Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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A limit involving the Hurwitz zeta function I want to show that
$$ \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) = - \psi(a)$$
where $\zeta(s,a)$ is the Hurwitz zeta function and $\psi(a)$ is the digamma function.
The only approach I can think of is to use the integral representation
$$\zeta(s,a) = 2 \int_{0}^{\infty} \frac{\sin (s \arctan \frac{t}{a} )}{(a^{2}+t^{2})^{s/2} (e^{2 \pi t}-1)} \, dt + \frac{1}{2a^{s}} + \frac{a^{1-s}}{s-1} $$ which is valid for all complex values of $s$.
$$ \begin{align} \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) &= 2 \int_{0}^{\infty} \frac{\sin (\arctan \frac{t}{a} )}{(a^{2}+t^{2})^{1/2} (e^{2 \pi t}-1)} \, dt + \frac{1}{2a} \\ &= 2\int_{0}^{\infty} \frac{t}{(a^{2}+t^{2})(e^{2 \pi t}-1)} \, dt + \frac{1}{2a} \\ &= - 2 \, \frac{d}{d a} \int_{0}^{\infty} \frac{\arctan (\frac{t}{a})}{e^{2 \pi t}-1} \, dt + \frac{1}{2a}.\end{align}$$
Binet's second log gamma formula states
$$ 2 \int_{0}^{\infty} \frac{\arctan \left( \frac{x}{z} \right)}{e^{2 \pi x} -1} \, dx = \ln \Gamma(z) - \left( z- \frac{1}{2} \right) \ln z + z - \frac{\ln (2 \pi)}{2}.$$
So
$$ \begin{align} \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) &= - \frac{d}{da} \left(
\ln \Gamma(a) - \left( a- \frac{1}{2} \right) \ln (a) + a - \frac{\ln (2 \pi)}{2} \right) + \frac{1}{2a} \\ &= -\psi(a) +\ln (a) +1 -\frac{1}{2a} -1 + \frac{1}{2a} \\ &= - \psi(a) \color{#C00000}{+ \ln (a)} . \end{align} $$
Why do I have $\ln (a)$ in my final answer?
| Everything is correct except in the beginning where you take a limit of the integral representation. You conclude that $\lim_{s\rightarrow 1}\frac{a^{1-s}-1}{s-1}=1$, whereas by l'Hopital's rule, the correct answer is $-\ln(a)$ which should cancel the other log that you get later.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find AB where A= matrix and B=matrix $A=\left[\begin{array}{ccc} 2&1&0\\0&3&-1
\end{array}\right]$
$B=\left[\begin{array}{cc}a&1\\1&b\\b&a\end{array}\right]$
Matrices
Find $AB$
| $$AB=\left[\begin{array}{ccc} 2&1&0\\0&3&-1
\end{array}\right]\left[\begin{array}{cc}a&1\\1&b\\b&a\end{array}\right]$$
If you know basic matrix multiplication, you should know
$$\begin{align}
AB =& \left[\begin{array}{cc}
A\left[\begin{array}{c}a\\1\\b\end{array}\right]&
A\left[\begin{array}{c}1\\b\\a\end{array}\right]
\end{array}\right]\\=& \left[\begin{array}{cc}
\left[\begin{array}{ccc}2&1&0\\0&3&-1\end{array}\right]\left[\begin{array}{c}a\\1\\b\end{array}\right]&
\left[\begin{array}{ccc}2&1&0\\0&3&-1\end{array}\right]\left[\begin{array}{c}1\\b\\a\end{array}\right]
\end{array}\right]\\
=&\left[\begin{array}{cc}
\left[\begin{array}{ccc}2&1&0\end{array}\right]\left[\begin{array}{c}a\\1\\b\end{array}\right]&
\left[\begin{array}{ccc}2&1&0\end{array}\right]\left[\begin{array}{c}1\\b\\a\end{array}\right]\\
\left[\begin{array}{ccc}0&3&-1\end{array}\right]\left[\begin{array}{c}a\\1\\b\end{array}\right]&
\left[\begin{array}{ccc}0&3&-1\end{array}\right]\left[\begin{array}{c}1\\b\\a\end{array}\right]
\end{array}\right]\\
=&\left[\begin{array}{cc}
2a+1&
\left[\begin{array}{ccc}2&1&0\end{array}\right]\left[\begin{array}{c}1\\b\\a\end{array}\right]\\
\left[\begin{array}{ccc}0&3&-1\end{array}\right]\left[\begin{array}{c}a\\1\\b\end{array}\right]&
\left[\begin{array}{ccc}0&3&-1\end{array}\right]\left[\begin{array}{c}1\\b\\a\end{array}\right]
\end{array}\right]
\end{align}$$
For each value in the resultant array, the row vector and column vector multiplication is simply the sum of product of corresponding values. For example
$$\begin{align}
\left[\begin{array}{ccc}2&1&0\end{array}\right]\left[\begin{array}{c}a\\1\\b\end{array}\right] =& 2\times a + 1\times 1 + 0\times b\\
=& 2a +1
\end{align}$$
You should be able to find all values in terms of $a$ and $b$ now.
| {
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$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6 The problem is as follows: Prove that $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0 \, \text{mod n} \ \text{if and only if } n \equiv \pm 1 \, \text{mod} 6.$$
My idea is to of course rewrite the summation. We have $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6}.$$
Now, in order for this to be an integer, must we have in particular that $6 \mid n-1 \Rightarrow n \equiv 1 \, \text{mod} 6$?
We also know that the sum $$1^2 + 2^2 + 3^2 + \ldots + (n-1)^2 + n^2 = \frac{n(n+1)(2n+1)}{6}.$$ Must we have in particular that then $6 \mid n+1 \Rightarrow n \equiv -1 \, \text{mod} 6$?
I think the statement holds if and only if the whole way so we are done with both ways?
| Hint: $6 \mid (n-1)(2n-1)$ implies that $n$ is odd and cannot be divisible by $3$.
| {
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Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$
To solve it I used induction but it is leading me nowhere my attempt was as follows:
Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$
so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} > 13/24$
I don't know where to go from here please help.
| $$f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}=H_{2n}-H_n $$
is an increasing function, since
$$ f(n+1)-f(n) = \frac{1}{(2n+1)(2n+2)}>0. $$
It follows that $\forall n>1,\; f(n)\geq f(2)=\frac{7}{12}>\frac{13}{24}$.
Additionally, by Riemann sums
$$ \lim_{n\to +\infty}f(n)=\lim_{n\to +\infty}\sum_{k=1}^{n}\frac{1}{n+k} = \int_{0}^{1}\frac{dx}{1+x}=\log(2)$$
and
$$ 0< \int_{0}^{1}\frac{x^3(1-x)^3}{1+x}\,dx = \frac{111}{20}-8\log(2) $$
leads to $\log(2)<\frac{111}{160}$, so:
$$\boxed{ \forall n\geq 2,\quad \color{red}{\frac{7}{12}}\leq\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}<\color{red}{\frac{111}{160}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Laurent series of a trigonometric function Find the Laurent Serie(and residue) aroud $z_0=0$ of the function
$f(z) = \frac{1}{1-\cos z}$.
Progress:
It looks very trivial but it seems to get complicated so I'll only try with 3-4 terms:
We can use that $\cos z = 1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0}^{\infty}\frac{(-1)^k z^{2k}}{(2k)!}$ and hence
$f(z) = \frac{1}{1-\cos z} = \frac{1}{1-(1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0}^{\infty}\frac{(-1)^k z^{2k}}{(2k)!})} \Rightarrow f(z)= \frac{1}{\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}}
$
This is not correct apparently so I tried to factor out $z^2$ and$\frac{z^2}{2!}$ but I was unable to proceed. I suspect that I'll arrive to a nestled sum.
| Since we're interested only around $\,z=0\,$ we can try the following using what you've already done:
$$\frac1{1-\cos z}=\frac1{\frac{z^2}2\left(1-\frac{z^2}{12}+\mathcal O(z^4)\right)}=\frac2{z^2}\left(1+\frac{z^2}{12}+\frac{z^4}{12^2}+\ldots\right)=$$
$$\frac2{z^2}+\frac16+\frac{2z^2}{12^2}+\ldots$$
We used, of course, the development
$$\frac1{1-z}=1+z+z^2+\ldots\;\;,\;\;|z|<1$$
| {
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Integral of $ \frac{\sqrt{x+1}}{x} $. Im looking for a way to calculate:
$$ \int{\frac{\sqrt{x+1}}{x}} dx $$
I substituted $ u = \sqrt{x+1} $ and then $ dx = 2 * \sqrt{x+1} du $ so I have:
$$ \int{\frac{u}{x} * 2 \sqrt{x+1} du } $$
$ x = u^2-1 $
and
$$ 2 \int{\frac{u^2}{u^2-1} } du = 2 \int{\frac{u^2}{(u-1)(u+1)} } du $$
I'm not sure if this is correct. Maybe I had some errors on the way here.
But if this is correct. How would I proceed then?
| What you did was correct. To finish it off, notice that
$$\dfrac{u^2}{u^2-1} = \dfrac{(u^2-1)+1}{u^2-1} = 1 + \dfrac{1}{u^2-1}$$
The integral can then be computed using partial fractions:
$$\dfrac{1}{u^2-1} = \dfrac{1}{(u-1)(u+1)} = \dfrac{1}{2} \left( \dfrac{1}{u-1} - \dfrac{1}{u+1} \right)$$
| {
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How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be only pick from the rest, so the number of choices only 5
Third digit can be only pick from the rest, so the number of choices only 4
The total number of choices is $6\cdot 4\cdot 5\cdot 4= 480$
So, is my solution true? Or I miss something?
Thanks
| Consider cases:
*
*The last digit is $0$. Then you have six choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $6 \cdot 5 \cdot 4 \cdot 1 = 120$ four digit even numbers that end in $0$.
*The last digit is $2$, $4$, or $6$. Since the first digit cannot be zero, you have five choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $5 \cdot 5 \cdot 4 \cdot 3 = 300$ four digit even numbers that end in $2$, $4$, or $6$.
That exhausts the possibilities, so there are $120 + 300 = 420$ even four digit numbers that can be formed using the digits $0, 1, 2, 3, 4, 5, 6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
| Suppose $gcd(a,b)=1$ then you have $ax+by=1$, cubing this, we get $(ax+by)^3=1$
i.e., $a^3x^3+b^3y^3+3a^2x^2by+3axb^2y^2=1$
i.e., $a^2(ax^3+3x^2by)+b^2(by^3+3axy^2)=1$
does this imply $gcd(a^2,b^2)=1$???
| {
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A basic doubt on a infinite series problem I see in Rudin the following statement is claimed for the following convergent series:
$$1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$
If $s$ is the sum of this series then $$s \lt 1 -\frac{1}{2} + \frac{1}{3}$$. How is that possible to tell without knowing $s$ ?
| $$S_{2k+1} = 1-\frac{1}{2} + \frac{1}{3} - \left(\frac{1}{4} - \frac{1}{5}\right) - \ldots -\left(\frac{1}{2k} - \frac{1}{2k+1}\right) < 1-\frac{1}{2} + \frac{1}{3}-\left(\frac{1}{4} - \frac{1}{5}\right)$$ and $S_{2k+1}\to s$ imply
$$s \leqslant 1-\frac{1}{2} + \frac{1}{3}-\left(\frac{1}{4} - \frac{1}{5}\right) < 1-\frac{1}{2} + \frac{1}{3}.$$
| {
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Distributing amounts of different objects to distinct people I have the following exercise and I am wondering if the subsequent solution of mine is correct.
Alan, Peter, Steve, and Donald have a total of 11 chocolate bars that
they are going to divide among themselves. They have 9 Snickers (S), 1
Mars (M) and 1 Bounty (B). In how many different ways can they
distribute the chocolates among themselves such that each person gets
at least one chocolate?
Each person shall have at least one chocolate each, i.e. a total of 4 chocolates. Call these compulsory. There are 4 different ways to choose compulsory chocolates:
*
*$\{S, S, S, S\}$
*$\{S, S, S, M\}$
*$\{S, S, S, B\}$
*$\{S, S, M, B\}$
Case 1: There is only one way to distribute the four compulsory Snickers among the boys. Then there are $\binom{4}{1}$ ways of distributing the Mars and $\binom{4}{1}$ ways of distributing the Bounty. The remaing five Snickers can be distributed in $\binom{4+5-1}{5}$ ways using the equation $a+b+c+d = 5$. In total: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+5-1}{5}$.
Case 2: There are $\binom{4}{1}$ ways to distribute the one compulsory Mars and one way to distribute the three compulsory Snickers. Then there are $\binom{4}{1}$ ways to distribute the non compulsory Bounty and $\binom{4+6-1}{6}$ ways to distribute the six non compulsory Snickers. In total: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{6}$.
Case 3: There are $\binom{4}{1}$ ways to distribute the one compulsory Bounty, $\binom{4}{1}$ ways to distribute the one non compulsory Mars and $\binom{4+6-1}{6}$ ways of distributing the six remaining Snickers. In total: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{1}$.
Case 4: There are $\binom{4}{1}$ ways of distributing the one compulsory Mars, then $\binom{3}{1}$ ways of distributing the compulsory Bounty, and one way of distributing the two compulsory Snickers. The remaining seven non compulsory Snickers can be distributed in $\binom{4+7-1}{7}$ different ways. In total: $\binom{4}{1}$$\binom{3}{1}$$\binom{4+7-1}{7}$.
All in all: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+5-1}{5}$ + $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{6}$ + $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{1}$ + $\binom{4}{1}$$\binom{3}{1}$$\binom{4+7-1}{7}$.
| You seem to get something like $3824$, which looks far too many: my answer is about half that.
So looking at possible issues with your answer:
*
*In case 3 you seem to have changed a 6 into a 1, so case 3 does not give the same answer as case 2; I am unclear about whether this is deliberate or not
*I think you may have double, triple or quadruple counted cases like $(SSSS, SSM, SB, SS)$
*To avoid double counting, you might consider cases where the compulsory Mars or Bounty are not accompanied by any Snickers. You would need to add $\{S,S,S,MB\}$ to your patterns and do the calculations where non-compulsory bars can only be given to people who already have a compulsory Snickers.
| {
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The Fejer kernel has this $\sin$ closed form. Let $D_N$ be the $N$th Dirichlet kernel, $D_N = \sum_{k = -N}^N w^k$, where $w = e^{ix}$. Define the Fejer kernel to be $F_N = \frac{1}{N}\sum_{k = 0}^{N-1}D_k$. Then $$F_N = \frac{1}{N}\frac{\sin^2(N x/2)}{\sin^2(x/2)}$$.
So far I have $D_k = \frac{w^{k+1} - w^{-k}}{w-1}$, and so
$$
\begin{align*}
F_N &= \frac{1}{N}\sum_{k=0}^{N-1} D_k \\
&= \frac{1}{N(w-1)}\sum_{k=0}^{N-1} (w^{k+1} - w^{-k}) \\
&= \frac{1}{N(w-1)}\left ( w\sum_{k=0}^{N-1} w^k - \sum_{k=-N+1}^0 w^k \right ) \\
&= \frac{1}{N(w-1)}\left ( \frac{w(w^N - 1)}{w-1} - \frac{1-w^{-N + 1}}{w-1} \right ) \\
&= \frac{1}{N(w-1)^2}\left ( w^{N+1} +w^{-N + 1} - (w + 1) \right )
\end{align*}
$$
| We can prove the equality for the Fejér kernel in the following way. Using the formula for the geometric progression and the fact that $e^{i\theta}-e^{-i\theta}=2i\sin\theta$ for each $\theta\in\mathbb R$,
\begin{align*}
D_k(x)
&=e^{-ikx}\sum_{s=0}^{2k}e^{isx}\\
&=e^{-ikx}\frac{1-e^{ix(2k+1)}}{1-e^{ix}}\\
&=\frac{e^{-ix(k+1/2)}-e^{ix(k+1/2)}}{e^{-ix/2}-e^{ix/2}}\\
&=\frac{\sin(x(k+1/2))}{\sin(x/2)}.
\end{align*}
Using the product-to-sum identity and the power reduction formula,
\begin{align*}
F_n(x)
&=\frac1{n\sin(x/2)}\sum_{k=0}^{n-1}\sin(x(k+1/2))\\
&=\frac1{2n\sin^2(x/2)}\sum_{k=0}^{n-1}2\sin(x(k+1/2))\sin(x/2)\\
&=\frac1{2n\sin^2(x/2)}\sum_{k=0}^{n-1}[\cos(kx)-\cos((k+1)x)]\\
&=\frac{1-\cos (nx)}{2n\sin^2(x/2)}\\
&=\frac1n\biggl[\frac{\sin(nx/2)}{\sin(x/2)}\biggr]^2.
\end{align*}
The trigonometric identities can be found here.
| {
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How do I prove by induction? For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
| Although this is not induction, I still find it quite amusing that Leibniz (inventing the differential calculus almost simultaneously with Newton) solved these kind of problems with some sort of difference method with great similarity to his differential calculus:
First we define the differences of a sequence: For instance the sequence $x^2$ for integers $x$ has (first order) differences
$$
dx^2=(x+1)^2-x^2=2x+1
$$
or more generally $f:\mathbb Z\rightarrow\mathbb Z$ has differences $df=f(x+1)-f(x)$ in modern function notation. With this definition it can be seen (using the binomial theorem) that
$$
\begin{align}
dx^4&=4x^3+6x^2+4x+1\\
dx^3&=3x^2+3x+1\\
dx^2&=2x+1\\
dx&=1\\
d(kf)&=k\ df\\
dk&=0
\end{align}
$$
where $k$ denotes some constant. One notices that taking differences of polynomial expressions yields polynomials of one degree less.
Now your problem corresponds to saying $df=(x+1)^2=x^2+2x+1$ and searching for a polynomial expression $f(x)$ solving this. From the degree it follows that
$$
f(x)=ax^3+bx^2+cx+d
$$
Therefore
$$
\begin{align}
x^2+2x+1&=df\\
&=a(3x^2+3x+1)+b(2x+1)+c\\
&=3ax^2+(3a+2b)x+(a+b+c)
\end{align}
$$
Now to match coefficients of these two expressions for the same quadratic we must have
$$
\begin{align}
3a&=1\\
3a+2b&=2\\
a+b+c&=1
\end{align}
$$
which is a system of linear equations that can be solved to get $a=\frac{1}{3}$, $b=\frac{1}{2}$ and $c=\frac{1}{6}$. The constant term $d$ can then be determined from the initial value $f(1)=1$ leading to $d=0$. So we have found that
$$
\begin{align}
f(x)&=\frac{1}{3}x^3+\frac{1}{2}x^2+\frac{1}{6}x\\
&{}\\
&{}\\
&=\frac{x(x+1)(2x+1)}{6}
\end{align}
$$
Both note how this method is constructive we do not need know the answer beforehand as it just jumps out from the method, but also how it has great similarities to integration of a function with a given initial value.
| {
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Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$ Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$.
You can use L'Hospitale, or Maclaurin, etc
| Hint: Use $\tan x = x+\dfrac{x^3}{3}+o(x^3)$ and $\sin x = x-\dfrac{x^3}{6}+o(x^3)$ to obtain $\tan(\tan x) = x+\dfrac{2x^3}{3}+o(x^3)$ and $\sin (\sin x) = x-\dfrac{x^3}{3}+o(x^3)$.
Hence $$\dfrac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x} = \dfrac{x^3+o(x^3)}{\frac{x^3}{2}+o(x^3)} \to 2.$$
| {
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Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x\equiv2\pmod3$$What $x$ value?
| The number which leaves (5, 4, 3, 2) mod (6, 5, 4, 3) is one less than the one that leaves a residue of (0, 0, 0, 0), mod (6, 5, 4, 3). So one finds $n=\operatorname{lcm}(6,5,4,3)-1$ to get 59, which is the desired answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| Well, I tried and failed to find an answer made of sugar and spice. Here is a frogs and snails answer.
As karafka noted, the AM-GM inequality easily implies that $xyz\leq 1$. So let $\epsilon\geq 0$ be such that $xyz=1-\epsilon$ .
Let us order the variables so that $x\leq y\leq z$. We must have $z\geq 1$ and $x\leq 1$.
