Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Polynomial with "symmetrical" coefficients, and $\gcd(p,p')$ Consider the following polynomial:
$p(x) = x^6+x^5+2x^4+x^3+2x^2+x+1$. I would like to calculate $\gcd(p,p')$.
With Euclid's GCD algorithm, the calculation becomes very complicated, very quickly. I was wondering if the fact that the coefficient vector is "pa... | $p(x)=x^6 + x^5 + x^4 +x^4+ x^3 + x^2 +x^2+ x + 1$
$p(x)=x^4(x^2+x+1)+x^2(x^2+x+1)+1(x^2+x+1)=(x^2+x+1)(x^4+x^2+1)=(x^2+x+1)[(x^4+2x^2+1)-x^2]=(x^2+x+1)[(x^2+1)^2-x^2]=(x^2+x+1)(x^2+x+1)(x^2-x+1)=(x^2+x+1)^2(x^2-x+1)$
$p'(x)=2(x^2+x+1)(2x+1)(x^2-x+1)+(x^2+x+1)^2(2x-1)=(x^2+x+1)(6 x^3-x^2+3 x+1)$
Let $q(x)=6x^3-x^2+3x+... | {
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"timestamp": "2023-03-29T00:00:00",
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Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substi... | it must be
$$\frac{2x}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$$
| {
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A small trouble understanding the AMGM inequality usage for some basic minimization problems? I'm reading Byrne's "A first course in optimization". After the presentation and proof of the AMGM inequality, the author gives two examples:
Here's the first:
I guess I understand what was done, it seems he writes in a con... | In both cases it was used:
$$a+b+c\ge 3\sqrt[3]{abc},$$
the equality occurs when $a=b=c$.
Example 1:
$$\frac{12}{x}+\frac{18}{y}+xy\ge 3\sqrt[3]{\frac{12}{x}\cdot \frac{18}{y}\cdot xy}=18,$$
$$\frac{12}{x}=\frac{18}{y}=xy \Rightarrow x=2, y=3$$
Example 2:
$$\frac{1}{12}\cdot (3x)(4y)(72-3x-4y)\le \frac{1}{12}\left(\fra... | {
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Stationary points and linearisation of non-linear system So, the problem is:
Find and discuss the behavior of the stationary points of the system :
$$ x'=-y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =f(x,y)$$
$$ y'=x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=g(x,y)$$
So in the beggining I Linearised the non-linear sys... | In order to get the critical points, we need to solve the system :
$$\begin{cases} -y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =0 \\ \space \space\space x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=0\end{cases}$$
The obvious solution is $x=0,y=0$, which means that $O(0,0)$ is a Stanionary Point of your system.
To prove ... | {
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Proving $ \lim\limits_{n\to\infty} \frac{8n^2-5}{4n^2+7} = 2$ using $\epsilon-\delta-$ definition. I'm trying to prove the limit of this sequence using the formal definition. I've looked at other questions on the site but from the ones I've seen, the $n^2$ term always seems to cancel out, making it simpler.
Show that ... | we have $$\left|\frac{8n^2-5}{4n^2+7} -2 \right| = \left|\frac{-19}{4n^2+7}\right| < \frac{19}{4n^2} = $$
Now let such that $$\frac{19}{4n^2} < \epsilon\Longleftrightarrow n>\sqrt{\frac{19}{4\epsilon}} $$ Now choosing,
$$N =\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1$$ then we have $$n > N=\left\lfloor \... | {
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Value of $e$ given that $\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}$ I have five real numbers $a,b,c,d,e$ and their arithmetic mean is $2$. I also know that the arithmetic mean of $a^2, b^2,c^2,d^2$, and $e^2$ is $4$. Is there a way by which I can prove that the range of $e$ (or any ONE of t... | By C-S $$(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)\geq(a+b+c+d)^2$$ or
$$4(a^2+b^2+c^2+d^2)\geq(a+b+c+d)^2$$ or
$$4(20-e^2)\geq(10-e)^2$$ or
$$(e-2)^2\leq0$$ or
$$e=2.$$
This method works in the general case.
Given: $a+b+c+d+e=k$ and $a^2+b^2+c^2+d^2+e^2=l$.
Find the range of $e$.
| {
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How to simplify: $f(x)=(2^a-1)(3^a-1)(4^a-1)...(x^a-1)\mod {m}$ How to simplify $f(x)$ for positive integers $a,m,x$ where $x>1$ and $m$ is a composition of two very large primes, such that factoring $m$ is not an option. $a$ is relatively small number that doesn't have common divider with $m$
$f(x)=(2^a-1)(3^a-1)(4^a-... | This expression will be $0 \mod m\;$ in specific cases, for example if $x\ge m,$ because then one of the terms $n+i$ will be 1 $\pmod m,$ or if $a$ is a multiple of the Euler phi function $\varphi(m),$ because then $(n+i)^a\equiv 1 \pmod m$ for all $i.$
| {
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Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}... | Why complicate it with complex coefficients?
$$\int_1^\infty\frac{1}{x(x^2+1)}dx=\int_1^\infty\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx=\left(\ln(x)-\frac{1}{2}\ln(x^2+1)\right)\bigg\vert_1^\infty\\=\ln\frac{x}{\sqrt{x^2+1}}\bigg\vert_1^\infty=\ln\frac{1}{\sqrt{1+1/x^2}}\bigg\vert_1^\infty=\ln 1-\ln\frac{1}{\sqrt{2}}=... | {
"language": "en",
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Sketching $(x^2-2)^2+(y^2-2)^2=4$ I was recently asked to sketch $(x^2-2)^2+(y^2-2)^2=2$, which did not prove to be too problematic, for establishing the range and domain of the expression gives nearly all of it away.
I then asked myself what would happen as I change the constant on the R.H.S. I recommend you try this... | More generally,
$(x^2-a)^2+(y^2-a)^2-b = (x^2-Axy+y^2-B)(x^2+Axy+y^2-B)$
iff
$A^2=2, B=a, b=a^2$
but like @Lubin I can't see this at once.
| {
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$a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $3$) holds for all $x$? Today I wrote a computer program that finds whole positive $a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $n$) holds for all $x$. For $n=2$ I get several results, such as $x^2+x$, $3x^2+x$ and more. However, I was not able to find $a,b,c$ for the case $n=... | I assume $a,b,c$ and $x$ are all restricted to integers.
Considering the three cases $x = 0,1,2 \mod 3$ we have:
$x=0 \mod 3 \Rightarrow c=0 \mod 3$
$x=1 \mod 3 \Rightarrow a+b+c=0 \mod 3$
$x=2 \mod 3 \Rightarrow a +2b+c = 0 \mod 3$
These three simultaneous equations are only all satisfied if $a=b=c=0 \mod 3$ i.e. if $... | {
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Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
My Try :
$$\log x=\log 2+\log y \\ \log x=\log 2y \\x=2y $$
So we have :
$$2^x+8^y=4\\2^{2y}+2^{3y}=4 \\t=2^y \\t^2+t^3=4$$
now what ?
| I will assume you want a real solution. $t^2+t^3=4$ can be rewritten as $t^3+t^2-4=0$. Using the cubic formula, the real solution for $t$ is $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$
We also know that $y=log_2(t)$ and that $x=2y$. So, using that,
$$x=2\cdot log_2\left(\frac{1}{... | {
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Taylor expansion of arctan x How can I prove that the tailor expansion of arctan x converges only at [-1,1]? I was able to prove that it converges at the given interval but I don't know how to prove that it does not converge anywhere else . Can someone help?
| The derivative of $\arctan x$ is
$$\frac d{dx} \arctan x =\frac 1{1+x^2}$$
From the formula for geometric series, we have that
$1+y+y^2+y^3+...=\frac 1{1−y}$ if and only if $|y|<1$.
