Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to prove that $42|a^7-a$? Suppose we are given a number $a \in \mathbb{Z}$ prove that $42|a^7-a$.
I'm not too sure how to start any ideas?
| In fact, $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.
$a^7-a=a\left(a^6-1\right)=a\left(a^3-1\right)\left(a^3+1\right)$
$=a(a-1)\left(a^2+a+1\right)(a+1)\left(a^2-a+1\right)$
$a-1,a,a+1$ are three consecutive integers, so $2,3\mid a^7-a$.
Also by Fermat's Little theorem $7\mid a^7-a$. Therefore $2\cdot 3\cdot 7=42\mid a^7-a$.
$2^7-2=42\cdot 3$ and $3^7-3=42\cdot 52$. But $\gcd(3,52)=1$,
so it follows that $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sequence from generating function $\frac{1}{(1 - \frac{x}{3})^2}$ I know that for
$$ \frac{1}{1 - \frac{x}{3}} $$
sequence would be $a_n = \frac{1}{3^n}$ for $n \geq 0$ ($\sum_{n \geq 0} \frac{1}{3^n} x^n$ ).
How should I approach with that power of two?
| $$\frac{1}{(1-\frac{x}{3})^2}=\frac{1}{1-\frac{x}{3}}\cdot\frac{1}{1-\frac{x}{3}}$$
$$=\sum\limits_{n=0}^\infty(\frac{x}{3})^n\cdot\sum\limits_{n=0}^\infty(\frac{x}{3})^n $$
$$=(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots)(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots).$$
Now we want a single sum in the form of $\sum a_nx^n$, and so we're going to find all of the terms resulting from the multiplication above that correspond to $x^0, x^1, x^2,...$ and match those coefficients to $a_0, a_1, a_2,...$. For example, the only way to get a constant term is from multiplying the first terms of both sequences (1) together, so $a_0=1$. Then, there are two ways to get $x^1$, either from multiplying the 1 in the first sequence to $\frac{x}{3}$ in the second or vice versa, so we get $a_1=2\cdot\frac{1}{3}$. In general, when we add 1 to our index $n$, there is one more way to get an $x^n$ term. And so $a_n=(n+1)\cdot(\frac{1}{3})^n$, for $n\ge0$. Thus, your series would be:
$$\sum\limits_{n=0}^\infty (n+1)\frac{x^n}{3^n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits. I would like to solve that limit solved using only rules of algebra of limits.
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do not fully address my question.
A challenging limit problem for the level of student who knows that:
$$\begin{align*}
\lim\limits_{x\to +\infty} e^x&=+\infty\tag1\\
\lim\limits_{x\to -\infty} e^x&=0\tag2\\
\lim\limits_{x\to +\infty} \frac{e^x}{x^n}&=+\infty\tag3\\
\lim\limits_{x\to -\infty} x^ne^x&=0\tag4\\
\lim\limits_{x\to 0} \frac{e^x-1}{x}&=1\tag5
\end{align*}$$
| $$\lim_{x\to0} \dfrac{e^x-1-x}{x^2}= \lim_{x\to0} \dfrac{e^{-x}-1+x}{x^2}= \lim_{x\to0} \dfrac{e^x+e^{-x}-2}{2x^2}=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-e^{\frac{-x}{2}}}{x}\Bigg)^2$$
$$=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-1}{x}-\dfrac{e^{\frac{-x}{2}}-1}{x}\Bigg)^2=\dfrac{1}{2} \Bigg(\dfrac{1}{2}-\dfrac{-1}{2}\Bigg)^2=\dfrac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give me some hints? Thanks in advance!
| All of the other approaches here are based on guessing that the expressions under the cube root signs are perfect cubes.
Here's an approach that doesn't assume that.
Let $$a = \sqrt[3]{7 + \sqrt{50}} = \sqrt[3]{7 + 5\sqrt{2}}, \quad b = \sqrt[3]{7 - 5\sqrt{2}}.$$
We are trying to evaluate $s = a + b$. We have
$$a^3 = 7 - 5\sqrt{2}, \quad b^3 = 7 + 5\sqrt{2}.$$
So
$$a^3 + b^3 = 14, \quad a^3 b^3 = -1.$$
Since $a$ and $b$ are real, the second equation implies $ab = -1$. Now we find
$$14 = a^3 + b^3 = (a+b)^3 - 3ab(a + b) = s^3 +3s.$$
This cubic equation has only one real solution, $s = 2$, which can be found using the rational root theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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The number of ways of writing an integer as a sum of two squares Given an integer $m=pq$, where $p,q$ are both primes such that $p\equiv 1 \pmod{4}, q\equiv 1 \pmod{4}$.
It is known that $p$ can be written as a sum of two squares (of positive integers) in a unique way, and the same for $q$. Prove that $m$ can be written as a sum of two squares (of positive integers) in exactly two distinct ways.
Attempt
Notice that for positive integers $u,v,A,B$, we have
\begin{align*}(u^2+v^2)(A^2+B^2)&=(uA+vB)^2+(vA-uB)^2=(vA+uB)^2+(uA-vB)^2\end{align*}
Therefore if $p=a^2+b^2,q=c^2+d^2$, then
\begin{align*}m&=(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(bc-ad)^2=(bc+ad)^2+(ac-bd)^2\end{align*}
meaning that $m$ can be written as a sum of two squares (of positive integers) in at least two distinct ways.
How do I prove that there doesn't exist a third way of writing $m$ as a sum of two squares (of positive integers)?
| I think we can always translate everything to the basics operations but it will take a huge number of pages! Fortunately, your question has an elementary answer. We denote by $S_n$ the set of representation of the integer $n$ as sum of squares, we have $(a,b)\in S_n$ if and only if $0\leq b<a$ and $ a^2+b^2=n$;
Given two primes $p$ and $q$ let:
$$ S_p=\{(a_p,b_p)\}\\
S_q=\{(a_q,b_q)\}
$$
Now we want to prove that $S_{pq}$ contains exactly two elements, or in other words we want to prove that the following equations on unknown $(a,b),(c,d)$ has exactly one solution :
*
*$a>b>0$ and $c>d> 0$ (it's clear that neither $d=0$ or $b=0$ occurs)
*$a^2+b^2=c^2+d^2=pq$
*$a>c$ ( different elements)
The equation in the middle is equivalent to:
$$(a-c)(a+c)=(d-b)(d+b)$$
this implies the existence of integers $x,y,z,t$ pairwise coprime such that:
$$\begin{align} a-c&=&2xy\\ a+c&=&2zt\\d-b&=&2xz\\d+b&=&2yt \end{align}$$
Note that the existence is not hard for example $x=gcd\big(\frac{a-c}{2},\frac{d+b}{2}\big),y=gcd\big(\frac{a-c}{2},\frac{d-b}{2}\big)\cdots$, so this will justify also that they are pairwise corprime.
The result is the fact that:
$$pq=(x^2+t^2)(y^2+z^2)$$
obviously $xyzt\neq 0$ therefore $\{p,q\}=\{x^2+t^2,y^2+z^2\}$ we have two cases:
*
*$p=x^2+t^2$ and $q=y^2+z^2$, because $a+c>d-b$ we have $t>x$, and $a+c > d+b$ implies $z>y$ so $x=b_p,t=a_p,y=b_q, z=a_q$ so $a=a_pa_q+b_qb_p,b=a_pb_q-b_pa_q,c=a_pa_q-b_qb_p, d=a_pb_q+b_pa_q$
*$q=x^2+t^2$ and $p=y^2+z^2$ because of the symetry it gives different $x,y,z,t$ but the same $a,b,c,d$ as the first case.
So there is only one solution to the equations hence $S_{pq}$ contains exactly two elements.
| {
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"url": "https://math.stackexchange.com/questions/1181336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $a_{n+1} - a_n = n^2$ using generating functions The Full Question
Using the method of generating functions, solve $a_{n+1} - a_n = n^2$ where $a_0 = 1$
My Research
Scanned the website for similar answers, reviewed the following links:
Solve the following recurrences using generating functions.
can we use generating functions to solve the recurrence relation $a_n = a_{n-1} + a_{n-2}$, $a_1=1$, $a_2=2$?
Solving a recurrence equation using generating functions
Using Generating Functions to Solve Recursions
Solve the following recursive relation by using generating functions
Solve a recursion using generating functions?
Solving recurrences using generating functions
None of them would make my question a duplicate and none of them really addressed what I was having difficulty with.
My Work
If we multiply each side by $x^{n+1}$ and sum each of the possible cases $a_0,a_1,\dots$
we have:
$$\sum\limits_{j=0}^{\infty}{a_{n+1}x^{n+1}} - x\sum\limits_{j=0}^{\infty}a_nx^n = \sum\limits_{j=0}^{\infty}n^2x^{n+1}$$
If we let $f(x) = \sum\limits_{j=0}^{\infty}a_nx^n$ our equation transforms into:
$$f(x)-a_0 -xf(x) = \sum\limits_{j=0}^{\infty}n^2x^{n+1}$$
Factoring $f(x)$ and observing the LHS is a well known generating functions and using the fact that $a_0 = 1$, we get:
$$f(x)(1-x) = \frac{x(1+x)}{(1-x)^3} + 1 \implies f(x) = \frac{x(1+x)}{(1-x)^4} + \frac{1}{1-x}$$
We now need the coefficients of $x^n$ of these two functions we are adding. The right function it is well known that is has coefficients of $1$. We need partial fraction decomposition to obtain the coefficients of the left hand side.
$$\frac{x(1+x)}{(1-x)^4}= \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{(1-x)^3}+\frac{D}{(1-x)^4}$$
$\implies x(1+x)= A(1-x)^3+B(1-x)^2+C(1-x)+D$
Expanding this, we have
$Ax^3+3Ax^2 - 3Ax + A + Bx^2 -2Bx + B +C -Cx +D$
Which gives us the system of equations:
1) $A+B+C+D = 0$
2) $-3A-2B-C =1$
3) $3A+B = 1$
4) $A = 0$
5) $D = 0$
$3(0)+B=1 \implies B=1$ Using equations 3&4
$-3(0) -2 - C=1\implies C=3$ Using equations Using equations 2&1 and the result above
$0+1+3+0 = 0$ plugging in our results to equation 1
that is obviously not correct. Where did I make my mistake. I feel I'm very close and must have messed up the partial fraction decomposition somewhere, but maybe the mistake is somewhere else. Can anyone find where I went wrong?
| Define the generating function $A(z) = \sum_{n \ge 0} a_n z^n$. Take the recurrence, multiply by $z^n$, sum over $n \ge 0$:
$$
\sum_{n \ge 0} a_{n + 1} z^n - \sum_{n \ge 0} a_n z^n = \sum_{n \ge 0} n^2 z^n
$$
Recognize some sums, note that:
$\begin{align}
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2} \\
\sum_{n \ge 0} n^2 z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{z}{(1 - z)^2} \\
&= \frac{z + z^2}{(1 - z)^3} \\
\end{align}$
$$
\frac{A(z) - a_0}{z} - A(z) = \frac{z + z^2}{(1 - z)^3}
$$
Plug in $a_0 = 1$, solve for $A(z)$, split into partial fractions:
$\begin{align}
A(z)
&= \frac{1 - 3 z + 4 z^2}{1 - 4 z + 6 z^2 - 4 z^3 + z^4} \\
&= \frac{4}{(1 - z)^2}
+ \frac{5}{(1 - z)^3}
+ \frac{2}{(1 - z)^4}
\end{align}$
Consider that:
$$
(1 - z)^{-r}
= \sum_{n \ge 0} (-1)^n \binom{-r}{n} z^n
= \sum_{n \ge 0} \binom{r + n - 1}{r - 1} z^n
$$
Note that $\binom{n + r - 1}{r - 1}$ is a polynomial of degree $r - 1$ in $n$.
Pick the coefficient of $z^n$ of $A(z)$:
$\begin{align}
a_n
&= 4 \binom{n + 1}{1} + 5 \binom{n + 2}{2} + 2 \binom{n + 3}{3} \\
&= \frac{2 n^3 + 27 n^2 + 91 n + 66}{6}
\end{align}$
| {
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"url": "https://math.stackexchange.com/questions/1181510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show $\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$ How to show
$$\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$$
| Stahl's approach basically works. A walk-through solution is also possible: Notice that the digamma function $\psi(z)$ satisfies
$$ \psi(z) = -\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right). $$
Consequently
\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(24n+5)(24n+19)}
&= \frac{1}{14} \sum_{n=0}^{\infty} \left( \frac{1}{24n+5} - \frac{1}{24n+19} \right) \\
&= \frac{1}{14 \cdot 24} \left( \psi\left( \frac{19}{24} \right) - \psi\left( \frac{5}{24} \right) \right).
\end{align*}
Now applying the Euler's reflection formula shows that
$$ \psi(z) - \psi(1-z) = \frac{d}{dz}\log ( \Gamma(z)\Gamma(1-z) ) = \frac{d}{dz}\log\left( \frac{\pi}{\sin \pi z} \right) = -\pi \cot(\pi z). $$
Plugging $z = 19/24$ gives the same answer as Jack D'Aurizio. (Basically, one solution implies the other so both solutions are essentially equivalent.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$
Here's what I did to solve this exercise:
$$\frac{a+b}2 \ge \sqrt{ab} \implies \frac12 \ge \sqrt{ab} \implies \frac14\ge ab \implies ab\le\frac14$$
$$\left(1+\frac1a\right)\left(1+\frac1b\right) = 1+\frac1b+\frac1a+\frac1{ab} \ge 9 \Rightarrow \frac{a+b+1}{ab}\ge8 \Rightarrow \frac2{ab}\ge8 \Rightarrow 2\le8ab \Rightarrow ab\le\frac14$$
I came to a known point, but I'm not sure that this is the right form. If there is any other more clear proof please show it to me.
| If you're only allowed to use $x+y\geq 2\sqrt{xy}$ (for $x,y\geq 0$), then
$$
1=a+b\geq2\sqrt{ab}\implies ab\leq\frac{1}{4}
$$
and
$$
(1+1/a)(1+1/b)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}\geq 1+\frac{2}{\sqrt{ab}}+\frac{1}{ab}\geq1+\frac{2}{\sqrt{1/4}}+\frac{1}{1/4}=9.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Strange things happening with definite integration by substitution There are two ways solving the definite integration by substitution. One is to solve the indefinite integration first, then use the evaluation theorem. Another is to use the substitution rule for definite integral. However, when I solved some problems, say calculating the following integral:$$\int_0^{2\pi}\frac{dx}{(2+\cos (x)) (3+\cos (x))},$$
the answer is 0 by the first method. But, the graph of the integrand implies that this is impossible. So, what is the problem?
| You got the wrong answer.
Use the method similar to partial fractions,
$$\int \frac{dx}{(2+\cos{x})(3+\cos{x})}
=\int \left(\frac{1}{2+\cos{x}}-\frac{1}{3+\cos{x}}\right)dx.$$
Draw the following figure,
we have
$\cos{\frac{x}{2}}=\frac{1}{\sqrt{1+u^2}}$ and
$\sin{\frac{x}{2}}=\frac{u}{\sqrt{1+u^2}}$.
Then $\sin{x}=2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=\frac{2u}{1+u^2}$
and $\cos{x}=\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}=\frac{1-u^2}{1+u^2}$.
In addition,
$u=\tan{\frac{x}{2}}$ and $x=2\arctan{u}$
and $dx=\frac{2}{1+u^2}du$.
Therefore,
$$\int \frac{dx}{(2+\cos{x})}
=\int \frac{\frac{2}{1+u^2}}{2+\frac{1-u^2}{1+u^2}}du
=\int \frac{2}{3+u^2}du
=\frac{2\arctan{\frac{u}{\sqrt{3}}}}{\sqrt{3}}+C
=\frac{2\arctan{\frac{\tan{\frac{x}{2}}}{\sqrt{3}}}}{\sqrt{3}}+C$$
Similarly,
$$\int \frac{dx}{(3+\cos{x})}
=\frac{\arctan{\frac{\tan{\frac{x}{2}}}{\sqrt{2}}}}{\sqrt{2}}+C$$
So
$$\int_{0}^{2\pi} \frac{dx}{(2+\cos{x})(3+\cos{x})}
=\left[\frac{2\arctan{\frac{\tan{\frac{x}{2}}}{\sqrt{3}}}}{\sqrt{3}}
-\frac{\arctan{\frac{\tan{\frac{x}{2}}}{\sqrt{2}}}}{\sqrt{2}}\right]_{0}^{2\pi}
=\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{2}}\right)
\neq 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\sqrt{2}\cos\theta = 1$ and $\sqrt{2}\sin\theta = 1$, so the angle $\theta$ is $\frac{\pi}{4}$.
However, if we do $k=0$, then we get that one of the roots is $1$, which is obviously not true since $1^4 \neq 1 + i$. The professor says that the solutions are: $k=1: \cos(\frac{9\pi}{16}) + i\sin(\frac{9\pi}{16})$, $k=2: \cos(\frac{17\pi}{16}) + i\sin(\frac{17\pi}{16})$, and $k=3: \cos(\frac{25\pi}{16}) + i\sin(\frac{25\pi}{16})$. I plugged these into WolfRamAlpha and rose them to the $4$th power, but none of them return the form $1+i$.
What is incorrect about these steps?
| The $n$ $n$th roots of a nonzero complex number $z=r(\cos \theta+i\sin\theta)$ are given by
$$
w_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right)\right],\tag{1}
$$
where $k=0,1,2,\ldots,n-1$.
Problem: Find the four fourth roots of $z=1+i$.
Solution. We have, in this case, $r=\sqrt{2}$ and $\theta=\arg(z)=\pi/4$. Using $(1)$ with $n=4$, we get
$$
w_k = \sqrt[8]{2}\left[\cos\left(\frac{\pi/4+2k\pi}{4}\right)+i\sin\left(\frac{\pi/4+2k\pi}{4}\right)\right],\quad k=0,1,2,3.
$$
Thus, we have the following:
*
*$k=0,\quad w_0=\sqrt[8]{2}\left[\cos\frac{\pi}{16}+i\sin\frac{\pi}{16}\right]$
*$k=1,\quad w_1=\sqrt[8]{2}\left[\cos\frac{9\pi}{16}+i\sin\frac{9\pi}{16}\right]$
*$k=2,\quad w_2=\sqrt[8]{2}\left[\cos\frac{17\pi}{16}+i\sin\frac{17\pi}{16}\right]$
*$k=3,\quad w_3=\sqrt[8]{2}\left[\cos\frac{25\pi}{16}+i\sin\frac{25\pi}{16}\right]$
| {
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"url": "https://math.stackexchange.com/questions/1191759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: integral $$\frac{1 + t^2}{2(t + 2)(t + 3)}$$ or $$\frac{1 + t^2}{12 + 2t^2 + 10t}.$$
Could someone give me a hint?
Many thanks.
| Use the universal substitution $t=\tan\frac{x}{2}$, we have
$$\sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad dx=\frac{2}{1+t^2}dt$$
So
$$\frac{1}{5\cos x+\sin x+7}dx=\frac{1}{\frac{5(1-t^2)+2t}{1+t^2}+7}\frac{2dt}{1+t^2}=\frac{2}{2t^2+2t+12}dt=\frac{1}{t^2+t+6}dt=\frac{1}{\left(t+\frac{1}{2}\right)^2+\frac{23}{4}}dt=\frac{4}{23}\cdot\frac{1}{\frac{4}{23}\left(t+\frac{1}{2}\right)^2+1}dt$$
Now recall that
$$\frac{d}{dt}\arctan(a(t+b))=\frac{a}{a^2(t+b)^2+1}$$
Thus
$$\frac{d}{dt}\arctan\left(\frac{2}{\sqrt{23}}\left(t+\frac{1}{2}\right)\right)=\frac{2}{\sqrt{23}}\cdot\frac{1}{\frac{4}{23}\left(t+\frac{1}{2}\right)^2+1}$$
It follows that
$$\int \frac{4}{23}\cdot\frac{1}{\frac{4}{23}\left(t+\frac{1}{2}\right)^2+1}dt=\frac{2}{\sqrt{23}}\arctan\left(\frac{2}{\sqrt{23}}\left(t+\frac{1}{2}\right)\right)+C$$
Finally write your answer in terms of $x$:
$$\int\frac{1}{5\cos x+\sin x+7}dx=\frac{2}{\sqrt{23}}\arctan\left(\frac{2}{\sqrt{23}}\left(\tan\frac{x}{2}+\frac{1}{2}\right)\right)+C=\frac{2}{\sqrt{23}}\arctan\left(\frac{1}{\sqrt{23}}\left(2\tan\frac{x}{2}+1\right)\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all nonnegative integers
Determine all nonnegative integers $x$ and $y$ so that $$3^x + 7^y$$
is a perfect
square and $y$ is even.
