Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Let $A$ be a complex $2$ by $2$ matrix having distinct eigenvalues $a, b$. Show that $A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)$.
Let $A\in\mathscr{M}_{2\times 2}(\mathbb{C})$ be a matrix having distinct eigenvalues $a\neq b$. Show that, for all $n > 0$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A ... | i think you can do this with cayley-hamilton theorem which says that a matrix satisfies its characteristic equation. that is $$A^2 -(a+b)A + abI = 0 $$ or $$A^2 = (a+b)A - abI$$ we can use this to write every power of $A$ as a lineay combination of $A, I$. we will the division algorithm to find the remainder. suppose... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving a Radical Equation $5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2}$ (squaring doesn't help) How should I approach this problem:
$$ 5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2} $$
I've tried squaring both sides but to get rid of all the radicals requires turning it into a quartic equation, which I don't know... | HINT:
Method$\#1:$
Let $2y=\arccos x\implies x=\cos2y,\sqrt{1-x^2}=+|\sin2y|$
As for real $a,\sqrt a\ge0,$
Using the definition of Principal values, $0\le2y\le\pi\implies\sqrt{1-x^2}=+\sin2y$
Again, $0\le2y\le\pi\iff0\le y\le\dfrac\pi2\implies\sin y,\cos y\ge0$
$\implies\sqrt{1-x}=+\sqrt2\sin y,\sqrt{1+x}=+\sqrt2\cos y... | {
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"timestamp": "2023-03-29T00:00:00",
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What will be the equation of side $BC$.
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{gre... | the two two lines that bisects the $\angle BAC$ are given by
$$\frac{7x-y-3}{\sqrt{50}} = \pm\frac{x+y-5}{\sqrt2}.$$ they are
$$x-3y+11 = 0, \quad 3x + y - 7=0 \text{ and } A = (1, 4).$$
we will pick a point $$D = (1+2k, 4-6k), k \text{ to be fixed later} $$ on the angle bisector $3x+y-7 = 0.$ the line through $D$ ... | {
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"source": "stackexchange",
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Is my algorithm correct? (Polar decomposition) I cant seem to find my mistake. Consider this matrix $T = $\begin{bmatrix}
2 & 1 & 1 \\[0.3em]
-1 & 2 & 0 \\[0.3em]
0 & 1 & -1
\end{bmatrix}
I need the polar decomposition of $T$ ($T=AS$, with $A$ orthogonal and $S$ symmetrical). So, here are my s... | What you have for $S$ is exactly correct. To make finding $S^{-1}$ easier, normalize your columns of your matrix of eigenvectors, so that $S = QDQ^{T}$, where $Q$ is an orthogonal matrix. Then $S^{-1} = QD^{-1}Q^{T}$ and all of the matrices in that product are easy to calculate.
In particular,
$$Q = \left[\begin{array}... | {
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How to apply the AM-GM inequality? What is the minimum value of $8x^3+36x+54/x+27/x^3 $ for positive real numbers x?
Express your answer in simplest radical form.
I attempted to make an equation between the product of the terms and the original expression, $ 8*36*54*27 = 8x^3+36x+54/x+27/x^3 $
but it seems more compli... | For numbers $a_1,\ldots,a_n > 0$, the AM-GM inequality is $\dfrac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}$.
Equality occurs iff $a_1 = a_2 = \cdots = a_n$.
The trick here is to apply AM-GM to the right number of terms at a time.
Since $x$ is positive, the $4$ terms $8x^3$, $36x$, $\dfrac{54}{x}$, and $... | {
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Decomposition into partial fractions to compute an integral I'm having problems with:
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}... | HINT:
As $y^3+1=(y+1)(y^2-y+1),$
$$\dfrac{x^4+1}{x^6+1}=\dfrac{x^4-x^2+1}{x^6+1}+\dfrac{x^2}{x^6+1}=\dfrac1{x^2+1}+\dfrac{x^2}{x^6+1}$$
Set $x^3=u$ for the second part
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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why $ \sin \theta = \frac{7}{8} \cos \theta$? I have an example:
$$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{7}{8} $$
and then this equation is true? why there is cos multiplied?:
$$ \sin \theta = \frac{7}{8} \cos \theta$$
| You have $\dfrac{\sin \theta}{\cos \theta} = \dfrac{7}{8}$
Multiply both side by $\cos \theta$
You get $\dfrac{\sin \theta}{\cos \theta} \times\cos \theta = \dfrac{7}{8} \cos \theta$
So you get $\sin \theta = \dfrac78 \cos \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difficult inverse tangent identity
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$
I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{... | Given that $$\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$$ $$=\tan^{-1}\left(\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}\right)$$ $$=\tan^{-1}\left(\frac{1+x+1-x-2(\sqrt{1+x})(\sqrt{1-x})}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)$$ $$=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you solve this quadratic equation? The number of values of a for which
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in x is?
Here's how much I was able to solve through:-
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
$$ ((a-1)(a-2))x^2 + ((a-3)(a-2))x + (a+2)(a-2) = 0$$
$$ (a-2)[(a-1)x^... | HINT:
If $Ax^2+Bx+C=ax^2+bx+c$ is an identity
$A=a,B=b,C=c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim_{n\to\infty}nI_n$ with $I_n=\int_0^1\frac{x^n}{x^2+3x+2}dx$ We have to evaluate: $$\lim_{n\to\infty}nI_n$$ with $$I_n=\int_0^1\frac{x^n}{x^2+3x+2}\:dx.$$
There is an elegant way to solve this problem?
Here is all my steps:
*
*My first ideea was to find a recurrence relation such that:
$$I_{n+2}+3... | Substitute $x\mapsto x^{1/(n+1)}$ and use Dominated Convergence:
$$
\begin{align}
n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x
&=\frac{n}{n+1}\int_0^1\frac1{x^2+3x+2}\,\mathrm{d}x^{n+1}\\
&=\frac{n}{n+1}\int_0^1\frac1{x^{2/(n+1)}+3x^{1/(n+1)}+2}\,\mathrm{d}x\\
&\to1\int_0^1\frac1{1+3\cdot1+2}\,\mathrm{d}x\\
&=\frac16
\en... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Quick Method to Calculate the Maclaurin Series of $\frac{1}{\sqrt{\cos{x}}} $ I am supposed to calculate the maclaurin series for $\frac{1}{\sqrt{\cos{x}}} $ but I can't seem to figure out an efficient way to go about doing this.
| Hint. You may use, as $x \to 0$,
$$
\cos x =1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \tag1
$$ and, as $u \to 0$,
$$
\frac 1{\sqrt{1-u}} =1+\frac{u}{2}+\frac{3}{8}u^2+O(u^3) \tag2
$$ giving $$
\begin{align}
\frac 1{\sqrt{\cos x}}&=\frac 1{\sqrt{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}}\\\\
&=1+\frac12\left(\frac{x^2}{2!}-\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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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:
$$2x^4 + x^2
= 2(82) + 9
= 1... | The commutativity property states that:
*
*For all $\color{red}{a},\color{green}{b}$ we have $\color{red}{a}+\color{green}{b} = \color{green}{b}+\color{red}{a}$
*For all $\color{red}{a},\color{green}{b}$ we have $\color{red}{a}\times\color{green}{b} = \color{green}{b}\times\color{red}{a}$
distributivity property ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $... | Hint: $A^TA$ is positive semidefinite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
| You should just multiply the inequality by (x-1) instead of $(x-1)^2$.
