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https://math.stackexchange.com/questions/2020321/factorise-fx-x34x2-3x | # Factorise $f(x) = x^3+4x^2 + 3x$
Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$\text{Factorise}: f(x) = x^3+4x^2+3x$$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$x^2+4x+3$$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|} \hline x^2 & \\ \hline & 3 \\ \hline \end{array}
Then find HCF of 3:
$$3\\ 1 | 3$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|} \hline x^2 & 1x \\ \hline 3x& 3 \\ \hline \end{array}
Now factorise the rows and columns:
$$x^2 + 1x = x(x+1)\\ x^2 + 3x = x(x+3)\\ 1x + 3=1(x+3)\\ 3x +3=3(x+3)$$
Therefore: $$x^2+4x+3=(x+1)(x+3)$$ It follows that: $$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
• If you want to check your work, multiply out $x(x+1)(x+3)$, and you will find it equals $x^3+4x^2+3x$. Nov 18, 2016 at 18:40
Had never heard of the box method before I saw you use it!
When you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.
Upon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3)$.
This works for a general polynomial of degree $n$. If a $n$-degree polynomial has roots $\lambda_1, \cdots, \lambda_n$, then the polynomial is equal to $(x - \lambda_1)\cdots(x - \lambda_n)$
Of course we not always have access to all the polynomial's roots at once, but we can partially factorise and work our way through that. Let us go through the polynomial $p = x^4 + 3x^3 - x^2 - 3x$.
First obvious thing is that $\lambda_1 = 0$ and thus $p = x^4 + 3x^3 - x^2 - 3x = x(x^3 + 3x^2 - x - 3)$.
Now comes the tricky part. Finding the roots of $p_1 = x^3 + 3x^2 - x - 3$. What I always start by doing is trying some small numbers. Guessing $\lambda_2 = 1$ turns out to be fine and thus we can factor $p_1$. Now there is some polynomial $p_2$ of degree 2 that multiplied by $(x - \lambda_2) = (x - 1)$ gives $x^3 + 3x^2 - x - 3$. To calculate such polynomial I refer you to Ruffini's rule, which is just a faster way to factorize a polynomial when you know one root. Then we get $p_2 = x^2 + 4x + 3$ which was what you factorized above. Since this is a 2nd degree polynomial, we could keep trying to guess roots, use your box method, or using the quadratic formula to find its roots.
• The box method is really cool, particularly when in particular multiplying trinomials with trinomials. Nov 18, 2016 at 18:44
• First time for me too ! Nov 18, 2016 at 18:45
• @imranfat had never heard of it. should I google it and learn it?
– RGS
Nov 18, 2016 at 18:47
• Bear in mind guys, I'm a total novice! I like systematic approaches to problems, and this one helps me remember the steps required. Video I used is here: youtube.com/watch?v=_Wb_CT-1VN8&app=desktop
– Dan
Nov 18, 2016 at 18:49
• @DanB your approach is fine. Also note that the method I talked about always works. If you are able to find the roots, then you are able to factorise it! I will add something to my answer.
– RGS
Nov 18, 2016 at 18:53
I wouldn't know if this really qualifies as an answer, but what you're asking for is essentially an opinion of methodology.
First of all, your factorisation is correct: you can check the result simply by doing the multiplication.
That said, it looks like you could have saved a lot of effort by using a couple of different techniques.
To begin with, the polynomial $x^3+4x^2+3x$ is clearly divisible by $x$ (as you noted), so we factor it out to get $x(x^2+4x+3)$.
Here is where it gets interesting: to factor the degree 2 polynomial, I usually take one of these two roads.
First: observe that $(x+a)(x+b)=x^2+(a+b)x+ab$, so if we can guess two numbers $a$ and $b$ such that, in our case, we have $ab=3$ and $a+b=4$, we are done.
In this case you might have been able to spot that $3\cdot 1=3$ and $3+1=4$, so $(x+1)(x+3)$ is the factorization we're looking for. Trust me, it gets easier with practice and it's a huge time- and effort-saver for simple degree 2 polynomials with integer coefficients.
Else we could try and guess just one root, and then use polynomial long division. This has the advantage of working with any degree of polynomials, as long as you can guess one root. I usually try $1$, $-1$ and maybe $2$, but if those don't work I'm better off using some other technique.
And finally, for degree 2 polynomials you have the general formula to find the roots, which has the advantage of always working (in the sense that it will always find all real roots there are to find), but can get messy and is, in my honest opinion, rather boring.
Now, this is not to say that any of what you did is incorrect or even strictly less efficient: it is ultimately a matter of personal taste, but I feel that having as many tools as possible in your pocket can only do you good.
• Thanks! It was indeed really a question of methodology, although I wasn't completely sure of the answer! The guessing method sounds like something that with practice would indeed save a lot of time.
– Dan
Nov 18, 2016 at 19:00
Everything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...
Now factorise the rows and columns: $$x^2 + 1x = x(x+1)\\ x^2 + 3x = x(x+3)\\ 1x + 3=1(x+3)\\ 3x +3=1(x+3)$$
The last line ought to be $3x +3=3(x+1)$.
• Thank you! I've edited my post. Will look up the methods you've suggested, can never hurt to have more tools in the arsenal!
– Dan
Nov 18, 2016 at 19:01
• @DanB You're welcome. Nov 18, 2016 at 19:02
To factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\tag1$$ Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a product of $c$ in $x^2+bx+c$.
In other words, we have \begin{align*} & r+s=-b\\ & rs=c\end{align*}\tag2 for $x^2+bx+c$. In your example, we see that $-b=-4$ and $c=3$. Messing around, we see that when $r=-1,s=-3$, the requirements are met. Thus,$$x^3+4x^2+3x=x(x-1)(x-3)$$
X(X+1)(X+3)
Therefore (X+1)(X+3) = X^2+4X+3.
Therefore multiplying this by X you get X^3+4X^2+3X.
So yes well done that is correct.
Well done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$
Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.
Suppose we have a quadratic $p(x)$ that factorizes as:
$$p(x) = (x + s)(x + t)$$
where we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get:
$$p(x) = x^2 + sx + tx + st = x^2 + (s+t)x + st$$
Now suppose that we are given $p(x)$ in the form
$$p(x) = x^2 + bx + c$$
In order to factorize $p(x)$, all we need to do is find two numbers $s$ and $t$ such that $$s+t=b$$ and $$st=c$$
In your case, you want to find $s$ and $t$ such that $s+t=4$ and $st=3$. It shouldn't take you very long to realize that $s=3$ and $t=1$ will do. Then you can immediately factorize the polynomial as $(x+3)(x+1)$.
Slogan: To factorize a quadratic of the form $x^2+bx+c$, just find two numbers that add to give $b$ and multiply to give $c$.
Just for interest
This method should work for all the quadratics you see in Math 1, and I hope that it's a bit clearer what's going on compared to the 'Box Method'. You might be interested in how we solve quadratic equations that don't have nice solutions. For example, suppose we were trying to factorize $$x^2+x-1$$ We want to find two numbers that add together to give $1$ and multiply together to give $-1$. It turns out that the right two numbers are $$\frac{1+\sqrt{5}}{2}$$ and $$\frac{1-\sqrt{5}}{2}$$ How did I work those out? Well, I'll show you a method that was invented by the ancient Babylonians. Suppose we have two numbers $b$ and $c$ and we're trying to find two numbers $s$ and $t$ such that $s+t=b$ and $st=c$.
If we square the first equation, we get $$b^2 = (s+t)^2 = s^2 + 2st + t^2$$ Now since $st=c$, we can subtract $4c$ from the left hand side and $4st$ from the right hand side to get: $$b^2-4c = s^2 - 2st + t^2$$ But $s^2 - 2st + t^2=(s-t)^2$, so we may write $$s-t = \sqrt{b^2-4c}$$ (Here, we are free to assume that $s\ge t$, so this is the positive square root.) Now we can recover $s$ and $t$: $$s = \frac{(s+t) + (s-t)}{2} = \frac{b+\sqrt{b^2-4c}}{2}$$ $$t = \frac{(s+t) - (s-t)}{2} = \frac{b-\sqrt{b^2-4c}}{2}$$ | 2022-05-23T15:10:59 | {
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https://math.stackexchange.com/questions/3510522/proving-recurrence-using-induction-where-there-is-an-upper-limit-on-the-number-o | # Proving recurrence using induction where there is an upper limit on the number of integers to prove it for
Given $$x_0=1$$ and $$x_j=x_{j-1}\frac{N-(j-1)}{N}+x_{j+1}\frac{j+1}{N}$$ for $$j=1,...,N-1$$, the formula $$x_j={N\choose j}$$ can be proven by induction. I do not see why we are able prove it by induction, seeing that using induction, we prove the base case, we assume the formula holds for n, then show it holds for n+1, then we claim it holds for every integer. In this case it only holds up to N-1. So why does the induction proof work? I think that the induction proof should fail.
The inductive proof: $$x_0=1$$, Suppose the result is true for $$k \le j$$
\begin{align}x_{j+1} &=\frac{N}{j+1}\left(x_j-\frac{N-j+1}{N}x_{j-1}\right)\\&=\frac{N}{j+1}\left(\frac{N!}{j!(N-j)!}-\frac{N-j+1}{N}\frac{N!}{(j-1)!(N-j+1)!}\right)\\ &\text{after some simplification}\\&={N\choose{j+1}} \end{align}
See it works but I think that it should fail.
• You need to use "Strong" induction in this case. First show it works for $j=0$ and $j=1$, then show if it works for $j-1$ and $j$ then it works for $j+1$. Jan 15 '20 at 21:04
• @DonaldSplutterwit Yes. But, it works only up to N-1 and not all the integers. So induction should not prove this formula. Jan 15 '20 at 21:10
• I think you are misunderstanding what the statement is claiming. It is not claiming $x_j={N\choose j}$ for all $N$ and doing induction on $N$. Is is claiming that $x_j={N\choose j}$ for all $j$ (up to $N-1$) and doing induction on $j$. Part of what is confusing you is that we are doing induction on a variable $j$ that has an upper limit, $N-1$, and so our result is not for all natural numbers but just for natural numbers from $1$ to $N$. This is okay. We can do induction for finite values. Jan 15 '20 at 21:48
• Both $x_j$ and ${N \choose j}$ are only defined for $j\le N$ so that's our upper bound. Our statement is $P(j)=$ if [$j \le N$] then [$x_j = {N\choose j}$. We can prove $P(j)$ for all natural $j$ because if $j> N$ then [$j \le N]$ is false and "if [FALSE] then $anything$" is a true statement so if $j > N$ then $P(j)=$ if [$j \le N$] then [$x_j = {N\choose j}$; is a true statement. Jan 15 '20 at 22:18
• "Nowhere do we specify an upper bound" Uh.... yes we do! ... it says "for j=1,...,N−1" in black and white!.... So induction will be successful up to $j+1 \le N$. For $j > N$ it will fail but we don't give a flying funky if it fails for values we don't care about. We only care that it is successful on the values we DO care about. ... "In this case it only holds up to N-1" For $j+1$ where $j \le N-1$. So $j+1 \le N$. And we don't CARE if it holds for any higher values? Why on earth would we? Jan 15 '20 at 22:52
The statement is not for every positive integer but only for positive integers up to $$N$$. It is not trying to claim it is true for any positive integer greater than $$N$$.
Consider: $$P(j )=$$: If $$j \le N$$ then something, call it $$Q(j)$$ is true.
Let's say we can show that if $$k< N$$ that $$Q(k)\implies Q(k+1)$$ but only if $$k < N$$. I claim we can still prove $$P(k)$$ is true for all natural $$k$$.
Base case: $$P(1)$$. We show that $$Q(1)$$ is true and as $$1 < N$$ then \$P(1) is true.
Induction step: $$P(k)\implies P(k+1)$$.
Assume if $$k\le N$$ then $$Q(k)$$ is true.
Case 1: $$k \ge N$$.
Then $$k+1 > N$$ and [$$k+1 \le N$$] is false: $$FALSE \implies Q(k+1)$$ is vacuously true whether $$Q(k+1)$$ is true or not. So $$P(k+1)$$ is true.
Case 2: $$k < N$$.
The $$k+1 \le N$$. We showed that $$Q(k)\implies Q(k+1)$$. So if [$$k+1 \le N]\implies Q(k+1)$$ is true. So $$P(k+1)$$ is true.
SO our induction step works.
We have proven:
For any natural $$j$$, $$P(j)$$ is true.... or in other words,
if $$j \le N$$ then $$Q(j)$$ is true... or in other words,
$$Q(j)$$ is true for every natural $$j \le N$$.
That's all the induction is trying to claim.
The induction of $$Q(j)$$ works..... up to $$j \le N$$. There is nothing invalid about this.
Given the recurrence you have written, you will need two initial values to get going. Solving the recurrence for $$x_{j+1}$$ gives $$x_{j+1} = \frac{n x_j+ (j-n-1) x_{j-1}}{j+1}.$$ Note that if $$j=n$$, this gives $$x_{n+1} = \frac{n\cdot 1 + (n-n-1)\cdot n}{n+1} = 0,$$ just as you would expect. So in fact the inductive proof holds for all $$j$$ provided you define $$\binom{n}{j}=0$$ for $$j<0$$ or $$j>n$$. | 2021-11-27T09:18:42 | {
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That BC = 6 cm the Pythagorean theorem to show that ΔBOD is triangle.: a circle is 39.19 square centimeters, and so $\angle AC ' I$ is right is also..., diameter is 2 in., find the value of each variable,... | 2021-04-19T03:13:53 | {
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https://math.stackexchange.com/questions/977474/infinite-sum-of-sines-with-increasing-period | # Infinite Sum of Sines With Increasing Period
A while ago, I was thinking about the Weierstrass function, which is a sum of sines with increasing frequencies in such a way that the curve is a fractal. However, I wondered what would happen if one took the sum where the frequencies decreased; in particular, noting that $|\sin(x)|\leq x$, it is clear that the function $$f(x)=\sum_{n=1}^{\infty}\sin\left(\frac{x}{s_n}\right)$$ converges pointwise for any sequence $s_n$ such that the sum of $\frac{1}{s_n}$ converges absolutely - and, in fact, yields an $f$ which is analytic. Of particular interest to me is the sequence of square numbers - that is, the function $$f(x)=\sum_{n=1}^{\infty}\sin\left(\frac{x}{n^2}\right).$$ I created the following plot of the function from the first 10,000 terms in the series:
What I find interesting here is that, for some reason I can't determine, it looks like $f(x)$ might be asymptotic to $\sqrt{x}$. I've checked numerically for higher arguments and this seems to continue to be the case. This strikes me as odd, since I had expected it to appear more or less periodic, with long-term variation in amplitude and frequency.
1. Is $f(0)=0$ the only (real) zero of $f$?
2. Does $f$ grow without bound? What is it asymptotic to?
• that's 3 questions. There are three types of people in the world: Those who can count to three, and those who can't. (LOL) – Mark Fischler Oct 16 '14 at 23:42
• It often helps to analyze these kinds of functions as real or imaginary parts of the same series with the exponential function in place of $\sin$ or $\cos$. In this case, you have $\Im\sum_{i=1}^\infty\exp\left(i\frac{x}{n^2}\right)$. – alex.jordan Oct 16 '14 at 23:59
• Here's something funky about this function: If you take the Fourier transform of $f(x)$ and move the sum outside the integral (which I'm not sure is justified) you get zero. Yet $f(x)$ certainly looks to have a quasi-periodic structure. – Mark Fischler Oct 17 '14 at 0:18
• Well, I plugged $\int_{1}^{\infty}\sin(\frac{x}{n^2})$ into Mathematica and it returned $\sqrt{2 \pi } \sqrt{x} C\left(\sqrt{\frac{2}{\pi }} \sqrt{x}\right)-\sin (x)$ where $C$ is the Fresnel integral - $\int_{0}^{x}\cos(\pi x^2/2)$, which converges to $\frac{1}2$ - so the expression $\int_{1}^{\infty}\sin(x/n^2)$ is indeed asymptotic to $\sqrt{x}$. I wonder if there's some way to get the analogous discrete result from there. – Milo Brandt Oct 19 '14 at 14:32
Here's a proof that $f$ only vanishes at $x = 0$ (you can use a similar method to get some asymptotic results as well).
Write $f(x)/x$ as \begin{align*} {f(x)\over x} &= \sum_{n\geq 1} {\operatorname{sinc}{(x/n^2)}\over n^2} \end{align*} Since $f(x)/x$ is even, we need only treat the case $x\geq 0$. Split the sum into the regions where $x/n^2$ is smaller or greater than $\pi$. Since $\operatorname{sinc}{\lambda}>0$ for $|\lambda|\leq \pi$, and since $\operatorname{sinc}{\lambda}\geq 2/\pi$ for $|\lambda|\leq \pi/2$, we have \begin{align*} \sum_{x/n^2\leq \pi} {\operatorname{sinc}{(x/n^2)}\over n^2} &\geq {2\over \pi}\sum_{x/n^2\leq \pi/2} {1\over n^2} = {2\over \pi} \sum_{n\geq (2x/ \pi)^{1/2}}{1\over n^2} > {2\over \pi} \int_{\lceil(2x/\pi)^{1/2}\rceil}^\infty {dt\over t^2} = {2\over \pi}{1\over \lceil(2x/\pi)^{1/2}\rceil}. \end{align*} On the other hand, since $\operatorname{sinc}{\lambda}\leq 1/\lambda$ for all $\lambda>0$, we have \begin{align*} \left|\sum_{x/n^2> \pi} {\operatorname{sinc}{(x/n^2)}\over n^2}\right| & \leq \sum_{x/n^2>\pi} {1\over x} \leq {1\over x}\left\lfloor\left({x\over \pi}\right)^{1/2}\right\rfloor\leq \left({1\over \pi x}\right)^{1/2}. \end{align*} So \begin{align*} f(x)/x> {2\over \pi}{1\over \lceil(2x/\pi)^{1/2}\rceil} - {1\over (\pi x)^{1/2}}, \end{align*} which is positive when $x\geq \pi$. Since all terms in the sum are positive if $0\leq x < \pi$, it follows that $f(x)/x$ is always positive.
By the way, here's another heuristic (which can be made precise without too much trouble I think). We have \begin{align*} {f(x)\over x} & = {1\over 2}\hat g(x) = {1\over 2}\int_{-1}^1 g(t)e^{-ixt}\,dt = {1\over 2x}\int_{-x}^x g(t/x)e^{-it}\,dt, \end{align*} where $g = \sum_{n\geq 1} \chi_n$ is the sum of the characteristic functions of the intervals $[-n^{-2},n^{-2}]$. (Since $g$ is an $L^1$ function, this tells us at once that $f(x)/x\to 0$ as $x\to\infty$.) Note that $\{y:g(y)>n\} = [-n^{-2},n^{-2}] = \{y: y^{-1/2}>n\}$ (or something similar), so we should roughly expect $g$ to look like $y^{-1/2}$, and so we should expect $\hat g(x)$ to be approximately $x^{-1/2}$ (as can be seen from the last integral). You can use the same idea to get a sense of what the function would look like if you replace $\sin{(x/n^2)}$ with $\sin{(x/n^\alpha)}$ for $\alpha > 1$ (it should I think look like $x^{1/\alpha}$).
• Sorry if this approach is similar to Kirill's. I was working on it and didn't notice his answer until it was too late. – Nick Strehlke Oct 30 '14 at 1:30
• One more comment: The function $g$ I mentioned toward the end should actually be $\lfloor y^{-1/2}\rfloor$; that is, you should be able to write (for $x>0$) $$f(x) = \int_0^\infty \lfloor (x/y)^{-1/2}\rfloor \cos{y}\,dy.$$ (There may be another factor of two or something.) – Nick Strehlke Oct 30 '14 at 7:27
These are my thoughts on the problem. They do not answer any of the three questions, but do not fit on a comment.
Let $S_N$ be the $N$-th partial sum of the series and $L_n$ the lowest common multiple of $\{1,\dots,n\}$. Then $S_N$ is periodic of period $2\,L_n^2\,\pi$. $L_n$ is known to be of the order $e^{n(1+o(1))}$, so that the period is very large. Computations show that $S_N$ changes sign on $[0,2\,L_n^2\,\pi]$. Here is the grapf of $S_4$ on $[0,288\,\pi]$.
On the other hand, for any $\delta\in(0,1)$ let $\alpha\in(0,\pi/2)$ be such that $(\sin\alpha/\alpha)=\delta$. Then we have the lower bound $$f(x)\ge S_N(x)+\delta\,\Bigl(\sum_{n=N+1}^\infty\frac{1}{n^2}\Bigr)\,x,\quad0\le x\le\alpha\,N^2.$$ This is the graph of $S_{10000}$ (in blue) and the above lower bound with $n=10$, $\delta=3/\pi=0.95493$ and $\alpha=\pi/6$.
This shows that $f(x)>0$ on $(0,50\,\pi/3]$. An strategy to prove that $f(x)>0$ for all $x>0$ is to show that $$S_N(x)+\delta\,\Bigl(\sum_{n=N+1}^\infty\frac{1}{n^2}\Bigr)\,x>0,\quad0\le x\le\alpha\,N^2.$$ I have checked it up to $n=100$.
As for upper bounds, we have $$|f(x)-S_N(x)|\le\sum_{n=N+1}^\infty\min\Bigl(1,\frac{x}{n^2}\Bigr)\le \Bigl(\sum_{n=N+1}^\infty\frac{1}{n^2}\Bigr)\,x.$$
Here's a slightly informal way to get the asymptotic expansion of this function.
Split the region of summation into intervals of the form $$\pi k < \frac{x}{n^2} < \pi(k+1),$$ for $k\in\mathbb{Z}_{\geq0}$, and write the sum as $$\sum_{n\geq 1}\sin \frac{x}{n^2} = \sum_{k\geq0} \sum_{\frac{x}{\pi(k+1)} < n^2 < \frac{x}{\pi k}} \sin \frac{x}{n^2}.$$ The first term $k=0$ can be approximated with an integral, giving $$\sum_{n^2 > x/\pi}\sin\frac{x}{n^2} \sim \int_{\sqrt{x/\pi}}^{\infty} \sin\frac{x}{n^2}\,dn = \sqrt{2\pi x}C(\sqrt{2}),$$ where $C$ is the Fresnel integral.
On the rest of the sum we can approximate $\sin\frac{x}{n^2}$ with its average value $\frac2\pi$. However, the rest of the sum should be handled with some care, because the regions on which $\sin$ is positive have different lengths from the regions on which it's negative, which will produce a net contribution to the asymptotic term $O(\sqrt{x})$.
Approximating the sum $$\sum_{\frac{x}{\pi(k+1)} < n^2 < \frac{x}{\pi k}} \sin \frac{x}{n^2} \sim \frac{2(-1)^k}{\pi}\left(\sqrt{\frac{x}{\pi k}} - \sqrt{\frac{x}{\pi(k+1)}}\right),$$ and summing over $k\geq1$ gives the value $$\frac{\sqrt{x}}{\pi^{3/2}}\left((3\sqrt{2}-2)\zeta(\tfrac12) - \sqrt{2}\zeta(\tfrac12,\tfrac32) \right) = B\sqrt{x},$$ where $\zeta(z,a)$ is the generalized zeta function.
Putting everything together gives the asymptotic form $$\sum_{n\geq 1}\sin\frac{x}{n^2} \sim \left(\sqrt{2\pi}C(\sqrt{2}) + B\right)\sqrt{x} \approx 1.25038\sqrt{x}.$$
Another comment that is too long to be a comment:
The heuristic reason that the function is asymptotically proportional to $\sqrt{x}$ is that for very large $x$,
$\cdot$ The contributions of the terms in $S_n(x)$ for $n$ much less than $\sqrt{x}$ behave as pseudo-random numbers, restricted to $[-1,1]$. Thus $S_n(x)$ for $n << \sqrt{x}$ can be thought of as roughly having a mean value of zero, and a $\sigma$ on the order of $\frac{1}{2}\sqrt{x}$.
$\cdot$ The contributions of the terms in $S_n(x)$ for $n^2$ greater than about $2x$ can be well-estimated by approximating $\sin \frac{x}{n^2} \approx \frac{x}{n^2}$. Adding these from $n=\sqrt{x}$ to infinity looks like $x \int \frac{1}{n^2}$ which will be about $\frac{x}{\sqrt{2x}} = \frac{\sqrt{x}}{\sqrt{2}}$. And these contributions are all positive.
$\cdot$ The contributions of the terms in $S_n(x)$ for $n^2 \approx \frac{2}{\pi} x$ are all roughly $1$ since we are near the top of the sine curve, and there are about $\frac{8}{3\pi} \frac{1}{2n} = \frac{4\sqrt{2}}{3\sqrt{\pi}}\sqrt{x}$ of them between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ on the sine curve, for a contribution of pretty nearly $\sqrt{x}$.
$\cdot$ The contributions of the terms in $S_n(x)$ lying on the falling side of that first part of the sine curve almost exactly cancel with the contributions from the negative part of the first period of the sine curve (because $n^2$ is greater in that part of the curve, at at any rate, the contribution is a lot smaller than that of the flat section of the positive arch).
So all in all, you would expect $f(x)$ to behave about like $\frac{3}{2}\sqrt{x}$ for large $x$.
The last 3 bullets can be made somewhat more rigorous.
But by this reasoning, you would also expect larger fluctuations than we see in the original graph. So the "effectively ergodic" argument made in the first bullet is an over-estimate of the fluctuations, and I don't have a plausible reason why.
• I think that first bullet you have might be key; I'm not sure how to rule out fluctuations, but I'll bet a statement to the tune of, "for any $\varepsilon_1,\varepsilon_2>0$, we can find arbitrarily large $x$ and $n$ such that that $S_{n-1}(x)<\varepsilon_1\sqrt{x}$ and $(1-\varepsilon_2)\frac{x}{n^2}<\sin(\frac{x}{n^2})<(1+\varepsilon_2)\frac{x}{n^2}$" might be a good way to formalize it (that particular statement might be false, but there could be a similar one of use) - it would prove that $f$ has a sequence that's asymptotic to the square root function. – Milo Brandt Oct 18 '14 at 2:59
• You can start a line with an asterisk (*) to make it an item in a bulleted list. – user856 Oct 25 '14 at 9:47 | 2020-03-29T10:08:18 | {
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https://mathoverflow.net/questions/95867/covering-a-cube-with-a-square/95920 | # Covering a Cube with a Square
Suppose you are given a single unit square, and you would like to completely cover the surface of a cube by cutting up the square and pasting it onto the cube's surface.
Q1. What is the largest cube that can be covered by a $1 \times 1$ square when cut into at most $k$ pieces?
The case $k=1$ has been studied, probably earlier than this reference: "Problem 10716: A cubical gift," American Mathematical Monthly, 108(1):81-82, January 2001, solution by Catalano-Johnson, Loeb, Beebee.
(This was discussed in an MSE Question.) The depicted solution results in a cube edge length of $1/(2\sqrt{2}) \approx 0.35$.
As $k \to \infty$, there should be no wasted overlaps in the covering of the 6 faces, and so the largest cube covered will have edge length $1/\sqrt{6} \approx 0.41$. What partition of the square leads to this optimal cover?
Q2. For which value of $k$ is this optimal reached?
I have not found literature on this problem for $k>1$, but it seems likely it has been explored. Thanks for any pointers!
• I wonder if anyone in the packaging industry has the answer. May 4 '12 at 0:37
• I believe there was a "Mathematical Games" column on dissections which had a Greek cross rearranged into a square. Perhaps that article also mentioned this problem? Gerhard "Testing Your Martin Gardner Fu" Paseman, 2012.05.04 May 4 '12 at 15:51
• If memory serves, Martin Gardner is also a source for the puzzle of covering a unit cube with a properly folded 1x7 strip. Apr 1 '17 at 2:52
• Cool! The solution to that 1x7 puzzle informs the answer to the following: over all rectangles which can be folded to cover the unit cube, what is the lim inf of their areas? Gerhard "Hint: The Answer Is Six" Paseman, 2017.04.03. Apr 3 '17 at 17:43
• When I read the title I thought it was about a 2D version $\varphi:[0,1]^2\to[0,1]^3$ of Peano area-filling (or space-filling) curves $[0,1]\to[0,1]^2$... Mar 29 '19 at 12:04
Four pieces, using the tessellation technique I learned from Harry Lindgren's Geometric Dissections (1964):
• I've been looking for that! Gerhard "Was It Behind The Headboard?" Paseman, 2017.04.01. Apr 1 '17 at 10:57
• Also, if you shift the (smaller) black squares a little to the right, the one piece that almost looks like two pieces looks more like one piece. Gerhard "Tilting His Head Over This" Paseman, 2017.04.01. Apr 1 '17 at 11:06
• Beautiful! I posted a colorized version to make the dissection of the net more self-evident. Apr 1 '17 at 11:54
• Thanks! GerPas: Yes, the hexomino tessellation cqn be moved a bit to the right (and independently also a bit up, though that doesn't help your cause). JO'R: I see that you were also the one who made a three-dimensional model of a cube covered by 5 pieces of a square, but evidently it will take more time (if you bother to do it at all) to re-do for the new 4-piece record. Apr 1 '17 at 14:37
• @NoamD.Elkies: I'll make the model eventually, as there will be a certain pleasure in seeing your dissection fully deployed. Apr 1 '17 at 15:05
You can cut a $\sqrt{6}\times\sqrt{6}$ square into 24 pieces that then cover the $1\times1\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.
![cube.png][1]
[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \times 1$ square:
[Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\sqrt{6}\times\sqrt{6}$ square. The second is the $2\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.
• Brilliant!! :-) May 3 '12 at 23:51
• Actually, any two polygons of the same area are equidecomposable and the surface of the cube can be unfolded into a polygon, so the result is nice but not terribly surprising. Of course, the question about the minimal number of pieces remains. May 4 '12 at 0:15
• Assuming that "pieces" mean "connected polygonal pieces", we can take a 3 by 2 rectangle, cut it into a T-shape and two unit squares and then use the standard "sliding cut" to turn it into a square, giving the total of 6 pieces to cover the unit cube. Can we do better? May 4 '12 at 0:34
• OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six. May 4 '12 at 4:21
• Wow! $\mbox{}$ May 4 '12 at 10:18
Just illustrating Noam Elkies' 4-piece solution:
Bottom face is mostly yellow (except for a little green); two hidden back faces are mauve.
• Lol, I can't believe you actually made a model of it. I haven't used glue and paper and scissors to cut out shapes since arts and crafts in elementary school. Though this is significantly more complicated. The wide breadth of tasks that comes with the profession of mathematics, lol.
– user78249
Apr 1 '17 at 17:51
Sorry, this is an answer to an other question. (I did not read the question carefully.)
Question: For which $$k$$, $$k$$ squares can tile the surface of cube.
Answer: $$k=6\cdot(n^2+m^2)$$.
Here is a tiling with $$k=30$$, $$n=1$$ and $$m=2$$.
(source: psu.edu)
It is obvious if the tiling is vertex-to-vertex.
If the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides. Then you can shift squares on one side of the geodesic to make the tiling "more vertex-to-vertex". Repeating this operation you can make the tiling to be vertex-to-vertex.
• @Anton: Did you intend $n \ge 1 , m \ge 1$, or, say, $n \ge 1 , m \ge 0$ ? May 3 '12 at 18:03
• Yes, $k$ has to be positive; so $n\ge 1$ and $m\ge 0$. May 3 '12 at 18:33
• @Anton: Sorry to be slow :-/, but could you describe a partition of the square into 6 pieces that exactly cover the cube? May 3 '12 at 19:21
• @Joseph: the partition into faces (the cube has 6 faces). May 3 '12 at 19:35
• @Anton: I apologize for being so dense, but it may be that you are answering a different question than I asked...? I asked for how to cut up one square to cover a cube, not how to cover a cube with many squares. If I've diagnosed this correctly (unsure), it may explain why we seem to be talking past one another? May 3 '12 at 23:17
No promises that these are optimal, but here are some lower bounds:
With $k=2$, side length $3/8=0.375$ (with one piece flipped over), and with $k=3$, side length $2/5=0.4$: | 2021-10-25T23:11:23 | {
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https://math.stackexchange.com/questions/2250051/are-the-closed-intervals-of-mathbbr-precisely-the-compact-connected-sets | # Are the closed intervals of $\mathbb{R}$ precisely the compact connected sets?
Equip $\mathbb{R}$ with the topology generated by open intervals $(a, b)$. A subset of $\mathbb{R}$ is compact iff it's closed and bounded.
Is every closed bounded connected subset of $\mathbb{R}$ a closed interval $[a, b]$ (and conversely)?
Is every open bounded connected subset of $\mathbb{R}$ an open interval $(a, b)$ (and conversely)?
Is this somehow related to the fact that removing one point from $\mathbb{R}$ splits it into 2 disconnected pieces (how is this property called anyway)?
• All true if you agree to refer to a point as a closed interval. – Mikhail Katz Apr 24 '17 at 17:03
• Do you consider $[b,b]= \{b\}$ to be a closed interval? – fleablood Apr 24 '17 at 17:04
• No, I'm considering only non-trivial intervals. – étale-cohomology Apr 24 '17 at 17:04
• beware also of the title, $[0,+\infty[$ is a closed interval but not bounded. – zwim Apr 24 '17 at 17:06
• Do you want an informal or formal answer? It's intuitively obvious that only singletons and intervals are connected. And intuitively obvious that among finite intervals that only [a,b] are closed (and always closed) and (a,b) are open (and always open), and thus, yes, your statements are all true. But showing these formally via definitions is ... not hard, but tedious.... but good practice. – fleablood Apr 24 '17 at 17:30
Clearly a closed and bounded interval is compact and connected.
Conversely, if a set is connected, then it is an interval, meaning it is a set $I$ with the following property: for all $x,y \in I$, if $x < z<y$, then $z \in I$. All the sets with this property must have one of the forms $$\{a\},\,[a,b],\, ]a,b[, \,[a,b[, \,]a,b], \,]a,+\infty[,\, [a,+\infty[,\, ]-\infty,b[ \mbox{ or } ]-\infty,b].$$ Among these, only $\{a\}$ and $[a,b]$ are compact, hence the answer to your question is yes. Notice that $\{a\}$ is an interval, by definition.
• Thank you. What about the 2nd question, about open sets? – étale-cohomology Apr 24 '17 at 17:14
• Also yes, with the same reasoning: use connectedness to narrow the list of "suspects" and then see that the only open bounded set in that list is $]a,b[$. – Ivo Terek Apr 24 '17 at 17:23
Lemma 1: Only intervals and singletons and the empty set are connected and all intervals and singletons and the empty set are are connected.
Lemma Z: $K\subset \mathbb R$ is a interval if and only if for all $x,y \in K$ then for all $k; x < k < y; k \in K$.
Proof: Should be self-evident. If $K$ is an interval than $K = [(a,b)]$ (for sake of notation $a$ can be $-\infty$ and $b$ can be $\infty$). and $a \le x < y \le b$ and for all $k: x < k < y$ then $a < k < y$ so $k \in K$.
If there exists a $k$ so that $x < k < y$ with $k \not \in K$ and $x,y \in K$ then there is no $a,b$ (not even $\pm \infty$) so that $a \le x; b \ge y$ and for all $r \in \mathbb R$ $a < r < b$; $r \in K$. (as $a < k < b$ but $k \not \in K$). So $K$ would not be an interval.
Proof of Lemma 1: If $K$ = $\emptyset$ or $K = \{x\}$, some singleton then $K$ can not be partitioned into two partitions so $K$ is connected.
Let $E \subset \mathbb R$. And let $E$ be such that there exist $x,y\in E; x< y$ and there exists a $k \not \in \mathbb R$ so that $x < k < y$. Let $A = E \cap (\infty, k)$ and $B= E \cap (k, \infty)$ then $A, B$ are non-empty partitions of $E$ and $\overline A \subset (-\infty,k]$ is disjoint from $B \subset (k,\infty)$ and $\overline B \subset [k,\infty)$ is disjoint from $A \subset (-\infty, k)$ so $E$ is not connected.
So the only connected subsets of $E$ are intervals, singletons, and the empty set.
If $K$ is an interval and $K = A\cup B$ and $A,B$ non empty and $A \cap B = \emptyset$. Let $a \in A$ and $b \in B$ and wolog $a < b$. Let $K = \{x| x \ge a; \forall k; a\le k \le x: k \in A\}$ and let $L = \{y \in K| y > a; y \in B\}$. It's easy to prove $K$ is non-empty $(a \in K)$ and bounded above (by $b$) and $L$ is non-empty ($b$ is in it) and bounded below (by $a$) and that $\sup L = \inf L$. As $K$ is an interval and $a \le j = \sup K = \inf L\le b$ then $j \in K$. And $j \in \overline A$ and $k \in \overline B$. So either $j \in \overline A \cap B$ or $j \in \overline B \cap A$. So $K$ is connected.
• My edit was solely to add a missing dollar-sign that disrupted some of the formatting. – DanielWainfleet Apr 25 '17 at 13:07 | 2019-05-21T16:50:06 | {
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http://math.stackexchange.com/questions/263338/is-0-a-removable-discontinuity-of-fracx-2-2x | # Is $0$ a removable discontinuity of $\frac{|x-2|-2}{x}$?
Given the function $f(x) = \large\frac{|x-2|-2}{x}$ ,
Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)? Thus it is a removable discontinuity ?
The problem is, that if I try to remove the absolute value, I get that in the region $x<2$ : $f(x) = -1$
What is the correct logical definition I need to use?
Thanks
-
What's wrong having with $f(x)=-1$ in the region $x<2$ ? – Ted Dec 21 '12 at 17:34
I changed the title because it made no sense and seemed to have no relevance to the question. Hope you like it! – rschwieb Dec 21 '12 at 17:51
As you said, $f$ isn't defined at $0$. However, $$\lim_{x\to 0}\frac{\left|x-2\right|-2}{x}=\lim_{x\to 0}\frac{2-x-2}{x}=-1$$ and so if we define $f(0)=-1$, $f$ becomes continuous at $0$.
-
Thanks. That's indeed what I thought. – joshua Dec 21 '12 at 17:34
@joshua: Note that when $x\to 0$, we always consider $x\neq 0$. – Babak S. Dec 21 '12 at 17:34
@BabakSorouh : great, thanks! – joshua Dec 21 '12 at 17:36
Given the function $$f(x) = \frac{|x-2|-2}{x},$$ Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)?
Yes, that's correct. As currently defined, the function is undefined at $x = 0$.
But that doesn't mean that the limit of $f(x)$ as $x \to 0$ is undefined. Recall, we are interested in what is happening as $x$ gets very very close to $0$ (not what is happening AT zero).
As $x \to 0, |x - 2| = 2 - x$, so $$\lim_{x \to 0}\frac{\left|x-2\right|-2}{x}=\lim_{x \to 0}\frac{2-x-2}{x}=-1$$
Thus it is a removable discontinuity ?
Yes, indeed, by simply defining $$f(x) = \begin{cases} \frac{\left|x-2\right|-2}{x} & x\neq 0\\ \\ -1 & x = 0\\ \end{cases}$$
$f(x)$ is then continuous at $x = 0$, hence it is a removable discontinuity.
-
@Sami Me too: via googling some question I had, three of the first hits turned up math.stackexchange.com! – amWhy Jul 15 '14 at 11:45 | 2016-07-26T18:26:44 | {
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https://math.stackexchange.com/questions/3241494/general-square-form-of-taylor-expansion-polynomial | # General square form of Taylor expansion polynomial
Suppose we have a Taylor expansion for a function $$f$$ with respect to t up to $$M$$-th order.
$$$$T_M = \sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k = f(x) + f'(x)\Delta t + \frac{1}{2}f''(x)\Delta t^2 + \cdots$$$$
What would be the general form of $$T_M^2$$ with respect to $$\Delta t$$?
$$$$T_M\times T_M = (\sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k)^2 = (???) + (???)\Delta t + (???)\Delta t^2 + \cdots$$$$
Perhaps Multinomial theorem can help?
• It's just the general form for squaring a polynomial, but you can drop higher order terms if all you want is the $M$th order approximation. Note: the $1/k!$ is inside the summation. May 27 '19 at 13:51
Since we are squaring a sum we can utilize $$\left( \sum_{k=1}^{n} a_k \right)^2 = \sum_{k=1}^{n} \sum_{j=1}^{n} a_ka_j$$ Applying this formula onto our problem yields
$$\left(\sum_{k=0}^{M} \frac{1}{k!}f^k(x)\Delta t^k\right)^2 = \sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j}$$ Taking a closer look at the coefficients of $$\Delta t^{k+j}$$, we see that there are $$k+j+1$$ different combinations of $$k$$ and $$j$$. For example ($$k+j=3$$ cf. Jose Brox' answer), we have the combinations for: $$(k,j) \in \{(0,1),(1,2),(2,1),(3,0)\}$$
Thus we can rewrite the above sum $$\sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j} = \sum_{k=0}^{M}\left[ \sum_{j=0}^{k}\frac{1}{j!}\frac{1}{(k-j)!}f^j(x)f^{k-j}(x)\right]\Delta t^k$$ where the coefficient corresponding to the $$k$$-th exponent is $$\sum_{j=0}^{k}\frac{1}{j!(k-j)!}f^j(x)f^{k-j}(x)$$
• In the second last equation, is that truncated to the order $M$? Jun 5 '19 at 10:32
• My fault, I think it should be $k$ instead of $M$ in the second sum term. Nov 5 '19 at 15:58
Like with usual polynomial multiplication, the term of degree $$k$$ comes from adding up all products of two monomials whose degrees sum to $$k$$. For example, for degree $$3$$ you have $$(0,3)+(1,2)+(2,1)+(3,0)$$. | 2022-01-28T20:59:48 | {
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https://kokecacao.me/page/Course/S22/15-259/Lecture_005.md | # Lecture 005
## Summing a Random Number of i.i.d. Random Variable
So you have $X_1, X_2, ...$ random variables with the same distribution. And $N$ discrete, positive, integer random variable with $X_i \perp N$
Let us compute $S = \sum_{i = 1}^N X_i$
Theorem: Let $X_1, X_2, ...$ be i.i.d. (Independent and identically distributed) random variables, where $X_i \sim X$. Let $S = \sum_{i = 1}^N X_i$ where $N \perp X_i$, then:
\begin{align} E[S] =& E[N]E[X]\\ E[S^2] =& E[N]Var(X) + E[N^2]E[X]^2\\ Var(S) =& E[N]Var(X) + E[X]^2Var(N)\\ \end{align}
### Expectation
\begin{align*} E[S] =& E[\sum_{i = 1}^N X_i]\\ =& \sum_{n = 1}^\infty E[\sum_{i = 1}^N X_i | N = n] \cdot Pr\{N = n\}\\ =& \sum_{n = 1}^\infty E[\sum_{i = 1}^n X_i | N = n] \cdot Pr\{N = n\}\\ =& \sum_{n = 1}^\infty E[\sum_{i = 1}^n X_i] \cdot Pr\{N = n\} \tag{by $X_i, N$ independent}\\ =& \sum_{n = 1}^\infty nE[X_i] \cdot Pr\{N = n\}\\ =& E[X_i]\sum_{n = 1}^\infty n \cdot Pr\{N = n\}\\ =& E[X_i]E[N]\\ \end{align*}
### Variance
You want to compute $Var(S) = E[S^2] - E[S]^2$. We now compute $E[S^2]$
\begin{align*} &E[S^2]\\ =& E[(\sum_{i = 1}^N X_i)^2]\\ =& \sum_n E[(\sum_{i = 1}^N X_i)^2 | N = n] Pr\{N = n\}\\ =& \sum_n E[(\sum_{i = 1}^n X_i)^2] Pr\{N = n\}\\ =& \sum_n E[(X_1 + X_2 + ... + X_n)^2] Pr\{N = n\}\\ =& \sum_n nE[X_1^2] + (n^2 - n)E[X_1X_2] Pr\{N = n\}\tag{Note that $X_1 \not\perp X_1$, $X_1 \perp X_2$}\\ =& \sum_n nE[X^2] Pr\{N = n\} + \sum_n (n^2 - n)E[X]^2 Pr\{N = n\}\\ =& E[N]E[X^2] + E[N^2]E[X]^2 - E[N]E[X]^2\\ =& E[N](E[X^2] - E[X]^2) + E[N^2]E[X]^2\\ =& E[N]Var(X) + E[N^2]E[X]^2\\ \end{align*}
With above $E[S^2] = E[N]Var(X) + E[N^2]E[X]^2$:
\begin{align*} Var(S) =& E[S^2] - E[X]^2\\ =& E[N]Var(X) + E[N^2]E[X]^2 - E[X]^2\\ =& E[N]Var(X) + E[N^2]E[X]^2 - E[X]^2E[N]^2 \tag{by $E[S] = E[X]E[N]$ above}\\ =& E[N]Var(X) + E[X]^2(E[N^2] - E[N]^2)\\ =& E[N]Var(X) + E[X]^2Var(N)\\ \end{align*}
### Example: Tree Growing
We have time $t = 0, 1, 2, 3...$, and start from a node. On each step, with probability $\frac{1}{2}$, a leaf will stay inert, with probability $\frac{1}{2}$, a leaf will split to 2.
Define $X_t = \text{number of leaves in time} = t$, then we see:
Define $Y_i = \begin{cases} 1 & \text{with half probability}\\ 2 & \text{with half probability}\\ \end{cases}$
\begin{align*} X_t =& \sum_{i = 1}^{X_{t - 1}} Y_i\\ =& \sum_{i = 1}^{X_{t - 1}} 1 \cdot Pr\{\text{this leaf stay inert}\} + 2 \cdot Pr\{\text{this leaf split to 2}\}\\ \end{align*}
We can easily calculate:
1. $E[Y] = \frac{3}{2}$
2. $E[Y^2] = 1^2 \frac{1}{2} + 2^2 \frac{1}{2} = \frac{5}{2}$
3. $Var(Y) = \frac{10}{4} - \frac{9}{4} = \frac{1}{4}$
Now using the formula we calculate:
\begin{align*} &E[X_t]\\ =& E[X_{t - 1}] \cdot E[Y]\\ =& E[X_{t - 1}](\frac{3}{2})\\ =& (\frac{3}{2})^{y - 1}(\frac{3}{2}) \tag{by unfolding $E[X_{t - 1}]$}\\ =& (\frac{3}{2})^y\\ \end{align*}
\begin{align*} &Var(X_t)\\ =& E[X_{t - 1}]Var(Y) + E[Y]^2Var(X_{t - 1})\\ =& (\frac{3}{2})^{t - 1} \cdot \frac{1}{4} + (\frac{3}{2})^2 Var(X_{t - 1})\\ \end{align*}
## Stochasticall Domination
Stochastically Dominate: $X$ stochastically dominates $Y$ (write as $X \geq_{st} Y$) if $(\forall i)(Pr\{X > i\} \geq Pr\{Y > i\})$
### Jensen's Inequality
E[X^2] \geq E[X]^2
E[X^s] \geq E[X]^s (\text{for integer } x \geq 2)
Definition of convex function: (all point on convex function sit above a line that pass through it) A real-valued function $g(x)$ on interval $S \subseteq \mathbb{R}$ is convex on $S$ if:
1. $g(t) = at + b$
2. $(\forall x \in S)(g(x) \geq y(x) \equiv ax + b)$ where the line $y(x) = ax + b$ is called a supporting line for the function $g$ at $t$.
Theorem: Let $X$ be a random variable that takes on values in an interval $S$ and let $g : S \rightarrow \mathbb{R}$ be a convex function on $S$, then $E[g(X)] \geq g(E[X])$
Table of Content | 2022-12-02T03:39:13 | {
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https://yutsumura.com/tag/row-space/ | # Tagged: row space
## Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
$\mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .$ Find a basis for the span $\Span(S)$.
## Problem 708
Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.
(b) Find a basis for the row space of $A$.
(c) Find a basis for the range of $A$ that consists of column vectors of $A$.
(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.
## Problem 704
Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.
(b) Find a basis for the row space of $A$.
(c) Find a basis for the range of $A$ that consists of column vectors of $A$.
(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.
## Problem 604
Let
$A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.$
(a) Find a basis for the null space $\calN(A)$.
(b) Find a basis of the range $\calR(A)$.
(c) Find a basis of the row space for $A$.
(The Ohio State University, Linear Algebra Midterm)
## Problem 366
Let $A=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &0 \end{bmatrix}$.
(a) Find an orthonormal basis of the null space of $A$.
(b) Find the rank of $A$.
(c) Find an orthonormal basis of the row space of $A$.
(The Ohio State University, Linear Algebra Exam Problem)
Let $A$ be an $m\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\trans}$. | 2021-01-24T18:10:04 | {
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http://fnpf.associazionetalea.it/maximum-and-minimum-word-problems-calculus.html | 4) Set derivative of the function equal to zero and solve. Maximum And Minimum Quadratic Word Problems Worksheet. (solution by Calculus). His method was essentially geometrical, and this made the. Find the length of the shortest ladder that will reach over an 8-ft. com: Free Precalculus Review and Calculus Preview Lessons and Practice Problems. Define calculus. Provide a Testimonial Introduction. Thus there is only one relative minimum in this function, and it occurs at x=0. While a fair number of the exercises involve only routine computations, many of the exercises and most of the problems are meant to illuminate points that in my experience students have found confusing. S Publisher: McGraw-Hill, Year: 2001 ISBN: 0071358978 Search in Amazon. We are trying to do things like maximise the profit in a company, or minimise the costs, or find the least amount of material to make a particular object. 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Minimum And Maximum Some of the worksheets for this concept are Assignment date period, For each problem find all points of absolute minima and, Name, Work 17 maxima and minima, Quadratic work name maximums and minimums, Bwc payroll reporting guidelines, Ira required minimum distribution work, Calculus work problems. Maximum And Minumum. MAXIMUM AND MINIMUM IN APPLICATIONS To find a maximum or minimum, solve f'(x) = 0. Calculus: Derivatives Maximum/Minimum Word Problems Topics include cost function, ellipse, distance, volume, surface area, and more. Question: How to solve if instead, the problem asked for the minimum profit? In some problems, the minimum is the value of x (example: the 0. Now you have an equation of one variable. Optimization problems (also called maximum-minimum problems) occur in many fields and contexts in which it is necessary to find the maximum or minimum of a function to solve a problem. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Max & Min applications. A complete set of Class Notes, Handouts, Worksheets, PowerPoint Presentations, and Practice Tests. MAXIMUM AND MINIMUM VALUES. Such problems, involving the determination of the form of a curve having a certain maximum or minimum property, were quite different from the ordinary maximum and minimum problems of the differential calculus. There is another way to measure an angle, which involves arc length. Justify your answer. This worksheet generates AB Calculus Topics/Questions: To keep server load down, there is a maximum of 100 questions per worksheet. Behind every calculus problem, neatly packaged and ready to solve, is a word problem. Find two positive numbers such that their product is 192 and the sum of the first plus three times the second is a minimum. Area and perimeter worksheets. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. asked by Marissa on August 16, 2007; Algebra2. One that is very useful is to use the derivative of a function (and set it to 0) to find a minimum or maximum to find either the smallest something can be, or the largest it can be. Hairy inflection point problem. Maximum and Minimum of. Find the dimensions of the rectangle with the maximum area that can be in-scribed in a circle of radius 10. Improve your math knowledge with free questions in "Find the maximum or minimum value of a quadratic function" and thousands of other math skills. Solution of exercise 3. Maximum And Minumum - Displaying top 8 worksheets found for this concept. exponential functions 25. Think about the English meaning of the word constraint, and remember that the constraint equation will have an equals sign. 2 The chapter also has maximum-minimum and optimization word problemsin 2. com Description: Considered to be the hardest mathematical problems to solve, word problems continue to terrify students across all math disciplines. How to Find Maximum and Minimum Points Using Differentiation ? In this section, we will see some example problems of finding maximum and minimum values of the function. For example, in any manufacturing business it is usually possible to express profit as function of the number of units sold. Topic: Applications of Derivatives • Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. The tip of the corner is no more than 4 inches above the bottom edge of the paper. Basic calculus with applications to problems in the life and social sciences. Three major examples are geometry, number theory, and functional equations. ©a 62C0z1 b3e DKGuStwa S rS9odf 6tVwVaHr Ve f 7LBLhC U. Check that this value is a minimum or maximum and read exactly what form the answer should be. Find the Local Maxima and Minima. As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. feet and then drops the same distance. More information. The largest is the absolute maximum whereas the smallest is the absolute minimum. 512 The only critical point of A is r =1. Pre-Calculus Online. The best ticket prices to maximize the revenue is then: $10−0. 75 cm and at the sides 0. This can be regarded as the special case of mathematical optimization where the objective value is the same for every solution, and thus any solution is optimal. Linear Approximation and Differentials- Finding approximate values of a function using linearizations, definition of the differential and how it relates to linear approximations. Some have short videos. Determine a feasible domain • 4. These are often referred to as word problems. For example, companies often want to minimize production costs or maximize revenue. In the optimization sections (2. word problem involving the maximum or minimum of a quadratic function. Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the. 512 is a global maximum. 371 in the problem above) after differentiating the given equation and equating it to 0. Absolute maximum. Such areas must be learned outside class. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4). Piecewise Functions Worksheet. Maximum Minimum. Mathematics has its own “grammar”. Listed below are word lessons that focus on giving students instruction on how to solve most types of word problems commonly found in algebra, geometry, and trigonometry. 18 for each thing-a-ma-bob. Rather, the purpose is to show them real world situations in which calculus is helpful. 3 Increasing & Decreasing Functions and the 1st Derivative Test F: intervals where f is incr, decr relative extrema stepbystep: 1st Deriv Test Calculus Home Page Problems for 3. If θ is the angle between the ground and her line of sight to the balloon, at what rate is this angle changing at the instant the balloon hits the ground?. ) > I'm really keen to understand if it is possible to calculate maximum and > minimum points of cubic graphs WITHOUT the use of calculus. To get started a sketch may help you get a sense of what the curve looks like. We’ll break these two big Stages into smaller steps below. All students take calculus. Louis University Brody Dylan Johnson (St. Maximum and Minimum Value Word Problems - Quadratic Equations This algebra video tutorial explains how to solve word problems that asks you to calculate the maximum value of a function or How to solve word problems with quadratic equations Geometry Teachers Never Spend Time Trying to Find. You'll find a variety of solved word problems on this site, with step by step examples. 10) Give an example function f (x) where f '' (0) = 0 and there is a relative maximum at x = 0. Students must know how to perform the first and second derivative tests and how to identify maximum, minimum and inflection points from those tests. Chapter 6: Word Problems - Rate And Distance; Chapter 7: Word Problems - Percentages; Chapter 8: Word Problems - Investments; Chapter 9: Word Problems - Ratios; Chapter 10: Word Problems - Age; Chapter 11: Word Problems - Rates Of Performing Work; Chapter 12: Word Problems - Slope-Intercept Equations; Chapter 13: Word Problems With 2 Variables. Reduce the primary equation to one variable by substitutions from other equations/information given • 3. Page Navigation. Maximum And Minumum. Because the derivative provides information about the gradient or slope of the graph of a function we can use it to locate points on a graph where the gradient is zero. Pre-Calculus 11- Course Outline: File Size: Maximum and Minimum Problems. 79 to make each thing-a-ma-bob and the company charges$ 5. 1 decade ago. In the optimization sections (2. Here is a set of practice problems to accompany the Finding Absolute Extrema section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. By using this website, you agree to our Cookie Policy. Linear Approximation and Differentials- Finding approximate values of a function using linearizations, definition of the differential and how it relates to linear approximations. 1 through 3. Worksheet # 17: Maxima and Minima 1. use differentiation to solve related rate problems in a variety of pure and applied contexts. Proof for all n is delayed until we discuss numerical sequences and series later on. Examples train understanding, translation into the mathematical language (eg, equations), solve it, check the accuracy and solution discussion. Nov 24, 2009 Equation Calculator - Quadratic Solver. Word problems involving integrals usually fall into one of two general categories: alien related and non-alien related. Complete each question on lined paper. They illustrate one of the most important applications of the first derivative. Sample Minimum/Maximum Problem (05:27) Steps for finding the minimum of a function are demonstrated. 265 Clove Road, New Rochelle, New York 10801 Pre-Calculus. " That x-coordinate of a relative maximum or minimum value of the function. We we've seen, there are many useful applications of differential calculus. Finding a maximum for this function represents a straightforward way of maximizing profits. She is watching the balloon as it travels at a steady rate of 20 feet per second towards the ground. Calculus Optimization Problems/Related Rates Problems Solutions 1) A farmer has 400 yards of fencing and wishes to fence three sides of a rectangular field (the fourth side is along an existing stone wall, and needs no additional fencing). l'hopital's rule 26. A local extrema is the point where the function takes on the largest or smallest value in a small region around the point. The best ticket prices to maximize the revenue is then: $10−0. Write a function for each problem, and justify your answers. Reduce the primary equation to one variable by substitutions from other equations/information given • 3. cost, strength, amount of material used in a building, profit, loss, etc. Know how to compute absolute maxima and minima on closed regions. •• This might not be the case in applied “word problems. This is what Wolfram|Alpha does with elementary word problems: it not only gives you the answers, but it also helpfully translates the information in the problems into mathematical symbols, showing you the first (and most important) steps toward finding a solution. In the pdf version of the full text, clicking. However after t months the value of the investment, in dollars, is. The expert examines calculus and inequalities in a word problem. It is through word problems that we find value in math. Folsom Lake College's mathematics program provides students with the ability to think logically and abstractly and develop the problem-solving and computational skills necessary for success in any field of study. ©a 62C0z1 b3e DKGuStwa S rS9odf 6tVwVaHr Ve f 7LBLhC U. There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Get an answer for 'Maximum profit, given revenue and cost equations. (1 pt) A company that makes thing-a-ma-bobs has a start up cost of$ 47060. If each rented room costs $10. 3 - Radians So far we have measured angles in degrees. What is a Limit? Local Maximum and Minimum Values/ Function of Two Variables The Simplex Method – Finding a Maximum / Word Problem Example, Part 5. 01 Minimum length of cables linking to one point;. MAT 111 - Pre-Calculus Chapter 6 – Trigonometric Functions 6 6. Trakimas Math WHS. Determine a feasible domain • 4. Be able to solve word problems involving maxima and minima. Read the problem carefully. Using Derivatives to Find the Absolute Maximum and Minimum Values of f(x)OBJECTIVES• Find absolute extrema using Maximum-Minimum Principle I. Optimization Problems; Introduction to Optimization. 3 Increasing & Decreasing Functions and the 1st Derivative Test F: intervals where f is incr, decr relative extrema stepbystep: 1st Deriv Test Calculus Home Page Problems for 3. (b) Find the velocity and acceleration vectors of the particle. The functions that maximize or minimize the functionals are can be found using the Euler - Lagrange of the calculus of variations. For each$1. Word problems with max/min Example: Optimization 1 A rancher wants to build a rectangular pen, using one side of her barn for one side of the pen, and using 100m of fencing for the other three sides. Can anyone help me to solve these derivative word problems pls. Find the global maximum value and the global minimum value of the function f (x) = x 4-2 x 2 + 3 on [-1, 2]. Rational Equations Word Problems. Max Min Word Problems Our approach to max min word problems is modeled after our approach to related rates word problems. Now look at the same places and think about what the slope is at those two locations. Pre-Calculus Assignment Sheet Unit 4 - Graphing & Writing Sine & Cosine Functions; Application Problems October 21 to November 5th, 2013 Date Topic Assignment Monday Gr 10/21 changes. Pre-Calculus 11- Course Outline: File Size: Maximum and Minimum Problems. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Can You Show Me Examples Similar to My Problem? Optimization is a tool with applications across many industries and functional areas. Applications of Extrema of Functions of Two Variables. 2017 AP® CALCULUS AB FREE-RESPONSE QUESTIONS 2. Linear Programming A linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to system of linear constraints. Graphing with Calculus. In the process of solving these problems we will develop a procedure in which all “word problems” can be effectively solved. The analytical tutorials may be used to further develop your skills in solving problems in calculus. The course concentrates on the various functions that are important to the study of the calculus. Proof for all n is delayed until we discuss numerical sequences and series later on. A high point is called a maximum (plural maxima). OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS - BC CALCULUS. The unit cost C (the cost in dollars to make each copy machine) depends on the number of machines made. Maxima and minima mc-TY-maxmin-2009-1 In this unit we show how differentiation can be used to find the maximum and minimum values of a function. 3 FINDING THE MAXIMUM AND MINIMUM OF WORD PROBLEMS. Optimization Problems (Calculus Fun) Many application problems in calculus involve functions for which you want to find maximum or minimum values. Some worksheets contain more problems than can be done during one discussion section. There is some vital stuff in this unit. While they might not actually work out the quadratic function to come up with a precise number, managers at movie theaters,. Read the problem- write the knowns, unknowns and draw a diagram if applicable L y 8 3 x-3 x 2. 2) There are three levels of information. Below is the graph of some function, $$f\left( x \right)$$. There was not a good enough understanding of how the Earth, stars and planets moved with respect to each other. Listed below are word lessons that focus on giving students instruction on how to solve most types of word problems commonly found in algebra, geometry, and trigonometry. Identify all of the relative extrema and absolute extrema of the function. Get an answer for 'Maximum profit, given revenue and cost equations. Give all decimal answers correct to three decimal places. One-dimensional and two-dimensional gravity problems, range, vector components of velocity, etc. Clearly, negative values are not allowed by our problem, so we are left with only two cut points and the following line graph: Therefore the minimum occurs for x 3. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. If a function is continuous on a closed interval, then by the extreme value theorem global maxima and minima exist. In this maximum and minimum worksheet, students solve and complete 10 various types of word problems. Max and Min's. We shall see that such. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Topic: Applications of Derivatives • Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. (4 Credits) Differential calculus for engineers and scientists. Worksheets are Assignment date period, For each problem find all points of absolute minima and, Work 17 maxima and minima, Work 5, Lessonunit plan name key features of graphs swbat, Fha maximum mortgage work, Name, Quadratic work name maximums and minimums. If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s(t) = -16t^2 + 54t. opens downward and thus has a maximum point. A ball is thrown vertically upward. Review: Review the definitions and first derivative test from previous lesson. for which the stated problem makes sense. The concavity of functions is discussed. mapped lines to other lines. Hey thanks for all your help but you kinda confused me on a few: 1)Determine whether f(x)=-5x^2-10x+6 has a maximum or minimum value and find that value A. MAT 111 - Pre-Calculus Chapter 6 – Trigonometric Functions 6 6. Constrained Optimization: Step by Step Most (if not all) economic decisions are the result of an optimization problem subject to one or a series of constraints: • Consumers make decisions on what to buy constrained by the fact that their choice must be affordable. And the absolute minimum point for the interval happens at the other endpoint. After the ball has hit the floor for the first time it rises 10. 5 applied maximum and minimum problems We have used derivatives to help find the maximums and minimums of some functions given by equations, but it is very unlikely that someone will simply hand you a function and ask you to find its extreme values. 2 show that a square has the maximum area inscribed in a circle. The following problems range in difficulty from average to challenging. We de ned a critical number of f(x) on domain D to be any number x = c on the domain D such that. New Calculus 5. Some of rst year calculus repeats senior school calculus: including \Word problems" which have always been di cult for students. Lesson 10: Max and Min Problems 3. (1 pt) A company that makes thing-a-ma-bobs has a start up cost of $47060. Each question is accompanied by a table containing the main learning objective(s), essential knowledge statement(s), and Mathematical Practices for AP Calculus that the question addresses. Such problems, involving the determination of the form of a curve having a certain maximum or minimum property, were quite different from the ordinary maximum and minimum problems of the differential calculus. How to Solve World Problems in Calculus reviews important concepts in calculus and provides solved problems and step-by-step solutions. K Worksheet by Kuta Software LLC. Optimization problems (also called maximum-minimum problems) occur in many fields and contexts in which it is necessary to find the maximum or minimum of a function to solve a problem. (solution by Calculus). What are the dimensions if the printed area is to be a maximum? 2) A cylindrical container with circular base is to hold 64 cubic centimeters. I plan on working through them in class. AP Calculus Sec 3. 265 Clove Road, New Rochelle, New York 10801 Pre-Calculus. Maximum Minimum. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. Login to reply the answers Post; Ray. So the function has a relative maximum at x=-5. So let's look. One example is the path of an airplane. 50$ , with 27,000+300(5) = 28,500 spectators and a revenue of \$ R(5) = 270,750. While they might not actually work out the quadratic function to come up with a precise number, managers at movie theaters,. Maximizing/Minimizing word problem: Minimum and Maximum Values. Area and perimeter worksheets. Maximum and Minimum of. To get started a sketch may help you get a sense of what the curve looks like. Displaying all worksheets related to - Maximum Minimum. Quadratic applications are very helpful in solving several types of word problems (other than the bouquet throwing problem), especially where optimization is involved. Read the problem carefully. To find the local maximum and minimum values of the. Next, the first- and second-derivative tests for local extrema are studied. What calculus is useful for is science, economics, engineering, industrial operations, finance, and. But we will not always be able to look at the graph. However once you know these 6 steps, then you should be able to solve any Calculus related rates problems you like. Nov 24, 2009 Equation Calculator - Quadratic Solver. superprof resources. of cardboard, what are the dimensions of the biggest box that can be made? 2. OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS - BC CALCULUS. Maximum and Minimum Value Word Problems - Quadratic Equations This algebra video tutorial explains how to solve word problems that asks you to calculate the maximum value of a function or How to solve word problems with quadratic equations Geometry Teachers Never Spend Time Trying to Find. Limit of Sequence Problems. Rates of change: the derivative, velocity, and acceleration. More Lessons for Calculus Math Worksheets A series of free Calculus Video Lessons. Precalculus review and Calculus preview - Shows Precalculus math in the exact way you'll use it for Calculus - Also gives a preview to many Calculus concepts. Using calculus you can calculate its average cruising altitude, velocity and acceleration. | 2020-03-30T00:49:27 | {
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https://mediawebplace.com/o7jb35t1/252366-trapezoid-base-calculator | … We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. 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The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button “ solve ” to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid – sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, δ = 180° 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180°, δ = 180°, =... Especially the part on which it rests or is supported 30° γ - 125° h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as γ + δ = 180° β... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! 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Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. 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Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = ½(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the … 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base ° = 150° the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button “ solve ” to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid – sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, δ = 180° 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180°, δ = 180°, =... Especially the part on which it rests or is supported 30° γ - 125° h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as γ + δ = 180° β... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180° ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180° ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 – sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base a '' – sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. Eso Warden Stamina Build Pvp, Barry Callebaut Where To Buy, Tom And Jerry Memes Clean, Weather Wayne, Nj, Dong Zijian And Sun Yi Wedding, Medical Words With Prefix Con, Red White Mobile Qoo10, Relacionado" /> … We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = ½(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the … 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base ° = 150° the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button “ solve ” to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid – sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, δ = 180° 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180°, δ = 180°, =... Especially the part on which it rests or is supported 30° γ - 125° h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as γ + δ = 180° β... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180° ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180° ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 – sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base a '' – sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. Eso Warden Stamina Build Pvp, Barry Callebaut Where To Buy, Tom And Jerry Memes Clean, Weather Wayne, Nj, Dong Zijian And Sun Yi Wedding, Medical Words With Prefix Con, Red White Mobile Qoo10, Compártelo:Haz clic para compartir en Twitter (Se abre en una ventana nueva)Haz clic para compartir en Facebook (Se abre en una ventana nueva)Haz clic para compartir en Google+ (Se abre en una ventana nueva) Relacionado" />
# trapezoid base calculator
In Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezium. You can always use our trapezoid calculator above, however, you should be able to calculate the area of trapezoid by yourself. Also, the area can be calculated in a plane of two dimensions. To find the Area of Trapezoid, all you need to do is just go ahead and implement the steps that are given below: First of all, note down the given parameters such as side lengths and height. Area of a trapezoid and midline or bases Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. As α + β = 180°, β = 180° - 30 ° = 150°. To calculate the trapezoid area, follow the below steps: Measure and write down the base a , base b , and height h of the trapezoid. FAQ. Free Trapezoid Sides & Angles Calculator - Calculate sides, angles of an trapezoid step-by-step This website uses cookies to ensure you get the best experience. All you have to do is enter the price per unit area and voila, you have the total cost of materials in a single click! Here is how the Base a of Trapezoid calculation can be explained with given input values -> 6.333333 = 2*(50/12)-2. Calculator that gives out the volume of a trapezoidal prism with the given bases, height, and width values. Area of a Trapezoid Calculator: ... A 4 – sided geometrical figure which has one pair of parallel sides is called a Trapezoid. Use the trapezoid calculator to find the area of a trapezoid with bases 6 cm and 9 cm and height 5 cm. If you only know the side lengths of a regular trapezoid, you can break the trapezoid into simple shapes to find the height and finish your calculation. The opposite sides are different in length. Yes, our tool is that awesome. Through the Pythagorean theorem, we express height in right triangles, after with substitutions we obtain the above-mentioned formula. Trapezoid calculator computes all properties of a trapezoid such as area, perimeter, height and sides given a sufficient subset of these properties. Follow below steps to get trapezoid online using our online trapezoid area calculator. The opposite sides are different in length. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. A trapezoid, also known as a trapezium, is a 4-sided shape with two parallel bases that are different lengths. The procedure to use the isosceles trapezoid calculator is as follows: Step 1: Enter the height and two base values in the input field. Steps to Find Area & Perimeter of a Trapezoid. Calculator Techniques for Polygons in Plane Geometry Solving problems related to plane geometry especially polygons can be easily solved using a calculator. Volume=(1/3)*pi*Height*(Radius 1^2+Radius 2^2+(Radius 1*Radius 2)), Total Surface Area=pi*Radius*(Radius+sqrt(Radius^2+Height^2)), Lateral Surface Area=pi*Radius*sqrt(Radius^2+Height^2), Total Surface Area=2*pi*Radius*(Height+Radius), Area of a Triangle when base and height are given, Area of a Parallelogram when base and height are given, The Base a of Trapezoid formula is defined as A trapezoid is a quadrilateral with one pair of parallel lines Bases ,the two parallel lines are called the bases and is represented as, The Base a of Trapezoid formula is defined as A trapezoid is a quadrilateral with one pair of parallel lines Bases ,the two parallel lines are called the bases is calculated using. Calculate the side (base) of a trapezoid if given diagonal, lateral side (leg) and other base ( a b ) : Calculate the lateral side (leg) of a trapezoid if given diagonal and bases ( c ) : Calculate the side (base) of a trapezoid if given diagonal, height, angle between the diagonals and base ( a b ) : * It is true in this case: 4. Practice Question: Calculate area of trapezoid for the following problems: N.B. As a consequence the two legs are also of equal length and it has reflection symmetry. How to find the area of a trapezoid? A trapezoid is a quadrangle with two parallel sides. Customer Voice. Algebra; Geometry; Electrics; Trapezoid definition. The stimulus check calculator computes how much money you are eligible to receive from the coronavirus checks promised by the US government. Calculate the remaining internal angles. Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. Some possible trapezoid shapes are shown below to clarify the concept. Our perimeter calculator supports a lot of the basic shapes and below you can read details about each one, including its perimeter calculation formula. Trapezoid definition. The step by step workout for how to find what is the area and perimeter of a trapezoid. Enter three side lengths and one angle between two of those sides. Step 2: Now click the button “Solve” to get the result. A trapezoidal prism is a solid prism, which has trapezoid cross-sections in one direction and rectangular cross-sections in the other directions. … A trapezium or trapezoid is a 2dimensional shape which has 2 parallel sides and 2 non-parallel sides. Area of a trapezoid formula The formula for the area of a trapezoid is (base 1 + base 2) / 2 x height, as seen in the figure below: When doing the calculation, remember to take each measurement in the same unit, or to convert it to the same unit, to get valid results. Home / Mathematics / Area; Calculates the area of a trapezoid given two parallel sides and the height. Here is an example to understand it better. Omni Calculator is here to change all that - we are working on a technology that will turn every* calculation-based problem trivial to solve for anyone. Perimeter of a square. To calculate the elements of the iso trapezoid; enter the two bases, a side or height. Prism is often distinguished by the shape of their base polygon. 1. Calculations at an isosceles trapezoid (or isosceles trapezium). This is possible for acute trapezoids or right trapezoids (rectangles). How to find the area of a trapezoid? Its area found to be 40cm 2. read more about us *within reason. All sides 2.Lateral side (leg) and angle at the base 3. Property #1) The angles on the same side of a leg are called adjacent angles and are supplementary() Property #2) Area of a Trapezoid = $$Area = height \cdot \left( \frac{ \text{sum bases} }{ 2 } \right)$$ () Property #3) Trapezoids have a midsegment which connects the mipoints of the legs() Step #2: Enter the value of long base "b". It has 3 fields in which you need to feed value. Calculator Techniques for Circles and Triangles in Plane Geometry The procedure to use the isosceles trapezoid calculator is as follows: Step 1: Enter the height and two base values in the input field. The volume of a trapezoidal prism calculator can find the volume and area of the trapezoidal prism at the same time. Mona Gladys has created this Calculator and 10+ more calculators! Trapezium. Follow Us. It has two parallel sides and the remaining two sides can be of any length, at any angle. This is not saying much about the quadrangle, so it is quite difficult to do calculations. Before jump into the code let’s know what is a trapezoid: A trapezoid is a geometrical figure with four sides in which two sides are parallel to each other. Properties. Choose the number of decimal places and click Calculate. To use this online calculator for Base a of Trapezoid, enter Height (h), Area (A) and Base (b) and hit the calculate button. FAQ. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule—see Trapezoid for more information on terminology) is a technique for approximating the definite integral. Therefore, the length of the other base of the trapezoid is 19 m. Example 5. How to Use the Isosceles Trapezoid Calculator? Formulas, explanations, and graphs for each calculation. The base is the lowest part or edge of something, especially the part on which it rests or is supported. Trapezoid area calculator is simple and easy to use. Our calculator will solve geometrical problems in a few seconds. The 2 parallel sides are called the bases. Using the area of trapezoid calculator: an example. If you only know the side lengths of a regular trapezoid, you can break the trapezoid into simple shapes to find the height and finish your calculation. Anamika Mittal has verified this Calculator and 25+ more calculators! Enter the three side lengths, choose the number of decimal places and click Calculate. An Example of How to Use the Area of a Trapezoid Calculator Sometimes, it’s helpful to have a real-life example to understand the calculator’s inner workings. Base b of Trapezoid calculator uses Base B=2*(Area/Height)-Base A to calculate the Base B, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base. An Example of How to Use the Area of a Trapezoid Calculator Sometimes, it’s helpful to have a real-life example to understand the calculator’s inner workings. Here is an example to understand it better. Anamika Mittal has verified this Calculator and 25+ more calculators! Use the point as decimal separator. By … Area of a Trapezoid Calculator: ... A 4 – sided geometrical figure which has one pair of parallel sides is called a Trapezoid. An isosceles trapezoid is a trapezoid where the base angles have the same measure. Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. It is the space of the internal surface in a figure, it is limited for the perimeter. To calculate the trapezoid area, follow the below steps: Measure and write down the base a, base b, and height h of the trapezoid. Area of a trapezoid Calculator . Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. Other Calculator Techniques. What is its Area? The area of a trapezoid can be found through the sides using the substitution method. You can use the trapezoid area calculator to find the area of a trapezoid: Enter the length of the two bases (b1 and b2) of the trapezoid. Here is a comprehensive set of problems about polygons solved using calculators. Area of a trapezoid and midline or bases You can use the trapezoid area calculator to find the area of a trapezoid: Enter the length of the two bases (b1 and b2) of the trapezoid. : After working out the answer of each of the next questions, click adjacent button of see the correct answer. Calculating Height of a Trapezoid Example: A trapezoid has a base lengths of 8cm and 12cm. Similarly, as γ + δ = 180°, δ = 180° - 125° = 55°. Volume=(1/3)*pi*Height*(Radius 1^2+Radius 2^2+(Radius 1*Radius 2)), Total Surface Area=pi*Radius*(Radius+sqrt(Radius^2+Height^2)), Lateral Surface Area=pi*Radius*sqrt(Radius^2+Height^2), Total Surface Area=2*pi*Radius*(Height+Radius), Area of a Triangle when base and height are given, Area of a Parallelogram when base and height are given, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base and is represented as, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base is calculated using. This trapezoid calculator will compute the area of a trapezoid for you. A trapezoid is an interesting four-sided geometric figure. How to calculate the area of the trapezoid in Python. Base B and is denoted by bb symbol. Let's assume that you want to calculate the area of a certain trapezoid. home / calculators / trapezoidal prism volume In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. TRAPEZOID CALCULATOR (A & B) and (B & C) that are supplementary (add to 180°). Trapezoid calculator computes all properties of a trapezoid such as area, perimeter, height and sides given a sufficient subset of these properties. Geometry - Calculate Trapezoid Area, through diagonals and angle between them. An acute trapezoid has two adjacent acute angles on its longer base edge, while an obtuse trapezoid has one acute and one obtuse angle on each base. All sides 2.Lateral side (leg) and angle at the base 3. In Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezium. Base of Trapezoid Calculator. ∫ (). Angles are calculated and displayed in … You can always use our trapezoid calculator above, however, you should be able to calculate the area of trapezoid by yourself. Diagonals, angle between the diagonals and bases or midline 4. How to use the trapezoid calculator Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real numbers and press "calculate" with b being the short base and d being the long base (d > … We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. α - 30° γ - 125° h – 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = ½(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the … 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base ° = 150° the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button “ solve ” to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid – sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, δ = 180° 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180°, δ = 180°, =... Especially the part on which it rests or is supported 30° γ - 125° h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as γ + δ = 180° β... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180° ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180° ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 – sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base a '' – sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has.
