Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How do you integrate: $\int_0^r\sqrt{x-x^2}.dx$ $$\int_0^r\sqrt{x-x^2}.dx$$
I only have basic calculus and would like to know how would one go about integrating an expression of this form.
I have tried substituting say $u=x-x^2$ but I'm still left with an $x$.
The method is not in my book and I can't find a similar exa... | $$\mathcal{I}(\text{r})=\int_0^\text{r}\sqrt{x-x^2}\space\text{d}x=\int_0^\text{r}\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}\space\text{d}x=$$
Substitute $u=x-\frac{1}{2}$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=0-\frac{1}{2}=-\frac{1}{2}$ and upper bound $u=\text{r}-\frac{1}{2}$:
$$\int_{-\fr... | {
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"timestamp": "2023-03-29T00:00:00",
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Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?
For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\fra... | HINT:
As $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$
$$\implies\dfrac{\tan\dfrac A2+\tan\dfrac B2}{1-\tan\dfrac A2\tan\dfrac B2}=\dfrac1{\tan\dfrac C2}$$
$$\implies\tan\dfrac A2\tan\dfrac B2+\tan\dfrac B2\tan\dfrac C2+\tan\dfrac C2\tan\dfrac A2=1$$
Now $$\left(\tan\dfrac A2-\tan\dfrac B2\right)^2+\... | {
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"url": "https://math.stackexchange.com/questions/1915376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the determinant of the matrix $A^2+B^2$
Let $A$ and $B$ be two $n\times n $ matrices such that $A\neq B$, $A^3=B^3$, and $A^2B=B^2A,$ then what is the value of $\det(A^2+B^2)$ ?
My attempt: $A^3=B^3 \Rightarrow (A-B)(A^2+AB+B^2)=0 \Rightarrow A^2+AB+B^2=0$
but I can't get any idea
| First a couple of notes on your attempt:
$$(A-B)(A^2+AB+B^2) = A^3+A^2B+AB^2-BA^2-BAB-B^3$$
and if $A$ and $B$ do not commute, it doesn't need to be $A^3-B^3$. You have to be careful about factorization in non-commutative rings, familiar formulas do not always work.
Even it were true, as I already said in comments, $$(... | {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying surd of degree 1/4 to 1/2 What will be the simplification of
$$\sqrt { 6 } \cdot \sqrt { 2-\sqrt { 3 } } $$
Thanks in advance!
|
$$\sqrt { 6 } \cdot \sqrt { 2-\sqrt { 3 } } =\sqrt { 3 } \cdot \sqrt { 2 } \sqrt { 2-\sqrt { 3 } } =\sqrt { 3 } \sqrt { 4-2\sqrt { 3 } } =\sqrt { 3 } \sqrt { { \left( \sqrt { 3 } -1 \right) }^{ 2 } } \\=\sqrt { 3 } \left( \sqrt { 3 } -1 \right) = 3-\sqrt { 3 } $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots
The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.
How would I find the minimum value of $a^2 + b^2$ ?
| In the $(a,b)$ plane, the point $(a,b)$ has to be under the parabola $a^2-4b =8$. The point of this domain closest to origin is the vertex of the parabola, $(0,-2)$. Hence the minimum of $a^2+b^2$ is $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a less trickier way to solve this differential equation? Solve the following differential equation:
$$ y^3 dy+(x + y^2)dx = 0$$
Solution:
The following solution uses the substitution $y^2=tx$.
$$y^3\frac{dy}{dx}+x+y^2=0\tag1$$
Differentiating $y^2 = tx$,
$$ 2y\frac {dy}{dx}= t + x\frac{dt}{dx}$$
Now substituti... | $y^3~dy+(x+y^2)~dx=0$
$y^3\dfrac{dy}{dx}=-x-y^2$
Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{y^2}{2}\dfrac{du}{dx}=-x-y^2$
$\dfrac{du}{dx}=-\dfrac{2x}{y^2}-2$
$\dfrac{du}{dx}=-\dfrac{2x}{u}-2$
Luckily this becomes a first-order homogeneous ODE.
| {
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Find the equation of the plane that has distance $1$ from line $X: (1,0,2) + \lambda (1,0,2)$ and contains points $P=(1,1,-1)$ and $Q = (2,1,1)$ I am asked the following problem:
Find the equation of the plane that has distance $1$ from line $X: (1,0,2) + \lambda (1,0,2)$ and contains points $P=(1,1,-1)$ and $Q = (2,1... | ($d=1$ can be taken arbitrarily here. If another value is chosen, the values for $a,b,c$ are scaled by a factor of $d$ and the same plane equations result.)
$$1=\frac{|a+2c+1|}{\sqrt{a^2+b^2+c^2}}\tag1$$
$$a+b-c=-1\tag2$$
$$2a+b+c=-1\tag3$$
$$a+2c=0\tag4$$
Substitute (4) into (1) twice:
$$1=\frac{|0+1|}{\sqrt{(-2c)^2+b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of the all possible values of $n$ such that $5\cdot 3^m+4=n^2$ $5\cdot 3^m+4=n^2$.
Find the sum of all possible values of $n$.
It is an question from prermo 2016 west Bengal exam. I try to do it using theory of congruence. But I can't proceed. I am disappointed, how do I find the sum? Can anybody can help... | Given equation can be re written as
$$5\cdot 3^m = (n-2)(n+2)$$
let $\gcd((n-2),(n+2))=d$ So $d|(n+2)-(n-2)=4$ Thus possible values for $d$ is $1,2,4$. As $d|(5\cdot 3^m), d=1$ as $5\cdot 3^m$ is odd for all values of $m$. This ensure that $5\cdot 3^m$ can only be factored as $5$ and $3^m$ , not as $5\cdot 3^k$ and $ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| if we divide both polynomials we get
$$xa+2\,a+8+1/4\,{\frac {27\,a+3\,b+78}{x-3}}+1/4\,{\frac {a+b-14}{x+1}}$$
and the remainder must be zero. And it must be
$$27a+3b+78=0$$ and $$a+b-14=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
| Splitting into partial fractions can also be done this way:
Let $u = (x+3)(x+5)$ and $v = (x+4)^2$. Then $v-u = 1$ and hence
\begin{align*}
\frac{1}{(x+3)(x+4)^2(x+5)} &= \frac{v-u}{uv} \\
&= \frac{1}{u} - \frac{1}{v} \\
&= \frac{1}{(x+3)(x+5)} - \frac{1}{(x+4)^2}\\
&= \frac{1}{2}\frac{1}{x+3} - \frac{1}{2}\frac{1}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2... | Using the identity
$$
\log(\cos(x))=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2kx)}k\tag{1}
$$
and the evaluations
$$
\begin{align}
\sin(3k\pi/4)-\sin(k\pi/4)
&=2\color{#C00000}{\cos(k\pi/2)}\color{#00A000}{\sin(k\pi/4)}\\
&=\left\{\begin{array}{}
\color{#C00000}{0}&\text{if $k$ is odd}\\
2(-1)^{\frac{k/2+1}2}&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 2
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CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the stor... | Here is another variation of the theme. It is convenient to use the coefficient of operator $[z^r]$ to denote the coefficient of $z^r$ of a series. This way we can write e.g.
\begin{align*}
[z^r](1+z)^t=\binom{t}{r}
\end{align*}
We observe LHS and RHS are polynomials in $p$ with lowest degree $r$ and highest deg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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On fifth powers $x_1^5+x_2^5+\dots = y_1^5+y_2^5+\dots$ There's a nice identity by Vandermergel. If,
$$a^3+b^3 = c^3+d^3\tag1$$
then,
$$(ac)^3+(bc)^3+(d^2)^3=(ad)^3+(bd)^3+(c^2)^3\tag2$$
Here's one by yours truly. If,
$$a^4+b^4 = c^4+d^4\tag3$$
then,
$$(a^2 + d^2)^2 - (a^2 - d^2)^2 + (2 b c)^2 = (b^2 + c^2)^2 - (b^2 - ... | Vandermergel's generalizes to any power.
For any $m$, $$a^m + b^m = c^m + d^m$$ implies
$$ (ac)^m + (bc)^m + (d^2)^m = (ad)^m + (bd)^m + (c^2)^m$$
since it's really just
$$AC + BC + D^2 - AD - BD - C^2 = (A+B-C-D)(C-D)$$
Similarly,
$$A^2+AC-AD-AE-AF-B^2-BC+BD+BE+BF = (A+B+C-D-E-F)(A-B)$$
so that $$a^m + b^m + c^m = d^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $m! (n − m)!$ divides $n!$ for all $m$, $n \in \Bbb{N}$ with $m \leq n$. I am studying Analysis by Amann and Escher by my own I am stuck at this exercise:
Show that $m! (n − m)!$ divides $n!$ for all $m$, $n\in\Bbb{N}$ with $m\leq n$.
