Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Infinite sum of reciprocal shifted Fibonacci numbers I found on Wikipedia the following infinite sum :
$$\sum_{k=0}^{\infty} \frac{1}{1+F_{2k+1}}=\frac{\sqrt{5}}{2}$$
There is no reference for this sum in the article and I couldn't find it anywhere else .I have no idea how to prove it so I'm asking for help .
Thank ... | We can use Binet's formula for $F_{2k+1}$:-
$$F_{2k+1}=\frac{\phi^{2k+1}-(-\phi)^{-(2k+1)}}{\sqrt{5}}$$
where $\phi=\frac{1+\sqrt{5}}{2}$.
The summation (after substitution of $F_{2k+1}$ and partial fraction decomposition) becomes a telescoping sum
$$\begin{align}\sum_{k=0}^{\infty} \frac{1}{1+F_{2k+1}}&=\sum_{k=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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$25$ colored dice $25$ colored dice are thrown and their results are put at random at a square matrix $5\times 5$. Dice colors are as follows: $5$ Red, $5$ Blue, $5$ Yellow, $5$ Green and $5$ Orange.
I am interested in calculating the following probabilities:
*
*The probability that a bar of length $3$ ($3$ conse... | The third question can be solved without a computer.
We have one condition for each row and one for each column. Multiple conditions can only be satisfied simultaneously if they correspond to the same direction.
The probablity for $k$ particular compatible conditions to be satisfied simultaneously is $\prod_{j=0}^{k-1}... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Evaluation of $\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$
Evaluation of $$\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \int\frac{\sin 2x}{(3+4\cos x)^3}dx = 2\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^3 x.$
So we get $$\displaystyle... | Going by your work so far we have
$$
2\int \frac{\sin \theta \cos \theta}{(3+4\cos \theta)^3}d\theta
$$
Lets set $u = 3+4\cos \theta\to du = -4\sin\theta d\theta$
thus your integral becomes
$$
-\frac{2}{16}\int \frac{u-3}{u^3}du=?
$$
This yields the result you desire! Though the interesting question is how to go from y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Forming a committee from $4$ gentlemen and $4$ ladies with certain conditions
From $4$ gentlemen and $4$ ladies a committee of $5$ is to be formed .
If the committee consists of $1$ president, $1$ vice president and $3$ secretaries.
What will be the number of ways of selecting the committee with at least $3$
wom... | As pointed out in a comment, there are numerous errors in your answer.
Pres-Veep . . . . . Secretaries
W-W . . . . . . . . . 2W,1M or 1W-2M: $\left[{4\choose1 }{3\choose 1}\right]\left[{2\choose2}{4\choose1} + {2\choose 1}{4\choose 2}\right]= 192$
W-M or M-W . . . . 3W or 2W,1M: $\left[2{4\choose1}{4\choose1}\right]\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430501",
"timestamp": "2023-03-29T00:00:00",
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If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work,
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$,
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\t... | Notice, we have $$\cos \theta+\sin\theta=1$$
$$(\cos \theta+\sin\theta)^2=1$$$$\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta=1$$ $$1+2\sin \theta\cos \theta=1$$
$$\iff \sin\theta\cos \theta=0\tag 1$$
Now, we have
$$(\cos \theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\sin \theta\cos \theta$$
$$=(\cos^2\theta+\sin^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Coin toss puzzle with 1 biased and 2 unbiased coins Three coins are given: one two-head coin and two fair coins. You randomly choose a coin and the first three tosses give heads. What is the probability that 4-th toss is a head.
I have two solutions that give different results. Please help me to find the error.
1st sol... | Your second solution is erroneous.
\begin{align}
& \Pr(\text{4th is head} \mid \text{first 3 tosses are heads})\\[10pt]
= {} & \Pr(\text{fair} \mid \text{1st 3 are heads}) \times \Pr(\text{4th is a head} \mid \text{fair & 1st 3 are heads}) \\[4pt]
& {} + \Pr(\text{two-headed}\mid \text{1st 3 are heads} ) \\
& \phantom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$ Problem :
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$
My approach :
Put $x = \tan\theta$
we get $$\int \frac{\sqrt{1+x^2}}{1-x^2}dx = \frac{\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta \cos\theta}}{\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta}} d\theta $$
$$= \frac{1}{(\cos^2\theta ... | Letting $t=\frac{x}{\sqrt{x^{2}+1}}$, then $$
x^{2}=\frac{t^{2}}{1-t^{2}}=\frac{1}{1-t^{2}}-1 \Rightarrow 2 x d x=\frac{2 t d t}{1-t^{2}} \Rightarrow\sqrt{x^{2}+1} d x=\frac{d t}{1-t^2}$$
Plugging them into the integral yields
\begin{aligned}
I &=\int \frac{1}{1-\frac{t^2}{1-t^{2}}} \cdot \frac{d t}{1-t^{2}} \\
&=\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding an angle in a triangle, given the angle bisector and some conditions.
$ABC$ is a triangle in which $\angle B= 2\angle C$. $D$ is a point on side $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$.
What is the measure of $\angle BAC? $
I tried using angle bisector theorem, similarity, but the sides, don'... | Notice, $$\angle ADB=180^\circ -\left(\angle B+\frac{\angle BAC}{2}\right)$$
$$\angle B+\angle C+\angle BAC=180^\circ\iff 2\angle C+\angle C+\angle BAC=180^\circ$$ $$\iff \angle C=\frac{180^\circ-\angle BAC}{3}\tag 1$$$$\iff \angle B=\frac{2(180^\circ-\angle BAC)}{3}\tag 2$$
apply sine rule in $\triangle ABD$ $$\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a vector function represented by the curve of intersection? I'm struggling with the following problem:
Given $\, z = \sqrt{x^2 + y^2}\,$ and $\, z = y+1\,$ find the vector function represented by the curve of intersection of the surfaces using the parametrization $\, x = t$.
I've done problems like this before... | First, equate $\,z\,$ and square both sides of obtained expression:
$$
z = \sqrt{x^2 + y^2} = y+1
\implies
x^2 + y^2 = y^2 + 2y + 1
\implies
\boxed{y = \dfrac{x^2 - 1}{2}}
$$
Second, apply parametrization $\, x = t$:
\begin{align}
\begin{cases}
y = \dfrac{x^2 - 1}{2}\\
z = y + 1 \\
z = \sqrt{x^2 + y^2}
\end{cases}
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number Theoretic Transform (NTT) example not working out I'm reading up on the NTT, which is a generalisation of the DFT. I'm working in $\mathbb{F}_5$ with primitive root $w=2 \mod 5$. Suppose I want to compute the NTT of $x=(1,4)$. So far I have obtained:
$$\hat{x}=\mathcal{N}(x)=\left(\sum_{j=0}^1 2^{jk}x_j\mod 5\ri... | I think this is right. It seems the problem was related to the size of the NNT, $n=2$ in my case, and the order of the primitive root of unity. Since $n=2$ then I needed to choose a $2$-th root of unity, which would mean switching to prime $p=3$, but then this would not work because my vector $(1,4)\equiv(1,1)$ so I lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Smallest no of balls in the box? A box contains white and black balls. When two balls are drawn without replacement suppose the probability that both are white is 1 /3.
a) Find smallest number of balls in the box ?
b) How small can the total number of balls be if black balls are even in number ?
I considered no o... | $$\frac{n}{m+n}\frac{n-1}{m+n-1} = \frac{1}{3}$$
$$\frac{n(n-1)}{(m+n)(m+n-1)} = \frac{1}{3}$$
$$\frac{n(n-1)}{m^2 + 2nm + n^2 - m - n} = \frac{1}{3}$$
$$\frac{n(n-1)}{(m+n)^2 - (m+n)} = \frac{1}{3}$$
Let us now set $m+n = B$, the number of total balls
$$3(n^2-n) = B^2 - B$$
$$B^2 - B - 3(n^2-n) = 0$$
$$B = \frac{1 \pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin... | Hint:
$${f(x)\over g(x)}={{f(x)\over x}\over{g(x)\over x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:... | $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$
substitute $t = 1-x^7$, $\,\,dt = -7x^6dx$, and by extension $\,\,x = \sqrt[7]{1-t}$
$$=-\frac{1}{7}\int_{0}^{1} \frac{\sqrt[3]{t}}{(1-t)^\frac{6}{7}}dx$$
Now integrate by parts choosing the factors $u = \sqrt[3]{t}$ $\,\, u' = \frac{1}{3t^{2/3}}$$\,\,dv = \frac{1}{(1-t)^{6/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate $\lim \limits_{x \to 0} \frac{x^2 \sin^2x}{x^2-\sin^2x}$ with $\lim \limits_{x \to 0} \frac{\sin x}{x}=1$? How to calculate
$$\lim \limits_{x \to 0} \frac{x^2 \sin^2x}{x^2-\sin^2x}$$
with
$$\lim \limits_{x \to 0} \frac{\sin x}{x}=1?$$
Yes I know the question has been asked, the answer is $3$, L'Hospit... | We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{x^{2}\sin^{2}x}{x^{2} - \sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{x^{4}}{x^{2} - \sin^{2}x}\cdot\frac{\sin^{2}x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{x^{4}}{x^{2} - \sin^{2}x}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{x}{x + \sin x}\cdot\frac{x^{3}}{x - \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Real part of $\sqrt{ai-1}$ Is there a way to find the real and imaginary parts of
$$
z=\sqrt{ai-1},\qquad a>0
$$
where $i=\sqrt{-1}$. Thanks. I do not know what to to do.