Now, it can be checked that \begin{eqnarray*} (xyz)^4(x^3+y^3+z^3) & = & (x+y+z)^3(xyz)^4-3(x+y+z)(xy+xz+yz)(xyz)^4+3(xyz)^5,\end{eqnarray*}
and since $x+y+z=3$ we deduce
\begin{eqnarray*} (xyz)^4(x^3+y^3+z^3) & = & 27(1-\epsilon)^4-9(xy+xz+yz)(1-\epsilon)^4+3(1-\epsilon)^5\\ &=& 3\left(10-\epsilon+3(xy+xz+yz)\right)(1-\epsilon)^4\end{eqnarray*}
Thus we're done if we can show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq \frac{1}{(1-\epsilon)^4}.$
Now, since $$1+\epsilon+\epsilon^2 = \frac{1-\epsilon^3}{1-\epsilon}\leq \frac{1}{1-\epsilon},$$ we can take the fourth power of both sides. Then note that $$1+4\epsilon\leq 1+4\epsilon+10\epsilon^2\leq \frac{1}{(1-\epsilon)^4}.$$
Case 1: $\frac{3}{5}\leq x \leq 1$.
In this case we aim to show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq 1+4\epsilon$, i.e, that $C\geq 0$, where $C=1+4\epsilon - \left(10-\epsilon-3(xy+xz+yz)\right)$.
Since $\frac{3}{5}\leq x\leq 1$ and $z\geq 1$ there exists $a\in[0,\frac{2}{5}]$ such that $x=1-a$ and $b\in[0,2)$ such that $1+b=z$. Hence $y=1+a-b$. Fix such an $a$.
Now, $C$ can be written in terms of $a$ and $b$ as $C=(2-5a)b^2+(5a^2-2a)b+2a^2$. Since $a\leq \frac{2 }{5}$, this corresponds to an upwards parabola in the $C-b$ plane with vertex $\left(\frac{a}{2}, \frac{a^2}{5}(6+5a)\right)$. Hence $C\geq0$ for every choice of $b$.
Case 2: $\frac{3}{25}\leq x\leq \frac{3}{5}$.
In this case, we'll show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq 1+4\epsilon+10\epsilon^2$. For this case, write the right side minus the left in terms of $x,y,z$ to obtain:
$$C=10x^2(yz)^2+(3-25x)(yz)+6+9x-3x^2.$$
Now fix an $x$ in the range. Then $C$ corresponds to upward parabola in "variable" $yz$. Its discriminant is $120x^4-360x^3+385x^2-150x+9.$ Using the quartic formula (...) , this polynomial has real roots $r_1\approx 0.08<\frac{3}{25}$ and $r_2\approx 0.78>\frac{3}{5}$, and is negative in between. Thus for any valid choice of $y,z$, we must have $C>0$.
Case 3: $0<x<\frac{3}{25}$.
The vertex of the parabola from Case $2$ is located at $yz=\frac{25x-3}{20x^2}$. However, this is negative since $x<\frac{3}{25}$. Since the parabola is upwards with $C$-intercept at $C=6+9x-3x^2>6-3>0$, this means that $C>0$ when $yz$ is positive (which it is).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
Justify $\gcd$ of $f(x) = x^3 - 6x^2 + x + 4$ and $g(x) = x^5 - 6x +1$ Let $f(x) = x^3 - 6x^2 + x + 4$ and $g(x) = x^5 - 6x +1$. Using Euclidean algorithm I find $\gcd[f(x), g(x)] = 1$. How could I JUSTIFY that $h(x) = 1$ is the ACTUAL $\gcd$ of $f(x)$ and $g(x)$?
Thanks.
| Actually Euclidean Algorithm is a really good proof.
One other way is to factorize one of the polynomials into linear polynomials. To do the find the zeroes of that polynomials. I'll use the polynomial $f(x)$ so the zeroes are at $x = 1$, $x=\frac 12 (5 \pm \sqrt{41})$.
So we can factorize it as:
$$x^3 - 6x^2 + x + 4 = (x-1)(\frac 12 (x-5+\sqrt{41}))(\frac 12 (x-5-\sqrt{41}))$$
$$x^3 - 6x^2 + x + 4 = \frac14(x-1)(x-5+\sqrt{41})(x-5-\sqrt{41})$$
So now we need to check which of this factors divide $g(x)$. Using the Polynomial remainder Theorem we know that a linear polynomial of the type $(x-a)$ divides some polynomial $p(x)$ if and only if $p(a) = 0$
So we have three cases to check those are $a = 1$, $a = 5 + \sqrt{41}$ and $a = 5 - \sqrt{41}$. And for all three cases we'll get that $g(a) \neq 0$, so neither of the factors of $f(x)$ is divisor of $g(x)$, i.e $gcd(f(x),g(x)) = 1$
Also sometimes to get rid of those ugly calculation you can use Sylvester matrix to check whether 2 polynomials are coprime. You can read how to generate one reading the intructions in the link so for this example the matrix would look like:
$$\begin{bmatrix}
1 & 0 & 0 & 0 & -6 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & -6 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & -6 & 1 \\
1 & -6 & 1 & 4 & 0 & 0 & 0 & 0 \\
0 & 1 & -6 & 1 & 4 & 0 & 0 & 0 \\
0 & 0 & 1 & -6 & 1 & 4 & 0 & 0 \\
0 & 0 & 0 & 1 & -6 & 1 & 4 & 0 \\
0 & 0 & 0 & 0 & 1 & -6 & 1 & 4
\end{bmatrix}$$
Take the determinant of it. If it's $0$ the two polynomials have a common root and a common factor. Otherwise they are coprime and you can stop your caclulation there. So for this two polynomials we have:
$$\begin{vmatrix}
1 & 0 & 0 & 0 & -6 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & -6 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & -6 & 1 \\
1 & -6 & 1 & 4 & 0 & 0 & 0 & 0 \\
0 & 1 & -6 & 1 & 4 & 0 & 0 & 0 \\
0 & 0 & 1 & -6 & 1 & 4 & 0 & 0 \\
0 & 0 & 0 & 1 & -6 & 1 & 4 & 0 \\
0 & 0 & 0 & 0 & 1 & -6 & 1 & 4
\end{vmatrix} = 120784$$
So clearly this isn't equal to $0$ and we can conclude that these two polynomails are coprime.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Convergence of a sequence $\frac{1}{1+n^3}$ How can I prove by integral test that the sequence $1+ \dfrac{1}{1+2^3} + \dfrac{1}{1+3^3} + \dots + \dfrac{1}{1+n^3}$ is convergent?
Thank you. Is there a way that I can integrate $\dfrac{1}{1+n^3}$ ?
| I will guess that the $n$-th term of the sequence is
$$\frac{1}{1+1^3}+\frac{1}{1+2^3}+\frac{1}{1+3^3}+\cdots+\frac{1}{1+n^3}.$$
There is indeed a closed form for $\int \frac{1}{1+t^3}\,dt$. However, the details are somewhat unpleasant. It is simpler to note that the $n$-th term is less than $\frac {1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots +\frac{1}{n^3}$
and use the fact that $\int_1^\infty \frac{1}{t^3}\,dt$ converges.
Alternately, we can use the fact that the $n$-th term is less than or equal to
$$\frac{1}{1+1^2}+\frac{1}{1+2^2}+\frac{1}{1+3^2}+\cdots+\frac{1}{1+n^2},$$
and use the fact that $\int_0^\infty \frac{1}{1+t^2}\,dt$ converges.
| {
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"url": "https://math.stackexchange.com/questions/523071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.
| Presented below is a self-contained evaluation. With
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$
\begin{align*}
I & = \int_0^1 \frac{\ln x \ln (1-x^4 )}{1+x^2}dx \\
& = \int_0^1 \ln (1-x^4 ) d\left(\int_1^x \frac{\ln t}{1+t^2}dt \right) \overset{IBP}=\int_0^1 \frac{ 4x^3}{1-x^4} \underset{t=xs }{\left(\int_0^x \frac{\ln t}{1+t^2}dt +G \right) } dx \\
& =4\int_0^1 \left( \int_0^1 \frac{x^4 \ln x+x^4\ln s }{(1-x^4 )(1+x^2s^2)}ds +\frac{Gx^3}{1-x^4} \right) dx\\
& =4\int_0^1 \int_0^1 \frac{x^4\ln x}{(1-x^4)(1+x^2s^2)}dsdx -4 \int_0^1\int_0^1 \frac{\ln s}{1+x^2s^2}dx ds \\
& \>\>\>\>\>+ 4 \int_0^1 \left(\int_0^1 \frac{\ln s }{(1-x^4 )(1+x^2s^2)}ds +\frac{Gx ^3}{1-x^4} \right) dx\\
\end{align*}
Integrate the 2nd integral
\begin{align*}
& \int_0^1\int_0^1 \frac{\ln s}{1+x^2s^2}dx ds =\int_0^1 \frac{\ln s\tan^{-1}s}sds
\overset{IBP}=-\frac12 \int_0^1 \frac{\ln^2s}{1+s^2}ds=-\frac{\pi^3}{32}
\end{align*}
and apply the decomposition below in the 3rd integral
$$\frac{1 }{(1-x^4 )(1+x^2s^2)}
= \frac{-s^4}{(1-s^4)(1+x^2s^2)} +\frac1{2(1-s^2)(1+x^2)}+ \frac1{2(1+s^2)(1-x^2)}
$$
Then, the 1st integral cancels and
\begin{align*}
I =& -4\left(-\frac{\pi^3}{32}\right)
+2\int_0^1 \int_0^1 \frac{\ln s }{(1-s^2 )(1+x^2)}dx ds \\
&\>\>\> + 2\int_0^1 \left( \int_0^1 \frac{\ln s }{(1+s^2 )(1-x^2)}ds +\frac{2Gx ^3}{1-x^4}\right) dx\\
= & \frac{\pi^3}8+ 2\int_0^1 \frac{\ln s ds }{1-s^2}\int_0^1\frac{dx }{1+x^2}
-2G \int_0^1 \left( \frac{1}{1-x^2} -\frac{2x^3}{1-x^4}\right) dx\\
= & \frac{\pi^3}8+ 2\left(-\frac{\pi^2}{8}\right) \frac\pi4
-2G \int_0^1 \left( \frac{x}{1+x^2} +\frac{1}{1+x}\right) dx\\
= & \frac{\pi^3}{16}
-3G\ln2\\
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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} |
If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem :
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach :
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get :
$ y = \sqrt{x + 2005}$
Now putting this value of y in (ii) we get :
$ \Rightarrow 2005 +\sqrt{x + 2005} =x^2$
$ \Rightarrow \sqrt{x + 2005} =x^2 -2005 $
Now squaring both sides we get :
$\Rightarrow (\sqrt{x + 2005})^2 =(x^2 - 2005)^2$
Is there any other way I can solve this problem please suggest... thanks.
| I am new here so don't know how to type math equation but I am providing the solution.
2005+x=y^2 ..(1)
2005+y=x^2 ..(2)
After subtracting 2nd equation from first
x-y=y^2-x^2
x-y=(x-y)(x+y)
(x-y)+(x-y)(x+y)=0 (we change the sign x-y to y-x here)
(x-y)(x+y+1)=o
because as mentioned x and y are different integers so x-y!=0
In this case
x+y+1=0 -> x+y=-1 ...(3)
Now adding equation (1) and second we got
4010+x+y=x^2+y^2
because x+y=-1
so 4010-1=x^2+y^2
4009=x^2+y^2
because x^2+y^=(x+y)^2-2xy
as (x+y)^2=x^2+y^2+2xy
so 4009=(x+y)^2-2xy
and 4009=(-1)^2-2xy
and 2xy=1-4009
xy=-2004
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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√2+√3 is irrational I’m trying to prove that $\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ is irrational.
I have already proved that $\sqrt{2}+\sqrt{3}$ is irrational. Should I use a similar approach as below or is there a different way?
Proof: $\sqrt{2}+\sqrt{3}$ is irrational
Assume $\sqrt{2}+\sqrt{3}$ is rational so $\sqrt{2}+\sqrt{3}=p/q$ where $p$ and $z$ are integers, $q\neq 0$, $\gcd(p,q)=1$
$r^2=5+2\sqrt{6}$
$r^2-5=2\sqrt{6}$ then I squared both sides…
$r^4-10r^2+25=24$
$r^4-10r^2+1=0$
$r=p/q$ so $p^4/q^4-10p^2/q^2+1=0$
then I got a common denominator
$p^4-10p^2q^2+q^4=0$
$p^4=10p^2q^2-q^4$
I found that $q=1$ or $-1$ and then I plugged it back into $r^4-10r^2+1=0$ and found that it can’t $=-1$ so there is a contradiction.
| You have a shorter proof: if $\sqrt{2}+\sqrt{3}=\frac{p}{q}$, where $p\in\mathbb Z$ and $q\in\mathbb N$, $q\neq 0$, then $5+\sqrt{6}=\frac{p^2}{q^2}$. So, $\sqrt{6}=\frac{p^2-5q^2}{q^2}$ is rational, which is known to be false.
For $\sqrt{n}$, with $n\in\mathbb N$, you have the following alternative:
1) $\sqrt{n}$ is an integer (it happens when $n$ is a perfect square)
2) $\sqrt{n}$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$ Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$.
This is what I've managed:
$x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}= 2^{k} \cdot \sin \frac{\pi}{2^{k}} \sqrt{\frac{2^{k+1} \cdot \sin \frac{\pi}{2^{k}}}{2^{k} \cdot \sin \frac{\pi}{2^{k}}+2^{k-1} \cdot \sin \frac{\pi}{2^{k}}}}=2^{k} \cdot \sin \frac{\pi}{2^{k}} \sqrt{\frac{2}{1+\cos \frac{\pi}{2^{k}}}}$
And I don't see how to proceed....
| You can use that
$$\cos (2x) = \cos^2 x - \sin^2 x,$$
and hence
$$1 - \cos \frac{\pi}{2^k} = 1 - \cos^2 \frac{\pi}{2^{k+1}} + \sin^2 \frac{\pi}{2^{k+1}} = 2\sin^2 \frac{\pi}{2^{k+1}}.$$
Then write
$$\sqrt{\frac{2}{1+\cos \frac{\pi}{2^k}}} = \sqrt{\frac{2(1 - \cos \frac{\pi}{2^k})}{1 - \cos^2 \frac{\pi}{2^{k}}}} = \sqrt{\frac{4\sin^2 \frac{\pi}{2^{k+1}}}{\sin^2 \frac{\pi}{2^k}}} = \frac{2\sin \frac{\pi}{2^{k+1}}}{\sin \frac{\pi}{2^k}},$$
since all the sines involved are non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area of a circle inscribed in a rhombus? Let's say we have a rhombus with diagonals $a$ and $b$, which contains an inscribed circle. How can we find the area of that circle in terms of $a$ and $b$?
|
Obviously, the radius of inscribed circle is also a height $h=OH$ of the right triangle $\triangle AOB$. To find it, use equations for triangle's area
$$
S_{\triangle AOB} = \frac 12 \frac a2 \frac b2 = \frac {ab}8 = \frac 12 ch
$$
where $c = AB$ is a hypotenuse. So $r = h = \frac {ab}{4c} = \frac {ab}{4\sqrt{\frac {a^2}4+\frac {b^2}4}} = \frac {ab}{2\sqrt{a^2+b^2}}$, and therefore
$$
S_c = \pi r^2 = \frac {\pi a^2 b^2}{4\left (a^2+b^2 \right )}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528877",
"timestamp": "2023-03-29T00:00:00",
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Partial Fractions and power of a factor with $x^2$ I just started working with partial fractions and hit a wall with splitting this one:
$$ \frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} $$
I get here:
$$ \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}$$
Then on to:
$$ (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2 $$
I find that $E = 1$ by using $x = -2$. I am unsure how to proceed from here.
| You don't have any equations there.
Presumably, you start with
$$\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}
= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}
$$
then multiply both sides
by $ (x + 2)(x^2 + x + 1)^2$
to get
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2
.$$
There are a variety of ways
to go from here.
As you did,
when you set
$x = -2$,
you get
$12-4+1=9E$
so $E = 1$.
The equation then becomes
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + (x^2 + x + 1)^2
.$$
You can do as
Michael Albanese
suggested to get equations for the
other coefficients.
This is probably the
best way.
You can look at the roots
of
$x^2 + x + 1=0$,
which are complex.
You can let $x=0$,
and this becomes
$1
=2B + 2D + 1
$,
or $B = -D$.
Looking at the coefficient of $x^4$,
you get
$Cx^4+x^4 = 0$
or $C = -1$.
I'll stop here.
| {
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Inequality $\sum^{k}_{i=1}\sum^{n}_{j=k}{{1}\over{j - i + 1}} \le n$ I was proving a statement and have finished all the other part, the remaining boiling down to proving the inequality below (if I proved the other part correctly):
$$\sum^{k}_{i=1}\sum^{n}_{j=k}{{1}\over{j - i + 1}} \le n$$
where $1 \le i \le k \le j \le n$.
I wrote a simple function taking $k$ and $n$ as parameters to compute the left side and the result never exceeds $n$. Also it seems that the maximum is reached when $k = n / 2$. Any ideas would be appreciated!
Thanks...
| Proof: Define $s=\sum^{k}_{i=1}\sum^{n}_{j=k}{{1}\over{j - i + 1}}$. We prove the required inequality by factoring out common and distinct summand $\frac{1}{d+1}$, where $d=j-i$. The region of summation $\{1,2,...,k\}\times\{k,k+1,...,n\}$ is partitioned into 3 parts, with 2 isosceles right triangles sandwiching 1 parallelogram with two 45 degree angles. We distinguish 2 cases:
1) $n\le 2k-1$
The summation over the 3 partitions are:
$$\begin{align}
s_1 &= \sum_{i=0}^{n-k}\frac{d+1}{d+1}, \\
s_2 &= \sum_{d=1}^{2k-n-1} \frac{n-k+1}{n-k+1+d}, \\
s_3 &= \sum_{d=1}^{n-k} \frac{n-k+1-d}{k+d} = (n+1)\sum_{d=1}^{n-k}\frac{1}{k+d} - (n-k).
\end{align}$$
Since the summands are decreasing function of d, we bound the summations by the integration of the summands.
$$s_2 < (n-k+1)\int_0^{2k-n-1}\frac{1}{n-k+1+x}=(n-k+1)\ln\frac{k}{n-k+1},$$
and
$$s_3+n-k < (n+1)\int_0^{n-k}\frac{1}{k+x}dx=(n+1)\ln\frac{n}{k}.$$
2) $n\ge 2k-1$
$$\begin{align}
s_1 &= \sum_{i=0}^{k-1} \frac{d+1}{d+1} = k, \\
s_2 &= k\sum_{d=1}^{n-2k+1} \frac{1}{k+d} <k\int_0^{n-2k+1}\frac{1}{k+x}dx = k\ln\frac{n+1-k}{k}, \\
s_3 &= \sum_{d=1}^{k-1} \frac{d}{n-d+1} = (n+1)\sum_{d=1}^{k-1}\frac{1}{n+1-i}-(k-1) \\
&<(n+1)\int_1^k \frac{1}{n+1-x}dx = (n+1)\ln\frac{n}{n+1-k}-(k-1).
\end{align}$$
For both cases, put the parts $s_1, s_2, s_3$ together, we get the same expression
$$\begin{align}s &=s_1+s_2+s_3 \\
&= (n+1)\ln n - \big(k\ln k + (n+1-k)\ln(n+1-k)\big)+1 \\
&< (n+1)\ln n - (n+1)\ln\frac{n+1}{2} +1 \\
&\text{by the convexity of the function } x\ln x \\
&< (n+1)\Big(\ln 2-\frac{1}{n+1} \Big)+1 \\
& = (n+1)\ln 2 \\
&< 0.7(n+1).
\end{align}$$
This actually gives a slightly tighter bound for $s$. We only need to check the inequality for $n\in\{1,2\}$ to complete the proof.
Also, this proof points the way to showing $s$ achieves maximum at $k \cong \frac{n+1}{2}$ for fixed $n$.
