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A series involving the Dirichlet Beta function; How to evaluate $\sum_{n=1}^\infty \frac{\beta(n)-1}{n}$? Let the beta and the zeta function be defined as usual:
\begin{align}
& \beta(s) & = & \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} & = & 1-\frac{1}{3^s}+\frac{1}{5^s}\dots +\frac{(-1)^n}{(2n+1)^s}+\dots
\\ & \zeta(s) & = & \sum_{n=1}^\infty \frac{1}{n^s} &= & 1+\frac{1}{2^s}+\frac{1}{3^s}+\dots +\frac{1}{n^s}+\dots
\end{align}
Question
Is there a closed form for
$$\sum_{n=1}^\infty \frac{1-\beta(n)}{n}$$
Exposition
I was hunting for some analogues to the zeta function while reading this. In the post we can find
$$ \begin{align}
& \sum_{n=2}^\infty \zeta(n)-1 & = & 1
\\ & \sum_{n=2}^\infty \zeta(2n)-1 & = & 3/4
\\ & \sum_{n=2}^\infty \frac{\zeta(n)-1}{n} & = & 1-\gamma \end{align} $$
Where $\gamma$ is the Euler-Mascheroni constant
So I found analogues for the first two but I can't quite see if this is possible for the third. I found using the techniques of the accepted answer in the link above (and it can also be found here):
$$\sum_{n=1}^\infty 1-\beta(n)= \ln(\sqrt2) $$ and $$ \sum_{n=1}^\infty 1-\beta(2n)= \ln(\sqrt2)-\frac{1}{4} $$
Possibly a red herring but hopefully not:
$$\sum_{n=1}^\infty \frac{1-\beta(n)}{n} \approx \ln (L/2)=\ln\bigg( \int_0^1 \frac{dx}{\sqrt{1-x^4}} \bigg) $$
These match up for at least the first 100 decimal places.
Where $L=2.62205755429 \dots $ is the Lemniscate constant. If these two are equal my question becomes: What is the connection between the Lemniscate constant and the $\beta$ function?
|
First of all : $$ \prod_{p=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{p=0}^{n}{\left(1-\frac{1}{4p+3}\right)}=\frac{4\Gamma\left(\frac{7}{4}\right)\Gamma\left(n+\frac{5}{4}\right)\Gamma\left(n+\frac{3}{2}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)\Gamma\left(n+1\right)\Gamma\left(n+\frac{7}{4}\right)}\underset{n\to +\infty}{\longrightarrow}\frac{4\Gamma\left(\frac{7}{4}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)} $$
Let $ n $ be a positive integer, we have the following : \begin{aligned}\sum_{p=1}^{2n+1}{\left(-1\right)^{p}\ln{\left(1+\frac{1}{2p}\right)}}&=\sum_{k=1}^{n}{\ln{\left(1+\frac{1}{4p}\right)}}-\sum_{k=0}^{n}{\ln{\left(1+\frac{1}{4p+2}\right)}}\\&=\ln{\left(\prod_{k=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{k=0}^{n}{\left(1-\frac{1}{4p+3}\right)}\right)} \end{aligned}
Thus, $ \sum\limits_{p\geq 0}{\left(-1\right)^{p}\ln{\left(1+\frac{1}{2p}\right)}} $ converges, and its sum values $ \ln{\left(\frac{4\Gamma\left(\frac{7}{4}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}\right)}\cdot $
Now, since $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n-1}}{\left(2n+1\right)^{s}}} $ converges absolutly, for any $ s>1 $, we can prove that the family $ \left(\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}\right)_{\left(p,n\right)\in\mathbb{N}^{*}\times\mathbb{N}^{*}\setminus\left\lbrace 1\right\rbrace} $ is summable, and hence, thanks to Fubini's theorem, we can write the following : \begin{aligned}\sum_{n=2}^{+\infty}{\frac{1-\beta\left(n\right)}{n}}=\sum_{n=2}^{+\infty}{\sum_{p=1}^{+\infty}{\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}}}=\sum_{p=1}^{+\infty}{\sum_{n=2}^{+\infty}{\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}}}\end{aligned}
After switching the summations, adding the first term of the sum, recognising logarithm's series expansion, we get that : $$ \sum_{n=1}^{+\infty}{\frac{1-\beta\left(n\right)}{n}}=\sum_{p=1}^{+\infty}{\left(-1\right)^{p-1}\ln{\left(1+\frac{1}{2p}\right)}}=\ln{\left(\frac{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}{4\Gamma\left(\frac{7}{4}\right)}\right)} $$
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"url": "https://math.stackexchange.com/questions/3639176",
"timestamp": "2023-03-29T00:00:00",
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|
Number of solvable lights out puzzles on m x n rectangle I'm trying to replicate the results on Dr. Brouwer website,
I consulted two of the papers referenced [1] and [2] both of them make use of the same recursively defined polynomial and GCD.
Let $p_n(x)$ be the mod 2 characteristic polynomial of the path of length n. The table has a zero in position (m,n) if and only if $gcd(p_m(x), p_n(x+1)) = 1$ [1]. More generally, $codim(m,n) = degr(gcd(p_m(x), p_n(x+1)))$ [2]
The $p_n(x)$ can be simply computed via $p_0(x) = 1, p_1(x) = x, p_{n+1}(x) = x p_n(x) + p_{n-1}(x)$.
Using a different definition, $Dimensions(Nullspace(Adjacency + I))$, I can verify the $codim(m,n)$ table
But when I try to use $degree(gcd(p_m(x), p_n(x+1)))$ I get weird results.
Here's a discreet example
$codim(5,2) = codim(2,5) = 1$
$codim(5,2) = degree(GCD(p_5(x), p_2(x+1))) = $
$degree(GCD(x(1+x^4), x^2) = degree(x) = 1$ (correct for this m,n and in general about half of m,n pairs)
but this doesn't match $codim(2,5)$
$codim(2,5) = degree(GCD(p_2(x), p_5(x+1))) = $
$degree(GCD(1+x^2, x*(x^3+x^4))) = degree(1) = 0$
Here's my math (double checked by Mathematica) on constructing the various mod 2 characteristic polynomials.
$p_1(x) = x$
$p_2(x) = x*p_1(x) + p_0(x) = x^2 + 1$
$p_3(x) = x*p_2(x) + p_1(x) = x^3 + x + x = x^3$
$p_4(x) = x*p_3(x) + p_2(x) = x^4 + x^2 + 1$
$p_5(x) = x*p_4(x) + p_3(x) = x^5 + x^3 + x + x^3 = x^5 + x$
$p_0(x+1) = 1$
$p_1(x+1) = x+1$
$p_2(x+1) = x*(x+1) + 1 = x^2$
$p_3(x+1) = (x+1)*x^2 + x+1 = x^3 + x^2 + x + 1$
$p_4(x+1) = (x+1)*(x^3+x^2+x+1) + x^2 = x^4 + x^2 + 1$
$p_5(x+1) = (x+1)*(x^4+x^2+1) + x^3+x^2+x+1 = x^5 + x^4$
[1] R. Barua & S. Ramakrishnan, σ-game, σ+-game, and two-dimensional cellular automata. Theor. Comput. Sci. 154 (1996) 349-366.
[2] Klaus Sutner, σ-Automata and Chebyshev-Polynomials, Theor. Comput. Sci. 230 (2000) 49-73. preprint.
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$$1+x^2 \equiv (1+x)^2 \mod 2$$
$$x(x^3+x^4) = x^4(1+x)$$
So $\gcd(1+x^2,x(x^3+x^4)) = 1+x$ and is also of degree $1$.
Note that if $\gcd(p(x),q(x+1))=r(x)$, then substituting $x=X+1$ gives $\gcd(p(X+1),q(X+2))=r(X+1)$ showing that when working modulo $2$ the degrees of $\gcd(p(x),q(x+1))$ and $\gcd(p(x+1),q(x))$ are equal.
P.S. Some of the links on Brouwer's site are very out of date. The link to my website should be updated to https://www.jaapsch.net/puzzles/lomath.htm
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evaluate it directly. And the LHS equals to
\begin{align}
\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
& = \frac{q}{1+q^2}+\frac{q^2}{1+q^4}\cdot\frac{q^3}{q^3}+\frac{q^3}{1+q^6}\cdot\frac{q}{q} \\
& = \frac{q}{1+q^2}+\frac{q^5}{1+q^3}+\frac{q^4}{1+q} \\
& = q\cdot\frac{(1+q)(1+q^3)+q^4(1+q)(1+q^2)+q^3(1+q^2)(1+q^3)}{(1+q)(1+q^2)(1+q^3)} \\
& = q\cdot\frac{1+q+q^3+q^4+q^4+q^5+q^6+1+q^3+q^5+q^6+q}{(1+q)(1+q^2)(1+q^3)} \\
& = \frac{-2q^3}{(1+q)(1+q^2)(1+q^3)} \\
\end{align}
And
$$(x-q)(x-q^2)(x-q^3)(x-q^4)(x-q^5)(x-q^6)=x^6+x^5+x^4+x^3+x^2+x+1$$
Let $x=-1$ I get that
$$(1+q)(1+q^2)(1+q^3)(1+q^4)(1+q^5)(1+q^6)=1$$
and
$$(1+q)(1+q^2)(1+q^3)\cdot q^4(q^3+1)\cdot q^5(q^2+1)\cdot q^6(q+1)=1$$
therefore
$$\left[(1+q)(1+q^2)(1+q^3)\right]^2=\frac{1}{q^{15}}=\frac{1}{q}$$
hence
$$\left[\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}\right]^2=\frac{q}{1}\cdot 4q^6=4$$
$$\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}=\pm 2$$
And I try for a solution as a polar-form method$.\\$Suppose $q=\cos\frac{2j\pi}{7}+i\sin\frac{2j\pi}{7}$
$$\frac{q^k}{1+q^{2k}}=\frac{\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}}{2\cos\frac{2jk\pi}{7}\left(\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}\right)}=\frac{1}{2\cos\frac{2jk\pi}{7}}$$
Am I going to the right direction? How I finish it? And please help to figure out what's wrong with my calculation at the first part. I appreciate for your help.
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Suppose that $x_1,x_2,\ldots,x_{n-1}$ are the roots of $z^{n-1}+z^{n-2}+\ldots+z+1=0$. We have
$$\frac{1}{x_j+x_j^{-1}}=\frac{x_j}{1+x_j^2}.$$
If $n$ is odd, then
$$1+z^{2n}=(1+z^2)(1-z^2+z^4-z^6+\ldots+z^{2(n-1)}).$$
Because $x_j^n=1$, we have
$$\frac{1}{1+x_j^2}=\frac{\sum_{k=0}^{n-1}(-1)^kx_j^{2k}}{1+x_j^{2n}}=\frac{\sum_{k=0}^{n-1}(-1)^kx_j^{2k}}{2}$$
so
$$\frac{x_j}{1+x_j^2}=\frac{1}{2}\sum_{k=0}^{n-1}(-1)^kx_j^{2k+1}.$$
Note that $\sum_{j=1}^{n-1}x_j^d=-1$ unless $d$ is a multiple of $n$, in which case $\sum_{j=1}^{n-1}x_j^d=n-1$. Therefore
\begin{align}\sum_{j=1}^{n-1}\frac{x_j}{1+x_j^2}&=\frac{1}{2}\sum_{j=1}^{n-1}\sum_{k=0}^{n-1}(-1)^kx_j^{2k+1}\\&=\frac{1}{2}\sum_{k=0}^{n-1}(-1)^k\sum_{j=1}^{n-1}x_j^{2k+1}=\frac{1}{2}\left((-1)^{\frac{n-1}{2}}n-1\right).\end{align}
If $n\equiv 2\pmod4$, then
$$1+z^n=(1+z^2)(1-z^2+z^4-z^6+\ldots+z^{n-2}).$$
Because $x_j^n=1$, we have
$$\frac{1}{1+x_j^2}=\frac{\sum_{k=0}^{\frac{n}{2}-1}(-1)^kx_j^{2k}}{1+x_j^{n}}=\frac{\sum_{k=0}^{\frac{n}{2}-1}(-1)^kx_j^{2k}}{2}$$
so\begin{align}\sum_{j=1}^{n-1}\frac{x_j}{1+x_j^2}&=\frac{1}{2}\sum_{j=1}^{n-1}\sum_{k=0}^{\frac{n}{2}-1}(-1)^kx_j^{2k+1}\\&=\frac{1}{2}\sum_{k=0}^{\frac{n}{2}-1}(-1)^k\sum_{j=1}^{n-1}x_j^{2k+1}=-\frac{1}{2}.\end{align}
(Alternatively, note that $\sec\theta+\sec(\pi+\theta)=0$ and $\pi$ is an integer multiple of $\frac{2\pi}{n}$.) If $n\equiv0\pmod{4}$, then clearly $\sum_{j=1}^{n-1}\frac{x_j}{1+x_j^2}$ is not defined.
Therefore, we have
$$\sum_{j=1}^{n-1}\sec\frac{2\pi j}{n}=2\sum_{j=1}^{n-1}\frac{x_j}{1+x_j^2}=\left\{\begin{array}{ll}
(-1)^{\frac{n-1}{2}}n-1&\text{if $n\equiv1\pmod{2}$},\\
-1&\text{if $n\equiv 2\pmod{4}$},\\
\text{undefined}&\text{if $n\equiv 0\pmod{4}$}.
\end{array}\right.$$
The sum in question is equal to
$$\frac{1}{4}\sum_{j=1}^6\sec\frac{2\pi j}{7}=\frac{(-1)^{\frac{7-1}{2}}\cdot 7-1}{4}=-2.$$
|
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|
Evaluating $\iint_{[0,1]^2} \frac{2-4xy}{(9-xy)(8+xy)}dxdy$ I am trying to compute the following double integral:
$$I=\iint_S \frac{2-4xy}{(9-xy)(8+xy)}dxdy$$
with $S=[0,1]\times[0,1].$
What I have tried:
*
*I have written the integral as follows:
$$I=I_1+I_2=-2\iint_S \frac{1}{(9-xy)}dxdy+2\iint_S \frac{1}{(8+xy)}dxdy$$
When trying to compute $I_1$, I encountered this integral $\int_0^1-\frac{1}{x}\ln(1-\frac{x}{9})dx$. I used an online calculator to solve it and the result involves the function $\operatorname{Li}(z)$, which I am not familiar with.
*Another thing I have tried is a change of variables:
$$u=9-xy$$ $$v=8+xy$$
The problem is that the Jacobian associated to this change of variables is null, so I cannot use it.
My question:
Could someone show me a way of computing this integral without having to use $\operatorname{Li}_2(z)$?
|
Let's generalize it, in particular the integral in the question is just the case $a=8$.
$$I(a)=\int_0^1\int_0^1\left(\frac{1}{a+xy}-\frac{1}{1+a-xy}\right)dxdy\overset{xy=t}=\int_0^1\int_0^y\frac{1}{y}\left(\frac{1}{a+t}-\frac{1}{1+a-t}\right)dtdy$$
$$=\int_0^1\int_t^1\frac{1}{y}\left(\frac{1}{a+t}-\frac{1}{1+a-t}\right)dydt=\int_0^1\ln t \left(\frac{1}{1+a-t}-\frac{1}{a+t}\right)dt$$
$$\overset{\large \frac{1+a-t}{a+t}=x}=\int_{a/(1+a)}^{(1+a)/a}\ln\left(\frac{(1+a)-ax}{1+x}\right)\frac{1-x}{1+x}\frac{dx}{x}\overset{x\to 1/x}=\int_{a/(1+a)}^{(1+a)/a}\ln\left(\frac{1+x}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
By averaging the two integrals from above we obtain:
$$I(a)=\frac12 \int_{a/(1+a)}^{(1+a)/a}\ln\left(\frac{(1+a)-ax}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
Similarly, via the substitution $x\to \frac{1}{x}$ we can see that:
$$\int_{a/(1+a)}^{1}\ln\left(\frac{(1+a)-ax}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}=\int_{1}^{(1+a)/a}\ln\left(\frac{(1+a)-ax}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
$$\Rightarrow I(a)=\int_{1}^{(1+a)/a}\ln\left(\frac{(1+a)/a-x}{(1+a)/a x-1}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
$$\overset{\large \frac{(1+a)/a-x}{(1+a)/a x-1}=t}=\int_0^1 \ln t \left(\frac{2}{1+t}-\frac{1}{(1+a)/a+t}-\frac{1}{a/(1+a)+t}\right)dt$$
Now we will use the substitution $t=k x$, where $k$ is the constant found in each denominator.
$$I(a)=2\int_0^1 \frac{\ln x}{1+x}dx-\int_0^{(1+a)/a}\frac{\ln\left(\frac{1+a}{a}x\right)}{1+x}dx-\int_0^{a/(1+a)}\frac{\ln\left(\frac{a}{1+a}x\right)}{1+x}dx$$
$$=-\ln\left(\frac{1+a}{a}\right)\ln\left(1+\frac{1+a}{a}\right)-\ln\left(\frac{a}{1+a}\right)\ln\left(1+\frac{a}{1+a}\right)$$
$$+2\int_0^1 \frac{\ln x}{1+x}dx-\int_0^{(1+a)/a}\frac{\ln x}{1+x}dx-\int_0^{a/(1+a)}\frac{\ln x}{1+x}dx$$
$$=\ln^2\left(\frac{1+a}{a}\right)+\int_{(1+a)/a}^1 \frac{\ln x}{1+x}dx+\int_{a/(1+a)}^1 \frac{\ln x}{1+x}dx$$
Again, the substitution $x\to \frac{1}{x}$ yields:
$$\int_{a/(1+a)}^1 \frac{\ln x}{1+x}dx=\int_{(1+a)/a}^1 \frac{\ln x}{x}dx-\int_{(1+a)/a}^1 \frac{\ln x}{1+x}dx$$
$$\Rightarrow I(a)=\ln^2\left(\frac{1+a}{a}\right)+\int_{(1+a)/a}^1 \frac{\ln x}{x}dx=\boxed{\frac12 \ln^2\left(\frac{1+a}{a}\right)},\ a>0$$
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Why does the difference equation $x_nx_{n+2}=w^5x_n-(w^2+w^3)x_{n+1}+x_{n+2}$ generate cyclic sequences? Let $w$ be a primitive 5th root of unity. Then the difference equation $$x_nx_{n+2}=x_n-(w^2+w^3)x_{n+1}+x_{n+2}$$ generates a cycle of period 5 for general initial values:
$$u,v,\frac{u-(w^2+w^3)v}{u-1},\frac{uv-(w^2+w^3)(u+v)}{(u-1)(v-1)},\frac{v-(w^2+w^3)u}{v-1},u,v, ...$$
For equations of the form $$x_nx_{n+2}=w^{a+b}x_n-(w^a+w^b)x_{n+1}+x_{n+2},\text{ for }w^a+w^b\ne 0$$ with the same globally periodic property, I can show that the only possible periods are 5,8,12,18 and 30.
Curiously, the 'same' equation works for all these periods:
$$x_nx_{n+2}=w^5x_n-(w^2+w^3)x_{n+1}+x_{n+2},$$
where $w$ is a primitive $p$th root of unity for $p=5,8,12,18,30$. Is this a fluke or is there a way of seeing why this 'family' of equations always generate cycles?
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Let us prove that for all nonzero $a,b\in \mathbb{C}$ the difference equation $$x_nx_{n+2}=ax_n+bx_{n+1}+x_{n+2}$$
never generates a cycle of length $8$.
Let us denote $u=x_0, y=x_1$.
Suppose the contrary. We know that
$$x_{n+2}=\frac{ax_n+bx_{n+1}}{x_n-1}, x_{n-1}=\frac{bx_n+x_{n+1}}{x_{n+1}-a},$$
so for every integer $k$ we can represent $x_k$ as rational function in variables $u,v$:
$$x_k=\frac{P_{k}(u,v)}{Q_k(u,v)}.$$
If the equality $x_{-4}=x_4$ is true then $P_{-4}(u,v)Q_{4}(u,v)=P_{4}(u,v)Q_{-4}(u,v)$.
Calculations in Wolfram Mathematica gives the next result for the left side
and for the right side
Looking at some coefficients at $u^iv^j$ we get
$$[u^0v^1]: 2a^2b-ab^3=-2a^6b+a^5b^3,$$
$$[u^1v^1]: -a^7+a^6b^2=-a^3+3a^2b^2-ab^4,$$
$$[u^3v^2]: b-ab+b^2-ab^2=-a^2b+a^3b-ab^2+a^2b^2.$$
These equations after excluding some terms turn into (we remember that $a,b\not =0$:
$$ (a^4+1)(2a-b^2)=0, \tag{1}$$
$$[u^1v^1]: b^4+(a^5-3a)b^2+a^2-a^6=0, \tag{2}$$
$$[u^3v^2]: (a-1)(b(a+1)+a^2+1)=0. \tag{3}$$
If $a=1$, then from $(1)$ we have that $b^2=2$. In this case we have
$$P_{-4}(u,v)Q_{4}(u,v)-P_{4}(u,v)Q_{-4}(u,v)\not =0,$$ as it equals to
So $a\not =1$, and equation $(3)$ gives us $a\not =-1$ and the next relationship
$$b=-\frac{a^2+1}{a+1}. \tag{4}$$
Let us consider the equation $(1)$ and suppose that $b^2=2a$. Then from $(2)$ we have that $a^2-a^6=0$. As $a\not =0,\pm 1$ then $a=\pm i$. But in that case $a=\pm i$, so $b=0$ by $(4)$, contradiction with our assumption.
Hence $a^4+1=0$, so for some odd $k$ we have
$$a=e^{ik\pi/4}=\frac{\pm 1 \pm i}{\sqrt{2}}. \tag{5}$$
Joining together $b^2=2a$ and equation $(4)$ we have that $a$ is also a root of polynomial $x^4-2x^3-2x^3-2x+1$ but that's not true, so we get final contradiction.
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|
Factorise $Σa^{2}(b^{4}-c^{4})$ I expanded this cyclic expression then found the three factors and assumed third to be constant but after that I got stuck at this question. The options are
(a)$(a-b)^{2}(b-c)^{2}(c-a)^{2}$
(b)$(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$
(c)$(a+b)^{2}(b+c)^{2}(c+a)^{2}$
(d)None of these
The answer given is option (b).
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Factorizing the difference of two squares comes up a lot here. Let $u=b^4-c^4,\,v=c^4-a^4$ so $a^4-b^4=-u-v$ and the sum is$$\begin{align}a^2u+b^2v-c^2(u+v)&=(a^2-c^2)u+(b^2-c^2)v\\&=(a^2-c^2)(b^2-c^2)(b^2+c^2)+(b^2-c^2)(c^2-a^2)(c^2+a^2)\\&=(b^2-c^2)(c^2-a^2)(-b^2-c^2+c^2+a^2)\\&=(a^2-b^2)(b^2-c^2)(c^2-a^2),\end{align}$$which is (b). Incidentally, (c) is obviously wrong because it doesn't in general vanish if $a=b=c$.
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"url": "https://math.stackexchange.com/questions/3654780",
"timestamp": "2023-03-29T00:00:00",
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For $a$, $b$, $c$ the sides of a triangle, show $\sum_{cyc}\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}\geq1$
Let $a,b,c$ be the side lengths of a triangle. How to prove the following inequality?
$$\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}+\frac{\sqrt{ac}\,ac}{b(a+b+c)(a+c-b)}+\frac{\sqrt{ab}\,ab}{c(a+b+c)(a+b-c)}\geq1$$
Via some numerical experiment and graphing techniques, I am sure it is true. Could we use AM-GM?
My goal is to find an analytic/conventional proof of this inequality. I have tried AM-GM, but it didn't work. Since it is a homogeneous function with degree 0, WLOG, let $a=1.$ Then I plotted the LHS of this inequality using the Maple command "plot3d" which clearly shows that the LHS ≥ 1.
|
We have to prove that (see @g.kov's result)
$$3\sqrt[3]{\frac{(abc)^2}{(a+b+c)^3(a+b-c)(b+c-a)(c+a-b)}} \ge 1.$$
By using Ravi Substitution $a = x + y, b = y+z, c = z + x$ for $x, y, z > 0$, the inequality becomes
$$3\sqrt[3]{\frac{[(x+y)(y+z)(z+x)]^2}{64(x+y+z)^3xyz}} \ge 1.$$
Let $p = x+y+z, q = xy + yz + zx, r = xyz$. The inequality becomes
$$3\sqrt[3]{\frac{(pq - r)^2}{64p^3 r}} \ge 1.$$
Since $pq \ge 9r$ and $q^2 \ge 3pr$, we have
$$3\sqrt[3]{\frac{(pq - r)^2}{64p^3 r}} \ge 3\sqrt[3]{\frac{(pq - \frac{pq}{9})^2}{64p^2q^2/3}} = 1.$$
We are done.
|
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|
Dividing polynomial $f(x^3)$ by $x^2+x+1$ Given two polynomials with real coefficients $g(x)$ and $h(x)$.
Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.
