Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.
Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(... | You have by distance from a point to a line formula,
$$|h-2|=|k-5|=\frac{|3h-4k-10|}{5}=r.$$
Consequently, $h=2\pm r$, and $k=5\pm r$. So we have four cases
*
*$h=2+r$, $k=5+r$ implies $|3(2+r)-4(5+r)-10|=|r+24|=5r \implies r=6$. So the equation of the circle is $$\color{red}{(x-8)^2+(y-11)^2=36}.$$
*$h=2+r$, $k=5-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Asymptotic notation question (logarithms) I am working through some problems and I have come across one I do not understand. Could someone clarify why
$$2x^3 + 3x^2\log(x) + 7x + 1$$
is $O(x^3\log(x))$ for $x>0$?
I guess I am missing some sort of knowledge to be able to answer this question :(
| if f(x) = $2x^3 + 3x^2\log(x) + 7x + 1$, f(x) = $x^3$$(2 + \frac{3\log(x)}{x} + \frac{7}{x^2} + \frac{1}{x^3})$. Then if x approaches the infinity, ($\frac{3\log(x)}{x} + \frac{7}{x^2} + \frac{1}{x^3}$) is negligible comparing to $x^3$.
And, if $f(x)$ = $2x^3 + 3x^3\log(x) + 7x + 1$, f(x) = $x^3log(x)$$(\frac{2}{log(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Evaluate the infinite product $\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$ $$\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$$
I've seen some similar questions asked. But this one is different from all these. Euler product does not apply. One cannot simply factorize $\left(1-\frac{2}{(2n+1)^2}\right)... | Considering the partial products
$$A_p=\prod_{n=1}^{p} \left(1-\frac{2}{(2n+1)^2}\right)$$ a CAS produced
$$A_p=-\cos \left(\frac{\pi }{\sqrt{2}}\right) \frac{\Gamma
\left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma
\left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma
\left(p+\frac{3}{2}\right)^2}$$ and u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answe... | $$\alpha'=\frac{\alpha}{\beta} ,\beta'=\frac{\beta}{\alpha}$$ now find sum and product of $\bf new$ roots
$$\quad{S'=\alpha'+\beta'\\p'=\alpha'\times \beta'\\
S'=\alpha'+\beta'=\frac{\alpha}{\beta} +\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{s^2-2p}{p}=\frac{(-\frac{-3}{2})^2-2.\frac{6}{2}}{\frac{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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How to prove $\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}$? $$
\quad{\forall x\in \mathbb{R}:\\
\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}}$$ I want to prove this.I tried to graph it and se... | There is a nice proof for $x\geq0$.
By Minkowski we obtain:
$$\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73}-\sqrt{157}=$$
$$=\sqrt 2\left(\sqrt{x^2} + \sqrt {{x^2} + x + \frac{1}{2}} + \sqrt {{x^2} - 5x + \frac{13}{2}} + \sqrt {{x^2} - 11x + \frac{73}{2}}-\sqrt{\frac{157}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| Vieta's theorem is the key, like in your previous question. It gives $\alpha+\beta=-\frac{5}{3}$ with no effort and
$$\alpha^3+\beta^3 = (\alpha+\beta)\left((\alpha+\beta)^2-3\alpha\beta\right) = -\frac{5}{3}\left(\frac{25}{9}-4\right) = \color{red}{\frac{55}{27}}$$
with very simple manipulations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do we use EEA and why do we multiply by 4 throughout? I am trying to understand the following example and solution, but I am confused by why we use EEA and why we need to multiply by $4$?
Example. Find a solution to
$$\begin{align}x &\equiv 88 \phantom{1}\mod 6 \\x &\equiv 100 \mod 15 \end{align}$$
Solution 1:
Fro... | This is how I would solve it.
$x \equiv 88 \pmod{6}$ implies $x = 88 + 6s$ for some integer $s$.
So $x \equiv 100 \pmod {15}$ becomes.
Eliminating $x$, we get
\begin{align}
x &\equiv 100 \pmod {15}\\
88 + 6s &\equiv 100 \pmod {15} \\
6s &\equiv 12 \pmod{15} &(\text{$3$ divides $6,12,$ and $15$})\\
2s &\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question about analytic geometry (vectors) Let $a , b, c$ be three vectors such that $a+3b+c = o$ where $o$ is $(0,0,0)$ vector. we also know $|a| = 3 , |b| = 4 , |c| =6$. we want to find $a.b + a.c + b.c$
I know we can solve it by saying $a+b+c = -2b$ and then squaring the sides. and at last we get the answer 3/2 . ... | Dot the equation with each of the vectors
\begin{eqnarray*}
\mid a \mid^2 +3 a \cdot b +a \cdot c =0 \\
a \cdot b +3 \mid b \mid^2 +b \cdot c =0 \\
a \cdot c +3 b \cdot c+\mid c \mid^2 =0 \\
\end{eqnarray*}
Now add the first and third equations and subtract the second $2(a \cdot b +a \cdot c +b \cdot c)+9+36-48=0$ ..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{4x}{x+7}I know how to solve the problem. The reason I post the problem here is to see whether there is a quick approach, rather than a traditional method, to solve the problem.
The problem is: find $x$ that satisfy $\frac{4x}{x+7}<x$
I considered two cases: $x+7>0$ and $x+7<0$, and then went through... | Transform $\frac{4x}{x+7}<x$ into the equivalent $-\frac{x^2+3x}{x+7}=\frac{4x}{x+7}-x<0$ or $\frac{x^2+3x}{x+7}>0$.
This reduces to question to finding the sign of $\frac{x^2+3x}{x+7}$, which can be solved by making a simple table:
$$
\begin{array}{c|ccccc}
& & -7 & & -3 & & 0 & \\
\hline
x & - & - & - & - & -& 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Conditional probability of multivariate gaussian I'm unsure regarding my (partial) solution/approach to the below problem. Any help/guidance regarding approach would be much appreciated.
Let $\mathbf{X} = (X_1, X_2)' \in N(\mu, \Lambda ) $ , where
$$\begin{align}
\mu &= \begin{pmatrix}
1 \\
1
... | https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions Just apply the result after you obtain the distribution of $\mathbf{Y}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find value of $\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$ Find value of
$$S=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$$
I started with $$S+\frac{1}{S}=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}+\frac{\sum... | Let
$$S_1=\sum_{k=1}^{m^2-1} \sqrt{m+\sqrt{k}}$$
and
$$S_2=\sum_{k=1}^{m^2-1} \sqrt{m-\sqrt{k}}.$$
(your sum is $S_1/S_2$ at $m=20$). Note that
$$\left(\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}\right)^2=2m+2\sqrt{m^2-k},$$
so
$$\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}=\sqrt{2}\sqrt{m+\sqrt{m^2-k}}.$$
Thus
$$S_1+S_2=\sum_{k=1}^{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find all points on which a function is discontinuous. $ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex] 0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $
The only point it could be discontinuous in is [0,0]. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \... | $x = r\cos \theta$, $y = r\sin \theta$
instead of $(x,y) \rightarrow (0,0)$ I can now use $r\rightarrow0$
$$\begin{align}
\lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2\cos^2\theta + r^2\sin^2\theta}
&=\, \lim_{r\to0} \frac{r (\cos^3\theta + \sin^3\theta)}{\cos^2\theta + \sin^2\theta} \\
&=\,0
\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer. Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer.
Well if this is an integer then $3x-1 \mid 2x^2-3$ so $2x^2-3=3x-1(k)$ such that $k\in \mathbb{Z}$ from here not sure where to go I know that it has no solutions I just can't see th... | Using the extended Euclidean algorithm in $\mathbb Q[x]$, we get
$$
25 = (-9)(2x^2-3)+(6x+2)(3x-1)
$$
Therefore, if $3x-1$ divides $2x^2-3$, then it divides $25$.
So, $3x-1 \in \{\pm 1, \pm 5, \pm 25 \}$ and there is not much to test.