(Hint: $(n+1)!=n!(n+1−m)+n!m$.)
Thanks in advance
| Take any prime number $p$ and observe that within
$$1 \cdot 2 \cdot 3 \cdot \ldots \cdot k$$
it is used exactly
$$
\left\lfloor \frac{k}{p} \right\rfloor
+\left\lfloor \frac{k}{p^2} \right\rfloor
+\left\lfloor \frac{k}{p^3} \right\rfloor
+\ldots
$$ times (this series becomes zero after $\log_p k$ steps).
So to prove y... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not.
| Let $$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}$$
$$B=\frac23\cdot\frac45\cdot\frac67\cdot...\frac{2n}{2n+1}$$
Then $$A<B$$
Then $$A^2<AB=\frac1{2n+1}$$
$$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940425",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Does $\rvert z \rvert +z= i$ have any solutions? $$z=a+i \ b$$
$$\vert z \rvert = \sqrt{a^2+b^2} \\$$
$$\sqrt{a^2+b^2}+a+i \ b=i \\$$
\begin{cases}b=1 \\
\sqrt{a^2+b^2}+a=0 \end{cases}
$$\sqrt{a^2+1}+a=0$$
$$\sqrt{a^2+1}=-a$$
$$a^2+1=a^2 \\ \\$$
The equation has no solution
Is it correct?
| with $$z=a+bi$$ you will get the equation
$$\sqrt{a^2+b^2}+a+bi-i=0$$ this is equivalent to
$$\sqrt{a^2+b^2}+a+i(b-1)=0$$ from here we get
$$b=1$$
and $$\sqrt{a^2+b^2}+a=0$$
from the second equation we get
$$\sqrt{a^2+b^2}=-a$$
squaring gives
$$a^2+b^2=a^2$$ substracting $a^2$ gives $$b=0$$
this is a contradiction to ... | {
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"timestamp": "2023-03-29T00:00:00",
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How can we calculate this limit How can i calculate this limit :
$$\lim_{x \rightarrow 1} \frac{\sqrt[3]{x^2}-\sqrt[3]{(x-1)^2}-1}{x-1}$$
I cannot calculate this square root
Please help
| Rewrite the expression as follows:
\begin{equation}
\lim_{x\to 1}\frac{x^{2/3}-1}{x-1} - \lim_{x\to 1} \frac{(x-1)^{2/3}}{x-1} \tag{*}
\end{equation}
The second limit is
$$\lim_{x\to 1^+} \frac{1}{(x-1)^{1/3}} = \infty,$$
$$\lim_{x\to 1^-} \frac{1}{(x-1)^{1/3}} = -\infty.$$
To simplify the first limit in (*), let $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the missing entry in determinant We have,
$\begin{vmatrix}
a^2+2a & 2a+1 & 1 \\
2a+1 & x & 1\\
3 & 3 & 1
\end{vmatrix}=(a-1)^3$ I am asked to prove $x=a+2$ by using properties of determinants.
I've no idea how to solve it. Though tried to add or substract one column or row with others. No combinations had wo... | $$\begin{vmatrix}
a^2+2a & 2a+1 & 1 \\
2a+1 & x & 1\\
3 & 3 & 1
\end{vmatrix}\underbrace{=}_{C_1-3C_3, C_2-3C_3} \begin{vmatrix}
a^2+2a-3 & 2a-2 & 1 \\
2a-2 & x-3 & 1\\
0 & 0 & 1
\end{vmatrix}=\begin{vmatrix}
a^2+2a-3 & 2a-2 \\
2a-2 & x-3\\
\end{vmatrix}$$
$$=(x-3)(a^2+2a-3)-4(a-1)^2=(a-1)^3 \iff x-3=\frac{(a-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopita... | $$ =\lim_{x \to 1} \frac{23 - 23x^{11} -11 + 11x^{23}}{1 - x^{11} - x^{23} + x^{34}} = \lim_{x \to 1} \frac{-23\cdot 11 x^{10} + 11\cdot 23 x^{22}}{-11x^{10} - 23x^{22} + 34x^{33}} =$$$$= \lim_{x \to 1} \frac{-23\cdot 11 \cdot 10 x^9 + 11\cdot 23 \cdot 22x^{21}}{-11\cdot 10 \cdot x^9 - 23\cdot 22x^{21} + 34\cdot 33 x^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 6
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Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numb... | Hint, given
$$n \equiv 0 \pmod{n}$$
$$n+1 \equiv 1 \pmod{n}$$
$$n+2 \equiv 2 \pmod{n}$$
$$n+3 \equiv 3 \pmod{n}$$
$$n+4 \equiv 4 \pmod{n}$$
$$n+5 \equiv 5 \pmod{n}$$
if there exists such a partition, $Q\cap S=\varnothing$ and $Q\cup S=\left \{ 0,1,2,3,4,5 \right \}$
$$\prod_{q \in Q} \left ( n+q \right ) = \prod_{s \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Proving that $\angle B\le60^\circ$
$A$ is the smallest angle of $\triangle ABC$. The height from $A$ over $BC$ is equal to the median from $B$. Prove that $\angle B\le 60^\circ$.
I tried extending the median to construct a parallelogram and expressed the area in terms of $BC$, the height over $BC$ and the median, but... | Note that the median partitions the triangle into two triangles with the same area. Hence if $M$ is the midpoint of $AC$ we have: $$[MBC] = \frac{[ABC]}{2} \implies \frac{BC \cdot BM \cdot \sin \angle MBC}{2} = \frac{BC \cdot h_a}{4} $$
$$\implies \sin \angle MBC = \frac 12 \implies \angle MBC = \frac{\pi}{6} $$
Simila... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove by induction that $\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $2^{n}$ Prove by induction that :
$\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $ 2^{n}$
My proof is :
At $n=1$
$$\frac{\left ( 3+\sqrt{5} \right ... | You're on the right track, but you need to use a slightly extended form of induction for this problem. Note that to prove $P(k+1)$, you don't have to limit yourself to $P(k)$; you can also use $P(i)$ for $i\leq k$. In this case, the proposition $P(k-1)$ will prove helpful for dealing with the last term. (But note th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to determine if vector b is in the span of matrix A? Given a matrix A = \begin{bmatrix}
1 &2 &3 \\
4 &5 &6 \\
7 &8 &9
\end{bmatrix}
Determine if vector $b$ is in $span(A)$
where
$$
b = \begin{bmatrix}
1 \\
2 \\
4
\end{bmatrix}
$$
| First we address $\mathrm{Span}(A)$,
$$\mathrm{Span}(A) = \left\{ \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right] : \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right] = a \left[\begin{array}{c} 1\\ 4\\ 7\\ \end{array}\right] + b \left[\begin{array}{c} 2\\ 5\\ 8\\ \end{array}\right] + c \left[\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate $\lim_{x\to 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$ without Taylor's theorem or L'Hospital rule Calculate this limit without using taylor or hopital
$$\lim_{x\rightarrow 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$$
I have no idea to start the problem please help
| We can proceed as follows
\begin{align}
L &= \lim_{x \to 0^{+}}\dfrac{\dfrac{4}{\pi}\arctan\left(\dfrac{\arctan x}{x}\right) - 1}{x}\notag\\
&= \lim_{x \to 0^{+}}\frac{4}{\pi}\cdot\dfrac{\arctan\left(\dfrac{\arctan x}{x}\right) - \arctan 1}{x}\notag\\
&= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\arctan\left(\frac{\arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Challenging Integral [indefinite] Integrate: $$\int \frac{x^2+n(n-1)}{(x\sin x+n\cos x )^2}dx$$
I've been beating my head around this problem for quite some time now, but I've got nowhere. I'd request the person writing the solution to please explain his thought process because I would like to learn how to appraoch suc... | Using $\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$
$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$ and $\displaystyle \cos \phi = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of the value of a function at a point in a Non-linear initial value problem Given $y''y^3=1,\quad y(0)=1,\quad y'(0)=2$. I have to show that $y(1)=\sqrt{10}$. I tried to substitute $y^2=z$ but did not succeed.
