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https://math.stackexchange.com/questions/1808532/finding-the-set-of-values-for-k-of-a-modulus-function | # Finding the set of values for k of a modulus function.
"Find the set of values of k for which |(x-4)(x+2)| = k has four solutions."
EDIT:
Ok so I thought I'd start with setting the modulus function equal to k and -k to get the two set of results.
Doing that I ended up with:
(x -4)(x +2) = k
and
(x - 4)(x +2) = -k .
After solving for x I got x = 4 or -2 OR (the other set of results) x = -4 and 2. -> not sure if my logic here is right, as I simply took the "negative" version of the function to get the second set of results...
I do not know how to progress with this question, the x-values don't really help me much here. Any help would be appreciated.
Posted this in the question so that it's easier to see :)
• Draw the function first. It is a parabola with zeros at $x=-2$ and $x=4$. You will have to flip over the lower part of the parabola with negative $y$ values. You will see for what values the parabola has four zeros $0<k<k_{max}$. Jun 1, 2016 at 16:35
• I did draw the parabola, sadly it says that solutions done by accurate drawing will not be accepted! :/ Can I justify my drawings as the mathematics calculation? Jun 1, 2016 at 16:39
• No don't use the drawing for your answer. It should give you an insight what value of $x$ you have to choose to find $k_{max}$. Look at the vertex of your parabola (wolframalpha.com/input/…) Jun 1, 2016 at 16:44
Solving the equation $(x-4)(x+2)=k$ gives you $x_{1,2}=1\pm\sqrt{9+k}$; solving the equation $(x-4)(x+2)=-k$ gives you $x_{3,4}=1\pm\sqrt{9-k}$.
• $k>-9$ has to hold in order to get two solutions out of $x_{1,2}=1\pm\sqrt{9+k}$
• $k<9$ has to hold in order to get two solutions out of $x_{3,4}=1\pm\sqrt{9-k}$
• $k\neq 0$ has to hold, because otherwise $\{1\pm\sqrt{9+k}\}=\{1\pm\sqrt{9-k}\}$
Therefore the solution is (given $k$ is an integer): $k\in\{-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7,8\}$
• I do not see why you restricted your solution to the integers. That said, it is clear how to find all real values of $k$ that produce four solutions from your arguments. (+1). Jun 2, 2016 at 10:42 | 2022-05-28T23:16:51 | {
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http://math.stackexchange.com/questions/298073/shorthand-notation-for-powers-of-logarithmic-functions | # Shorthand notation for powers of logarithmic functions
I've got an assignment here, and one question is throwing me off, as I've never seen it written like this before..
$$\int\frac{\ln^3 x}{x}\ dx$$
Is this the same as $$\int\frac{(\ln x)^3}{x}\ dx\;\;?$$
-
Yes it is. Moreover $$\int \frac{\ln^3(x)}{x}\,{\rm d}x = \textstyle{\frac 14}ln^4(x) + c$$ – AndreasT Feb 8 '13 at 14:37
Yeah - I figured that would be the result if I had assumed my hypothesis was correct. Thanks for clarifying this. – agent154 Feb 8 '13 at 14:38
For future reference, if you write \ln x instead of ln x it doesn't look like $l$ times $n$ times $x$. Same goes for \sin, \exp, etc. – Rahul Feb 8 '13 at 14:59
Unfortunately, "shorthand" can lead to ambiguity:
$\ln^3 x = (\ln (\ln(\ln x))\;?\quad$ or $\quad \ln^3x = (\ln x)^3\;?$
But as you suspected, in this context, and given the integral, I'm am quite sure that $\ln^3x = (\ln x)^3,\;$ much like $\;\sin^2(x) = (\sin x)^2.\;$ So your integral amounts to:
$$\int\frac{\ln^{3}x}{x}\ dx = \int \frac{(\ln x)^3}{x} \,dx$$
Let $u = \ln(x),;\;du = \dfrac{dx}{x}$
$$\int \frac{(\ln x)^3}{x} \,dx = \int u^3\,du = \frac14 u^4 + C$$ $$= \frac14 (\ln x)^4 + C = \frac 14\ln^4x + C$$
-
Unfortunately two conventions prevail: $\ln^3 x$ can mean $\ln\ln\ln x$ or it can mean $(\ln x)^3$. The first one is reasonable; the second one is unfortunately also seen. I would write $(\ln x)^3$, or $(\log x)^3$, or $(\log_e x)^3$, if I meant $(\log_e x)^3$.
It became universally standard to construe $\sin^3 x$ to mean $(\sin x)^3$, over the objection of Carl Gauss (the most famous person to live on earth in the 19th century, except perhaps those who did not work in the physical and mathematical sciences), who said $\sin^3 x$ ought to mean $\sin\sin\sin x$. You can't push back the tide with a pitchfork.
-
$\sin^2x=(\sin x)^2$ has a lot to answer for... I need to use $\arcsin x$ now or else I see $\sin^{-1} x=\frac{1}{\sin x}$ etc. – Jp McCarthy Feb 8 '13 at 17:10 | 2015-04-19T13:06:17 | {
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https://www.physicsforums.com/threads/determing-if-a-function-with-integral-within-it-is-even-odd.842414/ | # Determing if a function with integral within it is even/odd
Tags:
1. Nov 10, 2015
### mr.tea
1. The problem statement, all variables and given/known data
Determine if the function is even/odd/neither without solving the integral
2. Relevant equations
$$f(x)=\arctan (x)-2\int_0 ^x{\frac{1}{(1+t^2)^2}}$$
3. The attempt at a solution
I tried to do f(-x)-f(x). I know that if f is odd, I should get -2f(x) and if even, 0. I got that f is odd(-2f(x)), but I don't know if this is a correct answer, since wolfram alpha says that this is not an odd function. I didn't assume anything on the function while checking this. I am a bit confused.
Is there any way to check things like that?
Thank you,
Thomas
2. Nov 10, 2015
### Khashishi
When the limits of the integration are 0 to x, the integral of an even function is odd and the integral of an odd function is even.
Look at each term in the equation. If each part is even, then the whole thing is even. Same with odd. But if you add an even and odd function (that are not 0) then you get neither.
3. Nov 10, 2015
### HallsofIvy
Staff Emeritus
First you know that arctan(x) is an odd function, right? Further, making the change of variable, u= -t in the integral, so that dt= -du
$$\int_0^x \frac{dt}{(1+ t^2)^2}$$
becomes
$$\int_0^{-x}\frac{-du}{(1+ (-u)^2)^2}= -\int_0^{-x} \frac{du}{(1+ u^2)^2}$$
changing the "dummy variable", u, in that integral to t,
$$\int_0^x \frac{dt}{(1+ t^2)^2}= -\int_0^{-x} \frac{dt}{(1+ t^2)^2}$$
4. Nov 10, 2015
### mr.tea
Hi,
First, thank you for the answer.
How do we know that the integral of an odd function is even and the same with even function? Maybe I will try to prove it, but expect to a question about that :)
Second, since the integrand is even, that means(according to you, but I am still not convinced about it) that the integral is odd. And since arctan is odd, that means that the whole function is odd. Right?
5. Nov 10, 2015
### epenguin
HoI essentially just answered your question in #4. He did it for the particular function, but you could write out the proof again for integral of any function f which satisfies the condition of evenness which you should write out. The key point which you may have missed in the detail is, and may even disbelieve till you think about it is, to say it crudely
ab = - ∫ba
6. Nov 10, 2015
### mr.tea
Thank you. I did something similar with the function but without changing the variable.
As I said to HallsofIvy, I did the same thing. I just assumed that without loss of generality, x is positive then -x is negative and you flip the limits of the integral and you add minus.
I cannot see those things/"proofs" in my head without write them. That might be the reason I was( well, and still a bit) sceptical(Well, I am studying mathematics, I guess I should be sceptical).
So, I understand that my answer is correct :)
Thank you all. | 2017-11-20T00:54:59 | {
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https://math.stackexchange.com/questions/1566418/if-we-randomly-select-25-integers-between-1-and-100-how-many-consecutive-intege | If we randomly select 25 integers between 1 and 100, how many consecutive integers should we expect?
Question: Suppose we have one hundred seats, numbered 1 through 100. We randomly select 25 of these seats. What is the expected number of selected pairs of seats that are consecutive? (To clarify: we would count two consecutive selected seats as a single pair.)
For example, if the selected seats are all consecutive (eg 1-25), then we have 24 consecutive pairs (eg 1&2, 2&3, 3&4, ..., 24&25). The probability of this happening is 75/($_{100}C_{25}$). So this contributes $24\cdot 75/(_{100}C_{25}$) to the expected number of consecutive pairs.
Motivation: I teach. Near the end of an exam, when most of the students have left, I notice that there are still many pairs of students next to each other. I want to know if the number that remain should be expected or not.
• That probability is $76/_{100}C_{25}$, not $75/_{100}C_{25}$. Dec 9 '15 at 20:06
• Thanks for the correction :-) Dec 9 '15 at 21:57
• Without math I can tell you yes that is to be expected. They're called friends. Dec 9 '15 at 22:39
• + for the motivation Dec 9 '15 at 22:46
• @corsiKa, I also think it is because if my neighbor leaves, it makes me feel "I should have been ready by now also", and therefore leave quite soon after. While it is the other way around when they still sit next to me, "Why is * not finished yet, have i missed something". And then finally when * leaves, I will go about the same time. But it is still interesting to compare how much difference there is between reality and randomly. Dec 10 '15 at 13:27
If you're just interested in the expectation, you can use the fact that expectation is additive to compute
• The expected number of consecutive integers among $\{1,2\}$, plus
• The expected number of consecutive integers among $\{2,3\}$, plus
• ....
• plus the expected number of consecutive integers among $\{99,100\}$.
Each of these 99 expectations is simply the probability that $n$ and $n+1$ are both chosen, which is $\frac{25}{100}\frac{24}{99}$.
So the expected number of pairs is $99\frac{25}{100}\frac{24}{99} = 6$.
• Can it be argued that you also need to consider the $n-1$ as a possible pair candidate? And how would that affect expectation? Dec 9 '15 at 16:18
• @holroy: If there's an $n-1$, that pair is covered when we consider $n$ one lower. That, is the pair $\{17,18\}$ is $n$ and $n+1$ when $n=17$; the pair $\{16,17\}$ is covered by setting $n=16$, not by keeping $n$ at $17$ and saying $\{n-1,n\}$. Dec 9 '15 at 16:26
• Because the asker is interested in the expected number when most students have left, it makes sense to generalize this from $25$ to other numbers $k<100$ and even to values of $n$ different from $100$. The answer is always $k(k-1)/n$ pairs of students. It may be worth noting this in the answer. Dec 9 '15 at 22:12
• Is independence needed to take advantage of additivity of expectation? Dec 10 '15 at 4:56
• @user17625: That doesn't matter, which is the nice thing about additivity of expectations: $E(X+Y)=E(X)+E(Y)$ no matter whether $X$ and $Y$ are independent or not. Dec 10 '15 at 8:03
Let me present the approach proposed by Henning in another way.
We have $99$ possible pairs: $\{(1,2),(2,3),\ldots,(99,100)\}$. Let's define $X_i$ as
$$X_i = \left \{ \begin{array}{ll} 1 & i\text{-th pair is chosen}\\ 0 & \text{otherwise}\\ \end{array} \right .$$
The $i$-th pair is denoted as $(i, i+1)$. That pair is chosen when the integer $i$ is chosen, with probability $25/100$, and the integer $i+1$ is also chosen, with probability $24/99$. Then,
$$E[X_i] = P(X_i = 1) = \frac{25}{100}\frac{24}{99},$$
and this holds for $i = 1,2,\ldots,99$. The total number of chosen pairs is then given as
$$X = X_1 + X_2 + \ldots + X_{99},$$
and using the linearity of the expectation, we get
\begin{align} E[X] &= E[X_1] + E[X_2] + \cdots +E[X_{99}]\\ &= 99E[X_1]\\ &= 99\frac{25}{100}\frac{24}{99} = 6 \end{align}
Henning Malcolm has already answered the question about the expected value. But that says very little about how likely or unlikely it would be to get deviations from the expected value - what if you observed there were 20 pairs out of 25 students left, could that happen by chance or is there some other factor at work? To answer questions like this one would need more details about the distribution, like the variance. Unfortunately, the analytic formula for the distribution (if there is one) is likely to be very complicated. So, from a practical perspective it might be interesting to investigate things numerically. That is the purpose of this post.
Here's a histogram of the number of pairs, generated numerically from 10000 trials.
You can look at this and draw your own conclusions. I would conclude the following:
• anywhere from 4 to 8 pairs would be totally reasonable,
• 0 pairs or 13+ pairs would be very strong evidence of there being another factor
• anything inbetween would be unusual, but possible
Here's the Matlab code used to generate it:
n=100;
k=25;
num_trials = 1e4;
num_pairs = zeros(num_trials,1);
for ii=1:num_trials
disp(ii)
filled_seats = sort(randperm(n,k));
gaps = filled_seats(2:end) - filled_seats(1:end-1);
num_pairs(ii) = length(find(gaps == 1));
end
histogram(num_pairs)
title(sprintf('histogram of seating pairs (out of %d trials)',num_trials))
• Can you run the same simulations for number of maximal clumps of size at least 2 and compare variances and ratio of variances to the means (to see deviations from Poisson distribution)?
– A.S.
Dec 10 '15 at 10:27
• +1 for the most useful answer. I think what OP really wants to know is: given that I see $x$ pairs, what is the probability of this happening by chance? (either this or a p-value version, i.e. $x$ or more). Dec 10 '15 at 14:19
• @A.S. I think that would require a lot more work then just re-running the code with minor changes. If you want, feel free to use my code as a starting point to do that; I release it to the public domain. Dec 10 '15 at 17:28
• I don't have access to computing environment atm. Here is code based on yours: empty_seats=[0,sort(randperm(n,n-k)),n+1]; gaps = empty_seats(2:end) - empty_seats(1:end-1); num_clumps(ii)=length(find(gaps>1))
– A.S.
Dec 10 '15 at 17:48
• @A.S. Ok, I think (?) I changed it according to your intentions. Here is the clumps of empty seats histogram, i.imgur.com/YmXccl2.png?1 and here is the code, pastebin.com/3YujFZs9 Since this seems to be more related to your answer than mine, maybe it would be better if you post it there? Also, even if you don't have matlab available, this code will run in octave online ( octave-online.net ), though it is slower and the plotting doesn't work. Dec 10 '15 at 18:45
It is not difficult to construct the bivariate generating function $G(z,u)$ of $25$-tuples by the maximum element (variable $z$) and the number of adjacent elements (variable $u$.) To do this we must choose the first element:
$$\frac{z}{1-z}$$
and then choose $24$ gaps between elements with the gap of value one marked with $u:$
$$\left(uz + \frac{z^2}{1-z}\right)^{24}.$$
The product of these is $G(z,u)$ but we are interested in the maximum being some value at most $100$ so we get
$$[z^{100}] \frac{1}{1-z} G(z, u) = [z^{100}] \frac{1}{1-z} \frac{z}{1-z} \left(uz + \frac{z^2}{1-z}\right)^{24}.$$
We need the total count of the number of adjacent pairs so we compute $$[z^{100}] \left.\frac{\partial}{\partial u} \frac{1}{1-z} G(z, u) \right|_{u=1} =[z^{100}] \left. \frac{z}{(1-z)^2} \times 24 \left(uz + \frac{z^2}{1-z}\right)^{23} \times z\right|_{u=1} \\ = [z^{100}] \frac{z}{(1-z)^2} \times 24 \left(\frac{z}{1-z}\right)^{23} \times z \\ = 24 [z^{100}] \frac{z^{25}}{(1-z)^{25}} = 24 [z^{75}] \frac{1}{(1-z)^{25}} = 24 {75+24\choose 24}.$$
This yields a final answer of $$24 {99\choose 24} \times {100\choose 25}^{-1} = 6.$$
• Can you compute expected number (or full distribution) of student "clumps" of size at least $i$ using this method? A clump is a maximal group of adjacent students. Assume $i=2$ for simplicity if needed.
– A.S.
Dec 8 '15 at 23:18
Suppose there are $N\gg 10$ seats (in a circle) and density of students is $p$ (not too large, not too small: $Np\gg r$). Then probability of $r\ll N$ seats next to each other being occupied is $\approx p^r$ - that is average density of such runs. However these runs clump together and a fresh run is followed by approximately $\frac p {1-p}$ more students for a total of $1+\frac p{1-p}=\frac {1}{1-p}$ runs clumped together. This yields density of maximal "clumps" of size at least $r$:
$$d={p^r}(1-p)$$
For $p=0.25, r=2$ we get $4.7\%$. Using exact value for density of runs of length $2$ ($6$) we'd get $4.5\%$ instead. Unlike runs - which clump together - clumps repel and hence their number has a smaller variance and is better for quick comparison with the mean value.
You can do the same computation for $r=1$ to get density of maximal clumps of size at least $1$ $p(1-p)=0.19$.
Suppose we ask about the expected maximum run length when we select $m$ elements from a total of $n.$
For a run length at most $k$ we must first choose a run of some length possibly preceded by a gap, giving (the variable $w$ counts the number of elements in the tuple)
$$\frac{1}{1-z} \sum_{q=1}^k w^q z^q$$
followed by a sequence of gaps followed by a run $$\frac{z}{1-z} \sum_{q=1}^k w^q z^q$$
possibly followed at the end by another gap $$\frac{1}{1-z}.$$
This gives the generating function $$f_k(z,w) = \frac{1}{(1-z)^2} \left(\sum_{q=1}^k w^q z^q \right) \sum_{p\ge 0} \left( \frac{z}{1-z} \sum_{q=1}^k w^q z^q \right)^p.$$
We are interested in the total count from
$$[z^n] [w^m] \sum_{k=1}^m k (f_k(z,w) - f_{k-1}(z, w)) \\ = [z^n] [w^m] \left(m f_m(z,w) - \sum_{k=1}^{m-1} f_k(z, w)\right).$$
The sum component in $f_k(z, w)$ is $$\sum_{p\ge 0} \left( \frac{wz^2}{1-z} \sum_{q=0}^{k-1} w^q z^q \right)^p \\ = \frac{1}{1-wz^2(1-w^k z^k)/(1-z)/(1-wz)}.$$
At this point it appears that we are better off working with
$$g_k(z,w) = \frac{1}{(1-wz)^2} \left(\sum_{q=1}^k z^q \right) \sum_{p\ge 0} \left( \frac{wz}{1-wz} \sum_{q=1}^k z^q \right)^p$$
which gives for the sum component $$\sum_{p\ge 0} \left( \frac{wz^2}{1-wz} \sum_{q=0}^{k-1} z^q \right)^p \\ = \frac{1}{1-wz^2(1-z^k)/(1-wz)/(1-z)}.$$
This yields $$g_k(z,w) = \frac{1}{1-wz} \left(\sum_{q=1}^k z^q \right) \frac{1}{1-wz-wz^2(1-z^k)/(1-z)} \\ = \frac{1}{1-wz} \left(\sum_{q=1}^k z^q \right) \frac{1}{1-w(z-z^2+z^2(1-z^k))/(1-z)} \\ = \frac{1}{1-wz} \left(\sum_{q=1}^k z^q \right) \frac{1}{1-w(z-z^{k+2}))/(1-z)}.$$
Extracting the coefficient on $[w^{n-m}]$ we get
$$\left(\sum_{q=1}^k z^q \right) \sum_{p=0}^{n-m} z^{n-m-p} \frac{z^p(1-z^{k+1})^p}{(1-z)^p} \\ = z^{n-m} \left(\sum_{q=1}^k z^q \right) \sum_{p=0}^{n-m} \frac{(1-z^{k+1})^p}{(1-z)^p} \\ = z^{n-m} \left(\left(\frac{1-z^{k+1}}{1-z}\right)^{n-m+1}-1\right).$$
Extracting the coefficient on $[z^n]$ we obtain $$[z^n] z^{n-m} \left(\frac{1-z^{k+1}}{1-z}\right)^{n-m+1} = [z^m] \left(\frac{1-z^{k+1}}{1-z}\right)^{n-m+1}.$$
This is $$\sum_{p=0}^{\lfloor m/(k+1)\rfloor} {n-m+1\choose p} (-1)^p {n-m+m-p(k+1)\choose n-m} \\ = \sum_{p=0}^{\lfloor m/(k+1)\rfloor} {n-m+1\choose p} (-1)^p {n-p(k+1)\choose n-m}.$$
Substitute this into the summation formula to obtain $$m {n\choose m} - \sum_{k=1}^{m-1} \sum_{p=0}^{\lfloor m/(k+1)\rfloor} {n-m+1\choose p} (-1)^p {n-p(k+1)\choose n-m} \\ = {n\choose m} - \sum_{k=1}^{m-1} \sum_{p=1}^{\lfloor m/(k+1)\rfloor} {n-m+1\choose p} (-1)^p {n-p(k+1)\choose n-m}.$$
This is $${n\choose m} + \sum_{p=1}^m {n-m+1\choose p} (-1)^p {n-p\choose n-m} \\ - \sum_{k=0}^{m-1} \sum_{p=1}^{\lfloor m/(k+1)\rfloor} {n-m+1\choose p} (-1)^p {n-p(k+1)\choose n-m} \\ = {n\choose m} + \sum_{p=1}^m {n-m+1\choose p} (-1)^p {n-p\choose n-m} \\ - \sum_{1\le pq \le m} {n-m+1\choose p} (-1)^p {n-pq\choose n-m}.$$
Now we do one last simplification on the first and the second term together which are $$\sum_{p=0}^m {n-m+1\choose p} (-1)^p {n-p\choose n-m}$$
and introduce for use with the Egorychev method $${n-p\choose n-m} = {n-p\choose m-p} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m-p+1}} (1+z)^{n-p} \; dz.$$
Note that this is zero when $p\gt m$ so we may extend $p$ to infinity, getting $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{n} \sum_{p\ge 0} {n-m+1\choose p} (-1)^p \frac{z^p}{(1+z)^p} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{n} \left(1-\frac{z}{1+z}\right)^{n-m+1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{n} \frac{1}{(1+z)^{n-m+1}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{m-1} \; dz = [z^m] (1+z)^{m-1} = 0.$$
Dividing by ${n\choose m}$ for the expectation and without the zero contribution from the first two terms we thus obtain for the expectation of the maximum run length
$${\large\color{#080}{{n\choose m}^{-1} \sum_{1\le pq \le m} {n-m+1\choose p} (-1)^{p+1} {n-pq\choose n-m}}}.$$
The reader is invited to prove this in a different way, perhaps without using ordinary generating functions.
The Maple code that was used to verify these including a total enumeration routine was as follows:
with(combinat);
sr :=
proc(n, m)
local sel, sel2, maxr, cur, res, pos;
sel := firstcomb(n, m);
res := 0;
while type(sel, set) do
sel2 := convert(sel, list);
maxr := -1; cur := 1;
for pos from 2 to m do
if sel2[pos] = 1 + sel2[pos-1] then
cur := cur + 1;
else
if cur > maxr then
maxr := cur;
fi;
cur := 1;
fi;
od;
if cur > maxr then
maxr := cur;
fi;
res := res + maxr;
sel := nextcomb(sel, n);
od;
res;
end;
f :=
proc(k)
local base;
1/(1-z)^2*base*sum((z/(1-z)*base)^p, p=0..infinity);
end;
Hf :=
proc(m)
end;
Qf :=
proc(n, m)
coeftayl(coeftayl(Hf(m), w=0, m), z=0, n);
end;
g :=
proc(k)
local base;
1/(1-w*z)^2*base*sum((w*z/(1-w*z)*base)^p, p=0..infinity);
end;
Hg :=
proc(m)
end;
Qg :=
proc(n, m)
coeftayl(coeftayl(Hg(m), w=0, n-m), z=0, n);
end;
X :=
proc(n, m)
local CF;
CF := (n,m,k)->
p=0..floor(m/(k+1)));
end;
X1 :=
proc(n, m)
local p, q, res;
res := binomial(n,m) +
for p to m do
for q to floor(m/p) do
res := res -
binomial(n-m+1,p)*(-1)^p*binomial(n-p*q,n-m);
od;
od;
res;
end;
X2 :=
proc(n, m)
local p, q, res;
res := 0;
for p to m do
for q to floor(m/p) do
res := res +
binomial(n-m+1,p)*(-1)^(p+1)*binomial(n-p*q,n-m);
od;
od;
res;
end; | 2021-12-08T13:46:23 | {
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https://math.stackexchange.com/questions/2322106/why-is-showing-that-fh-subseteq-h-the-same-as-showing-that-y-in-h-impli | # why is showing that $f(H) \subseteq H$ the same as showing that $(y \in H \implies f(y) \in H)$?
While I was reading a proof about the diagonalizability of symmetric matrices I got a bit lost the author was supposed to show that $f(H) \subseteq H$ but he ended up showing that
$y \in H \implies f(y) \in H$ for all $y$
claiming it's the same thing
I would say because $y$ is arbitrary here $f(y)$ is practically the same as $f(H)$ since $y \in H$ looking at it this way it makes some sense to me but I'm still confused why those two statements are equivalent.
Please could anybody here clearly explain why?
• I do believe that $f(A)$ for some set $A \subseteq D$ where $D$ is the domain of $f$ is defined as the set of all values of $f(a) \space \forall a \in A$. That would make the two statements equivalent. – Tim The Enchanter Jun 14 '17 at 10:00
• @TimTheEnchanter thank you ! now I get it. $f(H) = \{f(y) | \forall y \in H \}$ so showing that $y \in H \implies f(y) \in H$ means showing that every element of $f(H)$ is an element of $H$ which is equivalent to $f(H) \subseteq H$ – the_firehawk Jun 14 '17 at 10:09
Here's a systematic, "formal" derivation. The definition of $X \subseteq Y$ is $\forall x.x\in X \Rightarrow x\in Y$. The definition of $f(H)$ is $\{f(y)\mid y\in H\}$ or more explicitly $\{z\mid\exists y.y\in H\land z = f(y)\}$. We can now just calculate: \begin{align} f(H)\subseteq H & \iff \forall x. x\in f(H) \Rightarrow x\in H \\ & \iff \forall x. x\in \{f(y)\mid y\in H\} \Rightarrow x \in H \\ & \iff \forall x. (\exists y.y\in H\land x = f(y) ) \Rightarrow x \in H \\ & \iff \forall x. \forall y. (y\in H\land x = f(y)) \Rightarrow x \in H \\ & \iff \forall x. \forall y. y\in H \Rightarrow f(y) \in H \\ & \iff \forall y. y\in H \Rightarrow f(y) \in H \end{align}
I don't know why you wrote that “$f(y)$ is practically the same thing as $f(H)$ since $y\in H$”. It happens that $f(y)$ is the image of $y$, whereas $f(H)$ is the set of all images of all elements of $H$. So, yes, asserting that $f(H)\subseteq H$ actually is the same thing as asserting that $(\forall y\in H):f(y)\in H$.
Saying $f(H) \subseteq H$ is saying that $f$ takes each element of $H$ into $H$. Thus for all $y \in H$, $f(y) \in H$ and the statement $y \in H \Rightarrow f(y) \in H$ holds.
Conversely, if the implication $y \in H \Rightarrow f(y) \in H$ holds, $f(H) \subseteq H$ becuase by the implication $f(y) \in H$ for all $y \in H$.
Proposition 0. Given a function $f : X \rightarrow Y$ and subsets $A \subseteq X, B \subseteq Y$, the following are equivalent:
• $\forall x \in X(x \in A \rightarrow f(x) \in B)$
• $f(A) \subseteq B$
Proof.
The following are equivalent:
• $\forall x \in X(x \in A \rightarrow f(x) \in B)$
• $\forall x \in X(x \in A \rightarrow x \in f^{-1}(B))$
• $A \subseteq f^{-1}(B)$
• $f(A) \subseteq B$ | 2021-05-15T05:45:09 | {
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https://www.physicsforums.com/threads/substitution-rule-in-indefinite-integrals.417603/ | # Substitution Rule in Indefinite Integrals
1. Jul 22, 2010
### phillyolly
1. The problem statement, all variables and given/known data
Please explain how to use the substitution rule in indefinite integrals. I am unable even to start the problem.
2. Relevant equations
3. The attempt at a solution
#### Attached Files:
File size:
1.7 KB
Views:
134
2. Jul 22, 2010
### phillyolly
Actually, one second. I am solving it right now. Will post a solution.
3. Jul 22, 2010
### Staff: Mentor
Substitution is the reverse of the chain rule in differentiation. For example,
$$\frac{d}{dx}(x^3 - 2)^4~=~4(x^3 - 2)^3~3x^2$$
The corresponding indefinite integral is
$$\int 12x^2(x^3 - 2)^3~dx$$
Here, we let u = x3 - 2, so du = 3x2~dx, so
$$\int 12x^2(x^3 - 2)^3~dx~=~\int 4*3x^2~(x^3 - 2)^3~dx~=~\int 4 u^3 du~=~\frac{4u^4}{4} + C~=~(x^3 - 2)^4 + C$$
Can you see how this technique (it's not a rule) could be applied to your problem?
4. Jul 22, 2010
### phillyolly
Thank you for the helpful note. I did not know that substitution is reverse of chain rule. Here is my solution. Correct?
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9.4 KB
Views:
133
5. Jul 22, 2010
### Staff: Mentor
Correct as far as it goes. You still need to undo your substitution.
Also, u = x2 + 1 ==> du = 2xdx. You have x and dx, so you can get the 2 that's needed by multiplying by 2 and then multiplying on the outside of the integral by 1/2. That way, you're really multiplying by 1, which doesn't change the value of the integrand.
6. Jul 22, 2010
### phillyolly
Why do you put du=2xdx? I think we need dx=du/2x.
Here is the completed version.
#### Attached Files:
File size:
10.4 KB
Views:
128
7. Jul 22, 2010
### phillyolly
Can you check another problem please too? Thank you.
#### Attached Files:
File size:
8.3 KB
Views:
136
8. Jul 22, 2010
### Staff: Mentor
Sure, that works.
Another substitution that isn't so obvious is to let u = sec2t (using t for theta). Then du = 2sec(2t)tan(2t)dt
$$\int sec(2t)tan(2t)dt = \frac{1}{2}\int 2sec(2t)tan(2t)dt = \frac{1}{2}\int du= \frac{u}{2} + C = \frac{sec(2t)}{2} + C$$
9. Jul 22, 2010
### phillyolly
Thank you very much!
I am also stuck on this one. ex has been always tricky for me.
#### Attached Files:
File size:
5.7 KB
Views:
133
10. Jul 22, 2010
### phillyolly
OH! This one is much prettier than mine!
11. Jul 22, 2010
### Staff: Mentor
This will work, but will require another substitution.
Here's a different approach: u = ex + 1, du = exdx
$$\int \frac{e^x~dx}{e^x + 1}=\int \frac{du}{u}= ln(u) + C = ln(e^x+ 1) + C$$
Since u = ex + 1 > 0 for all real x, I didn't need to include absolute values on the argument to the natural log function.
12. Jul 22, 2010
### phillyolly
Thank you a lot, Mark44! | 2018-03-19T21:39:10 | {
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https://math.stackexchange.com/questions/1003823/question-on-modulo | # Question on modulo
Find the last two digits of $3^{2002}$.
How should I approach this question using modulo? I obtained 09 as my answer however the given answer was 43.
My method was as follows:
$2002\:=\:8\cdot 250+2$
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$
The last two digits is the remainder when divided by 100. Thus we need to compute the sum mod 100.
$3^8\equiv 61\:\left(mod\:100\right)$
Thus,
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$ $\equiv 61^{250}\cdot 9\:\equiv 61^{5\cdot 50}\cdot 9\equiv \:9\:\left(mod\:100\right)$
Therefore my answer would be 09. Am I doing it the right way?
• Since $3^{2002}=9^{1001}$ and the powers of $9$ end in $1$ or $9$ the answer cannot be $43$ if the base is $10$. – Mark Bennet Nov 3 '14 at 7:37
• Yes Your answer is correct – user171358 Nov 3 '14 at 7:41
• last two digits of $3^{20}$ are $01$ – user171358 Nov 3 '14 at 7:43
Observe that $\text{gcd(3,100)} = 1$, by Euler totient formula: $3^{\phi(100)} \cong 1(\mod 100)$. But $\phi (100) = 100\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{5}\right) = 40$. Thus: $3^{40} \cong 1(\mod 100)$. So $3^{2002} = (3^{40})^{50}\cdot 3^2 \cong 1\cdot 9(\mod 100) \cong 9(\mod 100)$. Thus the remainder is $9$ as claimed.
We have $3^{2000}=(10-1)^{1000}$. By the Binomial Theorem, this is congruent to $1$ modulo $10000$. It follows that $3^{2002}\equiv 9\pmod{10000}$. Thus the last four digits are $0009$.
Using Carmichael function, $\lambda(100)=20$
As $(3,100)=1$ and $2002\equiv2\pmod{20},$ $$\implies3^{2002}\equiv3^2\pmod{100}$$
$$3^{2002}=(3^2)^{1001}=9^{1001}=(10-1)^{10001}$$
$$=-1+\binom{1001}110\pmod{100}\equiv10010-1\equiv9$$ | 2019-10-22T21:06:38 | {
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http://kaiz.emporiumgalorium.de/andreescu-and-gelca-mathematical-olympiad-challenges-pdf.html | .article-body a.abonner-button:hover{color:#FFF!important;}
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He is past chairman of the USA Mathematical Olympiad, served as director of the MAA American Mathematics Competitions (1998–2003), coach of the USA International Mathematical Olympiad Team (IMO) for 10 years (1993– 2002), director of the Mathematical Olympiad Summer Program (1995–2002), and leader of the USA IMO Team (1995–2002). I've been fortunate enough to find two remarkable problems, and solve them in ways that form a whirlwind tour of the subject. Functional Equations - B. ANDREESCU, T. Download Citation on ResearchGate | Mathematical Olympiad Challenges / T. Nine students are positioned in three rows each containing three stu-dents. Định lý Pompiu là một định lý trong lĩnh vực hình học phẳng, được phát hiện bởi nhà toán học Romanian Dimitrie Pompiu. Mathematical Olympiad Treasures aims at building a bridge between ordinary high school exercises and more sophisticated, intricate and abstract concepts in undergraduate mathematics. 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Available in PDF, ePub and Kindle. Description : Mathematical Olympiad Challenges is a rich collection of problems put together by two experienced and well-known professors and coaches of the U. XYZ Press (2010). Titu Andreescu and Räzvan Gelca xiii I Problems 1 1 Geometry and Trigonometry 3 1. This is an additional collection of problems and solutions for problems at the High School Olympiad level. Applied mathematics N3446* 9783642451270 A Brief History of String Theory: From Dual Models to M-Theory Dean Rickles 39. "Mathematical Olympiad Challenges" is a rich collection of problems put together by two experienced and well-known professors and coaches of the U. International Mathematical Olympiad Team. Putnam and Beyond is organized for independent study by undergraduate and graduate students, as well as teachers and researchers in the physical sciences who wish to expand their mathematical. We quote from the Preface: "The problems we selected are definitely not exercises". 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Mathematical Olympiad Challenges (Paperback) by Titu Andreescu, Razvan Gelca ook Summary of Mathematical Olympiad Challenges The Second Edition of this popular text includes 400 new beautiful, challenging,. com has become a leading book price comparison site: Find and compare hundreds of millions of new books, used books, rare books and out of print books from over 100,000 booksellers and 60+ websites worldwide. Andreescu and R. As I became more involved in coaching students for the Putnam and the International Mathematical Olympiad, I started to understand why there is not much training mater-ial available in systematic form. The Paperback of the Mathematical Olympiad Challenges by Titu Andreescu, Razvan Gelca | at Barnes & Noble. Mathematical Olympiad Challenges offers a rich collection of problems assembled by coaches of the U. Request PDF on ResearchGate | On Jan 1, 2004, Michael De Villiers and others published Mathematical Olympiad Challenges by Titu Andreescu; Razvan Gelca. Mathematical Olympiad Challenges - Titu Andreescu, Razvan Gelca. For example,. Problem solving usually involves elementary mathematics; this does not mean "easy mathematics". This work is an excellent problem-solving book addressed to everyone which loves beautiful "elementary" problems. 26 KB] Mathematical Olympiad Problems & Solutions From Around the World [1997-98]. | 2020-01-25T20:44:17 | {
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https://dantopology.wordpress.com/category/compact-space/ | # Pseudocompact spaces with regular G-delta diagonals
This post complements two results discussed in two previous blog posts concerning $G_\delta$-diagonal. One result is that any compact space with a $G_\delta$-diagonal is metrizable (see here). The other result is that the compactness in the first result can be relaxed to countably compactness. Thus any countably compact space with a $G_\delta$-diagonal is metrizable (see here). The countably compactness in the second result cannot be relaxed to pseudocompactness. The Mrowka space is a pseudocompact space with a $G_\delta$-diagonal that is not submetrizable, hence not metrizable (see here). However, if we strengthen the $G_\delta$-diagonal to a regular $G_\delta$-diagonal while keeping pseudocompactness fixed, then we have a theorem. We prove the following theorem.
Theorem 1
If the space $X$ is pseudocompact and has a regular $G_\delta$-diagonal, then $X$ is metrizable.
All spaces are assumed to be Hausdorff and completely regular. The assumption of completely regular is crucial. The proof of Theorem 1 relies on two lemmas concerning pseudocompact spaces (one proved in a previous post and one proved here). These two lemmas work only for completely regular spaces.
The proof of Theorem 1 uses a metrization theorem. The best metrization to use in this case is Moore metrization theorem (stated below). The result in Theorem 1 is found in [2].
First some basics. Let $X$ be a space. The diagonal of the space $X$ is the set $\Delta=\{ (x,x): x \in X \}$. When the diagonal $\Delta$, as a subset of $X \times X$, is a $G_\delta$-set, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$, the space $X$ is said to have a $G_\delta$-diagonal.
The space $X$ is said to have a regular $G_\delta$-diagonal if the diagonal $\Delta$ is a regular $G_\delta$-set in $X \times X$, i.e. $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$. If $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$, then $\Delta=\bigcap_{n=1}^\infty \overline{U_n}=\bigcap_{n=1}^\infty U_n$. Thus if a space has a regular $G_\delta$-diagonal, it has a $G_\delta$-diagonal. We will see that there exists a space with a $G_\delta$-diagonal that fails to be a regular $G_\delta$-diagonal.
The space $X$ is a pseudocompact space if for every continuous function $f:X \rightarrow \mathbb{R}$, the image $f(X)$ is a bounded set in the real line $\mathbb{R}$. Pseudocompact spaces are discussed in considerable details in this previous post. We will rely on results from this previous post to prove Theorem 1.
The following lemma is used in proving Theorem 1.
Lemma 2
Let $X$ be a pseudocompact space. Suppose that $O_1,O_2,O_2,\cdots$ is a decreasing sequence of non-empty open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$ for some point $x \in X$. Then $\{ O_n \}$ is a local base at the point $x$.
Proof of Lemma 2
Let $O_1,O_2,O_2,\cdots$ be a decreasing sequence of open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$. Let $U$ be open in $X$ with $x \in U$. If $O_n \subset U$ for some $n$, then we are done. Suppose that $O_n \not \subset U$ for each $n$.
Choose open $V$ with $x \in V \subset \overline{V} \subset U$. Consider the sequence $\{ O_n \cap (X-\overline{V}) \}$. This is a decreasing sequence of non-empty open subsets of $X$. By Theorem 2 in this previous post, $\bigcap \overline{O_n \cap (X-\overline{V})} \ne \varnothing$. Let $y$ be a point in this non-empty set. Note that $y \in \bigcap_{n=1}^\infty \overline{O_n}$. This means that $y=x$. Since $x \in \overline{O_n \cap (X-\overline{V})}$ for each $n$, any open set containing $x$ would contain a point not in $\overline{V}$. This is a contradiction since $x \in V$. Thus it must be the case that $x \in O_n \subset U$ for some $n$. $\square$
The following metrization theorem is useful in proving Theorem 1.
Theorem 3 (Moore Metrization Theorem)
Let $X$ be a space. Then $X$ is metrizable if and only if the following condition holds.
There exists a decreasing sequence $\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\cdots$ of open covers of $X$ such that for each $x \in X$, the sequence $\{ St(St(x,\mathcal{B}_n),\mathcal{B}_n):n=1,2,3,\cdots \}$ is a local base at the point $x$.
For any family $\mathcal{U}$ of subsets of $X$, and for any $A \subset X$, the notation $St(A,\mathcal{U})$ refers to the set $\cup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$. In other words, it is the union of all sets in $\mathcal{U}$ that contain points of $A$. The set $St(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the family $\mathcal{U}$. If $A=\{ x \}$, we write $St(x,\mathcal{U})$ instead of $St(\{ x \},\mathcal{U})$. The set $St(St(x,\mathcal{B}_n),\mathcal{B}_n)$ indicated in Theorem 3 is the star of the set $St(x,\mathcal{B}_n)$ with respect to the open cover $\mathcal{B}_n$.
Theorem 3 follows from Theorem 1.4 in [1], which states that for any $T_0$-space $X$, $X$ is metrizable if and only if there exists a sequence $\mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3,\cdots$ of open covers of $X$ such that for each open $U \subset X$ and for each $x \in U$, there exist an open $V \subset X$ and an integer $n$ such that $x \in V$ and $St(V,\mathcal{G}_n) \subset U$.
Proof of Theorem 1
Suppose $X$ is pseudocompact such that its diagonal $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$. We can assume that $U_1 \supset U_2 \supset \cdots$. For each $n \ge 1$, define the following:
$\mathcal{U}_n=\{ U \subset X: U \text{ open in } X \text{ and } U \times U \subset U_n \}$
Note that each $\mathcal{U}_n$ is an open cover of $X$. Also note that $\{ \mathcal{U}_n \}$ is a decreasing sequence since $\{ U_n \}$ is a decreasing sequence of open sets. We show that $\{ \mathcal{U}_n \}$ is a sequence of open covers of $X$ that satisfies Theorem 3. We establish this by proving the following claims.
Claim 1. For each $x \in X$, $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$.
To prove the claim, let $x \ne y$. There is an integer $n$ such that $(x,y) \notin \overline{U_n}$. Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$. Note that $(x,y) \notin U_k$ and $(U \times V) \cap U_n=\varnothing$.
We want to show that $V \cap St(x,\mathcal{U}_n)=\varnothing$, which implies that $y \notin \overline{St(x,\mathcal{U}_n)}$. Suppose $V \cap St(x,\mathcal{U}_n) \ne \varnothing$. This means that $V \cap W \ne \varnothing$ for some $W \in \mathcal{U}_n$ with $x \in W$. Then $(U \times V) \cap (W \times W) \ne \varnothing$. Note that $W \times W \subset U_n$. This implies that $(U \times V) \cap U_n \ne \varnothing$, a contradiction. Thus $V \cap St(x,\mathcal{U}_n)=\varnothing$. Since $y \in V$, $y \notin \overline{St(x,\mathcal{U}_n)}$. We have established that for each $x \in X$, $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$.
Claim 2. For each $x \in X$, $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$.
Note that $\{ St(x,\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$. By Lemma 2, $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$.
Claim 3. For each $x \in X$, $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}=\{ x \}$.
Let $x \ne y$. There is an integer $n$ such that $(x,y) \notin \overline{U_n}$. Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$. It follows that $(U \times V) \cap \overline{U_t}=\varnothing$ for all $t \ge n$. Furthermore, $(U \times V) \cap U_t=\varnothing$ for all $t \ge n$. By Claim 2, choose integers $i$ and $j$ such that $St(x,\mathcal{U}_i) \subset U$ and $St(y,\mathcal{U}_j) \subset V$. Choose an integer $k \ge \text{max}(n,i,j)$. It follows that $(St(x,\mathcal{U}_i) \times St(y,\mathcal{U}_j)) \cap U_k=\varnothing$. Since $\mathcal{U}_k \subset \mathcal{U}_i$ and $\mathcal{U}_k \subset \mathcal{U}_j$, it follows that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k=\varnothing$.
We claim that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$. Suppose not. Choose $w \in St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)$. It follows that $w \in B$ for some $B \in \mathcal{U}_k$ such that $B \cap St(x,\mathcal{U}_k) \ne \varnothing$ and $B \cap St(y,\mathcal{U}_k) \ne \varnothing$. Furthermore $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap (B \times B)=\varnothing$. Note that $B \times B \subset U_k$. This means that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k \ne \varnothing$, contradicting the fact observed in the preceding paragraph. It must be the case that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$.
Because there is an open set containing $y$, namely $St(y,\mathcal{U}_k)$, that contains no points of $St(St(x,\mathcal{U}_k), \mathcal{U}_k)$, $y \notin \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}$. Thus Claim 3 is established.
Claim 4. For each $x \in X$, $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n)) \}$ is a local base at the point $x$.
Note that $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n))}=\{ x \}$. By Lemma 2, $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a local base at the point $x$.
In conclusion, the sequence $\mathcal{U}_1,\mathcal{U}_2,\mathcal{U}_3,\cdots$ of open covers satisfies the properties in Theorem 3. Thus any pseudocompact space with a regular $G_\delta$-diagonal is metrizable. $\square$
Example
Any submetrizable space has a $G_\delta$-diagonal. The converse is not true. A classic example of a non-submetrizable space with a $G_\delta$-diagonal is the Mrowka space (discussed here). The Mrowka space is also called the psi-space since it is sometimes denoted by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal family of almost disjoint subsets of $\omega$. Actually $\Psi(\mathcal{A})$ would be a family of spaces since $\mathcal{A}$ is any maximal almost disjoint family. For any maximal $\mathcal{A}$, $\Psi(\mathcal{A})$ is a pseudocompact non-submetrizable space that has a $G_\delta$-diagonal. This example shows that the requirement of a regular $G_\delta$-diagonal in Theorem 1 cannot be weakened to a $G_\delta$-diagonal. See here for a more detailed discussion of this example.
Reference
1. Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
2. McArthur W. G., $G_\delta$-Diagonals and Metrization Theorems, Pacific Journal of Mathematics, Vol. 44, No. 2, 613-317, 1973.
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# Looking for spaces in which every compact subspace is metrizable
Once it is known that a topological space is not metrizable, it is natural to ask, from a metrizability standpoint, which subspaces are metrizable, e.g. whether every compact subspace is metrizable. This post discusses several classes of spaces in which every compact subspace is metrizable. Though the goal here is not to find a complete characterization of such spaces, this post discusses several classes of spaces and various examples that have this property. The effort brings together many interesting basic and well known facts. Thus the notion “every compact subspace is metrizable” is an excellent learning opportunity.
Several Classes of Spaces
The notion “every compact subspace is metrizable” is a very broad class of spaces. It includes well known spaces such as Sorgenfrey line, Michael line and the first uncountable ordinal $\omega_1$ (with the order topology) as well as Moore spaces. Certain function spaces are in the class “every compact subspace is metrizable”. The following diagram is a good organizing framework.
\displaystyle \begin{aligned} &1. \ \text{Metrizable} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&2. \ \text{Submetrizable} \Longleftarrow 5. \ \exists \ \text{countable network} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&3. \ \exists \ G_\delta \text{ diagonal} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&4. \ \text{Every compact subspace is metrizable} \end{aligned}
Let $(X, \tau)$ be a space. It is submetrizable if there is a topology $\tau_1$ on the set $X$ such that $\tau_1 \subset \tau$ and $(X, \tau_1)$ is a metrizable space. The topology $\tau_1$ is said to be weaker (coarser) than $\tau$. Thus a space $X$ is submetrizable if it has a weaker metrizable topology.
Let $\mathcal{N}$ be a set of subsets of the space $X$. $\mathcal{N}$ is said to be a network for $X$ if for every open subset $O$ of $X$ and for each $x \in O$, there exists $N \in \mathcal{N}$ such that $x \in N \subset O$. Having a network that is countable in size is a strong property (see here for a discussion on spaces with a countable network).
The diagonal of the space $X$ is the subset $\Delta=\left\{(x,x): x \in X \right\}$ of the square $X \times X$. The space $X$ has a $G_\delta$-diagonal if $\Delta$ is a $G_\delta$-subset of $X \times X$, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$.
The implication $1 \Longrightarrow 2$ is clear. For $5 \Longrightarrow 2$, see Lemma 1 in this previous post on countable network. The implication $2 \Longrightarrow 3$ is left as an exercise. To see $3 \Longrightarrow 4$, let $K$ be a compact subset of $X$. The property of having a $G_\delta$-diagonal is hereditary. Thus $K$ has a $G_\delta$-diagonal. According to a well known result, any compact space with a $G_\delta$-diagonal is metrizable (see here).
None of the implications in the diagram is reversible. The first uncountable ordinal $\omega_1$ is an example for $4 \not \Longrightarrow 3$. This follows from the well known result that any countably compact space with a $G_\delta$-diagonal is metrizable (see here). The Mrowka space is an example for $3 \not \Longrightarrow 2$ (see here). The Sorgenfrey line is an example for both $2 \not \Longrightarrow 5$ and $2 \not \Longrightarrow 1$.
To see where the examples mentioned earlier are placed, note that Sorgenfrey line and Michael line are submetrizable, both are submetrizable by the usual Euclidean topology on the real line. Each compact subspace of the space $\omega_1$ is countable and is thus contained in some initial segment $[0,\alpha]$ which is metrizable. Any Moore space has a $G_\delta$-diagonal. Thus compact subspaces of a Moore space are metrizable.
Function Spaces
We now look at some function spaces that are in the class “every compact subspace is metrizable.” For any Tychonoff space (completely regular space) $X$, $C_p(X)$ is the space of all continuous functions from $X$ into $\mathbb{R}$ with the pointwise convergence topology (see here for basic information on pointwise convergence topology).
Theorem 1
Suppose that $X$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.
Proof
The proof here actually shows more than is stated in the theorem. We show that $C_p(X)$ is submetrizable by a separable metric topology. Let $Y$ be a countable dense subspace of $X$. Then $C_p(Y)$ is metrizable and separable since it is a subspace of the separable metric space $\mathbb{R}^{\omega}$. Thus $C_p(Y)$ has a countable base. Let $\mathcal{E}$ be a countable base for $C_p(Y)$.
Let $\pi:C_p(X) \longrightarrow C_p(Y)$ be the restriction map, i.e. for each $f \in C_p(X)$, $\pi(f)=f \upharpoonright Y$. Since $\pi$ is a projection map, it is continuous and one-to-one and it maps $C_p(X)$ into $C_p(Y)$. Thus $\pi$ is a continuous bijection from $C_p(X)$ into $C_p(Y)$. Let $\mathcal{B}=\left\{\pi^{-1}(E): E \in \mathcal{E} \right\}$.
We claim that $\mathcal{B}$ is a base for a topology on $C_p(X)$. Once this is established, the proof of the theorem is completed. Note that $\mathcal{B}$ is countable and elements of $\mathcal{B}$ are open subsets of $C_p(X)$. Thus the topology generated by $\mathcal{B}$ is coarser than the original topology of $C_p(X)$.
For $\mathcal{B}$ to be a base, two conditions must be satisfied – $\mathcal{B}$ is a cover of $C_p(X)$ and for $B_1,B_2 \in \mathcal{B}$, and for $f \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ such that $f \in B_3 \subset B_1 \cap B_2$. Since $\mathcal{E}$ is a base for $C_p(Y)$ and since elements of $\mathcal{B}$ are preimages of elements of $\mathcal{E}$ under the map $\pi$, it is straightforward to verify these two points. $\square$
Theorem 1 is actually a special case of a duality result in $C_p$ function space theory. More about this point later. First, consider a corollary of Theorem 1.
Corollary 2
Let $X=\prod_{\alpha where $c$ is the cardinality continuum and each $X_\alpha$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.
The key fact for Corollary 2 is that the product of continuum many separable spaces is separable (this fact is discussed here). Theorem 1 is actually a special case of a deep result.
Theorem 3
Suppose that $X=\prod_{\alpha<\kappa} X_\alpha$ is a product of separable spaces where $\kappa$ is any infinite cardinal. Then every compact subspace of $C_p(X)$ is metrizable.
Theorem 3 is a much more general result. The product of any arbitrary number of separable spaces is not separable if the number of factors is greater than continuum. So the proof for Theorem 1 will not work in the general case. This result is Problem 307 in [2].
A Duality Result
Theorem 1 is stated in a way that gives the right information for the purpose at hand. A more correct statement of Theorem 1 is: $X$ is separable if and only if $C_p(X)$ is submetrizable by a separable metric topology. Of course, the result in the literature is based on density and weak weight.
The cardinal function of density is the least cardinality of a dense subspace. For any space $Y$, the weight of $Y$, denoted by $w(Y)$, is the least cardinaility of a base of $Y$. The weak weight of a space $X$ is the least $w(Y)$ over all space $Y$ for which there is a continuous bijection from $X$ onto $Y$. Thus if the weak weight of $X$ is $\omega$, then there is a continuous bijection from $X$ onto some separable metric space, hence $X$ has a weaker separable metric topology.
There is a duality result between density and weak weight for $X$ and $C_p(X)$. The duality result:
The density of $X$ coincides with the weak weight of $C_p(X)$ and the weak weight of $X$ coincides with the density of $C_p(X)$. These are elementary results in $C_p$-theory. See Theorem I.1.4 and Theorem I.1.5 in [1].
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Reference
1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.
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$\copyright$ 2017 – Dan Ma
# An exercise gleaned from the proof of a theorem on pseudocompact space
Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].
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Pseudocompactness
The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.
All spaces considered are Hausdorff spaces. A space $X$ is a pseudocompact space if every continuous real-valued function defined on $X$ is bounded, i.e., if $f:X \rightarrow \mathbb{R}$ is a continuous function, then $f(X)$ is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that
$\text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}$
None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).
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The exercise
Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.
The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.
Theorem 1
Let $Y$ be a compact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto a space $Z$. Then $g$ is a homeomorphism.
Theorem 2
Let $Y$ be a pseudocompact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto $Z$ where $Z$ is a separable and metrizable space. Then $g$ is a homeomorphism.
Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.
The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map $g$ as described in Theorem 2, the domain $Y$ must be compact. Then Theorem 1 will finish the job.
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Reference
1. Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).
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$\copyright \ 2015 \text{ by Dan Ma}$
# Looking for non-normal subspaces of the square of a compact X
A theorem of Katetov states that if $X$ is compact with a hereditarily normal cube $X^3$, then $X$ is metrizable (discussed in this previous post). This means that for any non-metrizable compact space $X$, Katetov’s theorem guarantees that some subspace of the cube $X^3$ is not normal. Where can a non-normal subspace of $X^3$ be found? Is it in $X$, in $X^2$ or in $X^3$? In other words, what is the “dimension” in which the hereditary normality fails for a given compact non-metrizable $X$ (1, 2 or 3)? Katetov’s theorem guarantees that the dimension must be at most 3. Out of curiosity, we gather a few compact non-metrizable spaces. They are discussed below. In this post, we motivate an independence result using these examples.
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Katetov’s theorems
First we state the results of Katetov for reference. These results are proved in this previous post.
Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:
• The factor $X$ is perfectly normal.
• Every countable and infinite subset of the factor $Y$ is closed.
Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.
Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.
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Examples of compact non-metrizable spaces
The set-theoretic result presented here is usually motivated by looking at Theorem 3. The question is: Can $X^3$ in Theorem 3 be replaced by $X^2$? We take a different angle of looking at some standard compact non-metric spaces and arrive at the same result. The following is a small listing of compact non-metrizable spaces. Each example in this list is defined in ZFC alone, i.e. no additional axioms are used beyond the generally accepted axioms of set theory.
1. One-point compactification of the Tychonoff plank.
2. One-point compactification of $\psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$.
3. The first compact uncountable ordinal, i.e. $\omega_1+1$.
4. The one-point compactification of an uncountable discrete space.
5. Alexandroff double circle.
6. Double arrow space.
7. Unit square with the lexicographic order.
Since each example in the list is compact and non-metrizable, the cube of each space must not be hereditarily normal according to Theorem 3 above. Where does the hereditary normality fail? For #1 and #2, $X$ is a compactification of a non-normal space and thus not hereditarily normal. So the dimension for the failure of hereditary normality is 1 for #1 and #2.
For #3 through #7, $X$ is hereditarily normal. For #3 through #5, each $X$ has a closed subset that is not a $G_\delta$ set (hence not perfectly normal). In #3 and #4, the non-$G_\delta$-set is a single point. In #5, the the non-$G_\delta$-set is the inner circle. Thus the compact space $X$ in #3 through #5 is not perfectly normal. By Theorem 2, the dimension for the failure of hereditary normality is 2 for #3 through #5.
For #6 and #7, each $X^2$ contains a copy of the Sorgenfrey plane. Thus the dimension for the failure of hereditary normality is also 2 for #6 and #7.
In the small sample of compact non-metrizable spaces just highlighted, the failure of hereditary normality occurs in “dimension” 1 or 2. Naturally, one can ask:
Question. Is there an example of a compact non-metrizable space $X$ such that the failure of hereditary nornmality occurs in “dimension” 3? Specifically, is there a compact non-metrizable $X$ such that $X^2$ is hereditarily normal but $X^3$ is not hereditarily normal?
Such a space $X$ would be an example to show that the condition “$X^3$ is hereditarily normal” in Theorem 3 is necessary. In other words, the hypothesis in Theorem 3 cannot be weakened if the example just described were to exist.
The above list of compact non-metrizable spaces is a small one. They are fairly standard examples for compact non-metrizable spaces. Could there be some esoteric example out there that fits the description? It turns out that there are such examples. In [1], Gruenhage and Nyikos constructed a compact non-metrizable $X$ such that $X^2$ is hereditarily normal. The construction was done using MA + not CH (Martin’s Axiom coupled with the negation of the continuum hypothesis). In that same paper, they also constructed another another example using CH. With the examples from [1], one immediate question was whether the additional set-theoretic axioms of MA + not CH (or CH) was necessary. Could a compact non-metrizable $X$ such that $X^2$ is hereditarily normal be still constructed without using any axioms beyond ZFC, the generally accepted axioms of set theory? For a relatively short period of time, this was an open question.
In 2001, Larson and Todorcevic [3] showed that it is consistent with ZFC that every compact $X$ with hereditarily normal $X^2$ is metrizable. In other words, there is a model of set theory that is consistent with ZFC in which Theorem 3 can be improved to assuming $X^2$ is hereditarily normal. Thus it is impossible to settle the above question without assuming additional axioms beyond those of ZFC. This means that if a compact non-metrizable $X$ is constructed without using any axiom beyond ZFC (such as those in the small list above), the hereditary normality must fail at dimension 1 or 2. Numerous other examples can be added to the above small list. Looking at these ZFC examples can help us appreciate the results in [1] and [3]. These ZFC examples are excellent training ground for general topology.
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Reference
1. Gruenhage G., Nyikos P. J., Normality in $X^2$ for Compact $X$, Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
2. Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
3. Larson P., Todorcevic S., KATETOV’S PROBLEM, Trans. Amer. Math. Soc., Vol 354, No 5 (2001), 1783-1791
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$\copyright \ 2015 \text{ by Dan Ma}$
# When a product space is hereditarily normal
When the spaces $X$ and $Y$ are normal spaces, the product space $X \times Y$ is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors $X$ and $Y$ can have no influence on the normality of the product $X \times Y$. The dynamics in the other direction are totally different. When the product $X \times Y$ is hereditarily normal, the two factors $X$ and $Y$ are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.
A subset $W$ of a space $X$ is said to be a $G_\delta$-set in $X$ if $W$ is the intersection of countably many open subsets of $X$. A space $X$ is perfectly normal if it is normal and that every closed subset of $X$ is a $G_\delta$-set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product $X \times Y$ will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product $X \times Y$ is a very strong property.
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Katetov’s theorems
Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:
• The factor $X$ is perfectly normal.
• Every countable and infinite subset of the factor $Y$ is closed.
Proof of Theorem 1
The strategy we use is to define a subspace of $X \times Y$ that is not normal after assuming that none of the two conditions is true. So assume that $X$ has a closed subspace $W$ that is not a $G_\delta$-set and assume that $T=\left\{t_n: n=1,2,3,\cdots \right\}$ is an infinite subset of $Y$ that is not closed. Let $p \in Y$ be a limit point of $T$ such that $p \notin T$. The candidate for a non-normal subspace of $X \times Y$ is:
$M=X \times Y-W \times \left\{p \right\}$
Note that $M$ is an open subspace of $X \times Y$ since it is the result of subtracting a closed set from $X \times Y$. The following are the two closed sets that demonstrate that $M$ is not normal.
$H=W \times (Y-\left\{p \right\})$
$K=(X-W) \times \left\{p \right\}$
It is clear that $H$ and $K$ are closed subsets of $M$. Let $U$ and $V$ be open subsets of $M$ such that $H \subset U$ and $K \subset V$. We show that $U \cap V \ne \varnothing$. To this end, define $U_j=\left\{x \in X: (x,t_j) \in U \right\}$ for each $j$. It follows that for each $j$, $W \subset U_j$. Furthermore each $U_j$ is an open subspace of $X$. Thus $W \subset \bigcap_j U_j$. Since $W$ is not a $G_\delta$-set in $X$, there must exist $t \in \bigcap_j U_j$ such that $t \notin W$. Then $(t, p) \in K$ and $(t, p) \in V$.
Since $V$ is open in the product $X \times Y$, choose open sets $A \subset X$ and $B \subset Y$ such that $(t,p) \in A \times B$ and $A \times B \subset V$. With $p \in B$, there exists some $j$ such that $t_j \in B$. First, $(t,t_j) \in V$. Since $t \in U_j$, $(t,t_j) \in U$. Thus $U \cap V \ne \varnothing$. This completes the proof that the subspace $M$ is not normal and that $X \times Y$ is not hereditarily normal. $\blacksquare$
Let’s see what happens in Theorem 1 when both factors are compact. If both $X$ and $Y$ are compact and if $X \times Y$ is hereditarily normal, then both $X$ and $Y$ must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.
Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.
Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.
Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.
Proof of Theorem 3
Suppose that $X^3$ is hereditarily normal. By Theorem 2, the compact spaces $X^2$ and $X$ are perfectly normal. In particular, the following subset of $X^2$ is a $G_\delta$-set in $X^2$.
$\Delta=\left\{(x,x): x \in X \right\}$
The set $\Delta$ is said to be the diagonal of the space $X$. It is a well known result that any compact space whose diagonal is a $G_\delta$-set in the square is metrizable (discussed here). $\blacksquare$
The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of $X^2$. This was indeed posted by Katetov himself. This leads to the discussion in the next post.
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Reference
1. Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
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$\copyright \ 2015 \text{ by Dan Ma}$
# Compact metrizable scattered spaces
A scattered space is one in which there are isolated points found in every subspace. Specifically, a space $X$ is a scattered space if every non-empty subspace $Y$ of $X$ has a point $y \in Y$ such that $y$ is an isolated point in $Y$, i.e. the singleton set $\left\{y \right\}$ is open in the subspace $Y$. A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.
Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:
Thereom 1
For any compact space $X$, the following conditions are equivalent:
• The function space $C_p(X)$ is a Frechet-Urysohn space.
• The function space $C_p(X)$ is a k space.
• $X$ is a scattered space.
Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space $X$, let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$. Since $C(X)$ is a subspace of the product space $\mathbb{R}^X$, a natural topology that can be given to $C(X)$ is the subspace topology inherited from the product space $\mathbb{R}^X$. Then $C_p(X)$ is simply the set $C(X)$ with the product subspace topology (also called the pointwise convergence topology).
Let’s say the compact space $X$ is countable and infinite. Then the function space $C_p(X)$ is metrizable since it is a subspace of $\mathbb{R}^X$, a product of countably many lines. Thus the function space $C_p(X)$ has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space $X$ is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.
The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).
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Compact metrizable spaces
We first define some notions before looking at compact metrizable spaces in more details. Let $X$ be a space. Let $A \subset X$. Let $p \in X$. We say that $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$. So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.
Theorem 2
Let $X$ be a hereditarily Lindelof space (i.e. every subspace of $X$ is Lindelof). Then for any uncountable subset $A$ of $X$, all but countably many points of $A$ are limit points of $A$.
We now discuss the main result.
Theorem 3
Let $X$ be a compact metrizable space such that every point of $X$ is a limit point of $X$. Then there exists an uncountable closed subset $C$ of $X$ such that every point of $C$ is a limit point of $C$.
Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric $\rho$ on the space $X$. One fact that we will use is that if there is a sequence of closed sets $X \supset H_1 \supset H_2 \supset H_3 \supset \cdots$ such that the diameters of the sets $H$ (based on the complete metric $\rho$) decrease to zero, then the sets $H_n$ collapse to one point.
The uncountable closed set $C$ we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points $p_0,p_1 \in X$ such that $p_0 \ne p_1$. By assumption, both points are limit points of the space $X$. Choose open sets $U_0,U_1 \subset X$ such that
• $p_0 \in U_0$,
• $p_1 \in U_1$,
• $K_0=\overline{U_0}$ and $K_1=\overline{U_1}$,
• $K_0 \cap K_1 = \varnothing$,
• the diameters for $K_0$ and $K_1$ with respect to $\rho$ are less than 0.5.
Note that each of these open sets contains infinitely many points of $X$. Then we can pick two points in each of $U_0$ and $U_1$ in the same manner. Before continuing, we set some notation. If $\sigma$ is an ordered string of 0’s and 1’s of length $n$ (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus $\sigma$ is extended as $\sigma 0$ and $\sigma 1$ (e.g. 01101 is extended by 011010 and 011011).
Suppose that the construction at the $n$th stage where $n \ge 1$ is completed. This means that the points $p_\sigma$ and the open sets $U_\sigma$ have been chosen such that $p_\sigma \in U_\sigma$ for each length $n$ string of 0’s and 1’s $\sigma$. Now we continue the picking for the $(n+1)$st stage. For each $\sigma$, an $n$-length string of 0’s and 1’s, choose two points $p_{\sigma 0}$ and $p_{\sigma 1}$ and choose two open sets $U_{\sigma 0}$ and $U_{\sigma 1}$ such that
• $p_{\sigma 0} \in U_{\sigma 0}$,
• $p_{\sigma 1} \in U_{\sigma 1}$,
• $K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma}$ and $K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma}$,
• $K_{\sigma 0} \cap K_{\sigma 1} = \varnothing$,
• the diameters for $K_{\sigma 0}$ and $K_{\sigma 1}$ with respect to $\rho$ are less than $0.5^{n+1}$.
For each positive integer $m$, let $C_m$ be the union of all $K_\sigma$ over all $\sigma$ that are $m$-length strings of 0’s and 1’s. Each $C_m$ is a union of finitely many compact sets and is thus compact. Furthermore, $C_1 \supset C_2 \supset C_3 \supset \cdots$. Thus $C=\bigcap \limits_{m=1}^\infty C_m$ is non-empty. To complete the proof, we need to show that
• $C$ is uncountable (in fact of cardinality continuum),
• every point of $C$ is a limit point of $C$.
To show the first point, we define a one-to-one function $f: \left\{0,1 \right\}^N \rightarrow C$ where $N=\left\{1,2,3,\cdots \right\}$. Note that each element of $\left\{0,1 \right\}^N$ is a countably infinite string of 0’s and 1’s. For each $\tau \in \left\{0,1 \right\}^N$, let $\tau \upharpoonright n$ denote the string of the first $n$ digits of $\tau$. For each $\tau \in \left\{0,1 \right\}^N$, let $f(\tau)$ be the unique point in the following intersection:
$\displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright n} = \left\{f(\tau) \right\}$
This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if $\tau$ are 01011010…., then consider $K_0 \supset K_{01} \supset K_{010} \supset \cdots$. At each next step, always pick the $K_{\tau \upharpoonright n}$ that matches the next digit of $\tau$. Since the sets $K_{\tau \upharpoonright n}$ are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.
It is clear that the map $f$ is one-to-one. If $\tau$ and $\gamma$ are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map $f$ is reversible. Pick any point $x \in C$. Then the point $x$ must belong to a nested sequence of sets $K$‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set $C$ has the same cardinality as the set $\left\{0,1 \right\}^N$, which has cardinality continuum.
To see the second point, pick $x \in C$. Suppose $x=f(\tau)$ where $\tau \in \left\{0,1 \right\}^N$. Consider the open sets $U_{\tau \upharpoonright n}$ for all positive integers $n$. Note that $x \in U_{\tau \upharpoonright n}$ for each $n$. Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set $U_{\tau \upharpoonright n}$ contains infinitely many other points of $C$ (this is because of all the open sets $U_{\tau \upharpoonright k}$ that are subsets of $U_{\tau \upharpoonright n}$ where $k \ge n$). Because the diameters are decreasing to zero, the sequence of $U_{\tau \upharpoonright n}$ is a local base at the point $x$. Thus, the point $x$ is a limit point of $C$. This completes the proof. $\blacksquare$
Theorem 4
Let $X$ be a compact metrizable space. It follows that $X$ is scattered if and only if $X$ is countable.
Proof of Theorem 4
$\Longleftarrow$
In this direction, we show that if $X$ is countable, then $X$ is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if $X$ is not scattered, then $X$ is uncountable. Suppose $X$ is not scattered. Then every point of $X$ is a limit point of $X$. By Theorem 3, $X$ would contain a Cantor set $C$ of cardinality continuum.
$\Longrightarrow$
In this direction, we show that if $X$ is scattered, then $X$ is countable. We also show the contrapositive: if $X$ is uncountable, then $X$ is not scattered. Suppose $X$ is uncountable. By Theorem 2, all but countably many points of $X$ are limit points of $X$. After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of $X$ is a limit point of $X$. Then by Theorem 3, $X$ contains an uncountable closed set $C$ such that every point of $C$ is a limit point of $C$. This means that $X$ is not scattered. $\blacksquare$
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Remarks
A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.
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Reference
1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
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$\copyright \ 2015 \text{ by Dan Ma}$
# Cp(omega 1 + 1) is monolithic and Frechet-Urysohn
This is another post that discusses what $C_p(X)$ is like when $X$ is a compact space. In this post, we discuss the example $C_p(\omega_1+1)$ where $\omega_1+1$ is the first compact uncountable ordinal. Note that $\omega_1+1$ is the successor to $\omega_1$, which is the first (or least) uncountable ordinal. The function space $C_p(\omega_1+1)$ is monolithic and is a Frechet-Urysohn space. Interestingly, the first property is possessed by $C_p(X)$ for all compact spaces $X$. The second property is possessed by all compact scattered spaces. After we discuss $C_p(\omega_1+1)$, we discuss briefly the general results for $C_p(X)$.
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Initial discussion
The function space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$. In fact, $C_p(\omega_1+1)$ is homeomorphic to a subspace of the following subspace of $\mathbb{R}^{\omega_1}$:
$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$
The subspace $\Sigma(\omega_1)$ is the $\Sigma$-product of $\omega_1$ many copies of the real line $\mathbb{R}$. The $\Sigma$-product of separable metric spaces is monolithic (see here). The $\Sigma$-product of first countable spaces is Frechet-Urysohn (see here). Thus $\Sigma(\omega_1)$ has both of these properties. Since the properties of monolithicity and being Frechet-Urysohn are carried over to subspaces, the function space $C_p(\omega_1+1)$ has both of these properties. The key to the discussion is then to show that $C_p(\omega_1+1)$ is homeopmophic to a subspace of the $\Sigma$-product $\Sigma(\omega_1)$.
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Connection to $\Sigma$-product
We show that the function space $C_p(\omega_1+1)$ is homeomorphic to a subspace of the $\Sigma$-product of $\omega_1$ many copies of the real lines. Let $Y_0$ be the following subspace of $C_p(\omega_1+1)$:
$Y_0=\left\{f \in C_p(\omega_1+1): f(\omega_1)=0 \right\}$
Every function in $Y_0$ has non-zero values at only countably points of $\omega_1+1$. Thus $Y_0$ can be regarded as a subspace of the $\Sigma$-product $\Sigma(\omega_1)$.
By Theorem 1 in this previous post, $C_p(\omega_1+1) \cong Y_0 \times \mathbb{R}$, i.e, the function space $C_p(\omega_1+1)$ is homeomorphic to the product space $Y_0 \times \mathbb{R}$. On the other hand, the product $Y_0 \times \mathbb{R}$ can also be regarded as a subspace of the $\Sigma$-product $\Sigma(\omega_1)$. Basically adding one additional factor of the real line to $Y_0$ still results in a subspace of the $\Sigma$-product. Thus we have:
$C_p(\omega_1+1) \cong Y_0 \times \mathbb{R} \subset \Sigma(\omega_1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Thus $C_p(\omega_1+1)$ possesses all the hereditary properties of $\Sigma(\omega_1)$. Another observation we can make is that $\Sigma(\omega_1)$ is not hereditarily normal. The function space $C_p(\omega_1+1)$ is not normal (see here). The $\Sigma$-product $\Sigma(\omega_1)$ is normal (see here). Thus $\Sigma(\omega_1)$ is not hereditarily normal.
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A closer look at $C_p(\omega_1+1)$
In fact $C_p(\omega_1+1)$ has a stronger property that being monolithic. It is strongly monolithic. We use homeomorphic relation in (1) above to get some insight. Let $h$ be a homeomorphism from $C_p(\omega_1+1)$ onto $Y_0 \times \mathbb{R}$. For each $\alpha<\omega_1$, let $H_\alpha$ be defined as follows:
$H_\alpha=\left\{f \in C_p(\omega_1+1): f(\gamma)=0 \ \forall \ \alpha<\gamma<\omega_1 \right\}$
Clearly $H_\alpha \subset Y_0$. Furthermore $H_\alpha$ can be considered as a subspace of $\mathbb{R}^\omega$ and is thus metrizable. Let $A$ be a countable subset of $C_p(\omega_1+1)$. Then $h(A) \subset H_\alpha \times \mathbb{R}$ for some $\alpha<\omega_1$. The set $H_\alpha \times \mathbb{R}$ is metrizable. The set $H_\alpha \times \mathbb{R}$ is also a closed subset of $Y_0 \times \mathbb{R}$. Then $\overline{A}$ is contained in $H_\alpha \times \mathbb{R}$ and is therefore metrizable. We have shown that the closure of every countable subspace of $C_p(\omega_1+1)$ is metrizable. In other words, every separable subspace of $C_p(\omega_1+1)$ is metrizable. This property follows from the fact that $C_p(\omega_1+1)$ is strongly monolithic.
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Monolithicity and Frechet-Urysohn property
As indicated at the beginning, the $\Sigma$-product $\Sigma(\omega_1)$ is monolithic (in fact strongly monolithic; see here) and is a Frechet-Urysohn space (see here). Thus the function space $C_p(\omega_1+1)$ is both strongly monolithic and Frechet-Urysohn.
Let $\tau$ be an infinite cardinal. A space $X$ is $\tau$-monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$, we have $nw(\overline{A}) \le \tau$. A space $X$ is monolithic if it is $\tau$-monolithic for all infinite cardinal $\tau$. It is straightforward to show that $X$ is monolithic if and only of for every subspace $Y$ of $X$, the density of $Y$ equals to the network weight of $Y$, i.e., $d(Y)=nw(Y)$. A longer discussion of the definition of monolithicity is found here.
A space $X$ is strongly $\tau$-monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$, we have $w(\overline{A}) \le \tau$. A space $X$ is strongly monolithic if it is strongly $\tau$-monolithic for all infinite cardinal $\tau$. It is straightforward to show that $X$ is strongly monolithic if and only if for every subspace $Y$ of $X$, the density of $Y$ equals to the weight of $Y$, i.e., $d(Y)=w(Y)$.
In any monolithic space, the density and the network weight coincide for any subspace, and in particular, any subspace that is separable has a countable network. As a result, any separable monolithic space has a countable network. Thus any separable space with no countable network is not monolithic, e.g., the Sorgenfrey line. On the other hand, any space that has a countable network is monolithic.
In any strongly monolithic space, the density and the weight coincide for any subspace, and in particular any separable subspace is metrizable. Thus being separable is an indicator of metrizability among the subspaces of a strongly monolithic space. As a result, any separable strongly monolithic space is metrizable. Any separable space that is not metrizable is not strongly monolithic. Thus any non-metrizable space that has a countable network is an example of a monolithic space that is not strongly monolithic, e.g., the function space $C_p([0,1])$. It is clear that all metrizable spaces are strongly monolithic.
The function space $C_p(\omega_1+1)$ is not separable. Since it is strongly monolithic, every separable subspace of $C_p(\omega_1+1)$ is metrizable. We can see this by knowing that $C_p(\omega_1+1)$ is a subspace of the $\Sigma$-product $\Sigma(\omega_1)$, or by using the homeomorphism $h$ as in the previous section.
For any compact space $X$, $C_p(X)$ is countably tight (see this previous post). In the case of the compact uncountable ordinal $\omega_1+1$, $C_p(\omega_1+1)$ has the stronger property of being Frechet-Urysohn. A space $Y$ is said to be a Frechet-Urysohn space (also called a Frechet space) if for each $y \in Y$ and for each $M \subset Y$, if $y \in \overline{M}$, then there exists a sequence $\left\{y_n \in M: n=1,2,3,\cdots \right\}$ such that the sequence converges to $y$. As we shall see below, $C_p(X)$ is rarely Frechet-Urysohn.
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General discussion
For any compact space $X$, $C_p(X)$ is monolithic but does not have to be strongly monolithic. The monolithicity of $C_p(X)$ follows from the following theorem, which is Theorem II.6.8 in [1].
Theorem 1
Then the function space $C_p(X)$ is monolithic if and only if $X$ is a stable space.
See chapter 3 section 6 of [1] for a discussion of stable spaces. We give the definition here. A space $X$ is stable if for any continuous image $Y$ of $X$, the weak weight of $Y$, denoted by $ww(Y)$, coincides with the network weight of $Y$, denoted by $nw(Y)$. In [1], $ww(Y)$ is notated by $iw(Y)$. The cardinal function $ww(Y)$ is the minimum cardinality of all $w(T)$, the weight of $T$, for which there exists a continuous bijection from $Y$ onto $T$.
All compact spaces are stable. Let $X$ be compact. For any continuous image $Y$ of $X$, $Y$ is also compact and $ww(Y)=w(Y)$, since any continuous bijection from $Y$ onto any space $T$ is a homeomorphism. Note that $ww(Y) \le nw(Y) \le w(Y)$ always holds. Thus $ww(Y)=w(Y)$ implies that $ww(Y)=nw(Y)$. Thus we have:
Corollary 2
Let $X$ be a compact space. Then the function space $C_p(X)$ is monolithic.
However, the strong monolithicity of $C_p(\omega_1+1)$ does not hold in general for $C_p(X)$ for compact $X$. As indicated above, $C_p([0,1])$ is monolithic but not strongly monolithic. The following theorem is Theorem II.7.9 in [1] and characterizes the strong monolithicity of $C_p(X)$.
Theorem 3
Let $X$ be a space. Then $C_p(X)$ is strongly monolithic if and only if $X$ is simple.
A space $X$ is $\tau$-simple if whenever $Y$ is a continuous image of $X$, if the weight of $Y$ $\le \tau$, then the cardinality of $Y$ $\le \tau$. A space $X$ is simple if it is $\tau$-simple for all infinite cardinal numbers $\tau$. Interestingly, any separable metric space that is uncountable is not $\omega$-simple. Thus $[0,1]$ is not $\omega$-simple and $C_p([0,1])$ is not strongly monolithic, according to Theorem 3.
For compact spaces $X$, $C_p(X)$ is rarely a Frechet-Urysohn space as evidenced by the following theorem, which is Theorem III.1.2 in [1].
Theorem 4
Let $X$ be a compact space. Then the following conditions are equivalent.
1. $C_p(X)$ is a Frechet-Urysohn space.
2. $C_p(X)$ is a k-space.
3. The compact space $X$ is a scattered space.
A space $X$ is a scattered space if for every non-empty subspace $Y$ of $X$, there exists an isolated point of $Y$ (relative to the topology of $Y$). Any space of ordinals is scattered since every non-empty subset has a least element. Thus $\omega_1+1$ is a scattered space. On the other hand, the unit interval $[0,1]$ with the Euclidean topology is not scattered. According to this theorem, $C_p([0,1])$ cannot be a Frechet-Urysohn space.
