(A)).11Some other sources call the adjugate the adjoint; however on PM the adjoint is reserved for the conjugate transpose. The inverse of a matrix is that matrix which when multiplied with the original matrix will give as an identity matrix. To solve this, we first find the LU decomposition of A, then iterate over the columns, solving Ly=Pbk and Uxk=y each time (k=1…n). The matrix A can be factorized as the product of an orthogonal matrix Q (m×n) and an upper triangular matrix R (n×n), thus, solving (1) is equivalent to solve Rx = Q^T b The inverse of a matrix does not always exist. Next, transpose the matrix by rewriting the first row as the first column, the middle row as the middle column, and the third row as the third column. Method 2: You may use the following formula when finding the inverse of n × n matrix. 2.5. computational complexity . Some caveats: computing the matrix inverse for ill-conditioned matrices is error-prone; special care must be taken and there are sometimes special algorithms to calculate the inverse of certain classes of matrices (for example, Hilbert matrices). There are really three possible issues here, so I'm going to try to deal with the question comprehensively. It should be stressed that only square matrices have inverses proper– however, a matrix of any size may have “left” and “right” inverses (which will not be discussed here). Typically the matrix elements are members of a field when we are speaking of inverses (i.e. The general form of the inverse of a matrix A is. If A cannot be reduced to the identity matrix, then A is singular. Instead of computing the matrix A-1 as part of an equation or expression, it is nearly always better to use a matrix factorization instead. First calculate deteminant of matrix. A precondition for the existence of the matrix inverse A-1 (i.e. The inverse of an n × n matrix A is denoted by A-1. Note that the indices on the left-hand side are swapped relative to the right-hand side. 0. Definition. A-1 A = AA-1 = I n. where I n is the n × n matrix. It can be proven that if a matrix A is invertible, then det(A) ≠ 0. An n × n matrix, A, is invertible if there exists an n × n matrix, A-1, called the inverse of A, such that. If this is the case, then the matrix B is uniquely determined by A, and is called the inverse of A, denoted by A−1. Definition and Examples. 3x3 identity matrices involves 3 rows and 3 columns. One can calculate the i,jth element of the inverse by using the general formula; i.e. Definition. However, the matrix inverse may exist in the case of the elements being members of a commutative ring, provided that the determinant of the matrix is a unit in the ring. where adj(A) is the adjugate of A (the matrix formed by the cofactors of A, i.e. The resulting values for xk are then the columns of A-1. inverse of n*n matrix. The matrix Y is called the inverse of X. Inverse of an identity [I] matrix is an identity matrix [I]. the reals, the complex numbers). Their product is the identity matrix—which does nothing to a vector, so A 1Ax D x. If A is invertible, then its inverse is unique. It looks like you are finding the inverse matrix by Cramer's rule. The inverse is defined so that. Multiply the inverse of the coefficient matrix in the front on both sides of the equation. 4. No matter what we do, we will never find a matrix B-1 that satisfies BB-1 = B-1B = I. Here you will get C and C++ program to find inverse of a matrix. the matrix is invertible) is that detA≠0 (the determinant is nonzero), the reason for which we will see in a second. We can even use this fact to speed up our calculation of the inverse by itself. A-1 A = AA-1 = I n. where I n is the n × n matrix. In this tutorial, we are going to learn about the matrix inversion. For example, when solving the system Ax=b, actually calculating A-1 to get x=A-1b is discouraged. For the 2×2 case, the general formula reduces to a memorable shortcut. So I am wondering if there is any solution with short run time? That is, multiplying a matrix … The n × n matrices that have an inverse form a group under matrix multiplication, the subgroups of which are called matrix groups. Example 1 Verify that matrices A and B given below are inverses of each other. It's more stable. Current time:0:00Total duration:18:40. We look for an “inverse matrix” A 1 of the same size, such that A 1 times A equals I. $$Take the … Theorem. The inverse matrix can be found for 2× 2, 3× 3, …n × n matrices. Instead, they form. Determining the inverse of a 3 × 3 matrix or larger matrix is more involved than determining the inverse of a 2 × 2. The need to find the matrix inverse depends on the situation– whether done by hand or by computer, and whether the matrix is simply a part of some equation or expression or not. You probably don't want the inverse. This can also be thought of as a generalization of the 2×2 formula given in the next section. Assuming that there is non-singular ( i.e. The inverse of a matrix Introduction In this leaflet we explain what is meant by an inverse matrix and how it is calculated. Matrix inversion is the process of finding the matrix B that satisfies the prior equation for a given invertible matrix A. Vote. LU-factorization is typically used instead. n x n determinant. Then the matrix equation A~x =~b can be easily solved as follows. The inverse matrix has the property that it is equal to the product of the reciprocal of the determinant and the adjugate matrix. If the determinant is 0, the matrix has no inverse. I'm betting that you really want to know how to solve a system of equations. Though the proof is not provided here, we can see that the above holds for our previous examples. f(g(x)) = g(f(x)) = x. where Cij(A) is the i,jth cofactor expansion of the matrix A. The proof has to do with the property that each row operation we use to get from A to rref(A) can only multiply the determinant by a nonzero number. As in Example 1, we form the augmented matrix [B|I], However, when we calculate rref([B|I]), we get, Notice that the first 3 columns do not form the identity matrix. The inverse of a matrix exists only if the matrix is non-singular i.e., determinant should not be 0. Search for: Home; The inverse is: The inverse of a general n × n matrix A can be found by using the following equation. Form an upper triangular matrix with integer entries, all of whose diagonal entries are ± 1. With this knowledge, we have the following: A square matrix An£n is said to be invertible if there exists a unique matrix Cn£n of the same size such that AC =CA =In: The matrix C is called the inverse of A; and is denoted by C =A¡1 Suppose now An£n is invertible and C =A¡1 is its inverse matrix. where In is the n × n matrix. The inverse of a matrix The inverse of a square n× n matrix A, is another n× n matrix denoted by A−1 such that AA−1 = A−1A = I where I is the n × n identity matrix. … 0 ⋮ Vote. When we calculate rref([A|I]), we are essentially solving the systems Ax1 = e1, Ax2 = e2, and Ax3 = e3, where e1, e2, and e3 are the standard basis vectors, simultaneously. We will denote the identity matrix simply as I from now on since it will be clear what size I should be in the context of each problem. Golub and Van Loan, “Matrix Computations,” Johns Hopkins Univ. In this method first, write A=IA if you are considering row operations, and A=AI if you are considering column operation. However, due to the inclusion of the determinant in the expression, it is impractical to actually use this to calculate inverses. A square matrix is singular only when its determinant is exactly zero. I'd recommend that you look at LU decomposition rather than inverse or Gaussian elimination. Subtract integer multiples of one row from another and swap rows to “jumble up” the matrix… Inverse Matrices 81 2.5 Inverse Matrices Suppose A is a square matrix. Rule of Sarrus of determinants. We then perform Gaussian elimination on this 3 × 6 augmented matrix to get, where rref([A|I]) stands for the "reduced row echelon form of [A|I]." Adjoint can be obtained by taking transpose of cofactor matrix of given square matrix. An n × n matrix, A, is invertible if there exists an n × n matrix, A-1, called the inverse of A, such that. Decide whether the matrix A is invertible (nonsingular). The inverse is defined so that. It may be worth nothing that given an n × n invertible matrix, A, the following conditions are equivalent (they are either all true, or all false): The inverse of a 2 × 2 matrix can be calculated using a formula, as shown below. Formula for 2x2 inverse. Use Woodbury matrix identity again to get$$ \star \; =\alpha (AA^{\rm T})^{-1} + A^{+ \rm T} G \Big( I-GH \big( \alpha I + HGGH \big)^{-1} HG \Big)GA^+. which has all 0's on the 3rd row. A square matrix that is not invertible is called singular or degenerate. The reciprocal or inverse of a nonzero number a is the number b which is characterized by the property that ab = 1. For the 2×2 matrix. Let A be an n × n (square) matrix. The converse is also true: if det(A) ≠ 0, then A is invertible. Inverse of a Matrix is important for matrix operations. Below are some examples. In this method first, write A=IA if you are considering row operations, and A=AI if you are considering column operation. The matrix A can be factorized as the product of an orthogonal matrix Q (m×n) and an upper triangular matrix R (n×n), thus, solving (1) is equivalent to solve Rx = Q^T b where the adj (A) denotes the adjoint of a matrix. Click here to know the properties of inverse … (We say B is an inverse of A.) Det (a) does not equal zero), then there exists an n × n matrix. The left 3 columns of rref([A|I]) form rref(A) which also happens to be the identity matrix, so rref(A) = I. We use this formulation to define the inverse of a matrix. 3 x 3 determinant. More determinant depth. Matrices are array of numbers or values represented in rows and columns. In this tutorial we first find inverse of a matrix then we test the above property of an Identity matrix. An easy way to calculate the inverse of a matrix by hand is to form an augmented matrix [A|I] from A and In, then use Gaussian elimination to transform the left half into I. Therefore, B is not invertible. Determinants along other rows/cols. We will denote the identity matrix simply as I from now on since it will be clear what size I should be in the context of each problem. The reciprocal or inverse of a nonzero number a is the number b which is characterized by the property that ab = 1. Remember that I is special because for any other matrix A. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). This method is suitable to find the inverse of the n*n matrix. Using the result A − 1 = adj (A)/det A, the inverse of a matrix with integer entries has integer entries. Therefore, we claim that the right 3 columns form the inverse A-1 of A, so. We can obtain matrix inverse by following method. We can cast the problem as finding X in. De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. Definition. We use this formulation to define the inverse of a matrix. 1. Remark When A is invertible, we denote its inverse as A 1. We say that A is invertible if there is an n × n matrix … This general form also explains why the determinant must be nonzero for invertibility; as we are dividing through by its value. was singular. The inverse of a matrix A is denoted by A −1 such that the following relationship holds −. Follow 2 views (last 30 days) meysam on 31 Jan 2014. The need to find the matrix inverse depends on the situation– whether done by hand or by computer, and whether the matrix is simply a part of some equation or expression or not. where a, b, c and d are numbers. Example of finding matrix inverse. [x1 x2 x3] satisfies A[x1 x2 x3] = [e1 e2 e3]. To calculate inverse matrix you need to do the following steps. Commented: the cyclist on 31 Jan 2014 hi i have a problem on inverse a matrix with high rank, at least 1000 or more. Let us take 3 matrices X, A, and B such that X = AB. which is matrix A coupled with the 3 × 3 identity matrix on its right. For instance, the inverse of 7 is 1 / 7. An inverse matrix times a matrix cancels out. Theorem. An n × n matrix, A, is invertible if there exists an n × n matrix, A-1, called the inverse of A, such that. Using determinant and adjoint, we can easily find the inverse of a square matrix using below formula, If det(A) != 0 A-1 = adj(A)/det(A) Else "Inverse doesn't exist" Inverse is used to find the solution to a system of linear equation. Let us take 3 matrices X, A, and B such that X = AB. We will denote the identity matrix simply as I from now on since it will be clear what size I should be in the context of each problem. We prove that the inverse matrix of A contains only integers if and only if the determinant of A is 1 or -1. Let A be an n × n (square) matrix. Next lesson. This method is suitable to find the inverse of the n*n matrix. Generated on Fri Feb 9 18:23:22 2018 by. Whatever A does, A 1 undoes. If A can be reduced to the identity matrix I n , then A − 1 is the matrix on the right of the transformed augmented matrix. Set the matrix (must be square) and append the identity matrix of the same dimension to it. We say that A is invertible if there is an n × n matrix … Problems in Mathematics. AA −1 = A −1 A = 1 . where In denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication. For instance, the inverse of 7 is 1 / 7. An invertible matrix is also said to be nonsingular. Inverse of a matrix A is the reverse of it, represented as A-1.Matrices, when multiplied by its inverse will give a resultant identity matrix. The need to find the matrix inverse depends on the situation– whether done by hand or by computer, and whether the matrix is simply a part of some equation or expression or not. This is the currently selected item. But since [e1 e2 e3] = I, A[x1 x2 x3] = [e1 e2 e3] = I, and by definition of inverse, [x1 x2 x3] = A-1. As a result you will get the inverse calculated on the right. A noninvertible matrix is usually called singular. But A 1 might not exist. We prove that the inverse matrix of A contains only integers if and only if the determinant of A is 1 or -1. When rref(A) = I, the solution vectors x1, x2 and x3 are uniquely defined and form a new matrix [x1 x2 x3] that appears on the right half of rref([A|I]). Inverse matrix. which is called the inverse of a such that:where i is the identity matrix. The inverse of a 2×2 matrix take for example an arbitrary 2×2 matrix a whose determinant (ad − bc) is not equal to zero. You’re left with . Finding the inverse of a 3×3 matrix is a bit more difficult than finding the inverses of a 2 ×2 matrix . Definition of The Inverse of a Matrix Let A be a square matrix of order n x n. If there exists a matrix B of the same order such that A B = I n = B A then B is called the inverse matrix of A and matrix A is the inverse matrix of B. To find the inverse of a 3x3 matrix, first calculate the determinant of the matrix. The inverse of an n × n matrix A is denoted by A-1. You now have the following equation: Cancel the matrix on the left and multiply the matrices on the right. For n×n matrices A, X, and B (where X=A-1 and B=In). Recall that functions f and g are inverses if . Inverse of matrix. Press, 1996. http://easyweb.easynet.co.uk/ mrmeanie/matrix/matrices.htm. It can be calculated by the following method: Given the n × n matrix A, define B = b ij to be the matrix whose coefficients are … Use the “inv” method of numpy’s linalg module to calculate inverse of a Matrix. First, since most others are assuming this, I will start with the definition of an inverse matrix. determinant(A) is not equal to zero) square matrix A, then an n × n matrix A-1 will exist, called the inverse of A such that: AA-1 = A-1 A = I, where I is the identity matrix. A matrix that has no inverse is singular. De &nition 7.1. The inverse is defined so that. Definition :-Assuming that we have a square matrix a, which is non-singular (i.e.
2020 inverse of n*n matrix | HuggingFaceTB/finemath | |
If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $R$ is $\pi R^2$ .
The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk.
The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings — sum them till whole disk is obtained.
Rings and Sections
Mathematically, we can imagine a ring of with radius $x$ and thickness $dx$ , anywhere in the disk having the same center as disk, calculate its area and then sum up (integrate) it from $x=0$ to $x=R$ . Area of a thin ring is since $\pi x dx$ . And after integrating we get, area of disk $A=2 \int_0^R \pi x dx$ or $A=\pi R^2$ .
There is another approach to achieve the area of a disk, A.
An inscribed Triangle
Imagine a disk is made up of a number equal sections or arcs. If there are $n$ number of arcs then interior angle of an arc is exactly $\frac{2\pi}{n}$ , since $2 \pi$ is the total angle at the center of disk and we are dividing this angle into $n$ equal parts. If we join two ends of each sections –we can get $n$ identical triangles in which an angle with vertex O is $\frac{2 \pi}{n}$ . Now, if we can calculate the area of one such section, we can approach to the area of the disk intuitively. This approach is called the method of exhaustion.
Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser.
Area remaining after the Triangle
So, if we increase the number of triangles in disk —-we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle —sum it up to n — make $n$ tending to infinity to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles.
Now, if we draw a radius-line OT’, perpendicular to AB at point T and intersecting the circumference at point T’, we can easily draw another triangle OA’B’ as shown in figure. AOB and A’OB’ are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A’OB’ is equal to the angle AOB. Triangle A’OB’ is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA’B’ is either greater than or equal to area of the disk.
Let disk be divided into n- such inscribed and (thus) superscribed triangles. Since, total angle at point O is 360° or 2π —-the angles AOB and A’OB’ are of $2 \pi/n$ . And also since OT and OT’ are normals at chord AB and line A’B’ respectively, then they must divide the angles AOB and A’OB’ in two equal parts, each of $\pi/n$ radians.
In triangle AOB, the area of triangle AOB is the sum of the area of triangles AOT and BOT. But since both are equal to each other in area, area of AOB must be twice of the area of triangle AOB (or BOT). Our next target is to find, the area of AOT in order to find the area of AOB.
From figure, area of $\bigtriangleup{AOT}= \frac{1}{2} \times AT \times OT$ ….(1)
$OA=R$
And, $\angle{AOT}= \frac{\pi}{n}$ .
Thus, $\frac{AT}{OA}=\sin {\frac{\pi}{n}}$
or, $AT=OA \sin {(\pi/n)}$
or, $AT=R \sin {(\pi/n)}$ …..(2)
Similarly, $OT=R \cos {(\pi/n)}$
Therefore, area of $\bigtriangleup {AOT}=\frac{1}{2} \times R \sin {(\pi/n)} \times R \cos {(\pi/n)}=\frac{1}{2} R^2 \sin{(\pi/n)} \cos {(\pi/n)}$
And thus, area of $\bigtriangleup{AOB}=2 \times \frac{1}{2} R^2 \sin{(\pi/n)} \cos{(\pi/n)}=R^2 \sin{(\pi/n)} \cos{(\pi/n)}$
Since there are $n$ such triangles: sum of areas of such triangles
$S_1=n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)}$ .
In triangle A’OB’, the total area of triangle A’OB’ is the sum of areas two identical triangles A’OT’ and B’OT’. Therefore, area of $\bigtriangleup{A’OB’}=2 \times \text{area of} \bigtriangleup{A’OT’}$ .
And area of $\bigtriangleup{A’OT’}=\frac{1}{2} \times AT’ \times OT’$ .
We have $OT’=R$
and angle A’OT’=$\frac{\pi}{n}$
Thus, A’T’/OT’= $\tan{\frac{\pi}{n}}$
or, $A’T’=OT’ \tan{\frac{\pi}{n}} =R \tan{\frac{\pi}{n}}$ .
Therefore, area of triangle A’OT’= $\frac{1}{2} \times R \times R \tan{\frac{\pi}{n}}=\frac{1}{2} R^2 \tan{\frac{\pi}{n}}$ .
Hence, area of $\bigtriangleup A’OB’=2 \times \frac{1}{2}R^2 \tan{\frac{\pi}{n}}$
As, there are $n$ such triangles: sum of areas of those triangles $S_2=n \times R^2 \tan{\frac{\pi}{n}}$ .
As it is clear that Sum of areas of triangles like $\bigtriangleup AOB$ is an approximation for the area of disk from below, i.e., $S_1 \le A$ when $n \to \infty$ or,
$\displaystyle{\lim_{n \to \infty}} n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)} \le A$
$\displaystyle{\lim_{n \to \infty}} n \times R^2 \frac{\pi}{n} \dfrac{\sin{(\pi/n)}}{\pi/n} \cos{(\pi/n)} \le A$
$\pi R^2 \le A \ldots (I)$
Similarly, $\displaystyle{\lim_{n\to \infty}} n \times R^2 \tan{\frac{\pi}{n}} \ge A$
or, $\pi R^2 \ge A \ldots (II)$
From (I) and (II), we have $A=\pi R^2$ .
So the area of a disk is $\pi R^2$ . | HuggingFaceTB/finemath | |
# MAS115 Mathematical Investigation Skills: Python
Dr Alexander Fletcher
Autumn 2021/22
## Lecture 2: Intended learning outcomes
By the end of this lecture, you will be able to:
• identify standard elements of a programming language;
• describe the process of writing a computer program;
• use this process to find factors by trial division.
## 2 Computers, languages and programming
### 2.1 Programs
A program is a sequence of instructions for the computer to follow.
1. If there is a crossing nearby use it!
2. Otherwise, look left and right until no traffic is coming.
3. Then walk across the road, looking and listening until you get to the other side.
These instructions are written in "natural language" – in this case English.
A computer program is similar but the language is more formalized.
The above illustrates basic features shared by many computer languages.
### 2.2 Standard elements of a programming language
Variables to store data and information, can be of various types `a = 7` Mathematical operations add, multiply, etc. `+`, `*` Input and output operations communicate with the user and the World `input`, `print` Loops for repeating a set of steps `while`, `for` Conditional statements for checking if you want to do one thing or another `if`, `else` Functions / procedures to encapsulate a sequence of steps you perform often `def`
### 2.3 Simplified schematic diagram of a computer ### 2.4 The process of writing a program
1. Understand the problem or task.
2. Design your solution. (In something like English.)
3. Write your program. (In Python.)
4. Test and debug your program.
Writing in Python does not come until step 3!
Often you will go back to previous steps.
Often you will program a little bit at a time.
Task. For each number from 1 to 50,000 find its factors by trial division.
[Trial division means that to find if x is a factor of n you try to divide n by x and see if you have any remainder.]
### 2.6 Part 1: Understand the problem or task
2. Do I understand what a factor is?
3. Do I understand what trial division means?
### 2.7 Part 2: Design your solution
1. Write down the mathematical problem.
2. Start to think how to address it algorithmically.
3. How can I perform trial division?
4. Will I need to repeat any actions?
5. Do I need a loop?
6. What do I need to print to screen?
### 2.8 Part 3: Write your program
```1 2 3 4 5 6 7 8 9``` ```n = 1 while n <= 50000: print("The factors of", n, "are:") i = 1 while i <= n: if n % i == 0: print(i) i = i + 1 n = n + 1 ```
Line 2 starts a loop that will repeat lines 3–9 from `n==1` until `n==50001`.
Line 5 starts a loop that will repeat lines 6–8 from `i==1` until `i==n+1`.
Line 7 prints `i` to screen if `i` is a factor of `n`.
### 2.9 Part 4: Test and debug your program
1. Does the program run?
2. Does the program print anything to screen?
3. Does the program print 1 and n as factors for each value of n?
4. Does the program give the correct answer for a smaller value than 50,000?
5. Does the program take a long time to run?
(We will discuss refactoring and debugging in the next two lectures.)
### 2.10 Conditional statements: the `if` command
```if (boolean_condition):
do this
followed by this
carry on here
```
For example
```password = input("Enter password: ") | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Subtract 3/56 from 8/6
1st number: 1 2/6, 2nd number: 3/56
8/6 - 3/56 is 215/168.
#### Steps for subtracting fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 6 and 56 is 168
Next, find the equivalent fraction of both fractional numbers with denominator 168
2. For the 1st fraction, since 6 × 28 = 168,
8/6 = 8 × 28/6 × 28 = 224/168
3. Likewise, for the 2nd fraction, since 56 × 3 = 168,
3/56 = 3 × 3/56 × 3 = 9/168
4. Subtract the two like fractions:
224/168 - 9/168 = 224 - 9/168 = 215/168
5. In mixed form: 147/168
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
# Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$
$$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
• Note that $\sqrt{a+b}+\sqrt{a-b}\neq\sqrt{2a},$ which is where I think you made your mistake. – Arcturus Jul 26 '16 at 19:04
• Observe that the minimal polynomial for roots $6±2\sqrt5$ is $x^2-12x+16$ – Zack Ni Jul 27 '16 at 11:36
• Consider $\sqrt{6.5 - 2.5} + \sqrt{6.5 + 2.5}$. The answer is not $\sqrt{13}$. – John Joy Jul 27 '16 at 17:05
Set $$t=\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$ $$t^2=6-2\sqrt5+2\sqrt{(6-2\sqrt5)(6+2\sqrt5)}+6+2\sqrt5$$ $$t^2=12+2\sqrt{36-20}=12+2(4)=20$$ $$t^2=20\implies t=2\sqrt5$$
Hint. Observe that $$(\sqrt{5}-1)^2=6-2\sqrt{5},\quad (\sqrt{5}+1)^2=6+2\sqrt{5}.$$
Hint: $(\sqrt{5}\pm 1)^2 = 6 \pm 2 \sqrt{5}$.
More generally, if $t =\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$,
$\begin{array}\\ t^2 &=(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}})^2\\ &=a+\sqrt{b}+2(\sqrt{a+\sqrt{b}}\sqrt{a-\sqrt{b}})+a-\sqrt{b}\\ &=2a+2\sqrt{(a+\sqrt{b})(a-\sqrt{b})}\\ &=2a+2\sqrt{a^2-b}\\ \text{so}\\ t &=\sqrt{2a+2\sqrt{a^2-b}}\\ \end{array}$
In this case, $a=6$ and $b=20$ so $t =\sqrt{2\cdot 6+2\sqrt{36-20}} =\sqrt{12+8} =\sqrt{20} =2\sqrt{5}$.
Hint $\,\ \sqrt a + \sqrt b\, = \sqrt{(\sqrt a + \sqrt b)^2} = \sqrt{a+b +2{\sqrt{ab}}}.\$ Here $\ a+b=12,\ \color{}{\sqrt{ab}} = 4.$
• Did you mean $\sqrt{a + b + 2\sqrt{ab}}$? – N. F. Taussig Jul 26 '16 at 15:54
Let $\alpha = \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}$. Then $$\alpha^2 = 12+2\sqrt{36-20} = 20$$ hence $\color{red}{\alpha=2\sqrt{5}}$. | HuggingFaceTB/finemath | |
# What is the standard form of y= (x-2)(3x+2)(-x+8)?
Dec 10, 2017
$- 3 {x}^{3} + 28 {x}^{2} - 28 x - 32$
#### Explanation:
$\text{expand the first 2 factors using FOIL then multiply the}$
$\text{result by the third factor}$
$\left(x - 2\right) \left(3 x + 2\right) \leftarrow \textcolor{b l u e}{\text{first 2 factors}}$
$= 3 {x}^{2} + 2 x - 6 x - 4$
$= 3 {x}^{2} - 4 x - 4$
$\Rightarrow \left(3 {x}^{2} - 4 x - 4\right) \left(- x + 8\right) \leftarrow \textcolor{b l u e}{\text{multiply by third factor}}$
$= - 3 {x}^{3} + 24 {x}^{2} + 4 {x}^{2} - 32 x + 4 x - 32$
$= - 3 {x}^{3} + 28 {x}^{2} - 28 x - 32 \leftarrow \textcolor{b l u e}{\text{in standard form}}$ | HuggingFaceTB/finemath | |
## Precalculus (6th Edition) Blitzer
$685$
Formula to calculate the nth term of an arithmetic sequence is: $a_n=a_1+d(n-1)$ Here, $a_1:$ First term and $d=$ common difference Since, we have $a_1=-60$ Then $a_{160}=-60+5(150-1)=-60+745=685$ | HuggingFaceTB/finemath | |
# Output impedance
ananthu
The concept of output impedance is highly confusing to me.
I will be thankful if some body gives simple explanations to the following.
1. I undertand that in the case of current amplifiers, the out put impedance should be low and the input one should be high. In the case of voltage amplifiers the opposite is true. If you take slope in the saturation region of the output characteristics of a transistor, the impedance will be very low and if you take it on the active region it will be enormously high. If you take it at a point close the knee point it will be intermediate. So it goes on varying. Then how can we judge the right value of output imp. and where it should be measured ?
