Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the recursive formula for calculating the solution of the linear equation system $A$x = b. Let $K$ be a field and n $\in$ $\mathbb{N}$. It's given that $A$ = $
\begin{bmatrix}
0 & p_{1} & 0 & 0 & \dots & 0 \\
q_{2} & 0 & p_{2} & 0 & \dots & 0 \\
0 & q_{3} & 0 & p_{3} & \dots & 0 \\
\vdots & & \ddots & \ddots... | So,
a) solve the 1st as $x_2=b_1/p_1$;
b) solve the 2nd in terms of $x_3=(b_2-q_2x_1)/p_2$, the $x_1$ is unknown: keep it as such and continue;
c) all the following $x_{2k}$ will be a function of the precedent even-index and are therefore determined;
d) all the following $x_{2k+1}$ will be a linear function of the prec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$... | You write the equation in the form
$$
[(9+2)^k-9^k]-[(7+1)^k-7^k]-[(5+1)^k-5^k]=0.
$$
If $k\ge 0$, then you can use that
$$
(9+2)^k-9^k=\int_9^{9+2}kt^{k-1}dt\ge 2k9^{k-1},\\
-[(7+1)^k-7^k]=-\int_7^{7+1}kt^{k-1}dt\ge -k8^{k-1},\\
-[(5+1)^k-5^k]\ge -k6^{k-1}.
$$
Your equation becomes
$$
0\ge k\cdot\left(2\cdot 9^{k-1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Determining in which quadrant an angle terminates. Let $A$ be in the fourth quadrant and let $\sec(A) = \dfrac{13}{5}$, Let $B$ be in the third quadrant and let $\csc(B)=\dfrac{-5}{3}$. Find $\sin(A + B)$ and determine in which quadrant $A + B$ terminates.
$\sec(A) = \dfrac{13}{5} \Rightarrow \cos(A) = \dfrac{5}{13}$
... | Hint: Your calculations above are correct (I cleaned them up a little bit). If sine is positive but cosine is negative, what quadrant are you in? Does this give a hint to $A + B$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to calculate $\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$ How to calculate $$\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$$
The limit equals a but any hints for the method?
| Let us consider $$A=(x+a)^{1+\frac1x}\qquad \text {and} \qquad B=(x) ^{1+\frac{1}{x+a}}$$
$$A=(x+a)^{1+\frac1x}\implies \log(A)=\left({1+\frac1x}\right)\log(x+a)$$ Using Taylor we then have
$$\log(A)=\log \left({x}\right)+\frac{a+\log
\left({x}\right)}{x}+\frac{a-\frac{a^2}{2}}{x^2}+O\left(\frac{1}{x^3
}\right)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer sho... | Using the same rationalization method, when $x\not\in\{0,1\}$,
$$
\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}
=\frac1{x+1}\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How to calculate $\int \frac{\cos^2 x}{1 + \sin^2 x}dx$ My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$\sin^2 x = \frac{1 - \cos(2x)}{2}$ and $\cos^2x = \frac{1 + \cos(2x)}{2}$
| Note that we can write
$$\begin{align}
\frac{\cos^2(x)}{1+\sin^2(x)}&=\frac{1+\cos(2x)}{3-\cos(2x)}\\\\
&=\frac{4-(3-\cos(2x))}{3-\cos(2x)}\\\\
&=-1+\frac{4}{3-\cos(2x)}
\end{align}$$
Enforce the substitution $y=\tan(x)$, so that $\cos(2x)=\frac{1-y^2}{1+y^2}$ and $dx=\frac1{1+y^2}\,dy$. Proceeding reveals
$$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably... | This fails for $a=1,$ $b=2,$ $c=3,$ & $d=4$. The left-hand side becomes $\frac{275}{504}=0.5456349...$ and the right-hand side is $\frac{16}{10}=1.6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Showing that if $p_1 + \cdots p_n = 1$ then $\displaystyle \sum_{k=1}^n \left(p_k + \dfrac {1}{p_k} \right)^2 \ge n^3+2n+\dfrac 1n$? This problem is from the book "Cauchy-Schwarz Masterclass":
Show that if $p_1 + \cdots p_n = 1$ with each $p_i$ positive, then $\displaystyle \sum_{k=1}^n \left(p_k + \dfrac {1}{p_k} \ri... | This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $\ge$ HM.
Apply CS to $n$ copies of $1$ and $\displaystyle\;p_k+\frac{1}{p_k}$, we obtain
$$n \sum_{k=1}^n \left(p_k + \frac{1}{p_k}\right)^2
= \left( \sum_{k=1}^n 1^2\right)\sum_{k=1}^n\left(p_k + \frac{1}{p_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Finding all measurable functions maximizing an expression given integral conditions I'd like to know how to answer the question located in this comprehensive exam from the 1990s. The question is:
Find the maximum value of $\int_{-1}^{1} x^3 g(x) dx$ for measurable functions g(x) satisfying
$$ \int_{-1}^{1} g(x) dx = \... | We have four conditions. Define
$$
g(x) = a_0+a_1 x+ a_2 x^2+ a_3 x^3
$$
and then calculate
$$
\int_{-1}^1 g(x)dx = 2\left(a_0+\frac 13 a_2\right)=0\\
\int_{-1}^1 g(x)x dx = 2\left(\frac 13 a_1+\frac 15 a_3\right) = 0\\
\int_{-1}^1 g(x)x^2 dx =2\left(\frac 13 a_0 +\frac 15 a_2\right) = 0\\
\int_{-1}^1 |g(x)|^2 dx = 2 a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the values of $a$ and $b$ such that the following matrices are similar
Find the values of $a$ and $b$ such that the following matrices are similar.
$$A=\begin{bmatrix} -2& 0 & 0 \\ 2 & a & 2 \\ 3& 1 & 1 \end{bmatrix}, \qquad B=\begin{bmatrix} -1& 0 & 0 \\ 0 & 2 & 0 \\ 0& 0 & b \end{bmatrix}$$
I know that simila... | Calculate the characteristic polynomials:
$$\det (A - \lambda I) = (-2-\lambda) \begin{vmatrix}a-2 & 2 \\ 1 & 1-\lambda\end{vmatrix} = (-2-\lambda)(\lambda^2 - \lambda(a+1) + (a-2)) = (-2 - \lambda)\left(\frac{1+a-\sqrt{a^2-2a+9}}2 - \lambda\right)\left(\frac{1+a+\sqrt{a^2-2a+9}}2 - \lambda\right)$$
$$\det (B - \lambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $ab \mid c(c^2-c+1)$ and $c^2+1 \mid a+b$ then prove that $\{a, b\}=\{c, c^2-c+1 \}$ If $ab \mid c(c^2-c+1)$ and $c^2+1 \mid a+b$ then prove that $\{a, b\}=\{c, c^2-c+1 \}$ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve t... | Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=\frac{2mc+(c^2-c+1)/2}{2m^2(c^2+1)^2}$
Now problem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Integration of $\frac{1}{x^2-a^2}$ by trigonometric substitution?
$$\int \frac{1}{x^2-a^2}dx$$
Now, I know this can be done by splitting the function into two integrable functions,
$\displaystyle\dfrac{1}{2a}\int \bigg(\dfrac{1}{x-a} - \dfrac{1}{x+a}\bigg)dx$
And then doing the usual stuff.
My question is, how can we... | You might use $x=a\sec\theta$: $dx=a\sec\theta\tan\theta\,d\theta$ and the integral becomes
$$
\int\frac{1}{a^2(\sec^2\theta-1)}a\sec\theta\tan\theta\,d\theta
=
\frac{1}{a}\int\frac{\cos^2\theta}{\sin^2\theta}\frac{1}{\cos\theta}\frac{\sin\theta}{\cos\theta}\,d\theta
=
\frac{1}{a}\int\frac{1}{\sin\theta}\,d\theta
$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Arithmetic series problem Given $\left\{a_n\right\}$ arithmetic progression, $a_1=2$, $a_{n+1}=a_n+2n$ $\left(n\:\ge \:1\right)$. $a_{50}=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_{50}=2+d\left(n-1\right)$$
$$a_{50}=2+2\left(n^2-n\right)$$
$$a_{50}=2+2\cdot 2450$$
$$a_{50}=4902$$
But this is wrong. Answers:
$$A=2452... | Alternatively, note: $a_{n+1}-a_n=2n$. So:
$$\begin{align} a_2-a_1&=2\cdot 1\\
a_3-a_2&=2\cdot 2\\
a_4-a_3&=2\cdot 3\\
&\vdots \\
a_{50}-a_{49}&=2\cdot 49 \end{align}$$
Summing all (midterms telescope):
$$a_{50}-a_1=2(1+2+3+\cdots +49) \Rightarrow \\
a_{50}=a_1+2\cdot \frac{1+49}{2}\cdot 49=2+2450=2452.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Convergence of the series $\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$.