| You have
$$y^{\prime} y^{\prime \prime} y^3= y^{\prime} \iff y^{\prime} y^{\prime \prime}= \frac{y^{\prime}}{y^3} \text{ (for } y^{\prime} \neq 0)$$ hence
$$\frac{1}{2}\left( y^{\prime} \right)^2 - \frac{1}{2}\left( y^{\prime}_0 \right)^2 = \frac{1}{2} \frac{1}{y_0^2} - \frac{1}{2} \frac{1}{y^2}$$ or
$$\left( y^{\prime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Fourier series for $|x|$ not in complex form $\mathbf{f(x)=|x|=\frac{1}{2}a_0+ \sum_{n=1}^{\infty}(a_n \cos{nx}+b_n \sin{nx})}$
for $-\pi <x< \pi$
Since $|x|$ is even $\rightarrow b_n=0$
f(x) =
\begin{cases}
-x, & {-\pi<x<0} \\
x, & {0<x<\pi}
\end{cases}
$a_0=\frac{1}{2\pi}[\int_{-\pi}^0 f(x)$ $dx$ + $\int^{\pi}_0 f(... | Note that
$$1+(-1)^n=\begin{cases}
2 & \text{if $n$ is even, i.e. } n=2k \\
0 & \text{if $n$ is odd, i.e. }n=2k+1
\end{cases}$$
And since $\frac{\cos nx}{n}$ is indeterminate for $n=0$:
$$\sum_{n=\color{red}{1}}^\infty\frac{1+(-1)^n}{n^2}\cos nx=\sum_{k=\color{red}{1}}^\infty\frac{2}{(2k)^2}\cos(2k x)+0=\frac{1}{2}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a closed-form expression for $\binom n1+3\binom n3+5\binom n5+\cdots ,$ where $n > 1$. Find a closed-form expression for
$$\binom n1+3\binom n3+5\binom n5+\cdots ,$$
where $n > 1$. You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful.
I got $2^{n-2}$ but it was wrong! I don't know where I miscal... | Consider
$$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\cdots$$
Differentiating with respect to $x$,
$$n(1+x)^{n-1}=\binom{n}{1}+2x\binom{n}{2}+3x^2\binom{n}{3}+\cdots$$
Put $x=1$,
$$n2^{n-1}=\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots\tag1$$
Put $x=-1$,
$$0=\binom{n}{1}-2\binom{n}{2}+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \bino... | $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$
Prove $$\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$$
Hardy uses this fact without proof in a monograph on different ways to evaluate $\int_0^{\infty}\frac{\sin(x)}{x} dx$.
| Presumably the principal value of the two-sided infinite sum is what was intended in the question.
We'll solve this just using Euler's product formula for the sine function:
\begin{align}
\sin x=x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2 \pi^2}\right).
\end{align}
Compute the logarithmic derivative:
\begin{align}
\cot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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Evaluate $\sum_{n=1}^\infty \frac{1}{(2n+1)^4}$ using Fourier Series I am given $f(x) = x$ for $0 \le x \le 2$. The question wants me to evaluate $\sum_{n=1}^\infty \frac{1}{(2n+1)^4}$ by first evaluating the Fourier sine of $f(x)$ by extending it outside the interval.
I get $f(x) = 2 + \sum_{n=1}^\infty \frac{-4}{\pi... | Using your function, $f(x)=x$ on $[0,2]$ then extended by periodicity (that is, the period is $T=2$), you get Fourier coefficients
$$a_0=\frac1T\int_0^T f(x)\mathrm{d}x=1$$
And for all $n>0$
$$a_n=\frac2T\int_0^T f(x)\cos \frac{2n\pi x}{T}\mathrm{d}x=\int_0^2 x\cos (n\pi x)\mathrm{d}x=\left[x\frac{\sin(n\pi x)}{n\pi}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Probability and Conditional Probability Could someone please provide a worked solution to this question?
A box contains 5 amber, 7 blue and 9 green balls. Six of the balls are removed from the box at random and without replacement.
Find the probability that
(a) three out of the six balls are blue;
(b) four of the ball... | (A) The probability that when selecting $6$ from $21$ balls, you select $3$ from $7$ blue and $3$ from $14$ non-blue is: $$\dfrac{\binom 73\binom {14}3}{\binom {21}6} \tag{$= \dfrac{\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}\cdot \frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1}}{\frac{21\cdot 20\cdot 19\cdot 18\cdot 17\cdot 16}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq\frac{3}{\sqrt2}$$
After substitution $a=\frac{y}{x}$... I tried C-S, but without success.
| Using Hölder's inequality, we have in general:
$$
\left( \frac{a}{\sqrt{X}}+\frac{b}{\sqrt{Y}}+\frac{c}{\sqrt{Z}}\right)^2(aX+bY+cZ) \geq (a+b+c)^3
$$
Substitute $\sqrt{a+b^2}, \sqrt{b+c^2}, \sqrt{c+a^2}$ for $X, Y, Z$:
$$
\left( \frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}} \right)^2 \geq \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Integration involving rational function and exponentials I hope I can find closed form solution for two following definite integrals. Unfortunately I don't have Mathematica and I can't find similar integrals in Tables. Can any one help me please?
\begin{equation}
\int_0^\infty \frac{e^{-x}}{x^{1/2}+a x^{3/2}} dx
\end{... | The second integral is calculated in the same fashion:
\begin{eqnarray}
&&\int\limits_0^\infty \frac{x e^{-x}}{\sqrt{x} (1+b x)^2} dx \underbrace{=}_{z=\sqrt{x}}\\
&&\int\limits_0^\infty \frac{z^2}{(1+b z^2)^2} 2 e^{-z^2} dz=\\
&&\int\limits_0^\infty \left(\frac{1}{b} \frac{1}{1+b z^2} - \frac{1}{b} \frac{1}{1+b z^2)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do I solve differential equation $\frac{d}{dt}x(t)=(2x(t)+8)(t^4+2t^2-t)$? How do I solve differential equation $\frac{dx}{dt}=(2x+8)(t^4+2t^2-t)$?
$\frac{dx}{2x(t)+8}=dt(t^4+2t^2-t)$
$\frac{1}{2}\ln(2x+8)=\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2}$
$e^{\ln(2x+8)}=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}$
... | $\frac{1}{2}\ln(2x+8)=\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2}+C$
I advocate waiting until the end to solve for $C.$ There is a lot of algebra to follow, and it is easier if there is one less thing to think about.
For example, $C$ is an arbitrary constant. Multiply by $2, 2C$ is equally arbitrary, so just leave it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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When does $\int_a^\infty g(x)/f(x)\,dx$ converge?
When does $\int_a^\infty g(x)/f(x) \,dx$ converge?
Where $g$ and $f$ are polynomials.
I am not so sure about this pattern. Consider,
$\int_1^\infty \frac{1}{x^3 + x^2 + x + 1} \, dx$, this integral converges, but $\int_1^\infty \frac{4x^2}{x^3 + x^2 + x + 1} \,dx$ doe... | In the integral $\displaystyle \int_1^\infty \frac{4x^2}{x^3 + x^2 + x + 1} \,dx$ the leading terms in the numerator and denominator are $4x^2$ and $x^3$, so as $x\to\infty$ it goes to $0$ at a rate comparable to that of $4x^2/x^3 = 4/x$, so the integral diverges.
$$
\frac{4x^2}{x^3+x^2+x+1} = \frac 4 {x + 1 + \dfrac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Investigate the convergence of $\sum_{n=1}^{\infty} u_{n}(x)$, where $u_{n}(x)=(-1)^n \frac{x^2+n}{n^2}$ Let $u_{n}(x)=(-1)^n \dfrac{x^2+n}{n^2}$. Show that the series $\sum_{n=1}^{\infty} u_{n}(x)$ converges uniformly on any interval $[a;b]$ but does not converge absolutely at any point $x\in R$.
