Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{... | These numbers may look pretty complicated but notice that if we square both sides a lot can be simplified.
$$ z^2 = \big((2 - \sqrt{3}) - (2 + \sqrt{3})\big) - 2\sqrt{(2 - \sqrt{3})(2 + \sqrt{3})}\,i
= -2\sqrt{3} - 2i = 4 \left(\cos \left(\frac{7\pi}{6}\right) + i\sin \left(\frac{7\pi}{6}\right) \right) $$
So now we kn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\b... | :( I have not enough reputation to even add a comment...
Yes, the cot<-->tan problem.
Comment: it should be equal to $e^{-\frac{1}{6}}$, if the cot<-->tan was not there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Using integrating factor to find a series solution for an ODE Consider the ODE
$$xy''-y=0$$
Find the solution $y_1$ in the form of a power series, and use an integrating factor in $$y_2=y_1\int\frac{exp(-\int P(x)dx)}{y_1^2}dx$$
to determine $y_2$
To find $y_1$, I put the ODE in standard form $y''-\frac{0y'}{x}-\frac{x... | You can factor out the $x^2$ term from the denominator, as you noticed, to get:
$$Q(x) = \frac{e^{-x}}{x^2 \, (1+P(x)) } \approx \\ \frac{1}{x^2} (1-x+x^2/2 - x^3/6 + x^4/24 + \ldots) \, (1-P+P^2-P^3+ \ldots ), $$
where $P(x) = x+11 x^2/12 + \ldots $ and use has been made of the Taylor expansion of both $e^{-x}$ and $1... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$... | Hint: $1+\cos \theta = 2\cos^2\frac{\theta}{2}$, and use:
$\cos \frac{\theta}{2} = \dfrac{\sin \theta}{2\sin \frac{\theta}{2}}$, or
try to create a quartic equation and use Viete's formula for the product of zeroes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Find Jordan basis of a given matrix $A=\begin{bmatrix}
-1 & 4 & -4 \\
0 & 1 & 0 \\
1 & -2 & 3
\end{bmatrix}$
The eigenvalues of the matrix are all $1$. The dimension of it's eigenspace is 2 so the Jordan normal form of the matrix is \begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
this is all confirmed... | The characteristic polynomial is given by
$$\begin{align} p (\lambda) &= (-1-\lambda)(1 - \lambda)(3 - \lambda) - (-4)(1- \lambda)(1) \\&= (1 - \lambda)[-(-1-\lambda)(3-\lambda) - 4]\\&= -(\lambda - 1)^3\end{align}$$
Which gives you eingenvalue $\lambda = 1$ of multiplicity $3$.
To find the eigenvectors, solve
$$(A - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+... | Up to $$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$$ your work was fine but, immediately after, you have some mistakes since, developing, you are supposed to get $$(4 x^4+4 x^3+6 x^2+6 x)+(6 x^4+12 x^3-45 x^2+6 x-24)=10 x^4+16 x^3-39 x^2+12 x-24$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me ... | You can write $\displaystyle\frac{x}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$; then multiply by $(x-1)^2$ to get $x=A(x-1)+B$.
Letting $x=1$ gives $B=1$, and comparing the coefficients of $x$ gives $A=1$.
(Notice that $\displaystyle\int\frac{1}{(x-1)^2}=-\frac{1}{x-1}+C$.)
You could also find $\displaystyle\int\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expressio... | Note that
\begin{align}
f(k)&=(a+k)(b+k)(c+k)(d+k) \\
&=k^4+(a+b+c+d)k^3+(a b+a c+a d+b c+b d+c d) k^2
+(a b c+a b d+a c d+b c d) k+a b cd \\
&=k^4 + k^3 e_1 + k^2 e_2 + k e_3 + e_4.
\end{align}
Now you have a linear system of 4 equations and 4 unknowns ($e_1,e_2,e_3,e_4$).
Solve it and find $f(5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 1
} |
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int... | Another aproach to the integral : substitution $x=t^2$
$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \int\frac{2t^2 \mathrm{d}t}{\sqrt{t^2+1}} $$
By per partes :
$$I = \int\frac{t^2 \mathrm{d}t}{\sqrt{t^2+1}} = t\sqrt{t^2+1}-\int\sqrt{t^2+1}\;\mathrm{d}t = t\sqrt{t^2+1}-\int\frac{t^2+1}{\sqrt{t^2+1}}\;\mathrm{d}t $$
Ergo ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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In which interval is the value of $\sum\limits_{k=1}^\infty \frac{1}{{\left(k + 3\right)}^{2}} $ The question is
But none of these possible answers makes any sense to me.
I know that $\displaystyle\sum_{k=1}^\infty \left( \displaystyle\frac{1}{{\left(k + 3\right)}^{2}} \right)$ is an over estimate of $\displaystyle\in... | We can actually find the value of this series:
Note that:
$$\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \dots + \frac{1}{n^2} = \frac{\pi^2}{6}$$
Now, take a look at the actual sum. Let $S$ be the value of our sum:
$$S = \sum_{k = 1}^{\infty} \frac{1}{(k+3)^2} = \frac{1}{16} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
| $$\cos(a+b)=\cos a \cos b -sin a sin b\\cos(x+x)=cos x \cos x -\sin x \sin x\\cos(2x)=cos^2x-sin^2x\\cos(2x)=cos^2x-(1-cos^2x)=2cos^2x-1\\or =(1-\sin^2x)-\sin^2x\\so\\1-cos(2x)=2sin^2x\\$$put there x instead of 2x $$1-cos(x)=2sin^2(\frac{x}{2})\\\frac{1-cos(x)}{sinx}=\frac{2sin^2(\frac{x}{2})}{2sin(\frac{x}{2})cos(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$ $$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$
Can someone help me to solve it?
result of online calculator: 3/5
| Hint
$$\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)\left(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}+\sqrt{x-3}}\right)=\frac{\big(\sqrt{x+1}-\sqrt{x-2}\big)\big(\sqrt{x+2}+\sqrt{x-3}\big)}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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$\int^{\infty}_{-\infty}u(x,y) \,d y$ independent of x I need to prove that
$$I = \int^{\infty}_{-\infty}u(x,y) \,dy$$
is independent of $x$ and find its value, where
$$u(x,y) = \frac{1}{2\pi}\exp\left(+x^2/2-y^2/2\right)K_0\left(\sqrt{(x-y)^2+(-x^2/2+y^2/2)^2}\right)$$
and $K_0$ is the modified Bessel function of the ... | Although I'm about 7 years late, here is an answer anyway for anyone interested:
Claim
$$I = \frac{e}{2} \sqrt{\pi} \, \text{erfc} (1)$$
and is thus independent of $x$.
Proof.
By https://dlmf.nist.gov/10.32#E10
$$K_0 (z) = \frac{1}{2} \int_{0}^{\infty} \exp \left(-t-\frac{z^2}{4t}\right) \, \frac{dt}{t}$$
This allows... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Identity with Harmonic and Catalan numbers Can anyone help me with this.
Prove that
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
The left side is equal to $$2\log(C(x))=2\log\left(\fr... | Note: Please note that according to the calculation below the factor $(H_{2n-1}-H_n)$ in OPs RHS should be removed.
The following identity is valid
\begin{align*}
\sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^n}{n}=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)
\end{align*}
The initial point for all calculations is the generati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
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Continuity of $f(x,y)=x^3\sin\left(\frac 1x\right)+y^2$ at $(0,0)$ Consider $$f(x,y)= \begin{cases} x^3\sin(1/x) + y^2\quad & x\ne 0 \\ y^2 &\text{otherwise}\end{cases}$$
Prove that $f$ is continuous at $(0,0)$.
