Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute 100th derivative A friend suggested me a rather tricky problem, namely find the $100^{th}$ derivative of
$$
f(x)=\frac{x^2+1}{x^3-x}.
$$
I have computed the zeroth derivative
$$
\frac{x^2+1}{x^3-x}
$$
and the first derivative
$$
\frac{2x(x^3-x)-(3x^2-1)(x^2+1)}{(x^3-x)^2}=\frac{1-x^4-4x^2}{(x^3-x)^2}
$$
but I ... | $$\frac{x^2+1}{x^3-x}=\frac{x^2-1+2}{x^3-x}=\frac{1}{x}+\frac{2}{x(x-1)(x+1)}=$$
$$=\frac{1}{x}+\frac{2}{x+1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x+1}+2\left(\frac{1}{x+1}-\frac{1}{x}\right)=$$
$$=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Proof by induction - stuck on simple question! Question:
(Part 1) Show that the inequality
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(n-1)\cdot n} < \frac{1}{2}
$$
works for all natural numbers $n > 2$.
(Part 2) Deduce that for all natural numbers $n$, the following inequality holds:
$$
\frac{1}{... | Hints:
$1.\qquad\dfrac1{2\cdot 3}=\dfrac12-\dfrac 13,\quad \dfrac1{3\cdot 4}=\dfrac 13-\dfrac14,\;\dots,\;\dfrac1{(n -1)n}=\dfrac1{n-1}-\dfrac 1n$.
$2.\qquad \dfrac1{3^2}<\dfrac 1{2\cdot 3},\quad \dfrac1{4^2}<\dfrac1{3\cdot4},\;\dots\;,\dfrac1{n^2}<\dfrac1{(n-1)n}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Inverse of $A^T Y A$ Is there a formula for the inverse of a matrix
$$ X = A^\top Y A $$
in terms of $A$ and $Y$, given that $A \in F^{m \times n}$ is full rank with $m >n$, and $Y$ is positive definite?
| Even with the restriction $m>n$, it need not be the case that $X=A^\top Y A$ be invertible. As a counterexample, suppose $m=2,n=1$. Then the choice $$Y=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\; A=\begin{pmatrix} 1 \\ 1\end{pmatrix}$$
yields $$X = A^\top Y A = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a + b = \frac{1}{4}$, what is $a^3 + b^3$? I just want to know how to solve this problem.
If $a + b = \dfrac{1}{4}$, what is $a^3 + b^3$ equals to?
My work - but still doesn't give me an answer.
$a^2 + b^2 = (a + b)^2 - 2ab = \dfrac{1}{16} - 2ab$
$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = \dfrac{1}{64} - \dfrac{3ab}... | The value of $y = a^3+b^3$ contains a family of solutions (like a curve). You cannot establish a value for $y$ unless something else is specified.
For example besides $a+b = \tfrac{1}{4}$, let us use some artbitrary value $a-b = \lambda$. The solve for $a$ and $b$ in terms of $\lambda$ and put them into $y$
$$ y = \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$S_n = \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{n^4 +r^4+2r^3+2n^2r^2+2n^2r+n^2+r^2}})$ find$S_{100}$ $$S_n = \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{n^4 +r^4+2r^3+2n^2 r^2+2n^2r+n^2+r^2}})$$ find$S_{100}$
Now the denominator of this expression is to big and confusing.I don't know how can we reso... | Notice that $\displaystyle n^4 +r^4+2r^3+2n^2 r^2+2n^2r+n^2+r^2 = (n^2 + r^2 + r)^2 + n^2$
Hence
$\displaystyle S_n=$
$\displaystyle \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{(n^2 + r^2 + r)^2 + n^2}})$
$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{n}{n^2+r^2+r} $
$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the initial value problem $xyy' + xy' = 1$ and $y(1) = 0$ I can solve the differential equation, which is $y + y^{2}/2 = \ln(x) + C$.
But I cannot solve the IVP because I can't isolate for $y$ and find the value of $C.$
| We have (just reproducing your work for completeness, and combining some of the comment ideas):
\begin{align*}
xyy'+xy'&=1\\
xy'(y+1)&=1\\
(y+1)\,dy&=\frac{dx}{x}\\
\frac{y^2}{2}+y&=\ln(x)+C\\
0&=\ln(1)+C\\
0&=C\\
\frac{y^2}{2}+y&=\ln(x)\\
y^2+2y&=2\ln(x)\\
y^2+2y+1&=2\ln(x)+1\\
(y+1)^2&=2\ln(x)+1\\
y+1&=\pm\sqrt{2\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without usin... | The nearest integer greater than $4m^2 +2m +1/4$ is $4m^2 +2m +1$..And if we consider $a=b=c =m $ and $d=m+1$.We can get this minimum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x... | In two variables the epsilon-delta definition for $\lim_{\substack{x\to a\\ y\to b}}f(x,y)=L$ means that for every $\epsilon >0$ there exists a $\delta>0$ such that $\big|f(x,y)-L\big|<\epsilon$ whenever $0<\sqrt{(x-a)^2+(y-b)^2}<\delta$.
In your case, you want to show that $\big|f(x,y)-0\big|<\epsilon$ whenever $0<\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
If $u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$ , find the maximum and minimum value of $u^2$.
If $u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$ , find the maximum and minimum value of $u^2$.
This problem was bothering me for a while. The minimum value of $u $ seemed relatively... | $u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$
Let
$p = a \cos^2 x + b \sin^2 x$
$q = b \cos^2 x + a \sin^2 x$
and
$u = \sqrt{p} + \sqrt{q}$
Then $u^2 = p + q + 2 \sqrt{pq}$
Now
$p + q = a + b$
and
$\displaystyle pq = \frac{(a+b)^2}{4} - \frac{(a-b)^2}{8} - \frac{(a-b)^2}{8} \cos 4x$
If $\cos 4x = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x \to \infty} x^2\big(\ln(x\cot^{-1}(x))$
$$\lim_{x \to \infty} x^2\big(\ln(x\cot^{-1}(x))$$
I tried using the Series Expansion of the $\ln(x)$ but then got stuck in between. I also tried using the L'Hopital but the expression got quickly messy.
After applying L'Hopital for the first time, I got
$$\lim_{... | Note that as $x\to +\infty$,
$$\cot^{-1}(x)=\arctan(1/x)=\frac{1}{x}-\frac{1}{3x^3}+o(1/x^3).$$
Hence, from your work,
$$\frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}(x)\bigg)=
\frac{-x^3}{2}\bigg(-\frac{1}{x}\frac{1}{1+\frac{1}{x^2}} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\\
=\frac{-x^3}{2}\bigg(-\frac{1}{x}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of three non-linear equations Can anyone help me solve this system of equations?
$$a_1 a_3 -a_2 ^2=0$$
$$a_1+a_3-2a_2-16=0$$
$$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$
After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Than... | Subtract the first row from the last row of the system: then you get
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1 -16 a_2 -64 &=0
\end{split}
$$
Then multiply by 8 the second row and subtract it again from the third row:
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1 -16 a_2 -64 &=0
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this Ramanujan series
Prove how this series: $\frac{1}{1^3}\cdot\frac{1}{2} + \frac{1}{2^3}\cdot\frac{1}{2^2} + \frac{1}{3^3}\cdot\frac{1}{2^3} + \frac{1}{4^3}\cdot\frac{1}{2^4} +...= \frac{1}{6}(\log 2)^3+\frac{\pi ^2}{12}(\log 2) + (\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+..)$
I don't know how to begin but ... | This is by no means a complete answer but really a long comment. We could also ask if for a fixed number $k$ is it true that
$$
\sum_{n=1}^{\infty}\frac{1}{n^3k^n}=\text{Li}_{3}\left( \frac{1}{k}\right);
$$
where $\text{Li}_{3}()$ is the trilogarithm. Really we are computing $\text{Li}_{3}(\frac{1}{2}).$
We have the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Throwing a die three times - check my proof "A teacher rolls a 6-sided die (numbers from 1 to 6) three times. The lowest number will be the grade of the student.
Calculate the probability of each grade."
My approach is:
Firstly we determine the sample space $\Omega$. So $\Omega:= \{\{\omega_1, \omega_2, \omega_3\},~wi... | The grade $1$ occurs when $1,2$ or $3$ dice show $1$.
