Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Maclaurin expansion of $\arcsin x$ I'm trying to find the first five terms of the Maclaurin expansion of $\arcsin x$, possibly using the fact that
$$\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}.$$
I can only see that I can interchange differentiation and integration but not sure how to go about this. Thanks!
| Considering the binomial expansion
\begin{align}
\frac 1{\sqrt{1-u}}&=1+\frac12 u+\frac12\cdot\frac32\frac{u^2}{2!}+\frac12\cdot\frac32\cdot\frac52\frac{u^3}{3!}++\frac12\cdot\frac32\cdot\frac52\cdot\frac 72\frac{u^4}{4!}+\dotsm \\
&=1+\frac12 u+\frac{1\cdot3}{2^2\,2!}u^2 +\frac{1\cdot3\cdot 5}{2^3\,3!}u^3 + +\frac{1\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/197874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 3
} |
Completing the square with negative x coefficients I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?
For example:
\begin{align*}
f(x) & = x^2 + 6x + 11 \\
& = (x^2 + 6x) + 11 \\
& = (x^2 + 6x + \mathbf{9}) + 11 - \mathbf{9} \\
& = (x+3... | $$(2ax+b)^2 = 4a^2x^2 + 4abx + b^2$$
So, multiplying by $4a$, or in this case $-4a$, will avoid fractions until the last step.
\begin{align}
f(x) &= -3x^2 + 5x + 1 \\
-12f(x) &= 36x^2 -60x - 12 \\
-12f(x) &= 36x^2 - 60x + 25 - 25 - 12 \\
-12f(x) &= (6x-5)^2 - 37 & \left(b = \frac{4ab}{4a} = \dfrac{-60}{12}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| $\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=\sqrt{6}$
Is an addition of complex conjugates of the form
$(a+b\sqrt{-1}) + (a-b\sqrt{-1})$
So then
$1+\sqrt{-3}=(a+b\sqrt{-1})^2$
$=a^2+2ab\sqrt{-1}-b^2$
If we equal the real and complex parts we get:
$1=a^2-b^2$ and $\sqrt{-3}=2ab\sqrt{-1}$
Solving for a in the complex equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 4
} |
Euclidean Algorithm in Polynomials $\Bbb Z_2[x]$ Express the greatest common divisor of the following pair of polynomials as a combination of polynomials:
$f(x) = x^3 + x^2 +x +1$ and $g(x) = x^4 + x^2 + 1$.
I've been trying to understand this, but still can't get how I should do it in $\mathbb{Z}_2[x]$.
| There’s really nothing different from solving such problems in $\Bbb Z$.
Using the Euclidean algorithm in its most straightforward form, not trying to be mechanically efficient:
$$\begin{align*}
x^4+x^2+1&=(x+1)(x^3+x^2+x+1)+x^2\\
x^3+x^2+x+1&=(x+1)x^2+(x+1)\\
x^2&=x(x+1)+x\\
x+1&=1\cdot x+\color{red}{1}
\end{align*}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $n>1$ such that $\frac{2^n+1}{n^2}$ is an integer.
Find all $n>1$ such that $\dfrac{2^n+1}{n^2}$ is an integer.
I know that $n$ must be odd, then I don't know how to carry on. Please help. Thank you.
| Let's consider $$\frac{2^n+1}{n^k}$$
If $p$ be the smallest prime that divide $n$
Let $$\operatorname{ord}_p2=d,d\mid(p-1,2n)\implies d\mid 2$$ as $p-1<$ all other primes, so it implies
$$ p\mid (2^2-1)\implies p=3.$$
Let $$3^r||n, 2^{2n}\equiv 1{\pmod {3^{kr}}}\implies \phi(3^{kr})|2n$$ as $2$ is a primitive root of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $\frac{1}{\sqrt{x+5}}$
I'm trying to find the derivative of $\dfrac{1}{\sqrt{x+5}}$
using $\displaystyle \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$
So,
$$\begin{align*}
\lim_{h\to 0} \frac{\dfrac{1}{\sqrt{x+h+5}}-\dfrac{1}{\sqrt{x+5}}}{h} &= \frac{\dfrac{\sqrt{x+5}-\sqrt{x+h+5}}{(\sqrt{x+h+5})(\sqrt{x+5})... | I would suggest, start it over..
We can introduce a variable, say $u:=x+5$. Then it's
$$\begin{align*} \lim_{h\to 0} \frac1h\cdot \left( \frac1{\sqrt{u+h}} -\frac1{\sqrt{u}} \right)
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \right) = \\
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
| Can you see by inspection that 2 is a root? If so, we can write your equation as $x^{3}-12x+16=(x-2)(ax^2+bx+c)=ax^{3}+(b-2a)x^{2}+(c-2b)x-2c$ by the factor theorem.
Thus, $a=1$, $b-2a=0$, $c-2b=-12 $ and $-2c=16$
Which gives $a=1$, $b=2$ and $c=-8$.
Therefore $x^{3}-12x+16=(x-2)(x^2+2x-8)=(x-2)(x-2)(x+4)=(x-2)^{2}(x+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the distance between the line and plane Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my ne... | If $z=t,$ for the given line $x=z+3=t+3$ and
$x+2y+4z=6$ or, $t+3+2y+4(t)=6$ or, $2y=3-5t$
So, any point on the given line is $(t+3,t,\frac{3-5t}2)$
The distance of $(t+3,t,\frac{3-5t}2)$ from $3x+2y+2z=5$ is $$\left|\frac{3(t+3)+(3-5t)+2(t)-5}{\sqrt{3^2+2^2+2^2}}\right|=\frac 7{\sqrt {17}}$$
Had it not been a constan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/209910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
I'm not familiar to factorial and I don't have much idea, can someone show me how to prove this? Thank you.
| Consider a prime $p$. The highest power of $p$ dividing the numerator is $$\alpha_{Nr} = \left(\left \lfloor \dfrac{2m}{p}\right \rfloor + \left \lfloor \dfrac{2m}{p^2}\right \rfloor + \left \lfloor \dfrac{2m}{p^3}\right \rfloor + \cdots \right) + \left(\left \lfloor \dfrac{2n}{p}\right \rfloor + \left \lfloor \dfrac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/215355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Why is there a pattern for $2^n=x \pmod {13}$ for all integers n? I was doing a question yesterday on very elementary number theory, and I came across this pattern.
$2^1=2 \pmod {13}$
$2^2=4 \pmod {13}$
$2^3=8 \pmod {13}$
$2^4=3 \pmod {13}$
$2^5=6 \pmod {13}$
$2^6=12 \pmod {13}$
$2^7=11 \pmod {13}$
$2^8=9 \pmod {13}$
$... | Simpler answer: For all these modulo questions, it is enough to find some exponent that gives a modulo of 1. Consider
$2^n = 1 (mod 13)$ for some n. Then $2^{n + 1} = 2 (mod 13)$ which is same as $2^{0 + 1} = 2 (mod 13)$ and $2^{n + 2} = 4 (mod 13)$ which is $2^{1 + 1} = 4 (mod 13)$, in other words, they repeat. And a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
| Note that using the AM-GM inequality, we have
$$\frac{a^3}{a+b}=a^2-\frac{a^2b}{a+b}\geq a^2-\frac{a^2b}{2\sqrt{ab}}=a^2-\frac 12\sqrt{a^3b},$$
And so summing up cyclically, it is enough to check that
$$2(a^2+b^2+c^2)\geq\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}+ab+bc+ca,$$
Which is obviously true since $\sqrt{a^3b}\leq \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Solving system of recurrence relations Base Case:
$$
\left\{
\begin{array}{c}
T(1) = 1 \\ T(2) = 1 \\T(3) = 4\end{array}
\right.
$$
I have the system:
$$
\left\{
\begin{array}{c}
T(N) = G(N-1) + F(N-1) \\
G(N) = F(N-1) + G(N-1) \\
F(N) = 2H(N-1) + F(N-2) \\
H(N) = H(N-1) + F(N-1)\end{array}
\right.
$$
I seems $$T(N... | Note that
$$
\begin{pmatrix}
T_n \\
F_n \\
F_{n - 1}
\end{pmatrix} = \begin{pmatrix}
1 & 1 & 0 \\
2 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
T_{n-1}\\
F_{n-1}\\
F_{n-2}
\end{pmatrix}.
$$
So starting with $(T_1, F_1, F_0)$ you can find $(T_n, F_n, F_{n-1})$ quickly by matrix exponentiation. The latter can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/223289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find formula for ${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27},\frac{8\ln6}{81} ,....}$ The question is to find a general formula for the nth term, $a_n$, of the sequence:
$${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27}, \frac{8\ln6}{81} ,\frac{10\ln7}{243},...}$$
Here is what I got, but when I plugged... | Each term has a form like $\pm \frac{x\ln y}{z}$.
