Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$ Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$
$y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,\frac{1}{4})$.I dont know how to solve further.
| $p(x,y)=(x,y)+(1,3)=(x+1,y+3), \quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix
$\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}$ which is $\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}\pmatrix{x \\ y \\ 1}=\pmatrix{x+1 \\ y+3 \\ 1}$
so for $y=(x-1)^2$ it is
$\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}\pmatrix{x \\ (x-1)^2 \\ 1}=\pmatrix{x+1 \\ (x-1)^2+3 \\ 1}$
so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Summation of series and Taylor series are giving different results $A = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots$
$B = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots$
$A + B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$
Using Taylor expansion for $\ln(1+x)$
$\ln(2) = 1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$
Where did I go wrong ?
| As others have already pointed out, the series $\sum_{n=1}^\infty \frac1n$ and $\sum_{n=1}^\infty \frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.
Another way to analyze this is to look at the partial sums $A_N=\sum_{n=1}^{N} \frac{1}{2n}$ and $B_N=\sum_{n=1}^N \frac{1}{2n-1}$.
While $A_N+B_N =\sum_{n=1}^{2N} \frac1n$, we have $A_N-B_N=-\sum_{n=N+1}^{2N}\frac1n \ne 0$ (In fact, we have $\lim_{N\to\infty}(A_N-B_N)=\log(1/2)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $p>1$ that ${\int\limits^p_1}\frac{1}{x}\,\mathrm{d}x={\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x$ Find $p>1$ that $${\displaystyle\int\limits^p_1}\dfrac{1}{x}\,\mathrm{d}x={\displaystyle\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x$$
\begin{align*}
&{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x=\ln\left(\mid x \mid \right) && \vert \ \text{general integral}
\end{align*}
$F_1(x)=\ln\left({\mid x \mid} \right)+C$
\begin{align*}
&{\displaystyle\int}1\cdot\ln\left(x\right )\,\mathrm{d}x && \vert \ 2. \text{ with } f'=1, g=\ln(x)\\
&=x\ln\left(x\right)-{\displaystyle\int}1\,\mathrm{d}x\\
&=x\ln\left(x\right)-x
\end{align*}
$F_2(x)=x\ln\left(x\right)-x+C$
\begin{align*}
\left[\ln\left(\mid{x}\mid\right)+C\right]^p_1&=\left[\ln({\mid p \mid})+C\right]-\left[\ln({\mid 1 \mid})+C\right]\\
&=\ln\left(p\right)
\end{align*}
\begin{align*}
&\left[x\ln\left(x \right)-x+C\right]^p_1=\left[p\ln\left(p \right)-p+C\right]-\left[1\ln\left(1\right)-1+C\right]
\end{align*}
It was already mentioned, that I had a typo. I corrected everything and contributed my own solution.
| Find $p>1$ that
$$
{\displaystyle\int\limits^p_1}\dfrac{1}{x}\,\mathrm{d}x={\displaystyle\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x
$$
\begin{align*}
&{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x=\ln\left(\mid x \mid \right) && \vert \ \text{general integr}
\end{align*}
$F_1(x)=\ln\left({\mid x \mid} \right)+C$
\begin{align*}
&{\displaystyle\int}1\cdot\ln\left(x\right )\,\mathrm{d}x && \vert \ 2. \text{ with } f'=1, g=\ln(x)\\
&=x\ln\left(x\right)-{\displaystyle\int}1\,\mathrm{d}x\\
&=x\ln\left(x\right)-x
\end{align*}
$F_2(x)=x\ln\left(x\right)-x+C$
\begin{align*}
\left[\ln\left(\mid{x}\mid\right)+C\right]^p_1&=\left[\ln({\mid p \mid})+C\right]-\left[\ln({\mid 0 \mid})+C\right]\\
&=\ln\left(p\right)
\end{align*}
\begin{align*}
\left[x\ln\left(x \right)-x+C\right]^p_1&=\left[p\ln\left(p \right)-p+C\right]-\left[1\ln\left(1 \right)-1+C\right]\\
&=\left[p\ln\left(p \right)-p+C\right]-[-1+C]\\
&=p\ln\left(p \right)-p+1
\end{align*}
Solve $\ln\left(p\right)=p\ln\left(p \right)-p+1$ for $p$:
\begin{align*}
&\ln\left(p\right)=p\ln\left(p \right)-p+1&&\vert \ -(1-p+p\ln(p)\\
&\iff-1+p+\ln(p)-p\ln(p)=0\\
&\iff-(p-1)\cdot (\ln(p)-1)=0&& \vert \ \cdot (-1)\\
&\iff (p-1)\cdot (\ln(p)-1)=0 && \vert \ p>1\\
&\iff (\ln(p)-1)=0\\
&\iff (\ln(p))=1&& \vert \ \exp{()}\\
&\iff \underline{\underline{p=e}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$ The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.
Prove $$\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$
I had small tries for it, such as writting:
$$I=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{ x\to \tan \frac{x}{2}}=-\frac12 {\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx}$$
And with Feynman's trick we obtain:
$$J(t)=\int_0^\frac{\pi}{2} \frac{x\ln(1-t\sin x)}{\sin x}dx\Rightarrow J'(t)=\int_0^\frac{\pi}{2} \frac{x}{1-t\sin x}dx$$ But I don't see a way to obtain a closed from for the above integral.
Also from here we have the following relation:
$$\int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$
Thus we can rewrite the integral as:
$$I=\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx -2\int_0^1 \frac{\arctan x \ln(1-x)}{x}dx$$
Another option might be to rewrite:
$$\ln\left(\frac{1+x^2}{(1-x)^2}\right)= \ln\left(\frac{1+x}{1-x}\right)+\ln\left(\frac{1+x^2}{1-x^2}\right)$$
$$\Rightarrow I= \int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x}{1-x}\right)dx+\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{1-x^2}\right)dx$$
And now to use the power expansion of the log functions to obtain:
$$\small I=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \frac{\arctan x}{x} \, \left(x^{2n+1}+x^{4n+2}\right)dx=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1\int_0^1 \frac{\left(x^{2n+1}+x^{4n+2}\right)}{1+y^2x^2}dydx$$
This seems like an awesome integral and I would like to learn more so I am searching for more approaches.
Would any of you who also already solved it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?
Edit:
In the meantime I found a nice solution by Roberto Tauraso here and another impressive approach due to Yaghoub Sharifi here.
| We have $\log\left(\frac{1+x^2}{(1-x)^2}\right)=\log(1-x^4)-\log(1-x^2)-2\log(1-x)$, hence by integration by parts
$$ I_1 = \frac{3\pi^3}{32}-2\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx\qquad \chi(n)=\left\{\begin{array}{rcl}1&\text{if}&n\equiv 1\pmod{2}\\ 2 & \text{if} & n\equiv 2\pmod{4}\\ 0 &\text{if}&n\equiv 0\pmod{4}\end{array}\right. $$
where
$$\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx=\iint_{(0,1)^2}\frac{-\log(y)}{y(1+x^2)}\sum_{n\geq 1}\chi(n)(xy)^n\,dx\,dy=\iint_{(0,1)^2}\frac{-\log(y)}{y(1+x^2)}\cdot\frac{(1+xy)^2 xy}{1-(xy)^4}\,dx\,dy$$
equals
$$ \iint_{(0,1)^2}\frac{-\log(y)x(1+xy)}{(1+x^2)(1-x y)(1+x^2 y^2)}\,dx\,dy. $$
By performing a partial fraction decomposition this integral is reduced to four integrals in the $y$-variable, with two of them (namely $\int_{0}^{1}\frac{y\log y}{1-y^4}\,dy = -\frac{\pi^3}{32}$ and $\int_{0}^{1}\frac{\log(2)}{1+y^2}\,dy = \frac{\pi}{4}\log(2)$) being elementary and the remaining ones being
$$ J_1 = \int_{0}^{1}\frac{\arctan(y)\log(y)}{1-y^2}\,dy,\qquad J_2=\int_{0}^{1}\frac{\log(y)\log(1-y)}{1+y^2}\,dy. $$
$J_2$ can be tackled by performing the substitution $y=\tan\theta$ and exploiting the Fourier series of $\log \sin$ and $\log\cos$. By performing the substitution $y\mapsto\frac{1+y}{1-y}$ $J_1$ is reduced to
$$ \int_{0}^{1}\frac{\arctan(x)\operatorname{arctanh}(x)}{x}\,dx $$
then, by computing $\int_{0}^{1}\frac{x^{2k}}{2k+1}\operatorname{arctanh}(x)\,dx$, to the series (also appearing here)
$$ \sum_{k\geq 0}\frac{(-1)^k H_k}{(2k+1)^2}=\int_{0}^{1}\frac{\log(1+z^2)\log(z)}{1+z^2}\,dx. $$
Via $z\to\tan\theta$ we have that both $J_1$ and $J_2$ are reduced to the integrals $\int_{0}^{\pi/4}\log^2(\sin\theta)\,d\theta$ and $\int_{0}^{\pi/4}\log(\sin\theta)\log(\cos\theta)\,d\theta$, which are well-known and related to Euler sums with weight $3$. Luckily the contributions related to $\pi\log^2(2),\pi^2\log(2),K\log(2)$ and $\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)$ cancel out and only leave a rational multiple of $\pi^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 4,
"answer_id": 3
} |
Denote $f(x)=\int_x^{x+1}\cos t^2 {\rm d}t.$Prove $\lim\limits_{x \to +\infty}f(x)=0.$ Problem
Denote $$f(x)=\int_x^{x+1}\cos t^2 {\rm d}t.$$Prove $\lim\limits_{x \to +\infty}f(x)=0.$
Proof
Assume $x>0$. Making a substitution $t=\sqrt{u}$，we have ${\rm d}t=\dfrac{1}{2\sqrt{u}}{\rm d}u.$ Therefore,
\begin{align*}
f(x)&=\int_x^{x+1}\cos t^2 {\rm d}t\\
&=\int_{x^2}^{(x+1)^2} \frac{\cos u}{2\sqrt{u}}{\rm d}u\\
&=\frac{1}{2\sqrt{\xi}}\int_{x^2}^{(x+1)^2}\cos u{\rm d}u\\
&=\frac{\sin(x+1)^2-\sin x^2}{2\sqrt{\xi}},
\end{align*}
where $x^2 \leq \xi\leq (x+1)^2$. Further,
$$0\leq |f(x)|\leq \frac{|\sin(x+1)^2|+|\sin x^2|}{2\sqrt{\xi}}\leq \frac{1}{\sqrt{\xi}}\leq \frac{1}{x}\to 0(x \to +\infty),$$
which implies $f(x)\to 0(x \to +\infty)$, according to the squeeze theorem.
| Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $\cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.
Another Proof
Likewise, assume $x>0$. Making a substitution $t=\sqrt{u}$，we have ${\rm d}t=\dfrac{1}{2\sqrt{u}}{\rm d}u.$ Therefore,
\begin{align*}
f(x)&=\int_x^{x+1}\cos t^2 {\rm d}t\\
&=\int_{x^2}^{(x+1)^2} \frac{\cos u}{2\sqrt{u}}{\rm d}u\\
&=\int_{x^2}^{(x+1)^2}\frac{1}{2\sqrt{u}}{\rm d}(\sin u)\\
&=\frac{\sin u}{2\sqrt{u}}\bigg|_{x^2}^{(x+1)^2}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}\sin u{\rm d}u\\
&=\frac{\sin(x+1)^2}{2(x+1)}-\frac{\sin x^2}{2x}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}\sin u{\rm d}u.\\
\end{align*}
Thus
\begin{align*}
|f(x)|&\leq \frac{|\sin(x+1)^2|}{2(x+1)}+\frac{|\sin x^2|}{2x}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}|\sin u|{\rm d}u\\
&\leq \frac{1}{2(x+1)}+\frac{1}{2x}+\frac{1}{4}\int_{x^2}^{(x+1)^2}u^{-\frac{3}{2}}{\rm d}u\\
&=\frac{1}{2(x+1)}+\frac{1}{2x}+\frac{1}{4}\cdot \left[-\frac{2}{\sqrt{u}}\right]_{x^2}^{(x+1)^2}\\
&=\frac{1}{x}\to 0(x \to +\infty),
\end{align*}
which implies $|f(x)|\to 0$ according to the squeeze theorem. It follows that $f(x) \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
power series for $\frac1{x^2+x+1}$ So I am trying to find the sequence $(a_n)_{n\geq0}$ such that
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$a_0(x^2+x+1)+\sum_{n\geq1}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$x=0$:
$$a_0=1$$
So
$$x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$\frac{x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)}x=\frac1x$$
$$x+1+a_1(x^2+x+1)+\sum_{n\geq2}a_n(x^{n+1}+x^{n}+x^{n-1})=0$$
$x=0$:
$$a_1=-1$$
One more time:
$$x^2+x+1-x^3-x^2-x+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$-x^3+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=0$$
Divide both sides by $x^2$:
$$-x+a_2(x^2+x+1)+\sum_{n\geq3}a_n(x^{n}+x^{n-1}+x^{n-2})=0$$
$x=0$:
$$a_2=0$$
How do I find $a_n$? Am I doing things right so far? Thanks.
| Another way of doing this is to let the roots of $1+x+x^2=0$ be $a$ and $b$ so that $1+x+x^2=(1-ax)(1-bx)$ with $ab=1$ and $a+b=-1$. Then use partial fractions $$\frac 1{1+x+x^2}=\frac A{1-ax}+\frac B{1-bx}$$ where clearing fractions gives $$1=A(1-bx)+B(1-ax)$$
Set $a=\frac 1b$ to obtain $1=B(1-\frac ab)$ so that $B=\frac b{b-a}$ and similarly $A=\frac a{a-b}$ so that $$\frac 1{1+x+x^2}=\sum_{n=0}^\infty\frac a{a-b}a^nx^n-\sum_{n=0}^\infty \frac b{a-b}b^bx^n=\sum_{n=0}^\infty\frac {a^{n+1}-b^{n+1}}{a-b}x^n$$
Now observe $a^2=-1-a=b$ and $a^3=ab=1$ and similarly $b^2=-1-b=a$ and $b^3=1$. For $n=3m$ the coefficient is $\frac {a-b}{a-b}=1$, for $n=3m+1$ the coefficient is $\frac {b-a}{a-b}=-1$ and for $n=3m+2$ the coefficient is $0$.
We can also identify $a=\omega$ and $b=\omega^2$ as being primitive cube roots of $1$. Note how the coefficients are periodic with period $3$. These facts are related, as another answer shows.
Note: I have tried to express this in such a way as to show how partial fractions can be used more generally - even where a convenient short cut is not available - it doesn't matter that you get complex roots, the answer ends up with real coefficients. The only slight trick is that we would normally factorise $$x^2+px+q=(x-c)(x-d)$$ but, setting $a=\frac 1c, b= \frac 1d$ we need the form $$x^2+px+q=cd(1- ax)(1-bx)=q(1-ax)(1-bx)$$In the case above we had $q=1$, but in general the factor must not be forgotten.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factoring $3^6-3^3 +1$ $3^6-3^3 +1$ factors?, 37 and 19, but how to do it using factoring, $3^3(3^3-1)+1$, can't somehow put the 1 inside
| Viewing this as a quadratic in $3^3$ leads you to try to factorise $x^2-x+1$ or $x^6-x^3+1$, neither of which splits into anything useful. However, you can pull out a factor of $9$ from $3^6$ and a factor of $3$ from $3^2$ to obtain
$$3^6-3^3+1 = 9 \cdot 3^4 - 3 \cdot 3^2 + 1$$
And the quartic $9x^4-3x^2+1$ does factorise into two quadratic factors:
$$9x^4-3x^2+1 = (3x^2+3x+1)(3x^2-3x+1)$$
Now set $x=3$ and voilà!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
For what values of n (where n is a natural number) is this statement true: $3^n - n - 1 ≡ 0\pmod5$ I'm clueless as where to start with this, $3^n$ seems to be of period 4, where $ - n - 1$ is of period 5.
| $\, n = 4q\!+\!r,\ 0\le r\le 3.\,$ $\bmod 5\!:\ \color{#0a0}1 \equiv 3^{\large n}\!-n\equiv 3^{\large r} \overbrace{(3^{\large 4})^{\large q}}^{\large 3^{\Large 4} \equiv\ 1}\!-4q\!-\!r\equiv \color{#0a0}{3^{\large r}\!+q-r}$
$\,r=0\,\Rightarrow\ \color{#0a0}{1\equiv 3^{\large 0}\!+q-0}\ $ so $\,\color{#c00}{q\equiv 0\pmod{\!5}}\,$ so $\,n = 4\color{#c00}q\!+\!r=4(\color{#c00}{0\!+\!5k})\!+\!0 = 20k$
$\,r = 1,2,3\,$ are done the same way. You'll find the $\rm\color{#c00}{solutions}$ enumerated below. $\!\!\begin{array}{|r|r|}
\hline
n & 0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\
\hline
n\!+\!1\bmod 5 &\color{#c00}1&2&3&4&0&1&2&3&4&0&1&\color{#c00}2&3&4&0&1&2&\color{#c00}3&\color{#c00}4&0\\
\hline
3^n\bmod 5 &\color{#c00}1&3&4&2&1&3&4&2&1&3&4&\color{#c00}2&1&3&4&2&1&\color{#c00}3&\color{#c00} 4&2\\
\hline
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Find other angles
Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Then find the other angles
My Attempt
Let $a=\sqrt{3}+1$, $b=\sqrt{3}-1$ and $C=60$
$$
c^2=a^2+b^2-2.a.b\cos C\\=(\sqrt{3}+1)^2
+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}-1).\frac{1}{2}
=8-2=6\\
\implies c=\sqrt{6}=\sqrt{2}\sqrt{3}\\
\frac{a}{\sin A}=\frac{c}{\sin C}\implies\sin A=\frac{\sqrt{3}+1}{\sqrt{2}\sqrt{3}}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}\\
A=75^\circ\quad\&\quad B=45^\circ
$$
But the solution given in my reference is $105^\circ$ and $15^\circ$, what is going wrong with my attempt ?
| Playing around with the angle sum and the Law of Sines ($|\angle A|>|\angle B|$):
$|\angle A|+|\angle B|=120°$
$(\sin A)/(\sin B)=(\sqrt{3}+1)/(\sqrt{3}-1)=2+\sqrt{3}$
Put $|\angle A|=120°-|\angle B|$ and apply the formula for the wine of a difference:
$((\sqrt{3}/2)(\cos B)-(1/2)(\sin B))/(\sin B)=2+\sqrt{3}$
$(\cos B)/(\sin B)=\cot B = 2+\sqrt{3}$
$\tan B=2-\sqrt{3}$
Then
$\tan A = \tan (120°-B)=\dfrac{-\sqrt{3}-(2-\sqrt{3})}{1+(-\sqrt{3})((2-\sqrt{3}))}=-(2+\sqrt{3})$
Then $\tan A \tan B=-1$ so the angle measures must differ by a right angle. So
$|\angle A|+|\angle B|=120°$
$|\angle A|-|\angle B|=90°$
$|\angle A|=105°, |\angle B|=15°$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
A Series for $\pi$
Question: Can we show that $$\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}=\frac{\pi}{8} $$ ?
According to wolfram alpha this result is true.
Just amateur curiosity, not sure of a starting point to show if this true.
| Note that after cancellation we have
$$\frac{(2n-3)!!}{(2n+3)!!}=\frac{1}{(2n-1)(2n+1)(2n+3)}$$
Taking partial fractions gives
$$
\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{8}\left(-\frac{1}{2n-1}+\frac{2}{2n+1}-\frac{1}{2n+3}\right)
$$
Now, the standard arctangent series for $\pi$ gives $\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}=\frac{\pi}{4}$. So
$$\sum_{n=0}^\infty (-1)^n \frac{1}{2n-1}=-\sum_{n=-1}^\infty (-1)^n \frac{1}{2n+1}=-1-\frac{\pi}{4}$$
$$
\sum_{n=0}^\infty (-1)^n \frac{1}{2n-3}=-\sum_{n=1}^\infty (-1)^n \frac{1}{2n+1}=1-\frac{\pi}{4}
$$
which means that when we combine all three infinite sums we get
$$
\frac{1}{8}\left[-\left(-1-\frac{\pi}{4}\right)+2\left(\frac{\pi}{4}\right)-\left(1-\frac{\pi}{4}\right)\right]=\frac{\pi}{8}
$$
as desired.
(This solution makes it obvious that the sum should be of the form $a+b\pi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $\pi$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right)\rightarrow \sin0 $. However I have also $x$ before $\sin x$ and I don't know how to calculate it.
| Letting $h=\frac1x$:
$$\begin{array}{cl}
&\displaystyle \lim_{x \to \infty} x \sin \left( \sqrt{x^2+3} - \sqrt{x^2+2} \right) \\
=&\displaystyle \lim_{x \to \infty} x \sin \left( \frac 1 {\sqrt{x^2+3} + \sqrt{x^2+2} } \right) \\
=&\displaystyle \lim_{h \to 0^+} \frac1h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\
=&\displaystyle \lim_{h \to 0^+} \frac 1 {\sqrt{1+3h^2} + \sqrt{1+2h^2}} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\
=&\displaystyle \frac12 \times \lim_{h \to 0^+} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\
=&\displaystyle \frac12 \times 1 \\
=&\dfrac12
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my question is then why does $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)} = -\frac{1}{6}$$
As I though that it would be $\frac{1}{2}$.
| Actually your finite sum is equal to $$\sum_{r=2}^n\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}-\frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! \approx \sqrt{2\pi\cdot 15}
\cdot \left(\frac{15}{e}\right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
| Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b \equiv b \pmod {10}$ and
$1*2*3*4*6*7*9*2*11*12*13*14*3\equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3\equiv$
$-2^9*3^4 \equiv -512*81\equiv -2 \equiv 8\pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0\le a \le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is? Question based on Permutations and Combinations
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?
A) $2^{20} + \binom{20}{10}$
B) $2^{20} - \binom{20}{9}$
C) $2^{19} + \binom{20}{9}$
D) $2^{19} - \binom{20}{10}$
| If you choose $n$ of the unique object, then there are $20\choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$\sum_{n=0}^{10} {20\choose n}$$
By symmetry, we have
$$2\sum_{n=0}^{10} {20\choose n} = \sum_{n=0}^{20} {20\choose n} + {20\choose 10} = 2^{20} + {20\choose 10}$$
So
$$\sum_{n=0}^{10} {20\choose n} = 2^{19} + \frac12{20\choose 10}$$
And ${19\choose10} = {19\choose9}$ and ${19\choose10} + {19\choose9} = {20\choose10}$ by Pascal's Identity, so this gives $\frac12 {20\choose10} = {19\choose9}$ so the final answer is
$$2^{19} + {19\choose 9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve this simple equation $\frac{46}{y} + y = 25$? How do I solve this simple equation? $\frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
| $$\frac{46}{y}+y = 25$$
Here, assuming $y \neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)(y+y_2) = 0$$
The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes
$$(y-23)(y-2) = 0$$
Setting either factor equal to $0$ yields
*
*$$y-23 = 0 \iff y = 23$$
*$$y-2 = 0 \iff y = 2$$
You could also use the Quadratic Formula if desired:
$$ax^2+bx+c = 0 \iff x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$y = \frac{-(-25)\pm\sqrt{(-25)^2-4(1)(46)}}{2(1)} = \frac{25\pm\sqrt{441}}{2} = \frac{25\pm 21}{2}$$
$$y = 23; \quad y = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Linear Transformation involving polynomial and matrix Determine if its linear the transformation
$f:\Re_{3} [x]\rightarrow M_{2x2}$
such as
$f(ax^3+bx^2+cx+d) = \begin{bmatrix}
a-c & 0 \\
0&b+d
\end{bmatrix}$
for any $ax^3+bx^2+cx+d \in \Re _{3} [x]$
Im having trouble associating the polynomial with the resulting matrix.