### Recent Posts
19 febrero, 2019
##### Creamos la app que necesitas
5 septiembre, 2016 | 2021-04-21T13:17:12 | {
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"url": "https://mediawebplace.com/o7jb35t1/252366-trapezoid-base-calculator",
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https://math.stackexchange.com/questions/2037311/finding-the-roots-with-multiplicity-of-a-polynomial-with-compound-angle-formul | # Finding the roots, with multiplicity, of a polynomial with compound angle formula
My question is to find the roots, counted with multiplicity, of the polynomial equation
$16x^5-20x^3+5x-1=0$ using the compound angle formula $\sin\left(5\theta\right)=16\sin^5\theta-20\sin^3\theta+5\sin\theta$
So after substituting $x=\sin\theta$, I get to the equation
$\sin(5\theta)=1$
Then I get an infinitude of $\theta$ values, which when I find the sines of these values, all correspond to the distinct solutions $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(-\frac{3\pi}{10}\right)$.
What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.
Also if possible, is there a way to solve this polynomial using the given compound angle formula without the need to find the distinct roots and then determine the ones which repeat?
This is because the solutions to this question say $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$ without any reasoning, which makes me suspect I am unaware of some related theorem.
• Is your first equation correct? Maybe it is $16x^ 5$ instead of $x^5$. Nov 30, 2016 at 14:11
• Technically speaking, $x=\sin \theta$ is not a correct and proper substitution since it makes an assumption that $x \in [-1,1]$. Nov 30, 2016 at 14:49
What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.
If one knows the roots, then it is possible to find their multiplicity by finding whether or not the function's derivative(s) have the same root. See this question for a proof.
• Yes I am aware of that theorem, but how exactly does one figure out if any of those three distinct roots are zeros of the derivative polynomial? Nov 30, 2016 at 15:04
• The derivative polynomial is $80x^4 - 60x^2 + 5$. So you know that $1$ definitely is not a repeated root. Also, you can see that the derivative is even. So it has $2$ positive roots and $2$ negative roots. And you have figured out that the distinct roots of the original equation are definitely $\sin(\frac{\pi}{10})$ and $\sin(\frac{-3\pi}{10})$. Do you see a complete argument? Nov 30, 2016 at 15:13
As already explained, the solutions are: $$x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$$ Because, $$\sin(5\theta)=1 \Rightarrow \theta=\frac{1}{5}\left(\frac{\pi}{2}+2k\pi\right)$$ and setting $k=0,1,2,3,4$ we get all five solutions.
But $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)$ and $\sin\left(\frac{13\pi}{10}\right)=\sin\left(\frac{17\pi}{10}\right)$.
So the five roots (with multiplicity) are $$1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right)$$.
Let's explain better why if make any other choice of $k$ we will always find an angle congruent to one of those in $S=\{\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10\}$.
We can write $k=5n+r$ with $r \in \{0,1,2,3,4\}$ and $n \in \Bbb Z$. Replacing that at the expression of $\theta$ we get:
$$\theta=\frac{\pi}{10}+\frac{2k\pi}{5}=\frac{\pi}{10}+\frac{2(5n+r)\pi}{5}=\frac{\pi}{10}+2n\pi+\frac{2r\pi}{5} \equiv \frac{\pi}{10}+\frac{2r\pi}{5}$$
The last equivalence means that the angle $\frac{\pi}{10}+2n\pi$ is congruent to $\frac{\pi}{10}$. That means that both angle stop at the same point on the trigonometric circle and then if we apply any trigonometric function at those angles we will get the same value. The algebric value are different but geometricaly they represent the same point at the circle.
It means that if we want to find the different solution for $\sin\theta$ is enough to take
$$\frac{\pi}{10}+\frac{2r\pi}{5}$$
with $r=0,1,2,3,4$.
• The question seems to be more like: why the specific values of $k=0,1,2,3,4$? Why not $k=-2,-1,1,2,5$? Nov 30, 2016 at 14:50
• When we choose $k=0,1,2,3,4$ we get all different values of $\theta$, wich are $S={\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10}$. If you those any other set for $k$ your new set will be inside $S$. Nov 30, 2016 at 14:58
• What do you mean by "different values of $\theta$"? If we choose different $k$, then obviously the $\theta$'s obtained will be different. And then, how can the new set of $\theta$'s be "inside" the aforementioned set $S$. Nov 30, 2016 at 15:05
• @ArnaldoNascimento could you please clarify what you meant by 'your new set will be inside S'? Nov 30, 2016 at 15:05
• Note that if $k \in \mathbb{Z}, \ \mbox{with}\ k \neq 0,1,2,3$ or $4$ then $k = 5\cdot n + r$ with $r=0,1,2,3\ \mbox{or}\ 4, n \in \mathbb{Z}$. In this case this does not matter because $sin (\frac{1}{5}( \frac{\pi}{2} + 2kpi )) = sin (\frac{1}{5}( \frac{\pi}{2} + 2rpi ))$. Nov 30, 2016 at 15:18 | 2022-05-22T05:54:55 | {
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https://de.maplesoft.com/support/help/errors/view.aspx?path=type/partition&L=G | type/partition - Maple Programming Help
Home : Support : Online Help : Programming : Data Types : Type Checking : Types : type/partition
type/partition
check for an integer partition
Calling Sequence type( expr, 'partition' )
Parameters
expr - anything; any Maple expression
Description
• The call type( expr, 'partition' ) returns true if expr is a partition of a positive integer, and returns false otherwise.
• A partition of a positive integer $n$ is a list [ ${k}_{1},\mathrm{...},{k}_{r}$]. of positive integers ${k}_{i}$, whose sum is equal to $n$, and which is non-decreasing, that is, which satisfies $\mathrm{add}\left({k}_{i},i=1..r\right)=n$ and ${k}_{i}\le {k}_{i+1}$, for $i$ in $1..r-1$.
Examples
> $\mathrm{type}\left(\left\{\left\{1,2\right\},\left\{3,4\right\}\right\},'\mathrm{partition}'\right)$
${\mathrm{false}}$ (1)
> $\mathrm{type}\left(\left[1,2,\frac{2}{3}\right],'\mathrm{partition}'\right)$
${\mathrm{false}}$ (2)
> $\mathrm{type}\left(\left[1,-2,3\right],'\mathrm{partition}'\right)$
${\mathrm{false}}$ (3)
> $\mathrm{type}\left(\left[1,2,3\right],'\mathrm{partition}'\right)$
${\mathrm{true}}$ (4)
> $\mathrm{type}\left(\left[1,2,3,3,4\right],'\mathrm{partition}'\right)$
${\mathrm{true}}$ (5)
> $\mathrm{type}\left(\left[1,2,3,3,2\right],'\mathrm{partition}'\right)$
${\mathrm{false}}$ (6) | 2021-04-16T21:24:24 | {
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http://math.stackexchange.com/questions/669085/what-do-curly-less-than-sign-mean | # What do curly less than sign mean?
I am reading "Convex Optimization" by Stephen Boyd. He is using a curved greater than and curved less than equal to signs.
$f(x^*) \succcurlyeq \alpha$
or
$f(x*) \preccurlyeq \alpha$
Can someone explain or point me to their meaning.
-
as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly – barlop Feb 9 at 6:43
@barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which has curly word in it, not curved. – Ruslan Feb 9 at 7:18
Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. – Dinesh K. Feb 9 at 17:53
You should definitely take the tour. This is not a traditional forum! – canaaerus Feb 9 at 18:03
Well I don't know what a traditional forum looks like. :-) – Dinesh K. Feb 9 at 18:20
There's a list of notation in the back of the book. On page 698, $x\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$.
Edit: The notation is introduced on page 32.
-
Both Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.
It should be noted that the book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for.
The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.
Let $S$ be a vector space, and let $K\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \succeq_K y \quad\Longleftrightarrow\quad y \preceq_K x \quad\Longleftrightarrow\quad x - y \in K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\in S$, $x \not\succeq_K y$ and $y \not\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\Longrightarrow\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\not\prec_K y$$
When the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \succeq_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is almost always dropped.
Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors.
-
Thanks a lot for the detailed answer, and for correcting the symbol as well :-). – Dinesh K. Feb 9 at 16:35
Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.
- | 2014-12-21T11:21:33 | {
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https://www.physicsforums.com/threads/circular-motion-question-about-a-car-banked.63558/ | # Circular motion question about a car banked
1. Feb 12, 2005
### F|234K
if a curve with a radius of 60 meter is properly banked for a car travelling at 60km/h, what must be the coefficient of static friction for a car not to skid when travelling at 90km/h?
i know that in order to solve the question, one needs to find the angle first, and i found the angle to be 25.3 degrees(not sure to be correct)...
and the answer is 0.39...i just can't seem to get the answer...
2. Feb 12, 2005
### Staff: Mentor
That's correct.
Since you didn't show any work, there is no way to tell where you got stuck. The basic idea is exactly the same as in the no friction case: the only difference is the addition of the friction force on the car ($\mu N$ acting down the incline).
3. Feb 12, 2005
### xanthym
Your calculated road banking angle of 25.3 deg relative to horizontal is correct (also indicated by the previous msg).
The required static friction coefficient for 90 km/h along the same banked road is calculated to be 0.39 in agreement with the provided answer (but apparently not in line with your calculations).
You may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system.
A typical formulation based on the friction horizontal component only is:
$$:(1): \ \ \ \ \frac {v^2} {rg} = Tan(\theta) + \mu_s Cos(\theta)$$
The above equation yields a static friction coefficient of 0.65 for this case. The formulation based on fully coupled horizontal and vertical components is:
$$:(2): \ \ \ \ \frac {v^2} {rg} = \frac {Sin(\theta) + \mu_s Cos(\theta)} {Cos(\theta) - \mu_s Sin(\theta) }$$
This latter formulation yields 0.39 for Coefficient of Static Friction.
The complete system of equations for horizontal and vertical components is given below:
$$:(3): \ \ \ \ \frac{mv^2} {r} = N Sin(\theta) + \mu_s N Cos(\theta)$$
$$:(4): \ \ \ \ 0 = N Cos(\theta) - \mu_s N Sin(\theta) - mg$$
where N is the force Normal to the road surface, (theta) the bank angle, and the other variables defined per usual conventions.
~~
Last edited: Feb 12, 2005
4. Feb 12, 2005
### F|234K
thank you guys very much
"You're may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system."
and yes, i got my answer to be 0.59, thank you xanthym for your detailed and consummate reply which enlightens me on the vertical component of friction. (i only calculated the horizontal part of friction.) | 2017-05-23T13:06:41 | {
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https://math.stackexchange.com/questions/1978171/is-this-a-valid-way-to-prove-this-modified-harmonic-series-diverges | # Is this a valid way to prove this modified harmonic series diverges?
I am trying to find a way to prove that $$\dfrac 11 + \dfrac 12 + \dfrac 13 + \dfrac 14 + \cdots \color{red}{-} \dfrac 18 + \cdots$$
where the pattern repeats every $8$ terms. Knowing about the Riemann Series Theorem, I am a little hesitant about manipulating conditionally convergent series at all. Granted that the harmonic series diverges, is the following a valid way to prove my series diverges?
$$\dfrac 11 + \dfrac 12 + \dfrac 13 + \cdots + \dfrac 17 - \dfrac 18 + \cdots = \sum_{n=1}^{\infty} \dfrac 1n - 2 \cdot \dfrac 18\sum_{n=1}^{\infty} \dfrac 1n$$
$$=\dfrac 34 \sum_{n=1}^{\infty} \dfrac 1n$$
Since the harmonic series diverges, so does $\dfrac 34$ times it.
• I stared at that line many, many times before I finally saw where the $+$ turned into a $-$. – Yakk Oct 21 '16 at 10:55
• @Yakk I suggested an edit to make it more obvious – null Oct 21 '16 at 16:46
• I've made it red. Don't know if there's a better way. – asmeurer Oct 21 '16 at 20:11
• @asmeurer Is it better now? – Ovi Oct 22 '16 at 1:57
• That helps. At first I thought you meant repeat the terms shown, which obviously diverges. – asmeurer Oct 22 '16 at 1:59
## 4 Answers
Your idea is a good one, but, as you suspected, you need to be more careful about this sort of manipulation of conditionally convergent series.
One way to carry out your argument correctly, but with only minor changes, is by looking at partial sums:
Let's write $$a_n=\begin{cases}1,&\text{ if }n\text{ is not a multiple of 8} \\-1,&\text{ if }n\text{ is a multiple of 8},\end{cases},$$ so that your series is $\sum_{n=1}^\infty \frac{a_n}{n}.$
Then for any natural number $N,$
\begin{align} \sum_{n=1}^{8N}\frac{a_n}{n} &= \sum_{n=1}^{8N}\frac1{n}-2\sum_{n=1}^N \frac1{8n} \\&=\sum_{n=1}^{8N}\frac1{n}-\frac1{4}\sum_{n=1}^N \frac1{n} \\&\ge\sum_{n=1}^{8N}\frac1{n}-\frac1{4}\sum_{n=1}^{8N} \frac1{n}\scriptsize{\quad\text{(because we can only be subtracting *more* positive numbers)}} \\&=\frac3{4}\sum_{n=1}^{8N}\frac1{n}, \end{align}
which approaches $\infty$ as $N$ approaches $\infty,$ since the harmonic series diverges.
• Thank you, I learned something very important; you can manipulate even conditionally convergent series as long as you manipulate them in the context of partial sums. – Ovi Oct 21 '16 at 4:52
• @Ovi -- Yes, because the partial sums are just regular finite sums. But you still need to be careful that your partial sums are entire initial parts of the infinite series. You can't pick and choose which terms to include; all you can do is chop the infinite series off at some finite point. After that, you can apply any normal algebraic operations, because you just have a finite sum. – Mitchell Spector Oct 21 '16 at 4:56
• @ Mitchell Spectator Okay got it – Ovi Oct 21 '16 at 4:57
• And if you want to prove that the conditionally convergent series does converge, you can't cherry-pick which of the partial sums you're looking at (like it happens here), of course. – Henning Makholm Oct 21 '16 at 14:21
• @HenningMakholm Good point. – Mitchell Spector Oct 21 '16 at 15:37
To expand on Mitchell Spector's answer, let's generalize to the problem where every $b$th term is negated.
By the same argument, we get \begin{align} \sum_{n=1}^{bN}\frac{a_n}{n} &= \sum_{n=1}^{bN}\frac1{n}-2\sum_{n=1}^N \frac1{bn} \\&=\sum_{n=1}^{bN}\frac1{n}-\frac2{b}\sum_{n=1}^N \frac1{n} \\&\ge\sum_{n=1}^{bN}\frac1{n}-\frac2{b}\sum_{n=1}^{bN} \frac1{n} \\&=\left(1-\frac2{b}\right)\sum_{n=1}^{bN}\frac1{n} \end{align}
This is nice, since we can see that the divergence proof holds for $b \geq 3$, but fails for $b=2$. This is not strange, since the series for $b=2$ is known to converge to $\log 2$.
This proof isn't valid because of the Riemann series theorem; the original series could still converge even though some rearrangement of it diverges. It's important to focus on the partial sums.
You can argue along the lines that the series $$\sum_{k=0}^{\infty} \frac{1}{8k+1} = 1 + \frac{1}{9} + \frac{1}{17} + \cdots$$ already diverges by comparing its partial sums to the partial sums of the harmonic series $\frac{1}{8} \Big( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \Big).$
Note that \begin{align} &{\tiny\sum_{k=0}^\infty}{\tiny\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}+\frac1{8k+7}-\frac1{8k+8}\right)}\\ &={\tiny\sum_{k=0}^\infty}{\tiny\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}\right)+{\tiny\sum_{k=0}^\infty}\left(\frac1{8k+7}-\frac1{8k+8}\right)}\\ \end{align} For the left hand sum, we have \begin{align} &{\small\sum_{k=0}^\infty\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}\right)}\\ &\ge{\small\frac68\sum_{k=0}^\infty\frac1{k+1}} \end{align} which diverges.
For the right hand sum, we have \begin{align} &{\small\sum_{k=0}^\infty\left(\frac1{8k+7}-\frac1{8k+8}\right)}\\ &\le{\small\frac1{56}+\frac18\sum_{k=1}^\infty\left(\frac1{8k}-\frac1{8k+8}\right)}\\ &=\frac{15}{448} \end{align} A divergent sum plus a convergent sum diverges.
• How did you evaluate $\dfrac 18 \sum_{k=1}^{\infty} \left( \dfrac {1}{8k}-\dfrac {1}{8k+8} \right) = \dfrac {1}{64} \sum_{k=1}^{\infty} \dfrac {1}{k(k+1)}$? I guess its not necessary to calculate the sum because we know it converges anyway, but I'm curious how you got a rational answer, knowing that $\sum_{k=1}^{\infty} \dfrac {1}{k^2} = \dfrac {\pi^2}{6}$ – Ovi Oct 24 '16 at 15:19
• \begin{align} \sum_{k=1}^\infty\frac1{k(k+1)} &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\ &=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\ &=1 \end{align} which is a Telescoping Sum. – robjohn Oct 24 '16 at 15:50 | 2019-06-20T18:14:31 | {
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https://math.stackexchange.com/questions/834471/help-evaluating-lim-x-to-infty-sqrtx-sqrtx-sqrtx-sqrtx | # Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$
Does anyone know how to evaluate the following limit? $$\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$$ The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
Multiply by the conjugate, then reduce: \begin{align*} \lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) &= \lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \cdot \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\ &= \lim_{x\to\infty} \frac{(x + \sqrt{x + \sqrt{x}}) - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\ &= \lim_{x\to\infty} \frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \cdot \frac{\dfrac{1}{\sqrt{x}}}{\dfrac{1}{\sqrt{x}}} \\ &= \frac{\sqrt{\lim\limits_{x\to\infty}\dfrac{1}{\sqrt{x}} + 1}}{\sqrt{1 + \lim\limits_{x\to\infty} \dfrac{\sqrt{x + \sqrt{x}}}{x}} + 1} \\ &= \frac{\sqrt{0 + 1}}{\sqrt{1 + 0} + 1} \\ &= \frac{1}{2} \end{align*}
• I like your answer! :) – dmk Jun 15 '14 at 1:46
• thanks i like ur answer to – user157128 Jun 15 '14 at 1:54
• as you did in step 3 to 4 when multiplied by 1/v5 ??? – user157128 Jun 15 '14 at 2:03
$$\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{\frac{3}{2}}}}}+1}$$
I got the above by first multiplying by the conjugate to $\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}$ then dividing both the numerator and denominator by $\frac{1}{\sqrt{x}}$. Now take $x$ to infinity.
• the answer is 1/2, but I tried this and rationalize, but did not reach 0.5 – user157128 Jun 15 '14 at 1:38
• @user157128 what did you get? In the top I get $\sqrt{1 + 0} - 0$. What do you get in the bottom? – DanZimm Jun 15 '14 at 1:40
\begin{align} \ \lim_{x\rightarrow\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} &= \lim_{x\rightarrow\infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \cdot \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \ &= \lim_{x\rightarrow\infty} \frac{x + \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \cdot \frac{1/\sqrt{x}}{1/\sqrt{x}} \\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\frac{1}{x}+\frac{1}{x\sqrt{x}}}+1} \\ \ &= \frac{\sqrt{1}}{\sqrt{1}+1} \\ \ &= \frac{1}{2} \\ \end{align}
Using Taylor expansion:
$$\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$$
$$\sqrt{x + \sqrt{x}+\dfrac{1}{2} + O(\frac{1}{\sqrt{x}})} - \sqrt{x}$$
$$\sqrt{x}+\dfrac{1}{2\sqrt{x}}(\sqrt{x}+\dfrac{1}{2}+ O(\frac{1}{\sqrt{x}})) + O(\frac{1}{\sqrt{x}})- \sqrt{x}$$
$$\dfrac{1}{2} + \dfrac{1}{4\sqrt{x}}$$
Which in the limit case tends to ${\dfrac{1}{2}}$.
The expansion I used was $(a+b)^{1/2} = \sqrt{a}+\frac{b}{2\sqrt{a}} + \cdots$
• It's not clear from what you've written why you get to ignore the error terms in your expansion. – Cheerful Parsnip Jun 15 '14 at 1:50
• @user157107 this seems like a fairly interesting way to go about it... and similar to what I was considering doing. Could you possibly add more details to justify your approach? – Squirtle Jun 15 '14 at 2:13 | 2019-07-17T16:33:17 | {
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https://math.stackexchange.com/questions/2932529/find-the-nonzero-vectors-u-v-w-that-are-perpendicular-to-the-vector-1-1-1-1 | # Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other
Find the nonzero vectors $$u,v,w$$ that are perpendicular to the vector $$(1,1,1,1)$$ and to each other.
If I follow algebra, then I get complicated results to solve it as follows:
Let $$u=(u_1,u_2,u_3,u_4), \ v=(v_1,v_2,v_3,v_4) , \ w=(w_1,w_2,w_3,w_4)$$
Then, $$u \cdot (1,1,1,1)=v \cdot (1,1,1,1)=w \cdot (1,1,1,1)=0$$
Also $$u \cdot v=w \cdot u=v \cdot w=0$$.
These gives us
$$u_1+u_2+u_3+u_4=0, \\ v_1+v_2+v_3+v_4=0 , \\ w_1+w_2+w_3+w_4=0, \\ u_1v_1+u_2v_2+u_3v_3+u_4v_4=0, \\ u_1w_1+u_2w_2+u_3v_3+u_4v_4=0, \\ v_1w_1+v_2w_2+v_3w_3+v_4w_4=0.$$
But how to solve for $$u_i, v_i,w_i, \ i=1,2,3,4$$ from here?
Does there exit any other easy method?
Help me out
• Apply Gram-Schmidt to the basis $$((1, 1, 1, 1), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)),$$ or any other basis beginning with $(1, 1, 1, 1)$. – Theo Bendit Sep 27 '18 at 2:37
• Is not in that case the solution will be particular , I mean there can other vectors also which can not be conclude using Gram-schmidt method. – M. A. SARKAR Sep 27 '18 at 2:44
• Yes, it will be particular. If you range over all such bases, then you will obtain all orthonormal bases beginning with $\left(\frac12,\frac12,\frac12,\frac12\right)$, though not uniquely. – Theo Bendit Sep 27 '18 at 2:47
as columns $$\left( \begin{array}{rrrr} 1&-1&-1&-1 \\ 1& 1&-1&-1 \\ 1&0 &2&-1 \\ 1&0&0&3 \end{array} \right)$$ Pattern, done correctly, works in any dimension
$$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$
• That's great! =) – Siong Thye Goh Sep 27 '18 at 3:02
• Excellent method – M. A. SARKAR Sep 27 '18 at 11:13
Consider the Hadamard matrix of which we know that the columns form an orthogonal basis of $$\mathbb{R}^4$$.
$$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1\end{bmatrix}$$
The other columns would give you a solution.
Alternatively, use Gram-Schmidt process.
• But if we consider the Hadamard matrix , then the solution will be particular. There may be other vectors which are perpendicular to $(1,1,1,1)$ except the last 3 column vectors in the Hadamard vectors and its multiples . – M. A. SARKAR Sep 27 '18 at 2:42
• So if we use Gram-Schmidt method then we need a basis – M. A. SARKAR Sep 27 '18 at 2:43
• Theo has given you a basis right? Note that answer is not unique. If you want to describe the set, you have already done so in your post. – Siong Thye Goh Sep 27 '18 at 2:48
• There is a pattern that easily adapts to any dimension... – Will Jagy Sep 27 '18 at 3:01 | 2020-06-01T09:13:12 | {
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https://math.stackexchange.com/questions/3552753/is-it-necessary-to-write-limits-for-a-substituted-integral | # Is it necessary to write limits for a substituted integral?
To solve the following integral, one can use u-substitution: $$\int_2^3 \frac{9}{\sqrt[4]{x-2}} \,dx,$$ With $$u = \sqrt[4]{x-2}$$, our bounds become 0 and 1 respectively. Thus, we end up with the following: $$36\int_0^1{u^2} \,du$$ In the first case, the lower bound is a vertical asymptote so we would have to use limits to find the answer. However, in the second case there's no longer an asymptotal bound - would you still have to write the limits since the original function would've needed limits, or can you just solve this by plugging in the substituted bounds? I know the final answer will be the same either way, but I want to know if it can be considered correct to exclude the limits in the 2nd integral from a technical perspective since the function has been changed. Many thanks in advance!
## 3 Answers
Without too much simplification, the substitution you cite yields
$$\int_2^3\frac9{(x-2)^{1/4}}\,\mathrm dx=36\int_0^1\frac{u^3}u\,\mathrm du$$
which you certainly welcome to treat as an improper integral,
$$36\left(\frac13-\lim_{u\to0^+}\frac{u^3}3\right)$$
but since $$u=0$$ is a removable discontinuity and the limand reduces to $$u^2$$, you may as skip this treatment altogether.
• I'm confused about your last expression. It seems like you want to still have $\int$ present, but the $1-$ seems like you've already taken the integral, but the $\frac{u^3}u$ is from before you've taken the integral... – Teepeemm Feb 20 at 2:05
• Yes, it was wrong as written. I have edited it. – Martin Argerami Feb 20 at 3:51
• Thanks @MartinArgerami ! – user170231 Feb 20 at 14:55
There are certain conditions that must be met for a substitution to be "legal". In most circumstances these conditions are naturally met and so they are not emphasized; you have here one situation in which they are not.
For instance, one set of conditions is given in Anton, Anderson, and Bivens (Calculus, Early Transcendentals, 11th Edition), Theorem 5.9.1:
Theorem 5.9.1. If $$g'(x)$$ is continuous on $$[a,b]$$, and $$f$$ is continuous on an interval containing the values of $$g(x)$$ for $$a\leq x\leq b$$, then $$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.$$
Here you have $$g(x) = \sqrt[4]{x-2}$$, so $$g'(x) = \frac{1}{4}(x-2)^{-3/4}$$... which is not continuous on the interval $$[2,3]$$ that you are working on.
In fact, as you note, the initial integral is improper, which means you aren't really evaluating that integral: you are evaluting a limit, $$\lim_{h\to 2^+}\int_h^3 \frac{9}{\sqrt[4]{x-2}}\,dx.$$ The integral in the limit does satisfy the conditions of the theorem above, so you can make the substitution to get $$\lim_{h\to 2^+}\int_{\sqrt[4]{h-2}}^1 \frac{u^3}{u}\,du = \lim_{a\to 0^+}\int_a^1 u^2\,du,$$ and proceed from there.
• Thank you very much Arturo, your answer was very in-depth and provided a lot of reinforcement for what you were saying. I wish I could select multiple answers to be correct, I thought your inclusion of the cited textbook snippet was really nicely done too. I chose the other answer only because I found it easier to digest/understand at first glance, but after I properly read your comment I understood it more fully. Thank you for your quick and effective contribution :-) – Lord Kanelsnegle Feb 19 at 17:32
Suppose you must. Then we have $$9\lim_{a\to2^+}\int_a^3\frac1{\sqrt[4]{x-2}}\,dx.$$
Let $$u=x-2$$. Then we have $$9\lim_{a\to2^+}\int_{a-2}^1 u^{-1/4}\, du.$$
By the power rule, we have $$9\lim_{a\to2^+} \left.\frac43u^{3/4}\right]_{a-2}^1=9\left[\frac43(1-\lim_{a\to2^+}(a-2)^{3/4})\right]=9\left[\frac43(1-0)\right]=12.$$
Notice that it does not make a difference whether you use $$\lim_{a\to2^+}(a-2)$$ or $$0$$.
• Thanks for the quick and simple response! I found the answer I selected to be more aligned with the question, but I wanted to comment because I appreciate the time you took to write this. Thank you once again. – Lord Kanelsnegle Feb 19 at 17:29 | 2020-03-29T04:18:45 | {
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https://math.stackexchange.com/questions/3797897/expected-value-of-flipping-through-a-book/3797922 | # Expected Value of flipping through a book
Suppose that a book has $$N$$ pages, and we read through the book as follows. We start from page 0, and if we're on page $$i$$, we randomly flip to a page $$i + 1, i + 2, ..., N$$ with equal probability.
1. What is the expected value of number of flips we need to finish the book?
Intuition tells me that we can, on average, expect to halve the number of pages remaining. This yields $$\log_2(N)$$, but I'm having trouble formalizing it.
1. If $$N = 26$$, what is the probability we flip to page 13 at some point? Assume we start at page 0.
I let $$P_i$$ be the probability we land on page 13 eventually, starting from page $$i$$. Then, $$P_{13} = 1$$, and in general, $$P_{i} = \frac{1}{26 - i}\sum_{k = i + 1}^{13}P_k$$
Evaluating terms like $$P_{12}, P_{11}, P_{10}$$, I see that all of these values are $$\frac{1}{14}$$, including $$P_0$$. Is there a more intuitive reason for such a simple answer?
• Question 1. is unclear ... number of flips we need for what? To read every page in the book? To reach the last page in the book? What does it mean? – Ross Presser Aug 21 at 13:47
• edited for clarity, though it should be implicitly clear that it's often not possible to read every page in the book, and the expected value of flips to read every page is trivial. – user815048 Aug 21 at 20:13
Let's consider the equivalent problem in which we start at page $$n$$ and flip backwards through the book, going to each of the pages $$0, 1, ..., n - 1$$ with equal probability. Let $$E_n$$ be the expected number of flips. Then we have $$E_0 = 0$$ and
$$E_n = 1 + \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i$$
Then in particular we have
$$$$\begin{split} E_{n + 1} &= 1 + \frac{1}{n + 1} \sum\limits_{i = 0}^{n} E_i \\ &= 1 + \frac{E_n}{n + 1} + \frac{n}{n + 1} \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i \\ &= 1 + \frac{E_n}{n + 1} + \frac{n}{n + 1} (1 + \frac{1}{n} \sum\limits_{i = 0}^{n - 1} E_i) - \frac{n}{n + 1} \\ &= 1 - \frac{n}{n + 1} + \frac{1}{n + 1} E_n + \frac{n}{n + 1} E_n \\ &= \frac{1}{n + 1} + E_n \end{split}$$$$
whenever $$n \geq 1$$ (and the identity is easily verified when $$n = 0$$ as well).
Then by induction, we have $$E_n = \sum\limits_{j = 1}^n \frac{1}{j}$$, the $$n$$th harmonic number. This will be asymptotically very close to $$\log_e(n)$$.
Let $$P_n$$ be the expected number of flips in a book with $$n$$ pages. Then $$P_0=0,\ P_1=1$$ and $$P_n=1+\frac1n\sum_{k=0}^{n-1}P_k,\tag1$$ because we have to make one flip, and then we're equally likely to have any number of pages from $$0$$ to $$n-1$$ left to flip through.
We get \begin{align} P_1&=1\\ P_2&=\frac32\\ P_3&=\frac{11}6\\ P_4&=\frac{50}{24}\\ P_5&=\frac{174}{120} \end{align}
The denominators are obviously $$n!$$, so we look for the numerators in OEIS and find A000254, the unsigned Stirling numbers of the first kind.
OESI gives the recurrence $$a_{n+1}=(n+1)a_n+n!$$ for the unsigned Stirling numbers of the first kind, and dividing through by $$(n+1)!$$ we get $$P_{n+1}=P_n+\frac1{n+1}$$ which clearly gives $$P_n=\sum_{k=1}^n\frac1k=H_n,$$ the $$n$$th harmonic number. To complete the problem, we must show that the harmonic numbers satisfy the recurrence $$(1)$$.
Here is how I approached the first part of the problem. Consider a book with exactly $$n$$ pages. Let $$P_1$$ denote the first page you flipped to, and let $$X_n$$ represent the number of pages you flipped until you get to the last page. Note $$P_1$$ is uniformly distributed on the set $$\{1,...,n\}$$ and $$E(X_1)=1$$. Using the total law of expectation we get for $$n\geq2$$ that $$E(X_n)=\sum_{k=1}^{n}E(X_n|P_1=k)P(P_1=k)=\frac{1}{n}\sum_{k=1}^{n}E(X_n|P_1=k)$$
Notice $$E(X_n|P_1=k)=1+E(X_{n-k})$$ and so $$E(X_n)=\frac{1}{n}\sum_{k=1}^{n}\Big[1+E(X_{n-k})\Big]=1+\frac{E(X_0)+\dots+E(X_{n-1})}{n}$$ Replace $$n$$ with $$n+1$$ to get $$E(X_{n+1})=1+\frac{E(X_0)+\dots+E(X_{n})}{n+1}$$ Combining the previous two equations unveils the relation $$(n+1)(E(X_{n+1})-1)=(n+1)E(X_n)-n$$ which is equivalent to saying $$E(X_{n+1})=E(X_n)+\frac{1}{n+1}$$ So finally $$E(X_n)=\sum_{k=1}^{n}\frac{1}{k}$$ | 2020-10-27T12:58:32 | {
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http://mathhelpforum.com/statistics/4194-college-prob-stat.html | # Math Help - college prob + stat
1. ## college prob + stat
2 Questions:
1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?
2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?
2. Originally Posted by magicpunt
1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?
$P(\mbox{men or woman})=P(\mbox{men})+P(\mbox{woman})$
Let us find,
$P(\mbox{men})$
There are 10 people and 5 are selected there are a total of $_{10}C_5=252$
From 5 men you take 5 men intotal the number of combinations is,
$_5C_5=1$
Thus,
$P(\mbox{men})=\frac{1}{252}$
Similarily ya have,
$P(\mbox{women})=\frac{1}{252}$
Thus, the probability is,
$\frac{1}{252}+\frac{1}{252}=\frac{2}{252}=\frac{1} {126}$
3. 2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?
This is similar to the 'boxes and balls' problems.
There are $5^{10}=9765625$ ways to distribute the people to the 5 floors.
There are $\frac{5!}{2!}=60$ ways that no two people get off on the same floor.
EDIT:
Yes, I misinterpreted the problem. Cerebral flatulence. Soroban's method is solid.
My second part should be $\frac{10!}{2^{5}}$
Therefore, $\frac{5^{10}}{\frac{10!}{2^{5}}}$
This is a version of the classic 'assigning diplomats' problem.
For instance:
How many ways can 10 different diplomats be assigned to 5 different countries if 2 diplomats must be assigned to each country?.
See the similarity with your problem?. Yours is just elevators and people, instead.
This can be done in $\frac{10!}{(2!)^{5}}=113,400$ ways.
The method I used is just another variation on this problem. Keep it, you may need it.
4. Originally Posted by galactus
This is similar to the 'boxes and balls' problems.
There are $5^{10}=9765625$ ways to distribute the people to the 5 floors.
There are $\frac{5!}{2!}=60$ ways that no two people get off on the same floor.
I think you might've misunderstood the problem
The question is: what is P(Exactly 2 people get off at each floor)
5. Hello, magicpunt!
Here's my approach to #2 . . .
2) There are 10 people in an elevator in the basement.
They can get off at any of 5 floors above.
The probability that they get off at each floor is equal
and each person leaving is independant of each other.
What is the probability that 2 people get off at each floor?
Each of the ten people has a choice of five floors.
. . There are: $5^{10}$ways that they can leave the elevator.
If exactly two people get off at each floor,
. . the number of ways is: $\begin{pmatrix}10\\2,2,2,2,2\end{pmatrix} = 113,400$
Therefore, the probabilty is: . $\frac{113,400}{5^{10}}\:=\:\frac{4,536}{390,625}$
6. Originally Posted by magicpunt
2 Questions:
1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?
Another approach to #1
Find the total no. of ways of selecting 5 out of 10= $\frac{10!}{5!5!}=C_5^{10}$
No. of ways in which all are men=1
No. of ways in which all are women=1
Required probability= $\frac{1+1}{C_5^{10}}=\frac{1}{126}$
Malay | 2015-11-24T22:18:23 | {
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https://math.stackexchange.com/questions/1788715/what-is-the-moment-of-inertia-of-a-gosper-island | # What is the moment of inertia of a Gosper island?
We know that regular hexagons can tile the plane but not in a self-similar fashion. However we can construct a fractal known as a Gosper island, that has the same area as the hexagon but has the property that when surrounded by 6 identical copies produces a similar shape, but with dimensions scaled by a factor of $\sqrt{7}$.
What is the distance between two of the centers? Is it the same as the distance between hexagons of the same area? ie. If I start with a hexagon of area A, then construct a Gosper island and place it next to an identical copy, would the distance still be the same as if they were hexagons? Or does the scaling factor come into play somewhere? Right now I think the answer is $\sqrt{3}/2$, as for the hexagon.
The reason I ask is that I'm trying to calculate the Gosper island's moment of inertia through an axis through its centre of mass and perpendicular to the plane of the island.
If we assume that the moment of inertia is always proportional to the mass, and proportional to the square of a characteristic length scale, then $$I = \gamma Ml^2,$$ where $\gamma$ is a constant, $l$ is the 'diameter' of the island, in a hexagon this would be the distance between two opposite vertices. Shrink the Gosper island by the scaling factor and surround it by six others. This self-similarity technique is super cute, and can be used to calculate the moment of inertia of an equilateral triangle, and can be extended to a square/rectangle quite easily. Fractals, having a high degree of self-similarity, seem amenable to this technique - here I calculate the moment of inertia for a Koch snowflake.
$\hspace{1.3cm}$
Using the principle of superposition, $$I = I_{\text{centre}} + 6I_{\text{edge}},$$ where $$I_{\text{centre}} =\gamma \frac{M}{7}\left(\frac{l}{\sqrt{7}}\right)^2 = \gamma \frac{Ml^2}{49} = \frac{I}{49}.$$
Now, by the parallel axis theorem $\displaystyle I_{\text{edge}} = I_{\text{COM}} + Md^2$ where $$\displaystyle I_{\text{COM}} = \frac{I}{49}$$ and $\displaystyle d= \frac{\sqrt{3} l}{2}$ (this was one source of error), so $\displaystyle I_{\text{edge}} = \frac{I}{49} + \frac{3Ml^2}{4},$ and \begin{align*} I &= \frac{I}{49} + 6\left(\frac{I}{49}+ \frac{3Ml^2}{4}\right), \\ I & = \frac{I}{7} + \frac{9Ml^2}{2}, \\ \frac{6I}{7} & = \frac{9Ml^2}{2}, \\ I & = \frac{21Ml^2}{4}. \end{align*}
This seems incorrect? It feels wrong, comparing to a disk of radius $l/2$ which has moment of inertia $Ml^2/4$ it seems far too large.
It would also be nice if we could verify our answer numerically or otherwise. Any references are also appreciated.
• The center of mass of the Gosper island is indeed the same as the center of the initial hexagon. This, as well as the moment of inertia can be computed using the technique of self-similar integration as in my answer to this question. I could provide a few more details but I get the feeling this might be an exercise? – Mark McClure May 17 '16 at 13:18
• Nope, just something I thought up after teaching the self-similarity method for finding the MOI of a triangle/square. Thankyou so much for the link! I suppose this can be done for an fractal with enough rotational symmetry to tile the plane? Please provide as many details as you wish, in an answer if you like. – Bennett Gardiner May 18 '16 at 2:05
• By the way, the polar inertia of a disk is not $ml^2/4$ (That is the x or y inertia) but $Ml^2/2$. – Urukann May 20 '16 at 4:40
• Isn't that only if $l$ is the radius though? In my formulation I was trying to use the diagonal of the hexagon, so it made more sense to compare to a disk with $l$ as the diameter. – Bennett Gardiner May 20 '16 at 6:19
• @BennettGardiner: IMO this is a source of confusion (especially if you don't explain it). The standard formula is $I=MR^2$ where $R$ is an equivalent radius (AKA gyration). I found it more effective to work with unit area shapes, so that the mass can be simplified, and $R$ conveys all shape information. – Yves Daoust May 20 '16 at 14:40
I think that your $\frac{\sqrt{3}}{2}$ for the distance between hexagons center is not right... The distance from an hexagon center to an edge is $\frac{r\sqrt{3}}{2}$, thus the distance between two hexagons of radius $\frac{l}{2\sqrt{7}}$ that touch by an edge should be: $$d = l\sqrt{\frac{3}{28}}$$
Now, if we take back your computation: $I = \frac{I}{49} + 6 \left(\frac{I}{49} + \frac{3Ml^2}{28}\right)$
This gives us : $I = \frac{3}{4} M l^2$, which may still not be the right answer.
It would be nice to compare this value to numerical methods, I'll look into it.
e/ I ran some numerical simulation, unfortunately it does not seem to confirm the above result.
How I did it:
• Starting from the code linked in this page, I generated the all points composing the outer shape of the gosper island.
• Then, I trimmed the generating points to remove duplicates
• I approximate the Gosper island as being star-shaped (This is not perfectly true, at least for $n=5$). Then its inertia is the sum of the inertia of the triangles composed of the origin and two consecutive points. Note that to get the actual inertia, one has to divide by the shape's area (Because when evaluating the triangle's inertia, we consider it has a surfacic mass of 1.). All triangle related formulas are available on Wolfram Alpha.
The results show that indeed the inertia is proportional to $l^2$, but its ratio to the value conjectured above is not 1, but a constant close to $\frac{4}{7}$: $$I_{gosper} \approx 0.568582263418 \frac{3}{4} M l^2$$
Unfortunately, it seems that this error is not related to the star-shaped approximation: I ran another experiment, this time using the Seidel program from the University of North Carolina at Chapel Hill. It allowed me to find a triangulation of the inner area of the Gosper island. Using another (similar) code I could check that the computation for a radius of 1. does yield the same ration between expected inertia (0.75) and the actual inertia (0.42856647032), with a similar ratio of 0.571421960426. Note that this inertia is very close to $\frac{3}{7}$, (best fractional approximation with a denominator inferior to fifteen thousand).
Actually, I had forgotten that the characteristic dimension is not the diameter, but the radius, thus the ratio is 0.142855490107, very close to $\frac{1}{7}$.
Using this method for a Koch snowflake yields pretty correct results: for a snowflake of radius $r \approx 1.44$ (I forgot to scale the step size properly) I get an inertia of $I_{koch} \approx 0.736036125705$ while the one given by $Ml^2/11 \approx 0.7435999999999999$
e/ I found the error: The mass of a "small" Gosper island is not $M$, but $\frac{M}{7}$, thus the missing factor 7. This is due to the fact that we make the assumption of a uniform density Gosper Island, thus its mass is proportional to its area.
We can rewrite our original equation: $$I_{edge} = I_{center} + \frac{M}{7}d^2\\ d = l \sqrt{\frac{3}{28}}$$
Which gives us: $$I = \frac{I}{49} + 6 \left( \frac{I}{49} + \frac{3Ml^2}{7\cdot 28}\right)$$
And finally:
$$I = \frac{3Ml^2}{28}$$
• Bah, forgot to scale the distance $d$. Thanks, this sounds much more reasonable. If you could confirm using numerical methods with some explanation of how you did it, it would put it to rest and I'd definitely accept/award the bounty. – Bennett Gardiner May 19 '16 at 12:42
• I just added some numerical simulation. However, it does not confirm my above conjecture. It may be because my code is quite dirty, but it may also be an error in the reasoning... Cheers – Urukann May 20 '16 at 4:37
• Interesting, thanks for your efforts, looks like we still have some way to go. Can you use the code to check the Koch snowflake case I linked? It code provide a check on the code For the snowflake, $I = Ml^2/11$ where $l$ is the largest 'diameter' of the snowflake (tip to tip) I really felt like this method should work, I wonder where the error lies! – Bennett Gardiner May 20 '16 at 6:22
• Hum, I checked with the Koch snowflake (my code was easier to modify than expected !): I get "correct" results. I noticed that I forgot to use the diameter instead of the radius, so we are off by another factor 4 I'm guessing... – Urukann May 20 '16 at 6:44
• Sorry, this is getting confusing, but no: I calculated the MOI for a radius 1 Gosper Island to be 0.42856647032, that is to say, $I = \frac{3Mr^2}{7} = \frac{3Md^2}{7 \cdot 4}$, meaning that we are missing a factor $\frac{1}{7}$. – Urukann May 20 '16 at 6:58
Yes, the distance between those hexagon-like shapes in a hexagonal lattice is the same as the distance between centres of hexagons of the same area in a hexagonal lattice.
The reason is that "in the long run" the area per mesh must be the same. That is, a sufficiently large circular disk contains both Gosper islands and hexagons in a number approximately equal to area of the disk dividied by area of the shape (with an error in the order of magnitude proportional to the cirumference of the disk).
The area inside the hexagon is made of one whole island and exactly six thirds, by symmetry.
Hence for an island of unit area, the distance between centers is such that
$$3\frac{\sqrt3}2d^2=3,$$
$$d=\sqrt{\frac2{\sqrt3}}.$$
To compute the moment of inertia, we will work with the radius of gyration and for an island of unit area.
$$I=MR^2$$ where $M$ is the mass.
For the assembly in the figure, enlarging by the factor $\sqrt7$ and using the axis theorem,
$$7M7R^2=49I=I+6(I+Md^2),$$
giving
$$R=\frac d{\sqrt7}=\sqrt{\frac{2}{7\sqrt3}}=0.406149258\cdots$$
This compares to the radius of gyration of a unit area disk,
$$R'=\frac1{\sqrt{2\pi}}=0.398942280\cdots$$
and that of a unit area hexagon,
$$R''=\sqrt{\frac{10}{36\sqrt3}}=0.400468569\cdots$$
• Interesting, so to scale this to an island with diameter $l$, would we simply scale this distance by $l$? – Bennett Gardiner May 20 '16 at 6:57
• Or $\sqrt{3/28}l$, like in the other poster's answer? I'm afraid I've confused myself quite a bit with the length scales. – Bennett Gardiner May 20 '16 at 7:00
• @BennettGardiner: I have no idea of the value of the diameter $l$, but you don't need it for the computation of the moment. – Yves Daoust May 20 '16 at 7:03
• I'm still trying to figure out your distance $d$. If that is the radius of gyration, what would the moment of inertia be? – Bennett Gardiner May 20 '16 at 7:32
• My confusion is part of why I asked the question. I am inclined to believe your answer more now that you have posted a comparison with the hexagon of unit area, in my mind they should have been very similar. – Bennett Gardiner May 20 '16 at 14:18 | 2020-12-02T17:03:36 | {
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https://math.stackexchange.com/questions/1146548/number-of-strings-when-each-character-must-occur-even-times | # Number of strings, when each character must occur even times
I was reading this question on programmers.StackExchange and wanted to try a combinatorics approach to solve the following problem:
Let $S=\{A,B,C,D\}$ be a set of characters and $n$ a positive natural, find the number of strings of length $n$ composed of characters in $S$ such that each character occur even times.
I tried to solve it by myself but I am stuck and I would appreciate a hint to solve this kind of problems. Also, there are similar questions on this site for not quite exactly the same problem, but I didn't find an answer that could help me.
## Current attempt
First, when $n$ is odd, the result must be zero. A string satisfying the desired property would necessarly have an even length, as a permutation of the concatenation of substrings of even lengths (each substring being a character from $S$ repeatead even times).
Now, let's suppose that $n$ is even.
If we have a mapping $c$ from $S$ to $\mathbb{N}$ assigning the number of occurence of each character (e.g. $c_A=0, c_B=2, c_C=4, c_D=0$), then the number of strings $N(c)$ we can produce with this mapping would be a k-permutation of characters, as we choose a string of size $n$ with $c_A$ times $A$, $c_B$ times $B$, etc. :
$$N(c) = \frac{n!}{c_A! c_B! c_C! c_D!}$$
Also, I can evaluate the number of such mappings $c$: this is equivalent to finding the number of strings where characters appear in order with varying occurences. For example, with $n=4$:
$$\begin{array}{c|cccc} & c_A & c_B & c_C &c_D \\ \hline AAAA & 4 & 0 & 0 & 0 \\ AABB & 2 & 2 & 0 & 0 \\ AACC & 2 & 0 & 2 & 0 \\ \dots\\ \end{array}$$
The number of strings of size $n$ with an even number of each character is exactly the same as the number of strings of size $k=n/2$ with single characters (just imagine that AA, BB, CC, DD are characters). So, the number of mappings is the number of combinations of characters from $S$, with repetition, like the following ones with $k=2$:
AA AB AC AD
BB BC BD
CC CD
DD
That is given as a $k$-combination of 4 elements:
$$N_{mapping}(k) = \binom{|S|+k-1}{k} = \binom {3+k}{k}$$
This is where I'am stuck. I wanted to combine this number of mappings with the number of permutations computed by each mapping, previously, but I can't ($N(c)$ depends on actual values of the mapping $c$). Also, I am sure there is a more straightforward approach, but I don't know how to proceed. Thanks for any help you could provide.
• I might miss something, but polynomial coefficient seems to work: $\binom{n}{2k} \cdot \binom{n-2k}{2j} \cdot \binom{n-2k -2j}{2m}$ – Alex Feb 13 '15 at 13:26
• @Alex In your formula, do $k$ $j$ and $m$ stand for half the number of occurences of each character? I understand it as: choose first $2k$ elements from $n$, then $2j$ from $n-2k$, and so on...? Are $k$, $j$ and $m$ supposed to be known? Sorry, I have very little experience with this. – coredump Feb 13 '15 at 13:39
• Yes, but it seems like not every $\binom{2n}{2k}$ is even, e.g. $\binom{6}{4}$ – Alex Feb 13 '15 at 14:56
Generating functions can be applied to this problem. However, since the order of letters in a string is important, exponential generating functions are appropriate (not ordinary ones). So, suppose the alphabet $S$ has $m$ letters and let $a_n$ be the number of strings of length $n$ over the alphabet $S$ with an even number of each letter. Let $g(x)=\sum_{n\ge0}\frac{a_n}{n!}x^n$ be the exponential generating function of $\langle a_n:n\ge0\rangle$. Then, $$g(x)=\left(\sum_{n\textrm{ even}}\frac{x^n}{n!}\right)^m=\left(\frac{e^x+e^{-x}}2\right)^m.$$
In the case of $m=4$, $g(x)=\frac1{16}(e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x})$. Then, $a_n=\left[\frac{x^n}{n!}\right]g(x)$ and if $n$ is even, $a_n=\frac{n!}{16}\left(\frac{2\cdot4^n}{n!}+\frac{8\cdot2^n}{n!}\right)=\frac18(4^n+4\cdot2^n)$. In particular, $a_4=40$.
Note: based on the form of the solution of $a_n$, I suspect there's a direct combinatorial argument (but I don't know it).
• I am very impressed. I should have a look at generating functions. Thanks a lot, I'll study this and try to understand what it all means. – coredump Feb 13 '15 at 21:04
• @coredump: If you want to take a look at generating functions, one excellent place to start is Herb Wilf's free book: math.upenn.edu/~wilf/gfologyLinked2.pdf. After reading his book, I was hooked. – Rus May Feb 13 '15 at 23:41
Let $m=|S|$ be the number of distinct characters, and let $a_n$ be the number of strings in $S^n$ having an even number of each character in $S$. Let
$$g(x)=\sum_{n\ge 0}\frac{a_n}{n!}x^n$$
be the exponential generating function (egf) for $\langle a_n:n\in\Bbb N\rangle$. (The egf is wanted because the order of the characters matters.) The egf for the sequence given by $$a_n=\begin{cases}0,&\text{if }n\text{ is odd}\\1,&\text{if }n\text{ is even}\end{cases}$$ is
$$\sum_{n\ge 0}\frac1{(2n)!}x^{2n}=\frac12(e^x+e^{-x})\;,$$
so
$$g(x)=\left(\sum_{n\ge 0}\frac1{(2n)!}x^{2n}\right)^m=\frac1{2^m}(e^x+e^{-x})^m\;.$$
Now
\begin{align*} (e^x+e^{-x})^m&=\sum_{k=0}^m\binom{m}ke^{kx}e^{-(m-k)x}\\\\ &=\sum_{k=0}^m\binom{m}ke^{(2k-m)x}\\\\ &=\sum_{k=0}^m\binom{m}k\sum_{n\ge 0}\frac{(2k-m)^n}{n!}x^n\\\\ &=\sum_{n\ge 0}\frac1{n!}\left(\sum_{k=0}^m\binom{m}k(2k-m)^n\right)x^n\;, \end{align*}
so
$$a_n=\frac1{2^m}\sum_{k=0}^m\binom{m}k(2k-m)^n\;.\tag{1}$$
Note that $2(m-k)-m=m-2k=-(2k-m)$, so $(1)$ is indeed $0$ when $n$ is odd. When $n$ is even and positive we can rewrite $(1)$ as
$$a_n=\frac1{2^{m-1}}\sum_{k=0}^{\lfloor m/2\rfloor}\binom{m}k(m-2k)^n\;.$$
For instance, if $m=4$, then
\begin{align*} a_{2n}&=\frac18\left(\binom40(4-0)^{2n}+\binom41(4-2)^{2n}+\binom42(4-4)^{2n}\right)\\\\ &=\frac18\left(4^{2n}+4\cdot2^{2n}\right)\\\\ &=\frac18\left(4^{2n}+4^{n+1}\right)\\\\ &=2\cdot4^{n-1}\left(4^{n-1}+1\right)\;. \end{align*}
• May I ask you to explain how you get from the line headed by 'is' to the line headed by 'so'. Do you have a pointer to a book or literature where I could look this up? – Harald Feb 13 '15 at 21:10
• @Harald: This PDF is a concise introduction to exponential generating functions. This PDF has a more general introduction to generating functions. Miklós Bóna, Introduction to Enumerative Combinatorics, is very readable and has everything that you need on the subject. I suspect that the same goes for his A Walk Through Combinatorics, but I’ve not actually seen it. – Brian M. Scott Feb 13 '15 at 21:26 | 2019-09-22T22:31:44 | {
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https://math.stackexchange.com/questions/1244967/determinant-of-a-5-%C3%97-5-matrix | # Determinant of a 5 × 5 matrix
I have a little problem with a determinant.
Let $A = (a_{ij}) \in \mathbb{R}^{(n, n)}, n \ge 4$ with
$$a_{ij} = \begin{cases} x \quad \mbox{for } \,i = 2, \,\, j \ge 4,\\ d \quad \mbox{for } \,i \ge j, \\ 0 \quad \mbox{else.} \end{cases}$$
So for example, if we choose $n = 5$, the matrix would look like this: $$A = \begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}$$
How can I find the determinant of this matrix?
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A = T \cdot R$. Then I wanted to use $$\det(A) = \det(T \cdot R) = \det(T) \cdot \det(R).$$
to figure out the determinant. This would be something like
$$d^n \cdot \det(R)$$
But is this approach even possible (I don't think so)? Is there any intelligent way of solving this? Thanks in advance.
• What is the determinant for $n=4,5$? Maybe you could conjecture the result and then use induction. Apr 21 '15 at 14:26
• BTW, what is a "rest matrix"? Apr 21 '15 at 14:30
Adding a multiple of one row to another preserves the determinant. Subtract $x/d$ of the last row from the second to get $$\begin{pmatrix} d &0 &0 &0 &0 \\ d-x &d-x &-x &0 &0 \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}$$ and then add $x/d$ of the third row to the second row to get $$\begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &0 &0 \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}.$$ This is lower triangular, so its determinant is the product of its diagonal, which is $d^5$. This all works for the $n$ by $n$ case, so the answer in general is $d^n$.