The sign alternates so it is $(-1)^n$.
$x$ is $2, 4, 6, 8, \dots$, so $x$ is $2n$.
$y$ is $3,4,5,\dots$ so $y$ is $n+2$.
$z$ is $3,9,27, 81, \dots$ so $z$ is $3^n$.
Combining all contributions, we get $a_n = (-1)^n \frac{2n \ln(n+2)}{3^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/224983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding an unknown angle Geometry: Auxiliary Lines
As shown in the figure:
| Let central point is $O$.
$\angle OBC = 180^\circ - (46^\circ+12^\circ) - (62^\circ+8^\circ) - X = 52^\circ-X$.
Using the sine formula, we have:
$$\frac{AO}{\sin 62^\circ} = \frac{CO}{\sin 12^\circ},$$
$$\frac{AO}{\sin X} = \frac{BO}{\sin 46^\circ},$$
$$\frac{CO}{\sin (52^\circ-X)} = \frac{BO}{\sin 8^\circ},$$
so,
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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prove, that following formula is correct As in the statement, I got problems with:$$\binom{n}{0} +\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2[\frac{n}{2}]}=2^{n-1}$$ I started with Newton conjecture, trying to work with $(1+1)^n=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}$ but actually it does not guide me anywhere.... | The root of the solution lies in the fact, $\binom n {2[\frac{n}{2}]}$ is the last term of $(1+1)^n+(1-1)^n$ as proved below:
for $n=2m, [\frac n2]=[\frac{2m}2]=m, \binom n {2[\frac{n}{2}]}=\binom{2m}{2m}$
for $n=2m+1, [\frac n2]=[\frac{2m+1}2]=m ,\binom n {2[\frac{n}{2}]}=\binom{2m+1}{2m}$
Now,
$$2^{2m}=(1+1)^{2m}=\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\frac{x-1}{x+3}>\frac{x}{x-2}$ I'm having little trouble solving$$\frac{x-1}{x+3}>\frac{x}{x-2}$$
What steps should I take?
Need this to write the topological spaces of the set defined by this inequation.
| $$\frac{x-1}{x+3}>\frac{x}{x-2}$$
$$\frac{x-1}{x+3}-\frac{x}{x-2}>0$$
$$
\frac{(x-1)(x-2)}{(x+3)(x-2)}-\frac{x(x+3)}{(x-2)(x+3)}>0
$$
$$
\frac{(x-1)(x-2)-x(x+3)}{(x-2)(x+3)}>0
$$
$$
\frac{(x^2 -3x +2)-(x^2+3x)}{(x-2)(x+3)}>0
$$
$$
\frac{-6x +2}{(x-2)(x+3)}>0
$$
$$
\frac{x -(1/3)}{(x-2)(x+3)}<0
$$
This fraction undergoe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Use Lagrange multiplier to find absolute maximum and minimum Use Lagrange multiplier to find absolute maximum and minimum of $f(x,y) =x^2+xy+y^2, x^2+y^2 =8$.
What i've done so far..
$f_x = \lambda g_x \Rightarrow 2x+y =\lambda2x, \\f_y = \lambda g_y \Rightarrow x+2y = \lambda 2y,\\g(x,y) = x^2+y^2 -8 =0$
May I kno... | On a side note, this problem can be solved very nicely with substitution since it's equivalent to:
$$f(\theta) = 8 \left ( 1 + \frac{1}{2}\sin 2\theta \right )$$
where $ x = r \cos\theta$, $y = r \sin \theta$ and $r = 2\sqrt{2}$.
Since $f(\theta)$ is the same as maximizing $\sin 2 \theta$, $f(\theta)$ is maximum and mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/234909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant. *help needed please* Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant.
This is what i've done so far....
Let $a,b,c$ be ... | BTW: you don't need Lagrange multipliers for this problem:
To find the minimum of $abc$ for $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$ you can use $AM-GM$:
$$\frac{1}{3}=\frac{\frac{1}{a}+\frac{2}{b}+\frac{3}{c}}{3} \geq \sqrt[3]{\frac{6}{abc}}$$
Thus
$$abc \geq 3^3 \cdot 6 \,,$$
with equality if and only if
$$\frac{1}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Jordan canonical form of a matrix for distinct eigenvalues How can i find the Jordan canonical form of this matrix $$A=\begin{pmatrix}1 &0 &0 &0 \\ 1& 2& 0& 0\\ 1 &0& 2& 0\\ 1 &1& 0& 2\end{pmatrix}.$$
In my book there are examples but all the matrices in these examples have only one eigenvalue repeated n times(for $n\t... | The properties listed here can help you.
In this case you have three possibilities for the Jordan Canonical Form.These are:
$$J_1=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 0\\ 0 &0& 0& 2\end{pmatrix} , \ J_2=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2\end{pmatrix}, \ J_3=\begin{pmatrix}1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement hold... | This can be shown using a variant of Lagrange Inversion. Introduce
$$T(z) = w = \sqrt{1-4z}$$ so that
$$z = \frac{1}{4} (1-w^2)$$ and
$$dz = -\frac{1}{2} w \; dw.$$
Then we seek to compute
$$[z^n] \frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{T(z)} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/237810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 4
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Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$
Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.
The answer is $Θ (n \log \log n)$.
Can anyone arrive at the solution.
| Yes, the Master Theorem can be applied. Here's how:
$$T(n) = \sqrt{n} T(\sqrt{n}) + n = \sqrt{n} T(\sqrt{n}) + \mathcal{O}(n)$$
Let $n = 2^k$, $\sqrt{n} = 2^{k/2}$, and $k = \log{n}$. Substituting in above equation, we get:
$$T(2^k) = 2^{k/2} T(2^{k/2}) + 2^k \tag{1}$$
Dividing (1) by $2^k$, we get:
$$\frac{T(2^k)}{2^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
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integration by parts from Apostol I working through Apostol's calculus, and I need to prove integrating by parts that :
$\int (a^2 - x^2)^n \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2 a^2 n}{2n+1} \int (a^2 - x^2)^{n-1} \,dx + C $
Now, using the integration by parts formula after first division the integral to par... | OK. Another problem. Miserably stuck again :/. Integration by parts. Need to show that:
If $I_n(x)=\int_{0} ^{x}t^n(t^2+a^2)^{-\frac{1}{2}}dt$
Then: $nI_n(x) = x^{n-1}\sqrt{x^2+a^2}-(n-1)a^2I_{n-2(x)}$ if $x\geq2$
I can get to the point where:
$nI_{n}=x^{n+1}(x^2+a^2)^{-\frac{1}{2}}-a^2I_{n-2}+a^4\int t^{n-2}(t^2+a^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? The numbers in the AP must be distinct.
For example if $n=6$ then number of AP's possible are $6$
*
*$1,2,3$
*$2,3,4$
*$3,4,5$
*$4,5,6$
... | Let the three term AP be $a,a+d,a+2d$ where $a,d$ are natural numbers.
So, $a\ge1$ and $a+2d\le n\implies 1\le a\le n-2d \implies d\le \frac{n-1}2$
If $d=1,a$ can assume $1,2,\cdots, n-2$ i.e., $n-2$ values.
If $d=2,a$ can assume $1,2,\cdots, n-4$ i.e., $n-4$ values.
If $n$ is even, $d_{max}=\frac{n-2}2$
for $d=\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/248703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to solve $ \sum_{i=1}^{n-1} i^2 \equiv \;? \pmod n$ I solved expressions below . it is ok?
