Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why can we use inspection for solving equation with multiple unknowns? In our algebra class, our teacher often does the following:
$a + b\sqrt{2} = 5 + 3\sqrt{2} \implies \;\text{(by inspection)}\; a=5, b = 3
$
I asked her why we can make this statement. She was unable to provide a satisfactory answer. So I tried prov... | We want to show that the only rational solutions $a,b,x,y$ of
$$a + b\sqrt{2} = x + y\sqrt{2}$$
are given by $a = x$ and $b = y$.
(Note that if you allow for real values of $a,b,x,y$, then for example $a = x + \sqrt{2}$, $b = y - 1$ would be a solution, too.)
If $y = b$, then obviously $a = x$.
Now assume $y \neq b$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3... | Hint:
The expansion of $(x+h)^\frac{4}{3}$ is
$$h\sqrt[3]{x+h} + x\sqrt[3]{x+h}$$
So then you will have
$$\lim_{x \rightarrow \infty}\frac{h\sqrt[3]{x+h} + x\sqrt[3]{x+h} - x^{\frac{4}{3}}}{h}$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $ \int_{25\pi/4}^{53\pi/4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx $ How to evaluate the integral
$$\int_{25\pi / 4}^{53\pi / 4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx\ ?$$
| For any integrable function $f:\mathbb{R}\to\mathbb{R}$ with period $T$ and any $a,b\in\mathbb{R}$ we have
$$
\int_{b}^{b+T} f(x)dx=\int_{a}^{a+T}f(x)dx\tag{1}
$$
And as the consequence for $k\in\mathbb{N}$
$$
k\int_{b}^{b+T} f(x)dx=\int_{a}^{a+kT}f(x)dx\tag{2}
$$
Also for any integrable $f$ we have
$$
\int\limits_{-c}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Splitting field of $f=x^6+x^5+1$, orders of elements, and minimal polynomials.
Let $F$ be the splitting field over $\Bbb{Z}_2$ of $f=x^6+x^5+1$, an irreducible polynomial over $\Bbb{Z}_2$.
We know that:
$\bullet$ The Galois group of $F$ over $\Bbb{Z}_2$ is cyclic and is generated by the Frobenius autormorphism.
$\bull... | $(x^9-1)/(x-1)$ is not $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$, but $x^8+x^7+\cdots +x+1$.
Indeed,
$$
x^9-1=(x^6+x^3+1)(x^2+x+1)(x-1),\quad \frac{x^9-1}{x-1}=(x^6+x^3+1)(x^2+x+1).
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solve the differential equation ${dy \over dx} + qy^4+y=0$ Solve the differential equation ${dy \over dx}+qy^4+y=0$
the initial condition is $y(x=1)=-\frac 12$ with considering $q \ll 1$.
don't consider the powers of $q$ which are higher than $2$.
You will use Expansion.(expand the function $y=f(x)$ with the powers o... | In the ODE $\displaystyle \quad\frac{dy}{dx} + qy^4 + y = 0\quad$ treat $x$ as a function of $y$, we have:
$$
x(y) = \int dx = - \int \frac{dy}{qy^4 + y} = -\frac13\int \frac{d(y^3)}{y^3(qy^3+1)}
= -\frac13 \int \left(\frac{1}{y^3} - \frac{q}{qy^3+1}\right)d(y^3)\\
= -\frac13 \left(\log(y^3)-\log(qy^3+1)\right) + K
= -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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finiding a ratio in Trapezoid
Given $ABCD$ a right angle trapezoid, $\measuredangle A=90^{0}$, $AB\parallel DC$.
$O$ is the point of intersection of the diagonals.
$E$ is a point on $BC$ such that: $AB=BE$, $DC=CE$.
$AD=12$, $BC=13$
Need to find the ratio $AO:OC$.
Tried to compute the area and divided it the partial ... | The given $ABCD$ trapazoid with $\angle BAD = 90^\circ$ is depicted in the following figure with given data:
It is given that $AD=12, \ BC=13$, and $E$ is situated on $BC$ in such a way that $AB=BE$ and $DC=CE$. If you draw a perpendicular line to $CD$ from the point $B$, which meets $CD$ at point $F$, then $AB = DF$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to ... | I provide one more solution, where we don't use sines and cosines.
First, some preparation.
We all know that
$$
\tan (x \pm y) = \frac {\tan (x) \pm \tan (y)} {1 \mp \tan (x)\tan (y)}.
$$
Then
$$
\cot (x - y) = \frac {\cot (x) \cot (y) + 1} { \cot (y) - \cot (x)},
$$
or
$$
\cot (x) \cot (y) = \cot(x-y) (\cot(y) -\... | {
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"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 4
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Prove that: $\sqrt{x^2+21}+\sqrt{2y^2+14}+\sqrt{z^2+91}\ge 19$ Let $x, y, z$ be real number such that $xy+yz+zx=11$. Prove the inequality:
$$\sqrt{x^2+21}+\sqrt{2y^2+14}+\sqrt{z^2+91}\ge 19$$
I think that inequality can be solved by Minkowski. Equality holds if only is $(x;y;z)=(2;1;3)$...But I couldn't continue...
| we can prove this inequality $x,y,z$ are positive numbers.
By cauchy-Schwarz inequality have
$$\sqrt{\dfrac{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}{n}}\ge\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}$$
where $a_{i}>0,i=1,2\cdots,n$\
then
$$\sqrt{x^2+21}=5\sqrt{\dfrac{\dfrac{x^2+1}{5}+1+1+1+1}{5}}\ge 5\dfrac{\sqrt{\dfrac{x^2+1}{5}}}{5}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Cyclic polynomial proof 2. So here's my second question for the day,
Q. Manipulate the given equality
$$a^2b^2(bc-a^2)+b^2c^2(ca-b^2)+c^2a^2(ab-c^2)$$$$ = a^2b^2(b^2-ac)+b^2c^2(c^2-ab)+c^2a^2(a^2-bc)$$
to obtain the following:
$$(a^2+b^2)(b^2+c^2)(c^2+a^2)=abc(a+b)(b+c)(c+a)$$
| Rearrange the LHS as
$$a^2b^2(bc)+b^2c^2(ca)+c^2a^2(ab) - (a^2b^2(a^2)+b^2c^2(b^2)+c^2a^2(c^2)) = A - B$$
Rearrange the RHS as
$$a^2b^2(b^2)+b^2c^2(c^2)+c^2a^2(a^2) - (a^2b^2(ac)+b^2c^2(ab)+c^2a^2(bc)) = C - D$$
So now we have
$$A - B = C - D$$
Exchange
$$A + D = B + C$$
First we have the following
$$(x+y)(y+z)(x+z) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluating $\int_0^\infty \frac{\cos(ax)-e^{-ax}}{x \left(x^4+b^4 \right)}dx$ How can we evaluate
$$\int_0^\infty \frac{\cos(ax)-e^{-ax}}{x \left(x^4+b^4\right)}dx \quad a,b>0$$
using Complex Analysis? This problem was given in a Complex Analysis book which I was reading. The answer given in it is
$$\frac{\pi}{4b^4}e^... | The following is an evaluation that does not use contour integration.
