Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Multiplying square roots How do I simplify the following types of question:
$ \sqrt{x^2+5} \times \sqrt{x^2+20}$
Do I need to get both answer out of their roots first or not? This is how I would do it:
$ \sqrt{x^2+5} \times \sqrt{x^2+20} = x^2 + x\sqrt{20} + x\sqrt{5} + \sqrt{100}$ but I'm very certain this isn't the ... | For positive real $a,b,c,d$
$\sqrt{a+b}\cdot\sqrt{c+d}=\sqrt{(a+b)(c+d)}=\sqrt{ac+ad+bc+bd}$ which is $\ne \sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating the following limit: $\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}} $ I am trying to calculate this limit:
$$
\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}
$$
I've tried using conjugate of both denominator and numerator but I can't get the right result.
| $$
\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}=\frac{x^2-x}{-x}\cdot\frac{1+\sqrt{x+1}}{\sqrt{x^2+1}+\sqrt{x+1}}\sim1\cdot\frac{2}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$ How can I solve integral of types
(1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$
(2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$
| (2) $$\int \frac{dx}{(x^4-1)^\frac14}=\int\frac{dx}{x(1-\frac1{x^4})^\frac14}=\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$
Let $1-\frac1{x^4}=y^4,4y^3dy=-4\frac{dx}{x^5}\implies \frac{dx}{x^5}=-y^3dy$ and $\frac1{x^4}=1-y^4\implies x^4=\frac1{1-y^4}$
$$\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$
$$=\int\frac{-y^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Showing that $ |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$ For every nonnegative integer $n$ and every real number $ x$ prove the inequality:
$$\sum_{k=0}^n|\cos(2^kx)|= |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$$
| This is terse and skips some important steps, but I speculate that it can be turned nicely into a clean proof!
We show:
$$|\cos 2^n x | + |\cos 2^{n+1} x| \geq \frac{1}{\sqrt{2}}$$
We have
$$
\sqrt{\cos^2 2^n x } + \sqrt{\cos^2 2^{n+1} x}.
$$ And since $\cos^2 x = \frac{1}{2}(\cos(2x)+1)$ we get
$$
\frac{1}{\sqrt{2}}\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 2
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Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? $$\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$$
Do you have an idea about this serie? If it converges what is the sum?
| Use the equality
$$
\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Finding the derivative of $x$ to the power something that is a function of $x$ if $y = x^{(x+1)^\frac12}$
then how can I get the first derivative of $y$?
| $$y=x^{(x+1)^{1/2}}=x^{\sqrt{x+1}}=e^{\sqrt{x+1}\log x}$$
and since
$$(\sqrt{x+1}\log x)'=\frac{\log x}{2\sqrt{x+1}}+\frac{\sqrt{x+1}}{x}=\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$
we get, applying as suggested the chaing rule, that
$$y'=y\frac{x\log x+2x+2}{2x\sqrt{x+1}}=x^{\sqrt{x+1}}\cdot\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does this sum go to $0$? http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf
In problem (2), at the very end it says
$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$
I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \fr... | One can forget about the square root part for a while. Note that $\dfrac{1}{k^2}\lt \dfrac{1}{(k-1)k}$.
But $\dfrac{1}{(k-1)k}=\dfrac{1}{k-1}-\dfrac{1}{k}$. Thus your sum is less than
$$\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}+\cdots.$$
Note the wholesale cancellation: the abov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Integrating a Rational Function I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will simplify it enough to solve in reasonably quick . I plugged it into wolfram and as usual it... | I think the integral can be solved in a very easy way.
$$
\begin{align}
\int\frac{x^3+x+2}{x^4+2x^2+1}dx&=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{x^4+2x^2+1}dx\\
&=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{(x^2+1)^2}dx.
\end{align}
$$
In the RHS part, for the left integral uses substitution $u=x^4+2x^2+1\;\Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\... | Update: Upps, that is the same as that of @Vincent, sorry didn't see that first. However, it's a bit more explicte
A nicer way to use your results and proceed frm there is the following. You've got that
$$ a^2+b^2+c^2 \ge ab+bc+ca $$
Now use the fact that $(a+b+c)=S=3$ and multiply each side of your original inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Cramer von Mises test statistic I am trying to derive the Cramer von Mises test statistic
$$nC_{n}=\frac{1}{12n}+\sum_{i=1}^{n}\left(U_{(i)}-\frac{2i-1}{2n}\right)^2$$
where $U_{(i)}=F_{0}(X_{(i)})$ the order statistics
from the original
$$C_{n}:=\int (\hat{F}_{n}(t)-F_{0}(t))^2dF_{0}(t)$$
Could anyone help me with t... | Let $X_1,\dots,X_n$ be a random sample of size $n$ from $f(x)$.
The empirical CDF is
\begin{equation}
\hat{F}_n(x) = \frac{1}{n} \sum_{i=1}^n I(X_i \leq x)
= \begin{cases}
0, & x < X_{1:n},\\
i/n, & X_{i:n} \leq x < X_{i+1:n},\\
1, & X_{n:n} \leq x.
\end{cases}
\end{equation}
Let $U_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Problem involving permutation matrices from Michael Artin's book. Let $p$ be the permutation $(3 4 2 1)$ of the four indices. The permutation matrix associated with it is
$$ P =
\begin{bmatrix}
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0
\end{bmatrix}
$$
This is the matrix that permutes the comp... | The permutation is indeed $(1342)$, but this decomposes as $(13)(14)(12)$, so you should compute
$$
P_{(13)} P_{(14)} P_{(12)}
=
\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Find $\lim\limits_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$.
Find $\displaystyle \lim_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$.
I tried to rationalize it, but doesn't help either. Please give me some hints. Thank you.
| Hint:
$$
\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}=\left(\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}\frac {\sqrt{x+3}+2}{\sqrt{x+8}+3}\right)\frac {\sqrt{x+8}+3}{\sqrt{x+3}+2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
| $(a^5+b^5) = (a+b)(a^4+b^4) - ab(a^3+b^3)$.
So for any $n$, if $n \mid a^3+b^3$ and $n \mid a^4+b^4$ then $n \mid a^5+b^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finding the area of a triangle using fractions?
To find the area of the triangle do you use Pythagorean theorem from what you have? Could this use similar triangles.
| The area of $\triangle PST$ is the sum of the areas of $\triangle PSV$ and $\triangle VST$.
$$\begin{align}
\text{The area of }\triangle PSV &= \frac12 PV\cdot SV\\
&=\frac12 PV\cdot\frac23 QV\\
&=\frac13 PV\cdot QV,
\end{align}$$
$$\begin{align}
\text{and the area of }\triangle VST &= \frac12VT\cdot SV\\
&=\frac12 VT\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the points on the given curve where the tangent line is horizontal or vertical $r=e^θ $ $($ Assume $0 ≤ θ ≤ 2π.)$
Apparently I keep getting this answer wrong. I dont know if i need to use $n $ in the answer or not...
| As $r=\sqrt{x^2+y^2}$ and $\theta=\arctan \frac yx$
Differentiating $$\sqrt{x^2+y^2}=e^{\arctan \frac yx}$$ wrt $x,$
$$\frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}}$$
$$=e^{\arctan \frac yx}\cdot\frac1{1+\left(\frac yx\right)^2}\cdot\frac{\left(x\frac{dy}{dx}-y\right)}{x^2}$$
$$=\sqrt{x^2+y^2}\frac{\left(x\frac{dy}{dx}-y\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/322705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Commuting Skew-symmetric Nilpotent 4x4 Matrices Suppose $A$ and $B$ are nonzero, commuting, skew-symmetric, nilpotent matrices in $M_4(k)$, $k$ a field (char $k\ne 2$). Must $A=\lambda B$ for some $\lambda\in k$? I have shown that this is true for $3\times 3$ matrices, and I believe it should also be true for $4\time... | It's not true. Consider
$$ A = \left[ \begin {array}{cccc} 0&0&1&-i\\ 0&0&i&1\\ -1&-i&0&0\\ i&-1&0&0
\end {array} \right],\ B = \left[ \begin {array}{cccc} 0&1&-i&0\\ -1&0&0&i\\ i&0&0&1\\ 0&-i&-1&0\end {array}
\right]$$
where $i$ is a square root of $-1$.