For the absolute conv... | (Rough Sketch of Proof): Fix an interval $[a, b]$, then there exists $N_1$ such that for all $n>N_1$ we have
\begin{align}
\frac{x^2+n}{n^2} \leq \frac{b^2+a^2+n}{n^2}
\end{align}
which tends to $0$ as $n\rightarrow \infty$. Hence $|u_n(x)|$ tends to $0$ uniformly for all $x \in [a, b]$. Thus, by the alternating series... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the determinant of $I + A$ What is the determinant of an $n \times n$ matrix $B=A+I$, where $I$ is the $n\times n$ identity matrix and $A$ is the $n\times n$ matrix
$$
A=\begin{pmatrix}
a_1 & a_2 & \dots & a_n \\
a_1 & a_2 & \dots & a_n \\
\vdots & \vdots & \ddots & \vdots \\
a_1 & a_2 & \dots & a_n \\
\end{pmatr... | First look at what happens when $n=2$, $n=3$, renaming our matrix $B_n(a_1,\cdots,a_n)$ one gets
$$\begin{align}\det{B_2(a_1,a_2)}&=1+a_1+a_2\\\det{B_3(a_1,a_2,a_3)}&=1+a_1+a_2+a_3\end{align}$$
Assume that $\det{B_{n-1}(a_1,\cdots,a_{n-1})}=1+a_1+\cdots a_{n-1}$ and consider
$$\begin{vmatrix}1+a_1 & a_2 & \cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
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Solving a complex equation with $z=x \pm iy$ I'm trying to solve part of the question below:
The solution is below:
I managed to get to the line $x-y=\pm 2$ , $y=\pm x$ but I can't understand how you can deduce $y=-x$ from these equations. I tried substituting them into each other but got nothing.
| The equation (b) becomes
$$
x^2-y^2+2xyi=(x^2+y^2-4)i
$$
translates into the system
$$
\begin{cases}
x^2-y^2=0 \\[6px]
2xy=x^2+y^2-4
\end{cases}
$$
and both equations should be satisfied. From the second one you get $(x-y)^2=4$, so $x-y\ne0$. The first equation tells us
$$
(x-y)(x+y)=0
$$
so we get $x+y=0$. Now we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
| Notice that $x+x^{-1}=1$ implies that
$$1 = (x+x^{-1})^3 = (x^3+x^{-3}) + 3(x+x^{-1}),$$
and thus
$$x^3+x^{-3} = -2.$$
Therefore,
$$1 = (x+x^{-1})^5 = (x^5+x^{-5}) + 5(x^3+x^{-3}) + 10(x+x^{-1}),$$
giving
$$x^5+x^{-5} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.
Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.
| Subtract $b^3$ from both sides.
$2b^2 + 1 -b^3 = 3^b-b^3$
$\Rightarrow b^2(2-b) = 3^b - (b^3+1)$
Now think about the expression
$3^b - (b^3+1)$.
Both $3^b$ and $b^3+1$ are monotonically increasing. Being an exponential function, $3^b$ grows faster than $b^3+1$ for positive b. These two facts ensure that if at any non... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How would you find the trigonometric roots of a cubic?
Question: How would you find the roots of the cubic$$x^3+x^2-10x-8=0\tag{1}$$
I'm not too sure where to begin. I'm thinking of somehow, implementing $\cos 3\theta=4\cos^3\theta-3\cos\theta$.
I've tried substituting $x$ with $t+t^{-1}$, but didn't get anywhere, a... | 1) Reduce the standard way your equation $x^3+x^2-10x-8=0$ to the form $$x^3+ax+b=0\qquad(*)$$
2) Since you have the identity $$4\cos^3\theta-3\cos \theta-\cos 3\theta=0\qquad(**)$$ make $x=u\cos\theta$ in $(*)$ so you get
$$u^3\cos^3\theta+au\cos\theta+b=0\iff4\cos^3\theta+\frac{4au}{u^3}\cos\theta+\frac{4b}{u^3}\qqua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
| Your problem is of the following form
$$a\sin x+b\cos x = c$$
where $a = 1$, $b = 2\sqrt{2}$ and $c = 3$.
Let $R = \sqrt{a^2 + b^2}$. We can define $$A=\dfrac{a}{R} =\cos\theta$$ and $$B=\dfrac{b}{R} =\sin\theta$$
Therefore
$$\begin{align*}
a\sin x+b\cos x&=R(A\sin x+B\cos x)=R(\cos\theta\sin x+\sin\theta\cos x)=R\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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How to solve this limit problem? $\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )$ I am stuck with the following question from my homework:
$\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )$
Using wolfram alpha gives me 1 for the solution. However, I would li... | $$\lim _{ n\to \infty } \left( \sqrt { n+\sqrt { n } } -\sqrt { n-\sqrt { n } } \right) =\lim _{ n\to \infty } \frac { \left( \sqrt { n+\sqrt { n } } -\sqrt { n-\sqrt { n } } \right) \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) }{ \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find the determinants below using the fact: $\det \left[\begin{smallmatrix}a&b&c\\d&e&f\\g&h&i\end{smallmatrix}\right]=7$ a. \begin{bmatrix}g&h&i\\2d&2e&2f\\3a&3b&3c\end{bmatrix}
b. \begin{bmatrix}a&b&c\\d-2a&e-2b&f-2c\\5g&5h&5i\end{bmatrix}
Hello, I am not sure how to go about answering this question. I don't need and... | Lets try solve a similar determinant:
$$\det\begin{pmatrix}2d & 2e & 2f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} $$
We have that:
\begin{align*}
\det\begin{pmatrix}2d & 2e & 2f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} & =
2\det\begin{pmatrix}d & e & f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Mathemathic induction proof I need to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$
Here is what I tried:
\begin{align}
& \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}+\frac{1}{n+1(n+2)} \\[10pt]
= {} & \frac{n}{n+1}+\f... | $$\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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roots of the equation $(abc^2)x^2+3a^2 cx+b^2cx-6a^2-ab+2b^2=0$ are rational $a,b,c$ are non zero , unequal rational numbers then prove that roots of the equation $$(abc^2)x^2+3a^2 cx+b^2cx-6a^2-ab+2b^2=0$$ are rational
quadratic eqn. in standard form $(abc^2)x^2+(3a^2c+b^2c)x-(6a^2+ab-2b^2) = 0$
$\displaystyle D=(3a^2... | $$B=3a^2c+b^2c=c(3a^2+b^2)$$
$$-C=6a^2+ab-2b^2=6a^2+4ab-3ab-2b^2=2a(3a+2b)-b(3a+2b)=(3a+2b)(2a-b)$$
$$A=abc^2$$
Break $$C\cdot A=(3a+2b)ac\cdot bc(2a-b)$$ as $$(3a+2b)ac-bc(2a-b)=B$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the conditions on $a, b, c$ so that $x^3+ax^2+bx+c$ is bijective? I would like to find the conditions on $a$, $b$, $c$ so that function $$f(x)=x^3+ax^2+bx+c$$ is bijective.
I thought about resolving the equation
$$x^3+ax^2+bx+c=y$$
but I didn't succeed. And our math teacher told us that we cannot prove that a... | Surjectivity is clear, because a third degree equation always has at least a real root.
Suppose $f(x)=f(y)$, with $x\ne y$; then
$$
(x^3-y^3)+a(x^2-y^2)+b(x-y)=0
$$
that becomes
$$
(x-y)(x^2+xy+y^2+a(x+y)+b)=0
$$
and so $x^2+xy+y^2+a(x+y)+b=0$.
Set $s=x+y$ and $p=xy$: then $s^2-4p>0$. We have $s^2+as+b-p=0$, so
$$
s^2-... | {
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"source": "stackexchange",
"question_score": "3",
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How to integrate $\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$ I am asked to solve the following integral:
$$\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$$
I've made a couple of guesses but none of them took me anywhere. One of my ideas was to do, at first, an u-substitution:
$$
u = 1+x^2\\
du = 2x... | Let us integrate by parts and put
$u'=\frac{x}{(1+x^2)^2}$
and
$v=arctan(x).$
thus
the integral becomes
$$I=[-\frac{1}{2(1+x^2)}arctan(x)]_0^\infty$$
$$+\frac{1}{2}\int_0^\infty\frac{dx}{(1+x^2)^2}$$
$$=\frac{1}{2}\int_0^\infty\frac{1+x^2-x^2}{(1+x^2)^2}dx$$
$$=\frac{\pi}{4}+\frac{1}{4}\int_0^\infty x \frac{-2x}{(1+x^2... | {
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"url": "https://math.stackexchange.com/questions/1985109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the expected determinant of a symmetric $2\times2$ matrix, whose three elements are distinct and draw from$ [-n,n]$
The symmetric matrix has three elements:
\begin{pmatrix} A&B\\ B&C \end{pmatrix}
$A,B$ and $C$ are integers draw randomly from $\{-n,-n+1,\dots,n-1,n\}$, $n\ge2$ and they are distinct with ea... | The problem is in the notation. When you write $B$ you actually mean $B$ conditional on $B\neq A=a$, which is not uniformly distributed on $[-n,n]$ but on $[-n,n]\backslash\{a\}$. Similarly, conditional on the values of $A=a,B=b$ the random variable $C$ is uniformly distributed on $[-n,n]\backslash\{a,b\}$. So, actuall... | {
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"source": "stackexchange",
"question_score": "3",
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Can somebody prove this infinite series? Transformation of the Leibniz formula for $\pi$ results in the infinite series:
$$
\frac 1 {1 \times 3} + \frac 1 {5 \times 7} + \frac 1 {9 \times 11} + \frac 1 {13 \times 15} +\cdots = \frac \pi 8
$$
If you recombine the numbers in the denominator you get e.g. the following se... | The same type of method as used to prove that $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \ldots$ can be used here, as soon as you write $\frac1{(2n + 1)(2n + 5)} = \frac1{4}(\frac1{2n + 1} - \frac1{2n + 5})$.