Let $ \displaystyle I(a) = \int_{0}^{\infty} \frac{\cos (ax) - e^{-ax}}{x(x^{4}+b^{4})} \ dx$.
Then differentiating under the integral sign, $$ \begin{align} I^{(4)}(a)+b^{4}I(a) &= \int_{0}^{\infty} \frac{\cos(ax)-e^{-ax}}{x} \ dx \\ &= \text{Ci}(ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Diophantine equation involving prime numbers : $p^3 - q^5 = (p+q)^2$ Find all pairs of prime nummbers $p,q$ such that $p^3 - q^5 = (p+q)^2$.
It's obvious that $p>q$ and $q=2$ doesn't work, then both $p,q$ are odd. Assuming $p = q + 2k$ we conclude, by the equation, that $k|q^3 - q - 4$ because $\gcd(k,q)=1$ (else $p$ i... | $$p^3-q^5=(p+q)^2=p^2+2pq+q^2$$
So, $p^3-q^5\ge 0$, so $p^3\ge q^5$, so $p>q$.
$$p^2(p-1)=p^3-p^2=q^5+2pq+q^2=q(q^4+2p+q)$$
Hence, $p^2$ divides $q^4+2p+q$ (as it can't divide q) and so $q$ divides (p-1).
So $p^2\le q^4+2p+q$, so $p\le q^2+1$. We can then verify easily that $q=2$ has no solution for $p$.
But $p$ is pri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Asymptotic expansion of a function $\frac{4}{\sqrt \pi} \int_0^\infty \frac{x^2}{1 + z^{-1} e^{x^2}}dx$ How to find the asymptotic expansion of the following function for large values of $z$.
$$f_{3/2}(z) = \frac{4}{\sqrt \pi} \int_0^\infty \frac{x^2}{1 + z^{-1} e^{x^2}}dx $$
I have to get something like (in the book)
... | For simplicity let's write $\ln z = \lambda$. The integral we're trying to approximate is
$$
I(\lambda) = \int_0^\infty \frac{x^2}{1+\exp(x^2-\lambda)}\,dx
$$
as $\lambda \to \infty$.
Begin by making the change of variables $x^2-\lambda = y$ to get
$$
\begin{align}
I(\lambda) &= \frac{1}{2} \int_{-\lambda}^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
| \begin{align}
&\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)}
=
{1 \over 9}\sum_{k = 0}^{\infty}{1 \over \left(k + 1/3\right)\left(k + 2/3\right)}
=
{1 \over 9}\,{\Psi\left(1/3\right) - \Psi\left(2/3\right) \over 1/3 - 2/3}
\\[3mm]&=
{{1 \over 3}\left\lbrack \Psi\left(2 \over 3\right) - \Psi\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4... | You could use generating functions. Put $$f(z) = \sum_{n\ge 1} a_n z^n.$$
Summing your recurrence for $n\ge 2$ and multiplying by $z^n,$ we get
$$ f(z) - \frac{1}{4} z = \frac{1}{4} \sum_{n\ge 2} n z^n
- \frac{1}{2} z \sum_{n\ge 2} z^{n-1} \sum_{k=1}^{n-1} a_k.$$
Simplify to obtain
$$ f(z) - \frac{1}{4} z = \frac{1}{4... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
| Hint: Write $$x+\sin \frac1x = x\left(1+\frac1x \sin \frac1x\right)$$ and use
$$\sin \frac1x=\frac1x - \frac{1}{3!x^3} + \frac{1}{5!x^5}+ \ldots, \quad x \to \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working :
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$
N... | Hint:
$$2^{x+2}\cdot3^{\frac{3x}{x-1}}=3^2\iff 2^{x+2}=3^{2-\frac{3x}{x-1}}=3^{-\frac{x+2}{x-1}}\iff \begin{cases}x+2=0\\{}\\x-1=-\log_23\end{cases} \;\;(\text{why?)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially.
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? ... | We are given $\frac{1}{x^2 + 6 x + 10}$ which we can easily rearrange to $\frac{1}{(x+3)^2 +1}$.
If we consider the maximum value of $(x+3)^2 + 1$ then clearly for large positive or negative numbers this can become arbitrarily large and will approach $+\infty$ for large positive or negative $x$. We are dividing 1 by ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Where's the mistake in this proof of Euler's reflection formula? $$\frac{\sin(\pi x)}{\pi}=x\prod_{r=1}^\infty\left(1-\frac{x^2}{r^2}\right)$$
$$\Gamma(x)=\frac{1}{x}e^{-\gamma x}\prod_{r=1}^\infty\left(\frac{r}{x+r}\right)e^{\frac{x}{r}}$$
$$\Gamma(x) \Gamma(1-x)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\fra... | In the line
$$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$
you have divergent products. If one wants to give meaning to it, the only somewhat sensible meaning is $0 = 0$. Of course dividing that equation by $0$ does not produce anyt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Alternating binomial sum with intervals of two Fix integer $n\geq 1$. Consider the number $$1-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$where the sum continues as long as the lower number in the binomial is $\leq n$. Is there a way to simplify this sum?
The first few values are $1, 0, -2, -4, -4$.
| Note that
$$(1+x)^n = \dbinom{n}0 + x \dbinom{n}1 + x^2 \dbinom{n}2 + x^3 \dbinom{n}3 + \cdots + x^n \dbinom{n}n$$
Hence,
$$(1+i)^n = \dbinom{n}0 + i \dbinom{n}1 + i^2 \dbinom{n}2 + i^3 \dbinom{n}3 + \cdots + i^n \dbinom{n}n$$
where $i^2=-1$.
Hence,
$$\text{Real part of }(1+i)^n = 1 - \dbinom{n}2 + \dbinom{n}4 - \dbino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486193",
"timestamp": "2023-03-29T00:00:00",
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If $a,b,c > 0$ and $a+b+c = 1$, then $(\tfrac{1}{a} −1)(\tfrac{1}{b} −1)(\tfrac{1}{c} −1) \geq 8$
If $a, b, c$ are positive real numbers and $a+b+c = 1$, prove that
$$\left(\frac{1}{a} −1\right)\left(\frac{1}{b} −1\right)\left(\frac{1}{c} −1\right) \geq 8.$$
Thank you.
| This following is my new methods:
let $$x=1-\dfrac{1}{a},y=1-\dfrac{1}{b},z=1-\dfrac{1}{c}$$
then
$$a+b+c=1\Longrightarrow \dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}=1$$
$$\dfrac{x}{x+1}=
\dfrac{1}{1+y}+\dfrac{1}{1+y}\ge2\sqrt{\dfrac{1}{(1+y)(1+z)}}\cdots (1)$$
$$\dfrac{y}{1+y}\ge2\sqrt{\dfrac{1}{(1+x)(1+z)}}\cdots (... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $ for $0Let $a,b,c$ be positive real numbers between $0$ and $1$ ,i.e., they lie in the closed interval $[0,1]$. Prove that :
$$ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $$
| Let $f(a,b,c)=\sum\limits_{cyc}\frac{a}{b+c+1}+\prod\limits_{cyc}(1-a)$.