Check whether the series
$$\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$$
is convergent.
I tried to sandwich the function by $\dfrac{1}{1+x^2}$ and $\dfrac{x}{1+x^2}$ , but this did not help at all.... | When $x\in [0,1], \frac {\sqrt x}{2} \le\frac {\sqrt x}{1+x^2} \le \sqrt x$
Which means that $\frac 12\int_0^\frac1n \sqrt x\ dx \le \int_0^\frac1n \frac {\sqrt x}{1+x^2} \ dx \le \int_0^\frac1n \sqrt x \ dx$
if $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \sqrt x \ dx $ converges then $\sum_\limits{n=1}^{\infty} \int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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The limits of integration for a paraboloid and plane.
Consider the volume, $V$, enclosed by the paraboloid
$$z=x^2+y^2$$
and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think ... | The two surfaces intersect when $$z=x^2+y^2=3-2x \Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 \Rightarrow \left(x+1\right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= \int{\int{\int dz}dy}dx=\int_{x=-3}^{x=1}{\int_{y=-\sqrt{4-(x+1)^2}}^{y=\sqrt{4-(x+1)^2}}{\int_{z=x^2+y^2}^{z=3-2x}1dz}dy}dx$$
In Cylindrical... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to integrate this indefinite integral? $$\int\frac{x^2\sec^2x}{(x \tan{x}+1)^{2}}\,\mathrm{d}x$$
I tried the online available calculators but they cannot calculate the answer or provide the solution.
| This is basically working backwards, like @Arthur suggests:
\begin{align}
\frac{x^2\sec^2 x}{(x\tan x+1)^2} &= \frac{x^2}{\left(x\frac{\sin x}{\cos x} + 1\right)^2\cos^2 x}\\
&= \frac{x^2}{(x\sin x + \cos x)^2}\\
&= \frac{x^2(\sin^2 x + \cos ^2 x) - x\cos x\sin x + x \cos x\sin x}{(x\sin x + \cos x)^2}\\
&= \frac{x\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the least value of $n\in N$ for which $(n-2)x^2+8x+n+4>\arcsin(\sin12)+\arccos(\cos12)$ for every $x \in \mathbb {R}$ Find the least value of $n\in N$ for which $(n-2)x^2+8x+n+4>\arcsin({\sin12})+\arccos({\cos12})$ for every $x \in \mathbb {R}$
$(n-2)x^2+8x+n+4>\arcsin({\sin12})+\arccos({\cos12}) $
$(n-2)x^2+8x+... | Consider $f(x)=\arcsin(\sin x)+\arccos(\cos x)$; then
$$
f'(x)=\frac{1}{\sqrt{1-\sin^2x}}\cos x+\frac{1}{\sqrt{1-\cos^2x}}\sin x=
\frac{\cos x}{\lvert\cos x\rvert}+\frac{\sin x}{\lvert\sin x\rvert}
$$
Note that $7\pi/2<12<4\pi$; one inequality is obvious; the other one is equivalent to $\pi<24/7$, but we know from Arch... | {
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"url": "https://math.stackexchange.com/questions/2874792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$ Solve the equation:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$
Things I have done so far:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$ Deducting 2 from both sides of equation
$$\Leftrightarrow (\sqrt[3]{x^2+4}-2)=(\sqrt{x-1}-1)+2x-4$$
$$\Leftrightarrow \frac{x^2+4-8}{\sqrt[3]{... | Actually, there is only one solution in this equation.
let $f(x)=(x^2+4)^{1/3}-\sqrt{x-1}-2x-3(x\geq 1)$,then $f'(x)=\frac{2x}{3(x^2+4)^{\frac{2}{3}}}-\frac{1}{2\sqrt{x-1}}-2 \leq \frac{2x}{3(x^2+4)^{\frac{2}{3}}} -2\leq \frac{2x}{3(x^2)^{\frac{2}{3}}} -2=\frac{2}{3x^{\frac{1}{3}}}-2 \leq -\frac{4}{3}$.
So $f(x)$ is st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}... | Your sum can be written as $\sum_{k=1}^n\frac{1}{k^{3/2}}=1+2^{-3/2}+\sum_{k=3}^n k^{-3/2}$. Since $f(x)=x^{-3/2}$ is decreasing, we can bound the sum from above by an integral, that is,
$$
\sum_{k=3}^n\frac{1}{k^{3/2}}\leq\int_2^n x^{-3/2}\,dx.
$$
Note specifically the lower bound of the integral.
Evaluating the integ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluate the Sum $S=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+\cdots \infty$ Evaluate the Sum
$$S=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+\cdots \infty$$
My try: We have the $n$ th term as
$$T_n=\frac{1.3.5. \cdots (2n-1)}{4.6.8 \cdots (2n+2)}$$ $\implies$
$$T_n=\frac{1.3.5. \cdots (2n-1)}{2^n \times (... | Hint: See Catalan Number with $x=\dfrac14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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You flip a coin $10$ times. How many ways can you get at least $7$ heads?
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$\binom{10}{10}+ \binom{10}9\cdot\binom{10}1 + \binom{10}8\cdot\binom{10}2+\binom{10}7\cdot\binom{10}3$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $\cdo... | the combinations notation ${10\choose 7} = \frac {10!}{7!3!}$ accounts for the fact that if you have 7 heads you also have 3 non-heads. ${10\choose 7}{10\choose 3}$ is effectively squaring the value that you need for that term. Similar for the other terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Find the oblique asymptote of $\sqrt{x^2+3x}$ for $x\rightarrow-\infty$ I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$
For $x\rightarrow+\infty$ we have:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x... | The problem is in
$$
f(x)-mx=\sqrt{x^2+3x}+x\underset{\begin{array}{c} \uparrow \\ \text{problem}\end{array}}{=}x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty
$$
which should be
$$
f(x)-mx=\sqrt{x^2+3x}+x=x\left(-\sqrt{\frac{x^2}{x^2}+\frac{3x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886545",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$
Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$.
I tried taking $2x$ as $a$, and then $2y$ as $b$, and then finding the possibilities. However, I am not sure about the approac... | Assume $x,y,z$ are required to be positive integers.
Consider two cases . . .
Case $(1)$:$\;n$ is even.
From the equation $2x+2y+z=n$, it follows that $z$ is also even.
Writing $z=2w$, and $n=2m$, for some positive integers $w,m$, we get
\begin{align*}
&2x+2y+z=n\\[4pt]
\iff\;&2x+2y+2w=2m\\[4pt]
\iff\;&x+y+w=m\\[4p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos... | Abbreviating $\sin\theta$ by $s$ and $\cos\theta$ by $s$ we have on base of $c^2+s^2=1$:$$1+2sc=c^2+2sc+s^2=(c+s)^2$$ so that:$$(1+2sc)(c-s)=(c+s)^2(c-s)=(c^2-s^2)(c+s)$$
Dividing boths sides by $c$ and abbreviating $\tan(\theta)$ by $t$ we get on base of $t=s/c$:$$(1+2sc)(1-t)=(c^2-s^2)(1+t)$$
Now divide both sides by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Value of a triangle inscribed in a rectangle I was doing a little trigonometry challenge given by my professor, which is not anything especially complicated. However, it is completely based on getting the area of the green triangle in relation to the side of the square and the triangle, equal to $x$.
Reference image: ... | I am not sure that this is a solution you want, since it uses vectors and not trigonometry. I'll put it here anyway since it is a simpler approach and gives the right answer. I will use $l$ instead of $x$ for the length of the sides of the square and triangle to avoid confusion with the variable $x$.