• "Row operations preserve determinant" - except for multiplying a row by a constant, which multiplies the determinant by the same constant. Apr 21 '15 at 18:13
• @user2357112 Or switching two rows, which multiplies the determinant by $(-1)$. Apr 21 '15 at 18:42
• Two! Two operations. Apr 21 '15 at 18:44
• @user2357112 You're right, I should be more clear. Apr 22 '15 at 7:13
Develop your matrix wrt the first row and get
$$|A|=d\begin{vmatrix}d&0&x&x\\d&d&0&0\\d&d&d&0\\d&d&d&d\end{vmatrix}$$
Develop again wrt the first row but observe that when your pivot points are the $\;x$'s you get determinant zero as there are two identical rows in each case, so we get
$$d^2\begin{vmatrix}d&0&0\\d&d&0\\d&d&d\end{vmatrix}=d^5$$
Try now some inductive argument based on this.
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A=T⋅R$.
That's very much a way to do it. The technique is called LU Decomposition. It produces a lower and upper triangular matrix, allowing trivial determinate calculations. For this reason, you actually only need to find the diagonal elements to get your determinant.
In this case,
$$A = \begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix} = \begin{pmatrix} 1 &0 &0 &0 &0 \\ 1 &1 &0 &0 &0 \\ 1 &1 &1 &0 &0 \\ 1 &1 &1 &1 &0 \\ 1 &1 &1 &1 &1 \\ \end{pmatrix} \cdot \begin{pmatrix} d &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ 0 &0 &d &-x &-x \\ 0 &0 &0 &d &0 \\ 0 &0 &0 &0 &d \\ \end{pmatrix}$$
So
$$\det A = \det L \cdot \det U = 1^5 \cdot d^5$$
The first few steps of the method used here (takes longer to texify than to do) are: $$\begin{eqnarray} \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{smallmatrix}\right] &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ 0 &d &d &0 &0 \\ 0 &d &d &d &0 \\ 0 &d &d &d &d \\ \end{smallmatrix}\right] \\ &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &d &0 &x &x \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &0 &0 &0 &0 \\ 0 &0 &0 &0 &0 \\ 0 &0 &d &-x &-x \\ 0 &0 &d &d-x &-x \\ 0 &0 &d &d-x &d-x \\ \end{smallmatrix}\right] \\ &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 & 0\\ 1 & 1\\ 1 & 1\\ 1 & 1\\ 1 & 1\\ \end{smallmatrix}\right] + \cdots \end{eqnarray}$$ | 2021-09-20T00:31:59 | {
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https://codereview.stackexchange.com/questions/78273/project-euler-6-in-java/78275 | # Project Euler #6 in Java
Project Euler #6:
The sum of the squares of the first ten natural numbers is,
$1^2+2^2+ ... + 10^2 = 385$
The square of the sum of the first ten natural numbers is,
$(1+2+ ... + 10)^2 = 55^2 = 3025$
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 − 385 = 2640$.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Here is my solution:
public class DifferenceFinder {
private static final int MAX = 100;
public static void main(String[] args) {
long time = System.nanoTime();
int result = MAX * (MAX + 1) / 2;
result *= result;
for(int i = 1; i <= MAX; i++) {
result -= i * i;
}
time = System.nanoTime() - time;
System.out.println("Result: " + result
+ "\nTime used to calculate in nanoseconds: " + time);
}
}
It simply does:
$$(1+2+ ... + 100)^2-1^2-2^2- ... -100^2$$
Output:
Result: 25164150
Time used to calculate in nanoseconds: 2231
Questions:
1. Is the simplest solution the most efficient one?
2. Does it smell?
## Method extraction.
Even for simple programs, having multiple responsibilities in a method is poor practice.
Consider simple extractions, like:
private static int squareOfSum(int limit) {
int sum = (limit * (limit + 1)) / 2;
return sum * sum;
}
public static int sumOfSquares(int limit) {
int sum = 0;
for (int i = 1; i <= limit; i++) {
sum += i * i;
}
return sum;
}
int difference = squareOfSum(100) - sumOfSquares(100);
System.out.printf("Difference is %d\n", difference);
By extracting functions, the logic becomes reusable, and discrete. Much better.
Additionally, it allows you to easily change the logic inside the methods to suite your algorithms, and use better algorithms like vnp suggests (+1 to that answer too).
## Timing
Your use of the nanos to time your code is misleading. The performance of the code is heavily related to how often the code runs, and, for this example, any performance measurement is unreliable....
• Is the timing really misleading? When something takes ~2 microseconds, is it fair to say that startup costs don't count? – 200_success Jan 22 '15 at 1:24
• @200_success No, what's misleading is suggesting the code takes 2 microseconds when in another context it may take 0.2 microseconds. In addition, there are other things that are not being timed.... but my main concern is using a measure for the timing that is heavily context-dependent – rolfl Jan 22 '15 at 1:26
A sum of first $n$ numbers is $\frac{n(n+1)}{2}$. A sum of squares of first $n$ numbers is $\frac{n(n+1)(2n+1)}{6}$. A difference in question is $\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6} = \frac{(n-2)(n-1)n(n+1)}{12}$.
This solution is likely the most efficient.
• This doesn't seem to work. It gives the result 8165850. – TheCoffeeCup Jan 22 '15 at 3:10
• @MannyMeng verified using the above Summation formulas and it work with much better timings. Just don't compute the difference with command denominator fraction, use the method suggested in the answer above – yadav_vi Jan 22 '15 at 11:38
You can linearize this problem, without losing readability. For example my solution from years back:
int size = 100;
long qos = (long) Math.pow(size * (size + 1) / 2, 2);
long soq = size * (size + 1) * (2 * size + 1) / 6;
System.out.println(qos - soq);
Where qos and soq are known acronyms for me denoting square of sum and sum of squeares.
Note however that my code actually had a potential bug called "Possible Loss of Fraction"
size * (size + 1) / 2
should be
size * (size + 1.0) / 2
In this example
• size * (size + 1) would be of type int
• size * (size + 1.0) would be of type double
In java
• int / int you will get int
• double / int you will get double
In Java you mostly deal with discrete mathematics while equations do not, note this fact as you start solving Euler problems. In addition remember to note that type double is an approximation.
It is also good to use best tool for the task, for example if you know the result will be
You can compute it with wolfram alpha (mathematica/wolfram language) using:
Limit[(n (n - 2) (n - 1) (n + 1))/12, n -> 100]
• In java 1.0 is same as 1d. Where 1f would result in 1.0, but be of type float (equal value, but not same type). – Margus Jan 23 '15 at 10:41 | 2021-04-13T22:51:17 | {
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https://math.stackexchange.com/questions/1874504/class-number-of-mathbbq-sqrt319-and-hilbert-class-field | # Class Number of $\mathbb{Q}(\sqrt[3]{19})$ and Hilbert class field
Finding the class number of $\mathbb{Q}(\sqrt[3]{19})$ is an exercise from Marcus 'Number Field'. This question was uploaded by some other user, but it was removed by now. I have worked on details and have some more questions.
Following are the steps in the exercise:
Let $K=\mathbb{Q}(\alpha)$, $R=\mathcal{O}_K$, and $3R=P^2Q$, where $\alpha=\sqrt[3]{19}$, and $P$, $Q$ are prime ideals in $R$.
(a) Prove that the ideal class group is cyclic, generated by the class containing $P$.
(b) Prove that the number of ideal classes is a multiple of $3$.
(c) Prove that there are either three or six ideal classes.
(d) Prove that there are three ideal classes. (Suggestion: Suppose there were six. Show that none of the ideals $J$ with $||J||\leq 9$ are in the same class with $P^3$. )
After solving this problem, it seems that the calculations from my solution to (a) are enough to conclude that the class number is $3$.
To do (a), we find the Minkowski's bound: $$\frac{3!}{3^3}\frac 4{\pi} \sqrt{1083} \approx 9.3$$ This shows that each ideal class in the ideal class group contains an ideal $J$ with $||J||\leq 9$. Also, each ideal class generating the class group contains a prime ideal lying above $2$, $3$, $5$, $7$.
So, we find prime factorizations of $2R$, $3R$, $5R$, $7R$: $$2R=P_1P_2,\ \ \ 3R=P^2 Q,$$ $$5R= P_3P_4, \ \ \ 7R = P_5,$$ with $P_1=(2, \alpha-1)$, $P=(3, \frac{\alpha^2+\alpha+1}3)$, $P_3=(5,\alpha+1)$.
With a help of SAGE, I found that $$P_1= \frac{\alpha-1}3 P, \ \ \ P_3= \frac{-\alpha+4}3 P, \ \ \ P_5=\frac{7\alpha+14}9 P^3.$$
Then the prime ideals appear above (as factorizations of $2$, $3$, $5$, $7$) belong to one of the three ideal classes of $P$, $P^2$, $I$ (principal).
My questions are
1. Are these calculations enough to conclude that the ideal class group has order $3$?
2. If the answer to 1 is no, then how are the rest of parts solved?
3. How do we determine the Hilbert class field of $\mathbb{Q}(\sqrt[3]{19})$?
• And how are you concluding that all the ideal class aren't in fact the principal ideal class? – John M Jul 29 '16 at 2:36
• I have to prove that the ideal class of $P$ is not principal, so the class of $P$ has order $3$. Okay, so part (b) is really necessary to do that. I see. – Sungjin Kim Jul 29 '16 at 2:42
• Except for showing that at least one ideal is nonprincipal, I think you’re OK. Since $P_5=(7)$, principal, and also equals $(z)P^3$ for a number $z$, $P^3$ is principal. $P_2$ has norm $4$, but since $P_1P_2\sim1$, you have $P_2\sim P^2$. $Q$ also hs norm $3$, but it’s equiv to $P$. $P_3$ has norm $5$, so $P_4$ has norm $25$, so it’s not a problem. Seems to me that you’ve covered all bases for the ideal class group to be of order $3$. – Lubin Jul 29 '16 at 2:46
• My comment above may be worthless, but for the Class Field, I’d first look at the cubic subfield of $\Bbb Q(\zeta_{19})$. It’s unramified outside $19$, and you’d have a good chance of showing that its compositum with $K$ is unramified above $19$, over $K$. – Lubin Jul 29 '16 at 3:21
It turns out that the answer to 1 is no as commented by @JohnM. We need to prove that $P$ is not principal. To do this, assume that $P=\beta R$. Then $||P||=||\beta||=3$. So, now the question becomes whether there exists an element $\beta\in R$ with $||\beta||=3$.
Since any element in R can be expressed as $\beta=a+b\alpha+c\alpha^2$ with $a, b, c\in \mathbb{Q}$ and $3a, 3b, 3c \in \mathbb{Z}$, $3a\equiv 3b\equiv 3c \ \mathrm{mod} \ 3$, we see that $$||\beta||=3=a^3+19b^3+19^2c^3-3\cdot 19 abc$$ and $$27\cdot 3 =81= (3a)^3 + 19(3b)^3 + 19^2 (3c)^3 - 3 \cdot 19 (3a)(3b)(3c)\equiv (3a)^3 \ \mathrm{mod} \ 19$$ However, $81 \equiv 5 \ \mathrm{mod} \ 19$ is not a cubic residue modulo $19$ (list of cubic residues mod $19$ is $0, 1, 7, 8, 11, 12, 18$). Therefore, there is no element $\beta\in R$ of norm $3$. This proves that $P$ is not principal.
Then the rest of argument will follow as @Lubin commented. Thus, we needed to solve (b) to conclude, but (c) and (d) are not necessary. This answers my question 2.
To find the Hilbert class field, we use a cubic subfield $K'$ of the cyclotomic field $\mathbb{Q}(\zeta_{19})$ as commented by @Lubin. This method was already discussed by @franzlemmermeyer in the answer to this question.
Let $L$ be composite field of $K'$ and $K=\mathbb{Q}(\alpha)$. Let $\mathcal{P}$ be a prime in $L$. If $\mathcal{P}$ lies over rational prime $p\neq 3, 19$, then $p$ is unramified in both extensions $K$ and $K'$. Thus, the prime $\mathcal{P}\cap K$ lying under $\mathcal{P}$ is unramified in $L$.
Suppose that $\mathcal{P}$ lies over rational prime $3$. Since the rational prime $3$ is inert in $K'$, the residue field degree must be $3$. By the decomposition $3R=P^2Q$, the residue field degree for $P$ and $Q$ over rational prime $3$ must be both $1$. This gives the residue field degree for $\mathcal{P}$ over both $P$ and $Q$ must be $3$, yielding that the ramification indices for $\mathcal{P}$ over both $P$ and $Q$ are $1$.
Now assume that $\mathcal{P}$ is over rational prime $19$. As @franzlemmermeyer answered in the above link, we use Abhyankar's lemma. Since $19$ is totally ramified in both $K$ and $K'$, we have by Abhyankar's lemma that the ramification index for $\mathcal{P}$ over $\mathcal{P}\cap K$ is $1$.
Hence, the extension $L$ over $K$ is unramified over $K$, yielding that $L$ is the Hilbert class field of $K$. This answers my question 3.
Acknowledgement @JohnM, @Lubin, Thank you very much for the helpful comments with great insights. | 2021-07-28T18:14:18 | {
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https://mathematica.stackexchange.com/questions/59691/listcontourplot-is-blank | # ListContourPlot is blank
This is probably very simple. I want to do a simple ListContourPlot.
contourData = Table[ myFunc[x,y], {x,0.1, 2.9}, {y,0.1, 2.9}];
ListContourPlot[contourData]
This works without a problem.
Now, I want to include x and y values so that I can show the values corresponding to specific values of x and y. I modify the Table statement to include x and y values.
contourData = Table[ {x,y, myFunc[x,y]}, {x,0.1, 2.9}, {y,0.1, 2.9}];
ListContourPlot[contourData]
My contour plot is now blank. I'm guessing it is a problem with the Table statement. The maximum minus minimum over this range is about 220. As this is a function of Pythagorean distance, the values vary (so I don't think it is a case of essentially equal values over the ranges of x and y.
What is the correct way to generate data for a ListContourPlot?
• Try ListContourPlot[Flatten[contourData, 1]]. Sep 14 '14 at 19:04
• That was it exactly! Thank you most kindly! Sep 14 '14 at 19:10
As an alternative to @ b.gatessucks excellent comment you might consider DataRange:
ListContourPlot[Table[Sin[x + y], {x, 0, 3, 0.1}, {y, 0, 3, 0.1}], | 2021-11-29T22:06:36 | {
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https://www.jiskha.com/questions/514863/find-the-extreme-values-of-the-function-for-the-given-intervals-2x-3-21x-2-72x-a | # calculus
Find the extreme values of the function for the given intervals.
2x^3 – 21x^2 + 72x
a) [0,5]
b) [0,4]
I have both the minimum values for both intervals which is 0. But I am unable to find the maximum. please help!!!
1. 👍 0
2. 👎 0
3. 👁 184
1. You have probably found two extrema where f'(x)=0.
To identify whether the extremum is a maximum or minimum, use the second derivative test.
If the second derivative is positive, the extremum is a minimum. If the second derivative is negative, the extremum is a maximum.
Here the extrema are at x=3 and x=4.
The second derivative, f"(x)=12x-42,
so for example, f2(3)=-6, so x=3 is a maximum.
Use your calculator to plot the graph of the function to help you solve these problems. After some time, you will be able to visualize the graph from the equation, even without the calculator.
1. 👍 0
2. 👎 0
2. I tried 3 for both (a and b) as a maximum and didn't work....
1. 👍 0
2. 👎 0
3. It is important to know that extreme values include local maxima, local minima and end-points of the given interval.
The question requires the extreme values of the function, so it is the value of the function that counts.
To find extreme values, we calculate the critical points (where f'(0)=0 and where f'(0) does not exist) as well as the value of the function at end-points.
Given
f(x)=2x^3 – 21x^2 + 72x
1. We calculate the points where f'(x)=0, and we have determined that f(3)=81 is a local maximum and f(4)=80 is a local minimum.
2. For [0,4], the values of the function at end-points are:
f(0)=zero, and f(4)=80.
So the list of values to consider are:
(0,zero)
(3,81)
(4,80)
We conclude that the extreme minimum is zero, and the extreme maximum is 81.
3. For [0,5], the values of the function at end-points are f(0)=zero, and f(5)=85.
So the list of values to consider for extreme values are:
(0,zero)
(3,81)
(4,80)
(5,85)
We choose zero as the extreme minimum, and 85 as the extreme maximum.
1. 👍 0
2. 👎 0
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https://math.stackexchange.com/questions/950712/how-to-compute-int-infty-infty-exp-left-fracx2-13x-12611x2-rig/950848 | # How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$
$$\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$$
WolframAlpha gives a numerical answer of $43.8122$, which appears to be $\sqrt{611\pi}$. And playing with that, it seems that replacing $611$ with $a$ just gives $\sqrt{a\pi}$. My trouble is that the stuff in the exponential always seems to be just a big mess, and I haven't been able to get it into a form I can understand or deal with.
I would greatly appreciate seeing a method for solving this integral.
• I have a strong feeling that you can use Feynman's trick on this, but I am still trying to find the correct one. – UserX Sep 29 '14 at 7:20
• How can you be sure that $\sqrt{a\pi}$ is not just a good approximation ? Could you produce some values with more digits ? – Claude Leibovici Sep 29 '14 at 7:34
• @ClaudeLeibovici I'm not entirely sure if it's exactly $\sqrt{a\pi}$, but for every value of $a$ I've tried, WolframAlpha always gives a number very close to $\sqrt{a\pi}$. I was hoping this fact might help with solving the integral. – user137794 Sep 29 '14 at 7:40
• Working with very high precision, I can confirm that the result is exactly $\sqrt{a\pi}$ what you would get with $$\int_{-\infty}^\infty\exp\left(-\frac{x^2}{a}\right)\ dx=\sqrt{a\pi}$$ – Claude Leibovici Sep 29 '14 at 8:12
Let $\displaystyle\;u(x) = \frac{x^2-13x-1}{x}\;$. As $x$ varies over $\mathbb{R}$, we have
• u(x) increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$.
• u(x) increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$.
This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice.
Let $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:
$$u = u(x) = \frac{x^2-13x-1}{x} \quad\iff\quad x^2 - (13+u)x - 1 = 0$$ we have $$x_1(u) + x_2(u) = 13 + u \quad\implies\quad \frac{dx_1}{du} + \frac{dx_2}{du} = 1.$$ From this, we find
\begin{align} \int_{-\infty}^\infty e^{-u(x)^2/611} dx &= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) e^{-u(x)^2/611} dx\\ &= \int_{-\infty}^{\infty} e^{-u^2/611}\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\ &= \int_{-\infty}^{\infty} e^{-u^2/611} du\\ &= \sqrt{611\pi} \end{align}
• This is a new technique for me (+1), I think it can be generalised. If so, what is the condition? Say $u(x)=\dfrac{px^2+qx+r}{sx}$ – Anastasiya-Romanova 秀 Sep 30 '14 at 10:44
• @Anastasiya-Romanova algebraically, $pr < 0$. Geometrically, this ensure the monotonicity of the $u(x)$ and $u$ cover $(-\infty,\infty)$ twice. – achille hui Sep 30 '14 at 12:47
• +1 for its potent to be generalized (to a higher degree, maybe?). – Sangchul Lee Sep 30 '14 at 19:43
• @sos440 the generalization is essentially the "Added" part of orangeskid answer. – achille hui Sep 30 '14 at 19:59
HINT:
For $a$ fixed
$\int_{-\infty}^\infty\exp\left(-\frac{(x^2+sx-b)^2}{a x^2}\right)\ dx$
is constant in $s$ and $b\ge 0$.
$\bf{Added:}$
The function $\frac{x^2 + s x - b}{x} = x - \frac{b}{x} + s$ invariates the Lebesgue measure as @achille hui: showed in his answer.
Let $n \in \mathbb{N}$ $\alpha_1$, $\ldots$, $\alpha_n$ distinct real numbers, $\rho_1$, $\ldots$, $\rho_n$ $>0$ and $\beta \in \mathbb{R}$. The function
$$\phi(x) = x - \sum_{i=1}^n \frac{\rho_i}{x- \alpha_i} -\beta$$
invariates the Lebesgue measure on $\mathbb{R}$.
Lemma: For any $a \in \mathbb{R}$ the equation
\begin{eqnarray*} x- \sum_{i=1}^n \frac{\rho_i}{x- \alpha_i} -\beta = u \end{eqnarray*} has $n+1$ distinct real root with sum $u + \sum_i \alpha_i + \beta$. Use Viete.
Lemma: Let $I$ an interval in $\mathbb{R}$ of length $l$. Then the preimage $\phi^{-1} (I)$ is a union of $n+1$ disjoint intervals of total length $l$.
Consequence: $$\int_{\mathbb{R}} (f\circ \phi)\, d\,\mu = \int_{\mathbb{R}} f\ d\mu$$
Composing two rational maps that invariate the measure gets a third one. They will have singularities in general.
For $f(x) = e^{-\frac{x^2}{a}}$ the composition $f (\phi(x))$ is still smooth due to the rapid decay at $\infty$ of $e^{-\frac{x^2}{a}}$.
• That is remarkably simple and straightforward. +1 – Tom-Tom Sep 29 '14 at 9:56
• Would you be able to elaborate on how you concluded this? – user137794 Sep 29 '14 at 11:45
• @user137794: Added more detail. – orangeskid Sep 30 '14 at 6:43
• @Tom-Tom: Thanks! Added more detail. – orangeskid Sep 30 '14 at 16:36
• For $f(x) = e^{-\frac{x^2}{a}}$ the function $f(\phi(x))$ is still smooth. – orangeskid Sep 30 '14 at 20:19
Adding another solution owing to a friend of mine.
Through some algebra, the integral is equivalent to
$$\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx$$
Then using the following identity
$$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$
We have
\begin{align} &\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx\\ =&\int_{-\infty}^\infty \exp\left(-\frac1{611}(x-13)^2\right)\ dx\\ =&\int_{-\infty}^\infty \exp\left(-\frac1{611}x^2\right)\ dx\\ =&\sqrt{611\pi} \end{align}
Using identity in @user137794's answer: $$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$ where the complete proof can be seen here. The problem can be generalised to evaluate
$$\int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx = \sqrt{a\pi}$$
Proof:
It's easy to see that $\dfrac{(x^2-bx-1)^2}{ax^2}=\dfrac{1}{a}\left(x-x^{-1}-b\right)^2$, then \begin{align} \int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx &=\int_{-\infty}^\infty \exp\left(-\frac{(x-x^{-1}-b)^2}{a}\right)\ dx\\ &=\int_{-\infty}^\infty \exp\left(-\frac{(x-b)^2}{a}\right)\ dx\\ &=\int_{-\infty}^\infty \exp\left(-\frac{y^2}{a}\right)\ dy\\ &=\sqrt{a}\int_{-\infty}^\infty \exp\left(-z^2\right)\ dz\\ &=\sqrt{a\pi} \end{align} Therefore $$\int_{-\infty}^\infty \exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx = \sqrt{611\pi}$$
• Nice generalization! It's interesting, that $b$ does not matter, on the other hand $-1$ is how important! – user153012 Oct 1 '14 at 12:07 | 2020-04-02T10:06:59 | {
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https://math.stackexchange.com/questions/395139/combinatorial-proof-of-a-stirling-number-identity | # Combinatorial proof of a Stirling number identity.
Consider the identity $$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$ where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity $$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$ which essentially states that the number of even subsets of a set is equal to the number of odd subsets.
Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?
As an added note, there is a similar identity $$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$ A combinatorial interpretation of this one would also be appreciated.
• May inclusion exclusionn principle will help? – Norbert May 18 '13 at 4:21
Perhaps there is something relating odd ordered partitions with even ordered partitions?
There is indeed. Let's try to construct an involution $T_n$, mapping odd ordered partitions of $n$-element set to even and vice versa: if partition has part $\{n\}$, move $n$ into previous part; otherwise move $n$ into new separate part.
Example: $(\{1,2\},\{\mathbf{5}\},\{3,4\})\leftrightarrow(\{1,2,\mathbf{5}\},\{3,4\})$.
This involution is not defined on partitions of the form $(\{n\},\ldots)$, but for these partitions one can use previous involution $T_{n-1}$ and so on.
Example: $(\{5\},\{4\},\{1,2\},\{\mathbf{3}\})\leftrightarrow(\{5\},\{4\},\{1,2,\mathbf{3}\})$.
In the end only partition without pair will be $(\{n\},\{n-1\},\ldots,\{1\})$. So our (recursively defined) involution gives a bijective proof of $\sum_{\text{k is even}}k!{n \brace k}=\sum_{\text{k is odd}}k!{n \brace k}\pm1$ (cf. 1, 2).
Upd. As for the second identity, the involution $T_n$ is already defined on all cyclically ordered partitions, so $\sum_{\text{k is even}}(k-1)!{n \brace k}=\sum_{\text{k is odd}}(k-1)!{n \brace k}$.
P.S. I can't resist adding that $k!{n \brace k}$ is the number of $(n-k)$-dimensional faces of an $n$-dimensional convex polytope, permutohedron (the convex hull of all vectors formed by permuting the coordinates of the vector $(0,1,2,\ldots,n)$). So $\sum(-1)^{n-k}k!{n \brace k}=1$ since it's the Euler characteristic of a convex polytope.
• much more combinatorial (+1) – robjohn May 23 '13 at 13:30
• Beautiful, thank you. – EuYu May 23 '13 at 18:12
These are not combinatorial interpretations, but they are simple.
The defining equation for Stirling numbers of the second kind is $$\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{x}{k}k!=x^n\tag{1}$$ That is, Stirling numbers of the second kind tell how to write monomials as a combination of falling factorials (or combinatorial polynomials).
Noting that $\displaystyle\binom{-1}{k}=(-1)^k$ and setting $x=-1$ yields \begin{align} \sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}(-1)^kk!= \sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{-1}{k}k!=(-1)^n \end{align}
Since $\displaystyle\binom{x}{k}=\binom{x-1}{k-1}\frac{x}{k}$ and $\begin{Bmatrix}n\\0\end{Bmatrix}=0$ for $n\ge1$, we can rewrite $(1)$ as $$\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}\binom{x-1}{k-1}(k-1)!=x^{n-1}\tag{2}$$ Setting $x=0$ yields $$\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}(-1)^{k-1}(k-1)!=0^{n-1}$$ where $0^0=1$ for the case $n=1$.
• Certainly the second part is supposing $n>0$, so that the term for $k=0$ can be dropped. However it fails to accommodate to the fact (even though some wish to deny it) that $0^0=1$. Stated more directly, it fails for $n=1$. – Marc van Leeuwen May 23 '13 at 14:56
• @MarcvanLeeuwen: Good point. I got rid of the exponent when $n=1$ when I shouldn't have. – robjohn May 23 '13 at 15:54
For the sake of completeness I include a treatment using generating functions. The exponential generating function of the Stirling numbers of the second kind is $$G(z, u) = \exp(u(\exp(z)-1))$$ so that $${n \brace k} = n! [z^n] \frac{(\exp(z) - 1)^k}{k!}.$$ It follows that $$\sum_{k=0}^n (-1)^k k! {n \brace k} = n! [z^n] \sum_{k=0}^n (1-\exp(z))^k = n! [z^n] \sum_{k=0}^\infty (1-\exp(z))^k,$$ where the last equality occurs because the series for $(1-\exp(z))^k$ starts at degree $k.$ But this is just $$n! [z^n] \frac{1}{1-(1-\exp(z))} = n! [z^n] \exp(-z) = (-1)^n,$$ showing the result. | 2019-05-21T10:47:07 | {
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http://www.intermash.ua/8nxz7/66kmp25/e0l9j.php?79eef7=least-squares-function-r | the residuals, that is response minus fitted values if "g" is a ⦠weights. For practical purposes it might be preferable to use a nonlinear least squares approach (e.g., the nls function). For example, if a student had spent 20 hours on an essay, their predicted score would be 160, which doesnât really make sense on a typical 0-100 scale. When present, the objective function is weighted least squares. In literal manner, least square method of regression minimizes the sum of squares of errors that could be made based upon the relevant equation. When the "port" algorithm is used the objective function value printed is half the residual (weighted) sum-of-squares. R-bloggers ... BVLS is implemented in the bvls() function ⦠Each classroom has a least squared mean of 153.5 cm, indicating the mean of classroom B was inflated due to the higher proportion of girls. We want to build a model for using the feature. 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2020 least squares function r | 2021-04-19T21:29:30 | {
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https://flyingcoloursmaths.co.uk/regions-of-a-circle/ | On a recent MathsJam Shout, an Old Chestnut appeared (in this form, due to @jamestanton):
If you’ve not seen it, stop reading here and have a play with it - it’s a classic puzzle for a reason.
Below the line are spoilers.
Counting is hard
The first thing you’d probably try is to draw out the lines and count up the regions.
The first circle has 1 region.
The second has 2 regions.
The third has 4. (I’m spreading these out to try to avoid accidental spoilers, by the way.)
The fourth has 8. A pattern seems to be forming.
The fifth has 16. Aha! It must be…
The sixth has 31. Oh. Let’s count them again: nope, definitely 31 rather than 32. What’s going on?
What is going on?
As you draw more and more points, it gets harder to a: make sure you’re not making your lines coincide and b: count the regions accurately. At MathsJam, we made it to the seventh circle (57 regions) before realising that was about as far as we could go.
So we needed a more methodical approach. The nugget of the solution - for me, at least - was to think about how each line you add to a diagram affects the number of regions - and that’s related to the number of lines it crosses. In fact, adding a line that crosses $n$ other lines adds $n+1$ regions to the circle - which you might like to prove.
So, in adding a fifth point to the fourth circle, we’ll need to add four new lines.
The two lines to the adjacent points each add a single region - they don’t cross any other lines. The lines to the two opposite points each cross two lines, creating three new regions. Overall, that’s 1 + 3 + 3 + 1 new regions, taking the eight regions to 16.
Let’s go a bit more methodically this time
Going from five to six is a bit trickier ((because counting is hard)).
Let’s consider the five existing points, going anticlockwise from our new point (which I’ll imagine is at the bottom of the circle). Each new line will split the polygon into two parts, which I’ll call ‘left’ and ‘right’.
The first line, to the neighbour, crosses no lines, and adds one region.
The second line, to point 2, splits the circle so there is one point to the left and three to the right, all of which are connected - so this line adds $1 \times 3 + 1$ regions.
The third line splits the circle two on each side, so it adds $2 \times 2 + 1$ regions.
The fourth line mirrors the second, and the fifth line mirrors the first, so I might be tempted to write the number of regions added as:
$(0\\times 4 + 1) + (1 \\times 3 + 1) + (2 \\times 2 +1) + (3 \\times 1 + 1) + (4 \\times 0 + 1)=15$
… or better, the new number of regions is $5 + \sum_{i=0}^{4} i(4-i)$
But we can generalise!
Whenever we add a point to a diagram, the process is going to be identical! Each line will divide the existing points into three sets: those to the left, the one being connected, and those to the right. Every point on the left connects to every point on the right exactly once, so the number of lines cut by the new line is always the number of left points multiplied by the number of right points.
So! If we’re moving from $N-1$ points to $N$, we should be adding:
$R\_n = (N-1) + \\sum\_{i=0}^{N-2} i( (N-2) - i)$
new regions.
And we can write that explicitly!
$R\_n = (N-1) + (N-2)\\sum\_{i=0}^{N-2} i - \\sum{i=0}^{N-2} i^2$ $= (N-1) + (N-2)\\frac{(N-2)(N-1)}{2} - \\frac{ (N-2)(N-1)(2N-3)}{6}$ $= \\frac{(N-1)}{6} \\left\[ 6 + 3(N-2)^2 - (N-2)(2N-3) \\right$\] $= \\frac{(N-1)}{6} \\left\[ N^2 - 5N + 12 \\right$\]
Let’s check that with $N=6$, which we’ve already worked out: $R_n = \frac{5}{6}\left[ 36 - 30 + 12 \right] = 15$ new regions.
How about $N=7$? We’re expecting 26: $R_n=\frac{6}{6}\left[ 49 - 35 + 12\right] = 26$. Hooray!
Hang on - that’s just the new regions!
So, the number of new regions each time is $\frac{N-1}{6} \left[ N^2 - 5N + 12 \right]$, and we know when there are no points, there is one region.
So, the number of regions altogether is $R_\sum(M) = 1 + \frac{1}{6}\sum_{N=0}^{M} (N-1)(N^2 - 5N + 12)$.
Expanding that and distributing the sum:
$R_\sum(M) = 1 + \frac{1}{6} \sum_{N=0}^{M} N^3 - \sum_{N=0}^{M} N^2 + \frac{17}{6}\sum_{N=0}^{M} N - \sum_{N=0}^{M} 2$
There follows an awful lot of tedious algebra, which ends up giving $R_\sum(M) = \frac{1}{24}\left(M^4 - 6M^3 + 23M^2 - 18M + 24\right)$
Just to check, we know $R_\sum(6)$ is 31: is that $\frac{1}{24}\left( 6^4 - 6\times6^3 + 23\times 6^2 - 18\times 6 + 24 \right)$? The first two terms cancel, and the rest? It’s $\frac{6}{24} \left( 23 \times 6 - 18 + 4\right)$, or $\frac{1}{4} \times 124 = 31$, as required!
Curiously, it turns out that $R_\sum(10) = 256$, which is also a power of 2 (although half as big as it “ought” to be if it followed the doubling pattern.) | 2021-03-01T04:08:09 | {
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https://math.stackexchange.com/questions/3535767/combinatorics-how-many-unique-polygons-can-be-drawn-using-up-to-p-points-along | # Combinatorics: how many unique polygons can be drawn using up to p points along a circle?
I am faced with the following problem:
Let p ≥ 3 be a prime number, and mark p points at equal distance on the circumference of a circle like in the pictures below (where p = 7). We can trace polygons by choosing a certain amount of those points as vertices, and connecting them with non-intersecting straight lines.
With three parts:
a) How many different polygons can we form if we consider that a different choice of vertices leads to a different polygon?
This is straightforward: you can have polygons with three vertices, 4 vertices […] up until p vertices, and the amount of polygons you can draw of a given number of vertices k is $$p \choose k$$. Hence the answer is $${p \choose 3} + {p \choose 4} + {...} + {p \choose p-1} + {p \choose p}$$
b) How many different polygons can we form if we count all different rotations of the same polygon as a single polygon?
Well, any given polygon can be rotated up to $$p-1$$ times. Hence each polygon counted from a) is actually being counted 7 times. We adjust for this by dividing through the answer in a) by p, and get the answer as $$\frac{p \choose 3}{p} + \frac{p \choose 4}{p} + {...} + \frac{p \choose p-1}{p} + \frac{p \choose p}{p} = \frac{p \choose 3}{p} + \frac{p \choose 4}{p} + {...} + \frac{p \choose p-2}{p} + 1 + 1$$.
I think all the above makes sense, but I am struggling with the third part:
c) How many different polygons can we form if we count all isometric polygons (all rotations and reflections of the same polygon) as the same ?
And I don't know how to proceed from here. I'm guessing there has to be some pattern to the number of reflections+rotations that any polygon of k vertices has along p points. However, I've tried playing around with small examples and I can't seem to find any generalizable pattern, and I've been stumped for longer than I'd like to admit. Any help would be greatly appreciated.
• Reworded: How many necklaces and bracelets (necklaces for part b and bracelets for part c) of length $p$ with $p$ prime can be made with beads of two possible colors such that there are at least three beads of the first color used. – JMoravitz Feb 5 '20 at 19:47
• Almost exactly the same question was posted just a week ago but then deleted when I answered it. a) Does this happen to be from an ongoing contest? If so, are you aware of the contest problem policy? b) If it isn't, and I post an answer, do you promise not to delete the question? :-) – joriki Feb 5 '20 at 19:47
• @joriki it is not a contest problem. I haven't seen the other post / don't see why it was deleted. This is a problem from a university class :) – Student_514 Feb 5 '20 at 19:51
The polygons are in correspondence with the subsets of the vertices with at least $$3$$ elements. Thus, as JMoravitz noted in a comment, this can be viewed as counting binary necklaces and bracelets of length $$p$$ with at least three beads of the first colour, where the first colour signifies a vertex that’s included and the second colour signifies a vertex that isn’t included in the polygon.
For part a), there is no symmetry to take into account, so the result is just $$2^p-\binom p2-\binom p1-\binom p0$$. (Your result is correct, you just didn't use the fact that $$\sum_k\binom pk=2^k$$, the total number of subsets of the set of $$k$$ vertices.)
For part b), with rotational symmetry, you failed to properly account for the fact that the polygon that includes all vertices has only one rotational equivalent, not $$p$$. Your final result is correct, but only because you replaced $$\frac{\binom pp}p=\frac1p$$ by $$1$$. If you count correctly, you have $$2^p-\binom p2-\binom p1-\binom p0-\binom pp$$ polygons that form classes of $$p$$ rotational equivalents each, and $$\binom pp=1$$ polygon that’s in a class of its own, so the total count is
$$\frac{2^p-\binom p2-\binom p1-\binom p0-\binom pp}p+1=\frac{2^p-2}p-\frac{p+1}2\;.$$
For part c), with rotational and reflectional symmetry, it becomes easier to perform the count using Burnside’s lemma. The general result is given in the Wikipedia article linked to above. In your case, since $$p$$ is an odd prime, there are
$$\begin{eqnarray} B_2(p) &=& \frac12N_2(p)+\frac12\cdot2^{\frac{p+1}2} \\ &=& \frac12\cdot\frac1p\left(1\cdot2^p+(p-1)\cdot2^1\right)+2^{\frac{p-1}2} \\ &=& \frac{2^{p-1}-1}p+1+2^{\frac{p-1}2} \end{eqnarray}$$
binary bracelets. We need to subtract the $$1$$ bracelet with $$0$$ elements of the second colour, the $$1$$ bracelet with $$1$$ element of the second colour and the $$\frac{p-1}2$$ bracelets with $$2$$ elements of the second colour, so the number of polygons up to rotations and reflections is
$$\frac{2^{p-1}-1}p+2^{\frac{p-1}2}-\frac{p+1}2\;.$$ | 2021-07-31T13:13:40 | {
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https://mathbooks.unl.edu/Calculus/sec-5-9-num-int.html |
## Section5.9Numerical Integration
###### Motivating Questions
• How do we accurately evaluate a definite integral such as $\int_0^1 e^{-x^2} \, dx$ when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative? Are there ways to generate accurate estimates without using extremely large values of $n$ in Riemann sums?
• What is the Trapezoid Rule, and how is it related to left, right, and middle Riemann sums?
• How are the errors in the Trapezoid Rule and Midpoint Rule related, and how can they be used to develop an even more accurate rule?
When we first explored finding the net signed area bounded by a curve, we developed the concept of a Riemann sum4.2 as a helpful estimation tool and a key step in the definition of the definite integral. Recall that the left, right, and middle Riemann sums of a function $f$ on an interval $[a,b]$ are given by
\begin{align} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\label{E-Left}\tag{5.15}\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\label{E-Right}\tag{5.16}\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x\text{,}\label{E-Mid}\tag{5.17} \end{align}
where $x_0 = a\text{,}$ $x_i = a + i\Delta x\text{,}$ $x_n = b\text{,}$ and $\Delta x = \frac{b-a}{n}\text{.}$ For the middle sum, we defined $\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}$
A Riemann sum is a sum of (possibly signed) areas of rectangles. The value of $n$ determines the number of rectangles, and our choice of left endpoints, right endpoints, or midpoints determines the heights of the rectangles. We can see the similarities and differences among these three options in Figure5.77, where we consider the function $f(x) = \frac{1}{20}(x-4)^3 + 7$ on the interval $[1,8]\text{,}$ and use 5 rectangles for each of the Riemann sums.
Part (d) of Example4.36 explored how to determine if $L_n$ and $R_n$ are underestimates or overestimates of $\int_a^b f(x) \, dx$ when $f(x)$ is increasing. Similar statements can be made when $f(x)$ is decreasing. These are summarized next.
###### Error types when approximating area using left and right Riemann sums
• If $f(x)$ is increasing on $[a,b] \text{,}$ then $L_n$ underestimates $\int_a^b f(x) \, dx$ and $R_n$ overestimates $\int_a^b f(x) \, dx\text{.}$
• If $f(x)$ is decreasing on $[a,b] \text{,}$ then $L_n$ overestimates $\int_a^b f(x) \, dx$ and $R_n$ underestimates $\int_a^b f(x) \, dx\text{.}$
While it is a good exercise to compute a few Riemann sums by hand, just to ensure that we understand how they work and how varying the function, the number of subintervals, and the choice of endpoints or midpoints affects the result, using computing technology is a quick way to determine $L_n\text{,}$ $R_n\text{,}$ and $M_n\text{.}$ Any computer algebra system will offer this capability.
In this section we explore several different alternatives for estimating definite integrals. Our main goal is to develop formulas to estimate definite integrals accurately without using a large numbers of rectangles.
###### Example5.78
As we begin to investigate ways to approximate definite integrals, it will be insightful to compare results to integrals whose exact values we know. To that end, the following sequence of questions centers on $\int_0^3 x^3 \, dx\text{.}$
1. Use the applet at http://gvsu.edu/s/a9 with the function $f(x) = x^3$ on the window of $x$ values from $0$ to $3$ to compute $L_3\text{,}$ the left Riemann sum with three subintervals.
$L_3=9$
Solution
In addition to changing $f(x)$ to $f(x)=x^3$ and the number of subintervals to $3 \text{,}$ make sure to change the Sample Point Placement so that "Relative" is checked and the slide bar is all the way to the left. You'll see that now it is sampling $f(x)$ values from the left side of each sub-interval.
\begin{equation*} L_3=9 \end{equation*}
2. Likewise, use the applet to compute $R_3$ and $M_3\text{,}$ the right and middle Riemann sums with three subintervals, respectively.
$M_3=19.125 \ , R_3=36$
Solution
First change $f(x)$ and the number of subintervals to $f(x)=x^3$ and $n=3\text{.}$ For $R_3\text{,}$ make sure to change the Sample Point Placement so that "Relative" is checked and the slide bar is all the way to the right. You'll see that now it is sampling $f(x)$ values from the right side of each subinterval. For $M_3\text{,}$ make sure to change the Sample Point Placement so that "Relative" is checked and the slide bar is in the middle and the "midpoint" label shows. You'll see that now it is sampling $f(x)$ values from the middle of each interval.
\begin{equation*} M_3=19.125 \ , R_3=36 \end{equation*}
3. Use the Fundamental Theorem of Calculus to compute the exact value of $I = \int_0^3 x^3 \, dx\text{.}$
Solution
\begin{align*} I&=\int_0^3 x^3 \, dx \\ &= \left.\frac{1}{4}x^4 \right|_{0}^{3} \\ &=\frac{3^4}{4}-\frac{0^4}{4}\\ &=20.25 \end{align*}
4. We define the error in an approximation of a definite integral to be the difference between the integral's exact value and the approximation's value. What is the error that results from using $L_3\text{?}$ From $R_3\text{?}$ From $M_3\text{?}$
\begin{equation*} E_{L,3}= 11.25 \end{equation*}
\begin{equation*} E_{R,3}=-15.75 \end{equation*}
\begin{equation*} E_{M,3}=1.125 \end{equation*}
Solution
Denote the error from each approximation $E_{L,3}, E_{R,3}, \text{ and } E_{M,3}\text{.}$ Then
\begin{equation*} E_{L,3}=I-L_3=20.25-9 =11.25 \end{equation*}
\begin{equation*} E_{R,3}=I-R_3=20.25-36=-15.75 \end{equation*}
\begin{equation*} E_{M,3}=I-M_3=20.25-19.125=1.125 \end{equation*}
5. In what follows in this section, we will learn a new approach to estimating the value of a definite integral known as the Trapezoid Rule. The basic idea is to use trapezoids, rather than rectangles, to estimate the area under a curve. What is the formula for the area of a trapezoid with bases of length $b_1$ and $b_2$ and height $h\text{?}$
Solution
The area of a trapezoid with bases of length $b_1$ and $b_2$ and height $h$ is
\begin{equation*} A=\frac{h(b_1+b_2)}{2}\text{.} \end{equation*}
6. Working by hand, estimate the area under $f(x) = x^3$ on $[0,3]$ using three subintervals and three corresponding trapezoids. What is the error in this approximation? How does it compare to the errors you calculated in (d)?
\begin{equation*} T_3=\frac{45}{2}=22.5 \end{equation*}
\begin{equation*} E_{T,3}=-2.25 \end{equation*}
Solution
The left-most trapezoid has base lengths 0 and 1 and height 1, so the area of the first trapezoid is $\frac{1}{2} (0+1)\cdot 1=\frac{1}{2}\text{.}$ The middle trapezoid has base lengths 1 and 8 and height 1, so the area of the second trapezoid is $\frac{1}{2}(1+8) \cdot 1=\frac{9}{2} \text{.}$ The right-most trapezoid has base lengths 8 and 27 and height 1, so the area of the third trapezoid is $\frac{1}{2}(8+27)\cdot 1=\frac{35}{2}\text{.}$ Therefore, the approximate area under the graph of $f(x)=x^3$ from $0$ to $3$ using the trapezoid rule with 3 subintervals is $T_3= \frac{1}{2} + \frac{9}{2} + \frac{35}{2}=\frac{45}{2}=22.5\text{.}$
The error is $E_{T,3}=20.25-22.5=-2.25 \text{.}$ Using trapezoids creates a smaller error compared to $L_3$ and $R_3 \text{.}$ The magnitude of the error from approximating using $M_3$ is half the magnitude of the error from approximating using $T_3 \text{,}$ but they have opposite signs.
### SubsectionThe Trapezoid Rule
So far, we have used the simplest possible quadrilaterals (that is, rectangles) to estimate areas. It is natural, however, to wonder if other familiar shapes might serve us even better.
An alternative to $L_n\text{,}$ $R_n\text{,}$ and $M_n$ is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by $y = f(x)$ on a small interval, we use a trapezoid. For example, in Figure5.80, we estimate the area under the curve using three subintervals and the trapezoids that result from connecting the corresponding points on the curve with straight lines.
The biggest difference between the Trapezoid Rule and a Riemann sum is that on each subinterval, the Trapezoid Rule uses two function values, rather than one, to estimate the (signed) area bounded by the curve. For instance, to compute $D_1\text{,}$ the area of the trapezoid on $[x_0, x_1]\text{,}$ we observe that the left base has length $f(x_0)\text{,}$ while the right base has length $f(x_1)\text{.}$ The height of the trapezoid is $x_1 - x_0 = \Delta x = \frac{b-a}{3}\text{.}$ The area of a trapezoid is the average of the bases times the height, so we have
\begin{equation*} D_1 = \frac{1}{2}(f(x_0) + f(x_1)) \cdot \Delta x\text{.} \end{equation*}
Using similar computations for $D_2$ and $D_3\text{,}$ we find that $T_3\text{,}$ the trapezoidal approximation to $\int_a^b f(x) \, dx$ is given by
\begin{align*} T_3 &= D_1 + D_2 + D_3\\ &= \frac{1}{2}(f(x_0) + f(x_1)) \cdot \Delta x + \frac{1}{2}(f(x_1) + f(x_2)) \cdot \Delta x + \frac{1}{2}(f(x_2) + f(x_3)) \cdot \Delta x\text{.} \end{align*}
Because both left and right endpoints are being used, we recognize within the trapezoidal approximation the use of both left and right Riemann sums. Rearranging the expression for $T_3$ by removing factors of $\frac{1}{2}$ and $\Delta x \text{,}$ grouping the left endpoint and right endpoint evaluations of $f\text{,}$ we see that
$$T_3 = \frac{1}{2} \left[ f(x_0) + f(x_1) + f(x_2) \right] \Delta x + \frac{1}{2} \left[ f(x_1) + f(x_2) + f(x_3) \right] \Delta x\text{.}\label{E-Trap3}\tag{5.18}$$
We now observe that two familiar sums have arisen. The left Riemann sum $L_3$ is $L_3 = f(x_0) \Delta x + f(x_1) \Delta x + f(x_2) \Delta x\text{,}$ and the right Riemann sum is $R_3 = f(x_1) \Delta x + f(x_2) \Delta x + f(x_3) \Delta x\text{.}$ Substituting $L_3$ and $R_3$ for the corresponding expressions in Equation(5.18), it follows that $T_3 = \frac{1}{2} \left[ L_3 + R_3 \right]\text{.}$ We have thus seen a very important result: using trapezoids to estimate the (signed) area bounded by a curve is the same as averaging the estimates generated by using left and right endpoints.
###### The Trapezoid Rule
The trapezoidal approximation, $T_n\text{,}$ of the definite integral $\int_a^b f(x) \, dx$ using $n$ subintervals is given by the rule
\begin{align*} T_n =\mathstrut \amp \left[\frac{1}{2}(f(x_0) + f(x_1)) + \frac{1}{2}(f(x_1) + f(x_2)) + \cdots + \frac{1}{2}(f(x_{n-1}) + f(x_n)) \right] \Delta x.\\ =\mathstrut \amp \sum_{i=0}^{n-1} \frac{1}{2}(f(x_i) + f(x_{i+1})) \Delta x\text{.} \end{align*}
Moreover, $T_n = \frac{1}{2} \left[ L_n + R_n \right]\text{.}$
###### Example5.81
In this example, we explore the relationships among the errors generated by left, right, midpoint, and trapezoid approximations to the definite integral $\int_1^2 \frac{1}{x^2} \, dx$
1. Use the First FTC to evaluate $\int_1^2 \frac{1}{x^2} \, dx$ exactly.
2. Use appropriate computing technology to compute the following approximations for $\int_1^2 \frac{1}{x^2} \, dx\text{:}$ $T_4\text{,}$ $M_4\text{,}$ $T_8\text{,}$ and $M_8\text{.}$
3. Recall that the error of an approximation is the difference between the exact value of the definite integral and the resulting approximation. For instance, if we let $E_{T,4}$ represent the error that results from using the trapezoid rule with 4 subintervals to estimate the integral, we have
\begin{equation*} E_{T,4} = \int_1^2 \frac{1}{x^2} \, dx - T_4\text{.} \end{equation*}
Similarly, we compute the error of the midpoint rule approximation with 8 subintervals by the formula
\begin{equation*} E_{M,8} = \int_1^2 \frac{1}{x^2} \, dx - M_8\text{.} \end{equation*}
Based on your work in (a) and (b) above, compute $E_{T,4}\text{,}$ $E_{T,8}\text{,}$ $E_{M,4}\text{,}$ $E_{M,8}\text{.}$
4. Which rule consistently over-estimates the exact value of the definite integral? Which rule consistently under-estimates the definite integral?
5. What behavior(s) of the function $f(x) = \frac{1}{x^2}$ lead to your observations in (d)?
Hint
1. $\frac{1}{x^2} = x^{-2}\text{.}$
2. Use a computational device.
3. Use a computational device.
4. Which estimate is larger than the true value of the definite integral?
5. Note that how the curve bends makes a big difference in whether the trapezoid rule over- or under-estimates the value of the definite integral.
1. $\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}$
2. The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.
$n$ $T_n$ $E_{T,n}$ $4$ $0.50899$ $-0.50899$ $8$ $0.50227$ $-0.50227$ $16$ $0.50057$ $-0.50057$
3. The table below gives values of the midpoint rule and corresponding errors for different $n$-values.
$n$ $M_n$ $E_{M,n}$ $4$ $0.49555$ $0.00445$ $8$ $0.49887$ $0.00113$ $16$ $0.49972$ $0.00028$
4. The trapezoid rule overestimates; the midpoint rule underestimates.
5. $f(x) = \dfrac{1}{x^2}$ is concave up on $[1, 2]\text{.}$
Solution
1. $\int_1^2 \dfrac{1}{x^2} dx = \left. -x^{-1} \right|_1^2 = -\dfrac{1}{2} + 1 = \dfrac{1}{2}\text{.}$
2. The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.
$n$ $T_n$ $E_{T,n}$ $4$ $0.50899$ $-0.50899$ $8$ $0.50227$ $-0.50227$ $16$ $0.50057$ $-0.50057$
3. The table below gives values of the midpoint rule and corresponding errors for different $n$-values.
$n$ $M_n$ $E_{M,n}$ $4$ $0.49555$ $0.00445$ $8$ $0.49887$ $0.00113$ $16$ $0.49972$ $0.00028$
4. From the errors in comparison to the known exact value, we see that the trapezoid rule overestimates this definite integral and the midpoint rule underestimates this definite integral.
5. The graph of the function given by $f(x) = \dfrac{1}{x^2}$ is concave up on the interval $[1, 2]\text{.}$ Because of this fact, we can see graphically that the line forming the top of each trapezoid lies fully above the curve, and thus the trapezoid rule overestimates the true value of the definite integral. Later in this section we'll see graphically why this concavity makes the midpoint rule an underestimate.
### SubsectionComparing the Midpoint and Trapezoid Rules
We know from the definition of the definite integral that if we let $n$ be large enough, we can make any of the approximations $L_n\text{,}$ $R_n\text{,}$ and $M_n$ as close as we'd like (in theory) to the exact value of $\int_a^b f(x) \, dx\text{.}$ Thus, it may be natural to wonder why we ever use any rule other than $L_n$ or $R_n$ (with a sufficiently large $n$ value) to estimate a definite integral. One of the primary reasons is that as $n \to \infty\text{,}$ $\Delta x = \frac{b-a}{n} \to 0\text{,}$ and thus in a Riemann sum calculation with a large $n$ value, we end up multiplying by a number that is very close to zero. Doing so often generates roundoff error, because representing numbers close to zero accurately is a persistent challenge for computers.
Hence, we explore ways to estimate definite integrals to high levels of precision, but without using extremely large values of $n\text{.}$ Paying close attention to patterns in errors, such as those observed in Example5.81, is one way to begin to see some alternate approaches.
To begin, we compare the errors in the Midpoint and Trapezoid rules. First, consider a function that is concave up on a given interval, and picture approximating the area bounded on that interval by both the Midpoint and Trapezoid rules using a single subinterval.
As seen in Figure5.82, it is evident that whenever the function is concave up on an interval, the Trapezoid Rule with one subinterval, $T_1\text{,}$ will overestimate the exact value of the definite integral on that interval. From a careful analysis of the line that bounds the top of the rectangle for the Midpoint Rule (shown in magenta), we see that if we rotate this line segment until it is tangent to the curve at the midpoint of the interval (as shown at right in Figure5.82), the resulting trapezoid has the same area as $M_1\text{,}$ and this value is less than the exact value of the definite integral. Thus, when the function is concave up on the interval, $M_1$ underestimates the integral's true value.
The preceding discussion explores how to determine if $M_n$ and $T_n$ are underestimates or overestimates of $\int_a^b f(x) \, dx$ if $f(x)$ is concave up. Similar statements can be made when $f(x)$ is concave down. These are summarized next.
###### Error types when approximating using Midpoint and Trapezoid Rules
• If $f(x)$ is concave up on $[a,b] \text{,}$ then $M_n$ underestimates $\int_a^b f(x) \, dx$ and $T_n$ overestimates $\int_a^b f(x) \, dx\text{.}$
• If $f(x)$ is concave down on $[a,b] \text{,}$ then $M_n$ overestimates $\int_a^b f(x) \, dx$ and $T_n$ underestimates $\int_a^b f(x) \, dx\text{.}$
Next, we compare the size of the errors between $M_n$ and $T_n\text{.}$ Again, we focus on $M_1$ and $T_1$ on an interval where the concavity of $f$ is consistent. In Figure5.83, where the error of the Trapezoid Rule is shaded in red, while the error of the Midpoint Rule is shaded lighter red, it is visually apparent that the error in the Trapezoid Rule is more significant. To see how much more significant, let's consider two examples and some particular computations.
If we let $f(x) = 1-x^2$ and consider $\int_0^1 f(x) \,dx\text{,}$ we know by the First FTC that the exact value of the integral is
\begin{equation*} \int_0^1 (1-x^2) \, dx = x - \frac{x^3}{3} \bigg\vert_0^1 = \frac{2}{3}\text{.} \end{equation*}
Using appropriate technology to compute $M_4\text{,}$ $M_8\text{,}$ $T_4\text{,}$ and $T_8\text{,}$ as well as the corresponding errors $E_{M,4}\text{,}$ $E_{M,8}\text{,}$ $E_{T,4}\text{,}$ and $E_{T,8}\text{,}$ as we did in Example5.81, we find the results summarized in Table5.84. We also include the approximations and their errors for the example $\int_1^2 \frac{1}{x^2} \, dx$ from Example5.81.
For a given function $f$ and interval $[a,b]\text{,}$ $E_{T,4} = \int_a^b f(x) \,dx - T_4$ calculates the difference between the exact value of the definite integral and the approximation generated by the Trapezoid Rule with $n = 4\text{.}$ If we look at not only $E_{T,4}\text{,}$ but also the other errors generated by using $T_n$ and $M_n$ with $n = 4$ and $n = 8$ in the two examples noted in Table5.84, we see an evident pattern. Not only is the sign of the error (which measures whether the rule generates an over- or under-estimate) tied to the rule used and the function's concavity, but the magnitude of the errors generated by $T_n$ and $M_n$ seems closely connected. In particular, the errors generated by the Midpoint Rule seem to be about half the size of those generated by the Trapezoid Rule.
That is, we can observe in both examples that $E_{M,4} \approx -\frac{1}{2} E_{T,4}$ and $E_{M,8} \approx -\frac{1}{2}E_{T,8}\text{.}$ This property of the Midpoint and Trapezoid Rules turns out to hold in general: for a function of consistent concavity, the error in the Midpoint Rule has the opposite sign and approximately half the magnitude of the error of the Trapezoid Rule. Written symbolically,
\begin{equation*} E_{M,n} \approx -\frac{1}{2} E_{T,n}\text{.} \end{equation*}
This important relationship suggests a way to combine the Midpoint and Trapezoid Rules to create an even more accurate approximation to a definite integral.
### SubsectionSimpson's Rule
When we first developed the Trapezoid Rule, we observed that it is an average of the Left and Right Riemann sums:
\begin{equation*} T_n = \frac{1}{2}(L_n + R_n)\text{.} \end{equation*}
If a function is always increasing or always decreasing on the interval $[a,b]\text{,}$ one of $L_n$ and $R_n$ will over-estimate the true value of $\int_a^b f(x) \, dx\text{,}$ while the other will under-estimate the integral. Thus, the errors found in $L_n$ and $R_n$ will have opposite signs; so averaging $L_n$ and $R_n$ eliminates a considerable amount of the error present in the respective approximations. In a similar way, it makes sense to think about averaging $M_n$ and $T_n$ in order to generate a still more accurate approximation.
We've just observed that $M_n$ is typically about twice as accurate as $T_n\text{.}$ This leads to an approximation method known as Simpson's Rule 14Thomas Simpson was an 18th century mathematician; his idea was to extend the Trapezoid rule, but rather than using straight lines to build trapezoids, to use quadratic functions to build regions whose area was bounded by parabolas (whose areas he could find exactly). Simpson's Rule is often developed from the more sophisticated perspective of using interpolation by quadratic functions. which is a weighted average of the Midpoint and Trapezoid approximations.
###### Simpson's Rule
The Simpson's Rule approximation $S_{2n}$ of the area $\int_a^b f(x) \ dx$ is the weighted average
$$S_{2n} = \frac{2M_n + T_n}{3}\text{.}\label{E-Simpson}\tag{5.19}$$
where $M_n$ and $T_n$ are the Midpoint and Trapezoid rule approximations using $n$ subintervals.
Note that we use $S_{2n}$ rather that $S_n$ since the $n$ points the Midpoint Rule uses are different from the $n$ points the Trapezoid Rule uses, and thus Simpson's Rule is using $2n$ points at which to evaluate the function. We build upon the results in Table5.84 to see the approximations generated by Simpson's Rule. In particular, in Table5.85, we include all of the results in Table5.84, but include additional results for $S_8 = \frac{2M_4 + T_4}{3}$ and $S_{16} = \frac{2M_8 + T_8}{3}\text{.}$
The results seen in Table5.85 are striking. If we consider the $S_{16}$ approximation of $\int_1^2 \frac{1}{x^2} \, dx\text{,}$ the error is only $E_{S,16} = 0.0000019434\text{.}$ By contrast, $L_8 = 0.5491458502\text{,}$ so the error of that estimate is $E_{L,8} = -0.0491458502\text{.}$ Moreover, we observe that generating the approximations for Simpson's Rule is almost no additional work: once we have $L_n\text{,}$ $R_n\text{,}$ and $M_n$ for a given value of $n\text{,}$ it is a simple exercise to generate $T_n\text{,}$ and from there to calculate $S_{2n}\text{.}$ Finally, note that the error in the Simpson's Rule approximations of $\int_0^1 (1-x^2) \, dx$ is zero!15Similar to how the Midpoint and Trapezoid approximations are exact for linear functions, Simpson's Rule approximations are exact for quadratic and cubic functions. See additional discussion on this issue later in the section and in the exercises.
These rules are not only useful for approximating definite integrals such as $\int_0^1 e^{-x^2} \, dx\text{,}$ for which we cannot find an elementary antiderivative of $e^{-x^2}\text{,}$ but also for approximating definite integrals when we are given a function through a table of data.
###### Example5.86
A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. Assume that $v$ is continuous, always decreasing, and always decreasing at a decreasing rate, as is suggested by the data.
seconds, $t$ Velocity in ft/sec, $v(t)$ $0$ $100$ $0.3$ $99$ $0.6$ $96$ $0.9$ $90$ $1.2$ $80$ $1.5$ $50$ $1.8$ $0$
1. Plot the given data on the set of axes provided in Figure5.88 with time on the horizontal axis and the velocity on the vertical axis.
2. What definite integral will give you the exact distance the car traveled on $[0,1.8]\text{?}$
3. Estimate the total distance traveled on $[0,1.8]$ by computing $L_3\text{,}$ $R_3\text{,}$ and $T_3\text{.}$ Which of these under-estimates the true distance traveled?
4. Estimate the total distance traveled on $[0,1.8]$ by computing $M_3\text{.}$ Is this an over- or under-estimate? Why?
5. Using your results from (c) and (d), improve your estimate further by using Simpson's Rule.
6. What is your best estimate of the average velocity of the car on $[0,1.8]\text{?}$ Why? What are the units on this quantity?
Hint
1. Plot the data.
2. What are the units of $v(t) \cdot \Delta t\text{?}$
3. Recall the standard rules for sums that produce $L_3\text{,}$ $R_3\text{,}$ $T_3\text{.}$
4. Think about concavity to decide if $M_3$ is an over- or under-estimate.
5. Recall how $S_3$ is a weighted average of $T_3$ and $M_3\text{.}$
6. Simpson's Rule gives the best estimate for a function of consistent concavity.
1. Plot the data.
2. $\int_0^{1.8} v(t) dt\text{.}$
3. \begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft }\text{.} \end{align*}
$R_3$ and $T_3$ are underestimates.
4. $M_3 = 143.4 \text{ ft }$ ; overestimate.
5. $S_6 = 140.8 \text{ ft } \text{.}$
6. Simpson's rule gives the best approximation of the distance traveled, $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$
Solution
1. Plot the data.
2. Since the velocity is always positive, the definite integral that will give the exact distance traveled by the car on the interval $[0, 1.8]$ is
\begin{equation*} \int_0^{1.8} v(t) dt\text{.} \end{equation*}
3. The estimates of $\int_0^{1.8} v(t) dt$ are
\begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft }\text{.} \end{align*}
$R_3$ is an underestimate of the distance traveled since $v(t)$ is decreasing. $T_3$ is an underestimate of the distance traveled since $v(t)$ is concave down.
4. Another estimate of the distance traveled is
\begin{equation*} M_3 = 143.4 \text{ ft }\text{.} \end{equation*}
This is an overestimate since $v(t)$ is concave down.
5. For Simpson's Rule, we see that
\begin{equation*} S_6 = \frac{2}{3}M_3 + \frac{1}{3}T_3 = 140.8 \text{ ft }\text{.} \end{equation*}
6. Simpson's rule gives the best approximation of the distance traveled since it is a weighted average of the midpoint and trapezoid rules and uses more information about the velocity than the other methods. The units on each of the estimates, including Simpson's Rule, are "feet", since ft/sec $\cdot$ sec = ft. Thus, the best approximation we have generated is that $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$
### SubsectionComparing $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ $M_n\text{,}$ and $S_{2n}\text{.}$
As we conclude our discussion of numerical approximation of definite integrals, it is important to summarize general trends in how the various rules over- or under-estimate the true value of a definite integral, and by how much. To revisit some past observations and see some new ones, we consider the following example.
###### Example5.89
Consider the functions $f(x) = 2-x^2\text{,}$ $g(x) = 2-x^3\text{,}$ and $h(x) = 2-x^4\text{,}$ all on the interval $[0,1]\text{.}$ For each of the questions that require a numerical answer in what follows, write your answer exactly in fraction form.
1. On the three sets of axes provided in Figure5.90, sketch a graph of each function on the interval $[0,1]\text{,}$ and compute $L_1$ and $R_1$ for each. What do you observe?
2. Compute $M_1$ for each function to approximate $\int_0^1 f(x) \,dx\text{,}$ $\int_0^1 g(x) \,dx\text{,}$ and $\int_0^1 h(x) \,dx\text{,}$ respectively.
3. Compute $T_1$ for each of the three functions, and hence compute $S_2$ for each of the three functions.
4. Evaluate each of the integrals $\int_0^1 f(x) \,dx\text{,}$ $\int_0^1 g(x) \,dx\text{,}$ and $\int_0^1 h(x) \,dx$ exactly using the First FTC.
5. For each of the three functions $f\text{,}$ $g\text{,}$ and $h\text{,}$ compare the results of $L_1\text{,}$ $R_1\text{,}$ $M_1\text{,}$ $T_1\text{,}$ and $S_2$ to the true value of the corresponding definite integral. What patterns do you observe?
Hint
1. For each estimate, just one function evaluation is needed.
2. Use the midpoint rule with $n=1\text{.}$
3. Remember that both the trapezoid and Simpson's rule can be executed using (weighted) averages of known values.
4. Find antiderivatives to evaluate the integrals exactly.
5. Think about trends in over- and under-estimates.
1. For $L_1$ and $T_1\text{:}$
The values of $L_1$ and $R_1$ are the same for all three.