Plugging in $-x^2$ for $y$, we get that
$$\begin{align*}
\frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\
&= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \... | {
"language": "en",
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Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$. $$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$
I thought that at $x=1$ the expansion is:
$$n+n(n-1)(x-1)+n(n-1)(n-2)(x-1)^2+...$$
but the answer is:
$$n+\frac{1}{2}n(n-1)(x-1)+\frac{1}{6}n(n-1)(n-2)(x-1)^2+...$$
Can you explain the origin of $$\frac{1}{2},\frac{1}{6... | Let $f(x)=\frac{x^n-1}{x-1}=\sum_{k=0}^{n-1}x^k$ for $x\ne 1$. $\lim_{x->1}f(x)=n$.
| {
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How to find Laurent expansion of $f(z)=\frac{1}{(z+i)z^2}$ on $0<|z-i|<1$ I think the way is doing the decomposition as follows, but I just don't know how to deal with the quadratic term:
$f(z)=\frac{1}{(z+i)z^2}=-\frac{1}{z+i}+\frac{1}{z}-\frac{i}{z^2}$
where, $-\frac{1}{z+i}=-\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}$ a... | With $z-i=w$ and geometric series
$$\dfrac{1}{z}=\dfrac{1}{w+i}=\dfrac{1}{i}\dfrac{1}{1+\frac{w}{i}}=\dfrac{1}{i}\left(\sum_{n=0}^\infty(-\frac{w}{i})^n\right)$$
then it's derivative
$$\dfrac{1}{(w+i)^2}=\sum_{n=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n$$
valid for $\Big|\dfrac{w}{i}\Big|<1$. Also
$$\dfrac{1}{z+i}=\d... | {
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Almost-half iterate of $x^2+1$ Because I can't find a function $h:\mathbb R\mapsto\mathbb R$ with the property
$$h^{\circ 2}(x)=x^2+1$$
I'm looking for a function that almost has that property - that is, I would like to find a closed-form (and preferably elementary) function $h:\mathbb R\mapsto\mathbb R$ satisfying
$$\... | Starting with
$$
g(x) = x^\sqrt{2} + C
$$
we have
$$
\begin{align}
g(g(x)) &= \left(x^\sqrt{2} + C\right)^\sqrt{2} + C \\
&= x^2 \left(1 + Cx^{-\sqrt{2}}\right)^\sqrt{2} + C \\
&\approx x^2 \left( 1 + C\sqrt{2}x^{-\sqrt{2}} \right) + C \qquad \text{(binomial theorem)} \\
&= x^2 + C\sqrt{2}x^{2-\sqrt{2}} + C.
\end{align... | {
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Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
I have tried to express $x$ in terms of $a$ and $b$, namely: $x = a+bi, |x|= \sqrt{a^2+b^2}$ and to evaluate this but this simply led me n... | So let’s evaluate the expression by plugging in $x = a + bi$. $\frac{x-1}{x+1} = \frac{(a-1)+bi}{(a+1)+bi}$.
Multiply by the conjugate of the denominator, which is $(a+1)-bi$, to get $\dfrac{((a-1)+bi)((a+1)-bi)}{((a+1)+bi)((a+1)-bi)}$.
We know that the resulting denominator must be real and equal to $(a+1)^2+b^2$. We ... | {
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Inequality with angles of a triangle Let $\alpha$, $\beta$ and $\gamma$ be the three angles in a non-right triangle. How can i prove the following inequality?
$\dfrac{1+\sin^2(\alpha)}{\cos^2(\alpha)}+\dfrac{1+\sin^2(\beta)}{\cos^2(\beta)}+\dfrac{1+\sin^2(\gamma)}{\cos^2(\gamma)}\ge \dfrac{1+\sin(\alpha)\sin(\beta)}{1-... | $$\sum_{cyc}\frac{(2bc)^2+16\Delta^2}{(b^2+c^2-a^2)^2}\geq \sum_{cyc}\frac{4 R^2+ab}{4R^2-ab}$$
is equivalent to
$$ 16\Delta^2\sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2} \geq \sum_{cyc}\frac{ab}{4R^2-ab} $$
or to
$$\sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2} \geq \sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2+4(a-b)bc^2}$$
which can be proved by ... | {
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Miller-Rabin: Showing non trivial divisors of $n$
Let's assume $n$ doesn't pass the Miller-Rabin test and $b$ is a witness. Meaning, $b^{\frac{n-1}{2^r}} \equiv 1 \pmod{n}$ where $\frac{n-1}{2^r}$ is even, but $c = b^{\frac{n-1}{2^{r+1}}} \not\equiv \pm1 \pmod{n}$. Show that $\gcd(c+1,n),\gcd(c-1, n)$ are non-trivial ... | The general statement is that whenever we have a nontrivial square root of $1$ modulo $n$, a value $c$ such that
$$
c^2 \equiv 1 \pmod{n}, \qquad c \not\equiv \pm1 \pmod{n},
$$
then we obtain a factorization of $n$.
If $c^2 \equiv 1 \pmod n$, then $(c+1)(c-1) \equiv 0 \pmod n$.
If $c+1$ were relatively prime to $n$,... | {
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How to find $\int \frac{e^{-x^2}}{x^2 + 1} dx$? I have a question about improper integrals:
How can we find $\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx$?
$\textbf{Some effort:}$
$\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1}{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx = \lim_{n \right... | Unfortunately, this integral has no closed form in terms of known functions, elementary or non-elementary. You'd have to integrate it numerically if you want to do anything practical with it.
The series expansion of the integral (at $x=0$), while by no means a closed form, is, according to W|A:
$$\int \frac{e^{-v^2}}{v... | {
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"url": "https://math.stackexchange.com/questions/2556615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Expansion of $S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$ Expand;
$S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$
See solution, Also I am looking for an advance method if exists.
| We can also see that $$S =\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} $$ $$\implies S(1-x)=\frac{(n+1)x^n-nx^{n+1}-1}{x-1}=\frac{(x-1)[1+x+x^2…x^{n-1}-nx^n]}{x-1}=[1+x+x^2+…x^{n-1}-nx^n]$$ $$S=\frac{1+x+x^2…x^{n-1}-nx^n}{1-x}=[1+x+x^2+…][1+x+x^2+…x^{n-1}-nx^n] = ?$$
| {
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"source": "stackexchange",
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Let $f(x)=x^3-x^2-x+a$ only have a real negative root then what is range of $a$
Let $f(x)=x^3-x^2-x+a$ only have a real negative root then what is range of $a$
I got it : $$g(x)=x^3-x^2-x \ \ \ h(x)=-a$$
Now we have :
so must $-a<-1 \to a>1$
But I want to be algebraic please help me !
| Discriminant $\Delta$ of the cubic function
\begin{align}
f(x)&=ax^3+bx^2+cx+d
\end{align}
is known to be
\begin{align}
\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2
,
\end{align}
and the cubic polynomial has only one real root
when its $\Delta<0$.
For
\begin{align}
f(x)=x^3-x^2-x+a
\end{align}
we have
... | {
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Geometric interpretation for median formula $m_c = \sqrt{\frac{2(a^2 + b^2)- c^2}{4}}$ for $a + b < c$? For a triangular with side lengths $a, b, c$, the length of the median of $c$ is $m_c = \sqrt{\frac{2(a^2 + b^2) - c^2}{4}}$.
If $a + b < c$, the configuration is obviously not a triangle anymore.
However, there a... | If $a+b>c$, then we can draw circles of radius $a$ and $b$ about endpoints of segment $\overline{AB}$ of length $c$, and these circles will meet at distinct points $C$ and $C^\prime$. The circle about $M$, the midpoint $\overline{AB}$, with radius $m:=\frac12 \sqrt{2(a^2+b^2)-c^2}$, meets those points, as well. Thus, $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Large Remainders with Mod Determine the remainder of $5^{2017}$ when divided by 7.
I know that we need to use mod 7 to find all of the different remainders but I am not sure what specific steps to take and how to finish it out.
Please use Mod 7 as this question is very specific in the way it needs to be worked out.
| The powers of $5$ modulo $7$ obviously form a finite set. So one can try and look at a few values in order to get the idea:
\begin{alignat}{2}
5^0&\equiv 1&&\pmod{7}\\
5^1&\equiv 5&&\pmod{7}\\
5^2&\equiv 5\cdot 5\equiv 25\equiv 4&&\pmod{7}\\
5^3&\equiv 4\cdot 5\equiv 20\equiv 6&&\pmod{7}\\
5^4&\equiv 6\cdot 5\equiv 30\... | {
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Find values of $p$ such that $p^2-p+1$ is cubic? Find all prime numbers such that $p^2-p+1$ is a cubic number. I've tried substituting values of $x$ for $p^2-p+1 = x^3$.
This gave me a few values, which were small. As values of $x$ ranged higher and higher, it was tedious to find values of $p$ that satisfied this cond... | Here is a solution using simpler tools.
Our starting point is
$$p^2 - p + 1 = x^3$$
with $p$ prime, and $x$ is an integer.
As the smallest prime is 2, note that
$$x^3 = p^2 - p + 1 \ge 2^2 - 2 + 1 = 3 \quad \rightarrow \quad x \ge 3^{1/3} > 1$$
$$p^2 > p^2 - p + 1 = x^3 \quad \rightarrow \quad p > x^{3/2} > x $$
Since... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving: $(x-a)(x-b)=x-c$, $(x-c)(x-b)=x-a$ and $(x-c)(x-a)=x-b$ Given three distinct real numbers $a$, $b$, and $c$, show that at least two of the three following equations
$$(x-a)(x-b)=x-c$$
$$(x-c)(x-b)=x-a$$
$$(x-c)(x-a)=x-b$$
have real solutions.