I do not have an experience in proving continuity for functions of several variables so any hint or out... | Here's what we know: For all $x$, $-1 \leq \sin{(\frac{1}{x})} \leq 1$, which means for all $x, y$ with $x > 0$:
$$-x^{3} + y^{2} \leq x^{3} \sin{(\frac{1}{x})} + y^{2} \leq x^{3} + y^{2}$$
Now both of the functions on the left and right are continuous at $(0,0)$, and taking the limit as $(x,y) \to (0,0)$ for each of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148541",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Let $A$ be a $3$X$3$ matrix whose eigenvalues are $1$, $2$, $3$. Find $\det(B)$ where $B = A^2 + A^T$. Let $A$ be a $3$X$3$ matrix whose eigenvalues are $1$, $2$, $3$. Find $\det(B)$ where $B = A^2 + A^T$.
I know that $\det(A) = 6$, but I cannot proceed after $|A^2 + A^T|$. Any hints as to how to approach the problem?
| The answer, it seems, is that it depends. If we have $$ A = \left[ \begin{array}{ccc}
1 & 0 & 1 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}
\right]$$
Then we have $\det(A^2 + A^T) = 120$.
On the other hand, if we have $$ A = \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}
\right]$$
Then we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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List the elements of $\mathbb Z_2 \times \mathbb Z_3$ and write its operation table (the notation is additive). $\mathbb Z_2 \times \mathbb Z_3 = \{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)\}.$
$$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, ... | As Omne Bonum mentions, adding two members of a group must yield a member of the group. Using the notation given in the question, in $\mathbb Z_2 \times \mathbb Z_3$,
$$(a, b) + (c, d) = ((a+c)\mod 2, \ (b+d)\mod 3)$$
Here is the corrected addition table.
$$\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149643",
"timestamp": "2023-03-29T00:00:00",
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Minimum value of $\displaystyle \left(x_{1}-x_{2}\right)^2+\left(12-\sqrt{1-x^2_{1}}-\sqrt{4x_{2}}\right)^2.$ The Minimum value of $\displaystyle \left(x_{1}-x_{2}\right)^2+\left(12-\sqrt{1-x^2_{1}}-\sqrt{4x_{2}}\right)^2\;,$ Where $x_{1}\;,x_{2}\in \mathbb{R}$.
$\bf{My\; Trial}::$ Here we have to find Distance between... | One point is in the circle with radius 1 and with center at (0,12) and the other point is in a parabola $x= y^2/4$. I used EXCEL and found that the shortest distance and the minimum value is 7.944277 and the points are (0.45,11.06971) on the circle and (4.0,4.0). As the other responder said, you could use calculus to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the coefficient of $x^{10}$ in $\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2$? I did (parcially) the following exercise:
There are $10$ identical gift boxes. Each one must be colored with a unique color and there are the colors red, blue, green and yellow. It's po... | You're almost there. Expand $(1-x^3)(1-x^4)$ as $1-x^3-x^4+x^7$; when you take the $x^{10}$ coefficient of the product of this with $1/(1-x)^4$, you'll get the four terms with $j=10,10-3,10-4,10-7$ (with appropriate signs):
$$ \binom{13}{10}-\binom{10}{7}-\binom{9}{6}+\binom{6}{3} = 102.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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If $\text{lcm}(a,b,c,d)=a+b+c+d$, then $\gcd(abcd,15)>1$ $\text{LCM}(a,b,c,d)=a+b+c+d$, where $a,b,c,d\in\mathbb Z^+$.
Prove that $abcd$ is divisible by at least one of $3$ and $5$.
I am not sure how to create this proof. Does it involve divisibility tests? How should I set it out?
| We suppose that $a\ge b\ge c\ge d\ge 1$ and $s=a+b+c+d$ is the lowest common multiple of $a$, $b$, $c$ and $d$.
Now, by contradiction, we suppose that $s$ is not a multiple of $3$ nor $5$.
Note that if $a=b=c=d$ then $4a=s$ (because $s$ is the sum) and $s=a$ (because $s$ is the lcm). This is not possible.
*
*As $a<s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0<r< \infty\}$ under the mapping $w=\frac{z}{z-1}$.
Here is what I got so far. First I got the reverse function
$$z= \frac{w}{w-1}=\frac{2u+2iv}{u-1+iv}$$
I did some algebr... | For the sector $\{z=re^{it}\colon 0<r<\infty\quad\text{and}\quad 0\leq t\leq \pi/4\}$, we have in Cartesian coordinates that $y > 0$ and $0<y\leq x$. Then let $z=x+iy$ so
$$
w = \frac{z}{z-1} = \frac{x+iy}{x-1+iy}.
$$
Since $x,y>0$ and $y\leq x$,
$$
(x-1)^2+y^2\leq x^2+y^2
$$
Therefore, we now know $x\geq 1/2$ and $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Summation of sine by considering the imaginary part of exp(ik*seta)
prove that
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac{\cos{\frac {1} {2}}\theta - \cos(n + {\frac 1 2}\theta)} {2\sin \frac{1} {2}\theta}
$$
I would like to solve this by considering the imaginary part of $\sum {\exp(ik\theta)}$.
This is an a... | HINT:
Try using
$$\begin{align}
S_n&=\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\\\\
&=\frac{(1-e^{i(n+1)\theta})(1-e^{-i\theta})}{2(1-\cos(\theta))}
\end{align}$$
The imaginary part is
$$\frac{(\sin(n\theta)+\sin(\theta)-\sin((n+1)\theta)}{2(1-\cos(\theta))}$$
Now, using $1-\cos(\theta)=2\sin^2(\frac{\theta}{2})$, $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometric proof of $\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$ It is well-known that $$\sin{20^\circ}\sin{40^\circ}\sin{80^\circ}=\frac{\sqrt{3}}{8}$$
It follows that $$\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$$
But how to prove this by geometry?
Thank you.
| By using Briggs formulas,
$$\sin A \sin B = \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right),$$
$$\sin C \cos D = \frac{1}{2}\left(\sin(C+D)+\sin(C-D)\right),$$
we have:
$$\begin{eqnarray*} \sin A \sin B \sin C &=& \frac{1}{2}\left(\cos(A-B)\sin C-\cos(A+B)\sin C\right)\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)+\sin(A+B-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Singular value decomposition for matrices that are not square? I understand that the Singular Value Decomposition is defined as SVD = $U\Sigma V^T$ , but I am slightly confused about the calculations when the matrix is not square.
For example, I have the matrix:
$$
\begin{bmatrix}
1 & -1 \\
-2 & 2 \\
2 & -2
\end{bmatr... | First, correct a simple error
$$
\mathbf{A}^{*}\mathbf{A} =
\left[
\begin{array}{rr}
9 & -9 \\
-9 & 9 \\
\end{array}
\right].
$$
This system matrix $\mathbf{A}$ is a rank one matrix (column 2 = $-$column 1) with singular value decomposition
$$
\begin{align}
\mathbf{A}
&=
\mathbf{U}\, \Sigma \, \mathbf{V}^{*} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is $53$ expressible in this form? It seems as if prime numbers may always be expressed in the form $a\cdot 2^b+c \cdot 3^d$ for some nonnegative integers $b,d$ and $a,c\in \{-1,0,1\}$.
Examples:
$2=1\cdot 2^1+0\cdot 3^d$
$3=0\cdot 2^b+1\cdot 3^1$
$5=1\cdot 2^1+1\cdot 3^1$
$7=1\cdot 2^2+1\cdot 3^1$
$11=1\cdot 2^3+1\cdot... | Let's count the positive integers up to $X$ of the shape $2^b+3^d$. It is easy to see that there are at most a constant times $\log^2 X$ of them. That the same statement is true for integers of the shape $|2^b-3^d|$ is much less obvious, but follows from what are called lower bounds for linear forms in logarithms (for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Finding matrix $A$ knowing that $A^2 = B$ Let $B$ be the $3\times3$ matrix
$$
\begin{pmatrix}
1&8&5\\
0&9&5\\
0&0&4
\end{pmatrix}.
$$
How can I find a triangular matrix $A$ with positive diagonal entries such that $A^2 = B$ ?
| As per my comments, we have
$$\begin{bmatrix}1&x&y\\0&3&z\\0&0&2\end{bmatrix}^2 = \begin{bmatrix}1&8&5\\0&9&5\\0&0&4\end{bmatrix}$$
and wish to find $x, y, z$. This gives
$$
\cases{4x = 8\\3y + xz = 5\\5z = 5}
$$
where we immediately get the solutions $x = 2, z = 1$, which again gives $y = 1$. So we have
$$
A = \begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What are some rigorous definitions for sine and cosine? Here are some of my ideas:
1. Addition Formula: $\sin{x}$ and $\cos{x}$ are the unique functions satisfying:
*
*$\sin(x + y) = \sin x \cos y + \cos x \sin y $
*$\cos(x + y) = \cos x \cos y - \sin x \sin y$
*$\sin 0 = 0\quad$ and $\quad\displaystyle{\lim_{x \... | My preferred definition is:
$\cos x$ and $\sin x$ are the real and imaginary parts of the
exponential function $\exp(ix)$.