EDIT: Or, a bit more generally, with the same $A$,
$$ B = \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Further clarification needed on proof invovling generating functions and partitions (or alternative proof)
Show with generating functions that every positive integer can be written as a unique sum of distinct powers of $2$.
There are 2 parts to the proof that I don't understand. I will point them out as I outline the... | For your second question, first note that
$$g^*(x)=\prod_{k\ge 0}\left(1+x^{2^k}\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)\dots$$
makes sense: if $n\le 2^m$, the coefficient of $x^n$ in $g^*(x)$ is its coefficient in the polynomial $$\prod_{k=0}^m\left(1+x^{2^k}\right)\;,$$ because the contribution of any factor $1+x^{2^k}$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of definite integral using residue theorem $$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$$
I need to evaluate the above integral .
My idea is to consider the same integral but with the $x$'s as $z$'s, over the complex plane, have a closed contour integral over $\gamma$, and then use the residue theorem. ... | Note that $\frac{x - 1}{x^3 - 1} = \frac{1}{x^2 + x + 1}$. So the integral is
$$
\int_{-\infty}^{+\infty} \frac{1}{x^2 + x + 1} \, dx.
$$
First take the integral from $-a$ to $a$ and then take the limit after. Just complete the square in the denominator and use substitution to get an integral in terms of $\arctan$.
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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General term of $a_{n}= \sqrt{ a_{n-1}+6}$ What is the general term of $a_{n}= \sqrt{ a_{n-1}+6}$,$a_{1}=4$?
| I doubt if such general closed form exists.
We have $\sqrt{a_{n-1}+6}$ to be $\sqrt{10}$ in the first step, and $\sqrt{6+\sqrt{10}}$ is hard to simplify as if one assume it has the form $a+\sqrt{b}$, then we must have $a^{2}+b=6,2a\sqrt{b}=\sqrt{10}$. This force $a^{2}b=5/2$, and one has to solve $x^{2}-6x+5/2$. This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"Fat" sets of integers and Fibonacci numbers Let us call a set of integers "fat" if each of its elements is at least as large as its cardinality. For example, the set $\{10,4,5\}$ is fat, $\{1,562,13,2\}$ is not.
Define $f(n)$ to count the number of fat sets of a set of integers $\{1...n\}$ where we count the empty set... | This problem has a simple solution using ordinary generating functions.
Observe that the generating function of subsets of $\{k, n\}$ indexed
by number of elements ($u$) and total sum ($x$) is by inspection seen
to be
$$\prod_{m=k}^n (1+ux^m).$$
Therefore the generating function of the sets being considered is
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/330123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
| Note that each term is of the form
$$\dfrac{df_k/dx}{f_k} = \dfrac{d(\log(f_k)))}{dx}$$
Hence,
$$\sum_{k=0}^n \dfrac{df_k/dx}{f} = \sum_{k=1}^n \dfrac{d(\log(f_k)))}{dx} = \dfrac{d(\sum_{k=0}^n \log(f_k)))}{dx} = \dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx}$$In your case, $f_0 f_1 f_2 \cdots f_n$ r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/332191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Logarithm simplification Simplify: $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$
Can we use the formula to solve this: $\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$
Therefore first term will become: $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{1}{2}}$
$\log_4$ can be written as $\frac{1}{2}\log_2$
Please guide further.... | Let $x=\sqrt{2+\sqrt 3}+\sqrt{2-\sqrt 3}$
$x^2=(2+\sqrt 3)+(2-\sqrt 3)+2\sqrt{(2+\sqrt 3)(2-\sqrt 3)}=6$
I think you get it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am hav... | Your work looks good so far. Here is a hint:
$$
\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}
$$
To elaborate, apply the hint to get:
$$
\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 8,
"answer_id": 0
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Proof $ab-cd \ge 3$ $a$, $b$, $c$, and $d$ are real number
$a\ge b\ge c\ge d$
$a+b+c+d = 13$
$a^2+b^2+c^2+d^2 =43$
Proof that $ab-cd\ge 3$
| Hint
$$ (a + b + c + d)^2 = 169 = \sum a^2 + 2 ( ab + bc + cd + da + ac + bd) $$
This gives: $ ab + bc + cd + da + ac + bd = 63 $. We already had $ ab + cd \leq \frac{43}{2} $ from $ (a - b)^2 + (c - d)^2 \geq 0 $.
Use the above results with $$ \left[ (a + b) - (c + d) \right]^2 \geq 0 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\frac{(5m)!(5n)!}{(m!)(n!)(3m+n)!(3n+m)!}$ is a natural number?
How to prove that $$\frac{(5m)! \cdot (5n)!}{m! \cdot n! \cdot (3m+n)! \cdot (3n+m)!}$$ is a natural number $\forall m,n\in\mathbb N$ , $m\geqslant 1$ and $n\geqslant 1$?
If $p$ is a prime, then the number of times $p$ divides $N!$ is... | I became aware of this question due to some recent edits. The shorter proofs seem to rely on
$$
\lfloor5x\rfloor+\lfloor5y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor3x+y\rfloor+\lfloor3y+x\rfloor
$$
My answer also uses this inequality. I have tried to provide a complete, yet shorter, proof of it.
Define
$$
f(x... | {
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"url": "https://math.stackexchange.com/questions/336208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{n \to \infty}\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$ Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$
For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge t... | $$ \frac{x + n}{x + n^{-1}} = 1 + \frac{n-1}{nx + 1}$$
Using the expansion of $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$
We see that the resulting expansion of the integrand is $$\frac{n - 1}{nx + 1} - \frac{(n - 1)^2}{2(n x+ 1)^2} + O\left (\left(\frac{n-1}{nx + 1} \right)\right)^3$$
Squaring this again,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find a polynomial of degree $4$ with the coefficient $x^2$ equal to $6$, and zeros $-3$, $ 2$, $ -1$, $-2$ I started off with:
$$f(x)= a(x-(-3)) (x-(2)) (x-(-1)) (x-(-2))$$
$$f(x)= a(x+3) (x-2) (x+1) (x+2)$$
| You are right that $f(x)=a(x+3)(x+2)(x+1)(x-2)$
Expanding gives $$f(x)=a(x^{2}+3x+2x+6)(x^{2}+x-2x-2)=a(x^{2}+5x+6)(x^{2}-x-2)$$
Multiplying the quadratics, $$f(x)=a(x^{4}+5x^{3}+6x^{2}-x^{3}-5x^{2}-6x-2x^{2}-10x-12)$$
Collecting terms gives $f(x)=a(x^{4}+4x^{3}-x^{2}-16x-12)$
Hence, the coefficient on $x^{2}$ is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What may be the ratio of the perimeter of the trapezium to its midline? The diagonals of a symmetric trapezium are perpendicular to each other. What may be the ratio of the perimeter of the trapezium to its midline?
| Let the perimeter $p = a+b+2c$, where $c$ are the sides and set $d = d_a + d_b$ to be the length of the diagonals with $d_a$ and $d_b$ being their parts. From the Pythagorean theorem we know that $d_a = \frac{a}{\sqrt{2}}$ and $d_b = \frac{b}{\sqrt{2}}$. Also, $c^2 = d_a^2+d_b^2$, hence $c = \frac{1}{\sqrt{2}}\sqrt{a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How prove this summation prove that:
$$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\dfrac{n}{2}$$
I think can prove by the probability
my idea:
$$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\displaystyle\sum_{k=1}^{n}\dfrac{(n!)^... | Here is another method using the convolution of two generating functions.
