Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
trying to solve $\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$ The equation is
$$\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$$
The system is
$$
\begin{cases}
\cos(x)-2\cos(2x)=2\cos^2(2x) \\
-\sqrt{2}\cos(2x)\ge 0 \iff \cos(2x)\le 0
\end{cases}
$$
The equation:
$$\cos(x)-2(2\cos^2(x)-1)=2(2\cos^2(x)-1)^2$$
$$\cos(x)-4\c... | HINT:
$$8\cos^3x-4\cos x-1=4\cos x(2\cos^2x-1)$$
$$\implies2(\cos3x+\cos x)=1\iff4\cos x\cos2x=1$$
If $\sin x=0,4\cos x\cos2x\ne1\implies\sin x\ne0$
$$4\cos x\cos2x=1\iff\sin x=2(2\sin x\cos x)\cos2x=2\sin2x\cos2x=\sin4x$$
$$\sin4x=\sin x\implies4x=n\pi+(-1)^nx$$ where $n$ is any integer
Also, $\cos2x\le0\implies2m\pi+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simultaneous orthogonal diagonalization of two matrices Let $A=\begin{pmatrix}
1 & -2\\
-2 & 5
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 & 6\\
6 & -10
\end{pmatrix}$. Obviously $A$ is positive-definite and thus we can simultaneously diagonalize $A$ and $B$. It's easy to check that $P=\begin{pmatrix}
1 & 2\\
0 & 1
\en... | Simultaneous congruence via an orthogonal matrix implies in particular simultaneous orthogonal diagonalization and in particular, simultaneous diagonalization. Assume that there exists an orthogonal matrix $Q$ such that $Q^tAQ = \mathrm{diag}(\lambda_1, \lambda_2)$ and $Q^tBQ = \mathrm{diag}(\mu_1, \mu_2)$. Since $Q^t ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the definite integral of a trigonometric expression Find the integral of $$ \int_0^{\frac{\pi}{2}}{{\sqrt{\sin(2\theta)}} \cdot \sin(\theta)d\theta}$$
I got $$I=\int_0^\frac{\pi}{4}{\sqrt{\sin(2\theta)} \cdot (\sin(\theta)+\cos(\theta))d\theta}$$
But, I'm stuck here.
| The integral
$$I = \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta$$
is evaluated by making use of the Beta function. This is seen as follows.
\begin{align}
I &= \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta \\
&= \sqrt{2} \, \int_{0}^{\pi/2} \sin^{3/2}(\theta) \, \cos^{1/2}(\theta) \, d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Simple Trig Equations - Why is it Wrong to Cancel Trig Terms? In the following problem, I first did it using a cancellation of $sin^2\theta$, working shown below, which gave the wrong answer. Having looked at the question again, I saw it could be solved by factoring, working again below.
My question then, is why is it ... | Here's a simpler way to get your answer. Remember what $sin(\theta)$ and $tan(\theta)$ actually represent on a right triangle. For a given right triangle, let $a$ be the length of the adjacent side, $b$ be the length of the opposite side, and $c$ be the length of the hypotenuse. This means:
$sin(\theta) = \frac{a}{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmat... | here is another way to find $e^A.$ we will use the interpretation that $x = e^{At}x_0$ is the unique solution of $$\frac{dx}{dt} = Ax, x(0) = x_0.$$
for $$\frac{d}{dt}\pmatrix{x\\y} = \pmatrix{0&1\\-4&0}\pmatrix{x\\y}$$ which in component form is $$\dot x = y, \dot y = -4x \to \ddot x=-4x, \ddot y= -4y \text{ and } x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,coul... | Observe that the equality fails if one of the $\cos$ is negative. We can assume that triangle is acute. By Jensen and followed by AM-GM inequality:
$x = \cos A, y = \cos B, z = \cos C\to x+y+z \leq \dfrac{3}{2}$:
$1 = 2xyz + xy+yz+zx \leq 2\cdot \dfrac{(x+y+z)^3}{27}+\dfrac{(x+y+z)^2}{3}\leq \dfrac{2}{27}\cdot \left(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculation of improper integral: $\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$ One of an exam's task was to calculate the following integral: $$I=\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$$
I tried integration by parts:
$$I=\frac{1}{4}\int_{0}^{\infty}\ln x \cdot (x^4+1)^{-3} \,d(x^4+1)$$
but then things g... | $$I = \int\frac{x^3\ln(x)}{(x^4+1)^3}dx=\frac{\ln x}4\int \frac{4x^3}{(x^4+1)^3}dx-\int\left(\frac1{4x}\int \frac{4x^3}{(x^4+1)^3}dx\right)dx $$
$$=-\frac{\ln x}{8(x^4+1)^2}+\frac 1{32}\int \frac{4x^3}{x^4(x^4+1)^2}dx$$
$$=-\frac{\ln x}{8(x^4+1)^2}+\frac1{32}\ln\left(\frac{x^4}{1+x^4}\right)+\frac 1{32(1+x^4)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove by Induction: $8^n - 3^n$ is divisible by $5$ for all $n \geq 1$ Prove by Induction that for all $n \geq 1$ we have
$$8^n - 3^n \text{is divisible by 5} ...(*)$$
My proof so far
Step 1: For $n=1$ we have $8^1 - 3^1 = 8 - 3 = 5$ which is divisible by 5.
Step 2: Suppose (*) is true for some $n=k\geq 1$ that is... | Hint: note that $8=5+3$ and then factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x<2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}<2$$
$$ 4t^2-2t>0$$
$$ 2t(2t-1)>0$$
$$ t(2t-1)>0$$
$$ (t>0 \wedge t>\dfrac{1}{2}) \vee (t<0 \wedge t<\dfrac{1}{2})$$
From this, I can only find $x<2\pi+2k\pi$, and, $x<2k\pi$, these are good (I think), ... | First, we solve $\sin x + 2 \cos x = 2$
Let $\theta$ be the angle corresponding to the point $(x,y) = (2,1)$ with amplitude $r = \sqrt 5.$ Then
$\cos \theta = \dfrac{2}{\sqrt 5}$ and
$\sin \theta = \dfrac{1}{\sqrt 5}$
Then
\begin{align}
\dfrac{2}{\sqrt 5}\cos x + \dfrac{1}{\sqrt 5} \sin x &= \dfrac{2}{\sqrt 5}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How to get formula for sums of powers? Assuming I have Bernoulli numbers:
$B = [\frac{1}{1},\frac{1}{2},\frac{1}{6},\frac{0}{1},-\frac{1}{30}, \frac{0}{1}, \frac{1}{42}, ...]$
How can I get the coefficients of the sums of powers formulas?
For example the sum of squares is $(1/3)n^3 + (1/2)n^2 + 1/6n$
| What you're looking for is Faulhaber's formula (Wikipedia link):
$$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p (-1)^j{p+1 \choose j} B_j n^{p+1-j},\qquad \mbox{where}~B_1 = -\frac{1}{2}$$
expressing the sum of the first $n$ $p$th powers as a polynomial in $n$ whose coefficients involve the Bernoulli numbers and some ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How to show that $\lim_{x\to 1}(x^3-1)/(x-1)=3$? How to show that
$$
\lim_{x\to 1}\frac{x^3-1}{x-1}=3?
$$
I tried to solve but couldn't. Please help me.
| Let $x = y+1$,
so we want to see what happens as $y \to 0$.
Note:
I almost always want to have
a variable that controls a limit
go to zero
instead of some other value
(like $1$ in this case).
\begin{align}
\frac{x^3-1}{x-1}
&=\frac{(y+1)^3-1}{(y+1)-1}\\
&=\frac{(y^3+3y^2+3y+1)-1}{y}\\
&=\frac{y^3+3y^2+3y}{y}\\
&=y^2+3y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Turning 500, 1100, 1800, 2600, 3500, etc into 1, 2, 3, 4, 5, etc I would love to have a formula of the form $n = F(m)$ for this progression ...
n m
---------
1 500
2 1100
3 1800
4 2600
5 3500
...
where the second-differences in $m$ are constant, on up to $n = 100$.
| Divide the elements by $100$ to get the sequence $5,11,18,26,35,...$, which is
1 5
2 5+6
3 5+6+7
4 5+6+7+8
...
which is
1 1+2+3+4+5 - 10
2 1+2+3+4+5+6 - 10
3 1+2+3+4+5+6+7 - 10
...