My attempt:
$$1)\;\;\;\;\; ?\cong\pmod n\sum_{i=1}^{n-1} i^2 \stackrel{?}= {(n-1)^3\over3}+{(n-1)^2\over2}+{(n-1)\over6}={n^3\over3}-{n^2\over6}+{n\over6}=n\left({n^2\over3}-{n\over2}+{1\over6}\right)
$$
(continue?)how to ... | It never hurts to gather some numerical data. For your first problem:
$$\begin{array}{cc|cc|cc}
n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n&n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n&n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n\\ \hline
2&1&8&4&14&7\\
3&2&9&6&15&10\\
4&2&10&5&16&8\\
5&0&11&0&17&0\\
6&1&12&2&18&3\\
7&0&13&0&1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving a system of linear equations to find two eigenvectors. So I have
\begin{align*}
x - 2y + z & = 0 \\
-2x + 4y - 2z & = 0 \\
x - 2y + z & = 0
\end{align*}
I know I need to find two eigenvectors for the eigenspace with eigenvalue 2 as I know the matrix is diagonalizable and I've already found the eigenvector for t... | $$ \begin{align*}
x - 2y + z & = 0 \\
-2x + 4y - 2z & = 0 \\
x - 2y + z & = 0
\end{align*} $$
$$ \implies \begin{align*}
x - 2y + z & = 0 \\
0 & = 0 \\
0 & = 0.
\end{align*}$$
So you have two free variables. Assuming $y=t, z=s,\, t,s\in \mathbb{R} $, then the solution is given by
$$ \begin{bmatrix}
x \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ converges. I view the orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ as the sequence $(x_{n})$. So
$$x_n = \frac{1}{2}, f\bigg(\frac{1}{2}\bigg), f\... | Note that $f(\frac{1}{2})=\frac{3}{4}$(as commented by bonext), $f(\frac{2}{3})=\frac{2}{3}$(as you have noticed) and $f(\frac{3}{4})=\frac{9}{16}\in(\frac{1}{2},\frac{2}{3})$. Also note that $f$ is strictly decreasing on $[\frac{1}{2},1]$. Then we have
$$f((\frac{1}{2},\frac{2}{3}))\subset(\frac{2}{3},\frac{3}{4})\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a quadratic equation via a tangent half-angle formula (Maybe I'll post my own answer here, but maybe others will make that redundant.)
This is a fun (?) trivia item that fell out of a bit of geometry I was thinking about.
One of the tangent half-angle formulas says
$$
\tan\frac\alpha2 = \frac{\sin\alpha}{1+\cos... | When you took $\sqrt{1+b^2}$ you did not consider the negative square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/252724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that every element of $a_{n+2013}=\frac{a_{n+1}a_{n+2}...a_{n+2012}+1}{a_n}$ is an integer
Given $\displaystyle a_1=a_2=\cdots=a_{2013}=1$ and $\displaystyle a_{n+2013}=\frac{a_{n+1}a_{n+2}\cdots a_{n+2012}+1}{a_n}$.
Prove that $a_{n+2013}\in\mathbb{N}$ for all $n\in\mathbb{N}$.
I tried to prove it with inducti... | For these kind of questions, it is clear that $2013$ is arbitrary. Let us replace $2013$ by $k \geq 2$, so
$$a_1, a_2, \ldots , a_k=1, a_{n+k}=\frac{a_{n+1}a_{n+2} \ldots a_{n+k-1}+1}{a_n}, n \geq 1$$
It is clear that $a_{n+k}>0$.
We prove by induction on $n \geq 1$ that $a_{n+k} \in \mathbb{Z}$.
When $1 \leq n \leq k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove the point is the midpoint of a segment Let a circle (O) and a point A outside of the circle. AB, AC are tangents of (O) (B,C $\in$ (O)). BD is an diameter of (O). CK perpendicular to BD (K $\in$ BD). Let I is intersection of CK and AD. Prove that I is midpoint of KC
|
Without any loss of generality we can assume $B(a,0),O(0,0)$ so that $D(-a,0)$ the equation of the circle $(x-0)^2+(y-0)^2=(a-0)^2\implies x^2+y^2=a^2$
Using this, the equation of the tangent at any point $P(a\cos\theta,a\sin\theta)$ will be
$xa\cos \theta+ya\sin\theta=a^2\implies x\cos \theta+y\sin \theta=a$
So, $AB:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots... | Your determinant is equal to
$$
2^n\det\begin{bmatrix}2x & 1 & 0 & 0 & 0 & \cdots & 0\\ 1 & 2x & 1 & 0 & 0 & \cdots & 0\\ 0 & 1 & 2x & 1 & 0 & \cdots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots &\vdots\\ 0 &
\cdots & 0 & 1 & 2x & 1 & 0\\ 0 & \cdots & 0 & 0 & 1 & 2x & 1\\ 0 & \cdots & 0 & 0 & 0 & 1 & 2x\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 0
} |
$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$ Let $a,b,c,d >0$ and $a+b+c+d=2$. Prove this:
$$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$$
| This is a brute force approach.
First, let's show it for $a,b,c,d\geq 0$, since it is true in that case, too, and this set is compact, so if there is a minimum, it is reached somwhere.
Let $f(x)=\frac{1}{3x^2+1}$. Then $f''(x)=\frac{6(9x^2-1)}{(3x^2+1)^3}$. So $f(x)$ is convex on $[1/3,2]$. In particular, if $a,b,c,d\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Listing all the subfield and corresponding subgroup Here is an exercise in the book of Dummit Foote :
Find the Galois group of the splitting field of $(x^2-2)(x^2-3)(x^2-5)$ over $\mathbb{Q}$. Then list all the subgroups and the corresponding subfield
Here is my argument :
It is not hard to see that $\mathbb{Q}(\sqrt... | For the first question, consider $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Note this relationship for "independent" automorphisms.
Alright, since I don't know how to do the larger tables in latex here (usually I use xymatrix!) lets (diagrammatically) look at the example at th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/270780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate value of expression $(\sin^6 x+\cos^6 x)/(\sin^4 x+\cos^4 x)$
Calculate the value of expresion:
$$
E(x)=\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}
$$
for $\tan(x) = 2$.
Here is the solution but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this:
Can you explain to me why t... | $$\tan x=2\implies \frac{\sin x}2=\frac{\cos x}1\implies \frac{\sin^2x}4=\frac{\cos^2x}1=\frac{\sin^2x+\cos^2x}{4+1}=\frac15$$
$$\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}=\frac{\left(\frac45\right)^3+\left(\frac15\right)^3}{\left(\frac45\right)^2+\left(\frac15\right)^2}=\frac{(4^3+1)5^2}{5^3(4^2+1)}=\frac{13}{17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| The problem can be restated as minimizing ${1 \over x^2} + {8 \over y^2} + {27 \over z^2}$ subject to $x + y + z = 1$. By Lagrange multipliers you seek an $(x,y,z)$ and $\lambda$ such that $-{2 \over x^3} = \lambda = -{16 \over y^3} = -{54 \over z^3}$. Dividing by $-2$ and taking cube roots leads to
$${1 \over x} = {2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Evaluating the series: $1 +(1/3)(1/4) +(1/5)(1/4^2)+(1/7)(1/4^3)+ \cdots$
Sum the series:
$$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots$$
How can I solve this? I am totally stuck on this problem.
| Using Infinite geometric series, $$1+x+x^2+\cdots=\frac1{1-x}$$ for $|x|<1$
Applying integration wrt to $x, \log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots+c$ where $c$ is an arbitrary constant of indefinite integral.
Putting $x=0,\log(1)=c\implies c=0$
So, $$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots$$
Putting $x=y$ we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ I am wondering if there exists any formula for the following power series :
$$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$
Is there any way to calculate the sum of above series (if $k$ is given) ?
| Consider the the sequence $a_n$ defined as:
If $n=1$ then $a_n=1$, else if $n>1$ then $a_n=-1$.
This is the sequence $a_n = 1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,...$
The Dirichlet inverse (call it $b_n$) of $a_n$ is the oeis sequence called
"Number of ordered factorizations of n" http://oeis.org/A074206 , starting: $b_n = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 5,
"answer_id": 4
} |
Definite integral of inverse cosine I want to calculate the definite integral $\Large{\int_0^{\frac \pi 2} \frac{1}{2 + \cos x} \; dx}$
I tried it with the properties of definite integrals but it was of no help.
| Firstly, rewrite integral as
$$\displaystyle \int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{2+\cos{x}}}=\int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{1+\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}+ \cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}}=\int\limits_{0}^{\frac{\pi}{2}}{\dfrac{dx}{1+2\cos^2\dfrac{x}{2}}}.$$
Substitute
\begin{gather... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
the value of :$ ((a-b)(b-c)(c-a))^2$ If the polynomial : $f(x)=x^3-3x+2$ have the roots :$a,b,c$
How to find the value of :$$ ((a-b)(b-c)(c-a))^2$$
| Notice that $f(x)=(x+2)(x-1)(x-1)$. This means the polynomial has three real roots which are $-2$, $1$ and $1$. Therefore you have that $((a-b)(b-c)(c-a))^2=(a-b)^2(b-c)^2(c-a)^2$. Notice that you have a cyclic product of squares. This means the product will be invariant of your choice of $a$, $b$ and $c$. You get $((a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$ Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$
| You labeled this as homework, but I am not sure what methods you have learnt. (Are you in HSGS of Vietnam for example?) I would assume that you know the tangent line method, and will add more explanation if you need it.
