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is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function? I have a function $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one). The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b
$\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$
$\sqrt{a^2 + 1} + b = \sqrt{b^2 + 1} + a$
$(\sqrt{a^2 + 1} + b)^2 = (\sqrt{b^2 + 1} + a)^2$
$a^2 + 1 + b^2 + 2b\sqrt{a^2 + 1} = b^2 + 1 + a^2 + 2a\sqrt{b^2 + 1}$
$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$
$b\sqrt{a^2 + 1} = a\sqrt{b^2 + 1}$
$(b\sqrt{a^2 + 1})^2 = (a\sqrt{b^2 + 1})^2$
$b^2(a^2 + 1) = a^2(b^2 + 1)$
$b^2a^2 + b^2 = a^2b^2 + a^2$
$b^2 = a^2$
$\sqrt{b^2} = \sqrt{a^2}$
$b = \pm a$
then the function is not injective because $b$ in not equal to $a$, but using online solver the function is injective.
|
When you wrote $$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$$ this gave you a new condition : $a$ and $b$ must have same sign. (The square roots are positive). Hence $a^2=b^2$ implies $a=b$ in this case.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding stagnation points and stream function
Sorry for the lack of latex. The question I want to ask would need all this info and it would take very long to write it.
(a) Irrotational flow means $\nabla \times \textbf u =0$ so we can define potential to be $\textbf u =\nabla \phi$ and since it is irrotational and inviscid, the continuity equation holds so $\nabla \cdot \textbf u =0$ so the laplacian of the potential holds, $\nabla \cdot \textbf u =\nabla \cdot \nabla \phi=\nabla^2 \phi=0$.
(b) $u_r = U\cos \theta (1-a^3 / r^3)$ and $u_{\theta}=-U\sin \theta (1+a^3/2r^3)$
(c) $(r=a, \theta = 0)$ and $(r=a, \theta = \pi)$
(d) $ψ=\frac U2 \sin^2 \theta (r^2 - a^3/r)$
(e) Would it be something like this:
I am mainly stuck on c, d, and e. I am a tiny bit unsure on part (a) but it seems to make sense. Please do say if not.
Please help.
|
For (b), we can take partial derivatives of $\phi$ to obtain
$$u_r = \frac{\partial \phi}{\partial r} = U \cos \theta - U\frac{a^3}{r^3} \cos \theta, \\ u_\theta = \frac{1}{r} \frac{\partial \phi}{\partial \theta}= -U \sin \theta - U\frac{a^3}{2r^3} \sin \theta.$$
For (c), on the surface of the sphere $r = a$ and the radial velocity component vanishes, $u_r(a, \theta) = 0,$ since this component is normal to the surface. This is apparent upon substituting $a$ for $r$ in the above expression for $u_r$. The tangential component is
$$u_\theta(a, \theta) = -\frac{3}{2}U \sin \theta.$$
We have a stagnation points on the surface $r = a$ where both velocity components vanish: $\theta = 0$ (fore) and $\theta = \pi$ (aft).
For (d), we solve for the streamfunction using
$$\frac{\partial \psi}{\partial \theta} = r^2 \sin \theta u_r = U\sin \theta \cos \theta \left(r^2 - \frac{a^3}{r} \right) = \frac{U}{2} \sin 2 \theta\left(r^2 - \frac{a^3}{r} \right)\\ \implies \psi = -\frac{U}{4} \cos 2 \theta\left(r^2 - \frac{a^3}{r} \right) + F(r) \\ \implies \psi = -\frac{U}{4} (1 - 2\sin^ 2 \theta)\left(r^2 - \frac{a^3}{r} \right) + F(r) \\ \implies \psi = \frac{U}{2} \sin^ 2 \theta\left(r^2 - \frac{a^3}{r} \right) - \frac{U}{4}\left(r^2 - \frac{a^3}{r} \right) + F(r)$$
and
$$\frac{\partial \psi}{\partial r} = -r \sin \theta u_\theta = U\sin^2 \theta\left(r + \frac{a^3}{2r^2} \right) \\ \implies \psi = U \sin^2 \theta \left(\frac{r^2}{2} - \frac{a^3}{2r} \right) + G(\theta),$$
We get a consistent solution by choosing
$$G(\theta) = 0, \\ F(r) = \frac{U}{4}\left(r^2 - \frac{a^3}{r} \right),$$
and it follows that
$$\psi = \frac{U}{2} \sin^2 \theta \left(r^2 - \frac{a^3}{r} \right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$
Find the value of $$\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$$
I know how to solve when all surds are of the same order, but what if they are different?
Technically, (as some users wanted to know exactly what is to be found), find:
$$\lim_{n\to\infty}\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots+\sqrt[n]{4}}}}} $$
|
Put $$y=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...}}}}\qquad (1)$$ stopping successively at the $n$-th root we have a sequence strictly increasing and bounded so the limit $y$ is well defined.
From $(1)$ we can easily get a sequence $\{P_n\}$ of polynomials such that its largest real roots $$\alpha_n=\sqrt{4+(4+(4+(4+….(4+\sqrt[n]4)^{1/n}….)^{1/5})^{1/4})^{1/3}}$$
form a sequence $\{\alpha_n\}$ converging to $y$. In fact we have
$$\begin{cases}P_2(x)=x^2-4\\P_3(x)=(x^2-4)^3-4\\P_4(x)=( (x^2-4)^3-4)^4-4\\.....\\.....\\P_n(x)=(P_{n-1}(x))^n-4\end{cases}$$
We can see the polynomial $P_n$ has degree $n!$ and $P_n(\alpha_{n-1})=-4$. Since $P_n(x)$ grows very quickly, this indicates that $\{\alpha_n\}$ rapidly converges to the limit $y$ because the arc of the graphic of $P_n(x)$ between the points $(\alpha_{n-1},-4)$ and $(\alpha_n,0)$ is almost a vertical line so $\alpha_{n-1}\approx \alpha_n$. This argument is enhanced by the equality
$$y=2\sqrt{1+\left(\frac{4}{4^3}+\left(\frac{4}{4^{12}}+\left(\frac{4}{4^{60}}+….\left(\frac{4}{4^n}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ hence the factor of $2$ is equal to $$\sqrt{1+\left(\frac{1}{2^{6-2}}+\left(\frac{1}{2^{24-2}}+\left(\frac{1}{2^{120-2}}+….\left(\frac{1}{2^{n!-2}}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ or
$$\sqrt{1+\left(\frac{1}{2^{4}}+\left(\frac{1}{2^{22}}+\left(\frac{1}{2^{118}}+….\left(\frac{1}{2^{n!-2}}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ This factor of $2$ grows very slowly obviously.
It seems to me there is not a closed form for $y$ (It could be an asymptotic one perhaps). Consequently I give here an (justified) approximation taking the above referred real root of, say, $P_6$; we have$$y\approx\sqrt{4+\left(4+\left(4+\left(4+\left(4+\sqrt[6]4\right)^{1/6}\right)^{1/5}\right)^{1/4}\right)^{1/3}}\approx \color{red}{2.40161550315}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating $\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}$ We want only the real 3rd root.
By calculation, $[\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}]^3= 4-3[\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}]$
Therefore, the answer is a root of $t^3=4-3t$ , which obviously has the real solution $t =1$.
But I want another way of showing $\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2} =1$, maybe by using simple algebraic formulas.
|
By observation,
$$(\sqrt5+1)^3=16+8\sqrt5=8(\sqrt5+2)$$
$$\implies\left(\dfrac{\sqrt5+1}2\right)^3=\sqrt5+2$$
Similarly,$$\left(\dfrac{\sqrt5-1}2\right)^3=\sqrt5-2$$
Motivation:
$$(\sqrt5+a)^3=a^3+15a+\sqrt5(3a^2+5)$$
Let us find $a$ such that $$\dfrac{a^3+15a}{3a^2+5}=\dfrac21\iff0=a^3-6a^2+15a-10=(a-1)(a^2-5a+10)$$
Observe that the only real root is $a=1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit of $\lim_{n\to \infty}\frac{2^n n^2}{n!}$ I am trying to find the limit of the following, $$\lim_{n\to \infty}\frac{2^n n^2}{n!}$$
L'Hospital is not going to work. Hoping for a squeeze, by the observation that $2^n<n!$ for $n\geq 4$ does not help either as one side of the limit goes to $\infty$. How can I solve this? Any hints?
|
$n!>2\cdot 3^{n-2}$ for $n\ge2$ is apparent from the product definition, so:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot 3^{n-2}}=\frac{9n^2}{4\left(\frac{3}{2}\right)^n}$$
Note that $\left(\frac{3}{2}\right)^n=\left(1+\frac{1}{2}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{1}{2}\right)^k\ge\binom{n}{3}\left(\frac{1}{2}\right)^3$, so we have:
$$\frac{2^nn^2}{n!}\le\frac{9n^2}{4\left(\frac{3}{2}\right)^n}\le\frac{9n^2}{4\binom{n}{3}\left(\frac{1}{2}\right)^3}=\frac{108n^2}{n(n-1)(n-2)}=\frac{108}{n}\frac{1}{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}\to0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+x^2} = \frac{1}{x^2} + \frac{2}{x} - \frac{1}{x+1}$$
Therefore:
$$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$
The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$
Where is my mistake?
|
Observe that
$$\frac{x^4+1}{x^3+x^2}=\frac{x^4+x^3-x^3-x^2+x^2+1}{x^3+x^2}=\frac{(x-1)(x^3+x^2)+x^2+1}{x^3+x^2}=x-1+\frac{x^2+1}{x^2(x+1)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form of $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$? I am trying to find a closed form for the integral $$I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$$ So far, my reasoning is thus: write, by symmetry through $x=\pi/2$, $$I=2\sum_{n=1}^{\infty}n\int_{\arctan n}^{\arctan (n+1)}\frac{dx}{|\tan x|}=2\sum_{n=1}^{\infty}n\ln\frac{\sin\arctan(n+1)}{\sin\arctan n}$$
Using $\sin{\arctan {x}}=\frac{x}{\sqrt{1+x^{2}}}$, we get: $$I=2\sum_{n=1}^{\infty}n\ln(\frac{(n+1)\sqrt{1+n^2}}{n\sqrt{1+(n+1)^2}})=\sum_{n=1}^{\infty}n\ln\frac{(n+1)^2(1+n^2)}{n^2(1+(n+1)^2)}=\sum_{n=1}^{\infty}n\ln(1+\frac{2n+1}{n^2(n+1)^2})$$ Expanding the logarithm into an infinite series we get $$I=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}(2n+1)^m}{mn^{2m-1}(n+1)^{2m}}$$ Here I am a bit stuck.. Does anyone have any suggestions to go further?
Thank you.
EDIT:
keeping in mind the nice answer below, applying summation by parts to $$I_N=2\sum_{n=1}^{N}n\ln\frac{\sin\arctan(n+1)}{\sin\arctan n}=2\sum_{n=1}^{N}n(\ln\sin\arctan(n+1)-\ln\sin\arctan n)$$ gives $$I_N=2((N+1)\ln\sin\arctan(N+1)+\frac{\ln 2}{2}-\sum_{n=1}^{N}\ln\sin\arctan(n+1))$$ hence: $$I-\ln2=-\sum_{n=2}^{\infty}\ln\frac{n^2}{1+n^2}=\sum_{n=2}^{\infty}\ln\frac{1+n^2}{n^2}=\sum_{n=2}^\infty\sum_{m=1}^\infty\frac{(-1)^{m+1}}{mn^{2m}}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{m}\sum_{n=2}^\infty n^{-2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}(\zeta(2m)-1)}{m}$$
Is this valid and helpful?
EDIT 2: Coming back to $$\sum_{n=2}^{\infty}\ln(1+\frac{1}{n^2})=\ln(\prod_{n=2}^{\infty}(1+\frac{1}{n^2}))=\ln(\prod_{n=2}^{\infty}(1-\frac{i^2}{n^2}))=\ln(\prod_{n=1}^{\infty}(1-\frac{i^2}{n^2}))-\ln2$$
$$=\ln(\frac{\sin(i\pi)}{i\pi})-\ln2=\ln\frac{\sinh\pi}{\pi}-\ln2$$ hence $I=\ln\frac{\sinh\pi}{\pi}$
|
$$\begin{align}
I
&=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor\left|\tan{\left(x\right)}\right|\rfloor}{\left|\tan{\left(x\right)}\right|}\,\mathrm{d}x\\
&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\lfloor\left|\cot{\left(y\right)}\right|\rfloor}{\left|\cot{\left(y\right)}\right|}\,\mathrm{d}y;~~~\small{\left[x-\frac{\pi}{2}=y\right]}\\
&=2\int_{0}^{\frac{\pi}{4}}\frac{\lfloor\left|\cot{\left(y\right)}\right|\rfloor}{\left|\cot{\left(y\right)}\right|}\,\mathrm{d}y\\
&=2\int_{0}^{\frac{\pi}{4}}\frac{\lfloor\cot{\left(y\right)}\rfloor}{\cot{\left(y\right)}}\,\mathrm{d}y\\
&=2\int_{1}^{\infty}\frac{\lfloor t\rfloor}{t\left(1+t^{2}\right)}\,\mathrm{d}t;~~~\small{\left[\cot{\left(y\right)}=t\right]}\\
&=2\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{\lfloor t\rfloor}{t\left(1+t^{2}\right)}\,\mathrm{d}t\\
&=2\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{n}{t\left(1+t^{2}\right)}\,\mathrm{d}t\\
&=2\sum_{n=1}^{\infty}n\int_{n}^{n+1}\frac{\mathrm{d}t}{t\left(1+t^{2}\right)}\\
&=2\sum_{n=1}^{\infty}n\int_{n}^{n+1}\left[\frac{1}{t}-\frac{t}{1+t^{2}}\right]\,\mathrm{d}t\\
&=\sum_{n=1}^{\infty}n\left[\ln{\left(\frac{t^{2}}{1+t^{2}}\right)}\right]_{n}^{n+1}\\
&=\ln{\left(2\right)}-\sum_{n=2}^{\infty}\ln{\left(\frac{n^{2}}{1+n^{2}}\right)}\\
&=\ln{\left(2\right)}-\ln{\left(\prod_{n=2}^{\infty}\frac{n^{2}}{1+n^{2}}\right)}\\
&=\ln{\left(2\right)}-\ln{\left(2\pi\operatorname{csch}{\left(\pi\right)}\right)}\\
&=\ln{\left(\frac{\operatorname{sinh}{\left(\pi\right)}}{\pi}\right)}.\blacksquare\\
\end{align}$$
|
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|
How to calculate the nullity for a matrix with variable? In order to find the number of (x,y) satisfying
$$f(x,y) =\begin{cases}
xy^2-y+x^2+1=0\\
x^2y^2+y-1=0\\
\end{cases}$$
I use the Jacobi-Darboux Theorem by fix 1.x; 2.y and get the Bezoutian matrix, but I don't know how to find the nullity of the both matrix...
*
*$$B =
\begin{pmatrix}
y^3-y^2 & y^3-y^2+y-1\\
y^3-y^2+y-1 & -y^4
\end{pmatrix}$$
*$H_{f_x} =
\begin{pmatrix}
-1 & x\\
x & 0 \end{pmatrix}$, $T_{f_x} =
\begin{pmatrix}
1+x^2 & -1\\
0 & 1+x^2 \end{pmatrix}$, $H_{g_x} =
\begin{pmatrix}
1 & -1-x\\
-1-x & 0 \end{pmatrix}$, $T_{g_x} =
\begin{pmatrix}
-1 & 1\\
0 & -1 \end{pmatrix}$
$$B = \begin{pmatrix}
1 & -1-x\\
-x & -x \end{pmatrix} - \begin{pmatrix}
1+x^2 & -1+x^2+x^4\\
x^2+x^4 & -x^2 \end{pmatrix}=
\begin{pmatrix}
-x^2 & -x^4-x^2-x\\
-x^4-x^2-x & x+x^2
\end{pmatrix}$$
|
We can solve the two polynomial equations using $S$-polynomials (from Buchberger's algorithm). Apart from the trivial solution $(x,y)=(0,1)$ this yields
$$
x=y^4 + y^3 - 2y^2 + y - 1,
$$
so that we obtain a polynomial in $y$, namely
$$
y^6 + y^5 - y^4 + 2y^3 - 2y^2 + y - 1=0.
$$
This has $6$ solutions over the complex numbers. Two of them are real solutions,
i.e., $y=0.863774039675$ and $y=- 2.15866388153$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving an identity involving floor function
Prove that :
$$\left \lfloor \dfrac{2 a^2}{b} \right \rfloor - 2 \left \lfloor \dfrac{a^2}{b} \right \rfloor = \left \lfloor \dfrac{2 (a^2 \bmod b)}{b} \right \rfloor $$
Where $a$ and $b$ are positive integers.
Please provide some hints/solutions. Thanks.
|
Let us assume that $b>0$. By division algorithm $a^2=bq+r$ with $0 \leq r <b$. Then
\begin{align*}
a^2&=bq+r\\
\frac{2a^2}{b} & =2q+\frac{2r}{b}\\
\left\lfloor \frac{2a^2}{b} \right\rfloor & = 2q+\left\lfloor \frac{2r}{b} \right\rfloor\\
\left\lfloor \frac{2a^2}{b} \right\rfloor -2q& =\left\lfloor \frac{2r}{b} \right\rfloor\\
\left\lfloor \frac{2a^2}{b} \right\rfloor -2\left\lfloor \frac{a^2}{b} \right\rfloor& =\left\lfloor \frac{2(a^2 \bmod b)}{b} \right\rfloor\\
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Can composite numbers be uniquely written as a sum of two squares?
Let $X = a^2 +b^2$ where all the terms are positive integers and $X$ is a composite number and $\gcd(a,b)=1$ . Do there exist positive integers $c$ and $d$ other than $a$ and $b$ such that $X = c^2+d^2$ ?
By Fermat's Two Square Theorem, since $\gcd(a,b) =1$ , all prime factors of $X$ (other than $2$) must be of the form $4k+1$. I can prove that primes of the form $4k+1$ can be uniquely written as a sum of two squares, but can composites also follow the same property?
Any help will be appreciated.
Thanks.
|
If $N$ is a product of two numbers of the form $4k+1$, then it usually has at least two representations as a sum of two squares, because:
$$\begin{align}
(a^2+b^2)(c^2+d^2) & = (ac-bd)^2 + (ad+bc)^2 \\
& = (ac+bd)^2 + (ad-bc)^2
\end{align}$$
This is known as
Brahmagupta's identity.
By applying Brahmagupta's identity repeatedly, you can find composite numbers that can be decomposed into two squares in arbitrarily many ways. For example,
$$\begin{align}
5^4\cdot 13 & = \\
8125 & = 5^2 + 90^2 \\
& = 27^2 + 86^2 \\
& = 30^2 + 85^2 \\
& = 50^2 + 75^2 \\
& = 58^2 + 69^2
\end{align}$$
The smallest such example is $25 = 3^2+4^2 = 0^2+5^2$.
|
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|
Given $f(z)=\dfrac{U(z)}{V(z)}=\dfrac{2z^3-3z^2+7z-8}{z^4-5z^3+4z^2-6z+1}$ find $f(1-\sqrt{2}i)$ without lots of complex arithmetic. Such a problem is usually done either by direct substitution (ugh!) or synthetic division.
Synthetic division after several complex products and additions gives $U(1-\sqrt{2}i)=-8-3\sqrt{2}i$
After several more complex operations one finds, barring errors, that $V(1-\sqrt{2}i)=9+7\sqrt{2}i$
leaving the solution to the straightforward evaluation of
\begin{equation}
f(1-\sqrt{2}i)=\frac{-8-3\sqrt{2}i}{9+7\sqrt{2}i}
\end{equation}
All standard stuff. The question is this: Can this be done without so many complex operations?
|
Since it has been 12 hours and no more answers have been suggested, I am posting a solution based upon a suggestion by @almagest.
\begin{equation}
[z-(1-\sqrt{2}i)]\cdot[z-(1+\sqrt{2}i)]=z^2-2z+3
\end{equation}
Using long division (or better yet, quadratic synthetic division) no complex arithmetic is required to obtain
\begin{equation}
U(z)=(2z-1)(z^2-2z+3)+3z-11
\end{equation}
and
\begin{equation}
V(z)=(z^2-3z-5)(z^2-2z+3)-7z+16
\end{equation}
Therefore
\begin{equation}
U(1-\sqrt{2}i)=0+3(1-\sqrt{2}i)-11=-8-3\sqrt{2}i
\end{equation}
and
\begin{equation}
V(1-\sqrt{2}i)=0-7(1-\sqrt{2}i)+16=9+7\sqrt{2}i
\end{equation}
Therefore
\begin{align}
f(1-\sqrt{2}i)&=\frac{-8-3\sqrt{2}i}{9+7\sqrt{2}i}\\
&=\frac{-8-3\sqrt{2}i}{9+7\sqrt{2}i}\cdot\frac{9-7\sqrt{2}i}{9-7\sqrt{2}i}\\
&=\frac{-114+29\sqrt{2}i}{179}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1787748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Inclusion–Exclusion Identical Computers Problem
Find the number of ways to distribute 19 identical computers to four schools, if School A must get at least three, School B must get
at least two and at most five, School C get at most four, and School D gets the rest.
a) Solve using inclusion-exclusion
b) Solve using generating functions
I've been tackling this question for a couple of days and I am pretty confused where to even start.
So I bet the answer is in the form $x_1 + x_2 + x_3 + x_4 = 19$,
where $x_1 \ge 3; 2 \le x_2 \le 5; x_3 \le 4$ and $x_4$ is whatever is left over.