But these sums are triangle numbers, so the terms are given by $\frac{(n+4)(n+5)}{2} -10$. Now multiply by $100$ to recover ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
... | Given $\displaystyle \lim_{x\rightarrow 1}\frac{1}{2}\left[\frac{x^{\frac{1}{3}}-1}{x^{\frac{1}{2}}-1}\right] = \frac{1}{2}\times \lim_{x\rightarrow 1}\left[\left(\frac{x^{\frac{1}{3}}-1}{x-1}\right)\times \left(\frac{x-1}{x^{\frac{1}{2}}-1}\right)\right]$
Now Using The formula $\displaystyle \lim_{y\rightarrow 1}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| Hint: $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)=18.$$
Let $$x+\frac{1}{x}=t.$$ Then your equation reduces to $(t(t^2-3))=18$. Solve this cubic and you get $x+\frac{1}{x}.$
Now I think you can carry on after that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore,
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any... | Hint:
Notice
\begin{align}
7+5\sqrt{2}&=2\sqrt{2}+3(2)(1)+3(\sqrt{2})(1)+1\\
&=(\sqrt{2})^3+3(\sqrt{2})^2(1)+3(\sqrt{2})(1)^2+(1)^3
\end{align}
Can you recognize this pattern?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Help factorising a sixth degree polynomial I have to factorise- $$x^6+5x^3+8$$Answer is $$(x^2−x+2)(x^4+x^3−x^2+2x+4)$$.I have also used factor theorem.Please help me.Thanks in advance.
| Break the equation $x^6+5x^3+8$ into $x^6+8-x^3+6x^3$. It then comes into the form $a^3+b^3+c^3-3abc$. Factorise it using the formula $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$? If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$?
At the end when you get $4ax=0$, can I divide by $4x$ to cancel out $4$ and $x$?
| $$(x-a)^2 = (x+a)^2$$
Expand squares:
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
Subtract $x^2+a^2$ from both sides:
$$-2ax = 2ax$$
add $2ax$ and divide by $4$:
$$0 = ax$$
So either 1) $a = 0$ or 2) $x = 0$.
*
*$$(x-0)^2 = (x+0)^2 \Leftrightarrow x^2 = x^2$$
Which is true for all x.
*$$(0-a)^2 = (0+a)^2 \Leftrightarrow ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
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Prove that $a^2+ab+b^2\ge 0$ How to prove that $a^2+ab+b^2\ge 0$?
Obviously the squares are positive, but how can I be sure that $ab$ doesn't become too negative with a certain combination of $a$ and $b$?
| There is yet another of seeing this.
Since $ab$ is between $-2ab$ and $2ab$, the quantity $a^2+ab+b^2$ must be between $a^2-2ab+b^2$ and $a^2+2ab+b^2$. But these two quantities obey
$$a^2-2ab+b^2=(a-b)^2\ge0 \tag{1}$$
and
$$a^2+2ab+b^2=(a+b)^2\ge0 \tag{2}$$
with equality only if $a=b$ for (1), $a=-b$ for (2),or $a=b=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
| Hint: notice that
$$y(x) = \frac{1+x}{1-x}\cdot\frac{1-x^3}{1+x^3}\tag{1}$$
and that:
$$ \frac{1+x}{1-x}=1+2x+2x^2+2x^3+\ldots,\tag{2}$$
from which:
$$ \frac{1-x^3}{1+x^3}=1-2x^3+2x^6-2x^9+2x^{12}-\ldots\tag{3}$$
Are you able to multiply $(2)$ and $(3)$ and prove that:
$$ y(x) = 1 + \left(2x+2x^2\right) - \left(2x^4+2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Limit of $f(n)^{g(n)}$ Given the limit:
$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n}$$
Is it possible to assume that
$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n} = L$$
and then take the natural log of both sides
$$\lim_ {n\to ∞}\left(n \cdot \ln\left(\frac{n^2+5}{3n^2+1}\right)\right) ... | First, consider the function
$$\mathrm{f}(x) = \frac{x^2+5}{3x^2+1}$$
The derivative is given by using the quotient rule:
\begin{eqnarray*}
\mathrm{f}'(x) &=& \frac{2x(3x^2+1)-(x^2+5)(6x)}{(3x^2+1)^2} \\ \\
&=& \frac{-28}{(3x+1)^2}
\end{eqnarray*}
This tells us that $\mathrm{f}'(x) < 0$ for all $x \ge 0$, i.e. $\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Integration by partial-fractions, I´m stuck in this one. I don´t really know how get the factors in the denominator which allow me to use a case
$\int\frac{x^2+1}{x^2-x} dx$
| $$\frac{x^2+1}{x^2-x}=\frac{x^2}{x^2-x}+\frac{1}{x^2-x}$$
$$=\frac{x^2}{x(x-1)}+\frac{1}{x(x-1)}$$
$$=\frac{x}{(x-1)}+\frac{1}{x(x-1)}$$
$$=\frac{x-1+1}{(x-1)}+\frac{1}{x(x-1)}$$
$$=1+\frac{1}{(x-1)}+\frac{1}{x(x-1)}$$
now let
$$\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$
equate the two sides to find
$$A=-1$$
$$B=1$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$ I'm trying to prove the following equation above. So far I have:
\begin{align}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} & \text{(By the Binomial Theorem)}
\end{align}
I know I have to use the following identity somehow... | I would suggest to use induction:
For $n=1$:
$2^2 = \binom{3}{0}+\binom{3}{1}$, true.
For $m=n+1$:
$2^{2n+2}=4\cdot 2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k} + \binom{2n+2}{n+1} = 2^{2n} + \binom{2n+2}{n+1} = 2^{2n} + \frac{(2n+2)!}{(n+1)!^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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How would i go about solving this exponential equation? $$2^{ x-2 }+\frac { 1 }{ \sqrt { { 4 }^{ x+2 } } } =\frac { 17 }{ 16 } $$
I can't seem to find any direction to go that will lead me to a solution. I decided to try to take the natural log all terms in the equation to see where that would lead me. Please do not g... | There's no need to use logarithms at all here-just indices will do.
Note that: $$\frac{1}{\sqrt{4^{x+2}}}= \frac{1}{2^{x+2}} = \frac{1}{4(2^x)}$$
$$ 2^{x-2} + \frac{1}{2^{x+2}} = \frac{17}{16} $$
$$ \frac{2^x}{4}+ \frac{1}{2^x(4)}= \frac{17}{16}$$
Let $2^x = a$
$$\frac{a}{4}+ \frac{1}{4a}= \frac{17}{16}$$
$$ 4a^2 -17a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
a conjectured continued-fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ that leads to a new limit for $\pi$ Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\cot\... | O. Perron, Die Lehre von den Kettenbrüchen, Chapter XI, Section 82.
Satz 5 on page 488 (2nd edition, 1927):
Continued fraction
$$
d+\underset{k=1}{\overset{\infty }{\mathbf K}}\;
\frac{a+bk+ck^2}{d+ek}
$$
with $c \ne 0, e \ne 0, e^2+4c \ne 0$ has value
$$
\frac{\sqrt{e^2+4c}\;{}_2F_1(\alpha,\beta;\gamma;x)\;\gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
iterated sine function on different arguments I want to evaluate the following:
$\lim_{n\rightarrow \infty} \sqrt{n} \sin^{(n)}(2/\sqrt{n})$, where $\sin^{(n)}$ is the iterated sine function.