Since LHS is homogeneous, we may assume that $a+b+c = 3$. So we need to prove that
$$\frac{(3-2a)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate the limit $\lim_{n\to\infty}\left[n+{n^2}\log{\frac{n}{n+1}}\right]$ Evaluate
$$\lim_{n\to\infty}\left[n+{n^2}\log{\frac{n}{n+1}}\right]$$
I know that this limit is equal to $\frac1{2}$ but I don't know how to do it.
Thanks!
| Recall that $\log(1-x) = -x - \dfrac{x^2}2 + \mathcal{O}(x^3)$. Hence,
\begin{align}
\log \left(\dfrac{n}{n+1}\right) & = \log \left(1-\dfrac1{n+1}\right)\\
& = -\dfrac1{n+1} - \dfrac1{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)^3}\right)\\
n^2\log \left(\dfrac{n}{n+1}\right) & = -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How would I find the area of a triangle given three sides and using either the sine/cosine laws? Triangle ABC has sides $8.5m$ (a), $7.1$ (b), and $9$ (c). I have been asked to find the area of the triangle using trigonometry.
| If you must use Trigonometry, we can use this formula:
$$ K = \frac{1}{2} ab \sin C $$
In order to find the $ \sin C $, we can use the law of cosines:
$$ c^2 = a^2 + b^2 - 2ab \cos C $$
$$ 9^2 = 8.5^2 + 7.1^2 - 2 \cdot 8.5 \cdot 7.1 \cos C $$
$$ -41.66 = -120.7 \cos C $$
$$ 0.34515327257 = \cos C $$
$$ C \approx 69.80^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/283216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find the order of a recurrence relation I have some homework that I'm working on where there is a whole section of problems I need to solve taking the following form:
"Assume that T(1) = 1, and find the order of function T(n)."
I have no idea what this means, really, and I'm having difficulty finding anything sp... | We want to find $T(n)$ given the recurrence $$T(n) = 3T(n/2) + n$$ Let us write $n= 2^m$ and call $g(m) = T(2^m)$. We then have
\begin{align}
g(m) & = 3 g(m-1) + 2^m = 2^m + 3 (2^{m-1} + 3g(m-2)) = 2^m + 3 \cdot 2^{m-1} + 3^2 g(m-2)\\
& = 2^m + 3 \cdot 2^{m-1} + 3^2 \cdot 2^{m-2} + 3^3 g(m-3)\\
& = 2^m + 3 \cdot 2^{m-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/283793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to ... | $$
\begin{align}
\int_0^1\frac{t+1}{t^2+1}\frac{t-1}{\log(t)}\,\mathrm{d}t
&=\int_0^1\frac{t+1}{t^2+1}\int_0^1t^x\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\int_0^1\frac{t^{x+1}+t^x}{t^2+1}\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^1\left(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}+\dots\right)\,\mathrm{d}x\\
&=\left(\log\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 1
} |
Solving lyapunov equation, Matlab has different solution, why? I need to solve the lyapunov equation i.e. $A^TP + PA = -Q$. With $A = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$ and $Q = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$.
Hence...
$$
\begin{bmatrix}
-2 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
P... | from MATLAB documents it actually solves $AP+PA'+Q=0$ so that the result is different than your equation $A'P+PA+Q=0$
to get the same answer in MATLAB use $A'$ instead of $A$ since $A'P+P(A')'+Q=0$ equals
$A'P+PA+Q=0$ which is equivalent to your equation. eventually the following
lyap(A',Q) matches your results
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simple question - Proof How is $\frac{1}{2}ln(2x+2) = \frac{1}{2}ln(x+1) $ ?
As $\frac{1}{2}ln(2x+2)$ = $\frac{1}{2}ln(2(x+1))$, how does this become$ \frac{1}{2}ln(x+1)$?
Initial question was $ \int \frac{1}{2x+2} $
What I done was $ \int \frac{1}{u}du $ with $u=2x+2$ which lead to $\frac{1}{2}ln(2x+2) $ but WolframA... | $\frac{1}{2}\ln(2(x+1))=\ln\sqrt{2}+\frac{1}{2}\ln(x+1)\neq \frac{1}{2}\ln(x+1)$
But as far as $$\int\frac{1}{(2x+2)}dx$$ is concerned , it comes out to be $$\frac{1}{2}\ln(2x+2)+c$$ or you can write it as $$\frac{1}{2}\int\frac{1}{(x+1)}dx$$ which comes out to be $$\frac{1}{2}\ln(x+1)+k$$
Here , the extra $\ln\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $B_{2\times 2}$ such that $A^{51}=B$ Assuming $$ A=\left(\begin{matrix}0 & \sqrt{3}\\\sqrt{3}\ &4\end{matrix}\right)$$ how to find B such that $$A^{51}=B$$
My attempts:
if find $P,D$ that $D$ be diagonal matrix $A=P^{-1} DP$ then $$A^{51}=P^{-1} D^{51}P$$ therefore $B=P^{-1} D^{51}P$ but how to find $D$ an... | Here's another way. The generating function of $A^n$ is $$G(t) = \sum_{n=0}^\infty t^n A^n = (I-tA)^{-1} =
\dfrac{1}{-1+4t+3t^2} \left( \begin {array}{cc} -1+4\,t&-t\sqrt {3}\\ -t
\sqrt {3}&-1\end {array} \right)
$$
Now using a partial fraction decomposition
$$
\eqalign{\frac{1}{-1+4t+3t^2} &={\frac {\sqrt {7}}{14(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Permutations of elements of a matrix I'm a bit confused about the second part of the question I'm working on. The question is as follows
Let A be the $4 \times 4$ matrix
$$A=\begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{... | I believe the following is meant:
The general formula (complete expansion) of the determinant is:
$$\det A = \sum_{\textrm{permutations p}} \left( \textrm{sign p} \right) a_{1, p_1} a_{2, p_2} \textrm{...} a_{n, p_n}.$$
In the case of your first permutation, the term would thus be:
$$\left( -1 \right) a_{1, 4} a_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF int... | It's probably easiest to expand the exponential in the exponent first, since that will lead to a finite number of terms to be evaluated:
$$\begin{align}e^{(e^x-1)} &= \exp\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right) +\frac{1}{2}\left(x+\frac{x^2}{2}+\frac{x^3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Factorization problem Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$
My reasoning:
Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$
It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$
Since: $$mn\gt0 \Rightarrow 16k^... | It is useful to write the stipulation of the problem in a few different ways:
$$\frac{49}{4}=\frac{(m+n)^2-mn}{m+n}=m+n-\frac{mn}{m+n},$$
$$\frac{49}{4}=\frac{(m+n)n+m^2}{m+n}=n+\frac{m^2}{m+n},$$
$$\frac{49}{4}=\frac{(m+n)m+n^2}{m+n}=m+\frac{n^2}{m+n}.$$
From the first, we glean that $m+n>12$ because $49/4=12+1/4$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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Help with $\large \intop\frac{\sqrt{x^2-4}}{x} dx$ I have to $\large \intop\frac{\sqrt{x^2-4}}{x} dx$. Can you tell me what to substitute?
Should I substitute $x$ or $\sqrt{x^2-4}$? Would it be better If I'd substitute $x=\tan u$
or should I substitute $\sqrt{x^2-4}= \tan u$?
| You may try several substitutions, perhaps more elementary and maybe easier:
$$(1)\;\;\text{First substitution :}\;\;u=x^2-4\Longrightarrow du=2xdx\Longrightarrow dx=\frac{du}{2\sqrt{u+4}}\;\;,\;\;\text{so :}$$
$$\int\frac{\sqrt{x^2-4}}{x}dx=\int\frac{\sqrt u}{2(u+4)}du$$
$$(2)\;\;\text{Second substitution :}\;\;\;\;\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Differentiation question? How would I solve the following problem?
