Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$ prove that
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<\dfrac{5}{3}$$
my idea: use
$$3^k-2^k>2^k(k\ge 2)$$
then
$$\Longleftrightarrow 1+\sum_{k=2}^{n}\dfrac{k}{3^k-2^k}<1+\sum_{k=2}^{n}\dfrac{k}{2^k}<1+\sum_{k=2}^{\infty}\dfrac{k}{2^k}=1+\dfrac{3}{2}>\dfrac{5}{3}$$
u... | By arithmetic/geometric means,
$$
3^k-2^k=3^{k-1}+3^{k-2}\cdot 2+\ldots +2^{k-1}< k\cdot 6^{(k-1)/2}.
$$
Hence
$$
\sum_{k\ge m}\frac{k}{3^k-2^k}<\sum_{k\ge m}\frac{1}{6^{(k-1)/2}}<\frac{1}{6^{(m-1)/2}}\frac{1}{1-\frac{1}{\sqrt{6}}}
$$
and it remains to choose $m$ as Raymond Manzoni advices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). ... | $(a^3+b^3+c^3) = (a+b+c)(a^2 +b^2+c^2-ab-bc-ac)$
$1+\omega + \omega^2 = 0 \implies \omega+\omega^2 = -1$
$= (a+b+c)(a^2 + b^2 + c^2 +ab(\omega+\omega^2)+bc(\omega+\omega^2)+ac(\omega+\omega^2)) = (a+b+c)(a(a+b\omega^2+c\omega)+b\omega(a+b\omega^2+ c\omega)+ c\omega^2 (a+ \dfrac b\omega+c\omega)$
Now, note that $\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 2
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Solve $(a^2-1)(b^2-1)=\frac{1}4 ,a,b\in \mathbb Q$ Does the equation $(a^2-1)(b^2-1)=\dfrac{1}4$ have solutions $a,b\in \mathbb Q$?
I search $0<p<1000,0<q<1000$, where $a=\dfrac{p}q$, but no solutions exist. I wonder is this equation solvable?
| If there are rational points on your curve than there are rational points on the curve:
$$ (x^2-4)(y^2-1)=1, $$
so $(y^2-1)=\frac{p}{q}$ and $(x^2-4)=\frac{q}{p}$, with $(p,q)=1$, must hold.
This implies that both $\frac{p+q}{q}$ and $\frac{4p+q}{p}$ are squares of rational numbers, so $p=u^2,q=v^2$ and both $u^2+v^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \... | Hint:
$$\begin{align}
0&=\cos x+\sin x\\
\sin x &= -\cos x \\
\text{Divide}&\text{ both sides by }\cos x\\
\tan x &= -1
\end{align}$$
Now you can solve for $x$ using inverse trigonometric functions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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proof that $f(x)=\sum_{n=1}^\infty \frac{\ln(x+n)}{x^2 + n^2}$ converges uniformily Prove that $f(x)=\sum_{n=1}^\infty \frac{\ln(x+n)}{x^2 + n^2}$ converges uniformly for $x\geq 0$.
This exercise is in a text about uniform convergence, I've tried to use Weierstrass test with no success. trying to quote with
$a_n=\sqrt... | Hint: $(x+n)^2 \le 2(x^2 + n^2)$.
$$\dfrac{\sqrt{x+n}}{x^2+n^2} \le \dfrac{\sqrt[4]{2}\sqrt[4]{x^2+n^2}}{x^2+n^2} = \dfrac{\sqrt[4]{2}}{(x^2+n^2)^{3/4}} \le \dfrac{\sqrt[4]{2}}{n^{3/2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a... | HINT: Let $b_n=\frac1{1\cdot3\cdot\ldots\cdot(2n-1)}$; for all $n\ge 1$ we have $b_{n+1}\le\frac13b_n$, so $$a_n=\sum_{k=1}^nb_k\le\sum_{k=1}^n\left(\frac13\right)^k\;.$$
Now use what you know about geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Projections on the Riemann Sphere are antipodal Prove that given two points $z,w\in\mathbb{C}$, we have that their projections on the Riemann sphere are antipodal if and only if: $z\bar{w}=-1$.
| We could of course just compute it using the explicit formula for the stereographic projection. But let's argue geometrically.
Let $z \in \mathbb{C}\setminus\{0\}$ and $\zeta$ the image of $z$ on the sphere under stereographic projection. Let $C$ be the great circle on the sphere passing through $\zeta$ and the north a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving exponential complex equation. I need assistance in solving the following equation:
$$ e^{z^2}=1, $$
where $z$ is a complex number.. I can't seem to get the answer of $z=\sqrt{k\pi}(1\pm i)$. Thank you in advanced for your help.
| We have $\displaystyle e^{z^2}=1=e^{2k\pi i}$ where $k$ is any integer
$\displaystyle\implies z^2=2k\pi i$
$\displaystyle\implies z=\sqrt{2k\pi i}=\sqrt{2k\pi }\sqrt i$
Method $1:$
By observation, $i=\frac{i^2+1+2i}2=\frac{(i+1)^2}2$
$\displaystyle\implies i^{\frac12}=\pm\frac{(1+i)}{\sqrt2}$
Method $2:$
Let $\sqrt{i}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1^3 + 2^3 + \cdots + n^3 < n^4$. I am trying to prove the following: $1^3 + 2^3 + \cdots + n^3 < n^4$ if $n \in \mathbb{N}, n>1$ by induction. From there, I am to prove that the sum is $< \frac{n^4}{2}$ if $n>2$. My attempt to do so is as follows:
First, we use $n = 2$ as a basis step since $n \in \mathbb{N... | As the others have said, you went wrong in the induction step. Here's one way to prove what we want without having to expand the binomials:
\begin{align*}
1^3 + 2^3 + \ldots + (n+1)^3 &= \left[1^3 + 2^3 + \ldots + n^3 \right] + (n+1)^3 \\
&< n^4 + (n+1)^3 \qquad\text{by the induction hypothesis} \\
&= n(n)^3 + 1(n+1)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to evaluate the sum $\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2n}}+\cdots+\frac{1}{\sqrt{n^2}}$ when $n$ grows? I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$
Thanks.
| Notice for large $n$, we expect $\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}}
\text{ behave like }\int_1^n \frac{dx}{\sqrt{x}} \sim 2\sqrt{n}$. This suggests
$$\frac{1}{\sqrt{k}} \sim \int_{k-1/2}^{k+1/2} \frac{dx}{\sqrt{x}} \sim 2\left( \sqrt{k+\frac12} - \sqrt{k-\frac12}\right)$$ and the terms $\displaystyle \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
$S_n$ be the sum of areas of $y=\sin x, y=\sin {nx}$. Then $\lim_{n \to \infty}S_n=8/{\pi}$? Let $n$ be a natural number. Also let $S_n$ be the sum of the areas of the regions enclosed with the two curves $y=\sin x$ and $y=\sin {nx}$ in $0\le x\le \pi$.
It's easy to find $S_2, S_3$. For smaller $n$, we can also use wo... | Since I wasted some time solving this last night, I might as well post it:
The integrand changes sign at the zeros, unless the zero is a double zero, so we need to find the zeros. For that, we bring it in a form where the zeros are more easily determined:
$$\sin (nx) - \sin x = 2\sin \frac{(n-1)x}{2}\cos \frac{(n+1)x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \inft... | $$F(a)=\int-\frac{1}{1+ax^2}dx=-\frac{\arctan(\sqrt{a}x)}{\sqrt a}+c$$
$$\int_{-\infty}^{+\infty}-\frac{1}{1+ax^2}dx=-\frac{\pi}{\sqrt a}$$
so we take the derivaitve of $-\frac{\pi}{\sqrt a}$ with respect to $"a"$ then put a=1 so
$$\int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how t... | HINT
The denominators in every term, except the first, are the alternating sums of even powers.
\begin{array}
.x^2-1 &\equiv& x^2 - x^0 \\
x^4-x^2+1 &\equiv& x^4 - x^2 + x^0 \\
x^6-x^4+x^2-1 &\equiv& x^6-x^4+x^2-x^0
\end{array}
Putting these in to summation notation:
\begin{array}
.x^2-x^0 &\equiv& \sum_{k=0}^1 (-1)^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is so... | It is useful, when approaching a problem like this, to eliminate the more complex square roots. So given:
$$\sqrt{4 + 2 \sqrt{3}} - \sqrt{3} = 1$$
Rearrange to (so as to isolate the more complex square root on the LHS):
$$\sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$$
Square both sides:
$$4 + 2 \sqrt{3} = (\sqrt{3} + 1)(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that t... | Not quite a proof to the original IMO problem, but there definately is a very easy way to compute all possible answers.