I do know the proof for $\lim_{n\rightarrow \infty} \sqrt{n} \sin^{(n)}(x_0) =\sqrt{3}$ for any non trivial $x_0 \neq k\pi$ (by ... |
Claim. For any $x \in \Bbb{R}$ the limit is given by
$$ \lim_{n\to\infty} \sqrt{n} \sin^{\circ n}\left( \frac{x}{\sqrt{n}} \right) = \frac{\sqrt{3}x}{\sqrt{x^2+3}}. $$
Proof. To prove this, WLOG we assume that $x > 0$. Let $f_n(x) = \sqrt{n}\sin(x/\sqrt{n})$ and define
$$ x_{n;0} = x, \qquad x_{n;k+1} = f_n(x_{n;k... | {
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"source": "stackexchange",
"question_score": "2",
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How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3... | this limit doesn't exist since $$\frac{1}{x-1}-\frac{2}{x^3-1}=\frac{x^2+x-1}{(x-1)(x^2+x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding the next limit How to calculate $\displaystyle\lim_{(x,y)\rightarrow(a,a)}\frac{(x-y)a^n+(a-x)y^n-(a-y)x^n}{(x-y)(a-x)(a-y)}$ where $a\neq 0?$
I have the following idea:the factors on the denominator are roots of the expression on numerator because nullify it. So, the polinomyal has these three roots and then,... | In order for the limit to exist it must have the same value along all paths that approach the point $(a,a)$. We will find two paths that disagree for the limit in question.
So first consider the limit along the parametric curve (a line) $x=a-h,y=a+h$, and thus let $h\to0$. We have:
*
*$x-y=-2h$
*$a-y=-h$
*$a-x=h$... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Is -2 is a root of the equation : $\sqrt {x^2 - 8} = \sqrt {3x + 2}$? Is -2 is a root of the equation : $\sqrt{x^2 - 8} = \sqrt{3x + 2}$ ?
Is it any limitation of the root of this equation?
If the root is -2, both sides of the equation are equal to 2i.
Is this acceptable since the variable x has no limitation.
| Notice, the given equation is $\sqrt{x^2-8}=\sqrt{3x+2}$, now squaring both the sides we get $$x^2-8=3x+2$$ $$ x^2-3x-10=0$$ $$(x+2)(x-5)=0$$ On solving we get $$x=-2, \ 5$$
Now, let's check out both the roots, if they are acceptable by satisfying the given (original) equation
Substituting $x=-2$, we get $LHS=RHS=\sq... | {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form
$$\int_{1/2}^... | It can be easily checked by differentiation that:
$${\large\int}\,\frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx=\\
\frac i4\left[\vphantom{\Large|}\operatorname{Li}_2\left(\left(-\tfrac12+\tfrac i6\right)(x-1)\right)-\operatorname{Li}_2\left(\left(-\tfrac12-\tfrac i6\right)(x-1)\right)\\
\,+\operato... | {
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"url": "https://math.stackexchange.com/questions/1453345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Proving tan((x + y)/2) = (sin x + sin y)/(cos x + cos y) with the angle sum and difference identities I've recently come across this problem of proving
$$ \tan \frac{x + y}{2} = \frac{\sin x + \sin y}{\cos x + \cos y} $$
Not a difficult problem, I thought. I would have rewritten the RHS using the sum-to-product identit... | \begin{align}
\frac{\sin x + \sin y}{\cos x + \cos y}
&= \frac{\sin\left(\frac{x+y}{2} + \frac{x-y}{2}\right)
+ \sin\left(\frac{x+y}{2} - \frac{x-y}{2}\right)}
{\cos\left(\frac{x+y}{2} + \frac{x-y}{2}\right)
+ \cos\left(\frac{x+y}{2} - \frac{x-y}{2}\right)}\\
&= \frac{2\sin\left(... | {
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"url": "https://math.stackexchange.com/questions/1453588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
inverse of $\arcsin (\frac{x}{x-1})$ determine the inverse of
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
b) $y= \dfrac{1-2e^{-x}}{4}$
I learned you the steps for finding the inverse are
1) get it in a form of $x= \dots$
2) change $x$ and $y$
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
is $y=\arcsin \left(\dfrac{x}{x-1}... | $\sin y(x-1)=x\Rightarrow \sin y x-x=siny \Rightarrow x(\sin y -1)=\sin y$ so $$x=\frac{\sin y}{\sin y -1}\Rightarrow y^{-1}=\frac{\sin x}{\sin x -1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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isomorphic equations W. Ross Ashby states in his book "An Introduction to Cybernetics" that the system:
$\ x' = 1/2(x^2+y^2) + x*y + y$
$\ y' = 1/2(x^2+y^2) + x*y + x$
is isomorphic to the system:
$\ u' = -u$
$\ v' = v + v^2$
under the transformation:
$\ u = x - y$
$\ v = x + y$
can someone explain this properly?... | Note that you can express $x$ and $y$ in terms of $u$ and $v$:
$$x=\frac{u+v}2\\y=\frac{-u+v}2$$
So, from the first equation of your system,
$$x' = \frac12(x^2+y^2) + xy + y,$$
we have
$$\left(\frac{u+v}2\right)' = \frac12\left(\left(\frac{u+v}2\right)^2+\left(\frac{-u+v}2\right)^2\right) + \left(\frac{u+v}2\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is the following limit? $\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$ What is the following limit?
$$\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$$ Should I calculate the exact value of $(1+x)^5$?
| It suffices to know the coefficient of $x^2$ in the numerator.
Note that $(1+x)^5-1-5x=\binom{5}{2}x^2+\mathcal{O}(x^3)$ and $x^2+x^5=x^2+\mathcal{O}(x^3)$ and hence the limit will be
$$\lim_{x \to 0} \frac{10x^2+\mathcal{O}(x^3)}{x^2+\mathcal{O}(x^3)}=10$$
Alternatively, if you don't want to use these symbols, just ca... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve absolute value inequality I have to show the inequality
$$
\left|\frac{1}{2 + a}\right| < 1.
$$
How do I do this?
I know that a fraction is less than 1 when the denominator is greater than the numerator, but I cannot just check if $2 + a > 1$ because of the absolute value sign.
Edit
If I use
$$
\left|\frac{1}{2 +... | Both sides are positive, so you can take their reciprocals (of course the 'less than' flips to 'greater than'):
$$\left|\frac 1{2+a}\right| < 1 \iff \frac 1{\left|\frac 1{2+a}\right|} = \frac {|2+a|}{|1|}= |2+a| > 1$$
That is equivalent to an alternative:
$$(2+a) < -1 \lor (2+a) > 1$$
which resolves to:
$$a < -3 \lor a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving a property of a prime number (Elementary number theory) Prove the following statement.
For all prime numbers $a$, $b$, and $c$, $a
^2 + b^2
\neq c^2$
.
What i tried
Proving by contradiction
Assume the negation of the statement
There exists prime numbers $a$, $b$, and $c$ such that $a
^2 + b^2
= c^2$
Since we kn... | We cannot find primes $a,b,c$ such that $a^{2} + b^{2} = c^{2}$ because of an elementary classical result: Take any integers $a,b,c$ such that $(a,b) = 1$; then $a^{2} + b^{2} = c^{2}$ if and only if $a = u^{2} - v^{2}$, $b = 2uv$, $c = u^{2} + v^{2}$ for some integers $u, v$ such that $(u,v) = 1$.
| {
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"url": "https://math.stackexchange.com/questions/1461241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule?
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$
Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is... | Set $x-1=u$ and rewrite the expression:
*
*$2\cos(x-1)-2=2\cos u-2=-u^2+o(u^2)$,
*$\begin{aligned}[t]x^2-2\sqrt x+1&=(1+u)^2-2\sqrt{1+u}+1=1+2u+u^2-2-u+\dfrac{u^2}4+o(u^2)+1\\&=u+\dfrac{5u^2}4+o(u^2)
\end{aligned}.$
Thus $2\cos(x-1)-2\sim_1 -(x-1)^2$, $\;x^2-2\sqrt x+1\sim_1 x-1$, and
$$\dfrac{2\cos(x-1)-2}{x^2-2... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
| This sum may be tackled using the residue theorem.
$$S=\sum_{k=0}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right ) = \frac12 \sum_{k=-\infty}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right )$$
Thus, consider the contour integral
$$\oint_{C_N} dz \frac{2 \pi \, \cot{\pi z}}{36 \left (z+\frac16\right ) \left... | {
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"source": "stackexchange",
"question_score": "17",
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Difficult integral evaluation I'm working through Vector Calculus by Marsden and Tromba to review for my GRE (and because it has really interesting historical snippets) and I ran into a wall on a problem where I have to evaluate the arc length of the vector valued function $<t, tsint, tcost>$ for $0 \leq t \leq \pi$. H... | You can use integration by parts, or you can use the hyperbolic substitution $t = \sqrt{2}\sinh(u)$. I'll use the latter.