And then I get confused. I've looked all over the internet, through textbooks and I'm not getting anywhere. Any help would be appreciated, thank you.
|
Addendum to the generating function part of the answer of @callculus. Assuming WA is not available, it's not too hard to calculate the coefficient by hand. In order to do so it's convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write e.g.
\begin{align*}
[x^j](1+x)^n=\binom{n}{j}
\end{align*}
We obtain
\begin{align*}
[x^{19}]&\left(\sum_{k=3}^{17}x^k\right)\left(\sum_{k=2}^5x^k\right)\left(\sum_{k=0}^4x^k\right)\left(\sum_{k=0}^{14}x^k\right)\\
&=[x^{19}]\left(\sum_{k=0}^{14}x^{k+3}\right)\left(\sum_{k=0}^3x^{k+2}\right)\frac{1-x^5}{1-x}\cdot\frac{1-x^{15}}{1-x}\tag{1}\\
&=[x^{19}]x^3\frac{1-x^{15}}{1-x}\cdot x^2\frac{1-x^4}{1-x}\cdot\frac{1-x^5}{1-x}\cdot\frac{1-x^{15}}{1-x}\tag{2}\\
&=[x^{14}]\left(1-x^{15}\right)^2\left(1-x^4\right)\left(1-x^5\right)\sum_{n=0}^{\infty}\binom{-4}{n}(-x)^n\tag{3}\\
&=[x^{14}]\left(1-x^4\right)\left(1-x^5\right)\sum_{n=0}^{\infty}\binom{-4}{n}(-x)^n\tag{4}\\
&=[x^{14}](1-x^4-x^5+x^9)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n\tag{5}\\
&=\left([x^{14}]-[x^{10}]-[x^9]+[x^5]\right)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n\tag{6}\\
&=\binom{17}{3}-\binom{10}{3}-\binom{9}{3}+\binom{8}{3}\tag{7}\\
&=680-286-220+56\\
&=230
\end{align*}
Comment:
*
*In (1) we shift the index of $k$ to start from $0$ and apply the finite geometric series formula.
*In (2) we apply the finite geometric series formula again.
*In (3) we use the linearity of the coefficient of operator and apply the rule $$[x^{n-m}]A(x)=[x^n]x^mA(x)$$ We also use the binomial series expansion of $\frac{1}{(1-x)^4}$.
*In (4) we observe that we can replace the binom $(1-x^{15})$ with $1$, since multiplication with $x^{15}$ does nothing contribute to the coefficient of $x^{14}$.
*In (5) we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
*In (6) we apply the rule $[x^{n-m}]A(x)=[x^n]x^mA(x)$ again.
*In (7) we select the coefficients accordingly.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to factor $a^3 - b^3$? I know the answer is $(a - b)(a^2 + ab + b^2)$, but how do I arrive there? The example in the book I'm following somehow broke down $a^3 - b^3$ into $a^3 - (a^2)b + (a^2)b - a(b^2) + a(b^2) - b^3$ and factored that into $(a-b)(a^2 + ab + b^2)$ from there, but I don't quite understand how it was done.
|
Long division makes it as easy as $1$, $2$, $3$:
$$\begin{align}
\frac{a^3-b^3}{a-b}&=a^2+\frac{ba^2-b^3}{a-b} \tag 1\\\\
&=a^2+ab+\frac{b^2a-b^3}{a-b} \tag 2\\\\
&=a^2+ab+b^2 \tag 3
\end{align}$$
as was to be shown!
Note that in $(1)$, the term $\frac{ba^2-b^3}{a-b}$ is the remainder of $a^2$ in $\frac{a^3-b^3}{a-b}$.
Note that in $(2)$, the term $\frac{b^2a-b^3}{a-b}$ is the remainder of $ab$ in $\frac{ba^2-b^3}{a-b}$.
Note that in $(3)$ the remainder is zero after dividing $b^2a-b^3$ by $a-b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1789286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Please help me compute this$ \sum_m\binom{n}{m}\sum_k\frac{\binom{a+bk}{m}\binom{k-n-1}{k}}{a+bk+1}$ Compute following:
$$
\sum_m\binom{n}{m}\sum_k\frac{\binom{a+bk}{m}\binom{k-n-1}{k}}{a+bk+1}
$$
Only consider real numbers a, b such that the denominators are never 0.
Now I simplify it into
$$
-\frac{1}{n}\sum_k\binom{n}{k}\binom{-n}{a+bk+1}(-1)^{a+bk+k}
$$
I have problem with this question in which I can't eliminate coefficient b.
But I can't find any formula to get answer.Please help me!
|
Here is a slightly different variation of the theme. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\sum_{m=0}^{n}&\binom{n}{m}\sum_{k=0}^{n}\frac{1}{a+bk+1}\binom{a+bk}{m}\binom{k-n-1}{k}\tag{1}\\
&=\sum_{k=0}^{n}\frac{(-1)^k}{a+bk+1}\binom{n}{k}\sum_{m=0}^{n}\binom{n}{m}\binom{a+bk}{m}\tag{2}\\
&=\sum_{k=0}^{n}\frac{(-1)^k}{a+bk+1}\binom{n}{k}\binom{a+bk+n}{n}\tag{3}\\
&=\frac{1}{n}\sum_{k=0}^{\infty}(-1)^k\binom{n}{k}\binom{a+bk+n}{n-1}\tag{4}\\
&=\frac{1}{n}\sum_{k=0}^{\infty}(-1)^k[z^{k}](1+z)^{n}[u^{n-1}](1+u)^{a+bk+n}\tag{5}\\
&=\frac{1}{n}[u^{n-1}](1+u)^{a+n}\sum_{k=0}^\infty(-1)^k(1+u)^{bk}[z^k](1+z)^n\tag{6}\\
&=\frac{1}{n}[u^{n-1}](1+u)^{a+n}(1-(1+u)^b)^n\tag{7}\\
&=\frac{(-1)^n}{n}[u^{n-1}](1+u)^{a+n}\left(\sum_{j=1}^\infty\binom{b}{j}u^j\right)^n\tag{8}\\
&=0
\end{align*}
Comment:
*
*In (1) we write lower and upper limits of the sum.
*In (2) we exchange the sums, do small rearrangements and use the identity
\begin{align*}
\binom{k-n-1}{k}=(-1)^k\binom{n}{k}
\end{align*}
*In (3) we apply Vandermonde's identity. With $q:= a+bk$ we get
\begin{align*}
\sum_{m=0}^{n}\binom{n}{m}\binom{q}{m}=\sum_{m=0}^{n}\binom{q}{m}\binom{n}{n-m}=\binom{q+n}{n}
\end{align*}
*In (4) we use the identity
\begin{align*}
\frac{q}{p-q+1}\binom{p}{q}=\binom{p}{q-1}\\
\end{align*}
and we also change the uppper limit of the series to $\infty$ without changing anything since we add only zeros.
*In (5) we apply the coefficient of operator twice
*In (6) we use the linearity of the coefficient of operator and do some rearrangements
*In (7) we apply the substitution rule of the coefficient of operator
\begin{align*}
A(u)=\sum_{k=0}^\infty a_ku^k=\sum_{k=0}^\infty u^k[z^k]A(z)
\end{align*}
*In (8) we use the binomial series expansion and observe the smallest power of $u$ is $n$ so that the coefficient $[u^{n-1}]$ is zero.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1789981",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Sketch the heart and indicate its orientation with arrows $ r = 1 - \cos(\theta)$. Find the area enclosed by the heart Hi all I am trying to figure out how to sketch the heart. Here is what I have tried so far:
$$r = 1 - \cos(\theta) \\
r(r = 1 - \cos(\theta)) \\
r^2 = r - r\cos(\theta) \\
$$
Use the fact that
$$r =\sqrt{x^2 + y^2} \text{ and } x=r\cos(\theta) \text{ to rewrite the equation as , } \\
x^2 + y^2 = \sqrt{x^2 + y^2} + x \\
x^2 - 2x + y^2 - y = 0 \\
\text{Completing the square gives } \\
(x-1)^2 + (y-\frac{1}{2})^2 = -\frac{5}{4}\\
\sqrt{(x-1)^2 + (y-\frac{1}{2})^2 } = \sqrt{\frac{5}{4}} \\
(x-1) + (y-\frac{1}{2}) = \sqrt{\frac{5}{4}} \\
(y-\frac{1}{2}) = -(x-1) + \sqrt{\frac{5}{4}} \\
y = -(x-1) + \sqrt{\frac{5}{4}} + \frac{1}{2} \\
y = -x + \frac{\sqrt{5} + 3}{2}
$$From here we can get a table of values and sketch the function. For the area the formula is
$$\frac{1}{2} \int f(\theta)^2 \,d\theta$$
So should it be
$$\frac{1}{2} \int y = -x + \frac{\sqrt{5} + 3}{2}$$
|
Drawing the figure is the hard part by far. For example, Matlab doesn't have a method for directly adding arrows to a curve like this. What I tried was to draw a little right-pointing arrowhead with its tip at $(0,0)$ and some reasonable size. Then at a selection of points along the path, I rotated that arrow counterclockwise by $\text{atan2}\left(\frac{dy}{d\theta},\frac{dx}{d\theta}\right)$ and then added $(x(\theta),y(\theta))$ to its coordinates. Here is my Matlab code:
% cardioid.m
% Plot main curve
theta = linspace(0,2*pi,500);
r = 1-cos(theta);
x = r.*cos(theta);
y = r.*sin(theta);
plot(x,y,'b-');
hold on;
% Plot arrows
asize = 0.07;
axy = [-1 0 -1;
-1 0 1]*asize/sqrt(2);
% We want fewer points
theta = linspace(0,2*pi,10);
% Avoid the cusp
theta = theta(2:end-1);
r = 1-cos(theta);
x = r.*cos(theta);
y = r.*sin(theta);
% Orientations needed
drdtheta = sin(theta);
phi = atan2(sin(theta).*drdtheta+r.*cos(theta), ...
cos(theta).*drdtheta-r.*sin(theta));
for k = 1:length(theta),
R = [cos(phi(k)) -sin(phi(k));
sin(phi(k)) cos(phi(k))];
xy = R*axy;
plot(xy(1,:)+x(k), xy(2,:)+y(k),'b-');
end
% Titles and labels
axis equal;
xlabel('x');
ylabel('y');
title('Cardioid with Direction');
hold off;
And here is the figure it drew:
Having drawn the cardioid, finding its area is anticlimactic:
$$A=\int_0^{2\pi}\int_0^{1-\cos\theta}r\,dr\,d\theta=\int_0^{2\pi}\frac12\left(1-2\cos\theta+\cos^2\theta\right)d\theta=\frac12(2\pi)\left(1-0+\frac12\right)=\frac{3\pi}2$$
Where we have used the average values for $1$ of $1$, for $\cos\theta$ of $0$ and for $\cos^2\theta$ of $\frac12$ over the interval of length $2\pi$ to zip through the azimuthal integral.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$ $a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$
I try to apply AM-GM
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3\cdot \sqrt[3]{\left(\frac{a+1}{a+b} \right)^{\frac25}\left(\frac{b+1}{b+c} \right)^{\frac25}\left(\frac{c+1}{c+a} \right)^{\frac25}}$$
Thus it remains to prove
$$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1 $$ with the condition $a+b+c=3.$
But I found the counter example for $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1 $$ :(
|
Only a partial answer.
Assuming $a\le b\le c$, then we have $0<a\le1$ and $1\le c<3$. Now we have two cases, $b\le1$ and $b>1$. The case $b\le1$ is easy to deal with.
Assuming $0<a\le b\le1\le c<3$, we have
\begin{align}
\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \ge 1
&\iff (a+1)(b+1)(4-a-b)\ge(a+b)(3-a)(3-b)\\
&\iff {\left(2-a-b\right)} {\left(1-a\right)} {\left(1-b\right)}\ge0.
\end{align}
Now assuming $0<a\le1\le b \le c<3$. Below is not an answer but only an analysis.
We want to know why this case is difficult, and why $\frac25$ is important.
Using the series expansion of $\exp$, and let $A = \operatorname{diag}\left(\ln\left(\frac{a+1}{a+b} \right),\ln\left(\frac{b+1}{b+c}\right),\ln\left(\frac{c+1}{c+a} \right)\right)$, then we have
\begin{align}
LHS
&= \sum_{n=0}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!}\\
&= 3 + \frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3 +R_3,
\end{align}
where
$R_3 = \sum_{n=4}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!} \ge 0$, since $e^x-(1+x+x^2/2+x^3/6)$ is positive for any $x\in\mathbb R$.
Then we can prove the inequality if we have
$$\frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3\ge\!\!\!?\;0,$$
which can be simplified to
$$75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge\!\!\!?\;0.\tag{1}$$
Numerical results suggest that using the 3 first terms is enough to prove the inequality. Note that in the first case where $b\le1$, using the first term $\operatorname{tr}A$ is enough (and what we did in the first part is in fact proving $\operatorname{tr}A\ge0$), that's why that case is easy.
So,
*
*Why the case $b\ge1$ is difficult?
Because we have 2 more terms, $\operatorname{tr}A^2$ and $\operatorname{tr}A^3$, to deal with.
*Why $\frac25$ is important?
Because $\frac25$ gives the coefficients 75, 15, and 2, which makes $75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c ^{\frac34}}{2^{\frac34}}$$
I tried the substitution $x=a^4,\ldots$ but I have no idea how to deal with the left- hand side. I tried some C-S but it goes nowhere. I think Bernoulli's inequality may be the only way to prove this inequality.
|
Partial answer :
Hint :
Using WRCF theorem see https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2011-101
We have $c=\min{(a,b,c)}$ and let $f(x)$ such that :
$$f\left(x\right)=\left(\frac{1}{1+x^{3}}\right)^{\frac{3}{4}}$$
It seems we have $f''(x)>0$ for $x\geq 1$
For and $a,b,c\in[1,2]$ it seems we have :
$$g\left(x\right)=\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}f\left(x\right)+\left(1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}\right)f\left(\frac{\left(1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}x\right)}{1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}}\right)-f\left(1\right)\geq 0$$
Where $0.5\leq x\leq 1$ .
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1793680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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|
Is $x^4 + 4$ irreducible in $\mathbb{Z}_5$? Well, I'm having doubts, isnt that $\mathbb{Z}_5$ has no zero divisors, and now you cant factor $x^4 + 4$ ?
|
\begin{align}
x^4+4&=x^4-1\\
&=(x^2-1)(x^2+1)\\
&=(x-1)(x+1)(x^2-4)\\
&=(x-1)(x+1)(x+2)(x-2)\\
&=(x-1)(x-2)(x-3)(x-4).
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1796350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solve summation expression For a probability problem, I ended up with the following expression
$$\sum_{k=0}^nk\ \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k$$
Using Mathematica I've found that the result should be $\frac{n}{3}$. However, I have no idea how to get there. Any ideas?
|
\begin{align*}
\sum_{k=0}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k &=\sum_{k=0}^n k\frac{n!}{k!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\
&=\sum_{k=0}^n \frac{n!}{(k-1)!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\
&=\frac{n}{3}\sum_{k=0}^n \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\
&=\frac{n}{3}\sum_{k=0}^n \binom{n-1}{k-1}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\
&=\frac{n}{3}\left(\frac{2}{3}+\frac{1}{3}\right)^{n-1}\\
&=\frac{n}{3}.
\end{align*}
The second to last line is a result of the binomial theorem.
Edit: It was pointed out that I need to be careful when $k=0$. I also made a mistake in applying the binomial theorem. Here is a revised proof. Note that when $k=0$, the term of the sum is $0$, so it is the same as starting at $k=1$.
\begin{align*}
\sum_{k=0}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k &= \sum_{k=1}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=
\sum_{k=1}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\ &=\sum_{k=1}^n k\frac{n!}{k!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\
&=\sum_{k=1}^n \frac{n!}{(k-1)!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\
&=\frac{n}{3}\sum_{k=1}^n \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\
&=\frac{n}{3}\sum_{k=1}^n \binom{n-1}{k-1}\left(\frac{2}{3}\right)^{(n-1)-(k-1)}\left(\frac{1}{3}\right)^{k-1}\\
&=\frac{n}{3}\sum_{k=0}^{n-1} \binom{n-1}{k}\left(\frac{2}{3}\right)^{(n-1)-k}\left(\frac{1}{3}\right)^{k}\\
&=\frac{n}{3}\left(\frac{2}{3}+\frac{1}{3}\right)^{n-1}\\
&=\frac{n}{3}.
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/1798085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Verify the correctness of $\sum_{n=1}^{\infty}\left(\frac{1}{x^n}-\frac{1}{1+x^n}+\frac{1}{2+x^n}-\frac{1}{3+x^n}+\cdots\right)=\frac{\gamma}{x-1}$ $x\ge2$
$\gamma=0.57725166...$
(1)
$$\sum_{n=1}^{\infty}\left(\frac{1}{x^n}-\frac{1}{1+x^n}+\frac{1}{2+x^n}-\frac{1}{3+x^n}+\cdots\right)=\frac{\gamma}{x-1}$$
Series (1) converges very slowly, we are not sure that (1) has closed form $\frac{\gamma}{x-1}$
Can anyone verify (1)?
|
For $x =2$ the identity is true.
Claim. $\displaystyle \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k} = \gamma. $
Proof. We first rearrange the given sum:
$$ S
:= \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k}
= \sum_{l = 2}^{\infty} \bigg( \sum_{\substack{(n,k) \ : \ 2^n + k = l \\ n \geq 1, k \geq 0}} \frac{(-1)^k}{2^n+k} \bigg)
= \sum_{l = 2}^{\infty} \frac{(-1)^l}{l} \lfloor \log_2 l \rfloor. \tag{1} $$
(For a careful reader: see below for a rigorous proof.) Then group each $2^n$ terms to write
$$ S = \sum_{n=1}^{\infty} n \bigg( \sum_{l=2^n}^{2^{n+1}-1} \frac{(-1)^l}{l} \bigg) = \sum_{n=1}^{\infty} n ( A_{n-1} - A_n ), $$
where $A_n = H(2^{n+1}-1) - H(2^n-1)$ and $H(n) = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$ is the $n$-th harmonic number. Now the $N$-th partial sum of the RHS is
\begin{align*}
\sum_{n=1}^{N} n(A_{n-1} - A_n)
&= (A_0 + \cdots + A_{N-1}) - N A_N \\
&= (N-1)H(2^N-1) - N H(2^{N+1}-1).
\end{align*}
Taking limit as $N \to \infty$ gives the desired limit $\gamma$. ////
Remark. Let $f(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k}$.
*
*Using the formula $f(z) = \psi_0(z) - \psi_0(z/2) - \log 2$ together with the asymptotic expansion of $\psi_0(z)$ gives a much shorter proof.
*We can check that $f(z) = \frac{1}{2z} + \mathcal{O}(z^{-2})$. Thus
$$ \sum_{n=1}^{\infty} f(x^n) = \frac{1}{2(x-1)} + \mathcal{O}\left(\frac{1}{x^2} \right). $$
This tells us that the proposed formula is false, which is already confirmed by other users.
Addendum 1 - A careful justification of $\text{(1)}$. The trick is to group each pair of successive terms to create a non-negative sum. Then by the Tonelli's theorem we can freely rearrange the order of summation.
\begin{align*}
\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k}
&= \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(2^n+2k)(2^n+2k+1)} \\
&= \sum_{l' = 1}^{\infty} \bigg( \sum_{\substack{(m,k) \ : \ 2^m + k = l' \\ m \geq 0, k \geq 0}} \frac{1}{(2^{m+1}+2k)(2^{m+1}+2k+1)} \bigg) \\
&= \sum_{l' = 1}^{\infty} \frac{1 + \lfloor \log_2 l' \rfloor}{2l'(2l'+1)} \\
&= \sum_{l' = 1}^{\infty} \bigg( \frac{\lfloor \log_2 (2l') \rfloor}{2l'} - \frac{\lfloor \log_2 (2l'+1) \rfloor}{2l'+1} \bigg)\\
&= \sum_{l = 1}^{\infty} \frac{(-1)^l}{l}\lfloor \log_2 l \rfloor.
\end{align*}
Addendum 2 - Heuristic. We use the following asymptotic expansion
$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k} = \psi_0(z) - \psi_0(z/2) - \log 2 = \frac{1}{2z} + \sum_{k=1}^{\infty} \frac{B_{2k}}{2k} \frac{2^{2k}-1}{z^{2k}}, $$
where $(B_k)$ are Bernoulli numbers. (Notice: This is only a formal sum because the asymptotic expansion above does not converge for any $z$. It is numerically meaningful only when we truncate the sum.) Then we formally have
\begin{align*}
\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{x^n + k}
&= \frac{1}{2(x-1)} + \sum_{n=1}^{\infty} \frac{B_{2n}}{2n} \frac{2^{2n}-1}{x^{2n}-1} \\
&= \frac{1}{2(x-1)} + 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \Gamma(2n)\zeta(2n)}{(2\pi)^{2n}} \frac{2^{2n}-1}{x^{2n}-1} \\
&= \frac{1}{2(x-1)} + 2 \int_{0}^{\infty} \bigg( \sum_{n=1}^{\infty} (-1)^{n-1}\zeta(2n) \frac{2^{2n}-1}{x^{2n}-1} t^{2n-1} \bigg) e^{-2\pi t} \, \mathrm{d}t.