So your series is
$$\frac1{4}\left(1 + \frac1{3} - \frac1{5} - \frac1{7} + \ldots\right)$$
where the expression inside p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sq... | For $n>0$ we have
(1). $ \sqrt {n^2+n}-n=\frac {n}{\sqrt {n^2+n}+n}.$
(2). $ n=\sqrt {n^2}<\sqrt {n^2+n}<\sqrt {n^2+n+\frac {1}{4}}=n+\frac {1}{2}.$
(3). Therefore $ \frac {1}{2}-\frac {1}{8n+2}=\frac {n}{( n+\frac {1}{2}) +n}<\frac {n}{\sqrt {n^2+n} +n}<\frac {n}{n+n}=\frac {1}{2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdo... | Using an Extension of Pascal's Rule
$$
\begin{align}
\sum_{m=0}^n\frac1{2^{2m}}\binom{2m}{m}
&=\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\tag{1a}\\
&=\sum_{m=0}^n\binom{m-\frac12}{m}\tag{1b}\\
&=\sum_{m=0}^n\left[\binom{m+\frac12}{m}-\binom{m-1+\frac12}{m-1}\right]\tag{1c}\\
&=\binom{n+\frac12}{n}\tag{1d}\\[6pt]
&=\frac{(2n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989966",
"timestamp": "2023-03-29T00:00:00",
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Point on surface closest to a plane using Lagrange multipliers Find the point on $z=1-2x^2-y^2$ closest to $2x+3y+z=12$ using Lagrange multipliers.
I recognize $z+2x^2+y^2=1$ as my constraint but am unable to recognize the distance squared I am trying to minimize in terms of 3 variables. May someone help please.
| A simpler approach for computing the distance between these objects: the given plane is orthogonal to the vector $(2,3,1)^T$, so the point(s) on the surface of minimal distance from the plane are the ones for which the tangent plane of the surface at such point(s) is orthogonal su $(2,3,1)^T$. If for some $k$
$$\left\{... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothi... | Hint
You can use the fact that
$$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$
so you can rewrite your inequation
$$\frac{\ln(3)}{\ln(2)}<\frac{\ln(6)}{\ln(3)}$$
if and only if
$$\ln^2(3)<\ln(2\times 3)\ln 2=\ln(2)(\ln(2)+\ln(3)).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to do SVD of the differentiation operator of $P_{2}$ Find the singular values of the differentiation operator $D \colon P_{2} \rightarrow P_{2}$ where $P_{2}$ is the vector space of real polynomials of degree less than or equal to $2$, and the inner product on $P_{2}$ is given by
$$ \left<f,g \right> :=\int_{-1}^{... | You can start with finding an orthonormal basis for $P_2$ by applying Gram-Schmidt to the standard basis $(1,x,x^2)$. Using the fact that the integral of an odd function on $[-1,1]$ vanishes, we get:
$$ e_1 = \frac{1}{\|1\|} = \frac{1}{\sqrt{2}}, \\
e_2 = \frac{x - \left< x, e_1 \right> e_1}{x - \left< x, e_1 \right> e... | {
"language": "en",
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Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
Given, for every $x>1$,
$$f(x)=4\arctan\frac{1}{\sqrt{x-1}+\sqrt{x}}$$
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
I have tried to use the fact that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2
}$
So I obtain: $f(x)=4(\frac{\pi}{2}-\arctan(\sqrt{x-1}+\sqrt{x})$
I am stuck here !
| Observe that
$$\frac1{\sqrt{x-1}+\sqrt x}=\sqrt x-\sqrt{x-1}$$
and thus
$$\left(4\arctan(\sqrt x-\sqrt{x-1})\right)'=4\left(\frac1{2\sqrt x}-\frac1{2\sqrt{x-1}}\right)\cdot\frac1{1+(\sqrt x-\sqrt{x-1})^2}=$$
$$=2\left(\frac1{\sqrt x}-\frac1{\sqrt{x-1}}\right)\frac1{2\sqrt x(\sqrt x-\sqrt{x-1})}=\frac{\sqrt{x-1}-\sqrt x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996649",
"timestamp": "2023-03-29T00:00:00",
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Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$ How one can prove the following.
Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:
$(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$
WLOG one can assume that $0\le a\le b\le c$.
It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le... | $a=0, b=0, c=2 \ \implies \ 1 \cdot 3 \cdot 3 \ge 3 \cdot 2^2$, not true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating limit of $\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$ As the title says we want to calculate:
$$\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$$
By multiplying nominator and denominator in their conjugates
$=\lim_{x\t... | You need one more conjugate-multiplication.
You already have
$$\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$$
First, note that
$$\dfrac{\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4}}{\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x}}=\frac{\sqrt{1+\frac 2x}+2+\sqrt{1-\frac 4... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that exactly half of $1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}$ are congruent to 1 modulo $p$
Let $p$ be an odd prime number. Look at the numbers in the set
\begin{align*}
S \in \{1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}\}
\end{align*}
Show that exactly half... | Recall that each of $1, \dots, p-1$ is a root of $X^{p-1}-1$.
Now as you alluded to $X^{p-1}-1 = (X^{(p-1)/2} -1)(X^{(p-1)/2} +1)$.
This means that for each $a=1, \dots, p-1$:
$$(a^{(p-1)/2} -1)(a^{(p-1)/2} +1)=0$$
thus at least one of the two $(a^{(p-1)/2} -1)$ and $(a^{(p-1)/2} +1)$ is zero, and of course not bo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\s... | Hint:
$$45\pm4\sqrt{41}=(2\pm\sqrt{41})^2.$$
| {
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Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
| The standard way is to make the substitution
$y=x+\sqrt{x^2+2x+2}$
so that $x=\frac{y^2-2}{2(y+1)}$
$\sqrt{x^2+2x+2}=\frac{y^2+2y+2}{2(y+1)}$
$dx=\frac{y^2+2y+2}{2(y+1)^2}dy$
which changes your integral into an integral of rational function, solvable by partial fraction expansion:
$\int\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)... | {
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Limit square roots of polynomials I am trying to find $\lim \limits_{n \to \infty} {\sqrt{n^3+1}-n\sqrt{n} \over \sqrt{n^2+1}-n}$. I rewrite the fraction as
$${(\sqrt{n^3+1}-n\sqrt{n})(\sqrt{n^3+1}+n\sqrt{n}) \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})} = {1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}$$
I notice... | Observe that $${\sqrt{n^3+1}-\sqrt{n^3} \over \sqrt{n^2+1}-n}=\dfrac{n^{3/2}}{n}\left({\sqrt{1+\dfrac{1}{n^3
}}-1 \over \sqrt{1+\dfrac{1}{n^2}}-1}\right)$$ and by multiplying conjugates of both top and bottom yields $${\sqrt{1+\dfrac{1}{n^3
}}-1 \over \sqrt{1+\dfrac{1}{n^2}}-1}=\dfrac{\dfrac1{n^3}}{\dfrac1{n^2}}\left({... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003994",
"timestamp": "2023-03-29T00:00:00",
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Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$ Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$
For this I think I should use De l'Hopital's rule but it takes a lot time and I can't get to answer.