Note
$$
i=e^{i\pi/2}=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}.
$$
| for $z=x+iy$ you can right its Polar Representation $z=re^{i\theta}=r(\cos\theta+i\sin\theta)=\color{red}{r\cos\theta}+i\color{blue}{r\sin\theta}$(respectively real part and imaginary part red and blue) where $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\frac{y}{x} $ for $x>0$ and $\theta=tan^{-1}\frac{y}{x}+\pi $ for $x<0$... | {
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Solving $\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$ without L'Hopital Without L'Hopital:
$$\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$$
Rationalize:
$$\frac{x}{\sqrt{1-3x}-1}\cdot \frac{\sqrt{1-3x}+1}{\sqrt{1-3x}+1}$$
$$\frac{x\cdot(\sqrt{1-3x}+1)}{(1-3x)-1}$$
This will still yield $\frac{0}{0}$. Maybe I should now try variable sub... | Let $\sqrt{1-3x}=1+h$. We know that when $x=0$ the LHS $=1 \Rightarrow h=0$
Then $1-3x=1+2h+h^2$
$-3x=2h+h^2$
$x=\frac{2h+h^2}{-3}$
The expression we want is $\frac x{\sqrt{1-3x}-1}=\frac {\frac{2h+h^2}{-3}}{1+h-1}=\frac{2+h}{-3}$.
As $h$ tends to $0$ this tends to $-\frac23$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
A... | $$\lim_{x\to0}\frac{x (\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}\color{red}x}{\color{red}2x} = \sqrt 3$$
Note the changes in red.
You didn't cancel correctly here.
| {
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Prove that using induction that $\binom22+\dots+\binom n2 = \binom{n+1}2$ so I have this math problem where I have to prove this using induction. $$\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}n\\2\end{pmatrix}=\begin{pmatrix}n+1\\3\end{pmatrix... | To prove $P(k+1)$, start with your left hand side, i.\,e.
$$ \def\P#1{\binom 22 + \binom 32 + \cdots + \binom{#1}2}\P{k+1}$$
Write it as
$$ \P k + \binom{k+1}2 \tag+ $$
Now use, that by the induction hypothesis, we have
$$ P(k): \P k = \binom{k+1}3 $$
Hence, we can write $(+)$ as
$$ \P k + \binom{k+1}2 = \binom{k+1}... | {
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"timestamp": "2023-03-29T00:00:00",
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Order of operations in polynomial with exponent I have a simple question about whether or not my approach is correct in simplifying a polynomial, here it is,
$(n(n+1)/2)^2 = ((n^2+n)/2)^2 = 1/4(n^4+2n^3+n^2)$
I apologize if you find that hard to read, I can't write the equations the way I would like.
Many thanks.
| It is correct, but if desired to write with no parenthesis you may opt to distribute the $\frac{1}{4}$ through the expression to arrive at $$\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}$$
A small piece of advice for if you are faced with exponents of higher power than $2$.
There is such a thing known as the Binomial Theor... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the solution of $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$ Is anyone able to help me with the following equation concerned the floor function $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$
I don't know how to deal with the floor terms properly.
| From the definition of the floor function we have
[correction here]
$$\begin{align}
x^2-1 &< &\lfloor x^2 \rfloor &\le x^2 \\
3x-1 &< &\lfloor 3x \rfloor &\le 3x
\end{align}$$
and so
$$\lfloor x^2 \rfloor - \lfloor 3x \rfloor + 2 > x^2-3x-1+2=x^2-3x+1$$
So all solutions to the floor equation must lie between the two ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \rig... | Hint: Taking logs, and using $u>\ln (1+u), u> 0,$ we see it suffices to show
$$\ln x \ge (x-1)/3\cdot (1/x + 1/x^2 + 1/x^3), \ \ x> 1.$$
Both sides are $0$ at $x=1,$ so it suffices to show the inequality for the derivatives of each side.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there $a,b,c,d \in \mathbb N$ such that $\frac{a + b}{a + b + c + d} < \frac{a}{a + c} < \frac{b}{b + d}$? Consider the following $2 \times 2$ contingency table:
\begin{array}{c|cc|c}
& C & \overline C & \Sigma \\ \hline
V & 4000 & 3500 & 7500 \\
\overline V & 2000 & 500 & 2500 \\ \hline
\Sigma & 6000 & 4000 & 100... | None of the two double-inequalities can hold for any $a,b,c,d\in\Bbb R^+$.
$$\frac{a + b}{a + b + c + d} < \frac{a}{a + c} \iff (a+b)(a+c)<a(a+b+c+d)$$
$$\iff \frac{a}{b}>\frac{c}{d}$$
The same way you get:
$$ \frac{a}{a+c} < \frac{b}{b + d} \iff \frac{a}{b}<\frac{c}{d}$$
The two inequalities contradict each other.
Use... | {
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \fra... | From the Basel Problem, we have
$$\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\dots$$
$$\frac{\pi^2}{24} = \frac{\pi^2}{6\cdot 2^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2}\dots$$
so that
$$\begin{align}\frac{\pi^2}{8} &= \frac{\pi^2}{6} - \frac{\pi^2}{24}\\&=\frac{1}{1^2} + \f... | {
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"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
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How many different values can $(x^2 + y^2, x^2 + 2y^2 )$ have mod 4? It is known that if a prime number $p = x^2 + y^2 $ is equivalent to $p \equiv 1 \mod 4$. This is Fermat's theorem on the sum of two squares.
My question is about the value of two simultaneous quadratic forms. Consider the values:
$$ (x,y) \in \mat... | $x^2$ and $y^2$ modulo 4 can each only take on 2 values. Thus, there can be at most four distinct combinations of those values, no matter how you arrange them.
Modulo 11, $x^2$ and $y^2$ can take on up to 6 distinct values, so you can have up to 36 possible combinations.
Note that if $(a,b)$ are independent variables,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area under curves functional analysis question Consider the functions defined inplicitly by te equation $y^{3}-3y+x=0$ on various intervals in the real line . If $x\in(-\infty,2)\cup(2,\infty)$ the equation implicitly defines a unique real valued differentiable function $y=f(x)$ . if $x\in (-2,2)$, the equation implici... | $(1)\;\;$ Given $$y^3-3y+x=0\;,...........(1)$$ Now Differentiate both side w. r to $x\;,$ We get
$$\displaystyle 3y^2\cdot y'-3y'+1=0....................(2)\Rightarrow y'=\frac{1}{3-3y^2}\;,$$ Where $$\displaystyle y=f(x)\;\;, y'=\frac{dy}{dx}=f'(x)\;\;,y''=f''(x)=\frac{d^2y}{dx^2}$$
Now Given $\displaystyle y=f\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integration of elementary function $\int \frac {\log x}{(1+x)^3}\,{\rm d}x $
The question is to find the integral
$$\int \frac {\log x}{(1+x)^3}\,{\rm d}x $$
It can be easily be solved by integration by parts, but I want to solve it without using integration by parts.
| It can be done by a sort of "undetermined coefficients". Suppose we can guess that the solution is of the form
$$ F(x) = a(x) + b(x) \ln(1+x) + c(x) \ln(x) $$
where $a, b, c$ are rational functions. Taking the derivative and comparing to $\ln(x)/(1+x)^3$, we see
$$ \eqalign{c' &= \dfrac{1}{(1+x)^3}\cr
b'... | {
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"url": "https://math.stackexchange.com/questions/1457716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin... | $$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x} =-\lim_{x\to0}\frac{\sin x(1-\cos x)}{2x^2\sin x\cos^2x} =-\lim_{x\to0}\frac{(1-\cos x)}{2x^2}\cdot\frac1{\lim_{x\to0}\cos^2x}$$
Now $$\lim_{x\to0}\frac{(1-\cos x)}{2x^2}=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{2x^2}\cdot\dfrac1{\lim_{x\to0}(1+\cos x)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluate Integral $\int\frac{1+x}{1+x^2}\ dx$ $\int\frac{1+x}{1+x^2}\ dx$
Let $u=1+x^2$
Then $du = 2x\ dx$
Here is my work.