The idea is to fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all $x,\ y,\ z$ that satisfy: $(x+y-z)^{5}+(x+z-y)^{5}+(z+y-x)^{5}=0$
Find all integers $x,\ y,\ z$ that satisfy:
$$(x+y-z)^{5}+(x+z-y)^{5}+(z+y-x)^{5}=0$$
I guessed that, considering the symmetry and the similar case $\sum x_{i}^{2n}=0$, that these three expressions must be equal to zero, which leads to $x=y... | Since $x, y, z$ are integers so are $a=x+y-z, b=x+z-y, c=z+y-x$.
We know from Fermat's last theorem that $a^5+b^5+c^5=0$ is impossible for non-zero integers $a, b, c$.
This means that $x+y-z, x+z-y$ or $z+y-x$ must be zero.
If $x+y-z=0$ then your equation gives $x+z-y=x-y-z$ which means that $z=0$ and $x=-y$.
If $x+z-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For $\triangle ABC$ with inradius $r$, if $\frac{c + a}{b} +\frac{c + b}{a} = \frac{c}{r}$, then which angle is $90^\circ$?
For $\triangle ABC$ with inradius $r$, if
$$\frac{c + a}{b} +\frac{c + b}{a} = \frac{c}{r}$$ then which angle is $90^\circ$?
My try :
I am stuck here I think my process is wrong please help ... | The equality on the $3^{rd}$ line can be written as:
$$r = \frac{abc}{a^2+b^2+(a+b)c}$$
The symmetry in $\,a,b\,$ suggests that the right angle is $\,C\,$, in which case $\,a^2+b^2=c^2\,$, then:
$$\require{cancel}
r = \frac{abc}{c^2+(a+b)c} = \frac{ab\cancel{c}}{(a+b+c)\cancel{c}} =\frac{ab}{a+b+c} = \frac{\cancel{2}S}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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On Lame's Theorem I was trying to study Lame's theorem but I'm a little bit stuck on one specific step... I write here the theorem and the proof:
Lame's theorem:
using the Euclidean algorithm to find the greatest common divisor of two positive integers has number of divisions less than or equal five times the number... | Let's chase through the rest of the proof without using the line $\log_{10}\alpha>1/5$. So instead of writing $\log_{10}b>(n-1)/5$, we'll just write $$\log_{10}b>(n-1)\log_{10}\alpha.$$ Continuing through the rest of the algebra, we end up with $$n-1<k/\log_{10}\alpha.$$
So, this actually gives a more precise conclus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to calculate the limit $\lim\limits_{n\to\infty} \int\limits_0^1 \frac{n(2nx^{n-1}-(1+x))}{2(1+x)}\,dx$? How to calculate the limit $\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$?
I have to calculate the limit when solving
Find $a,b$ for $\displaystyle\int_0^1 \dfrac{x^{n-1}}{... | \begin{align*}
2b &= \lim_n n \int_0^1 \frac{nx^{n-1}} {x+1} \mathrm dx - \frac n 2 \\
&= \lim_n n \int_0^1 \frac {\mathrm d (x^n)} {x+1} - \frac n2\\
&= \lim_n n \left. \frac {x^n} {1+x}\right|_0^1 + n \int_0^1 \frac {x^n \mathrm dx} {(1+x)^2} - \frac n 2\\
&= \lim_n \frac n {n+1} \cdot \left.\frac {x^{n+1}}{(x+1)^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Is the function is differentiable at 0? Is the function given by
$ f(x) = \begin{cases} \frac{1}{x \log2} - \frac{1}{2^x-1},\text{if} \ x\neq 0 ,\\\frac{ 1}{2} \ \text {if} \ x=0 \end{cases}$.
is differentiable at zero ?
My Attempts:
$f'(0) = \frac{f(x) - f(0)}{x-0} =\frac{\frac{1}{x \log2} - \frac{1}{2^x-1}- \... | $$\frac1{2^x-1}=\frac1{\exp(x\ln2)-1}=\frac1{x\ln 2+x^2(\ln 2)^2/2+O(x^3)}
=\frac1{x\ln 2}\left(1-\frac{\ln 2}2+O(x)\right)$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove that $ a^2+b^2+c^2=1$ implies $ab+bc+ca \in [-\frac12,1]$? If $a,b,c$ belong to set of real numbers and $ a^2+b^2+c^2=1$ then prove that $ab+bc+ca$ belongs to $[-\frac12,1]$
I have tried AM>GM>HM(progressions mean inequality) but I am unable to do anything.
I have even attempted to assume a,b and c to be s... | Let's denote $S = ab+bc+ca$.
For the lower bound:
$$(a+b+c)^2 \geq 0 \Leftrightarrow a^2+b^2+c^2 + 2S = 1 + 2S \geq 0 \Leftrightarrow \boxed{S \geq -\frac{1}{2}}$$
For the upper bound we use Cauchy-Schwarz:
$$1+2S = (a+b+c)^2 = (1\cdot a + 1\cdot b+1\cdot c)^2 \leq 3\cdot (a^2+b^2+c^2) = 3 $$$$\Leftrightarrow 1+2S \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How do I show that $\cos^4x=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x )+\frac{3}{8}$ I know how to prove that
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
by substituting $\cos(2x)$ with $2\cos^2x-1$ according to the double angle identity
$$\cos(2x)=2\cos^2x-1$$
However, how do I do that for $\cos^4x$?
Do I do it by wri... | Since
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
we have
$$\cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos(4x)$$
and thus, if we square first equation we get
$$\cos^4x=\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\cos ^2(2x)=$$
$$ =\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\Big{(}\frac{1}{2}+\frac{1}{2}\cos(4x)\Big{)}$$
$$ =\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A sequence of full rank matrices $A_i \in \mathbb{R}^{m\times n}$ such that $A_i \rightarrow A$ in $\|\cdot\|_2$ $\textbf{Question:}$
Prove that for every $A\in \mathbb{R}^{m\times n}$, $\exists$ a sequence of full rank matrices $A_i \in \mathbb{R}^{m\times n}$ such that $A_i \rightarrow A$ in $\|\cdot\|_2$.
$\textbf{... | They were by choice, with the goal of obtain $\|A-A_i\|_2=\frac1{i}$, in fact, you can replace $\frac1{i}$ by another positive function of $i$ that goes to $0$ as $i \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$... | We need to prove that
$$\sum_{cyc}\ln(a^2+3)\geq4\ln4$$ or
$$\sum_{cyc}\left(\ln(a^2+3)-\ln4-\frac{1}{2}(a-1)\right)\geq0.$$
Now, let $f(x)=\ln(x^2+3)-\ln4-\frac{1}{2}(x-1).$
Thus, $$f'(x)=\frac{(x-1)(3-x)}{2(x^2+3)},$$ which says that $f$ decreases on $[3,+\infty)$ and since $f(3)>0$, there is unique $x_0>3$, for whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Show that the equation represents an ellipsoid. Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.
Workings
$$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$
C... | We have $$\frac32 = (x+y)^2 + (y+z)^2 + (x+z)^2 = 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz + 2xy$$
so set $A =\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$ and the equation can be rewritten as $$\left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Partial Fraction Decomposition of quadratic factor I'm trying to break up the following equation into partial fractions:
$$\frac{1}{(x^2-1)^2}=\frac{1}{(x+1)^2(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
I'm ending up with two equations and four unknowns though:
$$A+C=0 \\ -A + B + C+D = 0... | We have
$$\frac{1}{(x^2-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
$$=\frac{A(x+1)(x^2-1)+B(x+1)^2+C(x-1)(x^2-1)+D(x-1)^2}{(x^2-1)^2}=$$
$$=\frac{A(x^3+x^2-x-1)+B(x^2+2x+1)+C(x^3-x^2-x+1)+D(x^2-2x+1)}{(x^2-1)^2}=$$
that is
*
*$A+C=0 \implies A=-C$
*$A+B-C+D=0\implies 2A+B+D=0$
*$-A+2B-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If x is a number between 0 and 1, what number results from dropping all odd powers from the binary representation of x? Let $x$ be a number in the interval $(0,1).$ $x$ has a representation in base 2, using the finite one if the sequence terminates. Clearly we can drop either all the even powers or the odd powers in th... | One way to represent the result of such dropping seems to be with the series:
$$\sum_{n=0}^\infty \frac{\lfloor4^{n+1}x\rfloor-\lfloor4^nx\rfloor}{4^{2n+2}}$$
But this, admittedly, is a rather cheap trick
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that the series converges and find its sum Show that
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial ... | The partial sum must be:
$$S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\\
\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...\color{red}{\frac1N-\frac{1}{N+1}}= \\
\left( \color{blue}{\frac{1}{1}}-\require{cancel}\cancel{\frac{1}{2}} \right)+\left( \cancel{\frac{1}{2}}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find the Matrix X and Y $\text { Find the matrix } X \text { and } Y , \text { if } X + Y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] \text { and } X - Y = \left[ \begin{array} { c c } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$
Adding 1 and 2
$x + y +x-y = \left[ \begin{... | You have the right ideas but I see multiple mistakes.