With $t = \sqrt{2}\sinh(u)$, we have $\sqrt{2 + t^2} = \sqrt{2}\cosh(u)$ and $dt = \sqrt{2}\cosh(u)\, du$, so
$$\int_0^\pi \sqrt{2+t^2}\, dt = \int_0^{\sin^{-1}(\pi/\sqrt{2})} 2\cosh^2(u)\, du.$$
U... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
| $$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\iff x^3=-1$$
$$x^{2000}=(x^3)^{666}\cdot x^2=(-1)^{666}\cdot x^2=x^2$$
Finally use $a^2+b^2=(a+b)^2-2ab$ for $x^2+\dfrac1{x^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What... | First let's figure out what $C$ is. Letting $x=0$, we see that $\arctan(-1) = C$ (i.e. $C = -\frac{\pi}{4}$ or $C = \frac{3\pi}{4}$). Since inverse trigonometric functions are hard to work with, let's take $\tan$ of both sides. Doing so we get
$$\tan\left(\arctan(x-\sqrt{1+x^2})\right) = \tan\left(\frac{1}{2}\arctan(x)... | {
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"source": "stackexchange",
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I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red... | The "usual" proof that $\Bbb Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+ \sqrt{3})$ ), the other inclusion is obvious.
It suffices to show that $\sqrt{2} \in \Bbb Q(\sqrt{2}+ \sqrt{3})$, since then $\sqrt{3} = (\sqrt{2} + \sqrt{3}) - \sqrt{2}$ must be as well.
Now $(\sqrt{2} + \sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$ A straight line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$,then square of th... | Broad Steps
(1) We have the dr's of the two given lines
(2) Let P & Q be points on the two lines so that PQ is the shortest distance.
Find equation of L through PQ
(3) Find distance of L from origin.
Details
Note that $\color{blue}{(2,3,-10)}$ and $\color{blue}{(4,-3,-2)}$ are direction ratios of the 2 given lines.
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$? Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$
Attempt:
intersect when:-
$x^3 - 4x = 0$
$x ( x² - 4 ) = 0 $
$x = 0 , x = \pm 2$
Area is given by :-
$$2 ∫_0^2 4x - x^3\; ... | The real problem is that $\displaystyle \int x^3 = \frac{x^4}{4}$, not $\displaystyle \frac{x^3}{3}$.
$$\int_0^2 4x-x^3$$
$$= \left[4 \cdot \frac{x^2}{2} - \frac{x^\color{red}{4}}{\color{red}{4}} \right]_0^2$$
$$= \left[ 2x^2 - \frac{x^4}{4} \right]_0^2$$
$$= \left(2\cdot 2^2 - \frac{2^4}{4} \right) - \left(2\cdot 0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473083",
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"source": "stackexchange",
"question_score": "1",
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Let $A=\left\{a\in R:\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\right\}$ has atleast one real root. Let $A=\left\{a\in R:\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\right\}$ has atleast one real root.Find the value of $\sum_{a\in A}a^2$.
What should i do in this question to find the possible value... | $2 a^2+x^3-6 x^2+i (2 x^3-2 x^2-4 x)+5 x = 0$
root of $2 x^3-2 x^2-4 x = 0$ are $-1, 0$ and $2$
$2 a^2+(-1)^3-6 (-1)^2+5 (-1) = 0$ imply $2a^2-12=0$ ie $a=\pm\sqrt{6}$
$2 a^2+(0)^3-6 (0)^2+5 (0) = 0$ imply $2a^2=0$ ie $a^2=0$
$2 a^2+(2)^3-6 (2)^2+5 (2) = 0$ imply $2a^2-6=0$ ie $a=\pm\sqrt{3}$
therefore $\sum a^2=18$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Maximizing $3 x^2+2 \sqrt{2} x y$ with $x^4+y^4=1$ I want to have maxizing value of $3 x^2+2 \sqrt{2} x y$ when $x^4+y^4=1$, $x>0,y>0$.
How can I solve it.
| I have got solution with precaculus method. thank you everyone advices.
$3 x^2+2 \sqrt{2} x y$
$\leq$ $3 x^2+\left(x^2+2 y^2\right)$
$=$ $4 x^2+2 y^2$ (when equal is $x=\sqrt{2} y$ with arithmetic geometric mean inequality )
$=$ $2\cdot \left(2 x^2\right)+1\cdot \left(2
y^2\right)$
$\leq$ $\sqrt{\left(2^2+1^2\r... | {
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"url": "https://math.stackexchange.com/questions/1474767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Domain of functions that involve composition Suppose the domain of $f$ is $(-1,1)$. Define the function $\ell$ by
$$\ell(x)=f\left(\frac{x+1}{x-1}\right).$$ What is the domain of l
| As pointed in the comments, you want to solve: $$-1< \frac{x+1}{x-1} < 1$$for $x$. Rewrite $\frac{x+1}{x-1} = 1 + \frac{2}{x-1}$, so that: $$-1< \frac{x+1}{x-1} < 1 \implies -1 < 1 + \frac{2}{x-1} < 1 \implies -1 < \frac{1}{x-1} < 0.$$Can you solve that last one now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\r... | Somehow, you missed a square.
$$
(\sqrt{4x^2-5x})^2-(2x)^2=4x^2-5x-4x^2=-5x
$$
which radically simplifies the ensuing limit calculation,
$$
\lim_{x\to-\infty}\frac{5|x|}{\sqrt{4x^2+5|x|}+2|x|}
=\lim_{x\to-\infty}\frac{5}{\sqrt{4+5/|x|}+2}=\frac54
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^... | Another approach uses Lagrangian multipliers. Our Lagrangian is $L=2\ln x+5\ln y+\lambda (35-x-y)$ so $0=\partial_x L = \frac{2}{x}-\lambda,\,0=\partial_y L=\frac{5}{y}-\lambda$ and $x=10,\,y=25$. This maximises $x^2y^5$ because the matrix of second derivatives is diagonal, with negative eigenvalues $-\frac{2}{x^2},\,-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Show that $\lim_{n \longrightarrow \infty} \frac{3n^2+2n}{4+3n^2}=1$ I need show that $\forall \, \epsilon > 0$, exist $M \in \mathbb N$ such that, $\left|\frac{3n^2+2n}{4+3n^2}-1\right| < \epsilon$ if $n \geq M$.
Let's consider
$\left|\frac{3n^2+2n}{4+3n^2}-1\right|=\left|\frac{2n-4}{4+3n^2}\right|<\left|\frac{2n-4}{3... | $$\lim_{n \to \infty} \frac{3n^2 + 2n}{3n^2 + 2} = \lim_{n \to \infty} \frac{3 + 2/n}{3+2/n^2} = \frac{3}{3} = 1$$
$\epsilon$-proof:
$$ \left|\frac{3n^2 + 2n}{3n^2 + 2} - 1\right| = |\frac{2n - 2}{3n^2 + 2}| < \epsilon \Rightarrow |2n-2| < 3\epsilon n^2 + 2\epsilon$$
The absolute value can be removed since both sides a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} $ $$\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} $$
This question was asked today in my maths exam,
It was one of those two questions which I couldn't answer, How do you go about answering it ?
| $$\begin{align}\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} dx
&=
\int \frac{4\tan(x)+5}{\cos^2(x)(\tan (x)+1) (\tan (x)+2)}dx\\
&=
\int \frac{4u+5}{(u+1) (u+2)}du
\end{align}$$
The last part is by using $u=\tan x$, $du = \frac{dx}{\cos^2 x}$.