QED
| {
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} |
Use L'Hôpital's rule to evaluate this limit $$\lim_{x\to0} \left(\frac{a^x + b^x}{2}\right)^{1/x} = (ab)^{1/2}$$
I tried taking logarithm to both sides, but got stuck
I got to
e^ (lim (a^x loga + b^x logb)/(a^x + b^x)
| Note that $ \ln \left( \left( \dfrac {a^x+b^x}{2} \right)^{1/x} \right) = \dfrac {1}{x} \cdot \ln \left( \dfrac {a^x+b^x}{2} \right) $.
Now, consider $ \lim \left( \dfrac {\ln \left( \dfrac {a^x+b^x}{2} \right)}{x} \right) $.
The derivative of the numerator is $ \dfrac {a^x \ln a + b^x \ln b}{a^x+b^x} $.
The derivative of the denominator is $ 1 $.
Now, our original limit is just $$ \exp \left( \lim_{x \to 0} \left( \dfrac {a^x \ln a + b^x \ln b}{a^x+b^x} \right) \right). $$
Can you do it from here?
| {
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$f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}\;\;,x,y\in \mathbb{R}$. Let $f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}\;\;,x,y\in \mathbb{R}$.
Then Max. of $f(x,y)$.
$\underline{\bf{My\;Try}}::$ We can convert into Complex no. form ...
Let $z=x+iy$, Then $f(z) = \left|z-i\right|+\left|z-3-4i\right|-\left|z\right|-\left|z-1\right|$
Now for Max., we use $\triangle$ Inequality,
But I did not understand in which pair i have used $\triangle$ inequality.
Help Required
Thanks
| let $A(0,1),B(3,4),C(0,0),D(1,0),P(x,y)$, we want $PA+PB-PC-PD$ have max ,but $PA-PC\le AC$, the "=" will hold when $P$ is on $AC$, $PB-PD \le BD$,the "=" will hold when $P$ is $BD$, so MAX is $AC+BD=1+\sqrt{(3-1)^2+(4-0)^2}=1+2\sqrt{5}$,when $P(0,-2)$
| {
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Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
| It seems like the way to attack this that doesn't require guessing is this: Start with $N (N-1) (N-2) (N-3) = N^4 + 6 N^3 + 11 N^2 + 6 N = M^2 - 1$ Since the product is "near" $N^4$, $M$ has to be "near" $N^2$. And it's very likely that $M$ is a polynomial in $N$. So set $M = N^2 + aN + b$. Then $M^2-1 = N^4 + 2aN^3 + (2b+a^2)N^2 + 2abN + (b^2-1)$. Setting that into the first equation and equating powers of $N$ gives $2a = 6, 2b+a^2 = 11, 2ab = 6, b^2-1 = 0$, which has the unique solution $a = 3, b = 1$. So $M = N^2 + 3N + 1$ and $N (N-1) (N-2) (N-3) = (N^2 + 3N + 1)^2 -1$.
| {
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Express $\sin\frac{\pi}{8}$ and $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ I've been trying with no success expressing this functions.
a) $\sin\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
b) $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
I've tried formula of the double angle ($\sin$ and $\cos$) and the ecuation $\cos^2X+\sin^2X=1$.
Can anyone point what to use or where to start?
I appreciate it, thanks.
UPDATE
I think this is achived with the half formula as njguliyev and nbubis said.
$$\sin\frac{\pi}{8} = \sin(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1-\cos\frac{\pi}{4}}{2}$$
$$\cos\frac{\pi}{8} = \cos(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1+\cos\frac{\pi}{4}}{2}$$
| Hint:
You my want to take a look at the half angle formulas.
| {
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How find the value of $\cos{x}+\cos{y}$ let
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$$
Find the value $$cos{x}+\cos{y}=?$$
this following My ugly solution: let
$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$$
then
$$\cos{x}=\dfrac{1-a^2}{1+b^2},\cos{y}=\dfrac{1-b^2}{1+b^2}$$
then
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\dfrac{1}{1+\tan{x}\tan{\dfrac{y}{2}}}=\dfrac{1-a^2}{1-a^2+2ab}$$
and
$$\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\dfrac{1-b^2}{1-b^2+2ab}$$
so
$$\dfrac{1-a^2}{1-a^2+2ab}+\dfrac{1-b^2}{1-b^2+2ab}=1$$
then
$$\Longrightarrow (1-a^2)(1-b^2+2ab)+(1-b^2)(1-a^2+2ab)=(1-a^2+2ab)(1-b^2+2ab)$$
then
$$(1-b^2)(1-a^2+2ab)=2ab(1-b^2+2ab)$$
$$a^2+b^2=1-3a^2b^2$$
so
$$\cos{x}+\cos{y}=\dfrac{2-2a^2b^2}{1+a^2+b^2+a^2b^2}=1$$
My question: this problem have nice methods? Thank you,because My methods is very ugly.Thank you
| $$\dfrac{\cos x\cos{\dfrac y2}}{\cos\left(x-\dfrac y2\right)}+\dfrac{\cos y\cos{\dfrac x2}}{\cos\left(y-\dfrac{x}{2}\right)}=1$$
$$\implies\dfrac{\cos x\cos{\dfrac y2}}{\cos\left(x-\dfrac y2\right)}=1-\dfrac{\cos y\cos{\dfrac x2}}{\cos\left(y-\dfrac{x}{2}\right)}$$
$$\implies\frac1{1+\tan x\tan\frac y2}=\frac{\sin y\sin \frac y2}{\cos\left(y-\dfrac{x}{2}\right)}=\frac1{\cot y\cot\frac x2+1}$$
$$\implies \tan x\tan\frac y2=\cot y\cot\frac x2\iff \tan x\tan\frac y2\tan y\tan\frac x2=1$$
Now, $\displaystyle \tan x\tan\frac x2=\frac{2\tan\frac x2}{1-\tan^2\frac x2}\tan\frac x2=\frac{2\sin^2\frac x2}{\cos^2\frac x2-\sin^2\frac x2}$ (multiplying the numerator & the denominator by $\cos^2\frac x2$)
$\displaystyle\implies \tan x\tan\frac x2=\frac{1-\cos x}{\cos x}$
and similarly for $\displaystyle \tan y\tan\frac y2$
| {
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How do I prove that $(1+\frac{1}{2})^{n} \ge 1 + \frac{n}{2}$ for every $n \ge 1$? How do I prove that $(1+\frac{1}{2})^{n} \ge 1 + \frac{n}{2}$ for every $n \ge 1$
My base case is $n=1$
Inductive step is $n=k$
Assume $n=k+1$
$(\frac{3}{2})^{k} \times \frac{3}{2} \ge (1 + \frac{k+1}{2})$
I'm not sure how to proceed.
| Let $P(n):(1+x)^n\ge 1+nx$
Clearly $P(n)$ is true for $n=1$
Let $P(n)$ is true for $n=m\implies (1+x)^m\ge 1+mx$
$\implies (1+x)^{m+1}\ge (1+mx)(1+x)=1+(m+1)x+mx^2\ge 1+(m+1)x$ if $m\ge0$ and $x$ is real
| {
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Use mathematical induction to prove that for any $k \in\mathbb N , \lim (1+k/n)^n = e^k$. Use mathematical induction to prove that for any $k \in \mathbb N, \lim (1+k/n)^n = e^k$.
I already used monotone Convergence Theorem to prove $k=1$ case. Do I just need to go through the same process to show $k$? If not, could you please help?
Thanks
| First a useful bound: Suppose $\theta_n \ge 0$ is such that $\lim_n n \theta_n = 0$, then $\lim_n (1+ \theta_n)^n = 1$. To see this, note $(1+\theta_n)^n = \sum_{k=0}^n \binom{n}{k} \theta^k \le \sum_{k=0}^n n^k \theta^k \le \frac{1}{1-n \theta_n}$.
Suppose the result is true for $k$ (I am taking $k=1$ as already proved). Then note that
$(1+ \frac{k}{n}) (1+ \frac{1}{n}) = (1+ \frac{k+1}{n}+ \frac{k}{n^2}) = (1+ \frac{k+1}{n})(1+ \frac{k}{n^2 (1+ \frac{k+1}{n})}) $, and let $\theta_n = \frac{k}{n^2 (1+ \frac{k+1}{n})}$. This gives
$ (1+ \frac{k+1}{n})^n = \frac{1}{(1+\theta_n)^n} (1+ \frac{k}{n})^n (1+ \frac{1}{n})^n$. Taking limits gives
$ \lim_n (1+ \frac{k+1}{n})^n = \frac{1}{1}e^k e^1 = e^{k+1}$, hence the result is true for $k+1$.
This was my original answer which is not correct for the reason Robjohn points out below:
Assuming that you have proved it for $k=1$, note that the function $x \mapsto x^k$ is continuous, and
$(1+ \frac{k}{n})^n = (1+ \frac{k}{n})^{\frac{n}{k}k} = \left ( (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k$.
Hence $\lim_n (1+ \frac{k}{n})^n = \lim_n \left ( (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k = \left( \lim_n (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k = \left( \lim_n (1+ \frac{1}{\frac{n}{k}} )^{\frac{n}{k}} \right)^k = e^k $.
| {
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Evaluate $\binom{12}0+\binom{12}2+\ldots+\binom{12}{12}$ using binomial theorem
Solve the sum:
$$
{12 \choose 0}+
{12 \choose 2}+
{12 \choose 4}+
{12 \choose 6}+
{12 \choose 8}+
{12 \choose 10}+
{12 \choose 12}
$$
using the binomial theorem.
I know the binomial theorem:
$$ \left(a+b\right)^2 = \sum_{k=0}^{n} {n \choose k} a^nb^{n-k}$$
However I fail to translate the sum into the theorem. I probably need to choose a variable ($b$?) to be negative to get half of the terms to disappear.
| With $a = 1, b = -1$ you get that
$$
\sum_{i= 0}^{12}(-1)^i\binom{12}{i} = (1-1)^{12} = 0
$$
and $a = 1, b = 1$ gives you
$$
\sum_{i = 0}^{12}\binom{12}{i} = (1 + 1)^{12} = 2^{12}
$$
Add these two together and see what terms are left.
Edit
Expanding, another way to write the $b = -1$ equation over is
$$
0 = \binom{12}{0} -\binom{12}{1} +\binom{12}{2} -\binom{12}{3}+ \binom{12}{4}- \binom{12}{5}+ \binom{12}{6}\\- \binom{12}{7} +\binom{12}{8} -\binom{12}{9} +\binom{12}{10} -\binom{12}{11} +\binom{12}{12}
$$
The $b = 1$ equation ca be written
$$
2^{12}= \binom{12}{0} +\binom{12}{1} +\binom{12}{2} +\binom{12}{3}+ \binom{12}{4}+ \binom{12}{5}+ \binom{12}{6}\\+ \binom{12}{7} +\binom{12}{8} +\binom{12}{9} +\binom{12}{10}+\binom{12}{11} +\binom{12}{12}
$$
If we add these two together, we get
$$
2^{12} = 2\binom{12}{0} +2\binom{12}{2} +2\binom{12}{4} +2\binom{12}{6}+ 2\binom{12}{8}+ 2\binom{12}{10}+ 2\binom{12}{12}
$$
Dividing by $2$ gets you the answer you're after.
| {
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Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| Write the integrand as a sum of fractions and use the polylogarithm function $\mathrm{Li}_2:$
$$f(x):=\int\frac{\ln(x)}{x^2-1}dx=\int\frac{1}{2}\left(\frac{\ln(x)}{x-1}- \frac{\ln(x)}{x+1}\right) dx \\
=\frac{1}{2} \int \frac{\ln(x) dx}{x-1}- \frac{1}{2} \int \frac{\ln(x)dx }{x+1}
=-\frac{1}{2}\mathrm{Li}_2(1-x) -\frac{1}{2} \left(\mathrm{Li}_2(-x) + \ln(x)\ln(x+1)\right)$$
Since $\mathrm{Li}_2(0)=0$ and $\ln(x)\ln(x+1)$ vanishes at $x=0$ and $x=1$, we have
$$f(0) = -\frac{1}{2}\mathrm{Li}_2(1)= -\frac{\pi^2}{12}$$
and
$$f(1) = -\frac{1}{2}\mathrm{Li}_2(-1)= \frac{\pi^2}{24}$$
and the value of the integral is
$$
\int_0^1\frac{\ln(x)}{x^2-1}dx=\frac{\pi^2}{24}+\frac{\pi^2}{12} = \frac{\pi^2}{8}
$$
| {
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How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
| We'll using the following notation:
$$s_1 = x + y + z$$
$$s_2 = x^2 + y^2 + z^2$$
$$s_3 = x^3 + y^3 + z^3$$
And we'll express $x,y,z$ as root of the cubic equation $f(x) = ax^3 + bx^2 + cx + d$
From the Vieta's formulas we know that $d = -xyz$
From the condition we have $d = - s_2$ and $s_1 = s_3$
Now use Newton's identitities and we have:
$$as_1 + b = 0$$
$$b = -as_1$$
Now repeat the process:
$$as_2 + bs_1 + 2c = 0$$
$$as_2 + -as_1^2 + 2c = 0$$
$$a(s_2 - s_1^2) + 2c = 0$$
$$2c = -a(s_2 - s_1^2)$$
$$c = \frac{-a(s_2 - s_1^2)}{2}$$
And one last time:
$$as_3 + bs_2 + cs_1 + 3d = 0$$
$$as_1 - as_1s_2 + \frac{-as_1(s_2 - s_1^2)}{2} - 3s_2 = 0$$
$$as_1\left( 1 - s_2 - \frac{s_2 - s_1^2}{2}\right) - 3s_2 = 0$$
$$as_1\left(\frac{-s_2 + s_1^2 + 2 - 2s_2}{2}\right) = 3s_2$$
$$a = \frac{6s_2}{s_1(-s_2 + s_1^2 + 2 - 2s_2)}$$
Now for arbitrary $s_1$ and $s_2$ you should be able to obtain a cubic equation, which roots will satisfy the condition.
| {
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Find the value of theta. Let $X$ have the density function $f(x) = \dfrac{3x^2}{\theta^3}$ for $0 < x < \theta$, and $f(x) = 0$ otherwise. If $P\{X > 1\} = 7/8$, find the value of $\theta$.
I don't know
| We know that $\Pr(X\le 1)=\dfrac{1}{8}$. But
$$\Pr(X\le 1)=\int_0^1 \frac{3x^2}{\theta^3}\,dx,$$
which is $\dfrac{1}{\theta^3}$. Thus we need to solve the equation $\dfrac{1}{\theta^3}=\dfrac{1}{8}$.
Remark: Alternately, we could have used the equation
$$\int_1^\theta \frac{2x^2}{\theta^3}\,=\frac{7}{8}.$$
The integral turns out to be $1-\dfrac{1}{\theta^3}$, so we solve $1-\dfrac{1}{\theta^3}=\dfrac{7}{8}$.
| {
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$\int_0^{\frac{\pi}{2}}x\cot(x)dx$ and $ \lim_{m \rightarrow \infty}\log\left( e^{2m}\left(\frac{(2m-1)!!}{(2m+1)^m}\right)^2\right)$. I'm trying to evaluate the integral, but in doing so have stumbled upon the limit, which I don't know whether it exists, and if so how to resolve it (and whether I've derived the relationship between the integral and limit correctly [see below]).
Derivation: First, use the cotangent formula: $$\cot(x)=\sum_{-\infty\le n\le\infty}\frac{1}{x+\pi n},$$
and apply it to the integral:
$$\int_0^{\frac{\pi}{2}}x\cot(x)dx=\sum_{-\infty\le n\le\infty}\int_0^{\frac{\pi}{2}}\frac{x}{x+\pi n}dx$$
$$=\sum_{-\infty\le n\le\infty}\int_{n \pi}^{\pi \left(n+\frac{1}{2}\right)}\frac{u- \pi n}{u}du$$
$$=\sum_{-\infty\le n\le\infty}[u-\pi n \log(u)]_{n \pi}^{\pi \left(n+\frac{1} {2}\right)},$$
and rearranging,
$$=\sum_{-\infty\le n\le\infty}\frac{\pi}{2}-\pi n \log\left(\frac{2n+1}{2n}\right)$$
$$=\frac{\pi}{2}\sum_{-\infty\le n\le\infty} \log\left( e\left(\frac{2n}{2n+1}\right)^{2n}\right)$$
$$=\frac{\pi}{2} \log\left( \prod_{-\infty\le n\le\infty} e\left(\frac{2n}{2n+1}\right)^{2n}\right).$$
Note that $$\prod_{-\infty\le n\le\infty} \left(\frac{2n}{2n+1}\right)^{2n}= \prod_{1\le n\le\infty} \left(\frac{2n}{2n+1}\right)^{2n}\left(\frac{-2n}{-2n+1}\right)^{-2n} $$
$$= \prod_{1\le n\le\infty} \left(\frac{2n-1}{2n+1}\right)^{2n} $$
$$=\left(\frac{1^1}{3^1}\cdot\frac{3^2}{5^2}\cdot\frac{5^3}{7^3}\cdots\right)^2=\lim_{m \rightarrow \infty}\left(\frac{(2m-1)!!}{(2m+1)^m}\right)^2.$$
Thus the original integral is equal to
$$\frac{\pi}{2} \lim_{m \rightarrow \infty}\log\left( e^{2m+1}\left(\frac{(2m-1)!!}{(2m+1)^m}\right)^2\right).$$
| IMHO, it will be simpler if one integrate the integral by parts. Using answers from the question Evaluate: $\int_0^{\pi} \ln \left( \sin \theta \right) d\theta$, we have
$$\int_0^{\frac{\pi}{2}} x \cot x dx = \int_0^{\frac{\pi}{2}} x (\log \sin x)' dx
= \Big[x \log \sin x\Big]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \log\sin x dx\\
= - \int_0^{\frac{\pi}{2}} \log\sin x dx
= \frac{\pi}{2}\log 2$$
| {
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Prove by induction that $3^{4n + 2} + 1$ is divisible by $10$ Prove by induction: $3^{(4n+2)} + 1$ is divisible by $10$.
My basic step: $3^{(4n+2)} + 1$, where $n = 1$ gives me $3^6 + 1 = 730$, which is divisible by $10$. However, then I have to do the induction hypothesis and I am kind of stuck because I do not have an equality. How do I finish proving this by induction?
Many thanks.
Edit: I am thinking of creating a formula which involves $10n$? Would this be correct?
| Suppose $3^{4k+2} +1$ is divided by 10, then we need to show $3^{4(k+1)+2} +1$ is also divided by 10. Note that:
\begin{align*}
\ 3^{4(k+1)+2} +1 &= 3^{4k+2+4}+(81-80)
\\ &= 81\cdot3^{4k+2} + 81 - 80
\\ &= 81\cdot(3^{4k+2}+1) - 80
\\ &= 81\cdot10m -80\ldots\ldots (\text{where}~3^{4k+2}+1 = 10m ~~\text{for some integer}~m)
\\ &= 10(81m-8)
\end{align*}
Hence, $3^{4(k+1)+2} +1$ is also divided by 10, which complete the proof.
| {
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Finding the limit of $( 1 + a + a^2 + \ldots+ a^n)/(1 + b + b^2 + \ldots+ b^n)$ I have some problems finding the limit of $$\frac{ 1 + a + a^2 +\cdots + a^n}{1 + b + b^2 + \cdots + b^n}.$$
$0\le a,b \le +∞$
Here is what I got :
Forcefuly factorize $a^n$ and $b^n$ :
$$ \frac{a^n ( 1 + \frac1{a} + \frac1{a^2} + ... + \frac1{a^n})}{b^n( 1 + \frac1{b} + ... + \frac1{b^n})}
= \left(\frac ab\right)^n\cdot \frac{\dfrac{1-\frac{1}a^{n+1}}{1-\frac{1}a}}{\dfrac{1-\frac{1}b^{n+1}}{1-\frac{1}b}} $$
From here I'm kinda stuck since I don't know the limit of $\left(\frac ab\right)^n$ unless I take all the possible cases. Same goes for the other part.