\end{align*}
Although the summation in the last line does not converge for $t \geq x/2$, it may have an analytic continuation on all of $t > 0$. Then with that continuation the last expression may make sense and even possibly give the correct formula for the original sum. Indeed, when $x = 2$ it follows that
$$ 2 \sum_{n=1}^{\infty} (-1)^{n-1}\zeta(2n) t^{2n-1} = \pi \coth \pi t - \frac{1}{t} $$
and hence
$$ \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n + k} \text{ “}=\text{" } \frac{1}{2} + \int_{0}^{\infty} \bigg( \pi \coth \pi t - \frac{1}{t} \bigg) e^{-2\pi t} \, \mathrm{d}t = \gamma. $$
Now we know that this alleged equality is indeed true. So the above formal computation seems to give the correct answer (at least for $x =2$, and hopefully for $x > 2$). But for $x > 2$, I have no idea what will come out.
|
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|
Evaluating the arc length integral $\int\sqrt{1+\frac{x^4-8x^2+16}{16x^2}} dx$
Find length of the arc from $2$ to $8$ of $$y = \frac18(x^2-8 \ln x)$$
First I find the derivative, which is equal to $$\frac{x^2-4}{4x} .$$
Plug it into the arc length formula $$\int\sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$ and get
$$\int\sqrt{1+\frac{x^4-8x^2+16}{16x^2}} dx.$$
I am not sure how to proceed from here, as I cant figure out a way to put the sqrt argument into a form that I can find the square root of
|
Hint
$$1 + \frac{x^4 - 8 x^2 + 16}{16 x^2} = \frac{16 x^2}{16 x^2} + \frac{x^4 - 8 x^2 + 16}{16 x^2} =\frac{x^4 + 8 x^2 + 16}{16 x^2}$$
Can you write the rightmost expression as a square of another expression?
|
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|
Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$. I've been working on this epsilon delta proof for the longest time now, and I can't quite get it.
Let $a>0$ and $b>0$. Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$.
I've found that I need to prove $|x-b|$ < $\epsilon$ $|a+x|$ $|a+b|$ and that I can do this by calling $\delta$ something greater than $|x-b|$, and that I also need to eliminate $|a+x|$ somehow, but that's where I'm struggling.
The concept of calling $\delta$ the minimum of a few values has been helpful, but I'm not sure how to apply it here.
Thanks for your thoughts.
|
Let $\varepsilon > 0$. Want some $\delta > 0$ such that $0 < |x-b| < \delta$ implies $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$. We start with the inequality $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$ and try to manipulate it to give us insight into what $\delta$ should be. We have
\begin{align}
\frac{1}{a + x} - \frac{1}{a + b} &= \frac{a+b - (a+x)}{(a+x)(a+b)} \\
&= \frac{1}{b+a} \cdot \frac{b-x}{a+x} \\
\implies \left| \frac{1}{a + x} - \frac{1}{a + b} \right| &= \frac{1}{b+a} \cdot \frac{\left| b-x \right|}{\left| a+x \right|}. \tag{1}
\end{align}
We want to establish an upper bound for the LHS, which is accomplished by picking the largest possible RHS. This is found by maximizing $|b - x|$ and minimizing $|a - x|$. Our selection of $\delta$ determines which values of $x$ are allowed:
$$
0 < |x - b| < \delta \iff x \in (b-\delta, b) \cup (b, b+\delta). \tag{2}
$$
Certainly $(2)$ implies the maximum of $|b - x|$ is $\delta$. The minimum of $a+x$ is $a + b - \delta$, and hence
$$
\text{minimum of } |a + x| = \begin{cases}
a + b - \delta & a + b - \delta > 0 \\
0 & \text{otherwise}
\end{cases}. \tag{3}
$$
Armed with this info, let's try to pick a reasonable $\delta$ to bound the LHS of $(1)$. By $(3)$ we had better
$$
\text{enforce $\quad \delta < a + b$}. \tag{4}
$$
If we do, by $(1)$ we have
\begin{align*}
\left| \frac{1}{a + x} - \frac{1}{a + b} \right| & \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b + \delta} \\
& \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b}. \tag{5}
\end{align*}
Now, if we also
$$
\text{enforce $\quad \delta < \varepsilon \cdot (a + b)^2$}, \tag{6}
$$
then $(5)$ becomes
$$
\left| \frac{1}{a + x} - \frac{1}{a + b} \right| \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b} < \frac{1}{(a + b)^2} \cdot \varepsilon \cdot (a + b)^2 = \varepsilon.
$$
That is, if we enforce $(4)$ and $(6)$, i.e., $\delta < \min\{a + b, \varepsilon \cdot (a+b)^2\}$, then $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$, as desired.
|
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|
Solve $ord_x(2) = 20$ Given that the (multiplicative) order of $2$ mod $x$ is $20$, how can I work out what $x$ is?
|
Method 1:-
From Euler Totient Function(or theorem) we know-
$$a^{\phi(n)}\equiv1\pmod n$$ for $\gcd(a,n)=1$
Here you have,$2^{20}\equiv1\pmod n$.
Comparing the two equations we have,$\phi(n)=20$.Try to find $n$ from it.
Method 2:-
We know that for all $n$,$(a-b)|(a^n-b^n)$ and $a^n-b^n$ is also divisible by $a+b$ if $n$ is odd.
So,We need to find $x$ such that $x|2^{20}-1$
Note that $2^{20}-1$ can be written as $2^{20}-1^{20}$.
This can be written in the following forms-
$(2^5)^4-(1^5)^4$ Hence,$n=2^5-1$ is a factor.
$(2^{10})^2-(1^{10})^2$ Hence,$n=2^{10}-1$ is a factor.
$(2^4)^5-(1^4)^5$ Hence,both $2^4-1$ and $2^4+1$ are factors.
Important note-Factors of the values of $n$ given are also factors of the number.
|
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|
Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot \sin^2 xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot (1-\cos^2 x)dx$$
$$I_{n} =I_{n-1}-\int_{0}^{\frac{\pi}{2}}\cos x\cdot \sin^{2n-2}\cdot \cos xdx$$
Now Using Integration by parts, We get $$I_{n} = I_{n-1}-\frac{I_{n}}{2n-1}\Rightarrow I_{n} = \frac{2n-1}{2n}I_{n-1}$$
Now Using Recursively, We get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I_{n-2} =\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}I_{n-3}$$
So we get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \frac{2n-7}{2n-6}\cdot \cdot \cdot \cdot \cdot \cdot \cdot\cdot \frac{3}{2}I_{0}$$
and we get $\displaystyle I_{0} = \frac{\pi}{2}$
So we get $$I_{n} = \frac{(2n)!}{4^n\cdot n!\cdot n!}\cdot \frac{\pi}{2}$$
Now I did not understand How can I calculate value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Help Required, Thanks.
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Here's another probabilistic approach. We don't evaluate any (Riemann) integrals.
Consider a simple symmetric random walk on ${\mathbb Z}$ starting from $0$ at time $0$. The probability that at time $2n$ the walk is at zero is equal to $\binom{2n}{n}2^{-2n}$, and the probability that at time $2n+1$ the walk is at zero is $0$.
From this it follows that the expression we wish to evaluate is
$$S= \sum_{j=0}^\infty E[ 2^{-T_j}].$$
where $0=T_0<T_1<\dots $ are the times the walk is at zero. Note that $(T_{j+1}-T_j)$ are IID and have the same distribution as $T_1$. Therefore, this is a geometric series. Its sum is
$$S = \frac{1}{1-E [2^{-T_1}]}.$$
To compute $E [ 2^{-T_1}]$ we consider first $\rho$, the time until the walk hits $1$, starting from $0$. Conditioning on the first step, we have
$$ E[ 2^{-\rho}] = 2^{-1} \left ( \frac 12 + \frac 12 E [2^{-\rho}]^2\right),$$
representing, either moving to the right first or moving to the left first. Therefore
$$ E [2^{-\rho} ] ^2 -4E [ 2^{-\rho}]+1=0,$$
or
$$(E[ 2^{-\rho}] -2)^2-3=0 \quad \Rightarrow \quad E[2^{-\rho}] = 2-\sqrt{3} $$
Let's go back to our original problem. Conditioning on the first step,
$$ E [2^{-T_1} ] = 2^{-1} E [ 2^{-\rho}]=1- \frac{\sqrt{3}}{2}.$$
Thus,
$$ S = \frac{2}{\sqrt{3}}.$$
|
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|
Why is this incorrect $\int_{0}^{1}{\ln(x)\over (1+x)^3}dx=-\sum_{n=0}^{\infty}{(-1)^n(n+2)\over 2(1+n)}$ $$I=\int_{0}^{1}{\ln(x)\over (1+x)^3}dx$$
Recall
$${1\over (1+x)^3}=\sum_{n=0}^{\infty}{(-1)^n(n+1)(n+2)\over 2}x^n$$
$$\int_{0}^{1}x^n\ln(x)dx=-{1\over (n+1)^2}$$
Substitute in
$$I=\sum_{0}^{\infty}{(-1)^n(n+1)(n+2)\over 2}\int_{0}^{1}x^n\ln(x)dx$$
$$I=-\sum_{n=0}^{\infty}{(-1)^n(n+1)(n+2)\over 2(1+n)^2}$$
$$\int_{0}^{1}{\ln(x)\over (1+x)^3}dx=-\sum_{n=0}^{\infty}{(-1)^n(n+2)\over 2(1+n)}$$
Can somebody help me here. I don't understand why it is incorrect here. Where did I go wrong?
I still don't get it, because these two work perfectly fine
$$\int_{0}^{1}{\ln(x)\over 1+x}dx=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^n\ln(x)dx=-\sum_{n=0}^{\infty}{(-1)^n\over (n+1)^2}$$
$$\int_{0}^{1}{\ln(x)\over (1+x)^2}dx=\sum_{n=0}^{\infty}(-1)^n(n+1)\int_{0}^{1}x^n\ln(x)dx=-\sum_{n=0}^{\infty}{(-1)^n(n+1)\over (n+1)^2}$$
|
You are working on the assumption that
\begin{align*}
&\int_{0}^{1} \left( \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+2)}{2} x^n \right) \log x \, \mathrm{d}x \\
&\hspace{5em} = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+2)}{2} \int_{0}^{1} x^n \log x \, \mathrm{d}x.
\tag{1}
\end{align*}
But this is not true because the latter sum does not converge in ordinary sense. We can, however, give an alternative meaning to this formula, which eventually leads to the correct answer. Notice that for $r \in (0, 1)$ the following holds
\begin{align*}
&\int_{0}^{r} \left( \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+2)}{2} x^n \right) \log x \, \mathrm{d}x \\
&\hspace{5em} = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+2)}{2} \int_{0}^{r} x^n \log x \, \mathrm{d}x
\end{align*}
by the Fubini's theorem. So the issue of convergence occurs when you take limit as $r \uparrow 1$. However, this also shows that $\text{(1)}$ becomes true in Abel sense. So we have
$$ \text{(1)}
= - \frac{1}{2} \sum_{n=0}^{\infty}(-1)^{n} \left(1 + \frac{1}{n+1} \right)
= - \frac{1}{2} \left( \frac{1}{2} + \log 2 \right) \quad \text{in Abel sense.} $$
Here, we utilized that
$$ 1 - 1 + 1 - 1 + \cdots = \frac{1}{2} \quad \text{in Abel sense.} $$
|
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|
$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}\cdot\frac{2t}{\sqrt{t^4-2t^2}}dt=2\int\frac{1-t^2}{\sqrt{t^4-2t^2}}dt$
$\ = 2\displaystyle\int\frac{1}{\sqrt{t^4-2t^2}}dt-2\displaystyle\int\frac{t}{\sqrt{t^2-2t}}dt$
$\ = 2\displaystyle\int t^{-1}(t^2-2)^{-\frac{1}{2}}dt-2\displaystyle\int t(t^2-2t)^{-\frac{1}{2}}dt$
But after that I don't know how to continue.
|
By setting $x=\frac{\pi}{2}-t$ the problem boils down to finding:
$$ \int \frac{\cos t}{\sqrt{1-\cos t}}\,dt = \frac{1}{\sqrt{2}}\int\frac{1-2\sin^2\frac{t}{2}}{\sin\frac{t}{2}}\,dt $$
where:
$$ \int \frac{1}{\sin\frac{t}{2}}\,dt = C + 2\log\left(\tan\frac{t}{4}\right).$$
|
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|
How we can show this ;$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$ Let be $\quad x+y+z=0$
show this:
$$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$$
I solved ,but Im interesting what are you thinking about this,how can we arrive to solution quickly?
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Your polynomials are invariant by the action of $S_3$ on $(x,y,z)$.
A standard result says that the graded ring $\Bbb Q[x,y,z]^{S_3}$ is $\Bbb Q[x+y+z;xy+yz+xz;xyz]$.
Furthermore here, quotienting by $(x+y+z)$ we can work in the ring $(\Bbb Q[x,y,z]/(x+y+z))^{S_3} \cong \Bbb Q[xy+yz+xz;xyz]$.
Now in this graded ring, the pieces of degree $2,5,7$ are all of dimension $1$ (there is only one decomposition of $2,5,7$ as a sum of $2$s and $3$s).
Picking $x,y,z = 1,1,-2$ shows that $x^k+y^k+z^k$ are nonzero in that ring for $k > 1$, which proves that $(x^2+y^2+z^2)(x^5+y^5+z^5) = q (x^7+y^7+z^7) \pmod {x+y+z}$ for a rational number $q$ (and also that $(x^2+y^2+z^2)(x^3+y^3+z^3) = q' (x^5+y^5+z^5)$ but we don't particularly care).
Then you can easily check that when replacing $z$ with $-x-y$ in your expressions, the dominant coefficient in your polynomials is $1$ after doing the
divisions (note that for an odd prime $p$, by Fermat's little theorem, $(x^p+y^p+(-x-y)^p)/p$ even has integral coefficients, ... and the case $p=2$ just works), and you only have to check $1 \times 1 = 1$ to finish the proof.
You can also evaluate $(x^2+y^2+z^2)(x^5+y^5+z^5) = q (x^7+y^7+z^7)$ at $x,y,z = 1,1,-2$ to get $6\times (-30) = q \times (-126)$ hence $q=10/7$.
|
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|
derivative with square root I have been trying to figure this equation for some time now, but have come up empty. I have tried multiple ways on solving it. Whether by using the Quotient Rule or some other method, I can't seem to figure it out. Any help would be appreciated.
Find the derivative of the function
$
y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}
$
The answer is:
$
y' = \frac{3x^{2}+ 8x - 3 }{2x^\frac{3}{2}}
$
I'm having trouble getting to that answer. So if someone could point me in the right direction by showing the steps, I would be grateful. Thanks!
|
While using the quotient rule for $f(x)=\dfrac{h(x)}{g(x)}\rightarrow f'(x)=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}$ is straightforward and gives
$y'=\dfrac{(2x+8)\sqrt{x}-\frac{1}{2}(x^2+8x+3)x^{-\frac{1}{2}}}{x}=(2x+8)x^{-\frac{1}{2}}-\frac{1}{2}(x^2+8x+3)x^{-\frac{3}{2}}=x^{-\frac{3}{2}}\left[(2x+8)x-\frac{1}{2}(x^2+8x+3)\right]=\dfrac{3x^2+8x-3}{2x^{-\frac{3}{2}}}$
we can also, with a simple manipulation, avoid the quotient rule:
$y=x^{\frac{3}{2}}+8\sqrt{x}+3x^{-\frac{1}{2}}$
$y'=\frac{3}{2}x^{\frac{1}{2}}+4x^{-\frac{1}{2}}-\frac{3}{2}x^{-\frac{3}{2}}=\frac{3x^2+8x-3}{2x^{-\frac{3}{2}}}$
|
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|
Problem solving a word problem using a generating function
How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions.
So the first couple steps are easy.
The coefficient is $x^{24}$
$g(x) = x^6(1+x^2+x^4+x^6+x^8)^3$ or what I got was $x^6 (1 + (x^2)^1 +...+ (x^2)^4)^3$
now finding the closed formula is where I am having problems
My answer: using the fact that $\dfrac{1-x^{n+1}}{1- x}$
I get $x^6\left(\dfrac{1-x^9}{1-x}\right)$ which is wrong
The correct answer: $x^6\left(\dfrac{1-x^{10}}{1-x^2}\right)$
If someone could explain in some detail on how to get the correct formula would be much appreciated. Thanks!
|
Here is variation of the theme which might be helpful when doing the calculation. It's convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
[x^{24}]&(x^2+x^4+x^6+x^8+x^{10})^3\tag{1}\\
&=[x^{24}]x^6(1+x^2+x^4+x^6+x^8)^3\\
&=[x^{18}]\left(\frac{1-x^{10}}{1-x^2}\right)^3\tag{2}\\
&=[x^{18}](1-x^{10})^{3}\sum_{n=0}^\infty\binom{-3}{n}(-x^2)^n\tag{3}\\
&=[x^{18}](1-3x^{10})\sum_{n=0}^\infty\binom{n+2}{2}x^{2n}\tag{4}\\
&=\left([x^{18}]-3[x^8]\right)\sum_{n=0}^\infty\binom{n+2}{2}x^{2n}\tag{5}\\
&=\binom{11}{2}-3\binom{6}{2}\tag{6}\\
&=10
\end{align*}
Comment:
*
*In (1) we respect that an even number of at least $2$ and at most $10$ cookies is to distribute to three children. The coefficient of operator selects the coefficient of $x^{24}$ which gives the wanted number.
*In (2) we use the rule
\begin{align*}
[x^{p+q}]A(x)=[x^p]x^{-q}A(x)
\end{align*}
and apply the finite geometric series with argument $x^2$.
*In (3) we use the binomial series expansion for $\frac{1}{(1-x^2)^3}$
*In (4) we expand $(1-x^{10})^3$ and take the first two terms only since all other therms have powers greater than $18$. We also use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
*In (5) we use the linearity of the coefficient of operator and apply the rule stated in the comment of (2).
*In (6) we select the coefficients of $x^{18}$ and $x^{8}$ by taking $n=9$ and $n=4$.
|
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|
Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I thought maybe it's better to consider odd and even cases for $n$ but there's no unified rule here, because for example: $8^2\equiv{4}\pmod{10}$ and $8^4\equiv{6}\pmod{10}$. Any ideas??
|
Modulo $10$ easy inductions show that
$$\begin{align*}
&1^n+9^n\equiv\begin{cases}
0,&\text{if }n\text{ is odd}\\
2,&\text{if }n\text{ is even}\;,
\end{cases}\\
&2^n+8^n\equiv\begin{cases}
0,&\text{if }n\text{ is odd}\\
8,&\text{if }n\equiv 2\pmod4\\
2,&\text{if }n\equiv 4\pmod4\;,
\end{cases}\\
&3^n+7^n\equiv\begin{cases}
0,&\text{if }n\text{ is odd}\\
8,&\text{if }n\equiv 2\pmod4\\
2,&\text{if }n\equiv 0\pmod4\;,
\end{cases}\\
&4^n+6^n\equiv\begin{cases}
0,&\text{if }n\text{ is odd}\\
2,&\text{if }n\text{ is even}\;,\text{ and}
\end{cases}\\
&5^n\equiv 5\;.
\end{align*}$$
Thus,
$$\sum_{k=1}^9k^n\equiv\begin{cases}
5,&\text{if }n\not\equiv0\pmod4\\
3,&\text{if }n\equiv 0\pmod4\;.
\end{cases}$$
And as a quick check of the second case above,
$$\sum_{k=1}^9k^4=1+16+81+256+625+1296+2401+4096+6561=15,333\;.$$
Note that the stated result is true only when $n$ is not a multiple of $4$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
A probability question that uses the binomial expansion The question is as follow:
(i) Find the binomial expansion of $(1-x)^{-3}$ up to and including $x^{4}$.
(ii)
A player throws a 6-sided fair die at random. If he gets an even number, he loses the game and the game ends. If he gets a "1", "3" or "5" he throws the die again. He wins the game if he gets either "3" or "5" thrice consecutively (eg. 335, 555, 353) and the game ends. Find the exact probability of him winning the game.
I have been thinking about this question for quite a while. Obviously, the author of the question wants us to solve part (ii) with the help of part (i). However, to solve part (ii), it looks more an infinite series to me (the possible combinations of winning the game).
Could anyone contribute to solve this question please?
|
Binomial expansion of
$$(1 + x)^a = 1 + ax + \frac{a(a-1)}{2!} x^2 + \frac{a(a-1)(a-2)}{3!} x^3 + \frac{a(a-1)(a-2)(a-3)}{4!} x^4 + {\cal O}(x^5)$$
when $|x| < 1$.
Part 2) I have just written out the pattern of him winning in 3 steps, 4 steps, 5 steps, etc.
$$ \Pr(Win) = \frac{1}{3^3} \bigg[ 1 + \Big(\frac{1}{6}\Big) + \Big(\frac{1}{6^2} + \frac{1}{6 \cdot 3}\Big) + \Big(\frac{1}{6^3} + 2 \frac{1}{6^2\cdot3} + \frac{1}{6\cdot 3^2}\Big) + \Big(\frac{1}{6^4} + 3 \frac{1}{6^3\cdot3} + 3\frac{1}{6^2\cdot 3^2}\Big) + \Big(\frac{1}{6^5} + 4 \frac{1}{6^4\cdot3} + 6\frac{1}{6^3\cdot 3^2}\Big) + \cdots \bigg]
$$
Do you see what can be done from here?
I wrote them as series which can be summed.
$$\Pr(Win) = \frac{1}{3^3}\bigg[\sum_{n=0}^\infty 6^{-n} + \frac{1}{3}\sum_{k=0}^\infty k 6^{-k} + \frac{1}{3^2} \sum_{h=1}^\infty {h+1 \choose 2}6^{-h} \bigg],$$
Each of which can be summed without too much difficulty.
|
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|
A curious algebraic fraction that converges to $\frac{\sqrt{2}}{2}$ I have noticed that the algebraic fraction
$\frac{3a+2b}{4a+3b} $
Gives better and better approximations to $\sin 45^\circ = \frac{\sqrt{2}}{2} $
For $ a = b = 1$ we get $5/7 \approx 0.714 $
Now, taking $ a = 5, b = 7$, we get $ 29/41 \approx 0.707$
All the times, take
$ a_{n+1} = 3a_n + 2b_n$
$ b_{n+1} = 4a_n+3b_n$
And this process converges to $\frac{\sqrt{2}}{2} $
I have arrived at this result by a casual consideration. When I take the approximation $5/7$, I wonder what happens if I rewrite the numbers $5$ and $7$ as a function of two quantities $a, b$ in order to get a better approximation by successive iterations, in which the first approximation is the case $ a = b = 1$. Thus I have written $ 5 = 3a + 2b$ and the number $7 = 4a+3b$ I have selected both numerator and denominator to be consecutive numbers, and surprisingly this gives better approximations to $\frac{\sqrt{2}}{2}$
The question now is: how to prove that, in fact, this algebraic fraction, whit the described iteration above, converges to the desired value?
|
I think we can split the ratio $(3a + 2b)/(4a + 3b)$ further. Let $a/b$ be a rational approximation to $1/\sqrt{2}$. Then I prove that $(a + b)/(2a + b)$ is a better approximation to $1/\sqrt{2}$ but in opposite direction.