Without trial-and-error of course.
$$3^x + 7^y = a^2$$ For some integer $a$.
$$\implies a = \sqrt{3^x + 7^y}$$
But is there really a way?
| Writing $y = 2z$ we have $3^x = (a + 7^z)(a - 7^z )$. Now $a + 7^x$ and $a - 7^x$ must be two powers of $3$, say $a + 7^x = 3^d$, $a - 7^x = 3^e$.
But $3^d - 3^e = 2 \times 7^x \equiv 2 \mod 3$ so we must have $e = 0$. Then
$a = 1 + 7^x$, and $3^d = a + 7^x = 1 + 2 \times 7^x$. One solution is $x = 0$, $d = 1$, corresponding to $3^1 + 7^0 = 2^2$. Otherwise, note that $3^d \equiv
1 \mod 7$ iff $d$ is divisible by $6$, but if $d$ is divisible by $6$ then $3^d \equiv 1 \mod 4$ as well, so $(3^d - 1)/2$ is divisible by $2$ and can't be a power of $7$. So there are no other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
show this $\gcd{(a_{1}a_{2}-b_{1}b_{2},a_{1}b_{2}+a_{2}b_{1})}=1$ let $a_{1},b_{1},a_{2},b_{2}$ be no-zero integer,and such $\gcd{(a_{i},b_{i})}=1,|a_{i}|\neq 1,|b_{i}|\neq 1,i=1,2$
show that
$$\gcd{(a_{1}a_{2}-b_{1}b_{2},a_{1}b_{2}+a_{2}b_{1})}=1$$
My attemp
Assmue that prime $p$ such
$$p|(a_{1}a_{2}-b_{1}b_{2}),p|(a_{1}b_{2}+a_{2}b_{1})$$
It follows that
$$p|[(a_{1}b_{2}+a_{2}b_{1})a_{2}-(a_{1}a_{2}-b_{1}b_{2})b_{2}]$$
$$p|b_{1}(a^2_{2}+b^2_{2})$$
What approaches do you think I could take to solving the next step?
| This is not true. For counterexample: $(3+4i)\overbrace{(11+2i)}^{(3-4i)(1+2i)}=25+50i$.
That is, take $\gcd(3,4)=1$ and $\gcd(11,2)=1$. Then
$$
\gcd(3\cdot11-4\cdot2,3\cdot2+4\cdot11)=\gcd(25,50)=25
$$
Another counterexample: $(5+2i)\overbrace{(7+3i)}^{(5-2i)(1+i)}=29+29i$.
That is, take $\gcd(5,2)=1$ and $\gcd(7,3)=1$. Then
$$
\gcd(5\cdot7-2\cdot3,5\cdot3+2\cdot7)=\gcd(29,29)=29
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that the series $\sum_{1}^{\infty}\frac{k}{(k+1)(k+2)(k+3)}$ converges and find its limit I try to split the summand into differences, but that seems to be a futile way in our case right here, because the numerator is $k$, instead of a given number.
A closely-related series, say $\sum_{1}^{\infty}\frac{2}{(k+1)(k+2)(k+3)}$, can however be directly tackled by writing its partial sum as
$$\sum_{1}^{n}\frac{1}{(k+1)(k+2)} - \frac{1}{(k+2)(k+3)}.$$
| Since:
$$-\frac{1}{k+1}+\frac{4}{k+2}-\frac{3}{k+3}=\frac{2k}{(k+1)(k+2)(k+3)}$$
we have:
$$S_N=\sum_{n=1}^{N}\frac{k}{(k+1)(k+2)(k+3)}=\sum_{k=1}^{N}\left(\frac{-1/2}{k+1}+\frac{2}{k+2}+\frac{-3/2}{k+3}\right) $$
and since $(-1/2)+(2)+(-3/2)=0$ we have:
$$ \lim_{N\to +\infty}S_N = \frac{-1/2}{1+1}+\frac{2}{1+2}+\frac{-1/2}{2+1}=\color{red}{\frac{1}{4}}.$$
The convergence of the original series is trivial since:
$$0\leq \frac{k}{(k+1)(k+2)(k+3)}\leq\frac{1}{(k+2)(k+3)}$$
and:
$$\sum_{k\geq 1}\frac{1}{(k+2)(k+3)}=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Are there any two positive integers such that the square root of the sum of their squares give a perfect square. Are there two positive integers $$ \text{{ (x,y) | $ \sqrt{x^2 + y^2} = z $ }}$$
where $ z$ is a perfect square.
| If you're looking for integers such that $\sqrt{x^2+y^2}=z$, then there's plenty, for example, $3,4,5$ or $12,5,13$.
If you're looking for integers such that $\sqrt{a^2+b^2}=c^2$, then take any solutions to the first equation and multiply everything by $z$, e.g. take $3,4,5$ and multiply by $5$.
Then, since
$$\sqrt{x^2+y^2}=z,$$
it follows that
$$\sqrt{(xz)^2+(yz)^2}=\sqrt{z^2(x^2+y^2)}$$
$$=z\sqrt{x^2+y^2}=z^2$$
For example, $15^2+20^2=25^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{5-12i}$ by square root definition I KNOW it can be solved by the trig formula, but I want to solve it by the square root definition, so please don't just post an alternative way to do it.
By the square root definition:
$$z = 5-12i$$
$$\sqrt{z} = w\implies w^2 = z$$
So if I suppose $w = a+bi$ we have:
$$w^2 = z \implies (a+bi)^2 = 5-12i\implies\\a^2-b^2+2abi = 5-12i\implies \\5 = a^2-b^2\\-12 = 2ab\implies$$
$$b^4-5b^2-36 = 0\implies b = \pm \sqrt{-4} = \pm 2i, b = \pm 3$$
Then I get $4$ solutions:
For $b = \pm 2i$ I get $a = \pm 3i$ then $\sqrt{z}$ is
$$\pm 2 + 3i$$
For $b = \pm 3$ I get $a = \pm 2$ and $\sqrt{z}$ is
$$2\pm 3i$$
But two of these, when squared, aren't $z$. What am I doing wrong?
| An easier way is to solve:
$$\begin{cases}a^2-b^2=5\ \ (1)\\ 2ab=-12\ \ (2)\\a^2+b^2=|5-12i|=13\ \ (3),\end{cases}$$
because you don't have to solve a equation of degree $4$.
Therefore, by $(1)$
$$a^2=b^2+5$$
By $(3)$
$$2b^2=8\implies b^2=4\implies b=\pm2$$
and by $(2)$
$$ab=-6\implies a=\mp3$$
Therefore $$\sqrt{5-12i}=\mp3\pm 2i$$
| {
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"url": "https://math.stackexchange.com/questions/1199681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Determining convergence of $\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$ I have the following infinite series:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$$
I want to examine its convergence. First thing that came to my mind was "unfolding" $e^{\frac{1}{n}}$ and see what will happen:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}=\sum_{n=1}^{+\infty}(\sum_{k=0}^{+\infty}\frac{1}{n^kk!}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}=\sum_{n=1}^{+\infty}(\sum_{k=3}^{+\infty}\frac{1}{n^kk!})^{\frac{1}{2}}$$
Now it kinda "looks" convergent but I'm clueless about how to prove it. Any hints?
| In general,
for $m \ge 1$,
let
$d_n
=e^{1/n}-\sum_{k=0}^{m-1} \frac1{n^k k!}
$.
Then
$d_n
=\sum_{k=m}^{\infty} \frac1{n^k k!}
=\frac1{m!}\sum_{k=m}^{\infty} \frac{m!}{n^k k!}
$.
Therefore,
$d_n > \frac1{m!n^m}$
and,
if $n \ge 2$,
$d_n
<\frac1{m!}\sum_{k=m}^{\infty} \frac{1}{n^k }
=\frac1{m!n^m(1-1/n)}
\le\frac{2}{m!n^m}
$.
For $m=3$ (your case),
$\frac1{6n^3}
< d_n
<\frac1{3n^3}
$,
so
$\frac1{\sqrt{6}n^{3/2}}
< d_n^{1/2}
<\frac1{\sqrt{3}n^{3/2}}
$,
and the sum of these converges.
Note that
if you use the cube root,
$\frac1{\sqrt[3]{6}n}
< d_n^{1/3}
<\frac1{\sqrt[3]{3}n}
$,
and the sum of these diverges.
In general,
$\sum d_n^r$
converges for
$r < \frac1{m}$
and diverges for
$r \ge \frac1{m}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} + 3
\\ -3 &= \sqrt{36x^2-36x^4}
\\ (-3)^2 &= (\sqrt{36x^2-36x^4})^2
\\ 0 &= -36x^4 + 36x^2 - 9
\\ \frac{0}{9} &= \frac{9(-4x^4 + 4x^2 - 1)}{9}
\\ 0 &= -4x^4 + 4x^2 - 1
\\ \end{align*}
$$
I've considered the rational root theorem at this point, through which I find possible roots to be $x = \{ \pm1, \pm\frac{1}{2}, \pm\frac{1}{4} \}$.
Clearly I've made an error, as the actual root is $x = -\dfrac{1}{\sqrt{2}}$.
| When you find yourself in such a situation, a good trick is to insert the root (any number will work but the root will often make the calculations easy) at each step and see when the value change.
Personally I think you are doing to much at once in the last step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Number of ways to set 3 queens to attack each other
We play chess and want to set 3 queens to attack each other. How many ways we can do it?
I know to solve this problem when I have 2 queens. I see the chess board as 4 squares, from an outer square (the 28 squares on the edges and corners) to the inner square, the 4 squares in the center of the board.
| $3$ queens on a chessboard attack each other if they are in the vertices of a right and isosceles triangle. There are three possibilities, according to the hypotenuse of such a triangle lying on a diagonal, a row or a column of the chessboard. Counting them is not so difficult:
*
*hypotenuse being on the $a7-b8$ diagonal: $2$ possibilities (third queen on $a8$ or $b7$);
*hypotenuse being on the $a6-c8$ diagonal: $2\cdot\binom{3}{2}$ possibilities;
*$\ldots$
*hypotenuse being on the $a1-h8$ diagonal: $2\cdot\binom{8}{2}$ possibilities;
*htpotenuse being on the $b1-h7$ diagonal: same number of possibilities as $a2-g8$;
*$\ldots$
So if the hypotenuse is along a diagonal we have
$$ 2\left(2\binom{2}{2}+2\binom{3}{2}+\ldots+2\binom{7}{2}+2\binom{8}{2}+2\binom{7}{2}+\ldots+2\binom{2}{2}\right) = 4\left(2\binom{9}{3}-\binom{8}{2}\right) = 560 $$
possibilities. Assume now that the hypotenuse is along a row: the endpoints of the hypotenuse must have the same colour, so we have:
*
*along the $A$ or $H$ line : $2\binom{4}{2}$ possibilities;
*along the $B$ or $G$ line : $2\binom{4}{2}+2\cdot 3$ possibilities;
*along the $C$ or $F$ line : $2\binom{4}{2}+2\cdot (3+2)$ possibilities;
*along the $D$ or $E$ line : $2\binom{4}{2}+2\cdot (3+2+1)$ possibilities;
so if the hypotenuse is along a row we have
$$ 16\binom{4}{2}+4\cdot 3+4\cdot(3+2)+4\cdot(3+2+1) = 152 $$
possibilities and the total number of ways to place three queens on a chessboard such that they attack each other is given by:
$$ 560+2\cdot 152 = \color{red}{864}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Elements of order $10$ in $S_6$
Find the number of elements of order 10 in $S_6$
I conjecture that there are no elements of order 10 in $S_{6}$.
If there were a $\sigma \in S_6$ s.t. $|\sigma| = 10$, then $\sigma$ is a 10 cycle, or $\sigma$ is the product of two disjoint 2 and 5-cycle. Since there is no 10-cycle or 5-cycle in $S_6$, then $\sigma$ can't be in $S_6$
| The order of a permutation is the LCM of its cycle lengths. Therefore we are interested in all possible partitions of $6$, and the corresponding LCMs:
$$
\begin{align*}
6 && 6 \\
5+1 && 5 \\
4+2 && 4 \\
4+1+1 && 4 \\
3+3 && 3 \\
3+2+1 && 6 \\
3+1+1+1 && 3 \\
2+2+2 && 2 \\
2+2+1+1 && 2 \\
2+1+1+1+1 && 2 \\
1+1+1+1+1+1 && 1
\end{align*}
$$
Therefore the possible orders are $1,2,3,4,5,6$.
There is no real need to look at the entire list in this case. If the LCM of the cycle sizes is 10, then one of the cycle lengths needs to be a multiple of 5. The only multiply of 5 which is at most 6 is 5, and the only corresponding partition is 5+1, with LCM 5.
Similarly, let us show that in $S_{17}$ there is no permutation of order 65. Since $65 = 13\times 5$, we would need to have a cycle of length 13, which leaves no room for a cycle whose length is a multiple of 5.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Taylor Series for $\frac{1}{1+e^z}$ and radius of convergence I have done some manipulation and got that $$\frac{1}{1+e^z} = \sum_{n=0}^\infty \frac{n!}{n!+z^n}$$ by the fact that:
$$\frac{1}{1+e^z}= \frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}=\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{2}}+\ldots = \frac{1}{2}+\frac{1}{1+z}+\frac{2!}{2!+z^2}+\frac{3!}{3!+z^3}+\ldots$$ $$= \sum_{n=0}^\infty\frac{n!}{n!+z^n}$$
Assuming I did the above right, I am having trouble finding the radius of convergence of the Taylor series given above. I tried the ratio test but got stuck.
Edit: I just realized I did this completely wrong, thinking that I was taking $$\sum\frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}$$.
Could anyone help me? My question wants me to compute the first four terms of the taylor series for $\frac{1}{1+e^z}$ and find the radius of convergence. Perhaps they do not want me to actually find the explicit form?
| I assume you want the Maclaurin series, i.e. the Taylor series about $0$.
Write $$1 + e^z = 2 (1 + Q(z))$$
where
$$Q(z) = \dfrac{z}{2\cdot 1!} + \dfrac{z^2}{2\cdot 2!} + \dfrac{z^3}{2\cdot 3!} + \ldots $$
So $$ \dfrac{1}{1+e^z} = \dfrac{1}{2(1+Q(z))} = \dfrac{1}{2} \left( 1 - Q(z) + Q(z)^2 - Q(z)^3 + \ldots \right) $$
For $n \ge 1$, the $z^n$ term in the Taylor series comes from the $Q(z)^j$ terms for $j = 1$ to $n$. Thus the series is $\sum_{n=0}^\infty a_n z^n$ where
$$ \eqalign{
a_0 &= \dfrac{1}{2}\cr
a_1 &= \dfrac{1}{2} \left( -\dfrac{1}{2\cdot 1!}\right) = -\dfrac{1}{4}\cr
a_2 &= \dfrac{1}{2} \left( - \dfrac{1}{2 \cdot 2!} + \dfrac{1}{(2 \cdot 1!)^2} \right) = 0\cr
a_3 &= \dfrac{1}{2} \left( -\dfrac{1}{2 \cdot 3!} + \dfrac{2}{(2\cdot 1!)(2\cdot 2!)} - \dfrac{1}{(2 \cdot 1!)^3}\right) = \dfrac{1}{48}\cr
\ldots}$$
The radius of convergence is the radius of the largest disk around $0$ in which $1/(1+e^z)$ is analytic, i.e. the distance to the closest points where $1 + e^z = 0$. These points are $\pm i\pi$, so the radius of convergence is $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Optimisation to solve for trigonometric expression? I have a question that requires the use of optimisation to solve for the following expression:
$$\cos ec{(\cos^{-1}{(-\frac{\sqrt{3}}{2})}+\sin^{-1}{(-\frac{\sqrt{3}}{2})})}$$
I'm a bit baffled, as I'm not sure whether this refers to finding the value of cosec or finding the value of cos(ec)(cos^(-1)...). No further clarity is given.
Without using optimisation, the expression can just be evaluated at face-value to:
$$\csc{(\frac{5\pi}{6}+\frac{4\pi}{3})} = \csc{(2\pi+\frac{\pi}{6})} = \frac{1}{\sin{\frac{\pi}{6}}} = \frac{1}{\frac{1}{2}} = 2$$
Any suggestions would be appreciated!
| Since $\left|-\dfrac{\sqrt{3}}{2}\right|\le 1$, we have that $\cos^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)+\sin^{-1}\left( -\dfrac{\sqrt{3}}{2}\right) =\dfrac{\pi}{2} $.
So $\operatorname{cosec} \left( \cos^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)+\sin^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)\right)=\operatorname{cosec} \left(\dfrac{\pi}{2}\right) =\dfrac{1}{\sin \left(\dfrac{\pi}{2}\right)}=1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $
Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$
I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another question when the $n^2$ is in the numerator, but no significant progress so far.
Any help will be appreciated.
Thanks!
| the taylor series of $$y=\frac{1}{x}\log(\frac{1}{1-x})$$ is
$$y=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+...\frac{x^{n-1}}{n}$$
$$\int_{0}^{0.5}ydx=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\frac{x^4}{4^2}+...\frac{x^n}{n^2}=\sum_{n=1}^{\infty }\frac{(.5)^n}{n^2}=\sum_{n=1}^{\infty }\frac{1}{2^nn^2}$$
$$\sum_{n=1}^{\infty }\frac{1}{2^nn^2}=\int_{0}^{0.5}\frac{1}{x}\log\frac{1}{1-x}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207908",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Show that $r_k^n/n \le \binom{kn}{n} < r_k^n$ where $r_k = \dfrac{k^k}{(k-1)^{k-1}}$ Show that
for $n \ge 2$,
$\dfrac{r_k^n}{n+1} \le \binom{kn}{n} < r_k^n$ where $r_k = \frac{k^k}{(k-1)^{k-1}}$.
This is a generalization of
How to prove through induction
which asks for a proof that
$\binom{2n}n<4^n$.
I wondered what was true for
$\binom{kn}{n}$
and this is what I came up with.
Note that
$r_k
=k(1+\frac1{k-1})^{k-1}
\sim ek
$
for large $k$.
The key is to get
bounds for
$\dfrac{\binom{k(n+1)}{n+1}}{\binom{kn}{n}}$
that are not too dependent on $n$.
The upper bound I get is not,
the lower bound is,
but telescopes nicely.
Note:
The algebra here
is uncomfortably messy.
I would definitely like
a simpler proof,
perhaps depending on
Stirling's theorem.
Here goes.