Where would the function $|2x-1|$ not be differentiable?
I am thinking it would not be differentiable at $x=1/2$ because there it would be zero.
| Let $f(x) = |2x-1|$. Then, if $x<\frac{1}{2}$, $f(x) = 1-2x$, if $x\geq \frac{1}{2}$, then $f(x) = 2x-1$.
Hence $\lim_{x \downarrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \downarrow \frac{1}{2}}\frac{2x-1}{x-\frac{1}{2}} = +2$, but
$\lim_{x \uparrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To get addition formula of $\tan (x)$ via analytic methods Assume that we only know $\tan (0)=0$ and also given the relation $\tan'(x)=1+\tan^2(x)$ about $\tan (x)$ and we do not know other $\tan (x)$ relations of trigonometry.
How can I get the additon formula $$ \tan (x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}$... | Here is my approach:
Let
$$
\arctan(x) := \int_0^x \frac{dt}{1+t^2}
$$
and $\tan(x)$ is defined as the inverse of $\arctan(x)$, so that $\tan(0) = 0$.
Consider the differential equation
$$
\frac{d x}{1+x^2} + \frac{dy}{1+y^2} = 0 \tag{DE},
$$
one solution is
$$
\arctan x + \arctan y = c,
$$
but the equation has also th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to
\infty} s_n$.
\begin{align}
s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx
\end{align}
I have written a solution of my own, but I would like to ... | We simplify the formulate for $s_n$ by integrating by parts.
\begin{align}
s_n &= \int\limits_0^1 \frac{nx^{n-1}}{1+x} d x \\
&= \left[
\frac{1}{1+x} \int nx^{n-1} d x
- \int \frac{1}{\left(1+x\right)^2} \left(\int nx^{n-1} d x\right) d x
\right]^1_0 \\
&= \left[\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 3
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Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $ How to evaluate the series:
$$ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.
| Recall that, formally,
$$
\left(\sum_{n=1}^{\infty} a_n\right)\left(\sum_{n=1}^{\infty} b_n\right) = \sum_{n=1}^{\infty} c_{n+1},$$
where
$$
c_n = \sum_{k=1}^{n-1} a_k b_{n-k}.
$$
If the series $\sum c_{n+1}$ converges, then the above equality is actually true. You seem to know how to show this, so I'll just demonstra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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"answer_id": 1
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Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$ This is what I've been able to do:
Base case: $n = 1$
$L.H.S: 1^3 = 1$
$R.H.S: (1)^2 = 1$
Therefore it's true for $n = 1$.
I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.
Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + ... | Consider the case where $n = 1$. We have $1^3 = 1^2$. Now suppose $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$. Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Rightmost digit of $ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor $ How could I find $$ 0 \leq a \leq 9 $$ such that
$$ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor \equiv a \mod 10 $$
?
|
This didn't work as a comment to Hagen von Eitzen's answer, so I post it here.
Writing the fraction as
$$
\small\begin{align}
&\frac{10^{20000}}{10^{100}}\frac1{1+3\cdot10^{-100}}\\
&=10^{19900}\left(1-3\cdot10^{-100}+\left(3\cdot10^{-100}\right)^2-\dots
\color{#00A000}{-\left(3\cdot10^{-100}\right)^{199}}\color{#C00... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum Value of Trig Expression What is the general method for finding the maximum and minimum value of a trig expression without the use of a calculator. For example, given the expression :
$$\sin(3x) + 2 \cos(3x) \text{ where } - \infty < x < \infty$$
How would one go about finding the maximum and minimum values ac... | $$f(x) = \sin{(3 x)} + 2 \cos{(3 x)}$$
$$f'(x) = 3 \cos{(3 x)} - 6 \sin{(3 x)} $$
Set $f'(x)$ equal to zero for maxima or minima.
$$f'(x) = 0 \implies 3 \cos{(3 x)} - 6 \sin{(3 x)} = 0 $$
or
$$\tan{(3 x)} = \frac{1}{2} \implies x = \frac{1}{3} \arctan{\left ( \frac{1}{2} \right )} + \frac{k \pi}{3}$$
where $k \in \mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Quadratic Diophantine Equations I note that the Diophantine equation, $x^2 + y^2 = z^2$, with $x, y, z \in \mathbb{N}$, has infinitely many solutions. Indeed, $(x, y, z) = (3,4,5)$ provides a solution, and for any $k \in \mathbb{N}$ : $(kx, ky, kz ) = (3k, 4k, 5k)$ provides a solution.
However, assuming $x, y, z \in \m... | For the special case when the number of is a square and is given us.
That is, in the equation:
$X^2+Y^2=Z^2+q^2$
where the number of $q$ - given us.
Then the solutions of the equation can be written ospolzovavshis Pell: $p^2-2k(k-1)s^2=\pm{q}$
$k$ - given us can be anything.
$X=p^2+2(k-1)ps+2k(k-1)s^2$
$Y=p^2+2kps+2k(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/298053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Inequality problem $(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$ This problem that I have been unable to solve is from the book "Introduction to Inequalities" by Beckenbach and Bellman, chapter 2, page 22, problem 4.
Problem 4. Show that $$(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$$ and $$(a^2+b^2)(a^4+b^4)\ge (a^3+b^3)^2$$ for all $a,b... | If $a = 0$ or $b = 0$, the two sides are equal, and hence the inequality holds. Now, let us assume $a \neq 0$ and $b \neq 0$:
$(a^2 - b^2)(a^4 - b^4) \leq (a^3 - b^3)^2$
iff $a^6 + b^6 - a^2b^2(a^2 + b^2) \leq a^6 + b^6 - 2a^3b^3$
iff $- a^2b^2(a^2 + b^2) \leq - 2a^3b^3$
iff $a^2b^2(a^2 + b^2) \geq 2a^3b^3$
iff $(a^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/298670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How many solutions to prime = $a^3+b^3+c^3 - 3abc$ Let $a,b,c$ be integers.
Let $p$ be a given prime.
How to find the number of solutions to $p = a^3+b^3+c^3 - 3abc$ ?
Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $a^3+b^3+c^3 - 3abc$ below $w$. How does the fu... | Andre's quadratic form $a^2 + b^2 + c^2 - bc - ca - ab$ is only positive semidefinite. It is 0 if $a=b=c.$ Meanwhile,
$$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (b-c)^2,$$
$$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (c-a)^2,$$
$$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$? I am finishing a proof. It seems like I can use $\cos^2 + \sin^2 = 1$ to figure this out, but I just can't see how it works. So I've got two questions.
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$?
And if it does, then how?
| Observe that
$$
\begin{align*}
\sin^2(x)-\cos^2(x)&=\sin^2(x)+\bigl(\cos^2(x)-\cos^2(x)\bigr)-\cos^2(x)\\
&= (\sin^2(x)+\cos^2(x))-2\cos^2(x)\\
&= 1-2\cos^2(x).
\end{align*}
$$
More easily, just subtract $2\cos^2(x)$ from both sides of $\sin^2(x)+\cos^2(x)=1$ to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Solving non-linear congruence $x^2+2x+2\equiv{0}\mod(5)$, $7x\equiv{3}\mod(11)$
My attempt:
$x^2+2x+2\equiv{0}\mod(5)$
$(x+1)^2\equiv-1\mod(5)$, we have $x+1\equiv-1\mod(5)$
since $5$ and $11$ are coprime. We have a solution in $\mathbb{Z}_{11}$
With $[3]$ represent $3$, $[13]$ works for $7x\equiv3\mod11$
so $[3]$ is ... | As you said, $(x+1)² \equiv -1 \equiv 4 \pmod 5$, hence $(x+1)\equiv 2 or -2 \pmod 5$. Now, if $(x+1)\equiv -2 \equiv 3 \pmod 5$, then, since also $x\equiv 2 \pmod {11}$ is a solution to the last congruence, one solution is $x\equiv 2 \pmod {55}$. On the other hand, for $x\equiv 1 \pmod {5}$, we should solve for $x\equ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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contour integration of logarithm I must compute the following integral
$$\displaystyle\int_{0}^{+\infty}\frac{\log x}{1+x^3}dx$$
Can someone suggest me the right circuit in the complex plane over which to do the integration? I tried different paths, avoiding the origin, but unsuccessfully
| Let
$$I = \int_0^{\infty} \dfrac{\log(x)}{1+x^3} dx = \underbrace{\int_0^1 \dfrac{\log(x)}{1+x^3} dx}_J + \underbrace{\int_1^{\infty} \dfrac{\log(x)}{1+x^3} dx}_K$$
$$K = \int_1^{\infty} \dfrac{\log(x)}{1+x^3} dx = \int_1^0 \dfrac{\log(1/x)}{1+1/x^3} \left(-\dfrac{dx}{x^2}\right) = - \int_0^1 \dfrac{x \log(x)}{1+x^3} d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/301573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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Pursuit curves solution For our math class we have to do some calculations with respect to pursuit curves. The chased object starts at point $(p,0)$. Chaser starts at $(0,0)$.(x,y) Speed of the chased object is $u$. Speed chaser = $v$.
