Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly ... | Hint: notice that
$$ (2n^2+n)^2 = 4n^4+4n^3+n^2, $$
$$ (2n^2+n+1)^2 = 4n^4+4n^3+5n^2+2n+1 $$
are two consecutive squares and $4(n^4+n^3+n^2+n+1)$ is just between them, with very few exceptions. An old motto says "too close to a square to be a square, it is not a square".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a)... | $$\det(A)= \det\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}= \det\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix} \\
=\det\begin{pmatrix}b-a&(b-a)(b+a)\\c-a&(c-a)(c+a)\end{pmatrix} =(b-a)(c-a)\det\begin{pmatrix}1&b+a\\1&c+a\end{pmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do you add two series together How do you add the series
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
=\... | This isn't wrong but if i were you i would write it as Laurent series. $\displaystyle\frac {1}{2}\left(\sum_{n=0}^\infty \frac{2^n}{(z-3)^{n+1}}+\sum_{n=0}^\infty \frac{(z-3)^n}{4^{n+1}}\right) = \frac {1}{2}\left(\sum_{n=-1}^{-\infty} \frac{(z-3)^n}{2^{n+1}}+\sum_{n=0}^\infty \frac{(z-3)^n}{4^{n+1}}\right)$
$\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the possible values of a in the cubic equation. Given that $(x-a)$ is a factor of $x^3-ax^2+2x^2-5x-3$, find the possible values of the constant $a$.
I believe you first have to find the $a$ in the cubic equation then the other $a$ in $(x-a)$, but I'm having problems finding the first $a$. So far all I have is:
$x... | Hint: If $x-a$ is a factor, then $a$ is a zero. Hence,
$$a^3-a^3+2a^2-5a-3=0\implies 2a^2-5a-3=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality with Algebra $\sup_{x,y,z\in A}\left(\left(13x+\frac{5}{x}\right)+\left(13y+\frac{5}{y}\right)+\left(13z+\frac{5}{z}\right)\right)= 63$ let $A=\{(x,y,z)|x,y,z\in [\frac{1}{2},2],xyz=1\}$
Maybe have
$$f=\sup_{x,y,z\in A}\left(\left(13x+\dfrac{5}{x}\right)+\left(13y+\dfrac{5}{y}\right)+\left(13z+\dfrac{5}{z}\r... | $p=x+y,xy=q, \dfrac{1}{2} \le z \le 2 \implies \dfrac{1}{2} \le q \le 2 $
$f=13p+\dfrac{5p+13}{q}+5q$
consider $\dfrac{A}{q}+Bq$ will get max at two ends, so $f(p,q)$ will get max at $q=2$ or $q= \dfrac{1}{2}$
when $q=2 ,y=\dfrac{2}{x}$, put it in $f$,it become $f(x)= \dfrac{A}{x}+Bx+C$,again, $f(x)$ will get max at tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}
*I tried to zero some elements and expand until I reach the Right hand side.
*Also tried C1-C3, C2-C3 then decompose the determinant into t... | \begin{gather*}
\begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\
= \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\
= \begin{bmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3 - a^3\end{bmatrix} \\
= (b^2-a^2)(c^3 - a^3)-(b^3-a^3)(c^2 - a^2)
= (b-a)(c-a)[(b+a)(c^2+ac+a^2)-(c+a)(b^2+ab+a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all values of positive integers $x,y,z$ so that $4^x+4^y+4^z$ is a perfect square I have to solve the equation
$$4^x+4^y+4^z=k^2$$
I posted my solution but I don't know if there are other proofs. How can I demonstrate that this expression is a perfect square?
Thanks in advance.
| The equation that I have to solve is:
$$4^x+4^y+4^z=k^2$$
Manipulating the equation it becomes: $$4^x(4^{|y-x|}+4^{|z-x|}+1)=k^2$$
Now $4^x$ is always a perfect square, therefore we have to find value of $x,y, z$ so that $4^{|y-x|}+4^{|z-x|}+1$ is a perfect square.
We put $u=|y-x|$ and $v=|z-x|$ and the equation become... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding out this combination In how many ways three non-empty strings of length less than or equal to $N$ using $k$ different characters can be selected so that in each case, among the three strings, no string is prefix (not necessarily proper prefix) of one of the other two strings.
Example:
Let $N=1,k=3$.
Result wil... | Too long for a comment:
If $N=1$, then your problem has $k(k-1)(k-2)$ solutions.
If $N=2$, then the string lengths are $(1,1,1),(1,1,2),(1,2,2)$ or $(2,2,2)$. The number of solutions is $$F(1,1,1)=k(k-1)(k-2)\\F(1,1,2)=k(k-1)(k-2)k\\F(1,2,2)=k(k-1)k[(k-1)k-1]\\F(2,2,2)=k^2(k^2-1)(k^2-2)$$
respectively. The total is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Limit with polylog How do you show the following limit?
$$\lim_{x\to\infty} x\log(-e^x + 1)+\operatorname{Li}_2(e^x)-\frac12x^2=\frac{\pi^2}3$$
Where $\operatorname{Li}_n(x)$ is the polylogarithm.
This question is inspired by a thread in the sagemath mailinglist.
| The asymptotic expansion of dilog is :
$$Li_2(X)=-\frac{1}{2}\ln^2\left(\frac{1}{X}\right)+i\pi\ln\left(\frac{1}{X}\right)+\frac{\pi^2}{3}+O\left(\frac{1}{X}\right)$$
$$Li_2(e^x)=-\frac{1}{2}\ln^2\left(\frac{1}{e^x}\right)+i\pi\ln\left(\frac{1}{e^x}\right)+\frac{\pi^2}{3}+O\left(e^{-x}\right)=-\frac{x^2}{2}-i\pi x+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.
Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed f... | Basis Step:
$P(1) = 12^1 + 2\cdot5^{1-1} = 12 + 2 = 14 = 7\cdot 2$
Thus $P(1)$ is true.
Inductive Step:
Assume $P(k)$ holds for an arbitrary positive integer $k$. Under this assumption, let us prove that $P(k+1)$ is true, namely that
$$12^{k+1} + 2\cdot5^{k} $$ is also a multiple of 7.
\begin{align}
12^{k+1}+2\cdot 5^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving equations including floor function. I got a little trouble solving equations that involve floor function in an efficient way.
For example :
$$
\left\lfloor\frac{x+3}{2}\right\rfloor = \frac{4x+5}{3}
$$
In the one above, I get that you basically let $$ \frac{4x+5}{3} = k $$ and then inserting $k$ in the left sid... | As the left hand side is integer, so should be $\dfrac{3x-5}2\iff2|3(x-1)\iff2|(x-1)\implies x$ is odd (assuming $x$ to be an integer)
Again as lcm$(3,6)$ we need to test for $x\equiv0,1,2,3,4,5\pmod6$
But as $x$ is odd, $x\equiv1,3,5\pmod6$
If $x=6b+1$
$$\left\lfloor\frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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number $e$, why $\sum_{k=0}^\infty \frac{1}{k!}=e$? The definition that I have for $e$ is
$$e:=\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n.$$
How can we get that
$$e=\sum_{k=0}^\infty \frac{1}{k!}\ \ ?$$
My try
By Newton formula
$$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{... | As you say, by the Binomial Theorem,
$$
\left(1+\frac1n\right)^n
=\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\tag{1}
$$
For each $k\le n$, by Bernoulli's Inequality,
$$
\begin{align}
\frac{\binom{n+1}{k}\frac1{(n+1)^k}}{\binom{n}{k}\frac1{n^k}}
&=\frac{n+1}{n-k+1}\left(1-\frac1{n+1}\right)^k\\
&\ge\frac{n+1}{n-k+1}\left(1-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Plea... | Substitute $t=1+x^3$ and $dt=3x^2$:
\begin{align*}
\int \frac{x^5 dx}{\sqrt{1+x^3}}&=\frac{1}{3}\int \frac{(t-1) dt}{\sqrt{t}}\\
&=\frac{1}{3}\int \left( \sqrt{t}-\frac{1}{\sqrt{t}}\right)dt\\
&=\frac{1}{3}\left( \frac{2}{3}t^\frac{3}{2}-2\sqrt{t}\right)\\
&=\frac{2}{9}(x^3-2)\sqrt{1+x^2}
\end{align*}
And of course don... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
| Suppose that $x$ is a solution. Flip things over. We get
$$\frac{1}{\sqrt{x-4}+\sqrt{x-7}}=1.$$
Rationalizing the denominator, and minor manipulation gives
$$\sqrt{x-4}-\sqrt{x-7}=3.$$
From this, addition gives $2\sqrt{x-4}=4$. The only possibility is $x=8$, which is not a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Seating arrangements with no 3 objects together.