Hence, $\frac{\partial^2f}{\partial a^2}=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\geq0$,
which says that $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $$\max_{\{a,b,c\}\subset[0,1]}f=\max_{\{a,b,c\}\subset\{0,1\}}f=f(1,1,1)=1.$$
Done!... | {
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find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$ Let $a,b,$ be real numbers such that $a\ge 4b$ .
Find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$.
This problem is http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=553680
| The minimum is either on the interior of the feasible set, or on the boundary.
The gradient of the objective is
$$\left[2(a-b)+2(a-1), -2(a-b)+2(b-1)\right] = \left[4a-2b-2, 4b-2a-2\right],$$
and so the unconstrained minimum is the infeasible point $(1,1)$. Therefore the minimum is on the boundary. Substituting $a=4b$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\gcd( m,n)$=1 If $a$ and $b$ are co prime and $n$ is a prime, show that:
$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a+b$ is a multiple of $n$
Also enlighten me why $n$ has to be prime so that $\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a$+$b$ is a multiple of $n$?
| If you like modular arithmetic, note that $$\tag 1 a\equiv -b\mod a+b$$
Now, since $n$ is odd $$\frac{{{a^n} + {{b }^n}}}{{a + b}}=\frac{{{a^n} - {{\left( { - b} \right)}^n}}}{{a - \left( { - b} \right)}} = \sum\limits_{k = 0}^{n - 1} {{a^k}{{\left( { - b} \right)}^{n - k-1}}} $$ and using $(1)$ $$\frac{{{a^n} + {{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
what to do next recurrence relation when solving exponential function? find gernal solution of :$a_n = 5a_{n– 1} – 6a_{n –2} + 7^n$
Homogeneous solution:
$$a_n -5a_{n– 1} + 6a_{n –2} = 7^n$$
put $a_n=b^n$:
$$b^n -5b^{n– 1} + 6b^{n –2} =0
\\b^{n-2} (b^2-5b^{} + 6b) =0
\\b^2-5b^{} + 6b =0
\\(b-2)(b-3)=0\\
b=2,3$$
$$a^h_{... | Use generating functions... Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence so there aren't subtractions in indices:
$$
a_{n + 2} = 5 a_{n + 1} - 6 a_n + 49 \cdot 7^n
$$
Multiplply by $z^n$, sum over valid indices ($n \ge 0$), and recognize the resulting sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 5 \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find $\int_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\,\mathrm dx$ $$\large\int \limits_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\mathrm dx$$
TL;DR: Is there an elegant way of integrating this? I've reduced it to a series, detailed below, but the closed form eludes me, and the only solution I've seen uses a ra... | Use the Fourier series
$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k \\
\ln(\cos(x)) = -\ln(2) - \sum_{k=1}^\infty (-1)^k \frac{\cos(2kx)}k$$
and expand the integrand to
$$\begin{align*}
\ln(\sin(x)) \ln(\cos(x)) &= \ln^2(2) + \ln(2) \sum_{k=1}^\infty \frac{1+(-1)^k}2 \frac{\cos(2kx)}k \\[1ex]
& \qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Solving $y^2(x^2+1) +x^2(y^2+16) =448$ $y^2(x^2+1) +x^2(y^2+16) =448$
The task is to find all solutions in integers $(x,y)$. This is the fourth question of rmo 1st stage.The solution here is not complete. I have tried to solve unable to.
enter link description here
| First, this equation is symmetric under $x\to-x$ and $y\to-y$, independently. Therefore we can restrict attention to $x,y>0$ and such a solution will represent four actual solutions (we easily see that $x=0$ or $y=0$ are not possible integer solutions).
Solve with the quadratic formula to find
$$y=4\frac{\sqrt{28-x^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding a non-recursive formula for a recursively defined sequence So I have a recursive definition for a sequence, which goes as follows:
$$s_0 = 1$$
$$s_1 = 2$$
$$s_n = 2s_{n-1} - s_{n-2} + 1$$
and I have to prove the following proposition: The $n$th term of the sequence defined above is $s_n = \frac{n(n+1)}{2} + 1$.... | Your work is fine.
$${s_{k + 1}} = \frac{{k(k + 3)}}{2} + 2 = \frac{{k(k + 3) + 2}}{2} + 1 = \frac{{{k^2} + 3k + 2}}{2} + 1 = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2} + 1$$
As GitGud has pointed out, you should assume that the formulas for $s_k$ and $s_{k-1}$ hold true, since you need them both for you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:
Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow 3|(n(n(n+3)) + 2)$
But then?
| To do it inductively, we also need a base case: $0 \times 1 \times 2=0$ is divisible by $3$. Then, if $3$ divides $k(k+1)(k+2)=k^3+3k^2+2k$ then
\begin{align*}
(k+1)(k+2)(k+3) &= k^3+6k^2+11k+6 \\
&= (k^3+3k^2+2k)+3(k^2+3k+2).
\end{align*}
We know $k^3+3k^2+2k$ is divisible by $3$, by the inductive hypothesis, and $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found t... | Although another answer is hardly needed at this point, here is yet another way of thinking about this. Note that, for any real numbers $x$ and $y$, one has
$$ \begin{pmatrix}
1 & x \\ 0 & 1 \\
\end{pmatrix} \cdot
\begin{pmatrix}
1 & y \\ 0 & 1 \\
\end{pmatrix} =
\begin{pmatrix}
1 & x+y \\ 0 & 1 \\
\end{pmatrix}.$$
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 2
} |
Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive ... | If $$f(x)=(x-a)^3+(x-b)^3+(x-c)^3$$ then $$f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2$$ Since $a,b,c$ are distinct real numbers $f'(x) > 0$ for all $x\in\mathbb{R}$ and therefore $f$ is strictly increasing and therefore it has only one real root.