Prove that $x-1|h(x)$ and $x-1|g(x)$
I have tried to solve it by doing this;
$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$
then;
$(x^3-1)|(x)(h(x^3)-g(x^3)) + (x^2)g(x^3) - h(x^3)$
consider degree of $3k, 3k+1, 3k+2$ separately, I got;
$(x^3-1)|(x^2)g(x^3)$ and $h(x^3)$
Lastly, $(x^3-1)|g(x^3), h(x^3)$ and so on finish proving.
I wonder, is it correct to separate three cases of degrees in this problem?
|
I'm not sure what you mean by that separate cases.
But here is how I did it.
Let $a$ and $b$ be roots for $x^2+x+1=0$, then $a^3=b^3=1$ and clearly $a\ne b$ since the discriminat is not $0$.
Since $$x^2+x+1\mid f(x)$$ we see that $f(a)= f(b)=0$
Now put
$$x=a:\;\;\; f(a^3) = h(a^3)+ag(a^3)$$
so $$ 0 = f(1) = h(1)+ag(1)$$ and similary for $b$:$$ 0 = h(1)+bg(1)$$
Sunbstract does eqautions: $$(a-b)g(1)=0\implies g(1)=0$$ and thus also $h(1)=0$. This means $x-1\mid g(x)$ and $x-1\mid h(x)$.
|
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|
Trying to find a general equation for an ellipse given the foci and sum of focal distances I'm trying to find an equation for the ellipse in the form $$Ax^2 + Bxy + Cy^2 +Dx +Ey +F = 0$$ given the foci $(a,b)$ and $(c,d)$ and the sum of focal distances $r$. I started from the definition $$\sqrt{(x-a)^2+(y-b)^2} + \sqrt{(x-c)^2+(y-d)^2} = r$$ squared both sides, moved not radical terms to the right side and squared again and after trudging through a lot of algebra I've arrived at the equation (easier reading of coefficients below)
\begin{align}
0 &= (r^2 + (a-c)^2)x^2 + 2(a-c)(b-d)xy + (r^2 + (b-d)^2)y^2\\
&\qquad + (r^2(a+c) - (a-c)(a^2+b^2-c^2-d^2))x \\
&\qquad +(r^2(b+d) - (b-d)(a^2+b^2-c^2-d^2))y \\
&\qquad+ \frac{1}{4}(r^4 + 2r^2(a^2+b^2+c^2+d^2) + (a^2+b^2-c^2-d^2)^2),
\end{align}
which I want to believe is close, but this does not produce a graph on Desmos. If anyone just has a reference for the equation that I might be able to look at and find my mistakes that would be much appreciated. When I looked on Wikipedia, they talked about using the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 $$ and rotating the major axis, but I have no idea how to translate those coefficients to be in terms of the foci. For easier reading my coefficients are
\begin{align}
A &= r^2+(a-c)^2 \\
B &= 2(a-c)(b-d) \\
C &= r^2 + (b-d)^2 \\
D &= r^2(a+c)-(a-c)(a^2+b^2-c^2-d^2) \\
E &= r^2(b+d)-(b-d)(a^2+b^2-c^2-d^2) \\
F &= \frac{1}{4}(r^4 +2r^2(a^2+b^2+c^2+d^2) + (a^2+b^2-c^2-d^2)^2).
\end{align}
If the middle term of F was $2r^2(a^2+b^2-c^2-d^2)$ I could factor it, but because the left hand side (after the second squaring round) has no $r$ term, there isn't an opportunity for it to change like all the other terms did. Sorry I can't be more specific, but I didn't think typing out the mountain of algebra I've done was a good idea.
|
Thanks to User Blue for helping find the errors. Just a couple of negative signs went awry. The correct coefficients are $$A=(a-c)^2-r^2$$ $$B=2(a-c)(b-d)$$ $$C=(b-d)^2-r^2$$ $$D = r^2(a+c)-(a-c)(a^2+b^2-c^2-d^2)$$ $$E=r^2(b+d)-(b-d)(a^2+b^2-c^2-d^2)$$ $$F = \frac{1}{4}(r^4-2r^2(a^2+b^2+c^2+d^2)+(a^2+b^2-c^2-d^2)^2)$$
|
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"url": "https://math.stackexchange.com/questions/3661068",
"timestamp": "2023-03-29T00:00:00",
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|
Sequence $\left\{ a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$. Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$. As title. Sequence $\left\{a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$.
Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$.
I tried the induction.
$a_{k+1}-3=\sqrt{a_k+6}-3<\sqrt{\frac{1}{3}\left(a_{k-1}-3\right)+6}-3=\frac{1}{3}\sqrt{a_{k-1}+15}-3$
But It seems not work. And I tryied trigonometrical substitution with letting $a_1=6\cos\theta$, then $a_2=\sqrt{6+6\cos\theta}=2\sqrt{3}\cos\frac{\theta}{2}$. And it seems not work either.
And I tried to square them to find a relation $\left(a_n-3\right)^2+6\left(a_n-3\right)=\left(a_{n-1}-3\right)$ or $\left(a_n-3\right)\left(a_n+3\right)=\left(a_{n-1}-3\right)$
It seems not work agian. Please help me, and thanks a lot.
|
Hint :
$$
a_{n+1}-3=\sqrt{a_n+6}-\sqrt{9}=\frac{a_n+6-9}{\sqrt{a_n+6}+\sqrt{9}}
$$
|
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|
Question about $I = \int x^3 \sqrt{x^2+2} \, dx$ I'm trying to find the result of the following integral:
$$I = \int x^3 \sqrt{x^2+2} \, dx$$
This integral appears in MIT OCW 18.01SC course
I try to make a substitution as follow :
Let $x = \sqrt{2}\tan(u), \quad dx = \sqrt{2}\sec^2(u)\, du$
If we substitute that in our integral and use the trig identity which tells as that $\sec^2(u) = 1 + \tan^2(u)$
We found that :
$$I = 4\sqrt{2} \int\tan^3(u)\sec^3(u) \, du$$
Using the previous trig identity we can write it as follow :
$$I = 4\sqrt{2} \int\sec^2(u)(\sec^2(u) - 1)\sec(u)\tan(u) \,du$$
Now we can make a substitution as follow :
Let $v = \sec(u),\quad dv = \sec(u)\tan(u)\,du$
After doing that the integral is easy to calculate and we can get the result :
$$I = 4\sqrt{2} \left(\frac{1}{5}v^5 - \frac{1}{3}v^3\right) + c$$
And after reverse all previous trig substitution we should get :
$$I = \frac{4}{5}(x^2+2)^{5/2} - \frac{4}{3}(x^2+2)^{3/2} + c$$
The professor uses another way to get the result using another substitution in the first which is :
$$u = x^2 + 2, x^2 = u - 2$$
$$du = 2x\,dx, x\,dx = du/2$$
And his answer is :
$$I = \frac{1}{5}(x^2+2)^{5/2} - \frac{2}{3}(x^2+2)^{3/2} + c$$
Can anyone tell me where the wrong is in my solution?
|
You have made an error in simplification after "reverse all previous trig substitution".
\begin{align*}
&\left. \left. 4\sqrt{2}\left( \frac{1}{5}v^5 - \frac{1}{3}v^3 \right) + C \right|_{v = \sec u} \right|_{u = \arctan(x/\sqrt{2})} \\
&= \left. 4\sqrt{2}\left( \frac{1}{5}\sec^5 u - \frac{1}{3}\sec^3 u \right) + C \right|_{u = \arctan(x/\sqrt{2})} \\
&= 4\sqrt{2}\left( \frac{1}{5} \left( \frac{x^2}{2}+1 \right)^{5/2} - \frac{1}{3}\left( \frac{x^2}{2}+1 \right)^{3/2} \right) + C \\
&= \frac{4\sqrt{2}}{5 \cdot 2^{5/2}} \left( x^2 + 2 \right)^{5/2} - \frac{4\sqrt{2}}{3 \cdot 2^{3/2}}\left( x^2 + 2 \right)^{3/2} + C \\
&= \frac{4\sqrt{2}}{5 \cdot 4 \sqrt{2}} \left( x^2 + 2 \right)^{5/2} - \frac{4\sqrt{2}}{3 \cdot 2\sqrt{2}}\left( x^2 + 2 \right)^{3/2} + C \\
&= \frac{1}{5} \left( x^2 + 2 \right)^{5/2} - \frac{2}{3}\left( x^2 + 2 \right)^{3/2} + C \text{.}
\end{align*}
|
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|
If $\omega$ is a cube root of unity $\not = 1$ then find the minimum value of $|a+b\omega +c\omega^2|$, where $a,b,c$ are integers but not all equal. Let $z=a+b\omega + c\omega^2$
$$z=a+b\omega -c (1+\omega)$$
$$z=a-c+\omega (b-c)$$
Therefore $$|a-c+\omega (b-c)| \ge ||a-c|-|b-c||$$
How should I proceed?
|
We want to minimize $|x+y\omega|^2=\bigg(x-\frac{y}{2}\bigg)^2+\frac{3}{4}y^2$ where $x$ and $y$ are nonzero integers.
If $y$ is even, we can take $x=\frac{y}{2}$ and hence $|x+y\omega|^2 \geq \frac{3}{4}y^2 \geq 3$.
If $y$ is odd, we can take $x=\frac{y+1}{2}$ and hence $|x+y\omega|^2 \geq \frac{1}{4}+\frac{3}{4}y^2 \geq 1$.
Conclusion : the minimum is $1$. It is reached when $y=\pm 1,x=y$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum_{i=1}^{n-1} \left[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \right] =(n-1)^2$? Apparently the following expression
$$
\sum_{i=1}^{n-1} \Bigg[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \Bigg] \\
$$
simplifies to $(n-1)^2$, where $H_i$ is the i-th harmonic number.
I tried to simplify but I'm not seeing the simplification
\begin{align}
\sum_{i=1}^{n-1} \Bigg[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \Bigg] \\
= n\sum_{i=1}^{n-1} \frac{1}{i(i+1)}\Bigg[1 + (n-1) (n(H_{n-2} - H_{n-i-1}) - (i-1))) \Bigg]
\end{align}
It's not obvious to me how I can further simplify, especially with the harmonic terms.
In addition, what kind of series is $\sum_i \frac{1}{i(i+1)}$?
I have a little Python script below in case anyone wants to see that the 2 expressions are equal:
def solution(n):
ans = 0
for i in range(1, n):
sum1 = 0
for r in range(n-i, n-1):
sum1 += 1/r
sum1 *= n
sum1 -= i-1
sum1 *= n*(n-1)/i/(i+1);
ans += sum1 + n/i/(i+1)
return ans
for n in range(100,1000):
print(solution(n) - (n-1)**2)
```
|
I got $-(n-1)^2$ and by calculating the first few cases, it seems to be the correct answer. I will assume that $n\geq 2$ and $H_0=0$. First note that
$$
\sum\limits_{i = 1}^{n - 1} {\frac{1}{{i(i + 1)}}} = \sum\limits_{i = 1}^{n - 1} {\left[ {\frac{1}{i} - \frac{1}{{i + 1}}} \right]} = 1 - \frac{1}{n}.
$$
We can decompose the sum into
\begin{align*}
& n\sum\limits_{i = 1}^{n - 1} {\frac{1}{{i(i + 1)}}} + n^2 (n - 1)H_{n - 2} \sum\limits_{i = 1}^{n - 1} {\frac{1}{{i(i + 1)}}} - n^2 (n - 1)\sum\limits_{i = 1}^{n - 1} {\frac{{H_{n - i - 1} }}{{i(i + 1)}}} \\ &\quad - n(n - 1)\sum\limits_{i = 1}^{n - 1} {\frac{1}{i}}
\\ &
= n - 1 + n(n - 1)^2 H_{n - 2} - n^2 (n - 1)\sum\limits_{i = 1}^{n - 1} {\frac{{H_{n - i - 1} }}{{i(i + 1)}}} - n(n - 1)H_{n - 1}
\\ &
= n - 1 + n(n - 1)^2 H_{n - 2} - n^2 (n - 1)\sum\limits_{i = 1}^{n - 1} {\frac{{H_{n - i - 1} }}{{i(i + 1)}}} - n(n - 1)\left( {H_{n - 2} + \frac{1}{{n - 1}}} \right)
\\ &
= - 1 + n(n - 1)(n - 2)H_{n - 2} - n^2 (n - 1)\sum\limits_{i = 1}^{n - 1} {\frac{{H_{n - i - 1} }}{{i(i + 1)}}} .
\end{align*}
Now
\begin{align*}
& \sum\limits_{i = 1}^{n - 1} {\frac{{H_{n - i - 1} }}{{i(i + 1)}}} = \sum\limits_{i = 1}^{n - 2} {\frac{{H_{n - i - 1} }}{{i(i + 1)}}} = \sum\limits_{i = 1}^{n - 2} {\left[ {\frac{{H_{n - i - 1} }}{i} - \frac{{H_{n - i - 1} }}{{i + 1}}} \right]} \\ & = \sum\limits_{i = 1}^{n - 2} {\left[ {\frac{{H_{n - i - 1} }}{i} - \frac{{H_{n - i - 2} + \frac{1}{{n - i - 1}}}}{{i + 1}}} \right]}
\\ &
= \sum\limits_{i = 1}^{n - 2} {\left[ {\frac{{H_{n - i - 1} }}{i} - \frac{{H_{n - i - 2} }}{{i + 1}}} \right]} + \sum\limits_{i = 1}^{n - 2} {\frac{1}{{(n - i - 1)(i + 1)}}} \\ & = H_{n - 2} + \frac{1}{n}\sum\limits_{i = 1}^{n - 2} {\left[ {\frac{1}{{n - i - 1}} + \frac{1}{{i + 1}}} \right]}
\\ &
= H_{n - 2} + \frac{1}{n}\left( {H_{n - 1} - 1 + H_{n - 2} } \right).
\end{align*}
Therefore, the original sum is
$$
- 1 + n(n - 1)(n - 2)H_{n - 2} - n^2 (n - 1)H_{n - 2} + n(n - 1)\left( {H_{n - 1} - 1 + H_{n - 2} } \right) \\ = - n^2 + 2n - 1 = - (n - 1)^2 .
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate the area of the surface I have to calculate the area of such surface:
$$(x^2+y^2+z^2)^2=x^2-y^2$$
I use the spherical coordinates transform, and get $r^2=\sin^2 \varphi \cos 2\theta$, but I don't know how to use this to calculate the area.
I'll be grateful if there's any help. :)
|
First notice that by even symmetry in each of the variables we can reduce the integral to just one in the first octant
$$\iint_S \:dS = 8 \iint_{S\:\cap\:\text{First octant}} \:dS$$
Next we will parametrize the surface like so:
$$x = \sin^2\theta \cos\phi \sqrt{\cos 2\phi}$$
$$y = \sin^2\theta \sin\phi \sqrt{\cos 2\phi}$$
$$z = \sin\theta\cos\theta \sqrt{\cos 2\phi}$$
with $\theta \in \left[0,\frac{\pi}{2}\right]$ and $\phi \in \left[0,\frac{\pi}{4}\right]$
Then we can use the following convenient fact about parametrizations in spherical coordinates:
$$\vec{r}(\theta,\phi) = f(\theta,\phi)\hat{r} \implies |\vec{r}_\theta\times\vec{r}_\phi| = f\sqrt{f_\phi^2 + f_\theta^2\sin^2\theta + f^2\sin^2\theta}$$
$$ = f\sin\theta\sqrt{\frac{f_\phi^2}{\sin^2\theta}+f_\theta^2+f^2}$$
Notice the similarity to the spherical coordinates Jacobian, $r^2\sin\theta$, when $f = \text{const}$
Since in this case $f = \sin\theta\sqrt{\cos2\phi}$ we get that
$$|\vec{r}_\theta\times\vec{r}_\phi| = \sin^2\theta\sqrt{\cos2\phi}\sqrt{\frac{\sin^22\phi}{\cos2\phi} + \cos^2\theta\cos2\phi+\sin^2\theta\cos2\phi}$$
$$ = \sin^2\theta\sqrt{\sin^22\phi + \cos^2\theta\cos^22\phi+\sin^2\theta\cos^22\phi} = \sin^2\theta$$
giving us the integral
$$S = 8\int_0^{\frac{\pi}{4}} \int_0^{\frac{\pi}{2}} \sin^2\theta\:d\theta\:d\phi = \frac{\pi^2}{2}$$
|
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|
Contest Math sum Partition question From IMO 2012 shortlist:
What is the maximum value of $k$ such that the set $S=\{1,2,\cdots,2018\}$ can be partitioned into $k$ disjoint pairs such that the sum of (the two numbers in) each pair is pairwise distinct and none of the sums exceeds $2018$.
I predict that the answer is $672$, that is when the pairs are $(1,2017),(2,2015),(3,2013),\cdots,(672,675)$. But I am not sure whether my prediction is right. Furthermore, I wanted to try proving that $673$ pairs is impossible, and I think some pigeonhole principle might be needed but I don't know how to make use of it. Any help is surely appreciated. Thanks!
N.B : If anyone got a solution without using the pigeonhole principle, that would be fine too!
|
This problem is in IMO 2012 shortlist, page 20.
The answer is $\left\lfloor \frac{2n-1}{5}\right\rfloor$. In this case $807$.
Original post in the shortlist:
C2. Let $n \geqslant 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{1,2, \ldots, n\}$ such that the sums of the different pairs are different integers not exceeding $n ?$
Solution. Consider $x$ such pairs in $\{1,2, \ldots, n\} .$ The sum $S$ of the $2 x$ numbers in them is at least $1+2+\cdots+2 x$ since the pairs are disjoint. On the other hand $S \leqslant n+(n-1)+\cdots+(n-x+1)$ because the sums of the pairs are different and do not exceed $n$. This gives the inequality
$$\frac{2 x(2 x+1)}{2} \leqslant n x-\frac{x(x-1)}{2}$$
which leads to $x \leqslant \frac{2 n-1}{5}$. Hence there are at most $\left\lfloor\frac{2 n-1}{5}\right\rfloor$ pairs with the given properties. We show a construction with exactly $\left\lfloor\frac{2 n-1}{5}\right\rfloor$ pairs. First consider the case $n=5 k+3$ with $k \geqslant 0,$ where $\left\lfloor\frac{2 n-1}{5}\right\rfloor=2 k+1 .$ The pairs are displayed in the following table.
$$\begin{array}{|c|c|c|c|}
\hline
\text{pairs}& 3k+1 & 3k &\cdots&2k+2&4k+2&4k+1&\cdots&3k+3&3k+2 \\ &2 & 4& \cdots&2k&1&3&\cdots&2k-1&2k+1\\ \hline
\text{sums} &3k+3 &3k+4&\cdots&4k+2&4k+3&4k+4&\cdots&5k+2&5k+3\\ \hline
\end{array}$$
The $2 k+1$ pairs involve all numbers from 1 to $4 k+2 ;$ their sums are all numbers from $3 k+3$ to $5 k+3 .$ The same construction works for $n=5 k+4$ and $n=5 k+5$ with $k \geqslant 0 .$ In these cases the required number $\left\lfloor\frac{2 n-1}{5}\right\rfloor$ of pairs equals $2 k+1$ again, and the numbers in the table do not exceed $5 k+3 .$ In the case $n=5 k+2$ with $k \geqslant 0$ one needs only $2 k$ pairs. They can be obtained by ignoring the last column of the table (thus removing $5 k+3$ ). Finally, $2 k$ pairs are also needed for the case $n=5 k+1$ with $k \geqslant 0 .$ Now it suffices to ignore the last column of the table and then subtract 1 from each number in the first row.
Comment. The construction above is not unique. For instance, the following table shows another set of $2 k+1$ pairs for the cases $n=5 k+3, n=5 k+4,$ and $n=5 k+5$
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \text{Pairs } & 1 & 2 & \cdots & k & k+1 & k+2 & \cdots & 2k+1\\
& 4k+1 & 4k-1 & \cdots & 2k+3 & 4k+2 & 4k & \cdots & 2k+2\\ \hline
\text{Sums} & 4k+2& 4k+1 & \cdots & 3k+3 & 5k+3& 5k+2 & \cdots & 4k+3\\ \hline
\end{array}
The table for the case $n=5 k+2$ would be the same, with the pair $(k+1,4 k+2)$ removed. For the case $n=5 k+1$ remove the last column and subtract 2 from each number in the second row.
|
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}
|
When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. When $ab/(a+b)$ is an integer, where $a,b$ are positive integers?
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
print a,b,q1 div q2;
end if;
end for;
end for;
the Magma code given above outputs the following.
2 2 1
3 6 2
4 4 2
4 12 3
5 20 4
6 6 3
6 12 4
6 30 5
8 8 4
8 24 6
9 18 6
10 10 5
10 15 6
12 12 6
12 24 8
14 14 7
15 30 10
16 16 8
18 18 9
20 20 10
20 30 12
21 28 12
22 22 11
24 24 12
26 26 13
28 28 14
30 30 15
So I conjectrue that $\frac{ab}{a+b}$ is an integer if and only if $\frac{a}{d}+\frac{b}{d} \mid d$, where $d=\gcd(a,b)$. It is true if $a=b$.
If $\frac{a}{d}+\frac{b}{d} \mid d$, then
$$\frac{ab}{a+b}=\frac{a}{d} \cdot \frac{b}{d}\cdot \frac{d}{\frac{a}{d}+\frac{b}{d}}$$
is a product of three positive integers and hence is an integer.
Conversely, i.e., $\frac{ab}{a+b}$ is an integer. Let $b=\frac{p}{q}\cdot a$ with $\gcd(p,q)=1$ where $p=\frac{b}{d}$, $q=\frac{a}{d}$ and $d=\gcd(a,b)$. Then
$$\frac{ab}{a+b}=\frac{a\cdot \frac{p}{q}\cdot a }{(1+\frac{p}{q})a}=\frac{ap}{p+q}$$
Since $\gcd(p,p+q)=\gcd(p,q)=1$, it has $p+q \mid a$. Similarly, it has $p+q \mid b$. As a result, $p+q \mid \gcd(a,b)$, i.e. $\frac{a}{d}+\frac{b}{d} \mid d$.
It completes the proof.
|
Let $\gcd(a,b)=d$, $a=du$ and $b=dv$.
Thus, $\gcd(u,v)=1$ and $$\frac{ab}{a+b}=\frac{ab+b^2-b^2}{a+b}=b-\frac{dv^2}{u+v},$$ which says $$d=k(u+v)$$ and we obtain:
$$(a,b)=(k(u^2+uv),k(v^2+uv)),$$ where $u,$ $v$ and $k$ are integers, $k\neq0$ and $\gcd(u,v)=1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3668069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
Proving $(1+x)f'(x)=a f(x)$ where $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$ A function $f$ is defined for $-1<x<1$ by $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$. Here $a$ is a real number which is neither zero nor a positive integer. Prove that $(1+x)f'(x)=a f(x)$.
I took the derivative of $f(x)$ which is $\sum _{n=0}^\infty {a\choose n} nx^{n-1}$. And then multiplied $(x+1)$ to that and didn't find it is $a f(x)$ then I gave up and looked at the book's answer.
And in book it says:
$$f'(x)=\sum _{n=0}^\infty {a\choose n+1} (n+1)x^{n}$$.
Now
" $\frac {n a(a-1)...(a-n+1)}{n!}+\frac{(n+1)a(a-1)...(a-n)}{(n+1)!}=\frac{a(a-1)...(a-n+1)} {(n-1)!}(1+\frac{(a-n)}{n})$.
So $(1+x)f'(x)=a f(x)$." This is the part where I don't understand. I don't know how after
writing this $\frac {n a(a-1)...(a-n+1)}{n!}+\frac{(n+1)a(a-1)...(a-n)}{(n+1)!}=\frac{a(a-1)...(a-n+1)} {(n-1)!}(1+\frac{(a-n)}{n})$ you can know that $(1+x)f'(x)=a f(x)$ is true. Can anyone explain?
|
For problems like these, I like to write out the first few terms in order to really understand what I've got.
We are working with $f$ defined by
$$f(x) = \binom a0 + \binom a1 x + \binom a2 x^2 + \cdots.$$
Your definition of $f'(x)$ is correct, except that the series should start at $n=1$ since the derivative of $\binom a0$ is simply $0$. Then,
$$f'(x) = 1\binom a1 + 2\binom a2 x + 3\binom a3 x^2 + \cdots = \sum_{n=1}^\infty n\binom{a}{n} x^{n-1}.$$
Now, in general, we have that
$$n \binom an = n \cdot \frac{a!}{n! (a-n)!} = n \cdot \frac{a \cdot (a-1)!}{n \cdot (n-1)! (a-n)!} = a \cdot \frac{(a-1)!}{(n-1)! (a-1 - (n-1))!} = a \binom{a-1}{n-1}.$$
Substituting this into our definition of $f'(x)$ shows that
$$f'(x) = a\binom {a-1}0 + a\binom {a-1}1 x + a\binom {a-1}2 x^2 + \cdots = \sum_{n=1}^\infty a\binom{a-1}{n-1} x^{n-1} = \sum_{n=0}^\infty a\binom{a-1}{n} x^{n}.$$
The last step above stems from the transformation $n \mapsto n+1$.