Only one die showing $1$:
$${3\choose 1}\cdot 5^2$$
Two dice showing $1$ each:
$${3\choose 2}\cdot 5$$
All three dice showing $1$:
$${3\choose 3}$$
Hence:
$$P(1)=\frac{{3\choose 1}5^2+{3\choose 2}5+{3\choose 3}}{6^3}\approx 0.421$$
Similarly:
$$P(2)=\frac{{3\choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Value of $(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}$ if $\beta$ is the root of $x^3-x-1=0$
If $\beta$ is the root of the equation $x^3-x-1=0$, find the value of
$$(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}.$$
This is what I tried:
$x=\beta$ is a root of $x^3-x-1=0,$
s... | The trick is that if we reduce $(x + a)^3$ modulo $x^3 - x - 1$, we'll get $3 a x^2 + (3 a^2 + 1) x + (a^3 + 1)$. Therefore, if $\beta$ is a root of $x^3 - x - 1$,
$$(1 + \beta)^3 = 3 \beta^2 + 4 \beta + 2, \\
(1 - \beta)^3 = 3 \beta^2 - 4 \beta.$$
Then we need to define the cube root in such a way that $(z^3)^{1/3} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\cos(x)=-\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$. Calculate $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$. If $\cos(x)= -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ a... | Your method is correct. Indeed, refer to the graphs:
$\hspace{1cm}$
$$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$
Also note:
$$\cos x=-\cos(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A double integral for $\frac{\pi}{2} \ln 2$.
Show that
\begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2.
\end{eqnarray*}
My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from th... | Through series expansions:
$$ \iint_{(0,1)^2}\frac{dx\,dy}{\sqrt{1-x^2 y^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\iint_{(0,1)^2}x^{2n}y^{2n}\,dx\,dy=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)^2}=\int_{0}^{1}\frac{\arcsin(x)}{x}\,dx $$
and this is
$$ \int_{0}^{\pi/2}x\cot(x)\,dx\stackrel{\text{IBP}}{=}\int_{0}^{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding the length of the arc for function $8y = x^4 +2x^{-2}$
Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$
I first isolated for $y$ and derived:
$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$
$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$
Then found the arc length:
$$L = \int ^2 _... | Hint:
$$2+x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Difficult integral involving trigonometric and hypertrigonometric functions This is definitely the most difficult integral that I've ever seen.
Of course, I'm not able to solve this.
Could you help me?
$$\int { \sin { x\cos { x } \cosh { \left( \ln { \sqrt { \frac { 1 }{ 1-\sin { x } } } +\tanh ^{ -1 }{ \left( \sin ... | Mathematica gives:
$$-\frac{\sqrt{\frac{1}{1-\sin (x)}} \sqrt{\sin ^2(2 x)} \csc^2(x) \\ \left(-90 \sin \left(\frac{x}{2}\right)+35 \sin \left(\frac{3 x}{2}\right)-3 \sin \left(\frac{5 x}{2}\right)+15 \cos \left(\frac{3 x}{2}\right)+3 \cos \left(\frac{5 x}{2}\right)+30 \cos \left(\frac{x}{2}\right) \left(4 \sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Simplify system of 3 equations of 3 recursive functions Is it possible to simplify this system into one equation of f only?
f(N) = f(N-1) + f(N-2) + 2*g(N) + h(N)
g(N) = f(N-2) + g(N-1)
h(N) = f(N-2) + h(N-2)
Initial values if needed:
f(1) = 2
f(2) = 5
g(2) = 1
g(3) = 2
h(2) = 1
h(3) = 1
I'm not sure it is possible. T... | We start with
$$\begin{align}
&f(n) = f(n - 1) + f(n - 2) + 2·g(n) + h(n)
\\&-\\
&f(n - 2) = f((n - 2) - 1) + f((n - 2) - 2) + 2·g(n - 2) + h(n - 2)
\end{align}$$
And obtain
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1) - 2·g(n - 2) + 2·g(n) - h(n - 2) + h(n)
$$
We can substitute $h(n)-h(n-2)$ in the equati... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Formula for calculating the odds per user of winning in a raffle each player can win once Am trying to write a program which gives each user in a raffle contest their odds so far at winning.
The rules of the game are simple:
A predefined number of tickets are sold for example 1000 each user can at most buy 50 tickets ... | I don't think you'll find a closed formula for this. If you're OK with only specifying the odds up to a couple of decimal places, the easiest way to do this would be to simulate the draw a couple of billion times and use the resulting relative frequencies. If you want the exact probabilities instead, you can keep track... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7},\cos\frac{5 \pi}{7}$
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7}\space \text{and}\space \cos\frac{5 \pi}{7}$.
How to even solve it ? I have no idea.
B... | You could prove it with just a couple of high-school trigonometric identities.
Let $\theta = \pi/7 $ and recognize
$$x_1=\cos \theta = -\cos 6\theta$$
$$x_2=\cos3\theta = -\cos 4\theta$$
$$x_3=\cos5\theta = -\cos 2\theta$$
(1). Evaluate their product by applying the identity $\sin 2x = 2\sin x \cos x$
$$ \cos2\theta \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$ How to prove
$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx\tag{1}$$
Here is how I came up with this relation:
In this solution @Kemono Chen elegantly proved
$$\... | In other words we want to show that:
$$\color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$
This can be seen via the substitution:
$$\frac{1-x}{1+a^2 x}=t\Rightarrow x=\frac{1-t}{1+a^2 t}\Rightarrow dx=-\frac{1+a^2}{(1+a^2t)^2}dt$$
$$\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Evaluate $\sum\limits_{n=0}^{\infty} \frac{\cos(nx)}{2^n}$ where $\cos x = \frac15$ Evaluate
$$\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{2^n}$$
where $\cos x = \frac{1}{5}$.
This is a complex number question. But I don’t know where to start. Maybe need to use the DeMoivre’s Theorem?
| Consider the following auxiliar sum: $\sum_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}$, and use that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ to get: $\sum_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+\sum_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+i\dfrac{\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\... | Translating up by $6$ units again, we find that the angle between the $x$-axis and $(4,-7)$ is $-\tan^{-1} \frac{7}{4}$. In addition, the angle between the line and the $x$-axis is $-\tan^{-1} 5$. So the reflected point must have an angle of:
$$\theta =-\tan^{-1} 5 -\tan^{-1} 5 + \tan^{-1} \frac{7}{4}$$
In addition, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it:
https://imgur.c... | It is $$4(a^6+b^6)-(a+b)(a^2+b^2)(a^3-b^3)=(a-b)^2(a^2+b^2)(3a^2+5ab+3b^2)\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
how do you differentiate x^(3/4) using first principle $$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac{3}{4}}-(x)^{\frac{3}{4}}\Bigr)}{h}$$
I understand the process till
$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac34}-(x)^{\frac{3}{4}}\Bigr)}{h} * \frac{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}{\Bigl((x+h)^{\frac{3}{4}}+(... | Why don't you expand from the start
$$\lim_{h\to0}\dfrac{(x+h)^n-x^n}h=x^n\cdot\lim_{h\to0}\dfrac{\left(1+\dfrac hx\right)^n-1}h$$
Now use Binomial series
Alternatively, set $$(x+h)^{1/4}=a,x+h=a^4; x^{1/4}=b, x=b^4$$
$$\lim_{h\to0}\dfrac{(x+h)^{3/4}-x^{1/4}}h=\lim_{a\to b}\dfrac{a^3-b^3}{a^4-b^4}=\lim_{a\to b}\dfrac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\d... | Hint: What is $\tfrac{-b+\sqrt{b^2-4ac}}{2a} - \frac{-b-\sqrt{b^2-4ac}}{2a}$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos... | For positives $A$, $B$ and $C$ there is the following way.
Let $A=\max\{A,B,C\},$ $\pi-A=\alpha,$ $\frac{\pi}{2}-B=\beta$ and $\frac{\pi}{2}-C=\gamma$.
Thus, $\alpha$, $\beta$ and $\gamma$ are measured angles of the triangle and let sides-lengths of the triangle be $a$, $b$ and $c$ respectively.
Thus, since by law of s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
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Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | Hint:
Let $$x^{81}+x^{49}+x^{25}+x^9+x=p(x)(x^3-x)+ax(x-1)+bx(x+1)+c(x-1)(x+1)$$
where $a,b,c$ are arbitrary constants
Put $x=0,-1,1$ to find $a,b,c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$?
If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:
$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.