I want to know the next step into solving this limit.
| $\displaystyle\lim_{n\rightarrow\infty}\frac{1+a+\cdots+a^n}{1+b+\cdots+b^n}=\displaystyle\lim_{n\rightarrow\infty}\frac{1-b}{1-a}.\frac{1-a^{n+1}}{1-b^{n+1}}=\frac{1-b}{1-a}$ if $1<a\leq b$.
| {
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Solution verification: $\lim\limits_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$ I am trying to find the following limit
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2}$$
and I did the following steps:
\begin{align}
\require{cancel}
&\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} \\
&\lim_{x\rightarrow \infty} \frac{x^2\left(\frac{\sqrt{x}}{x^2}+1\right)}{x^2\left(\frac{2x}{x^2}-1\right)} \\
& \lim_{x\rightarrow \infty} \frac{\cancel{x^2}\left(\frac{\sqrt{x}}{x^2}+1\right)}{\cancel{x^2}\left(\frac{2\cancel{x}}{\cancel{x^2}}-1\right)}\\
& \lim_{x\rightarrow \infty} \frac{\left(\frac{\sqrt{x}}{x^2}+1\right)}{\left(\frac{2}{x}-1\right)} \\
\end{align}
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only a $+1$ on top. On the bottom, the same thing happens, $\frac{2}{x}$ goes to $0$ and we left with $-1$ in the denominator. Therefore
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$$
Is my solution correct and did I take the right steps with the correct logic?
Thanks!
| Looks good! I would be a bit more precise about "the top portion goes to $0$," but your reasoning is just fine.
| {
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By definition, how is a prime number represented? Even numbers can be easily represented as $2n$. Odd numbers as $2n+1$. An exactly divisible operation can be defined as $n = dq$.
But, is there an specific way of representing a prime number, obtained by a proof of some sort?
| From Jones, J., Sato, D., Wada, H. and Wiens, D. (1976). Diophantine representation of the set of prime numbers. American Mathematical Monthly, 83, 449-464.
The set of prime numbers is identical with the set of positive values taken on by the polynomial
$(k+2)(1-(wz+h+j-q)^2-((gk+2g+k+1)\cdot(h+j)+h-z)^2-(2n+p+q+z-e)^2-(16(k+1)^3\cdot(k+2)\cdot(n+1)^2+1-f^2)^2-(e^3\cdot(e+2)(a+1)^2+1-o^2)^2-((a^2-1)y^2+1-x^2)^2-(16r^2y^4(a^2-1)+1-u^2)^2-(((a+u^2(u^2-a))^2-1)\cdot(n+4dy)^2+1-(x+cu)^2)^2-(n+l+v-y)^2-((a^2-1)l^2+1-m^2)^2-(ai+k+1-l-i)^2-(p+l(a-n-1)+b(2an+2a-n^2-2n-2)-m)^2-(q+y(a-p-1)+s(2ap+2a-p^2-2p-2)-x)^2-(z+pl(a-p)+t(2ap-p^2-1)-pm)^2)$
as the variables range over the nonnegative integers.
| {
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$xy=22$ and $yz=26$: What is $x+y+z $ equal to? Given the following: $$xy=22,\qquad yz=26,$$ where $x,y,z\in\mathbb{N}$. Which of the following is a possible value of $ x + y + z $?
$ \textbf {(A) } 22 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 26 \qquad \textbf {(D) } 48 $
| We know that $(x+z)y=48$. We are looking for $(x+z)+y$.
$$
x+z=1;y=48\implies (x+z)+y=49\\
x+z=2;y=24\implies (x+z)+y=26\\
x+z=3;y=16\implies (x+z)+y=19\\
x+z=4;y=12\implies (x+z)+y=16\\
x+z=6;y=8\implies (x+z)+y=14\\
$$
we could switch the values, and get the other choices, but the sums and products remain the same.
The only common element is $26$.
If we want the actual values, note that $(x+z)y=2\cdot24$. Since neither $xy=22$ nor $yz=26$ are divisible by $24$, we need $y=2$. Then we get $x=11$ and $z=13$.
| {
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Problems while solving the cubics Some time ago, I got a question of the form $\sqrt{a+(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3} }$, which after cubing I realized that $(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3} = -a $. That set my instincts, and I figured out that that all cubic of the form $x^3 = a + bx$ must have a solution of the form $(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3}$, so I came to the formula:
$$(\frac{a}{2} - \sqrt{\frac{27a^2-4b^3}{108}})^\frac{1}{3} + (\frac{a}{2} + \sqrt{\frac{27a^2-4b^3}{108}})^\frac{1}{3}$$
I have still not managed to verify the formula, especially because of my faulty labelling of $a, b$, and have been a bit lousy to fix that. Anyway, to check my formula i invented the random cubic, $2x^3-5x - 6$, just because it had the root $2$. Now I plug in the values, $b=\frac{5}{2}$ and $a=3$ into my formula, which gives:
$$(\frac{9\sqrt{6} - 19}{6\sqrt{6}})^\frac{1}{3} + (\frac{9\sqrt{6} + 19}{6\sqrt{6}})^\frac{1}{3}$$
After that I was not able to do a single manipulation, apart from noticing that the common denominator is $\sqrt{6}$, however Wolfram Alpha tells me that the value is just simply $2$. Which is my first question, how should I simplify the above expression such that it quickly gives the simplified answer. Infact, I would like a general sure-fire technique, since I encounter this kind of dilemma all the time.
Anyways, I did some Googling thereafter, and came to this: http://www.sosmath.com/algebra/factor/fac11/fac11.html , which showed me how to compute the solution for any kind of cubic. I very much liked the method, though I have some doubts on the substitution, $x=y-\frac{b}{3a}$. Not that its incorrect, I would like to know the motivation from which we obtain that substitution. I would like to know how do we find such beautiful results, which we can use for example to clear the cubic term of a quartic equation.
| Better late than never. This is strongly related to this answer of mine, when I proved a general form for all solutions of $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n$$ for rational numbers. Here, we should work backwards: we have $$N=\left(\frac{9\sqrt{6} - 19}{6\sqrt{6}}\right)^{1/3} + \left(\frac{9\sqrt{6} + 19}{6\sqrt{6}}\right)^{1/3}=\left(\frac{3}{2}-\frac{19\sqrt 6}{36}\right)^{1/3} + \left(\frac{3}{2}+\frac{ 19\sqrt{6}}{36}\right)^{1/3},$$ so we have $r=6$, $p=3/2$, $q=19/36$. Putting this in the solutions yields $$(p,q)=(3/2, 19/36)=\left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right)=\left(\frac{18t^2N+N^3}{8},\,\frac{3N^2t+6t^3}{8}\right)$$ so we have the system $$\begin{cases}
18t^2N+N^3=12 \\
27N^2t+54t^3=38
\end{cases}
$$ with a real solution which can be worked out (nontrivial!) as $(N, t)=(2, 1/3).$ However, I wonder whether solving this system requires essentially the original problem.
| {
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A problem on indefinite integration $$\int\frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx$$
I tried some substitutions, but none succeeded in simplifying the expression. Please help.
| Once we know the answer,
$$\displaystyle\int\frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx =\int\frac{x^4-2}{x^3\sqrt{\frac{x^4+x^2+2}{x^2}}}dx=\int\frac{x-\frac2{x^3}}{\sqrt{x^2+1+\frac2{x^2}}}dx$$
Put $\displaystyle x^2+1+\frac2{x^2}=u^2$
| {
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Infinite Series $\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$
$H_n$ denotes the harmonic numbers.
| using the following identity :
$$\displaystyle \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$
multiply both sides by $\frac{\ln^2x}{x}$ then integrate both sides w.r.t $x$ from $0$ to $1$, we have:
\begin{align*}
S&=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ln^2x\ dx=\color{blue}{2\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{n^3}}=\int_0^1\frac{\ln^2x\ln^2(1-x)}{x(1-x)}\ dx\\
&=\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx+\underbrace{\int_0^1\frac{\ln^2x\ln^2(1-x)}{1-x}\ dx}_{x\mapsto1-x}=2\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx\\
&=4\sum_{n=1}^{\infty}\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx=4\sum_{n=1}^{\infty}\left(\frac{H_n}{n}-\frac1{n^2}\right)\frac{2}{n^3}=\color{blue}{8\sum_{n=1}^{\infty}\frac{H_n}{n^4}-8\zeta(5)}
\end{align*}
where we used $\ln^2(1-x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)x^n$
rearranging the terms of the blue sides, we have:
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}&=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^3}+4\sum_{n=1}^{\infty}\frac{H_n}{n^4}-4\zeta(5)\\
&=\left(3\zeta(2)\zeta(3)-\frac92\zeta(5)\right)+4\left(3\zeta(5)-\zeta(2)\zeta(3)\right)-4\zeta(5)\\
&=\frac72\zeta(5)-\zeta(2)\zeta(3)
\end{align*}
The proof of $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$ can be found here.
| {
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How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy} \right)(x^2+y^2+z^2+3xyz)\ge (x+y+z)^2$$
then we only prove
$$\dfrac{9}{x^2+y^2+z^2+3xyz}\ge\dfrac{3}{2}$$
$$\Longleftrightarrow x^2+y^2+z^2+3xyz\le 6$$
Then I can't,and for $n$ how prove it?
| Starting with $x+y+z =3$
From the relation between $AM - GM$:
$\frac{x+yz}2 \ge \sqrt{xyz}$
$\Rightarrow \frac 2{x+yz} \le \frac 1{\sqrt{xyz}}$
$\Rightarrow \frac{2x}{x+yz} \le \frac x{\sqrt{xyz}}$
Similarly for others it can be shown that
$ \frac {2y}{y+xz} \le \frac y{\sqrt{xyz}}$
$\frac {2z}{z+xy} \le \frac z{\sqrt{xyz}}$
Adding all the above inequations we get
$2[\frac x{x+yz} + \frac y{y+zx} + \frac z{z+xy}] \le \frac {x+y+z}{\sqrt{xyz}}$
Also since $x+y+z=3$ i.e. $(xyz)^{\frac 13} \le \frac {x+y+z}3 = 1$
$\Rightarrow (xyz)^{\frac 13} \le 1$
$\Rightarrow xyz \le 1$
Also when $a \le 1$, $a^{\frac 1n} \le a^{\frac 1{n+1}}$ which can be proved from the assumption that when $a\gt 1$
$a^n \le a^{n+1}$
So $(xyz)^{\frac 12} \le (xyz)^{\frac 13}$
$\frac {x+y+z}{(xyz)^{\frac 12}} \le \frac {x+y+z}{(xyz)^{\frac 13}}$
I came upto this, someone try to reach the last step.
| {
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Probability Problem on Divisibility of Sum by 3 From the 3-element subsets of $\{1, 2, 3, \ldots , 100\}$ (the set of the first 100 positive integers), a subset $(x, y, z)$ is picked randomly. What is the probability that $x + y + z$ is divisible by 3?
This is a math Olympiad problem. I would welcome a good solution.
| Note that our set has $34$ members equivalent to $1$ modulo $3$ (that is, with remainder $1$ upon division by $3$), $33$ equivalent to $2$, and $33$ equivalent to $0$.
Suppose we draw three numbers at random to create our subset. In order to get a sum divisible by $3$, we could draw any of the following sets of remainders:
$$
2,2,2\\
1,1,1\\
0,0,0\\
0,1,2 \text{ (In each of }3! \text{ arrangements)}
$$
So, the number of possible drawings that will yield a multiple of $3$ (order matters here) is
$$
(34\times 33\times 32)+ (33\times 32\times 31)\times 2+(33\times 34 \times 33)\times (3\times 2 \times 1)
$$
There are $100\times 99\times 98$ possible ways of drawing three numbers, our probability is
$$
\frac{(34\times 33\times 32)+ (33\times 32\times 31)\times 2+(33\times 34 \times 33)\times (3\times 2 \times 1)}{100\times 99 \times 98}
$$
Which, upon simplification, yields an answer of $\frac{817}{2450}$.
It is also possible to approach this with combinations instead of permutations. That is, noting we must have either all in one category or one from each category, our total number of possibilities is
$$
\binom{33}{3} + \binom{33}{3} + \binom{34}{3} + 33\times 34 \times 33
$$
producing a probability of
$$
\frac{\binom{33}{3} + \binom{33}{3} + \binom{34}{3} + 33\times 34 \times 33}{\binom{100}{3}}
$$
| {
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if $f(\frac{x+y}{2})=\frac{1}{2}[f(x)+f(y)], f(0)=0,f(1)=1$ then$f(\frac{1}{22})=?$ let function $f:[0,1]\to [0,1]$,and such $f(0)=0,f(1)=1$,
and foy any $0\le x\le y\le 1$,then we have
$$f\left(\dfrac{x+y}{2}\right)=\dfrac{1}{2}[f(x)+f(y)]$$
Question 1
Find the value $f(\dfrac{1}{22})$
Qusetion 2: Find the
$$f(\dfrac{1}{n})=\dfrac{1}{n}?,n\in N^{+}$$
My try: Now I have solve question 1:
since
$$f\left(\dfrac{x+y}{2}\right)=\dfrac{1}{2}[f(x)+f(y)]$$
then let $x=0$,we have
$$f(y)=2f(\dfrac{y}{2})$$
let $f(\dfrac{1}{22})=a$,then
$$f(\dfrac{1}{11})=f(\dfrac{2}{22})=2f(\dfrac{1}{22})=2a,f(\dfrac{2}{11})=4f(\dfrac{1}{22})=4a,f(\dfrac{4}{11})=8f(\dfrac{1}{22})=8a$$
and $$f(\dfrac{8}{11})=16f(\dfrac{1}{22})=16a$$
and note
$$f(\dfrac{6}{11})=f(\dfrac{\dfrac{1}{11}+1}{2})=\dfrac{1}{2}(f(\dfrac{1}{11}+1)=\dfrac{1}{2}(2a+1)$$
and other hand
$$f(\dfrac{6}{11})=f(\dfrac{4}{11}+\dfrac{8}{11}/2)=\dfrac{1}{2}(8a+16a)$$
so
$$\dfrac{1}{2}(8a+16a)=\dfrac{1}{2}(2a+1)\Longrightarrow a=\dfrac{1}{22}$$
so
$$f(\dfrac{1}{22})=\dfrac{1}{22}$$
But question 2,How prove it? Thank you
| As long as you consider $f$ at rational places, no continuity assumption is needed. Since 1. is a special case of 2., let's solve that:
Let $n\in\mathbb N$ and $x_i=\frac in$, $0\le i\le n$.
Then for $0<i<n$ we have $x_i=\frac{x_{i-1}+x_{i+1}}2$ and hence $f(x_i)=\frac{f(x_{i-1})+f(x_{i+1})}2$, i.e. $$\tag1f(x_{i+1})=2f(x_i)-f(x_{i-1}).$$ Then by induction we find that $f(x_i)=if(x_1)$ for $0\le i\le n$. Indeed, this is trivially true for $i=0$ and for $i=1$ and the induction step follows immediately from $(1)$. From $nf(x_1)=f(x_n)=f(1)=1$ we then conclude that $f(x_1)=\frac1n$, as was to be shown.
(In fact, we have at the same time shown that $f(x)=x$ for all $x\in[0,1]\cap \mathbb Q$ as $\frac mn$ with $0\le m\le n$ occurs as $x_m$ in the above sequence and $f(x_m)=mf(x_1)=\frac mn$)
| {
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Elementary proof that $\pi < \sqrt{5} + 1$ I wanted to show that
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove that $\dfrac{\pi}{\phi^2}<\dfrac{6}5 $.
By squaring my inequality (valid since both sides are positive), and dividing by $6$ I obtain.
$$ \frac{\pi^2}{6} < 1 + \frac{\sqrt{5}}{3} $$
Where the left handside has a very neat series representation, alas the same does not hold for the right handside. However this is far from an elementary solution. Does someone have a relative simple proof for the equality? To be precise something that is not using advanced knowledge of series. =)
| Start with an unit circle inscribed inside a $2 \times 2$ square. One can chop off 4 right angled isosceles triangle whose shorter side has length $2 - \sqrt{2}$ from the four corners. This will turn the square into a octagon with the unit circle inscribed inside it. By comparing the area of the circle and the octagon, we have
$$\pi = \text{Area(circle)} < \text{Area(octagon)} = 4 - 2 (2 - \sqrt{2})^2 = 8(\sqrt{2}-1)$$
Since $2 \cdot 12^2 = 288 < 289 = 17^2$, we have
$$12 \sqrt{2} < 17
\implies 8 (\sqrt{2}-1) < \frac{10}{3}
\implies \frac{3\pi}{10} < 1
$$
For $x \in (0,1)$, if one look at the Taylor expansion of $\sin x$ at $x = 0$, we have
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
One will notice the magnitude of each term is monotonic decreasing while the sign is alternating. This implies
$$\sin x > x ( 1 - \frac{x^2}{6} ) > \frac{5x}{6}\quad\text{ for } x \in (0,1)$$
In particular, this is true for $ \displaystyle x = \frac{3\pi}{10}$. As a result,
$$ 1 + \sqrt{5} = 4 \cos\left(\frac{\pi}{5}\right) = 4 \sin\left(\frac{3\pi}{10}\right)
> 4 \left(\frac{5}{6}\right)\left(\frac{3\pi}{10}\right) = \pi$$
Update
An alternate proof without calculus.
Start with a circle centered at $O$ with radius $2$. Inscribe a regular pentagon inside it.
Let $AB$ be an edge of the pentagon and $C$ be its mid-point. Extend $OC$ until it hit the circle at $D$. Construct a line through $O$ perpendicular to $OC$ and let it hit the circle at $E$. Let $F$ be the point so that $OCFE$ forms a rectangle.
From elementary geometry, we know the area of rectangle $OCFE$ is $1 + \sqrt{5}$ and the area of the quarter circle (the red one as shown) $ODE$ is $\pi$. To prove the inequality
$1 + \sqrt{5} > \pi$, one just suffices to show the area of the green shape $BFE$ above
is larger than that of the shape $BCD$.
Extend $OB$ to the point $G$ where $BG = OB = 2$. Construct a line through $G$ parallel to $OD$. Let $C'$ be its intersection with the line $CF$. Let $D'$ be a point on the line at a distance $2$ from $G$. By symmetry, the shape $BC'D'$ enclosed by the red dashed lines
has the same area as the shape $BCD$.
Construct a line through $D'$ parallel to $OE$ and let it hit the line $EF$ at $H$.
The area of that portion of the shape $BC'D'$ outside the rectangle $OCFE$ is smaller than
that of the rectangle $FC'D'H$. Extend $D'H$ to the point $D''$ where $HD'' = D'H$. Extend $FH$ to the point $C''$ where $HC'' = 2FH$. The triangle $HC''D''$ has the same area as the rectangle $FC'D'H$'. It is clear this triangle lies completely in the green shape $BEF$ and
disjoint from the shape $BC'D'$. As a result, we have:
$$\text{Area}(BFE) > \text{Area}(BC'D') = \text{Area}(BCD)$$
and hence
$$1 + \sqrt{5} = \text{Area}(OCFE) > \text{Area}(ODE) = \pi$$
| {
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Which step is wrong in this proof
Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give
$$x=−1−\frac{1}{x}$$
Substitute this back into the $x$ term in the middle of the original equation, so
$$x^2+(−1−\frac{1}{x})+1=0$$
This reduces to
$$x^2=\frac{1}{x}$$
So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives
$1^2+1+1=0$
Therefore, $3=0$.
What happened?
| if $x^2+x+1=0$ then $(x-1)(x^2+x+1)=0$ thus $x^3-1=0$, thus: $x^3=1$.
if ... then .... is not a logical equivalence but only a logical implication.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$
are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)
I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $
$$
\begin{vmatrix}
f & g & h \\
f' & g' & h' \\
f'' & g'' & h'' \\
\end{vmatrix}
$$
If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and $h(x)$ are linearly independent.