Observe that when we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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When are both fractions integers? The sum of absolute values of all real numbers $x$, such that both of the fractions $\displaystyle \frac{x^2+4x−17}{x^2−6x−5}$ and $\displaystyle \frac{1−x}{1+x}$ are integers, can be written as $\displaystyle \frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the v... | Note that
$$\frac{1+x}{1-x}=\frac{2}{1+x}-1$$
and
$$\frac{x^2+4x-17}{x^2-6x-5} = \frac{10x-12}{x^2-6x-5}+1.$$
We need both of these to be integers. So $\frac{2}{1+x}$ must be an integer, and happens when, for example, $x=-3,-2,0,1$ (you also need to check rational values). You can plug these into the second fraction t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4... | Felix Marin's approach is nice and short, but I thought I'd add another approach that uses standard Binomial tools: Vandermonde's Identity, the symmetry of Pascal's Triangle, and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$.
$$
\hspace{-12pt}\begin{align}
&\phantom{={}}\sum_{k=0}^r(-1)^k(k+1)(k+2)\binom{n}{r-k}\\
&=\sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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$x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$. How do I prove that the diophantine equation $x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$.
| We have
$$x^4-4y^4=z^2\iff\left(2y^2\right)^2+z^2=\left(x^2\right)^2.$$
Therefore if a primitive solution exists then $2y^2=2nm, \ z=m^2-n^2, \ x^2=m^2+n^2.\;$ [Note: Primitive here means one such that $x^2, 2y^2$ and $z$ are coprime which has as a consequence that $n$ and $m$ are coprime.]
From $2y^2=2nm$ and because... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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proving or disproving that two tangent lines are parallel to a curve Im trying to prove or disprove that given the function, $f(x)=0.5\sqrt{1-x^{2}}$,
There are two different tangent lines to $f(x)$ so they are parallel.
I tried to derivative but with no success.
| $y=f(x)=0.5\sqrt{1-x^2}\implies 4y^2=1-x^2$ with $y\ge0$
Let us find the condition of $y=mx+c$ being a tangent to the curve
So, $4y^2=1-\left(\frac{y-c}m\right)^2\implies y^2(1-4m^2)+2cy+(m^2-c^2)=0$
This is a Quadratic Equation in $y$
For tangency, both the root must be same i..e, the discriminant must be $0$
So, $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$?
$Approach$:
$N$=$11^2$.$13^4$.$17^6$
$N^2$=$11^4$.$13^8$.$17^{12}$
This means $N... | N= 11^2 * 13^4 * 17^6
No of Factors of N = 3*5*7 = 105
There will be one factor which is N, So remaining = 105-1 = 104
No of factors of N^2= 5*9*13 = 585, of these N will be one of the factors.
Out of the remaining 584 factors, half will be less than N and half will be grater than N , So no. of factors of N^2 which are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/340533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof by induction for a recursive function $F(n) = F(n–1)+F(n–2)$ I'm having a lot of trouble with the following homework question:
Consider the following recursive function:
Base Case: $F(0) = 0,F(1) = 1$.
Recursive Step: $F(n)=F(n−1)+F(n−2)$ for all $n≥2$.
Prove that $F(n)\le \left ( \cfrac{1+ \sqrt5}{2} \right) ^{... | Hint: $\displaystyle \frac{1+\sqrt{5}}{2}+1= \frac{3+\sqrt{5}}{2}=(\frac{1+\sqrt{5}}{2})^2$
The above implies that,
$$\displaystyle F(k+1)=F(k)+F(k-1)\le(\frac{1+\sqrt{5}}{2})^{k-1}+(\frac{1+\sqrt{5}}{2})^{k-2} \text{ (By induction hypothesis)}$$
$$\displaystyle \Rightarrow (\frac{1+\sqrt{5}}{2})^{k-1}+(\frac{1+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3}... | Because $$\cos \frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{2\sin\frac{\pi}{7}\cos \frac{\pi}{7} - 2\sin\frac{\pi}{7}\cos\frac{2\pi}{7} + 2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{2\pi}{7}- \sin\frac{3\pi}{7}+\sin\frac{\pi}{7} + \sin\frac{4\pi}{7}-\sin\frac{2\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
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Integer solutions for $x^3+2=y^2$? I've heard a famous result that $26$ is the only integer, such that $26-1=25$ is a square number and $26+1=27$ is a cubic number.In other words, $(x,y)=(5,3)$ is the only solution for $x^2+2=y^3$.
How if we make it like this $x^3+2=y^2$? Are there any integral solutions? If so, finite... | Edit: Misread the question. Will leave the answer for a while, since the writeup of the solution to the wrong problem may be interesting to MSE users.
We need to take a moderately lengthy detour through the arithmetic of $\mathbb{Z}[\sqrt{-2}]$, that is, the arithmetic of numbers of the form $a+b\sqrt{-2}$, where $a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Proving $x\equiv 2^{11} \mod{97}$ In $\mathbb{N}$ we suppose the equation $(F): x^{35} \equiv 2 \mod{97}$, and $(97,x)=1$.
How could I prove that $x\equiv 2^{11} \mod{97}$ ?
| Here we look for relations ('tricks') that will solve the problem without using Fermat's Little Theorem.
We interpret the OP's question as a 'how to calculate' ${2^{11}}^{35} \pmod{97}$.
Well $2^{11} \equiv 2^8 \times 2^3 \equiv 62 \times 8 \equiv 11 \pmod{97}$ so we've found our first relation,
$\tag 1 {2}^{11} {11}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $1-\frac{1}{2}\sin(2x)=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}$ without factoring Is there a way to prove this identity without factoring?
$$1-\frac{1}{2}\sin(2x)=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}$$
| Motivated by the fact that $(a^2 + b^2 - ab)(a+b) =a^3 + b^3$, we see LHS = $1 - \frac{1}{2} \sin(2x) = \sin(x)^2 + \cos(x)^2 - \sin(x) \cos(x)$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding roots of complex equation? Determine all roots of the equation $x^6+(3+i)x^3 + 3i = 0$ in $\mathbb{C}$ Express answers in standard form
| Let $x^3 = z$. We then have that
$$z^2 + (3+i)z + 3i = 0 \implies (z+3)(z+i) = 0 \implies z = -3, -i \implies x^3 = -3,-i$$
Hence,
$$x = \sqrt[3]{-3},\sqrt[3]{-3} \omega, \sqrt[3]{-3} \omega^2, i, i\omega, i\omega^2$$where $\omega$ is the complex cube-root of $1$ and $\sqrt[3]{-3}$ is the real number $y$ such that $y^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find Rational Roots and Factor $x^4 + x^3 - 4x^2 + x + 1$ Find all rational roots of the polynomial $x^4 + x^3 - 4x^2 + x + 1$. Factor this polynomial into a product of monic irreducible polynomials over the rationals.