I know I'm supposed to check the two main Linear transformation conditions but I don't know how to aproach this example.
Here's my attempt:
Condition 1: $T(u+v)=T(u)+T(v)$
$f((ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d'))=f((ax^3+a'x^3)+(bx^2+b'x^2)+(cx+c'x)+(d+d'))=f((a+a')x^3+(b+b')x^2+(c+c')x+(d+d'))=\begin{bmatrix}
(a+a')-(c+c') & 0\\
0 & (b+b')+(d+d')
\end{bmatrix} = \begin{bmatrix}
a-c & 0\\
0 & b+d
\end{bmatrix} + \begin{bmatrix}
a'-c' & 0\\
0 & b'+d'
\end{bmatrix}=f(ax^3+bx^2+cx+d)+f(a'x^3+b'x^2+c'x+d')$
Is this correct?
| That looks fine. Another way to go is this:
Consider the standard basis $$\bigl\{A_{1,1},A_{2,1},A_{1,2},A_{2,2}\bigr\}$$ for $M_{2\times 2},$ where $A_{i,j}$ is the $2\times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis $\{x^3,x^2,x,1\}$ for $\mathfrak{R}_3[x],$ then the given transformation has the matrix representation $$\begin{bmatrix}1 & 0 & -1 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 1 & 0 & 1\end{bmatrix}.$$
Since it has a matrix representation, then it's a linear transformation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determinant of a Hankel matrix with sequence (1, 2, 3,..., n, 1, 2,..., n-1) I'm looking for a closed form of the determinant of matrices like
$\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\\ 3& 1 &2\end{bmatrix}$
or
$\begin{bmatrix}1 & 2 & 3 &4\\2 & 3 &4 & 1\\ 3&4& 1 &2\\ 4& 1&2&3\end{bmatrix}$,
that means Henkel matrices of the size n with sequence (1, 2, 3,..., n, 1, 2,..., n-1)
| These matrices are actually circulant matrices, whose determinant is well known.
For the matrices at hand, the determinant is
$
\Delta_n=\prod _{j=0}^{n-1}f(\omega _{j})
$,
where $f(z)=1+2z+3z^{2}+\dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.
We have $f(\omega_0)=f(1)=1+2+3+\cdots +n = \dfrac{n(n+1)}{2}$.
We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+\cdots+z^n=\dfrac{z^{n+1}-1}{z-1}$. Therefore,
$$
g'(z)=\dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
$$
and so, for $j\ne0$,
$$
f(\omega_j)=g'(\omega_j)=\dfrac{n \omega_j - n}{(\omega_j-1)^2}=\dfrac{n}{\omega_j-1}
$$
Thus,
$$
\Delta_n=\prod _{j=0}^{n-1}f(\omega _{j})
=\dfrac{n(n+1)}{2} \prod _{j=1}^{n-1}\dfrac{n}{\omega_j-1}
=\dfrac{n^n(n+1)}{2} \dfrac{1}{\prod _{j=1}^{n-1}(\omega_j-1)}
$$
Now $\prod _{j=1}^{n-1}(\omega_j-1)=h(1)$ for
$$
h(z)=\prod _{j=1}^{n-1}(\omega_j-z)
=(-1)^{n-1}\prod _{j=1}^{n-1}(z-\omega_j)
=(-1)^{n-1}\dfrac{z^{n}-1}{z-1}
=(-1)^{n-1}(1+z+z^2+z^3+\cdots+z^{n-1})
$$
and so $h(1)=(-1)^{n-1}n$. Thus
$$
\Delta_n
=\dfrac{n^n(n+1)}{2} \dfrac{1}{(-1)^{n-1}n}
=(-1)^{n-1}\dfrac{n^{n-1}(n+1)}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a value of $t$ above which these inequalities aren't jointly satisfiable Let $\{a_{1}, ...., a_{8}\} \in [0, 1]$ be such that $\sum \limits_{0 < i \leq 8} a_{i} = 1$ and let $t \in (\frac{1}{2}, 1]$. Is there a value of $t$ above which the following inequalities aren't jointly satisfiable? I know that they are all satisfiable for any value of $t \leq \frac{2}{3}$.
(1) $\frac{a_{1} + a_{2}}{a_{1} + a_{2} + a_{3} + a_{4}} \geq t$
(2) $\frac{a_{1} + a_{5}}{a_{1} + a_{2} + a_{5} + a_{6}} \geq t$
(3) $\frac{a_{4} + a_{8}}{a_{2} + a_{4} + a_{6} + a_{8}} \geq t$
(4) $\frac{a_{7} + a_{8}}{a_{3} + a_{4} + a_{7} + a_{8}} \geq t$
(5) $\frac{a_{1} + a_{3}}{a_{1} + a_{2} + a_{3} + a_{4}} \leq 1 - t$
(6) $\frac{a_{6} + a_{8}}{a_{2} + a_{4} + a_{6} + a_{8}} \leq 1 - t$
| Let’s start a quest for a largest $t<1$ for which there exist feasible $a_1,\dots, a_8$ that satisfy all six inequalities. In this quest we can drop the condition $\sum a_i=1$, because if non-negative numbers $a_i$ satisfy the inequalities for some $t$ then numbers $a’_i=\frac {a_i}{\sum a_j}$ also satisfy the inequalities for the same $t$ and $\sum a’_i=1$.
Now note that each of $a_5$ and $a_7$ occur in only one inequality, (2) and (4), respectively. Pick any of these $a_i$’s and fix all other $a_j$’s. Then when $a_i$ tends to infinity, the left hand side of the inequality, in which it occurs, tends to $1$. Since $t<1$, for all other $a_j$’s the inequality will be satisfied for sufficiently large $a_i$. Thus in our quest we can drop both $a_5$ and $a_7$ and the inequalities (2) and (4).
After routine transformations, we can simplify the remaining inequalities to the following
(1’) $a_1+a_2\ge s(a_3+a_4)$
(3’) $a_4+a_8\ge s(a_2+a_6)$
(5’) $a_2+a_4\ge s(a_1+a_3)$
(6’) $a_2+a_4\ge s(a_6+a_8)$.
Here $s=\frac {t}{1-t}>0,$ because $t>\frac 12$.
Since $a_3$ and $a_6$ occur only at the right hand sides of the inequalities, when we annulate them, the inequality still be satisfied.
Now the inequalities transform to
(1’’) $a_1+a_2\ge sa_4$
(3’’) $a_4+a_8\ge sa_2$
(5’’) $a_2+a_4\ge sa_1$
(6’’) $a_2+a_4\ge sa_8$.
Adding (1’’) and (3’’), we obtain $a_1+a_8\ge (s-1)(a_2+a_4)$. Adding (5’’) and (6’’), we obtain $2(a_2+a_4)\ge s(a_1+a_8)$. Since, by $1$, $a_1+a_2+a_3+a_4>0$, and $a_3=0$, we have that both $A=a_1+a_8$ and $B=a_2+a_4$ are positive. Since $2A\ge 2(s-1)B\ge s(s-1)A$, we have $s(s-1)\le 2$. This implies $s\le 2$ and $t\le\frac 23$.
In order to satisfy inequalities (1’’)-(6’’) with $s=2$ we may try to put $a_1=a_8=a$ and $a_2=a_4=b$. Then the inequalities will be satisfied iff $a=b$. If we put $a>0$, $a_3=a_6=0$, $a_5=a_7=a$, the inequalities (2) and (4) also will be satisfied. At last, to achieve $\sum a_i=1$, we put $a=\frac 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n}\cos\left(\frac{n\pi}{3}\right)=-\frac{\pi^2}{36}$ I learnt on www.pi314.net that $$\sum_{n=1}^{\infty}\frac{H_n}{n}\cos\left(\frac{n\pi}{3}\right)=-\frac{\pi^2}{36}$$
This result is hard to verify using Wolfram Alpha since the series converges very slowly.
I do not know how to prove this result.
I tried to rewrite the original series as
$$\sum_{k\geq1} \frac{1}{k}\sum_{n\geq k}\frac{\cos\left(\frac{n\pi}{3}\right)}{n}$$
I know that
$$\sum_{n\geq 1}\frac{\cos\left(\frac{n\pi}{3}\right)}{n}=0$$
But since $n$ starts from $k$, I cannot continue from here.
Any hint?
| Perhaps it is in the extraction of the values you are having trouble with. If so, here is one rather (tedious and laborious) way to do it.
From Eq. (590) in the link, one has
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \operatorname{Li}_2 (x) + \frac{1}{2} \ln^2 (1 - x).$$
Setting $x = e^{i\pi /3}$ we see that
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n} \cos \left (\frac{n \pi}{3} \right ) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n} e^{i \pi n/3}\\ &= \operatorname{Re} \left [\operatorname{Li}_2 \left (e^{i\pi/3} \right ) + \frac{1}{2} \ln^2 \left (1 - e^{i \pi/3} \right ) \right ].
\end{align}
The log term can be disposed of immediately. Here
$$\ln (1 - e^{i \pi/3}) = -\frac{i \pi}{3}.$$
For the dilogarithm term, $\operatorname{Re} \operatorname{Li}_2 (e^{i \theta})$ is an elementary function (for a reason why, see here) and means a value for this term can be found. What I present below is one way to find it (and there are no doubt other, simpler ways).
Starting with the series representation for the dilogarithm function, we have
\begin{align}
\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] &= \operatorname{Re} \sum_{n = 1}^\infty \frac{e^{in \pi/3}}{n^2}\\
&= \sum_{n = 1}^\infty \frac{\cos \left (\frac{n \pi}{3} \right )}{n^2}\\
&= \sum_{\substack{n = 1\\n \in 6,12,\ldots}}^\infty \frac{1}{n^2} - \sum_{\substack{n = 1\\n \in 3,9,\ldots}}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{\substack{n = 1\\n \in 1,7,\ldots}}^\infty \frac{1}{n^2}\\
& \quad - \frac{1}{2} \sum_{\substack{n = 1\\n \in 2,8,\ldots}}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{\substack{n = 1\\n \in 5,11,\ldots}}^\infty \frac{1}{n^2} - \frac{1}{2} \sum_{\substack{n = 1\\n \in 4,10,\ldots}}^\infty \frac{1}{n^2}.
\end{align}
Terms in the series can be rearranged as the series absolutely converges. Shifting the indices as follows: $n \mapsto 6n$, $n \mapsto 6n + 3$, $n \mapsto 6n +1$, $n \mapsto 6n + 2$, $n \mapsto 6n + 5$, $n \mapsto 6n + 4$ leads to
\begin{align}
\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] &= \frac{1}{36} \sum_{n =1}^\infty \frac{1}{n^2} - \frac{1}{36} \sum_{n = 0}^\infty \frac{1}{(n + 1/2)^2} + \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 1/6)^2}\\
& \quad - \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 1/3)^2} + \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 5/6)^2} - \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 2/3)^2}\\
&= \frac{1}{72} \left [2 \cdot \zeta (2) - 2 \psi^{(1)} \left (\frac{1}{2} \right ) + \psi^{(1)} \left (\frac{5}{6} \right ) + \psi^{(1)} \left (\frac{1}{6} \right ) - \psi^{(1)} \left (\frac{2}{3} \right ) - \psi^{(1)} \left (\frac{1}{3} \right ) \right ],
\end{align}
where $\psi^{(1)} (z)$ is the polygamma function (trigamma function).
Using the known values of $\zeta (2) = \pi^2/6$, $\psi^{(1)} (1/2) = \pi^2/2$, and making use of the reflection relation for the trigamma function of
$$\psi^{(1)} (1 - z) + \psi^{(1)} (z) = \pi^2 \csc^2 (\pi z),$$
we see that
$$\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] = \frac{1}{72} \left [2 \cdot \frac{\pi^2}{6} - 2 \cdot \frac{\pi^2}{2} + 4 \pi^2 - \frac{4 \pi^2}{3} \right ] = \frac{\pi^2}{36}.$$
Thus
$$\sum_{n = 1}^\infty \frac{H_n}{n} \cos \left (\frac{n \pi}{3} \right ) = \frac{\pi^2}{36} - \frac{\pi^2}{18} = -\frac{\pi^2}{36},$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Inequality. $\sum_{cyc}(\frac{1}{a+b+\sqrt{2a+2c}})^3 \le \frac{8}{9}$ Problem. When $a, b, c>0, a, b, c \in \Bbb R, 16(a+b+c)\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, Prove that
$$\sum_{cyc}(\frac{1}{a+b+\sqrt{2a+2c}})^3 \le \frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that
$$4(x+y+z) \ge -2\frac{x^2+y^2+z^2-2xy-2yz-2zx}{(-x+y+z)(x-y+z)(x+y-z)}$$
And the inequality that I have to prove will be:
$$\sum_{cyc}(\frac{1}{x+\sqrt{2y}})^3 \le \frac{8}{9}$$
But I cannot think further. Can anyone give me a hint?
| Let $a=\frac{x}{4},$ $b=\frac{y}{4}$ and $c=\frac{z}{4}.$
Thus, the condition gives $$x+y+z\geq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ or
$$1\leq\frac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$\sum_{cyc}\frac{1}{\left(x+y+2\sqrt{2(x+z)}\right)^3}\leq\frac{1}{72}.$$
Now, by AM-GM
$$\left(x+y+2\sqrt{2(x+z)}\right)^3\geq\left(3\sqrt[3]{(x+y)\left(\sqrt{2(x+z)}\right)^2}\right)^3=54(x+y)(x+z).$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{1}{(x+z)(y+z)}\leq\frac{3}{4}$$ or
$$8(x+y+z)\leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that
$$8(x+y+z)\cdot\frac{xyz(x+y+z)}{xy+xz+yz}\leq3(x+y)(x+z)(y+z),$$ or
$$3(x+y)(x+z)(y+z)(xy+xz+yz)\geq8xyz(x+y+z)^2.$$
Now, since $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that
$$\frac{8}{9}(x+y+z)(xy+xz+yz)\cdot3(xy+xz+yz)\geq8xyz(x+y+z)^2$$ or
$$(xy+xz+yz)^2\geq3xyz(x+y+z).$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 \implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 \implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 \implies yxz =1$. If we take $G$ to be the group of $2\times 2$ matrices and let $x = \left( \begin{array} { c c } { 1 } & { 2 } \\ { 0 } & { 2 } \end{array} \right)$, $y = \left( \begin{array} { l l } { 0 } & { 1 } \\ { 2 } & { 1 } \end{array} \right)$ and $z = \left( \begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \\ { 1 } & { - 1 } \end{array} \right)$. Then $x y z = \left( \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) = 1$ but $y x z = \left( \begin{array} { c c } { 2 } & { - 2 } \\ { 5 } & { - 9 / 2 } \end{array} \right) \neq 1$.
I don't understand where my proof went wrong.
| Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the inequality $\frac{2}{x} > 3 x$ I've been slowly losing my mind on how to do this question, any help would be greatly appreciated.
Solve the inequality $$\frac{2}{x} > 3 x.$$
| To solve the equation
$$
\frac{2}{x} > 3x
$$
we could multiply be $x$ to obtain
$$
2 > 3x^2.
$$
But you have to be careful!
If $x < 0$, the inequality is reversed.
In other words:
$$
\frac{2}{x} > 3x
\iff \begin{cases}
2 < 3x^2 & x < 0 \\
2 > 3x^2 & x > 0.
\end{cases}
$$
Notice the equation isn't defined for $x = 0$, because you can't divide by $0$.
Case 1: $x > 0$.
Then we have
$$
2> 3x^2
\iff x^2 < \frac{2}{3}
\iff x \in \left(- \sqrt{ \frac{2}{3}}, \sqrt{ \frac{2}{3}}\right).
$$
But, since $x > 0$, our first solution interval is $I_1 := \left(0, \sqrt{ \frac{2}{3}}\right)$.
Case 2: $x < 0$.
Then we have
$$
2 < 3x^2
\iff x^2 > \frac{3}{2}
\iff x > \sqrt{\frac{3}{2}} \quad \text{and} \qquad x < -\sqrt{\frac{3}{2}}
$$
But again, since $x < 0$, our second (and final) solution interval is $I_2 := \left(- \infty, -\sqrt{\frac{3}{2}}\right)$.
So our solution is $I_1 \cup I_2 =\left(- \infty, -\sqrt{\frac{3}{2}}\right) \cup \left(0, \sqrt{ \frac{2}{3}}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that a semigroup satisfying $a^pb^q=ba$ is commutative
Let $(S, \cdot)$ be a semigroup. There are natural numbers $p,q \geq 2$ such that $a^pb^q=ba$ for all $a,b \in S$. Prove that $S$ is commutative.
I wrote
$$\begin{align}
a^{p+1}b^{q+1} &=b^{(q+1)p}a^{(p+1)q} \\
&=b^{p}\cdot(b^q)^p \cdot (a^p)^q\cdot a^q \\
&=b^p\cdot a^p \cdot b^q \cdot a^q \\
&= b^p\cdot b \cdot a \cdot a^q \\
&=b^{p+1}a^{q+1}.
\end{align}$$
From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.
Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
| Let $a\in S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $d\mid p+q-2$. But we also have $$a^3=\color{red}{a^2}\cdot \color{blue}{a}=\color{blue}{a^p}\color{red}{(a^2)^q}=a^{p+2q}$$ which means $d\mid p+2q-3$. Thus $d$ divides $(p+2q-3)-(p+q-2)=q-1$. Similarly, writing $a^3=a\cdot a^2$ we find that $d$ divides $p-1$.
Thus we have shown that for any $a\in S$, there exists $N$ such that for any $n\geq N$, $$a^n=a^{n+p-1}=a^{n+q-1}.$$
Now let $a,b\in S$; we will show that $ab=ba$. Notice first that $$ba=a^pb^q=(b^q)^p(a^p)^q=b^{pq}a^{pq}.$$ Iterating this, we have $ba=b^{(pq)^k}a^{(pq)^k}$ for any $k$. In particular, let us choose $k$ such that $n=(pq)^k$ is large enough such that $a^n=a^{n+p-1}$ and $b^n=b^{n+q-1}$. Now observe that $$ba=b^na^n=(a^n)^p(b^n)^q=a^{np}b^{nq}.$$ But $a^{np}=a^n$, since $a^n=a^{n+p-1}$ and $np-n$ is divisible by $p-1$, and similarly $b^{nq}=b^n$. Thus $$ba=a^nb^n,$$ where $n=(pq)^k$ for some $k$. But swapping the roles of $a$ and $b$ from the start of the argument, we know $a^nb^n=ab$, and thus $$ba=ab.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
derivative with quotient rule and summations I want to differentiate this with respect to $\eta$:
$$C(\eta) = \frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\left(1-\frac{\eta}{\eta_c}\right)}$$
Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math
$$C'(\eta) = \frac{\left[3\left(1-\frac{\eta}{\eta_c}\right)\sum\limits_{k=1}^{3} B_kk\left(\frac{\eta}{\eta_c}\right)^{k-1}\frac{1}{\eta_c}\right]-\left[\left(1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k\right)\left(-\frac{3}{\eta_c}\right)\right]}{\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^2}$$
| Here's how you could differentiate this particular function:
$$
C'(\eta) = \left[\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\left(1-\frac{\eta}{\eta_c}\right)}\right]'=
\left[\frac{1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}}{3\left(1-\frac{\eta}{\eta_c}\right)}\right]'=\\
\left[\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-1}\right]'=\\
\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)'\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-1}+\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)\left(\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-1}\right)'=\\
\frac{B_{1}\frac{1}{\eta_c}+B_{2}\frac{2\eta}{\eta_c^2}+B_{3}\frac{3\eta^2}{\eta_c^3}}{3\left(1-\frac{\eta}{\eta_c}\right)}+\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)(-1)\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-2}\left(3-3\frac{\eta}{\eta_c}\right)'=\\
\frac{\sum\limits_{k=1}^{3}B_k\frac{k\eta^{k-1}}{\eta_c^k}}{3\left(1-\frac{\eta}{\eta_c}\right)}+\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^2}\cdot\frac{3}{\eta_c}=\\
\frac{\sum\limits_{k=1}^{3}B_k\frac{k\eta^{k-1}}{\eta_c^k}}{3\left(1-\frac{\eta}{\eta_c}\right)}+\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\eta_c\left(1-\frac{\eta}{\eta_c}\right)^2}.
$$
The answer I get appears to be equivalent to what you got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating the probability of obtaining exactly four distinct values when a die is rolled six times Please could someone help me with determining the probability of getting $4$ distinct numbers (no order in the outcome e.g. 1,2,3,4 or 4,5,6,2 etc) or $5$ distinct from rolling a die $6$ times. So far I was able to calculate the probability of getting $6$ distinct numbers from $6$ rolls of a dice
$$\frac{6!}{6^6}$$
But I am having issues determining if it's only $5$ or $4$ distinct numbers out of a $6$ rolls.
Additionally, is this type of probability binomial or hypergeometric?
| Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $\binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $\binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of
$$\binom{6}{5}\cdot 5\cdot\binom{6}{2}\cdot 4!$$
desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is
$$\frac{\binom{6}{5}\cdot 5\cdot\binom{6}{2}\cdot 4!}{6^{6}}$$
Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $\binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.
*
*Case 1: There are $4$ choices for which value will appear $3$ times, and it will appear in $\binom{6}{3}$ locations. We can then order the remaining rolls in $3!$ ways. So there are $4\cdot\binom{6}{3}\cdot 3!$ outcomes of this form.
*Case 2: There are $\binom{4}{2}$ choices for which values will appear twice, and we can place them in $\binom{6}{2}\cdot\binom{4}{2}$ ways. There are then $2$ possible orders for the remaining two values. So there are $\binom{4}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot 2$ outcomes of this type.
Combining these cases, the probability of obtaining $4$ distinct numbers is
$$\frac{\binom{6}{4}\left[4\cdot\binom{6}{3}\cdot 3! + \binom{4}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot 2\right]}{6^{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$
My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is not an easy inequality.
| Note that
$$a^2+b^2+c^2-\frac{1}{3}(a+b+c)^2=\frac{1}{3}((a-b)^2+(b-c)^2+(a-c)^2)$$
(this is useful if you read Mark Bennet's hint).
So it remains to prove that
$$-2(a-b+1)\leq \frac{1}{3}((a-b)^2+(b-c)^2+(a-c)^2).$$
Now you may use the real variables $x=a-b$, $y=b-c$ and $z=c-a$ (and you are right, the condition $a>1$, $b>1$ and $c>1$ is irrelevant!):
$$-6x-6\leq x^2+y^2+z^2.$$
That is equivalent to
$$\sqrt{3}\leq \sqrt{(x+3)^2+y^2+z^2}$$
where $x+y+z=0$. The above inequality holds because the distance of any point on the plane $x+y+z=0$ from the point $(-3,0,0)$ is greater or equal to $\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$. Where's my error?