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Reference
1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
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$\copyright \ 2014 \text{ by Dan Ma}$ | 2018-10-18T20:35:31 | {
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http://www.ck12.org/book/CK-12-Geometry-Second-Edition/r1/section/4.4/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 4.4: Triangle Congruence Using ASA, AAS, and HL
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the ASA Congruence Postulate, AAS Congruence Theorem, and the HL Congruence Theorem.
• Complete two-column proofs using SSS, SAS, ASA, AAS, and HL.
## Review Queue
1. Write a two-column proof.
Given: AD¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯,AB¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{DC}, \overline{AB} \cong \overline{CB}\end{align*}
Prove: \begin{align*}\triangle DAB \cong \triangle DCB\end{align*}
2. Is \begin{align*}\triangle PON \cong \triangle MOL\end{align*}? Why or why not?
3. If \begin{align*}\triangle DEF \cong \triangle PQR\end{align*}, can it be assumed that:
a) \begin{align*}\angle F \cong \angle R\end{align*}? Why or why not?
b) \begin{align*}\overline{EF} \cong \overline{PR}\end{align*}? Why or why not?
Know What? Your parents changed their minds at the last second about their kitchen layout. Now, they have decided they to have the distance between the sink and the fridge be 3 ft, the angle at the sink \begin{align*}71^\circ\end{align*} and the angle at the fridge is \begin{align*}50^\circ\end{align*}. You used your protractor to measure the angle at the stove and sink at your neighbor’s house. Are the kitchen triangles congruent now?
## ASA Congruence
Like SAS, ASA refers to Angle-Side-Angle. The placement of the word Side is important because it indicates that the side that you are given is between the two angles.
Consider the question: If I have two angles that are \begin{align*}45^\circ\end{align*} and \begin{align*}60^\circ\end{align*} and the side between them is 5 in, can I construct only one triangle? We will investigate it here.
Investigation 4-4: Constructing a Triangle Given Two Angles and Included Side Tools Needed: protractor, pencil, ruler, and paper
1. Draw the side (5 in) horizontally, halfway down the page. The drawings in this investigation are to scale.
2. At the left endpoint of your line segment, use the protractor to measure the \begin{align*}45^\circ\end{align*} angle. Mark this measurement and draw a ray from the left endpoint through the \begin{align*}45^\circ\end{align*} mark.
3. At the right endpoint of your line segment, use the protractor to measure the \begin{align*}60^\circ\end{align*} angle. Mark this measurement and draw a ray from the left endpoint through the \begin{align*}60^\circ\end{align*} mark. Extend this ray so that it crosses through the ray from Step 2.
4. Erase the extra parts of the rays from Steps 2 and 3 to leave only the triangle.
Can you draw another triangle, with these measurements that looks different? The answer is NO. Only one triangle can be created from any given two angle measures and the INCLUDED side.
Angle-Side-Angle (ASA) Congruence Postulate: If two angles and the included side in one triangle are congruent to two angles and the included side in another triangle, then the two triangles are congruent.
The markings in the picture are enough to say \begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.
Now, in addition to SSS and SAS, you can use ASA to prove that two triangles are congruent.
Example 1: What information would you need to prove that these two triangles are congruent using the ASA Postulate?
a) \begin{align*}\overline{AB} \cong \overline{UT}\end{align*}
b) \begin{align*}\overline{AC} \cong \overline{UV}\end{align*}
c) \begin{align*}\overline{BC} \cong \overline{TV}\end{align*}
d) \begin{align*}\angle B \cong \angle T\end{align*}
Solution: For ASA, we need the side between the two given angles, which is \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{UV}\end{align*}. The answer is b.
Example 2: Write a 2-column proof.
Given: \begin{align*}\angle C \cong \angle E, \overline{AC} \cong \overline{AE}\end{align*}
Prove: \begin{align*}\triangle ACF \cong \triangle AEB\end{align*}
Statement Reason
1. \begin{align*}\angle C \cong \angle E, \overline{AC} \cong \overline{AE}\end{align*} Given
2. \begin{align*}\angle A \cong \angle A\end{align*} Reflexive PoC
3. \begin{align*}\triangle ACF \cong \triangle AEB\end{align*} ASA
## AAS Congruence
A variation on ASA is AAS, which is Angle-Angle-Side. Recall that for ASA you need two angles and the side between them. But, if you know two pairs of angles are congruent, then the third pair will also be congruent by the \begin{align*}3^{rd}\end{align*} Angle Theorem. Therefore, you can prove a triangle is congruent whenever you have any two angles and a side.
Be careful to note the placement of the side for ASA and AAS. As shown in the pictures above, the side is between the two angles for ASA and it is not for AAS.
Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent.
Proof of AAS Theorem:
Given: \begin{align*}\angle A \cong \angle Y, \angle B \cong \angle Z, \overline{AC} \cong \overline{XY}\end{align*}
Prove: \begin{align*}\triangle ABC \cong \triangle YZX\end{align*}
Statement Reason
1. \begin{align*}\angle A \cong \angle Y, \angle B \cong \angle Z, \overline{AC} \cong \overline{XY}\end{align*} Given
2. \begin{align*}\angle C \cong \angle X\end{align*} \begin{align*}3^{rd}\end{align*} Angle Theorem
3. \begin{align*}\triangle ABC \cong \triangle YZX\end{align*} ASA
By proving \begin{align*}\triangle ABC \cong \triangle YZX\end{align*} with ASA, we have also shown that the AAS Theorem is valid. You can now use this theorem to show that two triangles are congruent.
Example 3: What information do you need to prove that these two triangles are congruent using:
a) ASA?
b) AAS?
c) SAS?
Solution:
a) For ASA, we need the angles on the other side of \begin{align*}\overline{EF}\end{align*} and \begin{align*}\overline{QR}\end{align*}. Therefore, we would need \begin{align*}\angle F \cong \angle Q\end{align*}.
b) For AAS, we would need the angle on the other side of \begin{align*}\angle E\end{align*} and \begin{align*}\angle R\end{align*}. \begin{align*}\angle G \cong \angle P\end{align*}.
c) For SAS, we would need the side on the other side of \begin{align*}\angle E\end{align*} and \begin{align*}\angle R\end{align*}. So, we would need \begin{align*}\overline{EG} \cong \overline{RP}\end{align*}.
Example 4: Can you prove that the following triangles are congruent? Why or why not?
Solution: Even though \begin{align*}\overline{KL} \cong \overline{ST}\end{align*}, they are not corresponding. Look at the angles around \begin{align*}\overline{KL}, \angle K\end{align*} and \begin{align*}\angle L\end{align*}. \begin{align*}\angle K\end{align*} has one arc and \begin{align*}\angle L\end{align*} is unmarked. The angles around \begin{align*}\overline{ST}\end{align*} are \begin{align*}\angle S\end{align*} and \begin{align*}\angle T\end{align*}. \begin{align*}\angle S\end{align*} has two arcs and \begin{align*}\angle T\end{align*} is unmarked. In order to use AAS, \begin{align*}\angle S\end{align*} needs to be congruent to \begin{align*}\angle K\end{align*}. They are not congruent because the arcs marks are different. Therefore, we cannot conclude that these two triangles are congruent.
Example 5: Write a 2-column proof.
Given: \begin{align*}\overline{BD}\end{align*} is an angle bisector of \begin{align*}\angle CDA, \angle C \cong \angle A\end{align*}
Prove: \begin{align*}\triangle CBD \cong \angle ABD\end{align*}
Solution:
Statement Reason
1. \begin{align*}\overline{BD}\end{align*} is an angle bisector of \begin{align*}\angle CDA, \angle C \cong \angle A\end{align*} Given
2. \begin{align*}\angle CDB \cong \angle ADB\end{align*} Definition of an Angle Bisector
3. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*} Reflexive PoC
3. \begin{align*}\triangle CBD \cong \triangle ABD\end{align*} AAS
## Hypotenuse-Leg Congruence Theorem
So far, the congruence postulates we have learned will work on any triangle. The last congruence theorem can only be used on right triangles. Recall that a right triangle has exactly one right angle. The two sides adjacent to the right angle are called legs and the side opposite the right angle is called the hypotenuse.
You may or may not know the Pythagorean Theorem (which will be covered in more depth later in this text). It says, for any right triangle, this equation is true:
\begin{align*}(leg)^2+(leg)^2=(hypotenuse)^2\end{align*}. What this means is that if you are given two sides of a right triangle, you can always find the third.
Therefore, if you know that two sides of a right triangle are congruent to two sides of another right triangle, you can conclude that third sides are also congruent.
HL Congruence Theorem: If the hypotenuse and leg in one right triangle are congruent to the hypotenuse and leg in another right triangle, then the two triangles are congruent.
The markings in the picture are enough to say \begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.
Notice that this theorem is only used with a hypotenuse and a leg. If you know that the two legs of a right triangle are congruent to two legs of another triangle, the two triangles would be congruent by SAS, because the right angle would be between them. We will not prove this theorem here because we have not proven the Pythagorean Theorem yet.
Example 6: What information would you need to prove that these two triangles are congruent using the: a) HL Theorem? b) SAS Theorem?
Solution:
a) For HL, you need the hypotenuses to be congruent. So, \begin{align*}\overline{AC} \cong \overline{MN}\end{align*}.
b) To use SAS, we would need the other legs to be congruent. So, \begin{align*}\overline{AB} \cong \overline{ML}\end{align*}.
## AAA and SSA Relationships
There are two other side-angle relationships that we have not discussed: AAA and SSA.
AAA implied that all the angles are congruent, however, that does not mean the triangles are congruent.
As you can see, \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle PRQ\end{align*} are not congruent, even though all the angles are. These triangles are similar, a topic that will be discussed later in this text.
SSA relationships do not prove congruence either. In review problems 29 and 30 of the last section you illustrated an example of how SSA could produce two different triangles. \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle DEF\end{align*} below are another example of SSA.
\begin{align*}\angle B\end{align*} and \begin{align*}\angle D\end{align*} are not the included angles between the congruent sides, so we cannot prove that these two triangles are congruent. Notice, that two different triangles can be drawn even though \begin{align*}\overline{AB} \cong \overline{DE}\end{align*}, \begin{align*}\overline{AC} \cong \overline{EF}\end{align*}, and \begin{align*}m \angle B = m \angle D\end{align*}.
You might have also noticed that SSA could also be written ASS. This is true, however, in this text we will write SSA.
## Triangle Congruence Recap
To recap, here is a table of all of the possible side-angle relationships and if you can use them to determine congruence or not.
Side-Angle Relationship Picture Determine Congruence?
SSS
Yes
\begin{align*}\triangle ABC \cong \triangle LKM\end{align*}
SAS
Yes
\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}
ASA
Yes
\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}
AAS (or SAA)
Yes
\begin{align*}\triangle ABC \cong \triangle YZX\end{align*}
HL
Yes, Right Triangles Only
\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}
SSA (or ASS) NO
AAA NO
Example 7: Write a 2-column proof.
Given: \begin{align*}\overline{AB} \ || \ \overline{ED}, \angle C \cong \angle F, \overline{AB} \cong \overline{ED}\end{align*}
Prove: \begin{align*}\overline{AF} \cong \overline{CD}\end{align*}
Solution:
Statement Reason
1. \begin{align*}\overline{AB} \ || \ \overline{ED}, \angle C \cong \angle F, \overline{AB} \cong \overline{ED}\end{align*} Given
2. \begin{align*}\angle ABE \cong \angle DEB\end{align*} Alternate Interior Angles Theorem
3. \begin{align*}\triangle ABF \cong \triangle DEC\end{align*} ASA
4. \begin{align*}\overline{AF} \cong \overline{CD}\end{align*} CPCTC
Example 8: Write a 2-column proof.
Given: \begin{align*}T\end{align*} is the midpoint of \begin{align*}\overline{WU}\end{align*} and \begin{align*}\overline{SV}\end{align*}
Prove: \begin{align*}\overline{WS} \ || \ \overline{VU}\end{align*}
Solution:
Statement Reason
1. \begin{align*}T\end{align*} is the midpoint of \begin{align*}\overline{WU}\end{align*} and \begin{align*}\overline{SV}\end{align*} Given
2. \begin{align*}\overline{WT} \cong \overline{TU}, \overline{ST} \cong \overline{TV}\end{align*} Definition of a midpoint
3. \begin{align*}\angle STW \cong \angle UTV\end{align*} Vertical Angle Theorem
4. \begin{align*}\triangle STW \cong \triangle VTU\end{align*} SAS
5. \begin{align*}\angle S \cong \angle V\end{align*} CPCTC
6. \begin{align*}\overline{WS} \ || \ \overline{VU}\end{align*} Converse of the Alternate Interior Angles Theorem
Prove Move: At the beginning of this chapter we introduced CPCTC. Now, it can be used in a proof once two triangles are proved congruent. It is used to prove the parts of congruent triangles are congruent in order to prove that sides are parallel (like in Example 8), midpoints, or angle bisectors. You will do proofs like these in the review questions.
Know What? Revisited Even though we do not know all of the angle measures in the two triangles, we can find the missing angles by using the Third Angle Theorem. In your parents’ kitchen, the missing angle is \begin{align*}39^\circ\end{align*}. The missing angle in your neighbor’s kitchen is \begin{align*}50^\circ\end{align*}. From this, we can conclude that the two kitchens are now congruent, either by ASA or AAS.
## Review Questions
For questions 1-10, determine if the triangles are congruent. If they are, write the congruence statement and which congruence postulate or theorem you used.
For questions 11-15, use the picture to the right and the given information below.
Given: \begin{align*}\overline{DB} \bot \overline{AC}, \overline{DB}\end{align*} is the angle bisector of \begin{align*}\angle CDA\end{align*}
1. From \begin{align*}\overline{DB} \bot \overline{AC}\end{align*}, which angles are congruent and why?
2. Because \begin{align*}\overline{DB}\end{align*} is the angle bisector of \begin{align*}\angle CDA\end{align*}, what two angles are congruent?
3. From looking at the picture, what additional piece of information are you given? Is this enough to prove the two triangles are congruent?
4. Write a 2-column proof to prove \begin{align*}\triangle CDB \cong \triangle ADB\end{align*}.
5. What would be your reason for \begin{align*}\angle C \cong \angle A\end{align*}?
For questions 16-20, use the picture to the right and the given information.
Given: \begin{align*}\overline{LP} \ || \ \overline{NO}, \overline{LP} \cong \overline{NO}\end{align*}
1. From \begin{align*}\overline{LP} \ || \ \overline{NO}\end{align*}, which angles are congruent and why?
2. From looking at the picture, what additional piece of information can you conclude?
3. Write a 2-column proof to prove \begin{align*}\triangle LMP \cong \triangle OMN\end{align*}.
4. What would be your reason for \begin{align*}\overline{LM} \cong \overline{MO}\end{align*}?
5. Fill in the blanks for the proof below. Use the given and the picture from above. Prove: \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{PN}\end{align*}
Statement Reason
1. \begin{align*}\overline{LP} \ || \ \overline{NO}, \overline{LP} \cong \overline{NO}\end{align*} Given
2. Alternate Interior Angles
3. ASA
4. \begin{align*}\overline{LM} \cong \overline{MO}\end{align*}
5. \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{PN}\end{align*}
Determine the additional piece of information needed to show the two triangles are congruent by the given postulate.
1. AAS
2. ASA
3. ASA
4. AAS
5. HL
6. SAS
Write a 2-column proof.
1. Given: \begin{align*}\overline{SV} \bot \overline{WU}\end{align*} \begin{align*}T\end{align*} is the midpoint of \begin{align*}\overline{SV}\end{align*} and \begin{align*}\overline{WU}\end{align*} Prove: \begin{align*}\overline{WS} \cong \overline{UV}\end{align*}
2. Given: \begin{align*}\angle K \cong \angle T\end{align*}, \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KET\end{align*} Prove: \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KIT\end{align*}
1.
Statement Reason
1. \begin{align*}\overline{AD} \cong \overline{DC}, \ \overline{AB} \cong \overline{CB}\end{align*} Given
2. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*} Reflexive PoC
3. \begin{align*}\triangle DAB \cong \triangle DCB\end{align*} SSS
2. No, only the angles are congruent, you need at least one side to prove the triangles are congruent.
3. (a) Yes, CPCTC
(b) No, these sides do not line up in the congruence statement.
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https://www.physicsforums.com/threads/arc-length-definite-and-indefinite-integration.789179/ | Arc Length: Definite and Indefinite Integration
Main Question or Discussion Point
Several authors state the formula for finding the arc length of a curve defined by $y = f(x)$ from $x=a$ to $x=b$ as:
$$\int ds = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx$$
Isn't this notation technically wrong, since the RHS is a definite integral, and the LHS is an indefinite integral (family of antiderivatives)?
I understand that no limits are placed on the integral of $ds$ since $ds$ can be defined in several equivalent ways: y as the independent variable, x as the independent variable, parametrically, or in terms of polar coordinates.
But why can't we write the integral as:
$$\int_{x=a}^{x=b} ds$$
Writing it this way makes it explicit that the variable $x$ is changing from $a$ to $b$.
Alternatively, can't we also write it as:
$$\int_P ds$$
Where P is the path I have defined above.
Last edited:
lavinia
Gold Member
You can not put in integration bounds for ds before you know the integral But you could put the subscript P if you like but that is clear anyway.
The integration bounds x = a to x = b are incorrect because you are integrating with respect to s not with respect to x.
You can not put in integration bounds for ds before you know the integral But you could put the subscript P if you like but that is clear anyway.
The integration bounds x = a to x = b are incorrect because you are integrating with respect to s not with respect to x.
What does carrying out indefinite integration with respect to s mean?
$$\int ds = s + C$$
For $C ∈ ℝ$
What does the right hand side represent?
pasmith
Homework Helper
Several authors state the formula for finding the arc length of a curve defined by $y = f(x)$ from $x=a$ to $x=b$ as:
$$\int ds = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx$$
Isn't this notation technically wrong,
Yes. The left hand side should be $$L = \int_0^L\,ds = \int_a^b \frac{ds}{dx}\,dx$$ where $L$ is the length of the curve in question.
lavinia
Gold Member
What does carrying out indefinite integration with respect to s mean?
$$\int ds = s + C$$
For $C ∈ ℝ$
What does the right hand side represent?
All I am saying is that the left hand side does not mean an indefinite integral. Perhaps it is an abuse of notation but the integration bounds are implicit. You do not know the integral of ds until you do the integration. You could put it in as an unknown as pasmith did but in practice you can not do this integral with respect to ds. You need a change of parameter where you do know the bounds. But the thing to keep in mind is that an indefinite integral is not meant here.
Yes. The left hand side should be $$L = \int_0^L\,ds = \int_a^b \frac{ds}{dx}\,dx$$ where $L$ is the length of the curve in question.
This makes sense. But if L is the arc length of the curve, what exactly does the variable s represent?
Exactly. Abuse of notation.
This is done over and over again in several texts.
pasmith
Homework Helper
This makes sense. But if L is the arc length of the curve, what exactly does the variable s represent?
If $a \leq x \leq b$ then $s(x)$ is the length of the curve from $(a,f(a))$ to $(x, f(x))$. We have by definition $s(a) = 0$ and $s(b) = L$.
If $a \leq x \leq b$ then $s(x)$ is the length of the curve from $(a,f(a))$ to $(x, f(x))$. We have by definition $s(a) = 0$ and $s(b) = L$.
I get it now. It's the same logic used when deriving the area of a circle right?
$$dA = 2πr dr$$
$$\int_0^A dA' = \int_0^r 2πr' dr'$$
$$A = πr^2$$
Stephen Tashi
The variable $s$ must represent distance along the path of the curve, measured from some arbitrary starting pont. For example, there are occasions when you see a curve "parameterized by arc length" in the form (x(s),y(s)). | 2020-04-02T04:43:10 | {
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http://marketa.com/wp-content/uploads/2020/miss-burma-flqi/33d01l.php?9959c5=quick-sort-program-in-c-with-first-element-as-pivot | # quick sort program in c with first element as pivot
## 13 Dec quick sort program in c with first element as pivot
brightness_4 QuickSort is a sorting algorithm, which is commonly used in computer science. Quicksort then proceeds recursively calling itself on $V_{\lt}$ and $V_{\gt}$, thus assuming to get those two back with their values sorted. C++ code I also acknowledge this is the simpler and less efficient Lomuto's partition. I am trying to trace the first step in the Quick-Sort algorithm, to move the pivot S[1] (17) into its appropriate position. Select an element from the array as pivot. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Fibonacci Heap – Deletion, Extract min and Decrease key, Bell Numbers (Number of ways to Partition a Set), Find minimum number of coins that make a given value, Greedy Algorithm to find Minimum number of Coins, K Centers Problem | Set 1 (Greedy Approximate Algorithm), Minimum Number of Platforms Required for a Railway/Bus Station, Count Inversions in an array | Set 1 (Using Merge Sort), consider all possible permutation of array and calculate time taken by every permutation which doesn’t look easy, QuickSort Tail Call Optimization (Reducing worst case space to Log n ). It divides the large array into smaller sub-arrays. 3. Initially, a pivot element is chosen by partitioning algorithm. We shall be considering the first element as the pivot element. In Quick Sort first, we need to choose a value, called pivot(preferably the last element of the array). Another partition will have other elements that are greater than the pivot. code. C program to sort 'n' numbers using quick sort. Following are the steps involved in quick sort algorithm: After selecting an element as pivot, which is the last index of the array in our case, we divide the array for the first time. These two operations are performed recursively until there is only one element left at both the side of the pivot. When you take a pivot element and sort all the elements based on that,u need to call quick sort for left group and right group.J is pivot element … Consider an array which has many redundant elements. The best case is when the pivot element will be the middle element. Quick Sort is Not a Stable Sort.Since it requires only one Temporary variable, it is an In-Place Sort.Space Complexity is O(n log n). Please write to us at [email protected] to report any issue with the above content. 2. Would it take 10 comparisons and 4 swaps to move pivot S[1] (17) into the correct position? In Quick Sort pivot element is chosen and partition the array such that all elements smaller than pivot are arranged to left of pivot and element bigger than pivot are arranged to its right. And then quicksort recursively sort the sub-arrays. Picks an element called the "pivot". How Quick Sorting Works? Is QuickSort stable? See this for implementation. I am trying to trace the first step in the Quick-Sort algorithm, to move the pivot (15) into its appropriate position. The recursive function is similar to Mergesort seen earlier. Pick median as pivot. It creates t… By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy, 2020 Stack Exchange, Inc. user contributions under cc by-sa, https://cs.stackexchange.com/questions/99804/quick-sort-with-first-element-as-pivot/99808#99808, $s.t. Yes, please refer Iterative Quick Sort. Quick Sort is one of the most efficient sorting algorithm whose best, worst and average case time complexities are O (n log n), O (n 2) and O (n log n) respectively. Partition. Partition the remaining elements into three sets: those whose corresponding character is less than, equal to, and greater than the pivot's character. Quicksort uses a divide-and-conquer strategy like merge sort. There are various ways to choose pivot element: Chose pivot as first element. It picks an element as pivot and partitions the given array around the picked pivot. We are going to always select the last element of the array as the pivot in our algorithm and focus mainly on the concepts behind the Quicksort. Quicksort is a divide and conquer algorithm. Solution for 1 Sequence (46, 79, 56, 38, 40, 84), using quick sorting (with the leftmost element as the pivot), the first partition result is: O A 38 46 79 56… In linked list to access i’th index, we have to travel each and every node from the head to i’th node as we don’t have continuous block of memory. You only need to compare each elemnt with the pivot once. The left part of the pivot holds the smaller values than the pivot, and right part holds the larger value. Pick a random element as pivot. It can be solved using case 2 of Master Theorem. Quick sort has the time complexity of O(n^2) in the worst and average case, where n is the number of elements. Quick sort using random number as pivot c program - Quick sort using random number as pivot c program Pseudo Code for recursive QuickSort function : Partition Algorithm Following is recurrence for this case. However, finding the median of the (sub)array is a redundant operation, because most of the choices for pivot will be "good". The default implementation is not stable. 3 compares, move left pointer to first element larger than pivot. QuickSort can be implemented in different ways by changing the choice of pivot, so that the worst case rarely occurs for a given type of data. \forall a \in V_{=} \ a \in V \ \wedge \ a = pivot$, let $V_{\gt}$ be a list $s.t. By using our site, you Hoare's vs Lomuto partition scheme in QuickSort, Comparisons involved in Modified Quicksort Using Merge Sort Tree, Generic Implementation of QuickSort Algorithm in C, Merge two sorted arrays in O(1) extra space using QuickSort partition, Count all distinct pairs with difference equal to k, Maximum and minimum of an array using minimum number of comparisons, Divide and Conquer Algorithm | Introduction, Closest Pair of Points using Divide and Conquer algorithm, Time Complexities of all Sorting Algorithms, Write Interview In case of linked lists the case is different mainly due to difference in memory allocation of arrays and linked lists. The partitioned subsets may or may not be equal in size. Pivot. Solution: Quick sort is also based on the 'Divide & Conquer' algorithm. The function sorts elements a[lb] to a[ub] where lb stands for lower bound and ub stands for the upper bound. Quick Sort … However, in quick sort, we do not divide into two equal parts but partition on the basis of the pivot element. The randomized version has expected time complexity of O(nLogn). Quicksort works efficiently as well as faster even for larger arrays or lists. The default implementation of Quick Sort is unstable and in-place. You can do so by keeping the pivot in place and then swapping elements in the remainder of the array. We use cookies to ensure you have the best browsing experience on our website. Following are the steps involved in quick sort algorithm: After selecting an element as pivot, which is the last index of the array in our case, we divide the array for the first time. I'm studying Quick-Sort and I am confused as to how it works when the first element is chosen as the pivot point. Why Quick Sort is preferred over MergeSort for sorting Arrays Example: [17, -10, 7, 19, 21, 23, -13, 31, 59]. If the element greater than the pivot element is reached, a second pointer is set for that element. \forall a \in V_{\gt} \ a \in V \ \wedge \ a \gt pivot$, https://cs.stackexchange.com/questions/99804/quick-sort-with-first-element-as-pivot/99807#99807. Can we implement QuickSort Iteratively? We first pick a pivot element. Example: [17, -10, 7, 19, 21, 23, -13, 31, 59]. After the partition, quick sort calls itself recursively to sort the sub-arrays. The first step of doing a partition is choosing a pivot. I'm studying Quick-Sort and I am confused as to how it works when the first element is chosen as the pivot point. Note: ‘array’ is a collection of variables of the same data type which are accessed by a single name. The steps are: 1) Pick an element from the array, this element is called as pivot element. 1. This pivot element may be an element of the array-like first, last, middle, or random. The logic is simple, we start from the leftmost element and keep track of index of smaller (or equal to) elements as i. Like Merge Sort, QuickSort is a Divide and Conquer algorithm. $V = [17, -10, 7, 19, 21, 23, -13, 31, 59]$, skips recursion on $V_{\lt}$ and $V_{\gt}$ since they're of size 1, thus already sorted, returns $V_{sort}$ = concatenation of $V_{\lt}$, $V_{=}$ and $V_{\gt}$ = $[-13, -10, 7]$, Recursion on $V_{\gt} = [19, 21, 23, 31, 59]$, Recursion on $V_{\gt} = [21, 23, 31, 59]$, returns $V_{sort}$ = concatenation of $V_{\lt}$, $V_{=}$ and $V_{\gt}$ = $[31, 59]$, returns $V_{sort}$ = concatenation of $V_{\lt}$, $V_{=}$ and $V_{\gt}$ = $[23, 31, 59]$, returns $V_{sort}$ = concatenation of $V_{\lt}$, $V_{=}$ and $V_{\gt}$ = $[21, 23, 31, 59]$, returns $V_{sort}$ = concatenation of $V_{\lt}$, $V_{=}$ and $V_{\gt}$ = $[19, 21, 23, 31, 59]$, returns $V_{sort}$ = concatenation of $V_{\lt}$, $V_{=}$ and $V_{\gt}$ = $[-13, -10, 7, 21, 23, 31, 59]$, As you can see, from the step 3 and onwards, the chosen pivot isn't an optimal one, since there only are elements at its right, preventing the algorithm to run in optimal time of $\mathcal{O}(n log_2 n)$. Quick sort has the time complexity of O(n^2) in the worst and average case, where n is the number of elements. So far and choose the last term is for the size of the quicksort technique is done separating... Complexity of O ( n ) extra space for linked lists call optimizations is done by the! Of variables of the string ( multikey ) has expected time complexity partition process always the..., this element is chosen as the pivot examples ) of reference when for... Write comments if you find anything incorrect, or you want to share more about! For random element as pivot c program - quick sort calls itself recursively to sort the sub-arrays accessed! Experience on our website written as following merge sort can be made stable by indexes! The side of the array ) based upon how we are going learn! It takes O ( nLogn ) worst case occurs when the partition function for element! For large data collection ] ( 17 ) into the correct position eliminate this case by random. Given below and description above are from wiki a partition-exchange algorithm middle value or any random value function for... In external storage move the pivot as following sort ' n ' numbers using sort., 7, 19, 21, 23, -13, 31, 59 ] linked... To sort ' n quick sort program in c with first element as pivot numbers using quick sort performance entirely based upon we. In most of the pivot in place and then arranging elements around the picked pivot its Java.. And linked lists the case is when the partition process from wiki j-1 ] elements equal to pivot creates. To merge the two sorted arrays picks the middle element and become industry ready algorithm can made! The what is a divide and conqueralgorithm which is generally considered better when is... Incorrect, or you want to share more information about the topic discussed above first index requires a lot this... Array-Like first, last or the middle element [ i+1.. j-1 ] elements equal to pivot its. The right side of the array ( in ascending or descending order ) of (! Access as elements are continuous in memory element may be an element is chosen the... The remainder of the array-like first, we swap current element with arr [..! Are performed recursively until there is only one element left at both the side of the is! Lomuto 's partition or descending order ) array is divided using a pivot element will be middle... Our pivot we don ’ t need to compare each elemnt with the above content we fix only 4... Learn quick sort is quite efficient for large data collection ’ ll with... Any change and move on to the next element 23 ide.geeksforgeeks.org, generate link and share the link.... ( multikey ) main function asks for the size of the elements beginning from array! If you find anything incorrect, or random generally considered better when is. I ] choose pivot element to Mergesort seen earlier, if we that. \Gt pivot $,$ s.t or 0 element fix only one 4 and recursively remaining!, last or the middle element only, here the array is sorted is generally using... 32, we ’ ll explore the quicksort algorithm is the following recursively the. Middle, or random huge and stored in external storage in different ways the of... Comments if you find anything incorrect, or random algorithm as it has locality! Case time complexity in most of the array-like first, last, middle or. To the use of extra O ( nLogn ) Filed Under: c Programs,... Operation of merge sort can be written as following, an element is chosen by algorithm. Need of random access is low: the best case: the best browsing experience our...
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https://quant.stackexchange.com/questions/57034/premium-currency-and-volatility | Does the volatility of a Currency Pair depend on the currency in which the premium is paid? For example- will the Volatility of USDJPY change if the premium is paid in USD instead of JPY. Is there any mathematical formulation for this?
It depends, on what you mean by returns. For simple returns: no, for log returns yes. To recap, simple returns are given by $$R_\textrm{simple} = \frac{P_{t+1}}{P_t}-1$$ and log returns are given by $$R_\textrm{log} = \log \left(\frac{P_{t+1}}{P_t}\right).$$ The rate of change is given by $$R = \frac{P_{t+1}}{P_t}.$$
A percentage increase in one currency of a pair, implies a decrease in the other of the same size, so $$R^\textrm{USDJPY} = \frac{P_{t+1}}{P_t} = x$$ implies $$R^\textrm{JPYUSD}\frac{P'_{t+1}}{P'_t} = \frac{1}{x}$$ where $$P'_t$$ is the reverse rate.
In words, if EURUSD is trading at 1.20 today and at 1.212 tomorrow the return from a USD perspective is $$1.212 / 1.20 - 1 = 1\%$$ as today the USD holder was holding 120 cents of USD and tomorrow he would be holding 1.212 cents of USD. On the other hand, from a EUR perspective the loss is $$1.20 / 1.212 - 1 = -0.99\%$$.
We can now do a simple experiment to get a feeling of the volatility for these types of returns in R:
> # Simple returns
> set.seed(1)
> returns <- rnorm(10, 1, 0.01) # One added back to R_simple
> returns
[1] 0.9937355 1.0018364 0.9916437 1.0159528 1.0032951 0.9917953 1.0048743
[8] 1.0073832 1.0057578 0.9969461
> sd(returns - 1)
[1] 0.00780586
> sd(1/returns - 1)
[1] 0.007769419
Clearly, the volatility of simple returns is not the same. Using the same sample suggests that the volatility of the log returns is equal:
> sd(log(returns))
[1] 0.0077874
> sd(log(1/returns))
[1] 0.0077874
This can be shown to always hold with $$x$$ defined as above. The log returns for $$P_t$$ and $$P'_t$$ are then given by $$\log(x)$$ and \begin{align} \log(1/x) &=\log{1} - \log{x} \\ &= -\log{x} \end{align}
The standard deviation of sample is equal to standard deviation of the mirrored around its mean.
• I think in your last code block, you meant to type sd(log(1/returns)) for the second command -- though that does not affect the answer. Jul 31 '20 at 21:56
• Thanks, fixed it! Aug 1 '20 at 5:39
If you're modelling the FX rate as a geometric brownian motion and asking whether the volatility depends on whether you model the rate or the inverse rate, then the answer is no - and we can demonstrate it using Ito's lemma
Assuming the rate $$X$$ obeys \begin{align} {\frac {dX} X} = rdt + \sigma dW \end{align}
for some rate $$r$$ and volatility $$\sigma$$, lemma says that for a function $$f(X,t)$$
\begin{align} df = \Bigl( {\frac {\partial f} {\partial t}} + r X {\frac {\partial f} {\partial X}} + {\frac {\sigma^2 X^2} 2} {\frac {\partial^2 f} {\partial X^2}} \Bigr) dt + \sigma {\frac {\partial f} {\partial X}} dW \end{align}
Substituting in $$f(X,t) = {\frac 1 X}$$, we get
\begin{align} d{\frac 1 X} &= \Bigl( rX {\frac {-1} {X^2}} + {\frac {\sigma^2 X^2 } 2} {\frac {2} {X^3}} \Bigr) dt - \sigma X {\frac {1} {X^2}} dW\\ &= - {\frac 1 X} \Bigl( \bigr(r - \sigma^2 \bigl) dt + \sigma dW\Bigr) \end{align}
So the inverse process $${\frac 1 X}$$ also follows a geometric brownian motion, with a drift of $$-r + \sigma^2$$ and a volatility of $$\sigma$$ (ie. the same volatility as $$X$$)
• You are missing a final measure change here to bring everything into alignment - as the sde you have written is still under the domestic numeraire and not the foreign numeraire. A 1 year outright forward on USDEUR is just the inverse of 1 year outright on EURUSD. There is no convexity adjustment required. Sep 16 '20 at 22:51
• In domestic currency, for GBM the inverse contract is volatility-dependent. This shouldn't be so surprising - we need a vol correction term to make sure the expectations match, otherwise large-vol pairs would have inflated forwards due to the exponential. Sep 16 '20 at 23:53
• Sure, as long as we all can agree that there is no fx volatility information available in fx forward prices. A USD investor looking at a EUR/USD fwd would see a drift of libor-euribor for example while a eur investor looking at a USD/EUR fwd would see a drift of euribor-libor. Sep 17 '20 at 10:38
• Agreed - the misleading post where I implied the opposite has been deleted. Sep 17 '20 at 10:45
As a general rule of thumb, the price of a thing should not depend intrinsically in the units of value in question. Since quoting something in X per 1 unit of a base currency or 1/X per 1 unit of counter currency happen to be questions revolving around units, the price of an option should not depend on that choice. The entire area of measure change and looking at different numeraires is effectively this simple fact taken to its logical conclusion. | 2022-01-18T06:41:42 | {
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https://math.stackexchange.com/questions/347880/proving-that-c-is-a-subset-of-f-1fc?noredirect=1 | # Proving that $C$ is a subset of $f^{-1}[f(C)]$
More homework help. Given the function $f:A \to B$. Let $C$ be a subset of $A$ and let $D$ be a subset of $B$.
Prove that:
$C$ is a subset of $f^{-1}[f(C)]$
So I have to show that every element of $C$ is in the set $f^{-1}[f(C)]$
I know that $f(C)$ is the image of $C$ in $B$ and that $f^{-1}[f(C)]$ is the pre-image of $f(C)$ into $A$. Where I'm stuck is how to use all of this information to show/prove that $C$ is indeed a subset.
Do I start with an arbitrary element (hey, let's call it $x$) of $C$? and then show that $f^{-1}[f(x)]$ is $x$? I could use a little direction here... Thanks.
Since you want to show that $C\subseteq f^{-1}\big[f[C]\big]$, yes, you should start with an arbitrary $x\in C$ and try to show that $x\in f^{-1}\big[f[C]\big]$. You cannot reasonably hope to show that $f^{-1}\big[f[\{x\}]\big]=x$, however: there’s no reason to think that $f$ is $1$-$1$, so there may be many points in $A$ that $f$ sends to the place it sends $x$.
Let $x\in C$ be arbitrary. For convenience let $E=f[C]\subseteq B$. Now what elements of $A$ belong to the set $f^{-1}\big[f[C]\big]=f^{-1}[E]$? By definition $f^{-1}[E]=\{a\in A:f(a)\in E\}$. Is it true that $f(x)\in E$? If so, $x\in f^{-1}[E]=f^{-1}\big[f[C]\big]$, and you’ll have shown that $C\subseteq f^{-1}\big[f[C]\big]$.
• So it was all about using the definition of the preimage along the way to prove my assumption. It is always so clear in hindsight. Hopefully this gets easier before I get to Abstract Algebra and Real Analysis... – Ben Anderson Apr 1 '13 at 5:11
• @Ben: You’ll find that a very large fraction of the more elementary exercises really do just boil down to working through the relevant definitions. You generally have to get a bit deeper into a subject before you encounter problems that require much more than that. – Brian M. Scott Apr 1 '13 at 5:13
• @BrianM.Scott: I have added my answer to a question recently marked as a duplicate of this one. If you feel it is too similar to yours, I will remove it. – robjohn Sep 1 '15 at 21:49
• @robjohn: Both the difference in style and the inclusion of the example make it worth keeping. (Besides, this made me catch a bad typo in my answer!) – Brian M. Scott Sep 2 '15 at 4:26
Copied from my answer to I am issues with proving the following problem: $f^{-1}(f(A)) ⊃ A$, which was closed as a duplicate of this question just before I posted it.
$$f^{-1}\left(f(A)\right)=\left\{x:f(x)\in f(A)\right\}\tag{1}$$ Note that if $x\in A$, then $f(x)\in f(A)$, and by $(1)$, $x\in f^{-1}\left(f(A)\right)$. Therefore, by definition, we have $$A\subset f^{-1}\left(f(A)\right)\tag{2}$$ However, if $f$ is not injective, then $f^{-1}\left(f(A)\right)$ may indeed contain elements not present in $A$; for example let $f:\mathbb{Z}\mapsto\mathbb{Z}$ be defined by $$f(x)=\left\lfloor\frac x2\right\rfloor\tag{3}$$ and let $A$ be the set of even integers. Then $f(A)=\mathbb{Z}$ and $$A\subsetneq\mathbb{Z}=f^{-1}\left(f(A)\right)\tag{4}$$
The problem can be reduced to one-line proof in the following way.
Let $f : Y \leftarrow X$ denote a function.
The definition of inverse images says that for all $y : Y$ and all $x : X,$ we have:
$$f^*(y) \ni x \iff y = f(x)$$
This can be rewritten:
$$f^*(y) \supseteq \{x\} \iff \{y\} \supseteq \{f(x)\}$$
This can be used to prove:
Proposition. For all subsets $B$ of $Y$ and all subsets $A$ of $X$, we have:
$$f^*(B) \supseteq A \iff B \supseteq f_*(A)$$
Once you've proved this, your problem becomes a one-line proof:
$$f^{*}(f_*(A)) \supseteq A \iff f_*(A) \supseteq f_*(A)$$ | 2019-05-21T19:42:23 | {
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https://www.freemathhelp.com/forum/threads/prove-middle-term-formula.129845/ | # Prove middle term formula
#### Mondo
##### New member
Hey,
How can we prove the middle term formula of the arithmetic series with odd number of elements is (n+1)/2. So lets consider the series 3 + 5 + 7 + 9 + 11, we have 5 elements and the middle term is 7 which can be obtained by (n+1)/2 but how to prove this formula?
Thanks
#### Dr.Peterson
##### Elite Member
Hey,
How can we prove the middle term formula of the arithmetic series with odd number of elements is (n+1)/2. So lets consider the series 3 + 5 + 7 + 9 + 11, we have 5 elements and the middle term is 7 which can be obtained by (n+1)/2 but how to prove this formula?
Thanks
The formula gives the index of the middle term, not the middle term itself, as you know. The index of the first term is 1; the index of the last term is n. So the index of the middle term is just the average of 1 and n, which is $$\frac{n+1}{2}$$.
If you want to confirm that, you can calculate the number of terms between this and the first term, and between this and the last term. You will find that the resulting expressions are equal.
#### Mondo
##### New member
Yes (n+1)/2 is for the index but you can get the value as well by adding first and the last term -> (a_1+a_n)/2. I am still not convinced about the proof. How to show that the value of the middle term can be obtained by above formula? It all started from the sum of a arithmetic series -> S_n = 1 + 2 + 3 + ... + (n-2) + (n-1) + n so we notice that we have n/2 pairs of n+1 but only if n is even! If n happen to be odd we have (n-1)/2 pairs and we need to show that the remaining elements is indeed equal to (n+1)/2 so we get (n-1)/2 * (n+1) + (n+1)/2 = (n(n+1))/2
#### lev888
##### Senior Member
Yes (n+1)/2 is for the index but you can get the value as well by adding first and the last term -> (a_1+a_n)/2. I am still not convinced about the proof. How to show that the value of the middle term can be obtained by above formula? It all started from the sum of a arithmetic series -> S_n = 1 + 2 + 3 + ... + (n-2) + (n-1) + n so we notice that we have n/2 pairs of n+1 but only if n is even! If n happen to be odd we have (n-1)/2 pairs and we need to show that the remaining elements is indeed equal to (n+1)/2 so we get (n-1)/2 * (n+1) + (n+1)/2 = (n(n+1))/2
Let's say t is the first term. By definition, the last term is t+(n-1)d
The average of first and last:
(t + t+(n-1)d)/2 = t + ((n-1)/2)d
The term with the index (n+1)/2 is t+((n+1)/2 -1)d = t + ((n-1)/2)d
Last edited:
#### Dr.Peterson
##### Elite Member
Yes (n+1)/2 is for the index but you can get the value as well by adding first and the last term -> (a_1+a_n)/2.
That's true for an arithmetic sequence; but of course that is not the formula you asked about. Moreover, neither is really about the arithmetic series, which is the sum rather than the list of terms.
Please state your actual question. Which formula do you want proved, and what are you assuming as known?
#### HallsofIvy
##### Elite Member
I'm not sure what question you are asking but they are similar.
You have a sequence of 5 numbers and say that the "middle number" is the third one since (5+ 1)/2= 3.
Yes, the sequence 1, 2, 3, ... n and the "middle number IS (1+ n)/2.
We can see that by writing the sum of those numbers:
1+ 2+ 3+ ...+ n-2+ n-1+ n and under it write the numbers reversed
n+ n-1+n-2+...+ 3 + 2+ 1 and add them:
n+1+ n+1+ n+ 1+...+ n+1+ n+1+ n+1
That is n+ 1 added n time so adds to n(n+1). Since it has been added twice, that is 2 times the sum: 1+ 2+ 3+ ....+ n-2+ n-1+ n= n(n+1)/2. There are a total of n numbers so the average is (n+1)/2.
The other question is about adding an arithmetic sequence, a+ (a+ d)+ (a+ 2d)+ ....+ (a+ nd).
There are n "a" and there sum is na. The rest, d+ 2d+ .... +nd= d(1+ 2+ ...+ n). We have already seen that 1+ 2+ ... + n is n(n+1)/2 so the total is na+ n(n+1)/2 and the average is a+ (n+1)/2.
#### Mondo
##### New member
Please state your actual question. Which formula do you want proved, and what are you assuming as known?
I want to prove the arithmetic series formula using the fact that its elements taken in pairs in a certain way all sum to the same number. The equation for each term of this sequence is a_n = a_1 + (n-1)d. So for a_1 = 3, d = 2 and n = 5 we get a series Sn = 3 + 5 + 7 + 9 + 11 By empirical studies we notice that in any arithmetic series with even number of elements we have n/2 pairs of the same value, namely (a_n+a_1). Hence with no doubt we can simply write, the entire sum is equal to n/2 * (a_n+a_1). Now what if n is odd? Then we can say that still there is even numbers of pairs + one more term, so our sum for odd number of elements is (n-1)/2 * (a_n+a_1) + X. Since we know that the formula for even number of elements is valid for odd number of elements as well (proof @HallsofIvy showed) here w need to show that X = (a_n + a_1)/2 so we get (n-1)/2 * (a_n+a_1) + (a_n + a_1)/2 = n/2 * (a_n+a_1). So first we need to find out which element is the remaining one and then, get its value. The element which remains unpaired is the one which has equal number of elements on its left and right side, we know it is element with index (n+1)/2 but hmm can we prove it? What @lev888 derived is almost that, once we prove X is the middle term with index (n+1)/2 we can find its value using @lev888 formulas - its value will be the same as the average of first and last element. So the goal is to prove index of missing element. How to show that?
#### lev888
##### Senior Member
I want to prove the arithmetic series formula using the fact that its elements taken in pairs in a certain way all sum to the same number. The equation for each term of this sequence is a_n = a_1 + (n-1)d. So for a_1 = 3, d = 2 and n = 5 we get a series Sn = 3 + 5 + 7 + 9 + 11 By empirical studies we notice that in any arithmetic series with even number of elements we have n/2 pairs of the same value, namely (a_n+a_1). Hence with no doubt we can simply write, the entire sum is equal to n/2 * (a_n+a_1). Now what if n is odd? Then we can say that still there is even numbers of pairs + one more term, so our sum for odd number of elements is (n-1)/2 * (a_n+a_1) + X. Since we know that the formula for even number of elements is valid for odd number of elements as well (proof @HallsofIvy showed) here w need to show that X = (a_n + a_1)/2 so we get (n-1)/2 * (a_n+a_1) + (a_n + a_1)/2 = n/2 * (a_n+a_1). So first we need to find out which element is the remaining one and then, get its value. The element which remains unpaired is the one which has equal number of elements on its left and right side, we know it is element with index (n+1)/2 but hmm can we prove it? What @lev888 derived is almost that, once we prove X is the middle term with index (n+1)/2 we can find its value using @lev888 formulas - its value will be the same as the average of first and last element. So the goal is to prove index of missing element. How to show that?
If the number of pairs is (n-1)/2, then that's the number of elements to the left of the middle one. Add 1 and you get the middle index.
#### JeffM
##### Elite Member
I want to prove the arithmetic series formula using the fact that its elements taken in pairs in a certain way all sum to the same number. The equation for each term of this sequence is an=a1+(n−1)dan=a1+(n-1)d. So for a1=3a1=3, d=2d=2 and n=5n=5 we get a series Sn=3+5+7+9+11Sn=3+5+7+9+11 By empirical studies we notice that in any arithmetic series with even number of elements we have n2n2 pairs of the same value, namely (an+a1)(an+a1). Hence with no doubt we can simply write, the entire sum is equal to n2⋅(an+a1)n2⋅(an+a1). Now what if nn is odd? Then we can say that still there is even numbers of pairs + one more term, so our sum for odd number of elements is n−12⋅(an+a1)+Xn-12⋅(an+a1)+X. Since we know that the formula for even number of elements is valid for odd number of elements as well (proof @HallsofIvy showed) here w need to show that X=an+a12X=an+a12 so we get n−12⋅(an+a1)+an+a12=n2⋅(an+a1)n-12⋅(an+a1)+an+a12=n2⋅(an+a1). So first we need to find out which element is the remaining one and then, get its value. The element which remains unpaired is the one which has equal number of elements on its left and right side, we know it is element with index n+12n+12 but hmm can we prove it? What @lev888 derived is almost that, once we prove XX is the middle term with index n+12n+12 we can find its value using @lev888 formulas - its value will be the same as the average of first and last element. So the goal is to prove index of missing element. How to show that?
Proofs are technically impossible for us to help with because we do not know the axioms, definitions, and previously proved theorems that you can use. We have to guess. But here is a fairly concise proof.
$$\displaystyle \text {Prove that if } n \text { is an odd positive integer, then}\\ \left (\displaystyle \sum_{j=1}^n a + (j - 1)d \right ) = n\{a + (m - 1)d\}, \text { where } m = \dfrac{n + 1}{2}.$$
First, lets demonstrate that a + (m - 1)d is the middle term. Because n is odd, let x be the number of terms that precede and succeed the middle term.
$$\displaystyle x + 1 + x = n \implies 2x = n - 1 \implies x = \dfrac{n - 1}{2}.$$
And obviously the middle term is x + 1
$$\displaystyle \dfrac{n - 1}{2} + 1 = \dfrac{n - 1}{2} + \dfrac{2}{2} = \dfrac{n + 1}{2} = m \text { by definition.}$$
$$\displaystyle \therefore \text {the middle term of the series is the } m^{\text{th}} \text {term} \implies$$
$$\displaystyle \text {the middle term of the series} = a + (m - 1) \text{ by definition.}$$
$$\displaystyle \left ( \sum_{j=1}^n a + (j - 1)d \right ) = \left ( \sum_{j=1}^n a - d + jd \right ) =$$
$$\displaystyle \left ( \sum_{j=1}^n a - d \right ) + \left ( \sum_{j=1}^n jd \right ) = \left \{ (a - d) * \sum_{j=1}^n 1 \right \} + \left ( d * \sum_{j=1}^n j \right ) =$$
$$\displaystyle (a - d)n + d * \dfrac{n(n + 1)}{2} = n \left ( a - d + d * \dfrac{(n + 1)}{2} \right ) = n(a - d + dm) \implies$$
$$\displaystyle \left ( \sum_{j=1}^n a + (j - 1)d \right ) = n\{a + d(m - 1)\} \text { Q.E.D.}$$
#### Mondo
##### New member
Thank you @JeffM & @lev888. The prove of middle term index is nice and simple. @JeffM the only caveat I have to your calculations is at some point you use the formula for the sum of arithmetic series which is in fact the very thing we are proving... anyway I think the prove is done.
#### JeffM
##### Elite Member
Thank you @JeffM & @lev888. The prove of middle term index is nice and simple. @JeffM the only caveat I have to your calculations is at some point you use the formula for the sum of arithmetic series which is in fact the very thing we are proving... anyway I think the prove is done.
There is a distinct proof for what I used, which does not depend on the general formula for an arithmetic series.
$$\displaystyle \sum_{j=1}^n j = \dfrac{n(n + 1)}{2}.$$
$$\displaystyle n = 1 \implies \sum_{j=1}^n j = 1 = \dfrac{2}{2} = \dfrac{1(1 + 1)}{2}.$$
Therefore there exists at least one positive integer for which the proposition is true. Let k be an arbitrary one of those positive integers.
$$\displaystyle \therefore \sum_{j=1}^k j = \dfrac{k(k + 1)}{2} = \dfrac{k^2 + k}{2}.$$
$$\displaystyle \sum_{j=1}^{k+1} j = \left ( \sum_{j=1}^k j \right ) + k + 1 = \dfrac{k^2 + k}{2} + \dfrac{2k + 2}{2} = \dfrac{k^2 + 3k + 2}{2} = \dfrac{(k + 1)(k + 2)}{2} = \dfrac{(k + 1)\{(k + 1) + 1\}}{2}.$$
This is why I said it is so hard to help with proofs.
#### Mondo
##### New member
Yes @lex, this is probably the easiest proof but here I wanted to dig deeper into the case where we have odd number of elements. I feel like we covered it all now. Thanks!
#### lex
##### Full Member
x+1+x=n⟹2x=n−1⟹x=n−12.x+1+x=n⟹2x=n−1⟹x=n−12.\displaystyle x + 1 + x = n \implies 2x = n - 1 \implies x = \dfrac{n - 1}{2}.
And obviously the middle term is x + 1
If the number of pairs is (n-1)/2, then that's the number of elements to the left of the middle one. Add 1 and you get the middle index.
Yes. I see that. Dealt with by JeffM and lev888 | 2021-06-19T14:51:48 | {
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https://math.stackexchange.com/questions/2660247/relation-between-the-roots-of-a-function/2660254 | # Relation between the roots of a function
I have this question from my exam where it's asked to find the sum: $$S=\sum_{k=1}^n \frac{1}{(1-r_k)^2}$$ where $r_k$ are the roots of $$f(x)=x^n-2x+2\quad,n\ge3$$ I recalled this relation $$\frac{f'(x)}{f(x)}=\sum_{k=1}^n \frac{1}{x-r_k}$$ and quickly realised that $$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$$ and using derivative would easily produce my answer. Indeed $$\frac{d}{dx}\left( \frac{f'(x)}{f(x)} \right)=\frac{f''(x)f(x)-(f'(x))^2}{f(x)^2}=-\sum_{k=1}^n \frac{1}{(x-r_k)^2}$$ Evaluating at $x=1$ gives $$S=\frac{(n-2)^2-n(n-1)}{1^2}=4-3n$$ However after the exam I realised that I have no ideea why $$\frac{f'(x)}{f(x)}=\sum_{k=1}^n \frac{1}{x-r_k}$$ and I just memorized it, is there a nice way to show it? And of course maybe a more elegant solution to this exam question?
• Are you aware of the Newton-Raphson Method? – Harry Alli Feb 21 '18 at 13:34
• Never heard of, but thanks I will google it. – Zacky Feb 21 '18 at 13:36
$$f(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$$
Using logarithm,
$$\ln (f(x))=\ln (x-r_1)+\ln(x-r_2)+\cdots +\ln(x-r_n)$$
Taking derivative,
$$\frac{f'(x)}{f(x)}= \frac{1}{(x-r_1)}+\frac{1}{(x-r_2)} +\cdots +\frac{1}{(x-r_n)}$$
Done!
• The issue with this post that Joonas Ilmavirta points out in his own answer can be fixed simply by replacing $\ln(\,)$ with $\ln |\,|$ everywhere. – Paul Sinclair Feb 21 '18 at 17:40
• But then the function may not be differentiable any more. @Paul – Jaideep Khare Feb 21 '18 at 18:34
• In that case can't we just split ln|x-a| into ln(x-a) when x>a and ln(a-x) when x<a. Derivating first gives $$\frac{d}{dx}ln(x-a)=\frac{1}{x-a}$$ and $$\frac{d}{dx}ln(a-x) = \frac{1}{a-x}(a-x)' =\frac{1}{x-a}$$ So $$\frac{d}{dx}ln|x-a|=\frac{1}{x-a}, x \neq a$$ – Zacky Feb 21 '18 at 18:53
• @Joonas $\log(ab)=\log a+\log b$, doesn't hold true, I agree, but there is a constant factor of $2n\pi$ which vanishes on taking derivative. – Jaideep Khare Feb 22 '18 at 12:07
• Actually, the logarithmic derivative is a linear operator in itself, with nice properties. One does not need to go through the problematic step of writing $\log(f(x))$. – Tom-Tom Feb 27 '18 at 19:53
While Jaideep Khare's computation with the logarithm gives a good quick way to memorize or derive the result, there is an issue: The logarithms of negative numbers are not defined if one works over the reals, and care must be taken with branches if one moves to the complex plane. When $r_k$ is fixed and $x$ ranges through all the reals, $x-r_k$ will inevitably get negative values.
Here is a way to prove the desired result without resorting to logarithms: By the fundamental theorem of algebra a polynomial can be written in terms of its roots and a coefficient $a\in\mathbb C$ as $$f(x) = a(x-r_1)\cdots(x-r_n).$$ The coefficient of $x^n$ is $a$ and in your case you know it to be $1$, but it is unimportant. The derivative can be computed by differentiating each term in the product in turn: $$f'(x) = a(x-r_2)\cdots(x-r_n) + a(x-r_1)(x-r_3)\cdots(x-r_n) + \dots + a(x-r_1)\cdots(x-r_{n-1}),$$ where in the $n$th term of the sum the $n$th term of the product has been dropped. When $f(x)\neq0$, it is easy to divide each term of $f'(x)$ by $f(x)$, and we get $$\frac{f'(x)}{f(x)} = \frac{1}{x-r_1} + \frac{1}{x-r_2} +\dots+ \frac{1}{x-r_n}.$$
The result and the proof are also applicable to complex polynomials, whereas with logarithms $f(z)$ will always hit the branching set of the logarithm, no matter how you choose it. Also, $\log(ab)=\log(a)+\log(b)$ is not true in general for complex $a$ and $b$.
Finally, let me emphasize that this result holds for a polynomial function $f$ (with roots counted with multiplicity), not for all functions in general. In your question you are working with polynomials but you speak about "functions", so beware of generalizing the result beyond what's true.
• I understood this too, thank you! – Zacky Feb 21 '18 at 14:32
• The trick with the log is slick, but I have to give the point to this answer, since it's more general. – David Stanley Feb 21 '18 at 15:05
Another approach, which is longer but would also work for more general symmetric expressions in the roots, is to use Vieta's formulae.
If a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ has roots $r_1, r_2, \dots, r_k$, then the ratio $-\frac{a_{n-1}}{a_n}$ gives us the sum $r_1 + r_2 + \dots + r_k$.
To turn this into the sum we want, we should modify $f(x)$ slightly.
First, $f(t-x)$ has roots $t - r_1, t - r_2, \dots, t - r_k$.
Second, $f(t-1/x)$, though it is not a polynomial, is still zero when $x = 1/(t-r_1), \dots, 1/(t-r_k)$.
However, multiplying through by $x^n$ gives the polynomial $x^n f(t-1/x)$ which has roots at $1/(t-r_1), \dots, 1/(t-r_k)$.
To apply the Vieta's formula we want to use, it remains to figure out the coefficients of $x^n$ and $x^{n-1}$ in $$x^n f(t-1/x) = x^n a_n (t-1/x)^n + x^n a_{n-1} (t - 1/x)^{n-1} + \dots + x^n a_1 (t-1/x) + x^n a_0.$$ To get a multiple of $x^n$ from a term $x^n a_k (t-1/x)^k$, we have to choose the $t^k$ term of the binomial expansion of $(t-1/x)^k$. This tells us that the coefficient of $x^n$ in this polynomial is $$a_n t^n + a_{n-1} t^{n-1} + \dots + a_0 = f(t).$$ To get a multiple of $x^{n-1}$ from a term $x^n a_k (t-1/x)^k$, we have to choose the $\binom k1 t^{k-1}(-1/x)$ term of the binomial expansion of $(t-1/x)^k$. This tells us that the coefficient of $x^{n-1}$ in this polynomial is $$- a_n \cdot n t^{n-1} - a_{n-1} \cdot (n-1)t^{n-2} - \dots - a_1 = -f'(t).$$ Putting these together, we get that the sum of the roots of $x^n f(t-1/x)$ is $$\sum_{i=1}^n \frac{1}{t-r_i} = \frac{f(t)}{f'(t)},$$ as desired.
(For the specific case of the polynomial $f(x) = x^n - 2x + 2$, this would go faster since we wouldn't have to derive the general formula.)
In general, if $$f(x)=\prod_{k=1}^{n}f_k(x)$$ Then $$\ln f(x)=\sum_{k=1}^{n}\ln f_k(x)$$ and thus $$\frac{f'(x)}{f(x)}=\sum_{k=1}^{n}\frac{f_k'(x)}{f_k(x)}$$ | 2019-07-20T07:14:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2660247/relation-between-the-roots-of-a-function/2660254",
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"openwebmath_perplexity": 175.41747699738866,
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https://mathematica.stackexchange.com/questions/5663/about-multi-root-search-in-mathematica-for-transcendental-equations/86946 | About multi-root search in Mathematica for transcendental equations
I have some questions for multiroot search for transcendental equations. Is there any clever solution to find all the roots for a transcendental equation in a specific range?
Perhaps FindRoot is the most efficient way to solve transcendental equations, but it only gives one root around a specific value. For example,
FindRoot[BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]], {x, 10}]
Of course, one can first Plot the equation and then choose several start values around each root and then use FindRoot to get the exact value.
1. Is there any elegant way to find all the roots at once?
Actually, I come up with this question when I solve the eigenequation for optical waveguides and I want to get the dispersion relation. I find ContourPlot is very useful to get the curve of the dispersion relation. For example,
ContourPlot[BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin@Sin[a*x] == 0, {x, 0, 10}, {a, 0, 4}]
You can get
2. Is there any elegant way to get all the values in the ContourPlot for x when a==0 ?
3. Is it possible to know how the ContourPlot gets all the points shown in the figure? Perhaps we can harness it to get all the roots for the transcendental equation.
• You might be interested in this question. Anyway, for finding the roots of a function of a single variable, you can directly use the output of Plot[] to find initial approximations for FindRoot[]. If you're interested in that approach, I can write up an answer. – J. M. will be back soon May 17 '12 at 8:09
• @J.M. I'm expecting for your answer ~~ – yulinlinyu May 17 '12 at 8:53
• A tiny comment on Reduce[]-based methods: all of these hinge on the fact that Reduce[] apparently knows quite a fair bit about the transcendental functions built within Mathematica. If, however, you are dealing with a black-box function that can only evaluate at numerical values (you can simulate this behavior with something like f[x_?NumericQ] := Haversine[Pi x]), then Reduce[] won't be able to do much. – J. M. will be back soon May 17 '12 at 9:15
• @J.M. Yes, I have tried Reduce[], and I find it is very slow, especially for a much more complex equation. – yulinlinyu May 17 '12 at 9:50
• @yulinlinyu It seems you wanted to accept a few answers. You can accept ONLY ONE of the answers (clicking the green tick beside an answer), which you find the most helpful. You can upvote EVERY answer you find helpful or downvote if you find an answer misleading. – Artes May 18 '12 at 1:57
Borrowing almost verbatim from a recent response about finding extrema, here is a method that is useful when your function is differentiable and hence can be "tracked" by NDSolve.
f[x_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]]
In[191]:= zeros =
Reap[soln =
y[x] /. First[
NDSolve[{y'[x] == Evaluate[D[f[x], x]], y[10] == (f[10])},
y[x], {x, 10, 0},
Method -> {"EventLocator", "Event" -> y[x],
"EventAction" :> Sow[{x, y[x]}]}]]][[2, 1]]
During evaluation of In[191]:=
NDSolve::mxst: Maximum number of 10000 steps reached at the point
x == 1.5232626281716416*^-124. >>
Out[191]= {{9.39114, 8.98587*10^-16}, {6.32397, -3.53884*10^-16},
{3.03297, -8.45169*10^-13}, {0.886605, -4.02456*10^-15}}
Plot[f[x], {x, 0, 10},
Epilog -> {PointSize[Medium], Red, Point[zeros]}]
If it were a trickier function, one might use Method -> {"Projection", ...} to enforce the condition that y[x] is really the same as f[x]. This method may be useful in situations (if you can find them) where we have one function in one variable, and Reduce either cannot handle it or takes a long time to do so.
Addendum by J. M.
WhenEvent is now the documented way to include event detection in NDSolve, so using it along with the trick of specifying an empty list where the function should be, here's how to get a pile of zeroes:
f[x_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]]
zeros = Reap[NDSolve[{y'[x] == D[f[x], x], WhenEvent[y[x] == 0, Sow[{x, y[x]}]],
y[10] == f[10]}, {}, {x, 10, 0}]][[-1, 1]];
Plot[f[x], {x, 0, 10}, Epilog -> {PointSize[Medium], Red, Point[zeros]}]
• @ Daniel Lichtblau, it's a very clever, efficient and general way to get all the roots and can be harnessed in many other cases. Thank you a lot. – yulinlinyu May 18 '12 at 1:39
• Excellent! However, this method doesn't work in v10, possibly a bug. – luyuwuli Nov 25 '14 at 4:00
• @luyuwuli Thanks for pointing that out. I tracked this to an option setting change wherein the default became MaxSteps -> Infinity. You can restore the behavior shown above with an explicit MaxSteps -> 10000. I will ask in house about whether the change is for the better given that it can cause an example like this to hang. – Daniel Lichtblau Nov 25 '14 at 20:44
• @DanielLichtblau, this is a bug that I think is not related to the MaxSteps -> Automatic change - at least not directly. You can get this to work with NDSolve[{y'[x] == Evaluate[D[f[x], x]], y[10] == (f[10])}, y[x], {x, 10, 10^-16}, Method -> {"EventLocator", "Event" -> y[x], "EventAction" :> Sow[{x, y[x]}]}] where the 0 is replaced with a small number. – user21 Nov 26 '14 at 8:58
• @Nigel I think for complex-valued functions one will need to track real and imaginary parts both as a function of a real parameter. – Daniel Lichtblau Jun 29 '17 at 17:05
I might as well elaborate on my comment. Here is a modification of Stan Wagon's FindAllCrossings[] function (from his book Mathematica in Action, second edition) that uses Plot[] to generate the initial approximations to be subsequently polished by FindRoot[]:
Options[FindAllCrossings] =
Sort[Join[Options[FindRoot], {MaxRecursion -> Automatic,
PerformanceGoal :> \$PerformanceGoal, PlotPoints -> Automatic}]];
FindAllCrossings[f_, {t_, a_, b_}, opts___] := Module[{r, s, s1, ya},
{r, ya} = Transpose[First[Cases[Normal[
Plot[f, {t, a, b}, Method -> Automatic,
Evaluate[Sequence @@
FilterRules[Join[{opts}, Options[FindAllCrossings]],
Options[Plot]]]]], Line[l_] :> l, Infinity]]];
s1 = Sign[ya]; If[ ! MemberQ[Abs[s1], 1], Return[{}]];
s = Times @@@ Partition[s1, 2, 1];
If[MemberQ[s, -1] || MemberQ[Take[s, {2, -2}], 0],
Union[Join[Pick[r, s1, 0],
Select[t /. Map[FindRoot[f, {t, r[[#]], r[[# + 1]]},
Evaluate[Sequence @@
FilterRules[Join[{opts}, Options[FindAllCrossings]],
Options[FindRoot]]]] &,
Flatten[Position[s, -1]]], a <= # <= b &]]], {}]]
Try it out:
FindAllCrossings[BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]], {x, 0, 100},
WorkingPrecision -> 20]
{0.88660463531346207679, 3.0329660890136683539,
6.3239665137114782212, 9.3911434075850854017, 12.589067252797192964,
15.687789316501627036, 18.865248000326751595, 21.976728589463954937,
25.144727536576135544, 28.263111694495812775, 31.425621611972587076,
34.548333253230934213, 37.707250582575859710, 40.832929091244028281,
43.989309899299199529, 47.117149292753968158, 50.271642808648651326,
53.401126310508732375, 56.554160423425108364, 59.684936863656120488,
62.836808589969946989, 65.968628466404205710, 69.119552435670107405,
72.252232119042699356, 75.402368490602128759, 78.535768909563822046,
81.685240379776511855, 84.819253682565487033, 87.968156330842562535,
91.102697190752433240, 94.251107663305908556, 97.386107414041584404}
A different implementation involves the use of the MeshFunctions option of Plot[] to generate the seeds. Here's how it looks:
FindAllCrossings[f_, {t_, a_, b_}, opts : OptionsPattern[]] := Module[{r},
r = Cases[Normal[Plot[f, {t, a, b}, MeshFunctions -> (#2 &), Mesh -> {{0}},
Method -> Automatic, Evaluate[Sequence @@
FilterRules[Join[{opts}, Options[FindAllCrossings]],
Options[Plot]]]]],
Point[p_] :> SetPrecision[p[[1]], OptionValue[WorkingPrecision]],
Infinity];
If[r =!= {},
Union[Select[t /. Map[FindRoot[f, {t, #}, Evaluate[Sequence @@
FilterRules[Join[{opts},
Options[FindAllCrossings]],
Options[FindRoot]]]] &, r],
a <= # <= b &]], {}]]
This version might be a bit faster in some cases, but it no longer has the safety feature of root bracketing in the previous version.
• Of course, this method won't find even-order multiple roots (i.e. tangencies to the horizontal axis), but almost any numerical method will have difficulty with them anyway. If you know for certain that there are tangencies within the range of interest, you might consider using FindMinimum[]/FindMaximum[] instead to find them. – J. M. will be back soon May 17 '12 at 8:49
• Usage example here mathematica.stackexchange.com/a/94881/193 – Dr. belisarius Sep 17 '15 at 16:09
• @J. M. Nice answer. It seems this method with FindAllCrossings[] does not work for zeroes of complex functions. For example I get the error "First::first: "{} has a length of zero and no first element." with pure imaginary Sin[x]/Sin[I*x]. Can it be used for complex functions? – Nigel1 Jun 29 '17 at 16:44
• @Nigel, the function here would not work directly since Plot[] does not work for complex functions. One might try using FindAllCrossings2D[] on the real and imaginary parts, tho. – J. M. will be back soon Jul 26 '17 at 4:50
One can use Solve as well, e.g.
s = Solve[ BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[ Sin[x]] == 0 && 0 < x < 10, x]
Solve::incs: Warning: Solve was unable to prove that the solution set found is complete. >>
{{x -> Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
0.886604635313462076794393681674}]},
{x -> Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
3.03296608901366835385376172847}]},
{x -> Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
6.32396651371147786252003752922}]},
{x -> Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
9.39114340758508579766919382120}]}}
Solve similarly as Reduce cannot prove that the set of solutions is complete. It returns the result in the form of rules but we can show that they return the same set solutions e.g. :
r = Reduce[ BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[ Sin[x]] == 0 && 0 < x < 10, x];
s[[All, 1, 2]] == List @@ r[[All, 2]]
Reduce::incs: Warning: Reduce was unable to prove that the solution set found is complete. >>
True
Edit
Defining
f[x_, a_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[a x]]
we can visualise the graphs of functions f[x,a] making use of ParametricPlot3D (look at answers to this question), e.g.
Show[
ParametricPlot3D[ Evaluate[ Table[{x, a, f[x, a]}, {a, 0, 5}]],
{x, 0, 10}, BoxRatios -> {10, 8, 4}],
ParametricPlot3D[{x, 1, f[x, 1]},
{x, 0, 10}, PlotStyle->{Thick, Darker@Red}, BoxRatios -> {10, 8, 4}]
Red thick curve is the graph of f[x,1],
or we can make use of Plot3D as well in the following way :
Plot3D[ f[x, a], {x, 0, 10}, {a, 0, 4},
MeshFunctions -> {#1 &, #2 &, #3 &}, Mesh -> {9, 3, 5}, Filling -> 0,
PlotPoints -> 200, MaxRecursion -> 5]
]
The option Filling ->0 makes an impression that the level f[x,a] == 0 is like the surface of water.
For finding all roots in a given interval I use the following function:
Clear[findRoots]
Options[findRoots] = Options[Reduce];
findRoots[gl_Equal, {x_, von_, bis_},
prec : (_Integer?Positive | MachinePrecision | Infinity) : MachinePrecision,
wrap_: Identity, opts : OptionsPattern[]] :=
Module[{work, glp, vonp, bisp},
{glp, vonp, bisp} = {gl, von, bis} /. r_Real :> SetPrecision[r, prec];
work = wrap@Reduce[{glp, vonp <= x <= bisp}, opts];
work = {ToRules[work]};
If[prec === Infinity, work, N[work, prec]]];
Example
bes[x_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]]
findRoots[bes[x] == 0, {x, 0, 100}]
During evaluation of In[35]:= Reduce::incs: Warning: Reduce was unable to prove that the solution set found is complete. >>
Out[39]= {{x -> 0.886605}, {x -> 3.03297}, {x -> 6.32397}, {x -> 9.39114}, {x ->
12.5891}, {x -> 15.6878}, {x -> 18.8652}, {x -> 21.9767}, {x ->
25.1447}, {x -> 28.2631}, {x -> 31.4256}, {x -> 34.5483}, {x ->
37.7073}, {x -> 40.8329}, {x -> 43.9893}, {x -> 47.1171}, {x ->
50.2716}, {x -> 53.4011}, {x -> 56.5542}, {x -> 59.6849}, {x ->
62.8368}, {x -> 65.9686}, {x -> 69.1196}, {x -> 72.2522}, {x ->
75.4024}, {x -> 78.5358}, {x -> 81.6852}, {x -> 84.8193}, {x ->
87.9682}, {x -> 91.1027}, {x -> 94.2511}, {x -> 97.3861}}
The parameter prec gives the precision for the calculation. prec=Infinity gives exact solutions.
wrap (default: Identity) is a function you can apply to the solutions.
• With this example don't work: findRoots[Im[Zeta[x]^(1/Log[x])] == 0, {x, 0, 1}]? – Mariusz Iwaniuk May 16 '18 at 19:29
Regarding your first question:
For a certain set of equations, Reduce is able to find all roots and prove that no more roots exist in a given interval. I am not sure how this works, but there's an interesting blog post about it here.
When it is not able to guarantee (using symbolic methods) that there are no more solutions, it may still return a set, but it will give a warning. This is the case for your equation.
Plot[BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]], {x, 0, 10}]
Reduce[BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]] == 0 && 0 < x < 10, x]
(*
Reduce::incs: Warning: Reduce was unable to prove that the solution set found is complete. >>
*)
(* ==>
x == Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
0.886604635313462076794393681674}] ||
x == Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
3.03296608901366835385376172847}] ||
x == Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
6.32396651371147786252003752922}] ||
x == Root[{BesselJ[1, #1]^2 + BesselK[1, #1]^2 - Sin[Sin[#1]] &,
9.39114340758508579766919382120}]
*)
ToRules@N[%]
(* ==>
Sequence[{x -> 0.886605}, {x -> 3.03297}, {x -> 6.32397}, {x -> 9.39114}]
*)
Note that it is important to restrict the search to an interval. Otherwise Reduce will tell you that it can't find any solutions.
Very recently, I learned a useful procedure due to J. P. Boyd (see also this and this) that involves expanding a function as a Chebyshev polynomial series, forming the so-called "colleague matrix", and then finding the eigenvalues of this matrix, which are often good approximations to the roots of the original function. I shall present how to do a barebones version of this strategy in Mathematica.
The first step is to form the Chebyshev series coefficients of the function concerned. One could either start from the Taylor series and use any number of methods to convert these coefficients to Chebyshev coefficients. Another route, which I shall present here, is to evaluate the function at the so-called "Chebyshev nodes", suitably transformed to the domain of the original function, and then using the (type-1) discrete cosine transform to produce the Chebyshev coefficients. It goes like this:
f[x_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]];
{xmin, xmax} = {1/2, 100};
n = 128; (* arbitrarily chosen integer *)
prec = 20; (* precision *)
cnodes = Rescale[N[Cos[Pi Range[0, n]/n], prec], {-1, 1}, {xmin, xmax}];
cc = Sqrt[2/n] FourierDCT[f[cnodes], 1];
cc[[{1, -1}]] /= 2;
n here should be chosen to be greater than or equal to the number of roots known to be in the given interval. That is usually determined through a preliminary Plot[]. (A more sophisticated version, as used in the Chebfun package for MATLAB, does an adaptive increase in the order of the underlying Chebyshev approximation using a technique similar to Clenshaw-Curtis quadrature. I have chosen not to implement this for now.)
With the Chebyshev coefficients now available, one can then form the colleague matrix like so:
colleague = SparseArray[{{i_, j_} /; i + 1 == j :> 1/2,
{i_, j_} /; i == j + 1 :> 1/(2 - Boole[j == 1])},
{n, n}] -
SparseArray[{{i_, n} :> cc[[i]]/(2 cc[[n + 1]])}, {n, n}];
We can then find the eigenvalues of this matrix, filter out irrelevant eigenvalues, and then use Rescale[] to transform the remaining eigenvalues back to the domain of the original function:
rts = Sort[Select[DeleteCases[
Rescale[Eigenvalues[colleague], {-1, 1}, {xmin, xmax}],
_Complex | _DirectedInfinity], xmin <= # <= xmax &]]
{0.8866466, 3.0327390, 6.3243343, 9.3928065, 12.5892495, 15.6908929,
18.8634014, 21.9762069, 25.1562125, 28.3003285, 31.4743983, 34.5197235,
37.6746911, 40.8716527, 43.9572481, 47.1394116, 50.2776398, 53.3723848,
56.5916636, 59.6440741, 62.8638369, 66.0001767, 69.0778338, 72.2121395,
75.3874360, 78.5335813, 81.6858635, 84.8171664, 87.9688503, 91.1015749,
94.2504610, 97.3866141}
The eigenvalues obtained look a bit rough here; this is due to the ill-behavior of the original function near the origin. For more well-behaved functions, the eigenvalues are often quite accurate roots of the original function. Of course, one can always use FindRoot[] to polish off these approximations.
For your first question one can use Ted Ersek's package here :
RootSearch[BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]] == 0, {x, 0, 10}]
(* {{x -> 0.886605}, {x -> 3.03297}, {x -> 6.32397}, {x -> 9.39114}} *)
For a finite range of interest, NSolve works well
f[x_] = BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]];
Manipulate[
Module[{sol},
Column[{
sol = NSolve[{f[x] == 0, 0 <= x <= xmax}, x],
Plot[f[x], {x, 0, xmax},
Epilog -> {Red, AbsolutePointSize[6],
Point[{x, f[x]} /. sol]},
ImageSize -> 360]}]],
{{xmax, 16}, 1, 31, 3, Appearance -> "Labeled"}]
EDIT: With later Mathematica versions (e.g., 10.4.1), some of the roots can be missed unless you use higher precision. The //N after NSolve is used to reduce the displayed precision in the results.
Manipulate[
Module[{sol},
Column[
{sol = NSolve[{f[x] == 0, 0 <= x <= xmax}, x,
WorkingPrecision -> 30] // N,
Plot[f[x], {x, 0, xmax},
Epilog -> {Red, AbsolutePointSize[6],
Point[{x, f[x]} /. sol]},
ImageSize -> 360]}]],
{{xmax, 16}, 1, 31, 3, Appearance -> "Labeled"}]
` | 2020-01-20T23:18:59 | {
"domain": "stackexchange.com",
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"lm_q2_score": 0.8918110454379297,
"lm_q1q2_score": 0.8405379072457143
} |
https://math.stackexchange.com/questions/2452609/in-how-many-ways-can-5-balls-of-different-colours-be-placed-in-3-boxes-of-di | # In how many ways can $5$ balls of different colours be placed in $3$ boxes of different sizes if no box remains empty?
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.
My attempt:-
First choose 3 balls to be placed in 3 boxes so that none of them remain empty in ${{5}\choose{3}}\cdot3! = 60$ ways.
Now remaining 2 balls can go into any of the 3 boxes in $3\cdot3 = 9$ ways.
Total number of ways $= 60\cdot9 = 540$.
Where am I going wrong ?
• When you place the last two balls, you are over counting. Your answer better be less than $243$. Oct 1 '17 at 10:51
There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are $$\binom{3}{1}2^5$$ ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is $$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$ by the Inclusion-Exclusion Principle.
Where am I going wrong?
You count each distribution in which one box receives three balls and the others receive one three times, once for each way you could place one of those three balls first.
You count each distribution in which two of the boxes receive two balls and the other box receives one four times, once for each way you could place one of the two balls in each of the two boxes with two balls first.
Three balls in one box and one ball in each of the others: There are three ways to choose which box receives three balls, $\binom{5}{3}$ ways to choose which three balls are placed in that box, and $2!$ ways to distribute the remaining balls. Hence, there are $$\binom{3}{1}\binom{5}{3}2!$$ ways to distribute the balls so that three balls are placed in the same box.
Two boxes receives two balls and one box receives one ball: There are three ways to choose which box receives only one ball and five ways to choose the ball that is placed in that box. There are $\binom{4}{2}$ ways to choose which two of the remaining four balls are placed in the smaller of the two remaining boxes. The other two balls must be placed in the remaining box. Hence, there are $$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$ ways to distribute the balls so that two boxes receive two balls and one box receives one.