$\begin{array}\\
s_k(n)
&=\dfrac{\binom{k(n+1)}{n+1}}{\binom{kn}{n}}\\
&=\dfrac{\frac{(kn+k)!}{(n+1)!(kn+k-n-1)!}}{\frac{(kn)!}{n!(kn-n)!}}\\
&=\dfrac{n!(kn-n)!(kn+k)!}{(n+1)!(kn)!(kn+k-n-1)!}\\
&=\dfrac{\prod_{j=0}^{n}(kn+k-j)}{(n+1)\prod_{j=0}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)\prod_{j=k}^{n}(kn+k-j)}{(n+1)\prod_{j=0}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)\prod_{j=0}^{n-k}(kn-j)}{(n+1)\prod_{j=0}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)}{(n+1)\prod_{j=n-k+1}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)}{(n+1)\prod_{j=0}^{k-2}(kn-(j+n-k+1)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)}{(n+1)\prod_{j=0}^{k-2}((k-1)n-(j-k+1)}\\
&=\dfrac{(kn+k)\prod_{j=1}^{k-1}(kn+k-j)}{(n+1)\prod_{j=0}^{k-2}((k-1)n+k-j-1)}\\
&=k\dfrac{\prod_{j=0}^{k-2}(kn+k-j-1)}{\prod_{j=0}^{k-2}((k-1)n+k-j-1)}\\
&=k\prod_{j=0}^{k-2}\dfrac{kn+k-j-1}{((k-1)n+k-j-1)}\\
&<k\left(\dfrac{k}{k-1}\right)^{k-1}\\
&=\dfrac{k^k}{(k-1)^{k-1}}\\
\end{array}
$
since,
if $1 \le a \le k-1$,
$\begin{array}\\
\dfrac{kn+a}{(k-1)n+a}-\dfrac{k}{k-1}
&=\dfrac{(kn+a)(k-1)-k((k-1)n+a)}{(k-1)((k-1)n+a)}\\
&=\dfrac{(k^2n-kn+ka-a)-(k^2n-kn+ak)}{(k-1)((k-1)n+a)}\\
&=\dfrac{-a}{(k-1)((k-1)n+a)}\\
&< 0\\
\end{array}
$
To get the lower bound,
$\begin{array}\\
\dfrac{kn+a}{(k-1)n+a}-\dfrac{k}{k-1}
&=\dfrac{-a}{(k-1)((k-1)n+a)}\\
&=\dfrac{-1}{(k-1)((k-1)n/a+1)}\\
&\ge\dfrac{-1}{(k-1)(n+1)}\\
&=-\dfrac{1}{(k-1)n+1}\\
\end{array}
$
or
$\begin{array}\\
\dfrac{kn+a}{(k-1)n+a}
&\ge\dfrac{k}{k-1}-\dfrac{1}{(k-1)n+1}\\
&=\dfrac{k((k-1)n+1)-(k-1)}{(k-1)((k-1)n+1)}\\
&=\dfrac{k(k-1)n+1}{(k-1)((k-1)n+1)}\\
&=\dfrac{k(k-1)n+1}{(k-1)^2n+k-1}\\
&>\dfrac{k(k-1)n}{(k-1)^2n+k-1}\\
&=\dfrac{kn}{(k-1)n+1}\\
&=\dfrac{k}{(k-1)+1/n}\\
&=\dfrac{k}{(k-1)(1+1/(n(k-1))}\\
&=\dfrac{k}{k-1}\dfrac1{1+1/(n(k-1))}\\
\end{array}
$
so that
$s_k(n)
\ge k\prod_{j=0}^{k-2}\dfrac{k}{k-1}\dfrac1{1+1/(n(k-1))}
= \dfrac{k^k}{(k-1)^{k-1}}\prod_{j=0}^{k-2}\dfrac1{1+1/(n(k-1))}
= \dfrac{k^k}{(k-1)^{k-1}}\dfrac1{(1+1/(n(k-1)))^{k-1}}
$.
In an earlier problem
I showed
(here: Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$)
that
if $0 < x < 1$
then
$(1+x/n)^n < \frac1{1-x}$.
Therefore
$(1+1/(n(k-1)))^{k-1}
<\frac1{1-1/n}
$
so that
$s_k(n)
>\dfrac{k^k}{(k-1)^{k-1}}(1-1/n)
=\dfrac{k^k}{(k-1)^{k-1}}\dfrac{n-1}{n}
=r_k\dfrac{n-1}{n}
$.
The products of
$\dfrac{n-1}{n}$
telescope
to give the lower bound stated.
| No one else has done anything,
so I thought I'd see what happens
when I follow my own suggestion
of using Stirling's formula.
Definitely easier
and, of course,
more precise.
Here we go.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{kn}{n}
&=\dfrac{(kn)!}{n!(kn-n)!}\\
&\sim \dfrac{\sqrt{2\pi kn}(kn/e)^{kn}}
{(\sqrt{2\pi n}(n/e)^n)(\sqrt{2\pi (kn-n)}((kn-n)/e)^{kn-n})}\\
&= \sqrt{\dfrac{2\pi kn}{(2\pi n)(2\pi (kn-n))}}\dfrac{(kn)^{kn}}
{((n)^n)(((kn-n))^{kn-n})}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}\dfrac{k^{kn}n^{kn}}
{(n)^n((n(k-1))^{nk-n)})}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}\dfrac{k^{kn}}
{(((k-1))^{kn-n})}\\
&= k^n\sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k}
{k-1}\right)^{n(k-1)}\\
&= k^n\sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k}
{k-1}\right)^{kn-n}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}k^n\left(\left(\dfrac{k}
{k-1}\right)^{k-1}\right)^{n}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k^k}
{(k-1)^{k-1}}\right)^{n}\\
\end{array}
$
This confirms my main result
of the ratio of consecutive
terms being
$\dfrac{k^k}
{(k-1)^{k-1}}
$,
with the multiplier of
$\sqrt{\dfrac{k}{2\pi n(k-1)}}$
being nicely between
$\dfrac1{n}$ and $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $1/(x^2+y^2)^a$. I'm curious about the convergence of the series:
\begin{align}
\sum_{x,y=1}^\infty \frac{1}{(x^2+y^2)^\alpha}\ ,\ \alpha \in \mathbb{N}
\end{align}
I'm wondering for what values of $\alpha$ this converges, and if so is there an analytic way to find the limit? I'm not really sure where to start, it's been a long time since I studied sequences and series.
| Let $N,M \in \Bbb N$. On the square $[N,N+1]\times [M,M+1]$,
$$\frac{1}{[(N+1)^2 + (M + 1)^2]^\alpha} \le \frac{1}{(x^2 + y^2)^\alpha} \le \frac{1}{(N^2 + M^2)^\alpha}.$$
Integrating the inequality over the square, we find
$$\frac{1}{[(N+1)^2 + (M+1)^2]^\alpha} \le \int_N^{N+1}\int_M^{M+1} \frac{dx\, dy}{(x^2 + y^2)^{\alpha}} \le \frac{1}{(N^2 + M^2)^\alpha}.$$
Summing from over all $N,M \ge 1$, we have
$$\sum_{N,M\ge 2} \frac{1}{(N^2 + M^2)^\alpha} \le \int_1^\infty \int_1^\infty \frac{dx\, dy}{(x^2 + y^2)^\alpha} \le \sum_{N,M\ge 1} \frac{1}{(N^2 + M^2)^\alpha}$$
Hence, the series $\sum_{x,y\ge 1} (x^2 + y^2)^{-\alpha}$ converges if and only if $\int_1^\infty \int_1^\infty (x^2 + y^2)^{-\alpha}dx\, dy$ converges. Using the transformation $x = u\tan \theta$, $y = u$, we obtain
$$\int_1^\infty \int_1^\infty \frac{dx\, dy}{(x^2 + y^2)^\alpha} = C_\alpha\int_1^\infty u^{1 - 2\alpha}\, du $$
where $C_\alpha = \int_{\pi/4}^{\pi/2} \cos^{2\alpha - 2}\theta\, d\theta$. Since $C_\alpha$ is finite (since $\alpha \ge 1$) and $\int_1^\infty u^{1 - 2\alpha}\, du$ converges if and only if $\alpha > 1$, it follows that $\sum_{x,y \ge 1} (x^2 + y^2)^{-\alpha}$ converges if and only if $\alpha > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$. To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.
$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.
$199x \equiv 1\pmod {10}$ has a solution $x =-1$ and $10x \equiv 1\pmod{199}$ has a solution $x = 20$.
Thus using Chinese Remainder Theorem we have,
$a \equiv 4(-1)199 + 29(20)10 \equiv 5004 \equiv 1024\pmod {1990}$.
Thus our $a = 1024$.
Is the solution correct? Is there any better proof?
| Yes, it is correct. More simply, by little Fermat $\,2^4\equiv 1\pmod 5\,$ and $\,2^{198}\equiv 1\pmod{199}\,$ so $\,4,198\mid 1980\,\Rightarrow\,2^{1980}\equiv 1\,\Rightarrow\,2^{1990}\equiv 2^{10}\,$ mod $5$ and mod $199$. Finally, combining these we conclude that $\,2^{1990}-2^{10}\,$ is divisible by $\,2,5,199,\,$ so also by their lcm = product $=1990$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Help evaluating $ \int \sqrt{{x}^{2} + 3} \; dx $ Can you help me evaluating the following indefinite integral?
$$
\int \sqrt{{x}^{2} + 3} \; dx
$$
Please, don't give a full solution, just some hint on which method to use...
** UPDATE **
Thank you very much to everybody for the useful comments and suggestions. I'm sorry for the delay with my reply, unfortunately do some mathematics as an hobby and often don't have time to work at it.
I tried to take Lucian suggestion on board and use trigonometric substitution as follows.
$$ x = \sqrt{3} \tan\theta $$
and
$$ \int \sqrt{{x}^{2} + 3} \; dx = \int \sqrt{{3\tan}^{2}\theta + 3} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{{3\sec}^{2}\theta} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{3}\sec\theta \; \sqrt{3}\sec^{2}\theta \; d\theta = 3 \int \sec^{3}\theta \; d\theta $$
which (according to common integral tables) is equal to
$$ 3 \left[ \frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \ln\left| \sec\theta + \tan\theta \right| + C \right] $$
Problem arises when I try to substitute back the variable from $ \theta $ to $ x $ because I know $ \tan\theta = \frac{x}{\sqrt{3}} $ but I don't know how to substitute back $ \sec\theta $, so basically I stopped here:
$$
\frac{3}{2}\sec\theta\;\frac{x}{\sqrt{3}} + \frac{3}{2}\ln\left| \sec\theta + \frac{x}{\sqrt{3}} \right| + 3C
$$
It looks close to the final answer but still not there...any suggestions?
| You have modified your question and now you want to determine the substitution for $\sec\theta$. So, here we go:
Given:
$$ \tan\theta = \frac{x}{\sqrt{3}} $$
knowing that $\tan\theta$ is positive for positive $x$, squaring both sides:
$$ tan^2\theta = \frac{x^2}{3} $$
$$ 1+ tan^2\theta = 1+ \frac{x^2}{3} $$
$$ 1+ \frac{sin^2\theta}{cos^2\theta} = 1+ \frac{x^2}{3} $$
$$ \frac{cos^2\theta + sin^2\theta}{cos^2\theta} = 1+ \frac{x^2}{3} $$
and since $ cos^2\theta + sin^2\theta = 1 $, we have:
$$ \frac{1}{cos^2\theta} = 1+ \frac{x^2}{3} $$
$$ \sec^2\theta = 1+ \frac{x^2}{3} $$
$$ \sec\theta = \sqrt{1+ \frac{x^2}{3}} $$
This is your substitution. One would argue that how did I remove squares from both sides, isn't that illegal? Well, not here - since I was the one who squared the sides in the first place, I know that the positive square root is the correct one, as the negative one was the result of me squaring the equation and creating an additional root of the equation. The positive root is consistent with the equation I started with, hence it's the one we are looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Simplify $\left(\sqrt{\left(\sqrt{2} - \frac{3}{2}\right)^2} - \sqrt[3]{\left(1 - \sqrt{2}\right)^3}\right)^2$ I was trying to solve this square root problem, but I seem not to understand some basics.
Here is the problem.
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2$$
The solution is as follows:
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2 = \Bigg(\frac{3}{2} - \sqrt{2} - 1 + \sqrt{2}\Bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \frac{1}{4}$$
Now, what I don't understand is how the left part of the problem becomes:
$$\frac{3}{2} - \sqrt{2}$$
Because I thought that $$\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2}$$
equals to $$\bigg(\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2\bigg)^{\frac{1}{2}}$$
Which becomes $$\sqrt{2} - \frac{3}{2}$$
But as you can see I'm wrong.
I think that there is a step involving absolute value that I oversee/don't understand.
So could you please explain by which property or rule of square root is this problem solved?
Thanks in advance
| A square root is always a non negative number so $\sqrt{x^2}=|x|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Bijection from $\{ (x,y) : 0 \leq x ,y \leq 1\}$ to $\left\lbrace (u,v) : u,v \geq 0, u+v \leq \frac{\pi}{2} \right\rbrace$ Suppose the set $M :=\{ (x,y) : 0 \leq x ,y \leq 1\}$. Now we define
$$ u := \arccos \sqrt{\frac{1-x^2}{1-x^2y^2}} \ \ \ \ \ \text{and} \ \ \ \ \ v:= \arccos \sqrt{\frac{1-y^2}{1-x^2 y^2}}.$$
How can I show, that this new coordinatens $(u,v)$ map each point $(x,y) \in M$ bijective to a point in the new set
$$ \widetilde{M} := \left\lbrace (u,v) : u,v \geq 0, u+v \leq \frac{\pi}{2} \right\rbrace?$$
I found out that we have $x = \frac{\sin u}{\cos v}$ and $y = \frac{\sin v }{\cos u}$ but I do not understand how to get the condition $u+v \leq \frac{\pi}{2} $.
| One can show that
$$
u := \arccos \sqrt{\frac{1-x^2}{1-x^2y^2}} \quad \text{and} \quad
v:= \arccos \sqrt{\frac{1-y^2}{1-x^2 y^2}} \quad \tag 1
$$
is a bijective mapping of the open rectangle
$M :=\{ (x,y) : 0 < x ,y < 1\}$ onto the open triangle
$\widetilde{M} := \left\lbrace (u,v) : u,v > 0, u+v < \frac{\pi}{2} \right\rbrace$.
The inverse mapping is
$$
x := \frac{\sin u}{\cos v} \quad \text{and} \quad y := \frac{\sin v }{\cos u} \tag 2
$$
as shown below.
Since
$$
(u(1, y), v(1, y)) = (\frac \pi 2, 0) \\
(u(x, 1), v(x, 1)) = (0, \frac \pi 2) \\
$$
the mapping cannot be extended one-to-one to the boundary of the square $M$.
Now let $(x, y) \in M$ and define $(u,v)$ by $(1)$. It is clear that
both $u$ and $v$ are in the range $(0, \frac \pi 2)$. To show that
$u + v < \frac \pi 2$, we compute
$$
\cos (u + v) = \cos u \cos v - \sin u \sin v =
\cos u \cos v - \sqrt{1 - \cos^2 u} \sqrt{1 - \cos^2 v} \\
= \sqrt{\frac{1-x^2}{1-x^2y^2}} \sqrt{\frac{1-y^2}{1-x^2 y^2}}
- \sqrt{\frac{x^2(1-y^2)}{1-x^2y^2}} \sqrt{\frac{y^2(1-x^2)}{1-x^2 y^2}} \\
= (1 - xy) \sqrt{\frac{1-x^2}{1-x^2y^2}} \sqrt{\frac{1-y^2}{1-x^2 y^2}}
> 0 \quad .
$$
It follows that $u + v < \frac \pi 2$.
So up to now we know that $(1)$ is a mapping from the square $M$
into the triangle $\widetilde{M}$. To show that the mapping is in fact
bijective, we verify that $(2)$ is the inverse mapping from
$\widetilde{M}$ to $M$.
Let $(u, v) \in \widetilde{M}$ and define $(x, y)$ by $(2)$.
It is clear that both $x$ and $y$ are positive. From
$u + v < \frac \pi 2$ it follows that
$$
\sin u < \sin(\frac \pi 2 - v) = \cos v \quad \Longrightarrow \quad x < 1 \, ,\\
\sin v < \sin(\frac \pi 2 - u) = \cos u \quad \Longrightarrow \quad y < 1 \, .$$
and therefore $(x, y) \in M$.
To verify that $(x, y)$ is mapped back exactly to the given $(u, v)$
is a straightforward calculation and I'll leave that final part to you :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof of identity: cross product of three vectors A book I'm reading contains the following (paraphrased)
\begin{equation}
(a \times b) \times c = (a \cdot c)b - (b \cdot c)a
\end{equation}
This is supposed to follow from:
\begin{equation}
(a \times b) \cdot (c \times d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c)
\end{equation}
Where in both equations $a, b, c, d$ are all vectors. The question is how to prove this?
| Chappers' proof is the way that was most likely intended (using the identity given). However, here is another approach (if only to show how much work is saved by using the identity given)
Since $(a\times b)\times a$ is perpendicular to $a$,
$$
((a\times b)\times a)\cdot a=0\tag{1}
$$
Since $|a\times b|=|a|\,|b|\,|\sin(\theta)|$ and $a\cdot b=|a|\,|b|\cos(\theta)$, we have
$$
|a\times b|^2+|a\cdot b|^2=|a|^2|b|^2\tag{2}
$$
Since $(a\times b)\cdot c=(b\times c)\cdot a$, we can use $(2)$ to get
$$
\begin{align}
(\color{#C00000}{(a\times b)}\times\color{#00A000}{a})\cdot\color{#0000F0}{b}
&=(\color{#00A000}{a}\times \color{#0000F0}{b})\cdot\color{#C00000}{(a\times b)}\\
&=(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)\tag{3}
\end{align}
$$
$(a\times b)\times a$ is perpendicular to $a\times b$ which is perpendicular to both $a$ and $b$. Therefore, $(a\times b)\times a$ is in the plane of $a$ and $b$. Thus, we can verify the following equation by taking dot products with $a$ and $b$ and using $(1)$ and $(3)$
$$
(a\times b)\times a=(a\cdot a)b-(a\cdot b)a\tag{4}
$$
Swapping $a$ and $b$ and negating $(4)$ gives
$$
(a\times b)\times b=(a\cdot b)b-(b\cdot b)a\tag{5}
$$
Suppose that $c$ is in the plane of $a$ and $b$. Writing $c=xa+yb$, we can take the dot product of this equation with $a$ and $b$ to get two equations to solve for $x$ and $y$. This gives
$$
c=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}a+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}b\tag{6}
$$
Furthermore, since $c$ only appears on the right side of $(6)$ in a dot product with $a$ or $b$, any component of $c$ perpendicular to the plane of $a$ and $b$ will not change the right side. Thus, the right side of $(6)$ is a formula for the perpendicular projection of $c$ onto the plane of $a$ and $b$.
Using $(4)$, $(5)$, and $(6)$, we have
$$
\begin{align}
(a\times b)\times c
&=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}((a\cdot a)b-(a\cdot b)a)\\
&+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}((a\cdot b)b-(b\cdot b)a)\\[4pt]
&=(a\cdot c)b-(b\cdot c)a\tag{7}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$
Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?
| You have basically put your matrix $A$ in Jordan normal form.
This means that you can split your matrix $J$ as follows:
$$J = D + N$$
where $D$ is diagonal and $N$ is nilpotent, i.e. there exists an a positive integer $n$ such that $N^n = 0$.
In your case
$$J = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}
=
\underbrace{
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}}_{=D}
+
\underbrace{
\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}}_{=N}
$$
and in your case you have $N^2 = 0$.