In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, point $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$.
I know the correct solution, but I'm more interested in where is the problem in my solution.
My solution :
Now in $\triangle ABD$, applying the sine rule:
$$\frac{AD}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{1}$$
In $\triangle BDC$:
$$\frac{BD}{\sin 80^\circ} = \frac{BC}{\sin(180^\circ-\beta)} \tag{2}$$
We know $AD= BC$; put in to $(1)$:
$$\frac{BC}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{3}$$
Comparing $(2)$ and $(3)$:
$$\frac{BC}{BD} = \frac{\sin\alpha}{\sin 20^\circ} = \frac{\sin(180^\circ-\beta)}{\sin 80^\circ} \tag{4}$$
$$\frac{\sin \alpha}{\sin(180^\circ-\beta)} = \frac{\sin 20^\circ}{\sin 80^\circ} \tag{5}$$
Now, $\alpha = 20^\circ$ and $\beta = 100^\circ$, but when I plug these values in $\triangle ABC$, it's not even triangle. oO
Where I am wrong? Thanks.
PS : sorry for poor editing, I don't have any clue about it.
| So we have that $\frac{\sin \alpha}{\sin (180-\beta)} = \frac{\sin 20}{\sin 80}$.
The first thing we use is that $\alpha + \beta = 160$ from the triangle $ABD$. From here, $180 - \beta = 180 - (160-\alpha) = 20 + \alpha$.
Next, we note that:
$$
\frac{\sin 20}{\sin 80} = \frac{\sin 20}{\cos(90-80)} = \frac{\sin 20}{\cos 10} = \frac{2 \sin 10 \cos 10} {\cos 10} = 2 \sin 10
$$
So, we have the equation :
$$
\frac{\sin \alpha}{\sin (\alpha + 20)} = 2 \sin 10\\ \implies \sin \alpha = 2 \sin 10 \sin (20+\alpha) = 2 \sin 10 \sin 20 \cos \alpha + 2 \sin 10 \cos 20 \sin \alpha
$$
Now, collecting terms of $\sin \alpha$ on one side, and facctorizing it out,
$$
\sin \alpha(1 - 2 \sin 10 \cos 20) = 2 \sin 10 \sin 20 \cos \alpha \\ \implies
\tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 \sin A\cos B = \sin(A+B) + \sin(A-B) \quad ; \quad \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
We will also use the fact that $\sin 30 = \frac 12$.
In our case,
$$
1 - 2 \sin 10 \cos 20 = 1- (\sin 30 + \sin (-10)) = 2 \sin 30 - (\sin 30 - \sin 10) \\ = \sin 30 +\sin 10 = 2 \sin 20 \cos 10
$$
Therefore,
$$
\tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20} = \frac{2\sin 10 \sin 20}{2 \cos 10 \sin 20} = \tan 10
$$
Now, since $0 < \alpha < 180$, we get that $\alpha = 10$. From here, $80-\alpha = 70$ is the desired angle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $
I made this far:
$$
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
| Hint
First, let $n=\frac 1x$ and compose Taylor series around $x=0$; back to $n$, get successively
$$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}+O\left(\frac{1}{n^4}\right)$$
$$\sqrt{n^2+\sqrt{n^4+1}}=\sqrt{2} n+\frac{1}{4 \sqrt{2} n^3}+O\left(\frac{1}{n^5}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Is $a$ or $b \equiv 0 \pmod{5}$ if $a^2-b^2 \equiv 0 \pmod{5}$? I have this question, knowing the last in title can i define that at least a or b is divisible by 5?
| Well, ... no because $a^2 - a^2 \equiv 0$ but neither $a$ need not be equivalent to $0 \mod 5$.
Or for that matter $a^2 - (-a)^2 \equiv 0$ but if $a \ne \equiv 0 \pmod 5$ then $-a\equiv 5-a\not \equiv 0$ either.
You can exhaust all cases. $0^2 - 0^2, 1^2 - 1^2, 1^2 - 4^2, 2^2 - 2^2, 2^2 - 3^2, 3^2 - 3^2, 3^2 - 2^2, 4^2 - 4^2, 4^2 - 1^2$ are all equiv $0$.
In fact we can see that if $a^2 - b^2 \equiv 0$ then either both are equivalent to zero or neither are.
Which makes perfect sense becuase if one of them, say $b$, was equivalent to $0$ then $a^2 - b^2 \equiv a^2 - 0^2 \equiv a^2\pmod 5$ (Or if $a\equiv 0$ then $a^2 - b^2 \equiv 0^2 - b^2 \equiv -b^2 \equiv 0\pmod 5$) and there's no reason to assume the other squared (or negative the other squared) would be $0$.
If we think about it is seems really weird that one would think it would imply one is equiv to $0$.
Why DID you think that?
...
Note also $a^2 - b^2 = (a+b)(a-b)\equiv 0 \pmod 5$. Because $5$ is prime that means either $5|a+b$ and $a+b \equiv 0 \pmod 5$ or $5|a-b$ and $a-b \equiv 0 \pmod 5$. Those would mean that if $a^2 - b^2 \equiv 0\pmod 5$ we MUST have either $a \equiv b\pmod 5$ or $a \equiv -b \pmod 5$.
That is almost the exact opposite of your assumption.
However it's worth noting that $a^2 -b^2 \equiv 0 \pmod n$ wont mean this if $n$ is not prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
In square $ABCD$ with side $1$, points $E$, $P$, $F$ are the midpoints of $AD$, $CE$, $BP$. What is the area of $\triangle BFD$? I'm not sure if my solution to this Olympiad Geometry question is valid.
Let $\square ABCD$ be a square with side length $1$. Let $E$, $P$, $F$ be the midpoints of $AD$, $CE$, $BP$, respectively. What is the area of $\triangle BFD$?
To solve this I used coordinate geometry but I'm not sure if it works.
I dropped a perpendicular from point P to side CD and called the point X. Then, since the area of triangle DPB is exactly double the desired area, I assumed point D to be (0, 0) and found the side length XP using the Pythagorean theorem. Since I have the three coordinates for triangle DPB, I used shoelace formula to find the area and divided by 2 to get {(19)^(1/2) - 1} / 8
| The equation of the line $\overleftrightarrow{BD}$ is $x+y-1=0$. The coordinates of the point $F$ are $\left(\dfrac 34, \dfrac 38 \right)$. So the distance from the point $F$ to the line $\overleftrightarrow{BD}$ is
$$h = \dfrac{\left|\dfrac 34 + \dfrac 38 - 1 \right|}{\sqrt{1^2+1^2}}
= \dfrac{1}{8 \sqrt 2}$$
The area of $\triangle BDF$ is therefore
$$\dfrac 12 \cdot BD \cdot h \
= \dfrac 12 \cdot \sqrt 2 \cdot \dfrac{1}{8 \sqrt 2}
= \dfrac{1}{16}$$
An easier-to-remember way to write @Rosenberg's answer is
$$BDF = \dfrac 12 \cdot
\left\| \begin{array}{c}
1 & 1 & 1 \\
1 & 0 & \dfrac 34 \\
0 & 1 & \dfrac 38
\end{array} \right\| = \dfrac{1}{16}$$
where $\| \cdot \|$ indicates the absolute value of the determinant.
Notes.
The distance from the point $(h,k)$ to the line $ax+by-c=0$ is
$\dfrac{|ah+bk-c|}{\sqrt{a^2+b^2}}$
The area of the triangle whose vertices have coordinates
$(a_1, a_2), (b_1, b_2),(c_1, c_2)$ is
$$\dfrac 12 \cdot
\left\| \begin{array}{c}
1 & 1 & 1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array} \right\|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximize the following equation under a constraint Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.
Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:
| Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
\begin{eqnarray} g(r,\theta)&=&(r\cos\theta)^4+(r\sin\theta)^4\\
&=&r^4(\cos^4\theta+\sin^4\theta)\\
\frac{dg}{dr}&=&4r^3(\cos^4\theta+\sin^4\theta)=0\\
\frac{dg}{d\theta}&=&-4r^4\cos^3\theta\sin\theta+4r^4\sin^3\theta\cos\theta\\
&=&-4r^4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)\\
&=&-2r^4(2\sin\theta\cos\theta)(\cos^2\theta-\sin^2\theta)\\
&=&-2r^4\sin(2\theta)\cos(2\theta)\\
&=&-r^4\sin(4\theta)=0
\end{eqnarray}
So extrema should occur when $\sin(4\theta)=0$. And $\sin(4\theta)=0$ only when $4\theta$ is a whole multiple of $\pi$. Thus you must check points of the circle corresponding to multiples of $\dfrac{\pi}{4}$.
You should find that the minima occur when $\theta$ is an odd multiple of $\dfrac{\pi}{4}$ and the maxima will occur when $\theta$ is an even multiple of $\dfrac{\pi}{4}$ (i.e. a multiple of $\dfrac{\pi}{2})$.
$$g\left(3,\frac{\pi}{4}\right)=g\left(3,\frac{3\pi}{4}\right)=g\left(3,\frac{5\pi}{4}\right)=g\left(3,\frac{7\pi}{4}\right)=\left(\pm\frac{3}{\sqrt{2}}\right)^4+\left(\pm\frac{3}{\sqrt{2}}\right)^4=\frac{81}{2}$$
$$g(3,0)=g\left(3,\frac{\pi}{2}\right)=g\left(3,\pi\right)=g\left(3,\frac{3\pi}{2}\right)=3^4=81$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Mixed Fractions and Multiplication (with Variables) I stumbled over this expression: $3 \frac{1}{x^3}$.
How should you interpret something like that?
While you could see that as implicitit multiplication ($3 * \frac{1}{x^3}$),
you could also argue that $3 \frac{1}{x^3}$ is a mixed fraction ($3 + \frac{1}{x^3}$).
I think in situations with only numbers or only variables everything should be clear:
$3 \frac{1}{2} = 3 + \frac{1}{2} = 3.5$
$a \frac{b}{c} = a * \frac{b}{c}$
This should also be true: $3 \frac{b}{c} = 3 * \frac{b}{c}$.
But what do our conventions say to something like $3 \frac{1}{x^3}$ or $6 \frac{x}{3}$ or $\frac{1}{x^2}5$? Is there any written standard that you should generally follow? Multiplication or addition?
| When there is no operator between expressions the usual interpretation is multiplication, so $3\frac 1{x^3}=3\cdot \frac 1{x^3}=\frac 3{x^3}$
Mixed fractions are an exception to this, so $3\frac 13 \neq 3 \cdot \frac 13$. Instead it is $3 \frac 13=3+\frac 13$.
If there is any possibility of confusion one should supply the operator, but sometimes people do not. In that case you need to figure it out from context.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k}\text{ } \text{ } \text{} \text{ } \text{ }\text{the } n^{th} \text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
| For $x \in [0,1]$ let
$$ f(x) = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n} x^{2n} \, . $$
Using the power series of $\arcsin$ we find
$$ x \frac{\mathrm{d}}{\mathrm{d} x} x \frac{\mathrm{d}}{\mathrm{d} x} f(x) = 4 \frac{\mathrm{d}}{\mathrm{d} x} [\arcsin(x) - x] = 4 \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] $$
for $x \in [0,1)$ . In particular,
$$ f'(1) = 4 \int \limits_0^1 \frac{1}{x} \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] \, \mathrm{d} x \stackrel{x=\sqrt{1-y^2}}{=} 4 \int \limits_0^1 \frac{\mathrm{d} y}{1+y} = 4 \ln(2) \, . $$
Now we can compute
\begin{align}
S &\equiv \sum \limits_{n=1}^\infty \frac{H_{2n} {2n \choose n}}{n^2 4^n} = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n} \int \limits_0^1 \frac{1-x^{2n}}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{f(1) - f(x)}{1-x} \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{- \ln(1-x)}{x} x f'(x) \, \mathrm{d} x
= \operatorname{Li}_2 (1) f'(1) - 4 \int \limits_0^1 \frac{\operatorname{Li}_2 (x)}{x} \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] \, \mathrm{d} x \\
&= \operatorname{Li}_2 (1) f'(1) + 4 \operatorname{Li}_3 (1) - 4 \int \limits_0^1 \frac{\operatorname{Li}_2 (x)}{x \sqrt{1-x^2}} \, \mathrm{d} x \equiv 4 \left[\frac{\pi^2}{6} \ln(2) + \zeta(3) - I\right] \, .
\end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
\begin{align}
I &= \int \limits_0^\infty t \int \limits_0^1 \frac{\mathrm{d} x}{(\mathrm{e}^t - x) \sqrt{1-x^2}} \, \mathrm{d} t \stackrel{(*)}{=} \int \limits_0^\infty \frac{t \left[\frac{\pi}{2} + \arcsin(\mathrm{e}^{-t})\right]}{\sqrt{\mathrm{e}^{2t}-1}} \, \mathrm{d} t \\
&\stackrel{\mathrm{e}^{-t} = \sin(u)}{=} \frac{1}{2} \int \limits_0^{\pi/2} -\ln[\sin(u)] (\pi + 2 u) \, \mathrm{d} u = \frac{1}{2} \int \limits_0^{\pi/2} u (\pi + u) \cot(u) \, \mathrm{d} u \\
&= \frac{1}{2} [\pi K_1^{(1)} + K_2^{(1)}] = \frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3) \, .
\end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ \boxed{S = \sum \limits_{n=1}^\infty \frac{H_{2n} {2n \choose n}}{n^2 4^n} = \frac{23}{4} \zeta(3) - \frac{5}{6} \pi^2 \ln(2)} \, . $$
Proof of $(*)$:
For $a \in [0,1]$ let
$$ g(a) = \int \limits_0^1 \frac{-\ln(1-a x)}{x \sqrt{1-x^2}} \, \mathrm{d} x= \sum \limits_{n=1}^\infty \frac{a^n}{n} \int \limits_0^{\pi/2} \sin^{n-1} (t) \, \mathrm{d} t \, .$$
Using Wallis' integrals we find
$$ g(a) = \frac{\pi}{2} \sum \limits_{k=0}^\infty \frac{{2k \choose k} a^{2k+1}}{4^k(2k+1)} + \frac{1}{4} \sum \limits_{m=1}^\infty \frac{4^k a^{2k}}{k^2 {2k \choose k}} = \frac{\pi}{2} \arcsin(a) + \frac{1}{2} \arcsin^2 (a) \, . $$
Therefore
$$ \int \limits_0^1 \frac{\mathrm{d} x}{(1-a x)\sqrt{1-x^2}} = g'(a) = \frac{\frac{\pi}{2} + \arcsin{a}}{\sqrt{1-a^2}} $$
holds for $a \in [0,1)$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Infinite series problem
The sum of $$\frac{2}{4-1}+\frac{2^2}{4^2-1}+\frac{2^4}{4^4-1}+\cdots \cdots $$
Try: write it as $$S = \sum^{\infty}_{r=0}\frac{2^{2^{r}}}{2^{2^{r+1}}-1}=\sum^{\infty}_{r=0}\frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
| You may prove by induction that $$\sum_{r=0}^n\frac{2^{2^r}}{2^{2^{r+1}}-1}=1-\frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$\sum_{r=0}^{k+1}\frac{2^{2^r}}{2^{2^{r+1}}-1}=1-\frac{1}{2^{2^{k+1}}-1}+\frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-\frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-\frac{1}{a-1}+\frac{a}{a^2-1}=1-\frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
On the evalution of an infinite sum
I wish to show that
$$\sum_{n = 0}^\infty (-1)^n \left [\frac{2n + 1/2}{(2n + 1/2)^2 + 1} + \frac{2n + 3/2}{(2n + 3/2)^2 + 1} \right] = \frac{\pi}{\sqrt{2}} \frac{\cosh \left (\frac{\pi}{2} \right )}{\cosh (\pi)}.$$
The reason I wish to find such a sum is as follows.
The question here called for the evaluation (I have added its value) of
$$\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx = \frac{\pi}{\sqrt{2}} \frac{\cosh \left (\frac{\pi}{2} \right )}{\cosh (\pi)}.$$
As one of the comments, the OP remarked that they would like to see different approaches to the evaluation of the integral so I thought I would try my hand at one that does not rely on contour integration and the residue theorem. My approach was as follows:
\begin{align}
\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx &= \int_0^1 \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx + \int_1^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx\\
&= \int_0^1 \frac{\cos (\ln x) (x + 1)}{\sqrt{x} (1 + x^2)} \, dx,
\end{align}
after a substitution of $x \mapsto 1/x$ has been enforced in the second of the integrals. Now if we enforce a substitution of $x \mapsto e^{-x}$ one arrives at
$$\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx = \int_0^\infty \frac{\cos x \cosh (x/2)}{\cosh x} \, dx.$$
Writing the hyperbolic functions in terms of exponentials we have
\begin{align}
\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx &= \int_0^\infty \frac{\cos x (e^{-x/2} + e^{-3x/2})}{1 + e^{-2x}} \, dx\\
&= \text{Re} \sum_{n = 0}^\infty (-1)^n \int_0^\infty \left [e^{-(2n + 1/2 - i) x} + e^{-(2n + 3/2 - i)x} \right ] \, dx\\
&= \text{Re} \sum_{n = 0}^\infty (-1)^n \left [\frac{1}{2n + 1/2 - i} + \frac{1}{2n + 3/2 - i} \right ] \tag1\\
&= \sum_{n = 0}^\infty (-1)^n \left [\frac{2n + 1/2}{(2n + 1/2)^2 + 1} + \frac{2n + 3/2}{(2n + 3/2)^2 + 1} \right],
\end{align}
which brings me to my sum.
Some thoughts on finding this sum
Rewriting the sum $S$ in (1) as follows:
\begin{align}
S &= \text{Re} \cdot \frac{1}{4} \sum_{n = 0}^\infty \left [\frac{1}{n + 1/8 - i/4} + \frac{1}{n + 3/8 - i/4} - \frac{1}{n + 5/8 - i/4} - \frac{1}{n + 7/8 - i/4} \right ]\\
&= \text{Re} \cdot \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 7/8 - i/4} \right ) + \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 5/8 - i/4} \right )\\
& \qquad - \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 3/8 - i/4} \right ) - \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 1/8 - i/4} \right )\\
&= \frac{1}{4} \text{Re} \left [\psi \left (\frac{7}{8} - \frac{i}{4} \right ) + \psi \left (\frac{5}{8} - \frac{i}{4} \right ) - \psi \left (\frac{3}{8} - \frac{i}{4} \right ) - \psi \left (\frac{1}{8} - \frac{i}{4} \right ) \right ].
\end{align}
Here $\psi (z)$ is the digamma function. I was rather hoping to use the reflexion formula for the digamma function, but alas it does not seem to take me any closer to a final real solution.
Final thought
While it would be nice to see how to evaluate this sum, perhaps my approach was not the best so alternative methods to evaluate the integral that avoid this sum and does not rely on contour integration would also be welcome.
| $$
\begin{align}\newcommand{\Re}{\operatorname{Re}}
&\sum_{n=0}^\infty(-1)^n\left[\frac{2n+1/2}{(2n+1/2)^2+1}+\frac{2n+3/2}{(2n+3/2)^2+1}\right]\tag1\\
&=\frac12\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac{2n+1/2}{(2n+1/2)^2+1}+\frac{2n+3/2}{(2n+3/2)^2+1}\right]\tag2\\
&=\frac14\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac{n+\frac14}{\left(n+\frac14\right)^2+\frac14}+\frac{n+\frac34}{\left(n+\frac34\right)^2+\frac14}\right]\tag3\\
&=\frac18\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac1{n+\frac14-\frac i2}+\frac1{n+\frac14+\frac i2}+\frac1{n+\frac34-\frac i2}+\frac1{n+\frac34+\frac i2}\right]\tag4\\
&=\frac18\left[\frac\pi{\sin\left(\pi\!\left(\frac14-\frac i2\right)\right)}+\frac\pi{\sin\left(\pi\!\left(\frac14+\frac i2\right)\right)}+\frac\pi{\sin\left(\pi\!\left(\frac34-\frac i2\right)\right)}+\frac\pi{\sin\left(\pi\!\left(\frac34+\frac i2\right)\right)}\right]\tag5\\
&=\frac{\pi\sqrt2}8\left[
\frac{\cosh\left(\frac\pi2\right)+i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}+
\frac{\cosh\left(\frac\pi2\right)-i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}+
\frac{\cosh\left(\frac\pi2\right)-i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}\right.\\
&\left.\phantom{=\frac{\pi\sqrt2}8}+
\frac{\cosh\left(\frac\pi2\right)+i\sinh\left(\frac\pi2\right)}{\cosh(\pi)}\right]\tag6\\
&=\frac\pi{\sqrt2}\frac{\cosh(\pi/2)}{\cosh(\pi)}\tag7
\end{align}
$$
Explanation:
$(2)$: use symmetry
$(3)$: pull factor of $\frac12$ out front
$(4)$: partial fractions
$(5)$: use $(3)$ from this answer
$(6)$: evaluate the sine of a complex number
$(7)$: simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 0
} |
Correctness of the induction step Prove by induction that for all $ n \geq 2$ that the following:
$\sum_{i = 1}^n \frac{i}{i+1} < \frac{n^2}{n+1}$
Looking at the $n+1$ step, is it safe to assume the following ?
$\sum_{i = 1}^{n+1} \frac{i}{i+1} < \frac{(n+1)^2}{n+2} \iff \frac{n+1}{n+2} \leq \frac{(n+1)^2}{n+2} - \frac{n^2}{n+1}$
Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.
We then solve the inequality on the right:
$(n+1)^2 \leq (n+1)^3 -n^2(n+2)$
$n^2+2n+1 \leq n^3+3n^2+3n+1 -n^3-2n^2$
$0 \leq n$
Since the questions assumes $ n \geq 2$ q.e.d.
| Hint: You must show that $$\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{n}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
$$\frac{n^2}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$$
And we get
$$\frac{(n+1)^2}{n+2}-\frac{n+1}{n+2}-\frac{n^2}{n+1}=\frac{n}{(n+1) (n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the complex numbers $z$ such that $w=\frac{2z-1}{2+iz}$ is real
Find the complex numbers $z$ such that $w=\dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one.
Trouble is that, by manipulating $w$, it just seems to get much worse.
Any path to follow?
| $$w=\frac{(2x-1)+iy}{(2-y)+ix}=\frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+(y^2-2y+1)+\frac{1}{8}-1=0$$
$$(y-1)^2-2\left(x-\frac{1}{4}\right)^2=\frac{7}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Rewriting $2x^2-12x+11$ by completing the square I want to rewrite $2x^2-12x+11$ in square form. According to my book, the answer is $2(x-3)^2-7$. However, I get $2(x-3)^2+2$ for some reason.
My Steps:
$2x^2-12x+11$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+9$
$2(x - 3)^2=-2$
$2(x-3)^2+2$
What am I doing wrong?
| As noted in the comments, in the fourth line, you have to add 18 to the right member and not 9.