Suppose that five $1$'s and six $0$'s need to be arranged in such a way that no three $0$'s are consecutive. How many different arrangements are possible?
This is a variation on a problem where some number of boys and girls need to be seated such that no two boys sit ... | user84413 has provided an elegant approach to solving this problem.
Here is another approach:
If we place five ones in a row, we create six spaces, four between successive ones and two at the ends, which we wish to fill with six zeros. If we let $x_k$ denote the number of zeros placed in the $k$th space from the lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the absolute maximum and minimum value of $f$ I have to find the absolute maximum and minimum value of $$f(x, y)=\sin x+\cos y$$ on the rectangle $[0, 2\pi] \times [0, 2\pi]$.
I have done the following:
$$\nabla f=(\cos x, -\sin y)$$
$$\nabla f=0 \Rightarrow \cos x=0 \text{ and } -\sin y=0 \Rightarrow x=\frac{\... | Answer to question 2. (only)
$f(x,y)=\sin x+\cos y.$ One has $\partial U=A\cup B\cup C\cup D$ with
\begin{eqnarray*}
A &=&\left\{ (x,0):0\leq x\leq 2\pi \right\} \\
B &=&\left\{ (2\pi ,y):0\leq y\leq 2\pi \right\} \\
C &=&\left\{ (x,2\pi ):0\leq x\leq 2\pi \right\} \\
D &=&\left\{ (0,y):0\leq y\leq 2\pi \right\} .
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the soluti... | Your denominator of $4$ in $(b)$ is wrong - $\log_2(4)=2$, not $4$. Then you'd get:
$$2^{\frac{\log_2(3)}{2}} = \left(2^{\log_2(3)}\right)^{1/2}=3^{1/2}$$
It might be easier to just use that $\log_4{x}=\frac{\log_2(x)}{2}$ and solve:
$$\log_2(x)\log_3(x)=\frac{\log_2(x)}{2}$$
This is only true when $\log_2(x)=0$ or whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv... | Notice
$$8^5=210\cdot{156}+8$$
So
$$8^5\equiv8\pmod{210}--(1)$$
Also
$$8^2\equiv64\pmod{210}--(2)$$
By (1),(2) we have
$$8^7\equiv8\cdot{64}\equiv92\pmod{210}$$
Futhermore
$$8^{35}=(8^5)^7\equiv8^7\equiv92\pmod{210}--(3)$$
Morever
$$8^3\equiv512\equiv92\pmod{210}--(4)$$
Therefore, by (3),(4),
$$8^{38}=8^{35}\cdot8^3\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 6
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What is the area of shaded region which is lies between outer and inner circle. There is a outer circle with radius 2r and another inner circle with radius r whose center is the middle of big circle.As depicted in the following figure.
Foo graph Image
There is a sector of 120 degree in inner circle which leads to shad... | You can also easily solve the problem by rotating the figure by $90^o$ clockwise so that the dotted line coincides with the x-axis & center of outer circle at the origin. Thus, equation of the outer circle: $x^2+y^2=4r^2$ & the equation radius above x-axis: $y-0=\tan60^o(x-r)$ or $y=\sqrt{3}(x-r)$ solving the equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Question regarding Proof by Contradiction This is taken out of Apostol's Calculus I. It's part of his retelling of Archimedes method for finding the integral of $x^2$.
At one point, the conclusion is reached
Therefore any number $A$ must also satisfy
$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$... | Note that "for all integers $n \geq 1$'' goes with the first sentence of the quote, not the second. The alternatives in the second sentence do not depend on $n$, but rather compare two constants.
In his version of Archimedes' proof, Apostol assumes $A > \frac{b^3}{3}$ and then uses the first statement of the quote -- ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Minimum number of circular segments. Let K be any natural number. Consider the unit square, and the circle of diameter 1 inside of the square. We then consider circular segments of area $\frac{1}{2K}$ and claim that there exists a constant $c$ such that $\#$(mutually disjoint circular segments of area $\frac{1}{2K}$)$\... | Forget about the unit square, it's irrelevant.
Let K be a positive real. The maximum area of disjoint segments arises if the segments bound a polygon inscribed in the circle, and since the areas of a set of segments for a given K are equal that polygon is regular.
Let $n$ be the number of segments. Let $\theta$ be the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the length of tangent $x$.
Two circles $C_1$ and $C_2$ of radius $2$ and $3$ respectively touch each other as shown in the figure .If $AD$ and $BD$ are tangents then the length of $BD$ is
$a.)3\sqrt6\\
b.)5\sqrt6\\
\color{green}{c.)\dfrac{7\sqrt6}{3}}\\
d.)6\\$
I did a construction of $CE$
And with help of P... | As shown in the second diagram, we have in right $\Delta ABD$ we have $$AD=\sqrt{x^2+10^2}=\sqrt{x^2+100}$$ Now, in right triangles $\Delta ABD$ & $\Delta AEC_{2}$, we have $$\sin \angle DAB=\sin \angle EAC_{2} \implies \frac{BD}{AD}=\frac{EC_{2}}{AC_{2}} \implies \frac{x}{\sqrt{x^2+100}}=\frac{3}{7}$$ $$49x^2-9x^2=900... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the sum of three series which may be telescoping Let $$\sum_{n=1}^\infty \frac{n-2}{n!}$$
$$\sum_{n=1}^\infty \frac{n+1}{n!}$$
$$\sum_{n=1}^\infty \frac{\sqrt{n+1} -\sqrt n}{\sqrt{n+n^2}}$$
I have to calculate their sums. So I guess they are telescoping. However, I've no idea about how to make the telescopy... | Lets begin with the first two since they represent a very simple class. Consider the function
$$ e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + ... = \sum_{n=0}^{\infty}{\frac{1}{n!}{x^n}} $$
From here it suffices to consider
$$ \frac{e^x}{x^2} = \sum_{n=0}^{\infty}{\frac{1}{n!}{x^{n-2}}} $$
And therefore
$$ \frac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Multiplicative Inverse Element in $\mathbb{Q}[\sqrt[3]{2}]$ So elements of this ring look like
$$a+b\sqrt[3]{2}+c\sqrt[3]{4}$$
If I want to find the multiplicative inverse element for the above general element, then I'm trying to find $x,y,z\in\mathbb{Q}$ such that
$$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(x+y\sqrt[3]{2}+z\sqrt... | First let me remark that there is a general abstract argument which shows that if $L/K$ is a field extension and $a \in L$ is algebraic over $K$, then the $K$-algebra $K[a]$ is a field. Namely, $K[a]$ is an integral domain which is also a finite-dimensional vector space over $K$. This implies for $0 \neq x \in K[a]$ t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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If $a_1,a_2,a_3$ are roots $x^3+7x^2-8x+3,$ find the polynomial with roots $a_1^2,a_2^2,a_3^2$ If $a_1,a_2,a_3$ are the roots of the cubic $x^3+7x^2-8x +3,$ find the cubic polynomial whose roots are:
$a_1^2,a_2^2,a_3^2$ and the polynomial whose roots are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.$ Not really sure wh... | For the second,
if
$f(x)
= x^3+ux^2+vx+w
=0
$
then,
if
$g(x)
=x^3f(1/x)
= x^3(1/x^3+u/x^2+v/x+w)
= 1+ux+vx^2+wx^3
$
$g(x)$
has the same roots as
$f(1/x)$
as long as none of the roots
are zero.