Now, there is another identity about binomials which is well-known, namely Pascal's Identity, which states that
$$\binom{a-1}n + \binom{a-1}{n-1} = \binom an.$$
Note that
$$\begin{align*}
f'(x) + xf'(x) &= \left(a\binom {a-1}0 + a\binom {a-1}1 x + a\binom {a-1}2 x^2 + \cdots\right) \\& \quad+ \left(a\binom {a-1}0x + a\binom {a-1}1 x^2 + \cdots\right), \\
&= a\binom {a-1}0 + \left(a\binom {a-1}0 + a\binom{a-1}1\right)x + \left(a\binom {a-1}1 + a\binom{a-1}2\right) x^2 + \cdots, \\
&= a\left(\binom {a-1}0 + \binom{a}1x + \binom{a}2 x^2 + \cdots\right), \\
&= af(x).
\end{align*}$$
In the above, we have $\binom{a-1}0 = \binom a0 = 1$, so our reasoning is valid. Hence, $f'(x) + xf'(x) = af(x)$, or
$$(1+x)f'(x) = af(x).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3669436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\lim_{n\to\infty} \lvert x_n \rvert ^\frac{1}{n}$ when $x_{n+1}=3x_n + \frac{2}{x_n^2},x_1=1$ $\lim_{n\to\infty} \lvert x_n \rvert ^\frac{1}{n}$ when $x_{n+1}=3x_n + \frac{2}{x_n^2}, x_1=1$
I figured out that $x_n$ increases as n approaches infinity.
However, the power approaches 0 as n approaches infinity.
So, I thought the limit does not approach 0 because the base approaches infinity while the power approaches 0.
Any hint that I can get?
|
Assuming that
$x_0 > 0$,
$x_{n+1}
=3x_n + \frac{2}{x_n^2}
\gt 3x_n
$
so
$x_n > 3^n x_0$.
Also
$x_{n+1}
=3x_n + \frac{2}{x_n^2}
\lt 3x_n+\frac{2}{(3^nx_0)^2}
= 3x_n+\frac{2}{9^nx_0^2}
$
so
$\dfrac{x_{n+1}}{3^{n+1}}
\lt \dfrac{x_n}{3^n}+\frac{2}{27^nx_0^2}
$.
Letting
$y_n = \dfrac{x_n}{3^n}$,
$y_{n+1}-y_n
\lt \frac{2}{27^nx_0^2}
$.
Summing
$y_m-y_0
=\sum_{n=0}^{m-1}(y_{n+1}-y_n)
\lt \sum_{n=0}^{m-1} \frac{2}{27^nx_0^2}
=\frac{2}{x_0^2} \sum_{n=0}^{m-1} \frac{1}{27^n}
\lt \frac{2}{x_0^2}\dfrac{1}{1-1/27}
= \frac{27}{13x_0^2}
=c$
so
$y_m
\lt y_0+c
$
or
$\dfrac{x_m}{3^m}
\lt y_0+c
$
or
$\dfrac{x_m^{1/m}}{3}
\lt (y_0+c)^{1/m}
\to 1
$.
Therefore
$\lim_{m \to \infty} x_m^{1/m}
=3
$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3671525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Finding the area of $\triangle ABC$ with $B$ and $C$ lying on the line $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line, $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, –1, 2)$, is: $5 \sqrt{17} /\sqrt{34}/ 6 / 2\sqrt{34}?$
My attempt: Let $B$ be $(3k+2,1,4k)$ and $C$ be $(3l+2,1,4l)$. So, using distance formula between $B$ and $C$, I get $k-l=1$. Now, using determinant, where the first row is coordinates of $A$, the second is of $B$, and the third $C$, I get area to be $5$. But the answer is given as $\sqrt{34}$. What is wrong in my method?
|
The problem with your method is that you're not measuring the area of triangle $\triangle ABC$, but the volume of the parallelepiped with vertices $0, A, B, C, A + B, A + C, B + C, A + B + C$. This is what the $3 \times 3$ determinant measures.
Instead, try computing
$$\frac{1}{2}\|(B - A) \times (C - A)\|.$$
The vectors $B - A$ and $C - A$ are sides of the triangle, and the length of their cross product is the area of the parallelogram formed by these vectors. Half of this quantity will be the area of $\triangle ABC$.
|
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"url": "https://math.stackexchange.com/questions/3674370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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|
continuity of function$f(x,y) = xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ in $(0,0)$? I try to find a value for $f(x,y)=xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}}$ in (0,0) which f to be continuous.
First we must show that the $\lim_{(x,y) \to (0,0)} xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ exits and then do the alignment $f(0,0) = \lim_{(x,y) \to (0,0)} xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}}$.
I try different ways to calculate the limit but I fail. I try to split the fraction in function to $ \dfrac {xy}{x^{2}+y^{2}}\times\dfrac{1}{x^{2}-y^{2}} $ or $xy \times \dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ to simplify the calculation of limit but I fail.
I even google this and it led to a website which I comment below and it had used the polar coordinates and I don't want to use polar coordinates.
Thanks.
|
For any given $\varepsilon>0$ we have
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&\le \frac12|x^2-y^2|\tag1\\\\
&\le \frac12(x^2+y^2)\\\\
&<\varepsilon
\end{align}$$
whenever $\sqrt{x^2+y^2}<\delta=\sqrt{2\varepsilon}$.
Note in arrving at $(1)$ we used the AM-GM inequality $x^2+y^2\ge 2|xy|$.
Alternatively, note that
$$\left|\frac{x^2-y^2}{x^2+y^2}\right|\le 1$$
Then,
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&\le |xy|\\\\
&\le \frac12(x^2+y^2)\\\\
&<\varepsilon
\end{align}$$
Finally, a transformation to polar coordinates $(x,y)\mapsto (\rho,\phi)$ reveals
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&=\rho^2|\cos(\phi)\sin(\phi)\,|\cos^2(\phi)-\sin^2(\phi)|\\\\
&\le \frac12 \rho^2\\\\
&<\varepsilon
\end{align}$$
whenever $\rho<\delta=\sqrt{2\varepsilon}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3676921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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|
Do we reduce the problem into the known $3^x+4^x=5^x $? I want to solve the following equations: \begin{align*}&(i) \ \ \ \ \ 3^x+4^x=5^x \\ &(ii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+3)^x, \ x>-1 \\ &(iii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+4)^x, \ x>-1\end{align*}
$$$$
I have done the following:
For (i):
We have that $$3^x+4^x=5^x\Rightarrow \left (\frac{3}{5}\right )^x+\left (\frac{4}{5}\right )^x=1$$ This equation of the form $f(x)=a^x+b^x$ with $0<a<b<1$. For this equation we have that: Since $0<a,b<1$ the function $a^x$ and $b^x$ and so also $f(x)$ are decreasing.
So there is at most one solution.
Since $a<b$ so $a^x<b^x$ for positive $x$ then we get $2a^x<f(x)<2b^x$. Both the lower and upper bound for $f(x)$ pass through 1, we know that f(x) must also pass through 1, so there is exactly one solution,.
We symbolize by $c$ the unique solution. From $2a^c<1<2b^c$ we get the inequality $-\log_a2<c<-\log_b2$. For $a=\frac{3}{5}$ and $b=\frac{4}{5}$ we get $1.35<c<3.10$. We consider the natural numers between $1.35$ and $3,10$.
We see that for $c=2$ the equation is satisfied, and so $x=2$ is the only solution.
Is everything correct?
As for (ii) and (iii) do we have to find an appropriate $x$ so that we can reduce it to a problem as in (i) ?
|
To answer your last question: no this would not work. Finding an appropriate $x$ would mean… that you have solved the equation.
You can try the same "monotonicity trick", but the function is more complicated:
$$\left (\frac{x+1}{x+3}\right )^x+\left (\frac{x+2}{x+3}\right )^x=1$$
Note that a function like
$$\left (\frac{x+1}{x+3}\right )^x=\left (1-\frac{2}{x+3}\right )^x$$ tends to a finite value ($e^{-2}$) and it is not completely obvious that it is decreasing.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3682819",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Prove that $\ f(x)=x^{3}$ is continuous at $\ x = -2$ using the delta-epsilon approach Prove that $\ f(x)=x^{3}$ is continuous at $\ x = -2$.
I am really struggling on how to choose an initial $\delta_1$ value, are we able to choose any value or are there specific values to target, in this case I chose $\delta_1=1$
The below is the work I had:
For any given $\epsilon > 0$, we want to find a $\delta >0$ such that $\ |f(x) - (-8)| < \epsilon$ whenever $\ 0<|x-(-2)|<\delta$ each simplifies to $\ |x^{3} +8| < \epsilon$ and $\ 0<|x+2|<\delta$
I expand the inequality so that I get
$$\ |(x+2)(x^{2}-2x+4)|<\epsilon$$
I temporarily set $\ |(x^{2}-2x+4)|<K$ so then the above inequality becomes:
$$\ K|x+2|<\epsilon\\
|x+2|<\frac{\epsilon}{K}$$
This is the part I am struggling with, now I need to find out K.
We know that $\ 0<|x+2|<\delta$, removing the absolute values results in $-\delta<x+2<\delta$ and getting to only x results in $-\delta-2<x<\delta-2$
If I make the assumption of $\delta_1=1$ then the resulting inequality is $-3<x<-1$
I also tried this with a $\delta=\frac{1}{2}$ just to make sure it fell inside these bounds.
Since we want to look at the upper bound here I plugged in -1 back into K and got the following
$$\
|x+2|<\frac{\epsilon}{|-1|^2+2|-1|+4|}\\
|x+2|<\frac{\epsilon}{7} \\
\delta=\frac{\epsilon}{7}\\
\delta=min(1,\frac{\epsilon}{7})
$$
Then I did the following to prove it
$$
|f(x) - (-8)| =|x^{3}+8|=|(x^{2}-2x+4)||x+2|\\
<|x+2|\cdot (|-1|^{2}+2|-1|+4)\\= |x+2|\cdot7\\
=7\delta \le 7 \cdot \frac{\epsilon}{7}=\epsilon
$$
|
Note that\begin{align}x^2-2x+4&=\bigl((x+2)-2\bigr)^2-2\bigl((x+2)-2\bigr)+4\\&=(x+2)^2-6(x+2)+12\end{align}and that therefore$$|x+2|<1\implies|x^2-2x+4|<1+6+12=19.$$So, take $\delta=\min\left\{1,\frac\varepsilon{19}\right\}$ and then$$|x+2|<\delta\implies\bigl|(x+2)(x^2-2x+4)\bigr|<\frac\varepsilon{19}\times19=\varepsilon.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3685485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
A sequence includes $a_p=\sqrt2$, $a_q=\sqrt3$, $a_r=\sqrt5$ for some $1\leq p
In a sequence $a_1, a_2,\dots$ of real numbers it is observed that $a_p=\sqrt{2}$, $a_q=\sqrt{3}$, and $a_r=\sqrt{5}$, where $1\leq p<q<r$ are positive integers. Then $a_p$, $a_q$, $a_r$ can be terms of
(A) an arithmetic progression
(B) a harmonic progression
(C) an arithmetic progression if and only if $p$, $q$, and $r$ are perfect squares
(D) neither an arithmetic progression nor a harmonic progression
I have tried using the definition of AP and argued that if the first option is true then $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{q-p}{r-q}$ is true which implies that the left hand side is an rational number. I don't know whether the left hand side is a rational number or not. I am stuck here. Also, I don't know how correct this approach is. How do I approach and solve this problem?
|
Suppose that $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ were rational.
Then, squaring the number would give the following:
$$\begin{align}
x^2
&=(\sqrt{15}-\sqrt{10}-\sqrt{6})(\sqrt{15}-\sqrt{10}-\sqrt{6})\\
&=15-\sqrt{150}-\sqrt{90}-\sqrt{150}+10+\sqrt{60}-\sqrt{90}+\sqrt{60}+6\\
&=31-2\sqrt{150}-2\sqrt{90}+2\sqrt{60}\\
&=31-2(5\sqrt{6})-2(3\sqrt{10})+2(2\sqrt{15})\\
&=31-10\sqrt{6}-6\sqrt{10}+4\sqrt{15}
\end{align}$$
Next, to get rid of the $\sqrt{6}$ term, one must multiply $x$ by $-10$ to cancel the $\sqrt{6}$ term from $x^2$. This gives $x^2-10x=(31-10\sqrt{6}-6\sqrt{10}+4\sqrt{15})+(-10\sqrt{15}+10\sqrt{10}+10\sqrt{6})=31+4\sqrt{10}-6\sqrt{15}$.
The above would then lead to the rationality of $31+4\sqrt{10}-6\sqrt{15}$. Since $31$ is clearly rational, we may ignore that term, leading to the rationality of $x^2-10x-31=4\sqrt{10}-6\sqrt{15}$.
Squaring one more time, we get the rationality of $(x^2-10x-31)^2=(4\sqrt{10}-6\sqrt{15})^2=160-48\sqrt{150}+540=700-48(5\sqrt{6})=700-240\sqrt{6}$.
Finally, subtracting $700$ and then dividing by $-240$ leads to the rationality of $\sqrt{6}$.
But we know that square roots of integers are either themselves integers or else irrational, leading to a contradiction. Hence, $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ must be irrational.
Alternatively, one could instead get rid of the $\sqrt{10}$ term by considering $x^2-6x$, or the $\sqrt{15}$ term by considering $x^2-4x$, and similar steps would show that the original term that one got rid of is rational.
|
{
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"url": "https://math.stackexchange.com/questions/3690033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Taylor Series of $f(x) =\frac{1}{x^2}$ I've found the Taylor Series for $f(x)=\frac{1}{x^2}$ centered at $a=-1$.
$f(-1)=1$, $f'(-1)=2$, $f"(-1)=6$, $f'''(-1)=24$, $f^4(-1)=120$
I used this formula to get each the first coefficients of the terms of the series: $c_n=\frac{f^n(a)}{n!}$
So I got the expansion:
$$f(x)=1+2(x+1)+3(x+1)^2+4(x+1)^3+5(x+1)^4+...$$
What is difficult for me is expressing this in summation notation. Is this right?
$$\sum\limits_{n=0}^\infty (n+1)(x+1)^n$$
|
You can verify it by computing this sum, because for $|x+1|<1$ this infinite sum is convergent:
$$
\frac{d}{dx}(x+1)\sum_{k=0}^{\infty} (x+1)^{k+1} = \frac{d}{dx}\bigg(-\frac{x+1}{x} - 1 \bigg) = \frac{d}{dx} \bigg( -\frac{1}{x} \bigg) = \frac{1}{x^2}
$$
|
{
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"url": "https://math.stackexchange.com/questions/3691493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& = \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\
& =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\
&=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$
I don't know how to continue from this.
Can someone help me?
|
Use AM-GM. $\frac{ad + bc}{2} \ge \sqrt{abcd}$. Squaring both sides, you get the answer. A tiny tip: If everything is positive, and you have an inequality, think about AM-GM once at least.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3691678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ . What is the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ ?
I have been stuck on this problem with no direction. I have tried multiplying the sequence with $x$ and trying out $S-Sx$ but have gotten nowhere. Any help?
Thanks.
|
Hint:
$$S=2x+6x^2+12x^3+20x^4+\cdots$$
$$S(1-x)=2x+(6-2)x^2+(12-6)x^3+(20-12)x^4+\cdots=V\text{(say)}$$
$$V(1-x)=2x+x^2(4-2)+x^3(6-4)+x^4(8-6)+\cdots=\dfrac{2x}{1-x}\iff|x|<1$$
Can you take it from here?
See : this
|
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}
|
If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$ If $z+\frac{1}{z}=2 \cos \theta,$ where $z$ is a complex number, show that
$$
\left|\frac{z^{2 n}-1}{z^{2 n}+1}\right|=|\tan n \theta|
$$
My Approach:
$$
\begin{array}{l}|\sin \theta|=\left|\sqrt{1-\cos ^{2} \theta}\right| \\ =\left|\sqrt{1-\left(\frac{z^{2}+1}{2z}\right)^{2}}\right| \\ =\left|\sqrt{\frac{4 z^{2}-z^{4}-2 z^{2}-1}{4 z^{2}}} \right|\\ =\left|\sqrt{\frac{-\left(z^{4}-2 z^{2}+1\right)}{4 z^{2}}}\right|=\left|\sqrt{\frac{\left(z^{2}-1\right)^{2}}{4 z^{2}}}\right| \\ =|\frac{z^{2}-1}{2 z}|\end{array}
$$
So $|\tan \theta|=\left|\frac{z^{2}-1}{z^{2}+1}\right|$ ( proven when $n = 1$)
Is there any way to prove directly by taking $n$?
|
You could express the equation directly as a quadratic:
$z^2 - 2 z \cos \theta + 1 = 0$
and use the quadratic formula, for further insight.
|
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|
Find all values at which the vector is linearly expressed through vectors I need to find all values $\lambda$ at which the vector $(7,-2, \lambda)$ is linearly expressed through vectors $(2,3,5), (3,7,8), (1,-6,1)$
I can't go to the next step after building a system:
$2x_{1} + 3x_{2} + x_{3} = 7$
$3x_{1} + 7x_{2} + 8x_{3} = -2$
$5x_{1} + 8x_{2} + x_{3} = \lambda$
|
Reduce the augmented matrix:
$$
\left[\begin{array}{ccc|c}
2&3&1&7\\
3&7&-6&-2\\
5&8&1&\lambda
\end{array}\right]
\sim
\left[\begin{array}{ccc|c}
1&0&5&11\\ 0&1&-3&-5\\ 0&0&0&15 - \lambda
\end{array}\right].
$$
Therefore, the only possible $\lambda$ that works is $\lambda = 15$, and it does in fact work: if $\lambda = 15$, then $(7,-2,15) = (-5z + 11)(2,3,5)+(3z-5)(3,7,8)+z(1,-6,1)$ where $z$ can be any number.
Why does this work? Reducing a matrix doesn't change the set of solutions, so this justifies the reduction. The system is inconsistent if the augmented matrix has a row of the form $\left[\begin{array}{cccc|c}0 & 0 & \cdots & 0 & x \end{array}\right]$ for some $x \neq 0$. So, we must have $\lambda = 15$. For $\lambda = 15$, the matrix has the last row as zero, with one free variable. For a vector $(x,y,z)$ to be a solution of the system means that $x + 5z = 11$ and $y -3z = -5$. This can be rewritten as $x = -5z + 11$ and $y = 3z - 5$. In other words, $z$ is free to be any number and $z$ determines $x$ and $y$ by those two equations. So any vector of the form $v = (-5z+11, 3z-5, z)$ is a solution. By definition of matrix-vector multiplication, $Av$ is a linear combination of the columns of $A$ with the coefficients coming from $v$. Therefore, we can write out the equation $Av = (7,-2,15)$ as $(-5z + 11)(2,3,5)+(3z-5)(3,7,8)+z(1,-6,1) = (7,-2,15)$.
|
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Measure of an angle "subtended by each pentagon" in a truncated icosahedron A soccer ball is a truncated icosahedron; it consists of 12 black regular pentagons and 20 white regular hexagons; the edge lengths of the pentagons and hexagons are congruent. What is the measure of the solid angle (in ste-radian) subtended by each pentagon of the soccer ball?
My Explanation:
If the edge length of the regular pentagon is $a$, according to mathworld.wolfram.com, the radius of the circle circumscribing the pentagon is
\begin{equation*}
R = \left(\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\right)a \approx 0.8507 a ,
\end{equation*}
and the radius of the circle inscribed in the pentagon is
\begin{equation*}
r = \left(\frac{1}{10}\sqrt{25 + 10\sqrt{5}}\right)a \approx 0.6882 a.
\end{equation*}
So, the length of the line segment between a vertex of a pentagon and the midpoint of the side of the same pentagon across from this vertex is
\begin{equation*}
R + r
= \left(\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\right)a + \left(\frac{1}{10}\sqrt{25 + 10\sqrt{5}}\right)a
\approx 1.5388 a.
\end{equation*}
According to wikipedia.org/wiki/Truncated_icosahedron, the radius of the sphere circumscribing the truncated icosahedron is
\begin{equation*}
\frac{a}{4} \sqrt{58 + 18\sqrt{5}} \approx 2.478 a.
\end{equation*}
An implementation of Heron's Formula yields the distance between the center of the sphere and the midpoint of a side of the pentagon of about $2.427a$. An implementation of the Law of Cosines shows that the angle with its vertex at the center of the icosahedron and its endpoints at a vertex of a pentagon and the midpoint of the side of the same pentagon across from this vertex is approximately $36.5^\circ$.
Is that correct? Is that what is meant by "the angle subtended by each pentagon"?
|
We know that the normal distance $H_p$ of each of 12 regular pentagonal faces from the center of a truncated icosahedron with edge length $a$ is given by following formula
$$H_p=\frac{a(2\sqrt5+1)}{\sqrt{10-2\sqrt5}}=\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}$$
The solid angle subtended by any regular polygonal plane with each side $a$ at any point lying on the perpendicular at a distance $h$ from the center is given by generalized formula from HCR's Theory of Polygon as follows
$$\omega=2\pi-2n\sin^{-1}\left(\frac{2h\sin\frac{\pi}{n}}{4h^2+a^2\cot^2\frac{\pi}{n}}\right)$$
Now, substituting the corresponding values $h=H_p=\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}$ and $n=5$ in above formula, the solid angle subtended by each regular pentagon $\omega_p$ at the center of truncated icosahedron (soccer ball) is
$$\omega_p=2\pi-2\cdot 5\sin^{-1}\left(\frac{2\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}\sin\frac{\pi}{5}}{4\left(\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}\right)^2+a^2\cot^2\frac{\pi}{n}}\right)$$
After simplifying, one will get
$$\boxed{\color{red}{\omega_p}=\color{blue}{2\pi-10\sin^{-1}\left(\frac{9-\sqrt5}{12}\right)\approx 0.295072263}\ sr}$$
|
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|
Number of ways of arranging 10 tiles in four colors such that any consecutive block of 5 tiles contain all four colors This problem is from Purple Comet High school contest, 2016.
Ten square tiles are placed in a row, and each can be painted with one of the four colors red (R), yellow (Y), blue (B), and white (W). Find the number of ways this can be done so that each block of five adjacent
tiles contains at least one tile of each color. That is, count the patterns RWBWYRRBWY and
WWBYRWYBWR but not RWBYYBWWRY because the five adjacent tiles colored BYYBW does not
include the color red.
It is easy to see that if a particular color to appear in any block of five tiles, there must be at least two tiles of that color and the two tiles must be at one of the following pairs of positions:
\begin{align*}
& 1,6 \\
& 2,6 \quad 2,7 \\
& 3,6 \quad 3,7 \quad 3,8 \\
& 4,6 \quad 4,7 \quad 4,8 \quad 4,9 \\
& 5,6 \quad 5,7 \quad 5,8 \quad 5,9 \quad 5,10 \\
\end{align*}
We need to choose 4 of the above pairs such that no two have the same first coordinate/second coordinate and assign the four colors one each to a pair. The remaining two tiles can be arbitrary color.
If we choose four from $(1,6), (2,7), (3,8), (4,9), (5,10)$, there are 24 ways to map the four colors and the number of colorings is
$5 \cdot 24 \cdot\left(\frac{4}{2} + \binom{4}{2} \cdot 2\right) = 1680$.
When we choose four pairs other than the above five, there are 26 ways to choose the four pairs and there are multiple countings in subtle ways:
For example, when we choose that pairs $(1,6), (3,7), (4,8), (5,9)$, the coloring $WWBRYWBRYY$ is counted 4 times: the other three occur from the pairs $((2,6), (3,7), (4,8), (5,9))$, $((2,6), (3,7), (4,8), (5,9))$, $((1,6), (3,7), (4,8), (5,10))$ and $((2,6), (3,7), (4,8), (5,10))$ and the colorings $WWBRYWBRYW, WWBRYWBRYB, WWBRYWBRYR$ are counted twice each.
I am not able to eliminate all the multiple countings. The answer is 7296.
|
There are four types of legitimate sequence of length $n$:
Type $A$: the last four colours are distinct,
Type $B$: the second from last colour is repeated later in the sequence,
Type $C$: the third from last colour is repeated later in the sequence,
Type $D$: the fourth from last colour is repeated later in the sequence.