My attempt: I tried to multiply all the equations side by side to ob... | Let (x-a) (x-b) / (x-c) = y
Solving, $ x^2 -(a+b+y)x + ab + cy = 0 $
D>0
$(a+b+y)^2 -4(ab+cy) > 0 $
$a^2 + b^2 + y^2 + 2ab + 2by + 2ay - 4ab - 4cy >0$
$ y^2 + (2b+2a-4c)y + (a-b)^2 > 0 $
Now for this Quadratic, D<0 gives the quadratic the whole Real values as range.
$ 4(b+a-2c)^2 - 4(a-b)^2 < 0$
$(a-c)(b-c)<0$
$Either... | {
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L'Hospital Rule with Trigonometry I've been working on this question for the final exam, yet I could not come to a closure. I'd appreciate if you can help me.
What is $$\lim_{x\to 0} \frac{\sin(\sin(\sin x)) - x}{x^3}\qquad?$$
| $$\lim_{x\rightarrow0}\frac{\sin\sin\sin{x}-x}{x^3}=\lim_{x\rightarrow0}\frac{\cos\sin\sin{x}\cos\sin{x}\cos{x}-1}{3x^2}=$$
$$=\lim_{x\rightarrow0}\tfrac{-\sin\sin\sin{x}\cos^2\sin{x}\cos^2x-\cos\sin\sin{x}\sin\sin{x}\cos^2x-\cos\sin\sin{x}\cos\sin{x}\sin{x}}{6x}=$$
$$=\lim_{x\rightarrow0}\tfrac{-\tfrac{\sin\sin\sin{x}... | {
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How to calculate last four digits of $2^{2017}$? can u suggest me any short trick of solving these kind of problems. How to find last four digits of any number raised to some power ?
| Since $10^4 = 2^4 \times 5^4$, my first reaction is one should use Chinese remainder theorem.
For $2^4$, it is clear $2^{2017} \equiv 0 \pmod {2^4}$.
For $5^4$, we use following theorem
For $a,b \in \mathbb{Z}_{+}$ such that $\gcd(a,b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(\cdot)$ is the Eul... | {
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$X\sim N(1,1)$. Find $\operatorname{Var}(X^2)$
Let $X\sim N(1,1)$. Find $\operatorname{Var}(X^2)$.
I know since, $X-1\sim N(0,1)$, then $(X-1)^2\sim \chi^2_1$, hence $\operatorname{Var}(X-1)^2=2$. But don't understand how to find $\operatorname{Var}(X^2)$. Any help appreciated. Thanks.
| $\newcommand{\e}{\operatorname E}$
$$
\operatorname{var}(X^2) = \e((X^2)^2) - (\e(X^2))^2.
$$
$$
\e(X^2) = \operatorname{var}(X) + (\e(X))^2 = 1 + 1^2 = 2.
$$
Let $Z=X-1$ so that $Z\sim N(0,1).$ Then
\begin{align}
\e(X^4) & = \e((Z+1)^4) = \e(Z^4) + 4\e(Z^3) + 6\e(Z^2) + 4\e(Z) + 1. \\[10pt]
\e(Z^2) & = \operatorn... | {
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Need help to understand the following number theory proof Prove that there exist infinitely many integers n such that $n, n + 1, n + 2$ are each the sum of the squares of two integers:
Solution: First solution: Let a be an even integer such that $a^2 + 1$ is not prime. (For example, choose a ≡ 2 (mod 5), so that $a^2 +... | Just follow the instructions:
*
*Pick some even $a$ such that $a^2+1$ is not a prime, for instance some $a\equiv 2\pmod{5}$.
Fine, let us pick $a=22$;
*Write $a^2+1$ as $r\cdot s$ with $r\geq s >1$.
Fine, $22^2+1 = 97\cdot 5$;
*Write $a^2+1$ as a difference of squares through the previous decomposition.
Fine, $22^... | {
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$f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$, $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$ Given $f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$
Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$
I have simplified the given to $f(n)=\sqrt{n}+\sqrt{n+1}$ but I am still not sure of how to solve th... | The hint.
Use $$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$$ and the telescopic sum.
Finally, we obtain $$\sqrt{100}-1=9.$$
| {
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$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$ Prove that $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$
I believe x and y could both be 0 as that satisfies the equations. Beyond that, I do not know how to prove this.
| By continuous rearranging and squaring:
$$\text{Equation} \Rightarrow x+\sqrt{x^2+1}=\frac{1}{y+\sqrt{y^2+1}} \Rightarrow x+y=\sqrt{y^2+1}-\sqrt{x^2+1} \Rightarrow$$
$$1-xy=\sqrt{x^2y^2+x^2+y^2+1} \Rightarrow (x+y)^2=0.$$
| {
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find the value of $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is a root of the polynomial? Find the value of $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is a root of the polynomial
$$x^3-5x^2 +ax -1$$
I got the answer and value of $a = -10 -5\sqrt{3} -1$.
Is my answer is correct or not?
| If $a \in \mathbb Z$, then if $2+\sqrt3$ is a root, then so is $2-\sqrt3$.
Therefore, $x^3-5x^2 +ax -1$ is divisible by $x^2 - 4 x + 1$, which has $2\pm\sqrt3$ as roots.
Now, $x^3-5x^2 +ax -1 = (x^2 - 4 x + 1)(x-1)+(a-5)x$ and so $a=5$.
| {
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Reverse continued fractions Basic continued fractions arise from recurrence relations such as:
$$
n = a + \frac{b}{n},
$$
This gives rise to the continued fraction:
$$
n = a + \frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}.
$$
What about relations such as:
$$
n = a + \frac{n}{b}?
$$
Do these give rise to things like:
$$
n = a... | Note that $$n = a + \frac{a+\frac{a+\frac{a+\frac{a+\frac{a+...}{b}}{b}}{b}}{b}}{b}=a+\frac ab + \frac a{b^2}+ \frac a{b^3} +\ldots$$
gives you a geometric series. If the constants vary it is still just another way of writing an infinite sum, so doesn't give anything new.
| {
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Is this a valid method of solving $3^x + 2^x = 35$? I want to know if this is a valid method of solving this equation:
$3^x + 2^x = 35$
$3^x+2^x = (7)(5)$
$3^x+2^x = (3+2^2)(2+3)$
$3^x+2^x = 3^2 + 2^3 + 18$
$3^x+2^x = 9 + 18 + 2^3 $
$3^x+2^x = 27 + 2^3 $
$3^x+2^x = 3^3 + 2^3 $
And now comes my problem. Is it correct to... | As others have noted, $x=3$ is the only solution in the real numbers. However, there are infinitely many complex solutions. For example, there is a solution near $3.40258654642902+5.90727700642090 i$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit without l'hopital $\lim_{x \to 0} \left ( \frac{1}{x} -\frac{1}{e^x-1}\right )$ I want to find the limit without L'hopital's rule and without Taylor series:
$
\lim_{x \to 0} \left ( \frac{1}{x} -\frac{1}{e^x-1}\right )
$
Is it possible?
Hints are more than welcome!
Thanks!
| Let us rewrite the expression in the limit as follows
$$ \frac{1}{x} - \frac{1}{e^x-1} = \frac{e^x - 1 - x}{xe^x - x} $$
Using the Taylor expansion of $e^x$, which we recall is
$$ e^x = \sum_{i = 0}^{\infty} \frac{x^i}{i!} \ , $$
we further rewrite the numerator and denominator as
$$ e^x - 1 - x = \frac{x^2}{2} + \m... | {
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Inequality with integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}>b\sqrt{7}$.
Prove that $(a\sqrt{3}-b\sqrt{7})(a+b)>1$.
Attempt. From the assumption we have $3a^2-7b^2\geq 1$. We try to argue that $3a^2-7b^2\geq 3$ to get
$$
a\sqrt{3}-b\sqrt{7}=\frac{3a^2-7b^2}{a\sqrt{3}+b\sqrt{7}}\geq\f... | $3a^2=7b^2+1$ is impossible for $b$ by modulo $3$.
$3a^2-2=7b^2$ is impossible for $a$ by modulo $7$.
Thus, $3a^2-7b^2\geq3$ and we are done!
The equality in the last inequality occurs for $a=55$ and $b=36$, which gives
$$S\geq91(55\sqrt3-36\sqrt7)=1.43....$$
Since $(55,36)$ is a minimal solution of the Pell $3a^2-7b^2... | {
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Equation of circle after rolling Q: A circle of radius $5$ units touches both the axes and lies in the first quadrant. If the circle makes one complete roll on $+ x$-axis, then what will be its equation in the new position $?$
I got my answer as $x^2 +y^2 -x(10+20\pi)-10y+100\pi^2+100\pi+25=0.$
Is it correct?
| The circle of radius $5$ has equation $(x-5)^2+(y-5)^2=5^2$.
One complete revolution covers a distance of $2\pi r$, or $10\pi$ in this case.
So the centre also shifts by $10\pi$ to the right.