2.This problem comes while we teach the higher secondary students in the practical class when they draw the graph for the same.
3. The textbooks demand that their output imp. should be lower than the input one. (in the case of NPN transistor in CE mode.
4. In this case I feel another contradiction. Because the theory part says that a transistor is operated in active region as only in that regions the collector current is independent of the coll-emit. voltage. But if you take impedance in that region the output imp. will be higher than the input one.
If I get a detailed explanation it will be useful to me.
yungman
I think it will be much more beneficial if you spend the time to understand how the output and input impedance behave. I don't think you can just have a universal statement.
For transistor, the input impedance $Z_{in}=\beta \; (r'_e+R_E)\;\hbox { where }\;r'_e=\frac {I_c}{V_t}\;\hbox { and } \;V_t=25mV \;\hbox { at 25 deg C.}$
I don't have the equation for output impedance, someone can give you this or I'll find that tomorrow if I have time. It is not important to answer your question. You just regard it to be very high.
I take that your "current amplifier" means common collector or emitter follower. You always have an emitter resistor $R_E\;$, so you can calculate the input impedance from the equation above. The output impedance is $r'_e\;$.
I take that your "voltage amplifier" means common emitter stage with a resistor $R_c\;$ from collector to the power supply. You calculate the input impedance using the equation given. Usually the collector resistor $R_c\;$ is much lower than the output impedance of the transistor, so the output impedance is very very close to just simple $R_c\;$.
You can make your generalization from this. | HuggingFaceTB/finemath | |
1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Examples
Transcript
Example 47 Find intervals in which the function given by f(𝑥) =310𝑥4 – 45𝑥3– 3𝑥2 + 365𝑥 + 11 is (a) strictly increasing (b) strictly decreasing f𝑥 = 310𝑥4 – 45𝑥3– 3𝑥2 + 365𝑥 + 11 Step 1: Finding f’𝑥 f’𝑥 = 310 × 4𝑥3 – 45 × 3𝑥2 – 3 × 2x + 365 + 0 f’𝑥 = 1210𝑥3– 125𝑥2– 6x + 365 f’𝑥 = 65𝑥3− 125𝑥2– 6x + 365 f’𝑥 = 6𝑥35−2𝑥25−𝑥+65 f’𝑥 = 6𝑥3 − 2𝑥2− 5𝑥 + 65 = 65 𝑥3−2𝑥2−5𝑥+6 = 65𝑥−1𝑥2−𝑥−6 = 65 𝑥−1𝑥2−3𝑥+2𝑥−6 = 65 𝑥−1𝑥𝑥−3+2𝑥−3 = 65 𝑥−1𝑥+2𝑥−3 Hence f’𝑥 = 65𝑥−1𝑥+2𝑥−3 Step 2: Putting f’𝑥=0 65𝑥−1𝑥+2𝑥−3 = 0 𝑥−1𝑥+2𝑥−3 = 0 Hence x = –2 , 1 & 3 Step 3: Plotting point on real line Thus, we get four disjoint intervals i.e. −−2 ,−2, 1 ,1 , 3, 3 , uc1 ⇒ f𝑥 is strictly decreasing on the interval 𝑥 ∈−−𝟐& 𝟏 , 𝟑 f𝑥 is strictly increasing on the interval 𝑥 ∈−𝟐,𝟏 & 𝟑 , uc1
Examples | HuggingFaceTB/finemath | |
# OptimizationExpression
Arithmetic or functional expression in terms of optimization variables
## Description
An OptimizationExpression is an arithmetic or functional expression in terms of optimization variables. Use an OptimizationExpression as an objective function, or as a part of an inequality or equality in a constraint or equation.
## Creation
Create an optimization expression by performing operations on OptimizationVariable objects. Use standard MATLAB® arithmetic including taking powers, indexing, and concatenation of optimization variables to create expressions. See Supported Operations for Optimization Variables and Expressions and Examples.
You can also create an optimization expression from a MATLAB function applied to optimization variables by using fcn2optimexpr. For examples, see Create Expression from Nonlinear Function and Problem-Based Nonlinear Optimization.
Create an empty optimization expression by using optimexpr. Typically, you then fill the expression in a loop. For examples, see Create Optimization Expression by Looping and the optimexpr function reference page.
After you create an expression, use it as either an objective function, or as part of a constraint or equation. For examples, see the solve function reference page.
## Properties
expand all
Index names, specified as a cell array of strings or character vectors. For information on using index names, see Named Index for Optimization Variables.
Data Types: cell
Optimization variables in the object, specified as a structure of OptimizationVariable objects.
Data Types: struct
## Object Functions
evaluate Evaluate optimization expression or objectives and constraints in problem show Display information about optimization object write Save optimization object description
## Examples
collapse all
Create optimization expressions by arithmetic operations on optimization variables.
x = optimvar('x',3,2);
expr = sum(sum(x))
expr =
Linear OptimizationExpression
x(1, 1) + x(2, 1) + x(3, 1) + x(1, 2) + x(2, 2) + x(3, 2)
f = [2,10,4];
w = f*x;
show(w)
(1, 1)
2*x(1, 1) + 10*x(2, 1) + 4*x(3, 1)
(1, 2)
2*x(1, 2) + 10*x(2, 2) + 4*x(3, 2)
Create an optimization expression by transposing an optimization variable.
x = optimvar('x',3,2);
y = x'
y =
2x3 Linear OptimizationExpression array with properties:
IndexNames: {{} {}}
Variables: [1x1 struct] containing 1 OptimizationVariable
See expression formulation with show.
Simply indexing into an optimization array does not create an expression, but instead creates an optimization variable that references the original variable. To see this, create a variable w that is the first and third row of x. Note that w is an optimization variable, not an optimization expression.
w = x([1,3],:)
w =
2x2 OptimizationVariable array with properties:
Name: 'x'
Type: 'continuous'
IndexNames: {{} {}}
Elementwise properties:
LowerBound: [2x2 double]
UpperBound: [2x2 double]
Reference to a subset of OptimizationVariable with Name 'x'.
See variables with show.
See bounds with showbounds.
Create an optimization expression by concatenating optimization variables.
y = optimvar('y',4,3);
z = optimvar('z',4,7);
f = [y,z]
f =
4x10 Linear OptimizationExpression array with properties:
IndexNames: {{} {}}
Variables: [1x1 struct] containing 2 OptimizationVariables
See expression formulation with show.
Use optimexpr to create an empty expression, then fill the expression in a loop.
y = optimvar('y',6,4);
expr = optimexpr(3,2);
for i = 1:3
for j = 1:2
expr(i,j) = y(2*i,j) - y(i,2*j);
end
end
show(expr)
(1, 1)
y(2, 1) - y(1, 2)
(2, 1)
y(4, 1) - y(2, 2)
(3, 1)
y(6, 1) - y(3, 2)
(1, 2)
y(2, 2) - y(1, 4)
(2, 2)
y(4, 2) - y(2, 4)
(3, 2)
y(6, 2) - y(3, 4)
Create an optimization expression corresponding to the objective function
$f\left(x\right)={x}^{2}/10+\mathrm{exp}\left(-\mathrm{exp}\left(-x\right)\right).$
x = optimvar('x');
f = x^2/10 + exp(-exp(-x))
f =
Nonlinear OptimizationExpression
((x.^2 ./ 10) + exp((-exp((-x)))))
Find the point that minimizes fun starting from the point x0 = 0.
x0 = struct('x',0);
prob = optimproblem('Objective',f);
[sol,fval] = solve(prob,x0)
Solving problem using fminunc.
Local minimum found.
Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
sol = struct with fields:
x: -0.9595
fval = 0.1656
If f were not a supported function, you would convert it using fcn2optimexpr. See Supported Operations for Optimization Variables and Expressions and Convert Nonlinear Function to Optimization Expression.
f = @(x)x^2/10 + exp(-exp(-x));
fun = fcn2optimexpr(f,x)
fun =
Nonlinear OptimizationExpression
((x.^2 ./ 10) + exp((-exp((-x)))))
prob = optimproblem('Objective',fun);
[sol,fval] = solve(prob,x0)
Solving problem using fminunc.
Local minimum found.
Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
sol = struct with fields:
x: -0.9595
fval = 0.1656
Create an optimization expression of two variables.
x = optimvar("x",3,2);
y = optimvar("y",1,2);
expr = sum(x,1) - 2*y;
Evaluate the expression at a point.
xmat = [3,-1;
0,1;
2,6];
sol.x = xmat;
sol.y = [4,-3];
val = evaluate(expr,sol)
val = 1×2
-3 12 | HuggingFaceTB/finemath | |
# GMAT Math : DSQ: Calculating discrete probability
## Example Questions
← Previous 1 3 4 5 6 7 8 9
### Example Question #1 : Dsq: Calculating Discrete Probability
Data sufficiency question- do not actually solve the question
A bag of marbles consist of a mixture of black and red marbles. What is the probability of choosing a red marble followed by a black marble?
1. The probability of choosing a black marble first is .
2. There are 10 black marbles in the bag.
Statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question
Statements 1 and 2 together are not sufficient, and additional information is necessary to answer the question
Each statement alone is sufficient to answer the question
Statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question
Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient
Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient
Explanation:
From statement 1, we know the probabilty of choosing the first marble. However, since the marble is not replaced, it is impossible to calculate the probability of choosing the second marble. By knowing the information in statement 2 combined with statement 1, we can calculate the total number of marbles initially present.
### Example Question #1 : Discrete Probability
A certain major league baseball player gets on base 25% of the time (once every 4 times at bat).
For any game where he comes to bat 5 times, what is the probability that he will get on base either 3 or 4 times? - Hint – add the probability of 3 to the probability of 4. Explanation: Binomial Table
### Example Question #1 : Dsq: Calculating Discrete Probability
A type 1 error (False Alarm or 'Convicting the innocent man') occurs when we incorrectly reject a true null hypothesis.
A type 2 error (failure to detect) occurs when we fail to reject a false null hypothesis.
Which one of the following 5 statements is false?
Note - only 1 of the statements is false.
A) For a given sample size (n=100), decreasing the significane level (from .05 to .01) will decrease the chance of a type 1 error.
B) For a given sample size (n=100), increasing the significane level (from .01 to .05) will decrease the chance of a type 2 error.
C) The ability to correctly detect a false null hypothesis is called the 'Power' of a test.
D) Increasing sample size (from 100 to 120) will always decrease the chance of both a type 1 error and a type 2 error.
E) None of the above statements are true.
A)
B)
C)
E) None of the above statements are true.
D)
E) None of the above statements are true.
Explanation:
Statements A, B, C, and D are all true - so -
The only false statement is E (the statement that declares that A and B and C and D are all false)
### Example Question #1 : Dsq: Calculating Discrete Probability
A marble is selected at random from a box of red, yellow, and blue marbles. What is the probability that the marble is yellow?
1) There are ten blue marbles in the box.
2) There are eight red marbles in the box.
Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is not sufficient.
BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.
EITHER Statement 1 or Statement 2 ALONE is sufficient to answer the question.
BOTH statements TOGETHER are NOT sufficient to answer the question.
Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is not sufficient.
BOTH statements TOGETHER are NOT sufficient to answer the question.
Explanation:
To determine the probability that the marble is yellow we need to know two things: the number of yellow marbles, and the number of marbles total. The first quantity divided by the last quantity is our probablility.
But the two given statements together only tell us that eighteen marbles are not yellow. This is not enough information. For example, if there are two yellow marbles, the probability of drawing a yellow marble is . But if there are twenty-two yellow marbles, the probability of drawing a yellow marble is Therefore, the answer is that both statements together are insufficient to answer the question.
### Example Question #5 : Discrete Probability
Several decks of playing cards are shuffled together. One card is drawn, shown, and put aside. Another card is dealt. What is the probability that the dealt card is red, assuming the first card is known?
1) The card removed before the deal was red.
2) The cards were shuffled again between the draw and the deal.
Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is not sufficient.
BOTH statements TOGETHER are insufficient to answer the question.
Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is not sufficient.
BOTH statements TOGETHER are insufficient to answer the question.
EITHER statement ALONE is sufficient to answer the question.
BOTH statements TOGETHER are insufficient to answer the question.
Explanation:
To answer this question you need to know two things: the number of red cards left and the number of total cards left. The second statement is irrelevant, as a reshuffle does not change the composition of the deck. The first statement tells you that there is one fewer red card than black cards, but it does not tell you how many of each there are, as you do not know how many decks of cards there were.
And that information, which is not given, affects the answer. For example, if there were four decks, there were 103 red cards out of 207; if there were six decks, there were 155 red cards out of 311. The probabilities would be, respectively, and a small difference, but nonetheless, a difference.
The correct answer is that both statements together are insufficient to answer the question.
### Example Question #1 : Dsq: Calculating Discrete Probability
Data sufficiency question
What is the probability of choosing a red marble at random from a bag filled with marbles?
1. There are only red and black marbles in the bag
2. of the marbles are black
statements 1 and 2 together are not sufficient, and additional data is needed to answer the question
statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question
both statements taken together are sufficient to answer the question, but neither statement alone is sufficient
statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question
each statement alone is sufficient
both statements taken together are sufficient to answer the question, but neither statement alone is sufficient
Explanation:
From statement 1, we learn that there are only two different colored marbles in the bag. From statement 2, we learn that are black which tells us that are red. Without statement 1, it is impossible to determine if there is another color of marble in the bag.
### Example Question #7 : Discrete Probability
Assume that we are the immortal gods of statistics and we know the following population statistics:
1) average driving speed for women=50 mph with a standard deviation of 12
2) average driving speed for men=45 mph with a standard deviation of 11
We look down from our statistical Mount Olympus and notice that the Earth mortals have randomly sampled 60 women and 65 men in an attempt to detect a significant difference in the average driving speed.
What is the probability that the Earth mortals will properly reject the assumption (i.e. the null hypothesis) that there is no significant difference between the average driving speeds. The Earth mortals have decided to use a 2-tailed 95% confidence test. Explanation:
standard deviation of the difference between the sample means = at 95% (2-tailed) = 1.96 the sample difference must be 4 or greater
(note: the probability of the sample difference being -4 or less is so small (4.5 standard deviations) that it will be ignored and we will only consider the probability that the difference is 4 or more.)
the probablity of the sample difference being 4 or greater (knowing that the population difference is 5) = the table shows that .3156 lies below -.48, so, .6844 lies above -.48
In English - there is a .6844 probability that the 2 sample means will yield a sample difference that is 1.96 or more standard deviations above 0.
### Example Question #1 : Discrete Probability
In a popular state lottery game, a player selects 5 numbers (on 1 ticket) out of a possible 39 numbers. There are 575,757 possible 5 number combinations. So, the odds are 575,757 to 1 against winning.
What are the odds of getting 4 of the 5 numbers correct on 1 ticket? Explanation:
575,757 must be divided by - ### Example Question #1041 : Data Sufficiency Questions
Your friend at work submits a bold hypothesis. He suggests that the number of sales per day follow the pattern - Monday-10%, Tuesday-10%, Wednesday-10%, Thursday 35% and Friday 35%.
You and he then record the number of sales for the following week: Monday-120, Tuesday-85, Wednesday-105, Thursday-325 and Friday-365.
After viewing the observed data, your friend expresses serious concern regarding his hypothesis.
You can help; you can tell him the probability of the observed data occuring if the hypothesis is true. Hint - Excel ChiTest. Explanation:
Use Excel ChiTest to get the .063 probability. If you are old-fashioned, you can also obtain the Chi-Squared number (8.928) by using ChiInv; but, it is not needed
### Example Question #1 : Dsq: Calculating Discrete Probability
A certain tutor boasts that his 2 week training program will increase a student's score on a 2400 point exam by at least 100 points (4.167%). A 10 student 'before-and-after' study was conducted to validate the claim. The following results were obtained - the 3 columns represent the before, afer and increase numbers for each of the 10 students:
1300 1340 40
1670 1790 120
1500 1710 210
1360 1660 300
1580 1730 150
1160 1320 160
1910 2100 190
1410 1490 80
1710 1880 170
1990 2060 70
Assume the null Hypothesis:
'The average increase is less than 100 points'
What is the highest level of significance (p-value) at which the null hypothesis will be rejected? Explanation:
Using Excel, the average increase (column 3) is 149 and Standard Deviation of the increases is 76.37 Using Excel - a t value of 2.03 for 9 degrees of freedom = .036
← Previous 1 3 4 5 6 7 8 9
### All GMAT Math Resources | HuggingFaceTB/finemath | |
Mandelbrot set approximation
Is there a function $f:\mathbb N\to \mathbb R$ such that $\lim_{n\to\infty} f(n) = 0$ and for every $c\in\mathbb C$:
If $z_0=0$, $z_{n+1}=z_n^2+c$ and $|z_k|<2$, then there exists a point $c'$ in the Mandelbrot set satisfying $|c-c'|<f(k)$?
Such fact would be very useful for rendering the Mandelbrot set, as it would give the exact count of iterations needed for given resolution.
-
What is the motivation for the condition $|z_k| < 2$? Also, do you mean that for any $k$ with $|z_k| < 2$, $|c - c'| < f(k)$, or do you mean that if $|z_k| < 2$ for all $k$< $|c - c'| < f(k)$ for all $k$? – 6005 Jun 12 '13 at 14:48
@Goos: I think neither? What the OP wants is probably "If the Mandelbrot iteration starting from $c$ stays within the critical circle $|z| < 2$ for $k$ steps, then there exists $c'$ in the Mandelbrot set such that $|c - c'| < f(k)$." – Willie Wong Jun 12 '13 at 14:59
On the contrary, you have Koebe $\frac14$ disks completely disjoint to $M$ for every point that does escape. – Hagen von Eitzen Jun 12 '13 at 15:00
@Goos: It's well known that if $|z_k|>2$, then $(z_n)$ diverges, that's why. I mean that for any k with $|z_k|<2$ you can find $c'$ (depending on that k) such that $|c-c'|<f(k)$. I don't get the second part of this question (from "or do you mean"), why is "for all $k$" twice and what's with "$k<|c-c'|$"? – janek37 Jun 12 '13 at 15:00
@HagenvonEitzen: can you elaborate? Is the radius of the Koebe disk independent of $k$? – janek37 Jun 12 '13 at 15:41
In principle, yes. The set of points whose orbits stay within $|z|< 4$ for the first $n$ iterations is a bounded open subset $M_n\subset\mathbb C$. We have $M_{n+1}\Subset M_n$ and $M=\bigcap _{n\in\mathbb N}M_n$. Let $f(n)=\max\{d(z,M)\mid z\in \overline{M_n}\}$. Then $z\in M_n$ implies there is a point $c\in M$ with $d(c,z)\le f(n)$. The sequence $(f(n))$ is nonincreasing, hence $a:=\lim_{n\to\infty} f(n)$ exists. Then for all $n$, we get a point $z_n\in M_n$ with $d(z_n,M)=a$. A subsequence of $(z_n)$ converges to some $\zeta$ with $d(\zeta,M)=a$, but we also have $\zeta\in\bigcap \overline{M_n}=M$. However, this mere existence of $f$ won't help you in practice.
You're right. I hoped that there are some upper bounds for the $f$ you defined. Hopefully in the form $f(n)<Cq^n$ for some $C>0$, $0<q<1$. – janek37 Jun 12 '13 at 17:15 | open-web-math/open-web-math | |
# Exposure Factor when calculating SLE
I am studying for an exam and came across a question asking to find what the single loss expectance would be, but I am failing to understand exposure factor in the calculation.
SLE = Asset Value * exposure factor
The question: Your company owns a machine that is worth \$100,000, if it was damaged in a fire it would only be worth \$8,000 in parts. What would the single loss expectance be?
My thought is that since the machine is \$100,000 and you can only salvage \$8,000 then the answer should be \$92,000. You are expected to lose \$92,000 when a fire occurs.
The answer they give is that SLE = \$8,000. This doesn't make sense because you aren't losing \$8,000, you are losing \$92,000.
Can someone please explain why SLE is \$8,000 and not \$92,000?
Exposure factor is a percentage: the percentage of the value that would be lost if a loss occurs. If the asset is completely lost, then the EF = 1.0
Let's redo your formula with a total loss:
``````SLE = AV * EF
SLE = \$100 * 1.0
SLE = \$100
``````
But in your book, you are expected to lose 92%, so:
``````SLE = AV * EF
SLE = \$100 * 0.92
SLE = \$92
``````
You can see that the more you "lose" in the event, the higher the exposure factor is, and the higher the SLE becomes.
The calculations are as you expect. I'd check your book's errata ....
In this example Exposure Factor (EF) is equal to the worth of the parts of the machine after the fire (\$8,000) divided by the value of the machine (\$100,000) or EF = 8,000/100,000 or .08 then using the formula Single Loss Expectancy (SLE) = Asset Value (\$100,000) * EF (.08) = \$8,000
Your SLE after the fire is \$8,000 as that will be the new value after the fire. But before the fire your SLE is \$92,000 as your machine value is expected to depreciate by 92% hence EF is 92% ans SLE = 100,000*0.92 | HuggingFaceTB/finemath | |
FromQuarkstoQuasars
# The Physics of a Spinning Top
###### / FromQuarkstoQuasars
News
Pictured here is a a spinning top. If you throw the top out and simultaneously pull on the string, the top spins. And you get to see this top spend a lot of time balancing on the sharp tip. (I have taken the liberty of choosing a traditional spinning top from my region of the world, instead of the ones you find elsewhere.)
When I was young, we would play with these tops. They were heavy, but the heavier ones where more fun. No, I am not an old fogey; computers and the Internet simply came to us a bit later, and the computers were so slow back then that we would play with real stuff.
Why am I bringing these tops to your attention?
It turns out that the mathematical solution to how a top spins, especially when it is slowing down and starts wobbling, is a particularly important physical example that explains precession and nutation motions. In fact, this is an example which you can find in many classical mechanics textbooks. Moreover, if you solve this problem in high detail, you’ll notice that the Earth’s motion is like a spinning top.
In this understanding, the spinning top exhibits the usual rotation. When it wobbles, you start seeing how precession and nutation come into play (when we speak of the “precession of the equinox” for the Earth’s motion, this is where the name comes from).
But why does a spinning top display this motion?
Yes, if you skip the equations a bit, we can get by with a borderline pass (the next two paragraphs, that is).
I hope that you still remember Newton’s 2nd Law, which simplifies to F = ma in many treatments. Using this law, which is strictly for motion in a line, we can derive what would happen for rotation. We can do this in two ways—either we hand wave and just cross product both sides with position, or we can really be detailed and check what every little bit of material is going to do and integrate over.
In the end, we find that the familiar equations of linear motion is translated, symbol for symbol, to get the equations of rotational motion. For example, the equation F = ma above turns into τ = I · α. In this case, τ (tau) is the torque, I is the moment of inertia that behaves just like the mass (but this time, the different directions get different moments of inertia), and α (alpha) is the angle θ (theta) differentiated twice with time (angular acceleration).
The energy and momentum equations are also translated and conserved, such that, if ω (omega) is the angular speed (rate of rotation), the angular momentum L would be L = I · ω. To know that this is conserved is enough to understand the motion of a wobbling spinning top!
When the top is initially spinning fast, it is pointing straight up and rotating around that axis. Its angular momentum is also pointing straight up and, through dissipative forces, it is decreasing a bit, but still pointing straight up. As it slows down even more, however, the spinning motion starts to fail to counteract the slight and random tilting forces. So, it wobbles a bit. Here, the interesting thing happens. It is tilting a bit, so the main rotation is now giving an angular momentum pointing in some other direction than straight up. But the initial angular momentum is pointing straight up, and this is conserved! So, what is going to happen is that there is some new rotation in the other two possible directions, just enough to vectorial add back to an angular momentum bivector that points straight up. Work out the maths, and this is exactly what explains the precession and nutation motions. Of course, this is quite strenuous, but this explanation is pretty beautiful, especially compared to the other treatments.
Spinning bodies do a lot of things that defy intuition. Back in the day, when we were still playing with these, the scientific community would routinely get letters claiming a way to use spinning bodies to create anti-gravity machines. Understanding their motion is quite amusing just in their own right. We do not need those letters back, but we do lament the loss of childhood for the current generations. | HuggingFaceTB/finemath | |
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# If x<y, is x(1+x)<y(1+y)?
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[GMAT math practice question]
If $$x<y$$, is $$x(1+x)<y(1+y)$$?
$$1) x>\frac{1}{2}$$
$$2) x+y>1$$
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Retired Moderator Joined: 22 Jun 2014 Posts: 1093 Location: India Concentration: General Management, Technology GMAT 1: 540 Q45 V20 GPA: 2.49 WE: Information Technology (Computer Software) Re: If x<y, is x(1+x)<y(1+y)? [#permalink] ### Show Tags 19 Mar 2018, 08:35 MathRevolution wrote: [GMAT math practice question] If $$x<y$$, is $$x(1+x)<y(1+y)$$? $$1) x>\frac{1}{2}$$ $$2) x+y>1$$ Let us re-arrange the equation: $$x (1+x) < y (1+y)$$ $$x + x^2 < y + y^2$$ $$x + x^2 - y - y^2 < 0$$ $$x - y + x^2 - y^2 < 0$$ $$(x-y) + (x-y) (x+y) < 0$$ $$(x-y) (x+y+1) < 0$$ - now this is actually what we need to prove. Already given: $$x<y$$ i.e. $$x-y < 0$$ Statement-1: $$x > 1/2$$ $$x > 1/2$$, it means $$x$$ is +ve. it is given that $$x<y$$, which means y is +ve too. Hence $$(x+y+1)$$ is +ve. with this, We are now left to prove $$(x-y) < 0$$ but it is already given. Sufficient. Statement-2: $$x+y > 1$$ it means $$x+y+1$$ is also positive. we are now left to prove $$(x-y) < 0$$ but it is already given. Sufficient. _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8009 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If x<y, is x(1+x)<y(1+y)? [#permalink] ### Show Tags 21 Mar 2018, 02:49 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Now, $$x(1+x)<y(1+y)$$ $$=> x+x^2 - y - y^2 < 0$$ $$=> (x-y) + (x^2 - y^2) < 0$$ $$=> (x-y) + (x-y)(x+y) < 0$$ $$=> (x-y)(1+x+y) < 0$$ $$=> 1+x+y > 0$$, since $$x < y$$. Condition 1) Since $$y > x > \frac{1}{2}$$, we have $$x + y + 1 > 0.$$ Thus, condition 1) is sufficient. Condition 2) Since $$x + y > 1$$, we have $$x + y > 0$$. Thus, condition 2) is sufficient too. Therefore, D is the answer. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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### Show Tags
21 Mar 2018, 04:19
MathRevolution wrote:
[GMAT math practice question]
If $$x<y$$, is $$x(1+x)<y(1+y)$$?
$$1) x>\frac{1}{2}$$
$$2) x+y>1$$
I find statement 2. to be insufficient
now, considering statement 2.
mandatory :
x < y (from first line of question)
x+y >1 (part 2 of the question)
Case I : x=2 , y = 3
2 / (2+1) & [align=][/align] 3 / (3+1)
2/3 & 3/4
.66 & .75
solution for is x part is less than y
Case II : x=-3 , y = 5
-3 / (-3+1) & 5 / (5+1)
-3/-2 & 5/6
1.5 & .83
solution for is x part is more y
need support of experts and friends to fly high.
pls assess and give your input.