The basic idea is that when you expand your power $J^k = (D+N)^k$ using the binomial theorem, terms with $N^i$ with $i \geq n$ will disappear, since $N^n = 0$. In your case:
$$(D+N)^k = {k \choose 0} D^k + {k \choose 1} D^{k-1}N + \dots$$
where the remaining terms will be zero, e.g. the next term will be
$${k \choose 2}D^{k-2} \underbrace{N^2}_{=0} = 0.$$
More information can be found in this math.SE post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}
\end{align}$$
But I don't know what I can do after this.
| $$\begin{align}(x + x^{1/2})^{1/2} &= x^{1/2} + \frac 12 x^{-1/2}x^{1/2} -\frac 1 8 x^{-3/2}x+\cdots\\&=\sqrt x+\frac 12-\frac1{8\sqrt x} +\cdots\\
(x-1)^{1/2} &=\sqrt x -\frac 1{2\sqrt x}+\cdots \end{align}$$
therefore $$(x + x^{1/2})^{1/2} - (x-1)^{1/2}=\frac 1 2 + \frac 3{8\sqrt x} + \cdots \rightarrow \frac 12 \text{ as } x \to \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| ![The method I was speaking of if anyone especially OP is interested.... ][1]
[1]: http://i.stack.imgur.com/r05sp.jpg $$
\text{ Let there be a right triangle with angle A (not with measurement 90 deg) } \\
\text{ whose adjacent has measurement } 3 \text{ and whose opposite has measurement } 2. \\
\text{ Therefore the hypotenuse of the triangle has measurement } \sqrt{13}. \\
\text{ So this means } \sin(A)=\frac{2}{\sqrt{13}} \text{ and } \cos(A)=\frac{3}{\sqrt{13}} . \\ \text{ } \\ \text{ } \\
3 \sin(x)+2 \cos(x)=\sqrt{13}(\frac{3}{\sqrt{13}} \sin(x)+\frac{2}{\sqrt{13}} \cos(x)) \\
=\sqrt{13}(\cos(A) \sin(x)+\sin(A) \cos(x)) =\sqrt{13} \sin(x+A) \\ \text{ } \\ \text{ }
2 \sin(x)+3 \cos(x) = \sqrt{13}(\frac{2}{\sqrt{13}} \sin(x)+\frac{3}{\sqrt{13}} \cos(x)) \\ \text{ } \\ \text{ }
=\sqrt{13}(\sin(A) \sin(x)+\cos(A) \cos(x)) =\sqrt{13} \cos(x-A) \\
\text{… } \\
\int \frac{ 3 \sin(x)+2 \cos(x)}{ 2 \sin(x)+3 \cos(x) } dx= \int \frac{\sqrt{13} \sin(x+A) }{\sqrt{13} \cos(x-A) } dx \\
=\int \frac{ \sin(x-A+2A)} {\cos(x-A)} dx=\int \frac{ \sin(x-A) \cos(2A)+ \sin(2A) \cos(x-A)}{\cos(x-A)} dx \\ =\cos(2A) \int \frac{ \sin(x-A)}{\cos(x-A)} dx+ \sin(2A) \int \frac{\cos(x-A)}{\cos(x-A)} dx\\
=(\cos^2(A)-\sin^2(A)) \int \frac{\sin(x-A)}{\cos(x-A)} dx+2 \sin(A) \cos(A) \int 1 dx \\
=((\frac{3}{\sqrt{13}})^2-(\frac{2}{\sqrt{13}})^2) \int \frac{-du}{u} +2 \frac{2}{\sqrt{13}} \frac{3}{\sqrt{13}} x +C \\
\text{ (note: where } u=\cos(x-A) \text{ and so } du=-\sin(x-A) dx \text{ ) } \\
=(\frac{9}{13}-\frac{4}{13}) (- \ln|u|)+\frac{12}{13} x+C \\
=- \frac{5}{13} \ln|\cos(x-A)| +\frac{12}{13}x+C \\
=-\frac{5}{13} \ln|\cos(x) \cos(A)+\sin(x) \sin(A)|+\frac{12}{13}x+C \\
=-\frac{5}{13} \ln|\cos(x) \frac{3}{\sqrt{13}}+ \sin(x) \frac{2}{\sqrt{13}}|+\frac{12}{13}x+C \\
=-\frac{5}{13} \ln| \frac{1}{\sqrt{13}} (3 \cos(x)+2 \sin(x))| +\frac{12}{13}x+C \\
=-\frac{5}{13} \ln|\frac{1}{\sqrt{13}}|-\frac{5}{13} \ln|3 \cos(x)+2 \sin(x)|+\frac{12}{13}x+C \\
=-\frac{5}{13} \ln|3 \cos(x)+2 \sin(x)|+\frac{12}{13}x-\frac{5}{13} \ln|\frac{1}{\sqrt{13}}|+C \\
=-\frac{5}{13} \ln|3 \cos(x)+2 \sin(x)|+\frac{12}{13}x+K \\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
} |
Help me to prove this statement about quadratic equations? (from Gelfand's Algebra). $ x^2+px+q=0 ${p,q are integers; a,b are roots}. Prove $a^n+b^n$(n is any natural number) is an integer.
This is the third part of the problem.I have previously proved that $a^2+b^2$ and $a^3+b^3$ are integers using the binomial theorem and vieta's laws$(a+b=-p, ab=q)$. Using the same approach here,i have:
$a^n+b^n = (a+b)^n-a^{n-1}b-a^{n-2}b^2...-ab^{n-1}$
$=(-p)^n-ab(a^{n-2}+a^{n-3}b+...+b^{n-2})$
$=(-p)^n-q((a+b)^{n-2}...))$
Here i am stumped.I think that depending on whether n is even or odd the equation reduces to the previous two cases i have already proven.How do i state this formally?
PS: i am an independent learner without much experience with proofs.Please give a concise proof if you can(or a really detailed hint).
| It can be proved by induction on $n$, as follows:
From
$x^2 + px + q = 0 \tag{1}$
it follows that the roots $a$, $b$ satisfy
$a + b = -p, \tag{2}$
$ab = q; \tag{3}$
from (2), (3) we have
$p^2 = (a + b)^2 = a^2 + 2ab + b^2 = a^2 + 2q + b^2, \tag{4}$
or
$a^2 + b^2 = p^2 - 2q \in \Bbb Z, \tag{5}$
since $p, q \in \Bbb Z$. Also, since
$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b), \tag{6}$
it follows that
$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = -p^3 + 3pq \in \Bbb Z. \tag{7}$
Now assume that $m \ge 3$ and that $a^k + b^k \in \Bbb Z$ for $1 \le k \le m$; note we have established the case $m = 3$ in the above. Then
$a^{m+ 1} + b^{m + 1} = (a + b)(a^m + b^m) - (a^mb + ab^m) = (a + b)(a^m + b^m) - ab(a^{m - 1} + b^{m - 1})$
$= -p(a^m + b^m) - q(a^{m - 1} + b^{m - 1}); \tag{8}$
by hypothesis, $a^m + b^m, a^{m - 1} + b^{m - 1} \in \Bbb Z$; since $p, q \in \Bbb Z$, it follows that $a^{m + 1} + b^{m + 1} \in \Bbb Z$ as well. This completes the induction and shows that $a^n + b^n \in \Bbb Z$ for all $n \ge 1$. QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Recovering the optimal primal solution from dual solution I'm having trouble finding the optimal primal solution of a particular problem from its dual solution.
Primal:
$\texttt{Maximize} \ \ 10 x_1 + 24 x_2 + 20 x_3 + 20 x_4 + 25 x_5$
Subject to
$x_1 + x_2 + 2 x_3 + 3 x_4 + 5 x_5 \leq 19$
$2 x_1 + 4 x_2 + 3 x_3 + 2 x_4 + x_5 \leq 57$
$x_1, x_2, x_3, x_4, x_5 \geq 0$
Dual:
$\texttt{Minimize} \ \ 19 u_1 + 57 u_2$
Subject to
$u_1 + 2 u_2 \geq 10$
$u_1 + 4 u_2 \geq 24$
$2 u_1 + 3 u_2 \geq 20$
$3 u_1 + 2 u_2 \geq 20$
$5 u_1 + u_2 \geq 25$
$u_1, u_2 \geq 0$
Optimal solution to the dual problem: (4,5).
Then, since all the slack variables except for second are not zero, $x_1=x_3=x_4=x_5=0$.
Substituting this we have:
$x_2 \leq 19$ and $4 x_2 \leq 57$
But I can't seem to find the answer to the maximization problem.
| The following condition must hold:
$(A^T\cdot u^*-c)^T\cdot x^*=0$
Inserting the given values:
$\left[ \left( \begin{array}{} 1&2 \\ 1&4 \\ 2&3 \\ 3&2 \\ 5&1 \end{array} \right) \cdot \left( \begin{array}{} 4 \\ 5 \end{array} \right)-\left( \begin{array}{} 10 \\ 24 \\ 20 \\ 20 \\ 25 \end{array} \right)\right]^T\cdot \left( \begin{array}{} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array} \right)=0 $
$\left[ \left( \begin{array}{} 14 \\ 24 \\ 23 \\ 22 \\ 25 \end{array} \right) -\left( \begin{array}{} 10 \\ 24 \\ 20 \\ 20 \\ 25 \end{array} \right)\right]^T\cdot \left( \begin{array}{} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array} \right)=0 $
$\left[ \left( \begin{array}{} 4 \\ 0 \\ 3 \\ 2 \\0 \end{array} \right)\right]^T\cdot \left( \begin{array}{} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array} \right)=0 $
Thus the equation is $4x_1+0x_2+3x_3+2x_4+0x_5=0$
All $x_i$ are greater or equal to zero. Then $x_1=x_3=x_4=0$.
And the condition $s_i\cdot u_i=0 \ \ \forall i \in \{1,2\}$ must hold. Both $u_i $ have a positive value. Thus $s_1=s_2=0$
$s_i$ are the slack variables of the primal problem.
The equations are:
$x_2+5x_5=19$
$4x_2+x_5=57$
I think you can solve this little equation system on your own.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{x\rightarrow0}\frac{e^{x^2} - \cos x}{\sin^2 x}$ Find the limit as $x$ approaches $0$ of
$\dfrac{e^{x^2} - \cos x}{\sin^2 x}$
What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\frac 12$.
Is this right? Any help would be appreciated.
| As Jean-Claude Arbaut showed, Taylor expansions are extremely useful for this kind of problems.
But, using them with a few more terms, you also see how is the limit approached $$A=\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{\Big(1+x^2+\frac{x^4}{2}+O\left(x^6\right)\Big)-\Big(1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)\Big)}{x^2-\frac{x^4}{3}+O\left(x^6\right)}$$ $$A\approx\frac{\frac{3 x^2}{2}+\frac{11 x^4}{24}+O\left(x^6\right)}{x^2-\frac{x^4}{3}+O\left(x^6\right)}=\frac{\frac{3 }{2}+\frac{11 x^2}{24}}{1-\frac{x^2}{3}}$$ Performing the long division, we have $$A\approx \frac{3}{2}+\frac{23 }{24}x^2$$ If you plot on the same graph the function and the above approximation, you will probably be amazed to see how close to each other are the two curves for $-\frac 12\leq x \leq \frac 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is even $x^2\equiv (1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
Thus, when $p\equiv 1\pmod{4}$, the congruence $x^2\equiv -1\pmod{p}$ is soluble.
Point of contention: I understand the general argument
I understand the relation
$(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
But I cant work out how $x^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
How is $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
| If the general case proves obfuscatory, it often proves enlightening to first examine a few small special cases in order to help grasp the general idea, e.g. let's examine the two smallest cases first $\,p=5,13.\,$ mod $\, p = 5\!:\,\ \color{#0a0}{4\equiv -1},\ 3\equiv \color{#c00}{ -2}\ $ so substituting these into Wilson's formula
$\qquad\qquad \begin{align} {\rm mod}\ 5\!:\,\ {-}1\, \equiv& \ \ \ \ \color{#0a0}{(4)}\cdot \color{}{(3)}\cdot 2\cdot 1\\
\equiv&\, \color{#0a0}{(-1)}(-\color{#c00}2)\cdot\color{#c00} 2\cdot 1\,\equiv\, \color{#c00}2!^2\end{align}$
$\qquad\quad\ \ \begin{align} {\rm mod}\ 13\!:\,\ {-}1\, \equiv& \ \ \color{#0a0}{(12)}\cdot\,(11)\,\cdots\ \ (7)\ \cdot 6\cdot 5\cdots 1\\
\equiv&\, \color{#0a0}{(-1)}\cdot(-2)\cdots (-\color{#c00}6)\cdot\color{#c00} 6\cdot 5\cdots 1\,\equiv\, \color{#c00}6!^2 \end{align}$
$\qquad\qquad\quad\overset{\vdots}{\phantom{c}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \overset{\ddots}{\phantom{c}}$
$\qquad\begin{align} {\rm mod}\ 4k\!+\!1\!:\,\ {-}1\, \equiv& \ \ \color{#0a0}{(4k)}(4k\!-\!1)\cdots (2k\!+\!1)\ \ (2k)(2k\!-\!1)\cdots 1\\
\equiv&\, \underbrace{\color{#0a0}{(-1)}\,\cdot\,(-2)\, \cdots\ \,(-\color{#c00}{2k})}_{\Large (-1)^{2k}(2k)!\ =\ (2k)!} \ \ \,\underbrace{(\color{#c00}{2k})(2k\!-\!1)\cdots 1}_{\Large(2k)!}\equiv (2k)!^2 \end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Equilateral Triangle equality Let ABC be an equilateral triangle, and P be an arbitrary point within the triangle.
Perpendiculars PD, PE, PF are drawn to the three sides of the triangle. Show
that, no matter where P is chosen,
PD + PE + PF / AB + BC + CA = 1/2√3
| Let $AB=BC=AC=a$.
It follows from the areas equality:
\begin{align}
S_{\triangle APB}+S_{\triangle APC}+S_{\triangle BPC}
&=
S_{\triangle ABC}
\\
\frac{1}{2}PD\cdot a+\frac{1}{2}PE\cdot a+\frac{1}{2}PF\cdot a
&=
a^2\frac{\sqrt{3}}{4}
\\
PD+PE+PF
&=
\frac{3a}{2\sqrt{3}}
\\
\frac{PD+PE+PF}{AB+BC+AC}
&=
\frac{1}{2\sqrt{3}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the limit $\lim\limits _{x\to 0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}$ $$\lim _{x\to \:0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}=\left|\frac{0}{0}\right|$$
I think you have to multiply by the conjugate. And then make the change equivalent small. Right?
| I am waiting for a simpler more succinct answer but in the meantime we can use Taylor's theorem
$$ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{4n} = 1 - \frac{x^4}{2} + \frac{x^8}{120} + O(x^{12})$$
$$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2} + \frac{x^4}{120} + O(x^{6}) $$
Now we have
$$\lim_{x \to 0} \frac{\sqrt{\frac{x^4}{2} - \frac{x^8}{120} + O(x^{12})}}{\frac{x^2}{2} - \frac{x^4}{120} + O(x^{6})} = \lim_{x \to 0} \frac{\sqrt{\frac{1}{2} - \frac{x^4}{120} + O(x^{8})}}{\frac{1}{2} - \frac{x^2}{120} + O(x^{4})} = \frac{\sqrt{\frac{1}{2}}}{\frac{1}{2}}$$
$$ \lim_{x \to 0} \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)} = \sqrt{2}$$
I don't really like using Taylor's theorem for "simple" limits, so I await a cleverer answer!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 5
} |
Integrating $ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $ I am trying to show that
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \frac{\pi}{a(a+b)} $$ where $ a,b > 0$.
I have tried a few things, but none have worked.
For example, one approach was
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \int^{\pi}_{0} \frac{\cos^2 x}{(a^2 - b^2)\cos^2 x + b^2} \ dx$$ and then divide by $ a^2 - b^2 $ throughout. However I am not given that $ a \not= b $, so this approach didn't work.
I also tried splitting up the integral, but then was not sure how to proceed:
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx + \int^{\pi}_{\frac{\pi}{2}} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$
Substituting $ x \to x + \frac{\pi}{2} $ followed by $ x \to \frac{\pi}{2} -x $ a in the second integral gives
$$ \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$
Hence $$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = 2 \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$
Next I thought I could substitute $ u = a^2 \cos^2 x + b^2 \sin^2 x $ which changes my limits from $[0, \frac{\pi}{2}]$ to $[a^2, b^2]$, however I could not complete the substitution as I didn't see how to express $ du = (b^2 - a^2) \sin 2 x \ dx $ or $ \cos^2 x$ in terms of $u$.
| Suppose we seek to evaluate
$$\int_0^\pi \frac{\cos^2 x}{a^2\cos^2x + b^2\sin^2x} dx
= \frac{1}{2}
\int_0^{2\pi} \frac{\cos^2 x}{a^2\cos^2x + b^2\sin^2x} dx$$
with $a,b>0.$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\frac{1}{2}\int_{|z|=1}
\frac{(z+1/z)^2}{a^2(z+1/z)^2-b^2(z-1/z)^2} \frac{dz}{iz}
\\ = \frac{1}{2}\int_{|z|=1}
\frac{(z^2+1)^2}{a^2(z^2+1)^2-b^2(z^2-1)^2} \frac{dz}{iz}.$$
This factors to give
$$\frac{1}{2}\int_{|z|=1}
\frac{(z^2+1)^2}{(a(z^2+1)+b(z^2-1))(a(z^2+1)-b(z^2-1)} \frac{dz}{iz}.$$
The roots of the first term are
$$\rho_{1,2} = \pm\sqrt{\frac{b-a}{a+b}}
= \pm \sqrt{1-\frac{2a}{a+b}} $$
and of the second term
$$\rho_{3,4} = \pm\sqrt{\frac{a+b}{b-a}}
= \pm \sqrt{1+\frac{2a}{b-a}}.$$
We now treat the case of $b>a$ which gives $|\rho_{1,2}|<1$ and
$|\rho_{3,4}|>1.$ We may therefore apply the Cauchy Residue Theorem,
taking into account the contributions from $\rho_{1,2}$ and from
$z=0.$
The residues at $\rho_{1,2}$ are
$$\left.
\frac{(z^2+1)^2}
{a^2\times 2(z^2+1)2z-b^2\times 2(z^2-1)2z}
\frac{1}{iz}\right|_{z=\rho_{1,2}.}$$
This is
$$\frac{1}{4i} \left.
\frac{(z^2+1)^2}
{a^2\times (z^2+1)z^2-b^2\times (z^2-1)z^2}
\right|_{z=\rho_{1,2}}
\\ = \frac{1}{4i} \left.
\frac{(z^2+1)^2}
{a^2\times (z^2+1)^2-b^2\times (z^2-1)^2
- a^2(z^2+1) - b^2(z^2-1)}
\right|_{z=\rho_{1,2}.}$$
Considering the definition of $\rho_{1,2}$ this becomes
$$-\frac{1}{4i} \left.
\frac{(z^2+1)^2}
{a^2(z^2+1) + b^2(z^2-1)}
\right|_{z=\rho_{1,2}}$$
which yields
$$-\frac{1}{4i}
\frac{4b^2}
{a^2\times 2b(a+b) + b^2\times (-2a)(a+b)}
\\ = -\frac{1}{i}
\frac{b^2}
{2ab(a-b)(a+b)}
= -\frac{1}{2i}
\frac{b}
{a(a-b)(a+b)}.$$
The residue at $z=0$ is
$$\frac{1}{i} \frac{1}{a^2-b^2}
= \frac{1}{i} \frac{1}{(a-b)(a+b)}.$$
Collecting everything we obtain
$$\frac{1}{2}\times 2\pi i\times
\left(\frac{1}{i} \frac{1}{(a-b)(a+b)}
- 2\times \frac{1}{2i} \frac{b}{a(a-b)(a+b)}\right)
\\ = \frac{1}{2}\times 2\pi i\times
\frac{1}{i} \frac{1}{(a-b)(a+b)}
\left(1 - \frac{b}{a}\right)
\\ = \frac{1}{2}\times 2\pi i\times
\frac{1}{i} \frac{1}{(a-b)(a+b)} \frac{a-b}{a}
\\ = \frac{\pi}{a(a+b)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
4 girls and 8 boys are randomly divided into 3 groups of equal size 4 girls and 8 boys are randomly divided into 3 groups of equal size
What is the probability that there will be at least one girl in each group?
| You can calculate the probability that all girls are in a only group, the probability that four girls are disposed in two groups and then you can use the inverse probability. The probability that all girls are in only group is: $$1\cdot \frac {3}{11}\cdot \frac{2}{10}\cdot {1}{9}$$ while the probability that all girls are disposed in two groups is:$$1\cdot 1\cdot \frac{6}{10}\cdot \frac{5}{9}.$$ Then you can use the inverse probability $$1-(1\cdot \frac{3}{11}\cdot \frac{2}{10}\cdot \frac{1}{9})-(1\cdot 1\cdot \frac{6}{10}\cdot \frac{5}{9}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rewrite also as: $p^2+pq\equiv 3 \pmod5$.
Now I analyzed five possible cases: $p^2\equiv 0$, $p^2\equiv 1$, $p^2\equiv 2$, $p^2\equiv 3$ and $p^2\equiv 4$.