What is the correct way... | The way to find the smallest denominator "from scratch" is with continued fractions.
Begin by rendering the proposed bounds thusly:
$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$
$\dfrac{11}{15}=\dfrac{1
}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$
The upper layers of the continued f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Sum of the series $1-2r\cos~\theta +3r^2 \cos~2\theta -4r^3\cos~3\theta+\dots$..
Sum of the series
$$1-2r\cos~\theta +3r^2 \cos~2\theta -4r^3\cos~3\theta+\dots\qquad |r|<1.$$
I don't know how to find sum other than for geometric series and this is not geometric.
| You'll need to combine$$\sum_{n\ge0}(n+1)z^n=\frac{d}{dz}\sum_{n\ge0}z^{n+1}=\frac{d}{dz}\frac{z}{1-z}=\frac{d}{dz}\frac{1}{1-z}=\frac{1}{(1-z)^2}$$with$$\sum_{n\ge0}(n+1)(-r)^n\cos n\theta=\Re\sum_{n\ge0}(n+1)(-r\exp i\theta)^n=\Re\frac{1}{(1+r\exp i\theta)^2}\\=\Re\frac{(1+r\exp -i\theta)^2}{(1+r^2+2r\cos\theta)^2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS?
$$\frac{d}{dt}\left(\frac{m \ve... | Using the product rule reveals
$$\begin{align}
\frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=m\vec u \frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)+\frac{1}{\sqrt{1-u^2/c^2}}\frac{d}{dt}\left(m\vec u\right)\\\\
&=m\vec u\left(-\frac12(1-u^2/c^2)^{-3/2}\left(\frac{-2u}{c^2}\right)\,\frac{du}{dt}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Is there a systematic way to derive constraints in a system of equations? I have this system of equations:
$x_1 = y_1+ y_2+ y_3$
$x_2 = y_1- y_2$
$x_3 = y_1 + y_2 - 2y_3$
And I have these constraints:
$y_1 \geq y_2 \geq 0 \geq y_3$
From the constraints I can derive these constraints for the $x$'s:
$x_2 \geq 0$
$x_... | Your answer is not true. Indeed,
\begin{align}
\left\{\begin{array}{l}
x_2 \ge 0\\
x_3 \ge x_1\\
x_3 \ge x_2 \\
\end{array}
\right. \Leftrightarrow
\left\{\begin{array}{l}
y_1-y_2 \ge 0\\
y_1+y_2 - 2y_3 \ge y_1+y_2 + y_3\\
y_1+y_2 - 2y_3 \ge y_1 - y_2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3358244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the number of solutions of an augmented matrix? Given a system of linear equations is represented as a augmented matrix:
$$M = \left[\begin{array}{cccc|c}1&0&8&0&1\\0&1&1&0&3\\0&0&0&1&1\end{array}\right]$$
How do I find the exact number of solutions in the system of equations?
If I'm not mistaken, it has un... | In case you mean to solve the under-determined system $AX=B:$
$$\begin{pmatrix} 1 & 0 & 8 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}
\begin{pmatrix} w \\x \\ y \\ z \\ \end{pmatrix}= \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}.$$ Multiply from let with $A^T$ to get
$$\begin{pmatrix} 1 & 0 & 8 & 0\\ 0 & 1 & 1 &0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the probability that the highest of the three numbers obtained from three dice will be exactly a $4$?
Question: You are going to toll three dice. What is the probability that the highest of the three numbers will be exactly a $4$?
My attempt:
The first die obtains number $4$ with probability $\frac{1}{6}.$
T... | You've summed the probabilities of the events that one particular die shows a $4$ and the other two show not more than $4$. Unfortunately, those events are not disjoint, so you've double-counted the cases where several $4$s appear. To fix that, it helps to separate your computations by the number of $4$s rolled:
The pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?
$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$
I used
$$\sin... | Let $\arctan\left(\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}\right)=y,-90^\circ<y<90^\circ$
$\implies\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}=\tan y=\dfrac{\sin y}{\cos y}$
Rearranging we get $$\cos(6^\circ+y)+\cos(18^\circ+y)=\sin y=\cos(90^\circ-y)$$
$$\iff\cos(90^\circ-y)=\cos(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + ... | HINT.-$$N=3^x\cdot5^y\cdot7^z\quad\text{ has (x+1)(y+1)(z+1) divisors }\\5N=3^x\cdot5^{y+1}\cdot7^z\quad\text{ has (x+1)(y+2)(z+1) divisors }\\27N=3^{x+3}\cdot5^{y}\cdot7^z\quad\text{ has (x+4)(y+1)(z+1) divisors }$$ You can get from this the system
$$(x+1)(z+1)[(y+2)-(y+1)]=8\\(y+1)(z+1)[(x+4)-(x+1)]=18$$ Then........... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
I think that $(0,0)$ , $(1,0)$ and $(0,-1)$ are the only solutions to the above equation, but I'm unable to prove it.
I tried all sorts of things like working $\mod 9$ (but ther... | Letting $x=y+k$, we are looking for the solutions of
$$ 2y^2+2yk+k^2 = k^3 $$
$$ (2y+k)^2+k^2 = 2k^3 $$
which depend on the integer points on the elliptic curve $w^2=2z^3-z^2=z^2(2z-1)$.
We may assume that $z=\frac{q^2+1}{2}$, leading to the solution $k=\frac{q^2+1}{2},w=q\frac{q^2+1}{2},y=(q-1)\frac{q^2+1}{4},x=(q+1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the maximum of the value $f(x)=\frac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$
Let $n$ be a given positive integer, find the maximum of the value of
$$f(x)=\dfrac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$$ where $x \le \dfrac{n^2}{4}$ and $x \in \mathbb N^{+}.$
If the $x$ is real positive number, the problem also not easy, because
$$f'... |
Let $\mathbb{N}$ denote the positive integers. Given $n\in\mathbb{N}$ fixed, we look for
\begin{align*}
&f:\mathbb{N}\to\mathbb{R}\\
&f(x)=\frac{n^4}{x}+4\left(n^2+3\right)x-4x^2-5n^2\qquad\longrightarrow\qquad\max\tag{1}\\
\end{align*}
provided that
$x\leq \frac{n^2}{4}$
*
*The additive constant $-5n^2$ in (1) i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Are there any examples of $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$? I am writing a short discussion of Farey numbers and was wondering if there are any examples of when the mediant function is ever actually equal to the sum of the two fractions in the usual sense? (Not to produce Farey numbers, I just thought it might... | Seems like straight forward algebra manipulation
$\frac ab + \frac cd = \frac {a+c}{b+d}$
$(ad+bc)(b+d) = bd(a+c)$
$abd + b^2c + ad^2+bcd = abd +bcd$
$b^2c +ad^2 = 0$
$b^2c = -ad^2$
Wolog $b,d$ are positive. Wolog $c$ is positive an $a$ negative (I assume you don't want $a=c =0$.)
If we let $c = c'(M^2)$ where $c'$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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A property of Nesbitt's inequality I think it's not new however I think it's interesting :
Let $a,b,c>0$ and $x>0$ then the function :
$$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{b^x+a^x}$$
$f(x)$ is increasing
My first try :
The derivative of $f(x)$ is :
$$f'(x)=\sum_{cyc}\frac{a^x\ln(a)(b^x+c^x)-a... | By your assumption
$$f'(x)=\sum_{cyc}\frac{a^x\ln{a}(b^x+c^x)-a^x(b^x\ln{b}+c^x\ln{c})}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}\frac{a^x(b^x(\ln{a}-\ln{b})-c^x(\ln{c}-\ln{a}))}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}(\ln{a}-\ln{b})\left(\frac{a^xb^x}{(b^x+c^2)^2}-\frac{a^xb^x}{(c^x+a^x)^2}\right)=$$
$$=\sum_{cyc}\frac{a^xb^x(\ln{a}-\ln{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find all n such that : $3 \mid (n2^{n}+1)$ Question :
Determine all natural numbers n such that 3
divides $n\cdot2^{n}+1$
Actually I don't have any ideas to approach but
my efforts :
I see $n=1,2,7,8,13,14$ so I think :
$n=6k+1$ and $n=6k+2$ $k\in \Bbb{N} $
If I'm not wrong but I don't know how ? I prove it ?
| $2^n \equiv (-1)^n \pmod 3$.