Clearly we have $$\left(\frac{a + b}{2a + b}\right)^{2} - \frac{1}{2} = \frac{b^{2} - 2a^{2}}{2(2a + b)^{2}} = \frac{b^{2}}{(2a + b)^{2}}\left(\frac{1}{2} - \frac{a^{2}}{b^{2}}\right)$$ so that $Y = \left(\dfrac{a + b}{2a + b}\right)^{2} - \dfrac{1}{2}$ and $X = \dfrac{a^{2}}{b^{2}} - \dfrac{1}{2}$ have opposite signs and $|Y| < |X|$ and thus the claim in previous paragraph is established.
Applying the transformation $a/b \to (a + b)/(2a + b)$ twice on $a/b$ we get $$\frac{3a + 2b}{4a + 3b}$$ and we are thus ensured that if $a/b$ is a rational approximation to $1/\sqrt{2}$ then $\dfrac{3a + 2b}{4a + 3b}$ is a better approximation to $1/\sqrt{2}$ in the same direction. Since a bounded and monotone sequence is convergent it follows that repeated application of the transformation $a/b \to (3a + 2b)/(4a + 3b)$ to any positive rational number $a/b$ converges to $1/\sqrt{2}$.
BTW this comes from an exercise in Hardy's A Course of Pure Mathematics (page 11, Examples III, problem 3) but Hardy uses it for $\sqrt{2}$:
Show that if $m/n$ is a good approximation to $\sqrt{2}$, then $(m + 2n)/(m + n)$ is a better one, and that the errors in the two cases are in opposite directions.
|
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|
Limit of gamma and digamma function In my answer of the previous OP, I'm able to prove that
\begin{align}
I(a)&=\int_0^\infty e^{-(a-2)x}\cdot\frac{1-e^{-x}(1+x)}{x(1-e^{x})(e^{x}+e^{-x})}dx\tag1\\[10pt]
&=\int_0^1\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy\tag2\\[10pt]
&=\log\Gamma\!\left(\frac{a+2}{4}\right)-\log\Gamma\!\left(\frac{a}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+1}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+2}{4}\right)\tag3
\end{align}
From the integral representations of $I(a)$ in $(1)$ and $(2)$, it's easy to show that
\begin{align}
\lim_{a\to\infty}I(a)&=\int_0^\infty \lim_{a\to\infty}e^{-(a-2)x}\cdot\frac{1-e^{-x}(1+x)}{x(1-e^{x})(e^{x}+e^{-x})}dx\\[10pt]
&=\int_0^1\lim_{a\to\infty}\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy\quad,\quad 0<y<1\\[10pt]
&=0
\end{align}
But I'm having trouble proving
\begin{equation}
\lim_{a\to\infty}\left[\log\Gamma\!\left(\frac{a+2}{4}\right)-\log\Gamma\!\left(\frac{a}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+1}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+2}{4}\right)\right]=0
\end{equation}
Indeed the above result is confirmed by Wolfram Alpha. How does one prove the above limit?
|
We have that $$\log\left(\Gamma\left(x\right)\right)\sim x\log\left(x\right)-x-\frac{1}{2}\log\left(\frac{x}{2\pi}\right)+O\left(\frac{1}{x}\right)
$$ and $$\psi\left(x\right)\sim\log\left(x\right)+O\left(\frac{1}{x}\right)
$$ as $x\rightarrow\infty$ (see here and here) hence we have to evaluate $$\begin{align}
&\frac{a+2}{4}\log\left(\frac{a+2}{4}\right)-\frac{a+2}{4}-\frac{1}{2}\log\left(\frac{a+2}{8\pi}\right)-\frac{a}{4}\log\left(\frac{a}{4}\right)+\frac{a}{4}+\frac{1}{2}\log\left(\frac{a}{8\pi}\right)\\
& -\frac{1}{4}\log\left(\frac{a+1}{4}\right)-\frac{1}{4}\log\left(\frac{a+2}{4}\right)+O\left(\frac{1}{a}\right)\\
&=\frac{1}{2}\log\left(\frac{a}{a+2}\right)+O\left(\frac{1}{a}\right)\rightarrow0.
\end{align}
$$
|
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|
Solve congruence Solve:$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod {2016}$$
So $ 2016 = 2^5 \cdot 3^2 \cdot 7$
And $$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2^5} \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{3^2} \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod 7 \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
What now?
|
The lambda-function : https://en.wikipedia.org/wiki/Carmichael_function
helps us to find a quick answer. Because of $\lambda(2016)=24$ and considering $2016=2^5\cdot 3^2\cdot 7$, we can conclude $a^k=a^{k+m\cdot 24}$ for all $a$ and $m$, when $k\ge 5$.
The given number can be written by $2^M$, where $M$ is a power tower of $2's$ with height $2015$. Obviously , $M$ is divisble by $8$ and the remainder modulo $3$ is $1$.
Applying the chinese remainde theorem , we get $M=16$ modulo $24$.
So, we can conclude $2^M=2^{16}=1024$ modulo $2016$.
Another approach :
The given number is obviously divisble by $2^5$. With induction, you can prove that $2^{2^m}=7$ modulo $9$ for every even $m\ge 2$.
The base case $m=2$ is easy to verify.
Then , $2^{2^{m+2}}=(2^{2^m})^4=7^4=7$ mod $9$ is the induction step.
Finally, you can show $2^{2^m}=2$ modulo $7$ for every even $m\ge 2$ analogue to the proof for $9$ (I will leave it to you as an exercise).
So, your number is congruent
$0$ modulo $32$
$7$ modulo $9$
$2$ modulo $7$
Now, apply the chinese remainder theorem.
|
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|
On the composition of formal power series I an attempt to compute the coefficients of the composition $f(g(x))$ of two power series $f(x) = \frac{1}{1-x}$ and $g(x) = \frac{1}{1-x}-1$, I used the definition of composition to get to
$$f(g(x)) = \sum_{n=0}^{\infty} \left(\frac{1}{1-x} - 1\right)^n = \sum_{n=0}^{\infty} \left(\frac{x}{1-x}\right)^n$$
but I don't seem to be able to find a way to rewrite this in a way I can get the coefficients out, i.e. something of the form $\sum_{n=0}^\infty c_n x^n$.
Another way to compute the coefficients I know of, could be to use that $f(x) = \sum_{n=0}^\infty x^n$ and $g(x) = \sum_{n=1}^\infty x^n$ and actually compute each $c_N(x) = \sum_{n=0}^N f_n (g(x))^n = \sum_{n=0}^N (g(x))^n$, but this seems cumbersome if you need a lot of coefficients.
Using the transfer principle, on the other hand, one easily can get to
$$ f(g(x)) = \frac{1}{1-\frac{1}{1-x} + 1} = \frac{1-x}{1-2x}$$
with expansion $1 + \sum_{n=1}^\infty 2^{n-1} x^n$.
Am I missing something obvious here or is it just not that easy to find the coefficients of the composition of two power series without leaving the "formal world"? I am wondering, because I personally am not the greatest fan of analysis and prefer to think in the formal world
|
We have $$f(g(x)) = \sum_{k = 0}^\infty \left( \dfrac x {1 - x} \right)^k .$$
Now, for $k \ge 1$, using $\dfrac 1 {(1 - x)^k} = \sum\limits_{m = 0}^\infty \binom{m + k- 1}{m} x^m$, we get
\begin{align*}
f(g(x)) & = 1 + \sum_{k=1}^\infty x^k \sum_m \binom{m + k - 1}{m} x^m\\
& = 1 + \sum_{k=1}^{\infty} \sum_m \binom{m + k - 1}{m} x^{m +k}.
\end{align*}
To collect together the coefficients of the same power of $x$ in the above summation, let $m + k = n$, to get
\begin{align*}
f(g(x)) & = 1 + \sum_{n=1}^\infty \sum_{m=0}^{n-1} \binom{n - 1}{m} x^n\\
& = 1 + \sum_{n = 1}^\infty 2^{n-1} x^n,
\end{align*}
since $\sum\limits_{m=0}^{n-1} \binom{n - 1}{m} = 2^{n-1}$.
|
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|
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $n=k+1$ I setup the problem like so:
$\frac{k}{(k+1)!}+\frac{(k+1)}{(k+2)!}=1-\frac{1}{(k+2)!}$
After trying to simplify it I got the following:
$\frac{k(k+2)!+(k+1)(k+1)!}{(k+1)!(k+2)!}=1-\frac{1}{(k+2)!}$
However, I'm having trouble simplifying it to match the RHS. Hints?
|
The equality you want to prove is
$$
\underbrace{\frac{1}{2!}+\dots+\frac{k}{(k+1)!}}_{*}+
\frac{k+1}{(k+2)!}=1-\frac{1}{(k+2)!}
$$
The term marked $*$ is equal, by the induction hypothesis, to
$$
1-\frac{1}{(k+1)!}
$$
and so you need to manipulate
$$
1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}
$$
Hint:
$$
1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}
=
1-\frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}
=\dots
$$
Do the necessary steps in order to finish up at $1-\dfrac{k+1}{(k+2)!}$.
|
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|
Solutions to a linear system of equations. I want to find the solution of this system when the parameter $a \in R$ varies.
\begin{cases} (a+2)x_2 + x_4 = 1 \\ -x_1 +x_3 = a+1 \\ (a+1)x_1 + 2x_2 -x_3 = 0 \\ x_1 -2x_2 -(a+1)x_3 = -2 \end{cases}
I notice that the matrix associated with the homogeneous system has rank $= 3$ for $a \ne 0,-2$ and never has $rank = 4$.
Now taking the matrix \begin{bmatrix} 0 & a+2 & 0 & 1 \\ -1 & 0 & 1 & a+1 \\
a+2 & 2 & -1 & 0 \\
0 & -2 & -a-1 & -2 \end{bmatrix}
this has rank $= 4 \quad \forall a \ne 3,2$
and taking \begin{bmatrix} 1 & a+2 & 0 & 1 \\ 0 & 0 & 1 & a+1 \\
0 & 2 & -1 & 0 \\
0 & -2 & -a-1 & -2 \end{bmatrix}
I get that this has rank $= 4 \quad \forall a \ne \frac{-2 + 2 \sqrt{11}}{4}, \frac{-2 - 2 \sqrt{11}}{4}$
So the original system has no solutions, correct?
|
Transforming the system of equations into an augmented matrix, we have:
$$\begin{bmatrix} 0 & (a+2) & 0 & 1 &|& 1 \\ -1 & 0 & 1 & 0 &|& (a+1) \\ (a+1) & 2 & -1 & 0 &|& 0 \\ 1 & -2 & -(a+1) & 0 &|& -2\end{bmatrix}$$
(note that the vertical lines between columns 4 and 5 seperate the sides of the equations). Row reduction (not all the way to the RREF form) yields:
$$\begin{bmatrix} 1 & 0 & -1 & 0 &|& -(a+1) \\ 0 & 1 & \frac{1}{2}a & 0 &|& -\frac{1}{2}(a-1) \\ 0 & 0 & a(a+2) & -2 &|& -a(a+1) \\ 0 & 0 & 0 & 0 &|& a(a+3) \end{bmatrix}$$
We can now see that the last row must have $0$'s on both sides of the vertical line (otherwise the system is inconsistent), which means $a(a+3)=0$ which gives $a=-3$ or $a=0$. The solution set for when $a$ equals either of those values can be easily determined by plugging-in the values for $a$ and row reducing further.
Hopefully this helps.
|
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|
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after getting here i got stuck
$$\lim_{x\to\infty } \left({\sqrt{x+1}\left(a+b\sqrt{1+\frac{1}{x+1}}+c\sqrt{1+\frac{2}{x+1}}\right)}\right)$$
Got here by substituting $\sqrt{x+2}$ with $\sqrt{(x+1)(1+\dfrac{1}{x+1})}$
Edit: x tends to infinity, not to 0. I transcribed wrongly.
|
Lessee...
first prove $\lim_{x\rightarrow \infty}\frac{\sqrt {x + j}}{\sqrt{x+k}}= 1$ which should be easy (albeit tedious) enough. [$\frac{\sqrt{x + j}}{\sqrt{x + k}} =\frac{ \sqrt{1 + j/x}}{\sqrt{1 + k/x}}$ so limit is 1].
$\lim_{x\rightarrow \infty}a\sqrt{x + 1} + b \sqrt{x+2} + c\sqrt{x+3} =$
$\lim_{x\rightarrow \infty}(a + b\frac{\sqrt{x+2}}{\sqrt{x+1}} + c\frac{\sqrt{x+3}}{\sqrt{x+1}})\sqrt{x+1}=$
$\lim_{x\rightarrow \infty}(a + \lim_{x\rightarrow \infty}b\frac{\sqrt{x+2}}{\sqrt{x+1}} + \lim_{x\rightarrow \infty}c\frac{\sqrt{x+3}}{\sqrt{x+1}})\sqrt{x+1}=$
$\lim_{x\rightarrow \infty}(a + b + c)\sqrt{x+1}=$
$(a + b + c)\lim_{x\rightarrow \infty}\sqrt{x+1}=$
$\{\sqrt{x+1}\}$ diverges. So If $a + b + c \ne 0$ then $\{(a + b + c)\sqrt{x+1}\}$ diverges.
If $a + b + c = 0$ then $\lim_{x\rightarrow \infty}(a + b + c)\sqrt{x+1}=\lim_{x \rightarrow \infty} 0 = 0.$
|
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|
Factoring the sequence ${10}^{2n}+10^{n}+1$ While I am waiting for the basketball NBA game between Cleveland Cavaliers and Golden State Warriors to begin I sort of played with the sequence $a_n={10}^{2n}+10^{n}+1$ in a way that I looked for the prime factors for small values of $n$.
For $n=1$ we have $111=3 \cdot 37$.
For $n=2$ we have $10101=3 \cdot 7 \cdot 13 \cdot 37$
For $n=3$ we have $1001001=3 \cdot 333667$
For $n=4$ we have $100010001=3 \cdot 7 \cdot 13 \cdot 37 \cdot 9901$
For $n=5$ we have $10000100001=3 \cdot 31 \cdot 37 \cdot 2906161$
For these small values of $n$ we see that $a_n={10}^{2n}+10^{n}+1$ have no repeated prime factors, in other words, for these small values of $n$ every prime number in the factorization occurs with an exponent $1$.
I am interested in the following:
What is the smallest value of $n$ for which $a_n={10}^{2n}+10^{n}+1$ has repeated prime factor?
|
$a_n$ is divisible by $p^2$ if $x = 10^n$ is a solution of $x^2 + x + 1 \equiv 0 \mod p^2$. We may assume $p$ is not $2$ or $5$, so $x$ is coprime to $p$. Now $4 (x^2 + x + 1) = (2x + 1)^2 + 3$, so $2x + 1$ must be a square root of $-3$ mod $p^2$. If $p$ is an odd prime (other than $3$) for which $-3$ is a quadratic residue, i.e. $p \equiv 1 \mod 6$, then there are two square roots of $-3$ mod $p^2$, and thus two solutions of $x^2 + x + 1 \equiv 0 \mod p^2$ in $[1, \ldots, p^2-1]$. There may or may not be $n$ for which $10^n$ is one of these values: if there is, such $n$ will repeat according to the order of $10$ in the multiplicative group mod $p^2$.
For example, $x^2 +x+1 \equiv 0 \mod 7^2$ iff $x \equiv 18, 30 \mod 7^2$; $10^n \equiv 18 \mod 7^2$ for $n \equiv 28 \mod 42$ and $10^n \equiv 30 \mod 7^2$ for $n \equiv 14 \mod 42$, where $42$ is the order of $10$ mod $7^2$.
The first few values of $n$ for which $10^{2n}+10^n+1$ is divisible by the square of a prime turn out to be $14$, $26$, $28$ and $37$, for which $10^{2n}+10^n + 1$ is divisible by $7^2$, $13^2$, $7^2$ and $37^2$ respectively.
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?
Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
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\begin{align}
\int_0^{\infty}\frac{\mathrm dx}{a^2x^4+bx^2+c^2}
&= \int_0^{\infty}\frac{\mathrm dx}{\left(ax-\frac{c}{x}\right)^2+b+2ac}\cdot \frac{1}{x^2}\\[9pt]
&=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{\left(ay-\frac{c}{y}\right)^2+b+2ac}}_{\large\color{blue}{x=\frac{c}{ay}}}\\[9pt]
&=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{y^2+b+2ac}}_{\large\color{blue}{y=z\sqrt{b+2ac}}}\tag{$\spadesuit$}\\[9pt]
&=\frac{c}{a\sqrt{b+2ac}} \int_0^{\infty}\frac{\mathrm dz}{z^2+1}\\[9pt]
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{c\pi}{2a\sqrt{b+2ac}}}}
\end{align}
Setting $a=b=c=1$, then
$$I=\int_0^{\infty}\frac{\mathrm dx}{x^4+x^2+1}=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi}{2\sqrt{3}}}}$$
$(\spadesuit)$ Cauchy-Schlomilch transformation for $f(\cdot)$ is a continuous function and $a,c>0$
$$\int_0^\infty f\left(\left(ay-\frac{c}{y}\right)^2\right)\ \mathrm dy=\frac{1}{a}\int_0^\infty f\left(y^2\right)\ \mathrm dy$$
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The largest root of a recursively defined polynomial Suppose that for all $x \in \mathbb{R}$, $f_1(x)=x^2$ and for all $k \in \mathbb{N}$,
$$
f_{k+1}(x) = f_k(x) - f_k'(x) x (1-x).
$$
Let $\underline{x}_k$ denote the largest root of $f_k(x)=0$.
I want to prove the following conjectures:
*
*$0=\underline{x}_1 < \underline{x}_2 < \underline{x}_3 < \dots < 1$.
*$\lim_{n \to \infty} \underline{x}_n = 1$.
*$f_k'(\underline{x}_k) \geq 0$ for all $k$.
These conjectures are the missing part of a larger proof and it seems to hold for any $f_k$ that I can compute by hand or numerically. The first few polynomials:
*
*$f_1(x)=x^2$, so that $\underline{x}_1=0$ and $f_1'(\underline{x}_1) = 0$
*$f_2(x)=x^2 (2x-1)$, so that $\underline{x}_2=\frac{1}{2} \in (\underline{x}_1,1)$ and $f_2'(1/2)=1/2>0$
*$f_3(x)=x^2 (6x^2-6x+1)$, so that $\underline{x}_3 = \frac{1}{2} + \frac{1}{2 \sqrt{3}} \approx 0.7887 \in (\underline{x}_2,1)$ and $f_3'(\underline{x}_3) \approx 2.1547>0$
*$f_4(x)=x^2 (2x-1) (12 x^2-12 x+1)$, so that $\underline{x}_4=\frac{1}{2}+\frac{1}{\sqrt{6}} \approx 0.9082 \in (\underline{x}_3,1)$ and $f_4'(\underline{x}_4) \approx 6.5993 > 0$
The polynomials $f_k$ have many properties that should help. For example it is easy to show that $f_k(1)=1$ for all $k$ and $f_k(x) > 1$ for all $x>1$ for all $k$. Therefore all real roots must be strictly below $1$.
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From
$f_{k+1}(x) = f_k(x) - f_k'(x) x (1-x)
$,
let
$f_k(x)
=x^2 g_k(x)
$,
so
$f_k'(x)
=x^2 g_k'(x)+2xg_k(x)
=x(x g_k'(x)+2g_k(x))
$.
Then
$\begin{array}\\
x^2g_{k+1}(x)
&= x^2g_k(x) - x (1-x)x(x g_k'(x)+2g_k(x))\\
&= x^2(g_k(x) - (1-x)(x g_k'(x)+2g_k(x)))\\
\text{so}\\
g_{k+1}(x)
&= g_k(x) - (1-x)(x g_k'(x)+2g_k(x))\\
&= g_k(x) - (1-x)x g_k'(x)-2(1-x)g_k(x)\\
&= (1-2(1-x))g_k(x) - (1-x)x g_k'(x)\\
&= (2x-1)g_k(x) - (1-x)x g_k'(x)\\
&= (2x-1)g_k(x) + x(x-1) g_k'(x)\\
\end{array}
$
As a check,
since
$g_1(x) = 1$,
$g_2(x)
=2x-1
$
and
$\begin{array}\\
g_3(x)
&=(2x-1)(2x-1)+x(x-1)(2)\\
&=4x^2-4x+1+2x(x-1)\\
&=6x^2-6x+1\\
\end{array}
$
Note that
$g_{k+1}(x)
= (2x-1)g_k(x) + x(x-1) g_k'(x)
=((x^2-x)g_k(x))'
$
so that
$\begin{array}\\
g_{k+2}(x)
&=((x^2-x)g_{k+1}(x))'\\
&=((x^2-x)((2x-1)g_k(x) + x(x-1) g_k'(x))'\\
&=((x^2-x)(2x-1)g_k(x) + x^2(x-1)^2 g_k'(x))'\\
&=((2x^3-3x^2-x)g_k(x) + (x^4-2x^3+x^2) g_k'(x))'\\
&= (6 x^2-6 x-1) g_k(x)+x (x (x-1)^2 g_k''(x)+(6 x^2-9 x+1) g_k'(x))\\
\end{array}
$
Let
$G(x, y)
=\sum_{k=1}^{\infty} y^kg_k(x)
$.