So the new equation is
$$(x-5-10\pi)^2+(y-5)^2=5^2$$
So:
$$x^2-2(5+10\pi)x+(5+10\pi)^2+y^2-10y+25=25$$
$$x^2-(10+20\pi)x+25+100... | {
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"timestamp": "2023-03-29T00:00:00",
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sequence of functions converging uniformly on compact sets but not on the real line Consider the sequence of functions
$$
f_n(x)=n\sin((4\pi^2 n^2+x^2)^{1/2}), 0\leq x\leq C, C>0
$$
I cannot prove $(f_n)$ converges uniformly on $[0,C]$ by finding f(x) such that
$$
\lim_n \sup_{0\leq x\leq C}||f_n(x)-f(x)||=0
$$
But is... | Let's study pointwise convergence first. For fixed $x$ you have
$$\lim_{n \to \infty} 2 \pi n \left( \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} -1\right) = \lim_{n \to \infty} 2 \pi n \left( \frac{x^2}{2(2 \pi n)^2}\right) = 0 $$
where we used the asymptotic approximation $\sqrt{1+t^2} \sim \frac{t^2}{2}$. Hence
$$n \sin \lef... | {
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Constrained stars and bars. I have been racking my brain to understand the solutions of constrained stars and bars problems like:
$a+b+c+d = 19$; $a,b,c,d \in \{0,1,2,3,4,5,6,7\}$
Finding the total number of solutions is really simple $\dbinom{22}{3}$.
What to do after that? I want to understand the entire procedure ... | Here's another method which applies generally to many such problems. You have the problem equivalent to searching for coefficient of $x^{19}$ in expansion of $(1+x+x^2 ... x^7)^4$
Each power of $x$ in each of the $4$ factors represents value of particular variable ( say, $a$) and power of $x$ in each term of expansion... | {
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Intersection points of two functions Find all intersection points (with integer coordinates) of $$y=p$$ and $$y=|x^2+2x-15|$$ if p is prime number.
Now what I tried:
$$|x^2+2x-15|=p$$
then
$$|(x+1)^2-16|=p$$
then
$$(x+1)^2=16-p$$ $$or$$ $$(x+1)^2=p+16$$
So $p+16$ and $16-p$ must be square numbers. I found that if $p... | You can instead write $y=x^2 + 2x - 15$ as $$y=(x+5)(x-3).$$ Then, we are searching for integers $x$ such that $$p=|(x+5)(x-3)| = |x+5| \cdot |x-3|$$ where $p$ is a prime. Thus, $$\begin{align}\text{either} \quad &|x+5| = 1 \quad\\ \text{or} \quad &|x-3|=1.\end{align}$$
Now,
$$
\begin{align}
|x+5| = 1 &\Leftrightarrow ... | {
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Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\in... | Consider $$A=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}\sqrt{1+\frac 1x}}=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}}\left(1+\frac 1x \right)^{-1/2}$$ Now, by Taylor (or generalized binomial theorem
$$\left(1+\frac 1x \right)^{-1/2}=1-\frac{1}{2 x}+\frac{3}{8 x^2}-\... | {
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If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$
$P,Q,R,S$ are polynomials such that: $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$
I thought a lot on this but no result!!
By the way,one idea is to insert some values f... | Firstly it is easy to see that the roots of $x^4+x^3+x^2+x+1 =0$ are all different. Also if $a$ is a root for $ x^4+x^3+x^2+x+1=0$ then $a^5=1$. The same holds for the rest of three roots: $b,c,d$.
Now if we plug in each root in
$P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$
we get:
$$P(1)+aQ(1)+a^2R(1)= 0$$
$$P(1)+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $a^4+b^4+c^4$ The problem:
The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?
Based on the above information, we have:
\begin{align}
a + b + c &= 6 \\
a^2 + b^2 + c^2 & = 8 \\
a^3 + b^3 + c^3 & = 5 \\
\end{ali... | $f(\alpha)=0$ implies that $\alpha f(\alpha)=0$.
So, $\alpha^4-6\alpha^3+14\alpha^2-\frac{41}{3}\alpha=0$
We have similar results for $\beta$ and $\gamma$.
So, $\alpha^4+\beta^4+\gamma^4=6(\alpha^3+\beta^3+\gamma^3)-14(\alpha^2+\beta^2+\gamma^2)+\frac{41}{3}(\alpha+\beta+\gamma)$
| {
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Manual calculation doesn't match Wolfram Alpha. Why? Say I want to evaluate this sum:
$$\sum_{x=2}^\infty \ln(x^3+1)-\ln(x^3-1)$$
We can rewrite the sum as
$$\ln\left(\prod_{x=2}^\infty \frac{x^3+1}{x^3-1}\right)$$
We can split the product into two products:
$$\ln\left(\prod_{x=2}^\infty \frac{x+1}{x-1}\right)+\ln\left... | When you split the product into two products, the first diverges to $\infty$ and the second goes to zero. This gives you the indeterminate $\infty-\infty$. All bets are off.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=\frac{t}{a}J_1(ax)$ Question:
Assume that $J_\alpha(x)$ is defined as below:
$$J_\alpha(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!\, \Gamma(n+\alpha+1)}\left(\frac{x}{2}\right)^{2n+\alpha}.$$
(a) Prove that $\frac{d}{dx}(x^{-\alpha} J_\alpha(x))=-x^{-\alpha}J_{\alpha+1}(x)$;... | Well, let's write:
$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{a}}\cdot\mathscr{L}_x\left[x\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\tag1$$
Using the 'frequency-domain derivative' proper... | {
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The integral of $\frac{x+1}{\sqrt {x}-1}$ Can anyone help me with this? I have been racking my brain for days trying to solve this. I tried splitting the equation into two integrals and multiplying by $\frac{\sqrt{x}}{\sqrt{x}}$ for the differential but nothing works. The only hint the book gives me is that $u = \sqrt{... | $$\int\frac{x+1}{\sqrt{x}-1}dx=\int\frac{x\sqrt{x}+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}dx=\int\frac{x\sqrt{x}+\sqrt{x}-2+2}{\sqrt{x}(\sqrt{x}-1)}dx=$$
$$=\int\frac{x\sqrt{x}-x+x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}dx+\int\frac{4}{2\sqrt{x}(\sqrt{x}-1)}dx=$$
$$=\int\frac{x+\sqrt{x}+2}{\sqrt{x}}dx+4\ln|\sqrt{x}-1|=\fra... | {
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A triangle in ellipse Coordinates of the vertices B and C of a triangle ABC are (2,0) and (8,0)respectively.The vertex A is varing in such a way that $4\tan(B/2)\tan(C/2)=1$.then the locus of A has to be find.
Now , i didnt get any idea about the same.
Pls help me solving this.
| Applying Componendo-Dividendo we get $\dfrac{\cos \left ( \dfrac{B+C}{2}\right)}{\cos \left ( \dfrac{B-C}{2}\right)} = \dfrac{3}{5}$
We can easily prove that $\dfrac{a}{b+c} = \dfrac{\cos \left ( \dfrac{B+C}{2}\right)}{\cos \left ( \dfrac{B-C}{2}\right)}$
Thus $b+c = 10$ and hence locus of $A$ is an ellipse with focii ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A counterintuitive problem in conditional probability/combinatorics Please help me to finish/validate my solution for the following problem:
Three players $A,B,C$ play a game. A game is played between two
players. After a game the winner plays against player who did not play
last time. Everyone has equal winning c... | The game tree won't repeat itself after 3 games because even though the 4th game is between $A$ and $B$, one of them will be coming in with one win under her belt and will now win with probability $\frac{1}{2}$, not $\frac{1}{4}$.
However, your solution idea works with a little modification, now that we understand that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show that any integer can be expressed as $S^{k}_{n}$ Consider the following sequence: $a_1=6$, $a_2=4$, $a_3=1$, $a_4=2$, and $a_n=a_{n-4}$ for $n \geqslant 5$.
Let $S^{k}_{n}$ be the sum of $k + 1$ elements from $a_n$ of the previous sequence. For example:
$
S^{0}_{3} = 1
\\
S^{0}_{4} = 2
\\
S^{1}_{3} = 2 + 1 =... | As you notice, the sequence $a_n$ is by definition periodic: $6,4,1,2,6,4,1,2,6,4,1,2,...$
The sum of a whole "period" is $6+4+1+2=13$.