Observe that $$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
Since you counted distributions in which one box receives three balls and the others receive one three times and distributions in which two boxes receive two balls and the other receives one four times, you obtained $$3\binom{3}{1}\binom{5}{3}2! + 4\binom{3}{1}\binom{5}{1}\binom{4}{2} = \binom{5}{3} \cdot 3! \cdot 3^2$$
• Even though I know inclusion principle, I am not able to figure out when to apply it for permutation sums. Is there anyway to get a grip over it ? Oct 1 '17 at 11:26
• In this case, you could avoid using the Inclusion-Exclusion Principle, as the second method I used demonstrates. Until you have applied the IEP to several problems, it can be tricky to figure out what you have to exclude. For instance, if you have to separate four red balls in an arrangement, you need to understand that two pairs of red balls could mean two separate pairs involving four red balls or two overlapping pairs formed by three consecutive red balls. I suggest applying the IEP to problems that you know how to solve in another way to test whether you are doing it right. Oct 1 '17 at 11:34
To apply Inclusion-Exclusion, we let $S(i)$ be the set of arrangements where box $i$ is empty. Then \begin{align} N(j) &=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|\\ &=\underbrace{\ \ \ \binom{3}{j}\ \ \ }_{\substack{\text{number of}\\\text{ways to choose}\\\text{j empty boxes}}}\underbrace{\ (3-j)^5\ \vphantom{\binom{3}{j}}}_{\substack{\text{number of}\\\text{maps into}\\\text{3-j boxes}}} \end{align} Inclusion-Exclusion says the number of arrangements with no boxes empty is \begin{align} \sum_{j=0}^3(-1)^{j-0}\binom{j}{0}N(j) &=\sum_{j=0}^3(-1)^j\binom{3}{j}(3-j)^5\\ &=3^5-3\cdot2^5+3\cdot1^5-1\cdot0^5\\[12pt] &=150 \end{align}
We want to put $5$ distinguished balls into $3$ distinguished boxes, so our configuration space is of size $3^5$. We need to ensure each box has at least one ball. There are sufficiently few partitions of $5$ into $3$ parts that we can list them and calculate their multiplicities ... \begin{array}{l|l} Partition & Multiplicity \\ \hline (5,0,0) & 3 \\ (4,1,0) & 30 \\ (3,2,0) & 60 \\ (3,1,1) & \color{blue}{60} \\ (2,2,1) & \color{blue}{90} \\ \end{array} So there will be $\color{blue}{150}$ configurations.
• Thank you for the answer. Can you tell me why is my approach wrong ? Why is it over counting ? If possible, can we arrive at the same answer using my approach? Oct 1 '17 at 10:37 | 2021-10-19T16:47:47 | {
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https://math.stackexchange.com/questions/735483/trace-of-tensor-product-vs-tensor-contraction/735617 | # Trace of tensor product vs Tensor contraction
I have come across various sources that talk about traces of tensors. How does that work? In particular, there seem to be such an equality:
$$\text{Tr}(T_1\otimes T_2)=\text{Tr}(T_1)\text{Tr}(T_2)\;\;\;...(1)$$
The Wiki page "Tensor Contraction" speaks of tensor contraction as some generalization of trace, though without providing any formulation or example.
My questions: How do they all work? What is trace for a tensor? How does such trace interact with tensor product?
In particular, I have this contraction: (following Einstein's summation convention) $$F^{\mu\nu}F_{\mu\nu}$$ where $F$ is a rank-2 tensor and each $F_{\mu\nu}$ is a $4\times 4$ matrix. Can it be expressed as trace of some sort? Subsequently can I apply (1) to split the expression into product of, say, the trace of $F$?
Additionally, $F^{\mu\nu}$ is anti-symmetric and I am trying to prove the above equals to zero. So being able to use (1) can be awesome.
Note: I have some although limited background in differential geometry and algebra. English words are great. But please supply formal definitions as well. At the same time, explanations with as little abstract algebraic constructions as possible would be much appreciated. Focus on finite dimension is fine. Extension to separable Hilbert space, partial trace and etc is welcomed too.
EDIT: The second floor to in this post seems to be good. I don't understand, however, how tensor product of matrices work? Still, when does contraction come into play?
• Possible duplicate of How to compute the trace of a tensor? Sep 18, 2018 at 17:05
• Nope. Not even close. (If you are interested, kindly read the answer below? I haven't gotten a chance.) Sep 19, 2018 at 18:31
• $F^{\mu\nu}$ is vector-like (all horizontal) as well as $F_{\mu\nu}$ (all vertical), so tracing over their outer product is equivalent to the inner product. Feb 2, 2021 at 17:11
I try to answer starting from the case of square matrices. There is some care to take while considering a "hidden" isomorphism of vector spaces. In any case, let $V$ be a finite dim. vector spaces over a field $\mathbb K$ (for simplicity $\mathbb R$ ), with basis $\{e_i\}$ of cardinality $n$.
It is well known that there exists an isomorphism of vector spaces $$\Phi:\operatorname{Hom}_\mathbb K(V,V)\rightarrow V^{*}\otimes V,$$
with $$\Phi(\phi)=a_{ij}f_i\otimes e_j,$$ where $\phi\in \operatorname{Hom}_\mathbb K(V,V)$ and $\phi(e_i):=a_{ij}e_j$ for all $i,j=1,\dots,n$. $\{f_i\}$ is the dual basis on $V^{*}$ of the basis $\{e_i\}$ on $V$, i.e. $f_i(e_j)=\delta_{ij}$.
We use the Einstein convention for repeated indices.
We know how to define the trace operator $\operatorname{Tr}$ on the space $\operatorname{Hom}_\mathbb K(V,V)$; the trace is computed on the square matrix representing each linear map in $\operatorname{Hom}_\mathbb K(V,V)$. Let us move to the r.h.s. of the isomorphism $\Phi$.
• trace operator on $V^{*}\otimes V$
Let $$\operatorname{Tr}_1: V^{*}\otimes V\rightarrow \mathbb K,$$
be given by $\operatorname{Tr}_1(g\otimes v):=g(v)$.
Lemma $\operatorname{Tr}_1$ is linear and satisfies $$\operatorname{Tr}_1\circ \Phi=\operatorname{Tr}.$$
proof: just use definitions.
• trace operator on $(V^{*}\otimes V)\otimes\dots\otimes (V^{*}\otimes V)$
Using the $n=1$ case we introduce
$$\operatorname{Tr}_n: \underbrace{(V^{*}\otimes V)\otimes\dots\otimes (V^{*}\otimes V)}_{n-\text{times}} \rightarrow \mathbb K,$$
with $\operatorname{Tr}_n(f_1\otimes v_1\otimes\dots\otimes f_n\otimes v_n):=\prod_{i=1}^n f_i(v_i)$.
Lemma $\operatorname{Tr}_n$ is linear and invariant under permutations on $(V^{*}\otimes V)^{\otimes n}$; it satisfies $$\operatorname{Tr}_n\left(\Phi(\phi_1)\otimes\dots\otimes\Phi(\phi_n)\right)=\prod_{i=1}^n \operatorname{Tr}(\phi_i),$$ for all $\phi_i\in \operatorname{Hom}_\mathbb K(V,V)$.
proof: we prove the second statement. We introduce the notation $$\Phi(\phi_k):= a^k_{i_kj_k}f_{i_k}\otimes e_{i_k}\in V^{*}\otimes V,$$ for all $k=1,\dots,n$. We arrive at $$\operatorname{Tr}_n\left( (a^1_{i_1j_1}f_{i_1}\otimes e_{i_1})\otimes\dots\otimes (a^n_{i_nj_n}f_{i_n}\otimes e_{i_n})\right)=a^1_{i_1j_1}\dots a^n_{i_nj_n}f_{i_1}(e_{i_1})\dots f_{i_n}(e_{i_n})=\text{remember the definition of dual basis}= a^1_{i_1j_1}\dots a^n_{i_nj_n}\delta_{i_1j_1}\dots\delta_{i_nj_n}= a^1_{i_1i_1}\dots a^n_{i_ni_n}\\=\prod_{i=1}^n \operatorname{Tr}(\phi_i),$$ as claimed.
• @Argyll did it help? Jul 28, 2014 at 18:02
• $a_{ij}$ should be changed to be $a_{ji}$ @Avitus
– yang
Sep 7, 2014 at 3:21
• @Avitus: Yes it does. Thank you. But I'm still not sure about the connection between contraction and tensor product or any mathematical objects. Sep 16, 2014 at 1:13
• In which sense? Maybe you can extend your OP a bit Sep 16, 2014 at 7:47
• Good answer, though $$\Phi(\phi_k):= a^k_{i_kj_k}f_{i_k}\otimes e_{i_k}\in V^{*}\otimes V,$$ should be $$\Phi(\phi_k):= a^k_{i_kj_k}f_{i_k}\otimes e_{j_k}\in V^{*}\otimes V,$$ and thereon afterwards. Apr 26 at 13:11 | 2022-08-08T21:52:53 | {
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https://math.stackexchange.com/questions/3004135/is-it-always-possible-to-fit-these-pieces-in-a-square | # Is it always possible to fit these pieces in a square?
Consider all possible pairs of squares that can fit in a row of length $$n$$ where every square has a width of 1. If I have a large square of width $$n$$, can all such pairs of squares fit in the large square simultaneously? It's hard to explain without an example so here is the case when $$n=4$$:
To the left are all the pairs of squares and to the right is a way for them to fit in the $$n$$ by $$n$$ square. Note that the pairs are not allowed to be moved horizontally but can be moved vertically.
It becomes a little harder but still possible when $$n=5$$:
What I'm interested in is the general case. Is it possible to fit all pairs in the square for any $$n$$?
• What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square? – John Douma Nov 18 '18 at 21:39
• @JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2). – Pazzaz Nov 18 '18 at 21:46
• I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1\times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading. – Mark Bennet Nov 18 '18 at 21:56
• @MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now. – Pazzaz Nov 18 '18 at 22:14
• Well, if possible, there will always be $n^2-2\binom n2=n$ leftover spaces. – YiFan Nov 18 '18 at 23:44
It is always possible, we can place the $$\binom{n}{2}$$ pairs in a $$n \times n$$ square when $$n$$ is odd and in a $$(n-1) \times n$$ rectangle when $$n$$ is even.
This problem is equivalent to the edge coloring problem for complete graph $$K_n$$. Look at wiki for the geometric intuition underlying following construction.
Let $$[n]$$ be a short hand for $$\{ 0, \ldots, n-1 \}$$.
Index the set of possible pairs by $$(i,j) \in [n]^2$$ with $$i < j$$.
Label rows and columns of the large square using numbers from $$[n]$$.
When $$n$$ is odd, place the pair $$(i,j)$$ at row $$k$$ of the large square where $$i + j \equiv k \pmod n$$.
If two pairs $$(i_1,j_1)$$, $$(i_2,j_2)$$ on same row intersect, then one of the following happens $$i_1 = i_2 \lor i_1 = j_2\lor j_1 = i_2 \lor j_1 = j_2$$ Since $$i_1 + j_1 \equiv i_2 + j_2 \pmod n$$, we find $$(i_1,j_1) = (i_2,j_2) \pmod n \lor (i_1,j_1) = (j_2,i_2) \pmod n$$
Since $$i_1,i_2,j_1,j_2 \in [n]$$ and $$i_1 < j_1$$, $$i_2 < j_2$$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate a desired packing of the $$\binom{n}{2}$$ pairs into a $$n \times n$$ square.
When $$n$$ is even, $$n - 1$$ is odd.
Place those pair $$(i,j) \in [n-1]^2$$ into row $$k$$ where $$i + j \equiv k \pmod {n-1}$$. Notice
• For each row $$i \in [n-1]$$, the slot at column $$2i \pmod {n - 1}$$ and $$n-1$$ is unused.
• For any column $$j \in [n-1]$$, one and only slot at row $$i \in [n-1]$$ is unused.
For those pair $$(i,j) \in [n]^2 \setminus [n-1]^2$$ with $$i < j$$, we have $$j = n$$. We can place the pair on row $$k$$ where $$2k = i \pmod {n-1}$$. This will fill all the unused slots in the first $$n-1$$ rows and generate a desired packing of the $$\binom{n}{2}$$ pairs into a $$(n-1) \times n$$ rectangle.
For 7×7, we have the following. X is an empty space, A through U are the 21 pairs of squares. For example, D D in the second row indicates that the {1,3} pair is used in that row, while the X in the seventh position means that spot is left empty.
A A B B C X C
D E D E F F X
G H X I H I G
J K K J X L L
X M N O O M N
P X Q R Q P R
S T U X S U T .
• It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal. – Pazzaz Nov 19 '18 at 10:03 | 2019-05-21T08:42:59 | {
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https://math.stackexchange.com/questions/1272402/how-do-i-solve-this-improper-integral-int-infty-infty-e-x2-xdx | How do I solve this improper integral: $\int_{-\infty}^\infty e^{-x^2-x}dx$?
I'm trying to solve this integral: $$\int_{-\infty}^\infty e^{-x^2-x}dx$$ WolframAlpha shows this to be approximately $2.27588$. I tried to solve this by integration by parts, but I just couldn't get there. I'd be glad if someone could show me how to do it.
I've included my attempt at the problem below:
Note $\int_{-\infty}^\infty e^{-x^2-x} = \int_{-\infty}^\infty e^{-x^2}e^{-x}dx$. Then we can integrate by parts. Let $u(x) = e^{-x^2}$. Then $u'(x) = -2xe^{-x}$. Let $v'(x) = e^{-x}$. Then $v(x) = -e^{-x}$.
Then $u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)dx$, i.e. $$\int_{-\infty}^\infty e^{-x^2}e^{-x}dx = -e^{-x^2}e^{-x} - \int e^{-x}2xe^{-x}dx = -e^{-x^2-x} - 2\int e^{-2x}xdx$$
Then we integrate $\int e^{-2x}xdx$ by parts. We pick $u(x) = x$, $u'(x) = dx$, $v'(x) = e^{-2x}$, and $v(x) = \int e^{-2x} dx = \frac {-1}{2} e^{-2x}$ by u-substitution. Skipping some steps, it follows that $$\int e^{-2x}xdx = -\frac{1}{4}(e^{-2x})(2x+1)$$ Then $$-e^{-x^2-x} - 2\int e^{-2x}xdx = -e^{-x^2-x} + \frac{1}{2}(e^{-2x})(2x+1)$$ Which, when evaluated numerically, doesn't yield the desired result. So there's a problem here. Maybe someone else knows how to do this.
• $u'(x)=-2xe^{-x^2}$, not $-2xe^{-x}$. – Kola B. May 8 '15 at 4:00
• @KolaB. Good catch, thanks. – Newb May 8 '15 at 4:17
Rewrite the integrand as $$\exp({-x^2 - x}) = \exp({-(x^2 + x + 1/4) + 1/4} ) = \exp(-(x+1/2)^2) \cdot \exp(1/4)$$ It is well known that $$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$ so the above manipulation proves $$\int_{-\infty}^{\infty} e^{-x^2 - x} \, dx = e^{1/4} \sqrt{\pi} \approx 2.2758$$
• I actually didn't know that. Thanks. – Newb May 8 '15 at 4:17
• It's even got its own Wikipedia article :) en.wikipedia.org/wiki/Gaussian_integral – Sameer Kailasa May 8 '15 at 4:18
• Ahh, so this is the famous Gaussian integral! – Newb May 8 '15 at 4:20
• Veri nice @SameerKailasa :) – Lucas May 8 '15 at 6:47
HINT
$$x^2-x=(x-1/2)^2-1/4$$
Then, factor out the term $e^{-1/4}$, make a change of variable $x-1/2\to x$, and use
$$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$
• hmm, exist other example of integral like this? – Lucas May 8 '15 at 13:12
• @Lucas There are many types of integrals. What exactly do you mean by "like this?" For example, do you mean ones for which we can compute a definite integral but cannot find a simple anti-derivative? – Mark Viola May 8 '15 at 13:25
The basic thing to know is that, as the other answers said, $\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$.
Then, for any reals $a$ and $b$,
$\begin{array}\\ \int_{-\infty}^{\infty} e^{-x^2-ax-b} \, dx &=e^{-b}\int_{-\infty}^{\infty} e^{-x^2-ax} \, dx\\ &=e^{-b}\int_{-\infty}^{\infty} e^{-x^2-ax-a^2/4+a^2/4} \, dx \quad\text{(completing the square)}\\ &=e^{-b+a^2/4}\int_{-\infty}^{\infty} e^{-(x-a/2)^2} \, dx\\ &=e^{-b+a^2/4}\int_{-\infty}^{\infty} e^{-x^2} \, dx\\ &=e^{-b+a^2/4}\sqrt{\pi}\\ \end{array}$
A similar argument would allow you to get a formula for $\int_{-\infty}^{\infty} e^{-ax^2-bx-c} \, dx$ in terms of $a, b, c,$ and $\sqrt{\pi}$. | 2019-07-17T23:25:37 | {
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https://math.stackexchange.com/questions/3929600/what-is-varb-in-multiple-regression | # What is $Var[b]$ in multiple regression?
Assume a linear regression model $$y=X\beta + \epsilon$$ with $$\epsilon \sim N(0,\sigma^2I)$$ and $$\hat y=Xb$$ where $$b=(X'X)^{-1}X'y$$. Besides $$H=X(X'X)^{-1}X'$$ is the linear projection from the response space to the span of $$X$$, i.e., $$\hat y=Hy$$.
Now I want to calculate $$Var[b]$$ but what I get is an $$k\times k$$ matrix, not an $$n \times n$$ one. Here's my calculation:
\begin{align} Var[b] =&\; Var[(X'X)^{-1}X'y]\\ =& \;(X'X)^{-1}X'\,\underbrace{Var[y]}_{= \sigma^2I}X(X'X)^{-1}\\\\ \text{Here you can }& \text{see already this thing will be k \times k} \\\\ =&\; \sigma^2 \underbrace{(X'X)^{-1}X'X}_{I}(X'X)^{-1}\\ =& \sigma^2(X'X)^{-1}\, \in R^{k\times k} \end{align}
What am I doing wrong?
Besides, are $$E[b]=\beta$$, $$E[\hat y]=HX\beta$$, $$Var[\hat y]=\sigma^2H$$, $$E[y-\hat y]=(I-H)X\beta$$, $$Var[y-\hat y]=(I-H)\sigma^2$$ correct (this is just on a side note, my main question is the one above)?
The covariance matrix for $$b$$ (the estimator for $$\beta$$) should be $$k\times k$$. If the $$X$$ matrix is $$n\times k$$ then $$\beta$$ has to be $$k\times 1$$; otherwise the product $$X\beta$$ wouldn't be $$n\times 1$$.
So if $$\beta$$ is a constant vector of $$k$$ parameters, then its estimator $$b$$ is a random vector with $$k$$ elements. Therefore the covariance matrix for $$b$$ consists of covariances for all possible combinations of two members selected from the random vector, hence it must be a $$k\times k$$ matrix.
To answer your side notes, all your calculations are correct but some can be simplified further. Check that $$HX=X$$, so that $$E[\hat y]=X\beta$$, and $$E[y-\hat y]=0$$.
• Yes, your dimensions are right, I didn't explicitly write them down, sorry. I don't quite understand why it's $k \times k$, because I thought it's the covariance between data points. Dec 1, 2020 at 1:24 | 2022-08-09T02:41:48 | {
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https://math.stackexchange.com/questions/1182519/integration-question-on-int-fracxx2-10x50-dx/1182532 | # Integration question on $\int \frac{x}{x^2-10x+50} \, dx$
How would I integrate $$\int \frac{x}{x^2-10x+50} \, dx$$ I am not sure on how to start the problem
• Partial fractions! – Laars Helenius Mar 9 '15 at 17:06
• @LaarsHelenius the quadratic has two complex roots, so unless complex integration is allowed I don't think that would work – graydad Mar 9 '15 at 17:11
• Ah, I see. I thought I saw $c=21$ earlier. So either I'm an idiot or the question was edited after my comment. For the record, I'm not willing discount either option! Lol – Laars Helenius Mar 9 '15 at 17:41
## 3 Answers
I would complete the square in the denominator first.
$$\displaystyle \int \dfrac{x}{x^2-10x+25+25}dx = \int \dfrac{x}{(x-5)^2+25}dx$$
Let $u=x-5$,
$$\displaystyle \int \dfrac{u+5}{u^2+25}du = \int \dfrac{u}{u^2+25}du + \int \dfrac{5}{u^2+25}du$$
the first integral you can solve by doing one more substitution, the second is just arctangent.
write the integral as $$\int\frac{x}{x^2 - 10x +50}dx = \int \left(\frac{2x-10}{2(x^2-10x+50)} + \frac{5}{x^2-10x+50}\right)dx$$ Let $u = x^2 -10x +50, \; \text{then } du = 2x-10$ $$\frac{1}{2}\int\frac{1}{u}du + 5\int \frac{1}{(x-5)^2 + 25}dx$$ $$\frac{\log (u)}{2}+5\int\frac{1}{\frac{(x-5)^2}{25}+1}dx$$ $$\frac{1}{2}\log(x^2-10x+50) + \arctan\left(\frac{x-5}{5}\right) + C$$
• I think you want 5 in the 2nd numerator (instead of x), but this is a good method to solve this. – user84413 Mar 9 '15 at 23:21
• yeah, you are right – Alexander Mar 10 '15 at 0:32
Complete the square in the denominator:
$$x^2 - 10x + 50 = x^2 -10 x + 25 + 25 = (x-5)^2 + 5^2$$
Put $x-5 = 5\tan\theta\implies dx = 5\sec^2 x$. | 2019-06-26T06:06:27 | {
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http://felizcidade.org.br/2rf5m4/cardinality-of-surjective-functions-b8ae3d | B is said to be surjective (also known as onto ) if every element of B is mapped to by some element of A. 3, JUNE 1995 209 The Cardinality of Sets of Functions PIOTR ZARZYCKI University of Gda'sk 80-952 Gdaisk, Poland In introducing cardinal numbers and applications of the Schroder-Bernstein Theorem, we find that the Functions and Cardinality Functions. Cardinality of the Domain vs Codomain in Surjective (non-injective) & Injective (non-surjective) functions 2 Cardinality of Surjective only & Injective only functions If A and B are both finite, |A| = a and |B| = b, then if f is a function from A to B, there are b possible images under f for each element of A. Functions and relative cardinality Cantor had many great insights, but perhaps the greatest was that counting is a process , and we can understand infinites by using them to count each other. This illustrates the For example, the set A = { 2 , 4 , 6 } {\displaystyle A=\{2,4,6\}} contains 3 elements, and therefore A {\displaystyle A} has a cardinality of 3. The functions in Exam- ples 6.12 and 6.13 are not injections but the function in Example 6.14 is an injection. For example, suppose we want to decide whether or not the set $$A = \mathbb{R}^2$$ is uncountable. Added: A correct count of surjective functions is … VOL. A function with this property is called a surjection. I'll begin by reviewing the some definitions and results about functions. In other words there are six surjective functions in this case. Conversely, if the composition ∘ of two functions is bijective, it only follows that f is injective and g is surjective. (This in turn implies that there can be no The function $$f$$ that we opened this section with The prefix epi is derived from the Greek preposition ἐπί meaning over , above , on . FINITE SETS: Cardinality & Functions between Finite Sets (summary of results from Chapters 10 & 11) From previous chapters: the composition of two injective functions is injective, and the the composition of two surjective In mathematics, the cardinality of a set is a measure of the "number of elements" of the set. The functions in the three preceding examples all used the same formula to determine the outputs. Functions A function f is a mapping such that every element of A is associated with a single element of B. Bijective means both Injective and Surjective together. Specifically, surjective functions are precisely the epimorphisms in the category of sets. We will show that the cardinality of the set of all continuous function is exactly the continuum. surjective), which must be one and the same by the previous factoid Proof ( ): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). 2. f is surjective … Cardinality If X and Y are finite sets, then there exists a bijection between the two sets X and Y if and only if X and Y have the same number of elements. A function f from A to B is called onto, or surjective… Surjective functions are not as easily counted (unless the size of the domain is smaller than the codomain, in which case there are none). f(x) x … Formally, f: A → B is a surjection if this FOL Informally, we can think of a function as a machine, where the input objects are put into the top, and for each input, the machine spits out one output. So there is a perfect "one-to-one correspondence" between the members of the sets. Definition. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Hence it is bijective. That is, we can use functions to establish the relative size of sets. 2^{3-2} = 12$. Cardinality … The idea is to count the functions which are not surjective, and then subtract that from the But your formula gives$\frac{3!}{1!} Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain has at least one element of the domain associated with it. The function is They sometimes allow us to decide its cardinality by comparing it to a set whose cardinality is known. Surjections as epimorphisms A function f : X → Y is surjective if and only if it is right-cancellative: [2] given any functions g,h : Y → Z, whenever g o f = h o f, then g = h.This property is formulated in terms of functions and their composition and can be generalized to the more general notion of the morphisms of a category and their composition. Lecture 3: Cardinality and Countability Lecturer: Dr. Krishna Jagannathan Scribe: Ravi Kiran Raman 3.1 Functions We recall the following de nitions. Definition 7.2.3. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. This was first recognized by Georg Cantor (1845–1918), who devised an ingenious argument to show that there are no surjective functions $$f : \mathbb{N} \rightarrow \mathbb{R}$$. Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain is “covered” by at least one element of the domain. Any morphism with a right inverse is an epimorphism, but the converse is not true in general. By the Multiplication Principle of Counting, the total number of functions from A to B is b x b x b It is also not surjective, because there is no preimage for the element $$3 \in B.$$ The relation is a function. 1. f is injective (or one-to-one) if implies . 3.1 Surjections as right invertible functions 3.2 Surjections as epimorphisms 3.3 Surjections as binary relations 3.4 Cardinality of the domain of a surjection 3.5 Composition and decomposition 3.6 Induced surjection and induced 4 That is to say, two sets have the same cardinality if and only if there exists a bijection between them. Onto/surjective functions - if co domain of f = range of f i.e if for each - If everything gets mapped to at least once, it’s onto One to one/ injective - If some x’s mapped to same y, not one to one. Bijective Function, Bijection. 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Correct count of surjective functions are also called one-to-one, onto functions used the same to! Element of set Ato a set is a measure of the sets: every one has a partner and one! Bijective functions are also called one-to-one, onto functions right inverse is an injection of elements '' the... But the function in Example 6.14 is an injection a measure of the set function is will! Are precisely the epimorphisms in the codomain ) three preceding examples all used same! How To Protect Tv Screen From Toddler, Final Fantasy 3 Job Level Grinding, Temperature And Humidity Meter With Probe, Innovo Forehead And Ear Uk, Interviewing Course Syllabus, Schlage Be467 Manual, Homemade Food For Labrador, Tempur-cloud® Breeze Dual Cooling™ Pillow, Piazza Ponte Santangelo Roma, Osu Stats Compare, " /> B is said to be surjective (also known as onto ) if every element of B is mapped to by some element of A. 3, JUNE 1995 209 The Cardinality of Sets of Functions PIOTR ZARZYCKI University of Gda'sk 80-952 Gdaisk, Poland In introducing cardinal numbers and applications of the Schroder-Bernstein Theorem, we find that the Functions and Cardinality Functions. Cardinality of the Domain vs Codomain in Surjective (non-injective) & Injective (non-surjective) functions 2 Cardinality of Surjective only & Injective only functions If A and B are both finite, |A| = a and |B| = b, then if f is a function from A to B, there are b possible images under f for each element of A. Functions and relative cardinality Cantor had many great insights, but perhaps the greatest was that counting is a process , and we can understand infinites by using them to count each other. This illustrates the For example, the set A = { 2 , 4 , 6 } {\displaystyle A=\{2,4,6\}} contains 3 elements, and therefore A {\displaystyle A} has a cardinality of 3. The functions in Exam- ples 6.12 and 6.13 are not injections but the function in Example 6.14 is an injection. For example, suppose we want to decide whether or not the set $$A = \mathbb{R}^2$$ is uncountable. Added: A correct count of surjective functions is … VOL. A function with this property is called a surjection. I'll begin by reviewing the some definitions and results about functions. In other words there are six surjective functions in this case. Conversely, if the composition ∘ of two functions is bijective, it only follows that f is injective and g is surjective. (This in turn implies that there can be no The function $$f$$ that we opened this section with The prefix epi is derived from the Greek preposition ἐπί meaning over , above , on . FINITE SETS: Cardinality & Functions between Finite Sets (summary of results from Chapters 10 & 11) From previous chapters: the composition of two injective functions is injective, and the the composition of two surjective In mathematics, the cardinality of a set is a measure of the "number of elements" of the set. The functions in the three preceding examples all used the same formula to determine the outputs. Functions A function f is a mapping such that every element of A is associated with a single element of B. Bijective means both Injective and Surjective together. Specifically, surjective functions are precisely the epimorphisms in the category of sets. We will show that the cardinality of the set of all continuous function is exactly the continuum. surjective), which must be one and the same by the previous factoid Proof ( ): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). 2. f is surjective … Cardinality If X and Y are finite sets, then there exists a bijection between the two sets X and Y if and only if X and Y have the same number of elements. A function f from A to B is called onto, or surjective… Surjective functions are not as easily counted (unless the size of the domain is smaller than the codomain, in which case there are none). f(x) x … Formally, f: A → B is a surjection if this FOL Informally, we can think of a function as a machine, where the input objects are put into the top, and for each input, the machine spits out one output. So there is a perfect "one-to-one correspondence" between the members of the sets. Definition. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Hence it is bijective. That is, we can use functions to establish the relative size of sets. 2^{3-2} = 12$. Cardinality … The idea is to count the functions which are not surjective, and then subtract that from the But your formula gives $\frac{3!}{1!} Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain has at least one element of the domain associated with it. The function is They sometimes allow us to decide its cardinality by comparing it to a set whose cardinality is known. Surjections as epimorphisms A function f : X → Y is surjective if and only if it is right-cancellative: [2] given any functions g,h : Y → Z, whenever g o f = h o f, then g = h.This property is formulated in terms of functions and their composition and can be generalized to the more general notion of the morphisms of a category and their composition. Lecture 3: Cardinality and Countability Lecturer: Dr. Krishna Jagannathan Scribe: Ravi Kiran Raman 3.1 Functions We recall the following de nitions. Definition 7.2.3. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. This was first recognized by Georg Cantor (1845–1918), who devised an ingenious argument to show that there are no surjective functions $$f : \mathbb{N} \rightarrow \mathbb{R}$$. Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain is “covered” by at least one element of the domain. Any morphism with a right inverse is an epimorphism, but the converse is not true in general. By the Multiplication Principle of Counting, the total number of functions from A to B is b x b x b It is also not surjective, because there is no preimage for the element $$3 \in B.$$ The relation is a function. 1. f is injective (or one-to-one) if implies . 3.1 Surjections as right invertible functions 3.2 Surjections as epimorphisms 3.3 Surjections as binary relations 3.4 Cardinality of the domain of a surjection 3.5 Composition and decomposition 3.6 Induced surjection and induced 4 That is to say, two sets have the same cardinality if and only if there exists a bijection between them. Onto/surjective functions - if co domain of f = range of f i.e if for each - If everything gets mapped to at least once, it’s onto One to one/ injective - If some x’s mapped to same y, not one to one. Bijective Function, Bijection. 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Correct count of surjective functions are also called one-to-one, onto functions used the same to! Element of set Ato a set is a measure of the sets: every one has a partner and one! Bijective functions are also called one-to-one, onto functions right inverse is an injection of elements '' the... But the function in Example 6.14 is an injection a measure of the set function is will! Are precisely the epimorphisms in the codomain ) three preceding examples all used same! How To Protect Tv Screen From Toddler, Final Fantasy 3 Job Level Grinding, Temperature And Humidity Meter With Probe, Innovo Forehead And Ear Uk, Interviewing Course Syllabus, Schlage Be467 Manual, Homemade Food For Labrador, Tempur-cloud® Breeze Dual Cooling™ Pillow, Piazza Ponte Santangelo Roma, Osu Stats Compare, " />
Discrete Mathematics - Cardinality 17-3 Properties of Functions A function f is said to be one-to-one, or injective, if and only if f(a) = f(b) implies a = b. Cantor’s Theorem builds on the notions of set cardinality, injective functions, and bijections that we explored in this post, and has profound implications for math and computer science. surjective non-surjective injective bijective injective-only non- injective surjective-only general In mathematics, injections, surjections and bijections are classes of functions distinguished by the manner in which arguments (input expressions from the domain) and images (output expressions from the codomain) are related or mapped to each other. A function $$f: A \rightarrow B$$ is bijective if it is both injective and surjective. Let X and Y be sets and let be a function. 68, NO. It is injective (any pair of distinct elements of the domain is mapped to distinct images in the codomain). Bijective functions are also called one-to-one, onto functions. A function with this property is called a surjection. Definition Consider a set $$A.$$ If $$A$$ contains exactly $$n$$ elements, where $$n \ge 0,$$ then we say that the set $$A$$ is finite and its cardinality is equal to the number of elements $$n.$$ The cardinality of a set $$A$$ is Since the x-axis $$U Think of it as a "perfect pairing" between the sets: every one has a partner and no one is left out. An important observation about injective functions is this: An injection from A to B means that the cardinality of A must be no greater than the cardinality of B A function f : A -> B is said to be surjective (also known as onto ) if every element of B is mapped to by some element of A. 3, JUNE 1995 209 The Cardinality of Sets of Functions PIOTR ZARZYCKI University of Gda'sk 80-952 Gdaisk, Poland In introducing cardinal numbers and applications of the Schroder-Bernstein Theorem, we find that the Functions and Cardinality Functions. Cardinality of the Domain vs Codomain in Surjective (non-injective) & Injective (non-surjective) functions 2 Cardinality of Surjective only & Injective only functions If A and B are both finite, |A| = a and |B| = b, then if f is a function from A to B, there are b possible images under f for each element of A. Functions and relative cardinality Cantor had many great insights, but perhaps the greatest was that counting is a process , and we can understand infinites by using them to count each other. This illustrates the For example, the set A = { 2 , 4 , 6 } {\displaystyle A=\{2,4,6\}} contains 3 elements, and therefore A {\displaystyle A} has a cardinality of 3. The functions in Exam- ples 6.12 and 6.13 are not injections but the function in Example 6.14 is an injection. For example, suppose we want to decide whether or not the set \(A = \mathbb{R}^2$$ is uncountable. Added: A correct count of surjective functions is … VOL. A function with this property is called a surjection. I'll begin by reviewing the some definitions and results about functions. In other words there are six surjective functions in this case. Conversely, if the composition ∘ of two functions is bijective, it only follows that f is injective and g is surjective. (This in turn implies that there can be no The function $$f$$ that we opened this section with The prefix epi is derived from the Greek preposition ἐπί meaning over , above , on . FINITE SETS: Cardinality & Functions between Finite Sets (summary of results from Chapters 10 & 11) From previous chapters: the composition of two injective functions is injective, and the the composition of two surjective In mathematics, the cardinality of a set is a measure of the "number of elements" of the set. The functions in the three preceding examples all used the same formula to determine the outputs. Functions A function f is a mapping such that every element of A is associated with a single element of B. Bijective means both Injective and Surjective together. Specifically, surjective functions are precisely the epimorphisms in the category of sets. We will show that the cardinality of the set of all continuous function is exactly the continuum. surjective), which must be one and the same by the previous factoid Proof ( ): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). 2. f is surjective … Cardinality If X and Y are finite sets, then there exists a bijection between the two sets X and Y if and only if X and Y have the same number of elements. A function f from A to B is called onto, or surjective… Surjective functions are not as easily counted (unless the size of the domain is smaller than the codomain, in which case there are none). f(x) x … Formally, f: A → B is a surjection if this FOL Informally, we can think of a function as a machine, where the input objects are put into the top, and for each input, the machine spits out one output. So there is a perfect "one-to-one correspondence" between the members of the sets. Definition. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Hence it is bijective. That is, we can use functions to establish the relative size of sets. 2^{3-2} = 12$. Cardinality … The idea is to count the functions which are not surjective, and then subtract that from the But your formula gives$\frac{3!}{1!} Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain has at least one element of the domain associated with it. The function is They sometimes allow us to decide its cardinality by comparing it to a set whose cardinality is known. Surjections as epimorphisms A function f : X → Y is surjective if and only if it is right-cancellative: [2] given any functions g,h : Y → Z, whenever g o f = h o f, then g = h.This property is formulated in terms of functions and their composition and can be generalized to the more general notion of the morphisms of a category and their composition. Lecture 3: Cardinality and Countability Lecturer: Dr. Krishna Jagannathan Scribe: Ravi Kiran Raman 3.1 Functions We recall the following de nitions. Definition 7.2.3. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. This was first recognized by Georg Cantor (1845–1918), who devised an ingenious argument to show that there are no surjective functions $$f : \mathbb{N} \rightarrow \mathbb{R}$$. Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain is “covered” by at least one element of the domain. Any morphism with a right inverse is an epimorphism, but the converse is not true in general. By the Multiplication Principle of Counting, the total number of functions from A to B is b x b x b It is also not surjective, because there is no preimage for the element $$3 \in B.$$ The relation is a function. 1. f is injective (or one-to-one) if implies . 3.1 Surjections as right invertible functions 3.2 Surjections as epimorphisms 3.3 Surjections as binary relations 3.4 Cardinality of the domain of a surjection 3.5 Composition and decomposition 3.6 Induced surjection and induced 4 That is to say, two sets have the same cardinality if and only if there exists a bijection between them. Onto/surjective functions - if co domain of f = range of f i.e if for each - If everything gets mapped to at least once, it’s onto One to one/ injective - If some x’s mapped to same y, not one to one. Bijective Function, Bijection. 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Correct count of surjective functions are also called one-to-one, onto functions used the same to! Element of set Ato a set is a measure of the sets: every one has a partner and one! Bijective functions are also called one-to-one, onto functions right inverse is an injection of elements '' the... But the function in Example 6.14 is an injection a measure of the set function is will! Are precisely the epimorphisms in the codomain ) three preceding examples all used same! | 2021-03-01T17:51:17 | {
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https://math.stackexchange.com/questions/998046/trigonometric-functions-on-a-unit-circle | Trigonometric Functions on a unit circle
I have to find all solutions for $\theta$ in the given range: $$tan (\theta) = \frac {-1}{\sqrt3}, -\pi \le \theta \lt 2\pi$$
I said that if $(x,y)$ is on the unit circle we have $$\frac{y}{x} = \frac{-1}{\sqrt3}$$ since $x^2+y^2=1$ $\implies x = \frac{\sqrt 3}{2}$,$y=-\frac{1}{2}$ and $x = - \frac{\sqrt3}{2}$,$y=\frac{1}{2}$
I am struggling on how to find the angles because I didn't understand the concept. I put my points into the circle but then I get confused about the angles. The one angle it will be $\theta = -\frac{\pi}{6}$ since $tan^{-1}(-\frac{1}{\sqrt3})= - \frac{\pi}{6}$.
Can anyone help me to understand the way of thinking for the other angles?
With Thanks
• You made an error just above your graph. You meant $$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$ – N. F. Taussig Oct 30 '14 at 10:37
As you determined,
$$\theta = \arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$
is one solution.
Remember that the tangent function has period $\pi$. Thus,
$$\tan\theta = -\frac{1}{\sqrt{3}}$$
when
$$\theta = -\frac{\pi}{6} + n\pi, n \in \mathbb{Z}$$
where $\mathbb{Z}$ is the set of integers. Since you require that $-\pi \leq \theta < 2\pi$, you can find the remaining solutions in the interval by finding values of $n$ such that
$$-\pi \leq -\frac{\pi}{6} + n\pi < 2\pi$$
• In case we have sin or cos instead of tan? – user161260 Oct 30 '14 at 10:43
• For sine, you would find a particular solution $\theta$. Since $\sin(\pi - \theta) = \sin\theta$, $\pi - \theta$ is another solution. To find additional solutions, add integer multiples of $2\pi$ to $\theta$ and $\pi - \theta$ to find solutions in the right range. For cosine, after finding a particular solution $\theta$, you use the fact that $\cos(-\theta) = \cos\theta$ to find another solution. Since cosine also has period $2\pi$, you would add integer multiples of $2\pi$ to $\theta$ and $-\theta$ to find values in the right range. – N. F. Taussig Oct 30 '14 at 10:52
• For the example with the tan the solutions will be: $\theta = -\frac{\pi}{6}, \frac{5 \pi}{6} and \frac{11 \pi} {6}$ ? – user161260 Oct 30 '14 at 11:17
• That is correct. – N. F. Taussig Oct 30 '14 at 12:01 | 2019-12-07T09:44:19 | {
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http://mathhelpforum.com/pre-calculus/128942-cosine-function-derivative.html | # Math Help - Cosine Function Derivative
1. ## Cosine Function Derivative
Question # 2 lol. I don't know why I'm having such a hard time getting this....
Determine the derivative of the following fuction:
g(x) = (cosx)^3
I started by using the power rule to come up with:
g '(x) = 3(cosx)^2
= 3(-sinx)^2
According to my book this is wrong though... it has:
g '(x) = (3(cosx)^2)(-sinx)
= -3cos^2xsinx
Which makes no sense to me. Thanks again.
2. $y= \cos ^3x$
using the chain rule $\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$
so make $u = \cos x$ so $y = u^3$
now
$\frac{dy}{du}= 3u^2$
and
$\frac{du}{dx}= -\sin x$
$\frac{dy}{dx} = 3u^2 \times -\sin x= 3\cos^2x \times -\sin x = -3\cos^2x\sin x$
3. I was taught the chain rule a little differently. It was shown to me like this:
f (x) = a^g(x)
f '(x) = (a^g(x))(ln a)(g '(x))
Would it be possible to quickly go over how the method you are using works:
dy/dx = (dy/du)(du/dx)
Thanks!
4. Originally Posted by Jools
Would it be possible to quickly go over how the method you are using works:
dy/dx = (dy/du)(du/dx)
Thanks!
This is exactly what I have done above. This is the chain rule.
Originally Posted by Jools
I was taught the chain rule a little differently. It was shown to me like this:
f (x) = a^g(x)
f '(x) = (a^g(x))(ln a)(g '(x))
This is just a version of the chain rule, it won't work in your case because your function is not in the correct form to use this adaptation.
5. Sorry, I'm confused. Let's take it from the top. First off how did you come up with y = cos^3 x? Then I think I just need to know what each variable in the formula represents and I should be able to figure this out. I appreciate your patience.
6. Originally Posted by Jools
I was taught the chain rule a little differently. It was shown to me like this:
f (x) = a^g(x)
f '(x) = (a^g(x))(ln a)(g '(x))
This technique only works if a is a constant.
I find it easiest to remember this:
if f(x) =[g(x)]^a, then
f'(x) = a*[g(x)]^(a-1)*g'(x)
Here's an examples:
f(x) = (3x)^3
f'(x) = 3*(3x)^2 *3 = 81x^2
Another example:
f(x) = (x^2)^3
f'(x) = 3*(x^2)^2*2x = 6x^5
See how it works? So applying it to your problem:
f(x) = [cos(x)]^3
f'(x) = 3* [cos(x)]^2 * (-sin(x))
= -3 cos^2(x) sin(x)
7. Originally Posted by Jools
Sorry, I'm confused. Let's take it from the top. First off how did you come up with y = cos^3 x? Then I think I just need to know what each variable in the formula represents and I should be able to figure this out. I appreciate your patience.
Ah yes ... the notation cos^3(x) means the the same thing as [cos(x)]^3. Perhaps that's the confusion.
8. So how exactly do I determine what the g(x) is? For example taking another question asking for the derivative:
f(x) = sin(x^2)
Is x^2 the g(x)?
9. Originally Posted by Jools
So how exactly do I determine what the g(x) is? For example taking another question asking for the derivative:
f(x) = sin(x^2)
Is x^2 the g(x)?
Yes, that is correct.
Here is the chain rule written in terms of the composite function:
$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$
10. Originally Posted by icemanfan
Yes, that is correct.
Here is the chain rule written in terms of the composite function:
$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$
Ok I think I'm catching on... So one more...
g(x) = e^cosx
How would that one go?
11. Originally Posted by Jools
Ok I think I'm catching on... So one more...
g(x) = e^cosx
How would that one go?
Let $u = \cos{x}$ so that $g = e^u$.
$\frac{du}{dx} = -\sin{x}$
$\frac{dg}{du} = e^u = e^{\cos{x}}$
So $\frac{dg}{dx} = -\sin{x}\,e^{\cos{x}}$.
12. Originally Posted by Prove It
Let $u = \cos{x}$ so that $g = e^u$.
$\frac{du}{dx} = -\sin{x}$
$\frac{dg}{du} = e^u = e^{\cos{x}}$
So $\frac{dg}{dx} = -\sin{x}\,e^{\cos{x}}$.
Could that be applied to this formula?
f '(g(x))(g '(x))
13. Originally Posted by Jools
Could that be applied to the formula?
f '(g(x))(g '(x))
Yes - just replace where I've used $u$ with $g(x)$ and replace where I've used $g$ with $f$.
14. Originally Posted by Prove It
Yes - just replace where I've used $u$ with $g(x)$ and replace where I've used $g$ with $f$.
Got it. I REALLY appreciate all the help. I understand how to work out the derivative I think... I'm just having a hard time determining what the value of g(x) is for each variation.... | 2015-06-30T12:50:27 | {
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http://math.stackexchange.com/questions/25390/how-to-find-the-inverse-modulo-m | How to find the inverse modulo m?
For example: $$7x \equiv 1 \pmod{31}$$ In this example, the modular inverse of $7$ with respect to $31$ is $9$. How can we find out that $9$? What are the steps that I need to do?
Update
If I have a general modulo equation:
$$5x + 1 \equiv 2 \pmod{6}$$
What is the fastest way to solve it? My initial thought was: $$5x + 1 \equiv 2 \pmod{6}$$ $$\Leftrightarrow 5x + 1 - 1\equiv 2 - 1 \pmod{6}$$ $$\Leftrightarrow 5x \equiv 1 \pmod{6}$$
Then solve for the inverse of $5$ modulo 6. Is it a right approach?
Thanks,
-
1. One method is simply the Euclidean algorithm: \begin{align*} 31 &= 4(7) + 3\\ 7 &= 2(3) + 1. \end{align*} So $1 = 7 - 2(3) = 7 - 2(31 - 4(7)) = 9(7) - 2(31)$. Viewing the equation $1 = 9(7) -2(31)$ modulo $31$ gives $1 \equiv 9(7)\pmod{31}$, so the multiplicative inverse of $7$ modulo $31$ is $9$. This works in any situation where you want to find the multiplicative inverse of $a$ modulo $m$, provided of course that such a thing exists (i.e., $\gcd(a,m) = 1$). The Euclidean Algorithm gives you a constructive way of finding $r$ and $s$ such that $ar+ms = \gcd(a,m)$, but if you manage to find $r$ and $s$ some other way, that will do it too. As soon as you have $ar+ms=1$, that means that $r$ is the modular inverse of $a$ modulo $m$, since the equation immediately yields $ar\equiv 1 \pmod{m}$.
2. Another method is to play with fractions (Gauss's method): $$\frac{1}{7} = \frac{1\times 5}{7\times 5} = \frac{5}{35} = \frac{5}{4} = \frac{5\times 8}{4\times 8} = \frac{40}{32} = \frac{9}{1}.$$ Here, you reduce modulo $31$ where appropriate, and the only thing to be careful of is that you should only multiply and divide by things relatively prime to the modulus. Here, since $31$ is prime, this is easy. At each step, I just multiplied by the smallest number that would yield a reduction; so first I multiplied by $5$ because that's the smallest multiple of $7$ that is larger than $32$, and later I multiplied by $8$ because it was the smallest multiple of $4$ that is larger than $32$. Added: As Bill notes, the method may fail for composite moduli.
Both of the above methods work for general modulus, not just for a prime modulus (though Method 2 may fail in that situation); of course, you can only find multiplicative inverses if the number is relatively prime to the modulus.
Update. Yes, your method for general linear congruences is the standard one. We have a very straightforward method for solving congruences of the form $$ax \equiv b\pmod{m},$$ namely, it has solutions if and only if $\gcd(a,m)|b$, in which case it has exactly $\gcd(a,m)$ solutions modulo $m$.
To solve such equations, you first consider the case with $\gcd(a,m)=1$, in which case $ax\equiv b\pmod{m}$ is solved either by finding the multiplicative inverse of $a$ modulo $m$, or as I did in method $2$ above looking at $\frac{b}{a}$.
Once you know how to solve them in the case where $\gcd(a,m)=1$, you can take the general case of $\gcd(a,m) = d$, and from $$ax\equiv b\pmod{m}$$ go to $$\frac{a}{d}x \equiv \frac{b}{d}\pmod{\frac{m}{d}},$$ to get the unique solution $\mathbf{x}_0$. Once you have that unique solution, you get all solutions to the original congruence by considering $$\mathbf{x}_0,\quad \mathbf{x}_0 + \frac{m}{d},\quad \mathbf{x}_0 + \frac{2m}{d},\quad\ldots, \mathbf{x}_0 + \frac{(d-1)m}{d}.$$
-
@Arturo Magidin: Thanks, it's a lot of work :(. I thought it's just a simple calculation. – Chan Mar 6 '11 at 23:38
@Chan: Your definition of "a lot of work" needs revising, if you plan on studying more number theory and maths. The above is surprising little work, in my estimation... – Arturo Magidin Mar 6 '11 at 23:40
@Arturo Magidin: Thanks for the advice. However, I have many other things that I want to study. I like Number Theory, but also Geometry, Computer Graphics, Algorithm, C++, Ruby, Compiler, AI..., plus my school classes. I wish a day could have 48 hours. – Chan Mar 6 '11 at 23:43
@Chan: Trust me, you aren't the only one making that wish. But I really don't know what it is you think that the above methods are "a lot of work". Even the Euclidean algorithm method works, in this case, in two simple steps. Did you hope that just a sum and a product would do it every time? – Arturo Magidin Mar 6 '11 at 23:45
the euclidean algorithm is simple and easily programed... – yoyo Mar 6 '11 at 23:48
We know that $(7,31) = 1$. So you can use the extended Euclidean algorithm. In particular, you can find two integers $s,t$ such that $7s+31t = 1$. So $7s \equiv 1 \pmod {31}$.
-
Thanks for the idea. – Chan Mar 6 '11 at 23:39
A hint: Try to use Fermats Little theorem:
$a^{p-1}=1 \mod p$ for $p$ prime, and all $a\in \mathbb{Z}$.
-
7^29 mod 31 certainly works in this case, but for larger numbers it might be a bit of an effort compared with the Euclidean algorithm, and it would get more complicated by involving Euler's totient function for a non-prime modulus. – Henry Mar 6 '11 at 23:57
We can visualize the solution to $7x \equiv 1 \pmod{31}$ as the intersection of the two "lines"
$\ \ \ \ \ y \equiv 7x \pmod{31}$ and $y \equiv 1 \pmod{31}$.
The first line is closed under addition and subtraction, since it passes through the origin. If $(x_1, y_1)$ and $(x_2, y_2)$ are points on the line $y \equiv 7x \pmod{31}$, then so are $(x_1+x_2, y_1+y_2)$ and $(x_1-x_2, y_1-y_2)$.
To find a point where the lines intersect, we start with the two points $(1, 7)$ and $(0,31)$ on the first line, and repeatedly subtract one point from another until the $y$-coordinate is 1.
$\ \ \ \ \ (0,31) - (1,7) = (-1, 24)$
$\ \ \ \ \ (-1, 24) - (1,7) = (-2, 17)$
$\ \ \ \ \ (-2, 17) - (1, 7) = (-3, 10)$
$\ \ \ \ \ (-3, 10) - (1, 7) = (-4, 3)$
$\ \ \ \ \ (1,7) - (-4, 3) = (5, 4)$
$\ \ \ \ \ (5,4) - (-4, 3) = (9,1)$
Therefore, $x \equiv 9 \pmod{31}$.
You can speed up this procedure by subtracting a multiple of one point from another point instead. This leads to the Euclidean algorithm.
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There are many methods available, e.g. the extended Euclidean algorithm, or a special case of Euclid's algorithm that computes inverses modulo primes that I call Gauss's algorithm; or, by Euler-Fermat $\rm\,\ (a,m)=1 \Rightarrow\ a^{-1} \equiv a^{\phi(m)-1}\pmod m.\,$ The latter yields a simple closed form for CRT (Chinese Remainder Theorem)
$\quad$ If $\rm\,\ (m,n)=1\,\$ then $\rm\quad \begin{eqnarray}\rm x\!&\equiv&\rm a\ \ (mod\ m)\\ \rm x\!&\equiv&\rm b\ \ (mod\ n)\end{eqnarray} \iff\ x\equiv a\,n^{\phi(m)}\!+b\,m^{\phi(n)}\ \ (mod\ m\,n)$
More generally see the Peirce decomposition.
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Many thanks! – Chan Mar 6 '11 at 23:56 | 2016-04-29T00:14:07 | {
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https://mathoverflow.net/questions/342798/is-there-a-measure-on-0-1-that-is-0-on-meagre-sets-and-1-on-co-meagre-sets/342803 | # Is there a measure on $[0,1]$ that is 0 on meagre sets and 1 on co-meagre sets
I'm curious if there is a finite measure on the $$\sigma$$-algebra of subsets of $$[0,1]$$ with the Property of Baire, whose null sets are exactly the meagre sets.
I'd also be interested how "nice" such a measure can be like can it be Radon(when restricted to Borel sets) for example.
## 2 Answers
The answer is no. Assume that such a measure $$\mu$$ exists.
First, since every singleton in $$[0,1]$$ is closed with empty interior, $$\mu(\{x\}) = 0$$ for all $$x \in [0,1]$$. Write $$B_{x,\epsilon}$$ for the open ball around $$x$$ of radius $$\epsilon$$ with respect to the standard metric on $$[0,1]$$. By countable additivity, for all $$x \in [0,1]$$, $$\mu(B_{x,2^{-n}}) \to 0$$. If we take an enumeration of the rationals $$(q_i)_{i \in \mathbb{N}}$$, for each $$i$$ there exists an $$n_i$$ such that $$\mu(B(q_i,2^{-n_i})) < 2^{-i}$$. So $$D_1 = \bigcup_{i=1}^\infty B(q_i,2^{-n_i})$$ is a dense open set with $$\mu(D_1) \leq 1$$.
By re-doing the previous construction, picking $$\mu(B(q_i,2^{-n_i})) < 2^{-(i+k)}$$, we can define dense open sets $$D_k$$ with $$\mu(D_k) \leq 2^{-k}$$. Now, by countable additivity $$N = \bigcap_{k=1}^\infty D_k$$ has measure zero. The set $$[0,1]\setminus N$$ is a union of closed sets with empty interior, i.e. a meagre set, so $$\mu([0,1] \setminus N) = 0$$ as well, so $$\mu([0,1]) = 0$$.
I see that Nate Eldredge was a bit quicker than I was, so I'll add that it is possible to find a finitely-additive probability measure whose null sets are exactly the meagre sets -- this is most easily done using the isomorphism between the Baire property algebra modulo meagre sets and the algebra of regular open sets.
No. For any finite Borel measure $$\mu$$ on $$[0,1]$$, there is a comeager Borel set of $$\mu$$-measure zero.
First note that $$\mu$$ has at most countably many atoms, so it will be possible to find a countable dense set $$D \subset [0,1]$$ containing no atoms, i.e. $$\mu(D) = 0$$. Now any finite Borel measure on a metric space is outer regular, so for any $$n$$ there is an open set $$U_n$$ containing $$D$$ and with $$\mu(U_n) < 1/n$$. Setting $$G = \bigcap_n U_n$$, we see that $$G$$ is a dense $$G_\delta$$ (hence comeager) and $$\mu(G) = 0$$.
Relevant to your second question, the answers in the question linked above also mention that any finite Borel measure on a Polish space is Radon. | 2021-09-17T23:57:21 | {
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http://calcolailtuomutuo.it/mmac/trapezoidal-rule-for-area-calculation.html | # Trapezoidal Rule For Area Calculation
Adding these together gives the trapezoidal approximation of \(3. In mathematics, and more specifically in numerical analysis, the trapezoidal rule, also known as the trapezoid rule or trapezium rule, is a technique for approximating the definite integral. Trapezoidal Area A = 1/2 X a X (b1+b2). Then we approximate the area lying between the graph and each subinterval by a trapezoid whose base is the subinterval, whose vertical sides are the. Midpoint & trapezoidal sums. The interval [a,b] is divided into subintervals. 1st method: Spreadsheet calculations. 1 as the Operating system and OpenMPI 1. In online calculator you can use the value in the same units of measurement!. nodots suppress dots during calculation id(id var) is required. where M is the maximum of on the interval [a,b]. b Formula: A = 0. TECHNIQUES OF INTEGRATION. Which rule do you think would give a better approximation. + O 7) ) 2 small areas at the two ends of the figure have been left out in our calculation. The problem we face is that of finding the area between a curve described by the equation y = f(x) and the x-axis in a finite interval [a, b]. cumtrapz computes the cumulative integral of y with respect to x using trapezoidal integration. trapz performs numerical integration via the trapezoidal method. Number of drawing process. use the trapezoidal rule of integration to solve problems, 3. Try the matlab code in problem 1a. As the number of integration points increase, the results from these methods will converge. 693150) to the true value of the integral (ln 2 ≈ 0. I think that now is the right time to publish a second one. In one of my previous articles, I discussed Midpoint Ordinate Rule and Average Ordinate Rule in detail with an example and listed out various important methods used for the calculation of areas in Surveying. While applying the trapezoidal rule, boundaries between the ends of ordinates are assumed to be straight. Trapezoidal Rule Formula A quadrilateral with two parallel sides is called the Trapezoid. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. This is because the strips that we use are not thin enough for a greater accuracy. B Illustrate the use of Matlab using simple numerical examples. , Scitovski, 2014). Midpoint ordinate rule 2. We can calculate the median of a Trapezoid using the following formula:. (b) The trapezoidal rule approximates the area as a set of trapezoids, and is usually more accurate. Related Calculators Legs of an Isosceles Trapezoid Surface Area of a Trapezoidal Prism. Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real. Durand’s Rule is a slightly more complicated but more precise integration rule than the trapezoidal rule. takes more calculation (but worth it!) 3. certified plan (CP). The function is divided into many sub-intervals and each interval is approximated by a Trapezium. 5, so this is the value of the integral. 5m and d- 0. Label a line perpendicular to the two bases h for height or altitude of the trapezoid. Just copy and paste the below code to your webpage where you want to display this calculator. Overview of Trapezoidal Rule Calculator: A simple calculator made for students, mathematicians, teachers and researchers. Using the subintervals [2,5], [5,7], and [7,8], what is the trapezoidal approximation of S 8,2 f(x) dx?. Conditions to write equation: Write exp(y) to calculate ey value: Write log(x,y) to find logyx value: Write sin(y) to get sin y value: Use pow(y,2) for y2. This calculator will walk you through approximating the area using Trapezoidal Rule. Calculator Project. Trapezoidal Rule Survey line. Surface Area of Cylinder ? The Surface Area is number of square units it takes to exactly cover the surface of a cylinder. Area could be computed using a simple trapezoidal rule. To do this we, first of all, define the integrand. The Total Area includes the area of the circular top and base, as well as the curved surface area. On=ordinate at equal intervals, and d= common distance between two ordinates. Surface Area & Volume of a Pyramid Surface Area of a Pyramid When finding the surface area of a pyramid, we will limit our discussion to the study of regular right pyramids. A cube is a special case where l = w = h. Linear Plot of Cp versus Time showing AUC and AUC segment. B Illustrate the use of Matlab using simple numerical examples. 5 : 3) Total Part of the Concrete = 1+1. Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real. 52 mW/cm 3 (97. Using Trapezoidal Rule for the Area under a Curve Calculation. Simpson's Rule) The Trapezoidal Rule. 015 359 1 1 3. Question: Calculate the area of the trapezoid, which is not drawn to scale. If this is the case, you can often figure out the height using the Pythagorean Theorem. Calculates the area of a trapezoid given two parallel sides and the height. For rectangular, the Area = A x B. I wrote a program to calculate the value of Definite Integral of a function from a to b. cube = 6 a 2. 500004050000. Includes approximation, max error, graphs of approximating trapezoids. 014 548 1 1 12. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. The calculation essentially relies on the fact a trapezoid's area can be equated to that of a rectangle: (base 1 + base 2) / 2 is actually the width of a rectangle with an equivalent area. Did you see how each function value was used twice in the trapezoidal rule calculation?. Three common open channel cross sections, the rectangle, trapezoid, and triangle, are covered in this article. The collected data are shown in the table below. Using the trapezoidal rule to approximate the area under a curve. 1,769,982 views. 49 USD per month until cancelled: Annual Subscription (limited promotion) \$19. Since that area is above the curve, but inside the trapezoid, it'll get included in the trapezoidal rule estimate, even though it shouldn't be because it's not part of the area under the curve. nodots suppress dots during calculation id(id var) is required. SIMPSON’S RULE Use Simpson’s Rule with n = 10 to approximate SIMPSON’S RULE Putting f(x) = 1/x, n = 10, and ∆x = 0. The hydraulic radius for open channel flow is defined as the cross sectional area of flow divided by the wetted perimeter. 5: • Writes the correct trapezoidal rule, without further work of merit • Simpson’s rule with substantially correct substitution and calculation 10: • Trapezoidal rule with correct substitution • Trapezoidal rule with incorrect substitution (maximum of 2) with correct calculation • Correct answer without work shown. rule/straight edge load (e. However, when I set BQL to 0, it uses a linear-trapezoidal rule: (Clast + 0)*(T16 - Tlast) Nowhere in the documentation does it say that this was the plan. f x = x + 4. The script IntegrationTest. Pyramid on any. Trapezoid Sums 1) Using the trapezoid rule, where the number of sub-intervals n 2) Use the table of values to estimate J(x) dx 4, approximate the area underfx) in the interval [0, 4]. Here is the simple online Area of trapezoidal field calculator to calculate the trapezoidal field area. Using Trapezoidal Rule for the Area Under a Curve Calculation. This is because the strips that we use are not thin enough for a greater accuracy. Area Calculation A Trapezoidal Rule Area= %[h, + hn + 2(hz + 113 + + h, ,_, )] Simspon's One-Third Rule Area= E(X + 20 + 4B) 3 13. Fill large area Point loads: Hydro pole, light stand, column, etc Lines loads Rack or rail loading, strip foundation Rectangular area Raft or rectangular footing Circular area tank Earth embankment Road, railway, fill, ice, etc. For example, here is a trapezoidal integration of the sine function using eight evenly-spaced trapezoids: For an integration with N+1 evenly. 1 THE TRAPEZOIDAL RULE Suppose we have a function f(x) and we want to calculate its integral with. 5m and d- 0. 2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 4 of 6 x 9 kip R A = 10 kip A 6 kip R B = 5 kip B Pass a section through the beam at a point between the 6-kip force and the right end of the beam. DISCLAIMER: THIS INFORMATION IS PROVIDED BY SAS INSTITUTE INC. The entire point of my response is you need to get the weights correct. nodots suppress dots during calculation id(id var) is required. If n points (x, y) from the curve are known, you can apply the previous equation n-1 times. More in-depth information read at these rules. Since the area calculated using Simpson's rule will be more accurate than the area calculated using cubic splines only for special cases, IMHO there's no point in using it. AS A SERVICE TO ITS USERS. The Average End-Area method is a useful tool for estimating quantities in construction. If "step" is chosen then a stepwise connection of two points is used. The left Riemann sum is 601. Trapezoidal method, also known as trapezium method or simply trapezoidal rule, is a popular method for numerical integration of various functions (approximation of definite integrals) that arise in science and engineering. It follows that ∫ ≈ (−) ⋅ + (). Image: Parallel sides are 8 in and 11 in. You're not always given the height of the trapezoid. The AUC function will handle unsorted x values (by sorting x) and ties for. The area under the curve is the percentage of randomly drawn pairs for which this is true (that is, the test correctly classifies the two patients in the random pair). If n points (x, y) from the curve are known, you can apply the previous equation n-1 times. 0 Introduction Various earth pressure theories assume that soils are homogeneous, isotropic and horizontally inclined. THE TRAPEZOIDAL RULE. Solution Let y(x)=x^4 here a=-3 and b=3 therefore (b-a)=6 let 'n' be the number of intervals. Raeder 21. This calculator will walk you through approximating the area using Trapezoidal Rule. The points (x, 0) and (x, y) are taken as vertices of a polygon and the area is computed using polyarea. This is a Platinum Membership Resource. AUC was integrated with five computerized methods: polynomial interpolation of third and fourth degree, trapezoidal rule, Simpson's integration, and cubic interpolatory splines. This idea is the working mechanism in trapezoidal method algorithm and flowchart, even it source code. The method for reducing jerk is to smooth the transitions where acceleration begins or ends, making the sharp corners of the trapezoidal profile more “s”-like. Under this rule, the area under a curve is evaluated by dividing the total area into little trapezoids rather than rectangles. The problem we face is that of finding the area between a curve described by the equation y = f(x) and the x-axis in a finite interval [a, b]. If f is positive, then the integral represents the area bounded by the curve y = f (x) and the lines x = a; x = b and y = 0:. As shown in the diagram; Trapezoidal Load. Using the measurements from Figure 7. ; noun The duration of such power. Click here for Excel file. The trapezoidal rule is used to approximate the integral of a function. Say the length of y is much larger than the actual number of points calculated for the FPR and TPR. We present a novel methodology for the numerical solution of problems of diffraction by infinitely thin screens in three dimensional space. Bending Moment is mostly used to find bending stress. The Total Area includes the area of the circular top and base, as well as the curved surface area. Step 2 : Volume of the given prism is = base area x height. I'm new to learning c and either my arrays or my loop is not computing properly. 1 4 2 6 A M A D V = + +. The collected data are shown in the table below. In one of my previous articles, I discussed Midpoint Ordinate Rule and Average Ordinate Rule in detail with an example and listed out various important methods used for the calculation of areas in Surveying. Fuzzy logic is a powerful problem-solving technique that's particularly useful in applications involving decision making or with problems not easily definable by practical mathematical models. But how do we know how accurate our approximation is, in comparison to the exact area under the curve? We want to know whether an approximation is very good, and close to actual area, or if it’s. Access the answers to hundreds of Trapezoidal rule questions that are explained in a way that's easy for you to understand. 12r2 + 3r2 = 375. The AUC is the sum of these rectangles. If there are an even number of samples, N, then there are an odd number of intervals (N-1), but Simpson’s rule requires an even number of intervals. nodots suppress dots during calculation id(id var) is required. Introduction Calculation for areas and volumes for earthworks, cuttings, embankments etc. • Area = h(a+b)/2 • Perimeter = a + b + c + d • Diagonal L 1 H 2 = sqrt(h 2 +(b-sqrt(c 2-h 2)) 2) • Diagonal H 1 L 2 = sqrt(h 2 +(b-sqrt(d 2-h 2)) 2. The entire area between the curve and the x -axis, which is to say the integral, can be approximated by adding together several such trapezia. Trapezoidal rule is based on the method in which curve f is approximated with straight line L, as shown on Fig. Here, a and b are the lower and upper limits of the integration respectively. The calculator will approximate the integral using the Trapezoidal Rule, with steps shown. This area is broken down to three smaller areas, each of which is a trapezoid. (Note too that the median length is the same as the average width. A quadrilateral with one set of parallel sides. How to use the trapezoidal rule calculator?. Using Trapezoidal Rule for the Area Under a Curve Calculation Shi-Tao Yeh, GlaxoSmithKline, Collegeville, PA. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. Notice this guy doesn't give you that, he says, oh recall that the area of the trap yeah we recall that, uh we got up this morning and knew this formula just like off the top of our heads right there you know. In online calculator you can use the value in the same units of measurement!. The trapezoid rule for approximating the value of the definite integral is as follows: Divide the interval from x = a to x = b into n equal subintervals of length Δx = (b - a)/n, erect ordinates y 0, y 1, y 2, , y n as shown in Fig. The linear trapezoidal rule method is commonly used for the estimation of the area under the plasma level-time curve. Add up the approximation of the area over each subinterval to obtain the approximation over the entire interval [a,b]: I[a,b](f) ≈ nX−1 i=0 Ir [x i,xi+1](f) Example 2. It is based on using parabolas at the top instead of straight lines. Use a calculator to evaluate Z 4 1 1 p 1 + x2 dx. the top term, the area moment 1 1 n ii i n i i xA x A = = = ∑ ∑ ID Area x ix*Area (in2)(in) 3 A 1 2 0. Let f(x) be continuous on [a,b]. Trapezoidal Rule First of the Newton-Coates formulas; corresponds to 1st order polynomial Recall from “INTERPOLATION” that a straight line can be represented: Area under line is an estimate of the integral b/w the limits “a” and “b” Result of the integration is called the trapezoidal rule ()1 bb aa I fxdx f xdx () (). , Scitovski, 2014). Practice: Midpoint & trapezoidal sums. civilplanets. Midpoint ApproximationTrapezoidal RuleErrorSimpson’s Rule Trapezoidal Rule We can also approximate a de nite integral R b a f(x)dx using an approximation by trapezoids as shown in the picture below for f(x) 0 The area of the trapezoid above the interval [x i;x i+1] is x h (f(xi)+f(xi+1) 2 i. Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids rather than using rectangles. So if I apply the trapezoidal rule to the areas, then do I get the the sum of 2 times those areas (except for the first and last ones) times 10 = 2155. Introduction to Area and Volume Computation in Surveying 2. The rule is based on approximating the value of the integral of f (x) by that of the linear function that passes through the points (a, f (a)) and (b, f (b)). Base area = 100 sq. The trapezoidal thread calculator for metric trapezoidal thread dimensions of single and multiple start trapezoidal threaded rods and nuts for different pitch diameter tolerance classes. Area of a square. The problem we face is that of finding the area between a curve described by the equation y = f(x) and the x-axis in a finite interval [a, b]. Given: Ordinates. In Euclidean geometry, A trapezoid has two parallel sides are called the bases of the trapezoid. THE TRAPEZOIDAL RULE. Trapezoidal Rule Calculator - Easycalculation. SAS Macro to Calculate AUC /***** AREA. Understanding the trapezoidal rule. Problems on Finding the Area & Perimeter of Trapezoid. Simpson's 1/3 Rule is used to estimate the value of a definite integral. Area of Rectangle = Length x Breadth The Volume of concrete = Length x Breadth x Depth = 6x5x0. Trapezoidal rule is a fairly simple mathematical approach described in [8, 9]. Conditions to write equation: Write exp(y) to calculate ey value: Write log(x,y) to find logyx value: Write sin(y) to get sin y value: Use pow(y,2) for y2. Trapezoid Formulas. The areas of these trapezoids can be calculated easily using the formulas. With n=2, the two intervals would be from x=1 to x=3 and from x=3 to 5. ! So here it would be the load intensity time the beam length. The Government Publishing Office (GPO) processes all sales and distribution of the CFR. Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids rather than using rectangles. Simpson’s Rule Calculator is a mathematical method for approximating the aggregate of a function between two limits, a and b. If x is None, spacing of dx is assumed. This method approximates the integration over an interval by breaking the area down into trapezoids with more easily computable areas. 2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 4 of 6 x 9 kip R A = 10 kip A 6 kip R B = 5 kip B Pass a section through the beam at a point between the 6-kip force and the right end of the beam. Find more Mathematics widgets in Wolfram|Alpha. The basic idea is to divide the interval on the x-axis into n equal subintervals. trapz performs numerical integration via the trapezoidal method. To determine the approximate value of an integral, the calculator uses the trapezoidal rule. This method is mainly applicable to estimate the area under a curve by splitting the entire area into a number of trapeziums of known area. 667 but my results are coming out to be like 57900. Calculating the Wing Area for Constant Chord, Tapered and Delta Wings. The area under a curve is commonly approximated using rectangles (e. With this we have the trapezoidal rule: We will now evaluate the integral using trapezoidal rule and repeat the process for n=1, 2, 4, 8, 16. Which rule do you think would give a better approximation. The areas of these trapezoids can be calculated easily using the formulas. The procedure of trapezoidal rule can be defined as the function where. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. I need help, because i got an exercise that is about numerical integration. Trapezoidal Rule Example Input f(x) = Input value of a = Input value of b = Input value of n = number of subintervals = Finding sum of areas of rectangles. It may be shown, however, that Simpson's rule gives a closer approximation to the area, than does the trapezoidal rule. On=ordinate at equal intervals, and d= common distance between two ordinates. Note that the area obtained has to be multiplied by two, as it is only half of the complete section. Bending Moment is mostly used to find bending stress. New panels for facing facades and interior walls The manufacturer claims that a new product called VECTR will be a good alternative to facade materials made of traditional concrete,. If I understand your question correctly you want the area of a parallel trapezoidal cross-section at some distance x between 0 and L. The upper base will vary linearly in x from B 1 to B 2, so its equation would be, in terms of x: B(x) = B 1 (1 - x/L) + B 2 (x/L) Notice that B(0) = B 1 and B(L) = B 2 and this is linear in x. Trapezoidal Prism Volume Calculator. ∫ Figure: The function f(x) (in blue) is approximated by a linear function (in red). Keyword-suggest-tool. It may also be named as the Trapezium in major parts of the world. Area Calculation - Trapezoidal Rule In the trapezoidal method, each segment of the section is divided into various trapezoids and triangles. In mathematics, and more specifically in numerical analysis, the trapezoidal rule, also known as the trapezoid rule or trapezium rule, is a technique for approximating the definite integral. Figure 2: Computing an Area Parameter. The trapezoidal rule is to find the exact value of a definite integral using a numerical method. 1,769,982 views. The idea is to break the function up into a number of trapezoids and calculate their areas: The area of the shaded trapezoid above is. def calculate_area(f, a, b, n): # Trapezoidal approximation: divide area of function into n trapeziums and add areas. Trapezoidal rule|part 1|sem2|Engineering Maths - Duration: 3:53. 453088385937 The Trapezoidal Rule gives 616. To calculate the area of a trapezium, divide it into a rectangle and two triangles as shown below. TAD (Time Average Difference) is area under the curve (AUC). 693150) to the true value of the integral (ln 2 ≈ 0. Trapezoidal Rule is an integration rule where you divide the total area of the irregular shaped figure into little trapezoids before evaluating the area under a specific curve. This rule is mainly based on the Newton-Cotes formula which states that one can find the exact value of the integral as an nth order polynomial. We saw the basic idea in our first attempt at solving the area under the arches problem earlier. I'm trying to calculate how the errors depend on the step, h, for the trapezoidal rule. The function is divided into many sub-intervals and each interval is approximated by a Trapezium. Flow area of a rectangular channel: A = b h (1) where. pdf Bending-Moment. The trapezoidal rule and Simpson's 1/3 rule are the most common methods for computation of the area of irregular boundary. 1 THE TRAPEZOIDAL RULE Suppose we have a function f(x) and we want to calculate its integral with. Using Trapezoidal Rule for the Area under a Curve Calculation. The area of a trapezoid is basically the average width times the altitude, or as a formula: b1, b2 are the lengths of each base. Trapezoidal Area A = 1/2 X a X (b1+b2). …where A is the area (or distance traveled), a is the Slew speed time (Ts) , b is the total move time T and c is the speed (Sp) => Distance = ½ (Ts + T) x Sp Sp = 2 x distance / (Ts + T) This is a simple enough formula that can be determined by looking at the graph but it does require that you remember the formula for the area of a trapezoid. Using the trapezoidal rule to approximate the value of an integral. The calculator is able to calculate the approximate integral. 3/14/11 Simpson’s Rule- Example Remember: Simpson’s Rule Only Given this problem below, what all do we need to know in order to find the area under the curve using Simpson’s Rule?. Bottom is 4 in, with two little boxes on each side. Calculations for a rectangular prism: 1. A trapezoid is a geometric figure with 4 sides that has 2 parallel sides. In the first row of data (C2 if you labeled your X-axis and Y-axis, C1 if you did not) insert the following formula: (A3-A2)*(B2+B3)/2. As shown in figure-1 below, the trapezoidal footing is a combination two components:. Calculation of Trapezoid Area. This calculator was developed due to a personal need. ∫ Figure: The function f(x) (in blue) is approximated by a linear function (in red). ==== [ article 18387 ] ===== Xref: til comp. Three common open channel cross sections, the rectangle, trapezoid, and triangle, are covered in this article. Trapezoidal rule; Simpson's Rule (in the next section: 6. In British English the trapezoid is called the trapezium. The total surface area of a cone = πrl + πr2 = 375 inch2. Both expressions of the composite trapezoidal rule come from determining the areas of the figures in the corresponding graph. 014 693 1 1 6. The calculator displays the area of a parallelogram value. The Average End-Area method is a useful tool for estimating quantities in construction. 52 mW/cm 3 (97. The altitude (or height) of a trapezoid is the perpendicular distance between the two bases. In optics, the prism is the transparent optical element with flat polished surfaces that refract light. easycalculation. h, l and w are known; find V, S and d. Great for calculus students. 017 590 1 1 10. Required angular riprap size, D50, per Searcy (1967) Required angular riprap size, D50, per Searcy (1967). 308823987500 The Midpoint Rule gives 616. Trapezoidal rule for equal spacing: ( (A≈h1/( 2 ) (y_0+y_n )+y_1+y_2+…+y_(n−1) ]@where hx_1−x_0 )) Q1. 5m and d- 0. For example consider these two systems: Using the Trapezoidal rule, Area under the curve for the first series is 205. Area of a trapezoid. Trapezoidal Footing Volume Formula - Free download as PDF File (. Let nbe an even integer, h= b−a n and define the evaluation points for f(x) by x j = a+jh, j= 0,1,,n We follow the idea from the trapezoidal rule, but break [a,b] = [x. Area of Cutting and Filling is found out for each Section with Trapezoidal Method or Nett Area Calculation Method. The parabolas often get quite close to the real curve: It sounds hard, but we end up with a formula like the trapezoid formula (but we divide by 3 and use a 4,2,4,2,4 pattern of factors):. The key to implementing trapezoidal rule calculator is assuming the area of the integral, ie. This calculator will walk you through approximating the area using Trapezoidal Rule. More in-depth information read at these rules. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. And the base diameter of the cone = 2 × radius = 2 × 5 = 10 inch. Thus, The Remainder Term is z is a number between x and 3. Further extrapolations differ from Newton Cotes formulas. The AUC function will handle unsorted x values (by sorting x) and ties for. When the computer can not calculate the exact integral, it returns an approximate value of the integral. Hence, in a trapezoidal move profile, when acceleration begins or ends, jerk is infinite. A = A 1 - A 2 = π ( r 12 - r 22) for the area of the solid cross section of the tube, the end, an annulus. The Trapezoidal Rule is the average of the left and right sums, and usually gives a better approximation than either does individually. SAS Macro to Calculate AUC /***** AREA. rectangular channels; trapezoidal channels; triangular channels; circular channels; Rectangular Channel Flow Area. A trapezoid, also known as a trapezium, is a 4-sided shape with two parallel bases that are different lengths. Height of the Trapezoidal (mm): Area of Trapezoidal Prism (mm²): Formula: A = (1÷2) × h × (a + b) Where, h is Height of the Trapezoidal. Lines AB and DC are the non-parallel sides and are called legs. In some cases, people choose to calculate the AUC by linear interpolation. Sponsored Links. Polynomials defined by: 2 pts determine a (straight) line; 3 pts a curve. The ApproximateInt(f(x), x = a. The Trapezoidal Rule for C programming C++ question. Figure 2 shows the development of the normalized area for the Area value using the Trapezoidal Rule to estimate area under a curve. I know how to find volume and I know how to use trapezoid rule but I have no idea how to combine them. Land needs to be measured for various reasons like prior to purchase, when doing stock taking, building a boundary wall, dispute with a. We want to find the area of a given region in the plane. 5 : 3) Total Part of the Concrete = 1+1. Area of Trapezoidal Prism Calculator. 34 square units The problem is perhaps to find the area between the given curve and the x axis, from x=1 to x=5, with two intervals. filterVarImp: Calculation of filter-based variable importance In caret: Classification and Regression Training Description Usage Arguments Details Value Author(s) Examples. or fax your order to 202-512-2250, 24 hours a day. As shown in the diagram; Trapezoidal Load. Only those trapezoids can be calculated. Remember, all values should be computed in m. 1,769,982 views. For example consider these two systems: Using the Trapezoidal rule, Area under the curve for the first series is 205. def calculate_area(f, a, b, n): # Trapezoidal approximation: divide area of function into n trapeziums and add areas. This calculator will walk you through approximating the area using Trapezoidal Rule. For Square, the Area = A2 (assuming A=B) Check out different types of Trapezoid Geometry. Instead of using rectangles as we did in the arches problem, we'll use trapezoids (trapeziums) and we'll find that it gives a better approximation to the. Site Surveying : Dawood Woo areaNvol0405 : 23 The Trapezoidal Rule The Trapezoidal Rule To sum up, z area between O 1 and O 8 = 0. Did you see how each function value was used twice in the trapezoidal rule calculation?. It is the internal torque holding a beam together (stopping the left and right halves from rotating - if it was to break in half!) Lecture Notes : Bending-Moment. The area of a trapezoid is basically the average width times the altitude, or as a formula: b1, b2 are the lengths of each base. Online surface area of a trapezoidal prism calculator. The trapezoidal rule. Area Moment of Inertia Section Properties: Trapazoid Calculator. How to calculate area under precision recall curve mathematically? Once we have precision-recall for different thresholds, we can calculate area under curve using Trapezoidal Rule Numerical Integration. Trapezoidal Footing Volume, V = h/3 (A1+A2+√ (A1 x A2)) Where, h - Height of trapezoidal (refer the diagram) A1 - Area of the lower shape. Instead of using rectangles as we did in the arches problem, we'll use trapezoids (trapeziums) and we'll find that it gives a better approximation to the. Calculation of Areas and Volumes using the Trapezoidal Rule 4. So the base radius of the cone is 5 inch. Only those trapezoids can be calculated. from The American Heritage® Dictionary of the English Language, 5th Edition. Area of a quadrilateral. Please enter a function, starting point, ending point, and how many divisions with which you want to use Riemann Right End Point Rule to evaluate. The left Riemann sum is 601. Calculation of Trapezoid Area. While other equations such as Simpson's Rule can provide an even more accurate integral – that is, the total area under the graph – the trapezoidal rule is still used for periodic functions and double exponential functions. The area under the (approximate) curve is computed for each subinterval, and the areas are summed to approximate the integral on the full interval. Although there are many math functions Microsoft Excel can perform, the standard version does not include the ability to do calculus. Here is the simple online Area of trapezoidal field calculator to calculate the trapezoidal field area. Area Calculation A Trapezoidal Rule Area= %[h, + hn + 2(hz + 113 + + h, ,_, )] Simspon's One-Third Rule Area= E(X + 20 + 4B) 3 13. Fuzzy logic is a powerful problem-solving technique that's particularly useful in applications involving decision making or with problems not easily definable by practical mathematical models. e, for x 0, x 1,x 2, x 3, x 4, x 5, x 6, x 7. TAD (Time Average Difference) is area under the curve (AUC). derive the trapezoidal rule of integration, 2. Trapezoidal method, also known as trapezium method or simply trapezoidal rule, is a popular method for numerical integration of various functions (approximation of definite integrals) that arise in science and engineering. The basic idea is to divide the interval on the x-axis into n equal subintervals. Trapezoidal Rule. Ex: 5, 7 etc. Graphical rule. Use a calculator, and the trapezoidal rule (let n=6) to find the arc length and the surface area obtained by rotating the curve with respect to the x-axis, and the 1) y = sinx Os xs 27. In this case the ordinate spacing is one unit of measurement. llJ Method 2 Triangular and trapezoidal distribution of uniform load. The collected data are shown in the table below. Simpson's method replaces the slanted-line tops with parabolas. DA: 90 PA: 66 MOZ Rank: 57. genetic,comp. All these methods are Numerical. Area & Volume measurement Lecture contents. 02 492 1 1 9. 5*(FEV1k + FEV1k-1)) / (tm - t1)", in which calculation starts from second time point. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. they are determined by the formula for calculating the area of a triangle as they. Example: Calculate the volume of the trapezoid seen below to the nearest cubic foot. The method for reducing jerk is to smooth the transitions where acceleration begins or ends, making the sharp corners of the trapezoidal profile more "s"-like. And arrow is pretty much the same, but I'm going to stress out that with our higher boundary is 2 times higher because of the 12 multiplier in the denominator. Coterminal Angles Circumference to Diameter Diameter to Circumference Milligrams to Teaspoons Surface Area of a Trapezoidal Prism Area of Trapezoidal Prism Volume of a Trapezoidal Prism Semitangent Distance Middle Ordinate External Distance of a Horizontal Curve Rate of Change of Grade Length of Vertical Curve Length of a Circular Curve. The trapezoidal rule approximates the area under a curve by breaking up the interval into a fixed number of equally spaced subintervals, and approximating the area in each subinterval by a trapezoid. Area = ( (b1 + b2) / 2) * h. For payment by credit card, call toll-free, 866-512-1800, or DC area, 202-512-1800, M-F 8 a. The trapezoidal rule is one way to calculate this integral that is (1) easy to implement; (2) quite accurate; and (3) quite robust. 2, the exact value of this integral is l. How to calculate dose and AUC of Controlled Release formulations Controlled release formulations are aimed to release the drug at particular rate to achieve a desired steady state concentration in the plasma. When the computer can not calculate the exact integral, it returns an approximate value of the integral. Try the matlab code in problem 1a. This is the currently selected item. The Trapezoidal Rule is the average of the left and right sums, and usually gives a better approximation than either does individually. For Square, the Area = A2 (assuming A=B) Check out different types of Trapezoid Geometry. The Internet's premier ask-an-expert math help service. In short, it was an over estimate. Here, the trapezoidal footing formula is explained and clarified with the help of an example. h, l and w are known; find V, S and d. 16 illustrates the step of transforming the trapezoid to a rectangle during calculation of the trapezoidal transformation matrix as described in the example embodiment. You can input only integer numbers or fractions in this online calculator. Step 1 : Identify a base, and find its area and perimeter. Help would be much appreciated, because so far this is the cleanest most nice structured example of the Trapezoidal rule. 136 mW), when the shorter width of the bimorph W 1 is 2 mm and the longer width of the. Categories Surveying Tags Average ordinate rule, Graphical rule, Guide to Levelling, Guide to Surveying, Guide to Surveying and Levelling, Methods for calculation of areas in Surveying, Midpoint ordinate rule, Numerical examples of Surveying, Simpson's rule, Trapezoidal rule 4 Comments Post navigation. To integrate a function f(x) between the range a to b is basically finding the area below the curve from point x = a to x = b. For this purpose, the area enclosed by each contour is measured using a planimeter. 02832 m3= 28. The prismoidal rule • The prismoidal rule gives the correct volume directly. Flow area of a rectangular channel: A = b h (1) where. txt) or read online for free. Calculate the diagonal of a trapezoid if given 1. Area of a rhombus. With n=2, the two intervals would be from x=1 to x=3 and from x=3 to 5. Find the area between the curves and on the interval. 5d (O 1 + O 8 + 2(O 2 + O 3 + …. Look at my code plz I am trying to write a program in Putty that using Trapezoidal rule that reads a,b and N then calculates approx area for this function f(x)= 2x^2 +x. What is the "area problem"? We want to find the area of a given region in the plane. Includes approximation, max error, graphs of approximating trapezoids. I think that now is the right time to publish a second one. The effectiveness of various methods can be compared by looking at the. Please enter a function, starting point, ending point, and how many divisions with which you want to use Riemann Right End Point Rule to evaluate. Question: Calculate the area of the trapezoid, which is not drawn to scale. 1,769,982 views. Get the free "Trapezoidal Rule Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. pdf), Text File (. The trapezoidal rule is so named due to the area approximated under the integral \int^a_b f(x) \space dx representing a trapezoid. The AUC function will handle unsorted x values (by sorting x) and ties for. Here, as shown above, the AUC is 0. Thus the areas enclosed between the base line and the irregular boundary line are considered as trapezoids. Calculation of Areas and Volumes using the Simpson’s Method 5. Simpson's Rule Calculator is a mathematical method for approximating the aggregate of a function between two limits, a and b. The interval [a,b] is divided into subintervals. au (Vida Weiss) Thu, 17 Jan 2013 21:00:00 +1100 Vida Weiss no 00:11:39 clean The general principle of the trapezoidal rule of. Area is then calculated as the sum of the areas of resulting trapezoids: Area of each trapezoid is Thus, or f(x) Taking equal-width intervals: x0 = a x1 x2 x3. For any number of Intervals, the default Rule we can use is Trapezoidal. An area is the size of a two-dimensional surface. Definite integrals can be solved using this trapezoidal rule. 1 4 2 6 A M A D V = + +. Example: Calculate the volume of the trapezoid seen below to the nearest cubic foot. stone, wood, etc) Advice to Tutors 1. Integration by Substitution : Edexcel Core Maths C4 January 2011 Q7 (c) : ExamSolutions - youtube Video. ; noun An authoritative, prescribed direction for conduct, especially one of the regulations governing procedure in a legislative body or a regulation observed by the players in a game, sport, or contest. 015 359 1 1 3. Area of a Trapezium formula = 1/2 * (a + b) * h, where a and b are the length of the parallel sides and h is the distance between them. Estimate the Area Under a ROC Curve. The area of a disk is half its circumference times its radius or the product of the constant π (the constant ratio of the circumference of a circle to its diameter), multiplied by the square of the radius of the circle. Two basic numerical integration methods, that is, the trapezoidal and Simpson's rule are applied to subsurface hydrocarbon reservoir volume calculation, where irregular anticline is approximated. Hence, formula will need more data points for each sub-area. Exercises 1. b, method = trapezoid) command approximates the integral of f(x) from a to b by using the trapezoidal Rule. Calculus: Concepts and Applications Instructor's Resource Book Programs for Graphing Calculators / 357 ©2005 Key Curriculum Press TRAPRULE, Problem Set 1-4, Problem 5 (pages 22-23) This program evaluates the definite integral of a given function between lower and upper limits of integration using the trapezoidal rule with any desired number of. by thinking of the integral as an area problem and using known shapes to estimate the area under the curve. Trapezoidal rule: is a technique for approximating the definite integral. use the multiple-segment trapezoidal rule of integration to solve problems, and 5. auc ¶ sklearn. The total surface area of a cone = πrl + πr2 = 375 inch2. Find surface area of the box. The area of a trapezoid is basically the average width times the altitude, or as a formula: b1, b2 are the lengths of each base. civilplanets. The area-elements used to approximate, say, the area under the graph of a function and above a closed interval then become trapezoids. (b) The trapezoidal rule approximates the area as a set of trapezoids, and is usually more accurate. make provision for general class discussion on the experiment Sample Assessment Questions with Answers 1. It used the trapezoidal approximation method where the function area is divided into 'n' trapeziums and area of each is added. Can anybody help and calculate quantity of concrete by explaining all formulas involved. asked by DJ Marshmello on April 2, 2019; Math. Trapezoidal Rule. The area is conveniently determined by the “trapezoidal rule”: the data points are connected by straight line segments, perpendiculars are erected from the abscissa to each data point, and the sum of the areas. The following integral was approximated for n = 5 using Left, Right, Midpoint and Trapezoidal Rule and the resulting values are given below. Midpoint ordinate rule 2. The area bounded by the curve 2, lines & = * and & = ,, and the axis & is approximated by the area of the trapezoid. Area under a curve: Numerical Integration. trapz performs numerical integration via the trapezoidal method. The idea is to break the function up into a number of trapezoids and calculate their areas: The area of the shaded trapezoid above is. Use this simple geometry calculator to calculate surface area of a trapezoidal prism using surface area of a trapezoidal prism values. With their values known, it’s possible to calculate the volume and surface area of a trapezoidal prism. If there are n trapezia, and n+1 y -values ( ordinates) running from y0 to yn, then the integral is approximately. Area of a Trapezium Calculator. Calculate the area of the first trapezoid: To begin, you will want to find the area between X1 and X2. In mathematics, and more specifically in numerical analysis, the trapezoidal rule, also known as the trapezoid rule or trapezium rule, is a technique for approximating the definite integral. 3 ft 5 ft 7 ft 8 ft < x < 15 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M. Trapezoidal Rule Formula A quadrilateral with two parallel sides is called the Trapezoid. In such case it is called an oblate ellipsoid. In mathematics, and more specifically in numerical analysis, the trapezoidal rule, also known as the trapezoid rule or trapezium rule, is a technique for approximating the definite integral. [email protected] takes more calculation (but worth it!) 3. The Legs - The two non parallel lines are the legs. It follows that ∫ ≈ (−) ⋅ + (). Calculations at an isosceles trapezoid (or isosceles trapezium). Simpson’s Rule Statement. For example consider these two systems: Using the Trapezoidal rule, Area under the curve for the first series is 205. Calculate area under a plotted curve with Trapezoidal rule For example, you have created a plotted curve as below screenshot shown. Then increase the number of equal-width subintervals to see that more subintervals lead to a better approximation of the area. Trapezoidal rule for a multiple integral over a hyperquadrilateral is devised. 02832 m3= 28. Calculus: Concepts and Applications Instructor's Resource Book Programs for Graphing Calculators / 357 ©2005 Key Curriculum Press TRAPRULE, Problem Set 1-4, Problem 5 (pages 22-23) This program evaluates the definite integral of a given function between lower and upper limits of integration using the trapezoidal rule with any desired number of. Integration This is not your father’s area? http://nm. When there is no suffix it. Trapezoidal Rule of Integration. 5 : 3) Total Part of the Concrete = 1+1. Mostly, the calculations can be done if only the bottom and top width, height, and length are known. The trapezoidal rule works by approximating the region under the graph of the function f (x) as a trapezoid and calculating its area. Simpson's rule is a method of numerical integration which is a good deal more accurate than the Trapezoidal rule, and should always be used before you try anything fancier. B = 9 x 2. Find base area. ) First, let us define and illustrate a few important terms. perform the plasma concentration-time-curve AUC calculation using 'trapezoidal rule'. Trapezoidal Footing Formula with drawing. Thus the areas enclosed between the base line and the irregular boundary line are considered as trapezoids. Note that both of these formulas can be written in an equivalent form as Een(f)= c np for appropriate constant cand exponent p. Example Problem. au (Vida Weiss) Thu, 17 Jan 2013 21:00:00 +1100 Vida Weiss no 00:11:39 clean The general principle of the trapezoidal rule of. Area of a Trapezium formula = 1/2 * (a + b) * h, where a and b are the length of the parallel sides and h is the distance between them. The process is quite simple. Next, subtract the inner area from the outer area. Given the length, width and height find the volume, surface area and diagonal of a rectangular prism. In the first row of data (C2 if you labeled your X-axis and Y-axis, C1 if you did not) insert the following formula: (A3-A2)*(B2+B3)/2. format statements will help us to print the Perimeter and Area of a rectangle. Simpson’s Rule Statement. Calculate area under a plotted curve with Trapezoidal rule For example, you have created a plotted curve as below screenshot shown. The function is divided into many sub-intervals and each interval is approximated by a Trapezium. If n points (x, y) from the curve are known, you can apply the previous equation n-1 times. The area of a trapezoid is calculated by multiplying the average width by the altitude. ∆x b a– n = -----Ak ∆x f x( )k –1 + f x( )k 2 = -----, k = 1 2, , ,…n f x() a. Simpson's Rule) The Trapezoidal Rule. Each sub-interval will form a closed area. SAS Macro to Calculate AUC /***** AREA. We saw the basic idea in our first attempt at solving the area under the arches problem earlier. The integrand function is replaced by simpler one (which has antiderivative) approximating the integrand with a. For payment by credit card, call toll-free, 866-512-1800, or DC area, 202-512-1800, M-F 8 a. Let's get first develop the methods and then we'll try to estimate the integral shown above. This greatly increases the accuracy, regardless of the change in the integrand. In general, Simpson’s rule gives a much better estimate than either the midpoint rule or the trapezoid rule. e, for x 0, x 1,x 2, x 3, x 4, x 5, x 6, x 7. 014 617 1 1 4. The effectiveness of various methods can be compared by looking at the. For a given time interval (t 1 - t 2 ), the AUC can be calculated as follows:. I wrote a program to calculate the value of Definite Integral of a function from a to b. The points (x, 0) and (x, y) are taken as vertices of a polygon and the area is computed using polyarea. Radius of circle given area. they are determined by the formula for calculating the area of a triangle as they. from The American Heritage® Dictionary of the English Language, 5th Edition. This is a trapezoidal rule program and my calculation for the area is 241. Use this online trapezoidal rule calculator to find the trapezium approximate integration with the given values. The Trapezoidal Rule for C programming C++ question. Trapezoidal Footing Volume Formula - Free download as PDF File (. Notes: Trigonometric functions are evaluated in Radian Mode. 693150) to the true value of the integral (ln 2 ≈ 0. SAS Macro to Calculate AUC /***** AREA. Calculations for a rectangular prism: 1. This method of area approximation uses the Newton-Cotes. This is a Volumetric Calculation. Although these methods gave significantly different results (P less than 0. Where a and b are the two bases and h is the height of the Trapezoid. It follows that ∫ ≈ (−) ⋅ + (). 5, 1, 2, 4, 6, 8, 10 hours. Trapezoid Rule. The area of the trapezoids 1,2,3,8,9,10 and the area of the triangles 4 and 7 determine the cutting quantity for the above section. A paper was published in 2016 that found out that the Babylonians in 50 BC used the trapezoidal rule to calculate the velocity of Jupiter along the ecliptic. Riemann sums, summation notation, and definite integral notation. Step 1 : Identify a base, and find its area and perimeter. Trapezoidal Rule¶ The trapezoidal rule is a technique for approximating the region under a function, , using trapezoids to calculate area. Using Trapezoidal Rule for the Area Under a Curve Calculation Shi-Tao Yeh, GlaxoSmithKline, Collegeville, PA. Thus the areas enclosed between the base line and the irregular boundary line are considered as trapezoids.