EDIT: The last statement is true since $f$ is a polynomial function of degree $3$ (... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What are the values for which the series converges? Determining the values of $a$ and $b$ so that the series $\sum a_n$ converges, where
$$a_n=\ln n-a\ln(n+1)+b\ln(n+2)$$
| \begin{align}
a_{n}
&=
\ln\left(n\right) - a\ln\left(n + 1\right) + b\ln\left(n + 2\right)
\\[3mm]&=
\ln\left(n\right)
-
a\left[
\ln\left(n\right) + {1 \over n} - {1 \over 2n^{2}} + \cdots
\right]
+
b
\left[
\ln\left(n\right) + {2 \over n} - {2 \over n^{2}} + \cdots
\right]
\\[3mm]&=
\left(1 - a + b\right)\ln\left(n\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expressing as a sum of squares. Suppose a number can be expressed as a sum of two squares. If I know it's prime factorization, what is the easiest way to find two such numbers?
For e.g., let us consider the number $97^5 \cdot 641^3 \cdot 3^8$, how can I find $a$ and $b$ such that $a^2 + b^2 = 97^5 \cdot 641^3 \cdot 3^8... | First you must do it for its prime factors by trial, then you can use Brahmagputa's identity.
$$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2+2abcd-2abcd$$
$$=(ac+bd)^2+(ad-bc)^2$$
Notice that if $p$ is a prime number and there exist $a$ and $b$ such that $a^2+b^2=p$ then without loss of generality we can assume that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The discriminant of a cubic extension Let $K=\mathbb{Q}[\sqrt[3]{7}]$. I want to find the discriminant $d_K$ of the number field $K$.
I have computed $\operatorname{disc}(1,\sqrt[3]{7},\sqrt[3]{7^2})=-3^3\cdot 7^2$. I know that $d_K$ must divide this quantity, and may differ form it by a square, hence we have 3 possibi... | Yes, it is possible to prove it independent of computing the discriminant.
It suffices to check that $\mathbb{Z}[\sqrt[3]{7}]$ is normal (why?). To see this, we a priori need to check that $M\mathbb{Z}[\sqrt[3]{7}]_M$ is principal for all $M$ maximal in $\mathbb{Z}[\sqrt[3]{7}]$ (note $\mathbb{Z}[\sqrt[3]{7}]$ is alre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the coefficient of $x^{4}$ in $(1+x)^{\frac{1}{3}}$ Find the coefficient of $x^{4}$ in $(1+x)^{\frac{1}{3}}$
I try to use the general binomial coefficient to do this
so I will first find ${{\frac{1}{3}} \choose k}$
which is equal to ${\frac{(\frac{1}{3})(\frac{1}{3}-1) ... ({\frac{1}{3}-k+1})}{k!}}$
and I take o... | We have $(1+x)^\alpha = \sum_{k=0}^\infty \binom{\frac{1}{3}}{k} x^k$, for $|x|<1$.
Hence the coefficient of $x^4$ is $\binom{\frac{1}{3}}{4} =\frac{(\frac{1}{3}) (\frac{-2}{3}) (\frac{-5}{3}) (\frac{-8}{3}) }{4!} = - \frac{10}{243}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...
$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$
How can I get that?
| Welcome to StackExchange! You may use LaTeX-Formulas by using dollar signs \$ x^2 \$ $\longrightarrow{}$ $x^2$.
As a first step I used the substitution $u:=\sqrt{2-x^2}$ which implies $ \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2x}{2\sqrt{1-x^2}}=\frac{x}{u}$. So we have to solve $\int \sqrt{2-u}\cdot u\ \mathrm{d}u$. This... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (... | Assume $x^2+xy+y^2<0$. Adding and subtracting $xy$ on the left-hand side gives $x^2+2xy+y^2-xy=(x+y)^2-xy<0$, and therefore $0\leq(x+y)^2<xy$. Conversely, $x^2+xy+y^2<0$ implies $xy<-(x^2+y^2)\leq0$. Combining these, we have $$xy<0<xy,$$ a clear contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
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What is wrong with this proof of: $2+2 = 5$ I have seen this image and surprised that we can prove $2 + 2 = 5$. can any one tell me what is wrong with this image.
Prove that, $2+2=5$.
We know that, $2+2=4$
$$\begin{align}\Rightarrow2+2&=4-\dfrac92+\dfrac92\\\,\\
&=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{16... | Hint: Given any real number $x,$ we have $$\sqrt{x^2}=|x|.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Divisibility of Fibonacci numbers This question is inspired by a Project Euler problem I was working on. Noticing something that did not make sense led me to the conclusion that for all primes $p$ ending in $1$ or $9$, the $(p-1)$st Fibonacci number is divisible by $p$. I haven't proven it, but it's the only conclusi... | It's true because $5$ is a quadratic residue modulo an odd prime $p \neq 5$ if and only if $p \equiv \pm 1 \pmod{5}$. The relevance of $5$ can be seen from Binet's formula
$$F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi},$$
where
$$\varphi = \frac{1+\sqrt{5}}{2};\quad \psi = \frac{1-\sqrt{5}}{2} = 1 - \varphi = -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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An awful factorisation question. If $a+b+c = 0$ show that $$(2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)$$
I have tried substituting the values but it gets too complicated.
Can anyone please help with the method? I have been trying for 30 minutes.
Thanks!
| Let $x=2a-b$, $y=2b-c$, and $z=2c-a$. Then $x+y+z=0$, so $LHS=x^3+y^3+z^3=x^3+y^3+(-x-y)^3=3xy(-x-y)=3xyz=RHS$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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prove that $\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$ How can I prove that if $x$ and $y$ are positive then
$$\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$$
| I am using the notation $\{x\}$ to mean the fractional part of $x$.
$$
x = \lfloor x \rfloor + \{x\}, \;\; y = \lfloor y \rfloor + \{y\} \\
xy = \lfloor x\rfloor \lfloor y \rfloor + \lfloor x\rfloor \{y\} + \{x\}\lfloor y\rfloor + \{x\} \{y\} \\
\lfloor xy\rfloor=\lfloor x\rfloor \lfloor y\rfloor + \lfloor\lfloor x\rf... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
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Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| $ax^2+bx+c = a(x^2+\frac{bx}{a}+\frac{c}{a})=a((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2})$
Equate this to zero
NOTE: $a\neq0, $ if it is then the equation will not be quadratic.
you get $(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$
which gives the desired $$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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How is this matrix called (two diagonals)? I need to write an algorithm for solving this matrix but I wanted to first make a search online and that's why I need its name.