You can easily check that every number between $1$ and $12$ can be obtained as a sum of CONSECUTIVE numbers in the sequence $a_n$ (for instance, $9=1+2+6$). Then you just write any des... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve $(x^4+y^4)\,\text{d}x-xy\,\text{d}y=0$ I was trying to solve
$$(x^4+y^4)\,\text{d}x-xy\,\text{d}y=0.$$
I ended up to the point of obtaining $$x^4\, \text{d}x+xy^4\frac{\text{d}(e^{1/2y^2}x)}{e^{1/2y^2}x}=0.$$ Couldn't proceed further. Can someone help me?.
| $$x^4+y^4=x^2\frac{2ydy}{2xdx}$$
Let $\quad\begin{cases}x^2=X\\y^2=Y\end{cases}$
$$X^2+Y^2=X\frac{dY}{dX}$$
$$\frac{dY}{dX}=X+\frac{Y^2}{X}$$
Riccati ODE. Let : $\quad Y(X)= -X\frac{F'(X)}{F(X)} \quad\implies\quad Y'=-\frac{F'}{F} -X\frac{F''}{F}+X\frac{(F')^2}{F^2} $
$\frac{dY}{dX}=X+\frac{Y^2}{X}=-\frac{F'}{F} -X\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Continuity And Derivative Of A Function
Let $
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$
a. Find all the points where $f(x)$ is continuous
b. Find $f'(x)$ for all the points $f(x)$ is continuous
c. Is $f'(x)$ continuous where it is defined? if not where and which discontinuity typ... | Here you check whether this limit exists: $\lim_{x\to 0}2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$. The first term approaches zero, so we care about second (forget $-$ sign for some minuts):
$$\lim_{x\to 0}\frac{\cos(\frac{1}{x})\sin^2x}{x^2} \equiv \lim_{x\to 0}\cos(\frac{1}{x})$$
which sure... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$
My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\fra... | A couple of alternative approaches to Euler's Beta function or contour integration.
1) The reflection formula for the digamma function
By parity we have $I=\int_{-\infty}^{+\infty}\frac{x^4}{1+x^8}\,dx = 2\int_{0}^{+\infty}\frac{x^4}{1+x^8}\,dx $ and by splitting $\mathbb{R}^+$ as $(0,1]\cup(1,+\infty)$, then enforcing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$ Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$
My Attempt:
Let $x=3\sec (\theta )$
$$dx=3\sec (\theta).\tan (\theta).d\theta$$
Then,
$$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$
$$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d... | Let $ x + t = \sqrt{x^2 - 9}$
$$I = \dfrac{-1}{16}\int \dfrac{(t^2 + 9)^4}{t^5} dt = -\dfrac{486\ln(|t|)+\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}}{16}+C$$
Then, $$\dfrac1{16}\left(\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}\right) =- \left(\dfrac{27}8 x \sqrt{x^2 - 9} + x^3\dfrac{1}4 \sqrt{x^2 - 9}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find all pairs of integers (x,y) such that $x(x+1)(x^2+x+2)=2y^2$ A former Olympiad exercise is to find all pairs of integers $(x,y)$ such that
$$x(x+1)(x^2+x+2)=2y^2.$$
I could not solve this equation. How can one find all solutions?
I tried to rewrite the above equation in a different form, but it did not help. For ... | Set $z=x^2+x$, then we want $z(z+2) = 2y^2$. So $z$ is even and with $w=z/2 \in\Bbb{Z}$, we get the new Diophantine equation
$$2w(w+1) =y^2.$$
Notice that since $x^2+x=z$ has no real solution for $z<\frac{1}{4}$, we can assume $w\ge0$.
*
*If $2|w$, then since $gcd(2w, w+1)=1$ we have that $w+1$ is a square. Then $2w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$.
My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here. Any help?
| Hint
You found
$$(x-1)^2+(y+1)^2=4$$
which is the equation of a circle centered in $(1,-1)$ and with radius $2$.
The function $x^2+y^2$ is the square of the distance of $(x,y)$ to the origin. Which point $(x,y)$ on this circle is located the furthest from the origin? It helps to make a simple sketch.
Alternatively, y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$
Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$?
For example... | $n+\sqrt{n}$ never rounds to a square
By way of introduction, note first that $N^{2}+N$ can never be a
perfect square (when $N$ is a whole number) -- this is because perfect squares
of whole numbers must differ by (at least) $2N+1$. If we let $n=N^{2}$ then
$N^{2}+N=n+\sqrt{n}$ and this can never be a perfect square; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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How can one show that $\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?$ How can one show that
$$\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?\tag1$$
I have tried IBP but it is seem too complicate. I am not sure what to do or how to tackle th... | Enforcing a substitution of $x \mapsto \sqrt{x}$ the integral becomes
$$I = \frac{1}{4} \int_0^1 \frac{\ln x \ln (1 - x)}{\sqrt{1 - x}} \, dx.$$
From the definition for Euler's beta function, namely
$$\text{B}(x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt,$$
we observe that
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Deriving the fact that the approximation $\log(n!) \approx n\log(n) - n + \frac{1}{2}\log(2\pi n)$ is $O(1/n)$. I get to begin with Stirling's approximation, for any $C \in \mathbb{Z}_{\geq 0}$, there exists some $N \in \mathbb{Z}_{\geq 0}$ such that $N > C$ and for all $n > N$:
\begin{align*}
&\quad \left|n! - \sq... | Everything you've written here is 'correct' — but let's take a closer look at what it means. Your definition of $\approx$ says: "$\log(n!)\approx f(n)$ means that $\left|f(n)-\log(n!)\right|\leq K (U(n)-L(n))$ for some $0\leq K\leq 1$". And we have $L(n) = n\log n-n+\log(2\pi n)+\log(1-\frac Cn)$ and $U(n)=n\log n-n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\... | I may contribute by multiplying $(1-\cos(x))/(1-\cos(x)) $ :
$$ \int \frac{1 + \sin(x)}{1 + \cos(x)} dx= \int \frac{1 + \sin(x) - \cos(x) - \sin(x)\cos(x)}{1 - \cos^{2}(x)} dx = \int \frac{1 + \sin(x) - \cos(x) - \sin(x)\cos(x)}{\sin^{2}(x)} dx = \int \frac{\sin^{2}(x) + \cos^{2}(x) + \sin(x) - \cos(x) - \sin(x)\cos(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Integrating $\int (t^2+7)^{\frac{1}{3}}$dt
I have to integrate $\int (t^2+7)^{\frac{1}{3}}$dt
I tried some trigonometric substitution and then rationalisation but didn't get anything
Please provide a hint on how to proceed with this question
Please help!!!
| Positioning your integral:
\begin{equation}
\int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt = \int_0^t \left(w^2 + 7 \right)^{\frac{1}{3}}\:dw = \int_0^t \frac{1}{ \left(w^2 + 7 \right)^{\frac{1}{3}}}\:dw
\end{equation}
We can employ the solution that I've spoken to here:
\begin{equation}
\int_0^x \frac{t^k}{\left(t^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Number of non-negative integral solutions of $2x+3y+z=19$ The question is to find the number of solutions of $2x+3y+z=19$.
There are posts like this with answers arrived by counting the possibilities. But I decided to approach it in a different way.
My Approach:
I learnt a theorem stating that
The number of integral... | You can try directly. We see that $0\leq y\leq 6$
case 1: $y=6$ then we have $2x+z=1$, so $z=1$ and $x=0$;
case 2: $y=5$ then we have $2x+z=4$, so $x=2,z=0$ or $x=1,z=2$ or $x=0, z=4$;
case 3: $y=4$ then we have $2x+z=7$, so $0\leq x\leq 3$ so we have 4 solution
case 4: $y=3$ then we have $2x+z=10$, so $0\leq x\leq 5$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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$3$ primitive pythagorean triples from 6 integers. Do there exist $3$ different primitive Pythagorean triples $(a,d,w), (a,b,z)$ and $(c,d,z)$?
Explicitly, we want $6$ different integers $a,b,c,d,w,z$ such that...
(1) $a^2 + d^2 = w^2$
(2) $a^2 + b^2 = c^2 + d^2 = z^2$
| This is surprisingly simple given a function derived from Euclid's formula:
$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$
To find a triple matching side-C,
$$C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$
The lower limit ensures $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Ratio of volumes of $n$-dimensional unit balls
Prove that $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le 2$$ where $S_i$ is the volume of the $i$th dimensional unit ball and $n\ge 2$.
I think we can use the fact that an $n$-dimensional unit ball is contained in a hypercube whose edge measures $2$ units, and the ball ... | Since the volume of a $n$-dimensional unit ball is (see volume of an $n$- ball)$$
S_n = \frac{π^{\frac{n}{2}}}{Γ\left( n + \frac{1}{2} \right)},
$$
then\begin{align*}
\left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2 &\Longleftrightarrow \frac{n \sqrt{π} \cdot Γ\left( n - \frac{1}{2} \right)}{Γ\left( n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove the convergence of $\sum\limits_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$ Hints preferred:
I need to prove convergence and determine the limit of following series: $$ \sum_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$$
I need to use partial fraction decomposition to simplify the term, which I attempted with following:
$$
\fra... | Yes, this sum telescopes. We have: $$S =\sum_{k=2}^{\infty}\frac14\left(\frac1{2k+1}-\frac1{2k+5}\right)$$ giving us $$4S =\left(\frac15-\frac19\right) +\left(\frac17-\frac1{11}\right)+\left(\frac19-\frac1{13}\right)+ \ldots$$ $$\implies 4S=\frac15+\frac17$$ $$\boxed{\therefore S=\frac3{35}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $\exists n \in \mathbb{Z+}$,s.t. $2^n + 1$ is a prime, prove $\exists k \in \mathbb{Z+}, n = 2^k$ The actual description of the question is:
Suppose that $2^n + 1$ is a prime where $n$ is a positive integer. Prove that
$n = 2^k$ for some positive integer $k$.