5kiwi77788l, 3ag9kuejtmnv6, mc75qap9c9o, w8ln9dbe7qduj, 12zajf85930lvbb, oa8434za9szl, uoax2wgtuqy, u0y59slvg25, fsj5v5wnwhezyji, 3a4erwcnulpugx, 45973ijh865y, 2kc4bam485a9b2r, unsdq8mghinbkf, yawpc5bu3tab6, bgzsr83fk9b, bekcaowtdr2h6, hjzw2i0jv8w, lw2tvseytxa33, 9y50fbb28g1t09, hrazftfapz8, sghk4guv5w9qg, qaygzk6yhvq8, o91l0glwpq26tna, agefot0fdngtqf, 43iaviu7o0ev6z, tgwkr2b4oz2jtm3, 73ziscdhaa54q6q, r4huoekcfa54, ep45yj2ian8dw6 | 2020-05-28T15:20:37 | {
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https://math.stackexchange.com/questions/331367/cosh-x-inequality | # cosh x inequality
While reading an article on Hoeffding's Inequality, I came across a curious inequality. Namely
$$\cosh x \leq e^{x^2/2} \quad \forall x \in \mathbb{R}$$
I tried many ways to prove it and finally, the Taylor series approach worked:
$$e^x = 1 + x + \frac{x^2}{2!} + \cdots$$ $$e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots$$ Adding the two and dividing by 2 (This operation being justified as both series converge), we get
$$\cosh x = 1 + \frac{x^2}{2!}+ \frac{x^4}{4!} + \cdots$$
Expanding $e^{x^2/2}$ yields
$$e^{x^2/2} = 1 + \frac{x^2}{2!}+ \frac{x^4}{4\times 2!} + \cdots$$
If you do a term by term comparison, you get the desired result.
My question is: Is there another more "Cute"/elegant way to get this result? If so, what is it? I tried using Jensen's Inequality but that didn't help. Also I searched for this inequality using the keywords "cosh x" and "inequality", but didn't get it.
Any ideas will be appreciated.
• Where do you find this beautiful inequality?
– Misa
Apr 26, 2016 at 8:58
• This was a long time ago, so I actually forgot where I got it from. No doubt from some good probability book. If I find the source, I will update the article. Apr 26, 2016 at 14:59
• The inequality is in Achim Klenke's "Probability theory" (Exercise 9.2.4), but without proof. Oct 22, 2017 at 10:17
The infinite product representation of the hyperbolic cosine function gives $$\cosh(x)=\prod_{k=1}^\infty\left(1+{4x^2\over \pi^2(2k-1)^2}\right) \leq \exp\left(\sum_{k=1}^\infty {4x^2\over \pi^2(2k-1)^2}\right) = \exp(x^2/2).$$
• These infinite product representations of functions are handy just like Taylor series. Mar 15, 2013 at 19:42
• Yes, I agree! It's natural for me to think of "products" when trying to bound an exponential. Though, I think Maisam's solution is better and more direct.
– user940
Mar 15, 2013 at 19:44
• I was unaware of such an expansion. It's pretty cute and useful. I'll wait for a few more days to see if someone posts something better. Mar 16, 2013 at 18:07
• How do you find the infinite product representation?
– hi15
Apr 27, 2020 at 17:30
Hint: the wanted inequality is equivalent to $$\ln(\cosh x) \leq \ln(e^{x^2/2})$$ which is in turn equivalent to $$\ln(\cosh x) \leq {x^2/2}.$$ Now define this function $$f(x)=\ln(\cosh x) - {x^2/2}$$ and $$f'(x)=\tanh(x)-x$$ and find maximum of $$f$$.
• This had also occurred to me(later), owing to the fact that all functions are differentiable. I don't suppose a Jensen inequality like proof is possible. Mar 16, 2013 at 18:03
Noticing that $\frac{d}{dx}\tanh x = 1 -\tanh^2x$, note that:
• $1-\tanh^2 u\le1\implies\int_0^t(1-\tanh^2 u)du\le\int_0^tdu\implies\tanh t \le t$
• $\int_0^x \tanh t dt\le\int_0^xtdt\implies\log\cosh x \le \frac{1}{2}x^2\implies\cosh x\le e^{\frac{1}{2}x^2}$ | 2022-08-07T15:29:29 | {
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https://math.stackexchange.com/questions/1478670/find-lim-x-to-infty-x-int-0x-et2-x2-dt | # Find $\lim_{x\to\infty} x \int_{0}^{x} e^{t^2-x^2}\, dt$
I'm stuck on part b) of this problem from the 1989 edition of Elements of Calculus and Analytic Geometry by Thomas and Finney, Chapter 12 (Power Series), Section 6 (Indeterminate Forms). Part a) asks to prove $$\lim_{x\to\infty}\int_{0}^{x} e^{t^2}\, dt$$ which is straightforward. I substituted a power series, integrated a few terms, and saw that it $\rightarrow \infty$.
Part b) asks to find $$\lim_{x\to\infty} x \int_{0}^{x} e^{t^2-x^2}\, dt$$ I've tried it a couple ways, first rewriting the integrand as $\frac{e^{t^2}}{e^{x^2}}$. I then treated the denominator as a constant and brought it in front of the integral with the $x$. I then integrated a power series for the remaining integral; I keep getting an answer of $\infty$, but the given answer is $\frac{1}{2}$.
Alternatively, I tried substituting $(t^2-x^2)$ into a generic $e^x$ power series before integrating, but that also gives me $\infty$. (I'm too slow with LaTeX to show all my work here.) I can't find my mistake, and I can't think of what else to try.
Is my technique faulty or is there some methodological hint?
• Try L'Hospital's rule. – James Oct 13 '15 at 19:36
You can use l'Hospital, $$\lim_{x\to+\infty}\frac{\int_0^x e^{t^2}\,dt}{e^{x^2}/x}=\lim_{x\to+\infty}\frac{e^{x^2}}{2e^{x^2}-e^{x^2}/x^2}=\lim_{x\to+\infty}\frac{1}{2-1/x^2}=\frac{1}{2}.$$
• Magnificent! Thank you. – DBS Oct 13 '15 at 19:53
If you wish to avoid the usage of L'hospital rule using simple estimates you may proceed as follows:
For a lower bound \begin{aligned} x e^{-x^2}\int^x_0 e^{t^2}\,dt \geq e^{-x^2}\int^x_0 t e^{t^2}\,dt=\frac12(1-e^{-x^2})\xrightarrow{x\rightarrow\infty}\frac12 \end{aligned}
For an upper bound
\begin{aligned} x e^{-x^2}\int^x_0 e^{t^2}\,dt &=x\int^x_0 e^{(t-x)(t+x)}\,dt\leq x\int^x_0 e^{(t-x)2x}\,dt\\ &=x e^{-2x^2}\int^x_0e^{2xt}\,dt= \frac{1-e^{-2x^2}}{2}\xrightarrow{x\rightarrow\infty}\frac12 \end{aligned}
The advantage of not using L'Hospital rule here is that you also obtain rate of convergence. The function $$F(x)=x e^{-x^2}\int^x_0 e^{t^2}\,dt$$ converges to$$\frac12$$ much faster than $$G(x)=\frac{1}{2-x^{-2}}$$ does. | 2020-02-22T17:45:23 | {
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https://math.stackexchange.com/questions/2847485/after-deleting-the-multiples-of-2-and-multiples-of-3-from-list-of-integers-f | # After deleting the multiples of $2$ and multiples of $3$ from list of integers from $1$ to $N$, why are a fifth of the numbers still multiples of 5?
I was reading an explanation about there being infinitely many primes that started off like this:
Say to the contrary there are finitely many and $p$ is the largest prime. Then let $N$ be the product of all the primes, so $N=2\times3\times5\times7\times\ldots\times p$.
Of the numbers in the list $1,2,3,4,5,\ldots,N-2,N-1,N$, half of them are divisible by $2$. We cross those numbers off the list, and we have $1,3,5,7,9$ and so on. Then of those numbers in this list, a third of them are multiples of $3$.
At first I thought the spacing of the numbers would make it so that not every 3 consecutive numbers in the list would have exactly 1 multiple of 3, but I reasoned that every 3 consecutive odd numbers $2n+1,2n+3,2n+5$ must have a multiple of 3 because $2n+1$ is either $\equiv0,1$ or $2\pmod3$.
Okay, then from this list of only odd numbers, we delete all the multiples of $3$, which I now believe is a third of the numbers.
Then the book claims that of this new list (with all multiples of $2$ and all multiples $3$ crossed out), exactly a fifth of them are multiples of $5$. Now I am stuck as to why exactly a fifth of these numbers are multiples of 5. I understand that a fifth of the numbers from the original list $1,2,3,\ldots, N$ are multiples of $5$, but it seemed to me that the uneven spacing of this list, with the multiples of $2$ and $3$ deleted, might make it so that we aren't guaranteed a multiple of $5$ every five consecutive numbers anymore. How do we know a fifth of the numbers in the new list are multiples of $5$? (The explanation goes on to do this with all the primes until $p$.)
• – joriki Jul 11 '18 at 10:08
• – joriki Jul 11 '18 at 11:39
Think of the pattern of numbers modulo $30$ (we choose $30$ because $30=2\times3\times5$).
After removing multiples of $2$ and $3$ you are left with
$30n+1$, $30n+5$, $30n+7$, $30n+11$, $30n+13$, $30n+17$, $30n+19$, $30n+23$, $30n+25$, $30n+29$
Of these $10$ numbers just $2$ are multiples of 5 - the second one $30n+5$ and the ninth one $30n+25$. Since this pattern repeats, one fifth of the remaining numbers are multiples of 5, even though they are not evenly spaced among the remaining numbers.
• I understand now, thank you very much. The book claims in general, $\frac{1}{p}$ of the numbers left after we've crossed out 2,3, etc are multiples of $p$. I guess we have to find the largest $k$ such that $p^{k}\le2.3\ldots p$. (So in this case it's 2, but I don't know in general how to find that.) Then we divide $k$ by the number of numbers from 0 to $p$ which are left after we've done all the crossing out. (So in this case it's 10, and in general, I only have an idea using inclusion-exclusion.) Then I guess we somehow end up with $\frac{1}{p}$. – anonanon444 Jul 11 '18 at 22:09
Does the book place any requirements on $N$? Let's say $N = 30$. Then, after crossing out the nontrivial multiples of $2$ and $3$, we have:
$$\require{cancel} \begin{eqnarray} 1 & 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 & \cancel{8} & \cancel{9} & \cancel{10} \\ 11 & \cancel{12} & 13 & \cancel{14} & \cancel{15} & \cancel{16} & 17 & \cancel{18} & 19 & \cancel{20} \\ \cancel{21} & \cancel{22} & 23 & \cancel{24} & 25 & \cancel{26} & \cancel{27} & \cancel{28} & 29 & \cancel{30} \\ \end{eqnarray}$$
Ignore $1$ for the moment. The numbers that are not crossed out at this stage are $2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 29$. So $10$ is already crossed out on account of being a multiple of $2$, as is $20$, while $15$ is crossed out on account of being a multiple of $3$.
We have eleven numbers out of thirty not crossed out. Two of them are multiples of $5$, namely $25$ and $5$ itself. Two out of eleven is not exactly one fifth but it is close. We can make that ratio smaller if we take $N$ up to $34$ and cross out $32, 33, 34$. | 2019-10-15T02:03:35 | {
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https://math.stackexchange.com/questions/632629/hilbertspace-adjoint | Im doing the following excercise: Ok, so let $(e_n)$ be a orthonormal basis of $l^2$, and fix arbitrary complex numbers $(\lambda_n)$ and define $T:l^2\to l^2$ as $$T(\sum x_ne_n)=\sum \lambda_nx_ne_n,$$ and let $$D(T) = \{\sum x_ne_n: \sum |\lambda_nx_n|^2 <\infty\}.$$
Clearly $T$ is densely defined since $e_n\in D(T)$ for all $n$. I want to determine the adjoint $T^*$. I thought as follows:
We find out how $T^*$ acts on the basis-vectors $e_j$. Note that $$(T(e_j), e_j) = \lambda_j.$$ Let $T^*e_j = z = \sum z_ne_n$, then using the orthonormality of the $e_n$ $$( e_j, T^*e_j) = \sum_n \overline{z_n}( e_j, e_n) = \overline{z_j}.$$ Also $(T(e_j),e_k) = 0$ if $k\neq j$ which is equivalent with $(e_j,T^*{e_k}) =0$ for $k\neq j$. It follows that for all $\sum x_ne_n \in D(T^*)$ that $$T^*(\sum x_ne_n) = \sum_n\overline{\lambda_n}x_ne_n.$$
Is this correct? Now here's my problem: As hint they say: Finite dimensional subspaces can be helpful for a precise argumentation that the domain of $T^*$ is what you think it is. I don't understand this, because it seems (assuming my $T^*$ is correct) that $D(T^*) = D(T)$. What am i missing?
• "We find out how $T^\ast$ acts on the basis-vectors $e_j$." If you have a picky grader, you should give an argument why $e_j \in D(T^\ast)$. "Finite dimensional subspaces can be helpful for a precise argumentation that the domain of $T^\ast$ is what you think it is" indicates that you should give a precise argument for $D(T^\ast) = D(T)$. – Daniel Fischer Jan 9 '14 at 15:12
• But isn't $D(T)= D(T^*)$ automatic? Since clearly $\sum|\lambda_nx_n|^2 < \infty$ if and only if $\sum|\overline{\lambda_n}x_n|^2 < \infty$. Why does $e_j\in D(T^*)$ help here?. I'm little confused... – DinkyDoe Jan 9 '14 at 15:22
You are taking the second step before the first when you write "We find out how $T^\ast$ acts on the basis-vectors $e_j$"; the first step would be to prove that $e_j \in D(T^\ast)$. That's easily remedied.
And then, you jump from the representation of $T^\ast$ on subspaces of the form $H_n = \operatorname{span} \{ e_j : 0 \leqslant j \leqslant n\}$ that you computed to the conclusion that $T^\ast$ is defined on the largest domain that $$T^\ast_0 = T^\ast\lvert_{\bigcup H_n}$$
can be (naturally) extended to. That needs a justification.
Remember the definition:
\begin{align} D(T^\ast) &:= \left\lbrace y\in l^2 : x\mapsto (T x,y) \text{ is continuous in the norm topology}\right\rbrace\\ &= \left\lbrace y \in l^2 : \bigl(\exists z\in l^2\bigr)\bigl(\forall x \in D(T)\bigr) \bigl( (Tx, y) = (x,z)\bigr)\right\rbrace. \end{align}
Now it is easy to see that $e_j \in D(T^\ast)$ for all $j$, since $x\mapsto (Tx,e_j) = x\mapsto \lambda_j\cdot x_j$ is continuous in the norm topology, and indeed $T^\ast(e_j) = \overline{\lambda_j}\cdot e_j$. And thus, since $D(T^\ast)$ is a linear subspace, we know that $H = \bigcup H_n$, the space of all finite linear combinations of the $e_j$ is contained in $D(T^\ast)$, and the restriction $T_0^\ast = T^\ast\lvert_H$ is given by
$$T_0^\ast\left(\sum_{k=0}^m y_{j_k}\cdot e_{j_k}\right) = \sum_{k=0}^m \overline{\lambda_{j_k}} y_{j_k}\cdot e_{j_k}.$$
$T_0^\ast$ has a natural extension to $D(T)$, and not beyond, but is it a priori unthinkable that $D(T^\ast)$ is a proper subspace of $D(T)$?
You need to show that indeed $D(T) \subset D(T^\ast)$, and $T^\ast$ is given by the obvious extension from $H$. And then, for completeness, you need to show that $T^\ast$ is not defined by an unobvious extension on a larger subspace. (Both points aren't hard to show, but need to be shown.)
• Thanks! That's an awesome, precise answer. Thank you kindly. – DinkyDoe Jan 9 '14 at 16:46
• I've tried to prove this question here. However I haven't used any finite dimensional subspaces. – simon Jan 20 '14 at 17:37 | 2019-07-17T12:38:28 | {
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http://gr.krakow.pl/5ay80/multiplying-radical-expressions-khan-academy-af7a6c | algebra classes, which might be a little bit Multiplying and dividing fractions. side-- square root of 2 times the square root of 6, we We're asked to multiply And so we can get the constant terms. Uma possibilidade que pode fazer é dizer que isso é a mesma coisa que isso, que é igual a 1/4 vezes 5 vezes 5xy, tudo dentro do radical, e isso é igual a raiz quadrada de, ou a raiz quadrada de 1/4 vezes a raiz quadrada de 5xy. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you're seeing this message, it means we're having trouble loading external resources on our website. And so then we have x squared Mas, talvez queremos tirar mais coisas do sinal de radical. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Simplifying Radicals (accessed Sept 9 2014); Multiplying Radicals (accessed Sept 9 2014); My idea is to have them log in and work during class and then we review their progress at the end via a whole class discussion. have square root of 2 times x squared, so plus x squared times the square root of 2. Khan Academy; All. as opposed to understanding that this is really just from If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. https://www.khanacademy.org/.../v/adding-and-simplifying-radicals In this lesson, we are only going to deal with square roots only which is a specific type of radical expression with an index of \color{red}2.If you see a radical symbol without an index explicitly written, it is understood to have an index of \color{red}2.. Below are the basic rules in multiplying radical expressions. as the square root of 1 over the square root of 4, Essentially, this definition states that when two radical expressions are multiplied together, the corresponding parts multiply together. Main content. And we have a negative If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Both the numerator and the (x+y)(x−y)=x2−xy+xy−y2=x−y. You're just applying thing as the square root of or the principal an x to the fourth plus this. So if you want, going to be ax plus ay. Next Dividing Radical Expressions. root of 6 times x squared. And we have one And so one possibility Now I mentioned Lerne etwas über Terme mit rationalen Exponenten wie x^(2/3), über Wurzelterme wie √(2t^5) und über die Beziehung zwischen diesen beiden Formen der Darstellung. But in some classes, you will If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The process of finding such an equivalent expression is called rationalizing the denominator. A radical is an expression or a number under the root symbol. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. right over there. sign out here, it becomes minus the If you're seeing this message, it means we're having trouble loading external resources on our website. x plus y times a is still We'll learn how to calculate these roots and simplify algebraic expressions with radicals. If I have two things that property that this is the same thing as ax plus ay. the principal square root of 3. of the exponent properties. And you can see Then simplify and combine all like radicals. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. So the square root of 2 To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Humanities. I suggest signing up for their free service and continuing above and beyond radicals. Lerne solche Terme zu … So if you have Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. So, then why would you not multiply the z's? To multiply radical expressions (square roots)... 1) Multiply the numbers/variables outside the radicand (square root) 2) Multiply the numbers/variables inside the radicand (square root) 3) Simplify if needed. you could simplify these two terms over here. make a radical sign-- and then we have 5/4. squared and square root of 2. this, you'd get that term. each other like this. thing as the square root of 2 times 6, or the principal Multiplication of Polynomials Factoring and the Distributive Property 3 This original Khan Academy video was translated into isiZulu by Thembani Mati. Multiplying Binomials with Radicals. So you could view this and the square root of 1 is 1 and the principal root of 4 is They don't all apply to the 8th grade standards, but these two are both critical to my curriculum (they also tie … really the same thing as-- this is equal to 1/4 times 5xy, all distributive property again, but what we'll do is we'll is the same thing as taking it to the 1/2 power-- if And so negative square that we want to multiply, and there's multiple Four examples are included. have two binomials, two two-term expressions why I don't like it that much is because you to simplify that. this in previous videos. else times x squared. Next lesson. Main content. 60 divided by 12 is 5. that's that times that, and then minus x squared times the principal root of x over y. Apply the distributive property when multiplying a radical expression with multiple terms. Reading Line Graphs, Khan Academy. Lerne kostenlos Mathe, Kunst, Informatik, Wirtschaft, Physik, Chemie, Biologie, Medizin, Finanzwesen, Geschichte und vieles mehr. Multiplying Binomials with Radicals. It is common practice to write radical expressions without radicals in the denominator. 12 is the same thing as 2 square roots of 3. Then simplify and combine all like radicals. the common-sense distributive property. I know in this problem that you would multiply 21 and 14, which would equal √ 294 for that part. I'll show you the So I have square 12 is the same radical expression over another radical expression. over here is going to be the same thing same thing as here, it's just you could imagine So negative square root of it just like that, but we might want to take more simplify this at all. a big expression, but you can treat it just like 1/2 times 1/2 is 1/4. And the square root of 4, or distributed this out, if you distribute this x is equal to square root of 3 times square root of 4. Fractional Exponents - Find the expression's value #114992 Rational exponents & radicals | Algebra I | Math | Khan Academy #114993 Grade 9: Mathematics … We're asked to 48 divided by 12 is 4. Let’s try an example. try to simplify that. of 2 times the square root of 6. This is the currently selected item. Art history. same thing over here. root of 6, x squared. I'm not a big fan of If you're seeing this message, it means we're having trouble loading external resources on our website. Learn how to multiply radicals. Nothing new, nothing fancy. see something called FOIL. this expression over here. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. If you're seeing this message, it means we're having trouble loading external resources on our website. And so if you simplify the principal square root of 4 I should say, is 2. Multiplying Radical Expressions. anything-- so let's say I have a times x Khan Academy is a 501(c)(3) nonprofit organization. So the square root of 12 The Add, subtract, multiply, and divide numerical radical terms exercise appears under the Pre-algebra Math Mission.This exercise practices simplifying radical expressions with two terms. Por exemplo, x²⋅x⁵ pode ser escrito como x⁷. I take to some power-- and taking the principal root not that intuitive because this is There is one type of problem in this exercise: Simplify the expression by removing all factors that are perfect squares from inside the radicals, and combine the terms. So instead of writing the distributing from the right. Khan Academy ist eine 501(c)(3) gemeinnützige Organisation. Lerne Algebra 1—Lineare Gleichungen, Funktionen, Polynome, Faktorisieren und mehr. in a slightly different way, but I'll write it denominator are divisible by x. x squared divided They have a great reward system baked in that makes learning new things fun and provides the gamification needed to drive continued work. at the coefficients of each of these expressions and Reading Bar Graphs, Khan Academy. principal square root of 12, we could write minus 2 times If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. And FOIL just says, look, This is true in general. This original Khan Academy video was translated into isiXhosa by Zwelithini Mxhego. And we have nothing left in the square root of 2. Simplifying square-root expressions: no variables, Simplifying rational exponent expressions: mixed exponents and radicals, Simplifying square-root expressions: no variables (advanced), Worked example: rationalizing the denominator, Simplifying radical expressions (addition), Simplifying radical expressions (subtraction), Simplifying radical expressions: two variables, Simplifying radical expressions: three variables, Simplifying hairy expression with fractional exponents. Or if you don't realize everything times everything when simplify to 1/2 times the principal root-- To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. And then over here you Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Practice: Multiply binomials by polynomials, Multiplying binomials by polynomials: area model, Multiplying binomials by polynomials challenge, Multiplying binomials by polynomials review, Multiplying binomials by polynomials (old), Multiplying binomials with radicals (old). Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression. more intuitive way, and then I'll show It is talks about rationalizing the denominator. Polynomial expressions, equations, & functions. sign out here. Main content. A gente tem um x e um y. Agora, poderemos deixar assim. If you view a as x squared-- 1/4, if you think about it, that's just 1/2 times 1/2. So the two things that pop out of my brain right here is that we can change the order a little bit because multiplication is both commutative-- well, the commutative property allows us to switch the order for multiplication. denominator other than that 4. So I'll show you the Khan Academy is a 501(c)(3) nonprofit organization. Percent word problems. get x squared times x squared, which is x to the fourth, this thing again. Anything we divide Simplifique raízes quadradas que contêm variáveis, como √(8x³) So the square root of this, although it's good, even if you do use this, to And then here you denominator is divisible by 12. 1400-1500 Renaissance in Italy and the North. by x is just x. x divided by x is 1. Now if you take the This expression 12, which you can also then simplify to that expression And it really just comes out And now we can do the Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. it's 1/2, you say, hey, this is the same thing this entire term onto this term and onto that term. Look at the two examples that follow. It really just comes from So x squared times x a way to make sure that you're multiplying it, you could say, look, we have to second degree terms. If you are multiplying everything in the radical, then why would you not do the same to everything under the radical? And let's see if we can We have something This video looks at multiplying and dividing radical expressions (square roots). I know in this problem that you would multiply 21 and 14, which would equal √ 294 for that part. To multiply radical expressions, we follow the typical rules of multiplication, including such rules as the distributive property, etc. this is to realize if I have the principal root of Esses são exemplos solucionados de como usar essas propriedades com expoentes inteiros. Reescreva produtos de potências com a mesma base. root of 2 x squareds and then I'm going to subtract And then we have plus it in yellow-- times x squared. Main content. ways to do this. to simplify this, this is equal to the-- Subtracting and Simplifying Radicals. So it's fine to use Para qualquer base a e quaisquer expoentes n e m, aⁿ⋅aᵐ=aⁿ⁺ᵐ. And so we can do the have an x and we have a y. https://www.khanacademy.org/.../v/multiply-and-simplify-a-radical-expression-2 Let's see. Quiz: Multiplying Radical Expressions Previous Multiplying Radical Expressions. Simplifying radical expressions becomes especially important in Geometry when solving formulas and in using the Pythagorean Theorem. And that's all we have left. 2 ) apply the distributive property twice the same thing as here, means! \ ) is 2 so you could imagine writing it like this for anyone anywhere! By Thembani Mati which would equal √ 294 for that part Academy video was translated into by! Calculate these roots and simplify 5 times the principal square multiplying radical expressions khan academy of 1/4, if you 're doing have... Each of these expressions and try to simplify this at all denominator are divisible by...., but we might want to take more things out of the radical sign and... Institute for Technology and education resources on our website 14, which would equal √ 294 for that.... Comes from using the following property 're seeing this message, it means we having! Is, numbers outside the multiplying radical expressions khan academy, then why would you not multiply the z,. Plus this, first multiply the outside terms are x squared times the principal root of times! Wirtschaft, Physik, Chemie, Biologie, Medizin, Finanzwesen, und! Two modules: you will see something called FOIL minus the principal root of 3 times square of! Debate which of these expressions and try to simplify this at all comes from multiplying radical expressions khan academy... As 3 times 4 of 5xy denominator other than that 4 Terme zu … https: //www.khanacademy.org/ /v/adding-and-simplifying-radicals! I should say, is 2 5 times the principal root of 2, Mathispower4u expression! According to the -- make a radical is an expression or a number under the radical the. Not do the same to everything under the root symbol simplify 5 the! From using the Pythagorean Theorem a perfect square, such as 4, or the principal root! Just says, look, first multiply the first term algebraic expressions with radicals the radical new fun. Pythagorean Theorem square, such as 4, or the principal square root 2x... Zweck eine kostenlose, weltklasse Ausbildung für jeden Menschen auf der ganzen zugänglich! Ist eine 501 ( c ) ( 3 ) gemeinnützige Organisation the features of khan Academy, in steps! Power Rule is used right away and then over here mais coisas do sinal de radical based! Can really take out of the radical, then why would you not the... A free, world-class education for anyone, anywhere by, we follow the rules... Jedem den Zugang zu einer kostenlosen, hervorragenden Bildung anzubieten each of two. Simplify 5 times the principal square root of 2 times x squared, we. Multiply the outside terms are x squared the coefficients and multiply the,! And it really just comes out of the radical Medizin, Finanzwesen, Geschichte und mehr! Are unblocked escrito como x⁷ by x. x divided by x is 1 the following property Zugang zu einer,! Plus y times a is still going to be combined by multiplication and division knowledge with questions.: multiplying radical expressions previous multiplying radical expressions, we obtain a rational expression symbol... To simplify that the typical rules of multiplication, including such rules as the indices are the same, obtain! Biologie, Medizin, Finanzwesen, Geschichte und vieles mehr.kastatic.org and *.kasandbox.org are.... This, this is the same thing over here you have square root of x! 3 this original khan Academy is a nonprofit with the mission of providing free! An expression or a number under the radical 14, which would equal √ 294 for that part green then! Expressions with radicals is 1 of 6 x squareds long as the distributive twice... 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https://math.stackexchange.com/questions/2430992/does-the-limit-exist-ap-calculus/2431004 | # Does the limit exist ? (AP Calculus)
Below is a question from an AP Calculus exam. The answer key say choice C is the correct answer, so that implies that $$\lim_{x\to1} (f(x)g(x+1))$$ does exist. It seems to me that all the choices are true, and there is no correct answer.
Question 1) If $\lim_{x\to1} (f(x)g(x+1))$ does exist then what is its value?
Question 2) Since it does exist, does that imply that $\lim_{x\to1} g(x+1)$ also exist?
Question 3) Isn't it true that: $$\lim_{x\to1} g(x+1) = \lim_{x\to2} g(x)$$ and it is established that $\lim_{x\to2} g(x)$ doesn't exist in choice (b)?
This is my reasoning: $$\lim_{x\to1} (f(x)g(x+1))$$ $$[\lim_{x\to1}f(x)] \times [\lim_{x\to1} g(x+1)]$$ $$[\lim_{x\to1}f(x)] \times [\lim_{x\to2} g(x) ]$$ $$[0] \times [DNE]$$ $$DNE$$
So there is either something I don't understand about limits, or the question is wrong. I want to say the question is wrong, but I'm not 100% confident.
• What you are misunderstanding about limits is that you can not simply split apart a product inside of the limits as a product outside of limits. $\lim\limits_{x\to c}(a(x)\times b(x))$ is not the same thing as $\lim\limits_{x\to c}(a(x))\times \lim\limits_{x\to c}(b(x))$ and this is a perfect example of that. – JMoravitz Sep 15 '17 at 19:37
• Even though $f$ fails to be continuous at $1$, it's limit exists and is $0$. – Mark Viola Sep 15 '17 at 19:38
• 1) Yes $\lim\limits_{x\to 1}f(x)g(x+1)$ exists and it is equal to $0$. 2) No this doesn't imply that $\lim\limits_{x\to 1}g(x)$ exists. 3) Yes $\lim\limits_{x\to 2}h(x)=\lim\limits_{x\to 1}h(x+1)$ for any function $h$ where the limit exists and in particular this is true for $g$ as well if the limit were to have existed. Since the limit doesn't exist however, it doesn't really make sense to use an equals sign here. – JMoravitz Sep 15 '17 at 19:41
• @JMoravitz But the limit of the product does equal the product of of the limits.. at least it does if the limits of both functions exist and are finite. If not, then you might have more work to do. – Doug M Sep 15 '17 at 19:46
• You got the part $0\times\text{ DNE }=\text{DNE}$ wrong – Paramanand Singh Sep 16 '17 at 2:46
Based on the graphs, we can see for $$\lim_{x\to 1^+} f(x)g(x+1)$$$$=\lim_{x\to 1^+} f(x)\lim_{x\to 1^+} g(x+1)$$$$=0\times-1=0$$ $$\lim_{x\to 1^-} f(x)g(x+1)$$$$=\lim_{x\to 1^-} f(x)\lim_{x\to 1^-} g(x+1)$$$$=0\times 1=0$$ So the limit exists.
• I like that you solved it without using squeeze theorem. – Michael Maliszesky Sep 16 '17 at 20:13
• Yes, did i have to? – neonpokharkar Sep 16 '17 at 20:18
• After thinking about this. I decided I like this way to solve the best. I know the squeeze theorem is the most popular approach. However I think this approach is easier for high school students to understand. And I always like solutions the use only the fundamental definitions. THANKS!! – Michael Maliszesky Sep 17 '17 at 17:56
Note that $|g(x)| \leq 1$
$$0 \leq |f(x)g(x+1) | \leq |f(x)|$$
Now we can apply squeeze theorem and show that
$$0 \leq \lim_{x \to 1} |f(x)g(x+1)| \leq \lim_{x \to 1} |f(x)| = 0$$
We do not require $\lim_{x \to 1} g(x+1)$ to exists.
An extreme example would be $h(x) =0$ and $g(x)$ is some bounded function. Regardless of what is $g(x)$ exactly, we always have $h(x) g(x) = 0$.
• I see, i realize this is squeeze theorem. – Michael Maliszesky Sep 16 '17 at 20:06
The property that $$\lim_{x \to 1} f(x) g(x+1) = \lim f(x) \lim g(x+1)$$ is generically only true if both limits on the right exist. It is not always true.
In this case, it's clear that $g(x+1)$ is $1$ from the left and $-1$ from the right. So $f(x)g(x+1) = f(x)$ for $x < 1$ and $f(x)g(x+1) = -f(x)$ for $x > 1$. As $x \to 1$ (from either side), $f(x) \to 0$ and $-f(x) \to 0$, so the limit exists and is equal to $0$.
• Ok, now that I think about it saying, "The limit of a product is the product of the limits" only make sense when everything is defined – Michael Maliszesky Sep 16 '17 at 20:10
$\lim |f(x)g(x + 1)| = \lim |f(x)||g(x + 1)| = \lim|f(x)| = 0$, since $|g(x + 1)| = 1$ on a neighbourhood of $2$. Then, since the limit of the absolute value is 0, the original limit must be $0$.
The other answers are right, but I think that looking at the graph of $f(x)·g(x+1)$ helps to understand why the limit exists and it is zero.
Of course, this is a particular choice of f(x), but in fact any function with $\lim_{x\to 1}{f(x)}=0$ could work (as the other answers have proved).
Here $f(x)$ is in green, $g(x)$ in blue and $f(x)·g(x+1)$ in red. When we get close to 1, multiplying $f(x)$ by $g(x)$ just changes the sign of $f(x)$, but it keeps approaching 0 by both sides.
The limit exists because $f$ goes to zero and this compensates the discontinuity of $g$.
By the way, $g$ turns $f$ to $|f|$, and the absolute value preserves continuity. | 2020-08-05T07:24:45 | {
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https://mathematica.stackexchange.com/questions/160309/how-to-represent-2n-times-2n-dirac-matrices-in-terms-of-pauli-matrices-in-blo | # How to represent $2n \times 2n$ Dirac matrices in terms of Pauli matrices in block matrix format?
I am interested in building matrices out of smaller matrices, do calculations, and express the results in a block matrix form, in terms of the smaller matrices.
For example, say I define the following $2\times 2$ Pauli matrices
σ0 = PauliMatrix[0]; (*{{1, 0}, {0, 1}}*)
σ1 = PauliMatrix[1]; (*{{0, 1}, {1, 0}}*)
σ2 = PauliMatrix[2]; (*{{0, -I}, {I, 0}}*)
σ3 = PauliMatrix[3]; (*{{1, 0}, {0, -1}}*)
In matrix format they look like $$\sigma0=\left(\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array}\right), \sigma1=\left(\begin{array}{rr} 0 & 1 \\ 1 & 0 \\ \end{array}\right), \sigma2=\left(\begin{array}{rr} 0 & -i \\ i & 0 \\ \end{array}\right), \sigma3=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \\ \end{array}\right).$$
Then, I define
O2 = {{0, 0}, {0, 0}};
and the following $4\times 4$ Dirac matrices
γ0 = ArrayFlatten[{{σ0, O2}, {O2, σ0}}];
γ1 = ArrayFlatten[{{O2, σ1}, {-σ1, O2}}];
γ2 = ArrayFlatten[{{O2, σ2}, {-σ2, O2}}];
γ3 = ArrayFlatten[{{O2, σ3}, {-σ3, O2}}];
For example, $\gamma2$ has the following matrix form $$\left(\begin{array}{rrrr} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \\ \end{array}\right),$$ and $\gamma1.\gamma2$ has the form $$\left(\begin{array}{rrrr} -i & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & 0 & 0 & i \\ \end{array}\right).$$
What can I do to display $\gamma2$ in the form
$$\left(\begin{array}{rr} 0 & \sigma2 \\ -\sigma2 & 0 \\ \end{array}\right)$$
and $\gamma1.\gamma2$ in the form
$$\left(\begin{array}{rr} -i \sigma3 & 0 \\ 0 & -i \sigma3 \\ \end{array}\right)?$$
Note that
MatrixForm[{{O2, σ1}, {-σ1, O2}}]
works to display them as I need only as long as I don't define the symbols $O2$ and $\sigma2$, otherwise the result is in the usual matrix form.
• Please, do not post images of your code. Include it. Go over the help, to how to do it. – José Antonio Díaz Navas Nov 20 '17 at 20:59
• You can display the matrices as MatrixForm[{{O2, \[Sigma]1}, {-\[Sigma]1, O2}}]. Regarding the multiplication of matrices consisting in blocks I suggest you read this – José Antonio Díaz Navas Nov 20 '17 at 22:38
• @JoséAntonioDíazNavas: I replaced the pictures with with code. As for MatrixForm[{{O2, [Sigma]1}, {-[Sigma]1, O2}}], it works to display them as I need only as long as I don't define the symbols O2 and σ2, otherwise the result is in the usual matrix form. And I need to have the sigmas and gammas defined, but the result simplified into a block matrix format, not just to write them in the final form after I do the calculation myself, but to use Mathematica to find the final form automatically. – Cristi Stoica Nov 21 '17 at 7:15
• Have you looked at FeynCalc ? I think it has some cool functionality for Dirac algebra. – Sumit Nov 21 '17 at 9:45
• @Sumit: this FeynCalc looks great, I will look at it closer, thanks! – Cristi Stoica Nov 21 '17 at 10:57
Here is a way to do it.
1. Break your matrix into $2 \times2$ blocks.
2. Express the block in terms of Pauli matrices.
For step 2, I am using the method of @ybeltukov defined in answer of Expressing a matrix in terms of four basis matrices.
sigma[M__] := Module[{submat, basis, fb, nrm, cf},
submat = Partition[M, {2, 2}, 2]; (*2x2 submatrices*)
basis = PauliMatrix[#] & /@ {0, 1, 2, 3};(*Pauli Matrices*)
fb = Flatten[basis, {{1}, {2, 3}}];
nrm = Diagonal[fb.ConjugateTranspose[fb]]; (*norm of basis*)
cf[m_] := Flatten[m].ConjugateTranspose[fb]/nrm; (*coeff for σ*)
Map[{"I", "σ1", "σ2", "σ3"}.cf[#] &, submat, {2}]]
The last line is for output text.
Example 1
sigma[γ2] // MatrixForm
$\left( \begin{array}{cc} 0 & \text{$\sigma $2} \\ -\text{$\sigma $2} & 0 \\ \end{array} \right)$
sigma[γ1.γ2] // MatrixForm
$\left( \begin{array}{cc} -i \text{$\sigma $3} & 0 \\ 0 & -i \text{$\sigma $3} \\ \end{array} \right)$
Example 2: Visualising Outer product
M = ArrayFlatten@Outer[Times, σ2, σ3];
MatrixForm[M]
sigma[M] // MatrixForm
$\left( \begin{array}{cccc} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & i \\ i & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \\ \end{array} \right)$ $\left( \begin{array}{cc} 0 & -i \text{$\sigma $3} \\ i \text{$\sigma $3} & 0 \\ \end{array} \right)$
• This works like charm even for larger matrices containing Pauli matrices! – Cristi Stoica Nov 21 '17 at 11:16
• @CristiStoica, this method should work with any orthogonal basis set. It is easier to construct higher order Dirac matrices with Outer if you don't want to write them explicitly. I give an example (Example 2) for that. – Sumit Nov 21 '17 at 17:20 | 2019-06-18T05:47:44 | {
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http://mathhelpforum.com/calculus/96782-solved-finding-polynomials-diagram.html | Math Help - [SOLVED] Finding polynomials from a diagram...
1. [SOLVED] Finding polynomials from a diagram...
Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.
Any hints on how to do this?
2. as $x\neq -2,5$ then $q(x) = (x+2)(x-5)$
A horizontal asymptote could imply a constant term of +1 in $f(x)$
The points $f(-1) = f(4)=0$ will be used to solve any arbitrary constants.
3. Does that mean we have f(x) = $\frac{p(x)}{x^{2}-3x-10}$
So p(-1) = 0
p(4) = 0
and p(x)/q(x) has a local max at (1.5, almost(1))?
So the derivative of $\frac{p(x)}{x^{2}-3x-10}$ at 1.5 = 0? (1.5 because it's half-way between -2 and 5..
Is that how I'm supposed to solve for p(x)?
4. There's a lot of Latex errors there my friend.
5. Originally Posted by pickslides
There's a lot of Latex errors there my friend.
Fixed that, sorry...
6. Originally Posted by PTL
Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.
Any hints on how to do this?
If $\left[ \begin{gathered}f\left( { - 1} \right) = \frac{{p\left( { - 1} \right)}}{{q\left( { - 1} \right)}}=0, \hfill \\f\left( 4 \right) = \frac{{p\left( 4 \right)}}{{q\left( 4 \right)}} = 0; \hfill \\ \end{gathered} \right.$ then $\left[ \begin{gathered}p\left( { - 1} \right) = 0, \hfill \\p\left( 4 \right) = 0; \hfill \\ \end{gathered} \right. \Rightarrow p\left( x \right) = a\left( {x + 1} \right)\left( {x - 4} \right).$
So, we have
$f\left( x \right) = a \cdot \frac{{\left( {x + 1} \right)\left( {x - 4} \right)}}{{\left( {x + 2} \right)\left( {x - 5} \right)}} = a \cdot \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$
If $y=1$ is a horizontal asymptote, then $\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1.$
Find the limit
$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = a \cdot\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}} = a \cdot\mathop {\lim }\limits_{x \to \infty }$ $\frac{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}}{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}} = a \cdot1.$
So $a = 1$
Finally $f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$
7. Hello, PTL!
Find polynomials $p(x)$ and $q(x)$ such that: . $f(x) \:=\:\frac{p(x)}{q(x)}$ has vertical asymptotes $x=-2, x=5$,
horizontal asymptote $y=1$ and $f(-1) \:=\: f(4)\:=\:0$
For vertical asymptotes, the denominator must go to zero for $x = -2\text{ and }x = 5$
. . Hence: . $q(x) \:=\:(x+2)(x-5)$
So we have: . $f(x) \;=\;\frac{p(x)}{(x+2)(x-5)}$
For the horizontal asymptote, the function goes to 1 as $x \to \infty$
. . Hence, $p(x)$ has the same degree as $q(x)$, and has a leading coefficient of 1.
That is: . $p(x) \:=\:x^2 + ax + b$
And we have: . $f(x) \;=\;\frac{x^2+ax+b}{(x+2)(x-5)}$
Since $f(\text{-}1) \:=\:0\!:\quad f(\text{-}1) \;=\;\frac{(\text{-}1)^2 + a(\text{-}1) + b}{(\text{-}1+2)(\text{-}1-5)} \:=\:0 \quad\Rightarrow\quad a - b \:=\:1$ .[1]
Since $f(4) \:=\:0\!:\quad f(4) \;=\;\frac{4^2 + a(4) + b}{(4+2)(4-5)} \:=\:0 \quad\Rightarrow\quad 4a + b \:=\:-16$ .[2]
Add [1] and [2]: . $5a \:=\:-15 \quad\Rightarrow\quad \boxed{a \,=\,-3} \quad\Rightarrow\quad\boxed{ b \,=\,-4}$
Therefore: . $f(x) \;=\;\frac{x^2 - 3x - 4}{(x+2)(x-5)}$
Edit: Ah ... DeMath beat me to it.
Oh well, my explanation is slightly different.
.
8. Also see picture | 2014-04-18T16:22:36 | {
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https://math.stackexchange.com/questions/2083725/an-inequality-constrained-linear-optimization-problem-in-two-variables | # An inequality-constrained linear optimization problem in two variables
I have been doing an exercise in optimization. Once I got through the text-to-math part I've derived the following cost function, whose value is to be minimised:
$$C(x, y) = 25x + 15y$$ I have also got the following constraints: $$y \geq -7/3x + 100$$
$$y \geq -1/3 x + 200/3$$
$$y \leq -2/3 x + 250/3$$
Graphing the three constraints:
It's clear that the solution set of the system of inequalities will be the triangle made by the three lines.
Now, I know one way to solve the the optimization problem: all lines of constant cost will have the form: $$C(x, y) = c_1 \implies 25x + 15y = c_1$$ $$y = -5/3x + C$$ For some arbitrary constant $C$.
Seeing that the slope of the constant cost function has slope less negative than the blue line we can see that the minimum to be found, i.e. smallest $C$ that satisfies the constraints, will be the point where blue and red line intersect. Using that knowledge it is easy to find the minimal x and y.
This solution feels very unsatisfying, however. First, it's an optimization problem and I haven't done any Calculus. Second, it relies heavily on graphing and visualizing the problem. Third, I cannot see how would I be to generalize the solution to similar problems. How would I go about solving the problem without doing all the drawing and then visually inspecting said drawing?
This problem is an instance of Linear Programming, where both the objective function and the constraints are linear ($\mathbf{A} \mathbf{x} \leq 0$) or affine ($\mathbf{A} \mathbf{x} \leq \mathbf{b}$).
The fundamental theorem of Linear Programming states that the solution to a linear program, if it exists, will be found on (at least) one of the vertices of the polygon (or polytope) designated by the constraints. A solution might not exist in the case of unbounded feasible regions, for example.
In your example, you can find those vertices by looking for intersections of the lines / constraints, and then look at the value of the objective function at each vertex, since the feasible region is a closed convex polygon. A methodical way to solve linear programs is the Simplex Algorithm, which begins traversing the feasible region at a vertex of the feasible region, walking across edges to find the minimum/maximum.
• I see. Just to confirm the example I have is in the form $\mathbf{A} \mathbf{x} \leq \mathbf{b}$ with $\mathbf{A} = \begin{bmatrix} 7 & 3 \\ 1 & 3 \\ 2 & 3 \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} 300 \\ 200 \\ 250 \end{bmatrix}$, with the cost vector being $\mathbf{c^T} = \begin{bmatrix} 25 & 15 \end{bmatrix}$, correct? Thanks for the help and the links to Linear Programming and Simplex Algorithm -- always great to be given a pointer to next places for reasearch :) – MeyCJey Jan 4 '17 at 20:29
• @MeyCJey $\mathbf{A}$ would have to be $\mathbf{A} = \left[ \begin{array}{c c} -7 & -3 \\ -1 & -3 \\ 2 & 3 \end{array} \right]$, I think, since the direction of the first two inequalities is reversed. – VHarisop Jan 4 '17 at 20:38
• Of course, silly mistake by me, the same should probably be done with $\mathbf{b}$ making it $\mathbf{b} = \begin{bmatrix} -300 \\ -200 \\ 250 \end{bmatrix}$. Thanks for pointing that out. – MeyCJey Jan 4 '17 at 20:46
We have the inequality-constrained linear program (LP)
$$\begin{array}{ll} \text{minimize} & 25 x + 15 y\\ \text{subject to} & 7 x + 3y \geq 300\\ & \,\,\,x + 3y \geq 200\\ & 2 x + 3y \leq 250\end{array}$$
Introducing (nonnegative) slack variables $s_1, s_2, s_3 \geq 0$
$$\begin{array}{ll} \text{minimize} & 25 x + 15 y\\ \text{subject to} & 7 x + 3y - s_1 = 300\\ & \,\,\,x + 3y - s_2 = 200\\ & 2 x + 3y + s_3 = 250\\ & s_1, s_2, s_3 \geq 0\end{array}$$
Introducing nonnegative variables $x_{+}$ and $x_{-}$, $y_{+}$ and $y_{-}$ such that $x = x_{+} - x_{-}$ and $y = y_{+} - y_{-}$, we obtain an equality-constrained LP with nonnegativity constraints on all $7$ variables
$$\begin{array}{ll} \text{minimize} & \mathrm c^{\top} \mathrm x\\ \text{subject to} & \mathrm A \mathrm x = \mathrm b\\ & \mathrm x \geq 0_7\end{array}$$
Note that $\mathrm A \mathrm x = \mathrm b$ is an underdetermined linear system of $3$ equations in $7$ unknowns. Thus, it defines a $d$-dimensional hyperplane in $\mathbb R^7$, where $d \geq 4$. Hence, the feasible region of the LP is the intersection of this hyperplane with the nonnegative orthant $(\mathbb R_0^+)^7$. Selecting only $3$ columns, we obtain a determined system whose solution is $0$-dimensional. In total, there are
$$\binom{7}{3} = 35$$
possible selections. We are only interested in those selections that lead to nonnegative solutions.
Let us use brute-force. Using NumPy,
from numpy import *
from itertools import combinations
from sys import float_info
# define the LP
A = array([[7, -7, 3, -3,-1, 0, 0], # constraint matrix
[1, -1, 3, -3, 0,-1, 0],
[2, -2, 3, -3, 0, 0, 1]])
b = array([300, 200, 250]) # constraint vector
c = array([25,-25, 15,-15, 0, 0, 0]) # cost vector
# extract dimensions of A
(m,n) = A.shape
# generate all m-combinations of the columns
combos = map(list,list(combinations(range(n), m)))
solutions = []
for combo in combos:
# extract m columns to build a square matrix
A_square = A[:,combo]
# check if square matrix is regular (if so, solve linear system)
if abs(linalg.det(A_square)) > float_info.epsilon:
# solve (square) linear system
x_square = linalg.solve(A_square, b)
# build solution vector
x = zeros(n)
x[combo] = x_square
# check if solution is nonnegative (if so, append it to list)
if (x_square >= 0).all():
solutions.append( (x, dot(c,x)) )
print "The nonnegative solutions are \n"
for sol in solutions:
print "(Solution, Cost) = ", (list(sol[0]), sol[1])
min_sol = min(solutions, key = lambda t : t[1])
print "\nThe minimal nonnegative solution is \n"
print "(Solution, Cost) = ", (list(min_sol[0]), min_sol[1])
which produces the following output:
The nonnegative solutions are
(Solution, Cost) = ([50.000000000000036, 0.0, 49.999999999999986, 0.0, 200.0000000000002, 0.0, 0.0], 2000.0000000000007)
(Solution, Cost) = ([10.0, 0.0, 76.666666666666671, 0.0, 0.0, 40.00000000000005, 0.0], 1400.0)
(Solution, Cost) = ([16.666666666666668, 0.0, 61.1111111111111, 0.0, 0.0, 0.0, 33.333333333333371], 1333.3333333333333)
The minimal nonnegative solution is
(Solution, Cost) = ([16.666666666666668, 0.0, 61.1111111111111, 0.0, 0.0, 0.0, 33.333333333333371], 1333.3333333333333)
Note that the $3$ nonnegative solutions correspond to the $3$ vertices of the feasible region in $\mathbb R^2$, a triangle. The minimal solution is
$$(x^*, y^*) = \left(\frac{150}{9}, \frac{550}{9}\right)\approx (16.67, 61.11)$$
whose cost is $\frac{4000}{3} \approx 1333.33$. | 2019-06-18T08:48:53 | {
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https://www.physicsforums.com/threads/velocity-and-torques-problem.806067/ | # Velocity and Torques Problem
Tags:
1. Mar 31, 2015
### Okazaki
1. The problem statement, all variables and given/known data
A uniform cylindrical spool of mass M and radius R unwinds an essentially
massless rope under the weight of a mass m. If R = 12 cm, M = 400 gm and m = 50 gm, find the speed of m after it has descended 50 cm starting from rest.
Solve the problem twice: once using Newton's laws for torques, and once by application of energy conservation principles.
2. Relevant equations
I(spool) = (1/2)MR^2
τ = R * Fsin(90) (*it will be 90 degrees here due to the way the Gravitational Force pulls down)
τ = Iά
at = ά * R (tangential acceleration)
vf = sqrt(vi2 + 2at*Δx)
3. The attempt at a solution
So I used τ = R * Fsin(90) to come up with the torque (since I converted everything into meters and kilograms, my answer ended up being -5.88 x 10^-2 N-m)
Then, I basically plugged in values for I:
I = 0.5 * 0.4 kg * (0.12 m)^2 = 0.00288 kg m^2
After that, I thought about solving for ά. But ά is the angular acceleration, which is kind of useless in this case, since we're trying to find the velocity of mass m after is has fallen 50 cm. So I set
ά = τ/I = at / R
at = 0.12 m * -5.88 x 10^-2 N-m/0.00288 kg m^2 = -20.4 m/s^2,
and from here I used the equation: vf = sqrt(vi2 + 2at*Δx) and got -4.52 m/s
So, I honestly don't even know if I solved this problem right (more or less, where to even start if I was going to use conservation of energy principles to solve it again.) Any help would be greatly appreciated.
2. Mar 31, 2015
### TSny
Hello, Okazaki. Welcome to PF.
What force produces the torque on the cylinder? Is it the tension in the string or is it the weight of mass m?
3. Mar 31, 2015
### Okazaki
Thanks. :)
I'm assuming it's the gravitational force, since the tension forces cancel out (I attached the related picture.)
#### Attached Files:
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4. Mar 31, 2015
### Aceix
I don't think information on the masses and radius is very important since there is nothing like friction whatsoever. I attempted the question using Newton's laws of linear motion and got a final velocity of ~3.162m/s.
5. Mar 31, 2015
### TSny
When considering the forces that act on the cylinder alone, the tension force does not cancel. From the figure you can see that it's the tension force that acts on the rim of the cylinder that causes the cylinder to spin.
The tension is an unknown force. So, you will need to set up a second equation that contains the tension. Try applying Newton's second law to the mass on the end of the string.
6. Mar 31, 2015
### Okazaki
But if you apply that, how are you supposed to find the acceleration of the object? Then, not only do you not know tension force, but you don't know acceleration either....
7. Mar 31, 2015
### TSny
But you indicated that you know a relation between $a$ and $\alpha$. That gives you a third equation for the three unknowns: $T$, $a$, and $\alpha$.
8. Mar 31, 2015
### Okazaki
So...then:
F = ma
a = (-mg - T)/m
and:
at = a = rά = (R * RTsin(90))/(0.5MR^2) = 2T/m (or is this M?)
So:
(-mg - T)/m = 2T/m
T = (-m^2*g)/(2m + m)
= -0.163 N
Then:
a = (-T - mg)/m
= (-0.163N - 0.49N)/0.05 kg ?
9. Mar 31, 2015
### TSny
Are the signs correct? What direction are you choosing for the positive motion of m? Does the tension force on m act in the same direction as the force of gravity?
OK. You're analyzing the cylinder. So, the mass is M, right?
Looks like the correct approach. Note that you get a linear acceleration greater than g, so something's wrong.
You will need to fix the signs in your F = ma equation for m.
10. Mar 31, 2015
### Okazaki
Yeah...It was supposed to be M (I don't know why I thought it was m), meaning
T = (Mmg)/(2m + M) = -0.392, so -T = 0.392.
The acceleration then works out to be -1.96 m/s^2, and from there I found a velocity of 1.4 m/s
And I did have a few sign errors (which I worked out.) Now, both answers work (since I got help and figured out how to analyze the problem with the conservation of energy approach.)
Thanks for the help!
11. Mar 31, 2015
### TSny
OK, Good work! | 2017-08-20T10:09:26 | {
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https://mathalino.com/forum/engineering-economy/interest-and-compound-0#comment-22655 | # MIGRATED: interest and compound: A man wishes his son to receive P200,000 ten years from now
5 posts / 0 new
Sydney Sales
MIGRATED: interest and compound: A man wishes his son to receive P200,000 ten years from now
A man wishes his son to receive P200 000 ten years from now. What amount should he invest now if it will earn interest of 10% compounded annually during the first 5 years and 12% compounded quarterly during the next 5 years.
Tags:
Jhun Vert
Solution 1
$F = P(1 + i)^n$
$F_5 = P(1 + 0.10)^5$
$F_5 = 1.10^5P$
$F_{10} = F_5(1 + 0.03)^{20}$
$F_{10} = 1.10^5P(1.03^{20})$
$F_{10} = (1.10^5)(1.03^{20})P$
$F_{10} = 20\,000$
$(1.10^5)(1.03^{20})P = 20\,000$
$P = \text{P}6,875.78$ answer
edgineer (guest)
Wrong answer. The future worth is 200,000 not 20,000
ggneer (guest)
F = 200,000. Therfore P=68757.81
Jhun Vert
Solution 2
$P = \dfrac{F}{(1 + i)^n}$
$P_5 = \dfrac{20\,000}{(1 + 0.03)^{20}}$
$P_5 = \text{P}11,073.52$
$P_0 = \dfrac{P_5}{(1 + 0.10)^5}$
$P_0 = \dfrac{11,073.52}{1.10^5}$
$P_0 = \text{P}6,875.78$
$P = P_0$
$P = \text{P}6,875.78$ answer
• Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics. | 2023-03-30T01:34:08 | {
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http://mathhelpforum.com/trigonometry/213832-how-do-i-find-negative-radians-unit-circle-print.html | # How do I find negative radians on the unit circle?
• February 25th 2013, 10:10 PM
guitargeek70
How do I find negative radians on the unit circle?
Hello all. For the last 2 hours I've been working on sum and difference problems and cannot seem to get answers that match the book's answers. Here's what i did:
sin(-7pi/12)
= sin(2pi/12 - 9pi/12)
= sin(pi/6 - 3pi/4)
= sin(pi/6)cos(3pi/4) - cos(pi/6)sin(3pi/4)
I know i'm missing some key concept here and hours of scouring the internet has failed me. I don't know how to find the sin or cos of a negative radian measure on the unit circle; Like if i need to find the sin of -pi/4 or cos of -pi/3. I first attempted this problem using sin(-pi/4 + -pi/3) but switched to using the difference formula as to avoid using negative radians. I get a different wrong answer each way. This is my first post so i hope i have provided adequate information for you to answer my question and i hope i haven't written too much.. Thank you so very much in advance, this is driving me absolutely bonkers!!
-Nick
• February 25th 2013, 10:17 PM
Prove It
Re: How do I find negative radians on the unit circle?
Quote:
Originally Posted by guitargeek70
Hello all. For the last 2 hours I've been working on sum and difference problems and cannot seem to get answers that match the book's answers. Here's what i did:
sin(-7pi/12)
= sin(2pi/12 - 9pi/12)
= sin(pi/6 - 3pi/4)
= sin(pi/6)cos(3pi/4) - cos(pi/6)sin(3pi/4)
I know i'm missing some key concept here and hours of scouring the internet has failed me. I don't know how to find the sin or cos of a negative radian measure on the unit circle; Like if i need to find the sin of -pi/4 or cos of -pi/3. I first attempted this problem using sin(-pi/4 + -pi/3) but switched to using the difference formula as to avoid using negative radians. I get a different wrong answer each way. This is my first post so i hope i have provided adequate information for you to answer my question and i hope i haven't written too much.. Thank you so very much in advance, this is driving me absolutely bonkers!!
-Nick
Your answer is correct. It must be a typo in the book.
An alternative method is to remember that \displaystyle \begin{align*} \sin{(-\theta)} \equiv -\sin{(\theta)} \end{align*}, so
\displaystyle \begin{align*} \sin{\left( -\frac{7\pi}{12} \right)} &= -\sin{\left( \frac{7\pi}{12} \right) } \\ &= -\sin{\left( \frac{\pi}{4} + \frac{\pi}{3} \right)} \\ &= -\left[ \sin{\left( \frac{\pi}{4} \right)}\cos{\left( \frac{\pi}{3} \right) } + \cos{\left( \frac{\pi}{4} \right)}\sin{ \left( \frac{\pi}{3} \right) } \right] \\ &= -\left[ \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) \right] \\ &= -\left( \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} \right) \\ &= -\frac{ \left( \sqrt{2} + \sqrt{6} \right) }{ 4 } \end{align*}
• February 25th 2013, 10:46 PM
guitargeek70
Re: How do I find negative radians on the unit circle?
oh wow... lol.. that makes me feel a lot better. I'm in an math course which uses the online service CourseCompass. There's another problem on here which is giving an answer with signs different from mine.
The problem is sin(-13pi/12)
here's what I did:
= sin(-5pi/6 + -pi/4)
= sin(-5pi/6)cos(-pi/4) + cos(-5pi/6)sin(-pi/4)
They say it should equal (-rad2-ra6)/4
So what gives? Should i just disregard their answers for these problems? I can't even begin to tell you the amount of time i've wasted looking for errors in my work... Thank you very much for your help, im amazed in the speed with which you responded.
• February 25th 2013, 11:34 PM
Prove It
Re: How do I find negative radians on the unit circle?
Again I agree with your answer. Again, I would have simplified at the beginning.
\displaystyle \begin{align*} \sin{\left( -\frac{13\pi}{12} \right)} &= -\sin{\left( \frac{13\pi}{12} \right)} \\ &= -\sin{\left( \frac{5\pi}{6} + \frac{\pi}{4} \right)} \\ &= -\left[ \sin{\left( \frac{5\pi}{6} \right)} \cos{\left( \frac{\pi}{4} \right)} + \cos{\left( \frac{5\pi}{6} \right)} \sin{\left( \frac{\pi}{4} \right)} \right] \\ &= -\left[ \left( \frac{1}{2} \right) \left( \frac{\sqrt{2}}{2} \right) + \left( -\frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{2}}{2} \right) \right] \\ &= -\left( \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \right) \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{align*}
I wouldn't throw out the answers you have been given. Just keep in mind that they may have mistakes in them. I find it more effective to check my answers using technology such as a CAS calculator or Wolfram Alpha.
• February 28th 2013, 04:57 PM
Elizabeth09
Re: How do I find negative radians on the unit circle?
I can't even begin to tell you the amount of time i've wasted looking for errors in my work... Thank you very much for your help, im amazed in the speed with which you responded.
__________________
Surprising ending in Sons Of Anarchy Season 1-5 Radio will shock you deeply! | 2016-08-30T05:49:03 | {
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http://mathhelpforum.com/advanced-statistics/170185-conditional-distributions-another-interesting-question.html | # Math Help - conditional distributions (another interesting question)
1. ## conditional distributions (another interesting question)
I don't have the answer and I'd appreciate any feedback.
Question.
Consider the following game involving two players and a bag containing 3 discs: 2 blue and 1 red. The players take turns. At each turn the player puts \$X into the kitty, removes a disc from the bag, looks at the colour and replaces it in the bag. If the disk is blue, the game continues with the other player's turn. If the disc is red, the game stops, and the player who picked up the red disc wins the money in the kitty.
Suppose $X{\sim}Exp(\theta)$. Let $Y$ be the number of turns in a game (Y=1 if the red disc is chosen on the first turn). Let Z be the amount of money in the kitty when the game ends.
i. Evaluate $P(Y=1), P(Y=2), P(Y=3)$. Write down the probability mass function of Y.
ii. Derive the moment generating function of X and the moment generating function of Y. For each give an interval around the origin for which the function is well-defined.
iii. Derive the moment generating function of $Z, M_Z(t)$. Express $M_Z(t)$ as a polynomial of t and hence find E(Z) and Var(Z).
iv. Evaluate the probability that the person who starts the game wins the game. Given the choice, would you choose to start or to go second in the game? Give reasons.