Then
$G_x(x, y)
=\sum_{k=1}^{\infty} y^kg_k'(x)
$
and
$\begin{array}\\
\dfrac1{y}G(x, y)
&=\sum_{k=1}^{\infty} y^{k-1}g_k(x)\\
&=\sum_{k=0}^{\infty} y^{k}g_{k+1}(x)\\
&=g_1(x)+\sum_{k=1}^{\infty} y^{k}g_{k+1}(x)\\
&=1+\sum_{k=1}^{\infty} y^{k}((2x-1)g_k(x) + x(x-1) g_k'(x))\\
&=1+\sum_{k=1}^{\infty} y^{k}(2x-1)g_k(x) + \sum_{k=1}^{\infty}x(x-1)y^k g_k'(x)\\
&=1+(2x-1)\sum_{k=1}^{\infty} y^{k}g_k(x) + x(x-1)\sum_{k=1}^{\infty}y^k g_k'(x)\\
&=1+(2x-1)\sum_{k=1}^{\infty} y^{k}g_k(x) + x(x-1)\sum_{k=1}^{\infty}y^k g_k'(x)\\
&=1+(2x-1)G(x, y) + x(x-1)G_x(x, y)\\
\text{or}\\
x(x-1)G_x(x, y)
&=G(x, y)(\dfrac1{y}-(2x-1))-1
\end{array}
$
Not sure where to go from here,
so I'll leave it at this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can't find minimum using Lagrange multipliers I want to find the minimum of the function $f(x,y) = x + y^2$ with the constraint $2x^2 +y^2 = 1$.
Here are my partial derivatives:
$$f_x = 1$$
$$f_y = 2y$$
$$g_x = 4x$$
$$g_y = 2y$$
I have the following system of equations:
\begin{align*}
1 = \lambda4x\\
2y = 2y\lambda\\
2x^2 + y^2 = 1
\end{align*}
and I found that $$\lambda = 1,\ x = 1/4,\ \text{and }y = +-(7/8)^{1/2}.$$ And Wolfram Alpha shows that the above is the maximum value, and that the minimum value is $$ x= -1/(2)^{1/2}\text{ and }y = 0.$$
How can I find this value? I missed something.
|
There's no need to use a Lagrange multiplier. The ellipse
$$\{ (x,y) \in \mathbb{R}^2 \mid 2 x^2 + y^2 = 1 \}$$
is parametrized as follows
$$x (\theta) = \frac{\sqrt 2}{2} \, \cos (\theta) \qquad \qquad \qquad y (\theta) = \sin (\theta)$$
Hence,
$$g (\theta) := f (x (\theta), y (\theta)) = \frac{\sqrt 2}{2} \, \cos (\theta) + \sin^2 (\theta)$$
Differentiating,
$$g' (\theta) = -\frac{\sqrt 2}{2} \, \sin (\theta) + 2 \sin (\theta) \cos (\theta) = \left( 2 \cos (\theta) - \frac{\sqrt 2}{2} \right) \sin (\theta)$$
which vanishes when $\sin (\theta) = 0$ or $\cos (\theta) = \frac{\sqrt 2}{4}$. If $\sin (\theta) = 0$, then
$$x = \pm \frac{\sqrt 2}{2} \qquad \qquad \qquad y = 0 \qquad \qquad \qquad f = \pm \frac{\sqrt 2}{2}$$
If $\cos (\theta) = \frac{\sqrt 2}{4}$, then
$$x = \frac{1}{4} \qquad \qquad \qquad y = \pm \sqrt{\frac{7}{8}} \qquad \qquad \qquad f = \frac{9}{8}$$
Thus, the minimum is $- \frac{\sqrt 2}{2}$, which is attained at $(x,y) = \left(- \frac{\sqrt 2}{2}, 0\right)$.
|
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|
Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$ I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible
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You simply write out a skeleton $(x^4+x^3+8x^2+ax+b)=(x^2+1)(cx^2+dx+e)$.
(1) comparing the $x^4$ term we must have $c=1$.
(2) comparing the $x^3$ term we have $d=1$.
(3) comparing the constant term $e=b$.
So we now have $(x^4+x^3+8x^2+ax+b)=(x^2+1)(x^2+x+b)$.
(4) comparing the $x^2$ term $8=b+1$, so $b=7$.
(5) comparing the $x$ term $a=1$.
Check: $x^4+x^3+8x^2+x+7=(x^2+1)(x^2+x+7)$.
With practice a lot of this is done in your head and it is all quite fast.
|
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|
If the sides of a triangle satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$, and if one angle is $48^\circ$, then find the other angles.
In triangle $ABC$ one angle of which is $48^{\circ}$, length of the sides satisfy the equality:
$$(a-c)(a+c)^2+bc(a+c)=ab^2$$
Find the value in degrees the other two angles of the triangle.
I have no idea how to solve this problem
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From $c^2-a^2=ab,$
using $(i)$Sine Law,
$(ii)$Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
and $(iii)\sin(C+A)=\cdots=\sin B$
$$\sin B\sin(C-A)=\sin A\sin B$$
As $\sin B\ne0,\implies\sin(C-A)=\sin A$
$\implies C-A=n\pi+(-1)^nA$ where $n$ is any integer
If $n$ is odd, $n=2m+1$(say) $\implies C-A=(2m+1)\pi+(-1)A$
$\iff C=(2m+1)\pi$ which is impossible as $0<C<\pi$
If $n$ is even, $n=2m$(say) $\implies C-A=2m\pi+A\iff C=2m\pi+2A$
As $0<C,A<\pi; C=2A$
We also have $A+B+C=\pi$
Now check for the three cases namely, $A=48^\circ, B=48^\circ,C=48^\circ,$
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|
For all positive integer $n$ prove the equality: $\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}=\frac{\binom{2n}{n}}{2n}$
For all positive integer $n$ prove the equality:
$$\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}=\frac{\binom{2n}{n}}{2n}$$
My work so far:
$$\frac{n\binom{n-1}{k}}{k+1}=\frac{n(n-1)!}{(k+1)k!(n-k-1)!}=\frac{n!}{(k+1)!(n-k-1)!}=\binom{n}{k+1}$$
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Another approach.
$$ \sum_{k=0}^{n-1}\binom{n-1}{k}^2\frac{1}{k+1}=\int_{0}^{1}\color{blue}{\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n-1}{n-1-k}x^k}\,dx \tag{1}$$
and the blue term is the coefficient of $y^{n-1}$ in the following product:
$$\left(\sum_{k=0}^{n-1}\binom{n-1}{k}x^k y^k\right)\cdot\left(\sum_{k=0}^{n-1}\binom{n-1}{k}y^k\right)=(1+xy)^{n-1}(1+y)^{n-1}\tag{2}$$
so the original sum is the coefficient of $y^{n-1}$ in:
$$ (1+y)^{n-1}\int_{0}^{1}(1+xy)^{n-1}\,dx =(1+y)^{n-1}\cdot\frac{-1+(1+y)^{n}}{ny}\tag{3}$$
or $\frac{1}{n}$ times the coefficient of $y^n$ in $(1+y)^{2n-1}-(1+y)^{n-1}$. That leads to:
$$ \sum_{k=0}^{n-1}\binom{n-1}{k}^2\frac{1}{k+1}=\frac{1}{n}\binom{2n-1}{n}=\color{red}{\frac{1}{2n}\binom{2n}{n}}\tag{4}$$
as wanted.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
|
Claim:
$$\frac {3\sqrt {3}-4}{7-2\sqrt {3}}<\frac {3\sqrt {3}-8}{1-2\sqrt {3}}$$
Proof:
Let $u=\sqrt 3-1$. Notice $1>u>\frac12$.
If
$$\frac{3u-1}{5-2u}<\frac{3u-5}{-2u-1}$$
$$\Leftarrow-6u^2-u+1>-6u^2+25u-25$$
$$\Leftarrow26>26u$$
$$\Leftarrow1>u=\sqrt3-1$$
which is true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \cdot \frac{1}{2\sqrt{x+2}} = \frac{1}{(2x+6)\sqrt{x+2}}$$
$$g'(x) = 1$$
$$g(x) = x$$
So:
$$\bigg[\arctan(\sqrt{x+2}) \cdot x\bigg]_{-1}^{1} - \int_{-1}^{1} \frac{x}{(2x-6)\sqrt{x+2}}\ dx$$
How should I proceed with the new integral?
|
It is sometimes better to take $x+a$ (for some constant $a$) as the antiderivative of $1$. In this case
$$
\int \arctan\sqrt{x+2}\,dx=(x+a)\arctan\sqrt{x+2}-\frac{1}{2}\int(x+a)\frac{1}{1+x+2}\cdot\frac{1}{\sqrt{x+2}}\,dx,
$$
so $a=3$ seems to be a good choice.
I think you can continue from here.
|
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|
There are 10 boxes, 15 balls; 10 red, 5 blue. Each is randomly placed in a box in an independent manner. What's E[X=the number of empty boxes?] There are 10 boxes, 15 balls; 10 red, 5 blue.
Each is randomly placed in a box in an independent manner. The red balls are placed in boxes 1-10, blue balls are placed in 1-6. What is the expected value of the number of empty boxes?
I have no real ideas as to how to approach this problem as it seems very "layered." A hint in the right direction would be great.
|
Let $A_i=1$ if box $i$ is empty, 0 if not. Consider the four boxes numbered 7-10. They can only get red balls. For each one we have $p(A_i=1)=\left(\frac{9}{10}\right)^{10}$. So $E(A_i)=\left(\frac{9}{10}\right)^{10}$. But expectation is linear, so $E(\sum_7^{10}A_i)=4\left(\frac{9}{10}\right)^{10}\approx1.395 $.
The remaining six boxes numbered 1-6 can get both red and blue balls. We have $p(A_i=1)=\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}$ and $E(A_i)=\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}$, and hence $E(\sum_1^6A_i)=6\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}\approx0.841$.
So the expected number of empty boxes is $4\left(\frac{9}{10}\right)^{10}+6\left(\frac{5}{6}\right)^{5}\left(\frac{9}{10}\right)^{10}\approx2.235$
|
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|
Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$ Is there any closed form for the infinite product $\prod_{l=1}^\infty \cos\dfrac{x}{3^l}$? I think it is convergent for any $x\in\mathbb{R}$.
I think there might be one because there is a closed form for $\prod_{l=1}^\infty\cos\dfrac{x}{2^l}$ if I'm not wrong.
|
We follow the method of this paper, using the Fourier transform.
Following his normalizations,
$$
\hat f (\omega) = \frac{1}{2\pi} \int_{\mathbb R} f(x) e^{-i \omega x} \, dx, \qquad f(x) = \int_{\mathbb R} \hat f(\omega) e^{i \omega x} \, d\omega,
$$
the key facts to be used are:
*
*The identity $\cos(bx) = \frac{e^{i b x} + e^{-i b x}}{2}$
*If $f(x) = e^{ibx}$, then $\hat f(\omega) = \delta_b(\omega)$, the Dirac delta centered at $\omega = b$.
*The convolution formula $\widehat{fg} = \hat f * \hat g$.
*The identity $\delta_a * \delta_b = \delta_{a+b}$.
Applying these to your problem,
\begin{align*}
\prod_{k=1}^n \cos\left(\frac{x}{3^k}\right) &= \prod_{k=1}^n \frac{e^{i \frac{x}{3^k}} + e^{-i \frac{x}{3^k}}}{2} \\
\implies \widehat{\prod_{k=1}^n \cos\left(\frac{x}{3^k}\right)} &= \frac{1}{2^n} \left(\delta_{\frac{1}{3^1}} + \delta_{-\frac{1}{3^1}} \right) * \cdots * \left(\delta_{\frac{1}{3^n}} + \delta_{-\frac{1}{3^n}} \right)(\omega) \\
&= \frac{1}{2^n} \sum_{p \in P_n} \delta_p(\omega)
\end{align*}
where $P_n$ is the set of $2^n$ points given by $P_n = \left\{\sum_{k=1}^n \frac{b_k}{3^k} \;\middle|\; b_k \in \{-1,1\} \; \forall k \right\}$. Note that the endpoints of $\bigcup_{n=1}^\infty P_n$ are bounded tightly by $\pm \sum_{k=1}^\infty \frac{1}{3^k} = \pm \frac{1/3}{1 - 1/3} = \pm \frac{1}{2}$, and every point in $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$ is the limit of some sequence $(p_n)_{n=1}^\infty$, where $p_n \in P_n$. Hence $\tfrac{1}{2^n} \sum_{p \in P_n} \delta_p(\omega)$ tends to the uniform density of total mass $1$ over $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$ as $n\to\infty$, which is given by the indicator function $\chi_{\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]}(\omega)$:
\begin{align*}
\widehat{\prod_{k=1}^\infty \cos\left(\frac{x}{3^k}\right)} &= \chi_{\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]}(\omega).
\end{align*}
Taking inverse Fourier transforms, we get
$$
\prod_{k=1}^\infty \cos\left(\frac{x}{3^k}\right) = \int_{\mathbb R} \chi_{\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]}(\omega) e^{i \omega x} \, d\omega = \int_{-1/2}^{1/2} e^{i \omega x} \, d\omega = \frac{\sin\left(\tfrac{x}{2}\right)}{\tfrac{x}{2}}.
$$
UPDATE: here's why the above is wrong. The statement that "every point in $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$ is the limit of some sequence $(p_n)_{n=1}^\infty$, where $p_n \in P_n$" is incorrect; in fact, the points belonging to $P = \bigcup_n P_n$ are those which have balanced ternary expansion consisting of the digits $1$ and $-1$ but not $0$. I'm not sure how to better describe $P$, but we'd proceed as before by integrating over $P$ (rather than all of $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$) to compute the inverse Fourier transform yielding the correct answer.
|
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|
Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$
Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$ if $n>0$.
I didn't see an easy way of proving this without doing a lot of algebra and rearranging. Is there an easier way?
|
It does not look so terrible to me! We have:
$$ \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \tag{1}$$
and:
$$ \frac{1}{2\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)^2}\tag{2}$$
so the inequality boils down to $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, quite trivial.
|
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|
How to get the correct angle of the ellipse after approximation I need to get the correct angle of rotation of the ellipses. These ellipses are examples. I have a canonical coefficients of the equation of the five points.
$$Ax ^ 2 + Bxy + Cy ^ 2 + Dx + Ey + F = 0$$
Ellipses:
Points:
Zero ellipse: [16,46] [44,19] [50,35] [31,61] [17,54]
First ellipse: [14,95] [47,71] [55,83] [23,107] [16,103]
Second ellipse: [12,128] [36,117] [58,128] [35,146] [13,136]
Third ellipse: [16,164] [29,157] [54,188] [40,195] [17,172]
Fourth ellipse: [22,236] [31,207] [50,240] [40,252] [26,244]
Coefficients:
Zero ellipse First ellipse Second ellipse Third ellipse Fourth ellipse
|
There're two principal axes in general, so
\begin{align*}
\theta &=\frac{1}{2} \tan^{-1} \frac{B}{A-C}+\frac{n\pi}{2} \\
&= \tan^{-1}
\left(
\frac{C-A}{B} \color{red}{\pm} \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \:
\right) \\
\end{align*}
The centre is given by $$(h,k)=
\left(
\frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC}
\right)$$
Transforming to
$$\frac{A+C \color{red}{\pm} \sqrt{(A-C)^{2}+B^{2}}}{2} x'^2+
\frac{A+C \color{red}{\mp} \sqrt{(A-C)^{2}+B^{2}}}{2} y'^2+
\frac
{\det
\begin{pmatrix}
A & \frac{B}{2} & \frac{D}{2} \\
\frac{B}{2} & C & \frac{E}{2} \\
\frac{D}{2} & \frac{E}{2} & F
\end{pmatrix}}
{\det
\begin{pmatrix}
A & \frac{B}{2} \\
\frac{B}{2} & C \\
\end{pmatrix}}=0$$
where $\begin{pmatrix} x' \\ y' \end{pmatrix}=
\begin{pmatrix}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}
\begin{pmatrix} x-h \\ y-k \end{pmatrix}$.
The axes will match, up to reflection about the axes of symmetry, when the $\color{red}{\text{case}}$ (upper or lower) agree.
Numerical example
Given five points: $(2,1)$, $(1,1)$, $(-2,-2)$, $(-1,-2)$, $(1,-1)$
$A=1$, $B=-2$, $C=2$, $D=-1$, $E=2$, $F=-2$
$$(h,k)=(0,-0.5)$$
$$\det
\begin{pmatrix}
A & \frac{B}{2} & \frac{D}{2} \\
\frac{B}{2} & C & \frac{E}{2} \\
\frac{D}{2} & \frac{E}{2} & F
\end{pmatrix} = ACF-\frac{A E^2+C D^2+F B^2-EDB}{4}=-\frac{5}{2}$$
$$\det
\begin{pmatrix}
A & \frac{B}{2} \\
\frac{B}{2} & C
\end{pmatrix} = -\frac{B^2}{4}+AC=1$$
$$\frac{A+C \pm \sqrt{(A-C)^{2}+B^{2}}}{2}=\frac{3 \pm \sqrt{5}}{2}$$
Using upper case convention:
$$\frac{3+\sqrt{5}}{2} x'^2+\frac{3-\sqrt{5}}{2} y'^2=\frac{5}{2}$$
$$\frac{x'^2}{a^2}+\frac{y'^2}{b^2}=1$$
$$(x',y')= (a\cos t,b\sin t)$$
where $\displaystyle \begin{pmatrix} a \\ b \end{pmatrix}=
\begin{pmatrix}
\sqrt{\frac{5}{3+\sqrt{5}}} \\
\sqrt{\frac{5}{3-\sqrt{5}}}
\end{pmatrix}$
$$\theta = \tan^{-1}
\left(
\frac{C-A}{B}+\frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \:
\right) =
\tan^{-1} \left( -\frac{\sqrt{5}+1}{2} \right)
\approx -58.28^{\circ} $$
\begin{align*}
\begin{pmatrix} x \\ y \end{pmatrix} &=
\begin{pmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{pmatrix}
\begin{pmatrix} x' \\ y' \end{pmatrix}+
\begin{pmatrix} h \\ k \end{pmatrix} \\ &&\\
&=
\begin{pmatrix}
h+x'\cos \theta-y'\sin \theta \\
k+x'\sin \theta+y'\cos \theta \end{pmatrix} \\ &&\\
&=
\begin{pmatrix}
\sqrt{\frac{5}{2}+\sqrt{5}\,} \, \sin t+
\sqrt{\frac{5}{2}-\sqrt{5}\,} \, \cos t \\
-\frac{1}{2}+
\frac{\sqrt{5+\sqrt{5}}}{2} \, \sin t-
\frac{\sqrt{5-\sqrt{5}}}{2} \, \cos t \end{pmatrix}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is it really necessary to learn how to write a quadratic in standard to vertex form? How crucial is this skill or form of writing and polynomial function? Can't we just always use the $-b/2a$ trick for the $x$ intercept and just plug it back into the function to find the $y$?
|
To answer your question, it isn't crucial in any serious way. However, it is as handy as, say, "point-slope form" or "slope-intercept form" for lines, it gives you a way to immediately recognize certain bits of information. Additionally, as Daniel M. mentions, it covers completing the square which is more often than not, the desired way to solve for the roots etc. of a quadratic (it's also just a handy algebraic manipulation.
Perhaps to demonstrate some utility of completing the square, we'll derive the quadratic formula starting from standard form:
$$ax^2+bx+c=0 \implies x^2+\frac{b}{a}x=-\frac{c}{a}$$
but now, we really need to complete the square:
$$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a} \implies(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2} \implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Notice how one could deduce the $\frac{-b}{2a}$ trick from the middle implication! You're right to notice that the two are intimately related, I think it's best to know both.
Let's use a similar technique to examine what's going on.
vertex form: $y=a_1(x-h)^2+k$. But, given a quadratic in standard form, $y=ax^2+bx+c$, we can deduce the following by adding and subtracting the same thing, while factoring out an $a$, and then completing the square:
\begin{align*}y&=ax^2+bx+c\\
&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)+\left(\frac{b^2}{4a}-\frac{b^2}{4a}\right)\\
&=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\\
&=a\left((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2} \right)\\
&=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}.
\end{align*}
Notice that this is a way to always get to "vertex form" and tells you exactly what $a,h,k$ are supposed to be in vertex form.
Namely:
$a_1=a$ (Hoorah!)
$h=-\frac{b}{2a}$ (As you have noticed)
$k=c-\frac{b^2}{4a}$
|
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|
How can I solve this nice rational equation I am trying solve this equation
$$\dfrac{3x^2 + 4x + 5}{\sqrt{5x^2 + 4x +3}}+\dfrac{8x^2 + 9x + 10}{\sqrt{10x^2 + 9x +8}} = 5.$$ Where $x \in \mathbb{R}$. I knew that $x=-1$ is a given solution. But I can't solve it. I tried
We rewrite the given equation in the form
$$\dfrac{3(x^2 + x + 1) + x + 2}{\sqrt{5(x^2 + x + 1) - (x+2)}}+\dfrac{8(x^2 + x + 1) + x + 2}{\sqrt{10(x^2 + x + 1)- (x+2)} } = 5.$$
Put $a = x^2 + x + 1$ and $b = x + 2$, we get
$$\dfrac{3a + b}{\sqrt{5a - b}}+\dfrac{8a + b}{\sqrt{10a- b} } = 5.$$
From here, I stoped.
|
Hint : Square the equation , isolate the term with the square-root and square the equation again.
|
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|
Summation Notation Question in McMillan's Theorem Proof Let me preface by saying that this question does not pertain as much to coding theory, as it does to mathematical notation. Every symbol in this question is a natural number.