*
*We can not that if $p^2\equiv 0$ also $pq\equiv 0$ therefore the equation is impossible
*$p^2\equiv 2$ is impossible because $2$ is a non-quadratic residue $\pmod5$
*also $p^2\equiv 3$ is impossible because $3$ is a non-quadratic residue $\pmod5$.
*If $p^2\equiv 1 \pmod5$, $p\equiv \pm 1$, therefore $p\equiv -1\equiv 4 \pmod5$ and $p2\equiv 1 \pmod5$. These values are not acceptable because if $p=9$ ($9\equiv 4 \pmod5$) $275\cdot 9>2008$, if $p=11$ ($11\equiv 1\pmod5$) $275\cdot 9>2008$ therefore two values aren't acceptable.
*If $p^2\equiv 4 \pmod5$, $p\equiv \pm2$ and if $p\equiv -2\equiv 3 \pmod5$ there not exist solutions for the same reason of previous example. If $p\equiv 2 \pmod5$ possible values of $p$ are $2$ or $7$ because $275p<2008$. Replacing $p=2$ we obtain a valor not integer of $q$ while replacing $p=7$ we obtain $q=2$ therefore this is the only solution of equation.
Is correct my process or are there other solutions?
Thanks:)
| Suppose that $p>7$. Then $p\ge 11$ so that $p^2+pq+275p+10q\ge 3188$ since $q\ge 2$. Hence $p\le 7$ and we have to check $p=2,3,5,7$. It follows that $(p,q)=(7,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Minimum value of function $f(x)=x+\log_2(2^{x+2}-5+2^{-x+2})$ out of 5 options Minimum value of function $f(x)=x+\log_2(2^{x+2}-5+2^{-x+2})$ out of 5 options
A : $\log_2(1/2)$
B : $\log_2(41/16)$
C : $39/16$
D : $\log_2(4.5)$
E : $\log_2(39/16)$
I just... don't know how to approach this.
| $$x=\log_2(2^x)$$
$$f(x)=\log_2[2^x(2^{x+2}-5+2^{-x+2})]=\log_2[2^{2x+2}-5\cdot2^x+4]$$
Now $4\cdot2^{2x}-5\cdot2^x+4=(2^{x+1})^2-2\cdot2^{x+1}\cdot\dfrac54+\left(\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $=\left(2^{x+1}-\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $\ge4-\left(\dfrac54\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Last nonzero digit of $2010!$ I have to calculate the last nonzero digit of $2010!$
Till now I couldn't find any pattern.
| $2010!$ ends with $501$ zeroes since:
$$\sum_{n=1}^{5}\left\lfloor\frac{2010}{5^n}\right\rfloor=501.\tag{1}$$
For the same reason,
$$ \nu_2(2010!)=\sum_{n=1}^{10}\left\lfloor\frac{2010}{2^n}\right\rfloor=2002\tag{2}$$
so $\frac{2010!}{10^{501}}$ is an even number, and since:
$$ 1\cdot 2\cdot 3\cdot 4\equiv -1\pmod{5} \tag{3}$$
it follows that:
$$\begin{eqnarray*}\frac{2010!}{5^{501}}\equiv \frac{(-1)^{\frac{2010}{5}}5^{402}402!}{5^{501}}&\equiv& \frac{5^{402}\cdot 2 \cdot (-1)^{\frac{400}{5}}\cdot 5^{80}\cdot 80!}{5^{501}}\equiv\frac{5^{482}\cdot 2\cdot 5^{16}\cdot 16!}{5^{501}}\\&\equiv&\frac{2\cdot 16!}{5^3}\equiv\frac{2\cdot(-1)^{\frac{15}{5}}\cdot 5^3\cdot 3!}{5^3}\equiv 3\pmod{5}\end{eqnarray*}$$
and since $2^{501}\equiv 2\pmod{5}$ we have:
$$ \frac{2010!}{10^{501}}\equiv 4\pmod{5}\tag{4} $$
so the last non-zero digit of $2010!$ is $\color{red}{4}$ by the Chinese theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}{a^2}+4+b^2+\frac{4}{b^2}+4\ge\frac{81}{2} \\ a^2+b^2+\frac{4}{a^2}+\frac{4}{b^2}\ge \frac{65}{2}$$ and then multiplied both sides by $a^2b^2$ to get $$a^2b^2(a^2+b^2)+4(a^2+b^2)\ge\frac{65}{2}a^2b^2 \\ (a^2+b^2)(a^2b^2+4)\ge\frac{65}{2}a^2b^2, $$ which, combining $a+b=1$ and $(a+b)^2=a^2+b^2+2ab$, can be rewritten as $$(1-2ab)(a^2b^2+4)\ge\frac{65}{2}a^2b^2.\tag{$\star$}$$ Setting $c=ab$ makes $(\star)$ an inequality in one variable, but it didn't help me. Is this totally the wrong track? And, I was wondering if there are other approaches besides what I came up with.
| your calculations are ok, we are looking at the inequality $$(1-2c)(c^2+4)\geq \frac{65}{2}c^2$$ with $c=ab$ for $c$ we get $$a^2+b^2=1-2ab\geq 2ab$$ from here we get $$\frac{1}{4}\geq ab=c$$ and our inequality is equivalent to $$\frac{-1}{2}(4c-1)(c^2+16c+8)\geq 0$$
ready.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots of the equation $z^{11}=1$
So $1+z_1+z_2+...+z_{10}=1$ (De Moivre
s Theorem)
So $z_1+z_2+...+z_{10}=-1$
$-1=i^2$
So $\frac{i^2}{i}=i$
| $\bf{My\; Solution::}$ Given $$\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right) = -i\cdot \left[\cos \left(\frac {2k\pi}{11}\right)+i\cdot \sin \left(\frac {2k\pi}{11}\right)\right]$$
Now We know that $$e^{i\phi} = \cos \phi+i\sin \phi$$ and $$e^{-i\phi}=\cos \phi+i\sin \phi$$
So We Write $$\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right)=-i\cdot e^{i\frac{2k\pi}{11}} = -i\cdot e^{ik\phi}\;,$$ Where $$\displaystyle \phi = \frac{2\pi}{11}$$
So We have $$\displaystyle \sum_{k=1}^{10}\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right) = -i\sum_{k=1}^{10}e^{ik\phi} = -i\left[e^{i\phi}+e^{2i\phi}+........+e^{10i\phi}\right]$$
Above is Sum of $\bf{10-}$ terms of $\bf{G.P}.$
So We get $$\displaystyle = -i\cdot \left[\frac{e^{i\phi}-e^{11i\phi}}{1-e^{i\phi}}\right] = -i\cdot \left[\frac{e^{i\phi}-1}{1-e^{i\phi}}\right]=+i$$
Bcz $$\displaystyle e^{11i\phi} = \cos 11\phi+i\sin 11 \phi = \cos \left(11\cdot \frac{2\pi}{11}\right)+i\sin \left(11\cdot \frac{2\pi}{11}\right)=1\;,$$ Bcz $$\displaystyle \phi = \frac{2\pi}{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many permutations How many permutations $\pi \in S_{2n} $ for which $\exists a\in [2n] $ such that set $\lbrace a,\pi (a),\pi ^2(a),\pi^3(a),... \rbrace $ has exactly $n$ elements.
I need help to solve this.
| By way of enrichment here is an alternate formulation using
combinatorial species. The species of permutations with cycles of size $n$ marked is
$$\mathfrak{P}(\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathfrak{C}_{=2}(\mathcal{Z}) + \cdots
+ \mathcal{U}\mathfrak{C}_{=n}(\mathcal{Z})
+ \mathfrak{C}_{=n+1}(\mathcal{Z}) + \cdots).$$
This gives the generating function
$$G(z, u) =
\exp\left(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots
+u\frac{z^n}{n}+\frac{z^{n+1}}{n+1}+\cdots\right)$$
which is
$$G(z, u) =
\exp\left(u\frac{z^n}{n} - \frac{z^n}{n} +\log\frac{1}{1-z}\right)$$
or
$$G(z, u) =
\frac{1}{1-z}\exp\left(-\frac{z^n}{n}\right)
\exp\left(u\frac{z^n}{n}\right).$$
We seek to compute
$$(2n)! [z^{2n}] [u] G(z, u)
+ (2n)! [z^{2n}] [u^2] G(z, u).$$
The first component is
$$(2n)! [z^{2n}] \frac{1}{1-z}
\exp\left(-\frac{z^n}{n}\right)
\frac{z^n}{n}.$$
Since we are extracting $[z^{2n}]$ this becomes
$$(2n)! [z^{2n}] \frac{1}{1-z}
\left(1-\frac{z^n}{n}+\frac{1}{2}\frac{z^{2n}}{n^2}\right)
\frac{z^n}{n}.$$
The second component is
$$(2n)! [z^{2n}] \frac{1}{1-z}
\exp\left(-\frac{z^n}{n}\right)
\frac{1}{2} \frac{z^{2n}}{n^2}.$$
Since we are extracting $[z^{2n}]$ this becomes
$$(2n)! [z^{2n}] \frac{1}{1-z}
\left(1-\frac{z^n}{n}+\frac{1}{2}\frac{z^{2n}}{n^2}\right)
\frac{1}{2} \frac{z^{2n}}{n^2}.$$
Adding these we obtain
$$(2n)! [z^{2n}] \frac{1}{1-z}
\left(1-\frac{z^n}{n}+\frac{1}{2}\frac{z^{2n}}{n^2}\right)
\left(\frac{z^n}{n} + \frac{1}{2} \frac{z^{2n}}{n^2}\right).$$
This produces
$$(2n)! [z^{2n}] \frac{1}{1-z}
\left(\frac{z^n}{n}-\frac{z^{2n}}{n^2}+
\frac{1}{2}\frac{z^{3n}}{n^3}
+\frac{1}{2}\frac{z^{2n}}{n^2}
-\frac{1}{2}\frac{z^{3n}}{n^3}
+\frac{1}{4}\frac{z^{4n}}{n^4}\right).$$
We may omit the terms in $z^{3n}$ and $z^{4n}$ to get
$$(2n)! [z^{2n}] \frac{1}{1-z}
\left(\frac{z^n}{n}-\frac{z^{2n}}{n^2}
+\frac{1}{2}\frac{z^{2n}}{n^2}\right)
\\ = (2n)! [z^{2n}] \frac{1}{1-z}
\left(\frac{z^n}{n}-\frac{1}{2}\frac{z^{2n}}{n^2}\right).$$
This finally yields
$$(2n)!\frac{1}{n} [z^n]\frac{1}{1-z}
- (2n)! \frac{1}{2}\frac{1}{n^2} [z^0] \frac{1}{1-z}
\\ = (2n)! \left(\frac{1}{n} - \frac{1}{2} \frac{1}{n^2}\right)
\\ = \frac{(2n)!}{n^2} \frac{2n-1}{2}
= \frac{(2n-1)!}{n} (2n-1).$$
This gives the sequence
$$1, 9, 200, 8820, 653184, 73180800, 11564467200, 2451889440000,\ldots$$
which is OEIS A052145.
| {
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"url": "https://math.stackexchange.com/questions/1241917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How do I calculate the limit for this multiplication? $$\lim_{n\to\infty}\left(1-\frac{2}{3}\right)^{\tfrac{3}{n}}\cdot\left(1-\frac{2}{4}\right)^{\tfrac{4}{n}}\cdot\left(1-\frac{2}{5}\right)^{\tfrac{5}{n}}\cdots\left(1-\frac{2}{n+2}\right)^{\tfrac{n+2}{n}}$$
(original image)
I mean,I tried to use Sandwitch rule but it didnt work.
| The $m^{th}$ term is
\begin{align}
a_m & = \prod_{n=1}^m \left(1-\dfrac2{n+2}\right)^{(n+2)/m} = \prod_{n=1}^m \left(\dfrac{n}{n+2}\right)^{(n+2)/m} = \left(\dfrac{\prod_{n=1}^m n^{(n+2)/m}}{\prod_{n=1}^m (n+2)}\right)^{(n+2)/m}\\
& = \left(\dfrac{\prod_{n=1}^m n^{n/m}}{\prod_{n=3}^{m+2} n^{n/m}}\right) \prod_{n=1}^m n^{2/m} = \dfrac{2^{2/m}(m!)^{2/m}}{(m+1)^{(m+1)/m}(m+2)^{(m+2)/m}}
\end{align}
This gives us that
\begin{align}
\log(a_m) & = \dfrac{2\log2}m + \dfrac{2\log(m!)}m - \dfrac{m+1}m \log(m+1) - \dfrac{m+2}{m+1} \log(m+2)\\
& = \dfrac{2\log2}m + \dfrac{2}m\left(m\log(m)-m + \mathcal{O}(\log(m))\right) - \dfrac{m+1}m \log(m+1) - \dfrac{m+2}{m+1} \log(m+2)\\
\end{align}
where the last step relies on DeMoivre's identity.
Finishing step $1$:
This gives us$$\log(a_m) = -2 + 2 \log(m) - \log(m+1)-\log(m+2) + \mathcal{O}\left(\dfrac{\log(m)}m \right)$$
Finishing step $2$:
Taking $m \to \infty$, we obtain$$\lim_{m \to \infty} \log(a_m) = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that if $a\neq 1$, then $\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$
Need to show that if $a\neq 1$, then
$$\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$$
Here is my attempt:
$$\begin{aligned}
S & =\sum_{k=0}^{n-1}ka^k \\
&= \sum_{k=0}^{n}(k-1)(a^{k-1}) \\
\end{aligned}$$
from here we can say:
$$\begin{aligned}
S & =(1-1)(a^{1-1})+(2-1)(a^{2-1})+(3-1)(a^{3-1})+...+(n-1)(a^{n-1})\\
& =(0)(a^{0})+(1)(a^{1})+(2)(a^{2})+(3)(a^{3})+...+(n-1)(a^{n-1})\\
& =a+2a^2+3a^3+(n-1)(a^{n-1})\\
\end{aligned}$$
now compute $(a)S$:
$$\begin{aligned}
(a)S & =(a)(a)+(a)(2a^2)+(a)(n-1)(a^{n-1})\\
& =a^2+2a^3+(n-1)(a^{n-1+1})\\
& =a^2+2a^3+(n-1)(a^{n})\\
\end{aligned}$$
now compute $S-(a)S$:
$$\begin{aligned}
S-(a)S & = a+2a^2+3a^3+(n-1)(a^{n-1})-[a^2+2a^3+(n-1)(a^{n})]\\
& = a+2a^2+3a^3+(n-1)(a^{n-1})-a^2-2a^3-(n-1)(a^{n})\\
& = a+a^2+a^3+(n-1)(a^{n-1})-(n-1)(a^{n})\\
\end{aligned}$$
re-writing above:
$$(1-a)S = a+a^2+a^3+(n-1)(a^{n-1})-(n-1)(a^{n})$$
dividing both-sides by $(1-a)$:
$$S = \frac{a+a^2+a^3+(n-1)(a^{n-1})-(n-1)(a^{n})}{1-a}$$
Am I on the right track so far? If not, please point out where I went wrong.
Also, any examples would be very appreciated.
Update:
A lot of people have provided answers, however, no one has pointed out what is wrong with my current approach? I am looking to learn not to copy any answer. Please consider my "solution" and help direct me. Honestly I don't even want the final solution, I just want help understanding how to answer this. I really appreciate all the effort and time put.
| for $1 \le m \le n-1$, set
$$
S_m = a^m + \cdots + a^{n-1} = \frac{a^m -a^n}{1-a}
$$
then
$$
\sum_{k=0}^{n-1} ka^k = \sum_{m=1}^{n-1} S_m =(1-a)^{-1}\sum_{m=1}^{n-1}(a^m-a^n) \\ =(1-a)^{-2}\left( a-a^n-(n-1)a^n(1-a) \right) \\
= \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it looks like it works out.
Base Case:
Let $n=1$ then
\begin{align}
&a_2 = \sqrt{2 + a_1} \\
&a_2 < 2 \\
&\sqrt{2 + \sqrt{2}} < 2
\end{align}
The base case holds.
Induction hypothesis:
Let $n=k$
$$a_1 = \sqrt{2} \quad a_{k+1} = \sqrt{2 + a_k} \quad a_{k+1} = \sqrt{2}$$
Induction Step:
Now we have to prove that $a_{k+2} < 2$. Let $n=k+1$.
\begin{align}
a_{k+2} &= \sqrt{2 + a_{k+1}} \\
\implies a_{k+2} &= \sqrt{2 + \sqrt{2 + a_k}} \\
\end{align}
Now we have to show that $a_{k+2} < 2$.
\begin{align}
a_{k+2} &< 2\\
\sqrt{2 + \sqrt{2 + a_k}} &< 2\\
2 + \sqrt{2 + a_k} &< 4 \\
\sqrt{2 + a_k} &< 2 \\
\end{align}
Q.E.D
Are my steps correct?
Thanks for your time!
| $b_i:= a_i^2$ Then $$ b_1=2,\ b_{n+1}=2+\sqrt{b_{n}}$$
Show that $b_i$ is an increasing sequence bounded by $4$ :
(1) $b_1< 4$ If $b_n<4$ then $b_{n+1}=2+\sqrt{b_n}< 2+2=4$.
(2) $b_2=2+\sqrt{2} > b_1$ If $b_n>b_{n-1}$ then $$
b_{n+1}=2+\sqrt{b_n} > 2+\sqrt{b_{n-1}}=b_n $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then How to prove this inequality by AM-GM or other ?
| By setting $a=\cos\theta,b=\sin\theta$ you just have to prove that for any $\theta\in\left(0,\frac{\pi}{2}\right)$ we have:
$$ f(\theta)=(5a+2)(5b+5)=(5\cos\theta+2)(5\sin\theta+5)\leq 48,$$
or, by using Weierstrass substitution $\theta=2\arctan t$,
$$\forall t\in(0,1),\quad g(t) = (7-3t^2)(t+1)^2- \frac{48}{5}(1+t^2)^2 \leq 0$$
that is straightforward to check since:
$$ g'(t) = \frac{2}{5}(1-3t)\left(35+29 t+42 t^2\right), $$
hence we have equality only for $t=\frac{1}{3}$, i.e. for $\theta=2\arctan\frac{1}{3}=\arctan\frac{3}{4}=\arcsin\frac{3}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Exercise about Matrix diagonalization Well I have to diagonalize this matrix :
$$
\begin{pmatrix} 5 & 0 & -1 \\ 1 & 4 & -1 \\ -1 & 0 & 5 \end{pmatrix}
$$
I find the polynome witch is $P=-(\lambda-4)^2(\lambda-6)$
Now I want to know eignevectors so I solve $AX=4X$ and $AX=6X$ with $X=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$but I have a problem with the first system !
In the correction they say "after an elementary calculus we have $E_4=Vect\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\right)$"
And I don't know how and why because with my system I just find that $x=z$.
Can you explain to me, please ?
| You calculated correctly. The set of all eigenvectors of eigenvalue $4$ is a $2$-dimensional subspace. The condition $x = z$ is equivalent to all matrices of the form
$$
\begin{pmatrix} x \\ y \\ x \end{pmatrix}
= \begin{pmatrix} x \\ 0 \\ x \end{pmatrix}
+ \begin{pmatrix} 0 \\ y \\ 0 \end{pmatrix}
= x \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}
+ y \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
$$
where $x,y \in \Bbb{R}$. There is your basis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the type of triangle from equation. In triangle $ABC$, the angle($BAC$) is a root of the equation
$$\sqrt{3}\cos x + \sin x = \frac{1}{2}.$$
Then the triangle $ABC$ is
a) obtuse angled
b) right angled
c) acute angled but not equilateral
d) equilateral.