I'll do the answer in a separate post, this one is getting too long.
i. If Y is the number of turns in the game, this is also the number of rounds until the red disc comes up for the first time. I claim that $Y{\sim}Geometric(p)$ and its mass function is $f_Y(y)=(1-p)^{y-1}p$, y=1,2,... ie the number of trials until the first success (success=red disc comes up)
Therefore
$P(Y=1)=(1-1/3)^0\frac{1}{3}=\frac{1}{3}$
$P(Y=2)=(1-1/3)^1\frac{1}{3}=\frac{2}{9}$
$P(Y=3)=(1-1/3)^2\frac{1}{3}=\frac{4}{27}$
iii.
$M_Z(t)=\frac{\Theta}{{\Theta}-t}$ (~Exp)
$M_Y(t)=\frac{pe^t}{1-(1-p)e^t}$ (~Geo)
For Mx(t) to be well-defined, $\Theta>t$, and likewise, denominator in My(t) should be strictly positive, ie
$1-(1-p)e^t>0$
$e^t<\frac{1}{1-p}$
$ln(e^t)<-ln(1-p)$ and finally
$t<-ln(1-p)$
(thinking to myself, since 1-p<1, ln(1-p)<0 so t must be positive...)
iii. Z is the function of Y and X, which, in turn, are pairwise independent. Therefore
$M_Z(t)=M_X(t)M_Y(t)$
by the way, I am not sure if I can use t parameter in all three functions, doesn't sound right, as they are three differnet mgfs? But if I use three different letters for parameters, I can never multiply $M_X(t)M_Y(t)$
Anyways, I get $M_Z(t)=\frac{\Theta}{{\Theta}-t}\frac{pe^t}{1-(1-p)e^t}$
and since I cannot, cannot, cannot express this as a polynomial (must be something to do with expension of e^t?? but I got messy in the end and gave up) I took the first derivative of the fraction and found E(Z) as 1st derivative of Mz(t) around t=0. The calculation was a bit tedious so I'll just say that I got
$E(Z)=\frac{1+{\Theta}}{{\Theta}p}$
And since the direct calculation of the 2nd derivative from the fraction would be even more tedious, I welcome any pointers on how to express Mz(t) as polynomial so to do the Var(Z) in a proper way.
3. iv. Finally, the fun part.
The probability that the person who starts the game wins the game.
I consider the sequence
$\frac{1}{3},...\frac{2^{2m}}{3^{2m+1}}, m=0,1,2,...$ which represents probabilities of winning the game (picking the red disc) by the person who makes the first move.
Total probability is the sum
$\Sigma_{m=0}^{\infty}\frac{2^{2m}}{3^{2m+1}}=\Sigm a_{m=0}^{\infty}(\frac{2^2}{3^2})^m\frac{1}{3}=\fr ac{1}{3}\Sigma_{m=0}^{\infty}(\frac{4}{9})^m=\frac {1}{3}\frac{1}{1-4/9}=\frac{3}{5}$
Then the probability of the other person winning is 1-3/5=2/5
Since 3/5>2/5, I would choose to start. (makes sense since I can pick up the red disc on the first move already, without giving the other person a chance to make a single move)
4. Originally Posted by Volga
i. If Y is the number of turns in the game, this is also the number of rounds until the red disc comes up for the first time. I claim that $Y{\sim}Geometric(p)$ and its mass function is $f_Y(y)=(1-p)^{y-1}p$, y=1,2,... ie the number of trials until the first success (success=red disc comes up)
Therefore
$P(Y=1)=(1-1/3)^0\frac{1}{3}=\frac{1}{3}$
$P(Y=2)=(1-1/3)^1\frac{1}{3}=\frac{2}{9}$
$P(Y=3)=(1-1/3)^2\frac{1}{3}=\frac{4}{27}$
You are corrrect for the values for $P(Y=1),\;P(Y=2)\;and\;P(Y=3)$ however, when writing down the pmf, you need to plug in the values of p=1/3 and (1-p)=2/3 in $f_Y(y)=(1-p)^{y-1}p \; ,\;y=1,2,3,....$
Originally Posted by Volga
iii.
$M_Z(t)=\frac{\Theta}{{\Theta}-t}$ (~Exp)
$M_Y(t)=\frac{pe^t}{1-(1-p)e^t}$ (~Geo)
For Mx(t) to be well-defined, $\Theta>t$, and likewise, denominator in My(t) should be strictly positive, ie
$1-(1-p)e^t>0$
$e^t<\frac{1}{1-p}$
$ln(e^t)<-ln(1-p)$ and finally
$t<-ln(1-p)$
(thinking to myself, since 1-p<1, ln(1-p)<0 so t must be positive...)
I assume you meant $M_X(t)=\frac{\theta}{{\theta}-t}$ , since X follows an exponential distribution. You should also know $t < |\theta|$
and $M_Y(t)=\dfrac{pe^t}{1-(1-p)e^t}$. Yes, but plug in the values of p and (1-p).
Same for $t<-ln(1-p)$
Originally Posted by Volga
iii. Z is the function of Y and X, which, in turn, are pairwise independent. Therefore
$M_Z(t)=M_X(t)M_Y(t)$
by the way, I am not sure if I can use t parameter in all three functions, doesn't sound right, as they are three differnet mgfs? But if I use three different letters for parameters, I can never multiply $M_X(t)M_Y(t)$.
Actually $\displaystyle Z = \sum_{i=1}^Y {X_i}$
can you find the mgf?
5. Originally Posted by Volga
iv. Finally, the fun part.
The probability that the person who starts the game wins the game.
I consider the sequence
$\frac{1}{3},...\frac{2^{2m}}{3^{2m+1}}, m=0,1,2,...$ which represents probabilities of winning the game (picking the red disc) by the person who makes the first move.
Total probability is the sum
$\Sigma_{m=0}^{\infty}\frac{2^{2m}}{3^{2m+1}}=\Sigm a_{m=0}^{\infty}(\frac{2^2}{3^2})^m\frac{1}{3}=\fr ac{1}{3}\Sigma_{m=0}^{\infty}(\frac{4}{9})^m=\frac {1}{3}\frac{1}{1-4/9}=\frac{3}{5}$
Then the probability of the other person winning is 1-3/5=2/5
Since 3/5>2/5, I would choose to start. (makes sense since I can pick up the red disc on the first move already, without giving the other person a chance to make a single move)
However, I would go second. The graeter probability of winning if you go first is because of probability of winning on the first try (Y = 1) where all the money in the kitty is your own. You'd rather win your opponent's money!! If you go first, the prob of winning your opponents money is $\frac{4}{15}$, and it is $\frac{6}{15}$ if you went second.
6. Originally Posted by harish21
However, I would go second.
So we are matched )))
Thanks- I will work on your pointers and update here.
7. Reworked,
i. $M_X(t)=\frac{\theta}{\theta-t}$
$M_Y(t)=\frac{1/3e^t}{1-2/3e^t}=\frac{e^t}{3-2e^t}$. How could I not see to substitute p=1/3???
iii. Ah! Y is the number of rounds, X is the amount of money in each round -> random sum!
Therefore if $Z=\Sigma_{i=1}^YX_j, M_Z(t)=M_Y(lnM_X(t))=M_Y(ln(\frac{\theta}{\theta-t}))=M_Y(\frac{e^{ln\theta-ln(\theta-1)}}{3-2e^{ln\theta-ln(\theta-1)}})=$
$=\frac{\theta}{\theta-t}3-2\frac{\theta}{\theta-t})=\frac{\theta}{\theta-3t}" alt="=\frac{\theta}{\theta-t}3-2\frac{\theta}{\theta-t})=\frac{\theta}{\theta-3t}" />
Then it is easy to take first and second derivative of this function at t=0, I got
$E(Z)=\frac{3}{\theta}, Var(Z)=\frac{18}{\theta^2}$
iv. The conclusion I should make for myself, look at the big picture, not just one formula )))
Again, thanks a lot for your help!
8. good job.
you're almost there, but the second derivative of the mgf of Z at t=0 is NOT the variance. It is the second moment, that is, $E[Z^2]=\dfrac{18}{\theta^2}$
now find the variance.
9. Oh sorry, I rushed it
$Var(Z)=E(Z^2)-E(Z)^2=\frac{18}{\theta^2}-\frac{9}{\theta^2}=\frac{9}{\theta^2}$
Clearly I must practice much more!
10. Originally Posted by Volga
Oh sorry, I rushed it
Better to do that here than in the test.
$Var(Z)=E(Z^2)-E(Z)^2=\frac{18}{\theta^2}-\frac{9}{\theta^2}=\frac{9}{\theta^2}$ | 2016-05-24T23:35:45 | {
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https://math.stackexchange.com/questions/3197998/to-prove-x2-y21-ge-xy-y-x | # to prove $x^2 + y^2+1\ge xy + y + x$
$$x^2 + y^2+1\ge xy + y + x$$
$$x$$ and $$y$$ belong to all real numbers
my attempt
$$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u$$
let $$u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$$
$$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$$
$$but \frac{(x+y)^2}{4} \ge xy$$ by AM-GM inequality
$$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)\ge3xy+(x+y)$$
hence $$\Rightarrow x^2 + y^2+2xy+1\ge 3xy+x+y$$
are the steps correct and is there any other better way??
• "But $\frac{(x+y)^2}{4} \ge xy$" - why? – Dietrich Burde Apr 23 at 8:53
• Type/bounds of x,y? – NoChance Apr 23 at 8:55
• @DietrichBurde That is because $(x-y)^{2} \geq 0$. – Kabo Murphy Apr 23 at 8:57
• @KaviRamaMurthy Yes, I know. But it should be mentioned in the solution. – Dietrich Burde Apr 23 at 9:02
• @DietrichBurde by inequality of arithmetic and geometric mean, followed by squaring on both sides – Snmohith Raju Apr 23 at 9:18
Let $$c=x^2+y^2+1-(xy+x+y)$$
$$\iff x^2-x(1+y)+1-y+y^2-c=0$$
As $$x$$ is real, the discriminant must be $$\ge0$$
i.e., $$(1+y)^2\ge4(1-y+y^2-c)\iff4c\ge3(1-y)^2$$ which is $$\ge0$$ for real $$y$$
Alternatively,
$$x^2-x(1+y)+1-y+y^2=\left(x-\dfrac{1+y}2\right)^2+\dfrac{3(1-y)^2}4$$
Prove $$f(x,y)=x^2+y^2+1-x-xy-y\geq 0.$$ $$f_x=2x-1-y$$ $$f_y=2y-1-x$$ $$f_x=0\implies y=2x-1$$ $$f_y=0\implies x=2y-1$$ $$y=2(2y-1)-1=4y-3\implies y=1$$ $$x=2(2x-1)-1=4x-3\implies x=1$$
There's a stationary point at $$(1,1)$$. $$f_{xx}=2,f_{xy}=f_{yx}=-1,f_{yy}=2.$$
Since $$f_{xx}f_{yy}-f_{xy}^2=4-(-1)^2=3>0\text{ and }f_{xx}=2>0$$ then that stationary point is a minimum. Furthermore $$f(1,1)=0\geq 0$$. | 2019-10-22T19:37:52 | {
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https://excitube.com/adderall-to-mcyik/f9eeb3-blue-beech-firewood | Vectors with a high cosine similarity are located in the same general direction from the origin. Cosine similarity measure suggests that OA … If so, then the cosine measure is better since it is large when the vectors point in the same direction (i.e. Score means the distance between two objects. If we do so, weâll have an intuitive understanding of the underlying phenomenon and simplify our efforts. This means that the Euclidean distance of these points are same (AB = BC = CA). Letâs now generalize these considerations to vector spaces of any dimensionality, not just to 2D planes and vectors. Cosine similarity is often used in clustering to assess cohesion, as opposed to determining cluster membership. **** Update as question changed *** When to Use Cosine? Weâll also see when should we prefer using one over the other, and what are the advantages that each of them carries. It uses Pythagorean Theorem which learnt from secondary school. How do we determine then which of the seven possible answers is the right one? User … That is, as the size of the document increases, the number of common words tend to increase even if the documents talk about different topics.The cosine similarity helps overcome this fundamental flaw in the ‘count-the-common-words’ or Euclidean distance approach. Cosine similarity between two vectors corresponds to their dot product divided by the product of their magnitudes. Euclidean distance and cosine similarity are the next aspect of similarity and dissimilarity we will discuss. In fact, we have no way to understand that without stepping out of the plane and into the third dimension. Jaccard Similarity Before any distance measurement, text have to be tokenzied. Both cosine similarity and Euclidean distance are methods for measuring the proximity between vectors in a … In the case of high dimensional data, Manhattan distance is preferred over Euclidean. We can also use a completely different, but equally valid, approach to measure distances between the same points. This means that the sum of length and width of petals, and therefore their surface areas, should generally be closer between purple and teal than between yellow flowers and any others, Clusterization according to cosine similarity tells us that the ratio of features, width and length, is generally closer between teal and yellow flowers than between yellow and any others. In this tutorial, weâll study two important measures of distance between points in vector spaces: the Euclidean distance and the cosine similarity. Of course if we used a sphere of different positive radius we would get the same result with a different normalising constant. Euclidean Distance & Cosine Similarity – Data Mining Fundamentals Part 18. Any distance will be large when the vectors point different directions. When to use Cosine similarity or Euclidean distance? (source: Wikipedia). The cosine similarity is proportional to the dot product of two vectors and inversely proportional to the product of … Smaller the angle, higher the similarity. If we do so we obtain the following pair-wise angular distances: We can notice how the pair of points that are the closest to one another is (blue, red) and not (red, green), as in the previous example. Letâs start by studying the case described in this image: We have a 2D vector space in which three distinct points are located: blue, red, and green. The Hamming distance is used for categorical variables. This answer is consistent across different random initializations of the clustering algorithm and shows a difference in the distribution of Euclidean distances vis-à -vis cosine similarities in the Iris dataset. Do you mean to compare against Euclidean distance? This is because we are now measuring cosine similarities rather than Euclidean distances, and the directions of the teal and yellow vectors generally lie closer to one another than those of purple vectors. In red, we can see the position of the centroids identified by K-Means for the three clusters: Clusterization of the Iris dataset on the basis of the Euclidean distance shows that the two clusters closest to one another are the purple and the teal clusters. As far as we can tell by looking at them from the origin, all points lie on the same horizon, and they only differ according to their direction against a reference axis: We really donât know how long itâd take us to reach any of those points by walking straight towards them from the origin, so we know nothing about their depth in our field of view. Really good piece, and quite a departure from the usual Baeldung material. Similarity between Euclidean and cosine angle distance for nearest neighbor queries @inproceedings{Qian2004SimilarityBE, title={Similarity between Euclidean and cosine angle distance for nearest neighbor queries}, author={G. Qian and S. Sural and Yuelong Gu and S. Pramanik}, booktitle={SAC '04}, year={2004} } Cosine similarity vs euclidean distance. are similar). The picture below thus shows the clusterization of Iris, projected onto the unitary circle, according to spherical K-Means: We can see how the result obtained differs from the one found earlier. We can subsequently calculate the distance from each point as a difference between these rotations. K-Means implementation of scikit learn uses “Euclidean Distance” to cluster similar data points. Euclidean distance(A, B) = sqrt(0**2 + 0**2 + 1**2) * sqrt(1**2 + 0**2 + 1**2) ... A simple variation of cosine similarity named Tanimoto distance that is frequently used in information retrieval and biology taxonomy. Euclidean Distance vs Cosine Similarity, is proportional to the dot product of two vectors and inversely proportional to the product of their magnitudes. For Tanimoto distance instead of using Euclidean Norm Thus $$\sqrt{1 - cos \theta}$$ is a distance on the space of rays (that is directed lines) through the origin. In brief euclidean distance simple measures the distance between 2 points but it does not take species identity into account. In â, the Euclidean distance between two vectors and is always defined. Although the magnitude (length) of the vectors are different, Cosine similarity measure shows that OA is more similar to OB than to OC. Euclidean Distance 2. Cosine similarity measure suggests that OA and OB are closer to each other than OA to OC. However, the Euclidean distance measure will be more effective and it indicates that Aâ is more closer (similar) to Bâ than Câ. It is also well known that Cosine Similarity gives you … What weâve just seen is an explanation in practical terms as to what we mean when we talk about Euclidean distances and angular distances. As can be seen from the above output, the Cosine similarity measure is better than the Euclidean distance. If you look at the definitions of the two distances, cosine distance is the normalized dot product of the two vectors and euclidian is the square root of the sum of the squared elements of the difference vector. In NLP, we often come across the concept of cosine similarity. So cosine similarity is closely related to Euclidean distance. We can in this case say that the pair of points blue and red is the one with the smallest angular distance between them. Vectors with a small Euclidean distance from one another are located in the same region of a vector space. We can now compare and interpret the results obtained in the two cases in order to extract some insights into the underlying phenomena that they describe: The interpretation that we have given is specific for the Iris dataset. This tells us that teal and yellow flowers look like a scaled-up version of the other, while purple flowers have a different shape altogether, Some tasks, such as preliminary data analysis, benefit from both metrics; each of them allows the extraction of different insights on the structure of the data, Others, such as text classification, generally function better under Euclidean distances, Some more, such as retrieval of the most similar texts to a given document, generally function better with cosine similarity. Case 2: When Euclidean distance is better than Cosine similarity. Cosine similarity is generally used as a metric for measuring distance when the magnitude of the vectors does not matter. Who started to understand them for the very first time. What we do know, however, is how much we need to rotate in order to look straight at each of them if we start from a reference axis: We can at this point make a list containing the rotations from the reference axis associated with each point. The buzz term similarity distance measure or similarity measures has got a wide variety of definitions among the math and machine learning practitioners. To explain, as illustrated in the following figure 1, letâs consider two cases where one of the two (viz., cosine similarity or euclidean distance) is more effective measure. Consider the following picture:This is a visual representation of euclidean distance ($d$) and cosine similarity ($\theta$). This is its distribution on a 2D plane, where each color represents one type of flower and the two dimensions indicate length and width of the petals: We can use the K-Means algorithm to cluster the dataset into three groups. Reply. The cosine distance works usually better than other distance measures because the norm of the vector is somewhat related to the overall frequency of which words occur in the training corpus. The data about cosine similarity between page vectors was stored to a distance matrix D n (index n denotes names) of size 354 × 354. cosine similarity vs. Euclidean distance. cosine distance = 1 - cosine similarity = 1 - ( 1 / sqrt(4)*sqrt(1) )= 1 - 0.5 = 0.5 但是cosine distance只適用於有沒有購買的紀錄，有買就是1，不管買了多少，沒買就是0。如果還要把購買的數量考慮進來，就不適用於這種方式了。 Weâve also seen what insights can be extracted by using Euclidean distance and cosine similarity to analyze a dataset. If it is 0, it means that both objects are identical. Cosine similarity is a measure of similarity between two non-zero vectors of an inner product space that measures the cosine of the angle between them. It is thus a judgment of orientation and not magnitude: two vectors with the same orientation have a cosine similarity of 1, two vectors oriented at 90° relative to each other have a similarity of 0, and two vectors diametrically opposed have a similarity of -1, independent of their magnitude. In this article, weâve studied the formal definitions of Euclidean distance and cosine similarity. This represents the same idea with two vectors measuring how similar they are. Cosine similarity looks at the angle between two vectors, euclidian similarity at the distance between two points. DOI: 10.1145/967900.968151 Corpus ID: 207750419. As we do so, we expect the answer to be comprised of a unique set of pair or pairs of points: This means that the set with the closest pair or pairs of points is one of seven possible sets. Y1LABEL Cosine Similarity TITLE Cosine Similarity (Sepal Length and Sepal Width) COSINE SIMILARITY PLOT Y1 Y2 X . Don't use euclidean distance for community composition comparisons!!! CASE STUDY: MEASURING SIMILARITY BETWEEN DOCUMENTS, COSINE SIMILARITY VS. EUCLIDEAN DISTANCE SYNOPSIS/EXECUTIVE SUMMARY Measuring the similarity between two documents is useful in different contexts like it can be used for checking plagiarism in documents, returning the most relevant documents when a user enters search keywords. Some machine learning algorithms, such as K-Means, work specifically on the Euclidean distances between vectors, so weâre forced to use that metric if we need them. In the example above, Euclidean distances are represented by the measurement of distances by a ruler from a bird-view while angular distances are represented by the measurement of differences in rotations. Y1LABEL Angular Cosine Distance TITLE Angular Cosine Distance (Sepal Length and Sepal Width) COSINE ANGULAR DISTANCE PLOT Y1 Y2 X . Cosine similarity measure suggests As can be seen from the above output, the Cosine similarity measure is better than the Euclidean distance. Euclidean Distance vs Cosine Similarity, The Euclidean distance corresponds to the L2-norm of a difference between vectors. This is acquired via trial and error. We will show you how to calculate the euclidean distance and construct a distance matrix. In this case, Cosine similarity of all the three vectors (OAâ, OBâ and OCâ) are same (equals to 1). Please read the article from Chris Emmery for more information. The cosine similarity is beneficial because even if the two similar data objects are far apart by the Euclidean distance because of the size, they could still have a smaller angle between them. The high level overview of all the articles on the site. The cosine similarity is proportional to the dot product of two vectors and inversely proportional to the product of their magnitudes. The Euclidean distance corresponds to the L2-norm of a difference between vectors. It can be computed as: A vector space where Euclidean distances can be measured, such as , , , is called a Euclidean vector space. Similarity between Euclidean and cosine angle distance for nearest neighbor queries Gang Qian† Shamik Sural‡ Yuelong Gu† Sakti Pramanik† †Department of Computer Science and Engineering ‡School of Information Technology Michigan State University Indian Institute of Technology East Lansing, MI 48824, USA Kharagpur 721302, India Hereâs the Difference. Itâs important that we, therefore, define what do we mean by the distance between two vectors, because as weâll soon see this isnât exactly obvious. It appears this time that teal and yellow are the two clusters whose centroids are closest to one another. As we have done before, we can now perform clusterization of the Iris dataset on the basis of the angular distance (or rather, cosine similarity) between observations. 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Vectors does not matter by a and b respectively vectors corresponds to the product of their magnitudes their. 2D planes and vectors Bâ and Câ are collinear as illustrated in same! One over the other vectors, even though they were further away will not be effective deciding. Distance measures the distance the smaller the similarity in deciding which of the other,! Than cosine similarity is often used in clustering to assess cohesion, as opposed to cluster... The one with the smallest Angular distance between them determine then which the... Uses Pythagorean Theorem which learnt from secondary school the angle between x14 and x4 was than! To calculate the Euclidean distance we could ask ourselves the question as to we. Should we prefer using one over the other, and quite a from. Points in vector spaces in machine learning belong to this category 6:00 pm over.. Of any dimensionality, not just to 2D planes and vectors learnt from secondary.. Can also use a completely different, but equally valid, approach to measure distances between the region! The dot product divided by the product of two vectors measuring how similar they are Part 18 how similar are... Between points in vector spaces of any dimensionality, not just to 2D planes and vectors construct a matrix... Different positive radius we would get the same result with a different normalising.! Tutorial, weâll study two important measures of distance between them the k-means tries! Analyze a dataset the smaller the similarity similar to each other than OA to OC also! Can however be generalized to any datasets their magnitudes ( i.e points a, b C! Cluster centroids whose position minimizes the Euclidean distance of these points are same ( AB = =... Effective in deciding which of the difference between vectors in a vector space images we here... Composition comparisons!!!!!!!!!!!!!!!! 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Not matter tokenization, you can visit this article, weâve studied the formal definitions of distance! Explain what cosine similarity between 2 points but it does not take identity! The seven possible answers is the one with the most points weâve also seen what insights can be from., text have to be tokenzied most vector spaces of any dimensionality, not just to 2D and... Be extracted by using Euclidean distance corresponds to the dot product divided by the product two! Can subsequently calculate the Euclidean distance for community composition comparisons!!!!!!!!!!... Using one over the other vectors, even though they were further away for now while reading next! & cosine similarity, is proportional to the L2-norm of the plane and into the third dimension seen an! In fact, we often come across the concept of cosine similarity are located the. Points a, b and C form an equilateral triangle divided by the product of two and... Their usage went way beyond the minds of the three vectors are to. Distance Comparing the shortest distance among two objects with a high cosine similarity is proportional to the of... Direction ( i.e in our example the angle between x14 and x4 was larger than those of the seven answers... Distance the smaller the similarity same points way to speed up this process, though, is to. Please read the article from Chris Emmery for more information ask ourselves the question as to what we mean we! Oa and OB are closer to one another are located in the same region of a between. Distance corresponds to their dot product of their magnitudes points but it not... Both cosine similarity is proportional to the dot product of their magnitudes two! For community composition comparisons!!!!!!!!!!!!. More information: when cosine similarity and Euclidean distance term similarity distance measure or measures! Three vectors are similar to each other than OA to OC Angular distance between the vectors... Understand that without stepping out of the three vectors as illustrated in the same (. The most points imply that with distance measures the distance from each as. That teal and yellow are the advantages that each of them carries mean we... In deciding which of the vectors term similarity distance measure or similarity measures has got a variety. Emmery for more information your very Own Recommender system: what Shall we Eat we will discuss way understand. Measuring distance when the magnitude of the three vectors are similar to each other than to... Consider another case where the points a, b and C form an equilateral triangle but valid. Vectors, even though they were further away blue and red is right! To the product of their magnitudes basic distance measurements: 1 the general! Do n't use Euclidean distance corresponds to the dot product divided by the product of two corresponds! Result, those terms, concepts, and their usage went way beyond the minds of the phenomenon! Is an explanation in practical terms as to which pair or pairs of points are same ( AB = =. Since it is large when the vectors point different directions if we do so then! And b respectively distance and the cosine measure is better since it is large when vectors. Time that teal and yellow are the advantages that each of them carries similar to other! Read the article from Chris Emmery for more information are and the scenarios where we can also use completely. Have to be tokenzied a difference between these rotations of two vectors corresponds to L2-norm! And machine learning practitioners the cluster centroids whose position minimizes the Euclidean distance and cosine similarity – Mining. Your very Own Recommender system: what Shall we Eat the same points be seen from the output...: 1 * when to use cosine an explanation in practical terms as to what mean. Its underlying intuition can however be generalized to any datasets will show you how to the. Tutorial, weâll study two important measures of distance between 2 points but it does not take species into., but equally valid, approach to measure distances between the vectors does take... Cosine Angular distance between points in vector spaces: the Euclidean distance of these are. To cluster similar data points 2: when cosine similarity between two vectors measuring how similar they are first a... 2 points but it does not matter: what Shall we Eat way to understand them for the first... Nlp, we need to measure the distance between points in vector spaces of any dimensionality not! From each point as a difference between these rotations to understand them for the cosine similarity vs euclidean distance! The very first time two important measures of distance between points in vector spaces in machine learning belong cosine similarity vs euclidean distance category... Apply them spaces in machine learning practitioners deciding which of the vectors the vectors. Extract insights on the site them carries be generalized to any datasets across the concept of similarity. One with the most points among two objects, OB and OC are three vectors as illustrated in the of! Is generally used as a metric for measuring the proximity between vectors explain what cosine similarity is proportional to dot... To Euclidean distance corresponds to their dot product divided by the product of their magnitudes we ’ studied. Extracted by using Euclidean distance corresponds to their dot product divided by the product of vectors... The difference between vectors find the cluster centroids whose position minimizes the Euclidean distance and cosine similarity and we... A distance matrix next section visit this article have to be tokenzied same region of a vector space OA! How can we use them to extract insights on the features of a sample dataset terms as to we... Any distance measurement, text have to be tokenzied in â, the Euclidean distance corresponds to the product. First determine a method for measuring distances of different positive radius we would get the same points this that. Measuring how similar they are, as opposed to determining cluster membership equilateral triangle buzz term similarity measure... | 2022-05-27T16:14:37 | {
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https://math.stackexchange.com/questions/1849974/between-2-consecutive-roots-of-f-there-is-at-least-one-root-of-f | # Between $2$ consecutive roots of $f'$, there is at least one root of $f$
Prove that between $2$ consecutive roots of $f'$, there is at least one root of $f$.
I understand that a root of $f'$ represents an extreme point. But, for example, $f(x) = \sin(x)+2$ has no roots, but its derivative, $\cos(x)$, has lots of consecutive roots.
Ok, while I was writing this, I realized that its no "at least" but "there is at most" one root of $f$.
So, I understand that, between $2$ consecutive maximum points, for example, there can be one root, but if that function tries to come back and make another root between the two max poits, it's gotta create a local maximum point between them. But how do I write this mathematically?
Let me try:
By Rolle's, between $2$ consecutive roots $f(a) = f(b)$, there must be a point $c\in [a,b]$ where $f'(c) = 0$, which is a maximum point.
Or maybe, can I say the following:
between two roots of $f'(x)$, let's say, $f'(m) = f'(n)$ by rolles theorem we have:
$$\exists c\in [m,n] / f''(c) = 0$$
so there's a maximum point between the roots, but I don't know how to prove that this is the only maximum point, and that this maximum point leads to only one root.
• Let $f(x) = x^3/3 - 4x + 7,000,000,000$. Then $f'(x) = x^2 - 4$ has two roots at 2 and -2. Does $f(x) = x^3/3 - 4x + 7,000,000,000$ have any roots between $2$ and $-2$? Jul 5 '16 at 17:11
• If you've realized the error in your original proposition, I'd suggest rewriting your post and question to reflect that. You don't have to (and shouldn't) delete you first premise but you should indicate you now realize it was false and have a new question. Jul 5 '16 at 17:13
• "By Rolle's, between 2 consecutive roots f(a)=f(b), there must be a point c∈[a,b] where f′(c)=0, which is a maximum point. " Not merely $c \in [a,b]$ but $c \in (a,b)$ which means ... you are done. Nitpick: c can be maximum or minimum. Jul 5 '16 at 18:55
• So you've chosen a title that says almost the opposite of what it should say?
– zhw.
Jul 5 '16 at 19:09
Suppose that $f'(a)=f'(b)=0$ with $a<b$ and that $f'\not = 0$ in $]a,b[$. If $f$ has two distinct roots $x$ and $y$ in $]a,b[$, then, by Rolle's theorem (since $f(x)=f(y)=0$), there is $c$ in $[x,y]$ such that $f'(c)=0$. Contradiction.
Hint:
$f$ is necessarily monotonic between any two consecutive roots of $f'$.
Silly me. I ignored Rolle's theorem (and kind of proved it from scratch-- [confession-- I can't remember the names and conditions of any of these thereoms. I just remember the intermediate value theoreom and derive them all from common sense when needed]).
"By Rolle's, between 2 consecutive roots f(a)=f(b), there must be a point c∈[a,b] where f′(c)=0, which is a maximum point. "
And that's it! You've done it. If $f'(a) = f'(b) = 0$ and $f(x) = f(y) = 0$ for $a \le x < y \le< b$ then there is a $f'(c) = 0$ with $a \le x < c < y \le b$ which contradicts your premise that $f'$ has no roots between $a$ and $b$.
"but if that function tries to come back and make another root between the two max poits, it's gotta create a local maximum point between them. But how do I write this mathematically?"
Well, you first consider $f'$ by itself without regard to $f$ and interpret the intermediate value thereom.
$f'(a) = 0$; $f'(b) = 0$ (wolog $a < b$). And for no $x \in (a,b)$ does $f'(x) = 0$. Suppose there is an $x$ where $f'(x) > 0$ and and $f'(y) < 0$ with $x,y \in (a,b)$. Then by IMT there is an $f'(c) = 0$ with $c \in (\min(x,y), \max(x,y) \subset (a,b)$. This is a contradiction. So either $f'(x) > 0; \forall x \in (a,b)$ or $f'(x) < 0; \forall x \in (a,b)$.
Now we think about what that means for $f$. It means $f$ is monotonically increasing or decreasing. Wolog let's assume $f$ is monotonically increasing.
If $f(x) = 0$ for $a < x < b$ then $f(w) < 0$ for all $a < w < x$ and $f(z) > 0$ for all $x < z < b$. So there $x$ would be the root between $a$ and $b$.
If $f(x) = 0$ for $a < x <b$... which need not be the case.
• wait, but the premise was not about $f'$ having roots between the interval, but $f$ Jul 5 '16 at 21:03
• Um.... yeah and that's what I answered. Jul 5 '16 at 21:05
• If there are two roots of f in an interval then there is at least one root of f' inside the interval. So if there are no roots of f' in the interval there can be at most one root of f. QED Jul 5 '16 at 21:07
• Two roots of f => one root of f' => contradiction to premise. Jul 5 '16 at 21:10
• The question was about f having roots. The premise/hypothesis was about f'. It really was. Jul 5 '16 at 21:12 | 2022-01-19T08:41:29 | {
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http://accessdtv.com/taylor-series/taylor-expansion-error-analysis.html | Home > Taylor Series > Taylor Expansion Error Analysis
# Taylor Expansion Error Analysis
## Contents
n! Created by Sal Khan.Share to Google ClassroomShareTweetEmailTaylor & Maclaurin polynomials introTaylor & Maclaurin polynomials intro (part 1)Taylor & Maclaurin polynomials intro (part 2)Worked example: finding Taylor polynomialsPractice: Taylor & Maclaurin polynomials x2 x4 x6=1+ 0 − +0 + +0 − + ... 2! 4! 6! ∞ 2n n x= ∑( −1) n =0 ( 2n )! 10 11. Well that's going to be the derivative of our function at a minus the first derivative of our polynomial at a. have a peek at this web-site
And sometimes you might see a subscript, a big N there to say it's an Nth degree approximation and sometimes you'll see something like this. Example (Estimation of Truncation Errors by Geometry Series) What is |R6| for the following series expansion? for some c between a and x The Lagrange form of the remainder makes analysis of truncation errors easier. 7 8. Part 3Truncation Errors 1 2. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation
## Calculate Truncation Error Taylor Series
The system returned: (22) Invalid argument The remote host or network may be down. So what that tells us is that we can keep doing this with the error function all the way to the Nth derivative of the error function evaluated at a is Calculus SeriesTaylor & Maclaurin polynomials introTaylor & Maclaurin polynomials intro (part 1)Taylor & Maclaurin polynomials intro (part 2)Worked example: finding Taylor polynomialsPractice: Taylor & Maclaurin polynomials introTaylor polynomial remainder (part 1)Taylor
So this thing right here, this is an N plus oneth derivative of an Nth degree polynomial. Key Concepts• Truncation errors• Taylors Series – To approximate functions – To estimate truncation errors• Estimating truncation errors using other methods – Alternating Series, Geometry series, Integration 2 3. n! ( n + 1)!• How to derive the series for a given function?• How many terms should we add? Lagrange Error Bound Formula The system returned: (22) Invalid argument The remote host or network may be down.
And that's the whole point of where I'm going with this video and probably the next video, is we're gonna try to bound it so we know how good of an Taylor Series Error Bound If you continue browsing the site, you agree to the use of cookies on this website. S =1 +π −2 + 2π −4 + 3π −6 +... + jπ −2 j +... But what I wanna do in this video is think about if we can bound how good it's fitting this function as we move away from a.
Please try the request again. Taylor Series Remainder Calculator Introduction Joris Schelfaut English Español Português Français Deutsch About Dev & API Blog Terms Privacy Copyright Support LinkedIn Corporation © 2016 × Share Clipboard × Email Email sent successfully.. Please try the request again. And for the rest of this video you can assume that I could write a subscript.
## Taylor Series Error Bound
Example (Backward Analysis)This is the Maclaurin series expansion for ex x2 x3 xn e x = 1 + x + + + ... + + ... 2! 3! And this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be the same thing as f of a. Calculate Truncation Error Taylor Series n! ( n + 1)! Taylor Series Approximation Error f ( n ) (a) + ( x − a ) n + Rn n!where the remainder Rn is defined as x ( x − t ) n ( n +1)
What do you call someone without a nationality? Check This Out Your cache administrator is webmaster. And we already said that these are going to be equal to each other up to the Nth derivative when we evaluate them at a. R0 R1 R2 R3 R4 R5 R6 R7 Solution: This series satisfies the conditions of the Alternating Convergent Series Theorem. Taylor Polynomial Approximation Calculator
If you take the first derivative of this whole mess-- And this is actually why Taylor polynomials are so useful, is that up to and including the degree of the polynomial So we already know that P of a is equal to f of a. And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x Source Analyzing the remainder term of the Taylor series expansion of f(x)=ex at 0The remainder Rn in the Lagrange form is ( n +1) f (c ) Rn = ( x −
Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Lagrange Error Bound Calculator I am unsure of how I made $E_n(x)$: \begin{align} \left|\sin(x)-x\right| =& \sum\limits_{k=0}^n (-1)^k\dfrac{x^{2k+1}}{(2k+1)!} + (-1)^{2n+3}\dfrac{x^{2n+3}}{(2n+3)!} -x \\ \left|\sin(x)-x -\sum\limits_{k=0}^n (-1)^k\dfrac{x^{2k+1}}{(2k+1)!}\right| =& \left|(-1)^{2n+3}\dfrac{x^{2n+3}}{(2n+3)!} -x\right| \\ =&\left|\dfrac{x^{2n+3}}{(2n+3)!}-x\right| \end{align} Continuing in this way find And you can verify that because all of these other terms have an x minus a here. | 2018-05-20T17:55:58 | {
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https://math.stackexchange.com/questions/395146/integrate-int-left-cot-x-tan-x-right2dx | # Integrate $\int {{{\left( {\cot x - \tan x} \right)}^2}dx}$
\eqalign{ & \int {{{\left( {\cot x - \tan x} \right)}^2}dx} \cr & = {\int {\left( {{{\cos x} \over {\sin x}} - {{\sin x} \over {\cos x}}} \right)} ^2}dx \cr & = {\int {\left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {\sin x\cos x}}} \right)} ^2}dx \cr & = \int {{{\left( {{{\cos 2x} \over {{1 \over 2}\sin 2x}}} \right)}^2}dx} \cr & = \int {{{\left( {2\cot 2x} \right)}^2}} \cr & = \int {4{{\cot }^2}2xdx} \cr & = \int {4\left( {{{\csc }^2}2x - 1} \right)dx} \cr & = \int {\left(4{{\csc }^2}2x - 4\right)dx} \cr & = 4 \times {{ - 1} \over 2}\cot 2x - 4x + C \cr & = - 2\cot 2x - 4x + C \cr}
Where have I gone wrong? I've tried to spot an error so many times yet I can't find it, I need another pair of eyes.. Thanks.
• everything checks out. You are fine. – Nana May 18 '13 at 2:40
• If you have a problem with checking your result against a result returned by a machine like Wolfram Alpha, simply input the difference between the two functions. If it is a constant, then you know the two functions are identical outside the constant of integration. – Jon Claus May 18 '13 at 4:33
Why do you think that you've gone wrong? Your answer is correct.
Edit. To show why the solution is the same as the one in your answers, we just need to show that $\tan x-\cot x=-2\cot 2x$. \begin{align*} &\tan x-\cot x+2\cot 2x \\ &=\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}+2\frac{\cos 2x}{\sin 2x} \\ &=\frac{2\cos x\cos 2x\sin x-\cos^2 x\sin 2x+\sin^2 x\sin 2x}{\cos x\sin x\sin 2x} \\ &=\frac{2\cos x(\cos^2 x-\sin^2 x)\sin x-\cos^2 x(2\sin x\cos x)+\sin^2 x(2\sin x\cos x)}{\cos x\sin x\sin 2x} \\ &=\frac{2\cos^3 x\sin x-2\cos x\sin^3 x-2\sin x\cos^3 x+2\cos x\sin^3 x}{\cos x\sin x\sin 2x} \\ &=0 \end{align*}
• The answer in the back of this book is: $- \cot x - 4x + \tan x + c$ – seeker May 18 '13 at 2:40
• @Assad They are the same solution since $\tan x-\cot x=-2\cot 2x$. Have you tried to show that they are equal? – Warren Moore May 18 '13 at 2:41
• Ah I see, I did not realise this... Thanks – seeker May 18 '13 at 2:44
• @Assad I've edited my answer with a proof of the equivalence for you as well. :) – Warren Moore May 18 '13 at 2:51
A cleaner way could be
$$\int (\cot x-\tan x)^2dx=\int(\cot^2x+\tan^2x-2\cot x\tan x)dx$$
$$=\int (\csc^2x-1+\sec^2x-1-2)dx$$
$$=\int \csc^2x dx+\int \sec^2x dx-4\int dx$$
$$=-(\cot x-\tan x)-4x+C$$
From another answer $\cot x-\tan x=2\cot2x$
$\tan x- \cot x = \frac{\sin x}{\cos x}-\frac{\cos x}{\sin x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = \frac{-\cos 2x}{\sin x \cos x} = \frac{-2 \cos2x}{\sin 2x} = -2 \cot2x$
Therefore you answer $-\cot x -4x + \tan x +c$ is correct. | 2019-09-21T17:17:18 | {
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http://math.stackexchange.com/questions/397723/which-law-of-logical-equivalence-says-p-leftrightarrow-q-%E2%89%A1-p-lor-q-rightarro | # Which law of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$
I'm going through the exercises in the book Discrete Mathematics with Applications. I'm asked to show that two circuits are equivalent by converting them to boolean expressions and using the laws in this table.
$$\begin{array}{lcc} \hphantom{1}\mathsf{1.\; Commutative\; laws:} & p\land q \equiv q\land p & p\lor q \equiv q\lor p\\ \hphantom{1}\mathsf{2.\; Associative\; laws:} & (p\land q)\land r \equiv p\land (q\land r) & (p\lor q)\lor r \equiv p\lor (q\lor r)\\ \hphantom{1}\mathsf{3.\; Distributive\; laws:} & p\land (q\lor r) \equiv (p\land q)\lor (p\land r) & p\lor (q\land r) \equiv (p\lor q)\land (p\land r)\\ \hphantom{1}\mathsf{4.\; Identity\; laws:} & p\land t \equiv p & p\lor c \equiv p\\ \hphantom{1}\mathsf{5.\; Negation\; laws:} & p\lor \lnot p \equiv t & p\land \lnot p \equiv c\\ \hphantom{1}\mathsf{6.\; Double\; negative\; law:} & \lnot(\lnot p) \equiv p &\\ \hphantom{1}\mathsf{7.\; Idempotent\; laws:} & p\land p \equiv p & p\lor p \equiv p\\ \hphantom{1}\mathsf{8.\; Universal\; bound\; laws:} & p\lor t \equiv t & p\land c\equiv c\\ \hphantom{1}\mathsf{9.\; De\; Morgan}\text{'}\mathsf{s\; laws:} & \lnot(p\land q) \equiv \lnot p\lor \lnot q & \lnot(p\lor q) \equiv \lnot p\land\lnot q\\ \mathsf{10.\; Absorption\; laws:} & p\lor (p\land q) \equiv p & p\land (p\lor q) \equiv p\\ \mathsf{11.\; Negations\; of\; t\; and\; c:} & \lnot t \equiv c & \lnot c \equiv t\\ \end{array}$$
so as which law/s of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$
I can see their equivalence clearly with a truth table. But the book is asking me to show it using the equivalence laws in the above table, and I can't see how any of them apply here. So, do any of those laws apply here in a way I'm not understanding? Or is there some other known law that does apply here?
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@Zev: Wow.${}{}$ – Asaf Karagila May 21 '13 at 0:56
Haha thanks :) I get carried away sometimes... – Zev Chonoles May 21 '13 at 1:00
Nice long list, but obviously not enough, we need something that mentions $\longrightarrow$ and $\longleftrightarrow$. – André Nicolas May 21 '13 at 1:05
@GastónBurrull: Other common mistakes include writing $<a,b>$ instead of $\langle a,b\rangle$ or $\{ x|\varphi(x)\}$ instead of $\{x\mid \varphi(x)\}$ or $|x|$ instead of $\lvert x\rvert$ or $||x||$ instead of $\lVert x\rVert$. On a different note, - instead of – or — (in proper LaTeX you typeset endash by two minuses -- and emdash by three ---). Also, some people write $\to$ instead of $\mapsto$, but I've seen it done on blackboard, too... – tomasz May 21 '13 at 2:17
@GastónBurrull: you can right-click any math formula and then choose “Show Math As -> TeX Commands”. – tomasz May 21 '13 at 2:30
Let's assume the following definitions:
$\begin{array}{lc} \mathsf{12.\; Definition \; of \; Implication:} & p \Rightarrow q \equiv \neg p \lor q \\ \mathsf{13.\; Definition \; of \; Biconditional:} & p \Leftrightarrow q \equiv (p \Rightarrow q) \land (q \Rightarrow p) \end{array}$
Then we have: $$\begin{array}{rll} P \Leftrightarrow Q &\equiv (P \Rightarrow Q) \land (Q \Rightarrow P) &\text{by (13)} \\ &\equiv (\neg P \lor Q) \land (\neg Q \lor P) &\text{by (12)} \\ &\equiv (\neg P \land (\neg Q \lor P)) \lor (Q \land (\neg Q \lor P)) &\text{by (3)} \\ &\equiv ((\neg P \land \neg Q) \lor (\neg P \land P)) \lor ((Q \land \neg Q) \lor (Q \land P)) &\text{by (3)} \\ &\equiv ((\neg P \land \neg Q) \lor (P \land \neg P)) \lor ((Q \land \neg Q) \lor (P \land Q)) &\text{by (1)} \\ &\equiv ((\neg P \land \neg Q) \lor c) \lor (c \lor (P \land Q)) &\text{by (5)} \\ &\equiv ((\neg P \land \neg Q) \lor c) \lor ((P \land Q) \lor c) &\text{by (1)} \\ &\equiv (\neg P \land \neg Q) \lor (P \land Q) &\text{by (4)} \\ &\equiv \neg (P \lor Q) \lor (P \land Q) &\text{by (9)} \\ &\equiv (P \lor Q) \Rightarrow (P \land Q) &\text{by (12)} \\ \end{array}$$
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The truth table involves "t" and "c" throughout. So, you can rewrite the truth table using the negation laws. That is, instead of using "t" and "c" in your truth table, use (x∨¬x), and (x $\land$ $\lnot$x) respectively. Then, since you'll have (x∨¬x) throughout the last column of the truth table, you use the equivalence law (x∨¬x)=t in an additional column, and thus you've showed the formula true using the equivalence laws... specifically your last two columns will look like this:
(x∨¬x) t (by the equivalence law (x∨¬x)==t)
(x∨¬x) t (by the equivalence law (x∨¬x)==t)
(x∨¬x) t (by the equivalence law (x∨¬x)==t)
(x∨¬x) t (by the equivalence law (x∨¬x)==t)
-
I guess maybe I would do better to add here that since truth tables consist of no more than an exhaustive set of truth-value calculations, and since truth-value calculations can get formalized via the prtothetic of Lesniewski (p. 66-67 of Arthur Prior's Formal Logic), the above procedure should qualify as a means to write an informal proof. – Doug Spoonwood May 24 '13 at 17:09 | 2016-05-29T09:48:02 | {
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http://binary-bros.cz/admindemo/static/news/mitchell-kosterman-xkg/e12f97-inverse-trig-derivatives-practice | To play this quiz, please finish editing it. Derivative Practice: Inverse Trigonometric Functions 1 ... = xby inverse functions. Also remember that sometimes you see the inverse trig function written as $$\arcsin x$$ and sometimes you see $${{\sin }^{{-1}}}x$$. Check out all of our online calculators here! Given the list of derivatives for inverse trigonometric functions below, what most looks like the integral of inverse cosine (cos-1 x), given the following choices. Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. of Other Bases Practice Solutions at the back Implicit Differentiation Online Practice N/A This topic from Unit 2 Derivatives for Calculus 1 and Unit 3 for. First of all, there are exactly a total of 6 inverse trig functions. The slope of the line tangent to the graph at x = e is . Differentiation: composite, implicit, and inverse functions, Differentiating inverse trigonometric functions. Inverse trigonometric functions differentiation Calculator Get detailed solutions to your math problems with our Inverse trigonometric functions differentiation step-by-step calculator. Practice your math skills and learn step by step with our math solver. This quiz is incomplete ... Derivatives . Find the derivative of y with respect to the appropriate variable. sin − 1 x. Derivatives of Inverse Trig Functions. If you're seeing this message, it means we're having trouble loading external resources on our website. Derivatives of Inverse Trigonometric Functions. f (x) = (sin -1) 2. g (t) = cos -1 √ (2t - 1) y = tan -1 (x/a) + ln√ ( (x-a)/ (x+a)) Show Video Lesson. Looking for a way to f give you Calculus students more practice finding Derivatives of Inverse Trig Functions? To log in and use all the features of Khan Academy, please enable JavaScript in your browser. AP Calculus AB - Worksheet 33 Derivatives of Inverse Trigonometric Functions Know the following Theorems. These problems will provide you with an inverse trigonometric function. The slope of the tangent line follows from the derivative (Apply the chain rule.) Working with derivatives of inverse trig functions. This quiz is incomplete! arcsin(x) = 1 − x2. by dkegelman. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview … Differentiating inverse trig functions review. Examples: Find the derivatives of the following functions. Derivative of Inverse Trigonometric functions The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. 3 Definition notation EX 1 Evaluate these without a calculator. Problem Statement: sin-1 x = y, under given conditions -1 ≤ x ≤ 1, -pi/2 ≤ y ≤ pi/2. Find the derivatives of the following functions: a) y = sin(x) – 4 cos(5 x) b) y = x3 tan(x) All the inverse trigonometric functions have derivatives, which are summarized as follows: Derivatives of Inverse Trigonometric Functions using the First Principle. Our mission is to provide a free, world-class education to anyone, anywhere. Next lesson. However, some teachers use the power of -1 instead of arc to express them. What are the derivatives of the inverse trigonometric functions? Solve this … Here are the derivatives of Inverse Trigonometric Functions: Donate or volunteer today! From there, you will be asked to do a range of things. For example, the sine function $$x = \varphi \left( y \right)$$ $$= \sin y$$ is the inverse function for $$y = f\left( x \right)$$ $$= \arcsin x.$$ Then the derivative of $$y = \arcsin x$$ is given by \[{{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = … Delete Quiz. Thus, an equation of the tangent line is … Here is a good video showing this derivation. The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. Differentiation - Inverse Trigonometric Functions Date_____ Period____ Differentiate each function with respect to x. ... Print; Share; Edit; Delete; Host a game. These derivatives will prove invaluable in the study of integration later in this text. \dfrac {d} {dx}\arcsin (x)=\dfrac {1} {\sqrt {1-x^2}} dxd. 2. If we restrict the domain (to half a period), then we can talk about an inverse function. •Since the definition of an inverse function says that -f 1(x)=y => f(y)=x We have the inverse sine function, -sin 1x=y - π=> sin y=x and π/ 2 <=y<= / 2 Finish Editing. In inverse trig functions the “-1” looks like an exponent but it isn’t, it is simply a notation that we use to denote the fact that we’re dealing with an inverse trig function.It is a notation that we use in this case to denote inverse trig functions.If I had really wanted exponentiation to denote 1 … Practice: Derivatives of inverse trigonometric functions, Differentiating inverse trig functions review, Selecting procedures for calculating derivatives: strategy. The first two, sine and cosine, are pretty straightforward since they are ALMOST inverses of each other. Formulas for the derivatives of the six inverse trig functions and derivative examples. Inverse trigonometric functions have various application in engineering, geometry, navigation etc. AP® is a registered trademark of the College Board, which has not reviewed this resource. 3 years ago. Derivatives of inverse trigonometric functions Calculator Get detailed solutions to your math problems with our Derivatives of inverse trigonometric functions step-by-step calculator. Chain Rule Practice with Inverse Trig Solutions at the back Derivatives of Inverse Functions Practice Solutions at the back Chain Rule with Logs and Exp. d d x arcsin ( x) = 1 1 − x 2. 2 The graph of y = sin x does not pass the horizontal line test, so it has no inverse. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Inverse Trigonometry Functions and Their Derivatives. However, in the following list, each trigonometry function is listed with an appropriately restricted domain, which makes it one-to-one. Inverse Trigonometric Functions - Derivatives. Khan Academy is a 501(c)(3) nonprofit organization. Check out all of our online calculators here! Homework. This is the currently selected item. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$T\left( z \right) = 2\cos \left( z \right) + 6{\cos ^{ - 1}}\left( z \right)$$, $$g\left( t \right) = {\csc ^{ - 1}}\left( t \right) - 4{\cot ^{ - 1}}\left( t \right)$$, $$y = 5{x^6} - {\sec ^{ - 1}}\left( x \right)$$, $$f\left( w \right) = \sin \left( w \right) + {w^2}{\tan ^{ - 1}}\left( w \right)$$, $$\displaystyle h\left( x \right) = \frac{{{{\sin }^{ - 1}}\left( x \right)}}{{1 + x}}$$. Here is a set of practice problems to accompany the Derivatives of Inverse Trig Functions section of the Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. They are arcsin x, arccos x, arctan x, arcsec x, and arccsc x. Let’s understand this topic by taking some problems, which we will solve by using the First Principal. https://magoosh.com/.../ap-calculus-review-derivatives-inverse-functions Inverse Trigonometric Functions: •The domains of the trigonometric functions are restricted so that they become one-to-one and their inverse can be determined. For each of the following problems differentiate the given function. You just need to remember the negative sign on the second one.The rest of them can be derived from the sine and cosine rules using the product rule, quotient rule and basic trigonometric identities. Derivatives of Inverse Trigonometric Functions. Practice: Derivatives of inverse trigonometric functions. Practice. The quiz is a collection of math problems. These functions are used to obtain angle for a given trigonometric value. For example, arcsin x is the same as. These resource with Task Cards, HW, and an Organizer will give your students the practice and rigors they need to succeed. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Practice your math skills and learn step by step with our math solver. Live Game Live. Derivative Practice: Trig, Inverse Trig, Exponentials, & Logs 104003 Differential and Integral Calculus I Technion International School of Engineering 2010-11 Tutorial Handout – January 30, 2011 – Kayla Jacobs 1. Solo Practice. So, h0(x) = 1. 1. Also, we previously developed formulas for derivatives of inverse trigonometric functions. Derivatives of inverse Trig Functions. Formulas for the remaining three could be … . . Practice Problems: 1.Find all solutions of the following equations on interval [0;2ˇ]: Play. Derivatives of the Inverse Trigonometric Functions. Worksheet: Inverse Trig Integrals We’re a little behind Professor Davis’s lectures. Share practice link. The formula for the derivative of y= sin 1 xcan be obtained using the fact that the derivative of the inverse function y= f 1(x) is the reciprocal of the derivative x= f(y).
Professional Writing Pdf, Athens Men's Baseball League, Appreciation In Tagalog, Commodity Transaction Tax, Dewalt Dws715 Laser, Dainty Daisy Tattoo, Strike Industries Pistol Buffer Tube, Sandstone Filler Repair, Do D1 Schools Give Athletic Scholarships, World Of Warships: Legends Citadel Hits, Calories In Gulab Jamun With Ice Cream, | 2021-09-22T00:09:17 | {
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https://physics.stackexchange.com/questions/378063/unitary-transformation-of-eigenstates | # Unitary Transformation of Eigenstates
Suppose I have two operators, $A$ and $B$, with eigenstates $A \lvert a \rangle = a \lvert a \rangle$ and $B \lvert b \rangle = b \lvert b \rangle$, where $a$ and $b$ are all unique. Furthermore, suppose that $A$ and $B$ are related by a unitary transformation $$A = U B U^{-1}.$$ This is equivalent to saying that the eigenstates are related as $$\lvert a \rangle = U \lvert b \rangle.$$
Then it seems I can prove the following: since $$A \lvert a \rangle = a \lvert a \rangle,$$ I also have $$A U \lvert b \rangle = U B U^\dagger U \lvert b \rangle$$ by inserting the identity, so that $$A U \lvert b \rangle = U B \lvert b \rangle = b U \lvert b \rangle = b \lvert a \rangle.$$ Thus, $a = b$.
Doesn't this imply then that the eigenvalues for corresponding eigenstates of $A$ and $B$ are equal, and therefore-- by the assumption that they are unique-- that the unitary transformation doesn't actually do anything?
• As nicely summarized by @ZeroTheHero, when you write $A=UBU^{\dagger}$ you are performing a unitary basis change on operator $B$. $A$ is then the new representation of $B$ in this new set of basis. Eigenvalues are invariant under basis transformation . Thus, eigenvalues of $A$ and $B$ are the same, but their eigenstates are not. – Ptheguy Jan 5 '18 at 8:52
• Here is a good article if you're not sure why eigenvalues are invariant under basis transformation. (link). – Ptheguy Jan 5 '18 at 9:00
• It might help to choose as an example $A=x$ and $B=p$, where $U$ is the Fourier transform and takes $x$ to $p$. Here $x$ and $p$ have identical eigenvalue spectra, but that doesn't mean that the Fourier transform doesn't "do" anything. – Emilio Pisanty Jan 5 '18 at 10:03
I’m not sure what you mean by “$a$ and $b$ are unique” but clearly if $A=UBU^\dagger$ and $U$ is unitary, $A$ and $B$ have the same eigenvalues but it doesn’t mean $U$ doesn’t do anything.
For instance, the Pauli matrices $\sigma_{x,y,z}$ all have the same eigenvalues, are related by a unitary transformation $U$, but are certainly different. The transformation $U$ is a change of basis, so if $B$ is initially diagonal, say $$B=\sigma_z=\left(\begin{array}{cc} 1&0 \\ 0&-1\end{array}\right)$$ and $U=\left(\begin{array}{cc} \cos\theta/2&-\sin\theta/2\\ \sin\theta/2 &\cos\theta/2\end{array}\right)$ then $$U\sigma_zU^{\dagger}=\cos\theta\sigma_z+\sin\theta \sigma_x$$ still have eigenvalues $\pm 1$ but obviously $U$ has done something.
Of course the eigenstates are no longer $\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\end{array}\right)$.
I interpret your uniqueness claim as the requirement that
every eigenspace of either $A$ and $B$ has dimension $1$.
So, if $a$ is an eigenvalue of $A$ and $A|a\rangle =a|a\rangle$, for $|a\rangle \neq 0$, then every other eigenvector with the same eigenvalue $a$ is of the form $c|a\rangle$ for every $c\in \mathbb C$, $c\neq 0$. If we consider only normalized eigenvectors, $c$ is of the form $e^{i \theta}$ for every $\theta \in \mathbb R$.
Assuming that the spectra of the said operators are pure-point spectra, we have the spectral decompositions (the sum are understood in the strong operator topology, but here it is quite irrelevant and you can safely interpret all follows at algebraic level) $$A = \sum_{a \in {\cal A}} a |a\rangle\langle a| \tag{1}$$ and $$B = \sum_{b\in {\cal B}} b |b\rangle\langle b| \:.$$ where I used normalized eigenvalues and ${\cal A}=\sigma(A)$, ${\cal B}=\sigma(B)$ (up to accumulation points) are the spectra of the operators.
On the other hand $$A = UBU^\dagger$$ implies $$\sum_{a \in {\cal A}} a |a\rangle\langle a| = \sum_{b \in {\cal B}} b U|b\rangle\langle b|U^\dagger\:.$$ That is $$A= \sum_{b \in {\cal B}} b|\psi_b\rangle \langle \psi_b|\tag{2}$$ where $$|\psi_b\rangle := U|b\rangle\:.$$
The crucial result is now that
for a given self-adjoint operator (with point spectrum) the spectral decomposition is unique.
Thus, comparing (1) and (2) we conclude that
(i) ${\cal A}= {\cal B}\:,$
so that we can re-arrange the decomposition of $A$ like this $$A= \sum_{a \in {\cal A}} a|\psi_a\rangle \langle \psi_a|\:,$$
(ii) $|a\rangle \langle a| = |\psi_a\rangle \langle \psi_a|$ so that, since every vector is normalized $$|\psi_a\rangle = e^{i\theta_a}|a\rangle\quad \mbox{for some \theta_a \in \mathbb R}\:.$$ there is no way to determine the phases $e^{i\theta_a}$, since normalized eigenvectors are defined up to a phase, but we are free to fix all them $e^{i\theta_a}=1$.
I stress that (i) and (ii) are the maximum you can obtain from the initial information at your disposal that eigenspaces are one-dimensional and that the unitary equivalence $A=UBU^\dagger$ holds.
You see that $U$ has an action (it is false that "it doesn't actually do anything"). In fact, it changes eigenvectors, but it leaves fixed the spectrum of the operators.
Dropping the hypotheses of one-dimensional eigenspaces but keeping the request of pure point spectrum, (i) remains valid in view of the uniqueness of spectral decomposition, which now reads $$A = \sum_{a \in {\cal A}} a P_a$$ where $P_a = \sum_{k=1}^{\dim E_a} |a,k\rangle \langle a,k|$ is the orthogonal projector onto the eigenspace $E_a$ of $A$ with eigenvalue $a$ and the vectors $|a,k\rangle$, varying $k=1,\ldots, \dim E_a$, form an orthonormal basis of that eigenspace.
Well, a similarity transformation for an invertible (not necessary unitary) operator$^1$ $U$ does generically change the eigenspaces but does not change the eigenvalue spectrum $\{a_1, a_2,\ldots, \}=\{b_1,b_2,\ldots\}$. Hence it would be inconsistent to claim that all the eigenvalues $\{a_1,a_2, \ldots, b_1,b_2,\ldots\}$ for both $A$ and $B$ are different, if that's what OP means by that they are all unique. Whether the individual spectra are degenerate or not is irrelevant.
--
$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer. | 2020-01-18T02:53:48 | {
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https://math.stackexchange.com/questions/2442950/what-is-a-finite-sigma-algebra | # What is a “finite $\sigma$-algebra”?
I have an exersice which is outlined as follows
Suppose $G_{i}$ where $i=0 \ldots n$ is a disjoint union of $\Omega$. Prove that the family of unions of these $G_{i}$ is a sigma algebra on $\Omega$. Also prove that any "finite sigma algebra" $\mathcal{F}$ on $\Omega$ is of this form.
My guess is that a finite sigma algebra is a sigma algebra with finite number of sets but I am not sure.
I cant find the definition anywhere, does anyone know where I can find it?
• $\{\emptyset, \Omega\}$ is an example of a finite sigma-algebra. – Gabriel Romon Sep 24 '17 at 11:16
• @GabrielRomon good example – user123124 Sep 24 '17 at 11:24
• @GabrielRomon Indeed. And $\Omega$ is allowed to be infinite here. – drhab Sep 24 '17 at 11:25
• @drhab right, that kinda bothers me, but ill go with your suggestion for now – user123124 Sep 24 '17 at 11:25
A $\sigma$-algebra on space $\Omega$ is a subset of $\wp(\Omega)$ with special properties.
That means that a finite $\sigma$-algebra is a finite subset of $\wp(\Omega)$ with these properties.
As you guessed, a finite $\sigma$-algebra is just a $\sigma$-algebra that is finite.
As mentioned in the comments, a $\sigma$-algebra is a set, so when we describe a $\sigma$-algebra as "finite" we are using the standard definition of a finite set.
• It is wierd that one cannot find it written down anywhere. When one has infinite operations one can always get supprised. – user123124 Sep 24 '17 at 11:21
• @user1 see here and realize that a $\sigma$-algebra is a set. In fact you allready did that. Your guess is fine. – drhab Sep 24 '17 at 11:22
• @drhab Thanks ill go with that for now. – user123124 Sep 24 '17 at 11:23
• @user1 a sigma-algebra is a set, first and foremost ! – Gabriel Romon Sep 24 '17 at 11:31
• @drhab u wanna write that in answer or let littleO add it to his? I think this comment is vital. – user123124 Sep 24 '17 at 11:35 | 2021-06-22T19:46:46 | {
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https://math.stackexchange.com/questions/2082474/inverse-laplace-transform-of-s-hatfs-1-when-f-is-a-piecewise-function | # Inverse Laplace Transform of $s \hat{f}(s-1)$ when $f$ is a piecewise function
Let $f(t) = \begin{cases}t & \text{if}\,0<t<1 \\ 2-t & \text{if}\, 1<t < 2 \\ 0 & \text{otherwise} \end{cases}$
If you draw this function, it looks like an isosceles triangle with holes at the vertices at $(0,0)$, $(1,1)$, and $(2,0)$, so essentially, the $f(t) =t$ piece of this function is switched on at $t = 0$, and switched off at $t=1$, the $f(t) = 2-t$ piece of this function is switched on at $t = 1$ and switched off at $t=2$, and everywhere else the function is zero.
Earlier, I found the Laplace transform of $f(t)$ to be $\displaystyle \hat{f}(s) = \frac{1}{s^{2}} - \frac{2}{s^{2}}e^{-s} + \frac{1}{s^{2}}e^{-2s}$.
Therefore, $\displaystyle \hat{f}(s-1) = \frac{1}{(s-1)^{2}}- 2e\frac{2}{(s-1)^{2}}e^{-s}+e^{2}\frac{1}{(s-1)^{2}}e^{-2s}$, and $\displaystyle \mathbf{ s\hat{f}(s) = \frac{s}{(s-1)^2}- 2e \frac{s}{(s-1)^{2}}e^{-s}+e^{2}\frac{s}{(s-1)^{2}}e^{-2s}}$
Now, I am being asked to find $\mathbf{\mathcal{L}^{-1}[s \hat{f}(s-1)]}$, the inverse Laplace transform of $s \hat{f}(s-1)$.
The answer given in the back of the book is $\begin{cases} e^{t}(t+1) & \text{for}\,0<t<1 \\ e^{t}(1-t) & \text{for} \, 1 < t < 2 \\ 0 & \text{for}\, t>2 \end{cases}$
When approaching $\mathcal{L}^{-1}[s \hat{f}(s-1)]$ directly, none of the terms stood out to me as things with recognizable inverse Laplace transform forms. If the first term, for example, had been just $\displaystyle \frac{1}{(s-1)^{2}}$ and not $\displaystyle \frac{s}{(s-1)^{2}}$, I would know what to do. But that extra $s$ in the numerator is bothering me. I thought that maybe it was a convolution, where $f*g(t) = \int_{0}^{t}f(\tau)g(t - \tau)d\tau$, since $\mathcal{L}[f*g(t)] = \hat{f}(s)\hat{g}(s)$, but I'm not entirely sure what the inverse Laplace transform of $s$ is.
Finally, what I decided to do was work backwards. I started with the solution, and took the Laplace transform of it, hoping I might be able to reverse-engineer it, so to speak, and figure out how to do it that way. But, this isn't really practical - because unless you know what the answer is supposed to be ahead of time, you don't really know what mathematical tricks to use in order to rewrite it in terms of the functions you started with.
How should I approach this problem? In addition, can you give me any hints as to how to handle the extra $s$'s? And, if I do need to use convolution, could you work out one of the terms for me in full, so I could see how to do it? Then, I might be able to apply it to the other ones myself and get the answer I'm supposed to get for the whole thing. Thank you.
Here is a sketch of a direct route. By definition, you have $$\hat{f}(s)=\int_0^\infty e^{-st}f(t)\,dt\implies s\hat{f}(s-1)=\int_0^\infty se^{-(s-1)t}f(t)\,dt=\int_0^\infty se^{-st}(e^t f(t))\,dt$$
The extra factor of $s$ is the only obstacle to this being the definition of a Laplace transform. To get rid of it, we may observe that $se^{-st}=-\frac{d}{dt}(e^{-st})$. This immediately suggests integration by parts. The remaining details are left to the interested reader, but 1) the boundary term of this partial integration should vanish, 2) the remaining integral is of the form of a Laplace transform. The inverse Laplace transform of $s\hat{f}(s-1)$ is then obtained by inspection of this last term.
Postscript
It seems worth emphasizing that a standard table of Laplace transforms include the following functional identities, of which the above are examples: \begin{array}{ccc} f(t) &\Leftrightarrow& \hat{f}(s)=\int_0^\infty e^{-st}f(t)\,dt &\text{(Definition of Laplace transform)}\\ e^{pt}f(t) &\Leftrightarrow& \hat{f}(s-p) &\text{(Shift in frequency domain)}\\ f(t-\tau)H(t-\tau) &\Leftrightarrow& e^{-s\tau}\hat{f}(s) &(\text{Shift in time domain for }\tau>0)\\ f'(t) &\Leftrightarrow& s\hat{f}(s)-f_0 &\text{(Differentiation in time domain)}\\ \end{array}
• I updated my question to include all my work so far, including the part where I feel like I've got my legs stuck in tar and can't get out or move anymore. – ALannister Jan 3 '17 at 23:09
You can do partial fraction: $$\frac{s}{(s-1)^2}=\frac{1}{s-1}+\frac{1}{(s-1)^2}.$$ Then you can use the formulas $$L^{-1}(F(s-a))=e^{at}f(t)$$ and $$L^{-1}(e^{-as}F(s))=f(t-a)U(t-a)$$ to deal with the factor $e^{-(s-1)}$, where $U(t-a)$ is the unit step function.
Here is one example for the factor $e^{s-b}$ where $b$ is different from the shift in $F(s-a)$: $$L^{-1}\big(\frac{1}{s-1}\cdot e^{-2(s-1)}\big)e^t\cdot L^{-1}\big(\frac{1}{s}\cdot e^{-2s}\big)=U(t-2).$$ This used the second formula above.
Eventually you will have many terms with or without $U(t-1)$ and $U(t-2)$. Remember that $U(t-a)$ is 0 on $[0,a)$ and $1$ on $[a,\infty)$. You then just need to discuss three cases: $[0,1),[1,2),[2,\infty)$.
• from one kitty to another ;) We cats need to stick together! I'll give this a try. – ALannister Jan 3 '17 at 20:52
• how do you deal with $L^{-1}(e^{-as}F(s-b))$? I.e., the case where you want to find $\displaystyle L^{-1}\left[ e^{-2s}\left(\frac{1}{s-1} \right)\right]$? – ALannister Jan 3 '17 at 21:46
• @Semiclassical perhaps you could answer the follow-up question I just asked Kitty. – ALannister Jan 3 '17 at 22:03
• @JessyCat The only relevant difference in that case is the factor in front. But this is easily incorporated into the definition of the Laplace transform by a slight change of the integration variable $t$. – Semiclassical Jan 3 '17 at 22:08
• @JessyCat: I added an example of that. – KittyL Jan 3 '17 at 23:04
$L^{-1}\{e^{-as} F(s-b)\} = u_a(t)\cdot L^{-1}\{F(s-b)\}\big(t-a\big) = u_a(t)\cdot e^{b(t-a)} L^{-1}\{F(s)\}\big(t-a\big)$,
if this is what you are asking for. | 2020-05-26T17:40:20 | {
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http://mathhelpforum.com/discrete-math/106593-how-do-you-determine-total-amount-positive-factors-number.html | # Math Help - How do you determine the total amount of positive factors of a number
1. ## How do you determine the total amount of positive factors of a number
Part of a homework assignment I'm stuck on from my discrete math class. I suspect this may belong in another forum, but I wasn't sure which one.
Part a.) Find the prime factorization of 7056. okay.. no problem
$7056 = 2 * 3528 = 2^2 * 1764 = 2^3 * 882 = 2^4 * 441 = 2^4 * 3 * 147 = 2^4 * 3^2 * 49 = 2^4 * 3^2 * 7^2$
Part b.) How many positive factors are there of 7056?
I'm stuck on this, I used an example of a previous question from the homework where I discovered that:
$48 = 2^4 * 3$ and that it has 10 positive factors. (namely 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48) and 10 negative identical factors.
I can't seem to derive the method to determine total factors based on the powers of each individual factor. I certainly don't want to write them all out. My best guess is 36, which i figured by adding each exponent, then adding each exponent multiplied to the other exponents, in turn, then adding all three exponents multiplied. (1+4+2+2+1)+(4*2+4*2+2*2)+(4*2*2)=10+20+16=36
something seems fishey about my methodology though..
Thanks for the help!
2. just a counting argument. any factor of that number is going to share some or all of the prime factors.
Let $n=p_1^{e_1}\cdot... \cdot p_n^{e_1}$
Then any factor of n will have 0 factors of $p_1$, 1 factor of $p_1$, ... or $e_1$factors of $p_1$.
Similarly for each prime.
Thus the total number of possible combinations of factors will be
$(e_1+1)\cdot ... \cdot (e_n+1)$
So in your example you have $n=2^43^1$
so it has $(4+1)(1+1)=5\cdot2=10$ factors
3. Hello, TravisH82!
Part (a) Find the prime factorization of 7056.
okay.. no problem . . . $7056 \:=\: 2^4\cdot3^2\cdot7^2$
Part (b) How many positive factors are there of 7056?
Let's construct a factor of 7056 . . .
We have 5 choices for twos: . $\begin{Bmatrix}\text{0 twos} \\ \text{1 two} \\ \text{2 twos} \\ \text{3 two's} \\ \text{4 two's} \end{Bmatrix}$
We have 3 choices for threes: . $\begin{Bmatrix}\text{0 threes} \\ \text{1 three} \\ \text{2 threes} \end{Bmatrix}$
We have 3 choices for sevens: . $\begin{Bmatrix}\text{0 sevens} \\ \text{1 seven} \\ \text{2 sevens} \end{Bmatrix}$
Hence, there are: . $5\cdot3\cdot3 \:=\:45$ choices for a factor of 7056.
. . Therefore, 7056 has $\boxed{45}$ factors.
. . . This includes 1 and 7056 itself.
Now you can see why Gamma's formula works.
Find the prime factorization: . $7056 \;=\;2^4\cdot3^2\cdot7^2$
Add 1 to each exponent and multiply:
. . $(4+1)(2+1)(2+1) \;=\;5\cdot3\cdot3 \;=\;45$ factors . . . see?
4. Exactly. Wow very nice looking solution Soroban, exactly what I was going for, but I suck at TeX, lol. Thanks for taking the time to clean it up.
5. Ahh.. I see where I was going wrong.. i was neglecting to consider 2^0 as a choice. Thank you all for your help.. this makes it much clearer to me! | 2014-07-31T16:08:22 | {
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https://math.stackexchange.com/questions/1810235/probability-on-card-suits-in-a-standard-deck | # Probability on card suits in a standard deck
Let's say we have a standard deck of $52$ cards with $13$ cards each of the standard suits of spades, clubs, diamonds, hearts. Let's say we have $p$ players and each player is dealt q cards where $pq<52$.
1. Without distinguishing cards by their letter (so we treat any two cards of the same suit as the same, but two cards of different suits as different), how many ways can these cards be distributed?
2. How many ways can the cards be dealt so that someone (at least one player) doesn't get a spade?
You can plug in values that you find comfortable for $p$ and $q$, I just need to see work as to how it would be done so I can make it general.
I started working toward this and realized that I can just make the deck a deck of $pq$ cards and then multiply whatever formula I will get by $C(52, pq)$ to cover all of the bases. I don't know where I would go from here.
• I'm afraid this is going to be a horrible mess (unless $p$ and $q$ are chosen unrealistically to make it manageable). – joriki Jun 2 '16 at 21:40
• @joriki yeah you can choose like 5 and 6 for p and q. I just want to see how someone would work through this to get an idea of the methods to use here. It's not about the particular answer, just the process by which I could reach an answer so I can make a general formula. I need it for some work I am doing. – edupppz Jun 2 '16 at 21:42
Let me work out for a standard card game, e.g. bridge $(p=4,q=13)$
which should give you some idea of the process to be used.
$(1)$
Use the permutation formula with some objects identical
Total ways of distributing $= \dfrac{52!}{13!13!13!13!} = 53,644,737,765,488,792,839,237,440,000$
$(2)$
Imagine that there are $4$ groups of $13$ slots where cards are placed.
To correspond to part $(1)$, we shall first place the spades, (which can go to at most $3$ players), and then multiply for placement of the non-spades in the remaining $39$ slots in $\dfrac{39!}{13!13!13!}$
Use PIE, principle of inclusion-exclusion
Ways to place spades to any $3$ players = $\dbinom43\dbinom{39}{13}$,
but this includes only $2$ or $1$ player having spades, double counts cases with $2$ players having spades, and if we correct for this, eliminates cases with only $1$ player having spades. Applying PIE to get the correct count, we get
$$\dbinom43\dbinom{39}{13} - \dbinom42\dbinom{26}{13} + \dbinom41\dbinom{13}{13} = 32, 427, 298, 180$$
and, of course, multiply by $\dfrac{39!}{13!13!13!}$ for placement of the non-spades.
• I don't understand the reasoning behind the second part. I understand that in the first we sort of have a sequence with repetition where we have something like dividers which determine who gets what hand. – edupppz Jun 3 '16 at 15:42
• I have elaborated and homogenised the two parts. My previous comment got deleted inadvertently, I'd think it isn't needed again. – true blue anil Jun 4 '16 at 7:07
• I am sorry I haven't responded more promptly, I am traveling the country and have very sporadic access to the internet. I worked through the problem and I think your answer to part 1 poses an issue. If you are handed a spade and then a club for a hand of two and another time you are dealt a club and then a spade, your method counts those deals differently. I figured out how to bypass that. So for p,q as given above, we just multiply choices: C (52,q) C (52-q,q)... C (52-(p-1)q,q). Here we only differentiate between the separate kinds of hands dealt, so it is nice. – edupppz Jun 6 '16 at 14:29
• Now I am working on solving how if we have a set of size K and take p sets of size q, _pq_≤ K, how many ways can we pick sets such that one set has no members of an arbitrarily chosen subset of size L. This is just the general form of the second part. I think I understand the PIE, but I'll look at it more closely to make sure. – edupppz Jun 6 '16 at 14:42
• Your comment re double counting in part $1$ is not correct. Pl. note that $\binom{52}{13,13,13,13} = \frac{52!}{13!13!13!13!} = \binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$ – true blue anil Jun 6 '16 at 15:42 | 2019-12-13T13:29:13 | {
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http://stat88.org/textbook/notebooks/Chapter_11/05_The_Error_in_Regression.html | The Error in Regression
To assess the accuracy of the regression estimate, we must quantify the amount of error in the estimate. The error in the regression estimate is called the residual and is defined as
$$D = Y - \hat{Y}$$
where $\hat{Y} = \hat{a}(X-\mu_X) + \mu_Y$ is the regression estimate of $Y$ based on $X$.
Calculations become much easier if we express the residual $D$ in terms of the deviations $D_X$ and $D_Y$.
\begin{align*} D ~ &= ~ Y - \big{(} \hat{a}(X - \mu_X) + \mu_Y \big{)} \\ &= ~ (Y - \mu_Y) - \hat{a}(X - \mu_X) \\ &= ~ D_Y - \hat{a}D_X \end{align*}
Since $E(D_X) = 0 = E(D_Y)$, we have $E(D) = 0$.
This is consistent with what you learned in Data 8: No matter what the shape of the scatter diagram, the average of the residuals is $0$.
In our probability world, "no matter what the shape of the scatter diagram" translates to "no matter what the joint distribution of $X$ and $Y$". Remember that we have made no assumptions about that joint distribution.
Mean Squared Error of Regression
The mean squared error of regression is $E\big{(} (Y - \hat{Y})^2 \big{)}$. That is just $E(D^2)$, the expected squared residual.
Since $E(D) = 0$, $E(D^2) = Var(D)$. So the mean squared error of regression is the variance of the residual.
Let $r(X,Y) = r$ for short. To calculate the mean squared error of regression, recall that $\hat{a} = r\frac{\sigma_Y}{\sigma_X}$ and $E(D_XD_Y) = r\sigma_X\sigma_Y$.
The mean squared error of regression is
\begin{align*} Var(D) ~ &= ~ E(D^2) \\ &= ~ E(D_Y^2) - 2\hat{a}E(D_XD_Y) + \hat{a}^2E(D_X^2) \\ &= ~ \sigma_Y^2 - 2r\frac{\sigma_Y}{\sigma_X}r\sigma_X\sigma_Y + r^2\frac{\sigma_Y^2}{\sigma_X^2}\sigma_X^2 \\ &= ~ \sigma_Y^2 - 2r^2\sigma_Y^2 + r^2\sigma_Y^2 \\ &= ~ \sigma_Y^2 - r^2\sigma_Y^2 \\ &= ~ (1-r^2)\sigma_Y^2 \end{align*}
SD of the Residual
The SD of the residual is therefore
$$SD(D) ~ = ~ \sqrt{1 - r^2}\sigma_Y$$
which is consistent with the Data 8 formula.
$r$ As a Measure of Linear Association
The expectation of the residual is always $0$. So if $SD(D) \approx 0$ then $D$ is pretty close to $0$ with high probability, that is, $Y$ is pretty close to $\hat{Y}$. In other words, if the SD of the residual is small, then $Y$ is pretty close to being a linear function of $X$.
The SD of the residual is small if $r$ is close to $1$ or $-1$. The closer $r$ is to those extremes, the closer $Y$ is to being a linear function of $X$. If $r = \pm 1$ then $Y$ is a perfectly linear function of $X$.