| You can rearrange terms, to get the following system
$$
\left [ \begin{array}{ccccccc}
d_1 & a_{2n+1} \\
a_1 & d_{2n+1} \\
& & d_2 & a_{2n} \\
& & a_2 & d_{2n} \\
& & & & \ddots \\
& & & & & d_n & a_{n+2} \\
& & & & & a_n & d_{n+2} \\
& & & & & & & d_{n+1}
\end{array}\right ] \left [ \begin{array}{c}
x_1 \\ x_{2n+1} \\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove: $3(a^4+b^4+c^4)+48\ge 8(a^2b+b^2c+c^2a)$ Let $a, b, c$ - real numbers. Prove that $3(a^4+b^4+c^4)+48\ge8(a^2b+b^2c+c^2a)$
| Consider the following inequality obtained by AM-GM inequality:
$$
2a^4+b^4+16=a^4+a^4+b^4+16\geq 4\sqrt[4]{16a^8b^4}=8ba^2
$$
Writing down similar inequalities for other pairs we get:
$$
2b^4+c^4+16\geq 8cb^2\\
2c^4+a^4+16\geq 8ac^2
$$
It is enough to sum up all these inequalities and we get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/521268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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clarification for the solution of $(z^3=|z|^2 +4)$!? $(z-2)(z^2+z+2)=0$
so the solutions are
$$z=2;$$$$z^2+z+2=0;$$
$$z=-(1+-\sqrt{-7})/2$$
I removed the absolute value for imposing a solution in the field of real, but only because I saw it on wolfram alpha, now I ask you! I can always impose a solution in the real nu... | Let us proceed in the conventional way:
Let $z=r(\cos\theta+i \sin\theta)$ where $r$ is real
$\implies |z|=r$ and $z^3=\{r(\cos\theta+i \sin\theta)\}^3=r^3\cos3\theta+ir^3\sin3\theta$ (using de Moivre's formula)
$\implies r^3\cos3\theta+ir^3\sin3\theta=r^2+4$
Clearly $r\ne0$
Equating the imaginary terms $r^3\sin3\theta... | {
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"timestamp": "2023-03-29T00:00:00",
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Spot my error in solving a linear system I almost always get the unit matrix if I try to get to an row reduced echelon form. I probably always make a mistake. Can you spot the error? What illegal operations could a beginner do while trying to solve a linear system?
\begin{array}{}
1 & 1 & 3 \\
-4 & -3 & -8 \\
-2 & -1 &... | Your very first step on the very first row operation, unfortunately
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the center of mass of the region bounded by $y=x^2$ and $y=4$?
English: That the centroid of the area bounded by $y=x^2$ and $y=4$? The answer is $(0,\frac{12}{5}$) I would like to understand the calculations!
$$A=\int_{-2}^{2}|4-x^2|dx=\frac{32}{3}$$
Em Português: Qual o centróide da região limitada por $y=... | Center of mass is the "average point". For example, if I just gave you 2 points, $p_1$ and $p_2$, and asked for the center of mass, the answer would be $\frac {p_1 + p_2} 2$.
I'm assuming that your object has uniform density; in other words, it's not made of steel on one side and cotton on the other.
Center of mass th... | {
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"url": "https://math.stackexchange.com/questions/526139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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sums of legendre symbol I am studying the sums of Legendre symbol. It's easy to prove that for $p\equiv 1 \pmod 4$,
$$\sum_{0<n<p/2}\left(\frac{n}{p}\right)=0.$$
by writing the sum over all residue classes and then using the fact that sum over $0$ to $p/2$ is same as $p/2$ to $p$ (by a change of variable). But the same... | (1) If $p \equiv 3 \pmod{8}$ then $(\frac{-1}{p})=(\frac{2}{p})=-1$. Let $p=8k+3$. Observe that
\begin{align}
\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})}& =\sum_{1 \leq n \leq 4k+1}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k}{(\frac{2n}{p})}+\sum_{1 \leq n \leq 2k+1}{(\frac{2n-1... | {
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} |
How to solve step by step $\frac{2n+1}{n+2} < 8$ Given something like $\frac{2n+1}{n+2} < 8$ how do you solve that step by step to get all the possible intervals?
| Observe that $\displaystyle \lim_{n+2\to0^-}\frac{2n+1}{n+2}\to+\infty$ and $\displaystyle \lim_{n+2\to0^+}\frac{2n+1}{n+2}\to-\infty$
If $n+2\ne0,(n+2)^2>0$ for real $n$
$$\frac{2n+1}{n+2}<8\iff \frac{2n+1}{n+2}-8<0$$
Now, $$\frac{2n+1}{n+2}-8=-\frac{6n+15}{n+2}=-3 \frac{(2n+5)}{n+2}$$
$$\implies -3 \frac{(2n+5)}{n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
from $1-\sin x $ to $2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$ How can you go from $1-\sin x $ to $2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$? I mean how to prove that $1-\sin x = 2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$?
| Use $\displaystyle \sin x=\cos\left(\frac\pi2-x\right)$ and $\displaystyle\cos2y=1-2\sin^2y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/533043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the following Diffrential Equation $(3y-7x+7)dx-(3x-7y-3)dy=0$ I want to solve the following equation:
$$(3y-7x+7)dx-(3x-7y-3)dy=0$$
I will need two new variables? or I can solve it with 1, for example set expression as $z$?
What you are suggesting?thanks.
| Answer taken from
http://in.answers.yahoo.com/question/index?qid=20080830232129AAdNlmN
$$(3x−7y−3)dy=(3y−7x+7)dx \\
\frac{dy}{dx}=\frac{3y−7x+7}{3x−7y−3} $$
Now, let $x=x’+h$ and $y=y’+k$
Therefore
$$\frac{dy’}{dx’} =\frac{3y’−7x’+3k−7h+7}{3x’−7y’+3h−7k−3} $$
Putting $h=1$ and $k=0$, we get
$$\frac{dy’}{dx’}=\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\... | Since $\left(1/\sqrt{x}\right)^{\prime\prime} = \frac{3}{4 x^{5/2}} > 0$ for $x > 0$, we have that $f(x)=\frac{1}{\sqrt{x}}$ is convex.
Then, by Jensen's inequality we have:
$$ \tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Let p, q, r be distinct primes greater than 3, and let n = pqr. Show that if $x \in \mathbb{Z}$ satisfies $x^{2} \equiv 9\mod{n}$
then $x \equiv ±3 \mod{p}$, $x \equiv ±3 \mod{q}$ and $x \equiv ±3 \mod {r}$.
I'm not sure what to do. Any help is appreciated.
Thanks.
| Let $x \in \mathbb{Z}$, and $n=pqr$, such that $x^{2} \equiv 9 \mod {n} $, ie, $\exists k\in \mathbb{Z}$ such that
\begin{eqnarray}
x^{2}-9 &=& nk\\
(x-3)(x+3)&=&nk
\end{eqnarray}
Now
*
*P.D. $x\equiv 3\mod{p}$
\begin{eqnarray}
(x-3)(x+3)&=&nk\\
(x-3)(x+3)&=&pqrk\\
(x-3)(x+3)&=&p(qrk)\\
x&\equiv&\pm3\mod{p}
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integration by parts, error on Primitive of $\sqrt{1-x^{2}}$ Problem: Integrate $\sqrt{1-x^{2}}$
Attempt:
| The usual method uses the substitution $x=\sin\theta$ followed by the identity $\cos^2\theta=\frac{1}{2}(\cos 2\theta+1)$. However, since the OP used integration by parts, let's do it that way. It is perhaps a little harder, but introduces a useful technique.