I am confused about this question, but isn't the rational root theorem tell us the roots are only $\... | \begin{align}
x^4+x^3-4x^2+x+1 & = (x^4 - 2x^2 + 1) + (x^3 - 2x^2 + x) = (x^2-1)^2 + x(x-1)^2\\
& = (x-1)^2 \left( (x+1)^2 + x \right) = (x-1)^2(x^2 + 3x +1)\\
& = (x-1)^2 \left( \left(x+\dfrac32\right)^2 + 1 - \dfrac94\right)\\
& = (x-1)^2 \left( \left(x+\dfrac32\right)^2 - \dfrac54\right) = (x-1)^2 \left( \left(x+\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$2\int_0^\pi \sin x + \sin x \cos x\,dx$. Where am I going wrong?
$2\displaystyle\int_0^\pi \sin x + \sin x \cos x\,dx$
I let $u=\sin x \implies \dfrac{du}{dx}=\cos x \implies dx=\dfrac{du}{\cos x}$
$2\displaystyle\int_{x=0}^{x=\pi} 2u\,du$
When I try to change the limits I just get 0 and 0: Lower limit $=\sin 0 = 0$... | As $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$
If $f(x)=\sin x+\sin x\cos x,$
$f(\pi+0-x)=\sin(\pi+0-x)+\sin(\pi+0-x)\cos(\pi+0-x)=\sin x-\sin x\cos x$ as $\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$
So, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi (\sin x - \sin x \cos x)dx=I\text{ say}$$
So, $$2I=\int_0^\pi (\sin x + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?
How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?
My thought:
$365 = 5 \times 73$ where $5$ and $73$ are prime numbers.
So we can obtain $x^2 \equiv {-1} \pmod 5$ and $x^2 \equiv {-1} \pmod {73}$.
For $x^2 \equiv {-1} \pmod 5$,
we checked $5 \equ... | Because $5$ and $73$ are relatively prime, we have $x^2\equiv -1\pmod{5\cdot 73}$ if and only if $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{73}$.
Because $5$ is a prime of the form $4k+1$, the congruence $x^2\equiv -1 \pmod{5}$ has a solution, and hence two solutions. The same applies to $x^2\equiv -1\pmod{73}$.
Su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Determine the definite limit The following limit
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
evaluates to 1/12.
This is my progress so far:
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
$$\lim_{x\to 1}\frac{1 + \sqrt{x}}{2(1 - x)} - \frac{1 + \sqrt[3]{x} + \sq... | If we pose $x=1-h$ we find
$$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{2(1 - \sqrt{1-h})} - \frac{1}{3(1 - \sqrt[3]{1-h})}$$
and we have $$\sqrt{1-h}=1-\frac{1}{2}h-\frac{1}{8}h^2+o(h^2)$$ and $$(1-h)^{1/3}=1-\frac{1}{3}h-\frac{1}{9}h^2+o(h^2)$$
hence we find
$$\frac{1}{2(1 - \sqrt{x})} - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Use Cauchy product to find a power series represenitation of $1 \over {(1-x)^3}$
Use Cauchy product to find a power series represenitation of
$$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$.
Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is ... | This is most easily done with the binomial theorem and the identity
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}
$$
which gives
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k
\end{align}
$$
To use the Cauchy product... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$? Integration by parts is of no success. What else to try?
$$\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$$
| Let $z=e^{ix}$ and then $\cos x=\frac{1}{2}(e^{ix}+e^{-ix})=\frac{1}{2}\left(z+\frac{1}{z}\right),dz=izdx$. We have
\begin{eqnarray*}
\int_0^\pi(\log(1-2a\cos x+a^2))^2dx&=&\frac{1}{2}\int_{-\pi}^{\pi}(\log(1-2a\cos x+a^2))^2dx\\
&=&\frac{1}{2}\int_{|z|=1}\left[\log(1-a\left(z+\frac{1}{z}\right)+a^2)\right]^2\frac{dx}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
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How to derive a closed form of a simple recursion? Consider: $$T(n) = 2 T(n-1) + 1$$ with $T(1)$ a positive integer constant $a$.
I just stuck in finding a closed form for this simple recursion function. I would appreciate it, if someone gives me a hint.
| A proof with generating functions: Let $$f(X) := \sum_{n=1}^{\infty} T(n) X^n$$ be a formal power series (we don't care about convergence); then you have $$f(X) = aX +\sum_{n=2}^{\infty} (2T(n-1) + 1)X^n$$ $$= aX + 2Xf(X) + \sum_{n=2}^{\infty} X^n$$ $$=2Xf(X) + \frac{X^2}{1-X} +aX.$$ Manipulate this to $$f(X) = \frac{X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding functional equation for generating function I'm given
$$
a_n = \sum_{i=2}^{n-2} a_ia_{n-i} \quad (n\geq 3), a_0 = a_1=a_2 = 1
$$
and I need to find the functional equation for the generating function satisfying the above equality. I obtained the correct answer, but I don't understand why some of the steps are a... | For this problem, I would go in the other direction:
$\begin{align}
f^2(x)
&= (\sum_{n=0}^{\infty} a_n x^n) (\sum_{m=0}^{\infty} a_m x^m)\\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n x^n a_m x^m\\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n a_m x^{n+m}\\
&= \sum_{i=0}^{\infty} \sum_{j=0}^{i}a_j a_{i-j} x^{i}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all values x, y and z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes. Find all positive integers x, y, z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes.
It seems trivial that the only... | Odd primes: We first take care of the case $x^2+1$, $y^2+1$ both odd primes. If $x^2+1$ and $y^2+1$ are odd, then $x$, $y$, and $z$ are even.
Factor in the Gaussian integers. We get
$$(x-i)(x+i)(y-i)(y+i)=(z-i)(z+i).$$
Note that $x\pm i$ and $y\pm i$ are Gaussian primes. And since $z$ is even, the Gaussian integers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/357970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Hint: Follow the example in lectures to show that $2^n -1$ is not divisible by 3.
In lectures, the example showed the $3 \mid 2^n -1 \iff n$ is even. So wha... | Considering $7|2^n-1$.
$2^n-1= (2-1)(2^{n-1}+2^{n-2}+ \dots1)$
$7k=(2^{3m}-1)=(2-1)(2^{3m-1}+2^{3m-2}+ \dots1)$
Let $7k=2^n-1=(2-1)(2^{n-1}+2^{n-2}+ \dots1)=(2-1)(2^{3m-1}+2^{3m-2}+ \dots1)$
$(2^{n-1}+2^{n-2}+ \dots1)=(2^{3m-1}+2^{3m-2}+ \dots1) \implies n=3m$. But $n$ is a prime greater $3$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is "prime factorisation" of polynomials? I have the following question:
Find the prime factorization in $\mathbb{Z}[x]$ of $x^3 - 1, x^4 - 1, x^6 - 1$ and $x^{12} - 1$. You will need to check the irreducibility in $\mathbb{Z}[x]$, of three quadratic polynomials and of one quartic. In the case of the quadratic, yo... | For quadratics and cubics, irreducibility is easy to test, for if there are no rational roots, the polynomial must be irreducible. And the polynomials $x^2+x+1$ and $x^2-x+1$ don't even have real roots. In more complicated cases, the Rational Roots Theorem could be helpful.
