Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate integral for $\int \sin^2 (x+\frac{\pi}{6}) dx$ Can someone walk me through how to evaluate the integral $$\int \sin^2 (x+\frac{\pi}{6}) dx?$$
I get as far as
$$\int \frac {1 - \cos(x + \frac{\pi}{6})}2dx,$$
but I am not sure how to proceed.
| Careful with formula
$$
\sin^2\left(t\right)=\frac{1-\cos\left(2t\right)}{2}
$$
Then using linearity
$$
\int \frac{1-\cos\left(2x+\frac{\pi}{3}\right)}{2}\text{d}x=\int \frac{1}{2}-\frac{1}{2}\int \cos\left(2x+\frac{\pi}{3}\right)
$$
Hence
$$
\int \frac{1-\displaystyle \cos\left(2x+\frac{\pi}{3}\right)}{2}\text{d}x=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown of... | From $10b^2=a^2+2ab$ you can see that $2\mid a^2$, hence $a=2m$ (for some $m$), from which you can conclude that $b^2=2m^2+2mb-4b^2$, which implies $2\mid b$, contradicting the (tacit) assumption $\gcd(a,b)=1$.
This is an interesting variant on the standard proof of irrationality in that it only invokes the implication... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Unclear calculus problem I don't know if that's called calculus or precalculus in the Anglosphere, so please pardon me if I got the terminology incorrect.
I've been given the second derivative of $f(x): f''\left(x\right)=\frac{2x-1}{\sqrt{-x^2+x+6}} $.
I was asked to find $f'(x)$, if it is known that the slopes of all ... | Just continue
So
For $-2 < x < 3$ we have $-2\sqrt{-x^2+x+6} + C \ge -1$
$2\sqrt{-x^2 + x + 6} \le 1 + C$
So $ C \ge \max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)} - 1$.
So must find $\max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)}$. As $-x^2 +x+6$ must be non-negative we may assume $\max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)}$ where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A tricky algebraic inequality This is an old inequality but I haven't seen a satisfactory solution yet and am hoping someone here can provide one. There are a couple of brute force solutions but they provide no insight into the inequality and I'd be surprised if there isn't a trick to it:
$$x\frac{(y+z)^2}{(1+yz)^2} + ... | We need to prove that
$$\sum_{cyc}\frac{x(y+z)^2}{(xy+xz+2yz)^2}\geq\frac{3}4\sqrt{\frac{3}{xy+xz+yz}}$$ or
$$\sum_{cyc}\frac{\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)^2}{\left(\frac{1}{xy}+\frac{1}{xz}+\frac{2}{yz}\right)^2}\geq\frac{3}{4}\sqrt{\frac{3}{\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}}}$$ or
$$\sum_{cyc}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for... | $x$ and $y$ need to have the same parity, so that we can predict that the probability will be close to $\frac12$.
For a given $z$, there are $11-z$ combinations of $x,y$, which are all of the same parity or of the opposite parity.
Hence
$$p=\frac{11+9+7+5+3+1}{11+9+7+5+3+1+10+8+6+4+2}=\frac{36}{66}.$$
Using summation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Express generic "penthagorean" triplet If asked to express the generic pythagorean triplets satisfying
$a^2+b^2=c^2$ you would answer
$$
a = k(r^2 - s^2), b = 2krs, c = k(r^2+s^2)
$$
with $k,r,s \in \Bbb N$ and $r>s$ and $r \ne s \mod 2$ and $\gcd(r,s) = 1$. This generates every pythagorean triplet exactly once as ... | from pages 124,125 of Magnus. If we have integers $a,b,c,d$ with $ad-bc=1$ and $a+b+c+d \equiv 0 \pmod 2,$ and we take
$$
\left(
\begin{array}{ccc}
\frac{1}{2} \left( a^2 + b^2 + c^2 + d^2 \right)&ab+cd&\frac{1}{2} \left( a^2 - b^2 + c^2 - d^2 \right) \\
ac+bd&ad+bc&ac-bd \\
\frac{1}{2} \left( a^2 + b^2 - c^2 - d^2 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do I find out the value of $p$ for $\sin p + \cos p = 0$? I tried doing this by putting $$\frac{ \sin p}{\sqrt2} +\frac{ \cos p}{\sqrt2} = 0 $$ which implies $$ \sin(p + π/4) = \sin 0 $$ which implies $p+ π/4 = nπ $. Now according to the question I'm solving $p = 2πt/T$. I need to get the relation $t = 3T/8$. How d... | $$\sin x + \cos x = 0$$
$$\Rightarrow \sin^2 x + 2 \sin x \cos x + \cos^2 x = 0$$
$$\Rightarrow \sin(2x) + 1 = 0$$
$$\Rightarrow \sin(2x) = -1$$
Since we know that $\sin x = -1$ when $x = \frac{3\pi}{2} + 2\pi n$, then $\sin(2x) = -1$ when $x = \frac{3\pi}{4} + \pi n $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Let $S = \{1,2,3,4,5,6\}$ be a sample space of equally likely outcomes Let $S = \{1,2,3,4,5,6\}$ be a sample space of equally likely outcomes such that $P(\{s\}) = \frac{1}{6}$, $\forall s \in S$ let X = $I_{(0,9,3,1)}, Y = I_{(2,4,4,2)}$ and $W = X + Y$
Compute $P(W = 1)$
My attempt:
$$W(1) = X(1) + Y(1) = I_{(0,9,3,1... | Not entirely:
For one you are correct in saying:
$$
W(1) = 1
$$
But notice also that
\begin{align}
W(2) &= X(2) +Y(2) = 0 + 1 = 1, \\
W(3) &= X(3) +Y(3) = 1 + 0 = 1, \\
W(4) &= X(4) +Y(4) = 0 + 1 = 1, \\
W(5) &= X(5) +Y(5) = 0 + 0 = 0, \\
W(6) &= X(6) +Y(6) = 0 + 0 = 0.
\end{align}
So we get that
\begin{align}
P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there any formula to find $\sin^{-1}a +\sin^{-1}b$ We known that$$\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right).$$
Now we derive the above formula. Let$$
\tan^{-1}a=\theta _1 \implies \tan\theta_1=a,\\
\tan^{-1}b=\theta _2 \implies \tan\theta_2=b,\\
\theta _1+\theta _2 = \tan^{-1}a+ \tan^{-1}b,\\
... | Assume that $\arcsin(a)+\arcsin(b)\in\left[-\frac\pi2,\frac\pi2\right]$. Then, since$$\sin\left(\arcsin(a)+\arcsin(b)\right)=a\sqrt{1-b^2}+b\sqrt{1-a^2},$$we have$$\arcsin(a)+\arcsin(b)=\arcsin\left(a\sqrt{1-b^2}+b\sqrt{1-a^2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral using the Substitution y = $\frac{1}{2}(\sqrt{a}x^2+\frac{\sqrt{b}}{x^2})$ Given that $\int_{-\infty}^{\infty} e^{(-\frac{1}{2}x^2)}dx = \sqrt{2\pi}$,
Find $I(a,b) = \int_{-\infty}^{\infty} e^{-\frac{1}{2}(ax^2+\frac{b}{x^2})}dx$.
The question suggests using the substitution: $$y = \frac{1}{2}(\sqrt{a}x-\frac{... | The key here is to note that this integral is improper at zero, so before performing the substitution notice the integrand is even so we may rewrite the integral as
$$I(a,b)=2\int_{0+}^\infty e^{-\frac{1}{2}(ax^2+\frac{b}{x^2})}dx$$
Let $y=\sqrt{a}x-\frac{\sqrt{b}}{x}$. Note the bounds become $-\infty$ to $\infty$, sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced ... | If $m=30a$ and $n=30b$, then
$${m\over n}={30a\over30b}\\{m+2\over n+2}={2(15a+2)\over2(15b+1)}\\{m+3\over n+3}={3(10a+1)\over3(10b+1)}\\{m+4\over n+4}={2(15a+2)\over2(15b+2)}\\\text{and}\\{m+5\over n+5}={5(6a+1)\over5(6b+1)}$$ are all reducible. It remains to choose $a$ and $b$ so that
$$30a+1\over30b+1$$
is reduci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Number of solutions to the congruence $x^2\equiv 121\pmod {1800}$ I'm trying to find the number of solutions to this congruence:
$$x^2\equiv 121\pmod {1800}$$
I thought about writing it as a system of congruences. As $1800=3^2 \cdot 5^2 \cdot 2^3$, we get:
$x^2\equiv 121\pmod {5^2} \;,\; x^2\equiv 121\pmod {3^2} \;,\;... | Hint:
We need $\left(\dfrac x{11}\right)^2\equiv1\pmod{3^2,5^2,2^3}$
Now we can prove $y^2\equiv1\pmod{p^n}$ has exactly two solutions for prime $p\ge3$ and integer $n\ge1$
Finally we can apply Chinese Remainder Theorem to find the number of in-congruent solutions to be $$4\cdot2^2$$
See also: Number of solutions of $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Convex sum order If I have a strictly convex function $f(x)$ with $f''(x)>0$
and if I know that for some $a\le b \le c$ and $x \le y \le z$ I have
$$a+b+c = x+y+z$$
$$f(a)+f(b)+f(c)=f(x)+f(y)+f(z)$$
can I conclude that at least one of the following must be true about the order?
$$ a \le x \le y \le b \le c \le z$$
$$ ... | For simplicity, let us assume that all the points are different, i.e., $a < b < c$, $x < y < z$ and the all the points $a,b,c$ are different from $x,y,z$.
