Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Small angle approximations - different answers I would like to approximate $$\frac{\cos^2{x}}{\sin(x) \tan(x)}$$ using the small angle approximations.
Throughout I will use $\sin(x) \approx x$, $\tan(x) \approx x$, $\cos(x) \approx 1 - \frac{x^2}{2}$.
Method 1:
$$\frac{\cos^2{x}}{\sin(x) \tan(x)} \approx \frac{\left(... | Because you are using less terms in your first expansion. What happens if you take $\tan(x) \approx \frac{x}{1 -x^2/2}$? It gives the second expansion. So the first one is a weaker approximation. In the first your are not keeping all terms of order $x^2$, but you do in the second. Note that the leading order terms for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\det\left(\begin{smallmatrix}a & 1 & 1\\1 & b & 1\\1 & 1 & c\end{smallmatrix}\right) > 0 $ then prove that $abc> -8$
If the value of the determinant $$\begin{vmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & c
\end{vmatrix} > 0 $$ then prove that $abc> -8$
I have calculated its determinant and got
$$abc - (a+b+c) > -2$$
... | Another counterexample to this claim is $(a,b,c) = (-1,-1,-k)$ for any $k\ge 8$. Then
$$
\begin{vmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & c
\end{vmatrix} = \begin{vmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -k
\end{vmatrix} = 4
$$
for any choice of $k$, but $abc = -k$ can be made as small as we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $\frac{2-\cos(x)}{\sin(x)}$ without differentiation I have to find the minimum value of the expression
$$\frac{2 - \cos x}{ \sin x}$$
Also $x$ lies between $0$ to $\pi$. One way is to find the minima using differentiation. But it is not taught in my grade so my teacher asked me to do it without differ... | Let $$k=\frac{2-\cos x}{\sin x}\Rightarrow 2-\cos x=k\sin x$$
So we have $$k\sin x+\cos x=2$$
Using $$|a\sin x+b\cos x|\leq \sqrt{a^2+b^2}$$
So we have $$|k\sin x+\cos x|\leq \sqrt{k^2+1}$$
$$2\leq \sqrt{k^2+1}\Rightarrow k^2+1\geq 4\Rightarrow k\geq \sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prime numbers of the form $p=m^2+n^2$ such that $p \mid m^3+n^3-4$
Find all prime numbers $p$, for which there are positive integers $m$ and $n$ such that $p=m^2+n^2$ and $p \mid m^3+n^3-4$.
I simplified this a little bit.
$$m^2+n^2 \mid m^3+n^3-4 =(m+n)(m^2+n^2-mn)-4 \\ \Longrightarrow m^2+n^2 \mid (m+n)mn+4 $$
The ... | This problem is from 2004 Silkroad Mathematical Competition. By the way, if you are into this problems, I think you should really visit AoPS forums, audience here at stack exchange are not very experienced with olympiadic problems.
Here is a short reasoning (similar to above). Check that, $p=2,5$ indeed works. Suppose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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infinitely many negative and infinitely many positive numbers Suppose that
$$x_1=\frac{1}{4}, \ x_{n+1}=x_{n}^3-3x_n.$$
Show that the sequence has infinitely many negative and infinitely many positive numbers.
My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be l... | The desired claim follows from the following two observations:
Claim. If $x_n \in \{-1/4, 1/4\}$, then $x_n \neq 0$ for all $n \geq 1$.
Proof. Let $p_1 = \operatorname{sign}(x_1)$ and $p_{n+1} = p_n^3 - 3p_n 4^{2 \cdot 3^{n-1}}$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^{3^{n-1}}$. Since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$
If $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is
Try: us... | We'll prove that
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq\frac{3\sqrt3}{4},$$ where the equality occurs for $a=b=c=d=\frac{1}{\sqrt3}$ only.
Indeed, let $a=\frac{x}{\sqrt3},$ $b=\frac{y}{\sqrt3},$ $c=\frac{z}{\sqrt3}$ and $d=\frac{t}{\sqrt3}.$
Thus, we need to prove that
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I determine the limit of this sequence? I need a refresher.
I would assume that the answer would be $\infty$, but I am not quite sure.
The problem reads:
Determine the limit of this sequence? $c_n$ = $\sqrt {n^2+n} - \sqrt {n^2 - n}$ .
I would need a refresher on this. That's all thank you.
| $$\begin{align*}
c_n &= \frac{\left(\sqrt{n^2+n}-\sqrt{n^2-n}\right)\left(\sqrt{n^2+n}+\sqrt{n^2-n}\right)}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{n^2+n-n^2+n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{2n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{2}{\sqrt{1+1/n}+\sqrt{1-1/n}}\\
&\to \frac{2}{\sqrt{1+0}+\sqrt{1-0}}\\
&= 1
\end{... | {
"language": "en",
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What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3... | You can easily prove by induction on $k$ that $2\times 4^k+1$ is divisible by $3$: It is for $k=1$, and if it is for some $k$, then
$$2\times 4^{k+1}+1=2\times 4^k\times 4+1=4(2\times 4^k+1)-3$$
Therefore $2^{2m+1}+1$ is divisible by $3$ for every natural $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\fr... | Other answers followed my suggestion in the comments.
Here's an alternative:
Let $z=\dfrac1x.$ Then we have $-z^2+z+4=0$, so, using the quadratic formula, $z=\dfrac{-1\pm\sqrt{17}}{-2}.$
Therefore $x=\dfrac1z=\dfrac{-2}{-1\pm\sqrt{17}}=\dfrac{2(-1\mp\sqrt{17})}{16}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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An algebra problem: Let $a+b+c=0$. Prove that $a^2+b^2+c^2 =6/5$ Let $a,b,c$ be nonzero real numbers such that $a+b+c=0$ and $a^3+b^3+c^3 = a^5 +b^5 +c^5$. Prove that $a^2+b^2+c^2 =6/5$
I tried to expand $(a+b+c)^5$ but I can't get term of $a^2+b^2+c^2$.
| When you see symmetric equations, try the following substitution $$a+b+c=s=0\qquad ab+bc+ac=q\qquad abc=p$$ And rewrite the equations in terms of $s,q,p$
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\implies a^3+b^3+c^3=3p$$
$$a^5+b^5+c^5\stackrel{(*)}=-5(ab+bc+ac)(abc)=-5qp$$
(*) means: in this case.
If you're wond... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding the Inverse Laplace Transform of $-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}$
I am trying to find the inverse Laplace transform of $$F(s)=-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}.$$
I proceeded as follows:
\begin{align}
F(s)&=-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+... | Your partial fractions are not correct for the second term:
\begin{equation}
\frac{(s+1)(s+2)}{s((s+1)^2+1)}=\frac{1}{(s+1)^2+1}+\frac{1}{s}.
\end{equation}
Then,
\begin{equation}
F(s) = \frac{1}{(s+1)}+\frac{1}{(s+1)^2+1}.
\end{equation}
Thus, you get the desired result!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The number of positive integers of 5 digits such that each digit is 1, 2 or 3, and all three of the digits appear atleast once. The number of positive integers of 5 digits such that each digit is 1, 2 or 3, and all three of the digits appear at least once.
Working:
Three 1, one 2, one 3, number of numbers = 5!/3! = 20 ... | Inclusion Exclusion is somewhat easier.
The total number of such integers, ignoring the requirement that all three digits appear, is $3^5$. We then subtract off the instances in which one specified digit does not appear, that's $3\times 2^5$. Then we add back the $3$ instances in which two specified digits do not app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I... | Hint: $x^2-11x+28=(x-7)(x-4)$. Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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Use numerical and graphical evidence to guess the value of the limit $\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$?
Use numerical and graphical evidence to guess the value of the limit $$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$
I am a Calculus 1 student, and I'm not sure what this problem in my textboo... | You did it well but it could be done faster.