Also it demonstrates that here, Vieta jumping is basically just using
symmetry to jump between the two solutions of the good old quadratic formula.
Nothing complicated below, but I haven't seen this e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 1
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Summation of $\cos A+\cos(A+B)+ \ldots \cos(A+(n-1)B)$ How would I prove the following result?
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (n-1)B) = \frac{\sin\left(\frac{nB}{2}\right) \cos\left[A+\frac{(n-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
| Induction on $n$.
Checking for $n=1$:
$$\cos A =\frac{\sin(\frac{B}{2})}{\sin(\frac{B}{2})}\cos A =\cos A $$
Induction hypothesis:
Suppose that the equality holds for $n=k$, thus
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a])
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.
I wrote out some $m$ to try to conceive... | For any positive integer, $m$, there will be some non-negative integer, $r$ such that $m < 2^r$. This implies that there must be some non negative integer $k \le r$ such that $2^k \mid m$ and $2^{k+1} \nmid m$.$
For example $m = 96 \lt 128 = 2^7$ and $m = 2^5 \times 3$ where $5 \le 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Summation of series. If $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$
Then find the value of $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+..$$
The answer given in the book is $\frac{\pi^2}{8}$.
How can I find this and also how to find summation of fractions? Thank you!
| Riemann's zeta function is given by, $$ \zeta (s)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+.... $$
Group even and odd number containing terms of this function as shown below,
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}....) $$
$$ \zeta (s)=(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$
My try:
and I find when $a=4,b=2$,then
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$
it maybe use Cauchy-Schwarz inequality
Thank you
| Setting $x = a$ and $b = \frac{8}{x}$ we find that we want to minimise $$\sqrt{x^2 + 64} + \sqrt{\frac{64}{x^2} + 1} = \left (\frac{1}{x} + 1 \right ) \sqrt{x^2 + 64}$$
Differentiating, we obtain $$ \frac{x^3 - 64}{x^2 (x^2 + 64)} $$ Setting this equal to zero, we obtain a real extremum at $x=4$, corresponding to $a=4,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying large exponents in modular arithmetic like $1007$ in $4^{1007} \pmod{5}$ How would I rigorously prove that $4^{1007} \pmod{5} = 4$ and $4^{1008} \pmod{5} = 1$?
I was simplifying a larger modular arithmetic problem ($2013^{2014} \pmod{5}$) and got it down to $4^{1007} \pmod{5}$ and am wondering if there's a ... | Another method for this is repeated squaring.
Take $n^d \mod b$ then
find all $a^{2^k} \mod b$ for all $2^k <= d$ where $d$ is $n^d$
$$\begin{align}
4^2 &= 16 = 1 \\
4^4 &= (4^2)^2 = 1^2 \\
4^8 &= (4^4)^2 = 1^2 \\
&... \\
4^{512} &= 1 \\
\end{align}$$
for $4^{1007}$ we have:
$$\begin{align}
4^{1007}
&= 4^{512} * 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
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Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
| Multiplying by 81 on the top and the bottom gives
$\displaystyle\frac{16x^4-81y^4}{27(2x+3y)}=\frac{(4x^2-9y^2)(4x^2+9y^2)}{27(2x+3y)}=\frac{(2x-3y)(2x+3y)(4x^2+9y^2)}{27(2x+3y)}=$
$\frac{1}{27}(2x-3y)(4x^2+9y^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers
| I will look at
$x(x+1) = y(y+1)(y^2+k)$
for integral $k \ge 1$.
This becomes the original question
when $k = 2$.
I will show that
there are no solutions in
positive integral $x$ and $y$
for $y \ge k+2$.
Note that
$(x, y)
=(k^2-k, k-1)$
and
$(k^2+3k+1, k+1)
$
are solutions to this,
and there is no solution
with $y = k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the tangent line to a curve Find an equation for the tangent line to the curve
$$x\sin(xy-y^2)=x^2-1$$
through the point $(1,1)$.
| When the curve is given by $y = f(x)$ then the slope of the tangent is $\frac{dy}{dx}$,then the equation of the tangent line can be found by the equation.$$y=\frac{dy}{dx}x+c$$the value of $c$ can be found with the help of the given point $(1,1)$. given:$$x\sin(xy-y^2)=x^2-1$$$$y^2-xy+arcsin\left(\frac{x^2-1}{x}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Increasing and bounded sequence proof Prove that the sequence $a_n= 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln(n)$ is increasing and bounded above. Conclude that it’s convergent.
This what I got so far
Proof:
Part 1: Proving $a_n$ is increasing by induction.
Base:
$a_1=1$
$a_2=1+\frac 12= \frac 32$
$a_1≤a_2$
So the b... | This sequence is NOT increasing and is in fact DECREASING.
Proof:
$$a_n = 1+\frac{1}{2} + \cdots + \frac{1}{n} - \ln(n)$$
$$a_{n+1}=1+\frac{1}{2} + \cdots + \frac{1}{n}+\frac{1}{n+1} - \ln(n+1)$$
so
$$a_{n+1}=a_n + (\ln(n) +\frac{1}{n+1} - \ln(n+1)) = a_n + \frac{1}{n+1} - (\ln(n+1) - \ln(n))$$
$$\ln(n+1)-\ln(n) = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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integration by parts !!!! PD: I did a little change in the denominator !!!!
I need to solve this integral using integration by parts.
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$
Thanks!
PS: I know that I can to do:
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}=\int\frac{(x-c)\,dx}{\sqrt{(a^2+... | You don't actually need integration by parts. First, observe the following:
$$\left[\sqrt{a^2+b^2+(x-c)^2}\right]'=\frac {x-c}{\sqrt{a^2+b^2+(x-c)^2}}$$
Integrating both sides and isolating the integral under question gives
$$\int\frac x{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx=\sqrt{a^2+b^2+(x-c)^2}+c\int\frac {\mathrm dx}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When is $2^n \pm 1$ a perfect power Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?
I know that Catalan's conjecture (or Mihăilescu's theorem) gives the result directly, but I'm hopefully for a more elementary method.
I can show that it is never a square, except... | I collected everything you may need about elementary methods solving special cases of Mihailescu theorem here (see cases (iii) and (vii) for this problem)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 3,
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Show $\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{k+1} $ I have been attempting to show $$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1} $$ and my work is
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{r!((k+1)-r)!} + \frac{(k+1)!}{(r+1)!((k+1)-(r+1))!}$$
$$=\frac{(k+1)!(r+1)}{(r+1)!((k+1)-r)!} + \frac{(k+1)!