Split integral $\int\frac{1}{1+x^2}\ dx$ + $\int\frac{x}{1+x^2}\ dx$
Integrate first integral term:
$\int\frac{1}{1+x^2}\ dx=tan^{-1}x$
$\int\frac{x}{1+x^2}\cdot\frac{du}{2x} $
I am stuck when it... | HINT:
$$\int \frac{1}{u}\,du=\log|u|+C$$
So, let $u=1+x^2$, $du=2x\,dx$ and ...
SPOILER ALERT Scroll over the highlighted area to reveal the answer
$$\int\frac{1+x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx+\int\frac{x}{1+x^2}\,dx=\arctan(x)+\frac12 \log(1+x^2)+C$$where for the second integral, we used the HINT and substitut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expansion of $\frac{1}{\sqrt{1-4x}}$ Expand $\frac{1}{\sqrt{1-4x}}$ in ascending of power of $x$, up to and including the term in $x^2$, simplifying the coefficient. Hence find the coefficient of $x^2$ in the expansion of $\frac{1+2x}{\sqrt{4-16x}}$
My attempt, $(1-4x)^{-\frac{1}{2}}$
$=1+(-\frac{1}{2})(-4x)+\frac{(-\... | $$ [x^2]\frac{1+2x}{\sqrt{4-16 x}} = \frac{1}{2}[x^2]\frac{1}{\sqrt{1-4x}}+[x]\frac{1}{\sqrt{1-4x}}\tag{1} $$
and since:
$$ \frac{1}{\sqrt{1-4x}} = 1+2x+6x^2+O(x^3)\tag{2} $$
we have:
$$ [x^2]\frac{1+2x}{\sqrt{4-16 x}} = \frac{1}{2}\cdot 6+2=\color{red}{5}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The product of $2×653×733×977$ has each digit exactly once except for one, which one is it? The product of $2×653×733×977$ is a number with nine digits and it contains every digit once except for one, which digit is that?
I noticed that it is a product of primes, but so far I cannot solve it without multiplying it out... | In mod 9 we have
2*653*733*977 = 2*5*4*5 = 2 mod 9
Now we make the following table :
0+1+2+3+4+5+6+7+8+() = 36 = 0 mod 9
0+1+2+3+4+5+6+7+()+9 = 37 = 1 mod 9
0+1+2+3+4+5+6+()+8+9 = 38 = ...
0+1+2+3+4+5+()+7+8+9 = 39 = ...
0+1+2+3+4+()+6+7+8+9 = 40 = ...
0+1+2+3+()+5+6+7+8+9 = 41 = ...
0+1+2+()+4+5+6+7+8+9 = 42 = ...
0+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $ I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx$
If $\displaystyle I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx\;,$ Then value of $100(I-\ln 2) =$
$\bf{My\; Try::}$ Let $\cot^{-1}(x)=t\;,$ Then $\displaystyle \frac{1}{1... | HINT:
$$I=\int\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx =\int\dfrac{\dfrac{2x}{(1+x^2)^2}-\cot^{-1}x}{\dfrac1{1+x^2}-\cot^{-1}(x)}dx$$
As $\dfrac{d\left(\dfrac1{1+x^2}-\cot^{-1}(x)\right)}{dx}=-\dfrac{2x}{(1+x^2)^2}+\dfrac1{1+x^2},$ write
$$I=\int\dfrac{\dfrac{2x}{(1+x^2)^2}-\dfrac1{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=$ If $n=12m$, where $m\in\Bbb{N}$, prove that
$$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=(-1)^m\left(\frac{2\sqr... | Using $\displaystyle (1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+............$
Now Put $\displaystyle x=\frac{i}{2+\sqrt{3}}$ and $\displaystyle x=-\frac{i}{2+\sqrt{3}}$ Respectively, We get
$\displaystyle \left(1+\frac{i}{2+\sqrt{3}}\right)^{n} = \binom{n}{0}+\binom{n}{1}\cdot \frac{i}{2+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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binomials product alternating sum calculation I need to somehow prove that $\sum\limits_{k = 0}^{n - 1} {n \choose k} {3 n - k - 1 \choose 2 n - k}(-1)^k = (-1)^{n + 1} {2 n - 1 \choose n}$.
I didn't manage to do it using induction or any combinatorial ideas. Could someone help me?
| Suppose we seek to verify that
$$\sum_{k=0}^{n-1}
{n\choose k} {3n-k-1\choose 2n-k} (-1)^k
= (-1)^{n+1} {2n-1\choose n}$$
which is the same as
$$\sum_{k=0}^{n}
{n\choose k} {3n-k-1\choose 2n-k} (-1)^k
= \left((-1)^n+(-1)^{n+1}\right) {2n-1\choose n} = 0.$$
Introduce
$${3n-k-1\choose 2n-k} = {3n-k-1\choose n-1} =
\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square. For all real numbers $x$,let the mapping $f(x)=\frac{1}{x-i},\text{where} i=\sqrt{-1}$.If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex ... | $$\displaystyle f(x)=\frac { 1 }{ x-i }=\frac { x+i }{ x^{ 2 }+1 } $$
Now in Argand plane $f(a),f(b),f(c),f(d)$ all lie on a curve whose parametric coordinates
is given by $$\displaystyle x=\frac { t }{ t^{ 2 }+1 } ,y=\frac { 1 }{ t^{ 2 }+1 }$$
Now eliminating variable $t\;,$ We get $$x^{ 2 }+y^{ 2 }=y\Rightarrow x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to compute the coefficients of this generating function Working on some combinatorial problem, I arrived at the following generating function
$$K_m(x) = \sum_{n\geq 0}K_{mn}x^n =\frac{x}{1-\sqrt{1+x^2}\cdot\frac{\displaystyle{y_+(x)^{m+1}+y_-(x)^{m+1}}}{\displaystyle{y_+(x)^{m+1}-y_-(x)^{m+1}}}}$$
with
$$y_\pm(x) =... | In terms of Lucas sequences with the parameters $(P,Q)=(2x,-1)$, we have
$$y_+(x)^{m+1} + y_-(x)^{m+1} = V_{m+1},$$
$$\frac{y_+(x)^{m+1} - y_-(x)^{m+1}}{\sqrt{1+x^2}} = 2U_{m+1}.$$
It follows that
$$K_m(x) = \frac{x}{1-\frac{V_{m+1}}{2U_{m+1}}} = \frac{xU_{m+1}}{U_{m+1}-\frac12V_{m+1}}.$$
Since the denominator value at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 1,
"answer_id": 0
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$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I poste... | Here is one of my favorites.
Proof that $1=0$
Let's consider for real $x$ the function $f(x)=xe^{-x^2}$. Note, the following integral representation of $f$ is valid (substitute: $u=x^2/y$).
\begin{align*}
\int_{0}^{1}\frac{x^3}{y^2}e^{-x^2/y}\,dy
=\left[xe^{-x^2/y}\right]_0^1
=xe^{-x^2}
\end{align*}
We obtain for al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 7
} |
Prove inequality $\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3$ How to prove the following inequality :
$$
\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3
$$
with $a>0,\ b>0$ and $c>0$.
| since $$9(a+b)(b+c)(a+c)\ge 8(ab+bc+ac)(a+b+c)$$
By Cauchy-Schwarz inequality we have
$$\sum_{cyc}\sqrt{\dfrac{2a}{a+b}}\le\sqrt{\left[\sum(c+a)\right]\left[\sum_{cyc}\dfrac{2a}{(a+b)(c+a)}\right]}
=\sqrt{\dfrac{8(a+b+c)(ab+bc+ac)}{(a+b)(b+c)(a+c)}}\le 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum number of fractions to be summed up to $\frac45$ What is the minimum number of fractions having numerator 1 and a natural number as denominator to be summed up to $\frac 45$?
I have tested with 2 fractions: $\frac1a + \frac1b = \frac45$ and get into the diophantine equation: $5(a+b)=4ab$ and it seems this shou... | Considering prime factorization gives that $5 \mid ab$ and, by relabeling if necessary, we can assume that $5 \mid a$, and in particular $\frac{1}{a} \leq \frac{1}{5}$. Substituting in the two-fraction equation gives $b \leq \frac{5}{3}$, so we must have $b = 1$, but this gives $\frac{1}{a} + 1 = \frac{4}{5}$, and the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\... | You reindexed the sum incorrectly when you changed from $n$ to $k$ as dummy variable.