Let $x=y^2$
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}=\underset{y \to 1} \lim \frac{y^6-1}{y-1}=1+y+y^2+y^3+y^4+y^5\,\, \to 6$$
If you want to also see how the limit is approached, let $x=1+t$
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}=\underset{t \to 0} \li... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sum_{k=0}^\infty \binom{2n+k+1}{n}/2^{2n+k+1}=1$. I was trying to find a closed form for the sum
$\sum_{k=0}^\infty \binom{2n+k+1}{n}/2^{2n+k+1}$.
According to Wolfram
https://www.wolframalpha.com/input/?i=sum+(2n%2Bk%2B1)!%2F(n!*n%2Bk%2B1)!*2%5E(2n%2Bk%2B1))+from+k%3D0+to+infinity
this sum evaluates to 1, but... | Preliminary
$$
\begin{align}
a_n
&=\sum_{k=0}^n\frac{\binom{k+n}{k}}{2^k}\tag{1a}\\
&=\sum_{k=0}^n\frac{\binom{k+n-1}{k-1}+\binom{k+n-1}{k}}{2^k}\tag{1b}\\
&=\sum_{k=0}^{n-1}\frac{\binom{k+n}{k}}{2^{k+1}}+\sum_{k=0}^n\frac{\binom{k+n-1}{k}}{2^k}\tag{1c}\\
&=\frac12a_n-\frac{\binom{2n}{n}}{2^{n+1}}+a_{n-1}+\frac{\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Algebra question/geometry solving the system 3x + 2y = 21, (7 - x)² + y² = 2².
what does this mean ?
How to get the answer (7−4/√13,6/√13)
| This means that you have a linear graph 3x+2y=21 and a circle, with origin at (7,0) and a radius of 2. To solve this system, you need to find the intersection point of these 2 graphs.
Edit:
My algebraic solution:
$3x+2y=21\\
2y=21-3x\\
y=\frac{21-3x}{2}\\
(7-x)^2+\frac{9}{4} \cdot (7-x)^2=4\\
\frac{13}{4} \cdot (7-x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorize the polynomial in the ring $\mathbb{Z}_5[x]$ I need to factor the polynomial $x^5+3x^4+x^3+x^2+3$ into the product of the irreducible ones in the ring $\mathbb{Z}_5[x]$.
The problem is I don't see any whole roots (I tried every possible divisor of 3) in the given polynomial. Does it mean that it is alr... | It is easy to check that $-1$ is a root, and by Horner's algorithm, you obtain
$$x^5+3x^4+x^3+x^2+3=(x+1)(x^4+2x^3-x^2+2x-2).$$
Similarly, the quotient has $2$ as a root and
$$ x^4+2x^3-x^2+2x-2=(x-2)(x^3-x^2+2x+1).$$
The latter quotient has $-2$ as a root, and
$$x^3-x^2+2x+=(x+2)(x^2+2x+3),$$
so that $$x^5+3x^4+x^3... | {
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"timestamp": "2023-03-29T00:00:00",
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Any easy way to evaluate this double integral? $$ \iint_S \frac{x^2+xy+3y^2}{2x^2+xy+2y^2}dxdy$$
where
$$S=\left\{(x,y)\vert x^2+y^2\leq1 \right\}$$
I tried
$$x=r\cos \theta, \quad y=r\sin\theta$$ substitution but it still not clear.
Is polar-coordinates the only solution for this?
Or how can I solve this easier?
|
I tried
$$x=r\cos \theta, \quad y=r\sin\theta$$ substitution but it still not clear.
This is a good idea. After simplifying (see below), you get:
$$\frac{x^2+xy+3y^2}{2x^2+xy+2y^2} \xrightarrow{\color{red}{(\star)}} 1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}$$
Don't forget the extra $\color{blue}{r}$ when con... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer How many integer numbers, $x$, verify that the following
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}
is an integer?
I managed to do:
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
... | You're off to a good start. Now note that the denominator $x^2+4x+5$ is quickly larger than the numerator $3x+19$; you can quickly reduce the problem to only finitely many values for $x$ to check.
More details: (Hover to show)
The fraction is certainly not an integer if the denominator is greater than the numerator, ... | {
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"timestamp": "2023-03-29T00:00:00",
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Definite integral containing 2 trig functions and a square root function $$\int_{-\pi/4}^{\pi/4}\bigl(\cos x+ \sqrt {1+x^2}\sin^3x \cos^3x \bigr)\, dx $$
This question is from a math GRE practice test
I've tried to solve this integral for 2 days... starting to think it is a typo
The answer is $\sqrt2$ but I need to k... | Let $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\cos(x)+\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$ $$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$
For the second integral, notice $$f(x)=\sqrt{1+x^2}\sin^3(x)\cos^3(x)$$ is an odd fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left... | In response to reading some of the comments, I just want to say that the L'hospital calculation is not horrible, if one works a little strategically. Of course, this is not entriely tedium free...
Let $$g(x) = \frac{\log(1+x)}{x}.$$ We have $g \to 1,$ and $g' \to -1/2$ as $x \to 0.$ (you'll have to show this for $g'$... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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If $x, y, z$ are three distinct positive integers, where $x + y + z = 13$ and $xy, xz, yz$ form an arithmetic... If $x, y, z$ are three distinct positive integers, where $x + y + z = 13$ and $xy, xz, yz$ form an increasing arithmetic sequence, what is the value of $(x + y)^z$ ?
I've been trying to solve this by forming... | This is actually not too bad to brute force since $x + y + z = 13$ and they are all positive integers. There are actually only $11$ possible values for $x$: $1, 2, 3, \ldots, 11$.
However, with a bit of analysis, we can reduce the number of cases.
Since $xz - xy = yz - xz$, we have that $2xz = xy + yz = y(x+z) = y(13-y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that the largest eigenvalue of $A$ lies in the given interval
Show that if the given matrix $A$ is positive semi-definite then the largest eigen value of $A$ lies in the interval $(6,7)$.
$$A=\begin{bmatrix}
5&1&1&1&1&1\\
1&2&1&0&0&0\\
1&1&2&0&0&0\\
1&0&0&1&0&0\\
1&0&0&0&1&0\\
1&0&0&0&0&1
\end{bmatrix}$$
My try... | There exist $2\times 2$ orthogonal matrix $Q_2$ and $3\times 3$ orthogonal matrix $Q_3$ such that
$$Q_2\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}\sqrt{2}\\0\end{bmatrix}\quad\text{and}\quad Q_3\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}\,,$$
e.g. Householder reflections. Defin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can the number $2^n + n^2$ be divisible by $5$ for some natural number $n$? Can the number $2^n + n^2$ be divisible by $5$ for some natural number $n$?
I found some solutions, i.e., $n=6, n=8, n=12, n=14$ but not 18 or 20 by trial and error and somewhere I realized that the last digit of either of $2^n$ or $n^2$ should... | Hint:
First observe that if $n\equiv 0\mod 5$, there are no solutions. Next, the non-zero squares modulo $5$ are congruent to $\pm 1$, hence if $2^n+n^2\equiv 0\bmod 5$, $\;2^n\equiv\mp 1$.
Now, by lil' Fermat, $\;2^n\equiv 2^{n\bmod 4}\mod 5$, so that
*
*either $n\equiv \pm 1\mod 5$, which implies $n^2\equiv 1\mod... | {
"language": "en",
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"source": "stackexchange",
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can re... | 1) The second method is wrong because of a silly mistake.
2) The first method is wrong because apart from discriminant, it is also important to note that there are restrictions on the values of $ B $ and $ C $ , and thus their are restrictions on $ tan C $, and $ tan B $, and thus their are restrictions on $ p $.