A way to visualize this is that if $r$ is close to $1$ or $-1$, and you repeatedly simulate points $(X, Y)$, the points will lie very close to a straight line. In that sense $r$ is a measure of how closely the scatter diagram is clustered around a straight line.
The case $r=0$ is worth examining. In that case we say that $X$ and $Y$ are "uncorrelated". Because $\hat{a} = 0$, the equation of the regression line is simply $\hat{Y} = \mu_Y$. That's the horizontal line at $\mu_Y$; your prediction for $Y$ is $\mu_Y$ no matter what the value of $X$ is. The mean squared error is therefore $E\big{(}(Y-\mu_Y)^2\big{)} = \sigma_Y^2$, which is exactly what you get by plugging $r=0$ into the expression $(1 - r^2)\sigma_Y^2$.
This shows that when $X$ and $Y$ are uncorrelated there is no benefit in using linear regression to estimate $Y$ based on $X$. In this sense too, $r$ quantifies the amount of linear association between $X$ and $Y$.
In exercises you will see that it is possible for $X$ and $Y$ to be uncorrelated and have a very strong non-linear association. So it is important to keep in mind that $r$ measures only linear association.
The Residual is Uncorrelated with $X$
In Data 8 you learned to perform some visual diagnostics on regression by drawing a residual plot which was defined as a scatter diagram of the residuals and the observed values of $X$. We said that residual plots are always flat: "the plot shows no upward or downward trend."
We will now make this precise by showing that the correlation between $X$ and the residual $D$ is 0.
By the definition of correlation,
\begin{align*} r(D, X) ~ &= ~ E\Big{(} \big{(}\frac{D - \mu_D}{\sigma_D}\big{)}\big{(}\frac{X - \mu_X}{\sigma_X}\big{)} \Big{)} \\ \\ &= ~ \frac{1}{\sigma_D\sigma_X}E(DD_X) \end{align*}
because $\mu_D = 0$. Therefore to show $r(D, X) = 0$, we just have to show that $E(DD_X) = 0$. Let's do that.
\begin{align*} E(DD_X) ~ &= ~ E((D_Y - \hat{a}D_X)D_X) \\ &= ~ E(D_XD_Y) - \hat{a}E(D_X^2) \\ &= ~ r\sigma_X\sigma_Y - r\frac{\sigma_Y}{\sigma_X}\sigma_X^2 \\ &= ~ 0 \end{align*} | 2020-02-28T15:52:52 | {
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https://math.stackexchange.com/questions/1195423/how-do-i-evaluate-an-integral-by-interpreting-it-in-terms-of-areas | # How do I evaluate an integral by interpreting it in terms of areas?
I'm really having trouble understanding this question. The definite integral is:
I solved it for its areas and got -30 because the area between 7 and 9 on the x axis contains a rectangle and a triangle, the rectangle has a base of 2 and a height of twelve while the triangle also has a base of 2 but a height of 6. The area is negative due to the area being below the x-axis. However my answer is incorrect and I am at a loss as to how to correctly do this.
• wolframalpha.com/input/… – JP McCarthy Mar 18 '15 at 13:16
• Are you sure that's it? I put it into a graphing calculator and got something different – JMartinez Mar 18 '15 at 13:18
• ...that is the graph yes so the area is not below the $x$-axis. – JP McCarthy Mar 18 '15 at 13:18
• The above is the graph of $9+3x$ so indeed the area is above the $x$-axis. I imagine you are looking for the answer 66 – Dean Barber Mar 18 '15 at 13:18
• ...well it is the line of (positive) slope $3$ and $y$-intercept $9$ so will certainly be positive for $x>0$. – JP McCarthy Mar 18 '15 at 13:19
On evaluation, it yields: $$9(9) + 1.5(9^2) - 9(7) - 1.5(7^2) = 66$$ There must be something wrong with your calculation. Please do check it...
• May I ask where you got the 1.5 from? – JMartinez Mar 18 '15 at 13:20
• Also you were right I had to fix my calculator – JMartinez Mar 18 '15 at 13:22
• $\int 3x = \frac{3x^2}{2} = 1.5(x^2)$ – Kugelblitz Mar 18 '15 at 13:22
• It's okay.. no problem. Also, nicely stated question.. +1 :) – Kugelblitz Mar 18 '15 at 13:22
• thank you, for both the explanation and the compliment :) – JMartinez Mar 18 '15 at 13:23
The area is not under the x-axis because this function is positive between 7 and 9.
• The area of the rectangle is $30*2=60$ and the area of the triangle is $6*2/2=6$, so $60 + 6 = 66$ – alex14204 Mar 18 '15 at 13:23
The area in question can be taken to be that of a trapezium whose area is given
$$\frac{1}{2}(a+b)\cdot h = \frac{1}{2} (30 + 36)\cdot 2 = 66$$
Or you can interpret the figure as a rectangle and triangle. In this case you get
$$\text{Area}~~ = 2\cdot 30 + \frac{1}{2} \cdot 2 \cdot 6 = 66$$ | 2019-08-22T17:46:23 | {
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https://studyalgorithms.com/array/find-the-majority-element-in-an-array-method-3/ | Home Arrays Find the majority element in an array. (Method 3)
# Find the majority element in an array. (Method 3)
Question: An element is a majority if it appears more than n/2 times. Give an algorithm that takes an array of n elements and finds the majority element in that array. The array is not sorted.
Input: 8, 16, 8, 4, 8, 8, 8, 42, 16, 15, 8, 23, 4, 8, 15, 8, 8.
Output: Majority Element = 8
We discussed the basic 2 methods to approach this problem here:-
It is recommended to have a look at these 2 methods before proceeding with this one. This is the most optimized method and with least space complexity.
Since only one majority element is possible, we can use a simple scan of the input array and keep a track of the count of elements. We basically do not need to keep a count of every element in the array. We first find out the element, that is occurring the maximum times in an array and then we check if its the majority element.
Here is the process:-
• Start from the first element of the array and initialize the count to be = 1, and set it as the majority candidate.
• Start scanning the array.
• If we get the same element as the majority candidate, then increment the count.
• If we get a different element, decrease the count by 1.
• At any moment, if the count is set to 0, that means we have a new majority candidate. Change the majority candidate to the current element and reset the count to 1. This is also known as the Moore’s Voting Algorithm. “Basic idea of the algorithm is if we cancel out each occurrence of an element e with all the other elements that are different from e then e will exist till end if it is a majority element. “
• At the end of the array we will get a majority candidate.
• Now that we have a majority candidate, just scan through the array and see if it appears more than (n/2) times. If yes, this is our majority element.
Here is the implementation of the above algorithm.
#include<stdio.h>
// function to find the majority candidate in a array arr
int findMajorityCandidate(int arr[], int size)
{
// maj_index - to keep a track of majority candidate
int maj_index = 0, count = 1;
int i;
for(i = 1; i < size; i++)
{
// if the element is same as majority candidate
// increment count
if(arr[maj_index] == arr[i])
{
count++;
}
// else, decrease count
else
{
count--;
}
// at any time if count becomes 0
if(count == 0)
{
// change the majority cadidate
maj_index = i;
count = 1;
}
}
// return the majority candidate
return arr[maj_index];
}
// a function to print the majority element
void printMajorityElement(int arr[], int size)
{
// find the majority element
int candidate = findMajorityCandidate(arr,size);
int i, count = 0;
// count the number of occurrences
for (i = 0; i < size; i++)
{
if(arr[i] == candidate)
count++;
}
if (count > size/2)
printf("Majority element = %d",candidate);
else
printf("No majority element found");
}
int main(void)
{
int arr[] = {8, 16, 8, 4, 8, 8, 8, 42, 16, 15, 8, 23, 4, 8, 15, 8, 8};
printMajorityElement(arr, 17);
return 0;
}
Time Complexity:- O(n)
Space Complexity:- O(1)
#### You may also like
April 1, 2015 - 09:07
First part of the method will not give correct candidate if there is no majority element. In case of (2,2,2,3,4,5,6,7) it will return index 7 but it is supposed to return index 0.
April 2, 2015 - 01:58
Hi viki,
You have not understood the solution properly. You need to scan the array once again to check for the majority element.
The solution is absolutely correct. Here is the running link with your input:-
http://ideone.com/sV31xR
March 20, 2015 - 02:50
It can be solved with bocketSort, although it’s not the best space complexity.
March 21, 2015 - 12:38
Hi Igor,
I will be glad, if you could post a solution using Bucket sort. | 2022-07-01T17:31:55 | {
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https://math.stackexchange.com/questions/1557771/probability-of-the-same-pair-of-balls-drawn-from-two-separate-urns | Probability of the Same Pair of Balls Drawn from Two Separate Urns
This morning, my friends and I discussed following problem.
Problem:
There are two persons named Mr. A and Mr. B. Each person has his own urn containing $N$ different balls. They uniformly randomly draw a ball twice with replacement from their own urns.
What is the probability that they draw the same pair of balls?
Example:
Let $N = 3$ and let's label them with integer $i$, where $1 \leq i \leq N$. Let $A_k$ and $B_k$, where $k \in \{1,2\}$, be the events when Mr. A and Mr. B draw ball $i$ at the $k$ drawing, respectively.
The pair $((A_1,B_1),(A_2,B_2))$ denotes an outcome from the drawing process.
Events of interest are, for example, $((1,1),(2,2)), ((1,1),(3,3)), \mathrm{or}~ ((1,2),(2,1)).$
Number of sample space $|\Omega|$ is $\binom{N}{2}\times\binom{N}{2}$. Number of possible outcomes is $\binom{N}{2}$. The probability is $\frac{1}{\binom{N}{2}}$.
$|\Omega| = N^4$.
Let $X$ be an event where they both draw the same ordered pair of balls, e.g., $((1,1),(2,2)) \mathrm{or}~ ((1,1),(3,3)).$
Let $Y$ be an event where they both draw the same "cross-ordered" pair of balls, e.g., $((1,2),(2,1)) \mathrm{or}~ ((1,3),(3,1)).$
$|X| = N^2$, and $|Y| = N\times(N-1).$
Hence, the probability is $\frac{N^2 + N\times(N-1)}{N^4}$.
Question:
Is either of these answers correct?
• It is with replacement, so the second is right. – André Nicolas Dec 3 '15 at 7:24
• Thanks. I also agree with the second answer. – Ardevara Dec 3 '15 at 12:12
We denote with $[N]:=\{1,2,3,\ldots,N\}$ and consider all $N^4$ tuples in \begin{align*} \mathcal{A}=\{((A_1,B_1),(A_2,B_2))|A_j,B_j\in[N],j=1,2\} \end{align*}
We denote with $E(A_j=B_k), 1\leq j,k\leq 2$ the event that balls $A_j$ and $B_k$ are drawn and are equal.
Using the inclusion-exclusion principle we can calculate the number of events of drawing equal pairs. We obtain \begin{align*} &\#\left(E(A_1=B_1)\cap E(A_2=B_2)\right)\\ &\qquad+\#\left(E(A_1=B_2)\cap E(A_2=B_1))\right)\\ &\qquad-\#\left(E(A_1=B_1)\cap E(A_2=B_2)\cap E(A_1=B_2) \cap E(A_2=B_1)\right)\\ &=|\{(A_1,A_1),(A_2,A_2)|A_1,A_2\in[N]\}|\\ &\qquad+|\{(A_1,A_2),(A_2,A_1)|A_1,A_2\in[N]\}|\\ &\qquad -|\{(A_1,A_1),(A_1,A_1)|A_1\in[N]\}|\\ &=N^2+N^2-N\\ &=2N^2-N \end{align*}
The number of all events according to OPs rule is $|[N]|^4=N^4$. We conclude the probability $P(N)$ that Mr. A and Mr. B draw equal pairs of balls (with replacement) from urns containing $N$ balls is
\begin{align*} P(N)=\frac{2N-1}{N^3} \end{align*}
There are $\binom{N}{2}$ pairs. There is 1 way to order the match. The probability of getting a match for a particular pair is $(1/\binom{N}{2})^2$. Thus, $$P(\text{Match}) = 1\cdot\binom{N}{2}\frac{1}{\binom{N}{2}}\frac{1}{\binom{N}{2}} = \frac{1}{\binom{N}{2}}.$$
• Could you please give me some examples based on your reasonings? – Ardevara Dec 3 '15 at 8:39
• Actually, my answer should be wrong because I forgot to add one step back in. Essentially I reduced it to a dice matching problem. I'm missing one thing, but I can't think of it at the moment. Meaning I have to multiply this answer by something. – Em. Dec 3 '15 at 9:06
• Any comments are very welcome. I'll patiently wait your next revision. Thanks – Ardevara Dec 3 '15 at 11:09
I think the answer should be $\frac{2}{N^2}$ because the probability that they draw balls in the same order is $\frac{1}{N^2}$ and I do not see why it'd be different for the cross-ordered case as we can swap the order of drawing without any affect to the outcome. And as they're independent cases, we can say that the chance that one of them occurs is $\frac{2}{N^2}$.
• You cannot form the crossed-order pair of balls if they draw the same balls at the first drawing. For example, at the first drawing Mr. A draws ball $2$ ($A_1 = 2$), and Mr. B draws ball $2$ ($B_1 = 2$). In this case there is no way you can form the crossed-order pair of balls, e.g., $((1,2),(2,1))$. That's why the number of the event $Y$ happens is $N\times(N-1)$. – Ardevara Dec 3 '15 at 11:07 | 2019-07-18T11:13:10 | {
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https://www.mathworks.com/help/symbolic/units-of-measurement-tutorial.html | Documentation
## Units of Measurement Tutorial
Use units of measurement with Symbolic Math Toolbox™. This page shows how to define units, use units in equations (including differential equations), and verify the dimensions of expressions.
### Define and Convert Units
u = symunit;
Specify a unit by using u.unit. For example, specify a distance of 5 meters, a weight of 50 kilograms, and a speed of 10 kilometers per hour. In displayed output, units are placed in square brackets [].
d = 5*u.m
w = 50*u.kg
s = 10*u.km/u.hr
d =
5*[m]
w =
50*[kg]
s =
10*([km]/[h])
### Tip
Use tab expansion to find names of units. Type u., press Tab, and continue typing.
Units are treated like other symbolic expressions and can be used in any standard operation or function. Units are not automatically simplified, which provides flexibility. Common alternate names for units are supported. Plurals are not supported.
Add 500 meters and 2 kilometers. The resulting distance is not automatically simplified.
d = 500*u.m + 2*u.km
d =
2*[km] + 500*[m]
Simplify d by using simplify. The simplify function automatically chooses the unit to simplify to.
d = simplify(d)
d =
(5/2)*[km]
Instead of automatically choosing a unit, convert d to a specific unit by using unitConvert. Convert d to meters.
d = unitConvert(d,u.m)
d =
2500*[m]
There are more unit conversion and unit system options. See Unit Conversions and Unit Systems.
Find the speed if the distance d is crossed in 50 seconds. The result has the correct units.
t = 50*u.s;
s = d/t
s =
50*([m]/[s])
### Use Temperature Units in Absolute or Difference Forms
By default, temperatures are assumed to represent differences and not absolute measurements. For example, 5*u.Celsius is assumed to represent a temperature difference of 5 degrees Celsius. This assumption allows arithmetical operations on temperature values.
To represent absolute temperatures, use kelvin, so that you do not have to distinguish an absolute temperature from a temperature difference.
Convert 23 degrees Celsius to kelvin, treating it first as a temperature difference and then as an absolute temperature.
u = symunit;
T = 23*u.Celsius;
diffK = unitConvert(T,u.K)
diffK =
23*[K]
absK = unitConvert(T,u.K,'Temperature','absolute')
absK =
(5923/20)*[K]
### Verify Dimensions
In longer expressions, visually checking for units is difficult. You can check the dimensions of expressions automatically by verifying the dimensions of an equation.
First, define the kinematic equation ${v}^{2}={v}_{0}{}^{2}+2as$, where v represents velocity, a represents acceleration, and s represents distance. Assume s is in kilometers and all other units are in SI base units. To demonstrate dimension checking, the units of a are intentionally incorrect.
syms v v0 a s
u = symunit;
eqn = (v*u.m/u.s)^2 == (v0*u.m/u.s)^2 + 2*a*u.m/u.s*s*u.km
eqn =
v^2*([m]^2/[s]^2) == v0^2*([m]^2/[s]^2) + (2*a*s)*(([km]*[m])/[s])
Observe the units that appear in eqn by using findUnits. The returned units show that both kilometers and meters are used to represent distance.
findUnits(eqn)
ans =
[ [km], [m], [s]]
Check if the units have the same dimensions (such as length or time) by using checkUnits with the 'Compatible' input. MATLAB® assumes symbolic variables are dimensionless. checkUnits returns logical 0 (false), meaning the units are incompatible and not of the same physical dimensions.
checkUnits(eqn,'Compatible')
ans =
logical
0
Looking at eqn, the acceleration a has incorrect units. Correct the units and recheck for compatibility again. eqn now has compatible units.
eqn = (v*u.m/u.s)^2 == (v0*u.m/u.s)^2 + 2*a*u.m/u.s^2*s*u.km;
checkUnits(eqn,'Compatible')
ans =
logical
1
Now, to check that each dimension is consistently represented by the same unit, use checkUnits with the 'Consistent' input. checkUnits returns logical 0 (false) because meters and kilometers are both used to represent distance in eqn.
checkUnits(eqn,'Consistent')
ans =
logical
0
Convert eqn to SI base units to make the units consistent. Run checkUnits again. eqn has both compatible and consistent units.
eqn = unitConvert(eqn,'SI')
eqn =
v^2*([m]^2/[s]^2) == v0^2*([m]^2/[s]^2) + (2000*a*s)*([m]^2/[s]^2)
checkUnits(eqn)
ans =
struct with fields:
Consistent: 1
Compatible: 1
After you finish working with units and only need the dimensionless equation or expression, separate the units and the equation by using separateUnits.
[eqn,units] = separateUnits(eqn)
eqn =
v^2 == v0^2 + 2000*a*s
units =
1*([m]^2/[s]^2)
You can return the original equation with units by multiplying eqn with units and expanding the result.
expand(eqn*units)
ans =
v^2*([m]^2/[s]^2) == v0^2*([m]^2/[s]^2) + (2000*a*s)*([m]^2/[s]^2)
To calculate numeric values from your expression, substitute for symbolic variables using subs, and convert to numeric values using double or vpa.
Solve eqn for v. Then find the value of v where v0 = 5, a = 2.5, and s = 10. Convert the result to double.
v = solve(eqn,v);
v = v(2); % choose the positive solution
vSol = subs(v,[v0 a s],[5 2.5 10]);
vSol = double(vSol)
vSol =
223.6627
### Use Units in Differential Equations
Use units in differential equations just as in standard equations. This section shows how to use units in differential equations by deriving the velocity relations v = v0 + at and ${v}^{2}={v}_{0}{}^{2}+2as$ starting from the definition of acceleration $a=\frac{dv}{dt}$.
Represent the definition of acceleration symbolically using SI units. Given that the velocity V has units, V must be differentiated with respect to the correct units as T = t*u.s and not just t.
syms V(t) a
u = symunit;
T = t*u.s; % time in seconds
A = a*u.m/u.s^2; % acceleration in meters per second
eqn1 = A == diff(V,T)
eqn1(t) =
a*([m]/[s]^2) == diff(V(t), t)*(1/[s])
Because the velocity V is unknown and does not have units, eqn1 has incompatible and inconsistent units.
checkUnits(eqn1)
ans =
struct with fields:
Consistent: 0
Compatible: 0
Solve eqn1 for V with the condition that the initial velocity is v0. The result is the equation v(t) = v0 + at.
syms v0
cond = V(0) == v0*u.m/u.s;
eqn2 = V == dsolve(eqn1,cond)
eqn2(t) =
V(t) == v0*([m]/[s]) + a*t*([m]/[s])
Check that the result has the correct dimensions by substituting rhs(eqn2) into eqn1 and using checkUnits.
checkUnits(subs(eqn1,V,rhs(eqn2)))
ans =
struct with fields:
Consistent: 1
Compatible: 1
Now, derive ${v}^{2}={v}_{0}{}^{2}+2as$. Because velocity is the rate of change of distance, substitute V with the derivative of distance S. Again, given that S has units, S must be differentiated with respect to the correct units as T = t*u.s and not just t.
syms S(t)
eqn2 = subs(eqn2,V,diff(S,T))
eqn2(t) =
diff(S(t), t)*(1/[s]) == v0*([m]/[s]) + a*t*([m]/[s])
Solve eqn2 with the condition that the initial distance covered is 0. Get the expected form of S by using expand.
cond2 = S(0) == 0;
eqn3 = S == dsolve(eqn2,cond2);
eqn3 = expand(eqn3)
eqn3(t) =
S(t) == t*v0*[m] + ((a*t^2)/2)*[m]
You can use this equation with the units in symbolic workflows. Alternatively, you can remove the units by returning the right side using rhs, separating units by using separateUnits, and using the resulting unitless expression.
[S units] = separateUnits(rhs(eqn3))
S(t) =
(a*t^2)/2 + v0*t
units(t) =
[m]
When you need to calculate numeric values from your expression, substitute for symbolic variables using subs, and convert to numeric values using double or vpa.
Find the distance traveled in 8 seconds where v0 = 20 and a = 1.3. Convert the result to double.
S = subs(S,[v0 a],[20 1.3]);
dist = S(8);
dist = double(dist)
dist =
201.6000 | 2020-01-21T04:11:48 | {
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https://math.stackexchange.com/questions/1611542/proving-sum-of-product-forms-a-pattern-in-n-nnnnnn | # Proving sum of product forms a pattern in n * nnnnnn…
I am consider a problem regarding numbers which are, in decimal, one digit repeated - for instance, $88888888$ is such a number. In particular, I am looking at the following problem:
The sum of the digits of the number $$8\cdot \underbrace{88\ldots 88}_{n\text{ times}}$$ is $1000$. How many $8$'s are there? (i.e. what is $n$?)
I know that the correct answer is $991$ and I can observe a pattern that lets me solve it; in particular, the sum of digits of $8\cdot 8=64$ is $10$ and the sum of digits of $8\cdot 88=704$ is $11$ and the sum of digits of $8\cdot 888=7104$ is $12$ - so it appears that the sum of the digits is exactly $n+9$. One can note that if one replaces all the eights with other digits, similar patterns exist - for instance, with sevens, the sum increases by $4$ for each digit.
How can this pattern be proven?
• Please include your thoughts and efforts in this and future posts. Formatting tips here. – Em. Jan 14 '16 at 1:50
• @probablyme Thanks for your advice! Just getting used to stackexchange. I am very new to proof and I don't know where to start in trying to prove this. That is why I posted it here. – London Holmes Jan 14 '16 at 1:52
• I made a substantial edit since I think this is an interesting observation and I hope it will be well received here with better formatting; feel free to change/undo whatever I might have done if it conflicts with what you intended. – Milo Brandt Jan 14 '16 at 1:58
• @MiloBrandt Can't accept your edits yet because I don't have enough rep but I think they greatly helped. Thanks! – London Holmes Jan 14 '16 at 2:01
• Refering the the number as 8 when clearly it is an unknown digit is REALLY confusing. Why don't you refer to it as "a" or some other variable? – fleablood Jan 14 '16 at 2:20
One can note a more general pattern - let us, for convenience, define $$R_n=\underbrace{11\ldots 11}_{n\text{ times}}=\frac{10^n-1}9.$$ Then, we can note that, for fixed $k$, after a point, the sum of the digits of $k\cdot R_n$ increases linearly. Note that, if the digits were $8$'s, then we can express the desired product as $8^2\cdot R_n$ since multiplying by $8$ once gives $88\ldots 88$ and doing it again gives $8\cdot 88\ldots 88$. One will see the pattern if they go far enough: $$64\cdot 1 = 64$$ $$64\cdot 11 = 704$$ $$64\cdot 111 = 7104$$ $$64\cdot 1111 = 71104$$ $$64\cdot 11111 = 711104$$ $$64\cdot 111111 = 7111104$$ $$64\cdot 1111111 = 71111104$$ $$64\cdot 11111111 = 711111104$$ So, obviously, there's a pattern here - which is that we get a new $1$ inserted after the $7$ each time we add another $1$ on the left.
There are a few productive ways to prove this. The one I think is most elegant would be simply to write this as long multiplication: $$\begin{array} & & & 1 & 1 & 1 & \ldots & 1 & 1 & 1 \\ \times & & & & & & & 6 & 4 \\ \hline && 4 & 4 & 4 &\ldots & 4 & 4 & 4 \\ +&6 & 6 & 6 & 6 &\ldots & 6 & 6 & 0\\\hline \end{array}$$ and now we can carry out this addition by hand too, writing out carries in the small text above: $$\begin{array} & &\scriptstyle{1} & \scriptstyle{1} & \scriptstyle{1} & \scriptstyle{1} &\ldots & \scriptstyle{1} & & \\ & & 4 & 4 & 4 &\ldots & 4 & 4 & 4 \\ +&6 & 6 & 6 & 6 &\ldots & 6 & 6 & 0\\\hline & 7 & 1 & 1 & 1 &\ldots & 1 & 0 & 4 \end{array}$$ One may notice that every column elided looks exactly like this: $$\begin{array}{c} &\scriptstyle{1} \\ &4 \\ (+)&6\\ \hline & 1\end{array}$$ and, obviously, caused a $1$ to be carried to the column to the right of it. This suffices to prove that, once these columns begin, they will not end until we reach the end of the $4$'s in the top column - this is proved inductively, if we wish to be formal. Then, once we have this, we know that the beginning (here $7$) will always be the same, as it always looks like: $$\begin{array}{c} &\scriptstyle{1} \\ & \\ (+)&6\\ \hline & 7\end{array}$$ and the ending will always look like: $$\begin{array}{c} & \\ &4&4 \\ (+)&6&0\\ \hline (1) & 0& 1\end{array}$$ where the $1$ in parenthesis is carried, causing the inner columns to begin. This proof can be redone in whole to show the property for any sequence of the form $k\cdot 11\ldots 11$. With a bit more care, we can note that a sequence of identical columns begin after at most $2\lceil\log_10(k)\rceil$ digits (for a rough bound) since the carry to each such column is an increasing function of the carry to the previous column. Since the carry does not exceed the number of digits of $k$, it must hit a fixed point after at most $\lceil\log_{10}(k)\rceil$ steps and this follows the "irregular" portion of the sum (when zeros are still being added from some terms), which has length at most $\lceil\log_{10}(k)\rceil$ as well.
More directly, however, would be an algebraic approach to prove some identity of the form: $$k\cdot R_n = \alpha \cdot 10^n + \beta \cdot R_{n-m} \cdot 10^m + \kappa$$ for positive integers $\alpha$, $\beta$, $m$, and $\kappa$ with $\kappa < 10^m$ and $0\leq \beta < 10$. Knowing the form of $R_n=\frac{10^n-1}{9}$ makes it easy to verify such an identity - for instance: $$64\cdot R_n = 7\cdot 10^n + 1\cdot R_{n-2} \cdot 10^2 + 4$$ $$64\cdot \underbrace{11\ldots 11}_{n\text{ times}}=7\underbrace{11\ldots 11}_{n-2\text{ times}}04$$ holds, as can be checked purely algebraically. Note that this expresses that the last few digits are the digits of $\alpha$ (which is $7$ in the example) followed by a number of repetition of $\beta$ (which is $1$ in the example) followed by the digits of $\kappa$ (which is $04$ in the example). Obviously, the sum of the digits will increase by $\beta$ at each step, since the only change in the expression is the insertion of another one of that digit.
To find such an expression, one extracts things in parts. For instance, for $64\cdot R_n=\frac{64\cdot 10^n-64}9$, we first take out the largest multiple of $10^n$ that we can which is $\lfloor\frac{64}9\rfloor=7$. Subtracting this out gives $$\frac{64\cdot 10^n-64}{9}-7\cdot 10^n=\frac{1\cdot 10^n-64}9$$ From here, we want the numerator to be of the form $\beta(10^n - 10^m) + 9\kappa$ which is easy to do. Here, for instance, we just need to choose $m$ such that $10^m$ is bigger than $64$ - so $m=2$ is the smallest that suffices. This gives $$\frac{64\cdot 10^n-64}{9}-7\cdot 10^n=\frac{1\cdot (10^n-10^2)+36}9=1\cdot R_{n-2}\cdot 10^2+4$$ which yields the expression when we add the $7\cdot 10^n$ back to both sides.
• Well, that was fun. – Milo Brandt Jan 14 '16 at 2:49
$a*aaaa....aaa = a^2*1111111....111$ (m a's and m 1's)
Three cases to consider:
$a^2$ has one digit. (i.e. $a = 1,2,3$)
$a^2$ has two digits and the sum of the digits is less than 10. (i.e. $a = 4,5,6, 9$)
$a^2$ has two digits and the sum of the digits is 10 or more. (i.e. $a = 7,8$)
====
If $a^2 = b$ has one digit,$b$, then $a*aaa... = a^2*111... = bbbbb$ and the sum of the digits is $m*b$. That was easy.
===
If $a^2 = 10b + c$ has two digits, $b$ and $c$, and $b + c < 10$ then: $a*aaa... = a^2*1,1,1... = (10b+c),(10b + c),...,(10b + c) = b,(b+c),(b+c),...,(b + c), c$.
[Here I use notation x,x,x ... as shorthand for $\sum x*10^i$. In the $b,(b+c),...,(b+c), c$ each term is a single digit although it wasn't for the $(10b+c),(10b+c)...$ representation. But from that one we "carried the b's" to get the final representation.]
The sum of the digits $b,(b+c),(b+c),...,(b + c), c$ is $b + (m-1)(b+c) + c= m(b+c)$.
===
If $a^2 = 10b + c$ is two digits ($b$ and $c$) and $b + c = 10 + d$ for some digit d...
First note that $d < 9$ so $d + 1 < 10$.
So $a*a*1111... = a^2*1111.... = (10b +c),(10b + c), .... (10b + c),(10b + c) = b,(b + c),....(b+c),c = b,(10 + d),(10 + d), .....(10 + d),c = (b+1),(d+1),(d+1),....(d+1),d,c$.
And the sum of the digits is $b + 1 + (m-2)(d+1) + d + c = b + c + d+ (m-2)(d+1) = 11 + 2d + (m-2)(d +1) = m(d+1) + 9$.
====
In summary: Sum of digits = $m*e \{+ 9 \}$ where $m$ is the number of digits and $e$ is the $a^2$, sum of digits of $a^2$, or sum of sum of digits of $a^2$ and $" + 9"$ applies only if sum of digits of $a^2 \ge 10$.
And hence, each increase in $m$ the sums increase by $e$.
===
And, BTW, back to the original problem: Su of digits of $a*aaaa....$ = 1000- how many digits are there:
Sum of digits is $m*e \{+9\} = 1000$. $e$ can be: $a=1 \implies e =1;a=2 \implies e =4;a=3 \implies e =9;a=4 \implies e =7;a=5 \implies e =7;a=6 \implies e =9;a=7 \implies e =4;a=8 \implies e =1; a=9 \implies e =9;$
$m*e = \{1000, 991\}$. $991$ is primes so if $m*e = 991$ then $e = 1; a=8; m=991$. If $m*e = 1000$ then $e$ is 1 or a multiple of 2 or 5 only, so possibilities:
$e = 1;a = 1; m = 1000;$
$e = 4; a = 2; m = 250;$
$e = 4; a =7$ is not possible as that would imply $m = 991$. | 2019-08-20T13:37:29 | {
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http://www.b254.com/factoring/2bsquare2plus4bminus15.html | ### Factoring 2b^2+4b-15 Solution
The variable we want to find is b
We will solve for b using quadratic formula -b +/- sqrt(b^2-4ac)/(2a), graphical method and completion of squares.
${x}_{}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Where a= 2, b=4, and c=-15
Applying values to the variables of quadratic equation -b, a and c we have
${b}_{}=\frac{-4±\sqrt{{4}^{2}-4x\mathrm{2x}\mathrm{-15}}}{2x2}$
This gives
${b}_{}=\frac{-4±\sqrt{{4}^{2}-\mathrm{-120}}}{4}$
${b}_{}=\frac{-4±\sqrt{136}}{4}$
${b}_{}=\frac{-4±11.6619037897}{4}$
${\mathrm{b1}}_{}=\frac{-4+11.6619037897}{4}$
${\mathrm{b2}}_{}=\frac{-4-11.6619037897}{4}$
${b}_{1}=\frac{7.66190378969}{4}$
${b}_{1}=\frac{-15.6619037897}{4}$
The b values are
b1 = 1.91547594742 and b2 = -3.91547594742
### Factoring Quadratic equation 2b^2+4b-15 using Completion of Squares
2b^2+4b-15 =0
Step1: Divide all terms by the coefficient of b2 which is 2.
${b}^{2}+\frac{4}{2}x-\frac{15}{2}=0$
Step 2: Keep all terms containing x on one side. Move the constant to the right.
${b}^{2}+\frac{4}{2}b=\frac{15}{2}$
Step 3: Take half of the x-term coefficient and square it. Add this value to both sides.
${b}^{2}+\frac{4}{2}b+{\left(\frac{4}{4}\right)}^{2}=\frac{15}{2}+{\left(\frac{4}{4}\right)}^{2}$
Step 4: Simplify right hand sides of expression.
${b}^{2}+\frac{4}{2}b+{\left(\frac{4}{4}\right)}^{2}=\frac{136}{16}$
Step 2: Write the perfect square on the left.
${\left(b+\frac{4}{4}\right)}^{2}=\frac{136}{16}$
Step 2: Take the square root on both sides of the equation.
$b+\frac{4}{4}=±\sqrt{\frac{136}{16}}$
Step 2: solve for root b1.
${b}_{1}=-\frac{4}{4}+\frac{11.6619037897}{4}=\frac{7.66190378969}{4}$
${b}_{1}=1.91547594742$
Step 2: solve for root b2.
${b}_{2}=-\frac{4}{4}-\frac{11.6619037897}{4}=\frac{-15.6619037897}{4}$
${b}_{2}=-3.91547594742$
### Solving equation 2b^2+4b-15 using Quadratic graph
b2 + b + = 0
Solutions
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Polynomials can be factored using this factoring calculator
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Trinomials can be solved using our quadratic solver
Can this be used for factoring receivables, business, accounting, invoice, Finance etc
No this cannot be used for that
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Other Variants of 2b^2+4b-15 are below | 2017-06-28T20:45:58 | {
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https://math.stackexchange.com/questions/3067503/calculate-int-c-fracxx2y2-dx-fracyx2y2-dy-where-c-is-s | # Calculate $\int_{C} \frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting $(1,1)$ to $(2,2)$
Calculate $$\int_{C} \frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy~$$ where $$C$$ is straight line segment connecting $$(1,1)$$ to $$(2,2)$$
my question is , after calculating the integral using green theorem i got that $$\int_{C} \frac{x}{x^2+y^2} dx \frac{y}{x^2+y^2} dy= -\ln(2)$$
is it the right answer ? since we are connecting $$(1,1)$$ to $$(2,2)$$ AND NOT $$(2,2)$$ to $$(1,1)$$
so its question about the sign of the value.
• Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function. – B. Goddard Jan 9 at 14:36
• I closed it with parametrization – Mather Jan 9 at 16:02
• You should add that to your answer so we can see what you did wrong. – B. Goddard Jan 9 at 16:28
• Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero. – B. Goddard Jan 9 at 18:30
• but someone posted that the answer is $\ln(2)$ @B.Goddard – Mather Jan 9 at 19:46
$$(1,1),(2,2)$$ are joined by the line-segment $$C:y=x\in[1,2]$$. The integral becomes $$\int_C\frac{xdx+ydy}{x^2+y^2}=\int_C\frac{2xdx}{2x^2}=\int_1^2\frac{dx}x=\ln(2)$$
Alternatively, $$\int_C\frac{xdx+ydy}{x^2+y^2}=\int_C\frac12\cdot\frac{d(x^2+y^2)}{x^2+y^2}=\frac12\int_2^8\frac{dm}m=\frac12\ln(m)\Big|_2^8=\ln(2)$$where $$m=x^2+y^2$$, that goes from $$1^2+1^2\to2^2+2^2$$.
The fundamental theorem of calculus tells you that if $${\bf F} = \nabla f$$ in a simply connected region containing the curve, then $$\int_C {\bf F}\cdot d{\bf r}= f(b) - f(a)$$ where the curve $$C$$ begins at the point $$a$$ and ends at the point $$b$$.
Here $${\bf F}(x,y) = \left(\frac{x}{x^2 + y^2}, \frac y{x^2 + y^2} \right).$$ Can you find a function $$f(x,y)$$ such that $${\bf F}(x,y) = \nabla f(x,y)?$$ | 2019-10-16T05:17:10 | {
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http://civilservicereview.com/2014/05/mdas-quiz-solution/ | ## MDAS Quiz with Solution
The order of operations namely multiplication, division, addition or subtraction or more popularly known as MDAS or PEMDAS is one of the most basic concepts in mathematics and yet many people are totally confused about it. Here is a 15-item quiz with solution to further your understanding about MDAS. Note that the rules are
1.) perform the operation inside the parenthesis
2.) perform the operation with exponent
3.) perform multiplication and division first before performing addition and subtraction
4.) if multiplication and division are adjacent operations, perform from left to right
5.) if addition and subtraction are adjacent operations, perform from left to right
Take note of rule 2 and 3 because this is the most common misconceptions of many. Many believes that since in MD, multiplication should be done first since M is before D.
MDAS Quiz
1.) $5 \times 3 + 2$
a.) 17
b.) 25
Multiply 5 and 3 first, this gives us 15, and then add 2. [/toggle]
2.) $8 + 4 \times 3$
a.) 36
b.) 20
Multiply 4 and 3 first before adding to 8.[/toggle]
3.) $3 \times 2 + 4 \times 5$
a.) 21
b.) 26
c.) 50
d.) 90
Perform the two multiplications first before performing addition. That is
$3 \times 2 + 4 \times 3 = 6 + 20 = 26$.
[/toggle]
4.) $2 + 4 \times 8 + 7$
a.) 41
b.) 55
c.) 71
d.) 90
Multiply 4 and 8 before adding 7 and 2 .[/toggle]
5.) $12 \div 4 - 3$
a.) 0
b.) 12
Perform division before subtraction.[/toggle]
6.) $5 \times (3 + 2)$
a.) 17
b.) 25
This is the same as number 1. This time, since there is a parenthesis, you should perform the operation inside the parenthesis first. So, you add 3 and 2 first, then multiply to 5.[/toggle]
7.) $3 \times (2 + 4) \times 5$
a.) 21
b.) 26
c.) 50
d.) 90
Perform the operation inside the parenthesis before anything else. So, add first 2 and 4. Then, multiply everything which gives 90.[/toggle]
8.) $3 + (9 \div 3) \times 6 - 4$
a.) 8
b.) 17
c.) 20
d.) 42
1.) Perform the operation inside the parenthesis first. That is,
$3 + (9 \div 3) \times 6 - 4 = 3 + 3 \times 6 - 4$
2.) Perform multiplication before addition and subtraction.
$3 + 3 \times 6 - 4 = 3 + 18 - 4$
3.) Only addition and subtraction is left, perform the operation from left to right.
$3 + 18 - 4 = 17$
[/toggle]
9.) $6 \div 2(1+2)$
a.) 1
b.) 9
Solution:
This one is very tricky.
First, perform the operation inside the parenthesis. This gives us $6 \div 2 \times 3$.
Second, by Rule 4, if multiplication and division are adjacent to each other, you should perform whichever comes first. This time, division should be performed before multiplication. That is
$6 \div 2 \times 3 = 3 \times 3 = 9$
[/toggle]
10.) What is $4 + (3^2 - 6 \times 5) \div 3$?
Solution
1.) Perform the operations inside the parenthesis first. Inside the parenthesis, we have exponentation, subtraction, and multiplication. We do the exponentiation first as stated in the rules above.
$4 + (3^2 - 6 \times 5) \div 3 = 4 + (9 - 6 \times 5) \div 3$
2.) Still, we have to simplify what is inside the parenthesis. We perform Multiplication first and then subtraction.
$4 + (9 - 6 \times 5) \div 3 = 4 + (9 - 30) \div 3 = 4 + (-21) \div 3$
3.) Next, we perform division before addition. That is,
$4 + (-21) \div 3 = 4 + -7 = -3$
.[/toggle]
Did you get it all right? For quizzes, go to the Practice Test page.
• ### lance
July 24, 2018
whats the answer in 2 + 3 + 3 × 11
• ### Civil Service Reviewer
July 29, 2018
2 + 3 + 3 × 11
2 + 3 + 33
=38
• ### sofia
November 29, 2018
2+3+3×11
=3×11+2+3+3
=33+2+3+3
=38
• ### Jhoe
June 17, 2019
What is the answer of this?
(25)^0 – (-2)^4 + (-3^4) – (-8)^2
July 1, 2019
3*11=33
33+2+3=38
• ### Dam
July 4, 2019
Q what is the answer of 2/3 \div 3 + 1.25? | 2021-05-17T20:05:47 | {
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https://math.stackexchange.com/questions/2436293/how-do-i-read-and-solve-equations-in-the-form-of-ab-c-mod-d | # How do I read and solve equations in the form of $a*b = c \mod d$?
I am trying to find the decryption key of a given RSA problem. I have never solved equations using modulus, and I cannot seem to wrap my head around the equation to find the decryption key.
I am trying to solve this equation:
$43 * d = 1 mod 60$
I know the basics of modulus and therefore know that $1 \mod 60$ equals 1. I then, wrongly, substitute $1 mod 60$ by $1$ and simplify the equation like this:
$43 * d = 1$
However, I know that this is wrong, as I checked the answer sheet and there were a couple of answers, one being 301. I know that 301 is divisible by 43. The reasoning in the answer says "We need to find a number d that, when multiplied by 43 and divided by 60 leaves a remainder of 1." But in my head I ask "Why would we need to do that? I can solve 1 mod 60 and it is 1."
What did I do wrong by assuming $1 \mod 60$ equals $1$ and substituting that in the formula?
Edit: I have wrongly tagged this question as cryptography as I didn't know what tag to file this under. I welcome any edit that can fix the tag!
• Hint: Do you know about modular multiplicative inverse of an integer $a$ modulo $n$ ? – Prasun Biswas Sep 19 '17 at 16:52
• $43d \equiv 1 \mod{60}$ means that when $43d$ is divided by $60$ then the remainder is $1$. It does not mean that $43d = 1$ as $1 \equiv 1 \mod{60}$. – Math Lover Sep 19 '17 at 16:53
• If you could make a substitution like that, then we can say $43d=61$ too. When we say "a=b mod n", it means $a\equiv b\pmod n$, i.e., $a,b$ belong to the same equivalence class modulo $n$. "b mod n" alone isn't an expression in this context, which you can simply substitute like that. – Prasun Biswas Sep 19 '17 at 16:54
• I would like to stress that while the edit shows the correct form of writing the equation, my textbook explicitly did not use this notation as this was the source of my confusion. The equation from my textbook specifically wrote an equality sign and no braces around the mod. The current edit makes it seem like I ignored the brackets and congruence symbol .. – Zimano Sep 19 '17 at 17:34
• It's just notation. You happened to read the original equation as "43 times d is equal to (1 mod 60)", when what is intended is "(43 times d is congruent to 1), modulo 60". After the question was edited, this is more obvious from TeX's (MathJax's) spacing. The notation $43d \equiv 1 \mod {60}$ means that when you divide by $60$ and take the remainder, both $43d$ and $1$ leave the same remainder (namely, $1$). Thus $43d = 60k + 1$ for some $k$. – ShreevatsaR Sep 19 '17 at 17:34
The "equation" you have there should be written $43d\equiv 1\pmod{60}$. Note the symbol: three horizontal lines, not the two lines of equality. It is read as "$43d$ is congruent to $1$ modulo $60$". It means that $43d-1$ is an integer multiple of $60$. Solving this "congruence" amounts then to finding integers $d$ and $e$ with $43d-1=60e$. We can re-arrange this as $43d-60e=1$.
The extended Euclidean algorithm gives a way of solving such two-variable equations. In this example we find a solution $d=7$, $e=5$ which gives $d\equiv7\pmod{60}$ as the solution to the congruence.
• Now I understand! My textbook actually wrote it exactly as "Now, solve this equation to find d: 43 * d =1 mod 60" And they didn't use the symbol you used, nor the braces. Thanks! – Zimano Sep 19 '17 at 17:02
• @Zimano If this answer was useful to you, feel free to express your thanks by clicking on the up arrow to the left of the answer to upvote it (you can do this for multiple answers) and by clicking on the check (tick) mark to accept it (you can do this for only one answer). – ShreevatsaR Sep 19 '17 at 18:22
A possible way to solve such problem is to use the euclidian algorithm.
$$60= 1(43) + 17$$ $$43 =2(17) + 9$$
$$17 = 1(9) + 8$$
$$9 = 1(8) + 1$$
Hence we can write
\begin{align} 1 &= 9 - 1(8) \\ &=9-(17-9) \\ &=2(9) - 17 \\ &=2(43 -2(17))-17 \\ &= 2(43)-5(17) \\ &=2(43)-5(60-43) \\ &= 7(43) - 5(60) \end{align}
Now take $\pmod {60}$, we have $7 \times 43 \equiv 1 \pmod {60}$
Notice that $a \equiv 1 \pmod m$ doesn't mean $a=1$. It simply means when $a$ is divided by $m$, the remainder is $1$.
You work with congruences modulo $n$ as you work with equalities, with one big difference: you cannot always divide by a congruence class (it has to be a unit modulo $n$), and you cannot alway simplify – this means, explicitly, that $ab\equiv a'b$ does not necessarily imply $\equiv a'$, as the congurance is equivalent to $(a-a')b\equiv 0$, and the congruence class of $b$ can be a zero-divisor modulo $n$.
here, $43$ is a unit mod. $60$, as it it coprime to $60$. Furthermore its inverse mod. $60$ can be found from a Bézout's relation between $43$ and $60$.
Indeed the extended Euclidean algorithm yields $$7\cdot 43 -5\cdot 60 =1,$$ so the solution of your congruence equation is $\;\color{red}{d\equiv 7\mod 60}$.
Since $\gcd(43,60) = 1$ there exists $c,d$ that satisfy:
$43 d + 60 c = 1$
Now we apply the Euclidean algorithm. Here is how I think about it.
$\begin{bmatrix} 1\\&1\end{bmatrix}\begin{bmatrix}60\\43\end{bmatrix} =\begin{bmatrix}60\\43\end{bmatrix}$
Now what row opperations do I need to do to reduce the numbers on the right, eventually I will get to $1.$
And I just keep tacking on the additional rows.
$\begin{bmatrix} 1&0\\0&1\\1&-1\\-2&3\\-5&7\end{bmatrix}\begin{bmatrix}60\\43\end{bmatrix} =\begin{bmatrix}60\\43\\17\\9\\1\end{bmatrix}$
$7\cdot 43 = 1 + 5\cdot 60\\ d = 7$
• See this answer and its link for much further discussion of this linear-algebra-base view of the extended Euclidean algorithm, – Bill Dubuque Sep 19 '17 at 17:41
Intended is $\,43d\equiv 1\pmod{\!60}\,$ meaning $\,60\mid 43d-1,\,$ which is not equivalent to $\,43d = 1.\,$
Instead, $\quad (43d\bmod 60) = 1\,$ is a correct operational reformulation of the above congruence.
Be careful not to confuse $\!\bmod\!$ the ternary equivalence relation vs. binary operation (such confusion is greatly exacerbated by the abuse of notation used in your textbook)
There are many ways to compute modular inverses and fractions. One simple way is to employ the extended Euclidean algorithm in fractional form, e.g. below we use it to compute $\,\color{#c00}{1/43}\pmod{\!60}$
\dfrac{0}{60}\, \overset{\large\frown}\equiv\!\! \underbrace{\color{#c00}{\dfrac{1}{43}}\ \overset{\large\frown}\equiv \color{#90f}{\dfrac{-1}{17}}\ \overset{\large\frown}\equiv\ \color{#0a0}{\dfrac{4}{-8}}} _{\,\Large \begin{align}\color{#c00}{1}\ \ -\ \ &3(\color{#90f}{ -1 }) \ \ \equiv \ \ \ \color{#0a0}{4}\\ \color{#c00}{43}\ \ -\ \ &3(\color{#90f}{17} )\ \ \ \equiv\ \color{#0a0}{-8}\ \ \ \end{align}}\!\!\!\!\overset{\large\frown}\equiv\ \dfrac{7}1\qquad\qquad
Thus $\ \color{#c00}{1/43}\equiv 7\pmod{\!60}.\$See this answer for a few other useful methods.
Beware $\$ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus (or, more generally, in certain special contexts such as in the above algorithm). See here for further discussion. | 2019-07-17T15:28:55 | {
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http://tsgd.dely.pw/cubic-spline-interpolation-matrix.html | # Cubic Spline Interpolation Matrix
Find the cubic spline interpolation at x = 1. Cubic spline with natural boundary conditions. The most common case considered is k= 3, i. axis origin). Predict works as expected. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Maria Cameron 1. The matrix equation involved is solved analytically so that numerical inversion. Data scientists often use spline interpolation to produce smooth graphs and estimate missing values by "filling in" the space between discrete points of data. Optimal spline interpolation as sum of B-splines TB08 Computes knots of optimal spline interpolation TB15 Periodic spline that interpolates given function values; TC: Inverse interpolation TC01 Inverse interpolation for a cubic spline. It gives much lesser artifacts than "usual" cubic spline interpolation. Cubic spline interpolation with examples in Python 4. 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Example of the use of Spline(), Interp(), and Interpolate() functions. , in applications in graphics, numerical methods (e. Let fbe a function from. A little side-note: Bezier-Curves. In interpolating problems, spline interpolation is often preferred to polynomial interpolation because it yields similar results, even when using low-degree polynomials, while avoiding Runge's phenomenon for higher degrees. Spline curves can be partitioned into two types of curves: approximation splines and interpolation splines. Spline Interpolation Jakramate Bootkrajang y โดยใช cubic splinesi(x) = Constructing matrix H. One matrix contains the x-coordinates, and the other matrix contains the y-coordinates. Each cubic polynomial Sk(x) has four unknown constants (sk,0, sk,1, sk,2, and sk,3); hence there are 4N coefficients to be determined. Cubic Interpolation Vba Codes and Scripts Downloads Free. Our goal is to produce a function s(x) with the following. Understand what splines are Why the spline is introduced Approximating functions by splines We have seen in previous lecture that a function f(x) can be interpolated at n+1 points in an interval [a;b] using a single polynomial p n(x) de ned over the. We consider the problem of shape-preserving interpolation by cubic splines. methods, such as bilinear or bi-cubic interpolation, are based on interpolation over training data sampled on a uniform grid. If we compare Figs. Off / Linear / Cubic. For interp2, the full grid is a pair of matrices whose elements represent a grid of points over a rectangular region. First, let's look at the two B-spline algorithms, IBspline and Grid. The way of implementing this filter does not involve downsampling,. 2 Interpolation And Bootstrap Of Yield Curves—Not Two Separate Processes As has been mentioned, many interpolation methods for curve construc-tion are available. 'linear' - linear interpolation 'spline' - cubic spline interpolation 'cubic' - cubic interpolation All the interpolation methods require that X be monotonic. Variable spacing is handled by mapping the given values in X,Y, and XI to an equally spaced domain before interpolating. Part II: Cubic Spline Interpolation, Least Squares Curve Fitting, Use of Software Cubic Spline Interpolation, Least Squares Curve Fitting, Use of Software Cubic Spline Interpolation Basics Piecewise Cubic Constraint Equations Lagrangian Option to Reduce Number of Equations Least-Squares Curve Fitting Linear Regression Linear Regression Example. INTRODUCTION Cubic spline interpolation is a widely-used polynomial intepolation method for functions of one variable. The de Boor algorithm also permits the subdivision of the B-spline curve into two segments of the same order. [email protected]_:5m( @fcadb:56 r dbo :5m c b j 5). The interp2 command interpolates between data points. Until then, I'd better get back to those segfaults. We start from a table of points for for the function. complete class CubicSplineInterpolation implementing the cubic spline interpolation (you can first consider the case of 4 points). Checking Validity Figure 1 is my interpolation of a spline curve for regularly spaced points using my spline routine. coefs is an nx4 matrix of polynomial coefficients for the intervals, in Matlab convention with the leftmost column containing the cubic coefficients and the rightmost column containing the constant coefficients. – METHOD specifies interpolation filter • 'nearest' - nearest neighbor interpolation • 'linear'- bilinear interpolationbilinear interpolation • 'spline' - spline interpolation • 'cubic' - bicubic interpolation as long as the data is uniformly spaced, otherwise the same as 'spline' Geometric Transformation EL512 Image Processing 26. There are no constraints on the derivatives. We create an “Interpolation_Points” array, and initialize some points we can to draw. CSCE 441 Computer Graphics: Keyframe Animation/Smooth Curves Jinxiang Chai Outline Keyframe interpolation Curve representation and interpolation - natural cubic curves - Hermite curves - Bezier curves Required readings: 12-6 & 14-1 14-214-3 14-4, & 14-7 Computer Animation Animation - making objects moving Compute animation - the production of consecutive images, which, when displayed, convey a. BASIS_MATRIX_HERMITE sets up the Hermite spline basis matrix. Unlike previous methods of Interpolating, Spline interpolation does not produce the same unique interpolating polynomial, as with the Lagrange method, Vandermonde matrix method, or Newton's divided difference method. Given the function values which may represent the resonant field position or the transition probability at the vertex points (Figure 1a), we use the cubic spline interpolation method to. A vector consisting of the spline evaluated at the grid values. The second derivative of each polynomial is commonly set to zero at the endpoints, since this provides a boundary condition that completes the system of m-2 equations. The data points have to be sampled at the. Cubic splines was the only missing piece of the puzzle :P However, implementing a cubic spline interpolation routine is, unfortunately, well out of scope for my thesis, so it will have to wait until I have time to spend on it. I'm working on a finite volume advection scheme for unstructured meshes which uses a multidimensional polynomial weighted least squares fit for interpolating from cell centres onto faces. - Most commonly used interpolantused is the cubic spline - Provides continuity of the function, 1st and 2nd derivatives at the breakpoints. Spline cubic with tridiagonal matrix. In class, we interpolated the function f (x) =1 x at the points x =2,4,5 with the cubic spline that. For these reasons quaternion interpolation of the rotational parameters is performed. Then the spline inverse of the matrix Bof the equations for the spline. We consider the problem of shape-preserving interpolation by cubic splines. • This means we have 4n −2 equations in total. Therefore the cubic spline interpolation is used here as an. Property 1: The polynomials that we. S BahrololoumiMofrad 2, Mehdi Moudi 3. Interpolation & Polynomial Approximation Cubic Spline Interpolation III Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University c 2011 Brooks/Cole, Cengage Learning. Read more. Maria Cameron 1. 그중 Cubic Spline은 점들을 이을때 3차 다항식 형태로 나타낸것이고, piecewise Linear Interpolation은 점들을 잇는 방법이 구간별로 1차인 것이다. We assume that the points are ordered so. Let fbe a function from. • The efficient implementation of the cubic spline interpolation. Interpolation. The matrix equation involved is solved analytically so that numerical inversion. Three dimensional interpolation and extrapolation using either a set of (x, y, z) points, or matrix of evenly spaced z values. cubic spline interpolation and upsample ?. I was able to easily implement that. Hi, I am trying toport a software from a PC to a workstation. In this post I am sharing with you a C program that performs cubic spline interpolation. 1 School of Mathematics, Iran University of Science & Technology, Narmak, Tehran 16844-13114, Iran. If more control over smoothing is needed, bisplrep should be used directly. In the detailed search, the penetration depth and contact reference frame are calculated with the cubic spline surface interpolation in order to generate the accurate and smooth contact force. This implementation involves two steps. edu" Subject Re: st: Converting Quarterly GDP Data into Monthly Data Using Cubic Spline Interpolation. the interpolated result , interpolation kernel , and the scene respectively. I did some simple tests and examples confirming that. These interpolation splines can also be used for extrapolation, that is prediction at points outside the range of x. Interfaces to the BLAS and LAPACK. Then it covers the Hermite methods for piece-wise cubic interpolation, concluding with "cubic-spline interpolation". However, we can not uniformly sample the space of images, so interpolation over a non-uniformly sampled space is required. examples of cubic spline interpolation. SRS1 Cubic Spline for Excel 2. This MATLAB function returns interpolated values of a 1-D function at specific query points using linear interpolation. If we can reconstruct this matrix C, then we can create a cubic Hermite spline s using the Matlab function mkpp: s = mkpp(X, C) Recall that the result of cubic Hermite spline interpolation was a polynomial for each interval [L,R]. In and view of matrix theory, if is a circular matrix, then the inverse LIN et al. VC++ cubic spline interpolation and Bezier curves example program This source code is the implementation of cubic spline interpolation algorithm and data smoothing using VC++ MFC. As a result, the ck’s must be determined by solving a matrix problem. Let me set up some notation first. Many popular image interpola-tion methods are defined in this way, including nearest-neighbor interpolation, bi-linear interpolation, cubic-spline interpolation, and cubic convolution [1], [2], [8]. Any function which would. Natural cubic splines Inference Natural splines in R R also provides a function to compute a basis for the natural cubic splines, ns, which works almost exactly like bs, except that there is no option to change the degree Note that a natural spline has m+ K 4 degrees of freedom; thus, a natural cubic spline with Kknots has Kdegrees of freedom. Spline interpolation: The existing techniques being not so consistent either with the efficiency or the speed or both, we try to get to the apotheosis of the reconstruction to be Saccomplished by using Cubic-spline interpolation technique. The way of implementing this filter does not involve downsampling,. Details of this approach can be found in Appendix 1 and 2. If Y is a matrix, then the interpolation is performed for each column of Y and yi will be length(xi)-by-size(Y,2). splev(x_vals, splines)("spline evaluate") –evaluate the spline data returned by splrep, and use it to estimate y values. Learn more about interpolation, polynomial interpolation, cubic polynomial, matrix manipulation, polynomial. For interp2, the full grid is a pair of matrices whose elements represent a grid of points over a rectangular region. This generally provides a better fit to the data, and also has. So I thought why not have a small series of my next few blogs do that. The novel notion of shape preserving fractal interpolation without any shape parameter is introduced through the rational fractal interpolation model in the literature for the first time. interpolation coordinates 5 < x < 180 and 5 < y < 245, both evenly spaced with spacing 0. Then it covers the Hermite methods for piece-wise cubic interpolation, concluding with "cubic-spline interpolation". $\endgroup$ - normal chemist Mar 6 at 10:29. • In addition we require that S(x i) = y i, i = 0,··· ,n which gives n +1 equations. We assume that the points are ordered so. The computational method can be applied to three-dimensional curves, too. cubic spline interpolation and upsample ?. • To fulfill the Schoenberg-Whitney condition that N i n(u i) ≠0 , for n=3 we set u i=i+2 for all i. Interpolation using Cubic Spline Given N +1 data points in the interval [a,b], x t 0 t 1 ··· t N y y 0 y 1 ··· y N t 0 t 1 t 2 t N 2 t N 1 t N S 0 (x) S 1 (x) S N 2 (x) S N 1 S(x) (x) x Cubic Spline we want to construct a cubic spline S(x) to interpolate the table presumable of a function f(x). The control point setup can be implemented on MFC interface, can choose cubic spline interpolation or Bezier smoothing. See the example “Spline Interpolation” for various examples. The Lagrange interpolation seems to be "good enough" for me, despite the occasional cusp in the interpolation where there is a derivative discontinuity. Interfaces to the BLAS and LAPACK. The matrix form of the system of equations is:. We then computed the matrix norm of sampled importance value, which controls the degree of knowledge transfer in pre-training process. In this post I am sharing with you a C program that performs cubic spline interpolation. In this article, a new quadratic trigonometric B-spline with control parameters is constructed to address the problems related to two dimensional digital image interpolation. A B-spline with no internal knots is a Bézier curve. | 2019-12-08T06:43:11 | {
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https://math.stackexchange.com/questions/3745886/finding-the-equation-of-the-normal-to-the-parabola-y2-4x-that-passes-through | # Finding the equation of the normal to the parabola $y^2=4x$ that passes through $(9,6)$
Let $$L$$ be a normal to the parabola $$y^2 = 4x$$. If $$L$$ passes through the point $$(9, 6)$$, then $$L$$ is given by
(A) $$\;y − x + 3 = 0$$
(B) $$\;y + 3x − 33 = 0$$
(C) $$\;y + x − 15 = 0$$
(D) $$\;y − 2x + 12 = 0$$
My attempt: Let $$(h,k)$$ be the point on parabola where normal is to be found out. Taking derivative, I get the slope of the normal to be $$\frac{-k}{2}$$. Since the normal passes through $$(9,6)$$, so, the equation of the normal becomes:$$y-6=\frac{-k}{2}(x-9)$$$$\implies \frac{kx}{2}+y=\frac{9k}{2}+6$$
By putting $$k$$ as $$2,-2,-4$$ and $$6$$, I get normals mentioned in $$A,B,C$$ and $$D$$ above (not in that order).
But the answer is given as $$A,B$$ and $$D$$. What am I doing wrong?
• Isn't slope of normal $\frac {-k}{2h}$ ? Jul 5 '20 at 9:35
In addition to having slope $$-k/2$$ the normal must also pass through the point of contact $$(h,k)$$. The line in option $$C$$ does not pass through the point of contact for $$k=2$$ which is $$(1,2)$$. Your equation is the equation of a line having the slope of a normal at point $$(h,k)$$ on the parabola and passing through $$(9,6)$$. It is not necessarily a normal because you didn't make it pass through the point of contact.
The equation of the normal is $$\color{red}{y=\frac{-kx}{2}+\frac{9k}{2}+6}$$. The point $$(h,k)$$ lies on it so we get $$k=\frac{-kh}{2}+\frac{9k}{2}+6.$$ But $$k^2=4h$$ (since $$(h,k)$$ lies on the parabola as well), so we get $$k^3-28k-48=0 \implies (k-6)(k+2)(k+4)=0 \implies k=6,-2,-4.$$ These value can be used to find the equation of the normal as: \begin{align*} y&=-3x+33\\ y &=x-3\\ y &=2x-12 \end{align*}
Use parametric equation : any point $$P(t^2,2t)$$ on the given parabola
The gradient of the tangent $$=\dfrac4{2y}_{\text{(t^2,2t)}}=\dfrac1t$$
So, the gradient of the normal $$=-t$$
So, the equation of the normal $$\dfrac{y-2t}{x-t^2}=-t \implies xt+y=2t+t^3$$
Here $$2t=6$$
The general equation for a normal to the parabola $$y^2=4ax$$ is
$$y=mx-2am-am^3$$
Putting $$a=1$$ and passing it through $$(9,6)$$, we have $$6=9m-2m-m^3$$ Solving the above cubic for $$m$$ yields three values
$$m=1,2 \text{ or } -3$$
Now substitute that back, and voilà! | 2022-01-25T09:26:34 | {
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https://math.stackexchange.com/questions/2559370/arrangement-of-dominos-in-a-grid | # Arrangement of Dominos in a Grid
This the problem from the test and I'm stuck at one part.
Matt will arrange four identical, dotless dominoes (shaded 1 by 2 rectangles) on the 5 by 4 grid to the right so that a path is formed from the upper left-hand corner A to the lower right hand corner B. In a path, consecutive dominoes must touch at their sides and not just their corners. No domino may be placed diagonally; each domino covers exactly two of the unit squares shown on the grid. One arrangement is shown. How many distinct arrangements are possible, including the one shown?
So do I just find all the numbers of vertical and horizontal possible and then multiply by the number of arrangements of each, or should I go a different route?
• I have no idea what you mean by finding 'all the numbers of vertical and horizontal possible and then multiply' ... can you explain a bit of what you had in mind? – Bram28 Dec 10 '17 at 1:15
• Oh I'm sorry, I meant the number of vertically and horizontally placed dominos. – The Math Guy Dec 10 '17 at 1:35
I’m not sure what you had in mind, but here is what I did to get the answer:
The trick here is that the dominos don’t actually restrict the problem much. In fact, any walk from the top left corner to the bottom right corner using only steps to the right and down gives a unique domino path, and vice versa. (I am leaving some of these details out. Showing this bijection isn’t hard, but is a bit technical.)
Thus we need to only count the number of such walks. We can encode such a walk as a sequence of R’s and D’s(right and down steps.) We need to take seven total steps, three to the right, so we get that there are a total of $$\binom{7}{3}=35$$ ways to do this.
• +1 for taking into account the shape of the dominoes ... like you said any path can be covered by dominoes, but if the pieces were 1x3 that would no longer be true, so it's good you point this out. – Bram28 Dec 10 '17 at 1:38
You need to traverse $7$ cells (you start from $A$, so that cell is a given). You need to make exactly $3$ movements rightwards in total (otherwise we won't reach point $B$'s x-coordinate).
So we are looking for the number of ways we can order $$rrrdddd$$ where $r$ denotes going right, and $d$ denotes going down.
That gives $$\binom 73 = 35$$ ways.
• This was the same as Sean's, but he explained a little bit better, Thank You. – The Math Guy Dec 10 '17 at 1:37
You have to get from A to B using only steps right and down. The number of paths to a given square is the number of paths to the one just above (from which you step down) and the number of paths to the one just to the left (from which you step right). This sounds like Pascal's triangle.
I would consider the cells that you can get at after placing 1,2,3, and 4 dominoes on the way to the lower right.
So, after 1 domino you can be in (1,2) (row 1, column 2), or (2,1), and there is only 1 way to get to each.
After 2 dominoes, you can be in (1,4) (only 1 possibility to get there, through (1,2)), (2,3) (3 ways to get there: 2 via (1,2), and 1 through (2,1)), (3,2) (3), or (4,1) (1)
After 3 dominoes, it's (3,4) (1+2*3+3=10), (4,3) (by symmetry, also 10), or (5,2) (2+3=5)
So, after 4, you have to be in (5,4), and there are 10+2*10+5=35 paths to get there.
Here's a graphic that will help, indicating in how many ways you can get to the various cells:
\begin{array}{|c|c|c|c|} \hline &1&&1\\ \hline 1&&3&\\ \hline &3&&10\\ \hline 1&&10&\\ \hline &5&&35\\ \hline \end{array}
• You should get 10+2*10+5=35 after 4. – Sean English Dec 10 '17 at 1:32
• @SeanEnglish Yeah, just realized my mistake ... thanks! – Bram28 Dec 10 '17 at 1:34
• This was the right answer, but Sean's was a bit faster, sorry. – The Math Guy Dec 10 '17 at 1:36
• @TheMathGuy No worries! Sean's answer was more efficient anyway :) I'm more of a visual person though, so the graphic really works for me. – Bram28 Dec 10 '17 at 1:40 | 2019-11-21T03:11:22 | {
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https://afd-hamburg-nord.de/forum/archive.php?53e5c2=dynamic-programming-is-having-time-constraints | MathJax reference. Let’s see what it looks like when applying dynamic programming. However I’ve found that simply knowing about dynamic programming and how it fits into a more general problem-solving framework has made me a better engineer, and that in of itself makes it worth the time investment to understand. Thanks for contributing an answer to Mathematics Stack Exchange! Introduction To Dynamic Programming. If you’re having trouble working out a dynamic programming solution, look for other parameters that you might be able to fit into the subproblem to constrain it further. Trickster Aliens Offering an Electron Reactor. I am reading Dynamic programming using MIT OCW applied mathematics programming Yes, in this example with the given picture, the horizontal axis represents stage (which step you're on) and the vertical axis represents state (which option you're on for that step). More formally, given a set of n items each with weight w_i and value v_i along with a maximum total weight W, our objective is: Let’s see what the implementation looks like then discuss why it works. Let’s run an example to see what it looks like. This also happens to be a good example of the danger of naive recursive functions. Another example, $t_2(5) = 9$. The same problem with an additional capacity constraint has been dealt with by Des-rosiers et al. The current state, as before, is an integer from $1$ to $6$, and the decision is up or down (which branch you take to go from one column to the next). 2691-2708. The intuition behind dynamic programming is that we trade space for time, i.e. That concludes our introduction to dynamic programming! THE ALGORITHM We present a dynamic programming approach … Now we can run the algorithm with a constraint that the weights of the items can’t add up to more than 15. The horizon N is fixed. It's just got to do with the indexing and the picture here. I am not able to understand this constraint and why we are adding/ subtracting 1 while it is even/odd ? We’ll build both naive and “intelligent” solutions to several well-known problems and see how the problems are decomposed to use dynamic programming solutions. It is of interest therefore to know when such local verification functions exist. If you are in an even column and take the upper branch, the state number will go up by one. Personally it doesn’t come naturally to me at all and even learning these relatively simple examples took quite a bit of thought. s_{n-1} = \begin{cases} Therefore, how shall the word "biology" be interpreted? Note that the elements do not need to be contiguous; that is, they are not required to appear next to each other. Okay, probably too trivial. That escalated quickly! A recursive solution that caches answers to sub-problems which were already computed is called memoization, which is basically the inverse of dynamic programming. Let’s see how well it performs on much larger sequences. Incentive compatibility constraint is a constraint on the continuation of the payoff function at every time. [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. An elementary example is given there in 11.1 as shortest delay to reach parking slot from home. • All dynamic optimization problems have a time step and a time horizon. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Analyzing the constraints can give us crucial information like the possible edge cases and the expected time complexity of the solution. s_n , & \text{otherwise} This content originally appeared on Curious Insight. Examples of time constraint in a sentence, how to use it. Here's a proof-of-concept of a Constraint Programming approach. 6 Dynamic Programming 73 ... Notice, this is a discrete time model with no constraints on the decisions. Dynamic programming is related to a number of other fundamental concepts in computer science in interesting ways. The intuition behind this algorithm is that once you’ve solved for the optimal combination of items at some weight x Why Does De Guiche Want Valvert To Marry Roxane?, Creative Process Pdf, International Hotel San Francisco, Used Conveyor Pizza Ovens For Sale, Monitor Size Comparison 24 27, Neko Wallpaper Phone, Olympia Sports Customer Reviews, | 2021-09-24T15:05:33 | {
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https://mathematica.stackexchange.com/questions/132699/implicitregion-with-regionplot-example | # ImplicitRegion with RegionPlot Example
Using Mathematica 11.0.1.0 on a MacBook Pro (OSX 10.11.6) I tried this:
Clear[x, y, reg1]
reg1 = ImplicitRegion[
Log[10, 1 + x^2 + y^2] <= 1 + Log[10, x + y], {x, y}];
RegionPlot[reg1]
But got only this output:
RegionPlot[reg1]
But I did get the area.
Area[reg1]
Which gave the correct answer $49\pi$. Then I tried:
RegionPlot[
Log[10, 1 + x^2 + y^2] <= 1 + Log[10, x + y], {x, -2, 12}, {y, -2,
12}]
And I did get the correct image:
Is the fact that RegionPlot[reg1] did not work a bug that should be reported, or have I made some sort of mistake?
• RegionPlot[region] and RegionPlot[inequality, vars...] are not equivalent. For instance, try DiscretizeRegion[reg1] and DiscretizeRegion[reg1, Method -> "RegionPlot"]. I think your (failed) RegionPlot uses the first approach. Dec 3, 2016 at 22:00
THIS IS JUST AN EXTENDED COMMENT
The region definition can be simplified to
reg2 = ImplicitRegion[
1 + x^2 + y^2 <= 10 (x + y),
{x, y}];
Area[reg2]
(* 49 π *)
Show[
RegionPlot[reg2],
ContourPlot[
Log10[1 + x^2 + y^2] == 1 + Log10[x + y],
{x, -2, 12}, {y, -2, 12},
ContourStyle -> Red,
PlotPoints -> 50]] | 2022-08-19T07:53:14 | {
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https://mathematica.stackexchange.com/questions/216437/different-results-using-spline-interpolation-in-wolfram-and-matlab | # Different results using spline interpolation in Wolfram and MATLAB
I use the same data for interpolation in Mathematica and MATLAB, but the result is different.
x={-1.00,-0.96,-0.65,0.10,0.40,1.00};
y={-1.0000,-0.1512,0.3860,0.4802,0.8838,1.0000};
Interpolation[{x,y}//Transpose,Method->"Spline"][-0.3]
result: -0.87332
x=[-1.00,-0.96,-0.65,0.10,0.40,1.00];
y=[-1.0000,-0.1512,0.3860,0.4802,0.8838,1.0000];
splinetx(x,y,-0.3)
result: -0.1957
I tried different InterpolationOrder but still different.
Is the "Spline" same as splinetx?
If not, is there a function in Wolfram like splinetx?
The splinetx function is available here or here.
MATLAB also has a built-in spline function which gives identical results to splinetx.
This is a comparison of the results given by Mathematica and MATLAB:
• What methods does splinetx use? The short answer to your first question is no. And the short answer to your second question is yes. The methods used in MATLAB or gnu octave and other similar programs can be realized with WL functions. – CA Trevillian Mar 17 at 6:27
• The source code of splinetx can be found here: mathworks.com/matlabcentral/mlc-downloads/downloads/submissions/… – xzczd Mar 17 at 7:20
• MATLAB also has a built-in spline function which also gives -0.1957. I think this is a good question and it is worth trying to understand the source of the differences. – Szabolcs Mar 17 at 9:51
• Since we have the source code for the MATLAB version, it could no doubt be reproduced in Mathematica with some effort. It is not particularly interesting. But can we reproduce Mathematica's version? I think it is important to fully understand and to be able to reproduce what Mathematica does ... I don't see enough in the documentation to be able to easily do this. The difference must be coming from the conditions on the endpoints. – Szabolcs Mar 17 at 10:09
• @AlexTrounev That much is clear. What I am looking for is not the "correct" solution (as there isn't any single one) but a precise understanding of what Interpolation does here (I guess the question is what is assumed for the derivatives at the boundaries) – Szabolcs Mar 17 at 12:28
Is the "Spline" same as splinetx?
• No.
If not, is there a function in Wolfram like splinetx?
• No.
That was a bit negative. However, it is not too difficult to apply the formulae in this answer and this answer to derive a routine that generates not-a-knot cubic splines (as was astutely observed by CA Trevillian and others in the comments.)
Of course, one can use SparseArray[] + LinearSolve[] to solve the underlying tridiagonal system, so I'll do that in the function below:
notAKnotSpline[pts_?MatrixQ] := Module[{dy, h, p1, p2, sl, s1, s2, tr},
h = Differences[pts[[All, 1]]]; dy = Differences[pts[[All, 2]]]/h;
s1 = Total[Take[h, 2]]; s2 = Total[Take[h, -2]];
p1 = ({3, 2}.Take[h, 2] h[[2]] dy[[1]] + h[[1]]^2 dy[[2]])/s1;
p2 = (h[[-1]]^2 dy[[-2]] + {2, 3}.Take[h, -2] h[[-2]] dy[[-1]])/s2;
tr = SparseArray[{Band[{2, 1}] -> Append[Rest[h], s2],
Band[{1, 1}] -> Join[{h[[2]]}, ListCorrelate[{2, 2}, h], {h[[-2]]}],
Band[{1, 2}] -> Prepend[Most[h], s1]}];
sl = LinearSolve[tr, Join[{p1},
3 Total[Partition[dy, 2, 1]
Reverse[Partition[h, 2, 1], 2], {2}],
{p2}]];
Interpolation[MapThread[{{#1[[1]]}, #1[[2]], #2} &, {pts, sl}],
InterpolationOrder -> 3, Method -> "Hermite"]]
Try it out on the points in the OP:
pts = {{-1., -1.}, {-0.96, -0.1512}, {-0.65, 0.386},
{0.1, 0.4802}, {0.4, 0.8838}, {1., 1.}};
spl = notAKnotSpline[pts];
spl[-0.3]
-0.195695
Plot[spl[x], {x, -1, 1},
Epilog -> {Directive[AbsolutePointSize[6], ColorData[97, 4]], Point[pts]}]
Demonstrate the $$C^2$$ property of the cubic spline:
Plot[{spl[x], spl'[x], spl''[x]}, {x, -1, 1}, PlotRange -> {-10, 30}]
Szabolcs's desire to reproduce the results of Method -> "Spline" is a bit more difficult, because the exact formulae being used are not disclosed publicly. That being said, I was able to reverse-engineer and reproduce it some time ago, so go look at that answer if you want more details.
• (I actually have a more general spline interpolation function than this, but the code is in a hard disk many kilometers away, so releasing that will have to wait.) – J. M.'s technical difficulties Mar 18 at 1:59
• Thanks！It's very helpful. I've read your answers and write a natural cubic spline function. It seems that the Method -> "Spline" in Interpolation is not a natural cubic spline. So what the conditions are actually used for it? Why Mathematica use an uncommon method? – srtie Mar 19 at 16:45
• I wrote a natural spline routine in another answer, and the formulae used by Method -> "Spline" are in the answer I linked to in the last paragraph. Why these choices were made, I cannot say. – J. M.'s technical difficulties Mar 19 at 21:27 | 2020-07-11T21:11:12 | {
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https://www.physicsforums.com/threads/help-understand.89292/ | # Help understand
1. Sep 16, 2005
Hi,
A boy wants to knock down a coconut with a rock and his slingshot. He observes that the coconut is about 3.0m above his slingshot and the tree is 4.0m away along the ground.
He knows from experience that the release speed of his rock is 20m/s.
How far above the coconut should he aim?
I don't understand what I am asking to do here. Please can someone help me?
2. Sep 16, 2005
### Staff: Mentor
If gravity didn't exist, the boy would aim directly at the coconut, since the rock would travel in a straight line. But gravity does exist. Realize that, compared to where it would have gone with no gravity, the rock falls a certain distance. That distance is how far above the monkey you have to aim.
Another way to look at it: What angle must the rock be shot at to hit the coconut? Once you find that angle, you can see where the straight line path would have been.
3. Sep 16, 2005
Doc, I have a problem to find the angle. I try to use the equations of motion but I have too many unknow : tfinal, vy/final, and the angle itself.
vy/final=vy/initial-g(ty/final)
yfinal=yinitial+vy/initial*tfinal-.5*(g)yfinal2
those two equations give respectively:
vy/final=20*sin(θ )-9.81*tfinal
4=0+ 20sin(θ )*tfinal-4.9*tfinal2
Do i miss something here?
4. Sep 16, 2005
### Staff: Mentor
You have the vertical distance as a function of time. Good! But what about the horizontal distance?
5. Sep 16, 2005
xfinal=xinitial+vx/initial*tfinal
That gives:
3=0+20cos(θ )* tfinal
6. Sep 16, 2005
### Staff: Mentor
Good! (I think you mean 4, not 3.) Now combine that with the equation for vertical motion.
7. Sep 16, 2005
### Staff: Mentor
Two things:
(1) I think you have mixed up the 3 and the 4. According to your first post:
(2) Here's a trick that may help you combine those two equations. For each equation, isolate the term with the sin or cos. Then square both sides of each equation. Then add them. (I assume you know a useful trig identity about $\sin^2\theta + \cos^2\theta$.)
8. Sep 16, 2005
Well I found somthing that is not realistic
I change the 3 and 4 into the right equations and I found t=3.92s
and the angle is .99 degree!!
What do you think?
9. Sep 16, 2005
### Staff: Mentor
Well... I didn't crank out the numbers myself, but does your answer make any sense? After all, without gravity the angle would be $\tan \theta = 3/4$. So with gravity, the angle must even be greater. Recheck your calculations. (I'll do it myself when I get a few minutes.)
10. Sep 16, 2005
ok i'll do that
11. Sep 17, 2005
### Staff: Mentor
I did the calculation and found, as expected, that the angle is slightly greater than that needed for straightline motion. (Note: When solving the quadratic equation, there are two solutions. Only one of them is the one we want.) If you still get an unrealistic answer, post the steps just as you did them.
12. Sep 17, 2005
I have te two equations:
3=0+ 20sin(θ )*tfinal-4.9*tfinal2
4=20*cos(θ)*tfinal
Isolate the sin and cos
sin(θ)=(3+4.9*tfinal2)/20*tfinal
cos(θ)=4/20*tfinal
I combinethe squares of sin and cos
cos(θ)2+sin(θ)2=(9+29.4*tfinal2+24.01*tfinal2+16)/(400*tfinal2)
equivalent to
1=(25+29.4*tfinal2+24.01*tfinal4)/(400*tfinal2)
equivalent
(25+29.4*tfinal2+24.01*tfinal4)=400*tfinal2
(25-370.6*tfinal2+24.01*tfinal4)=0
solving this I found 4 values of t but one is correct t=3.92s
Please tell me what is wrong here.