We have that for the chaser
$$ \frac{dy}{dx}=\frac{ut-y}{p-x} $$
Then the length of... | You have some wrong signs in the last two equations.
Based on the equation of the derivative of the pursuit curve $y=f(x)$ described by the chaser object that you indicate
$$
\frac{dy}{dx}=\frac{ut-y}{p-x}\tag{1}
$$
I assume that the chased object moves along the straight line $x=p$, as I commented above. Assume furt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to solve equation $ \frac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$? $$ \dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$
How can I solve this equation in the easiest way?
| We have the equation
$$
\frac{1}{2}(\sqrt{x^2-16} + \sqrt{x^2-9}) = 1
$$
Let's multiply it by $\sqrt{x^2-16} - \sqrt{x^2-9}$ to get
$$
-\frac{7}{2}=\sqrt{x^2-16} - \sqrt{x^2-9}
$$
Hence
$$
2\sqrt{x^2-16} = (\sqrt{x^2-16} + \sqrt{x^2-9}) + (\sqrt{x^2-16} - \sqrt{x^2-9})=2-\frac{7}{2}<0
$$
This is imossible so there is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/304144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Polynomial factorization over finite fields How can i factorize the polynomial $x^{12}-1$ as product of irreducibles polynomials over $\mathbb{F}_4$? Anyone can help me?
| Let $\alpha$ and $\beta = \alpha^2 = \alpha+1$ denote the two elements of $\mathbb F_4-\mathbb F_2$. Then, $\alpha$ and $\beta$ are roots of $x^2+x+1$ and
we have that
$$x^3-1 = (x-1)(x^2+x+1) = (x-1)(x-\alpha)(x-\beta).$$
In a field of characteristic $p$, $(a-b)^{p^i} = a^{p^i} - b^{p^i}$ and so we have
that
$$x^{12} ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Is this simple proof about symmetric sums correct? I'm asked to prove
$$x \gt 0, y \gt 0, z \gt 0 \rightarrow$$
$$\left(\frac{x+y}{x+y+z}\right)^\frac{1}{2}+\left(\frac{x+z}{x+y+z}\right)^\frac{1}{2} + \left(\frac{y+z}{x+y+z}\right)^\frac{1}{2} \le 6^\frac{1}{2}$$
I rewrite the summands and say that it is sufficien... | Another approach is to use Lagrange multipliers. With $A$, $B$, and $C$ as you chose, the minimum value of $f(A,B,C) = \sqrt{A} + \sqrt{B} + \sqrt{C}$ subject to the constraint $A+B+C=2$ must occur at a local minimum of $F(A,B,C,\lambda) = \sqrt{A} + \sqrt{B} + \sqrt{C} + \lambda (A+B+C-2)$.
This can only occur if ${\... | {
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"url": "https://math.stackexchange.com/questions/306344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want ... | There's actually a quicker way to do this. As an exercise, prove the following lemma:
Lemma. Suppose that $\alpha$ is an arbitrary permutation and $\beta$ is an $n$-cycle, written $\beta=(\beta_1\hspace{5pt} \beta_2 \hspace{5pt}\ldots \hspace{5pt}\beta_n)$. Then $$\alpha^{-1}\beta \alpha=(\alpha(\beta_1)\hspace{5pt}... | {
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Absolute value inequality - Please guide further
Prove that if the numbers $x$, $y$ are of one sign, then $\left|\frac{x+y}{2}-\sqrt{xy}\right|+\left|\frac{x+y}{2}+\sqrt{xy}\right|=|x|+|y|$.
Expanding the LHS,
$$\left|\frac{x+y}{2}-\sqrt{xy}\right|= \left|\frac{x+y -2\sqrt{xy}}{2}\right| = \frac{(\sqrt{x}-\sqrt{y})^2... | Use AM-GM to conclude $\frac{x+y}{2} \ge \sqrt{xy}$ if $x,y > 0$ and $\frac{x+y}{2} \le -\sqrt{xy}$ if $x,y < 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Least value of $a$ for which at least one solution exists? What is the least value of $a$ for which
$$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a$$
has atleast one solution in the interval $(0,\frac{\pi}{2})$?
I first calculate $f'(x)$ and put it equal to $0$ to find out the critical points.
This gives
$$\sin(x)=\frac{2}{3... | My Solution:: Using the Cauchy-Schwarz inequality:: $\displaystyle \frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}$
and equality holds when $\displaystyle \frac{a}{x} = \frac{b}{y}.$
So here $\displaystyle \frac{2^2}{\sin x}+\frac{1^2}{1-\sin x}\geq \frac{(2+1)^2}{\sin x+1-\sin x}\Rightarrow a\geq 9$
and equality h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
integrating $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ I am doing $\oint_C \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $... | There are 3 poles: $z=1$ and $z=\pm 3 i$.
$$\mathrm{Res}_{z=1} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{5}{10} = \frac{1}{2}$$
$$\mathrm{Res}_{z=3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2-i 81}{(-1+3 i)(6 i)}$$
$$\mathrm{Res}_{z=-3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2+i 81}{(-1-3 i)(-6 i)}$$
The integral value is $i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that ... | Here's a rigorous derivation.
$$
\frac{n}{2} - \sum_{k=1}^n \frac{n^2}{(n+k)^2} = n \int_0^1 \frac{dx}{(1+x)^2} - \sum_{k=1}^n \frac{1}{(1+\frac{k}{n})^2} \\
= \sum_{k=1}^n \left(\int_{\frac{k-1}{n}}^{\frac{k}{n}} \frac{ n\, dx}{(1+x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\
= \sum_{k=1}^n \left(\int_{-\frac{1}{n}}^0... | {
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"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(... | Instead of maximizing $|\,\cdot\,|$ you can equivalently maximize $|\,\cdot\,|^2$, which has the advantage of being smooth. Then you can set
\begin{align*}
f(a,b) &= a^2 - 2ab + b^2\\
g(a,b) &= a^2 + 2ab + 2b^2\\
c &= 7
\end{align*}
and solve the corresponding Lagrange problem
\begin{align*}
(d/da):&& a - b + L*(a + b)... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Finding the affine transformation that will change a given ellipse into the unit circle $x^2$ + $y^2$ =1. We are given that ellipse $E$ is given by $x^2+4y^2-2x+16y+1=0$ and we are asked to find $t_2 \in A(2)$ such that $t_2(E)$ is the unit circle.
| Transformation is - tranlsation + scaling
$$
x^2+4y^2-2x+16y+1=x^2-2x+1+4(y^2+4y+4)-16=(x-1)^2+4(y+2)^2-16=0 \\
\frac{(x-1)^2}{16}+\frac{(y+2)^2}4=1
$$
So if you change coordinates
$$
\xi = \frac{x-1}4 \\
\eta= \frac{y+2}2
$$
your ellipse will become $\xi^2+\eta^2=1$
Update
Transformation itself can be represented in v... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A simultaneous system of equations Solve for $a,b,c$:
\begin{align}
2ab+a+2b=24\\
2bc+b+c=52\\
2ac+2c+a=74\\
\end{align}
Solving them simultaneously is leading to very difficult situation. Plz help.
| Here is a partial answer: User the first equation and third equation (add and substract) to obtain
\begin{align}
c-b&=\frac{25}{a+1} \\
c+b&=\frac{49-a}{a+1}
\end{align}
Adding and subtracting should again give
\begin{align}
c=&\frac{1}{2}\frac{74-a}{a+1} \\
b=&\frac{1}{2}\frac{24-a}{a+1}
\end{align}
Substituting thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \mathcal{O}(\log(n)) $. Prove that $ 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} = \mathcal{O}(\log(n)) $, with induction.