There is only one approach that know of: To show that t... | We have:
$$2^{ab} + 1 = (2^{a} + 1)(2^{a(b-1)} - 2^{a(b-2)} + \cdots + 1)$$
where $b$ is an odd integer. It's obvious that both of the factors are bigger than 1 and so if $n$ has an odd factor we $2^n + 1$ has a nontrivial factor.
If you're more interested in primes of that form, they are called Fermat's Primes and til... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving a biquadratic.
If $x$ is a real number and satisfies $$x+ \sqrt[4] {5-x^4}=2$$ then find the value of $$x\sqrt[4] {5-x^4}$$
My try :
The question is significantly asking for the value of $-x(x-2)$ if we get the root of the equation $$2x^4-8x^3+24x^2-32x+11=0$$
Using this I have reached till
$$-2x(x-2)(x^2-2... | $2x^4-8x^3+24^2-32x+11=0$
Let $u = x -1 $
$11-32(u+1)+24(u+1)^2-8(u+1)^3+2(u+1)^4=0$
$2u^4+12u^2-3=0$
Let $v = u^2$
$2v^2+12v-3=0$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
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differential equation $y''-2y'+y=\frac{2(1-x^2) e^x}{(1+x^2)^2}$ using variation of parameter $y''-2y'+y=\frac{2(1-x^2) e^x}{(1+x^2)^2}$
so i got the homogeneous part
$y=Ae^x+Be^x . x$
i tried to find the non homogeneous part by using variation of parameter:
$W = \begin{vmatrix}
e^x& x.e^x \\
e^x &e^x+x.e^x \\
\end... | I think the idea is correct, but
$$u'_2=\frac{gy_1}W=\frac{2(1-x^2)e^x\cdot e^x}{e^{2x}(1+x^2)^2}=2\frac{1-x^2}{(1+x^2)^2}$$
and the integral of the above doesn't look so daunting.
Also
$$u'_1=-\frac{gy_2}W=-\frac{2(1-x^2)e^x\cdot xe^x}{e^{2x}(1+x^2)^2}=-2\frac{x(1-x^2)}{(1+x^2)^2}$$
and again: not so terrible an integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Dividing polynomials such that we get the desired remainder?
Find $a,b,c$ such that the remainder of
$$\frac{a(x^{11}-x^4+1)^{15}-x^2+bx-c}{x^3+x}$$ is equal to
$x^2-4x+1$.
I can solve it by writing $$a(x^{11}-x^4+1)^{15}-x^2+bx-c=(x^3+x)q(x)+x^2-4x+1$$
And solve the system of linear equations by looking at $x\in... | Not sure if its easier, but here is another way:
Note we have $x^{11}-x^4+1 \equiv 1 \pmod x$ and $x^{11}-x^4+1 \equiv -x \pmod {x^2+1} $
So we have $(x^{11}-x^4+1)^{15} \equiv 1 \pmod x $ and $(x^{11}-x^4+1)^{15} \equiv x \pmod {x^2+1} $
Also as $(x^2+1)^{-1} \equiv 1 \pmod x$ and $x^{-1} \equiv -x \pmod {x^2+1}$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $\sin (y)=x\sin (a+y)$, prove that: If $\sin (y)=x\sin (a+y)$, prove that:
$$\dfrac {dy}{dx}=\dfrac {\sin^2 (a+y)}{\sin (a)}$$
My Attempt:
$$\sin (y)=x\sin (a+y)$$
$$\dfrac {d}{dy} \sin (y)=\dfrac {d}{dy} (x\sin (a+y))$$
$$\cos (y)=x\cdot \cos (a+y)+\sin (a+y)\cdot\dfrac {dx}{dy}$$
$$\cos (y)-x\cdot\cos (a+y)=\sin (... | By implicit differentiation, you should get
\begin{align}
\cos(y) y' = \sin(a+y) + x\cos(a+y)y' \ \ \implies \ \ y' = \frac{\sin(a+y)}{\cos(y)-x\cos(a+y)}.
\end{align}
However, since
\begin{align}
x= \frac{\sin(y)}{\sin(a+y)} \ \ \implies \ \ y' = \frac{\sin^2(a+y)}{\sin(a+y)\cos(y) - \sin(y)\cos(a+y)} = \frac{\sin^2(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $A_{n+1}=3A_n+2^n$ Let's say I want to find a formula for the following expression given $n$ number of threes
$$\ldots(3(3(3(3(3(3+1)+2)+4)+8)+16)+\ldots$$
If $A_0=1$, then
$$A_{n+1}=3A_n+2^n$$
Plugging in values to see the pattern,
$$A_2 = 3+1$$
$$A_3 = 3^2+3+2^1$$
$$A_4 = 3^3+3^2+3\cdot2+2^2$$
But I don't kno... | One way to see the correct answer is to use that:
$$x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}\right)$$
Putting in $x=3,y=2$ you get that:
$$3^n-2^n = 3^{n-1}+3^{n-2}\cdot2+\cdots+3\cdot 2^{n-2}+2^{n-1}$$
Now add $3^n$ to both sides, and you get:
$$2\cdot 3^n -2^n = 3^{n}+3^{n-1}+3^{n-2}\cdot2+\cdots+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that the series $\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}$ has a sum of $\frac{1}{2(x+1)(x+2)}$ I am trying to prove that the following series:
$$
\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}
$$
has a sum of:
$$
\frac{1}{2(x+1)(x+2)}
$$
What I've tried:
Using the criteria that $S_n = b_1 - l$:
$$
\fra... | I would suggest that you try to make what i mention below instead. I think it's more simple ( and it's a general method to find the sum of this type of series, later in this post i give you the general form of it)
The Trick:
To find the terms of telescoping sum we can add in the numerator the following difference: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finite difference approximation of $u''(x)+u(x)=0, u(0)=1, u(\pi)=-1$ Finite difference approximation of $u''(x)+u(x)=0, u(0)=1, u(\pi)=-1$.
For the approximation for this, do I get:
$$\frac{1}{h^2}(U_{j-1}-2U_j+U_{j+1})+U_j$$
$$=\frac{1}{h^2}(U_{j-1}+(h^2-2)U_j+U_{j+1})$$
Then the matrix A would have $h^2-2$ on the ma... | If you let $h^2-2-\lambda=x$, then the eigenvalue equation becomes $P_n(x)=0$, where $P_n(x)$ is the determinant of the $n\times n$ matrix with $x$ on the main diagonal and $1$ on the adjacent diagonals. For example,
$$P_4(x)=\det\begin{bmatrix}x&1&0&0\\
1&x&1&0\\
0&1&x&1\\
0&0&1&x\end{bmatrix}$$
If we expand the deter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of the generating function of $1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots$ The ordinary generating function for the sequence $\{a_n\}_{n\geq0}$ where $a_n = (-1)^n\,n$ is
$$1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots = \frac{1}{(x+1)^2}$$
I can see from the geometric formula, and even from long division that
$$1 -x +x^2 -x^3+x^4-... | $$S_n = 1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots$$ $$\implies
xS_n = x - 2x^2 + 3x^3 -\cdots$$
$$(1+x)S_n=1-x+x^2-x^3\cdots=\frac{1}{1+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is $\int \frac{ \sqrt{(a^2 - x^2)}}{x^2}$? I tried the following trig substitution:
$x = a\sin \theta$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta} = \int \frac{a \cos \theta}{a^2 \sin^2}$$
Setting $u = \sin x$ yields:
$$\frac{1}{a} * \int u^{-2} = - \fra... | You forgot to substitute dx. dx = acosθdθ
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$ Prove
$$
\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}
$$
and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.
My Attempt:
$$
\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}... | The method you used to eliminate $\frac{-4-\sqrt7}{3}$ is correct. Here I want to explain why it comes out to be a root of the equation so that things become more clear.
Consider the following
$$\tan(n\pi +2\theta)=\tan2\theta$$
$$\frac{\tan(\frac{n\pi}{2}+\theta)}{1-\tan^2(\frac{n\pi}{2}+\theta)}=\tan2\theta$$
Now it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Given $\triangle ABC$ with $\angle C = 60^\circ$, show $a^2+b^2-ab=c^2$ without trigonometry I have the following problem:
In $\triangle ABC$, the angle at vertex $C$ is $60^{\circ}$. Prove that $a^2+b^2-ab=c^2$.