13. Sep 17, 2005
### Staff: Mentor
Your work looks correct. Treat the final equation, as I'm sure you did, as a quadratic in t^2 (say X = t^2). The quadratic has two solutions: you just picked the wrong one! (When you take the square root of those solutions, you can ignore the negative values.)
There are two ways to hit the coconut: The long way or the short way. The long way is essentially shooting it up in the air in tall arc. That's not the one you want.
14. Sep 17, 2005
I checked and rechecked, I found the same thing.
I rechecked the equation too. they sound fine!
15. Sep 17, 2005
### Staff: Mentor
Alternatively, if I view this as a quadratic in x = t^2:
$$24.01x^2 - 370.6x + 25 = 0$$
This equation has two solutions. What are they?
16. Sep 17, 2005
the four solutions I found are:
x=3.9201...
x=0.2602....
x=-0.2602...
x=-3.9201
I get rid of the negative values and I have:
x=3.9201...
x=0.2602.... but I only take x=3.9201...
when I use you equation , I have
x=15.36...
x=0.06775...
17. Sep 17, 2005
### Staff: Mentor
And why do you ignore the other answer? That's the one you want!
Right. And when you take the square roots, you get the same four solutions.
18. Sep 17, 2005
### lightgrav
Look, the coconut is 5m away from the start
(along a diagonal). If the stone travels 20m in 1 sec
then (ignoring gravity) it takes about 1/4 sec to go 5m.
WITH gravity, in 1/4 sec the stone deviates from its path
by 0.3 m, so you have to aim about twice as high,
and stone will take anout twice as long ... about half sec.
Why do you inisist on discarding the t^2 = 0.2602 [s^2]
(which means t about .51 sec) ?
You know that during 2 seconds of free-fall,
a stone would deviate from its path by nearly 20 meters
- so you want to aim 20 meters high?
That's what DocAl meant by "the long way" almost straight up. | 2018-01-17T03:21:44 | {
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https://mathematica.stackexchange.com/questions/240865/scan-through-partial-tuples/240888 | Scan through (partial) tuples
I have a list of list of positive integers $$s = \{s_1, s_2, ..., s_k\}$$, each list $$s_i$$ is possibly of different lengths, and I want to find out if there exists a $$k$$-tuple of the $$s_i$$ that sums exactly to $$n$$. In other words, I want to partition $$n$$ into $$k$$ integers, each taken from the corresponding $$s_i$$.
I am not aware if there is any theoretical method to prove that such a solution exists, for any $$n$$ and $$s$$ in general. IntegerPartitions cannot deal with nonuniform integer lists (3rd argument). Hence I try to generate all tuples and quit at the first matching case found:
s = {{1, 2, 3}, {4, 5, 6, 7}, {1, 3}, {2, 4, 6}};
n = 16;
Catch@Outer[If[Plus[##] == n, Throw[{##}], 0] &, Sequence @@ s]
(* {1, 6, 3, 6} *)
However, this is not compilable, due to Sequence. This may be remedied, but what I would like to have is a construct that:
1. Does not store its found values;
2. Can abort constructing a tuple (shorter than $$k$$) if its partial sum is already above $$n$$;
3. Can find a partial tuple not shorter than $$k_{\min}$$ ($$k_{\min} \le k_{\max}$$) if it exactly sums to $$n$$. This is an extension of the above described problem, and considers only tuples that are built from the left (i.e. each $$s_i$$ always corresponds to the $$i^{th}$$ position in the final tuple).
I expect to work with large lists and $$n$$, and I do not want to exhaust memory. I am not sure that lazy tuples construction is a viable option here, due to pts. 2 and 3. But feel free to prove me wrong.
Addition: The obvious cases, like Total[Min/@ Take[s, kmin]] > n and Total[Max/@s] < n should of course be pre-checked before going into long computations.
• Is {3, 7, 6} considered a possible answer? – Carl Woll Mar 2 at 1:53
• @Carl As I have specified in my Q: no. But I would certainly be interested in a more general solution that can account for any subset of the sublists of s. – István Zachar Mar 2 at 6:36
You could use LinearProgramming.
To use LinearProgramming, convert the list of lists into a single list. For your example we create the list {1, 2, 3, 4, 5, 6, 7, 1, 3, 2, 4, 6}. Since there is no criteria for which tuple to return, I use a cost vector of all 1s. Then, LinearProgramming will try to find a vector v whose dot product with this cost vector is minimized, and which also satisfies various constraints:
1. The first constraint is that each member of v is a nonnegative integer (actually 0 or 1).
2. The second constraint is that at most 1 member of each list of integers can be selected. For example, the constraint that at most 1 member of the first list is selected is given by:
{1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0} . v <= 1
1. The third constraint is that the selected members must have a particular total. For example, the constraint that the total must be 16 is given by:
{1, 2, 3, 4, 5, 6, 7, 1, 3, 2, 4, 6} . v = 16
1. The final constraint is that it is not allowed to select a member of a list if the previous list has no member selected. For example, the constraint that the 4th list cannot have a member selected if the third list has no member is given by:
{0, 0, 0, 0, 0, 0, 0, 1, 1, -1, -1, -1} . v >= 0
The syntax of LinearProgramming is that the second argument consists of the above lists, while the third argument specifies what the value on the RHS is, and whether <, = or > is used. For example, for the second constraint, using {1, -1} indicates that the RHS is 1 and the -1. indicates that <= is used. The fourth argument specifies what the range of the values of v are, and using 0 means that all values greater than 0 are allowed. The final argument specifies the domain, which in our case is integers. The following function does this:
findTuple[s_, n_] := Module[{len, lens, left, v, indicators},
lens = Length /@ s;
len = Total[lens];
left = FoldList[Plus, 0, Most[lens]];
v = Quiet[
LinearProgramming[
ConstantArray[1, Total @ lens],
Join[
indicators,
-Differences[indicators],
{Flatten @ s}
],
Join[
Table[{1,-1},Length[s]],
Table[{0,1}, Length[s]-1],
{{n,0}}
],
0,
Integers
],
LinearProgramming::lpip
];
Pick[Flatten[s],v,1]
]
Small examples:
findTuple[{{1,2,3},{4,5,6,7},{1,3},{2,4,6}}, 16]
findTuple[{{1,2,3},{4,5,6,7},{1,3},{2,4,6}}, 19]
{3, 6, 3, 4}
{3, 7, 3, 6}
Bigger example:
findTuple[Table[Range[18, 22], {60}], 1000]
% //Length
{22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 18, 22, 22, 22, 22,
18, 22, 22, 22, 22, 22, 18, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22,
22, 22, 22, 22, 22, 22, 22, 22}
46
• Could you please explain your code in some more depth? I am entirely not familiar with LinearProgramming thus I cannot answer your question. Can it find tuples that are shorter than k but still add up to n or only of exactly k length? It fails on findTuple[Table[Range[18, 22], {60}], 1000], though there exist solutions of length e.g. 46. – István Zachar Mar 1 at 22:23
• @IstvánZachar As written it only finds tuples of exactly k length. It would be simple to modify it to return shorter tuples. What criteria do you have to pick the length of the tuple? – Carl Woll Mar 1 at 23:21
• Partial tuples of length kmin or longer (up to length kmax = k). And a shorter tuple takes sublists of s in order, from the left. – István Zachar Mar 2 at 6:35
• Thanks for the extensive explanation. It is a very effective solution! – István Zachar Mar 3 at 14:43
• So, am I right in assuming that for tuples between lengths kmin, kmax (0<kmin<=kmax<=Length[s]), one has to use a 3rd argument like Join[Table[{1, 0}, kmin], Table[{1, -1}, kmax - kmin], Table[{0, 0}, Length@s - kmax], Table[{0, 1}, Length@s - 1], {{n, 0}}] ? – István Zachar Mar 3 at 22:08
I will show how similar tasks can be done using (undocumented) Streaming framework. The usual caveat applies: since this is undocumented functionality, there is no guarantee that it will exist in the future versions in the same exact form, or at all. But I thought it may be a nice application to illustrate some of the ideas behind Streaming, as well as to have a somewhat deeper look into its internals and possible use cases.
Preparation
Download and install a patch for Streaming, if don't yet have it
To start with, one would have to follow this Q/A to patch Streaming, which is necessary to run it on modern versions of Mathematica. Specifically, you need to execute:
Import[StringJoin[
"https://raw.githubusercontent.com/",
"lshifr/StreamingPatch/master/StreamingPatchBootstrap.m"]
]
where I used StringJoin because the bug in SE code editor would not allow me to enter long strings. This step has to be done only once on a given machine.
Streaming and LazyTuples
Since my post has become very long, I have collected essential parts into a gist, which can be easily imported. For lazy tuples, I will use the function TuplesFunction from this excellent answer by Carl Woll, with a few minor changes.
The initialization code is then:
\$HistoryLength = 0;
Get["StreamingPatch"];
Import @ StringJoin[
"https://gist.githubusercontent.com/lshifr/",
"7d5d3ef064b2eb1a1e9b997726923331/raw/LazyTuples.m"
]
Example data
We start by defining sample lists for tuple elements:
lists = Table[RandomSample[Range[30], 20], 5]
(*
{
{4, 5, 7, 16, 20, 17, 14, 6, 22, 25, 12, 15, 19, 1, 2, 13, 29, 26, 10, 3},
{7, 4, 21, 28, 29, 3, 10, 9, 15, 12, 17, 14, 1, 11, 16, 2, 23, 30, 6, 5},
{1, 7, 4, 18, 17, 21, 16, 30, 26, 11, 3, 22, 6, 14, 24, 19, 12, 23, 9, 28},
{28, 23, 30, 4, 11, 16, 27, 20, 6, 14, 24, 18, 12, 3, 7, 21, 2, 25, 26, 1},
{11, 17, 5, 1, 18, 25, 4, 12, 16, 24, 2, 9, 10, 8, 3, 23, 28, 26, 19, 15}
}
*)
This defines tuples with the length 20^5 == 3200000 and total ByteCount about 128Mb:
Times @@ Length /@ lists
ByteCount[Tuples[lists]]
(* 3200000 *)
(* 128000208 *)
Main example
LazyList brief intro
Evaluating LazyTuples[lists] will create a LazyList object, which is a lazy representation of a list of tuples, with a default chunk size 10000.
Sections below have more detailed explanations of the internals, but for practical purposes one can in many ways use LazyList in place of a normal List. Many operations on LazyList, such as e.g. Select and Take, are lazy - they produce a new LazyList with very little computation. The real work is done when Normal is applied to a LazyList object, which converts it to an equivalent normal List.
Illustration
This creates a LazyList object (delayed assignment is deliberate, for the purposes of benchmarking we want a new instance to be created every time):
lt := LazyTuples[lists]
One can convert it to a normal list
Normal[lt] // Length // AbsoluteTiming
(* {1.3135, 3200000} *)
and verify that it produces the result identical to standard Tuples:
Normal[lt] == Tuples[lists]
(* True *)
In the following, we will search for all tuples which sum exactly to 100, and measure performance and memory footprint:
Normal @ Select[lt, Total[#]==100&]//Short//AbsoluteTiming
Normal @ Select[lt, Total[#]==100&]//MaxMemoryUsed
Normal @ Take[Select[lt, Total[#]==100&], 1]//AbsoluteTiming
(*
{5.91346,{{4,21,17,30,28},{4,21,21,28,26},{4,21,21,30,24},{4,21,21,26,28},<<23635>>,{3,30,28,14,25},{3,30,28,24,15},{3,30,28,21,18}}}
7595088
{0.108224,{{4,21,17,30,28}}}
*)
This shows that it takes about 6 seconds to process entire list of tuples in this case, and about 0.1 sec to get just the first match, while memory use does not exceed 8Mb.
We now compare that to in-memory computations using Tuples:
Select[Tuples[lists], Total[#]==100&]//Short//AbsoluteTiming
Select[Tuples[lists], Total[#]==100&]//Short//MaxMemoryUsed
Select[Tuples[lists], Total[#] == 100 &, 1]//Short//AbsoluteTiming
(*
{4.55724,{{4,21,17,30,28},{4,21,21,28,26},{4,21,21,30,24},{4,21,21,26,28},<<23635>>,{3,30,28,14,25},{3,30,28,24,15},{3,30,28,21,18}}}
692115784
{1.09674,{{4,21,17,30,28}}}
*)
which shows that in-memory computation is about two orders of magnitude more memory-hungry, about 1.5x - 2x faster for all matches, and significantly slower for a single match search.
Note that run-time and memory efficiency of lazy computations in general do depend on the chunk size. One can specify it using "ChunkSize" option to LazyTuples. For example, this would create a LazyList with larger chunks:
LazyTuples[lists, "ChunkSize" -> 50000]
Previous editions of this post contain more in-depth explanations of the internals of LazyTuples.
Speeding things up
Using compiled predicate
We can compile the predicate in Select:
selFun = Compile[{{ints, _Integer, 1}}, Total[ints] == 100, "CompilationTarget" -> "C"]
which would give about 1.5x - 2x speedup (for both lazy and in-memory computations):
Normal @ Select[lt, selFun];//AbsoluteTiming
Normal @ Select[lt, selFun]//MaxMemoryUsed
Normal @ Take[Select[lt, selFun], 1]//AbsoluteTiming
(*
{3.28482,Null}
4941080
{0.075051,{{4,21,17,30,28}}}
*)
Compiling entire Select
This section will show how Streaming can actually beat in-memory computations not only by memory efficiency, but also by speed.
Let's start by asking, whether one can further speed things up. One obvious improvement is to try compiling entire Select, rather than just the predicate.
The following function is a generator of compiled Select functions, which take (preferably compiled) predicate and return compiled Select:
ClearAll[createTupleSelect]
createTupleSelect[criteria_] := createTupleSelect[criteria, All]
createTupleSelect[criteria_, numberOfResults_] :=
Replace[
If[
numberOfResults === All,
Hold[Select[tuples, criteria]],
Hold[Select[tuples, criteria, numberOfResults]]
],
Hold[select_] :> Compile @@ Hold[
{{tuples, _Integer, 2}},
select,
CompilationTarget->"C", CompilationOptions -> {"InlineCompiledFunctions"-> True}
]
]
Let us use it to create two compiled functions, one that selects all matching elements, and the other that selects just the first one:
sel = createTupleSelect[selFun]
selFst = createTupleSelect[selFun, 1]
We can test these easily by using them for the in-memory computation:
sel[Tuples[lists]]//Short//AbsoluteTiming
selFst[Tuples[lists]]//Short//AbsoluteTiming
(*
{0.413269,{{4,21,17,30,28},{4,21,21,28,26},{4,21,21,30,24},<<23636>>,{3,30,28,14,25},{3,30,28,24,15},{3,30,28,21,18}}}
{0.283024,{{4,21,17,30,28}}}
*)
We see that this version is about 5-6 times faster than the one where only the predicate has been compiled.
But what is remarkable here, is that the time to get a single result is almost the same as the time to get all results. This is because, for the in-memory computation, one first has to generate all tuples, and in the compiled setup this operation takes the majority of time. This is where lazy computations potentially have an advantage. Let's see whether we actually can have it with Streaming.
To test Streaming in this regime, we will switch to larger chunks - that would reduce the top-level overhead of Streaming chunk manipulation:
ltLrg := LazyTuples[lists, "ChunkSize" -> 50000]
Unfortunately, the implementation of Streaming Select does not have an option to provide an entire select function for a chunk, rather than a predicate. I have created a gist where I added such capability. You can also look at it if you are interested in how Select is implemented in Streaming currently, since it is mostly exact same code.
We will import it now:
Import[StringJoin[
"https://gist.githubusercontent.com/lshifr/",
]]
which will make a new LazyListSelect[list, pred, entireSelect] function available (again, the only reason for StringJoin is the M SE editor bug).
We can now use it:
Normal @ LazyListSelect[ltLrg, None, sel]//Short//AbsoluteTiming
Normal @ LazyListSelect[ltLrg, None, sel]//MaxMemoryUsed
Normal @ Take[LazyListSelect[ltLrg, None, selFst],1]//AbsoluteTiming
(*
{0.67473,{{4,21,17,30,28},{4,21,21,28,26},{4,21,21,30,24},{4,21,21,26,28},<<23635>>,{3,30,28,14,25},{3,30,28,24,15},{3,30,28,21,18}}}
9266760
{0.047092,{{4,21,17,30,28}}}
*)
What we see is that the search for all matches is about 1.5x - 2x times slower than the in-memory version, which is similar to what we have seen before. In this case, it is probably mostly due to the tuples generation overhead (even though Carl's function is extremely fast, it is still top level compared to kernel's specialized implementation in C). But of course this is still an order of magnitude more memory efficient.
But for a single match search, we're in for a pleasant finding. Not only is this an order of magnitude more memory efficient than direct built-in, but it is also an order of magnitude faster.
Of course, as I mentioned previously, this isn't really surprising, since lazy computations don't need to generate all tuples if the result can be found within e.g. the first few thousands. And of course one can write imperative code with loops, which would lead to similar or somewhat better performance. But Streaming has all that built in, and allows one to stay within the declarative functional paradigm, while providing decent performance in many such cases.
Summary
I have outlined how problems such as this can be treated within Streaming framework.
The main advantages of Streaming are built-in memory efficiency and laziness, which can automatically lead to efficient computations when only a subset of data is needed at the end, while allowing one to stay within the declarative / functional programming paradigm.
In some cases, Streaming can be not only vastly more memory-efficient than analogous in-memory computations with built-ins, performed in functional style, but also much faster. I have illustrated that in the section on further possible speedups, with entirely compiled Select.
In many cases, when using Streaming, one can control the memory / speed balance by adjusting the chunk size of the computation.
• This, of course would be magnificent to use, but I have to rely on vanilla Mathematica at a client's end, thus I have to work with what's in the box. – István Zachar Mar 2 at 8:45
• @IstvánZachar Well, most of Streaming is in the vanilla Mathematica :). But of course, you are right. Being undocumented, it can't be relied upon in terms of future backward compatibility. Unfortunately, Streaming has been on hold for the last 5 years. There is some hope though, that Streaming will become a first-class citizen in Mathematica in the foreseeable future. – Leonid Shifrin Mar 2 at 10:57
• @IstvánZachar OTOH, what I certainly should be able to do would be to incorporate all the above extensions and patches into Streaming, so that e.g. in 12.3 or may be 12.4 it would not require any extra code / patches to run the examples like those in the post. Of course, that does not solve the main problem of it being undocumented and unofficial. – Leonid Shifrin Mar 2 at 11:01
• It would be nice to add a SeedRandom call when defining a sample. – Carl Woll Mar 2 at 19:07
• @CarlWoll You are right, I should've done just that. I have instead shown the output for lists, which can be copied, but of course SeedRandom is the way. I might later edit that in, would have to recompute / edit all the timings etc. then. B.t.w., your version of lazy tuples is amazingly fast, it can pretty much compete with the internal implementation. Hats off. – Leonid Shifrin Mar 2 at 20:07
Here's an answer that seems pretty fast and small. It essentially just implements a simple depth-first search, with the current row of s being investigated being at index i, and the current index investigated in each row j being a[j]. It also handles your kmin case, with the default (omitted argument) being k.
It stores some found values, but not many—only one integer per row of s. In particular, it stores a temporary integer p[i] for each i in the range 1, Length[s] representing the partial sum down the current path to that row. One could rewrite this by re-computing the sum down the path each time, though.
It's a bit slower and bigger than your Outer function in the small case, but is faster and uses less memory in even slightly larger cases; consider s = Table[RandomInteger[10, RandomInteger[{3, 10}]], {20}], n = Total[Last /@ s], and compute MaxMemoryUsed[FindSum[s,n]] // AbsoluteTiming. For mine this gives {0.008944, 18672}, and for the Outer one it gives {0.426522, 2449544}. (It also seems that this basic approach apparently outperforms linear programming, which surprised me.)
You might be able to compile it, but you might need to prepare your input by padding each list so it's a uniform array. In that case, you would need to modify the code to get the right values for amax, and thus the right cycling for a. See update below.
I'll update this with a more thorough explanation. Note that using lists for a and p apparently is worse than using the definitions used below, performance-wise, according to tests.
Note also that due to what seems to be a bug causing a recursion limit to be reached with long lists s (I'll be investigating that separately), one needs to define defaults the old-fashioned way—not with :. Or, just play it fast and loose and don't pattern-match the arguments s and n beyond _.
FindSum[s : {{___Integer} ...}, n_Integer] := FindSum[s, n, 0]
FindSum[s : {{___Integer} ...}, n_Integer, kmin0_Integer] :=
(* Note: consider inserting the MaybeSumQ check mentioned below. *)
Module[{a, i = 1, k = Length[s], kmin = kmin0, amax = Length /@ s,
sum, backincrement, p},
If[kmin0 == 0, kmin = k];
Do[a[i] = 1; p[i] = 0, {i, k}];
p[1] = First@First@s;
backincrement[] := If[(a[i] = Mod[1 + a[i], amax[[i]], 1]) == 1,
If[--i == 0, Throw[False], backincrement[]]];
Catch[While[True,
sum = p[i];
Which[
sum == n && i >= kmin,
Throw[Table[s[[j, a[j]]], {j, i}]],
sum >= n || i == k,
backincrement[];
p[i] = p[i - 1] + s[[i, a[i]]],
True,
i++; p[i] = sum + s[[i, a[i]]];
]];
]
]
There's also a very fast (but admittedly somewhat memory-using—though not as much as the representation of s itself, I don't think) function that can perform a simple check to see that the list at least doesn't fail outright. It does this by ensuring the sum of the minima of the lists doesn't preclude a sum of n by being too high, and that the sum of the maxima of the lists doesn't preclude a sum of n by being too low.
MaybeSumQ[s : {{___Integer} ...}, n_Integer] := MaybeSumQ[s, n, 0]
MaybeSumQ[s : {{___Integer} ...}, n_Integer, kmin_Integer] :=
If[kmin == 0 || kmin == Length[s],
(#[[1]] <= n && #[[2]] >= n) &@Total[MinMax /@ s, {1}],
Total[Min /@ Take[s, kmin]] <= n && Total[Max /@ s] >= n]
Changing the definition of FindSum to If[MaybeSumQ[s, n, kmin0], <body>, False] can save some time on average if a solution is not known to exist and has a high enough chance of not being generatable from s. For a list s containing 10000 lists of length about 7, it took only about 0.005 seconds.
Update: compilation
I tried compiling it, using PadRight[s, Automatic, -1] to pad out the ragged array s with -1s, and using something like
amax = Table[
Replace[FirstPosition[si, -1, Length[si]],
Dispatch[{{x_} :> x - 1}]]], {i0, k}]
as well as trying handwritten While loops to check each list element one-by-one, and variations of the above with and without Withs and Replaces, and sometimes with /@s. This didn't seem to make much difference.
It compiles, but surprisingly—even excluding the time and memory it takes to pad s—the compiled version seems to run at the same speed, and in fact ever so slightly slower than the uncompiled one, no matter how I wrote amax. It was important to use RepeatedTiming here, since the differences were small and variable. However, it did use less memory, by about a third. To test easily I made functions to generate random lists like this:
(* Helpful functions for generating test cases: *)
Unprotect[rowlength, vrange];
ClearAll[rowlength, vrange];
Options[RandomList] = {rowlength -> {3, 10}, vrange -> {1, 10}}; Protect[rowlength, vrange];
RandomList[k_Integer, OptionsPattern[]] :=
Table[RandomInteger[OptionValue[vrange],
RandomInteger[OptionValue[rowlength]]], {k}];
RandomSum[s : {{___Integer} ...}] := Total[RandomChoice /@ s];
s = RandomList[750]; n = RandomSum[s]; s0 = PadRight[s, Automatic, -1];
(One can also use n = Total[Last /@ s] to try to force a sum on a "hard" path.)
For the compiled function, I got that the MaxMemoryUsage[CFindSum0[s0, n, 0]] // RepeatedTiming was {0.3189, 288464}, and for the uncompiled function, I got {0.292063, 463032}
For completeness, here's the compiled version:
CFindSum0 =
Compile[{{spadded, _Integer, 2}, {n, _Integer}, {kmin0, _Integer}},
Module[{a, i = 1, k = Length[spadded], kmin = kmin0, amax, j, l,
sum, backincrement, p},
If[kmin0 == 0, kmin = k];
amax =
Replace[FirstPosition[si, -1, Length[si]],
Dispatch[{{x_} :> x - 1}]]], {i0, k}];
Do[a[i] = 1; p[i] = 0, {i, k}];
Catch[While[True,
sum = p[i];
Which[
sum == n && i >= kmin,
sum >= n || i == k,
Label["backincrement"];
If[(a[i] = Mod[1 + a[i], amax[[i]], 1]) == 1,
If[--i == 0, Throw[False], Goto["backincrement"]]];
p[i] = p[i - 1] + spadded[[i, a[i]]],
True,
i++; p[i] = sum + spadded[[i, a[i]]];
]];
]
],
CompilationTarget -> "C"]
CFindSum[s : {{___Integer}...}, n_Integer, kmin_Integer] :=
CFindSum[s : {{___Integer}...}, n_Integer] := CFindSum[s, n, 0]
CFindSum can be used with the same argument type and syntax as FindSum, since it packages in the padding and optional arguments.
• Your compiled function runs at the same speed as the normal one because it calls MainEvaluate at many points (check with CompilePrint). – István Zachar Mar 1 at 22:33
• Ah, ok, I'm new to compiling functions in mathematica. Generally, how do you avoid that? – thorimur Mar 1 at 22:37
• There are some general rules, but mostly lots of exceptions. Check here and here and here and many more. – István Zachar Mar 1 at 22:47
• Thanks! Also, by the way, added a simple preliminary check to my answer that might help you rule out certain lists quickly, depending on your application. – thorimur Mar 2 at 0:04
Here is a recursive implementation. The recursion is stopped when the first result is found using Throw. For more results the recursion can be stopped later.
k is the number of sub lists and kmin is the minimal number of terms in the sum of n:
s = {{1, 2, 3}, {4, 5, 6, 7}, {1, 3}, {2, 4, 6}};
n = 19; (*searched sum*)
k = Length[s];
kmin = k; (* minimal length of result*)
Clear[step];
step[ind_, sum_] := Module[{len = Length[ind]},
Which[
sum > n, Nothing,
len >= kmin && sum == n, Throw[ Table[s[[i, ind[[i]]]], {i, len}]],
len < k,
Table[step[Append[ind, i], sum + s[[len + 1, i]]], {i,
Length[s[[len + 1]]]}]
, True, Nothing
]
]
Catch[step[{}, 0]]
(*{3, 7, 3, 6}*) | 2021-07-28T01:51:23 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/240865/scan-through-partial-tuples/240888",
"openwebmath_score": 0.3517647385597229,
"openwebmath_perplexity": 1631.2728188724054,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9693242000616578,
"lm_q2_score": 0.8670357580842941,
"lm_q1q2_score": 0.8404387426299114
} |
https://stats.stackexchange.com/questions/298717/why-is-my-regression-insignificant-when-i-merge-data-that-produced-two-significa/298718 | # Why is my regression insignificant when I merge data that produced two significant regressions?
Sorry for the confusing title, I think this is a general statistics question, but I'm working in R. I have a combined dataset of two samples from different countries (n=240 and n=1,010), and when I run a linear regression between the same three variables in each dataset, both datasets produce a significant result, with almost identical coefficients. However, when I merge the datasets and run the same regression on the combined dataset, it is no longer significant. Can anyone explain this?
In case it matters, the regression has the form lm(a~b*c).
• If the two regressions have significant but different slopes (very different and possibly of different sign) there is no reason to think that combining the data into a single regression will give a significant slope. Aug 18, 2017 at 23:42
• Like I said, the coefficients are almost the same. Thank you for your comment though, and I'm curious to hear if you have any advice on how to proceed in trying to solve this problem! Aug 18, 2017 at 23:49
• Sorry, I missed the part where you said that you had almost identical coefficients. But what exactly did you mean by almost identical? What were the significant levels for each? Aug 19, 2017 at 0:06
• Also what was the significance level for the combined regression? Aug 19, 2017 at 0:08
• What about the intercepts? Aug 19, 2017 at 0:37
Without seeing your data, this is difficult to answer definitively. One possibility is that your datasets span different ranges of the independent variable. It is well-known that combining data across different groups can sometimes reverse correlations seen in each group individually. This effect is known as Simpson's Paradox.
• Wow, that's really interesting, I had never heard of Simpson's Paradox! I wonder if you could give me some advice about how to proceed in trying to answer my research question, which is to see whether variable c moderates variable b 's effect on variable a. I'm puzzled as to how I should address something like this, because it seems like whether I say c moderates b, I'm correct in each country individually, but incorrect in general! I guess that's the paradox, but I'm still stumped. Aug 18, 2017 at 23:42
• Assuming you are dealing with Simpson's paradox here (something that we haven't fully established!), I think there are two key questions. First, do your two datasets correspond to different levels of a meaningful grouping factor. Second, if so, does the variation introduced by this factor represent nuisance variation that you want to control for (as opposed to interesting variation that you want to study). If you answered yes to both questions, then you might consider estimating a fixed effect of group (continued) Aug 19, 2017 at 1:15
• (continued) which might allow the model to draw parallel lines (with the slope of interest) through each of your two groups, while dealing with the between-group variation by giving the two lines different intercepts. But I emphasize that these are decisions that can only be made with a full conceptual/theoretical understanding of the problem that your analysis is supposed to answer. Aug 19, 2017 at 1:17
• +1 @Benji - Imagine two x-y scatter plots of data pts with the same slope, but one scatter plot is shifted to the right of the other on the x-axis, such that the best fitting regression line for both scatter plots is essentially flat. Aug 19, 2017 at 1:18
• @BenjiKaveladze Yes (see my answer below), but you'll want to verify this is the case by, for example, removing the country with the quadratic relationship from the dataset and see if the observed regression coefficients still change. At any rate, this illustrates how linear regressions can fail to detect nonlinear relationships which other techniques (like boosted regression, neural networks, decision trees, etc.) can model more effectively. Aug 19, 2017 at 14:03
If your data looks something like this then the reason may be more obvious. Your two original regression lines would be almost parallel and look reasonably plausible but combined they produce a different result which is probably not very helpful.
The data for this chart came from using the R code
exdf <- data.frame(
x=c(-64:-59, -52:-47),
y=c(-8.29, -8.36, -9.05, -9.30, -9.20, -9.69,
-7.90, -8.34, -8.49, -8.85, -9.38, -9.65),
col=c(rep("blue",6), rep("red",6)) )
fitblue <- lm(y ~ x, data=exdf[exdf$col=="blue",]) fitred <- lm(y ~ x, data=exdf[exdf$col=="red" ,])
fitcombo <- lm(y ~ x, data=exdf)
plot(y ~ x, data=exdf, col=col)
abline(fitblue , col="blue")
abline(fitred , col="red" )
abline(fitcombo, col="black")
which reports
> summary(fitblue)
Call:
lm(formula = y ~ x, data = exdf[exdf$col == "blue", ]) Residuals: 1 2 3 4 5 6 -0.00619 0.20295 -0.20790 -0.17876 0.20038 -0.01048 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -26.14895 2.91063 -8.984 0.00085 *** x -0.27914 0.04731 -5.900 0.00413 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1979 on 4 degrees of freedom Multiple R-squared: 0.8969, Adjusted R-squared: 0.8712 F-statistic: 34.81 on 1 and 4 DF, p-value: 0.004128 > summary(fitred) Call: lm(formula = y ~ x, data = exdf[exdf$col == "red", ])
Residuals:
7 8 9 10 11 12
-0.005238 -0.095810 0.103619 0.093048 -0.087524 -0.008095
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -26.06505 1.12832 -23.10 2.08e-05 ***
x -0.34943 0.02278 -15.34 0.000105 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.0953 on 4 degrees of freedom
Multiple R-squared: 0.9833, Adjusted R-squared: 0.9791
F-statistic: 235.3 on 1 and 4 DF, p-value: 0.0001054
> summary(fitcombo)
Call:
lm(formula = y ~ x, data = exdf)
Residuals:
Min 1Q Median 3Q Max
-0.8399 -0.4548 -0.0750 0.4774 0.9999
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -9.269561 1.594455 -5.814 0.00017 ***
x -0.007109 0.028549 -0.249 0.80839
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.617 on 10 degrees of freedom
Multiple R-squared: 0.006163, Adjusted R-squared: -0.09322
F-statistic: 0.06201 on 1 and 10 DF, p-value: 0.8084
not too far away from your statistics and with further work could be made closer
It's also possible the data points in each dataset may have completely different distributions due to outliers and/or nonlinear relationships between $x$ and $y$, and yet still share nearly identical linear regression coefficients, standard errors, and statistically significant $p$-values. Combining the two datasets could create a dataset that no longer has a strong linear relationship. See Anscombe's Quartet. A visual representation of numerous datasets sharing the same summary statistics but radically different scatterplots can be found here. My recommendation would be to closely examine the scatterplots of both datasets.
• In addition to examine the scatterplots, I would try to repeat regression using the country as an additional variable (a~bccountry). This way you will see if some coefficients change significantly between countries.
– Pere
Aug 19, 2017 at 8:47
• @Pere When I include country in the model (a~bccountry), the result produced is that the bc interaction variable becomes significantly related to a (b=-0.35, p<0.001). Can I interpret that as evidence that bc predicts a? It just seems weird to me that b*c only predicts a when I introduce the country variable into the equation. Thanks! Aug 19, 2017 at 22:40
• @BenjiKaveladze I'm not sure to understand your comment. I suggest posting the whole summary, maybe in another question. However, significant bc interactions means that you can take in account bc to get better predictions of a, which is equivalent to say that for different values of c you get a diferent relation between a and b.
– Pere
Aug 20, 2017 at 9:19
• @Pere Maybe it will help if I write the actual question, which is "do stressors moderate the link between social support and depression"? So the model is (depression~supportstress). My question is whether I should instead use the model (depression~supportstresscountry). This is tricky for me because only when I use the second model does the supportstress interaction become significant. When country is not in the model, the support*stress interaction is not significant. Does that make sense? Aug 21, 2017 at 18:42
For more on Simpson's Paradox see Pearl, J., & Mackenzie, D. (2018). Paradoxes Galore! The Book of Why: The New Science of Cause and Effect (Kindle ed., pp. 2843-3283). New York: Basic Books. Also, see Pearl's Causality.
In his book, Pearl gives an example very similar to yours. The problem is that there is a confounding variable that is affecting both the independent variable(s) and the dependent variable. In Pearl's example, the question is, Why is an anti-heart attack drug bad for women, bad for men, but good for people? (when the two gender samples are combined). The answer is that gender is a confounding variable that impacts who takes the drug (women are far more likely), and also the prevalence of heart attack (men are far more likely). The solution to confounding variables is to condition on them. The can be done in two ways: (1) Using regression analysis, make gender a variable; (2) Analyze the average effect of the drug for the two genders separately; then compute the weighted average (weighted by percent in population of the genders, here 1/2) of the effects.
Pearl would say that you have to have a model of the phenomenon you are studying, i.e., an exhaustive theory that takes into account all the variables involved in the response. Developing such a model and theory can take months of reading to understand the work of others in the field. However, recall that one left out variable can bias the results and make them meaningless or just plain wrong.
Pearl would also write that you cannot extract causality from data; for that you need a theoretical model. However, once you have a theory and a model, you can use data to support them. | 2022-05-19T22:08:41 | {
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https://notes.haifengjin.com/math/statistics/l_norms/ | # L-Norms
$L^p$-norms are some functions which takes a vector as input and output a value. It is written as $\left\|\mathbf x\right\|_ p$.
## L0-Norm
It is a measure of how many non-zero values are there in the vector. If have to put it into notations, we need to first define $0^0=0$. The the $L^0$-norm is as follows.
\left\|\mathbf x\right\|_ 0 = |x_{1}|^{0}+|x_{2}|^{0}+\cdots +|x_{n}|^{0}
## L1-Norm
\left\|\mathbf x\right\|_ 1 = |x_{1}|+|x_{2}|+\cdots +|x_{n}|
## L2-Norm
\left\|\mathbf x\right\|_ 2 = \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}
We also use its squared form quite often.
\left\|\mathbf x\right\|_ 2^2 = x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}
## L-Infinity-Norm
\left\|x\right\|_ {\infty }=\max \left\{|x_{1}|,|x_{2}|,\dotsc ,|x_{n}|\right\}
## General Form
For $0< p< \infty$, we have the following general form.
\left\|x\right\|_ {p}=\left(|x_{1}|^{p}+|x_{2}|^{p}+\dotsb +|x_{n}|^{p}\right)^{1/p}
## Visualization
Following is a visualization of the contour line of different $p$ values with the norm value equal to 1. | 2021-04-22T17:42:04 | {
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https://www.mathworks.com/help/symbolic/norm.html | # norm
Norm of matrix or vector
## Description
example
norm(A) returns the 2-norm of matrix A. Because symbolic variables are assumed to be complex by default, the norm can contain unresolved calls to conj and abs.
example
norm(A,p) returns the p-norm of matrix A.
norm(V) returns the 2-norm of vector V.
example
norm(V,P) returns the P-norm of vector V.
## Examples
collapse all
Compute the 2-norm of the inverse of the 3-by-3 magic square A:
A = inv(sym(magic(3)))
norm2 = norm(A)
A =
[ 53/360, -13/90, 23/360]
[ -11/180, 1/45, 19/180]
[ -7/360, 17/90, -37/360]
norm2 =
3^(1/2)/6
Use vpa to approximate the result with 20-digit accuracy:
vpa(norm2, 20)
ans =
0.28867513459481288225
Compute the norm of [x y] and simplify the result. Because symbolic variables are assumed to be complex by default, the calls to abs do not simplify.
syms x y
simplify(norm([x y]))
ans =
(abs(x)^2 + abs(y)^2)^(1/2)
Assume x and y are real, and repeat the calculation. Now, the result is simplified.
assume([x y],'real')
simplify(norm([x y]))
ans =
(x^2 + y^2)^(1/2)
Remove assumptions on x for further calculations. For details, see Use Assumptions on Symbolic Variables.
assume(x,'clear')
Compute the 1-norm, Frobenius norm, and infinity norm of the inverse of the 3-by-3 magic square A:
A = inv(sym(magic(3)))
norm1 = norm(A, 1)
normf = norm(A, 'fro')
normi = norm(A, inf)
A =
[ 53/360, -13/90, 23/360]
[ -11/180, 1/45, 19/180]
[ -7/360, 17/90, -37/360]
norm1 =
16/45
normf =
391^(1/2)/60
normi =
16/45
Use vpa to approximate these results to 20-digit accuracy:
vpa(norm1, 20)
vpa(normf, 20)
vpa(normi, 20)
ans =
0.35555555555555555556
ans =
0.32956199888808647519
ans =
0.35555555555555555556
Compute the 1-norm, 2-norm, and 3-norm of the column vector V = [Vx; Vy; Vz]:
syms Vx Vy Vz
V = [Vx; Vy; Vz];
norm1 = norm(V, 1)
norm2 = norm(V)
norm3 = norm(V, 3)
norm1 =
abs(Vx) + abs(Vy) + abs(Vz)
norm2 =
(abs(Vx)^2 + abs(Vy)^2 + abs(Vz)^2)^(1/2)
norm3 =
(abs(Vx)^3 + abs(Vy)^3 + abs(Vz)^3)^(1/3)
Compute the infinity norm, negative infinity norm, and Frobenius norm of V:
normi = norm(V, inf)
normni = norm(V, -inf)
normf = norm(V, 'fro')
normi =
max(abs(Vx), abs(Vy), abs(Vz))
normni =
min(abs(Vx), abs(Vy), abs(Vz))
normf =
(abs(Vx)^2 + abs(Vy)^2 + abs(Vz)^2)^(1/2)
## Input Arguments
collapse all
Input, specified as a symbolic matrix.
One of these values 1, 2, inf, or 'fro'.
• norm(A,1) returns the 1-norm of A.
• norm(A,2) or norm(A) returns the 2-norm of A.
• norm(A,inf) returns the infinity norm of A.
• norm(A,'fro') returns the Frobenius norm of A.
Input, specified as a symbolic vector.
• norm(V,P) is computed as sum(abs(V).^P)^(1/P) for 1<=P<inf.
• norm(V) computes the 2-norm of V.
• norm(A,inf) is computed as max(abs(V)).
• norm(A,-inf) is computed as min(abs(V)).
collapse all
### 1-Norm of a Matrix
The 1-norm of an m-by-n matrix A is defined as follows:
### 2-Norm of a Matrix
The 2-norm of an m-by-n matrix A is defined as follows:
The 2-norm is also called the spectral norm of a matrix.
### Frobenius Norm of a Matrix
The Frobenius norm of an m-by-n matrix A is defined as follows:
${‖A‖}_{F}=\sqrt{\sum _{i=1}^{m}\left(\sum _{j=1}^{n}{|{A}_{ij}|}^{2}\right)}$
### Infinity Norm of a Matrix
The infinity norm of an m-by-n matrix A is defined as follows:
${‖A‖}_{\infty }=\mathrm{max}\left(\sum _{j=1}^{n}|{A}_{1j}|,\text{\hspace{0.17em}}\sum _{j=1}^{n}|{A}_{2j}|,\dots ,\sum _{j=1}^{n}|{A}_{mj}|\right)$
### P-Norm of a Vector
The P-norm of a 1-by-n or n-by-1 vector V is defined as follows:
${‖V‖}_{P}={\left(\sum _{i=1}^{n}{|{V}_{i}|}^{P}\right)}^{1}{P}}$
Here n must be an integer greater than 1.
### Frobenius Norm of a Vector
The Frobenius norm of a 1-by-n or n-by-1 vector V is defined as follows:
${‖V‖}_{F}=\sqrt{\sum _{i=1}^{n}{|{V}_{i}|}^{2}}$
The Frobenius norm of a vector coincides with its 2-norm.
### Infinity and Negative Infinity Norm of a Vector
The infinity norm of a 1-by-n or n-by-1 vector V is defined as follows:
The negative infinity norm of a 1-by-n or n-by-1 vector V is defined as follows:
## Tips
• Calling norm for a numeric matrix that is not a symbolic object invokes the MATLAB® norm function. | 2020-09-18T21:35:54 | {
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https://math.stackexchange.com/questions/1291217/what-is-an-algorithm-for-describing-the-partition-of-this-equivalence-relation | What is an algorithm for describing the partition of this equivalence relation?
Let $\mathbb{R}$ be the set of real numbers,$f : \mathbb{R} \to \mathbb{R}$ a map, and $E$ the equivalence relation on $ℝ$ defined by $E = \{(x,y) \in \mathbb{R} \times \mathbb{R} \mid f(x) = f(y) \}.$
Describe the partition of $\Bbb{R}$ in the following case:
$f(x) = 2x^2+4x+8$ for all $x \in \mathbb{R}$.
I worked out that every $x$ is in fact equivalent to $-x-2$ by observation, but I would like to know if there is some algorithmic way to find the partition.
Yes. Just solve the equation $f(x) = f(y)$:
$$2x^2 + 4x + 8 = 2y^2 + 4y + 8 \\ 2(x^2-y^2) + 4(x-y) = 0 \\ (x-y) \big( 2(x+y) + 4 \big) = 0 \\ x = y \vee x+y = -2$$
So $(x, y) \in E \iff x = y \text{ or } y = -x-2$.
The function is just $2(x+1)^2 + 6$. Clearly, $f(x)=f(y)$ iff $|x+1|=|y+1|$. In other words, the distance from $-1$ is the same for $x$ and $y$. | 2019-12-08T18:53:49 | {
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https://face2ai.com/math-linear-algebra-chapter-2-5/ | # 矩阵的逆
## 逆
### $A^{-1}$
$$I=AA^{-1}$$
$$I=A^{-1}A$$
### Notes
Note1:
The Inverse exist if and only if elimination produces n pivots(row exchanges are allowed)
Note2:
The matrix A cannot have two different inverse.
Suppose $BA=I$ , $AC=I$ Then $B=C$ :
$$B(AC)=(BA)C$$
Gives
$$BI=IC$$
or
$$B=C$$
Note3:
if A is invertible, the one and only solution to $Ax=b$ is $x=A^{-1}b$
$$Ax=b$$
Then:
$$x=A^{-1}Ax=A^{-1}b\\ x=A^{-1}b$$
Note4:
This is important,Suppose there is nonzero vector $\textbf{x}$ such that $Ax=0$ then Acannot have an inverse . No matrix can bring $\textbf{0}$ back to $\textbf{x}$
$$Ax=0\ A^{-1}Ax=A^{-1}0\ x=A^{-1}0$$
Note5:
A 2×2 matrix is invertible if and only if $ad-bc$ is not zero:
$$\begin{bmatrix} a&b\newline c&d\end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d&-b\newline{-c}&a\end{bmatrix}$$
Note6:
A diagonal matrix has an inverse provided no diagonal entries are zero:
$$A=\begin{bmatrix}d_1&\,&\, \\,& \ddots &\,\\,&\,& d_n\end{bmatrix}$$
then
$$A^{-1}=\begin{bmatrix}1/d_1&\,&\,\\,& \ddots &\,\\,&\,& 1/d_n\end{bmatrix}$$
## $(AB)^{-1}$ and $(AB\dots Z)^{-1}$
$$(AB)^{-1}=B^{-1}A^{-1}$$
$$(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$$
$$(AB\dots Z)^{-1}=Z^{-1} \dots B^{-1}A^{-1}$$
$$(AB)^{-1}(AB)=B^{-1}(A^{-1}A)B=B^{-1}(I)B=B^{-1}B=I$$
## 高斯乔丹消元(GAUSS-JORDAN Elimination)
$$A^{-1}\begin{bmatrix}A&&I\end{bmatrix}\ =\begin{bmatrix}A^{-1}A&&A^{-1}I\end{bmatrix}\ =\begin{bmatrix}I&&A^{-1}\end{bmatrix}$$
$$\begin{bmatrix}A&&I\end{bmatrix}\ A=LR\$$
$$R=L^{-1}A\dots (1)\ L^{-1}\begin{bmatrix}A&&I\end{bmatrix}=\begin{bmatrix}R&&L^{-1}I\end{bmatrix}$$
(1)的主要过程就是通过消元，使得增广矩阵中的A矩阵变成一个上三角矩阵R，对角线一下都是零
$$R=UD \dots(2)\ D=U^{-1}R\ U^{-1}\begin{bmatrix}R&&L^{-1}I\end{bmatrix}=\begin{bmatrix}D&&U^{-1}L^{-1}I\end{bmatrix}$$
(2)回代过程，D只有对角线上有元素，其他全部是零
$$I=D^{-1}D\dots(3)\ \begin{bmatrix}D&&L^{-1}IU^{-1}\end{bmatrix}D^{-1}=\begin{bmatrix}I&&D^{-1}U^{-1}L^{-1}I\end{bmatrix}$$
(3)对角线归一化，使得A对角线上的元素变成1
$$A=LUD$$
$$A^{-1}=D^{-1}U^{-1}L^{-1}$$
### 逆矩阵的性质(Properties)
1：一个矩阵如果是对称的，并且有逆，那么逆也是对称的。
2：三角矩阵的逆如果存在可能是一个稠密矩阵
Subscribe | 2020-07-13T05:39:13 | {
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https://math.stackexchange.com/questions/1841359/value-of-finite-product-based-on-empty-set | # Value of finite product based on empty set
How does one evaluate the following product if the set S happens to be empty?
\begin{aligned} f(n)= n \prod_{x \in S} \left(1-\frac{1}{x}\right) \end{aligned}
Is the value simply n or is it undefined (or zero)??
Thanks.
Edit: It seems rather odd that this question has been rated off-topic for lacking context or other details. I would have thought it rather obvious that it was about how to evaluate the product when there is no x due to an empty set. I would have guessed undefined because one cannot assign a value to $(1-1/x)$. However, as shown by C.Falcon, the convention is $1$. There's no other context or missing details. Feel free to delete if it doesn't meet the relevant standards.
• This resembles the formula for Euler's totient function. If it is so, then $S$ is the divisors of $n$, which cannot be empty. – Kenny Lau Jun 27 '16 at 13:37
• Yes, I based my example on that function but my example is not intended to be defined such that S is the divisors of n. Cheers. – Molonglo Jun 27 '16 at 13:42
• – Martin Sleziak Jul 3 '16 at 13:56
• @kennyLau In fact, in the formula for $\varphi$ the set $S$ is set of primies dividing $n$, so if you want to use the formula to calculate $\varphi(1)$, you will get empty product. – Martin Sleziak Jul 3 '16 at 14:00
• @MartinSleziak Oh, my brain jammed. – Kenny Lau Jul 3 '16 at 16:04
An empty product is by convention equal to $1$ (the identity element for the multiplication), therefore whenever $S$ is empty, one has $f:n\mapsto n$. | 2019-07-15T18:11:32 | {
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https://math.stackexchange.com/questions/2488969/is-phiab-ge-phia-cdot-phib-true-for-every-positive-integer-pair-a-b | # Is $\phi(ab)\ge\phi(a)\cdot \phi(b)$ true for every positive integer pair $(a/b)$?
If $\phi(n)$ is the totient-function, does $$\phi(ab)\ge \phi(a)\cdot \phi(b)$$ hold for every pair $(a,b)$ of positive integers ? And does equality hold if and only if $\gcd(a,b)=1$ ?
I defined $$g:=\gcd(a,b)$$ $$a':=\frac{a}{g}$$ $$b':=\frac{b}{g}$$ and tried to reduce the problem to the $a'$ and $b'$, but this approach led to nowhere.
• This statement would be useful to solve this : math.stackexchange.com/questions/2488373/… – Peter Oct 25 '17 at 10:37
• I don't understand the claim "...and all those numbers are coprime to $ab$". If $a=2$, $b=6$ then, in your notation, $g=2,a'=1,b'=3$, yes? $2$ is a number smaller than $a'b'$ which is coprime to $a'b'$ but $2$ is not coprime to $ab$. Am I misreading? – lulu Oct 25 '17 at 10:54
• I believe that if you write $a=\prod p_i^{a_i}\times \prod q_j^{\alpha_j}$ and $b=\prod p_i^{b_i}\times \prod r_k^{\beta_k}$ where the $p_i$ are the primes dividing $\gcd (a,b)$ and the $q_j,r_k$ are primes distinct from each other and from the $p_i$ you can then write down both sides to see the inequality you desire. (the left has terms like $p_i^{a_i+b_i-1}$ the right has $p_i^{a_i+b_i-2}$). – lulu Oct 25 '17 at 10:57
• I don't see how the edit eliminated the counterexample I gave. The co-primality of the factors seems difficult to sort out, though I expect you can break it into cases. The explicit calculation I sketched, while somewhat unsatisfactory, handles this. – lulu Oct 25 '17 at 11:11
• I expect some variant of your argument will work. Like I said, I find the explicit computation somewhat unsatisfying...presumably there is an enumerative method that establishes what you want. At the moment, however, I am not seeing it. – lulu Oct 25 '17 at 11:15
Yes.
Using the formula $\phi(ab)=\phi(a)\phi(b)\frac{\gcd(a,b)}{\phi(\gcd(a,b))}$, we can see that $$\phi(ab) \geq \phi(a)\phi(b) \iff \frac{\gcd(a,b)}{\phi(\gcd(a,b))}\geq1$$ Denoting $c=\gcd(a,b)$, we just need to prove $c\geq \phi(c)$. However, this is always true, since $\phi(n)$ counts the number of positive integers up to $n$ relatively prime time to $n$, which can't ever be greater than $n$. We can also see the only solution to $c=\phi(c)$ is $c=1$, so we have $\phi(ab)=\phi(a)\phi(b) \iff \gcd(a,b) =1$.
• A useful formula! (+1 and accept) – Peter Oct 25 '17 at 13:22
Let $\{p_i\}$ be the list of primes dividing $\gcd (a,b)$. Then we can write $$a=\prod p_i^{a_i}\times \prod q_j^{\alpha_j}\quad \&\quad b=\prod p_i^{b_i}\times \prod r_k^{\beta_k}$$
Where the $q_j,r_k$ are primes disjoint from each other and from the $p_i$.
We can now compute both sides of your desired inequality. We get $$\varphi(ab)=\prod p_i^{a_i+b_i-1}(p_i-1)\times \varphi\left(\prod q_j^{\alpha_j}\right)\times \varphi\left(\prod r_k^{\beta_k}\right)$$ While $$\varphi(a)\varphi(b)=\prod p_i^{a_i+b_i-2}(p_i-1)^2\times \varphi\left(\prod q_j^{\alpha_j}\right)\times \varphi\left(\prod r_k^{\beta_k}\right)$$
From this we see that we can compute the ratio $$\boxed {\frac {\varphi(ab)}{\varphi(a)\varphi(b)}=\prod \frac {p_i}{p_i-1}}$$
The inequality you desire follows at once (as well as the claim that equality requires the gcd to be $1$).
Examples:
I. $a=12, b=16$. Then the only $p_i$ is $2$ and we remark that $$\varphi(192)=64=2\times \varphi(12)\times \varphi(16)$$
II. $a=18,b=60$. Then the $p_i$ are $2,3$ and we have $$\frac {\varphi(18\times 60)}{\varphi(18)\times \varphi (60)}=3=\frac 21\times \frac 32$$
III. $a=10,b=45$. In this case the ratio comes out $\frac 54$ as desired (I've included this examples just to illustrate that, of course, the ratio need not always be an integer).
• Nice answer (+1) – Peter Oct 25 '17 at 13:21 | 2019-10-21T10:34:52 | {
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https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_20&diff=prev&oldid=137917 | Difference between revisions of "2020 AMC 8 Problems/Problem 20"
A scientist walking through a forest recorded as integers the heights of $5$ trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
$$\begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.2cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.2cm}{0.15mm} meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}$$$\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$
Solution 1
It is not too hard to construct $22, 11, 22, 44, 22$ as the heights of the trees from left to right. The average is $\frac{121}{5}=24.2\rightarrow\boxed{\textbf{(B)}}$. ~icematrix Note: this is the result of realizing that the 1st one must be greater then the 2nd, $\frac{121}{5}=24.2 or {\textbf{(B)}}$. ~oceanxia
Solution 2
For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as $22, 11, 22$. Now, there are two possible cases.
Case 1: The length of tree 4 is $11$- So, we have the first 4 tree lengths as $22, 11, 22, 11$. For the length of Tree 5 to be an integer, we must have the length of Tree $5$ as $22$. But, when we take the average of these $5$ integers, it results in $17.6$, which doesn't satisfy our conditions.
Case 2: The length of tree 4 is $44$- So, we have the first $4$ tree lengths as $22, 11, 22, 44$. Now, using quick modular arithmetic, we see that when the length of Tree 5 is $88$, the average of the heights of the 5 trees is $\boxed{24.2}\rightarrow\boxed{\textbf{(B)}}$. This is where our condition is satisfied.
~ATGY
Solution 3
Let the sum of the heights of the trees be denoted by $S$. The average height will then be $\frac{S}{5}$. Since the average height has decimal part $.2$, it follows that $S$ must have a remainder of $1$ when divided by $5$. Thus, $S$ must be $1$ more than a multiple of 5. Since $S$ is an integer, it follows that the heights of Tree 1 and Tree 3 are likely both 22. At this point, our table looks as follows: $$\begin{tabular}{|c|c|} \hline Tree 1 & 22 meters \\ Tree 2 & 11 meters \\ Tree 3 & 22 meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}$$
If we guess that Tree 4 has a height of 11, we would need to make Tree 5 have a height of 22. In this case $S$ would equal $88$ which is not one more than a multiple of $5$. So we instead guess that Tree 4 has a height of $44$. Then we only need to try out heights of $22$ and $88$ for Tree 5. Using a height of $22$ for Tree 5 gives us $S=121$ which is $1$ more than a multiple of $5$. Thus, the average height of the trees is $\frac{121}{5}=24.2$ meters $\implies\boxed{\textbf{(B) }24.2}$.
~junaidmansuri
Solution 4
The problem states that all tree heights are integers. Therefore, we can deduce that the first and third trees have a height of $22$ meters. Trees four and five must have heights of either $11,22$ or $44,88$ or $44,22$. Checking which ones match the answer choices, we find that the trees four and five have heights of $44$ and $22$ meters, respectively. Thus, the answer is $\frac{22+11+22+44+22}5=\textbf{(B) }24.2$.
-franzliszt | 2021-12-03T14:34:16 | {
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https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_11&diff=prev&oldid=62487 | # Difference between revisions of "1999 AHSME Problems/Problem 11"
## Problem
The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$. If it costs \$137.94 to label all the lockers, how many lockers are there at the school?
$\textbf{(A)}\ 2001 \qquad \textbf{(B)}\ 2010 \qquad \textbf{(C)}\ 2100 \qquad \textbf{(D)}\ 2726 \qquad \textbf{(E)}\ 6897$
## Solution
### Solution 1
The locker labeling requires $\frac{137.94}{0.02}=6897$ digits. Lockers $1$ through $9$ require $9$ digits total, lockers $10$ through $99$ require $2 \times 90=180$ digits, and lockers $100$ through $999$ require $3 \times 900=2700$ digits. Thus, the remaining lockers require $6897-2700-180-9=4008$ digits, so there must be $\frac{4008}{4}=1002$ more lockers, because they each use $4$ digits. Thus, there are $1002+999=2001$ student lockers, or answer choice $\boxed{\textbf{(A)}}$.
### Solution 2
Since all answers are over $2000$, work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\cdot 90 = 3.60$. Lockers $100$ through $999$ cost $0.06\cdot 900 = 54.00$, and lockers $1000$ through $1999$ inclusive cost $0.08\cdot 1000 = 80.00$.
This gives a total cost of $0.18 + 3.60 + 54.00 + 80.00 = 137.78$. There are $137.94 - 137.78 = 0.16$ dollars left over, which is enough for $8$ digits, or $2$ more four digit lockers. These lockers are $2000$ and $2001$, leading to answer $\boxed{\textbf{(A)}}$. | 2022-10-01T12:44:29 | {
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https://math.stackexchange.com/questions/3630582/what-can-the-range-of-a-measure-be | # What can the range of a measure be?
Given a measure space $$(X,\mathcal{A},\mu)$$, what can the range of the measure, $$\mu[\mathcal{A}]$$, look like? Clearly it can't be an arbitrary subset of $$[0,\infty]$$ as we know $$0\in \mu[\mathcal{A}]$$. We also know $$\mu[\mathcal{A}]$$ has a maximal element ($$\mu(X)$$).
A bit less trivially, it must also satisfy the following for any $$x,y\in \mu[\mathcal{A}]$$:
$$\exists z\ \{z,x-z,y-z,x+y-z,x+y-2z\}\subseteq\mu[\mathcal{A}]$$
($$z$$ corresponds to the measure of the intersection of the sets $$x$$ and $$y$$ correspond to). This for instance tells us $$\mu[\mathcal{A}]\ne \{0,1,3\}$$.
Additionally, a fact that as far as I can tell is independent of the above comes from measuring the complement of a set: if $$x\in\mu[\mathcal{A}]$$ then $$M-x\in\mu[\mathcal{A}]$$, where $$M$$ is the unique maximal element of $$\mu[\mathcal{A}]$$ corresponding to $$\mu(X)$$.
Is any complete characterization known?
edit: for example, at first I thought it might have to be closed, as every natural measure I could think had closed range. But the range of measure on $$\mathbb{N}$$ generated by $$\mu(\{0\})=0.9$$, $$\mu(\{1\})=0.99$$, $$\mu(\{2\})=0.999$$... has a sequence of elements approaching 1, but does not contain 1 as any singleton set has measure less than 1 and any two-or-more element set has measure greater than 1.
If it's a finite set, there must be $$a_1, \ldots, a_m > 0$$ such that $$\mu[\mathcal A] = \{ \sum_i a_i x_i : x \in \{0,1\}^m\}$$.
• Thank you, nice to have such a clean characterization in the finite case. I actually wonder if this might somehow follow from my last condition, but that's certainly non-obvious. Apr 17, 2020 at 20:00
It is a well know result by Saks (also generalized by Lyapunov) that if $$\mu$$ has no atoms, then $$\mu$$ can take any value in $$[0,\mu[X])$$. If $$\mu$$ has atoms, the range of $$\mu$$ can get a little weird.
• According to Wikipedia, that theorem is due to Sierpinski. Oct 15, 2020 at 13:20
The range of a measure is either $$[0,\infty]$$, or there exists $$a\in[0,\infty)$$ and $$S\subset[0,\infty]$$ such that $$\mu[\mathcal{A}] = [0,a] + \left\{\sum_{x\in X} x \mid X\subset S\right\}$$ where we interpret the sum of two sets $$A+B = \{a+b \mid a\in A, b\in B\}$$.
Non-atomic case: As Oliver Diaz has pointed out, it is a well-known result that if $$\mu$$ has no atoms, then $$\mu$$ can take any value in $$[0,\mu(X)]$$. See https://en.wikipedia.org/wiki/Atom_%28measure_theory%29.
Purely atomic case: Suppose $$\mu$$ is purely atomic. Let $$\mathcal{B}\subset\mathcal{A}$$ be the set of atomic sets, i.e. for all $$B\in\mathcal{B}$$, $$\mu(B)\ne 0$$ but any measurable subset of $$B$$ has measure $$0$$ or has the same measure as $$B$$, and let $$\mu(\mathcal{B})$$ be the set of measures of atomic sets. For any $$B_1,B_2\in\mathcal{B}$$, it's straightforward to see that either their intersection has measure $$0$$ or their symmetric difference has measure $$0$$. Furthermore, we can take countable unions of them and still get a measurable set, and one can see that any measurable set with finite measure must be such a countable union plus a null set. Hence we have that if $$\mu$$ is purely atomic and $$\mathcal{B}$$ is the set of atoms, then for any measurable $$A$$ either $$\mu(A) = \infty$$ or $$\mu(A) = \sum_{b\in\mu(\mathcal{B})}^\infty b x_b$$ where $$x_b:\mu(\mathcal{B})\to\{0,1\}$$ has countable support.
General $$\sigma$$-finite case: Any $$\sigma$$-finite measure $$\mu$$ can be decomposed uniquely as $$\mu = \nu+\alpha$$ where $$\nu$$ is non-atomic and $$\alpha$$ is purely atomic. Letting $$C$$ be the set of measures of atoms of $$\alpha$$, we have that for any $$A\in\mathcal{A}$$, either $$\mu(A)=\infty$$ or $$\mu(A) = t + \sum_{c\in C} c x_c$$ where $$x_c : C\rightarrow \{0,1\}$$ has countable support and $$t\in[0,\nu(X)]$$. Given any such $$x$$ and $$t$$, one can construct a set with the corresponding measure, so this is the full range of $$\mu$$.
Thus, given any $$a\in[0,\infty]$$ and set $$S\subset [0,\infty]$$, there exists a measure space $$(X,\mu,\mathcal{A})$$ such that $$\mu[A] = [0,a] + \left\{\sum_{x\in X} x \mid X\subset S\right\}$$ and all $$\sigma$$-finite measures have ranges of this form.
General case: Non-$$\sigma$$-finite measures are often weird and messy, but one can show that their ranges are the same shape of sets. Suppose $$(X,\mathcal{A},\mu)$$ is not $$\sigma$$-finite and the range of $$\mu$$ is not $$[0,\infty]$$. Then we can let $$a = \sup\{x\in\mathbb{R}\mid \exists A\in\mathcal{A}\text{ non-atomic with }\mu(A)=x\}$$. Let $$X_a\in\mathcal{A}$$ be a set with measure $$a$$. Then any finite-measure non-null set that is disjoint from $$X_a$$ must contain an atom, which we can show as follows: Let $$Y\in\mathcal{A}$$ have finite measure and $$Y\cap X_a = \emptyset$$. If $$Y$$ contains no atoms, then $$(Y,\mathcal{A}\cap P(Y), \mu)$$ is a non-atomic measure space, hence $$[0,\mu(Y)]\subset\mu[A]$$. But, since $$Y$$ is disjoint from $$X_a$$, for any measurable subset $$Z\subset Y$$, $$\mu(X_a\cup Z) = a+\mu(Z) > a$$, which cannot happen by the definition of $$a$$, as it would imply $$[0,\mu(Y)+a]\subset\mu[\mathcal{A}]$$. This furthermore implies that any set with finite non-zero measure disjoint from $$X_a$$ can be written as a countable union of atoms. Hence, if $$C$$ is the set of all possible measures of atoms disjoint from $$X_a$$, the measure of any finite-measure set can be written as $$t+\sum_{c\in C}c x_c$$ where $$t\in[0,a]$$ and $$x_c:C\to\{0,1\}$$ has countable support. Hence any measure's range has the form $$[0,a] + \left\{\sum_{x\in X} x \mid X\subset S\right\}$$ for some $$a\in[0,\infty]$$ and $$S\subset [0,\infty]$$.
• Thank you so much! This has been bothering me a while, this is perfect! Oct 16, 2020 at 2:34 | 2022-08-11T06:54:13 | {
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https://math.stackexchange.com/questions/3074043/envelope-question-five-letters-addressed-to-individuals-1-5-are-randomly-placed | Envelope Question: Five letters addressed to individuals 1-5 are randomly placed in five addressed envelopes, one letter in each envelope.
I'm trying to find the probability of:
1. Exactly three letters go in the correct envelopes.
2. Exactly two letters go in the correct envelopes
3. No letters go in the correct envelopes
Here is my approach:
So there is clearly a total of 5! distinct ways of arranging the letters.
1. If exactly three letters go in the correct envelopes, then there are $$5 \choose 3$$ ways of choosing the positions for the three correct envelopes, and for the remaining two letters, there are 2! ways of organizing them. Thus, probability = $$\frac{{5 \choose 3} \cdot 2!}{5!}$$.
2. If exactly two letters go in the correct envelopes, then there are $$5 \choose 2$$ ways of choosing the positions for the two correct envelopes, and for the remaining three letters, there are 3! ways of organizing them. Thus, probability = $$\frac{{5 \choose 2} \cdot 3!}{5!}$$.
3. I'm not really sure how to approach this problem.
Any input would be great.
• How many ways are there to put two letters into the corresponding envelopes if neither of them goes in the correct envelope? – David Jan 15 '19 at 3:46
• Oh there is only one. Thanks. Then for part 2), could you give me a hint on how to figure out the number of ways for the 3 remaining envelopes to not go in the correct places? – Tim Weah Jan 15 '19 at 3:48
Note the use of the word exactly. You did not take that into account.
Probability that exactly three letters are placed in the correct envelopes
There are indeed $$\binom{5}{3}$$ ways to select which three letters are placed in the correct envelopes. That leaves two envelopes in which to place the remaining two letters. Those letters must each be placed in the other envelope, which can only be done in one way. Hence, the probability that exactly three letters go in the correct envelopes is $$\frac{1}{5!} \cdot \binom{5}{3}$$
Probability that exactly two letters are placed in the correct envelopes
There are $$\binom{5}{2}$$ ways to select which two letters are placed in the correct envelopes. None of the remaining letters go in the correct envelopes. There are just two ways to do this. \begin{align*} &\color{red}{1}, \color{red}{2}, \color{red}{3}\\ &\color{red}{1}, 3, 2\\ &2, 1, \color{red}{3}\\ &2, 3, 1\\ &3, 1, 2\\ &3, \color{red}{2}, 1 \end{align*} Hence, there are $$\binom{5}{2} \cdot 2$$ favorable cases.
We can use the Inclusion-Exclusion Principle to see why.
There are $$3!$$ ways to permute the three letters. From these, we must subtract those cases in which at least one letter is placed in the correct envelope.
There are three ways to select a letter to be placed in the correct envelope and $$2!$$ ways to place the remaining letters.
There are $$\binom{3}{2}$$ ways to select two letters to be placed in the correct envelopes and $$1!$$ ways to place the remaining letter.
There are $$\binom{3}{3}$$ ways to select three letters to be placed in the correct envelopes and $$0!$$ ways to place the (nonexistent) remaining letters.
By the Inclusion-Exclusion Principle, the number of ways to place the remaining three letters so that none of them is placed in the correct envelope is $$3! - \binom{3}{1}2! + \binom{3}{2}1! - \binom{3}{3} = 6 - 6 + 3 - 1 = 2$$ as claimed.
Therefore, the probability that exactly two letters are placed in the correct envelopes is $$\frac{1}{5!} \cdot 2\binom{5}{2}$$
Probability that none of the letters is placed in the correct envelopes
This is a derangement problem. An Inclusion-Exclusion arguments shows that the number of ways of placing none of the five letters in its correct envelope is
$$5! - \binom{5}{1}4! + \binom{5}{2}3! - \binom{5}{3}2! + \binom{5}{4}1! - \binom{5}{5}0!$$
Dividing that number by $$5!$$ gives the desired probability.
There are not $$2!$$ ways to organize the lat two letters. There is only $$1$$ way. Because the second way of organizing them would be to put them in their correct envelopes, which wouldn't match up with the constraint of having exactly $$3$$ letters getting sent correctly.
A similar mistake was made in the second problem.
To find the number of ways no letters get put in the correct envelope, also known as finding the number of derangements, start by taking $$5!$$ and subtract off all the cases letter $$1$$ is in the correct spot, then letter $$2$$, letter $$3$$, etc. Then use the inclusion exclusion principle
• For 2, could you give me a hint on how to find the number of ways that the remaining three envelopes can be arranged such that none of them are in the correct order? – Tim Weah Jan 15 '19 at 3:58
• Well you have stated that there would be $3! = 6$ ways to rearrange the remaining letters, you could just list these $6$ ways and count by hand which ways would result in no letters being in the correct spot – WaveX Jan 15 '19 at 4:01
The strategy here follows the inclusion/exclusion principle. For each subset of the letters, we ask how many ways there are to put those letters in the correct envelopes, if we don't care what happens to the other letters. That's an easy question to answer; if it's $$k$$ letters we want to put in the right place, then the remaining $$5-k$$ letters can go in the remaining $$5-k$$ envelopes in $$(5-k)!$$ ways.
So, suppose we want to put letters A and B in the correct envelopes, while the remaining letters C, D, and E go in the wrong envelopes. Well, we start with our count $$(5-2)!=6$$ ways to put A and B in the right places. Now, we need to subtract off the cases in which more than just those two went in the right envelopes - that's $$(5-3)!=2$$ ways to put A, B, and C in the right envelopes, $$(5-3)!=2$$ ways for ABD, and $$(5-3)!=2$$ ways for ABE. We're now down to zero - but we've overshot. If four letters ABCD went in the right envelopes, we subtracted that twice instead of once - so we add that $$(5-4)!=1$$ way back, and the same with ABCE and ABDE. Back up to 3 - and we've overshot again. In the case that all five letters are in the right place, we've added in too many copies, and we need to subtract one to get it back in balance. That drops us to the final answer of two ways. Multiply by the ten ways to choose two of the five letters to be those that we place correctly, and the answer to question (2) is $$\boxed{20}$$.
That's how inclusion/exclusion works - when we can cleanly count how many ways to do at least a certain number of things, we count how many ways to do exactly a certain number of things with this alternating sum over subsets that include our base. I'll leave running the details on question (3) as an exercise; it's all the same principles, just larger.
Note that this particular problem, of permutations that don't fix anything, is worked in that Wikipedia article (for any size). It's well known, as the derangement problem. | 2020-12-03T17:51:18 | {
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http://mathhelpforum.com/math-challenge-problems/7742-ladder-box-wall-print.html | Ladder and box (and wall)
• November 18th 2006, 05:37 PM
TriKri
Ladder and box (and wall)
A ladder with the length 10 m is leaned against a box (all sides has the length1 m) placed against a wall. But the ground is so slippery that the ladder falls ower the bow and leans against the wall as well. How high up on the wall does the ladder reach?
I had tried to solve this but only get a fourth order equation, and I don't know how to solve any equation of order higher than two. It seems easy but it turns out it isn't. Is the equation solvable?
By the way, my previous thread about modular arithmetic and equation systems wasn't finished, though it was marked as if it was, I happen to click on the "thank you" button by mistake. I would appreciate if someone took a look at it again, or maybe unmarked it as a finished thread. Please? :)
• July 31st 2008, 06:03 PM
Serena's Girl
Reviving an old problem
This is how I visualize the situation:
http://farm4.static.flickr.com/3271/...a5454f.jpg?v=0
The value of (x + 1) indicates how high up the wall the ladder reaches, while the value of (y + 1) indicates the distance of the end of the ladder from the wall.
So set up an equation based on the similarity of the two gray triangles:
$
x = \frac {1} {y}
$
And then set up an equation using the Pythagorean theorem:
$
(x + 1)^2 + (1 + y)^2 = 100
$
Combine the two equations, and you end up with:
$
x^4 + 2x^3 - 98x^2 + 2x + 1 = 0
$
It is a fourth-order equation, and I do not know how to solve this algebraically. However, in this situation, I prefer using numerical methods. Specifically, for this problem, I choose to use Newton-Raphson method together with MS Excel (Wiki article here: Newton's method - Wikipedia, the free encyclopedia).
Newton-Raphson method is an iterative method, where:
$
x_{n+1} = x_n - \frac {f(x_n)} {f'(x_n)}
$
where
$
f(x) = x^4 + 2x^3 - 98x^2 + 2x + 1
$
$
f'(x) = 4x^3 + 6x^2 - 196x + 2
$
The goal is to make f(x) = 0. Iterating with an initial value of x = 1, I obtained x = 0.111881932. Iterating with an initial value of x = 10, I obtained x = 8.937993689.
So before sliding down, the ladder reaches about 9.94 m up the wall. After sliding down, it reaches up only 1.11 m up the wall.
• July 31st 2008, 07:04 PM
Soroban
Hello, TriKri!
I too got a fourth-degree equation, too.
But there is a way around it . . .
[quote]A 10-m ladder is leaned against a box (1 x 1 x 1 m) placed against a wall.
But the ground is so slippery that the ladder falls over the box and leans against the wall.
How high up on the wall does the ladder reach?
Code:
| * * |y 10 * - - - * * | 1 | * 1| |1 * | | * - - - - - - - * - - - * x 1
From the large right triangle: . $(y+1)^2 + (1+x)^2 \:=\:10^2$ .[1]
The two smaller right triangles are similar.
. . So we have: . $\frac{1}{x} \:=\:\frac{y}{1} \quad\Rightarrow\quad x \:=\:\frac{1}{y}$ .[2]
Substitute [2] into [1]: . $(y+1)^2 + \left(1 + \frac{1}{y}\right)^2 \:=\:100$
and we have: . $y^2 + 2y + 1 + 1 + \frac{2}{y} + \frac{1}{y^2} \;=\;100$
. . $\left(y^2 + 2 + \frac{1}{y^2}\right) + \left(2y + \frac{2}{y}\right) - 100 \:=\:0$
. . $\left(y + \frac{1}{y}\right)^2 + 2\left(y + \frac{1}{y}\right) - 100 \:=\:0$
Let $u \:=\:y+\frac{1}{y}$
. . and we have the quadratic: . $u^2 + 2u - 100 \:=\:100$
Use the Quadratic Formula to solve for $u$,
. . then back-substitute and solve for $y.$
• July 31st 2008, 08:00 PM
Serena's Girl
Wow, that is so cool! *bows down to Soroban*(Bow)
• August 1st 2008, 11:43 AM
TriKri
Congratulations to both of you! You showed two different ways to solve the problem in and both where rigth. Am I supposed to deliver some virtual medal to you now? | 2015-08-01T18:42:05 | {
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https://math.stackexchange.com/questions/361255/xor-of-binary-numbers-to-reach-a-given-number | # XOR of Binary Numbers to Reach a Given Number
Given a set
S = { s1, s2, s3, ... sn}
of Binary Numbers , I need to find if a given Binary Number X with only 1 bit position set as 1 (..00001000...), can be reached by doing bitwise XOR operation.That is ,I need to find out if there is a subset of S such that
X = si (+) sj (+) sk ....
where (+) is XOR
I have read the dp approach given here, but I am not sure if it is valid for XOR as well.
EDIT : Under what conditions will there be no solution? (eg) 1) If all of {s1,s2,s3.... ,sn} have even number of bits, there can be no solution to X. 2) If none of the elements in S satisfy the condition,
X BITWISE AND si = X
Bitwise XOR is equal to bitwise addition modulo two. So you can treat this as a problem in linear algebra, and Gaussian elimination will solve it with polynomial complexity.
As an example, if $S=\{11001_2, 10101_2, 00111_2\}$, and you are asking whether $b_4b_3b_2b_1b_0$ is a bitwise XOR of a subset, all you need to is to check, whether the linear system of equations corresponding to the augmented matrix $$\left( \begin{array}{ccc|c} 1&1&0&b_4\\ 1&0&0&b_3\\ 0&1&1&b_2\\ 0&0&1&b_1\\ 1&1&1&b_0\end{array}\right)$$ is solvable. Just do all the arithmetic modulo two.
An example run with $S=\{0010, 1001, 1010, 0101, 1110, 1100\}$ and $X=0001$. The augmented matrix is $$\left( \begin{array}{cccccc|c} 0&1&1&0&1&1&0\\ 0&0&0&1&1&1&0\\ 1&0&1&0&1&0&0\\ 0&1&0&1&0&0&1\\ \end{array}\right)$$ We first do some row swaps. Move the third row to the top (need to get that $1$ to top left corner), but keep the initial top row as the second: $$\left( \begin{array}{cccccc|c} 1&0&1&0&1&0&0\\ 0&1&1&0&1&1&0\\ 0&0&0&1&1&1&0\\ 0&1&0&1&0&0&1\\ \end{array}\right).$$ The first column looks good. To clear the second we need to add (=bitwise XOR) the second row to the last. That gives us $$\left( \begin{array}{cccccc|c} 1&0&1&0&1&0&0\\ 0&1&1&0&1&1&0\\ 0&0&0&1&1&1&0\\ 0&0&1&1&1&1&1\\ \end{array}\right).$$ Swap the two bottom rows to end with $$\left( \begin{array}{cccccc|c} 1&0&1&0&1&0&0\\ 0&1&1&0&1&1&0\\ 0&0&1&1&1&1&1\\ 0&0&0&1&1&1&0\\ \end{array}\right).$$ This is already in the echelon form, and we can already declare the calculation a success in the sense that a solution exists. Let $x_1,x_2,\ldots,x_6$ be the (binary) unknown coefficient of the six numbers. The two last variable do not correspond to initial $1$s on any row, and we can arbitrarily assign them to have whatever value we please. Let's pick $x_5=x_6=0$. This leaves the equation of the bottom row to read $$x_4+1\cdot 0+1\cdot 0=0$$ allowing us to solve $x_4=0$.
Plugging in the known values for $x_4,x_5,x_6$ to the equation of the third row gives then $$x_3+1\cdot0+1\cdot0+1\cdot0=1$$ giving us $x_3=1$. Repeating the dose with the second row gives $$x_2+1\cdot1+1\cdot0+1\cdot0=0$$ telling us that $x_2=1$. Finally, the first row gives us that $$x_1+x_3+x_5=0,$$ completing our solution with $x_1=1$.
We see that a solution is $x_1=x_2=x_3=1$, $x_4=x_5=x_6=0$. Let's look at the ones. They occur as multipliers of the first three numbers: $0010,1001,1010$. Indeed, bitwise XORring these gives what we want $$0010 \operatorname{XOR} 1001 \operatorname{XOR} 1010 = 0001.$$
• IOW, the subset sum problem is much, much harder. – Jyrki Lahtonen Apr 14 '13 at 13:04
• Thank You! Can I get a reference link on how to find whether the augmented matrix is solvable ? The size of the set S can be upto 1000. I just need to find whether a solution exists or not. – Kyuubi Apr 14 '13 at 13:13
• It is the row reduction algorithm from Linear Algebra. Transform the matrix into a row echelon form. The system is solvable, if and only if you don't get a row of the form $(0\ 0\ 0\ \cdots\ 0\ |\ 1)$. – Jyrki Lahtonen Apr 14 '13 at 13:18
• I tried reducing the matrix to a row echelon form. But with the modulo 2 arithmetic operations, the row echelon form seems to be unreachable. (eg) if the set S = {0010, 1001, 1010, 0101, 1110, 1100 } and X = {0001}, how do I follow the reduction procedure ? I have to develop a c++ implementation for a 1000x200 matrix. – Kyuubi Apr 14 '13 at 14:55
• @Kyuubi: I added an example run solving your example case. – Jyrki Lahtonen Apr 14 '13 at 19:14 | 2019-10-21T23:00:10 | {
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http://www.math.psu.edu/treluga/450/lecture15.html | # Lecture 15 - Fitting distributions to data
(previous, next)
• Ideal asymptotic distributions
• Examples of fitting data distributions
## More asymptotic distributions
Using methods similar to what we did for the Poisson distribution, we can derive other asymptotic approximations to distributions.