Therefore,
just reverse the order of the coefficients:
$3x^3-8x^2+7x+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$
the way my teacher wants us to solve is by substitution values for x,
I set it up like this:
(after setting the variables to the common denominator and getting rid of the denominator in the original equat... | After $B=\frac25$, you may obtain $C$ and $D$ similarly with
$$\frac{x^2-2x+3}{(x-1)^2}=\frac{A(x^2+4)}{x-1}+\frac{B(x^2+4)}{(x-1)^2}+{Cx+D}$$
by setting $x^2=-4$ to get $1=-8C+3D$, $2=3C+2D$, yielding
$C=\frac4{25}$ and $D=\frac{19}{25}$. On the other hand, $A$ can be obtained by examining the limit $x\to \infty$ of a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i... | How's about...
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
$$
\left(-\frac{4}{x}\right)u = (-2)x^{-1} -\frac{du}{dx}
$$
$$
u = \frac{2}{4x} + \frac{x}{4}\frac{du}{dx}
$$
As $u=y^{1-n} = y^{-2}$
$$\frac{1}{y^2} = \frac{2}{4x} + \frac{x}{4}\frac{du}{dx} $$
$$ y^2 = = \frac{4x}{2} + \frac{4}{x}\frac{... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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What are the facts used in each step of this proof? What are the facts used in each step of this proof ?
Suppose that $A\in F^{nm}$ and $B\in F^{ml}$
$$\begin{align}rank A + rank B &= rank\begin{bmatrix}0 & A\\B & 0\\ \end{bmatrix}\\
&\le rank\begin{bmatrix}0 & A\\B & I\\ \end{bmatrix}\\ &= rank\begin{bmatrix}I & A\\0 ... | Pick rank-many independent rows from $\begin{pmatrix}-AB&0\\B&1\end{pmatrix}$. Some of these rows live in $\begin{pmatrix}-AB&0\end{pmatrix}$ and some in $\begin{pmatrix}B&1\end{pmatrix}$, where they are still linearly independent hence certainly not more than $\operatorname{rank}\begin{pmatrix}-AB&0\end{pmatrix}$ and ... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving for a three dimensional vector. Let
$a = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$ and $b = \begin{pmatrix} -6 \\ 4 \\ -3 \end{pmatrix}$
There exists two nonzero three-dimensional vectors
${v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$
that are orthogonal to both ${a}$ and ${b}$, such that its entries $x$... | \begin{align}
2x + 5y - z = 0 \rightarrow x + \frac{5}{2}y - \frac{1}{2}z = 0\\
-6x + 4y - 3z = 0 \rightarrow x - \frac{2}{3}y + \frac{1}{2}z = 0\\
\rightarrow \\
\left(\frac{5}{2} + \frac{2}{3}\right)y - z = 0
\end{align}
This gives that $z = \frac{19}{6}y$ which gives that $x = -\frac{5}{2}y + \frac{1}{2}\frac{19}{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Is it possible to make a dataset given these following mean, range, and standard variance? Given that the:
$Mean = 30$
Range $= X_n - X_1 = 10$
$S^2 = Variance = 40$
is it possible to construct a dataset with those values?
So, to be honest I have no idea how to properly approach this and I have tried "brute forcing" th... | No, it is impossible.
Given any list of $n$ numbers $X_k$ without any constraint of its order. Let
$$\begin{cases}
\overline{X} &= \frac{1}{n}\sum_{k=1}^n X_k\\
\overline{X^2} &= \frac{1}{n}\sum_{k=1}^n X_k^2
\end{cases}
\quad\text{ and }\quad
\begin{cases}
P &= \max\{ X_k : 1 \le k \le n \}\\
Q &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which $n$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ imply $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$ I'm having trouble finishing a problem on an old national competition.
As the title states, the question says asks:
Given $a,b,c \neq 0,a+b=c$ such that $\frac{1}{a}+\frac{1}{... | Lemma: if $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$,then we have
$$\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}$$
where $n$ is odd.
proof:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}
\Longrightarrow \dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}$$
$$\Longrightarrow (a+b+c)(a... | {
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"url": "https://math.stackexchange.com/questions/1305121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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The biggest and smallest integer solution of $\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$ are? $$\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$$
I noticed that $5+2\sqrt6=(\sqrt2+\sqrt3)^2$ but that hardly helps.
After cancelling the square root and the power of two, I tried multiplying the whole thin... | Firstly note that
$$(5+2\sqrt{6})(5-2\sqrt{6}) = 25-24=1$$
so $(5+2\sqrt{6})^{-1}= (5-2\sqrt{6})$.
Putting $t=(5+2\sqrt{6})^x$ you get the inequality
$$t + \frac{1}{t} \leq 98$$
which has solution
$$(5+2\sqrt{6})^{-2} = 49 - 20 \sqrt{6} \le t \le 49 + 20 \sqrt{6} = (5+2\sqrt{6})^2$$
which means exactly
$$-2 \le x \le 2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Finding the Laurent representation of a complex function How can i find the Laurent representation fot the function:
$$f(z)=\dfrac{1}{1-z^2}+\dfrac{1}{3-z}$$
In the region of:
a) $\{z\in\mathbb C:1<|z|<3\}$
b) $\{z\in\mathbb C:1<|z-2|<3\}$
| $\frac{1}{1 - z} = \sum z^n$--the simplest function. You have the following:
$$
\frac{1}{2}\frac{1}{1 - (-z)} + \frac{1}{2}\frac{1}{1 - z} + \frac{1}{3}\frac{1}{1 - \frac{z}{3}}
$$
When $1 < |z| < 3$, we need the Laurent series for $\frac{1}{1 \pm z}$ but the Taylor series for $\frac{1}{3 - z}$:
$$
\frac{1}{1 - z} = \... | {
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"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
... | We can do a slick AM-GM "pairwise" token that I picked up from the "Cauchy masters":
$\dfrac{ab}{c} + \dfrac{bc}{a} \geq 2\sqrt{\dfrac{ab}{c}\cdot\dfrac{bc}{a}}= 2b$, and similarly: $\dfrac{bc}{a}+\dfrac{ca}{b} \geq 2c$, and $\dfrac{ca}{b} + \dfrac{ab}{c} \geq 2a$. Add them up and divide by $2$ to get the answer.