Also, the solution would then be $\frac{3}{2}\le x\lt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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What is wrong with this integral reasoning? $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$
We start by multiplying by $1=\frac{x}{x}$.
$$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$
Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$.
$$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+... | To explicitly check that they differ by a constant, subtract them:
\begin{align*}
&\frac{1}{2}\log \frac{x^4+x^2\sqrt{x^4+1}}{1+\sqrt{x^4+1}} - \log \frac{x^2-1+\sqrt{x^4+1}}{x}\\
&= \frac{1}{2} \log \frac{1-x^2+x^4-\sqrt{1+x^4} + x^2\sqrt{1+x^4}}{x^2}\\
&\qquad - \frac{1}{2} \log \frac{2-2x^2+2x^4-2\sqrt{1+x^4}+2x^2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
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$(1+i)^6$ 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... | Hint (you made an arithmetic error):
$$ (1+i)^6 \neq \left(\frac{\sqrt2}{2}\right)^6\left(\frac{\sqrt2}{2} +\frac{\sqrt2}{2}i\right)^6$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
| Notice that the limit is
$$
L=\lim_{x \to 0}\frac{1}{\sin x} \left(\frac{1}{\arctan x} - \frac{\cos x}{\arcsin x} \right)
$$
from the series expansion for small values of $x$ one has $\cos x =1- \frac{x^2}{2} + O(x^4)$ and $(\mbox{arc})\sin x = x + O(x^3) = \arctan x$ hence
$$L = \lim_{x \to 0}\frac{1}{(x + O(x^3))} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Find a unit tangent vector to a curve that is an intersection of two surfaces. The intersection of the two surfaces given by the Cartesian equations $2x^2+3y^2-z^2=25$ and $x^2+y^2=z^2$ contains a curve $C$ passing through the point $P=(\sqrt{7},3,4)$. These equations may be solved for $x$ and $y$ in terms of $z$ to gi... | For $(b)$, by
$$
\begin{cases}
2x^2+3y^2-z^2=25 \\
x^2+y^2=z^2 \\
\end{cases}
$$
and working on the first:
$$\underbrace{2x^2+2y^2}_{2 z^2}+y^2-z^2=25\ \Rightarrow \ z^2+y^2=25$$
Then the curve is a circumference in the y-z plane and its parametric representation is:
$$
\begin{cases}
x=5 \cos t \\
y=5 \sin t \\
\end{ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Probability of a rectangular card intersecting the lines of a floor I have attempted this problem, but do not understand how the answer was achieved. The question is as follows (coming from Henk Tijms 'Understanding Probability' (3rd edn.) book):
Problem 7.16
Consider the following variant of Buffon's needle problem.... | I think the problem is the range of integration for $x$. Assume WLOG that $a\leq b$. There being two diagonals and thus two choices for $x$, we want $x$ to be the angle between a diagonal and the parallel lines that is closest to a right-angle. Then we have $\theta\leq x\leq \theta+\pi/2$ where $\theta$ is the angle be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Coefficient Problem (polynomial expansion)
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$
But expanding on like this take too long.
In the end th... | We get the coefficient of $x^2$ by adding the product of pairs of coefficients of $x$ in the binomials, since the other coefficients would be $1$.
All the coefficients that come from the $1-x$ term multiplied by other $x$ terms add up to
$$-1\cdot 2 + -1\cdot -3 + \ldots + -1\cdot -15$$
$$=-1(2-3+\ldots -15)$$
$$=-1[(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$ So I've been trying to solve this problem for a couple of days now. What I've come up with is this:
By way of contradiction, assume that there are positive integers a and b such that $b^4 + b + 1 = a^4$.
Consider the case when $b \geq ... | Indeed, we can show that $b^4+b+1=c^2$ has no solution in positive integers.
This is because if $b>0$ is an integer, then $0<b+1<2b+1\leq 2b^2+1$. Adding $b^4$ to this inequality gives: so $$b^4<c^2=b^4+b+1<b^4+2b^2+1=(b^2+1)^2$$ so $$b^2<c<b^2+1.$$
There are also no solutions with $b\leq -2$, since then:
$$1-2b^2<b+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2... | First use the double angle formulas to lower the degree
$$3\frac{\cos(2x)+1}2-\frac82\sqrt3 \sin(2x)+5\frac{1-\cos(2x)}2-7
=-\cos(2x)-4\sqrt3 \sin(2x)-3.$$
The dot product $$(\cos(2x),\sin(2x))\cdot(-1,-4\sqrt3)$$ equals $$1\cdot\sqrt{(-1)^2+(-4\sqrt3)^2}\cdot\cos(\phi)$$ where $\phi$ is the angle between the vectors, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do you solve the summation of $2-4+8-16+32- \dots 2^{48}$? This is a summation problem but I can't seem to figure out how to solve this with the mix of subtraction and addition.
| Split it into a positive series and a negative series:
$\color\red{2}-\color\green{4}+\color\red{8}-\color\green{16}+\color\red{32}-\ldots-\color\green{2^{48}}=$
$\color\red{2^1}-\color\green{2^2}+\color\red{2^3}-\color\green{2^4}+\color\red{2^5}-\ldots-\color\green{2^{48}}=$
$\color\red{\sum\limits_{n=0}^{23}2^{2n+1}}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2... | Break this up into cases.
Case 1: $x > 0$.
Multiplying both sides by $x$, we have $1 > \dfrac{2x^2}{x^2 + 2}$ (since $x$ is positive, the sign doesn't change), and so $x^2 + 2 > 2x^2$ (since $x^2 + 2$ is positive, the sign doesn't change).
Then $2 > x^2$. Now, just find the positive $x$ for which this is true.
Case 2: ... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\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} -... | $$(\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) =(x \sqrt{1 + 3/x} - x\sqrt{1+ x^{-2} })$$
$$
=x\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right)
$$
$$
\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right) =1 +\frac{3}{2}\frac{1}{x} + O(x^{-2}) - (1+ O(x^{-2}))
$$
using Taylor's theorem.
So we get
$$
\frac{3}{2} + O(x^{-1})
$$
which ... | {
"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": 4
} |
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... | Multiply each term inside the parenthesis by $z$. You'll find it is simply:
$$\frac{(1-\cos z)^2}{z^2}.$$
| {
"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": 3
} |
Calculating determinant of a Vandermonde type matrix of order n Det$
\begin{bmatrix}
1 & 2 & 3 &\ldots &n\\
1& 2^3& 3^3& \ldots & n^3\\
1 &2^5& 3^5& \ldots & n^5\\
\vdots & \vdots& & \vdots \\
1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1}
\end{bmatrix}
$
If the powers were consecutively increasing down the rows than we... | Call $\Delta$ your determinant. It seems to me that $$\Delta = n! \det \left( \begin{array}{ccccc}
1 & 1 & 1 & \dots & 1 \\
1 & 4 & 9 & \cdots & n^2 \\
1 & 4^2 & 9^2 & \cdots & (n^2)^2 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
1 & 4^{n-1} & 9^{n-1} & \cdots & {(n^2)}^{(n-1)}
\end{array} \right) = n! \left( \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ a^2-4b \neq2$ if $ a,b \in \mathbb{ Z}$ My solution :
We suppose that is true. Then by contradiction:
$a^2-4b-2=0$
$a^2=4b+2$
$a=2(b+1/2) ^{0.5}$
then $(b+1/2)$ is fraction and rooted by $0.5$ so the square root of any fraction $+$ any-Integer will give fraction so then $a$ must be fraction, not an integer... | Suppose
$a^2=4b+2
$.
$a$ must be of the form
$2c$ or $2c+1$
(i.e., even or odd).