An easier (but equivalent) argument is $2x^2-3 = (3x-1)q(x) + r$, where $r$ is the va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is $\int \frac{x}{x^5-7} dx$? I have tried out many trigonometric substitution like $x=\sin^{\frac{2}{5}}z$. But it did not work.
| Substitute $x=\sqrt[5]{7}u$, and you have
$$
\begin{align}
\int\frac{\sqrt[5]{7}u}{7u^5-7}\,\sqrt[5]{7}du
&=\frac{\sqrt[5]{49}}{7}\int\frac{u}{u^5-1}\,du\\
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{Bu+C}{u^2-2hu+1}+\frac{Du+E}{u^2-2ku+1}\right)\,du\\
\end{align}
$$
where $h=\cos(2\pi/5)=\frac{\sqrt{5}-1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Verifying a long polynomial equation in (the reciprocal of) the Golden Ratio I'm trying to show that the following equation holds true:
$$4\sigma^{12}+11\sigma^{11}+11\sigma^{10}+9\sigma^9+7\sigma^8+5\sigma^7+3\sigma^6+\sigma^5+\sigma^4+\sigma^3+\sigma^2+\sigma = 1 + 2\sigma$$
where $\sigma$ is the reciprocal of the g... | As $\sigma$ verifies the identity $\sigma^2=-\sigma+1$, you can perform the following reductions (the powers of $\sigma$ have been omitted for conciseness):
$$4+11+11+9+7+5+3+1+1+1+1-1-1,\\
7+15+9+7+5+3+1+1+1+1-1-1,\\
8+16+7+5+3+1+1+1+1-1-1,\\
8+15+5+3+1+1+1+1-1-1,\\
7+13+3+1+1+1+1-1-1,\\
6+10+1+1+1+1-1-1,\\
4+7+1+1+1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x... | Since
\begin{align}
\left|\frac{x^3+y^3}{x^2+y^2}\right|&=\left|(x+y)\cdot\frac{x^2-xy+y^2}{x^2+y^2}\right|\\
&=|x+y|\left|1-\frac{xy}{x^2+y^2}\right|\\
&\le|x+y|\left(1+\frac{|xy|}{x^2+y^2}\right)\quad(\because\text{the triangle inequality})\\
&\le|x+y|\left(1+\frac{1}{2}\right)\quad(\because\text{the AM-GM inequality... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\... | You are almost done. Note that the three series on the LHS of your last line are convergent and they can be written as
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$
which is equal to
$$
{1\over 3}\sum_{n=1}^{2}{1\over 2^n-1}-\sum_{n=2}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Can someone tell me what should i do next? Should I use the inequality between Arithmetic and Geometric mean? Problem 1: Let $a,b,c> 0$
$ab+3bc+2ca\leqslant18$
Prove that:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$.
I started on this way:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$
$\frac{3bc+2ac+ab}{... | By the AM-GM inequality, we have $$\left(\frac{ab+3bc+2ac}{3}\right)^3\ge 6{(abc)}^2$$
$$\Rightarrow \frac 1{abc}\ge \sqrt{\frac{3^3.6}{(ab+3bc+2ac)^3}}$$
$$\Rightarrow \frac{ab+3bc+2ac}{abc}\ge \sqrt{\frac {3^3.6}{ab+3bc+2ac}}\ge\sqrt{\frac{3^3.6}{18}}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Summing up $3+5+9+17+...$
Find the sum of sum of $3
+5+9+17+...$ till $n$ terms.
Using Method of differences, the sum of the series is
$$\sum\limits_{j=1}^n 2^{j-1}+n$$
I am facing difficulty in evaluating $$\sum\limits_{j=1}^n 2^{j-1}$$. How do I do that?
Now I have $2^0 + 2^1 + 2^2 ... 2^{n-1}$
The sum of this ser... | Maybe do you mean the following?
$$3+5+9+17+...=\sum_{k=1}^n2^k+n=\frac{2(2^n-1)}{2-1}+n=2^{n+1}-2+n.$$
If so let $S_n=3+5+...+(2^n+1).$
Thus, $$S_n-S_1=\sum_{k=2}^n(S_k-S_{k-1})=\sum_{k=2}^n(2^k+1)=\frac{4(2^{n-1}-1)}{2-1}+n-1,$$
which gives $$S_n=2^{n+1}+n-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solve integral $\int\frac{dx}{\sin x+ \cos x+\tan x +\cot x}$ I need to find:
$$\int\frac{1}{\sin x+ \cos x+\tan x +\cot x}\ dx$$
My attempts:
I have tried the conventional substitutions. I have tried the $\tan(x/2)$ substitutions, tried to solve it by quadratic but nothing has worked so far.
| Let $t=x-\frac\pi4$ and recognize $2\sin x\cos x= 2\cos^2t-1$
$$I= \int\frac{dx}{\sin x+ \cos x+\tan x +\cot x}
=\frac1{\sqrt2}\int \frac{\cos^2t-\frac12}{\cos^3t-\frac12\cos t+\frac1{\sqrt2}}dt
$$
Factorize the denominator ($y=\cos t$)
$$y^3-\frac12y+\frac1{\sqrt2}
=(y+a)\left(y-\frac {a-ib}2 \right) \left(y-\frac {a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Laurent series of algebraic function Consider function
$$f(z)=\frac{1}{\sqrt{a^2-z^2}+b\sqrt{c^2-z^2}}.$$
It has pole
$$z = \frac{\sqrt{b^2 c^2 - a^2}}{\sqrt{b^2-1}}$$.
I wanna find Laurent series of $f$ around singular point.
For $b=1$, I put $f$ in partial fraction to proced.
How to find Laurent series for $b \ne 1$... | Hint:
$$
\frac{1}{\sqrt{a^2-z^2}+b\sqrt{c^2-z^2}}=\frac{\sqrt{a^2-z^2}-b\sqrt{c^2-z^2}}{a^2-z^2-b^2(c^2-z^2)}=\frac{\sqrt{a^2-z_0^2-(z^2-z_0^2)}-b\sqrt{c^2-z_0^2-(z^2-z_0^2)}}{(b^2-1)(z^2-z_0^2)},
$$
where
$$
z_0 = \frac{\sqrt{b^2 c^2 - a^2}}{\sqrt{b^2-1}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square. Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the... | Here is a solution using less machinery than in Will Jagy's answer.
As Daniel Robert-Nicoud noted in comments, the fact that $3^a+7^b\equiv(-1)^a+(-1)^b$ mod $4$ implies $a$ and $b$ must have opposite parity in order for the (even) sum $3^a+7^b$ to be a perfect square. So we seek to show that $(a,b)=(1,0)$ is the only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe ... | Hint
$${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.=\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j $$
$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} x^{j+1} \binom n j =\int \sum_{j=0}^{n} (-1)^j x^{j}\binom n j =\int (1-x)^n=- \frac { (1-x)^{n+1}} {n+1} $$
$$\sum_{j=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$\varepsilon$-$N$ proof question
Use the definition of limits to prove $$\lim_{n\to\infty} \frac{n^3-n^2+9}{n^3-3n} =1 $$
So far I got
$$ \left| \frac{n^3-n^2+9}{n^3-3n} -1 \right| < \epsilon, $$
the left hand side is the same as
$$ \left| - \frac{n^2-3n-9}{n(n^2-3)} \right|$$
which we can then drop the absolute v... | Since you choose $n$ large anyway, you may assume $n^2 - 3 > \frac{1}{2} n^2$ (that is, you choose $n>\sqrt 2$, or simply $n>2$). Then
$$\frac{n}{n^2 - 3}< \frac{n}{\frac 12 n^2} = \frac{2}{n}.$$
Now we want $\frac 2n<\epsilon$. So we need $n >\frac {2}{\epsilon}$.
Combining the two information, one choose
$$N = \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Basic modular inverses I know how to do modular inverses in a hypothetical sense with the Euclidian method, and have been trying to do the, but I seem to keep getting the incorrect answer.
I'm trying to find the inverse of $\;5\pmod {13}$, for example.