A length $n$ sequence of type $A$ may be extended uniquely to one sequence each of type $A,B,C,D$ of length $n+1$. Sequences of the other types may be extended uniquely to just one sequence of length $n+1$ of the following types:$$B\to C\to D\to A.$$
Let $A_n,B_n,C_n,D_n$ denote the number of length $n$ sequences of type $A,B,C,D$ respectively, each divided by 24 (to keep the numbers small). From the above we have:
\begin{eqnarray*}
A_{n+1}&=&A_n+D_n\\
B_{n+1}&=&A_n\\
C_{n+1}&=&A_n+B_n\\
D_{n+1}&=&A_n+C_n\\
\end{eqnarray*}
Writing $A_n,B_n,C_n,D_n$ as a column vector and starting at $n=4$ we get:
$$
\left(\begin{array}{c}
1\\1\\2\\3
\end{array}
\right)
\to
\left(\begin{array}{c}
4\\1\\2\\3
\end{array}
\right)
\to
\left(\begin{array}{c}
7\\4\\5\\6
\end{array}
\right)
\to
\left(\begin{array}{c}
13\\7\\11\\12
\end{array}
\right)
\to
\left(\begin{array}{c}
25\\13\\20\\24
\end{array}
\right)
\to
\left(\begin{array}{c}
49\\25\\38\\45
\end{array}
\right)
\to
\left(\begin{array}{c}
94\\49\\74\\87
\end{array}
\right)
$$
Adding the values for $n=10$ and returning the factor of 24 we get:$$24(94+49+74+87)=24*304=7296.$$
This was quick to do by hand for $n=10$. However the general solution will be a linear combination of $n$'th powers of the roots of the polynomial $$t^4-t^3-t^2-t-1.$$
|
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|
Find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ I'm trying to find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ but nothing works for me.
I made the plot and it looks like this:
How could you start? Any recommendation?
|
Try checking where the two surfaces intersect. Solve the given equations for $z$ and set them equal:
$$\begin{align*}
1-x^2-y^2&=x^2+(y-1)^2\\
0&=2x^2+2y^2-2y\\
0&=x^2+y^2-y\\
0&=x^2+\left(y-\frac12\right)^2-\frac14\\
x^2+\left(y-\frac12\right)^2&=\frac1{2^2}
\end{align*}$$
which corresponds to the cylinder of radius $\frac12$ with cross sections parallel to the $(x,y)$ plane and centered over the point $\left(0,\frac12,0\right)$.
With this in mind, a change to cylindrical coordinates is the "obvious" move. Take
$$\begin{cases}x=r\cos\theta\\y=\frac12+r\sin\theta\\z=z\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$$
The solid (call it $S$) is then given by the set
$$S=\left\{(r,\theta,z)\mid0\le r\le\frac12,0\le\theta\le2\pi,r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta\right\}$$
where the last inequality follows from
$$\begin{align*}
&x^2+(y-1)^2\le z\le1-x^2-y^2\\
&\implies r^2\cos^2\theta+\left(r\sin\theta-\frac12\right)^2\le z\le1-r^2\cos^2\theta-\left(\frac12+r\sin\theta\right)^2\\
&\implies r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta
\end{align*}$$
The volume is then
$$\iiint_S\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\frac12}\int_{r^2-r\sin\theta+\frac14}^{\frac34-r^2-r\sin\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$
$$=\frac\pi{16}$$
|
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|
Proving limits in terms of epsilon delta questions $ f(x)= {x^3}-2x+1. $
We want to show that $ \lim_{x\to 2} ({x^3}-2x+1)= 5 $.
So here,
$ \lvert f(x)-f(2)\rvert = \lvert {x^3}-2x+1-5\rvert$ = $ \lvert {x^3}-2x-4\rvert $.
$ \lvert ({x^3}-2x)+(-4)\rvert \leq \lvert {x^3}-2x\rvert + \lvert -4\rvert = \lvert {x^3}-2x\rvert + 4 $, by triangle inequality.
I'm confused on what to do next for the $ \epsilon$ and $\delta$.
|
As you have computed, we have
$$|f(x)-f(2)|=|x^3-2x-4|.\tag{0}$$
Now, we have
$$x^3-2x-4=(x-2)(x^2+2x+2)=(x-2)[(x+1)^2+1].\tag{1}$$
If $|x-2|<1$, then $1<x<3$ and we can bound
$$0<(x+1)^2+1\leq 16+1=17.\tag{2}$$
Therefore, for any $\epsilon>0$, we can choose $\delta=\min\{ 1, \epsilon/17\}>0$
such that if $|x-2|<\delta$, then
$$|f(x)-f(2)|=|x^3-2x-4|~~\mbox{ by }(0)\\
=|x-2|\cdot|(x+1)^2+1|~~\mbox{ by }(1)\\
\leq \frac{\epsilon}{17}\cdot 17 ~~\mbox{ by }(2)\\
=\epsilon.
$$
|
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|
Spivak Calculus chapter 1 problem 13 proof critique Question paraphrased: Prove that the maximum between two numbers $x$ and $y$ is given by:
$$\max(x,y)=\frac{x+y+|y-x|}{2}$$
Proof: Let $x$ and $y$ be two arbitrary numbers. Then, one and only one holds true: $x \ge y$ or $x \le y$
If $x \ge y$, $x=x+(y-y)=(x-y)+y=|x-y|+y$
Also, $$x=\frac {x+x}{2}=\frac{|x-y|+y+x}{2}=\frac{|x-y|+x+y}{2}$$
If $y \ge x$, $y=y+(x-x)=(y-x)+y=|y-x|+y$
Also, $$y=\frac {y+y}{2}=\frac{|y-x|+x+y}{2}=\frac{|x-y|+x+y}{2}$$
Thus, $$\max(x,y)=\frac{x+y+|y-x|}{2}$$
How can I improve this proof? I am not very convinced with $y=\frac {y+y}{2}$ and $x=\frac {x+x}{2}$ technique because it feels as if I have done this to satisfy the answer.
|
A quicker way to the proof would be to show LHS=RHS. If $x\geq y,$ then LHS$=x$ and $|y-x|=-(y-x)=x-y$ which gives $$\text{RHS }=\frac{x+y+(x-y)}{2}=\frac{x+x}{2}=x.$$
The case $y> x$ is similar.
|
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Prove that $\exists !c \in \mathbb{R} \exists ! x \in \mathbb{R} (x^2 + 3x + c = 0)$ This is an exercise from Velleman's "How To Prove It". I am struggling with how to finish the final part of the uniqueness proof, so any hints would be appreciated!
*a. Prove that there is a unique real number $c$ such that there is a unique real number $x$ such that $x^2 + 3x + c = 0$.
Proof: Let $c = \frac{9}{4}$. Let $x = -\frac{3}{2}$. It follows that $x^2 + 3x + c = \frac{9}{4} - \frac{9}{2} + \frac{9}{4} = 0$. To show that $x$ is unique, let $y \in \mathbb{R}$ be arbitrary such that $y^2 + 3y + c = 0$. So $y^2 + 3y + \frac{9}{4} = 0$, and $(y+\frac{3}{2})^2 = 0$. It immediately follows that $y = -\frac{3}{2} = x$.
Now to show that $c$ is unique, let $d, e \in \mathbb{R}$ be arbitrary such that $\exists ! x \in \mathbb{R} (x^2 + 3x + d = 0)$ and $\exists ! x \in \mathbb{R} (x^2 + 3x + e = 0)$. This means that $\exists x \in \mathbb{R}(x^2 + 3x + d = 0)$, $\exists x \in \mathbb{R}(x^2 + 3x + e = 0)$, $\forall y \in \mathbb{R} \forall z \in \mathbb{R} ((y^2 + 3y +d = 0 \wedge z^2 + 3z + d = 0 )\rightarrow y =z)$, and $\forall y \in \mathbb{R} \forall z \in \mathbb{R} ((y^2 + 3y +e = 0 \wedge z^2 + 3z + e = 0 )\rightarrow y =z)$. (How can we show that $d = e$ to finish this?)
|
Your approach is complicated and I do not recommend it. Instead, consider the quadratic formula applied here:
$$
x = \frac{-3 \pm \sqrt{9 - 4c}}{2}
$$
Let $d$ be another number such that there exists a unique $x$ in which $x^2 + 3x + d = 0$. We note that $d \not> \frac{9}{4}$, as otherwise $x$ it not real (so no such real $x$ exists). We also see that $d \not< \frac{9}{4}$, as otherwise this $x$ is not unique. Thus, $d = \frac{9}{4} = c$ necessarily.
|
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Standard limits not working on this particular question I have this question and its been troubling me for so long. I try to use the standard trig. limits but that just fails everytime and I get the answer as $\infty$.
$$\lim_{x \to 0} \frac{2x+x\cos(x)-3\sin(x)}{x^4\sin(x)}$$
note: I have posted the question by already taking the lcm since i didnt think it would matter (hopefully).
I even checked out some limit calculators but all they show is l hopital rule which is very tiring, but they end up with the right answer which is $1/60$.
More than the answer im trying to figure out why standard limits fail here?
Thank you in advance
|
One way you can find this is by using the Taylor series expansions for $\cos x$ and $\sin x$.
Using the first few terms in each,
\begin{align*}
\frac{2x + x \cos x - 3 \sin x}{x^4 \cdot \sin x} &= \frac{2x + x(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)) - 3(x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^7))}{x^4(x + O(x^3))}\\
&= \frac{2x + x - \frac{x^3}{2} + \frac{x^5}{24} + O(x^7) - 3x + \frac{x^3}{2} - \frac{x^5}{40} + O(x^7)}{x^5 + O(x^7)}\\
&= \frac{\frac{x^5}{60} + O(x^7)}{x^5 + O(x^7)}\\
&= \frac{1}{60} + O(x^2)
\end{align*}
As $x \to 0$, we get that the limit is $\boxed{\frac{1}{60}}$.
|
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If $ f(x) = f(x+1) $ and $ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $ then $ f(x) = f( x+ \frac{1}{4}) $ I was trying to solve following problem.
Let $ f: \Bbb R \rightarrow \Bbb R $ be a twice differentiable function such that $ f(x) = f(x+1) $ for all $x$. If $$ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $$ holds for all x . Then prove that $ f(x) = f( x+ \frac{1}{4}) $ for all $x$ To sove this problem i replaced $x$ by $x + \frac {1}{4} $ in $ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $ so i got $ f''(x + \frac{1}{4}) + f(x + \frac{1}{4}) = \frac {1} { f( x+ 1) }$ =$ \frac{1}{ f( x) } $. I got stuck here how to proceed further. Thanx for your help.
|
Let me try it : We put $x=\ln(y)$ and $y\neq 0$
The equality :$$ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $$
Becomes :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{1}{f(\ln(y)+0.75)}$$
Now we assume $$f''(\ln(y))-f'(\ln(y))\neq 0\quad \forall y\neq 0$$
But :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{1}{f(\ln(y)+0.75)}=\frac{1}{f(\ln(y)+0.75+1)}=\frac{f''(\ln(y)+1)-f'(\ln(y)+1)}{y^2e^2}+f(\ln(y)+1)$$
Or :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{f''(\ln(y)+1)-f'(\ln(y)+1)}{y^2e^2}+f(\ln(y)+1)$$
Wich implies $$\frac{1}{y^2e^2}=\frac{1}{y^2}$$
Wich is a contradiction .
So we deduce that :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}=0 \quad \forall y\neq 0$$
Wich implies $f(\ln(y))=Constant$
|
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If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that
$$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$
I'm struggling with that problem. I've recognized that the given equation is one of a circle, namely; $$(x-2)^2 + (y+5)^2 = 9$$
And what needs to be proven can be derived to the form: $$7-3\sqrt{2}\le x-y\le 7 + 3\sqrt{2} \quad\Leftrightarrow\quad (x-y)^2 - 14(x-y) + 31 \le 0$$
But I really can't find the correlation between what's given and what has to be proven! Also, I'm wondering if this problem can be generalized for any circle parameters, something like an universal interval of $x-y$ for any circle. Any help is apprecatiated! Thank you in advance!
|
WLOG $x=2+3\cos t, y=-5+3\sin t$
$$x-y=7+3(\cos t-\sin t)$$
Now $$(\cos t-\sin t)^2+(\cos t+\sin t)^2=2$$
$$\implies(\cos t-\sin t)^2\le2$$
|
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"url": "https://math.stackexchange.com/questions/3705594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do I find the sum of a power series $\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}$? I have found the area of convergence to be $ x \in (-\infty, \infty)$, and this is how far I had gotten before getting stuck:
$$
\begin{aligned}
\sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}} &=\sum_{k=0}^{\infty} \frac{x^{k+3}}{(k+4) !(k+3) 3^{k+1}} \\
&=\sum_{k=0}^{\infty} \frac{x^{3} x^{k}}{(k+4) !(k+3) 3 \cdot 3^{k}} \\
&=\frac{x^{3}}{3} \sum_{k=0}^{\infty} \frac{1}{(k+4) !(k+3)}\left(\frac{x}{3}\right)^{k} \\
\text{substituting }u=& \frac{x}{3} \\
&=\frac{x^{3}}{3} \sum_{k=0}^{\infty} \frac{1}{(k+4) !(k+3)} u^{k}
\end{aligned}$$
I do not know how to proceed from here.
|
$$f(x)= \sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}}$$
$$f^{'}(x)= \sum_{n=3}^{\infty} \frac{nx^{n-1}}{(n+1) ! n 3^{n-2}}= \frac{3^3}{x^2}\sum_{n=3}^{\infty} \frac{x^{n+1}}{(n+1) ! 3^{n+1}} = \frac{3^3}{x^2} \left( e^{\frac{x}{3}}-1- \frac{x}{3}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3706197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
|
After writing our equation in the form $$\sqrt{(x+1)(x+7)}+\sqrt{(x+1)(x+2)}=\sqrt{(x+1)(6x+13)}$$ we obtain the domain:
$$(-\infty-7]\cup[-1,+\infty).$$
1. $x\geq-1.$
We see that $-1$ is a root and it remains to solve here
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$ or
$$\sqrt{(x+7)(x+2)}=2(x+1)$$ or
$$3x^2-x-10=0,$$ which gives also $x=2.$
*$x<-1$.
Thus, $x\leq-7$ and we need to solve
$$\sqrt{-x-7}+\sqrt{-x-2}=\sqrt{-6x-13}$$ or
$$\sqrt{(x+7)(x+2)}=-2(x+1),$$ which gives $$3x^2-x-10=0$$ again or
$$(x-2)(3x+5)=0$$ and we see that this equation has no roots for $x\leq-7.$
Id est, we got the answer: $$\{-1,2\}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3707727",
"timestamp": "2023-03-29T00:00:00",
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|
Let $a,b,c>0$ then prove this inequality holds $$\left( \frac{a}{b+c}+\frac{b}{c+a} \right)\left( \frac{b}{c+a}+\frac{c}{a+b} \right)\left( \frac{c}{a+b}+\frac{a}{b+c} \right)\ge 1+\frac{1}{8}{{\left( \frac{a-b}{a+b} \right)}^{2}}{{\left( \frac{b-c}{b+c} \right)}^{2}}{{\left( \frac{c-a}{c+a} \right)}^{2}}$$
I broke the whole expression then expressed it like this
\begin{align*}
LHS-RHS & =\frac{\sum \limits_{cyc}{c(10a^2b+10ab^2+11abc+c^3)(a-b)^2}+\sum \limits_{cyc}{ab[(a^2-6ab+ac+2b^2+5bc-3c^2)^2+(2a^2-6ab+5ac+b^2+bc-3c^2)^2]}}{5(b+c)^2(c+a)^2(a+b)^2} \ge {0}
\end{align*}
But this proof is very ugly. Do you have any better easy and beautiful proof
|
Let $\frac{a}{b+c}=\frac{x}{2},$ $\frac{b}{a+c}=\frac{y}{2}$ and $\frac{c}{a+b}=\frac{z}{2}.$
Thus, since $$\frac{x-y}{x+2}=\frac{\frac{2a}{b+c}-\frac{2b}{a+c}}{\frac{2a}{b+c}+2}=\frac{a-b}{a+c},$$ we need to prove that:
$$\prod_{cyc}(x+y)\geq8+\frac{\prod\limits_{cyc}(x-y)^2}{\prod\limits_{cyc}(x+2)^2}.$$
Also, $$2=\sum_{cyc}\frac{b+c}{a+b+c}=\sum_{cyc}\frac{1}{1+\frac{a}{b+c}}=\sum_{cyc}\frac{1}{1+\frac{x}{2}}=2\sum_{cyc}\frac{1}{x+2},$$ which gives
$$xy+xz+yz+xyz=4.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $u$ and we need to prove
$$9uv^2-w^3\geq8+\frac{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}{(w^3+6v^2+12u+8)^2}$$ or $f(u)\geq0,$ where
$$f(u)=(9uv^2-w^3-8)(w^3+6v^2+12u+8)^2-27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6).$$
But $$f'(u)=9v^2(w^3+6v^2+12u+8)^2+$$
$$+24(9uv^2-w^3-8)(w^3+6v^2+12u+8)-27(6uv^4-12u^2w^3+6v^2w^3)\geq$$
$$\geq9(v^2(w^3+6v^2+12u+8)^2-18uv^4)\geq0,$$ which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $u$, which by $uvw$ happens for equality case of two variables.
Let $x=y$.
Thus, $a=b$ and since in this case our inequality is obvious, we are done!
|
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"url": "https://math.stackexchange.com/questions/3709287",
"timestamp": "2023-03-29T00:00:00",
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|
Rotating $(3,3)$ by $45^\circ$ about center $(2,1)$
With a two dimensional surface, if we take $(2,1)$ as the center point and consider a transformation with a rotation axis around $45^\circ$, then point $(3,3)$ is transformed into point $(?,?)$
I am a bit stumped on how to do a $45^\circ$ rotation. I'd prefer answers that steer clear of using a rotation matrix.
|
While the rotation matrix that rotates $(x, y)$ by angle $\theta$ counterclockwise around a center $(c_x, c_y)$ may look scary,
$$\left[\begin{matrix} x^\prime \\ y^\prime \end{matrix}\right] =
\left[\begin{matrix} c_x \\ c_y \end{matrix}\right] + \left[\begin{matrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta \\
\end{matrix}\right] \left[ \begin{matrix} x - c_x \\ y - c_y \end{matrix} \right]$$
it is only a way to write
$$\left\lbrace ~ \begin{align}
x^\prime &= c_x + (\cos\theta)(x - c_x) - (\sin\theta)(y - c_y) \\
y^\prime &= c_y + (\sin\theta)(x - c_x) + (\cos\theta)(y - c_y) \\
\end{align} \right.$$
where $(x^\prime, y^\prime)$ are the coordinates of the rotated point.
When $\theta = 45°$, $\cos\theta = \sqrt{1/2}$ and $\sin\theta = \sqrt{1/2}$. So, to rotate point $(3, 3)$ by $45°$ around point $(2,1)$:
$$\left\lbrace ~ \begin{align}
x &= 2 + \sqrt{\frac{1}{2}}(3 - 2) - \sqrt{\frac{1}{2}}(3 - 1) \\
y &= 1 + \sqrt{\frac{1}{2}}(3 - 2) + \sqrt{\frac{1}{2}}(3 - 1) \\
\end{align} \right.$$
which is approximately $(1.293, 3.121)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3711932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving $\frac{1}{16} \sum \frac{(b+c)(c+a)}{ab} +\frac{9}{4} \geq 4\sum \frac{ab}{(b+c)(c+a)}$ For $a,b,c>0$. Prove: $$\frac{1}{16} \sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+\frac{9}{4} \geq 4\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}$$
My SOS's proof is:
It's equivalent to: $$\frac{1}{27}\sum\limits_{cyc} ab \left( a+b-8\,c \right) ^{2} \left( a+b-2\,c \right) ^{2}+\frac{26}{27}\sum\limits_{cyc}ab \left( a-b \right) ^{2} \left( a+b-2\,c \right) ^{2} +{\frac{50}{27}} \Big[\sum\limits_{cyc} a(b-c)^2\Big]^2 \geq 0$$
However, it's hard to find without computer.
So I'm looking for alternative solution without $uvw$. Thanks for a real lot!
|
We have$:$\begin{align*}\sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+36 - 64\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}
=\frac{\prod \,(a+b)\sum\limits_{cyc} a(a-b)(a-c)+\Big[\sum\limits_{cyc} c(a-b)^2\Big]^2}{a\,b\,c\,(a+b)\,(b+c)\,(c+a)}\geq 0\end{align*}
The last inequality is Schur degree $3,$ you can see many proof by SOS here.
Or this one is stronger$:$ here
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\int_0^{2\pi} \frac{x \cos x}{2 - \cos^2 x} dx$. I have to find the integral:
$$\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx$$
I rewrote it as:
$$\int_0^{2\pi} \frac{x \cos x}{1 + \sin^ 2 x} dx$$
But nothing further. I plugged it in a calculator and the result was $0$. I can see that the following relation holds:
$$f(-x) = -f(x)$$
For
$$ f: [0, 2\pi] \rightarrow \mathbb{R} \hspace{2cm} f(x) = \frac{x \cos x}{1 + \sin^2 x}$$
so that means that the function is an odd function. So if the interval $[0, 2\pi]$ is a symmetric interval for $f(x)$ then the result would be $0$.
I can see that the interval $[0, 2\pi]$ is symmetric for $\sin x$ and for $\cos x$, so it is not far fetched to believe it is symmetric for $\dfrac{\cos x}{1 + \sin^2 x}$, but wouldn't multiplying it with $x$ interfere with that symmetry? I don't see why $[0, 2\pi]$ is symmetric for the function
$$f(x) = \frac{x\cos x}{1 + \sin^2 x}$$
How come that $x$ doesn't ruin the symmetry?
|
$$I=\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx\tag 1$$
$$I=\int_0^{2\pi} \frac{(2\pi-x) \cos x}{2 - \cos^ 2 x} dx\tag 2$$
Adding (1) & (2) $$2I=\int_0^{2\pi} \frac{2\pi \cos x}{2 - \cos^ 2 x} dx$$
$$I=2\pi \int_0^{\pi} \frac{\cos x}{2-\cos^ 2 x} dx\tag 3$$
$$I=2\pi \int_0^{\pi} \frac{\cos (\pi-x)}{2-\cos^ 2(\pi- x)} dx$$
$$I=-2\pi \int_0^{\pi} \frac{\cos x}{2-\cos^ 2x} dx\tag 4$$
Adding (3) & (4), we get $$I=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3713569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed{n\cdot2^{n-2}}.$ Am I on the right track?
|
Spoilered answer:
$$\sum_{i} 2i \binom{n}{2i}=n\sum_{i} \binom{n-1}{2i-1}= n2^{n-2}$$
To get the second equaltiy note that $\sum_{i}\binom{n-1}{2i-1}=\sum_{i} \binom{n-1}{2i}$ as results from writing the binomial formula for $(1-1)^{n-1}$, and recall that
$$\sum_{i}\binom{n-1}{2i-1}+\sum_{i} \binom{n-1}{2i}= \sum_{i} \binom{n-1}{i}= 2^{n-1}$$
as follows from using binomial formula for $(1+1)^{n-1}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Showing this integral is always greater than $\pi/2$ Define $$I(b)= \int_1^\infty {(u+b)^{1/2} du\over (u^2+b)}$$
I want to show that $\mathit I(b)$ is greater than $\pi/2$, for every $\mathit b >0. $
|
The following is a more-or-less elementary approach:
Changing variables $u = bv$ in the original $I(b)$ integral, we get
$$ I(b) = \int_{1/b}^{\infty} \frac{b^{1/2} \sqrt{v+1}}{(b^{1/2}v)^2 +1} \, dv = \int_{1/b}^{\infty} \sqrt{v+1} \, \partial_v(\arctan(b^{1/2}v)-\pi/2) \, dv.$$
Integrating by parts, we get
$$I(b) = \sqrt{1/b + 1} (\pi/2 - \arctan(1/b^{1/2})) + \int_{1/b}^{\infty}\frac{(\pi/2 - \arctan(b^{1/2}v)}{2\sqrt{v+1}} \, dv.$$
As mentioned before, $\limsup_{b \to \infty} I(b) \ge \pi/2,$ which also follows from this last identity. In order to prove the result, it suffices to show that $I(b)$ is nonincreasing in $b>0.$ In order to do so, it suffices to prove that the function
$$g(t) = \sqrt{\frac{1}{t^2} + 1} (\pi/2 - \arctan(1/t)) + \int_{1/t^2}^{\infty} \frac{\pi/2 - \arctan(tv)}{2\sqrt{v+1}} \, dv $$
is nonincreasing. Before we do that, we remark that a function
$$\eta(t) = \int_{a(t)}^{\infty} F(x,t) \, dx$$
is, given $a,F$ are smooth, is differentiable, and its derivative is
$$\eta'(t) = -a'(t) F(a(t),t) + \int_{a(t)}^{\infty} \partial_t F(x,t) \, dx.$$
The derivative of $\int_{1/t^2}^{\infty} \frac{\pi/2 - \arctan(tx)}{2\sqrt{x+1}} \, dx$ is then
$$ 2 t^{-3} \frac{\pi/2 - \arctan(1/t)}{2\sqrt{\frac{1}{t^2} + 1}} - \int_{1/t^2}^{\infty} \frac{x}{2\sqrt{x+1}} \frac{1}{(xt)^2+1} \, dx,$$
while that of $\sqrt{\frac{1}{t^2}+1}(\pi/2 - \arctan(1/t))$ is
$$ -2 t^{-3} \frac{1}{2\sqrt{\frac{1}{t^2} + 1}} (\pi/2 - \arctan(1/t)) + \frac{1}{t(1+t^2)^{1/2}}.$$
Therefore,
$$g'(t) = \frac{1}{t} \frac{1}{(1+t^2)^{1/2}} -\int_{1/t^2}^{\infty} \frac{x}{2\sqrt{x+1}} \frac{1}{(xt)^2+1} \, dx. $$
Change once more variables $x = v/t^2$ in the integral on the right. This yields
$$g'(t) = \frac{1}{t(1+t^2)^{1/2}} - \frac{1}{t} \int_1^{\infty} \frac{v}{2\sqrt{v+t^2}} \frac{1}{v^2 + t^2} \, dv.$$
We then use the Cauchy-Schwarz inequality $\sqrt{v+t^2} \le (v^2 + t^2)^{1/4} (1+t^2)^{1/4}$ in the integral, which yields
$$g'(t) \le \frac{1}{t} \left( \frac{1}{(1+t^2)^{1/2}} -\frac{1}{(1+t^2)^{1/4}} \int_1^{\infty} \frac{v}{2(v^2+t^2)^{5/4}} \right) \, dv.$$
But then $-\frac{v}{2(v^2 + t^2)^{5/4}} = \partial_v ((v^2+t^2)^{-1/4}),$ so the integral evaluates explicitly to $\frac{1}{(1+t^2)^{1/4}}.$ Thus $g'(t) \le 0.$ Notice, moreover, that the inequality is strict (due to the Cauchy-Schwartz inequality) unless $t=0.$ Therefore, $g$ is strictly decreasing on $t>0,$ and so is $I(b),$ finishing the problem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake? I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so that:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = \int\limits_\gamma \frac{dz}{\left(10+\frac{3}{i}(z-\frac{1}{z})\right)iz} = \int\limits_\gamma \frac{dz}{\left(10iz+3z^2-3\right)}
$$
Roots of denominator are $-3i$ and $\frac{-i}{3}$ but since the winding number of $-3i$ is equal to 0 we have:
$$
\int\limits_\gamma \frac{dz}{10iz+3z^2-3} = 2 \pi\, i\, Res\left(f,\frac{-i}{3}\right)\cdot1
$$
Calculating residue:
$$
Res\left(f,\frac{-i}{3}\right) = \lim_{\large z \to \frac{-i}{3}} \frac{1}{(z+3i)}=\frac{3}{8i}
$$
summing up:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{3}{8i} = \frac{3\pi}{4}
$$
But wolfram says it is equal to $\dfrac{\pi}{4}$. Could you help me spot my mistake?
|
The mistake comes from calculating the residue.