If $a=2c$,
then
$2 = (2c)^2-4b
= 4c^2-4b
= 4(c^2-b)
$.
But 4 divides the right side but not the left,
so this is impossible.
If $a=2c+1$,
then
$2 = (2c+1)^2-4b
= 4c^2+4c+1-4b
= 4(c^2+c-b)+1
$.
But the left side is even
and the right side i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2... | Correct, but too complicated (and missing several $\lim_{x\to0}$).
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=
\lim_{x\to0}\frac{2\sin^2(x/2)}{4(x/2)^2}=
\lim_{x\to0}\frac{1}{2}\left(\frac{\sin(x/2)}{(x/2)}\right)^{\!2}=
\frac{1}{2}
$$
Alternative way:
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=
\lim_{x\to0}\frac{1-\cos^2 x}{x^2(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
$a^2+b$ and $a+b^2$ prime implies $\gcd(ab+1,a+b)=1$ Let $a,b>1$ be integers such that $a^2+b$ and $a+b^2$ are prime. Prove that $\gcd(ab+1,a+b)=1$.
Clearly $a$ and $b$ are of different parities; suppose $a$ is odd and $b$ even. If a prime $p\neq 2$ divides $ab+1$ and $a+b$, then it also divides $(ab+1)+(a+b)=(a+1)(b+1... | Suppose a prime $p$ divides both $ab+1$ and $a+b$. Then, writing $b\equiv -a \pmod p$, and plugging this in $ab+1$, we get $p| 1-a^2$, which (since $p$ is prime) implies $a\equiv 1$ or $-1$ modulo $p$.
i) $a\equiv 1 \pmod p$: Then, $b\equiv -1 \pmod p$, which implies $a^2+b\equiv 0 \pmod p$, so $a^2+b=p$, but since the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A trig ratio integral Consider the two integrals, where $a$ and $n$ are integers,
\begin{align}
I_{1} &= \int_{-n \pi}^{n \pi} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x} \\
I_{2} &= \int_{-(2n+1)\pi/2}^{(2n+1)\pi/2} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x}.
\end{align}
It would seem that $I_{2}$ would hav... | inside $I_1$ it is even function so integral will be
\begin{align}
I_{1} &=2 \int_{0}^{n \pi} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x} \\
\end{align}
then what function written inside the integral holds $f(x)=f(n\pi-x)$
then integral $I_1$ will be
\begin{align}
I_{1} &=4 \int_{0}^{\frac{n \pi}{2}} \frac{\tan^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$xy + yz + zx + 2xyz = 1$ implies $4x+y+z\geq 2$ Let $x,y,z>0$ satisfy $$xy + yz + zx + 2xyz = 1.$$ Prove that $4x+y+z\geq 2$.
The condition invites the factoring $(1+x)(1+y)(1+z)+xyz-2=x+y+z$, but having the factor $4$ in the desired inequality makes things more difficult.
| $x(y+z)+yz(1+2x)=1,yz\le (\dfrac{y+z}{2})^2 ,p=y+z \implies xp+(1+2x)\dfrac{p^2}{4} \ge 1 \\ \implies x \ge \dfrac{1-\frac{p^2}{4}}{p+\frac{p^2}{2}}=\dfrac{1}{p}-\dfrac{1}{2}$
$4x+y+z \ge \dfrac{4}{p}-2+p \ge 2\sqrt{\dfrac{4}{p}*p}-2=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
When is $a \space \sin^2(x) + b \space \cos^2(x) \le 1$? When is the above expression less than or equal to $1$, meaning for what values of $a$ and $b$ will the above expression be less than or equal to $1$?
| Well, clearly it equals 1 when $a=b=1$. Rewrite $b=a+c$. Then you are asking when
$$a\sin^2 x+b\cos^2x=a\sin^2x+(a+c)\cos^2x=a(\sin^2x+\cos^2x)+c\cos^2x=a+c\cos^2x>1$$
or when
$$\cos^2x>\frac{1-a}{c}=\frac{1-a}{b-a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt... | For the first one, rewrite the integral as follows:
$$ \int \frac{1}{\sqrt{9+4x^2}} \, \mathrm{d} x = \frac{1}{3} \int \frac{1}{\sqrt{1+4x^2/9}} \, \mathrm{d} x = \frac{1}{3} \int \frac{1}{\sqrt{1+(2x/3)^2}} \, \mathrm{d} x, $$ now let $v = 2x/3$ so we have:
$$ \int \frac{1}{\sqrt{9+4x^2}} \, \mathrm{d} x = \frac{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": 3
} |
calculate the $1/6+1/12+1/24+1/48 \ldots $. Wolfram is wrong? I am trying to calculate the following sum
$$
S = \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots
$$
so
$$
S+\frac{1}{3} = \frac{1}{3} + \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots = \frac{1}{3} \cdot \sum_{k=0}^{\infty} \frac{1}{... | You seem to have found a bug in Wolfram Alpha. It appears to first interpret the sum $$\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{48} + \ldots $$ as $$\sum_{n=1}^\infty \dfrac{1}{n((n-3)n+8)} \approx 0.3450320299$$
(which does start $ \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{48}$, but the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Product of $A$ with the adjoint of $A$: why are all nondiagonal elements zero? Let \begin{align*} A = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ 6 & 8 & 2 \end{pmatrix}. \end{align*} We have $\det(A) = 44$. The cofactor matrix corresponding with $A$ is \begin{align*} C = \begin{pmatrix} -4 & 0 & 12 \\ 28 & -22 & 4 \\ -6... | Because this is the Laplace expansion for the determinant of a matrix with two identical rows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Substituting a value of sine function in a trigonometric equation I am trying to really understand trigonometric equations and I've stumbled upon a rather confusing example. Solve the following equation:
$\sin x= 2|\sin x|+ {\sqrt{3}}\cos x$
First step is to define the absolute $\sin x$:
$$|\sin x| =
\begin{cases}
\... | the function $y = 2|\sin x| + \sqrt 3 \cos x$ is an even $2\pi$-periocic function and is also symmetric about $x = \pi.$ the equation $$\sin x = 2|\sin x| + \sqrt 3 \cos x = \begin{cases} 2\sin x + \sqrt 3 \cos x & \text{ if } 0 \le x \le \pi \\
-2\sin x + \sqrt 3 \cos x & \text{ if } \pi \le x \le 2\pi \\\end{cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find all complex numbers so that $Im(\frac{z+2}{2-i})=1$ and $Re(z^2+1)=1$ and for $z$ which is in the first quadrant find $\sqrt{z}$
Find all complex numbers so that $Im(\frac{z+2}{2-i})=1$ and $Re(z^2+1)=1$ and for $z$ which is in the first quadrant find $\sqrt{z}$.
If $z=x+iy$ then $$\frac{z+2}{2-i}=\frac{x+2+iy}{... | Note that
$$z^2+1=(x+iy)^2+1=x^2+2xyi-y^2+1=x^2-y^2\color{red}{+1}+2xyi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^2=17\pmod{128}$ I'm attempring to solve a congruence $x^2 \equiv 17\pmod{128}$ but not quite sure how to go about it. I see that $128 = 2^7$, but the Chinese Remainder Theorem doesn't apply to $\gcd > 1$. I found one solution quite easily by finding solutiong to $x^2 \equiv\pmod{32}$ which was $x \equiv 23\p... | The series for the square root is
\begin{eqnarray}
\sqrt{1+16 t} =1 + 8 t - 32 t^2 + 256 t^3 - 2560 t^4 + 28672 t^5 - 344064 t^6+\\ +
4325376 t^7 - 56229888 t^8 + 749731840 t^9 - 10196353024 t^{10}+ \cdots
\end{eqnarray}
The first three terms truncation $1 + 8 t - 32t^2$ has square $1 + 16 t - 512 t^3 + 1024 t^4$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Calculus 2 - $\int(\sqrt{72+36x^2}dx$ I have done this problem several times and this is the only answer i ever come to. My schools webwork gives me incorrect for my answer (answer is not simplified but it should be accepted in this format). Did i do this correctly?