Thanks in advance.
| given that
$$\sqrt{3}cosx+sinx=\frac{1}{2}$$ Now, diving the above equation by $\sqrt{(\sqrt{3})^2+(1)^2}=2$ we get
$$\begin{align}
\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=\frac{1}{4}\\
cosxcos\frac{\pi}{6}+sinxsin\frac{\pi}{6}=\frac{1}{4}\\
cos\left(x-\frac{\pi}{6}\right)=\frac{1}{4}\\
x-\frac{\pi}{6}=cos^{-1}\frac{1}{4}\\
x=\frac{\pi}{6}+cos^{-1}\frac{1}{4}\\
\end{align}$$
$$x\approx1.841714847... rad\approx105.52^{o}>90^{o}$$
Thus, the triangle is an obtuse angled triangle. Hence, the option (a) is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proving $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$ by induction How can I prove by induction that $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$? My guess is that there must be another form to express the sum of nested square roots, but I don't know how to find it.
| We can actually prove a more general version of what you hope to prove (set $a=0$ for your problem, specifically):
Claim: For every $n\in\mathbb{Z^+}$ and every non-negative real number $a$
$$
\sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+n}}}< a+3.
$$
Proof. For $n\geq 1$, let $S(n)$ denote the statement that for any non-negative real $a$,
$$
S(n) : \sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+n}}}< a+3.
$$
Base step: $S(1)$ says that $\sqrt{a+1}<a+3$, and this is verifiable since
$$
a+1<(a+3)^2\Leftrightarrow 0<a^2+5a+8,
$$
which is true for $a\geq 0$.
Inductive step: Fix some $k\geq 0$, and suppose that $S(k)$ is true for any non-negative $a$ where
$$
S(k) : \sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+k}}}< a+3.
$$
To be shown is that $S(k+1)$ follows for any non-negative $b$ where
$$
S(k+1) : \underbrace{\sqrt{b+1+\sqrt{b+2+\cdots+\sqrt{b+k+\sqrt{b+k+1}}}}}_{\text{LHS}}<\underbrace{b+3}_{\text{RHS}}.
$$
Using $a=b+1$,
\begin{align}
\text{LHS} &= \sqrt{b+1+\sqrt{b+2+\cdots+\sqrt{b+k+\sqrt{b+k+1}}}}\\[1em]
&= \sqrt{b+1+\sqrt{a+1+\cdots+\sqrt{a+k-1+\sqrt{a+k}}}}\\[1em]
&< \sqrt{b+1+a+3}\tag{by $S(k)$}\\[0.5em]
&= \sqrt{2b+5}\\[0.5em]
&< b+3\\[0.5em]
&= \text{RHS},
\end{align}
where the last inequality follows since
$$
2b+5<(b+3)^2=b^2+6b+9
$$
and $b^2\geq 0$. This proves $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
Your particular problem holds for $S(n)$ where $a=0$.
| {
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"url": "https://math.stackexchange.com/questions/1248962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3 \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4 \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can make what's inside of the square root equal to $50$?
| Sometimes it might help to take things a little more symbolically. Without worrying about what $x$ is for now, think about $3x + 5x$. Clearly that's $8x$, just as you thought.
For the next problem you have $x - 3x$. The answer is $-2x$. Let's say $x = 1$. Then this boils down to $1 - 3 = -2$. When we put $x = \sqrt{11}$ back into the picture, we get $\sqrt{11} - 3\sqrt{11} = -2\sqrt{11}$.
With the third problem, think about $x \neq y$. We have $6x + 4y$ and you're pretty much stuck unless you can figure out some way to restate $x$ in terms of $y$ or vice versa. It turns out that $\sqrt{50} = 5\sqrt{2}$ (so $y = 5x$). Therefore $$6\sqrt{2} + 4\sqrt{50} = 6\sqrt{2} + 4(5\sqrt{2}) = 6 + 20\sqrt{2} = 26\sqrt{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~~x^{c-a}=y^{b-c}$$
$$2.~~y^{a-b}=z^{c-a}$$
$$3.~~z^{b-c}=x^{a-b}$$
How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.
Thank you.
| Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}=\lambda$$
we have:
$$ x = e^{\lambda(b-c)},\quad y=e^{\lambda(c-a)},\quad z=e^{\lambda(a-b)}, $$
hence:
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = \exp\left(\lambda\cdot\sum_{cyc}\left(b^2-c^2-a(b-c)\right)\right)=\exp(0)=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
differential equation power series solution I am trying to solve this equation using power series
$$
(1-x)y"-xy'+y=0
$$
Knowing that $y(0)=-2$ and $y'(0)=6$.
Please I need someone's help, I get a relation between
$c(n)$,$c(n+1)$, and $c(n+2)$.
| taking
$$\begin{align}
y&=\sum_{n=0}^{+\infty}a_n x^n\\
y'&=\sum_{n=0}^{+\infty}na_n x^{n-1}\\
y''&=\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}
\end{align}$$
we have
$$\begin{align}
(1-x)y''-xy'+y&=0\\
(1-x)\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-x\sum_{n=0}^{+\infty}na_n x^{n-1}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-x\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-1}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=2}^{+\infty}n(n-1)a_nx^{n-2}-\sum_{n=1}^{+\infty}n(n-1)a_nx^{n-1}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}(n+2)(n+1)a_{n+2}x^{n}-\sum_{n=0}^{+\infty}(n+1)na_{n+1}x^{n}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}\left[(n+2)(n+1)a_{n+2}x^{n}-(n+1)na_{n+1}x^{n}-na_n x^{n}+a_n x^n\right]&=0\\
\sum_{n=0}^{+\infty}\left[(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}-na_n+a_n \right]x^n&=0
\end{align}$$
then you get
$$\begin{align}
(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}-na_n+a_n&=0\\
(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}+(1-n)a_n&=0\\
a_{n+2}=\frac{(n+1)na_{n+1}+(n-1)a_n}{(n+2)(n+1)}
\end{align}$$
since you got $y(0)=-2$ and $y'(0)=6$ you got $a_0$ and $a_1$ because
$$\begin{align}
y(0)&=\sum_{n=0}^{+\infty}a_n x^n\\
&=a_0\\
y'(0)&=\sum_{n=0}^{+\infty}na_n x^{n-1}\\
&=a_1
\end{align}$$
then with the value of $a_0$ and $a_1$ you can get $a_2,a_3,\cdots$ and then get the solution (valid within the series convergence radius)
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the equation of a circle given two points on the circle
11. Find the equation of the circle which touches $x^{2} + y^{2} - 6x + 2y + 5 = 0$ at $(4, -3)$ and passes through $(0, 7)$.
My textbook has a worked example for obtaining the equation of a circle from three points on the circle. It also talks you through obtaining the equation of a circle if you're given two points on the circle and if it touches an axis (in this case you know a coordinate of the centre will be $\pm r$, where $r$ is your radius.)
I am reasonably well-practised at using these two techniques. I have also been practising finding the length of a tangent from a given point.
However, I have so far been unable to make the "jump" to this question, I suspect there's something I'm not seeing. So I was hoping for a hint that would help me work out how to approach this question.
| I've come back to this question after a while and have found a solution which agrees with that in the textbook. My method is based primarily on the tips given by @Mufasa.
Let the circle which touches $x^{2} + y^{2} − 6x + 2y + 5 = 0$ at $(4,−3)$ and passes through $(0,7)$ be $C_{1}$.
Let the centre of $C_{1}$ be $(p, q)$ and let the radius be $a$.
$(4 - p)^{2} + (-3 - q)^{2} = a^{2}$
$p^{2} + (7 - q)^{2} = a^{2}$
Let the circle $x^{2} + y^{2} − 6x + 2y + 5 = 0$ be $C_{2}$.
The centre of $C_{2}$ is $(3, -1)$.
The gradient of the radius of $C_{2}$ to $(4, -3)$ is $-2$.
$\therefore$ The gradient of the tangent to $C_{2}$ at $(4, -3)$ is $\frac{1}{2}$.
$\therefore$ The gradient of the tangent to $C_{1}$ at $(4, -3)$ is $\frac{1}{2}$.
$\therefore$ The gradient of the radius of $C_{1}$ to $(4, -3)$ is $-2$.
$\therefore \dfrac{q - (-3)}{p - 4} = -2$
$2p + q - 5 = 0 \qquad (1)$
$(4 - p)^{2} + (-3 - q)^{2} = a^{2} \qquad (2)$
$p^{2} + (7 - q)^{2} = a^{2} \qquad (3)$
Equating (2) and (3), we ultimately get:
$p = \dfrac{5}{2}q - 3 \qquad (4)$
Substituting $p$ into (1), we get:
$q = \dfrac{11}{6}$
Substituting $q$ into (4), we get:
$p = \dfrac{19}{12}$
Substituting $p$ and $q$ into (3) we ultimately get:
$a^{2} = \dfrac{4,205}{144}$
This gives us our equation of $C_{1}$:
$$\left(x - \frac{19}{12}\right)^{2} + \left(y - \frac{11}{6}\right)^{2} = \frac{4,205}{144}$$
Expanding and simplifying, we get:
$$6x^{2} + 6y^{2} - 19x - 22y - 140 = 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2^{12^{7}+3}\equiv x \pmod{36}$, then what is the value of $x$?
If $2^{12^{7} + 3} \equiv x \pmod{36}$, then what is the value of $x$?
We have:
$$
\begin{align}
2^5 & \equiv - 4 \pmod{36} \\
2^{10} & \equiv 16 \pmod{36} \\
2^{12} & \equiv - 8 \pmod{36} \\
(2^{12})^{12} & \equiv 8^{12} \pmod{36} \\
2^{12^2} & \equiv - 8 \pmod{36} \text{, as } 8^{12} \equiv - 8 \pmod{36} \\
(2^{12^2})^{12^5} & \equiv - 8 \pmod{36} \\
2^{12^7} & \equiv - 8 \pmod{36} \\
2^{{12^7} + 3} & \equiv - 64 \pmod{36} \\
2^{{12^7} + 3} & \equiv 8 \pmod{36} \\
\end{align}
$$
Am I right or wrong? Is there any simpler way to find it?
| It's probably easier to work modulo $4$ and modulo $9$, and use CRT.
First we have
$$x \equiv 2^{12^{7}+3}\equiv 0 \pmod{4}$$
To compute
$$x \equiv 2^{12^{7}+3} \pmod{9},$$
it suffices to compute $12^{7} + 3 \equiv 3 \pmod{6}$, as $6 = \varphi(9)$. (Or, if you prefer, as $2^{6} \equiv 1 \pmod{9}$.) So
$$2^{12^{7}+3}\equiv 2^{3} \equiv 8 \pmod{9}.$$
Clearly, then, the solution is indeed $x = 8$ as you found it, as $8 \equiv 0 \pmod{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
| Since
$$\left(\frac{x}{1+x^2}\right)'=\frac{2-(1+x^2)}{(1+x^2)^2}=\frac{2}{(1+x^2)^2}-\frac{1}{1+x^2}$$
one has
$$2\int \frac{1}{(1+x^2)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$$
$$\Rightarrow \int\frac{1}{(1+x^2)^2}dx=\frac{x}{2(1+x^2)}+\frac{\tan^{-1} x}{2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove a sum of series How do I prove that for any natural number $n$ we have
$$\sum_{i=0}^n i^4 \neq \left(\sum_{i=0}^n i\right)^3?$$
Any help would be greatly appreciated.
| Recall that $$\sum_{i=1}^n i^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
and
$$\sum_{i=1}^n i = \dfrac{n(n+1)}2$$
We hence need
$$\left(\dfrac{n(n+1)}2\right)^3 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
This gives us either $n(n+1)=0$ or
$$15(n(n+1))^2 = 4(2n+1)(3n^2+3n-1) \,\,\,\, (\spadesuit)$$
$(\spadesuit)$ can be simplified as
$$15n^4+6n^3-21n^2-4n+4 = 0 \implies (n-1)(15n^3+21n^2-4) = 0 \,\,\,\, (\clubsuit)$$
Now consider the function $f(n) = (15n^3+21n^2-4)$. Note that we have $f'(n) = 45n^2+42n >0 $ for all $n \geq 1$. Hence, the function $f(n)$ is increasing for all positive integers. Further, $f(1) = 32 > 0$. Hence, $f(n)$ has no non-negative integer as its roots. The only solution to your initial problem is $n=0$ and $n=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Orthogonal circles What is the equation of the circle that is orthogonal to the circles $x^2 + y^2 - 8x +5 =0$ and $x^2 + y^2 +6x +5 = 0$ and passes through the point $(3,4)$? I've spent hours trying to figure this out - help please
| you have $$C_1:(x-4)^2+y^2=11,\quad C_2:(x+3)^2+y^2= 4$$ we will look for a point $(a,0)$ on the $x$-axis so that the tangents to the circle have the same length.
we have $$(4-a)^2 -11=(a+3)^2-4 \to-8a+5=6a+5\to a = 0.$$
we will use the fact that if a circle is orthogonal to both $C_1, C_2,$ then the $x$-cdt of the center is $0.$
let the the center of $C$ that is orthogonal to both $C_1, C_2$ and goes through $(3,4)$ be $(0,b)$.
$$x^2 + (y-b)^2 = b^2 + 9-4 \to x^2+y^2 - 2by=5$$ making this go through $(3,4)$ requires $$3^2 + 4^2-8b=5\to b= \frac 52 $$ andn the circle is $$x^2 + y^2-5y=5. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to\infty}(\frac{x^2}{x^2+x})^x = 1 $ or $0$? $$\lim_{x\to\infty}\left(\frac{x^2}{x^2+x}\right)^x$$
as $\lim_{x\to\infty}\frac{x^2}{x^2+x} = 1$ but we all know that $\frac{x^2}{x^2+x} < 1$ and for $a<1$ lim $\lim_{x\to\infty} a^x = 0$
I don't know how to find the limit in this case
| Mathlove privided an efficient way forward. Here is a slightly different approach. Take the logarithm of the function of interest. Then, we have
$$\log\left(\frac{x^2}{x+x^2}\right)^x=x\log\left(\frac{x}{x+1}\right)=\frac{\log\left(\frac{x}{1+x}\right)}{\frac{1}{x}}=\frac{\log x- \log (1+x)}{\frac{1}{x}}$$
Now, apply L'Hospital's Rule to find
$$\lim_{x\to \infty} \log\left(\frac{x^2}{x+x^2}\right)^x=\lim_{x\to \infty} \frac{\frac{1}{x}- \frac{1}{1+x}}{-\frac{1}{x^2}}=-1$$
Since the exponential function is continuous, then we have
$$\lim_{x\to \infty} \left(\frac{x^2}{x+x^2}\right)^x=e^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I need help proving the following: Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I know that the smallest $n$ is 8 by testing $n=1,2,3,...$. I need help proving that $n=8$ is the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$.
| Hint:
$$ x^4+4m^4 = (x^4+4x^2m^2+4m^4)-(4x^2m^2)=(x^2+2m^2)^2-(2mx)^2 = (x^2-2mx+2m^2)(x^2+2mx+2m^2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the interval on which $ x^{2} - \lfloor x \rfloor - 3 < 0 $ holds. On what interval does the equation $ x^{2} - \lfloor x \rfloor - 3 < 0 $ hold?
My attempt: I tried sketching the graph, but it’s a bit complicated.
Is there any other approach?
| Probably graphing $y=x^2-3$ and $y=\lfloor x\rfloor$ and seeing where the first graph is below the second is the easiest way to solve this, but here is another method:
$\textbf{1)}$ Since $\lfloor x\rfloor \le x$,
$x^2-3<\lfloor x\rfloor\implies x^2-3<x\implies x^2-x-3<0\implies x^2-x+\frac{1}{4}<\frac{13}{4}\implies$
$\left(x-\frac{1}{2}\right)^2<\frac{13}{4}\implies|x-\frac{1}{2}|<\frac{\sqrt{13}}{2}<2\implies-\frac{3}{2}<x<\frac{5}{2}$.
$\textbf{2)}$ Since $x-1<\lfloor x\rfloor$, the inequality will be satisfied if $x^2-3\le x-1$:
$x^2-3\le x-1 \iff x^2-x-2\le0\iff (x-2)(x+1)\le 0\iff -1\le x\le2$.
$\textbf{3)}$ a) If $-\frac{3}{2}<x<-1$, then $\lfloor x\rfloor=-2$,
so $x^2-3<\lfloor x\rfloor \iff x^2<1\iff -1<x<1$; so there is no solution in this case.
b) If $2<x<\frac{5}{2}$, then $\lfloor x\rfloor=2$, so $x^2-3<\lfloor x\rfloor \iff x^2<5\iff-\sqrt{5}<x<\sqrt{5}$;
and therefore if $2<x<\sqrt{5}$, the inequality is satisfied.
Therefore the inequality is satisfied for $-1\le x<\sqrt{5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Residue $\frac{e^z}{z^3\sin(z)}$ I want to find the residue of $$\frac{e^z}{z^3\sin(z)}$$ and get $$ \frac 1 {3!} \lim_{z \to 0} \left( \frac{d^3}{dz^3} \left(\frac{ze^z}{\sin(z)} \right)\right) = \frac{1}{3}$$
Can anyone confirm this? I tried using the Laurent Series, but I didn't know how to compute it.
Oh, I was able to compute the Laurent Series and confirm it was $\frac{1}{3}$
I don't know how to close this question though!
| $$\begin{align}\frac{e^z}{z^3 \sin{z}} &= \frac1{z^4} \frac{\displaystyle 1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots}{\displaystyle 1-\frac{z^2}{6} + \frac{z^4}{120}+\cdots} \\ &= \frac1{z^4} \left (1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots \right )\left [1-\left (-\frac{z^2}{6} + \frac{z^4}{120}+\cdots\right )+ \left (-\frac{z^2}{6} + \frac{z^4}{120}+\cdots\right )^2+\cdots\right ] \end{align}$$
Now, we need to pluck the coefficient of $1/z$ from the above. To this effect, you should be able to see that we do not need the squared term in the brackets on the RHS, as its first term is $O(z^4)$, which cancels out the $1/z^4$ in front of everything.
Thus, the residue is simply
$$\frac16 + \frac16 = \frac13 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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show $p$ is divisible by $(x^2 +y^2 +1)$ Show that, for any prime $p$, there are integers $x$, and $y$ such that $p$ is divisible by $(x^2+y^2+1)$ Can you show me what to start with? do I prove $p$ is divisible by $x^2$ and $y^2$ separately?
| Take $x=y=0$, then $x^2+y^2+1=1$, then $p$ is divisible by $x^2+y^2+1=1$.
| {
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Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$ Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$. I put $(2+2x)^3=3-2\cdot3^{x+1}$. But I don't know how to go on.
| Hint: $3-2\cdot 3^{x+1}>0$
So we'll have: $\:3^{2+2x}+2\cdot 3^{x+1}=3\:\:\rightarrow 9\cdot 3^{2x}+6\cdot 3^x=3\:\:\rightarrow \:3\cdot 3^{2x}+2\cdot 3^x-1=0\:$
let $3^x=a\:\:\rightarrow \:3a^2+2a-1=0$ with $a_1=-1$ and $a_2=\frac{1}{3}$, and only $a_2=\frac{1}{3}$ is good. Therefore, comeback at substitution we'll obtain: $3^x=\frac{1}{3}$ with only solution $x=-1$.
| {
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$17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$, & excluding $(x+y)$ the sum of the digits $(3+5+5+6+4+2+8+9+6) =48$.Thus excluding option d) the ans can be any of a),b) or c).
Please help me to find the exact answer.
Thank you
| First use the fact that the factorial is divisible by $9$, and therefore the digit sum is divisible by $9$.
Then use the fact that the number with decimal representation $abcdefg\dots$ is divisible by $11$ if and only if $a-b+c-d+e-f+\cdots$ is divisible by $11$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that
$$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$
I conjecture:
Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then
$$n(a_{1}+a_{2}+\cdots+a_{n})-(a^2_{1}+a^2_{2}+\cdots+a^2_{n}-a_{1}a_{2}-a_{2}a_{3}-\cdots-a_{n}a_{1})\ge (\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}})^2$$
This inequality is stronger than Cauchy-Schwarz inequality,because
$$(a^2+b^2+c^2-ab-bc-ac)\ge 0$$
Applying Cauchy-Schwarz inequality
$$3(a+b+c)\ge(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$
How to prove the required statement (1)?
| We need to prove that
$$\sum_{cyc}(3a-a^2+ab)\geq\sum_{cyc}(a+2\sqrt{ab})$$ or
$$2\sum_{cyc}(a-\sqrt{ab})\geq\sum_{cyc}(a^2-ab)$$ or
$$2\sum_{cyc}\left(\sqrt{a}-\sqrt{b}\right)^2\geq\sum_{cyc}(a-b)^2$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2-(\sqrt{a}+\sqrt{b})^2\right)\geq0,$$
for which it's enough to prove that
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2(a+b+c)-(\sqrt{a}+\sqrt{b})^2\right)\geq0$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left((\sqrt{a}-\sqrt{b})^2+2c\right)\geq0,$$
which is obvious.