We have $Res(f, -\frac{i}{3})=lim_{z\rightarrow\frac{-i}{3}}\frac{z+\frac{i}{3}}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{3z+i}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{1}{(z+3i)}=\frac{1}{3}\frac{1}{\frac{-i}{3}+3i}=\frac{1}{3}\frac{1}{\frac{8i}{3}}=\frac{1}{3}\frac{-3i}{8}=\frac{-i}{8}$.
|
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|
Solving $(D^2-1)y=e^x(1+x)^2$ I did like this:
$$\text{Let,} y=e^{mx} \text{ be a trial solution of } (D^2-1)y=0$$
$$\therefore \text{The auxiliary equation is } m^2-1=0$$
$$\therefore m=\pm1\\
\text{C.F.} = c_1e^x+c_2e^{-x}$$
$$\begin{align}
\text{P.I.}& =\frac{1}{D^2-1}e^x(1+x)^2\\
& =e^x\frac{1}{(D+1)^2-1}(1+2x+x^2)\\
& =e^x\frac{1}{D^2+2D}(1+2x+x^2)\\
& =\frac{e^x}{2}\left[\frac{1}{D}-\frac{1}{D+2}\right](1+2x+x^2)\\
& =\begin{aligned}
\frac{e^x}{2}\frac{1}{D}(1+2x+x^2)-\frac{e^x}{2}\frac{1}{D+2}(1+2x& +x^2)\\
\end{aligned}\\
& =\frac{e^x}{2}\left(x+x^2+\frac{x^3}{3}\right)-\frac{e^x}{4}\left(x^2+x+\frac{1}{2}\right)\\
& =e^x\left(\frac{x^3}{6}+\frac{x^2}{4}+\frac{x}{4}-\frac{1}{8}\right)\\
\end{align}$$
$$\therefore \text{The solution is}$$
$$y=c_1e^x+c_2e^{-x}+e^x\left(\frac{x^3}{6}+\frac{x^2}{4}+\frac{x}{4}-\frac{1}{8}\right)$$
But, in my book the answer is:
$$y=c_1e^x+c_2e^{-x}+\frac{xe^x}{12}(2x^2+3x+3)$$
$$\text{Please, check if there is any } \color{red}{mistake}.$$
|
The answers are essentially the same. $-\frac18e^x$ can be absorbed in $c_1e^x$.
|
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|
T,N,B and the curvature and the equation of the osculating plane at a given point Given a vector valued function defined by $r(t):=(2t)i+(t^2)j+(\ln t)k$,Find the unit vectors $\vec T,\vec N,\vec B$ at the point $P(2,1,0)$,then find the curvature and the equation of the osculating plane at the point.
Using the definitions we see that :
$$\vec T=\frac{dr\left(t\right)}{dt}=2i+2tj+\frac{1}{t}k$$At the point we have $2i+2j+k$
$$\vec N=\frac{\frac{d\vec T}{dt}}{\left|\frac{d\vec T}{dt}\right|}=\frac{1}{\sqrt{4+\frac{1}{t^{4}}}}(2j-\frac{1}{t^2}k)$$
At the point we have $\frac{1}{\sqrt{5}}(2j-k)$
$$\vec B=\vec T\times \vec N=\begin{vmatrix}
i & j & k\\
2 & 2t& \frac{1}{t}\\
0 & \frac{2}{\sqrt{4+\frac{1}{t^{4}}}}&\frac{-1}{t^2\sqrt{4+\frac{1}{t^{4}}}}
\end{vmatrix}$$
At the point we have $\frac{1}{\sqrt{5}}\left(-4i-2j+4k\right)$
For the osculating plane we need a normal to the plane and a point on the plane:
The normal vector is given by:
$$\frac{dr\left(t\right)}{dt}\times\frac{d^{2}r\left(t\right)}{dt^{2}}=\begin{vmatrix}
i & j & k\\
2 & 2t& \frac{1}{t}\\
0 & 2&\frac{-1}{t^2}
\end{vmatrix}$$
Setting $t=1$ gives the equation of the plane:
$$-4\left(x-2\right)-2\left(y-1\right)+4\left(z-0\right)=0$$
Or in a more compact form:
$$2x+y-2z=5$$
And the curvature at the point is $\sqrt{5}$.
Can someone please check the process?
|
These three vectors are unit vectors, so they must have magnitude $1$. To begin, we will find $\vec{\mathbf{r}}\,'(t)$.
$$\vec{\mathbf{r}}(t)=\left\langle2t,t^2,\ln(t)\right\rangle\implies\vec{\mathbf{r}}\,'(t)=\left\langle2,2t,\frac{1}{t}\right\rangle$$
We are assuming $t\gt0$. Next, we will find the magnitude of $\vec{\mathbf{r}}\,'(t)$.
$$\left\lVert\vec{\mathbf{r}}\,'(t)\right\rVert=\sqrt{4+4t^2+\frac{1}{t^2}}=\sqrt{\frac{4t^4+4t^2+1}{t^2}}=\frac{2t^2+1}{t}$$
We can now find $\hat{\mathbf{T}}(t)$. This will be the first of the three unit vectors we will find.
$$\hat{\mathbf{T}}(t)=\frac{\vec{\mathbf{r}}\,'(t)}{\left\lVert\vec{\mathbf{r}}\,'(t)\right\rVert}=\frac{t}{2t^2+1}\cdot\left\langle2,2t,\frac{1}{t}\right\rangle=\left\langle\frac{2t}{2t^2+1},\frac{2t^2}{2t^2+1},\frac{1}{2t^2+1}\right\rangle$$
We are interested in the point $P(2,1,0)$. We can clearly see that the vector function $\vec{\mathbf{r}}(t)=\langle2,1,0\rangle$ when $t=1$.
$$\hat{\mathbf{T}}(1)=\left\langle\frac{2}{3},\frac{2}{3},\frac{1}{3}\right\rangle$$
Next, Let us find $\hat{\mathbf{T}}\,'(t)$. This is the first step to finding the unit normal vector $\hat{\mathbf{N}}(t)$.
$$\hat{\mathbf{T}}\,'(t)=\left\langle-\frac{4t^2-2}{\left(2t^2+1\right)^2},\frac{4t}{\left(2t^2+1\right)^2},-\frac{4t}{\left(2t^2+1\right)^2}\right\rangle$$
Let us find its magnitude.
$$\left\lVert\hat{\mathbf{T}}\,'(t)\right\rVert=\sqrt{\frac{\left(4t^2-2\right)^2+32t^2}{\left(2t^2+1\right)^4}}=\sqrt{\frac{4\left(2t^2+1\right)^2}{\left(2t^2+1\right)^4}}=\frac{2}{2t^2+1}$$
Now, let us find $\hat{\mathbf{N}}(t)$.
$$\hat{\mathbf{N}}(t)=\frac{\hat{\mathbf{T}}\,'(t)}{\left\lVert\hat{\mathbf{T}}\,'(t)\right\rVert}=\frac{2t^2+1}{2}\cdot\left\langle-\frac{4t^2-2}{\left(2t^2+1\right)^2},\frac{4t}{\left(2t^2+1\right)^2},-\frac{4t}{\left(2t^2+1\right)^2}\right\rangle$$
This easily simplifies.
$$\hat{\mathbf{N}}(t)=\left\langle-\frac{2t^2-1}{2t^2+1},\frac{2t}{2t^2+1},-\frac{2t}{2t^2+1}\right\rangle$$
Let us calculate $\hat{\mathbf{N}}(1)$.
$$\hat{\mathbf{N}}(1)=\left\langle-\frac{1}{3},\frac{2}{3},-\frac{2}{3}\right\rangle$$
The final unit vector that we will find is $\hat{\mathbf{B}}(1)$.
$$\hat{\mathbf{B}}(1)=\hat{\mathbf{T}}(1)\times\hat{\mathbf{N}}(1)=\left\langle-\frac{2}{3},\frac{1}{3},\frac{2}{3}\right\rangle$$
Next, let us find the curvature of the space curve when $t=1$.
$$k(1)=\frac{\left\lVert\hat{\mathbf{T}}\,'(1)\right\rVert}{\left\lVert\vec{\mathbf{r}}\,'(1)\right\rVert}=\frac{2}{2(1)^2+1}\cdot\frac{1}{2(1)^2+1}=\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$$
The general equation of the osculating plane at the point $P(2,1,0)$ can be found with the unit binormal vector we found earlier as it is a normal vector to the plane [1].
$$ax+by+cz+d=0\implies-\frac{2}{3}x+\frac{1}{3}y+\frac{2}{3}z+d=0$$
We can easily solve for $d$ by using the point in question.
$$-\frac{2}{3}(2)+\frac{1}{3}(1)+\frac{2}{3}(0)+d=0\implies d=1$$
We now have the general equation of the osculating plane and have finished solving the problem in its entirety.
$$-\frac{2}{3}x+\frac{1}{3}y+\frac{2}{3}z+1=0$$
|
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|
Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.
My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}})+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{63}{-16}+\tan^{-1}\frac{63}{16}$$
$$=-\tan^{-1}\frac{63}{16}+\tan^{-1}\frac{63}{16}$$
$$=0$$
But the answer is given as $\pi$. What is my mistake?
|
Actually
$\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}=$
$=\pi+\tan^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}\right)$
We can notice that
$\frac{\pi}{2}<\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}<\pi$.
I am going to prove that
if $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{2}$ then $\alpha+\beta=\pi+\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right).$
Proof:
$\tan(\alpha+\beta-\pi)=\frac{\tan\alpha+\tan(\beta-\pi)}{1-\tan\alpha\cdot\tan(\beta-\pi)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}$
As $|\alpha+\beta-\pi|<\frac{\pi}{2}$ and the function tangent is invertible in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that
$\alpha+\beta-\pi=\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right)$, therefore:
$\alpha+\beta=\pi+\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right)$.
|
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|
Rewrite a term as sum of squares At the moment I am trying to rewrite this term:
$2x^2-2xy+5y^2-4x+2y+2$
as a sum of squares.
So I am trying to find an experession of $2x^2-2xy+5y^2-4x+2y+2=a^2+b^2+c^2$ (for example)
It looks easy, but everything I have tried failed so far. So I wonder if there is such expression. I know that this term is nonnegativ for every pair $(x,y)$.
One try might look like this:
$2x^2-2xy+5y^2-4x+2y+2=x^2-2xy+y^2+x^2+4y^2-4x+2y+2=(x-y)^2+(x-2)^2+4y^2+2y-2$
Here $4y^2+2y-2=4(y-\frac12)(y+1)$
Do you see a nice sequence of calculations?
Thanks in advance.
|
The second matrix identity below says
$$ \frac{1}{2} (2x-y-2)^2 + \frac{9}{2} y^2 $$
The method is discussed at
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 \\
1 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & - 1 & - 2 \\
- 1 & 5 & 1 \\
- 2 & 1 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 1 }{ 2 } & 1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & \frac{ 9 }{ 2 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 1 }{ 2 } & 1 & 0 \\
- 1 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & \frac{ 9 }{ 2 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 1 }{ 2 } & - 1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
2 & - 1 & - 2 \\
- 1 & 5 & 1 \\
- 2 & 1 & 2 \\
\end{array}
\right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the value of $p^2q + q^2r + r^2p$ for the given cubic equation?
If $p$, $q$, $r$ are the real roots of the equation $x^3-6x^2+3x+1=0$, then find the value of $p^2q + q^2r + r^2p$.
My Attempt:
I tried $(p+q+r)(pq+qr+rp)$ but couldn't really figure out what to do with the extra terms. The roots are also not trivial to find.
Any help would be appreciated.
|
As already noted, $A(p,q,r)=p^2 q + q^2 r + r^2 p$ is not a symmetric function of the roots, but if we couple it with $B(p,q,r)=q^2 p + r^2 q + p^2 r$ we have that
$$ A+B = pq(p+q)+qr(q+r)+pr(p+r) = (pq+qr+pr)(p+q+r)-3pqr$$
$$ AB = pqr(p^3+q^3+r^3)+3 p^2 q^2 r^2+\frac{1}{2}(p^3+q^3+r^3)^2-\frac{1}{2}(p^6+q^6+r^6). $$
By the Cayley-Hamilton and Jordan's theorems
$$ p^n+q^n+r^n = \text{Tr}\begin{pmatrix}0 &0 &-1 \\ 1 & 0 & -3 \\ 0 & 1 & 6\end{pmatrix}^n$$
so $p^3+q^3+r^3=159$ and $p^6+q^6+r^6=25113$. This leads to
$$ A+B = 6\cdot 3+3 = 21$$
$$ AB = (-1)\cdot 159 + 3(-1)^2 + \frac{1}{2}(159)^2-\frac{1}{2}(25113)=-72 $$
so the value of $p^2 q+q^2 r + r^2 p$ is either $\color{red}{-3}$ or $\color{red}{24}$.
All the roots are real since the discriminant equals $729=3^6$.
I will outline a trigonometric solutions since it might be interesting to know and apply in similar circumstances. The equation $x^3-6x^2+3x+1=0$ is equivalent to
$$ 6\sqrt{3}\,T_3\left(\frac{x-2}{2\sqrt{3}}\right) = 9,\qquad T_3\left(\frac{x-2}{2\sqrt{3}}\right) = \frac{\sqrt{3}}{2} $$
hence by setting $x=2+2\sqrt{3}\cos\theta $ we have that the roots are given by
$$ 2+2\sqrt{3}\cos 10^\circ,\quad 2+2\sqrt{3}\cos 130^\circ,\qquad 2+2\sqrt{3}\cos 250^\circ $$
which are not difficult to approximate numerically. We know in advance that both $A+B$ and $AB$ are integers, so we may recover $A+B=21$ and $AB=-72$ also from approximated roots.
|
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|
Evaluate $\int_{-\pi}^{\pi} \frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$ I came across this integral:$$\int_{-\pi}^{\pi} \frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
I tried $u=x+\pi$
$$\int_{-\pi}^{\pi} \frac{(x+\pi)^2}{1-\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
but had no success.
I also tried $u=-x$:
$$\int_{-\pi}^{\pi} \frac{x^2}{1-\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
Does this help? Any suggestions.
Answer is $\dfrac{\pi^3}{3}$ by the way.
|
Rationalizing the denominator makes it much easier to proceed as follows
$$\frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}}=\frac{(1+\sin x-\sqrt{1+\sin^2x})x^2}{(1+\sin{x}+\sqrt{1+\sin^2{x}})(1+\sin{x}-\sqrt{1+\sin^2{x}})}$$
$$=\frac{(1+\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}$$
$$\therefore I=\int_{-\pi}^{\pi}\frac{(1+\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}\ dx\tag 1$$
Now, substituting $x=-x$, we get
$$I=-\int_{-\pi}^{\pi}\frac{(1-\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}\ dx\tag 2$$
Adding (1) & (2), we get
$$2I=\int_{-\pi}^{\pi} \frac{2x^2\sin x}{2\sin x} \mathop{dx}$$
$$I=\int_{0}^{\pi} x^2dx$$$$=\color{blue}{\frac{\pi^3}{3}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:$ If $a+b+c \neq 0$ then $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c} = \frac {\left [\{xb+(1-x)c\} + \{xc + (1-x)a\} + \{xa+(1-x)b\} \right]} {a+b+c} =1.$$ Therefore $$x = \frac {a-c}{b-c} = \frac {b-a} {c-a} = \frac {c-b} {a-b}.$$
Comparing the first two expressions of $x$ and simplifying we get \begin{align*} a^2+b^2+c^2 - ab - bc -ca & = 0 \\ \implies (a-b)^2+(b-c)^2 +(c-a)^2 & = 0. \end{align*} Therefore we have $a=b=c.$ But then $x$ would be an indeterminant form. Does it imply that $a+b+c = 0$? How to proceed further? Any help will be highly appreciated. Thank you very much.
|
I don't think that's true. Let $x=1/2$. Then:
$$\frac {b+c} {2a} = \frac {c + a}{2b} = \frac {a+b}{2c}.$$
Any $a,b,c\ne 0$ such that $c=-a-b$ and $a+b\ne 0$ are a solution for that:
$$\frac {b-a-b} {2a} = \frac {-a-b + a}{2b} = \frac {a+b}{-2a-2b} \implies \frac {-a} {2a} = \frac {-b}{2b} = \frac {a+b}{-2(a+b)} \implies -\frac {1} {2} = -\frac {1}{2} = -\frac {1}{2}.$$
Moreover, if $a=b=c$, then any value of $x$ satisfies the identity:
$$\frac {xa+(1-x)a} {a} = \frac {xa + (1-x)a}{a} = \frac {xa+(1-x)a}{a} \implies \frac {1} {a} = \frac {1}{a} = \frac {1}{a} \implies 1=1=1.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $a^4+2a^3 b+2ab^3+b^4 ≥ 6a^2b^2$ Prove that for any positive real numbers $a$ and $b$,$$a^4 + 2a^3b + 2ab^3 + b^4 ≥ 6a^2b^2.$$I tried using Vieta's formula to show the product of the LHS is greater than the RHS, but I don't think I am correct.
|
Rewrite the inequality and employ AM-GM:
$$a^4+a^3b+a^3b+ab^3+ab^3+b^4 \ge 6\sqrt[6]{a^{12}b^{12}} =6a^2b^2$$
|
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|
If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.
|
Th proof by Bjkjdz. The inequality equivalent to
$${x^2}z + {y^2}z - {z^3} > 6z(z - x)(z - y),$$
or
$${x^2}(z - x) + {y^2}(z - y) > 6z(z - x)(z - y),$$
or
$$\dfrac{{{x^2}}}{{z - y}} + \dfrac{{{y^2}}}{{z - x}} > 6z.$$
But
$$\left\{ \begin{array}{l} {x^2} = \dfrac{{{x^3}}}{x} = \dfrac{{{z^3} - {y^3}}}{x} = \dfrac{{(z - y)({z^2} + yz + {y^2})}}{x} \\ {y^2} = \dfrac{{{y^3}}}{y} = \dfrac{{{z^3} - {x^3}}}{y} = \dfrac{{(z - x)({z^2} + zx + {x^2})}}{y} \\ \end{array} \right.$$
Thefore, we will show that
$$\dfrac{{{y^2} + yz+ {z^2}}}{x} + \dfrac{{{z^2} + zx + {x^2}}}{y} > 6z,$$
equivalent to
$$\left( {{x^3} + y{z^2} + {y^2}z} \right) + \left( {{y^3} + {x^2}z + x{z^2}} \right) > 6xyz.$$
Which is true by the AM-GM inequality.
|
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|
Determine basis for solution space when there are more unknowns than equations Let's say we have
$$3x_1+x_2+x_3+x_4=0$$
$$5x_1-x_2+x_3-x_4=0$$
Then I would have matrix $\begin{bmatrix} 3 &1&1&1\cr8&0&2&0 \end{bmatrix}$ which simplifies to $\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}$
then $4x_1+x_3=0$ and $-x_1+x_2+x_4=0$ after substitution we get $\frac{1}{4}x_3+x_2+x_4=0$
and I'm not sure what to do from here.
|
Take it to reduced row-echelon form. You have
$$\begin{bmatrix}-1&1&0&1\\4&0&1&0\end{bmatrix} \sim \begin{bmatrix}\color{red}1&0&\frac{1}{4}&0\\0&\color{red}1&\frac{1}{4}&1\end{bmatrix}.$$
From here, identify the pivots, coloured in red, and the columns without pivots in them. In this case, we have the third and fourth columns. This means it is safe to consider the third and fourth variables (i.e. $x_3$ and $x_4$) as free variables. The other two variables, $x_1$ and $x_2$, will have a unique solution given fixed values of $x_3$ and $x_4$.
Our equations are
\begin{align*}
x_1 + \frac{1}{4} x_3 &= 0 \\
x_2 + \frac{1}{4} x_3 + x_4 &= 0,
\end{align*}
or equivalently,
\begin{align*}
x_1 &= -\frac{1}{4} x_3 \\
x_2 &= -\frac{1}{4} x_3 - x_4 \\
x_3 &= x_3 \\
x_4 &= x_4.
\end{align*}
This gives us
\begin{align*}
(x_1, x_2, x_3, x_4) &= \left(-\frac{1}{4}x_3, -\frac{1}{4}x_3 - x_4, x_3, x_4\right) \\
&= \left(-\frac{1}{4}x_3, -\frac{1}{4}x_3, x_3, 0\right) + (0, -x_4, 0, x_4) \\
&= x_3\left(-\frac{1}{4}, -\frac{1}{4}, 1, 0\right) + x_4(0, -1, 0, 1).
\end{align*}
If you're up to bases yet, this tells you that
$$\left\{\left(-\frac{1}{4}, -\frac{1}{4}, 1, 0\right) + (0, -1, 0, 1)\right\}$$
forms a basis for your solution space.
|
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|
Find affine function to approximate local inverse Let
$$
f(x,y) = (x^2 - y^2, 2xy)
$$
then by the inverse function theorem $f$ is invertible locally at any point $(x,y) \neq (0,0)$ because
$$
\det Df(x,y) = \det \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix} = 4(x^2+y^2) > 0 \iff (x,y) \neq (0,0)
$$
it's also not globally invertible given the counterexample
$$
f(1,1) = (0,2) = f(-1,-1)
$$
But how would I find a linear affine function that approximates $f^{-1}$ near $f(1,-1)$?
|
As a function $f:\Bbb{C}\to\Bbb{C}$ it is imply
$$f(z)=z^2$$
with some algebra we can find an inverse
$$f^{-1}(z)=\sqrt{z} = r^{\frac{1}{2}}e^{\frac{i\theta}{2}} = r^{\frac{1}{2}}\left[\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right)\right]$$
And using the half angle identities gives us that the inverses are of the form
$$f^{-1}(x,y) = \left(\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},\pm\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\right)$$
which is four possible options. Since we want a solution around $(1,-1)$, we can narrow this down to just
$$f^{-1}(x,y) = \left(\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},-\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\right)$$
$f(1,-1) = (0,-2)$, which is where we center our linear approximation like so:
$$\frac{d}{dz}\sqrt{z} = \frac{1}{2\sqrt{z}} = \frac{1}{2(1-i)} = \frac{1+i}{4}$$
which means as a function $f:\Bbb{R}^2\to\Bbb{R}^2$
$$f^{-1}(x,y) \approx (1,-1) + \begin{pmatrix}\frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4}\end{pmatrix}\left(x,y+2\right)$$
|
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|
Evaluate $\cos \frac\pi7\cos \frac{3\pi}7\cos\frac{9\pi}7 $ I need help evaluating the expression
$$\cos \frac\pi7\cos \frac{3\pi}7\cos\frac{9\pi}7 $$
Can someone show the steps he or she used to arrive at the answer?
|
Note
\begin{align}
& \cos \frac\pi7\cos \frac{3\pi}7\cos\frac{9\pi}7\\
= &-\cos \frac{8\pi}7\cos \frac{4\pi}7\cos\frac{2\pi}7\\
=& -\cos \frac{8\pi}7\cos \frac{4\pi}7\sin\frac{4\pi}7\cdot\frac1{2\sin\frac{2\pi}7}\\
=& -\cos \frac{8\pi}7\sin\frac{8\pi}7\cdot\frac1{4\sin\frac{2\pi}7}\\
=& -\frac{\sin\frac{16\pi}7}{8\sin\frac{2\pi}7}=-\frac18
\end{align}
|
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|
Find all functions which satisfy $f(m^2+n^2)=f(m)^2+f(n)^2$ $\forall\space m,n\in\Bbb{N}$ and $f(1)>0$ Remark. Dear voters, this question is not a duplicate of the following old question. Please refrain from closing it for being a duplicate.