Just look at the first terms: when $n=2$, the term is $(\frac12)^2\cdot(\frac12)^2=\frac1{16}$; but when $k=0$, your first term is $(\frac12)^0=1$. Your first index for $k$ should be $4$, not $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove by induction: $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$ $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$
Base Case: For $n=1$, $\frac{1}{2}\tan\frac{x}{2}=\frac{1}{2}\... | Proving that $$\frac{1}{2^k}\cot\frac{x}{2^k}-\cot x+\frac{1}{2^{k+1}}\tan\frac{x}{2^{k+1}}=\frac{1}{2^{k+1}}\cot\frac{x}{2^{k+1}}-\cot x$$ is equivalent to prove that $$\frac{1}{2^k}\cot\frac{x}{2^k}+\frac{1}{2^{k+1}}\tan\frac{x}{2^{k+1}}=\frac{1}{2^{k+1}}\cot\frac{x}{2^{k+1}}$$ by omitting $-\cot x$. Then if you mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the last three digits of $17^{102}$.
Find the last three digits of $17^{102}$.
That is we have to find out $17^{102}\equiv ?\pmod{1000}$.
Now if we solve it through general congruence relation , then the process becomes very difficult , computational and laborious. Again from Euler's theorem , $$17^{\phi(1000)}\... | The units digit of $17^1$ is $7$.
The units digit of $17^2$ is $9$ ($7 \times 7 = 49$).
The units digit of $17^3$ is $3$ ($7 \times 9 = 63$).
The units digit of $17^4$ is $1$ ($7 \times 3 = 21$).
Calculating,
$\; 17^4 = 289 \cdot 289 = (280+9)^2 = 280^2 + 18 \cdot 280 + 81 \equiv$
$\quad\quad\quad 400 + 800 + 600 + 640... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Closed form for ${\large\int}_0^1x\,\operatorname{li}\!\left(\frac1x\right)\ln^{1/4}\!\left(\frac1x\right)dx$ Let $\operatorname{li}(x)$ denote the logarithmic integral:
$$\operatorname{li}(x)=\int_0^x\frac{dt}{\ln t}.$$
How can we prove the following conjectured closed form?
$${\large\int}_0^1x\,\operatorname{li}\!\le... | Let $x = e^{-y}$ and denote the value of the integral by $I$. Then we have
\begin{equation}
I = \int_0^\infty y^{1/4} e^{-2y} \text{ li}(e^y) \, dy.
\end{equation}
Consider the parameter
\begin{equation}
I(b) = \int_0^\infty y^{1/4} e^{-2y} \text{ li}(e^{by}) \, dy.
\end{equation}
Differentiating we obtain
\begin{equat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
... | Using the following rules, with $a,b \in \mathbb{R}$:
*
*$\left|a^b\right|=\left|a\right|^b$;
*$\arg\left(a^b\right)=\tan^{-1}\left(\cos(b\cdot\arg(a)),\sin(b\cdot\arg(a))\right)$
$$(-8)^{\frac{4}{3}}=$$
$$\left|(-8)^{\frac{4}{3}}\right|e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$
$$\left|-8\right|^{\frac{4}{3}}e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$
If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$.
I don't know how to prove this statement.
$p=4m+3$, so $(2m+1)! \equiv \pm1\pmod p$
This is all I did.
| $\left( \frac{p-1}{2}! \right)^2 = \prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}n=\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}-n \equiv\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=\frac{p+1}{2}}^{p-1}n=(p-1)! \equiv 1$
You can change $n$ for $-n$ because t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more si... | Here is another way to pursue this (just for fun!).
Recall that for $|x| < 1$ we have:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
Multiplying both sides by $x^6$ yields:
$$\frac{x^6}{1-x} = x^6 + x^7 + x^8 + x^9 + \cdots$$
Now we subtract the former line from the latter:
$$\frac{x^6 - 1}{1-x} = -(1 + x + x^2 + x^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 1
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Formula for how many combinations of powers of 2 sum to $2^n$ Given a number $2^n, n\in\mathbb{Z}\gt 0$, I would like to find a formula for how many unique sets of powers of $2$ sum to that number. This is related to the triangular numbers but excludes non-power-of-$2$ terms.
For example,
$$\begin{align}2^3
&= 1 + 1 +... | Define $g(m)$ to be the number of ways of writing $m$ as sums of powers. Then $f(n)=g(2^n)$. The sequence $g$ is OEIS sequence A018819.
There's a little more at the OEIS page, but it doesn't list a closed form.
The generating function for $g$ is:
$$G(x)=\sum_{m=0}^\infty g(m)x^m = \prod_{k=0}^\infty \frac{1}{1-x^{2^k}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Let $p$ be prime not equal to 2 or 5. Show $p^2+1$ or $p^2-1$ is divisible by 10. I can do half the proof but can not think of a way to finish. If $p$ is prime then both $p-1$ and $p+1$ are even. In cases where either $p-1$ or $p+1$ is divisible by 5 that implies $(p-1)(p+1)=p^2-1=10n$ for some positive integer $n$. I ... | Since $p \neq 5$ and $p$ is prime, $p$ cannot be divisible by $5$, so $p \neq 0 \mod 5$. Now,
$1^2 = 1 \mod 5$
$2^2 = 4 \mod 5$
$3^2 = 4 \mod 5$
$4^2 = 1 \mod 5$
That is, $p^2 \mod 5$ can be either $1$ or $4$. If $p^2=1\mod 5$, then $p^2-1=0\mod 5$. If $p^2=4\mod 5$,then $p^2+1=0\mod 5$. That is, either $p^2-1$ or $p^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
If $|ax^2+bx+c|\le 1\ \forall |x|\le 1$, then what is the maximum possible value of $\frac 83a^2+2b^2$? Let $f(x) = ax^2 + bx + c$ ; $a,b,c\in\mathbb R$
It is given that $|f(x)| \le 1$ $\forall |x| \le 1$
Q1) The possible value of $|a+c|$, if $\displaystyle \frac{8}{3} a^2 + 2b^2$ is maximum, is given by:
a) $0$
b) $1... | I'm going to write an answer because the given answer seems to have some errors.
First of all, noting that $a,b,c$ can be written as
$$a=\frac 12\left(f(-1)+f(1)-2f(0)\right),\quad b=\frac 12\left(f(1)-f(-1)\right),\quad c=f(0)$$
might make things easy.
Let us consider first Q3).
$$\begin{align}\frac 83a^2+2b^2&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do I integrate $x^5(4+x^2)^{-1/2}$? I just started learning substitution and I can't seem to solve this exercise.
I'm using $x = 2sinh(t)$
Full Solution:
$32\int(sinh(t)*(cosh^2(t)-1)^2=$
$32\int(sinh(t)*(cosh^4(t)-2cosh^2(t)+1)=$
$32\int(cosh^4(t)sinh(t)-2cosh^2(t)sinh(t)+sinh(t))=$
$32(\int(cosh^4(arcsinx)sinh(ar... | Notice,
let, $x=2\tan \theta\ \implies dx=2\sec^2\theta \ d\theta $
$$\int \frac{x^5}{\sqrt{4+x^2}}\ dx$$$$=\int \frac{(2\tan\theta)^5}{\sqrt{4+4\tan^2\theta}}(2\sec^2\theta \ d\theta)$$
$$=32\int \frac{\tan^5\theta \sec^2\theta}{|\sec\theta|}\ d\theta$$
Assuming $0<\theta<\pi/2$
$$=32\int \frac{\tan^5\theta \sec^2\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Probability of hitting a target number on x number of dice with y number of sides I have a complicated probability question. I want to know how to calculate the probability of reaching a target number (or greater) on a variable number of dice with each die having a variable number of sides. For example, the chance of r... | One way is to use generating functions.
If you have $n$ dice, with sides $a_1,\dots,a_n$, then the coefficient on $x^k$ of the polynomial
$$
\prod_{i=1}^n \left( \frac{1}{a_i} \sum_{j=1}^{a_i} x^i \right)
$$
yields the probability of a sum of $k$ with these dice.