So if $n$ is even then $n2^n + 1\equiv n+1 \equiv 0 \pmod 3$ and $n\equiv -1\pmod 3$.
And so $n\equiv 0 \pmod 2$ and $n\equiv 2\pmod 3$ so $n\equiv 2\pmod 6$.
And indeed $(6k+2)2^{6k+2}+1=(6k+2)4^{3k+1}+1\equiv (6k+2)*1^{3k+1} +1\equiv 6k + 2+ 1\equiv 0 \pmod 3$.
If $n$ is odd then $n2^n+1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
| Using the Extended Euclidean Algorithm (which can be applied to $\mathbb{Q}[x]$ as well as $\mathbb{Z}$), we get
$$
\left(36n^2+24n+16\right)\left(n^4+n+1\right)-\left(18n^3+12n^2+8n-27\right)\left(2n^3+5\right)=151
$$
if $2n^3+5\mid n^4+n+1$, we must also have $2n^3+5\mid151$. Since $151$ is prime and there is no inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Maximum value of $(1 − x)(2 − y)^ 2 (x + y)$ Additional Info
$x < 1, y<2, x+y>0$
I tried doing AM-GM (since all of the terms are positive.
$(1 − x)(2 − y)^ 2 (x + y) = (2-2x)(2-y)(2-y)(2x+2y)*1/4$
this way we can cancel the x and y's
thus the maximum value is
$((2-2x+2-y+2-y+2x+2y+1/4)/5)^5$
but the answer is said to... | A much more Logical way to solve this without algebraic manipulations:
Since we want to maximize $(1-x)(2-y)^2(x+y)$ , and all the terms are positive, we can see that we might want to maximize $(2-y)^2$ in particular as it is the biggest among all terms.
$$\max \space(2-y)^2 = (2-0)^2 = 4 \space\space\space\space\space... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5... | You can get a faster result by using a different way to complete the square: split the middle term
$$
z^4-3z^2+1=(z^2-1)^2-z^2=(z^2-z-1)(z^2+z-1)
$$
Now the remaining two quadratic equations are easy to solve.
$$
0=z^2-z-1\implies z=\frac12(1\pm\sqrt{5})\\
0=z^2+z-1\implies z=\frac12(-1\pm\sqrt{5})\\
$$
The fifth unit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as
$(0, -7)$; no $x$ intercepts.
I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3... | For $$x\geq 2$$ we get $$3(x-2)-1=3x-6-1=3x-7$$ .For $$x<2$$ we have $$3(2-x)-1=-3x+5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary
Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary.
My attempt is as follows:-
Let $t=\frac{1}{1+x^2}$, and le... | In your proof you have excluded the interval $(0,4/9)$. Why? For example if $a=1/3\in(0,4/9)$ then the equation becomes
$$\frac{1+3x^2+3x^4}{(1+x^2)^2}=0$$
which has not real roots because the l.h.s. is always positive. Hence $1/3$ should be included in the required set.
If $z(x)=1/(1+x^2)$ then $z(\mathbb{R})=(0,1]$. ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Taylor Series for $\dfrac{x}{x ^ 2 + x + 1}$. I'm trying to calculate the taylor series of $\dfrac{x}{x ^ 2 + x + 1}$.
Algebraic manipulation didn't get me anywhere, since the roots of $ x ^ 2 + x + 1 $ are complex.
Integrate or derive made the problem worse
Any tips on how to proceed?
| There seem to be many ways to go about this, so here is one: put $\dfrac{x}{x ^ 2 + x + 1}=\sum_{n=0}^{\infty} a_nx^n$, then $$x=\sum_{n=0}^{\infty} a_nx^n(x^2+x+1)=a_0+(a_1+a_0)x+\sum_{n=2}^{\infty}(a_n+a_{n-1}+a_{n-2})x^n,$$
and by comparing the coefficients we get $a_0=0$, $a_1=1$, and $a_n+a_{n-1}+a_{n-2}=0$ for $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Compute $\lim_{N\to \infty} A_N$ where $\{A_N\}$ is a sequence of matrices
Given matrix, $$A_N = \begin{bmatrix} 1 & \frac{\pi}{2N}\\ \frac{-\pi}{2N} & 1\end{bmatrix}^N$$ compute $$\lim_{N \rightarrow \infty} A_N$$
I took the logarithm of both sides but was not able to figure out the limit. Any suggestions on how to ... | The characteristic polynomial of $A$ is given by:
$$ \det \begin{bmatrix} x-1 & \frac{-\pi}{2N} \\ \frac{\pi}{2N} & x-1\end{bmatrix} =(x-1)^2+\frac{\pi^2}{4N^2}$$
To find the eigenvalues of $A$, set the characteristic polynomial equal to zero, and solve for $x$. This gives us the complex conjugate eigenvalues $x=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to write this as a multiplication: $4 x^2 y^2 - (x^2 + y^2 - z^2)^2$? I have the following polynomial:
$$ 4 x^2 y^2 - (x^2 + y^2 - z^2)^2$$
(which comes up, for example, in computing the area of a triangle using the cosine law).
I would like to convert this to a product.
Wolfram tells me it's
$$ -(x - y - z) (x ... | $$ \begin{align} \color{magenta}{4 x^2 y^2 - (x^2 + y^2 - z^2)^2} & = \color{brown}{(2xy +x^2+y^2-z^2)(2xy -x^2-y^2+z^2)} \\ &= \color{blue}{((x+y)^2-z^2)(-(x-y)^2 -z^2)}\\
& = \boxed{\color{red}{-(x+y+z)(x+y-z)(x-y-z)(x-y+z)}} \end {align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
| Not really that different, but this avoids doing most of the multiplications. It is probably a bit longer though.
First a simplification of your computations. Write $x^4=x^4-1+1$. Then
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{(x^4-1)(1-x)^4}{1+x^2}+\frac{(1-x)^4}{1+x^2}=(x^2+1)(1-x)^4+\frac{(1-x)^4}{1+x^2}$$
Now, do the substi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
He... | That does work. What would be easier is to notice that $5^n$ dominates the numerator while $4^n$ dominates the denominator. Therefore $$\lim_{n \to \infty} \frac{2^n+5^n}{3^n+4^n} = \lim_{n \to \infty} \frac{5^n}{4^n} $$
This clearly diverges.
By the $n$th term test, you can see that the series also diverges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Compute the determinant of $A$
How to prove that the determinant of the following matrix
$$A = \begin{bmatrix}
a&-b&-c&-d\\
b&a&d&-c\\
c&-d&a&b\\
d&c&-b&a
\end{bmatrix}$$
is $\det A=(a^2+b^2+c^2+d^2)^2$?
Note that $AA^t=(a^2+b^2+c^2+d^2)I_4$, but we just have to $$|\det A|=(a^2+b^2+c^2+d^2)^2$$
Any hint would be appr... | A different approach: If $A,B,C,D$ are square matrices and $CD=DC$, then
$$\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(AD-BC).$$
Letting
$$A=\begin{pmatrix}a&-b\\b&a\end{pmatrix},B=\begin{pmatrix}-c&-d\\d&-c\end{pmatrix},C=\begin{pmatrix}c&-d\\d&c\end{pmatrix},D=\begin{pmatrix}a&b\\-b&a\end{pmatrix},$$
you can check ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a$, $b$, $c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are all integers, then $|a|=|b|=|c|$
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.
Well this is wha... | You have the $2$ values, stated to be integers, of
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{a^2c + ab^2 + bc^2}{abc} \tag{1}\label{eq1A}$$
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} = \frac{a^2b + b^2c + ac^2}{abc} \tag{2}\label{eq2A}$$
Consider any prime $p$ where $p \mid abc$. Note the $p$-adic valuation func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
$f(x+1) + f(x-1) = x^2$ ; $f(x+4) + f(x-4) = 2\sin x$ , then $f(x) =$? Given $f$ is a complex valued function satisfying $$f(x+1) + f(x-1) = x^2 \\
f(x+4) + f(x-4) = 2\sin x$$
what is $f(x)$ ?