When... | {
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"source": "stackexchange",
"question_score": "6",
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$S=\left \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \right \}$ $S=?$ If $n\in\mathbb{N}, n\geq2$ and $S=\big \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \big \}$ then $S=?$ The right answer is
$$
S=\left \{\operatorname{ctg}\frac{k\pi}{n} |1\leq k\leq n-1;k\in\mathbb{N}\right \}
$$
I started like this $\left(\frac{z+i}{z-i}\righ... | $$\frac{z+i}{z-i} = \frac{x + (y+1)i}{x + (y-1)i} = \frac{x^2 - y^2 + 1}{x^2 + (y-1)^2} + \frac{2xy}{x^2 + (y-1)^2}i = r e^{i\theta}$$
so
$$r = \vert \frac{z+i}{z-i} \vert = \sqrt{ \frac{(x^2 - y^2 + 1)^2}{(x^2 + (y-1)^2)^2} + \frac{4x^2y^2}{(x^2 + (y-1)^2)^2} } = 1$$
which is
$$ (x^2 - y^2 + 1)^2+4x^2y^2 = (x^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can this binomial expansion result in two different approximations of root 2? I have been working on a problem on approximating $\sqrt{2}$ using the first three terms of the following binomial expansion, and a substitution of $x = -\frac{1}{10}$ :
$$(4 - 5x)^.5 = 2 - \frac{5x}{4} - \frac{25x^2}{64}$$
After substit... | Let $f(x) = 2 - \frac{5x}{4} - \frac{25x^2}{64}$.
Let $c$ denote the value $f(-\frac{1}{10}) = \frac{543}{256}$.
We can write
$$\tag 1 \frac{3}{\sqrt 2} = c + \varepsilon$$
If we take the OP's route 1 (but keeping $\varepsilon$), then
$$\tag 2 \sqrt 2 = \frac{3}{c+\varepsilon}$$
We can check that the 'route 1 approxima... | {
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"source": "stackexchange",
"question_score": "1",
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Knowing that $a+b\equiv 1 \pmod{7^{n+1}}$ show that $a^7+b^7\equiv 1 \pmod{7^{n+2}}$ Knowing that $a,b$ are prime integers and $a+b\equiv 1 \pmod{7^{n+1}}$ show that $a^7+b^7\equiv 1 \pmod{7^{n+2}}$
I used $a^7+b^7=(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$ and tried to show that $(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^... | It's not true. Take $a=7, b=43$. Then $a+b\equiv 1\bmod 49$, but $a^7+b^7\equiv 295\bmod 343$.
In fact, I would think that the primality or otherwise of $a$ and $b$ is irrelevant, given that we are only interested in their values mod $7^{n+1}$; but perhaps if you require that they be co-prime to $7$, your result might ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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let $a_{k}=\frac{k^4-17k^2+16}{k^4-8k^2+16}$ real number for $5≤k\in N$ If
$a_{k}=\frac{k^4-17k^2+16}{k^4-8k^2+16}$ real
number for $5≤k\in N$
Then find :
$\lim_{n\to +\infty} a_{5}a_{6}a_{7}...a_{n}$
My try :
$k^4-17k^2+16=(k^2-1)(k^2-16)$
And
$k^4-8k^2+16=(k-4)^2$
But how I complete
Answer is $\frac{1}{14}$
| Let $a_k = p_k/q_k$, where $p_k = k^4 - 17k^2 + 16 = (k-4)(k-1)(k+1)(k+4)$, and $q_k = k^4 - 8k^2 + 16 = (k-2)(k-2)(k+2)(k+2)$. Then $$\prod_{k=5}^n a_k = \prod_{k=5}^n \frac{(k-4)(k-1)(k+1)(k+4)}{(k-2)(k-2)(k+2)(k+2)}.$$ This is a telescoping product; for instance, $$\prod_{k=5}^n \frac{k-4}{k-2} = \frac{\prod_{k=1}... | {
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"source": "stackexchange",
"question_score": "2",
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Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$ I have happened to have proved this sum while attempting to prove another summation. Let
$$S_n=\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$
${2k \choose k} $ is the coeff... | We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. Recalling the generating function of the central binomial coefficients we can write
\begin{align*}
[z^n]\frac{1}{\sqrt{1-4z}}=\binom{2n}{n}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| By the mean value theorem and since $(\sqrt{1+x})'=\frac{1}{2\sqrt{1+x}},$ we obtain: $$\frac{\sqrt{1+b}-\sqrt{1+a}}{b-a}=\frac{1}{2\sqrt{1+c}},$$ where $3<c<8$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\l... | i) Consider $$ A_n :=\int^{\frac{1}{n-1}}_{ \frac{1}{n} } \
\frac{1}{x}\ dx - (n-1) \bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg] $$
Hence $F(\frac{1}{n-1})=A_n +A_{n+1}+\cdots $
So $$ A_n\bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg]^{-1} \rightarrow
\frac{1}{2}$$
so that $\frac{F(\frac{1}{n-1}) }{ \frac{1}{n-1} } \rightarrow
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234229",
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"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Prove by mathematical induction that $\sum_{i=0}^n (2i-1)^2 = \frac{n(2n+1)(2n-1)}{3}$ One of my homework problems is to prove that $\sum_{i=1}^n (2i-1)^2 = \frac{n(2n+1)(2n-1)}{3}$
I already completed the basis step
$[2(1)-1]^2 = 1 $
$\frac{(1)[2(1)+1][2(1)-1]}{3} = 1$
Then I assumed that the proposition was true for ... | What you have to prove is $$\frac{k(2k+1)(2k-1)}{3}+(2k+1)^2=\frac{(k+1)(2k+3)(2k+1)}{3}$$
The left-hand side is $$\frac{k(2k+1)(2k-1)+3(2k+1)^2}{3}=\frac{(2k+1)(2k+3)(k+1)}{3}$$
$$(2k+1)(2k^2-k+6k+3)$$ and $$(2k+3)(k+1)=2k^2-k+6k+3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the condition on $p$ such that equation $x^2+3 \equiv 0 \pmod{4p^2}$ has roots
Problem: Find the condition on $p$ such that equation $x^2+3 \equiv 0 \mod 4p^2$ has roots with $p$ is a prime and find the number of roots of this equation.
My solution: $x^2+3 \equiv 0 \mod 4p^2 \Rightarrow x^2 \equiv -3 \mod 4p^2 \... | Here we go. I leave the $p=2$ case to you, and will assume $p>2$.
Now, $x^2+3\equiv 0\pmod{4p^2}$ if and only if $x^2\equiv 1\pmod{4}$ and $x^2\equiv -3\pmod{p^2}$. For the former, it suffices to have $x$ to be odd, that is, $x\equiv 1\pmod{2}$.
Now, we turn our attention to the latter. If $p=3$, observe that we don't ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Showing that generating function coefficients agree with recurrence relation I have the recurrence relation $a_n = 3a_{n-1} + 4a_{n-2}$ with $a_0 = 1, a_1 = 0$. The solution to this is $a_n = \frac{4^n}{5} + \frac{4(-1)^n}{5}$. The generating function is $g(x) = \frac{1-3x}{1-3x-4x^2} = (1-3x)(\sum_{i=0}^{\infty}(-1)^i... | One way is expanding $\sum_{n=0}^\infty a_nx^n$. Note that $a_n=\frac{4^n}{5} + \frac{4(-1)^n}{5}$ evaluated at $n=0$ and $n=1$ coincides with the stated $a_0=1, a_1=0$, so that we can focus at
\begin{align*}
a_n=\frac{4^n}{5} + \frac{4(-1)^n}{5}\qquad\qquad n\geq 0
\end{align*}
Recalling the geometric series formula ... | {
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Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2)... | The first equation is equivalent to $(x+y)^2=2xy(1+xy)$.