Summing the appropriate values will give you the proba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}$ I have to prove that if a function $f$ is differentiable on $(a,b)$, then
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}
\end{align*}
Using the fact that $f'(x) = \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{... | \begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}
\end{align*}
but also
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x)-f(x-h)}{h}
\end{align*}
sum them up and divide by 2 to get
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0}\frac{ \dfrac{f(x+h)-f(x)}{h} + \dfrac{f(x)-f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Show difference between values of turning points of $ f(x) = (c-\frac{1}{c}-x)(4-3x^2) $ How would you show that the difference between the values of the turning points of
$$
f(x) = (c-\frac{1}{c}-x)(4-3x^2)
$$
is
$$
\frac{4}{9}(c+\frac{1}{c})^3
$$
$c>0$
I have attempted to compute the derivative and set it to $0$ (wh... | Starting with
$$f(x) = (c-\frac{1}{c}-x)(4-3x^2)$$
to make computation a bit more manageable, set $\boxed{\gamma=c-\frac{1}{c}}$, so
$$f(x)=4\gamma-4x-3\gamma x^2+3x^3$$
You had the correct idea to set $f'(x)=0$ at the turning points. So
$$f'(x)=-4-6\gamma x+9x^2=0\quad(\text{at turning points})$$
By the quadratic form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think i... | Here's my attempt at a solution using complex exponentials...
$$\begin{align*}
\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}} &=
\sqrt[6]{\frac{\sqrt{2}+\left(e^{i3\pi/4}\right)^7}{\left(e^{i\pi/4}\right)^{11}}}
\\
& = \sqrt[6]{\frac{\sqrt{2}+e^{i2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(1 + \sqrt2)^{2n} + (1 - \sqrt{2})^{2n}$ is an even integer. Prove that $(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n}$ is an even integer.
I'm not sure how to prove that it is an even integer. What would I do for the Inductive Step? And for the basic step, can I plug in zero and prove something from that?
| Note that
$(1+\sqrt2)(1-\sqrt2)
=-1
$
and
$(1+\sqrt2)^2
=3+2\sqrt{2}
$.
Therefore,
if $a
=3+2\sqrt{2}
$,
then
$1/a = 3-2\sqrt{2}
$,
so that
$a+1/a
=6
$
and
$(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n}
=a^n+1/a^n
$.
We now use the identity
true for any $a$
that
$a^{n+1}+1/a^{n+1}
=(a+1/a)(a^n+1/a^n)-(a^{n-1}+a^{n-1})
$.
Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$
| The first step is dividing $x^6$ by $x^2$ to get $x^4$.
Then you subtract $x^6-x^4(x^2-4)=4x^4$ (as you also got). But you need to continue.
Second step is to divide $4x^4$ by $x^2$ to get $4x^2$.
Then you subtract $4x^4-4x^2(x^2-4)=16x^2$. Again you need to continue and dividing $16x^2$ by $x^2$ to get $16$.
Then you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $x^2-3x+9=0$, can we say $(x+3)(x^2-3x+9)$ is also $0$ hence $x^3=-27$? I was solving this question
If $\dfrac x3 + \dfrac 3x = 1$ then find the value of $x^3$.
I solved it as. Cube both sides and substitute $x^3$ with $t$,
$$ \dfrac{t}{27} + \dfrac{27}{t}=-2$$
$$\implies t^2 + {27}^2 + 2\cdot 27 \cdot t = 0$$
$... | Yes there is nothing as such wrong with this process since it really helps in identifying the roots.
BUT one thing to keep in mind is that when you multiply the expression with $(x+3)$, you are introducing extraneous roots. That is, you had a quadratic equation with 2 roots and you have now made it cubic with 3 roots. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Am I correctly finding the standard matrix?
Question:
Let $F: \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying
$F(1,0,1)=(-3,-3,1)$,$F(0,1,0)=(0,1,1)$, and $F(0,1,1)=(2,-2,1)$. Find
the standard matrix $A$ of $F$.
My Approach:
I used a method that I haven't been taught, so I'm not sure if I am corre... | Your matrix does not map $(0,1,0)^t$ to $(0,1,1)^t$, so I believe it is wrong.
Instead you could use
$$
A X = Y \iff \\
A = Y X^{-1}
$$
where $X$ has the given argument vectors and $Y$ the corresponding image vectors, assuming the argument vectors are linear independent.
Applying the above I get
$$
X =
\left(
\begin{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~... | Here is one suggestion. Let $a=x+y, b=xy$ then $a+b=44, ab=448$
This gives $a=16, b=28, x^2+y^2=a^2-2b=200$ or $a=28, b=16, x^2+y^2= 752$
The intermediate step is to formulate and solve the quadratic satisfied by $a$ and $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Which natural numbers satisfy $2^n > n^2$?
Which natural numbers satisfy $2^n > n^2$ ?
My work. Step 1: $n = 1 $, $2^1 > 1^2$. True.
For $n = k$, $2^k > k^2$. For $n = k+1$,
$$ 2^{(k+1)} > (k+1)^2 \\
2\cdot 2^k>k^2+2k+1 \\
2^k+2^k > k^2+2k+1$$
*
*$2^k > k^2 \text{ - from step 1}$
*$2^k > k^2+2k+1$
How I can find... | For $n = 1$ is true, but for $n = 2,3,4$ is not. but for $n \geq 5$ is true.
Hence, the solution set is $\{1\}\cup[5,\infty)$.
Now we assume is true for $n = k$, that is
$2^k > k^2$, we will have to show that it is true for $n = k+1$.
$2^{k+1} = 2(2^k)$, but $2^k > k^2$ by induction hypothesis then $2^{k+1} > 2k^2 > ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Mathematical induction $\sum_{i=0}^{n}{n \choose i} = 2^n $ Prove by mathematical induction:
$\sum_{i=0}^{n}{n \choose i} = 2^n ; n \ge 0$
Step 1: n = 0
${0 \choose 0}=2^0$
Step 2:
for n = k
$\sum_{i=0}^{k}{k \choose i} = 2^k$
assumption: for n = k+1
$\sum_{i=0}^{k+1}{k+1 \choose i} = 2^{k+1}$
Step 3:
$\sum_{i=0}^{k+... | Your base case is essentially correct, although I would prefer to see the justification that both sides are equal to $1$. In step 2, your assumption should be $$\sum_{i = 0}^{k} \binom{k}{i} = 2^k$$ since you must show that $P(k + 1)$ holds whenever $P(k)$ holds. In step 3, the statement
$$\sum_{i = 0}^{k + 1} \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this limit without using L'Hospital's Rule? $$
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{|\arctan \frac{2}{x}|}
$$
Can anybody help me to solve this one ?
I ve done somethig like this but im not sure if it is the correct aproach.
$$
\lim\limits_{x\to{\infty}}\frac{\arctan\frac{3}{x}}{|\arctan ... | Your approach is correct assuming you already know $\lim \frac{\sin x}{x}=1$. Make sure you justify your steps: you can only separate the big limit in a product/quotient of separate limits assuming those exist (which you prove at the end), and you can say that there exists a $y\in(0,\frac \pi2)$ such that $\tan y = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A field with an element of order $12$: $(a+a^{-1})^2=3.$
Let $K$ be a field and $a\in K^*$ of order $12.$ I need to prove that $(a+a^{-1})^2=3.$
'Progress': I expand $(a+a^{-1})^2=a^2+a^{-2}+2$ so equivalently I need to prove that $a^2+a^{-2}=1$, or the inverse of $a^2$ is $1+a^{-4}.$ I tried also Binomial theorem a... | We have $(a^6-1)(a^6+1)=0$. But $a^6\ne 1$, so $a^6+1=0$. Thus $(a^2+1)(a^4-a^2+1)=0$. Because the order of $a$ does not divide $4$, we have $a^4-a^2+1=0$. This is equivalent to what you want to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $3^n > n^2$ by induction Prove that $$3^n > n^2$$
I am using induction and I understand that when $n=1$ it is true. The induction hypothesis is when $n=k$ so $3^k>k^2$. So for the induction step we have $n=k+1$ so
$3^{k+1} > (k+1)^2$ which is equal to $3\cdot3^k > k^2+2k+1$. I know you multiple both sides of the... | You start with knowing that $3^k > k^2$ and you prove that if you know that then you prove that $3^{k+1} > (k+1)^2$.
One straightforward and Doy! way is to note:
$3^k > k^2$
$3^k + 3^k + 3^k> k^2 + k^2+k^2$.
Since $k \ge 2$ then $k^2 \ge 2k$ and as $k \ge 2$ then $k^2 > 1$
So
$3^k + 3^k + 3^k > k^2 + 2k + 1$
$3^{k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Conditions for symmetry for pair of commuting matrices Let A = $\left(\begin{array}{ccc}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2
\end{array} \right)$
Show that every real matrix $B$ such that $AB = BA$ has the form $B = aI + bA + cA^2$
My attempt: If we assume that $AB = BA$ are simultaneously diagonalizable by $P$,... | We can prove the statement by brute force.