Here is my work:
\begin{align}
\int \sqrt{72+36x^2}\,... | I believe that your work is correct. However, at the end you are most likely asked to put this in a nicer form. The way to simplify $trig_1(trig_2^{-1}(\frac{a}{b}))$ is form a right triangle that fits your $trig^{-1}$ conditions and then compute $trig_1(angle)$.
As an example, simplifying $\tan{(\arctan{\frac{x}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the logic/theorem/derivation behind finding the exponent of p in n! By [n/p] + [n/p^2] + [n/p^3] + ....? The exponent of prime number of 3 in 100! is 48.
It means 100! is divisible by $3^48$
$$E_3(100!) = \left\lfloor\frac{100}3\right\rfloor + \left\lfloor\frac{100}{3^2}\right\rfloor + \left\lfloor\frac{100}{3^... | suppose $p$ is prime number
$$1,2,3,4,5,...,p,p+1,p+2,....,2p,....3p,3p+1,...,p^2,p^2+1,....2p^2,...p^3,p^3+1,...$$ obviously : every p number has multiple of p
every $p^2 $number has multiple of $p^2$
every $p^3 $number has multiple of $p^3$
and so on
sum of the p multiple is :sum of $$\left \lfloor \frac{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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... | The sum of the first and fourth terms is
\begin{align*}
\frac{\omega}{1 - \omega^2} + \frac{\omega^4}{1 - \omega^3} &= \frac{\omega (1 - \omega^3) + \omega^4 (1 - \omega^2)}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega^6}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \... | {
"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": 2
} |
System of equations in a,b,c,d $a,b,c,d$ are complex numbers satisfying
\begin{cases}
a+b+c+d=3 \\
a^2+ b^2+ c^2+ d^2=5 \\
a^3+ b^3+ c^3+ d^3=3 \\
a^4+ b^4+ c^4+ d^4=9
\end{cases}
Find the value of the following: $$a^{2015} + b^{2015} + c^{2015} + d^{2015}$$.
| Hint
Let $a_{n}=a^n+b^n+c^n+d^n$,then
$$a_{n+3}=(a+b+c+d)a_{n+2}-(ab+ac+ad+bc+bd+cd)a_{n+1}+(abc+abd+acd+bcd)a_{n}-abcd\cdot a_{n-1}$$
and you have only find this $ab+ac+ad+bc+bd+cd,abc+abd+acd+bcd,abcd$ this is not hard to find it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the area of the part of the surface $z=yx$ bounded by $x^2+y^2=1$? A parametrization of the part of the surface $z=yx$ bounded by $x^2+y^2=1$ is
\begin{align}
x &= u \cos v \\
y &= u \sin v \\
z &= \frac12 u^2 \sin 2v,
\end{align}
or
$$r(u,v)=u \cos v \, {\bf i} + u \sin v \, {\bf j} + \frac12 u^2 \sin 2v \, {\... | $I=\int \int_Vxy\ dx\ dy = \int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^\sqrt{1-x^2} xy\ dy\ dx$
Change of coordinates -
$x=r\cos \theta$
$y=r\sin \theta$
Jacobian -
$I=\int \int r^2\cos\theta \sin\theta\ \left\|d\frac{x,y}{r,\theta}\right\|\ d\theta\ dr$
$\left\|d\frac{x,y}{r,\theta}\right\|=\left|\begin{array}{c}\cos \theta&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question:
If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...
Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$
My Attempt:
The new equation can be made into a quadratic as:
$$(a-b+c)x^... | Let $\alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$.
Note that $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$.
We want to calculate:
$$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} $$
In terms of $\alpha$ and $\beta$.
$$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How do I prove that $\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$ $$\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$$
I believe this is correct since I couldn't find a directional limit that won't validate this.
From what I know, I have to prove that
$$\forall\epsilon\gt 0, \exists\delta\g... | Here's yet another way
$$\lim\limits_{(x,y)→(0,0)}\frac{1-\cos\left(x^2+y^2\right)}{\sqrt{x^2+y^2}} $$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}} $$
$$=\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\left(\cos^2\phi+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Side limits of the derivative of this function $f:\mathbb{R}\to \mathbb{R}$ with $f\left(x\right)=\left(x^3+3x^2-4\right)^{\frac{1}{3}}$
Calculate side limits of this function's derivative, $f'_s\:and\:f'_d$, in $x_o=-2$
The answer key says I should get $\infty $ and $-\infty$ but I'm not getting that. The derivative I... | It might help to observe that $f(x) \; = \; (x+2)^{2/3}(x-1)^{1/3}.$ To obtain this factorization, note that $x^3 + 3x^2 - 2$ equals zero when $x = -2,$ so we know $x+2$ is a factor of $x^3 + 3x^2 - 2.$ Use long division, or use synthetic division, or note that $x^3 + 3x^2 - 2 = x^3 + 2x^2 + x^2 - 2$ (and factor by gro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to show that that the following three consecutive numbers $3^{2^{10}} − 1$, $3^{2^{10}}$,$3^{2^{10}}+ 1$ are the sum of two squares? Show that the following three consecutive numbers:
$$
3^{2^{10}} − 1, 3^{2^{10}} , 3^{2^{10}} + 1
$$
can be represented as sums of two integer squares.
| the second and third can be expressed as $(3^{2^{9}})^2+0^2$ and $(3^{2^9})^2+1^2$
For the first:
$3^{2^{10}}-1=$
$(3^{2^{9}}+1)(3^{2^8}+1)(3^{2^7}+1)(3^{2^6}+1)(3^{2^5}+1)(3^{2^4}+1)(3^{2^3}+1)(3^{2^2}+1)(3^{2}+1)8$
Each of the factors is trivially a sum of two squares, So all we need to conclude is the fact that the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic ... | the matrix you have has rank one and can be written as $$A = uu^\top \text{ where } u = \pmatrix{1,1,1,1}^\top.$$ now, $$Av = uu^\top v=(u^\top v)u.$$ note that $u^\top v$ is a scalar, therefore $A$ has eigenvalue $0$ of multiplicity $3$ corresponding eigenvector any vector orthogonal to $u$ which has dimension $3.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find this limit : $x\sin{f(x)}$ How to find the limit
$$\lim_{x\to\infty}x\sin f(x)$$
where $$f(x)=\left(\sqrt[3]{x^3+4x^2}-\sqrt[3]{x^3+x^2}\right)\pi\ ?$$
Is it possible to solve without L'Hospital's rule ?