Now multiplying the second equation by the previous relation yields: $$(x+y)^3=(2xy)^3$$ thus $x+y=2xy$ and then: $$(x+y)^2=2\times 2x^2y^2=2(x^2+y^2)\iff (x-y)^2=0$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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$(X^2-1)^2+(X^2-1)-12=0$ has 2 solutions only or 3? I am asked to solve for X using substitution:
$(X^2-1)^2+(X^2-1)-12=0$
Let $U = X^2-1$
$U^2+U-12=0$
$U^2+4U-3U-12=0$
$U(U+4)-3(U+4)=0$
$(U+4)(U-3)$
Then:
$U-3=0$
$U=3$
$X^2-1=3$
$X^2=4$
$X=\pm2$
This is the provided solution by my textbook.
However:
$U+4=0$
$U=-4$
$X^... | Note that if $X^2=-3$, this does not mean that $X=i\sqrt 3$. There is another solution $X=-i\sqrt 3$ that you have neglected. In either case, the reason your textbook might not have included $X=\pm 2,\pm i\sqrt 3$ as all four solutions is because it was only seeking for real solutions, of which there are only $\pm 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $\lim\limits_{n \to \infty}n\int_{0}^{1}(\cos x-\sin x)^ndx$? I want to compute
$$\lim\limits_{n \to \infty}n\int_{0}^{1}(\cos x-\sin x)^ndx$$
Someone helped me find the limit of the integral, which is $0$, but now I can't figure out this one. Also tried squeeze theorem but I only get only one side of it t... | Alternative solution
For $n\ge 4$, we have
\begin{align*}
I_n &= n\int_0^{1/\sqrt{n}} (\cos x - \sin x)^n \mathrm{d} x
+ n\int_{1/\sqrt{n}}^{\pi/4} (\cos x - \sin x)^n \mathrm{d} x\\
&\quad + n\int_{\pi/4}^1 (\cos x - \sin x)^n \mathrm{d} x\\
&= I_{n, 1} + I_{n, 2} + I_{n, 3}.
\end{align*}
Let us prove that $\lim_{n\to... | {
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"source": "stackexchange",
"question_score": "9",
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IMO 1984: Prove that $0 ≤ yz + zx +xy −2xyz ≤ \frac {7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$ I tried to solve this inequality because I find exotic. Actually, I didn't look at the right solution. Because before I look at the right solution, I want to know if my solution i... | Assume $x= \max\{\,x,\,y,\,z\,\}\,\therefore\,3\,x\geqq 1$.
For $z= 1- x- y$, we need to prove
$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$
We have
$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$
$$= \left \{ 4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,... | {
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"url": "https://math.stackexchange.com/questions/3245030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expre... | $$c^2=(a^2-b^2)^2+(2ab)^2$$ $$=(a^4-2a^2b^2+b^4)+(4a^2b^2)$$ $$=a^4+2a^2b^2+b^4=(a^2+b^2)^2$$
$$\Rightarrow c=a^2+b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Determinant of a specific block matrix Let $A$ be the following block matrix
$$A = \begin{pmatrix}
0 & A_{12} & A_{13} \\
A_{21} & A_{22} & 0 \\
A_{31} & 0 & A_{33}\\
\end{pmatrix}$$
I am finding difficulty to show that
$$\det A = \det \begin{pmatrix}
0 & A_{12} \\
A_{21} & A_{22} \\
\end{p... | This is not true. Counterexample:
$$
A=\left[\begin{array}{cc|cc|cc}
0&0&1&0&0&0\\
0&0&0&0&0&1\\
\hline
1&0&0&0&0&0\\
0&0&0&1&0&0\\
\hline
0&0&0&0&1&0\\
0&1&0&0&0&0
\end{array}\right].
$$
We have $\det(A)\ne0$ because $A$ is a permutation matrix, but as both $A_{22}$ and $A_{33}$ are singular,
$$
\det\pmatrix{0&A_{12}\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Closed form for $\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}$ By Mathematica, we find $$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$
How to find the closed form for general series:
$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \ (p\ge 3)$$
| We can make use of the following representation
$$\sf 2\arcsin^2z=\sum\limits_{n\geq1}\frac {(2z)^{2n}}{n^2\binom {2n}n}, \ z\in[-1,1]$$
Which gives integrating once with respect to $\sf z$ from $\sf 0$ to $\sf x$:
$$\sf 4\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3 \binom{2n}{n}}$$
So the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Jensen gap for a real-valued random variable $X \geq 0$ For a real-valued random variable $X \geq 0$,
We have $1 - \frac{1}{1+E \left[x\right]} \geq E \left[ 1 - \frac{1}{1+x} \right]$ (Jensen's inequality).
We want to get a tight constant gap between $1 - \frac{1}{1+E \left[x\right]}$ and $E \left[ 1 - \frac{1}{1+x} \... | Taylor expanding $\frac{1}{1+x}$ around $x = E[x]$ gives us
\begin{align*}
\text{Gap} &= \left|1 - \frac{1}{1+E[x]} - E\left[1 - \frac{1}{1+x}\right] \right| \\&= \left|E\left[\frac{1}{1+x}\right] - \frac{1}{1+E[x]} \right| \\
&= \left|\frac{1}{1+E[x]} - E\left[\frac{x - E[x]}{(1 + E[x])^2}\right] + E\left[\frac{(x - ... | {
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"url": "https://math.stackexchange.com/questions/3249677",
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"source": "stackexchange",
"question_score": "1",
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Advanced algebra-precalculus Find the value of $\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}$ when $\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}$ = 1, where $w,x,y,z \in \mathbb R $ .
I have tried setting one variable equal to zero, two variables equal to zero, and ... | Let's try your method! Set $z=w=0$, then:
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}=1 \Rightarrow \\
\frac{x}{y}+\frac{y}{x}=1 \Rightarrow x^2+y^2=xy\Rightarrow x^2+y^2\ge 2|xy|>xy$$
So, there is no real solution.
However, for complex numbers:
$$\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$. Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$.
I started from considering
$$\begin{align}\sin A+\sin B+\sin (180^o-A-B) &= \sin A+\sin B+\sin(A+B)
\\&=\sin A+\sin B+\sin A\cos ... | Also, we can make the following.
Let $b^2+c^2-a^2=x$, $a^2+c^2-b^2=y$ and $a^2+b^2-c^2=z$.
Thus, $x$, $y$ and $z$ are positives, $a=\sqrt{\frac{y+z}{2}},$ $b=\sqrt{\frac{x+z}{2}},$ $c=\sqrt{\frac{x+y}{2}}$ and in the standard notation
$$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{2S}{bc}=\sum_{cyc}\frac{\frac{1}{2}\sqrt{\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Elementary Second partial derivative If $V=\frac{xy} {(x^2+y^2)^2}$ and $x=r\cos\theta$, $y=r\sin\theta$ show that $\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=0$
My attempt :
$\frac{\partial {V}} {\partial{x}} =y\cdot\frac{y^2-3... | Hint: Write $$V=\frac{\sin2\theta}{2r^2}.$$Now take partial derivative with respect to $r$ and $\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Do I have a chance to get a closed form for this integral? I conjecture that
$$\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x \sim \frac 1 {z^2}$$
when $z\to \infty$.
Maybe there should be also equality.
| As Paul Enta already answered, the integral can be exactly computed.
$$J=\int_z^\infty \exp\left({- \frac { \pi^2}{16}(y-x)^2}\right) \,d y=\frac{2 }{\sqrt{\pi }}\left(1+\text{erf}\left(\frac{\pi}{4} (x-z)\right)\right)$$
$$I=\int \frac{\cos \left(\frac{\pi x}{2 z}\right)}{z}J \,dx=\frac{2 e^{-\frac{1}{z^2}}}{\pi ^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is it possible to have 2 different but equal size real number sets that have the same mean and standard deviation? By inspection I notice that
*
*Shifting does not change the standard deviation but change mean.
{1,3,4} has the same standard deviation as {11,13,14} for example.