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 1... | Hint:
$$
(A+B)^2= (A+B)(A+B)=A^2+AB+BA+B^2=A^2+B^2+2AB
$$
Use: $A=2x^2$ and $B=x$ and you find the right result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Homogenous equation to higher order ODE Hello I have a quick question in regard to general form of the solution to
$$y^{(4)}-2y^{(3)}+y''=0$$
I had thought to find this solution we would consider
$r^{4}-2r^{3}+r^{2}=0$
which factors as $r^2(r-1)^{2}$
that is we have $r_{1,2}=0$ and $r_{3,4}=1$
So what from what I had t... | For the equation $y^{(4)} - a y^{(3)} + y^{(2)} = c_{0}x$ the solution is as follows. Integrate twice to obtain $y^{(2)} - a y^{(1)} + y = c_{0} \frac{x^{3}}{6} + c_{1} x + c_{2}$. Now let $y(x) = f(x) + b_{1} x^{3} + b_{2} x^{2} + b_{3} x + b_{4}$ to obtain
\begin{align}
f'' - a f' + f + [b_{1} x^{3} + (b_{2}-3 a b_{1... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
| Multiplying $(x-1)^2\gt 0$ is a good idea, but you made mistakes after that.
$$2(x-1)^2\lt x(x-1)\iff x^2-3x+2\lt 0$$$$\iff (x-2)(x-1)\lt 0\iff 1\lt x\lt 2$$
and
$$x(x-1)\le 3(x-1)^2\iff 2x^2-5x+3\ge 0$$$$\iff (x-1)(2x-3)\ge 0\iff x\le 1\ \text{or}\ x\ge\frac 32.$$
Hence, the answer is
$$\frac 32\le x\lt 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where t... | We have that $|1+i| = \sqrt{2}$ and ${\rm Arg}(1+i) = \pi/4$. So: $$1+i = \sqrt{2}\left(\cos \frac{\pi}{4}+i \sin\frac{\pi}{4}\right).$$By De Moivre's formula: $$(1+i)^6 = (\sqrt{2})^6 \left(\cos \frac{6\pi}{4}+i\sin\frac{6\pi}{4}\right).$$But $\sqrt{2}^6 = (2^{1/2})^6 = 2^3=8$. Simplifying: $$(1+i)^6 = 8\left(\cos \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.
What are all integral solutions of this equation?
I can only find $x=y=z=1$.
| Using the Power Mean Inequality we have that:
$$ 1=\frac{1}{3}(x^3 + y^3 + z^3) \leq \left[\frac{1}{3}(x+y+z)\right]^3 =1$$
Equality only holds when $x=y=z=1$.
Let's replace $z \mapsto -z$ and try to solve the system of equations for $x,y,z \geq 0$:
$$ x^3 + y^3 = 3 + z^3 \text{ and } x + y = z + 3 $$
We can still ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Difficulty proving $x_n$ satisfaction in the inequality I am trying to prove that $$x_n = \int_{0}^{1}\left(\frac {n}{x}\right)^n \,dx$$ satisfies the inequalities
$$\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
... | $$x_n = n \int_{n}^{\infty} \frac {t^n}{t^2}\,dt = n \sum_{k=n}^{\infty}\int_{k}^{k+1} \frac {t^n}{t^2} \,dt = n \sum_{k=n}^{\infty}\int_{k}^{k+1} \frac {(t-k)^n}{t^2} \,dy = n \sum_{k=n}^{\infty} \int_{0}^{1} \frac {y^n}{(k+y)^2}\,dy = n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{(k+y)^2}\right) \,dy $$
Sin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can one evaluate the integral $\int\sqrt{\frac{1-ax}{1-x}}\ dx$? How can I evaluate the following integral:
$$\int\sqrt{\frac{1-ax}{1-x}}\ dx?$$
Here $a$ is a positive constant. I know that the function
$$\sqrt{\frac{1-ax}{1-x}}$$
has a primitive, but I don't know how to find it.
| The standard way for these integrals is to use substitution: set
$$y^2=\frac{1-ax}{1-x}=a+\frac{1-a}{1-x},\enspace y>0,\enspace\text{whence}\enspace \mathrm 2y\, d\mkern1mu y=\frac{1-a}{(1-x)^2}\mathrm d\mkern1mu x$$
so that
$$\mathrm d\mkern1mu x =2y\frac{(1-x)^2}{1-a}y=\frac{2(1-a)y}{(y^2-a)^2}\mathrm d\mkern1mu y$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go furth... | $$\int\left(\sin^3(x)\cos^2(2x)\right)dx=$$
$$\int \left(\frac{7}{16}\sin(x)-\frac{5}{16}\sin(3x)+\frac{3}{16}\sin(5x)-\frac{1}{16}\sin(7x)\right)dx=$$
$$\int \frac{7}{16}\sin(x)dx-\int \frac{5}{16}\sin(3x)dx+\int \frac{3}{16}\sin(5x)dx-\int \frac{1}{16}\sin(7x)dx =$$
$$\frac{7}{16}\int \sin(x)dx-\frac{5}{16}\int \sin(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $T(n)= T(n-1) + 2T(n-2)$ If $T(n)= T(n-1) + 2T(n-2)$ with $T(0)=0$ and $T(1) = 1$
What is $T(n)$ (in $Θ$–notation) in terms of $n$?
I am trying to solve by substitution, but I am not sure if I am doing this right, as I get stuck. Can anybody tell me where to go from here?
$T(n)= T(n-1) + 2T(n-2)$
$T(n-1)= T(n-2) ... | Generating functions for the win. Let $T(x) = \sum_{n=0}^{\infty}T_nx^n = \sum_{n=1}^{\infty}T_nx^n$, since $T_0 = 0$.
Then:
$$\begin{split}
T(x) &= x + \sum_{n=2}^{\infty}T_nx^n \\
&= x + \sum_{n=2}^{\infty}(T_{n-1} + 2T_{n-2})x^n \\
&= x + \sum_{n=2}^{\infty}T_{n-1}x^n + \sum_{n=2}^{\infty}2T_{n-2}x^n
\end{split}$$
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$
Can anyone show me a numerical method step-by-st... | $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}= \frac{2^n}{4^n} +\frac{1^n}{4^n} +\frac{3^n}{4^n} = \frac{1^n+2^n+3^n}{4^n}=1,\: \therefore 1^n+2^n+3^n=4^n $
I don't think this can be solved analytically,
you just have to use trial and error with n
| {
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"url": "https://math.stackexchange.com/questions/1316957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
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How many strings of $8$ English letters are there (repetition allowed)? a) at least one vowel
b) start with $x$ and at least one vowel
c) start and end with $x$ and at least one vowel
I can solve them easily by considering $total-no$ $vowel$. So,
a) $26^8 -21^8$
b) $26^7 -21^7$
c) $26^6 -21^6$
But, can I try
a) choos... | Your calculation has a lot of overcounting, whenever there is more than one vowel present.
If you really want to avoid (or, say, cross-check) the "negative space" method of simply excluding options with no vowels, you could perhaps sum through the possibilities of where the first vowel is:
*
*Vowel in first place: $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to find expansion of powers of multinomials without any coefficients? For example, $(a + b + c)^3 = a^3 + b^3 + c^3 + 3ab^2 + 3ac^2 + 3a^2b + 3a^2c + 3bc^2 + 3b^2c + 6abc$
Knowing the value of a, b and c, is there a way to find this without the coefficients i.e. $a^3 + b^3 + c^3 + ab^2 + ac^2 + a^2b + a^... | The answer to this question gives the formula you are looking for ($m=3$).
$$\frac{a^{m+2}}{(a-b)(a-c)}+\frac{b^{m+2}}{(b-a)(b-c)}+\frac{c^{m+2}}{(c-a)(c-b)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$
let $a,b,c>0$, such $ab+bc+ac=1$,show that
$$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{4}$$
by Macavity C-S:with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\f... | Using Cauchy-Schwarz in a different way:
$$\begin{align}
2(3a+5b+7c) &= 15(a+b+c)-(9a+5b+c) \\
&= \sqrt{(118+107)(2+a^2+b^2+c^2)} - (9a+5b+c)\\
&\ge \sqrt{118\cdot2}+\sqrt{(9^2+5^2+1^2)(a^2+b^2+c^2)} -(9a+5b+c)\\
&\ge 2\sqrt{59}
\end{align}
$$
$$\implies \sum_{cyc} \frac1{3a+5b+7c} \le \frac3{\sqrt{59}} < \frac{\sqrt3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the Range of a Trigonometric function The range of $$f(x)=3\cos^2x-8\sqrt3 \cos x\cdot\sin x+5\sin^2x-7$$is given by:(1)$[-7,7]$(2)$[-10,4]$(3)$[-4,4]$(4)$[-10,7]$
ANS: (2)
My Solution
The equation can be written as: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2... | Let's do this.
\begin{align}
f(x)&=3\cos^2 x−8\sqrt3 \cos x\sin x + 5\sin^2x−7 =\\
&= 3(\cos^2x + \sin^2 x) - 4\sqrt3\cdot 2\sin x\cos x + 2\sin^2 x - 7=\\
&=3 - 4\sqrt 3\sin 2x + 2\sin^2 x - 7 = \\
&=-4 -4\sqrt 3\sin 2x + 1 - \cos 2x=\\
&=-\cos2x -4\sqrt 3\sin 2x-3.