$2x^2-12x+11=0$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+18$
$2(x - 3)^2=7$
$2(x-3)^2-7=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $9x^2+2y^2-8xy-6x+11$ For any $x, y$ in real, find the minimum value of $9x^2+2y^2-8xy-6x+11$
| Let $$9x^2+2y^2-8xy-6x+11=k$$
Thus, the the following quadratic equation of $y$
$$2y^2-8xy+9x^2-16x+11-k=0$$ has real solutions, which says
$$16x^2-2(9x^2-6x+11-k)\geq0$$ or
$$(x-3)^2\leq k-2,$$ which gives $$k\geq2.$$
The equality occurs for $x=3$ and $y=6,$ which says that $2$ a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Is there any other way to establish this trig identity? $\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $ I needed to verify this trig identity: $$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $$
what I did is I worked on both sides individually:
LHS: $$\frac{\frac1{\cos(x)} + 1}{\frac {\sin(x)}{\cos (x)}} =\frac{1 + \cos(x)}{\sin(x)}$$
RHS: $$\frac{\sin(x)}{1 - \cos(x)}=\frac {\sin(x)(1-\cos^2(x))}{1 -\cos^2(x)}=\frac {\sin(x)(1-\cos^2(x))}{\sin^2(x)}= \frac{1 +\cos(x)}{\sin(x)} $$
QED.
I can't think of other way to verify this, like just working on one side.
Is there any?
| $$\begin{align*} \frac{\sec x + 1}{\tan x} &= \frac{1 + \cos x}{\sin x} = \frac{(1 + \cos x)(1 - \cos x)}{\sin x (1 - \cos x)} = \frac{1 - \cos^2 x}{\sin x (1 - \cos x)} = \frac{\sin^2 x}{\sin x (1 - \cos x)} \\ &= \frac{\sin x}{1 - \cos x}. \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Solving equations involving square roots I am a student and I often encounter these type of equations:
$$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6$$
I usually solve these by taking one term ($\sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.
Thanks
| A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.
Given that
$$\begin{align}\sqrt{x^2 +(y-2)^2} + \sqrt{x^2+(y+2)^2} &= 6 &[1]\\ \end{align}$$
and that
$$\begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\\ \end{align}$$
we have (by $\frac{[2]}{[1]}$)
$$\begin{align}\sqrt{x^2 +(y-2)^2} - \sqrt{x^2+(y+2)^2} = \frac{-8y}{6} &= -\frac{4}{3}y &[3]\\ \end{align}$$
Adding $[1]$ and $[3]$ gives us
$$\begin{align}
2\sqrt{x^2 +(y-2)^2} &= 6-\frac{4}{3}y &[4]\\
4(x^2 +(y-2)^2) &= \bigg(6-\frac{4}{3}y\bigg)^2\\
4x^2 +4y^2-16y+16 &= \frac{16}{9}y^2-16y+36\\
x^2 +y^2+4 &= \frac{4}{9}y^2+9\\
x^2 +\frac{5}{9}y^2 &= 5\\
\frac{x^2}{5} +\frac{y^2}{9} &= 1\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$a + b = c + c$ So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get:
(
| In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=\frac{a^2+b^2}{2}$$
$$X=\frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=\frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=\frac{(a^2+b^2)^2}{2}$$
$$X=\frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=\frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$
Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$
and that the remainder of the division of polynomial $P(x)$ by $x-3$ is $18$. Evaluate the constant term of polynomial $P(x-1)$.
All I could see so far is that the polynomial $P(x)$ should be quadratic because $Q(x)$ is multiplied by quadratic term, which is $x^2$.
|
the remainder of the division of polynomial $P(x)$ by $x−3$ is $18$
implies:
$$P(x)=(x-3)R(x)+18 \Rightarrow \\
P(x+1)=(x-2)R(x+1)+18=(x^2 -4)Q(x)+3ax+6\\
P(2+1)=18=6a+6 \Rightarrow a=2.$$
Hence:
$$P(\color{red}x-1)=P(x-2+1)=((x-2)^2-4)Q(x-2)+6(x-2)+6 =\\
(x^2-4x)Q(x-2)+6x-6 \Rightarrow \\
P(\color{red}0-1)=-6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mess with the constant edits, I will stop editing and not touch this so people can review the question. Sorry
a) Prove that P(1) is true
$~$
$~$
$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$
Show that P(k+1) is true as well
$$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$
$~~$
$$ = \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$
| [You can use the trick of observing that it's a telescoping sum, or you just blast ahead with a proof by induction. I think you're asking about the second approach, so here's an answer.]
It's like any other induction proof. You get to assume "P(k)", which means you get to assume
$$
\sum_{i=1}^{k} \frac{1}{i(i+1)} = \frac{k}{k+1}.
$$
Now you want to show that "P(k+1)" is true: you want to show
$$
\sum_{i=1}^{k+1} \frac{1}{i(i+1)} \overset{?}{=} \frac{k+1}{k+2}.
$$
Expand the left side:
$$
\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \sum_{i=1}^{k} \frac{1}{i(i+1)} + \frac{1}{(k+1)(k+2)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)},
$$
where the last equality is by the inductive hypothesis. Now do some algebra:
$$
= \frac{k}{k+1} + \frac{1}{k+1} - \frac{1}{k+2} = 1 - \frac{1}{k+2} = \frac{k+1}{k+2}.
$$
That completes the inductive step, and hence the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$.
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
| Proof by induction that $\sum_{k=1}^{n} (2k-1)(n-k+1) = \sum_{k=1}^{n} k^2 \space \forall n \ge 1$:
Base case: For $n=1$ we have $1 \times 1 = 1$
Assume true for $n=k$.
For $n=k+1$:
$\sum_{k=1}^{n+1} (2k-1)((n+1)-k+1) \\
=\sum_{k=1}^{n+1} (2k-1)((n-k+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} (2k-1)((n-k+1) + (2(n+1)-1)(n-(n+1)+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} k^2 +(2n+1)\times0 + (n+1)^2 \\
=\sum_{k=1}^{n+1} k^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
determining if sequence has upper bound I am somewhat stuck in my calculations when determining if sequence has an upper bound.
The sequence $$x_n = \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n-1}+\frac{1}{2n}$$
Is equal to $$\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}})$$
And so I notice that all the denominators are greater than 1, which means that all terms in the parentheses are less than 1.
But how can I determine further if there is an upper bound?
| By C-S $$\sum_{i=1}^n\frac{1}{n+i}=1+\sum_{i=1}^n\left(\frac{1}{n+i}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{i=1}^n\frac{i}{n+i}=$$
$$=1-\frac{1}{n}\sum_{i=1}^n\frac{i^2}{ni+i^2}\leq1-\frac{1}{n}\frac{\left(\sum\limits_{i=1}^ni\right)^2}{\sum\limits_{i=1}^n(ni+i^2)}=1-\frac{1}{n}\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=$$
$$=1-\frac{3(n+1)}{2(5n+1)}=\frac{7n-1}{10n+2}<\frac{7}{10}.$$
Actually, $$\ln2=0.6931...$$
Cauchy-Schwarz forever!
Actually, by calculus we can show that $$\lim_{n\rightarrow+\infty}\sum_{i=1}^n\frac{1}{n+i}=\ln2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $y=\frac {a+bz}{c+dz}$, $z=\frac{a+bx}{c+dx}$, $x=\frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$ I need to solve this problem and I don’t know how.
If $x, y, z$ are unequal and $y = \frac {a + bz}{c + dz}, z = \frac {a + bx}{c + dx}, x = \frac {a + by}{c + dy},$ then $ad + bc + b^2 + c^2 = 0$.
I see that I need to eliminate $x$, $y$, $z$ but I don’t know how to start.
I appreciate your help.
| $$
\begin{bmatrix}
y\\
1
\end{bmatrix}
=
\alpha
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}
\begin{bmatrix}
z\\
1
\end{bmatrix}, \qquad
\begin{bmatrix}
z\\
1
\end{bmatrix}
=
\beta
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}
\begin{bmatrix}
x\\
1
\end{bmatrix}, \qquad
\begin{bmatrix}
x\\
1
\end{bmatrix}
=
\gamma
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}
\begin{bmatrix}
y\\
1
\end{bmatrix}
$$
$$
\Rightarrow
\begin{bmatrix}
y\\
1
\end{bmatrix}
=
\alpha\beta\gamma
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}^{3}
\begin{bmatrix}
y\\
1
\end{bmatrix}, \qquad
\begin{bmatrix}
z\\
1
\end{bmatrix}
=
\alpha\beta\gamma
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}^{3}
\begin{bmatrix}
z\\
1
\end{bmatrix}, \qquad
\begin{bmatrix}
x\\
1
\end{bmatrix}
=
\alpha\beta\gamma
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}^{3}
\begin{bmatrix}
x\\
1
\end{bmatrix}
$$
Since $x$, $y$, $z$ are different, hence this is only possible iff
$$
\alpha\beta\gamma
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}^{3}
=
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}
$$
$$
\alpha\beta\gamma
\begin{bmatrix}
b^{3}+ad(2b+c) & a(b^{2}+c^{2} +ad+bc) \\
d(b^{2}+c^{2} +ad+bc) & c^{3}+ad(b+2c)
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}
$$
so either {$a=0$ and $d=0$} or $b^{2}+c^{2} +ad+bc =0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that planes $x+2y+3z=8$ and $2x+3y+4z=11$ intersect in a line coplanar with $\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$ Question is
Show that the line of intersection of the planes
$$ x + 2y + 3z = 8 \quad\text{and}\quad 2x + 3y + 4z = 11 $$
is coplanar with the line
$$\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$$
Also find the equation of the plane containing them.
$\text{Any hint how could I proceed ?}$
I know how to find the vector the line of intersection would be parallel to , which is given by cross product of the normal vectors of the two planes .
But that does not help in finding the equation of the line of intersection .
What I know is that to prove that two lines are coplanar we have to show that they intersect i.e the shortest distance between them is zero .
But unfortunately this formula again requires one point through which the line of intersection would pass which is unknown.
ps - I am in school yet :)
| The line of intersection between the two planes can be represented parametrically. Let $z = t$:
$$
\left\{
\begin{array}{c}
x + 2y = 8 - 3t \\
2x + 3y = 11 - 4t \\
\end{array}
\right.
$$
Then $2 (1) - (2)$ gives $y = (16 - 6t) - (11 - 4t) = 5 - 2t$, and so $x = (8 - 3t) - 2(5 - 2t) = -2 + t$.
We have to check that these lines are not skew. This is a sufficient condition for two lines in 3D space to be coplanar, as two lines can at most span $\mathbb R^2$ (imagine rotating a plane around one line until it intersects the other line).
The other line can be rewritten parametrically as $(\lambda - 1, 2 \lambda - 1, 3 \lambda - 1)$. Thus we have:
$$
\left\{
\begin{array}{c}
t - 2 = \lambda - 1 \\
-2t + 5 = 2 \lambda - 1 \\
t = 3 \lambda - 1 \\
\end{array}
\right.
$$
Substituting the third into the first, we have $(3 \lambda - 1) - 2 = \lambda - 1 \implies 3 \lambda - 3 = \lambda - 1$, thus $\lambda = 1$ and $t = 2$. Now we have to verify that $-2 \cdot 2 + 5 = 2 \cdot1 - 1$. This is true, so the lines are coplanar.
The equation of the plane is given by the normal vector to the direction vectors of the lines. This means we have to compute $(1, -2, 1)^T \times (1, 2, 3)^T$, which is $(-6-2, 1-3, 2--2)^T$ or $(-8, -2, 4)^T$. A multiple of this normal vector is $(4, 1, -2)^T$, so we need:
$$\begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} $$
or $4x + y - 2z = -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
PMF of throwing a die 4 times We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
*
*What is the probability that $X \ge 4$?
*Compute the PMF of $X$.
*Determine the mean and variance of $X$.
My attempt:
*
*$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.
*for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$
*I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?
| 1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.
*
*$1$ one: ${4 \choose 1} (\frac{1}{6})^1(\frac{5}{6})^3$
*$2$ ones: ${4 \choose 2} (\frac{1}{6})^2(\frac{5}{6})^2$
*$3$ ones: ${4 \choose 3} (\frac{1}{6})^3(\frac{5}{6})^1$
*$4$ ones: ${4 \choose 4} (\frac{1}{6})^4(\frac{5}{6})^0 = (\frac{1}{6})^4$
The sum of the above will give you the $P(X=1)$.
Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (\frac{1}{2})^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
computing the adjoint of operator $T$ on the space $P_2(\mathbb{R})$ Suppose that the inner product on $P_2(\mathbb{R})$ is defined by $$\langle f,g \rangle:= f(-1)g(-1)+f(0)g(0)+f(1)g(1).$$
Consider the operator $T \in B(P_2(\mathbb{R}))$ which is defined as $Tf=f'$, so the derivative of $f$. Find the adjoint of $T$.
I am trying to find the adjoint in the following way.
Since $\langle Tf, g\rangle = \langle f, T^*g\rangle$ then we have that
$\langle f, T^*g\rangle = f'(-1)g(-1) + f'(0)g(0) + f'(1)g(1)$. How do I continue now since I don't know how $T^*$ looks like? I appreciate your help.
| Every $f\in P_2$ can be written as
$$
f=f(-1)\frac{1}{2}x(x-1)+f(0)(1-x^2)+f(1)\frac{1}{2}x(x+1)$$
Therefore,
$$
f'(x)=f(-1)(x-\frac{1}{2})-2f(0)x+f(1)(x+\frac{1}{2})
$$
which gives
\begin{align}
f'(-1)&=-\frac{3}{2}f(-1)+2f(0)-\frac{1}{2}f(1) \\
f'(0)&=-\frac{1}{2}f(-1)+\frac{1}{2}f(1) \\
f'(1)&=\frac{1}{2}f(-1)-2f(0)+\frac{3}{2}f(1)
\end{align}
So
$$
\left[\begin{array}{c}(Tf)(-1)\\ (Tf)(0)\\ (Tf)(1)\end{array}\right]
= \left[\begin{array}{ccc}-\frac{3}{2} & 2 & -\frac{1}{2} \\
-\frac{1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & -2 & \frac{3}{2}\end{array}\right]\left[\begin{array}{c} f(-1)\\f(0)\\f(1)\end{array}\right]
$$
Therefore, $T^*$ is represented by the transpose
$$
\left[\begin{array}{c}(T^*f)(-1)\\ (T^*f)(0)\\ (T^*f)(1)\end{array}\right]
= \left[\begin{array}{ccc}-\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} \\
2 & 0 & -2 \\
-\frac{1}{2} & \frac{1}{2} & \frac{3}{2}\end{array}\right]\left[\begin{array}{c} f(-1)\\f(0)\\f(1)\end{array}\right]
$$
Therefore,
\begin{align}
T^*f &= (T^*f)(-1)\frac{1}{2}x(x-1) \\ & +(T^*f)(0)(1-x^2)\\ &+(T^*f)(1)\frac{1}{2}x(x+1) \\
&= \left(-\frac{3}{2}f(-1)-\frac{1}{2}f(0)+\frac{1}{2}f(1)\right)\frac{1}{2}x(x+1) \\
&+2\left(f(-1)-f(1)\right)(1-x^2) \\
&+\left(-\frac{1}{2}f(-1)+\frac{1}{2}f(0)+\frac{3}{2}f(1)\right)\frac{1}{2}x(x-1) \\
&= f(-1)A(x)+f(0)B(x)+f(1)C(x).
\end{align}
I'll let you write $A,B,C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\lim_{n->\infty}\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{3}{n}}}{\sqrt[6]{n^2+\sin{\frac{2}{n}}}-\sqrt[3]{n}}?$ I tried to get rid off cube root as written below but still can not get throught the next steps. What should be the right step to take after the steps below? Did I start as I should or do I have to take completely different approach?
$\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}=$
$=\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}\dfrac{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}$
$=\lim_{n->\infty}\dfrac{1-\cos\dfrac{3}{n}}{(\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n})((\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2)}$
| As said in comments, using Taylor series would make life much easier.
Start dividing all terms by $n^{\frac 13}$ and let $x=\frac 1n$. Now, use the well known Taylor series for the sine and the cosine. Continue with Taylor or with the binomial expansion.
Using three terms in the expansions, you should end with the limit itself.
Adding an extra term in the expansion, you will end with
$$\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{3}{n}}}{\sqrt[6]{n^2+\sin{\frac{2}{n}}}-\sqrt[3]{n}}=\frac{9}{2}-\frac{3}{n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
If you make $n=10$ (quite far away from $\infty$), using you pocket calculator (I made it on my phone), you should get an exact value of $\approx 4.22868$ while the above truncated expression would give $\frac{21}{5}=4.2$.
Making the problem more general as in a now deleted answer, the same process would give
$$\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{a}{n}}}{\sqrt[6]{n^2+\sin{\frac{b}{n}}}-\sqrt[3]{n}}=\frac{a^2}{b}-\frac{2 a^2}{3 b n}+O\left(\frac{1}{n^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\int_{0}^{1} x^2 \sqrt{1+x^2} dx$? I'm trying to calculate the following integral:
$\int_{0}^{1} x^2 \sqrt{1+x^2} dx$
I tried solving by parts but i'm getting nowhere close. I feel like some substitution will be good here, however neither $x=\cos(u)$ nor $x=\sin(u)$ get me anywhere.
| First let $x = \tan u$, we get
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
={\displaystyle\int}\sec^2\left(u\right)\tan^2\left(u\right)\sqrt{\tan^2\left(u\right)+1}\,\mathrm{d}u
\end{equation}
Using the fact that
$\tan^2 u + 1 =\sec^2 u$
we get
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
={\displaystyle\int}\sec^3\left(u\right)\tan^2\left(u\right)\,\mathrm{d}u
={\displaystyle\int}\sec^3\left(u\right)\left(\sec^2\left(u\right)-1\right)\,\mathrm{d}u
=
A_5 - A_3
\end{equation}
Expand to get
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
={\displaystyle\int}\sec^5\left(u\right)\,\mathrm{d}u-{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u
\end{equation}
Lets work with $A_5$, using the reduction formula,
\begin{equation}
\small{{\displaystyle\int}\sec^{n}\left(u\right)\,\mathrm{d}u={{\dfrac{n-2}{n-1}}}\int \sec^{n-2}\left(u\right)\,\mathrm{d}u+\dfrac{\sec^{n-2}\left(u\right)\tan\left(u\right)}{n-1}}
\end{equation}
which gives us
\begin{equation}
A_5 =\dfrac{\sec^3\left(u\right)\tan\left(u\right)}{4}+{{\dfrac{3}{4}}}A_3
\end{equation}
and
\begin{equation}
A_3 =\dfrac{\sec\left(u\right)\tan\left(u\right)}{2}+\frac{1}{2} A_1
\end{equation}
Now $A_1$ is easy and is known to be $A_1 = \ln\left(\tan\left(u\right)+\sec\left(u\right)\right)$. So plugging back upwards we get
\begin{equation}
A_3 =\dfrac{\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{2}+\dfrac{\sec\left(u\right)\tan\left(u\right)}{2}
\end{equation}
and. hence the original integral becomes
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
=-\dfrac{\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{8}+\dfrac{\sec^3\left(u\right)\tan\left(u\right)}{4}-\dfrac{\sec\left(u\right)\tan\left(u\right)}{8}
\end{equation}
Plugging back the initial change of variable and simplifying we get
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
=\dfrac{\sqrt{x^2+1}\left(2x^3+x\right)-\ln\left(\left|\sqrt{x^2+1}+x\right|\right)}{8}+C
\end{equation}
Now using the integration limits, we have in the simplest form
\begin{equation}
\int_0^1 x^2\sqrt{x^2+1}\,\mathrm{d}x
=
-\dfrac{\operatorname{arsinh}\left(1\right)-3\cdot\sqrt{2}}{8}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solution to a differential equation involving inseparable variables. What is the solution for the following DE
$\frac{dy}{dx} - \epsilon{y} = x$
Where $0 \leq x \leq 1$ and initial condition y = 1 when x = 0 and where $\epsilon$ is any positive parameter
| $$\frac{dy}{dx} - \epsilon y = x$$
$$\frac{dy}{dx} = x + \epsilon y$$
Let $v = x + \epsilon y$
$$\frac{dy}{dx} = v$$
Differentiating $v = x + \epsilon y$
$$\frac{dv}{dx} = 1 +\epsilon\frac{dy}{dx}$$
$$\frac{dy}{dx} = \frac{\frac{dv}{dx} - 1}{\epsilon}$$
Substituting $\frac{dy}{dx}$ we have $$ \frac{\frac{dv}{dx} - 1}{\epsilon} = v$$
$$ \frac{dv}{dx} = \epsilon v + 1$$
Dividing by $\epsilon v + 1$ and multiplying by $dx$
$$\frac{1}{\epsilon v + 1} {dv} = dx$$
Integrating both sides
$$\int\frac{1}{\epsilon v + 1} {dv} = \int dx$$
$$\frac{1}{\epsilon}ln(\epsilon v + 1) = x + C $$
$$ln(\epsilon v + 1) = \epsilon (x + C) $$
$$ln(\epsilon v + 1) = \epsilon x + \epsilon C $$
$\epsilon C$ is just another constant (C)
$$ln(\epsilon v + 1) = \epsilon x + C $$
Raising both sides to the power $e$
$$e^{ln(\epsilon v + 1)} = e^{\epsilon x + C}$$
$$e^{ln(\epsilon v + 1)} = e^{\epsilon x} e^C$$
$e^C$ is just another constant $C$.
$$e^{ln(\epsilon v + 1)} = Ce^{\epsilon x} $$
$$\epsilon v + 1 = Ce^{\epsilon x} $$
Substiuting $v = x + \epsilon y$
$$\epsilon(x + \epsilon y) + 1 = Ce^{\epsilon x}$$
$$\epsilon x + \epsilon^2 y + 1 = Ce^{\epsilon x} $$
$$\epsilon x + \epsilon^2 y = Ce^{\epsilon x} - 1$$
$$\epsilon^2 y = Ce^{\epsilon x} - {\epsilon x} - 1$$
Generic solution
$$y = C\frac{e^{\epsilon x}}{\epsilon^2 } - \frac{x}{\epsilon} - \frac{1}{\epsilon^2} $$
Using the initial condition $y= 1$ when $x = 0$ we calculate for $C$
$$C\frac{e^0}{\epsilon^2 } - \frac{0}{\epsilon} - \frac{1}{\epsilon^2} = 1$$
$$\frac{C - 1}{\epsilon^2 } = 1$$
$$C= 1 + {\epsilon^2 }$$
By substituting C the exact solution is
$$y = (1 + \frac{1}{\epsilon^2 })e^{\epsilon x} - \frac{x}{\epsilon} - \frac{1}{\epsilon^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
question about calculus I want to find the difference in volume between two balls, one with radius of x while the other with radius of $x+dx$. I find two answers, $\frac{4}{3}\pi(x+dx)^3-\frac{4}{3}\pi x^3$ and $4πx^2 dx$. Please tell me what I did wrong. Thanks.
Pardon my bad English.
| Your answers are the same if we treat higher powers of $dx$ as being equal to $0$.
$$\begin{align}
\frac{4}{3}\pi(x+dx)^3-\frac{4}{3}\pi x^3
&= \frac{4}{3}\pi(x^3+3x^2dx + 3x dx^2 + dx^3)-\frac{4}{3}\pi x^3 \\
&= \frac{4}{3}\pi(x^3+3x^2dx)-\frac{4}{3}\pi x^3 \\
&=\frac{4}{3}\pi x^3+4 \pi x^2dx-\frac{4}{3}\pi x^3 \\
&= 4 \pi x^2dx \\
\end{align}$$
Essentially, you're using the dual numbers approach to nonstandard analysis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.