(Maybe some arguments can be transferred to the general case).
Now, the order of the points can be described by a permutation of $abcxyz$ which preserves the order o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof... | This follows from the Cauchy–Schwarz inequality
$$
|\langle \mathbf{u},\mathbf{v}\rangle| ^2 \leq \langle \mathbf{u},\mathbf{u}\rangle \cdot \langle \mathbf{v},\mathbf{v}\rangle,
$$
with
$
\mathbf{u} = (1,1)$
and
$
\mathbf{v} = (a,b)
$
in $\mathbb R^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Spivak's Calculus: Finding the formula for the sum of a series. In Spivak's Calculus, Chapter 2, the second problem of the set asks you to find a formula for the following series:
$$\sum_{i=1}^n(2i-1)$$ and $$\sum_{i=1}^n(2i-1)^2$$
Now, for the former, this was fairly straightforward: it sums to $i^2$ and I have a proo... | $\sum_\limits{k=1}^n (2k - 1)^2 = \sum_\limits{k=1}^n 4k^2 - \sum_\limits{k=1}^n 4k + \sum_\limits{k=1}^n 1 = 4\left(\sum_\limits{k=1}^n k^2\right) - 2n(n+1) + n$
You need a way to find $\sum_\limits{k=1}^n k^2$
There are lots of other ways to derive this... but this one is nice.
Write the numbers in a triangle
$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Solution of the functional equation $f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right) $
If
$$f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$
prove that
$$f(4x^3 -3x) + 3f(x) =0\text.$$
I started by substituting $x = y$, in the expression and I get
$$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$... | I'll assume the arguments of $f$ are always between $-1$ and $1$. Then
$$2f(x)=f(2x\sqrt{1-x^2})$$
and
$$3f(x)=f(x)+f(2x\sqrt{1-x^2})=f(y)$$
where
$$
\begin{split}
y &= x\sqrt{1-4x^2(1-x^2)}+\sqrt{1-x^2}2x\sqrt{1-x^2}\\
&= x\sqrt{1-4x^2+4x^4}+2x\left(1-x^2\right)\\
&= x\left(1-2x^2\right)+2x\left(1-x^2\right)=3x-4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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integral of $ \int (x^3+1)^{1/3} / x^2 dx $ What do you think about
$$ \int \frac {(x^3+1)^{1/3}}{x^2}\, dx $$
how do you compute this ? Is it possible to use the Euler substitution?
In fact, I don't know if the integral on my sheet is $$ \int \frac {(x^3+1)^{1/3}} {x^2} \, dx $$
or
$$ \int \frac{x^2 + 1}{x \sqrt {... |
Hint : Divide numerator & denominator by $x$ & write the integrand
as $I= \int\frac {(1+\frac1 {x^3})^{1/3}}x dx $
The whole calculus :
\begin{align}
I =& \int \frac{(x^3 +1)^{1/3} }{x^2} dx \\
=& \int \frac{(1+ x^{-3})^{1/3} }{x} dx
\end{align}
let $y= x^{-3} \implies dy = \frac{-1}{3} x^{-4} dx $
$$ I =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding the value of a sum using Riemann sum theorem
Question: Find the value of $\sum_{i=1}^{n}(\frac{1}{n-i})^{c}$ for large $n$.
\begin{align}
\sum_{i=1}^{n}(\frac{1}{n-i})^{c}
& = \sum_{i=1}^{n}(\frac{1}{n})^{c}(\frac{1}{1-\frac{i}{n}})^{c}
\\ & = \frac{n}{n} \times \sum_{i=1}^{n}(\frac{1}{n})^{c}(\frac{1}{1-\f... | \begin{align}\label{eq:7777}
& \frac{2}{\sqrt{n-i} + \sqrt{n-i+1}} \leq \frac{1}{\sqrt{n-i}} \leq \frac{2}{\sqrt{n-i} + \sqrt{n-i-1}} \nonumber\\
& \qquad \Rightarrow 2(\sqrt{n-i+1} - \sqrt{n-i}) \leq \frac{1}{\sqrt{n-i}} \leq 2(\sqrt{n-i} - \sqrt{n-i-1}) \nonumber\\
& \qquad \Rightarrow 2 \sum_{i=1}^{n-1}(\sqrt{n-i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if ${x_1, x_2, x_3}$ are roots of ${x^3 + px + q = 0}$ then ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?
I've tried the following approach:
If $x_1$, $x_2$ and $x_3$ are roots then
$$(x-x_1)(x-x_2)(x-x... | Rewrite $x_1,x_2,x_3$ with $a,b,c$. From first Vieta formula we have $$a+b+c=0$$
so $a+b=-c$ and so on...
Now $$a^3+b^3+c^3= (a+b)(a^2-ab+b^2)+c^3 = c(\underbrace{-a^2+ab-b^2+c^2}_I)$$
Since
$$I = -a^2+ab-b^2+c^2 = a(b-a)+(c-b)(c+b) = $$ $$a(b-a)-a(c-b) = a(2b-a-c)=a(2b+b)=3ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Simplifying nested radicals with higher-order radicals I've seen that $$\sin1^{\circ}=\frac{1}{2i}\sqrt[3]{\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}+\frac{i}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}}-\frac{1}{2i}\sqrt[3]{\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}-\frac{i}{4}\sq... | $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$
$\begin{align*}\sin 1^\circ & = \sin\left(\dfrac{\pi}{180} \right) \\ & = \dfrac{e^{i\tfrac{\pi}{180}}-e^{-i\tfrac{\pi}{180}}}{2i} \\ & =\dfrac{\left(e^{i\tfrac{\pi}{6}}\right)^{1/30}-\left(e^{-i\tfrac{\pi}{6}}\right)^{1/30}}{2i} \\ & = \dfrac{\sqrt[30]{\cos \left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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I am stuck with some details when graphing this function
Let $f(x) = \frac{x^3 + 1}{x}. $ Plot the function using derivatives criteria.
First of all, $f$ is indeterminate at $x=0.$ So I would have to deal with 2 intervals, separately: $(- \infty, 0)$, and $(0, \infty)$.
I would also have to consider:
$\lim_{\to \inft... | A qualitative approach.
The function can be written as $x^2+\dfrac1x$, and is the sum of a parabola and an equilateral hyperbola.
The hyperbola has a vertical asymptote at $x=0$ and the horizontal asymptote $y=0$. Hence for small $|x|$ it is dominant, and for large $|x|$, it is neglectible and the function gets closer ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1... | $ \frac{d}{dx}(x^3+2x^2+x)^4 = 4(x^3+2x^2+x)^3\cdot(\frac{d}{dx}(x^3+2x^2+x))$ by the chain rule, then we get: $ \frac{d}{dx}(x^3+2x^2+x)^4 = 4(x^3+2x^2+x)^3\cdot(3x^2+4x+1) = 4x^3(x^2+2x+1)^3\cdot(3x^2+4x+1) $.
Now the rest should just be algebra: (note that $(x+1)^2=(x^2+2x+1)$)
$ 4x^3((x+1)^2)^3(3x^2+4x+1) = 4x^3(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Intersection spheroid -plane I have a spheroid $S$ with $a$ is the equatorial radius, and $b$ is the polar radius, and ($a>b$)
I would get the intersection between $S$ and a plane $P: ux+vy+wz+d=0$
Then calculating the semi-axis major and minor and the center of this intersection.