If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$
My attempt
Let
$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$
$W(f, g, h) =$
$$
\begin{vmatrix}
\sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\
\cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\
-\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\
\end{vmatrix}
$$
$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)] - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$
$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$
By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$
Not sure if this argument is sound?
| Other solution:
If you know that the set of diffrential equation: $$(E) \quad y''+y=0$$ solutions is a two dimensinal space, then since the three applications : $x \mapsto \sin(x+a)$ where $a \in\{\alpha,\beta,\gamma \}$ are solutions of $(E)$, there are linearly dependantes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
$a;b;c\in \mathbb{R}^+$. Prove : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$ $a;b;c\in \mathbb{R}^+$. Prove : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$
Thanks :)
I have proved that :
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{abc}{bca}}=3$
And : $\frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\leq \frac{\sqrt{3}(a+b+c)}{a+b+c}=\sqrt{3}$
!?
| use this known lema:
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{9(a^2+b^2+c^2)}{(a+b+c)^2}$$
proof:By applying the know inequality
$$(x+y+z)^3\ge\dfrac{27}{4}(x^2y+y^2z+z^2x+xyz)$$
for
$x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$,we get
$$\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^3\ge\dfrac{27}{4}\left(\dfrac{a^3+b^3+c^3}{abc}+1\right)$$
it suffices to prove that
$$\dfrac{a^3+b^3+c^3}{abc}+1\ge\dfrac{108(a^2+b^2+c^2)^3}{(a+b+c)^6}$$
or
$$\dfrac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+4\ge\dfrac{108(a^2+b^2+c^2)^3}{(a+b+c)^6}$$
Using now the obvious inequality $3abc(a+b+c)\le (ab+bc+ac)^2$,we have
$$\dfrac{a+b+c}{abc}=\dfrac{3(a+b+c)^2}{3abc(a+b+c)}\ge\dfrac{3(a+b+c)^2}{(ab+bc+ac)^2}$$
and hence,it is enough to check that
$$\dfrac{3(a+b+c)^2(a^2+b^2+c^2-ab-bc-ac)}{(ab+bc+ac)^2}+4\ge\dfrac{108(a^2+b^2+c^2)^3}{(a+b+c)^6}$$
let
$$t=\dfrac{3(a^2+b^2+c^2)}{(a+b+c)^2},1\le t<3$$
then inequality is equivalent to
$$\dfrac{54(t-1)}{(3-t)^2}+4\ge 4t^3$$
or
$$(t-1)(9-6t-8t^2+10t^3-2t^4)\ge 0$$
this is true because
$$9-6t-8t^2+10t^3-2t^4=2(3+3t-t^2)(t-1)^2+3>0$$
This lema prove by done.
so you can let
$$x=\dfrac{a^2+b^2+c^2}{(a+b+c)^2}\ge\dfrac{1}{3}$$
so use this lemma, it suffices to prove that
$$9x+\dfrac{1}{\sqrt{x}}\ge 3+\sqrt{3}$$
let $f(x)=9x+\dfrac{1}{\sqrt{x}},x\ge\dfrac{1}{3}$
then we have
$$f'(x)=9-\dfrac{1}{2x^{\frac{3}{2}}}>0$$
so
$$f(x)\ge f(\dfrac{1}{\sqrt{3}})=3+\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Concave Convex $e^{-x}-e^{-2x}$ My main goal is to see here if my discusison on concave/convex is good enough.
Problem
Find the extreme points (max and/or min) for $$f(x)=e^{-x}-e^{-2x}.$$ Also discuss concavity/convexity.
Attempt
Extreme points
Solve $f'(x) = 0 -e^{-x} + 2e^{-2x} = 0$
Multiply by $-e^{2x}$ and get $e^x - 2 = 0 \quad \Rightarrow \quad e^x = 2 \quad \Rightarrow \quad x = \ln(2).$
Then: $f(\ln(2)) = e^{-\ln 2} - e^{-2\ln(2)} = e^{\ln(2^{-1})} - e^{\ln(2^{-2})} = e^{\ln(\frac{1}{2})} - e^{\ln(\frac{1}{4})} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$
There is an extreme point at $\left(\ln(2), \frac{1}{4}\right).$
Second derivative test: $f''(x) = e^{-x} - 4e^{-2x}.$
$f''(\ln(2)) = e^{-\ln(2)} - 4e^{-2\ln 2} = e^{\ln(2^{-1})} - 4e^{\ln(2^{-2})} = e^{\ln(\frac{1}{2})} - 4e^{\ln(\frac{1}{4})}.$
$= \frac{1}{2} - 4\left(\frac{1}{4}\right) = \frac{1}{2} - 1 = -\frac{1}{2}.$
In $x_0=\ln(2)$ we have $f'(x_0)=0$ and $f''(x_0)<0.$
Thus, $\left(\ln(2), \frac{1}{4}\right)$ is a maximum point.
Concave/Convex
$f''(\ln(4)) = 0$. And $f''$ is negative to the left of $\ln(4)$, and positive to the right, so $f$ is concave for $x<\ln(4)$ and conve for $f>\ln(4)$.
| !
$f′′(\ln(4))=0$
$f′′$ is negative to the left of $\ln(4)$
and positive to the right
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/563726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the Unknown Coordinate when given Distances from two points I have this problem that I don't know how to do.
Locate the two points that are $5$ units from $(5,-3)$ and square root of $41$ units from$(-2,6)$
| You need the intersection of two circles.
Let's say $A=(5,-3)$ and $B=(-2,6)$, $r_1=5$ and $r_2=\sqrt{41}$.
Now
$$c_1:(x-5)^2+(y+3)^2=5^2 \quad and \quad c_2:(x+2)^2+(y-6)^2=\sqrt{41}^2$$
Now you have to find you radical axis of $c_1,c_2$:
\begin{eqnarray*}
c_1-c_2&:x^2-10x+y^2+6y+9=0 \\
&-[x^2+4x+y^2-12y-1=0]\\
&\Rightarrow z(x)=\frac 7 9 x - \frac 5 9
\end{eqnarray*}
With $z(x) \cap c_1$ you'll get
$$(x-5)^2+(z(x)+3)^2=25$$ with $L=\lbrace \frac{121}{65}, 2 \rbrace$ for $x$. So your two points are $S_1(\frac{121}{65};\frac{58}{65})$ and $S_2(2;1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/565325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fast way to integrate $\frac{x^2-y^2}{(x^2+y^2)^2} dx \,dy$ in unit square I am looking for a fast way to integrate $$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx \,dy$$ using standard techniques ( no complex analysis and no functional analysis).
I am aware that wolframalpha spits out a solution, but this one is quite long, I assume that there is a faster way to do this. The result will be $\frac{\pm \pi}{2}$(depending on which integral you do first). By the way: Please be aware of the fact, that fubini's theorem does not hold in this case.
| Start with the inside integral:
$$
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx
$$
Let $x=y\tan\theta$, so that $dx = y\sec^2\theta\,d\theta$ and $x^2+y^2 = y^2\tan^2\theta+y^2 = y^2\sec^2\theta$. As $x$ goes from $0$ to $1$, $\theta$ goes from $0$ to $\arctan(1/y)$, so the integral becomes
\begin{align}
& \phantom{={}}\int_0^{\arctan(1/y)} \frac{y^2\tan^2\theta-y^2}{(y^2\tan^2\theta+y^2)^2} y\sec^2\theta\,d\theta \\[12pt]
& = \frac1y\int_0^{\arctan(1/y)} \frac{\tan^2\theta-1}{\sec^4\theta}\sec^2\theta\,d\theta \\[12pt]
& =\frac1y\int_0^{\arctan(1/y)}(\sin^2\theta-\cos^2\theta)\,d\theta \\[12pt]
& = \frac1y\int_0^{\arctan(1/y)} -\cos(2\theta)\,d\theta \\[12pt]
& = \frac1y\left[\frac{-\sin(2\theta)}{2}\right]_{\theta=0}^{\theta=\arctan(1/y)} = \frac1y\left[-\sin\theta\cos\theta\right]_{\theta=0}^{\theta=\arctan(1/y)}. \tag 1
\end{align}
Since tangent${}={}$opposite$/$adjacent, draw a right triangle in which the "opposite" side is $1$ and the "adjacent" side is $y$. By the Pythagorean theorem, the hypotenuse is $\sqrt{1+y^2}$, so the sine is opposite$/$hypotenuse $=\dfrac{1}{\sqrt{1+y^2}}$ and the cosine is adjacent$/$hypotenuse $=\dfrac{y}{\sqrt{1+y^2}}$. Hence $(1)$ becomes
$$
\frac{-1}{1+y^2}.
$$
It's easy to integrate that from $0$ to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/567412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$
Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help.
| HINT:
Use $\log a+\log b+\log c=\log(abc) $
and then put $a=\tan A$ etc. to find
$\displaystyle \sum\tan A=\prod\tan A\implies A+B+C=n\pi$ where $n$ is any integer (See here)
and $\displaystyle\frac{2a}{1-a^2}=\frac{2\tan A}{1-\tan^2A}=\tan2A$
$\displaystyle\implies \sum\tan2A=\prod\tan2A$ as $2A+2B+2C=2n\pi$
Now apply logarithm now
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Hard contest type trigonometry proof Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
| I prefer @math110's solution, but here's a brute force method using complex exponentials,
with
$$\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \qquad \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$
We define
$$a := e^{ix} \qquad b := e^{iy} \qquad c := e^{iz}$$
so that
$$p = \frac{\cos x + \cos y + \cos z}{\cos(x+y+z)} \implies p(a^2b^2c^2 +1) = abc (a+b+c) + bc + ca + ab \qquad (1)$$
$$p = \frac{\sin x + \sin y + \sin z}{\sin(x+y+z)} \implies p(a^2b^2c^2 - 1 ) = abc (a+b+c) - bc - ca - ab \qquad (2)$$
Thus, from $(1)-(2)$ and $(1)+(2)$, we have
$$p = bc + ca + ab \qquad\qquad p = \frac{a+b+c}{abc} = \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$$
whereupon
$$2p = bc+\frac{1}{bc}\;+\;ca+\frac{1}{ca}\;+\;ab+\frac{1}{ab} = 2\left( \cos(y+z)+\cos(z+x)+\cos(x+y)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Show $f(x) = (x^4+x^2+1)/(x^3+1) $ is $O(x)$ How would I find the witnesses $C$ and $k$ such that $f(x)$ is $O(x)$?
What I tried was $$(x^4+x^2+1)/(x^3+1) ≤ (x^4+x^4+x^4)/(x^3+x^3) = (3/2)x $$
for values $x>1$. $C = 3/2, k = 1$
Is this right?
| We have
$$
f(x)= \frac{x^4 + x^2 +1 }{x^3 +1} \leq C x,
$$
for $x \geq \sqrt{\frac{C}{3(C-1)}}$ and $C>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question on Convergence of Improper integrals Question is to check which of the following improper integrals are convergent?
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I was having a stupid mindset that :
"$\textbf{Improper integrals should have at least one limit infinite}$"
and so i came to conclusion that given two integrals with finite limits are convergent and thus question is nonsense.
But sooner i realized that for two integrals $$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
denominators have zeros in given interval $[0,5]$
So, then I decided that these integrals are actually improper :)
Now, I tried seeing $$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$ as limit of $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}$$ with $b\rightarrow \infty$
$$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=2\sqrt{x^2+2x+2}.\frac{1}{2x+2}=\sqrt{x^2+2x+2}.\frac{1}{x+1}=\sqrt{\frac{(x+1)^2+1}{(x+1)^2}}=\sqrt{1+\frac{1}{(x+1)^2}}$$
sorry for forgetting limits
which would imply that $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=\sqrt{1+\frac{1}{(b+1)^2}}-\sqrt{1+\frac{1}{(1+1)^2}}$$
and when $b\rightarrow \infty$ as $b$ is in denominator... this limit exists and so,
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}=1-\sqrt{\frac{5}{4}}$$
As $x^2+2x+2$ is positive in $[1,\infty]$, the integral should be positive, I might have missed some signs in between but i am sure that this is convergent.. (I would be thankful if some one can help me to see where the problem is with signs)
Coming to second problem, $$\frac{1}{x^2-5x-6}=\frac{1}{x-3}-\frac{1}{x-2}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}=\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$
$$\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)=\int _0^2 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _2^3 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _3^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$
$$=\log (2-3)- \log(2-2) \dots$$ but then $\log 0$ is not defined
SO, I would like to confirm that this integral is divergent.
Coming to third integral $$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I would like to see this as limit of $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
and $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}=\frac{3}{2}(7x+2x^4)^{\frac{2}{3}}.\frac{1}{8x^3+7}$$
with similar calculations as i have done for first integral, I have seen that this integral also converges (I do not want to write that and kill readers patience :D)
I am sure if my idea for first integral is correct then third integral should also converge.
So, I would say that first and third integrals converge where as second integral do not converge.
I would be thankful if someone can verify my solution and please let me know if there are any gaps.
Thank You.
P.S : I would like to remind for users (who are as dumb as i am) that
$\textbf{Improper integrals need not necessarily have infinite end points}$
| Hint: For the issue concerning convergence it sufficient and necessary to understand only the two following types of integrals for $\alpha \in \mathbb{R}$:
$$ \int_{0}^1 x^\alpha d x , \qquad \int_{1}^\infty x^\alpha d x.$$
E.g.
$$ \int_{1}^\infty ( (x+1)^2 + 1)^{-1/2} d x = \int_{0}^\infty ( x^2 + 1)^{-1/2} d x \geq C * \int_{1}^\infty x^{-1} dx = \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/577096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to speedup evaluation of hypergeometric ${}_3 F_2(1)$? I need to make a table of ${}_3 F_2\left(\frac{a}2+\frac14, \frac{a}2+\frac34, \frac{a+b}2;\; a+1,\frac{a+b}2+1;\;1\right)$ for integer $a, b,$ $0\le a\le N_1$, $0\le b\le N_2$, with precision of 50 decimal places in mantissa.
Currently I'm trying to use Wolfram Mathematica for this task, but it appears to evaluate a single value for over a minute when $a$ and $b$ are $\sim50$. And as $N_1\approx50$ and $N_2\approx 300$, this appears to be too long a computation.
So I'm looking for ways to speed up this calculation. Namely, is there a way to reduce this function to something simpler for these particular arguments? Are there some relations which would allow to easily compute it for given $a_i, b_i$, having computed for several other values (like the relation $\Gamma(x+1)=x\Gamma(x)$, which lets one easily find $\Gamma(x+1)$ knowing $\Gamma(x)$ without computing $\Gamma$ once more)?
| Hint:
$_3F_2\left(\dfrac{a}{2}+\dfrac{1}{4},\dfrac{a}{2}+\dfrac{3}{4},\dfrac{a+b}{2};a+1,\dfrac{a+b}{2}+1;1\right)$
$=\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}\left(\dfrac{a+b}{2}\right)^{(n)}}{(a+1)^{(n)}\left(\dfrac{a+b}{2}+1\right)^{(n)}n!}$
$=\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}\dfrac{a+b}{2}}{(a+1)^{(n)}\left(n+\dfrac{a+b}{2}\right)n!}$
$=\dfrac{a+b}{2}\int_0^1\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}x^{n+\frac{a+b}{2}-1}}{(a+1)^{(n)}n!}~dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.
Thus, any help to get me out of this rift is more than welcome!
| Using Aron D'souza's idea further we can get:
$$L^2-\frac{1}{2}< \sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}$$
$$\sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}=\frac{1}{2} \left(1+\sqrt{\frac{7}{3}} \right)$$
$$L<\sqrt{1+\frac{1}{2} \sqrt{\frac{7}{3}}}=1.328067$$
To find the next bound we will need to solve:
$$c^4-c-\frac{1}{4}=0$$
The exact solution is too complex to write here (see Wolframalpha), so I'll just write it numerically:
$$\sqrt[4]{\frac{1}{4}+\sqrt[4]{\frac{1}{4}+\dots}}=1.0723501510383$$
The bound will become:
$$L<\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.0723501510383}}=1.272871$$
Which is three correct digits of the numerical value of the limit.
To make my answer more complete, the exact value of $c_4$ is:
$$c=\frac{1}{2} \left( b+\sqrt{\frac{2}{b}-b^2} \right)$$
$$b=\sqrt{ \sqrt[3]{ \frac{a}{18} }-\sqrt[3]{ \frac{2}{3a} } }$$
$$a=9+\sqrt{93}$$
And solving the quintic equation:
$$c^5-c-\frac{1}{5}=0$$
We get the upper bound for the limit with four correct digits:
$$L<1.272282$$
Taking into account the corresponding lower boundary:
$$L>\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\sqrt[5]{\frac{1}{5}+\sqrt[6]{\frac{1}{6}}}}}}=1.271035$$
We see that truncating the limit gives less accurate solutions than the method in this answer.
However, truncating at $\frac{1}{7}$ we can finally get very good boundaries:
$$1.27207<L<1.27228$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
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} |
Recursive sequence of functions Recursive sequence of functions: $f_{n+1}= \sqrt{x+f_n}$, $f_1(x)= \sqrt{x}$.
this sequence is monotonic, but what is bounding it? thanks
| So you have defined $$f_n(x) = \underbrace {\sqrt{x + \sqrt {x + \dots \sqrt x}}}_{\text {n square roots}}$$ and you say that it is "monotonic", suggesting some comparison like
$$f_{n+1} \ge f_n$$
which, correct me if I'm wrong, I'm going to assume means $\forall x\, f_{n+1}(x) > f_n(x)$.
A bound (..but not a least upper bound...) then would imply that you are looking for a function $g(x)$ such that
$$\forall n,\,x,\, g(x) \ge f_n(x)$$
This is a bit more than I would expect in a regular infinitessimal calculus class. The upper bound obviously has to be larger than $g(x) = \sqrt x$, but it doesn't seem like it has to be much more. Let's try $g(x) = \sqrt {2x}$. To use induction though, we have to prove a stronger result:
$$h_n(x) = \sqrt{\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^{n-1}}\right)\cdot x}$$
$$h_n(x) \ge f_n(x)$$
Base case:
$$\sqrt{x} \ge \sqrt{x}$$
Recursive assumptions:
$$h_n(x) \ge f_n(x)$$
Recursive cave:
$$h_{n+1}(x) \ge f_{n+1}(x)$$
$$\sqrt{\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^n}\right)\cdot x} \ge \sqrt{x + f_n(x)}$$
$$\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^n}\right)\cdot x \ge x + f_n(x)$$
$$\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^{n - 1}}\right)\cdot x + \frac x {2^n} \ge x + f_n(x)$$
$$h_n(x)^2 + \frac x {2^n} \ge f_n(x) + x $$
$$\text{Is implied by}$$
$$f_n(x)^2 + \frac x {2^n} \ge f_n(x) + x $$
$$f_n(x)^2 - f_n(x) \ge x - \frac x {2^n} $$
This last statement....sorry...we have to prove with induction over n.
Base case:
$$x - \sqrt{x} > x - \frac x 1$$
For $x\ge 1$ this holds.
Recursive assumption:
$$f_n(x)^2 - f_n(x) \ge x - \frac x {2^n} $$
Recursive case:
$$f_{n+1}(x)^2 - f_{n+1}(x) \ge x - \frac x {2^{n+1}} $$
$$x+f_n(x) - \sqrt{x+f_n(x)} \ge x - \frac x {2^{n+1}} $$
$$f_n(x) - \sqrt{x+f_n(x)} \ge \frac x {2^{n+1}} $$
$$f_n(x) + \frac x {2^{n+1}} \ge \sqrt{x+f_n(x)} $$
$$f_n(x)^2 + 2f_n(x)\frac x {2^{n+1}} + \frac {x^2} {2^{n+2}} \ge x+f_n(x) $$
$$\text{Is implied by: (using the recursive assumption)}$$
$$x - \frac x {2^n} + 2f_n(x)\frac x {2^{n+1}} + \frac {x^2} {2^{n+2}} \ge x$$
$$x^2 - 4 x + 4f_n(x)\ge 0$$
Which holds for $x \ge 1$.
Since $\forall n\, g(x) \ge h_n(x)$, $g$ is an upper bound for all $f_n$, for $x \ge 1$.
| {
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Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
| Heres a nice combinatorial proof: Lets say you have $n+1$ kids, and want to form a committee of three. Order the kids $a_1, a_2, \cdots, a_{n+1}$. There are $\dbinom{n+1}{3}$ ways to form the committee. On the other hand, if $a_1$ is the first person on the committee, we need to choose two more, in $\dbinom{n}{2}$ ways. If $a_2$ is the first person on the committee, we can choose two more in $\dbinom{n-1}{2}$ ways. In general, if $a_n$ is the first person on the committee, we can choosse two more in $\dbinom{n-k+1}{2}$ ways. Therefore, we have $$ \dbinom{n+1}{3} = \sum_{i=1}^{n+1} \dbinom{n-k+1}{2} = \dbinom{n}{2} + \dbinom{n-1}{2} + \cdots + \dbinom22$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fixed sum of combinations When you have combinations where numbers are $0,1,2,\dots,m$, meaning we have $n=m+1$ and $k$, is there a way to see how $k$ of them sum up to a given number?