Since we have:
$$
\begin{split}
e^{ix}= \sum_{k=0}^\infty\dfrac{(ix)^k}{k!}&=1+ix+\dfrac{(ix)^2}{2!}+\dfrac{(ix)^3}{3!}+\cdots+\dfrac{(ix)^n}{n!}+\cdots\\
&=1-\dfrac{(x)^2}{2!}+\dfrac{(x)^4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
| This inequality seems to be somehow too weak, so I'll prove the much stronger inequality:
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$1 + \frac ab + \frac ac \ge 3\sqrt[3]\frac{a^2}{bc} = \frac{3a}{\sqrt[3]{abc}}$$
$$1 + \frac ba ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give ... | First, let us cube the number at hand:
$$\begin{align*}
x^3
&=7+\sqrt{50}+7-5\sqrt{2}+3(7+\sqrt{50})^\frac{2}{3}(7-5\sqrt{2})^\frac{1}{3}+3(7+\sqrt{50})^\frac{1}{3}(7-5\sqrt{2})^\frac{2}{3}\\
&=14+3((49-50)(7+\sqrt{50}))^\frac{1}{3}+3((49-50)(7-5\sqrt{2}))^\frac{1}{3}\\
&=14-3x
\end{align*}$$
Then, $x^3=14-3x \implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
| $x=1$ is a double root of $x^4-4x+3$ (it is also a root of the derivative), hence it is divisible by $(x-1)^2$, hence you'll get $x^4-4x+3=(x-1)^2(x^2+ax+b)$, and partial fractions decomposition has the form:
$$\frac{-2x^2+6x+8}{x^4-4x+3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac{Cx+D}{x^2+ax+b}.$$
You'll obtain $A\ln\lvert ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Erroneously Finding the Lagrange Error Bound Consider $f(x) = \sin(5x + \pi/4)$ and let $P(x)$ be the third-degree Taylor polynomial for $f$ about $0$. I am asked to find the Lagrange error bound to show that $|(f(1/10) - P(1/10))| < 1/100$. Because $P(x)$ is a third-degree polynomial, I know the difference is in the f... | We have:
$$f(x) = \sin\left(5x + \dfrac{\pi}{4}\right)$$
The third degree Taylor Polynomial is given by:
$$T_3(x) = -\frac{125 x^3}{6 \sqrt{2}}-\frac{25 x^2}{2 \sqrt{2}}+\frac{5 x}{\sqrt{2}}+\frac{1}{\sqrt{2}}$$
The error term is given by:
$$R_{n+1} = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \le \dfrac{M}{(n+1)!}(x-a)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $\arg(-1/z)=-2\pi/3$ and $|1-\frac2z|=1$ $\arg(\frac{-1}z)=\frac{-2\pi}3$
what does this mean? how to i get the $\arg(z)$ from this.
I'm thinking of reciprocals.
$\left|1-\frac2z\right|=1$ how do i solve for this as well.. i'm confused when i negative sign appears
| Set
$z = r(\cos \theta + i \sin \theta); \tag{1}$
then $\arg(z) = \theta$, and
$z^{-1} = r^{-1}(\cos \theta - i \sin \theta), \tag{2}$
as may easily be checked by a simple complex multiplication:
$(r(\cos \theta + i \sin \theta))(r^{-1}(\cos \theta - i \sin \theta)) = rr^{-1}(\cos^2 \theta + \sin^2 \theta) = rr^{-1}(1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A conjecture about quadratic residues given $p \equiv 5 \pmod 8$ Original Problem
$p$ is a prime that is congruent to $5$ modulo $8$ and $a$ is a quadratic residue modulo $p$.
Prove that excactly one of $x_1=a^{\frac{p+3}{8}},x_2=(2a)(4a)^{\frac{p-5}{8}}$ is the solution to the congruence $x^2 \equiv a \pmod p$.
What I... | Let $p=8k+5$. By Euler's Criterion, $a^{4k+2}\equiv 1\pmod{p}$ and therefore $a^{2k+1}\equiv \pm 1\pmod{p}$.
Suppose that $a^{2k+1}\equiv 1\pmod{p}$. Then $a^{2k+2}\equiv a\pmod{p}$, and $a$ is congruent modulo $p$ to the square of $a^{k+1}$.
Now suppose that $a^{2k+1}\equiv -1\pmod{p}$. Since $p$ is of the form $8k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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algorithm for positive integer solutions of equation $a^3+b^3=22c^3$ This is a look-a-like to Fermat's last theorem for $n=3$, but it has solutions! I believe that its solution requires knowledge of the techniques of algebraic or analytic number theory which I don't have.
| Might have found something to generate more from an initial solution.
Suppose $a^3+b^3=22c^3$. If want the smallest such $a$ and $b$, they have to be relatively prime to each other and to $c$. $a^3+b^3=(a+b)(a^2+b^2-ab)$. Via trial and error, $a^2+b^2-ab$ cannot be divisible by 11 unless both $a$ and $b$ are, which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Can anyone tell me if this is correct? Suppose that the temperature of a metal plate is given by $T(x; y) = x^2 +2x+y^2$, for points $(x, y)$ on the elliptical plate defined by $x^2 + 4y^2 <= 24$.
Find the maximum and minimum temperatures on the plate.
This is what i have done so far.
Finding critical point:
$T(x)=2x+2... | hint:Equation $(2)$ gives $A = 1/4$, or $y = 0$. If $A = 1/4 \to 2x+2 = \dfrac{x}{2}\to x = -4/3$.
If $A = 2, (2) \to y = 0, (1) \to x = 1 \to x^2+4y^2 = 1^2 + 4\cdot 0^2 = 1 \neq 24 \to A \neq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right)$ $$\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) = \;?$$
I have been trying to see if it can be written as sum of two telescope terms but it looks tricky. Any help ?
| Hint: The sum is a rearrangement of the terms of the alternating harmonic series.
Can you find a series to compare term-wise to? Note that $-\frac{1}{2n} = - \frac{1}{4n} - \frac{1}{4n}$.
Since everyone is very hectic in the comments, let me be clear here:
The sum converges. The rearrangement itself is not enough to p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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critical numbers of a complex function $ f(x) = 2 x - 5 x^{\frac{2}{5}} $ Critical numbers of $ f(x) = 2 x - 5 x^{2/5} $ are 1,$0$
How come $0$ is also a critical number?
I found the critical number 1 this way:
$f'(x) = 2 - 5 \ \frac {2}{5} \ x^{-\frac{3}{5}}$
$ x^{\frac{3}{5}} = 1$
| Critical values of a function occur where the derivative is zero or where the derivative is undefined.
Since $f'(x)=2 - 2x^{\frac{-3}{5}} = 2 - \frac{2}{x^{\frac{3}{5}}} = 2 - \frac{2}{\sqrt[5]{x^3}}$
Hence, $f'(0)$ is undefined since we are dividing by zero.
Also, $f'(1) =0,$ therefore the critical values are 0 and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why is quadratic integer ring defined in that way? Quadratic integer ring $\mathcal{O}$ is defined by
\begin{equation}
\mathcal{O}=\begin{cases}
\mathbb{Z}[\sqrt{D}] & \text{if}\ D\equiv2,3\ \pmod 4\\
\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]\ & \text{if}\ D\equiv1\pmod 4
\end{cases}
\end{equation}... | Have you studied norms yet? If $x$ is an algebraic number in a given quadratic domain such that the leading coefficient of its minimal polynomial is 1, then its norm in that domain is an integer from $\mathbb{Z}$ and $x$ is called an algebraic integer.