There is always an intersection (resu... | Equation of spheroid:
$$\frac{x^2+y^2}{a^2}+\frac{z^2}{b^2}=1 \tag{1}$$
Let $(X,Y,Z)$ be the centre of the section, then equation of plane section is
$$\frac{Xx+Yy}{a^2}+\frac{Zz}{b^2}=\frac{X^2+Y^2}{a^2}+\frac{Z^2}{b^2} \tag{2}$$
Comparing coefficients,
$$-\frac{(u,v,w)}{d}=
\frac{\left( \dfrac{X}{a^2}, \dfrac{Y}{a^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ where $x,y\in [0,\pi]$.
I am trying to solve this but I am stuck. I know that $x=y=\pi/3$ is a solution but how do I show this is the only one? I think there are no others! Hints would be appreciated
| For $x,y \in [0,\pi]$
\begin{align}
\cos x &= u \\
\cos y &= v \\
\sin x &= \sqrt{1-u^2} \\
& \ge 0 \\
\sin y &=\sqrt{1-v^2} \\
& \ge 0 \\
u+v-uv+\sqrt{1-u^2} \sqrt{1-v^2} &= \frac{3}{2} \\
\sqrt{1-u^2} \sqrt{1-v^2} &= \frac{3}{2}+uv-u-v
\end{align}
By AM $\ge$ GM,
$$\frac{1-u^2+1-v^2}{2} \ge \frac{3}{2... | {
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"url": "https://math.stackexchange.com/questions/2827457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^... | Here is a way to prove it without induction. Notice that $$n^4-4n^2=n^2(n-2)(n+2)$$
If $3\not\mid n$ then $3\mid (n+1)-3$, that is $3\mid n-2$, or $3\mid (n+2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Choose $y\in\mathbb{R}$ so that $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. I know the identity $\sin^2y+\cos^2y=1$. Also, I notice that
$$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
without $u,v$ vanishing simultaneously. But I am not sure whether $\... | The identity $$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
means that the point $\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}}\right)$ is on the unit circle $x^2+y^2=1$. By definition, there exists an angle $y$ (actually, an infinity of angles) such that $\cos(y)=\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding the mean of all 9-digit numbers formed from four $4$s and five $5$s
I need to find the mean of the numbers ($9$-digit) formed of four $4$s and five $5$s .
MY WORK:
In order to find the sum I do the following and find sum of digits :
$$25\times5!=3000$$
$$16\times4!=384$$
So, sum of all possible numbers :
$$3... | For the reader with an interest in generating functions we have
from first principles the closed form
$${9\choose 4}^{-1}
\left. \frac{\partial}{\partial w} [z^4]
\prod_{q=0}^8 (z \times w^{4 \times 10^q} +
w^{5 \times 10^q}) \right|_{w=1.}$$
We may treat the derivative first and obtain
$${9\choose 4}^{-1}
\left. [z^4]... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Does hypergeometric distribution apply in this case? A box contains 4 red balls, 5 blue balls and 2 green balls.
Six balls are drawn without replacement, find the chance that exactly one of the first two is blue AND exactly two of the last four are blue.
The probability that I came up with is:
$\frac {{\binom {2}{1}}\b... |
A box contains 4 red balls, 5 blue balls and 2 green balls.
Six balls are drawn without replacement, find the chance that exactly one of the first two is blue AND exactly two of the last four are blue.
The total ways to draw groups of $2$ and $4$ balls from a heap of $4+5+2$ (ie $11$) is $\binom {11}2\binom{11-2}{4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$ I would like to evaluate this integral, $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}$$
An approach:
1. Using integration by parts
$u=\ln\left(\frac{a-x^2}{a+x^2}\right)$, $du=\frac{x... | Assuming $a>1$, the original integral equals
$$ \frac{1}{2}\int_{0}^{1}\log\left(\frac{a-x}{a+x}\right)\frac{dx}{x\sqrt{x(1-x)}}\,dx=-\int_{0}^{1}\frac{\text{arctanh}(x/a)}{x\sqrt{x(1-x)}}\,dx \tag{1}$$
or
$$ -\sum_{n\geq 0}\int_{0}^{1}\frac{x^{2n-1/2}}{(2n+1)a^{2n+1}\sqrt{1-x}}\,dx=-\sum_{n\geq 0}\frac{\pi\binom{4n}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is going wrong in this log expansion? I am getting a weird result here:
Let $p_1 = q_1 + \Delta$ and Let $p_2 = q_2 - \Delta$
I use the expansion $\log(1-x) = -x -x^2/2 -x^3/3 - ...$ in the third step. This expansion is valid for $x<1$ so there should be no problem with that.
\begin{align*}
D(\vec{p}||\vec{q})... | There are at least two errors:
$$p_1\log{\frac{q_1 + \Delta}{q_1}} = p_1\left(\frac{\Delta}{q_1} - \frac{\Delta^2}{\color{red}{2}q_1^2} - \frac{\Delta^3}{\color{red}{3}q_1^3} - \cdots\right)$$
Similar for the second term.
And even if the low order terms cancel, the result is not zero. Using Maple I get (if I did not ... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a... | You can square the given equation:
$$a_{1}+a_{2}+a_{3}+a_{4}=0 \Rightarrow \\
a_1^2+a_2^2+a_3^2+a_4^2+2(a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4)=0 \Rightarrow \\
2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})=-1-2a_1a_3-2a_2a_4.$$
Hence:
$$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2=\\
=2(a^2_{1}+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Need hint for equation system I stumbled upon the following in an old math book and could use a hint:
The sum of the volumes of two cubes is 35, whereas the sum of the
surface area of the two cubes is 78. Calculate the lengths of the
sides of the two cubes.
So, if we let $x$ be the length of the side of one cube,... | $35=x^3+y^3=(x+y)(x^2-xy+y^2)=$
$=(x+y)(13-xy)$
But
$(x+y)^2=x^2+y^2+2xy=13+2xy$ and so
$xy=\frac{1}{2}[(x+y)^2-13]$ and it can be replaced to first equation to get that
$35=(x+y)(13-\frac{1}{2}((x+y)^2-13))$
and if you define $z:=x+y$ you must resolve
$70=z(39-z^2)$
(You can observe that the solutions of the equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Does this pattern continue $\lfloor\sqrt{44}\rfloor=6, \lfloor\sqrt{4444}\rfloor=66,\dots$? By observing the following I have a feeling that the pattern continues.
$$\lfloor \sqrt{44} \rfloor=6$$
$$\lfloor \sqrt{4444} \rfloor=66$$
$$\lfloor \sqrt{444444} \rfloor=666$$
$$\lfloor \sqrt{44444444} \rfloor=6666$$
But I'm un... | This can be proved a bit more simple.
The general term can be written as $4 \frac{10^{2n} - 1}{9}$. Taking square root will give you $\frac{2}{3} \sqrt{10^{2n} - 1}$. As $\frac{2}{3} \approx 0.66666666666$ and $\sqrt{10^{2n} - 1} \approx \sqrt{10^{2n}} = 10^n$, this explains why the result is $6;66;666;...$ etc. To pro... | {
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"url": "https://math.stackexchange.com/questions/2849609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 3,
"answer_id": 0
} |
Does the equation $q^2\, n(n+1) = p^2\, m(m+1)$ have solutions only if $m(m+1) = q^2\, k(k+1)$, $kI have the following equation
\begin{equation}
q^2\, n(n+1) = p^2\, m(m+1), \qquad(1)
\end{equation}
where $m,n,p,q \in \mathbb{N}$ and $p>q>2$ and $\text{gcd}(p,q)=1$.
I could find numerically solutions only when $m(m+1) ... | First of all, a number $N$ is of the form $m(m+1)$ if and only if $4N+1$ is a perfect square.