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Re: If x<y, is x(1+x)<y(1+y)? [#permalink]
### Show Tags
21 Mar 2018, 07:10
GMAT215 wrote:
[/size]2 / (2+1) & [align=][/align] 3 / (3+1)
2/3 & 3/4
.66 & .75[b]
Case II : x=-3 , y = 5
-3 / (-3+1) & 5 / (5+1)
-3/-2 & 5/6
1.5 & .83[b]
.
you are dividing instead of multiplying as mentioned in question.
so it is x * (1+x) NOT x / (1+x)
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Re: If x<y, is x(1+x)<y(1+y)? [#permalink] 21 Mar 2018, 07:10
Display posts from previous: Sort by | HuggingFaceTB/finemath | |
## What is Hi-Lo card counting?
The Hi-Lo card counting system assigns a tag of +1 for the low cards (2, 3, 4, 5, and 6) and –1 for the high cards (10, J, Q, K, and A). The 7, 8, and 9 is assigned a tag of 0. The Hi-Lo is a balanced card counting system because there are equal numbers of +1 and –1 cards per deck.
How do you count cards for true count?
To calculate the True Count at any given time, divide the Running Count by the number of decks remaining. For example, if we have a Running Count of +9 and there are 3 decks remaining, the True Count is +3.
What does a high true count mean?
After all, it’s a sucker bet. But when the true count is +3 or greater, insurance becomes a positive expectation bet. That’s because as the ratio of high cards changes, the dealer has a better probability of getting a blackjack, too. When that probability gets high enough, insurance becomes a positive expectation bet.
### Is counting cards illegal in Australia?
Australian casino operators are on the lookout for people using new iPhone applications to cheat at blackjack by counting cards. Many casinos throw players out if they suspect them of counting cards but using a device to do the maths is strictly outlawed.
Which card counting system is best?
You might be able to use a more complicated system without mistakes, but most gamblers can’t. I recommend using either the Knock Out or Red 7 system. These are both good systems and they don’t use a conversion for true count. If you don’t struggle with a conversion, the hi lo is the system I recommend.
When should you count your high card counting?
You assign each card a value when counting and then add to the running “count” when cards are played. You will tend to bet high when the “count” is high, and bet lower when the “count” is lower. Even with different counting systems, counting always points in the same direction.
## How do you beat hi low?
Whilst high low may be a simple game, there is no easy way to win….Here is what the strategy consists of:
1. Your first bet should be the lowest possible stake.
2. Place the same bet each time until you lose.
3. After a loss, double your bet.
4. If you have successive losses, you should continue to double your bet each round.
Is it cheating to count cards?
Counting cards is not considered cheating. When counting a player uses their brain to gain an advantage. No “devices” are used. Counting is 100% legal but casinos frown on counting because it is an effective way to win.
What is the best card counting system?
Hi-Lo System. This is probably the most common system and is very popular with people beginning to learn how to card count.
• REKO System. This is a more efficient system generally than the Hi-Lo system.
• Zen Count System. This is an older system and dates back to 1983.
• Wong Halves System.
• ### How do you count cards?
Step 1. Assign a value to every card. Step 2. Keep a “Running Count” based off of the values of the card dealt. Step 3. Use this information to calculate the count per deck or “true count”. Step 4. Change your bets as the true count rises.
How to count cards?
Step 1:. Assign a value to every card. Remember you are not just counting your cards; you’re counting the other players…
• Step 2:. Keep a “running count based off the values of the card dealt. Normally the player would add 1, subtract 1, or…
• Step 3:. Use this information to calculate the count per deck or also “known as true count”. At this point we…
• What is a card counting method?
Card Counting. Thus, card counting is simply a method of determining when the remaining deck is player-favorable. When the deck is “rich” in tens and aces, the card counter places correspondingly larger wagers. When the deck instead has more small cards such as twos through sixes, the card counter will bet as little as possible or leave the table. | HuggingFaceTB/finemath | |
# Homework Help: Darn integral
1. Nov 16, 2009
### Void123
1. The problem statement, all variables and given/known data
I just can't crack the integral of (sin(x))^6 for some reason.
What is the exact solution to this? This is not really a homework question, as an immediate reference to an integral table would be sufficient. But I just need it right away. Thanks.
2. Relevant equations
...
3. The attempt at a solution
...
2. Nov 16, 2009
### jgens
Use integration by parts repeatedly.
3. Nov 16, 2009
### HallsofIvy
Or use trig identities. $cos(2x)= cos^2(x)- sin^2(x)= 1- 2sin^2(x)$ so $sin^2(x)= (1/2)(1- cos(2x))$. Then $sin^6(x)= (sin^2(x))^3= (1/2)^3(1- cos(2x))^3$$= (1/8)(1- 3cos(2x)$$+ 3cos^2(2x)+ cos^3(2x))$.
The integral of 1- 3cos(2x) is straightforward. The integral of $cos^3(2x)$ can be done by writing it as $cos^3(2x)= cos(2x)(1- sin^2(2x))$ and using the substitution u= sin(2x). The integral of $cos^2(2x)$ can be done by using the trig identity $cos^2(2x)= (1/2)(1+ cos(4x))$. | HuggingFaceTB/finemath | |
## Find the nth of each sequence a1 = 8,d = 3,n n = 16
Question
Find the nth of each sequence a1 = 8,d = 3,n n = 16
in progress 0
11 mins 2021-11-26T00:07:37+00:00 1 Answer 0
16th term of the given sequence is 53.
Step-by-step explanation:
Here, the first term (a 1) = 8
Common Difference (d) = 3
n = 16
Now in ARITHMETIC PROGRESSION:
The nth term of any sequence is given as
⇒ a 16 is given as:
⇒ a (16) = 53
Hence, 16th term of the given sequence is 53. | HuggingFaceTB/finemath | |
BrainDen.com - Brain Teasers
• 2
# Groundhog in a Hole - Yet Again
## Question
There is an infinite line of holes in an infinite field. You know there is a groundhog hiding in one of the holes. You can check a hole once a day. At night the groundhog will move one hole to the left or one hole to the right. Knowing you cannot find the groundhog yourself, you enlist a friend to help. Now you can check 2 holes a day. Can you find the groundhog? If so, how would you do it?
## Recommended Posts
• 1
To solve the question without knowing parity, we follow almost the same steps, except we change hypothesis of parity alternatively each time we go to next round. And it may take us one more round to find it, because we find it only when in the same parity.
Edited by xp2008
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Interesting! Does the line of boxes have an origin? Like 1 to infinity? Or is it from negative infinity to positive infinity?
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Negative infinity to positive infinity. You could pick any hole to be the origin.
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Barring blind luck, I don't think it can be caught with two hunters. Thoughts....
Spoiler
Think minimum required is two teams of 3. From origin one team heads -ve, other goes +ve. Teams can check two new holes each day, whilst preventing the hog from backtracking. They will be moving twice speed of hog so will eventually capture.
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8 hours ago, Wilson said:
Barring blind luck, I don't think it can be caught with two hunters. Thoughts....
Reveal hidden contents
Think minimum required is two teams of 3. From origin one team heads -ve, other goes +ve. Teams can check two new holes each day, whilst preventing the hog from backtracking. They will be moving twice speed of hog so will eventually capture.
Spoiler
You can cut one hunter off your strategy by alternating which direction checks two new holes and which only checks one.
Can anyone find the groundhog using less than 5 hunters?
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Perhaps this question will help move things along...
Spoiler
If I were to tell you on the first day that the groundhog was an odd number of holes away from the origin, could you then find the groundhog with 2 hunters?
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If they started from either end worked to the middle they would eventually find it... provided they lived for eternity
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20 hours ago, HotRodCow said:
If they started from either end worked to the middle they would eventually find it... provided they lived for eternity
There is no end to start from. Every hole has an infinite number of holes on both sides of it.
And, yes, both you and the groundhog are immortal. Your friend, not so much. Luckily, you'll always be able to enlist a member of his posterity to help check one hole each day, so close enough.
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On 5/12/2020 at 3:04 AM, EventHorizon said:
Perhaps this question will help move things along...
Hide contents
If I were to tell you on the first day that the groundhog was an odd number of holes away from the origin, could you then find the groundhog with 2 hunters?
If we know first day's parity then we can. Supposing it was odd, then I will check
1. 1 3
2. 4 6
3. 1 -1
4. -2 -4
5. 6 8
6. 9 11
7. ...
Which is to say,
day 4k+1, I check 5K+1, 5K+3
day 4k+2, I check 5K+4, 5K+6
day 4K+3, I check -5K+1, -5K-1
day 4K+4, I check -5K-2, -5K-4
We can see my range is ( -5K-4, 5K+6 ) , while lim ( 5K - 4K) = lim (K) -> infinity, thus we could always find the groundhog in this case.
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16 hours ago, xp2008 said:
If we know first day's parity then we can. Supposing it was odd, then I will check
1. 1 3
2. 4 6
3. 1 -1
4. -2 -4
5. 6 8
6. 9 11
7. ...
Which is to say,
day 4k+1, I check 5K+1, 5K+3
day 4k+2, I check 5K+4, 5K+6
day 4K+3, I check -5K+1, -5K-1
day 4K+4, I check -5K-2, -5K-4
We can see my range is ( -5K-4, 5K+6 ) , while lim ( 5K - 4K) = lim (K) -> infinity, thus we could always find the groundhog in this case.
Unfortunately, that does not work for the known parity case.
Notice day 4 checks even holes, and day 5 also checks even holes. So days 5-8 are checking the wrong parity.
Fixing this, the numbers change such that day 4k+x checks hole 4k+y, so the groundhog can escape by fleeing away from hole 0.
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3 hours ago, EventHorizon said:
Unfortunately, that does not work for the known parity case.
Notice day 4 checks even holes, and day 5 also checks even holes. So days 5-8 are checking the wrong parity.
Fixing this, the numbers change such that day 4k+x checks hole 4k+y, so the groundhog can escape by fleeing away from hole 0.
Apologise I considered it to be 2 days after when they come back to positive half axis, yeah, after fixing this 4K hole doesn't work for 4k day.
Is this an open question or you know there's an answer ?
Edited by xp2008
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I tried again like this, supposing we know the parity case, for example odd on first day
By checking (1,3),(4,6),(7,9),..., we can check a range of (3k-1) in k days. Now I construct a sequence of numbers ai, such that
an >= 3*sumi<nai
we can easily verify that ai = 4i qualifies.
The strategy now becomes,
in the first a1=4 days, we check (-5,-3),(-2,0),(1,3),(4,6), which is a range of -5 -> 6
in the next a2 = 16 days, we check (-23, -21),(-20,-18),...(-2,0),(1,3),..,(22,24), which is a range of -23 -> 24
etc.
Supposing groundhog is at P, while |6P|<ap = 4p
Then after ( a1 + a2 + .. +ap ) days, the range we just checked is the range of last ap days, which is about -1.5*ap -> 1.5*ap, while the range of groundhog could be is
-|P| - ( a1 + a2 + .. +ap ) -> |P|+( a1 + a2 + .. +ap ),
on the other hand, we have
|P|+( a1 + a2 + .. +ap ) - 1.5*ap = |P|+( a1 + a2 + .. +ap-1) - 0.5*ap < ap/6 +ap/3-0.5*ap = 0,
which is to say, groundhog's range is included by our last searching range.
Edited by xp2008
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15 minutes ago, xp2008 said:
To solve the question without knowing parity, we follow almost the same steps, except we change hypothesis of parity alternatively each time we go to next round. And it may take us one more round to find it, because we find it only when in the same parity.
You got it. Well done. I'll post my solutions later.
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6 minutes ago, EventHorizon said:
You got it. Well done. I'll post my solutions later.
Thanks !
We can see the days we need is
sumi<=P+1ap<3*ap+2=3*43*ap-1<3*43*6P, which is O(P).
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My thoughts and solutions:
Spoiler
Solution 1: If you have one person check, for instance, holes 0 and 3 on alternating days, you can keep the groundhog trapped in holes 1-2. Using this, you can have the other scan over the area being guarded. And once the scan is done, trap progressively larger areas alternating the assumed parity for each time a new area is selected. You can easily choose a multiplicative factor to increase the size of the areas by to make the time spent on the previous areas practically meaningless compared to the new area size, so the groundhog couldn't escape by running away.
Solution 2: Not that the groundhog, if the correct parity, cannot get past the person making the scan of the area. Then you can have one person start at one end of a chosen area and the other at the other end, and if the groundhog has the right parity and is in the area, it will be caught. This is simpler than solution 1 since both checkers have the same job. It also doubles the speed of solution 1.
Solution 1 was what I initially thought of when coming up with this puzzle. Using this idea you could also have a puzzle like two squares of holes with a shared edge, have one person alternate at the two intersection points and then have the other scan the 3 connected segments for the assumed parity, then clear it again with the other parity.
xp2008's solution was rather interesting. He showed that you didn't even need to have an area trapped if you scan in one direction using both checkers.
Looking at solution 2, you can see that if the line of holes has an end on one side (ie, one sided infinity), you only need 1 person to find the groundhog.
The solutions for my original additions back in the day all had solutions that were rather elegant extensions of the original problem's solution. This one turns out to be the same. Using solution 2, if you center the checking around hole 0 (ie, when one person checks hole x, the other checks -x), start with checking an odd number, and have the scale factor an odd number 5 or more (3?)... it seems much like a simple extension of the solution to the original problem. The requirements of starting at odd hole and having the scaling factor be odd makes the parity naturally alternate with each new area. Also, like how you never need to check holes 1 or 5 in the original, you never check hole 0 in this solution.
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On 6/6/2020 at 12:13 PM, EventHorizon said:
My thoughts and solutions:
Hide contents
Solution 1: ....You can easily choose a multiplicative factor to increase the size of the areas by to make the time spent on the previous areas practically meaningless compared to the new area size, so the groundhog couldn't escape by running away.
Never mind how big the area you chose, I always can say "Bad luck, the groundhog is somewhere about 17 holes outside the defined area."
I agree that the size of the cleared area will -> inf. Just the size is expressed as a number. No matter how big it is, there always is a bigger number.
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While the hole that the groundhog starts in is unknown, it is one specific hole and its movements are as described. No matter what hole it starts in, given its described movement, the strategy given will eventually find it. You make it seem like the groundhog can teleport just because it's starting hole was unknown.
Given your "about 17 holes outside" example, it would take just one or two (depending on parity) rounds of area clearing/expansion to get it. It does not matter that there are more holes outside the area, it will eventually be trapped and caught.
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Thinking it over and over again, I always finish with a problem I cannot solve.
The holes are numbered 1, 2, 3....
Three hunters check:
1 2 3
3 4 5
5 6 7
....
On day n, they will have checked up to the hole 2 * n +1
The groundhog starts in the hole k and moves to the right.
On day n, he will be in the hole k+n
Question 1: Will the hunters catch the groundhog (and if so, when)?
2 * n +1 grows faster than k + 1, I already have the answer. Nevertheless:
2 * n + 1 = k + n
n ⁼ k + 1
No matter how high the number of the starting hole the groundhog choses, it will be caught.
Question 2: Is there a starting hole so that the groundhog is not discovered on day n?
k + n 2 * n + 1
> n + 1
No matter how long the hunters hunt, it always is possible that the groundhog started in a hole leading him outside the checked area.
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I'm afraid I might be missing something, because it seems like whenever you switch parity you end up losing all the ground you previously covered.
Consider xp2008's solution...
Spoiler
If you assume the initial parity is odd and you search (-5,-3),(-2,0),(1,3),(4,6), then where could the groundhog be on day 4 if he's not caught? If he started from hole -7 then he could have reached hole -4 by day 4, and he could be at hole +8 on day 4, or anywhere outside that range.
If you then switch to cover the even parity case and search for four days, in the odd parity case the groundhog could have moved from hole -4 on day 4 to hole 0 on day 8, and could have moved from hole +8 on day 4 to hole +4 on day 8. So after you've searched both parities, now all you know for the odd parity case is that the groundhog is not in the range from hole 1 to hole 3. So when you resume your odd parity search on the next day, the groundhog could be in any odd hole. Have you really made any progress?
Similarly for EventHorizon's first solution
Spoiler
Take the case of having bounds of 0 and N (where N is odd) for an initial scan. If you assume initial even parity, then you could scan (0,2), (N,3), (0,4), (N,5), ... (0,N-1) and it would take you N-1 days. You could then re-scan the area covering the initial odd parity case, but after scanning for another N-1 days the groundhog could get anywhere within the even parity case again by the time you start the next scan.
You could argue that if you start off covering a finite area, then cover a slightly larger area, then cover a slightly larger area, etc. then you will eventually catch the groundhog because the groundhog must be at some finitely numbered hole which you will eventually reach. My counterargument to that is:
Spoiler
xp2008's strategy would cover holes -5 through 6 for the odd parity case and finish after 4 days. If you then cover the even parity cases, you could cover holes -6 through 5 for the even parity case and finish both the odd and even parity cases after 8 days.
Similarly, it would cover holes -23 to 24 for the odd parity case over the next 16 days, so if you start doing that on day 9 (after finishing the odd and even parity cases for -6 to 6) then you would finish on day 24. Then you would cover the even parity cases for that range over the next 16 days and finish on day 40.
If I'm a groundhog and I don't want to be caught, I start on hole +2 and move to the next higher hole every day. You'll never catch me.
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Spoiler
In both xp2008's and my solutions, you don't scan over the same range twice, once for each parity. You increase the range by the scale factor every time you switch the parity.
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On 11/1/2020 at 2:16 PM, EventHorizon said:
Reveal hidden contents
In both xp2008's and my solutions, you don't scan over the same range twice, once for each parity. You increase the range by the scale factor every time you switch the parity.
Ah, I see. So in that spirit you could present your solution #2 this way (not particularly efficient, but crystal clear that it works)
Spoiler
Suppose the groundhog started at hole #0. Check hole #0 on the first day, and if he’s there then you’re done.
Suppose the groundhog started at hole -1 or hole 1. By the second day (after you’ve checked hole 0) he’ll be at hole -2, 0, or 2. Have the two of you check holes ±2 on the second day, and if you don’t find him and he started at hole ±1 then he must have gone to hole 0 for the second day and will be at hole ±1 on the third day. Check holes ±1 on the third day and you’ll find him.
Suppose the groundhog started at hole -2 or hole 2 on the first day. By the fourth day (after you’ve taken care of the cases where he starts at -1, 0, or 1) he’ll be at an odd numbered hole from -5 to 5. Search holes ±5 on the fourth day and if he’s not found then he was in an odd numbered hole from -3 to 3 and will be in an even numbered hole from -4 to 4 on the next day. Check holes ±4 on the fifth day, then holes ±3 on the sixth day, etc. until you’ve taken care of all the cases where he started at hole -2 or 2.
Suppose the groundhog started at hole -3 or hole 3, or in general hole -N or N where you’ve already taken care of all of the smaller numbers. If you’ve taken D days to search, then he must be in hole -(N+D) to (N+D), so search holes ±(N+D) and work your way inward.
After taking care of that case, move on to the next higher N. No matter where he initially started, you’ll eventually find him.
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# Curiosities of linearly ordered sets
Andrei Chekmasov explores order and infinity You might look at the title and ask yourself: what is a linearly ordered set? A good example of such a set would be counting the number of years that have passed since the Big Bang—the age of the universe. It will look something like this:
$$\overset{\text{Bang!}}{0} \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow \dots \rightarrow \overset{\text{Now?}}{\text{13,772,000,000}} \rightarrow \cdots$$
More formally, a set $S$ is a collection of objects $x$, and if an object is in a set it is called an element of the set: $x \in S$. Simply put, a linearly ordered set is an arrangement of the elements of a set one after another. If we assume that the universe will exist forever, our example represents a linear arrangement for the set $\mathbb{N}$ of non-negative integers. This specific order is called the natural order on $\mathbb{N}=\mathbb{Z}^+\cup\{0\}$. Keep this example in mind while we define what a linear order is.
We begin by introducing a relation $\rightarrow$ on a set $S$, which determines the arrangement of elements of $S$; for convenience we shall use the symbol $\rightarrow$ whenever there are elements $x,y \in S$, and $x$ comes before (or to the left of) $y$ in the order of elements, writing $x \rightarrow y$.
We need the relation $\rightarrow$ to have some useful properties to become a linear order on $S$. First, the order needs to be consistent in some sense across the whole set, so we require the relation to have the transitivity property: if $x \rightarrow y$ and $y \rightarrow z$, then $x \rightarrow z$. Second, we want the order to apply to every element of the set $S$, so it is necessary for the relation to have the trichotomy property: for all $x,y \in S$, exactly one of the following holds: \begin{align*} x &\rightarrow y& x&=y& y &\rightarrow x \end{align*} Enforcing transitivity and trichotomy properties allows us to avoid ‘loops’ in the linear order like $x \rightarrow y \rightarrow z \rightarrow x$ If $\rightarrow$ is a linear order on $S$, then $S$ together with the relation $\rightarrow$ is called a linearly ordered set. Applying this definition to the natural order on $\mathbb{N}$, we see that the relation $\rightarrow$ is represented by the more familiar ‘less than’ relation $<_{\mathbb{N}}$ on the elements of $\mathbb{N}$.
## Dense orders
Broadly speaking, there are two types of linear orders: dense and non-dense orders. A linear order on a set is dense if between any two elements of the set one can always find a third element of the same set. Otherwise, a linear order is non-dense. We challenge the reader to answer this question: is it possible to make a dense order on a set with a finite number of elements?
An example of a dense linear order is well known to all of us. It is formed by the set of positive rational numbers, $\mathbb{Q}^+$, and the relation $<_{\mathbb{Q}^+}$. With respect to $<_{\mathbb{Q}^+}$, between any two positive rational numbers, one can find a third positive rational number. For example, if $x,y \in \mathbb{Q}^+$ and $x <_{\mathbb{Q}^+} y$, then their average is also a rational number and is between them with respect to $<_{\mathbb{Q}^+}$:
$$x <_{\mathbb{Q}^+} \frac{x+y}{2} <_{\mathbb{Q}^+} y$$
You need an infinite set to have a dense order, but this does not mean that every linear order on an infinite set is dense! The natural order of non-negative integers is not dense, despite being infinite. For example, there is no integer between 2 and 3.
Though we are limited to integers here we can still construct a dense linear order on the set $\mathbb{N}$! Here is an example. On a straight line we start step $0$ by marking out $0$ and $1$. Then at step $1$ we place $2$ at a midpoint between $0$ and $1$ on the straight line. We continue from left to right and at step $2$ find midpoints between $0$ and $2$, $2$ and $1$ and placing $3$ and $4$ at those midpoints respectively. We continue like this forever. From step $1$ onwards, for each $i\in\mathbb{Z}^+$ we fit the numbers $2^{i-1}+1$ through to $2^i$ into all the gaps between numbers in the previous step. The relative location of the first $17$ elements of $\mathbb{N}$ according to the linear order is shown in the figure below.
So is this linear order dense? At any step imagine two natural numbers $a$ and $b$ which have no numbers between them. On the next step we will put a number at the midpoint between $a$ and $b$ so that $a$ and $b$ are not ‘immediate’ neighbours any more. We built a dense order indeed! By applying this algorithm, we ‘packed’ all the integers inside the interval $[0,1]$ in order to construct a dense order on $\mathbb{N}$.
## Cantorian order
Looking at how many more rationals there seem to be in comparison with the natural numbers (after all, there is an infinite number of rationals between $0$ and $1$), you might expect that every linear order on $\mathbb{Q}^+$ is a dense order. Not at all! German mathematician Georg Cantor showed that there is a linear order on $\mathbb{Q}^+$ that is not dense. Can you come up with one?
Let us introduce the Cantorian order $\overset{\mathbb{Q}^+}{\longrightarrow}$ on $\mathbb{Q}^+$ via a diagram: The order is created by forming a plane of fractions where numerators increase as one goes diagonally right and denominators increase as one goes diagonally left. Then, by travelling along a path back and forth, one creates a linear order on $\mathbb{Q}^+$. Fractions which simplify are skipped: for example, $1/2$ is present, but $2/4$, $3/6$ and so forth are skipped (they are crossed out on the diagram). We can see that there are no elements in between adjacent fractions, yet this linear order contains every possible fraction, thus covering the whole set $\mathbb{Q}^+$.
Relative to the Cantorian order one can speak of the first positive rational number, the second positive rational number and so on. In other words, one can construct a one-to-one mapping from $\mathbb{Z}^+$, the set of positive integers, to $\mathbb{Q}^+$ since two different positive rational numbers are always in two different positions in the Cantorian order. So there are the same number of positive rational numbers as there are positive integers! If you think of the apparent abundance of positive rational numbers relative to the apparent sparseness of positive integers on the number line, you will find this conclusion a bit mind-bending. Unsurprisingly, Cantor’s contemporaries were shocked by his findings, and many rejected his research. Cantor’s thinking was clearly ahead of his time!
## A bigger picture
Cantor had grander ideas than just defining a weird linear order. After all, at the time he was developing what is now known as modern set theory. One of the problems Cantor faced was how to define infinity; he thought of an elegant approach to grasp this ethereal concept. We’ll walk ourselves through it. To start off, we know that every positive integer has a successor—just add $1$ to it. Thus you can count positive integers. The number of elements in a set like $\mathbb{Z}^+$ is called the cardinality of the set. The infinite set whose elements can be counted is called countably infinite. Other sets that can be put into one-to-one mapping with positive integers are also countably infinite. Cantorian order proves that $\mathbb{Q}^+$ is countably infinite.