Search a matrix $B$ the commute with $A$:
$$
\begin{bmatrix}
2&-1&0\\
-1&2&-1\\
0&-1&2
\end{bmatrix}
\begin{bmatrix}
x_1&y_1&z_1\\
x_2&y_2&z_2\\
x_3&y_3&z_3
\end{bmatrix}=
\begin{bmatrix}
x_1&y_1&z_1\\
x_2&y_2&z_2\\
x_3&y_3&z_3
\end{bmatrix}
\begin{bmatrix}
2&-1&0\\
-1&2&-1\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
| Another ad hoc solution is the following:
$$
32 = \Big((x+1)-(x-1)\Big)^5 \\
= (x+1)^5 -5(x+1)^4(x-1) +10(x+1)^3(x-1)^2 \qquad \\ \qquad -10(x+1)^2(x-1)^3 +5(x+1)(x-1)^4
-(x-1)^5 \\
= (x+1)^3\Big((x+1)^2-5(x+1)(x-1)+10(x-1)^2\Big) \qquad \\ \qquad
-(x-1)^3\Big(10(x+1)^2-5(x+1)(x-1)+(x-1)^2\Big)
$$
so
$$
f(x) =
-\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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Is the numerator of $\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}$ a power of $2$? I stumbled on something numerically, and was just starting to work on it, but it seemed fun enough to share.
Let $$f(n)=\sum_{k=0}^{n} \frac{(-1)^{k}}{2k+1}\binom{n}{k}$$
It appears, from the first few values, that $f(n)$ always has num... | $$\begin{eqnarray*} f(n) = \sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k}&=&\int_{0}^{1}\sum_{k=0}^{n}(-1)^k x^{2k}\binom{n}{k}\,dx\\&=&\int_{0}^{1}(1-x^2)^n\,dx\\&=&\frac{1}{2}\int_{0}^{1}z^{-1/2}(1-z)^{n}\,dz\\&=&\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2\,\Gamma\left(n+\frac{3}{2}\right)}\\&=&\frac{4^n\,n!^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
} |
convergence of series $ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$ $$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$$
determine the convergence of the series
$$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\} < {n}^{p} \cdot \frac{1}{\sqrt{n-1}\s... | $$\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n-1}}=\frac{1}{\sqrt{n^2-n}(\sqrt{n}+\sqrt{n-1})}=\frac{1}{n^{3/2}}\frac{1}{\sqrt{1-\frac{1}{n}}\left(1+\sqrt{1-\frac{1}{n}}\right)}\approx \frac{1}{2n^{3/2}}$$
So use the limit comparison test between $n^p\left(\frac{1}{\sqrt{n-1}}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Question on inequality Let $a,b,c$ be real numbers such that $a+b+c = 0$ and $a^2 + b^2 + c^2 = 1$. Prove that $$a^2b^2c^2\le \frac1{54}$$
| By calculation $$ab+bc+ca=-1/2$$ and
$$1/4=(ab+bc+ca)^2=2abc(a+b+c)+a^2 b^2+b^2 c^2+c^2 a^2=a^2 b^2+b^2 c^2+c^2 a^2$$.
Let $K=a^2b^2c^2$, then $a^2,b^2,c^2$ are the solutions of
$$t^3-t^2+\frac{1}{4}t-K=0$$
Thus
$$K=t^3-t^2+\frac{1}{4}t$$
By taking derivative, we know the maximal value $K=1/54$ is achieved at $t=1/6$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Spectral Decomposition for matrices with eigenvalue(s) 0 When expressing the spectral decomposition of a matrix that has eigenvalues of 0, do you express the corresponding matrices with factor '0', or leave them out altogether?
EDIT: let's consider some matrix M, where
$$
M =
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1... | What your book calls spectral decomposition is just the equation you obtain after carrying out the matrix product in what is usually called a spectral decomposition. This seems very odd to me, because usually the term decomposition is used to refer as a way to write a matrix as a product of other matrices, not as a sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_\limits{x\to 1 }\bigl( (2^x x + 1)/(3^x x)\bigr)^{\tan(\pi x/2)}$ I have to calculate limit
$$\lim_{x\to 1 } \left(\frac{2^x x + 1}{3^x x}\right)^{\tan(\frac{\pi x}{2})}.$$
I know $\tan(\frac{\pi x}{2})$ is undefined in $x = 1$, but can I just put $x = 1$ into $\frac{x\cdot 2^x + 1}{x\cdot3^x}$ and ge... | Let $$y = \left( \frac{x 2^x + 1}{x 3^x}\right)^{\tan\left(\frac{\pi x}{2}\right)}$$
Then $$\ln(y) = \tan\left(\frac{\pi x}{2}\right)\ln\left( \frac{x 2^x + 1}{x 3^x}\right) = \frac{\ln\left( \frac{x 2^x + 1}{x 3^x}\right)}{\frac{1}{\tan\left(\frac{\pi x}{2}\right)}}$$
Apply L'Hôpital's rule to obtain your answer, it i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Least squares and QR factorization I have a full-column-rank matrix $A \in \mathbb{R}^{N \times n} $ ($N >> n$):
$Q^{T} A = \begin{bmatrix}
R & w \\
0 & v \\
\end{bmatrix} , Q^{T} = \begin{bmatrix}
c \\
d \\
\end{bmatrix} $,
with $R \in \mathbb{R}^{(n-1)\times(n-1)}, w \in \mathbb{R}^{n-1}, v \in \mathbb{R}^{N-n+1}, c ... | $x \in R^n$ must be true, otherwise the matrices cannot be subtracted form each other.
\begin{equation}
\begin{split}
\displaystyle \min_{x} ||Ax - b||_2^2 = \displaystyle \min_{x} ||Q^T(Ax - b)||_2^2 \\
= \displaystyle \min_{x} ||Q^TAx - Q^Tb||_2^2 \\
= \displaystyle \min_{x} || \begin{pmatrix} Rx & wx \\ 0 & vx \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)... | You can simplify this by noting $$x^2+\frac1{x^2}=x^2+x^{-2}=e^{2\ln(x)}+e^{-2\ln(x)}=\cos(2i\ln(x))$$
Now you can solve for $x$ very easily.
$$x^2+\frac1{x^2}=y$$$$\cos(2i\ln(x))=y$$$$x=e^{\frac{\arccos(y)}{2i}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Mathematics Olympiad Question $a+b+c=7$, ... Given $a+b+c=7$ and $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} = 0.7$, need to find $\frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{a+c}$.
I have noted that these two differ by a factor of $10$. So I divided the first equation by $10$ and equated the two. But that did not lead... | $$\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}$$
$$=\frac{7-(a+b)}{a+b}+\frac{7-(b+c)}{b+c}+\frac{7-(a+c)}{a+c}$$
$$=\frac{7}{a+b}+\frac{7}{b+c}+\frac{7}{a+c}-3$$
$$=7\cdot 0.7-3=1.9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
put the following in standard form how do I compute the following by putting it in standad form?
\begin{equation}
(2-2\sqrt{3i})^{20}
\end{equation}
what I have tried to do is use de Moivre theorem
but that requires me to put it in polar form, as in z=...
but I don't know how to do that either.
| Let $2 - 2\sqrt 3 i = 4(\frac 1 2 - \frac {\sqrt 3} 2 i)$.
Now a neat thing happens when we raise $(\frac 1 2 - \frac {\sqrt 3} 2 i)$ to powers.
$(\frac 1 2 - \frac {\sqrt 3} 2 i)^2 = \frac 1 4 - \frac 3 4 - \frac {\sqrt 3} 2 i = - \frac 1 2 - \frac {\sqrt 3} 2 i$
So $(\frac 1 2 - \frac {\sqrt 3} 2 i)^3 = (- \frac 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
quadratic equation problem - proving a statement I was given that $ax^2+2bx+c=0$
Using $y=x+\frac{1}{x}$ I need to prove that $acy^2+2b(c+a)y+(a-c)^2+4b^2=0$
Tried to make the pattern $x+\frac{1}{x}$ and to substitute $y$, but couldn't prove it.
| Given that $y = x + \frac1x,$ if you make this substitution for $y$ in
$acy^2+2b(c+a)y+(a-c)^2+4b^2,$ you get
$$ac\left(x + \frac1x\right)^2
+ 2b(c+a)\left(x + \frac1x\right) + (a-c)^2 + 4b^2.$$
Use the binomial theorem, multiplication of polynomials, the distributive law,
or any other methods you know in order to con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Summation $1-4x+9x^2-16x^3 + \cdots ?$ Taylor series gives $$\frac 1 {(1+x)^2}=1-2x+3x^2-4x^3+\cdots$$
is there a nice expression for $1-4x+9x^2-16x^3 + \cdots ?$
It would be helpful for a problem I am trying to solve.
| For $|x| < 1$, we have
$$\begin{align}
\sum_{n=0}^\infty (n+1)^2 (-x)^n
&= \sum_{n=0}^\infty \left(x\frac{d}{dx} + 1 \right)^2 (-x)^n
= \left(x\frac{d}{dx} + 1 \right)^2 \sum_{n=0}^\infty (-x)^n\\
&= \left(x\frac{d}{dx} + 1 \right)^2 \frac{1}{1+x}
= \frac{d}{dx}\left[ x \frac{d}{dx}\left(\frac{x}{1+x}\right)\right]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Write $1, 2, \dots, n^2$ into a $n \times n$ square grid such that sum of each row and column is a power of 2 Let $n$ be a positive integer. Show that one cannot fill a $n \times n$ square grid with numbers $1, 2, \dots, n^2$ such that sum of numbers on each row and column is a power of 2.