Anyhow, I am currently reading through a proof of McMillan's Theorem. In the proof, there is the statement that
\begin{equation}\tag{1}
\left( \sum_{j=1}^m \dfrac{\alpha_j}{r^j} \right)^u = \left( \dfrac{\alpha_1}{r} + \cdots + \dfrac{\alpha_m}{r^m} \right)^u,
\end{equation}
and so far, so good. However, the next line states that $(1)$ out can be multiplied out to yield
\begin{equation}\tag{2}
\sum_{\substack{1 \leq i_j \leq m \\ i_1, i_2, \cdots, i_u}} \dfrac{\alpha_{i_1}}{r^{i_1}} \cdots \dfrac{\alpha_{i_u}}{r^{i_u}} = \sum_{\substack{1 \leq i_j \leq m \\ i_1, i_2, \cdots, i_u}} \dfrac{\alpha_{i_1} \alpha_{i_2} \cdots \alpha_{i_u}}{r^{i_1 + \cdots + i_u}},
\end{equation}
which confuses me. Can someone explain to me what the subscripts on the sigmas in equation $(2)$ are meant to indicate?
Also, I apologize if this question has already been answered somewhere else. I tried to look through the archives and didn't come across it.
|
Would this help?
\begin{align}
\left( \sum_{j=1}^2 \dfrac{\alpha_j}{r^j} \right)^3
&= \left( \dfrac{\alpha_1}{r}+\dfrac{\alpha_2}{r^2} \right)^3 \\
&= \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right)
\left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right)
\left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \\
&= \dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}+\\
&\phantom{=} \left(
\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}+
\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}+
\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}
\right) +\\
&\phantom{=} \left(
\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}+
\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}+
\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}
\right) + \\
&\phantom{=}
\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\\
&= \dfrac{\alpha_1\alpha_1\alpha_1}{r^{1+1+1}} + \\
&\phantom{=} \left(
\dfrac{\alpha_1\alpha_1\alpha_2}{r^{1+1+2}}+
\dfrac{\alpha_1\alpha_2\alpha_1}{r^{1+2+1}}+
\dfrac{\alpha_2\alpha_1\alpha_1}{r^{2+1+1}}
\right) +\\
&\phantom{=} \left(
\dfrac{\alpha_1\alpha_2\alpha_2}{r^{1+2+2}}+
\dfrac{\alpha_2\alpha_1\alpha_2}{r^{r^2+1+2}}+
\dfrac{\alpha_2\alpha_2\alpha_1}{r^{r^2+2+1}}
\right) + \\
&\phantom{=} \dfrac{\alpha_2\alpha_2\alpha_2}{r^{2+2+2}} \\
\end{align}
|
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|
Summing $3+7+14+24+37...$ up to $10$ terms
What is $3+7+14+24+37...$ up to $10$ terms?
I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?
|
This answer seeks to expand on the "one can check that" comment in Foobaz's answer. First, let $A=\{3,7,14,24,37,\ldots\}$. The pattern for the difference, $d_n$, between terms $n$ and $n+1$ is clear enough:
$$
d_n=4+(n-1)3=1+3n.
$$
Hence,
$$
a_{n+1}=\frac{n}{2}(5+3n)+3
$$
or, more usefully,
\begin{align}
a_n&=\frac{n-1}{2}(2+3n)+3\\[1em]
&=(n-1)+\frac{3n^2-3n}{2}+3\\[1em]
&= \frac{2n-2+3n^2-3n+6}{2}\\[1em]
&= \frac{3n^2-n+4}{2}.
\end{align}
Thus, we have the following:
\begin{align}
S_{10}&= \sum_{i=1}^{10}\frac{3i^2-i+4}{2}\\[1em]
&= \frac{3}{2}\sum_{i=1}^{10}i^2-\frac{1}{2}\sum_{i=1}^{10}i+2\sum_{i=1}^{10}1\\[1em]
&= \frac{3}{2}\left[\frac{10(10+1)(2(10)+1)}{6}\right]-\frac{1}{2}\left[\frac{10(10+1)}{2}\right]+2(10)\\[1em]
&= 570.
\end{align}
|
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|
Questions regarding dot product (possible textbook mistake) I am given the following exercise:
Show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
\overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ if and
only if $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel and
point to the same direction. Also, show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert \overrightarrow{a} - \overrightarrow{b} \Vert$ if and only if $\overrightarrow{a} \cdot \overrightarrow{b} = 0$
Regarding the first question:
Firstly, i feel like the first question is wrong since $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
\overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ for $\theta = 0$ or $\pi$ (so they must not necessarily point on the same direction). Is that correct?
if so, I proceeded the following way:
$$
\Vert \overrightarrow{a} + \overrightarrow{b} \Vert^2 = \Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
\Vert
\overrightarrow{a} \Vert^2 + 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert +\Vert \overrightarrow{b} \Vert^2 = \Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
\cos(\theta) = \pm 1\\
\theta = 0 \text{ or } \theta = \pi
$$
Regarding the second question:
$$
\Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 +2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta) = \Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 -2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
- \cos(\theta) = \cos(\theta)\\
\cos(\theta) = 0\\
\theta = \pi/2
$$
Can someone help me with these?
|
The first question is valid. For instance, if we let $0\neq b = -a$, then $||a+b|| = 0 \neq ||a||+||b||.$ So, $a$ and $b$ must point in the same direction.
As for a proof of the first statement: going backwards, assume $a$ and $b$ point in the same direction, i.e. (without loss of generality) $b = ca$ for some scalar $c\geq 0$, and so $$||a+b|| = ||a+ca|| = (1+c)\cdot||a|| = ||a||+||ca|| = ||a||+||b||$$
Now assume $||a+b||=||a||+||b||.$ The left hand side is equal to $\sqrt{||a||^2+||b||^2+2(a\cdot b)}$. So squaring both sides of the original equation and cancelling like terms, we get $a\cdot b = ||a||\cdot ||b||$, and so $b = ca$ where $c \geq 0$.
As for the second question, we can similarly expand $||a+b||=||a-b||$ to obtain $\sqrt{||a||^2+||b||^2+2(a\cdot b)} = \sqrt{||a||^2+||b||^2-2(a\cdot b)}$, and so $a\cdot b = 0$. And for proving this in the other direction, simply follow those steps backwards.
Your approach is more or less valid (depending on what sort of math class you're in), but the angles are unnecessary and should be avoided if possible.
|
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|
How to find the determinant of a 5 x 5 matrix? Finding a 3x3 matrix is easy, but how can I find the determinant of this 5x5 matrix?? I just need an example of the first couple steps to mimic
$A =$
$\begin{bmatrix} 7&1&9&-4&3\\0&-3&4&9&-6\\0&0&-6&-6&-9\\0&0&0&7&6\\0&0&0&0&2\end{bmatrix}$
then the $\det(A) = ?$
By the way, I did put the matrix in REF form and tried to multiply the diagonal and it didn't work at all.
I ended up getting
$\begin{bmatrix} 464,679,936&0&0&0&0\\0&-14,112&0&0&0\\0&0&-168&0&0\\0&0&0&14&0\\0&0&0&0&2\end{bmatrix}$
|
Developing with respect the last row, you get that the determinant equals
$$2\cdot\begin{vmatrix}7&1&9&\!-4\\0&\!-3&4&9\\0&0&\!-6&\!-6\\0&0&0&7\end{vmatrix}$$
Again deloping the last row we get
$$2\cdot7\cdot\begin{vmatrix}7&1&9\\0&\!-3&4\\0&0&\!-6\end{vmatrix}=2\cdot7\cdot(-6)\cdot\begin{vmatrix}7&1\\0&\!-3\end{vmatrix}=-84(-21)=1,764$$
|
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|
Solving differential equation of third degree If differential equation of the curves $$c(y+c)^2 = x^3$$ where 'c' is an arbitary constant is $$12y(y')^2 + ax = bx(y')^3$$
What is the value of a+b?
I tried differentiating the curve given and obtain a linear relation with the equation.
I got $$cyy'y'' + c(y')^3 + c^2y'y'' = 3x$$
But in the equation 'x' appears alongwith $(y')^3$.
I'm stuck here.
|
$$c(y+c)^2=x^3$$
Differentiate :
$$2c(y+c)y'=3x^2$$
$2c(y+c)^2y'=3x^2(y+c)=2x^3y'$
$$y+c=\frac{2}{3}xy'\quad\to\quad c=\frac{2}{3}xy'-y$$
$c(y+c)^2=x^3=(\frac{2}{3}xy'-y)(\frac{2}{3}xy')^2$
$x=\frac{8}{27}xy'^3-\frac{4}{9}yy'^2$
$-\frac{8}{27}xy'^3+\frac{4}{9}yy'^2+x=0$
$$12yy'^2+27x-8xy'^3=0$$
Compared to
$$12yy'^2+ax-bxy'^3=0$$
$$a=27\text{ and }b=8 \quad a+b=35$$
|
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|
Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image?
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is:
$90^{\circ}+ 720^{\circ}k$ where $k=1,2,3,...$
Intuitively, I understand the answer, but that's not the point, I would like to understand the correct mathematical solution. Thank you very much!
|
Note that: $$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2} \cos \left(\frac{\theta}{2} - \frac{\pi}{4} \right) \quad \quad \quad \quad (\star)$$
So your equation reduces down to $\cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right) = 1 \iff \frac{\theta}{2} - \frac{\pi}{4} = 2\pi k$ where $k$ ranges over the integers.
This can be re-written equivalently as $\theta = \frac{\pi}{2} + 4\pi k$ or in degrees: $\theta = 90^{\circ} + 720^{\circ} k$.
$(\star):$ Assume that $\sin x + \cos x = R\cos (x - \alpha)$ then expanding the left hand, we have $\sin x + \cos x = R\cos x \cos \alpha + R\sin x \sin \alpha$. Now we have $R\cos \alpha = 1$ and $R\sin \alpha = 1$, squaring and adding gives us $R^2 (\cos^2 \alpha + \sin^2 \alpha) = 2 \Rightarrow R = \sqrt{2}$. Dividing the two equations gives $\frac{R\sin \alpha}{R\cos \alpha} = \tan \alpha = 1\Rightarrow \alpha = \frac{\pi}{4}$, whence $(\star)$.
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Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get
$$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$
So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$
So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$
My question is how can we solve it in some shorter way, Help required, Thanks
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your change of variable is wrong.
$$\int\frac{1}{t(1+t^{2})}dt=\int\frac{1}{t^{3}\left(1+\frac{1}{t^{2}}\right)}dt
$$
and now we use the change of variable $u=\frac{1}{t^{2}} $ => $du=-\frac{dt}{t^{3}}$,then the integral take the form:
$$-\int\frac{1}{1+u}du=-\ln(1+u)=\ln\left(\frac{1}{1+u}\right)$$
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|
$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod n}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/n\rfloor}}{2k+1}=\sqrt{n}$ for $n\ge 2$ I found tasks in an old script. For the following problem I have no idea how to solve.
How can one prove $$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod n}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/n\rfloor}}{2k+1}=\sqrt{n}$$ for $n\in\mathbb{N}$ with $n\ge 2$?
It generalizes
How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$ .
Examples $n=2$ and $n=3$:
$$\prod\limits_{k=0}^\infty \frac{2k+1+(-1)^k}{2k+1}= \frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11}…=\sqrt{2}$$
$$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod 3}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/3\rfloor}}{2k+1}= \frac{2}{1}\frac{4}{5}\frac{8}{7}\frac{10}{11}\frac{14}{13}\frac{16}{17}…=\sqrt{3}$$
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Literature: Sierpinski Oeuvres Choisies I (page 222-226)
http://www.plouffe.fr/simon/math/Sierpinski%20Oeuvres%20Choisies%20I.pdf
Gamma function product: $\displaystyle \Gamma(1+x)=\lim\limits_{m\to\infty} m^x/\prod\limits_{k=1}^m \left(1+\frac{x}{k}\right)$
or written äquivalent $\displaystyle m^x/\prod\limits_{k=1}^m \left(1+\frac{x}{k}\right) \sim\Gamma(1+x)$ .
Therefore one get
$(1)$ $\displaystyle m^x\prod\limits_{k=1}^m \frac{k-x}{k}\sim\Gamma(1-x)$
and general
$(2)$ $\displaystyle (nm)^x\prod\limits_{k=1}^{nm} \frac{k-x}{k}\sim\Gamma(1-x)$
for $n\in\mathbb{N}$ and $x\in\mathbb{R}\setminus\mathbb{N}$.
It’s $\dfrac{nk}{nk-x}\dfrac{k-x}{k} =1+\dfrac{(1-n)x}{nk-x}$ and with $\dfrac{(2)}{(1)}$ follows therefore
$$ \prod\limits_{k=1}^m\frac{k-x}{k} \prod\limits_{k=1}^{nm}\frac{k}{k-x} =
\prod\limits_{k=1}^m \left(\frac{nk}{nk-x}\frac{k-x}{k}\right) \prod\limits_{ k=1 \atop k\ne 0 \bmod n }^{nm}\left(1+\frac{x}{k-x}\right) $$ $$=
\prod\limits_{k=1}^{nm}\left(1+\frac{x(1-0^{k-n\lfloor\frac{k}{n}\rfloor}n)}{k-x}\right) \sim n^x$$
with $0^0:=1$.
With $x:=\frac{1}{2}$ the formula of the question is proofed.
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|
Concyclicity of $4$ points using algebraic geometry
Consider the line $L_1:ax+4y-1=0$ and a circle $S:x^2+y^2-10x+2y+10=0$. The line intersects the circle at $2$ distinct points $A$ and $B$. Another line $5x-12y-67=0$ intersects the circle $x^2+y^2+6x+14y-28=0$ at $2$ distinct points $C$ and $D$. Find the value of $a$ for which the $4$ points $A,B,C,D$ are concyclic. Also, find the equation of the circle passing through these $4$ points.
How do I use the concyclic condition? If I make equations using opposite angles sum $180°$, the equations are pretty nasty. Some hints, please. Thanks.
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HINT:
The equation of the circle passing through the intersection of $L_1, S$
$$x^2+y^2-10x+2y+10+A(ax+4y-1)=0\ \ \ \ (1)$$
Similarly for $5x-12y-67=0, x^2+y^2+6x+14y-28=0$
$$x^2+y^2+6x+14y-28+B(5x-12y-67)=0\ \ \ \ (2)$$
$A,B,C,D$ will be concyclic if $(1)$ & $(2)$ become identical.
Can you take it from here?
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number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
Then number of points where $g(x)$ is not differentiable.
$\bf{My\; Try::}$ We can write $$f(x) =\lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}= \left\{\begin{matrix}
\frac{2}{x+1}\;\;,&x+1<-1\Rightarrow x<-2 \\
\frac{x^2}{x^2+1}\;\;, & -1<x+1<1\Rightarrow -2<x<0\\
\frac{2}{x+1}\;\;, & x+1>1\rightarrow x>0 \\
1\;\;, & x+1=1\Rightarrow x=0\\
\frac{3}{2}\;\;,& x+1=-1\Rightarrow x=-2
\end{matrix}\right.$$
Now How can i solve after that help Required, Thanks
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if you plot function f, you will see $-2 \lt f \lt 2$ with 2 discontinuity points at -2, 0, which are not differentiable.
Next you can say the following x are not differentiable, $$\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right) = \pm n\pi$$
This leads to the points satifying the following equation.
$$\frac{2f(x)}{1+(f(x))^2} = 0$$ or $$f(x) = 0$$
There is only one, i.e. x=0.
So the answer is two non differentiable points -2 and 0.
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|
If $x^2+\frac{1}{2x}=\cos \theta$, evaluate $x^6+\frac{1}{2x^3}$. If $x^2+\frac{1}{2x}=\cos \theta$, then find the value of $x^6+\frac{1}{2x^3}$.
If we cube both sides, then we get $x^6+\frac{1}{8x^3}+\frac{3x}{2} \cdot \cos \theta=\cos ^3 \theta$ but how can we use it to deduce required value?
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Let $c = \cos \theta$. You have $x^3 = c x - 1/2$, so $x^6 = (c x - 1/2)^2$ while $1/x = 2 c - 2 x^2$. Then I get
$$\eqalign{x^6 + \dfrac{1}{2x^3} &= - 3 c^2 x^2 - 3 c x + 4 c^3 - \frac{3}{4}\cr & =
-\dfrac{3}{2} (1 + \cos(2\theta)) x^2 - 3 \cos(\theta) x + \cos(3\theta) + 3 \cos(\theta) - \dfrac{3}{4}}$$
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|
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
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$$\int \frac { x^{ 2 }-2 }{ \left( x^{ 2 }+2 \right) ^{ 3 } } dx=\int { \frac { { x }^{ 2 }+2-4 }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } dx=\underbrace { \int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 2 } } } }_{ { I }_{ 1 } } -4\underbrace { \int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } }_{ { I }_{ 2 } } =\\ x=\sqrt { 2 } \tan { t } \\ dx=\sqrt { 2 } \frac { dt }{ \cos ^{ 2 }{ t } } \\ \\ { I }_{ 1 }=\int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 2 } } } =\sqrt { 2 } \int { \frac { dt }{ 4{ \left( \tan ^{ 2 }{ t } +1 \right) }^{ 2 }\cos ^{ 2 }{ t } } } =\frac { \sqrt { 2 } }{ 4 } \int { \cos ^{ 2 }{ t } dt } =\frac { \sqrt { 2 } }{ 4 } \int { \frac { 1+\cos { 2t } }{ 2 } } dt=\\ =\frac { \sqrt { 2 } }{ 8 } \left( t+\frac { \sin { 2t } }{ 2 } \right) =\frac { \sqrt { 2 } }{ 8 } \left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } +\frac { \sin { 2\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 2 } \right) \\ \\ { I }_{ 2 }=\int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } =\sqrt { 2 } \int { \frac { dx }{ { 8\left( \tan ^{ 2 }{ t } +1 \right) }^{ 3 }\cos ^{ 2 }{ t } } } =\frac { \sqrt { 2 } }{ 8 } \int { \cos ^{ 4 }{ t } dt } =\frac { \sqrt { 2 } }{ 8 } \int { { \left( \frac { 1+\cos { 2t } }{ 2 } \right) }^{ 2 } } dt=\\ =\frac { \sqrt { 2 } }{ 32 } \int { \left( 1+2\cos { 2t+\cos ^{ 2 }{ 2t } } \right) } dt=\frac { \sqrt { 2 } }{ 32 } \int { \left( 1+2\cos { 2t } +\frac { 1+\cos { 4t } }{ 2 } \right) } dt=\\ =\frac { \sqrt { 2 } }{ 32 } \left( \frac { 3t }{ 2 } +\sin { 2t+\frac { \sin { 4t } }{ 8 } } \right) =\frac { \sqrt { 2 } }{ 32 } \left( \frac { 3\arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } }{ 2 } +\sin { 2\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } +\frac { \sin { 4\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 8 } \right) $$
so the anwer will be $$\int \frac { x^{ 2 }-2 }{ \left( x^{ 2 }+2 \right) ^{ 3 } } dx=\frac { \sqrt { 2 } }{ 8 } \left( \frac { 3\sin { 2\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 2 } -\frac { \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } }{ 2 } +\frac { \sin { 4\left( \arctan { \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } }{ 8 } \right) $$
further you can simplify this "monster"
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Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$ Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$
Solutions are $x=2$ or $x=-1$.
But $x=2$ does not satisfy $\sqrt{x+2}+x=0$.Because $\sqrt{4}+2 \neq0$
So does it mean that they are different ? Why ?
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Yes. By convention $\sqrt{x+2}$ is non-negative.
You can multiply the first by $\sqrt{x+2}-x$ to get $x+2-x^2=0$. You introduced the solution $x=2$ when you assumed $\sqrt{x+2}\sqrt{x+2}=x+2$. This is not true, since $x+2$ can be negative.
Note that this does not violate the fundamental theorem of algebra since you did not begin with a polynomial.
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Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff & x^3+y^3\geq \frac{1}{4}(x+y)^3
\end{split}$$
is the proof valid? is there a shorter way?