Since $p$ and $q$ are coprime, we have $q^2|m(m+1)$, so $$\frac{m(m+1)}{q^2}$$ is a positive integer and we have $$S:=\frac{m(m+1)}{q^2}=\frac{n(n+1)}{p^2}$$
Using the criterion above, we get that $$4Sq^2+1$$ and $$4Sp^2+1$$ m... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | Without Calculus:
Multiplying both sides of $a+b+\dfrac 1a + \dfrac 1b \ge 3 \sqrt 2$ by $ab$ we get
$$a^2b+b^2a+a+b \ge 3 \sqrt2 ab$$
which factors as
$$(a+b)(ab+1) \ge 3 \sqrt2 ab$$
and squaring both sides yields
$$(a+b)^2(ab+1)^2 \ge 18(ab)^2$$
But $(a+b)^2 = 1+2ab$, and substituting $x = ab$ get
$$(1+2x)(x+1)^... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 2
} |
Find the Determinant when $p$ and $q$ are roots of $x^2-2x+5=0$ If $p$ and $q$ are roots of $x^2-2x+5=0$ Then find value of
$$\Delta=\begin{vmatrix}
1 & 1+p^2+q^2 & 1+p^3+q^3\\
1+p^2+q^2& 1+p^4+q^4 & 1+p^5+q^5\\
1+p^3+q^3 & 1+p^5+q^5 & 1+p^6+q^6
\end{vmatrix}$$
My try: I tried to express the determinant as product o... | Use Viète's formulas:
$$p+q=2\qquad pq=5$$
$$p^2+q^2=(p+q)^2-2pq=-6$$
$$p^3+q^3=(p+q)(p^2+q^2-pq)=-22$$
$$p^4+q^4=(p^2+q^2)^2-2(pq)^2=-14$$
$$p^5+q^5=(p+q)(p^4+q^4)-pq(p^3+q^3)=82$$
$$p^6+q^6=(p^3+q^3)^2-2(pq)^3=234$$
from which we may easily work out the determinant as
$$\begin{vmatrix}
1 & 1-6 & 1-22\\
1-6& 1-14 & 1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Number of ways of forming 10 student committee from 5 classes of 30 students each There are 5 classes with 30 students each. How many ways can a committee of 10 students be formed if each class has to have at least one student on the committee?
I figured that we first have to choose 5 people from each class, so there a... | Here is how it works without inclusion-exclusion.
There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284
The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.
For each kind of "hand" we have to do the specific calculus.
... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got a... | Try with $b=\cos 2x$ and $a= \sin 2x$.
\begin{eqnarray}{b+1\over a+b-2}&=& {2\cos^2 x\over -\cos^2x+2\sin x \cos x -3\sin^2x}\\
&=& {2\over -1+2\tan x -3\tan^2x}\\
&=& {2\over -1+2t -3t^2}
\end{eqnarray}
where $t= \tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
a tough sum of binomial coefficients
Find the sum: $$\sum_{i=0}^{2}\sum_{j=0}^{2}\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j},\space\space 0\leq k,l\leq 6$$
I know to find $\sum_{i=0}^{2}\binom{2}{i}\binom{2}{2-i}$, I need to find the coefficient of $x^2$ of $(1+x)^4$ (which is $\binom{4}{2}$). But I fa... | Here is a variant which could be seen as generalisation of OP's example. We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
and we also use Iverson brackets which are defined as
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by ... | Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
... | With $z = \log_2(3)$:
$$L = \log_{24}(48) = \frac{\ln(48)}{\ln(24)} = \frac{\ln(2^4\times 3)}{\ln(2^3 \times 3)} = \frac{4\ln(2) + \ln(3)}{3\ln(2) + \ln(3)} = \frac{4+z}{3 + z}$$
$$K = \log_{12}{(36)} = \frac{\ln(36)}{\ln(12)} = \frac{\ln(2^2\times 3^2)}{\ln(2^2 \times 3)} = \frac{2\ln(2) + 2\ln(3)}{2\ln(2) + \ln(3)} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Inverse matrix of matrix (all rows equal) plus identity matrix Let $A$ be a matrix where all rows are equal, for example,
$$A=\left[\begin{array}{ccc}
a_{1} & a_{2} & a_{3} \\
a_{1} & a_{2} & a_{3} \\
a_{1} & a_{2} & a_{3}
\end{array}\right]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity ma... | The inverse of $B$:
$$B^{-1}=\frac{\text{adj}{(B)}}{\det(B)}.$$
The determinant of $B$:
$$\det(B)=\begin{vmatrix}a_{1}+1 & a_{2} & a_{3} \\
a_{1} & a_{2}+1 & a_{3} \\
a_{1} & a_{2} & a_{3}+1\end{vmatrix}=
\begin{vmatrix}a_{1}+1 & a_{2} & a_{3} \\
-1 & 1 & 0 \\
-1 & 0 & 1\end{vmatrix}=a_1+a_2+a_3+1.$$
The adjugate of $B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi... | So, you have
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}+\sin\frac{\pi}{20})=2\sin\frac{2\pi}{20}\left(\frac{\cos^2\frac{\pi}{20}-\sin^2\frac{\pi}{20}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}}\right)= \frac{\sqrt2}{2},$$
or
$$ = \frac{2 \sin\frac{2\pi}{20}\cos\frac{2\pi}{20}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
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Is there a simpler way to determine m, n, p, such that the following holds for all reals? I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ \sin^4x + \cos^4x + m(\sin^6x + \cos^6x) + n(\sin^8x + \cos^8x) + p(\sin^{10}x + \cos^{10}x) = 1, \space \fora... | Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5\,.$$
If $t=\sin^2(x)$ is a root for every $x\in\mathbb{R}$, then $P\equiv 0$ identically. In particular,
$$0=[t^4]\,P(t)=2n+5p\,,$$
$$0=[t^2]\,P(t)=2+3m+6n+10p\,,$$
and
$$0=[t^0]\,P(t)=P(0)=m+n+p\,,$$
where $[t^k]\,P(t)$ denotes t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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$\lim_{n \to \infty} \sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}{\binom{n-k}{k}\frac{1}{2^{n-k}}}$?
Consider the following limit:
$$\lim_{n \to \infty} \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}{\binom{n-k}{k}\frac{1}{2^{n-k}}}.$$
I can find the limit numerically, but is it possible to compute it anal... | Let $a_n$ denote the expression in the limit. Using the identity $n! = \int_{0}^{\infty} x^n e^{-x} \, dx$, we have
\begin{align*}
a_n
&= \sum_{k=0}^{\lfloor n/2\rfloor} \frac{1}{k!(n-2k)!2^{n-k}} \int_{0}^{\infty} x^{n-k} e^{-x} \, dx \\
&= \int_{0}^{\infty} \underbrace{ \sum_{k=0}^{\lfloor n/2\rfloor} \frac{(x/2)^k}{... | {
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} |
$\langle 3 \rangle$ in $\textbf{Z}[\omega]$ ramifies, not splits, right? Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.
As you know, $$\omega = \frac{-1 + \sqrt{-3}}{2}$$ is a complex cubit root of unity, that is, $\omega^3 = 1$. This slightly obs... | It seems fine, although I must admit I had trouble understanding what you want.
Anyway, first to conclude that $\langle 3 \rangle $ ramifies you need to prove that $\left \langle \sqrt{-3} \right \rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.
Then you can establish the... | {
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Determining the last column so that the resulting matrix is an orthogonal matrix
Determine the last column so that the resulting matrix is an orthogonal
matrix $$\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & ? \\ \dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{6}} & ? \\ 0 & \dfrac{2}{\sqrt{6}} & ? \end{bmatrix... | Let $[ x \ y \ z]$ be the vector you are looking for, then it must satisfy
$$x \frac{1}{\sqrt{2}} +y \frac{1}{\sqrt{2}}+z (0)= 0$$
which is $y = - x$ and
$$x \frac{1}{\sqrt{6}} -y \frac{1}{\sqrt{6}}+z \frac{2}{\sqrt{6}} = 0$$
which is $z = \frac{1}{2}y - \frac{1}{2}x$. Using $y = -x$ in $z = \frac{1}{2}y - \frac{1}{2}x... | {
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"answer_count": 4,
"answer_id": 0
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How to determine if a set of five $2\times2$ matrices is independent $$S=\bigg\{\left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\-1&2\end{matrix}\right], \left[\begin{matrix}0&1\\1&2\end{matrix}\right],\left[\begin{matrix}1&0\\1&1\end{matrix}\right],
\left[\begin{matrix}1&4\\0&3\end{matrix}\ri... | As has been pointed out, four matrices form a basis for the $2\times2$ matrices (the easiest would be
$$
\left[\begin{matrix}1&0\\0&0\end{matrix}\right], \left[\begin{matrix}0&1\\0&0\end{matrix}\right], \left[\begin{matrix}0&0\\1&0\end{matrix}\right], \left[\begin{matrix}0&0\\0&1\end{matrix}\right]
$$) so five matrices... | {
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"source": "stackexchange",
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State the possible Cartesian equations of $p_2$. Given a plane $p_1$ to have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.
I think I used a wrong way but got the correct answer,... | The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.
The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.