Countably infinite sets have some interesting properties. All of their subsets are either finite or countably infinite. For example, you can map one-to-one positive integers to positive even integers, which is a subset of $\mathbb{Z}^+$. Another property is that if you combine two countably infinite sets, the resulting set would be countably infinite. Let’s combine the set $\mathbb{N}$ with the set of negative integers $\{\dots,-3,-2,-1\}$ and place the combined set in the following linear order \begin{equation*} 0 \rightarrow -1 \rightarrow 1 \rightarrow -2 \rightarrow 2 \rightarrow -3 \rightarrow 3 \rightarrow \dots \end{equation*} In other words, odd terms are defined by $(i-1)/2$ and even terms are defined by $-i/2$ where $i \in \mathbb{Z}^+$ is the position of the number in the linear order. This linear order shows that the combined set is countably infinite.
But there is more to infinite sets! Cantor showed that there are at least two types of infinity: countable as we have discussed above and uncountable. With uncountably infinite sets it is impossible to construct a one-to-one mapping between their elements and the positive integers. The set of real numbers $\mathbb{R}$ is uncountably infinite. In fact, its subset $[0,1) \subset \mathbb{R}$ turns out to be uncountably infinite too. Cantor showed that there is no one-to-one mapping between $\mathbb{Z}^+$ and $[0,1)$.
So let’s illustrate Cantor’s idea by trying to construct some linear order for the real numbers from $[0,1)$ which is equivalent to the natural order on $\mathbb{Z}^+$ and see where this goes wrong. For our linear order we will be continuously generating random numbers $x \in [0,1)$ and placing them one after another, skipping numbers which were already generated. We might get a linear order similar to this example:
\begin{equation*}
0.739\ldots \rightarrow 0.055\ldots \rightarrow 0.349\ldots \rightarrow \cdots \rightarrow 0.281\ldots7\ldots \rightarrow \cdots
\end{equation*}
Now Cantor’s diagonal argument will help us to construct a mysterious real number from $[0,1)$ that does not fit into the linear order! In our approach we plan to construct the new real number so that it differs from every number in the linear order in at least one digit. The new number has its first digit different from the first digit of the first number in the linear order, its second digit different from the second digit of the second number in the linear order, and so on. We change digits by taking the original digit and adding $1$ to it if the digit is from $0$ to $8$ and substitute the digit $9$ with $0$. In other words, we take diagonal digits of elements from the original linear order and add $1$ to each of them or substitute $9$ with $0$ and construct the new real number from this array of digits. For our example the new real number would be \begin{equation*} 0.860\!\ldots\!8\!\ldots \end{equation*} This new number cannot appear in the linear order: it differs from the first number in the linear order with its first digit, with the second number in the order with its second digit and so on. Yet it is a perfectly good real number! So there are more real numbers in $[0,1)$ than positive integers. We come to a conclusion that these sets do not have equal sizes.
There are plenty more surprises in set theory! We have seen that there are at least two different sizes of infinity, but are there any infinities which are bigger than countable, but smaller than uncountable? For that question Cantor formulated his continuum hypothesis which reads: ‘There is no set whose cardinality is strictly between that of the positive integers and the real numbers.’ The continuum hypothesis can neither be proven nor disproven using the standard axioms of set theory; it stands independent of those axioms. In other words, you can create a set theory where the continuum hypothesis is true, or you can create a set theory where it is not. Cantor’s work creating set theory led to captivating results by other mathematicians, like Russell’s paradox—a famous logical contradiction which has profound implications for the foundations of mathematics. So dive in to learn more! Andrei Chekmasov is a 10th grader at Cypress Bay High School in Weston, Florida. At school, he enjoys studying calculus and computer science, as well as hanging out with friends. When not wrestling with schoolwork, Andrei enjoys building Lego models and competing in maths and coding competitions.
+ More articles by Andrei | HuggingFaceTB/finemath | |
The square is multiply the number by itself. The integer is one type of number and it is defined both positive and negative numbers. The sum of squares is add the squares of integers and it is used in statistics. A sum of squares is an integer that is used to determine a variance and standard deviation for a group of integers.
The formula for sum of squares is $S$ = $x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + . . . x_{n}^{2}$
Where, $S$ = sum of squares and $x_{1}, x_{2}, x_{3}. . . x_{n}$ are the integers.
## How to Find Sum of Squares
Below you could see the steps:
Step 1 :Take the following set of integers.
Step 2 : Square each of the number and add all by using the formula $S$ = $x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + . . . x_{n}^{2}$ for calculation.
Step 3 : Finally you get the answer $S$.
This calculator works on positive and negative numbers.
### Difference of Squares Calculator
Sum of Perfect Squares Calculate Riemann Sum Calculate Perimeter of a Square Calculate Square Root Upper Sum and Lower Sum Angle Sum Formula
Find Sum of Series Calculator Geometric Series Sum Calculator Left Riemann Sum Calculator Sigma Sum Calculator Sum of Geometric Series Calculator 3 Square Root Calculator | HuggingFaceTB/finemath | |
Question
1. # In Which Quadrant Is The Number 6 – 8I Located On The Complex Plane?
## Introduction
Have you ever wanted to know where the number 6-8i is located on the complex plane? If so, you’ve come to the right place. The complex plane is a way of visualizing numbers in terms of their real and imaginary parts. It’s used to graphically represent complex numbers as points on a two-dimensional surface, which can help us better understand how they interact with each other. In this blog post, we will take a look at exactly what quadrant the number 6-8i is located in on the complex plane, and why it matters.
## The complex plane
The complex plane is a two-dimensional coordinate system in which the real part and the imaginary part of a complex number are represented by orthogonal axes. The point at which these two axes intersect is called the origin, and the distance from the origin to any other point on the plane is called the modulus or magnitude of the complex number. The angle between the real axis and the line joining the origin to any other point on the plane is called the argument or phase of the complex number.
On the complex plane, the four quadrants are often referred to as the Argand plane. In each quadrant, the x-axis is represented by a real number line and the y-axis is represented by an imaginary number line. The origin (0,0) is located in the center of the plane.
The first quadrant is located in the upper right hand corner of the plane. It is often referred to as the positive real axis because all numbers in this quadrant are positive real numbers. The second quadrant is located in the upper left hand corner of the plane. It is often referred to as the negative real axis because all numbers in this quadrant are negative real numbers.
The third quadrant is located in the lower left hand corner of the plane. It is often referred to as the imaginary axis because all numbers in this quadrant are imaginary numbers. The fourth quadrant is located in the lower right hand corner of the plane. It is sometimes referred to as the complex axis because it contains both real and imaginary numbers.
## Where is the number 6 – 8i located?
The number 6 – 8i is located in the fourth quadrant of the complex plane. This can be seen by looking at the real and imaginary components of the number. The real component, 6, is positive and the imaginary component, – 8i, is negative. This means that the number is located in the fourth quadrant.
## Conclusion
To conclude, the complex number 6-8i is located in the third quadrant of the complex plane. This means that it has a negative real part and a negative imaginary part. While this may seem daunting to understand at first, with practice and understanding of complex numbers you will find it much easier to plot them on the complex plane. Hopefully this article has helped clear up any confusion about where 6-8i lies on the complex plane so that you can better understand how these numbers are placed in relation to each other! | HuggingFaceTB/finemath | |
## Question 455:
1
We need to find the z-score for the 25th percentile. We can use the percentile to z-score calculator using the 1-sided area. This will provide us with the z-score corresponding to 25% of the area under the curve. Enter .25 and you should get a z-score of -0.675. Now we just setup an equation and solve for the unknonw IQ. The equation for a z-score is (x-mean)/standard deviation.
1. (x-10)/2 = -0.675
2. x-10 = -1.35
3. x = 8.65
So an IQ of 8.65 represents the 25th percentile given the mean of 10 and sd of 2. 75% of IQ scores would be above 8.65. | HuggingFaceTB/finemath | |
# math
posted by on .
If you have a class of 45 students and 15 are boys what is the percentage of boys in the class
• math - ,
15/45
Divide and multiply by 100
15/45 = 0.3333 = 33.33% are boys
• math - ,
45 | HuggingFaceTB/finemath | |
A contract is to paint 3 houses. Mr brown takes 6 days to
paint a house. Mr black takes 8 days and Mr blue 12 days. Mr
brown leaves for a vacation after 8 days. Mr black works for
6 days from then.How many days would Mr blue require to
complete the contract?
Answers were Sorted based on User's Feedback
A contract is to paint 3 houses. Mr brown takes 6 days to paint a house. Mr black takes 8 days and ..
11 days
brown completes 1+1/3 house in 8 days
black completes 3/4 house in 6 days
let the work done by blue=x
then
1+1/3+3/4+x=3 total houses to paint
x=11/12
so blue will complete the remaining work in 11 days
Is This Answer Correct ? 9 Yes 0 No
A contract is to paint 3 houses. Mr brown takes 6 days to paint a house. Mr black takes 8 days and ..
2 days
Is This Answer Correct ? 2 Yes 1 No
A contract is to paint 3 houses. Mr brown takes 6 days to paint a house. Mr black takes 8 days and ..
a tiny bit of modification in the first answer.
Mr. Brown completes 4/3 rd of a house
Mr. Black completes 3/4 th of a house
Mr. Blue completes x/12 th of a house
It is evident that 4/3+3/4+x/12=3
i.e. (x+25)/12=3
i.e. x=11
So Mr. Blue takes 11 days
Is This Answer Correct ? 0 Yes 0 No
A contract is to paint 3 houses. Mr brown takes 6 days to paint a house. Mr black takes 8 days and ..
4 days
Is This Answer Correct ? 0 Yes 1 No
A contract is to paint 3 houses. Mr brown takes 6 days to paint a house. Mr black takes 8 days and ..
I think it is 12.
Mr brown completes 1 house + 1/4 th a house in 8 days.
Mr black completes 3/4 th a house in 6 days. 1/4 +3/4 = 1
thus one more house left to be painted. Mr blue takes 12
days to paint 1 house( given in the question ).
Is This Answer Correct ? 0 Yes 3 No
More General Aptitude Interview Questions
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The largest rubber producer in the world__________
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There are four dogs/ants/people at four corners of a square of unit distance. At the same instant all of them start running with unit speed towards the person on their clockwise direction and will always run towards that target. How long does it take for them to meet and where?
plz send me last 5 years solved question paper of sbi clerk post on my yahoo id.my yahoo id is ven_mohanan2006@yahoo.co.in thanking you Venkatesh.M
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plese send me the sbi po model papers
what is the difference between sports and games?
2 Oranges,3 bananas and 4 apples cost rs.15 . 3 Oranges 2 bananas 1 apple costs rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples
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# Working with sequences
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Working with sequences [#permalink] 12 Jul 2011, 04:00
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Difficulty:
55% (hard)
Question Stats:
65% (02:05) correct 35% (02:13) wrong based on 17 sessions
Hi everybody,
I'm having a hard time going through sequences for some reason, so your explanations will be greatly appreciated. It goes like this:
The sequece S1, S2, S3..., Sn ... is such that Sn= 1/n - 1/n+1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?
1. k>10
2. k<19
Thanks.
-R
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Re: Working with sequences [#permalink] 12 Jul 2011, 11:51
First see the sum of first 3 terms: (1 - 1/2) + (1/2 - 1/3 ) + (1/3 - 1/4) = 3/4
similarly first sum of first 2 terms: 2/3
therefore it will be of the form, sum of first n terms = n/n+1
since this is a proper fraction, adding one to Numerator and denominator increases the overall value.
consider stmt 1: k > 10; the sum of first 10 terms is = 10/11 which is (9+1)/(10+1) that means greater than 9/10.
but in stmt 2: sum of fist 9 (or fewer terms) is less than 9/10 and sum of terms more than 9 is greater than 9/10.
therefore answer will be A
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Re: Working with sequences [#permalink] 14 Jul 2011, 05:06
I could not do this in 2mins. picked a wrong strategy.. damn
Sn = 1/n - 1/(n+1)
Sum of first k terms = S1 + S2 ... Sk = 1/1 -1/2 + 1/2 - 1/3 ... + 1/(k-1) - 1/k + 1/k - 1(k+1) = 1 - 1/(k+1) = k/(k+1)
St#1 : k>10 , assume k=11, then Sum = 11/12, which is > 9/10, (also subsequent sums for k = 12, 13, 14 etc. will be greater that 9/10). Suff. there possible options AD
St #2: k<19, assume k=2, then Sum = 2/3, which is < 9/10 and we already saw that for k=11, sum > 9/10 so not suff. eliminate D => A.
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Re: Working with sequences [#permalink] 14 Jul 2011, 05:06
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Re: Solving simple equations
• To: mathgroup at smc.vnet.net
• Subject: [mg122737] Re: Solving simple equations
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 8 Nov 2011 07:16:16 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201111071054.FAA03874@smc.vnet.net>
```I think one should note that when using PowerExpand one can get a
solution that is only valid for some values of the parameters or even
one that is never valid at all. As an example, consider the equation:
eq = Sqrt[1/z] - 1/Sqrt[z] + Sqrt[1/a] - 1/Sqrt[a] + a + z == 0
It actually has no solutions over the reals as you can tell with
Reduce[eq, z, Reals]
False
But if you use PowerExpand you will get a "solution":
sol = Solve[PowerExpand[eq], z, Reals][]
{z->-a}
The solution is, of course, wrong. If you use PowerExpand with Assumptions you can see this fact:
Solve[PowerExpand[eq,
Assumptions -> a >= 0], z, Reals]
{{z -> ConditionalExpression[-a, a < 0]}}
Solve[PowerExpand[eq, Assumptions -> a >= 0], z, Reals]
{{z->ConditionalExpression[-a,a<0]}}
Andrzej Kozlowski
On 7 Nov 2011, at 11:54, Dana DeLouis wrote:
>> ...Reduce (& Solve) doesn't work...
>> ...Solve[(A*n^a)^b/n == c, n]
>
> Hi. The only way I found to get a real solution for this particular problem is to use PowerExpand:
>
> data = {A -> 2, a -> 3, b -> 4, c -> 5.};
>
> equ = PowerExpand[(A*n^a)^b/n == c] ;
>
> Solve[equ, n]
>
> Which gives:
>
> n -> (c/A^b) ^ (1/(a*b - 1))
>
> % /. data
> 0.8996576447849666
>
> Check with Solve using the same data...
>
> (A*n^a)^b/n == c /. data
>
> 16*n^11 == 5.
>
> NSolve[%, n, Reals]
>
> n -> 0.8996576447849665
>
> = = = = = = = = = =
> HTH
> Dana DeLouis
> = = = = = = = = = =
>
>
>
> On Nov 5, 4:57 am, Mathieu <mat... at gmail.com> wrote:
>> Mathematica seems to struggle with very simple equations:
>>
>> Solve[(A*n^a)^b/n == c, n]
>> Solve::nsmet: This system cannot be solved with the methods available
>> to Solve
>>
>> Reduce doesn't work either. Even if I add assumptions:
>> Assuming[A > 0 && 1 > a > 0 && 1 > b > 0 && c > 0 && n > 0, Solve[(A
>> n^a)^b/n == c, n]]
>> Solve::nsmet: This system cannot be solved with the methods available
>> to Solve. >>
>>
>> Is there a way for Mathematica to solve these equations?
>>
>> Many thanks,
>> Mathieu
>
>
>
```
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# a. 2x = sqr root 12x+72 b. sqr root x+5 = 5 - sqr root x c. |2x| = -|x+6| d. 5 = |x+4|+|x-1| e. |x-2| = 4 -|x-3|Please Help!!!
giorgiana1976 | Student
a) Supposing that you want to solve:
2x = sqr root (12x+72)
First you have to check for what x values, the square roots exists. To check, you have to solve the inequality:
12x+72>=0
If you divide the inequality with 4, you'll have:
3x+18>=0
3x>=-18
x>=-18/3
x>=-6
That means that, solving the equation, you have to keep all x values which belong to the interval set by the condition of existence of the square root.
Now, let's solve it:
(2x)^2=(sqr root 12x+72)^2
4x^2=12x+72
4x^2-12x-72=0
x^2-3x-18=0
x1=[3+sqrt(9+72)]/2
x1=(3+9)/2
x1=6
x2=(3-9)/2
x2=-3
Though both solutions belongs to the interval, we have to verify them into the equation.
If we put x=-3 in equation, we'll have:
-6=sqrt(-36+72)
-6=sqrt36
-6=6, which is not true.
b) sqr root (x+5) = 5 - sqr root x
x+5 = 25-10sqrt x+x
20-10sqrt x=0
2-sqrt x=0
sqrt x = 2
x=4
c) |2x| = -|x+6|
In order to solve this equation, first let's see for what value of x, 2x>=0
|2x|=2x for x>=0
|2x|=-2x, for x<0.
|x+6|=x+6, for x>=-6
|x+6|=-x-6, for x<6
From these condition, occure 3 cases:
1) x belongs to (-inf., -6)
-2x=-(-x-6)
-2x=x+6
-3x=6
x=-2 which is not in the interval (-inf., -6).
2) x belongs to [-6,0)
-2x=-(x+6)
-2x+x=-6
-x=-6
x=6 which belongs to the interval [-6,0).
3) x belongs to [0,+inf.)
2x=-x-6
3x=-6
x=-2, which is not i the interval [0,+inf.).
The only solution of the equation is x=6.
d) 5= |x+4|+|x-1|
In order to solve this equation, first let's see for what value of x, |x+4|>0
|x+4|=x+4, for x>=-4
|x+4|=-x-4, for x<-4
Now, let's see, for what values of x,|x-1|>0.
|x-1|=x-1, for x>=1
|x-1|=-x+1, for x<1
From these condition, occure 3 cases:
1)x belongs to (-inf., -4)
5=-x-4-x+1
8=-2x
x=-4 which is not in the interval.
2) x belongs to [-4,1)
5=x+4-x+1
5=5, for any value from [-4,1).
3) x belongs to [1,+inf.)
5=x+4+x-1
2=2x
x=1, which belongs to [1, +inf.)
e) |x-2| = 4 -|x-3|
In order to solve this equation, first let's see for what value of x, |x-2|>0
|x-2|=x-2, for x>=2
|x-2|=-x+2, for x<2
Now, let's see, for what values of x,|x-3|>0.
|x-3|=x-3, for x>=3
|x-3|=-x+3, for x<3
From these condition, occure 3 cases:
1) x belongs to (-inf., 2)
-x+2=4+x-3
2x=1
x=1/2 which belongs to (-inf., 2).
2)x belongs to [2,3)
x-2=4+x-3
-2=1, is not true!
3) x belongs to [3, +inf.)
x-2=4-x+3
2x=9
x=4.5which belongs to [3, +inf.)
The solutions of the equation are:
x={0.5,4.5}
neela | Student
a)
2x = sqrt12x+72. Hope sqrt 12x means (sqrt12)x and not sqrt(12x). So,
2x-sqr12 x = 72. Or
(2-sqrt12)x = 72 Or
x = 72/(2-sqrt12) = 72(2+sqrt12)/(2-12) , after rationalising the denominator,
= 72(2+2sqrt3)/(-10)
=-7.2(2+2sqrt3)
b) sqrtx+5 = 5-sqrtx. Sqrtx+5 and sqrt(x+5) are different. Hope you do not mean sqrt(x+5) on LHS.
sqrtx=-sqrtx . Adding sqrtx to both sides,
2sqrtx = 0.
sqrtx = 0 . Or x = 0.
c)
|2x| = -|x+6|
Case (i) x<-6
|2x| = -(6-x) when x<-6 Or
-2x = 6-x. Or
-2x = 6. Or
-2x+x =-6.
-x = -6
So x = 6 and x<-6 which is a contradiction.
case(2) when 6< x <0
-2x = -(x+6). Or
-2x+x = -6. Or
-x = -6. Or x = 6 and x<0 is a contradiction.
Case (iii) x>0
2x = -(x+6) Or
2x+x = -6. Or
3x=-6 Or x =-2 and x>0 is a contradiction.
So there is no solution
d) 5 = |x+4|+|x-1|
When x>=1, 5 = x+4+x-1 Or 5 = 2x+3. Or x= (5-3)/2 = 1 is a contradiction as x>=1 and x=1. So x =1 is a solution.
When 0 = < x < 1, 5 = x+4 +(1-x). Or
5=5+2x . Or x = 0. Is a solution.
When -4 =< x<0, 5 = x+4+1-x, 5 =5 is an identity.So -4<=x< 0 is a solution.
When x <4, 5 = 4-x+(1-x) . Or 5 = 5-2x Or x=0 is a contradiction as x<-4 and x=5/2 are inconsistent.
So x=1 or (-4<x<=0) are the solution
e)
|x-2| = 4-|x-3|
case (i) x>=3
x-2 =4-(x-3). Or 2x = 4+3+2 or x =9/2 is consistent with x>=3. So x= 4.5 is a solution.
Case(ii) 2<x<3
x-2 = 4-(3-x) . Or x -2 = 3+x Or -2 = 3 is inconsistent result. So 2<x<3 cannot be a solution.
When x<2,
2-x =4-(3-x). Or
2-x=4+x-3. Or
2-4+3 = 2x Or
1 = 2x . Or x = 1/2 is consistent with x<2. So x=1/2 is a solution. | HuggingFaceTB/finemath | |
UT PHY 302L - DC circuits
Unformatted text preview:
familiari (nf3795) – DC circuits – yeazell – (58395) 1This print-out should have 16 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 (part 1 of 2) 10.0 pointsYou can obtain only four 20 Ω resistor s fromthe stockroom.How can you achieve a resistance of 50 Ωunder these circumstances?1. 1 in series with 3 in parallel2. 2 in series with 2 in parallel correct3. 4 in parallel4. 2 in series5. 3 in parallel6. 4 in series7. None of these8. 2 in parallel9. 3 in seriesExplanation:Let : R = 20 Ω .Placing two resistors of the same si ze in par-allel halves the original individual resistance:1Rnew=1R+1R=2RThus we needReq= R + R +R2Two resistors in series with two parallel resis-tors:Req= 20 Ω + 20 Ω +20 Ω2=50 Ω .002 (part 2 of 2) 10.0 pointsWhat can you do if you need a 5 Ω resistor?1. 2 in series with 2 in parallel2. None of these3. 3 in series4. 2 in parallel5. 3 in parallel6. 2 in series7. 1 in series with 3 in parallel8. 4 in parallel correct9. 4 in seriesExplanation:Placing 4 resistors of the same value inparallel will quarter the or iginal individualresistance:1Req=1R+1R+1R+1R=4R.Four parallel resistors:Req=R4=20 Ω4=5 Ω .003 10.0 pointsAfter a 6.67 Ω resistor is connected acrossa battery with a 0.23 Ω internal resistance,the electric potential between the physicalbattery terminals is 12 V.What is the rated emf of the battery?Correct answer: 12.4138 V.Explanation:Let : R = 6.67 Ω ,r = 0.23 Ω , andV = 12 V .The current drawn by the ex ternal resistoris given byfamiliari (nf3795) – DC circuits – yeazell – (58395) 2I =VR=12 V6.67 Ω= 1.7991 A .The output voltag e is reduced by the inter-nal resistance of the battery byV = E − I r ,so the electromotive force isE = V + I r= 12 V + (1.7991 A) (0.23 Ω)=12.4138 V .004 10.0 pointsFour resistors are connected as shown in thefigure.21 Ω53 Ω86 Ω91 V47 ΩS1abcdFind the resistance between points a and b.Correct answer: 37.8222 Ω.Explanation:R1R3R4EBR2S1abcdLet : R1= 21 Ω ,R2= 47 Ω ,R3= 53 Ω ,R4= 86 Ω , andEB= 91 V .Ohm’s law is V = I R .A good rule of thumb is to eliminate junc-tions connected by zero resistance.R2R3R1R4abcdThe parallel connection of R1and R2givesthe equivalent resistance1R12=1R1+1R2=R2+ R1R1R2R12=R1R2R1+ R2=(21 Ω) (47 Ω)21 Ω + 47 Ω= 14.5147 Ω .R12R3R4abThe series connection of R12and R3givesthe equivalent resistanceR123= R12+ R3= 14.5147 Ω + 53 Ω= 67.5147 Ω .R123R4abfamiliari (nf3795) – DC circuits – yeazell – (58395) 3The parallel connection of R123and R4gives the equivalent resistance1Rab=1R123+1R4=R4+ R123R123R4Rab=R123R4R123+ R4=(67.5147 Ω) (86 Ω)67.5147 Ω + 86 Ω= 37.8222 Ω .or combining the above steps, the equivalentresistance isRab=R1R2R1+ R2+ R3R4R1R2R1+ R2+ R3+ R4=(21 Ω) (47 Ω)21 Ω + 47 Ω+ 53 Ω(86 Ω)(21 Ω) (47 Ω)21 Ω + 47 Ω+ 53 Ω + 86 Ω=37.8222 Ω .005 (part 1 of 2) 10.0 pointsTwo identical light bulbs A and B are con-nected in series to a constant voltage source.Suppose a wire is connected across bulb B asshown.EA BBulb A1. will burn more brightly. correct2. will burn less brightly.3. will burn as brightly as before.4. will go out.Explanation:Because the wire is added in parallel tolight bulb B, the equivalent resistance nowbecomes just the resistance of light bulb A.Since the resistance dropped, the current hadto increase which means that light bulb A willget brighter.006 (part 2 of 2) 10.0 pointsand bulb B1. will burn as brightly as before.2. will burn more brightly.3. will burn less brightly.4. will go out. correctExplanation:Since the wire is the path of zero resistance,all the current wil l flow through the wire andnone will flow through light bulb B. With nocurrent to light the light bulb, bulb B will goout.007 10.0 pointsConsider resistors R1and R2connected inseriesER1R2and in parallelER1R2to a source of emf E that has no internalresistance.How does the power dissipated by the resis-tors in these two cases compare?1. It is greater for the series connection.2. It is different for each connection, but onefamiliari (nf3795) – DC circuits – yeazell – (58395) 4must know the values of R1and R2to knowwhich is greater.3. It is greater for the parallel connection.correct4. It is the same for both connections5. It is different for each connection, but onemust know the values of E to know which isgreater.Explanation:The power dissipated by the resistors isP =E2Req.The equivalent resistance for a series con-nection isRs= R1+ R2.The equival ent resista nce for a parallel con-nection isRp=R1R2R1+ R2.Regardless of the values of R1and R2, Rp<Rs, so more power is dissipated in the parallelconnection.008 (part 1 of 5) 10.0 pointsAssume the battery is ideal (it has no in-ternal resistance) and connecting wires haveno resistance. Unlike most real bulbs, the re-sistances of the bulbs in the questions belowdo not change as the current through themchanges. Three identical bulbs are in the cir-cuit as shown below in the figure. (The switchS is initially closed.)EABCSWhich of the following correctly ranks thebulbs in brightness?1. Bulb A is the brightest, B is next bright-est, and C is the least brightest.2. None of these is correct.3. Bulb B and C are equally bright, andeach is brighter than A.4. Bulb A is the brightest, and B and C areequally bright. correct5. All bulbs are equally bright.Explanation:Bulb B and C are connected parallel, theyhave the same potential difference, t hus theyare equally bright. IA= IB+ IC, so A thebrighter than B and C.009 (part 2 of 5) 10.0 pointsWhich of the following correctly ranks thecurrent flowing through the bulbs?1. B ul b B and C have the same current, andeach has more current than A.2. B ul b A has the largest current, and C hasthe smallest current.3. Bulb A has the largest current, and B andC have the same current. correct4. None of these is correct.5. All bulbs have the same current flowingthrough them.Explanation:As mentioned in t he last part, B and Chave the same potential difference and currentI =VR, while the current of A is the sumof that of B and C.010 (part 3 of 5) 10.0 pointsWhich of the following correctly ranks thepotential difference across these bulbs?1. The potential difference is l a r gest acrossfamiliari (nf3795) – DC circuits – yeazell – (58395) 5A, and smallest across C.2. None of these is correct.3. All bulbs have the same potential acrossthem.4. Bulb B and C
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# UT PHY 302L - DC circuits
Pages: 7
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4 pages | HuggingFaceTB/finemath | |
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• Print publication year: 2009
• Online publication date: June 2012
# Introduction
## Summary
Graph theory
This section presents the basic definitions, terminology and notation of graph theory, along with some fundamental results. Further information can be found in the many standard books on the subject – for example, Chartrand and Lesniak [1], Gross and Yellen [2], West [3] or (for a simpler treatment)Wilson [4].