For the sake of simplicity I have the numbers $0,1,2...,7$ (so $n=8$), and $k=3$. I need to find how much of these combinations with repetition sum up to $7$. By sum up, I mean the sum of all $3$ digits in each combination needs to be equal to $7$.
Is there a formula for this?
| Write $m$ as
$$\underbrace{1 + 1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m 1s}}$$
This isn't the only sum, however. Other possible sums are $$3 + 3 + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-6 1s}}$$ or $$2 + 1 + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-5 1s}} + 2$$ Notice, however, if we decompose all our non-unit integers into ones, they all end up as $$\underbrace{1 + 1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m 1s}}$$ In other words, we can represent $3 + 3 + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-6 1s}}$ as $$(1 + 1 + 1) + (1 + 1 + 1) + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-6 1s}}$$
and $2 + 1 + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-5 1s}} + 2$ as $$(1 + 1) + (1) + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-5 1s}} + (1 + 1)$$
Instead of using parenthesis, let us re-write this as
$$1 + 1 \ \Big\vert \ + 1 \ \Big\vert + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-5 1s}} \ \Big\vert + 1 + 1$$
and re-write the other sum as $$1 + 1 + 1 \ \Big\vert + 1 + 1 + 1 \ \Big\vert + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-6 1s}}$$
Notice that every single sum can be represented with such "bar notation". For sums with zeroes, such as $3 + 0 + 2 + 5$, we could write $$ 1 + 1 + 1 \ \Big\vert \ \Big\vert + 1 + 1 \ \Big\vert + 1 + 1 + 1 + 1 + 1$$ (there are no ones in between the first two bars) and for sums such as $0 + 0 + 0 + 1 + 1 + 2 + 3$ we could do $$ \ \Big\vert \ \Big\vert \ \Big\vert 1 \ \Big\vert + 1 \ \Big\vert + 1 + 1 \ \Big\vert + 1 + 1 + 1$$
as there is nothing to the left of the first divider, representing the first zero, nothing in between the first two dividers, and nothing in between the second and third dividers.
So suppose we have $k$ summands. This would require $k-1$ bars, just as $1 \ \Big\vert + 1 \ \Big\vert + 1$ requires two bars instead of three. There are $m$ ones, which must be arranged with the $k-1$ bars. Therefore, in total, there are $\dbinom{m + k - 1}{m}$ ways to do the arrangement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/585712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+1)}$
Now I Did not understand how can i solve after that
Help Required
Thanks
| Note that\begin{align}
&\frac{x^5}{x^5+x+1}
=1-\frac{2x+3}{7(x^2+x+1)}+
\frac{2x^2-x-4}{7(x^3-x^2+1)}\\
=& \ 1-\frac{2x+1}{7(x^2+x+1)}-\frac{2}{7(x^2+x+1)} +
\frac{2(3x-2x)}{21(x^3-x^2+1)} +
\frac{x-12}{21(x^3-x^2+1)}
\end{align}
Then
$$\int \frac{x^5}{x^5+x+1} dx
=x -\frac1{21}\ln\frac{(x^2+x+1)^3}{(x^3-x^2+1)^2}
-\frac4{7\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3}+\frac1{21}K
$$
where
$$K = \int \frac{x-12}{x^3-x^2+1}dx$$
$x^3-x^2+1= (x-a)(x^2-\frac x{a^2}-\frac1a) $ has a single real root $a=-0.7549$, or
$$a =\frac13\bigg( 1-\sqrt[3]{\frac{25+3\sqrt{69}}2}-\sqrt[3]{\frac{25-3\sqrt{69}}2}\bigg)
$$
which allows $K$ to be integrated below with partial fractions
\begin{align}
K= &\int\frac{x-12}{(x-a)(x^2-\frac x{a^2}-\frac1a)}dx\\
=& \ \frac1{a^3-2}\int
\frac{a(a-12)}{x-a}+ \frac{a(12-a)(2x-\frac1{a^2})}{2(x^2-\frac x{a^2}-\frac1a)}+ \frac{12a^2-\frac6a-\frac32}{x^2-\frac x{a^2}-\frac1a}\ dx\\
=& \ \frac1{a^3-2}\bigg(\frac{a(a-12)}{2}\ln\frac{(x-a)^2}{x^2-\frac x{a^2}-\frac1a}
+\frac {a^2-16a-4}{\sqrt{-4a^3-1}}\tan^{-1}\frac{2a^2x-1}{\sqrt{-4a^3-1}}\bigg)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
let sequence $\{a_{n}\}$ such $a_{1}=1$,and
$$a_{n+1}a_{n}=n,n\ge 1$$
show that
$$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$
(2): I consider we can find this limit
$$\lim_{n\to+\infty}\dfrac{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}{\sqrt{n}}=?$$
My try:since
$$a_{n+2}a_{n+1}-a_{n+1}a_{n}=n+1-n=1$$
so
$$a_{n+2}=\dfrac{1}{a_{n+1}}+a_{n}$$
so
$$\dfrac{1}{a_{n+1}}=a_{n+2}-a_{n}$$
so
\begin{align*}\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}&=a_{1}+(a_{3}-a_{1})+(a_{4}-a_{2})+(a_{5}-a_{3})+\cdots+(a_{n+1}-a_{n-1})\\
&=a_{1}+a_{n+1}+a_{n}-a_{1}-a_{2}\\
&=a_{n+1}+a_{n}-a_{2}
\end{align*}
since $$a_{1}=1,a_{1}a_{2}=1\Longrightarrow a_{2}=1$$
so
$$a_{n+1}+a_{n}-a_{2}\ge2\sqrt{a_{n+1}a_{n}}-1\ge 2\sqrt{n}-1$$
so left hand inequality is prove it.Now consider Right hand inequality,we only prove this
$$a_{n}+a_{n+1}<\dfrac{5}{2}\sqrt{n}$$
since
$$a_{n}a_{n+1}=n\Longrightarrow a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$$
so maybe we can $a_{n}<2\sqrt{n}$?
becasue $$\dfrac{5}{2}\sqrt{n}=2\sqrt{n}+\dfrac{n}{2\sqrt{n}}$$
so let $$f(x)=x+\dfrac{n}{x}$$
if we can prove $a_{n}<2\sqrt{n}$,Then
$$f(x)\le f(2\sqrt{n})=\dfrac{5}{2}\sqrt{n}$$
Thank you very much
| observe that for $a_{n}$ to satisfy the inequality $a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$
we need $\dfrac{\sqrt{n}}{2} <a_{n}<2\sqrt{n}$.
this can be proved by induction: first observe that it is true for $a_{1}$ and $a_{2}$ then suppose it is true for any n ($a_{n}$) then prove it is true for n+1 ($a_{n+1}$) and you are done
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of the expression. Find the minimum value of the expression. $x,y,z \in R$
$\sqrt{x^2+1}+ \sqrt {4+(y-z)^2} + \sqrt{1+ (z-x)^2} + \sqrt{9+(10-y)^2}$
| $S = d(E, A) + d(A,B) + d(B, C) + d(C,D) \\ \therefore S = \sqrt{10^2+7^2}=\boxed{\sqrt{149}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let n be a positive integer. Prove that: Let n be a positive integer. Prove that:
$\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$
| Case $1$: Let $n=m^2$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m$$
We now have $\sqrt{4n+2} = \sqrt{4m^2+2} \in [2m,2m+1]$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m$
Case $2$: Let $n+1=m^2$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m-1$$
We now have $\sqrt{4n+2} = \sqrt{4m^2-2} \in [2m-1,2m]$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m-1$
Case $3$: Neither are perfect squares, i.e., $m < \sqrt{n} < \sqrt{n+1} < (m+1)$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m$$
We now have $\sqrt{4n+2} \in \left(\sqrt{4m^2+2},\sqrt{4m^2+8m+2} \right)$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m$
| {
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"timestamp": "2023-03-29T00:00:00",
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Need Help With Strong Induction Let $(X_n)$ be a sequence given by the following recursion formula:
$$X_1 = 3, X_2 = 7,\text{ and }X_{n+1} = 5X_n - 6X_{n-1}$$
Prove that for all $n\in\Bbb N$, $X_n = 2^n + 3^{n-1}$.
Attempt:
For $n = 1$, we have $2^1 + 3^0 = 3 = a_1$ TRUE
For $n = 2$, we have $2^2 + 3^1 = 7 = a_2$ TRUE
Assume $X_k = 2^k + 3^{k-1}$ for some $k\in\Bbb N$.
Now, for $n = k+1$:
$$\begin{align*}
X_{k+1} &= 5X_k - 6X_{k-1}\\
&= 5\left(2^k + 3^{k-1}\right) - 6\left(2^{k-1} + 3^{k-2}\right)
\end{align*}$$
| Yes, go on. Use $2^k=2\cdot 2^{k-1}$ and associate the terms:
$$X_{k+1}=5\cdot(2^k+3^{k-1})\ -\ 6\cdot(2^{k-1}+3^{k-2}) = 2^{k-1}\cdot(10-6)+3^{k-2}\cdot(15-6)=\\
=4\cdot 2^{k-1}+9\cdot 3^{k-2}=2^{k+1}+3^k\,.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But even with that I don't get that result.
| $$\sqrt{3+2\sqrt2}=\sqrt{3+\sqrt8}=\sqrt\frac{3+\sqrt{9-8}}{2}+\sqrt\frac{3-\sqrt{9-8}}{2}=\sqrt2+1$$
$$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt12}=\sqrt\frac{4+\sqrt{16-12}}{2}-\sqrt\frac{4-\sqrt{16-12}}{2}=\sqrt3-1$$
Now we have:
$$\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=\sqrt2+2-\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is it always possible to find a $t>0$, such that $\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|dxIs it always possible to find a $t>0$, such that
$$\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|\,dx<C~~~?$$
where $C$ is independent of $n$.
Here is my idea:
We know that
\begin{align}
\sum_{k=1}^{n}\cos kx&=\frac{\sin(n+\frac{1}{2})x}{2\sin\frac{1}{2}x}-\frac{1}{2}\\
&=\frac{\sin nx \cos\frac{1}{2}x+\cos nx\sin\frac{1}{2}x}{2\sin\frac{1}{2}x}-\frac{1}{2}\\
&=\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}+\frac{\cos nx}{2}-\frac{1}{2}
\end{align}
So we have
\begin{align}
\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|\,dx\leq\int_{0}^{t}|\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}|+1\,dt
\end{align}
The key is that, could we find a constant $C$(just independent of $n$),such that
\begin{align}
\int_{0}^{t}|\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}|\,dt<C~?
\end{align}
It is an improper integral. If
$$\lim_{\varepsilon \to 0^+}\int_{\varepsilon}^{t}|\frac{\sin nx\cos\frac{1}{2}x}{2\sin\frac{1}{2}x}|\,dt \to \infty$$
Then could we find an interval $[a,b](a>0)$, such that
$$\int_{a}^{b}|\sum_{k=1}^{n}\cos kx|\,dx<C~~~?$$
| $$\vert \cos kx \vert \leq 1 \implies \bigg \vert \sum_{k = 1}^n \cos kx \bigg \vert \leq n \implies \int_0^t \bigg \vert \sum_{k = 1}^n \cos kx \bigg \vert \leq nt \implies nt < C \implies t < \frac{C}{n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x^4+y^4$ and $x^3+y^3$ if $x+y=2$ and $x^2+y^2=8$ Find $x^4+y^4$ if $x+y=2$ and $x^2+y^2=8$
So i started the problem by nothing that $x^2+y^2=(x+y)^2 - 2xy$ but that doesn't help!
I also seen that $x+y=2^1$ and $x^2+y^2=2^3$ so maybe $x^3+y^3=2^5$ and $x^4+y^4=2^7$ but i think this is just coincidence
So how can i solve this problem?
PLEASE i need some help and thanks for all!!
| Notice
$$(x^2 + y^2)^2 = 64 \implies x^4 + y^4 + 2(xy)^2 = 64$$
and
$$ (x + y )^2 = 4 \implies x^2 + y^2 + 2xy = 4 \implies 2xy = 4 - 8 = -4 \implies xy = -2
$$
$$ \therefore x^4 + y^4 = 64 - 2(xy)^2 = 64 - 2(-2)^2 = 56 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequalities In Algebra So the problems ask to find Find all values of $x$ for which
$$\frac{x}{x-4}<\frac{x-5}{x+1}.$$
So the solution requires moving both fractions to one side, finding a common denominator, combining, then factoring. Then you have a set of possible solutions for $x$ and you go on from there.
(If anyone is interested, the solution is $x < -1$, $2 < x < 4$.)
Why can I simply multiply both side by $(x+1)$ and $(x-4)$ WITHOUT shifting everything to one side?
I suspect it's because I don't really know if $(x-4)$ and $(x+1)$ are positive or negative. Since I don't know, I cannot appropriately flip the inequality. Hence, a tactic applicable for equalities simply doesn't work in this case.
Is it really just that or is there's a more fundamental reason?
| $$\frac x{x-4}<\frac{x-5}{x+1}\iff \frac x{x-4}-\frac{x-5}{x+1}<0\iff 10\frac{x-2}{(x-4)(x+1)}<0$$
$$\text{ Multiplying the numerator & the denominator by } \frac{(x-4)^2(x+1)^2}{10} \text{ which is}>0$$
$$\frac x{x-4}<\frac{x-5}{x+1}\iff (x-4)(x+1)(x-2)<0$$
so we need odd number of factors $<0$
If all of the three are $<0$ we need $x<$min$(4,-1,2)=-1$
If $x-4<0\iff x<4,$ we need $x>$max $(-1,2)=2\implies 2<x<4$
If $x+1<0\iff x<-1,$ we need $x>$max $(4,2)=4$ which is impossible
If $x-2<2\iff x<2,$ we need $x>$max $(4,-1)=4 $ which is impossible
| {
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Joint distribution of RVs involving rolls of die We roll a die until we get $4$ fives. Let $X$ be the number of rolls needed for the first $5$ and let $Y$ be the number of rolls needed to get the fourth five. What is the joint probability mass function of $X,Y.$
My attempt: $P(X=x \cap Y=y)=$ ${y-1}\choose{3}$$\left({\frac{1}{6}}\right)^4$$\left({\frac{5}{6}}\right)^{y-4}$
My reasoning: There are ${y-1}\choose{3}$ ways of choosing 3 spots for fives among the first $y-1$ rolls. Then we have $\left({\frac{1}{6}}\right)^4$= probability of rolling four fives. And $\left({\frac{5}{6}}\right)^{y-4}$= probability of rolling $y-4$ non-fives.
Does this look correct?
| You need to do it in this way . Let $x$ be the number of tries required to get the first $5$ then $p_X(x) = (\frac{5}{6})^{x-1}\frac{1}{6}$ After this you need $3$ more $5$ . You can treat this as a completely new event , because it doesn't depend on what happened earlier .
Now let $y$ be the total number of trials needed to get four $5$ , then you have already wasted $x$ tries . You are left with only $y-x$ tries now . Now the ways of getting two $5$s will be ${y-x-1}\choose{2}$ . Therefore the probability of this event will be ${y-x-1}\choose{2}$$(\frac{1}{6})^2 (\frac{5}{6})^{y-x-3}(\frac{1}{6})$
Therefore the joint PMF $p_{X,Y}(x,y)$ = $(\frac{5}{6})^{x-1}\frac{1}{6}$${y-x-1}\choose{2}$$(\frac{1}{6})^2 (\frac{5}{6})^{y-x-3}(\frac{1}{6}) $
| {
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Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}{n^4}}} = \frac{1}{\sqrt{n+\frac{1}{n^4}}}$
So $\lim\limits_{n \rightarrow \infty}{{\frac{1}{\sqrt{n+\frac{1}{n^4}}} = 0}} $
$1>|\frac{a_{n+1}}{a_n}|= \frac{(n+1)^2}{\sqrt{(n+1)^5+1}} \frac{\sqrt{n^5+1}}{n^2}= \frac{2n+1 \sqrt{n^5+1}}{\sqrt{(n+1)^5+1}}= \frac{n^2+2n+1 \sqrt{n+\frac {1}{n^4}}}{(n+1)^2\sqrt{n+1+1}} = \frac {\sqrt{n+\frac {1}{n^4}}}{\sqrt{n+2}}$
So $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges.
Could somebody please check my solution?
| The convergence of $$\sum_{n=1}^N (-1)^n \dfrac{n^2}{\sqrt{n^5+1}}$$ can be concluded based on alternating test. The general result is termed as generalized alternating test or Dirichlet test and is based on Abel partial summation. We will prove the generalized statement, though this is a bit of a overkill for this problem it is as easy to prove as the alternating test.
Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.
First note that from Abel summation, we have that
$$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\
= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that
$$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$
Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ converges absolutely. Now since
$$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges.
In your case, $a(n) = (-1)^n$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N (-1)^n$$which is clearly bounded.
Also, $b(n) = \dfrac{n^2}{\sqrt{n^5+1}}$ is a monotone decreasing sequence converging to $0$.
Hence, we have that $$\sum_{n=1}^N (-1)^n \dfrac{n^2}{\sqrt{n^5+1}}$$ converges.
| {
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Power series of trigonometric functions Problem statement: Determine those $x$, for which the power series is convergent and determine the sum.
$$f(x)=x+\sum_{n=2}^{\infty}(-1)^{n-1}2n\frac{x^{2n-1}}{(2n-1)!}$$.
Progress:
I have difficulty to handle the $2n$ inside the sum. I can clearly see, since $\sin x = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\large x^{2n-1}}{(2n-1)!}$, that $\sin x-x$ is involved. The term $2n$ is always even so I'm sure $\cos x$ will occur in the sum.
| This is similar to Andre's observation but more explicit.
We have $f(x)=x+\sum_{n=2}^{\infty}(-1)^{n-1}2n\frac{x^{2n-1}}{(2n-1)!}$,
thus we can kill the $2n$ term by integrating under summation, giving us
$$
\begin{align*}
\int f(x) dx &= C + \frac{x^2}{2} + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n}}{(2n-1)!}
\\
&= C + \frac{x^2}{2} + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n}}{(2n-1)!}\cdot\frac{x}{x}
\\
&= C + \frac{x^2}{2} + x\sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}
\\
&= C + \frac{x^2}{2} + x\left( \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}-x \right)
\\
&= C + \frac{x^2}{2} + x\left( \sin x-x \right)
\\
&= C - \frac{x^2}{2} + x\sin x.
\end{align*}
$$
Taking the derivative of the integral returns $f$, thus $f(x)=-x +\sin x + x\cos x$.
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Prove using mathematical induction pt 2 Assumed that i asked a question like 30 min ago thinking i got the hang of this, seems not.
So $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac12n(6n^2-3n-1) \text{ for all } n\in\mathbb N$$
This time it seems way harder with the squares.
so i did the steps and got stuck on the 3rd step(Again).
Step 1: prove LHS = RHS which it does for n=1
Step 2: Assume $n=k$ is true $$1^2+4^2+7^2+\dots+(3k-2)^2=\frac12k(6k^2-3k-1)$$
Step 3: would $n = k+1$? And would $n = k+1$ work for all equations?could someone help me with the last step, would be appreciated thanks
EDIT: Cheers for the help, i know where i went wrong!
| $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1), \forall n\in N$$
$$1^2+4^2+7^2+\dots+(3n-2)^2+(3n+1)^2=\frac{1}{2}n(6n^2-3n-1))+(3n+1)^2 $$
$$=\frac{1}{2}(6n^3-3n^2-n+2(3n+1)^2)=\frac{1}{2}(6n^3+15n^2+11n+2))= $$
$$\frac{1}{2}(6n^3+12n^2+6n+3n^2+5n+2)=$$
$$\frac{1}{2}(6n(n^2+2n+1)+3n^2+3n+2n+2)=$$
$$\frac{1}{2}(6n(n+1)^2+3n(n+1)+2(n+1))=$$
$$\frac{1}{2}(n+1)(6n(n+1)+3n+2)=\frac{1}{2}(n+1)(6n^2+9n+2)=$$
$$=\frac{1}{2}(n+1)(6n^2+12n+6-3n-3-1)=$$
$$=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)$$
| {
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$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$
$$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$
| $$T=\sqrt{2\sqrt{2\sqrt{2}}}...\\
\frac{T^2}{2}=T\\$$
T is a nonzero real number so:$$
T=2$$
| {
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How to find the limit of this equation? I have a Calc final coming up soon and am currently reviewing. I am currently stumped on finding the limit of as $x \rightarrow 1$, of $(1-\sqrt{2x^2 -1}) / (x-1)$.