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Using AM-GM Inequality:
$$\frac{x+y}2\ge\sqrt{xy}\iff xy\le \left(\frac{x+y}2\right)^2$$
We have:
\begin{align}
(x+y)^3&=x^3+y^3+3xy(x+y)\\
&\le x^3+y^3+3\left(\frac{x+y}2\right)^2(x+y)\\
&=x^3+y^3+\frac 34 (x+y)^3
\end{align}
So
$$x^3+y^3\ge \frac14 (x+y)^3$$
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|
Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
they use in solution manual :
My questions:
*
*I don't know how to achieve ( * ) could someone complete my attempts for ( * ) and is it correct if i use (**) to prove that the series is convergent :
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}-\dfrac{(-1)^{n}}{n}+O\left(\dfrac{1}{n^{\frac{4}{3}}}\right)$}\quad (*)$$
My thoughts :
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}
\end{align}
note that :
$$(1+x)^{\alpha}=1+\alpha x+O(x^{2})$$
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1-\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)+O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} -\dfrac{(-1)^{n}}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}} +\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\times O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \\
&=\ldots\ldots \\
&= \mbox{ I'm stuc here i hope someone complete my attempts }
\end{align}
Or i should use :
note that :
$$(1+x)^{\alpha}=1+O(x)$$
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+O\left( \dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\right) \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{1}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
\end{align}
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} =\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right)$}\quad (**) $$
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Let $f(n)=\frac{1}{n^{2/3}+n^{1/3}+A}$ and consider $\phi_m(n)=n^{-m/3}$ for $m=2,3,...$ Then you should be after an asymptotic expnasion for $f$ of the form $$f(n)=\sum a_m\phi_m(n).$$ We have $$a_2=\lim_{n\to\infty}\frac{f(n)}{\phi_2(n)}=1$$ and $$a_3=\lim_{n\to\infty}\frac{f(n)-a_2\phi_2(n)}{\phi_3(n)}=-1$$ and $$a_3=\lim_{n\to\infty}\frac{f(n)-a_2\phi_2(n)-a_3\phi_3(n)}{\phi_4(n)}=1-A$$
etc...
and as such we obtain $$f(n)=\frac{1}{n^{2/3}}-\frac1n+(1-A)\frac{1}{n^{4/3}}+O(n^{5/3})$$
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|
Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
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$$\begin{align*} \sin^2 x \cos 4x &= \sin x (\sin x \cos 4x) \\ &= \frac{1}{2} \sin x (\sin 5x - \sin 3x) \\ &= \frac{1}{4}(\cos 4x - \cos 6x + \cos 4x - \cos 2x) \\ &= \frac{1}{2} \cos 4x - \frac{1}{4} \cos 6x - \frac{1}{4} \cos 2x. \end{align*}$$
Consequently, we immediately and rather trivially obtain $$\int \sin^2 x \cos 4x \, dx = \frac{1}{8} \sin 4x - \frac{1}{24} \sin 6x - \frac{1}{8} \sin 2x + C.$$
Alternatively, observe $$\begin{align*}
\sin^2 x \cos 4x &= \sin^2 x (\cos 3x \cos x - \sin 3x \sin x) \\
&= \cos 3x \sin^2 x \cos x - \sin 3x \sin^3 x \\
&= \frac{1}{3} \left( \frac{d}{dx}\left[\sin^3 x\right] \cos 3x + \frac{d}{dx} \left[ \cos 3x \right] \sin^3 x \right) \\ &= \frac{1}{3} \frac{d}{dx}\left[\cos 3x \sin^3 x\right], \end{align*}$$ the last step due to the product rule applied to the functions $\sin^3 x$ and $\cos 3x$; thus $$\int \sin^2 x \cos 4x \, dx = \frac{1}{3} \cos 3x \sin^3 x + C.$$ It is quite straightforward to demonstrate that these antiderivatives are equivalent.
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|
Can $x^2+y^2,y^2+z^2,z^2+x^2$ and $x^2+y^2+z^2$ all be square numbers? I know that if we want $x^2+y^2$ to be square number, we are looking for pythagorean triple; if we want $x^2+y^2+z^2$ to be a square number, we are looking for pythagorean quadruple. But have we ever found any positive integers $x,y,z$ such that $x^2+y^2,y^2+z^2,z^2+x^2,x^2+y^2+z^2$ are all square numbers?
|
It's a famous open problem : the perfect cuboid problem.
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$f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.
My Attempt:
Let $f(x)=ax^2+bx+c$ and $a,c \neq 0 $
I got $a+b+c=3$
And by the functional equation
$a[ax^2+(b+1)x+c]^2+b[ax^2+(b+1)x+c]+c= [ax^2+bx+c][x^2+4x-7]$
Then by putting $x=0$ , $ac^2 +bc+c=-7c$
Since $c \neq 0 $ , we have $ac+b+8=0$
Then comparing the coeffiecient of $x^4$ , we get $a^3=a$
Since $a \neq 0$ , $a=-1 $ or $a =1 $
Then how to proceed with two values of $a$ ?
or Is there a polynomial satisfying these conditions?
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$ac + b + 8=0$, $ac - a - c + 11 = 0$. If $a = 1$, then $10 = 0$. So $a = -1$, hence $c = 6$, $b = -2$, $f(x) = -x^2 -2x + 6$
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|
How to solve this inequality problem? Given that $a^2 + b^2 = 1$, $c^2 + d^2 = 1$, $p^2 + q^2 = 1$, where $a$, $b$, $c$, $d$, $p$, $q$ are all real numbers, prove that $ab + cd + pq\le \frac{3}{2}$.
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Subtracting 2ab, 2cd and 2pq from the three equations gives us:
$$ a^2+b^2-2ab = 1 - 2ab$$$$
c^2 + d^2 - 2cd = 1 - 2cd$$$$p^2+q^2 - 2pq = 1-2pq$$
Noting that the LHS of each of these equations is a perfect square and all perfect squares are non-negative we have:$$1-2ab≥0$$$$1-2cd≥0$$$$1-2pq≥0$$
Adding the three equations and dividing by 2 across we obtain
$$\frac32 - (ab + cd + pq) ≥ 0$$$$\therefore ab + bc + cd ≤\frac32 \qquad\qquad \Box$$
Hence Proved
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|
Doubt in finding number of integral solutions Problem :
writing $5$ as a sum of at least $2$ positive integers.
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $${7\choose3}= 35$$
but we can count the cases and say that answer must be $6$ :
$4 + 1 $
$3 + 2 $
$3 + 1 + 1$
$2 + 2 + 1 $
$2 + 1 + 1 + 1$
$1 + 1 + 1 + 1 + 1$
At which stage am I making a mistake ? Thanks
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There are $2^{5-1} -2^{(2-1)-1}=15$ compositions of $5$ into at least $2$ positive integers while there are $6$ partitions of $5$ into at least $2$ positive integers. With compositions, $1+1+3$ is distinct from $3+1+1$, while with partitions they are the same.
But your first attempt counts even more than compositions. It actually counts cases where the first two elements are positive and the next three are non-negative, including $3+1+0+0+1$. I doubt this is what you want to count.
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|
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-sinxdx$
$$-\int \frac{(1-u^2 )u^2}{2-u^2}du=\int \frac{u^4-u^2 }{2-u^2}du$$
Polynomial division give us
$$\int-u^2-1+\frac{2}{u^2-2}du=-\frac{u^3}{3}-u=2\int \frac{1}{u^2-2}du$$
Using partial fraction we get to:
$$\frac{1}{u^2-2}=\frac{1}{(u+\sqrt{2})(u^2-\sqrt{2})}=\frac{A}{u^2+\sqrt{2}}+\frac{B}{u^2-\sqrt{2}}$$
So we get to:
$$1=(A+B)u^2+\sqrt{2}(A-B)$$
So $A=B$ but $0*\sqrt{2}\neq 1$, where did it go wrong?
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The decomposition in partial fractions should be
$$\frac{1}{u^2-2}==\frac{A}{u +\sqrt{2}}+\frac{B}{u -\sqrt{2}}.$$
But every one should know that
$$\int\frac1{a^2-x^2}\,\mathrm d x=\frac1{2a}\ln\biggl\lvert\frac{x+a}{x-a}\biggr\rvert=\frac1a\,\operatorname{argtanh}\Bigl(\frac xa\Bigr).$$
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|
$3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ Prove that $3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ for all $x,y\geq 0$.
Expanding, the inequality becomes
$$3x^2+3xy+3y^2-6x\sqrt{xy}-6y\sqrt{xy}+6xy\geq x^2+xy+y^2$$
which is
$$x^2+4xy+y^2\geq3\sqrt{xy}(x+y)$$
We can try using AM-GM:
$$x^2+xy+xy\geq 3\sqrt[3]{x^4y^2}$$
This is close to the right-hand side but still different.
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$\Leftrightarrow (\sqrt{x}-\sqrt{y})^2(x-\sqrt{xy}+y)\geq 0$.
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|
How does $\int_{\pi/3}^{\pi/2} \frac{1-\cos^2x}{\sqrt{\sin^2(x/2)}}dx$ simplify to $\int_{\pi/3}^{\pi/2} 4\sin(x/2)\cos^2(x/2)dx$? $$\int_{\large{\frac{\pi}{3}}}^{\large{\frac{\pi}{2}}} \frac{1-\cos^2x}{\sqrt{\sin^2\left(\frac x2\right)}}dx$$
How does the above simplify to the below?
$$\int_{\large{\frac{\pi}{3}}}^{\large{\frac{\pi}{2}}} 4\sin\left(\frac x2\right)\cos^2\left(\frac x2\right)dx$$
I suspect it has something to do with a double angle or half-angle identity, but I don't know where the $4$ came from, and I am not sure how the $\sin\left(\frac x2\right)$ made its way to the numerator.
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$$\frac{1-\cos^2 x}{\sin\frac{x}{2}}=\frac{\sin^2 x}{\sin\frac{x}{2}}=\frac{\left(\color{red}{2}\sin\frac{x}{2}\cos\frac{x}{2}\right)^2}{\sin\frac{x}{2}}=\color{red}{4}\cos^2\frac{x}{2}\sin\frac{x}{2}.$$
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|
Evaluate $\int \frac{x}{x^4+5}dx$
$$\int \frac{x}{x^4+5}dx$$
$u=x^2$
$du=2xdx\Rightarrow \frac{du}{2}=xdx$
$$\int \frac{x}{x^4+5}=\frac{1}{2}\int \frac{du}{u^2+5}$$
I want to get to the expression in the form of $\frac{da}{a^2+1}$ so I factor out $5$ to get to:
$$\frac{1}{2}\int \frac{du}{5[(\frac{u}{\sqrt{5}})^2+1]}$$
$\frac{du}{5[(\frac{u}{\sqrt{5}})^2+1]}$ is in the form of $\frac{\frac{a}{b}}{\frac{5c}{d}}$ so if I factor out the $5$ should it be $$\int \frac{1}{2}*\frac{1}{5}\frac{du}{[(\frac{u}{\sqrt{5}})^2+1]}$$
or
$$\int \frac{1}{2}*5\frac{du}{[(\frac{u}{\sqrt{5}})^2+1]}$$
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The First is correct. Let $t=u/\sqrt5$ then $dt=du/\sqrt5$:
$$\frac 12 \frac 1{\sqrt5}\int\frac {dt}{1+t^2}=\frac 12 \frac 1{\sqrt5}\tan^{-1}(t)+c=\frac 12 \frac 1{\sqrt5}\tan^{-1}(\frac {x^2}{\sqrt5})+c$$
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|
Combinatorial identity's algebraic proof without induction. How would you prove this combinatorial idenetity algebraically without induction?
$$\sum_{k=0}^n { x+k \choose k} = { x+n+1\choose n }$$
Thanks.
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Suppose we seek to verify that
$$\sum_{k=0}^n {q+k\choose k} = {q+n+1\choose n}.$$
The difficulty here lies in the fact that the binomial coefficients on
the LHS do not have an upper bound for the sum wired into them. We use
an Iverson bracket to get around this:
$$[[0\le k\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{w^k}{w^{n+1}} \frac{1}{1-w} \; dw.$$
Introduce furthermore
$${q+k\choose k} = {q+k\choose q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q+k} \; dz.$$
With the Iverson bracket in place we can let the sum range to
infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\sum_{k\ge 0} w^k (1+z)^k
\; dw\; dz.$$
This converges when $|w(1+z)| < 1.$ Simplifying we have
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{1-w(1+z)}
\; dw\; dz.$$
Now the residues of the inner integral sum to zero so we can evaluate
it by computing the negative of the residues at $w=1$ and at
$w=1/(1+z).$ We get for the first one
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+2}}
(1+z)^{q} \; dz = - [z^{q+1}] (1+z)^q = 0.$$
For the second one we have
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q-1}
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{1/(1+z)-w}
\; dw\; dz.$$
This yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q-1}
(1+z)^{n+1}
\frac{1}{1-1/(z+1)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+2}}
(1+z)^{q+n+1} \; dz
= {q+n+1\choose q+1}
= {q+n+1\choose n}.$$
This concludes the argument.
In order to be rigorous we must show that the residue at infinity
of the inner integral is zero. We get
$$\mathrm{Res}_{w=\infty}
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{1-w(1+z)}
\\ = - \mathrm{Res}_{w=0} \frac{1}{w^2}
w^{n+1} \frac{1}{1-1/w} \frac{1}{1-(1+z)/w}
\\ = - \mathrm{Res}_{w=0}
w^{n+1} \frac{1}{w-1} \frac{1}{w-(1+z)}.$$
There is most certainly no pole here an the residue is zero as
claimed (note that $1+z$ circles the value one.)
Addendum. I just realized that it is actually a lot simpler
without the Iverson bracket. We get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q}
\sum_{k=0}^n (1+z)^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
(1+z)^{q}
\frac{(1+z)^{n+1}-1}{1+z-1}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+2}}
(1+z)^{q}
((1+z)^{n+1}-1)
\; dz.$$
This has two pieces, the second is
$$-[z^{q+1}] (1+z)^q = 0$$
and the first is
$$[z^{q+1}] (1+z)^{q+n+1} = {q+n+1\choose q+1}
= {q+n+1\choose n}.$$
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|
Find all triples satisfying an equation Another question I saw recently:
Find all triples of positive integers $(a,b,c)$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.
Can someone help me with it?
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As $\frac{1}{n}>0$ for all natural $n$, all of $a,b,c$ have to be at least two.
Let $a\leq b\leq c$. Unless $a=b=c=3$, $a=2$. Then we have $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$, so similarly either $b=c=4$ or $b=3$. In the latter case, we get $c=6$.
So, $(a,b,c)=(3,3,3),(2,4,4),(2,3,6)$ are all the solutions.
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|
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$
$$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$
I think, I have made the things more difficult. How can I proceed further? Is there any better substitution for it?
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A trigonometric formula can be used：
\begin{align}
&(\sin x+\cos x+1)(\sin x+\cos x-1)=\sin 2x\\
I&=\int\frac{\sin x\cos x}{\sin x+\cos x+1}dx=\int\frac{1}{2}(\sin x+\cos x-1)dx=\frac{1}{2}(-\cos x+\sin x-x)+C\\
\end{align}
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|
Show that $a_n$ is decreasing
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.
First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_n) - 1}{3 - a_n} = \frac{8 - 3a_{n}}{3 - a_n} > 0$, which means showing $8 > 3a_{n}$, but I'm having trouble showing it.
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First compute, $a_2=\frac{1}{3-2}=1\leq a_1<3$. Then, notice that
$$
a_{n+1}-a_{n}=\frac{1}{3-a_n}-\frac{1}{3-a_{n-1}}=\frac{a_n-a_{n-1}}{(3-a_n)(3-a_{n-1})}\leq 0
$$
which gives you the induction step for a strong induction argument for a dual hypothesis: $a_n< 3$ and $a_{n}\leq a_{n+1}$ for all $n\geq 1$. Note that if you already have $a_n<3$, then $a_{n+1}\leq a_n$ will automatically give $a_{n+1}<3$.
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|
How to solve $y'' + y = -2\sin(x)$? I don't know how to find the particular solution of $$y'' + y = -2\sin(x)$$
I started with $$y'' + y = 0$$ to find the homogeneous form $$A\cos(x)+B\sin(x)$$
But now i am stuck.
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You need to find the particular solution.
Let $y_{p}=x(a\cos x+b\sin x)$.
\begin{align*}
y'_{p} &= a\cos x+b\sin x+x(-a\sin x+b\cos x) \\
y''_{p} &= 2(-a\sin x+b\cos x)-x(a\cos x+b\sin x) \\
y''_{p}+y_{p} &= 2(-a\sin x+b\cos x) \\
-2\sin x &= 2(-a\sin x+b\cos x) \\
(a,b) &= (1,0) \\
y(x) &= A\cos x+B\sin x+x\cos x
\end{align*}
Further points to be noticed
Note that $-2\sin x$ is one of the possibilities of the general solution $A\cos x+B\sin x$, the form $y_{p}=a\cos x+b\sin x$ always makes the LHS of the ODE to be zero. It's simpler to try the form of $x(a\cos x+b\sin x)$ and we found that it's doable. It's in fact a resonance term.
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|
Quadratic Inequality in terms of variable $x$
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help required, Thanks
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We are interested in $f(x)=x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$$f(1)=1+a+a^2+6a \leq 0$$
That is $$a^2+7a+1\leq 0$$
$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-7+\sqrt{45}}{2}$$
Also, we want, $$f(2)=4+2a+a^2+6a \leq 0$$
$$a^2+8a+4 \leq 0$$
$$\frac{-8-\sqrt{48}}{2} \leq a \leq \frac{-8+\sqrt{48}}{2}$$
Hence, overall, we need to take the intersection of the two constraint.$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-8+\sqrt{48}}{2}$$
|
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|
Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
|
For $n \geq 6$ we have
$$\frac{(n+3)^n}{(n+2)^n} = \left( 1+\frac{1}{n+2} \right)^n = \left( 1+\frac{1}{n+2} \right)^{n+2} \cdot \left(\frac{n+2}{n+3}\right)^2$$
Both factors are monotonely increasing, hence this is at least its value for $n=6$, which is $\left(\frac{9}{8}\right)^6 \approx 2.072\dotsc > 2$.
Since $k \mapsto \frac{k+1}{k}$ is monotonely decreasing, we obtain
$$\frac{(k+1)^n}{k^n} \geq \frac{(n+3)^n}{(n+2)^n} > 2$$
for all $k = 1, \dotsc, n+2$, if $n \geq 6$. We can write this as
$$(k+1)^n-k^n > k^n$$
for all $k = 1, \dotsc, n+2$, if $n \geq 6$. This also holds for $k=0$.
In particular we have $$(n+3)^n = \sum_{k=0}^{n+2} ((k+1)^n-k^n) > \sum_{k=0}^{n+2} k^n = 1^n + 2^n + 3^n + \dotsb + (n+2)^n$$
for $n\geq 6$, an even stronger result.
|
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|
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac{\pi}{12} $
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$
$$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$
Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$
How to solve this ?
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Using the fact that $\tan { \left( \alpha -\beta \right) =\frac { \tan { \alpha -\tan { \beta } } }{ 1+\tan { \alpha \tan { \beta } } } } ,\tan { \left( \tan ^{ -1 }{ \alpha } \right) =\alpha } $$$\tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) =t\ \tan { \left( \tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) \right) =\tan { t } } \ \frac { \tan { \left( \tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) \right) -\tan { \left( \tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) \right) } } }{ 1+\tan { \left( \tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) \right) \tan { \left( \tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) \right) } } } =\tan { t } \ \tan { t } =\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } }{ 1+\frac { 1 }{ \sqrt { 2 } } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) } =\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \sqrt { 3 } -\sqrt { 2 } }{ 1+\sqrt { 6 } } }{ 1+\frac { 1 }{ \sqrt { 2 } } \left( \frac { \sqrt { 3 } -\sqrt { 2 } }{ 1+\sqrt { 6 } } \right) } =\frac { 3 }{ 3\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } \ $$
so
$$\tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } } } }{ 1+\sqrt { 6 } } \right) =\frac { \pi }{ 6 } $$
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}[\ln(\tan \ x)^{\frac{2}{3}}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}-\ln(1)^\frac{2}{3}]=\frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}] = \bf \color{red}{\ln(\sqrt{3})}$$
However, the answer given is not this value, but instead $= \frac{3\sqrt[3]{3}-3}{2} \approx 0.663... $ while $\ln(\sqrt{3}) \approx 0.549...$
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Put $u = \tan x\implies I = \displaystyle \int_{1}^{\sqrt{3}} u^{-\frac{1}{3}}du= \left[\frac{3}{2}u^{\frac{2}{3}}\right]|_{u=1}^{u = \sqrt{3}}= \dfrac{3}{2}\left(\sqrt[3]{3}-1\right)$
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{
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"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$ $$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$
$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$
$$1=Ax+A+Bx-2B$$
$$1=(A+B)x+A-2B$$
$A+B=0\iff A=-B$
$-3B=1$
$B=-\frac{1}{3}$, $A=\frac{1}{3}$
$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\frac{1}{3}\int_{3}^{\infty}\frac{dx}{x-2}-\frac{1}{3}\int_{3}^{\infty}\frac{dx}{(x+1)}=|\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1)|_{3}^{\infty}$$
$$=lim_{t\to \infty}(\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1))-\frac{1}{3}ln(1)+\frac{1}{3}ln(4)=\infty-\infty-0+\frac{ln(4)}{3}$$
But the answer is $\frac{2ln(2)}{3}$, What have I done wrong?
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Observe $\infty-\infty$ is indeterminate form, so you need to get rid from it by writing:
$$\lim_{t\to \infty}(\frac{1}{3}\ln(t-2)-\frac{1}{3}\ln(t+1))=\lim_{t\to \infty}\frac{1}{3}\ln{\frac{(t-2)}{(t+1)}}=\frac{1}{3}\ln{1}=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040} \right)$.
How do we proceed further?
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For example, very close to zero we have
$$\frac{z^2+1}{\sin z}=\frac{z^2+1}{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots}=\frac{z^2+1}{z\left(1-\frac{z^2}6+\frac{z^4}{120}-\ldots\right)}\stackrel{\text{Devel. of geom. series}}=$$
$$=\left(z+\frac1z\right)\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=\frac1z+\frac76z+\frac7{36}z^3+\ldots$$
With the above you already have the pole at $\;z=0\;$ is a simple one and its residue is $\;1\;$ .
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Overcounting when doing Counting Problems Problem:
How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?
Attempted Solution:
$$N_\text{Teams} = {12 \choose 2}{10 \choose 5}{ 5 \choose 5} = {12 \choose 2}{10 \choose 5}$$
Arriving at this solution by first picking the team of 2 and then the two teams of 5, however the provided solution requires that my answer be divided by 2 due to "Overcounting" which is briefly mentioned but not explained. How does one gain an intuition for overcounting in this problem and also for counting problems in general?
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There's one more (cool, in my opinion) way to look at such problems.