HINT: The answer is $-2\pm \sqrt5$.
What I have tried so far:
$$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$
$$\tan{\frac{x}{2}} = \frac{1-\sin^2 x}{\sin x}$$
$$\tan{\frac{x}{2}} = \frac{\cos^2 x}{\sin x}$$
I don't know how to solve this problem. Pls help. Thank you :)
| Hint: Use that $$\frac{1}{\sin(x)}-\sin(x)=1/2\,{\frac {1+ \left( \tan \left( x/2 \right) \right) ^{2}}{\tan
\left( x/2 \right) }}-2\,{\frac {\tan \left( x/2 \right) }{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Sum of square of binomial coeffcient with positive and negative terms Finding $\displaystyle \binom{2n}{1}^2-2\binom{2n}{2}^2+3\binom{2n}{3}^2-\cdots \cdots -2n\binom{2n}{2n}^2.$
What I've tried:
$$(1-x)^{2n}=\binom{2n}{0}-\binom{2n}{1}x+\binom{2n}{2}x^2+\cdots \cdots +\binom{2n}{2n}x^{2n}$$
$$-2n(1-x)^{2n-1}=-\binom{2n}{1}+2\binom{2n}{2}x-3\binom{2n}{3}x^2+\cdots +n\binom{2n}{2n}x^{2n-1}$$
Sum notation:
$$\sum_{k=0}^{2n} (-1)^{k-1}k\binom{2n}{k}^2$$
| Starting from
$$\sum_{k=0}^{2n} (-1)^{k-1} k {2n\choose k}^2
= \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
k {2n\choose k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose k-1}
= 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose 2n-k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
[z^{2n-k}] (1+z)^{2n-1}
\\ = 2n [z^{2n}] (1+z)^{2n-1}
\sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} z^k
\\ = 2n [z^{2n}] (1+z)^{2n-1}
\left(1+\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k\right).$$
Now $2n [z^{2n}] (1+z)^{2n-1}$ is zero, so we may continue with
$$2n [z^{2n}] (1+z)^{2n-1}
\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k
\\ = - 2n [z^{2n}] (1+z)^{2n-1} (1-z)^{2n}
= - 2n [z^{2n}] (1-z^2)^{2n-1} (1-z)
\\ = - 2n [z^{2n}] (1-z^2)^{2n-1}
= - 2n [z^{n}] (1-z)^{2n-1}.$$
This is
$$(-1)^{n+1} \times 2n \times {2n-1\choose n}
= (-1)^{n+1} \times 2n \times {2n\choose n}
\frac{n}{2n}$$
for an answer of
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{n+1} \times n \times {2n\choose n}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Show an identity involving a sum of arctangents of algebraic expressions
Show that
$$
2\tan^{-1}\frac{\sqrt{x^2+a^2} - x + b}{\sqrt{a^2-b^2}} + \tan^{-1}\frac{x\sqrt{a^2-b^2}}{b\sqrt{x^2+a^2} + a^2} + \tan^{-1}\frac{\sqrt{a^2-b^2}}{b} = n\pi .
$$
I tried using $$ x= a \tan \theta ,\; b= a \sin\phi,$$
but then calculations are not working out; that is,
I am not able to further simplify.
| Hints:
*
*Write the equation as $\tan^{-1}\frac{x\sqrt{a^2-b^2}}{b\sqrt{x^2+a^2} + a^2} + \tan^{-1}\frac{\sqrt{a^2-b^2}}{b} = n\pi - 2\tan^{-1}\frac{\sqrt{x^2+a^2} - x + b}{\sqrt{a^2-b^2}} $
*Apply $\tan$ on both sides and use $\tan (n \pi - 2\alpha) = -\tan 2\alpha$
*Resolve the tangents on the left using $\tan \beta + \tan \gamma =\frac{\tan \beta + \tan \gamma}{1- \tan \beta \tan \gamma}$ and on the right using $\tan(2\alpha) = \frac{2\tan\alpha}{1-\tan^2\alpha}$
*Now you are left with an equation consisting of only algebraic fractions in the variables $a,b,x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration, get stuck at x=tan($\theta$) when calculating arc length I am learning to calculate the arc length by reading a textbook, and there is a question
However, I get stuck at calculating
$$\int^{\arctan{\sqrt15}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta$$ How can I continue calculating it?
Update 1:
$$\int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta = \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} (\csc{(\theta)} + \sec{(\theta)} \tan{(\theta)}) d\theta \\
= \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta) d\theta + \frac{1}{\cos{(\theta)}}} |^{arctan{\sqrt{15}}}_{arctan{\sqrt3}} \\
= \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta) d\theta + \frac{1}{\cos{(\sqrt{15})}} - \frac{1}{\cos{(\sqrt3)}}}$$
But how can I get the final result?
Update 2:
Because $\frac{1}{\cos{(x)}} = \sqrt{ \frac{\cos^2{(x)} + \sin^2{(x)}}{cos^2{(x)}}} = \sqrt{1+\tan^2{(x)}}$, I get
$$\frac{1}{\cos{(\sqrt{15})}} - \frac{1}{\cos{(\sqrt3)}} = \sqrt{1+15} - \sqrt{1+3} = 2$$
However, for the first part $\int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta)} d\theta$, I get
$$ \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta)} d\theta = \log \tan{\frac{\theta}{2}} |^{arctan{\sqrt{15}}}_{arctan{\sqrt3}}$$
How can I continue it?
| Think of the identities that you have available:
$$\begin{align}\cos^2\theta+\sin^2\theta&=1\\
1+\tan^2\theta&=\sec^2\theta\\
\cot^2\theta+1&=\csc^2\theta\\
\cosh^2\theta-\sinh^2\theta&=1\\
1-\tanh^2\theta&=\text{sech}^2\,\theta\\
\coth^2\theta-1&=\text{csch}^2\,\theta\end{align}$$
Of the $4$ that give you a formula for $1+x^2$, the one that seems to work best in the present context is the last. Accordingly we let $x=\text{csch}\,\theta$, $\sqrt{1+x^2}=\coth\theta$, $dx=-\text{csch}\,\theta\coth\theta\,d\theta$, so
$$\begin{align}\int\frac{\sqrt{1+x^2}}xdx&=-\int\coth^2\theta\,d\theta=-\int\left(1+\text{csch}^2\,\theta\right)d\theta=-\theta+\coth\theta+C\\
&=-\sinh^{-1}\left(\frac1x\right)+\sqrt{1+x^2}+C\\
&=-\ln\left(\frac1x+\sqrt{1+\frac1{x^2}}\right)+\sqrt{1+x^2}+C\\
&=\ln x-\ln\left(1+\sqrt{1+x^2}\right)+\sqrt{1+x^2}+C\end{align}$$
So
$$\int_{\sqrt3}^{\sqrt{15}}\frac{\sqrt{1+x^2}}xdx=\ln\sqrt{15}-\ln5+4-\ln{\sqrt3}+\ln3-2$$
As promised in the original question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Ellipse equations paradox We know that an ellipse can be plotted in cartesian coordinates using the following parametric function:
$$ ellipsePoint(\theta)=\left[\begin{array}{c}a \cdot \cos(\theta)\\b \cdot \sin(\theta)\end{array}\right], 0\leqslant \theta \lt 2\pi $$
I have rewritten it as an equation in polar coordinates:
$$ r=distance(ellipsePoint(\theta), \left[\begin{array}{c}0\\0\end{array}\right])=distance(\left[\begin{array}{c}a \cdot \cos(\theta)\\b \cdot \sin(\theta)\end{array}\right], \left[\begin{array}{c}0\\0\end{array}\right])=\sqrt{a^2 \cdot \cos(\theta)^2+b^2 \cdot \sin(\theta)^2} $$
However, plotting values of the $ ellipsePoint $ function and plotting the solutions of the second equation result in different graphs:
Why is it so? Have I done something incorrectly?
| Switching to polar coordinates,
$$\rho=\sqrt{a^2\cos^2t+b^2\sin^2t},\\\tan\theta=\frac ba\tan t.$$
After elimination of $t$,
$$\rho=\sqrt{a^2\cos^2\left(\arctan\left(\frac ab\tan\theta\right)\right)+b^2\sin^2\left(\arctan\left(\frac ab\tan\theta\right)\right)}
\\=\sqrt{a^2\frac1{\dfrac{a^2}{b^2}\tan^2\theta+1}+b^2\frac{\dfrac{a^2}{b^2}\tan^2\theta}{\dfrac{a^2}{b^2}\tan^2\theta+1}}
\\=\sqrt{\frac{a^2b^2}{a^2\sin^2\theta+b^2\cos^2\theta}}
.$$
Note that with $a=b=r$, this simplifies to $r$, as expected. The values at multiples of $\dfrac\pi2$ are also correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| Recall Taylor series : $$\sin^2 y =y^2+O(y^3) ,\ \ln\
(1+x^2)=x^2+O(x^3)
$$ so that there is $\delta_i$ s.t. $$ 1-\epsilon < \frac{y^2}{\sin^2y},\ \frac{x^2}{\ln\ (1+x^2)}<1+\epsilon $$ for $|y|<\delta_1$ and $|x|<\delta_2$ \begin{align*}\frac{x^2+y^2}{\sin^2y +\ln\ (1+x^2)} &=
\frac{x^2+y^2}{y^2 + O(y^3) + x^2 +O(x^3) } \\&<
\frac{x^2+y^2}{(1-\epsilon) x^2 + (1-\epsilon)y^2} \\&=
\frac{1}{1-\epsilon} \end{align*} for $0<\epsilon <1$, where $0<|x|,\ |y|<1\ \ast$
are small.
Similarly, we have $\frac{1}{1+\epsilon}< \frac{x^2+y^2}{\sin^2y
+\ln\ (1+x^2)}$. In $\ast$, consider the case $x=0$ or $y=0$ so that
we have the limit $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How many square matrices are there with given columns, rows and diagonals Suppose we have $n\times n$ matrix that contains numbers from $1$ to $n^2$.
How many matrices are there that their $n$ columns, $n$ rows and $2n$ diagonals contain given numbers?
For example $3\times 3$ matrix:
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{pmatrix}
Its rows are $(1,2,3),(4,5,6),(7,8,9)$.
Its columns are $(1,4,7),(2,5,8),(3,6,9)$.
Its diagonals are $(1,5,9),(2,6,7),(3,4,8),(1,6,8),(2,4,9),(3,5,7)$.
If we join them all together sorted inside and sorted overall we get:
$$\big((1,2,3),(1,4,7),(1,5,9),(1,6,8),(2,4,9),(2,5,8),(2,6,7),$$$$(3,4,8),(3,5,7),(3,6,9),(4,5,6),(7,8,9)\big)$$
Question is how many $3\times 3$ matrices there are with the same sorted set?
I brute-forced it for $n=3$. And the number is $432$. But I am not able to figure out a formula for it. The number $432$ is divisible by $9$ which is of course because the original matrix can be shifted to $3\cdot 3=9$ different positions preserving the ordered set. And also by $16$ as we can rotate whole matrix $4$ times by $90$ degrees and there are $4$ mirror images for each one. $432/9/16=3$ so there must be $3$ fundamental matrices from which we can generate all $432$ by shifting, rotating and mirroring. (If I am not mistaken.)
Just one example out of $432$ that has the same sorted set as original matrix:
\begin{pmatrix}
4 & 2 & 9 \\
1 & 8 & 6 \\
7 & 5 & 3 \\
\end{pmatrix}
...and perhaps additional question: How to produce all the required matrices algorithmically one by one?
| I believe the previous answers have sufficiently addressed the $3\times3$ matrix. In this post I will consider the $4\times4$ matrix below.
\begin{pmatrix}
0&1&2&3\\
4&5&6&7\\
8&9&A&B\\
C&D&E&F
\end{pmatrix}
Consider first the groups containing $0$:
$$(0,1,2,3),(0,4,8,C),(0,5,A,F),(0,7,A,D)$$
We see that $A$ appears in two of these groups. As such, there is only one position (relative to the position of $0$) in which $A$ can be placed, as shown here with $0$ in the upper left corner, where it will be fixed for the purpose of counting:
\begin{pmatrix}
0&?&?&?\\
?&?&?&?\\
?&?&A&?\\
?&?&?&?
\end{pmatrix}
Now there are $8$ ways to complete the two diagonals containing both $0$ and $A$:
\begin{matrix}
\begin{pmatrix}
0&?&?&?\\
?&5&?&7\\
?&?&A&?\\
?&D&?&F
\end{pmatrix}
&
\begin{pmatrix}
0&?&?&?\\
?&5&?&D\\
?&?&A&?\\
?&7&?&F
\end{pmatrix}
&
\begin{pmatrix}
0&?&?&?\\
?&7&?&5\\
?&?&A&?\\
?&F&?&D
\end{pmatrix}
&
\begin{pmatrix}
0&?&?&?\\
?&7&?&F\\
?&?&A&?\\
?&5&?&D
\end{pmatrix}
\\[2ex]
\begin{pmatrix}
0&?&?&?\\
?&D&?&5\\
?&?&A&?\\
?&F&?&7
\end{pmatrix}
&
\begin{pmatrix}
0&?&?&?\\
?&D&?&F\\
?&?&A&?\\
?&5&?&7
\end{pmatrix}
&
\begin{pmatrix}
0&?&?&?\\
?&F&?&7\\
?&?&A&?\\
?&D&?&5
\end{pmatrix}
&
\begin{pmatrix}
0&?&?&?\\
?&F&?&D\\
?&?&A&?\\
?&7&?&5
\end{pmatrix}
\end{matrix}
Next we will take one of these and fill in the remaining elements, first noting that the column containing both $D$ and $5$ must also contain both $1$ and $9$, and that $9$ does not appear in any group with $0$, so this column must be $(1,D,9,5)$. Likewise, the column $(?,F,?,7)$ must be $(3,F,B,7)$. The second and fourth rows can also be completed in this manner, and this leads to exactly one solution for each of the eight partial solutions:
\begin{matrix}
\begin{pmatrix}
0&?&?&?\\
?&D&?&F\\
?&?&A&?\\
?&5&?&7
\end{pmatrix}
&\implies&
\begin{pmatrix}
0&1&2&3\\
C&D&E&F\\
8&9&A&B\\
4&5&6&7
\end{pmatrix}
\end{matrix}
With $8$ solutions for each of the $4^2$ possible positions for the $0$, there are a total of $8\cdot4^2=128$ qualifying matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int \frac{dt}{(t^2-1)^2}$ I have this problem of evaluating the indefinite integral of a rational function
$$\int \frac{dt}{(t^2-1)^2}$$
and I'm a bit unsure about how to proceed. I could use partial fractions:
$$\frac{1}{(t+1)(t-1)(t+1)(t-1)}$$
$$\frac{1}{(t+1)^2(t-1)^2}$$
$$\frac{A}{t+1} + \frac{B}{t-1} + \frac{C}{t+1} + \frac{D}{(t+1)^2}$$
and go from there... but that seems long and complicated. Is there a trick I'm missing?
| $$\frac{1}{t^2-1} = \frac{1}{2}\left(\frac{1}{t-1} - \frac{1}{t+1}\right)\tag{1}$$
Square both sides:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} + \frac{1}{(t+1)^2}\right) - \frac{1}{2} \frac{1}{t^2-1}$$
The last term's partial fraction expansion is given by (1), therefore we have:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} - \frac{1}{t-1} +\frac{1}{(t+1)^2}+\frac{1}{t+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
show this inequality $\sum\frac{x}{2+xy}\ge\frac{1}{2}$
Let $x,y,z\ge 0$ such that $$x+y^2+z^3=1.$$
Show that
$$\dfrac{x}{2+xy}+\dfrac{y}{2+yz}+\dfrac{z}{2+zx}\ge\dfrac{1}{2}$$
I try do
$$\sum_{cyc}\dfrac{x}{2+xy}=\sum_{cyc}\dfrac{x^2}{2x+x^2y}\ge\dfrac{(x+y+z)^2}{(2x+2y+2z)+(x^2y+y^2z+z^2x)}$$
it have to prove
$$2(x+y+z)^2\ge (2x+2y+2z)+(x^2y+y^2z+z^2x)$$
it seem this is hold,But I can't prove it
| I don't know if this would help you but here is a hint for you.
$\left \{ \begin{aligned} x, y, z \ge 0\\ x + y^2 + z^3 = 1 \end{aligned} \right. \implies 0 \le x, y, z \le 1 \implies \left\{ \begin{aligned} x = x\\ y \ge y^2\\ z \ge z^3 \end{aligned} \right. \implies x + y + z \ge x + y^2 + z^3 = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$ Let $a,b,c$ be reals with $a+b+c=1$. Show that :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$
I have tried to something like this:
$$18a^4-24a^3+6a^2-12a+12\geq 0$$
$$18b^4-24b^3+6b^2-12b+12\geq 0$$
$$18c^4-24c^3+6c^2-12c+12\geq 0$$
After summing I get :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+24\geq24(a^3+b^3+c^3).$$
If someone has an ideea I would gratefully appreciate.
| Writing your inequality in the form $$\frac{18(a^4+b^4+c^4)}{a+b+c}+6(a^2+b^2+c^2)(a+b+c)+(a+b+c)^3-24(a^3+b^3+c^3)\geq 0$$ so we get that this is equivalent to
$$\frac{\left(a^2-4 a b-4 a c+b^2-4 b
c+c^2\right)^2}{a+b+c}\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$
since $x+y+z = 6$, $z = 6-x-y$, so
\begin{align*}
d &= \sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \\
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
\end{align*}
Find partial derivative $f_x$ and $f_y$ and critical points
\begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \\
&= 24-2y \quad (\text{set }= 0) \\
&= \text{critical point }y = 4 \\
f_y &= 2y + 2(-x-y+12) \\
&= 24 - 2x \quad (\text{set }= 0) \\
&= \text{critical point }x = 12 \\
\end{align*}
Plug in $x = 12$ and $y = 4$ to original equation
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
| Option:
Normal of the plane: $(1,1,1)$,
Normalized(i.e of unit length): $(1/√3)(1,1,1)$
Line through $(8,0,-6)$:
$\vec r= (8,0,-6)+t (1/√3)(1,1,1)$.
Determine $t$ when line intersects plane:
$(8+t/√3)+t/√3+(-6+t/√3)=6;$
$3(t/√3)=4$;
$t= 4/√3.$
Distance = $4/√3$ (Why?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the angle in a quadrilateral
This is the picture, and we are aiming for the angle $x$
It's easy to see that $\angle DGA = \angle CGB = 100°$, $\angle CGD = \angle AGB = 80°$, $\angle CBG = 50°$, but now i'm missing $\angle GBA = ? $ and $\angle GAB = x$
Any hints?
| Let $y$ denote the angle $\angle GBA$. By the law of sines applied to the four triangles with vertex $G$, we obtain the relations
$$\frac{AB}{\sin 80^\circ}=\frac{BG}{\sin x}=\frac{AG}{\sin y}.$$
$$\frac{AD}{\sin 100^\circ}=\frac{AG}{\sin 40^\circ}=\frac{DG}{\sin 40^\circ},$$
$$\frac{BC}{\sin 100^\circ}=\frac{BG}{\sin 30^\circ}=\frac{CG}{\sin 50^\circ},$$
$$\frac{CD}{\sin 50^\circ}=\frac{DG}{\sin 70^\circ}=\frac{CG}{\sin 30^\circ}.$$
Dividing the fourth equation by the third one, we get (cancelling out $CG$)
$$\frac{\sin^2 30^\circ}{\sin 50^\circ\sin 70^\circ}=\frac{BG}{DG}.$$
Now, from the second, we get that $DG=AG$, so
$$\frac{\sin^2 30^\circ}{\sin 50^\circ\sin 70^\circ}=\frac{BG}{AG}.$$
Now, we apply the law of cosines to get
$$AB^2=BG^2+AG^2-2BG\cdot AG \cos 80^\circ$$
Hence, by the first equation we gave by the law of sines
$$\frac{1}{\sin^2x} =\frac{1}{\sin^2 80^\circ}\frac{AB^2}{BG^2}=\frac{1}{\sin^2 80^\circ}\left(1+\frac{AG^2}{BG^2}-2\frac{AG}{BG} \cos 80^\circ\right)=\frac{1}{\sin^2 80^\circ}\left(1+\left(\frac{\sin 50^\circ\sin 70^\circ}{\sin^2 30^\circ}\right)^2-2\frac{\sin 50^\circ\sin 70^\circ}{\sin^2 30^\circ} \cos 80^\circ\right)$$
Easily you can check that $x=20^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Functional equation to determine $f(-1)$
There's a function $f$ satisfying:
$$f\left(\frac{1}{1−x}\right)+2\cdot f\left(\frac{x−1}{x}\right)=3x$$
Find the value of $f(−1)$.
I have no idea how to solve this one. If anyone could help me to understand it, I would be grateful.
| Substitute $1-\frac{1}{x}$ instead of $x$.
We obtain:
$$f\left(\frac{1}{1-1+\frac{1}{x}}\right)+2f\left(1-\frac{1}{1-\frac{1}{x}}\right)=3\left(1-\frac{1}{x}\right)$$ or
$$f(x)+2f\left(\frac{1}{1-x}\right)=3\left(1-\frac{1}{x}\right).$$
Now, substitute $1-\frac{1}{x}$ instead of $x$ in the last equation.
We obtain:
$$f\left(1-\frac{1}{x}\right)+2f\left(\frac{1}{1-1+\frac{1}{x}}\right)=3\left(1-\frac{1}{1-\frac{1}{x}}\right)$$ or
$$f\left(1-\frac{1}{x}\right)+2f(x)=\frac{3}{1-x}.$$
Id est,
$$f(x)=3\left(1-\frac{1}{x}\right)-2f\left(\frac{1}{1-x}\right)=\frac{3(x-1)}{x}-2\left(3x-2f\left(1-\frac{1}{x}\right)\right)=$$
$$=\frac{3(x-1)}{x}-2\left(3x-2\left(\frac{3}{1-x}-2f(x)\right)\right),$$
which gives
$$9f(x)=\frac{3(x-1)}{x}-6x+\frac{12}{1-x}$$ or
$$f(x)=\frac{x-1}{3x}-\frac{2x}{3}+\frac{4}{3(1-x)}$$ and $$f(-1)=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the integral $\int \frac{(\ln(x))^2}{x^3} \, dx$ $$\int \frac{(\ln(x))^2}{x^3} \, dx $$
Starting off with Integration by Parts
$$
\begin{align}
u = \ln(x)^2 &~~~ dv = x^{-3} \\\\
du = 2\ln(x)dx &~~~ v = \frac{x^{-2}}{-2}
\end{align}
$$
$$
\begin{align}
\int \frac{(\ln(x))^2}{x^3} &= (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\frac{2}{2} \int \frac{\ln(x)}{x^2}
\end{align}
$$
Integration by parts again...
$$
\begin{align}
u = \ln(x) &\hspace{10mm} dv = \frac{1}{x^2} \\\\
du = \frac{1}{x}dx &\hspace{10mm} v = -\frac{1}{x} \\\\
-\frac{2}{2} \int \frac{\ln(x)}{x^2} &= -\ln(x) \left(\frac{1}{x} \right) + \int \frac{1}{x^2}dx \\\\
\end{align}
$$
Integration by parts several times $\int \frac{1}{x^2}dx \rightarrow \int \frac{1}{x} \rightarrow \ln(x) + C$
Combining everything together, I get
$$
\int \frac{(\ln(x))^2}{x^3} = (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\ln(x) \left(\frac{1}{x} \right) - \ln(x) + C\\\\
= -\frac{(\ln(x))^2}{2x^2} - \frac{\ln(x)}{x} - \ln(x) + C
$$
Is there a shorter way I could have done this? I think I got the right answer, but I am not really sure either. Checking my answer via differentiate doesn't seem like a feasible test strategy and even without time constraints still seems too complex for my level right now (but then maybe this is why I need the practice)
| Let's prove by induction $$\displaystyle I_p(n)=\int\dfrac{\ln(x)^n}{x^p}\mathop{dx}=\dfrac{P_n(\ln(x))}{x^{p-1}}$$ where $P_n$ is a polynomial of degree $n$.