... | Without assuming much, we have (just by definition):
$\binom{n}{r}+\binom{n}{r-1}=\frac{n!}{(n-r)! r!}+\frac{n!}{(n-(r-1))! (r-1)!}$
$\frac{n!}{(n-r)! r!}+\frac{n!}{(n-(r-1))! (r-1)!}=\frac{n!}{(n-r)! r!}+\frac{n!}
{(n-r+1))! (r-1)!}$
$\frac{n!}{(n-r)! r!}+\frac{n!}
{(n-r+1))! (r-1)!}=\frac{n!}{(n-r)!(r-1)!}(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I get the integral of $\frac{1}{(x^2 - x -2)}$ I'm working with this problem $$ \int \frac{1}{x^2 - x - 2}$$
I'm thinking I break up the bottom so that it looks like this $$\int \frac{1}{(x-2)(x+1)} $$
Then I do $$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiple both sides by the common denominator and co... | In your partial fraction,
$\dfrac{1}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}$
So,
$\dfrac{1}{(x-2)(x+1)} = \dfrac{A(x+1) + B(x-2)}{(x-2)(x+1)}$
and hence,
$1 = (A+B)x + (A-2B)$
I think you can take it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$ $\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: Let $\tan x = t$. Then $\sec^2 xdx = dt\Rightarrow \displaystyle dx = \frac{1}{1+\tan^2 t}dt\Rightarrow dx = \frac{1}{1+t^2}dt$
So $\displaystyle \int\frac{t+... | Mathematica evaluates it as
$$\frac{\tan ^{-1}\left(\frac{2 \tan (x)-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{1}{6} \log \left(\tan ^2(x)-\tan (x)+1\right)-\frac{1}{3} \log (\tan (x)+1)$$
This can be done via partial fractions.
\begin{align*}
\int \frac{t}{1 + t^3} \; dt &= \frac 1 3 \int \frac{t+1}{t^2 - t + 1} dt - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Calculus limit homework problem $$ \lim_{n→\infty} \frac1 n \left(\left(a + \frac 1 n\right)^2 + \left(a + \frac 2 n\right)^2 + ... + \left(a + \frac{n-1}{n}\right)^2\right)$$
$$ \text{hint: }\ 1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} $$
I can't figure out how to find this problem's limit. Does anybody have ide... | \begin{align*}
& \; \; \; \lim_{n\to \infty}\frac 1 n \left( \left(a + \frac 1 n \right )^2 + \dots +\left(a + \frac {n-1} n \right )^2 \right ) \\
&= \lim_{n\to \infty} \frac 1 n \left( (n-1)a^2 + 2 a \left( \frac 1 n + \dots + \frac{n-1}{n} \right ) + \left( \frac {1^n}{n^2} + \dots + \frac{(n-1)^2}{n^2}\right ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Finding Volume using triple integrals Find the volume common to $y^2+z^2=4ax$ and $x^2+y^2=2ax$
I have problem establishing the common region. I know the region but unable to visualize it to get to the limits of the integral.
| $x^2 + y^2 = 2 ax \implies (x-a)^2 + y^2 = a^2$ is a cylinder centered at $(a,0)$ and parallel to $z-$ axis.
$y^2+z^2=4ax \implies \frac 1 {4a} (y^2 + z^2) = x $ is a parabloid opening towards $+$ ve $x$-axis. Putting $z=0$, gives two curves are $(x-a)^2 + y^2 = a^2$ and $y^2 = 4ax$, and the circle completely lines i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotic expansion of $(1+\frac{t}{n})^{-n-1}$ at $n \to \infty$ I'm reading through a proof in Analytic Combinatorics by Flajolet/Sedgewick and I have come across this:
We have the asymptotic expansion:
$(1+\frac{t}{n})^{-n-1}=e^{-(n+1)\log(1+\frac{t}{n})}=e^{-t}[1+\frac{t^2-2t}{2n}+\frac{3t^4-20t^3+24t^2}{24n^2}+..... | Using the Taylor expansion of the logarithm, we obtain
$$\begin{align}
(n+1)\log \left(1 + \frac{t}{n}\right) &= (n+1)\left(\frac{t}{n} - \frac{t^2}{2n^2} + \frac{t^3}{3n^3} + O\left(\frac{1}{n^4}\right)\right)\\
&= t + \frac{t}{n} - \frac{t^2}{2n} - \frac{t^2}{2n^2} + \frac{t^3}{3n^2} + \frac{t^3}{3n^3} - \frac{t^4}{4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Breaking up an integral using substitution? I am faced with the following integral, and am required to use substitution to solve it, using $u = \frac{1}{x}$.
$$\int \dfrac{1}{x^2\sqrt{x^2+1}} \, dx$$
I know I can break up the equation in:
$$\int \dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{\sqrt{x^2+1}} \, dx.$$
How would I go a... | $x=\tan u$ $\Rightarrow$ $dx=(1+\tan^{2}u)du$
So
\begin{align*}
\int\frac{1}{x^{2}\sqrt{x^{2}+1}} & =\int\frac{1}{\tan^{2}u\sqrt{1+\tan^{2}u}}(1+\tan^{2}u)du=\int\frac{\sqrt{1+\tan^{2}u}}{\tan^{2}u}du\\
& =\int\frac{\sec u}{\tan^{2}u}du
\end{align*}
$$
$$
after some simplification
\begin{align*}
\int\frac{\sec u}{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Writing a proposition as the conjunction of two conditional statements For each integer a, a is congruent to 3(mod 7) if and only if (a^2 + 5) is congruent to 3(mod 7).
A)Writing a proposition as the conjunction of two conditional statements
B) determine if the conditionals are true or false
| For completeness and because any deadline has probably passed for this homework-type quesiton, I will construct a full solution.
A) Let $A_a$ be defined as "$a$ is congruent to $3\bmod 7$" and $B_a$ defined as "$a^2+5$ is congruent to $3\bmod 7$." We are given $(\forall a \in \mathbb{Z})A_a \equiv B_a$, which by defini... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real... | We need to prove that
$$\sum_{cyc}\frac{1}{2a^2-6a+9}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or
$$\sum_{cyc}\left(\frac{18(a^2-3a+2)}{2a^2-6a+9}+1\right)\geq3$$ or
$$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}\geq\frac{3}{5}.$$
Now, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
| $a+b, c+a$ is positive real,
consider $\frac{1+a}{2}= \frac{(a+b)+(c+a)}{2} \geq \sqrt{(a+b)(c+a)}$
i.e$(1+a)\geq2\cdot\sqrt{(1-c)(1-b)}$
and so..
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far.
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true
Induction: Assum... | Set $f(n) = n^3 + (n+1)^3 + (n+2)^3$.
Hint: $f(n+1)-f(n) = (n+3)^3 - n^3 = 3 * 3 * n^2 + 3 * 3^2 * n + 3^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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On Shanks' quartic approximation $\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)$ In Mathworld's "Pi Approximations", (line 58), Weisstein mentions one by the mathematician Daniel Shanks that differs by a mere $10^{-82}$,
$$\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)\color{blue}{+10^{-82}}\tag{1}$$
and says that $u$ is a produc... | Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then
$$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$
As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressi... | {
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"url": "https://math.stackexchange.com/questions/515172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$ Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I don't have any idea on... | If @Cocopops is correct, in that the +,- signs go like +,+,-,+,+,+,-,+,+,+, ... and the aperiodicity is just at the beginning, this is far less impressive.
Then if
$$x= \sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}} $$
then $$
y = \sqrt{5+x} = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$ Let
$$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$
its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$
$S$ can be represented in terms of the generalized hypergeometric function:
$$S=... | We show that the sum equals
$$
\int_0^1 \frac{2-3x}{1-x^2+x^3} dx.
$$
This integral is "elementary", but requires expanding the integrand
in partial fractions, which in turn requires all the solutions of
the cubic polynomial in the denominator; so if one insists on
writing everything in radicals then the answer is boun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
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If both the sum and sum of squares of two rationals are integers, the two rationals are integers too There are two rational numbers $\alpha, \beta$ such that $\alpha + \beta,\ \alpha^2 + \beta^2$ are both integers. Prove that $\alpha, \beta$ are integers.
I started off by assuming that $\alpha = \frac{a}{b}, \beta = \f... | Since $\alpha + \beta$ is an integer, $\alpha$ and $\beta$ have the same denominator; i.e. $\alpha = \frac{a}{b}$, $\beta = \frac{c}{b}$, in reduced form.