Of course, it is easy if you use cosine rule. I believe there exists a beautiful proof as well without u... | Here's a proof using Ptolemy's Theorem:
$$\begin{align}
|\overline{AB}| |\overline{A^\prime B^\prime}| &= |\overline{AA^\prime}||\overline{BB^\prime}|+|\overline{AB^\prime}||\overline{A^\prime B}| \tag{1} \\[4pt]
c^2 &= a b + ( a - b )^2 \tag{2} \\[4pt]
&= a^2 + b^2 - a b \tag{3}
\end{align}$$
Incidentally, the Wikip... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
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Differential equation $y''=y^2$ I must solve the differential equation $y''=y^2$. Clearly, the function $y=0$ is a solution. So, assume that $y$ is not identically zero. Unless I'm not mistaken, there is a trick to solve equations of this kind. If I multiply for $2y'$, the left-hand member is $2y'y''$, that is the deri... | We must solve $y'' = y^2$. Assuming the independent variable is $x$, let us write this as:
$\frac{d^2y}{dx^2} = y^2$. Then, using the chain rule, we can write the left hand side as:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dy} \left(\frac{dy}{dx}\right) \frac{dy}{dx}$.
But, this is equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Finding the Sum of $a,b,c,d$ that satisfy the following conditions Let $a,b,c,d$ represent $4$ different non-zero integers such that the absolute value of each integer is less than $11$. If $c$ and $d$ are the solutions for $x$ of $x^2+ax+b=0$ and if $a$ and $b$ are the solutions for $x$ of $2x^2-cx-20d=0$, find the va... | From Vieta's formulas we know that:
$$
\left\{ \begin{array}{rrrrrr}
c+d &= &-\dfrac{a}{1}\\
cd &= &\dfrac{b}{1}\\
a+b &= &-\dfrac{-c}{2}\\
ab &= &\dfrac{-20d}{2}\\
\end{array} \right.
$$
Then
$$
\left\{ \begin{array}{rrrrrr}
a &= &-c -d\\
b &= &cd\\
-c-d+cd &= &\dfrac{c}{2}\\
-(c+d)cd &= &-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding $\int^{\pi}_{0}\frac{1}{\sqrt{a-\cos x}}dx$
Finding $$\int^{\pi}_{0}\frac{1}{\sqrt{a-\cos x}}dx$$
Try: using $$\cos x =\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$\int^{\pi}_{0}\frac{\sec \frac{x}{2}}{\sqrt{(a-1)+(a+1)\tan^2\frac{x}{2}}}dx$$
Could some help me to solve it , thanks
| Starting from
$$I(a)=\int^{\pi}_{0}\frac{\sec \frac{x}{2}}{\sqrt{(a-1)+(a+1)\tan^2\frac{x}{2}}}\mathrm{d}x$$
Let us consider the map $x\mapsto 2x$ and factor out the $(a-1)$ term, so that
$$I(a)=\frac{2}{\sqrt{a-1}}\int_{0}^{\pi/2}\frac{\sec x}{\sqrt{1+\dfrac{a+1}{a-1}\tan^2 x}}\mathrm{d}x \\
=\frac{2}{\sqrt{a-1}}\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $x+y=1$ then $x^{n+1}\sum\limits_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum\limits_{k=0}^n \binom{n+k}{k}x^k=1$ $$x^{n+1}\sum_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum_{k=0}^n \binom{n+k}{k}x^k=1$$
with one property such that x+y=1
I tried using the binomial equation to replace $y^k$ with $(1-x)^k$ but that will not cancel o... | I just copy my answer of this question.
$$\forall m,n\in\Bbb N,\ \forall x\in\Bbb R,\ \sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$$
Let $m,n\in\Bbb N$, $f:x\mapsto\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}$, we have $f(0)=0$ and for all $x\in\Bbb R$,
\begin{align}
f'(x)&=(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Differentiating w.r.t x Please help me understand this question.
\begin{gathered}
{(\sqrt x - \frac{1}{{\sqrt x }})^2} = y \hfill \\
{\text{Differentiating w}}{\text{.r}}{\text{.t x}} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left[ {(\sqrt x ) - \frac{1}{{\sqrt x }}} \right]^2} \hfill \\
\hf... | it is by the power and Quotient rule
$$2\left(\frac{x-1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-(x-1)\cdot \frac{1}{2}x^{-1/2}}{x}\right)$$
it can be simplified ton $$1-\frac{1}{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Arithmetic progression and polynomials Suppose the quadratic polynomial $$p(x)=ax^2 + bx + c$$ has positive coefficients $a, b, c$ such that these are in AP in the given order. If $m$ and $n$ are the integer zeros of the polynomial then what is the value of $m+n+mn$? I have tried quadratic formula but ain't getting the... | Since $m$ and $n$ are roots of the polynomial we have
\begin{align*}
ax^2+bx+c&=a\left(x-m\right)\left(x-n\right)\\
&=ax^2-a(m+n)x+amn
\end{align*}
Then, $$m+n=-\frac ba\quad\text{and}\quad mn=\frac ca$$
So, since $a,b$ and $c$ are in AP we get
$$m+n+mn=-\frac ba+\frac ca=\frac{c-b}a=\frac{b+b-a-b}a=\frac{b-a}a=\frac b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the real part of the following analytic function Given $v = \frac{-y}{x^2+y^2}$, find the real component of the function $u$.
I'm not sure why I'm not getting this. Here's my work so far:
$$v_x = \frac{2xy}{(x^2+y^2)^2} = -u_y \Rightarrow u(x,y) = \frac{x}{x^2+y^2} + \phi(x) $$
where $\phi(x)$ is (as of yet) unde... | Hint:
$$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \iff$$
$$\iff \frac{2}{x^2+y^2} = \frac{2(x^2+y^2)}{(x^2+y^2)^2} $$
So your work seems correct !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfr... | Continuing from where you stopped
We get
$$1-x^2\tan\alpha= \sqrt {1+x^2} .\sqrt {1-x^2}$$
Squaring both sides we get
$$1+x^4\tan^2\alpha-2x^2\tan\alpha=1-x^4$$
$$x^4(1+\tan^2\alpha)=2x^2\tan\alpha$$
$$x^2=\frac {2\tan\alpha}{1+\tan^2\alpha}$$
Hence $$x^2=\sin(2\alpha)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Prove that: $\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$
Prove that:
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
My work so far:
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
$$\frac{(n+0)!}{n!0!}+\fr... | You can calculate directly:
$$\begin{align} & \frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}=\frac{n!(1+n+1)}{1!}; \\
& \frac{n!(n+2)}{1!}+\frac{(n+2)!}{2!}=\frac{n!(n+2)(2+n+1)}{2!}; \\
& \cdots \\
& \frac{n!(n+2)(n+3)\cdots ((n-1)+n+1)}{(n-1)!}+\frac{(n+n)!}{n!}= \\
& \frac{n!(n+2)(n+3)\cdots (n+n+1)}{n!}=\frac{(2n+1)!}{(n+1)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How to show $\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}$ I am trying to show $\displaystyle{\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}}.$
Any help?
(I am having troubles using the half circle infinite contour)
Or more specifically, what is the residue $\text{res} \left(\frac{1}{z^3+1},... | Hint
Substitute $u=1/x \ $ then $ \ \text{d}u=(-1/x^2)\text{d}x=-u^2\text{d}x$
$$
I=\int_{0}^{+\infty}\frac{\text{d}x}{1+x^3}=-\int_{+\infty}^{0}\frac{1}{1+(1/u)^3}\left(\frac{\text{d}u}{u^2}\right)
$$
Then
$$
\int_{0}^{+\infty}\frac{\text{d}x}{1+x^3}=\int_{0}^{+\infty}\frac{u}{1+u^3}\text{d}u$$
What is beautiful here ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Real part of complex root of $X^5-X-k^5+k+1$ is $\frac{1}{23k}$ away from any integer ($k\geq 2$) Let $k\geq 2$ be an integer. The polynomial $P_k=X^5-X-k^5+k+1$ is easily
seen to have exactly one real root and two pair of conjugate non-real roots. Is it true
that if $\alpha_k + i \beta_k$ any non-real root of $P_k$, t... | Not a full solution, but for large $k$ this should explain it.
Let $X = r e^{i \phi}$. Then
$$
P_k=r^5 e^{5 i \phi}-r e^{i \phi}- k^5+k+1
$$
Let $a = r^{-4}$.Then
$$
P_k=r^5 (e^{5 i \phi}-a e^{i \phi})- k^5+k+1
$$
For the roots, the imaginary part of this has to vanish, so we have
$$
0 = \sin(5 \phi)-a \sin(\phi)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$? How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?
| It's true even for all reals $x$, $y$ and $z$.