$$X+Y+X-Y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] + \left[ \begin{array} { l l } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$$
$$2 X = \color{red}{1} \cdot \left[ \begin{array} { l l } { 8 } & { 8 } \\ { 0 } & { 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | By l'Hopital we have
$$\lim_{x \to 0}\frac{1}{x^2} - \frac{1} {\sin^2 x} =\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{\sin 2x-2x}{2x\sin^2 x+x^2\sin 2x }$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{2\cos 2x-2}{2\sin^2 x+2x\sin 2x+2x\sin 2x +2x^2\cos 2x}$$
$$\stackrel{H.R.}=\lim_{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
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Find sum of all integer solution of $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ Given $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ , which $a,b,c ,d$ are an positive integer , all solution of $a,b,c,d$ are members of Set "S" ,then find sum all of members in se... | Hint: For first question you can use following algorithm; it is a bit long operation.
$2/a+3/b+4/c+5/d=1/14$
we rewrite the equation as:
$$\frac{1}{\frac{a}{28}}+\frac{1}{\frac{b}{42}}+\frac{1}{\frac{c}{56}}+\frac{1}{\frac{d}{70}}=1$$
Now we solve equation:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1$$
Number o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $n \in \mathbb{N}$ such that n is a non-perfect square. Show that $\sqrt{n}$ is irrational. Can someone please check my proof?
Let $n \in \mathbb{N}$ such that n is a non-perfect square. Show that $\sqrt{n}$ is irrational.
Let's prove it by contradiction, that is, suppose there are $n,a,b \in \mathbb{N}$ such that ... | As the comments suggest, you overanalyze. If $\frac{a}{b}=\sqrt{n}$ then $\frac{a^2}{b^2}=n$ and $a^2=nb^2$. Thus, as with the cases of $2$ and $3$, $n\mid a^2$. Now, for every prime factor $p_i$ of $n$, if $p_i\mid a^2$, then $p_i\mid a$. Thus $(p_i)^2\mid a^2$. Remove that $p_i$ from each occurrence of $n$ and $a$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the three roots of the equation $\left(z+1\right)^3=-1$
Give your answers in the form $x+ iy$, where $x\in \mathbb{R}\:and\:y\in \mathbb{R}$.
I suspect I'll have to use de Moivre's theorem to solve this, but I don't know how to factor in the $+1$, because increasing the real part of a complex number by one doesn... | Consider instead
$$c^3 = -1$$
where $c = z + 1$. Denoting in geometric form $c = re^{i\theta}$ and $-1 = e^{i \pi}$, we get
$$r^3 e^{i3\theta} = e^{i\pi + 2k\pi}, \qquad k \in \lbrace 0,1,2 \rbrace$$
We get $r = 1$ and three different angles
\begin{align}
\theta_1 &= \frac{\pi}{3} = \\
\theta_2 &= \frac{\pi}{3} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Functions satisfying $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\}.$ I have used this identity: if $g(x)=1/(1-x),$ then
$$g^{-1}(x)=1-\frac{1}{x},$$
to get all functions satisfying: $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\},$ but I didn't get a general form of its solution. My que... | We will prove that $$f(x)=x-\frac{1}{2(1-x)}$$ when
$$f(x)+f\left(\frac{1}{1-x}\right)=x$$
At first we get $$f\left(\frac{1}{1-x}\right)+f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}$$ and we get
$$f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}$$
Now we get
$$f\left(\frac{x-1}{x}\right)+f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the inverse of a matrix using elementary matrixes This should be fairly simple as I know matrix A can be found by $$A = E_1^{-1} E_2^{-1}...E_k^{-1}$$
So it should go that $$A^-$$ can be found by $$E_kE_(k-1)...E_1$$
But for some reason my numbers aren't matching up with the book.
I have a matrix $$\begin{bmatr... | If $E_1\cdots E_n A = I$, then $A^{-1} = E_1\cdots E_n I$.
So you can transform $A$ into $I$ and simultaneously transform $I$ to $A^{-1}$ using the same transformations:
$$
\left[\begin{array}{rrr|rrr}
1&0&-1 & 1 & 0 & 0\\0 &6&-1& 0 & 1 & 0\\0&0&4& 0 & 0& 1
\end{array}\right]
\sim \left[\begin{array}{rrr|rrr}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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"Binomial coefficients" generalized via polynomial iteration This is a question I will answer myself immediately by repeating one of my old AoPS posts, since the latter post no longer renders on AoPS.
Convention. In the following, whenever $A$ is a commutative ring with $1$, and $f$ and $g$ are two polynomials over $A... | Proof of Theorem 2
The proof of Theorem 2 I shall give below is merely a generalization of GreenKeeper's proof at AoPS, with all uses of specific properties of $A$ excised.
If $a_1, a_2, \ldots, a_m$ are some elements of a commutative ring $R$, then we let $\left< a_1, a_2, \ldots, a_m \right>$ denote the ideal of $R$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
I was asked to evaluate the determinant of the I was asked to evaluate the determinant of the $n$x$n$ matrix
$$ A=
\begin{bmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{bmatrix}
$$
I tried starting from ... | First subtract the last row from the first $n-1$ rows. Then add first $n-1$ columns to the last one.
\begin{align}
\det A &= \begin{vmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{vmatrix} \\
&= \begin{v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the sum of squares of the real roots
Let $r_1,r_2,r_3,\cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+\cdots+r_n^2$ is $$(A)\,3\quad(B)\,14\quad(C)\,8\quad(D)\,16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't kno... | \begin{align}
x^8-14x^4-8x^3-x^2+1&=0
\tag{1}\label{1}
.
\end{align}
\eqref{1} can be factored as
\begin{align}
f_1(x)f_2(x)=0
,\\
f_1(x)&=x^4+4x^2+x+1
\tag{2}\label{2}
,\\
f_2(x)&=x^4-4x^2-x+1
\tag{3}\label{3}
.
\end{align}
\begin{align}
f_1(x)&=x^4+3x^2+(x^2+x+\frac14)-\tfrac14+1
\\
&=x^4+3x^2+\tfrac14(2x+1)^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$ as $n \rightarrow \infty$ The question tells me find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$.
I thought that I would use an idea similar to the one given below:
Exercise 1.8.4. $a_n=\sqrt[n]n.$
Solution. If we take $b=\sqrt[n]n$ and apply $(1.14)$, we obtain... | Assuming that you could be interested by more than the limit itself.
Considering Faulhaber polynomials
$$S_n=\sum_{i=1}^n i^7=\frac {a^2(6a^2-4a+1)} 3 \qquad \text{where} \qquad a=\frac 12 n (n+1)$$ you can write
$$S_n=\frac{n^8}8 \left(1+\frac{4}{n}+\frac{14}{3 n^2}-\frac{7}{3 n^4}+\frac{2}{3
n^6} \right)$$
$$\log(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the maximum value of $f(z)=|z^3-z+2|$ on the unit circle $|z|=1$
Let $f:\mathbb{C} \rightarrow \mathbb{R}$ be defined by $$f(z)=|z^3-z+2|\,.$$ What is the maximum value on the unit circle $|z|=1$ ?
My approach is as follows:
$z=e^{i\theta}$ as it is mentioned that the point lie on the unit circle.
$$f(z)=|1+\cos... | Let $z:=\exp(\text{i}\theta)$ and $t:=\cos(\theta)$. Then,
$$\begin{align}|z^3-z+2|^2&=\Big|\big(\cos(3\theta)-\cos(\theta)+2\big)^2+\text{i}\big(\sin(3\theta)-\sin(\theta)\big)\Big|^2
\\&=4\cos(3\theta)-2\cos(2\theta)-4\cos(\theta)+6
\\&=4(4t^3-3t)-2(2t^2-1)-4t+6
\\&=16t^3-4t^2-16t+8=:g(t)\,.