| Write
$\begin{array}\\
(x^3+ax^2)^{1/3}
&=x(1+a/x)^{1/3}\\
&=x(1+(a/3x)+(1/3)(-2/3)(a/x)^2/2 + O(1/x^3))\\
&=x+(a/3)-(a^2/9x) + O(1/x^2))\\
\end{array}
$
so
$\begin{array}\\
(x^3+ax^2)^{1/3}-(x^3+bx^2)^{1/3}
&=(x+(a/3)-(a^2/9x) + O(1/x^2)))-(x+(b/3)-(b^2/9x) + O(1/x^2)))\\
&=(a-b)/3-(a^2-b^2)/(9x) + O(1/x^2)\\
\end{arr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find variance of k+1 elements if variance of k elements is known? I need to find the variance of k+1 elements given the variance of k elements. I can also store other features for k elements like mean ($\mu_n$) etc. So, given the below function's value,
$$
\frac{1}{n}\sum\limits_{i=1}^n(a_i-\mu_{n})^2
$$
I nee... | $$\sigma_{n+1}^2 = \frac{1}{n+1}\sum_{k=1}^{n+1}((a_{k} - \mu_{n}) + (\mu_{n} - \mu_{n+1}))^2 $$
$$ = \frac{1}{n+1}\sum_{k=1}^{n+1}[(a_k-\mu_n)^2 - 2(a_k-\mu_n)(\mu_{n+1} - \mu_n) + (\mu_{n+1} - \mu_n)^2]$$
$$ = \frac{1}{n+1}[\sum_{k=1}^{n}[(a_k-\mu_n)^2] + (a_{n+1} - \mu_n)^2 - 2 (\mu_{n+1} - \mu_n)\sum_{k=1}^{n+1}[(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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... | $$\left(\frac{48}{10}\right)^x=1000 \Longleftrightarrow $$
$$\left(\frac{24}{5}\right)^x=1000 \Longleftrightarrow x=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}$$
$$\left(\frac{8}{10}\right)^x=1000 \Longleftrightarrow $$
$$\left(\frac{4}{5}\right)^x=1000 \Longleftrightarrow y=\frac{\log_{10}(1000)}{\log_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Trigonometric integration 3. $\int \frac{\sqrt{3cos2x-1}}{cosx}$
ATTEMPT:
I did the following substitution
Let $sinx=t$
$cosxdx=dt$
$I=\frac{\sqrt{3(1-2t^2)-1}}{1-t^2}=\frac{\sqrt{2-6t^2}}{1-t^2}$
Now let $t=\frac{1}{z}$
$dt=-\frac{1}{z^2}dz$
Now $I=\color{red}{-} \frac{\sqrt{2z^2-6}}{z(z^2-1)}$
Lastly substitute $2z^... | Not sure I can follow your last substitution, but I think the mistake was in dropping a sign when you changed variables from $t$ to $z$. It appears your answer is off by a sign.
Draw a right triangle and label one degree $\alpha$.Make the side opposite $\alpha$ a length of $\sqrt3\sin x$ and the hypoteneuse a length o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Faulhaber Formula Identity I have to show the following identity:
$$ S_n^p := 1^p+2^p+...+n^p $$
$$ (p+1)S_n^p+\binom{p+1}{2}S_n^{p-1}+\binom{p+1}{3}S_n^{p-2}+...+S_n^0=(n+1)^{p+1}-1 $$
What I did first is to use the binomial theorem on the term the right side of the equation, which results in:
$$ (n+1)^{p+1}-1=\binom{... | Suppose we have
$$S_n^p = \sum_{q=1}^n q^p$$
and we seek to evaluate
$$\sum_{q=1}^{p+1} {p+1\choose q} S_n^{p+1-q}
= - S_n^{p+1}
+ \sum_{q=0}^{p+1} {p+1\choose q} S_n^{p+1-q}.$$
Observe that
$$q^p = \frac{p!}{2\pi i}
\int_{|z|=\epsilon} z^{p-1} \exp(q/z) \; dz.$$
Introduce the generating function
$$f(w) = \sum_{p\ge 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\cos(2x+\frac{\pi}{4})= -1/2 $ My suggestion:
$$\cos\left(2x+\frac{\pi}{4}\right)= -\frac{1}{2}$$
$$ 2x+\frac{\pi}{4} = \frac{2\pi}{3} \pm 2\pi n, n\in\mathbb{Z}$$
$$ x= \frac{\left( \frac{2\pi}{3} - \frac{\pi}{4} \right)}{2} \pm 2\pi n, n\in\mathbb{Z}$$
My answer:
$$ x = \frac{5\pi}{24} \pm 2\pi n, n\in\math... | HINT:
$\dfrac12=\cos\dfrac\pi3\implies-\dfrac12=\cos\left(\pi-\dfrac\pi3\right)$ as $\cos(\pi-u)=-\cos u$
$\implies2x+\dfrac\pi4=2m\pi\pm\dfrac{2\pi}3$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why does this sum converge? I know that the following sum converges to 2 via WolframAlpha, but I am not sure why.
$$\sum_{k=1}^\infty k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = 2$$
WolframAlpha gives the following partial sum formula:
$$\sum_{k=1}^n k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2... | Let the sum of series be denoted by $S$, the term corresponding to each k be $s_k$. Thus each $s_k$ is made of three terms.
$S=\sum s_k=\sum k(\frac{2}{k}-\frac{4}{k+1}+\frac{2}{k+2})$
$=\sum\limits_{k=1}^{\infty} (2-\frac{4k}{k+1}+\frac{2k}{k+2})$
Notice that last term of each $s_k$ and first 2 terms of $s_{k+1}$ have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Show that $4$ does not divide $x^3-2$ Show that $4$ does not divide $x^3-2$ is what I need to prove.
I think I should put $4k$ is $x^3-2$ and then contradict it somehow.
Alternatively is to factor it out as $x^3$ is $x(x+2)(x-2)$ but I am not sure of that.
Do you know how to show this?
| *
*Using modular arithmetic:
If $x \equiv 0 \mod 4$, then $x^3 \equiv 0 \mod 4$ and $x^3-2 \equiv 2 \mod 4$.
If $x \equiv 1 \mod 4$, then $x^3 \equiv 1 \mod 4$ and $x^3-2 \equiv 3 \mod 4$.
If $x \equiv 2 \mod 4$, then $x^3 \equiv 8 \equiv 0 \mod 4$ and $x^3-2 \equiv 2 \mod 4$.
If $x \equiv 3 \mod 4$, then $x^3 \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Derive Cartesian cubic Möbius strip from parametric The following link:
http://mathworld.wolfram.com/MoebiusStrip.html
shows the Möbius strip parametrized as
\begin{eqnarray}
x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\
y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\
z = s \sin \left ( \frac1... | I'd just substitute the given coordinates into the equation and show that it's satisfied. The terms can be grouped according to the powers of $R$ and $s$ they contain, with the two exponents always summing to $3$, and the equation has to be satisfied for all four groups separately; that makes the calculation more manag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And... | $$6\cos { x } -5\sin { x } =8\\ 6\cos ^{ 2 }{ \frac { x }{ 2 } -6\sin ^{ 2 }{ \frac { x }{ 2 } -10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } =8 } \cos ^{ 2 }{ \frac { x }{ 2 } } +8\sin ^{ 2 }{ \frac { x }{ 2 } } } } } \\ 14\sin ^{ 2 }{ \frac { x }{ 2 } } +10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
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.