We will assume in the following $p>1$, so we can carry on integration by parts.
$\displaystyle I_p(0)=\int \dfrac{\mathop{dx}}{x^p}=\dfrac{1-p}{x^{p-1}}$ thus $P_0(x)=1-p$ is a polynomial of degree $0$
$\displaystyle I_p(n+1)=\int\dfrac{\ln(x)^{n+1}}{x^p}\mathop{dx}=\left[\ln(x)^{n+1}\times\dfrac{1-p}{x^{p-1}}\right]-\int \dfrac{(n+1)\ln(x)^n}{x}\dfrac {1-p}{x^{p-1}}\mathop{dx}=\alpha\dfrac{\ln(x)^{n+1}}{x^{p-1}}+\beta I_p(n)$
By induction hypothesis $P_{n+1}(x)=\alpha x^{n+1}+\beta P_n(x)$ is a polynomial of degree $n+1$ and the induction is verified.
Of course we could have calculated the exact coefficients, but it makes it harder to remember the formula. In fact we were just interested in the general form of the result.
In our case we have $\displaystyle I_3(2)=\int\dfrac{\ln(x)^2}{x^3}\mathop{dx}=\dfrac{a\ln(x)^2+b\ln(x)+c}{x^2}$
Derivate it and identify the coefficients : $x^3\times {I_3}'(2)=-2a\ln(x)^2-(2b-2a)\ln(x)-(2c-b)\iff\begin{cases}-2a=1\\2b-2a=0\\2c-b=0\end{cases}\iff \begin{cases}a=-\frac 12\\b=-\frac 12\\c=-\frac 14\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
3-variable symmetric inequality Given $a,b,c>0$ satisfying $a^2+b^2+c^2=3$. Prove that
$$2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+3(a+b+c)\geq 15.$$
I've tried to use the inequality $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{9}{a+b+c}$ as well as AM-HM and the condition $a+b+c \leq 3$ but it still doesn't work.
| There is also the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, and $abc=w^3$.
Thus, by AM-GM $v\geq w$ and the condition gives $3u^2-2v^2=1,$ which gives $u=\sqrt{\frac{1+2v^2}{3}}.$
Thus, $$\sum_{cyc}\frac{1}{a}=\frac{ab+ac+bc}{abc}=\frac{3v^2}{w^3}\geq\frac{3}{v}.$$
Id est, it's enough to prove that
$$\frac{2}{v}+3u\geq5$$ or
$$\frac{2}{v}+\sqrt{3(1+2v^2)}\geq5,$$ which is obvious for $\frac{2}{v}\geq5$ or $5v-2\leq0.$
But for $5v-2>0$ we need to prove that
$$v\sqrt{3(1+2v^2)}\geq5u-2$$ or after squaring of the both sides
$$(v-1)^2(3v^2+6v-2)\geq0,$$
which is true for $5v-2>0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find some digits of $17!$
$17!$ is equal to $$35568x428096y00$$
Both $x$ and $y$, are digits. Find $x$ and $y$.
So, $$17!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13\times 17=(2^3\times 5^3)\times 2^{12}\times 3^6\times 7^2\times 11\times 13\times 17$$
If there`s a product of $(2\times 5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
| HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$. I am new to multivariable limits and I am having a hard time arguing this limit:
Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$.
I can work out the overarching context of the proof as follows: Let $\epsilon>0$ be given. We need a $\delta$ so that whenever $\|(x,y)-(2,1)\|<\delta$ we have $\|f(x,y)-f(2,1)\|<\varepsilon.$ The difficulty I always have with these problems is that I know I need to rewrite $\|\sqrt{x+y-2}-1\|$ in terms of the given $\delta$ estimate but for some reason when it comes to these limit problems algebra turns into the most difficult thing in the world for me.
| Break down the problem
This function is a compositition of the multivariate affine function $(x, y) \mapsto x + y - 2$, and the univariate square root function $x \mapsto \sqrt{x}$. Both of these functions are continuous on their domains. Affine functions are very straightforward functions, and are easier to prove continuous. Try first proving the continuity of this function. You may find it helpful to remember the triangle inequality and that
\begin{align*}
|x| &= \sqrt{x^2} \le \sqrt{x^2 + y^2} = \|(x, y)\| \\
|y| &= \sqrt{y^2} \le \sqrt{x^2 + y^2} = \|(x, y)\|.
\end{align*}
The square root function is a little more complicated, but it is a single-variable function. Think about how to prove this function is continuous at $x = 1$ (or any $x > 0$). It involves some rationalisation, but hopefully if you're comfortable with $\varepsilon$-$\delta$ proofs in one variable, it shouldn't be an issue.
Then, putting it together, composing two continuous functions produces a continuous function. Think about the proof of this; essentially the $\delta$ in one function becomes the $\varepsilon$ in the other.
Figure out the steps
As with most limit questions, your proof will be figured out in reverse order to the way it should be written. People often find it's easier to start with the conclusion $|f(x, y) - f(x_0, y_0)| < \varepsilon$, working backwards, until they reach $\|(x, y) - (x_0, y_0)\| < \delta$, for some positive number $\delta$. Remember, unlike in many other mathematics proofs, it's important that each step implies the previous step, not the next step. We want a full proof to start from the assumption $\|(x, y) - (x_0, y_0)\| < \delta$ and reach the conclusion $|f(x, y) - f(x_0, y_0)| < \varepsilon$, in other words, to be presented in the reverse order to how it was written.
So, let's begin with $|\sqrt{x + y - 2} - 1| < \varepsilon$. Note how this is reminiscent of the (univariate) square root function continuity proof. At this point, we may want to perform the same rationalisation trick. We have
\begin{align*}
|\sqrt{x + y - 2} - 1| &= |\sqrt{x + y - 2} - 1| \cdot \frac{|\sqrt{x + y - 2} + 1|}{|\sqrt{x + y - 2} + 1|} \\
&= \frac{\left|(\sqrt{x + y - 2})^2 - 1^2\right|}{\sqrt{x + y - 2} + 1} \\
&= \frac{\left|x + y - 3\right|}{\sqrt{x + y - 2} + 1}.
\end{align*}
We are trying to make this expression small. By looking at the numerator, this is exactly $|x + y - 2 - 1|$, which is the expression you'd be making $\varepsilon$ small in the proof that $(x, y) \mapsto x + y - 2$ is continuous at $(2, 1)$. So, we shouldn't have any trouble making this numerator small.
The only thing we need to ensure is that our denominator doesn't become small in the process too. Fortunately, our denominator is $1$ plus a positive number; the denominator is at least $1$ (provided the expression is defined). So, for all such $x, y$, we have
$$|\sqrt{x + y - 2} - 1| = \frac{\left|x + y - 3\right|}{\sqrt{x + y - 2} + 1} \le \left|x + y - 3\right|,$$
hence if we ensure $\left|x + y - 3\right| < \varepsilon$, then we have what we need.
There was the caveat that $\sqrt{x + y - 2}$ must be well-defined. This shouldn't be overlooked. If our $\delta$ is too large, then certain $(x, y)$ satisfying $\|(x, y) - (2, 1)\| < \delta$ may make $f(x, y)$ undefined. For example, if we had $\delta = 3$, then $\|(0, 1) - (2, 1)\| = 2 < \delta$, but $f(0, 1)$ is not defined. All we need to do is put a hard limit on how large $\delta$ can be, so that we are absolutely sure that none of these undefined points sneak in. We can see that $\delta \le \frac{1}{2}$ works, as
\begin{align*}
\|(x, y) - (2, 1)\| < \frac{1}{2} &\implies \begin{Bmatrix}|x - 2| < \frac{1}{2} \implies x > \frac{3}{2} \\ |y - 1| < \frac{1}{2} \implies y > \frac{1}{2} \end{Bmatrix} \\
&\implies x + y - 2 > \frac{3}{2} + \frac{1}{2} - 2 = 0.
\end{align*}
(There are larger bounds on $\delta$ we could use, but $\frac{1}{2}$ works, so let's press on.)
Finally, we look at making $|x + y - 3| < \varepsilon$. If you could prove $(x, y) \mapsto x + y - 2$ is continuous, there should be no problem here. If you couldn't, apply triangle inequality on $\Bbb{R}$ (triangle inequality is a really, really useful tool in multivariate limits):
$$|x + y - 3| = |(x - 2) + (y - 1)| \le |x - 2| + |y - 1|.$$
As above, we have $|x - 2|, |y - 1| \le \|(x, y) - (2, 1)\|$, so if we force $\|(x, y) - (2, 1)\| < \frac{\varepsilon}{2}$, we should be ok. So, we choose $\delta = \frac{\varepsilon}{2}$ provided $\varepsilon$ is small enough, but never let it be more than $\frac{1}{2}$. Our choice of $\delta$ is therefore
$$\delta = \min\left\{\frac{1}{2}, \frac{\varepsilon}{2}\right\}.$$
Compile the final proof
(We now write up all our steps in proper order. This bit is the only bit you need to write up when showing the proof to other people.)
Suppose $\varepsilon > 0$, and
$$\|(x, y) - (2, 1)\| < \min\left\{\frac{1}{2}, \frac{\varepsilon}{2}\right\}.$$
Note that this implies that
\begin{align*}
\|(x, y) - (2, 1)\| < \frac{1}{2} &\implies \begin{Bmatrix}|x - 2| < \frac{1}{2} \implies x > \frac{3}{2} \\ |y - 1| < \frac{1}{2} \implies y > \frac{1}{2} \end{Bmatrix} \\
&\implies x + y - 2 > \frac{3}{2} + \frac{1}{2} - 2 = 0,
\end{align*}
hence $(x, y) \in \operatorname{Domain}(f)$. Further,
$$\|(x, y) - (2, 1)\| < \frac{\varepsilon}{2},$$
and so
$$|x + y - 3| \le |x - 2| + |y - 1| \le 2\|(x, y) - (2, 1)\| < \varepsilon.$$
Therefore,
$$|\sqrt{x + y - 2} - 1| = \frac{|x + y - 3|}{\sqrt{x + y - 2} + 1} \le |x + y - 3| < \varepsilon.$$
Hence, $\lim_{(x, y) \to (2, 1)} \sqrt{x + y - 2} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Inverse image of an open set with a function Let $f: (\mathbb{R},|.|)\to (\mathbb{R},|.|)$ defined by $f(x)=\dfrac{1}{1+x^2}$
I want to find $f^{-1}(]a,b[)$
What I do:
$$f^{-1}(]a,b[)=\{y\in \mathbb{R}, \frac{1}{1+y^2}\in ]a,b[\}$$
that is : $$a<\frac{1}{1+y^2}<b$$
*
*If $a>0$ then $$\frac1b-1<y^2<\frac1a-1$$
after that we must suppose two sases $\frac1b\leq 1 $ or $\frac1b>1$
Is there an other method ?
i suppose that $a<b$
$$
f^{-1}(]a,b[)=
\begin{cases}
\emptyset,~ b<0\\
]-\infty,-\sqrt{\frac1b-1}[\cup]\sqrt{\frac1b-1},+\infty[,~ a\leq0 ~\text{and}~b<1\\
\mathbb{R},~ b>1,a\leq0\\
\mathbb{R}\setminus{0},~ b=1,a\leq0\\
\emptyset,~b\geq1,a>1\\
]-\sqrt{\frac1a-1}, \sqrt{\frac1a-1}[,~b\geq1,a<1\\
]\sqrt{\frac1b-1},\sqrt{\frac1a-1}[,~ b<1, a>0
\end{cases}
$$
Thank you
| I think a good way to approach the problem is to think first about the function $f$:
First of all, since $x^{2} \geq 0$ then $1+x^{2} \geq 1$ which leads to $ f(x) = \frac{1}{1+x^{2}} \leq 1$. This means that if $a > 1$ then $f^{-1}(]a,b[) = \emptyset$.
We also have that $1+x^{2} \geq 1 > 0$ so $f(x) = \frac{1}{1+x^{2}} >0$. This means that if $b \leq 0$ then we also have $f^{-1}(]a,b[) = \emptyset$.
Furthermore, $\forall a,a' \in ]\infty, 0]$ and $b>0$ we have $f^{-1}(]a,b[) = f^{-1}(]a',b[)$ and $\forall a < 1$ and $b \in ]1, \infty]$ we have $f^{-1}(]a,b[) = f^{-1}(]a,b'[)$
With this results, we can then separate the problem in appropriate cases:
$\underline{\text{Case} \,\, 1:}$ $0<a<b \leq 1$
$$a < f(x) < b \,\,\Leftrightarrow \,\, a < \frac{1}{1+x^{2}} < b \,\,\overset{a,b >0}{\Longleftrightarrow}\,\, \frac{1}{b} < 1+x^{2} < \frac{1}{a} \,\,\Leftrightarrow \,\,\frac{1}{b}-1 < x^{2} < \frac{1}{a}-1$$
Now, since $a, b\leq 1$, we have that $\frac{1}{b}-1 \geq 0$ and $\frac{1}{a}-1 \geq 0$ and we can take square roots to have
$$ \frac{1}{b}-1 < x^{2} < \frac{1}{a}-1 \,\, \Leftrightarrow \,\, \sqrt{\frac{1}{b}-1} < |x| < \sqrt{\frac{1}{a}-1}$$
(notice that the fact of the square root being an increasing injective function allows us to take square roots and not changing the grater-than and less-than symbols and to have the equivalence symbol)
We finally have that
$$ x \in f^{-1}(]a,b[) \,\,\Leftrightarrow \,\, x\in \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[ \quad \text{or} \quad x\in \,\, ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[$$
We put now our attention on the limit cases:
$\underline{\text{Case} \,\, 2:}$
$\underline{\text{Subcase} \,\, 1:}$ $a=0$ and $b \leq 1$
$a=0 < \frac{1}{1+x^{2}}$ so the condition $a<f(x)<b$ is equivalent to $f(x)<b$.
$\underline{\text{Subcase} \,\, 2:}$ $a>0$ and $b > 1$
We have $\frac{1}{1+x^{2}} \leq 1<b$ so the condition $a<f(x)<b$ is equivalent to $a<f(x)$.
We can summarize the problem in the next cases:
The empty cases:
$a > 1\, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$
$b \leq 0 \, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$
The whole case:
$a \leq 0$ and $b >1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = \mathbb{R}$
The partial cases:
$a \leq 0$ and $0 < b \leq 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\infty,-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\infty[$
$1 > a > 0$ and $ b > 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},\sqrt{\frac{1}{a}-1}[ $
The restrictive case:
$0<a<b \leq 1$ and $0 < b \leq 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[$
Notice that the good choice of the limit points ($0$ for $a$ and $1$ for $b$) is the key.
The proof may appear very long but if you draw the function (maybe with WolframAlpha) you will understand it much better.
I hope it helps you!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove $\tan1 < \frac{\pi}{2}$? Prove that the equation
$$\sin x \sin({\sin x}) = \frac{\pi}{2} \cos({\sin x})$$
has no real solutions.
Let $t=\sin x$, $-1\leq t\leq 1$. Then the expression above is equvalent to $t\sin t = \frac{\pi}{2} \cos t$. As the function $f(t)=t\sin t - \frac{\pi}{2} \cos t$ is even, and $t=0$ is not a solution, I have to prove that $f(t)$ has no positive roots ($t>0$). So, for the left side $0<t\leq 1$ and $0<\sin t\leq \sin1$, then $t\sin t\leq \sin1$. For the right side $\cos t\geq \cos1$, so $\frac{\pi}{2} \cos t\geq \frac{\pi}{2} \cos1$. The objective is to prove that $\sin1<\frac{\pi}{2} \cos1$, or, equivalently, $\tan1 < \frac{\pi}{2}$.
I don't know how to approach this inequality. The arguments and the values are mixed up.
| This is a Community Wiki repost of one of several deleted answers to a deleted question from last year: Compare $\arcsin (1)$ and $\tan (1)$.
[I have taken this opportunity to incorporate a simplification - but only a slight one - I made later in a comment.]
I would be pleased if we were allowed to revisit the other deleted answers, too.
One can write:
$$
\frac{1}{\sqrt{1 + \tan^21}} = \cos1 = 1 - 2\sin^2\frac{1}{2} >
\frac{17639}{32768} > \frac{17632}{32768} =
\frac{551}{1024},
$$
because
$$
\sin\frac{1}{2} <
\frac{1}{2} - \frac{1}{2^3\cdot3!} + \frac{1}{2^5\cdot5!} =
\frac{1920 - 80 + 1}{3840} < \frac{1845}{3840} = \frac{123}{256}.
$$
On the other hand, using Archimedes's lower bound,
$\pi > 3\tfrac{10}{71}$:
$$
\frac{1}{\sqrt{1 + \left(\sin^{-1}1\right)^2}} =
\frac{1}{\sqrt{1 + \left(\frac{\pi}{2}\right)^2}} <
\frac{1}{\sqrt{1 + \left(\frac{223}{142}\right)^2}} =
\frac{142}{\sqrt{69893}} <
\frac{142}{\sqrt{69696}} =
\frac{142}{264} = \frac{71}{132}.
$$
So, one can prove that $\tan1 < \sin^{-1}1$ by proving that:
$$
\frac{551}{1024} > \frac{71}{132},
$$
which simplifies to $33 \times 551 > 71 \times 256$, that is,
$18183 > 18176$ - which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The inductive proof of $\sum_{k=1}^n \frac k{k+1} \leq n - \frac1{n+1}$ is unclear Prove by induction of $n$
$$\sum_{k=1}^n \frac k{k+1} \leq n - \frac1{n+1}$$
\begin{align}\sum_1^{n+1}\frac k{k+1}&\leq n-\frac 1{n+1}+\frac{n+1}{n+2}\\&=n-\frac 1{n+1}+1-\frac 1{n+2}\\&=(n+1)-\frac{2(n+2)-1}{(n+1)(n+2)}\\&=(n+1)-\frac 2{n+1}+\frac 1{(n+1)(n+2)}\\&\leq (n+1)-\frac 2{n+2}+\frac 1{n+2}=(n+1)-\frac 1{n+2}\end{align}
Now I'm a beginner at induction, and couldn't follow this solution very well.I was hoping someone could help break down the steps and explain them.
Questions
*
*How the inequality works
Wouldn't
$$\sum_1^{n+1}\frac k{k+1}\leq n-\frac 1{n+1}$$
become
$$\sum_1^{n+1}\frac k{k+1}\ +\frac{n+1}{n+2} \leq n-\frac 1{n+1}$$
and then
$$\sum_1^{n+1}\frac k{k+1}\leq n-\frac 1{n+1}-\frac{n+1}{n+2}$$
instead of
$$\sum_1^{n+1}\frac k{k+1}\leq n-\frac 1{n+1}+\frac{n+1}{n+2}$$
*My largest issue with induction, is when the inequalities change like in the first and last step. I don't understand how that works. Any explanation, or good resources to help with my understanding of how the inequality changes when performing induction would be helpful.
| Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
$$\sum_{k = 1}^{n} \frac{k}{k + 1} \leq n - \frac{1}{n + 1},$$
and want to end with the conclusion that $$\sum_{k = 1}^{n + 1} \frac{k}{k + 1} \leq n + 1 - \frac{1}{n + 2} .$$
Here's the argument written out in a bit more detail with commentary on each step.
\begin{align*}
\sum_{k = 1}^{n + 1} \frac{k}{k + 1} & = \frac{n + 1}{n + 2} + \sum_{k = 1}^{n} \frac{k}{k + 1} & \textrm{ (just writing out the sum)} \\
& \leq \frac{n + 1}{n + 2} + n - \frac{1}{n + 1} & \textrm{ (applying the induction hypothesis)} \\
& = 1 - \frac{1}{n + 2} + n - \frac{1}{n + 1} & \textrm{ (rewriting $\frac{n+ 1}{n + 2}$ as $\frac{n + 2 - 1}{n + 2} = 1 - \frac{1}{n + 2}$)} \\
& = n + 1 - \frac{1}{n + 1} - \frac{1}{n + 2} & \textrm{ (regrouping)} \\
& = n + 1 - \frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & \textrm{ (combining fractions)} \\
& = n + 1 - \frac{2(n + 2) - 1}{(n + 1)(n + 2)} & \textrm{ (regrouping the numerator)} \\
& = n + 1 - \frac{2(n + 2)}{(n + 1)(n + 2)} + \frac{1}{(n + 1)(n + 2)} & \textrm{ (breaking the fraction back apart)} \\
& = n + 1 - \frac{2}{n + 1} + \frac{1}{(n + 1)(n + 2)} & \textrm{ (simplifying the fraction)} \\
& \leq n + 1 - \frac{2}{n + 2} + \frac{1}{(n + 1)(n + 2)} & \textrm{ (we slightly modified the second-to-last summand)} \\
& \leq n + 1 - \frac{2}{n + 2} + \frac{1}{n + 2} & \textrm{ (modifying the last summand)} \\
& = n + 1 - \frac{1}{n + 2} & \textrm{ (combining the fractions)} .
\end{align*}
So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please.
| Let $a, b, c\in [0, 1]$ satisfying all the hypothesis and $a=\max\{a, b, c\}>b$, it's clear that this expression must have a maximum. Set
$$
a'=\frac{a+b}2<a\\
b'=\frac{a+b}2>b
$$
then $a'+b'+c=1$ and
$$
a'b'+a'c+b'c=\frac{(a+b)^2}4+ab+bc=ab+bc+ac+\frac{(a-b)^2}{4}\\
2a'b'c=2c\frac{(a+b)^2}{4}=2abc+2c\frac{(a-b)^2}4\\
a'b'+a'c+b'c-2a'b'c=ab+bc+ac-2abc+(1-2c)\frac{(a-b)^2}{4}
$$
If $c\geq \frac{1}{2}$ then $a=c=\frac{1}{2}$ and $b=0$ so $ac=\frac 14$. Otherwise
$$
ab+bc+ac-2abc<a'b'+b'c+a'c-2a'b'c
$$
and the maximum should be $a=b=c=\frac 13$. The proof is concluded observing that
$$
\frac{7}{27}>\frac 14
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Understanding a technique mentioned in 2016 IMO Shortlist The following question is there in the IMO shortlist 2016:
Let $a, b$ and $c$ be positive real numbers such that min $ (ab, bc, ca) > 1$. Prove that
$$((a^2 + 1)(b^2 + 1)(c^2 + 1))^{1/3} \leq \left(\frac{a+b+c}{3}\right)^2 + 1$$
In the shortlist document https://www.imo-official.org/problems/IMO2016SL.pdf, after the solution, in the comments section they have stated that the given inequality can also be obtained by first proving the stated lemma and then by "mixing variables".