\end{align}$$
Thus, we w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $\sqrt{1}+\sqrt{2}+...+\sqrt{n} \geq \frac{2}{3}n\sqrt{n}$ The base step is pretty obvious: $1 \geq \frac{2}{3}$.
Then we assume that $P(k)$ is true for some $k \in \mathbb{Z}^{+}$ and try to prove $P(k+1)$. So I have
$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{... | For the induction step, you want to show that:
$$
\frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\
2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\
$$
Working backwards:
$$
2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\
4k^2 \times k \geq (4k^2 - 4k+1)(k+1) = 4k^3 - 4k^2 + k + 4k^2 - 4k+ 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Given $(m-2)x^2-4x+3=3-x^2+2nx$, compute the value of $m$
Given $$(m-2)x^2-4x+3=3-x^2+2nx$$ Compute the value of $m$.
$$(m-2)x^2-4x+3=3-x^2+2nx
\\mx^2+(-4m-4)x+7=-x^2+2nx+3
\\m=-1, n=0$$
The answer is $m=1,n=-2$, please tell me where did I go wrong
| Hint:
The coefficient of $x^2$ on the left of the equal sign is $(m-2)$ and on the right it is $-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Partial decomposition of $\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$ Recently I have come arcross the following fraction
$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$$
Hence I have encountered this fraction within a task of integration I want to do a partial decomposition. First of all I rewrote it as following
$$\dfrac{z^2}{(2z^2... | Based on your demand, here's another method using residues
Let
\begin{equation}
F(z)= \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
=
\frac{\frac{z^2}{2(s^2+1)}}{(z^2 + \frac{3}{2})(z^2 + \frac{1}{s^2+1})}
\end{equation}
\begin{equation}
F(z)
= \frac{\frac{z^2}{2(s^2+1)}}{\Pi_{i=0}^3(z-z_i)}
=
\frac{r_0}{z-z_0}
+
\frac{r_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that
$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$
My try:
First I wrote the inequality as
$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac... | Another way.
By AM-GM for all $x\geq2$ we obtain:
$$x^3+16=x^3+8+8\geq3\sqrt[3]{x^3\cdot8^2}=12x.$$
Thus, $$\prod_{cyc}(a^3+b)\geq\prod_{cyc}(12a-16+b)=\prod_{cyc}(5a+7a+b-16)\geq\prod_{cyc}5a=125abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
The probability of getting a 7 This problem is probably too easy for all you math geniuses out there, but I am having trouble understanding it, so here it is:
An athlete has 10 cards, numbered 1 to 10. Every day, he picks a card randomly and does that number of pushups. He doesn't replace the card, and picks until all... | $$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$
*
*The number of decks for which the first $2$ cards sum to $12$ is $(4)(2!)(8!)$, where
*
*The factor $4$ counts the ordered pairs $(2,10),\;(3,9),\;(4,8),\;(5,7)$.
*The factor $2!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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A special case of the Fermat-Torricelli point in a triangle $\triangle{ABC}$ is an isosceles right triangle; its legs are of length $s = 30\sqrt{5}$ and hypotenuse is of length $30\sqrt{10}$. The Fermat-Torricelli point $P$ must be along the median from the vertex $C$ of the right angle of the triangle.
$\left\vert\ove... | I've found no errors in your calculations.
If we can use the fact that the point $P$ satisfies
$$\angle{APB}=\angle{BPC}=\angle{CPA}=120^\circ$$
then, we can easily see that
$$|\overline{CP}|=x=15\sqrt{10}-5\sqrt{30},\qquad|\overline{AP}|=|\overline{BP}|=y=10\sqrt{30}$$
which are the same as what you've got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to prove that this sequence converges to $0$? $0 \le x_0 \le \frac{1}{2}$ , and $x_{n+1}=x_n-\dfrac{4x_n^3}{n+1}$
When I take $x_0=\sqrt{\frac{1}{12}}$, it converges very very slow. I can see it is monotonic decreasing but don't know how to find its limit.
| Let's study
$$x_{n+1}^{-2} - x_n^{-2} = \left( x_n - \frac{4x_n^3}{n+1} \right)^{-2} - x_n^{-2} = x_n^{-2} \left( \left( 1- \frac{4x_n^2}{n+1}\right)^{-2} -1\right)$$
So
$$x_{n+1}^{-2} - x_n^{-2} = x_n^{-2} \left( \frac{8x_n^2}{n+1} + o\left(\frac{x_n^2}{n+1} \right) \right) = \frac{8}{n+1} + o\left( \frac{1}{n+1} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Roots of $2x^3-4x+1$ I'm having difficulty getting the solution to the cubic equation $2x^3-4x+1=0$ and from http://www2.trinity.unimelb.edu.au/~rbroekst/MathX/Cubic%20Formula.pdf it claims that the general solution to $Ax^3+Bx^2+Cx+D=0$ is $$p+q+r=-B/A$$
$$pq+qr+rp=C/A$$
$$pqr=-D/A$$
where $p,q,r$ are the roots.
I als... | We have a cubic equation $2x^3 - 4x + 1 = 0$, what $x$ is its root ?
May you can solve this by formula $\cos 3t = 4\cos^3 t - 3\cos t$. But I hope that you understood the means and enjoys from apply them on many cubic equations. Indeed, you will solve this cubic equation.
The cubic equation is $x^3 - 2x + \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$E(X^2)$ of discrete uniform distribution I have a discrete uniform distribution (from IFoA formulae) with parameters a, b, h, where a and b are the start/end points, and h is the interval between each value. The p.d.f. is given as $\frac{h}{b-a+h}$. I manage to find out $$E[X] = \sum_{x\in X} xf(x) = \sum_{x\in X} \fr... | With your edit, you are almost there. You can simplify
$$\frac{h}{b-a+h}\sum_{k=0}^{b-a}(hk+a)^2$$
$$=\frac{h}{b-a+h}\sum_{k=0}^{b-a}\left((hk)^2+2ahk+a^2 \right)$$
$$=\frac{h}{b-a+h}\sum_{k=0}^{b-a}(hk)^2+\frac{h}{b-a+h}\sum_{k=0}^{b-a}(2ahk)+\frac{h}{b-a+h}\sum_{k=0}^{b-a}a^2$$
$$=\frac{h^3}{b-a+h}\sum_{k=0}^{b-a}k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find height of a wall if at the beginning it exceeds $10$ meters and then $8$ meters When the foot of a staircase is $5$ meters from the base of a wall, it protrudes $10$ meters above the wall; and if it is $9$ meters from the base, it stands $8$ meters.
Find the height of the wall.
Using the Pythagorean theorem for ... | The hypotenuse is $x - 10$ and $x - 8$ where $x$ is the total length of the staircase with overhang and is the same for both.
\begin{cases}(x-10)^2=5^2+y^2\\(x-8)^2=9^2+y^2\end{cases}
$(x-8)^2 - (x-10)^2 = 56$
$(x^2 - 16x + 64)-(x^2 - 20x +100) = 56$
$4x -36 = 56$
$x = 23$
$y = \sqrt{13^2 - 5^2} = 12$
Also $y = \sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is there any mistake in my approach for solving $ \int_0^{\pi/2} \frac{ \cos x}{3 \cos x + \sin x} \, dx $ ?? I had to evaluate this integral .
$$
\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx
$$
Here is how I proceeded
Dividing $N^r$ And $D^r$ by $\cos^3 x$
$$
\int_0^{\pi/2} \frac{ \sec^2 x}{3 \sec^2 x + \tan... | A simple method to avoid using partial fractions and improper integrals. Let
$$ A=\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx, B=\int_0^{\pi/2} \frac{\sin x}{3 \cos x + \sin x} \, dx. $$
Clearly
$$ 3A+B=\frac{\pi}{2}. \tag{1}$$
Noting that
$$ A=\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d\sin x, B=-\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqr... | we have to show,
$$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$$
Let,$2+\sqrt{3}=p$
$$R.H.S.= -3-2\sqrt{3}$$
$$=-2(\frac{3}{2}+\sqrt{3})$$
$$=-2\bigg(\frac{3}{2}+\frac{1}{2}-\frac{1}{2}+\sqrt{3}\bigg)$$
$$=-2(2+\sqrt{3}-\frac{1}{2})$$
$$=-2(p-\frac{1}{2})$$
$$=1-2p$$
$$=\dfrac{(2-p)(1-2p)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that ... | One of the approaches may be as follows:
Suppose there exists some $a$ which is the minimum. Then:
$$
\begin{align}
x + 2\sqrt{x^2 - \sqrt2 x+1} &= a \\
2\sqrt{x^2 - \sqrt2 x+1} &= a-x
\end{align}
$$
Square both sides:
$$
(a-x)^2 = 4x^2 - 4\sqrt2x + 4
$$
Applying some transformations you can get:
$$
3x^2 + x(2a-4\sqrt2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to ... | One way is to write $z=r^2 e^{2\theta}$ and roots will be $re^\theta$ and $re^{\pi -\theta}$.