I get the intuition behind this question. Clearly, the given function isn’t even growing at a linear rate, but what i... | I'm expanding the answer by xan:
Define $H_n=\displaystyle\sum_{1\le k\le n} {1\over k}$, let's prove by induction that $H_{2^n}\le n+1$. This is true for $n=0$ since $H_{2^0}=H_1=1\le 1$.
Now suppose $H_{2^n}\le n+1$. We have:
$$\begin{align}
H_{2^{n+1}} &= \sum_{1\le k\le 2^{n+1}} {1\over k} \\
H_{2^{n+1}} &= \sum_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $... | $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$=ab^2-ac^2+bc^2-a^2b+a^2c-cb^2$$
Now the method to factor cyclic expressions is to arrange the expression with the highest powers of the first variable, i.e:
We take powers of $a$ $$a^2c-a^2b+ab^2-ac^2+bc^2-cb^2$$
$$=a^2(c-b)-a(c^2-b^2)+bc(c-b)$$
$$=(c-b)(a^2-ac-ab+bc)$$
Now we lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/314105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma... | To conclude this we actually do the simplification in terms of the $\Gamma$ function. We have
$$\prod_{k=0}^{b-a-1}\frac{1}{2a+2k+1} =
\frac{1}{2^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)}.$$
Furthermore $$\prod_{k=0}^{b-a} (a+k) = \frac{\Gamma(b+1)}{\Gamma(a)}.$$
Finally, $$ \frac{1}{2^{b-a}} {2(b-a) - 1 \choose b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/314856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
general solution of the differential equation $\frac{dy}{dx} + \frac{x+y+a}{x+y+b}=0$ I'm trying to find the general solution of the differential equation $\frac{dy}{dx} + \frac{x+y+a}{x+y+b}=0$ where a and b are constants.
I have tried puttting z=x+y thus, $\frac{dz}{dx} = 1 + \frac{dy}{dx}$. I subbed this into the eq... | Set $x+y+a=u$ so, $1+y'=u'$ and then $~~x+y+b=u-a+b~~$ so $~~~y'=-\frac{u}{u+b-a}$ or $$u'-1=\frac{-u}{u+b-a}$$ or $$u'=\frac{b-a}{u+b-a}$$ or $$(u+b-a)du=(b-a)dx$$ or $$\frac{u^2}{2}+(b-a)u=(b-a)x+C$$ wherein $x+y+a=u$. Now, by substiting $u$ we have: $$\frac{(x+y)^2}{2}+\frac{a^2}2+(x+y)b=(b-a)x+C$$ which is $$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& ... | There is no need to ask whether it is right. You can differentiate to check. A little problem: the arbitrary constant of integration is missing,
It is good to get accustomed to the approach you took, it is useful to master it. Here is an alternative. My guess is that the answer will look like $(A\cos 2x+B\sin 2x)e^x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Change of basis (to a more structured one) in a dynamic system, wrong result? Given a time-invariant homogeneous dynamic system $x(k +1) = Ax(k)$, where the system matrix is:
$$
A = \begin{bmatrix}
1 & -1 \\
2 & 4 \\
\end{bmatrix}
$$
And an initial state vector $x(0) = \begin{bmatrix}1 \... | You've written:
$$z(2) = \begin{bmatrix}4 & 0 \\ 0 & 9\end{bmatrix}\begin{bmatrix}-4 \\ 6\end{bmatrix} = \begin{bmatrix}-8 \\ 27\end{bmatrix}$$
However, it should be:
$$z(2) = \begin{bmatrix}4 & 0 \\ 0 & 9\end{bmatrix}\begin{bmatrix}-4 \\ 6\end{bmatrix} = \begin{bmatrix}-16 \\ 54\end{bmatrix}$$
after which $x(2)$ works... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $(e+x)^{e-x}>(e-x)^{e+x}$ I get stuck with proving that $$(e+x)^{e-x}>(e-x)^{e+x}$$ for $x \in (0, e)$. All I know, is that it is doable with Jensen inequality, and I started with defining $$f(x)=(e+x)^{e-x}$$ and further $$g(x)=\ln \cdot f(x)$$ and... nothing more come to my mind, I kindly ask for any help ... | Using power series,
$$
\begin{align}
(e-x)\log(e+x)
&=e-x+(e-x)\log\left(1+\frac xe\right)\\
&=e-x+(e-x)\left(\frac xe-\frac12\frac{x^2}{e^2}+\frac13\frac{x^3}{e^3}-\dots\right)\\
&=e-x+x-\frac32\frac{x^2}{e}+\frac56\frac{x^3}{e^2}-\dots\\
&=e-\frac32\frac{x^2}{e}+\frac56\frac{x^3}{e^2}-\frac7{12}\frac{x^4}{e^3}+\dots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Rank of matrix products Suppose $A_1$, $A_2$ and $A_3$ are $3$ by $3$ matrices of rank $2$ such that their kernels are linearly independent. Is the following true?
Define: $V_1=A_2A_1$, $V_2=A_3A_2$ and $V_3=A_1A_3$. Then each $V_i$ is rank $1$ and $V_iV_j=0$ for $i\neq j$.
Thanks for your help!
Stan
| If I interpret the question correctly the assumption is wrong, take
\begin{align*}
A_1&= \begin{pmatrix} 0& 1 & 0\\ 0&0 &1\\ 0 & 0 & 0\\ \end{pmatrix} \qquad \operatorname{kernel} \begin{pmatrix} \alpha \\0 \\ 0 \end{pmatrix} \\
A_2&= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 &0 \\ 0 & 0 &0 \end{pmatrix}
\qquad \operatorname... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
minima of $\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}$ Can anyone help me in finding minima of $\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}$ where $k$ and $k'$, are constants. I found the differential but it was too big to be equated.
$\frac{dy}{dx}=\frac{(1-k)\log(x^2-x)[\ln x^2(x-k')x+2(1-k')(x-1)]-(1-k,)\ln x^2... | Wolfram gave an easier derivative.
$$
\dfrac{d}{dx}\left(\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right) = \dfrac{((k-1) ((2 x-1) \log(x^2)-\log((x-1) x) (\log(x^2)+2 x-2)))}{((k'-1) (x-1)^2 \log^2(x^2))}
$$
To see this result visit this link.
Setting the numerator of the derivative equal to $0$,
$$
(k-1) ((2 x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$ My understanding would lead me to believe that:
$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$
But it actually equals: $\sqrt{1 + e^4}$
Can you explain why?
| Your mistake is to say that $\sqrt{4(a+b)}=2(a+b)$, which is not true. For instance $\sqrt{4(3^2+4^2)}=10\not=2(3^2+4^2)$
The correct equalities you need is
$$\sqrt{(c^2a+c^2b)^2}=\sqrt{c^2(a+b)}=\sqrt{c^2}\sqrt{a+b}=c\sqrt{a+b}$$
Now $\frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{2^2(1+e^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ How to prove:$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ ?
| $$\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$$
$$=\sum_{0\le a\le \frac{9-1}2}\{a^n+(9-a)^n\}-3\{1^n+8^n\}-3\cdot6\cdot6^{n-1}\text{ for }n\ge1$$
If $n$ is odd, $9\mid \{a^n+(9-a)^n\}$ and $9\mid \{1^n+8^n\}$
$$\implies 9\mid \{\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\}\text{ if } n \text{ is odd} $$
If $n$ is even, $8^n\equiv(-1)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proof of Wolstenholme's theorem According to the theorem, if
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}$$
then we have to prove that $r\equiv0 \pmod{p^2}$.
(Given $p>3$, otherwise $1+\dfrac{1}{2}=\dfrac{3}{2}$, $3 \not\equiv 0 \pmod 9$.)
I guess there's a $(\bmod p)$ solution... | $ (1 + \frac 1 {p-1}) + (\frac 1 2 + \frac 1 {p-2}) + \ldots + (\frac 1 {(p-1)/2} + \frac 1 {(p+1)/2}) = \frac p {p-1} + \frac p {2(p-2)} + \ldots + \frac p {((p-1)(p+1)/4)}
\\ = p \sum_{k=1}^{(p-1)/2} \frac 1 {k(p-k)} = \frac p 2 \sum_{k=1}^{p-1} \frac 1 {k(p-k)} $.