| Although the answer is straight forward but here are some steps if you find helpful
Notice, $$2\cos^2 x\sin^2 x+\cos^4 x+\sin^4 x$$ $$=(\cos^2 x)^2+(\sin^2 x)^2+2(\cos^2 x)(\sin^2 x)$$ Now, assume $\alpha =\cos^2 x$ & $\beta=\sin^2 x$ & apply $\alpha^2+\beta^2+2\alpha \beta=(\alpha+\beta)^2$$$=(\cos^2x+\sin^2 x)^2$$$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$ Prove
$$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$$
I got this problem in Harro Heuser's "Lehrbuch der Analysis Teil 1". It is surely smaller than 1 because $\sqrt{9n^2 + 2n + 1} < \sqrt{9n^2 + 6n + 1} = 3n + 1$, but I cannot get closer... | Notice, $$\lim_{n\to \infty}(\sqrt{9n^2+2n+1}-3n)$$ Let, $n=\frac{1}{t} \implies t\to 0\ as\ n\to \infty$ $$=\lim_{t\to 0}\left(\sqrt{9\left(\frac{1}{t}\right)^2+2\left(\frac{1}{t}\right)+1}-3\left(\frac{1}{t}\right)\right)$$ $$=\lim_{t\to 0}\frac{\sqrt{t^2+2t+9}-3}{t}$$ Now, applying L-hospital's rule for $\frac{0}{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$ Let, $b> \max\{a_1,a_2,...,a_n\}.$ Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$
f is convex if $f(t_1x_1+t_2x_2+...+t_nx_n)\leq t_1f(... | I would use the function $g(x)=\frac{1}{x}$ defined on $(0,\infty)$; you can readily verify that $g''(x)>0$ on its domain. (The issue with your $f$ is that $f''(x)>0$ only for $x<b$.) The rest of the argument proceeds similarly: convexity gives
$$
\frac{1}{n}\times\text{LHS}=\frac{1}{n}\sum_{i=1}^ng(b-a_i)\geq g\left(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=... | Using the law of sines, one can get
$$s=\frac{1}{2}bc \sin(\alpha)=\frac{1}{2}bc \frac{a}{2R}=\frac{abc}{4R}$$
where $R$ is the radius of the circumscribed circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 4
} |
$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$
find $\dfrac {x^2} {x-1}.$
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $... | Hint:
$$a^2 + b^2 = (a+b)^2-2ab$$
Going forward,
$$ \left(x + \frac{x}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$
$$ \left(\frac{x^2}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$
Let $\frac{x^2}{x-1} = n$
$$ n^2 -2n -2010 =0 $$
You should be able to solve this using the quadratic formula or otherwise to get a value for $ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac... | multiplying (2) by $x$ $(x \ne 0)$ we obtain
$$\frac{1}{z}=\frac{1}{4}-\frac{1}{xy}$$
with (3) we obtain $$y=\frac{12(x^2-1)}{4x^2-3x}$$ (I)
from (1) and (3) we get
$$\frac{x^2y}{3}-x^2+1=\frac{xy}{4}$$
plugging (I) in this equation and simplifying we get
$$-60 x^5+119 x^4+156 x^3-288 x^2-72 x+144=0$$
with five real... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\... | We have $$\int\frac{1}{x^{2}\left(x^{4}+1\right)^{3/4}}dx\overset{u=1/x^{4}}{=}-\frac{1}{4}\int\frac{1}{\left(u+1\right)^{3/4}}du\overset{u+1=v}{=}-\frac{1}{4}\int\frac{1}{v^{3/4}}d=
$$ $$=-\sqrt[4]{v}+C=-\frac{\sqrt[4]{x^{4}+1}}{x}+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the Integral: $\int(x^5+5^x)\ dx$ $\int(x^5+5^x)\ dx$
I made the the terms within the parenthesis u
$u=x^5+5^x$
$du=5x^4+5^xln\ 5$
$du=5x^4+5^x\ ln\ 5 dx$
$\frac{u}{5x^4+5^xln\ 5}\ du$
I am stuck at this point. Is there a better way to attack this problem? I know I have to use this formula
$\int\ a^x dx=\fr... | If you want to evaluate this integral, you should simplify like this:
$\int{(x^5 + 5^x) dx} = \int{x^5 dx} + \int{5^x dx}$. You can always split up an expression of the same variable like this to simplify your computation. Now, generally, $\int{x^ndx} = \frac{x^{n+1}}{n+1} +C$, and $\int{a^xdx} = \frac{a^x}{ln{(a)}} +C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluate the Integral: $\int \frac{\log_{10}\ x}{x}\ dx$ $\int \frac{\log_{10}\ x}{x}\ dx$
$du=x\ln10\ dx$
$\log_{10}x\ \ln10+ C$
Is this answer correct? If not what step should I take to convert the log into a term I can manipulate?
| $$\int { \frac { \log _{ 10 }{ x } }{ x } dx } =\int { \log _{ 10 }{ x } d\ln { x } =\log _{ 10 }{ x } \ln { x } -\int { \frac { \ln { x } }{ x\ln { 10 } } +C= } } \\ =\log _{ 10 }{ x } \ln { x } -\frac { 1 }{ \ln { 10 } } \int { \ln { x } d\ln { x } +C= } \log _{ 10 }{ x } \ln { x } -\frac { 1 }{ \ln { 10 } } \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. ... | Since you need to find the relation between $a$ and $m$, you can consider that you face a problem of three equations for three unknowns $x,y,a$ for which the solution is unique $$\left\{x= \frac{m+1}{m},y= m+1,a= \frac{(m+1)^2}{m}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But ... | Hint:
$$f(x)=K(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)+x$$
For some constant K. Plug in $x=0$ to get K and then plug $x=7$ to get answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
} |
Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that
$$\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\... | Actually, mathematical induction is not necessary for this formula. A direct calculation with a telescoping decomposition into partial fractions will show where the formula comes from.
Indeed one checks that
$$\frac1{k(k+1)(k+2)}=\frac12\frac1k-\frac1{k+1}+\frac12\frac1{k+2}.$$
Thus the sum is:
\begin{align*}&\Bigl(\f... | {
"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": 2
} |
How to rotate a whole rectangle by an arbitrary angle around the origin using a transformation matrix? Suppose, I have a 2D rectangle ABCD like the following:
$A(0,0)$, $B(140,0)$, $C(140,100)$, $D(0,100)$.
I want to rotate the whole rectangle by $\theta = 50°$.
I want to rotate it around the Z-axis by an arbitrary an... | When you show $B'$ as $(5,7)$ it is rotated by about 54.5^\circ and the length of the side is now $\sqrt{74} \approx 8.6$, not $7$ This is a problem with your expectation, not the code. You also have the sign of the sine backwards if you are using row vectors-rotate $(1,0,1)$ by $45^\circ$ and it should be $(\frac 12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that a trigonometric equation has six distinct roots Show that,in general,the equation $A \sin^3x+B\cos^3x+c=0 $has six distinct roots,no two of which differ by $2\pi$,and that the tangent of their semi-sum is $-\frac{A}{B}$.
My attempt:
I tried to express it as sixth degree equation.
$A \sin^3x+B\cos^3x+C=0 $
$A... | By setting $u=\tan\frac{x}{2}$, we have to solve:
$$ A\left(\frac{2u}{1+u^2}\right)^3+B\left(\frac{1-u^2}{1+u^2}\right)^3 = -c $$
that is equivalent to finding the roots of the sixth-degree polynomial:
$$ p(u)=(c+B)+3(c-B)u^2+8A u^3+3(c+B)u^4+(c-B)u^6. $$
I do not agree that this polynomial has in general six real root... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying transfer functions in Z domain I have difficulties to check whether the below transfer function is recursive or non-recursive:
$$H(z)=\frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})}$$
I know that I have to multiply the num and denum by ${z^3}$ but the problem here is the denum. I think it must be first sim... | Starting from
$$H(z)=\frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})}$$
then
\begin{align}
H(z) &= \frac{z^{3}}{z^{3}} \cdot \frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})} \\
&= \frac{z^{3} - z^{2} + z - 3}{z-1} = \frac{z^{2} (z-1) + (z-1) -2}{ z-1} \\
&= 1 + z^{2} + \frac{2}{1-z} = 1 + z^{2} + 2 \, \sum_{n=0}^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the differential equation $\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$ given that $y(1)=0$. Solve the differential equation $\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$ given that $y(1)=0$.