The answer should be 8, but I can't seem to get that. These ar... | Ok, after applying the extended Euclidean algorithm applied to $13$ (modulus) and $5$ (the number you want to invert)
you will find that
$$1 = 13\cdot 2 + -5 \cdot 5$$
as stated. But taking this whole equation modulo $13$, we get
$$1 \equiv 13\cdot 2 + -5 \cdot 5 \equiv -5 \cdot 5 \equiv 8 \cdot 5 \pmod{13}$$
using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find two arithmetic progressions of three square numbers I want to know if it is possible to find two arithmetic progressions of three square numbers, with the same common difference:
\begin{align}
\ & a^2 +r = b^2 \\
& b^2 +r = c^2 \\
& a^2 +c^2 = 2\,b^2 \\
\end{align}
and
\begin{align}
\ & d^2 +r = e^2 \\
& e... | $$(a,b,c,d,e,f,r)=(1,29,41,23,37,47,840)$$
satisfies
$$a^2 +r = b^2,\quad b^2 +r = c^2,\quad a^2 +c^2 = 2b^2$$
$$d^2 +r = e^2,\quad e^2 +r = f^2,\quad d^2 +f^2 = 2e^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the pdf of $Y=X^2-X$ given that the pdf of $X$ is $f_X(x) = 3x^2I_{(0,1)}(x)$
Consider a random variable $X$ having pdf
$$f_X(x) = 3x^2I_{(0,1)}(x)$$
Give the pdf of $Y=X^2−X$ and the value of $E(Y)$.
I have from an inverse calculator that if $Y=X^2−X$ then $X = {1\over2}(1\pm \sqrt{4Y+1})$. We have,
$F_Y(y)... | Sketch the graph $Y=X^2-X$, one can see that
$$0 \ge Y \ge -\frac{1}{4}$$
and there are two values of $X$ corresponding to each $Y$, except for $Y=-\frac{1}{4}$.
By transformation method formula
$$f_Y(y)=f_X(x_1)\bigg|\frac{dx_1}{dy}\bigg|+f_X(x_2)\bigg|\frac{dx_2}{dy}\bigg|$$
$$=\frac{3}{8}(1-\sqrt{4y+1})^2\bigg|\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find a + b + c + bc. My math trainer friend ask my help to look for the shortest possible solution for this problem:
Let $a$, $b$ and $c$ be positive integers such that
$$\left \{\begin{matrix}a + b + ab = 15 \\
b + c + bc = 99 \\
c + a + ca = 399\end{matrix}\right. $$
Find $a + b + c + bc.$
I tried elimination but... | Adding $1$ to each equation, we get that
$$\left \{\begin{matrix}a + b + ab +1 = 16 \\
b + c + bc+1 = 100 \\
c + a + ca+1 = 400\end{matrix}\right. $$
Factoring, we get
$$\left \{\begin{matrix}(b+1)(a+1) = 16 \\
(b+1)(c+1) = 100 \\
(c+1)(a+1) = 400\end{matrix}\right. $$
Dividing equation 2 from equation 3 and multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How should one proceed in this trigonometric simplification involving non integer angles? The problem is as follows:
Find the value of this function
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$
when $\omega=33^{\circ}{20}'$ and $\phi=... | There is a mistake in your expansion of the second bracket.
We should have
$A=\cos^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2\frac{\phi}{2}$
$=2+2(\cos\frac{\omega}{2} \cos\frac{\phi}{2} -\sin\frac{\omega}{2}\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Need Clarification on Trig Substitution $$\int\frac{x}{(1+x^2)^\frac{3}{2}}\ \mathrm dx$$
becomes
$$\sin\theta = \frac{x}{\sqrt{x^2+1}}$$
I am aware that we would use $x = a^2\tan^2\theta$. and that our answer before we plug in from the triangle becomes $\sin\theta$.
I just took a test and got the answer of
$$\sin\th... | I'd write $\displaystyle \int \frac 1 {(x^2+1)^{3/2}} \Big( x\, dx\Big) = \int \frac 1 {u^{3/2}} \left( \frac 1 2\,du\right) = \cdots,$ etc.
But if you must use a trigonometric substitution you can do this:
\begin{align}
x & = \tan\theta \\
dx & = \sec^2\theta\,d\theta \\[10pt]
\int \frac x {(1+x^2)^{3/2}} \,dx & = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $f(x)=\sqrt{x^2+2x+3}.$ Completing the square and letting $t=x+1$, I obtain $$\int\sqrt{(x+1)^2+2} \ dx=\int\sqrt{t^2+2}\ dt.$$
Letting $u=t+\sqrt{t^2+2},$ I get
\begin{array}{lcl}
u-t & = & \sqrt{t^2+2} \\
u^2-2ut+t^2 & = & t^2+2 \\
t & = & \frac{u^2-2}{2u} \\
dt &=& \frac{u^2+2}{2u^2}du
\end{array}
Thus ... | Compute
\begin{align}
\frac{(t+\sqrt{t^2+2})^2}{8}-\frac{1}{2(t+\sqrt{t^2+2})^2}
&=
\frac{(t+\sqrt{t^2+2})^2}{8}-\frac{(t-\sqrt{t^2+2})^2}{2(t^2-(t^2+2))^2}
\\[6px]
&=
\frac{t^2+2t\sqrt{t^2+2}+(t^2+2)-t^2+2t\sqrt{t^2+2}-(t^2+2)}{8}
\\[6px]
&=\frac{t\sqrt{t^2+2}}{2}
\end{align}
Do you see it, now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Implicit Differentiation of this function [If
$x^2 + xy + y^3 = 1$,
find the value of
$y'''$
at the point where
x = 1.]1
Am i going the right direction, if so how will know what is the values of y' and y''
| It's easier, I think, to avoid division as much a possible, thus:
With
$x^2 + xy + y^3 = 1, \tag 1$
and $x = 1$,
$1 + y + y^3 = 1; \tag 2$
thus
$y(y^2 + 1) = y^3 + y = 0; \tag 3$
thus, if we assume $y \in \Bbb R$,
$y = 0; \tag 4$
that is, we have
$(x, y) = (1, 0). \tag 5$
We find, taking $d/dx$ of (1),
$2x + y + xy' + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac... | For modest requirements of accuracy, we can use a quite basic and easy to understand convergence acceleration technique to transform the Gregory/Leibniz series into something that yields the result with little computation.
If we have an alternating series
$$\sum_{n = 0}^{\infty} (-1)^n a_n$$
where the $a_n$ are slowly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 11,
"answer_id": 8
} |
Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$
| For $\epsilon >0$ sufficiently small $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon \Leftrightarrow \left|\dfrac{-7}{9n^2+3}\right|<\epsilon $$ $$ \Leftrightarrow \dfrac{7}{9n^2+3}<\epsilon \Leftrightarrow \frac{7}{\epsilon}-3<9n^2\Leftrightarrow \frac{7-3\epsilon}{9\epsilon}<n^2\Leftrightarrow \sqrt{\frac{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to prove that $n=x^2+y^2$ if it is a product of primes congruent 3 mod 4 and 1,2 mod 4 I am stuck trying to show that if
\begin{equation}
n=\prod p_i^{2e_i}\prod q_j^{f_j}
\end{equation}
where each $p_i\equiv3(mod~4)$ and each $q_j\equiv 1,2(mod 4)$ then each $p_i^{2e_i}$ and each $q_j^{f_j}$ is the norm of an ele... | This is a simple model to show the solution.