From $z =-1+4\sqrt{3}i$, we obtain $r=7$ and $\tan{2\theta} = \frac{2\tan\theta}{1-\tan^2\theta} = -4\sqrt{3}$. Second expression gives you a quadratic equation, $2\sqrt{3}\tan^2\theta -2\sqrt{3} + \tan\theta =0$.
Roots of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Determine linear map so that equation is correct Let $f:\mathbb R^3→\mathbb R^2$, $(x,y,z)^T\mapsto(x^2y-5z,2x+4yz-3z^3)^T$,
and $(x_0, y_0, z_0)^T=(-1,0,1)^T$.
I need to determine the linear map $A∈L(\mathbb R^3,\mathbb R^2)\cong\mathbb R^{2×3}$ so that$$
f \begin{pmatrix} x \\ y \\ z \end{pmatrix}=f\begin{pmatrix} x_... | Hint.
$$
f(p) = f(p_0) +\frac{\partial f}{\partial p}|_{p_0}(p-p_0) + O(|p-p_p|^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
| Method 1:
$y = -4x^2 -2x -4$
$\frac {y}{-4} = x^2 + \frac 12x + 1$
$-\frac {y}{4}- 1 = x^2 + 2*\frac 14 x$
$-1-\frac y4 +(\frac 14)^2= x^2 + 2*\frac 14 x+ (\frac 14)^2$
$-1+\frac 1{16}-\frac y4 = (x + \frac 14)^2$
$(x+ \frac 14)^2 = -\frac {15}{16} - \frac y4 = \frac {-15 - 4y}{16}$
$x + \frac 14 = \pm \sqrt {\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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First year college question on divisibility of integers I'm having a hard time with a practice question.
Given $n$ is an integer, prove $2$ divides $(n^4 -3)$ iff $4$ divides $(n^2 +3)$.
So I know since it's an iff statement, I have to show the implication going both ways. Let's start with the left side first.
There ... | Trick: If $2|n^4 -3$ then $n$ is odd. So let $n = 2r + 1$ then $n^2 +3 = (2r+1)^2 + 3 = 4r^2 + 4r + 1 + 3 = 4r^2 + 4r + 4 = 4(r^2 + r + 1)$ and $4|n^2 + 3$.
Trick: If $4|n^2 + 3$ then $n$ is odd. So if $n$ is odd then $n^4$ is odd and $n^4 -3$ is even. So $2|n^4 -3$.
.....
In general to prove $4|n^2 +3$ we want to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Stuck solving a logarithmic calculation I'm preparing for my further studies (last year of high school, preparing so I can try and join the academy that I want), and just solving problems. Got stuck on this one:
What is the value of:
$$log_4log_3log_28 + log_{\sqrt{7}+1}(8+2\sqrt{7})+log_{\sqrt[3]{7}}7\sqrt{7}$$
Th... | Hint: $$8+2\sqrt{7}=(\sqrt{7}+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does Wolfram|Alpha make a mistake here? We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sq... | If you take the complex roots of $\sqrt[3]{x}$ you get $0$ as the limit, because the denominator is different from zero in this case.
So, Wolfram|Alpha did not make a mistake but just uses a different root of $\sqrt[3]{x}$.
For the real root you get $-2$:
*
*$t^3 = -x \Rightarrow \lim_{x \to -8}\frac{\sqrt{1-x}-3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 2
} |
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\df... | $ z - \frac{1}{z} = \frac{z^2-1}{z} $
So that, we get $\left| z - \frac{1}{z} \right| $ = $\frac{|z^2-1|}{2} $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ \frac{|x^2-y^2-1 + i(2xy)|}{2} = \frac{\sqrt{(x^2-y^2-1)^2 + (2xy)^2}}{2} $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$ here what I did .
$$f(0)=f(0)^2$$
so $f(0)=1$ or $f(0)=0$
IF $f(0)=1$
we have $f(2y)=f(y)$
$$f(1)=f(2)=\ldots=f(2^n)=a$$
the equation $f(x)-a=0$ has infinitly many solutions , so $f(x)=a$
since f(0)=1 , $f(x)=1$.
i don't ... | This is related to the Frobenius problem (or coins or Mcnuggets problem) of two numbers:
If $a,b\in\mathbb{N}$ are two numbers such that $\gcd(a,b)=1$, then all numbers $n\geq(a-1)(b-1)$ are representable as a nonnegative combination of $a,b$, i.e., $n=ax+by$ with $x,y\in\mathbb{N}$.
1) If $f(0)=0$:
Since $\gcd(4,3)=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Uniform convergence of $\sum\limits_{n=1}^∞n^{-x}(e^{\frac{x}{n^2}}-1)$
Pointwise and uniform convergence of the following series of functions:
$$\sum_{n=1}^{\infty} n^{-x}\left(e^{\frac{x}{n^2}}-1\right).$$
Now, the series of function converges pointwise as $x \in (-1, +\infty)$, because, as $n \to \infty$, $$f_n(... | $\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$First, since$$
\frac{1}{n^x} \left(\exp\left( \frac{x}{n^2} \right) - 1 \right) \sim \frac{x}{n^{x + 2}}\quad (n → ∞)
$$
for any fixed $x$, then $\displaystyle \sum_{n = 1}^∞ \frac{1}{n^x} \left( \exp\left( \frac{x}{n^2} \right) - 1 \right)$ converges iff $x + 2 > 1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Sum of $\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - ...$ I have to find the sum of :
$$\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - \frac{x^5}{5*4} +\cdots$$
So far I have :
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)}$$
which is very close to $\ln(1+x)$... but I just can't figure o... | \begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)} &= \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) \, (-1)^{n-1} \, x^{n+1} \\
&= x \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} + \sum_{n=1}^{\infty} \frac{(-1)^n \, x^{n+1}}{n+1} \\
&= x \, \sum_{n=1}^{\infty} \frac{(-1)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find the remainder when $13^{13}$ is divided by $25$. Find the remainder when $13^{13}$ is divided by $25$.
Here is my attempt, which I think is too tedious:
Since $13^{2} \equiv 19 (\text{mod} \ 25),$ we have $13^{4} \equiv 19^{2} \equiv 11 (\text{mod} \ 25)$ and $13^{8} \equiv 121 \equiv 21 (\text{mod} \ 25).$
Finall... | $13$ is coprime with $25$, so Euler-Fermat tells us that
$$
13^{\varphi(25)}\equiv1\pmod{25}
$$
Since $\varphi(25)=20$, this is not much of a help.