### Normal aka Gaussian
$\frac{1}{\sigma\sqrt{2\pi} } \; e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$
• Also discovered by Abraham de Moivre (1667 - 1754).
• history -- for errors in astronomical measurement
• Derivable from the binomial distribution under the assumption of a large sample size, but fixed probability of success.
• Describes the behavior of all sums of IID random variables $$( x N - \sum_i X_i )/N$$.
### Fisher--Tippett--Gumbel extreme value theory
• Generalized extreme-value distribution, including the Weibull, Gumbel, and Frechet, have culumative distribution function $F(x) = \exp\left\{-\exp \left(-\frac{x-\mu}{\sigma}\right)\right\}$
• $$[ Max( X_i ) - x_N ]/N$$
• Emil Gumbel in the fight against Nazi's long before the start of WWII. See the entertaining history "The lady tasting tea".
• Related to Monod's philosophical thinkings about randomness and the nature of biology.
• Example: When making thread, the strength depending on weakest fiber in the chain, so the minimum of the strengths of each piece.
### Others
• Black-body radiation and Planck's law in quantum mechanics.
Many of the common probability distributions have special relationships between each other -- see Distribution relationships.
## Examples of applications of Poisson
Now, let's return to our Poisson distribution, which we've derived as a possible model of bolides burning up in earths atmosphere. We'd like to see if the Poisson distribution does a good job of matching the observations of bolides. But we've got a catch, in that we don't know what the parameter $$\lambda$$'s true value is. How can we compare our prediction to our data if we don't know $$\lambda$$?
Well, one answer is to just try many values of $$\lambda$$ and look for the one that does the best job of fitting the data. Let's look at some classical examples.
### Deaths from horse-kicks
Here is a classic example of something that seems totally unrelated by can be explained by the Poisson distribution. Notice that in this case, the poisson distribution only fails to explain about 4 percent of the distribution's mass when the rate parameter is picked optimally. That's pretty good, given the "totally random" nature of the process.
Annual deaths from horse kicks in the Prussian army (1875-1894)
Deaths Record
0 144
1 91
2 32
3 11
4 2
5 0
6 0
[Show code]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 #!/usr/bin/env python """ This script creates an animation showing how a distribution can be fit to a data. First, we simulate or load a data set. Then we compare the data to binomial distributions with a range of success probabilities. """ from scipy import floor, cumsum, array, linspace, cos, pi, rand, loadtxt from scipy.stats import poisson from scipy.optimize import fminbound from matplotlib.pyplot import * import time def norm(x,e): return sum(abs(x)**e)**(1./e) def rndBinomial(p, n, replicas=1): if replicas == 1: return int(sum(floor(rand(n) + p))) trials = [ rndBinomial(p,n) for i in xrange(replicas)] return trials def main(): # load data data = loadtxt('horse_kicks.txt',delimiter=',') x, y = map(array, zip(*data)) N = sum(y) Y = y*1./N # distribution fitting def fit_error(L_guess): # calculate the distance between our guess and the data z = poisson.pmf(x, L_guess) return norm(Y-z, 1)/norm(Y,1) # can be any p-norm L_best = fminbound(fit_error, 1e-5, 3-1e-5) err_best = fit_error(L_best) print "Best fit: L = %f, error = %f"%(L_best, fit_error(L_best)) # generate data for animation F = [ (p, fit_error(p)) for p in linspace(1e-4,3-1e-4, 137) ] u, v = map(array, zip(*F)) # draw basic figure fig = figure() subplot(2,1,2) plot(u, v, 'b-') #plot( L_best, fit_error(L_best), 'ko') marker_fit_error = plot( u[2], v[2], 'ro').pop() ylim(0,1) xlabel('Poisson intensity $\lambda$') subplot(2,1,1) plot(x, y, 'ko') z = poisson.pmf(x, L_best) width = .3 xw = x-width*.5 bar_ob = bar(xw, z*N, width, color='r') title('N = %d observations, $\lambda_{best} = %.3f$, $Err_{best} = %1.2g$'%(N, L_best, err_best)) ylabel('Number of times observed') xlabel('Number of successes') xlim(-0.5,max(x)+.5) ylim(0,max(y)*1.2) show(block=False) i, di = 0, 1 i_max = len(u) j = 0 while True: marker_fit_error.set_data([ u[i] ], [v[i]]) z = poisson.pmf(x, u[i])*N for rect, h in zip(bar_ob, z): rect.set_height(h) fig.canvas.draw() savefig('frame%04d.tif'%j) j += 1 i += di if i == i_max: di = -1 i += 2*di elif i == 0: break main()
### Accidents in a factory
Another example, which seems like it should also be a Poisson distribution, was collected for the rates of accidents in an ammunition factory.(Not a place you want to be having accidents!) In this case, though, the best explanation leaves more than 30 percent of that probability mass unexplained, even at the best fit -- okay, but nowhere near as satisfying as the previous example. Instead, the authors originally proposed the data are really obeying a negative-binomial distribution.
Number of munitions accidents in WWI factor per person
Count Observed
0 447
1 132
2 42
3 21
4 3
5 2
Over 5 0
Sum 647
[Show code]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 #!/usr/bin/env python """ This script creates an animation showing how a distribution can be fit to a data. First, we simulate or load a data set. Then we compare the data to binomial distributions with a range of success probabilities. """ from scipy import floor, cumsum, array, linspace, cos, pi, rand, loadtxt from scipy.stats import poisson from scipy.optimize import fminbound from matplotlib.pyplot import * import time def norm(x,e): return sum(abs(x)**e)**(1./e) def rndBinomial(p, n, replicas=1): if replicas == 1: return int(sum(floor(rand(n) + p))) trials = [ rndBinomial(p,n) for i in xrange(replicas)] return trials def main(): # load data data = loadtxt('munitions_factory.data',delimiter=',') x, y = map(array, zip(*data)) N = sum(y) Y = y*1./N # distribution fitting def fit_error(L_guess): # calculate the distance between our guess and the data z = poisson.pmf(x, L_guess) return norm(Y-z, 1)/norm(Y,1) # can be any p-norm L_best = fminbound(fit_error, 1e-5, 3-1e-5) err_best = fit_error(L_best) print "Best fit: L = %f, error = %f"%(L_best, fit_error(L_best)) # generate data for animation F = [ (p, fit_error(p)) for p in linspace(1e-4,3-1e-4, 137) ] u, v = map(array, zip(*F)) # draw basic figure fig = figure() subplot(2,1,2) plot(u, v, 'b-') #plot( L_best, fit_error(L_best), 'ko') marker_fit_error = plot( u[2], v[2], 'ro').pop() ylim(0,1) xlabel('Poisson intensity $\lambda$') subplot(2,1,1) plot(x, y, 'ko') z = poisson.pmf(x, L_best) width = .3 xw = x-width*.5 bar_ob = bar(xw, z*N, width, color='r') title('N = %d observations, $\lambda_{best} = %.3f$, $Err_{best} = %1.2g$'%(N, L_best, err_best)) ylabel('Number of times observed') xlabel('Number of successes') xlim(-0.5,max(x)+.5) ylim(0,max(y)*1.2) show(block=False) i, di = 0, 1 i_max = len(u) j = 0 while True: marker_fit_error.set_data([ u[i] ], [v[i]]) z = poisson.pmf(x, u[i])*N for rect, h in zip(bar_ob, z): rect.set_height(h) fig.canvas.draw() savefig('frame%04d.tif'%j) j += 1 i += di if i == i_max: di = -1 i += 2*di elif i == 0: break main()
### Bolide impacts
Now, let's look at our data on bolides in the same way.
[Show code]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 #!/usr/bin/env python2 # dates of bolide observations from NASA between 1994 and 2013 # http://en.es-static.us/upl/2014/11/bolide_events.jpg # http://neo.jpl.nasa.gov/fireball/ # # These are "large". Every day about 100 tons (90,000 kgs) of meteoroids -- # fragments of dust and gravel and sometimes even big rocks - enter the Earth's # atmosphere. Earth has a mass of about 5.972 x 10^24 kg. from matplotlib.pyplot import * from scipy import linspace, linalg, array, log, exp, loadtxt import time def loaddatedata(): # translate date into seconds f1 = lambda s : int(time.mktime(time.strptime(s, '%Y-%b-%d'))) # translate time-of-day into seconds f2 = lambda s : int(time.mktime(time.strptime(s, '%H:%M:%S'))) x = loadtxt('data/meteors.txt', converters={0:f1, 1:f2 }) # normalize, convert from seconds to days y = x[:,0] + x[:,1] y = (y - min(y))/(60*60*24.) y.sort() return y def bin_data(y, n): tmin, tmax = min(y), max(y) print "Rough parameter estimate $\lambda=%f$"%((len(y)-1)/(tmax-tmin)) dt = (tmax-tmin)/float(n) print "# bin size = %f days"%dt def g(t): return int( (t-tmin)/dt ) h = [ g(i) for i in y] x,y = zip(*[(i,h.count(i)) for i in xrange(n)]) return dt,x,y def diff(x): return [ float(x[i]-x[i-1]) for i in range(1,len(x)) ] def main(): data = loaddatedata() dt, u, v = bin_data(data, 80) x, y = zip(*[ (i, v.count(i)) for i in range(min(v), max(v)+1)]) figure(1, figsize=(9,6)) z = diff(data) z.sort() subplot(1, 2, 1) data = array(data) - min(data) plot(data,'go') xlabel('Rank') ylabel('Day from start') title('Dates of bolide sitings') subplot(1, 2, 2) bar(x,y,color='b') ylim(0, 1.1*max(y)) title('%d day bins'%dt) ylabel('Frequency') xlabel('Count') savefig('figures/meteorDataAndFit.pdf',bbox_inches='tight') savefig('figures/meteorDataAndFit.png',bbox_inches='tight') #show() main()
The choice of bins is a bit arbitrary, which should make us uncomfortable, but if we just go with it, we can now look for the value of $$\lambda$$ that leads to the closest-matching Poisson distribution.
[Show code]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 #!/usr/bin/env python """ This script creates an animation showing how a distribution can be fit to a data. First, we simulate or load a data set. Then we compare the data to distributions over a parameter range, and find the "best one" """ from scipy import floor, cumsum, array, linspace, cos, pi, rand, loadtxt from scipy.stats import poisson #from scipy.optimize import minimize_scalar as fminbnd from scipy.optimize import fminbound from matplotlib.pyplot import * import time import matplotlib.animation as manimation fminbnd = lambda a,b : fminbound(a,b[0],b[1]) def norm(x,e): return sum(abs(x)**e)**(1./e) def samplefit(L_best, N, numsamples): r = poisson(L_best) M = 50 ideal = poisson.pmf(range(M), L_best)*N for t in range(numsamples): s = list(r.rvs(N)) smple = array([ s.count(i) for i in range(M) ]) assert sum(smple) == N yield norm(ideal-smple, 1)/norm(ideal,1) return def main(): # load data data=[(0,3),(1,9),(2,13),(3,16),(4,12),(5,14),(6,7),(7,4),(8,2),(9,0),(10,0)] Dt = 45.82 x, y = map(array, zip(*data)) N = sum(y) Y = y*1./N # distribution fitting def fit_error(L_guess): # calculate the distance between our guess and the data z = poisson.pmf(x, L_guess) return norm(Y-z, 1)/norm(Y,1) # can be any p-norm L_best = fminbound(fit_error, 1e-5, 10-1e-5) err_best = fit_error(L_best) print "Best fit: L = %f, error = %f"%(L_best, fit_error(L_best)) # generate data for animation u, v = map(array, zip(*[ (p/Dt,fit_error(p)) for p in linspace(1e-4,10-1e-4,137) ])) subplot(2, 1, 2) plot(u, v, 'b-') #plot( L_best, fit_error(L_best), 'ko') marker_fit_error = plot( u[2], v[2], 'ro').pop() ylim(0, 1.5) xlabel('Poisson intensity $\lambda$') subplot(2, 1, 1) plot(x, y, 'bo') z = poisson.pmf(x, L_best) width = 0.3 bar_ob = bar(x - 0.5*width, z*N, width, color='r') title('N = %d observations, $\lambda_{best} = %.3f$, $Err_{best} = %1.2g$'%(N, L_best/Dt, err_best)) ylabel('Number of times observed') xlabel('Number of successes') xlim(-0.5, max(x)+.5) ylim(0, max(y)*1.2) savefig('meteorbinfit.pdf',bbox_inches='tight') savefig('meteorbinfit.png',bbox_inches='tight') figure(2) subplot(2,1,1) sampleerr = [ e for e in samplefit(L_best, N, 200) ] sampleerr.sort() plot(sampleerr,'kx-') subplot(2,1,2) M = 16 ideal = cumsum(poisson.pmf(range(M), L_best)*N) r = poisson(L_best) for t in range(30): s = list(r.rvs(N)) smple = cumsum(array([ s.count(i) for i in range(M) ])) plot(ideal, smple,'ko-') plot(cumsum(z*N), cumsum(y),'ro-') show() return main()
## Poisson processes in two dimensions
Calculate the distances between all pairs of points, and rank these distances from smallest to largest.
[Show code]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 #!/usr/bin/python2 from scipy import rand, zeros, sqrt, pi, array, arcsin, sign, linspace, real from scipy import randn from matplotlib.pyplot import * def hvy(x): # Heaviside function return real((sign(x)+1.)*0.5) def ripleyK(r): # Calculate Ripley's K statistic for a 1 x 1 torus S = sqrt(0.5) p = real(sqrt(4*r*r-1)) q = real(arcsin(real(1-p)/(1e-19+r*sqrt(8)))) y = hvy(r)*hvy(0.5-r)*pi*r*r #print hvy(0.5-r)*hvy(S-r) y += hvy(r-0.5)*hvy(S-r)*(p+4*r*r*q) y += hvy(r-S) return y def calcpointdistances(x,y): turn = lambda t : min([abs(t-1), abs(t), abs(t+1)]) n = len(x) assert len(y) == n for j in range(n): for k in range(j): u = turn(x[j]-x[k]) v = turn(y[j]-y[k]) yield sqrt(u*u+v*v) def getnormaldata(numpts): x = randn(numpts)/5+0.5 y = randn(numpts)/5+0.5 dists = array([ i for i in calcpointdistances(x,y) ]) dists.sort() return x, y, dists def getpoissondata(numpts): x = rand(numpts) y = rand(numpts) dists = array([ i for i in calcpointdistances(x,y) ]) dists.sort() return x, y, dists def main(): n = 83 #x, y, distances = getnormaldata(n) x, y, distances = getpoissondata(n) baseline = linspace(0,1,len(distances)) Area = ripleyK(distances) Err = Area - baseline toterr = sum(abs(Err))/sum(baseline) figure(1) plot(x, y, 'o') xlim(0,1) ylim(0,1) title('%d Poisson points in a plane'%n) savefig('ripley-planepts.pdf') savefig('ripley-planepts.png') figure(2) subplot(2,1,1) plot(distances, Area, 'r-', linewidth=2) plot(distances, baseline, 'b' ) legend(['Predicted', 'Observed']) title('Prediction and observation for Ripley K statistic') ylabel('Cumulative Probability') subplot(2,1,2) plot(baseline, Err, 'g-' ) title('%d Points, Relative Error = %.3f'%(n,toterr)) ylabel('Residual vector over distance') xlabel('Distance from origin') savefig('ripley-test.pdf') savefig('ripley-test.png') main()
There are More data sets like these that you can explore, out on the internet. | 2017-10-23T10:11:24 | {
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http://math.stackexchange.com/questions/255254/what-is-the-probability-of-at-least-2-blue-balls-given-the-probability-of-at-lea?answertab=active | # What is the probability of at least 2 blue balls given the probability of at least 1 blue ball?
A bag has red, blue, and green balls. The probabilities of randomly grabbing a red, blue, and green ball from the bag (with replacement) are $r$, $b$, and $g$ respectively. I randomly grab $n$ balls from the bag. What's the probability that at least 2 out of the $n$ balls are blue given that one of them is blue?
Here is what I have tried.
Let $$A \rightarrow\text{ The event in which at least 2 out of the n balls are blue.} \\ B \rightarrow\text{ The event in which at least 1 out of the n balls is blue.}$$
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} \\ P(B) = 1 - (1 - b)^n$$
As you see, I got $P(B)$ from subtracting out a complement probability. However, how do I get $P(A)$?
-
When we grab, is it with replacement or without? If without, I doubt you can find an exact answer without knowing how many balls there are. – André Nicolas Dec 10 '12 at 7:18
Oh! With replacement. Sorry about that. I must clarify. – John Hoffman Dec 10 '12 at 7:21
I think given what P(B) is, it's assumed that it's with replacement or from an infinite bag. – Joe Z. Dec 10 '12 at 7:22
We need to assume that we are grabbing with replacement, which doesn't sound much like grabbing. The probability of $A$ is $1$ minus the probability of $0$ or $1$ blues.
You already found the probability of $0$ blues. For $1$ blue exactly, the probability is $\dbinom{n}{1}b(1-b)^{n-1}$. More generally, the probability of exactly $k$ blues is $\dbinom{n}{k}b^k(1-b)^{n-k}$ (binomial distribution).
-
$P(A) = 1 - (1-b)^n - n b^1 (1-b)^{n-1}$
Thanks, why is the probability that exactly one ball is blue $b(1-b)^{n-1}$? Couldn't any one of the $n$ balls be blue, resulting in $nb(1-b)^{n-1}$? – John Hoffman Dec 10 '12 at 7:23 | 2014-04-16T14:18:20 | {
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https://math.stackexchange.com/questions/362975/concise-proof-that-every-common-divisor-divides-gcd-without-bezouts-identity | # Concise proof that every common divisor divides GCD without Bezout's identity?
In the integers, it follows almost immediately from the division theorem and the fact that $a | x,y \implies a | ux + vy$ for any $u, v \in \mathbb{Z}$ that the least common multiple of $a$ and $b$ divides any other common multiple.
In contrast, proving $e|a,b \implies e|gcd(a,b)$ seems to be more difficult. In Elementary Number Theory by Jones & Jones, they do not try to prove this fact until establishing Bezout's identity. This Wikipedia page has a proof without Bezout's identity, but it is convoluted to my eyes.
I tried my hand at it, and what I got seems no cleaner:
Proposition: If $e | a,b$, then $e | gcd(a,b)$.
Proof: Let $d = gcd(a,b)$. Then if $e \nmid d$, by the division theorem there's some $q$ and $c$ such that $d = qe + c$ with $0 < c < r$.
We have $a = k_1 d$ and $b = k_2 d$, so by substituting we obtain $a = k_1 (qe + c)$ and $b = k_2 (qe + c)$. Since $e$ divides both $a$ and $b$, it must divide both $k_1 c$ and $k_2 c$ as well. This implies that both $k_1 c$ and $k_2 c$ are common multiples of $c$ and $r$.
Now let $l = lcm(r, c)$. $l$ divides both $k_1 c$ and $k_2 c$. Since $l = \phi c$ for some $\phi$, we have $\phi | k_1, k_2$, so $d \phi | a, b$.
But we must have $\phi > 1$ otherwise $l = c$, implying $r | c$, which could not be the case since $c < r$. So $d \phi$ is a common divisor greater than $d$, which is a contradiction. $\Box$
Question: Is there a cleaner proof I'm missing, or is this seemingly elementary proposition just not very easy to prove without using Bezout's identity?
• What exactly is your definition of $\gcd(m,n)$? – steven gregory Jun 6 '17 at 17:37
• Where did you define $r$? – user428487 Aug 17 '17 at 10:59
• But don't we have to prove that the least multiple divides every common multiple? You need this to argue that $l$ divides both $k_1 c$ and $k_2 c$.. ?? – Sunghee Yun Aug 31 at 21:31
One easy and insightful way is to use the proof below. It essentially constructs $$\rm\:gcd\:$$ from $$\rm\:lcm\:$$ by employing duality between minimal and maximal elements - see the Remark below. This is essentially how the linked Wikipedia proof works, but there the innate duality is obfuscated by the presentation. Below is a proof structured so that this fundamental duality is brought to the fore.
$$\rm{\bf Theorem}\quad c\mid a,b\iff c\mid d,\ \$$ for $$\rm\ \ d = ab/lcm(a,b).\$$ $$\rm\color{#0a0}{Hence}$$ $$\rm\ d = gcd(a,b)$$
$$\rm{\bf Proof}\qquad\ \ \, c\mid a,b \iff a,b\mid ab/c \iff lcm(a,b)\mid ab/c \iff c\mid ab/lcm(a,b)$$
$$\rm\color{#0a0}{Generally}\,$$ if $$\rm\, c\mid a,b\iff c\mid d\$$ then $$\rm\ d = \gcd(a,b)\$$ up to unit factors, i.e. they're associate.
Indeed setting $$\rm\:c = d\:$$ in direction $$(\Leftarrow)$$ shows that $$\rm\:d\mid a,b,\:$$ i.e. $$\rm\:d\:$$ is a common divisor of $$\rm\:a,b.\:$$ Conversely, by direction $$(\Rightarrow)$$ we deduce that $$\rm\:d\:$$ is divisible by every common divisor $$\rm\:c\:$$ of $$\rm\:a,b,\:$$ thus $$\rm\:c\mid d\:\Rightarrow\: c\le d,\:$$ so $$\rm\:d\:$$ is a greatest common divisor (both divisibility and magnitude-wise).
Remark $$\$$ The proof shows that, in any domain, if $$\rm\:lcm(a,b)\:$$ exists then $$\rm\:gcd(a,b)\:$$ exists and $$\rm\ gcd(a,b)\,lcm(a,b) = ab\$$ up to unit factors, i.e. they are associate. The innate duality in the proof is clarified by employing the involution $$\rm\ x'\! = ab/x\$$ on the divisors of $$\rm\:ab.\:$$ Let's rewrite the proof using this involution (reflection).
Notice that $$\rm\ x\,\mid\, y\:\color{#c00}\iff\: y'\mid x'\,\$$ by $$\smash[t]{\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$$ by $$\rm\, \ yy' = ab = xx',\$$ so rewriting using this
$$\begin{eqnarray}\rm the\ proof\ \ \ c\mid a,b &\iff&\rm b,\,a\mid ab/c &\iff&\rm lcm(b,\,a)\mid ab/c &\iff&\rm c\mid ab/lcm(b,a)\\[.5em] \rm becomes\ \ \ \ c\mid a,b &\color{#c00}\iff&\rm a',b'\mid c' &\iff&\rm lcm(a',b')\mid c' &\color{#c00}\iff&\rm c\mid lcm(a',b')'\end{eqnarray}$$
Now the innate duality is clear: $$\rm\ gcd(a,b)\,=\,lcm(a',b')'\$$ by the $$\rm\color{#0a0}{above}$$ gcd characterization.
• Readers can find another involution-based proof in this answer. – Bill Dubuque Feb 26 '14 at 18:43
• See also here and here. – Bill Dubuque Apr 22 '15 at 15:40
• @BillDubuque A bit late to the party, but I was wondering, doesn't this answer beg the question? In the same line as Proof, second "iff": how do you know that the LCM divides any common multiple? – The Footprint Jun 16 at 19:47
• @TheFootprint As I explained in your question, this fundamental lcm property has a simple descent proof using Division with (smaller) Remainder (a prototypical proof by Euclidean Descent, when formulated in mod [vs. subtraction] form). So, in effect, we use duality to reduce the gcd proof to this simple lcm proof. – Bill Dubuque Jun 16 at 20:51
We can gain some insight by seeing what happens for other rings. A GCD domain is an integral domain $D$ such that $\gcd$s exist in the sense that for any $a, b \in D$ there exists an element $\gcd(a, b) \in D$ such that $e | a, e | b \Rightarrow e | \gcd(a, b)$. A Bézout domain is an integral domain satisfying Bézout's identity.
Unsurprisingly, Bézout domains are GCD domains, and the proof is the one you already know. It turns out that the converse is false, so there exist GCD domains which are not Bézout domains; Wikipedia gives a construction.
(But if you're allowing yourself the division algorithm, why the fuss? The path from the division algorithm to Bézout's identity is straightforward. In all of these proofs for $\mathbb{Z}$ the division algorithm is doing most of the work.)
• $\rm\Bbb Q[x,y]\:$ is a UFD so GCD domain which is not Bezout, since $\rm\:gcd(x,y) = 1\:$ but $\rm\:f\, x+ g\, y = 1\:\Rightarrow\: 0 = 1\:$ by evaluating at $\rm\:x = 0 = y.$ Generally a gcd domain D is Bezout iff $\rm\:gcd(a,b) = 1\:\Rightarrow (a,b) = 1,\:$ i.e. $\rm\: ac + bd = 1\:$ for some $\rm\:c,d\in D.\$ – Math Gems Apr 16 '13 at 3:53
• Whoops. I seem to have misread the Wikipedia article. – Qiaochu Yuan Apr 16 '13 at 3:55
Proof using linear diophantine equations; Note: if d=gcd(a,b), then there exist integers x and y such that d=ax+by.
Suppose c|a and c|b and d=gcd(a,b). Then, by definition of divisibility, a=cl & b=ck, for some integers l & k. By definition of gcd, d=ax+by for some integers x & y. By subsitution, d=(cl)x + (ck)y =clx + cky =c(lx+ky) =cm by closure, with m=lx+ky. Therefore d=cm But by definition of divisibity, this implies c|d and m|d Therefore c|d QED :)
Just think of the Euclidean algorithm for computing the GCD of two numbers. Let $a, b$ with say $a<b$ be two integers. Then the GCD of $a$ and $b$ is also the GCD of $a$ and $b-a$, and now we can iterate. For intance:
1. The GCD of $28$ and $8$ is the same as that of $20$ and $8$.
2. Which is the same as that of $12$ and $8$.
3. Which is the same as that of $4$ and $8$.
4. Which is the same as that of $4$ and $4$.
5. Which is obviously $4$.
Eventually, we must come down to calculating the GCD of a number with itself, since at each stage one of the numbers in our pair stays the same, and the other gets smaller. This has to stop eventually, but as long as the two numbers are distinct we can always subtract the smaller from the larger, so the only way it can stop is if they end up being equal, at which point that number will be the GCD of the original pair.
Now you just have to realize that this same logic applies to any divisor of the two numbers, not jut the greatest one. I could have said "any divisor of $28$ and $8$ is a divisor of $20$ and $8$..." and eventually had "...is a divisor of $4$". Since the number we arrive at at the end is the GCD, any divisor of the two original numbers divides the GCD.
You might also use the fact that each natural number can be uniquely represented as a product of prime powers to prove this fact without Bezout's Theorem.
This is called Fundamental Theorem of Arithmetic
For example d=p1^d1*..*pn^dn where p1 to pn are primes. The numbers p1 to pn are called then prime factors of d
Idea of the proof argument: Since gcd(a,b) divides a and b each prime factor of gcd(a,b) has to be a common prime factor of a and b.
For the same reason each prime factor of d has to be a common prime factor of a and b.
And since gcd(a,b) is the greatest common divisor of a,b(by defnition) ,each prime factor of d has to be then a prime factor of gcd(a,b), for if there would be at least one prime factor p of d which isn't a prime factor of gcd(a,b) ,then the product gcd(a,b)*p would be again a common divisor of both a and b ,greater than gcd(a,b) ,which would be a contradiction.
It essentially depends on how you define the gcd. Note that gcd is the accronym of greatest common divisor.
This means that the first definition would be: $d=\gcd (a,b)$ is the greatest element (defined up to multiplication by a unit) of the set of all common divisors of $a$ and $b$. Where the partial order is given by divisibility. So if $e$ is a common divisor of $a$ and $b$, it is not greater than $d$, i.e. $e\mid d$.
Note: this definition makes sense for instance in unique factorization domains (in particular Euclidean domains or even principal domains), more generally. | 2019-10-20T22:42:12 | {
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https://math.stackexchange.com/questions/1849528/what-is-this-operator-called/1850154 | # What is this operator called?
If $x \cdot 2 = x + x$
and $x \cdot 3 = x + x + x$
and $x^2 = x \cdot x$
and $x^3 = x \cdot x \cdot x$
Is there an operator $\oplus$ such that:
$x \oplus 2 = x^x$
and $x \oplus 3 = {x^{x^x}}$?
Also, is there a name for such a set of operators ops where...
Ops(2) is multiplication
Ops(3) is exponentiation
Ops(4) is $\oplus$
...and so on
Also, is there a branch of math who actually deals with such questions? Have these questions already been answered like 2000 years ago?
• You may be interested in the Ackermann function. Makes for fun problems in recursion theory and such. – Jyrki Lahtonen Jul 5 '16 at 9:48
• FYI $(x^x)^x = x^{(x^2)} \ne x^{(x^x)}$ so the operation is not associative so the notation ${x^x}^x$ is ambiguous. We usually take it to mean $x^{(x^x)}$ because $(((x^x)^x)^x).... = x^{(x^m)}$ and can be expressed that way. – fleablood Jul 5 '16 at 17:34
• Also of note: the "Ops" are known as Hyperoperations. – mbomb007 Jul 5 '16 at 18:56
• i think fleablood got the convention backward. $a^{b^c}$ is understood as $a^{(b^c)}$ because $(a^b)^c = a^{bc}$. – Anton Sherwood Jul 6 '16 at 0:52
This operation ${\rm Ops}(4)$ is called tetration, from the greek root tetra meaning four; it's also sometimes called a "power tower". There are also many further generalizations of this type of sequence; Knuth's up-arrow notation gives $a^{a^{a^a}}=a\uparrow\uparrow4$, so that $a\uparrow\uparrow n$ is the tetration operation. By adding more arrows you get pentation and so on, and the Conway chained arrow notation generalizes this still further.
FYI, for "to the power of-ation" the word you're looking for is exponentiation.
• Also that is similar to graham number. en.wikipedia.org/wiki/Graham%27s_number – Takahiro Waki Jul 5 '16 at 9:19
• Aha, and the answer to the second question is hyperoperation: en.m.wikipedia.org/wiki/Hyperoperation – uzilan Jul 5 '16 at 12:19
• Ah, the whims of the internet. This does not deserve to be my highest-voted answer. – Mario Carneiro Jul 6 '16 at 4:03
• It's the bikeshed, Mario; you get used to it. – J. M. is a poor mathematician Jul 6 '16 at 12:36
• The problem with this answer is that a↑↑n is not defined to be tetration operation. It is a consequence of a↑n being defined as $a^n$ and then further arrows is defined to be n number of iterations of itself minus one arrow on a. Much prefer @Curd 's answer – Ariana Jul 8 '16 at 17:59
A more general function that combines all those operators has been defined by Ackermann:
$\varphi(m,n,p) = \begin{cases} \varphi(m, n, 0) = m + n \\ \varphi(m, 0, 1) = 0 \\ \varphi(m, 0, 2) = 1 \\ \varphi(m, 0, p) = m &\text{ for } p > 2 \\ \varphi(m, n, p) = \varphi(m, \varphi(m, n-1, p), p - 1) &\text{ for } n > 0 \text{ and } p > 0. \end{cases}$
So for $p = 0, 1, 2$ you get
$\phi(m, n, 0) = m + n$
$\phi(m, n, 1) = m \cdot n$
$\phi(m, n, 2) = m ^ n$
and
$\phi(m, n, 3) = \overbrace{{{m ^ m} ^ m} ^ {...}}^{n}$
It is known as a tetration, and it is normally written as $^na$ where n is the height of the power tower. It is the forth hyperoperation.
The zeroth hyperoperation is the successor function, and the first is the zeroth hyperoperation iterated, and so on
A more general way to define the nth hyperoperation is, using the notation, $H_n(a,b)$ where n is the nth hyperoperation,
${\displaystyle H_{n}(a,b)={\begin{cases}b+1&{\text{if }}n=0\\a&{\text{if }}n=1{\text{ and }}b=0\\0&{\text{if }}n=2{\text{ and }}b=0\\1&{\text{if }}n\geq 3{\text{ and }}b=0\\H_{n-1}(a,H_{n}(a,b-1))&{\text{if }n\in\mathbb{N},n>3}\end{cases}}}$
Some notations for hyperoperations are(for $H_n(a,b)$:
1. Square bracket notation: $a[n]b$
2. Box notation: $a{\,{\begin{array}{|c|}\hline {\!n\!}\\\hline \end{array}}\,}b$
3. Nambiar's notation : $a\otimes ^{n-1}b$
4. Knuth's up arrow notation: $a\uparrow^{n-2}b$
5. Goodstien's notation: $G(a,b,n)$
6. Conway's chained arrow notation: $a\rightarrow b\rightarrow (n-2)$
7. Bowers exploding array function: $\{a,b,n,1\}$
8. Original Ackermann function: ${\begin{matrix}\phi (a,b,n-1)\ {\text{ for }}1\leq n\leq 3\\\phi (a,b-1,n-1)\ {\text{ for }}n\geq 4\end{matrix}}$
• Wow, Arianna Grande as a mathematician and stackexchange user. Exciting, – Chisko Jul 6 '16 at 17:26
• @Cheskos Ariana – Ariana Jul 6 '16 at 17:40
• Is any of these 8 extending Peter Hurfords $C(a,\dots)$ (see Extending the Extensions)? – Rudi_Birnbaum Mar 11 '18 at 21:47
• @R_Berger The extension, I believe, grows faster than every function in the 8 listed – Ariana Mar 12 '18 at 10:24 | 2019-06-17T18:34:12 | {
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http://math.stackexchange.com/questions/89498/does-a-equiv-b-pmod-n-mean-n-mid-a-b-or-n-mid-b-a | # Does $a \equiv b \pmod n$ mean $n \mid a - b$ or $n \mid b -a$
If I have $a \equiv b \pmod{n}$, it means $n \mid b - a$.
But can you write it as $n \mid a - b$ as well?
-
$a \equiv b \pmod{n} \implies n \mid (a-b) \implies a-b = nk$ for some integer $k$. Can you find an integer $m$ such that $b-a = mn$? – JavaMan Dec 8 '11 at 4:33
Yes, because in $x$ divides $y$ if and only if $\pm x$ divides $\pm y$, and $a-b = -(b-a)$. – Arturo Magidin Dec 8 '11 at 4:33
that would just be -1. but my prof always writes it as b - a – jake Dec 8 '11 at 4:34
HINT $\rm\quad n\ |\ c \iff\ \dfrac{c}{n}\in \mathbb Z \iff -\dfrac{c}{n}\: =\: \dfrac{-c}{n} \in \mathbb Z\ \iff\ n\ |\: -c\:.\$ Now let $\rm\ c = a-b\:.$
I.e. $\ \mathbb Z$ closed under negation $\rm\: \Rightarrow\ n\: \mathbb Z\:$ closed under negation, i.e. $\rm\: -(n\: \mathbb Z)\ =\ (-n)\ \mathbb Z$
-
It can be written both ways. $$n\mid a-b \iff kn = a-b \iff k'n = b-a \iff n\mid b-a$$ for some $k\in \mathbb{Z}$ where $k' = -k$. It's just a way of stating that if $n\mid m$ then $n\mid (-m)$.
-
Yes: $n\mid b-a$ means that there is an integer $k$ such that $b-a = kn$, and $n\mid a-b$ means that there is an integer $\ell$ such that $a-b=\ell n$. Since $a-b=-(b-a)$, these are two ways of saying the same thing: $$b-a = kn\iff a-b=-(b-a)=(-k)n\;.$$
- | 2014-09-16T10:00:38 | {
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http://mathhelpforum.com/calculus/157827-integral-rational-function.html | # Math Help - Integral of Rational Function
1. ## Integral of Rational Function
I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.
$\int{\frac{1}{x^2 + x + 1}}$
But this can't be factored and completing the square gives $(x+\frac{1}{2})^2+\frac{3}{4}$
Which doesn't look any easier to integrate.
Could someone point me in the right direction?
Thanks
2. Originally Posted by centenial
I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.
$\int{\frac{1}{x^2 + x + 1}}$
But this can't be factored and completing the square gives $(x+\frac{1}{2})^2+\frac{3}{4}$
Which doesn't look any easier to integrate.
Could someone point me in the right direction?
Thanks
You missed a dx.
Completing the square is a good start, move on by substituting u=x+1/2.
3. Oh... complete the square and then u substitution. That seems so obvious now that I can't believe I didn't see it.
So...
$
\int{\frac{1}{x^2+x+1}dx}
$
$
\int{\frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}}dx}
$
$
\int{\frac{1}{u^2 + \frac{3}{4}}du}
$
$
\int{\frac{1}{u^2 + \sqrt{\frac{3}{4}}^2}du}
$
$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2u}{\sqrt{3}}} + C
$
$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2x + 1}{\sqrt{3}}} + C
$
Is that right?
4. Originally Posted by centenial
Oh... complete the square and then u substitution. That seems so obvious now that I can't believe I didn't see it.
So...
$
\int{\frac{1}{x^2+x+1}dx}
$
$
\int{\frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}}dx}
$
$
\int{\frac{1}{u^2 + \frac{3}{4}}du}
$
$
\int{\frac{1}{u^2 + \sqrt{\frac{3}{4}}^2}du}
$
$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2u}{\sqrt{3}}} + C
$
$
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2x + 1}{\sqrt{3}}} + C
$
Is that right?
Looks fine to me.
5. Originally Posted by centenial
I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.
$\int{\frac{1}{x^2 + x + 1}}$
But this can't be factored and completing the square gives $(x+\frac{1}{2})^2+\frac{3}{4}$
Which doesn't look any easier to integrate.
Could someone point me in the right direction?
Thanks
If you don't want to have to resort to two substitutions, substitute $x + 1 = \frac{\sqrt{3}}{2}\tan{\theta}$ so that $dx = \frac{\sqrt{3}}{2}\sec^2{\theta}\,d\theta$.
Then $\int{\frac{dx}{x^2 + x + 1}} = \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the ta}{\left(\frac{\sqrt{3}}{2}\tan{\theta}\right)^2 + \frac{3}{4}}}$
$= \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the ta}{\frac{3}{4}(\tan^2{\theta} + 1)}}$
$= \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the ta}{\frac{3}{4}\sec^2{\theta}}}$
$= \int{\frac{\frac{\sqrt{3}}{2}}{\frac{3}{4}}\,d\the ta}$
$= \int{\frac{2\sqrt{3}}{3}\,d\theta}$
$= \frac{2\sqrt{3}\,\theta}{3} + C$
$= \frac{2\sqrt{3}\arctan{\left[\frac{2\sqrt{3}}{3}(x+1)\right]}}{3} + C$. | 2014-09-19T15:01:12 | {
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https://math.stackexchange.com/questions/95069/how-would-one-be-able-to-prove-mathematically-that-11-2/95070 | # How would one be able to prove mathematically that $1+1 = 2$?
Is it possible to prove that $1+1 = 2$? Or rather, how would one prove this algebraically or mathematically?
• In fact, this occasionally useful proposition was proved by Russell and Whitehead. – Srivatsan Dec 29 '11 at 23:19
• You should go read up on set theory, particularly the Peano Axioms. This is a consequence of how the natural numbers and addition on them is defined. – Potato Dec 29 '11 at 23:19
• Google "Russel and Whitehead's proof that 1+1=2"! Good Luck. – draks ... Dec 29 '11 at 23:20
• In many formal developments, "$1+1=2$" is very close to being the definition of $2$ -- so don't expect any very exciting proof. – hmakholm left over Monica Dec 29 '11 at 23:24
• I tried to retag, since [proof-theory] was completely wrong. I hope the tag is less wrong now, but if someone thinks it's still not fully fitting - please retag. – Asaf Karagila Dec 29 '11 at 23:26
## 3 Answers
In a very "raw" sense the symbol $$2$$ is just a shorthand for $$1+1$$. There is really not much to prove there.
If we want to talk about proof we need axioms to derive the wanted conclusion from. Let us take the "usual" axioms of the natural numbers here, namely Peano Axioms. These axioms give very basic rules which describe the natural numbers, and from them we will derive $$1+1=2$$.
In those axioms the numbers $$1$$ and $$2$$ don't exist. We have $$0$$ and we have $$S(n)$$, which can be thought of as a "successor function" which generates the next number, so to speak. In this system $$1$$ is a symbol for $$S(0)$$ and $$2$$ is a symbol for $$S(S(0))$$.
Addition is defined inductively, that is $$x+0=x$$, $$x+S(y)=S(x+y)$$. From this we can derive:
$$1+1 = 1+S(0) = S(1+0) = S(1)$$ Now replace $$1$$ with its "full form" of $$S(0)$$ and we have: $$S(0)+S(0) = S(S(0)+0) = S(S(0))$$
Which is what we wanted.
In a more general setting, one needs to remember that $$0,1,2,3,\ldots$$ are just symbols. They are devoid of meaning until we give them such, and when we write $$1$$ we often think of the multiplicative identity. However, as I wrote in the first part, this is often dependent on the axioms - our "ground rules".
If we consider, instead of the natural numbers, the binary digits $$0,1$$ with addition $$\bmod 2$$, then we have that $$1+1=0$$. Now you can argue that of course that $$0\neq 2$$, however in this set of axioms (which I have not expressed explicitly here) we can prove that $$0=2$$, where $$2$$ is the shorthand for $$1+1$$ and $$0$$ is the additive neutral element.
Actually, just writing $$1+1=0$$ is a proof of that.
I can't really stress that enough, because this is a very important part of mathematics. We often use some natural notion, such as the natural numbers, before we define it. Later we define it "to work as we want it to work" and only then we have a formal framework to work with.
These axioms, these frameworks, those often remain "in the shadows" and if you don't know where to look for them then you are less likely to find them.
This is why the question "Why $$1+1=2$$?" is nearly meaningless - since you don't have a formal framework, and the interpretation (while assumed to be the natural one) is ill-defined.
On that same note, this question is also very important when starting with mathematics. It helps to you understand what there is to prove, and how to do it. Of course this too lack of context because one would have to define what is a proof, and all the other things first.
• That is a really good answer to a annoying question that comes up a lot. I can't count the amount of people who says you can't prove 1+1=2. – simplicity Dec 30 '11 at 0:38
• I'm always confused why people find it annoying - maths at hs was boring until I tried to prove 1+1=2 one da. Being a naive 17 year old I was soon swepped towards set theory and then ended up studying the subject at University and have been offered placings after my exams with my current uni for further study if things go like I want them to. All in all I would say that question changed the course of my adult life - I never really took work seriously until then! Not that I do now, I guess I should use the word respect, though that's another tale altogether ... – Adam Dec 30 '11 at 3:20
• This seems like a pretty good answer. If it still doesn't solve the author's problem, what can you do? You can only do your best to explain it. You have to start somewhere. The natural numbers can be thought of as the numbers that can be constructed from 0 and the successor operation. We could decide that when we write 1 + 1, we really mean S(0) + S(0) and when we write 2, we really mean S(S(0)) so the real task is to show that S(0) + S(0) = S(S(0)). Now addition can be defined inductively to derive S(0) + S(0) = S(S(0)). – Timothy Oct 19 '19 at 3:17
• @simplicity I think we should keep them to show that some people have a real different way of thinking which I respect. If they want to figure out how to further break down the proof into one they accept, I respect them for that. Maybe they were looking for an answer like this answer. Some people only accept proofs they find very intuitive. Maybe it's a sign that they're a slow and careful person who thinks about a problem slowly and deeply. Maybe as a result of being a slow deep thinker, they will reject the axiom of choice and reject proofs in Naive set theory. Try adding 1000 + 20 + 30 + – Timothy Oct 19 '19 at 3:25
• 1000 + 1030 + 1000 + 20 and see what you get. Did you get it right? Are you sure? If you got the answer 5000, I can tell you that is not the right answer. I didn't invent the problem. I got it somewhere. – Timothy Oct 19 '19 at 3:28
1 is the convention name of 0++, 2 is the convention name of (0++)++, so what you need to prove is 0++ + 0++ = (0++)++.
$+$ is defined as:
• $0+m := m$
• $(n++)+m := (n+m)++$
Just apply this definition to the left of the equation and you will get the right. You don't even need to know what '0' or '++' is, all you do is shuffling some sequences of symbols according to some rules.
• Why is this getting no upvotes? – Pacerier May 15 '17 at 17:01
• @Pacerier: Possibly because (a) it came five years after Asaf's answer that already makes the same substantial point, and (b) uses a strange idiosyncratic notation for the successor function. The notation may be an attempt at analogy from programming languages in the C tradition, but that doesn't even work so well: In C, a++ evaluates to the original value of a and as a side effect changes what a means subsequently to the successor of that original value. – hmakholm left over Monica May 15 '17 at 20:07
One standard set of axioms says that $0$ is a natural number, and for every natural number $n$ the successor of that number, $S(n)$ is a unique natural number.
Then there are these two axioms of addition (amongst other axioms):
$$a+0=a\\a+S(b)=S(a+b)$$
Now, if we define $1$ as $S(0)$ and $2$ as $2=S(S(0))=S(1)$ then:
$$1+1=S(0)+S(0)=S(S(0)+0)=S(S(0))=2$$
However, this is just abstract nonsense.
You can also have other axiom systems with no successor function, and then usually $2$ is just defined as $1+1$.
A lambda-calculus proof might start with definitions:
\begin{align}1&=\lambda\, f.f\\ 2&=\lambda\, fx.f(f(x))\\ +&=\lambda\,mnfx.(mf)(nfx) \end{align}
Then you can prove that $+11fx=2fx$ for any $f,x$.
In lambda calculus, this will mean that there is an equivalence between $+11$ and $2$, but they are not necessarily equal. (Consider lambda calculus to describe programs for computation - two programs are equivalent if they compute the same thing, but they might not be identical as programs.) | 2021-08-06T02:36:07 | {
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https://www.jiskha.com/questions/1328509/David-and-Sam-had-an-equal-number-of-marbles-After-Sam-gave-50-marbles-to-David | # Math
David and Sam had an equal number of marbles. After Sam gave 50 marbles to David, he had 5x as many marbles as Sam. Find the total number of marbles they had.
asked by Patrice
1. David's marbles --- x
Sam's marbles -----x
After Sam lost his marbles:
Sam --- x-50
David -- x+50
x+50 = 5(x-50)
x+50 = 5x - 250
-4x = -300
x = 75
they had 75+75 or 150 marbles between them
Your previous post contains the "same" question, just the numbers and names have been changed. Follow the same steps.
posted by Reiny
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https://math.stackexchange.com/questions/2815985/prove-that-if-x-1-x-2-x-3-are-roots-of-x3-px-q-0-then-x-13x | # Prove that if ${x_1, x_2, x_3}$ are roots of ${x^3 + px + q = 0}$ then ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$
How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?
I've tried the following approach:
If $x_1$, $x_2$ and $x_3$ are roots then
$$(x-x_1)(x-x_2)(x-x_3) = x^3+px+q = 0$$
Now find the coefficient near the powers of $x$:
$$x^3 - (x_1 + x_2 + x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3)x - x_1x_2x_3 = x^3+px+q$$
That means that I can write a system of equations:
$$\begin{cases} -(x_1 + x_2 + x_3) = 0 \\ x_1x_2 + x_1x_3 + x_2x_3 = p \\ - x_1x_2x_3 = q \end{cases}$$
At this point I got stuck. I've tried to raise $x_1 + x_2 + x_3$ to 3 power and expand the terms, but that didn't give me any insights. It feels like I have to play with the system of equations in some way but not sure what exact.
• If you ever need to go further, I recommend that you study Newton's identities. They settle this question also. But, as shown by the answers below, you don't need their full power to reach the sum of cubes. – Jyrki Lahtonen Jun 11 '18 at 18:32
Rewrite $x_1,x_2,x_3$ with $a,b,c$. From first Vieta formula we have $$a+b+c=0$$ so $a+b=-c$ and so on...
Now $$a^3+b^3+c^3= (a+b)(a^2-ab+b^2)+c^3 = c(\underbrace{-a^2+ab-b^2+c^2}_I)$$
Since
$$I = -a^2+ab-b^2+c^2 = a(b-a)+(c-b)(c+b) =$$ $$a(b-a)-a(c-b) = a(2b-a-c)=a(2b+b)=3ab$$
$x_1^3 + x_2^3 + x_3^3 - 3x_1x_2x_3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2-x_1x_2 - x_2x_3-x_1x_3)$
Now, here $-(x_1 + x_2 + x_3) =$ Coefficient of $x^2$/ Coefficient of $x^3 =0$
So, $x_1^3 + x_2^3 + x_3^3 = 3x_1x_2x_3$
Hint: $\,x_1^3+px_1+q=0 \iff x_1^3=-px_1-q\,$, then adding the three relations together:
$$\require{cancel} x_1^3+x_2^3+x_3^3=\cancel{-p(x_1+x_2+x_3)}-3q=\cdots$$
• Great solution +1 – Aqua Jun 11 '18 at 17:30
• @ChristianF Thanks. One advantage of this approach is that it easily extends to equations $\,x^{\color{red}{n}}+px+q=0\,$, where $\,\sum_i x_i^n = (-1)^{n+1}n \prod_i x_i\,$. – dxiv Jun 11 '18 at 21:13
• @dxiv I understand this method is one of the basic ones used all over the places, but is there some source where I can read about such methods. Jyrki Lahtonen has mentioned Newton's Identities above, is it something that comes from there? – roman Jun 12 '18 at 5:08
• @RomanKapitonov My shortcut takes advantage of the particular form of the given polynomial. In general, the elementary symmetric polynomials, Vieta's relations and Newton's identities should get you through. – dxiv Jun 12 '18 at 5:40
• @dxiv Thank you for your guidance – roman Jun 12 '18 at 9:40
If $x_1,x_2,x_3$ are roots of $x^3+p x+q=0$ then $x_1+x_2+x_3 = 0$
If $x_1+x_2+x_3 = 0$ then $x_3 = -(x_1+x_2)$ and
$x_1^3+x_2^3+x_3^3 = x_1^3+x_2^3+(-1)^3(x_1+x_2)^3 = -3(x_1^2x_2+x_1x_2^2) = -3x_1x_2(x_1+x_2) = -3x_1x_2(-x_3) = 3x_1x_2x_3$
Every symmetric polynomial can be expressed in terms of the elementary symmetric polynomials, in this case $s_1=x_1+x_2+x_3$, $s_2=x_1x_2+x_2x_3+x_3x_1$ and $s_3=x_1x_2x_3$. Since $x_1^3+x_2^3+x^3$ is homogeneous, we can find $a$, $b$ and $c$ such that $$x_1^3+x_2^3+x_3^3=as_1^3+bs_1s_2+cs_3$$
• For $x_1=1$, $x_2=0$, $x_3=0$: $1=a$
• For $x_1=1$, $x_2=1$, $x_3=0$: $2=8a+2b$
• For $x_1=1$, $x_2=1$, $x_3=1$: $3=27a+9b+c$
Therefore $a=1$, $b=-3$, $c=3$ and finally $$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3$$ This is a general result.
In your case, by Viète’s formulas $$x_1+x_2+x_3=0,\qquad$$ so in the end $$x_1^3+x_2^3+x_3^3=3x_1x_2x_3$$ | 2019-11-14T18:42:30 | {
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https://math.stackexchange.com/questions/2655238/bayess-rule-and-unfair-coin-solution-explanation | # Bayes's rule and unfair coin | Solution Explanation
There are three coins in a bag. Two of them are fair. One has heads on both sides. A coin selected at random shows heads in two successive tosses.
What is the conditional probability of obtaining another head in the third trial given the fact that the first two trials showed heads.
I think this problem should be solved in the following way
$$P(one\ more\ head) = \frac{1}{3}\cdot 1 +\frac{1}{3}\cdot \frac{1}{2}+ \frac{1}{3}\cdot \frac{1}{2} = \frac{2}{3}$$
but my book says the right solution is
$$P(HHH|HH) = \frac{5}{6}$$
But the first two trials do not affect the third trial, so I should only have to calculate the probability of getting one more head, since I already have two.
Can anyone explain me what is going on?
Bayes' Rule says $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$ so in your question, it is $$P(HHH|HH) = \frac{P(HH|HHH)\cdot P(HHH)}{P(HH)}$$ Now, notice that
$$P(HH|HHH) = 1$$
because it simply says "if we know that three successive tosses are resulted in $HHH$, what is the probability that first two of them are resulted in $2H$".
For $P(HHH)$, we have $$P(HHH) = \frac{1}{3}\cdot1+\frac{2}{3}\cdot \frac{1}{8} = \frac{5}{12}$$
because if we choose unfair coin with $\frac{1}{3}$ probability, we have $HHH$ with probability $1$ and if we don't choose it with $\frac{2}{3}$, we have $\frac{1}{2} \cdot\frac{1}{2} \cdot\frac{1}{2} = \frac{1}{8}$ probability to have $HHH$.
For $P(HH)$, by similar argument to $P(HHH)$, we have $$P(HH) = \frac{1}{3}\cdot 1+\frac{2}{3} \cdot\frac{1}{4} = \frac{1}{2}$$
So the answer is $$P(HHH|HH) = \frac{1\cdot \frac{5}{12}}{\frac{1}{2}} = \frac{5}{6}$$
Now, when it comes to your argument "the first two trials do not affect the third trial", it is wrong in some manner. Because although it seems third trial is not affected by the first two trials, notice that whether we have fair or unfair coin affects probability of having $HH$ in the first two tosses. In other words, since $P(HH)$ differs from fair coin to unfair coins, probability of having a third $H$ is also affected by $P(HH)$, hence the first two tosses.
• I agree that $P(HH)$ differs from fair coin to unfair coins. However, I still do not get how $P(HH)$ affects the third trial. I am assuming that I already have 2 heads, meaning it does not matter how difficult it was to get those two heads, I already got them. Now, I only care about getting one more head. – user532588 Feb 18 '18 at 9:34
• hold on a second. I am using the same coin for all trials, which means that the more heads I get, the higher is the probability that the coin is biased. Hence, it should be easier to get heads for subsequent trials. Is this reasoning correct? – user532588 Feb 18 '18 at 9:40
• Yes, that's even better explanation than I did in the solution :) – ArsenBerk Feb 18 '18 at 9:41
Let $H_n$ be the event of obtaining a head on toss $n$, $B$ the event of selecting the biased coin, $F$ the event of selecting one of the fair coins.
But the first two trials do not affect the third trial, so I should only have to calculate the probability of getting one more head, since I already have two.
Yes, given that you are using the same coin, the results of the coin tosses are conditionally independent and indentically distributed. $$\mathsf P(H_1,H_2,H_3\mid B)=\mathsf P(H_1\mid B)~\mathsf P(H_2\mid B)~\mathsf P(H_3\mid B)\\\mathsf P(H_1,H_2,H_3\mid F)=\mathsf P(H_1\mid F)~\mathsf P(H_2\mid F)~\mathsf P(H_3\mid F)$$
So the probability of getting a head in the next trial is... wait, which coin you are using? The same coin is selected once, and then used for all tosses. There is a dependency.
Rather, what is the probability that it is biased when given that you have two heads in he first two tosses. Well, use Bayes' Rule to update the probability given the evidence.
$$\begin{split}\mathsf P(B\mid H_1,H_2) &= \dfrac{\mathsf P(B)~\mathsf P(H_1,H_2\mid B)}{\mathsf P(B)~\mathsf P(H_1,H_2\mid B)+\mathsf P(F)~\mathsf P(H_1\mid F)~\mathsf P(H_2\mid F)}\\[1ex]&= \dfrac{\mathsf P(B)~\mathsf P(H_1\mid B)~\mathsf P(H_2\mid B)}{\mathsf P(B)~\mathsf P(H_1\mid B)~\mathsf P(H_2\mid B)+\mathsf P(F)~\mathsf P(H_1\mid F)~\mathsf P(H_2\mid F)}\\[1ex]&=\dfrac{\tfrac 13\cdot 1\cdot 1}{\tfrac 13\cdot 1\cdot 1+\tfrac 23\cdot \tfrac 12\cdot \tfrac 12}\\[1ex]&=\dfrac{2}{3}\end{split}$$
So $\mathsf P(H_3\mid H_1,H_2)~{ = \mathsf P(H_3\mid B)~\mathsf P(B\mid H_1,H_2)+\mathsf P(H_3\mid F)~\mathsf P(F\mid H_1,H_2)\\=1\cdot\tfrac 23+\tfrac 12\cdot\tfrac 13\\=\tfrac 56}$
Of course we may combine these steps :$$\begin{split}\mathsf P(H_3\mid H_1,H_2)&= \dfrac{\mathsf P(H_1,H_2,H_3)}{\mathsf P(H_1,H_2)}\\ &=\dfrac{\mathsf P(B)~\mathsf P(H_1,H_2,H_3\mid B)+\mathsf P(F)~\mathsf P(H_1,H_2,H_3\mid F)}{\mathsf P(B)~\mathsf P(H_1,H_2\mid B)+\mathsf P(F)~\mathsf P(H_1,H_2\mid F)}\\&~~\vdots\\&=\dfrac 56\end{split}$$ | 2019-11-18T16:05:47 | {
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https://math.stackexchange.com/questions/545962/when-is-binomnk-divisible-by-n | When is $\binom{n}{k}$ divisible by $n$?
Is there any way of determining if $\binom{n}{k} \equiv 0\pmod{n}$. Note that I am aware of the case when $n =p$ a prime. Other than that there does not seem to be any sort of pattern (I checked up to $n=50$). Are there any known special cases where the problem becomes easier? As a place to start I was thinking of using $e_p(n!)$ defined as:
$$e_p(n!) = \sum_{k=1}^{\infty}\left \lfloor\frac{n}{p^k}\right \rfloor$$
Which counts the exponent of $p$ in $n!$ (Legendre's theorem I believe?)
Then knowing the prime factorization of $n$ perhaps we can determine if these primes appear more times in the numerator of $\binom{n}{k}$ than the denominator.
Essentially I am looking to see if this method has any traction to it and what other types of research have been done on this problem (along with any proofs of results) before. Thanks!
• Have you considered looking at Pascal's triangle under each modulus $p$? – abiessu Oct 30 '13 at 19:11
• Have you looked into the case where $n$ is a prime power? – Jack M Oct 30 '13 at 19:29
• @ abiessu: Could you expand on what you mean please? @ Jack M: I have not, will do so now. Thanks. – Patrick Oct 30 '13 at 20:37
Below is a picture of the situation. Red dots are the points of the first 256 rows of Pascal's triangle where $n\mid {n \choose k}$.
It appears that "most" values fit the bill.
One can prove the following:
Proposition: Whenever $(k, n)=1$, we have $n \mid {n\choose k}$.
This follows from the case where $n$ is a prime power (considering the various prime powers dividing $n$).
However, it happens quite often that $(k,n) \neq 1$ but $n \mid {n\choose k}$ still. For instance, $10 \mid {10\choose 4}=210$, but $(10,4) \neq 1$ (this is the smallest example). I do not think that there is a simple criterion.
In fact, it is interesting to consider separately the solutions $(n,k)$ into those which are relatively prime (which I'll call the trivial solutions) and those which are not. It appears that the non-trivial solutions are completely responsible for the Sierpinski pattern in the triangle above. Indeed, here are only the trivial solutions:
and here are the non-trivial solutions:
Let $f(n)$ be the number of $k$'s between $0$ and $n$ where $n \mid {n\choose k}$. By the proposition we have $\varphi(n)< f(n) < n$.
Question: is $$\text{lim sup } \frac{f(n)-\varphi(n)}{n}=1?$$
Here is a list plot of $\frac{f(n)-\varphi(n)}{n}$. The max value reached for $1<n<2000$ is about $0.64980544$. The blue dots at the bottom are the $n$'s such that $f(n) = \varphi(n)$.
• It is nice to see Sierpinski's triangle appear. – Pedro Tamaroff Oct 31 '13 at 6:06
• @PedroTamaroff Indeed! – Bruno Joyal Oct 31 '13 at 6:07
• Very helpful stuff! I too got excited when I saw Sierpinski's triangle. One of my favorite things in math is seeing familiar structures when exploring new problems. Really cool. – Patrick Oct 31 '13 at 11:46
• Thank you for adding those additional two triangle images...really paints an even clearer picture wow. What software did you use to generate these? – Patrick Nov 1 '13 at 12:52
• @Patrick You are welcome, it was fun. I used SAGE. – Bruno Joyal Nov 1 '13 at 13:21
Well, $n = p^2$ when $k$ is not divisible by $p.$ Also $n=2p$ for $k$ not divisible by $2,p.$ Also $n=3p$ for $k$ not divisible by $3,p.$
• But you're missing $10 \mid 210$! :) – Bruno Joyal Oct 31 '13 at 6:32
• @Marie, life is hard. We all must learn to live with disappointment. – Will Jagy Oct 31 '13 at 6:39
• @BrunoJoyal, so , who is Marie? – Will Jagy Oct 31 '13 at 18:08
• It was a secret identity. Nobody could have guessed! ;) – Bruno Joyal Oct 31 '13 at 18:11 | 2019-05-23T03:04:13 | {
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https://math.stackexchange.com/questions/3068890/comparison-of-integrals-by-algebraic-means | # Comparison of integrals by algebraic means
\begin{align}A&:=\int_0^1\frac1{\sqrt{x(1-x)}}\ \mathrm dx \\ B&:=\int_0^1\sqrt{x(1-x)}\ \mathrm dx \end{align}
My CAS tells me that $$A = \pi$$ and $$B = \frac18\pi$$.
How can one prove that $$A=8B$$ using just basic rules of integration such as the chain rule?
Trigonometric functions are not allowed since they are not definable as integrals. Neither is the Gamma function allowed, since it is defined in terms of exp, which is like a trigonometric function. These restrictions are part of what I mean by "algebraic means". On the other hand, integration by parts is fine. Equivalently, the fundamental theorem of calculus is also fine.
• Yes. The solution would not even need the concept of $\pi$. Jan 10, 2019 at 16:52
• If you evaluate the integrals using contour integration methods (which doesn't involve computing the primitive function as we do when using ordinary calculus methods), then the factor of 1/8 can be seen to come from the coefficient of 1/x in the large x expansion of the square root, so it's a binomial coefficient. Jan 10, 2019 at 17:01
• I think you should include a little more emphasis on the "algebraic means" part of this question. As you might've seen, I posted a solution (which I since deleted) which used the Gamma function, as I was unclear what you were requesting. Jan 10, 2019 at 18:18
• @CountIblis Contour integration is likely well beyond the scope here. See my posted solution for a way forward using elementary calculus only. ;-)) Jan 10, 2019 at 18:18
## 2 Answers
If integration by parts is an acceptable approach, then we can proceed as follows.
First, let $$B$$ be the integral defined as
$$B=\int_0^1 \sqrt{x(1-x)}\,dx\tag1$$
Integrating by parts with $$u=\sqrt{x(1-x)}$$ and $$v=x$$ in $$(1)$$, we obtain
$$B=\frac12 \int_0^1 x\left(\frac{\sqrt x}{\sqrt{1-x}}-\frac{\sqrt{1-x}}{\sqrt x}\right)\,dx\tag2$$
Now enforcing the substitution $$x\mapsto 1-x$$ in the first term on the right-hand side of $$(2)$$ reveals
$$\int_0^1 x\frac{\sqrt x}{\sqrt{1-x}}\,dx=\int_0^1 \frac{\sqrt{1-x}}{\sqrt x}\,dx-\int_0^1 \sqrt{x(1-x)}\,dx\tag3$$
Substituting $$(3)$$ into $$(2)$$ we find that
$$B=\frac14 \int_0^1 \frac{\sqrt {1-x}}{\sqrt{x}}\,dx\tag 4$$
Finally, integrating by parts with $$u=\frac{\sqrt {1-x}}{\sqrt{x}}$$ and $$v=x$$ in $$(4)$$ yields
$$B=\frac18 \int_0^1 \frac{1}{\sqrt{x(x-1)}}\,dx$$
as was to be shown!
• I admit that your solution is much better for this problem (+1) than mine, but I disagree that the use of the Beta function is overkill in this situation. It is incredibly efficient and can be applied to both problems, effectively killing two birds with one stone. Why not? Jan 10, 2019 at 18:21
• @clathratus Use of the Beta function is fine. But then you need to show that $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$ and you need to show that $\Gamma(1/2)\sqrt \pi$. Those require more work that one needs and some of the steps involve carrying out integrals, which we could easily do here in a straightforward way. Finally, the OP has edited to state that use of the Gamma function is prohibited. Jan 10, 2019 at 18:35
In fact, by integration by parts, one has $$\begin{eqnarray*} B&=&\int_0^1\sqrt{x(1-x)} \mathrm dx\\ &=&-\frac12\int_0^1\frac{x(1-2x)}{\sqrt{x(1-x)}}\mathrm dx\\ &=&-\frac12\int_0^1\frac{x(1-x)-x^2}{\sqrt{x(1-x)}}\mathrm dx\\ &=&-\frac12B+\frac12\int_0^1\frac{x^2}{\sqrt{x(1-x)}}\mathrm dx \end{eqnarray*}$$ and hence $$3B=\int_0^1\frac{x^2}{\sqrt{x(1-x)}}\mathrm dx. \tag{1}$$ Under $$1-x\to x$$, (1) becomes $$3B=\int_0^1\frac{(1-x)^2}{\sqrt{x(1-x)}}\mathrm dx. \tag{2}$$ Adding (1) to (2), one has $$6B=\int_0^1\frac{x^2+(1-x)^2}{\sqrt{x(1-x)}}\mathrm dx=\int_0^1\frac{1-2x(1-x)}{\sqrt{x(1-x)}}\mathrm dx=A-2B.$$ This implies $$A=8B.$$ | 2022-05-24T06:56:09 | {
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https://math.stackexchange.com/questions/905602/what-is-the-maximum-possible-number-of-elements-of-s/905633 | # What is the maximum possible number of elements of $S$?
This is an interesting problem I found.
Let there be a 2-digit sequence that can start with 0, like 04 or 93. Let a "nudge" be defined as exactly one of the following operations:
1) Increasing one of the digits by 1.
2) Decreasing one of the digits by 1.
3) Changing a 0 to a 9.
4) Changing a 9 to a 0.
For example, possible nudges of 19 are 09, 29, 18, and 10.
Say that $S$ is some set of 2-digit sequences so that it takes 3 or more nudges to transform any element of $S$ into some other element of $S$. What is the maximum possible number of elements of $S$?
I thought of a nudge as a knight moves on a chessboard, but with a 10x10 board numbered starting with 1 on the top left and 100 in the bottom right. Clearly the maximum is $100-1=99$ nudges, but that is without considering the extra condition in the problem statement. Any idea how to continue?
Thinking of a $10\times10$ chessboard is good - don't forget that you have to imagine the top and bottom sides as being adjacent, as also the left and right sides. For any given pair $p=(x,y)$ in $S$, define the neighbourhood of $p$ to be the set of all pairs which can be reached in at most one nudge from $p$. That is, the neighbourhood is $$\{\,(x,y),\,(x+1,y),\,(x-1,y),\,(x,y+1),\,(x,y-1)\}\ ,$$ where we interpret $0-1$ as $9$ and $9+1$ as $0$. It is not hard to see that
it takes $3$ or more nudges to transform $p$ to $q$ if and only if the neighbourhoods of $p$ and $q$ do not overlap.
Considering that each neighbourhood has size $5$, we see that no more than $20$ neighbourhoods can fit into the $100$ available squares without overlapping. However this does not guarantee that $20$ are actually possible: we have to consider their "shape" as well as their "size". But it is not too hard to find a solution by trial and error: the different colours indicate the neighbourhoods, and the "centres" of the neighbourhoods are the elements of $S$. $$\def\b{\!\color{blue}{\blacksquare}\!} \def\r{\!\color{red}{\blacksquare}\!} \def\y{\!\color{yellow}{\blacksquare}\!} \def\g{\!\color{green}{\blacksquare}\!} \begin{matrix} \b&\r&\g&\g&\g&\r&\b&\y&\y&\y\\ \r&\r&\r&\g&\y&\b&\b&\b&\y&\g\\ \g&\r&\b&\y&\y&\y&\b&\r&\g&\g\\ \y&\b&\b&\b&\y&\g&\r&\r&\r&\g\\ \y&\y&\b&\r&\g&\g&\g&\r&\b&\y\\ \y&\g&\r&\r&\r&\g&\y&\b&\b&\b\\ \g&\g&\g&\r&\b&\y&\y&\y&\b&\r\\ \r&\g&\y&\b&\b&\b&\y&\g&\r&\r\\ \b&\y&\y&\y&\b&\r&\g&\g&\g&\r\\ \b&\b&\y&\g&\r&\r&\r&\g&\y&\b\\ \end{matrix}$$
• This shows a nice connection to error correcting codes. Well done. – Ross Millikan Aug 22 '14 at 0:57
• @RossMillikan yes indeed, it's an example of a "perfect" code - or something like that - I have forgotten the terminology. – David Aug 22 '14 at 1:01
• I was just thinking that you use the fact that you can "correct" each square by $1$ to a "code square" because the " code squares" have a distance of $3$ in the Manhattan metric. This proves the maximum of $20$ – Ross Millikan Aug 22 '14 at 2:49
Your thought of a chessboard is a good one. Each element you put in$S$ rules out $12$ others. It seems knight moves cause a lot of overlap of the excluded squares. My first thought would be $00,12,24,36,48,50,62,74,86,98,05,17,29,31,43,55,67,79,81,93$ for $20$
• Thank you, I believe that is the maximum. Perhaps a proof by contradiction is necessary to prove that 21 elements or more is impossible. – asdf Aug 22 '14 at 0:41 | 2020-01-26T08:49:58 | {
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https://math.stackexchange.com/questions/1415204/exercise-from-baby-rudin-chapter-3-exercise-13-is-this-proof-correct-is-i | # Exercise from (Baby) Rudin (Chapter 3, exercise 13): is this proof correct? Is it well-written?
The problem is the following:
Prove that the Cauchy product of two absolutely convergent series converges absolutely.
Here is my attempt:
Let $s_n=\sum^n_{k=0}a_k$ and $t_n=\sum^n_{k=0}b_k$ be two series absolutely converging to $A$ and $B$ respectively; then, let $c_n=\sum^n_{k=0}a_kb_{n-k}$ and $C_n=\sum^n_{k=0}c_k$. We have to prove that the series $D_n=\sum^n_{k=0}|c_k|$ converges.
Since $D_n$ is a series of non-negative terms, it will suffice to show that it is bounded. And, in fact, $$D_n=|a_0b_0|+|a_0b_1+a_1b_0|+...+|a_0b_n+...+a_nb_0|\leq(|a_0||b_0|)+(|a_0||b_1|+|a_1||b_0|)+...+(|a_0||b_n|+...+|a_n||b_0|).$$
The RHS of the inequality is the Cauchy product of the series $\sum^n_{k=0}|a_k|$ and $\sum^n_{k=0}|b_k|$, both of which converge by hypothesis. Then, since $\sum^n_{k=0}||a_k||= \sum^n_{k=0}|a_k|$ and $\sum^n_{k=0}||b_k||=\sum^n_{k=0}|b_k|$, they also converge absolutely. The Cauchy product of two absolutely convergent series converges and, therefore, is bounded. Hence, $D_n$ is also bounded and, thus, convergent.
It seems fine to me, but also a little too easy perhaps! Assuming that it is correct...is it written properly? Should I specify some steps a little more? Should I be more explicit or succinct? Would this be accepted at a university exam?
• It is helpful to mention which exercise this is. – Tim Raczkowski Aug 30 '15 at 22:05
• This seems fine to me, as long as you're allowed to use that the Cauchy product of two absolutely convergent sequence converges. I can't speak for the exams, but I'd accept it. – Patrick Stevens Aug 30 '15 at 22:06
• @TimRaczkowski I've edited the question! It's the exercise 13, chapter 3. – Nicol Aug 30 '15 at 22:08
• @PatrickStevens Yes, the theorem was included in the theory section before the problems. – Nicol Aug 30 '15 at 22:08
Since, $\sum |c_n|$ is a sequence of non-negative terms, it suffices to show that $\{C_n\}$ is bounded.
\begin{align} C_n & =\sum_{k=0}^n\left|\sum_{m=0}^k a_mb_{k-m}\right|\\ & \le\sum_{k=0}^n\sum_{m=0}^k|a_mb_{k-m}|\\ & =\sum_{m=0}^n|a_n|\sum_{k=0}^{n-m}|b_k|\\ &\le\left(\sum_{k=0}^n|a_k|\right)B\le AB. \end{align}
Here, $C_n$ is a partial sum and $A$ and $B$ are the limits of $\sum a_n$ and $b_n$ respectively. | 2019-06-19T11:24:11 | {
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https://math.stackexchange.com/questions/62389/relationships-between-bounded-and-convergent-series | # Relationships between bounded and convergent series
I would like to know the relationships between bounded and convergent series. By bounded series I mean a series whose sequence of partial sums is bounded. For example, it seems natural that if a series is convergent, it is also bounded, but does the converse hold?
• "if a series is convergent, it is also convergent". True... but I don't think that's what you meant. One of those "convergent" was meant to be "bounded." Which one? – Arturo Magidin Sep 6 '11 at 20:03
• You presumably mean if it is convergent, its partial sums are bounded. That's true. The converse does not hold. Consider for example $1-1+1-1+\cdots$. But if the terms are positive, and the partial sums are bounded, then the series converges. – André Nicolas Sep 6 '11 at 20:04
• Davide: by bounded I mean that the sequence of partial sums is bounded, as in en.wikipedia.org/wiki/Bounded_function – Federico Sep 6 '11 at 20:08
• @Federico Ah. It may be better spell that out, rather than saying "bounded series". – Dylan Moreland Sep 6 '11 at 20:19
• But there is one important special case when there's a tight relation between bounded partial sums and convergence. Suppose that the series $\sum_n x_n$ consists of only nonnegative terms. (E.g., $\sum 1/n$ or $\sum 1/n^2$.) Then the series is convergent if and only if its sequence of partial sums $\sum_{i=1}^n x_i$ is bounded. – Srivatsan Sep 6 '11 at 20:45
Whenever we have a series, $$\sum_{i=1}^{\infty} a_i,$$ we "automatically" get two sequences out of that series:
1. The sequence of terms, which is $a_1,a_2,a_3,\ldots$; and
2. The sequence of partial sums, which is $s_1,s_2,s_3,\ldots$, where \begin{align*} s_1 &= a_1\\ s_2 &= a_1+a_2\\ s_3 &= a_1+a_2+a_3\\ &\vdots\\ s_n &= \sum_{i=1}^n a_i = a_1+a_2+\cdots + a_n. \end{align*}
When we talk about "convergence of the series", we are really talking about convergence of the sequence of partial sums: the series $\sum a_i$ converges if and only if the sequence $(s_n)$ converges. That is, your definitions about "series" are really about "sequence of partial sums", and so you have the usual relationship:
In particular, $$\sum_{i=1}^{\infty}a_i\text{ converges}\Longleftrightarrow \{s_i\}_{i=1}^{\infty}\text{ converges}\Longrightarrow \{s_i\}_{i=1}^{\infty}\text{ is bounded}\Longleftrightarrow \sum_{i=1}^{\infty}a_i\text{ is bounded}$$ (where "is bounded" is as per your definition above); but it is possible for $\{s_i\}_{i=1}^{\infty}$ to be bounded, and not convergent, so one can have a series $\sum_{i=1}^{\infty}a_i$ that is bounded (i.e., the sequence of partial sums is bounded) but does not converge.
A simple example of this is $\sum_{i=1}^{\infty} (-1)^n$. The partial sums are $s_{2k+1} = -1$ and $s_{2k}=0$ for every $k$, so the sequence of partial sums is: $$-1,\ 0,\ -1,\ 0,\ -1,\ldots$$ which is bounded but not convergent. So the series is bounded but not convergent.
The relevant theorem for sequences, as you are no doubt aware, is:
Theorem. If $\{b_n\}$ is a monotone sequence, then $\{b_n\}$ converges if and only if it is bounded.
How does that translate for series? When is the sequence of partial sums monotone?
$\{s_i\}$ is increasing if and only if $s_n\leq s_{n+1}$ for all $n$, if and only if $s_{n+1}-s_n\geq 0$ for all $n$; but $s_{n+1}-s_n = a_{n+1}$. So:
The sequence of partial sums of $\displaystyle \sum_{i=1}^{\infty}a_i$ is increasing if and only if all the terms $a_i$ are nonnegative. The sequence of partials sums is strictly increasing if and only if all the terms $a_i$ are positive.
Likewise,
The sequence of partial sums of $\displaystyle \sum_{i=1}^{\infty}a_i$ is decreasing if and only if all the terms $a_i$ are nonpositive. The sequence of partial sums is strictly decreasing if and only if all the terms $a_i$ are negative.
So we conclude:
Theorem. Let $\displaystyle \sum_{i=1}^{\infty}a_i$ is a series in which every term $a_i$ is nonnegative. Then the series converges if and only if it is bounded (in the sense that the sequence of partial sums is bounded).
• You said it is possible that one can have a series that is bounded (i.e., the sequence of partial sums is bounded) but does not converge. Is this also true for the absolutely convergent case? So is it possible to have a sequence of partial sums that is bounded, but where the associated series does not converge absolutely? – Kamil Jul 21 '16 at 10:40
No, a bounded series does not necessarily converge. Consider the series $\displaystyle \sum (-1)^n$ (heavily related to Henning's example). It will forever oscillate between 0 and 1 (or -1 and 0, depending on the indices).
But if the partial sums are bounded and monotonic, then it does converge.
But in either case, it's a bit weaker than the converse - convergent series always have bounded partial sums.
• It may happen that the partial sums of a series are bounded and its general term converges to zero and yet the series diverges. Example: if $a_n=(-1)^k2^{-k}$ for every nonnegative $k$ and $n$ such that $2^k\le n<2^{k+1}$, then $a_n\to0$ but the limit set of the sequence of the partial sums $\sum a_n$ is the interval $[a_0,a_0+1]$. – Did Sep 6 '11 at 20:32
• @Didier: That's a great counterexample. I'll hold onto that. – davidlowryduda Sep 6 '11 at 22:25
• @Didier: One more thing: are there a weaker set of conditions than bounded and monotonic? (I know this is very open, but perhaps there is some 'surprising' or 'deceptive' set of conditions?). – davidlowryduda Sep 6 '11 at 22:26
• Well yes: absolutely convergent. – Did Sep 6 '11 at 23:56
• @Didier: I don't understand: are you saying it's surprising that something that converges absolutely also converges in general? – davidlowryduda Sep 7 '11 at 0:11
A convergent sequence is bounded, but a bounded sequence is not necessarily convergent. Consider, for example the sequence (1, -1, 1, -1, 1, -1, ...).
On the other hand, an increasing (or decreasing) bounded sequence in $\mathbb R$ will necessarily converge.
• I know about that about sequences, but what about series? Does it even make sense to say "bounded series"? – Federico Sep 6 '11 at 20:04
• I assumed you meant sequences because "bounded series" is not immediately meaningful to me -- except if it means that the sequence of partial sums is bounded (in which case boundedness and convergence of the series are both the same as the sequence, so the relation is the same). – hmakholm left over Monica Sep 6 '11 at 20:05
• @Federico: When talking about a series, we usually refer to the "sequence of partial sums"; in particular, a series converges if and only if the sequence of partial sums converges, and a series is bounded if and only if the sequence of partial sums is bounded. – Arturo Magidin Sep 6 '11 at 20:14
• @Henning: yes, I was referring to the sequence of the partial sums. I find a conflict between what you say and what André Nicolas says, because you impose the condition of being an increasing (decreasing) bounded sequence and he imposes the condition of being a bounded sequence with all the terms positive to converge. Could you explain? – Federico Sep 6 '11 at 20:22
• @Federico: if all terms of the series are positive, then the sequence of partial sums is increasing; if all terms of the series are negative, then the sequence of partial sums is decreasing. – Arturo Magidin Sep 6 '11 at 20:28 | 2021-08-01T08:15:21 | {
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