Don't you have to multiply by the conjugate? Any help is seriously appreciated. I know the answer is $-2$, but I cannot seem to get that.
| Here's a way to compute the limit without using derivatives but by multiplying by the conjugate:
\begin{align*}
\lim_{x\to1} \frac{1 - \sqrt{2x^2-1}}{x-1} \cdot \frac{1 + \sqrt{2x^2-1}}{1 + \sqrt{2x^2-1}}
&= \lim_{x\to1} \frac{1 - (2x^2-1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2x^2 + 2}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2(x^2 - 1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2(x - 1)(x + 1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2(x + 1)}{1 + \sqrt{2x^2-1}} \\
&= \frac{-2(1 + 1)}{1 + \sqrt{2(1)^2-1}} \\
&= -2
\end{align*}
| {
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Calculate the determinant of $3\times 3$ matrix with $\sin x$ and powers of $\cos x$ How to calculate the determinant of this matrix
$A=\begin{bmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{bmatrix}$
$$\left[A\right]=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{vmatrix}=\\=\sin x\begin{vmatrix}
\cos x& 0\\
1 & 1
\end{vmatrix}-\cos^2x\begin{vmatrix}
\sin x & 0 \\
\sin x & 1
\end{vmatrix}+\begin{vmatrix}
\sin x & \cos x\\
\sin x & 1\\\end{vmatrix}=\\=\sin x\cos x-\cos^2x\sin x+\sin x-\sin x\cos x =\\=\sin x\left(\cos x-\cos^2x+1-\cos x\right)=\sin x \left(1-\cos ^2x\right)=\\=\sin x\cdot \sin^2x=\sin^3x$$
The path is something like this? I'm using the wrong rule?
| The calculation will be easier if you showed zeros:
Subtract the first row from the second and third row and develop relative the first column we find
$$\det A=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{vmatrix}=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
0& \cos x(1-\cos x) & -1 \\
0& 1-\cos^2x & 0
\end{vmatrix}=\sin x(1-\cos^2x).$$
| {
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Find $\sum_{r=0}^{2n}\frac{r}{r+2}\binom{2n}r\;.$ Find
$$\sum_{r=0}^{2n}\frac{r}{r+2}\binom{2n}r\;.$$
I got
$$\frac{2^{2n+1}(2n^2+n+1)-1}{(2n+1)(2n+2)}$$
but the answer is
$$\frac{2^{2n+1}(2n^2-n+1)-2}{(2n+1)(2n+2)}$$
Thanks for the help...
| $$\frac r{r+2}\binom{2n}r=\left(1-\frac2{r+2}\right)\binom{2n}r=\binom{2n}r-2\cdot\underbrace{\frac1{r+2}\binom{2n}r}_1$$
$$\text{Now, }\underbrace{\frac1{r+2}\binom{2n}r}_1=(r+1)\cdot\frac{(2n)!}{(2n-r)!\cdot(r+2)(r+1) \cdot r!}$$
$$=\frac{r+1}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{\{2n+2-(r+2)\}!\cdot(r+2)!} =\underbrace{\frac{r+1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}}_2$$
$$\text{Again, }\underbrace{\frac{r+1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}}_2=\frac{r+2-1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}$$
$$=\underbrace{\frac{r+2}{2n+2)(2n+1)}\binom{2n+2}{r+2}}_3-\frac{1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}$$
by distributive law.
Now $\displaystyle\underbrace{(r+2)\binom{2n+2}{r+2}}_3=(r+2)\frac{(2n+2)!}{(2n-r)! (r+2)!}=(2n+2)\frac{(2n+1)!}{\{(2n+1)-(r+1)\}! (r+1)!}=(2n+2)\binom{2n+1}{r+1}$
$\displaystyle\implies\frac r{r+2}\binom{2n}r$
$\displaystyle=\binom{2n}r-\frac2{(2n+2)(2n+1)}\left[(2n+2)\cdot\binom{2n+1}{r+1}-\binom{2n+2}{r+2}\right]\ \ \ \ (4)$
Now using $\displaystyle2^m=(1+1)^m=\sum_{0\le r\le m}\binom mr,$
$\displaystyle(i)\sum_{0\le r\le 2n}\binom{2n}r=2^{2n}$
$\displaystyle(ii)\sum_{0\le r\le 2n}\binom{2n+1}{r+1}=\sum_{0\le u\le 2n+1}\binom{2n+1}u-\binom{2n+1}0=2^{2n+1}-1$
$\displaystyle(iii)\sum_{0\le r\le 2n}\binom{2n+2}{r+2}=\sum_{0\le v\le 2n+2}\binom{2n+2}v-\binom{2n+2}0-\binom{2n+2}1$
$\displaystyle=2^{2n+2}-1-(2n+2)$
Use $(i),(ii),(iii)$ in $(4)$
| {
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Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$ Evaluate the integral $$\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$$
| The answer given by Gordon is very good for computing the integral.
But it does not provide much why it works. The following figure might help with that
This is the integral for $\alpha=\sqrt{2}$. It looks as if the area under the function
is exactly half of the dashed rectangle... A good guess is therefore
$$
\int_0^{\pi/2} \frac{\mathrm{d}x}{1 + \tan(x)^\sqrt{2}} = \frac{1}{2}\left(\frac{\pi}{2}\right)
$$
This striking symmetry can be shown true for any choice of $\alpha$, play around in geogebra or any ploting tool you fancy =)
The symmetry of such integrals is studied further in the paper
Symmetri and Integration by Roger Nielsen
Where it is shown that if $f(x) + f(a+b-x)$ is constant for all $x\in[a,b]$ - meaning it has this nice symmetric property.
Then the area can calculated as
$$
\int_a^b f(x) \mathrm{d}x = \frac{f(a)+f(b)}{2}(b-a) = f\left(\frac{a+b}{2}\right)(b-a)\,.
$$
I will leave it to you to check that $f(x)+f(a+b-x)$ is constant.
Alternatively the integral can also be computed as follows
\begin{align*}
\int_0^{\pi/2} \frac{\mathrm{d}\theta}{(1 + (\tan \theta)^b}
= \int_0^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)}
= \frac{1}{2} \int_0^\infty \frac{\mathrm{d}x}{1 + x^2}
= \frac{\pi}{4}
\end{align*}
Where the substitution $u \mapsto \tan \theta$ was used and that
$$
\int_0^\infty \frac{R(x)}{x^b+1}\mathrm{d}x = \frac{1}{2} \int_0^\infty R(x)\,\mathrm{d}x
$$
Given that $R(x) = R(1/x)/x^2$, again check that this holds for $1/(1+x^2)$. A proof of the latter can be found on page $27$ here, section 1.9. More directly one has by the same methods
\begin{align*}
\int_0^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)}
& = \int_0^1 \frac{\mathrm{d}x}{(1+x^2)(1+x^b)} +
\int_1^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)} \\
& = \int_0^1 \frac{1}{(1+x^2)(1+x^b)} +
\frac{x^b}{(1+x^2)(1+x^b)} \mathrm{d}x \\
& = \int_0^1 \frac{\mathrm{d}x}{1+x^2}
= \frac{1}{2} \int_0^\infty \frac{\mathrm{d}x}{1+x^2}
= \frac{\pi}{4}
\end{align*}
Where the substitution $x \mapsto 1/x$ was used in the last integral.
These theorems and ideas are not the simplest way to attack the problem. But it might give you some broader insight further down the road.
| {
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Prove that :$\frac{1}{a+b} +\frac{1}{b+c} +\frac{1}{c+a}\ge \frac{4}{a^2+7} +\frac{4}{b^2+7} +\frac{4}{c^2+7}$ Let $a,b,c>0$ and satisfying $a^2+b^2+c^2=3$.
Prove that :$\dfrac{1}{a+b} +\dfrac{1}{b+c} +\dfrac{1}{c+a}\ge \dfrac{4}{a^2+7} +\dfrac{4}{b^2+7} +\dfrac{4}{c^2+7}$
| Use AM-GM we have :
$\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{2}{\sqrt{(a+b)(b+c)}}\ge\frac{4}{a+2b+c}\ge\frac{4}{\frac{a^2+1}{2}+b^2+1+\frac{c^2+1}{2}}=\frac{8}{(a^2+b^2+c^2)+4+b^2}=\frac{8}{b^2+7}$
Similar, the inequality is proved.
| {
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Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
| Alternatively $\rm\ \dfrac{2}{a-5} = \color{#c00}4+\color{#0a0}{3\sqrt2}\, <\, \color{#c00}5+\color{#0a0}{3\sqrt3} = \dfrac{2}{b-5}\ $ since $\rm\, \color{#c00}{4<5},\,\ \color{#0a0}{3\sqrt2 < 3\sqrt 3} $
Flipping gives $\rm\ a-5\, >\, b-5 \ \Rightarrow\ a > b\ \ $ QED
Remark $\ $ This is not pulled out of a hat. Rather, it arises from the continued fraction algorithm for comparing real numbers. Namely, first we compare the integer parts of $\rm\,a,b.\,$ Both are easily seen to be $\,5,\,$ so we subtract $5$ from both to get their fractional parts $\rm\,a-5,\ b-5,\,$ then recurse, comparing their reciprocals $\rm\,1/(a-5),\ 1/(b-5),\,$ etc, etc.
| {
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Mod calculation
$$10x + 20 \equiv 11 \pmod{23}$$
So I know the answer is $6 \mod{23}$. I understand that we subtract $20$ from each side. But how do we do extended gcd on this?
Can you explain the steps involved?
| $$10x+20 \equiv 11 \pmod{23} \implies 10x \equiv -9 \pmod{23} 10x \equiv -9+23 \pmod{23}$$
$$10x \equiv -9+23 \pmod{23} \implies 10x \equiv 14 \pmod{23} \implies 5x \equiv 7 \pmod{23}$$
$$5x \equiv 7 \pmod{23} \implies 5x \equiv 7+23 \pmod{23} \implies 5x \equiv 30 \pmod{23}$$
Hence,
$$x \equiv 6 \pmod{23}$$
| {
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prove statement. cyclic sum complex numbers If $(a+b)^3 = (b+c)^3 = (c+a)³\,\,\,\,a,b,c\,\,\in\mathbb{C},\,\,a\neq b\neq c$, show that $a^3 = b^3 = c^3$.
Some hints would be great.
| As $\displaystyle a\ne c, a+b\ne b+c\implies \frac{a+c}{b+c}\ne1$
So, from $\displaystyle(a+b)^3 = (b+c)^3 = (c+a)^3\implies a+b=\omega(b+c)=\omega^2(c+a)$ where $\omega$ is one of the two complex cube root of unity
From $\displaystyle a+b=\omega(b+c),a+b(1-\omega)-c \omega=0\ \ \ \ (1)$
From $\displaystyle\omega(b+c)=\omega^2(c+a),-a\omega^2+b\omega+c(\omega-\omega^2)=0\ \ \ \ (2)$
Solve for $a,b$ in terms of $c,$ then take the cubes
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find equation of a tangent on $y= \sin2x$ Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$
What I calculated:
Intersections: $$\sin2x= \frac{1}{2}$$
...
$$0=tg^2x-4tgx-1$$
$$tgx_{1}=2+\sqrt{3}$$
$$x_{1}=75+k\pi$$
$$tgx_{2}=2-\sqrt{3}$$
$$x_{2}=15+ k\pi$$
$$T_{1}(75+k\pi,\frac{1}{2})$$ and $$T_{2}(15+k\pi,\frac{1}{2})$$
Derivative of $y= sin2x$ is: $y'= 2cos2x$
$$ y'(75)= -\sqrt {3} =k_{t_{1}} $$
$$y'(15)= \sqrt {3} =k_{t_{2}}$$
formula for calculating tangent that I used: $y-y_{1}=k_{t}(x-x_{1})$
1.)$T_{1}(75+k\pi,\frac{1}{2})$
$k_{t_{1}}=-\sqrt {3}$
$y_{1}-\frac{1}{2}=-\sqrt {3}(x-75-k\pi)$
$y_{1}=-\sqrt {3}x+75\sqrt {3}+\sqrt {3}k\pi+\frac{1}{2}$
2.)$T_{2}(15+k\pi,\frac{1}{2})$
$k_{t_{2}}=\sqrt {3}$
$y_{2}-\frac{1}{2}=\sqrt {3}(x-15 -k\pi)$
$y_{2}=\sqrt {3}x-15\sqrt {3}-\sqrt {3}k\pi+\frac{1}{2}$
Both solutions are wrong; what I should actually get:
$y_{1}=\sqrt{3}x-\frac{\sqrt{3}pi-6+12kpi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}pi+6+12kpi\sqrt{3}}{12};k=Z $
How could I possibly get that ?!
| $$\sin2x= \frac{1}{2}$$
$$2x=n\pi+(-1)^n\pi/6$$
Therefore
$$x=\frac{n \pi}{2} + (-1)^n \frac{ \pi}{12} \text{ , } n=Z$$
$f(x)=\sin2x$
$f'(x)=2\cos2x$
where $m=f'(x)$ and c can be find by putting x and $y=\frac{1}{2}$
$y_{1}=\sqrt{3}x-\frac{\sqrt{3}\pi-6+12k\pi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}\pi+6+12k \pi\sqrt{3}}{12};k=Z $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cantor set in base 3. I'm trying to prove the cantor set $C$ is equivalent to the set of all numbers with ternary expansion of $2$'s and $0$'s. That is:
Let $A_0=[0,1]$. $A_n$ is defined to equal $A_{n-1}$ with it's middle third removed. let $C=\bigcap_{n\in\mathbb{N}} A_n$.
Prove $C =\{x=0.a_1a_2a_3...| a_n\in\{0,2\} \text{ for all }n\in\mathbb{N} \} $ where the decimal expansion is in base $3$.
Let $x=0.a_1a_2a_3...$ $\space$ be a ternary expansion of $x$
My strategy is this:
Prove that $a_n=1 \iff x\notin A_n$
(a) Proceed by induction on $n$. For $n=1$ we have:
$a_1 = 1 \iff \frac{1}{3} \le x < \frac{2}{3} \iff x \notin A_1$
Assume it's true for $n$ that is:$\space a_n = 1 \iff x \notin A_n$.
For $n+1$ we have: $a_{n+1}=1 \iff \text{there's some integer $m$ such that:} \frac{1}{3} \le 3^{n}x - m < \frac{2}{3} $
Furthermore $m = a_1a_2a_3...a_{n}$ (in base 3 digit expansion).
Here i got stuck.
| Let
$F_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$
$F_2 = [0,\frac{1}{3^2}] \cup [\frac{2}{3^2},\frac{3}{3^2}] \cup [\frac{6}{3^2},\frac{7}{3^2}] \cup [\frac{8}{3^2},1]$
and so on.
The Cantor set is then $C=\bigcap^{\infty}_{k=1} F_k$.
Each $x \in C$ can be written in the form $ x = \frac{a_0}{3} + \frac{a_1}{3^2} + \frac{a_2}{3^3} + ...+ \frac{a_n}{3^{n+1}}+ ...$
By Construction, if $x\in F_k$ and $0 \leq j < k$, then $3^jx=a+y$ where $a \in \mathbb{N} \cup \{0\}$ (*) and $y \in F_{k-j}$.
Noting this pattern suppose $x\in C$ therefore $x \in F_k$ for all $k \in \mathbb{N}$. Now assume that for some $m\in \mathbb{N}$ we have $a_m = 1$ Then we have $3^mx=a+\frac{1}{3}+\frac{a_{m+1}}{3^2}+\frac{a_{m+2}}{3^3}+...\ \ \ \Leftrightarrow$ $ \ \ \ \ y=3^mx-a \notin F_1 \Leftrightarrow x \notin F_{m+1} \Leftrightarrow x \notin C$ contradiction.
To see the pattern more clearly, consider $F_3$,
$F_3= [0,\frac{1}{27}] \cup [\frac{2}{27},\frac{3}{27}] \cup [\frac{6}{27},\frac{7}{27}]\cup [\frac{8}{27},\frac{9}{27}]\cup [\frac{18}{27},\frac{19}{27}]\cup [\frac{20}{27},\frac{21}{27}] \cup [\frac{24}{27},\frac{25}{27}] \cup [\frac{26}{27},1]$
Define the "multiplication" of an interval by an scalar, naturally by multiplying the endpoints by that scalar, for example $3F_2= [0,\frac{3}{9}] \cup [\frac{6}{9},1] \cup [\frac{18}{9},\frac{21}{9}] \cup [\frac{24}{9},3]=[0,\frac{1}{3}] \cup [\frac{2}{3},1] \cup [2,\frac{7}{3}] \cup [\frac{8}{3},3]$
This multiplication is important, since it gives us 2 copies of $F_1$, (the 2nd copy is a translation of $F_1$ by an integer, look at (*) )
More generally, $3^jF_k$ gives us $2^j$ copies of $F_{k-j}$
let $x \in F_3$
For $j=0 \ \ \ $, the statement is clear, i.e. $3^0x \in F_{3-0} \Rightarrow x \in F_3$ Trivial !
For $j=1 \ \ \ $Obviously $3^1x \in 3F_3$, i.e. $3^1x$ lies in a translated (by a number $a \in \mathbb{N} \cup \{0\}$) copy of $F_{3-1}=F_2$ Therefore $3x=a+y$ where a is the natural number of translation, and $y\in F_2$
For $j=2 \ \ \ $ Similarly $3^2x \in 3^2F_3 \Rightarrow 3^2x $ lies in a translated copy of $F_{3-2}=F_1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove Trig Identity
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$
This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\right)}^y2\cos\left(\frac{B-C}2\right)\\
\sin\left(\frac{A-C}2+\frac{A-B}2\right)+\sin\left(\frac{A-C}2-\frac{A-B}2\right)\\
\left(\frac{B-C}2\right)\left[\sin\left(\frac{2A-B-C}2\right)+\sin\left(\frac{B-C}2\right)\right]\\
\cos\left(\frac{B-C}2\right)\sin\left(\frac{2A-B-C}2\right)+\cos\left(\frac{B-C}2\right)\sin\left(\frac{B-C}2\right)$$
I know it's basically turning sum into product and I previously tried to derive the LHS the same way the normal sum to product identity works but I couldn't figure out how. I tried to use the sum-product identity on RHS but couldn't quite make it work.
| Now take out the common factor $\displaystyle\cos\frac{B-C}2$
and apply Prosthaphaeresis Formulas on $\displaystyle\sin\frac{2A-B-C}2+\sin\frac{B-C}2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.
I thought of it this way: $$f(x)=\begin{cases}2x-1-3(2x+4)+7 \,(\text{then I simplify)} & \text{if $x>0$}\\
-(2x-1-3(2x+4)+7)\,(\text{then I simplify)} & \text{if $x\le0$}\end{cases}$$
But is there some way without having to use the cases?
Edit: NEW work on this problem! I found three cases;
If $x\in ]-\infty,-2]$ then f(x)=$4x+20$
If $x\in]-2,1/2]$ then f(x)=$-8x-4$
If $x\in]1/2,+\infty[$ then f(x)=$-4x-6$
IS THIS TRUE?
Thank you very much!
| The absolute value has a discontinuous slope change at zero. Hence, f(x) has discontinuous slope changes at -2 from $|2x+4|$ and at 1/2 from $|2x-1|$. So there are three cases: $-\infty < x \le -2$, $-2 < x \le 1/2 $ and $1/2 < x < \infty$.