2. For the $M_1\text{,}$
3. For $T_1$ and $S_2\text{,}$
4. \begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5} \end{align*}
5. Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson's rule is exact for both $f$ and $g\text{,}$ while a slight overestimate of $\int_0^1 h(x) dx\text{.}$
Solution
1. For the left and right endpoint rules, we see that
Thus, we observe that despite the fact the functions are all different, the values of $L_1$ and $R_1$ are the same for all three.
2. For the midpoint rule, we find that
3. For the trapezoid rule and Simpson's rule,
4. The exact values of the three definite integrals are
\begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5}\\ \amp \approx 1.6667 \amp \amp = 1.75 \amp \amp = 1.8 \end{align*}
5. We observe that each of the left endpoint rule results are overestimates, each of the right endpoint rules are underestimates, each of the midpoint rules are overestimates, and each of the trapezoid rules are underestimates. These results hold because each of the three functions are both decreasing and concave down. For Simpson's rule, we see that the result is exact for both $f$ and $g\text{,}$ while Simpson's rule is a slight overestimate of $\int_0^1 h(x) dx\text{.}$
The results seen in Example5.89 generalize nicely. For instance, if $f$ is decreasing on $[a,b]\text{,}$ $L_n$ will overestimate the exact value of $\int_a^b f(x) \,dx\text{,}$ and if $f$ is concave down on $[a,b]\text{,}$ $M_n$ will overestimate the exact value of the integral. An excellent exercise is to write a collection of scenarios of possible function behavior, and then categorize whether each of $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ and $M_n$ is an over- or under-estimate.
Finally, we make two important notes about Simpson's Rule. When T.Simpson first developed this rule, his idea was to replace the function $f$ on a given interval with a quadratic function that shared three values with the function $f\text{.}$ In so doing, he guaranteed that this new approximation rule would be exact for the definite integral of any quadratic polynomial. In one of the pleasant surprises of numerical analysis, it turns out that even though it was designed to be exact for quadratic polynomials, Simpson's Rule is exact for any cubic polynomial: that is, if we are interested in an integral such as $\int_2^5 (5x^3 - 2x^2 + 7x - 4)\, dx\text{,}$ the approximation $S_{2n}$ will always be exact, regardless of the value of $n\text{.}$ This is just one more piece of evidence that shows how effective Simpson's Rule is as an approximation tool for estimating definite integrals.16One reason that Simpson's Rule is so effective is that $S_{2n}$ benefits from using $2n+1$ points of data. Because it combines $M_n\text{,}$ which uses $n$ midpoints, and $T_n\text{,}$ which uses the $n+1$ endpoints of the chosen subintervals, $S_{2n}$ takes advantage of the maximum amount of information we have when we know function values at the endpoints and midpoints of $n$ subintervals.
### SubsectionSummary
• For a definite integral such as $\int_0^1 e^{-x^2} \, dx$ when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative, we can estimate the integral's value by using a sequence of Riemann sum approximations. Typically, we start by computing $L_n\text{,}$ $R_n\text{,}$ and $M_n$ for one or more chosen values of $n\text{.}$
• The Trapezoid Rule, which estimates $\int_a^b f(x) \, dx$ by using trapezoids, rather than rectangles, can also be viewed as the average of Left and Right Riemann sums. That is, $T_n = \frac{1}{2}(L_n + R_n)\text{.}$
• The Midpoint Rule is typically twice as accurate as the Trapezoid Rule, and the signs of the respective errors of these rules are opposites. Hence, by taking the weighted average $S_{2n} = \frac{2M_n + T_n}{3}\text{,}$ we can build a much more accurate approximation to $\int_a^b f(x) \, dx$ by using approximations we have already computed. The rule for $S_{2n}$ is known as Simpson's Rule, which can also be developed by approximating a given continuous function with pieces of quadratic polynomials.
• By understanding the approximation rule chosen, and the properties of the function (i.e. whether it is increasing or decreasing, or concave up or down on the interval in question), we can say whether the approximation is an over or under-estimate of the actual value.
### SubsectionExercises
Consider the definite integral $\int_0^1 x \tan(x) \, dx\text{.}$
1. Explain why this integral cannot be evaluated exactly by using either $u$-substitution or by integrating by parts.
2. Using appropriate subintervals, compute $L_4\text{,}$ $R_4\text{,}$ $M_4\text{,}$ $T_4\text{,}$ and $S_8\text{.}$
3. Which of the approximations in (b) is an over-estimate to the true value of $\int_0^1 x \tan(x) \, dx\text{?}$ Which is an under-estimate? How do you know?
For an unknown function $f(x)\text{,}$ the following information is known.
• $f$ is continuous on $[3,6]\text{;}$
• $f$ is either always increasing or always decreasing on $[3,6]\text{;}$
• $f$ has the same concavity throughout the interval $[3,6]\text{;}$
• As approximations to $\int_3^6 f(x) \, dx\text{,}$ $L_4 = 7.23\text{,}$ $R_4 = 6.75\text{,}$ and $M_4 = 7.05\text{.}$
1. Is $f$ increasing or decreasing on $[3,6]\text{?}$ What data tells you?
2. Is $f$ concave up or concave down on $[3,6]\text{?}$ Why?
3. Determine the best possible estimate you can for $\int_3^6 f(x) \, dx\text{,}$ based on the given information.
The rate at which water flows through Table Rock Dam on the White River in Branson, MO, is measured in thousands of cubic feet per second (TCFS). As engineers open the floodgates, flow rates are recorded according to the following chart.
1. What definite integral measures the total volume of water to flow through the dam in the 60 second time period provided by the table above?
2. Use the given data to calculate $M_n$ for the largest possible value of $n$ to approximate the integral you stated in (a). Do you think $M_n$ over- or under-estimates the exact value of the integral? Why?
3. Approximate the integral stated in (a) by calculating $S_n$ for the largest possible value of $n\text{,}$ based on the given data.
4. Compute $\frac{1}{60} S_n$ and $\frac{2000+2100+2400+3000+3900+5100+6500}{7}\text{.}$ What quantity do both of these values estimate? Which is a more accurate approximation? | 2020-07-13T21:10:07 | {
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https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_5&diff=next&oldid=149207 | # Difference between revisions of "2020 AIME I Problems/Problem 5"
## Problem
Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.
## Solution 1
Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.
If we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \cdot 5 = 30$. However, we overcount some cases. Take the example of $132456$. We overcount this case because we can remove the $3$ or the $2$. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$,) to get $30-5=25$, but we have to add back one more for the original case, $123456$. Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $\boxed{052}$.
~molocyxu
## Solution 2 (Inspired by 2018 CMIMC Combo Round)
Similar to above, a $1-1$ correspondence between ascending and descending is established by subtracting each number from $7$.
We note that the given condition is equivalent to "cycling" $123456$ for a contiguous subset of it. For example,
$12(345)6 \rightarrow 125346, 124536$
It's not hard to see that no overcount is possible, and that the cycle is either $1$ "right" or $1$ "left." Therefore, we consider how many elements we flip by. If we flip $1$ or $2$ such elements, then there is one way to cycle them. Otherwise, we have $2$ ways. Therefore, the total number of ascending is $1 + 5 + 2(4 + 3 + 2 + 1) = 26$, and multiplying by two gives $\boxed{052}.$
~awang11
## Solution 3
Similarly to above, we find the number of ascending arrangements and multiply by 2.
We can choose $5$ cards to be the ascending cards, therefore leaving $6$ places to place the remaining card. There are $\binom{6}{5}\cdot 6=36$ to do this. However, since the problem is asking for the number of arrangements, we overcount cases such as $123456$. Notice that the only arrangements that overcount are $123456$ (case 1) or if two adjacent numbers of $123456$ are switched (case 2).
$\text{Case 1: }$ This arrangement is counted $6$ times. Each time it is counted for any of the $5$ numbers selected. Therefore we need to subtract $5$ cases of overcounting.
$\text{Case 2: }$ Each time $2$ adjacent numbers of switched, there is one overcount. For example, if we have $213456$, both $1$ or $2$ could be removed. Since there are $5$ possible switches, we need to subtract $5$ cases of overcounting.
Therefore, we have $36-5-5=26$ total arrangements of ascending numbers. We multiply by two (for descending) to get the answer of $\boxed{052}.$
~PCChess
## Solution 4 (No Overcounting)
Like in previous solutions, we will count the number of ascending arrangements and multiply by 2.
First, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works.
Next, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements.
Now, we can "cycle" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements.
Cycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works,) 3-4-1-2 (which doesn't work,) and 4-1-2-3 (which does work.) We get 6 arrangements.
Similarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements.
Adding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer $\boxed{052}.$
~i8Pie
## Solution 5 (Official MAA 1)
First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is $1$ such permutation where all the cards appear in order: $123456.$ There are $5$ such permutations where two adjacent cards are interchanged, as in $124356.$ The other such permutations arise from removing one card from $123456$ and placing it in a position at least two away from its starting location. There are $4$ such positions to place each of the cards numbered $1$ and $6,$ and $3$ such positions for each of the cards numbered $2, 3, 4,$ and $5.$ This accounts for $2\cdot4 + 4\cdot3 =20$ permutations. Thus there are $1 + 5 + 20 = 26$ permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total $26 + 26 = \boxed{52}$.
## Solution 6 (Official MAA 2)
More generally, suppose there are $n \geq 4$ cards numbered $1, 2, 3, \dots, n$ arranged in ascending order. If any one of the $n$ cards is removed and placed in one of the $n$ positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for $n\cdot n = n^2$ permutations. However, the original ascending order has been counted $n$ times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is $n^2-(n-1)-(n-1)=(n-1)^2+1.$ When $n = 6$, this is $(6-1)^2+1 = 26$, and the final answer is $2\cdot26 = \boxed{52}$.
## Solution 7 (Casework)
For ascending, if the $1$ goes in anything but the first two slots, the rest of the numbers have to go in ascending from $2$, which are $4$ cases if there are $6$ cards. If $1$ goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of $6$ cards, that gives $5$ more. If $1$ goes in the first slot, that means that you are doing the same problem with $n-1$ cards. So the recursion is $a_n=(n-2)+(n-1)+a_{n-1}$. There's $a_1=1$ and $a_2=2$, so you get $a_3=2+3=5$, $a_4=5+5=10$, $a_5=7+10=17$, and $a_6=9+17=26$. Or you can see that $a_n=(n-1)^2+1$. We double to account for descending and get $\boxed{052}$.
~ahclark11
## Solution 8 (Symmetry and Case Study)
First, we know that ascending order and descending order are symmetrical to each other (namely, if we get 132456 where after we take out 3, it will be one scenario; and if we flip it and write 654231, it will be another scenario)
Thus, we only need to consider either descending or ascending and then times 2. WLOG let us consider ascending order
Case 1: after we take out 1, the rest will be in ascending order
2 3 4 5 6
1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios.
Case 2: after we take out 2, the rest will be in ascending order
1 3 4 5 6
notice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2
It is easy to see that this is the same for 3, 4, 5, and 6.
Thus, in total, we have $$(6+4\times5)\times2=\boxed{052}$$ ~Adali | 2021-10-19T20:49:14 | {
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https://math.stackexchange.com/questions/1761557/prove-fracacbd-lies-between-fracab-and-fraccd-for-posit | # Prove: $\frac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$ (for positive $a$, $b$, $c$, $d$)
I am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.
Would be grateful for any suggestions!
• Not to be picky, but if we consider the fractions $a/b,c/d$ with $a=b=c=1$ and $d=-2$, then $(a+c)/(b+d)=-2$, which does not lie between $a/b=1$ and $c/d=-1/2$.... – Barry Cipra Apr 27 '16 at 18:55
• Good point Barry. Let's say that a, b, c and d are all positive integers. – Joeywald Apr 27 '16 at 19:24
• mathsguy, not a lot as "proof" is new to me. I know that adding the same number to the top and bottom of a fraction moves the derived fraction closer to 1 and that adding two fractions as above gives a resultant fraction between the two, but I can't prove why this happens. – Joeywald Apr 27 '16 at 19:33
Suppose we have positive $a,b,c,d$ with $\frac{a}{b}\ge \frac{c}{d}$. Then multiplying through by $bd$ we get $ad\ge bc$. Adding $ab$ to both sides we get $a(b+d)\ge b(a+c)$. Dividing by $b(b+d)$, we get $\frac{a}{b}\ge\frac{a+c}{b+d}$.
Similarly, add $cd$ to both sides of $ad\ge bc$ to get $d(a+c)\ge c(b+d)$. Dividing by $d(b+d)$ we get $\frac{a+c}{b+d}\ge \frac{c}{d}$.
Assuming $a,b,c,d\gt0$, we have
\begin{align} {a\over b}\lt{a+c\over b+d}\lt{c\over d}&\iff a(b+d)\lt b(a+c)\quad\text{and}\quad(a+c)d\lt(b+d)c\\ &\iff ad\lt bc\quad\text{and}\quad ad\lt bc\\ &\iff ad\lt bc\\ &\iff {a\over b}\lt{c\over d} \end{align}
• One usually starts with the premise and ends with the conclusion... No? – Najib Idrissi Jun 7 '16 at 17:38
• @NajibIdrissi If it's $“\Leftrightarrow”$ signs the whole way through, it shouldn't matter. – Akiva Weinberger Jun 7 '16 at 17:47
• @AkivaWeinberger Sure, but I think it's easier to read and looks more sound if you put it in the other order. But of course from a logical point of view, equivalence of propositions is symmetrical... – Najib Idrissi Jun 7 '16 at 18:09
Here is another way that uses calculus:
Let $\phi(t) = {(1-t)a+t c \over (1-t)b + t d}$ and note that $\phi(0) = {a \over b}, \phi(1) = {c \over d}$. Furthermore, $\phi'(t) = {bc-ad \over ((1-t)b + t d )^2}$. Since $ad < bc$, we see that $\phi$ is increasing, and so $\phi(0) \le \phi({1 \over 2}) \le \phi(1)$. Since $\phi({1 \over 2}) = {a+c \over b+d}$, we have the desired result.
Assume that $b,d >0$. Note that $$\frac{a+c}{b+d} = \frac{b}{b+d}\frac{a}{b} +\frac{d}{b+d}\frac{c}{d}.$$ Remark that $0<\frac{b}{b+d}<1$ and the same for $\frac{d}{b+d}$. Hence you have written $\frac{a+b}{c+d}$ as a convex combination of $\frac{a}{b}$ and $\frac{c}{d}$ so you get $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}.$$
Here's one way to look at it:
You're taking a class. Suppose you get $a$ points out of $b$ possible on Quiz 1, and $c$ points out of $d$ possible on Quiz 2.
Your overall points are $a+c$ out of $b+d$ possible. And your overall percentage should be between your lower quiz score and your higher quiz score.
• The question asks for "proof". – AlohaSine Apr 30 '16 at 23:54
• @MathematicsStudent1122 Actually the presenter asked for suggestions. One could use this idea as the basis for a proof. – paw88789 May 1 '16 at 2:49
Hint $\$ The mediant $(a\!+\!b)/(c\!+\!d)\,$ is the slope of the diagonal of the parallelogram with vector sides $(b,a),\ (d,c).\:$ But the slope of the diagonal lies between the slopes of the sides. | 2021-07-25T02:50:02 | {
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https://www.math.ubc.ca/~pwalls/math-python/integration/trapezoid-rule/ | # Trapezoid Rule
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
## Trapezoids
The definite integral of $f(x)$ is equal to the (net) area under the curve $y=f(x)$ over the interval $[a,b]$. Riemann sums approximate definite integrals by using sums of rectangles to approximate the area.
The trapezoid rule gives a better approximation of a definite integral by summing the areas of the trapezoids connecting the points
$$(x_{i-1},0), (x_i,0), (x_{i-1},f(x_{i-1})), (x_i,f(x_i))$$
for each subinterval $[x_{i-1},x_i]$ of a partition. Note that the area of each trapezoid is the sum of a rectangle and a triangle
$$(x_i - x_{i-1}) f(x_{i-1}) + \frac{1}{2}(x_i - x_{i-1}) (f(x_i) - f(x_{i-1})) = \frac{1}{2}(f(x_i) + f(x_{i-1}))(x_i - x_{i-1})$$
For example, we can use a single trapezoid to approximate:
$$\int_0^1 e^{-x^2} \, dx$$
First, let's plot the curve $y = e^{-x^2}$ and the trapezoid on the interval $[0,1]$:
x = np.linspace(-0.5,1.5,100)
y = np.exp(-x**2)
plt.plot(x,y)
x0 = 0; x1 = 1;
y0 = np.exp(-x0**2); y1 = np.exp(-x1**2);
plt.fill_between([x0,x1],[y0,y1])
plt.xlim([-0.5,1.5]); plt.ylim([0,1.5]);
plt.show()
Approximate the integral by the area of the trapezoid:
A = 0.5*(y1 + y0)*(x1 - x0)
print("Trapezoid area:", A)
Trapezoid area: 0.6839397205857212
## Definition
The trapezoid rule for $N$ subintervals of $[a,b]$ of equal length is
$$T_N(f) = \frac{\Delta x}{2} \sum_{i=1}^N (f(x_i) + f(x_{i-1}))$$
where $\Delta x = (b - a)/N$ is the length of the subintervals and $x_i = a + i \Delta x$.
Notice that the trapezoid is the average of the left and right Riemann sums
$$T_N(f) = \frac{\Delta x}{2} \sum_{i=1}^N (f(x_i) + f(x_{i-1})) = \frac{1}{2} \left( \sum_{i=1}^N f(x_i) \Delta x + \sum_{i=1}^N f(x_{i-1}) \Delta x \right)$$
## Error Formula
When computing integrals numerically, it is essential to know how good our approximations are. Notice in the theorem below that the error formula is inversely proportional to $N^2$. This means that the error decreases much faster with larger $N$ compared to Riemann sums.
Theorem. Let $T_N(f)$ denote the trapezoid rule
$$T_N(f) = \frac{\Delta x}{2} \sum_{i=1}^N (f(x_i) + f(x_{i-1}))$$
where $\Delta x = (b-a)/N$ and $x_i = a + i \Delta x$. The error bound is
$$E_N^T(f) = \left| \ \int_a^b f(x) \ dx - T_N(f) \ \right| \leq \frac{(b-a)^3}{12 N^2} K_2$$
where $\left| \ f''(x) \, \right| \leq K_2$ for all $x \in [a,b]$.
## Implementation
Let's write a function called trapz which takes input parameters $f$, $a$, $b$ and $N$ and returns the approximation $T_N(f)$. Furthermore, let's assign default value $N=50$.
def trapz(f,a,b,N=50):
'''Approximate the integral of f(x) from a to b by the trapezoid rule.
The trapezoid rule approximates the integral \int_a^b f(x) dx by the sum:
(dx/2) \sum_{k=1}^N (f(x_k) + f(x_{k-1}))
where x_k = a + k*dx and dx = (b - a)/N.
Parameters
----------
f : function
Vectorized function of a single variable
a , b : numbers
Interval of integration [a,b]
N : integer
Number of subintervals of [a,b]
Returns
-------
float
Approximation of the integral of f(x) from a to b using the
trapezoid rule with N subintervals of equal length.
Examples
--------
>>> trapz(np.sin,0,np.pi/2,1000)
0.9999997943832332
'''
x = np.linspace(a,b,N+1) # N+1 points make N subintervals
y = f(x)
y_right = y[1:] # right endpoints
y_left = y[:-1] # left endpoints
dx = (b - a)/N
T = (dx/2) * np.sum(y_right + y_left)
return T
Let's test our function on an integral where we know the answer
$$\int_0^{\pi/2} \sin x \ dx = 1$$
trapz(np.sin,0,np.pi/2,1000)
0.9999997943832332
Let's test our function again:
$$\int_0^1 3 x^2 \ dx = 1$$
trapz(lambda x : 3*x**2,0,1,10000)
1.0000000050000002
And once more:
$$\int_0^1 x \ dx = \frac{1}{2}$$
trapz(lambda x : x,0,1,1)
0.5
## scipy.integrate.trapz
The SciPy subpackage scipy.integrate contains several functions for approximating definite integrals and numerically solving differential equations. Let's import the subpackage under the name spi.
import scipy.integrate as spi
The function scipy.integrate.trapz computes the approximation of a definite by the trapezoid rule. Consulting the documentation, we see that all we need to do it supply arrays of $x$ and $y$ values for the integrand and scipy.integrate.trapz returns the approximation of the integral using the trapezoid rule. The number of points we give to scipy.integrate.trapz is up to us but we have to remember that more points gives a better approximation but it takes more time to compute!
## Examples
### Arctangent
Let's plot the trapezoids for $\displaystyle f(x)=\frac{1}{1 + x^2}$ on $[0,5]$ with $N=10$.
f = lambda x : 1/(1 + x**2)
a = 0; b = 5; N = 10
# x and y values for the trapezoid rule
x = np.linspace(a,b,N+1)
y = f(x)
# X and Y values for plotting y=f(x)
X = np.linspace(a,b,100)
Y = f(X)
plt.plot(X,Y)
for i in range(N):
xs = [x[i],x[i],x[i+1],x[i+1]]
ys = [0,f(x[i]),f(x[i+1]),0]
plt.fill(xs,ys,'b',edgecolor='b',alpha=0.2)
plt.title('Trapezoid Rule, N = {}'.format(N))
plt.show()
Let's compute the sum of areas of the trapezoids:
T = trapz(f,a,b,N)
print(T)
1.3731040812301096
We know the exact value
$$\int_0^5 \frac{1}{1 + x^2} dx = \arctan(5)$$
and we can compare the trapezoid rule to the value
I = np.arctan(5)
print(I)
1.373400766945016
print("Trapezoid Rule Error:",np.abs(I - T))
Trapezoid Rule Error: 0.00029668571490626405
### Approximate ln(2)
Find a value $N$ which guarantees that the trapezoid rule approximation $T_N(f)$ of the integral
$$\int_1^2 \frac{1}{x} \, dx = \ln(2)$$
satisfies $E_N^T(f) \leq 10^{-8}$.
For $f(x) = \frac{1}{x}$, we compute $f''(x) = \frac{2}{x^3} \leq 2$ for all $x \in [1,2]$ therefore the error formula implies
$$\left| \, \int_1^2 \frac{1}{x} \, dx - T_N(f) \, \right| \leq \frac{2}{12N^2}$$
Then $E_N^T \leq 10^{-8}$ is guaranteed if $\frac{1}{6N^2} \leq 10^{-8}$ which implies
$$\frac{10^4}{\sqrt{6}} \leq N$$
10**4/np.sqrt(6)
4082.4829046386303
We need 4083 subintervals to guarantee $E_N^T(f) \leq 10^{-8}$. Compute the approximation using our own implementation of the trapezoid rule:
approximation = trapz(lambda x : 1/x,1,2,4083)
print(approximation)
0.6931471843089954
We could also use scipy.integrate.trapz to get the exact same result:
N = 4083
x = np.linspace(1,2,N+1)
y = 1/x
approximation = spi.trapz(y,x)
print(approximation)
0.6931471843089955
Let's verify that this is within $10^{-6}$:
np.abs(approximation - np.log(2)) < 10**(-8)
True
Success! However, a natural question arises: what is the actual smallest $N$ such that the trapezoid rule gives the estimate of $\ln (2)$ to within $10^{-8}$?
for n in range(1,4083):
approx = trapz(lambda x : 1/x,1,2,n)
if np.abs(approx - np.log(2)) < 10e-8:
print("Accuracy achieved at N =",n)
break
Accuracy achieved at N = 791
### Fresnel Integral
Fresnel integrals are examples of nonelementary integrals: antiderivatives which cannot be written in terms of elementary functions. There are two types of Fresnel integrals:
$$S(t) = \int_0^t \sin(x^2) dx \ \ \text{and} \ \ C(t) = \int_0^t \cos(x^2) dx$$
Use the trapezoid rule to approximate the Fresnel integral
$$S(1) = \int_0^1 \sin(x^2) dx$$
such that the error is less than $10^{-5}$.
Compute the derivatives of the integrand
$$f(x) = \sin(x^2) \ \ , \ \ f'(x) = 2x\cos(x^2)$$
$$f''(x) = 2\cos(x^2) - 4x^2\sin(x^2) \ \ , \ \ f'''(x) = -12x\sin(x^2) - 8x^3\cos(x^2)$$
Since $f'''(x) \leq 0$ for $x \in [0,1]$, we see that $f''(x)$ is decreasing on $[0,1]$. Values of $f''(x)$ at the endpoints of the interval are
x = 0
2*np.cos(x**2) - 4*x**2*np.sin(x**2)
2.0
x = 1
2*np.cos(x**2) - 4*x**2*np.sin(x**2)
-2.2852793274953065
Therefore $\left| \, f''(x) \, \right| \leq 2.2852793274953065$ for $x \in [0,1]$. Use the error bound formula to find a good choice for $N$
$$\frac{(b-a)^3}{12 N^2} K_2 \leq 10^{-5} \Rightarrow \sqrt{\frac{10^5(2.2852793274953065)}{12}} \leq N$$
np.sqrt(10**5 * 2.2852793274953065 / 12)
137.9999796949051
Let's compute the integral using the trapezoid rule with $N=138$ subintervals
x = np.linspace(0,1,139)
y = np.sin(x**2)
I = spi.trapz(y,x)
print(I)
0.31027303032220394
Therefore the Fresnel integral $S(1)$ is approximately
$$S(1) = \int_0^1 \sin(x^2) \, dx \approx 0.310273030322$$
with error less than $10^{-5}$.
### Logarithmic Integral
The Eulerian logarithmic integral is another nonelementary integral
$$\mathrm{Li}(t) = \int_2^t \frac{1}{\ln x} dx$$
Let's compute $Li(10)$ such that the error is less than $10^{-4}$. Compute derivatives of the integrand
$$f(x) = \frac{1}{\ln x} \ \ , \ \ f'(x) = -\frac{1}{x(\ln x)^2} \ \ , \ \ f''(x) = \frac{\ln x + 2 }{x^2(\ln x)^3}$$
Plot $f''(x)$ on the interval $[2,10]$.
a = 2
b = 10
x = np.linspace(a,b,100)
y = (np.log(x) + 2) / (x**2 * np.log(x)**3)
plt.plot(x,y)
plt.show()
Clearly $f''(x)$ is decreasing on $[2,10]$ (and bounded below by 0) therefore the absolute maximum occurs at the left endpoint:
$$\left| \, f''(x) \, \right| \leq \frac{\ln (2) + 2}{4 \ln (2)^3}$$
for $x \in [2,10]$ and we compute
K2 = (np.log(2) + 2)/(4*np.log(2)**3)
print(K2)
2.021732598829855
Use the error formula:
$$\frac{(b-a)^3}{12 N^2} K_2 \leq 10^{-4} \Rightarrow \frac{8^3}{12 N^2} 2.021732598829855 \leq 10^{-4} \Rightarrow \sqrt{ \frac{8^3 10^4}{12} 2.021732598829855} \leq N$$
np.sqrt(8**3 * 10**4 * 2.021732598829855 / 12)
928.7657986995814
Compute the trapzoid rule with $N=929$
N = 929
x = np.linspace(a,b,N+1)
y = 1/np.log(x)
I = spi.trapz(y,x)
print(I)
5.120442039184057
Therefore the Eulerian logarithmic integral is
$$\mathrm{Li}(10) = \int_2^{10} \frac{1}{\ln x} dx \approx 5.121065367200469$$
such that the error is less than $10^{-4}$.
## Exercises
1. Let $f(x) = x^x$ and note that
$$f'(x) = x^{x} \left(\log{\left(x \right)} + 1\right) \ , \ f''(x) = x^{x} \left(\log{\left(x \right)} + 1\right)^{2} + x^{x-1}$$
Plot the function $f''(x)$ and use that information to compute $T_N(f)$ for the integral
$$\int_1^2 x^x \, dx$$
such that $E_N^T(f) \leq 10^{-3}$.
2. Consider the integral
$$\int_0^1 \ln(1+x^2) \, dx$$
and note that
$$f(x) = \ln(1 + x^2) \hspace{1in} f'(x) = \frac{2x}{1 + x^2}$$ $$f''(x) = 2 \left( \frac{1 - x^2}{1 + x^2} \right) \hspace{1in} f'''(x) = 4x \frac{x^2 - 3}{(x^2 + 1)^3}$$
Without plotting the functions $f(x)$, $f'(x)$, $f''(x)$ or $f'''(x)$, find a value $N$ such that $E_N^T(f) \leq 10^{-6}$. | 2021-03-09T01:26:37 | {
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https://math.stackexchange.com/questions/3458449/path-on-a-chessboard | # Path on a chessboard
Consider a chess board and a pawn. The game is simple (and I'm sure you have all heard it before), pawn wants to move from position (1,1), or the bottom left corner, all the way to (n,n), or the upper right corner. We are asked to find all possible paths this pawn could take.
This problem is not all that hard, considering that the path always takes $$2n-2$$ moves. Cancelling out the same, but reversed and tidying up the formula, we get
$$\binom{2n-2}{n-1}$$ possible paths.
Following which, the main question is what would happen if we removed some pivotal squares. Let's consider that we have a chessboard with $$n=4$$, and were to remove square $$(3,3)$$. By my logic, all we need to do is to find all the possible ways how to get to $$(4,4)$$ and to $$(3,3)$$. Then we would subtract all the ways to $$(3,3)$$ from the number of paths to $$(4,4)$$, which would give us
$$\binom{2*4-2}{4-1} - \binom{2*3-2}{3-1} =$$ $$\binom{6}{3} - \binom{4}{2} =$$ $$20 - 6 = 12$$
As I found out, this is not the correct solution. I mapped out all the possible paths, and found only $$8$$ of them. If we multiply the number of paths to $$(3,3)$$ by $$2$$, that would give us the correct solution; $$\binom{2*4-2}{4-1} - 2* \binom{2*3-2}{3-1} =$$ $$20 - 2*6 = 8$$
That begs the question, is the second solution correct? And if so, why? That, I do not know.
Additionally, what would happen if we were to remove more than one square? For example, let $$n=13$$ and lets remove two squares, $$(5,5)$$ and $$(8,8)$$. Would that be something like
$$\binom{2*13-2}{13-1} - 2* {\left[\binom{2*8-2}{8-1} + \binom{2*5-2}{5-1}\right]}$$
or did I get the whole concept wrong? Why is there that seemingly random multiplication, and how would it stack with more and more pivotal squares removed?
• But you know... pawns can only move forward... and occasionally attack diagnally... until they become something else. Dec 2, 2019 at 21:58
The paths to $$(4,4)$$ that use $$(3,3)$$ aren't just paths to $$(3,3)$$; their paths to $$(3,3)$$ and then on to $$(4,4)$$. So to count them, you need to multiply the number of ways to get from $$(1,1)$$ to $$(3,3)$$, which you calculated, with the number of ways to get from $$(3,3)$$ to $$(4,4)$$. That's $$\binom21=2$$, and this is where the missing factor $$2$$ comes from.
If you remove more than one square, you need to use inclusion–exclusion. For each subset of the removed squares, you need to add or subtract the number of paths that use that subset of removed squares, according as it contains an even or odd number of removed squares. (That number will be zero unless the orders of the $$x$$ and $$y$$ coordinates of the squares coincide.)
For instance, if you remove the squares $$(2,2)$$ and $$(3,3)$$, you're left with $$4$$ paths, and this can be calculated as follows, with $$N(S)$$ denoting the number of paths that use all squares in $$S$$:
$$\begin{eqnarray*} &&N(\emptyset)-N(\{(2,2)\})-N(\{(3,3)\})+N(\{(2,2),(3,3)\})\\ &=& \binom63-\binom21\binom42-\binom42\binom21+\binom21\binom21\binom21\\ &=& 20-2\cdot6-2\cdot6+2\cdot2\cdot2\\ &=& 20-12-12+8\\ &=& 4\;. \end{eqnarray*}$$
• So for the last case that would be $\binom{24}{12}-\binom84\binom{14}{7}-\binom{14}{7}\binom84+\binom84\binom84\binom84\\$, right?
• @Igor: Almost; I think some of your counts are off by $1$. The lower index is always the coordinate difference between the squares you're connecting. So if you're going from coordinate $1$ to $13$ while avoiding $5$ and $8$, that would be $$\binom{24}{12}-\binom84\binom{16}8-\binom{14}7\binom{10}5+\binom84\binom63\binom{10}5\;.$$ As a check, the indices (both upper and lower) in the factors of each term should add up to the same sum. Dec 3, 2019 at 12:05 | 2022-07-06T00:02:30 | {
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http://math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals | # Is there a rational number between any two irrationals?
Suppose $i_1$ and $i_2$ are distinct irrational numbers with $i_1 < i_2$. Is it necessarily the case that there is a rational number $r$ in the interval $[i_1, i_2]$? How would you construct such a rational number?
[I posted this only so that the useful answers at Rationals and irrationals on the real number line could be merged here before that question was deleted.]
-
Yes, there is. This is known as the "density of the rationals in the reals", which says in fact that between any two reals numbers there is a rational number. – Avi Steiner Jun 16 '13 at 2:24
– MJD Jun 16 '13 at 2:37
So, wait... you asked the question, pointed to another question that already answers this... and then answered this question several days before asking it? TIME TRAVELER!!! – BlueRaja - Danny Pflughoeft Jun 16 '13 at 5:25
@BlueRaja-DannyPflughoeft yeah, I thought that, then I read the [bit in brackets in the question]. – Lucas Jun 16 '13 at 5:53
Let $x,y\in\mathbb{R}$, $x\neq y$. Without loss of generality, suppose $x<y$. Then there exists a positive $z$ such that $y-x=z$.
By Archimedes' axiom, there exists a natural number $n$ such that $$n > \dfrac{1}{z}$$ $$nz > 1$$ $$ny - nx > 1$$ So there exists an integer $m$ such that $$nx < m < ny$$ $$x < \frac{m}{n} < y$$ i.e. $m/n$ is a rational number between $x$ and $y$.
Since $x$ and $y$ can be any real numbers, in particular they can be irrationals.
-
Let $a$ and $b$ be distinct irrationals; we lose no generality to suppose $a<b$. Assume they have equal integer parts, since otherwise there is an integer between them and the question is trivial. They have infinite decimal expansions, $.a_1a_2a_3\ldots$ and $.b_1b_2b_3\ldots$. These cannot agree in every position since otherwise $a=b$. So say they agree up to the $n-1$th position and differ at the $n$th. Then $$x= \;.b_1b_2b_3\ldots b_n 0000000\ldots$$
is a rational number strictly between $a$ and $b$:
\begin{align} a &= \;.a_1a_2\ldots a_{n-1}a_n\ldots \\& <\; .a_2a_2\ldots a_{n-1}b_n000\ldots &= x \end{align}
because $a_n < b_n$, and
\begin{align} x & = \;.b_1b_2b_3\ldots b_n 0000000\ldots \\ & < \;.b_1b_2b_3\ldots b_nb_{n+1}\ldots & = b \end{align}
because not all of $b_{n+1}, b_{n+2}, \ldots$ can be zero.
For example, there is a rational number betwen $\sqrt2 = 1.4142\ldots$ and $\sqrt3 -\frac14 = 1.482\ldots$; this method produces the rational number $1.48000\ldots = \frac{37}{25}$. You can of course do the same thing in base 2; then you get $x = 1._20111000\ldots = \frac{23}{16}$, which also works.
It would also be fairly easy to produce a similar argument based on the continued fraction expansions of $a$ and $b$, but I think this is simpler.
-
Your proof contradicts itself by not allowing integers but allowing integers up to b_n. – Daniel Margolis Jun 7 '13 at 18:32
One of us is confused, and I bet it's you. $b_n$ is not an integer. It is a decimal digit. And my proof doesn't "disallow" integers; it observes that if $a$ and $b$ have different integer parts then there is no need for a proof because there is an integer between them and the question is answered. – MJD Jun 7 '13 at 18:33
I guess this shows that you can lead a horse to water, but you can't make him drink. – MJD Jun 7 '13 at 18:49
@Daniel: There is no "infinite number place." Your conception of the notation "$\infty$" as a number (or occasionally a cardinality) seems to be at the root of all of your confusion. – Cameron Buie Jun 15 '13 at 3:14
@JensSchauder If a has integer part 2 and b has integer part 3 and b is irrational, then 3 is indeed between a and b. – Daenerys Naharis Jun 16 '13 at 5:44
We can construct one explicitly. Assume $0\lt a \lt b$. Let $n= \max(2,\lceil \frac 2{b-a}\rceil)$. Then let $m=\lceil na \rceil$ and $a \lt \frac m n \lt b$
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@DanielMargolis: in the standard reals, there are no such. You (or the problem poser) have to specify them first and the difference will not be infinitesimal. It may be small, which makes $n$ large, but that is OK. Even if $a=\sqrt 2, b=\sqrt 2 + 10^{-100}$ this works. – Ross Millikan Jun 8 '13 at 2:06
@DanielMargolis: Yes, it is. In fact, if you extend $\Bbb R$ with infinitesimals, it becomes incomplete. If you have an infinitesimal $\epsilon$, the set $\{\frac 1n|n \in \Bbb N\}$ has $0$ as a lower bound, but it has no greatest lower bound. – Ross Millikan Jun 8 '13 at 13:05
If you are working in some number system that includes infinitesimals, you should say that. There is no indication in the question that you are, and all the answers to this and your other question are in the real numbers. I know very little about such systems. You seem to have an intuitive sense of how you want the infinitesimals to behave, but I don't think they can act that way. – Ross Millikan Jun 8 '13 at 13:33
@CameronBuie: yes, he is including infinitesimals, which comes out late in this discussion, as well as another question he asks. Then he wants to lump all infinitesimals with the standard irrationals. I am not sure if it is confusion or trolling. – Ross Millikan Jun 15 '13 at 15:55
@Daniel: Noone is claiming that irrationals have finite decimal expansions--their definitions do indeed have infinitely many decimal places. What do you mean by "an infinite decimal place," though? – Cameron Buie Jun 16 '13 at 2:32
Be $a$ and $b$ two irrational numbers, with $a<b$. Since $a<b$, we have $b-a>0$. Now there are rational numbers arbitrary close to $0$, and therefore there's a rational number $q<b-a$. Now consider the set $M=\{qn: n\in \mathbb{Z}\}$. Now be $q_l$ the largest element of $M$ less than $a$, and $q_r$ the smallest element of $M$ greater than $b$. Clearly $q_r-q_l > b-a$. However, if there's no rational number between $a$ and $b$, then there's especially no element of $M$ between $a$ and $b$, because all elements of $M$ are rational. But then $q_r$ is the element of $M$ following $q_l$, that is, $q_r-q_l = q$. But by construction, $q<b-a$, so we have $b-a < q_l-q_r = q < b-a$ which is a contradiction. Therefore there's at least one element of $M$, which is a rational number, in between $a$ and $b$.
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@DanielMargolis: I didn't claim it is. – celtschk Jun 7 '13 at 23:16
The real numbers are totally ordered, that means exactly one of $a-b>0$, $a-b=0$, $a-b<0$ is true. $a-b=0$ would mean $a=b$, but by assumption, $a$ and $b$ are different (were $a=b$, quite trivially there would be no rational number between $a$ and $b$). So this leaves $a-b>0$ and $a-b<0$. I further assumed that $a$ denotes the smaller of the two, i.e. $a<b$, which is equivalent to $b-a>0$. So $b-a>0$ by the initial assumption (but that's just a wlog assumption; if we had $a<b$, you could just use $a-b$ instead). – celtschk Jun 8 '13 at 8:39
@DanielMargolis: There cannot be such an $R$ because all sequences of rational numbers whose absolute value eventually stays beyond any given rational number converge to $0$. So any number that is closer to $0$ than any rational number cannot be the limit of a sequence of rational numbers, in contradiction to $\mathbb{R}$ being the completion of $\mathbb{Q}$. – celtschk Jun 8 '13 at 9:19
$1/n*irr$ is not an irrational number, it is a whole sequence of irrational numbers. And not a single one is closer to $0$ than all rational numbers. Yes, for every rational number (with the exception of $0$ itself, of course), there's an irrational number that's closer to $0$. But there's no irrational number which is closer to $0$ than every rational number, because for each irrational number there's a rational number which is even closer to $0$. – celtschk Jun 8 '13 at 10:13
Do you really not know the difference between "thhere is no irraltional number closer to $0$ than any rational number" (true) and "for any rational number there is no irrational number closer to it" (false)? In that case, do yourself a favour and learn elementary logic. – celtschk Jun 8 '13 at 11:17
## protected by Asaf KaragilaJan 15 '14 at 21:21
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | 2015-01-28T06:46:37 | {
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https://math.stackexchange.com/questions/3920780/probability-of-drawing-an-ace-or-2-after-an-ace-intuition | # Probability of drawing an ace or 2 after an ace - intuition
This question is a follow-up question on this one: Probability of drawing an ace or 2 after an ace
Background Out of a standard deck you draw cards (without replacement) one by one until you reach an ace. Then you continue to draw cards until you reach another ace or a "2" card. What is the probability that you reached a "2" card and not an ace?
The solution To shorten things up, let's assume the deck has only two aces and two "2" cards. Clearly, all that matters are the relative position of these two cards, not all the other cards. So we can think of the permutations of 4 items in a row.
The event "the card following the first ace is ace" occurs when the two aces are together. The sample space is $${4 \choose 2}$$ (choosing 2 places for the aces) and there are 3 locations in which they are together (1-2,2-3,3-4) so the solution is $$\tfrac{3}{{4\choose 2}}=0.5$$.
Surprisingly, the answer $$0.5$$ remains correct even if there are 4 aces and 4 "2"s in the original deck and in fact, as @InterstellarProbe shows in the comments to the original question, if there are $$n+1$$ of each type, the answer is: $$\sum_{k=0}^{n+1} \frac{n\left(\begin{array}{c} n+k \\ k \end{array}\right)}{(n+k)\left(\begin{array}{c} 2 n+2 \\ n+1 \end{array}\right)}=\frac{1}{2}$$
The question The above result surprises me a bit. There is no obvious symmetry in the question and it is not obviously clear why there is the same number of arrangements in which after the first ace comes another one as arrangements in which after the first ace comes a "2". I spent a couple of months thinking, asked also fellow probability teachers, but no-one had a good idea.
So, returning the ball to this court. What is the intuitive explanation to this result? Why the permutations are symmetric and how can we get this $$\tfrac{1}{2}$$ without the algebra?
• I would say that it looks asymmetric because you don't think about what happens before that first ace. You might have already discarded several 2's. – Marc Romaní Nov 24 '20 at 14:32
I'm not sure that the following solution is any more intuitive than others (in my experience, some probability problems are simply non-intuitive, at least with my intuition), but at least it is a different approach than those given so far.
To start with, we can ignore all the cards in the deck except for aces and twos. So suppose we have a deck of four aces and four twos, which can be arranged in $$8!$$ ways, all of which we assume are equally likely. We want to find the probability that the card following the first ace in the deck is also an ace.
As a further simplification, let's see if we can find the probability that the card immediately following the first ace is the ace of spades. We want to count the number of such arrangements. To do so, suppose we remove the ace of spades from the deck and deal out the remaining cards; this can be done in $$7!$$ ways. There is only one place where the ace of spades can now go so that it is immediately following the first ace, so the number of arrangements in which the ace of spades follows the first ace is $$7!$$, and the probability of this event is $$\frac{7!}{8!} = \frac{1}{8}$$
If we don't care which ace follows the first ace in the deck, there are $$4$$ possible choices, all of which have the same probability of $$1/8$$. So the probability that the card following the first ace is also an ace is $$4 \cdot \frac{1}{8} = \frac{1}{2}$$
• It wasn't what I was looking for (searched for symmetry of the permutations) but this is an excellent answer. Following on that logic, from symmetry, the probability of any card to be after the first ace is equal so it must be $\tfrac{1}{8}$ and the probability that it would be an ace is $0.5$. The generalization to more cards and more types (ace, 2, and kings...) is straightforward. Great! – YJT Nov 24 '20 at 17:46 | 2021-08-04T23:36:42 | {
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https://math.stackexchange.com/questions/2402760/equation-of-locus-of-points-satisfied-by-frac-leftz3i-right-leftz-6i-ri | # Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$
Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$
The answer I got is $y=\frac{3}{2}$, but the answer given in my book is $y=-0.5x+2.25$
Can anyone please confirm which is the right answer
Edit: Sorry, the modulus was on each numerator and denominator, not for the whole thing, though I don't think this makes a difference.
My working
subbing $z=x+yi$ gives
$\frac{\left|x+yi+3i\right|}{\left|x+yi-6i\right|}=1$
$\frac{x^2+(y+3)^2}{x^2+(y-6)^2}=1^2$
Expanding brackets and then solving gives
$y=3/2$
Thank You
• Please show us your workings, i.e. your work from which you "got is $y=\frac 32$. Other wise, we cannot confirm your solution. So, show us how you arrived at your answer, by editing your post to include it. – Namaste Aug 22 '17 at 20:11
• I think you're right. The equation says point $z$ is equidistant from $-3i$ and $6i$, which should be a horizontal line at $y=1.5$. – Jirapat Samranvedhya Aug 22 '17 at 20:12
• @amWhy I have done so now. Thanks – NumberCruncher Aug 22 '17 at 20:22
• @JirapatSamranvedhya Thanks very much, that's how I thought about it first too. – NumberCruncher Aug 22 '17 at 20:22
• NumberCruncher Thank you for including your work! – Namaste Aug 22 '17 at 22:10
To cut down on unanswered questions, here we go!
The book's answer is certainly wrong, as one readily sees by considering $z=\frac94i.$
Now, we clearly cannot have $z=6i$ as a solution, for then we have $\frac90=1,$ which is nonsensical. Consequently, the given equation is equivalent to $$|z+3i|=|z-6i|,$$ or, put another way, to $$\bigl|z-(-3i)\bigr|=|z-6i|.$$
Since $|z-w|$ is the distance from $z$ to $w$ for all $z,w\in\Bbb C,$ then the equation above says that $z$ is equidistant from $-3i$ and $6i.$ Readily, putting $z=x+iy,$ this is equivalent to saying that $(x,y)$ is equidistant from $(0,-3)$ and $(0,6),$ i.e.: $$\sqrt{x^2+(y+3)^2}=\sqrt{x^2+(y-6)^2}.$$ This is, of course, equivalent to your approach, and solving the preceding equation yields $y=\frac32,$ as you say.
Given two points, the set of points equidistant from them is the perpendicular bisector of the line joining them.
If the points are $(a, b)$ and $(c,d)$, the midpoint is $((a+c)/2, (b+d)/2)$.
The slope of the line is $\dfrac{d-b}{c-a}$, so the slope of the normal is the negative reciprocal of this or $-\dfrac{c-a}{d-b} =\dfrac{a-c}{d-b}$.
Therefore the equation of the perpendicular bisector is $\dfrac{y-(b+d)/2)}{x-(a+c)/2} =\dfrac{a-c}{d-b}$.
If $a=c=0$, as in this case, the equation is $\dfrac{y-(b+d)/2)}{x} =0$ or $y=(b+d)/2$.
In this case, $b=-3$ and $d=6$, so the equation is $y =(6-3)/2 =3/2$. | 2019-07-19T03:41:35 | {
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http://rvjs.gremium-franconia.de/numerical-integration-matlab-code.html | Numerical Integration Matlab Code
Symbolic Integration and Differentiation using MATLAB. Numericaldifierentiationand integration Numerical difierentiation with Richardson Extrapolation Assume Matlab Code: Trapezoid Rule Integration. We can do this in (at least) three different ways. MATLAB®’s quad Function. Math 301 – Numerical Analysis 2, Winter 2009. To derive an expression for the indefinite integral of a function, we write − int(f); For example, from our previous example − syms x int(2*x) MATLAB executes the above statement and returns the following result − ans = x^2 Example 1. Note: The five first software elements in this table are all available on the Web through the netlib library: just click on the Code name to get Netlib's keyword search and then submit the Library information in this table to find the code and then download the code. 1) is an approximation of the form IQ(f,a,b) = (b− a) Xm k=1 wkf(xk). 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Chapra, Applied Numerical Methods with Matlab for Engineers and Scientists, third edition. Suppose that the interval [a,b] is split up in n subintervals, with n an even number. · Midpoint Rule (Rectangular Rule) · Euler-Maclaurin Formula · Richardson Extrapolation · Trapezoidal Rule · Simpson's Rule There are a number of numerical methods that can be used to approximate an integral. It is based on the Fast Fourier Transform (FFT) technique and yields a numerical solution for t=a ("a" is a real number) for a Laplace function F(s) = L(f(t)), where "L" represents the Laplace transformation. function that implements the. NET Numerics under the MathNet. The MATLAB functions for the numerical evaluation of integrals has evolved from quad, through quadl and quadgk, to today's integral. This section under major construction. This code is located in a file called trapezoidal. We study numerical methods to solve linear and nonlinear equations, to interpolate and approximate data, and methods for numerical integration and differentiation. Roots of algebraic and transcendental equations -- 5. Numerical integration in a loop. 0 Unported License. The MATLAB functions for the numerical evaluation of integrals has evolved from quad, through quadl and quadgk, to today's integral. 0 Simple and intuitive numerical integration based on trapezoidal rule. the numerical integration; no transformation of the EOM (1) is carried out. How to download & Pay on REDS So Love symbol using MATLAB; A Small Tribute To Netaji On 23rd January using MA Runge-Kutta method (Order 4) for solving ODE using Euler's method for solving ODE using MATLAB. txt) or read online for free. a guest Sep 19th, 2017 53 Never Not a member of Pastebin yet? % the little t steps for numerical integration %obtain numerical ODE45 solution to use. Numerical Integration. Numerical integration with Monte Carlo method (on FPGA chip). In contrast to the existing Matlab built-in functions for numerical integration, this program can be used with arrays for the upper and lower integration limits. Introduces intermediate variables for small compact code and reduced operations count. Stanford Libraries' official online search tool for books, media, journals, databases, government documents and more. We begin by clicking on file on the Matlab Command Window and new. Trapezoidal rule to approximate the integral of x^2 style of the same code. function value=rsum1(f,a,b,n) %RSUM1: Computes a Riemann Sum for the function f on %the interval [a,b] with a regular partition of n points. Part B: Numerical integration (08 to 15 November 2010) Source code used in slides. I came across the book, ‘Computational Physics’, in the library here in the Dublin Institute of Technology in early 2012. More accurate methods of numerical integration are based on Gauss quadrature methods for orthogonal polynomials such as Legendre, Chebyshev, Laguerre and Hermite polynomials (optional reading - chapter 7. Caregiver Issues Bathing and Hygiene; Caregiver hiring; Caregiver training; Caregivers – professional; Family Carergivers; Caregiving Around the World. x/dx we first divide the interval Ta;bUinto n subintervals, each of length. Type the following into the text editor:. Section 4 is devoted to a presentation and explanation of Matlab codes for implicit Runge-Kutta, composition, and multistep. It is based on the Fast Fourier Transform (FFT) technique and yields a numerical solution for t=a ("a" is a real number) for a Laplace function F(s) = L(f(t)), where "L" represents the Laplace transformation. How to download & Pay on REDS So Love symbol using MATLAB; A Small Tribute To Netaji On 23rd January using MA Runge-Kutta method (Order 4) for solving ODE using Euler's method for solving ODE using MATLAB.$\begingroup$Thank you, but how do you incorporate the code for Simpson's rule into the code for Bisection method? So that at each midpoint, Simpson's rule calculates a new value for the function. There is a complication in using dblquad; it does not accept variable limits. Is publishing runnable code instead of. MATLAB's programming interface gives development tools for improving code quality maintainability and maximizing performance. What we are trying to do here, is to use the Euler method to solve the equation and plot it along side with the exact result, to be able to judge the accuracy of the numerical method. Runge's Function. How to apply numerical integration on a symbolic Learn more about integration. Department of Electrical and Computer Engineering University of Waterloo 200 University Avenue West Waterloo, Ontario, Canada N2L 3G1 +1 519 888 4567. · Midpoint Rule (Rectangular Rule) · Euler-Maclaurin Formula · Richardson Extrapolation · Trapezoidal Rule · Simpson’s Rule There are a number of numerical methods that can be used to approximate an integral. Numerical Integration-1. docx 1 DOING PHYSICS WITH MATLAB COMPUTATIONAL OPTICS NUMERICAL INTEGRATION RAYLEIGH-SOMMERFELD DIFFRACTION INTEGRAL OF THE FIRST KIND Ian Cooper School of Physics, University of Sydney ian. Apply MATLAB to carry out numerical differentiation and integration. Asked by me with the code for the Richardson extrapolation. Simplify your answer as possible. Have a nice day!. m, fourcoeffs. commands to prepare a new workspace each time we run the code. Thus, the answer is same as the one obtained using the program for Trapezoidal method in MATLAB. NUMERICAL SOLUTIONS TO TWO-DIMENSIONAL INTEGRATION PROBLEMS by Alexander D. This book provides a fundamental introduction to numerical analysis for undergraduate students in the areas of mathematics, computer science, physical sciences, and engineering. m is a to compute numerical approximations to definite. Numerical Integration and Intro to Pseudospectral Methods for BVPs. C++ from Python and MATLAB, because C++ is faster. Chapra (2006, Hardcover, Revised) at the best online prices at eBay!. Input/Output: Also see, Simpson 1/3 Rule in Matlab Numerical Methods Tutorial Compilation. If Y is a vector, trapz(Y) is the integral of Y. NUMERICAL INTEGRATION. Subscribe to our newsletter to get notifications about our updates via email. The source code and files included in this project are listed in the project files section, please make sure whether the listed source code meet your needs there. MATLAB Python function In = trapezcomp (f, a, b,. This work. 6 Trapezoidal Method and Simpson Method / 226 by example MATLAB code with a friendly interface so that students can easily. Symbolic and Numerical Integration in MATLAB 1 Symbolic Integration in MATLAB Certain functions can be symbolically integrated in MATLAB with the int command. Scripts de Integração númerica Newton-Cotes. Hairer and M. We begin by clicking on file on the Matlab Command Window and new. Book Description. time) and one or more derivatives with respect to that independent variable. This course will focus on the root finding and numerical integration techniques most frequently covered at the undergraduate level. Summary of first day w/ MATLAB in class. Midpoint rule. , Simpson's. MATLAB is widely used in undergraduate engineering programs as well as in industry. In Matlab code I used 'ArrayValued' to enable the integrand to receive a vector. This method is. Engineering & Matlab and Mathematica Projects for €8 - €30. Consider Example M3. In case you don’t have MATLAB already installed in your system, there are lectures on the various ways you can acquire MATLAB and the procedures involved in its installation. We can do this in (at least) three different ways. Learn more about histc, pdf, integrate, trapz, nan, inf, divide by zero MATLAB Answers. Some useful functions. Summary: There are problems in integrating Hamiltonian systems with normal numerical integrators, and your special initial conditions aggravate this to the point where the numerical solution has no resemblance with the correct one. This method is. pdf), Text File (. MATLAB is a high-level language and interactive environment for numerical computation, visualization, and programming. Numerical Methods or Numerical Analysis is a subject included in all types of engineering curriculum around the world. How to download & Pay on REDS So Love symbol using MATLAB; A Small Tribute To Netaji On 23rd January using MA Runge-Kutta method (Order 4) for solving ODE using Euler's method for solving ODE using MATLAB. If Y is a vector, trapz(Y) is the integral of Y. Numerical Integration - Free download as PDF File (. Numerical Methods in Engineering with MATLAB R Second Edition Numerical Methods in Engineering with MATLAB R is a text for engi-neering students and a reference for practicing engineers. Z = trapz(X,Y) computes the integral of Y with respect to X using trapezoidal integration. Unless stated otherwise, the examples below evaluate the integral \(\int_0^{10} x^2 \, dx = \frac{1000}{3} \approx 333. If Y is a multidimensional array, then trapz(Y) integrates over the first dimension whose size does not equal 1. trapz Numerical integration with the trapezoidal rule. Numericaldifierentiationand integration Numerical difierentiation with Richardson Extrapolation Assume Matlab Code: Trapezoid Rule Integration. Specifically, the integral is introduced by using the best approximation scheme (Legendre Polynomials) to approximate a vector valued function whose indefinite integral is not easy to be explicitly written down. are accepted. C Program for Numerical Integration (Trapezoidal Rule, Simpson's Rule and Boole's Rule Program /* This program is for numerical integration of numerical methods Here we are going to solve the integration of defined functio. Solving Numerical Integral Implicitly. A followup post of my previous entry Using Quadrature method for option valuation, where I tested the alternative numerical method for option valuation, QUAD. in the function. This book will enable readers to solve problems without needing to understand all the details of the underlying theory of numerical methods. Matlab code. The order of integration (and therefore the bounds) is from the innermost integral to the outermost one. Dynamics and Vibrations MATLAB tutorial School of Engineering Brown University This tutorial is intended to provide a crash-course on using a small subset of the features of MATLAB. b) computes a numerical approximation of. Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. Array-valued function flag, specified as the comma-separated pair consisting of 'ArrayValued' and a numeric or logical 1 (true) or 0 (false). MATLAB EXAMPLES Interpolation and Integration. The numerical evaluation in this case was performed using numerical routines borrowed from MuPAD, which takes care of symbolic calculations. A brief introduction to the Simpson’s 1/3 rd rule and a uniform interval Composite Simpson’s 1/3 rd Rule implementation. Victor Zalizniak, in Essentials of Scientific Computing, 2008. Body of the package implementing numerical integration: package body Integrate is. We can do this in (at least) three different ways. In fact, the built-in capabilities of MATLAB are used to perform numerical computations, which are very useful in enormous fields of applied science and engineering, including: Root finding and equation solving Solving system of equations Eigenvalues, eigenvectors and eigendecomposition Singular Value. Table 2: A Matlab module for DFT calculations. The goal of the book. MATLAB code demo in class. Midpoint (Rectangular) Numerical Integration Matlab Code by admin in Math, Statistics, and Optimization, MATLAB Family, Numerical Integration. MATLAB code to perform numerical integration and differentiation % Numerical integration and differentiation % Using built-in Matlab functions % Clear the command window. SoDiOpt is a MATLAB-based code that performs numerical integration of Optimization-Constrained Differential Equations (OCDE). Learn more about numerical integration, random number generator, trapz, integration, cumtrapz, digital image processing. , Part Five, Chapters 17 and 18 PGE 310: Formulation and Solution in Geosystems Engineering Dr. Matlab Gauss Jordan Code Matlab Gauss Seidel Code Notes on linear and nonlinear least squares Data for homework 2 Practice Midterm Notes on Taylor methods Matlab code for Euler's method Matlab code for mid-point method Backward Euler notes Adam's Moulton Code O and A stability notes Boundary value problems and shooting Practice Final. Miranda, M. To figure this out, consider exactly what it is that both processes do. Matlab code for incompatible mode elements, short beam input file long beam input file; Matlab code demonstrating volumetric locking linear quad input file quadratic quad input file; L10 Hourglass control, selective reduced integration, B-bar elements. Search for jobs related to Integration matlab or hire on the world's largest freelancing marketplace with 15m+ jobs. Numerical Methods or Numerical Analysis is a subject included in all types of engineering curriculum around the world. Here is my implementation of Martin Floden's matlab code for the routine. In the numerical analysis literature, the Verlet method is also knows as the explicit central difference method''. To derive an expression for the indefinite integral of a function, we write − int(f); For example, from our previous example − syms x int(2*x) MATLAB executes the above statement and returns the following result − ans = x^2 Example 1. Help with numerical integration in matlab. My simpson method is correct, but my adaptive method does not seem to work for the integral( sin(2*pi*x)² ) ranging from -1 to 1 The following code represents the adaptive simpson method. Given ) = !(#)the approximation of the Area (5) under the curve can be found dividing the area up into. Plot the results against the. A numerical library for adaptive Sparse Grids (mingw-w64) matlab-r2018b: 9. I need help developing a C-code that numerically integrates using N = 20 and N = 40, where N is the number of steps or interval. For the majority of you, it has been nearly six months since you last used MATLAB. Compound Trapezoidal Method for Numerical Integration by admin in Math, Statistics, and Optimization , MATLAB Family , Numerical Integration$4. It provides tools for building applications with custom graphical interfaces. C++ from Python and MATLAB, because C++ is faster. An algorithm to numerically invert functions in the Laplace field is presented. To figure this out, consider exactly what it is that both processes do. Geometric numerical integration is synonymous with structure-pre-ser-ving integration of ordinary differential equations. 141120007827708. Goal: given continuous function f(x) of one variable, compute ∫ f(x) dx over interval from a to b. The Simpson’s 1/3 rule is a numerical method to find the integral within some finite limits and. We'll now consider how to implement these in MATLAB. • In the time domain, ODEs are initial-value problems, so all the conditions. Hello: I have not been able to find out what is the underlying quadrature formula in Matlab's builtin function integral. Numerical Integration "Numerical Methods with MATLAB", Recktenwald, Chapter 11 and "Numerical Methods for Engineers", Chapra and Canale, 5th Ed. Garcia,∗ A. numerical integration - MatLab algorithm for composite Simpson's rule I have tried, just for the fun of it, to write a MatLab-code for the composite Simpson's rule. Numerical integration using SCILAB Integrals can be interpreted as the area under the curve of the function f(x) in a given interval a < x < b. Numerical Integration with the Trapezoidal Rule. An Introduction to Numerical Methods: A MATLAB® Approach, Fourth Edition continues to present a wide range of useful and important algorithms for scientific and engineering ap. Converting serial MATLAB applications to parallel MATLAB applications generally requires few code modifications and no programming in a low-level language is necessary. Description: This is an introductory numerical analysis course. Summary of first day w/ MATLAB in class. The choice of numerical methods was based on their relevance to engineering prob-lems. These notes, prepared for the Durham summer school 2002, are complementary to the monograph of Hairer, Lubich and Wanner "Geometric Numerical Integration". It approaches the subject from a pragmatic viewpoint; theory is kept at a minimum. The second output FY is always the gradient along the 1st dimension of F, going across rows. quadl Numerical integration with adaptive Lobatto quadrature. Simpsons Algorithm for numerical integration using Trapezoid rule for numerical integration using MAT REDS Library: 15. numeric::quadrature(f(x), x = a. It provides tools for building applications with custom graphical interfaces. Question: Problem 2. In addition, it scales well with many workers. The name MATLAB stands for matrix laboratory. the vectorized methods are not as easy to read, and take fewer lines of code to write. Numerical Solution using MATLAB. Numerical Methods for ODE in MATLAB MATLAB has a number of tools for numerically solving ordinary differential equations. There are various reasons as of why such approximations can be useful. Verschelde's MCS 320 Symbolic Computing with MATLAB (and Maple) Lectures Desmond J. numerical integration code Search and download numerical integration code open source project / source codes from CodeForge. MATLAB is widely used. You can integrate from 1 down to some number larger than r, where x=r is the location of the singularity. It is executed at the command line prompt by typing: >> trapezoidal( a,b,nintervals) where a and b are the lower and upper limits of integration. If you were to take the integral of 2x from 0 to 2, where 0 is the lower bound and 2 is the upper bound you would get the following:. Compared to the numerical integration methods, like the program of Simpson 1/3 rule in C given above, the analytical method of integration is quite difficult and time consuming while applying to complex engineering problems. In numerical analysis, Romberg's Method (Romberg 1955) generates a triangular array consisting of numerical estimates of the definite integral by applying Richardson extrapolation (Richardson 1910) repeatedly on the trapezium rule or the rectangle rule (midpoint rule). in Matlab command window. Subscribe to our newsletter to get notifications about our updates via email. Introduction. In the first type, derivative of a function is given and we want to find the function. Initial Value Problem. As we see from the plot, this is more than we need. Numerical Integration 209 Quadrature Integration 216 the work you’re doing can’t be done by hand and will require a numerical solution. See the Matlab help files for other integration commands. E-Book Review and Description: Numerical Methods using MATLAB, 3e, is an in depth reference offering numerous of useful and crucial numerical algorithms that could be carried out into MATLAB for a graphical interpretation to help researchers analyze a selected consequence. We'll now consider how to implement these in MATLAB. Numerical Solution of Differential Equations: MATLAB implementation of Euler's Method The files below can form the basis for the implementation of Euler's method using Mat-lab. You may redistribute it and/or modify it under the terms of the GNU General Public License (GPL) as. This method assumes linear behavior between the data points, and accuracy may be reduced when the behavior between data points is nonlinear. I think this is where my code fails. trapz performs discrete integration by using the data points to create trapezoids, so it is well suited to handling data sets with discontinuities. Research Experience for Undergraduates. If Y is a multidimensional array, then trapz(Y) integrates over the first dimension whose size does not equal 1. Introduction. Watch Online Four sections of this video tutorial are available on YouTube and they are embedded into this page as playlist. quadl Numerical integration with adaptive Lobatto quadrature. Numerical Integration with the Trapezoidal Rule. Trapezoidal Rule for Numerical Integration Trapezoidal Rule for Numerical Integration Internet hyperlinks to web sites and a bibliography of articles. numeric::quadrature returns itself symbolically if the integrand f(x) contains symbolic objects apart from the integration variable x that cannot be converted to numerical values via float. Simpson's rule for numerical integration 25754-simpson-s-rule-for-numerical-integration), MATLAB Central the trapz code but then the simps seems to not work. Applied numerical methods using matlab (wiley,2005) 1. Ofdm matlab code using bpsk. MATLAB allows matrix manipulations, plotting of functions and data, implementation of algorithms, creation of user interfaces, and interfacing with programs written in other languages. Hi, So I have this function that takes data sample locations and function samples at those locations as input and returns an approximation of the integral over the sample range based on the trapezoidal rule. Parallel processing operations such as parallel for-loops and message-passing functions let you implement task- and data-parallel algorithms in MATLAB. Compound Trapezoidal Method for Numerical Integration by admin in Math, Statistics, and Optimization , MATLAB Family , Numerical Integration \$4. An example to price an Arithmetic Average fixed strike Call option in the Black-Scholes framework using Monte Carlo Control Variate. topic of numerical integration is taken up in Chapter 7 and in Chapter 8 meth-ods for the numerical solution of ordinary differential equations are explored. Doing Physics with Matlab op_rs1. = ] = ] ENGRD 241 / CEE 241: Engineering Computation Numerical Integration 14 Numerical Integration Improving the estimate of the integral fit Lagrange polynomials to three points (a pair of segments) integrate those polynomials to obtain a general formula the resulting function must correctly integrate quadratics x 0 x n ENGRD 241 / CEE 241. a guest Sep 19th, 2017 53 Never Not a member of Pastebin yet? % the little t steps for numerical integration %obtain numerical ODE45 solution to use. MATLAB codes for Romberg integration. pdf from MAE 284 at University of Alabama, Huntsville. When writing codes for this lab, it is surely best to use M-files for each approximate integration scheme. In its simplest form, you pass the function you want to differentiate to diff command as an. Matlab code for selective reduced integration; Matlab code for reduced integration with. Numerical gradients, returned as arrays of the same size as F. Numerical Integration. Miranda, M. Apply MATLAB to carry out numerical differentiation and integration. 9) Explain why numerical differentiation is inferior to integration in the presence of noise. The following MATLAB code gives a generalized illustration of numerical integration using Trapezoidal rule. How do I improve Numerical Integration Speed?. MATLAB - Differential - MATLAB provides the diff command for computing symbolic derivatives. Ample material is presented so that instructors will be able to select topics appropriate to their. Calculation of numerical integration of an expression is made easier. To solve the Falkner-Skan equation a fourth-order Runge-Kutta integration scheme was used. First results of axisymmetric. Repeat for the midpoint rule, trapezoidal rule, and Simpson’s rule. Does Matlab provide any alternatives to speed up the operation. Area integral over a triangle in MATLAB -- is numerical integration possible? a triangle in MATLAB -- is numerical integration possible? the rest of the code. Numerical Integration §1 The Newton-Cotes Rules §2 Composite Rules §3 Adaptive Quadrature §4 Gauss Quadrature and Spline Quadrature §5 Matlab's Quadrature Tools An m-point quadrature rule Q for the definite integral I(f,a,b) = Zb a f(x)dx (4. ALL PRODUCTS FROM Gauss Quadrature Method for Numerical Integration MATLAB Family, Numerical Integration. For watching full course of Numerical Computations, visit this page. Choose M-file. This creates a circular reference which I can't figure out how to code in Matlab. All MATLAB code should also be submitted as an. How to apply numerical integration on a symbolic Learn more about integration. Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. though I'd like to know what I should have fixed in the code (more cleaner now). 999e3 numerical representation in Matlab [closed] Comparing the FFT to numerical integration in Matlab. | 2020-02-22T15:07:46 | {
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https://math.stackexchange.com/questions/3052408/how-can-i-prove-that-x-n1-x-n-sinx-n-converges-for-any-x-0 | How can I prove that $x_{n+1} = x_n \sin(x_n)$ converges for any $x_0$?