Indeed, we need to prove that
$$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or
$$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or
$$\sum_{cyc}(2x^4+4x^2y^2-6x^2yz)\geq0$$ or
$$\sum_{cyc}(x^4-2x^2y^2+y^4+3z^2x^2-6z^2xy+3z^2y^2)\geq0$$ or
$$\sum_{cyc}(x-y)^2((x+y)^2+3z^2)\geq0,$$
which is obvi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute the dihedral angle of a regular pyramid Given a regular pyramid, defined as a right pyramid with a base which is a regular polygon, with the vertex above the centroid of the base, I would like to compute the dihedral angle between adjacent faces, as well as the angle between a lateral face and the base, in term... | An alternate approach to obtaining the formula for $\delta$ is to take a unit normal $\underline{\mathbf{u}}$ for a given slope and then rotate this by the angle $\beta$ about the $z$-axis, ie about the vector $\underline{\mathbf{k}}$, to produce a unit normal $\underline{\mathbf{v}}$ for the adjacent slope, as shown i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$ $\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$
$\implies \arccos (x)=\frac{3\pi}{4}-\arccos (2x)$
$\implies \cos(\arccos (x))=\cos(\frac{3\pi}{4}-\arccos (2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arccos(2x))$
$\implies x=\cos(\frac... | use that $$\arccos(x)+\arccos(y)=\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ for $$x+y\geq 0$$ and
$$2\pi -\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ for $$x+y<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that the sequence is bounded and subsequences converge Sequence is defined in the following way:
$x_1 = \frac{2}{3}$, $x_{2n+1} = \frac{x_{2n}}{3} + \frac{2}{3} $, $x_{2n} = \frac{x_{2n-1}}{3}$
I need to show that the sequence is bounded, i.e $0<x_{n}<1$, and it does not converge.
This is what I have done so fa... | We know that $$x_1=\dfrac{2}{3}\\x_{2n+1}=\dfrac{x_{2n}}{3}+\dfrac{2}{3}\qquad\qquad (I)\\x_{2n-1}=\dfrac{x_{2n-2}}{3}+\dfrac{2}{3}\qquad\qquad (II)\\x_{2n}=\dfrac{x_{2n-1}}{3}\qquad\qquad (III)$$by substituting $(II)$ in $(III)$ and $(III)$ in $(I)$ we respectively obtain$$x_{2n}=\dfrac{x_{2n-2}+2}{9}\\x_{2n+1}=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Collinearity when $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$
Let $\mathbf{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $... | If the vectors are collinear, they are of the form
$$
\mathbf{a}=\mathbf{u}+\alpha\mathbf{v},
\quad
\mathbf{b}=\mathbf{u}+\beta\mathbf{v},
\quad
\mathbf{c}=\mathbf{u}+\gamma\mathbf{v}
$$
Then
$$
\mathbf{a}\times\mathbf{b}=
\mathbf{u}\times\mathbf{u}+\alpha\mathbf{v}\times\mathbf{u}+
\beta\mathbf{u}\times\mathbf{v}+\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Homogeneous fourth degree inequality : $\sum x_i^2x_j^2 +6x_1x_2x_3x_4 \geq\sum x_ix_jx_k^2$ Let $x_1,x_2,x_3,x_4$ be four real numbers. The inequality
$$
\sum_{i,j} x_i^2x_j^2 +6x_1x_2x_3x_4 \geq \sum_{i,j,k} x_ix_jx_k^2
$$
(where the sums are over unordered uples of distinct indices) is true because if one views t... | It's $$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2\geq0.$$
Because if $x_1=a$, $x_2=b$, $x_3=c$ and $x_4=d$ then:
$$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2=$$
$$=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2=$$
$$=(a^2-2ab+b^2)(c^2-2cd+d^2)+$$
$$+(a^2-2ac+c^2)(b^2-2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simpiflying Exponential Expressions
Note that $x = 5$
$$\frac {x^{x+3}+ x^{x+2}}{x^{x+1}+x^{x+2}} = ?$$
This is what I'm struggling with. There's my attempt as seen below.
$$\frac {x^x.x^3+x^x.x^2}{x^x . x + x^x .x^2}$$
Factoring $x^x$ and we get
$$\frac {x^x (x^3 + x^2)}{x^x(x . x^2)} = \frac {x^3}{x} = x^2 = (5)^2... | write $$\frac{x^{x+2}(1+x^{x+3-x-2})}{x^{x+1}(1+x^{x+2-x-1})}$$
can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integer solutions for $x^3=y^2+5$ I have been able to solve the equation $x^3=y^2+a$ for integers where $1 \leq a \leq 4$ by splitting in $\Bbb Z [i\sqrt a]$. However as a natural continuation I would like to know whether the equation $x^3=y^2+5$ can be solved using elementary methods. Seems this case is much tougher t... | Write $$ (x-1)(x^2+x+1) = y^2+4$$
If $x\equiv_4 0$ then $x-1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x-1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!
If $x\equiv_4 1$ then $x^2+x+1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x^2+x+1$. So $p\mid y^2+2^2$ so $p\mid 2$, a con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $ax^2+bx+c=0$ has $2$ roots, one root is twice the other, find expression for $b$ $ax^2+bx+c=0$, has two roots with one being twice the other, so $x$ and $2x$
I need to find an expression for $b$ (in terms of $a$ and $c$)
I know $b = \dfrac{-ax^2-c}x$, but I don't know how to use the roots and how it would affect th... | Without using the Viète-formulas:
$$x_1=2x_2$$
$$\frac{-b+\sqrt{b^2-4ac}}{2a}=2\frac{-b-\sqrt{b^2-4ac}}{2a}$$
$$\frac{-b+\sqrt{b^2-4ac}}{2}=-b+\sqrt{b^2-4ac}$$
$$\frac{b}{2}=-\frac{3}{2}\sqrt{b^2-4ac}$$
$$b^2=9(b^2-4ac)$$
$$8b^2=36ac$$
$$b=\pm \sqrt{\frac{36ac}{8}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculating the Inverse of a matrix. $A)$
Solve:
$x_1-x_2=1$
$-x_1+2x_2-x_3=0$
$\vdots$
$-x_{99}+2x_{100}=0$
$B$)Deduce the inverse of
A=\begin{pmatrix}
1 & -1 & 0 & \cdots& \cdots & 0 \\
-1 & 2 & -1 & \ddots& \cdots &\vdots\\
0 & -1 & 2 &-1& \ddots& \vdots\\
\vdots & \ddots & \ddots... | $\begin{pmatrix}
1 & -1 \\
-1 & 2
\end{pmatrix}^{-1}=\begin{pmatrix}
2 & 1 \\
1 & 1
\end{pmatrix}$
$\begin{pmatrix}
1 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 2
\end{pmatrix}^{-1}=\begin{pmatrix}
3 & 2 & 1\\
2 & 2 & 1\\
1 & 1 & 1
\end{pmatrix}$
$\begin{pmatrix}
1 & -1 & 0 & 0\\
-1 & 2 & -1 & 0\\
0 & -1 & 2 & -1\\
0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
complex number written in $re^{ix}$ form where $r<0$
The complex number $\left(\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2}i}\right)^{1/3}$ can be written in the polar form as $r(\cos x + i\sin x)$. If $r<0$, find the smallest positive value of $x$ in degrees.
$\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i\pi/3}$. Thus $\left(\... | Just do it.
$z= re^{i\theta} = r(\cos \theta + i\sin \theta)$ where $r = \pm |z|$.
As $z = e^{i\frac {\pi}{18}}$ and $|z| = 1$ then $r = -1$ and
$e^{i\frac{\pi}{18}} = -1(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18})$.
We are asked to find the smallest positive $x$ so that -$1(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Minimum value of slope of line tangent to ellipse I would like to solve the following problem.
Given that the point $p$ lies on the ellipse $$\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1$$ and the line $y = ax$ passes through $p$, let $n$ be the smallest positive value of $a$. Compute $n$.
What first came to my mind w... | Not the best in math way answer because it is not general solution but it is based on my comment:
My idea is (for easy solution without hard work) to find the line
$y=αx$ that is tangent in the circle that contains the ellipse and
then test next few values with the real ellipse. [I know it is a hack
but also kno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the roots of $\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}$ I attempted the problem in two ways
Method 1
Resolved $\mathrm{ Z = X + iY}$ this led to a very big cubic equation and handling that became too difficult for me .
Method 2
I believe factorizing the equation is a far better method, I don't know how ... | \begin{align}
z &= x + iy \\
\hline
\overline z &= x - iy \\
z^2 &= (x^2-y^2)+2ixy \\
|z|^2 &= x^2+y^2
\end{align}
\begin{align}
\overline z + 1 &= iz^2 + |z|^2 \\
(x-iy) + 1 &= i(x^2-y^2 + 2ixy)+(x^2+y^2) \\
(x+1) - iy &= (x^2-2xy+y^2) + i(x^2-y^2) \\
x+1-x^2+2xy-y^2 &= i(x^2-y^2+y) \\
\hlin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"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.