Distance of a plane $Ax+By+Cz = D$ from the origin is given by $\frac{|D|}{\sqrt{A^2+B^2+C^2}}$ so in this case we have
$$... | {
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how to prove the below statement If $$\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac2w$$ where $a,b,c,d$ are real numbers and $w$ is a non real complex cube root of unity then prove that $$\frac{1}{a+w^2}+\frac{1}{b+w^2}+\frac{1}{c+w^2}+\frac{1}{d+w^2}=2w$$
I have no idea as how to start the problem. Ple... | Hint:
$ω^2$ is the conjugate of $ω$, so
$$\frac{1}{a+ω^2}+\frac{1}{b+ω^2}+\frac{1}{c+ω^2}+\frac{1}{d+ω^2}=\overline{\,\frac{1}{a+ω}+\frac{1}{b+ω}+\frac{1}{c+ω}+\frac{1}{d+ω}\,}.$$
| {
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"url": "https://math.stackexchange.com/questions/2864323",
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"source": "stackexchange",
"question_score": "1",
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Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1... | Hint.
$$
2^{2x}\sum_{k=0}^{99}2^{-2k} =2^x\sum_{k=0}^{99}2^{-k}
$$
hence
$$
2^x = \frac{\sum_{k=0}^{99}2^{-k}}{\sum_{k=0}^{99}2^{-2k}} = 1.5\to x = 0.5849625007211562
$$
| {
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"source": "stackexchange",
"question_score": "4",
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How do I solve this logarithm problem? I'm trying to solve this problem:
If $\log_{27}(a)=b$, find $\log_{\sqrt[6]{a}}\sqrt{3}$
However, I'm unable to see any connection in those given information. How can I solve this logarithm?
| Given $\log_{27}(a) = b$ then $a = 27^b$. Then:
$$\sqrt[6]{a} = (27^b)^\frac{1}{6} = (3^{3 \cdot \frac{1}{6}})^b = 3^\frac{b}{2}$$
$$\log_{\sqrt[6]{a}}x = \log_{3^\frac{b}{2}}x = \frac{\log_{3}x}{\log_{3}3^\frac{b}{2}} = \frac{2}{b}\log_{3}x$$
Since $x = \sqrt{3}$ then
$$\frac{2}{b}\log_{3}x = \frac{2}{b}\cdot\frac{1}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$ Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$
There are a few ways to approach it, one of the way i encountered is that using the $\tan2\theta$ formula, we get $$\tan\theta = \frac{2t}{1-t^2}$$
By trigonometry, we know t... | I think the second "solution" comes from squaring then unsquaring. This process can often add solutions to an equation. Using a much simpler example:
$$
\begin{align}
& &x-1 &= 2
\\ &\Rightarrow &(x-1)^2 &= 4
\\ &\Rightarrow &x^2-2x+1 &= 4
\\ &\Rightarrow &x^2-2x-3 &= 0
\\ &\Rightarrow &(x-3)(x+1) &= 0
\\ &\Rightarrow ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Log inequalities Solve the inequality
$\log(5^{1/x} +5^3) < \log 6 + \log 5^{1+\frac{1}{2x}}$
I came up with an equation
$5^{\frac{4x-1}{2x}} + 5^{\frac{-2x+1}{2x}}-6<0$
Which I couldn't solve, i tried to make it quadratic but I couldn't
| Yes, since $\log$ function is monotonic increasing we have
$$\log(5^{1/x}+5^3)<\log(6\cdot 5^{1+1/2x})\iff 5^{1/x}+5^3<6\cdot 5^{1+1/2x}$$
that is
$$5^{1/x-1-1/2x}+5^{3-1-1/2x}<6$$
and then your result
$$5^{\frac{4x-1}{2x}}+5^{\frac{1-2x}{2x}}-6<0$$
then
$$5^{2-\frac{1}{2x}}+5^{\frac{1}{2x}-1}-6<0$$
$$25\cdot 5^{-\fr... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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For $a> 0$, determine $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)$? For $a> 0$, determine $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)$
My attempts : $\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1) = ( a^\frac{1}{n^2}- 1) + ( a^\frac{2}{n^2}- 1)+......... +(a^\frac{n}{n^2}- 1)=a^\frac{1}{... | Intuition:
since $e^u = 1+u + o(u)$ when $u\to 0$ and
$0 \leq \frac{k}{n^2}\ln a \leq \frac{\ln a}{n} \xrightarrow[n\to\infty]{} 0$ for every $1\leq k\leq n$, we "expect" to have
$$
\sum_{k=1}^{n}( a^{\frac{k}{n^2}}-1)
= \sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1) \approx \sum_{k=1}^{n} \frac{k}{n^2}\ln a = \ln a\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868581",
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"source": "stackexchange",
"question_score": "3",
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How to simplify $\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$? I have been asked about the following integral:
$$\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$$
I think this is a joke of bad taste. I have tried every elementar... | Hint:
Let $\text{arcsec}x=t, x=\sec t$
Using Principal values, $0\le t\le\pi,t\ne\dfrac\pi2$
$\sin t=\sqrt{\left(1-\dfrac1x\right)^2}=\dfrac{\sqrt{x^2-1}}{|x|}$ as $\sin t\ge0$
$\tan t=\sin t\sec t=?$
$$1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}=\dfrac{\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t}{\cos^4t}$$
Now $\cos^4t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
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Gauss elimination. Where did I go wrong?
Gaussian elimination with back sub:
So my starting matrix:
\begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2
\end{bmatrix}
multiply the 2nd and 3rd row by -1 * (first row):
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 2 & 0 & 4
\end{bmatrix}
then ad... | Yes, your answer is correct.
Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.
$10z=0 \implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 \implies x-2=-1 \implies x=1$.
| {
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"source": "stackexchange",
"question_score": "3",
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compute the summation $\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$ compute the summation
$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$
My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$
Now
\begin{align}
& = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}... | It is wrong since
$$
2\times 4\times 6\times\dotsb (2n)=2^n n!\quad (n\geq 1).
$$
Write
$$
\sum_{n=1}^\infty\frac{2n-1}{2^n n!}=\sum_{n=1}^\infty\frac{1}{2^{n-1}(n-1)!}-\sum_{n=1}^\infty\frac{(1/2)^n}{n!}
=\sum_{n=0}^\infty\frac{(1/2)^n}{n!}-\sum_{n=1}^\infty \frac{(1/2)^n}{n!}=1.
$$
Note that we don't need to know the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Can we solve $A+D\sin^{2}x=B\sin x+C\cos x$ without having to solve a quartic polynomial? Suppose the following equation
$$
A+D\sin^{2}x=B\sin x+C\cos x,
$$
where $A,B,C,D\in\mathbb{R}$ are the real constants. Initially, I tried to find its solution from a simple substitution
\begin{align*}
A-B\sin x+D\sin^{2}x & =\pm ... | If you set $X=\cos x$ and $Y=\sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $D\ne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation betwe... | {
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"source": "stackexchange",
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Calculating $\int_{-\infty}^{\infty}\left( \frac{\cos{\left (x \right )}}{x^{4} + 1} \right)dx$ via the Residue Theorem? In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$\text{Proposition} \, \, \, (1) $
$$\int_{-\infty}^{\... | Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $\exp\left(\frac{\pi i}4\right)$, $\exp\left(\frac{3\pi i}4\right)$, $\exp\left(\frac{5\pi i}4\right)$, and $\exp\left(\frac{7\pi i}4\right)$. But only the first two matter, since they'r... | {
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"source": "stackexchange",
"question_score": "1",
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Inequality $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$ Let $k$ be a positive integer and $x$ a positive real number. Prove that $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$.
This looks similar to Bernoulli's inequality. If we write $X=x^k$, the inequality is equivalent to $(1+X)^{\frac{k+1}{k}}\geq 1+X^{\frac{k+1}{k}}$, but this is not exa... | Case 1. $x\le 1$.
Then $x^k\ge x^{k+1}$ and therefore
$$(1+x^k)^{k+1} = (1+x^k)(1+x^k)^k \ge (1+x^k)^k \ge (1+x^{k+1})^k.$$
Case 2. $x>1$.
We divide the inequality by $x^{k(k+1)}$ and get the following equivalent form
$$\left(1+\left(\frac 1x \right)^k\right)^{k+1} \ge \left(1+\left(\frac 1x \right)^{k+1}\right)^k.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Convergence of $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$ Define a sequence $a_n$ as follows: $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$
Determine if it's convergent and find its limit.
The sequence satisfies $a_n=\sqrt{7a_{n-1}}$. If it's converge... | Here's a totally different approach.
Let $T(a)=\sqrt{7a}$. For $a \geq 2$,
$$
T(a) \geq \sqrt{14} > \sqrt{4} = 2
\qquad
\text{and}
\qquad
|T^{\prime}(a)| = \frac{\sqrt{7}}{2\sqrt{a}} \leq \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4} < \frac{\sqrt{16}}{4} = 1.
$$
Therefore, $T$ is a contraction on $[2, \infty)$.
By ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Sum of n terms of this series $\frac{1}{1.3} + \frac{2}{1.3.5} +\frac {3}{1.3.5.7} + \frac{4}{1.3.5.7.9}........ n $ Terms.
I Know the answer to this problem but I couldn't find any proper way to actually solve this question. I thought the denominators were the product of n odd natural numbers. So I wrote the nth te... | the general term is given by $$a_n=\frac{n}{1*3*5....(2n+1)}$$
$$a_n=\frac{\frac{2n+1-1}{2}}{1*3*5....(2n+1)}$$
$$a_n=\frac{1}{2}*({\frac{2n+1}{1*3*5....(2n+1)}-\frac{1}{1*3*5...(2n+1)})}$$
$$a_n=\frac{1}{2}*({\frac{1}{1*3*5....(2n-1)}-\frac{1}{1*3*5...(2n+1)})}$$
for $n=1$ $$\frac{1}{2}*({\frac{1}{1}-\frac{1}{1*3})}$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating... | Your solution is fine but it can be shortened. By Lagrange's theorem $(x+2)\arctan(x+2)-x\arctan(x)$ is twice the value of the derivative of $z\arctan z$ at some $z\in(x,x+2)$. Since $\frac{d}{dz}z\arctan(z) = \underbrace{\frac{z}{1+z^2}}_{\to 0}+\underbrace{\arctan(z)}_{\to \pi/2}$, the wanted limit is $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Algebra - Solving for three unknowns. Find all possible solutions of
$$2^x + 3^y = z^2.$$
My approach.
First I substituted $x = 0$, and got the solution, then for $y = 0$. And for $x > 0$ and $y > 0$ , I just know the Pythagorean triple and got the solution as $(4,2,5)$ and $(4,2,-5)$. Please help me to solve it pro... | If $x,y>0$ then working mod $3$ we have $2^x\equiv 1$ so $x$ is even. Working mod $4$ gives $y$ even. So $(2^{x/2})^2+(3^{y/2})^2=z^2$, meaning $(2^{x/2},3^{y/2},|z|)$ form a pythagorean triple.
Any pythagorean triple can be written as $(a(b^2-c^2), 2abc, a(b^2+c^2))$ for some positive integers $a,b,c$ (where the first... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what values of $a$ and $b$ does $\lim_{x\rightarrow \infty}(\sqrt{x^2+x+1}-ax-b)=1$? I have a question about limits tending to infinity. I need to find the constants $a$ and $b$ for which this limit takes the value 1. Please, help! Thank you!
$$\lim_{x\rightarrow \infty}(\sqrt{x^2+x+1}-ax-b)=1.$$
I've tried vario... | $\lim_{x\to \infty} \frac{\sqrt{x^2+x+1}}{x} = \lim_{x\to \infty} \frac{x\sqrt{1+{\frac{1}{x}}+\frac{1}{x^2}}}{x} = 1=a$
$\lim_{x\to \infty} \sqrt{x^2+x+1}-ax = \lim_{x\to \infty} \sqrt{x^2+x+1}-x= \lim_{x\to \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\lim_{x\to \infty} \sqrt{x^2+x+1}-ax+b=1 \implies 1-b=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $ \lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$ where $[\cdot]$ denotes the greatest integer function. Evaluate $$\lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$$ where $[\cdot]$ denotes the greatest integer function.
Can anyone give me a hint to proceed?
I know that $$\frac {\sin x}{x}... | Can you just use the known expansions for $\sin x, \tan x$ for small $x$? Then have $$\frac{x^2}{\sin x \tan x} = \frac{x^2}{(x - \frac{x^3}{6} + \cdots)(x + \frac{x^3}{3} + \cdots)}.$$ The rhs only including terms up to $x^2$ can be written $$\frac{1}{(1 - x^2/6)(1 + x^2/3)}$$ and the denominator to order $x^2$ is $1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Difference of numbers in a unit interval $x,y,z\in \mathbb{R}$ are chosen at random from the unit interval $[0, 1]$.
What is the probability that
$$\max(x,y,z) - \min(x,y,z) \leq \frac{2}{3}$$
EDIT- Solutions not using calculus would be appreciated, as this problem appeared on a test where calculus was not meant to be... | Let $E$ be the event $\{ \max(x, y, z) - \min(x, y, z) \le \frac{2}{3}\}$. Our goal is to find $\Pr[E]$. Now let's split $E$ into 6 events $E_{xy}, E_{yx}, E_{xz}, E_{zx}, E_{yz}, E_{zy}$, where
$$E_{xy} = \{ (x = \max(x, y, z)) \land (y = \min(x, y, z)) \land (x - y \le 2/3)\},$$
and the rest are defined similarly. No... | {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$ is not uniformly convergent on $[0,\;\frac{\pi}{2})$? Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$
is not uniformly convergent on $[0,\;\frac{\pi}{2})$?
I was thinking that we need to show partial sums
\begin{equation}
\left|S_{2n}\... | First, your proof looks fine to me. Below I sketch a different kind of proof.
Let $f(x)$ be the sum of this series. Note that each summand increases on $[0,\pi/2).$ Hence so does $f(x).$ It follows that $\lim_{x\to \pi/2^-} f(x)=L$ for some $L\in (0,\infty].$
Suppose the series converges uniformly to $f$ on $[0,\pi/2).... | {
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"answer_count": 2,
"answer_id": 0
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Finding closed form for $\sum_{k=1}^n k2^{k-1}$ I am trying to use the perturbation method to find a closed form for:
$$
S_n = \sum_{k=1}^n k2^{k-1}
$$
This is what I’ve tried so far:
$$
S_n + (n+1)2^n = 1 + \sum_{k=2}^{n+1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=1}^{n} (k+1)2^{k}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{... | Note the mistake: $\sum_{k=1}^{n} (k+1)2^{k}\ne \sum_{k=0}^{n-1} k2^{k-1}$.
Here is the right way:
$$\begin{align}S_n + (n+1)2^n &= 1 + \sum_{k=1}^{n} (k+1)2^{k}=\\
&=1+\sum_{k=0}^{n-1} (k+2)2^{k+1}=\\
&=1+\color{blue}{\sum_{k=0}^{n-1} k\cdot 2^{k+1}}+\color{red}{\sum_{k=0}^{n-1}2^{k+2}}=\\
&=1+\color{blue}{\sum_{k=0}^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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From $5$ apples, $4$ mangoes and $3$ bananas, in how many ways we can select at least two fruits of each variety?
From $5$ apples, $4$ mangoes and $3$ bananas, in how many ways we can select at least two fruits of each variety if fruits of same species are different?
My attempts:
$$\underbrace{\bigg({5\choose 2}+{5\c... | Your solution is correct. We can confirm it with another approach.
The number of subsets of a set with $n$ elements is $2^n$ since each element is either included in a subset or it is not.
Since there are $5$ apples, there are $2^5$ subsets of apples. Of these, $\binom{5}{0} + \binom{5}{1} = 1 + 5 = 6$ have fewer t... | {
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"url": "https://math.stackexchange.com/questions/2891092",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Summation problem: $f(x)=1+\sum_{n=1}^{\infty}\frac{x^n}{n}$ I want to evaluate this summation:
$$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...+...$$
where, $|x|<1$
Here it is my approach
$$S=1+\sum_{n=1}^{\infty}\frac{x^n}{n}=f(x)$$
$$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$
$$f(x)=\int f'(x)dx=\int \frac{1}{... | It's well-known that $$\ln(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots,~~~\forall y \in (-1,1].\tag1$$
Thus, let $y=-x.$ we have $$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots,~~~\forall x \in [-1,1).\tag2$$
As result, $$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots=1-\left(-x-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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Binomial expansion lower bound $A^n + B^n \le (A+B)^n$ for non-integer $n$ By the Binomial expansion for integer powers,
$$ (A+B)^n = \sum_{k=0}^n {n\choose{k}} A^{n-k} B^{k}$$
(I'm assuming $A,B\ge 0$) and so we get the easy estimate $A^n + B^n \le (A+B)^n$ for any positive integer $n$.
Now what happens when we take ... | If $A=B$,
$A^n + B^n \le (A+B)^n$
is
$2A^n \le (2A)^n
=2^nA^n$
or
$2 \le 2^n$\which is true since
$n \gt 1$.
Assuming
$A \lt B$,
dividing by
$B^n$
and letting
$A/B = x$,
$A^n + B^n \le (A+B)^n$
becomes
$1+x^n \le (1+x)^n$
with
$0 < x < 1$.