You can try with repeated squares: $13=1+4+8$; since
$$
13^2\equiv 19,\quad
13^4\equiv 19^2\equiv11\quad
13^8\equiv 11^2\equiv21
$$
we get
$$
13^{13}\equiv 13\cdot 11\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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An analytical solution of a tricky integral Can anyone propose a method (some methods) to determine the following indefinite integral?
$$I=\int\frac{\textrm{d}x}{\sqrt[3]{\sqrt{1+x^{2}}-x}}.$$
I think an analytical solution should be possible...
| Using the substitution $x = \sinh t$, we have $dx = \cosh t dt$.
\begin{align}
I &= \int\frac{dx}{\sqrt[3]{\sqrt{x^2 + 1}-x}}\\
&= \int\frac{\cosh t}{\sqrt[3]{\cosh t - \sinh t}}\,dt\\
&= \int e^{t/3}\cosh t \,dt\\
&= \frac{1}{2}\int\left(e^{4t/3} + e^{-2t/3}\right)\,dt\\
&= \frac{1}{2}\left(\frac{3}{4}e^{4t/3} - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $A=PDP^T$, does $P$ have to be orthogonal. A matrix A is called orthogonally diagonalizable if $A=PDP^{-1}$ and $A=PDP^{T}$, where $D$ is diagonal. Therefore, $P^{-1}=P^T$ and thus $P$ is an orthogonal matrix. If you are only given the fact that $A=PDP^T$, and $D$ is diagonal, is it guaranteed that $P$ is orthogonal... | No, your assertion is not true. Choose $$
P = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix},
$$ and let $$D = I_2 = \begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}$$ be the identity matrix (which is obviously diagonal) and $A$ defined by $$A = PDP^T = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix} \begin{pmatrix}
1 & 1 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
| It's not hard to see that $x^4+x^3-1=0$ has two real and two complex roots. If we let these be $a$, $b$, $c+di$, and $c-di$, then we have $a+b+2c=-1$ and $ab(c^2+d^2)=-1$ from the $x^3$ and constant coefficients, $ab+(c^2+d^2)+2c(a+b)=0$ from the (missing) $x^2$ coefficient, and
$${1\over a}+{1\over b}+{1\over c+di}+{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding the minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$ Minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$.
$(2+\sqrt3)^2$= $7+4\sqrt3$
$\implies$ $((2+\sqrt3)^2-7)^2$=$48$
$\implies$ $((2+\sqrt3)^4-14(2+\sqrt3)^2+1=0$
So $2+\sqrt3$ is a root of $x^4-14x^2+1$. Did i do this correct so far? Any tips on factoring ... | This strategy can be applied in lots of situations:
\begin{align}
\alpha = 2+\sqrt 3 &\implies \alpha - 2 = \sqrt 3\\
&\implies (\alpha-2)^2 = 3\\
&\implies \alpha^2 - 4\alpha + 1 = 0
\end{align}
so $\alpha$ vanishes on $x^2-4x+1$. It is irreducible since it has no rational roots, so it is the minimal polynomial.
Your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n... | \begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
&& 1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9} + \cdots\\
&=&1 + \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9} + \cdots\\
&=& \int_0^1\frac{1+x^2-2x^3}{1-x^4}dx=\int_0^1\frac{(1-x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Prove that $2z^4-3z^3+3z^2-z+1=0$ has exactly one complex root in each of the four quadrants.
I am trying to show that $$p(z)=2z^4-3z^3+3z^2-z+1=0$$ only has a single root in all four quadrants.
From two previously related posts, I have shown that $p(z)$ does not have a root on neither the imaginary or real axes. We ... | Let's do a proof by contradiction. I'm going to start from Batominovski's (1-4),
\begin{eqnarray}
a+c &=& \frac{3}{4}\tag{1}\\
x+y &=& \frac{3}{2}\tag{2} -4m\\
cx + ay &=& \frac{1}{4}\tag{3} \\
xy &=& \frac{1}{2}\tag{4}
\end{eqnarray}
and assume $m = ac \ge 0$. Since $m \ge 0$, it follows from (2) that $x + y \le 3/2$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solving $e^{2z}+e^z +1=0$ Am I doing this right?
$e^{2z}+e^z +1=0$. Let $x=e^z$ so the original equation translates to \begin{align}x^2+x+1=0\end{align}
Using the quadratic formula for real numbers, we get $\frac{-1}{2}+\frac{i\sqrt{3}}{2}, \frac{-1}{2}+\frac{i\sqrt{3}}{2}$. Equating $e^z$ with both values, let $z=r(\c... | Beware the notations ($x$) clash.
$\color{red}{\text{Suppose there exists }} z \in \mathbb{C} \text{ such that } e^{2z}+e^z+1=0.$
Let $u=e^z$. $$u^2+u+1=0$$Then, $e^z=u$, with $u\in\{e^{i\frac{2\pi}{3}},e^{i\frac{4\pi}{3}}\}$. Then, $\color{red}{\text{necessarily, }}z=i\left(\frac{2\pi}{3}+2n\pi\right) \text{ or } i\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
how to find a, b that satisfy $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ How can I find those $a$ and $b$ in $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ ?
[my attempt] Since the denominator is $x^2$, it will be $0$. And $e^{-2x}$ is 1 so I get $1-\frac{1+ax}{1+b... | As already notice the case $\lim_{x \to \infty} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ is trivial for $$\lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$$
we have by $e^{-2x}=1-2x+2x^2+o(x^2)$
$$\frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}
=\frac{e^{-2x}(1+bx) -(1+ax)}{x^2(1+bx)}
=\frac{(1-2x+2x^2+o(x^2))(1+bx) -(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^25^2\cdots (4n-3)^2}{3^27^2\cdots(4n-1)^2}$ I need to study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}$.
Now, I think we can do it by using the fact that if we have a series $\sum_{n=1}^{\infty}a_n$ ... | We are interested in
$$ \sum_{n\geq 1}\left[\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(n+\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(n+\frac{3}{4}\right)}\right]^2 $$
and by Gautschi's inequality the main term of this series behaves like $\frac{K}{n}$ as $n\to +\infty$, hence the given series is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to calculate $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$ How do I calculate the $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$ ?
[my attempt]:
I could calculate that $nI_n = 2cos^{n-1}\theta sin\theta+2(n-1)\int_0^\frac{\pi}{... | \begin{align}
I_n
&= \int_{-\pi/2}^{\pi/2}\cos^n\theta\ d\theta \\
&= 2\int_{0}^{\pi/2}\cos^n\theta\ d\theta \\
&= {\bf B}\left(\dfrac{1}{2},\dfrac{n+1}{2}\right) \\
&= \dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+1}{2}\right)}{\Gamma\left(\dfrac{n}{2}+1\right)} \\
\dfrac{I_{n+2}}{I_{n}}
&= \dfrac{\Gamma\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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find all integers $n$ such that $2^{n-1}*n+1$ is a perfect square. Clearly $n=0$
and i found that also $n=5$ gives a perfect square
And By representing the two functions , we found that there are only two solutions that are $n=0,5$
But I don't know how to prove that using elementary number theory.
| Suppose $2^{n-1} n + 1 = x^2$ with $n > 5$. Then $2^{n-1} n = (x-1)(x+1)$ is a product of two integers that differ by $2$. Corresponding to this we must have $x-1 = 2^k u$ and $x+1 = 2^{n-1-k} v$ where $uv = n$. Now $\min(k,n-1-k) \le 2$, otherwise $x+1$ and $x-1$ would differ by at least $4$.
If $k \le 2$, $x+1 \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$ Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$
My working out so far:
$|x-1|$ =
$x-1$ when $x≥1$
$1-x$ when $x<1$
Then we compute $x^2 + |x-1|$ as follows:
$x^2 + |x-1|$ =
$x^2 +x -1$ when $x≥1$
$x^2 - x + 1$ when $x<1$
So we have the following 2 ... | You have almost got it right but the final conclusion is as follows: for $x \geq 1$ we get $x=1$ and for $x <1$ the answer is $0\leq x <1$. Hence the final answer is $0\leq x \leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Geometrically proving the half-angle formula for sine from a particular diagram
I have attached a picture of the diagram I am using to prove the trig identity $\sin(\frac{\alpha}{2})= \sqrt{(\frac{1-\cos\alpha}{2})}$. I have that $\sin \alpha = \frac{DG}{OD}$ and $\sin\frac{\alpha}{2}=\frac{DE}{OD}$ as well as $\sin\f... | Construct $DF$
let $DF$ intersect $OE$ at $P$
$\triangle OPF \cong \triangle OPD$
$\angle OPD\cong \angle OPF$ is a right angle.