I hope you know to continue from here.
Note:
$$
\sin^2(x)+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$ I have to show that the series
$\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$
converges.
I have tried Ratio Test and Cauchy Condensation Test but it didn't work for me. I tried using Comparison Test but I couldn't mak... | Since $\log (1 + x) < x$ for non-negative $x$, we have
$$\begin {eqnarray}
0 < \sum_{n = 1}^{\infty} \log \left (\frac {(2n)^2} {(2n - 1) (2n + 1)} \right) & = & \sum_{n = 1}^{\infty} \log \left (1 + \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & < & \sum_{n = 1}^{\infty} \left ( \frac {1} {(2n - 1) (2n + 1)} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to prove this $\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$ Let $x,y>0$, and $x+y=2$, show that
$$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$$
I tried using Minkowski inequality
$$\sqrt{x^2+3}+\sqrt{y^2+3}\ge\sqrt{(x+y)^2+(\sqrt{3}+\sqrt{3})^2}=\sqrt{4+12}=4$$
But
$$\sqrt{xy+3}\le\sqrt{\dfrac{(x+y)^2}{4}+3}=2$$
so i... | To prove that
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$,
We use Jensen's inequality.
$1*\sqrt{x^2+3}+1*\sqrt{y^2+3}+1*\sqrt{xy+3}\ge (1+1+1) \sqrt{\frac{x^2+y^2+xy+9}
{1+1+1}}$
Then, since $x+y=2$, we have:
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2+(2-x)^2+x(2-x)+9}{3}}$
$\sqrt{x^2+3}+\sqrt{y^2+3}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the sum from the system of equations
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x +3y + 4z$.
This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x... | You can add up the three equations to get
$2x+2y+2z = x^2+y^2+z^2 +3
= (x^2+1)+ (y^2+1)+ (z^2+1)\geq 2x+2y+2z$
Hence, we know that $x=y=z=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
proving montonity and convergence of sequence en = (1 + 1/n)^n Prove the following. What would be the summation formula be for the first part?
| From the Binomial Theorem, we have
$$\begin{align}
e_n&=\left(1+\frac1n\right)^n\\\\
&=\sum_{k=0}^n\binom{n}{k}\left(\frac1n\right)^k\\\\
&=1+n\left(\frac1n\right)+\frac{n(n-1)}{2!}\left(\frac1n\right)^2+\cdots+\frac{(n)(n-1)(n-2)\cdots(n-(n-1))}{n!}\left(\frac1n\right)^n\\\\
&=1+1+\frac1{2!}\left(1-\frac1n\right)+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Logarithmic inequality and properties of logarithms So the inequality:
$$2 \cdot \log_{\sqrt3}{(1-x)} - \log_\sqrt3 {(3-x)} \lt 2$$
Can be written as:
$$\log_{\sqrt3}{(1-x)^2} - \log_\sqrt3 {(3-x)} \lt 2$$
?????????????
I have tried both on Wolfram|Alpha and both gave different intervals of x, although the logarithm pr... | Note that throughout, $x\neq 1$ and $x\neq 3$ applies.
$$\begin{align}
2\log_\sqrt{3}(1-x)-\log_\sqrt{3}(3-x)&<2\\
\log_\sqrt{3}\frac{(1-x)^2}{3-x}&<\log_\sqrt{3}3\\
\frac {(1-x)^2}{3-x}&<3\\
\frac {(1-x)^2}{3-x}-3&<0\\
\frac {x^2+x-8}{3-x}&<0\\
\frac {x^2+x-8}{x-3}&>0\\
\frac {(x-\alpha)(x-\beta)}{x-3}&>0\\
\text{whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\lim... | Notice, in the third step of induction, you should follow the right procedure
substituting $n=m+1$ in the equality, we get
$$\sum_{k=1}^{m+1}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
$$\sum_{k=1}^{m}\sin(kx)+\sin(m+1)x=\frac{\sin\left(\frac{(m+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
question related with half angle from a given equation If $\sec x + \tan x = 3$ , then what is the value of $\tan \frac x2?$
i squared both sides and then converted the $\tan x $ into $\sec x$ and then $\sec x $ into $\cos x$ which gave me value $\sqrt\frac15$. Kinda confused after this step.. help is appreciated
| HINT:
$$\sec(x)+\tan(x)=3\Longleftrightarrow$$
$$-3+\sec(x)+\tan(x)=0\Longleftrightarrow$$
Substitute $y=\tan\left(\frac{x}{2}\right)$. Than $\sin(x)=\frac{2y}{y^2+1}$ and $\cos(x)=\frac{1-y^2}{y^2+1}$:
$$\frac{2}{y-1}-\frac{4y}{y-1}=0\Longleftrightarrow$$
$$-\frac{2(2y-1)}{y-1}=0\Longleftrightarrow$$
$$\frac{2y-1}{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$2^\text{nd}$ Derivative of normal distribution, evaluated at one standard deviation What is the $2^{nd}$ derivative of the normal distribution at one standard deviation?
The normal distribution is given by $N(x,\mu ,\sigma)=\frac{1}{\sigma\sqrt{2\pi }}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$. To make this problem easier, let... | Your way of applying the chain rule is wrong. Here's the right way:
$$
\frac d {dx} e^{-x^2/2} = e^{-x^2/2} \cdot \frac d {dx} \left( \frac{-x^2} 2 \right) = e^{-x^2/2} \cdot(-x).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show trigonometric identity $\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}} = \sec\theta - \tan\theta$ I need to show that $\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}} = \sec\theta - \tan\theta$
So far, I have changed the fraction to (where $t=\tan\frac{\theta}{2}$), $\frac{1-2t+t^2}{1-t^2}$(*)
I know ... | Notice, $$\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}=\frac{\left(1-\tan\frac{\theta}{2}\right)\left(1+\tan\frac{\theta}{2}\right)}{\left(1+\tan\frac{\theta}{2}\right)^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}+2\tan\frac{\theta}{2}}$$
$$=\frac{1}{\frac{1+\tan^2\frac{\theta}{2}}{1-\tan^2\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^... | I assume that you have checked that $(x,y) = (-2,-5)$ is indeed a solution.
Why is it the only one?
It is, because you have shown that your three equations must imply that $(x+2)^2=0$, which in turn implies $x=-2$. In other words, whatever solutions $(x,y)$ the three original equations have, you have shown that they mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Why test of divisible by $12$ works with $3$ and $4$ but not with $2$ and $6 ?$ Test of divisible by $4 ,$ last two digit must be divisible by $4 ,$ since $100$ is always divisible by $4$ remaining two digit $,$ we need to check $.$
Test of divisible by $3 ,$ sum digits must be divisible by $3 .$ But I don't know why $... | The reason the divisibility by $3$ test works is that $10$ and $1$ have the same remainder when divided by $3$. Say your number is (as digits) $x = a_n a_{n-1} \ldots a_1 a_0$. Then you could write it as $x = a_n \cdot 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10 + a_0$. If you mod out by $3$:
$$
\begin{align*}
x \bmod 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A serie about $\sum_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}}$ How to prove $$\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.$$
Add: Maybe we can follow this!
| \begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding equation of chord of hyperbola. Equation of chord of hyperbola joining points $(a\sec\phi,b\tan\phi)$ and $(a\sec\phi_1,b\tan\phi_1) $ $$y-b\tan\phi=\frac{b\tan\phi-b\tan\phi_1}{a\sec\phi-asec \phi_1}(x-a\sec\phi) $$ This reduces to $$\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)-\frac{x}{a}\cos\Big(\frac{\ph... | Another method which does not require ingenious use of trigonometric identities goes as follows
Equation of tangent to hyperbola at point $(asec \ A,btan \ A)$ is $$\frac{x}{a}sec \ A-\frac{y}{b}tan\ A=1 $$
Equation of tangent to hyperbola at point $(asec \ B,btan \ B)$ is $$\frac{x}{a}sec \ B-\frac{y}{b}tan\ B=1 $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Of sum of cosines and the $7$th roots of unity In my solution here, it was shown that
$$\omega+\omega^2+\omega^4=-\frac 12\pm\frac{\sqrt7}2\qquad\qquad (\omega=e^{i2\pi/7})$$
from which we know that
$$\sin \frac{2\pi}7+\sin \frac{4\pi}7-\sin \frac{6\pi}7=\Im (\omega+\omega^2+\omega^4)=\frac{\sqrt7}2$$
This can also b... | You're starting from the wrong assumption that $-\omega^3=\omega^4$, which is obviously wrong, because it would imply $\omega=-1$.