Consider following product of primes in forms $(4k+1) and (4k+3)$:
$$n=(4 k_1+1)(4 k_2+3)(4 k_3+1)(4 k_4+3). . .$$
$$(4 k_1+1)(4 k_2+3)≡ 3 mod 4$$
$$(4 k_3+1)(4 k_4+3)≡ 3 mod 4$$
$$⇒ (4 k_1+1)(4 k_2+3)(4 k_3+1)(4 k_4+3)≡ 1 mod 4$$
Or $$(4 k_1+1)(4 k_2+3)(4 k_3+1)(4 k_4+3) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486942",
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"source": "stackexchange",
"question_score": "1",
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How to prove $1+{x\over 2}-{x^2 \over 8}<\sqrt{1+x}$ for all $x>0$? How to prove $1+(1/2)x-(1/8)x^2<\sqrt{1+x}$ for $x>0$ by Taylor Expansions up to $n$ order or Mean Value Theorem? I tried to apply MVT on $\sqrt{1+x}$ and get $\displaystyle \sqrt{1+x}=1+\frac{1}{2\sqrt{1+\xi}}x$ for $\xi\in (0,x)$. How to do next?
| $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{\left(\sqrt{x+1}\right)'''_{x=\xi}x^3}{6}$$ and since $\left(\sqrt{x+1}\right)'''>0$ for all $x>0$, we are done!
A proof without Taylor series.
If $1+\frac{x}{2}-\frac{x^2}{8}<0$ or $x>2+2\sqrt3$ then it's obvious.
But for $0<x\leq2+2\sqrt3$ it's enough to prove that
$$(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadra... | You can find roots of $z^4+4=0$. For example, one easy way is to multiply the equation by $z^4-4$ to get $z^8 - 16 = 0$, so we conclude that roots of $z^4+4$ are $8$th roots of unity (times $\sqrt 2$) that are not $4$th roots of unity, i.e. the roots are $$\{\omega\sqrt 2,\omega^3\sqrt 2,\omega^5\sqrt 2,\omega^7\sqrt 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
... | At your step
$$3=a^2+2\sqrt 2 ab +2b^2$$, you can rearrange it to be
$$3-a^2-2b^2=2\sqrt 2 ab$$, the left hand side is a rational number while the right hand side is irrational, which you can't do unless a and b are zero, which does not satisfy the first condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this? I tried expressing the 2nd equation into 2 fractions which contain the first summation but i could... | The usual trick: separate even and odd indices of the sum (things converge absolutely, so you can).
We have $$\frac{\pi^2}{6}
= \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n)^2}
+\sum_{n=0}^\infty \frac{1}{(2n+1)^2}
$$
but
$$
\sum_{n=1}^\infty \frac{1}{(2n)^2} = \sum_{n=1}^\infty \frac{1}{4n^2}
= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Combinatoric problem with balls
Suppose there are $10$ balls in an urn, $4$ blue, $4$ red, and $2$ green. The balls are also numbered $1$ to $10$. How many ways are there to select an ordered sample of four balls without replacement such that the number B $\geq 0$ of blue balls, the number $R \geq 0$ of red balls, an... | You've got the color combinations correct: $RBG = [(3,1,0), (1,3,0), (3,0,1), (0,3,1)]$.
For the first color combination, you have $4$ sets of numbers you can draw for the reds, and for those numbers, $6$ orders to draw them in. For the one blue, you have $4$ choices for the number, and $4$ choices for when you draw it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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l'Hôpital vs Other Methods Consider the first example using repeated l'Hôpital:
$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} =... | You haven't checked whether L'Hopital could be applied each time.
| {
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"url": "https://math.stackexchange.com/questions/2499036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 1
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Modular arithmetic and probability $a$, $b$ and $c$ are chosen from $1$-$999$ with order and replacement, what is the probability that $a^2 + bc$ is divisible by $3$? (each choice is equally likely)
I split it up into 2 cases
$a^2$ has a remainder of $1$ while $bc$ has a remainder of $2$
$a^2$ has a remainder of $1$, $... | The error is here:
$a^2$ has a remainder of $0$ 1/3 of the time while $bc$ has a remainder of $0$ if $b$ or $c$ have remainders of $0$. probability $1/3 \times 1/3 \times 2 = 2/9$
You multiplied these numbers, which would be appropriate if you had "...if $b$ and $c$ have remainders of $0$." Since it's or instead, thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the value of ''a'' and ''b'' for which function is always increasing
Question Let f(x)= x$^{3}+ax^{2}+bx+5sin^{2}x$ be an increasing
function$\forall$ x $\in$$\mathbb{R}$$ Then$
(a) $a^2 -3b-15$<0
(b) $a^2 -3b-15$>0
(c)$a^2 -3b+15<$0
(d)a>0,b>0
My Approach i tried by making perfect square of the derivative
$\lef... | When we differentiate the function we get
\begin{align*}
3x^2+2ax+b+10\sin(x)\cos(x)=3x^2+2ax+b+5\sin(2x)
\end{align*}
Now as the function should be increasing, its differential should be everywhere nonnegative. But if it is everywhere nonnegative, so will
\begin{align*}
3x^2+2ax+b-5
\end{align*}
as $\sin(2x)$ is betwe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Olympiad inequality
Let $a,b,c,d \in [0,1]$.
Prove
$$ \frac {a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}+abcd\le3$$
Hi. Im a school student so ive been unable to look for ways to apply this problem. I attempted to use AM-GN on 1+a etc but the bound $2\sqrt{a}$ was too large and was easily larger than 3. Th... | For $a,b,c,d \in [0,1]$, let
$$
f(a,b,c,d)
=
\frac{a}{1+b}
+
\frac{b}{1+c}
+
\frac{c}{1+d}
+
\frac{d}{1+a}
+
abcd
$$
The goal is to prove $f(a,b,c,d) \le 3$, for all $a,b,c,d \in [0,1]$.
For fixed $b,c,d \in [0,1]$, let $g(a) = f(a,b,c,d)$.
Then
$$g''(a) = \frac{2d}{(a+1)^3}$$
hence $g''(a) \ge 0$, for all $a \in [0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$. How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?
I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be... | The product $p(p+1)$ of two consecutive numbers is divisible by $2$
because either $p$ or $p+1$ is even.
The product $p(p+1)(p+2)$ of three consecutive numbers is divisible by $3$
because either $p,\ p+1$ or $p+2$ is a multiple of $3$.
$f(n)=n(n^2-1)=(n-1)n(n+1)$ is then divisible by $3$.
Also for $n=2p+1$ odd then $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $|\sin x−\cos x|≤ 2$ for all $x.$ Please help me out in answering this:
Show that $|\sin x−\cos x|≤ 2$ for all $x.$
I don't know where to start and I think we have to use the mean value theorem to show this.
| I have another method for you if you want to use some straight-up calculus optimization. Let us first determine the extrema of $\sin(x)-\cos(x)$. To do this, we take the derivative and set it equal to $0$. In this case, the derivative is $\cos(x) +\sin(x)$, so we want to solve $\cos(x) + \sin(x) = 0$, or $\cos(x) = -\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was havi... | Let $F(a,b,c,d) = a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac25$.
With help of a CAS, one can verify
$$\begin{align}
F\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)
&= p^2 - pq + q^2 - qr + r^2 -rs + s^2\\
&= \frac12\left(p^2 + (p-q)^2 + (q-r)^2 + (r-s)^2 + s^2\right)\end{align}$$
If one set $(a,b,c,d)$ to $\left(\frac15+p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Splitting field of $x^3-3x+1$ I know there is a trick in that you can use $\omega + \omega ^{-1}$ where $\omega$ is a 9th root of unity but I can't see how this relates. I know this question has already been asked on here but there was no solution given using the trick.
| $x = z + \frac 1z\\
x^3 - 3x + 1 = z^3 + \frac 1{z^3} + 1 = \frac {1}{z^3} (z^3 + e^\frac {2\pi i}{3})(z^3 + e^\frac {-2\pi i}{3})\\
x^3 - 3x + 1 = (x + 2\cos \frac {2\pi}{9})(x + 2\cos \frac {4\pi}{9})(x + 2\cos \frac {8\pi}{9})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Completing squares with three variables. I want to complete the squares for this polynomial
$2x^2+2y^2-z^2+2xy+3xz-4yz$
Is there any kind of easy and non-confusing way to solve it?