*Sets with the same (or reversed) sequ... | You seem to like the set $\{1,3,4\}$. Here are six more three element sets having the same mean and standard deviation as your given set. \begin{align*}
&\frac{202}{171} & &\frac{1}{171} \left(583+\sqrt{19842}\right) & &\frac{1}{171}
\left(583-\sqrt{19842}\right) \\
&\frac{688}{171} & &\frac{1}{171} \left(340+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it... | We have
\begin{split}\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac1k - \frac1{k+1}&=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\dots+\left(\frac1n-\frac1{n+1}\right)\\&=1-\frac1{n+1}=\frac n{n+1}.\end{split}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove that $HK$ passes through a fixed point.
Point $A$ lies on the perpendicular bisector of $BC$. $M$ and $N$ are points respectively in line segments $AB$ and $AC$ such that $MN$ is tangent to the incircle of $ABC$ at point $H$. $MP, NQ \perp BC$ ($P,Q \in BC$). The intersection of $MQ$ and $NP$ is point $K$. Prov... | We can analytically prove that the straight line spanned by $HK$ passes through the midpoint of $BC$ as follows. Let the side $BC$ of the triangle lies along $x$-axis, and the origin $O$ is the midpoint of $BC$. Let ther vertex $A$ is a point $(0,a)$ and the vertices $B$ and $C$ are points $(-b,0)$ and $(b,0)$ respect... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Investigate $\sum_{n\geq1}\frac{3^nx^n\left(\frac{1}{3}-x\right)}{n}$ for uniform convergence on $x\in [0, \frac{1}{3}]$: is my solution correct?
Exercise:
Investigate $$\sum_{n\geq1}\frac{3^nx^n\left(\frac{1}{3}-x\right)}{n}$$ for uniform convergence on $x\in [0, \frac{1}{3}].$
Graphs:
From a few graphs, it seems ... | Your solution shows pointwise convergence, but not uniform. To prove uniform convergence, we can use some calculus and the Weierstrass M-test. Define
$$g_n(x) = \frac{3^nx^n(\frac13-x)}{n}.$$
The function $g_n$ is continuous and non-negative on $[0,\frac13]$, and $g_n(0)=g_n(\frac13)=0$, so $g_n$ will take a maximum so... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x+y$ is a factor of $x^2$ prove it's also a factor of $y^2$ If $X+Y$ is a factor of $X^2$ prove $X+Y$ is a factor of $Y^2$.
I have tried the rmainder theorem, attempted factorisation but those don't work.
| $x,y$ and $k$ are integers.
$x^2=k(x+y)$;
$((x+y)-y)^2=$
$ (x+y)^2-2y(x+y)+y^2=k(x+y);$
$y^2=k(x+y) -(x+y)^2 +2(x+y)y;$
Hence?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\ 4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx$ How to prove that
$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$
Where $\chi_2(x)=\sum_{n=1}... | This approach is pretty identical to Cornel's solution posted on his FB page.
using the fact that $\quad\displaystyle \sum_{n=1}^\infty a_{2n}=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty (-1)^na_n\right),\ $ we have
\begin{align}
\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}&=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Counterexample to the primality test This is a generalization of this claim .
Can you provide a counterexample to the following claim?
Let $n$ be a natural number greater than two . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $P_n^{(a)}(x)=\left(\frac{1}{2}\righ... | The claim is true.
It is true that if $n$ is a prime number, then $P_n^{(a)}(x)\equiv x^n\pmod{x^r-1,n}$.
Proof :
We have, by the binomial theorem,
$$\begin{align}P_n^{(a)}(x)&=\frac 12\left(\left(x-\sqrt{x^2+a}\right)^n+\left(x+\sqrt{x^2+a}\right)^n\right)
\\\\&=\frac 12\sum_{i=0}^{n}\binom nix^{n-i}\bigg(\bigg(-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.
The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....
I tried the polynomial remainder theorem but I am not sure if I did it correctly.
By factor theorem definition, provided by W... | (1) Since $x^3=(x^3+1)-1=(x+1)(x^2-x+1)-1$, the remainder is $-1$.
(2) $\displaystyle x^{2007}=(x^3)^{669}=[(x+1)(x^2-x+1)-1]^{669}=-1+\sum_{k=1}^{669}(x+1)^k(x^2-x+1)^k(-1)^{669-k}$
The remainder is $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$ does not exist over $0 \leq x \leq 1 , 0 \leq y \leq 1$ To prove that double integral does not exist:
$$\iint\frac{x-y}{(x+y)^2}\,dx\,dy$$ over region $0 \leq x \leq 1 , 0 \leq y \leq 1$.
I put $x - y = u$ and $x+y = v$ and I got integral as
$$\iint \frac{u}{v^3}\... | Given Integral $$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^2} dx dy \\
& = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^2} dx dy \\ & = \int^1_0 \int^1_0 \cfrac{1}{(x+y)} dx dy - 2 \int^1_0 \int^1_0 \cfrac{y}{(x+y)^2} dx dy \\ & = \int^1_0 \ln(x+y) \bigg|_0^1 dy + 2 \int^1_0 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you come up with great and ingenious questions In mathematics how are the professionals and great authors able to come up with such ingenious questions which which are so difficult yet elegant.
Like in integral calculus it involve a substitution or adding and subtracting sometimes or a trick which would make you... | I can show, how it can work in creating of inequalities.
We know that the $uvw$ method (see here: https://artofproblemsolving.com/community/c6h278791 or here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mo&paperid=194&option_lang=rus ) is very useful.
Sometimes this method gives a proof without great eff... | {
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Development of the sine function into the power series I would like to check my solution for the development of the real sine function, $f(x) = \sin x$ into the power series. Here is my solution:
First, we have that $\mathcal{D}_f = \mathbb{R}$. Now, we have to check if the given function is infinitely diferentiable an... | Using Taylor's theorem: $f(x)=\sum_{n=0}^{\infty}\dfrac {f^{(n)}(0)}{n!}x^n$, as you appear to have done, you should get $\sin x=\sum_{n=0}^\infty(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.
Note that $\sin n\frac{\pi}2$ takes values $0,1,0,-1$ in succession.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence and divergence of three series Which of the following is true.
A)$ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$ does not converge.
B) $ \sum_{n=1}^{\infty} \dfrac{1}{n}$ converges.
C) $ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(n+m)^2}$ converges.
D) $ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{... | Note that
\begin{align*}
\sum_{m = \mathbf{0}}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(n + m)^2} &= \sum_{n=1}^{\infty} \frac{1}{(n + 0 )^2} + \sum_{n=1}^{\infty} \frac{1}{(n + 1)^2} + \sum_{n=1}^{\infty} \frac{1}{(n + 2 )^2} + \ldots \\\\ &= \sum_{n=1}^{\infty} \frac{1}{(n )^2} + \sum_{n=2}^{\infty} \frac{1}{(n )^2} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum\limits_{cyc}\sqrt{a+11bc+6}\geq9\sqrt2.$
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc+abc=4.$ Prove that:
$$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}+\sqrt{c+11ab+6}\geq 9\sqrt2.$$
The equality occurs for $(a,b,c)=(1,1,1)$ and again for $(a,b,c)=(2,2,0)$ and for the cyclic permutations of... | Michael Rozenberg actually gave a proof. I do a little bit by the Buffalo Way to prove that
\begin{align}
&\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\\
\geq\ & 81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3.
\end{align}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | Your first step is wrong. It should be $$\sqrt{3x+7}-\sqrt{x+2}=1\implies\sqrt{3x+7}=1+\sqrt{x+2}$$ so we have $$3x+7=(1+\sqrt{x+2})^2$$ from which I think you can continue.