The true fact is that
$$
-\sin\frac{6\pi}{7}=\sin\frac{8\pi}{7}
$$
but
$$
\cos\frac{6\pi}{7}=\cos\frac{8\pi}{7}
$$
because
$$
\frac{6\pi}{7}+\frac{8\pi}{7}=2\pi
$$
and
$$
\sin(2\pi-\alpha)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I convert this tricky complex number into a real number: $\int_0^{\infty}\frac{x^α}{x^3+1}dx$? The problem statement is:
$$\int_0^{\infty}\frac{x^α}{x^3+1}dx$$
for α in the range −1<α<2.
$$\huge \frac{2\pi i}{1-e^{\frac{i2\pi (\alpha+1)}{3}}} \frac {e^{\frac{i \pi \alpha}{3}}} { 3e^{\frac{2\pi i}{3}}}$$
$\alpha... | HINT:
Assuming $x\in\mathbb{R}$:
$$\frac{2\pi i}{1-e^{\frac{i2\pi x}{3}}} \frac {e^{\frac{i \pi x}{3}}} { 3e^{\frac{2\pi i}{3}}}=$$
$$\frac{\left(2\pi i\right)\cdot\left(e^{\frac{x\pi i}{3}}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$
$$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .
| Denote $a=x/y$, $b=xy$.
Then
$$a+1+\frac{1}{a}=\frac{4}{b},$$
$$a^2+1+\frac{1}{a^2}=\frac{8}{b^2},$$
$$a^3+1+\frac{1}{a^3}=?$$
Then from the $1$st equation:
$$
\left(a+\frac{1}{a}\right)^2 = \left(\frac{4}{b}-1\right)^2,
$$
combining it with $2$nd equation, $\Rightarrow$ $b=1$.
Then, $a+\frac{1}{a}=3$, $a^2+\frac{1}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Needs to show Monotone While doing one of my works I need to show this expression is monotone(increasing).
$$e(2n+2)!\sum_{k=2n+2}^{\infty} \frac{(-1)^k}{k!}$$
I am sure this is increasing on $n$ because I explicitly computed for $n=1$ to $10$. And observed it is increasing but I need to prove it.
I tried many proce... | Let
$$
a_n
=(2n)! \sum_{k=2n}^{\infty} \frac{(-1)^k}{k!}
=(2n)! \sum_{k=n}^{\infty} \left(\frac{1}{(2k)!} - \frac{1}{(2k+1)!} \right)
$$
Clearly, $a_n > 0$ and
\begin{align}
\frac{a_n}{(2n)!} = \frac{2n}{(2n+1)!} + \frac{a_{n+1}}{(2n+2)!}
\end{align}
So
\begin{align}
a_n =\frac{2n}{2n+1} + \frac{a_{n+1}}{(2n+2)(2n+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Power series method to solve this ODE: what did I do wrong? Problem: Use the power series method to solve the ODE $$ 2(x - 1) y' = 3y. $$
Attempt: I solved this with power series method, and then compared with the technique of separation of variables, and I'm not getting the same answer. Here is what I did:
We look for... | There is a mistake in your computation of the coefficients $a_n$ (see below) :
One can reconize the binomial coefficients
\begin{pmatrix}
\frac{3}{2} \\
n
\end{pmatrix}
wich allows to link with $(1-x)^{3/2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\frac{1}{(1+x^4)^\frac{1}{4}}$ This question has been puzzling me a lot. This is in the middle list of question(meaning its moderately tough). I have tried everything i could think of.
My try: $$\int \frac{1}{x(1+\frac{1}{x^4})^\frac{1}{4}}dx$$
$$-\frac{1}{4}\int \frac{-4x^4}{x^5(1+ \frac{1}{x^4})^\fra... | Try this let $t^4 = 1 + x^{-4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$ Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$
For $n=1$ inequality holds.
For $n=k$
$$\left(\frac{1}{k+1}+\frac{1}{k+2}+...\frac{1}{2k}\right)^2<\frac{1}{2}$$
For $n=... | I have Cauchy-Schwarz inequality prove your inequality
we have
\begin{align*}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n}&\le\sqrt{n\left(\dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\cdots+\dfrac{1}{(n+n)^2}\right)}\\
&<\sqrt{n\left(\dfrac{1}{n(n+1)}+\dfrac{1}{(n+1)(n+2)}+\cdots+\dfrac{1}{(n+n-1)(n+n)}\right)}\\
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which of the constants A,B,C,D does T depend on? Let $f(x)=cos(5x)+Acos(4x)+Bcos(3x)+Ccos(2x)+Dcos(x)+E$ and $T=f(0)-f(\pi/5)+f(2\pi/5)-f(3\pi/5)+..-f(9\pi/5)$.Then out of A,B,C,D which does T depend on?
Hints please!
P.S:KVPY 2011 question
| Given $$f(x) = \cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+d\cos x+E$$
Using $$f(\pi-x) = f(\pi+x)\Rightarrow f(x) = f(2\pi-x).$$
So we get $$\displaystyle f\left(\frac{\pi}{5}\right)=f\left(\frac{9\pi}{5}\right)\;\;\;,\;\;\; \displaystyle f\left(\frac{2\pi}{5}\right)=f\left(\frac{8\pi}{5}\right)$$
and $$\displaystyle f\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How to solve $\lvert{x}\rvert - \lvert{2+x}\rvert = x$? How do I solve the following equation?
$$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$
I was thinking about dividing it into 4 cases: plus plus, plus minus, minus plus and minus minus. What is the best way to solve this?
| Another way is to use $|x| = \sqrt{x^2}$. If $|x| - |2+x| = x$ then
$$\begin{align*}
\sqrt{x^2} - \sqrt{(x+2)^2} &= x\\
x^2 + (x+2)^2 -2\sqrt{x^2}\sqrt{(x+2)^2} &= x^2\\
(x+2)^2 &= 2\sqrt{x^2} \sqrt{(x+2)^2}\\
(x+2)^4 &= 4x^2(x+2)^2
\end{align*}$$
So either $x= -2$, which doesn't satisfy the original equation, or else... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How should I go about solving this system of equations using Gauss Elimination? $${ x }_{ 1 }+2{ x }_{ 2 }-{ x }_{ 3 }+2{ x }_{ 4 }=0\\ { x }_{ 2 }+{ x }_{ 3 }-2{ x }_{ 4 }+2{ x }_{ 5 }=0\\ 2{ x }_{ 1 }+{ x }_{ 2 }-5{ x }_{ 3 }-4{ x }_{ 5 }=0$$
Steps I took:
$$\left[\begin{array}{rrrrr|r}
1 & 2 & -1 & 2 & 0 & 0 \\
... | multiplying the first equation by $-2$ and adding to the third equation we get
$$x_1+2x_2-x_3+3x_4=0$$
$$x_2+x_3+2x_4+2x_5=0$$
$$-3x_2-2x_3-x_4-4x_5=0$$
multiplying the second equation by $3$ and adding to the third we obtain
$$x_2+2x_5=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The sum of non real roots of the polynomial equation $x^3+3x^2+3x+3=0$ Problem :
The sum of non real roots of the polynomial equation $x^3+3x^2+3x+3=0$
(a) equals 0
(b) lies between 0 and 1
(c)lies between -1 and 0
(d) has absolute value bigger than 1
My approach :
The discriminant of cubic equation $ax^3+bx^2+... | Hint: The equation can be written as
$$(x+1)^3=-2.$$
Hence the three roots are $\sqrt[3]2-1$, $\sqrt[3]2\omega-1$, $-\sqrt[3]2\omega-1$, where $\omega$ is a cubic root of unit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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if $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has repeated roots prove... if the equation $$a(b-c)x^2+b(c-a)x+c(a-b)=0$$ has repeated roots prove that $${1\over a}, {1\over b},{1\over c} $$ are in Arithmetic Progression
Any idea about how to go about solving this ? Thanks is advance!