I’ve done this up until now:
$$2x^2+2y^2-z^2+2xy+3xz-4yz$$
$$2x^2+2xy+3xz-4yz+2y^2-z^2$$
$$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$
$$2(x^2+... | If you mean writing polynomial as the sum of squares, here is one solution:
$p=2x^2+2y^2-z^2+2xy+3xz-4yz$
polynomial p can be written as following forms:
$p=(x+y)^2+(x+z)^2+(y-2z)^2-5z^2+xz$
$p=(x+y)^2 +(x+2z)^2+(y-2z)^2-5z^2 -xz$
Summing these relations and dividing the result by 2 we get:
$p= (x+y)^2 +\frac{1}{2}(x+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$
I have to solve this system of equations \begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}
From the third equation I have done this to manipulate it:
$$ x^3+y^3-z^3=43-3xy \implies (x+y)(x^2-xy+y^2)-z^3=43... | $$\begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}\Rightarrow z^2+5=z^2+2z+1\Rightarrow z=2$$
Because of $x^3+y^3=(x+y)(x^2-xy+y^2$ it follows $$\begin{cases} x+y=3 \\ x^2+y^2=17 \end{cases}\Rightarrow (x,y)=(-1,4),(4,-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simple algebra derivation I am reading a paper and came across, what the author claims, is simple algebra. I made a few attempts, but have struggled. The equivalence claim is
$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
| What exactly is your problem with that expression? You can't see why the two statements on the left and right sides of the equals sign are equivalent?
\begin{align}\require{cancel}
\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}
&=\frac{\cancel{n}y}{\cancel{n}(n+4)} + \left(\frac{n+4}{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Why is the limit of $\frac{1}{n}$ is $0$ however the series $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent? As I recall, according to the test for divergence, if you have a series $\sum_{n=1}^{\infty}a_{n}$ and if the limit of $a_{n}$ is $0$, then the series is convergent. the limit of $\frac{1}{n}$ is $0$. However, if... | Actually, if $a_{n}$ does not tend to $0$, then $\sum_{n=1}^{\infty}a_{n}$ is divergent. The opposite is not true.
A good example is $a_n = 1/n$. To visualize this, consider the series,
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $A^2=B$ where $B$ is the $3\times3$ matrix whose only nonzero entry is the top right entry Find all the matrices $A$ such that $$A^2= \left( \begin {array}{ccc} 0&0&1\\ 0&0&0 \\ 0&0&0\end {array} \right) $$ where $A$ is a $3\times 3$ matrix.
$A= \left( \begin {array}{ccc} 0&1&1\\ 0&0&1 \\ 0&0&0\end {array} \ri... | You have $A^4=0$, and since it is a $3\times 3$ matrix it follows that $A^3=0$. This means that
$$A\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0\end{pmatrix}=0$$
and
$$\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0\end{pmatrix}A=0$$
Giving that
$$A=\begin{pmatrix}
0&a&b\\
0&c&d\\
0&0&0\end{pmatrix}$$ Now we have
$$A^2=\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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A tangent line to $y=\frac1{x^2}$ cuts the $x$-axis at $A$ and $y$ at $B$, minimize $AB$. The problem:
A tangent line to $y= \frac{1}{x^2}$ intersects the x-axis at the point A and the y-axis at the point B.
What is the length of the shortest such line segment AB?
I know that the graph of $y= \frac{1}{x^2}$ looks lik... | Some preliminary theory: If a line tangent to a given function intersects the x-axis at a point A and the y-axis at a point B, then it must form a right triangle with the right angle situated at the origin and the two sides with vertices located at $(A, 0)$ and $(0, B)$ of lengths $A$ and $B$ respectively. The length o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2517748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$ Calculate the following limit :
$$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$
This is what I have tried:
Using Maclaurin series for $ (1+x)^a $:
$$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$
Usi... | Hint. You need a longer expansion of $\sqrt{1+x^2}$ and $\cos(x)$:
$$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\quad,\quad
\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4).$$
Then
\begin{align}\sqrt{1+x^2}\cos(x)&=\left(1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\right)\left(1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2... | By I.H. we have $2^{3n}-3^{n}=5k$ for some integer $k$ and thus $2^{3n}=3^{n}+5k$.
Now we have:
$$2^{3(n+1)}-3^{n+1} = 8\cdot 2^{3n}-3^{n+1}= 8(3^{n}+5k)-3^{n+1} = 3^n(8-3)+40k = 5\cdot (3^n+8k)$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Write formula for matrix in terms of Fibonacci numbers How could I express this matrix in terms of Fibonacci numbers? It seems like I'd have to use induction once I have a candidate for a formula but I'm unsure of where to start with expressing the matrix in terms of Fibonacci numbers.
Thanks in advance!
Let $T:\math... | Let us try to do it for $n = 1$. In that case, $T(e_1) = e_2$ and $T(e_2) = e_1 + e_2$, so this gives the matrix $\begin{pmatrix}0 \quad 1 \\ 1 \quad 1\end{pmatrix}$ for $T$.
If we have to find $T^n$ now, the first thing that we do is to find some elementary powers of $T$. Let's try to find $T^2$:
$$
T^2 = \begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify the expression: $(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$ Simplify the expression:
$$(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$$
I used the general method:
$(2x+1)(2x^2+2x+1)(2x^4+4x^3+6x^2+4x+1)×...... | A method like this can work:
$$\left( x+1-x\right)×\left(x+1+x \right)×\left((x+1)^2+x^2 \right)×\left( (x+1)^4+x^4\right)×...×\left((x+1)^{64}+x^{64}\right)=(x+1)^{128}-x^{128}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.
The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$
I ... | Let's do it without any tricks using standard results from the usual theory of polynomial equations. If $\alpha$ is a root of $f(x) =0$ then $\beta=\alpha-1$ is a root of $f(x+1)=0$. Thus it follows that $y_{i} =(x_{i} - 1)$ are the roots of equation $$(y+1)^{n}+(y+1)^{n-1}+\dots+(y+1)+1=0$$ ie $$y^{n} +(n+1)y^{n-1}+\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Approximation of exponential function by power series Let $x \in (-\frac{1}{2},\frac{1}{2}), n \in \mathbb{N}$
How can I choose a $n$ that the the inequality is valid?
$$\left|e^x-\sum_{k=0}^n \frac{x^k}{k!}\right| \leq \frac{|e^x|}{10^{16}}$$
My ideas:
Try some values for $n$ and verify the inequality for value greate... | $$e^x-\sum_{k=0}^n \frac{x^k}{k!}= \sum_{k=n+1}^\infty \frac{x^k}{k!}= \frac{x^{n+1}}{(n+1)!}+ \frac{x^{n+2}}{(n+2)!}+\cdots = x^{n+1} {\left[ \frac{1}{(n+1)!}+ \frac{x}{(n+2)!}+\frac{x^2}{(n+3)!}+\cdots\right]} $$
$$\left|x^{n+1} {\left[ \frac{1}{(n+1)!}+ \frac{x}{(n+2)!}+\frac{x^2}{(n+3)!+\cdots}\right]}+\cdots\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2534119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Rational series for positive integer root of $x>1$ With reference to the method @MarkViola used to prove a formula for finding the square root of $x>1$ in Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$
I have now found a formula for generating a rational series for ... | $\def\peq{\mathrm{\phantom{=}}{}}$Note that for $0 < y < 1$, using the binomial series and $Γ(z + 1) = zΓ(z)$,\begin{align*}
(1 - y)^{-\frac{1}{r}} &= \sum_{n = 0}^∞ \binom{-\frac{1}{r}}{n} (-y)^n = \sum_{n = 0}^∞ \frac{\left( -\dfrac{1}{r} \right) \cdots \left( -\dfrac{1}{r} - n + 1 \right)}{n!} (-y)^n\\
&= \sum_{n = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
$$\cot A = 2\\
\frac{\cos A }{ \sin A }= 2\\
\cot B = \frac34\\
\frac{\cos B }{ \sin B} = \frac34$$
What should I do with $A+B+C=180^\circ$?