Note that as a check to your textbook solution, at $x=-2$, we get $$\sqrt{3(-2)+7}-\sqrt{-2+2}$$ which is indeed equal to $1$.
| {
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"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy$ where $A=\{(x,y)|4x^2+3y^2\leq1\}$
Find $\displaystyle\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy$ where $A=\{(x,y)|4x^2+3y^2\leq1\}.$
I got an answer, however the computer marks it as wrong and I can't find my mistake (if there is one, because sometimes it dee... | There is an error in the computation of the Jacobian: it is equal to $\frac r{2\sqrt3}$.
| {
"language": "en",
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Proofs involving strict inequalities let $a,b \in R$
Prove that if $3 \lt a \lt 5$ and $b= 2 + \sqrt{a-2}$ then, $3 \lt b \lt a$
My approach was simply to start with the first inequality and transform it into b and see what happens.
Subtracting 2 gives: $1 \lt a-2 \lt 3$
Taking the square root gives: $1 \lt \sqrt{a-2} ... | Let $b=f(a)=2+\sqrt{a-2}$, where $a\in (3,5)$. You have proved that $3<f(a)=b$. Now consider $$g(a)=a-f(a)=a-2-\sqrt{a-2}.$$
Then $g(a)$ is a increasing function since $g'(a)=1-\frac{1}{2\sqrt{a-2}}>1-\frac{1}{2\sqrt{3-2}}=\frac{1}{2}>0$. So $$a-b=a-f(a)=g(a)> g(3)= 3-2-\sqrt{3-2}=0.$$
More precisely, we have $0 < a-b ... | {
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Finding A point $Q$ on a surface such that Line $PQ$ get tangent to that surface
Consider the surface $f: x^2 + y^2 - 2z^2 = 1$ and Point $P= (1,1,1)$.
We want to find all points $Q$ on the surface such that line $PQ$ get
tangent to the surface. Also we want to find the point $Q_0$ with
above conditions such tha... | Given point $P=(1,1,1)$, find set $S$ of all $Q=(x,y,z)$ on the surface $x^2+y^2-2z^2=1$
subject to the constraint that the line $PQ$ is tangent to the surface at $Q$. Then locate the $Q$ in that set closest to $P$.
OP has already found that the set $S$ satisfies the equations
$$x+y-2z =1 $$
$$x^2 + y^2 -2z^2 = 1 $$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275112",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$
$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able t... | You noticed a pattern about the denominators, they are straight forward. How do you figure out what's going on with the numerators?
It's good to check ratios between consecutive terms if they are increasing, but that doesn't get you anywhere here. Anther good rule of thumb, take consecutive differences, perhaps differ... | {
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Convergence of Newton Method for monotonic polynomials Consider a polynomial $p : \mathbb R \to \mathbb R$ with $p'(x) > 0$ for all $x \in \mathbb R$. The function $p$ has exactly one real zero. Will the Newton method
$$x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)}$$
converge for all $x_0 \in \mathbb R$?
Intuitively I think t... | No, not necessarily. In particular, consider the polynomial,
$$p(x) = \frac{7}{2}x - \frac{5}{2}x^3 + x^5.$$
Note that
$$p'(x) = \frac{7}{2} - \frac{15}{2}x^2 + 5x^4,$$
a positive quadratic in $x^2$ with a negative discriminant $-\frac{55}{4}$, and hence is strictly positive everywhere. This means $p$ is strictly incre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Let $f(x) =x^3+x+1 \in \mathbb{Q} [x] $. If ideal $I=(f(x)) $, find the inverse of $x^2+x+1$ in quotient $\mathbb{Q} [x] /I$. Let $f(x) =x^3+x+1 \in \mathbb{Q} [x] $. If ideal $I=(f(x)) $, find the inverse of $x^2+x+1$ in quotient $\mathbb{Q} [x] /I$.
I am having trouble with this. What exactly is the quotient $\mathbb... | A polynomial in the quotient ring can be represented by $ax^2+bx+c.$
We want $(ax^2+bx+c)(x^2+x+1)\equiv 1 \pmod {x^3+x+1}$.
Note that $\color{blue}{x^3\equiv-x-1}$ and $\color{green}{x^4\equiv -x^2-x}$.
Thus $(ax^2+bx+c)(x^2+x+1)\equiv \color{green}{ax^4}+\color{blue}{(a+b)x^3}+(a+b+c)x^2+(b+c)x+c$
$\equiv \color{gree... | {
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"timestamp": "2023-03-29T00:00:00",
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$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$ My book has used the equality $$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$$ But its proof is not given.
When I open the LHS, I get $b^2c-bc^2+c^2a-ca^2+a^2b-ab^2$. I do not know how to proceed next.
| $$bc(b-c)+ca(c-a)+ab(a-b)=b^2c-a^2c-bc^2+ac^2+ab(a-b)=$$
$$=c^2(a-b)-c(a-b)(a+b)+ab(a-b)=(a-b)(c^2-ac-bc+ab)=$$
$$=(a-b)(c-a)(c-b).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that all multiples of 4 can appear in a Primitive Pythagorean Triple I have attempted by using the fact that the even numbers $b$ in a PPT $(a,b,c)$ can be generated using $ b = \frac{s^2 - t^2}{2} $ for any odd integers $s >t \ge 1$ that share no common factors.
We can express any odd numbers as $2k+1$ for some ... | $(4n^2-1)^2+(4n)^2=(4n^2+1)^2.$
All Pyth. triplets are of the form $(\,k(a^2-b^2),\,2kab,\, k(a^2+b^2)\,)$ with positive integers $k,a,b.$
All primitive triplets have $k=1$, with $a,b$ co-prime and such that $a-b$ is odd. These conditions are necessary but are also sufficient. (Along with, obviously, $a>b.)$
If $2ab=4... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving simple equation I can't seem to solve/simplify step by step to get from equation 3) to 4) as they do in this paper. As per the paper:
3) $p = \frac{p' + (r-R)p}{1+r}$
Because both sides of equation 3) involve the current price, $p$, the equation can be solved as follows:
4) $p = \frac{p'}{1+R}$
Could somebody... | We start with the original equation:
\begin{align*}
p = \frac{p' + (r-R)p}{1+r}
\end{align*}
We want to obtain $p$ as a function of the other values. So, we first want to collect all terms with $p$ on the same side. We do this by splitting the fraction and bringing one term over.
\begin{align*}
p &= \frac{p'}{1+r} + \f... | {
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"timestamp": "2023-03-29T00:00:00",
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Compute the integral in a closed form : $\int_0^{1}\operatorname{Li}_2(1-x)dx$ How I can find the closed form of the following integration :
$I=\int_0^{1}\operatorname{Li}_2(1-x)dx$
$J=\int_0^{1}\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-\frac{1}{x})dx$
$K=\int_0^{1}\ln(x)\operatorname{Li}_2(1-\frac{1}{x})dx$
Wolfr... | The integrals $J$ and $K$ can be computed using the polylogarithm identity
$$ \operatorname{Li}_2 \left(1 - \frac{1}{x}\right) = - \operatorname{Li}_2(1-x) - \frac{1}{2} \log^2(x) \, , \, x > 0 \, , \tag{1} $$
and the integrals
$$ \int \limits_0^1 \frac{[-\log(x)]^n}{1-x} \, \mathrm{d} x = n! \zeta(n+1) \, , \, n \in \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| The symmetry of the problem suggests to substitute $x = u+v$, $y = u-v$. Then the given constraint is
$$
3 = x^2 + xy + y^2 = 3u^2 + v^2
$$
and in particular $-1 \le u \le 1$.