Here, only for the first part MathWolfram alpha is showing $f(x)$ to be of type $c_1(i)^x + c_2(-i)^x + \dfrac{x^2 - 1}{2}$ b... | Let's do two change of variables. First, $u = x-1$:
$$f(u+2) = (u+1)^2 - f(u)$$
Let's apply that multiple times:
$$\begin{align}
f(u+4) &= (u+3)^2 - f(u+2) = (u+3)^2 - (u+1)^2 + f(u)\\
f(u+6) &= (u+5)^2 - f(u+4) = (u+5)^2 - (u+3)^2 + (u+1)^2 - f(u)\\
f(u+8) &= (u+7)^2 - f(u+6) = (u+7)^2 - (u+5)^2 + (u+3)^2 - (u+1)^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I solve $2^{x}+1=y^{2}$? I found an exponential equation with two variables $x$ and $y$, and I can't find the solutions. I want to solve it for $x$ and $y$. I tried using logarithms but I couldn't solve it. This is the equation:
$$2^{x}+1=y^{2}\qquad\text{with }\big(x,y\big)\in\mathbb{N}.$$
| Observe that $2^x=y^2−1=(y+1)(y−1) \Rightarrow y+1 = 2^a$ and $y-1 = 2^b$, where $x=a+b$. Then, we have $a \geq b$, and $2^a-2^b= (y+1) - (y-1) = 2$, which gives us $2^b(2^{a-b}-1)=2$. So, we have the following cases:
(I) $\left\{\begin{array}[rcl]
22^b& = & 1 \\
2^{a-b}-1 & = & 2
\end{array} \right. $ $\quad$ (II) ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$\int \frac{\sin x}{1+2\sin x}dx$ calculate:
$$\int \frac{\sin x}{1+2\sin x}dx$$
I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification:
$$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$
and I am not able to calulate that.
| Before doing that substitution I might say
$\int \frac{\sin x}{1+2\sin x} \ dx\\
\int \frac 12 \frac{2\sin x + 1 - 1}{1+2\sin x} \ dx\\
\int \frac 12 - \frac{1}{1+2\sin x} \ dx\\
\int \frac 12 \ dx - \frac 12 \int \frac{1}{1+2\sin x} \ dx$
Now when we do the substitution it isn't as messy.
$\frac x2 - \int \frac{... | {
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"source": "stackexchange",
"question_score": "2",
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Find $n$ for a $\sum_{x=1}^n \left[(x+1)^3-x^3\right]$ $$ \sum_{x=1}^n \left[(x+1)^3-x^3\right]$$
This is my sum, I tried simplfifying and got $3x^2+3x+1$ but Im stuck on how to resolve the sum for $n$.
| As you stated:
$$ \sum_{x=1}^{n} [(x+1)^3-x^3] = \sum_{x=1}^{n} 3x^2+3x+1 = 3\frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n = $$ $$ = n(n+1)(n+2)+n = (n+1)^3-1$$
Keep in mind that:
$$ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6} $$
$$ \sum_{x=1}^{n} x = \frac{n(n+1)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find limit of sequence $a_1=1$, $a_{n+1}=\left( 1-\frac{1}{(n+1)^2} \right)a_n$ for $n\geq 1$? I need some help finding this limit.
Define a sequence by $a_1=1$ and
$a_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)a_n$ for $n\geq1$. Show that the limit exists and find the limit.
I've shown that the limit $L$ exists ... | $1-\frac 1{k^2} = \frac{(k-1)(k+1)}{k^2}$, so the product is
$$
\frac{(2-1)(2+1)}{2\cdot 2}\cdot\frac{(3-1)(3+1)}{3\cdot 3}\cdot\frac{(4-1)(4+1)}{4\cdot 4}\cdot\frac{(5-1)(5+1)}{5\cdot 5}\cdot\dots
$$
i.e.,
$$
\frac{3}{2\cdot 2}\cdot\frac{2\cdot 4}{3\cdot 3}\cdot\frac{3\cdot 5}{4\cdot 4}\cdot\frac{4\cdot 6}{5\cdot 5}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For all $x$ in $\mathbb{R}$, there exists $n$ such that the below inequality holds (Proof)
Prove that for all $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ such that:
$$ \frac{3}{3x^2 - 3x + 2} < \frac{4}{n} < \frac{5}{5x^2 - 5x +2} .$$
Here's how I went about it:
For the first part, I reasoned that if we choo... | Since $5x^2-5x+2>0$ and $3x^2-3x+2>0$, we have $$\frac{4}{5}(5x^2-5x+2)<n<\frac{4}{3}(3x^2-3x+2).$$ Now, we see that it's enough to prove that:
$$\frac{4}{3}(3x^2-3x+2)-\frac{4}{5}(5x^2-5x+2)>1.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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What is the coefficient of x^2 in the expansion of (x+2)^4(x+3)^5 I'm missing something here.
I've calculated the $x^2$ coefficient of $(x+2)^4$ as 24 with constant term 16. And $x^2$ term coefficient of $(x+3)^5$ as 270 with constant term 243.
if I'm correct here then the answer should be $(16*270)+(24*243)$? but this... | \begin{array}{r}
(x+2)^4 = && \star x^4 &+ \star x^3 &+ 24x^2 &+ 32x &+ 16 \\
(x+3)^5 = &\star x^5 &+ \star x^4 &+ \star x^3 &+ 270x^2 &+ 405x &+ 243
\end{array}
\begin{array}{c|c}
\times &16 & 32x &24x^2 &\cdots \\
\hline
243 & \star & \star x & 5832x^2 &\cdots \\
405x & \star x & 12960x^2 & \star... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Formula for length of the diagonal of a parallelepiped Let $a,b,c$ and $\alpha, \beta, \gamma $ are sides and angles ($\alpha$ is the angle between the sides $b$ and $c$ and so on) of a parallelepiped. By using the vector algebra it is easу to prove the formula for the length of the diagonal $d$ of this parallelepi... | The formula can be derived from two geometric results/theorems:
*
*Parallelogram law
In a parallelogram $ABCD$ with $AB = CD = a, BC = DA = b$, one has $$AC^2 + BD^2 = 2(AB^2 + BC^2) = 2(a^2+b^2)$$
Since in a parallelogram, $\angle A = \angle C$ and $\angle B = \angle D = \pi - \angle A$, one can easily derive thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $ Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $
My attempt is as follows:-
$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$
$$\left(\sin\theta-\cos... | As $1+2+3=1\cdot2\cdot3$
$\arctan1+\arctan2+\arctan 3=n\pi$
using Show $\tan(x)+\tan(y)+\tan(z) = \tan(x) \tan(y) \tan(z)$
As $\arctan(u)<\dfrac\pi2$
$n=1$
Now $\arctan1=\dfrac\pi4$
$\arctan2+\arctan\dfrac12=\arctan2+$arccot$2=\dfrac\pi2$
See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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proof the given limit using epsilon delta definition
$$\lim_{x \to 1} \frac{\sqrt{x}-1}{x-1}=\frac{1}{2}$$
let $$f(x)=\frac{\sqrt{x}-1}{x-1}$$ then
$\forall \epsilon>0, \exists\delta>0,\forall x\in D_f \left(\left|x-1 \right|<\delta\Longrightarrow \large\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|<\epsilon\right)$... | The proof looks fine, there is a typo for
$$\delta <\color{red}{\frac12} \implies0.5<x<1.5$$
and here, since $\sqrt x$ is an increasing function, we can take
$$\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{\left|x-1\right|}{2\left(0+1\right)^{2}}<\epsilon$$
which leads to a simpler expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\cos(\pi/7)$ is a root of $8x^3-4x^2-4x+1=0$. How is the polynomial generated? According to Wolfram MathWorld, $\cos(\frac{\pi}{7})$ is a root of $8x^3-4x^2-4x+1=0$. Similarly, $\cos(\frac{2\pi}{7})$ is a root of $8x^3+4x^2-4x-1=0$. What's the procedure to generate these polynomials? I understand you can solve the cub... | By DeMoivre's theorem,
$$(\cos x+i\sin x)^7=\cos7x+i\sin7x\quad\quad(*)$$
With $x=\frac\pi7$, the right hand side of $(*)$ reduces to $-1$. Expand the binomial on the left hand side and match up the real and imaginary parts (the latter of which is $0$), leaving us with
$$\cos^7x-21\cos^5x\sin^2x+35\cos^3x\sin^4x-7\cos ... | {
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"url": "https://math.stackexchange.com/questions/3412677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Division of 2 different Infinite Nested square roots Has anyone come across the following type of nested square roots problem?