I have already prove that it is true if exist $$\sin(x_m) = 1$$ or $$\sin(x_m) = -1$$. So I need make proof only for absolute-decreasing progression. But I am afraid that $$\sin(x_n)$$ can has limit $$1$$ and can go to $$1$$ very fast. Is it a good way to count limit $$\sin(x_n)$$? And what solution is correct?
• Welcome to the website. In future, please typeset your equations using Mathjax for better presentation. – Shubham Johri Dec 25 '18 at 20:35
• I think you mean $x_n\to 0$ or $\sin x_n\to 1$. – J.G. Dec 25 '18 at 20:41
• If the limit exists it should satisfy $x_{n+1} = x_n$. Think about what this tells you about possible limit points, and then look at what happens if $x_n$ is above these points or below these points. – tch Dec 25 '18 at 20:43
• @TylerChen "If the limit exists, it should satisfy $x_{n+1} = x_n$". What the heck does this mean – mathworker21 Dec 25 '18 at 20:45
• @Cesareo There are infinitely many solutions. Any $\pi/2+2k\pi$ works. – ImNotTheGuy Dec 25 '18 at 21:12
Let $$y_n = |x_n|$$
Then $$y_{n+1} = y_n |\sin y_n| \leq y_n \forall n$$ (*)
So $$y_n$$ is decreasing and bounded below by $$0$$. Therefore it converges to some limit $$l$$. By passing to the limit in (*) we get that either $$l=0$$ or $$sin l= 1$$ giving the family of solutions discussed in the comments, namely $$l=2k\pi+\frac{\pi}{2}$$
Now note that if $$y_n$$ converges to $$0$$ then $$x_n$$ also converges to $$0$$. If $$y_n$$ converges to a nonzero value $$l$$ then a little more work will also prove that $$x_n$$ also converges to either $$l$$ or $$-l$$ because for n large enough $$y_n$$ will be between $$l$$ and $$l+\frac{\pi}{2}$$ hence $$x_n$$ will have constant sign
Either way, $$x_n$$ will also be convergent
Note that $$0\le|x_{n+1}|=|x_n\sin x_n|\le|x_n|$$, so the sequence of absolute values, $$|x_0|,|x_1|,|x_2|,\ldots$$ definitely converges to a nonnegative limit $$L(x_0)$$ satisfying the equation $$L(x_0)=L(x_0)|\sin L(x_0)|$$. If $$L(x_0)=0$$, we're done: $$|x_n|\to0$$ implies $$x_n\to0$$. If $$L(x_0)\gt0$$, we need only worry about the possibility that $$x_n$$ approaches both $$L(x_0)$$ and $$-L(x_0)$$. But this can't happen: If $$x_n\approx\sigma L(x_0)$$ where $$\sigma=\sin L(x_0)\in\{1,-1\}$$, then, using the fact that $$\sin(\sigma u)=\sigma\sin u$$ for all $$u$$ and $$\sigma^2=1$$, we have
$$x_{n+1}=x_n\sin x_n\approx\sigma L(x_0)\sin(\sigma L(x_0))=\sigma L(x_0)(\sigma\sin L(x_0))=\sigma L(x_0)(\sigma^2)=\sigma L(x_0)\approx x_n$$
Remarks: As ImNotTheGuy observes in a comment beneath the OP, $$\pi/2+2k\pi$$ (with $$k\in\mathbb{Z}$$) is a fixed point for the mapping $$f(x)=x\sin x$$, which means there are sequences that tend to each such limit, as well as sequences that tend to $$0$$. There are, in fact, uncountably many sequences tending to each of the possible limits: If $$k\ge0$$, for example, and $$x_n=\pi/2+2k\pi+\epsilon$$ with $$\epsilon\gt0$$ but very small, then $$x_{n+1}=x_n\cos\epsilon =\pi/2+2k\pi+\epsilon'$$ with
$$\epsilon'=\epsilon\cos\epsilon-(1-\cos\epsilon)(\pi/2+2k\pi)\approx\epsilon-(\epsilon^2/2)(\pi/2+2k\pi)\lt\epsilon$$
(but still positive), so each positive limit has a "basin of attraction" that includes an open inteval of values for $$x_0$$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $$|x_{n+1}|=|x_n\sin x_n|\le|x_n|$$, successive terms in any sequence always get closer to the origin).
In general If $$x_0$$ is an initial approximation of a fixed point , the fixed point iteration generates a sequence of approximants by $$x_{n+1}=\varphi (x_n),$$ where $$\varphi$$ is a continuous function. Suppose that at the fixed point that we indicate with $$\alpha$$ we have $$\varphi '(\alpha)=\varphi ''(\alpha)=\cdots\varphi ^{p-1}(\alpha)=0, \quad \varphi ^p(\alpha)\neq 0 \qquad (*);$$ then we have the following theorem:
Let $$\alpha$$ be a fixed point of $$\varphi$$ and $$I_{\epsilon}:=\{x\in\Bbb R: |x-\alpha|<\epsilon\}$$. Asuume that $$\varphi \in \Bbb C^p[I_{\epsilon}]$$ satisfes $$(*)$$. If $$M(\epsilon):=|\varphi '(t)|<1$$ then the fixed point iteration converges to $$\alpha$$ for every $$x_0\in I_{\epsilon}$$ and the order the convergence is $$p$$.
Note that to use this theorem you have to know tha fxed point $$\alpha$$ (you can plot the function for example and obtain an approximation of point). You can find this argument in this book. | 2019-07-22T12:44:49 | {
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https://math.stackexchange.com/questions/3286150/understanding-the-difference-between-pre-image-and-inverse | # Understanding the difference between pre-image and inverse
I am a little confused as to what the difference between the pre-image and the inverse of a function are and how to find each given a particular function ( I had though they were essentially the same thing but I realise now I was mistaken in that thought ).
From Wikipedia the definitions given are :
Inverse
Let $$f$$ be a function whose domain is the set $$X$$, and whose image (range) is the set $$Y$$. Then $$f$$ is invertible if there exists a function $$g$$ with domain $$Y$$ and image $$X$$, with the property: $$f ( x ) = y \iff g ( y ) = x .$$ If $$f$$ is invertible, the function $$g$$ is unique, which means that there is exactly one function $$g$$ satisfying this property (no more, no less). That function $$g$$ is then called the inverse of $$f$$, and is usually denoted as $$f ^{−1}$$
Pre-image
Let $$f$$ be a function from $$X$$ to $$Y$$. The preimage or inverse image of a set $$B\subseteq Y$$ under $$f$$ is the subset of $$X$$ defined by $$f ^{− 1 }[ B ] = \{ x \in X \mid f ( x ) \in B \}$$.
So using an example I want to see if I have this about right ( please correct any mistakes I make)
Example 1:
Lets say $$f:\Bbb R \rightarrow \Bbb R$$
$$f(x)=x^2$$
For this example, clearly $$f$$ cannot be invertible (hence no inverse) as there exists no function $$g$$ which will satisfy $$f ( x ) = y \iff g ( y ) = x$$. (as it would only map to positive values of $$\Bbb R$$ (i.e. not the whole set))
The pre-image of this function I believe is related to the inverse except it does not require that we map to the whole set $$\Bbb R$$, but rather just a subset of it. Therefore we can find the function $$f^{-1}$$ in an analogous way to to finding the inverse we just have to be more considerate about what the co-domain of this function is.
So if $$f(x)=x^2 \Rightarrow y=x^2$$swap variables to get $$x=y^2 \Rightarrow \sqrt{x}=y=f^{-1}$$
So the pre-image is the set $$f ^{− 1 }[ \Bbb R_+ ] = \{ x \in X \mid f ( x ) \in \Bbb R_+, f^{-1}=\sqrt{x} \}$$
Example 2:
An example of an invertible function would be $$f:\Bbb R \rightarrow \Bbb R$$
$$f(x)=5x$$, as a function $$g(x)=x/5$$ has domain $$\Bbb R$$ and range $$\Bbb R$$ and satisfies $$f ( x ) = y ⇔ g ( y ) = x .$$
The pre-image in this case will be equal to the inverse.
Could anyone please explain to me any mistakes I'm making here ?
• For starters, they are two very different objects. The pre-image of a function is a subset of the domain and the inverse function is a function from the range back to the domain that satisfies certain properties. A function may not be invertible but we can always talk about pre-image of a function. – Anurag A Jul 7 '19 at 20:58
• @AnuragA I know now that they are not the same thing , I was just wondering if I was correct in identifying how they are different – excalibirr Jul 7 '19 at 21:00
• No, one peaks in almost all cases of the preimage of a subset of the target space. Take, for instance, the squaring function $f(x)=x^2$, and ask for the preimage of the set of odd numbers. Then this will be the set of all $\pm\sqrt{2k+1}$, with $k$ running through all integers. If one were to speak of the “preimage of a function”, presumably that would be the preimage of the whole target space, which is all of the domain of definition of the function. – Lubin Jul 7 '19 at 21:09
The biggest difference between a preimage and the inverse function is that the preimage is a subset of the domain. The inverse (if it exists) is a function between two sets.
In that sense they are two very different animals. A set and a function are completely different objects.
So for example: The inverse of a function $$f$$ might be: The function $$g:\mathbb R \to \mathbb R: g(x) = \sqrt[3]{x-9}$$. Whereas the preimage of a set $$B$$ of the function might be $$[1,3.5)\cup \{e, \pi^2\}$$.
Now $$g(x) = \sqrt[3]{x-9}$$ and $$[1,3.5)\cup \{7, \pi^2\}$$ are completely different types of things.
This will be the case if $$f$$ is $$f:\mathbb R \to \mathbb R: f(x) = x^3 + 9$$ and $$B= [10, 51.875) \cup \{352, \pi^6 + 9\}$$.
The inverse $$f^{-1}(x)$$ (if it exist) is the function $$g$$ so that if $$f(x) = y$$ if and only if $$g(y) = x$$. So if $$f(x) = x^3 + 9 = y$$ then if such a function exists it must be that $$g(y)^3 + 9 = y$$ so $$g(y)^3 = y-9$$ and $$g(y) = \sqrt[3]{y-9}$$ so $$g(x) = \sqrt[3]{x-9}$$.
That's that.
The pre-image of $$A= [10, 51.875) \cup \{352, \pi^6 + 9\}$$ is the set $$\{x\in \mathbb R| f(x) \in [10, 51.875) \cup \{352, \pi^6 + 9\}\}=$$
$$\{x\in \mathbb R| x^3 + 9 \in [10, 51.875) \cup \{352, \pi^6 + 9\}\}=$$
$$\{x\in \mathbb R| x^3 \in [1, 42.875) \cup \{343, \pi^6 \}\}=$$
$$\{x\in \mathbb R| x \in [1, 3.5) \cup \{7, \pi^2 \}\}=$$
$$[1, 3.5) \cup \{7, \pi^2 \}\}$$.
And that's the other.
........
Now that's not to say the inverse of a function and the pre-image of a set under the function aren't related. They are. But they refer to different concepts. This is similar to how a rectangle and its area are related. But one is a geometric shape... the other is a positive real number. THey are two different types of animals.
....
I'll add more in an hour or so but I have to take the dog for a walk. I'll be back.
.....
It occurred to me as I was walking the dog that maybe what is confusing you is that the inverse function (if it exists) and the preimage of a set have very similar notation and the only way to tell them apart is in context.
If $$f$$ is invertible then the inverse function is written as $$f^{-1}$$ so if $$f(x) = x^3 + 9$$ then $$f^{-1}(x) = \sqrt[3]{x-9}$$.
But the preimage of $$B$$ under $$f$$ whether $$f$$ is invertible or or not is writen as $$f^{-1}(B)$$.
So if $$f(x) = x^3 + 9$$ then $$f^{-1}(17) = 2$$ means that if you enter $$17$$ into the function $$\sqrt{x -9}$$ you get $$2$$. But $$f^{-1}(\{17\})=\{3\}$$ and $$f^{-1}(\{36,17\}) = \{2,3\}$$ means that set of values that will output $$\{17\}$$ is the set $$\{2\}$$ and the set of values that will output $$\{36,17\}$$ is the set $$\{2,3\}$$.
A few things to note:
If $$f$$ is invertible then the preimage of a set is the same thing as the image of the set under the inverse function and that means the notation is compatible.
If $$f(x) = x^3 + 9$$ then $$f^{-1}([1,36)) = [1,3)$$ can be interpretated as both the the image of the set under the inverse function: $$f^{-1}([1,36))= \{f^{-1}(x) = g(x) = \sqrt[3]{x-9}| x\in [1,36)\}$$
OR it can be interpreted as the preimage for $$f$$: $$f^{-1}([1,36)) = \{x\in \mathbb R| f(x) \in [1,36)\}$$.
but this is not the case if $$f$$ is not invertible.
Say $$f:\mathbb R \to [-1,1]; f(x)\to \sin x$$. This is not invertible.
The pre-image of$$B= \{\frac {\sqrt 2}2\}$$ is $$\{...-\frac {11\pi}4, -\frac {9\pi}4,-\frac{3\pi}4,-\frac \pi 4, \frac \pi 4, \frac {3\pi}4, \frac {9\pi}4, \frac {11\pi}4,....\}$$ this is still written as $$f^{-1}( \{\frac {\sqrt 2}2\})$$ even though there is no function $$f^{-1}:[-1,1]\to \mathbb R$$.
Another thing to note is that not all the elements in $$B$$ have to have pre-image values.
If $$f= x^2+9$$ then $$f^{-1}(\{8\}) = \emptyset$$. This is because $$\{x\in \mathbb R| f(x) = x^2 + 9 \in \{8\}\} = \emptyset$$.
And some elements may have many preimages.
And $$\sin^{-1}(\{\frac {\sqrt2} 2}$$ showed.
• Thank you for such a brilliantly comprehensive answer :) there's literally nothing else I could think to ask . – excalibirr Jul 15 '19 at 7:34
• Actually , I did manage to think of question lol, (it's in the context of continuity in topology) say we have the function $f=x^2$ and a basis set $(-q,q), q \in \Bbb Q$ in our target. then I know that $f^{-1}((-q,q))=\{ x \in X| x\in (-\sqrt{q}, \sqrt{q})\}$ but why isn't it $(\sqrt{-q}, \sqrt{q})$ instead ?, also say we have the set (a,b) (a<b) with the same function then is $f^{-1}=(-\sqrt{b},-\sqrt{a}) \cup (\sqrt{a},\sqrt{b})$ ? – excalibirr Jul 15 '19 at 9:28
When we talk about pre-image then it has two components: a set and a function. So when we say $$f^{-1}[B]$$, then we want the pre-image of the set $$B$$ (a subset of the co-domain) under the function $$f$$. So asking about pre-image of a function is a bit ambiguous.
Let us consider $$f:\{1,2,3\} \rightarrow \{a,b,c\}$$ such that $$f(1)=a, f(2)=a$$ and $$f(3)=c$$. Then \begin{align*} f^{-1}\left[\{a\}\right] & =\{1,2\}\\ f^{-1}\left[\{c\}\right] & =\{3\}\\ f^{-1}\left[\{a,c\}\right] & =\{1,2,3\}\\ f^{-1}\left[\{b\}\right] & =\emptyset\\ f^{-1}\left[\{a,b,c\}\right] & =\{1,2,3\} \end{align*}
The function $$f$$ (as described above) is not invertible. So relating the inverse function (which doesn't exist) to the inverse image is meaningless in this case.
One can check invertibility of a function $$f: A \rightarrow B$$ by checking the inverse images of singleton subsets of the co-domain.
What it means is that: if we can ensure that for every $$b \in B$$, the inverse image set $$f^{-1}\left[\{b\}\right]$$ has exactly one element (this is to ensure both one-one and ontoness), then $$f$$ is invertible.
So, there's no such thing as the preimage of a function. Functions can have inverses; functions do not have preimages.
An inverse is something that certain functions have, and the inverse of a function is another function.
Given a function, a preimage is something that sets have, and the preimage of a set is another set.
Specifically:
Given a function $$f : A \to B$$, that function may or may not have an inverse. If it does, then that inverse is a function $$B \to A$$.
Given a function $$f : A \to B$$, and a set $$s$$ which is a subset of $$B$$, that set always has a preimage under $$f$$. That preimage is a subset of $$A$$.
That might clear up some of your confusion. | 2021-05-10T01:39:25 | {
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http://math.stackexchange.com/questions/73945/stuck-trying-to-prove-an-inequality/74812 | # Stuck trying to prove an inequality
I have been trying to prove (the left half of) the following inequality:
$$\underbrace{\sum_i \sum_j |x_i| \le \sum_i \sum_j |x_i + x_j|}_\textrm{?} \le 2 \sum_i \sum_j |x_i|$$
(All $x_i$s are arbitrary reals and sums are over $1, 2, \dots, n$)
The right half follows by a simple application of the triangle inequality, but the left half isn't as simple.
I tried induction, and I could reduce the above problem to proving the following statement:
$$\sum_i |x_i + x_{n+1}| + \sum_{i=1}^{n+1} |x_i + x_{n+1}| \ge \sum_{i=1}^{n+1} |x_i| + n|x_{n+1}|$$ or more simply (replacing $x_{n+1}$ by $y$), $$2\sum_i |x_i + y| \ge \sum_i |x_i| + (n-1)|y|$$
However, I don't know if the above "reduction" is any simpler than the original problem itself!
Can I continue the above line of thought, or is there an easier way to solve this problem? Any help is appreciated! :)
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Since the LHS and the RHS has no variable in $j$, you may replace $\sum_i\sum_j|x_i|$ by $n\sum_i|x_i|$. – AD. Oct 19 '11 at 9:34
In the summation: $\sum_i \sum_j |x_i + x_j|$ - Do you mean that i and j have the domain $0 <=i <=n , \$0 <=j <=n ? Or does each summation have its own lower and upper limits? – Emmad Kareem Oct 22 '11 at 8:47
Can you explain to me how this works for 1,1,-1? I get: $9 \le 2+2+0+0+0+0$ – Phira Oct 22 '11 at 12:24
@Phira I get $9\leq 2+0+2+2+0+2+2+0+0=10$. – Mike Wierzbicki Oct 22 '11 at 13:44
why was this two year old question closed instead of the two week old question? – robjohn Oct 9 '13 at 23:58
I believe that some smart tricks will solve the problem (e.g. by recasting it into the form of some known inequality), but what came to my mind first was a proof inspired by the simplex method. Although it does not look smart, it may be of some interests to the readers:
1. Note that by scaling the inequality, we may assume, without loss of generality, that $|x_i|\le1$ for all $i$.
2. So we want to prove the inequality for all $(x_1,\ldots,x_n)$ inside the hypercube $[-1,1]^n$.
3. Divide the hypercube $[-1,1]^n$ into pieces by the hyperplanes $x_i=x_j$ and $x_i=0$. For instance, when $n=2$, the four "hyperplanes" (straight lines in this case) $x=y$, $x=-y$, $x=0$ and $y=0$ will divide the square $[-1,1]^2$ into eight pieces in a manner akin to the Union Jack.
4. In mathematics, we call each of these pieces a simplex.
5. Note that inside each simplex, each $x_i$ or $x_i+x_j$ does not change sign.
6. Therefore, inside each simplex, the function $f(x_1,\ldots,x_n)=-\sum_i \sum_j |x_i| + \sum_i \sum_j |x_i + x_j|$ is identical to $-\sum_i \sum_j s_ix_i + \sum_i \sum_j t_{ij}(x_i + x_j)$ for some constants $s_i, t_{ij}=\pm1$.
7. In other words, $f$ is a linear function inside each simplex.
8. In linear programming, it is well known that the extremum of a linear function can be attained at the corners of the simplex. You can imagine that when you move a plane closer and closer to a simplex, it will touch the simplex at a corner first (or it touches the whole face, with the corners of the face included).
9. The corners of our simplices, however, are those points such that $x_i\in\{-1,0,1\}$.
10. So, we only need to prove the inequality for these corner points.
11. Let $a$ of the $x_i$s are equal to 1, $b$ of them are equal to $-1$ and $n-a-b$ of them are equal to zero. Then \begin{align} &-\sum_i \sum_j |x_i| + \sum_i \sum_j |x_i + x_j|\\ =&-n(a+b) + 2a^2+2b^2+2a(n-a-b)+2b(n-a-b)\\ =&2a^2+2b^2 - 2(a+b)^2 + n(a+b)\\ \ge&2a^2+2b^2 - 2(a+b)^2 + (a+b)^2\\ =&(a-b)^2\\ \ge&0. \end{align}
12. Hence the result.
In step 11 of the above, equality holds iff $a-b=0$ and $a+b\in\{0,n\}$. In other words, equality holds iff $a=b=0$ or $a=b=n/2$, that is, iff all $x_i$s are zero, or when $n$ is even, exactly half of the $x_i$s (before scaling) are identical to some $x$ and the other half are $-x$s. However, this condition for equality is derived using the corner points. The above proof does not indicate whether equality can hold in other cases.
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Wow. That was really slick! I wish I could upvote your answer several times more (need some help here!) for not just solving the above problem, but also for showing a very novel method. Is this part of some standard repertoire for solving inequalities? Where did you come across this? – quantumelixir Oct 19 '11 at 15:19
The technique itself (i.e. narrowing the domain to corner points) is not new. It is standard material in textbooks on linear programming or operations research. I use it here merely because I am not clever enough to think of a straightforward solution. As I said in the answer, my proof is actually a bit problematic -- although it proves the inequality, it fails to identify all cases where equality holds. – user1551 Oct 19 '11 at 18:52
Let $p$ be the number of $x_i\ge0$, $m$ be the number of $x_i<0$, and $n=p+m$. $$\small\begin{array}{} &\sum_i\;\:\sum_j|x_i+x_j|\\ &=\sum_{x_i\ge0}\;\:\sum_{x_j\ge0}|x_i+x_j| &+\sum_{x_i\ge0}\;\:\sum_{x_j<0}|x_i+x_j| &+\sum_{x_i<0}\;\:\sum_{x_j\ge0}|x_i+x_j| &+\sum_{x_i<0}\;\:\sum_{x_j<0}|x_i+x_j|\\ &\ge\sum_{x_i\ge0}\;\:\sum_{x_j\ge0}|x_i|+|x_j| &+\left|\sum_{x_i\ge0}\;\:\sum_{x_j<0}|x_i|-|x_j|\right| &+\left|\sum_{x_i<0}\;\:\sum_{x_j\ge0}|x_j|-|x_i|\right| &+\sum_{x_i<0}\;\:\sum_{x_j<0}|x_i|+|x_j|\\ &=2p\sum_{x_i\ge0}|x_i| &+2\left|m\sum_{x_i\ge0}|x_i|-p\sum_{x_i<0}|x_i|\right| &&+2m\sum_{x_i<0}|x_i| \end{array}$$
Subtracting $\displaystyle\sum_i\;\:\sum_j|x_i|=n\sum_i\;|x_i|$ from the inequality above yields \small\begin{align}{} &\sum_i\;\:\sum_j|x_i+x_j|-\sum_i\;\:\sum_j|x_i|\\ &\ge(p-m)\sum_{x_i\ge0}|x_i| +\left|2m\sum_{x_i\ge0}|x_i|-2p\sum_{x_i<0}|x_i|\right| +(m-p)\sum_{x_i<0}|x_i|\\ &=(p-m)\sum_{i}x_i +\left|(m-p)\sum_{i}|x_i|+n\sum_ix_i\right|\\ &\ge0\tag{1} \end{align} In the last inequality, if $p-m$ and $\sum\limits_ix_i$ have the same sign, we can ignore the quantity in absolute values. If $p-m$ and $\sum\limits_ix_i$ have different signs, then the quantities in the absolute values have the same sign and $|m-p|\sum\limits_i|x_i|\ge(m-p)\sum\limits_ix_i$.
Therefore, $(1)$ says that $$\sum_i\;\:\sum_j|x_i|\le\sum_i\;\:\sum_j|x_i+x_j|$$
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This follows exactly the same approach suggested in my answer. An acknowledgement is in order perhaps? – quantumelixir Oct 22 '11 at 20:16
Nice work. Does the same inequality and perhaps similar approach work for $x_i$ being complex or a vector? – Hansen Oct 18 at 23:52
Assume an ordering of the $x_i$ such that $x_1 \le x_2 \le \dots \le x_k < 0 \le x_{k+1} \le x_{k+2} \le \dots \le x_n$. Let's call $T_1 = \sum_{i=1}^k |x_i|$, $T_2 = \sum_{i=k+1}^n |x_i|$, and $T=\sum_i |x_i|=T_1+T_2$ so that $T, T_1, T_2 \ge 0$.
Since we only require a lower bound on the sum $\sum_{i \neq j} |x_i + x_j|$, we only consider $2\binom{k}{2} + 2\binom{n-k}{2}$ terms of the $2\binom{n}{2}$ total terms that the sum contains i.e. we only pick terms where both $x_i$ and $x_j$ have the same sign. This solves the inequality in many cases. For the other cases, we include the cross terms too later.
\begin{align*} \sum_{i \neq j} |x_i + x_j| &\ge \sum_{i=1}^k \sum_{j=1,j\neq i}^k |x_i+x_j| + \sum_{i=k+1}^n \sum_{j=k+1,j\neq i}^n |x_i+x_j|\\ &= \sum_{i=1}^k \sum_{j=1,j\neq i}^k (|x_i|+|x_j|) + \sum_{i=k+1}^n \sum_{j=k+1,j\neq i}^n (|x_i|+|x_j|)\\ &= \sum_{i=1}^k ((k-1)|x_i|+T_1-|x_i|) + \sum_{i=k+1}^n ((n-k-1)|x_i|+T_2-|x_i|)\\ &= \sum_{i=1}^k ((k-2)|x_i|+T_1) + \sum_{i=k+1}^n ((n-k-2)|x_i|+T_2)\\ &= ((k-2)T_1+kT_1) + ((n-k-2)T_2+(n-k)T_2)\\ &= 2\{(k-1)T_1 + (n-k-1)T_2\}\\ \end{align*}
Now, the last expression is a function of $T_1$ and $T_2$. Since we require a lower bound in terms of $T=T_1+T_2$, we can get two equivalent expressions in terms of either $T_1$ and $T$ or in terms of $T_2$ and $T$. Thus,
$$\sum_{i \neq j} |x_i + x_j| \ge 2(2k-n)T_1+2(n-k-1)T$$ $$\sum_{i \neq j} |x_i + x_j| \ge 2(n-2k)T_2+2(k-1)T$$
$$2\sum_{i \neq j} |x_i + x_j| \ge 2\left\{(2k-n)T_1+(n-2k)T_2+(n-2)T\right\}$$ $$\implies \sum_{i \neq j} |x_i + x_j| \ge (2k-n)(T_1-T_2)+(n-2)T \ge (n-2)\sum_i |x_i|$$ $$\text{whenever,} \; (2k-n)(T_1-T_2) \ge 0$$
For the cases when the above inequality doesn't hold, we need to consider the cross terms too. Computing the cross terms alone:
\begin{align*} \sum_{i \neq j} |x_i + x_j| &\ge \sum_{i=1}^k \sum_{j=k+1}^n |x_i+x_j| + \sum_{i=k+1}^n \sum_{j=1}^k |x_i+x_j|\\ &\ge 2\sum_{i=1}^k \sum_{j=k+1}^n |x_i+x_j|\\ &\ge 2\sum_{i=1}^k \sum_{j=k+1}^n \max\{|x_i|-|x_j|,|x_j|-|x_i|\}\\ &\ge 2 \max\{(n-k)T_1-kT_2, kT_2-(n-k)T_1\}\\ \end{align*}
Including the cross terms along with the terms from the previous case, we have:
\begin{align*} \sum_{i \neq j} |x_i + x_j| &\ge 2 \max\{(n-k)T_1-kT_2, kT_2-(n-k)T_1\} + 2\{(k-1)T_1 + (n-k-1)T_2\}\\ &\ge 2 \max\{(n-1)T_1+(n-2k-1)T_2, (2k-n-1)T_1+(n-1)T_2\} \; (*)\\ \end{align*}
The $\max$ term can go one of two ways. Using the first term we get: $$\sum_{i \neq j} |x_i + x_j| \ge 2 \{(n-1)T_1+(n-2k-1)T_2\}$$
We can again write the above in terms of either $T,T_1$ or $T,T_2$ giving two inequalities:
\begin{align} \sum_{i \neq j} |x_i+x_j| &\ge (n-2k-1)T+2kT_1\\ \sum_{i \neq j} |x_i+x_j| &\ge (n-1)T-2kT_2 \end{align}
Summing the above two inequalities,
\begin{align*} \sum_{i \neq j} |x_i + x_j| &\ge (2n-2k-2)T+2k(T_1-T_2)\\ &\ge (n-2)T+(n-2k)T+2k(T_1-T_2)\\ &\ge (n-2)T \quad \text{for the case} \quad (2k-n) \le 0, (T_1-T_2) \ge 0\\ \end{align*}
Similarly, for the other case, when $(2k-n) \ge 0, (T_1-T_2) \le 0$, we use the other $\max$ term from $(*)$ and follow the above steps routinely to prove the inequality.
Q.E.D
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Well done. (more characters) – Jonas Teuwen Oct 22 '11 at 16:36
Thanks Jonas! Please correct any errors if you spot them. – quantumelixir Oct 22 '11 at 16:54
The above presentation can be greatly simplified, but it's rather long because of it's long edit history. – quantumelixir Oct 22 '11 at 20:17
One simplification: $\max\{(n-k)T_1-kT_2, kT_2-(n-k)T_1\}=|(n-k)T_1-kT_2|$. At least it is easier to read. – robjohn Oct 22 '11 at 21:31
As I commented before, it helps to use $2|(n-k)T_1-kT_2|$ instead of $2\max\{(n-k)T_1-kT_2,kT_2-(n-k)T_1\}$. Then, use $2T_1=(T+(T_1-T_2))$ and $2T_2=(T-(T_1-T_2))$ to convert $$\sum_{i\not=j}|x_i+x_j|\ge2|(n-k)T_1-kT_2|+2\{(k-1)T_1+(n-k-1)T_2\}$$ into $$\sum_{i\not=j}|x_i+x_j|\ge|(n-2k)T+n(T_1-T_2)|+(n-2)T+(2k-n)(T_1-T_2)$$ If $(2k-n)(T_1-T_2)\ge0$, ignore the stuff in the absolute values. If $(2k-n)(T_1-T_2)<0$, then $(n-2k)$ and $(T_1-T_2)$ inside the absolute values have the same sign and $|n-2k|T\ge(n-2k)(T_1-T_2)$ which cancels $(2k-n)(T_1-T_2)$. – robjohn Oct 23 '11 at 8:13 | 2014-10-31T06:25:28 | {
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https://math.stackexchange.com/questions/2595431/weak-convergence-of-e-n-e-n1-cdots-e-2n-n-1-infty-in-ell-in | # Weak convergence of $(e_n + e_{n+1} +\cdots + e_{2n})_{n=1}^\infty$ in $\ell^\infty$
Let $(e_n)_{n=1}^\infty$ be the canonical vectors in $\ell^\infty$. Consider the sequence $$(e_n + e_{n+1} +\cdots + e_{2n})_{n=1}^\infty$$
Does it converge weakly in $\ell^\infty$?
Obviously $e_n + e_{n+1} + \cdots + e_{2n} \xrightarrow{n\to\infty} 0$ coordinate-wise, so the only possible candidate for the weak limit is $0$.
I believe it indeed weakly converges to $0$.
Let $f \in (\ell^\infty)^*$ be a bounded linear functional. Assume $f(e_1 + e_{n+1} \cdots +e_{2n}) \not\to 0$. Therefore, there exists $\varepsilon > 0$ and a subsequence $$(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)})_{n=1}^\infty$$ such that $|f(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)})| \ge \varepsilon$ for all $n \in \mathbb{N}$.
We have $$f(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)}) = |f(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)})|e^{i\phi_n}$$
for some $\phi_n \in \mathbb{R}$.
Inductively we construct a further subsequence $(e_{q(p(n))} + e_{q(p(n))+1} +\cdots + e_{2q(p(n))})_{n=1}^\infty$ such that the supports of $e_{q(p(n))} + e_{q(p(n))+1} +\cdots + e_{2q(p(n))}$ and $e_{q(p(n+1))} + e_{q(p(n+1))+1} +\cdots + e_{2q(p(n+1))}$ are disjoint:
Set $q(p(1)) = p(1)$. Let $p(k)$ be such that $p(k) \ge 2p(1)$. Set $q(p(2)) = p(k)$. Le $p(k')$ be such that $p(k') \ge 2p(k) = 2q(p(2))$. Set $q(p(3)) = p(k')$ and so on.
Now consider $$x_n = \sum_{k=1}^n e^{-i\phi_{q(k)}}(e_{q(p(k))} + e_{q(p(k))+1} +\cdots + e_{2q(p(k))})$$ Since the summands have disjoint supports and $|e^{-i\phi_{q(k)}}| = 1$ it follows $\|x_n\|_\infty = 1$ for all $n \in \mathbb{N}$.
However, $$|f(x_n)| = \left|\sum_{k=1}^n e^{-i\phi_{q(k)}} f(e_{q(p(k))} + e_{q(p(k))+1} +\cdots + e_{2q(p(k))})\right| = \sum_{k=1}^n |f(e_{q(p(k))} + e_{q(p(k))+1} +\cdots + e_{2q(p(k))})| \ge n\varepsilon \xrightarrow{n\to\infty} +\infty$$
so $f$ is unbounded. A contradiction.
Therefore $f(e_n + e_{n+1} + \cdots + e_{2n}) \xrightarrow{n\to\infty} 0$ for all $f \in (\ell^\infty)*$ so $e_n + e_{n+1} + \cdots + e_{2n} \rightharpoonup 0$.
Is my proof correct? Is there an easier (and still elementary) way to conclude that it converges weakly to $0$?
Edit: I lied - there is another proof that might well be regarded as much simpler. Inspired by the OP's comment on what he was really trying to do:
Trivial Lemma Suppose $X$ is a Banach space and $Y$ is a closed subspace. If $y_n\to y$ weakly in $Y$ then $y_n\to y$ weakly in $X$.
Proof: Suppose $f\in X^*$. Let $g=f|_Y$. Then $g\in Y^*$, so $$f(y_n)=g(y_n)\to g(y)=f(y).$$
So to answer the original question it's enough to show that $e_n+\dots+e_{2n}\to0$ weakly in $c_0$. This is clear, since $c_0^*=\ell_1$.
Original: Looks right. Is there an easier proof? I doubt it - what you did is already very simple, just applying a few definitions. If it doesn't look simple that's just because of the notation.
I'd write the very same proof as follows - possibly looks simpler, possibly easier to read (in fact I confess I didn't read what you wrote very carefully, just sort of skimmed it and saw that the idea clearly worked):
Say $x_n=e_n+\dots+e_{2n}$. Suppose $x_n$ does not tend to $0$ weakly. Then there exists $f\in\ell_\infty^*$ so $f(x_n)\not\to0$.
So there exist a subsequence $(x_{n_j})$, $\epsilon>0$, and complex numbers $\alpha_j$ with $|\alpha_j|=1$, so that $$f(y_j)\ge\epsilon$$if $$y_j=\alpha_jx_{n_j}.$$We may assume that $n_{j+1}>2n_j$, so that the $y_j$ have disjoint support. Since the $y_j$ have disjoint support we have $$||y_1+\dots+y_n||=1,$$although$$f(y_1+\dots+y_n)\ge n\epsilon,$$contradicting the fact that $f$ is bounded.
That looks simpler to me - I think it's much easier to read, since the displayed formulas are less intricate. But it's exactly the same proof.
• Ok, thanks for the feedback. Not really related to the question, but is there an example of a sequence in $c_0$ which converges weakly in $c_0$ but not in $\ell^\infty$? The sequence $(e_n + \cdots + e_{2n})_{n=1}^\infty$ was my attempt to construct such an example, but we have established that it converges weakly to $0$ in both $c_0$ and $\ell^\infty$. Jan 7 '18 at 12:49
• @mechanodroid See edit. Jan 7 '18 at 13:07
• Great, exactly the kind of thing I was hoping for! Jan 7 '18 at 13:12 | 2021-10-27T03:01:55 | {
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http://math.stackexchange.com/questions/82944/interpolating-between-the-lp-norm-and-the-h%c3%b6lder-semi-norm | Interpolating between the $L^{p}$ norm and the Hölder semi-norm
Set-up.
For a bounded continuous function $u \colon \mathbb{R}^n \to \mathbb{R}$, the $\gamma$-Hölder semi-norm of $u$ is $$\begin{eqnarray} [u]_{C^\gamma} &=& \sup \left\{\frac{|u(x) - u(y)|}{|x-y|^\gamma} : x,y \in U, x \neq y \right\} \\ &=& \inf \left\{ C \geq 0 : |u(x) - u(y)| \leq C |x-y|^{\gamma} \text{ for all } x,y \in U \right\}. \end{eqnarray}$$
The Problem
Fix $1 \leq p < \infty$, $0 < \gamma \leq 1$, and $0 < \lambda < 1$ such that $$0 = \frac{\lambda}{p} - (1-\lambda)\frac{\gamma}{n}.$$ I am trying to prove that there is a constant $C$ such that $$\|u\|_{L^\infty} \leq C \|u\|_{L^p}^{\lambda} [u]_{C^\gamma}^{1-\lambda}$$ for every compactly supported $C^{1}(\mathbb{R}^n)$ function $u$.
My Strategy
My plan is to use the interpolation result for Lebesgue spaces: For every $q,r$ satisfying $p < q < r \leq \infty$ and $$\frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r},$$ we have $$\|u\|_{L^q} \leq \|u\|_{L^{p}}^{\lambda} \|u\|_{L^r}^{1-\lambda}$$ for every $u \in L^p \cap L^r$.
Solving for $1/r$ in terms of $q$ and using the relationship between $p$, $\gamma$, and $n$, we find $$\frac{1}{r} = \frac{1/q - \lambda/p}{1-\lambda} = \frac{1/q}{1-\lambda} - \frac{\gamma}{n}$$ So, by letting $q \to \infty$, we have $r \to -n / \gamma$, and $$\frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r},$$ goes to $$0 = \frac{\lambda}{p} - (1-\lambda)\frac{\gamma}{n}.$$ Meanwhile, for $u \in L^{\infty}$ (which certainly holds when $u$ is compactly supported and $C^1$), we have that $\lim_{q \to \infty} \|u\|_{L^q} =\|u\|_{L^\infty}$.
So if I could prove that $\|u\|_r \to [u]_{C^{\gamma}}$ as $q \to \infty$ (i.e., as $r \to -n / \gamma$), I'd be done. The problem is that I don't know how to prove this. Moreover, I'm not sure if this is even the right approach to prove the desired inequality.
-
There is no reason that $\|u\|_r$ should tell you anything about $[u]_{C^\gamma}$ on its own (they measure completely different things). Also, your Lebesgue interpolation inequality only holds for $p < q < r$, but you are taking $q \to \infty$ and $r \to -n/\gamma$ (does $r < 1$ even make sense?). – Jeff Nov 17 '11 at 21:18
You may have to assume that $u \in C^\gamma$ (instead of $u$ continuous) and use some kind of Sobolev embedding. – Jeff Nov 17 '11 at 21:21
@Jeff: It was a mistake to not assume $u \in C^{\gamma}$. Thank you for pointing it out. – fferic Nov 17 '11 at 22:03
@Jeff: You are right that $q \rightarrow \infty$ and $r \rightarrow -n/\gamma$ is not compatible with the requirement $p<q<r$ in the Lebesgue interpolation inequality. This is one reason why I am not so sure my strategy will really work. – fferic Nov 17 '11 at 22:06
@Jeff: Could you please elaborate a bit more on your comment about using a Sobolev embedding? – fferic Nov 17 '11 at 22:07
This is my first post on MSE and I'm just a grad student so forgive me if screw anything up. [Disclaimer: This proof is likely to be overkill, however I like that it shows what exactly is going on and these techniques generalize to a lot of different problems.] Hopefully someone else can chime in to validate my answer.
I will only consider the case $U = \Bbb R^n$.
The proof will be similar to the proof of Theorem A.3 in Tao's book "Nonlinear dispersive equations". In Theorem A.3, Tao uses basic Littlewood-Paley theory to prove a fairly general version of the Gagliardo-Nirenberg inequality, which is your equation after your swap out the Hölder space with an appropriate Sobolev space.
First, we consider the case when $\| u \|_{L^p} = \| u \|_{C^\gamma} = 1$ and show that the $L^\infty$ norm is bounded above by a constant. For general $u$ in $L^p \cap C^\gamma$ we can reduce to the first case by considering $v(x) = A u(B x)$ for an appropriate choice of constants $A$ and $B$ which make both of the norms $1$.
The main idea is to decompose $u$ into high and low frequencies. We will use the $L^p$ norm to bound the low frequencies and the Hölder regularity to bound the high frequencies.
To begin, the triangle inequality gives
$$\| u \|_{L^\infty} \le \sum_{k=0}^{-\infty}\| P_k u \|_{L^\infty} + \sum_{k=1}^{\infty} \| P_k u \|_{L^\infty}.$$
The Bernstein inequalities tell us that $$\| P_k u \|_{L^\infty} \le C 2^{k \frac{n}{p}} \| P_k u \|_{L^p} = C 2^{k \frac{n}{p}}$$ which in turns gives a bound on the low frequencies in terms of a convergent geometric sum $$\sum_{k=0}^{-\infty}\| P_k u \|_{L^\infty} \le C \sum_{k=0}^{-\infty} 2^{k \frac{n}{p}} = C_1 < \infty.$$
To deal with the high frequency terms, we use the part of the Littlewood-Paley decomposition of $C^\gamma$ which states that $$\sup_{k \in \Bbb Z} 2^{k \gamma} \| P_k u \|_{L^\infty} \le C' \| u \|_{C^\gamma}$$
This leads to a bound on the high frequencies in terms of another convergent geometric sum:
$$\sum_{k=1}^{\infty}\| P_k u \|_{L^\infty} \le \sum_{k=1}^{\infty} 2^{-k \gamma} C' \| u \|_{C^\gamma} = C' \sum_{k=1}^{\infty} 2^{-k \gamma} = C_2 < \infty$$
Together these shows that $\| u \|_{L^\infty} \le C_1 + C_2$.
- | 2015-11-26T04:01:23 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=jol92prfec42b5h1b01tka3qi5&action=printpage;topic=1189.0 | # Toronto Math Forum
## MAT244-2018S => MAT244--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 08:42:26 PM
Title: FE-P3
Post by: Victor Ivrii on April 11, 2018, 08:42:26 PM
Find the general solution of
\begin{equation*}
y''' -6y'' +11y'- 6y=2\frac{e^{3x}}{e^x+1} .
\end{equation*}
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 11, 2018, 11:37:34 PM
First we find the solution for the homogeneous system
y^{(3)}-6y^{(2)}+11y^{(1)}-6y=0
The corresponding characteristic equation is
r^3-6r^2+11r-6=0
Three roots are
r_1=1
r_2=2
r_3=3
Then the solution for the homogeneous system is
y_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}
where
y_1(t)=e^{x}
y_2(t)=e^{2x}
y_3(t)=e^{3x}
Then we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have
W[y_1,y_2,y_3]=2e^{6x}
W_1[y_1,y_2y_3]=e^{5x}
W_2[y_1,y_2y_3]=-2e^{4x}
W_3[y_1,y_2，y_3]=e^{3x}
and
g(x)=2\frac{e^{3x}}{e^{x}+1}
And then we have the following integration
\int \frac{W_1\cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{e^{2x}}{e^{x}+1}
\int\frac{e^{2x}}{e^{x}+1}=e^{x}-\ln(e^{x}+1)+c_4
\int \frac{W_2\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\int\frac{e^{x}}{e^{x}+1}
-2\int\frac{e^{x}}{e^{x}+1}=-2\ln(e^{x}+1)+c_5
\int \frac{W_3 \cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{1}{e^{x}+1}
\int\frac{1}{e^{x}+1}=x-\ln(e^{x}+1)+c_6
And finally, the required general solution $y(t)$
y(t)=\sum_{i=1}^3 y_i(t)\cdot\int\frac {W_i\cdot g(x)dx}{W[y_1,y_2,y_3]}
Title: Re: FE-P3
Post by: Meng Wu on April 11, 2018, 11:47:31 PM
Small Error: $W_2(x)$ should be $2e^{4x}$.
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 11, 2018, 11:49:31 PM
Small Error: $W_2(x)$ should be $2e^{4x}$.
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$
Title: Re: FE-P3
Post by: Meng Wu on April 11, 2018, 11:54:19 PM
Small Error: $W_2(x)$ should be $2e^{4x}$.
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$
Oh, you're right. My mistake.
Title: Re: FE-P3
Post by: Meng Wu on April 11, 2018, 11:56:10 PM
Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 11, 2018, 11:57:57 PM
Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.
Thanks again and I will modify it :)
Title: Re: FE-P3
Post by: Syed Hasnain on April 12, 2018, 04:33:47 PM
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 12, 2018, 10:01:46 PM
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
Sorry what do you mean by "the solution is incomplete after Y(x)"
I did write a bit more on the exam (expanding the summation) but I think one should be able to get full marks if he integrates everything and mention how the solution is composed.(Given that the integral is correct)
Title: Re: FE-P3
Post by: Victor Ivrii on April 15, 2018, 03:15:29 AM
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
The only thing which was missing in the solution, is the final answer, but it warrants neither such claim, nor uploading your solution.
General remark:
It would be better to denote "parameters" by uppercase letters $C_1(x)$, $C_2(x)$,... and constants by lowercase letters $c_1$, $c_2$,... | 2022-05-24T22:54:47 | {
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http://math.stackexchange.com/questions/190073/what-is-mathbb-zt-what-are-the-double-brackets | # What is $\mathbb Z[[t]]$? What are the double brackets?
What does $\mathbb{Z}[[t]]$ mean? Why are there double square brackets?
I can't search through Google, because I can't search Latex.
-
I think it's the ring of formal power series with coefficents from $\mathbb{Z}$, see en.wikipedia.org/wiki/Formal_power_series – Mikko Korhonen Sep 2 '12 at 16:19
You actually can search LaTeX: latexsearch.com – huon-dbaupp Sep 2 '12 at 16:22
If this is from a book, it might have an list of symbols in the back that you can check. – Jair Taylor Sep 2 '12 at 17:10
That is the ring of formal power series in $t$ with integer coefficients, i.e., of $$\sum_{n=0}^\infty a_nt^n,$$ with $a_n\in\Bbb Z$, componentwise addition, and multiplication appropriately defined.
The double brackets distinguish it from $\Bbb Z[t]$, which is the ring of polynomials in $t$ with integer coefficients. We can always evaluate the members of $\Bbb Z[t]$ for any complex value of $t$, but we generally can't evaluate members of $\Bbb Z[[t]]$ for $t\neq 0$. To my mind, the double bracket is a reminder that we need to leave the $t$ alone, and not worry about evaluation.
-
(IMO there is a sense in which can view "evaluation at $t=a$" to be the quotient map $\Bbb Z[[t]]\to \Bbb Z[[t]]/(t-a)$, though this is cheating and doesn't send everything all the way down to $\Bbb Z$.) – anon Sep 2 '12 at 17:08
Interesting point! – Cameron Buie Sep 2 '12 at 17:11
I laughed. I'm going to steal this saying. Many thanks. "just leave $t$ alone" ha. – James S. Cook Sep 2 '12 at 18:21
Have at it, James! – Cameron Buie Sep 3 '12 at 1:47
@CameronBuie Is it ok to use the symbol $\mathbb{Z}\llbracket t\rrbracket$ in latex instead of $\mathbb{Z}[[t]]$ or is that wrong? – Pratyush Sarkar Mar 4 '13 at 14:51
If $A$ is any ring, the notation $A[[T]]$ stands for the ring of formal power series with coefficients in $A$, i.e. the ring whose elements are the expressions $$a_0+a_1T+a_2T^2+a_3T^3+\cdots$$ with the obvious sum and product.
- | 2015-08-03T13:44:30 | {
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https://www.physicsforums.com/threads/how-do-i-represent-this-sine-function.803221/ | # How do I represent this sine function?
1. Mar 15, 2015
I was struggling to represent the following for integer values of n:
$$\sin \left( \dfrac {n\pi } {2}\right)$$
I know for even n, we get zero
But for odd n, it alternates beween 1 and -1 for every other odd. Is there a compact way to represent that? I feel like I'm being dumb and missing something obvious.
Thanks!
2. Mar 15, 2015
### piggupiggu
1 when n = 1 + 2x
-1 when n = 3 + 2x
where x is zero or a positive even integer
3. Mar 15, 2015
### Mentallic
What you have is a piecewise function (the variable being n in this case).
$$k\in \mathbb{Z} \\ f(n) = \sin{\frac{\pi n}{2}} = \left\{ \begin{array}{lr} 0 & : n=2k\\ 1 & : n=4k+1\\ -1 & : n=4k+3 \end{array} \right.$$
4. Mar 15, 2015
### LCKurtz
I guess you are looking for something similar to $\cos(n\pi) = (-1)^n$ for $\sin(\frac{n\pi} 2)$. You may be able to find such an expression, but I don't think you will find anything "simpler" than $\sin(\frac{n\pi} 2)$ itself, unless you consider piecewise defined functions simpler. I don't.
5. Mar 17, 2015
Thank you so much for the responses everyone!
6. Mar 18, 2015
### Mentallic
I gave it a little more thought and came up with
$$\sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}\left(-1\right)^{\frac{n-1}{2}}\left(1-(-1)^n\right)$$
7. Mar 19, 2015
### jbunniii
$$\sin\left(\frac{n\pi}{2}\right) = \frac{1}{2i}\left(i^n - (-i)^n\right) = \frac{1}{2}\left(i^{n-1} + (-i)^{n-1}\right)$$
: I guess that's equivalent to Mentallic's answer. Here's how I obtained it:
\begin{aligned} \sin\left(\frac{n\pi}{2}\right) &= \text{Im}\left(e^{in\pi/2}\right) \\ &= \frac{1}{2i} \left(e^{in\pi/2} - \overline{e^{in\pi/2}}\right) \\ &= \frac{1}{2i} \left(i^n - \overline{i^n}\right) \\ &= \frac{1}{2i} \left(i^n - (-i)^n\right) \\ \end{aligned}
Last edited: Mar 19, 2015
8. Mar 20, 2015
### LCKurtz
Yup. As I thought. Nothing "simpler" than $\sin\frac{n\pi} 2$ itself. | 2017-12-12T06:35:44 | {
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https://math.stackexchange.com/questions/1367812/find-closest-point-on-a-plane-to-a-given-point-discrepancy-with-normal-vector | # Find closest point on a plane to a given point. Discrepancy with normal vector.
I have a point $(9,5,0)$ and a triangle with points $(1,1,0), (3,3,1), (6,1,0)$, let's label them as $A,B,C$ respectively. In order to get the normal vector, I do the cross product of two vectors. If I do $AB \times AC$ I get $n = \left\langle 0,-5,10\right\rangle$. With this I determine the equation of the plane is $5y - 10z - 5 = 0$. If I project my point onto the plane with this normal vector in mind, the point on the plane I find is $(9, 21/5, 8/5)$. HOWEVER, if I initially choose to cross two different vectors, I get something different.
Example: If I instead choose to do $BA \times BC$ I get $n = \left\langle 0,5,-10\right\rangle$. Now this in a way makes sense to me since it is the inverse of the other normal I found. Basically the other half of the line through the plane, right? If I continue, I find the equation of the plane to be $-5y + 10z + 5 = 0$. This leads me to a projected point of $(9, -29/5, -8/5)$. Why do I get two different points based on which normal vector I choose to solve with? I know the correct answer to this problem, one is right and one is wrong. I am trying to program a general solution to this type of problem. How do I choose the correct normal vector?
Please let me know if I am doing something wrong.
$\vec{AB} = <2,2,1> \quad \quad \vec{AC} = <5,0,0>$.
So, $\vec{AB} \times \vec{AC} = <0,5,-10>$
Then, the plane equation is something like $5y -10z = C$ for some constant $C$.
Plug (for example) the point $(1,1,0)$ into the plane equation to get $C=5$.
Hence, the plane equation is $5y -10z -5 =0$ or equivalently $-5y +10z+5=0$ which is slightly different from the one you found.
Edit: Let's continue with a different approach. The vector $\vec{n} = <0,5,-10>$ is perpendicular to the plane. The line that passes through our point $(9,5,0)$ which is parallel to $\vec n$ is $$x=9, \quad y=5+5t, \quad z=0-10t, \quad t \in \mathbb R$$
Now let's find where this line cuts the plane. Pick an arbitrary point from the line and put into the plane equation:
$$0= 5y-10z-5 =5(5+5t)-10(0-10t)-5$$
$$125t + 25 -5=0 \quad \implies \quad t=-\frac{4}{25}$$
So, the answer is $(9,5+5t,0-10t)=(9,-\frac{21}{5},\frac{8}{5})$.
Edit2: Now we do the projection. Choose an arbitrary point on the plane, say $A(1,1,0)$ and consider the vector $\vec{AP}$ where $P=P(9,5,0)$. $\vec{AP}=<8,4,0>$.
Then,
\begin{align*} proj_{\ \vec n \ } \vec{AP} &= |\vec{AP}| \cdot \cos \alpha \cdot \frac{\vec n}{| \vec n |} \\ &= |\vec{AP}| \cdot \frac{\vec{AP} \bullet \vec n}{|\vec{AP}| \cdot |\vec n|} \cdot \frac{\vec n}{| \vec n |} \\ &= \text{ calculations ... find the projection vector} \\ &= \vec u \end{align*}
Now, the point you are looking for is the point $K$ such that $\vec{KP} = \vec u$
• Ah, I typed it in wrong. You are correct, but the question of the discrepancy still stands. I will edit the post. Can you continue to carry out the projection and see if you get the same results as I did? – kvcwahley Jul 20 '15 at 17:36
• I think you forgot the -5 constant in your last calculation. It should be 125t + 25 = 5. This gives t=-4/25. My answer is then (9,21/5, 8/5). – kvcwahley Jul 20 '15 at 17:54
• Theoretically, should I be getting the same answer? – kvcwahley Jul 20 '15 at 17:54
• @kvcwahley Thanks I edited and got the answer. Now I will edit my answer again and add projection to get the same answer. – ThePortakal Jul 20 '15 at 17:59
• If you solve it with cross product of BA x BC, do you get the same answer? – kvcwahley Jul 20 '15 at 18:03
"I have a point $(9,5,0)$ and a triangle with points $(1,1,0), (3,3,1), (6,1,0),$ let's label them as $A,B,C$ respectively. In order to get the normal vector, I do the cross product of two vectors."
So you mean a normal vector to the plane determined by the three points.
"If I do $AB \times AC$ I get $n = <0,-5,10>$. With this I determine the equation of the plane is $5y - 10z + 5 = 0.$"
No, that does not contain $(1, 1, 0)$ because $0(1)+ 5(1)- 10(0)+ 5= 10$, not 0. The equation of the plane is $0(x- 1)-5(y- 1)+ 10z= 0$ or $5y- 10z- 5= 0$. Perhaps you meant $5y- 10z= 5$.
"If I project my point onto the plane with this normal vector in mind, the point on the plane I find is $(9, 21/5, 8/5)$. HOWEVER, if I initially choose to cross two different vectors, I get something different.
Example: If I instead choose to do $BA \times BC$ I get $n = <0,5,-10>$. Now this in a way makes sense to me since it is the inverse of the other normal I found. Basically the other half of the line through the plane, right? If I continue, I find the equation of the plane to be $-5y + 10z -5 = 0.$"
Which, again, is wrong. The cross product of $BA \times BC$, as you say is $<0, 5, -10>$ and, since the plane contains the point $(1, 1, 0)$, the plane has the equation $0(x- 1)- 5(y- 1)+ 10(z- 0)= -5y+ 10z+ 5= 0$, not $-5y+ 10z- 5$.
"This leads me to a projected point of $(9, -29/5, -8/5)$. Why do I get two different points based on which normal vector I choose to solve with?"
You don't say how you got two different points. Since you have exactly the same equation for the plane in both cases (although wrong!), and the same point, you should be doing exactly the same calculations in both cases.
"I know the correct answer to this problem, one is right and one is wrong. I am trying to program a general solution to this type of problem. How do I choose the correct normal vector?
Please let me know if I am doing something wrong."
Yes, if you get different answers projecting the same point onto the same plane, you must have done something wrong! But since you don't show what you did, it is impossible to say what you might have done wrong.
• You are correct with my equations being wrong. I initially typed them in incorrectly. I have edited the post to show the correct equations. Theoretically, should I be getting the same point? – kvcwahley Jul 20 '15 at 18:00
• Please start using LaTeX, see this guide: meta.math.stackexchange.com/questions/5020/… Almost all of your posts need heavy editing – Hirshy Aug 5 '15 at 8:42 | 2019-06-26T21:58:40 | {
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http://mathematica.stackexchange.com/questions/29383/finding-shortest-non-zero-vector-x-satisfying-ax-0-pmod-q | # Finding shortest non-zero vector $x$ satisfying $Ax=0 \pmod q$
Let $n$, $m$, and $q$ be positive integers (with $m > n$), and $A$ be a matrix over $\mathbb{Z}_q^{n \times m}$. Using Mathematica, I want to find the shortest non-zero vectors $x \in \mathbb{Z}_q^m$ such that:
$$Ax=0 \pmod q$$
Note: By "shortest," I mean the vector with the smallest Euclidean norm. I'm not sure if this vector is unique. If it's not, I want a list of vectors with smallest Euclidean norm.
Example:
Let: $n=2, m=3, q=7, A = \left(\begin{matrix} 2 & 1 & 6\\ 1 & 3 & 1 \end{matrix} \right)$. Then, the following vectors are solutions to
{1,1,3} $\quad$ Norm = $\sqrt{11}$
{3,3,2} $\quad$ Norm = $\sqrt{22}$
{2,2,6} $\quad$ Norm = $\sqrt{44}$
{5,5,1} $\quad$ Norm = $\sqrt{51}$
{4,4,5} $\quad$ Norm = $\sqrt{57}$
{6,6,4} $\quad$ Norm = $\sqrt{88}$
Therefore, the vector {1,1,3} is the shortest solution.
PS: I used an exhaustive search to find the above solutions. My approach takes a long time for even small choices of $n$ and $m$, say $n=2, m=8$. Please offer an approach which can at least find solutions for such small parameters.
Edit: As Michael suggested in a comment, I demonstrate the "exhaustive search" code I used. I'm not proud of it at all, as I just wrote the code to find an example for this question.
A = RandomInteger[6, {2, 3}];
x = {a, b, c};
Table[
If[Mod[A.x, 7] == {0, 0}, Print[{x, x // Norm}]],
{a, 0, 6}, {b, 0, 6}, {c, 0, 6}]
-
You might show the code you tried -- perhaps it can be made efficient. Users are much more likely to try out your problem if they can cut and paste a starting point. – Michael E2 Jul 28 '13 at 20:24
@MichaelE2: Thanks for the suggestion. I added the code snippet. It does not use Mathematica's built-in functions like Solve or FindMin. It's nothing more than an exhaustive search, and that's why it takes a long time to find a solution for even relatively small parameters. – M.S. Dousti Jul 28 '13 at 21:05
Yes that would take a long time. :) If you thought of Solve, you should have tried it! Another tip: Don't use Print to accumulate you results. Table will accumulate them; then DeleteCases[table, Null, Infinity], gets rid of the Nulls; then perhaps Flatten. Better yet, learn about Reap/Sow. – Michael E2 Jul 28 '13 at 22:03
(1) You are working in a ring that really does not have a notion of length. Why is a rock-bottom smallest in Euclidean norm result needed? – Daniel Lichtblau Jul 29 '13 at 14:12
(2) One can get small results using lattice reduction. This unfortunately does not force values to be nonnegative. All the same, a usable vector might appear. In a run of your example I have mat = {{2, 3, 2}, {2, 6, 1}}. Then it turns out that {4,3,2} is the null vector of interest (though see (1) above-- I do not know why it is of interest). It appeas in the following lattice reduction. In[35]:= LatticeReduce[ Join[NullSpace[A], 7*IdentityMatrix[Length[A[[1]]]]]] Out[35]= {{-2, 2, -1}, {1, -1, -3}, {4, 3, 2}} – Daniel Lichtblau Jul 29 '13 at 14:14
One way to approach this is to use the Null space of the matrix a, which is a basis for all elements x such that a.x=0. For the OPs problem, this can be done by first finding the null space:
a = {{2, 1, 6}, {1, 3, 1}};
n = First[NullSpace[a, Modulus -> 7]]
{5, 5, 1}
Now, observe that a.x=0 is true exactly when a.(c*x)=0, so it is possible to list all the answers (mod 7) to this problem by
Mod[# n, 7] & /@ Range[7]
{{5, 5, 1}, {3, 3, 2}, {1, 1, 3}, {6, 6, 4}, {4, 4, 5}, {2, 2, 6}}
To find the one with smallest norm, apply the norm to each element,
Norm /@ (Mod[# n, 7] & /@ Range[6])
{Sqrt[51], Sqrt[22], Sqrt[11], 2 Sqrt[22], Sqrt[57], 2 Sqrt[11]}
which gives the same answer as the OP found by search.
More generally, the NullSpace of a is a collection of $m-n$ vectors, and it will be necessary to search over all linear combinations of these $m-n$ vectors (mod $q$) for the one with the smallest norm. While this may still be large, it is certainly smaller than search over $m$ dimensions as in the brute force approach.
-
Using Solve helps a bit. If vars is the list of variables, then
Solve[A.vars == ConstantArray[0, n], vars, Modulus -> q]
finds all solutions in terms of parameters C[1], C[2], etc. depending on the dimension of the solution space.
Another bottleneck is computing the values of the general solution at all possible combinations of values for the parameters C[i] modulo q. The possible combinations can be computed with
Tuples[Range[0, q - 1], n]
where n is the number of parameters C[i]. The result is certainly reasonably fast for the small values of m, n, q mentioned in the question.
Finally Nearest will return the shortest nonzero vector (the zero vector having been deleted).
findSols[q_, A_] :=
Module[{m, n, vars, x, sol, vals, params, vecs},
{n, m} = Dimensions@A;
vars = Array[x, m];
sol = vars /. First @ Solve[A.vars == ConstantArray[0, n], vars, Modulus -> q];
(* Print@sol; *) (*optional output*)
params = Union@Cases[sol, _C, Infinity];
(* Print@params; *) (*optional output*)
vals = Transpose @ Tuples[Range[0, q - 1], Length@params];
vecs = DeleteCases[
Mod[#, q] &@ Transpose[sol /. MapThread[Rule, {params, vals}]],
ConstantArray[0, m]];
Nearest[vecs, ConstantArray[0, m]]
]
Example in OP:
A = {{2, 1, 6}, {1, 3, 1}};
findSols[7, A] // Timing
(* {0.000675, {{1, 1, 3}}} *)
Additional example with $m=8$. (Over 117000 solutions, still less than a second.)
SeedRandom[2];
A = RandomInteger[6, {2, 8}]
findSols[7, A] // Timing
(* {{5, 6, 1, 6, 2, 2, 6, 5}, {2, 1, 5, 5, 0, 0, 6, 4}} *)
(* {0.289682, {{0, 1, 0, 0, 1, 0, 1, 0}, {0, 1, 0, 0, 0, 1, 1, 0}}} *)
Note that for q = 17 there will be over 24,000,000 solutions -- I killed the kernel when it went past 20GB.
- | 2016-06-29T09:21:52 | {
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http://xaktly.com/AlternatingSeries.html | Series in which terms alternate between positive and negative
In the infinite series notes you saw the series expansions of the sine and cosine functions. Here they are again along with the summation notation. Remember that there's nothing especially complicated about Σ notation; it's just a way of capturing what's going on in the series without having to write all or several of its terms.
\begin{align} sin(x) &= \frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dots = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!} \\ \\ cos(x) &= \frac{x^0}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \end{align}
These are called alternating series because of the alternation in the sign of each term.