Then $2\alpha\beta = (\alpha + \beta)^2 - (\alpha^2 + \beta^2)$ is an integer, so $\frac{2ac}{b^2}$ is an integer. This means $b^2 \;\big| \; 2$, which means $b = 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
| It can be seen that $\sqrt{x^2+\frac{1}{x^2}}$ is convex. Therefore we can use Jensen inequality:
$$
\frac{1}{3}\sqrt{a^2+1/a^2}+\frac{1}{3}\sqrt{b^2+1/b^2}+\frac{1}{3}\sqrt{c^2+1/c^2} \ge \sqrt{(\frac{a+b+c}{3})^2+(\frac{3}{a+b+c})^2}.
$$
On the other hand because $a+b+c<1$, we can see that $f(x)=\sqrt{\frac{x^2}{9}+... | {
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"timestamp": "2023-03-29T00:00:00",
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Functional equation $f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$ Find all of the functions defined on the set of integers and receiving the integers value, satisfying the condition:
$$f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$$
for each pair of integers $(a,b)$.
| An easier answer.
Using $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ and putting $x=f(a+b),y=-f(a),z=-f(b)$ we get that $x^3+y^3+z^3-3xyz=0$ so we have that either $x+y+z=0$ or $x^2+y^2+z^2-xy-yz-zx=0$. If $x+y+z=0$ then we have that $f(a+b)=f(a)+f(b)$ which gives us $f(a)=af(1)$. If the second one is true we get f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove algorithm for solving a square congruence when p ≡ 5 (mod 8) I'm having trouble understanding why this algorithm works and where it comes from:
"Suppose p ≡ 5 (mod 8) is a prime and y is a square (mod p); that is, for some $ x, x^2
≡ y\ (mod\ p)$. This can be solved by the following algorithm
*
*Comput... | The $d^2 \equiv -1 \pmod{p}$ part is wrong. Presumably, the author mixed some parts of the $p \equiv 1 \pmod{8}$ case in, where one computes $d = y^{(p-1)/8}$.
For $p \equiv 5 \pmod{8}$, we always have $1 = \left(\frac{y}{p}\right) \equiv d^2 \pmod{p}$.
So if $d \equiv 1 \pmod{p}$, we choose $r \equiv y^{(p+3)/8}\pmod{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you factor $x^3 - 8 = 0$? I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$?
| First of all, just a comment about the terminology. You can factor a polynomial, you can't factor an equation. What you are trying to do is solve the equation $x^3 - 8 = 0$.
As $2$ is a zero of the polynomial, $(x-2)$ is a factor by the factor theorem. So $x^3 - 8 = (x-2)p(x)$ for some polynomial. As the degree of a pr... | {
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"timestamp": "2023-03-29T00:00:00",
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How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
| Let $b^2=1-\delta$ where $\delta\ge 0$. Then, $a^2=2-c^2+\delta$. Now, $(a+b)^2=a^2+b^2+2ab=3-c^2+2\sqrt{1-\delta}\sqrt{2-c^2+\delta}$. The product of square roots is decreasing in $\delta$ (we can explicitly differentiate or just note that $1-\delta<1$ while $2-c^2+\delta>1$ so a small increase in $\delta$ decreases f... | {
"language": "en",
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Inclusion-Exclusion How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 32$ with $0 \leq x_i \leq 10$ for $i = 1, 2, 3, 4$?
How many integer solutions are there to the inequality
$$
y_1 + y_2 + y_3 + y_4 < 184
$$
with $y_1 > 0$, $0 < y_2 \leq 10$, $0 \leq y_3 \leq 17$, and $0 \leq y_4 < 19$?
H... | Generating functions seem to give a cleaner looking solution for the second question.
The generating function for number of ways to sum to $k$ is
$$
\overbrace{\ \ \frac{x\vphantom{x^1}}{1-x}\ \ }^{y_1\gt0}\overbrace{\frac{x-x^{11}}{1-x}}^{0\lt y_2\le10}\overbrace{\frac{1-x^{18}}{1-x}}^{0\le y_3\le17}\overbrace{\frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
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If $x=\prod^{27}_{n=1}(1+\frac{2}{n})$ then find $13x$ - Ramanujan Mathematics olmpiad 2013 I tried this: $$x=\prod^{27}_{n=1}(1+\frac{2}{n})=(1+\frac{2}{1})(1+\frac{2}{2})(1+\frac{2}{3})\ldots(1+\frac{2}{27})=\frac{3}{1}\cdot\frac{4}{2}\cdot\frac{5}{3}\cdots\frac{29}{27}$$ Then the terms cancel out. But I am not getti... | As you noted, $$x=\frac31\cdot\frac42\cdot\frac53\cdots\frac{29}{27}=\frac11\cdot\frac12\cdot\frac{28}1\cdot\frac{29}1=406$$hence $$13x=13\cdot406=5278$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form,... | Hint: $y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
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Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
| $${ \left( \frac { a }{ b } \right) }^{ 2 }+1={ \left( \frac { c }{ b } \right) }^{ 2 }\\ \frac { c }{ b } +\frac { a }{ b } =k\\ \frac { c }{ b } -\frac { a }{ b } =\frac { 1 }{ k } \\ \frac { c }{ b } =\frac { { k }^{ 2 }+1 }{ 2k } \\ \frac { a }{ b } =\frac { { k }^{ 2 }-1 }{ 2k } \\ $$Say:$$a=({ k }^{ 2 }-1)n\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ Prove that
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$
What should I do for this equation? Should I focus on proving $\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}... | I view that the purpose on the lefthand side of the eqn as:
Let $A = {x_{1},x_{2},...,x_{n}}$ and $B={y_{1},y_{2},...,y_{m}}$. where $n \ge m$
$ \left( \begin{array}{cc}
m \\
0 \\
\end{array} \right) $
$ \left( \begin{array}{cc}
n \\
m \\
\end{array} \right) $ accounts the number of ways to choose $m$ elements f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How ... | Here's another approach you might use, a sort of brute-force approach. You said you got as far as $$(n^2 + 3n)(n^2 + 3n + 2).$$ Continuing to multiply, you get: $$ n^4 + 6n^3 + 11n^2 + 6n.$$
The claim is that that this is one less than a perfect square, or equivalently, that $$n^4 + 6n^3 + 11n^2 + 6n + 1$$ is a perfe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
For what values of a and b does the following limit equal 0?
I understand I need to make the sum of the individual limits equal 0 - but I'm a little lost. I computed the limit of the first term to be -4/3 via L'Hospitals Rule but Wolframalpha contradicts me (http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin%28... | First let us put this into a better form, with one variable term and one constant term:
$$ \lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a = 0$$
well if we evaluate the limit using L'Hopitals we get:
$$\lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} = \lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2} $$
$$\lim\limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove : $\dfrac{a}{ac+1}+ \dfrac{b}{ab+1}+ \dfrac{c}{bc+1} \le \frac 12 (a^2+b^2+c^2)$ $a;b;c\in \mathbb{R}^{+}$ such that $abc=1$
Prove : $\frac{a}{ac+1}+ \frac{b}{ab+1}+ \frac{c}{bc+1} \leq \frac{1}{2}(a^2+b^2+c^2)$
| $\frac{a}{ac+1} \leq \frac{a}{2\sqrt{ac}} = \frac{1}{2} a \sqrt{b}$. So it suffices to show that
$$a^2+b^2+c^2 \ge (abc)^{1/6} (a\sqrt{b} + b\sqrt{c} + c\sqrt{a}) = \sum_{cyc} a^{7/6}b^{2/3}c^{1/6}$$
This follows from adding up the AM-GM inequalities:
$$7a^2+4b^2+c^2 \ge 12 a^{7/6}b^{2/3}c^{1/6}$$
and its cyclic var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\new... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
} |
Derivative Arc-tangens We have to show that $(\arctan(x))' = \frac{1}{1+x^2}$, derived with the chain rule.