Consider these two numbers: $$\frac{1}{2} + \frac{\sqrt{-7}}{2}$$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 6,
"answer_id": 3
} |
Prove that the series $\sum_{1}^{\infty}\frac{k}{(k+1)(k+2)(k+3)}$ converges and find its limit I try to split the summand into differences, but that seems to be a futile way in our case right here, because the numerator is $k$, instead of a given number.
A closely-related series, say $\sum_{1}^{\infty}\frac{2}{(k+1)(... | $$\begin{align}\sum \limits_{k=1}^{\infty}\frac k{(k+1)(k+2)(k+3)}&= \sum \limits_{k=1}^{\infty}\dfrac{2}{k+2} - \dfrac{3}{2(k+3)} - \dfrac{1}{2(k+1)} \\~\\&=\frac{3}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2} -\dfrac{1}{k+3}\right) + \frac{1}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2}-\frac{1}{k+1}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Unexplicit sum evaluation (Putnam)
For positive integers $n,$ let the numbers $c(n)$ be determined by the rules $c(1)=1,c(2n)=c(n),$ and $c(2n+1)=(-1)^nc(n).$ Find the value of $$S = \sum_{n=1}^{2013}c(n)c(n+2).$$
Let $S_k$ represent the $k$th partial sum.
$$S_1 = c(1)c(3)$$
Since, $c(2n+1) = (-1)^nc(n)$ it follows: ... | Split into even and odd terms.
$$\begin{align} S &= \sum_{n=1}^{1006} c(2 n) c(2 n+2) + \sum_{n=1}^{1006} c(2 n+1) c(2 n+3) + c(1) c(3) \\ &= \sum_{n=1}^{1006} c(n) c(n+1) + \sum_{n=1}^{1006} (-1)^n c(n) (-1)^{n+1} c(n+1) - 1 \\ &= \sum_{n=1}^{1006} c(n) c(n+1) - \sum_{n=1}^{1006} c(n) c(n+1) - 1 \\ &= -1\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1}{z^2}$ is holomorphic
I have to show that $z\mapsto\frac1{z^2}$ is holomorpic on $\mathbb C\setminus\{0\}$ and compute its $n$-th derivative
I know that $\frac{1}{z^2}=\sum\limits_{n\ge0}(-1)^n(n+1)(z-1)^n$, so it has a power series representation. Is that sufficient to conclude that the function is analyti... | Given$$f(z) = {z^{ - 2}}$$
then, for$$z = x + iy$$
we may write$$f(x + iy) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} + i\frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}}$$
For$$\begin{gathered}
u(x,y) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} \hfill \\
v(x,y) = \frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}} \hfill... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Number theory: $a\in A\iff \frac{1}{2}-a\in A$. Let $A$ be the set of all $a\in \mathbb{Q}$ for which there exist $x,y,z\in \mathbb{Z}$ not all $=0$ such that $$a=\frac{xy+yz+zx}{x^2+y^2+z^2}.$$ Prove that $$a\in A\iff \frac{1}{2}-a\in A.$$
Note: This is a special case of a more general result I discovered while invest... | Let
$$
a = \frac{xy+yz+zx}{x^2+y^2+z^2}
$$
as above, note that this is equivalent to
$$
1+2a = \frac{(x+y+z)^2}{x^2+y^2+z^2}
$$ and let $a'=\frac12-a$.
If $x=y=z$ then $a=1$ and
$$
a' = -\frac12 = \frac{-1\cdot 2 -1\cdot 2 + 1}{(-1)^2+(-1)^2+2^2} \in A
$$
Otherwise, if $x,y,z$ are not all equal then
$$
1+2a' = 2-2a =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
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Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pic... | There's another (exhaustive) way of seeing it, and it's counting.
Let us represent the act of picking a number $a$ and then a number $b$ by $(a,b)$, we then have $5\times 4=20$ of this possible "actions". Now, the cases that favors your situation are:
$$(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,5),(4,1),(4,3),(4,5),(5,1),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Differentiation of Power Series Let $$f(x)= \sum_{n=0}^{\infty} \frac{x^n}{n!} $$
for $x\in \mathbb R$.
Show $f′ =f$.
Note: Do not use the fact that $f(x) = e^x$. This is true but has not been established at this point in the text.
| Here is a proof which avoids uniform convergence.
Let $$f(x) = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\tag{1}$$ Using multiplication of infinite series it is easy to show that $$f(x)f(y) = f(x + y)\tag{2}$$ for all real $x, y$ (the proof of this also requires binomial theorem for positive integer index).
Next we can see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integral of rational function with trigonometric functions $$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}
$$
I saw this problem online and it looked like an interesting/difficult problem to try and tackle. My attempt so far is to use tangent half-angle substitution.
Let $t= \tan^2 (\frac{x}{2})$, then $dt= \frac{... | You may write
$$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}=\int \frac{1}{(1+ \sqrt{\tan x})^4}\frac{dx}{\cos^2 x}
$$ then make the change of variable $$\displaystyle u=\sqrt{\tan x},\quad \frac{1}{\cos^2 x}=1+u^4, \quad dx=\frac{2u}{1+u^4}du,$$ to just obtain
$$
\begin{align}
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proving that a sequence is increasing Problem: A sequence $(a_n)$ is defined recursively as follows, where $0<\alpha\leqslant 2$:
$$
a_1=\alpha,\quad a_{n+1}=\frac{6(1+a_n)}{7+a_n}.
$$
Prove that this sequence is increasing and bounded above by $2$. What is its limit?
Ideas: How should I go about starting this proof? ... | First, $0\leq a_1\leq2$.
Second, $a_{n+1}=\frac{6(1+a_n)}{7+a_n}=6-\frac{36}{7+a_n}$.
So, if $0\leq a_n\leq2$,
$\qquad$then $7\leq 7+a_n\leq 9$,
$\qquad$then $\frac{1}{7}\geq\frac{1}{7+a_n}\geq\frac{1}{9}$,
$\qquad$then $\frac{36}{7}\geq\frac{36}{7+a_n}\geq\frac{36}{9}$,
$\qquad$then $-\frac{36}{7}\leq-\frac{36}{7+a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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General solution of $(1-x^2)y''-2y=0$ about $x_0=0$? I've expanded this differential equation as a series to obtain the recurrence relation $$a_{n+2}=\frac{a_n(n^2-n+2)}{n^2+3n+2}.$$ I don't know how to find $a_n$ in terms of $a_0$ and $a_1$ so that I can get the general solution. How do I do this?
| This is not an answer but it is too long for a comment.
As commented by Winther, finding something nicer than the general recurence seem to be almost impossible.
Playing with a CAS and starting from what Winther wrote, what I obtained is the following monster for $a_{2n}$ $$\frac{ 4^{n-1} \cosh \left(\frac{\sqrt{7} \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\big((1+\cos(x))\sin(x)\big)^2$ What is
$$\int \big((1+\cos(x))\sin(x)\big)^2dx$$
?
| use $$\begin{align}(1 + \cos t)^2\sin^2 t &= \sin^2 t + 2\sin^2 t \cos t + \sin^2 t \cos^2 t\\
&=\frac 12 - \frac 12 \cos 2t + 2\sin^2 t \cos t + \frac 18 - \frac 18 \cos 4t \\
&=\frac 58 - \frac 12 \cos 2t + 2\sin^2 t \cos t - \frac 18 \cos 4t \end{align}$$
now we can integrate $$\int (1 + \cos t)^2\sin^2 t\, dt = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Does $a \leq b + c $ imply $a^2 \leq (b+c)^2 + (b-c)^2$? Givens
$$
a \leq b + c
$$
or
$$
a^2 \leq b^2 + c^2+2bc
$$
Can we prove that?:
$$
a^2 \leq (b+c)^2 + (b-c)^2 = 2b^2 + 2c^2
$$
| If
$$a^2 \leq b^2+c^2 + 2bc= (b+c)^2,$$
then
$$a^2 \leq (b+c)^2 + (b-c)^2$$
as $(b-c)^2 \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2... | Note: Here is a clear, clean way of going about proving your result going from $k$ to $k+1$:
For $n\geq 1$ let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^{2n}(-1)^i\cdot i^2 = (2n+1)\cdot n.
$$
To prove this, we argue by induction on $n$.