My attempt: Assume the contra... | The sum is $\frac{n^2(n^2+1)}{2}$. For $n\gt 1$, $n$ odd is impossible.
Let $2^k$ be the highest power of $2$ that divides $n$. Then the highest power of $2$ that divides our sum is $2^{2k-1}$, which is $\le \frac{n^2}{2}$.
Consider the smallest row or column sum. Say it is a row sum, and is equal to $2^l$. Then al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Exponential generating function and number of balls
Use exponential generating functions to determine the number $a_n$ of ordered choices of $n$ balls such that there are $2$ or $4$ red balls, an even number of green balls, and an arbitrary number of blue balls.
If we denote red with $r$, green with $g$, and blue wit... | Exponential generating functions are useful for this sort of problem because we can find the answer by multiplying the EGF for each of the types of balls:
$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$
This expands to
$$\frac12\cdot\frac{x^2}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
| For any $a,b > 0$ and as $x \to 0$:
$\def\l{\!\left}$
$\def\r{\right}$
$a^x = \exp(x\ln(a)) \in 1 + x\ln(a) + \frac{1}{2}(x\ln(a))^2 + o(x^2)$.
$b^x = \exp(x\ln(b)) \in 1 + x\ln(b) + \frac{1}{2}(x\ln(b))^2 + o(x^2)$.
$\l( \dfrac{a^x - x\ln(a)}{b^x - x\ln(b)} \r)^\dfrac{1}{x^2} = \exp\l( \dfrac{1}{x^2}\ln\l( \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to solve the following pde? How to solve the following PDE?
For an arbitrary continuously differentiable function $f$ , which of the following is a general solution of $\;$
$z(px-qy)=y^2-x^2$?
1)$\;$ $x^2+y^2+z^2=f(xy)$
2)$\;$ $(x+y)^2+z^2=f(xy)$
3)$\;$ $x^2+y^2+z^2=f(y-x)$
4)$\;$ $x^2+y^2+z^2=f((x+y)^2+z^2)$
Her... | $z(xz_x-yz_y)=y^2-x^2$
$xz_x-yz_y=\dfrac{y^2-x^2}{z}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$
$\dfrac{dz}{dt}=\dfrac{y^2-x^2}{z}=\dfrac{y_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove congruence rule Say you have integers $a$ and $b$
If $a \equiv b \pmod 5$, we know that $a \pmod 5 = b \pmod 5$.
Let $a \pmod 5 = c$ and $x,y$ be some whole numbers satisfying
$$a = 5x + c \quad \text{and} \quad b = 5y + c$$
Then, $a-b = 5x + c -(5y + c) = 5(x-y)$, so $5$ is a factor in $a-b$.
Now my question is,... | You correctly showed that $a\bmod 5=b\bmod 5\implies 5\mid a-b$ and now want to prove $5\mid a-b\implies a\bmod 5=b\bmod 5$.
Let $a=5x+(a\bmod 5)$, $b=5y+(b\bmod 5)$. If $5\mid a-b$, then $$5\mid 5(x-y)+(a\bmod 5-b\bmod 5)\iff 5\mid a\bmod 5-b\bmod 5$$
But $a\bmod 5,b\bmod 5\in\{0,1,2,3,4\}$, so $a\bmod 5=b\bmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integral of $\frac{1}{x^2+4}$ Different approach underneath is a brief method of partial fractions integration on the problem given in the title
Using a standard trigonometric result it is known that:
$$ \int \frac{1}{x^2+4}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})+C$$
But also:
$$\frac{1}{x^2+4}=\frac{A}{x+2i}+\frac{B}{x-2... | For those kinds of integrals, for how many ideas one might have, the simplest way to proceed is to collect first the constant you have in the denominator:
$$\frac{1}{x^2 + 4} = \frac{1}{4\cdot \left(\frac{x^2}{4} + 1\right)}$$
Remember about the $\frac{1}{4}$ factor, that you bring out of the integral.
Now use the subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How do I choose a free variable? I have a question regarding the Gaussian method for solving linear equations. I had to solve 2 equations with 3 unknowns and naturally with the elimination process I had 2 variables left. I thought that it didn't matter which variable I chose as a free variable, but apparently after inp... | As noted by Bye_World in the comments, in Reduced Row Echelon Form, the matrix specified becomes:
\begin{pmatrix}
1 & 0 & \frac{13}{14} & | & \frac{1}{7} \\
0 & 1 & -\frac{5}{14} & | & -\frac{9}{7}
\end{pmatrix}
You then note that you can use $x_3$ as the free variable. So you get:
$$x_1 = -\frac{13}{14}x_3 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finite Element Method for the 1d wave equation I'm solving the 1D wave equation
\begin{equation}
\frac{\partial^2 \eta}{\partial t ^2} - \frac{\partial^2 \eta}{\partial x ^2} = 0
\end{equation}
with boundary conditions
\begin{equation}
\frac{\partial \eta}{\partial x} = 0 \qquad \qquad \text{on} \qquad \qquad x = 0, 1... | I think I found a way...
Rearranging the Crank-Nicolson formulation as follows:
\begin{align*}
\eta_j^{n+1} - \frac{\Delta t }{2} p_j^{n+1} &= \eta_j^{n} + \frac{\Delta t }{2} p_j^{n} \\
M_{ij} p_j^{n+1} +\frac{\Delta t}{2} S_{ij} \eta_j^{n+1} &= M_{ij} p_j^{n} -\frac{\Delta t}{2} S_{ij} \eta_j^{n}
\end{align*}
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$ Problem:
If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$
My work:
$a^3b-ab^3=ab(a+b) (a-b)... | Let $p,q$ be remainders when $5$ divides $a,b$ respectively. If any of the following holds:
*
*$p=0$
*$q=0$
*$p=q$
*$p+q = 5$
Then, $5|a$, $5|b$, $5|(a-b)$ or $5|(a+b)$ respectively.
So, now check for $(p,q)=$
*
*$(1,2)$ Now, if $c = 5k+1$, then $5|(a-c)$, so repeat the question taking the two numbers $a,c$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Intersection of three period functions Let $f(x)=\frac{1}{2}-|\frac{\sqrt{3}}{2}x-1/2|$ for $x\in [0,\frac{2}{\sqrt{3}}]$ and $f(x+\frac{2}{\sqrt{3}})=f(x)$ for all $x\in\mathbb{R}$.
$g(x)=\frac{1}{2}-|\frac{1}{2}x-1/2|$ for $x\in [0,2]$ and $g(x+2)=g(x)$ for all $x\in\mathbb{R}$.
$h(x)=\frac{1}{2}-|\sqrt{2}x-1/2|$ for... | Let's assume there exists such $t>0$. Then for some $p,q,r\in\mathbb N$,
$$t-\frac2{\sqrt3} p\in\left[0,\frac2{\sqrt3}\right],\quad t-2q\in\left[0,2\right],\quad t-\frac1{\sqrt2}q\in\left[0,\frac1{\sqrt2}\right]$$
then
$$f\left(t-\frac2{\sqrt3} p\right)=g\left(t-2q\right)=h\left(t-\frac1{\sqrt2}r\right)$$
And
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How can we evaluate the following limit?
How can this problem be solved?
$$
\lim_{(n,r) \rightarrow (\infty, \infty)} \frac{\prod\limits_{k=1}^{r} \left( \sum\limits_{i=1}^{n} i^{2k-1} \right)}{n^{r+1} \prod\limits_{k=1}^{r-1} \left( \sum\limits_{i=1}^{n} i^{2k} \right)}
$$
| The sum of the $k$ powers of the first $n$ natural numbers is a polynomial of degree $k+1$ with leading coefficient $\frac{1}{k+1}$, that is,
$$
\sum_{i=1}^n i^k
=
\frac{1}{k+1}n^{k+1} + p_k(n)
$$
where $p_k(n) \in O(n^k)$ is a polynomial of degree (at most) $k$.