Graphs
A graph G is a pair of sets (V, E), where V is a finite non-empty set of elements called vertices, and E is a finite set of elements called edges, each of which has two associated vertices (which may be the same). The sets V and E are the vertex-set and edge-set of G, and are sometimes denoted by V (G) and E(G). The order of G is the number of vertices, usually denoted by n, and the number of edges is denoted by m.
An edge whose vertices coincide is called a loop, and if two vertices are joined by more than one edge, these are called multiple edges. A graph with no loops or multiple edges is a simple graph. In many areas of graph theory there is little need for graphs that are not simple, in which case an edge e can be considered as a pair of vertices, e = {v, w}, or vw for simplicity. However, in topological graph theory, it is often useful, and sometimes necessary, to allow loops and multiple edges. A graph of order 4 and its underlying simple graph are shown in Fig. 1. | HuggingFaceTB/finemath | |
## Nothing Short of a Miracle!
That’s what happened today. The occasion was a cultural night arranged by the Staff Club of NITC (no one knew one even existed, till today!). Some of the faculty got together and wrote a drama to be staged tonight. I learnt about it from Deepak sir’s blog a few days ago, and that in itself seemed like a miracle. But the actual performance was nothing short of unbelievable, to say the least! It was fantastic. About the drama itself, I keep that for another day. I’m just too dazed by that performance to analyze it critically. Besides, Deepak sir has promised to make the script and video available.
I always knew that the faculty members were fantastic people outside the classroom, but hats off to them for this wonderful performance!
## DSP Lab – Week 1
### Constructing the Complex Plane
Suppose we have a sampled signal defined by the sequence $h(n)$, $n=0,1,2,...,N-1$
Its Z- transform is given by $H(z) = \sum_{n=0}^{N-1} h(n)z^{-n}$ .
It maps the original sequence into a new domain, which is the complex plane $z=e^{sT}$ where $s=\sigma+j\omega$ is the parameter in the Laplace domain and $T$ is the sampling period.
The $j\omega$ axis in the $s$-plane maps onto the unit circle with centre at the origin in the $z$-plane. So the value of $H(z)$ at different points on the unit circle actually gives the contribution of the frequency component given by $\angle z$, in the original signal.
This, in effect, gives the Discrete Fourier Transform of the sequence. Consider the following example:
%original sequence
h = [1,2,3,4];
%number of chosen points on the unit circle
N = 64;
%define the chosen points
z = complex(cos(2*pi/N*(0:N-1)),sin(2*pi/N*(0:N-1)));
%evaluate H(z) at each point
for i = 1:N
H(i) = 1+2*z(i)^-1+3*z(i)^-2+4*z(i)^-3;
end
%plot the unit circle
plot(z)
%plot the value of H(z) along the unit circle
figure
plot(abs(H))
%plot the N-point DFT of h(n)
figure
plot(abs(fft(h,64)))
This example computes the value of $H(z)$ at 64 uniformly spaced points on the unit circle and compares it with the 64 point DFT. We can see that both (fig. b & c) are identical.
unit circle
value of H(z)
abs(fft(h))
## Freedom Walk at NITC
It was almost midnight when the Freedom Walkers arrived here on Wednesday. They were thouroughly worn out from the long long walk from Thamarassery. They had dinner from our mini canteen, and then I led them to the rooms in PG-2 hostel which Sandeep sir from the Electrical department had booked.
Next morning, a few S3 guys and I went to meet them. We had a small gathering in their room and discussed what all we could do to spread Free Software here at NITC. Since all of us except one were from Electronics, Jemshid suggested that we could get started on some Embedded GNU/Linux work. We can think of conducting workshops to get people interested in it.
Prasad talked about the Freedom Toaster they had made, and suggested that we could try to make one with a vending machine, as a project. They said we would have their support if someone is ready to take it up.
It’s too bad we couldn’t organize a more elaborate meeting with the Freedom Walkers, because of the exams. But we’ve got some pointers to think about, when we sit down to make a concrete plan regarding the FOSS Cell activities.
## GNU/Linux Install Fest at NITC
As the first activity of the upcoming FOSS Cell NITC, we organized a GNU/Linux install fest on the occasion of Software Freedom Day. Around twenty people turned up during the day. The only undesirable part was that a couple of laptops, after installing Ubuntu, couldn’t boot Windows. Got to sort out their issues soon. We’ve set up a technical support mailing list for people to post their problems.Considering that it was the first ever event by our FOSS Cell, it didn’t go too badly.
## Software Freedom Day at Kozhikode
Software Freedom Day 2008 was celebrated today at Malabar Christian College, Kozhikode. The event was organized by Swatantra Malayalam Computing, in association with Malabar Christian College. The main attraction was a seminar on Language Computing, led by SMC.
There was also a GNU/Linux install fest and demo in parallel. We installed GNU/Linux on around 6-7 systems, apart from the 5 in the computer lab of Malabar Christian College. There was also an installation demo for hardware technicians. I couldn’t attend the seminar, and I’m looking forward to reading other reports about it.
One of the highlights of the day was the revival of Free Software Users’ Group Calicut, which had been dormant for over two years. Jemshid, of Ascent Engineers, the team from KSEB led by Mohammed Unais, have taken the initiative to kick start the community’s activities.
There were a few representatives from GEC West Hill and AWH Engg. College, and we’ve decided to organize a few workshops, to get some people from those colleges involved in FOSS as well, and try to create a network of college FOSS communities. Shyam put forward the necessity of a common platform for engineering colleges throughout Kerala, based on Free Software. We also have to explore the possibility of encouraging students to take up Free Software development as their projects.
Jemshid and his team had managed to contact the Malabar IT dealers’ association, and their representatives had turned up. They expressed genuine interest in migrating to GNU/Linux for the default installation in new systems they sell. They would thus be able to avoid distributing so called “pirated” software. We have proposed to arrange a basic GNU/Linux workshop for the hardware vendors. This move has the potential to start a revolution. If they can show their customers that they can do almost anything on GNU/Linux that they normally use a computer for, they’ll be encouraged to switch to it. And the customers will have someone to turn to for support.
On the whole, the event was a great success. Tomorrow, we have a small event planned in our campus. More on that later.
See other blogs and photos of the event:
Hiran
## Software Freedom Day at NITC
We are planning to celebrate the Software Freedom Day through an install fest and demos. There was a meeting today to get some volunteers for the event, and around twenty S3 students turned up. Only some of them have used GNU/Linux before, and they have been given the task of familiarising the others with it before the event. We are also hopeful of launching our FOSS Cell officially on that day. More about the event as it materializes…
## Liberation of Environment Knowledge Repository
The Centre for Science and Environment, in association with the National Knowledge Commission, has set up a National Portal on Environment. Read more here
I became a keen reader of CSE’s Down to Earth magazine, while I was at IUAC. It’s very informative and covers stories of development and environment from a rural perspective- many things which never appear in the mainstream media. It is great to know that the Environment Portal will make the whole Down to Earth archive freely available. | HuggingFaceTB/finemath | |
0
Q:
# An amount of 5,000 is invested at a fixed rate of 8 per cent per annum. What amount will be the value of the investment in five years time, if the interest is compounded every six months?
A) 7401.22 B) 3456 C) 4567 D) 7890
Explanation:
With slight modifications, the basic formula can be made to deal with compounding at intervals other than annually.
Since the compounding is done at six-monthly intervals, 4 per cent (half of 8 per cent) will be added to the value on each occasion.
Hence we use r = 0.04. Further, there will be ten additions of interest during the five years, and so n = 10. The formula now gives:
V = P(1 + r)10 = 5,000 x (1.04)10 = 7,401.22
Thus the value in this instance will be £7,401.22.
In a case such as this, the 8 per cent is called a nominal annual
rate, and we are actually referring to 4 per cent per six months.
Q:
Gopal borrowed some money at 12% simple interest. If he had to pay back Rs. 1280 after 5 years, in order to clear off the loan. How much did he borrow?
A) Rs. 800 B) Rs. 620 C) Rs. 560 D) Rs. 480
Explanation:
Let the principle amount be Rs. P
Interest rate = 12%
Total amount he paid after 5 years = Rs. 1280
ATQ,
Hence, the amount he borrowed = P = Rs. 800.
1 152
Q:
The simple interest on ₹ 10 for 4 months at the rate of 3 paise per month is
A) 0.3 Paise B) 1.2 Paise C) 30 Paise D) 3 Paise
Explanation:
Given Principal amount P = Rs. 10
Time T = 4 months
Rate of interest R = 3 ps
Interest I = PTR/100 = 10 x 4 x 3/100 = 12/10 = 1.2 paise.
1 1460
Q:
Avinash gave a sum of 5400 at the simple rate of interest of 8% to Rajeev for 4 years and Rajeev gave this amount to Chanukya at the rate of 6% for 4 years. Find how much extra amount has to be paid by Rajeev to Avinash after 4 years?
A) Rs. 364 B) Rs. 432 C) Rs. 498 D) Rs. 554
Explanation:
Here Avinash gave to Rajeev and Rajeev gave the same to Chanukya
Principal amount and Time is same but the only difference is Rate of interest.
Rajeev took @ 8% and gave it to Chanukya @ 6%
Here the difference in interest rate = 2%
2% of 5400 should be the extra amount paid by Rajeev to Avinash
Required amount =
Hence, Rs. 432 is the extra amount has to be paid by Rajeev to Avinash after 4 years.
1 1239
Q:
On retirement, a woman gets Rs. 1,45,440 of her provident fund which she invests in a scheme at 20% per annum. Her monthly income from this scheme will be
A) Rs. 2550 B) Rs. 2424 C) Rs. 2224 D) Rs. 2380
Explanation:
Now, the principal amount P = Rs. 1,45,440
Rate of interest R = 20%/annum
Now monthly income I = PTR/100 = 1,45,440 x 20 x 1/100 x 12
= 2908800/1200
= Rs. 2424.
Hence, her monthly income = Rs. 2424.
3 1334
Q:
In what time will Rs. 1860 amount to Rs. 2641.20 at simple interest 12% per annum?
A) 2.9 years B) 3.5 years C) 4.2 years D) 4.7 years
Explanation:
Given that Rs. 1860 will become Rs. 2641.20 at 12%
=> Simple Interest = 2641.20 - 1860 = Rs. 781.20
We know I = PTR/100
=> 781.20 x 100 = 1860 x T x 12
=> T = 78120/1860x12
=> T = 78120/22320
=> T = 3.5 years.
17 2568
Q:
A sum of money triples itself in 8 years simple interest. Find the rate of percent per annum?
A) 30% B) 25% C) 22% D) 18%
Explanation:
The rate of percent R is given by,
30 7340
Q:
A certain amount earns simple interest of Rs. 2260 after 3 years. Had the interest been 1 % more how much more interest would it have earned ?
A) Rs. 175 B) Rs. 220.75 C) Rs. 126 D) Can't be determined
Explanation:
Here given Interest earned = Rs. 2260
Time = 3 years
Rate of interest = ?
Principal Amount = ?
So, it can't be determined.
16 1867
Q:
A person lends Rs. 1540 for five years and Rs. 1800 for four years. If he gets Rs. 1788 as interest on both amounts, what is the rate of interest ?
A) 14.5% B) 11% C) 12% D) 10.5%
Explanation:
Let the interest rate be r%
We know that,
S.I = PTR/100
=> (1540 x 5 x r)/100 + (1800 x 4 x r)/100 = 1788
=> r = 178800/14900 = 12% | HuggingFaceTB/finemath | |
# How does lack of deadlock relate to computability in process calculi?
I'm interested in knowing things about the computability of concurrent programs. If you had a Turing complete language that also let you branch off new programs but had no means of communication between them there would be programs that you couldn't write. Namely those that required communication between concurrently running programs. At the other extreme it seems like the $\pi$-calculus can pretty well compute any program and has the power to implement almost any synchronization primitive I've ever heard of. But it also has deadlock much like Turing complete programs have infinite loops. So there seem to be degrees of power in synchronization/communication primitives although all I have figured out are extremes. This however seems analogous to programs without any kind of iteration or recursion and full Turing complete programs. One in-between would be mutexs that have levels to them at the type level. Every mutex could have levels to them at the type level. Every mutex could only be acquired if a higher mutex level had already been acquired by that thread. This would prevent deadlock but it would probably restrict some kinds of communication from occurring because the type system couldn't encode all well orderings on the mutexs (this is a bet, I have no proof). This feels analogous to something like primitive recursion or some other limited form of repeating computation (in fact, it's the kinda same problem of finding a well-ordering on objects in a program)
It's also easy to prove that a sufficiently strong language with a way to cause deadlock would not have decidable deadlock. Just replace infinite loops for dead lock in the standard proof of the halting problem. So yet again deadlock seems to have the kinds of properties that infinite looping has. On the other hand there is no way (that I know of, please tell me if this is possible) to detect infinite loops at run time but you can detect deadlock at run time. So dead lock also seems like an easier problem in a way as well.
Say I define a notion of total-program equality that goes something like "two languages are equal IFF every total program that can be written in one can be written in the other" where "total program" here is a function from naturals to naturals. More clearly stated if the set of total computable functions two languages can compute are the same then I'll call them "total-program equal". Every Turing complete language is total program equal and another neat fact is that have language that is total-program equal to a Turing complete language is not a total language! Assume that such a language existed, then an interpreter for it would be a total program that could be interpreted by a Turing complete language. Thus the supposed language could interpret itself. But such a total language (namely one which definitely has composition and paring and such) can't interpret itself so we have a contradiction. This is to say, there is no way to define a language that isn't going to contain non-total members.
I'd like to define something similar but for deadlock and the pi-calculus. Deadlock is tricky however. Separating deadlock from livelock is tricky to work with. Moreover I don't have a notion of the semantics of elements of a concurrent language. Certainly $\mathbb{N} \to \mathbb{N}$ doesn't really capture this because the programs were interested in keep outputting data on various channels as new data comes in. So there are some things that need to be formalized but basically I want a notion of "deadlock-free equivalent" languages. That is two languages are deadlock-free equivalent if the set of programs that they can define that never dead lock is the same set. Then I'd like to ask "is there a dead-lock free language that is deadlock-free equivalent to the pi-calculus?".
I don't really expect this exact question to have been answered but I was wondering if anyone has done work in this vain that I could look at? What are good models of process calculi like how $\mathbb{N} \to \mathbb{N}$ models computation. How do you formalize the difference between deadlock lock and livelock? Does being deadlock free cause such a language to not be able to interpret itself? Is there some other reason a dead lock free language as good as the pi-calculus can't exist? Answers to these questions here would be fantastic but I mainly just want to be pointed in the right direction to read further.
In a nutshell, lack of infinite loops means that you can't compute some programs even though they don't have infinite loops in them. Is the same true for deadlock? That is, will any language that never deadlocks have some programs that it can't compute?
• Pi calculus is Turing Complete: cnd.mcgill.ca/~ivan/functions_as_processes_BFb0032030.pdf – jmite Oct 3 '16 at 3:22
• Sure I was aware of that fact when writing this. I'm not sure that it matters or at least how it matters is very obscure to me. I'm specifically interested in programs in which the computation is a continuous reaction to a set of streams not a function on naturals. I'd wager this falls under higher type computability just for that reason alone. Either way I'd need more than just that fact. – Jake Oct 3 '16 at 3:28
• The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Oct 3 '16 at 7:15
I think you are asking about expressivity of concurrent programming languages. This is a deep and not well-understood field. For example you say that "the $\pi$-calculus [...] has the power to implement almost any synchronization primitive I've ever heard of". It is well known that the $\pi$-calculus cannot implement broadcasting (see e.g. here). The $\pi$-calculus can also not implement $n$-ary synchronisation (think Petri nets) for all $n > 2$, and $\pi$ can also not implement Ambient-calculus. Moreover, full mixed choice cannot be implemented by the asynchrnous $\pi$-calcululs, this is an old result by Palamidessi. There are more such results.
• @Jake it really depends on what you mean by deadlock, for example is $\mathbf{Nil}$, the process that does nothing, deadlocked for you or not? There are elaborate conceptions of the role of deadlocks and livelocks in process calculi. – Martin Berger Oct 4 '16 at 8:36
• @Jake There is terminological confusion between the concurrent/distributed systems community and the process theorists. The former thinks of a deadlock as a cycle in the waits-for graph, while the latter see a process $P$ as deadlocked if $P$ has no transitions whatsoever. Hence $Nil$ is deadlocked in process theory but not for the concurrent/distributed systems community. You can model cycle in the waits-for graph with processes, e.g. $(\nu ab)(a.\overline{b}.P \ |\ b.\overline{a}.Q)$. – Martin Berger Oct 4 '16 at 14:57
• @Jake There has been a lot of work on types for concurrent processes where types gurantee certain liveness properties of processes. In such typing systems, the notion of linear (output) channel is key. If a typing systems guarantees linearity at (output) channel $x$, then an output will eventually be sent on this channel. This is a good generalisation of absence-of-deadlock in my experience. Note that in this approach it's not processes that are or are not deadlocked, but (output) channels. – Martin Berger Oct 4 '16 at 14:59 | open-web-math/open-web-math | |
## Conversion formula
The conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:
1 oz = 0.0625 lb
To convert 1525 ounces into pounds we have to multiply 1525 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:
1 oz → 0.0625 lb
1525 oz → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 1525 oz × 0.0625 lb
M(lb) = 95.3125 lb
The final result is:
1525 oz → 95.3125 lb
We conclude that 1525 ounces is equivalent to 95.3125 pounds:
1525 ounces = 95.3125 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.010491803278689 × 1525 ounces.
Another way is saying that 1525 ounces is equal to 1 ÷ 0.010491803278689 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand five hundred twenty-five ounces is approximately ninety-five point three one three pounds:
1525 oz ≅ 95.313 lb
An alternative is also that one pound is approximately zero point zero one times one thousand five hundred twenty-five ounces.
## Conversion table
### ounces to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to pounds
ounces (oz) pounds (lb)
1526 ounces 95.375 pounds
1527 ounces 95.438 pounds
1528 ounces 95.5 pounds
1529 ounces 95.563 pounds
1530 ounces 95.625 pounds
1531 ounces 95.688 pounds
1532 ounces 95.75 pounds
1533 ounces 95.813 pounds
1534 ounces 95.875 pounds
1535 ounces 95.938 pounds | HuggingFaceTB/finemath | |
# Finding the correct formula for a sequence
As already discussed a little bit in the chatroom the following question came to my mind while studying some sequences of natural numbers.
If you need to find closed formulas (not piecewise function or polynomials which interpolate the numbers) for sequences like
(i) 2-4-8-16-...
(ii) 1-8-64-512-...
the first thing you'll probably try is to find a rule how the numbers are build up. In (i) it seems to be that the next number is the previous number multiplied by the factor 2, in (ii) we always multiply the previous number by 8.
My question is; What is the minimum number of numbers you need to find an unique closed formula expression? Lets take for simplicity only the the natural numbers.
Is there for exmaple a sequence $a(n)=?$ where $a(1)=2,a(2)=4, a(3)=8,a(4)=16$ but $a(5)\not=32$ and if yes is there also a sequence where $a(5)=32$ but $a(6)\not=64$ and so on?
-
You will need more restrictions. Here are two examples: 2, 4, 8, 16, 42, 42, 42, 42 2, 4, 8, 16, 32, 2, 3, 5, 7 , 11, 13, 17 – PyRulez Jun 10 '13 at 12:59
(i) is obviously part of the sequence $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 56953,\ldots$ of numbers $n$ such that $n$ and $n^3$ have the same number of ones in binary expansion – Hagen von Eitzen Jun 10 '13 at 14:54
Whatever "rule" of the "formula" $f(n)$ you come up with, you can always say that $$h(n):=\begin{cases} f(n),& n\leq N \\ g(n),& n>N \end{cases}$$ is a new "rule" where $N$ and $g$ are completely arbitrary. Even more, for any finite sequence $a(1)$, $a(2)$, $\dots, a(N)$ you can come up with a formula $f$ (for example, a polynomial) that agrees with all $a(n)$ up to $N$, but you can force $f(N+1)$ to be arbitrary.
In a way, IQ test "continue the sequence" are mathematically ill-posed: there are infinitely many continuations of any sequence, and some of them may be equally "natural". However, sometimes you may think of finding the rule for the sequence with the smallest complexity, e.g. Kolmogorov complexity.
-
Sure. Given $n$ values, you can always construct a $n-1$ degree polynomial which matches those values. Since the first five numbers in your example are in GP, i.e. are increasing exponentially ($2^n$), it is unlikely that the resulting polynomial will grow fast enough to match $2^n$ for the next value of $n$ as well.
If $p(x)$ is a $n-1$ polynomial which satisfies $p(n)=2^n$ for $n=1,2,3,4$, then $p$ is:
$$p(x)=\frac 13(x^3-3x^2+8x)$$
(which comes from a simple WolframAlpha computation., taking $p(x) = ax^3+bx^2+cx+d$)
This can be done for any number of terms of the sequence you want to match.
Of course, the polynomial is only interesting if you can get it to match $n$ terms without requiring an $n-1$ polynomial, as then there are lesser chances of over-fitting occurring.
- | HuggingFaceTB/finemath | |
# Mutual Inductance
Mutual inductance is the main operating principle of generators, motors, and transformers. Any electrical device having components that tend to interact with another magnetic field also follows the same principle. The interaction is usually brought about by a mutual induction where the current flowing in one coil generates a voltage in a secondary coil.
## What Is Mutual Inductance?
When two coils are brought in proximity with each other the magnetic field in one of the coils tend to link with the other. This further leads to the generation of voltage in the second coil. This property of a coil which affects or changes the current and voltage in a secondary coil is called mutual inductance. Changing I1 produces changing magnetic flux in coil 2.
In the first coil of N1 turns, when a current I1 passes through it, magnetic field B is produced. As the two coils are closer to each other, few magnetic field lines will also pass through coil 2.
${{\phi }_{21}}\to$ magnetic flux in one turn of coil 2 due to current I1.
If we vary the current with respect to time, then there will be an induced emf in coil 2.
${{\varepsilon }_{ind}}=-\frac{d\phi }{dt}$ [According to Faraday’s law] ${{\varepsilon }_{21}}=-{{N}_{2}}\frac{d{{\phi }_{21}}}{dt}$ ${{\varepsilon }_{21}}=-{{N}_{2}}\frac{d}{dt}\left( \overline{B}.\,\overline{A} \right)$
The induced emf is coil 2 directly proportional to the current passes through the coil 1.
${{N}_{2}}\,{{\phi }_{21}}\propto {{I}_{1}}$ ${{N}_{2}}\,{{\phi }_{21}}={{M}_{21}}{{I}_{1}}$ … (1)
The constant of proportionality is called as mutual inductance. It can be written as
${{M}_{21}}=\frac{{{N}_{2}}{{\phi }_{21}}}{{{I}_{1}}}$ … (2)
The SI unit of inductance is henry (H)
$1H=\frac{1(\text{Tesla}).1\left( {{m}^{2}} \right)}{1\,\,A}$
In a similar manner, the current in coil 2, I2 can produce an induced emf in coil 1 when I2 is varying with respect to time. Then,
${{\varepsilon }_{12}}=-{{N}_{1}}\frac{d{{\phi }_{12}}}{dt}$ ${{N}_{1}}{{\phi }_{12}}\propto {{I}_{2}}$ ${{N}_{1}}{{\phi }_{12}}={{M}_{12}}{{I}_{2}}$ … (3)
${{M}_{12}}=\frac{{{N}_{1}}{{\phi }_{12}}}{{{I}_{2}}}$ … (4)
This constant of proportionality is another mutual inductance. Changing I2 produces changing magnetic flux in coil 1.
## Reciprocity Theorem
Experiments and calculations that combine Ampere’s law and Biot-Savart law confirm that the two constants, M21 and M12 are equal in the absence of material medium between the two coils.
M12 = M21 … (5)
This property is called reciprocity and by using reciprocity theorem, we can simply write the mutual inductance between two coils as;
${{\text{M}}_{12}}\text{=}\,{{\text{M}}_{21}}\equiv \text{M}$
## EMF of Mutual Inductance
Considering the mutual inductance between two coil we just discussed, we defined mutual inductance M21 of coil 2 with respect to 1 as,
${{\text{M}}_{21}}\text{=}\,\frac{{{N}_{2}}{{\phi }_{21}}}{{{I}_{1}}}$ ${{\text{M}}_{21}}{{I}_{1}}={{N}_{2}}{{\phi }_{21}}$
If I1 changes with time,
${{M}_{21}}\frac{d{{I}_{1}}}{dt}={{N}_{2}}\frac{d{{\phi }_{21}}}{dt}$ … (6)
According to Faraday’s law of induction,
${{\varepsilon }_{ind}}=-\frac{d\phi }{dt}$
We can write (6) as,
${{M}_{2}}\frac{d{{I}_{1}}}{dt}=-{{\varepsilon }_{21}}$
Thus induced emf in coil 2 due to current in coil 1 is given by
${{\varepsilon }_{2}}=-{{M}_{21}}\frac{d{{I}_{1}}}{dt}$ … (7)
Similarly, induced emf in coil 1 due to changing current in coil 2 can be given as,
${{\varepsilon }_{1}}=-{{M}_{12}}\frac{d{{I}_{2}}}{dt}$ … (8)
From experiments (equation (5)),
M21 = M12 = M
Therefore ${{\varepsilon }_{1}}=-M\frac{d{{I}_{2}}}{dt}\,\,\,\,\,\,{{\varepsilon }_{2}}=-M\frac{d{{I}_{1}}}{dt}$
The coefficient of mutual induction – mutual inductance depends only on the geometrical factor of the two coils such as the number of turns, radii of two coils and on the properties of a material medium such as magnetic permeability of the medium surrounding the coils.