We look for the minimal value of
$$
(5+x)(5+y) = (5+u)^2 - v^2 = (5+u)^2 - (3-3u^2) \\
= 4u^2 + 10 u + 22 \, .
$$
This expression is increasi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.
| You can use Cauchy-Schwarz inequality. $X+y=1-z$, thus $(1-z)^2 = (x+y)^2$ is less than or equal to $(1+1)(x^2+y^2)=2 (1-z^2)$. I get $z$ more than or equal to $-1/3$. $x^3+y^3+z^3= (x^3+y^3)+z^3=(x+y)(x^2-xy+y^2)+z^3= (3(x^2+y^2)-(x+y)^2)/2 + z^3$ doing the necessary substitutions I get $1-3 z^2 +3 z^3$, equals $5/9$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if so... | On applying Taylor series, and binomial expansion and taking linear degree terms in numerator to evaluate the limit,(all constant terms are powers of 2, they cancel out as they are same for both terms in numerator)
We get the limit as $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}. = -1 + \frac{{3} \choose {1} * 2^2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe... | Make an exponential generating function with $I_k$,
$$I_k: = \frac{1}{2} (1 + (-1)^k) \int_0^\infty \frac{ \log^{k}(u) }{u^2+1} du $$
Then
$$I(x)=\sum_{k=0}^\infty I_k\,\frac{x^k}{k!} = \frac{1}{2}\int_0^\infty \frac{ u^x + u^{-x} }{u^2+1} du $$
where an interchange of $\sum$ and $\int$ has been made. The integral ca... | {
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"url": "https://math.stackexchange.com/questions/3294446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Real values of $k$ such that $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots
There is only one real values of $k$ for which the quadratic equation $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots. Then the product of these two solutions is
What i tried.
$$kx^2+(k+3)x+k-3=0$$
for $k=0,$ i am getting $x=1$(only on... | There is no such $k.$
Since the product we're looking for is $$\frac {k-3}{k},$$ first we must have $k\ne 0.$ Also, since the roots are positive integers, it follows that their product is also a positive integer. Thus, $k\ne 3$ and we also have that $k-3$ and $k$ must have the same sign.
Having noticed this, we find t... | {
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"url": "https://math.stackexchange.com/questions/3294564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that if $a$ and $b$ are real numbers with $0 < a < b$ then $\frac{1}{b} < \frac{1}{a}$
Suppose a and b are real numbers. Prove that if $0 < a < b$
then $\frac{1}{b} < \frac{1}{a}$.
My attempt:
Given that $0<<$
We can write $b$ as $b = an$, where $n>1$
$$\tag1\frac{1}{a} = \frac{n}{an} = \frac{n}{b} $$
$$\tag... | Yes, your solution is correct.
Of course you could have multiplied both sides of your inequality by $$\frac {1}{ab}$$ to get the result in one shot.
| {
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"url": "https://math.stackexchange.com/questions/3295820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ dividing by $x^2-1$ have remainder $2x$ I need help in this problem.
Problem: Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ divided by $x^2-1$ has remainder $2x$ a... | Hint:
$f(x)=(x^2-1)(x-a)+2x$ and $f(2)=0$.
Solve for $a$.
$f(x)=x^3-ax^2+x+a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find modulus and argument of $\omega = {\frac {\sin (P + Q) + i (1 - \cos (P + Q))} {(\cos P + \cos Q) + i (\sin P - \sin Q) }} $ A past examination paper had the following question that I found somewhat difficult. I tried having a go at it but haven't come around with any possible double angle identities. How would on... | We have
\begin{align}
N
& := \sin^2(P+Q) + (1-\cos(P+Q))^2 = \sin^2(P+Q) + \cos^2(P+Q) + 1 - 2\cos(P+Q) \\
& = 2 (1-\cos(P+Q))
= 2\cdot2\sin^2\frac{P+Q}{2}
= 4\sin^2\frac{P+Q}{2}
\end{align}
and
\begin{align}
D
& = \cos^2P +\cos^2Q + \sin^2P + \sin^2Q + 2(\cos P\cos Q - \sin P \sin Q) \\
&= 2 +2(\cos(P+Q))
= 2(1+\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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If $\alpha$ and $\beta$ are the solutions of $a\cos \theta+ b\sin \theta= c$ show that If $\alpha$ and $\beta$ are the solutions of $a\cos \theta+ b\sin \theta= c$ show that
1) $\sin \alpha + \sin \beta = \dfrac{2bc}{a^2 + b^2}$
2) $\sin \alpha \sin \beta = \dfrac{c^2-a^2}{a^2+b^2}$
I couldn't even start the problem, ... | $$a\cos\theta = c - b\sin\theta \implies a^2\cos^2\theta = c^2-2bc\sin\theta+b^2\sin^2\theta \\ \implies (a^2-a^2\sin^2\theta) = c^2-2bc\sin\theta+b^2\sin^2\theta $$
$$(a^2+b^2)\sin^2\theta - (2bc)\sin\theta+(c^2-a^2)=0 \equiv Ax^2 + Bx +C = 0$$
As $\alpha$ and $\beta$ are the solutions of the given equation, $\sin\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) ... | A higher brow way of writing the same thing is to say the $n^{th}$ term is
$2^{\lfloor \log_2 n\rfloor}$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^{10}=1024$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
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How to derive this Bessel function relationship? $$J_n(x) = \frac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt$$
| Show that
$J_n(x) = \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt
$
I did my usual
naive plugging in of definitions
and, with some help from Wolfy,
it worked out.
A pleasant surprise.
Since
$J_a(x)
=\sum_{m=0}^{\infty} \dfrac{(-1)^m}{m!(m+a)!}\dfrac{x^{2m+a}}{2^{2m+a}}
$
and,
according to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find range of $m$ such that $|x^2-3x+2 | = mx$ has 4 distinct solutions
$|x^2-3x+2 | = mx$
I think that $(x^2-3x+2) = mx$ or $-(x^2-3x+2) = mx$
For $-(x^2-3x+2) = mx$, $D > 0$,
$m = 3 \pm \sqrt{8}$
For $(x^2-3x+2) = mx$, $D > 0$,
$m = -3 \pm \sqrt{8}$
Now. What do you think it's the range?
Note that $ |x^2-3x+2 | = m... | Solving $|x^2-3x+2|=mx$ is same as finding the intersection of the graph of $y=|x^2-3x+2|$ and the lines $y=mx$ passing through the origin. Here is a figure to better illustrate the idea.
By changing the value of $m$ we are changing the slope of the line passing through the origin. Unless this line is tangent or pass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$
using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
ho... | My solution without a calculator:
Since $2^2 = 4$ and $3^2=9$, we know that $\sqrt{7} = 2 +b $ and therefore
$$
\frac{a}{b} = \frac{2}{b} = \frac{2}{\sqrt{7}-2} = \frac{2}{(\sqrt{7}-2)} \frac{(\sqrt{7}+2)}{(\sqrt{7}+2)} = \frac{2(\sqrt{7}+2)}{7-4} = \frac{2}{3} \left(\sqrt{7}+2 \right)
$$
And since $2 < \sqrt{7} < 3$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the Range of $y=\frac{x+a}{x^2+bx+c^2}$ Given that $x^2-4cx+b^2 \gt 0$ $\:$ $\forall$ $x \in \mathbb{R}$ and $a^2+c^2-ab \lt 0$
Then find the Range of $$y=\frac{x+a}{x^2+bx+c^2}$$
My try:
Since $$x^2-4cx+b^2 \gt 0$$ we have Discriminant
$$D \lt 0$$ $\implies$
$$b^2-4c^2 \gt 0$$
Also
$$x^2+bx+c^2=(x+a)^2+(b-2a)(x+a... | This is not a complete solution. Just my opinion on the possible solution.
Cross multiplying and subtracting $x+a$ gives an equation:
$x^2y+x(by-1)+(yc^2-a)=0$
The discriminant turns out to be:
$y^2(b^2-4c^2)+y(4a-2b)+1$
Further the discriminant of the above equation is:
$16(a^2+c^2-ab)$
Which is certainly less than 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h... | The top simplifies
$$ (x-a)^2 -[(x-a)-h]^2 =$$
$$(x-a)^2 -[(x-a)^2-2h(x-a)+h^2]$$
$$=2h(x-a)-h^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} =... | By differentiating and setting to $0$ or plotting the graph, the minimum value is actually $49$, so there must be something wrong in your solution.