QUESTION: Find all functions $f:\Bbb{N}→\Bbb{N}$ which satisfy
$$f(m^2+n^2)=f(m)^2+f(n)^2\,,\forall\space m,n\in\Bbb{N}\,.$$
Here, $\mathbb{N}=\{1,2,3,\ldots\}$.
MY APPROACH: Set $m=n$.. we obtain- $$f(2n^2)=2f(n)^2$$ Studying this equation I found out that $f(x)=\sqrt{\frac{x}2}$ satisfies the condition. But I could not think any further. If my claim is true, how do I prove it? And how do I make sure that there aren't any other functions satisfying the property?
Thank you for your help in advance :)
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The only such function is $f(n)=n$.
The hard part is to show $f(1)=1$. After that, one just follows the answer to
Find all $f$ such that $f\left(m^{2}+n^{2}\right)=f(m)^{2}+f(n)^{2},$ plus some small values of $f(n)=n$ that will be proved below.
Let $f(1)=a$.
The given formula applies directly to any number that is the sum of two squares. Thus \begin{align*}
f(2)&=f(1+1)=2a^2\\
f(5)&=f(4+1)=(2a^2)^2+a^2=4a^4+a^2\\
f(8)&=f(4+4)=8a^4\\
f(10)&=f(9+1)=f(3)^2+a^2\\
f(13)&=f(9+4)=f(3)^2+4a^4
\end{align*}
However, some numbers are the sum of two squares in more than one way, and this is the key to the proof.
\begin{align}
7^2+1&=5^2+5^2,&f(7)^2&=2f(5)^2-a^2\\
2^2+11^2&=5^2+10^2,&f(11)^2&=f(10)^2+f(5)^2-f(2)^2\\
11^2+7^2&=13^2+1,&f(11)^2&=f(13)^2-f(7)^2+a^2\\
\hline
5^2+14^2&=10^2+11^2,&f(14)^2&=f(10)^2+f(11)^2-f(5)^2\\
6^2+13^2&=3^2+14^2,&f(6)^2&=f(14)^2+f(3)^2-f(13)^2\\
\end{align}
The first three equations involve the variables $f(3)^2$, $f(7)^2$, $f(11)^2$. Eliminating $f(11)^2$ and $f(7)^2$ gives $$a^2 (4 a^2-1)f(3)^2 = a^2 (2a^2+1)^2(4a^2-1)$$ so $f(3)=2a^2+1$.
We can now use the last two equations to find $f(14)^2$, then $$f(6)^2=2(1+8a^2+17a^4+8a^6-16a^8)$$
This forces $a=1$ since the polynomial takes negative values for $a>1$.
It then follows that $f(n)=n$ for small values of $n$, by substitution, and hence by induction.
|
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|
inverse of a funtion I'm trying to calculate the inverse of the function:
$$y= \tan \left( \cos^{-1} \frac{x}{x+2} \right) $$
In my opinion it is $y= \frac{2 \cos ( \tan^{-1} x )}{1-\cos ( \tan^{-1} x )} $ but in my book it is reported :
$f(x)\begin{cases} -2\left( \frac{\sqrt{1+x^2}-1}{x^2} \right) & x < 0\\
-1 & x = 0\\2 \left( \frac{\sqrt{1+x^2}+1}{x^2} \right) & x>0 \end{cases}$
Can someone explain me the passages?
|
Hint
$$y= \tan\big[ \arccos(f(x)) \big]=\frac{\sqrt{1-f(x)^2}}{f(x)}$$ could help you.
|
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|
Convergence of series with negative terms Among the series $\sum\limits_{n=1}^{\infty}\frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}, \sum\limits_{n=1}^{\infty} \frac{(-1)^{\lfloor\log n\rfloor}}{n}$, and $\sum\limits_{n=1}^{\infty}\frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$; which are convergent?
The leibniz test fails here as the series are not alternating and the series are not absolutely convergent. I think we have to consider separately the sum of positive and negative terms and then check for their convergence. Any hints? Thanks beforehand.
|
1. Denote by $S_m = \sum_{n=1}^{m} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}$ the $m$th partial sum. Define $N_0 = 1$ and let $N_k$ be the $k$-th index at which the sign of the sequence $\{(-1)^{\lfloor \sqrt{n}\rfloor}\}_{n=1}^{\infty}$ flips. In fact, we have $N_k = (k+1)^2$. Then $S_m$ lies between $S_{N_{k-1}-1}$ and $S_{N_k-1}$ whenever $N_{k-1} \leq m < N_k$ holds. So by applying the Squeezing Theorem, we can easily prove that $\{S_m\}_{m=1}^{\infty}$ converges if and only if $\{S_{N_k - 1}\}_{k=1}^{\infty}$ converges.
Now, for each given $K \geq 1$,
$$ S_{N_k-1}
= \sum_{n=1}^{N_K-1} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}
= \sum_{k=1}^{K} (-1)^{k} \sum_{n = N_{k-1}}^{N_k - 1} \frac{1}{n}. $$
Then by using that $ \left| \frac{1}{n} - \log\left(1+\frac{1}{n}\right) \right| \leq \frac{C}{n^2} $ for some $C > 0$ and $\log\left(\frac{N_k}{N_{k-1}}\right) = 2\log\left(1+\frac{1}{k}\right)$,
\begin{align*}
\left| \left( \sum_{n = N_{k-1}}^{N_k - 1} \frac{1}{n} \right) - \frac{2}{k} \right|
&\leq 2 \left| \log\left(1+\frac{1}{k}\right) - \frac{1}{k} \right| +
\sum_{n = N_{k-1}}^{N_k - 1} \left| \frac{1}{n} - \log\left(1+\frac{1}{n}\right) \right| \\
&\leq \frac{2C}{k^2} + \sum_{n = N_{k-1}}^{N_k - 1} \frac{C}{n^2}
\end{align*}
From this, we deduce:
$$ \sum_{k=1}^{\infty} \left| (-1)^k \left( \sum_{n = N_{k-1}}^{N_k - 1} \frac{1}{n} \right) - \frac{2}{k} \right| < \sum_{k=1}^{\infty} \frac{2C}{k^2} + \sum_{n=1}^{\infty} \frac{C}{n^2} < \infty $$
This reveals that $S_{N_k-1}$ is written as the sum of $\sum_{k=1}^{K} (-1)^k \frac{2}{k}$ and a convergent term, and hence $S_{N_k-1}$ converges by the alternating series test.
2. Now we consider $S_m = \sum_{n=1}^{m} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$. Let $N_k$ be as in the previous part. Then
$$ \left| S_{N_k-1} - S_{N_{k-1}-1} \right|
= \sum_{n = N_{k-1}}^{N_k - 1} \frac{1}{\sqrt{n}}
\geq \int_{N_{k-1}}^{N_k} \frac{\mathrm{d}x}{\sqrt{x}}
= 2\left( \sqrt{N_k} - \sqrt{N_{k-1}} \right) = 2. $$
Therefore $S_m$ does not converge.
3. Let $S_m = \sum_{n=1}^{m} \frac{(-1)^{\lfloor \log n \rfloor}}{n}$. Similarly as before, define $N_0 = 1$ and let $N_k$ be the $k$-th index at which the sign of the sequence $\{(-1)^{\lfloor \log n \rfloor}\}_{n=1}^{\infty}$ flips. It is easy to find that $N_k = \lceil e^k \rceil $. Then
$$ \left| S_{N_k-1} - S_{N_{k-1}-1} \right|
= \sum_{n = N_{k-1}}^{N_k - 1} \frac{1}{n}
\geq \int_{N_{k-1}}^{N_k} \frac{\mathrm{d}x}{x}
= \log N_k - \log N_{k-1} \xrightarrow[k\to\infty]{} \log e = 1, $$
and therefore $S_m$ does not converge.
|
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|
Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=\frac1{2\sqrt2}$$
$$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac1{2\sqrt2}$$
$$\alpha-\frac{\pi}{4}=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)$$
I calculated the value of $\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)\approx 20.705^\circ$,
so I got $\alpha\approx 45^\circ+20.705^\circ=65.705^\circ$
I calculated
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{1}{\sin^365.705^\circ}-\frac{1}{\cos^3 65.705^\circ}\approx -13.0373576$$
My question: Can I find the value of above trigonometric expression without using calculator? Please help me solve it by simpler method without solving for $\alpha$. Thanks
|
$$\sin\alpha-\cos\alpha=\frac{1}{2}$$
$$(\sin\alpha-\cos\alpha)^2=\frac{1}{4}$$,
$$\sin\alpha\cos\alpha=\frac38$$
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\left(\frac 1{\sin\alpha}-\frac 1{\cos\alpha}\right)\left(\frac 1{\sin^2\alpha}+\frac 1{\cos^2\alpha}+\frac 1{\sin\alpha \cos\alpha}\right)$$
$$=\left(\frac {-(\sin\alpha-\cos\alpha)}{\sin\alpha\cos\alpha}\right)\left(\frac {\cos^2\alpha+\sin^2\alpha+\sin\alpha \cos\alpha}{\sin^2\alpha\cos^2\alpha}\right)$$
$$=\frac {-(\sin\alpha-\cos\alpha)(1+\sin\alpha\cos\alpha)}{(\sin\alpha\cos\alpha)^3}$$
$$=\frac{-{1\over2}(1+ {3\over8})}{({3\over8})^3}$$
$$=\frac{-352}{27}$$
|
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|
An inequality involving real numbers Let $x,y,z$ be real numbers such that $xyz=-1$. Prove that
$$\sqrt[3/2]{\frac{3}{2}}\geq E:=\frac{4(x^3+y^3+z^3)}{(x^2+y^2+z^2)^2}$$
I tried to imitate an idea by River Li but it does not work. The idea is to find a function $f$ such that for all $x,y>0$,
$$E\leq f(x+y)$$
and then use calculus to show $\displaystyle f_{\max}=\sqrt[3/2]{\frac{3}{2}}$.
For instance, $$x^2+y^2\geq\frac{(x+y)^2}{2},\qquad z^2=\frac{1}{x^2y^2}\geq\frac{16}{(x+y)^4},\qquad z^3=-\frac{1}{x^3y^3}\leq-\frac{64}{(x+y)^6}$$
Unfortunately, there does not exist any function $g$ such that $x^3+y^3\leq g(x+y)$.
|
Let $x+y+z=3u$, $xy+xz+yz=3v^2,$ where $v^2$ can be negative, and $xyz=w^3$.
Thus, we need to prove that:
$$\left(\frac{3}{2}\right)^{\frac{2}{3}}(x^2+y^2+z^2)^2\geq-4\sqrt[3]{xyz}(x^3+y^3+z^3).$$
Now, we can assumme that $x^3+y^3+z^3>0,$ otherwise the inequality is obviously true.
But, if so $$x^3+y^3+z^3-3xyz>0$$ or
$$(x+y+z)\sum_{cyc}(x-y)^2>0,$$ which gives $u>0$.
Id est, we need to prove that:
$$3\left(\frac{3}{2}\right)^{\frac{2}{3}}(3u^2-2v^2)^2\geq-4w(9u^3-9uv^2+w^3)$$ or
$$12\left(\frac{3}{2}\right)^{\frac{2}{3}}v^4-\left(36uw+36\left(\frac{3}{2}\right)^{\frac{2}{3}}u^2\right)v^2+27\left(\frac{3}{2}\right)^{\frac{2}{3}}u^4+36u^3w+4w^4\geq0,$$ for which it's enough to prove that
$$324u^2\left(w+\left(\frac{3}{2}\right)^{\frac{2}{3}}u\right)^2-12\left(\frac{3}{2}\right)^{\frac{2}{3}}\left(27\left(\frac{3}{2}\right)^{\frac{2}{3}}u^4+36u^3w+4w^4\right)\leq0$$ or
$$18\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3w+27u^2w^2-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^4\leq0$$ or
$$18\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3+27u^2w-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^3\geq0,$$ which is true by AM-GM:
$$18\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^3+27u^2w\geq$$
$$\geq3\sqrt[3]{\left(9\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3\right)^2\cdot\left(-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^3\right)}+27u^2w=0$$ and we are done!
|
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|
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
|
You can use the partial fraction decomposition $$\dfrac{x^3}{(x^2+1)^3}=\dfrac{3i}{16}\left(\dfrac{1}{(x+i)^2}-\dfrac{1}{(x-i)^2}\right)+\dfrac{1}{8}\left(\dfrac{1}{(x+i)^3}+\dfrac{1}{(x-i)^3}\right)$$ to integrate this function. Then simplify the solution to get rid of complex unit $i.$
|
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|
Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
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Let $\frac{b}{a}=q$ and $1+q^2=2uq.$
Thus, $|u|\geq1$, $$a=-\frac{40}{1+q+q^2+q^3}$$ and $$a^2(1+q^2+q^4+q^6)=3280,$$ which gives
$$\frac{1600}{(1+q)^2(1+q^2)^2}\cdot(1+q^2)(1+q^4)=3280$$ or
$$20(1+q^4)=41(1+q)^2(1+q^2)$$ or
$$10(2u^2-1)=41(u+1)u$$ or
$$21u^2+41u+10=0,$$ which gives $$u=-\frac{5}{3},$$
$$1+q^2=-\frac{10}{3}q$$ and $$q\in\left\{-3,-\frac{1}{3}\right\}.$$
Can you end it now?
|
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|
How can I find the roots of the polynomial $12x^{4}+2x^3+10x^2+2x-2$? It's clear that I can divide by $2$, but I don't know what can I do with $$6x^{4}+x^3+5x^2+x-1$$
Is there any algorithm for it or a trick? I have found the roots by an online calculator but I don't know how can I calculate them. Thank you for your help.
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Here, I try to give a way of factorization, which isn't too hard to be noticed:
$6x^4+x^3+5x^2+x-1$
$=5x^4+x^3+5x^2+x+x^4-1$
$=x^3(5x+1)+x(5x+1)+(x^2+1)(x^2-1)$
$=x(x^2+1)(5x+1)+(x^2+1)(x^2-1)$
$=(x^2+1)(6x^2+x-1)$
$=(x^2+1)(3x-1)(2x+1)$
|
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|
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
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Apparently I can evaluate integrals but not powers. To the ones who stumble upon this, when something is raised to a power, it is multiplied to that not added to it, which was the mistake that I was making. So for my problem above
$$(\sec^{2} \theta)^{\frac{3}{2}} = ((\sec \theta)^{2})^{\frac{3}{2}} = (\sec \theta)^{2 \times \frac{3}{2}} = \sec^3 \theta $$
|
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|
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
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Here's an approach :
Let :
$$L=\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}$$
\begin{align}
\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}&=\lim_{x\to 0}\frac{\bigg(\sqrt{1+x\sin x}-\sqrt{\cos x}\bigg)\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}\\
&=\lim_{x\to 0}\frac{1+x\sin x-\cos x}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}\\
&=\frac{1}2\Bigg(\lim_{x\to 0}\frac{x\sin x}{x\tan x}+\lim_{x\to 0}\frac{1-\cos x}{x\tan x}\Bigg)\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{\cos x(1-\cos x)}{x\sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{1-\cos x}{x\sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{x(1-\cos x)}{x^2 \sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{x}{\sin x}\frac{1-\cos x}{x^2}\\
&=\frac{1}2+\frac{1}4=\frac{3}4
\end{align}
Hence :
$$L=\frac{3}4$$
|
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|
Evaluating $\int_{0}^{2\pi}\frac{1}{\cos^2(\theta)+1}\, d\theta$ What would be the values of this definite Integral?
$$\int_{0}^{2\pi}\frac{1}{\cos^2(\theta)+1}\, d\theta$$
So, I have solved this definite integral using the substitution method, taking $u=\tan(\theta)$.
After some simplification, the solution to the definite integral I get is follows:
$$\frac{1}{\sqrt{2}} \, \tan^{-1}\left(\frac{\tan{\theta}}{\sqrt{2}}\right) \Biggr|_{0}^{2\pi}$$
Whenever I am evaluating the above result at the limits of the integration I am getting an answer of $0$.
My simplification,
$$=\frac{1}{\sqrt{2}} \, \left[ \tan^{-1}\left(\frac{\tan{2\pi}}{\sqrt{2}}\right) - \tan^{-1}\left(\frac{\tan{0}}{\sqrt{2}}\right) \right]$$
$$=\frac{1}{\sqrt{2}} \, \big[ \tan^{-1}(0) - \tan^{-1}(0) \big]$$
$$=\frac{1}{\sqrt{2}} \, \big[ 0-0]$$
$$=0$$
However, using Mathematical/Integral calculator, the value of this Integral is
$$2\pi$$
I am probably doing something very silly, as I can't figure out what I am doing wrong. Any help would be appreciated. Thanks!
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Here is an alternate way with contour integration:
$$\begin{aligned}
J= \int_{0}^{2\pi} \frac{d\theta}{\cos^2 \theta+1} &=
\oint_{|z|=1} \frac{\frac{dz}{iz}}{ \left[ \frac{(z+z^{-1})^2}{4}+1 \right]}\\
&= \frac{4}{i}\oint \frac{ z \, dz}{ z^4 + 6 z^2 +1 }\\
\end{aligned}$$
Take the integral in the positive (counter-clockwise) direction.
There are four roots to the denominator of the integrand:
$$z_k \in \left\{ \mp i \sqrt{3 \mp 2\sqrt{2}} \right\}, \quad k=1,\cdots,4 $$
The roots inside the circle are
$z_1 = - i \sqrt{3 - 2\sqrt{2}} $ and $z_2 = i \sqrt{3 - 2\sqrt{2}} $.
$$J=2\pi i \cdot \frac{4}{i} \left[ \text{Res}_{z=z_1} \frac{z}{z^4+6z^2+1}+\text{Res}_{z=z_2} \frac{z}{z^4+6z^2+1}\right]
= 8\pi \left[\frac{1}{8\sqrt{2}}+ \frac{1}{8\sqrt{2}} \right]=\sqrt{2}\pi.$$
Even easier is to recognize that the substitution $w=z^2$ inside the integral leads to
$$J=\frac{4}{i} \oint_{|w|=1} \frac{dw}{w^2+6w+1}$$
We either have to wind around the circle twice in the $w$-plane or multiply by $2$ so that the constant out in front is again ${4}/{i}$.
and so
$$J=8\pi \text{ Res}_{z=2\sqrt{2}-3} \, \frac{1}{w^2+6w+1}=8\pi\cdot \frac{1}{4\sqrt{2}}=\sqrt{2}\pi.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
|
Substitute $t=x-2$
\begin{align}
\int \dfrac{dx}{(x^2-4x+13)^2}
& =\int \dfrac{dt}{(t^2+9)^2}= \int \frac1{18t}d\left( \frac{t^2}{t^2+9}\right)\\
&= \frac t{18(t^2+9)}+\frac1{18}\int \frac{dt}{t^2+9}\\
&= \frac t{18(t^2+9)}+\frac1{54}\tan^{-1}\frac t3+C
\end{align}
|
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|
Find the matrix $A^{15}$. Let $I=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$ and $O=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$.
1.Let $A=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}$ and $ B=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}$. Find the value of $x$ which satisfies $AB=BA$.
$AB=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}\cdot\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}=\begin{pmatrix}
x+9 & 21 \\
3x+15 & 39 \\
\end{pmatrix}$
$BA=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}\cdot\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}=\begin{pmatrix}
x+9 & 3x+15 \\
21 & 39 \\
\end{pmatrix}$
.So that we get $3x+15=21 \Rightarrow x=2$
2.Let $A=\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}$ and $ B=\begin{pmatrix}
-2 & x \\
4 & y \\
\end{pmatrix}$.Find the values of x and y which satify $BA=O$.
$BA=\begin{pmatrix}
-2 & x \\
4 & y \\
\end{pmatrix}\cdot\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}=\begin{pmatrix}
-2+2x & 0 \\
4+2y & 8+4y \\
\end{pmatrix}=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$
S0 that we can get $x=1,y=-2$.
3.Let $A$ satisfying $A^2=A-I$. Find $A^{15}$.
Please help to show me about this. Thank you in advance!
|
You may proceed as follows:
From the given condition we have
$$A-A^2 = A(I-A) =I$$
It follows
\begin{eqnarray} A^{15}
& = & A(A^2)^7 \\
& = & A(A-I)^7 \\
& = & -(A-I)^6 \\
& = & -(A^2-2A + I)^3 \\
& = & -(-A)^3 \\
& = & A(A-I) \\
& = & -I
\end{eqnarray}
Another way would be seeing
$$A^3+I = (A+I)(A^2-A+I)=O \Rightarrow A^3 = -I$$
|
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|
Strange Cube Root Offense in an Inequality I don't know how to tackle the unusual cube root present in this inequality-
$1.$For real numbers $a,b,c > 0$ and $n\le3$ prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)\ge 3+n$$
Here is another question with the same lesser side (and of course I couldn't prove)-
$2.$Let $a, b, c$ be positive real numbers such that $a + b + c = ab + bc + ca$
and $n ≤ 3$. Prove that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
What I attempted was this-
$$\left(a+b+c\right)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\right)\ge \left(a+b+c\right)\left(3+n\right)$$
Avoiding the RHS for some time-
$$\left(a+b+c\right)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\right)\ge (a+b+c)^2+\frac{3n(a+b+c)}{a^2+b^2+c^2}$$
After this step I don't know where to use $a+b+c=ab+bc+ca$.
These are very basic. I need a solution using AM-GM Inequality.
Any help will be appreciated.
|
The first inequality for $n=3$.
By AM-GM $$\sum_{cyc}\frac{a}{b}+\frac{9\sqrt[3]{abc}}{a+b+c}=\frac{1}{3}\sum_{cyc}\left(\frac{2a}{b}+\frac{b}{c}\right)+\frac{9\sqrt[3]{abc}}{a+b+c}\geq$$
$$\geq\sum_{cyc}\sqrt[3]{\frac{a^2}{b^2}\cdot\frac{b}{c}}+\frac{9\sqrt[3]{abc}}{a+b+c}=\sum_{cyc}\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq$$
$$\geq6\sqrt[6]{\left(\frac{a+b+c}{3\sqrt[3]{abc}}\right)^3\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)^3}=6.$$
|
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|
find primes p,q,and r
Find all primes $p, q, r$ such that $pq+r$, $pq+r^2$, $qr+p$, $qr+p^2$, $rp+q$,
$rp+q^2$ are all primes.
|
We know that $pq+r>2$. If all of $p,q,r$ are odd, then $pq+r$ would be even, which would make is impossible for the number to be prime. Now, without loss of generality, let $p=2$. We then have $qr+2$ and $qr+4$ to be prime. We know that both these numbers are greater than $3$ and prime, thus cannot be divisible by $3$.
If $qr$ leaves remainder $1$ modulo $3$, then $qr+2$ would be divisible by $3$. If $qr$ leaves remainder $2$ modulo $3$, then $qr+4$ would be divisible by $3$. Thus, we must have $qr$ divisible by $3$. Without loss of generality, let $q=3$. Now, we need the following numbers to be prime:
$$r+6,r^2+6,2r+3,2r+9,3r+2,3r+4$$
All of the above primes must be greater than $5$, and thus, none of them are divisible by $5$. Consider the possible remainders when $r$ is divided by $5$. If $r \equiv 1 \pmod{5}$, then $2r+3$ would be divisible by $5$. If $r \equiv 3 \pmod{5}$, then $2r+9$ would be divisible by $5$. If $r \equiv 2 \pmod{5}$, then $3r+4$ would be divisible by $5$. Finally, if $r \equiv 4 \pmod{5}$, then $r+6$ would be divisible by $5$. As none of these are the case, $r$ is divisible by $5$, and thus, $r=5$. We can verify that this solution works.