## How To Find Mutual Inductance?
Steps for finding mutual inductance (M).
(i) Assume current in one of the coils (say I1 in coil 1)
(ii) Deduce the expression for the magnetic field in the neighbouring coil (2) due to I1.
(iii) Write the flux linkage equation.
${{N}_{2}}{{\phi }_{2}}=M{{I}_{1}}$
(iv) Obtain the magnetic flux linked,
${{N}_{2}}\left( {{\phi }_{2}} \right)={{N}_{2}}\left( {{B}_{1}}.\,{{A}_{2}} \right)$
(v) Compare the above two equations and find mutual inductance, M.
### Mutual Inductance Problems
(1) Mutual induction between circular coils. Consider two circular coils (closely packed) coaxially placed to each other. The coil with a larger radius has N1 turns and that with smaller radius has N2 turns. Also, assume that R1>> R2.
The above-mentioned calculation is the same for the following case as well.
• Mutual inductance of two concentric coplanar loops. • Mutual induction between two solenoids.
## Mutual Inductance of a Coaxial Solenoid Consider two coaxial solenoids of which the outer solenoid S2 has radius r2 and N2 turns whereas the inner solenoid S1 has radius r1 and N1 turns. Both the solenoids are of equal length.
When there is a current I2 in the solenoid S2, the magnetic induction due to I2 is given by,
${{B}_{2}}={{\mu }_{0}}{{n}_{2}}{{I}_{2}}={{\mu }_{0}}\frac{{{N}_{2}}}{l}{{I}_{2}}.$
The corresponding flux linkage with solenoid S1 is,
${{N}_{1}}{{\phi }_{12}}={{N}_{1}}{{B}_{2}}.\,{{A}_{1}}$ ${{N}_{1}}{{\phi }_{12}}={{N}_{1}}{{\mu }_{0}}\frac{{{N}_{2}}}{l}{{I}_{2}}.\,\pi r_{1}^{2}$ … (11)
Also, ${{N}_{1}}{{\phi }_{1}}={{M}_{12}}{{I}_{2}}$ … (12)
On comparing (11) and (12),
${{M}_{12}}=\frac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi \,\,r_{1}^{2}}{l}$
Similarly when a current I1 is set up through S1, then the magnetic flux linked in S2 is given by,
${{N}_{2}}{{\phi }_{21}}={{M}_{21}}{{I}_{1}}$ … (13)
Note that the magnetic flux due to current I1 in S1 is assumed to be confined only inside the solenoid S1. Also the solenoids are very long compared to their radii, the flux linkage in S2 is
${{N}_{2}}{{\phi }_{21}}={{N}_{2}}\left( {{B}_{1}}\,.\,\,{{A}_{1}} \right)$ → (Area confined within S1.)
${{N}_{2}}{{\phi }_{21}}={{N}_{2}}{{\mu }_{0}}\frac{{{N}_{1}}}{l}.\,{{I}_{1}}\,.\,\,\pi \,r_{1}^{2}$ … (14)
From (13) and (14),
${{M}_{21}}={{\mu }_{0}}\frac{{{N}_{1}}{{N}_{2}}}{l}\,\pi \,\,r_{1}^{2}={{M}_{12}}=M$
Example Problem: Calculate the mutual inductance between a solenoid of length l and cross-sectional area A with N1 turns and a circular coil of N2 turns that is wound at the centre of the solenoid. Solution:
(i) Assume current I1 in the solenoid. The magnetic field B at its centre is given by
$B={{\mu }_{0}}n\,i={{\mu }_{0}}\frac{{{N}_{1}}}{l}{{I}_{1}}$
(ii) Magnetic flux linked with the small coil of area A is due to magnetic field B, then,
${{N}_{2}}{{\phi }_{21}}={{N}_{2}}\left( B.\,A \right)$ ${{N}_{2}}{{\phi }_{21}}={{N}_{2}}.\,{{\mu}_{0}}\frac{{{N}_{1}}}{l}{{I}_{1}}\,\,.\,\,A$ … (a)
(iii) ${{N}_{2}}{{\phi }_{21}}=M\,\,{{I}_{1}}$ … (b)
Equating (a) and (b),
$M={{\mu }_{0}}\frac{{{N}_{1}}{{N}_{2}}}{l}A$
## Applications of Mutual Inductance
The principle of mutual inductance is followed in various electronic devices. Some of them are as follows;
### Motors Note the inductors (Lf and La) in the dc motor circuit that are mutually inducted.
### Transformer ### Generators
The induced EMG in a generator by electromagnetic induction is shown below. The direction of induced emf is given by Lenz law. When an electrical component (coil) is interacting or being influenced by the magnetic field in the neighbouring component, mutual inductance arises. The current flowing in one coil induces an emf in the neighbouring coil.
Solved Problems:
1. Two circular coils can be arranged in any of the three situations in the figure. The mutual inductance will be
(a) maximum in the situation (C)
(b) maximum in the situation (B)
(c) maximum in the situation (A)
(d) Same in all situations
Solution:
The mutual inductance will be maximum when maximum field lines due to one coil pass through the other. In such a situation, magnetic flux linked will be maximum.
Therefore, in situation A both coils are parallel to each other. Hence more flux will be linked. Hence mutual inductance will be larger.
2. Two coaxial coils are very closer to each other and their mutual inductance is 5mH. If a current (50 A)sin 500t is passed in one of the coils, then find the peak value of induced emf in the secondary coil.
(a) 50V (b) 500V (c) 125V (d) 250V
Solution:
Given:
$i=50\,\sin 500t$
emf of mutual induction is given by
$\varepsilon =-M\frac{di}{dt}$ $=-\left( 5\times {{10}^{-3}} \right)\frac{d}{dt}\left( 50\sin 500t \right)$ $=-5\times {{10}^{-3}}\times 50\cos 500t\times 500$ $\varepsilon =-125\cos 500t$
The peak value is 125V. Option c is correct.
## Frequently Asked Questions On Mutual Inductance
### What is the induced emf in a stationary circular coil kept in a uniform magnetic field?
As the coil is stationary in a uniform magnetic field, there is no change in magnetic flux generated. Hence there is no induced emf in the coil.
### Explain how mutual inductance between a pair of circular coil changes when a sheet of iron is placed in between the coils. Assume all the other factors remain constant.
We know that the mutual inductance depends (directly proportional) on the permeability of the medium surrounding the coils. When the permeability of the medium is increased by inserting a sheet of iron, then the mutual inductance between the coils also increased.
### When two spherical balls, one made up of copper and the other made up of glass is allowed to fall freely under gravity from the same height above the ground. Which one of the balls will reach the ground fast? Why?
Having known that acceleration due to gravity does not depend on the mass the falling objects, the glass ball will reach the ground first. Because being an insulator, glass is not affected by the earth’s magnetic field. Hence there is no induced emf in the glass ball. Whereas in a copper ball the induced emf by earth’s magnetic field will oppose the change that produced it. Thus falls slower.
### What are the factors that affect mutual inductance?
Mutual inductance between two coils is affected by;
-Area of cross-section
-Number of turns in each coil
-Space between the two coils
-Permeability of medium between the two coils
-Length (in case of the solenoid) | HuggingFaceTB/finemath | |
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## Choosing between simple and systematic sampling - Quiz
(go to Outline)
1
Could you do systematic random sampling in the camp pictured above?
a) Yes b) No Correct. The shelters are in neat rows. Incorrect. You certainly could do systematic random sampling. Just calculate the sampling interval, choose a starting random number. Then start counting and sampling at one corner of the camp and continue until you reach the other end of the camp.Your answer has been saved.
You can do systematic random sampling even if households are not numbered. As a result, if you are faced with a large population of unnumbered but systematically organized households, systematic random sampling may be much easier than trying to list and number each household before doing simple random sampling.
2
##### True or false?
In the camp pictured above, there is no list of all the households. Nonetheless, you could easily do systematic random sampling.
a) True b) False Correct. It would be very difficult to do systematic random sampling by counting households. Shelters are not arranged in any order, so it would be very difficult to count households while being sure all households were counted once and only once.Incorrect. It would be very difficult to do systematic random sampling. The shelters are not arranged in any order, so it would be very difficult to count down the households while being sure all households were counted once and only once.Your answer has been saved.
So what do you do in the situation pictured above? You have no list of households and the shelters or households are not arranged in an orderly way. This is the real situation in many populations of displaced people.
One thing you could do is pick any eligible child or adult who happens to walk by.
3
##### True or false?
This is a valid method to sample the population to estimate some health outcome.
a) True b) False Correct. This is convenience sampling and should never be used in an assessment survey.Incorrect. This is convenience sampling and should never be used in an assessment survey. It is probably not at all representative of the larger population. If you use convenience sampling, you risk making seriously wrong conclusions and misdirecting your program resources based on these invalid conclusions. Your answer has been saved.
4
Which of the following is convenience sampling? (Tick all correct answers.)
a) Choosing all the households in which the family names begin with A through E b) Numbering a list of households and selecting them with a random number table c) Selecting the first house at random and then counting down the sampling interval d) Asking anyone who wishes to participate in the survey to come to the health centre on a certain date e) Randomly selecting children who come to the clinic for illness a) Correct. This is convenience sampling. This is not random sampling because a large part of the population, those with names beginning with F-Z, are not eligible to be selected for the sample.a) Incorrect. This is convenience sampling. This is not random sampling because a large part of the population, those with names beginning with F-Z, are not eligible to be selected for the sample.b) Correct. This is random sampling.b) Incorrect. This is random sampling in which each household selected is selected randomly.c) Correct. This is random sampling.c) Incorrect. This is random sampling in which each household selected is selected randomly.d) Correct. This is convenience sampling. Persons in such a sample are "self-selected" meaning that they choose to be selected; they are not randomly selected.d) Incorrect. This is convenience sampling. Persons in the sample are "self-selected" meaning that they choose to be selected; they are not randomly selected.e) Correct. This is convenience sampling. Children coming to the clinic for illness are very different from all children in the population. They are ill, and by definition, have poorer health status than average at the time they come to the clinic. Moreover, their mothers are motivated to seek health care and are therefore may be different from the average mother.e) Incorrect. This is convenience sampling. Children coming to the clinic for illness are very different from all children in the population. They are ill, and by definition, have poorer health status than average at the time they come to the clinic. Moreover, their mothers are motivated to seek health care and are therefore may be different from the average mother. | HuggingFaceTB/finemath | |
Recent Submissions
• Symmetric Ethers as Bioderived Fuels: Reactivity with OH Radicals
(Energy & Fuels, American Chemical Society (ACS), 2021-09-27) [Article]
Environmental pollution and greenhouse gas emissions are major challenges faced by our society. One possible way to mitigate global warming is to cut CO2 emissions by taking a shift from conventional fuels to renewable fuels for future sustainability. Carbon neutral fuels produced in a sustainable carbon cycle can close the carbon cycle and reach net zero-carbon emission. To this end, ethers are promising renewable fuels and/or additives for future advanced combustion engines. Therefore, understanding the oxidation behavior of ethers under engine-relevant conditions is of utmost importance. In this work, the reaction kinetics of hydroxyl radicals with dimethyl ether (DME), diethyl ether (DEE), di-n-propyl ether (DPE), and di-n-butyl ether (DBE) were investigated behind reflected shock waves over the temperature range of 865–1381 K and the pressure range of 0.96–5.56 bar using a shock tube and a UV laser diagnostic technique. Hydroxyl radicals were monitored near 306.7 nm to follow the reaction kinetics. These reactions did not exhibit discernible pressure effects. The temperature dependence of the measured rate coefficients can be expressed by the following modified Arrhenius equations in units of cm3 mol–1 s–1: k1(DME+OH) = 1.19 × 1014 exp(−2469.8/T), k2(DEE+OH) = 1.27 × 107T2 exp(327.8/T), k3(DPE+OH) = 1.64 × 107T2 exp(368.4/T), k4(DBE+OH) = 9.12 × 1011T0.65 exp(−843.5/T). Our measured rate data were analyzed to obtain site-specific rates and branching ratios. Our results are compared with the available literature data wherever applicable. Furthermore, the ability of Atkinson’s structure–activity relationship (SAR) to predict the kinetic behavior of the reactions of dialkyl ethers with OH radicals was examined.
• Single-Particle Spectroscopy as a Versatile Tool to Explore Lower-Dimensional Structures of Inorganic Perovskites
(ACS Energy Letters, American Chemical Society (ACS), 2021-09-27) [Article]
The remarkable defect-tolerant nature of inorganic cesium halide perovskites, leading to near unity photoluminescence (PL) quantum yield and narrow emission line width across the entire visible spectrum, has provided a tantalizing platform for the development of a plethora of light-emitting applications. Recently, lower-dimensional (2D, 1D, and 0D) perovskites have attracted further attention due to their enhanced thermal, photo, and chemical stability as compared to their three-dimensional (3D) analogues. The combination of external size quantization and internal octahedral organization provides a unique opportunity to study and harness “multi-dimensional” electronic properties engineered on both atomic scale and nanoscale. However, crucial research to understand the elementary charge carrier dynamics in lower-dimensional perovskites lags far behind the enormous effort to incorporate them into optoelectronic devices. In this Perspective, we provide a review of recent developments that focus on studies of the dynamics of excitonic complexes in Cs-based perovskite nanocrystals using single-particle time-resolved PL spectroscopy and photon correlation measurements. Single-photon statistical studies not only offer an unprecedented level of detail to directly assess various recombination pathways, but also provide insights into specifics of the charge carriers' localization. We discuss the underlying physicochemical processes that govern PL emission and draw attention to a number of attributes within this class of the materials, especially lower-dimensional perovskites, that may indicate the common origin of the PL emission as well as provide a route map for the vast unexplored territories where single-particle spectroscopy can be a powerful tool to unravel crucial information.
• On the lubricity mechanism of carbon-based nanofluid fuels
(Fuel, Elsevier BV, 2021-09-27) [Article]
Utilizing fuels blended with nanofluid particles may enhance fuel delivery and combustion in engines. However, the underlying tribochemistry related to fuel delivery when using nanofluids remains unclear. In this study, we investigate fuel lubricity over low-sulfur diesel (D100), diesel fuel containing 10 wt% ethanol (DE10), and DE10 blended with 50 to 200 ppm surface modified graphene oxide (mGO), i.e., G50, G100, and G200. The fuel lubricity experiment shows that as compared to D100, the DE10 fuel produced 50% larger wear volumes on rubbed balls, while lubrication with the G200 fuel reduced wear by 6%. The tribochemical reaction kinetic model developed in this work unravels the lubrication mechanism. The blended mGO reduces direct metal-to-metal contacts, produces graphitic tribofilms on wear tracks, and serves as tribo-active sources to grow frictional products.
• Seasonal Simulations of Summer Aerosol Optical Depth over the Arabian Peninsula using WRF-Chem : Validation, Climatology, and Variability
(International Journal of Climatology, Wiley, 2021-09-27) [Article]
This study investigates the climatology and variability of summer Aerosol Optical Depth (AOD) over the Arabian Peninsula (AP) using a long-term high-resolution Weather Research and Forecasting model coupled with the chemistry module (WRF-Chem) simulation, available ground-based and satellite observations, and reanalysis products from 2008 to 2018. The simulated spatial distribution of the summer AOD agrees well with the satellite observations and reanalysis over the AP, with spatial correlation coefficients of 0.81/0.83/0.89 with MODIS-A/MODIS-T/MERRA-2, respectively. Higher values of summertime AOD are broadly found over the eastern AP regions and the southern Red Sea and minima over the northern Red Sea and northwest AP, consistent with observational datasets. The WRF-Chem simulation suggests that the two regions of high AOD are associated with dust advected from the Tigris–Euphrates by the northwesterly summer Shamal wind in the eastern AP and from the African Sahara via Sudan by westerly winds through the Tokar Gap for the southern AP. The high AOD over the south-central east AP is due to locally generated dust by the action of northerly winds, modulated by variations in relative humidity, vertical motion, soil moisture, and soil temperature over the desert regions. The vertical extent of this dust is primarily driven by upward motion triggered by thermal convection over the local source region. In terms of interannual variability, summer AOD exhibits significant year-to-year variations over the AP region. In particular, enhanced (reduced) AOD over the southern AP (Persian Gulf) is observed during La Niña conditions, favored by stronger (weaker) Tokar westerly (northwesterly summer Shamal) winds.
• Nanofibrous membranes comprising intrinsically microporous polyimides with embedded metal–organic frameworks for capturing volatile organic compounds
(Journal of Hazardous Materials, Elsevier, 2021-09-25) [Article]
Here, we report the fabrication of nanofibrous air-filtration membranes of intrinsically microporous polyimide with metal–organic frameworks (MOFs). The membranes successfully captured VOCs from air. Two polyimides with surface areas up to 500 m2 g −1 were synthesized, and the impact of the porosity on the sorption kinetics and capacity of the nanofibers was investigated. Two Zr-based MOFs, namely pristine UiO-66 (1071 m2 g −1 ) and defective UiO-66 (1582 m2 g −1 ), were embedded into the nanofibers to produce nanocomposite materials. The nanofibers could remove polar formaldehyde and non-polar toluene, xylene, and mesitylene from air. The highest sorption capacity with 214 mg g−1 was observed for xylene, followed by mesitylene (201 mg g−1 ), toluene (142 mg g−1 ), and formaldehyde (124 mg g−1 ). The incorporation of MOFs drastically improved the sorption performance of the fibers produced from low-surface-area polyimide. Time-dependent sorption tests revealed the rapid sequestration of air pollutants owing to the intrinsic porosity of the polyimides and the MOF fillers. The porosity allowed the rapid diffusion of pollutants into the inner fiber matrix. The molecular level interactions between VOCs and polymer/MOFs were clarified by molecular modeling studies. The practicality of material fabrication and the applicability of the material were assessed through the modification of industrial N95 dust masks. To the best of our knowledge, this is the first successful demonstration of the synergistic combination of intrinsically microporous polyimides and MOFs in the form of electrospun nanofibrous membranes and their application for VOC removal.
• Accordion-Like Carbon with High Nitrogen Doping for Fast and Stable K Ion Storage
(Advanced Energy Materials, Wiley, 2021-09-24) [Article]
Potassium ion battery (PIB) is a potential candidate for future large-scale energy storage. A key challenge is that the (de)potassiation stability of graphitic carbon anodes is hampered by the limited (002) interlayer spacing. Amorphous carbon with a hierarchical structure can buffer the volume change during repeated (de)potassiation and enable stable cycling. Herein, a direct pyrolysis approach is demonstrated to synthesize a highly nitrogen-doped (26.7 at.%) accordion-like carbon anode composed of thin carbon nanosheets and a turbostratic crystalline structure. The hierarchical structure of accordion-like carbon is endowed by a self-assembly process during pyrolysis carbonization. The hierarchical nitrogen-doped accordion structure enables a high reversible capacity of 346 mAh g−1 and superior cycling stability. This work constitutes a general synthesis methodology that can be used to prepare hierarchical carbon anodes for advanced PIBs.
• Efficient and Spectrally Stable Blue Perovskite Light-Emitting Diodes Employing a Cationic π-Conjugated Polymer
Metal halide perovskite semiconductors have demonstrated remarkable potentials in solution-processed blue light-emitting diodes (LEDs). However, the unsatisfied efficiency and spectral stability responsible for trap-mediated non-radiative losses and halide phase segregation remain the primary unsolved challenges for blue perovskite LEDs. In this study, it is reported that a fluorene-based π-conjugated cationic polymer can be blended with the perovskite semiconductor to control film formation and optoelectronic properties. As a result, sky-blue and true-blue perovskite LEDs with Commission Internationale de l'Eclairage coordinates of (0.08, 0.22) and (0.12, 0.13) at the record external quantum efficiencies of 11.2% and 8.0% were achieved. In addition, the mixed halide perovskites with the conjugated cationic polymer exhibit excellent spectral stability under external bias. This result illustrates that π-conjugated cationic polymers have a great potential to realize efficient blue mixed-halide perovskite LEDs with stable electroluminescence.
• Topological Aspects of Antiferromagnets
(Journal of Physics D: Applied Physics, IOP Publishing, 2021-09-22) [Article]
The long fascination antiferromagnetic materials have exerted on the scientific community over about a century has been entirely renewed recently with the discovery of several unexpected phenomena including various classes of anomalous spin and charge Hall effects and unconventional magnonic transport, but also homochiral magnetic entities such as skyrmions. With these breakthroughs, antiferromagnets standout as a rich playground for the investigation of novel topological behaviors, and as promising candidate materials for disruptive low-power microelectronic applications. Remarkably, the newly discovered phenomena are all related to the topology of the magnetic, electronic or magnonic ground state of the antiferromagnets. This review exposes how non-trivial topology emerges at different levels in antiferromagnets and explores the novel mechanisms that have been discovered recently. We also discuss how novel classes of quantum magnets could enrich the currently expanding field of antiferromagnetic spintronics and how spin transport can in turn favor a better understanding of exotic quantum excitations.
• 3D Printing of Hydrogels for Stretchable Ionotronic Devices
(Advanced Functional Materials, Wiley, 2021-09-21) [Article]
In the booming development of flexible electronics represented by electronic skins, soft robots, and human–machine interfaces, 3D printing of hydrogels, an approach used by the biofabrication community, is drawing attention from researchers working on hydrogel-based stretchable ionotronic devices. Such devices can greatly benefit from the excellent patterning capability of 3D printing in three dimensions, as well as the free design complexity and easy upscale potential. Compared to the advanced stage of 3D bioprinting, 3D printing of hydrogel ionotronic devices is in its infancy due to the difficulty in balancing printability, ionic conductivity, shape fidelity, stretchability, and other functionalities. In this review, a guideline is provided on how to utilize the power of 3D printing in building high-performance hydrogel-based stretchable ionotronic devices mainly from a materials’ point of view, highlighting the systematic approach to balancing the printability, printing quality, and performance of printed devices. Various 3D printing methods for hydrogels are introduced, and then the ink design principles, balancing printing quality, printed functions, such as elastic conductivity, self-healing ability, and device (e.g., flexible sensors, shape-morphing actuators, soft robots, electroluminescent devices, and electrochemical biosensors) performances are discussed. In conclusion, perspectives on the future directions of this exciting field are presented.
• Revealing the Side-Chain Dependent Ordering Transition of Highly-Crystalline Double-Cable Conjugated Polymers
(Angewandte Chemie International Edition, Wiley, 2021-09-21) [Article]
• Influence of pressure, temperature and organic surface concentration on hydrogen wettability of caprock; implications for hydrogen geo-storage
(Energy Reports, Elsevier BV, 2021-09-21) [Article]
Hydrogen (H2) as a cleaner fuel has been suggested as a viable method of achieving the decarbonization objectives and meeting increasing global energy demand. However, successful implementation of a full-scale hydrogen economy requires large-scale hydrogen storage (as hydrogen is highly compressible). A potential solution to this challenge is injecting hydrogen into geologic formations from where it can be withdrawn again at later stages for utilization purposes. The geostorage capacity of a porous formation is a function of its wetting characteristics, which strongly influence residual saturations, fluid flow, rate of injection, rate of withdrawal, and containment security. However, literature severely lacks information on hydrogen wettability in realistic geological and caprock formations, which contain organic matter (due to the prevailing reducing atmosphere). We, therefore, measured advancing (θa) and receding (θr) contact angles of mica substrates at various representative thermo-physical conditions (pressures 0.1-25 MPa, temperatures 308–343 K, and stearic acid concentrations of 10−9- 10−2 mol/L). The mica exhibited an increasing tendency to become weakly water-wet at higher temperatures, lower pressures, and very low stearic acid concentration. However, it turned intermediate-wet at higher pressures, lower temperatures, and increasing stearic acid concentrations. The study suggests that the structural H2 trapping capacities in geological formations and sealing potentials of caprock highly depend on the specific thermo-physical condition. Thus, this novel data provides a significant advancement in literature and will aid in the implementation of hydrogen geo-storage at an industrial scale
• Preconditioned BFGS-based Uncertainty Quantification in Elastic Full Waveform Inversion
(Geophysical Journal International, Oxford University Press (OUP), 2021-09-21) [Article]
Full Waveform Inversion (FWI) has become an essential technique for mapping geophysical subsurface structures. However, proper uncertainty quantification is often lacking in current applications. In theory, uncertainty quantification is related to the inverse Hessian (or the posterior covariance matrix). Even for common geophysical inverse problems its calculation is beyond the computational and storage capacities of the largest high-performance computing systems. In this study, we amend the Broyden-Fletcher-Goldfarb-Shanno (BFGS) algorithm to perform uncertainty quantification for large-scale applications. For seismic inverse problems, the limited-memory BFGS (L-BFGS) method prevails as the most efficient quasi-Newton method. We aim to augment it further to obtain an approximate inverse Hessian for uncertainty quantification in FWI. To facilitate retrieval of the inverse Hessian, we combine BFGS (essentially a full-history L-BFGS) with randomized singular value decomposition to determine a low-rank approximation of the inverse Hessian. Setting the rank number equal to the number of iterations makes this solution efficient and memory-affordable even for large-scale problems. Furthermore, based on the Gauss-Newton method, we formulate different initial, diagonal Hessian matrices as preconditioners for the inverse scheme and compare their performances in elastic FWI applications. We highlight our approach with the elastic Marmousi benchmark model, demonstrating the applicability of preconditioned BFGS for large-scale FWI and uncertainty quantification.
• On Linear Time-Invariant Systems Analysis via A Single Trajectory: A Linear Programming Approach
(arXiv, 2021-09-21) [Preprint]
In this note, a novel methodology that can extract a number of analysis results for linear time-invariant systems (LTI) given only a single trajectory of the considered system is proposed. The superiority of the proposed technique relies on the fact that it provides an automatic and formal way to obtain valuable information about the controlled system by only having access to a single trajectory over a finite period of time (i.e., the system dynamics is assumed to be unknown). At first, we characterize the stability region of LTI systems given only a single trajectory dataset by constructing the associated Lyapunov function of the system. The Lyapunov function is found by formulating and solving a linear programming (LP) problem. Then, we extend the same methodology to a variety of essential analysis results for LTI systems such as deriving bounds on the output energy, deriving bounds on output peak, deriving $\mathbf{L}_2$ and RMS gains. To illustrate the efficacy of the proposed data-driven paradigm, a comparison analysis between the learned LTI system metrics and the true ones is provided.