| You have
$$ \sin C = \sin \big(180^\circ -(A+B)\big) = \sin (A+B) $$
Then, using angle-sum formulas to obtain
$$ \cot (A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} = \frac{2}{11} $$
And
$$ \csc^2 (A+B) = \cot^2 (A + B) + 1 = \frac{125}{121} $$
You know what to do next.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$ The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, I n... | For your question about how $\sum{\frac{2}{n(n+1)}}$ is split,
$$\frac{2}{n(n+1)} = \frac{2[(n+1)-n]}{n(n+1)}$$
$$\frac{2[(n+1)-n]}{n(n+1)} = \frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)}$$
$$\frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$$
So your question about you would know this in an exam sit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$
My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\... | $$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(\frac{x}{1-x^2}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}$$
Now for Taylor's expansion we have that
$$\lim\limits_{x \to 0}\frac{1}{1-x^2}=1-x^2+o(x^2)$$
So
$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Getting x in terms of y I have this equation:
$$\dfrac{x}{y} = \dfrac{y-x}{x}$$
How would I separate $x$ and $y$ in $x^2+xy-y^2=0$ ?
| If
$\dfrac{x}{y} = \dfrac{y-x}{x}, \tag 1$
then, assuming $x \ne 0 \ne y$, as we must for the equation to make sense, we may set
$\alpha = \dfrac{x}{y}, \tag 2$
and writing (1) as
$\dfrac{x}{y} = \dfrac{y}{x} - 1, \tag 3$
we find
$\alpha = \alpha^{-1} - 1, \tag 4$
or
$\alpha^2 = 1 - \alpha, \tag 5$
or
$\alpha^2 + \alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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find a basis for $R^4$ given the vectors. i need help with this excercise..
find a basis for $R^4$ given the vectors.
$$\begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}\quad \text{and}\quad \begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix}$$
What... | Clearly, $v_1$ and $v_2$ are linearly independent, then you can add a new vector $v_3$ that is not in the span of
$$v_1=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}\quad \text{and}\quad v_2=\begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad... | $$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.$$
From first equation,
$$2x + 2\sqrt{x^2 - y^2} = a^2 \tag 2$$
$$(x- a^2/2)^2 = x^2 - y^2$$
$$x^2 + \dfrac{a^4}4 - xa^2 = x^2 - y^2 $$
$$-\dfrac{a^4}4 + xa^2 = y^2 \tag 3$$
Add (2) to second equation
$$ x... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$ The question is to evaluate $$\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$$ if $x^2 \neq (n \pi-1) \forall n\in N$
I tried to rewrite the integral as $$\int \frac{x\sqrt{(2\sin(x^2+1))(1-\cos... | Note that the substitution $x^2+1=y$ and $2x\,\mathrm dx=\mathrm dy$ leads to\begin{align}\frac12\int\frac{\sqrt{2\sin(y)-2\sin(y)\cos(y)}}{2\bigl(\sin(y)+2\sin(y)\cos(y)\bigr)}\,\mathrm dy&=\frac{\sqrt2}4\int\frac{\sqrt{\sin(y)-\sin(y)\cos(y)}}{\sin(y)+2\sin(y)\cos(y)}\,\mathrm dy\\&=\frac{\sqrt2}4\int\frac{\sqrt{\csc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| The unpopular limit formula $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}\tag{1}$$ is your friend here. First divide the numerator and denominator of the expression by $x-7$ and note that the denominator itself becomes a fraction $$\frac{\sqrt[4]{x+9}-2} {x-7}=\frac{t^{1/4}-16^{1/4}}{t-16}$$ where $t=x+9\to 16$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Finding the turning points of $f(x)=\left(x-a+\frac1{ax}\right)^a-\left(\frac1x-\frac1a+ax\right)^x$ I've just come across this function when playing with the Desmos graphing calculator and it seems that it has turning points for many values of $a$.
So I pose the following problem:
Given $a \in \mathbb{R}-\{0\}$, fin... | I've made some progress on the question, but I'm nowhere close to solving it.
For ease of reading, I will repeat the boxed equation below: $$\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Computing an infinite product Let $a_0=5/2$ and $a_k=a^2_{k-1}-2$ for all $k\geq 1$.The question is to compute $$\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$
I tried to calculate few terms.$a_0=5/2$, $a_1=17/4,a_2=257/16$ it seems that $a_k$ is of the form $2^{2^k}+2^{-2^{k}}$ however I am not sure about it.How ... | The explicit formula $a_k=2^{2^k}+2^{-2^{k}}$ is straightforward to prove by induction.
We also have
$$ a_k+1 = a_{k-1}^2-1 = (a_{k-1}-1)(a_{k-1}+1) $$
hence
$$ a_{k-1}-1 = \frac{a_k+1}{a_{k-1}+1} $$
and
$$ \prod_{k=1}^{N}\left(a_k-1\right)=\prod_{k=2}^{N+1}\left(a_{k-1}-1\right)=\frac{a_{N+1}+1}{a_1+1}. $$
Since
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
eigen vector that orthogonal to each other, symmetry matrix symmetry matrix $\left(\begin{array}{ccc} 0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{array}\right) $
one of eigen value is $\lambda_1=2$ and one of eigen vector is $x_1=\left(\begin{array}{ccc} \frac{1}{\sqrt3} \\
\frac{1}{\sqrt3} \\
... | \begin{eqnarray*}
\begin{bmatrix}
\frac{2}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\
\end{bmatrix},
\begin{bmatrix}
0 \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \\
\end{bmatrix}
\end{eqnarray*}
Edit: Note that
\begin{eqnarray*}
\begin{bmatrix}
2 \\ -1 \\ -1 \\
\end{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is The Value of This Definite Integral $0$?
Q. Evaluate -$$\int\limits_0^\pi {{{x\,dx} \over {{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} $$
My Try -
$$\eqalign{
& Let\,I\, = \,\int\limits_0^\pi {\frac{{x\,dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \int\limits_0^\pi {\frac{{... | As @José Carlos Santos points out, using a substitution of $t = \tan x$ on the interval $[0,\pi]$ introduces a discontinuity at the point $x = \pi/2$.
One way this can be avoided, as you correctly identified, is by first changing the interval of integration. If we note that the integrand in the integral
$$I = \int^\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Definite Integral $\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$ for Logarithm and Algebraic.
Evaluate
$$\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$$
I have been thinking for this question quite some time, I've tried all the methods I have learnt but still getting nowhere. Hope that someone can explain it for me. Thanks ... | If you want to avoid series altogether one could always try Feynman's trick of differentiating under the integral sign.
Consider
$$I(a) = \int^1_0 \frac{\ln \left [a(x^2 - x) + 1 \right ]}{x - x^2} \, dx, \quad a \geqslant 0.$$
Note that $I(0) = 0$ and we require $I(1)$.
Differentiating under the integral sign with re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove the associative la of multiplication for 2x2 matrices, show that (AB)C=A(BC)? Let
$$A= \left[
\begin{array}{cc|c}
a_{11}&a_{12}\\
a_{21}&a_{22}
\end{array}
\right] , B= \left[
\begin{array}{cc|c}
b_{11}&b_{12}\\
b_{21}&b_{22}
\end{array}
\right],C= \left[
\begin{array}{cc|c}
c_{11}&c_{12}\\
c_{21}&c_{... | It's wrong, this is the correct solution for $(AB)C$:
$$ (AB)C=\left[
\begin{array}{cc|c}
(a_{11}b_{11}+a_{12}b_{21})C_{11}+(a_{11}b_{12}+a_{12}b_{22})C_{21}&(a_{11}b_{11}+a_{12}b_{21})C_{12}+(a_{11}b_{12}+a_{12}b_{22})C_{22}\\
(a_{21}b_{21}+a_{22}b_{21})C_{11}+(a_{21}b_{12}+a_{22}b_{22})C_{21}&
(a_{21}b_{21}+a_{22... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Substitution in integral gives $0$ to $0$ bounds $\require{cancel}$
I must find the length of the curve given by those 2 equations:
$x = \cos^3t$
and
$y = \sin^3t$ between $0$ and $2\pi$.