Base case ($n=1$): For $S(1)$, we have that
$$
\sum_{i=1}^{2}(-1)^i\cdot i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$, sum, area of convergence & center I want to find the
$$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
now this is how I thought about doing it but I get stuck.
$$\sum _{n=0}^{\infty } x^n=\frac{1}{1-x}$$ given that absolute value of x is less then zero.
$$\int x^n \, dx=\frac{x^{n+1}... | I approached it in a slightly different way using Taylor expansions.
The sum for $x=0$ is trivial.
For $x \in (-1,1)$
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots$$
$$-\log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\dots$$
$$-\log(1-x)-x-\frac{x^2}{2}=\frac{x^3}{3}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the substitution $t=\tan\theta$ transforms the integral ${\int}\frac{d\theta}{9\cos^2\theta+\sin^2\theta}$, into ${\int}\frac{dt}{9+t^2}$ To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:
$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\... | $\sec^2\theta = \frac{1}{\cos^2\theta} \implies \cos^2\theta\sec^2\theta = \frac{\cos^2\theta}{\cos^2\theta} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Odd divisibility induction proof Prove that for odd $n>3$
$$64\ | \ n^4-18n^2+17$$
I checked that for $n=5$ it works. I think I need to assume that for $2n+1$ it holds and show that $2n+3$ also holds. Any ideas?
| we have:
$$n^4-18n^2+17+64=(n^2-9)^2 $$
and because $(n^2-9)=(n-3)(n+3)$ is divisible by $8$ for odd numbers we can conclude.
By induction:
Assume that $n^4-18n^2+17=(n^2-9)^2-64$ is divisible by $64$ for an odd number $n$ we want to prove that this still true for $n+2$, we have
$$\begin{align}(n+2)^4-18(n+2)^2+17+64&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is the sequence defined by the recurrence $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ convergent?
Let $ a _ 0 = 1 $, $ a _ 1 = 1 $ and $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ for every natural number $ n $. How can I prove that this sequence is convergent?
I know that ... | Graphs, illustrating asymptotic behavior of the sequence $\{a_n\}$. The graphs suggest that $$(a_n-\sqrt{2})\sqrt{2}^n=O(1).$$
Added:
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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calculate the serie $\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$ The following serie is obviously convergent, but I can not figure out how to calculate its sum :
$$\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$$
| You can use one of the following identities to telescope the series: $$\dfrac{3^{n+1}}{3^{n+1}-2^{n+1}}-\dfrac{3^n}{3^n-2^n}=\dfrac{2^n}{3^n-2^n}-\dfrac{2^{n+1}}{3^{n+1}-2^{n+1}}=\dfrac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is e... | You can see it the following way:
$$\begin{align}(1\cdot 2\cdot...\cdot r)\underbrace{((p-1)\cdot(p-2)\cdot...\cdot(p-r))}_{\equiv (-1)(-2)(-3)\cdots(-r)}\pmod{p}
\end{align}$$
so every $i$ appears two times in this product so this product is equivalent to $r!(-1)^r r!$ and this gives you $r!^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $ Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $
I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.
In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx ... | While partial fraction expansion is certainly the natural choice, here is a slightly different approach that continues from the OP's attempt.
$$\begin{align}
\frac{x^2}{x^2+x-2}&=1-\frac{x-2}{x^2+x-2}\\\\
&=1-\frac12 \frac{2x-4}{x^2+x-2}\\\\
&=1-\frac12 \frac{2x+1-5}{x^2+x-2}\\\\
&1-\frac12 \frac{2x+1}{x^2+x-2}+ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can $\oint_{|z|=2}z^3 \bar {z} e^\frac{1}{(z-1)} dz$ be solved? How we can calculate the result of following Integral?
$$\oint_{|z|=2}z^3 \bar {z} e^\frac{1}{z-1} \mathrm{d}z$$
| $$\begin{align}
\oint_{|z=2|} z^3\bar z e^{\frac{1}{(z-1)}}dz&=\oint_{|z=2|} 4z^2 e^{\frac{1}{(z-1)}}dz\\
&=8\pi i \text{Res}_{z=1}\left(z^2 e^{\frac{1}{(z-1)}}\right)\\
\end{align}$$
The residue $\text{Res}_{z=1} \left(z^2 e^{\frac{1}{(z-1)}}\right)$ can be found as follows
Note that $z^2=(z-1)^2+2(z-1)+1$ and the Lau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Question about matrix multiplication notation I have the following matrices:
$A=\begin{pmatrix}
-\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{2} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\end{pmatrix}$
; $B=\begin{pmatrix}
2 & 3\\
2 & 0 \\
... | $a_{ij}$ denotes the $i^{th}$ row and $j^{th}$ column of $A$. The first summation means keeping $i$ fixed, increment the $j$ parameter as below for each $i$.
$$\sum_j a_{ij}u_j = a_{i1}u_1+a_{i2}u_2+a_{i3}u_3$$
This translates to the product $A.U$.
Similarly for the second sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
| $$[x^{30}]\left(\frac{1-x^{13}}{1-x}\right)^5 = [x^{30}]\sum_{k=0}^{5}\binom{5}{k}(-1)^k x^{13k}\sum_{n\geq 0}\binom{n+4}{4}x^n \tag{1}$$
hence the LHS of $(1)$ equals:
$$\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}=\color{red}{17151.}\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Evaluate $\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$ My question is:
Evaluate $$\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$$ for $k=0,1,...,(n-1)$ and $n \in \mathbb{N}$.
I've tried integration by parts but without much success. Any help/trick for this integral? Thanks!
| For every $n\in \mathbb{N}$ we have:
\begin{eqnarray}
\cos(kx)\cos^n(x)&=&\frac{e^{ikx}+e^{-ikx}}{2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n\\
&=&\frac{e^{ikx}+e^{-ikx}}{2^{n+1}}\sum_{p=0}^n{n\choose p}e^{i(n-p)x}e^{-ipx}\\
&=&\frac{1}{2^{n+1}}\sum_{p=0}^n{n\choose p}\left[e^{i(n+k-2p)x}+e^{i(n-k-2p)x}\right]\\
&=&\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maya Lists the Positive Divisors
Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$... | The perfect squares must have an even power of each prime factor. Therefore, you could count the divisors of $2010$; the square of each of these is a square divisor of $2010^2$.
More precisely, since $2010^2$ has 4 distinct pairs of divisors, if $a$ is a square divisor of $2010^2$ and $2$ is a factor of $a$, then $2^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
| your differential equation $$Mdx + Ndy = (x^3 + xy^2 + y)\, dx +(y^3 + yx^2 - x)\, dy = 0.$$ then we have $$M_y = 2xy + 1, N_x = 2xy - 1 $$ therefore is not exact. we can make it exact by multiplying by $a$ and demanding that $$(aM)_y = (aN)_x \to a_y M + aM_y = a_x N + aN_x \to a_xN- a_y M = a(M_y - N_x) = 2a $$ tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Binomial Theorem of Differentiation? I noticed that $$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ and it's had me scratching for a little bit. It's easy to see how the cross terms add up but can anyone draw a direct comparison between the behavior being exhibited here and the binom... | Taylor series expansion of $f(x+h)$
$$
f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....
$$
Taylor series expansion of $g(x+h)$
$$
g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$? I need to compute $$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}\ dx.$$ I tried it on wolfram but it timed out, maybe because I am on a mobile device. Any hint is appreciated.
| $\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x^2+2x+1)}dx$$
Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$
$$\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{x^2\cdot \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.
| It all comes from the algebraic identity:
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\dots+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Proof that $0.33333... = \frac{1}{3}$ using $\epsilon-N$ method This proof is quite prevalent on the web, yet I struggle using this particular method.