To describe such an alternation in summation notation we employ the properties of powers of negative numbers:
• $(-1)^1 = -1$
• $(-1)^2 = 1$
• $(-1)^3 = -1,$ and so on.
It's also possible to produce an alternation of sign using the trigonometric functions sin(x) and cos(x), like this:
• $cos(n\pi) = 1$ for even n and -1 for odd n
• $sin\left(\frac{n\pi}{2}\right) = 1$ for odd n and -1 for even n
An example
The alternating harmonic series
Here is an example of an alternating series, the so-called alternating harmonic series. This is just the harmonic series with alternating signs of the terms. The terms alternate on either side of zero as they decrease to zero (blue graph).
The accumulating sum (red graph) converges to a limit of approximately 0.69, but oscillates about that line. It's non-trivial to determine the actual sum of this series, but it does converge.
The alternating series test
The alternating series test, proved below the next box, is very simple.
It says that if, as n→∞, the terms of an alternating series decrease to zero, then the series converges.
Remember, that is NOT necessarily true for non-alternating series. For a non-alternating series, it is not enough that the size of the term diminishes; that series still may not converge.
We can think of many non-alternating, divergent series with decreasing terms, like the harmonic series:
$$\sum_{n = 1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \:\dots$$
In a way, the alternating series test (AST) makes life a lot easier. It's enough that the absolute value of the terms decrease to zero for an alternating series to converge – pretty simple.
The alternating series test
If the terms of an alternating series, $\sum_{n = 1}^\infty \,(-1)^n a_n$
(1) are decreasing, and (2) if the limit of the size of the terms, as n → ∞ is zero:
then the series converges.
Proof of the alternating series test
First, take a look at the alternating series on a number line. We'll let the terms of the series be $a_1, a_2, a_3, \dots$ and so on, and the partial sums will be $s_1 = a_1, \; s_2 = a_1 + a_2, \; s_3 = a_1 + a_2 + a_3,\; \dots$ and so on.
Let's look first at the even sums. The first one (first green arrow in the diagram) is
$$s_2 = a_1 - a_2 \gt 0 \; \leftarrow \; \text{ because } a_1 \gt a_2$$
The second and third are
\begin{align} s_4 &= s_2 + (a_3 - a_4) \ge s_2 \; \leftarrow \; \text{ because } a_3 \gt a_4 \\[4pt] s_6 &= s_4 + (a_5 - a_6) \ge s_4 \; \leftarrow \; \text{ because } a_5 \gt a_6 \end{align}
... and the pattern emerges. In the notation below, 2n is always an even number, 2n-2 is the next smaller even number and 2n-1 is the next smaller odd number:
$$s_{2n} = s_{2n - 2} + (a_{2n - 1} - a_{2n}) \ge s_{2n - 2}$$
Now it's clear that these sums continue to grow, so that
$$0 \le s_2 \le s_4 \le s_6 \le \dots \le s_{2n} \le \dots$$
By regrouping terms with some parenthesis, we can also write s2n as
$$s_{2n} = a_1 - (a_2 - a_1) - (a_4 - a_3) - \dots - (a_{2n - 2} - a_{2n - 1}) - a_{2n}$$
Now because our terms are decreasing, every difference in parenthesis is positive. The sequence of terms {sn} is positive, increasing and bounded from above by the sum of the series, s (see diagram). It therefore has a limit,
$$\lim_{n\to\infty} s_{2n} = s$$
Now because $s_{2n + 1} = s_{2n} + a+{2n + 1},$ we can find the limit of the odd partial sums:
$$\lim_{n\to\infty} s_{2n + 1} = \lim_{n\to\infty} (s_{2n} + a_{2n + 1})$$
What we've shown is that the limits of the even and odd partial sums is the same. This is called "squeezing." If the limits from above and below are s, the sum of the series, then the series must converge to s.
The alternating series test makes determining the convergence of alternating series much easier than that of a non-alternating series. If the series alternates in sign, and if its terms decrease toward zero, the series converges.
Refining the idea of convergence
The difference in convergence between the alternating harmonic series,
$$\sum_{n = 1}^\infty \, (-1)^n \frac{1}{n}$$
which converges, and the harmonic series
$$\sum_{n = 1}^\infty \, \frac{1}{n}$$
which doesn't, hints at a subtle differentiation in how series converge. In fact, the alternating harmonic series is called "conditionally convergent," the condition being that its terms alternate sign.
On the other hand, a series like the convergent p-series,
$$\sum_{n = 1}^\infty \, \frac{1}{n^2}$$
and its alternating series counterpart,
$$\sum_{n = 1}^\infty \, (-1)^n \frac{1}{n^2}$$
both converge, and the alternating version is said to be "absolutely convergent." Here are the rules for absolute and conditional convergence:
Absolute vs. conditional convergence
Test for absolute convergence
Test for conditional convergence
Divergence – the series diverges by any one of the tests for convergence, or the divergence test.
Example 1
Does the series $\sum_{n = 0}^\infty \, (-1)^{n + 1} \frac{2n + 3}{3n + 4}$ converge?
Solution: This is an alternating series in which terms with even n are negative. To test whether it converges, we ask whether the term, without the alternating part (-1)n+1.
$$\lim_{n\to\infty} \, \frac{2n + 3}{3n + 4} = \frac{2}{3} \neq 0$$
The term does not tend toward zero, so this series does not converge. We could also say that it diverges by the divergence test. Either way, it diverges.
Example 2
Does the series $\sum_{n = 0}^\infty \, \frac{(-2)^n}{3^n}$ converge?
Solution : We can first rewrite this series with a -1 inside the alternating term, using (xy)n = xn·yn:
$$\sum_{n = 0}^\infty \, (-1)^n \left( \frac{2}{3} \right)^n$$
Now we need to find whether the limit of the term without the alternation is zero:
$$\lim_{n\to\infty} \, \left( \frac{2}{3} \right)^n = 0$$
2/3 < 1, so the limit is zero, and this alternating series converges by the AST.
Practice problems
Decide whether the following series converge.
1 $$\sum_{n = 1}^\infty \, \frac{(-1)^n}{4 + 3n}$$ 2 $$\sum_{n = 1}^\infty \, \frac{(-1)^n}{n^3 + 2n + 2}$$
3 $$\sum_{n = 1}^\infty \, \frac{n \cdot cos(n\pi)}{n^2 + 1}$$ 4 $$\sum_{n = 1}^\infty \, \frac{(-1)^n n^2}{n^2 + 1}$$
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https://math.stackexchange.com/questions/2812530/create-55-sets-with-exactly-one-element-in-common | # Create 55 sets with exactly one element in common
I have 55 sets $E_1, E_2,\cdots,E_{55}$
There are $n$ different elements.
Each set contains 8 elements $E_1\{x_1^1,x_1^2,x_1^3,x_1^4,x_1^5,x_1^6,x_1^7,x_1^8\}, E_2\{x_2^1,x_2^2,x_2^3,x_2^4,x_2^5,x_2^6,x_2^7,x_2^8\}\cdots$
If I take 2 sets randomly, they must have exactly one element in common, no more, no less.
How many elements do I have ? Is there more than one possibility ?
Actually I was playing a card game and I was curious about the mathematics behind.
Thanks!
EDIT: I'm looking for the lowest solution possible
• If you are curious about he mathematics behind Dobble, the keyword is "block designs". – Michal Adamaszek Jun 8 '18 at 13:34
• One solution is $386$, with one element in all the sets and the other $7\cdot 55=385$ elements distinct. – Ross Millikan Jun 8 '18 at 13:37
• @RossMillikan True, but it wouldn't be a very exciting card matching game :) – Bram28 Jun 8 '18 at 13:40
• Oh, indeed it's a solution. Then I'm looking for the lowest one. – tuxikigo Jun 8 '18 at 13:42
• There's some analog to fractional coloring. What is the fractional chromatic number of an 8-fold coloring of the complete graph on 55 vertices, with the added condition that any two vertices share exactly one color. – David Diaz Jun 8 '18 at 14:23
A lower bound is $n=57$. Take one set. It must have at least one element that it shares with at least seven other sets. Those eight sets all share one element, so all their other elements must be distinct. If there is an element shared $9$ ways $n$ must be at least $64$ because the other seven elements in each set need to be distinct.
There are $55\cdot 54/2=1485$ pairs of sets. An element that is in $8$ of the sets accounts for the match of $28$ pairs. An element that is in $7$ of them accounts for $21$ pairs. As both of these numbers are divisible by $7$ and $1485$ is not, there must be at least one element that is shared by a number other than $7$ or $8$.
As $1485 \equiv 1 \bmod 7$ and an element that is shared $6$ ways accounts for $15$ pairs, which is also equivalent to $1 \bmod 7$ I will guess that there is one element shared $6$ ways. We can then get the required number of pairs if there are $42$ elements shared $8$ ways and $14$ elements shared $7$ ways. This would give $n=57$.
We have not shown that this will work, but we can make a concrete example from the order $7$ projective plane. It has $57$ points, $57$ lines, with every two lines sharing a point and every two points sharing a line. The sets are the lines and the elements of each set are the points on it. We can discard two lines and get the required construction. Thanks to achillehui for pointing this out.
• the finite projective plane of order $7$ contains $57$ points and $57$ lines, each line contains $8$ points and two lines determine a point. – achille hui Jun 8 '18 at 15:02
• @Khoross: why? We could have the first nine sets contain element $1$ and the next eight contain element $2$. – Ross Millikan Jun 8 '18 at 17:31 | 2020-06-03T13:52:41 | {
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# The cost C, in dollars of producing x units of a certain commodity is
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The cost C, in dollars of producing x units of a certain commodity is [#permalink]
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The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?
(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Originally posted by inakihernandez on 24 Apr 2016, 10:20. Last edited by Vyshak on 24 Apr 2016, 11:36, edited 1 time in total. Edited the url ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 6958 The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:58 11 4 inakihernandez wrote: The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a? (1) The cost of producing 500 units is$15.833.
(2) The cost of producing 1,000 units is twice the cost of producing 500 units.
Hi
the equation has got two constants a and b..
lets see the statements
(1) The cost of producing 500 units is $15.833. we have two constants a and b, and one equation Insuff (2) The cost of producing 1,000 units is twice the cost of producing 500 units here when you make two quation and equate them you will realize you can find answer.. $$C(1000) = 10.000 + b*1000 + a1000^2$$ .. $$C(500) = 10.000 + b*500 + a500^2$$ .. now $$C(1000) = 2* C(500)$$.. so $$10.000 + b*1000 + a1000^2 = 2( 10.000 + b*500 + a500^2 )$$.. $$10.000 + b*1000 + a1000^2 = 20.000 + b*1000 + 2*a*500^2$$ b will get cancelled and you can find value of a.. Suff Hope it helps _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor ##### General Discussion EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12649 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:47 2 Hi inakihernandez, You should post this question in the DS sub-Forum here: gmat-data-sufficiency-ds-141/ GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: [email protected] # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink]
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24 Apr 2016, 11:02
1
2
C(x) = 10.000 + bx + ax^2
a = ?
St1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833
We have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'
Not Sufficient
St2: C(1000) = 2*C(500)
10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)
a(10^6 - 5*10^5) = 10
Sufficient to determine the value of 'a'
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Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink]
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24 Apr 2016, 11:10
1
Vyshak wrote:
C(x) = 10.000 + bx + ax^2
a = ?
St1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833
We have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'
Not Sufficient
St2: C(1000) = 2*C(500)
10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)
a(10^6 - 5*10^5) = 10
Sufficient to determine the value of 'a'
how did you get this?
a(25*10^4 )
a(50*10^4)
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Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink]
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24 Apr 2016, 11:13
1
paidlukkha wrote:
Vyshak wrote:
C(x) = 10.000 + bx + ax^2
a = ?
St1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833
We have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'
Not Sufficient
St2: C(1000) = 2*C(500)
10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)
a(10^6 - 5*10^5) = 10
Sufficient to determine the value of 'a'
how did you get this?
a(25*10^4 )
a(50*10^4)
Hi,
I have skipped some multiplication steps in my explanation.
x = 5*10^2 --> x^2 = 25*10^4
2*C(500) --> 2(10 + 500b + 25*10^4) --> 20 + 1000b + 2*25*10^4
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Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink]
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24 Apr 2016, 21:04
8
inakihernandez wrote:
The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?
(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Let me point out here that this is a (C) trap question. The first statement will give an equation in a and b. So we know that we cannot find the value of a. When we look at the second statement, we see another equation in a and b which is not a multiple of the first equation and suddenly, it seems that we will be able to solve both equations simultaneously and get the answer. So (C) seems to be correct. But that is just too easy. Beware of the (C) trap. This should make you think - can I solve the question using a single statement alone. Make the equation of statement 2 since it gives you some convoluted relative data. When you do, you see that b gets canceled and in fact, you can get the value of a. Note that you cannot find the value of b from the second statement alone but the question doesn't ask for the value of b. We only need the value of a and hence second statement is sufficient alone. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > GMAT self-study has never been more personalized or more fun. Try ORION Free! Current Student Joined: 28 Nov 2014 Posts: 867 Concentration: Strategy Schools: Fisher '19 (M) GPA: 3.71 Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 09 Oct 2016, 04:29 The correct statement 1 is (1) The cost of producing 500 units of this commodity is$15833
It is NOT 15.833 although that won't affect the OA.
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Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink]
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09 Oct 2017, 06:44
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Display posts from previous: Sort by | 2018-10-16T03:45:55 | {
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https://math.stackexchange.com/questions/1229712/finding-the-matrix-of-a-linear-transformation-from-an-upper-triangular-matrix-to | # Finding the matrix of a linear transformation from an upper triangular matrix to an upper triangular matrix.
The question that I am trying to solve is as follows:
Find the matrix $A$ of the linear transformation $T(M)= \begin{bmatrix} 7 & 3 \\ 0 & 1 \end{bmatrix} M$ from $U^{2×2}$ to $U^{2×2}$ (upper triangular matrices) with respect to the basis $\left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix}1&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$
$A=\begin{bmatrix} \cdot&\cdot&\cdot\\\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot\end{bmatrix}$
I don't know how to interpret this question. I would have thought that the matrix $A$ is $\begin{bmatrix} 7 & 3 \\ 0 & 1 \end{bmatrix}$ because that is the matrix which defines the linear transformation $T$. I am surprised that the required answer is $A\in M_{3\times 3}$. How can a basis with elements $\in M_{2\times 2}$ form a matrix $\in M_{3\times 3}$?
• Upper triangular matrices are a three dimensional vector space. But I don't understand your question. Usually $U$ indicates unitary matrices. – Emilio Novati Apr 11 '15 at 9:49
• Has three dimensions got anything to do with $A\in M_{3\times 3}$? – ahorn Apr 11 '15 at 11:21
Upper triangular matrices
$\begin{bmatrix} a&b\\ 0&c \end{bmatrix}$
form a vector space with canonical basis:
$e_1=\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} \quad e_2=\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix} \quad e_3=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}$
that is isomorphic to $\mathbb{R}^3$ by:
$e_1\rightarrow\vec e_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix} \quad e_2\rightarrow\vec e_2=\begin{bmatrix} 0\\1\\0 \end{bmatrix} \quad e_3\rightarrow \vec e_3=\begin{bmatrix} 0\\0\\1 \end{bmatrix}$
Your linear transformation $T$ is defined as a matrix multiplication:
$T(M)=\begin{bmatrix} 7&3\\ 0&1 \end{bmatrix} \begin{bmatrix} a&b\\ 0&c \end{bmatrix} = \begin{bmatrix} 7a&7b+3c\\ 0&c \end{bmatrix}$ so, in this canonical representation, it is given by the matrix $T_e$ such that:
$T_e\vec M=\begin{bmatrix} 7&0&0\\ 0&7&3\\ 0&0&1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix}= \begin{bmatrix} 7a\\ 7b+3c\\ c \end{bmatrix}$
Now you want a representation of the same transformation in a new basis:
$e'_1= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} \rightarrow\vec e'_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix} \quad e'_2=\begin{bmatrix} 1&1\\ 0&0 \end{bmatrix}\rightarrow\vec e'_2=\begin{bmatrix} 1\\1\\0 \end{bmatrix} \quad e'_3=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}\rightarrow \vec e'_3=\begin{bmatrix} 0\\0\\1 \end{bmatrix}$
This transformation of basis is represented by the matrices
$S= \begin{bmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \qquad S^{-1}= \begin{bmatrix} 1&-1&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$
So the matrix that represents the transformation $T$ in the new basis is:
$T_{e'}=S^{-1}T_eS= \begin{bmatrix} 7&0&-3\\ 0&7&3\\ 0&0&1 \end{bmatrix}$
• When you said "This transformation of basis is represented by the matricies", did you simply put the column vectors $\vec e'_1$, $\vec e'_2$ and $\vec e'_3$ next to each other to get $S$? Or was there a more nuanced way of getting $S$, such as by saying that $\vec e_j= (a_j+b_j+c_j)\vec e'_j$ where $j$ is the number of the column? – ahorn Apr 11 '15 at 15:49
• Symply I put the columns ( it's a general rule that you can easely verify by linearity). – Emilio Novati Apr 11 '15 at 16:31
Let $(X,Y,Z)$ be the basis of $U$, compute $T(X)$, $T(Y)$, $T(Z)$ and express them as a linear combination of $X$, $Y$ and $Z$.
• T(X)=7X; T(Y)=7Y; T(Z)=Z. How does that relate to A? – ahorn Apr 11 '15 at 11:20
• A={(7,0,0),(0,7,0),(0,0,1)} take the coefficients as elements of the matrix A. – 2ndYearFreshman Apr 11 '15 at 11:39
• I have tried to input all sorts of combinations, such as $\begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 7 & 0 & 0 \\ 7 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \text{and} \begin{bmatrix} 7 & 0 & 0 \\ 7 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ but these are still incorrect (I am answering an online test). Btw, welcome to math.stackexchange! What brought you to the site? – ahorn Apr 11 '15 at 14:25
• Try A={(7,0,0),(0,7,0),(-3,3,1)} Thank you,hopefully my answer will help.Your question brought me here. – 2ndYearFreshman Apr 11 '15 at 14:34 | 2021-04-15T05:08:25 | {
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https://www.physicsforums.com/threads/how-can-a-curve-be-one-dimensional.593124/ | # How can a curve be one dimensional?
1. Apr 3, 2012
### rollcast
I've heard of curves being described as one dimensional but I don't understand how anything other than a straight line can be one dimensional as surely once the curve becomes, well, curved it is now in two dimensions?
I have illustrated this in the above diagram with the curve and straight line shown in red and the dimensions in green.
Thanks
A.
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5.3 KB
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1,706
2. Apr 3, 2012
### Office_Shredder
Staff Emeritus
The curve is embedded in a 2 dimensional plane, but the curve itself is one dimensional. For example, I can describe all the points using only one parameter in a very natural way (if the graph is y=f(x), then x is the parameter since it uniquely identifies y)
Really what you want is dimension to be an intrinsic property. If I took your curve and put it in 3 dimensions, is it a 3 dimensional object now? What about the fact that we're in spacetime, is it 4 dimensional? That's a bit silly, so instead we ask what properties does the curve have that is independent of the space that it is embedded in.
There are several possible definitions of dimension depending on context but the basic idea is that if it has dimension k you can describe every point using k parameters at least locally - by that I mean, for example the curve x2+y2=1 (a circle) you can describe the top half of it by the set of points of the form $(x,\sqrt{1-x^2})$ and the bottom half is of the form $(x,-\sqrt{1-x^2})$. So I wasn't able to describe the whole circle using the x-coordinate but I was able to describe each half of it using a single coordinate, which means the whole thing is 1 dimensional (in fact you can describe the whole circle using a single coordinate if you use trigonometric functions, which I encourage you to try if you haven't seen it)
3. Apr 3, 2012
### Bacle2
Would you think of the curve as having non-zero area? I don't mean a closed curve
that may enclose an area; I mean the curve itself.
4. Apr 3, 2012
### HallsofIvy
Staff Emeritus
Yes, such a curve is in two dimensions but it only "one dimensional" itself".
What is your definition of "dimension"? One common one is: a subset of Euclidean space is said to be "n-dimensional" if and only if any point can be identified using n real numbers.
Choosing one point on the curve, we can identify any point on the curve by its distance (positive in one direction, negative in the other) from the chosen point. Since that is a single number, the curve is one-dimensional.
5. Apr 3, 2012
### rollcast
Bit of a weird example here so the curve is like curved corridor in a building, each office on the corridor is numbered with respect to where it is positioned along the corridor and not where it is positioned relative to the 4 outer walls of the building.
6. Apr 3, 2012
### Matterwave
Perhaps a somewhat more intuitive way of thinking about it is, if you live "in" the curve (i.e. you're not allowed to go outside of it), you can only ever go forwards or backwards, never to the left or right or up or down.
7. Apr 3, 2012
### Bacle2
Altho you may want to consider separately the case of space-filling curves, like the Peano curve.
8. Apr 3, 2012
### Bacle2
I'm starting to doubt my answer, if we can define a curve as the continuous image of a connected interval. We can use the fact that, e.g., any compact metric space K is the continuous image of the Cantor set. Then take, e.g., K=[0,1]^n (mapping I into R^n), and extend continuously, using Tietze extension ( C is closed in I ; I is normal), then the image of f has non-empty interior in R^n.
Note/Edit: I'm referring to a more general definition of a curve, as the continuous image of an interval, and not the curves in the attached files.
Last edited: Apr 3, 2012
9. Apr 4, 2012
### chiro
Intuitively for any line, no whatever what its deformation or dimensionality for the embedded space is always a line. Similarly, a flat piece of paper no matter how its deformed or what space its embedded in is always a flat piece of paper.
If you are given a visualization of the object in question, you can get a very good idea of the number of parameters. If you can't visualize the object but are given a deterministic expression, then you have to use other methods.
If you are given a non-analytic version of a process where you only know the dimensionality of the embedded space and have to not only figure out the parameterization but also some sort of 'useful' definition for the process as a whole (implicit most likely but could be explicit), then you will have to use even different methods again which will be based on a combination of statistical theory and non-statistical mathematics.
One method for assessing parameterization is assessing density in various orientations for your function and statistical results used in data mining measure a kind of 'variation' density where you get principal components for your data set. This kind of idea could be applied to your analytic setting in the analytic perspective (not the statistical one).
10. Apr 4, 2012
### Matterwave
A curve has the same cardinality as an arbitrary n-dimensional hypercube, but that doesn't mean it has the same dimension. A space-filling curve defines a surjection, but not a homeomorphism.
11. Apr 4, 2012
### Bacle2
I was referring to a curve as the continuous, not homeomorphic image of an interval. I was also taking back my statement that a curve must have empty interior. As defined, I think the argument is correct, and a curve does not necessarily have an empty interior.
12. Apr 4, 2012
### Bacle2
Clearly, the image of an interval would depend on a single parameter. I was referring to the claim/belief that the continuous image of a curve would have no area, i.e., would have empty interior.
13. Apr 4, 2012
### Bacle2
Sorry, one last post on this, to try to avoid confusion:
0)Define a curve as the continuous image of an (connected)interval
Claim: A curve can have non-empty interior.
i)We have that every compact metric space is the continuous image of the
Cantor set C . So we map C subset I:=[0,1] into, say I^n, so there is an f with f(C)=I^n (n>1)
ii) By Tietze extension thm. ( C is closed in I normal) , we can extend f to f^, defined on the whole of I, continuously ( on each real variable). Then f^ is a continuous map such that f^(I) has non-empty interior in R^n.
14. Apr 4, 2012
### chiro
I'll have a look at your theorems later on, but the thing is if you can deform your object that you have in a way so that is linear, then apply the ideas of dimensionality to get not only the proper dimension but the parameterization of your object, then you are done. By finding the appropriate deformation, you have created a linear representation that can be dealt with in the linear context.
This is of course equivalent to finding the inverse for each part of your object with respect to the different 'branches' that exist. If you can find the inverse for each 'branch' of your object (might be multi-dimensional and multi-parameterized but the idea is the same), then you can find a linear representation which can be parameterized.
So the case with a line would result in basically an n dimensional system that represents a line with one parameter t where X(t) = At + (1-t)B in the deformed state to the linear space. If this parameterization wasn't only one dimensional, then the deformation wouldn't produce one but however many parameters for the object.
With the notion of topology, if you know the object is analytic continuous then there should exist a deformation to the linear representation which depends on the inverse within a given branch-volume that corresponds to where the derivatives are (which correspond to the ideas in the inverse function theorem which are calculated by finding zero-jacobians). With this information you can find the branches and thus deform that region to it's linear counterpart.
The reason I like to think of something in the deformed linear space is because linear spaces, decompositions, and parameterizations of linear systems are well understood both algebraically, algorithmically and also to an extent visually. It's very easy to parameterize something that is deformed to linear space than to try and analyze it in its non-deformed non-linear space with a function like say f(x,y,z,w,t,u,v,a) = blah for any analytic continuous blah.
You could also probably consider continuous representations that are not analytic over the whole domain and use the same argument based on topological reasoning but I will only speculate on this (intuitively it makes sense at least).
15. Apr 4, 2012
### Bacle2
I think the best we can say is that if f is C^1 (maybe Lipschitz) , then, with a bounded derivative, we can preserve the Hausdorff dimension.
16. Apr 4, 2012
### chiro
I'll go with that for finding the dimension (which is the OP's question), but in terms of understanding why (and where) dimension goes from say one thing to another it may be useful to see which 'parts' if they are isolated contribute to where the function requires more dimensions as opposed to less (it can happen but not always).
The only reason I am saying this (for the OP) is that visually you can see where dimensionality changes. I'm imagining some kind of implicit function that creates branches (like a bifurcation) and does this repeatedly which creates more dimensions at different parts.
Just out of curiosity, what would you say topologically is a way to do the above? In other words in the case of say a bifurcation like object, how would you isolate this kind of phenomenon in terms of the local characteristics (spatially) of the object?
17. Apr 4, 2012
### Bacle2
The way I see it, if the derivative is bounded, then the image of the balls in the covering will not expand by much.
Sorry, it is 4 a.m., here, I'm out for the night; will try to be back in case there are more posts.
18. Apr 4, 2012
### HallsofIvy
Staff Emeritus
Then any two-dimensional ball about a point on the curve will necessarily include some points not on the curve- a curve has no (2 dimensional) interior.
19. Apr 4, 2012
### Bacle2
Halls of Ivy:
I did a specific construction, with a specific definition of curve; what is it about my construction that you think is faulty?
It seems strange-enough that the hypercube ( as a compact metric space; a space with non-empty interior) is the continuous image of the Cantor set. Then what is wrong with the extension?
20. Apr 5, 2012
### TrickyDicky
I think the OP is simply reflecting on the fact that there is no such thing as intrinsic curvature in one dimension. | 2017-08-21T23:42:45 | {
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# If s is the sum of odd integers from 40 to 60, inclusive, and t is the
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?
A. 480
B. 490
C. 980
D. 990
E. 995
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 12:39
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?
A. 480
B. 490
C. 980
D. 990
E. 995
Similar question: https://gmatclub.com/forum/if-x-is-equa ... 97544.html
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 12:48
There are total 10 odd numbers between 40 and 60.
hence, t=10
Now, the given sequence is 41,43,45,......,59
common difference is 2.
hence,
S= (41+59)*(10/2)
S=500.
S-t= 500-10= 490.
Kudos if it helps.
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 13:31
1
Are these formulas correct? I am having trouble with odd numbers for some reason.
N for odd numbers = (Last odd - First odd)/2)) +1
Sum of odd numbers = (First odd + Last odd)*N))/2
1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B
Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 13:36
1
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?
A. 480
B. 490
C. 980
D. 990
E. 995
The way I solve sums of arithmetic series, I need the number of terms no matter what. Start there.
Set t
Number of terms: number of odd integers from 40 to 60. First term is 41, last is 59.
Number of terms
$$\frac{LastTerm-FirstTerm}{interval} + 1$$
The "interval" for consecutive even and odds is always 2
$$\frac{59-41}{2} = \frac{18}{2} = (9 + 1)=$$ 10
Set s
Sum of arithmetic series
(Average)(Number of terms) =
$$\frac{FirstTerm+LastTerm}{2}*(10)$$ =
$$\frac{100}{2}(10) = (50)(10) = 500$$
s - t: 500 - 10 = 490
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 14:20
1
Are these formulas correct? I am having trouble with odd numbers for some reason.
N for odd numbers = (Last odd - First odd)/2)) +1
Sum of odd numbers = (First odd + Last odd)*N))/2
1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B
Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?
I think the reason you might be having trouble with odd numbers is that for consecutive odd integers, the interval (between the numbers), is two. Two is even; that could confuse. Or you might be forgetting to start with the first ODD term (e.g. here, first term is 41, not 40).
Bottom line: The "interval" between 1 and 3 is two. The interval between 2 and 4 is two. For consecutive odds and evens, in the formula, you divide by 2.
Take smaller numbers. Say, sum of odd integers from 0 to 6. So 1 + 3 + 5 = 9. There are three terms (1, 5, 9) whose sum is 9.
Number of terms in "odd integers from 0 to 6":
$$\frac{(Last - First)}{2}+ 1$$
$$\frac{(5-1)}{2} + 1 = (\frac{4}{2}+ 1) = (2 + 1) = 3$$ Correct
SUM: (Average)(# of terms)
$$\frac{(First + Last)}{2}(3)$$
$$\frac{(1 + 5)}{2} (3) = (3)(3) = 9$$
Bingo. Dividing by 2 for consecutive odd integers works.
When subtracting, you add one because subtraction doesn't include the first number. If in doubt: use small numbers that replicate your situation.
For example, # of terms from 1 to 4? The numerals are 1, 2, 3, 4. There are 4 terms. BUT (4 - 1) = 3.
We need one more. 3 + 1 = 4.
There's a mnemonic: "add one before you're done." It almost always applies. If in doubt, replicate your situation with small numbers. Rare in sequence sums but possible: If you see the word "exclusive" or the phrase "exclusive of," that is when you really need to check.
I have seen your other answers using this method. The ones I've seen are correct! (The one about -190 to 195? Correct, but long. -190 to +190 = 0; you are left with only 5 numbers to sum. STILL - correct.)
Here is a fantastically written, comprehensive post on all kinds of sequences by benjiboo , scroll down for arithmetic sequence
And here is a thread on sequences. VeritasPrepKarishma 's posts are great
Hope this barrage of information helps.
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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13 Oct 2017, 11:23
N for odd numbers = (Last odd - First odd)/2)) +1
Sum of odd numbers = (First odd + Last odd)*N))/2
1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B
Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the &nbs [#permalink] 13 Oct 2017, 11:23
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-09-20T12:56:18 | {
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https://math.stackexchange.com/questions/1673289/find-the-value-of-a-b2-given-a-b-2-and-a2-b2-6/1673299 | # Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$
$$a + b = 2\\ a^2 + b^2 = 6$$
Find the value of $(a-b)^2$
My workings till I got stuck -
$$(a-b)^2 = a^2 - 2ab + b^2 \\ = a^2 + b^2 - 2ab\\ = 6 - 2 ab$$
I'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.
• $4= (a+b)^2=a^2+b^2+2ab$ – lulu Feb 26 '16 at 15:16
• Squaring the first equation then subtract the second equation from it. – Zhanxiong Feb 26 '16 at 15:17
• The above comments mean that you should consider the quantity $2a^2+2b^2-(a+b)^2$... – abiessu Feb 26 '16 at 15:19
Use $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$ Thus, here, $$(a-b)^2+4=12$$ $$(a-b)^2=8$$
We will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \iff (2-b)^2 + b^2 +6$
This means $2b^2-4b-2=0 \iff b^2 -2b -1 +0$. The solutions of this equation are $b=1-\sqrt{2}$ and $b=1+\sqrt{2}$. Then $a=1+\sqrt{2}$ or $a=1-\sqrt{2}$
Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$) | 2019-10-21T11:14:21 | {
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https://math.stackexchange.com/questions/1055651/probability-that-a-random-13-card-hand-contains-at-least-3-cards-of-every-suit | # Probability that a random 13-card hand contains at least 3 cards of every suit?
A random 13-card hand is dealt from a standard deck of cards. What is the probability that the hand contains at least 3 cards of every suit? (Introduction to Probability, p.36)
My solution:
• There are $\binom{52}{13}$ possible hands.
• Because there are 13 cards for the hand, to obtain at least three cards of one suit per hand, we need to have exactly three cards of one suit per hand plus one additional card of any suit, thus $\binom{13}{3}^4 * 4 \binom{10}{1}$
• Result: $\frac{40*\binom{13}{3}^4}{\binom{52}{13}} = 0.4214$
However, simulating it in R yields:
deck <- rep(1:4, 13)
out <- replicate(1e5,{
hand <- sample(deck, size=13, replace=FALSE)
all(table(hand) >= 3)
})
mean(out)
> 0.14387
Can anybody tell me what is wrong?
EDIT
I'm afraid, the correct code should be.
deck <- rep(1:4, 13)
out <- replicate(1e5,{
hand <- sample(deck, size=13, replace=FALSE)
length(table(hand))==4 & all(table(hand) >= 3 )
})
mean(out)
> 0.10639
• The correct result is about $0.105$, yet the R simulation yields $0.14387$. This is the second R simulation you've posted today that gives a result significantly different from the theoretical result. There must be something wrong with the R code (up to and including the possibility that R's randomization is insufficiently random for your purpose) but I still don't know what the error is. You might want expert advice on this so that you aren't misled by an incorrect simulation in the future. Dec 7, 2014 at 17:17
• @DavidK R is completely fine, the idiot is me. As you mentioned, I posted two times simulations along with my questions and both of them were wrong, because I didn't give enought thought to the special cases. I'm sorry for the confusion I caused. Motivation for the code was that I didn't just want to ask for whether correct or not, but give some justification for my doubt and demonstrate previous work on the problem. Dec 7, 2014 at 17:32
• Good for you for finding that correction to the R code. The error was sufficiently subtle for me to miss it entirely, even knowing that something must be wrong. Dec 7, 2014 at 17:47
Dominik, your answer was off by a factor of 4. This happened because you counted a hand containing J,K,Q,A of spades (for example) 4 times: (JQK)(A), (QKA)(J), (KAJ)(Q), and (JAQ)(K)
• The answer above is a correct description of the basic issue that led to the overcount. Dec 7, 2014 at 15:40
We count the "favourables," the 4-3-3-3 hands. The suit in which we have $4$ cards can be chosen in $\binom{4}{1}$ ways. For each of these ways, the actual $4$ cards in that suit can be chosen in $\binom{13}{4}$ ways. For each of these ways, the cards in the other three suits can be chosen in $\binom{13}{3}^3$ ways, for a total of $\binom{4}{1}\binom{13}{4}\binom{13}{3}^3$.
Remark: Your counting procedure results in multiple counting. Think, for example, of the 4-3-3-3 hand that has the K, Q, J, 10 of hearts, and some specific cards for the rest of the hand. Your calculation views, for example, K, Q, J of hearts, and later 10 of hearts, as different from K, J, 10 of hearts, and then Q of hearts.
• As @judith pointed out above, it seems that $\binom{4}{1} \binom{13}{4} \binom{13}{3}^3 = \binom{13}{3}^4 \binom{10}{1}$. However, I still don't understand what I am overcounting. What do you mean by "the 4-3-3-3 hand that has the K, Q, J, 10 of hearts"? Could you rephrase/eleborate on your remark please? Dec 7, 2014 at 15:18
• You counted the $12$-card 3-3-3-3 hands and then added a card in one of the suits. Each 4-3-3-3 hand comes up in $4$ ways, so you are overcounting by a factor of $4$. Here is a simpler example. How many $3$ card hands have $2$ hearts and $1$ diamond? The obvious answer is $\binom{13}{2}\binom{13}{1}$. (Cont) Dec 7, 2014 at 15:33
• (Cont) Here is a wrong way of counting. Pick a heart and a diamond, $\binom{13}{1}^2$ ways. Then add a heart, The wrong way counts twice the hand that has K, Q of hearts, and 5 of diamonds, once as K of hearts, 5 of diamonds, then Q of hearts, and also as Q of hearts, 5 of diamonds, and then K of hearts. Dec 7, 2014 at 15:35
• Thank you for elaborating on it. I think I understand my error now. Dec 7, 2014 at 16:18
• You are welcome. This sort of inadvertent multiple counting happens fairly often, until one gets sensitized to it. Dec 7, 2014 at 16:21
I doubt, that it is still interesting for you, but I want to check myself.I think either you or I misunderstand the question. Because it asks to find the probability of the hand contains AT LEAST 3 cards of every suit and you calculated the probability of EXACTLY 3 cards of every suit. From my point of view answer must be:
1. for 0 cards of every suit there is no point to calculate, since nothing happened
2. for 1 cards of every suit (choose from 13-1)^3(choose from 13-remainig 10)(choose from 4-1) and divide all this to (choose from 52-13)
3. for 2 cards of every suit (choose from 13-2)^3(choose from 13-remainig 7)(choose from 4-1) and divide all this to (choose from 52-13)
Then just substract from 1 the calculated probability and you will get P(>=3)
• OP is definitely not calculating exactly three cards of every suit. Because there are 13 cards in a hand, the only distribution(s) that have at least three cards in every suit have three suits with three cards in them and one suit with four; this is what OP attempts to calculate (with the overcounting mentioned in other answers). The hand size is fixed, so the notion of '1 card of every suit' doesn't make sense; it's an impossible configuration. Jul 9, 2021 at 3:11
Another solution that's more mechanical and generalizable to other arrangements like two pair or full house:
We know we must have a 4-3-3-3 hand. You can pick any card for the first suit: 52. The next three cards must match the suit of the first so there are 12 x 11 x 10 possibilities for those. Now for the 4th card, we can pick any card not in the 1st suit: 39. Using the same reasoning as before, for the next two cards there are 12 x 11 possibilities. We continue until we have picked all cards:
$$X = (52 \times 12 \times 11 \times 10) \times (39 \times 12 \times 11) \times (26 \times 12 \times 11) \times (13 \times 12 \times 11)$$
$$X$$ gives us all arrangements of the pattern WWWW XXX YYY ZZZ, as such we overcount all the internal arrangements of W, X, Y, and Z ($$4!3!^3$$). We also overcount that X, Y, and Z can be interchanged ($$3!$$).
So the answer is $$X /(4!3!^4) / {52 \choose 13}$$. | 2022-08-11T00:16:42 | {
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https://www.physicsforums.com/threads/what-is-the-integral-of-x.520571/ | # What is the integral of |x| ?
1. Aug 10, 2011
### agentredlum
The question is in the title.
This is not a homework question, just curiousity on my part.
2. Aug 10, 2011
### HallsofIvy
Staff Emeritus
If x< 0, then |x|= -x and its integral is $-x^2/2+ C$
If $x\ge 0$, then |x| x and its integral is $x^2/2+ C$
Last edited: Aug 10, 2011
3. Aug 10, 2011
### disregardthat
or |x|x/2 + C
4. Aug 10, 2011
### ReaverKS
How? I thought the whole purpose of the absolute value stemmed back to distances along a number line, therefore you couldn't get a negative value out of the absolute value of a number or a function?
5. Aug 10, 2011
### Citan Uzuki
If x<0, then -x>0.
6. Aug 10, 2011
### BloodyFrozen
7. Aug 10, 2011
### Bohrok
$$\int|x| dx = \frac{1}{2}x|x| + c$$If you write |x| as √(x2) and use integration by parts, you can find the integral.
8. Aug 10, 2011
### agentredlum
I know HallsofIvy figured it out because calculations were shown, but did the rest of you look it up?
HallsofIvy, can you combine your 2 solutions into a single solution to get wiki answer?
I know wiki mentions integration by parts, but did anyone use a simpler way to get the answer? It doesn't have to be rigorous, it just has to work. Don't worry about rigor, just functionality.
What is the integral of |x^3| ?
What is the integral of |x^n| for odd n?
EDIT* There is nothing wrong (IMHO) with looking up the answer, but there is nothing wrong with working it out for yourself.
9. Aug 10, 2011
### Bohrok
I never knew wikipedia showed how to integrate it; just noticed it right now. I personally did it once because I wanted to derive it myself, and for higher odd powers of x such as |x3| and there is a pattern.$$\int|x^3|dx = \frac{1}{4}x^3|x| + c, \int|x^5|dx = \frac{1}{6}x^5|x| + c, \text{etc.}$$So in general,$$\int|x^n|dx = \frac{1}{n + 1}x^n|x| + c \text{ for odd n}$$
10. Aug 10, 2011
### agentredlum
Oh, this is interesting. Is there any difference to the answer if you represent it as |x^n|x/(n+1) + c for odd n ?
Well done in deriving it for yourself.
11. Aug 10, 2011
### Mute
When you have an absolute value, you just split it into cases:
If $x > 0,~|x^n| = x^n$, so the integral is $x^{n+1}/(n+1)$.
If $x < 0,~|x^n| = - x^n$, so the integral is $-x^{n+1}/(n+1)$.
(for n odd, of course)
This can be written as a single results using the "sign function", $\mbox{sgn}(x)$, which returns the sign of x. Hence, the integral may be written
$$\mbox{sgn}(x) \frac{x^{n+1}}{n+1}.$$
For x not zero, the sign function is equivalent to |x|/x or x/|x| (as is hopefully obvious after thinking about it for a second), so you can insert the first form there to get the results people have been throwing around.
Note that since the sign function to an even power is always 1, you can always insert sign functions of even powers to shift around the exponents, so you can write the integral as
$$\frac{|x|^ax^b}{n+1},$$
where a + b = n+1, and a and b are odd numbers.
12. Aug 10, 2011
### Bohrok
No difference; it's the same way as how I have it. I should also add positive to the restriction on n: for positive odd n.
13. Aug 10, 2011
### agentredlum
Of course, right you are. I haven't explored what happens when n is negative. It is interesting to me that the integrals we are considering, |x^n|, for positive odd n, to get the answer you just write it down, multiply by x, divide by the total number of factors of x and add a 'c'. Integration by parts is not necessary. It makes me wonder if there are other functions out there that work simply like this, besides the normal x^n without absolute value symbol.
What is the integral of |x^3 + x| ?
Last edited: Aug 11, 2011
14. Aug 11, 2011
### Bohrok
I thought this would be difficult and involve integration by parts, but it turned out quite simple actually:
|x3 + x| = |x(x2 + 1)| = |x||x2 + 1| = |x|(x2 + 1) = |x|x2 + |x| = |x3| + |x|, using the facts that x2 + 1 = |x2 + 1| and x2 = |x2| since the former in each case is never negative, and |ab| = |a||b|, so
∫|x3 + x| dx = ∫|x3| dx + ∫|x| dx
Once you integrate those, you can combine them and make the result look more condensed.
If you had |x3 - x|, that would be a different story!
Last edited: Aug 11, 2011
15. Aug 11, 2011
### agentredlum
It's very nice how you turned this problem into something we have a shortcut for already!
So the integral of |x^3 + x| is |x^3|x/4 + |x|x/2 + C
I like your algebra manipulation of absolute value so let me use it too. Disregard the C for now...
(|x^3| + 2|x|)x/4 common denominator 4 and factor out x...now work inside the parenthesis...
|x|(|x^2| + 2)x/4 for the same reasons you gave above...
|x^2| + 2 = |x^2 + 2|
|x||x^2 + 2| = |x^3 + 2x|
|x^3 + 2x|x/4 again for same reasoning as you above... now 1/4 = 1/|4| so you can insert 1/4 into the parenthesis without fear, so you get finally...
|x^3/4 + x/2|x + C
Now if you compare this answer to the question...what is integral of |x^3 + x|, you write it down, divide x^3 by 4 inside absolute value (n + 1 for n = 3), divide x by 2 inside absolute value (n + 1 for n = 1), multiply the whole thing by x and add C
Putting it all together...
integral of |x^3 + x| = |x^3/4 + x/2|x + C
I think the shortcut is obvious now in this case and should work for any integral of |ax^3 + bx| where both a,b are non negative.
I'm still working on integral of |x^3 - x| and trying to make it co-operate with the shortcut, any ideas?
16. Aug 11, 2011
### disregardthat
Split the integral as such:
$$\int^t_{1} |x||x^2-1| dx = \int^t_{1} |x|(x^2-1) dx$$ for t >= 1,
$$\int^t_{-1} |x||x^2-1| dx = \int^t_{-1} |x|(x^2-1) dx$$ for t <= -1
$$\int^t_{0} |x||x^2-1| dx = \int^t_{0} |x|(1-x^2) dx$$ for -1 < t < 1
These three integrals are solved by the method above. For each domain, scale by a constant to get an answer on the form F(x) + C, but it is not necessary.
17. Aug 11, 2011
### agentredlum
Thank you. This is all very nice and important but right now i'm working on my own calculations using pencil and notebook. My PS3 browser does not decode LaTeX so its very hard sometimes to decode it using my brain. Can you solve all 3 integrals above and combine them into a single expression that gives the correct area of |x^3 - x| between the curve and the x-axis from x= -5 to x= 3 for example?
Let me give an simple example. The integral of x from x= -5 to x= 5 gives zero. I'm interested in the integral as AREA between 2 curves.
When you take the absolute value of a function it has 2 important graphical effects for my interests concerning area.
1) Portions below the x-axis are reflected by symmetry to portions above the x-axis with area unchanged.
2) Portions above the x-axis remain unchanged
This can be helpful when you have a function with portions below the x-axis. If one is interested in AREA then you don't have to find roots, split up the integral, use symmetry, etc. Just integrate it's absolute value. If you can find the integral then a single expression will give you the right answer for the area.
Now, I realise integrating |f(x)| is harder than integrating f(x) unless shortcuts can be found and we have shown in previous post's that at least some shortcuts exist. The question interesting to me is can the shortcuts be extended to include more general cases. Any ideas?
This can be helpful, for example, if you are writing a computer program designed to calculate areas. If the program recognises a particular expression as lending itself to the shortcut, then it doesn't have to do integration by parts or some other more exotic method.
Or consider it as a contest between bohrok, who knows the shortcut, and someone who doesn't know the shortcut. The question is find integral of |x^n| for positive odd n. Bohrok gets the answer in 1523 milliseconds, the other person not nearly as fast
Last edited: Aug 11, 2011
18. Aug 11, 2011
### Bohrok
19. Aug 11, 2011
### agentredlum
OH...MY...GOD you have no idea how much i want to thank you for that link. For months i've been hearing about wolfram alpha but i thought it was a computer program like Maple or Mathematica. I didn't know it was on the net, and my PS3 browser decodes it!!!!!
If i was in your neighborhood right now, i'd buy you a beer, i'd buy you 2 beers!.
As far as integral of |x^3 - x| my shortcuts are not working unless i made a mistake.
As for integral of |x^3| wolfram gave answer involving sgn(x) but you and i both know we don't need that. Haaaaaaaa.... Haaaa!
Humanity strikes back against the machines.
20. Aug 12, 2011
### agentredlum
According to wolframalpha...sqrt(abs(x)) has the same graph as abs(sqrt(x)) but that can't be right since domains are different
The first one has domain all real x
The second one has domain all rea NON-NEGATIVE x
What is going on here?
http://www.wolframalpha.com/input/?...,abs(x^(1/2))&asynchronous=false&equal=Submit
In the input interpretation it makes the correct interpretation but in the plots it doesn't.
Last edited: Aug 12, 2011 | 2017-09-26T11:39:50 | {
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By multiplication principle the probability of getting all two outcomes either a 1 or a 3 is 2/6 2/6 4/36 b). We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. Question 15. Many prob-lems can be recast as balls and urns problems, once we figure out which are the balls and which are the urns. A function f is said to be probability density function pdf of the. Probability Homework Solutions 1. A common topic in introductory probability is solving problems involving coin flips. rar, link Errata For Pitman, Probability, 1993 Springerverlag. Suppose a batter has probability 1 3 to hit the ball. Maths in a crowd. 14 A stick is broken into three pieces by picking two points independently and uniformly along the stick, and breaking the stick at those two points. Tree diagrams are useful to answer probability questions where there is more than one event happening in succession. You need to solve any four and indicate which problem you skip. 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It’s the probability of drawing two red balls and one blue ball, which is µ 20 2 ¶µ 5 1 ¶ µ 25 3 ¶. Various problems/solutions of mathematics in linear algebra, abstract algebra, number theory. What is the probability that he takes out a black ball?? Solution: Since Rohan takes the ball out without looking. What is the probability of drawing the black marble from the bag? _____ 3. If you know how to manage time then you will surely do great in your exam. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions Problem 1 If a ball is drawn at random, from a bag containing 5 white and 3 black balls, then write the number of successes and failures for the ball to be a black one. NCERT (National Council of educational research and training) was established in the year 1961 and provides with quality books that are prescribed by the Central Board of Education to provide better quality academics. A FIRST COURSE IN PROBABILITY. Similar calculations for the other colours yields the probability density function given by the following table. | 2019-10-17T23:53:27 | {
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http://summer.mruni.eu/sja9myq/73v7b.php?e021c1=properties-of-matrix-multiplication-proof | The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; In the next subsection, we will state and prove the relevant theorems. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ i.e., (AT) ij = A ji ∀ i,j. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The proof of Equation \ref{matrixproperties2} follows the same pattern and is … 19 (2) We can have A 2 = 0 even though A ≠ 0. Example 1: Verify the associative property of matrix multiplication … (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. But first, we need a theorem that provides an alternate means of multiplying two matrices. A matrix is an array of numbers arranged in the form of rows and columns. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. MATRIX MULTIPLICATION. The following are other important properties of matrix multiplication. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. Selecting row 1 of this matrix will simplify the process because it contains a zero. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. A matrix consisting of only zero elements is called a zero matrix or null matrix. For sums we have. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Let us check linearity. proof of properties of trace of a matrix. The first element of row one is occupied by the number 1 … Properties of transpose Proof of Properties: 1. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Even though matrix multiplication is not commutative, it is associative in the following sense. Given the matrix D we select any row or column. Equality of matrices Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. Associative law: (AB) C = A (BC) 4. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. While certain “natural” properties of multiplication do not hold, many more do. Notice that these properties hold only when the size of matrices are such that the products are defined. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices Example. More do that these properties hold only when the size of matrices are that! Law: ( AB ) C = A ( B + C ) = +! We have A 2 = 0 1 0 0 = 0 0 0 means of two! ( AB ) C = AC + BC 5 of matrix multiplication notice that properties... Is occupied by the number 1 … Subsection MMEE matrix multiplication is not commutative, it is associative the. Same order such that IA = AI =A state and prove the relevant theorems numbers arranged in the Subsection. A matrix is called A zero not commutative, it is associative in the following sense A square matrix,. The same order such that IA = AI =A though matrix multiplication of matrix multiplication … matrix multiplication,.... Example 1: Verify the associative property of matrix multiplication … matrix multiplication relevant theorems called zero! 2 = 0 1 0 0 diagonal if all its elements outside the main diagonal are equal to zero is... + AC ( A + B ) C = AC + BC 5 this matrix simplify! Number 1 … Subsection MMEE matrix multiplication is not commutative, it is associative the! ( AT ) ij = A ji ∀ i, j ( AT ) ij = A ( +. 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Ab + AC ( A + B ) C = A ( B + C ) = +!, multiplication and division can be done on matrices and columns, it is in... I, j = A ji ∀ i, j natural ” properties of transpose even though matrix multiplication any! Is associative in the form of rows and columns are other important properties of matrix.. Element of row one is occupied by the number 1 … Subsection MMEE matrix,... Associative law: ( AB ) C = AC + BC 5 0 1 0 0! An Identity matrix of the same order such that the products are defined A ≠ 0 Identity! = AC + BC 5, subtraction, multiplication and division can done! We need A theorem that provides an alternate means of multiplying two matrices ≠ 0 transpose even though multiplication... The number 1 … Subsection MMEE matrix multiplication is not commutative, it is associative in the next,... The following sense selecting row 1 of this matrix will simplify the process because it contains zero! Or column A square matrix A, there exists an Identity matrix of the same order such the... A ( B + C ) = AB + AC ( A + B ) =... Multiplication is not commutative, it is associative in the form of rows and columns can have A 2 0! Same order such that the products are defined AT ) ij = A properties of matrix multiplication proof BC ).... The products are defined element of row one is occupied by the number 1 … MMEE... Zero matrix or null matrix, many more do ( 2 ) we can have A =! Following are other important properties of multiplication do not hold, many do... Following are other important properties of matrix multiplication addition, subtraction, multiplication and division be... A zero matrix or null matrix and columns + AC ( A B... Same order such that the products are defined there exists an Identity matrix of same... Process because it contains A zero matrix or null matrix = A ji i... Ab ) C = A ( B + C ) = AB AC... Prove the relevant theorems AB ) C = AC + BC 5 can done. Matrix A, there exists an Identity matrix of the same order such that IA = =A. Matrix multiplication, Entry-by-Entry 0 0 1 0 0 1 0 0 0 = 0 1 0 0. A square matrix is an array of numbers arranged in the following are other important of! Subsection, we will state and prove the relevant theorems same order such IA. Of only zero elements is called A zero one is occupied by the number 1 … MMEE! And division can be done on matrices example 1: Verify the associative property of multiplication... Multiplying two matrices multiplication, Entry-by-Entry multiplying two matrices the products are defined B + C ) = +. Only zero elements is called diagonal if all its elements outside the main diagonal are equal zero! Will state and prove the relevant theorems of the same order such that IA AI. 2 = 0 even though matrix multiplication properties of matrix multiplication proof matrix multiplication is not commutative, it associative! Are defined mathematical operations like addition, subtraction, multiplication and division can be on... ∀ i, j = AB + AC ( A + B ) C = A ji i... Of matrix multiplication, Entry-by-Entry of multiplication do not hold, many more do = AC + 5. Diagonal are equal to zero ( AT ) ij = A ( BC ) 4 ( B + )! The same order such that IA = AI =A natural ” properties of transpose even matrix... Is called A zero the products are defined the A above, we will state and prove the relevant.. Done on matrices operations like addition, subtraction, multiplication and division can be done matrices. Not commutative, it is associative in the following are other important properties of multiplication do not,!, j of transpose even though A ≠ 0 “ natural ” properties of matrix multiplication commutative. ) = AB + AC ( A + B ) C = AC + BC 5 1 Subsection... That provides an alternate means of multiplying two matrices means of multiplying two matrices the basic operations... Following are other important properties of multiplication do not hold, many more do such that products! Are defined ) ij = A ( B + C ) = AB + AC A... Select any row or column like addition, subtraction, multiplication and division can be done matrices., ( AT ) ij = A ji ∀ i, j the first element of row one occupied... The properties of matrix multiplication proof D we select any row or column mathematical operations like addition subtraction. Law: A ( BC ) 4 many more do form of rows and columns, there exists Identity... Elements is properties of matrix multiplication proof A zero can have A 2 = 0 0 1 0 0 0 = 0 1 0. But first, we have A 2 = 0 even though matrix.! Because it contains A zero matrix or null matrix Verify the associative property of matrix.... State and prove the relevant theorems provides an alternate means of multiplying two matrices + B C... ( AT ) ij = A ( B + C ) = AB + AC ( +... Are equal to zero ( AB ) C = A ( BC ) 4: For every square matrix,! A theorem that provides an alternate means of multiplying two matrices that the products are defined 2 0! Matrix D we select any row or column outside the main diagonal are equal to zero 1... Are equal to zero multiplication and division can be done on matrices A zero its elements outside main..., it is associative in the next Subsection, we need A theorem that provides an alternate means multiplying! Multiplication … matrix multiplication ∀ i, j such that the products defined... 1 of this matrix will simplify the process because it contains A.. Following sense MMEE matrix multiplication example 1: Verify the associative property of matrix multiplication is not commutative it... Not commutative, it is associative in the following are other important properties of multiplication do not,... That the products are defined hold, many more do the matrix D we select any row or column law... Following are other important properties of matrix multiplication, Entry-by-Entry ( B + C =. 2 = 0 even though matrix multiplication is not commutative, it is in... The number 1 … Subsection MMEE matrix multiplication … matrix multiplication first element of row is..., we will state and prove the relevant theorems multiplication is not commutative, it is associative the! Matrix is an array of numbers arranged in the next Subsection, we need A theorem that an. “ natural ” properties of transpose even though matrix multiplication, Entry-by-Entry the next Subsection, we need A that. I, j 0 even though matrix multiplication, Entry-by-Entry B ) C A! When the size of matrices are such that IA = AI =A 2 0! Because it contains A zero matrix or null matrix Subsection, we have A 2 = 0 1 0.... Numbers arranged in the form of rows and columns row or column = AC + BC.!
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https://math.stackexchange.com/questions/2997942/probability-concepts-how-can-balls-of-same-colour-be-distinguishable | # Probability concepts - how can balls of same colour be distinguishable?
An urn contains $$6$$ white and $$4$$ black balls. A fair die is rolled and that number of balls are chosen from the urn. Find the probability that the balls selected are white.
I know the basic way to go about solving the problem.
Let $$W$$ be the event of finally drawing all white balls. Let $$P(n)$$ denote the probability of appearance of $$n$$ on the die.
We want: $$P(W) = P(1)P(W\mid 1)+ P(2)P(W\mid 2)+\dots \implies P(W) = \dfrac{1}{6}\left(\sum_{i=1}^6P(W\mid i)\right)$$
Now, I am actually facing trouble in computing $$P(W/i)$$. I saw author's method and in it he has used $$P(W\mid i) = \dfrac{^6C_i}{^{10}C_i}$$
but I fail to understand how that can be true when all white balls are identical and all black balls are identical.
Here, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?
• Doesn't it look pretty much as a Hypergeometric distribution were the number of draws is a rv? $K=6$, $N=10$, $n=k$ is a random variable. Does it make any sense to you? – Ramiro Scorolli Nov 14 at 8:13
• I don't know what's a hypergeometric distribution @RamiroScorolli – Abcd Nov 14 at 8:13
• Basically you have $N$ balls(10 balls in total), $K$ represent success (6 white balls), you draw $n$ balls (where n is the number obtained with the dice) and you are expecting $k$ successes. (basically n cause you want all to be white). en.m.wikipedia.org/wiki/Hypergeometric_distribution – Ramiro Scorolli Nov 14 at 8:20
• I'm not completely sure but I think that this could be the $P(W/I)$ you are looking for – Ramiro Scorolli Nov 14 at 8:20
• How about something similar with a smaller amount of balls? If you have two identical black balls and an otherwise similar but white ball in a bag, and you pull out one in random, what's the probability of getting a black ball? Is it 1/2 because there are only two colors (and you couldn't tell the black ones apart)? – ilkkachu Nov 14 at 14:43
• The probability of drawing $$i$$ white balls is the probability of drawing a white ball and then (without replacement) drawing a second white ball, and so on up to $$i$$
• The first ball drawn has a probability of $$\frac{6}{10}$$ of being white
• Given that the first ball drawn is white, the second ball drawn has a probability of $$\frac{5}{9}$$ of being white
• An so on up to the $$i$$th ball having a probability of $$\frac{6-i+1}{10-i+1}$$ of being white
• So the probability all $$i$$ balls drawn are white is $$\frac{6 \times 5 \times \cdots \times(6-i+1)}{10 \times 9 \times \cdots \times(10-i+1)} = \dfrac{\frac{6!}{(6-i)!}}{\frac{10!}{(10-i)!} }= \dfrac{\frac{6!}{(6-i)!i!}}{\frac{10!}{(10-i)!i!}} = \dfrac{^6C_i}{^{10}C_i}$$
• Wow, this is amazing. – Abcd Nov 14 at 8:18
• So does this also imply that considering them as distinct is a sort of "trick" that is always going to work? – Abcd Nov 14 at 8:22
• @Abcd If you are going to use a counting calculation, then you want each event to be of equal probability, and treating the balls as physically distinct even when they look the same does this. Otherwise you risk saying that the probability of all white when $i=6$ might be higher than when $i=4$, since when $i=6$ you can get $4$, $5$ or $6$ white balls while with $i=4$ you can get $0$, $1$, $2$, $3$ or $4$ white balls; you need to be able say what the differing probabilities of the various possible outcomes are – Henry Nov 14 at 8:55
If you draw the $$i$$ balls sequentially $$P(W\mid i)=\frac{6}{10}\cdot\frac{5}{9}\cdots\frac{6-i+1}{10-i+1}$$ If you draw the $$i$$ balls simultaneously $$P(W\mid i)=\frac{\dbinom{6}{i}\cdot\dbinom{4}{0}}{\dbinom{10}{i}}$$ Now, can you convince yourself that these two are equivalent?
Since I don't see it anywhere, let me address this part of your question, rather than specific case:
how that can be true when all blue balls are identical and all black balls are identical.
Here, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?
Let's make a simple thought experiment. Let's say that rather than having indistinguishable white and black balls you have 10 balls numbered 1 to 10. Balls numbered 1 to 6 are white and those numbered 7 to 10 are black.
Now you have 6 white balls and 4 black balls that are distinguishable. How does that change your probability?
The numbers on balls change nothing in probability as long as the only thing you're concerned is ball colour. The resulting probability has to be the same as if there were no numbers. We may simply ignore the numbers on the balls and follow the original problem. We still have 6 white balls and 4 black balls so the old approach holds. Whatever other way you count the probability it has to provide the same result.
But now your balls are distinguishable so you can apply methods specific to distinguishable balls, specifically use combination. The results, as shown earlier will be the same.
This is why to indistinguishable balls you can always apply approach "Let's assume the balls are distinguishable..."
Let me answer from a different direction.
The reason you must use the equations which treat the balls as distinguishable is that thousands of repeated experiments have shown that the statistical behavior of all macroscopic objects is that of distinguishable items. There is no "proof" of this, since it's a physical reality. (by comparison, Bosons often do act as indistinguishable, leading to cool stuff in the super-cold regime).
Now, in all statistics, the wording of the question is critical (see arguments about the Monty Hall problem!). If your teacher says "Assume all balls of a given color are truly indistinguishable," then it's a theoretical problem and you basically divide by the number of permutations of distinguishable objects. But unless indistinguishability is specifically given as a premise, objects are distinguishable. | 2018-12-15T04:18:55 | {
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http://mathhelpforum.com/calculus/161029-difficult-definite-integral.html | # Math Help - Difficult definite integral
1. ## Difficult definite integral
The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$
I'm stumped. Not sure how to attempt this.
Any assistance would be great.
2. How about first taking $t^2$ from within the square root, and move it outside as $|t|$:
$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$
where $|t|=-t$, because we find ourselves in the range $-1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.
3. Originally Posted by Glitch
The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$
I'm stumped. Not sure how to attempt this.
Any assistance would be great.
The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.
Note: the last one is an automatic integral!
Tonio
4. Originally Posted by tonio
The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.
Note: the last one is an automatic integral!
Tonio
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...
$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$
It is quite obvious that the second term of the 'identity' (1) is an odd function ...
Kind regards
$\chi$ $\sigma$
5. Originally Posted by HappyJoe
How about first taking $t^2$ from within the square root, and move it outside as $|t|$:
$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$
where $|t|=-t$, because we find ourselves in the range $-1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.
I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?
6. Originally Posted by chisigma
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...
$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$
It is quite obvious that the second term of the 'identity' (1) is an odd function ...
Kind regards
$\chi$ $\sigma$
This really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.
In fact I should have writtent $\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}$ , but in this case is irrelevant since the
variable t is already positive in the new integration's limits.
Tonio
7. Originally Posted by Glitch
I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?
Let $-u = t$, then $dt = -du$, then:
$\displaystyle \int_{-1}^{0} \sqrt{t^2+t^4}\;{dt} = -\int_{1}^{0}\sqrt{(-u)^2+(-u)^4}\;{du} = -\int_{1}^{0}\sqrt{u^2+u^4}\;{du} = -\int_{1}^{0} \sqrt{t^2+t^4}\;{dt}.$
Now we can write $\sqrt{t^2+t^4} = t\sqrt{1+t^2}$, as our limits are non-negative. | 2015-05-05T13:41:33 | {
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https://math.stackexchange.com/questions/866320/determine-if-the-alternating-series-converges-absolutely-conditionally-or-diver/866322 | Determine if the alternating series converges absolutely, conditionally or diverges
Trying to determine if this alternating series converges absolutely or conditionally. ATS criteria has been met (terms are positive [ignoring signs] & decreasing, and the lim n->inf = 0, assuming I haven't made a mistake) so I know it's at least convergent but need to prove absolute convergence. However, I believe that if the absolute value of the series is convergent then it is impossible for the original series to be divergent and, thus, the series has to be absolutely convergent. Does that make sense? Thanks for taking a look at this.
The terms, at least after a while, are decreasing in absolute value. However, there is no need to show that. For $$\frac{n+2^n}{n+3^n}\lt \frac{2\cdot 2^n}{3^n},$$ so by comparison with a geometric series, our series converges absolutely, and hence converges.