Can we use the De l'Hopital's rule twice?or thr... | The limit above gives: $$\lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}=\lim\limits_{x\to 0}\frac{x^4+10x^3+35x^2+50x}{x^2+5x}$$
Now, it's enough to do the classical polynomial division, so we have: $$\lim\limits_{x\to 0}\frac{x^4+10x^3+35x^2+50x}{x^2+5x}=\lim\limits_{x\to 0}x^2+5x+10=10\hspace{75pt} (*1)$$... | {
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Solve the Diophantine equation $a+2b=2ab$, where $(a,b)$ are positive integers. Let $a$ be a positive integer. Show that $\gcd(a, a-1) = 1.$
Let $d$ be the greatest common divisor of $a$ and $a-1.$ I.e. $\gcd(a, a-1) = d$
Therefore, $d$ must divide $a-(a-1),$ following the rule that if any number, when some number d di... | Note we can rewrite your equation as
$$a + 2b = 2ab \iff a = 2b (a - 1) $$
Both sides of the equation are integers so it follows that $a $ divides $2b(a -1) $. But because of your lemma we know $a $ and $a-1$ are coprime so $a \mid 2b \rightarrow 2b = ak $ for some integer $k $. Substituting we get that
$$a = 2b (a - 1... | {
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Minimizing $|a+bw+cw^2|$ where $a,b,c\in\mathbb Z$ and $w = \zeta_3$.
If $a,b,c\in\mathbb Z$ are not all equal and $w$ is a cube root of unity $(w\neq 1$), then the minimum value of:
$$|a+bw+cw^2|$$
is what?
I'm pretty stuck with the above problem. Could someone help me out?
| Using $|z|^2 = z\bar{z}$
and Using $\displaystyle \bar{\omega} = \omega^2$ and $\bar{\omega^2} = \omega$ and $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$
So $$|a+b\omega+c\omega^2|^2 = (a+b\omega+c\omega^2)\cdot \bar{(a+b\omega+c\bar{\omega^2})} = (a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
So we get $$|a+b\omega+c\omeg... | {
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Prove $27195^8 - 10887^8 + 10152^8\equiv 0\pmod{26460}$ So far I've tried:
1) $26460 = 2^2 * 3^3 * 5 * 7^2$
2) $10152 = 2^3 * 3^3 * 47$
3) $27195 - 10887 = 16308 = 2^2 * 3^3 * 151$ (I know $a^8 - b^8\equiv0\pmod{a - b}$)
Therefore I conclude that $27195^8 - 10887^8 + 10152^8$ is divisible by $2^2 * 3^3$, as well is 1).... | Let $a=27195, b=10887, c=10152$ and consider this:
$$
\matrix{
m &a \bmod m &b \bmod m &c \bmod m \\
2^2 &3 &3 &0 \\
3^3 &6 &6 &0 \\
5\hphantom{^1} &0 &2 &2 \\
7^2 &0 &9 &9 \\
}
$$
Note how for each $m$, we have $b \equiv... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Induction divisibility question Q. Prove by induction that $2^{3n-1} + 5(3^n)$ is divisible by $11$ for any even number $n$, where $n$ is an element of natural numbers.
What is have so far:
(base case): $p(2) = 77$, $77/11 = 7$. so base case holds
$p(k) = 2^{3k-1} +5(3^k) $
$p(k+2) = 2^{3k+5} + 5(3^{k+2}) $
$p(k+2) = 2... | The next step I would take would be to write this in terms of $p(k)$ somehow:
$$p(k+2)=(2^{3k-1}+5(3^k))3^2+(2^6-3^2)2^{3k-1},$$
and see what the remainder is.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the mistake? $$1=1$$
$$\Rightarrow\frac{-1}{1}=\frac{1}{-1}$$
$$\Rightarrow \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$
$$\Rightarrow\frac{i}{1}=\frac{1}{i}$$
$$\Rightarrow\frac{i}{2}=\frac{1}{2i}$$
$$\Rightarrow\frac{i}{2}+\frac{3}{2i} = \frac{1}{2i} +\frac{3}{2i}$$
$$\Rightarrow i(\frac{i}{2}+\frac{3}{2i} ) = ... | $$\sqrt{\frac{a}{b}} \neq \frac{\sqrt{a}}{\sqrt{b}} $$ in general unless both $a$ and $b$ are positive.
| {
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Throwing a dice: specify possible combinations of $(i, j, k) \in$ $\Omega = \{1, 2, 3, 4, 5, 6\}^3$ with $i - j + k = l$
Let $(\Omega, P)$ be a laplacian probability space with $\Omega = \{1, 2, 3, 4, 5, 6\}^3$, and let $X: \Omega \rightarrow \Bbb Z$, $(i, j, k) \rightarrow i - j + k$ be a random variable. Specify $P_... | Generating functions work nicely here.
Let $p=\frac{1}{6}\left(x+x^2+x^3+x^4+x^5+x^6\right)$.
Let $q=\frac{1}{6}\left(x^{-1}+x^{-2}+x^{-3}+x^{-4}+x^{-5}+x^{-6}\right)=\frac{p}{x^7}$.
Then the distribution you seek is given by $p^2q = \frac{p^3}{x^7}$:
\begin{align*}
p^2q =&\frac{1}{216}\left(x^{11}
+ 3 x^{10}
+ 6 x^{... | {
"language": "en",
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"answer_id": 0
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Prove that $a+b\ge6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$ if equation $2x^3+ax^2+bx+4=0$ has $3$ real roots
Peoblem Staement:-
If equation $2x^3+ax^2+bx+4=0$ has $3$ real roots ($a,b\gt0$), then prove that $a+b\ge6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$.
Attempt at a solution:-
Let the roots of the equation $2x^3+ax^2+bx+4=... | Since $a, b > 0$, the polynomial is strictly positive for any nonnegative $x$. But this means that all $\alpha_i$ are negative. Now for example
$$2^{1/3} = (|\alpha_1| |\alpha_2| |\alpha_3|)^{1/3} \le \frac{1}{3}(|\alpha_1| + |\alpha_2| + |\alpha_3|) = -\frac{1}{3}(\alpha_1 + \alpha_2 + \alpha_3) = \frac{a}{2}.$$
Note ... | {
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"answer_count": 1,
"answer_id": 0
} |
Mac-Laurin-series for $\psi(\frac{2x+1}{2x})-\psi(\frac{x+1}{2x})$? Consider $$f(x)=\psi(\frac{2x+1}{2x})-\psi(\frac{x+1}{2x})$$
$\psi(x)$ is the digamma-function. This function occurs in the calculation of the definite integral
$$\int_0^1 \ln(x^n+1)dx=\ln(2)-\frac{f(n)}{2}$$ for $n>0$. Wolfram alpha gives a series exp... | Let's take a look at the chapter about polygamma functions in Abramowitz and Stegun.
The multiplication formula (6.4.8) shows:
$$\begin{align*}
\psi(2x) &= \ln(2) + \frac{1}{2}(\psi(x) + \psi(x+1/2)) \\
\iff \psi(x + 1/2) &= 2\psi(2x) - \ln(4) - \psi(x)
\end{align*}$$
The recurrence formula (6.4.6) shows:
$$\psi(1 + x)... | {
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"answer_count": 2,
"answer_id": 0
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factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, a... | "How are $(x+2)(x^2−2x+4)(x−2)(x^2+2x+4)$ and $(x+2)(x−2)(x^4+4x^2+16)$ equivalent?"
Because $(x^2−2x+4)(x^2+2x+4)=x^4+4x^2+16$
===
Nother way of looking at it.
$a^3-b^3=(a-b)(a^2+ab+b^2) $
$a^2-b^2=(a-b)(a+b) $
So therefore $a^6-b^6=(a^2-b^2)(a^4+a^2b^2+b^4) =(a-b)(a+b) (a^4+a^2b^2+b^4)$
Yet, $a^6-b^6=(a^3-b^3)(a^3+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Estimates for solution of Laplace equation The function
$\Phi(x):= \left\{\begin{array}{lr}
- \frac{1}{2\pi} log(|x|), & \text{for } n=2\\
\frac{1}{n*(n-2)\alpha(n)} \frac{1}{|x|^{n-2}}, & \text{for } n \ge 3
\end{array}\right\}
$ defines for $x \in \mathbb{R}^n, x \not=0$ is the fundamental solution ... | You did the calculation wrong for $n\geq 3$. We have
$$\nabla |x|^{2-n} = (2-n)|x|^{2-n-1}\frac{x}{|x|} = (2-n)\frac{x}{|x|^{n}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it true that $\frac1n<\sum\limits_{j=n}^{m}\frac{1}{j^2}<\frac1n+\frac1{m^2}$ for $n\geq2$ and $m=\lceil 2n^2-\frac{2}{3}n \rceil$?