I was able to find the solution to it by the method of solving homogeneous differential equati... | $$\left(xy\log\frac{x}{y}\right)dx+\left(y^2-x^2\log\frac{x}{y}\right)=0$$
$$\left(xy\log\frac{y}{x}\right)dx=\left(y^2+x^2\log\frac{y}{x}\right)dy$$ $$\left(\frac{y}{x}\log\frac{y}{x}\right)dx=\left(\left(\frac{y}{x}\right)^2+\log\frac{y}{x}\right)dy$$ Let $\frac{y}{x}=u \implies y=ux \implies \frac{dy}{dx}=u+x\frac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Behaviour of $\zeta(s)$ near $s=1$ I would appreciate if somebody could run this over and see if it works out? any suggestions or pointers would be appreciated. I denote the standard eta function $\eta$ by $\zeta^{*}$. I have not used big O notation and just used general well behaved functions. I do not wish to express... | It looks fine to me. Maybe it's interesting to note that from $$\zeta\left(s\right)=s\int_{1}^{\infty}\frac{\frac{1}{2}-\left\{ x\right\} }{x^{s+1}}dx+\frac{1}{s-1}+\frac{1}{2},\,\,\textrm{Re}\left(s\right)>0$$ where $\left\{ x\right\}$ is the fractional part of $x$, we can easily get $$\zeta\left(s\right)=\frac{1}{s-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means t... | If $x^2 + y^2 = 1$ then
$$
f(x,y) = 1 + x + y.
$$
Now, let $x=\cos t$, $y=\sin t$. So,
$$
f(t) = 1 + \cos t + \sin t = 1 + \sqrt2 \sin(t + \pi/4).
$$
Could you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
10th derivative of a function I want to find $f^{(10)}(0)$ where $f(x)=\ln(2+x^2)$.
I know that it can be done "by hand", but I believe there is a smarter way.
I think I should use Taylor series and the fact that $f^{(n)}(0)=a_n*n!$ , but I'm not sure how.
| We have
$$
\frac{2x}{x^2+2} = \frac{x}{1+\frac{x^2}{2}} = x\bigg(1-\frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{8} + \frac{x^8}{16}+\cdots \bigg).
$$
Then integrating we get
$$
\log(x^2+2)-\log2 = \int_0^x \frac{2t}{t^2+2}\; dt = \frac{x^2}{2}-\frac{x^4}{2\cdot 4} + \frac{x^6}{4\cdot 6} - \frac{x^8}{8\cdot 8} + \frac{x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises sectio... | To solve this problem, first of all think about the x values for which the following function is undefined.
We have,
$$f(x) = \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}$$
which can be written as,
$$f(x) = \frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)}$$
Note that the above function is undefined for
$$x = \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
If $\alpha, \beta, \gamma$ solutions for the equation If $\alpha, \beta, \gamma$ solutions for the equation
$$x^{3}+x^{2}+1=0$$
Then
$$\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma }=??$$
I know that the answer is $\frac{9}{13}$, It's just for sharing a new ideas, thanks :)
| Using the root-coefficient relationship, we have the following:
$$\alpha + \beta + \gamma = -1,$$
$$\alpha\beta\gamma = -1,$$
$$\alpha\beta + \beta\gamma + \gamma\alpha = 0.$$
The sum in question can be evaluated easily now.
\begin{align*}
\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ ... | Let us define two series. The first is
\begin{align}
a_1 &= 3 \\
a_2 &= 3 - \frac{2}{3} \\
a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\
a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\
&\vdots \\
a_{n+1} &= 3 - \frac{2}{a_n} \quad (*)
\end{align}
and
\begin{align}
b_1 &= 3 - 2 \\
b_2 &= 3 - \frac{2}{3-2} \\
b_3 &= 3 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
| Given $5x^2 −x−4=0 $ you must use, $a=5, b=-1, c=-4$ in the formula$$\begin{align}x & =\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[2ex] & = \frac{-(-1)\pm\sqrt{(-1)^2-4(5)(-4)}}{2(5)} \\[2ex] & = \frac{1\pm\sqrt{1+80}}{10}\\[2ex] & = \frac{1\pm 9}{10}\\ \therefore x\in \{-\tfrac {4}{5}, 1\}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Is there another way to compute this Laurent series? $\frac{1}{(1+z)^2}$ Consider $g(z) = \frac{1}{(1+z)^2}$. To compute a Laurent series of $g$ at $0$ on the region $|z| > 1$, I let $w = \frac{1}{z}$, so $$g(z) = \frac{w^2}{(1+z)^2w^2} = w^2 \frac{1}{(1+w)^2} $$ Since $|w| < 1$, we have $\frac{1}{1+w} = 1 - w + w^2 -... | HINT:
Note that $\frac{d}{dz}\left(\frac{1}{1+z}\right)=-\frac{1}{(1+z)^2}$. Find the Laurent series for $\frac{1}{1+z}$ (this is pretty easy), differentiate term by term, and multiply by $-1$ should do it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
evaluate the integral Evaluate the integral:
$$\int_0^{\pi} \frac{\cos 2\theta}{1 -2a\cos \theta +a^2}d\theta$$
The way I approach this problem is:
since, $\cos \theta = \frac{e^{it} + e^{-it}}{2}$; and $cos 2\theta = Re(z^2)$. Then, the integral will be written as follow:
$$\frac{1}{2}\int_0^{2 \pi} \frac{Re(z^2)}{... | Notice, the following expression
$$\frac{\cos 2\theta}{1-2a\cos \theta+a^2}=\frac{2\cos^2 \theta-1}{1-2a\cos \theta+a^2}=\frac{A\cos \theta(1-2a\cos \theta+a^2)+B(1-2a\cos \theta+a^2)+C}{1-2a\cos \theta+a^2}$$$$=A\cos \theta+B+\frac{C}{1-2a\cos \theta+a^2}$$
solving for $A, B, C$, we get $$A=-\frac{1}{a}, \ B=-\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$ Find the range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$.
By simply looking at the problem and simplifying trigonometrically,it looks as if range is zero but not.I think there is... | Using $\displaystyle |\sin x|=\left\{\begin{matrix}
\displaystyle \sin x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle \sin x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\sin x\;, & \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle -\sin x \;,& \displaystyle\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understa... | Let $$p(n):\displaystyle\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
Put $n=1\;,$ We get $$\displaystyle \frac{1}{1\cdot 3} = \frac{3}{4}-\frac{5}{2\cdot 2\cdot 3} = \frac{4}{12}$$
So it is true for $n=1$
Now Put $n=k\;,$ We get $$\displaystyle \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $ Prove that:
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $
What I've tried:
I solved the improper integral: $\int_0^\infty \frac{1}{1+x^2} = \lim_{b\to \infty} \arctan b -\arctan 0 = \frac{\pi}2 $.
Now, everything in the sum (and in the int... | An advanced approach using zeta function values $\zeta(2)$ and $\zeta(4)$ to get even better bounds.