Let's deal with $x^{12}-1$. This immediately... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$
| You can use the Law of cosine and the area formula of a triangle to solve this problem. Suppose the three angles are $A,B,C$ opposite to the sides $a,b,c$, respectively. Then
$$ c^2=a^2+b^2-2ab\cos C, S=\frac{1}{2}ab\sin C $$
and hence
\begin{eqnarray*}
a^2+b^2+c^2-4S\sqrt{3}&=&2[a^2+b^2-ab(\cos C+\sqrt{3}\sin C)]\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Summation of sequence $a_n - a_{n-1} = 2n$ $(a_1,a_2,a_3,..)$ be a sequence such that $a_1$ =2 and $a_n - a_{n-1} = 2n$ for $n \geq 2$. Then $a_1 + a_2 + .. + a_{20}$ is equal to?
$a_1$ = 2
$a_2$ = 2 + 2x2
$a_3$ = 6 + 2x3
$a_4$ = 12 + 2x4
$a_5$ = 20 + 2x5
$a_n$ = $b_n$ + $g_n$ ,here $g_n$=2xn
$b_3$-$b_2$=4
$b_4$-$b_... | I have 2 methods of doing this:
Method one:
$a_2-a_1=4$, i.e. $a_2=6$
$a_3-a_2=6$, i.e. $a_3=12$
$a_4-a_3=8$, i.e. $a_4=20$
You will notice that $a_k = 2+4+6+8+ \cdots +2k = \frac{k}{2}(2k+2)=k^2 + k$ (using summation formula)
$a_1+a_2+ \cdots + a_{20} = \displaystyle\sum_{k=1}^{20} (k^2+k)=\frac{(20)(20+1)(2\times20+1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Elementary Diophantine equation Solve $(x+y)(xy+1)=2^z$ in positive integres. My attempts is to use $x+y=2^a$, $xy=2^b-1$ and therefore $x,y$ are the roots of the quadratic equation $w^2-2^aw+2^b-1=0$. I try to analyze its dicriminant but it seems to be a dead end...
| The idea of discriminant works nicely, as @Jyrki, @Andre and @Mike answered. Yet, I think of another (perhaps more "number theory"ish :) way.
First, the equation is symmetric in $x$ and $y$, hence without loss of generality, we assume $x\ge y$.
Since $xy+1$ divides $2^z$ and is not equal to $1$, then it should be eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
| Following answer is for general case.
Denote by $c_m(1,n)=\binom{n-1}{m-1}$ the number of composition of $n$ into $m$ positive parts, and by $c_m(2,n)$ number of compositions of $n$ into $m$ parts greater than $1$.
Each composition of $n$ into $m$ parts greater than $1$, if all parts decrease by 1, becomes a compositi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$ The question is basically in the title: Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$
I get how to do it from $7\mid $ and $7\mid y$ to $7\mid x^2+y^2$, but not the other way around.
Help is appreciated! Thanks.
| $x^2+y^2 \equiv \mod 7 \implies x^2 \equiv k \mod 7$ and $y^2 \equiv 7-k \mod 7$
And any $a^2 \equiv 0,1,4,2\mod 7$(Why?) $\implies k=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Determine the values of $a,b,c$ for which the function is continuous at $x=0$
Determine the values of $a,b,c$ for which the function is continuous at $x=0$
$$
f(x) =
\begin{cases}
\frac{\sin(a+1)x+sinx}{x} \qquad \text{if} \ x<0 \ ; \\ \\
c \qquad\quad\qquad \text{if} \ x=0 \ ; \\ \\
\frac{\sqrt{x+bx^2}-\sqrt{x}}{b... | Hint: $$\frac{\sqrt{1+bx}-1}{bx}=\frac{\sqrt{1+bx}-1}{bx}\cdot\frac{\sqrt{1+bx}+1}{\sqrt{1+bx}+1}=\frac{1}{\sqrt{1+bx}+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$
Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$
No idea if I'm on the right track. May... | Take a Weierstrass substitution $u = \tan \frac{\theta}{2}$ after reducing the integration range using $\cos(2\pi - \theta) = \cos(\theta)$ in the second integral below:
$$
\int_0^{2\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} = \int_0^{\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} + \int_{\pi}^{2\pi} \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/370711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
For positive numbers a,b,c ,$abc=1$ prove$ \frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$ For positive numbers $a$,$b$,$c$ , with $abc=1$ prove $$\frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$$
| The problem as stated is false. Take $a=b=0.1$, $c=100$. $\frac{1}{a+b}=5$.
However, if $\min(a,b,c)\ge 1$ (for example, if they are integers), then it's true, by the other posted solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the locus of the midpoint of chord that subtends a right angle at $(\alpha,\beta)$ There is a circle $x^2+y^2=a^2$. On any line that cuts the circle in two distinct points(it is a secant), the points of intersection with circle are taken and at those two points I draw the tangents that intersect at some point, ... | Parameterize circle as $(a \cos \alpha, a \sin \alpha)$ with $\alpha \in \left[ 0 , 2 \pi \right]$, now the tangent vector is given as $(-a \sin \alpha , a \cos \alpha)$. Let the two points correspond to end of chord be at angle $\alpha$ and at angle $\beta$, then:
$$ (a \cos \alpha, a \sin \alpha) \cdot ( a \cos \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding a coefficient using generating functions I need to find a coefficient of $x^{21}$ inside the following expression:
$$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}$$
I think the only way (using generating functions) is to express the parentheses content with a generating function.
The generating function for the $... | $$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}=(x^2(1+x+x^2+x^3+x^4))^8=$$
$$=x^{16}\left(\frac{1-x^5}{1-x}\right)^8=x^{16}(1-x^5)^8(1-x)^{-8}=$$
$$=x^{16}\sum_{k=0}^{8}\binom{8}{k}(-x)^{5k}\sum_{l=0}^{\infty}\binom{8+l-1}{l}x^l=$$
$$=\sum_{l=0}^{\infty}\sum_{k=0}^{8}(-1)^{5k}\binom{8+l-1}{l}\binom{8}{k}x^{16+5k+l}$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
| Let $p = a+b+c$, then $ab+bc+ca = \frac{1}{2}(p^2 - p)$ and $abc=1$. So $a,b,c$ are solutions to the equation
$$x^3 - px^2 + \frac{1}{2} (p^2-p)x - 1 = 0$$
The discriminant is $(\frac{1}{2}(p^2-p))^2 - 4(\frac{1}{2}(p^2-p))^3 - 4p^3 - 27 + 18 p (\frac{1}{2}(p^2-p)) \ge 0$, since the roots are real.
This is equivalent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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The number $ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}$ For a real number $a$ and a positive integer $k$, denote by $(a)^{(k)}$ the number $a(a+1)\cdots (a+k-1)$ and $(a)_k$ the number
$a(a-1)\cdots (a-k+1)$. Let $m$ be a positive integer $\ge k$. Can anyone show me, or point me to a reference, why the number
$$ \frac{(m)... | Using Knuth's notation:
$$
x^{\underline{k}} = x (x - 1) \ldots (x - k + 1) = (x)_{(k)} \qquad
x^{\overline{k}} = x (x + 1) \ldots (x + k - 1) = (x)^{(k)}
$$
What you are looking for is:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
$$
As if $k > m$ then $m^{\underline{k}} = 0$, while all other ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to solve a polynomial with power fractions like $a-ax+x^{0.8}-x^{0.2}=0$ I have something like
$a-ax+x^{0.8}-x^{0.2}=0$
with parameter a>0 and variable x>0.