Let $n\geq 2$ and $m=\lceil 2n^2-\frac{2}{3}n \rceil$. Is it true
that
$$
\sum_{j=n}^{m}\frac{1}{j^2} > \frac{1}{n} > \sum_{j=n}^{m-1}\frac{1}{j^2}\ ?
$$
I have checked this for $... | Using the Euler-Maclaurin Sum Formula, we get that
$$
\begin{align}
f(n)
&=\sum_{k=n}^\infty\frac1{k^2}\\
&=\frac1n+\frac1{2n^2}+\frac1{6n^3}-\frac1{30n^5}+\frac1{42n^7}-\frac1{30n^9}+O\left(\frac1{n^{11}}\right)
\end{align}
$$
Then
$$
\begin{align}
\sum_{k=n}^{\left\lceil2n^2-\frac23n\right\rceil}\frac1{k^2}
&=f(n)-f\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the equation of an ellipse using eccentricity and directrix with focus at (0,0) The ellipse $\varepsilon$ has eccentricty $\frac{1}{2}$, focus $(0,0)$ and the line $x=-1$ as the corresponding directrix. Find the equation of $\varepsilon$. Find the other focus and directrix of $\varepsilon$.
I'm confused by this... | From your information:
\begin{align*}
\frac{\sqrt{x^2+y^2}}{x+1} &= \varepsilon \\
x^2+y^2 &= \varepsilon^2(x+1)^2 \\
(1-\varepsilon^2)x^2-2\varepsilon^2 x+y^2 &= \varepsilon^2 \\
\left[
(1-\varepsilon^2)x^2-2\varepsilon^2 x+
\frac{\varepsilon^4}{1-\varepsilon^2}
\right]+
y^2 &= \varepsilon^2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Gauss-Jordan Elimination Trying to solve this using Gauss-Jordan Elimination.
$x_1 + x_2 - x_3 = -4$
$-x_1 - 2x_2 + x_3 = 3+t$
$2x_1 + x_2 + (s-3)x_3 = st-9$
$2x_1 + (s-3)x_3 = st+2t-10$
I came across:
$t=0$
And
$s\neq 1$
Need to find the values of s and t that the equations will have: no solution, 1 solution, ... | From your final system, $0=-t$, hence if $t \neq 0$, there is no solution.
If $t=0$, we can drop the very last equation as it is just $0=0$.
The $3 \times 3 $ matrix on the LHS is non-singular, hence the system has a unique solution.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a,b,c$ be the roots of $x^3 - x - 1= 0$ find $a^5 + b^5 + c^5$ Consider a cubic polynomial $x^3 - x - 1 = 0$ I want the sum of the fifth powers of the roots $\sum a^5$. I know that
\begin{eqnarray*}
a + b + c &=& 0 \\
ab + bc + ca &=& 1 \\
-abc &=& 1
\end{eqnarray*}
but I have no way of combining this information... | $a^3=a+1$ implies that $a^5=a^3+a^2=a+1+a^2$.
$a^5+b^5+c^5=3+a+b+c+a^2+b^2+c^2=3+(a+b+c)^2-2(ab+ac+bc)=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{8^{n}}= \sqrt{2}$ using $f(x) = \frac{1}{\sqrt{1-4x}}$ Prove that $\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{8^{n}}= \sqrt{2}$ using $f(x) = \frac{1}{\sqrt{1-4x}}$
For the past 2 days I've tried to prove this but with no results.
I've done some research but with no re... | Us the extended binomial theorem to show that:
$$(1-x)^{-1/2} = 1 +\frac{1}{1!}\frac{1}{2}x+\frac{1}{2!}\frac{1}{2}\cdot\frac{3}{2}x^2+\frac{1}{3!}\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}x^3\dots$$
And then prove (by induction or just directly) that:
$$\frac{1}{n!}\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}...\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024655",
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"source": "stackexchange",
"question_score": "1",
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Cubics with one real root
$(1) \quad f(x) = x^3- 3ax + b$
$(4) \quad f(x) = x^3 - 3hkx - (h^3 + k^3)$
By comparing the coefficients in equations $(1)$ and $(4)$, obtain two equations that relate $h$ and $k$ to $a$ and $b$.
One of these is solved to give $k$ in terms of $a$ and $h$.
Substitute this into the other equa... | Obviously,
$$a=hk,\\b=-h^3-k^3.$$
Then
$$k=\frac ah,\\b=-h^3-\frac{a^3}{h^3}=-t-\frac{a^3}t$$
or
$$t^2+bt+a^3=0.$$
From the Vieta formulas, the sum of the roots of this quadratic equation is $-b$, and from the definition,
$$t_1+t_2=-b=h^3+k^3.$$ So if $t_1=h^3$, then $t_2=k^3$.
Now $x=h+k$ is a root of the cubic becau... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 ... | The pattern is
$$
\frac{1}{x^2+2x+2}=\frac12+\sum_{k\in \Bbb{Z}^+ | k-3\not\in \Bbb{Z}} (-1)^\left\lfloor \frac34 k\right\rfloor2^{-\left\lfloor \frac{k}2+1\right\rfloor}
$$
where $\lfloor s \rfloor$ means the greatest integer not exceeding $s$, and the sum on $k$ skips numbers of the form $4n+3$.
Thus
$$
\frac12 - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Arithmetic proof of absolute value function of complex numbers I am looking for the arithmetic proof that:
$ |z| = \sqrt{(x^2 + y^2)} $ where $ z = x + i y $
Previously I assumed squaring a function then square rooting it would be analogous to the absolute value function (modulus) but it seems not to be the case i... | First,
the absolute value
of a complex number
is a real number.
The standard definition is
$|z|^2
=z\, \bar{z}
$.
From this,
if $z = x+iy$,
$|z|^2
=z\, \bar{z}
=(x+iy)(x-iy)
=x^2+y^2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What will be the 49th derivative? Let be the function $f=(x^3+3x) \cdot sin(x)$
$f'(x)=\left(3x^2+3\right)\sin\left(x\right)+\left(x^3+3x\right)\cos\left(x\right)$
$f''(x)=\left(3x-x^3\right)\sin\left(x\right)+\left(6x^2+6\right)\cos\left(x\right)$
$f'''(x)=\left(-9x^2-3\right)\sin\left(x\right)+\left(15x-x^3\right)\co... | Going off A.Riesen's comment, the General Leibniz Rule states that
$$(uv)^{(n)} = \sum_{k=0}^n {n \choose k} u^{(k)} v^{(n-k)} %seriously, did nobody notice that I had written down the formula incorrectly the first time?$$
where $u,v$ are functions of $x$ and $f^{(n)}$ is the $n^{th}$ derivative of $f$ with respect to ... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find $\int_{|z|=3} {1 \over P(z)}dz$ where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$
Evaluate
$$\int_{|z|=3} {1 \over P(z)}dz$$
where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$.
I proved that $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, except for z=9, so all the poles of ${1 \over P(z)}$ except $z=9$ are i... | Let
$$
P(z)=2z^5-16z^4-17z^3-11z^2+20z-18
$$
Since
$$
\begin{align}
&(-218840155-162129667x-67092610x^2+13150850x^3)P(z)\\
&-(3927717-68602966x-46831817x^2-17626794x^3+2630170x^4)P'(z)\\
&=3860568450
\end{align}
$$
$\gcd(P,P')=1$. Therefore, $P$ has no repeated roots.
$P(9)=0$ and $\frac{P(x)}{x-9}=2x^4+2x^3+x^2-2x+2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031768",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Given $a^2+b^2=2$ prove $a+b\le2$
*
*Given $a^2+b^2=2$ prove $a+b\le2$
*Given $a+b=2$ prove $a^4+b^4\ge2$
I was trying to prove these using the fact that we know $a^2+b^2\ge2ab$ but not sure where to start.
| Using AM-GM inequality:
$\displaystyle\frac{a^2+b^2}{2} \geq \sqrt {a^2b^2}$
$\displaystyle{a^2+b^2}\geq {2ab}$
We know :
$\displaystyle\ a^2+b^2=2$
Equating the above equations, we get
$\displaystyle\ 2ab \leq 2$
Adding two equations above we get:
$\displaystyle\ a^2+b^2+2ab \leq 4$
Taking Square Roots both the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Suppose we have two positive integers $a$ and $b$ which satisfy the condition $a^3 − 2b^3 = 2$. Suppose we have two positive integers $a$ and $b$ which satisfy the condition
$a^3 − 2b^3 = 2$.
It then follows that the greatest common divisor of $a$ and $b$ must be either $1$ or $2$. True or false?