Done!
| {
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A difficult integral: $\int_0^{+\infty} e^{ - x}\left(\frac1{x( e^{ - x} - 1 )} + \frac1{x^2} + \frac1{2x} \right) \, dx$. Could you help me calculate the integral?
$$\int_0^{+\infty} e^{-x} \left(\frac1{x(e^{-x}-1)} + \frac1{x^2} + \frac1{2x} \right) \, dx .$$
| One may recall Binet's formula, for $\Re z >0$,
$$
\log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) +
\int_0^{\infty} \!
\left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x. \tag1
$$ Observing that
$$
{\left( {\frac{1}{{x\left( {{e^{ - x}} - 1} \right)}} + \frac{1}{{{x^2}}} + \frac{1}{{2x}}} \right)} =-\left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{1}{x}
$$ then, using $(1)$ with $z:=1$, readily gives
$$
\int_0^{ + \infty } {{e^{ - x}}\left( {\frac{1}{{x\left( {{e^{ - x}} - 1} \right)}} + \frac{1}{{{x^2}}} + \frac{1}{{2x}}} \right)dx}= \frac{1}{2}\log(2\pi)-1.
$$
| {
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"question_score": "6",
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How to compute the series $\sum_{n=0}^{\infty}{\frac{1}{(n+2)7^n}}$ I have to compute the series
$$\sum_{n=0}^{\infty}{\frac{1}{(n+2)7^n}}$$
It's easy to prove that the series converges using comparison test but how to compute it?
| Consider
$$ S(x) = \sum_{n=0}^{\infty} \frac{x^n}{n+2}. $$
Your sum is $S(1/7)$. Then
$$ x S'(x)+2 S(x) = \sum_{n=0}^{\infty} \frac{(n+2)x^{n}}{n+2} = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} $$
An integrating factor for this differential equation is $x$, so we find that
$$ (x^2S)' = \frac{x}{1-x} = \frac{1}{1-x}-1 \\
x^2 S = C-x-\log{(1-x)}, $$
and setting $x=0$ implies that $C=0$. Hence
$$ S(x) = \frac{-1}{x^2}(x+\log{(1-x)}), $$
so
$$ S(1/7) = -49\left( \frac{1}{7} + \log{\left( \frac{6}{7} \right)} \right) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Boolean Algebra - proof without associativity? I would like to prove the following:
$(x\cdot y) + (\overline{x} + \overline{y}) = 1$
without the Associativity Property. I can't seem to do this algebraically (without truth tables).
| First note that
\begin{align*}
x + (\overline x + \overline y)
&= 1\cdot (x + (\overline x + \overline y)) \\
&= (x + \overline x)\cdot (x + (\overline x + \overline y)) \\
&= x + (\overline x\cdot (\overline x + \overline y)) \\
&= x + ((\overline x + 0)\cdot (\overline x + \overline y)) \\
&= x + (\overline x + (0\cdot \overline y)) \\
&= x + (\overline x + 0) \\
&= x + \overline x \\
&= 1
\end{align*}
Similarly,
$$ y + (\overline y + \overline x) = 1 $$
So
\begin{align*}
(x\cdot y) + (\overline x + \overline y)
&= (x + (\overline x + \overline y))\cdot (y + (\overline x + \overline y))
\\
&= (x + (\overline x + \overline y))\cdot (y + (\overline y + \overline x))
\\
&= 1\cdot 1 \\
&= 1
\end{align*}
This uses distributivity and the properties of $0$ and $1$, but not associativity and not De Morgan. (I also used the commutativity of $+$, but it's not essential.)
| {
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Application of the Banach fixed point theorem Let $a > 0$. We consider the function: $f: (0, \infty) \to (0, \infty)$, defined by $f(x) = \frac{1}{2}(x + \frac{a}{x})$.
Let $(x_n)_{n \in \mathbb{N}_0}$ be defined by:
$x_0 \in (0, \infty)$, $x_{n+1} := f(x_n)$
What is the smallest $b > 0$ so that f is contracting on $[b, \infty)$, and what is $f's$ Lipschitz-constant $L > 0$?
Also, I want to prove using the Banach fixed point theorem, that $(x_n)$ as defined above converges against $\sqrt{a}$.
Finally, why is it that $|f(x_n) - \sqrt{a}| ≤ \frac{1}{2^n}|\frac{a}{x_0} - x_0|$?
Thanks in advance. I'm not very familiar with the Banach fixed point theorem, so I've been struggling with these questions so far.
| *
*To answer that question, it is most convenient to solve $|f'(x)| < 1$. Since $f'(x) = \frac{1}{2} \left[ 1 - \frac{a}{x^2} \right],$ the solution for positive $x$ is $x \in \left( \sqrt{ \frac{a}{3} }, \infty \right)$.
So:
*
*for every $b > \sqrt{\frac{a}{3}}$ there is such $L \in (0, 1)$ that $|f'(x)| \leqslant L$ for $x \geqslant b$ and so by the mean value theorem $|f(x) - f(y)| \leqslant L|x-y|$ for $x, y \in [b, \infty)$, thus $f$ is Lipschitz on $[b, \infty)$
*$f$ is not a contraction on $\left[ \sqrt{\frac{a}{3}}, \infty \right)$, since $f' \left( \sqrt{\frac{a}{3}} \right) = -1$, so $\displaystyle \frac{\left| f \left( \sqrt{\frac{a}{3}} \right) - f(x) \right|}{\left| \sqrt{\frac{a}{3}} - x \right|}$ is arbitrarily close to $1$ for $x$ sufficiently close to $\sqrt{\frac{a}{3}}$.
Therefore there is no smallest $b$, but "a limit value" is $\sqrt{\frac{a}{3}}$.
*Fine. For any $x > 0$ we have $$f(x) = \frac{x+\frac{a}{x}}{2} \geqslant \sqrt{ x \cdot \frac{a}{x} } = \sqrt{a},$$ so we have $x_1 \in [ \sqrt{a}, \infty)$ and $x_{n+1} = f(x_n)$ and $f : [\sqrt{a}, \infty) \to [\sqrt{a} \to \infty)$ is a contraction (as shown above). Just apply Banach theorem and show that $x = \sqrt{a}$ is the only solution of $f(x) = x$.
*Since $$\left| f(x) - \sqrt{a} \right| = \left| \frac{1}{2} \left( x + \frac{a}{x} \right) - \sqrt{a} \right| = \frac{1}{2} \left| x - \sqrt{a} \right| \cdot \left| 1 - \frac{\sqrt{a}}{x} \right| \leqslant \frac{1}{2} \left| x - \sqrt{a} \right|$$ for $x \geqslant \sqrt{a}$, verify for $n = 0$ and use induction.
| {
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What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S = \hspace{8.5pt} \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256} + \ldots$
$\displaystyle \frac{3}{4}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \qquad \leftarrow S- \frac{1}{4}S$
For the Infinite Sum on the RHS $\displaystyle \left(S = \frac{a}{1-r}\right)$:
$\displaystyle a = 1$
$\displaystyle r = \frac{1}{2}$
Then
$\displaystyle \frac{3}{4}S = \frac{1}{1-\frac{1}{2}}$
$\displaystyle \frac{3}{4}S = 2$
$\displaystyle S = \frac{8}{3}$
Using Excel the answer is $\displaystyle \frac{5}{3}$, but I don't know where is the issue.
Thanks!!
| Your answer seems to be correct. We see that Excel's answer is wrong by adding just the first 2 terms in the series, $1, \frac 34$ and we see that $1.75>1.66..$, so we can see that Excel's answer is wrong.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Limit and Integral problem work verification-2 I have to calculate the following:
$$\large\lim_{x \to \infty}\left(\frac {\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
My attempt:
Let $F(x)=\displaystyle\int\limits_0^xt^4e^{t^2}dt$. Then,
$$\large\lim_{x\to\infty}\left(\frac{\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)=\lim_{x \to \infty}\left(\frac {F(2x) - F(x^2)}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
Applying L'Hôpital's rule, we have,
$$\large\begin{align}\lim_{x \to \infty}\left(\frac {32x^4e^{4x^2} - 2x^9e^{x^4}}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}}\right) &= \lim_{x \to \infty}(32x^4e^{4x^2-x} - 2x^9e^{x^4-x}) \\&= \lim_{x \to \infty}\bigg(2x^4e^{4x^2-x}(16-x^5e^{x^4-4x^2})\bigg) = -\infty\end{align}$$
Am I right?
| For large $x, x^2 > 2x.$ So it seems like a good idea to write the numerator as $-\int_{2x}^{x^2}t^4e^{t^2}\,dt.$ Notice that the integrand is positive and increasing. Thus the absolute value of the numerator is $\ge (2x)^4 e^{4x^2}(x^2-2x).$ The denominator is $< e^x$ for large $x.$ So in absolute value, it is clear, nay blatantly obvious, that our expression in absolute value $\to \infty.$ Now put the minus sign in to see the limit in question is $-\infty.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrate $\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}$using partial fractions How do I integrate the improper integral $$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}$$ using partial fraction decomposition?
I am restricted to use only the principles taught in Calculus 2, which entails partial fraction decomposition. I have tried factoring the denominator but was quickly stumped at the complicated roots. I am unsure of how to proceed. I even thought of looking at it as a series but I don't think I can do that.
Thank you very much for your help. I greatly appreciate it.
| Decompose the integrand
$$ \frac{6}{1+x^{12}}
=\frac2{x^4+1}-\frac{\sqrt3x^2-2}{x^4-\sqrt3x^2+1}
+ \frac{\sqrt3x^2+2}{x^4+\sqrt3x^2+1} $$
Then
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}
=\frac23 J(0) +\frac{2+\sqrt3}3 J(\sqrt3)+ \frac{2-\sqrt3}3 J(-\sqrt3)\tag1$$
where
\begin{align}
J(a) &= \int_{0}^{\infty} \frac{dx}{x^4+ax^2+1}
\overset{x\to \frac1x} =\int_{0}^{\infty} \frac{x^2dx}{x^4+ax^2+1} \\
& =\frac12 \int_{0}^{\infty} \frac{(x^2+1)dx}{x^4+ax^2+1}
=\frac12 \int_{0}^{\infty} \frac{d(x-\frac1x)}{(x-\frac1x)^2+2+a}
=\frac\pi{2\sqrt{2+a}}\\
\end{align}
Let $a=0, \pm \sqrt3$ and plug the results into (1) to obtain
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}} = \frac{\sqrt3+1}{3\sqrt2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integration of $\frac{1}{\sin x+\cos x}$ I'm given this $\int\frac{1}{\sin x+\cos x}dx$.
My attempt,
$\sin x+\cos x=R\cos (x-\alpha)$
$R\cos \alpha=1$ and $R\sin \alpha=1$
$R=\sqrt{1^2+1^2}=\sqrt{2}$,
$\tan\alpha=1$
$\alpha=\frac{\pi}{4}$
So, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi}{4})$
$\int\frac{1}{\sin x+\cos x}dx=\int \frac{1}{\sqrt{2}\cos (x-\frac{\pi}{4})}dx$
$=\frac{1}{\sqrt{2}}\int \sec (x-\frac{\pi}{4})dx$
$=\frac{1}{\sqrt{2}} \ln \left | \sec (x-\frac{\pi}{4})+\tan (x-\frac{\pi}{4}) \right |+c$
Am I correct? Is there another way to solve this integral? Thanks in advance.
| Try using the substitution $t=\tan\frac x2$
| {
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Is it possible to find the sum of all integer values that $x$ can take? Is it possible to find the sum of all integer values that $x$ can take? In:
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
| hint: $x+3-4\sqrt{x-1}=(\sqrt{x-1}-2)^2, x+8-6\sqrt{x-1}=(\sqrt{x-1}-3)^2$
| {
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A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive one for demonstration
$\frac{dy}{dx}=\frac{1}{2}x(\frac{1}{2}x^2-\frac{1}{2})^{-\frac{1}{2}}=\frac{x}{2\sqrt{\frac{1}{2}x^2-\frac{1}{2}}}$
$\implies(1-2m^2)x^2=-2m^2$
Since the tangent touches the curve, we can make $x^2-2(mx+2)^2=1$, we then get $(1-2m^2)x^2=9+8mx$
$\implies(1-2m^2)x^2=-2m^2$ and $(1-2m^2)x^2=9+8mx$ are two equations with two unknowns, then we should be able to find the values of $m$, but I couldn't find any easy way to solve those 2 simultaneous equations. Is there any easier method?
I tried solving $9+8mx=-2m^2$ but we still have two unknowns in one equation?
Also, if we don't use those two simultaneous equations, can we solve this question with a different method?
I am trying to solve WITHOUT implicit differentiation.
Many thanks for the help!
| You have $$(1-2m^2)x^2=-2m^2\\(1-2m^2)x^2=9+8mx\\-2m^2=9+8mx\\x=\frac{-2m^2-9}{8m}\\x^2=\frac{-2m^2}{1-2m^2}\\\frac{4m^4+36m^2+81}{64m^2}=\frac{-2m^2}{1-2m^2}$$Now you can use the quadratic equation in $m^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $ Cauchy and Ramanujan both gave the formula:
$$ \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} = (2\pi)^{4p+3}\sum_{k=0}^{2p+2} (-1)^{k+1}
\frac{B_{2k}}{(2k)!}\frac{B_{4(p+1)-2k}}{(4(p+1)-2k)!}
$$
If we lightly manipulate things, we can use a bit of umbral calculus to simplify the formula
\begin{eqnarray} -\frac{[4(p+1)]!}{(2\pi)^{4p+3}}\sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} &=& \sum_{k=0}^{2p+2} (-1)^{k}
\binom{4(p+1)}{2k}B_{2k}B_{4(p+1)-2k} \\
&=& \frac{1}{2}\left[ (B+iB)^{2(p+1)}+(B-iB)^{2(p+1)}\right]
\end{eqnarray}
It's still a mess, but at least we have normalization. Prove LHS = RHS
| There is also a way to obtain this formula using the partial fraction expansion of cotangent:
$$ \frac{\pi}{2x} \coth(\pi x) - \frac{1}{2x^2} = \sum_{k=1}^{\infty} \frac{1}{k^2+x^2} $$
Let $m\geq 0$ be an integer, set $x=n$, divide by $n^{4m+2}$ and then sum from $n=1$ to infinity to get
$$ \sum_{n=1}^{\infty} \frac{\pi}{2n^{4m+3}} \coth(\pi n) - \sum_{n=1}^{\infty} \frac{1}{2n^{4m+4}} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} $$
The sum on the right may be handled by switching the order of summation (justified by absolute convergence) and then rewriting things as follows:
$$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}} \frac{1}{n^2(k^2+n^2)} = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}k^2} \left(\frac{1}{n^2}-\frac{1}{k^2+n^2} \right) = \zeta(4m+2)\zeta(2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}k^2(k^2+n^2)} $$
Repeating the procedure results in
$$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} = \zeta(4m+2)\zeta(2) - \zeta(4m)\zeta(4) + \cdots + \zeta(2) \zeta(4m+2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^{4m+2}(k^2+n^2)}$$
The final sum is the same as the first if we switch the dummy variables, and so we have that
$$ 2\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} = 2 \sum_{k=0}^{m-1} (-1)^k \zeta(4m+2-2k) \zeta(2k+2) +(-1)^m \zeta^2(2m+2) $$
Replacing this in the equality above and cleaning up a bit gives the result
$$ \sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^{4m+3}} = \frac{\zeta(4m+4)}{\pi} + \frac{2}{\pi} \sum_{k=0}^{m-1} (-1)^k \zeta(4m+2-2k)\zeta(2k+2) + \frac{(-1)^m}{\pi} \zeta^2(2m+2) $$
This may be expressed in the form you wrote by using the well known formula for zeta at the even integers.
| {
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Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$
I know that
\begin{equation*}
(1 + x)^n = 1 + nx +\frac {n(n-1)}2!\cdot x^2 +\frac {n(n-1)(n-2)}3! \cdot x^3 +...
\end{equation*}
but how can I use it to solve the above problem>Is there any other easier way to do it ?
| $$
(x+1)^n=((x-1)+2)^n=2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2+(x-1)^3 \cdot F(x)
$$
So the remainder is $$2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $(G,*)$ is a group? Let $G=\mathbb{R_0}\times\mathbb{R}$ where $\mathbb{R_0}=\mathbb{R}\setminus\{{0}\}$. Define operation $*$ on $G$ by $(a,b)*(x,y)=(ax,a^2y+b)$.
I'd like to prove that $(G,*)$ is a group. Immediately $(a,b), (x,y)\in G$ implies $(ax,a^2y+b)\in G$.
I need help to prove associativity of $*$.
Thanks for replies.
| *
*First note that $*$ is well-defined. If $(a,b), (x,y) \in G$, then $a,x \in \Bbb R^*$, so $ax \in \Bbb R^*$, which means that $(ax, a^2 y + b) \in \Bbb R^* \times \Bbb R = G$.
*We show that $(1, 0) \in G$ is the neutral element of $G$:
$$\begin{align*} (1,0) * (x,y) &= (1\cdot x, 1^2 y + 0) = (x,y) \\ (x,y) * (1,0) &= (x\cdot 1, x^2 \cdot 0 + y) = (x,y) \end{align*}$$
*Next we show, that for $(x,y) \in G$, $\left(\frac 1 x, -\frac{y}{x^2} \right) \in G$ is the inverse of $(x,y)$:
$$\begin{align*}(x,y) * \left( \frac{1}{x}, - \frac{y}{x^2} \right) &= \left( x \cdot \frac 1 x, x^2 \cdot \left(-\frac{y}{x^2} \right) + y \right) = (1,0) \\ \left( \frac 1 x, - \frac{y}{x^2} \right) * (x,y) &= \left( \frac{1}{x} \cdot x, \left( \frac{1}{x}\right)^2 \cdot y - \frac{y}{x^2} \right) = (1,0) \end{align*}$$
*It remains to show, that the operation $*$ is associative. Let $(a,b), (c,d), (e,f) \in G$. Then
$$\begin{align*} \left( (a,b) * (c,d) \right) * (e,f) &= \left( ac, a^2d+b) *(e,f) \right) = \left( ace, a^2c^2f + a^2d + b\right) \; , \\ (a,b) * \left( (c,d) * (e,f) \right) &= (a,b) * \left( ce, c^2f + d\right) = \left(ace, a^2 c^2 f+a^2 d + b\right) \; . \end{align*}$$
So we have shown, that $G$ is a group.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $f(x,y)$ is continuous in $(0,0)$ Prove that $f(x,y)$ is continuous in $(0,0)$, where
\begin{equation}
f(x,y) = \begin{cases}
\frac{x^2y}{x^4+y^2}, & (x,y)\neq 0\\
0, & (x,y) = (0,0)
\end{cases}
\end{equation}
The solution I have is that f is not continuous in $(0,0)$. (The solution doesn't say more than that.)