The hint given is that we should start with deriving $\tan(\arctan(x)) = x$; I am not sure though how this is helpful, since the derivative of $\arctan(x)$ is what we are looking for yet..
| You are given $\tan(\tan^{-1} x) = x$. Take a derivative of that to get $\displaystyle \sec^2(\tan^{-1} x) \cdot \frac{d}{dx} \left(\tan^{-1} x\right) = 1$.
Now, consider a right triangle with adjacent length $1$ and opposite length $x$, so that the triangle has an angle $\theta$ such that $\tan \theta = x$. Now we tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it... | You may also use the following formula valid for $x\neq 1$:
$\sum_{k=1}^{n-1}\frac{1}{(x-1)^2+4x\sin^2(\frac{k\pi}{n})}=\frac{n(x^n+1)}{(x^2-1)(x^n-1)}-\frac{1}{(x-1)^2}$.
The left hand side of the equation can be written as $\frac{x^{n+1}(n-1)-(n+1)x^n+x(n+1)-n+1}{(x-1)^3}\cdot \frac{1}{(x^{n-1}+x^{n-2}+\ldots +1)(x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 8,
"answer_id": 7
} |
Finding a,b,c,d in a quartic expression Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far.
\begin{align}
a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51
\end{align}
Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)... | You have $p(x) = x^4 + ax^3 + bx^2 + cx + d$, and you're given that $a + b + c + d = 9$, $8a + 4b + 2c + d = 4$, and $27a + 9b + 3c + d = -51$
Now, $p(12) + p(-8) = 12^4 + 8^4 + (12^3 - 8^3) a + (12^2 + 8^2) b + (12 - 8) c + 2d = 24832 + 1216 a + 208 b + 4c + 2d = 24832 + 1216 a + 208 b + 2 (2c+d)$. You note that $2c +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/547966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Let $a,b,c \in \mathbb{R^+}$, does this inequality holds $\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$? Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours an... | Assume $n,k\in\mathbb R^+$.
Necessary condition.- Notice that LHS-RHS is equal to $$\frac{(c-1)^2k(k-n)}{(k+n)(kc+n)(k+nc)}$$ for $a=1,b=1$. Therefore it is mandatory that $k\geq n$ for the inequality to hold.
Sufficient condition.- ...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Vector norm Inequality proof Does anyone know how to start proving this inequality
$$
\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{4 \|x-y\|}{\|x\|+ \|y\|}
$$
The norm is a random norm on a vector space $V$
| Here is a somewhat complicated proof, I do not know if there is a simpler
and more natural one.
Let $a=||x||,b=||y||,c=||x-y||$. The inequality is easy if $a=b$, so
assume that it is not the case. Since the problem is symmetric in
$x$ and $y$, we can assume $a < b$.
Then $2a < 2b$, so $a+b < 3b-a$, $ \frac{a+b}{3b-a}<1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute the exact length of the curve $y = \frac{x^2}{4}-\frac{1}{2}\ln(x)$ from $x=1, x=2$ So here is my problem: $y = \frac{x^2}{4}-\frac{1}{2}\ln(x)$
Taking the derivative:
$$\frac{\textrm{d}y}{\textrm{d}x}=\frac{x}{2}-\frac{1}{2x}$$
And that simplifies further to:
$$\frac{\textrm{d}y}{\textrm{d}x}=\frac{x^2-1}{2x... | \begin{align}
&\int_{1}^{2}\sqrt{1 + \left(\frac{{\rm d}y}{{\rm d}x}\right)^2}\,{\rm d}x
=
\int_{1}^{2}\sqrt{1 + \left(\frac{x^{2} - 1}{2x}\right)^2}\,{\rm d}x
=
\int_{1}^{2}\sqrt{1 + \frac{x^{4} - 2x^{2} + 1}{4x^{2}}}\,{\rm d}x
\\[3mm]&=
\int_{1}^{2}\sqrt{\frac{x^{4} + 2x^{2} + 1}{4x^{2}}}\,{\rm d}x
=
\int_{1}^{2}\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/553837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
| This result was a problem in American mathematical monthly in 50s.
Source:
H. F. Sandham and Martin Kneser, The American mathematical monthly, Advanced problem 4305, Vol. 57, No. 4 (Apr., 1950), pp. 267-268
\begin{align}S&=\sum_{n=1}^\infty \left(\frac{\text{H}_n}{n}\right)^2\\
&=\sum_{n=1}^\infty \frac{1}{n^2}\left(\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 8,
"answer_id": 4
} |
How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfra... | let $$x=\dfrac{bc}{a^2},y=\dfrac{ca}{b^2},z=\dfrac{ab}{c^2}$$
then we only prove follow inequality
$$\dfrac{a^4}{3a^4+2b^2c^2+4a^2bc}+\dfrac{b^4}{3b^4+2c^2a^2+4b^2ca}+\dfrac{c^4}{4c^4+2a^2b^2+4c^2ab}\ge\dfrac{1}{3}$$
By Cauchy-Schwarz inequality
\begin{align*}
\sum_{cyc}\dfrac{a^4}{3a^4+2b^2c^2+4a^2bc}&\ge\dfrac{(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Expected number of rolls required to get sum greater than n for n faced die? Suppose a guy has a die with $n$ faces. He can go on rolling it as many times as possible and add the sum of each outcome. What is the expected number of rolls after which the sum is at least $n$?
| I'm assuming the OP means what is the expected number of rolls until the sum is at least $n$ (as Alecos assumes in his answer).
Let $E(m)$ be the expected number of rolls until the sum is at least $n$, starting with a sum of $m$.
Then we have, for a start:
\begin{align*}
E(n) &=0 \\
E(n-1) &=1 \\
E(n-2) &=1+\frac{1}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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France Olympiad Team Selection Test 2005 In an international meeting of $n ≥ 3$ participants, $14$ languages are spoken. We know that:
- Any $3$ participants speak a common language.
- No language is spoken by more than half of the participants.
What is the least value of $n$?
| Set up the standard incidence matrix.
Let the row be the $n$ participants and the 14 columns be the languages. Fill in the entry of 1 if the participant can speak the language.
We wish to count the number of column triples $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. Since every 3 participants speak a common language,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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problem on solving equations with three variables Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously.
$x^3+y^3+z^3=x+y+z$
$x^2+y^2+z^2=xyz$
| The equations do not have positive solutions. It follows directly from AM-GM inequalities.
Recall that $x,y,z> 0$ then
$x+y+z=x^3+y^3+z^3\geq3xyz=3(x^2+y^2+z^2)$
$3(x^2+y^2+z^2)-(x+y+z)^2=(x-y)^2+(y-z)^2+(z-x)^2\geq 0$
Hence $x+y+z\geq (x+y+z)^2$
then $x+y+z\leq 1$.
But $x,y,z>0$, so $x,y,z$ are strictly smaller than 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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The largest value of $f(n) $ Let us consider a function $ f:\mathbb{N}_0 \to \mathbb{N}_0 $ following the relations:
*
*$ f(0)=0$
*$f(n) = n+f\left(\left\lfloor \frac{n}{p} \right\rfloor\right)$ when the $n$ is not divisible by $p$
*$ f(np) = f(n) $
Here $p>1$ is a positive integer .$ \mathbb{N}_0 $ is the set... | Proof by induction on $n$:
$$\begin{eqnarray}
f(p^{n+1}-1)=\\
f((p-1)(1+p+p^2+ \cdots p^{n-1}+p^n))=\\
f((p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1) p^{n-1} + (p-1)p^n)=\\
((p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1) p^{n-1} + (p-1)p^n)
+ f( \lfloor \frac{p-1}{p} + (p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1)p^{n-1} \rfloor)=\\
p^{n+1}-1 + f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/558005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Exponentiation by squaring I need to calculate $7^{2012} \mod {13}$ by hand using exponentiation by squaring, but I cant seem to figure it out.