Wikipedia (http://en.wikipedia.org/wiki/Limit_of_a_sequence) tells us:
We call $x$ the limit of the sequence $\{ x_n\}$ if:
for each real number $\epsilon> 0$ , there e... | You did no justify your value for $x_n$.
The value of a decimal (base 10) string with infinite number of digits 3 behind the period is:
$$
(0.333\cdots)_{10}
= \sum_{k=1}^{\infty} 3\cdot 10^{-k}
= 3 \sum_{k=1}^{\infty} 10^{-k}
= 3 \sum_{k=1}^{\infty} \left(\frac{1}{10}\right)^k
= 3 \lim_{n\to \infty} S_n
$$
for the pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers n such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients. I am unsure of how to approach this problem. I have thought about using the Rational root theorem, but I am unsure if this answers the question being asked.
Using the theorem, I ... | If a quadratic polynomial $ax^2 + bx + c$ with integer coefficients, $a \neq 0$, has factorization
$$ax^2 + bx + c = (rx + s)(tx + u) \tag{1}$$
then
\begin{align*}
a & = rt\\
b & = ru + st\\
c & = su
\end{align*}
which can be demonstrated by substituting $0$, $1$, and $-1$ in equation 1, then solving the resulting sy... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Solve the recurrence of the alternating sum $R_n=R_{n-1}+(-1)^{n}(n+1)^{2}$ I have been trying to solve this recurrence for a few hours, but I haven't been able to find the solution yet:
$R_0=1$
$R_n=R_{n-1}+(-1)^{n}*(n+1)^{2}$.
I have been trying to substitute $T_n=(-1)^{n}*R_n$ and then solving for $T_n$ and got the ... | $$\begin{align}
R_n-R_{n-1}&=(-1)^n(n+1)^2;\qquad R_0=1\\
R_n-\underbrace{R_0}_{=1}&=\sum_{r=1}^{n}(-1)^r(r+1)^2\qquad\text{by telescoping}\\
R_n&=\sum_{r=0}^{n}(-1)^r(r+1)^2\\
&=\sum_{r=1}^{n+1}(-1)^{r-1}r^2\\
\end{align}$$
Note that $-r^2+(r+1)^2=r+(r+1)$.
Hence, for even $n$,
$$\begin{align}
R_n&=1^2
\underbrace{-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
| For inductive step :
$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{k^2}&=\color{red}{\sum_{k=1}^{n}\frac{1}{k^2}}+\frac{1}{(n+1)^2}\\&\lt\color{red}{2-\frac 1n}+\frac{1}{(n+1)^2}\\&=2-\frac{(n+1)^2-n}{n(n+1)^2}\\&=2-\frac{1}{n+1}\left(\frac{n^2+n+1}{n(n+1)}\right)\\&=2-\frac{1}{n+1}\left(1+\frac{1}{n(n+1)}\right)\\&\lt 2-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $n^a-n^b$ is divisible by 10 Let $n$ be positive integer. Prove that there exists positive integers $a$ and $b$, with $a \neq b$, such that $n^a-n^b$ is divisible by $10$.
I have tried using mathematical induction and logs but I am really stuck .
| Here is way of proving that $n^5-n$ works by induction, just so you know it can be done. It does rely on knowing the answer. The pigeonhole answer is the easiest elementary argument, but doesn't tell you which powers to use, and a more sophisticated approach tells you in general which powers will work for numbers other... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}... | Hint: $$\log (t\times C) = \log t + \log C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Why does the equation $x^2\equiv2 \pmod 5$ have no solutions? Why does the equation $x^2\equiv2 \pmod 5$ have no solutions?
I did a remainders table and found that $$x^2\equiv0;1;4\pmod 5$$
But is there any way to justify this besides that?
The original equation was $2x^2\equiv9\pmod 5$ but I got it to the form above.
| The best way is what you have done.
*
*First note that given any $x$, we have
$x\equiv 0\pmod5 \text{ or }x \equiv 1\pmod5 \text{ or }x \equiv 2\pmod5 \text{ or }x \equiv 3\pmod5 \text{ or }x \equiv 4\pmod5$
*Next note that if $x \equiv y\pmod{n}$, then $x^2 \equiv y^2\pmod{n}$. Hence, we have that
*
*If $x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How would you convert this particular polar equation to cartesian? How would you go about converting the polar equation $r^2 = 4cos(2\theta)$ into a cartesian equation in terms of y?
I have just started working on polar-cartesian equations, but do not yet have sufficient pattern recognition. My current approach of try... | if you multiply the equation by $r^2,$ you get $$r^4 = 4r^2\cos 2\theta = 4r^2(\cos^2 \theta - \sin^2 \theta) = 4(x^2 - y^2) $$ that is $$(x^2 + y^2 )^2 = 4x^2 - 4y^2 \to y^4+2y^2(x^2+2) +(x^4-4x^2).$$ use the quadratic formula to get $$y^2 = -x^2 - 2\pm\sqrt{(x^2+2)^2 -x^4 +4x^2}=-2-x^2 \pm 2\sqrt{2x^2+1} $$ finally... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equations with Sinus. How to find equation solution? I don't understand how to get a solution for sinus equation.
I have:
$$\sin x = \frac{1}{2}$$
\begin{align*}
x & = (-1)^k \cdot \arcsin\left(\frac{1}{2}\right) + 180^\circ \cdot k, k \in \mathbb{Z}\\
& = (-1)^k \cdot 30^\circ + 180^\circ \cdot k, k \in \mathbb{Z}
\... | what you have $x = (-1)^k 30^\circ+180^\circ \cdot k$ is correct.
but can also be done the following way.
the solutions of $$\sin x = \frac 12$$ are $$x = \sin^{-1}(1/2) + k\cdot 360^\circ, 180 - \sin^{-1}(1/2) + k\cdot 360^\circ \\
x = 30^\circ + k\cdot 360^\circ, 150^\circ + k\cdot 360^\circ, \quad k \text{ any int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $M = \int_{o}^{\pi/2}\frac{\cos x}{x+2}dx$ and $N = \int_{0}^{4}\frac{\sin x\cos x}{(x+1)^2}dx$ then the value of $M - N$ is I don't think we can directly solve both the definite integrals and get the answer because I checked these two with WolframAlpha and integration is too difficult.
What I tried is if I execute ... | If the question is Like this way...
$$\displaystyle M=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{x+2}dx$$ and $$\displaystyle N=\int_{0}^{\frac{\pi}{4}}\frac{\sin x\cdot \cos x}{(x+1)^2}dx\;,$$ Then find $M-N$
$\bf{Then\; Solution::}$ Given $$\displaystyle M=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{x+2}dx$$ and $$\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Indefinite Integral of a function $$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$
I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.
| $\int cx^n=c\int x^n=c\frac{x^{n+1}}{n+1}$ when $n\neq -1$
Thus, $\int \frac{1}{5}x^3=\frac{1}{20}x^4$
Also, by the same rule, $\int -2x=-x^2$
$\int \frac{1}{x}=\ln(x)$
Thus, $\int \frac{3}{x}=3\ln(x)$
$\int e^x=e^x$
Thus, by above, the required integral is
$F(x)=\frac{1}{20}x^4-x^2+3\ln(x)+e^x+c$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$ How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$?
I try
$x=2r \cos \phi$,
$y=2r \sin \phi$,
$z=z$, but but probably not the wa... | Letting $r=\sqrt{x^2+y^2}$ gives $r^2+4z^2=4$ in the 1st equation and $z=r-2$ in the 2nd equation, so
substituting gives $r^2+4(r-2)^2=4$ and so $5r^2-16r+12=0$,
which gives $(5r-6)(r-2)=0$ so $r=\frac{6}{5}$ or $r=2$.