Now for $x$ sufficiently negative $|2x+4| = -2x-4$ and we know that $|2x+4|$ had discontinuous slope at $2x+4=0$. Similarly for sufficiently negative $x$ $|2x-1| = -2x+1$ and changes slope when $2x-1=0$. Hence
$$
f(x) = \left{
\begin{array}{ll}
+3(2x+4)-(2x-1)+7, & x \le -2; \\
-3(2x+4) -(2x-1)+7, & -2 <x \le 1/2; \\
-3(2x+4)+(2x-1)+7, & 1/2 < x
\right.
$$
Sorry! Don't see why the formatting is messed up. Hope it makes sense! I tried to use plain vanilla latex commands. I need to learn MathJaX
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $1^{2007}+2^{2007}+\cdots+n^{2007}$ is not divisible by $n+2$ Prove that for any odd natural number $n$, the number $1^{2007}+2^{2007}+\cdots+n^{2007}$ is not divisible by $n+2$.
| By taking modulo $n + 2$, and partition the terms as groups of two
each,
$$\begin{align*}
1^{2007}+2^{2007}+3^{2007}+\cdots n^{2007} \\
\equiv1+\big\lfloor 2^{2007}+(n)^{2007}\big\rfloor+\cdots +\cdots \big\lfloor \ \Big(\frac{n+1}{2}\Big)^{2007}+\Big(\frac{n+3}{2}\Big)^{2007}\big\rfloor\\ \equiv1+\big\lfloor 2^{2007}+(-2)^{2007}\big\rfloor+\cdots +\cdots \big\lfloor \ \Big(\frac{n+1}{2}\Big)^{2007}+\Big(-\frac{n+1}{2}\Big)^{2007}\big\rfloor\\ \equiv 1\pmod{n+2}
\end{align*}
$$
Thus, the conclusion is proven
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont know how to continue:
$2x^2+7x+5 = -1 \text{ and } -2.5$
The solution should be: $(-2.5;-1) \cup (-\infty;-3.25)$
| HINT:
$$\frac{4x+13}{2x^2+7x+5}=0\implies 4x+13=0$$
$$\frac{4x+13}{2x^2+7x+5}<0 \iff (4x+13)(2x^2+7x+5)<0$$
$$\iff(4x+13)(2x+5)(x+1)<0\iff\left(x+\frac{13}4\right)\left(x+\frac52\right)(x+1)<0$$
So, we need odd number of factor(s) to be $<0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Solving inequation with two absoulte values I need to solve the following inequation:
$$
|x| \cdot |x-1|-1>-x\\
$$
I cant get the correct result.
I tried to solve it like this:
$$
|x| \cdot |x-1|-1>-x
$$
I know that I can write $|x \cdot y|=|x| \cdot |y|$, so:
$$
\begin{eqnarray}
|x(x-1)|-1&>&-x\\
|x^2-x|-1&>&-x \qquad \text{for }\; x < 0 \\
-(x^2-x)-1&>&-x\\
-x^2+x-1&>&-x\\
-x^2+x+x-1&>&0\\
-x^2+2x-1&>&0
\end{eqnarray}
$$
but I think that this is incorrect, because when I solve this inequality I get that $x<1$ and $x>1$, and that doesn't make a lot of sense
Any suggestion is helpful. Than you very much!!!
| You should distinguish three cases:
*
*when $x<0$, $|x|=-x$ and $|x-1|=1-x$;
*when $0\le x\le 1$, $|x|=x$ and $|x-1|=1-x$;
*when $x>1$, $|x|=x$ and $|x-1|=x-1$.
Thus you have to solve
\begin{align}
&\begin{cases}
(-x)(1-x)-1>-x\\
x<0
\end{cases}\\
&\begin{cases}
x(1-x)-1>-x\\
0\le x\le 1
\end{cases}\\
&\begin{cases}
x(x-1)-1>-x\\
x>1
\end{cases}
\end{align}
and put the solutions sets together.
Alternative solution, which, however, could bring to more complicated computations in other cases.
Write the inequation as
$$
|x^2-x|>1-x
$$
We can see that any value of $x$ such that $1-x<0$ is a solution. So $x>1$ is always a solution.
In case $1-x>0$, we can square, so we get
$$
\begin{cases}
x^2(1-x)^2>(x-1)^2\\
1-x>0
\end{cases}
$$
This simplifies to
$$
\begin{cases}
x^2>1\\
x<1
\end{cases}
$$
which is satisfied for $x<-1$.
For $x=1$ the inequality doesn't hold
Putting together the two solution sets, we conclude that the inequality is satisfied for $x<-1$ or $x>1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve a 2nd order non-homogeneous linear recurrence? I have a problem in solving this equation :
$x_{n+2} + 3\ x_{n+1} + 2\ x_{n} = 5 \times 3^n $
given that $x_{0} = 0$ and $x_{1} = 1$.
I solved the homogeneous associated equation and got $v_{n} = c_{1} \times (-1)^{n} + c_{2} \times (-2)^{n}$ (where $c_{1}$ and $c_{2}$ are constants).
Could somebody explain the general method for solving second order non-homogeneous linear recurrence ?
| Use generating functions directly. Define $G(z) = \sum_{n \ge 0} x_n z^n$, multiply by $z^n$ and sum over $n \ge 0$, recognize a few sums:
$$
\frac{G(z) - x_0 - x_1 z}{z^2} + 3 \frac{G(z) - x_0}{z} + 2 G(z)
= 5 \frac{1}{1 - 3 z}
$$
As partial fractions:
$$
A(z) = \frac{1}{4} \cdot \frac{1}{1 - 3 z} - \frac{1}{4} \cdot \frac{1}{1 + z}
$$
Everything in sight is just geometric series;
$$
a_n
= \frac{3^n - (-1)^n}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving quadratic system If $a,b,c\in \mathbb{R}$ satisfy the system
$a^2+ab+b^2=9$;
$b^2+bc+c^2=16$;.
$c^2+ac+a^2=25$.
Find $ab+ac+bc$
| If you subtract the first equation from the second, you get $c^2-a^2 + bc - ab = 7,$ so $(c-a)(c+a +b) = 7.$ You get similar equations (all with an $a+b+c$ factor) when you subtract the second from the third, etc, which gives you a linear system in $a, b, c$ (you know $a+b+c \neq 0,$ from the first sentence).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the possible values of $0\pmod{2}$ and $1\pmod{3}$?
$$
1\pmod{3} = 1, 4, 7, 10
$$
so on.
Since only $4, 10$ and so on satisfy $\pmod{2}$, only values that satisfy both criteria can be used.
In general, can we do this for $\pmod{n}$, $n$ being any integer?
| As $\displaystyle10\equiv1\pmod3, 10^n\equiv1\pmod3$ for integer $n\ge0$
As $\displaystyle a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c }$ where $a,b,m,c$ are integers
$\displaystyle10\cdot10^n\equiv10\pmod{3\cdot10}\equiv10\pmod6$ as $6|30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620994",
"timestamp": "2023-03-29T00:00:00",
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Determine the smallest positive value of x(in degrees) for which: $\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$ Determine the smallest positive value of x(in degrees) for which:
$\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$
I tried to apply the formula of $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$ but that led me nowhere resulting in a huge equation.
Please help.
| Given $\displaystyle \tan(x+100^0) = \tan(x+50^0)\cdot \tan (x)\cdot \tan(x-50^0)$
$\displaystyle \Rightarrow \frac{\tan(x+100^0)}{\tan(x-50^0)}\Rightarrow =\tan(x+50^0)\cdot \tan(x^0)$
$\displaystyle \Rightarrow \frac{\sin(x+100^0)\cdot\cos(x-50^0)}{\cos(x+100^0)\cdot \sin (x-50^0)}=\frac{\sin(x+50^0)\cdot\sin(x)}{\cos(x+50^0)\cdot \cos (x)}$
Using Componendo and dividendo,
$\displaystyle\Rightarrow \frac{\sin(2x+50^0)}{\sin (150^0)} = -\frac{\cos(50^0)}{\cos(2x+50^0)}$
$\displaystyle \Rightarrow \sin(4x+100^0)=-\cos(50^0) = -\sin (50^0)=\sin(180^0+40^0)=\sin(360^0-40^0)$
$\displaystyle \Rightarrow (4x+100^0)=220^0 = 320^0$
So $x=30^0$ or $x=55^0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621213",
"timestamp": "2023-03-29T00:00:00",
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probability using permutations what is the probability that if we pick n natural numbers then their product's unit digit contains $2,4,6,8$.and i have no idea how to start ....so please mention all the steps...
i thought that last digit may be $0-9$ and fav cases are $4$..so prob should be $4/10=2/5$..but that is wrong .
correct answer is $(4^n-2^n)/5^n$..so in a way my answer was true for $n=1$.
please help
| We assume that our $n$ numbers are independently chosen, and that each has final digit equally likely to be $0,1,2,3,\dots,9$.
The product ends in $2$, $4$, $6$, or $8$ precisely if (i) the digits $0$ or $5$ are not chosen and (ii) the digits are not all odd.
The probability $0$ or $5$ are never chosen is $\left(\frac{8}{10}\right)^n$. Given that these digits are never chosen, the probability they are all odd is $\left(\frac{1}{2}\right)^n$, so the probability they are not all odd is $1-\left(\frac{1}{2}\right)^n$. Our probability is therefore
$$\left(\frac{8}{10}\right)^n\left(1-\left(\frac{1}{2}\right)^n\right).$$
We can simplify this in various ways. For example, we can multiply through and get $\left(\frac{4}{5}\right)^n-\left(\frac{2}{5}\right)^n$.
Another way: There are $10^n$ equally likely patterns of last digits of our $n$ numbers. Of these, $8^n$ patterns give us a product whose end digit is neither $0$ nor $5$. Out of these $8^n$ patterns, $4^n$ give final digit one of $1,3,7,9$. So the number of patterns that give us end digit $2$, $4$, $6$, or $8$ is $8^n-4^n$.
Our probability is therefore $\frac{8^n-4^n}{10^n}$.
Remark: It is not possible to have a random variable $X$ which takes on all non-negative integer values with equal probabilities. That is why we interpreted the question to mean that our numbers are chosen to have all final digits equally likely.
| {
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How prove this inequality $x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$ let $x,y,z\ge 0$,and such
$$xy+yz+xz=xyz+2$$
show that
$$x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$$
my try: let $x+y+z=p,xy+yz+xz=q, xyz=r$
then
$$q=r+2$$
show that
$$p^2-2q+r(p-2)\ge 4$$
then I can't,Thank you
| WLOG, $z=$Min{$x,y,z$},$z=\dfrac{2-xy}{x+y-xy}, t^2=xy,p=x+y, \implies p^2 \ge 4t^2,p\ge2t$
$z\ge 0 \implies (2-t^2)(p-t^2)\ge 0 \implies p\ge 2, t^2 \le2 $.
(if $p<2, t^2\le \dfrac{1}{2} \implies z>1$, but $z\le \dfrac{x+y}{2} < 1, $ contradict.)
(if $t^2>p\ge2 ,z=\dfrac{t^2-2}{t^2-p}<\dfrac{p}{2} \iff t^2 >\dfrac{p^3-4}{p-2} \iff \dfrac{p^2}{4} > \dfrac{p^3-4}{p-2} \iff 3p^3+2p^2<16$, but $3p^3+2p^2\ge40$ ,contradict)
replace $x,y,z$ with $p,t$ we have the inequality:
$f(p)=p^4-2t^2p^3-4p^2+(t^6+4t^4+4t^2)p-3t^6-3t^4+4 \ge 0$
$f'(p)=4p^3-6t^2p^2-8p+t^6+4t^4+4t^2 , f''(p)=12p^2-12tp-8 ,\dfrac{12t}{2*12}=\dfrac{t}{2}<1<2 \implies 2t\ge 2 ,f''_{min}=f(2t)=24t^2-8>0 \\ 2t<2,f''_{min}=f(2)=40-24t>0$
so $f'(p)$ is mono increasing function $\implies$
$2t>2,f'_{min}=f(2t)=t(t^5-20t^3+32t^2+4t-16)=t(13(2-t^2)(t-1)+(t-1)^2(t^3+2t^2-4x+9)+1)\ge t > 0 \\ 2t<2,f'_{min}=f(2)=t^6+4t^4-20t^2+16=t^6+4t^2(1-t)(4-t)>0$
so $f(p)$ is mono increasing function $\implies$
$2t \ge 2, f_{min}=f(2t)=(t-1)^2(t^2-2)^2(2t+1) \ge 0,t=1 $ or $t^2=2$ get min zero.
$2t<2,f_{min}=f(2)=-t^6+5t^4-8t^2+4=(1-t^2)(t^2-2)^2 \ge 0$
so $f(p) \ge 0$, the "=" will hold when $p=2t, t=1 $ or $t=\sqrt{2} \implies x=y=1=z$ or $x=y=\sqrt{2},z=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
how to integrate : $\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$ Need to compute:
$$\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$$
I tried to use partial fraction but got a messy algebra.
thanks.
| For any $x \in [0,\infty]$, let $I(x)$ and $J(x)$ be the integrals
$$
I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}
\quad\text{ and }\quad
J(x) = \int_{x^2}^{x^3} \frac{dt}{\sqrt{1+t^4}} = I(x^3) - I(x^2)
$$
Notice under the substitution $t = \frac{1}{u}$,
$$\frac{dt}{\sqrt{1+t^4}} = - \frac{du}{\sqrt{1+u^4}}$$
We find $I(x)$ and $J(x)$ satisfy following identities:
$$I(x) + I(\frac{1}{x}) = I(\infty)\quad\text{ and }\quad J(x) = -J(\frac{1}{x})$$
This means we only need to figure out what $I(x)$ is for $0 < x < 1$.
Part 1 : As an Hypergeometric function
For $|t| < 1$, $\frac{1}{\sqrt{1+t^4}}$ has an absolute convergence power series expansion. We can integrate the expansion term by term to get
$$I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}
= \int_0^x \left[ \sum_{k=0}^{\infty} \frac{(\frac12)_k}{k!}(-t^4)^k\right] dt
= \sum_{k=0}^\infty \frac{(\frac12)_k}{k!}\frac{(-1)^k x^{4k+1}}{4k+1}
$$
where $(\gamma)_k = \gamma(\gamma+1)\cdots(\gamma+k-1)$ is the rising
Pochhammer symbol. Notice
$$\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma(\gamma+1)\cdots(\gamma+k-1)}{(\gamma+1)(\gamma+2)\cdots(\gamma+k)} = \frac{\gamma}{\gamma+k}$$
We have $\frac{1}{4k+1} = \frac{(\frac14)_k}{(\frac54)_k}$ and $I(x)$ becomes
$$I(x) = x \sum_{k=0}^\infty \frac{(\frac14)_k (\frac12)_k}{k!(\frac54)_k}(-x^4)^k$$
The expansion of the RHS is that for a
Hypergeometric function with
argument $-x^4$. More precisely,
$$\begin{cases}
I(x) &= x\cdot{}_2F_1( \frac14, \frac12; \frac54 ; -x^4)\\
J(x) &= I(x^3) - I(x^2) = x^3\cdot{}_2F_1( \frac14, \frac12; \frac54 ; -x^{12})
- x^2\cdot{}_2F_1( \frac14, \frac12; \frac54 ; -x^8)
\end{cases}$$
Even though it is sort of hard to extract analytic relations from hypergeometric functions,
there are highly efficient numeric routines around to evaluate them accurately. If what you
want is to evaluate it and get a number, this will be a good representation for $I(x)$ and hence $J(x)$.
On WolframAlpha, you can access the hypergeometric function though the function
$\bf\text{HypergeometricPFQ}$ and evaluate $I(x)$ using the expression:
$$\bf x*\text{HypergeometricPFQ}[\{1/4,1/2\},\{5/4\},-x^4]$$
Part 2: As an elliptic integral
As pointed out by others, the integrals can be recasted as elliptical integrals.
Introduce variable $c$ and $s$ such that
$$t^2 = \frac12 \left(\frac{1}{c^2} - c^2\right)\quad\text{ and }\quad s = \sqrt{1-c^2}$$
We have $\quad\displaystyle \sqrt{1+t^4} = \sqrt{ 1 + \left(\frac12 \left(\frac{1}{c^2} - c^2\right)\right)^2 } = \frac12 \left(\frac{1}{c^2} + c^2\right).$
This implies
$\quad\displaystyle 2tdt = dt^2 = -\left(\frac{1}{c^3} + c\right) dc = -2\sqrt{1+t^4}\frac{dc}{c}$ and as a result,
$$\begin{align}
\frac{dt}{\sqrt{1+t^4}}
&= -\frac{dc}{tc} = -\frac{cdc}{tc^2} = \frac{sds}{tc^2}
= \frac{sds}{c\sqrt{\frac12\left(1-c^4\right)}} = \frac{ds}{c\sqrt{\frac12 (1+c^2)}}\\
& = \frac{ds}{\sqrt{(1-s^2)(1 - \frac12 s^2})}
\end{align}$$
Notice $s^2 = 1 - c^2 = 1 - \left(\frac12(\frac{1}{c^2} + c^2) -
\frac12(\frac{1}{c^2} - c^2)\right) = 1 + t^2 - \sqrt{1+t^4}$, we obtain
an alternate representation of $I(x)$:
$$I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}} = F( \sqrt{1 + x^2 - \sqrt{1+x^4}} \;; \frac12 )$$
where $\quad\displaystyle F(y \;; m ) = \int_0^y \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}}\quad$
is the Jacobi's form of incomplete elliptic integral of the first kind.
Please note that there is more than one form/convention of elliptic integrals. A lot of authors use the same term "incomplete elliptic integral of the first kind" for following integral:
$$F( \phi \mid m ) = \int_0^\phi \frac{d\theta}{\sqrt{1-m\sin^2\theta}}
= \int_0^{\sin\phi} \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}} = F( \sin\phi \;; m)
$$
On WolframAlpha, you can access the second form using the function $\bf \text{EllipticF}$ and
evaluate $I(x)$ using the expression:
$$\bf \text{EllipticF}[ \text{ArcSin}[ \text{Sqrt}[1+x^2-\text{Sqrt}[1+x^4]] ], m ]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find this $DF=?$ In diamond $ABCD$,such $\angle B=\dfrac{\pi}{3}$,and the point $E$ in on $BC$.such
$BE=3CE$,and the point $F$ is on $DE$,such $\angle AFC=\dfrac{2\pi}{3}$
Find $$DF=?$$
My try: since
$$\angle B+\angle AFC=\pi$$
so
$A,B,C,F$ is cyclic
and follow I can't
| Let's draw another equilateral triangle $BCG.$
Let $O$ be the center of that triangle.
Let's show that points $D, E$ and $O$ are collinear $\Longleftrightarrow$ points $D, F$ and $O$ are collinear.
$\overrightarrow{DE}=\overrightarrow{DC}+\overrightarrow{CE}=\overrightarrow{DC}+\dfrac{1}{4}\overrightarrow{CB},$
$\overrightarrow{DO}=\dfrac{1}{3}\left(\overrightarrow{DB}+\overrightarrow{DC}+\overrightarrow{DG}\right) = \dfrac{1}{3}\left(\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{DC}+2\overrightarrow{DC}\right)=\dfrac{1}{3}\left(\overrightarrow{CB}+4\overrightarrow{DC}\right) = \dfrac{4}{3}\overrightarrow{DE}.$
Thus, $D, F, E$ and $O$ are collinear.
Since $\angle COB + \angle BAC = \dfrac{2\pi}{3} + \dfrac{\pi}{3} = \pi,$ all five points $B, A, F, C$ and $O$ are concyclic.
The circumscribed circle of $\triangle ABC$ is tangent to line $DC$, which follows from the fact that $\angle BAC = \dfrac{\pi}{3} = \angle BCG.$ As an alternative, it can be observed from the symmetry of the picture with respect to the perpendicular bisector of $AB.$
It follows from the tangent-secant theorem that $DC^2 = DF\cdot DO.$ We know $DC,$ we want $DF,$ so the only thing that's left to calculate is $DO.$
We use cosine formula for that. Assuming the side of any of the equilateral triangles is $a$, $DO^2 = CD^2+CO^2-2\cdot CD \cdot CO \cos \cfrac{5\pi}{6}=a^2+\dfrac{a^2}{3} + 2a\dfrac{a}{\sqrt{3}}\dfrac{\sqrt{3}}{2} = \dfrac{7}{3}a^2.$
So $DF = a^2\bigg/\sqrt{\dfrac{7}{3}a^2}=\sqrt{\dfrac{3}{7}}a = \sqrt{\dfrac{3}{7}}AB$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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