I started with this but I don't know for sure if its correct or where it's going.
\begin{align}
(7^2)^{1006} &\equiv 49^{1006} \pmod {13} \\
(10^2)^{503} &\equiv 100^{503} \pm... | Looks good so far. Continuing, we find that:
\begin{align*}
100^{503} \pmod {13}
&\equiv 9^{503} \pmod {13} \\
&\equiv 81^{251} \cdot 9 \pmod {13} \\
&\equiv 3^{251} \cdot 9 \pmod {13} \\
&\equiv 9^{125} \cdot 3 \cdot 9 \pmod {13} \\
&\equiv 81^{62} \cdot 9 \cdot 1 \pmod {13} & \text{since $3 \cdot 9 = 27 = 2\cdot 13 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find a minium value of a function Given x,y,z are positive real numbers such that
$$
x^2+y^2+6z^2=4z(x+y).
$$
Find the minimum value of the following function
$$
P=\frac{x^3}{y(x+z)^2}+\frac{y^3}{x(y+z)^2}+\frac{\sqrt{x^2+y^2}}{z}
$$
| let $$a=\dfrac{x}{z},b=\dfrac{y}{z}$$
then $$a^2+b^2+6=4(a+b)$$
and we only find this minimum
$$\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}+\sqrt{a^2+b^2}$$
use Cauchy-Schwarz inequality,we have
$$\left(\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}\right)(ab(a+1)^2+ba(b+1)^2)\ge (a^2+b^2)^2$$
then we only find this follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Power series help please! $$ \frac{2}{3-x} $$
I need to find a power series representation for this. I figured i'd pull the two out, but I can't figure out what to do with the three.
| We know that $1 + r + r^2 + r^3......= \dfrac{1}{1-r}$. So we can write $\dfrac{2}{3-x} = \frac{2}{3}\dfrac{1}{1-\frac{x}{3}} = \frac{2}{3}(1+\frac{x}{3}+\frac{x^2}{9} + \frac{x^3}{27} + \frac{x^4}{81}.......)= \frac{2}{3}\sum_{n=0}^\infty(\frac{x}{3})^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/564341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Average of 3 consecutive odd numbers
The average of $3$ consecutive odd numbers is $14$ more than one third of the first of these numbers, what is the last of these numbers?
$17/19/15/$data inadequate/none of these
Let three consecutive odd numbers be $a-2,a,a+2$. Their average is $a.$ So, $a=\frac13(a-2)+14\implies ... | We take $3$ odd numbers $a,b,c$ such that $a= (2n+1), b=(2n+3), c=(2n+5)$ for some integer $n$. From information given, we know $$\frac{(2n+1)+(2n+3)+(2n+5)}{3} = 14 + \frac{1}{3}(2n + 1) \\ \frac{6n+9}{3}=\frac{43+2n}{3} \\ \frac{4n-34}{3} = 0 \\ n = \frac{34}{4} \\ n=8.5$$ Since n is not an integer, the solution cann... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Someone who is calculating $43434343^2$. The answer is $18865ab151841649$, where the two digits $a$ and $b$ were lost. So I used congruence $\bmod 9$ and $\bmod 11$. First, $43434343^2 \pmod{11}$ is $5$ and I did the answer $18865ab15184169 \pmod{11}$ and got $-b+a+3 \pmod{11}$. I also did $\bmod 9$ and got $a+b+67 \pm... | The simplest is probably to look modulo $99$.
Since $100 \equiv 1 \pmod {99}$,
$43434343^2 \equiv (43+43+43+43)^2 \equiv 172^2 \equiv 73^2 \equiv 5329 \equiv 82 \pmod {99} $
$18865ab151841649 \equiv 18+86+(50+a)+(10 b+1)+51+84+16+49 \equiv 355+ba \equiv 58+(10b+a) \pmod {99}$.
Hence $(10b+a) \equiv 24 \pmod {99}$. Luc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int\limits_0^\pi \frac{\sin^2x}{2-\cos x}\ \mathrm dx$ by complex methods find integral $$\int\limits_0^\pi \frac{\sin^2x}{2-\cos x}dx$$
what I had in mind is to use Euler formula, to turn it into a complex integral and change the limits of integration from $ -\pi$ to $\pi$ so that any odd parts of the in... | Hint: Let $\gamma\colon [-\pi, \pi]\to \Bbb C, \theta \mapsto e^{i\theta}$. It is true that $\displaystyle \int \limits_0^ \pi\frac{(\sin(x))^2}{2-\cos(x)}\mathrm dx=\dfrac 1{4i}\int _\gamma \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}\mathrm dz$.
Further details: Note that for all $t\in \Bbb R$, $$\sin(t)=\dfrac{e^{it}-e^{-it}}{2... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there are no such positive integers $a,b,c,d$ such that $a^2 + b^2 = 3(c^2 + d^2)$ Prove that there are no positive integers a, b, c, d such that $a^2 + b^2 = 3(c^2 + d^2)$. Hint: What can you say about divisibility of a and b by 3? Look at solution with smallest possible a.
| By way of contradiction, suppose there exist a,b,c,d $\in \mathbb N$ such that $a^2 + b^2 = 3(c^2 + d^2)$.
(Note: see that the square of a natural number is also natural and the sum of natural numbers is also natural.)
Now let $a_o , b_o , c_o , d_o $ be the smallest solution set to the equation.
Then 3|$a_o$$^2 + b_... | {
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"url": "https://math.stackexchange.com/questions/570922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many 5 digits numbers are there, whose digits sum to 22?
How many 5 digits numbers are there, whose digits sum to 22? Of course the first digit has to be larger than 0.
| Using generating functions:
$$[x^{22}](x+x^2+\cdots+x^9)(1+x+\cdots +x^9)^4=\\
[x^{21}]\frac{1-x^9}{1-x}\cdot \frac{(1-x^8)^4}{(1-x)^4}=\\
[x^{21}](1-x^9)(1-x^8)^4(1-x)^{-5}=\\
[x^{21}]\sum_{i=0}^1 {1\choose i}(-x^9)^i\sum_{j=0}^4 {4\choose j}(-x^8)^j\sum_{k=0}^{\infty}{4+k\choose k}x^k=\\
\{(i,j,k)=(0,0,21),(0,1,11),(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/572321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $a=2,b=2,c=2$ for a system of three equations. Let us consider a system equations :
\begin{gather}
a^3+b=3a+4\tag{i}\\
2b^3+c=6b+6 \tag{ii}\\
3c^3+a=9c+8\tag{iii}
\end{gather}
I have tried with the following steps :
$6\times(i)+3\times(ii)+2\times$(iii) gives :
$6(a^3+b^3+c^3)+6b+3c+2a=18(a+b+c)+58$. But... | Write the equations as
\begin{align*}
(a-2)(a+1)^2 = 2-b\\
2(b-2)(b+1)^2 = 2-c\\
3(c-2)(c+1)^2 = 2-a
\end{align*}
Note that $(a,b,c) = (2,2,2)$ is a valid solution.
$a>2 \Rightarrow b<2\Rightarrow c>2 \Rightarrow a<2. $ A similar logic precludes $a<2$. This forces $a=b=c=2$ which is the unique solution
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how to solve trigonometric inequalities? how does one solve trigonometric inequalities? Is there a method to this or is every solution done ad hoc?
simple equations of the type: $cos3x \leq 0$ when: $0\leq x \leq 2π$
The attempt at a solution: equating $cos 3x = 0$ yields $$ π /6 + 2\frac13πk\leq x \leq 2π -π /6 -... | I use Nghi H Nguyen's method to visually solve trig inequalities by the number unit circle.
Solve: $ F(x) = \cos (3x) < 0$
There are totally 6 end points at: $ \frac{\pi}{6}, \frac{\pi}{2}, 5\frac{\pi}{6}, 7\frac{\pi}{6}, 3\frac{\pi}{2} $ and $11\frac{\pi}{6} $that divide the unit circle into $6$ equal arc lengths.