Let
$f(x)
= (1+x)^n-1-x^n$.
$f(0) = 0$.
$f'(x)
=n(1+x)^{n-1}-nx^{n-1}
=n((1+x)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$.
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \c... | As an alternative we have by binomial expansion
$$\sqrt \frac{x^3+x}{x+1}=x\sqrt \frac{x+1/x}{x+1}=x\sqrt \frac{x^2+1}{x^2+x}=x\sqrt \frac{(x+1)^2-2x}{x(x+1)}=x\sqrt{\frac{x+1}{x}-\frac{2}{x+1}}=x\sqrt{1+\frac1x-\frac{2}{x+1}} =x\left( 1+\frac1{2x}-\frac{1}{(x+1)}+o\left(\frac1x\right)\right)=x+\frac12-\frac{x}{(x+1)}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Given $f(x) = \arcsin\left(\frac{2x}{1+x^2}\right)$, find $f'(1)$.
Let
$$
f(x) = \arcsin\left(\frac{2x}{1+x^2}\right).
$$
What is the value of $f'(1)$?
The function splits into
$$
f(x) = \begin{cases}
\phantom{\pi-\,}2\arctan(x), & \text{if $-1 \leq x \leq 1$},\\
\pi - 2\arctan(x), & \text{if $x > 1$}.
\end{ca... | Notice that
\begin{align*}
f'(x) &= \frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)
\\&= \frac{1}{\sqrt{\left(\frac{1+x^2}{1+x^2}\right)^2-\left(\frac{2x}{1+x^2}\right)^2}}\cdot\frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\\&= \sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}}\cdot \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the shortest distance from a point to curved surface
Find the shortest distance from a point $(0,0,0)$ to curved surface $x^2+2y^2-z^2=5.$
What I have done is,
Let a point in the curved surface be $(a,b,c)$, and $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix}$ be vector v.
A vector w perpendicu... | Alternatively, you can set up an optimization (minimization) problem.
Let the point $(a,b,c)$ belong to the curve $x^2+2y^2-z^2=5$. The squared distance from the point to the origin is:
$$d^2(a,b,c)=(a-0)^2+(b-0)^2+(c-0)^2=a^2+b^2+c^2,$$
which needs to be minimized subject to the constraint: $a^2+2b^2-c^2=5$.
Using the... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+... | $$60n^2+10n \gt 6n^2+13n+7 \iff 54n^2>3n+7$$
Note that $$ 54n^2 \ge 54n=3n+51n\ge 3n+51>3n+7$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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$3a4b \equiv 0\pmod{3}$, $3a4b \equiv 3\pmod{5}$
When the four digit number $3a4b$ is divided by $5$, the remainder is $3$. This number can also be divided by $3$ without remainder. Evaluate $a$ and $b$.
We have two conditions as illustrated below
$$3a4b \equiv 0\pmod{3} \tag{1}$$
$$3a4b \equiv 3\pmod{5} \tag{2}$$
... | $$4a+b \equiv 0\pmod{3}$$
This step is wrong
$$3000 + 100a +40 +b \equiv 0\pmod{3}$$
$$3000 + 99a+a +39+1 +b \equiv 0\pmod{3}$$
$$a+1 +b \equiv 0\pmod{3}$$
$$a +b \equiv 2\pmod{3}$$
Also given that
$$3000 + 100a +40 +b \equiv 3\pmod{5}$$
$$b \equiv 3\pmod{5}$$
Now what can you say about $b$ where $0\le b\le9$ and
$b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How can I find the limit of the following sequence $\sin ^2 (\pi \sqrt{n^2 + n})$? How can I find the limit of the following sequence:
$$\sin ^2 (\pi \sqrt{n^2 + n})$$
I feel that I will use the identity $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n})), $$
But then what? how can I deal with... | Intuitively, $n^2+n$ is almost exactly halfway between $n^2$ and $(n+1)^2$, and so $\sqrt{n^2+n}$ is very close to a half-integer. This means $\sin(\pi\sqrt{n^2+n})$ should be very close to $\pm 1$, and in either case $\sin^2(\pi\sqrt{n^2+n})$ will be very nearly $1$.
More formally, we have
$$\sin^2(\pi\sqrt{n^2+n})=\s... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
$\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
Clearly, we can see that $$T_r=\frac{n-r+1}{r(r+1)(r+2)}$$
Now, somehow, we have to make this telescoping.
But we do not have $n$ as a factor o... | Let $$u_r= \frac 1 r - \frac 1{r+1}$$
Then
$$\begin{split}
u_r-u_{r+1} &= \frac 1 r - \frac 1{r+1} -\bigg( \frac 1 {r+1} - \frac 1{r+2}\bigg) \\
&= \frac 1 r - \frac 2{r+1} + \frac 1 {r+2} \\
&= \frac{ (r+2)(r+1) -2r(r+2) + r(r+1)}{r(r+1)(r+2)}\\
&= \frac 2 {r(r+1)(r+2)}\\
\end{split}$$
With that, the sum to compute i... | {
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"url": "https://math.stackexchange.com/questions/2903555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n... | To do the inductive step, observe that:
$(n+1)3^{n+1}+\frac{3(3^n(2n-1)+1)}4=\frac{4(n+1)3^{n+1}+3(3^n(2n-1)+1)}4=\frac{(6n+3)3^{n+1}+3}4=\frac{3(2n+1)3^{n+1}+3}4=\frac{3(3^{n+1}(2(n+1)-1)+1)}4$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$ Problem
Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$
Attempt to solve
if i set $z=x+iy$ when $(x,y) \in \mathbb{R}, z \in \mathbb{C}$
$$ |\frac{1}{x+iy}|<1 $$
Trying to multiply denominator and nominator with complex conjugate.
$$ |\frac{x-iy}... | Why not just write
$$ |1/z| < 1 \quad\iff\quad |z| > 1 \quad\iff\quad x^2+y^2 > 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } ... | Hint : Note that $2-x^2 = (\sqrt{2-x^2}) * (\sqrt{2-x^2})$ and $(\sqrt{2-x^2}) = \frac{2-x^2}{(\sqrt{2-x^2})}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Monotonicity of the function $(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x$. Let $f(x)=(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x, 0<x\leq 1.$
Prove that $f$ is strictly increasing and $e<f(x)\leq 4.$
In order to study the Monotonicity of $f$, let
$$g(x)=\log f(x)=\frac{1}{x}\log (1+x)+x\log \left(1+\frac{1}{x... | The right inequality.
We can use the TL method here.
We need to prove that $$(1+a)^b(1+b)^a\leq4$$
for $a>0$, $b>0$ such that $ab=1$, which is $$\frac{\ln(1+a)}{a}+\frac{\ln(1+b)}{b}\leq2\ln2.$$
But $$\sum_{cyc}\left(\ln2-\frac{\ln(1+a)}{a}\right)=\sum_{cyc}f(a),$$
where $f(a)=\ln2-\frac{\ln(1+a)}{a}-(\ln2-0.5)\ln a$.... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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If $2^t = a^b \pm 1$, what are all the possible values of $t$? Let $t$ be a positive integer such that $2^t = a^b \pm 1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$?
The question is taken from here. I know the answer is $3$, so $a = 3$ and $b = 2$. But how can I prove t... | The answer is $t=3$.
This answer proves that the following two claims are true :
Claim 1 : If $t,a,b$ are integers such that $t\ge 1,a\ge 2$ and $b\ge 2$, then $2^t=a^b+1$ has no solutions.
Claim 2 : If $t,a,b$ are integers such that $t\ge 1,a\ge 2$ and $b\ge 2$, then $2^t=a^b-1$ has the only one solution $(t,a,b)=(3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Calculate $\int \frac{x^4+1}{x^{12}-1} dx$ So I found this problem:
Calculate $$\int \frac{x^4+1}{x^{12}-1} dx$$ where $x\in(1, +\infty)$
and I don't have any ideea how to solve it. I tried to write $$x^{12}-1=(x^6+1)(x+1)(x-1)(x^2+x+1)(x^2-x+1)$$ but with $x^4+1$ I had no idea what to do, an obvious thing would be t... | Use $$x^6+1=(x^2+1)(x^4-x^2+1)$$
$$I={\displaystyle\int}\dfrac{x^4+1}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)}\,\mathrm{d}x$$
Apply partial Factor decomposition you get
$$I={\displaystyle\int}\left(-\dfrac{x^2+1}{6\left(x^4-x^2+1\right)}-\dfrac{2x... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Arithmetic sequence problem $\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$ Choose such x that the following
$$\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$$
forms finite arithmetical sequence.