For $n>1$, $$\frac{1}{1+n^2}=\frac{1}{n^2}\frac{1}{1+\frac{1}{n^2}} = \frac1{n^2}-\frac{1}{n^4}+\frac{1}{n^6}-\frac{1}{n^8}+...$$
So $$\frac{1}{n^2}-\frac{1}{n^4}<\frac{1}{n^2+1}<\frac{1}{n^2}$$
So:
$$\frac{3}{2}+\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
solving integral How can I solve this integral to get the result as follow:
$${\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r= {1\over{\sqrt{\pi t}}} \cos({\alpha\over2t}) $$
Thanks!
| Let
$$I(t) = {\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r $$
then let $x = (\alpha/2 r)$ to obtain
\begin{align}
I(t) &= \frac{\alpha \, \sqrt{\alpha}}{4 \pi} \left(\frac{\alpha}{2}\right)^{3/2} \, \int_{\alpha/2t}^{\infty} \left(\sin(x) + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)... | It's two equations, with two variables. We can do what we always do. You can write $y= 8/x$, and substitute into the first equation, obtaining $x^3 + 8^3/x^3 =72.$
From here, we want to write this as a polynomial, so we multiply through $x^3$ so there are no negative powers, and then move everything to the left. This g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Triangle Area problem I've been trying to solve the following:
Let $ABC$ be a triangle with sides $a, b $ and $ c$, inradius $r$ and exradii $r_a, r_b$ and $r_c$. If $A'B'C'$ is another triangle with sides $\sqrt{a}, \sqrt{b}$ and $\sqrt{c}$ show that $Area(A'B'C')=\frac {\sqrt{r(r_a+r_b+r_c)}} {2}$.
I tried to combin... | Using
$$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$
$$\Rightarrow \frac{a+b+c}{2}r\cdot\frac{1}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac 14$$
and
$$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{-a+b+c}{2}r_a=\frac{a-b+c}{2}r_b=\frac{a+b-c}{2}r_c$$$$\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve trigonometric equation $ \cot x + \cos x = 1 + \cot x \cos x $ Solve trigonometric equation: $$ \cot (x) + \cos (x) = 1 + \cot (x) \cos (x) $$
I tried to multiply both sides with $\sin x$ (which I'm not sure if I can multiply with sin).
| Notice, we have $$\cot x+\cos x=1+\cot x\cos x$$
$$\frac{\cos x}{\sin x}+\cos x=1+\frac{\cos x}{\sin x}\cos x $$
$$\cos x+\sin x\cos x=\sin x+\cos^2 x $$
$$\cos x+\sin x\cos x-\sin x-\cos^2 x=0 $$
$$\underbrace{\cos x-\sin x}+\underbrace{\sin x\cos x-\cos^2 x}=0 $$
$$(\cos x-\sin x)-\cos x(\cos x-\sin x)=0 $$
$$(1-\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why is this sum zero? I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$... | The partial sums are
$$\begin{eqnarray}
1/2&0\\
2/3&4/9&0\\
3/4&12/16&18/64&0\\
4/5&24/25&72/125&96/625&0\\
5/6&40/36&180/216&480/1296&600/6^5&0\end{eqnarray}$$
The ratios from one partial sum to the next are
$$\begin{eqnarray}0\\
2/3&0\\
4/4&(3/2)/4&0\\
6/5&(6/2)/5&(4/3)/5&0\\
8/6&(9/2)/6&(8/3)/6&(5/4)/6&0\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
Intersection between two three-dimensional planes The intersection of the planes defined by
$x \bullet \begin{pmatrix} 8 \\ 1 \\ -12 \end{pmatrix} = 35$
and
$x \bullet \begin{pmatrix} 6 \\ 7 \\ -9 \end{pmatrix} = 70$
is a line. Find an equation of this line.
I've been attempting this problem for hours and cannot solve ... | Let $x=(a,b,c)$. Then we can rewrite your equations as
$$\begin{align}
8a+1b-12c&=35 \\[2 ex]
6a+7b-9c&=70
\end{align}$$
Now we want to solve those simultaneous equations. Subtracting $3/4$ of the first equation from the second we get
$$\begin{align}
8a+1b-12c&=35 \\[2 ex]
0a+\frac{25}4b-0c&=\frac{175}4
\end{align}$$
D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What techniques would be use to prove $6x^2 +12x +8$ cannot be perfect cube for integer x > 0 I'm wondering if there are any basic techniques.
| Let $(x,y)$ be any integer solution to $y^3 = 6x^2 + 12x + 8$.
Since RHS is even, so does LHS and $y$. This implies $\text{RHS} \equiv 0 \pmod 8$.
It is easy to check
$$\text{LHS} \equiv
\begin{cases} 0,& x \text{ even}\\2, & x \text{ odd}\end{cases}
\pmod 8$$
So $x$ is also even. This implies existence of integers $X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
To compute improper integral $\int_3^{5}\frac{x^{2}\, dx}{\sqrt{x-3}{\sqrt{5-x}}}$ I am given improper integral as
$$\int_3^{5}\frac{x^{2}}{\sqrt{x-3}{\sqrt{5-x}}}dx$$
DOUBT
I see that problem is at both the end points, so i need to split up the integral. But problem seems to me that when i split up integral one of te... | Beta Function Approach
Substituting $x\mapsto2x+3$,
$$
\begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=\int_0^1\frac{(2x+3)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=\int_0^1\frac{(3(1-x)+5x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Modular maths: How do I find the remainder? How do I find the remainder of $5^{22} \pmod{25}$?
And also how do I find the remainder of $3^{16} + 7 \pmod{5}$?
| Try to see which power of $5$ can give $1$ or $-1$ modulo $7$. This will reduce your computations quite significantly. For example, $5^2 \equiv 4 \pmod{7}$ so $5^3 \equiv 20 \equiv 6 \equiv -1 \pmod{7}$. Thus
$$5^{22} \equiv 5^{21} \cdot 5 \equiv (5^{3})^7 \cdot 5 \equiv -5 \equiv 2 \pmod{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to express $\phi$ in terms of $R\text{, }x\text{ and }\theta$ Let $S$ be a circle with radius $R$ and center at $O$.
Let $P$ be any arbitrary point inside circle such that its distance from $O$ is $x$ and the ray $\overrightarrow{OP}$ cuts the circle $S$ at $M$.
Let $N$ be any other point on the circle such that $... | First, it's important to realise that we can choose a coordinate system such that $P$ and $M$ are on the horizontal axis passing through the centre $O$ (see figure below). This can be done without any loss of generality.
With respect to the centre $O$, then, the coordinates of the various points of interest are:
$$
P ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ ... | work Translation(unnecessary once you know how to do this.)
| 88 1 0 88 = 1(88) + 0(15)
-6 | 15 0 1 15 = 0(88) + 1(15), add -6 x this row to the above row
7 | -2 1 -6 -2 = 1(88) - 6(15), add 7 x this row to the above row
| 1 7 -41 1 = 7(88) - 41(15)
Hence -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$ Find:
$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$
I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What met... | If you know that $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0\tag{1}$$ then it is easy to give a simple evaluation for the limit in question
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve the sequences inequality If $a_1=1$ and $a_n=a_{n-1}+\dfrac{1}{a_{n-1}}$ for $n≥2$ , then prove that $12 < a_{75} < 15$ ?
I have tried solving this by:
$$a_{75} - a_1 = \frac{1}{a_1}+\frac{1}{a_2}....+\frac{1}{a_{74}}$$
| Use the hint given by Did:
We have that $a_n^2=a_{n-1}^2+\frac{1}{a_{n-1}^2}+2$. Thus $a_n^2\geq a_{n-1}^2+2$ for all $n\geq 2$. Hence $a_n^2\geq 2\cdot (n-1) +1$. Hence $a_{75}\geq \sqrt{2\cdot 74+1}\geq \sqrt{2\cdot 72}=\sqrt{144}=12$.
For the other inequality, use the following reasoning:
$$a_n^2=a_{n-1}^2+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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Fractions in Questions and Answers
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