For each possible allocation of people into groups, let's take the same example as in the OP, there are a total of
$$
\binom{12}{2}\binom{10}{5}\binom{5}{5} =16632
$$
allocations, such as (letters are people, numbers are groups):
$$
(1)AB|(2)CDEFG|(3)HIJKL
$$
Now, $\mathbf{each}$ such allocation $\textbf{enters the list exactly twice}$:
$$
(1)AB|(2)CDEFG|(3)HIJKL\\
(1)AB|(2)HIJKL|(3)CDEFG
$$
There are no other $\mathbf{identical}$ allocation, i,e. no other way to produce an identical allocation. Hence, the number must be halved, or remove ecery second entry from the list:
$$
\binom{12}{2}\binom{10}{5}\binom{5}{5} - \frac{\binom{12}{2}\binom{10}{5}\binom{5}{5}}{2} = 8316
$$
Here's another example: also 12 people, but 2 groups of 4 and 2 groups of 2. What do we have? The list size is
$$
\binom{12}{4}\binom{8}{4}\binom{4}{2}\binom{2}{2}
$$
Base allocation is
$$
(1)ABCD|(2)EFGH|(3)IJ|(4)KL
$$
how many of such identical allocations?
$$
(1)ABCD|(2)EFGH|(3)IJ|(4)KL\\
(1)ABCD|(2)EFGH|(3)KL|(4)IJ\\
(1)EFGH|(2)ABCD|(3)IJ|(4)KL\\
(1)ABCD|(2)EFGH|(3)KL|(4)IJ
$$
Hence $\frac{3}{4}$ of all allocations are repeated, so you need to multiply
$$
\binom{12}{4}\binom{8}{4}\binom{4}{2}\binom{2}{2} - \frac{3}{4} \times \binom{12}{4}\binom{8}{4}\binom{4}{2}\binom{2}{2} = 51975
$$
|
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"url": "https://math.stackexchange.com/questions/1884818",
"timestamp": "2023-03-29T00:00:00",
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|
Question on the inequality Question.
prove that if ${ a }_{ 1 },{ a }_{ 2 },...{ a }_{ n }>0$ then $$ \frac { { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } } $$
Proof
$$\\ \left( { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } \right) \left( \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } \right) =\underset { n\left( n-1 \right) /2\quad terms }{ \underbrace { \left( \frac { { a }_{ 1 } }{ { a }_{ 2 } } +\frac { { a }_{ 2 } }{ { a }_{ 1 } } \right) +...+\left( \frac { { a }_{ n-1 } }{ { a }_{ n } } +\frac { { a }_{ n } }{ { a }_{ n-1 } } \right) + } \quad n\ge } \\ \ge n+2\cdot \frac { n\left( n-1 \right) }{ 2 } ={ n }^{ 2 }$$
in this proof i didn't understand this step "$n\left( n-1 \right) /2\quad ?
terms$" I mean how the number of terms can be $n\left( n-1 \right) /2\quad $ .Can anybody explain it.
Thanks in advance!
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$$(a_1+a_2+...+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})$$ gives clearly $n^2$ terms but each $a_i$ gives $$1+\sum_{i\ne j}\frac{a_i}{a_j}$$ so you have
$$(a_1+a_2+...+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})=(1+1+...+1)+\sum_{i=1}^{n-1}(\frac{a_i}{a_{i+1}}+\frac{a_{i+1}}{a_i})$$
$$(a_1+a_2+...+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})=n+\sum_{i=1}^{n-1}(\frac{a_i}{a_{i+1}}+\frac{a_{i+1}}{a_i})$$ Since there are $n^2$ terms in total (you know this, said at the beginning) there are $n^2-n$ fractional terms and $\frac{n^2-n}{2}=\frac{n(n-1)}{2}$ terms each of two fractions inside the parenthesis. This is the explanation you wanted to have.
|
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|
Prove a geometric inequality, if 3 numbers are satisfying the condition I just got this problem, but I have no idea on how to prove that.
Prove that if $x,y,z\in\mathbb{R},\ x,y,z\ge 0$ and $2\cdot(x\cdot z+x\cdot y+y\cdot z)+3\cdot x\cdot y\cdot z = 9$, then $(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72$. This is a geometric inequality.
Can anyone help me, please? Any kind of help (solutions, hints etc) is really appreciated. Thank you!
NOTE: I REALLY DON'T KNOW WHAT TITLE SHOULD I WRITE FOR THIS POST, SO PLEASE LEAVE A COMMENT IF YOU HAVE AN IDEA FOR THE POST TITLE. THANK YOU.
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We need to prove that $x+y+z\geq\sqrt[4]{72}$, where $x$, $y$ and $z$ are non-negatives such that
$$\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=9$$
Let $x+y+z<\sqrt[4]{72}$, $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=\sqrt[4]{72}$.
Hence, $k<1$ and $9=\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=k^4\sum\limits_{cyc}(2a^2b^2+k^2a^2b^2c^2)<$
$<\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)$, which is a contradiction because we'll prove now that
$$\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)\leq9$$
Indeed, we need to prove that
$$9\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^6\geq2(a^2b^2+a^2c^2+b^2c^2)\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^2+3a^2b^2c^2$$ or
$$(a+b+c)^6\geq16(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+144\sqrt2a^2b^2c^2$$
Since by AM-GM
$(a+b+c)^4=(a^2+b^2+c^2+2(ab+ac+bc))^2\geq$
$\geq\left(2\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\right)^2=8(a^2+b^2+c^2)(ab+ac+bc)$,
it remains to prove that
$$(a+b+c)^2(a^2+b^2+c^2)(ab+ac+bc)\geq2(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2((a^2+b^2+c^2)(ab+ac+bc)-2(a^2b^2+a^2c^2+b^2c^2))\geq18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2\sum\limits_{cyc}(a^3b+a^3c-2a^2b^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2\sum\limits_{cyc}(ab(a-b)^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$
Thus, it remains to prove that $(a+b+c)^3\geq18\sqrt2abc$, which is AM-GM:
$(a+b+c)^3\geq27abc\geq18\sqrt2abc$. Done!
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"timestamp": "2023-03-29T00:00:00",
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|
A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:
$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$
What he thought that since $x^2 > x$ and $x^2 > 3x$ when $x \to \infty$ so the first square root must be $x$ and same for the second. Hence, the limit must be $0$.
It is obviously problematic.
And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.
After perfect-squaring, $$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$ I assert that since there is a perfect square and a square root. As $x \to \infty$, the constant does not matter. So
$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1 $$
But when I conquer this limit: this post.
Here is my argument:
$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$ when $x \to \infty$
$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x} = \frac{3}{2}$$
Very concise get this answer but it gets downvoted.
I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:
$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$
My solution is:
$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$
So the limit becomes: $$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty} (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3} $$
This result gets verified by wolframalpha.
To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.
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Let our problem be of the form $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)}$, where $p$ and $q$ are $n$th degree monic(coefficient of $x^n$ is $1$) polynomials. First of all, without loss of generality, we will rewrite $p$ and $q$ in the forms:
$$
p = (x + a)^n + \sum_{i=0}^{n-1} a_ix^i, q = (x + b)^n + \sum_{i=0}^{n-1} b_ix^i
$$
where $a_i$ and $b_i$ are real constants which may be zero.(This we did in the cubic and quadratic case).
Now, note that $x+a \leq \sqrt[n]{(x + a)^n + \sum_{i=0}^{n-1} a_ix^i} \leq \sqrt[n]{(x + a)^n} + \sqrt[n]{\sum_{i=0}^{n-1} a_ix^i}$.Note that $$
\sqrt[n]{\sum_{i=0}^{n-1} a_ix^i} \leq \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}
$$
However, the term on the left goes to $0$ as $x \to \infty$, because $i < n, $ so $x^i \to 0$. A similar logic follows for $q$. To complete the argument,
However, this is better expressed by: $x+a - ((x+b) + \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) \leq \sqrt[n]{p(x)} - \sqrt[n]{q(x)} \leq (x+a +\sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) - (x+b))$.
On taking limits on both sides, we get $a-b$ on both sides, and using squeeze theorem, we get the result: $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)} = a-b$.
|
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|
Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
|
Integrating by parts we have $$I_{n}=\int_{0}^{1}x^{n}\arctan\left(x\right)dx=\frac{\pi}{4\left(n+1\right)}-\frac{1}{\left(n+1\right)}\int_{0}^{1}\frac{x^{n+1}}{1+x^{2}}dx
$$ $$ \stackrel{x^{2}=u}{=}\frac{\pi}{4\left(n+1\right)}-\frac{1}{2\left(n+1\right)}\int_{0}^{1}\frac{u^{n/2}}{1+u}du
$$ and now since $$ \frac{u^{n/2}}{2}\leq\frac{u^{n/2}}{1+u}\leq\frac{u^{n/2-1}}{2}
$$ we get $$\frac{\pi}{4\left(n+1\right)}-\frac{1}{2\left(n+1\right)n}\leq I_{n}\leq\frac{\pi}{4\left(n+1\right)}-\frac{1}{2\left(n+1\right)\left(n+2\right)}
$$ and so $$\lim_{n\rightarrow\infty}n\left(\left(n+1\right)I_{n}-\frac{\pi}{4}\right)=\color{red}{-\frac{1}{2}}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can we prove this inequality? Let $a$ , $b$ and $c$ be positive real numbers and $a+b+c=1$ How can we show this inequality?
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
Thanks.
|
We have that
$$1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$$
so it suffices to show that
$$ab+bc+ca\geq \sqrt{3abc}.$$
Now $X^2+Y^2+Z^2\geq XY+YZ+XZ$ implies
$$ab+bc+ca=(\sqrt{ab})^2+(\sqrt{bc})^2+(\sqrt{ca})^2
\\\geq b\sqrt{ac}+c\sqrt{ab}+a\sqrt{bc}
=(\sqrt{a}+\sqrt{b}+\sqrt{c})\sqrt{abc}.$$
Hence, we still have to prove that
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{3}$$
which holds because $\sqrt{x}$ is concave and
$$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\geq \sqrt{\frac{a+b+c}{3}}=\frac{1}{\sqrt{3}}.$$
|
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|
An integral involving a the dilogarithmic function Consider a following definite integral:
\begin{equation}
\phi_a(x):=\int\limits_0^x \frac{Li_2(\xi)}{\xi+a} d \xi
\end{equation}
By using the integral representation of the dilogarithmic function and swapping the integration order we have shown that our integral satisfies the following functional equation:
\begin{equation}
\phi_a(x)=
\text{Li}_3\left(\frac{a+x}{a+1}\right)-\text{Li}_3\left(\frac{a}{a+1}\right)
-2 \text{Li}_3\left(\frac{1}{a+1}\right)+2 \text{Li}_2\left(\frac{1}{a+1}\right) \log
\left(\frac{1}{a+1}\right)+
\text{Li}_2(-a) \log \left(\frac{a+x}{a}\right)+\text{Li}_2\left(\frac{a}{a+1}\right) \log \left(\frac{1}{a+1}\right)+\log \left(\frac{a}{a+1}\right) \log ^2\left(\frac{1}{a+1}\right)+2 \zeta (3)
-\phi_{-\frac{a+x}{a}}(\frac{x+a}{1+a})
\end{equation}
Now, the question is how do we solve this equation?
|
Here we are computing the antiderivative in closed form rather than solving the functional equation above. In our calculations we will be transforming the integrand by adding and subtracting terms to it in order to -- at every step-- extract terms which are full derivatives. In here we are using the following identity:
\begin{equation}
\left[Li_3(f(x)) - \log(f(x)) Li_2(f(x))\right]^{'} =
\log(1-f(x)) \log(f(x)) \cdot \frac{f^{'}(x)}{f(x)}
\end{equation}
valid for any function $f(x)$ that is smooth enough.
We define ${\mathcal I}_a(x) := Li_2(x)/(x+a)$ and
\begin{eqnarray}
{\mathcal J}_0(x) &:=& \log(x+a) \\
{\mathcal J}_1(x) &:=& Li_3(1-x) + \log(\frac{x+a}{a-a x}) Li_2(1-x)\\
{\mathcal J}_2(x) &:=& \log(\frac{x+a}{a})^2 - \log(\frac{a-a x}{x+a})^2\\
{\mathcal J}_3(x) &:=& Li_3\left(\frac{x+a}{a-a x}\right)-\log(\frac{x+a}{a-a x}) Li_2\left(\frac{x+a}{a-a x}\right)\\
{\mathcal J}_4(x) &:=& Li_3\left(\frac{x+a}{-1+ x}\right)-\log(\frac{x+a}{a-a x}) Li_2\left(\frac{x+a}{-1+ x}\right)\\
{\mathcal J}_5(x) &:=& Li_3\left(\frac{x+a}{a}\right) - \log(\frac{x+a}{a})Li_2\left(\frac{x+a}{a}\right)
\end{eqnarray}
Now we start from Euler's reflection formula for the dilogarithmic function:
\begin{eqnarray}
{\mathcal I}_a(x) &=& -\frac{\log(1-x) \log(x)}{x+a}-\frac{Li_2(1-x)}{x+a} + \frac{\pi^2}{6} \frac{1}{x+a} \\
&=&
-\frac{\log(1-x) \log(x)}{x+a} + \frac{\log(x) \log(\frac{x+a}{a-a x})}{1-x} - {\mathcal J}_1^{'}(x)+ \frac{\pi^2}{6} {\mathcal J}_0^{'}(x)
\end{eqnarray}
In here we have just added and subtracted the second term on the right hand side and then we have recognized that the $Li_2$ term along with the subtracted term form a full derivative. Now we have:
\begin{eqnarray}
&&{\mathcal I}_a(x)=\\
&&\log(-\frac{1}{a})
\left(-\frac{(1+a)\log(\frac{a- a x}{a+x})}{(x+a)(1-x)} -\frac{\log(\frac{x+a}{a})}{x+a}
+\frac{\log(\frac{x+a}{a})}{x+a}
+\frac{(1+a)\log(\frac{a- a x}{a+x})}{(x+a)(1-x)}\right)\\
&&
-\frac{\log(1-x) \log(x)}{x+a} + \frac{\log(x) \log(\frac{x+a}{a-a x})}{1-x} - {\mathcal J}_1^{'}(x)+ \frac{\pi^2}{6} {\mathcal J}_0^{'}(x)
\end{eqnarray}
Note that the last two terms in the parentheses on the right hand side form a full derivative. Therefore we have:
\begin{eqnarray}
&&{\mathcal I}_a(x)=\\
&&\log(-\frac{1}{a})
\left(-\frac{(1+a)\log(\frac{a- a x}{a+x})}{(x+a)(1-x)} -\frac{\log(\frac{x+a}{a})}{x+a}
\right)+
\frac{1}{2}\log(-\frac{1}{a})
{\mathcal J}^{'}_2(x)\\
&&
-\frac{\log(1-x) \log(x)}{x+a} + \frac{\log(x) \log(\frac{x+a}{a-a x})}{1-x} - {\mathcal J}_1^{'}(x)+ \frac{\pi^2}{6} {\mathcal J}_0^{'}(x)
\end{eqnarray}
At this stage we intuitively know what to do. To each of the logarithms in the parentheses on the right hand side we need to attach "a reflected log" meaning that we end up with a product of logs of arguments that sum up to one. If we achieve that we end up with another full derivative (see the first line from the top). We skip the tedious, but straightforward, intermediate calculations and we state the result:
\begin{eqnarray}
&&{\mathcal I}_a(x)=\\
&&
\frac{\log \left(\frac{a+x}{a-a x}\right) \log \left(1-\
\frac{a+x}{a-a x}\right)}{\frac{x+a}{a-a x}} \frac{1+a}{a(x-1)^2}
-
\frac{ \log \left(\frac{a+1}{1-x}\right) \log \
\left(\frac{a+x}{a-a x}\right)}{\frac{x+a}{a-a x}} \frac{1+a}{a(x-1)^2}
-
\frac{\log \left(-\frac{x}{a}\right) \log \left(\frac{x}{a}+1\right)}{\frac{a+x}{a}} \frac{1}{a}
- {\mathcal J}_1^{'}(x)+ \frac{\pi^2}{6} {\mathcal J}_0^{'}(x)+\frac{1}{2}\log(-\frac{1}{a})
{\mathcal J}^{'}_2(x)
\end{eqnarray}
Note that the only thing we did in here was to rewrite the terms which are not yet full derivatives . Now clearly the first and the third term on the right hand side are clearly full derivatives . The second term needs to be manipulated a bit but is conceivable that it can be transformed into a full derivative as well. Therefore we have:
\begin{eqnarray}
&&{\mathcal I}_a(x)=\\
&&{\mathcal J}_3^{'}(x)- {\mathcal J}_4^{'}(x)-{\mathcal J}_5^{'}(x)- {\mathcal J}_1^{'}(x)+ \frac{\pi^2}{6} {\mathcal J}_0^{'}(x)+\frac{1}{2}\log(-\frac{1}{a})
{\mathcal J}^{'}_2(x)
\end{eqnarray}
Now integrating the above we get the final result:
\begin{eqnarray}
&&\phi_a(x)=\\
&&{\mathcal J}_3(x)- {\mathcal J}_4(x)-{\mathcal J}_5(x)- {\mathcal J}_1(x)+ \frac{\pi^2}{6} {\mathcal J}_0(x)+\frac{1}{2}\log(-\frac{1}{a})
{\mathcal J}_2(x) +\\
&& Li_3(-a) + Li_3(1) - \log(a) \frac{\pi^2}{6}
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1893279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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|
Given a matrix $B$, what is $det(B^4)$? My task is this:
Compute det$(B^4)$, where
$$B =\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix}$$
My work so far:
It can be shown that
$B =
\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\end{pmatrix}\implies B$ is invertable.
Applying theorem for determinants to triangular matrices we get det$(B) = 1*1*(-4)=-4$.
Appling the multiplicative property theorem, we should get that det$(B^4)=$det$(B)$det$(B)$det$(B)$det$(B)=[$det$(B)]^4=(-4)^4= 256.$
Now I need help from the community to verify if the reasoning is correct, because I've already had a fellow student telling me that the answer is 16, and indeed the textbook I'm using states the same, which makes me abit confused. So please point out my errors or verify that each step holds true.
Thanks in advance.
|
The issue here is that you are computing the determinant wrong. $\det(B) = -2$, therefore $\det(B^4) = 16$. To check it, you can use Sarrus rule:
$$\det(B) = b_{11}b_{22}b_{33}+b_{12}b_{23}b_{31}+b_{13}b_{21}b_{32}-b_{11}b_{23}b_{32}-b_{12}b_{21}b_{33}-b_{13}b_{22}b_{31}$$
$$= 1+0+2-4-0-1 = 3-5 = -2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1895160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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|
$5$ divides one of $x,y,z$ in $x^2+y^2=z^2$? 1) If $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, show that $3$$∣$$xy$.
2) Now again if $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, can a similar result to the one in (part 1) be said mod $5$? Not that $5$$∣$$xy$, but that $5$ divides one of $x,y,z$?
I believe I have 1) down:
We note that $3$ is a prime, so $3∣xy⟺3∣x$ or $3∣y$. Suppose to the contrary that $3∤x$, and $3∤y$, then:$x=1$, $2$ mod $3$ and $y=1$, $2$ mod $3$. Then:
$x^2+y^2=2$ mod $3$, and $z^2=0$, $1$ mod $3$. Contradiction, proving the claim.
But I am having trouble with 2). Can anyone show a solution for 2)?
|
Squares are $0,1$ or $-1\bmod 5$
If both $x$ and $y$ are not multiples of $5$ then $x^2\equiv \pm 1$ and $y^2\equiv \pm 1\bmod 5$
This implies $x^2+y^2\equiv -2,0$ or $2\bmod 5$.
Of course, since $x^2+y^2$ is a square it must be $0\bmod 5$ given the previous options.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1896761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter example would work.
Note if $\mathfrak a(n) =\binom{2^n}{2}+1$ then $7$ divides $\mathfrak a(n)$ for $n=2,5,8,11,14,17,\ldots$ and from this I am guessing that $n=3k+2$
|
Something slightly more general: we have ${k\choose 2} \equiv -1 \pmod{7}$ if and only if $k\equiv 4\pmod{7}$.
To see this, write ${k\choose 2} = \frac{k(k-1)}{2}$, so the equation is equivalent to $k^2 - k \equiv -2\pmod{7}$. But $k^2 - k + 2 \equiv (k-4)^2 \pmod{7}$, so this is satisfied exactly when $k\equiv 4\pmod{7}$.
Finally, it is easy to see that $2^n$ cycles through the values $1,2,4\pmod{7}$, and the result follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1897494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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|
Prove that if $a+b+c=1$ then $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\le\frac{9\sqrt{2}}{2}$ Let $a,b,c>0$,and such $a+b+c=1$,prove or disprove
$$\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\le\dfrac{9\sqrt{2}}{2}\tag{1}$$
My try:since
$$\sqrt{a^2+b^2}\ge\dfrac{\sqrt{2}}{2}(a+b)$$
it suffices to prove that
$$\sum_{cyc}\dfrac{1}{a+b}\le\dfrac{9}{2}$$
But since Cauchy-Schwarz inequality we have
$$2\sum_{cyc}(a+b)\sum_{cyc}\dfrac{1}{a+b}\ge 9$$
or
$$\sum_{cyc}\dfrac{1}{a+b}\ge \dfrac{9}{2}$$
so this try can't works.so How to prove $(1)$
|
It is false. Try $a=\frac{9}{10},\, b=c=\frac{1}{20}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1897844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.
|
So thanks to @Henning Makholm, I realized that it does hold for every integer $n >2$.
Case 1. Suppose $n$ is an odd integer, say $n=2k+1$. ($k\geq1$)
\begin{align*}
n&=2k+1\\
c&=2k^2+2k+1\\
b&=2k^2+2k\\
c^2&=b^2+n^2
\end{align*}
Case 2. Suppose $n$ is two times an odd integer, say $n=4k+2$.($k\geq1$)
\begin{align*}
n&=4k+2\\
c&=4k^2+4k+2\\
b&=4k^2+4k\\
c^2&=b^2+n^2
\end{align*}
Case 3. If $n$ is an arbitrary even number then you can write is as $n=2^\alpha t$, where $t$ is an odd number. If $\alpha$ is odd you can use Case 2 to generate $b,c$, otherwise you can use Case 1.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1898127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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.