Use... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substi... | Putting $\displaystyle x=a\sin t,t=\arcsin\frac xa \implies -\frac\pi2\le t\le \frac\pi2$ as the principal value of inverse sine ratio lies $\displaystyle\left[-\frac\pi2,\frac\pi2\right]$
$\displaystyle\implies \cos t\ge0$
$\displaystyle\implies\sqrt{a^2-x^2}=\sqrt{a^2\cos^2t}=|a\cos t|=|a|\cos t$
$\displaystyle \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Laurent Series of $(z^2-1)^{-2}$ I have problems to determine the Laurent series of the function $(z^2-1)^{-2}$ in the regions:
$$0<|z-1|<2$$
and
$$|z+1|>2$$.
My idea was as follows:
$$\frac{1}{z^2-1}=\frac{1}{4z(z-1)^2}-\frac{1}{4z(z+1)^2}$$
However, this procedure only makes it harder to make the series, and tried to... | Make a partial fraction decomposition:
$$\begin{align}
\frac{1}{z^2-1} &= \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right)\\
\frac{1}{(z^2-1)^2} &= \frac{1}{4}\left(\frac{1}{(z-1)^2} - \frac{2}{z^2-1} + \frac{1}{(z+1)^2}\right)\\
&= \frac{1}{4}\left(\frac{1}{(z-1)^2} - \frac{1}{z-1} + \frac{1}{z+1} + \frac{1}{(z+1... | {
"language": "en",
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"source": "stackexchange",
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Verifying a solution from Strauss' "Partial Differential Equations - An Introduction", 2nd edition In section 9.2 on page 241, question #12 is given as follows:
"Solve the three-dimensional wave equation in $\{r\ne0,t>0\}$ with zero initial conditions and with the limiting condition
\begin{equation*}
\lim_{r\to 0}4\pi ... | I prefer to write $u_{rr}+\frac{2}{r}u_r$ as $\frac{1}{r^2}(r^2 u_r)_r$, which has more meaning ($r^2 u_r$ is the flux through a sphere of radius $r$) and is easier to calculate with. Given
$$u=-\frac{1}{4\pi} r^{-1} g(t-r/c)$$
we get
$$u_r= \frac{1}{4\pi} r^{-2} g(t-r/c)+\frac{1}{4\pi c} r^{-1} g'(t-r/c)$$
hence
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(... | Let us utilize the fact that $a,b,c$ are real
Using this,
either $\displaystyle(i) ab+bc+ca=0$
$\displaystyle\implies bc-a^2=-(ab+ca)-a^2=-a(a+b+c)$
$\displaystyle\implies \frac a{(bc-a^2)^2}=\frac a{a^2(a+b+c)^2}=\frac{bc}{abc(a+b+c)^2} $
$\displaystyle\implies \sum_{\text{cyc}}\frac a{(bc-a^2)^2}=\frac{bc+ca+ab}{ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581884",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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On a ring $R$ where $x^2 = x$, we must have that $2x = 0$ and $R$ is commutative I want to show that on a ring $R$ where $x^2 = x$ for all elements $x$, we must have that $2x = 0$ and $R$ is commutative. Does my answer look sound?
Well, if $x = 0$ we have that $x^2 = x$ is trivially true.
If $x \neq 0$ then $x^2 = x$
$... | The party line on this problem, since I am such a good party member!: since for all $x \in R$ we have $x^2 =x$, it follows that
$(x + y)^2 = x + y, \tag{1}$
or
$x^2 + xy + yx + y^2 = x + y, \tag{2}$
or
$x + xy + yx + y = x + y, \tag{3}$
or
$xy + yx = 0, \tag{4}$
whence taking $y = x$,
$2x = x + x = x^2 + x^2 = 0, \tag... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -... | P(n) = (2^3n)-1
p(1) = (2^3x1)-1
= 8-1 = 7 is divisible by 7
Let n =k
then p(k) =( 2^3k)-1 = 7x [assume it is divisible by 7]
therefore 2^3k = 7x +1
Let n =k+1
then p(k+1) = (2^3(k+1))-1
= (2^3k+3)-1
= (2^3k x 2^3)-1
= ((7x +1)2^3 ) -1
= ((7x +1)8 ) -1
= (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/584686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Calculation of a square root of a big number
How can I calculate the following number:
$$ \sqrt{444 \cdots (2n \text{ digits}) + 111 \cdots (n+1 \text{ digits}) - 666 \cdots (n \text{ digits})}.$$
My trying :
I have tried to calculate these data by observing the pattern of $ 7^2 , 67 ^2 ,667^2 , 6667^2 , \dots$.... | Hint:
Use the fact that $111\dots 111 = \frac{10^n-1}{9}$ (where there are $n$ ones)
Then your number is
$$4\frac{10^{2n}-1}{9} + \frac{10^{n+1}-1}{9} - 6\frac{10^n-1}{9} = \frac{4\times 10^{2n}-4 +10^{n+1}-1-6\times 10^n+6}{9}$$
$$=\frac{4\times 10^{2n}+4\times 10^n+1}{9}$$
$$= \frac{(2\times 10^n+1)^2}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/585525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Inequality with cube roots $$(\sqrt{n}+1)^{1/3}-(\sqrt{n}-2)^{1/3} \geq (\sqrt{n+1}+1)^{1/3}-(\sqrt{n+1}-2)^{1/3}$$
$$n\in \mathbb{N}$$
I come upon this inequality when trying to use prove series declension for the Leibnit'z criterion.
I haven't been able to find a good solution to this.
| Hint: Consider the function :$f(x)=(\sqrt{x}+1)^{\frac{1}{3}}-(\sqrt{x}-2)^{\frac{1}{3}}$ ,$ \forall x >4$.
$f'(x)=\frac{1}{6\sqrt{x}}(\frac{1}{\sqrt[3]{(\sqrt{x}+1)^2}} -\frac{1}{\sqrt[3]{(\sqrt{x}-2)^2}})=\frac{1}{6\sqrt{x}}\frac{\sqrt[3]{(\sqrt{x}-2)^2}-\sqrt[3]{(\sqrt{x}+1)^2}}{\sqrt[3]{(\sqrt{x}+1)^2}\times\sqrt[3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
| Let $m$ be the slope. Then,
$$m = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Find $m'(x)$ and set this to $0$.
Well, to find $m'(x)$, we use the Quotient Rule and we have $$ m'(x) = \dfrac {(1+x^4)(2x) - (x^2)(4x^3)}{(1+x^4)^2} = \dfrac {2x^5+2x-4x^5=2x-2x^5}{(1+x^4)^2}. $$Well, so the nu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing $(a+b+c)(x+y+z)=ax+by+cz$ given other facts $$x^2-yz/a=y^2-zx/b=z^2-xy/c$$
None of these fractions are equal to 0.We need to show that,
$(a+b+c)(x+y+z)=ax+by+cz$
This question comes from a chapter that wholly deals with factoring homogeneous cyclic polynomials.I multiplied the three sides of the first equalit... | Using Addendo formula $\displaystyle \frac Aa=\frac Bb=\cdots=\frac{A+B+\cdots}{a+b+\cdots},$
$$\frac{x^2-yz}a=\frac{y^2-zx}b=\frac{z^2-xy}c=\frac{x^2-yz+y^2-zx+z^2-xy}{a+b+c}\ \ \ \ (1)$$
Multiplying the numerator & the denominator of the first term by $x,$ and those of second by $y$ and those of third by $z$
$$\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ratio of binomial coefficients If the ratio of the coefficients of $x^6$ and $x^7$ in the expansion of $(2+ax)^{11}$ is $14:25$, find the value of $a$.
What I have so far:
$$\binom{11}{6} \cdot a^6 \cdot 2^5$$
And
$$\binom{11}{7} \cdot a^7 \cdot 2^4$$
| You calculated the coefficients correctly. If $A$ is the coefficient of $x^6$ and $B$ is the coefficient of $x^7$, then you want to find $A/B$, and set it equal to $14/25$.