I know by trial and error that the equation has three real roots for parameter a greater than certain value, otherwise it has only one root. x=1 is one obviou... | Let $y=x^{0.2}>0$, then the problem reduces to finding all $a>0$ such that the polynomial equation $a-ay^5+y^4-y=0$ has only 1 positive real root $y=1$.
For $y=1$ to be a double root, we must have $-5a(1)^4+4(1)^3-(1)=0$, so $a=\frac{3}{5}$. Putting this into the equation gives $$0=\frac{3}{5}-\frac{3}{5}y^5+y^4-y=-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\... | When going from line
$$\ln y = \ln x + x \ln a$$
to
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + (x \times \frac{1}{a} + \ln a \times 1)$$
you made a mistake: when you tried to differentiate $x \ln a$ with respect to x via the product rule, you differentiated $\ln a$ with respect to a, turning it into $\frac{1}{a}$, whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Write the expression in terms of $\log{a}$ and $\log{b}$ Write in terms of $\log{a}$ and $\log{b}$ : $\log{\frac{a^3}{b^4}}$?
using log laws $log(\frac{x}{y})$ = $\log{x} - \log{y}$
$\log{\frac{a^3}{b^4}} = \log{a^3} - \log{b^4}$
and using log law that $\log{x^y} = y\log{x}$
$\log{a^3} = 3 \log{a}$ & $\log{b^4} = 4\log... | It seems correct, you used well the logarithm laws.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/376857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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finding a primitive root. It says for part A to Find a primitive root r of 38? Im not sure if I did it right.
I first calculated $\phi(38)=\phi(19*2)=18$. So there are 18 numbers that are relatively prime to 38. Listing them out we get 1,3,5,7,9,11,13,15,17,21,... so on. So I decided to test out 1.
But $ord_1 38$ doe... | As $\phi(19)=18$ and $ord_{19}2$ must divide $\phi(19),$ we only need to test for the powers $1,2,3,6,9,18$
$3^2=9;3^3=27\equiv-11\pmod{38};3^6\equiv(-11)^2\equiv121\equiv7\pmod{38}$
So, $3^9=3^3\cdot3^6\equiv(-11)(7)\equiv-77\equiv-1\pmod{38}$
$\implies 3 $ is a primitive root $\pmod{38}$
Alternatively, let's start w... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving/disproving $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}$ Let $a\in\mathbb{R}\setminus\mathbb{Z}$. Prove or disprove that there do not exist three distinct $k_1, k_2, k_3\in\mathbb{N}$ such that $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}\neq 0$, where $\{x\}=x-\lfloor x \rfloor$.
| This is not a complete answer, but I thought the approach might help someone find a solution. We can ask when the following condition holds for positive integers $a>b>c$ and $a>d\ge c$:
$$
x^a - x^d - m = P(x)(x^b - x^c - n),
$$
where $P(x)$ is some polynomial over $x$, and $m,n\in\mathbb{Z}$. If it does, then $\{x^a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
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Prove by mathematical induction that $1 + 1/4 +\ldots + 1/4^n \to 4/3$ Please help. I haven't found any text on how to prove by induction this sort of problem:
$$
\lim_{n\to +\infty}1 + \frac{1}{4} + \frac{1}{4^2} + \cdots+ \frac{1}{4^n} = \frac{4}{3}
$$
I can't quite get how one can prove such. I can prove basic di... | i have a solution, but not seem mathematical induction
$s = 1 + \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \cdots$
$\dfrac1{4}s = \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \dfrac1{4^5} + \cdots$
$\dfrac1{4}s = - 1 + (1 + \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \dfrac1{4^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 0
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Distance between point and a line - problems simplifying the minimised distance equation Someone asked how to prove the distance between a point $(x_1,y_1)$ and a line $Ax + By + C = 0$ is:$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$ The currently accepted answer shows that a poin... | Since you wanted to see how this is done algebraically, we'll need to look at a particular property of quadratic functions. As I've said in a comment, if $r > 0$, then $ f(x) = rx^2 + sx + t \ $ represents an "upward-opening" parabola. By completing the square, we can write this as
$$f(x) \ = \ r \cdot (x^2 \ + \ \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find integral $\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$ (most likely substitution) $$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$
I tried letting $x^2=\tan \theta$ but it didn't work. What should I do?
Please don't give full solution, just a hint and I will continue.
| Integrate as follows
\begin{align}
&\int_0^1\frac{\ln(1+x^2)}{1+x^2} dx\\
=&\int_0^1\int_0^1 \frac{2x^2y}{(1+x^2)(1+x^2y^2)}dy\ dx=\int_0^1\frac{\frac\pi2 y-2\tan^{-1}y}{y^2-1}
\ \overset{y\to \frac{1-y}{1+y}}{dy}\\
= &\ \frac\pi2\int_0^1 \frac{1}{1+y}dy
-\int_0^1\frac{\tan^{-1}y}{y}dy=\frac\pi2\ln2-G
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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To find the x and y-intercepts of the line $ax+by+c=0$ Please check if I've solved the problem in the correct way:
The problem is as follows:
Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.
My solution:
We have to find the x and y-intercepts of the line.... | easier way is this equation of line intercept form
$$ \frac xa+\frac yb=1$$
so write the given eqn in this form
$$ax+by+c=0$$
$$ax+by=-c$$
$$\frac {ax}{-c}+\frac {by}{-c}=1$$
$$\frac {x}{\frac{-c}{a}}+\frac {y}{\frac {-c}{b}}=1$$
so intercept at X axis is (-c/a,0) and Y axis is (0,-c/b)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + ... | Note that $f_n(x)(x^2-1) = x^{2n+2} - 1 = (x^{n+1}-1)(x^{n+1}+1)$. Now use unique factorization...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5... | I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.
| {
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"source": "stackexchange",
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Class Group of $\mathbb Q(\sqrt{-35})$ As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$.
We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$.
$-35\equiv 5$ mod $8$, so $2$ is ... | Note that the ring of integers is $\mathbb Z[(1+\sqrt{-35})/2]$.
If you compute $(3, 1 + \sqrt{-35})^2$, you get $$(9,3 + 3\sqrt{-35}, -34 + 2 \sqrt{-35} ) = (9, 1 + \sqrt{-35}) = ( \dfrac{1-\sqrt{-35}}{2} \dfrac{1 + \sqrt{-35}}{2}, 2 \dfrac{1+\sqrt{-35}}{2}) = ((1 + \sqrt{-35})/2 )$$
(because
$\dfrac{1-\sqrt{-35}}{2}... | {
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How to prove the existence of odd numbers $a$ and $b$ in $2m\equiv a^{20}+b^{11}\pmod{2^n}$
Show that for any natural numbers $m,n$, there exist odd numbers $a,b$ such that
$$2m\equiv a^{20}+b^{11}\pmod{2^n}$$
Thank you everyone.
| We proceed by induction on $n$.