| With $a,b,c\ne 0$, observe that $1/a,1/b,1/c$ is an Arithmetic Progression iff $$K=0\text {, where}$$ $$K=a b+b c-2a c.$$ The condition that the quadratic has a repeated root is that the discriminant is zero. We have $$0=b^2(c-a)^2-4a c(a-b)(b-c)\iff$$ $$0=b^2c^2+b^2a^2-2b^2a c-4a c(-b^2+b c+b a-ac)\iff$$ $$0=b^2c^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4... | The discriminant is non-positive because of the geometric-quadratic means inequality. Indeed the reduced discriminant is
$$\Delta'=4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)$$
Sert $A=a^2, B=b^2, \&c.$ We have to prove $\;4ABCD\ge (A^2+B^2)(C^2+D^2)$
Now by the GQM inequality, we have
\begin{alignat*}{2}\sqrt{AB}&\le\sqrt\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove inequality $\left| \frac{x+yz}{x^2+y^2} \right| \leq 1$ for $x^2+y^2-z^2=1$ Prove this:
If $x,y,z \in \mathbb R$ and $x^2+y^2-z^2=1$, then $$\left| \frac{x+yz}{x^2+y^2} \right| \leq 1 $$ holds.
Own ideas:
If we eliminate $z$ the inequality is equivalent to $$\left| \frac{x \pm y\sqrt{x^2+y^2-1}}{x^2+y^2} \right|... | WLOG, let $z=\tan C$ and $x=\sec C\cos B,y=\sec C\sin B$
$$\dfrac{x+yz}{x^2+y^2}=\cos^2C(\sec C\cos B+\sec C\sin B\tan C)$$
$$=\cos C\cos B+\sin B\sin C=\cos(B-C)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find the least squares solution of the linear system...... Hi I'm looking for the least squares solutions of....
$$ \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
So Assuming this goes by $ A\vec{x}=\vec{b}$
Then using.... $A^T A \vec{... | Since you are only asking for the solution...
$$ \left( \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix}^\intercal \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix} \right) \begin{pmatrix} x \\ y \end{pmatrix} = \left( \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix}^\intercal \begin{pmatrix} 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\ri... | You might get some easy simplifications by extending the fraction with the factor $(1-\frac1n)^n$ so that all terms transform to the form
$$
\left(1+\frac{c_n}{n^2}\right)^n=1+\frac{c_n}{n}+O(n^{-2}).
$$
to obtain
\begin{align}
&\ \lim_{n \to \infty}
\frac{\left(1 - \frac{1}{n^3}\right)^n -
\left(1 - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\... | I like to write it this way: $p$ is assumed to be a prime that divides both $m^2 - n^2$ and $m^2 + n^2.$ Then $p$ divides their sum, so $p | 2 m^2$ and $p|2m.$ Either $p=2$ or $p |m.$
Also $p$ divides their difference, so $p | 2 n^2$ and $p|2n.$ Either $p=2$ or $p |n.$
If $p \neq 2,$ then $p$ divides both $m,n.$ We kno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Computing Two-Norm for interpolation of functions $$ f(x) = x^3, p(x) = (3/2)x^2 − (1/2)x $$
The two-norm of f(x) - p(x) is: $$( \int_0^1 (f(x) - p(x))^2 dx )^{1/2} $$
p(x) interpolates f(x) at $$x=0, x=1/2, x=1$$
The result of the two-norm computation should be $$\sqrt{210}/420$$
I can't seem to calculate the numerato... | I suppose some minor mistakes (probably a sign error in the development of the square).
If $$f(x)=x^3 \quad , \quad p(x)=\frac{3 }{2}x^2-\frac{1}{2}x$$ then $$f(x)-p(x)=x^3-\frac{3 }{2}x^2+\frac{1}{2}x$$ Now square carefuly and group terms for same powers of $x$ to get $$\big(f(x)-p(x)\big)^2=x^6-3 x^5+\frac{13 }{4}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
| $$a^3+a=b^3+b\iff a^3-b^3=b-a$$
$$\iff (a-b)\left(a^2+ab+b^2\right)=b-a$$
If $a=b$, then we're done. For contradiction, assume $a\neq b$. Then $a-b\neq 0$ and we can divide both sides by $a-b$:
$$a^2+ab+b^2=-1\iff 4a^2+4ab+4b^2=-4$$
$$\iff (2a+b)^2+3b^2=-4,$$
contradiction, because $(2a+b)^2+3b^2\ge 0$ for all $a,b\in\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find all the real solutions to the equation: $(x+i)^n-(x-i)^n=0$ Find all the real solutions to the equation: $$(x+i)^n-(x-i)^n=0$$
The answer is given, I will type it out until the line which is unclear to me (meaning I understand all the steps leading up to the last line).
$$\left(\frac{x+i}{x-i}\right)^n=1 \implies ... | I think that you could go one step further $$\frac{x+i}{x-i}= a + i b \implies x=i\frac{a+i b+1}{a+i b-1}=i\frac{a+i b+1}{a+i b-1}\times\frac{a-i b-1}{a-i b-1}$$ Develop and isolate the real and imaginary parts to get $$x=\frac{2 b}{(a-1)^2+b^2}+ i\frac{a^2+b^2-1}{(a-1)^2+b^2}$$ But, an here starts the beauty in your p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$ How can one evaluate the below sum? Any help would be greatly appreciated.
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$$
| $$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}=\sum_{k=1}^\infty\frac{k(k+1)}{15^k}+\sum_{k=1}^\infty\frac{k}{15^k}$$
start with the first term
depending on the geometric series
$$\frac{1}{1-x}=\sum_{k=0}^{\infty }x^k$$
$$\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k-1}$$
$$\frac{d^2}{dx^2}(\frac{1}{1-x})=\sum_{k=2}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
| By substitution: $\DeclareMathOperator\ach{arg\,cosh}$
Set $ x=\dfrac32\cosh t,\enspace t\ge0$, whence $\mathrm d\mkern1mu x=\dfrac32\sinh t\,\mathrm d\mkern1mu t$. With some hyperbolic trigonometry, we get
\begin{align*}
\int \frac{\sqrt{4x^2-9}}{x^2}\,\mathrm d\mkern1mu x&=2\int \frac{\sinh t}{\cosh^2t}\sinh t\,\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Taylor Series and Differentiation with Sigma notation $f(x) = \frac{x}{(2-3x)^2}$ Use Term By Term Differentiation to Find the Taylor Series about $x$=3 for
Give The Open Interval of Convergence and express as sigma notation
$\sum A_n(x-3)^n$
$f(x) = \frac{x}{(2-3x)^2}$
So I have Found the Taylor Series for 1/(2-3x)... | you can make a change of variable $x = 3 + h.$ then we have
$$\begin{align} f(x) &= \frac{x}{(2-3x)^2} = \frac{3+h}{(2 - 9-3h)^2}=\frac 1{49}(3+h)\left(1 + \frac{3h}7\right)^{-2} \\
&=\frac 1{49}(3+h)\left(1 -\frac 21 \frac{3h}7 + \frac{2\cdot3}{1\cdot2}\left(\frac{3h}{7}\right)^2 - \frac{2\cdot3 \cdot 4}{1\cdot2\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How many solutions this a multidimensional system have? How many solutions have this system of equations:
\begin{array}{l}
\ 3x^2+y+2xy^2-3 = 0 \\
\ x + 2yx = 0
\end{array}
The more general case is when we want to solve two nonlinear algebraic equations for two unknowns
\begin{array}{l}
\ {A_1}{x^2} + {B_1}xy + {C_1}{y... | There are two roots of the second equation.