If I use the equation: $$2a_2=a_3+a_1$$
I always get the wrong answer.
| Let
$$
a_1=\frac{x+4}{x-3}, \quad a_2=\frac{x+6}{2},\quad a_3=\frac{4}{x-2}
$$
If $ a_1 $, $ a_2 $ and $ a_3 $ are in arithmetic progression the possible values for $ x $ are
$$
4,\quad \frac{-4+\sqrt{72}}{2} \quad\mbox{and}\quad \frac{-4-\sqrt{72}}{2}
$$
We have $2a_2={a_3+a_1}$, that is,
$$
{x+6}=\frac{x+4}{x-3}+\fr... | {
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"source": "stackexchange",
"question_score": "3",
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Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$. Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous... | Notice that $x^2+1$ is irreducible over $\mathbb{Q}$. Let $d = \gcd(x^2+1, x^3+1)$. Then $d \mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 \not\mid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $\gcd$ is usually supposed to be monic.
Finally, notice that
$$1=... | {
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"question_score": "2",
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Please show detailed steps of double integration of absolute difference Please show detailed steps of integration
$$\int_0^2 \int_0^1 0.5|x-y| dxdy$$
$$=\frac{1}{2}\int_0^2\int_0^y(y-x)dxdy + \frac{1}{2}\int_0^2\int_y^1(x-y)dxdy$$
$$=\frac{1}{2}\int_0^2\{y\int_0^ydx - \int_0^yxdx\}dy + \frac{1}{2}\int_0^2\{\int_y^1xdx ... | The problem is with your integration limits. You integrate $y$ from $0$ to $2$ and $x$ from $y$ to $1$. But For say $y=1.5$ it's the wrong way around. If you integrate $x$ from $0$ to $1$ and then you split the $y$ integration from $0$ to $x$ and from $x$ to $2$, you get the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding value of $\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n^3}}{\ln(n)}$
Find the value of
$$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n^3}}{\ln(n)}$$
My Try: Using Stolz-Cesaro,
Let $\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots ... | Your $a_n$ should go up to $\frac1{n^3}$. Then $a_{n+1}-a_n=\frac{1}{(n+1)^3}+\frac{1}{(n+1)^3-1}\dots+\frac1{n^3+2}+\frac{1}{n^3+1}$.
I don't know whether you can arrive there by your method, but you can find the limit by using the asymptotics of harmonic numbers / definition of the euler $\gamma$ constant. Then the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)... | We have for $x\neq 0$
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x$$
$$ (1 - x)^3 + x(1 - x)^2\color{red}{+ x(1 - x)-(1-x)}=0 $$
$$(1 - x)^3 + x(1 - x)^2\color{red}{+ (1 - x)(x-1)}=0$$
$$(1 - x)^3 \color{blue}{+ x(1 - x)^2- (1 - x)^2=0}$$
$$(1 - x)^3 \color{blue}{+ (1 - x)^2(x-1)}=0$$
$$(1 - x)^3 \color{blue}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\... | *
*No, it's not valid, it is not defined. Consider two sequences: $(a_n)_{n\in\mathbb{N}} = \frac{1}{2^{-n}}$ and $(b_n)_{n\in\mathbb{N}} = \frac{1}{-2^{-n}}$. You could say that limit of both of them is $\frac{1}{0}$, since $\pm2^{-n} \to 0$, but does this make sense? Do these sequences have the same limit?
*Saying ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 4
} |
A double sum or a definite integral. I am trying to evaluate the following double sum
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(3n+m)}.
\end{eqnarray*}
Using the integral trick
\begin{eqnarray*}
\frac{1}{3n+m} =\int_0^1 y^{3n+m-1} dy,
\end{eqnarray*}
the sum can be transformed into... | This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $\operatorname {Li_2}(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$\int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy = -2\int_0^1 \frac{ (y-\frac12)\ln(1+y)}{(y-\frac12)^2 +\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Proving Altitudes of Triangle can never form a Triangle Prove that Altitudes of Triangle can never form a Triangle
My try: we have altitudes proportional to reciprocals of sides of given triangle
Let $a,b,c$ are sides we have
$$a+b \gt c $$
$$b+c \gt a$$ and $$c+a \gt b$$
Now if $\frac{1}{a}$, $\frac{1}{b}$ and $\fra... | The condition for a triangle is
$c(a+b) > ab$.
For the 3-4-5 triangle,
this is
$5(3+4) > 3\cdot 4$
which is true.
So the altitudes do form a triangle
in this case.
Note that you can write
the condition as
$\begin{array}\\
0
&\gt ab-c(a+b)\\
&= ab-c(a+b)+c^2-c^2\\
&= (a-c)(b-c)-c^2\\
\text{or}\\
c^2
&\gt (a-c)(b-c)\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the limit by using the definition of derivative. Here is the problem.
Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of
$$ L = \lim_{x \to 1} {\frac{\arctan{\sqrt{f(x)}-\arctan{f(x)}}}{ \left (\arcsin{\sqrt{f(x)}}-\arcsin{f(x)}\right)^2}} $$
I have tried using
$$
L=\lim... | Probably the derivation is wrong because arcsin x is not differentiable at $x=1$, therefore your first step is not allowed.
As an alternative, since $f(x)$ is continuos at $x=1$ we have that
$$L = \lim_{y \to 1^-} {\frac{\arctan{\sqrt{y}-\arctan{y}}}{ \left (\arcsin{\sqrt{y}}-\arcsin{y}\right)^2}}$$
then we can use tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus) $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ I'll use induction to solve prove this. Then
$$\sum_{k=0}^n \binom{n}{k} = 2^n = \binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n}$$
First prove with n = 1
$$\binom{1}{0} + \binom{1}{1} = 2^1... | A better mathematician than you or I once struggled as mightily with this question as you have, and I told him to expand $(1+1)^n$ with the binomial expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Question about Paul Erdös' proof on the Sylvester-Schur Theorem I'm going through the proof and I am not clear the sufficiency of Erdös's argument at one step.
Here's the set up to the argument:
(1) Let $\{x\}$ be the least integer greater or equal to $x$.
(2) Let $a_i$ be shorthand for $\left\{\dfrac{n}{2^i}\right\}$ ... | From the observations, we can say that
$$\prod\limits_{p_i \le \sqrt[k]n}p_i\le \left(\prod_{\left\lfloor\sqrt[k]{a_1}\right\rfloor < p_i\le \left\lfloor\sqrt[k]{2a_1}\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt[k]{a_2}\right\rfloor < p_i\le \left\lfloor\sqrt[k]{2a_2}\right\rfloor}p_i\right)\cdots \left(\prod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Prove formula $\operatorname{arctanh} x = \frac12\,\ln \left(\frac{1+x}{1-x}\right)$ Problem
Prove formula $\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Attempt to solve
To start off with definition of functions $\sinh(x)$ and $\cosh(x)$
$$ \cosh(x)=\frac{e^x+e^{-x}}{2} $$
$$ \sinh(x) = \f... | Looks fine to me. Well done!
Just a small thing, a matter of taste: I'd prefer to do this:
\begin{align*}
e^x(1-y) &= e^{-x}(1+y)\\
e^{2x} &=\dfrac{1+y}{1-y}\\
x &= \dfrac{1}{2}\ln\left(\dfrac{1+y}{1-y}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How i can reduce this polynomial in $\mathbb{Z}_2 [x]$? How i can reduce the polynomial $f(x) = x^9+x^8+x^6+x^5+x^4+x^3+1 \in \mathbb{Z}_2 [x]$. I've tried use the reduction modulo a prime, but that don't solve the problem totally.
| Using sage the factorization is easy:
sage: R.<x> = PolynomialRing(GF(2))
sage: f = x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1
sage: f.factor()
(x^3 + x^2 + 1) * (x^3 + x + 1)^2
(Assuming that "reduce" means "factor" in the OP.)
Manually, as a human, we expect factors - if any - till degree four. So it is a good try to see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\sum\limits_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$, is $\sum\limits_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}$? Suppose $p(n)$ is a polynomial with rational coefficients and rational roots of degree at least $3$. If we know
$$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$
are we able to infer that
$$\sum_{n=... | $\newcommand\Q{\mathbf{Q}}$
I object a little bit to the "accepted" answer. It would require proving that $\sqrt{3} \pi - 3 \log(2)$ is not a rational number. I'm not sure this is so obvious.
Here is an alternate solution which is less random:
Let~$b > a$ be distinct non-zero rational numbers such that $2a$ and $2b$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
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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.