• Shear wave velocity structure beneath Northeast China from joint inversion of receiver functions and Rayleigh wave group velocities: Implications for intraplate volcanism
(Wiley, 2021-09-17) [Preprint]
A high-resolution 3-D crustal and upper-mantle shear-wave velocity model of Northeast China is established by joint inversion of receiver functions and Rayleigh wave group velocities. The teleseismic data for obtaining receiver functions are collected from 107 CEA permanent sites and 118 NECESSArray portable stations. Rayleigh wave dispersion measurements are extracted from an independent tomographic study. Our model exhibits unprecedented detail in S-velocity structure. Particularly, we discover a low S-velocity belt at 7.5-12.5 km depth covering entire Northeast China (except the Songliao basin), which is attributed to a combination of anomalous temperature, partial melts and fluid-filled faults related to Cenozoic volcanism. Localized crustal fast S-velocity anomaly under the Songliao basin is imaged and interpreted as late-Mesozoic mafic intrusions. In the upper mantle, our model confirms the presence of low velocity zones below the Changbai mountains and Lesser Xing’an mountain range, which agree with models invoking sub-lithospheric mantle upwellings. We observe a positive S-velocity anomaly at 50-90 km depth under the Songliao basin, which may represent a depleted and more refractory lithosphere inducing the absence of Cenozoic volcanism. Additionally, the average lithosphere-asthenosphere boundary depth increases from 50-70 km under the Changbai mountains to 100 km below the Songliao basin, and exceeds 125 km beneath the Greater Xing’an mountain range in the west. Furthermore, compared with other Precambrian lithospheres, Northeast China likely has a rather warm crust (~480-970 °C) and a slightly warm uppermost mantle (~1200 °C), probably associated with active volcanism. The Songliao basin possesses a moderately warm uppermost mantle (~1080 °C).
• Chemical speciation and soot measurements in laminar counterflow diffusion flames of ethylene and ammonia mixtures
(Fuel, Elsevier BV, 2021-09-17) [Article]
Ammonia is considered as one of the most promising alternative fuels due to its carbon neutrality and the existing infrastructure for its mass production and delivery. However, burning neat ammonia has the issue of poor flame stability and high NOx emissions, making co-firing ammonia with conventional fuel a more feasible approach. The present work investigated the sooting characteristics of counterflow diffusion flames of ethylene/ammonia mixtures. Experimentally, soot volume fraction (SVF) and average soot particle diameter in the neat ethylene, ammonia- and nitrogen- doped flames were non-intrusively measured. Both SVF and average soot particles diameter were found to decrease with the addition of ammonia. Flame temperature were measured with tunable diode laser absorption spectroscopy and the results suggested that the inhibiting effect of ammonia on soot formation was chemical instead of thermal. For further kinetic insights, numerical simulation with newly-constructed reaction mechanisms were performed and the results were compared against chemical speciation data from gas chromatography (GC) measurements; the results showed that ammonia doping would lead to more significant reduction of benzene concentration than nitrogen doping. Kinetic pathways of the chemical suppressing effect of ammonia addition on soot and its precursor formation were then explained based on numerical results. The major contribution of the present work can be summarized in the following aspects: 1) New comprehensive experimental data on sooting characteristics, important intermediate species concentrations of diffusion counterflow flames of ethylene/ammonia mixtures were provided for model validation; 2) One coupled mechanism with detailed hydrocarbon-nitrogen interactions was established to predict PAHs formation and soot formation; 3) Detailed chemical kinetics insight of ammonia effect on soot formation was presented in the counterflow flame.
• Low-Defect, High Molecular Weight Indacenodithiophene (IDT) Polymers Via a C–H Activation: Evaluation of a Simpler and Greener Approach to Organic Electronic Materials
(ACS Materials Letters, American Chemical Society (ACS), 2021-09-16) [Article]
The development, optimization, and assessment of new methods for the preparation of conjugated materials is key to the continued progress of organic electronics. Direct C–H activation methods have emerged and developed over the last 10 years to become an invaluable synthetic tool for the preparation of conjugated polymers for both redox-active and solid-state applications. Here, we evaluate direct (hetero)arylation polymerization (DHAP) methods for the synthesis of indaceno[1,2-b:5,6-b′]dithiophene-based polymers. We demonstrate, using a range of techniques, including direct visualization of individual polymer chains via high-resolution scanning tunneling microscopy, that DHAP can produce polymers with a high degree of regularity and purity that subsequently perform in organic thin-film transistors comparably to those made by other cross-coupling polymerizations that require increased synthetic complexity. Ultimately, this work results in an improved atom economy by reducing the number of synthetic steps to access high-performance molecular and polymeric materials.
• Crystallization and Morphology of Triple Crystalline Polyethylene-b-poly(ethylene oxide)-b-poly(ε-caprolactone) PE-b-PEO-b-PCL Triblock Terpolymers
(Polymers, MDPI AG, 2021-09-16) [Article]
The morphology and crystallization behavior of two triblock terpolymers of polymethylene, equivalent to polyethylene (PE), poly (ethylene oxide) (PEO), and poly (ε-caprolactone) (PCL) are studied: PE227.1-b-PEO4615.1-b-PCL3210.4 (T1) and PE379.5-b-PEO348.8-b-PCL297.6 (T2) (superscripts give number average molecular weights in kg/mol and subscripts composition in wt %). The three blocks are potentially crystallizable, and the triple crystalline nature of the samples is investigated. Polyhomologation (C1 polymerization), ring-opening polymerization, and catalyst-switch strategies were combined to synthesize the triblock terpolymers. In addition, the corresponding PE-b-PEO diblock copolymers and PE homopolymers were also analyzed. The crystallization sequence of the blocks was determined via three independent but complementary techniques: differential scanning calorimetry (DSC), in situ SAXS/WAXS (small angle X-ray scattering/wide angle X-ray scattering), and polarized light optical microscopy (PLOM). The two terpolymers (T1 and T2) are weakly phase segregated in the melt according to SAXS. DSC and WAXS results demonstrate that in both triblock terpolymers the crystallization process starts with the PE block, continues with the PCL block, and ends with the PEO block. Hence triple crystalline materials are obtained. The crystallization of the PCL and the PEO block is coincident (i.e., it overlaps); however, WAXS and PLOM experiments can identify both transitions. In addition, PLOM shows a spherulitic morphology for the PE homopolymer and the T1 precursor diblock copolymer, while the other systems appear as non-spherulitic or microspherulitic at the last stage of the crystallization process. The complicated crystallization of tricrystalline triblock terpolymers can only be fully grasped when DSC, WAXS, and PLOM experiments are combined. This knowledge is fundamental to tailor the properties of these complex but fascinating materials.
• Evaluation of Thermoacoustic Applications Using Waste Heat to Reduce Carbon Footprint
(American Society of Mechanical Engineers, 2021-09-16) [Conference Paper]
Abstract Thermoacoustics (TA) engines and refrigerators typically run on the Stirling cycle with acoustic networks and resonators replacing the physical pistons. Without moving parts, these TA machines achieve a reasonable fraction of Carnot’s efficiency. They are also scalable, from fractions of a Watt up to kW of cooling. Despite their apparent promise, TA devices are not in widespread use, because outside of a few niche applications, their advantages are not quite compelling enough to dislodge established technology. In the present study, the authors have evaluated a selected group of applications that appear suitable for utilization of industrial waste heat using TA devices and have arrived at a ranked order. The principal thought is to appraise whether thermoacoustics can be a viable path, from both an economic and energy standpoint, for carbon mitigation in those applications. The applications considered include cryogenic carbon capture for power plant exhaust gases, waste-heat powered air conditioning/water chilling for factories and office buildings, hydrogen liquefaction, and zero-boiloff liquid hydrogen (LH2) storage. Although the criteria used for evaluating the applications are somewhat subjective, the overall approach has been consistent, with the same set of criteria applied to each of them. Thermoeconomic analysis is performed to evaluate the system viability, together with overall consideration of a thermoacoustic device’s general nature, advantages, and limitations. Our study convincingly demonstrates that the most promising application is zero-boiloff liquid hydrogen storage, which is physically well-suited to thermoacoustic refrigeration and requires cooling at a temperature and magnitude not ideal for standard refrigeration methods. Waste-heat powered air conditioning ranks next in its potential to be a viable commercial application. The rest of the applications have been found to have relatively lower potentials to enter the existing commercial space.
• Capturing 3D atomic defects and phonon localization at the 2D heterostructure interface
(Science Advances, American Association for the Advancement of Science (AAAS), 2021-09-15) [Article]
The three-dimensional (3D) local atomic structures and crystal defects at the interfaces of heterostructures control their electronic, magnetic, optical, catalytic, and topological quantum properties but have thus far eluded any direct experimental determination. Here, we use atomic electron tomography to determine the 3D local atomic positions at the interface of a MoS2-WSe2 heterojunction with picometer precision and correlate 3D atomic defects with localized vibrational properties at the epitaxial interface. We observe point defects, bond distortion, and atomic-scale ripples and measure the full 3D strain tensor at the heterointerface. By using the experimental 3D atomic coordinates as direct input to first-principles calculations, we reveal new phonon modes localized at the interface, which are corroborated by spatially resolved electron energy-loss spectroscopy. We expect that this work will pave the way for correlating structure-property relationships of a wide range of heterostructure interfaces at the single-atom level.
• Two-Dimensional TiO2/TiS2 Hybrid Nanosheet Anodes for High-Rate Sodium-Ion Batteries
(ACS Applied Energy Materials, American Chemical Society (ACS), 2021-09-15) [Article]
The sodium-ion battery (NIB) is promising for next-generation energy storage systems. One promising anode material is titanium dioxide (TiO2). However, the sluggish sodiation/desodiation kinetics of TiO2 hinders its application in NIBs. Herein, we converted TiO2 into a two-dimensional (2D) TiO2/TiS2 hybrid to improve its sodium storage capability. The 2D TiO2/TiS2 hybrid nanosheet electrode provides high kinetics and excellent cycling performance for sodium-ion storage. This work provides a promising strategy to develop 2D hybrid nanomaterials for high-performance sodium storage devices. | open-web-math/open-web-math | |
# Investigating sampling Part 2: Confidence intervals Teacher’s notes
A downloadable version of this activity is available in the following format:
### Purpose
The purpose of this activity is to investigate the relationship between the standard error and the population standard deviation.
### Outcomes
This activity will teach students how to
• apply characteristics of normal distributions
• demonstrate an understanding of the role of the central limit theorem in the development of confidence intervals
• demonstrate an understanding of how the size of a sample affects the variation in sample results
• graph and interpret sample distributions of the sample mean and sample distributions of the sample proportion
### Format
This activity will take the form of a teacher-led full-class discussion. Students are encouraged to share information. To facilitate discussion, divide students into small groups.
### Introduction
It is important that students realize that we are usually unable to collect information about a total population. The goal of sampling is to draw reasonable conclusions about a population by obtaining information from a relatively small part (a sample) of that population. In order to do that, we need to know what formulas to use and what degree of confidence we can claim in the resulting information.
The previous lesson on investigating sampling was a first step toward developing the concept of confidence intervals.
### Classroom instruction
1. Normal distribution
Students should recall the percentages associated with normal distributions:
Roughly: 68% of the data fall within one standard deviation of the mean
95% of the data fall within two standard deviations of the mean
99% of the data fall within three standard deviations of the mean
The concept of ‘within’ may require some discussion (between one standard deviation above the mean and one standard deviation below the mean). A diagram can help in visualizing this.
Using 164.5 as the mean and 2.28 as the standard deviation, students can examine the sample means in Appendix A. ‘Within one standard deviation of the mean’ gives the following calculation:
Students can examine the sample means given in Appendix A to determine how many of them fall within the interval from 162.2 to 166.8. (Answer: 67 sample means fall in that interval, with 97 sample means falling within the interval 159.9 to 169.1.) The results are close to what one would expect from a normal distribution. From the histogram, we recognized that the distribution was not perfectly normal.
2. The central limit theorem
Have the students discuss their answers in small groups or as a class.
Sample 2: The interval 157.6 to 165.0 contains the population mean.
Sample 3: The interval 164.6 to 172.0 contains the population mean.
Sample 4: The interval 165.5 to 172.9 does not contain the population mean.
In working through this exercise, students recognize that when an interval is calculated as described, it will ‘very likely’ contain the population mean.
3. Project time
Ensure that students recognize the characteristics of the population from which they have chosen a sample. For example, is the population all Grade 11 students in the school, or is it all students taking Grade 11 Mathematics in the school? Is it all students in the school district, in the province, or in the whole of Canada? It may also be useful to discuss alternative ways of describing the confidence level. Here are some examples:
. . .with 95% confidence.”
or
” . . . 19 times out of 20.”
Contributed by Anna Spanik, Math teacher, Halifax West High School, Nova Scotia.
This entry was posted in Teacher Resources. Bookmark the permalink. | HuggingFaceTB/finemath | |
# proof of property
#### mathodman25
Let a 3 × 3 matrix A be such that for any vector of a column v ∈ R3 the vectors Av and v are orthogonal. Prove that At + A = 0, where At is the transposed matrix.
#### romsek
Math Team
do you mean $A^TA=0$
Consider the 3x3 identity matrix.
Clearly all the column vectors are orthogonal, and $Iv = v$
$I^T=I$
$I+I= 2I \neq 0$
#### Greens
I may have read this wrong, but I don't believe $I_{3 \times 3}$ can be $A$ since it would mean $Iv$ is orthogonal to $v$.
#### mathodman25
I may have read this wrong, but I don't believe $I_{3 \times 3}$ can be $A$ since it would mean $Iv$ is orthogonal to $v$.
Yeah, A is definitely not an identity matrix. I think that maybe A is zero matrix. Or maybe I have forgotten some theorem or criteria.
Last edited by a moderator:
#### Greens
This was somewhat long so I wouldn't be surprised if there's an error in here somewhere so read critically.
Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$
and $v= \begin{bmatrix} v_1 \\ v_2\\ v_3 \end{bmatrix}$
Two vectors are othogonal if their dot product is $0$, so we know $Av \cdot v = 0$
$Av = \begin{bmatrix} a_1 v_1 + a_2 v_2 + a_3 v_3 \\ b_1 v_1 + b_2 v_2 + b_3 v_3 \\ c_1 v_1 + c_2 v_2 + c_3 v_3 \end{bmatrix}$
$Av \cdot v = v_1 (a_1 v_1 + a_2 v_2 + a_3 v_3)\\ + v_2 (b_1 v_1 + b_2 v_2 + b_3 v_3)\\ + v_3 (c_1 v_1 + c_2 v_2 + c_3 v_3) = 0$
after a bit of simplifying:
$Av \cdot v = a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} + v_1 v_2 (a_2 + b_1) + v_1 v_3 (a_3 + c_1) + v_2 v_3 (b_3 + c_2) = 0$
So , if $a_1 , b_2 , c_3 = 0$ and $b_1 = -a_2$ , $c_1 = -a_3$ , and $c_2 = -b_3$ then $Av \cdot v = 0$
So $A=\begin{bmatrix} 0 & a_2 & a_3 \\ -a_2 & 0 & b_3 \\ -a_3 & -b_3 & 0 \end{bmatrix}$
And you then have $A^{t} + A = 0$
2 people
#### mathodman25
But how do you get that a1 , b2 , c2= 0?
#### Greens
$Av \cdot v = a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} + v_1 v_2 (a_2 + b_1) + v_1 v_3 (a_3 + c_1) + v_2 v_3 (b_3 + c_2) = 0$
If $a_1 , b_2 , c_3 = 0$ , then $a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} =0$ for any $v$ | HuggingFaceTB/finemath | |
# Electrodynamics/Electromagnetic Field Tensors
Before discussing Maxwell's Equations, it's useful to define two tensors for the Electromagnetic field. These two tensors will be used in the remainder of the book for several applications, including Maxwell's Equations.
## Field Tensors
We will define two field tensors, F and G, and we will define another tensor, σ, that we can use to generate one field tensor from the other.
$\mathbb {F} ={\begin{bmatrix}0&B_{z}&-B_{y}&-iE_{x}\\-B_{z}&0&B_{x}&-iE_{y}\\B_{y}&-B_{x}&0&-iE_{z}\\iE_{x}&iE_{y}&iE_{z}&0\end{bmatrix}}$
$\mathbb {G} ={\begin{bmatrix}0&E_{z}&-E_{y}&-iB_{x}\\-E_{z}&0&E_{x}&-iB_{y}\\E_{y}&-E_{x}&0&-iB_{z}\\iB_{x}&iB_{y}&iB_{z}&0\end{bmatrix}}$
## Completely Antisymmetrical Tensor
We can define a completely antisymmetrical tensor of the fourth rank (4 dimensional, so we cannot visualize it) as follows:
$\sigma _{\mu \nu \lambda \rho }=0$ if any two indices are the same.
$\sigma _{1234}=1$
$\sigma _{\mu \nu \lambda \rho }=1$ if the indices are all different, and if the total difference in the indices from 1234 is even.
$\sigma _{\mu \nu \lambda \rho }=-1$ if the indices are all different and if the total difference in the indices from 1234 is odd.
We can define the two field tensors, F and G in terms of this new tensor as:
$\mathbb {G} ={\frac {i}{2}}\sum \sigma _{\mu \nu \lambda \rho }\mathbb {F} _{\lambda \rho }$ | HuggingFaceTB/finemath | |
# Given a = log2(x-7) and b = log2(3x-5) solve a=6-b?
giorgiana1976 | Student
We'll replace a and b by the given logarithms:
log2(x-7) = 6 - log2(3x-5)
log2(x-7) + log2(3x-5) = 6
We notice that the bases of logarithms from the left side are equal, therefore, we'll transform the sum into a product:
log2 (x-7)(3x-5) = 6 <=> (x-7)(3x-5) = 2^6
We'll eliminate the brackets using FOIL method:
3x^2 - 5x - 21x + 35 = 64
We'll combine like terms:
3x^2 - 26x + 35 = 64
We'll subtract 64 both sides:
3x^2 - 26x + 35 - 64 = 0
3x^2 - 26x - 29 = 0
x1 = (26+32)/6
x1 = 58/6 => x1 = 9.(6)
x2 = -1
We'll impose constraints of existence of the logarithms:
x - 7 > 0
x > 7
3x - 5 > 0
3x > 5
x > 5/3
The common interval of admissible values for x is (7 ; +`oo` ).
We notice that only one value belongs to this interval, therfore the equation will have only one solution, namely x = 9.(6). | HuggingFaceTB/finemath | |
# 微分方程：线性代数和NxN微分方程组
## Differential Equations: Linear Algebra and NxN Systems of Differential Equations
Learn how to use linear algebra and MATLAB to solve large systems of differential equations.
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edX
• 完成时间大约为 8
• 中级
• 英语
### 你将学到什么
Model and solve different real world phenomena with systems of differential equations.
Find the dimension and a basis for a (finite dimensional) vector space.
Formulate and solve eigenvalue and eigenvector problems.
Use MATLAB to explore solutions to large systems of equations.
### 课程概况
Differential equations are the mathematical language we use to describe the world around us. Most phenomena can be modeled not by single differential equations, but by systems of interacting differential equations. These systems may consist of many equations. In this course, we will learn how to use linear algebra to solve systems of more than 2 differential equations. We will also learn to use MATLAB to assist us.
We will use systems of equations and matrices to explore:
The original page ranking systems used by Google,
Balancing chemical reaction equations,
Tuned mass dampers and other coupled oscillators,
Three or more species competing for resources in an ecosystem,
The trajectory of a rider on a zipline.
The five modules in this series are being offered as an XSeries on edX. Please visit the Differential Equations XSeries Program Page to learn more and to enroll in the modules.
*Zipline photo by teanitiki on Flickr (CC BY-SA 2.0)
### 课程大纲
Unit 1: Linear Algebra
Solving linear systems: elimination and RREF
Nullspace, vector space
Column space, determinants, and inverses
eigenvalues, eigenvectors, and diagonalization
Unit 2: Systems of Differential Equations
Solving homogeneous NxN systems of differential equations
Matrix exponential and diagonalization
Decoupling and solving inhomogeneous systems of equations
Solving nonlinear systems with MATLAB
### 预备知识
18.031x Introduction to Differential Equations (Scalar equations), 18.032x Differential Equations: 2x2 Systems (2x2 first order differential equations)
### 常见问题
What are the prerequisites for this course?
We assume familiarity with both 18.031x Introduction to Differential Equations and 18.032x Differential Equations: 2x2 systems. Knowledge from 18.03Lx is not required to succeed in this course.
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• gacco | HuggingFaceTB/finemath | |
# PI Tank Calculations
This document investigates the theory behind the PI-tank, which is a subject dating back more than a century. The challenge is to find a working design that is understandable. Most amateur solutions consist of tables or equations without explanation. Some on-line calculators refer to equations stated in historic documents. However, I have an aversion to using equations without understanding derivation and without knowing the source. The purpose of this document is to investigate the derivation of the equations and check them against other sources.
The Problem
The following diagram is a simple PI arrangement. The approach is to use complex analysis to find the impedance looking into the network Zin. The first stage is to combine C2 and R as a parallel circuit. The next stage is to add L in series and then convert L, C2 and R into an equivalent parallel network. It should be possible to find the dynamic impedance of the parallel circuit for a given Q. The wanted impedance at the input is Rp Derivation of the above equations
Taking C2 and R in parallel, the combined impedance is the product over sum:
Taking into account the series impedance due to ‘L’:
Convert this into a series circuit by multiplying the second term top and bottom by the complex conjugate. This removes the ‘j’ terms in the denominator:
After multiplying out:
After tidying up:
Comparing Equation 3 against the standard series complex impedance:
Hence:
In a series circuit, the circuit Q is defined as the ratio of reactance to resistance. Some texts complicate the issue by referring to loaded and unloaded Q. Here, Q is assumed to be the loaded Q which is the Q of the tank circuit with the load R connected (This is 50Ω in this case). An assumption is made that the coil and capacitor is constructed from low loss components and that the unloaded Q (The Q without R connected) is much higher than the loaded Q.
Hence the value of Q is:
After substituting:
Equation 4 is needed later.
The next stage is to convert the series circuit, Equation 3, into a parallel circuit. When converted, the result will be a parallel resistance known as the dynamic impedance in parallel with an inductive reactance. The capacitor C1 is then placed in parallel and is used to tune the inductance to resonance.
The next stage is converting Equation 3 into an equivalent parallel equivalent circuit using the following equations. The subscript ‘s’ means ‘series’ and ‘p’ means ‘parallel’. These results are derived in the appendix to this document.
and
Hence, substituting for the series components:
and
Rearrange Equation 5 to give:
Also, from Equation 7:
Strictly speaking the square root is plus or minus. However, the correct sign in this application is positive. Now substitute Equation 7 into Equation 4:
Substitute Equation 8 into Equation 9:
Equation 10 is now solved for L: Notes
• In equation 11, the only terms are the anode load impedance Rp, L, R and the circuit Q.
• The value of equation 12 does not depend upon L.
• The capacitor C1 is only required to tune out the inductive reactance which then completes the match. This is probably why, for the last 100 years, the capacitor C2 has been called the load capacitor and why C1 has been called the tune capacitor!
Practical Example
To provide an example, Figure 1 shows a MathCad worksheet for 2MHz, a Q of 11, a load of 50Ω and a target anode load impedance of 5000Ω. It is best to calculate the parameters in the order: equations 11, 12, 13 and then 14. The Mathcad sheet is WYSIWYG and calculates the formulas numerically as shown below. The phase plot shows that, at resonance, the impedance looking into the PI-tank is indeed 5000Ω resistive. Figure 1 – Pi-Tank Impedance Measured against Frequency (MHz) – Example
Very often, we need to calculate C1, C2 and the associated Q for a given inductor value. This happens when a practical coil is wound and the resulting inductance is close to the idea value but not exact. In this case, you will need to know if the tank will tune without having to actually try it for real. Rearranging equation 10 gives the Q:
This rearrangement is quite tricky to calculate. Q ends up as a quadratic equation with two solutions. The above is the correct version. The derivation of this equation is provided in Appendix B. Equation 15 is used to calculate the Q before evaluating equations 12, 13 and 14 to determine the other parameters. The components are assumed ideal.
Comparison with other Calculators and Designs
An objective of this document is to compare the values predicted by equations 11, 12, 13 and 14 against other designs and on line calculators. For this comparison I assumed an anode impedance of 5000Ω to allow comparison against the G2DAF design. I included amateur bands all the way down to 475kHz.
The load is assumed 50Ω, anode load 5000Ω the Q is assumed 12.
BandElementThis
Document
On Line
Calculator
Note 1
On Line
Calculator
Note 2
G2DAF
Note 3
ARRL
Handbook
1983
Note 4
0.475 MHzC1804.1 pF804 pF804.1 pF
L146.4 μH146 μH146.4 μH
C24495 pF4500 pF4495.3 pF
1.9 MHzC1201 pF201 pF201 pF
L36.6 μH36.6 μH36.6 μH
C21124 pF1120 pF1123.8 pF
3.7 MHzC1103.2 pF103 pF103.2 pF116 pF118 pF
L18.8 μH18.8 μH18.8 μH20 μH18.6 μH
C2577.1 pF577 pF577.1 pF900 pF766 pF
7.1 MHzC153.8 pF53.8 pF53.8 pF58 pF59 pF
L9.8 μH9.8 μH9.8 μH10 μH9.28 μH
C2300.7 pF301 pF300.7 pF450 pF383 pF
14.2 MHzC126.9 pF26.9 pF26.9 pF29 pF30 pF
L4.9 μH4.9 μH4.9 μH5 μH4.6 μH
C2150.3 pF150 pF150.4 pF225 pF192 pF
21.2 MHzC118.0 pF18 pF18 pF20 pF20 pF
L3.3 μH3.3 μH3.3 μH3.5 μH3.1 μH
C2100.7 pF101 pF100.7 pF160 pF128 pF
29 MHzC113.2 pF13.2 pF13.1 pF15 pF15 pF
L2.4 μH2.4 μH2.4 μH2.5 μH2.32 μH
C273.6 pF73.6 pF73.6 pF113 pF96 pF
Note 1: This is taken from the excellent OwenDuffy.net site that presents an on-line calculator based on the Eimac “Care and Feeding of Power Grid Tubes 5th edition 2003” book. On inspection, the equations presented in the Eimac document turn out to be (essentially) identical to the equations derived here. There are some differences in appearance but these can be shown identical by suitable substitutions. The results are identical. https://owenduffy.net/calc/pi.htm
Note 2: https://www.changpuak.ch/electronics/calc_18.php The web site does not state the equations or the source of the equations but the results are the same numerically. I suspect the calculator uses the same source as the OwenDuffy.net calculator.