You calculated the following:
$$A = \binom{11}{6} \cdot a^6 \cdot 2^5$$
$$B = \binom{11}{7} \cdot a^7 \cdot 2^4$$
We can write out the $\binom{11}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to proceed to show this hold by induction?
Show that $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{n+2}{n(n+1)(n+3)}=\frac16\left[\frac{29}6-\frac4{n+1}-\frac1{n+2}-\frac1{n+3}\right],\text{ for $n\in\mathbb N$}.$$
I try induction. For $n=1$, it is trivial and then let it is true for some $k\in\mathbb... | $$\begin{eqnarray*}\frac{n+2}{n(n+1)(n+3)}&=&\frac{1}{n(n+1)}\left(1-\frac{1}{n+3}\right)\\&=&\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(1-\frac{1}{n+3}\right)\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n(n+3)}+\frac{1}{(n+1)(n+3)}\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Apply the Frobenius theorem to a problem Given
$x^2y''+x(x-3)y'+3y=0$
In standard form, we have
$y''+\frac{x-3}{x}y'+\frac{3}{x^2}y=0$
So $p(x)=\frac{x-3}{x}$ and $q(x)=\frac{3}{x^2}$
Find the first and second solutions. I'm not sure how to start this.. I know I have to find the indicial roots and indicial equation.. ... | Note that your differential equation is of the form
$$x^2y^{\prime\prime} + xp(x)y^{\prime}+q(x)y=0$$
So it follows that $p(x) = x-3$ and $q(x) = 3$ (the important thing to remember in these types of problems is that $p(x)$ and $q(x)$ are polynomials). Hence $p_0 = p(0) = -3$ and $q_0=q(0)=3$; thus the indicial equati... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Logarithm / exponential equation, not sure what to make of this, (simple) Solve for $a:(2 \log_a x)(3 \log_{x^2} 4) = 3$
No idea how to approach this problem other than moving the 2 and the 3 into an exponent..
| $(2 \log_a x)(3 \log_{x^2} 4) = 3\iff( \log_a x^2)( \log_{x^2} 4) = 1$. Then, because of the well known formula $\log_xy\log_y x=1$ one has $$\log_4 x^2\log_{x^2}4=1\Rightarrow\log_4 x^2=\frac{1}{\log_{x^2}4}=\log_a x^2\Rightarrow4^{x^2}=a^{x^2}\iff\left(\frac4a\right)^{x^2}=1$$
Thus (since $g(x)=b^x$ is injective) we ... | {
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calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\c... | $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Series expansion of a function at infinity I know it is possible to expand an expandable function for a real, and for infinite by setting $x=\dfrac1y$ and then expanding for $0$.
But my question is, how do we do if the evaluation of the new function and its derivatives is not possible ?
I mean I find things like $\sq... | If you want $\sqrt{x^2 - x +1}\;$ for $x\rightarrow \infty,\;$ you have $x > 0\;$ and write
$$\sqrt{x^2 - x +1} = x \sqrt{1 - \frac{1}{x} +\frac{1}{x^2}}.$$
Now substitute $y=\frac{1}{x}$ and compute the series for $y\rightarrow 0$
$$\sqrt{1 - y + y^2} = 1-\frac{1}{2}y+\frac{3}{8}y^2 + \frac{3}{16}y^3 + O(y^4)$$
Reve... | {
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"source": "stackexchange",
"question_score": "12",
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How to find $\arccos(\cos\frac{15\pi}{11})$? How to find $\arccos(\cos\frac{15\pi}{11})$?
I'm completely lost. $\frac{15\pi}{11}$ isn't on the unit circle so how do I find the cosine of it?
| As $\cos x=\cos A\implies x=2n\pi\pm A$ where $n$ is any integer
The general value of $$\arccos\left(\cos\frac{15\pi}{11}\right)=2n\pi\pm \frac{15\pi}{11}$$
Based on the definition of principal value inverse cosine ratio, $\displaystyle0\le \arccos x\le\pi$
Taking '+' sign, $\displaystyle0\le 2n\pi+\frac{15\pi}{11}\le... | {
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"timestamp": "2023-03-29T00:00:00",
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definite integral to evaluation (what am I doing wrong here 3) I was given the below definite integral to evaluate. below is the equation and my answer after I plug in the values but it is wrong. correct answer should be 29/6 what am I doing wrong? thank you
$$\int_1^2 \left(x + \frac 1x\right)^2 \,dx = \int_1^2 x^2 +... | You expanded incorrectly: Recall
$$(a + b)^2 = a^2 + 2ab + b^2$$
That gives us
$$\left(x + \frac 1x\right)^2 = x^2 + 2\left(x \cdot \frac 1x\right) + \frac 1{x^2} = x^2 + 2 + \frac 1{x^2}$$
Now go ahead and apply the power rule to evaluate the integral.
$$\begin{align} \int_1^2 \left(x + \frac 1x\right)^2 \,dx & = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the Following First Order PDE Solve
$$F\frac{\partial F}{\partial x} - \frac{\partial F}{\partial y} = y$$
subject to $F(s,0) = s^2$.
This is the first time I am using the method of characteristics, so I would like to know if I have made any errors in my working. I have
$$
\frac{dx}{dt} = z, \quad \frac{dy}{dt} =... | Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=-1$ , letting $y(0)=0$ , we have $y=-t$
$\dfrac{dF}{dt}=y=-t$ , letting $F(0)=F_0$ , we have $F=F_0-\dfrac{t^2}{2}=F_0-\dfrac{y^2}{2}$
$\dfrac{dx}{dt}=F=F_0-\dfrac{t^2}{2}$ , letting $x(0)=f_1(F_0)$ , we have $x=f_1(F_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A pack contains $n$ card numbered from $1$ to $n$ A pack contains $n$ card numbered from $1$ to $n$. Two consecutive numbered cards are removed from
the pack and sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on
the removing cards is $k$, Then $k$ is.
$\bf{My\; Try}::$ Let two consec... | $(n, k) = (50, 25)$
I think you are almost there. If I understand your question right you just solve for $k$ in terms of $n$.
\begin{align*}
& n(n+1) - (4k + 2) = 2448 \\
\implies & n(n+1) - 2 - 2448 = 4k \\
\implies & n(n+1) - 2450 = 4k \\
\implies & \frac{n(n+1) - 2450} 4 = k
\end{align*}
So $(n, k)$ can be written... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/599394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$ This is my attempt:
$$
\begin{align}
& \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt]
& = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt]
& = (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8p... | You got off to a good start:
$$
\sin(A+B)\sin(A-B) = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B)-\cos(A)\sin(B))
$$
This is of the form $(x+y)(x-y)$ so
$$
\sin(A+B)\sin(A-B) = \sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B)
$$
Eliminate the cosines (since $\sin^2(x)+\cos^2(x)=1$, so $\cos^2(x)=1-\sin^2(x)$) and expand:
$$
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Why does $r = \cos \theta$ produce a circle? I am trying to do a double integral over the following region in polar coordinates:
I know that the limits of integration are:
$$\theta = -\frac{\pi}{2} \quad \to \quad \theta = \frac{\pi}{2} \\ r = 0 \quad \to \quad r = \cos \theta$$
However, I don't understand how $r = 0 ... | $$r = \cos \theta$$
$$\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}}$$
$$x^2 + y^2 = x$$
$$x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}$$
$$\left(x - \frac{1}{2}\right)^2 + (y - 0)^2 = \left(\frac{1}{2}\right)^2$$
This is the equation for a circle in Cartesian coordinates $(x, y)$ with center $({1 \over 2}, 0)$ and radius ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 10,
"answer_id": 2
} |
Prove $1 + \tan^2\theta = \sec^2\theta$ Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$
I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
| Here is an alternative using exponential forms:
$$
\begin{align*}
1+\tan^2 \theta &=1+\left( \frac{e^{i \theta}-e^{-i \theta}}{i\left( e^{i \theta}+e^{-i\theta} \right)} \right)^2 \\
&=1-\frac{\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{\left( e^{i \theta}+e^{-i\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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.