$\sqrt{2-{\sqrt{2+{\sqrt{2+...n times {\sqrt{2}}}}}}}$ divided by
$\sqrt{2-{\sqrt{2+{\sqrt{2+...(n+1)times {\sqrt{3}}}}}}}$
Converging towards 3 as the 'n' increases
Are there any theorem or f... | Well, you seem to have basically proven it. We want to show that
$$
\lim_{n\to\infty}\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}} = 3
$$
We would like to use that $\sin x \approx x$ for small $x$. To be rigorous, we can write $\sin x = x + xo(1)$, where $o(1)$ is a function that goes to $0$ as $x\to0$. Then
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do I find all solutions of $X^2 = I_2$ in $M_2(\mathbb{N})$? I know that the solutions of the equation:
$$X^2 = I_2$$
in $M_2(\mathbb{N})$ are the $2$x$2$, natural, involutory matrices:
$$ X_1 =
\begin{pmatrix}
1 & 0\\
0 & 1\\
\end{pmatrix}
$$
and
$$ X_2 =
\begin{pmatrix}
0 & 1\\
1 & 0... | Take $$X=\begin{bmatrix}a& b \\ c &d
\end{bmatrix} $$
with $a,b,c,d \in \mathbb{N}$
Now
$$X^2=\begin{bmatrix}a^2+bc & ab+bd \\ ac+cd & bc+d^2
\end{bmatrix} $$
and we have a system of equations in $\mathbb{N}$:
$$\begin{cases}
a^2+bc=1\\
ab+bd=0 \\
ac+cd=0 \\
bc+d^2=1
\end{cases}$$
For $b(a+d)0=0$, we have either $b=0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Inverse of polynomial over a finite field The question is the following:
Can you deduce if ${2x + 1}$ is invertible in $\mathbb{Z}_3[x]/(x^2 + 2x + 2)$? In case of a positive answer, give its inverse.
Following the Wilson's theorem, for $K[X]/(f)$, any polynomial of degree $1 \leq deg < n$ will admit an inverse of degr... | Hint: $0=x^2 + 2x + 2$ implies $0 = 2x^2+4x+4 = 2x^2+x+1 = x(2x+1)+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact. From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\b... | This may not be the solution you are looking for:
Let $\mathcal C=\{x^2+y^2=1\}$ be the unit circle. Let $O' = (\alpha, \beta)$ be any point outside of this circle. Let $O'P'$ and $O'Q'$ be two tangent line to $\mathcal C$. One can check that the midpoint $m' = (x, y)$ of $P'Q'$ is given by (why?)
$$(x, y) = m' = \frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that if p and q are both prime numbers, with p > q > 2, then $p^4− q^4$ is divisible by 16 Prove that if p and q are both prime numbers, with p > q > 2, then $p^4 − q^4$ is divisible by 16.
This is my attempt so far:
Since p and q are both prime numbers greater than 2, then they must be odd and hence can be writt... | Alternatively you may show that any odd integer power $4$ leaves the same remainder $1$ when divided by $16$:
$(2n+1)^4 \\
= (2n)^4 + \binom{4}{1}(2n)^3 + \binom{4}{2}(2n)^2+\binom{4}{3}(2n)+1
\\= 16(n^4+2n^3+n^2) + 8n(n+1)+1\\\equiv 1\pmod{16}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Determine all prime numbers $p,q,r$ such that : $p^{2}+1=r^{2}+q^{2}$ Problem :
find all prime numbers such that :
$$p^{2}+1=q^{2}+r^{2}$$
My attempt :
equation equivalent :
$p^{2}-q^{2}=r^{2}-1$
so :
$(p+q)(p-q)=(r+1)(r-1)$
now :
$r=2$ we find : $p+q=2$ or $p-q=1$ $×$
$r=3$ we find : $p+q=3,2$ or $p-q=2,3$ $... | COMMENT.- One way could be to consider the general solution of the equation $x^ 2 + y^ 2 = z^ 2 + w^ 2$ that gives a parameterization with four independent coprime parameters with which one could look for prime values for $r$ and $q$. For example by making $at-bs = 1$ and $as + bt = p$, try to see if primes $r$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding a Lyapunov function for a specific problem I have tried for a long time to find a Lyapunov function for the specific problem
$$\begin{align} x' &= -x - 2y + xy^2 \\ y' &= 3x - 3y + y^3 \end{align}$$
Do you know any Lyapunov function what is going to work to determine the stability of the origin? According to a ... | Hint.
We have
$$
\cases{
3x \dot x = -3 x^2-6 x y + 3x^2y^2\\
2y \dot y = 6x y-6 y^2+2y^4
}
$$
and after addition
$$
\frac 12\frac{d}{dt}(3x^2+2y^2) = -3x^2-6y^2+3x^2y^2+2y^4
$$
now there exist $\rho_m$ such that $x^2+y^2 < \rho_m^2$ makes
$$
-3x^2-6y^2+3x^2y^2+2y^4 < 0
$$
in fact, $\rho_m = 1$ suffices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a Difficult Definite Integral in One Variable Let $t > 0 $, $N \in \mathbb{N}$, $q \geq 1 + \frac{1}{N}$.
Also let $r > 0 $, $r' = \frac{r}{r-1}$, such that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$.
We wish to prove the following integral equation:
$ \displaystyle \int^{t}_{0} \large (t-s)^{ -\frac{N}{2... | Only a partial answer:
For $a>-1,b>-1,x\in[0,1]$, let $$B(x,a+1,b+1):=\int_0^x u^a \cdot (1-u)^b \,\mathrm du$$ denote the Incomplete Beta function.
For $a,b>0$ and $c>0$, we can give an anti-derivative on $]0,c[$ of the function $$f:[0,c]\to \mathbb R, x\mapsto(c+x)^a\cdot(c-x)^b$$ in terms of the incomplete Beta fun... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometric determinants I've recently obtained my University entrance papers from 1967 (yes,52 years ago!) and I found the question below difficult. I presume the answer is a symmetric expression in the differences between alpha,beta and gamma.Am I missing some obvious trick? Any help would be appreciated.
Simplify ... | To make it a bit better readable, you may substitute
*
*$u = \theta + \alpha, v= \theta + \beta, w = \theta + \gamma$
So, the first row of your determinant looks like
$$\begin{pmatrix} \sin^2 u & \sin u \cos u & \cos^2 u
\end{pmatrix}$$
Now, you may use
*
*$\cos^2 u - \sin^2 u = \cos 2u$ and
*$2 \sin u \cos u ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the equation has only one real root. Prove that $(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only one real root.
It's easy to show that the equation has a real root using Rolle's theorem. But how to show that the real root is unique? By Descartes' rule of sign, it can be shown that it has 3 or 1 real root.
But i... | Here is an elementary way that uses only
*
*$(1)$: $a^3+b^3 = (a+b)(a^2-ab+b^2)$ and
*$(2)$: $a^2+b^2 >ab$ for $|a|+|b| > 0$
Note, that $|x-1|+|x-4|\geq |x-1 +(4-x)| =3 > 0$ and $|x-2|+|x-3|\geq |x-2 +(3-x)| =1 > 0$
Let's call $p(x) = (x-1)^3+ (x-2)^3 + (x-3)^3 + (x-4)^3$.
Now, using $(1)$ write
$$(x-1)^3 + (x-4)^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 6
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Use the epsilon-delta definition to show that $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$ $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$
by using the epsilon-delta method.
I have been trying so solve this for the last hour, but I'm completely stuck.
Help, anyone?
| Fix $\epsilon > 0$.
Find $\delta_1$ such that $|\frac{2}{x} - \sqrt{2}| < \epsilon$ for any $x$ satisfying $|x-\sqrt{2}| < \delta_1$.
Find $\delta_2$ such that $|x - \sqrt{2}| < \epsilon$ for any $x$ satisfying $|x - \sqrt{2}| < \delta_2$.
Then let $\delta = \min\{\delta_1, \delta_2\}$.
We then have $$|\frac{1}{2} (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for $f: [0, 1] \to \mathbb{R} , f_n(x) = \frac{nx}{1+n^2 x^2}$ is uniform convergent. Prove that for $f: [0, 1] \to \mathbb{R} , f_n(x) = \frac{nx}{1+n^2 x^2}$ is uniform convergent.
I found the pointwise limit to be 0.