\end{align}
Now we have
$$1\cdot\cos2x +4\sqrt 3\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} -... | I do not agree with your result after factorizing in the last step. I get
$$\frac{x(3-1/x)}{x\left(\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}\right)}=\frac{3-1/x}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}}\rightarrow 3/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First ter... | Here is another way to square the series without multiplying out the terms.
Since $\cos z = 1 - \dfrac{z^2}{2!} + \dfrac{z^4}{4!} - \dfrac{z^6}{6!} + \cdots$, we have $\dfrac{1-\cos z}{z} = \dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots$.
Thus, $\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to retrieve the expression of $a^3+b^3+c^3$ in terms of symmetric polynomials? I recently had to find without any resources the expression of $a^3+b^3+c^3$ in terms of $a+b+c$, $ab+ac+bc$ and $abc$.
Although it's easy to see that $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$, I couldn't come up myself with $a^3+b^3+c^3=(a+b+... | Let $E$ be the expression equal to $a^3+b^3+c^3$ in terms of symmetric polynomials. Since the reduced expression has no mixed terms, you must have $E = (a+b+c)^3 + E'$, where
$$ E' = a^3+b^3+c^3 - (a+b+c)^3 = -3(a^2b +ab^2 +a^2 c +2abc +b^2c +ac^2 +bc^2) $$
To get all of the terms like $a^2b$, we must have $(a+b+c)(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Values $a,b,c$ are all positive reals. I tried Muirhead and a few AM $\geq$ GM.
This problem is equivalent to proving $a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5$.
| Earth lended me it's energy. Without any assumption on the size of $a,b,c$:
$a^7 + a^7 + a^7 + a^7 + b^7 + b^7 + b^7 \geq 7a^4b^3$
$b^7 + b^7 + b^7 + b^7 + c^7 + c^7 + c^7 \geq 7b^4c^3$
$c^7 + c^7 + c^7 + c^7 + a^7 + a^7 + a^7 \geq 7c^4a^3$
adding up I get what I wanted
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are... | Take the integral in the form $$I=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{4}+u^{2}\right)^{1/4}\left(1-u^{2}\right)^{1/2}}=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{2}+1\right)^{1/4}\left(1-u^{2}\right)^{1/2}\left(u^{2}\right)^{1/4}}
$$ then put $u^{2}=s
$ $$=\frac{1}{2^{5/4}}\int_{0}^{1}\frac{ds}{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 4,
"answer_id": 3
} |
Points of intersection of two functions Through the following steps I found the x-coordinates of the intersection points of two functions:
$(x)= -x^{2}+3x+1\: and\: g(x)=3/x $
The numbers I found are x=(1, 2, 3)
But on the graph, one of the points has a negative x value, could you guys point me to anything I have mi... | The first equation is a cubic, $ -(x+1)(x-1) (x-3) = 0, $
with roots $ x = \pm 1, 3 $ at points of intersection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Mathematical Reflections J339: Solving $\frac{x-1}{y+1} + \frac{y-1}{z+1} + \frac{z-1}{x+1} = 1$ in positive integers I have been trying to solve $\frac{x-1}{y+1} + \frac{y-1}{z+1} + \frac{z-1}{x+1} = 1$ in positive integers and I'm shaky on one specific step. I solved the problem through case-by-case analysis and I wo... | observe that the quadratic $(z-3)(z+1)$ has a minimum where $z=1$ and its value there is $-4$, so we get $-2(y-1)\ge -4$, that is $y\le 3$. since $y\ge 1$, we get $(z-3)(z+1)\le 0$, so $z\le 3$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equa... | Well you have:
$$1^3+\ldots+n^3=\frac{n^2(n+1)^2}{4} $$
You can prove it easily by induction. Now your proposition could not work without even knowing the formula for the following reason:
$$\frac{n^3+(n+1)^3}{2}\text{ is equivalent to } n^3$$
Whereas I claim that:
$$\int_0^{n}x^3dx \leq \sum_{k=1}^nk^3\leq \int_0^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$?
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?
I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \... | One can also integrate by parts. To that end, we have
$$\begin{align}
I=\int_0^1x\sqrt{1+px}dx&=\left.\left(\frac{2x}{3p}(1+px)^{3/2}\right)\right|_0^1-\frac{2}{3p}\int_0^1(1+px)^{3/2}dx\\\\
&=\frac{2}{3p}(1+p)^{3/2}-\frac{2}{3p}\left.\left(\frac{2}{5p}(1+px)^{5/2}\right)\right|_0^1\\\\
&=\frac{2}{3p}(1+p)^{3/2}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove that a root $z_2=\frac{-a-\sqrt{a^2+b^2}}{2b} <1$ I want to prove that that the integral $I= \int_{0}^{2π} \frac{1}{a+b \cos\theta}\,d \theta$, where $a>0$, $b>0$ and $a>b$, is equal to $ I=\dfrac{2πb}{\sqrt{a^2+b^2}}$ with the method of residues.
My approach is:
Let $z=e^{i\theta}$ be the parametrizatio... | Without loss of generality, we can divide everything by $b$, and so look at $c=a/b$:
$$ -\frac{a/b}{2} \pm \frac{\sqrt{(a/b)^2+1}}{2} = -\frac{c}{2} \pm \frac{\sqrt{1+c^2}}{2}. $$
The conditions give $c>1$. Now, $\sqrt{1+c^2}>\sqrt{c^2}=c$, so
$$ -\frac{c}{2} - \frac{\sqrt{1+c^2}}{2} < -\frac{c}{2}-\frac{c}{2} = -c < -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to?
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?
Other conditions ... | Lethal weapon #2: Multiply and divide by the same thing:
$$
\frac{1+i\sin{\alpha}}{1-i\sin{3\alpha}}=\frac{i}{i}\frac{1+i\sin{\alpha}}{1-i\sin{3\alpha}}=\frac{i-\sin\alpha}{i+\sin 3\alpha}.
$$
Thus $2x=-\sin\alpha$ and $y=\sin3\alpha$, so
$$
\frac{x}{y}=-\frac{\sin\alpha}{2\sin3\alpha}=-\frac{1}{2+4\cos2\alpha}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $ without using L'hopital's rule Find $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $$ without using L'hopital's rule. Using the rule, I got it as 1/12. What's a way to do it without L'hopital?
| Let's first analyze the limit when $x \to 0^{+}$. If $0 < t < 1$ then we know that $$t - t^{2} = t(1 - t) < \frac{t}{1 + t} < \log(1 + t) < t$$ Multiplying by $t$ we get $$t^{2} - t^{3} < t\log(1 + t) < t^{2}\tag{1}$$ Also we can easily see that $$\frac{1}{4}\left(1 - \frac{t}{4}\right) < \frac{1}{t + 4} < \frac{1}{t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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How to solve this type of exercises $\sqrt{x^6+x^5-2x^3+O(x^2)}$ I have a simulation test with this type of exercise, asymptotic expansion:
$$\sqrt{x^6+x^5-2x^3+O(x^2)}$$
with
$$ x\rightarrow \infty$$
I have studied the theory of Landau's symbols but I have no idea about how to solve.