Note 3: The document by G2DAF states that the method of calculating the PI tank components is an approximation. Surprisingly, the results are similar particularly the inductors. With variable capacitors, the tank circuits would be perfectly acceptable. Note that the G2DAF results were calculated for a Q of 12 and output impedance of 75Ω and not 50Ω.
Note 4: This data is presented as a table for Q=12 only. Note the frequencies are approximate. No explanation for the data or derivation were given. Later, in the 1997 ARRL handbook, the method of calculation changed and equations are quoted. Unfortunately, the tables now only covered loads from 1500Ω to 2500Ω. The 1997 book provided software to calculate other values.
Conclusion
Metaphorically speaking, I have reinvented the wheel and taught grandma how to suck eggs. Most of the existing calculations provide perfectly satisfactory results. For me it has been an interesting investigation and I now understand the PI-tank a little better. The only limitation I can see in these calculations is the condition:
If you attempt a large impedance transformation, the loaded Q must be sufficient. For example, transforming 50Ω to 5000Ω, the Q must be 9.95 or greater. If you try to calculate results for lower Q, then C1 and C2 become complex which is nonsense.
I conclude that the OwenDuffy.net site provides identical results and involves no approximations. I recommend using that site with confidence.
Appendix A – Convert a Series Impedance to an Equivalent Parallel Circuit
This appendix demonstrates how to convert from a series to a parallel tuned circuit of known Q. The circuit Q is identical in both arrangements because the circuits are equivalent. In the above example, the reactance is inductive but expressions can easily be calculated for capacitive reactance (not done here).
First, find the parallel impedance by taking the product over sum and then multiply top and bottom by the conjugate of the denominator:
Express this in series form:
Here, the equivalent series impedance takes the form:
so
However, for a parallel connection, Q is defined as the parallel resistance divided by the reactance:
So, rearranging equations A1 and A2:
To convert in the other direction from series to parallel, rearrange equations A3 and A4:
==========================================
Some engineers avoid using secondary parameters such as Q because it can cause confusion. In this case, Q removes unnecessary complexity and is quite useful.
End of Appendix A
Appendix B – Convert Equation 10 into Equation 15
This is quite tricky and took several attempts so I recorded the arithmetic for posterity. The task is to express equation 10 making Q the subject:
First eliminate the square root by rearranging and then squaring both sides:
Multiply out and add 1 to each side:
divide through by:
giving:
Arrange as a quadratic in Q:
Divide through by:
To give:
Tidy up
so
Hence, taking the positive option (this is the one that works – found by trial and error)
Sorry that was so tedious!
73 Chris | HuggingFaceTB/finemath | |
# [SOLVED] Epsilon-N Proof
• October 28th 2009, 04:45 PM
Roam
[SOLVED] Epsilon-N Proof
Prove that the sequence $(\frac{1}{1+n+n^4})$ converges to 0.
Here's the solution:
http://img517.imageshack.us/img517/7235/81768116.gif
Here's my problem: I don't understand why they wrote $N=\frac{1}{\epsilon}$ !
We know that $n+1 > \frac{1}{\epsilon}$
Therefore $n> \frac{1}{\epsilon} -1$
And so we should take: $N= \frac{1}{\epsilon} -1$
Isn't that right???
Also, I don't understand why they first omitted $n^4$ from the denominator when they could omit $n+1$ instead. Because I think $\frac{1}{n^4}$ goes to zero a lot quicker. I appreciate some explanation.
• October 28th 2009, 05:22 PM
proscientia
Quote:
Originally Posted by Roam
We know that $n+1 > \frac{1}{\epsilon}$
Therefore $n> \frac{1}{\epsilon} -1$
And so we should take: $N= \frac{1}{\epsilon} -1$
No. Firstly, $N$ must be an integer. $\frac1\epsilon-1$ may not be an integer. Secondly, $n+1>\frac1\epsilon$ is what we want, not what we know. By choosing $N=\left\lceil\frac1\epsilon\right\rceil,$ we can be assured that whenever $n>N,$ we have $n+1>\frac1\epsilon.$
• October 29th 2009, 12:54 AM
Roam
Quote:
Originally Posted by proscientia
$\frac1\epsilon-1$ may not be an integer.
Why is that?
And, what about my second question? Why couldn't we start out by omitting $1+n$ instead of $n^4$?
$\frac{1}{1+n+n^4} < \frac{1}{n^4} < \epsilon$
Thus $N= \frac{1}{\epsilon ^4}$
• October 29th 2009, 02:44 AM
Plato
Quote:
Originally Posted by Roam
Thus $N= \frac{1}{\epsilon ^4}$
That does not work.
There is no reason that $\frac{1}{\epsilon ^4}$ is an integer.
• October 29th 2009, 10:30 AM
Roam
Quote:
Originally Posted by Plato
That does not work.
There is no reason that $\frac{1}{\epsilon ^4}$ is an integer.
Then how do you know that $\frac{1}{\epsilon}$ is an integer?
• October 29th 2009, 02:35 PM
proscientia
Where is it stated that $\frac1\epsilon$ is an integer? $N$ is $\left\lceil\frac1\epsilon\right\rceil,$ not $\frac1\epsilon.$
• October 30th 2009, 06:41 AM
HallsofIvy
Quote:
Originally Posted by Roam
Then how do you know that $\frac{1}{\epsilon}$ is an integer?
You don't! If, for example, $\epsilon= .00000003= \frac{3}{100000000}$ then $\frac{1}{\epsilon}= \frac{100000000}{3}$ which is not an integer.
If you look in any text book you should see that they say something like $N> \frac{1}{\epsilon} -1$, not $N= \frac{1}{\epsilon} -1$ | open-web-math/open-web-math | |
# Objectives: Derivation Transform Pairs Response of LTI Systems Transforms of Periodic Signals Examples - PowerPoint PPT Presentation
Objectives: Derivation Transform Pairs Response of LTI Systems Transforms of Periodic Signals Examples
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Objectives: Derivation Transform Pairs Response of LTI Systems Transforms of Periodic Signals Examples
## Objectives: Derivation Transform Pairs Response of LTI Systems Transforms of Periodic Signals Examples
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##### Presentation Transcript
1. LECTURE 10: THE FOURIER TRANSFORM • Objectives:DerivationTransform PairsResponse of LTI SystemsTransforms of Periodic SignalsExamples • Resources:Wiki: The Fourier TransformCNX: DerivationMIT 6.003: Lecture 8 Wikibooks: Fourier Transform TablesRBF: Image Transforms (Advanced) • URL: .../publications/courses/ece_3163/lectures/current/lecture_10.ppt • MP3: .../publications/courses/ece_3163/lectures/current/lecture_10.mp3
2. Motivation • We have introduced a Fourier Series for analyzing periodic signals. What about aperiodic signals? (e.g., a pulse instead of a pulse train) • We can view an aperiodic signal as the limit of a periodic signal as T . • The harmonic components are spaced apart. • As T ,0 0, then k0 becomes continuous. • The Fourier Series becomes the Fourier Transform.
3. Derivation of Analysis Equation • Assume x(t) has a finite duration. • Define as a periodic extensionof x(t): • As • Recall our Fourier series pair: • Since x(t) andare identical over this interval: • As
4. Derivation of the Synthesis Equation • Recall: • We can substitute for ckthe sampled value of : • As • and we arrive at our Fourier Transform pair: • Note the presence of the eigenfunction: • Also note the symmetry of these equations (e.g., integrals over time and frequency, change in the sign of the exponential, difference in scale factors). (synthesis) (analysis)
5. Frequency Response of a CT LTI System • Recall that the impulse response ofa CT system, h(t), defines the properties of that system. • We apply the Fourier Transform toobtain the system’s frequency response: • except that now this is valid for finite duration (energy) signals as well as periodic signals! • How does this relate to what you have learned in circuit theory? CT LTI CT LTI
6. Existence of the Fourier Transform • Under what conditions does this transform exist? • x(t) can be infinite duration but must satisfy these conditions:
7. Example: Impulse Function
8. Example: Exponential Function
9. Example: Square Pulse
10. Example: Gaussian Pulse
11. CT Fourier Transforms of Periodic Signals
12. Example: Cosine Function
13. Example: Periodic Pulse Train
14. Summary • Motivated the derivation of the CT Fourier Transform by starting with the Fourier Series and increasing the period: T • Derived the analysis and synthesis equations (Fourier Transform pairs). • Applied the Fourier Transform to CT LTI systems and showed that we can obtain the frequency response of an LTI system by taking the Fourier Transform of its impulse response. • Discussed the conditions under which the Fourier Transform exists. Demonstrated that it can be applied to periodic signals and infinite duration signals as well as finite duration signals. • Worked several examples of important finite duration signals. • Introduced the Fourier Transform of a periodic signal. • Applied this to a cosinewave and a pulse train. | HuggingFaceTB/finemath | |
$$V_0$$ Initial investment
$$V_f$$ Final value of investment
$$r$$ Rate of return
# The rule of 72, and may I introduce the rule of 108? Photo credit to Pixabay
Here we will answer the question, where does the rule of 72 come from? And how can we derive similar rules for other tripling or other multiplicative factors?
## Origin of 72 for doubling time of an investment
The rule of 72 is an estimate for the doubling time of an investment at a given rate of return.
$$\left[ \text{Doubling time} \right] = \frac{72}{\left[ \text{Rate of return as a percent}\right]}$$
As an example, for a 6% return, the rule of 72 says an investment will double in 12 years.
$$\left[ \text{Doubling time}\right] = \frac{72}{6\text{% per year}} = 12 \text{ years}$$
An investment at a fixed rate of return grows exponentially according to the simple differential equation from the article on what it takes to save \$1 million.
\begin{align} \frac{dV}{dt} &= rV\\ \int^{V_f}_{V_0} \frac{dV}{rV} &= \int^t_0 t\\ V_f = V_0 e^{rt} \end{align}
To find the doubling time, set $$V_f = 2V_0$$.
$$t_\text{double} = \frac{\log(2)}{r}$$
Notice that $$\log($$2$$) =$$ 0.693147 and if we want a formula that does not require the conversion from percent to dimensionless fraction, we have to multiply by 100.
$$t_\text{double} = \frac{69.3}{r_\text{as percent}}$$
The only trouble is that no reasonable rate of return divides cleanly into 69, except for 1 and 3. However, 72 can be cleanly divided by 1, 2, 3, 4, 6, 8, 9, and 12 making it a much more convenient number for a back-of-the-envelope estimation. That is the story of how 69 became 72.
This begets the question - "if the rule of 72 is actually the rule of 69.3, what multiplicative factor does 72 correspond to?" The answer is "pretty much 2."
\begin{align} \left. rt \right| _\text{72rule} = \log\left(\frac{V_f}{V_0}\right) &= 0.72\\ \frac{V_f}{V_0} &= e^{0.72}\\ \frac{V_f}{V_0} &= 2.0544 \approx 2 \\ \end{align}
The purpose of the rule of 72 is for back of the envelope calculations and for that, a <3% error is acceptable.
## Rules for the doubling, tripling, and quadrupling of investments
Using this logic we can extend the rule. How long does it take our money to triple?
$$t_\text{triple} = \frac{\log(3)}{r} = \frac{1.0986}{r} \approx \frac{108}{r_\text{as percent}}$$
Approximate 109.86 as 108 since 109 and 110 are only evenly divisible by 1 and 5 of the first 12 integers respectively. However, 108 is evenly divisible by 1, 2, 3, 4, 6, 9, and 12; the result being 108, 54, 36, 27, 18, 12, and 9 respectively.
This gives us the following table. The quadrupling time is just double the doubling time.
Doubling time Tripling time Quadrupling time
$$\displaystyle \frac{72}{r_\text{as percent}}$$ $$\displaystyle\frac{108}{r_\text{as percent}}$$ $$\displaystyle\frac{144}{r_\text{as percent}}$$ | HuggingFaceTB/finemath | |
# Würfeln
Würfeln ist Einstellungssache is a game of luck and high precision dice rolling. Tweak statistics to your advantage with the right mindset. Is your dice roll determination stronger than the one of your opponents? Play a game, bet on your roll attempts and show that you have mastered the art of rolling the die.
## How to play
### Setup
Challenge up to three opponents, three standard D6 dice (numbers 1-6) per player (idealy visually different), an einhorn and plenty of smooth rocks.
The poor mans option that works is any object that will resemble the einhorn and something to keep track of the score e.g. rocks, coins, etc. Pen and paper works but is a little more work and kills the flow.
Everyone starts with 6 rocks (=points). Roll all three die and the player with highest total number starts the game.
### Play
The current player decides to either predict the upcoming roll or not (passive roll) - then rolls all three dice once.
Based on the result add/remove points as follows:
Result Rule / Dice combination Correct Wrong Passive
⚰️ Das Unvermeidliche n & n+1 & (!n / !n+1)
Example: ⚀⚁⚃ or ⚂⚃⚀
+2 -2 -1
🎁 Wunsch 2x n & !n
Example: ⚀⚀⚁
+1 -1 0
🦄 Einhorn n & n+2 & (n+4 / n+5)
n & n+3 & n+5
Example: ⚀⚂⚄ or ⚀⚃⚅
+5 -5 -5
☢️ Dreifaltigkeit 3x n
Example: ⚁⚁⚁ or ⚅⚅⚅
+3 -3 +3
### Example round
If it is your turn, you bet on rolling 🎁(Wunsch) and you rolled 1,2,5 which is ⚰️(Das Unvermeidliche). Therefore, you lose 2 points. If this is the first round of the game, you started with 6 points and now have 4 points. Now it is the next players turn. The player choses to not predict anything and roll a 2,2,2 which is ☢️(Dreifaltigkeit). In this case the player gains 3 points and it is the next players turn. The third player predicts an 🦄(Einhorn) and roll a 1,3,6 and gains 5 points. (this never happens IRL, lol :D)
### 🦄 Einhorn
In case a player rolls an 🦄(Einhorn) the player then recieves the Einhorn. It doesn't matter if the prediction was correct, wrong or passiv. From this moment on the player may choose from where to take and give points: either the main rock stash from any other player. If the player with the Einhorn needs to give rocks back, those are given to any other player of choice. The 🦄(Einhorn) is passed once another player roles an 🦄(Einhorn) and returned once the round ends.
### How to win
The first player to reach 0 points (or less) loses the game. The game ends at this point and the player with the most points wins the game.
### Multiple games
Should you play multiple games note down all remaining points from the current game and add them to the total score.
Agree on how many games shall be played and after that game the player with the highest overall score wins.
On every new round you start again with 6 points but can add already gained points from previous matches into this game if you like. You can add your entire point stach if you want.
You do not keep the 🦄(Einhorn) over multiple games. After each game is over the 🦄(Einhorn) returns to the middle of the table and is called once an 🦄(Einhorn) is rolled.
The looser of the last round begins the new game.
### Multiple games example
This is the score board of an example session with 4 games and 3 players:
Game Player 1
Points -> Score
Player 2
Points -> Score
Player 3
Points -> Score
1 11 -> 11+0 = 11 0 -> 0+0 = 0 2 -> 2+0 = 2
2 5 -> 5+11 = 16 6 -> 6+0 = 6 0 -> 0+2 = 2
3 0 -> 0+16 = 16 8 -> 8+6 = 14 16 -> 16+2 = 18
4 3 -> 3+16 = 19 2 -> 2+14 = 16 0 -> 0+18 = 18
Player 1 takes the lead in round one with 11 points and can continue the lead in round 2.
Player 3 gets 16 points in round three adding to 18 points in total and taking the lead.
Player 1 wins the game at the end of round 4 with a total of 19 points.
### Extras
Have a look at the math part to see all possible combinations and possibly the rules without words are interesting as well :)
Warning: Math
With 3x D6 we get a total of 216 (6x6x6 or 6^3) possible outcomes.
Let us now look at unique roles. To get all possible combinations we can use this formula for combination with repetition:
\begin{align} \frac{(n+k-1)!}{k!*(n-1)!} \end{align}
With 3x D6 we get n = 6, k = 3 resulting in 56 combinations.
\begin{align} \frac{(6+3-1)!}{3!*(6-1)!} = 56 \end{align}
You can copy (6+3-1)!/(3!*(6-1)!) to wolframalpha to calculate it or use this link.
These 56 combinations distribute as follows:
Result Amount Percent
⚰️ Das Unvermeidliche ⚀⚁⚃, ⚀⚁⚄, ⚀⚁⚅, ⚁⚂⚀
⚁⚂⚄, ⚁⚂⚅, ⚂⚃⚀, ⚂⚃⚁
⚂⚃⚅, ⚃⚄⚀, ⚃⚄⚁, ⚃⚄⚂,
⚄⚅⚀, ⚄⚅⚁, ⚄⚅⚂, ⚄⚅⚃
16 29 %
🎁 Wunsch ⚀⚀⚁, ⚀⚀⚂, ⚀⚀⚃, ⚀⚀⚄, ⚀⚀⚅
⚁⚁⚀, ⚁⚁⚂, ⚁⚁⚃, ⚁⚁⚄, ⚁⚁⚅
⚂⚂⚀, ⚂⚂⚁, ⚂⚂⚃, ⚂⚂⚄, ⚂⚂⚅
⚃⚃⚀, ⚃⚃⚁, ⚃⚃⚂, ⚃⚃⚄, ⚃⚃⚅
⚄⚄⚀, ⚄⚄⚁, ⚄⚄⚂, ⚄⚄⚃, ⚄⚄⚅
⚅⚅⚀, ⚅⚅⚁, ⚅⚅⚂, ⚅⚅⚃, ⚅⚅⚄
30 54 %
🦄 Einhorn ⚀⚂⚄, ⚀⚂⚅, ⚀⚃⚅, ⚁⚃⚅ 4 7 %
☢️ Dreifaltigkeit ⚀⚀⚀, ⚁⚁⚁, ⚂⚂⚂
⚃⚃⚃, ⚄⚄⚄, ⚅⚅⚅
6 11 %
Rules without words
👱🧔🧕
👱 🎲
2-4 3 +6
🎲🎲🎲
= 🎲∑ 🎲🎲🎲 # %
⚰️ n & n+1 & (!n / !n+1) ⚀⚁⚃, ⚀⚁⚄, ⚀⚁⚅, ⚁⚂⚀
⚁⚂⚄, ⚁⚂⚅, ⚂⚃⚀, ⚂⚃⚁
⚂⚃⚅, ⚃⚄⚀, ⚃⚄⚁, ⚃⚄⚂,
⚄⚅⚀, ⚄⚅⚁, ⚄⚅⚂, ⚄⚅⚃
16 29
🎁 2x n & !n ⚀⚀⚁, ⚀⚀⚂, ⚀⚀⚃, ⚀⚀⚄, ⚀⚀⚅
⚁⚁⚀, ⚁⚁⚂, ⚁⚁⚃, ⚁⚁⚄, ⚁⚁⚅
⚂⚂⚀, ⚂⚂⚁, ⚂⚂⚃, ⚂⚂⚄, ⚂⚂⚅
⚃⚃⚀, ⚃⚃⚁, ⚃⚃⚂, ⚃⚃⚄, ⚃⚃⚅
⚄⚄⚀, ⚄⚄⚁, ⚄⚄⚂, ⚄⚄⚃, ⚄⚄⚅
⚅⚅⚀, ⚅⚅⚁, ⚅⚅⚂, ⚅⚅⚃, ⚅⚅⚄
30 54
🦄 n & n+2 & (n+4 / n+5)
n & n+3 & n+5
⚀⚂⚄, ⚀⚂⚅, ⚀⚃⚅, ⚁⚃⚅ 4 7
☢️ 3x n ⚀⚀⚀, ⚁⚁⚁, ⚂⚂⚂
⚃⚃⚃, ⚄⚄⚄, ⚅⚅⚅
6 11
= 🎲∑
⚰️ n & n+1 & (!n / !n+1) +2 -2 0
🎁 2x n & !n +1 -1 0
🦄 n & n+2 & (n+4 / n+5)
n & n+3 & n+5
+5 -5 -5
☢️ 3x n +3 -3 +3
🏁
🏁 P1 👱 P2 🧔 P3 🧕
1 11 -> 11+0 = 11 0 -> 0+0 = 0 2 -> 2+0 = 2
2 5 -> 5+11 = 16 6 -> 6+0 = 6 0 -> 0+2 = 2
3 0 -> 0+16 = 16 8 -> 8+6 = 14 16 -> 16+2 = 18
4 3 -> 3+16 = 19 2 -> 2+14 = 16 0 -> 0+18 = 18
P1👱=19 > P3🧕=18 > P2🧔=16
P1👱 = 🥇 | open-web-math/open-web-math | |
# What is a prime factor?
WHAT IS A PRIME FACTOR
Answers
A prime factor is a factor of a number, and that factor is also prime. A prime number is one that cannot be neatly divided into by any other number than itself and 1. Factors are numbers that do wholly divide into another number. Here "wholly divide" means "leaves no remainder". An example, then. The number 60 can be wholly divided by 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. These are the factors of 60. The Prime factors are 2, 3 and 5, and 60 is 2235 ( used here to mean "times by"). From the prime factors we can find all the other factors by using all the possible combinations of prime factors, e.g. 22 gives 4 and 35 gives 15, and 415 = 60, while 23 = 6 and 2*5 = 10 gives another pair of factors. And so on. Prime factors are useful for a number of reasons. Firstly, when working with awkward fractions it can be useful to know all the prime factors of the numbers invlved to help with simplifying the numbers. Suppose you had to work out (14/15) * (130/231). Doing this by hand could take a while and leave lots of chances to mess up. However if we write out all the numbers by prime factors we can find any possible simplfying before we do the sum. In this example, we would get 14 = 27 and 130 = 2513, while 15 = 35 and 231 = 3711, so the sum could have been written (225713)/(335711). We see that 5 and 7 appear on top and bottom so we cross these out, leave (2213)/(3311), which is much easier to deal with and turns out to be 52/99. Also, if you want to find out whether a number is prime, or if not what are its factors, you simply check whether any prime numbers divide into it. Take 887 as an example. You could test whether this is prime by checking every number below it to see if it divides into 887, but it's easier to just try 2, 3, 5, 7, 11, 13... the prime numbers, in other words, which reduces the amount of effort involved in discovering that 887 is indeed prime.
jim360
14 September 2012 | HuggingFaceTB/finemath | |
## Lecture on The Circle
Subject: Mathematic | Topics:
major objective of this lecture is to present on The Circle. In a plane,each point of the circle is at equal distance from a fixed point.The fixed point is called the centre of the circle. The distance from centre to any point on the circle is called radius of the circle. A Line segment passing through the centre of the circle and whose end points lie on the circle is called the diameter of the circle. The length of the circle or the distance around it is called circumference of the circle. The distance from centre to any point on the circle is called radius of the circle.
### Mental Math with Tricks and Shortcuts
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### Define and Discuss on Central Limit Theorem
principle purpose of this article is to Define and Discuss on Central Limit Theorem. Here explain Central Limit Theorem with mathematical examples. The central limit theorem states that even if a population distribution is usually strongly non‐normal, its sampling distribution of means is goi.....
### Discuss on Keywords for Mathematical Operations
Primary objective of this article is to Discuss on Keywords for Mathematical Operations. Here explain Keywords for Mathematical Operations in different language point of view. The initial step in solving the mathematical word problem is usually always in order to read the problem. Every one nee.....
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Primary objective of this lecture is to analysis Discrete Mathematics and its Applications based on Trees. A tree is often a connected undirected graph without any simple circuits. Brief hypothesis: An undirected graph is often a tree if and only if there is a unique simple way between any two ..... | HuggingFaceTB/finemath | |
# HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5
Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Exercise Questions and Answers.
## Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.5
Solve the following linear equations
Question 1.
$$\frac{x}{2}-\frac{1}{5}$$ = $$\frac{x}{3}+\frac{1}{4}$$
Solution:
Question 2.
$$\frac{n}{2}$$ – $$\frac{3n}{4}$$ + $$\frac{5n}{6}$$ = 21
Solution:
Question 3.
x + 7 – $$\frac{8x}{3}$$ = $$\frac{17}{6}$$ – $$\frac{5x}{2}$$
Solution:
Question 4.
$$\frac{x-5}{3}$$ = $$\frac{x-3}{5}$$
Solution:
$$\frac{x-5}{3}$$ = $$\frac{x-3}{5}$$
or, 5(x – 5) = 3 (x – 3)
or, 5x – 25 = 3x – 9
or, 5x – 3x = -9 + 25
or, 2x = 16
or, x = $$\frac{16}{2}$$
∴ x = 8
Question 5.
$$\frac{3t-2}{4}$$ – $$\frac{2t-3}{3}$$ = $$\frac{2}{3}$$ – t
Solution:
or, 3 (13t – 18) = 24
or, 39t – 54 = 24
or, 39t = 24 + 54 = 78
or, 39t = 78
or, t = 2
Question 6.
m – $$\frac{m-1}{2}$$ = 1 – $$\frac{m-2}{3}$$
Solution:
or, 5m – 1 = 6
or, 5m = 6 + 1
or, 5m = 7
∴ m = $$\frac{7}{5}$$
Simplify and solve the following linear equations:
Question 7.
3 (t – 3) = 5 (2t + 1)
Solution:
3 (t – 3) = 5 (2t + 1)
or, 3t – 9 = 10t + 5
or, 3t – 10t = 5 + 9
or, -7t = 14
or, t = $$-\frac{14}{7}$$ =-2
t = -2
Question 8.
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
or, 15y – 60 – 2y + 18 + 5y + 30 = 0
or, 20y – 2y – 60 + 48 = 0
or, 18y – 12 = 0
or, 18y = 12
or, y = $$\frac{12}{18}$$ = $$\frac{2}{3}$$
∴ y = $$\frac{2}{3}$$
Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
or, 15z – 21 – 18z + 22 = 32z – 52 – 17
or, 15z – 18z – 32z = -52 – 17 + 21 – 22
or, 15z – 50z = 21 – 91
or, -35z = -70
or, 35z = 70
or, z = 2
∴ z = 2
Question 10.
0.25 (4f – 3) = 0.05 (10f – 9)
Solution:
0.25 (4f – 3) = 0.05 (10f – 9)
or, f – 0.75 = 0.5f – 0.45
or, f – 0.5f = -0.45 + 0.75
or, 0.5f = 0.3
f = $$\frac{0.3}{0.5}$$ = $$\frac{3}{5}$$ = 0.6
∴ f = 0.6 | HuggingFaceTB/finemath |
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