Taking the first derivative, $$f_n '(x) = \frac{n(1-n^2 x^2)}{(1+n^2 x^2)^2} $$ $f_n '(... | Clearly, the convergence is not uniform on $[0,1]$ since $1/n \in [0,1]$ and as $n \to \infty,$
$$\sup_{x \in [0,1]} |f_n(x)-0| = \sup_{x \in [0,1]} \frac{nx}{1 + n^2x^2} \geqslant \frac{n \cdot \frac{1}{n}}{1 + n^2 \cdot \frac{1}{n^2}} = \frac{1}{2} \not\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to calculate this integral using Cauchy's Formula?
Calculate $\int_{C} \sinh(z+1)/(z^2+1)dz$ where $C$ is $x^{2/3}+y^{2/3}=3^{2/3}$
My try is redefine $\int_{C} \sinh(z+1)/(z^2+1)dz$ as
$
\begin{align}
\int_{C} \sinh(z+1)/(z^2+1)dz &= \int_{C} \sinh(z+1)/((z-i)(z+i))dz \\
&= \int_{C} \sinh(z+1)/(2i(z-i))dz-\int_{... | Your answer is actually equivalent after applying some identities!
There is a formula for the difference of hyperbolic sines (Wikipedia article)
$$\sinh(x) - \sinh(y) = 2 \cosh \left( \frac{x+y}{2} \right) \sinh \left( \frac{x-y}{2}\right)$$
Another identity of importance is
$$\sinh(x) = -i \sin(ix) \;\;\; \text{or, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the following limit: $=\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $ I am studying MIT OpenCourseware 18.01 Single Variable Calculus on my own and am stuck on a final exam question.
Evaluate the following limit:
$$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} \sqrt{1... | Just for your curiosity (without Riemann sum).
Assuming that you know about generalized harmonic numbers, you could approximate quite well the partial sums since
$$ S_n= \frac{2}{n} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}}=2 \sqrt{2}\,\frac{ H_{\frac{3
n}{2}}^{\left(-\frac{1}{2}\right)}-H_{\frac{n}{2}}^{\left(-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
CNF or DNF, $p(x)=0 \implies f=1$ Write either a $CNF$ or a$\;DNF$ of the expression $f$ such that
$f=0\iff (x_2x_1x_0)_{(2)}$ is a zero point of the polynomial:
$$p(x)=x^3-8x^2+12x=x(x-2)(x-6)$$
$$p(x)=0\iff x\in \{0,2,6\}$$
$$\begin{array}{|c|c|c|c|}
\hline
x_2& x_1 & x_0 & f=1 \\ \hline
0 &0 &0 & 1\\ \hline
0 &0 &1... | It's correct, and if you want to find the Minimal form, you can use a k-map:
$$\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&\color{orange}1&\color{orange}1&1&0\\
x_0&0&0&0&0\end{array}}
\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&\color{red}1&\color{red}1&0\\
x_0&0&0&0&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\le... | Given
\begin{align}
\tan\tfrac12A&=u-d
,\quad
\tan\tfrac12B=u
,\quad
\tan\tfrac12C=u+d
,\quad
u,d\in\mathbb{R}
\tag{1}\label{1}
.
\end{align}
We can express $d$ in terms of $u$ using
known identity for triangles
\begin{align}
\tan\tfrac A2\tan\tfrac B2+
\tan\tfrac B2\tan\tfrac C2+
\tan\tfrac C2\tan\tfrac A2&=1
,\\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solving limit for $\lim_{n \to \infty} \frac{3n^3 + 5n^2 + (-1)^n \cdot n}{5n^2 - 13}$ Trying to solve:
$$\lim_{n \to \infty} \frac{3n^3 + 5n^2 + (-1)^n \cdot n}{5n^2 - 13}$$
I tried to divide with $n^3$ and $n^2$, but the limit of $(-1)^{n}$ is not existed, so i am wondering if the limit of $(-1)^{n}\cdot n$ does exi... | We have indeed that $\lim_{n \to \infty} (-1)^{n}\cdot n$ doesn't exist but we have that
$$ \frac{3n^3 + 5n^2 + (-1)^n \cdot n}{5n^2 - 13} = \frac{n^3}{n^2}\cdot\frac{3 + \frac 5 n +\frac{(-1)^n}{n^2} }{5- \frac{13}{n^2}}$$
and
$$\left|\frac{(-1)^n}{n^2}\right|= \frac{1}{n^2}\to 0 \implies \frac{(-1)^n}{n^2}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the family of functions given an equation I played with the solution for the problem $$\text{if }\; x + \frac{1}{x} = a$$ what is $$x^5 + \frac{1}{x^5}$$
I tried different exponents other than 5 and tried finding the solution to it. I defined $f(x) = a^x + \frac{1}{a^x}$. I got $f(x + y)=f(x)f(y) - f(x-y).$ I t... | Hint:
$$\left(x^3+\dfrac1{x^3}\right)\left(x^2+\dfrac1{x^2}\right)=x^5+\dfrac1{x^5}+x+\dfrac1x$$
Now $x^3+\dfrac1{x^3}=\left(x+\dfrac1{x}\right)^3-3\left(x+\dfrac1{x}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solution attempt $xuu_x+yuu_y=u^2-1$ Solve $$
\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}
$$
I have got using Lagrange method:
$$F\left(\frac{x}{y}\right)=\frac{x^2}{u^2-1}$$
Applying $u(x,x^2)=x^3$:
$$u^2=\frac{y^6-x^6}{x^2y^2}+1$$
But plug in it to the PDE show that there is a mistake
| For the given PDE, Lagrange auxiliary equation $$\dfrac{dx}{xu}=\dfrac{dy}{yu}=\dfrac{du}{u^2-1}\tag1$$
From the first two ratio, $$\dfrac{dx}{xu}=\dfrac{dy}{yu}\implies \dfrac{dx}{x}=\dfrac{dy}{y}$$
Integrating we have $$\log x~=~\log y~+~\log c_1\implies\dfrac xy=c_1$$where $~c_1~$is a constant.
Again from the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \fra... | $$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}= \frac {(3x^2+3x +3)-(x+2)}{(x+2)(x^2+x+1)^2}=$$
$$ \frac {3}{(x+2)(x^2+x+1)}-\frac { 1}{(x^2+x+1)^2} $$
Now you may proceed with the first fraction.
$$ \frac {3}{(x+2)(x^2+x+1)}=\frac {1}{x+2} -\frac {x-1}{x^2+x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Understanding the equation $|x+1|=x^2 -1$ I want to understand the equation
$$|x+1|=x^2 -1$$
$$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$
$$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \... | When $x + 1 < 0$ then the initial equation becomes $$\begin{align*} && -(x+1) &= x^2 - 1 \\ &\iff&-x -1 &= x^2 - 1 \\ &\iff& 0 &= x^2 + x\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $ \arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2} $ I want to prove that: $$\arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2}, \forall x \in \mathbb{R} $$
I know that $$ \arctan x + arctan \frac{1}{x} = \frac{\pi}{2} $$
But that doesn't seem to be helping. How should I proceed?
| You may set $x=\tan\theta$ (for some $\theta\in\left(-\pi/2,\pi/2\right)$) and check that
$$2\arctan\left(\frac{1-\sin\theta}{\cos\theta}\right) = \frac{\pi}{2}-\theta $$
holds, or that
$$ 2\arctan\left(\frac{1-\cos\theta}{\sin\theta}\right)=\theta $$
holds for any $\theta\in(0,\pi)$. Since $1-\cos\theta=2\sin^2\frac{\... | {
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"url": "https://math.stackexchange.com/questions/3465969",
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"source": "stackexchange",
"question_score": "6",
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$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=... | If $ \{ a, b \} = \{c, d \}$ then $a+b+c+d = 2(a+b)$ is not a prime.
Henceforth, we have distinct pairs. If $ a = c$, then $b \neq d$ are roots to $x^2 + ax + a^2 = 0$, so $b+d = -a$ contradicting the requirement that all of the terms are positive.
WLOG, $ a > c \geq d > b$.