I can find their derivative: $\frac{dx}{dt} = -3\cos^2t\sin t$ and $\frac{dy}{dt} = 3\sin^2t\cos t$.
Then I can apply the formula (r... | The problem is that $\sqrt{\cos^2 t\sin^2 t}$ is only $\cos t \sin t$ when $\cos t \sin t \ge 0$. Otherwise it's $-\cos t \sin t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$; $a,b,$ are positive integers. Is it true: $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$?
Somehow I can't find counterexample nor to prove it. I try to write it $a=gx$ and $b=gy$ where $g=\gcd(a,b)$ but didn't help. It seems that there is no $a\ne b$ such that $a+b\mid a^4+b^4$. O... | Well,
\begin{align}
&& a+b &\mid a^4 + b^4 \\
&\iff & a+b &\mid a^4 + b^4 - (a+b)^4 + 4ab(a+b)^2 \\
&\iff & a+b &\mid 2a^2b^2 \\
&\iff & a+b &\mid ab\bigl((a+b)^2 - (a^2+b^2)\bigr) \\
&\iff & a+b &\mid ab(a^2+b^2)\,
\end{align}
so for coprime $a,b$ it follows that $a+b \mid a^4+b^4 \iff a+b \mid a^2+b^2$.
Writing $a = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac... | Let me give a general method which is useful
for this sum $\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}$
we have
$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}=\sum_{n=0}^\infty\dfrac{1}{(2n+1)(2n+3)}$
we can use this method
$$\sum_{n\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find out all solutions for the system Given the system
$$ \left[
\begin{array}{ccc|c}
x_1&x_2&x_3&k\\
x_1&x_2&kx_3&1\\
x_1&kx_2&x_3&1\\
kx_1&x_2&x_3&1\\
\end{array}
\right] $$
I tried to solve this...It looks simple but I found a problem at the end...
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
1&1&k&1\\
1&k... | Let us add the last three equations to gather to get: $$(k+2)(x_1+x_2+x_3) =3$$
Then we use the first one to get, $$(k+2)(k)=3 \implies k=1, -3$$
So, ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions a... | Let's put $x=1/t$ so that $t\to 0^{+}$ and the expression under limit is transformed into $$\frac{\sqrt{1+1/t^{2}}-\sqrt[3]{1+1/t^{3}}}{t}=\frac{\sqrt{1+t^{2}}-\sqrt[3]{1+t^{3}}}{t^{2}}$$ and this can be expressed as a difference of two standard limits $$\lim_{t\to 0^{+}}\frac{(1+t^{2})^{1/2}-1}{t^{2}}-t\cdot \frac{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Residue calculations involving logarithms. I am trying to solve a definite integral using residue theory. At some step, I wish to calculate the residue of the function $\frac{\log{(z)}}{z^3+8}$ at $-2e^{2\pi i/3}$. The log function is defined for $0 < \arg z < 2\pi$.
I'm having difficulty doing this simple looking calc... | $$\frac{\log(z)}{z^3+8}= \frac{\log(z)}{(z+2e^{\frac{2i\pi}3})(z^2-2e^{\frac{2i\pi}{3}}z+4e^{\frac{-2i\pi}{3}})}$$
Then the residue at $\displaystyle z=-2e^{\frac{2\pi i}{3}}$ should be:
$$\frac{\log(-2e^{\frac{2\pi i}{3}})}{(-2e^{\frac{2\pi i}{3}})^2-2(-2e^{\frac{2\pi i}{3}})(e^{\frac{2i\pi}{3}})+4e^{\frac{-2i\pi}{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $z^2 - 8(1-i)z + 63 - 16i = 0$ with reduced discriminant $z^2 - 8(1-i)z + 63 - 16i = 0$
I have come across a problem in the book Complex Numbers from A to Z and I do not understand the thought process of the author when solving:
The start to the solution is as follows:
$\Delta$ = discriminant
$\Delta ' = (4 - 4... | Since $∆' = -63-16i$
$$ z = \frac{ -b±\sqrt{∆'}}{a} $$
$$ z= 8 - 8i ± \sqrt{-63-16i} $$
Using Quadratic Formula, instead of reduced determinant
$$ z= \frac{8(1-i)±\sqrt{\left[8(1-i)\right]^2-4\cdot1\cdot(63+16i}}{2\cdot1} $$
$$ = \frac{8-8i±\sqrt{\left[64(2i)\right]-\cdot(4×63+4×16i}}{2} $$
$$ = \frac{8-8i±\sqrt{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Sum based on stolz theorem The Sum is as follows:
$$ \lim_{n\to \infty} \frac{1} {\sqrt n} \left( \frac {1} {\sqrt n}+ \frac {1} {\sqrt {n+1}} +...+ \frac{1}{\sqrt{2n}} \right) $$.
I solved as follows using stolz theorem
$$ \lim_{n\to \infty} \frac{x_n-x_{n-1}} {y_n-y_{n-1} } = \lim_{n\to \infty} \frac {\frac {1} {\... | You've made a mistake in evaluating $x_n-x_{n-1}$ \begin{align}x_{n}=&&\frac1{\sqrt n}+\cdots + \frac{1}{\sqrt{2n-2}}&+\frac1{\sqrt{2n-1}}+\frac1{\sqrt{2n}}\\x_{n-1}=&\frac1{\sqrt{n-1}}&+\frac1{\sqrt n}+\cdots + \frac{1}{\sqrt{2n-2}}\end{align}
They didn't.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& =... | Because $$6^n+8^n=(6+8)(6^{n-1}-6^{n-2}\cdot8+...-6\cdot8^{n-2}+8^{n-1}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.... | Let the LHS be $A(n)$ and let the RHS be $2B(n)^4.$
Then $A(n+1)-A(n)=(n+1)^7+(n+1)^5=(n+1)^5(n^2+2n+2).$
$$\text {We have }\quad 2B(n+1)^4-2B(n)^4=$$ $$(*)\quad =2(B(n+1)^2+B(n)^2)\cdot (B(n+1)+B(n))\cdot (B(n+1)-B(n)).$$ Since $B(n)=n(n+1)/2$ we have $$B(n+1)^2+B(n)^2=(n+1)^2((n+2)^2+n^2)/4=(n+1)^2(n^2+2n+2)/2$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
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How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| By power series we need the following two:
$$\cos x=1-\frac{x^2}{2}+o(x^2)$$
$$(1+x)^n=1+nx+o(x)$$
Thus
$$\sqrt {\cos 2x}=\left( 1-\frac{4x^2}{2}+o(x^2)\right)^\frac12=1-x^2+o(x^2)$$
$$\sqrt[3] {\cos 3x}=\left( 1-\frac{9x^2}{2}+o(x^2)\right)^\frac13=1-\frac{3x^2}{2}+o(x^2)$$
Thus
$$\sqrt {\cos 2x}×\sqrt[3] {\cos 3x}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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derivative of $\frac 2x \sin(x^3)$ by definition The function: $$\frac 2x \sin(x^3)$$ when $x\neq0$, and $0$ while $x=0$.
I need to find if the function is derivation at $x=1$.
First step was to check if the function is continuous. there is just 1 side of limit to check (since its the same function around $x=1$, so I c... | I did not answer your exact question, I looked at the case of $x=0$ rather than $x=1$, but the argument is analogous.
Evaluate the derivative of $f : \mathbb{R} \to \mathbb{R}$, where $$f(x) : = \begin{cases}
\frac{2}{x} \sin(x^3), & x \in \mathbb{R} \backslash \{ 0 \}, \\
0, & \text{otherwise}.