$\;\;$Using $z=r-2$, we have that
if $0\le r\le\frac{6}{5}, \;\;-\frac{1}{2}\sqrt{4-r^2}\le z\le \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I finish integrating $\int {\sqrt{x^2-49} \over x} $ using trig substitution? $$\int {\sqrt{x^2-49} \over x}\,dx $$
$$ x = 7\sec\theta$$
$$ dx = 7\tan\theta \sec\theta \,d\theta$$
$$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - ... | If it matters, $\sec^{-1}(x) = \cos^{-1}\left(\frac 1x\right)$. No getting away from inverse trig, though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Inverse of piecewise function I've the following function:
$$f(x)=
\begin{cases}
12x+3, & \text{if $x\ge0$} \\
x+3, & \text{if $x\lt0$}
\end{cases}
$$
What will be its inverse?
For me is $f(x)^{-1}= \frac{x-3}{12}$ per $x\ge 0$ and $3-x$ for $x\lt 0$. Right?
| To find the inverse of a function, we set $y=f(x)$ and try to solve for $x$ in terms of $y$. You can do this by direct manipulation of each case:
$$
\begin{align}
&y=
\begin{cases}
12x+3, & \text{if $x\ge0$} \\
x+3, & \text{if $x\lt0$}
\end{cases}\\
&\implies
\begin{cases}
y=12x+3, & \text{if $x\ge0$} \\
y = x+3, & \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can't find solution for equation $100z=a(z+i)(z-i)^2(z-2)+b(z-i)^2(z-2)+c(z+i)^2(z-i)(z-2)+d(z+i)^2(z-2)+e(z+i)^2(z-i)^2$
I already got $b=5+10i, d=-5+10i $ and $e =8$ by eliminating the factors using $z=i, z=-i, z=-2$ but I cant get $a$ and $c$ with that method. What am I doing wrong?
| $100z=a(z+i)(z-i)^2(z-2)+b(z-i)^2(z-2)+c(z+i)^2(z-i)(z-2)+d(z+i)^2(z-2)+e(z+i)^2(z-i)^2$
Setting $z=i$ gives
$$100i=d(i+i)^2(i-2)\quad\Rightarrow\quad d=-5+10i$$
Setting$z=-i$ gives
$$100(-i)=b(-i-i)^2(-i-2)\quad\Rightarrow\quad b=\color{red}{-}5\color{red}{-}10i$$
Setting $z=\color{red}{+}2$ gives
$$100\cdot 2=e(2+i)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a recurrence relation using generating functions My recurrence relation is
D(n) = D(n - 1) + D(n - 2) + 5(n - 1);
with the initial conditions D(2),D(3) being 6, 17 respectively.
The generating function G(z) for the sequence D(n) is given
I don't know how i got this. Please can anyone give explanation for this... | Note that we can work backwards from $D(2)$ and $D(3)$ to discover what $D(1)$ and then $D(0)$ should be if the recurrence is to hold for $D(2)$ and $D(3)$ as well. We find that if $D(0)=0$ and $D(1)=1$, the recurrence does indeed yield $D(2)=6$ and $D(3)=17$, so we can simplify matters by taking $D(0)=0$ and $D(1)=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}$ I'm trying to solve evaluate this limit
$$\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}.$$
I've tried to rationalize the denominator but this is what I've got
$$\lim_{x\to\infty}(\sqrt{x-1} - \sqrt{... | Note that
$$\sqrt{x-1}-\sqrt{x-2} = \dfrac1{\sqrt{x-1}+\sqrt{x-2}}$$
and
$$\dfrac1{\sqrt{x-2}-\sqrt{x-3}} = \sqrt{x-2}+\sqrt{x-3}$$
We hence have
$$\dfrac{\sqrt{x-1}-\sqrt{x-2}}{\sqrt{x-2}-\sqrt{x-3}} = \dfrac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}}$$
We have
$$\dfrac{\sqrt{x-3}}{\sqrt{x-2}}<\dfrac{\sqrt{x-2}+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Showing $s_n = \left(\frac{1}{2}\right)(s_{n-1} + s_{n-2})$ is Cauchy. Let $(s_n)$ be a sequence defined as $s_1 = 1, s_2 = 2$ , and $s_n = \left(\frac{1}{2}\right)(s_{n-1} + s_{n-2})$. Prove that $(s_n)$ is Cauchy.
I can see how it is convergent and Cauchy but not sure how to put it into a formal proof.
| We have
\begin{align}
s_n - s_{n-1} &= \dfrac{s_{n-1}+s_{n-2}}{2} - s_{n-1}
\\
&= -\dfrac{1}{2}(s_{n-1}-s_{n-2})
\\
&= \cdots
\\
&=(-\dfrac{1}{2})^{n-2}(s_2-s_1)
\\
\end{align}
Let $M=\dfrac{4}{3}|(s_2-s_1)|$, for any $\epsilon$, set $\space N=\log_2(\dfrac{M}{\epsilon})+1$, then for any $n>m, \space m>N,\space m,n\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I know when the limit of a function at a certain point doesn't exist? $$\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8}$$
I have this limit. I can't use L'Hopital. After rationalizing the numerator I get:
$$\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(x-8)(\sqrt{7+\sqrt[3]{x}}+3)}$$
Isn't there anything left to do aft... | \begin{align}
\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8}
&=\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8} \cdot \frac{\sqrt{7+\sqrt[3]{x}}+3}{\sqrt{7+\sqrt[3]{x}}+3}\tag{1}\\
&=\lim_{x \to 8} \frac{(7+\sqrt[3]{x})-9}{(x-8)(7+\sqrt[3]{x}+3)}\tag{2}\\
&=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(x-8)(7+\sqrt[3]{x}+3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What are two easiest numerical method to calculate $\sqrt[23]{123456789}$ by hand? What are two easiest numerical methods to calculate $\sqrt[23]{123456789}$ by hand?
I want to compare that two methods myself
Just name the methods
| Note that $2^{23}=8388608$ and that $10^{23}=100000000000000000000000$
Divide by $8388608$ until a value less than $8388608$ is achieved. Then multiply by $100000000000000000000000$.
Repeat as long as you want until you achieve a value close to $1$.
I discovered that $123456789 = k \times 2^{23} \times \frac 1 {10^{23}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.
Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed f... | For $n=k$, we have
$$12^k+2\cdot 5^{k-1} = 7M$$
Multiplying by $12$ throughout, we obtain
$$12^{k+1}+24\cdot 5^{k-1} = 84M \implies 12^{k+1} + (2\cdot 5 + 14)\cdot 5^{k-1} = 84 M$$
which in-turn implies
$$12^{k+1} + 2\cdot 5^k +14\cdot5^{k-1} = 84M \implies 12^{k+1}+2\cdot 5^k = 14\left(6M-5^{k-1}\right)$$
Hence, the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
Limit of square root where x approaches infinity I have to calculate the following limit, and I wondered if my solution to the question was true. Here it is:
$$\lim _{x \to -\infty} (\sqrt{(1+x+x^2)}-\sqrt{1-x+x^2})$$
Now I divide by $x^2$ and get:
$$\lim _{x \to -\infty} (\sqrt{\frac{1}{x^2}+\frac{x}{x^2}+\frac{x^2}{x... | $$\sqrt{1+x+x^2}-\sqrt{1-x+x^2}=\frac{(1+x+x^2)-(1-x+x^2)}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$
$$=\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}\stackrel{(1)}=\frac{-2}{\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}+\sqrt{\frac{1}{x^2}-\frac{1}{x}+1}}$$
$$\stackrel{x\to -\infty}\to \frac{-2}{\sqrt{0+0+1}+\sqrt{0-0+1}}=-1$$
$(1)\!\! :\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Progression from indefinite integral to definite integral - $\int_{0}^{2\pi}\frac{1}{5-3\cos x} dx$ I'm trying to evaluate the following integral:
$$\int_{0}^{2\pi}\frac{1}{5-3\cos x} dx$$
We can evaluate indefinite one first - $\int\frac{1}{5-3\cos x}dx = \frac{1}{2}\tan^{-1}(2\tan(\frac{x}{2})) + C$. The problem is t... | the graph of $\cos x$ is symmetric about $x = \pi,$ therefore $$\int_0^{2\pi}\frac{dx}{5-3\cos x} = 2 \int_0^{\pi}\frac{dx}{5-3\cos x} = \tan^{-1}\left(2\tan(x/2)\right)\Big|_0^{\pi} =\frac{\pi}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Need a more compact formula This is a part of solution of a programming contest problem
$$\sum_{i=0}^{k} {x-i \choose 2} $$ given $x-i \ge 2$ is always true.
for
$k=1$,$(x-1)^2$
$k=2$, $(x-1)^2+((x-2)*(x-3)/2)$
$k=3$, $(x-1)^2 + (x-3)^2$
and so on.