The stated lemma is as follows:
For any positive real numbers $x, y$ with $xy\geq 1$, we have $$(x^2 + 1)(y^2 + 1)\leq \left(\left(\frac{x + y}{2}\right)^2 + 1\right)^2$$
What does this mean? Is it any specific technique which might prove to be useful in olympiad problems?
| I understood the question being about the "mixing variables" technique, not about proving the lemma. While I haven't heard the term, the technique itself is familiar, though I could rarely apply it in an olympiad setting.
Using the lemma on $(a,b), (a,c)$ and $(b,c)$ and multiplying the result we get
$$(a^2+1)^2(b^2+1)^2(c^2+1)^2 \le \left(\left(\frac{a+b}2\right)^2+1\right)^2\left(\left(\frac{a+c}2\right)^2+1\right)^2\left(\left(\frac{b+c}2\right)^2+1\right)^2.$$
Notice that the term on the right has the same form as the term on the left. Let's shorten this with
$$f(x,y,z)=(x^2+1)^2(y^2+1)^2(z^2+1)^2.$$
and the above inequality simply becomes
$$f(a,b,c) \le f(\frac{a+b}2, \frac{a+c}2,\frac{b+c}2).$$
Once we've established that $\frac{a+b}2,\frac{a+c}2$ and $\frac{b+c}2$ also fulfill the property that any product of 2 (distinct) of them is at least $1$ (which I'll do below to not disturb the flow of the argument), we can apply the above inequality again, using $a'=\frac{a+b}2,b'=\frac{a+c}2, c'=\frac{b+c}2$:
$$f(a,b,c) \le f(\frac{a+b}2, \frac{a+c}2,\frac{b+c}2) \le f(\frac{b+2a+c}4,\frac{a+2b+c}4,\frac{a+2c+b}4).$$
We can continue to do this, in each step increasing the RHS of the previous inequality by replacing it with function $f$ applied the the pairwise arithmetic mean of the previous arguments.
The important part is that if you take 3 numbers and repeatedly replace them with their pairwise arithmetic mean, you get a series of tuples that converges on their "3-way" arithmetic mean. That can be seen by noticing that the sum of elements for all those tuples stays the same ($a+b+c$) and the difference between highest and lowest element is halved in each step.
So we have a series of inequalities $f(a,b,c) \le f(a_1,b_1,c_1) \le f(a_2,b_2,c_2)\le \ldots$, where $\lim_{n\to\infty}a_n = \lim_{n\to\infty}b_n = \lim_{n\to\infty}c_i = {a+b+c \over 3}$.
Since $f$ is continuous, we finally get
$$f(a,b,c) \le f({a+b+c \over 3},{a+b+c \over 3},{a+b+c \over 3}),$$
which is the statement to prove in the problem, raised to the 6th power.
Now comes the proof omited above that the "min product condition" is 'inherited' from $(a,b,c)$ to $(\frac{a+b}2,\frac{a+c}2,\frac{b+c}2)$. Since the condition is symmtrical, we can assume $a \le b \le c$ w.l.o.g. We know that $ac,ab \ge 1$. The two smallest among $(\frac{a+b}2,\frac{a+c}2,\frac{b+c}2)$ are $\frac{a+b}2$ and $\frac{a+c}2$, and we get
$$\frac{a+b}2\frac{a+c}2=\frac{a^2+ac+ab+bc}4 \ge \frac{a^2+1+1+\frac1{a^2}}4 \ge 1,$$
using $b,c \ge \frac1a$ and the well known $x+\frac1x \ge 2$ for $x=a^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $ with Taylor I have to calculate some limits and try to solve them in use of taylor.
$$
\lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right)
$$
In taylor pattern I have $x_0$ to put, but there $x_0$ is $\infty$ so I want to replace it with something other
$$
y = \frac{1}{x} \\
\lim_{y\to 0^+} \left(\frac{1}{y}-\frac{1}{y^2} \ln (1+y)\right) $$
Let $$ f(y) = \frac{1}{y}-\frac{1}{y^2} \ln (1+y) $$
$$f'(y) = -\frac{1}{y^2} + \left(-\frac{2}{y^3}\ln (1+y) - \frac{y^2}{1+y}\right) $$
but $f'(0)$ does not exists because I have $0$ in denominator.
| You just have to write the Taylor expansion of $\ln \left( 1 + \frac{1}{x} \right)$ when $x$ tends to $+\infty$ :
$$x - x^2 \ln \left( 1 + \frac{1}{x} \right) = x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + o\left( \frac{1}{x^2} \right) \right) = \frac{1}{2} + o(1)$$
So the limit is equal to $\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How obtain $\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$ when integrating $\sqrt{\dfrac{x-1}{x+2}}, x > 1$ The book I'm using states two different answers, one being
$\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$
I have tried two different approaches.
Approach one was to rewrite the integral like this:
$\int \sqrt{\dfrac{x-1}{x+2}} dx = \int \dfrac{x-1}{\sqrt{x^2+x-2}} dx = \dfrac{1}{2}\int \dfrac{2x + 1 - 3}{\sqrt{x^2+x-2}} dx = \\ \dfrac{1}{2}\int \dfrac{2x + 1}{\sqrt{x^2+x-2}} dx - \dfrac{3}{2}\int \dfrac{1}{\sqrt{x^2+x-2}} dx = \\
\dfrac{1}{2}[2\sqrt{x^2+x-2} +C_1] - \dfrac{3}{2} [\ln |x + \frac{1}{2} + \sqrt{x^2+x-2}| + C_2] = \\
\sqrt{x^2+x-2} - \dfrac{3}{2} \ln |x + \frac{1}{2} + \sqrt{x^2+x-2}|+ C$
Where in the last step I used the fact that $x^2+x-2 = (x+\frac{1}{2})^2 - \frac{9}{4}$ together with the rule that $\int \dfrac{1}{\sqrt{x^2 + a}} dx = \ln|x + \sqrt{x^2+a}|$
The result is correct according to the book, but not the form I want to achieve.
Apprach two was to make the substitution $y = \sqrt{\dfrac{x-1}{x+2}} \implies dx = \dfrac{6y}{(y^2-1)^2} dy$
rewriting as
$\int \dfrac{6y^2}{(y^2-1)^2} dy = 6 \int \dfrac{y^2 - 1 + 1}{(y^2-1)^2} dy = 6\int \dfrac{1}{y^2-1} dy + 6 \int \dfrac{1}{(y^2-1)^2} dy$
And using partial fraction decomposition on both integrals. This gave me the answer $\sqrt{x^2+x-2} + \dfrac{3}{2} \ln |x + \frac{1}{2} - \sqrt{x^2+x-2}| + C$
Should I try a different approach or am I maybe missing some step in the approaches I've already tried?
| A slightly different substitution from yours gives the result stated in your book:
$$u=\sqrt{x+2}\implies x=u^2-2\implies\mathrm dx=2u\,\mathrm du$$
$$\begin{align*}
\int\sqrt{\frac{x-1}{x+2}}\,\mathrm dx&=\int\frac{\sqrt{u^2-3}}u(2u\,\mathrm du)\\[1ex]
&=2\int\sqrt{u^2-3}\,\mathrm du\\[1ex]
&=u\sqrt{u^2-3}-3\ln\left|\sqrt{u^2-3}+u\right|+C&(*)
\end{align*}$$
Computing $\int\sqrt{u^2-3}\,\mathrm du$ can be indeed be done by parts:
$$\begin{cases}f=\sqrt{u^2-3}\implies\mathrm df=\dfrac u{\sqrt{u^2-3}}\,\mathrm du\\[1ex]\mathrm dg=\mathrm du\implies g=u\end{cases}$$
$$\int\sqrt{u^2-3}\,\mathrm du=u\sqrt{u^2-3}-\int\frac{u^2}{\sqrt{u^2-3}}\,\mathrm du$$
and the remaining integral can be approached with the substitution $u=\sqrt 3\sec t$. (Then again, you could have used this substitution to compute the antiderivative of $\sqrt{u^2-3}$; up to you.)
The result $(*)$ above follows, and replacing $u=\sqrt{x+2}$ gives the final antiderivative,
$$=\underbrace{\sqrt{x+2}\sqrt{x-1}}_{\sqrt{x^2+x-2}}-3\ln\left|\sqrt{x+1}+\sqrt{x+2}\right|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 + 1$$
$$1\cdot 1000 + (a-11)\cdot 100+(2-a-b)+1-a-b=0$$
So, our number will be something like this: $1(a-11)(2-a-b)(1-a-b)=0$. This is contradiction since the first digit is $1$, and on the right we have $0$.
Does that mean there are no $a$ and $b$ to satisfy this equality?
Alternatively, I did the following: since the last digit in $1a21$ is $1$, it has to be $a+b=L1$. $a$ and $b$ are digits, so there are couple of cases: $a=0$ and $b=1$, $a=1$ and $b=0$, $a=5$ and $b=6$, $a=6$ and $b=5$... None of those cases satisfy what we want, so there are no $a$ and $b$.
Are those two ways to do it the right ones? I don't have any solutions therefore I'm not sure.
Thank you.
| You can also bring in the divisibility test for $11$ into the problem. First note that $700+400=1100$ and $ab+ba=11(a+b)$, forcing your sum $1a21$ to be a multiple of $11$.
By the divisibility test for $11$, then, $1-a+2-1$ must be a multiple if $11$, forcing $a=2$. The ones digit sum then must be $b+2=11(\ge 2)$, thus $b=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3145603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Pattern in power towers of 2 involving last digits We have
\begin{align}
2^{2^{2}} &\mod 10 = 6 \\
2^{2^{2^2}} &\mod 100 = 36 \\
2^{2^{2^{2^2}}} &\mod 1000 = 736 \\
2^{2^{2^{2^{2^{2}}}}} &\mod 10000 = 8736 \\
2^{2^{2^{2^{2^{2^2}}}}} &\mod 100000 = 48736
\end{align}
I think you get the point. Basically, it seems like $^n2 \equiv ^{n+1}2 \mod 10^{n-2}$ for $n \geq 3$, where $^n 2$ represents tetration. How could one go about proving this?
| Let $x_0=4$ and $x_{n+1}=2^{x_n}$.
The claim is $x_{n+1}\equiv x_n\pmod{10^n}$.
By induction on $n$, we assume $x_{n+1}\equiv x_n\pmod{10^n}$ and we prove $x_{n+2}\equiv x_{n+1}\pmod{10^{n+1}}$.
We have
$$x_{n+2}-x_{n+1}=2^{x_{n+1}}-2^{x_n}=2^{x_n}(2^{x_{n+1}-x_n}-1)$$
Since $x_n\geq n+1$ we get $x_{n+2}\equiv x_{n+1}\pmod{2^{n+1}}$.
On the other hand for $n\geq 2$ we have $x_{n+1}\equiv x_n\pmod{4}$ and $x_{n+1}\equiv x_n\pmod{5^n}$ by assumption, so that $x_{n+1}-x_n\equiv 0\pmod{4\cdot 5^n}$.
Since $\varphi(5^{n+1})=4\cdot 5^n$, this gives $2^{x_{n+1}-x_n}\equiv 1\pmod{5^{n+1}}$, thus giving $x_{n+2}\equiv x_{n+1}\pmod{5^{n+1}}$.
This, together with $x_{n+2}\equiv x_{n+1}\pmod{2^{n+1}}$ gives $x_{n+2}\equiv x_{n+1}\pmod{10^{n+1}}$ concluding the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$y-xy' =2(x+yy'), y(1) =1 $ is..
Question: $y-xy' =2(x+yy'), y(1) =1 $ is..
Firstly, I do not even know what the question is asking for.
Secondly, Why does this question put $y(1)=1$? Isn't it obvious or is there any special meaning to it.
Surprisingly, the answer to it is also very unique.
The answer is: $\tan^{-1} \frac{y}{x} + \log (x^2 + y^2) = \frac{\pi}{4} + \log 2$.
Also, another thing is that this question was solved by trigonometry, which is confusing me further.
P.S.: Please do not downvote as I completely do not understand the question even a bit. Thats why I am asking it here.
| How the equation can be solved
The equation can be rewritten as
$$
y' = \frac{y-2x}{2y+x}.
$$
Introduce $u = y/x$ so that $y = xu$ and $y' = xu' + u.$ This makes the equation take the form
$$
xu' + u = \frac{xu-2x}{2xu+x} = \frac{u-2}{2u+1}
$$
i.e.
$$
xu' = \frac{u-2}{2u+1} - u = \frac{(u-2)-u(2u+1)}{2u+1} = -2 \frac{u^2+1}{2u+1}.
$$
or, separated,
$$
\frac{2u+1}{u^2+1}u' = -\frac{2}{x}.
$$
The left hand side can be split into two terms:
$$
\frac{2u+1}{u^2+1}u' = \frac{2uu'}{u^2+1} + \frac{u'}{u^2+1} = \left( \ln(u^2+1) + \arctan(u) \right)'.
$$
Thus,
$$
\left( \ln(u^2+1) + \arctan(u) \right)' = -\frac{2}{x} = \left( -2 \ln(x) \right)' = \left( \ln\frac{1}{x^2} \right)'
$$
so
$$
\ln(u^2+1) + \arctan(u) = \ln\frac{1}{x^2} + C,
$$
which after backsubstitution of $u=y/x$ gives
$$
\ln((y/x)^2+1) + \arctan(y/x) = \ln\frac{1}{x^2} + C
$$
or
$$
\ln(x^2+y^2) + \arctan(y/x) = C.
$$
We now get $C$ from inserting $(x,y)=(1,1)$:
$$
C = \ln(1^2+1^2) + \arctan(1/1) = \ln 2 + \pi/4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that
$$a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$$
I found a point at which the equality is attended, say $a=0,b=1,c=2$. But I have no idea how to prove it. I tried to use the AM-GM inequality but then I obtained the more difficult one. Please help me. Thank you very much.
| Start out as before:
$$
a\sqrt{b^{3}+1}+b\sqrt{c^{3}+1}+c\sqrt{a^{3}+1}\leq5
$$
$$
a\cdot\sqrt{\left(b+1\right)\left(b^{2}-b+1\right)}\leq a\cdot\frac{b^{2}+2}{2}
$$
$$
\frac{a\left(b^{2}+1\right)+b\left(c^{2}+1\right)+c\left(a^{2}+1\right)}{2}=3+\frac{ab^{2}+bc^{2}+ca^{2}}{2}
$$
Continue by lagrange multipliers:
$$
\Lambda=ab^{2}+bc^{2}+ca^{2}+\lambda\left(a+b+c-3\right)
$$
$$
\begin{cases}
\partial_{a}\Lambda=b^{2}+2ac-\lambda & \Rightarrow\lambda=b^{2}+2ac\\
\partial_{b}\Lambda=c^{2}+2ab-\lambda & \Rightarrow\lambda=c^{2}+2ab\\
\partial_{c}\Lambda=a^{2}+2bc-\lambda & \Rightarrow\lambda=a^{2}+2bc
\end{cases}
$$
$$
H=\left(\begin{array}{cccc}
2c & 2b & 2a & 1\\
2b & 2a & 2c & 1\\
2a & 2c & 2b & 1\\
1 & 1 & 1 & 0
\end{array}\right)
$$
$$
b^{2}+2ac=c^{2}+2ab\Rightarrow(b-c)\left(b+c-2a\right)=0
$$
$$
(b-c)\left(b+c-2a\right)=(a-c)\left(a+c-2b\right)=(a-b)\left(a+b-2c\right)=0
$$
If we have no two equal, then we arrive at a contradiction:
$$
b+c-2a=a+c-2b=a+b-2c=0\Rightarrow a=b=c
$$
Hence assume $a-b=0$, then either $a-c=0$ and getting $a=b=c=1$,
or: $a+c-2a=0\Rightarrow a=c$ yielding the same results. Hence the
only candidate where $abc\neq0$ is:
$$
a=b=c=1
$$
Here, we get:
$$
ab^{2}+bc^{2}+ca^{2}=3
$$
Assume $b=0$ if we are on the boundary, then by AM-GM:
$$
ab^{2}+bc^{2}+ca^{2}=ca^{2}=\frac{1}{2}\left(2\left(3-a\right)a^{2}\right)\leq\frac{1}{2}\left(\frac{2\left(3-a\right)+a+a}{3}\right)^{3}=4
$$
Hence we have
$$
\frac{ab^{2}+bc^{2}+ca^{2}}{2}\leq2
$$
As needed (maybe not too detailed why the $a=b=c=1$ is a saddle point). THe surface is shown below:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove this identity nicely?
Show that$$\sum_{1\le i<j\le n}\left((x_j-x_i)-(x_j-x_i)^2\right)=\left(\sum_{i=1}^n{x_i}\right)^2-n\sum_{i=1}^n{x_i^2}-\sum_{i=1}^n{(n-2i+1)x_i}\\=-n\sum_{i=1}^n\left(x_i-\frac1n\sum_{j=1}^n{x_j}+\frac{n-2i+1}{2n}\right)^2+\frac1{4n}{\sum_{i=1}^n(n-2i+1)^2}.$$
I wonder how this identity comes up with the first and second recipes? Can anyone explain this in detail? I got this identity when I was reading the problem here. Thanks.
| $\def\peq{\mathrel{\phantom{=}}{}}$For the first identity, because$$
\sum_{i < j} (x_j - x_i) = \sum_{k = 1}^n x_k \left( \sum_{l < k} 1 + \sum_{l > k} (-1) \right) = \sum_{k = 1}^n (2k - n - 1) x_k,
$$\begin{gather*}
\sum_{i < j} (x_j - x_i)^2 = \sum_{k = 1}^n x_k^2 \left( \sum_{l < k} 1 + \sum_{l > k} 1 \right) - 2 \sum_{i < j} x_i x_j\\
= (n - 1) \sum_{k = 1}^n x_k^2 - \left( \left( \sum_{k = 1}^n x_k \right)^2 - \sum_{k = 1}^n x_k^2 \right) = n \sum_{k = 1}^n x_k^2 - \left( \sum_{k = 1}^n x_k \right)^2,
\end{gather*}
then$$
\sum_{i < j} ((x_j - x_i) - (x_j - x_i)^2) = \left( \sum_{k = 1}^n x_k \right)^2 - n \sum_{k = 1}^n x_k^2 - \sum_{k = 1}^n (n - 2k + 1) x_k.
$$
For the second identity, denoting $\bar{x} = \dfrac{1}{n} \sum\limits_{k = 1}^n x_k$,\begin{align*}
&\peq -n \sum_{k = 1}^n \left( x_k - \bar{x} + \frac{1}{2n} (n - 2k + 1) \right)^2 + \frac{1}{4n} \sum_{k = 1}^n (n - 2k + 1)^2\\
&= -\sum_{k = 1}^n \left( n(x_k - \bar{x})^2 + (n - 2k + 1)(x_k - \bar{x}) + \frac{1}{4n} (n - 2k + 1)^2 \right) + \frac{1}{4n} \sum_{k = 1}^n (n - 2k + 1)^2\\
&= -\sum_{k = 1}^n \left( n(x_k - \bar{x})^2 + (n - 2k + 1)(x_k - \bar{x}) \right)\\
&= -n \sum_{k = 1}^n (x_k - \bar{x})^2 - \sum_{k = 1}^n (n - 2k + 1) x_k + \bar{x} \sum_{k = 1}^n (n - 2k + 1)\\
&= -n \left( \sum_{k = 1}^n x_k^2 - n\bar{x}^2 \right) - \sum_{k = 1}^n (n - 2k + 1) x_k + 0\\
&= \left( \sum_{k = 1}^n x_k \right)^2 - n \sum_{k = 1}^n x_k^2 - \sum_{k = 1}^n (n - 2k + 1) x_k.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Convergence of the series $1+1/2 -1/3 -1/4 -1/5 +1/6 +1/7 - 1/8 -1/9 - 1/10 + 1/11 + 1/12-...$ I would like to prove that $1+1/2 -1/3 -1/4 -1/5 +1/6 +1/7 - 1/8 -1/9 - 1/10 + 1/11 + 1/12 -...$ converges or diverges
I first showed that $1-1/2 -1/3 +1/4+1/5-1/6-1/7+1/8+1/9-1/10...$ converges by writing it as $1+\Sigma(-1)^n(4n+1)/((2n)(2n+1))$
which converges by A.S.T. However I do not see how to do something similar with this series. The sign has a pattern of 5 and thus I don't think it can be written as an alternating series and thus I do not see how it can converge. I was also unable to find a regrouping which obviously diverges.
Any help would be appreciated
| Note that you can group all but the first two terms into groups of five terms of the form
$$
-\frac{1}{5k+3}-\frac{1}{5k+4}-\frac{1}{5k+5}+\frac{1}{5k+6} + \frac{1}{5k+7}
< -\frac{1}{5k+3}
$$
This means that the $5k+7$-th partial sum will be bounded above by
$$
1 + \frac{1}{2} - \frac{1}{3}-\frac{1}{8} - \frac{1}{13} - ... - \frac{1}{5k+3}
$$Hopefully from here you can show that this implies divergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.
Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.
I have been doing similar problems, but can't get my head around this one. I'd rather get a tip instead of the full answer. Thanks to anyone who contributes.
My Solution After Tips:
Let $u = x + y$ and $ v = x - y$. Then $ x = u - y $ and $ y = x - v $. From these equations, we have that
$ x = u - y = u - (x - v) \rightarrow x = u - x + v \rightarrow 2x = u + v \Rightarrow x = (u + v)/2 $.
Similarly, $u - v = (x + y) - (x - y) = x + y - x + y = 2y \Rightarrow y = (u - v)/2$.
Substituting, we get
$$ f(u, v) = (\frac{u + v}{2})(\frac{u - v}{2}) + (\frac{u + v}{2})^2 $$
$$ f(u, v) = \frac{u^2 - uv}{2} $$
| You can solve it directly as follows:
*
*$f(x+y,x-y) = xy+y^2 = (x+y)y$
Now, write
*
*$y = \frac{1}{2}((x+y)-(x-y))$
It follows:
$$f(x+y,x-y)= (x+y)\cdot \frac{1}{2}((x+y)-(x-y)) = \frac{1}{2}\left( (x+y)^2 - (x+y)(x-y)\right)$$ $$\Rightarrow f(x,y) = \frac{1}{2}(x^2-xy)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
| Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^{\prime\prime} + y^\prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
\begin{align}\label{eq:1}\tag{1}
y^\prime(x) = 2Ax + B \quad \text{and} \quad y^{\prime\prime}(x) = 2A.
\end{align}
So, if $y$ solves the differential equation, then
\begin{align}
x^2 = y^{\prime\prime} + y^\prime - 2y &= 2A + (2Ax+B) -2\left(Ax^2 + Bx + C \right)\label{eq:2}\tag{2}\\
&= -2Ax^2 + (2A-2B)x + \left( 2A + B -2C\right)\label{eq:3}\tag{3}.