\end{cases}$$
Let us ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Steiner inellipse Hello it's related to my answer for Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq 3\frac{a+b+c}{ax+by+cz}$
My answer fails but I don't know why ... So I was thinking a generalization of the following formula:
$$\frac{IA^2}{CA\cdot AB}+\frac{IB^2}{BC\cdot AB}+\frac... | A triangle $\triangle ABC$ with its centroid at the origin is the image of an origin-centered equilateral under a linear transformation. The transformation carries the equilateral's incircle to $\triangle ABC$'s Steiner inellipse. With an appropriate rotation, and by considering similar triangles (and/or ellipses) equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why is $\int_0^{\pi/4} 5(1+\tan(x))^3\sec^2(x)\,dx$ equal to $18.75$ and not $3.75$? Why is $\int_0^{\pi/4} 5(1+\tan(x))^3\sec^2(x)\,dx$ equal to $18.75$ and not $3.75$?
I know the indefinite integral $\int 5(1+\tan(x))^3\sec^2(x)\,dx= \frac {5(1+\tan(x))^4} {4}+c$ by using $u$ substitution. Then shouldn't I evaluate t... | Note that we have $\tan\left(\frac{\pi}{4} \right)=1$ and $\tan(0)=0$
Hence,
\begin{align}
\frac{5(1+\tan\left(\frac\pi4 \right))^4}{4}-\frac{5(1+0)^4}{4}&=\frac{5(2^4)}{4}-\frac{5}{4}\\&=\frac{5(15)}{4}\\&= \color{blue}5(3.75)\\&=18.75
\end{align}
A potential mistake is that you have forgotten to multiply by $5$. We h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that if $c^2 = a^2 + b^2$ then $c < a + b$ I am having some trouble in proving a conjecture that occurred to me some time ago, based on the Pythagorean theorem.
If, for a non-degenerate triangle, $$c^2 = a^2 + b^2$$
Then can the following be proven?
$$c < a + b$$
Is this statement always true?
| I guess your doubt is:
does a triple $(a,b,c)$ of positive numbers with $c^2=a^2+b^2$ also satisfy $c<a+b$, so that the three numbers are the sides of a triangle?
In particular, does a Pythagorean triple define a (right) triangle?
Indeed, the other two conditions, namely $a<b+c$ and $b<a+c$, are obviously satisfied ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ Is it possible to prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ for $n \geq 1 $ . I know that it is true because $ \frac{3^n - 4(2^n) + (-1)^n + 6}{24}$ represents the number of ways to uniquely $4$-colour an n-cycle , excluding permutations of colours... | Let us try modulo $\pmod3,\pmod8$ separately.
As $4\equiv1\pmod3,2\equiv-1\implies2^n\equiv(-1)^n$
$$3^n-4\cdot2^n+(-1)^n+6\equiv-(-1)^n+(-1)^n\equiv0\pmod3$$
For $n\ge1,8\mid4\cdot2^n$
$$3^n-4\cdot2^n+(-1)^n+6\equiv3^n+(-1)^n+6\pmod8$$
Now as $3^2\equiv1\pmod8,(3^2)^m\equiv1$
If $n$ is even, $$3^n+(-1)^n+6\equiv9^m+1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Invertible 4x4 matrix $$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
$$
Is this matrix invertible? I would like to show that it is invertible but first I should find the det(Matrix) which should not be equal to zero. To find the determinant, ... | Proof only by determinant
We have $$\begin{vmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{vmatrix}=A-B+C-D$$ where $$A=5\begin{vmatrix}
2 & 2 & 8 \\
6 & 2 & 8 \\
3 & 6 & 7 \\
\end{vmatrix}=5\left(2\begin{vmatrix}
2 & 8 \\
6 & 7 \\
\end{v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that
$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$
I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\l... | HINT
$$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)=\frac{1}{n^2}\log\Big(1+\frac 1{n^2}\Big)^{n^2}+ \frac{2}{n^2}\log\Big(1+\frac 2{n^2}\Big)^\frac{n^2}{2}+...+ \frac{n}{n^2}\log\Big(1+\frac n{n^2}\Big)^\frac{n^2}{n}=\frac{1}{n^2}\sum_1^n k\log\left(1+\frac{k}{n^2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Show that this polynomial has three different real roots There is polynomial $f(x)=x^3 +ax^2+bx+c$ and $ab=9c$ and $b<0$. Like in the title. I've only come up so far with $9x^3 + 9ax^2 + 9bx+ab=0$ and factoring here does nothing.
| Regarding the requirement that $a,c$ be real. Set $a = 3\mathrm{i}$, $b = -3$, $c = -\mathrm{i}$. Then $b < 0$, $ab = -9\mathrm{i} = 9c$, and $f(x) = x^3 + 3 \mathrm{i} x^2 - 3 x - \mathrm{i} = (x+\mathrm{i})^3$, so all three roots are the same and none are real. Consequently, for the problem to be correct, another ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \f... | We have
$$\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2\,\text{AGM}(a,b)}=\frac{\pi}{2\,\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)}\tag{1} $$
as a consequence of Lagrange's identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$.
On the other hand
$$ K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum... | Since $|\sin t|\le |t|$ we have
$$|\sin(a)-\sin(b)|= 2\left|\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\right|\le |a-b|= \frac{1}{2n}-\frac{1}{2n+1}\sim \frac{1}{n^2}$$
with $a= \frac{1}{2n}$ and $b=\frac{1}{2n+1}$
Hence, the series is absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$z ≤ x + y$ implies $z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$
Suppose $x,y,z $ be nonnegative reals. Show that $z ≤ x + y\implies z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$.
My Proof: If $z=0$, then we are done. So, suppose $z>0$. Since $z ≤ x + y$, and $x$ and $y$ nonnegative, $z ≤ x + y+2xy+xyz$, which leads to $z(1+x+y+xy)\le (... | If
$z \le \sum_{i=1}^n x_i
$
then
we want to show that
$\dfrac{z}{1+z}
\le \sum_{i=1}^n \dfrac{x_i}{1+x_i}
$.
Let
$f(z)
=\dfrac{z}{1+z}
$.
We want to show that
$f(z)
\le \sum_{i=1}^n f(x_i)
$.
$f(z)
=\dfrac{z}{1+z}
=1-\dfrac{1}{1+z}
$
so
$f'(z)
=\dfrac1{(1+z)^2}
> 0
$.
$f$ is increasing so
$f(z)
\le f(\sum_{i=1}^n x_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Constant of integration change So, sometimes the constant of integration changes, and it confuses me a bit when and why it does. So for example, we have a simple antiderivative such as $$\int \frac{1}{x} dx $$ and we know that the result is $$\log|x| + C$$ and the domain is $$x\in\mathbb R \backslash \{0\} $$ If we wan... | "Do we need to do change the constant every time there is a gap in the domain [...]?" The answer is Yes: the generic C stands for a locally constant function - a function constant on intervals. The values on disjoint intervals may be different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Elementary Inequality from a local olympiad test The problem I'm stuck with is the following :
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$a+x \ge b+y \ge c+z \ge 0 $ and $ a+ b+c = x+y+z$
Prove that $ay+bx \ge ac+xz$
As easy as it sounds, I didn't achieve any significant progress after several tri... | Let us try to minimize
$$\varphi = a y + b x - ac - x z$$
We suppose that $a,b,c,x,y,z$ are given, satisfying the required conditions.
Without any computation one can decrease $\varphi$ by decreasing $b$ and $y$ by the same amount. Indeed, it doesn't change the condition
$a+b+c = x+y+z$ and it decreases $\varphi$ becau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplify this equation if $x$ is negative I am trying to simplify $7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}$, given that $x$ is negative. The answer is $-210x^3$, but I am getting $210x^3$. Below is my reasoning:
For a $ k > 0 $, let $ k = - x $. Then, $7\cdot\sqrt{5x}\cdot\sqrt{180x^5}$ = $7\cdot5\cdot6\cdot i\sqrt{k} \cdot... | Since $k=-x$, we have $$7\cdot\sqrt{5x}\cdot\sqrt{180x^5}=7\cdot i\cdot\sqrt{5k}\cdot i\sqrt{180k^5}=-7\cdot\sqrt{900k^6}=-7\cdot\sqrt{900x^6}=-210x^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.