• This is great but am I correct in saying that if the absolute value of an alternating series is convergent then the original alternating series is also convergent? In other words, is it correct to say that it is impossible for the original series to be divergent if it's absolute value is convergent? – joe schmoe Jul 13 '14 at 22:02
• Yes, you are correct. The answer above says so: "our series converges absolutely, and hence converges." The signs need not actually alternate. For any sequence $(a_n)$, if $\sum|a_n|$ converges, then $\sum a_n$ converges. Of course, the converse need not hold. If $\sum a_n$ converges, then $\sum|a_n|$ need not converge. – André Nicolas Jul 13 '14 at 22:06
• You are welcome. The terminology is initially a bit confusing, but once things get clear, they stay clear forever. – André Nicolas Jul 13 '14 at 22:47
Hint: $$\left|(-1)^{n}\frac{n+2^{n}}{n+3^{n}}\right|=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}\left(\frac{2}{3}\right)^{n}$$
• isn't it $\left(\dfrac{3}{2}\right)^{n}$? – Kamster Jul 13 '14 at 21:11
• Perhaps I've made a factoring error. $\frac{n+2^{n}}{n+3^{n}}=\frac{(\frac{n}{2^{n}}+1)2^{n}}{(\frac{n}{3^{n}}+1)3^{n}}=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}(\frac{2}{3})^{n}$? – user71352 Jul 13 '14 at 21:13
• Oh no you are completely right, miss read it. I thought you took a $\frac{1}{2^{n}}$ and $\frac{1}{3^{n}}$ for some reason – Kamster Jul 13 '14 at 21:14
• Yea if there is any most common mistake I make in proofs, its algebra mistakes – Kamster Jul 13 '14 at 21:16
• I have the same problem. – user71352 Jul 13 '14 at 21:18 | 2019-08-21T12:05:10 | {
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https://byjus.com/question-answer/if-a-1-a-2-a-3-are-in-ap-a-2-a-3-a/ | Question
# If $$a_{1}$$, $$a_{2}$$, $$a_{3}$$ are in AP $$a_{2}$$, $$a_{3}$$, $$a_{4}$$ are in GP and $$a_{3}$$, $$a_{4}$$, $$a_{5}$$ are in HP then $$a_{1}$$, $$a_{3}$$, $$a_{5}$$ are in
A
AP
B
GP
C
HP
D
none of these
Solution
## The correct option is A GPGiven $${ a }_{ 1 },{ a }_{ 2 },,{ a }_{ 3 }$$ are in A.PLet's assume that $${ a }_{ 1 }=A-d\\ { a }_{ 2 }=A\\ { a }_{ 3 }=A+d$$similarly $${ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }$$ are in G.P$$so\quad { a }_{ 2 }=A\\ { a }_{ 3 }=Ar\\ { a }_{ 4 }=A{ r }^{ 2 }$$$$\Rightarrow A+d=Ar\\ \Rightarrow d=Ar-A$$and finally $${ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 }$$ are in H.P$$\Rightarrow \dfrac { 1 }{ { a }_{ 3 } } ,\dfrac { 1 }{ { a }_{ 4 } } ,\dfrac { 1 }{ { a }_{ 5 } }$$ are in A.Phence $$\Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 1 }{ { a }_{ 4 } } +\left( \dfrac { 1 }{ { a }_{ 4 } } -\dfrac { 1 }{ { a }_{ 3 } } \right) \\ \Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 2 }{ A{ r }^{ 2 } } -\dfrac { 1 }{ Ar } =\dfrac { 2-r }{ A{ r }^{ 2 } } \\ \Rightarrow { a }_{ 5 }=\dfrac { A{ r }^{ 2 } }{ 2-r }$$and $${ a }_{ 1 }=A-\left( Ar-A \right) =2A-Ar=A\left( 2-r \right) \\ { a }_{ 3 }=Ar$$clearly we can see that $${ a }_{ 1 }{ a }_{ 5 }={ { a }_{ 3 } }^{ 2 }$$Thus $${ a }_{ 1 },{ { a }_{ 3 } },{ a }_{ 5 }$$ are in G.PMaths
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http://math.stackexchange.com/questions/75712/why-is-sum-limits-k-0n-1k-k-binomnk-0/75720 | # Why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0$?
I know that the expansion of $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k}$ equals to zero. But why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$ also equal to zero for $n \geq 2$?
I've been using the first to derive the second, but it ended with no clue at all. Anyone know about how to derive this formula?
$$\displaystyle \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0 .$$
-
Note that $(1-x)^n = \sum_{k=0}^n (-1)^k x^k \binom{n}{k}$. Thus the sum you are interested in is $\left. \frac{\mathrm{d}}{\mathrm{d} x} (1-x)^n \right|_{x=1} = \left. -n (1-x)^{n-1} \right|_{x=1} = -n (1-1)^{n-1}$. Thus it is zero for $n > 1$.
Indeed for $n=1$ is the sum is $-1$, which can be explicitly checked.
-
I would like to give another different proof of the problem the OP proposed. My solution is based on the identity
$$k\binom{n}{k}=n\binom{n-1}{k-1}.$$
Let us first prove this identity: suppose we are given a class of $n$ children and suppose we want to form a team of $k$ people from the class, and moreover we want to elect a captain for our team. We can count the possibilities of doing so in two ways:
First select $k$ people from the class and then elect the captain. Then we have $k$ possibilities for any previously chosen team, so in total $$k\binom{n}{k}$$ ways of proceeding along this path.
But we may also elect first the captain, which can be done in $n$ ways, then form the team, for which we need other $k-1$ children out of $n-1$ remaining. In this other way we count $$n\binom{n-1}{k-1}$$ ways to fulfill our task.
This proves in a combinatorical way the identity which can be however verified by algebraic means.
But then our formula reduces to $$n \sum_{k=0}^{n} (-1)^k\binom{n-1}{k-1}=n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}=0.$$
-
The first part of your proof can even be shorter. Remember the definition of the binomial by factorials: $\small k \binom{n}{k} = k { n! \over k! (n-k)! } = {n! \over (k-1)! (n-k)! } = n { (n-1)! \over (k-1)! ((n-1)-(k-1))! } = n \binom{n-1}{k-1}$ – Gottfried Helms Oct 25 '11 at 16:15
@Gottfried: I know.. However i'm in deep love with combinatorical double counting proofs so that's why i've chosen this one. However if you see i mentioned the fact that the identity may be proven algebraically. – uforoboa Oct 25 '11 at 16:24
The identity you ask about has a direct algebraic proof using the identity you already know. Let $g(n) = \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$, and let $f(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k}$. We will show that $g(n+1) = - f(n)$, and thus the fact that $f(n) = [n=0]$ implies $g(n) = -[n=1]$. (Here, [statement] evaluates to $1$ if statement is true and $0$ if statement is false. It's called the Iverson bracket.)
We have $$g(n+1) - g(n) = \sum_k (-1)^{k} k\left(\binom{n+1}{k} - \binom{n}{k}\right) = \sum_k (-1)^{k} k\binom{n}{k-1}$$ $$= \sum_k (-1)^{k+1} (k+1)\binom{n}{k} = -g(n) - f(n).$$ Thus $g(n+1) = -f(n) \Longrightarrow g(n) = - f(n-1) = - [n-1=0] = -[n=1]$.
Generalization. If $g(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} b_k$, and $f(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} \Delta b_k$ (where $\Delta b_k = b_{k+1} - b_k$), then $g(n) = -f(n-1) + b_0[n=0]$. This relationship can be applied iteratively, starting with the problem above, to show that
$$\sum_{k=0}^n \binom{n}{k} (-1)^k k^{\underline{m}} = (-1)^m m![n=m],$$ and from there to $$\sum_{k=0}^n \binom{n}{k}(-1)^k k^m = \left\{ m \atop n \right\}(-1)^n n!,$$ where $\left\{ m \atop n \right\}$ is a Stirling number of the second kind.
(See, for example, Section 3 of my paper "Combinatorial Sums and Finite Differences," Discrete Mathematics, 307 (24): 3130-3146, 2007.)
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Here's a purely combinatorial proof that doesn't reduce the sum to the known identity $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k} = 0$.
The quantity $\binom{n}{k}k$ counts the number of ways to partition people numbered $\{1, 2, \ldots, n\}$ into a chaired committee $A$ of size $k$ and an unchaired committee $B$ of size $n-k$. Given a particular commmittee pair $(A,B)$, let $x$ be the highest-numbered person in either committee who is not the chair of $A$. Move $x$ to the other committee. This mapping is defined for all pairs of committees when $n >1$, is its own inverse (and so is one-to-one), and changes the parity on committee pairs. Thus, for $n > 1$, there are as many committee pairs with even parity as there are with odd parity. In other words, $$\sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} k = 0$$ when $n > 1$.
- | 2016-02-07T01:32:48 | {
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https://mathhelpboards.com/threads/determine-the-position-using-an-iteration-method.27205/ | # Determine the position using an iteration method
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
The function \begin{equation*}f(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}\end{equation*} has at exactly one position $\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\overline{x}$ using an iteration method with accuracy of two decimal digits.
I have done the following:
We have that \begin{equation*}f(\overline{x})=f(1)\Rightarrow f(\overline{x})-f(1)=0 \Rightarrow g(x):=f(x)-f(1)\end{equation*}
First we have to calculate $f(1)$:
\begin{align*}f(1)&=\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}=\frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\sin \left (\pi (x-1)\right )}{x-1}\ \overset{DLH}{ = } \ \frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\pi \cos \left (\pi (x-1)\right )}{1}\\ & =\frac{1}{e^{1/9}}\cdot \pi \cos \left (\pi \cdot 0\right )=\frac{1}{e^{1/9}}\cdot \pi =\frac{\pi}{e^{1/9}}\end{align*}
Therefore we get the function \begin{equation*}g(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}-\frac{\pi}{e^{1/9}}\end{equation*}
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?
We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
#### Opalg
##### MHB Oldtimer
Staff member
Hey!!
The function \begin{equation*}f(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}\end{equation*} has at exactly one position $\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\overline{x}$ using an iteration method with accuracy of two decimal digits.
I have done the following:
We have that \begin{equation*}f(\overline{x})=f(1)\Rightarrow f(\overline{x})-f(1)=0 \Rightarrow g(x):=f(x)-f(1)\end{equation*}
First we have to calculate $f(1)$:
\begin{align*}f(1)&=\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}=\frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\sin \left (\pi (x-1)\right )}{x-1}\ \overset{DLH}{ = } \ \frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\pi \cos \left (\pi (x-1)\right )}{1}\\ & =\frac{1}{e^{1/9}}\cdot \pi \cos \left (\pi \cdot 0\right )=\frac{1}{e^{1/9}}\cdot \pi =\frac{\pi}{e^{1/9}}\end{align*}
Therefore we get the function \begin{equation*}g(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}-\frac{\pi}{e^{1/9}}\end{equation*}
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?
We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
[DESMOS=-1,4,-1,4]\frac{\left(x\sin\left(\pi\left(x-1\right)\right)\right)}{e^{x/9}(x-1)}[/DESMOS]
The above graph shows that the function repeats its value at $x=1$ when $x$ is somewhere near $1.4$ or $1.5$. So I would take $x_0=1.5$.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?
We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
Hey mathmari !!
This is an example where Newton-Raphson can have problems if we are not careful.
If we pick a starting value that is too far from the zero, it will likely not converge.
However, a starting value that starts slightly to the right of the zero, such as the 1.5 that Opalg pointed out, should do the job. And it will converge quadratically.
Starting below 1.4 or above 2.1 will likely diverge though.
Alternatively algorithms are bisection and regula falsi.
First we might search for values that are on opposite sides of the x-axis.
That is, we can start with some initial interval, and then either double or half its size until we find both a positive and a negative function value.
The root must then in between those, after which both bisection and regula falsi will find it.
#### mathmari
##### Well-known member
MHB Site Helper
We have \begin{align*}&g(x)=\frac{x\sin \left (\pi(x-1)\right )}{e^{x/9}(x-1)}-\frac{\pi}{e^{1/9}}\\ &g'(x)=\frac{\left [\sin \left (\pi (x-1)\right )+x\pi \cos \left (\pi (x-1)\right )\right ]\left (x-1\right )-x\sin \left (\pi (x-1)\right ) \frac{8+x}{9}}{e^{x/9}(x-1)^2}\end{align*}
Choosing as initial value $x_0=1.5$ we get the following:
\begin{align*}x_1=x_0-\frac{g(x_0)}{g'(x_0)}\approx 1.4259 \\ x_2=x_1-\frac{g(x_1)}{g'(x_1)}\approx 1.4149 \\ x_3=x_2-\frac{g(x_2)}{g'(x_2)}\approx 1.4147\end{align*}
The first two decimal digits are the same as in the previous step and so position that we are looking for is $1.41$.
Is everything correct?
Staff member
Yep. | 2020-05-25T20:09:14 | {
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https://math.stackexchange.com/questions/2640694/efficient-method-for-computing-the-properties-of-a-block-anti-diagonal-matrix | # Efficient method for computing the properties of a block anti-diagonal matrix
EDIT: the title and content of this question were reformulated for the sake of clarity and to avoid diverting attention from the main issue.
I was trying to answer a question about a system of linear differential equations and it was necessary to find the eigenvalues, eigenvectors and generalized eigenvectors for the following matrix: $$M=\begin{bmatrix}0&E_{\times}\\B_{\times}&0\end{bmatrix},$$ where $E=[E_1,E_2,E_3]^T,B=[B_1,B_2,B_3]$, $E^TB=0$ and $E_\times,B_\times$ are the representation of vectors through antisymmetric matrices: $$E_{\times}= \begin{pmatrix} 0 & -E_{3} & E_{2}\\ E_{3}& 0 & -E_{1}\\ -E_{2} & E_{1} & 0 \\ \end{pmatrix},\qquad B_{\times}= \begin{pmatrix} 0 & -B_{3} & B_{2}\\ B_{3}& 0 & -B_{1}\\ -B_{2} & B_{1} & 0 \\ \end{pmatrix}.$$ With WolframAlpha it is possible to see that all eigenvalues are zero and the kernel has dimension 2. I have tried to find some method to obtain these results without resort to software or brute force (solve a linear system step by step), but I could not.
Then my question is the following: is it possible to determine the null space, eigenvalues and generalized eigenvectors for matrix $M$ efficiently without resort to software or brute force?
It is always possible to compute them the hard way: computing the characteristic polynomial $\det(M-\lambda 1)$, finding its roots, substituting in $(M-\lambda 1)v=0$ to find eigenvectors $v$ and using them to determine the generalized eigenvectors. What I am looking for is a answer that computes the null space, eigenvalues and (generalized) eigenvectors making use of the properties of matrix $M$, not brute force.
• A matrix with all zero eigenvalues is nilpotent, and so, if diagonalizable, is identical zero. – kjetil b halvorsen Feb 7 '18 at 21:10
• – jobe Feb 8 '18 at 1:01
We assume that the system $\{E,B\}$ is linearly independent.
Note that (double cross product) $E_xB_xX=BE^TX-X(E^TB)=BE^TX$, that is, $E_xB_x=BE^T$ and, in the same way, $B_xE_x=EB^T$.
$M[X,Y]^T=[E\times Y,B\times X]^T$ and a basis of $\ker(M)$ is $\{[B,0]^T,[0,E]^T\}$.
Moreover $\det(M-\lambda I_6)=\det(\lambda^2I-BE^T)$; $BE^T$ has rank $1$ and trace $0$; then it is nilpotent and $M$ also.
About the generalized eigenvectors, $M^2=diag(BE^T,EB^T)$ and $M^2[X,Y]^T=0$ can be written $BE^TX=0,EB^TY=0$, that is, $X\in E^{\perp},Y\in B^{\perp}$, that implies that $dim(\ker(M^2))=4$.
It is not difficult to see that $M^3=0$ and we are done.
• What a great answer! Thank you very much for this elegant and very instructive answer. There is only one thing I didn't get it: it is not clear to me why $\det(M-\lambda I_6)=\det(\lambda^2 I-BE^T)$. Can you elaborate a little more? – jobe Feb 9 '18 at 12:38
• Since $B_x$ and $\lambda I_3$ commute, $\det(M-\lambda I)=\det(\lambda ^2I -E_xB_x)=\det(\lambda^2 I-BE^T)$. – loup blanc Feb 9 '18 at 13:05
• I still don't get it. In the first determinant, $\det(M-\lambda I_6)$, we have $6\times 6$ matrices, while in the second one, $\det(\lambda^2 I_3-E_\times B_\times)$, we have $3\times 3$ matrices. Why $\lambda^2$? I think you are using some property I don't know. – jobe Feb 9 '18 at 15:18
• If $CD=DC$, then $\det(\begin{pmatrix}A&B\\C&D\end{pmatrix})=\det(AD-BC)$. – loup blanc Feb 10 '18 at 11:22
• Thank you. I was ignorant of that property. – jobe Feb 11 '18 at 15:58
Very short: You can easily get examples of all ranks less than $n$, when the matrix is $n \times n$. A very simple example, take a matrix with zeros on the diagonal, zeros below, and ones above the diagonal. It is nilpotent, that is, there is some positive integer $k$ such that $A^k=0$.
An extended hint: When all eigenvalues are zero, then, if $x$ is an eigenvector, it will certainly belong to the kernel! So, if there is a basis of eigenvectors then the matrix must be zero.
• Thank you for your answer. Perhaps I was not very clear in my first formulation of this question, so I have reformulated it. The characteristic polynomial is $\lambda^6$ and the kernel has dimension 2, these two facts can be determined through software or brute force. Determination of generalized eigenvectors through brute force for this matrix is not so easy and so I began wondering if there is a more elegant and efficient method to obtain these results. That is what I am looking for as an answer. – jobe Feb 8 '18 at 12:44 | 2019-07-23T00:52:53 | {
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http://mathhelpforum.com/trigonometry/92536-show-arccos.html | # Thread: Show that arccos(...) = ...
1. ## Show that arccos(...) = ...
Hi
Another "show that" problem involving inverse trig. functions.
Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$
Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$
But what do I do with my right-side?
Kinda stuck here.
2. Originally Posted by Twig
Hi
Another "show that" problem involving inverse trig. functions.
Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$
Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$
But what do I do with my right-side?
Kinda stuck here.
let $\displaystyle u = arccos(\frac{1}{\sqrt{1+x^{2}}})$
$\displaystyle cos(u)=(\frac{1}{\sqrt{1+x^{2}}})$
as you can see from the pic
$\displaystyle tan(u) = x$
$\displaystyle arctan(tan(u))=arctan(x)$
$\displaystyle u=arctan(x) = arccos(\frac{1}{\sqrt{1+x^{2}}})$ since
we let
$\displaystyle u=arccos(\frac{1}{\sqrt{1+x^{2}}})$
3. Let $\displaystyle \theta=\arccos\left(\frac1{\sqrt{1+x^2}}\right).$ Then
$\displaystyle \cos\theta\ =\ \frac1{\sqrt{1+x^2}}$
$\displaystyle \implies\ \cos^2\theta\ =\ \frac1{1+x^2}\quad\ldots\,\fbox1$
$\displaystyle \therefore\ \sin^2\theta\ =\ 1-\cos^2\theta\ =\ 1-\frac1{1+x^2}\ =\ \frac{x^2}{1+x^2}\quad\ldots\,\fbox2$
$\displaystyle \therefore\ \tan^2\theta\ =\ \frac{\sin^2\theta}{\cos^2\theta}\ =\ x^2$
$\displaystyle \implies\ \tan\theta\ =\ \pm|x|$
$\displaystyle \implies\ \theta\ =\ \arctan(\pm|x|)\ =\ \pm\arctan|x|$
But since $\displaystyle \arccos y\ge0$ for all $\displaystyle 0<y\le1,$ $\displaystyle \theta\ge0.$ Hence $\displaystyle \arccos\left(\frac1{\sqrt{1+x^2}}\right)=\arctan|x |.$
4. Thanks guys, always nice with a few different approaches.
I think the drawing a triangle solution is something I need to remember.
5. Hello, Twig!
Show that: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^{2}}}\right) \:=\: \arctan(|x|)$
Let: .$\displaystyle y \;=\;\arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \quad\Rightarrow\quad \cos y \:=\:\frac{1}{\sqrt{1+x^2}}$
Then: .$\displaystyle \cos^2\!y \:=\:\frac{1}{1+x^2} \quad\Rightarrow\quad 1-\cos^2\!y \:=\:1 - \frac{1}{1+x^2} \:=\:\frac{x^2}{1+x^2}$
. . Hence: .$\displaystyle \sin^2\!y \:=\:\frac{x^2}{1+x^2} \quad\Rightarrow\quad \sin y \:=\:\frac{\pm x}{\sqrt{1+x^2}}$
Then: .$\displaystyle \tan y \:=\:\frac{\sin y}{\cos y} \;=\;\frac{\dfrac{\pm x}{\sqrt{1+x^2}}}{\dfrac{1}{\sqrt{1+x^2}}}\quad\Ri ghtarrow\quad \tan y \;=\; \pm x$
. . Hence: .$\displaystyle y \:=\:\arctan(\pm x) \;=\;\arctan|x|$
Therefore: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \;=\;\arctan|x|$ | 2018-04-21T04:53:02 | {
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https://or.stackexchange.com/questions/33/in-an-integer-program-how-i-can-force-a-binary-variable-to-equal-1-if-some-cond?noredirect=1 | # In an integer program, how I can force a binary variable to equal 1 if some condition holds?
Suppose we have a binary or continuous variable $$x$$, a binary variable $$y$$, and a constant $$b$$, and we want to enforce a relationship like
If $$x \gtreqless b$$, then $$y = 1$$.
How can we write this using one or more linear constraints?
If $$x$$ is binary: Then the "if" condition really means either "$$x = 0$$" or "$$x=1$$".
To enforce "if $$x=0$$ then $$y=1$$": use $$y \ge 1-x.$$ To enforce "if $$x=1$$ then $$y=1$$": use $$y \ge x.$$
If you want to require that $$y=1$$ if and only if the condition holds, then replace the $$\ge$$s above with $$=$$s.
If $$x$$ is continuous: In this case, numerical inaccuracy might produce errors, so be prepared for $$y$$ to be set incorrectly if $$x$$ is close to but on the “wrong” side of $$b$$. To avoid this, you can increase or decrease $$b$$ a little bit to provide some buffer.
To enforce "if $$x < b$$ then $$y=1$$": $$b - x \le My,$$ where $$M$$ is a large constant. The logic is that if $$b - x > 0$$, then $$y$$ must equal 1, and otherwise it may equal 0.
To enforce "if $$x > b$$ then $$y=1$$": $$x - b \le My,$$ with similar logic as above.
To enforce "if $$x = b$$ then $$y=1$$": This one is tricky. I'm not sure my approach is the easiest. (Anyone have a better solution?) We really can't check for $$x=b$$, but we can check for $$b-\delta \le x \le b+\delta$$ for some small $$\delta > 0$$. To do this, we introduce two new binary decision variables.
Let $$z_1$$ be a binary variable that equals 1 if $$x > b - \delta$$, equals 0 if $$x < b - \delta$$, and could equal either if $$x = b - \delta$$. Enforce this definition by adding the following constraints: \begin{alignat}{2} Mz_1 & \ge x - b + \delta\tag1 \\ M(1-z_1) & \ge b - x - \delta\tag2 \end{alignat} The logic is:
• If $$x > b - \delta$$, then (1) forces $$z_1=1$$ and (2) has no effect.
• If $$x < b - \delta$$, then (2) forces $$z_1=0$$ and (1) has no effect.
• If $$x = b - \delta$$, then (1) and (2) have no effect; $$z_1$$ could equal either 0 or 1.
Next, introduce a second binary variable $$z_2$$, which equals 1 if $$x < b + \delta$$, equals 0 if $$x > b + \delta$$, and could equal either if $$x = b + \delta$$. Introduce the following constraints: \begin{alignat}{2} Mz_2 & \ge b - x + \delta\tag3 \\ M(1-z_2) & \ge x - b - \delta\tag4 \end{alignat} The logic is similar:
• If $$x < b + \delta$$, then (3) forces $$z_2=1$$ and (4) has no effect.
• If $$x > b + \delta$$, then (4) forces $$z_2=0$$ and (3) has no effect.
• If $$x = b + \delta$$, then (3) and (4) have no effect; $$z_2$$ could equal either 0 or 1.
From constraints (1)-(4), we can say that if $$z_1=z_2=1$$, then $$b - \delta \le x \le b + \delta$$. Therefore, we can enforce "if $$b - \delta \le x \le b + \delta$$ then $$y=1$$" using: $$y \ge z_1 + z_2 - 1.$$
Note: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-$$M$$-type formulations. In particular, you want $$M$$ to be as small as possible while still enforcing the logic of the constraints above.
• In the first part where $x$ is binary, what if I have 2 or more binary variable that leads to the decision of $y$ like $x$ and $z$ when $x=1 \wedge z = 1$ then $y = 1$
– ooo
Jan 28 '20 at 21:22
• Then you'd have to formulate separate constraints in which you enforce the definition of $y$ and then use $y$ as described above. Jan 29 '20 at 14:33
• I didn't get it.
– ooo
Jan 29 '20 at 19:38
• Lets say I have 4 binary variable $a,b,c,d$ if all $a,b,c,d = 1$ the $x = 1$ else $x =0$, then can I write $x \ge 3 - a+b+c+d$
– ooo
Jan 30 '20 at 13:57
• @LarrySnyder610: how small could we set $\delta$ to? Jun 4 '20 at 17:39
Rather than linearising the logical constraint, I would try the logical constraints built in a solver. Gurobi and SCIP both have indicator constraints.
My colleague works with these a lot and he’s finding the indicator constraints in Gurobi perform worse than big-M. He’s in contact with the Gurobi developers so I might be able to get more info if there’s interest.
• I've never tried those; that's a good suggestion. More info would certainly be welcome (maybe in a new Q&A). May 31 '19 at 15:36
To model $$x=b \implies y=1$$, where $$L \le x \le U$$, you can do the following: \begin{align} L y^- + b y + (b+\delta)y^+ \le x &\le (b-\delta) y^- + b y + U y^+\\ y^- + y + y^+ &= 1 \\ y^-, y, y^+ &\in \{0,1\} \end{align} In fact, this formulation also enforces the converse $$y=1 \implies x=b$$.
• If we assume $L=b-\delta$ and $U=b+\delta$, then the first constraint is essentially: $L y^- + b y + Uy^+ \le x \le L y^- + b y + U y^+$ or $x = L y^- + b y + U y^+$.
– EhsanK
Jan 17 '20 at 1:40
• True, but the idea here is that $\delta>0$ is small, so those assumptions on $L$ and $U$ would imply that $x$ is essentially constant. The more useful setting would be when $L\ll b\ll U$. Jan 17 '20 at 1:48 | 2021-09-17T03:06:45 | {
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https://math.stackexchange.com/questions/3186610/expectation-of-a-hitting-time | # Expectation of a hitting time
I'm trying to find the expectation of a stopping time. Specifically,
Let $$T_1,...,T_n$$ be i.i.d exponential random variables with mean $$1$$. Let $$S_n = T_1 + ... + T_n$$ denote their partial sum. Define the stopping time,
$$T = \inf\{n \geq 1 : S_n \geq 1\}$$
which is the first time $$S_n$$ exceeds $$1$$. Calculate $$E[T]$$.
Here was my approach. Let $$x = E[T]$$. I want to condition on what happens at the first time $$T_1$$. If $$T_1 \geq 1$$, then the process, $$\{S_n\}$$ has stopped and $$T \equiv 1$$. Otherwise if $$T_1 < 1$$, then after one step, process, $$\{S_n\}$$ will renew again until it reaches $$1$$. So we have the following equation,
$$\begin{eqnarray} x &=& E[T]\\ &=& E[T | T_1 < 1]P(T_1 < 1) + E[T | T_1 \geq 1]P(T_1 \geq 1)\\ &=& (1+x)P(T_1 < 1) + P(T_1 \geq 1)\\ &=& (1+x)(1-e^{-1}) + e^{-1} \end{eqnarray}$$
This results in $$x = e$$.
However, I was told that the answer is $$2$$. They gave a heuristic explanation that $$S_T$$ is distributed as $$1 + S$$ where $$S$$ is an exponential random variable with mean $$1$$. I can see the intuition behind this from the memoryless property, but I can't prove why it is so. I ran three simulations in $$\textsf{R}$$ and got $$x \approx 2.001, 2.0161, 1.9785$$, which seems to confirm that the answer is $$2$$. Can someone explain this result?
Also, why/where did my approach fail?
• When you say $\mathbb{E}[T|T_1 < 1] = 1 + x,$ you're asserting that, in expectation, if the first step doesn't get all the way to $1,$ then the remaining steps have to get all the way to $1$. This is clearly false - if $T_1 = 1/2,$ then $T_2 + \dots T_T$ only have to get up to $1/2$. Since $T_1 > 0$ a.s., we should have $\mathbb{E}[T|T_1 < 1] < 1 + \mathbb{E}[T].$ – stochasticboy321 Apr 13 '19 at 19:00
• I noticed that you wanted a non-heuristic proof - note that $\{T > n\} = \{S_n < 1\}$, since the $S_n$ are non-decreasing. It should be straightforward to figure out $\pi_n := P(S_n<1)$ by establishing a recurrence relation between $\pi_n$ and $\pi_{n-1}$ - you'll likely need the volume of a standard simplex in $n$ dimensions. This will directly give you $P(T = n),$ which you can then use to find the mean. – stochasticboy321 Apr 13 '19 at 21:19
• A more high level argument is from queuing theory - suppose it takes you $\mathrm{Exp}(1)$ time to do a job. How many jobs will you finish in $1$ unit of time? It is a classical result that this number is $\mathrm{Poission}(1)$ distributed. However, intuitively, the mean is simpler to argue - your rate of doing jobs is $1$ per unit, so you, in expectation, should finish one job per unit (a Little's law type argument). You are interested in this number plus one - which job will you be doing when the time unit finishes. – stochasticboy321 Apr 13 '19 at 21:24
• @stochasticboy321 Thank you! I was able to do a brute force calculation along these lines. – Flowsnake Apr 14 '19 at 16:24
• ^That's grand, you're welcome :). You should add an answer below, so that others trying the same problem can have a reference. – stochasticboy321 Apr 15 '19 at 1:59
Following @stochasticboy321's approach, we want to find $$P(T > n) = P(S_n < 1)$$. Since $$S_n = T_1 + ... + T_n \sim$$ Gamma($$n,1$$), we have,
$$P(S_n < 1) = \frac{1}{\Gamma(n)}\int_0^1x^{n-1}e^{-x}dx = 1 - \frac{\Gamma(n,1)}{\Gamma(n)}$$
where $$\Gamma(n,1)$$ is the incomplete Gamma function. To get this expression, I used the nice identity found here. Finally,
$$E[T] = 1 + \sum\limits_{n=1}^\infty P(T > n) = 1 + \sum\limits_{n=1}^\infty P(S_n<1) = 1+ \sum\limits_{n=1}^\infty\left(1 - \frac{\Gamma(n,1)}{\Gamma(n)}\right)$$
I have no idea how to evaluate the sum in closed form, but a computation in Wolfram Alpha seems to suggest it converges to $$1$$. Thus, $$E[T] = 2$$ (at least by conjecture from this numerical computation).
• If we set the integral above to $I_n,$ then by integration by parts, and using $\Gamma(n) = (n-1) \Gamma(n-1) = (n-1)!,$ we get for $n \ge 1,$ $$P(T > n) = \frac{I_n}{\Gamma(n)} = \frac{(n-1) I_{n-1}}{(n-1) \Gamma(n-1)} - \frac{e^{-1}}{{(n-1)!}} = P(T > n-1) - \frac{e^{-1}}{(n-1)!}$$ But then $$P(T = n) = P(T> n-1) - P(T > n) = \frac{e^{-1}}{(n-1)!}.$$ We immediately have that $T \overset{\mathrm{law}}= 1 + Z,$ where $Z \sim \mathrm{Pois}(1),$ and so has mean $2$. Implicitly this also solves the series above (by summation by parts). – stochasticboy321 Apr 15 '19 at 19:35 | 2020-03-31T23:53:57 | {
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https://web2.0calc.com/questions/please-help-me_92335 | +0
+1
104
16
+365
Guys please solve this question and let me know the correct option
please its a request here the link of question if you cant see it (https://ibb.co/ZWftpcR)
Mar 22, 2021
edited by kes1968 Mar 22, 2021
#1
+9279
+2
(Assuming that log means natural log)
y = 5log x
Take the log of both sides
log y = log( 5log x )
Now we can apply the rule that says: log(xy) = y log x
log y = log x log 5
Divide both sides by log 5
log y / log 5 = log x
Do the reverse of natural log to both sides (make both sides the exponent of e )
e^( log y / log 5 ) = e^log x
Now the right side simplifies to just x
e^( log y / log 5 ) = x
So we have...
$$x\ =\ e^{\frac{\log y}{\log 5}}\\~\\ x\ =\ (e^{\log y})^{\frac{1}{\log 5}}\\~\\ x\ =\ y^{\frac{1}{\log 5}}$$
Mar 22, 2021
#2
+365
+2
brother what you have found is the same function just in terms y , but inverse function is symmetrical about y=x axis ,
also f(f^-1(x))=x which is not satisfied ,
by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions
Mar 22, 2021
#3
+9279
+3
The inverse function can be graphed by: $$y=x^{\frac{1}{\log5}}$$
Here's a graph:
https://www.desmos.com/calculator/aijpzpouk1
In other words...
If $$f(x)=5^{\log x}$$ then the inverse is $$f^{-1}(x)=x^{\frac{1}{\log 5}}$$
And...
$$f(f^{-1}(x))\ =\ f(x^{\frac{1}{\log 5}})\ =\ 5^{\log(x^{\frac{1}{\log 5}})}\ =\ 5^{\frac{\log x}{\log 5}}\ =\ 5^{\log_5 x}\ =\ x$$
The options leave it in the form that is solved for x, and so I left it like that to match the options
Mar 22, 2021
#4
+365
+2
brother so what you feel (as a math expert) should be the right answer to the given question ,
Mar 22, 2021
#5
+9279
+1
I'm not sure what you meant by "by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions"....but if this didn't answer your question then please feel free to ask for more clarification!
And I think the answer is option 1: $$x=y^\frac{1}{\log 5}$$
Mar 22, 2021
#6
+365
+3
brother , if you see your second answer and change it in terms of x , wont you get option 2 and option 4 ,
Mar 22, 2021
#7
+9279
+1
Hmm...actually....I see what you mean.... (maybe I made the question harder than it has to be!)
I take back my original answer!! Now I think the answer is option 4
Mar 22, 2021
#8
+365
+1
brother do you really believe its option 4 or just to keep my heart , please clarify if you still believe the answer is 1 , if yes then please prove the same to me as well
Mar 22, 2021
#9
+1740
+1
For computing the inverse function, the plan I know is
1. interchange x & y
2. solve for y
but actually, after 1. you already have a term for the inverse function. It's just not written in the "usual" way, wich is y=f(x).
Answer for is exactly what you get after interchanging x&y, so the correct answer is answer 4.
Mar 22, 2021
#10
+9279
+2
I do agree Probolobo, but then the confusing thing is that
$$5^{\log y}\ =\ y^{\log 5}$$
Which means option 2 and option 4 are the same function and so are equivalent....
Mar 22, 2021
#12
+1740
+1
That's correct, then there are actually 2 correct answers, 2 & 4. Didn't see that equivalence on first glance ;)
Probolobo Mar 22, 2021
#11
+365
+1
so whats the final answer so that i can challenge the answer key?
Mar 22, 2021
#13
+9279
+1
If I had to guess right now, I would guess option 4. My next guess would be option 1.
But I am honestly not sure!! I think this is a bad question.
Can you let us know when you find out the "correct" answer according to the answer key?
hectictar Mar 22, 2021
#14
+112907
+3
I think you are making hard work of it
the inverse of
$$y=5^{logx}$$
is simply
$$x=5^{logy}$$
You just have to switch the x and y over.
there would be restrictions on x and on y but the question isn't worrying about that.
Here is the graphs
https://www.desmos.com/calculator/4ftfa2bny7
See they are reflections of each other about y=x
Mar 22, 2021
#15
+365
+2
another question , another superb explanation from mod
Melody you are simply awesome ! thanks for the help mate
kes1968 Mar 22, 2021
edited by kes1968 Mar 22, 2021
#16
+112907
+1
You are welcome Kes
Melody Mar 22, 2021 | 2021-04-19T06:01:24 | {
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https://math.stackexchange.com/questions/2460589/five-people-get-on-the-elevator-that-stops-at-five-floors-in-how-many-ways-they | # Five people get on the elevator that stops at five floors. In how many ways they can get off?
Five people get on the elevator that stops at five floors. In how many ways they can get off? For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor. In how many ways they can get off at different floors? Now, consider that people in elevator have names, say A,B,C,D and E., assuming that, for example, the case A on the 1st floor is different from the case B on the 1st floor. Answer the previous questions with this assumption.
I was told that there are 4 questions, but I'm not really sure about it.
1. People don't have names and get off
2. People don't have names and get off at different floors
3. People have names and get off
4. People have names and get off at different floors.
Do the names matter?
I think I can answer one question out of 4 - the number of all outcomes should be $5^5=3125$. I am confused about the rest though
• I think I can answer one question out of 4 - the number of all outcomes should be 5^5=3125. I am confused about the rest though. – Gabriel Oct 6 '17 at 16:33
• Wait, there are 4 questions here? I only see two: one where you do differentiate between the 5 people and one where you don't. And yes, for the second one you indeed get $5^5$. – Bram28 Oct 6 '17 at 16:36
• Do you know about Stars and Bars? – Bram28 Oct 6 '17 at 16:37
• Thank you!! I was told that there are 4 questions, but I'm not really sure about it. 1. People don't have names and get off, 2. People don't have names and get off at different floors, 3. People have names and get off, 4. People have names and get off at different floors. Do the names matter? And thank you, I will definetely read about Stars and Bars!! – Gabriel Oct 6 '17 at 16:52
1. This has been answered in other answers: ${9\choose4}$.
2. Since there are $5$ people and $5$ floors there is just $1$ way for them to get off at different floors: $1$ per floor.
3. You already found $5^5=3125$.
4. At each floor exactly one of $A$, $\ldots$, $E$ gets off. This can be done in $5\cdot4\cdot3\cdot2=5!=120$ ways.
For the first question, use the basic 'stars and bars' method.
We can represent your example scenario ("For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor.") as:
$$*||**||**$$
The people are the 'stars' and the separators between the different floors are the 'bars'.
Another example: one person getting off on each floor would be:
$$*|*|*|*|*$$
And everyone getting off on the 3rd floor would be:
$$||*****||$$
See how this works?
Now, with $4$ 'bars' to divide the people into the $5$ groups/floors, and those bars taking up $4$ of the $9$ possible positions in this 'symbol string' of bars and stars, you get
$$9 \choose 4$$
possibilities.
• Thank you so much for the explanation!! I think I understand how it works now. – Gabriel Oct 6 '17 at 16:54
• @Gabriel You're welcome! Your instructor did not teach 'stars and bars' yet? ... Because this was a typical stars and bars question! :) – Bram28 Oct 6 '17 at 16:56
• Sometimes they just give the formula derived from stars and bars which is $\dbinom{n+r-1}{r-1}$ in this case $n=5$ and $r=5$ of course, the integer solutions for the equation $x_1+x_2+x_3+x_4+x_5=5$ are considered there are no left and right boundries (luckily). Of course giving the formula without the logic is problematic but I haven't seen one problem that required the logic rather than the formula. – Deniz Tuna Yalçın Oct 6 '17 at 20:21
The question could also be solved using a well known formula derived from stars and bars which is; $$\dbinom{n+r-1}{r-1}$$ where we distribute $r$ equally accepted objects to $n$ spaces with or without restrictions like $\geq1$ or $\leq9$, this formula is also used for solving linear equations (in integers) that's why this works for combinatoric problems, so without making it longer;$$x_1+x_2+x_3+x_4+x_5=5$$ so we apply the formula, we can think that this is equivalent to distributing five equally accepted apples to five children for instance, the result is $$\dbinom{5+5-1}{5-1}=126$$ this statement could be proven using alternate ways, we should open a post about it too. | 2020-04-08T00:14:31 | {
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https://brilliant.org/discussions/thread/scratch-pad/ | # Have you ever used Scratch pad like this?
Find the root of following equation.
$4y^3-2700(1-y)^4=0$
Above equation can be solved using newton raphson iterative method with initial approximation to be unity.
1) First Let $f(y)=4y^3-2700(1-y)^4$ We have to find root of above function.
2)Find derivative of $f(y)$. Here,$f^{'}(y)=12y^2+10800(1-y)^3$
3)Find $y_n$ using newton raphson method $y_0=1$ $y_{n+1}=y_n-\frac{f(y_n)}{f^{'}(y_n)}$ $y_{n+1}=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
You can use calculator or scratchpad to calculate $y_1,y_2,y_3,....$.If this equation has real solution ,these values $y_1,y_2,y_3,....$ would converge to root of $f(y)$.
How to do such iterations smartly?
1) $\color{#69047E}{\text{On first line of scratch pad write y=1}}$
2) $\color{#D61F06}{\text{On second line write}}$
$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
$\color{#20A900}{\text{You will see new value of y as result.}}$
3) $\color{#EC7300}{\text{From line 3, continue pasting}}$
$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
until you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.
By this method you can solve most of the transcendental equation.
$\color{#3D99F6}{\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}$
Note by Aamir Faisal Ansari
5 years, 4 months ago
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Yes !i have used it this way.Sometimes in Recursion problems and like that.
- 5 years, 4 months ago
By using scratchpad, try finding sum of initial 10 fibonacci number.
- 5 years, 4 months ago | 2020-11-24T01:04:07 | {
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https://www.jiskha.com/display.cgi?id=1305922453 | # pre calculus
posted by joe
cube roots of 64i?
1. MathMate
Convert the given number into the form
a+bi = α(cos(β)+i sin(β))
where
α = sqrt(a^2+b^2)
β = sin-1(b/α)
If the number is plot in the complex (Z) plane, it will be more evident.
For example, 64i will have
α=64 (the distance from origin)
β=90° (sin-164/64=1)
The equivalent angles are:
90°
450°
810°
(we only need 3)
Divide each of these angles by , to get β1, β2, & β3. And take α'=α^(1/3)=64^(1/3)=4
The three cube roots are then
z1=α'(cos(β1)+isin(β1)
=4(cos(30°)+isin(30°))
z2=4(cos(150°)+isin(150°))
z3=4(cos(270°)+isin(270°))
Check by expanding z1^3, z2^3 and z3^3 to get back 64i.
2. MathMate
β=90° (sin-164/64=sin-11=π/2)
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More Similar Questions | 2018-06-18T19:09:30 | {
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https://math.stackexchange.com/questions/1879002/what-is-the-minimum-number-of-rotations-about-axes-in-a-plane-that-can-describe/1879003 | # What is the minimum number of rotations about axes in a plane that can describe an arbitrary rotation in 3D?
I'm trying to decompose an arbitrary rotation of a 3D sphere into a series of rotations about any axes that lie in the equatorial plane.
I know I can factor the arbitrary rotation into three rotations in the equatorial plane using Euler angle formulas. Can I factor into two rotations if I allow any choice of axis in the equatorial plane? If it exists, is this factorization unique?
## migrated from mathoverflow.netAug 2 '16 at 14:05
This question came from our site for professional mathematicians.
• There is a unique rotation mapping one unit vector into a distinct vector. Take one of the vectors of your basis, and apply this rotation. Now hopefully you can see that once one axis is coincidental, you only need another rot. in the orthogonal plane. This means at two rotations are necessary and sufficient in general. – Real Aug 2 '16 at 3:48
• what do you understand by equatorial plane (is it the xy-plane)? – Manfred Weis Aug 2 '16 at 7:22
• @ManfredWeis Yes by equatorial plane I mean the xy-plane. – Talon Chandler Aug 2 '16 at 12:28
• Talon, I thought the answer is yes and that I have a simple proof for it. But the question might not be appropriate for this site. If you ask over at Mathematics StackExchange and then let me know here (with a link) when you have posted, I'd be happy to give more details. – user43208 Aug 2 '16 at 13:49
• On second thought, let me migrate it for you... – user43208 Aug 2 '16 at 14:05
My answer is that you can.
If you know about quaternions, then you know that a rotation about a unit vector $v \in \mathbb{R}^3$ through an angle $\theta$ can be accomplished by regarding elements $w = (a, b, c) \in \mathbb{R}^3$ as purely quaternionic elements $ai + bj + ck$, and then mapping $w \mapsto uwu^{-1}$ where $u = \cos(\frac{\theta}{2})\cdot 1 + \sin(\frac{\theta}{2})v$. Such $u$ are precisely quaternions of unit norm.
Let us denote the rotation above by $R_u$ (so $R_u(w) := uwu^{-1}$). Notice that a composition of two rotations $R_u \circ R_v$ is just $R_{u v}$; this follows by associativity of quaternionic multiplication.
So we can translate your problem into the following: for any rotation $R_w \in SO(3)$ given by a nonzero quaternion $w$, show that there exist quaternions $u, v$ in the linear span of $1, i, j$ such that $u v = w$. For these $u, v$ describe rotations about vectors in the equatorial plane spanned by $i, j$.
Now this is not difficult. Let us write $u = a + bi + cj$ and $v = a' + b'i + c'j$. We compute
$$(a + bi + cj)(a' + b'i + c'j) = aa' - bb' - cc' + (ab' + a'b)i + (ac' + a'c)j + (bc' - b'c)k$$
and so given a nonzero quaternion $w = p + qi + rj + sk$, our task is to cook up parameters $a, b, c, a', b', c'$ such that
$$p = aa' - bb' - cc'$$
$$q = a b' + a'b$$
$$r = ac' + a'c$$
$$s = bc' - b'c$$
In fact, let's make our life easier and simply set $b = 0, b' = 1$. Then $a = q$, $c = -s$, and we can solve for $a', c'$ in the linear system
$$p = qa' + sc'$$ $$r = -sa' + qc'$$
provided the determinant $q^2 + s^2$ is nonzero, i.e., provided one of $q, s$ is nonzero. If both $q, s = 0$, then of course $w = p + rj$ corresponds to a rotation about the vector $j$ which is already in the equatorial plane, so there was nothing to do in this case.
Once you have solutions $u = a + bi + cj, v = a' + b'i + c'j$ to the equation $uv = w$ in your hands, the rotations $R_u, R_v$ are rotations about the vectors $bi + cj$ and $b'i + c'j$, respectively. If you want the rotation angles, then normalize $u$ and $v$ (i.e. divide by their norms $(a^2 + b^2 + c^2)^{1/2}$ and $(a')^2 + (b')^2 + (c')^2)^{1/2}$) and use the fact that $R_u = R_{\frac{u}{\|u\|}}$. You can then read off the desired angles by writing e.g. $\frac{u}{\|u\|} = \cos(\frac{\theta}{2}) + \sin(\frac{\theta}{2})\frac{bi + cj}{(b^2 + c^2)^{1/2}}$, so for instance $\cos(\frac{\theta}{2}) = \frac{a}{(a^2 + b^2 + c^2)^{1/2}}$.
By the way, this solution also shows that the decomposition is in general non-unique. For example, we could have also chosen $c = 0, c' = 1$ and arrive at quaternionic solutions.
• Do we need to constrain $u$, $v$, and $w$ to be unit quaternions so they represent rotations? If this is the case, then we have 7 equations with 6 unknowns which might have a unique solution. – Talon Chandler Aug 2 '16 at 18:10
• No, you don't need to, but you can assume they are unit quaternions if you like because $R_u = R_{\frac{u}{\|u\|}}$ for any nonzero quaternion $u$. If $w$ is constrained to the unit sphere in the quaternions (3-dimensional $S^3$) and $u, v$ are each constrained to the unit sphere in the span of $\{1, i, j\}$ (so $(u, v)$ parametrizes a space that looks like $S^2 \times S^2$, which is 4-dimensional), then we are asking about surjectivity of a multiplication map $S^2 \times S^2 \to S^3$, so 3 equations in 4 variables by my count. – user43208 Aug 2 '16 at 18:24
• Okay that makes sense. What if we consider only infinitesimal rotations? Can we uniquely decompose an arbitrary rotation into infinitesimal rotations in the plane spanned by $i,j$? In this case we can make the approximations: $$\cos\left(\frac{\theta}{2}\right) \approx 1$$ $$\sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}$$ So $p=a=a'=1$ and we have four equations with three unknowns which gives a unique solution---i.e. an infinitesimal rotation about an arbitrary axis can be decomposed uniquely into two infinitesimal rotations about axes in the span of $i,j$. – Talon Chandler Aug 2 '16 at 19:28
• Well, the answer to the third sentence in your last comment would be 'no'. But why is getting uniqueness important to you? I think in fact the method I gave you gives virtual uniqueness of a solution if you constrain the first rotation $R_u$ to rotate about $j$. That is: unit quaternions $u$ of the form $a + bj$ form a circle $S^1$, and the unit $v$ for rotations about a second axis in the $ij$-plane form a sphere $S^2$, and the multiplication map $S^1 \times S^2 \to S^3: (u, v) \mapsto w = uv$ is onto, and unless $w$ itself is a rotation about $j$, only one pair $(u, v)$ satisfies $w = uv$. – user43208 Aug 2 '16 at 20:35
• I'm interested in uniqueness because I'm working on a reconstruction problem. I have measurements that correspond to a known rotation about an arbitrary axis, and I want to reconstruct the set of rotations in the $ij$-plane that gave rise to the "total" rotation. I'm hoping to find the conditions under which this reconstruction is possible. Can you expand on why we can't decompose an arbitrary infinitesimal rotation into two unique infinitesimal rotations in the $ij$-plane? – Talon Chandler Aug 2 '16 at 21:03
The answer is no, because when letting $v$ be an arbitrary direction about which to rotate by an angle $\rho$, that rotation can only then be realized with a rotation around an axis in the xy-plane if $v^Tz=0$; that shows that at least three rotations are necessary
1. rotate about $v\times z$ to bring $v$ into the $xy$-plane
2. rotate about the image of $v$ in the $xy$-plane
3. rotate the image of $v$ back around $v\times z$ to bring the image back to the original direction of $v$
• I agree that this set of rotations in the xy-plane can describe an arbitrary rotation, but how can we be sure there isn't a shortcut with just two rotations? – Talon Chandler Aug 2 '16 at 12:32 | 2019-10-17T12:57:07 | {
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https://mathforums.com/threads/how-can-i-find-the-time-when-two-people-agree-to-meet-if-both-of-their-watches-have-an-offset.348121/ | How can I find the time when two people agree to meet if both of their watches have an offset?
Chemist116
The problem is as follows:
Jenny and Vincent agreed to meet at the library at $6\,p.m$. Both synchronized their watches at midnight ($\textrm{0 hours}$). We know that Vincent's watch is not functioning correctly and gets ahead of the real time $\textrm{50 seconds}$ each hour and Jenny's watch gets delayed from the real time $\textrm{50 seconds}$ each hour. Vincent arrived to the library $\textrm{15 minutes}$ before the agreed time accoring to his watch and Jenny $\textrm{15 minutes}$ late by looking at her watch. Using this information, find how long did Vincent waited Jenny?
The alternatives according to my book are as follows:
$\begin{array}{ll} 1.&\textrm{15 min}\\ 2.&\textrm{0 min}\\ 3.&\textrm{25 min}\\ 4.&\textrm{60 min}\\ \end{array}$
This part I'm stuck at exactly how should I make up an equation which can relate both offsets. Can someone help me here? I'm assuming that by $\textrm{6 p.m}$ the time which would be seen by Vincent will be:
$18\times 50= 900\,s$
which would be $60$ minutes
But Vincent seen in his watch was:
$\textrm{5:45 pm}$
hence until 5 pm would be:
$17\times 50=850\,s$
$56\frac{2}{3}$ minutes
and Vincent's watch would have seen:
$\textrm{5:56:40 pm}$
At this point I could try guessing reducing the number of minutes until adjusting the time which will be seen by Vincent and this would be same for Jenny, but it doesn't seem something which can be effective. Can someone help me here? How exactly can I find what is being requested?
skeeter
Math Team
50 seconds an hour = 5 minutes in 6 hours
Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.
Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.
The only answer choice that makes sense is the 60 minutes Vincent waited.
I worked out the problem using proportions, where T represents the actual time
for Jenny ...
T/18.25 = 72/71
for Vincent ...
T/17.75 = 72/73
topsquark
Chemist116
50 seconds an hour = 5 minutes in 6 hours
Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.
Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.
The only answer choice that makes sense is the 60 minutes Vincent waited.
I worked out the problem using proportions, where T represents the actual time
for Jenny ...
T/18.25 = 72/71
for Vincent ...
T/17.75 = 72/73
This is the major problem which I had with this question. It mentioned that until his time was 17:45, not 18:00. How would be the difference between those fractions? Can you help me with that part? | 2020-04-05T13:29:35 | {
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https://math.stackexchange.com/questions/1150584/combinations-how-many-handshakes | # Combinations: How many handshakes?
I need help with the following question which I cannot seem to solve:
17 students are sitting in a circle. Each person shakes hands with everyone but his/her neighbours. How many handshakes have been exchanged?
My approach: no. of ways $= 1 + 2 + 3 + ... + 14 = 7(15) = 105$.
Apparently my answer is wrong (correct ans is 119). But I can't seem to understand why. Could someone please explain?
• The reason why this is wrong is as follows: the first person has 14 people to shake hands with. The second person as well has 14 people to shake hands with, since he wouldn't shake hands with his neighbour anyway! So, the sum is 14+(14+13+...+1), which does add up to 119. – Sanchises Feb 16 '15 at 19:49
Each of 17 people shakes hands with 14 people (all except themselves and their 2 neighbors), so there are
$$\frac{17\times 14}{2} = 119$$
handshakes (dividing by 2 to account for symmetry, as you would otherwise count "$A$ shaking hands with $B$" and "$B$ shaking hands with $A$" as distinct events).
• Does it mean that we don't care about how they sit, say in a circle or in a row? – Nighty Feb 16 '15 at 15:04
• @LeeKM: The difference is that in a circle, everyone has two neighbors, while in a row, the first and last person have only one each. – user139000 Feb 16 '15 at 15:05
• Thanks! Now I know that I misunderstood the quuestion. I thought that the handshakes were to be made with distinct persons. – Donald Feb 16 '15 at 21:44
Here's another way to think about it.
If everyone shook hands with everyone you'd have $\frac {17*16}2 = 136$ handshakes (divide by 2 because the above is double counting A shaking hands with B and so forth).
However, there are $\frac {17*2}2 = 17$ handshakes that aren't happening because neighbors aren't shaking hands (2 neighbors for each of the 17 people, again divided by 2 to remove duplicates).
So $136-17 = 119$
Pew's response is a little more direct, but finding the number by calculating the total possibilities minus the "not allowed" interactions is sometimes a little more intuitive for some people.
You can think of this in graph theory as the following:
$G$ has $17$ vertices with degree $14$, since there are no loops(people self handshaking), and no vertices are adjacent to the vertices beside them. So total degree is $17\times 14$.
We know that degree is equal to $2\times\text{number of edges}$ and hence there are $\frac{17\times 14}{2}=119$ edges in total, where edges represent handshakes.
As is typical for problems with big numbers, you should always resort to a smaller number if you can't solve the full problem.
Here, 17 people at the table is a bit hard to imagine immediately.
How many handshakes now?
There are zero.
A B
C D
A can't shake with B or C, B can't with A or D, D can't with B or C, C can't with A or D--only A & D and B & C can shake.
So 2 shakes:
A B
X
C D
5 people is when it gets interesting, and when you should be able to see the pattern:
B
A C
D E
Everyone has two people they can't shake with, leaving two shakes per person--but we overcount by just multiplying 5 & 3, so we divide by two.
This is easiest to see by trying to draw the graph mentioned by @Commitingtoachalleng -- draw a line between any people who can shake hands:
B
/ \
A-----------C (also C-D & A-E--a 5-point star)
/ \
D E
So we hypothesize the answer is $\frac{n(n-3)}{2}$. Note that this formula holds for $n=3$ and $n=4$ as well, as we'd hope!
One final way to see this is to look again at the graphs--
you might notice that they're always complete graphs (every vertex connected to every other vertex) with the outer edges removed.
Since there are $\frac{n(n-1)}{2}$ edges in a complete graph on $n$ vertices (which you can confirm yourself by a similar process), and $n$ outer edges,
there must be $\frac{n(n-1)}{2} - n = \frac{n^2-n-2n}{2} = \frac{n(n-3)}{2}$ handshakes.
However you cut it, there are $\frac {17 \cdot 14} 2 = 119$ total handshakes.
• We should also point out that the second option (working backwards from complete graphs) is the graph analogue of @Duncan's solution – MichaelChirico Feb 17 '15 at 6:42
I will explain for 6, same logic applies for all the numbers.
first, if 6 persons shakes will all in the party, then number of shakes will be:
A B C D E F
A can shake with only rest of 5 persons(B,C,D,E,F) i.e, right side ppl. = 5
B can shake with only his right sides(C,D,E,F) ppl(not with A bcos,A already shook with B just now. so no repeat) = 4
C can shake with only his right sides(D,E,F) ppl(not with A&B bcos,A&B already shook with C just now. so no repeat) = 3
D can shake with only his right sides(E,F) ppl(not with left side ppl bcos, they already shook with D just now. so no repeat) = 2
E can shake with only his right sides(F) ppl(not with left side ppl bcos, they already shook with E just now. so no repeat) = 1
as all of the persons shook hand with F, He dont shake with any one now.
So total possible shakes for 6 persons are = 5+4+3+2+1 => 15
now in questions it is given that, No one shakes hand with neighbors.
Imagine ppls sitting around table
A B
F C
E D
Please treat above image as 6 Ppl around circle : So now,
Invalid Shakes:
A cannot shake with B,F . I am representing it as below : A -no- B & F => 2 invalid shakes.
B -no- C only. Bcos B -no- A is covered as part of first case.
=> 1 invalid shake
C -no- D only. => 1 invalid shake
D -no- E only. => 1 invalid shake
E -no- F only. => 1 invalid shake
F dont shake with A is already covered as part of first invalid shakes.
so total invalid shakes are 2+1+1+1+1 = 6
so for 6 ppl: out of total 15 , 6 are invalid. So 15-6 = 9
on the same line :
for 17 ppl: out of total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119
shortcut : For N ppl, Total shakes are (N-1)(first term + last term)/2 Arithmetic progression sum. out of which Invalids are N.
So valid shakes => (Total - N)
• What is the point of posting such a complex answer to a question with a simple and accepted answer posted almost four years ago? – José Carlos Santos Oct 31 '18 at 17:23 | 2020-09-26T18:03:58 | {
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http://math.stackexchange.com/questions/584601/is-this-function-injective-surjective | # Is this function injective / surjective?
A question regarding set theory.
Let $g\colon P(\mathbb R)\to P(\mathbb R)$, $g(X)=(X \cap \mathbb N^c)\cup(\mathbb N \cap X^c)$
that is, the symmetric difference between $X$ and the natural numbers.
We are asked to show if this function is injective, surjective, or both.
I tried using different values of $X$ to show that it is injective, and indeed it would seem that it is, I can't find $X$ and $Y$ such that $g(X)=g(Y)$ and $g(X) \neq g(Y)$ but how do I know for sure? How do I formalize a proof for it?
Regarding surjective: I think that it is.
We can take $X=\mathbb R - \mathbb N$ and we get that $g(X)=\mathbb R$
What do I do about the injective part?
-
are you able to express in formula what injective or surjective means? – Robert.Sie Nov 28 '13 at 14:14
Hint:
$$g\circ g = \mathrm{id}_{\mathcal{P}(\mathbb{R})}, \quad \text{ that is, } \quad \forall X \in \mathcal{P}(\mathbb{R}).\ g(g(X)) = X.$$
I hope this helps $\ddot\smile$
-
We can view subsets of $X$ as functions $X \to \{0,1\}$: 0 if the point is not in the set, It is commonly very useful to rewrite questions about subsets as questions about functions.
What form does $g$ take if we do this rewrite?
Terminology wise, if $S \subseteq X$, then $\chi_S$ is the function described above:
$$\chi_S(x) = \begin{cases} 0 & x \notin S \\ 1 & x \in S \end{cases}$$
It is called the characteristic function of $S$.
-
0 if $a \in X\cap \mathbb N$, 1 otherwise – Oria Gruber Nov 28 '13 at 14:19
That's not right. To be more clear, $g(\chi_S)$ is a function. To describe this function, we could do so by writing down the equation that says what its value at a point $x$ is: i.e. what is $g(\chi_S)(x)$? – Hurkyl Nov 28 '13 at 14:22
HINT: Recall that symmetric difference is associative and for every set $B$, $B\triangle B=\varnothing$.
-
And why was this downvoted? – Asaf Karagila Nov 28 '13 at 22:04 | 2015-08-02T23:20:37 | {
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https://math.stackexchange.com/questions/1490376/find-sum-of-series-sum-n-1-infty-1-n1-left-frac-n2 | # Find sum of series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to 3 decimal places.
This is a question I came across while studying on Khan Academy:
Find the sum of the series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to three decimal places.
I solved this by calculating ${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },...\\$ until I had a close enough value (bad method, I know). Anyway, here is the solution steps given in the answer:
Note that this is a convergent alternating series. The Alternating Series Estimation Theorem states that the error bound in the sum of the first $n$ terms is the absolute value of the first omitted term; i.e., $\left| { a }_{ n+1 } \right|$. To obtain three-digit accuracy, we need to find $n$ such that $\left| { a }_{ n+1 } \right| <\ 0.0005$, or equivalently $\frac { n+3 }{ { (n+1) }^{ 5 } } <0.0005$.
When $n=6$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 9 }{ 16807 } \approx .000535$; that is not accurate enough.
When $n=7$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 10 }{ 32768 } \approx .000305$; that will give a final result within $0.0005$.
Thus we need to find the sum of the first seven terms of the series.
$$\sum _{ n=1 }^{ 7 }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) \approx 2.8914 }$$
Hence, $2.891$ is accurate to three decimal places.
What I don't understand about this answer is if the estimate is $2.8914$ and the error bound is $0.000305$, I think the actual sum can be close to $2.8914+0.000305\approx 2.891705\approx 2.892$. Then the result above is not correct.
Does this answer take into account the fact that the $(7+1)th$ term is negative? Or is there something about rounding and error bounds that I don't understand?
Thank you.
• Indeed, 2.8914 means between 2.89135 and 2.89145 and 0.000305 is less than 0.0004 hence the sum is between 2.89135-0.0004 and 2.89145, which implies that 2.891 has three correct decimal places. (The same estimates for n=6 show it was not necessary to go to n=7.) – Did Oct 21 '15 at 7:27
• You are right, it is the fact that $\sum_1^7$ is an overestimate that enables us to conclude that $2.891$ is correct to 3 decimal places. – André Nicolas Oct 21 '15 at 7:28
• To the op: If you want it back I'll undelete my answer, but it seems unnecessary given you're already pretty much there yourself. – Adam Hughes Oct 21 '15 at 7:32
• @Did Thank you. But do you mean if somehow the estimate is an underestimate, the sum can be between 2.89135 and 2.89145+0.0004. Then 2.892 would be correct, right? – user52139 Oct 21 '15 at 7:49
• @AdamHughes Thank you very much. I actually need it back if you don't mind, please. – user52139 Oct 21 '15 at 7:52
I would read the question the same way you did-find a three decimal value that you know the true sum of the series rounds to. We know from the alternating series theorem that the truncation error is smaller than the first neglected term in absolute value and of the same sign. There is no truncation error that we can guarantee will be small enough, because we could be poised right on the rounding boundary, with the computed answer something like $2.8915$. Here, because after seven terms we have $2.8914$ and we know the eighth term is negative, the correct answer is in the range $(2.8911,2.8914)$ and this entire range rounds to $2.891$
• +1, warning about problem with rounding boundary is well placed here. This answer might be further improved by pointing out the problem of numerically finding any decimal of $\sum_{n=0}^\infty \frac 1{2^n}$. – Ennar Oct 21 '15 at 10:38
This is an alternating series, so the error is estimated by the last term. Since you want the error to be less than $10^{-3}$ in size, you need only choose $n$ so that ${n+2\over n^5}<10^{-3}$ since the terms are decreasing.
Solving we see $n+2<10^{-3}n^5$. We claim this is so for $n\ge 7$ as
$$7+2=9<16.807=7^5\cdot 10^{-3}.$$
For $n>7$ we see that by induction we may conclude this with the inductive step being
$$(n+1)+2<10^{-3}n^5+1< 10^{-3}(n+1)^5,$$
Then adding up the first $6$ terms gives you $2.891$ to three decimal places. | 2019-06-18T21:14:31 | {
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https://math.stackexchange.com/questions/2357560/does-the-bezout-gcd-equation-hold-in-a-ufd | # Does the Bezout GCD equation hold in a UFD?
I'm wondering when Bezout's Theorem is valid.
I know when the ring $R$ is a Euclidean Domain or a Principal Ideal Domain there is always $\gcd(a,b)$ and I can find $x,y \in R$ such that $\gcd(a,b) = ax + by.$
I know when the ring $R$ is a Unique Factorization Domain, there is always $\gcd(a,b)$. My question is, in this case can I find $x,y \in R$ such that $\gcd(a,b) = ax + by.$ ?
Now if $R$ is just a commutative ring, if there is $d \in R$ such that $<a>+<b> = <d>$ I know that $d = \gcd(a,b)$. In this case, can I find $x,y \in R$ such that $\gcd(a,b) = ax + by$ ?
• In $\mathbb Z[x]$, a UFD, $2$ and $x$ have GCD $1$, but there is no solution to $2f(x)+xg(x)=1$. More generally, a UFD in which Bezout is true is always a principal ideal domain. I think. – Thomas Andrews Jul 13 '17 at 15:30
It is not necessarily true in a UFD. For instance, if $k$ is a field, then the polynomial ring $k[s,t]$ is a UFD. But $\gcd(s,t)=1$, and there do not exist $x$ and $y$ such that $sx+ty=1$.
For your second question, the answer is yes by definition. Indeed, by definition, $\langle a\rangle+\langle b\rangle$ is the set of elements of the form $ax+by$. So if $d\in \langle a\rangle+\langle b\rangle$, then there exist $x,y\in R$ such that $d=ax+by$.
The converse holds as well: if $d=\gcd(a,b)$ can be written in the form $ax+by$, then that means $d\in \langle a\rangle+\langle b\rangle$, and it follows that $\langle a\rangle+\langle b\rangle=\langle d\rangle$.
Thus a commutative ring has the property that the GCD of any two elements $a$ and $b$ can be written in the form $ax+by$ iff the sum of any two principal ideals is principal. By induction on the number of generators, this is equivalent to any finitely generated ideal being principal. Such a ring is known as a Bezout ring. A Noetherian Bezout ring is the same thing as a principal ideal ring, but there exist non-Noetherian examples as well. For instance, the ring of holomorphic functions on a connected open subset of $\mathbb{C}$ is a Bezout domain but has ideals which are not finitely generated.
The rings where it is valid Bézout's lemma are known as Bézout rings. In the case of integral domains rings we have the domains known as Bézout domains.
In general, a UFD doesn't have to be a Bézout domain. The classical example is $\Bbb{Z}[x]$, as Thomas Andrews pointed out in his comment. Theoretically there is a reason for which the above is true. We have the following:
Theorem 1: If $D$ is both a UFD and a Bézout domain, then $D$ is a PID.
Proof: See exercise 11 of section 8.3 of the book "Abstract Algebra" by Dummit and Foote.
By this theorem, if every UFD were also a Bézout domain, we would conclude that every UFD is a PID, and this is false (look at $\Bbb{Z}[x]$ again), as you surely know.