I tried solving by su... | Let be $d=gcd(a,b)$ so $a=d.a_1$ and $b=d.b_1$, with $gcd(a_1,b_1)=1$ and backing to the equation we have:
$$a^3-2b^3=2 \Rightarrow d^3(a_1^3-2b_1^3)=2 \Rightarrow d^3|2 \Rightarrow d=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$ I am checking for convergence of
$$\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$$
First, we notice
$${n+1 \over n-1}= {1+{2 \over n-1}}$$
Then, we use $\ln(1+x) \le x$ and get
$$\ln\left({1 + {2 \over n-1}}\right) \le {2 \over n-1}$$
$$... | If we use the well-known equivalence,
$$\ln(1+X)\sim X\;\;(X\to 0),$$
we get that
$$\ln(1+\frac{2}{n-1})\sim\frac{2}{n-1}\sim\frac{2}{n}\;\;(n\to +\infty)$$
$$\implies \frac{1}{\sqrt{n}}\ln(\frac{n+1}{n-1})\sim\frac{2}{n^\frac{3}{2}}\;(n\to+\infty)$$
the general terms are positive and equivalent, the series are both co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Error computing determinant of a $4\times 4$ matrix I am trying to compute the determinant of
$$C = \begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ -2 & -2 & -1 & 2 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
I first did the row operation $R_3 \leftarrow R_1-R_3$ so it doesn't change the determinant. So you want the determina... | Recall that the determinant does not change if you add to a row a linear combination of the others.
You should replace $R_3$ with $R_3-R_1$. Then you obtain
$$\begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ 0 & -5 & -3 & 1 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
and the determinant of
$$\begin{bmatrix} 3 & 1 & -3 \\ -5 ... | {
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"source": "stackexchange",
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Evaluate the sum $\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}$ Evaluate the sum $\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}$
I don't to know how to processed this problem. Can any one help with problem please. thanks
| Complex analysis gives a very clear explanation of the identity
$$ \sum_{n=1}^{\infty} \frac{1}{n^2+z^2} = \frac{1}{z}\bigg( \frac{\pi}{2}\coth(\pi z) - \frac{1}{2z} \bigg) $$
where $z = a^2$. This equality follows from the observation that both sides have exactly the same poles and decay to $0$ as $\Re(z) \to \pm \inf... | {
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"url": "https://math.stackexchange.com/questions/2042931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show a sequence is increasing
How can you show that :
$$D_n=\sum_{k=1}^{n } \frac{1}{k}-\int_{1}^{n+1} \frac{1}{x} \ dx $$ is increasing and bounded (and hence convergent). I'm having trouble.
| Assuming you meant
$$D_n=\sum_{k=1}^n \frac{1}{k}-\int_1^{n+1}\frac{1}{x}\ dx$$
we have
\begin{align*}
D_{n+1}-D_n &= \left(\sum_{k=1}^{n+1} \frac{1}{k}-\int_1^{n+2}\frac{1}{x}\ dx\right)-\left(\sum_{k=1}^n \frac{1}{k}-\int_1^{n+1}\frac{1}{x}\ dx\right)\\
&= \frac{1}{n+1}-\int_1^{n+2}\frac{1}{x} \ dx+\int_1^{n+1}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+... | So it doesn't explicitly use generating functions as I supposed, but it is pretty close.
You have (for ease of visualization):
$$\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)^2=\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)=\sum_{n=0}^{\infty}\left[\f... | {
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"url": "https://math.stackexchange.com/questions/2044953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove that $\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-0.0064$ (Motivation) As homework, we have been asked to prove that the following series converges: $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}$$
I did it in two ways:
*
*Using the alternating series test (Leibniz criterion), proving that $\frac{n^3}{4^n}$ is ... | You can get this by clever sum manipulations.
Start from
$$
\sum \frac1{(-4)^n}=-\frac14+\frac1{16}-\frac1{64} + \cdots = -\frac15
$$
Then look at
$$
\sum \frac{n}{(-4)^n}=\begin{array}{lllll}
-\frac14&+\frac1{16}&-\frac1{64} &+ \cdots \\
&+\frac1{16}&-\frac1{64}&- \cdots \\
&&-\frac1{64}&+ \cdots
\end{array}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Need help evaluating $ \int_{0}^{2\pi} \frac{d\theta}{5 + 4\sin{\theta}} $. From Churchill and Brown's Complex Analysis I am trying to conclude:
$$ \int_{0}^{2\pi} \frac{d\theta}{5 + 4\sin{\theta}} = \frac{2\pi}{3}$$
I have tried substituting in:
$$\sin{\theta} = \frac{z + z^{-1}}{2i}$$
$$d\theta = \frac{dz}{zi}$$
To g... | Note that
$$\left(5+4\left(\frac{z-z^{-1}}{2i}\right)\right)iz=2z^2+i5z-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find general sequence equation linear algebra Suppose that the sequence $x_0, x_1, x_2,\dots$ is defined by $x_0 = 7$, $x_1 = 2$, and $x_{k+2} = −x_{k+1}+2x_k$ for $k\geq0$. Find a general formula for $x_k$. Be sure to include parentheses where necessary, e.g. to distinguish $1/(2k)$ from $1/2k$.
I have no idea how t... | Just to show an alternative way, through generating function.
Allow me to change notation from $x$ to $a$ to avoid confusion in the following.
Starting from your recurrence
$$
\left\{ \begin{gathered}
a_{\,0} = 7 \hfill \\
a_{\,1} = 2 \hfill \\
a_{\,k + 2} = - a_{\,k + 1} + 2a_{\,k} \hfill \\
\end{gathered... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
I have tried this
$$f(x)=x^2\ln(x+1)$$
$$ f'(x) = 2x\ln(x+1) + \frac{x^2}{x+1}$$
$$ f''(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
However I do not see any pattern :(
| Let $u=x^2$ and $v=\ln (1+x)$ then,
$$u'=2x,u''=2,u'''=0$$
And
$v^{(n)}=\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}$ for $n \geq 1$.
Let's compute $(uv)^{(3)}=\sum_{k=0}^{3} {3 \choose k} u^{(k)}v^{(n-k)}={3 \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{3 \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{3 \choose 2}2\frac{(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I find all prime solutions $p, q, r$ of the equation $\displaystyle p(p+1)+q(q+1) = r(r+1)$?
Find primes $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$
I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.
Thank you for any help
| May this lead to a simple proof for your problem according to your unic example of solution . There is only one
solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
$n$ is a positive integer. Our equation yields
$p(p+1) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Solution of Logarithmic Inequalities: $\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1$
If $\log_{0.5}(\log_5 (x^2-4)) >\log_{0.5}1$ then x lies in the interval:
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3)$
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3\sqrt{5})$
(c) $(\sqrt{5}, 3\sqrt{5})$
(d) $\phi $
I have solved quite a few lo... | From the inequality we get
$x^2-4>1$
Or$x^2>5$
Which provide one condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $g(u)= \frac{E [ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} ] }{E [ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} ]}$ decreasing in $u$ Let $X$ be a positive random variable, let us define a function
\begin{align}
g(u,a)= \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\fra... | This is false in general.
Change the variables as in my comment: $1/(2X) \to X$, $a^2 \to a$, $u^2 \to u$. Then the problem is to show that
$$
f(u) = \frac{\mathsf{E}[\sqrt{X}e^{-auX}]}{\mathsf{E}[\sqrt{X}e^{-uX}]}.
$$
decreases.
Set $a=1.1$ and let $X=1$ or $100$ with probability $1/2$. Then
$$
f(u) = \frac{e^{-1.1u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
what is the value of $ x$? If $\log_{2}3^4\cdot\log_{3}4^5\cdot\log_{4}5^6\cdot....\log_{63} {64}^{65}=x!$, what is the value of $ x$?
I've tried
$$\log_2 3^4\cdot\log_3 4^5\cdot\log_4 5^6\cdot.... \log_{63} 64^{65}$$
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log ... | \begin{align}\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}&=\frac{65!}{3!}\frac{\log 64}{\log 2}
\\&=\frac{65!}{3!}\frac{\log 2^6}{\log 2} \\
&=\frac{65!}{3!}\frac{\log 2}{\log 2}.6
\\&=65!
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Strange limit problem to be solved without Hospital's Rule...? Having trouble solving this limit problem without L'Hôpital's Rule...
$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
Tried multiplying the function by the conjugate/inverse-conjugate, of both the numerator and denominator... but no avail.... any ide... | $$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)(\sqrt{3-x}+1)}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$$
$$=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}{(2-x)(\sqrt{6-x}+2)}=\lim_{x \to 2} \frac{(\sqrt{3-x}+1)(2-x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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