Now, working on this second sum in the field $\Bbb Z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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need help Jordan base Need help, how to find Jordan base for matrix:
$A=\begin{pmatrix}
-1&-1 &-2 &4 \\
1&-3 &1 &-2 \\
0&0&2&-8\\ 0&0 & 2&-6
\end{pmatrix}$
I found the Minimal polynomial: $(x+2)^3$
and normal Jordan is $Aj=\begin{pmatrix}
-2&0 &0 &0 \\
0&-2 &1 &0 \\
0&0&-2&1\\ 0&0 & 0&-2
\end{pmatrix}$
Then I... | Ok i have as jordan basis
$$\frac{1}{3} \cdot \begin{pmatrix}
2 \\ 2\\ -4 \\ -2 \end{pmatrix}; \qquad \begin{pmatrix} 6 \\ 6\\0 \\ 0\\ \end{pmatrix}; \qquad
\begin{pmatrix} 4 \\ -2 \\ -8 \\ -4 \end{pmatrix}; \qquad \begin{pmatrix} 0 \\ 0\\ 0\\ 1 \end{pmatrix} $$
Got them with that ugly algorithm.
At first we compute $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$ suppose $n\in \Bbb Z $ then how to prove this statement:
$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$
| tnx for Gerry Myerson.
it's equal with this statement:
$x^2-y^2=n$ iff $ n$ is odd or $4|n$
$(x-y)(x+y)=x^2-y^2=n=ab , \exists a,b \in \Bbb Z$
so we have ($x+y=n$,$x-y=1$ or $x-y=n$,$x+y=1$ ) that $n$ must be odd or $x=\frac{a+b}{2}, y=\frac{a-b}{2}$ that so two case exists :
$I)$both $a,b$ is even so $4|ab=n$
$II)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/328195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solution to a linear recurrence What is the general solution to the recurrence: $x(n + 2) = 6x(n + 1) - 9x(n)$ for $n \geq 0$;
with $x(0) = 0; x(1) = 1$?
Solution. The first few values of $x(n)$ are $0,1,6,27,...$ The auxiliary equation for the
recurrence is $r^2-6r+9$ which factors as $(r-3)^2$. Thus, we have repeated... | The best technique for solving recurrences I've seen is given in Wilf's "generatingfunctionology".
Define the ordinary generating function:
$$
X(z) = \sum_{n \ge 0} x_n z^n
$$
From the recurrence, by the properties of ordinary generating functions (see section 2.2 in the cited book):
$$
\begin{align*}
\frac{X(z) - x_0 ... | {
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"answer_id": 0
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Show that $(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$ I'm having a really hard time showing this equality is true, I've tried several ways of going about it and I just can't seem to make it work. Help!
$(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$
Thanks!
| $$\begin{align}\color{darkred}{\bf(n - 1)}\color{blue}{\bf 2^{n+1}} + 2 + \color{darkred}{\bf (n+1)}\color{blue}{\bf 2^{n+1}} & = \color{darkred}{\bf(n - 1 + n+1)}\color{blue}{\bf 2^{n+1}} + 2 \\ \\
& = 2n\cdot2^{n+1} +2 \\ \\
& = n\cdot 2\cdot 2^{n+1} + 2 \\ \\
& = n\cdot 2^{n+2} + 2 \\ \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluate $\sum_{k=1}^{n}(k^2 \cdot (k+1)!)$ We have to evaluate the following:
$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$
Any hints ?
| Using k2=(k+3)(k+2)−5(k+2)+4 we write
fk=k2(k+1)!=(k+3)!−5(k+2)!+4(k+1)!=Gk+1−Gk
where Gk=(k+2)!−4(k+1)!, therefore
∑k=1nfk=∑k=1n(Gk+1−Gk)=Gn+1−G1
Since G1=−2, and Gn+1=(n+3)!−4(n+2)!=(n+2)!(n−1) we get
∑k=1nk2(k+1)!=(n−1)(n+2)!+2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Series expansion $(1-\cos{x})^{-1}$ How do i get the series expansion $(1-\cos{x})^{-1} = \frac{2}{x^2}+\frac{1}{6}+\frac{x^2}{120}+o(x^4)$ ?
| $$
\begin{align}
\frac{1}{1 - \cos x} &= \left(\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{720} + o(x^6)\right)^{-1}\\
& = \frac{2}{x^2}\left(1-\frac{x^2}{12} + \frac{x^4}{360} + o(x^4)\right)^{-1}\\
& = \frac{2}{x^2}\left(1 + \frac{x^2}{12} - \frac{x^4}{360} + \frac{x^4}{144} + o(x^4)\right)\\
& = \frac{2}{x^2} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum up to number $N$ using $1,2$ and $3$ So the question asked was finding out the number of ways(combinations), a given number $N$ can be formed using the sum of $1,2$ or $3$.
(eg)
For $n = 8$, the answer is $10$
The given solution for this is simply
$$\left\lfloor\frac{(N+3)^2}{12}\right\rfloor$$
But I don't understa... | One approach to this sort of problem is generating functions. If $f(n)$ is the answer for $n$, then we have the formula:
$$\sum_{n=0}^\infty f(n)x^n =\frac{1}{1-x}\frac{1}{1-x^2}\frac{1}{1-x^3} = \frac{1}{(1-x)^3(1+x)(1+x+x^2)}$$
This in turns lets us see that there must be $a,b,c,d,e,f$ such that:
$$f(n)=an^2 + bn + c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | Cauchy-Schwarz inequality is not the way to go here. Notice that at $a = b = c =1$ your original inequality becomes exact, whereas your relaxed inequality is no longer satisfied. Moreover, you have lost a factor of 2 there.
Your new edits are very interesting, but still lead to a dead-end: you can check that when $a=b=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 1
} |
Calculate:$y'$ for $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$ and $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$ (1) If $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$
(2) If $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$
then find $y'$ in both cases
(3)If $ y= \sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\sqrt{\... | $y = x^{x^{x^{x^{{.^{.^{.^{}}}}}}}} = x^{y}$
Take the log of both sides.
Then $\ln y = y \ln x$
$ \implies\frac{1}{y} \frac{dy}{dx} = \ln x \frac{dy}{dx} + \frac{y}{x}$
$ \implies \frac{dy}{dx} = \frac{y^{2}}{x(1-y \ln x)}$ and then substitute for $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| \begin{gather*}
|x|<1:\quad f(x)=\sum_{n=1}^{\infty} x^n=\frac{x}{1-x} \\
xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}
\end{gather*}
Let $x=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 1
} |
Complex numbers and trig identities: $\cos(3\theta) + i \sin(3\theta)$ Using the equally rule $a + bi = c + di$ and trigonometric identities how do I make...
$$\cos^3(\theta) - 3\sin^2(\theta)\ \cos(\theta) + 3i\ \sin(\theta)\ \cos^2(\theta) - i\ \sin^3(\theta)=
\cos(3\theta) + i\ \sin(3\theta)$$
Apparently it's easy b... | $(\cos x+i\sin x)^3=(\cos 3x+i\sin 3x)$(by De Moivre's theorem)
But $(\cos x+i\sin x)^3=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i\cos^2 x\sin x-3\cos x\sin^2x=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i(1-\sin^2x) \sin x-3\cos x(1-\cos^2 x)=4\cos ^3x-3\cos x+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\r... | You shouldn't change direction of your inequalities in a chain--for example, $$|2(x+1)-1|\le 2|x+1|+1>1$$ doesn't allow you to conclude that $|2(x+1)-1|>1,$ as transitivity breaks down when you switch directions.
Instead, we can use triangle inequality (why?) to say $$|2(x+1)-1|\ge1-2|x+1|,$$ so whenever $|x+1|<\frac14... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
generating functions / combinatorics Calculate number of solutions of the following equations:
$$ x_1 + x_2 + x_3 + x_4 = 15 $$
where $ 0 \le x_i < i + 4 $
I try to solve it using generating functions/enumerators :
$$ (1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6+x^7)$$
and ta... | The expression you have is
$$
\frac{1-x^5}{1-x}\frac{1-x^6}{1-x}\frac{1-x^7}{1-x}\frac{1-x^8}{1-x}
$$
Then treat this as
$$
(1-x^5)(1-x^6)(1-x^7)(1-x^8)(1-x)^{-4}
$$
The $(1-x)^{-4}$ can be treated by taking derivatives for the geometric series $(1-x)^{-1}$, and you can easily compute the product of the first four te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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.