$x=0$ substituted into the first equation gives $ y=3.$
$y=-\frac12$ substituted into the first equation gives two solutions from the quadratic resulting in:
$$ x = \dfrac{-3 \pm \sqrt{82}}{2} $$
So, in all 3 different roots and 3 solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ find the ratio of $x$, $y$ and $ z$ I have this question from higher algebra by Hall and Knight:
if $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $\frac x4 =\frac y2 =\frac ... | y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Where did I mistake to integrate $I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx\; ?$ It was given to integrate $$I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx.$$
Attempt:
\begin{align}I&= \int\sqrt{\frac{\sin((x+\alpha)-2\alpha)}{\sin(x+\alpha)}}\; dx\\&= \sqrt{\sin 2\alpha}\int\sqrt{\cot 2\alpha ... | \begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ \\ &= \int\frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\... | Given that $n^2/(n^2+1) < 1$, and $1>\frac{n-1}{n+1} >0$ for $n>1$, we can upper bound $\left(\frac{n^2}{n^2+1} \right)^{\frac{n-1}{n+1}} \leq 1^{\frac{n-1}{n+1}} \leq 1$. So, the limit is upper bounded by $1$.
Now, note that we have the lower bound $\left(\frac{n^2}{n^2+1} \right)^{\frac{n-1}{n+1}} \geq \frac{n^2}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fibonacci sequence, number of trees, probability How can I prove that the numbers of orderings described below are Fibonacci numbers?
We are given $p_1, \dots , p_n > 0$ such that each $p_i= \frac{1}{2^k}$ for $i \in \{1, \dots , n\}, \ \ k \in \mathbb{N}$ and $$p_1 + \dots +p_n = 1$$ and the sequence $(p_1, ..., p_n)... | It is not the Fibonacci sequence, as it starts $1, 1, 1, 2, 3, 5, 9, 16, \ldots $ so has too many examples when $n \ge 7$. For $7$ parts, the examples are:
*
*$ \frac1{2} + \frac1{4} + \frac1{8} + \frac1{16} + \frac1{32} + \frac1{64} + \frac1{64} $
*$ \frac1{2} + \frac1{4} + \frac1{8} + \frac1{32} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum up trigonometric series $$\cos \frac{2π}{2013} +\cos \frac{4π}{2013} +\cdots+\cos \frac{2010π}{2013} + \cos \frac{2012π}{2013}$$
How to sum it up?
*Calculator is not allowed.
| Original Approach: Using the formula for the sum of a geometric series, we get
$$
\sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}=\frac{e^{2\pi i}-e^0}{e^{\frac{2\pi i}n}-1}=0\tag{1}
$$
and using Euler's Formula
$$
\begin{align}
\sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}
&=\sum_{k=0}^{n-1}\cos\left(\frac{2\pi}nk\right)+i\sum_{k=0}^{n-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$ $\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$
I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?
| Since $x \approx \sin x$ when $x \rightarrow 0$, the denominator is of order $x^4$.
The numerator needs more careful analysis of $\sin x$. Using $\sin x = x - \frac{x^3}{6} + o(x^3)$, we get $\sin^2 x = x^2 - \frac{x^4}{3} + o(x^4)$, so the numerator is $\sin^2 x - x^2 \approx -\frac{x^4}{3}$.
Dividing and ignoring ter... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $y^2 = x^3 - 4$ where $x,y \in \mathbb{Z}$ Solve $y^2 = x^3 - 4$
So far i've tried to look at this:
$x^3 = (y-2i)(y+2i)$.
I think we need to prove that $y+-2i$ are coprime, so suppose not then:
$d = u+iv$ divides both numbers. So also d divides the sum and difference: $d|2y$ and $d|4i$.
If one applies the euclide... | Your computation are done in the ring of Gaussian integers, which is a PID. Let $p$ be a prime in this ring dividing both $y+2i$ and $y-2i$. Then $p$ divides $2y$ and $4i$, so it is $1+i$ (up to invertible elements), because this is the only prime dividing $2$.
From $(1+i)(h+ki)=y+2i$ we get $h-k=y$, $h+k=2$, so $h=(2+... | {
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Is it possible to find solution of this system of equations? Following is augmented matrix which has been reduced to row echelon form by using row operations. So when I convert it to system of equations I would get 3 equations with 5 unknowns. Is it possible to find values of 5 unknowns in 3 equations? Is it true that ... |
Is it possible to find values of 5 unknowns in 3 equations?
Yes, the solutions of your system form a 2 dimensional subspace of $F^5$, where $F$ is your field, e.g. $F = \mathbb{R}$.
$$
\left[
\begin{array}{rrrrr|r}
1 & 7 & -2 & 0 & -8 & -3 \\
0 & 0 & 1 & 1 & 6 & 5 \\
0 & 0 & 0 & 1 & 3 & 9 \\
0 &... | {
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"url": "https://math.stackexchange.com/questions/1534338",
"timestamp": "2023-03-29T00:00:00",
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Show that $x^2 \equiv 1 \pmod{n}$ can have arbitrarily many solutions? Equivalently, show that for every $m$, there exists some $n$ such that $x^2 \equiv 1 \pmod{n}$ has at least $m$ solutions.
Would appreciate any help or hints. I have a feeling this may be a pretty simple proof, but right now I am stuck. Thanks!
| Expanding on my comment:
Let $k$ be some positive integer such that $2^k \geq m$. Let $p_1, p_2, p_3, \cdots, p_k$ be $k$ distinct odd primes. (It's possible to choose $k$ distinct odd primes since there are infinitely many primes.)
Consider $n = p_1 p_2 p_3 \cdots p_k$, and look at the system of congruences
$$\begin{a... | {
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Evaluation of $ \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$ $\bf{My\; Try:}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx = \underbrace{\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{1}}+\underbrace{\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{2}}.$$
Now Here $$\displaystyle I_{2} = \int_... | Your solution is simple and straight to the point. Here is a longer way:
You could take the indefinite integral then apply limits to find the definite integral.
Let $v=\log x$, $u'=\frac{x}{(1+x^2)^2}$
So $v'=\frac{1}{x}$, $u=-\frac{1}{2(1+x^2)}$
$$\int\frac{{x\log x}}{(1+x^2)^2}dx=-\frac{\log x}{2(1+x^2)}+\int\frac{1}... | {
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How to calculate this infinite sum? $$ \sum_ {n=0}^\infty \frac {1}{(4n+1)^2} $$
I am not sure how to calculate the value of this summation. My working so far is as follows:
Let $S=\sum_ {n=0}^\infty \frac {1}{(4n+1)^2}$.
$\Longrightarrow S=\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{9^2}+...$
$\Longrightarrow S=(\frac{1}{1^2... | This has no closed form, unless you want to use Catalan's constant. Then the answer is $\frac{K}{2}+\pi^2/16$. See here: http://mathworld.wolfram.com/CatalansConstant.html
| {
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"question_score": "2",
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The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the ... | Sides AB and AC make an angle of 45∘ with the given line.
Let B=(−6+5t,−10+3t,−14+8t)
AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^
cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√
Squaring and simplifying t2–5t+6=0⟹t=2or3
Taking 2 for B and 3 for C.
AB→=... | {
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"source": "stackexchange",
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Prove that if $A\ge B$ then $\left[ {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right]\ge0$. Let $A$ and $B$ be $n \times n$ matrices, i.e., $A, B \in M_n$. Also, $A \ge 0$, $B \ge 0$, and $A-B \ge 0$ which mean all these matrices are semi-positive-definite.
Why does
$\left[ {\begin{array}{*{20}{c}}
A... | First, write
$$
\begin{pmatrix}
x' & y'
\end{pmatrix}
\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
=x'Ax+y'Bx+x'By+y'Ay.
$$
Next, consider the RHS above
\begin{align*}
\text{RHS}&=x'(A-B)x+x'Bx+y'Bx+x'By+y'By+y'(A-B)y\\
&=x'(A-B)x+(x+y)'Bx+(x+y)'By+y'(A-B)y\\
&=x'(A-B)x+(x+y)'B(x+y)+... | {
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"url": "https://math.stackexchange.com/questions/1539605",
"timestamp": "2023-03-29T00:00:00",
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