From $9 \sec^2 x = 16 \csc^2 x$ you cannot deduce $\tan x = \frac{4}{3}$. You have $\tan^2 x = \frac{16}{9}$, but $\tan x$ can either be $\frac{4}{3}$ or $-\frac{4}{3}$.
In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding a function with given conditions I want to find the equation of a function $f(x)$ such that the following conditions are met:
$(i)\ f(3) = 0, f(0) =1$.
$(ii)\ f(x)$ is an even function.
$(iii)\ f(x)$ has vertical asymptodes at $x=\pm 4$.
$(iv)\ f(x)$ has horizontal asymptode at $ y=2$.
Now since the function is... | We use building blocks having one or more of the wanted properties and put them together conveniently so that the final function $f$ has all wanted properties.
*
*We start with (iii) and take as function with asymptote at $x=\pm 4$ the rational function
\begin{align*}
x\to\frac{1}{(x-4)(x+4)}=\frac{1}{x^2-16}... | {
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"url": "https://math.stackexchange.com/questions/3305372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$
If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$
This problem was bothering me for a while, and I finally could not solve it. I tried taking the whole equation as $k$ but the ... | Consider (first & second terms) $$x\sin A=y\sin\left(A+\frac{2\pi}{3}\right)$$
$$x\sin A=y\left(\sin A\cos \frac{2\pi}{3}+\cos A\sin\frac{2\pi}{3}\right)$$
$$x=y\left(\frac{\sin A\cos \frac{2\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{2\pi}{3}}{\sin A}\right)$$
$$x=y\left(-\frac12+\cot A \frac{\sqrt3}{2}\right)$$
$$\cot A=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 1... | Because $y$ should be positive.
By your work it's:
$$2\int\limits_0^2\left(2-\frac{x^2}{2}\right)dx+\frac{1}{4}\cdot\pi(2\sqrt2)^2-\frac{1}{2}(2\sqrt2)^2.$$
I got $\frac{4}{3}+2\pi.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$
Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3... | Your expression can be verified by the following identity:
\begin{align}
& \; \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} \\
= & \:
\dfrac
{(a+b+c)^3-27abc}
{\left(
\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-
\sqrt[3]{bc}-\sqrt[3]{ca}-\sqrt[3]{ab} \,
\right)
\left[
(a+b+c)^2+
3(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find limit of $2x+ (4x^2 +3x - 2)^{1/2}$ when $x\to-\infty$ I have to find a limit $\lim _{x \to - \infty}{(2x + \sqrt{4x^2 +3x - 2})}$. I know, that the correct answer is $-3/4$, I've checked it in a graphing tool.
But could someone explain it to me, how to find it analytically, preferably without the Lhopital ... | $\begin{array}\\
ax-\sqrt{a^2x^2+bx+c}
&=(ax-\sqrt{a^2x^2+bx+c})\dfrac{ax+\sqrt{a^2x^2+bx+c}
}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{a^2x^2-(a^2x^2+bx+c)
}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{-bx-c}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{-b-c/x}{a+\sqrt{a^2+b/x+c/x^2}}\\
&=\dfrac{-b-c/x}{a+a\sqrt{1+b/(a^2x)+c/(a^2x^2)}}\\
&\to\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Calculating Decimal points of modulo operations I am not too sure of the step that has occurred here (I found this form a cryptography book but I do not understand how he finds the second 9).
The function found in the image is
$(2 \cdot 1)^{-1}(3 \cdot 5^2 + 2) = 2^{-1} \cdot 9 \equiv 9 \cdot 9 \equiv 13\,\mod\,17$
My... | In modulo arithimetic then expression $(2^{-1})$ does not mean $0.5$ which is completely out of the purview of modular arithmetic (which is about equivalence classes of integers).
Instead $(2^{-1})$ refers to the the solution to $2x \equiv 1\pmod n$
And $2x \equiv 1 \pmod {17}\iff x \equiv 9 \pmod {17}$[1]
So $2^{-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the area under a curve when the area is bounded by 3 curves. The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney.
Problem:
Find the area of the region bounded by the given curves and lines.
$$ y = x, y = \frac{1}{ \sqrt{x} }, x = 2 $$
Answer:
Let $A$ be the area we seek... | You have the bounds correct, but since the area is bounded above by $x$ and below by $\frac{1}{\sqrt{x}}$ the integral should be
$$\int_1^2 (x-\frac{1}{\sqrt{x}})dx $$
$$=(\frac{1}{2}x^2-2\sqrt{x} )|^2_1$$
$$=2 - 2\sqrt{2} - (\frac{1}{2}-2)$$
$$=\frac{7}{2}-2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/... | Well, in the first case, there is a sign error:
$$\frac{5y-1+12}{3} = \frac{-8y+4}{6}$$
$$5y-11 = \frac{-8y+4}{2}$$
It should be
$$5y+11 = \frac{-8y+4}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Choosing two numbers from set $\{10,11...,99\}$ that satisfy the given conditions Given the set of numbers $\{10,11,...,99\}$, with no repetitions and no order significance.
Let $A$ be the set of options of choosing pairs with the same tens number.
Let $B$ be the set of options of choosing only two even numbers.
Let... | In your attempt, you used the Multiplication Principle. Since the order of selection does not matter in the first two parts, I will use combinations.
The number of ways of choosing a subset with $k$ elements from a set with $n$ elements is
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
where $n!$, read "$n$ factorial", is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having problems with Exponents $1^n =1$ where n is a positive integer I understand.
But $1^{1/2}$ or $1^{0.5}$ is also $1$, I have difficulty in understanding that.
And does that mean $1^n$ where n is any positive real no equals $1$?
$2^{-5}$ = $1 / 2^5$ = $1/32 =0.03125 $
I can compute this but I really don't understa... | $1^N = 1$ for any real value of N.
You are OK with the case of N being integer. This is good start.
The case where $N$ is not integer, as in your example: $1^{0.5}$ (also referred to as $\sqrt{1}$) could be justified as follows:
What is the number $x$ that if you multiply it by itself, you get 1?
This calls for solving... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Sum of products of combinatorials In the proof of some proposition, it appears that the following statement should hold:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =2^{2n}. $$
However, using the definition of combinatorials does not help:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\be... | We have
$$\sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta=0}^{r-1} {n\choose \beta}
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta\ge 0} {n\choose \beta} [[0\le \beta\le r-1]]
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta\ge 0} {n\choose \beta}
[z^{r-1}] \frac{z^{\beta}}{1-z}
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
[z^{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
| $$a+b+c=4\,\Rightarrow\,(a+b+c)^2=16$$
$$(a^2+b^2+c^2)+2(ab+bc+ac)=16\Rightarrow2(ab+bc+ac)=9$$
Now:
$$(a+b+c)^3=(a^3+b^3+c^3)+3\left[(a+b+c)(ab+ac+bc)-abc\right]$$
$$64=28+3\left[4\times9-abc\right]\therefore\,abc=24$$
Now you need to try and simplify:
$${256=(a^4+b^4+c^4)+6(a^2b^2+a^2c^2+b^2c^2)+4(ab^3+ac^3+ba^3+bc^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Pair of straight lines parallel vs coincident
The equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel
lines if $\dfrac{a}{h}=\dfrac{h}{b}=\dfrac{g}{f}$ and the distance between the parallel lines is $2\sqrt{\dfrac{g^2-ac}{a(a+b)}}$ or $2\sqrt{\dfrac{f^2-bc}{b(a+b)}}$
This is given in many reference... | I should have read your links and their comments before starting an answer.
That would have saved a big waste of time.
I have posted an alternative proof in an answer to
Deriving conditions for a pair of straight lines to be parallel..
I do not think the approach using partial derivatives in that question is a good one... | {
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"url": "https://math.stackexchange.com/questions/3325860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.