When $n=1$, we have $1^{20}+1^{11} \equiv 0 \equiv 2m \pmod{2^1}$.
Suppose that the statement holds for $n=k$, so $\exists a, b \in \mathbb{Z}$, $a, b$ odd, such that $2m \equiv a^{20}+b^{11} \pmod{2^k}$. Thus $a^{20}+b^{11} \equiv 2m, 2m+2^k \pmod{2^{k+1}}$. If $a^{20}+b^{11} \equiv 2m ... | {
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finding nth term Let
3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.
| Note: I just added a derivation of an explicit formula
for the terms as the OP requested.
From 3,8,17,32,57,
if the term is n,
the next term is 2n+k
where k = 2, 1, -2, -7.
The differences of k are
-1, -3, -5.
If we assume that the next difference is -7,
the next k is -7-7=-14
and the next term is 2*57-14 = 100.
To get... | {
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Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$ If $n=7+4\sqrt3$,then what is the simplified value of
$$\sqrt n+\frac {1}{\sqrt n}$$
I was taking LCM but how to get rid of $\sqrt n$ in denominator
| You could as OP suggests, add 2, and take the square root.
For example, $\dfrac{1}{(7+4\sqrt3)}$ is $(7-4 \sqrt3)$, so the sum of these numbers is $14$. You add $2$ to it to get $16$, and take the square root.
The new number is $4 = x + \dfrac {1}x$, which leads to $x = \dfrac{\sqrt{4+2} + \sqrt{4-2}}{2}$ = 1.931... | {
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"question_score": "3",
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Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$
I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.
Any help is app... | A general technique is taught by Wilf's "generatingfunctionology". Define $S(z) = \sum_{n \ge 0} s_n z^n$ and write ($s_0$ you get from the recurrence "backwards", mostly for not having to mess around with indices):
$$
s_{n + 1} = 3 s_n + 2^{n - 1} - 1 \qquad s_0 = \frac{1}{6}
$$
Multiply the recurrence by $z^n$, add o... | {
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Formulas for calculating pythagorean triples I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is:
$$\begin{align}
&a = m^2-n^2 \\
&b = 2mn\\
&c = m^2+n^2
\end{align}$$
With numerous parameters.
So are there other fo... | Here is the method I use:
*
*Take a number $c >= 1$
*Find all pairs $a, b$ such that $ab = 2c^2$
*For every pair of $a, b$ you get an unique solution $x = a + 2c, y = b + 2c, z = a + b + 2c$. It can be easily proven that $x^2 + y^2 = z^2$ is reduced to $4c^2 = 2ab$ which is satisfied at point 2.
The advantage of... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 4
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Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to b... | In this answer, the substitution $x=\frac{1-y}{1+y}$ is used to get
$$
\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x=\frac\pi8\log(2)\tag{1}
$$
We can use the substitution $x\mapsto1/x$ to get
$$
\begin{align}
\int_1^\infty\frac{\log(1+x)}{1+x^2}\mathrm{d}x
=\int_0^1\frac{\log(1+x)-\log(x)}{1+x^2}\mathrm{d}x\tag{2}
\end{a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
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linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q)... | Note that you're hiding the core of the proof, which is simply that if $p,q$ are polynomials, we define their sum, which is also a polynomial, and $(p+q)(x)=\sum_{i=1}^n (a_i+b_i)x^i=p(x)+q(x)$.
Thus, if $T$ is your transformation, $$T((p+q)(x))=(p+q)(x+1)=p(x+1)+q(x+1)=T(p(x))+T(q(x))$$ and $$T(\lambda p(x))=(\lambda ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$
However, the answer is 3
| 3 √x+13 = x + 9
Squaring both sides, we get
9 (x + 13) = x^2 + 81 + 18x
9x + 117 = x^2 + 81 + 18x
x^2 + 18x - 9x + 81 - 117 = 0
x^2 + 9x - 36 = 0
x^2 + 12x - 3x - 36 = 0
x(x + 12) - 3(x + 12) = 0
(x + 12)(x - 3) = 0
Therefore, x = -12 or 3
However, when the problem equation does not get satisfied when x is substituted ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Behavior of $|\Gamma(z)|$ as $\text{Im} (z) \to \pm \infty$ Let $\Gamma(z)$ be the gamma function.
In a paper I'm reading, the author states that $$ |\Gamma(z)| = |\Gamma(a+ib)| \sim \sqrt{2 \pi} |b|^{a-\frac{1}{2}} e^{-|b|\frac{\pi}{2}}$$ as $|b| \to \infty$.
Can this asymptotic behavior be derived from Stirling's fo... | Assume $|y|\gg|x|$. Note that
$$
\begin{align}
y\arctan\!2(y,x)
&=|y|\arctan\!2(|y|,x)\\[6pt]
&=\frac\pi2|y|-|y|\arctan\!2(x,|y|)\\
&\sim\frac\pi2|y|-x
\end{align}
$$
and
$$
x^2+y^2\sim|y|^2
$$
Therefore,
$$
\begin{align}
\left|\sqrt{2\pi}\,\frac{\color{#C00}{z^{z-1/2}}}{\color{#090}{e^z}}\right|
&=\sqrt{2\pi}\color{#C... | {
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About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime The number $30$ has a curious property:
All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!
I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not f... | Let $n>30$ be such a number. Observe that $p^2<n$ implies $p|n$.
Thus $2|n$, $3|n$, $5|n$ follow directly from $n>25>9>4$.
Thus $n$ is a multiple of $30$ and $n>30$, which implies $n\ge 60>49$ and hence $7|n$.
Up to now, $n$ is divisible by the four smallest primes.
Lemma. The product of four consecutive primes is gre... | {
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"source": "stackexchange",
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how to solve this multivariate quadratic equation? Any hope to acquire an analytic solution to such equations:
Solve:
$$\sum_{j=1}^n a_{ij} x_i x_j = b_i$$
for $i=1,\ldots,n$, where $a_{ij}$'s and $b_i$'s are known constants and $x_i$'s are unknowns to be solved.
Thanks a lot!
P.S. Thanks for alex.jordan's interesting ... | Let $A=\left( a_{ij} \right)_{n \times n} $, $B=\mathrm{diag}\{b_1, b_2, \cdots, b_n\}=\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}$, $C=A^{-1}B$.
There is a possible numerical solution when $A$ is invertible and $b_i\neq 0$ ($i$ = 1, 2, $\cdots$, $n$ ).
From the conditions, it will result
$$
\begin{pmatr... | {
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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education... | You want to integrate
$$J = \int_0^{\pi/2} \dfrac{dx}{\left(a^2 +(b^2-a^2) \sin^2(x) \right)^2} = \dfrac1{a^4}\int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$$
where $c = \dfrac{b^2-a^2}{a^2}$.
We now want to integrate $I = \displaystyle \int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$. From Taylor ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving that sequence $a_n = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}} = x^{1-2^{-n}}$ Let $x>0$.