Thus, the possible triples are:
$$(p,q,r)=(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2)$$
|
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|
Questions about counting the number of triples arranged in geometric progression
Problem: Three tickets are chosen from a set of $100$ tickets numbered $1,2,3,\ldots,100$. Find the number of choices such that the numbers on the three tickets are in geometric progression.
Solution: Let $k, n \in \mathbb Z_+$ s.t. $n \ge 2, \ kn^2 \le 100.$ Then the number of possibilities for $k$ is $\lfloor \frac{100}{n^2} \rfloor$. Hence total number of $k, kn, kn^2$ with integer common ratio is $\sum_{n = 2}^{10}\lfloor \frac{100}{n^2} \rfloor = 25 + 11 + 6 + 4 + 2 + 2 + 1 + 1 + 1 = 53.$ Now the number of $k, k\frac mn, k\frac{m^2}{n^2}$ where $n > 1, \ n^2 \mid k, \ m > n, \ \gcd(m, n) = 1$ is given by $f(n) = \sum\lfloor \frac{100}{m^2} \rfloor $. So the number of GPs with non-integer common ratio $\sum_{n= 2}^9f(n)$. Thus the total number of GP's in question is $53 + \sum_{n= 2}^9f(n) = 105.$
My questions:
*
*When $n \in \mathbb Z_+$, we have the following
$k, 2k, 4k \iff 25, 50, 100$
$k, 3k, 9k \iff 11, 33, 99$
$k, 4k, 16k \iff 6, 24, 96$
$\ldots$
$k, 10k, 100k \iff 1, 10, 100$
That is, nine triples in geometric order. I am confused as to what exactly $\sum_{n = 2}^{10}\lfloor \frac{100}{n^2} \rfloor$ counts.
*When the common ratio is rational, we have $km^2 \le 100n^2$. Why, then, is the number of $k$ is given by $\lfloor \frac{100}{m^2} \rfloor$ instead of, say, $\lfloor \frac{100n^2}{m^2} \rfloor$?
|
Your first patterN, with a common factor of $2$, can start with any number from $1$ through $25$. $\lfloor \frac {100}{n^2} \rfloor$ is the number of progressions with common factor $n$. The third number is $n^2$ times the first, so the progression must start with a number small enough that the third term is no greater than $100$. That is all numbers from $1$ through $\lfloor \frac {100}{n^2} \rfloor$, where the brackets represent the floor function, so $\lfloor \frac {100}{9} \rfloor=11$
For the second, $k$ must be a multiple of $n^2$, so you should not have $n^2$ in the numerator.
|
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|
Existence of limit for sequence $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right)$ with initial values $x_0=5,x_1=10$ Let $x_0=5,x_1=10,$ and for all integers $n\ge2$ let $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right).$ By induction, we have $\forall m\in\mathbb Z_{\ge0}\enspace x_m>0,$ so we can avoid division by $0$ and the sequence is well-defined.
According to a Math GRE practice problem, the limit exists. How can we prove that? Note that, if we assume the limit exists, then we can show it equals $\sqrt8,$ but finding the value of the limit is not my goal here.
My work: We can compute $x_2=5.8,x_3=3.3,$ which are strictly between $4/3$ and $6,$ and then, assuming an inductive hypothesis, for all integers $n\ge4$ we have $4/3<x_{n-1}<6$ and $4/3<8/x_{n-2}<6,$ so that $4/3<x_n<6.$ We can probably compute more values of $x_n$ to get tighter bounds, but I don't see how to actually show convergence.
|
it is easy to show that $|x_{n}-A|<\epsilon$ for all the $x_{n}$ for a given $A$.
such that
$$A-\epsilon<x_{n}<A+\epsilon$$
$$-A-\epsilon<-x_{n}<-A+\epsilon$$
$$\frac{8}{A+\epsilon}<\frac{8}{x_{n}}<\frac{8}{A-\epsilon}$$
use equation $$x_{n}=\frac{1}{2}(x_{n-1}+\frac{8}{x_{n-2}})$$
we have $$x_{n}-x_{n-1}=\frac{1}{2}(-x_{n-1}+\frac{8}{x_{n-2}})$$
using inequatity (2)(3) we have
$$\frac{1}{2}(-A-\epsilon+\frac{8}{A+\epsilon})<x_{n}-x_{n-1}<\frac{1}{2}(-A+\epsilon+\frac{8}{A-\epsilon})$$
since $\epsilon<<A$ we use the expansion and keep the first term we get:
$$\frac{1}{1+\epsilon}=1-\epsilon+O(\epsilon^{2})$$
we get:
$$\frac{1}{2}(-A-\epsilon+\frac{8}{A(1+\frac{\epsilon}{A})})<x_{n}-x_{n-1}<\frac{1}{2}(-A+\epsilon+\frac{8}{A(1-\frac{\epsilon}{A})})$$
$$\frac{1}{2}(-A-\epsilon+\frac{8}{A}(1-\frac{\epsilon}{A}))<x_{n}-x_{n-1}<\frac{1}{2}(-A+\epsilon+\frac{8}{A}(1+\frac{\epsilon}{A}))$$
$$\frac{1}{2}(-A+\frac{8}{A}-\epsilon-8\frac{\epsilon}{A^{2}})<x_{n}-x_{n-1}<\frac{1}{2}(-A+\frac{8}{A}+\epsilon+8\frac{\epsilon}{A^{2}})$$
Also, we know that $|a+b|<|a|+|b|$ and $|a-b|<|a|+|b|$, such that
$$|x_{n}-x_{n-1}|<\frac{1}{2}(|-A+\frac{8}{A}|+|\epsilon+8\frac{\epsilon}{A^{2}}|)$$
if we set $A=\sqrt{8}$
we get:
$$|x_{n}-x_{n-1}|<\frac{1}{2}(|-\sqrt{8}+\frac{8}{\sqrt{8}}|+|\epsilon+8\frac{\epsilon}{8}|)$$
$$|x_{n}-x_{n-1}|<|\epsilon|$$
which means that $x_{n}$ converge to A
|
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|
How can I study the convergence of the improper integral $\int_{0}^{ \infty} \frac{\sin(x)}{x+1} \, \mathrm dx\,$? I need to study the convergence of the following improper integral:
$$\int_{0}^{\infty} \dfrac{\sin(x)}{x+1}\, \mathrm dx$$
I did the following:
$$ -1 \leq \sin(x) \leq 1 \\
\implies \dfrac{-1}{x+1} \leq \dfrac{\sin(x)}{x+1} \leq \dfrac{1}{x+1} \\
\implies \left|\dfrac{\sin(x)}{x+1}\right| \leq \dfrac{1}{x+1} \\
\implies \int_{0}^{\infty} \left|\dfrac{\sin(x)}{x+1}\right| \, \mathrm dx \leq \int_{0}^{\infty}\dfrac{1}{x+1}\, \mathrm dx = \infty $$
I planned to use the comparison criterion and then the absolute convergence criterion. However, the idea did not work for me.
|
Notice that
$$\int_0^\infty \frac{\sin x}{x+1}\,dx = \frac{-\cos x}{x+1}\Bigg|_0^\infty - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx = 1 - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx$$
and the last integral converges absolutely since
$$\int_0^\infty \frac{\left|\cos x\right|}{(x+1)^2}\,dx \le \int_0^\infty \frac{dx}{(x+1)^2} = \int_1^\infty \frac{dx}{x^2} < +\infty.$$
The original integral however does not converge absolutely. Namely, we have $$x \in \bigcup_{k \in \mathbb{N}_0} \left[\frac\pi6+k\pi,\frac{5\pi}6+k\pi\right] \implies \left|\sin x\right| \ge \frac12$$
so
$$\int_0^\infty \frac{\left|\sin x\right|}{x+1}\,dx \ge \frac12\sum_{k=0}^\infty \int_{\frac\pi6+k\pi}^{\frac{5\pi}6+k\pi} \frac{dx}{x+1} = \frac12\sum_{k=0}^\infty \ln \frac{\frac{5\pi}6+k\pi+1}{\frac\pi6+k\pi+1} = +\infty.$$
|
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|
Find the values of $k$ that satisfy $\gcd(a,b)=5$ Let $a$,$b$ and $k$ be integers such that
$$a=6k+4$$
$$b=11k+4$$
Find $k$ values that satisfy $\gcd(a,b)=5$
Note: $(a,b)$ are solutions to the equation : $11a-6b=20$
My try :
$\gcd(6k+4,11k+4)=\gcd(k-16;20)=5$
Possible values of $\gcd(a,b)$ are ${1,2,5,10,20}$ so we want $(k-16)$ to be divisible by $5$ but not by $10,20$
So we pose the following :
$k-16 ≡ 5 \pmod{20}$
$k ≡ 1 \pmod{20}$
This means that $k$'s ones number is $1$ so
$k≡ 1\pmod{20}$
$k=10p+1$ , $p$ is an integer.
|
Use Euclids algorithm
$\gcd( 6k + 4,11k+4) = \gcd(6k+4, 5k)=\gcd(k+4,5k)=\gcd(k+4, 5k-5(k+4))=\gcd(k+4, -20)=\gcd(k+4,20)$
so $5|k+4$ but $k+4$ is odd.
That is $k+4 \equiv 0 \pmod 5$ or $k\equiv 1 \pmod 5$. And $k+4\equiv 1 \pmod 2$ so $k\equiv 1\pmod 2$ and (we know by Chinese remainder theorem that there is one unique solution $\mod 10$) so $k\equiv 1 \pmod {10}$.
So as long as $k = 10m + 1$ we have
$\gcd(6(10m + 1)+4,11(10m+1)+4)=\gcd(60m+10, 110m + 15)=$
$\gcd(5(12m+2), 5(22m+3)) = 5\gcd(12m+2, 22m+3)=$
$5\gcd(12m+1, 10m+1)=5\gcd(2m,10m+1)=5\gcd(2m, 1)=5$
And if $k = 10m + i; i\ne 1;0\le i \le 9$ we have
$\gcd(6(10m + i)+4,11(10m+i)+4)=\gcd(60m+5i+(4+i), 110m + 10i+(i+4))$
$=\gcd(5[12m+i] + (4+i),5[22m+2i] + (4+i))$.
If that is to equal to $5$ we must have $5|i+4$ but $i\ne 1$ so $i$ must equal $6$
But $\gcd(5[12m+6] + (4+6),5[22m+12] + (4+6))=$
$\gcd(10[6m+3] + 10, 10[11m+6] + 10)=10\gcd(6m+4,11m+7) > 5$.
|
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|
Is the sum of an alternating series satisfies the following inequality
Question if $s$ is sum of the alternating series $\sum (-1)^{n+1}z_n$ and if $s_n$ is the nth partial sum then, $|s-s_{n}|≤z_{n+1}$
My attempt:
$|s-s_n|=|s-s_{n+1}+s_{n+1}-s_n|$
$$≤|s-s_{n+1}|+|s_{n+1}-s_n|$$
$$≤\epsilon + z_{n+1}$$
($s_n\rightarrow s$ so that, $s_{n+1}\rightarrow s$)
Now as $\epsilon >0$ is arbitrary, letting $\epsilon\rightarrow 0$, we havve $|s-s_{n}|≤z_{n+1}$
Is my attempt is correct? Please help....
Is there is any other way?
|
Here, two cases may arrise,one is when " $(z_n)$ is not monotone " and another is "$(z_n)$ is monotone ".
Case $1$, when $(z_n)$ is not monotone, then your statement is wrong.
Take the example, $z_n = \begin{cases}2^{-n} & \text{ n is even} \\ 3^{-n} & \text{ n is odd }\end{cases}$. ,
The sum, $\sum_{n=1}^{\infty} (-1)^{n+1} z_{n} = \frac{1}{24} $.
Clearly, $|s - s_1|=| \frac{1}{24} - \frac{1}{3} | = \frac{21}{72} > \frac{1}{4} = |z_2| $.
Case $ 2 $ , when $(z_n) $ is monotone,
Since, $ \sum_{k=1}^{\infty} (-1)^{k+1} z_{k} = s $,
Just do, $(s - s_{n}) = \sum_{k=1}^{\infty} (-1)^{k+1} z_{k} -\sum_{k=1}^{n} (-1)^{k+1} z_{k} =\pm z_{n+1} \mp( ( z_{n+2} - z_{n+3}) + (z_{n+4} - z_{n+5})+( z_{n+6} - z_{n+7}) + ............................ ) $ .
If $(z_{n}) $ is monotone,then , the sum = $( ( z_{n+2} - z_{n+3}) + (z_{n+4} - z_{n+5})+( z_{n+6} - z_{n+7}) + .........) $
must be positive ,
Now take $( ( z_{n+2} - z_{n+3}) + (z_{n+4} - z_{n+5})+( z_{n+6} - z_{n+7}) + .........) =A $
A is some positive real.
Now, $(s - s_{n}) = \pm z_{n+1} \mp A = \pm(z_{n+1} - A) $.
Hence, $|s - s_{n}| \le z_{n+1} $
|
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How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$$
At first, I assumed $x^2+y^2=z^2$. But, it is so complicated. Then, I assumed $x=r\cos\theta$ and $y=r\sin\theta$. But, I can't calculate the limits. Solving the equations I got three values of $\theta$ i.e. $\theta=0$, $\theta=\frac{\pi}{4}$ and $\theta=\frac{\pi}{2}$. I am just confused. Please help.
|
Hint:
$$\begin{align}\int\dfrac{x^2 - y^2}{\left(x^2 + y^2\right)^2}\,\mathrm dy&\equiv \int\dfrac{2x^2}{\left(x^2 + y^2\right)^2} - \dfrac{x^2 + y^2}{\left(x^2 + y^2\right)^2}\,\mathrm dy\\ &= 2x^2\int\dfrac1{\left(x^2 + y^2\right)^2}\,\mathrm dy - \int\dfrac1{x^2 + y^2}\,\mathrm dy\end{align}$$
The first integral can be solved with the help of a reduction formula. The second integral is straightforward to solve (substitute $t = \dfrac yx$ if you get stuck).
|
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|
System of congurences and the Chinese Remainder Theorem I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $N_1=35, N_2=21, N_3=15$.
From this, we get the following
\begin{align*}
35x_1 &\equiv 1 \pmod{3} \\
21x_2 &\equiv 1 \pmod{5} \\
15x_3 &\equiv 1 \pmod{7}
\end{align*}
and this will result in
\begin{align*}
2x_1 &\equiv 1 \pmod{3} \\
x_2 &\equiv 1 \pmod{5} \\
x_3 &\equiv 1 \pmod{7}
\end{align*}
so from CRT $x =x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3 = 2 \cdot 35 \cdot3 + 1 \cdot 21 \cdot 5 + 1 \cdot 15 \cdot7 = 420 $.
However $420$ doesn't seem to satisfy the given system, what would be the problem here?
|
Here's how I would solve this particular system.
$ x \equiv 4 \pmod{5}$ and $x \equiv 6 \pmod{7}$ means $x\equiv-1\pmod5$ and $x\equiv-1\pmod7$,
which means $x\equiv-1\pmod{35}$, i.e., $x\equiv34\pmod{35}$,
which means $x\equiv34, 69$, or $104\pmod{105}$.
By the Chinese remainder theorem, only one of these satisfies $x\equiv1\pmod3$.
|
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|
How can I prove that $\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$? $$\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$$
Have no clue how to prove it.
Possibly rewrite $\sin(7n)$ as $\frac{1}{2\sin(7)}\left(\cos(7n-7)-\cos(7n+7)\right)$.
But what next?
|
Here is a method which does not use complex numbers. Let $x=\sum_{n=0}^\infty\frac{\sin(7n)}{13^n}$ and $y=\sum_{n=0}^\infty\frac{\cos(7n)}{13^n}$.
Then, we have
\begin{align}
x&=\sum_{n=1}^\infty\frac{\sin(7n)}{13^n}=\sum_{n=0}^\infty\frac{\sin(7n+7)}{13^{n+1}}\\
&=\frac1{13}\sum_{n=0}^\infty\frac{\sin(7)\cos(7n)+\cos(7)\sin(7n)}{13^n}=\frac{\sin(7)}{13}y+\frac{\cos(7)}{13}x.
\end{align}
Similarly,
\begin{align}
y&=1+\sum_{n=1}^\infty\frac{\cos(7n)}{13^n}=1+\sum_{n=0}^\infty\frac{\cos(7n+7)}{13^{n+1}}\\
&=1+\frac1{13}\sum_{n=0}^\infty\frac{\cos(7n)\cos(7)-\sin(7n)\sin(7)}{13^n}=1+\frac{\cos(7)}{13}y-\frac{\sin(7)}{13}x.
\end{align}
Now, all that is left is to solve for $x$.
|
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|
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
|
A polynomial $f(x)$ is a multiple of $x^2-x-1$ (the characteristic polynomial of the Fibonacci sequence) iff $f(\varphi)=f(\bar{\varphi})=0$, with $\varphi=\frac{1+\sqrt{5}}{2}$, $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$ being algebraic conjugates. Since $\varphi^2=\varphi+1$ we have by induction $\varphi^k=F_{k}\varphi+F_{k-1}$ and the same holds for $\bar{\varphi}$. It follows that $ax^9+bx^8+1$ is a multiple of $x^2-x-1$ iff
$$\begin{eqnarray*} 0 &=& a\varphi^9+b\varphi^8+1 = a(34\varphi+21)+b(21\varphi+13)+1\\&=&(34a+21b)\varphi+(21a+13b+1)\end{eqnarray*}$$
and
$$ 0 = (34a+21b)\bar{\varphi}+(21a+13b+1). $$
It follows that we must have $21a+13b=-1$ and $34a+21b=0$, so $\color{red}{a=21,b=-34}$ works.
You may also invoke the more general identity
$$ (x^2-x-1)\sum_{k=0}^{n}(-1)^{k+1}F_k x^k = (-1)^{n+1}F_n x^{n+2}+(-1)^n F_{n+1} x^{n+1}-x.$$
|
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|
If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + c} \geq \frac {1}{3}$.
So I've been trying to solve this problem, and I've been trying to find a way to modify it into using AM-GM. The issue is that the $(a+c)(b+d) = 1$ is really throwing me off, as I haven't dealt with any inequalities that have used that as a condition yet (most other conditions I have seen go along the lines of $abcd = 1$ or something like that), and I'm not sure how exactly to deal with this inequality. Does anyone have any ideas?
|
Another way.
Since by AM-GM
$$1=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2,$$ we obtain $$a+b+c+d\geq2.$$
Now, let $a=kx$, $b=ky$, $c=kz$ and $d=kt$ such that $k>0$ and $x+y+z+t=4$.
Thus, $$k(x+y+z+t)\geq2,$$ which gives $$k\geq\frac{1}{2}.$$
But, $$\sum_{cyc}\frac{a^3}{b+c+d}=\sum_{cyc}\frac{k^2x^3}{y+z+t}\geq\frac{1}{4}\sum_{cyc}\frac{x^3}{y+z+t}$$ and it's enough to prove that
$$\sum_{cyc}\frac{x^3}{y+z+t}\geq\frac{4}{3}$$ or
$$\sum_{cyc}\left(\frac{x^3}{4-x}-\frac{1}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(3x^2+3x+4)}{4-x}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(3x^2+3x+4)}{4-x}-\frac{10}{3}(x-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(9x+28)}{4-x}\geq0$$ and we are done!
|
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|
The existence of a point in a geometric configuration Consider $ABC$ a right triangle in $A$ and let $M$ be the middle of the height from the vertex $A$. Construct $B'=f(B)$ and $C'=g(C)$, where $f$ is the reflection about line $CM$ and $g$ is the reflection about line $BM$. Decide when do lines $BC'$ and $CB'$ intersect? When $BC'\cap CB'={E}$, prove that $BE+CE=\frac{5}{3}BC$.
I tried to find a length or an angle that takes some values when the lines intersect each other and other values when the lines do not intersect. I do not get anything concrete.
For the second part I tried to apply bisector theorem but I didn't obtain anything.
|
Let $D$ be the feet of altitude from $A$ to $BC$. Then $AD=\frac{bc}a$, $DB=\frac{c^2}a$ so $\tan\angle MBC=\frac{b}{2c}$ and hence
$$
\tan\angle C'BC=\frac{4bc}{4c^2-b^2},
$$
and similarly
$$
\tan\angle B'CB=\frac{4bc}{4b^2-c^2}.
$$
The two lines $C'B$ and $B'C$ does not intersects iff $\angle C'BC+\angle B'CB=180^\circ$, i.e., $4b^2-c^2=-(4c^2-b^2)$, which is impossible. So the two lines always intersect.
Now use $\sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}$ and similar to calculate
$$
\sin\angle ECB=\frac{4bc}{4b^2+c^2},\quad
\sin\angle EBC=\frac{4bc}{4c^2+b^2},\quad
\sin\angle BEC=\frac{12a^2bc}{(4b^2+c^2)(4c^2+b^2)}.
$$
So sine rule gives
\begin{align*}
BE+CE&=\frac{a}{\sin\angle BEC}\cdot(\sin\angle ECB+\sin\angle EBC)\\
&=a\cdot\frac{(4b^2+c^2)(4c^2+b^2)}{12a^2bc}\left(\frac{4bc}{4b^2+c^2}+\frac{4bc}{4c^2+b^2}\right)\\
&=\frac{4abc}{12a^2bc}((4b^2+c^2)+(4c^2+b^2))\\
&=\frac{5(b^2+c^2)}{3a}=\frac53a
\end{align*}
as desired.
|
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|
Find the derivative of $f(x)= \int_{\sin x}^{\tan x} \sqrt{t^{2}+t+1}\, \mathrm d t$
Find the derivative of $$f(x)=\int_{\sin x}^{\tan x} \sqrt{t^{2}+t+1}\,
\mathrm d t$$ with respect to $x$
So from may understanding, I need to apply the fundamental theorem of calculus and then differentiate. I think the upper and lower limits are throwing me off.
|
Let $g(t)$ denote the integrand $\sqrt{t^2+t+1}$. On the one hand, the FTC guarantees
$$
\frac{d}{dx}\int_{\sin x}^{\tan x} g(t)\, dt
$$
$$
=g(\tan(x))\cdot (\tan(x))' - g(\sin(x))\cdot (\sin(x))'
$$
$$
=g(\tan(x))\cdot \sec^2(x) - g(\sin(x))\cdot \cos(x)
$$
$$
=\sqrt{\tan^{2} (x)+\tan(x)+1}\cdot \sec^2(x) - \sqrt{\sin^{2} (x)+\sin(x)+1}\cdot \cos(x)
$$Were we masochisitic, we could compute the antiderivative using the substitution $(t+1/2)^2= (3/4)\tan^2(\theta)$ (note this cannot always be done, which is part of the power of the FTC), back-substitute, and then differentiate to verify we get the same result.
$$
\int \sqrt{t^2+t+1}\,dt = \int \sqrt{(t+1/2)^2+3/4}\,dt
$$
$$
=\frac{1}{2} t\sqrt{t^2+t+1} +\frac{1}{4} \sqrt{t^2+t+1}+\frac{3}{8} \log \left(\frac{2
t+1}{\sqrt{3}}+\sqrt{\frac{1}{3} (2 t+1)^2+1}\right)
$$For instance, replacing $t$ with $\tan(x)$ at the upper limit and differentiating gives:
$$
\frac{d}{dx}\left(\frac{1}{2} \tan (x) \sqrt{\tan ^2(x)+\tan (x)+1}+\frac{1}{4} \sqrt{\tan ^2(x)+\tan
(x)+1}+\frac{3}{8} \log \left(\frac{2 \tan (x)+1}{\sqrt{3}}+\sqrt{\frac{1}{3} (2 \tan
(x)+1)^2+1}\right)\right)
$$
$$
=\frac{1}{2} \sqrt{\tan ^2(x)+\tan (x)+1} \sec ^2(x)+\frac{\tan (x) \left(\sec ^2(x)+2 \tan (x)
\sec ^2(x)\right)}{4 \sqrt{\tan ^2(x)+\tan (x)+1}}+\frac{\sec ^2(x)+2 \tan (x) \sec ^2(x)}{8
\sqrt{\tan ^2(x)+\tan (x)+1}}+\frac{3 \left(\frac{2 \sec ^2(x)}{\sqrt{3}}+\frac{2 (2 \tan
(x)+1) \sec ^2(x)}{3 \sqrt{\frac{1}{3} (2 \tan (x)+1)^2+1}}\right)}{8 \left(\frac{2 \tan
(x)+1}{\sqrt{3}}+\sqrt{\frac{1}{3} (2 \tan (x)+1)^2+1}\right)}$$
$$
=\sqrt{\tan^2(x)+\tan(x)+1}\cdot \sec^2(x),
$$as promised. If you want, you can try the lower limit.
|
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|
Show that by means of the transformation $w=\frac 1z$ the circle $C$ given by $|z-3|=5$ is mapped into the circle $|w+\frac{3}{16}|=\frac{5}{16}$
Show that by means of the transformation $w=\frac 1z$ the circle C given by $|z-3|=5$ is mapped into the circle $\left|w+\frac{3}{16}\right|=\frac{5}{16}$
My try:
$$\begin{align}\\
&w=\frac 1z\\
&\implies 3w=\frac 3z\\
&\implies1-3w=1-\frac 3z=\frac{z-3}{z}\\
&\implies|1-3w|=\left|\frac{z-3}{z}\right|\\
&\implies|1-3w|=|z-3|\left|\frac 1z\right|\\
&\implies|1-3w|=5|w|\\
\end{align}\\$$
But, I can't understand what to do next and how can I proof that. Any suggestions? or, you can add an answer.
|
We need to prove that if $|z-3|=5$ so $$\left|\frac{1}{z}+\frac{3}{16}\right|=\frac{5}{16}.$$
Let $z=x+yi,$ where $x$ and $y$ are reals.