Can someone please explain me how ... | Write
$$\sqrt{x^6+x^5-2x^3+O(x^2)}=$$
$$=\sqrt{x^6+x^5-2x^3}+\sqrt{x^6+x^5-2x^3+O(x^2)}-\sqrt{x^6+x^5-2x^3}=$$
$$=\sqrt{x^6+x^5-2x^3}+\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}.$$
But $\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}\sim 2x^3$, hence
$$\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Understanding the solution to a basic annuity problem involving an unknown interest rate The following is the problem and the solution:
Before looking at the solution, here is how I approached the problem:
Let $X$ be the amount that each child receives. (i) and (ii) imply that $Xa_{20} = v^{20}\cdot{}2X\cdot\frac{1}{i... | First of all it is assumed that john´s estate is 1. It works also with other values. Thus all three, children and charity, receive $\frac{1}{3}$.
Now you use the formula for the present value. The present value have to be 1/3 for both children.
The present value for one child is
$C_0=r\cdot v\cdot \frac{1-v^n}{1-v}\Ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt... | The first integral is correct.
The second is easier! Substute $3x=t$ and note that $\tan^2 t= \dfrac{1}{\cos^2t}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Substituting a value of sine function in a trigonometric equation I am trying to really understand trigonometric equations and I've stumbled upon a rather confusing example. Solve the following equation:
$\sin x= 2|\sin x|+ {\sqrt{3}}\cos x$
First step is to define the absolute $\sin x$:
$$|\sin x| =
\begin{cases}
\... | For simplicity, let's find the solutions in $[0,2\pi)$.
1) If $\sin x\ge0$, we get $\sin x=2\sin x+\sqrt{3}\cos x$, which yields $-\sin x=\sqrt{3}\cos x$.
$\hspace{.2 in}$Dividing by $-\cos x$ gives $\tan x=-\sqrt{3}$, so $x=\pi-\frac{\pi}{3}=\color{red}{\frac{2\pi}{3}}$
$\hspace{.2 in}$(since $\sin x>0, \tan x<0\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit? The limit is
$$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$
which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using t... | Since you already received good answers, let me add another point of view.
Rewriting the expression
$$A=\frac1{\sin^2(x)}-\frac1{x^2}=\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}$$ and knowing that, close to $0$, $\sin(x)\approx x$, then the denominator looks like $x^4$ that is to say that, if the limit exist the numerator shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 2
} |
Volume of the intersection of two cylinders I have two infinite cylinders of unit radius
in $\mathbb{R}^3$, whose axes are skew lines.
Say that the axis of one is centered on the $x$-axis, and the axis of the
other is determined by the two points $a$ and $b$. Is there a formula for the
volume of their intersection, as ... | Let
*
*$r = 1$ be the common radii of the two cylinders.
*$2d$ be the nearest distance between the two axis of the cylinders.
*$\alpha = 2\beta$ be the angle between the two axis of the cylinders.
Choose the coordinate system so that the axis of the cylinders pass through
$(0, \pm d, 0)$ with tangent vectors $( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of cosA+cosB+cosC in a triangle ABC I have used jensen's inequality but couldn't move on.
| $$
\begin{align}
\sqrt[3]{\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})}
&
\le \frac{\sin(\frac{A}{2}) + \sin(\frac{B}{2}) + \sin(\frac{C}{2}) }{3} \text{ (AM-GM inequality)}\\
&
\le \sin \left( \frac{ \frac{A}{2} + \frac{B}{2} + \frac{C}{2}}{3} \right) \text{ (Jensen's inequality)}\\
&
= \sin(\frac{\pi}{6})\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:
$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one... | I would like to add some point to answer posted user @mixedmath.
For equation
$$
ax^2 + bx + c = 0
$$
roots are
$$
x = \frac {-b\ \pm \ \sqrt {b^2 - 4ac} }{2a}
$$
Lets put
$$
y = \frac {1}{x}
$$
Then we get converted equation as,
$$
cy^2 + by + a = 0
$$
Equivalent roots are,
$$
y = \frac {-b\ \mp \ \sqrt {b^2 - 4ac} }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in... | $$
\frac1a+\frac1b=\frac1c\implies ac+bc=ab
$$
$a\mid bc$. Therefore, each prime that divides $a$ must divide one and only one of $b$ or $c$ (since $(a,b,c)=1$). Therefore, $a=(a,b)(a,c)$. Similarly, $b=(a,b)(b,c)$ and $c=(a,c)(b,c)$.
Thus,
$$
a+b=\frac{ab}c=\frac{(a,b)(a,c)(a,b)(b,c)}{(a,c)(b,c)}=(a,b)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Need hint to solve a nasty integral.
Let $f(x)=\frac{x+2}{2x+3}$, $x>0$. If
$$\int \left( \frac{f(x)}{x^2} \right)^{1/2}dx=\frac{1}{\sqrt{2}}g \left(\frac{1+\sqrt{2f(x)}}{ 1-\sqrt{2f(x)}} \right) -\sqrt{\frac{2}{3}}h \left(\frac{\sqrt{3f(x)}+\sqrt{2}}{\sqrt{3f(x)}-\sqrt{2} } \right)+C$$
where $C$ is the constant of in... | if you find this integral, you can
Hint: Let
$$\sqrt{\dfrac{x+2}{2x+3}}=t,\Longrightarrow x=\dfrac{3t^2-2}{1-2t^2}$$
$$-\int \dfrac{1-2t^2}{3t^2-2}\cdot t\cdot\dfrac{2t}{(2t^2-1)^2}dt=\int\dfrac{2t^2}{(3t^2-2)(2t^2-1)}dt$$
and Note
$$\dfrac{2t^2}{(3t^2-2)(2t^2-1)}=\dfrac{4}{3t^2-2}-\dfrac{2}{2t^2-1}$$
maybe as this re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ t... | $$(33)^{100}\equiv(33)^{5\cdot5\cdot4}\equiv(93)^{5\cdot4}\equiv(93)^4\equiv 01\pmod{100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
| $$
\begin{align}
(1-\omega+\omega^2)(1+\omega-\omega^2)
&=1-\omega^2+2\omega^3-\omega^4\\
&=1-\omega^2+2-\omega\\
&=4-(1+\omega+\omega^2)\\
&=4
\end{align}
$$
The last step is true since $(1+\omega+\omega^2)\overbrace{(1-\omega)}^{\text{not $0$}}=1-\omega^3=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Why are the coefficients always equal? Take the equation $ax^{2} + bx + c = 3x^{2} + 4x + 53$.
Why is it always true that $a = 3, b = 4$ and $c = 53$?
I've seen many examples like this where the coefficients are equated, and was just wondering why that is always true.
| Suppose that $ax^{2} + bx + c = 3x^{2} + 4x + 53$ for all $x$, or $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ for all $x$.
$ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ is a polynomial with at most degree $2$ so by the fundamental theorem of algebra it has at most $2$ roots if it is not the zero polynomial.