For sequence $a_n$, such that $n$ denotes the $n$th term:
$$\begin{align} a_1&= \sqrt{x}\\
a_2&= \sqrt{x \sqrt{x}}\\
a_3&= \sqrt{x \sqrt{x \sqrt{x}}}\\
a_4&= \sqrt{x \sqrt{x \sqrt{x \sqrt{x}}}}\\
&\vdots\\
a_{n-1}&= \sqrt{... | Hint:
\begin{align}
a_4&=\sqrt{x \sqrt{x\sqrt{{x \sqrt{x}}}}}\\&=\sqrt{x \sqrt{x\sqrt{{x^{\frac{3}{2}} }}}}\\&=\sqrt{x \sqrt{x^{\frac{7}{4}}}}\\&=\sqrt{x^{\frac{15}{8}}}\\&=x^{\frac{15}{16}}\\&=x^{1-2^{-4}}
\end{align}
Can you use induction in these footsteps?
| {
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Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\di... | $$\dfrac{(6x^3-2x^2)}{x(9x^3-x)} = \dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} $$
Cancel common factor of $x$ in numerator and denominator gives us:
$$\dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} = \frac{2x(3x-1)}{\left[9x^2 - 1\right]}$$
Now we have a difference of squares in the denominator, and can factor it:
$$\frac{2x(3x-1)}{\color{bl... | {
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Subtracting roots of unity. Specifically $\omega^3 - \omega^2$ This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headwa... | $$\omega=\cos(\pi/5)+i\sin(\pi/5)$$
$$\omega^3-\omega^2=\cos(3\pi/5)+i\sin(3\pi/5)-(\cos(2\pi/5)+i\sin(2\pi/5))$$
$$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$
$$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$
$$\omega^3-\omega^2=-2\sin(\pi/2)\cdot\sin(\pi/10)-2... | {
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$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up ge... | Having rewritten the recurrence, we get an equation:
$$10R_{n-2}-7R_{n-1}+R_n=0$$
that can easily be solved.
The roots of a characteristical polymomial $10-7k+k^2=0$ are $k_1=2,k_2=5$.
So general solution is $R_n=c_12^{n}+c_25^{n}$, where $c_1$ and $c_2$ are arbitrary constants.
| {
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"timestamp": "2023-03-29T00:00:00",
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$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| If you type $1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$ into most pocket calculators you will get the answer $1$. This is because the calculator applies the next operation to the result of the previous one, effectively computing$$(1+1+1+1+1+1+1+1+1+1+1+1)⋅0+1=0+1=1$$
When using a calculator you have to respect the logic of the machi... | {
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Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$ Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test... | Ratio test is not enough.
Example 1:
$$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$
$$\dfrac{u_n}{u_{n+1}}=\dfrac{\left(\dfrac{1}{n}\right)}{\left(\dfrac{1}{n+1}\right)}=\frac{n+1}{n}=1+\frac{1}{n}$$
Ratio will be $1$ if $n--->\infty$
But we know that S series is divergent.
So the ratio test says us If the ratio... | {
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"source": "stackexchange",
"question_score": "6",
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How to simplify $a^n - b^n$? How to simplify $a^n - b^n$?
If it would be $(a+b)^n$, then I could use the binomial theorem, but it's a bit different, and I have no idea how to solve it.
Thanks in advance.
| $$
a^n - b^n = (a-b+b)^n - b^n = \sum_{k=0}^n {n \choose k}(a-b)^{k}b^{n-k} - b^n = (a-b)b^{n-1} + {n \choose 1}(a-b)^2b^{n-2} + ... + {n \choose n-1} (a-b)^{n-1}b + (a-b)^n = (a-b)(b^{n-1} + {n \choose 1}(a-b)b^{n-2} + {n \choose 2}(a-b)^2b^{n-3} + {n \choose 3}(a-b)^3b^{n-4} + ... + {n \choose n-1} (a-b)^{n-2}b + (a... | {
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What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to? What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.
| without L-hospital law
$$\lim_{x\to\dfrac \pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\frac{1-\sqrt{3}\dfrac {\sin x}{\cos x}}{6\left(\dfrac\pi 6-x\right)}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac{1\cdot\cos x-\sqrt{3} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0... | {
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Why dividing by zero still works Today, I was at a class. There was a question:
If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.
What my teacher did was this:
$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he... | The division your teacher did is polynomial division. He did not divide by zero; he divided by $x^2 - 4x + 5$.
The long division he did is just an algorithm that allows you to get the following identity:
$$x^3 - 3x^2 + 2x - 1 = (x^2 - 4x + 5)(x+1) + x-6$$
This is an identity involving multiplication, not division. Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
} |
How find $\int(x^7/8+x^5/4+x^3/2+x)\big((1-x^2/2)^2-x^2\big)^{-\frac{3}{2}}dx$ How can I compute the following integral:
$$\int \dfrac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{\left(\left(1-\frac{x^2}{2}\right)^2-x^2\right)^{\frac{3}{2}}}dx$$
According to Wolfram Alpha, the answer is
$$\frac{x^4 - 32x^2 + 20}{2\sqr... | Perform the substitutions $y=\frac{x^2}{2}, y=2+\sqrt{3}\sec{\theta}$ to get:
\begin{align}
&\int{\frac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{((1-\frac{x^2}{2})^2-x^2)^{\frac{3}{2}}} dx} \\
& =\int{\frac{y^3+y^2+y+1}{(y^2-4y+1)^{\frac{3}{2}}} dy} \\
&=\int{\left(\frac{y+5}{\sqrt{(y-2)^2-3}}+\frac{20y-4}{\sqrt{((... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\frac{\cos^4\theta}{\cos^2\phi}+\frac{\sin^4\theta}{\sin^2\phi}=1$, show $\frac{\cos^4\phi}{\cos^2\theta} +\frac{\sin^4\phi}{\sin^2\theta}=1$
If $\dfrac{\cos^4\theta}{\cos^2\phi}+\dfrac{\sin^4\theta}{\sin^2\phi}=1$, prove that $\dfrac{\cos^4\phi}{\cos^2\theta} +\dfrac{\sin^4\phi}{\sin^2\theta}=1$.
Unable to move ... | Hint: Let $ x = \cos \theta$, $y = \cos \phi$.
Show by expansion (and clearing denominators) that both equations are equivalent to $x^4 - 2x^2 y^2 + y^4 =0$, hence these statements are equivalent to each other.
Note: This shows that the condition is satisfied iff $x = \pm y$. This is not required, but very strongly hin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate: $\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$ How to evaluate
$$\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$$
I tried trigonometric substitution $x + a = b \sec \theta$ and I encountered
$$\int \frac{\tan \theta}{ (b - a \cos \theta) \sqrt{\tan^2 \theta}}d\theta$$
how to handle this term $\displaystyle \frac{\tan \the... | I substituted $x=1/y$ and got
$$-\int \frac{dy}{\sqrt{1+2 a y+(a^2-b^2) y^2}}$$
I then completed the square in the square root to get
$$-\frac{1}{\sqrt{a^2-b^2}} \int \frac{dy}{\displaystyle\sqrt{\left(y+\frac{a}{a^2-b^2}\right)^2-\left(\frac{a}{a^2-b^2}\right)^2}}$$
Now let
$$y=\frac{a}{a^2-b^2} (\cosh{u}-1)$$
Then t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What are the odds of rolling 3 pairs with six dice? Given a roll of six standard six-side dice, how do you calculate the odds of rolling 3 pairs? This should included non-unique pairs like 2,2,2,2,3,3 or even 5,5,5,5,5,5.
| Let's split into $3$ cases:
Case 1: All dice show the same number. There are $6$ ways this can happen.
Case 2: $4$ dice show the same number, and $2$ dice show a different number. There are $\binom{6}{2}=15$ different ways to choose the $2$ dice that show a different number, and $\binom{6}{2}=15$ ways to choose which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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