Observe that $$ -(a-b+c-d)(a-b-c+d) =... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding the limit $\lim_{x\to\infty} (x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}$ Finding the limit $\lim_{x\to\infty} (x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}$
I have tried using Taylor series. On a side not, I am not sure what to write when expanding both the $\sqrt{x^6+1}$ and the $e^{1/x}$. Do I write ... | Let $(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}=(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6(1+\frac{1}{x^6})}$
This is $(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -x^3\sqrt{1+\frac{1}{x^6}}$
$(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}+\frac{1}{4!x^4}+o(\frac{1}{x^5}))-x^3(1+\frac{1}{2x^6}+o(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with this multi variable limit. $ \lim_{x,y\to (1,0)} \frac{2x^4+x^2y^2-2x^2-y^2}{2x^2y^2-x^2-2y^2+1} $
By just plugging in $(1,0)$ into the limit you get $-2$, but I don't know if this enough to solve the question
Ive tried approaching along $x=y+1$ but it got extremely messy.
| You have
$$
\frac{2x^4+x^2y^2-2x^2-y^2}{2x^2y^2-x^2-2y^2+1}= \frac{(2x^2+y^2)(x^2-1)}{(2y^2-1)(x^2-1)}= \frac{2x^2+y^2}{2y^2-1}\to\frac2{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^{2}y''-3xy'+4y=x^{2}lnx$.
Solve $x^{2}y''-3xy'+4y=x^{2}lnx$.
I wanted to try and follow my professor's notes but I was confused when she wrote $\eta=lnx$ and $\frac{dy}{dx}=\frac{1}{x}\frac{dy}{d\eta}$. Something about changing the equation to have constant coefficients.
Is there a step-by-step way of solvin... | Let $D \equiv \frac{d}{dx}, D^2 \equiv \frac{d^2}{dx^2}$ and $\theta \equiv\frac{d}{d\eta}, \theta^2 \equiv\frac{d^2}{d\eta^2}$.
Once you substitute $\ln x = \eta \implies xD = \theta, x^2D^2 = \theta (\theta -1)$
In general
$x^nD^n = \theta(\theta-1)\cdots(\theta -n +1)$
This works only in case you have differentia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
MixColumn AES manually with GF(2^8) In the step of MixColumn in AES, calculate d0,0 if given c0,0 = c2,0 =
1100 0101, and c1,0 = c3,0 = 0100 1100.
Converting to polynomials:
c0,0 = c2,0 = x^7 + x^6 + x^2 + 1
c1,0 = c3,0 = x^6 + x^3 + x^2
now using the Galois field to get d0,0 we will get 4 results and then XOR all of t... | The second product is incorrect. There is no overflow to degree $8$ in that product, and it comes out as
$$
\begin{aligned}
c_{1,0}\cdot 11&=\overline{(x^6+x^3+x^2)(x+1)}\\
&=\overline{x^7+x^6+x^4+2x^3+x^2}\\
&=\overline{x^7+x^6+x^4+x^2}=11010100.
\end{aligned}
$$
I use overline to denote the coset of a polynomial beca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the $\int \frac{dx}{\sin^5x}$ $\int \frac{dx}{\sin^5x}$
I did $t= \cos(x)$, $-\frac{dt}{\sin(x)}=dx$:
$$\int \frac{dx}{\sin^5x} = \int \frac{dt}{\sin^6x} = -\int \frac{dt}{(1-t^2)^3}=-\int (1-t^2)^{-3} = \frac{(1-t^2)^{-3}}{-3} = -\frac{1}{3(1-\cos^2x)}+C$$
Is this correct?
| With $t=\cos x$
\begin{align}
\int \frac{1}{\sin^5x} dx= &-\int \frac{1}{(1-t^2)^3}dt
=-\int \frac1{4t^3}\ d\left( \frac{t^4}{(1-t^2)^2}\right)\\
\overset{ibp}= & -\frac t{4(1-t^2)^2}-\frac38\int \frac1{t}\ d\left( \frac{t^2}{1-t^2}\right)\\
\overset{ibp}= & - \frac t{4(1-t^2)^2}-\frac38 \left(
\frac t{1-t^2}+\tanh^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The set of differences of square rationals At first, we observe that $A:=\{ p^2-m^2 : p,m\in \Bbb{Z}\}=\mathbb{Z}\setminus (4\mathbb{Z}+2)$ (because an integer $a$ can be written
as the form $a=p^2-m^2$ if and only if $a\neq 4k+2$, for every integer $k$) and so $A^c=4\mathbb{Z}+2$. Hence, $B:=\left\{ p^2-m^2 : p,m\in \... | We have $4m = (m+1)^2-(m-1)^2$ for all $m \in \mathbb Z$. Therefore,
$$
\frac{a}{b}
=
\frac{4ab}{4b^2}
=
\frac{(ab+1)^2-(ab-1)^2}{(2b)^2}
=
\left(\frac{a b + 1}{2 b}\right)^2 - \left(\frac{a b - 1}{2 b}\right)^2
$$
When $ab=\pm 1$, one of the terms is zero. In this case, $\frac{a}{b}=\pm1$ and we can use
$$
1 = \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve a system by putting new variables
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$$
The first step is to determine the domain: $\begin{array}{|l} x \ne 0 \\ y \ne 0 \end{array}$
We can simplify the first equation of the system, and we get: $\begin{arr... | Rewrite the first equation as
$$
\frac{x^2}{xy}+\frac{y^2}{xy}=\frac{7}{25}.
$$
The second equation then gives $\frac{625}{7}=xy$. Substituting gives
$$
49x^4 - 1225x^2 + 390625=0,
$$
which is a biquadratic equation with no real solution. This seems short enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$ \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} = \frac { 2^{\ln n \sqr... | Your use of $\sim$ is not correct although this small mistake does not change the limit. You have
$$2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}} \stackrel{n\to\infty}{\sim}\color{blue}{2\cdot}2^{\ln n}$$
This is so because
$$\frac{2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{2^{\ln n}}= 2^{\ln n\left(\sqrt{ 1+ \frac {2}{\ln n}}-1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
In triangle $ABC$, find maximum value of $\sin A \cos B + \sin B \cos C + \sin C \cos A$ In triangle $ABC$, find maximum value of $$\sin A \cos B + \sin B \cos C + \sin C \cos A$$
We could make $\cos C = - \cos(A+B)$ and $\sin C = \sin(A+B)$.
But then we have a rather awkward expression that doesn't share the same pow... | You can use Lagrange multipliers. You are maximizing
$$
\sin A\cos B+\sin B\cos C+\sin C\cos A
$$
under
$$
A+B+C=\pi.
$$
Using Lagrange multipliers and the identity $\cos (x+y)=\cos x\cos y-\sin x\sin y$, you easily get
$$
\cos(A+B)-\lambda=\cos(B+C)-\lambda=\cos(A+C)-\lambda=0,
$$
from where it follows that $\cos C=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int \frac{dx}{x\sqrt{x^2-2}}$ $\int \frac{dx}{x\sqrt{x^2-2}}$
My book says to solve this using $x = 1/t$ so I did $dx = -1/t^2$
$$-\int\frac{\frac{1}{t^2}}{t\sqrt{1-2t^2}} = -\int \frac{1}{\sqrt{1-2t^2}} = -\arcsin(\sqrt{2t})=-\arcsin(\sqrt{2}/x)$$
My book says the solution is $\frac{1}{\sqrt{2}}\arccos(\frac{\s... | First, remember integration constants: $\arcsin y=\frac{\pi}{2}-\arccos y$. Secondly, be careful with coefficients: $\sqrt{1/t^2-2}=\sqrt{2}\sqrt{1/(2t^2)-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots... | Assuming $n\neq 0$ (to make the expression under limit meaningful) the given expression can be written as $$\dfrac{\dfrac{x} {\sin x} - 1}{\left(\dfrac{x}{\sin x}\right)^n-1}\cdot\frac{x^3}{x-\sin x} \cdot x^{n-2}$$ The first factor tend to $1/n$ and second one tends to $6$ and hence the limiting behavior of the expres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Rational solutions of quadratic Diophantine equation $ax^2+by^2+cz^2+du^2=v^2$? What do we know about the rational solutions of quadratic Diophantine equation $ax^2+by^2+cz^2+du^2=v^2$ in five variables $x,y,z,u,v$?
I am looking for references/papers related to this equation.
| $ax^2+by^2+cz^2+du^2=v^2$
Let assume $a+b+c+d=r^2.$
$p,q$ are arbitrary.
Substitute $x=pt+1, y=qt+1, z=pt-1, u=qt-1, v=t+r$ to above equation, then we get $$t = \frac{2(ap-qr^2+qa+2bq+qc-cp-r)}{(-ap^2-q^2r^2+q^2a+q^2c-cp^2+1)}.$$
Thus, we get a parametric solution below.
\begin{eqnarray}
&x& = (a-3c)p^2+(2qc-2qr^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.