But $ax^{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then ad... | HINT : $$\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^3}{1-\omega}+\frac{\omega^4}{1-\omega^3}$$
$$=\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^7}{\omega^4-\omega^5}+\frac{\omega^6}{\omega^2-\omega^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Definite integral with logarithm and arctangent inside of arctangent
How to prove $$\int_0^1 \left[ \frac{2}{\pi }\arctan \left(\frac 2 \pi \arctan \frac{1}{x} + \frac{1}{\pi }\ln \frac{1 + x}{1 - x}\right) - \frac{1}{2} \right]\frac{\mathrm{d}x} x = \frac{1}{2} \ln \left( \frac \pi {2\sqrt 2 } \right).$$
I have trie... | One may show that this integral is equivalent to the integral in this problem as follows:
First, rewrite
$$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$
Then note that
$$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x} $$
The integrand is then equal to
$$\left [\frac{2}{\pi} \tan^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Closed form for a binomial series I am wondering if any knows how to compute a closed form for the following two series.
*
*$$\sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}$$
*$$\sum_{m=1}^{n}\frac{(-1)^m}{m^4}\binom{2n}{n+m}$$
Mathematica gave a result in terms of the Hypergeometric function, which I don't und... | The first expression:
$\begin{align}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}
&= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{n}{m}\frac{m!n!}{(m+n)!}\\
&= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{n}{m}\frac{(m-1)!n!}{(m+n)!}\\
&= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
What are the ways to solve trig equations of the form $\sin(f(x)) = \cos(g(x))$? if I have the following trig equation:
$$\sin(10x) = \cos(2x)$$
I take the following steps to solve it:
*
*I rewrite $\cos(2x)$ as $\sin\left(\frac{\pi}{2} + 2x\right)$ or as $\sin\left(\frac{\pi}{2} - 2x\right)$ cause $\sin\left(\frac{... | Otherwise as you ask could be as follows (although it is almost the same).
You have $$sin (f(x))-sin (\frac\pi2-g(x))=0$$ so from the identity
$$sin(a)-sin(b)=2sin\frac {a-b}{2}cos\frac{a+b}{2}$$
it follows $$2sin\frac{(f(x)+g(x)- \frac{\pi}{2})}{2}cos\frac{(f(x)-g(x)+\frac{\pi}{2})}{2}=0$$ hence $$f(x)+g(x)=2n\pi+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find this infinite product How to find this infinite product ?
$$\prod_{n=0}^\infty \left(1-\dfrac{2}{4(2n+1)^2+1}\right)$$
I try to use infinite product of $\cos{x}$ but it doesn't work.
Thank you.
| $$\begin{eqnarray*}\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) &=& \prod_{n\geq 0}\left(\frac{4(2n+1)^2-1}{4(2n+1)^2+1}\right)\\&=&\color{purple}{\prod_{n\geq 0}\left(1-\frac{1}{4(2n+1)^2}\right)}\color{blue}{\prod_{n\geq 0}\left(1+\frac{1}{4(2n+1)^2}\right)^{-1}}\end{eqnarray*}$$
hence by exploiting the Weiers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Solve $x+y+z=1; x^2+y^2+z^2=35; x^3+y^3+z^3=97$ It may be surprising that I can't get any analytical way of verifying that one of the solutions of $$x+y+z=1$$ $$x^2+y^2+z^2=35$$ $$x^3+y^3+z^3=97$$ is $x=-1, y=-3$ and $z=5$. Although it may be possible that I have to revisit my elementary arithmetic.
| First note that
$$x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)=35$$
Now substituting $x+y+z=1$, we have $1^{2}-2(xy+yz+zx)=35$, and thus $xy+yz+zx=-17$.
Then, $$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)+3xyz=97$$ and substituting with $1$ and our above result once again leads us to have $xyz=15$. Now we need a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make i... | HINT :
$$4.8=1000^{\frac 1x},\ \ \ 0.8=1000^{\frac 1y}$$
So, $$1000^{\frac 1x-\frac 1y}=\frac{1000^{1/x}}{1000^{1/y}}=\frac{4.8}{0.8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Limit and factorization I have the following very interesting homework exercise:
Let $$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}$$ Find the
following limit, if it exists: $$\lim_{x\to 2}f(x)$$
I understand that I need to "delete" the $x-2$ factor from both nominator and denominator and then evaluate the li... | $$\begin{align}\lim_{x\to 2}\frac{x^2-4}{2-x\sqrt{x+2}+\sqrt{x+2}}&=\lim_{x\to 2}\frac{(x-2)(x+2)}{2-(x-1)\sqrt{x+2}}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)}{2-(x-1)\sqrt{x+2}}\cdot\frac{2+(x-1)\sqrt{x+2}}{2+(x-1)\sqrt{x+2}}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)(2+(x-1)\sqrt{x+2})}{4-(x-1)^2(x+2)}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
proof by induction $2^n \leq 2^{n+1}-2^{n−1}-1$ My question is prove by induction for all $n\in\mathbb{N}$, $2^n \leq 2 ^{n+1}-2^{n−1}-1$
My proof
$1+2+3+4+....+2^n \leq 2^{n+1}-2^{n−1}-1$
Assume $n=1$,$1 ≤ 2$
Induction step
Assume statement is true for $n=k$, show true for $n=k + 1$
$1+2+3+4+....+2^k+2^k+1 ≤ 2 ^{k+1... | In the case that $n=1$ we have $$2 \leq 2^2 - 2^0 - 1 \iff 2 \leq 2$$ which is true. Now, assuming that $$2^k \leq 2^{k+1} - 2^{k-1} - 1$$ holds. We proceed by multiplying our inductive hypothesis above by $2$ to get $$2 \cdot 2^{k} \leq 2(2^{k+1} - 2^{k-1} - 1)$$ Which simplifies to $$2^{k+1} \leq 2^{k+2} - 2^{k} - 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Quadratic approximation of $9^{1/3}$ Find a quadratic approximation of the cube root of $9$ by using the equation $9=8(1+\frac 18)$, and estimate the difference between the exact value and the approximation.
How am I supposed to find the quadratic approximation of this formula? If there is no $x$, and the formula for q... | Since $9=8(1+\frac{1}{8})$,
$\;\;\sqrt[3]9=2\big(1+\frac{1}{8}\big)^{1/3}=2\big(1+\frac{1}{3}(\frac{1}{8})-\frac{1}{9}(\frac{1}{8})^2+\frac{5}{81}(\frac{1}{8})^3+\cdots\big)$
since $\displaystyle(1+x)^{1/3}=1+\frac{1}{3}x+\frac{(\frac{1}{3})(\frac{1}{3}-1)}{2!}x^2+\frac{(\frac{1}{3})(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Simplifying $ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$ who can simplify the following term in a simplest way?