Is there a reduced form of this?
| Since
$$(x-i)(x-i-1)=\frac13\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)$$
one has
$$\begin{align}\sum_{i=0}^{k}\binom{x-i}{2}&=\sum_{i=0}^{k}\frac{(x-i)(x-i-1)}{2}\\&=\frac 12\sum_{i=0}^{k}\frac 13\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)\\&=\frac 16\sum_{i=0}^{k}\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
finding determinants using different properties The equation is as follows:
$\operatorname{det}(2A^{-1} + 7\operatorname{adj}(A))$
Here I know that $\operatorname{det}(A^{-1}) = (\operatorname{det}(A))^{-1}$ and $\operatorname{det}(kA) = k^n \operatorname{det}(A)$
using these, we know that $\operatorname{det}((2A)^{-1}... | Look at this:
since
$A\operatorname{adj}(A) = \det(A)I, \tag{1}$
we have
$\operatorname{adj}(A) = \det(A) A^{-1} \tag{2}$
provided $A^{-1}$ exists. Well, since we are given that (see comments above)
$\det(A) = 2, \tag{3}$
we know $A^{-1}$ does indeed exist, so using (2) we have
$2A^{-1} + 7\operatorname{adj}(A) = 2A^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the soluti... | Your mistake in part (b) is on line three where you replaced $\log_24$ with $4$ when it should be replaced by $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
For which primes is $-2$ a quadratic residue? For which primes is $-2$ a quadratic residue?
We are trying to find primes that have solution for $x^2 \equiv -2 \mod p.$ Using the Lagrange symbol I know that $2$ is a quadratic residue when $p \equiv 1$ or $7 \mod 8,$ but when is $-2$ a quadratic residue?
Work so far:
$\... | Hint:
$$
\left(\frac{-1}{p}\right) = \begin{cases} 1 & : p \equiv 1 \pmod{8} \text{ or } p \equiv 5 \pmod{8} \\ -1 & : p \equiv 3 \pmod{8} \text{ or } p \equiv 7 \pmod{8} \end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
General Solution of ODE (complex eigenversion) I am trying to figure out the general solution to the following matrix:
$ \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} -3 & -5 \\ 3 & 1 \end{pmatrix}\mathbf{Y}$
I got a solution, but it is so complex I am not sure, if it's even remotely right....:
My Solution:
$x = 1/3*k_1*(-\... | The coefficient matrix $A$ has characteristic polynomial
\begin{align}
\left|\begin{array}{cc}
\lambda+3 & 5 \\
-3 & \lambda-1
\end{array}\right|
& = (\lambda+3)(\lambda-1)+15 \\
& = \lambda^{2}+2\lambda+12 \\
& = (\lambda+1)^{2}+11 \\
& = (\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not abl... | L'Hospital once the fraction becomes
$$\frac{2\sin^2{x}+\tan^2{x}-3x^2}{7x^6}$$
Still an indeterminate
L'hospital 6 more times and rearranging the fraction becomes
$$\frac{4\cos{2x}}{315}+\sec^8{x}-\frac{4\sec^6{x}}{3}+\frac{2\sec^4{x}}{5}-\frac{4\sec^2{x}}{315}$$
And the limit for $x\to 0$ is $1-\frac{4}{3}+\frac{2}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Problems with a Simple Differential Equation I am trying to solve the following: $y' = (y-5)(y+5)$ if $y(4) = 0$.
So far, I have tried separating the variables and then use partial fractions and have followed these steps:
(1) $A(y+5) + B (y-5) =1$.
(2) For $y = -5$ we have $B = \dfrac{-1}{10}$.
(3) For $ y= 5$, we have... | Let's start by decomposing $\frac{1}{(y-5)(y+5)}$ into partial fractions.
We want to to find $A$ and $B$ such that
$$\frac{1}{(y-5)(y+5)} = \frac{A}{(y-5)} + \frac{B}{(y+5)} = \frac{A(y+5)}{(y-5)(y+5)} + \frac{B(y-5)}{(y-5)(y+5)}.$$
Since we have the same denominator for all terms, the numerators of the fractions must ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $Z$ transform of given signal Given the discrete signal $h(n)=r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}$ if $n \geq 0$ and $h(n)=0$ otherwise, find the $Z$ transform of $h(n)$.
What I did:
We know that $H(z)=\sum_{n=0}^{\infty}h(n)z^{-n}=\sum_{n=0}^{\infty}r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}z^{-n}$ which... | The fastest way to find the Z-transform of $h(n)=r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}$ is to utilize the Chebyshev polynomials of the second kind. These polynomials are given as
\begin{align}
U_{n}(\cos \theta) = \frac{\sin(n+1)\theta}{\sin \theta}
\end{align}
with the generating function
\begin{align}
\sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the length of angle bisector $AD$.
In $\triangle ABC$ , the internal bisector of the angle $\angle A$ meets $BC$ at $D$. If $AB=4$, $AC=3$ and $\angle A=60^{\circ}$, then the length of $AD$ is
$a.)\ 2\sqrt3\\
\color{green}{b.)\ \dfrac{12\sqrt3}{7}}\\
c.)\ \dfrac{15\sqrt3}{8}\\
d.)\ \dfrac{6\sqrt3}{7}\\$
With t... | Let the length of AD be $x$ & $\angle ADC=\theta$
Apply law of sine in $\Delta ABD$ $$\frac{\sin(\pi-\theta)}{AB}=\frac{\sin 30^o}{BD} \implies BD=\frac{4}{2\sin\theta}=\frac{2}{\sin\theta}$$
Apply law of cosine in $\Delta ABD$ $$BD=\sqrt{4^2+x^2-2(4)(x)\cos 30^o}\implies \frac{2}{\sin\theta}=\sqrt{x^2-4x\sqrt{3}+16}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $12-\sin(\theta)=\cos(2\theta)$? $$12-\sin(\theta)=\cos(2\theta)$$
What's the correct answer on the $[0,2\pi]$?
I started with $12-\sin(\theta)=1-2\sin^2(\theta)$ and then i cant get anything sensible as i end up with $12=\sin(\theta)+2\sin^2(\theta)$
| $$12-\sin(x)=\cos(2x)\Longleftrightarrow$$
$$12-\cos(2x)-\sin(x)=0\Longleftrightarrow $$
$$11-\sin(x)+2\sin^2(x)=0\Longleftrightarrow $$
$$\frac{11}{2}-\frac{\sin(x)}{2}+\sin^2(x)=0\Longleftrightarrow $$
$$\sin^2(x)-\frac{\sin(x)}{2}=-\frac{11}{2}\Longleftrightarrow $$
$$\frac{1}{16}-\frac{\sin(x)}{2}+\sin^2(x)=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Compute $\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^3+1}} + \frac{1}{\sqrt{n^3+4}} + \cdots + \frac{1}{\sqrt{n^3+n^2}}\right)$ How do I evaluate the following limit?
I guess I should do a comparison, but I've got no clue about what to do. Could you give me a hand?
$$\lim_{n \to \infty}\left( \frac{1}{\sqrt{n^3+1}} + ... | Easily we can find that, $$\frac{1}{\sqrt{n^3+n^2}}\le \frac{1}{\sqrt{n^3+k^2}}\le\frac{1}{\sqrt{n^3}}.$$for all $k=1,2,...,n$.
Taking summation, $$\sum_{k=1}^n\frac{1}{\sqrt{n^3+n^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3}}$$
$$\implies \frac{n}{\sqrt{n^3+n^2}}\le \sum_{k=1}^n\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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.