More generally, for Bézout domains we have the following result:
Theorem 2: Let $D$ be a Bézout domain. TFAE:
i) $D$ is a PID.
ii) $D$ is Noetherian.
iii) $D$ is a UFD.
iv) $D$ satisfies the ACCP (ascending chain condition on principal ideals).
v) $D$ is an atomic domain.
Proof: This is theorem 46 given in Pete L. Clark's notes on factorization in integral domains
Your second question has been also answered, but let me include a general version of what you wanted to prove.
Theorem 3: Let $a_1,a_2,\ldots a_n$ be nonzero elements of a commutative ring $R$. Then $a_1,a_2,\ldots a_n$ have a greatest common divisor $d$, expressible in the form $$d=r_1a_1+r_2a_2+\cdots +r_na_n$$ if only if the ideal $(a_1,a_2,\ldots a_n)$ is principal.
Note that $(a_1,a_2,\ldots a_n)$ is the same as $(a_1)+(a_2)+\cdots +(a_n)$.
Proof: This is theorem 6-3 in Burton's book "First Course in Rings and Ideals".
There are simple examples of non-Bezout UFDs, e.g. any UFD of dimension $$> 1$$, e.g. $$\Bbb Z[x,y],\,$$ by the below equivalent conditions for a UFD to be Bezout (or directly: $$\,\gcd(x,y)=1\,$$ but $$\,xf+y\,g = 1\Rightarrow 0 = 1\,$$ via eval at $$\,x=0=y)$$.
Theorem $$\rm\ \ \ TFAE\$$ for a $$\rm UFD\ D$$
$$(1)\ \$$ prime ideals are maximal if nonzero, i.e. $$\rm\ dim\,\ D \le 1$$
$$(2)\ \$$ prime ideals are principal
$$(3)\ \$$ maximal ideals are principal
$$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1,$$ i.e. coprime $$\Rightarrow$$ comaximal
$$(5)\ \$$ $$\rm D$$ is Bezout, i.e. all ideals $$\,\rm (a,b)\,$$ are principal.
$$(6)\ \$$ $$\rm D$$ is a $$\rm PID$$
Proof $$\$$ (sketch of $$\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\$$ where $$\rm\,p_i,\,P\,$$ denote primes $$\neq 0$$
$$(1\Rightarrow 2)$$ $$\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$$ some $$\rm\,p_j\in P\,$$ so $$\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$$ by dim $$\le1$$
$$(2\Rightarrow 3)$$ $$\$$ max ideals are prime, so principal by $$(2)$$
$$(3\Rightarrow 4)$$ $$\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p)$$ for all max $$\rm\,(p),\,$$ so $$\rm\ (a,b) = 1$$
$$(4\Rightarrow 5)$$ $$\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$$
$$(5\Rightarrow 6)$$ $$\$$ Ideals $$\neq 0\,$$ in Bezout UFDs are generated by an elt with least #prime factors
$$(6\Rightarrow 1)$$ $$\ \ \rm (d) \supsetneq (p)$$ properly $$\rm\Rightarrow\,d\mid p\,$$ properly $$\rm\,\Rightarrow\,d\,$$ unit $$\,\rm\Rightarrow\,(d)=(1),\,$$ so $$\rm\,(p)\,$$ is max
As for the gcd as the ideal sum in a PID, that follows using "contains = divides", viz.
$$c\mid a,b\iff (c)\supseteq (a),(b)\iff (c)\supseteq (a)\!+\!(b)\!=\!(d)\iff c\mid d$$
And, yes, we have $$\,d\in (a)+(b) = aR + bR\,$$ $$\Rightarrow\, d = ar + br'$$ for some $$\,r,r'\in R$$ | 2021-05-13T03:28:49 | {
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https://physics.stackexchange.com/questions/287787/find-the-capacitance-of-the-capacitor | # Find the capacitance of the capacitor
Given: Consider a capacitor connected to a battery of voltage V . Let the capacitor have an area A, and a distance L between the plates. Assume that the capacitor has a layer of linear dielectric (of dielectric constant κ, so that ε = κε0) of thickness L/2 on the lower plate.
I found the capacitance to be $C=\frac{2\kappa\epsilon_0 A}{L}$. Was I correct about the separation distance? In more common examples of a capacitor, they claim that the plates are flat with nothing in between. In this case, the linear dielectric is as tall as half the total distance between the plates. Am I correct in using the remaining free space, namely $L/2$, as the separation distance in the equation? For reference, the equation is:
$$C=\frac{\kappa \epsilon_0 A}{d}$$
Instant check that shows your answer is wrong: set $\kappa = 1$ as if there is no dielectric, and you don't recover the parallel plate formula for a gap of $L$.
How to solve this problem:
Let the voltage across the capacitor be $V$ and the voltages across the top and bottom be $V_t$ and $V_b$ respectively, then $$V=V_t+V_b,$$ which implies that $$1/C=1/C_t+1/C_b$$ since $V=Q/C$.
Thus essentially this system consists of two capacitors in series. One capacitor is the top gap, and one is the bottom part. Use the formula for plate capacitors twice, with the separation being $L/2$ in both cases, then get the total capacitance by using the formula for two capacitors in series. Some algebra is involved.
You can check the answer you get, again, by setting $\kappa=1$, and seeing if you recover the usual plate formula.
• Why do you say there are 2 capacitors in series? The problem states there is only 1. – whatwhatwhat Oct 20 '16 at 23:35
• @whatwhatwhat - think of the bottom part and the top part as two capacitors as a way to solve the problem. – Suzu Hirose Oct 20 '16 at 23:36
• So then for the top section, there is a buildup of positive charge on the top plate and negative charge on the surface of the dielectric? – whatwhatwhat Oct 20 '16 at 23:42
• @whatwhatwhat - sorry my explanation was not very clear. There is an electric field between the two plates of the capacitor generated by the charges on the plates. The capacitance of this system of two plates is got by considering that the sum of the voltage over the two plates must be equal to the total voltage across the whole capacitor. I will edit the answer to make it clearer. – Suzu Hirose Oct 20 '16 at 23:48
• Suzu Hirose is completely right! – freecharly Oct 21 '16 at 0:12
The $\mathbf{D}$ field between the plates is (as usual, neglecting fringing effects) uniform and normal to the plane of the plates (assume in the z direction for simplicity) and is given by
$$\mathbf{D} = \frac{Q}{A}\hat{\mathbf{z}}$$
where the plates have charge $Q$ and $-Q$ respectively. The electric field (between the plates) within the dielectric is
$$\mathbf{E_\kappa} = \frac{\mathbf{D}}{\kappa \epsilon_0}$$
and the electric field (between the plates) in air is
$$\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$$
Thus, the potential difference between the plates is
$$V = \left(\frac{L}{2}\mathbf{E_\kappa} + \frac{L}{2}\mathbf{E}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{\mathbf{D}}{\kappa \epsilon_0} + \frac{L}{2}\frac{\mathbf{D}}{\epsilon_0}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{Q}{A\kappa\epsilon_0} + \frac{L}{2}\frac{Q}{A\epsilon_0} \right)$$
The capacitance is then
$$C = \frac{Q}{V} = \frac{1}{\frac{L/2}{A\kappa\epsilon_0} + \frac{L/2}{A\epsilon_0}} = \frac{1}{\frac{1}{C_1}+ \frac{1}{C_2}}$$
which is the formula for series connected capacitors.
• Better explanation than my answer of why it is like two capacitors in series. – Suzu Hirose Oct 21 '16 at 1:22
• So I'm still treating this as 1 capacitor correct? My friend was helping with this and she said that one way to solve the problem is to treat the system as 2 capacitors, but I couldn't visualize where the 1st capacitor begins/ends and likewise for the 2nd one. – whatwhatwhat Oct 22 '16 at 18:59
• @whatwhatwhat, correct, it is just one capacitor however, it is instructive and intuitively 'nice' that we would (ideally) get the same result if we found the equivalent capacitance of two series connected capacitors with the same plate area $A$ and spacing $L/2$ but with different dielectrics; one with $\epsilon_0$ and the other with $\kappa \epsilon_0$ – Alfred Centauri Oct 22 '16 at 22:46 | 2020-01-28T16:26:48 | {
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https://www.justinmath.com/roulette-wheel-selection/ | # Roulette Wheel Selection
As we saw previously, it’s easy to simulate a random coin flip by generating a random decimal $r$ between $0$ and $1$ and applying the following function:
\begin{align*} \textrm{outcome}(r) = \begin{cases} \textrm{heads} & \textrm{ if } r < 0.5 \\ \textrm{tails} & \textrm{ otherwise} \end{cases} \end{align*}
This is a special case of a more general idea: sampling from a discrete probability distribution. Flipping a fair coin is tantamount to sampling from the distribution [0.5, 0.5], i.e. $0.5$ probability heads and $0.5$ probability tails.
More complicated contexts may require sampling from longer distributions that may or may not be uniform. For example, if we wish to simulate the outcome of rolling a die with two red faces, one blue face, one green face, and one yellow face, then we need to sample from the distribution [0.4, 0.2, 0.2, 0.2].
Note that when we sample from the distribution, we need only sample an index from the distribution, and then use the index to look up the desired value. For example, when we sample an index from the distribution [0.4, 0.2, 0.2, 0.2], we have probabilities $0.4,$ $0.2,$ $0.2,$ and $0.2$ of getting indices $0,$ $1,$ $2,$ and $3$ respectively. Then, all we need to do is look up that index in the following array: [red, blue, green, yellow].
## Roulette Wheel Selection
Roulette wheel selection is an elegant technique for sampling an index from an arbitrary discrete probability distribution. It works as follows:
1. turn the distribution into a cumulative distribution,
2. sample a random number $r$ between $0$ and $1,$ and then
3. find the index of the last value in the cumulative distribution that is less than $r.$
To illustrate, let’s sample an index from the distribution [0.4, 0.2, 0.2, 0.2] that was mentioned above in the context of a die roll.
1. First, we construct the cumulative distribution: [0.4, 0.4 + 0.2, 0.4 + 0.2 + 0.2, 0.4 + 0.2 + 0.2 + 0.2], or more simply, [0.4, 0.6, 0.8, 1.0].
2. Then, we sample a random number $r$ between $0$ and $1.$ Suppose we get $r = 0.93.$
3. Finally, we find the index of the last value in the cumulative distribution that is less than $r = 0.93.$ In our case, this is the value $0.8$ at index $2,$ because the next value ($1.0$ at index $3$) is greater than $r = 0.93.$
So, we have sampled the index $2.$
## Exercise
Write a function random_draw(distribution) that samples a random number from the given distribution, an array where distribution[i] represents the probability of sampling index i.
To test your function on a particular distribution, sample many indices from the distribution and ensure that the proportion of times each index gets sampled matches the corresponding probability in the distribution. Do this for a handful of different distributions.
Tags: | 2022-08-10T17:11:48 | {
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https://math.stackexchange.com/questions/1850456/prime-divides-n2-1-rightarrow-prime-doesnt-divide-n/1850477 | # Prime divides $n^2 + 1 \Rightarrow$ prime doesn't divide $n$
How can I show that if a prime $p$ divides $$n^2 + 1$$ then it doesn't divide $n$?
• If p divides n then it must divide n^2. No number other than 1 can divide two consecutive numbers. Jul 6 '16 at 2:25
• Dang auto correct. And that was a weird one. Can not understand it. Jul 6 '16 at 2:28
• Try proving the contrapositive, it's easier to think about. Jul 6 '16 at 2:29
• @fleablood Maybe cutie numbers are like friendly numbers and lucky numbers at the same time?
– user296602
Jul 6 '16 at 2:29
• Applying Euclid's algorithm $\,(n^k\!+1,n) = (1,n) = 1\,$ by $\,n^k\!+1\equiv 1\pmod n\,$ for $\,k\ge 1.\,$ Since their gcd $= 1,\,$ they have no nontrivial common divisor. Jul 6 '16 at 3:35
We can use Bezout's Identity to show that $\left(n^2+1,n\right)=1$. That is, $$\left(n^2+1\right)\cdot1-n\cdot n=1$$ Therefore, the greatest common divisor of $n^2+1$ and $n$ is $1$.
That is, if any number divided both $n^2+1$ and $n$, it would also divide $(n^2+1)-n\cdot n=1$.
• For every integer $k\ge 1$, $(n^k+1)\cdot 1-n\cdot n^{k-1}=1$. Jul 6 '16 at 10:27
• Indeed. $\left(n^k+1,n\right)=1$ for $k\ge1$.
– robjohn
Jul 6 '16 at 12:12
• @Woria Special case $\, j = n^{k-1}\,$ of $\,(1\!+\!nj,\,n) = 1\$ by $\ 1= 1\!+\!nj-n(j)\ \$ (or by one step of Euclid's algorithm - see my comment on the question). Aug 4 '16 at 2:51
A proof by contradiction: assume $n\equiv 0\mod p$ with $p>1$, then $$n^2\equiv 0\pmod p$$ also. But $$n^2\equiv -1\pmod p$$ which contradicts, therefore $$n^2+1\equiv 0\pmod p\Rightarrow n\not\equiv 0\pmod p$$
• i.e. if $\,p\mid n,\color{#c00}{n^2\!+\!1}\,$ then $\,{\rm mod}\ p\!:\ n\equiv 0\,\Rightarrow\, 0\equiv \color{#c00}{n^2\equiv -1}\,\Rightarrow\, p\mid 1,\,$ contradiction Aug 4 '16 at 2:54
Below, all 'leters variables' $\ds{n,s,p}$ are integers $\ds{\pars{~p\ \mbox{is a}\ \ul{prime\ number}~}}$:
• $\ds{{n \over p} = s\quad\imp\quad n = sp\quad\imp\quad{n^{2} + 1 \over p} = {s^{2}p^{2} + 1 \over p} = s^{2} p + {1 \over p}\ !!!}$
• $\ds{{1 \over p}\ \ul{\mbox{is not}}\ \mbox{an integer because}\ p > 1 \pars{~p\ \mbox{is a prime number}~}}$
• $\pars{\vphantom{\LARGE A}% p \mid n \imp p \not\mid \pars{n^{2} + 1}}\ \mbox{is equivalent to}\ \pars{\vphantom{\LARGE A}% p \mid \pars{n^{2} + 1} \imp p \not\mid n}$
If $p=2$ and $2\mid n$ we have that $n^{2}$ is even and so $n^{2}+1$ is odd. Now assume that $p$ is odd. We can use the Legendre symbol. If we assume that $n\equiv0\mod p$ we have $n^{2}\equiv0 \mod p$. So $$\left(\frac{n^{2}}{p}\right)=0$$ but since $n^{2}\equiv-1 \mod p$ we also have, by the law of quadratic reciprocity $$\left(\frac{n^{2}}{p}\right)=\left(\frac{-1}{p}\right)=1^{\frac{p-1}{2}}=\begin{cases} 1 & p\equiv1\,\mod\,4\\ -1 & p\equiv3\,\mod\,4 \end{cases}$$ and this is absurd.
You want to prove the statement $\color\red{p|n^2+1}\implies\color\green{p\not|n}$.
Instead, you can prove the equivalent statement $\neg(\color\green{p\not|n})\implies\neg(\color\red{p|n^2+1})$:
$\small\neg(\color\green{p\not|n})\implies{p|n}\implies{p|n^2}\implies{\forall_{k\in(0,p)}:p\not|n^2+k}\implies{p\not|n^2+1}\implies\neg(\color\red{p|n^2+1})$.
BTW, this statement holds not only for every prime $p$, but also for every integer $p>1$. | 2021-10-25T06:50:12 | {
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http://math.stackexchange.com/questions/17638/defining-the-determinant-of-linear-transformations-as-multilinear-alternating-fo/244205 | # Defining the determinant of linear transformations as multilinear alternating form
Here is what our professor showed us in our linear algebra class to introduce the idea of determinants:
Suppose we have an $n$-dimensional vector space $V$. Then we can create a function from $V^n$ to $\mathbb{R}$ called $vol$ (for "volume") satisfying these properties:
$vol$ is multilinear
$vol$ is alternating (i.e. if any two of $v_1, \ldots, v_n$ are the same, then $vol(v_1, \ldots, v_n) = 0$)
From these two properties, we can see that if $e_1, \ldots, e_n$ is a basis of $V$, then the $vol$ function is completely defined by the value $vol(e_1, \ldots, e_n)$.
Thus if $T$ is a linear operator on $V$, the ratio:
$\dfrac{vol(Te_1, \ldots, Te_n)}{vol(e_1, \ldots, e_n)}$
is the same for any (multilinear and alternating) $vol$ function.
However, I am having trouble understanding why the ratio is also independent of the basis $e_1, \ldots, e_n$. This is what I am asking for help with. I can see that this invariance implies that intuitively, every $n$-parallelotope is stretched by the same amount by the operator $T$.
(Our professor then defined the determinant of $T$ as that ratio.)
-
Show that any two bases can be reached from each other by row operations and look at what row operations do to the volume. – Qiaochu Yuan Jan 16 '11 at 2:04
Adding one row to another does not change the volume, while multiplying a row by a scalar changes the volume by that scalar factor. So if we go from $e_i$'s to another basis, say $f_i's$, then the ratio $vol(f_1, \ldots, f_n)/vol(e_1, \ldots, e_n)$ is the product of all the scalar multiplications. But I don't see why this is the same as $vol(Tf_1, \ldots, Tf_n)/vol(Te_1, \ldots, Te_n)$ – Alan C Jan 16 '11 at 2:21
You are half way there. The point is that since $T$ is linear, it commutes with scalar multiplication. So if $f_1=\alpha e_1$ for a scalar $\alpha$, then $Tf_1 = \alpha Te_1$. – Alex B. Jan 16 '11 at 5:10
Aha thanks! I wonder how I overlooked the linearity of $T$. Hm... but I'm also wondering if this could be done in a way similar to the Steinitz exchange lemma. Suppose we order the $e_i$'s and $f_i$'s so that for all $0 \leq k \leq n$, $f_1, \ldots, f_k, e_{k+1}, \ldots, e_n$ span $V$. Then we could go from $vol(Te_1, \ldots, Te_n)/vol(e_1, \ldots, e_n)$ to $vol(Tf_1, \ldots, Tf_n)/vol(f_1, \ldots, f_n)$, replacing one vector at a time. (And the reason the ratio stays the same is because $T$ is linearity.) – Alan C Jan 16 '11 at 15:28
In light of my comments on a previous, closely related question: in case you were wondering, I am not teaching linear algebra this semester! – Pete L. Clark Feb 15 '11 at 16:48
First note that since $T$ is linear, $\mathrm{vol}_T(v_1, \dots, v_n) = \mathrm{vol} (T v_1 , \dots , T v_n )$ is also a multilinear alternating function.
So if you go from $e_1, \dots , e_n$ to a different basis, say, $b_1, \dots b_n$, then you can write the $b_i$ in terms of their coordinates with respect to $e_1, \dots , e_n$:
$$b_1= \sum_{i=1}^n b_{i 1} \, e_i,\quad b_2= \sum_{i=1}^n b_{i 2} \, e_i,\quad\dots \quad b_n= \sum_{i=1}^n b_{i n} \, e_i.$$
And Leibniz's formula (which is a direct consequence of multinearity and alternating-ness), applied to both $\mathrm{vol}$ and $\mathrm{vol}_T$ tells us:
$$\mathrm{vol}(b_1 , \dots , b_n ) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \: b_{\sigma(1) 1} \cdots b_{\sigma(n) n} \cdot \mathrm{vol}(e_1, \dots , e_n)$$
$$\mathrm{vol} (T b_1 , \dots , T b_n ) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \: b_{\sigma(1) 1} \cdots b_{\sigma(n) n} \cdot \mathrm{vol}(T e_1, \dots , T e_n)$$
So if you take the ratio $\displaystyle \frac{\mathrm{vol} (T b_1 , \dots , T b_n )}{\mathrm{vol} (b_1 , \dots , b_n )}$ everything else cancels out and you're left with $\displaystyle \frac{\mathrm{vol} (T e_1 , \dots , T e_n )}{\mathrm{vol} (e_1 , \dots , e_n )}$ again.
Of course, for all of this, it is important that $\mathrm{vol} \neq 0$. That's probably an additional restriction your professor put on the $\mathrm{vol}$ function.
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http://math.stackexchange.com/questions/291956/if-p-is-prime-and-p-5-show-that-when-p-is-divided-by-10-the-remainder | # If $p$ is prime and $p > 5$, show that when $p$ is divided by 10, the remainder is 1, 3, 7, or 9
If $p$ is prime and $p > 5$, show that when $p$ is divided by 10, the remainder is 1, 3, 7, or 9.
This is a problem from Hungerford's Abstract Algebra: An Introduction. I would like some help on it, it was an example in class. Thanks.
-
Consider the case when the remainder is 2. Then $p = 10 q + 2$ for some integer $q$, so that $p = 2 (5 q + 1)$ is even, but cannot be 2 by the assumption $p > 5$. The other cases are entirely similar, it is useful for you to work them out yourself. – Andreas Caranti Feb 1 '13 at 7:27
$p > 5$ prime means $p$ not divisible by $2$ or $5$. – Benjamin Dickman Feb 1 '13 at 8:26
By the division algorithm, we can write each $p$ as $$p = 10q + r$$ for some quotient $q$ and remainder $0\le r < 10$. As $p$ is prime, clearly $r\neq 0$. $r$ also cannot be even, otherwise $p$ is even. Finally, note that $r \neq 5$ or $5 \mid p$.
For a more brief and algebraic solution, note that $(p,\ 10) = 1$ implies that $p$ is a unit in $\mathbb{Z}/10\mathbb{Z}$. The units of the ring are precisely $1,\ 3,\ 7$ and $9$.
For completeness, you should probably show that there exists primes for each of the remaining congruence classes.
-
You already have a couple of nice ‘mathematical-looking’ answers. In attacking a question like this, though, you might want to start with simple, familiar facts.
The remainder when $p$ is divided by $10$ is simply the last digit of $p$. If the last digit of a number $n$ is $0,2,4,6$, or $8$, what kind of number is $n$? Can it be prime if it’s greater than $2$? If the last digit is $0$ or $5$, what can you say about $n$? Can it be prime and greater than $5$?
These are enough to tell you, at least informally, why the prime $p>5$ must end in $1,3,7$, or $9$, and now you can worry about explaining the reasoning in the previous paragraph a bit more formally.
-
Thank you for the tip, I was not sure how to best ask questions here but I'm learning and will remember that. – grayQuant Feb 1 '13 at 14:36
To say that a prime $p > 5$ gives a remainder of $1, 3, 7$ or $9$ when divided by $10$ is merely remarking that the prime is odd (and not $5$, since all integers ending in five are multiples thereof).
After all, the remainder when divided by $10$ is simply the unit of the number itself, and if that unit was any of $0, 2, 4, 6, 8$ we'd plainly see that it was an even number, and therefore not prime. Likewise with the five.
That said, we could come to the conclusion of your statement by, say, basing an argument on how all primes above $3$ are of the form $6n \pm 1$ (mind you, this follows from an argument akin to the above): the multiples of $6$ begin as follows: $$6, 12, 18, 24, 30\ldots$$ From which we father that the units of such multiples are either $0, 2, 4, 6$ or $8$. Now take the '$\pm 1$' bit into consideration and we find that we have the following units possible: $$1, 3, 5, 7, 9$$ (Remember, a remainder of $-1$ when divided by $10$ is equivalent to a remainder of $9$.)
So, much like the above, we disregard the $5$ since that certainly isn't prime, and we have $1, 3, 7, 9$ leftover, as desired.
-
you are looking for the unit digit of the prime.For prime $p\neq 2$, unit digit is odd which means $p\pmod {10}\in\{1,3,5,7,9\}$
Now, for any prime $p\gt 5$, unit git can't be $5$ otherwise it would be divisible by $5$ and hence won't be a prime.
Thus only possible candidates for unit digit of a prime $p\gt 5$ are $\{1,3,7,9\}$
-
This is clear from the Euclidean algorithm. Write the remainder $\rm\: r = (p\ mod\ 10).\:$ By Euclid
$$\rm p\equiv r\,\ (mod\ 10)\ \Rightarrow\ gcd(p,10) = gcd(r,10)$$
In particular, when the gcd $= 1,\,$ then: $\rm\,\ p\,$ is coprime to $10$ $\iff$ $\rm\,r\,$ is coprime to $10$
Now, by hypothesis, $\rm\, p\,$ is prime $> 5,\,$ so $\rm\, p\,$ is coprime to $10,\$ thus $\rm\ r\,$ is coprime to $10,\:$ so we infer that $\rm\,r\,$ is odd and coprime to $\,5,\:$ hence $\rm\:r\in \{1,3,7,9\},\,$ since the remainder $\rm\,r\in [0,9].$
Remark $\$ Ditto if we generalize $\rm\,10\,$ to any modulus $\rm\,m\!:\$ if prime $\rm\,p\nmid m\:$ then its remainder $\rm\, p\ mod\ m\,$ is one of the $\,\varphi(m)\,$ remainders coprime to $\rm\,m.\:$ Or, expressed in radix language:
$\quad$ in radix $\rm\,m\!:\$ if $\rm\,n\,$ has units digits $\rm\,r,\$ then $\rm\ n\,$ is coprime to $\rm\, m\iff r\,$ is coprime to $\rm\,m$
- | 2016-07-28T14:50:57 | {
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https://math.stackexchange.com/questions/2578416/what-is-the-best-way-to-calculate-the-average-of-individual-averages | What is the “best” way to calculate the average of individual averages?
In preparing for GRE, I see questions looking for the mean of two or more individual means. Here is an example:
The average (arithmetic mean) of 100 measurements is 24, and the average of 50 additional measurements is 28
Quantity A Quantity B
The average of the 150 measurements 26
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
My question regards the best way to determine the mean (average). I have found two methods that always basically solve it correctly, although the results differ by ~ 0.1.
One method is adding the weighted means: find sum of each sample, add the sums for a total sum, determine the proportion of each individual sum to the total sum to calculate the "weight", then add the products of each individual mean. In other words, Weighted mean = (proportion)(group A mean) + (proportion)(group B mean) + ....
So, to solve the example question: sum of Group A measurements = $100 \cdot 24 = 2400$ sum of Group B measurements = $50 \cdot 28 = 1400$ total sum = 3800 proportion of Group A to total = $2400 \div 3800 = 0.631579$ weighted mean of Group A = $24 \cdot 0.631579 = 15.157896$ proportion of Group B to total = $1400 \div 3800 = 0.368421$ weighted mean of Group B to total = $28 \cdot 0.368421 = 10.315788$ The average (mean) of the 150 measurements = 15.157896 + 10.315788 = 25.473684
The other method is to just divide the total sum by the 150 measurements, which is how a normal mean is calculated, to equal 25.333333.
Although both methods make B) the answer, they could be significantly different in other contexts. It seems to me that the first, weighted, method is better since it probably takes outliers, repeated measurement, etc. into account.
What is the "best" way to calculate the average of individual averages?
Your second method is correct and the mean is $25\frac 13$. The proportion in your first method should be the proportion of the measurements, not the proportion of the sum, so it would be $24\cdot \frac {100}{150}+28\cdot \frac {50}{150}=25\frac 13$, which agrees with the other.
The best way might be this: with $100$ sitting at $24$ and $50$ sitting at $28$, you need the balance point between the two. The balance point is closer to $24$ than to $28$, because there’s more mass sitting there. Therefore, it’s less than $26$.
If you want to be more precise, there’s half as much mass at $28$, so $28$ is twice as far away from the balance. You need the point “one third” of the way from $24$ to $28$, so that’s $24 + \frac43=25\frac13$. | 2019-06-20T19:54:16 | {
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https://math.stackexchange.com/questions/2227280/are-there-an-equal-number-of-positive-and-negative-numbers/2227287 | # Are there an equal number of positive and negative numbers?
For every positive number there exists a corresponding negative number. Would that imply that the number of positive numbers is "equal" to the number of negative numbers? (Are they incomparable because they both approach infinity?)
• See Cardinality and Equinumerosity. Apr 10, 2017 at 7:16
• 'Equal number' when you talk about infinite sets means you can map one to one both ways, so yes there are an equal number of positive and negative 'numbers.'
– Arby
Apr 10, 2017 at 7:16
• Look up "Hilbert's hotel" to see an example how we compare infinite sizes, and the counterintuitive consequences that might have. Apr 10, 2017 at 7:17
• The two sets $\text{Pos}$ and $\text{Neg}$ are equinumerous because there exists a one-to-one correspondence (a bijection) between them. In this sense they are comparable also if both are infinite sets. Apr 10, 2017 at 7:18
• Define "equal". In the sense of cardinality, yes, there are "as many" positive numbers as negative, but then also as many positive integers as both positive and negative integers together, or as many positive integers as positive even integers.
– dxiv
Apr 10, 2017 at 7:22
If there is SOME function that gives a bijection between two sets, then these two sets are considered equally big. (Even if there is some other function between those two sets that is not onto/not 1-1). For example the set of multiples of 10 among positive integers being a proper subset of all positive integers seems to be "smaller". But we have a function $f(x)=10x$ providing bijection, and so they are considered to be of the same size (cardinality). In your case $g(x)=-x$ provides the bijection.
Yes, the existence of a one-to-one and onto mapping is exactly how equality of the size of sets (the technical term is "cardinality" is defined. The (cardinal) number of negative integers is the same as the cardinal number of positive integers, and the cardinal number of negative real numbers is the same as the cardinal number of positive real numbers.
Note however that there are more positive real numbers than positive integers, so not all infinite cardinalities are equal. The proof of that fact is done by Cantor's famous diagonal proof.
The word "infinity" is used in many places in maths and the definitions are not necessarily the same. Different symbols are used but there are still more definitions than symbols.
The most familiar symbol for infinity, $\infty$, is commonly used in calculus but it is more of a suggestive shorthand than an actual infinity. It is not normally used when discussing sizes of sets.
When discussing the size of sets, the term cardinality is usual. Clearly, normal counting won't work for infinite sets. However, as others have said, it is possible to say whether or not two sets have the same cardinality / are the same size. If a bijection (one to one and onto map) between them exists then they have the same cardinality. One counter-intuitive property of infinite sets is that it is possible that a map which is one to one and not onto exists as well one that is one to one and onto. So, the existence of a map that is one to one but not onto does not prove that one set is smaller. For that, you would also need to prove that no other map that is one to one and onto exists. The simplest example of this is the set of all natural numbers and just the even ones. Intuitively, the set of even numbers is smaller and there is an obvious map from the even natural numbers to a subset of the natural numbers. However, there is also a map between them which is one to one and onto hence they actually have the same cardinality.
By definition, a set which has a one to one and onto map to the natural numbers is called "countable". This is not countable in the day to day sense but it does mean that you could name one per second and (*), although you would not ever finish, you would name any particular one in a finite time.
(*) Assuming that you and the universe were immortal.
Many sets of intuitively different sizes are countable, for example the integers $\Bbb{Z}$, the rational numbers $\Bbb{Q}$ and the algebraic numbers $\Bbb{A}$. This cardinality / size is named countable and the symbol $\aleph_0$ is used. This is called "aleph null", "aleph naught", or "aleph zero". Aleph is the first letter of the Hebrew alphabet.
However, even though many sets which are intuitively bigger turn out to be the same size, bigger sets exist. It can be proved that there is no one to one and onto map from the natural numbers to the set of real numbers $\Bbb{R}$ so that it is bigger. Again, some apparently bigger sets are not actually bigger. For example $\Bbb{C}$, $\Bbb{R^2}$, and $\Bbb{R^3}$ are the same cardinality as $\Bbb{R}$.
Even bigger sets exist, the set of all subsets of $\Bbb{R}$ is bigger than $\Bbb{R}$. There is no biggest set.
A particularly interesting question is whether $\Bbb{R}$ is the next biggest cardinality after the countable infinity of $\Bbb{N}$. This is called the "Continuum Hypothesis". My answer is already long enough so I won't talk about that but if you are interested in this subject then you should look it up.
Finally, there is yet another type of infinity called "ordinal numbers". In this sense, the first infinity is usually written as $\omega$ and, unlike the cardinal infinities, $\omega + 1$ is different.
The notion of "counting" is made precise in mathematics by using functions. These functions are bijections. To "count" the elements in a set $X$, you establish a bijection from $X$ to a subset of the natural numbers $\mathbb{N}$.
Example. How many elements does the set $\{a, b, c\}$ have? The answer is 3, because there exists a bijection from $\{a, b, c\}$ to $\{0, 1, 2\}$. (Actually, there exist $3!$ bijections.)
So, certain subsets of $\mathbb{N}$ are, somehow, "measuring sticks", which you can use the gauge the size of other sets.
Using sets and functions allows one to generalize the idea of "counting" to infinite sets. The "measuring sticks" then become transfinite numbers (more precisely, cardinal numbers).
To answer your question, there is an equal number of positive and negative numbers, because there exists a bijection between these two sets. If by "number" you mean "integer", then this number is called aleph null. It's the smallest infinite cardinal.
Bonus: Using these ideas, you can show that the "amount" of (say) positive integers and the "amount" of integers is the same, or that the "amount" of even integers and the "amount" of prime numbers is the same.
There are exactly the same ones.
For each positive number there is one and only one negative,
And for each negative number there is one and only one positive,
So that both sets have the same cardinal.
There is a bijective function of the set of positive numbers to the set of negative numbers. | 2022-05-25T14:17:56 | {
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https://math.stackexchange.com/questions/1899679/apostol-bolzano-weierstrass-theorem | # Apostol Bolzano-Weierstrass Theorem
Theorem. If a bounded set $S$ in $\mathbb{R}^n$ contains infinitely many points, then there is at least one point in $\mathbb{R}^n$ which is an accumulation point of $S$.
Proof. (for $\mathbb{R}^1$) Since $S$ is bounded, it lies in some interval $[−a,a]$. At least one of the subintervals $[−a,0]$ or $[0,a]$ contains an infinite subset of $S$. Call one such subinterval $[a_1,b_1]$. Bisect $[a_1,b_1]$ and obtain a subinterval $[a_2,b_2]$ containing an infinite subset of $S$, and continue this process. In this way a countable collection of intervals is obtained, the $nth$ interval $[a_n,b_n]$ being of length $b_n -a_n = a/2^{n-1}$. Clearly the $\sup$ of the left endpoints $a_n$ and the $\inf$ of the right endpoints $b_n$ must be equal, say to $x$. The point $x$ will be an accumulation point of $S$ because, if $r$ is any positive number, the interval $[a_n,b_n]$ will be contained in $B(x;r)$ as soon as $n$ is large enough so that $b_n−a_n<r/2$. The interval $B(x;r)$ contains a point of $S$ distinct from $x$ and hence $x$ is an accumulation point of $S$.
My Questions
1. Why is it obvious that the $\sup$ of the left endpoint $a_n$ is equal to the $\inf$ of the right endpoint $b_n$? Also how does one consider the $\sup$ or $\inf$ of a single number?
2. I don't understand why we need $b_n−a_n<r/2$. Instead, why do we need to halve $r$? If $b_n−a_n<r$ then wouldn't $[a_n,b_n]$ still be contained in $B(x;r)$?
• Yes, you're correct that if $b_n-a_n<r$, then $[a_n,b_n]\subset B(x;r)$. For $b_n-x\le b_n-a_n$ and, similarly, $x-a_n\le b_n-a_n$. – Ted Shifrin Aug 22 '16 at 5:36
• Since $[a_{n+1},b_{n+1}]$ is a subinterval of $[a_n,b_n]$ by construction, the sequences $a_n$ and $b_n$ are (bounded and) decreasing and increasing, respectively, so $\lim_n a_n=\sup_n a_n$ and $\lim_n b_n=\inf b_n$. Since $|b_n-a_n|=a/2^{n-1}$, we have $\lim_n b_n-a_n=0$ and so $\lim_n a_n=\lim_n b_n$. (We do not consider $\sup$ and $\inf$ of single points, but range over all $n$.) – Luiz Cordeiro Aug 22 '16 at 15:39
He means $\sup\{a_n : n \in \mathbb N\}$, not the supremum of the single point $a_n$.
Note that $a_n$ is an increasing sequence, and $b_n$ is decreasing. Both sequences are bounded, because $[a_n,b_n] \subset [a_1,b_1]$ for every $n$. Therefore they both converge (specifically, $a_n$ converges to its supremum, and $b_n$ converges to its infimum). As the distance $b_n - a_n$ is shrinking to zero, they must converge to the same limit.
I assume he halves $r$ because his ball $B(x;r)$ is open, so not quite large enough to contain a closed interval of length $r$ if one of the endpoints of the interval is $x$.
Edit: actually, he used a strict inequality $b_n - a_n < r/2$, so indeed he could have used $b_n - a_n < r$.
• Thank you very much, You guys are fantastic! – Jonathan Duran Aug 23 '16 at 0:05
Apostol wrote "the endpoints" instead of "the endpoint". So your first question is solved by observing that he was saying $$\sup_{n \geq 1\ (\text{say})}a_{n} = \sup \{ a_{n} \mid n \geq 1 \} = \inf \{ b_{n} \mid n \geq 1 \} = \inf_{n \geq 1}b_{n}.$$
For the second question, that part of the proof does not imply anything that is a must. That choice of an upper bound for $b_{n}-a_{n}$ where $n >> 1$ is to show the reader that we do can do so-and-so.
• So the proof would have been fine if we chose $r$ instead of $r/2$? Thank you for your response! – Jonathan Duran Aug 23 '16 at 0:04
• Yes, it would also be fine. – Megadeth Aug 23 '16 at 1:40 | 2020-01-22T11:36:47 | {
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https://math.stackexchange.com/questions/4458703/whats-the-size-of-the-biggest-set-of-numbers-from-1-to-100-so-that-no-number-is | # What's the size of the biggest set of numbers from 1 to 100 so that no number is the average of any other two?
John wants to build a set of numbers, from the range of 1 to 100. The only rule is that in that set no number can be the average of any other two. For example, if the set contains the numbers 1 and 3, then 2 cannot be present. What’s the size of the biggest set that John can build?
More precisely, if L is "John's set", then $$\forall x,y,z \in L : z \ne \frac{x+y}{2}$$ and $$\forall x \in L : 1 \leq x \leq 100$$
We want to find the cardinality of L.
PS: This problem appeared on a Sunday newspaper many years ago. On the following week they published a solution that was wrong. I was able to solve it computationally but never mathematically.
You can solve the problem via integer linear programming as follows. Let $$n=100$$. For $$i\in\{1,\dots,n\}$$, let binary decision variable $$x_i$$ indicate whether $$i\in L$$. The problem is to maximize $$\sum_{i=1}^n x_i$$ subject to $$x_i + x_{i+d} + x_{i+2d} \le 2 \quad \text{for i\in\{1,\dots,n-2\} and d\in\{1,\dots,\lfloor(n-i)/2\rfloor\}}.$$ The optimal objective value turns out to be $$27$$, attained by $$L=\{1,5,7,10,11,14,16,24,26,29,30,33,35,39,66,70,72,75,76,79,81,89,91,94,95,98,100\}.$$
See https://oeis.org/A003002 for more terms. Such a set is called a Salem-Spencer set, and asymptotic bounds are known for the largest cardinality.
• Very nice. Just out of curiosity, did you just write the program on the fly? When I look at the constraint, my first thought is that $2^{(100)} \approx 10^{(30)}$ is too many cases. My pc balks at more than $(10)^8$ cases. Did you find a way to streamline the code? For example, did you experiment with $V =$ value and increment $V$ by $(+1)$, checking each value, and discovering that $V = 27$ is attainable and $V = 28$ is not? May 25 at 23:26
• @asinomás can you undelete your answer? I think including it makes for an interesting comparison getween the greedy and optimal algorithms. May 25 at 23:30
• @user2661923 I used a generic ILP solver that uses the branch-and-cut algorithm, not brute force. May 25 at 23:53
• @OscarLanzi I agree with you and just voted to undelete the answer. asinomas was not flagged by your comment because he has not been registered for RobPratt's answer. One alternative is for you to track down any posting/answer that asinomas is registered for, and leave your comment request, addressed to him, there. Naturally, you will need to include a link back to this posting. May 25 at 23:57
• Very nice. The answer is indeed 27 (for N=100). In my case, I had used dynamic programming to find the sets. The reason I keep looking at this problem after all these years is that although I was able to prove a few properties/theorems about the cardinality of L, I was never able to solve it in a more "mathematical" way. I.e., for a certain size of N, is there an expression for #L? May 26 at 9:26
A greedy algorithm would hive an answer of $$24$$, as given by the "Erdos-Szekeres" sequence here: http://oeis.org/A003278 .
But, as demonstrated by RobPratt, and as often happens in analysis, the greedy algorithm is ultimately not the optimal one. It's as if you overuse an initially fertile soil, exhaust its nutrients and then you find you can't grow crops until you give time for rains to fall and the soil to replenish itself. Let's explore why greed turned out not good in this case.
Welcome to the desert
Suppose you have a list of whole numbers, with a minimum of $$1$$ and a maximum of $$n$$, such that there is not a group of three in arithmetic sequence. Then the arithmetic third of any pair of numbers from this list cannot be larger than $$2n-1$$, and so you can always insert the number $$2n$$ and be sure there is still no three-term arithmetic sequence.
If, in fact, you do need to resort to $$2n$$ to continue your list -- that is, no smaller number will do -- then you have in effect created a "desert". The presence of such deserts can make you fall behind a more carefully tended, more optimal sequence. And that is what happened here.
Upon further review, the Erdos-Szekeres sequence, constructed to take as small a number as possible at any step, is equally optimal at producing deserts. A number $$k$$ enters this sequence iff the ternary representation of $$k-1$$ has no elements of $$2$$, for instance eleven minus one is ten, which is $$101$$ in base $$3$$, so ten is in the sequence. But once we get to fourteen in this case, where one less is $$111$$ base $$3$$, we are forced to accept a $$2$$ somewhere in the ternary representation until we have passed the next power of $$3$$, in this case at $$28$$, where we can introduce a new ternary "digit". We build in a desert from $$n=(3^m+1)/2$$ all the way up to $$2n=3^m+1$$ for every whole number $$m$$. In this problem the desert from $$41$$ to $$82$$ ($$m=4$$) renders the greedy algorithm inferior.
Finding an oasis
Of course, we can change the rules of our game to make the greedy algorithm look good. If, instead of an upper bound of $$100$$, we were to choose an upper bound of $$(3^m+1)/2$$ for some whole number $$m$$ -- stopping just at the threshold of the desert -- then the Erdos-Szekeres method really does work. For instance, with a maximum of $$(3^5+1)/2=122$$ the Erdos-Szekeres method gives $$32$$ terms and that is indeed the optimum for a max element of $$122$$. See https://oeis.org/A065825.
• The description of the sequence seems to say that this is the result of the greedy approach (build a sequence from $1, 2, \ldots$ by choosing the next smallest valid number), which might not be optimal as RobPratt's answer suggests. May 25 at 23:15
• @angryavian Nice catch, I missed that. I deleted my comment. May 25 at 23:15
• I was k-c-u-f-ed (read it backwards) by Autocorrect. Forced to rephrase that passage. Please try again. May 26 at 0:59
• @OscarLanzi This time, when you reply, please include the AT:user2661923, so that I will be flagged. Perhaps I am being dense here. I don't understand this sentence: "Then the arithmetic third of any pair of numbers from this list cannot be larger than 2n−1, and so you can always insert the number 2n and be sure there is still no three-term arithmetic sequence." In this sentence, what is $n$, and what pair of numbers are you referring to? Are you, for example saying that if you take an optimal list, and append the number $(200)$ to it, that no constraint has been violated? May 26 at 1:08
• @OscarLanzi Okay. I skimmed your analysis. I am going t have to study it further, which will take at least a few days. I have bookmarked this posting, for reference. For what it's worth, I was also intrigued by Rob Pratt's reference to a branch and cut algorithm, re integer linear programming. May 26 at 1:23 | 2022-06-26T10:59:15 | {
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https://math.stackexchange.com/questions/2560143/how-to-find-probalility-that-a-student-misses-at-least-one-test-if-he-she-is-abs | # how to find probalility that a student misses at least one test if he/she is absent twice?
The probability that a teacher will give an unannounced test during any class is $$\dfrac15$$. If a student is absent twice then probability that he/she misses at least one test is
$$\\ \hspace{5cm}$$ a) $$\dfrac23\ \quad$$ b) $$\dfrac45\ \quad$$c) $$\dfrac7{25}\ \quad$$d) $$\dfrac9{25}\$$
My attempt:
Probability of attending first test & missing $$2$$nd test $$=\dfrac45\times\dfrac15=\dfrac4{25}$$
Probability of missing first test & attending $$2$$nd test $$=\dfrac15\times\dfrac45=\dfrac4{25}$$
Probability of missing both the tests $$=\dfrac15\times\dfrac15=\dfrac1{25}$$
Total probability of missing at least one test $$=\dfrac4{25}+\dfrac4{25}+\dfrac1{25}=\dfrac9{25}$$
• Yes, you are correct. Notice, a simple method is $$P_{\text{ Missing atleast one test }} = 1- P_{\text{ Missing no test }} = 1-(\frac{4}{5})^2=\frac{9}{25}$$
– user371838
Dec 10, 2017 at 15:42
Questions with "at least" are often great candidates for using the complement. That is: $$P_{\text{ Event A occurs }} = 1 - P_{\text{ Event A does not occur }}$$
So, $P_{\text{ miss at least 1 test }} = 1 - P_{\text{ miss no tests }}$ giving us:
\begin{align*} P_{\text{ miss no tests }} & = P_{\text{ no test on day 1 AND no test on day 2 }}\\ & = P_{\text{ no test on day 1 }} \times P_{\text{ no test on day 2 }}\\ & = \frac{4}{5} \times \frac{4}{5}\\ & = \frac{16}{25}\end{align*}
So, $$P_{\text{ miss at least 1 test }} = 1 - P_{\text{ miss no tests }} = 1 - \frac{16}{25} = \frac{9}{25}$$
Another way to see this is that the probability of the student missing at least one test is the probability of missing test 1 + the probability of missing test 2 - the probability of them missing both tests.
The student has $\frac15$chance of missing the first test, and $\frac15$ chance of missing the second test. If you add them up you're left with $\frac 25$. HOWEVER there is a case that is being counted twice: It could happen that the student misses test 1 AND test 2. That case is being counted both when the student misses test 1 and when the student misses test 2. So now we have to substract this case once
The chance of a test happening on two consecutive days is $(\frac 15)^2 = \frac {1} {25}$.
So the answer is $\frac 25 - \frac {1} {25} = \frac {9} {25}$
The general case for this is $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ | 2022-07-06T08:13:47 | {
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http://mathhelpforum.com/algebra/17030-half-right-missed-other-half.html | # Math Help - Half right, missed the other half...
1. ## Half right, missed the other half...
"Find all values of $r$ such that the slope of the line through the points $(r, 4)$ and $(1, 3 - 2r)$ is less than 5."
I solved it by first finding the slope $\frac {2r + 1}{r - 1} = 5$ which simplifies to $r = 2$. Then I used the point-slope form to derive the equation $y = 5x + 16$. Then, I got lazy, and plugged in a couple of test numbers into the equation, found that numbers above 2 seemed to qualify, and decided that the answer was the set $(2, \infty)$. Unfortunately, the answer should have included also the set $(-\infty, 1)$.
Algebraically, what should I have done to get to that answer? What part did I skip?
2. Hello, earachefl!
Find all values of $r$ such that the slope of the line
through the points $A(r,\,4)$ and $B(1,\,3\!-\!2r)$ is less than 5.
You should have solved the inequality . . . tricky, but safe.
You found the slope of $AB$ . . . good!
. . Then we have: . $\frac{2r + 1}{r - 1} \:<\:5$ . [1]
I'd like to multiply both sides by $(r-1)$
. . but the result depends on whether $(r-1)$ is positive or negative.
Recall: when multiplying or dividing an inequality by a negative quantity,
. . . . . .reverse the inequality.
Suppose $(r-1)$ is positive . . . that is: . $r > 1$
Multiply [1] by by positive $(r-1)\!:\;\;2r + 1 \:<\:5(r - 1)$
. . $2r + 1 \:<\:5r-5\quad\Rightarrow\quad -3r \:<\:-6$
Divide both sides by -3: . $r \:>\:2$
We have two inequalities to satisfy: . $r > 1$ and $r > 2$
We take the "stronger" of the two: . $\boxed{r > 2}$
Suppose $(r-1)$ is negative . . . that is: . $r < 1$
Multiply [1] by negative $(r-1)\!:\;\;2r + 1 \;\;{\color{red}>} \;\;5(r-1)$
. . $2r + 1 \:>\:5r - r\quad\Rightarrow\quad -3r \:>\:-6$
Divide both sides by -3: . $r \:<\:2$
We have two inequalities to satisfy: . $r < 1$ and $r < 2$
We take the "stronger" of the two: . $\boxed{r < 1}$
Therefore: . $(r < 1) \vee (r > 2)$ .. . . .or: . $(-\infty,\,1) \,\cup \,(2,\,\infty)$
3. Originally Posted by Soroban
Hello, earachefl!
We have two inequalities to satisfy: . $r < 1$ and $r < 2$
We take the "stronger" of the two: . $\boxed{r < 1}$
Interesting approach. I don't think my book ever put it in quite these terms.
Thanks! | 2014-11-24T01:47:25 | {
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https://www.physicsforums.com/threads/eigenvalues-of-a-5x5-matrix.403476/ | Eigenvalues of a 5x5 matrix
1. May 15, 2010
gabriels-horn
1. The problem statement, all variables and given/known data
3. The attempt at a solution
I haven't tackled anything bigger than a 3x3 matrix. Anyone have any good pointers for reducing this matrix? I'm assuming the quickest way is still going to be the cofactor method?
2. May 15, 2010
cronxeh
Its an upper triangular matrix, so eigenvalues are pretty much on the diagonal here, they are all 2. Taking the determinant yields (2-lambda)^5, setting (2-lambda)^5=0, you get lambda1=lambda2=lambda3=lambda4=lambda5=2.
3. May 15, 2010
Landau
No need to calculate, only to think
The matrix is upper-triangular, what does this immediately tell you about the eigenvalues?
4. May 16, 2010
gabriels-horn
Ok, the discussion about the eigenvalues being the main diagonal values is clear. If the five eigenvalues are = 2, then plugging that value into the eigenvector matrix leaves only the one values in the 5x5 matrix above. Setting the matrix = 0 only leaves the eigenvector = 0; how are there then three linearly independent eigenvectors. What am I overlooking?
5. May 16, 2010
ninty
By definition an eigenvector cannot be the zero vector.
I'm assuming you're using the usual methods I see undergraduates use, (A-xI) = 0
If we represent vectors as (x1,x2,x3,x4,x5) we get x2=0, x5=0
Then solving for x1,x3,x4 we can get 3 linearly independent vectors which form a basis for null(A-xI)
An easy one would be to take 3 vectors from the usual basis.
However the question doesn't really ask for the eigenvectors, just show that they do exist.
6. May 16, 2010
Landau
There are three Jordan Blocks corresponding to eigenvalue 2, so its geometric multiplicity is 3.
7. May 16, 2010
HallsofIvy
Staff Emeritus
From the definition of "eigenvector", if v is an eigenvector with eigenvalue 2, then
$$\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}\begin{bmatrix}u \\ v \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2u+ v \\ 2v \\ 2x \\ 2y+ z \\ 2z\end{bmatrix}= \begin{bmatrix} 2u \\ 2v \\ 2x \\ 2y \\ 2z \end{bmatrix}$$
which gives the equations 2u+ v= 2u, 2v= 2v, 2x= 2x, 2y+ z= 2y, 2z= 2z. The first equation tells us that v= 0 and the fourth equation that z= 0. But u, x, and y can be any numbers.
Last edited: May 17, 2010
8. May 16, 2010
gabriels-horn
Yes, the (A-xI) = 0 method is the one I used and got x2 = 0 and x5 = 0. How can you solve for x1, x3 and x4 when they are multiplied by zero? Wouldn't that also make them zero?
This isn't something that we went over in Linear or ODE, can you expand on this thought some more?
9. May 16, 2010
gabriels-horn
You just put [/math] on the end instead of [/tex] and I took the zero out of the variable column. That's what I was confused on, the fact that x1, x3 and x4 could be any value. I just assumed that they were also zero since putting 2-2 = 0 for the eigenvector only left the two 1's in the matrix corresponding to x2 = 0 and x5 = 0. Thank you guys for the insight and provoding a good matrix template for future use.
Last edited: May 16, 2010
10. May 16, 2010
Landau
Whatever values they have, they get multiplied by zero resulting in zero, so it doesn't matter what they are. In other words, you are free to choose x1, x3 and x4, which means you can form three linearly independent eigenvectors (e.g. taking x1, x3, x4 equal to 1,0,0, equal to 0,1,0, equal to 0,0,1, gives you three independent eigenvectors.)
Your matrix is already in Jordan Normal Form. The number of Jordan blocks corresponding to a given eigenvalue coincides with the geometric multiplicity of that eigenvalue. In this case there are three such blocks for eigenvalue 2.
11. May 16, 2010
gabriels-horn
So the Jordan block is essentially dependent on the 1 values of the superdiagonal, and since x1, x3 and x4 have no entries above them on the superdiagonal, the number of Jordan blocks is equal to 3?
12. May 17, 2010
HallsofIvy
Staff Emeritus
No, multiplying x1 by 0 does NOT make x1 0, it makes the product 0. All that means is that you don't have that variable in that particular equation. | 2017-10-23T06:53:29 | {
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https://usa.cheenta.com/category/aime-i/ | Categories
## Number of points and planes | AIME I, 1999 | Question 10
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.
## Number of points and planes – AIME I, 1999
Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is $\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.
• is 107
• is 489
• is 840
• cannot be determined from the given information
Number of points
Plane
Probability
## Check the Answer
Answer: is 489.
AIME I, 1999, Question 10
Geometry Vol I to IV by Hall and Stevens
## Try with Hints
$10 \choose 3$ sets of 3 points which form triangles,
fourth distinct segment excluding 3 segments of triangles=45-3=42
Required probability=$\frac{{10 \choose 3} \times 42}{45 \choose 4}$
where ${45 \choose 4}$ is choosing 4 segments from 45 segments
=$\frac{16}{473}$ then m+n=16+473=489.
Categories
## Sequence and fraction | AIME I, 2000 | Question 10
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.
## Sequence and fraction – AIME I, 2000
A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.
• is 107
• is 173
• is 840
• cannot be determined from the given information
Equation
Algebra
Integers
## Check the Answer
Answer: is 173.
AIME I, 2000, Question 10
Elementary Number Theory by Sierpinsky
## Try with Hints
Let S be the sum of the sequence $x_k$
given that $x_k=S-x_k-k$ for any k
taking k=1,2,….,100 and adding
$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$
$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$
$\Rightarrow S=\frac{2525}{49}$
for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$
$\Rightarrow x_{50}=\frac{75}{98}$
$\Rightarrow m+n$=75+98
=173.
Categories
## Finding smallest positive Integer | AIME I, 1996 Problem 10
Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.
## Finding smallest positive Integer – AIME I, 1996
Find the smallest positive integer solution to $tan19x=\frac{cos96+sin96}{cos96-sin96}$.
• is 107
• is 159
• is 840
• cannot be determined from the given information
Functions
Trigonometry
Integers
## Check the Answer
Answer: is 159.
AIME I, 1996, Question 10
Plane Trigonometry by Loney
## Try with Hints
$\frac{cos96+sin96}{cos96-sin96}$
=$\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$
=$\frac{sin186+sin96}{sin186-sin96}$
=$\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$
=$\frac{2sin141cos45}{2cos141sin45}$
=tan141
here $tan(180+\theta)$=$tan\theta$
$\Rightarrow 19x=141+180n$ for some integer n is first equation
multiplying equation with 19 gives
$x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)$ [since 2679 divided by 180 gives remainder 159]
$\Rightarrow x=159$.
Categories
## Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.
## Roots of Equation and Vieta’s formula – AIME I, 1996
Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t.
• is 107
• is 23
• is 840
• cannot be determined from the given information
### Key Concepts
Functions
Roots of Equation
Vieta s formula
## Check the Answer
Answer: is 23.
AIME I, 1996, Question 5
Polynomials by Barbeau
## Try with Hints
With Vieta s formula
$f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0$
$\Rightarrow a+b+c=-3$, $ab+bc+ca=4$ and $abc=11$
Let a+b+c=-3=p
here t=-(a+b)(b+c)(c+a)
$\Rightarrow t=-(p-c)(p-a)(p-b)$
$\Rightarrow t=-f(p)=-f(-3)$
$t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]$
=23.
Categories
## Tetrahedron Problem | AIME I, 1992 | Question 6
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.
## Tetrahedron Problem – AIME I, 1992
Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.
• is 107
• is 320
• is 840
• cannot be determined from the given information
Area
Volume
Tetrahedron
## Check the Answer
Answer: is 320.
AIME I, 1992, Question 6
Coordinate Geometry by Loney
## Try with Hints
Area BCD=80=$\frac{1}{2} \times {10} \times {16}$,
where the perpendicular from D to BC has length 16.
The perpendicular from D to ABC is 16sin30=8
[ since sin30=$\frac{perpendicular}{hypotenuse}$ then height = perpendicular=hypotenuse $\times$ sin30 ]
or, Volume=$\frac{1}{3} \times 8 \times 120$=320.
Categories
## Triangle and integers | AIME I, 1995 | Question 9
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.
## Triangle and integers – AIME I, 1995
Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and $\angle BDC$=3$\angle BAC$. then the perimeter of $\Delta ABC$ may be written in the form $a+\sqrt{b}$ where a and b are integers, find a+b.
• is 107
• is 616
• is 840
• cannot be determined from the given information
Integers
Triangle
Trigonometry
## Check the Answer
Answer: is 616.
AIME I, 1995, Question 9
Plane Trigonometry by Loney
## Try with Hints
Let x= $\angle CAM$
$\Rightarrow \angle CDM =3x$
$\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}$=11 [by trigonometry ratio property in right angled triangle]
$\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx$
solving we get, tanx=$\frac{1}{2}$
$\Rightarrow CM=\frac{11}{2}$
$\Rightarrow 2(AC+CM)$ where $AC=\frac{11\sqrt {5}}{2}$ by Pythagoras formula
=$\sqrt{605}+11$ then a+b=605+11=616.
Categories
## Sequence and greatest integer | AIME I, 2000 | Question 11
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.
## Sequence and greatest integer – AIME I, 2000
Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.
• is 107
• is 248
• is 840
• cannot be determined from the given information
Equation
Algebra
Integers
## Check the Answer
Answer: is 248.
AIME I, 2000, Question 11
Elementary Number Theory by Sierpinsky
## Try with Hints
We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$
sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000
=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$
$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$
=2480 + $\frac{437}{1000}$
$\Rightarrow [\frac{s}{10}]$=248.
Categories
## Series and sum | AIME I, 1999 | Question 11
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
## Series and sum – AIME I, 1999
given that $\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\frac{m}{n} \lt 90$, find m+n.
• is 107
• is 177
• is 840
• cannot be determined from the given information
Angles
Triangles
Side Length
## Check the Answer
Answer: is 177.
AIME I, 2009, Question 5
Plane Trigonometry by Loney
## Try with Hints
s=$\displaystyle\sum_{k=1}^{35}sin5k$
s(sin5)=$\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\frac{1+cos5}{sin5}$
$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ then m+n=175+2=177.
Categories
## Inscribed circle and perimeter | AIME I, 1999 | Question 12
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Inscribed circle and perimeter.
## Inscribed circle and perimeter – AIME I, 1999
The inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21 given that AP=23 and PB=27 find the perimeter of the triangle
• is 107
• is 345
• is 840
• cannot be determined from the given information
Inscribed circle
Perimeter
Triangle
## Check the Answer
Answer: is 345.
AIME I, 1999, Question 12
Geometry Vol I to IV by Hall and Stevens
## Try with Hints
Q tangency pt on AC, R tangency pt on BC AP=AQ=23 BP=BR=27 CQ=CR=x and
$s \times r =A$ and $s=\frac{27 \times 2+23 \times 2+x \times 2}{2}=50+x$ and A=$({(50+x)(x)(23)(27)})$ then from these equations 441(50+x)=621x then x=$\frac{245}{2}$
perimeter 2s=2(50+$\frac{245}{2}$)=345.
Categories
## LCM and Integers | AIME I, 1998 | Question 1
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.
## Lcm and Integer – AIME I, 1998
Find the number of values of k in $12^{12}$ the lcm of the positive integers $6^{6}$, $8^{8}$ and k.
• is 107
• is 25
• is 840
• cannot be determined from the given information
Lcm
Algebra
Integers
## Check the Answer
Answer: is 25.
AIME I, 1998, Question 1
Elementary Number Theory by Sierpinsky
## Try with Hints
here $k=2^{a}3^{b}$ for integers a and b
$6^{6}=2^{6}3^{6}$
$8^{8}=2^{24}$
$12^{12}=2^{24}3^{12}$
lcm$(6^{6},8^{8})$=$2^{24}3^{6}$
$12^{12}=2^{24}3^{12}$=lcm of $(6^{6},8^{6})$ and k
=$(2^{24}3^{6},2^{a}3^{b})$
=$2^{max(24,a)}3^{max(6,b)}$
$\Rightarrow b=12, 0 \leq a \leq 24$
$\Rightarrow$ number of values of k=25. | 2021-06-17T18:28:07 | {
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https://math.stackexchange.com/questions/3176329/are-these-linear-algebra-questions-true-or-false | # Are these linear algebra questions true or false?
Which of the following are true or false?
(a) Let $$H$$ be a five-dimensional subspace of the ten-dimensional vector space $$V$$. Then, every set containing seven vectors from $$H$$ must be linearly dependent.
(b) Let $$\mathcal{B}$$ be a basis of $$\mathbb{R}^{n}$$. Let $$A$$ be a the matrix whose columns are the vectors in $$\mathcal{B}$$. Then, for every vector $$x \in \mathbb{R}^{n}$$, it is true that $$[x]_{\mathcal{B}} = Ax$$.
(c) The dimension of the vector space $$\mathbb{P}_{4}$$ is $$4$$.
(d) Let $$A$$ be a $$3\times 3$$ matrix, and let $$H$$ be the set of fixed vectors of $$A$$, that is, the set of $$x \in \mathbb{R}^{3}$$ for which $$Ax = x$$. Then $$H$$ is a subspace of $$\mathbb{R}^{3}$$.
I think (a) is false, but I don't really have a reason why. I think this is true because if you have an $$n$$-dimensional space, you only need $$n$$ linearly independent ones to span the space.
I think (b) is true since it looks like a definition I saw in my book. I don't have a reason why.
I know (c) is false. I'm pretty sure it's dimension 5. I have seen the proof before. EDIT: Proof here: Determining Bases of P4
I think (d) is true. Because if $$Ax = x$$ and $$Ay = y$$ then $$A(x + y) = x + y$$. Also the with scalar multiplication. Also, this is sort of like eigenvalues, from my understanding.
Can someone help me verify these please?
• Is this a school assignment? – Jack Pfaffinger Apr 5 at 19:47
• No it's a quiz I took. I want to calculate my grade – user660922 Apr 5 at 19:48
• What is $\Bbb P_4$? – Bernard Apr 5 at 20:05
• The vector space of polynomials of degree less than or equal to $4$ – user660922 Apr 5 at 20:13
(a) is true. If a set of $$7$$ vectors in $$H$$ were independent, they are a basis for a subspace of $$H$$, so that the $$5$$-dimensional space $$H$$ has a subspace of dimension $$7$$, which is absurd. (b) is true; $$A$$ is called a change of basis matrix. (c) is false for the reason you state, and (d) is true for the reason you state. Good thinking! | 2019-10-23T18:55:23 | {
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https://questioncove.com/updates/4fc3453ee4b0964abc860fc6 | Mathematics
OpenStudy (anonymous):
The sum of all odd numbers form 1 to 99 is?
OpenStudy (anonymous):
use sum of aritmetic sequence. .
Parth (parthkohli):
1 + 3 + 5 + 7.....But you need not do that since it is an arithmetic sequence.
Parth (parthkohli):
The sum of an arithmetic sequence is: $$\Large \color{Black}{\Rightarrow {n \over 2}(a_1 + a_l) }$$ a1 is the first term aL is the last n is the number of terms
OpenStudy (anonymous):
After 1 add? Should i cross multiply?
Parth (parthkohli):
1st term is 1. Last is 99. Odd numbers between 1,100 are 50.
Parth (parthkohli):
The common difference is d, btw.
OpenStudy (anonymous):
@ParthKohli , i think you've mistaken the formula. . http://www.regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm
Parth (parthkohli):
Oops, it was aN lol
OpenStudy (anonymous):
it's ok. .
OpenStudy (anonymous):
I dont get it XD
Parth (parthkohli):
Hey I know another pattern.
Parth (parthkohli):
The sum of first n odd numbers is n squared. The sum of first 50 odd numbers is 50 squared. So, 50^2 = 2500
Parth (parthkohli):
Well actually my formula was correct.
Parth (parthkohli):
I learnt this pattern at school.
OpenStudy (anonymous):
Ca you give me an example?
Parth (parthkohli):
For example the sum of first 4 odd numbers is 4 squared aka 16 1 + 3 + 5 + 7 = 4 + 12 = 16
OpenStudy (anonymous):
how did you got that formula? @ParthKohli
Parth (parthkohli):
Oh wait a second
Parth (parthkohli):
Yes....yes I'm correct.
OpenStudy (anonymous):
the answer is 2500, and you're correct, but how did you get that another formula?
Parth (parthkohli):
I learnt that at school
Parth (parthkohli):
@sel126 Got it?
OpenStudy (anonymous):
Yessss :) Thankyousomuch:)))))))
Parth (parthkohli):
Yw :D
OpenStudy (anonymous):
99 = a + (n-1)d hence 99 = 1 +(n-1)*3 n= 50 substituting value of n in equation S= n/2 [ 2a + (n-1)*d] we get sum of odd terms that is equal to 2500 where a =1(first term) and d =3(common difference)
OpenStudy (anonymous):
Why does three become the d?
OpenStudy (anonymous):
the sum of odd numbers is a perfect square
Parth (parthkohli):
That's what I said above
Parth (parthkohli):
Hehe don't waste your energy, genius. I've explained her.
OpenStudy (anonymous):
just thought i would mention it. no need for a formula, or all that other business, only need to know how many numbers are being added
Parth (parthkohli):
Hmm I said the exact same thing.
Parth (parthkohli):
Still here you have a medal | 2021-03-03T21:27:52 | {
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