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
(The answer is 1). Thanks for any suggestions.
|
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
$$(a+b)^2=2ab+a^2+b^2$$
$$=(\cos^2x+\sin^2x)^2$$
$$=1^2$$
$$=\boxed{\color{red}1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$
(D) none of the fore... | $a^n – b^n = (a – b)(a^{n–1} +a^{n–2}b + a^{n–3}b^2 + ··· + ab^{n–2} + b^{n–1})$
If $n$ is odd,$a^n+b^n = (a+b)(a^{n–1}-a^{n–2}b + a^{n–3}b^2 - ··· - ab^{n–2} + b^{n–1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=... | I have a little more:)
$$
S = 4R^2(\sin A + \sin B + \sin C)\sin\frac A2\sin\frac B2\sin\frac C2\\
S = \frac{r^2}{4}\frac{\sin A + \sin B + \sin C}{\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}\\
S = r^2\left(\cot\frac A2+\cot\frac B2 + \cot\frac C2\right)\\
S = r^2\cot\frac A2 \cot\frac B2 \cot\frac C2\\
S = 2p^2 \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 2
} |
$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$
find $\dfrac {x^2} {x-1}.$
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $... | $${ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( x-1 \right) }^{ 2 } } =2010\\ { \left( x+\frac { x }{ x-1 } \right) }^{ 2 }-2\frac { { x }^{ 2 } }{ x-1 } =2010\\ { \left( \frac { { x }^{ 2 } }{ x-1 } \right) }^{ 2 }-2\left( \frac { { x }^{ 2 } }{ x-1 } \right) -2010=0\\ \left( \frac { { x }^{ 2 } }{ x-1 } \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac... | \begin{align}
\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}
\end{align}
Is equivalent to the system
\begin{align}
xy+xz&=2(x+y+z)
\\
xy+yz&=3(x+y+z)
\\
xz+yz&=4(x+y+z),
\end{align}
Solving this system as a linear syst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Using complex variables to find sums of Fourier series Use complex variables to find the sum of the Fourier Series:
$$\sin(\theta) + \frac{\sin(2\theta)}{2^{2}} + \frac{\sin(3\theta)}{2^{3}}+\cdots$$
where $\theta$ is a real variable.
| We have
$$\begin{align}
f(\theta)&=\sin\theta + \sum_{k=2}^{\infty}\frac{\sin k\theta}{2^k} \tag 1\\\\
&=\sin\theta+\text{Im}\left( \sum_{k=2}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^k\right) \tag 2\\\\
&=\sin\theta+\text{Im}\left(\frac{e^{i2\theta}}{2-e^{i\theta}}\right) \tag 3\\\\
&=\sin\theta+\frac12 \frac{2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But ... | let $g(x) = f(x) - x$
Then:
$$g(1)=g(2)=g(3)=g(4)=g(5)=g(6)=0$$
So:
$$g(x) = A(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$$
We also know that $g(0)=f(0)-0=f(0)=h$ so we can find $A$.
$h=A\times 1\times 2\times 3\times 4\times 5\times 6$ so $A=\frac{h}{6!}$
$f(7) = g(7) + 7 = 6\times 5\times 4\times 3\times 2\times 1\times \frac{h}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$$
I tried taking $n=k$,
so it makes
$$\frac{1}{1\... | You actually copied it over incorrectly; the general term should be $\frac{1}{n(n+1)(n+2)}$, not $\frac{1}{n(n+1)(n+3)}$. That in mind, try to see if you can follow this argument:
For each $n\geq 1$, let $S(n)$ denote the statement
$$
\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin... | HINT:
$$\sin^2y-\sin^4y=\sin^2y(1-\sin^2y)=\dfrac{\sin^22y}4$$
$y=x\implies$ $$\sin^2x-\sin^4x=\dfrac{\sin^22x}4$$
$y=2x\implies$ $$\dfrac{\sin^22x-\sin^42x}{4^1}=\dfrac{\sin^22x}{4^2}$$
Set $y=4x,2^rx$ and add to get
$$\sin^2x-\sum_{r=0}^n\dfrac{\sin^4(2^rx)}{4^r}=\dfrac{\sin^4(2^{n+1}x)}{4^{n+1}}$$
$$\lim_{n\to\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
$\lim_{x\to 0}\frac{e^x-1}{\sin x}$ equal to $\lim_{x\to 0}\frac{e^x-1}{x}$ because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$ I'm stuck in this limit:
$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}$$
I tried to solve it using special limits, so:
$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}=$$
$$=\lim_{x\to 0}(e^x-1)\frac{... | $$\lim _{ x\rightarrow 0 }{ \frac { x\left( { e }^{ x }-1 \right) }{ \cos { x } -1 } = } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 }\left( { e }^{ x }-1 \right) }{ x\left( \cos { x } -1 \right) } = } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| from here, separating it so that they have same thing
$$\frac 1{1+(\sqrt2+\sqrt3)} + \frac 1{1-(\sqrt2-\sqrt3)} + \frac 1{1+(\sqrt2-\sqrt3)} + \frac 1{1-(\sqrt2+\sqrt3)}$$
let $x = \sqrt 2 + \sqrt 3$, $y = \sqrt 2 - \sqrt 3$
$$\frac 1{1+x} + \frac 1{1-y} + \frac 1{1+y} + \frac 1{1-x}$$
combine fraction with x into one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
| A few manipulations:
$$B=A^2-2A+2I$$
$$A^3 = 2I$$
then,
$$AB = A^3 -2A^2+2A \\= 2I -2A^2+2A \\= -2(A^2-2A+2I)-2A+6I \\= -2B -2A+6I$$
Now, the fact that $A^3 = 2I$ means $A$ is invertible, because $\det(A) \neq 0$
Thus,
$$B = -2A^{-1}B-2I+6A^{-1} $$
Take $B$ to one side:
$$B(I+2A^{-1}) = -2I+6A^{-1}$$
Multiply both side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 2
} |
Finding the equation of a circle. A circle of radius $2$ lies in the first quadrant touching both axis. Find the equation of the circle centered at $(6,5)$ and touching the above circle externally.
Let me share how I answered this question with your suggestions.
Since the radius of the first circle is $2$ and touches b... | The first circle is centered at $(2,2)$, so the distance between two centers of both the circles would be $\sqrt{(6-2)^2 +(5-2)^2} =5$. Hence the radius of circle centred at(6,5) is 3 units. So the final equation of circle is $(x-6)^2+(y-5)^2 = 9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Angle between medians in right triangle In a right angled triangle,medians are drawn from the acute-angles to the opposite sides.If maximum acute angle between these medians can be expressed as $tan^{-1}(\frac{p}{q})$ where p and q are relatively prime positive integers.Find $p+q$.
Let triangle is right angled at B.Le... | I think the slope of $AD$ is $\frac{2c}{-a}$ and that the slope of $CE$ is $\frac{-c}{2a}$.
Then, the acute angle between the medians can be expressed as
$$\arctan\frac{\frac{-c}{2a}-\frac{2c}{-a}}{1+\frac{2c}{-a}\cdot\frac{-c}{2a}}=\arctan\frac{3ac}{2a^2+2c^2}=\arctan\frac{\frac{3ac}{a^2}}{\frac{2a^2}{a^2}+\frac{2c^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.... | We can drop the factor of $\frac{1}{2}$, so we have $\sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}$ and $a+b$. Squaring gives $ab + \frac{a^2+b^2}{2}+\sqrt{ab\frac{a^2+b^2}{2}}$ and $a^2+b^2+ab$. They both have a factor $ab$ and $\frac{a^2+b^2}{2}$, so we have to compare $\sqrt{ab\frac{a^2+b^2}{2}}$ and $\frac{a^2+b^2}{2}$. Rewri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding $P$ such that $P^TAP$ is a diagonal matrix
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_2(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
So here's the solution:
$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{ar... | When you parameterize conics with rotation, its usual make the variable change $\mathbf {X=P^TY}$ so that $P^T\mathbf{A}P=D$ where $D$ is a diagonal matrix and $\mathbf{A}$ is the matrix associate to the conics. The goal is eliminate cross terms $Bxy$.
The change of variable $\mathbf {X=P^TY}$ is deduced by compl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Matrix representation with non-standard bases. In chaprer 2.2 of Fiedberg's Linear Algebra is wroten about matrix representation. But all examples are only with standard ordered bases. I made a task to understand it. Please, could you show me matrix representation on this task or your own example?
Let $$ T\begin{pmatr... | Problem: Let
$$ T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+y\\x-y\end{pmatrix}$$
is transformation from vector space V with a basis
$$ v= \{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}3\\4\end{pmatrix}\} $$ to vector space W with a basis
$$ w= \{ \begin{pmatrix}1\\4\end{pmatrix}, \begin{pmatrix}2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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.