$\sin \frac a2 = \frac {\text {opposite}}{\text{hypotenuse}} = \frac {PF}{OF} = \frac {PD}{OD}$
$OF = OD = \cos \frac a2\\
DP = PF = \cos \frac a2\sin \frac a2$
$DF= DP + PF = 2\cos \frac a2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^... | You were close to a shorter proof after you expanded $(k+5)^5-(k+1)$, since
\begin{align*}
(k+5)^5 - (k+1) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k \\
&= (k^5-k) + (5k^4 + 10k^3 + 10k^2 + 5k),
\end{align*}
and by induction hypothesis, $5$ divides $k^5-k$ and clearly $5$ divides each term in $5k^4 + 10k^3 + 10k^2 + 5k$, so $5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have n... | It's not out of the question to compute
$$(1-\sqrt[3]2+\sqrt[3]4)^2=1+\sqrt[3]4+2\sqrt[3]2+2(\sqrt[3]4-\sqrt[3]2-2)=3(\sqrt[3]4-1)$$
so that
$$c^2=\left(1-\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9 \right)^2={3(\sqrt[3]4-1)\over3\sqrt[3]3}={\sqrt[3]4-1\over\sqrt[3]3}$$
and, since $\sqrt[3]{54}=3\sqrt[3]2$,
$$c+\sqrt[3]6={1-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that any root $z$ of $z^4+z+3=0$ satisfies $|z|>1$ Show that any root $z$ of $z^4+z+3=0$ satisfies $|z| > 1$
My working is as such:
Let $|z| \leq 1$.
Then consider $|z^4+z| \leq |z^4| + |z| \leq 1 + 1 = 2$.
So, $|z^4 + z| \leq 2$.
Thus $-2 \leq z^4 + z \leq 2$. So $1 \leq z^4+z+3 \leq 5$.
And so $z^4+z+3 \neq 0$.
... | It's better to use Rouché's theorem, with $f(z)=z^4+z$ and $g(z)=3$ then on $z=1$ we have
$$|f(z)|=|z^4+z^3|\leq2<3$$
then the number of zeros $z^4+z+3$ and $3$ are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2989455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
The distances from a point to the corners of a rectangle are $6$, $7$, $9$, and (integer) $d$. Find $d$.
Christina is standing in a rectangular garden. Her distances from the corners of the garden are $6$ meters, $7$ meters, $9$ meters, and $d$ meters, where $d$ is an integer. How to find $d$?
Can someone lend me yo... | Let $P$ be a point inside a rectangle $XYZW$ such that $\{PX,PY,PZ\}=\{6,7,9\}$ and $PW=d$ is an integer. Suppose that the projections of $P$ onto $XY$, $YZ$, $ZW$, and $WX$ are $A$, $B$, $C$, and $D$, respectively. Write
$$x:=XA=WC\,,\,\,y:=YA=ZC\,,\,\,z:=YB=XD\,,\text{ and }w:=ZB=WD\,.$$
Therefore,
$$PX^2=x^2+z^2\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2989607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve equation in prime numbers
Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$
I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1\leq p^2 q^2$ for $p,q >2 $ but how to prove it.
| My solution
Case 1. Let $p \mod 3=q \mod 3=1$. Then $p^3+q^3+1=3=0 \mod 3 \neq p^2 q^2=1 \mod 3$
Case 2 $p \mod 3=q \mod 3=2$. Then $p^3+q^3+1=2+2+1=2 \mod 3 \neq p^2 q^2=1 \mod 3$
Сase 3. $p \mod 3=1, q \mod 3=2$. Then $p^3+q^3+1=1+2+1=1 \mod 3 \neq p^2 q^2=1 \cdot 2 =2 \mod 3$
So, must be, say $p=0 \mod 3 \implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove the matrix is positive Consider the matrix $A=\begin{bmatrix}
1 & 1/2 & 1/3 &\dots &1/n \\
1/2 & 1/3 & 1/4 &\dots &1/(n+1) \\
\vdots & \vdots & \vdots & \vdots & \vdots\\
1/n & 1/(n+1) & 1/(n+2) & \dots& 1/(2n-1)
\end{bmatrix}$
Prove that $A$ is positive.
My work: $A$ is diagonalisable, symm... | tried two , Sylvester Inertia
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 \\
\frac{ 1 }{ 3 } & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
60 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{ 1 }{ 3 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
maximal surface of a cone inscribed in a sphere with radius 1 Amongst all the cones inscribed in a sphere of radius 1, I have to find that one of maximal surface.
Every cone is characterized by a value for α, the opening angle of the cone,
with $\alpha \in [0, \frac{\pi}{2}]$ (the angle between the vertical line pass... | I'll start from the equation $S'(\alpha)=0$. $$\begin{aligned}S'(\alpha)&=\cos \alpha (-4 (\sin \alpha)^3-3 (\sin \alpha)^2 +2\sin \alpha +1)\\&=\cos\alpha(\sin\alpha+1)(1+\sin\alpha-4\sin^2\alpha)\end{aligned}$$
Thus, equation $S'(\alpha)=0$ is equal to $\cos\alpha=0$ or $\sin\alpha=-1$ or $1+\sin\alpha-4\sin^2\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problems with proof by induction $\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$?
$$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Prove for $n=1$: $$\frac1{1\times2}=\frac1{1+1}=\frac12$$
Hip: $$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = ... | Actually, this is incorrect. The correct answer should be $1-\frac1{n+1}$.
We can show that the above is true for $n=1$ easily. Now let us show $1-\frac1{n+1}+\frac1{(n+1)(n+2)}=1-\frac1{n+2}$
We can prove that $\frac1{n+1}-\frac1{n+2}=\frac {n+2}{(n+1)(n+2)}-\frac {n+1}{(n+1)(n+2)}=\frac{n+2-n-1}{(n+1)(n+2)}=\frac1{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$ Can anyone give me a proof or a reference for a proof that there exists a prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$, for $n$ sufficiently large?
I am reading a lecture where the professor uses this fact but does not provide any reference.
Thank you!
| I think this gets you in the ball park.
Number of primes $<x\approx \frac{x}{ln(x)}$
$\frac{\sqrt{x}}{ln(\sqrt{x})}=\frac{2}{\sqrt{x}}\frac{x}{ln(x)}$
$\frac{x^{1/3}}{ln(x^{1/3})}=\frac{3}{x^{2/3}}\frac{x}{ln(x)}$
So it suffices to prove $(\frac{2}{\sqrt{x}}-\frac{3}{x^{2/3}})\frac{x}{ln(x)}>1$
for sufficient large x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$
Prove that
$$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$
So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum... | $$k \binom nk^2=\binom nk\cdot k\binom nk$$
For $k\ge1,$ $$k\binom nk=k\cdot\dfrac{n!}{k!\cdot(n-k)!}=n\dfrac{(n-1)!}{(k-1)!\{n-1-(k-1)\}!}=n\binom{n-1}{k-1}$$
Now in the identity $(1+x)^{2n-1}=(x+1)^n(1+x)^{n-1},$ compare coefficients of $x^n$
$$\binom{2n-1}n=\sum_{k=0}^n\binom nk\binom{n-1}{k-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikip... | Let $S_n=\prod_{k=1}^n\frac{2k-1}{2k}$. From Wallis' product, we have
$$\prod_{k=1}^\infty\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)=\frac{\pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$\prod_{k=1}^n\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)<\frac{\pi}{2}$$
for all $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$ In a proof, the author states that it is clear that:
Given $x\geq 1$ and $ n-x \geq 1$ and finally also $n\geq 2$
$${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$$
This is not immediately clear to me. Of course, If I have $n$ objects and I split them up in $x$ and $n-... | Writing them out,
${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$
becomes
$x(x-1)+(n-x)(n-x-1) \le (n-1)(n-2)
$
or
$x^2-x+n^2-n(x+x+1)+x^2-x
\le n^2-3n+2
$
or
$2x^2-2x
\le n(2x+1-3)+2
$
or
$2x^2-2x
\le n(2x-2)+2
$
or
$x^2-x
\le n(x-1)+1
$
or
$x(x-1)
\le n(x-1)+1
$
or
$(x-n)(x-1)
\le 1
$
which is true for
$x \ge 1$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Find all the intervals in which $ -x^4 + x + 3 \ge 0 $ How do I find all the intervals in which $$-x^4 + x + 3 \ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f... | You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0\implies x^3=\frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $\dfrac1{\sqrt[3]4}$ we find
$$-\dfrac1{4\sqrt[3]4}+\dfrac1{\sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"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.