Let our integral be $I$, and let $g'(x)=1$ and $f(x)=\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of getting Required Sum
What is the probability of when four dice rolled together once,and getting a sum of Thirteen
If we do by just calculating all possible values of sum, then it will take more time; so we can solve the above problem as Multinomial Coefficents of sum, i.e.:
$$
x_1+x_2+x_3+x_4 = 13\;\te... | I assume that you already know how to calculate the probability for the sum of two dices. One gets the following table.
$$
\begin{array}{r|c}
k & p(x_1+x_2=k) \\
\hline{} \\
2 & 1/36 \\
3 & 2/36 \\
4 & 3/36 \\
5 & 4/36 \\
6 & 5/36 \\
7 & 6/36 \\
8 & 5/36 \\
9 & 4/36 \\
10 & 3/36 \\
11 & 2/36 \\
12 & 1/36 \\
\end{array... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof with mathematical induction that $ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $ Proof with mathematical induction.
I have the following induction problem:
$ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $
This property applies to all $n \g... | You are almost there. At the end, you wrote the difference between the two sums as a single complicated fraction. Maybe you have miscalculated something there, maybe not, anyway, it is not easy to see through. It becomes much easier if we proceed in steps, pairing like things with like things. The difference is
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Does there exist a prime number within the interval? Conjecture
$\forall p_{n}\in \mathbb{P} : n\geq3, \: \exists p_{m}\in \mathbb{P} : 3p_{n} - 4 \geq p_{m} > \sqrt{2(p^2_{n+1} - 1)} $
How would you go about proving/disproving this?
| This is true indeed. First consider $n$ s.t. $p_n \geq 29$. It has been proven that for $m \geq 3$, there is a prime in the interval $(m, \frac{4(m+2)}{3})$, see here, Corollary 2.2.
Since $p_n \geq 3$, there is a prime in $(p_n, \frac{4(p_n+2)}{3})$, so $p_{n+1}<\frac{4(p_n+2)}{3}$. Now $$\sqrt{2(p_{n+1}^2-1)}<\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Geometric mean proof with two tangents There are two tangent lines on $f(x) = \sqrt{x}$ each with the $x$-value $a$ and $b$ respectively.
I need to prove that $c$, the $x$ value of the point at which the two lines intersect each other, is equal to $\sqrt{ab}$, the geometric mean of $a$ and $b$.
I have been trying man... | The tangent line at $(a,\sqrt{a})$ has equation $y=\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}$.
The tangent line at $(b,\sqrt{b})$ has equation $y=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$.
Set $\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$ and solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/542454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ I am trying to prove that
$$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 2 +G\log 2$$
where $G$ is the Catalan's Constant. Numerically, it's value is $-0.199739$.
Using the substitution $x... | Expanding the inverse tangent in logarithms, writing $\frac{x}{1+x^2}=\Re\frac1{x-i}$, and expanding $\log(1-x^2)=\log(1-x)+\log(1+x)$, each of the resulting four indefinite integrals has a closed form. Each term is amenable to automatic integration, (an example), which means that after taking limits, slogging through ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 1
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How to find the equation of a line, tangent to a circle, that passes through a given external point I am given the equation of a circle: $(x + 2)^2 + (y + 7)^2 = 25$. The radius is $5$. Center of the circle: $(-2, -7)$.
Two lines tangent to this circle pass through point $(4, -3)$, which is outside of said circle. How ... | We can use the formula for the distance from a point to a line. The distance between a line $ax + by + c = 0$ and a point $(x_0, y_0)$ is
$$\left| \frac{a x_0 + b y_0 + c}{\sqrt{a^2+b^2}} \right|.$$
The line of tangency that passes the point $(4, -3)$ has equation $y+3 = m(x-4)$ for some slope $m$. So we just need to f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 6
} |
Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
| Consider the polynomial
$$(\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - a\lambda^2+b\lambda-c\tag{*1}$$
We know
$$\begin{cases}a = x + y +z\\ b = xy + yz + xz \\ c = x y z\end{cases}$$
Substitute $x, y, z$ for $\lambda$ in $(*1)$ and sum, we get
$$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$
This i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 3
} |
True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?
$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.
Can you find a counterexample for this or not?
| The following substitution transforms the constrained inequality into an unconstrained one:
$$a = t^2 \gt 0$$
$$b = u^2 \gt 0$$
$$c = v^2 \gt 0$$
The original inequality becomes:
$$F(t, u, v) = t^4+u^4+v^4 +2t^2u^2v^2+1 - 2*(t^2u^2 + t^2v^2 + u^2v^2) \ge 0$$
The new expression is fully symmetric in $t, u $ and $v$.
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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How to efficiently compute $17^{23} (\mod 31)$ by hand? I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 3... | Modulo $31$ we have:
$$17^1 \equiv 17$$
$$17^2 \equiv 10 \text{ (because $289$ mod $31$ is $10$)}$$
$$17^4 \equiv (10)^2 \equiv 7 \text{ (because $100$ mod $31$ is $7$)}$$
$$17^8 \equiv (7)^2 \equiv 18 \text{ (Similar)}$$
$$17^{16} \equiv (18)^2 \equiv 14 $$
And $17^{23} \equiv 17^{16} \cdot 17^4 \cdot 17^2 \cdot 17^1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Proving $\lim\limits_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right) =-\frac12$ How can I prove that
$$\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)=-\frac{1}{2}$$
| $$\begin{aligned}\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)
&=\lim _{x\to 0}\left(\frac{\ln \left(x+1\right)-\ln \left(x+\sqrt{x^2+1}\right)}{\ln \left(x+1\right)\ln \left(x+\sqrt{x^2+1}\right)}\right)
\\&=\lim _{x\to 0}\left(\frac{x-\frac{1}{2}x^2+o\left(x^2\right)-\left(x+o\left(x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find a matrix $X$ given $X^4$ Find the matrix $X$ such that
$$X^4=\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
This problem I can't work,and I think let the matrix the eigenvalue is $\lambda$,then
$\lambda^4$ is $$\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
eigenvalue?Thank you for your help.
| You can look for a solution of the same form as $X$: calculating $$\begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix} \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix} = a \cdot \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}$$ shows that $$\begin{pmatrix} a & 0 & 0 \\ 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| let $x=z+u,y=z+v,\to u\ge v \ge0$
$x^3y^2+y^3z^2+z^3x^2- xyz(x^2+y^2+z^2)=(u^2-uv+v^2)z^3+3u^2vz^2+uv(u^2-v^2)z+3u^2v^2z+u^3v^2 \ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the... | Standard conversion to hypergeometric series and use of the duplication
and triplication formulas for the $\Gamma$-function yields
\begin{equation}
\sum_{n\ge 0}\binom{3n}{n}z^n
=
\sum_{n\ge 0}\frac{\Gamma(3n+1)}{\Gamma(2n+1)}\frac{z^n}{n!}
=
\sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(n+1)}
{\Gamma(n+1/2)\Gamm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
} |
Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
| I just want to give two alternate forms of
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\frac34-\Im\operatorname{Li}_2\frac{3i}4.$$
First by using $(12)$ from here we get
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\frac{3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 4,
"answer_id": 3
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Residue/Contour integration problem Supposedly,
$\displaystyle\int_{-\infty}^\infty \frac{\cos ax}{x^4+1}dx=\frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left(\cos\frac{a}{\sqrt{2}}+\sin\frac{a}{\sqrt{2}}\right)$, $a>0$.