\end{align}
Comparing the coefficients of this equation, we find that
\begin{align*}
\begin{cases}
-2A = 1,\\
2A-2B = 0,\\
2A+B-2C = 0.
\end{cases}
\end{align*}
Solving this system will give the explicit values of $A,B$ and $C$ required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$
If $a,b,c\in\mathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is
what i try
Cauchy schwarz inequality
$$(a^2+(2b)^2+c^2)(2^2+\frac{1}{2^2}+3^2)\geq (2a+b+3c)^2$$
How do i solve it without Cauchy schwarz inequality Help me please
| Let $(a,b,c)=\left(\frac{160}{53},\frac{20}{53},\frac{240}{53}\right).$
Thus, we get a value $\frac{1600}{53}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$a^2+4b^2+c^2\geq\frac{1600}{53}\cdot\left(\frac{2a+b+3c}{20}\right)^2$$ or
$$37a^2-16(b+3c)a+208b^2-24bc+17c^2\geq0,$$ for which it's enough to prove that
$$64(b+3c)^2-37(208b^2-24bc+17c^2)\leq0,$$ which is
$$(12b-c)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.
Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.
Approach:
*
*I defined a function $f: \mathbb{R} \to \mathbb{R}$ with $$ f(x) = \cosh(x \log{2}) - \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b}.
$$ In order to show the inequality, it suffices to show $f(x) \geq 0$ for all $x \geq 0$.
*I tried to plot the function for some chosen parameters like $a=1$ and $b=2$. In all those cases the function was non-negative, so I suppose that this inequality is true.
*It is $f(0) = 1-1 = 0 \geq 0$.
*Now $f$ is differentiable, so I computed
$$ f'(x) = \frac{\log{2}}{2(a+b)} \big(a (2^x - 2^{-x} - 2^{x+1}) + b(2^x-2^{-x}+2^{-x+1} ) \big). $$
*I would be done if I could show that $f'(x) \geq 0$ for all $x \geq 0$, so $f$ is monotonically increasing and we get our desired result. However, I can not see how this can be shown.
Could you please help me with this problem? That would be nice, thank you in advance!
| Consider the following:
$$2^{x}-2^{-x}-2^{x+1}=2^x-2^{-x}-2\cdot2^x= -2^x-2^{-x}=-2(2^{x-1}+2^{-x-1}) = - 2 \left(\frac{2^x+2^{-x}}{2}\right) $$
Hence
$$2^{x}-2^{-x}-2^{x+1} = - 2 \left(\frac{2^x+2^{-x}}{2}\right) = -2 \cosh(x \log 2)$$
Similarly
$$2^{x}-2^{-x}+2^{1-x}=2^x-2^{-x}+2\cdot2^{-x}= 2^x+2^{-x}=2(2^{x-1}+2^{-x-1}) = 2 \left(\frac{2^x+2^{-x}}{2}\right) $$
Hence
$$2^{x}-2^{-x}+2^{1-x}= 2 \left(\frac{2^x+2^{-x}}{2}\right) = 2 \cosh(x\log 2)$$
So with this the derivative is:
\begin{align}
f'(x)&= \frac{\log 2}{2 (a+b)} \left[-2a \cosh(x \log 2) + 2b \cosh(x \log 2)\right] \\
&= \frac{\log 2}{ a+b} \left[-a \cosh(x \log 2) + b \cosh(x \log 2)\right]\\
&=\log 2\frac{b-a}{a+b} \cosh(x\log2)
\end{align}
with $\log 2 >0$, $\frac{b-a}{a+b}>0$ since $a<b$, and $\cosh(x \log 2)>0$ since $\cosh$ is non-negative.
Therefore
$$f'(x)>0$$
as you're expecting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding a limit of a matrix raised to the $n$-th power There is a stochastic matrix given
$$\mathbb{P} = \begin{bmatrix} 0&\frac{1}{2}&\frac{1}{2}&0 \\ 0&1&0&0 \\ \frac{1}{2}&0&0&\frac{1}{2} \\ \frac{1}{3}&0&\frac{1}{3}&\frac{1}{3} \end{bmatrix}$$
I am to find $\mathbb{P}^n$. I know a theorem which says that if $\exists n$ $[p_{i,j}^n] >0$ then the limit exists and can be easily calculated (of course by $[p_{i,j}^n]$ I mean elements of $\mathbb{P}^n$).
In my case however $[p_{2,1}^n] = [p_{2,3}^n] = [p_{2,4}^n] =0, \forall n$. How can I solve this problem? Is there another theorem?
| We have $P=SDS^{-1}$, where
$$
S=\pmatrix{
-1 & -1 & \frac{9}{16} & 1 \\
0 & 0 & 0 & 1 \\
1 & 0 & \frac{15}{16} & 1 \\
0 & 1 & 1 & 1 \\
}
,\quad
D=\pmatrix{
-\frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & \frac{5}{6} & 0 \\
0 & 0 & 0 & 1 \\
}
$$
Therefore, $P^\infty = S D^\infty S^{-1}$.
Since
$$
D^\infty = \pmatrix{
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
}
$$
we have
$$
P^\infty = \pmatrix{
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction decomposition of even function I need to do partial fraction decomposition of this function (to solve its integral):
$\frac{t^2}{t^4+4}$
Since $t^4+4=(t^2+2t+2)(t^2-2t+2)$ I would do:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}+C\frac{2t-2}{t^2-2t+2}+D\frac{1}{t^2-2t+2}$
And then continue with calculations.
According to the teacher's notes though, I can instead write it like this because the function is even:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}-A\frac{2t-2}{t^2-2t+2}+B\frac{1}{t^2-2t+2}$
But I can't understand how to get it, knowing that the function is even.
Can you please help me?
| Since the function is even we have that
$$A\frac{2t+2}{(t+1)^2}+B\frac{1}{(t+1)^2}+C\frac{2t-2}{(t-1)^2}+D\frac{1}{(t-1)^2}$$ is equal to
$$A\frac{-2t+2}{(-t+1)^2}+B\frac{1}{(-t+1)^2}-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}.$$ Reording terms we get
$$-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}-A\frac{2t-2}{(-t+1)^2}+B\frac{1}{(-t+1)^2}.$$
Comparing the first and third expressions we get $A=-C, B=D.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Bounded recursive sequence - Proof by induction
Given the sequence $(x_n)$ defined by
\begin{cases}
x_1 &= 1\\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right),
\end{cases}
prove that $1 \leq x_n \leq \frac{3}{2}, \forall n \in N$.
I verified the base case for $n=1$ and $n=2$.
Assumed that the boundaries hold for all $k \leq n$. Used the induction hypothesis to show that $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{1}{2} \leq \frac{x_n}{2} \leq \frac{3}{4} $ and $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{2}{3} \leq \frac{1}{x_n} \leq 1$.
Adding term by term I got $1 \leq \frac{7}{6} \leq \frac{1}{2}(x_n + \frac{2}{x_n}) \leq \frac{7}{4} $.
Is my reasoning correct? How could I show that $\frac{3}{2}$ is also an upper boundary, since it is smaller than $\frac{7}{4}$?
PS: I found similar questions for which $x_n \leq \sqrt{2}$.
| For induction step we have to prove
$1\leqslant x\leqslant\frac32\Rightarrow 2\leqslant x+\frac2x\leqslant3$
that is $x^2-3x+2\leqslant0$ and $x^2-2x+2\geqslant0$ for $1\leqslant x\leqslant\frac32$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\lambda$ in $\frac{3-\tan^2 {\pi\over 7}}{1-\tan^2 {\pi\over 7}}=\lambda\cos{\pi\over 7}$
Find the value of $\lambda$ in $$\dfrac{3-\tan^2 {\pi\over 7}}{1-\tan^2 {\pi\over 7}}=\lambda\cos{\pi\over 7}$$
The numerator looks similar to expansion of $\tan 3x$, so I tried this
$$\dfrac{3\tan {\pi\over 7}-\tan^3 {\pi\over 7}}{\tan {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)}=\lambda\cos{\pi\over 7}$$
$$\dfrac{\left(3\tan {\pi\over 7}-\tan^3 {\pi\over 7}\right)\left(1-3\tan^2 {\pi\over 7}\right)}{\tan {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)\left(1-3\tan^2 {\pi\over 7}\right)}=\lambda\cos{\pi\over 7}$$
$$\dfrac{\tan {3\pi\over 7}\left(1-3\tan^2 {\pi\over 7}\right)}{\sin {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)}=\lambda$$
But I'm stuck here. Need help. Thanks in advance.
| Let $7x=\pi$
$$\lambda\cos x=\dfrac{3-\tan^2x}{1-\tan^2x}=\dfrac{3\cos^2x-(1-\cos^2x)}{\cos^2x-(1-\cos^2x)}$$
$$\iff2\lambda\cos^3x-4\cos^2x-\lambda\cos x+1=0$$
From How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$,
$\cos\dfrac{(2n+1)\pi}7; n=0,1,2$ are the roots of $8t^3-4t^2-4t+1=0$
Comparing with the coefficients, $$\dfrac{2\lambda}8=\dfrac44=\dfrac\lambda4=\dfrac11$$
$\implies\lambda=?$
So, we can replace $\dfrac\pi7$ with $\dfrac{3\pi}7,\dfrac{5\pi}7$ in the given expression
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a closed form for coefficients in $x^{3n}=x_0\left(a_nx+b_n+\frac {c_n}{x}\right)$ Consider,
$$
x^3=x+1
$$
Let $x_0$ be a solution to the above equation. Now consider $x^{3n}$. For $n=2$ we have:
$$
x^6=(x+1)^2
$$
$$
=x^2+2x+1
$$
$$
=x\left(x+2+\frac {1}{x}\right)
$$
$$
=x_0\left(x+2+\frac {1}{x}\right)
$$
For $n=3$ we have:
$$
x^9=(x+1)^3
$$
$$
=x^3+3x^2+3x+1
$$
$$
=3x^2+4x+2
$$
$$
=x\left(3x+4+\frac {2}{x}\right)
$$
$$
=x_0\left(3x+4+\frac {2}{x}\right)
$$
Similarly for $n=4$ we have:
$$
x^{12}=x_0\left(7x+9+\frac {5}{x}\right)
$$
In general we have:
$$
x^{3n}=x_0\left(a_nx+b_n+\frac {c_n}{x}\right),
$$
Where,
$$
a_{n+1}=a_n+b_n, a_2=1,
$$
$$
b_{n+1}=a_n+b_n+c_n, b_2=2,
$$
$$
c_{n+1}=a_n+c_n, c_2=1.
$$
My question is: is there a closed form for $a_n,b_n$ and $c_n$?
Any help would be appreciated.
| We derive generating functions for the recurrence relation:
\begin{align*}
a_{n+1}&=a_n+b_n\tag{1}\\
b_{n+1}&=a_n+b_n+c_n\qquad\qquad (n\geq 2)\tag{2}\\
c_{n+1}&=a_n+c_n\tag{3}\\
a_2&=1,b_2=2,c_2=1\\
\end{align*}
Let $A(x)=\sum_{n\geq 2} a_nx^n, B(x)=\sum_{n\geq 2} b_nx^n, C(x)=\sum_{n\geq 2} c_n x^n$.
We obtain from (1)
\begin{align*}
\sum_{n\geq 2}a_{n+1}x^n&=\sum_{n\geq 2}a_nx^n+\sum_{n\geq 2}b_nx^n\\
\frac{1}{x}\sum_{n\geq 3}a_nx^n&=A(x)+B(x)\\
A(x)-x^2&=xA(x)+xB(x)\\
\color{blue}{(1-x)A(x)-xB(x)}&\color{blue}{=x^2}\tag{4}\\
\end{align*}
Since (3) and (1) have the same structure and initial condition, we get
\begin{align*}
\color{blue}{(1-x)C(x)-xA(x)}&\color{blue}{=x^2}\qquad\qquad\qquad\qquad\tag{5}\\
\end{align*}
The relationship (2):
\begin{align*}
\sum_{n\geq 2}b_{n+1}x^n&=\sum_{n\geq 2}\left(a_n+b_n+c_n\right)x^n\\
\frac{1}{x}\sum_{n\geq 3}b_nx^n&=A(x)+B(x)+C(x)\\
B(x)-2x^2&=xA(x)+xB(x)+xC(x)\\
\color{blue}{(1-x)B(x)-xA(x)-xC(x)}&\color{blue}{=2x^2}\tag{6}
\end{align*}
We take (4) - (6) and derive from them the generating functions.
\begin{align*}
(1-x)A(x)-xB(x)&=x^2\\
-xA(x)+(1-x)C(x)&=x^2\\
-xA(x)+(1-x)B(x)-xC(x)&=2x^2
\end{align*}
Solving the equations above we obtain
\begin{align*}
\color{blue}{A(x)}&\color{blue}{=\frac{x^2}{1-3x+2x^2-x^3}}\\
&=x^2 + 3 x^3 + 7 x^4 + 16 x^5 + 37 x^6 + 86 x^7+\cdots\\
\color{blue}{B(x)}&\color{blue}{=\frac{1-x}{x}A(x)-x}\\
&=2 x^2 + 4 x^3 + 9 x^4 + 21 x^5 + 49 x^6 + 114 x^7 +\cdots\\
\color{blue}{C(x)}&\color{blue}{=\frac{x}{1-x}A(x)+\frac{x^2}{1-x}}\\
&=x^2 + 2 x^3 + 5 x^4 + 12 x^5 + 28 x^6 + 65 x^7+\cdots
\end{align*}
where the expansion was done with some help of Wolfram Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How can I prove a curve lies on an ellipsoid? I am trying to prove a curve parameterized by
$$\mathbf{r} (t) = \cos(t) \, \mathbf{i} + \sqrt{2} \sin(t) \, \mathbf{j}-\sin(t) \, \mathbf{k}$$
lies on an ellipsoid. How do I do this?
| The general equation of an ellipsoid in $\Bbb R^3$ is
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1, \tag 1$
where
$abc \ne 0; \tag 2$
in order for the curve
$r(t) = \cos t \mathbf i + \sqrt 2 \sin t \mathbf j - \sin t \mathbf k \tag 3$
to lie on a surface (1), it's components
$x(t) = \cos t, \; y(t) = \sqrt 2 \sin t, \; z(t) = -\sin t, \tag 4$
must satisfy (1) for suitable $a, b, c$; thus we have
$\dfrac{\cos^2 t}{a^2} + \dfrac{2\sin^2 t}{b^2} + \dfrac{\sin^2 t}{c^2} = 1; \tag 5$
if we take
$a = 1, \; b = c = \sqrt 3, \tag 6$
then (5) becomes
$\dfrac{\cos^2 t}{1} + \dfrac{2\sin^2 t}{3} + \dfrac{\sin^2 t}{3} = \cos^2 t + \dfrac{3\sin^2 t}{3} = \cos^2 t + \sin^2 t = 1, \tag 7$
which shows that $r(t)$ lies on the ellipsoid
$x^2 + \dfrac{y^2}{3} + \dfrac{z^2}{3} = 1. \tag 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving an algebraic binomial identity related to Bertrand's ballot theorem I am trying to answer the math.stackexchange question found here by developing all the theory from scratch and not using Bertrand's ballot theorem. My logic boils down to being able to prove the following identity:
$$ \frac{n-m+1}{n+1} \binom{n+m}{n} = \frac{n-m}{n} \binom{n+m-1}{n-1} + \frac{n-m+2}{n+1} \binom{n+m-1}{n}$$
where $0 < m < n$.
I tried using Wolfram but not an expert (or a subscriber). I thought of multiplying both sides to clear the denominator and see what happens, but it seems a bit daunting.
So I post the problem here hoping that 'cranking it out' doesn't involve too many algebraic/binomial tricks of the trade.
| Well, we just need to simplify the right hand side:
\begin{align}
RHS &= \frac{n-1-m+1}{n} {n+m-1 \choose n-1} + \frac{n-m+2}{n+1} {n+m-1 \choose n} \\
&= \frac{n-m}{n} \frac{(n+m-1) !}{(n-1)! m!} + \frac{n-m+2}{n+1} \frac{(n+m-1) !}{n! (m-1)!} \\
&= \Big[\frac{n-m}{n+m} + \frac{(n-m+2)m}{(n+1)(n+m)}\Big] {n+m \choose n}
\end{align}
So now we just need to show the following equality:
$$ \frac{n-m+1}{n+1} = \frac{n-m}{n+m} + \frac{(n-m+2)m}{(n+1)(n+m)}$$
The right hand side can be written as:
\begin{align}
\frac{n-m}{n+m} + \frac{(n-m+2)m}{(n+1)(n+m)} &=
\frac{(n-m)(n+1) + (n-m+2)m}{(n+m)(n+1)} \\
&= \frac{n^2+n+m-m^2}{(n+m)(n+1)}\\
&= \frac{(n+m)(n-m+1)}{(n+m)(n+1)}\\
&= \frac{n-m+1}{n+1}\\
\end{align}
Which gives you the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| $$3x+2y=36\tag1$$
$$5x+4y=64\rightarrow \rightarrow \rightarrow \rightarrow 3x+2y+3x+2y-x=64$$
$$36+36-x=64$$
$$x=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 5
} |
For positive $a$, $b$, $c$, $d$ with $a+b+c+d=4$, show that $\sum_{cyc} \left(\frac{1}{a^2}-a^2\right)\ge0$ From MOP 2012:
If $a,b,c,d>0$, $a+b+c+d=4$, show
$$\sum_{cyc} \left(\frac{1}{a^2}-a^2\right)\ge0$$
My attempt: Using the derivative and tangent line, I obtain
$$(x-1)(-x^3+7x^2-x-1)=0$$
Does using $n-1$ EV work?
| EV works of course, but there is something nicer:
$$\sum_{cyc}\frac{1}{a^2}=\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)+\sum_{cyc}\frac{1}{ab}=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\sum_{cyc}\frac{1}{ab}=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2+\sum_{cyc}\frac{1}{ab}\geq$$
$$\geq\sum_{cyc}\frac{1}{ab}=\left(\frac{1}{ab}+\frac{1}{cd}\right)+\left(\frac{1}{ad}+\frac{1}{bc}\right)$$ and see here:
Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $a,b$ are even integers I'm trying to prove the following theorem:
Let $a,b\in\mathbb{Z}$ . Then $$a^{2}+b^{2}\equiv0\pmod 4 \iff a \;\text{and}\; b\;\text{are even}$$
I always struggle to prove some number is odd or even. How to prove it?
I thought of using the $(a+b)^2=a^2+2ab+b^2$ formula but not sure how.
| $\Rightarrow$:
By the rules of modular arithmetic we have
\begin{align}
0
&= (a^2 + b^2) \bmod 4 \\
&= \left( \left( a^2 \bmod 4 \right) + \left( b^2 \bmod 4 \right) \right) \bmod 4
\end{align}
For each summand we have
\begin{align}
x^2 \bmod 4
&= \left( \left( x \bmod 4 \right) \left( x \bmod 4 \right) \right) \bmod 4 \\
&\in
\{ y^2 \bmod 4 \mid y \in \{0,1,2,3\} \}
= \{ 0, 1 \} \quad (*)
\end{align}
So for the expression
$$
\left( \left( a^2 \bmod 4 \right) + \left( b^2 \bmod 4 \right) \right) \bmod 4
$$
we have the possible cases $0+0$, $0+1$, $1+0$ and $1+1$. This vanishes modulo $4$ only for the case $0+0$.
Now $x^2 \bmod 4$ only vanishes for $x \bmod 4 \in \{ 0, 2 \}$, see display $(*)$, which means
$x = 4k$ or $x = 4k + 2$ for some integer $k$ and both cases can be divided by $2$, thus are even.
$\Leftarrow$:
$a = 2p$ and $b=2q$ for some integers $p, q$. Then we have
\begin{align}
(a^2 + b^2) \bmod 4
&= (4p^2 + 4q^2) \bmod 4 \\
&= \left( \left(4p^2 \bmod 4\right) + \left(4q^2 \bmod 4 \right)\right) \bmod 4 \\
&= (0 + 0) \bmod 4 \\
&= 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to solve $|a+b|+|a-b|=c$? It is intuitive that $a=\pm \frac{c}{2}$, with $-\frac{c}{2}\leq b\leq \frac{c}{2}$ or vice-versa are solutions to the problem. Can I get to these solutions without dividing the expression in all the cases (i.e. $a$ and $b$ being positive or negative and $|a|$ being greater than or less than $|b|$)? Nothing wrong with that, I'm just looking for other way of solving this problem.
I tried using the fact that $|x|=\sqrt{x^2}$ but the expression quickly becomes very complex with the two variables.
| There are several standard methods: intervals (cases), squaring, graphical.
Let's do squaring:
$$(|a+b|+|a-b|)^2=c^2 \iff \\
2a^2+2b^2+2|a^2-b^2|=c^2 \iff \\
(2|a^2-b^2|)^2=(c^2-2(a^2+b^2))^2 \iff \\
4a^2+4b^2-8a^2b^2=c^4-4(a^2+b^2)c^2+4a^2+4b^2+8a^2b^2 \iff \\
c^4-4(a^2+b^2)c^2+16a^2b^2=0 \Rightarrow \\
c^2=2(a^2+b^2)\pm \sqrt{(2(a^2+b^2))^2-16a^2b^2}=2a^2+2b^2\pm \sqrt{4(a^2-b^2)^2} \Rightarrow \\
c^2=2a^2+2b^2\pm 2|a^2-b^2|\stackrel{WLOG: \ |a|\ge |b|}{=} 4a^2;4b^2 \Rightarrow \\
a=\pm \frac c2; |b|\le |a| \Rightarrow -\frac c2\le b\le \frac c2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show With Induction that $2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$ Show with induction that
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$
My solution:
Base case 1: n = 1
LHS = $(1+1)\cdot2^{1-1} = 2$
RHS = $1\cdot2^{1}= 2$
Case 2: n = p
When
$LHS_{P}$ =
$RHS_{P}$
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(p+1)\cdot2^{p-1}=p\cdot2^{p}$
Case 3: n = p + 1
$LHS_{P+1}$ = $LHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = (p+1)$\cdot2^{p+1}$
So i need to to prove that:
$RHS_{P+1}$ = $RHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$
Am I thinking right here?
$RHS_{P+1}$ =
$(p+1)2^{p+1}=$
$(p+1)\cdot2^{p}\cdot2$ = ?
(Can I get this equal to $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$ ? )
| The solution without induction:
$$2\cdot2^0+3\cdot2^1+...+(n+1)2^{n-1}=\frac{1}{2}\left(2\cdot2^1+3\cdot2^2+...+(n+1)2^n\right)=$$
$$=\frac{1}{2}\left(x^2+x^3+...+x^{n+1}\right)'_{x=2}=\frac{1}{2}\left(\frac{x^2(x^n-1)}{x-1}\right)'_{x=2}=$$
$$=\frac{1}{2}\left(\frac{\left((n+2)x^{n+1}-2x\right)(x-1)-\left(x^{n+2}-x^2\right)}{(x-1)^2}\right)_{x=2}=$$
$$=\frac{1}{2}\left((n+2)2^{n+1}-4-2^{n+2}+4\right)=\frac{1}{2}\cdot n2^{n+1}=n2^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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.