However, the result I got is that $f$ is continuous in $(0,0)$. Here's my approach:
Lets transform $x$ and $y$ into their polar coordinates, so that we can approach $(0,0)$ from any direction by varying $\theta$:
$(x,y) = (r\cos(\theta),r\sin(\theta)) \qquad r\in\mathbb{R}^+_0 \quad \theta\in[0,2\pi)$
Then $f$ is continuous iff
\begin{equation}
\lim_{(x,y)\to (0,0)} f(x,y) = 0 = f(0,0)
\end{equation}
By using the polar coordinates and letting $r\to 0$ we get:
\begin{equation}
\lim_{(x,y)\to (0,0)} f(x,y)
=
\lim_{r\to 0} \frac{r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta + r^2\sin^2\theta}
= \lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
\end{equation}
By a case distinction by $\theta$ we get:
*
*$\theta\in[0,2\pi)\backslash\{0,\pi\}$:
\begin{equation}
\lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
=
\lim_{r\to 0} \frac{0}{\sin^2\theta} = 0 = f(0,0)
\end{equation}
*$\theta\in\{0,\pi\}$: $\Longrightarrow \sin\theta = 0$
\begin{equation}
\lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
=
\lim_{r\to 0} \frac{0}{r^2\cos^4\theta} = 0 = f(0,0)
\end{equation}
Form 1. and 2. we can conclude that $f$ is continuous in $(0,0)$.
What am I doing wrong in my approach?
| You have shown the limit is $0$ along all straight lines through $(0,0).$ That's not enough: There are tons of paths to $(0,0)$ that are not straight lines. For example, the path $(x,x^2)$ as $x\to 0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the vertex of a rhombus? I am unable to solve this question.
If the area of a rhombus is 10 sq.unit . It's diagonals intersect at (0,0) if one vertex of the rhombus is (3,4) , then one of the other vertices can be ?
I took a rhombus as ABCD . I took A as (3,4) and took O(0,0) as the point of the intersections of the diagonals . I found out OA as 5 and OB as 1 . I found out C as(-3,-4) . Now , the problem is that I am unable to find the vertex B . Please tell me how do I find out the vertex B . Thank you!
| As we know that the diagonals of rhombus are equal in length & intersect each other normally. Distance of each vertex from the origin is $\sqrt{3^2+4^2}=5$.
Thus assuming the rhombus ABCD, the vertex C opposite to $A\equiv (3, 4)$ can be easily determined as the moi-point of AC is $(0, 0)$ Hence, $C(-3, -4)$. Now, assume any vertex say $(x, y)$ on the other diagonal BD. Since, the semi-diagonal OB is normal to OA, we get $$\left(\frac{y-0}{x-0}\right)\left(\frac{4-0}{3-0}\right)=-1 \implies y=-\frac{3}{4}x \tag 1$$ Since $OB=OA=\sqrt{3^2+4^2}=5$, we get
$$\sqrt{(x-0)^2+(y-0)^2}=5 \implies x^2+y^2=25$$ Setting the value of $y$, we get $$x^2+\left(-\frac{3}{4}x\right)^2=25 \implies x^2=16 \implies x=\pm 4$$$$ x=4\quad \implies y=-\frac{3}{4}(4)=-3$$ $$ x=-4\quad \implies y=-\frac{3}{4}(-4)=3$$ Thus, all the unknown vertices of rhombus ABCD are $B\equiv (4, -3)$, $C\equiv(-3, -4)$ & $D\equiv(-4, 3)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \binom{n-i}{k}=0$ I would like to prove that:
\begin{equation*}
\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \binom{n-i}{k}=0;~k\geq0 ; n\geq1.
\end{equation*}
Can any one help me how to do that? Thanks
| Just for fun, here's a generating function answer. Fix $k \geq 0$ and let $a_n = (-1)^n$ and $b_n = \binom{n}{k}$. Letting $a$ and $b$ be the exponential generating functions for these sequences, then
\begin{align*}
a(x) = \sum_{n \geq 0} (-1)^n \frac{x^n}{n!} = \sum_{n \geq 0} \frac{(-x)^n}{n!} = e^{-x}
\end{align*}
and
\begin{align*}
b(x) &= \sum_{n \geq 0} \binom{n}{k} \frac{x^n}{n!} = \sum_{n \geq k} \binom{n}{k} \frac{x^n}{n!} = \sum_{n \geq k} \frac{n!}{k!(n-k)!} \frac{x^n}{n!} = \sum_{n \geq k} \frac{1}{k!(n-k)!} x^n\\
&= \frac{x^k}{k!}\sum_{n \geq k} \frac{1}{(n-k)!} x^{n-k} = \frac{x^k}{k!}\sum_{j \geq 0} \frac{1}{j!} x^{j} = \frac{x^k}{k!} e^x
\end{align*}
where we have made the change of index $j = n-k$. By the convolution formula for exponential generating functions, then
\begin{align*}
\sum_{i=0}^n \binom{n}{i} (-1)^i \binom{n-i}{k} &= \left[\frac{x^n}{n!}\right] a(x) b(x) = \left[\frac{x^n}{n!}\right] e^{-x} \frac{x^k}{k!} e^{x}\\
&= \left[\frac{x^n}{n!}\right] \frac{x^k}{k!} =
\begin{cases}
1 & \text{if } \ n=k\\
0 & \text{otherwise}
\end{cases}
\end{align*}
where $\left[\frac{x^n}{n!}\right]$ is the coefficient of $\frac{x^n}{n!}$ in the formal series.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$\lfloor 2x \rfloor \lfloor 3x \rfloor$ From $1$ to $10000$ including both, how many of those integers can be written as:
$$\lfloor 2x \rfloor \lfloor 3x \rfloor$$
Where $x$ is a real number?
| To help you get you started:
If $n \le x < n+\dfrac{1}{3}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = 2n \cdot 3n = 6n^2$.
If $n+\dfrac{1}{3} \le x < n+\dfrac{1}{2}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = 2n(3n+1) = 6n^2+2n$.
If $n+\dfrac{1}{2} \le x < n+\dfrac{2}{3}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = (2n+1)(3n+1) = 6n^2+5n+1$.
If $n+\dfrac{2}{3} \le x < n+1$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = (2n+1)(3n+2) = 6n^2+7n+2$.
Now, you need to figure out how many integers between $1$ and $10000$ can be written in one of these forms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$ Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
This is my try
Let $t=\sqrt{2x+3}-2\sqrt{x+1}$ or $t^2=6x+7-4\sqrt{2x^2+5x+3}=1$
The equation is equivalent to: $t^2-4(x+2)t-6x-3=0\qquad(*)$
I have no idea how to solve the equation $(*)$. Who can help me?
| $$
(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1
$$
Let $t=x+1$,
$$2x+3=2t+1$$
$$(t+1)(\sqrt{2t+1}-2\sqrt{t})+\sqrt{2t^2+t}=1$$
$$(t+1)(\sqrt{2t+1}-2\sqrt{t})=1-\sqrt{2t^2+t}$$
Squaring both sides,
$$(t^2+2t+1)(2t+1+4t-4\sqrt{2t^2+t})=1+2t^2+t-2\sqrt{2t^2+t}$$
$$6t^3+12t^2+6t+t^2+2t+1-\sqrt{2t^2+t}(-4t^2-8t-2)=1+2t^2+t$$
$$t(6t^2+11t+7)=\sqrt{t}\sqrt{2t+1}(-4t^2-8t-2)$$
Again Squaring both sides,
$$
4t^6-12t^5-19t^4+10t^3+9t^2-4t=0
$$
Therefore , $t=-1,0,\frac{1}{2},4$
$$
x=-2,-1,-\frac{1}{2},3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$
How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?
*
*I squared the whole denominator, but that didn't help.
*Also I searched for a propriety or identity like $A^2-B^2$, but I didn't see one that could fit.
Any help is appreciated.
| To get rid of the square roots of the denominator, you may use $a^2-b^2=(a+b)(a-b)$.
$$\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{7-4\sqrt{35}+20-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{24-4\sqrt{35}}\\=\dfrac{(\sqrt{7}-2\sqrt{5}-\sqrt{3})(24+4\sqrt{35})}{24^2-16*35}\\=\dfrac{(\sqrt{7}-2\sqrt{5}-\sqrt{3})(24+4\sqrt{35})}{16}\\=\dfrac{-24\sqrt3-20\sqrt5-16\sqrt7-4\sqrt{105}}{16}\\=\dfrac{-6\sqrt3-5\sqrt5-4\sqrt7-\sqrt{105}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sin(a)$ + $\cos(a)\leq\sqrt{2}$ $$\begin{align*}
\sin (a) + \cos(a) &\leq \sqrt{2}\\
(\sin(a)+ \cos(a))^2 &\leq (\sqrt{2})^2\\
\sin^2(a) + 2\sin(a)\cos(a) + \cos^2(a) &\leq \text{2}
\end{align*}$$
Am I doing it right? I need help.
| Alternative path:
$$
\sin a+\cos a=
\sqrt{2}\left(\sin a\cos\frac{\pi}{4}+\cos a\sin\frac{\pi}{4}\right)=
\sqrt{2}\sin\left(a+\frac{\pi}{4}\right)\le\sqrt{2}
$$
(and also $\ge-\sqrt{2}$, of course).
However your reasoning is basically correct; only you need to do it backwards:
$$
\sin^2a+2\sin a\cos a+\cos^2a\le2
$$
because $\sin^2a+\cos^2a=1$ and $2\sin a\cos a=\sin 2a\le 1$; therefore
$$
(\sin a+\cos a)^2\le 2
$$
and so
$$
\sin a+\cos a\le \sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$ Let $a,b,c\in \mathbb{R^+}$.
Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$
PS: I don't have ay ideas about this problem :(
Thanks
| By Holder $\left(\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b^2+c^2}}\right)^3\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3\geq\left(\sum\limits_{cyc}(a^2+ab)\right)^4$.
Thus, it remains to prove that $\left(\sum\limits_{cyc}(a^2+ab)\right)^4(a+b+c)^3\geq729abc\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3$, which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304942",
"timestamp": "2023-03-29T00:00:00",
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Limit Problem-To find the value of a and b,when the value of the limit is given. Here,is a limit problem: $\lim \limits _{x \to 0} {x^3 \over {\sqrt {a+x}} (bx - \sin x)} = 1$. Here, $a \in \mathbb R _+$. The question is to find the values of $a$ and $b$.
Here is my workout. Dividing the numerator and denominator by $x$, $\lim \limits _{x \to 0} {x^2 \over {b \sqrt {a+x} - \sqrt {a+x}}} = 1$. Rationalising and simplifying, $\lim \limits _{x \to 0} {x^2 \over {\sqrt a (b-1)}} = 1$.
I could not do beyond this, please anyone help.
| Suppose that $b \neq 1$.
Concerning $a$, there are two possibilities. Assume first that $a \neq 0$. Then $\sqrt {a+x} \to \sqrt a$. The rest of your expression becomes ${x^3 \over {bx - \sin x}} = {x^2 \over {b - {\sin x \over x}}}$, which will tend to $0$, so your whole limit will be $0$, not $1$. If $a=0$, then your expresion becomes ${x^{3 \over 2} \over {b - {\sin x \over x}}}$, which again tends to $0$.
This shows that necessarily $b=1$.
Assume now that $a=0$. Your expression becomes ${x^{5 \over 2} \over {x - \sin x}}$; apply l'Hopital's theorem twice, reducing it to ${5 \over 2} {x^{3 \over 2} \over {1 - \cos x}}$, then to ${15 \over 4} {x^{1 \over 2} \over {\sin x}}$, which tends to $\infty$ (also, because of the $1 \over 2$ in the power, the previous expression is defined only for $x>0$). We conclude that $a \neq 0$.
Finally, $\lim \limits _{x \to 0} {x^3 \over {\sqrt {a+x}} (x - \sin x)} = {1 \over {\sqrt a}} \lim \limits _{x \to 0} {x^3 \over {x - \sin x}} = {3 \over {\sqrt a}} \lim \limits _{x \to 0} {x^2 \over {1 - \cos x}} = {6 \over {\sqrt a}} \lim \limits _{x \to 0} {x \over {\sin x}} = {6 \over {\sqrt a}}$. Since this must be $1$, necessarily $a=36$.
To conclude: $a=36, \space b=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show algebraically that $x^3 +3x +1$ is injective?
How to show algebraically that $$x^3 +3x +1$$ is injective?
Working with the usual method of assuming that $f(c)=f(d)$ and then seeing if $c=d$. I've tried several approaches, including factoring by the difference of cubes, followed by use of the quadratic formula. I'm stumped. Any help greatly appreciated!
Thanks, Bob
| Notice $$\begin{align}
f(x) - f(y) &= (x^3 + 3x + 1) - (y^3 + 3y+1)\\
&= (x-y)(x^2 + xy + y^2 + 3)\\
&= \frac12 (x-y)( x^2 + (x+y)^2 + y^2 + 6)\end{align}$$
Since $x^2 + (x+y)^2 + y^2 + 6 \ne 0$ for any pair of real numbers $x, y$;
we have $f(x) \ne f(y)$ whenever $x \ne y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$
where real $x,y,z > 0 $ and $xy + yz + zx = 1$
Thank you for your help .
| By the conditions, we have:
$$y =\frac{1-xz}{x+z}\ge0\quad\longrightarrow \quad xz\le1, z \le 1/x$$
Now, without loss of generality, assume $z\ge y\ge x$. We therefore have that the expression $f(x,y,z)$ in question satisfies:
$$f(x,y,z) \ge \frac{3}{2} \frac{x^n}{z^m}\ge \frac{3}{2}x^{n-m}$$
Now it remains to be shown that $x\ge 1/\sqrt{3}$. This is true since the minimal distance from the origin to the hyperboloid is exactly $1$, so that:
$$x^2+y^2+z^2 \ge 1 \quad\longrightarrow \quad 3x^2 \ge 1$$
Thus completing the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Show that $\sin^7 x + \cos^7 x < 1$ if $0 < x < \frac{\pi}{2}$. Could anybody help me solve this Problem? I don't quite get it.
This is what I got.
$\sin^7x+\cos^7x<1$
$0<x<\frac{\pi}{2}$
$0<\cos x$
$1>\sin x$
What else?
| Using the Pythagorean theorem on the unit circle, we can easily figure out the identity of $\sin^2 x + \cos^2 x = 1$. We also know that if $0<x<\frac{\pi}{2}$, then $0 < \sin x < 1$ and $0 < \cos x < 1$. We know that if $0<\alpha<1$ and $n>1$ are true, then $\alpha ^n < 1$. If $\sin^2 x + \cos^2 x = 1$, then $\boxed{\sin^7 x + \cos^7 x < 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{1}{2}(a+b)^2+\frac{1}{4}(a+b)\ge a\sqrt{b}+b\sqrt{a}$ How to prove
\begin{equation*}
\frac{1}{2}(a+b)^2+\frac{1}{4}(a+b)\ge a\sqrt{b}+b\sqrt{a}\ ,where \ a,b\ge 0
\end{equation*}
I applied AM-GM inequality but obtained only
\begin{equation*}
\frac{3}{2}(a+b)\ge a\sqrt{b}+b\sqrt{a}+1
\end{equation*}
which is not true for, example, $a=0, \ b=\frac{1}{3}$
| From first principles:
$$(\sqrt{x}-\sqrt{y})^2 \geq 0$$
$$x+y-2\sqrt{xy} \geq 0$$
$$\sqrt{xy} \leq \frac{x+y}{2}$$
And:
$$(\sqrt{x}-\frac12)^2 \geq 0$$
$$x+\frac14-\sqrt{x} \geq 0$$
$$\sqrt{x} \leq x + \frac14$$
Given that:
$$\begin{split}
a\sqrt{b}+b\sqrt{a} &= \sqrt{ab}(\sqrt{a}+\sqrt{b}) \\
&\leq\frac{a+b}2(\sqrt{a}+\sqrt{b}) \\
&\leq\frac{a+b}2(a+b+\frac12) \\
&=\frac12(a+b)^2+\frac14(a+b)
\end{split}$$
$$$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$. Hint: $x^2-y^2 = (x+y)(x-y)$. This is the exercise verbatim:
An integer n is a square modulo p if there exists another integer x
such that $n \equiv x^2 \pmod p$. Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.
Hint: $x^2-y^2 = (x+y)(x-y)$.
This is my attempt to solve it:
If $x \equiv y \pmod p$ then $\displaystyle \frac{x-y}{p} = q_1 \Rightarrow (x-y)=pq_1$.
If $x \equiv -y$ then $\displaystyle \frac {x+y}{p}=q_2 \Rightarrow (x +y)=pq_2$.
Then $x^2-y^2 = (x+y)(x-y)=(q_1p)\cdot(q_2p)=q_1q_2p^2$
If $p$ is to divide $ x^2-y^2 $ evenly, then it must also divide $q_1q_2p^2$, therefore $\displaystyle \frac {x^2-y^2}{p}=q_1q_2p$.
To satisfy the condition of divisibility by $p$, $q_1q_2p$ must be an integer. Since the assertion to be proven is "Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.", we must show that the premise is true if either $q_1$ or $q_2$ is an integer for sure.
Is this line of reasoning correct up until this point?
How to guarantee that $q_1q_2p$ is an integer?
| You can use the theorem : p( a prime) divides the product $ab$ if and only if p divides $a$ or $p$ divides $b$. Now $x^{2} \equiv y^{2}$ mod $p$ if and only if $p$ divides $x^{2}-y^{2}=(x-y)(x+y)$, if and only if $p$ divides $x+y$ or $p$ divides $x-y$ if and only if $x \equiv y$ mod $p$ o $x \equiv -y$ mod p.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Part of an induction question I might have done or not realize something stupid, but I can't seem to prove the following...
Inductive hypothesis
Assume $\exists$k$\in$N such that P(k) is true.
P(k): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1)}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k)}$ $\leq$ $\frac{1}{\sqrt{k + 1}}$
Inductive step
Prove P(k+1) is true.
P(k+1): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$ $\leq$ $\frac{1}{\sqrt{(k + 1) + 1}}$
$\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$
= $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1) \cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k) \cdot [2(k + 1)]}$
$\leq$ $\frac{1}{\sqrt{k + 1}}$ $\cdot$ $\frac{2(k + 1) - 1}{2(k + 1)}$ #by IH
= $\frac{2(k + 1) - 1}{2(k + 1) \cdot \sqrt{k + 1}}$
$\leq$ $\frac{2(k + 1)}{2(k + 1) \cdot \sqrt{k + 1}}$
= $\frac{1}{\sqrt{k + 1}}$
So what I seem to have done is proven that $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$ $\leq$ $\frac{1}{\sqrt{k + 1}}$ and not $\frac{1}{\sqrt{(k + 1) + 1}}$. I seem to need a minor adjustment somewhere.
| You have given away too much. You want to prove that
$$\frac{2k+1}{2(k+1)\sqrt{k+1}}\lt \frac{1}{\sqrt{k+2}}.$$
Equivalently, you want to prove that $(2k+1)^2(k+2)\lt 4(k+1)^3$. Expand, and the result will drop out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $\implies$ $(a^2+1)v=0$ $\implies$ $v=0$.
Case $n=2$: Write $A=\begin{pmatrix}a&b \\ c&d\end{pmatrix}$. Then, $A^T A=\begin{pmatrix}a^2+c^2&ab+cd\\ ab+cd&b^2+d^2\end{pmatrix}$, so
$$
\begin{align}
\det(A^TA+1)
&= \begin{vmatrix}a^2+c^2+1&ab+cd\\ ab+cd&b^2+d^2+1\end{vmatrix}\\
&= 1+a^2+b^2+c^2+d^2+(ad-bc)^2\\
&\neq 0.
\end{align}
$$
Now, can we generalize to all $n$?
| You may try to show that $A^TA+I$ is invertible. Hint: if $(A^TA+I)x=0$, consider $x^T(A^TA+I)x$. Try to rewrite it as a sum of squares and infer that $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial. Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?
| I started out by observing that $2+3 = 5$; I thought that might make it easier. Then
$$
x-\sqrt{5} = \sqrt{2}+\sqrt{3}
$$
$$
x^2-2\sqrt{5}x+5 = 2\sqrt{6}+5
$$
$$
x^2-2\sqrt{5}x = 2\sqrt{6}
$$
$$
x^4-4\sqrt{5}x^3+20x^2 = 24
$$
$$
x^4+20x^2-24 = 4\sqrt{5}x^3
$$
$$
x^8+40x^6+352x^4-960x^2+576 = 80x^6
$$
$$
x^8-40x^6+352x^4-960x^2+576 = 0
$$
(Or, of course, you can do it more systematically via the Galois theory!)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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