Using Residues/Contour integrals, I have
$\displaystyle\int_{-\infty}^\infty \frac{e^{iax}}{x^4+1}dx=\int_{... | Here is an answer that uses a quarter-circle instead of a semicircle. Start by observing that
$$\int_{-\infty}^\infty \frac{\cos(ax)}{x^4+1} dx
= 2\int_0^\infty \frac{\cos(ax)}{x^4+1} dx
= 2 \times\Re\left(\int_0^\infty \frac{e^{iax}}{x^4+1} dx\right).$$
Now use a contour consisting of three segments to evaluate the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Imaginary numbers and polynomials question I have a task which I do not understand:
Consider $w = \frac{z}{z^2+1}$ where $z = x + iy$, $y \not= 0$ and $z^2 + 1 \not= 0$.
Given that Im $w = 0$, show that $| z | = 1$.
Partial solution (thanks to @ABC and @aranya):
If I substitute $z$ with $x + iy$ then we have $w = \frac... | $$w=\frac z{z^2+1}=\frac{|z|^2\overline z+z}{|z^2+1|^2}$$
and thus
$$\text{Im}\,w=0\iff \text{Im}\,z\cdot\left(|z|^2-1\right)=0\stackrel{\text{since Im}\,z=y\neq 0}\iff |z|^2=1\;\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Infinite continued fraction expansion How can we find the first six partial quotients of the infinite continued fraction expansion of $\sqrt[3]2$?
I know how to do this by expanding when we have a square root function... but I"m not sure what to do with a cubic root.
| In response to André Nicolas's post, here's one way to use exact rational arithmetic. Let $x_0 = 2^{1/3}$.
First note that for rationals $a, b, c$, we can check the sign of $a x_0^2 + b x_0 + c$. The nontrivial case is where $a \ne 0$ and $b^2 - 4 a c > 0$. Let
$r_1 < r_2$ be the roots of $a x^2 + b x + c$, which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3... | Note that
$$
\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}=\left(\frac{1}{4}\right)^n 2^k
$$
So the sum in question simplifies to
$$
\left(\frac{1}{4}\right)^n\left(\sum_{k=1}^{n+3} 2^k\right)
$$
Now by the geometric series formula
$$
\sum_{k=1}^{n+3} 2^k=2\sum_{k=0}^{n+2} 2^k=2(2^{n+3}-1)
$$
So the total... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A square matrix A is invertible if and only if det A ≠ 0. Use the theorem above to find all values of k for which A is invertible $$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}?$$
I did a sample question before this one:
$$\begin{pmatrix} k & k & 0 \\ k^2 & 16 & k^2 \\ 0 & k & k \end{pmatrix}?... | Do the same thing you did before. First, take the determinant of
$$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}$$
to obtain the equation $25k^2-2k^4$. So if this equation is zero, then the matrix is not invertible. Then
$$
\begin{align}
25k^2-2k^4&=0 \\
k^2(25-2k^2)&=0 \\
k^2(5+\sqrt{2}k)(5-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Infinite sums - order of terms say we had an infinite sum $ \displaystyle \sum_{n=0}^\infty a_n $ where $ a_n = \dfrac{1}{2} +1 + \dfrac{1}{8} + \dfrac{1}{4} + \dfrac{1}{32} + ... $ this is obviously a rearrangment of $\displaystyle \sum_{n=0}^\infty \dfrac{1}{2^n} $ but does the order of the sum matter? I know by th... | If a series involves only positive (or at least non-negative) terms, then any rearrangement of the series has the same limit in the sum. One way to prove this is to note that if $a_n \ge 0$ for all $n$, then
$$\sum\limits_{n = 0}^{\infty} a_n = \sup_{A \text{ finite}} \sum_{n \in A} a_n$$
To prove that this series con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$ Regarding this problem, I conjectured that
$$ I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \ma... | So, following the procedure I outlined here, I get for the transformed integral:
$$I(r,s) = \int_0^{\infty} dv \frac{4 s \left(v^2-1\right) \left(v^4-(4 r-6) v^2+1\right)}{v^8+4 \left(2 r-s^2-1\right) v^6 +2 \left(8 r^2-8 r-4 s^2+3\right) v^4 +4 \left(2 r-s^2-1\right) v^2 +1} \log{v} $$
Note that this reduces to the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 2,
"answer_id": 0
} |
Find all real solutions for $16^{x^2 + y} + 16^{x + y^2} = 1$ Find all $x, y \in \mathbb{R}$ such that:
$$16^{x^2 + y} + 16^{x + y^2} = 1$$
The first obvious approach was to take the log base $16$ of both sides:
$$\log_{16}(16^{x^2 + y} + 16^{x + y^2}) = 0$$
manipulating did not give any useful result. The next thing I... | Using $$\bf{A.M\geq G.M}$$,
$\displaystyle \frac{16^{x^2+y}+16^{x+y^2}}{2}\geq \left(16^{x^2+y}\cdot 16^{x+y^2}\right)^{\frac{1}{2}}$
$\displaystyle 16^{x^2+y}+16^{x+y^2}\geq 2\cdot \left\{2^{4\left(x^2+y+x+y^2\right)}\right\}^{\frac{1}{2}} = 2\cdot \left\{2^{(2x+1)^2+(2y+1)^2-2}\right\}^{\frac{1}{2}}\geq 1$
and equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Strange mistakes when calculate limits I have difficulties with calculating the following limits. W|A gives the correct answers for both of them:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \... | In the first line of the first one:
$\sqrt{x}\sqrt{x + \sqrt{x}} = \sqrt{x^2 + x\sqrt{x}}$, not $\sqrt{x^2 + x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determining if a vector is in the column space of a matrix Hi there I'm having some trouble with the following problem:
I have a $3\times3$ symmetric matrix
$$
A=\pmatrix{1+t&1&1\\ 1&1+t&1\\ 1&1&1+t}.
$$
I am trying to determine the values of $t$ for which the vector $b = (1,t,t^2)^\top$ (this is a column vector) is in... | Gaussian elimination is not difficult in this case:
\begin{align}
\left[\begin{array}{ccc|c}
1+t & 1 & 1 & 1 \\
1 & 1+t & 1 & t \\
1 & 1 & 1+t & t^2
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
1 & 1+t & 1 & t \\
1+t & 1 & 1 & 1
\end{array}\right]
\\
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Calculate the limit of the function Please help me to calculate the limit of the function. I do not know where to start.