Thus, $$(x-3)^2+y^2=25$$ and we need to prove that:
$$|3z+16|=5|z|$$ or
$$(3x+16)^2+9y^2=25(x^2+y^2)$$ or
$$16(x^2+y^2)=96x+256$$ or
$$x^2-6x+y^2=16$$ or
$$(x-3)^2+y^2=25$$ and we are done!
|
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|
Identifying relation between numbers based on equation relating them
If $a^2 + b^2+16c^2=2(3ab+6bc + 4ac)$ , where $a,b,c$ are non zero numbers. Then $a,b,c$ are in __________?
1. Harmonic progression 2. Geometric progression 3. Arithmetic progression 4. None of these
My attempt:
$ a \rightarrow 4a'$
$ b \rightarrow 4b'$
$a'^2 +b'^2 +c^2 - 6a'b' -3b'c -2a'c=0$
Ok, this looks nicer than the original thing but still the solution doesn't seem in sight. Further, I started wondering, would the relationship between numbers be preserved under transformations to the equation?
|
There is a way to make this appear simpler. You are using letters a,b,c. Alright, take any $x,y,z$ you like such that
$2x^2 - 16 y^2 + 81 z^2 = 0.$ This is just a point on a cone. Then still to avoid fractions, let
$$ a = 4x+12y - 11z \; , \; \; b = 4y - 9 z \; , \; \; c = 4z $$
For infinitely many examples, we can introduce variables $u,v$ and take
$$ x = 576 u^2 + 72uv \; , \; \; y = 594 u^2 + 144 uv + 9 v^2 \; , \; \; z = 248 u^2 + 64 uv + 4 v^2 $$
For instance, let $u=1, v=1$ to get $x=648,y=747, z=316,$ then $a=8080, b= 144, c = 1264.$ These are all multiples of $16,$ we can divide out to get $a = 505, b= 9, c=79$
Here is a good one, as a single step:
$$ a = 16u^2 + 36 uv + 19 v^2 \; , \; \; b = 9 v^2 \; , \; \; c = 4 u^2 - 2 v^2 $$
With $u=0, v=1$ we get $a=19,b=9, c=-2,$ or with $u=1,v=1$ we get $a= 71, b=9,c=2$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
3 & 1 & 0 \\
- \frac{ 11 }{ 4 } & - \frac{ 9 }{ 4 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - 3 & - 4 \\
- 3 & 1 & - 6 \\
- 4 & - 6 & 16 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 3 & - \frac{ 11 }{ 4 } \\
0 & 1 & - \frac{ 9 }{ 4 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & - 8 & 0 \\
0 & 0 & \frac{ 81 }{ 2 } \\
\end{array}
\right)
$$
|
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|
what loops and points of numbers are possible when you take the alternating sum of the digits of squared? what loops of numbers are possible when you take the alternating sum of the digits of squared?
I've heard about the happy numbers and the sad numbers. if you don't know the happy numbers are numbers when you add it's digits$^2$ and do that n times it hits the floor value of $1$. sad numbers are numbers that go into the same endless loop of sadness $(20,4,16,37,58,89,145,42,20,...)$
$$(\space2^2+1^2=4,\space4^2=16,\space1^2+6^2=37,\space3^2+7^2=58,\space5^2+8^2=89,\space8^2+9^2=145,\space1^2+4^2+5^2=42,\space4^2+2^2=20)$$
my question is slightly different instead of the sum mine uses the absolute value of the alternating sum.
so far I've found $2$ loops and to $2$ points $$(\space9^2=81,\space8^2-1^2=63,\space6^2-3^2=27,\space|2^2-7^2|=45,\space|4^2-5^2|=9,\dots),(\space|1^2-6^2|=35,\space|3^2-5^2|=16)$$
$$the\space points\space are\space0\space and\space 1 \space . 1^2-0^2=1,1^2-1^2=0$$
doing this with a $3$ digit number looks like this you start with $125$ do this$|1^2-2^2+5^2|=23 $ and you get $22$.
My question is how many points and loops are there are?
|
Partial answer: the only points are $0,1$ and $48$.
Let $f()$ be the function that performs one step of your transformation, e.g. $f(125) = |1^2 - 2^2 + 5^2| = 22$ (not $23$ like you said).
If $x$ is a $k$-digit number, we have $f(x) \le \lceil k/2 \rceil \times 9^2$ because half the digits count positively and half count negatively, and each digit is at most $9$. This bound immediately shows that $f(x) < x$ for these cases ($d$ denotes leading digit):
*
*All $k \ge 4$.
*$k = 3, d \ge 2: f(x) \le 2 \times 81 $ but $x \ge 200$.
*$k = 3, d = 1:$ let the number be $x = (1bc)_{10}$ and $f(x) = |1 - b^2 + c^2| \le 1 + 81$ but $x \ge 100$.
Now consider $k=2, x = (ab)_{10} = 10a + b$.
*
*If $a \ge b$, we have $f(x) \le a^2 < 10a \le x$.
*If $a < b$, for $x$ to be a point we need $f(x) = b^2 - a^2 = 10a + b$ or $b(b-1) = (10+a)a.$ Now it is just a matter of trying each $a = 1, 2, 3, \dots$ and we find that the only solution is $x = 48, a = 4, b = 8:$
$$f(48) = |4^2 - 8^2| = 64 - 16 = 48$$
For $k=1$ we have $f(x) = x^2$ so $f(x) = x$ iff $x = 0,1$.
Incidentally, since $f(x) < x$ for sufficiently large $x$, this also means the process "collapses" onto the small integers, and there are only a finite number of loops.
|
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|
Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx
\end{align*}
\begin{align*}
I_x(y)&=\int_0^1\frac{-x\ln(xy)}{2(1-xy)}-\frac{x\ln(xy)}{2(1+xy)} \, dx\\[6pt]
&=\frac{1}{2}\int_0^1\frac{-x\ln(xy)}{1-xy}\, dx+\frac{1}{2} \int_0^1\frac{-x\ln(xy)}{1+xy} \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n \int_0^1x^{n+1}\ln(xy)\,dx+\frac{1}{2} \sum_{n=0}^\infty(-1)^{n+1} \int_0^1y^nx^{n+1}\ln(xy) \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n\left(\frac{\ln(y)}{n+2}-\int_0^1 \frac{x^{n+1}}{(n+2)y} \, dx\right)+\frac{1}{2} \sum_{n=0}^\infty (-1)^{n+1} y^n \left(\frac{\ln(y)}{n+2}-\int_0^1\frac{x^{n+1}}{(n+2)y} \, dx\right)\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty \frac{y^n\ln(y)}{n+2} ((-1)^{n+1}-1) +\frac{1}{2} \sum_{n=0}^\infty \frac{y^{n-1}}{(n+2)^2}(1-(-1)^{n+1})
\end{align*}
|
\begin{align}
-\int_0^1\int_0^1\frac{x\ln{(xy)}}{1-x^2y^2}\,\mathrm{d}x\,\mathrm{d}y
&=-\sum_{n=0}^\infty \int_0^1 y^{2n} \int_0^1x^{2n+1} \ln{(xy)} \, \mathrm{d}x \, \mathrm{d}y \\[6pt]
&=-\sum_{n=0}^\infty\int_0^1y^{2n}\left[\frac{x^{2(n+1)}\ln{(xy)}}{2(n+1)}-\frac{x^{2(n+1)}}{4(n+1)^2}\right]_0^1\,\mathrm{d}y\\[6pt]
&=-\sum_{n=0}^\infty\int_0^1y^{2n}\left(\frac{\ln{(y)}}{2(n+1)}-\frac1{4(n+1)^2}\right)\,\mathrm{d}y\\[6pt]
&=-\sum_{n=0}^\infty\left[\frac{y^{2n+1}\ln{(y)}}{2(n+1)(2n+1)}-\frac{(4n+3)y^{2n+1}}{4(n+1)^2(2n+1)^2}\right]_0^1\\[6pt]
&=\sum_{n=0}^\infty\frac{4n+3}{4(n+1)^2(2n+1)^2}\\[6pt]
&=\sum_{n=0}^\infty\left(\frac1{(2n+1)^2}-\frac1{(2(n+1))^2}\right)\\[6pt]
&=\sum_{n=1}^\infty\left(\frac1{(2n-1)^2}-\frac1{(2n)^2}\right)\\[6pt]
&=\eta(2)\\[6pt]
&=\frac{\pi^2}{12}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the greatest integer less than $\frac{1}{\sin^2(\sin(1))}$ without calculator. Find the greatest integer less than $$\frac{1}{\sin^2(\sin(1))}$$
This was on one of my tests. All angles in radians. Here's my work:
$$0<1<\frac{\pi}{3}<\frac{\pi}{2}$$
Since $\sin(x)$ is increasing in the first quadrant,
$$0<\sin(1)<\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}<\sin\left(\frac{\pi}{2}\right)=1<\frac{\pi}{2}$$
$\sin(1)$ radians and $\frac{\sqrt{3}}{2}$ radians also lie in the first quadrant, and $\sin^2(x)$ is increasing in the first quadrant
$$\sin^2(\sin(1))<\sin^2\left(\frac{\sqrt{3}}{2}\right)<\frac{3}{4}$$
$$\frac{1}{\sin^2(\sin(1))}>\frac{4}{3}$$
But I cannot find an upper bound on my expression. Any help?
|
$$1 > \frac{7}{24}\pi \approx 0.916$$
$$\sin(1) > \sin\left(\frac{7}{24}\pi\right) = \frac12\sqrt{2+\sqrt{2-\sqrt{3}}} \approx 0.793 > \frac{\pi}{4} \approx 0.785$$
$$\frac{1}{\sin^2(\sin(1))} < \frac{1}{\sin^2(\frac{\pi}{4})}=2$$
so $$1 < \frac{1}{\sin^2(\sin(1))} < 2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.
Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$
I have no clue on how this kind of matrices can be solved. Can anyone give a general strategy on how to solve matrices whose size are greater that $3\times 3$?
That would be really helpful.
|
Performing a Laplace expansion along the first column, we get
$$
\begin{vmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{vmatrix}
=
1\cdot
\begin{vmatrix}1&0&0&2&0\\0&1&2&0&0\\0&2&1&0&0\\2&0&0&1&0\\0&0&0&0&1\end{vmatrix}
- 2\cdot
\begin{vmatrix}0&0&0&0&2\\1&0&0&2&0\\0&1&2&0&0\\0&2&1&0&0\\2&0&0&1&0\end{vmatrix}
$$
Now both $5\times 5$ determinants can be Laplace expanded along the last column to get
$$
1\cdot 1\cdot
\begin{vmatrix}1&0&0&2\\0&1&2&0\\0&2&1&0\\2&0&0&1\end{vmatrix}
- 2\cdot 2\cdot
\begin{vmatrix}1&0&0&2\\0&1&2&0\\0&2&1&0\\2&0&0&1\end{vmatrix}
=
-3\cdot \begin{vmatrix}1&0&0&2\\0&1&2&0\\0&2&1&0\\2&0&0&1\end{vmatrix}.
$$
Now you can either repeat this procedure one more time to end up with a $2\times 2$ determinant, or notice the general pattern and prove a more general statement by induction:
Let $A_n$ be the $2n\times 2n$ matrix with ones on the main diagonal and twos on the antidiagonal. What we did above to $A_3$ works out in general as
\begin{align*}
\det(A_n) &= 1\cdot\begin{vmatrix} A_{n-1} & 0 \\ 0 & 1\end{vmatrix}
- 2 \cdot\begin{vmatrix} 0 & 2 \\ A_{n-1} & 0\end{vmatrix} \\
&= 1\cdot 1 \cdot \det(A_{n-1})
- 2 \cdot 2\cdot \det(A_{n-1}) \\
&= -3\cdot\det(A_{n-1}).
\end{align*}
Applied to your matrix this yields
$$
\det(A_3) = -3\det(A_2)=(-3)^2\det(A_1) = (-3)^3 = -27
$$
and in general you get $\det(A_n)=(-3)^n$.
|
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|
Proving: $\int_0^1 \int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$
Proving:$$\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$$
I tried using variable switching
$\ln(xy)=t$ But I did not reach any results after the calculation
\begin{align*}
k&=\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dx\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{(1+e^t)^2y^2}dtdy\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{y^2(1+e^t)}\displaystyle\sum_{n=0}^{\infty}(-e^t)dtdy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4(-e^{2t})}{1+e^t}dt\right)dy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{\ln(y)}^{\infty}\frac{t^4e^{2t}}{1+e^t}dt\right)dy\\
\end{align*}
|
\begin{align}J&=\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy\\
&=\int_0^1 \frac{1}{x}\left(\int_0^x \frac{\ln^4 u}{(1+u)^2 du}\right)\\
&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^4 u}{(1+u)^2} du\right)\ln x\right]_0^1-\int_0^1 \frac{\ln^5 x}{(1+x)^2}dx\\
&=-\int_0^1 \frac{\ln^5 x}{(1+x)^2}dx\\
&\overset{\text{IBP}}=\left[\ln^5 x\left(\frac{1}{1+x}-1\right)\right]_0^1-5\int_0^1 \frac{\left(\frac{1}{1+x}-1\right)\ln^4 x}{x}dx\\
&=5\int_0^1 \frac{\ln^4 x}{1+x}dx\\
&=5\left(\int_0^1 \frac{\ln^4 x}{1-x}dx-\int_0^1 \frac{2t\ln^4 t}{1-t^2}dt\right)\\
&\overset{x=t^2}=5\left(\int_0^1 \frac{\ln^4 x}{1-x}dx-\frac{1}{16}\int_0^1 \frac{\ln^4 x}{1-x}dx\right)\\
&=\frac{5\times 15}{16}\int_0^1 \frac{\ln^4 x}{1-x}dx\\
&=\frac{5\times 15}{16}\times 24\zeta(5)\\
&=\boxed{\frac{225}{2}\zeta(5)}\\
\end{align}
I assume that, $\displaystyle \int_0^1 \frac{\ln^4 x}{1-x}dx=24\zeta(5)$
|
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|
How can we take the derivative of this function: $y = \frac{x}{x^2+1}$ from first principles (using the limit definition of the derivative)? I was taking the derivative of the function: $y = \frac{x}{x^2+1}$. I know that we can solve it by the quotient rule. But I tried using the limit definition of differentiation. This is how I did it:
$$\lim_{h \to 0} \frac{(x+h)/\left((x+h)^2+1\right) - x/(x^2+1)}{h}$$
$$=\lim_{h \to 0}\frac{(x+h)/(x^2+h^2+2xh+1) - x/(x^2+1)}{h}$$
Then I multiplied both the denominator and numerator by the common factor $(x^2+h^2+2xh+1)(x^2+1)$ and expanded:
$$=\lim_{h \to 0} \ \Biggl(\frac{x^3+x+x^2h+h - x^3-xh^2-2x^2h-x}{x^4+x^2h^2+2x^3h+x^2+x^2+h^2+2xh+1}\Biggr) \cdot \frac{1}{h}$$
I then simplified the expression:
$$=\lim_{h \to 0} \ \frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}$$
Now what is to be done? It seems that I am very far from the actual derivative.
|
$$\frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}=\\ \frac{1-x^2-xh}{x^4 + x^2h^2+2x^3h^2+2x^2+h^2+2xh+1} \to \frac{1-x^2}{(1+x^2)^2}$$
when $h \to 0$.
|
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|
Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}= \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)}$$
but I don't know how to reduce this further.
|
\begin{align}
& \frac{\left(x\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\right)-\left(x + \frac{x^3}{3!} + \cdots\right)}{x\left(x + \frac{x^3}{3!}+\cdots\right)} \\ {} \\
= {} & \frac{\left(x + \frac{x^3}{2!} + \frac{x^5}{4!}+ \cdots\right) -\left( x + \frac{x^3}{3!} \right)}{x^2 + \frac{x^4}{3!}+ \cdots} \\ {} \\
= {} & \frac{\left(\frac{x^3}{2!} + \frac{x^5}{4!}+ \cdots\right) -\left(\frac{x^3}{3!} + \cdots \right)}{x^2 + \frac{x^4}{3!}+ \cdots} \\ {} \\
= {} & \frac{\left(\frac{x}{2!} + \frac{x^3}{4!}+ \cdots\right) -\left(\frac x {3!}+ \cdots \right)}{1 + \frac{x^2}{3!}+ \cdots} \to \frac 0 1
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ Notice that $\frac{n \sin x}{1+n^2x^2}\to 0$ pointwise.
And we have,$$\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx=\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx+\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$$
Then for $\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ portion we have,
$$\left|\frac{n \sin x}{1+n^2x^2}\right|\leq\left|\frac{n}{n^2x^2}\right|=\left|\frac{1}{nx^2}\right|\leq\frac{1}{x^2}$$
And $\frac{1}{x^2}$ is integrable on $(1,\infty)$
So by the dominated convergence theorem:
$$\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx\to\int\limits_0^{\infty}0=0$$
But how should I do the $\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx$?
Appreciate your help
|
And idea: substitute
$$x=\frac1u\implies dx=-\frac1{u^2}du\implies\int_0^1\frac{n\sin x}{1+n^2x^2}dx=\int_\infty^1-\frac{du}{u^2}\cdot\frac{n\sin\frac1u}{1+\frac{n^2}{u^2}}=$$
$$=\int_1^\infty\frac{n\sin\frac1u}{u^2+n^2}du$$
and now estimate
$$\left|\frac{n\sin\frac1u}{u^2+n^2}\right|\le\frac n{u^2+n^2}\,,\,\,\text{and}\;\int_1^\infty\frac n{n^2+u^2}du=\int_1^\infty\frac{d\left(\frac un\right)}{1+\left(\frac un\right)^2}=$$
$$=\left.\arctan\frac un\right|_1^\infty=\frac\pi2-\arctan\frac1n\xrightarrow[n\to\infty]{}\frac\pi2$$
Which means your integral is bounded above...
|
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|
Factorize: $x^3 + x + 2$. How do I factorize the term $x^3 + x +2$?
What I have previously tried is the middle term factor method but it didn't work...
$x^3 + x + 2$
$\Rightarrow x^3 + 2x - x + 2$
$\Rightarrow x(x^2 + x) - 2(x - 1)$
This doesn't work.
What should I do?
|
Another solution$:$
$$x^3+x+2=(x^3+1)+(x+1)=(x+1)(x^2-x+1)+(x+1)=(x+1)(x^2-x+2)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{2} \geqslant \sqrt{x} \sqrt{y}
$$
$$
\sqrt{x} \sqrt{y}(x-y) \geqslant 2 \sqrt{x} \sqrt{y}
\\ \therefore (x-y) \geqslant 2
$$
So, I have been able to arrive at this conclusion. But I am stuck here. Any help ?
Thanks
|
put $x=r^2{cos}^2a$ and $y=r^2{sin}^2a$ also let $a$ belong to $[0,\frac{\pi}{2}]$
thus we have to find max value of $r^2$
plugging the values in the given equation and simplifying using basic trig formulas we have $r^4(cosa)(sina)(cos2a)=r^2$ or
$ r^2=\frac{4}{sin(4a)} \ge 4$
|
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|
Olympiads Number Theory Problem: Show that there are no positive $a,b,c\in\Bbb Z$ such that $\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\in\Bbb Z$ I have been struggling to solve this problem, could someone give me an idea/solution? please!
Show that there are no positive $a,b,c\in\Bbb Z$ such that $$\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\in\Bbb Z$$
Edit source image: https://i.stack.imgur.com/mk9cl.png
|
I thought that I had a solution using Vieta Jumping but this was not a complete solution so we can use quadratic residues to solve it...
We can choose WLOG $\text{gcd} (a,b,c)=1$.
After letting $\frac{a^2+b^2+c^2}{ab+bc+ca}=3k$, we have $(a+b+c)^2=(3k+2)(ab+bc+ca)$.
Choose a prime $p \equiv 2 \pmod{3}$ with $v_p(3k+2) \equiv 1 \pmod{2}$
(clearly there exists one since $3k+2$ is not a square).
Then $p|a+b+c$, and we have $p|ab+bc+ca$ using our condition on $v_p$.
$\implies c \equiv -a-b \pmod{p}$, so $p|ab+(a+b)(-a-b)$, so $p|a^2+ab+b^2$.
If $p=2$, we check that $a$ and $b$ must be even.
Otherwise, we have $p \equiv 5 \pmod{6}$, and $\left( \frac{-3}{p} \right)=-1$.
This gives us $p|a,b,c$, contradiction.
|
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|
Finding Fourier series and evaluating at a point Ive been asked to find the fourier series of the following
Need help with finishing the question, i have included my work so far below:
and following that i need help with the final part of the question:
|
You wrote the correct expression for the coefficients $a_n$, $n\ge1$. However, there must have been an error in your arithmetic for carrying out the integral
$$a_n=\frac1\pi\int_{-\pi}^\pi (x^2-\pi^2)^2\cos(nx)\,dx=\frac{48(-1)^{n-1}}{n^4}$$
Also, we need the term $a_0$, which is
$$a_0=\frac1{2\pi}\int_{-\pi}^\pi (x^2-\pi^2)^2\,dx=\frac{8\pi^4}{15}$$
So, we have
$$\begin{align}
f(x)&=(x^2-\pi^2)^2\\\\
&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48(-1)^{n}}{n^4}\cos(nx)
\end{align}$$
Now, for $x=\pi$, $f(x)=0$. Furthermore, $\cos(n\pi)=(-1)^n$. Hence, we have
$$\begin{align}
f(\pi)&=0\\\\
&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48(-1)^{n}}{n^4}\cos(n\pi)\\\\
&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48(-1)^{n}}{n^4}(-1)^n\\\\
0&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48}{n^4}\tag1
\end{align}$$
From $(1)$, we see that
$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$$
as was to be determined!
|
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|
Recursive squaring algorithm by reducing to squaring 5 numbers
*(c) Describe a recursive algorithm that squares any $n$-digit number in $O(n^{log_35})$ time, by reducing to squaring only five $(n/3+O(1))$-digit numbers.
[Hint: What is $(a+b+c)^2+(a−b+c)^2$?]
I am currently studying the Toom-Cook algorithm which is highly related to this question. I came across a related post here but could not understand it sufficiently. For example, why the form of $10^{2m}x+10^my+z$ and the discovery of the last missing square.
Any help is appreciated.
|
Wikipedia's article on Toom–Cook multiplication has a nice explanation that doesn't involve guessing missing squares; let me simplify it for your case.
If you're writing your number as $10^{2m}a + 10^m b + c$ (essentially, breaking up the digits into three segments $a, b, c$) then you're writing your number as $p(10^m)$ where $p(x)$ is the quadratic polynomial $ax^2+bx+c$.
So the square of your number will be $p(10^m)^2$, which is the quartic polynomial $q(x) = p(x)^2 = (ax^2+bx+c)^2$ evaluated at $x=10^m$.
Now, we're going to try to find $q(x)$ in a different way. Specifically, we know the following:
*
*The constant term of $q(x)$ is $c^2$, or in other words $q(0)=p(0)^2=c^2$.
*The coefficient of $x^4$ in $q(x)$ is $a^2$, which Wikipedia thinks of as $q(\infty) = p(\infty)^2 = a^2$.
*$q(1) = p(1)^2 = (a+b+c)^2$.
*$q(-1) = p(-1)^2 = (a-b+c)^2$.
*$q(-2) = p(-2)^2 = (4a-2b+c)^2$.
The above squares are the five squares we're going to evaluate. Then, if $q(x) = q_4 x^4 + q_3 x^3 + q_2 x^2 + q_1 x + q_0$, we have a system of five equations in five variables:
\begin{align}
q_0 &= c^2 \\
q_4 &= a^2 \\
q_4 + q_3 + q_2 + q_1 + q_0 &= (a+b+c)^2 \\
q_4 - q_3 + q_2 - q_1 + q_0 &= (a-b+c)^2 \\
16q_4 - 8q_3 + 4q_2 - 2q_1 + q_0 &= (4a-2b+c)^2
\end{align}
Solve these for the $q_i$'s and then evaluate $q(10^m)$ to find $(10^{2m}a+10^mb+c)^2$.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.