It follows that
\begin{align}
\frac{
\prod\limits_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding a diagonalizing matrix associated with Jordan normal form Find the Jordan normal form $J$ of the upper triangular matrix
$$A = \begin{pmatrix}2 & 0 & 1 & 2 \\ 0 & 2 & 2 & 1 \\ 0 & 0 & 2 & 1\\ 0 &0 & 0& 3 \end{pmatrix}$$
and find a matrix $M$ such that $M^{-1}AM = J$.
Note that the characteristic polynomial is $... | You have found that there are two Jordan blocks for the eigenvalue $2$, so $\dim\ker(A-2I)=2$. There is a $1\times1$-block and a $2\times2$-block, which means that $\dim\ker(A-2I)^2=3$. Now pick a vector $v_3\in\ker(A-2I)^2$ that is not containd in $\ker(A-2I)$. Then $v_2=(A-2I)v_3$ is an eigenvector with eigenvalue $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Dividing the squares $1^2,2^2,\ldots,54^2$ into three equal groups with the same total sum Is it possible to divide the squares $1^2,2^2,\ldots,54^2$ into three groups, each of which contains $18$ squares, such that the sum of squares within each group is the same for all three groups?
| We can partition $9$ consecutive squares $n^2, (n+1)^2, \ldots,(n+8)^2$ in an almost equal fashion:
$$\begin{align}
(n+0)^2+(n+4)^2+(n+8)^2&=3n^2+24n+80\\
(n+1)^2+(n+5)^2+(n+6)^2&=3n^2+24n+62\\
(n+2)^2+(n+3)^2+(n+7)^2&=3n^2+24n+62\end{align} $$
Each group has three members and only one group has a sum too large by $18$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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exponential difference inequality Im asked to prove the inequality when $0\leq a<b$ and $x>0$:
$$
a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a)
$$
So far I have seen that obviously:
$$a^x(b-a)<b^x(b-a)$$
and that
$$b^{x+1}-a^{x+1} = (b-a)(b^x+b^{x-1}a+...+ba^{x-1}+a^x) > a^x(b-a)$$
This has done no good for me yet but ... | Take user math110's idea:
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x>a^x+a^x+...+a^x=(x+1)a^x$$
and
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x<b^x+b^x+\cdots+b^x=(x+1)b^x$$
From this and since $0\leq a<b$
$$
(x+1)a^x<(b^x+b^{x-1}a+...+ba^{x-1}+a^x)<(x+1)b^x
$$
Divide by $x+1$ and we get,
$$
a^x<{b^x+b^{x-1}a+...+ba^{x-1}+a^x\over{x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Is this a sufficient proof of a math contest problem? Problem:
If a,b,c,d are real, prove that
$$a^2+b^2=2$$
$$c^2+d^2=2$$
$$ac=bd$$
Is true if and only if
$$a^2+c^2=2$$
$$b^2+d^2=2$$
$$ab=cd$$
My proof is as follows:
Note that each of the second set of equations nearly corresponds to the lengths of the sides of a... | The difficulty with the geometric argument is that you cannot actually conclude that $b=c$ and $a=d$. For example, if the second set of equations is true, then you can have $$a = \sqrt{1/2}, c = \sqrt{3/2}, \quad b = -\sqrt{3/2}, d = -\sqrt{1/2},$$ which satisfies the conditions but it is clear that $b \ne c$. This i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
Partial integration can't ... | The integral $$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$$ converges.
Indeed $$\sum_{n=1}^\infty {\frac{1}{n\sqrt[3]{n^2+1}}} $$ converges. Indeed $$ {\frac{1}{n\sqrt[3]{n^2+1}}}\sim_{\infty} {\frac{1}{n\sqrt[3]{n^2}}} $$ that converges
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int 1/(x^3-25x) \mathrm{d}x$ Integrate $\displaystyle \frac{1}{x^3-25x}$
Between limits of $4,3$.
Think it involves the 'Difference in squares' substitution.
I ended up with an answer of $0.3242$ which is wrong as it is $0.023$.
Any help and pointers would be greatly appreciated.
Matt
| John's approach is correct.
$\frac{1}{x^3-25x} = \frac{1}{x(x+5)(x-5)} = \frac{A}{x} + \frac{B}{x+5} + \frac{C}{x-5}$.
Continuing, we generate the following equation: $$A(x+5)(x-5)+Bx(x-5)+Cx(x+5)=1$$
By strategically choosing our $x$ values (I recommend $x=-5,0,$ and $5$), we can derive that $A=\frac {-1}{25}, B= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Area bounded by $y=-x^2+4$, $y=x+2$ I set the two polynomials equal to each other and after multiplying everything by $-1$ I got $x^2+x-2=0$
My points of intersection are $x=1$ and $x=-2$
However, the graph of the polynomial is concave up. Why?
| Update: Setting your new equations equal yields $x^2+x-2=0$, which factors as $(x+2)(x-1)$. Now, your intersection points are correct. Also, $-x^2+4>x+2$ from $-2$ to $1$. Now, take
$$\int_{-2}^{1} -x^2-4-(x+2) \mathrm{d}x= \frac{-39}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\frac{1-2\cos(x)+\cos^2(2x)}{x^2}$ I tried to find the value of this limit without L'Hopital , but no luck
$$\lim_{x\to 0}\frac{1-2\cos(x)+\cos^2(2x)}{x^2}$$
| $$\frac{1-2\cos(x)+\cos^2(2x)}{x^2}=\frac{4\sin^2\frac{x}{2}-\sin^2(2x)}{x^2}$$
and since $\lim_{x\to 0}\frac{\sin x}{x}=1$, your limit equals $1-4=\color{red}{-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing the Integral $\int \frac{\sqrt{x}}{x^2+x} dx$
Find $\displaystyle \int \dfrac{\sqrt{x}}{x^2+x} dx$.
What would be the best way to integrate this? I saw the answer to this and it looked simple so that might mean the steps would be too?
| Method 1. Algebraic substitution:
let $\sqrt x=u\implies \frac{dx}{2\sqrt x}=du$ or $dx=2u\ du$, hence $$\int \frac{\sqrt x}{x^2+x}\ dx=\int \frac{u}{u^4+u^2}(2u\ du)=2\int \frac{1}{u^2+1}\ du=2\tan^{-1}(u)+C=2\tan^{-1}(\sqrt x)+C$$
Method 2. Trigonometric substitution:
let $\sqrt x=\tan\alpha\implies \frac{dx}{2\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$.
Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.
My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $... | Notice, $$\int \frac{1}{2\sin x-3\cos x}\ dx$$
$$=\int \frac{1}{2\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}-3\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}\ dx$$
$$=\int \frac{1+\tan^2\frac{x}{2}}{3\tan^2\frac{x}{2}+4\tan\frac{x}{2}-3}\ dx$$
$$=\frac13\int \frac{\sec^2\frac{x}{2}}{\tan^2\frac{x}{2}+\frac{4}{3}\tan\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I want to calculate the limit of $\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x)$ I want to calculate the limit of: $$\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x) $$ $$(a,b ∈ R)$$
Now I know the result is $\frac{1}{2}(a+b)$, but I am having trouble getting to it.
| Notice, $$\lim_{x\to \infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
$$=\lim_{x\to +\infty}\frac{\left(\sqrt{(x+a)(x+b)}-x\right)\left(\sqrt{(x+a)(x+b)}+x\right)}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$
$$=\lim_{x\to +\infty}\frac{x^2+(a+b)x+ab-x^2}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$
$$=\lim_{x\to +\infty}\frac{(a+b)+\frac{ab}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached:
| I prefer to use this substitution
$$\begin{align}
x &= 4 \sinh u \\
dx &= 4 \cosh u \, du
\end{align}$$
and hence the integral becomes
$$\begin{align}
I &= \int \frac{1}{\sqrt{16(1+\sinh^2u)}} 4 \cosh u du\\
&= \int \frac{4 \cosh u}{4\sqrt{\cosh^2u}} du \\
&= \int 1 \, du \\
&= u+C \\
&= \sinh^{-1} \frac{x}{4} + C\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Is there a name for a binomial expansion without coefficients? I am investigating a problem from George E. Andrews Number Theory (Dover, 1971), discussed previously here:
Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$
I was led astray for a bit, because I misunderstood the rhs to be the ... |
There is no specific name for the bivariate polynomial
\begin{align*}
x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1}
\end{align*}
but the nice expression
\begin{align*}
(x-&y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\
\end{align*}
is called a telescoping sum since all terms ... | {
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"url": "https://math.stackexchange.com/questions/1583607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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