Thank you.
| Let
$$
f(x)= \left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]
$$
and $$g(x)=-\frac{1}{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}.$$
We have to find the limit for $x\to0$ of the function $\varphi(x)=[f(x)]^{-\frac{1}{g(x)}}$
$$
\lim_{x\to 0}\varphi(x)=\lim_{x\to 0}[f(x)]^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Triangle ratio of areas This is a photo that was originally posted on Google Plus. I would like to know how to solve for S. I started by splitting S into two parts S1 and S2 by drawing a line from A to M.
I also know that I should use the fact that since the triangles share the same base, the ratio of areas = ratio of... | Let $x$ be the area of the triangle $AML$, and $y$ of $AMK$. Note that as $MBC$ and $LBC$ have the same bases, their heights must be in the ratio of $2/3$ and so must be the segments
$$\frac{MC}{LC}=\frac{2}{3}\Rightarrow\frac{LM}{LC}=\frac{1}{3}.$$
By a similar argument, we get
$$\frac{MB}{KB}=\frac{10}{18}=\frac{5}{9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/573933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_... | In a sense "Polar coordinates does not help much since the integrand likes them while the domain does not".
It helps to know about partial integration, that is
$$\int f'(x)g(x)dx = f(t)g(t)- \int f(x)g'(x)dx +C $$
In addition one should know about the derivatives of $\arcsin$ and $\arctan$.
Now to the integral, first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Squaring both sides leads to a complete different result I came across with this equation:
$v^2 \cos (2 \alpha) \sqrt{v^2 \sin^2 \alpha+2gh}+v \sin \alpha [v^2 \cos (2 \alpha)-2gh]=0$
or
$2 g h \sin \alpha - v^2 \cos(2\alpha) \sin \alpha = v \cos(2 \alpha) \sqrt{v^2 \sin^2 \alpha + 2gh}$
Now this equation before squari... | We have $$v^2\cos2\alpha(\sqrt{v^2 \sin^2 \alpha+2gh}+v\sin\alpha)=2gh v\sin\alpha$$
Rationalizing the numerator & cancelling $v\ne0$
$$v\cos2\alpha\frac{2gh}{\sqrt{v^2 \sin^2 \alpha+2gh}-v\sin\alpha}=2gh \sin\alpha$$
Cancelling $2gh\ne0$ & using $\cos2\alpha=\cos^2\alpha-\sin^2\alpha$
$$v(\cos^2\alpha-\sin^2\alpha)=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
| $$y^9+3y^6+3y^3-16y+9=0\iff(y-1)(y^2 + y - 1)(y^6 + 2y^4 + 2y^3 + 4y^2 + 2y +9) = 0$$
Three real-valued roots: $$y = 1, y = \dfrac{-1 \pm \sqrt 5}{2}$$
I noticed by inspection that $y = 1$ solves the equation. Then, using polynomial division, and manipulation with the resulting quotient, was able to find the quadratic ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Fixed points of group action Let $G=SL_2(Z)$ work on $\{z \in C | Im(z) \gt 0\}$ by $\begin{pmatrix}a&b \\ c&d\end{pmatrix}(z)=\frac{az+b}{cz+d}$.
First of all I want to find which elements of G leave $\zeta_3$ fixed.
Writing it out gives me $c\zeta_3^2+(d-a)\zeta_3-b=0$. If I now use $\zeta_3=e^{\frac{2\pi i}{3}}$ i f... | You don't need a computer. From $a+c-d=0$, you can substitute $d = a+c$ into $-a+2b+c+d = 0$ to get $2(b+c) = 0$. Thus you are at
$$\begin{pmatrix}a&b\\-b&a-b \end{pmatrix}.$$
Now the determinant gives you $a(a-b) + b^2 = 1$, or $(2a-b)^2 + 3b^2 = 4$, so you have
*
*$b = 0$, and $a = \pm 1$, which gives the identity... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \... | If we assume that $a\leqslant b\leqslant c$, then it suffices to show$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}.$$
*
*If $c-b\geqslant b-a$, denote $c-b=y,b-a=x$, then $y\geqslant x\geqslant 0$, we just ned to show$$x^2\leqslant \frac{a^2+(a+x)^2+(a+x+y)^2}{2}\iff 3a^2+(4x+2y)a+(y^2+2xy)\geqslant 0.$$Notic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Verifying identities, trigonometry I am completely stuck. I cannot come up with a proof for this identity. I am ready to pull my hair out. $$\sin(x) \left(\tan(x)+ \frac 1{\tan(x)}\right) = \sec(x)$$
| Hint: Remember $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $1 + \tan^2(x) = \sec^2(x)$.
Edit to reflect full answer:
Because we know that both $\tan(x)$ and $\sec (x)$ can be written in terms of $\sin(x)$ and $\cos(x)$, we will rewrite them. Remembering:
$$\tan(x) = \frac{\sin(x)}{\cos(x)} \ \ \mathrm{and } \ \ 1 + \tan^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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An Urn with 60 marbles A urn contains 60 marbles 10 red 10 white 10 blue 10 green 10 brown and 10 orange. I draw from the urn 6 times. What is the probability I have exactly 5 varieties?
Here's what I've got,
There are $10^5$ ways I can draw 5 different marbles. So the number of ways I am unable to draw the 6th variety... | Think of the marbles as distinct (write an ID number on each). We assume the drawing is done without replacement.
There are $\binom{60}{6}$ equally likely ways to choose $6$ marbles.
Now we count the "favourables." If there are $5$ types, we must have $2$ of one type, and $1$ each of $4$ other types.
There are $\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
| Solution from Thomas Mildorf notes
first we homogenize inequality with multipliying $(abc)^\frac{1}{3}$ and we get the following inequality $$a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{4}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{1}{3}c^\frac{4}{3}\leq a^2+b^2+c^2$$ then we can apply $A.O\geq H.O$ $$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Proving a non-linear diophantine equation has no solutions How can I prove that
$x^2 + y^2 + z^2 + 3(x+ y + z) + 5 = 0$
Has no integer solutions?
| Assume that $(x,y,z) \in \mathbb{Z}^3$ satisfies $x^2+y^2+z^2 + 3(x+y+z) + 5 = 0$. For simplicity, let $f(X,Y,Z) = X^2+Y^2+Z^2 + 3(X+Y+Z) + 5$. Then $f(x,y,z) \equiv 0 \pmod{2}$. Notice that $a^2 \equiv a \pmod{2}$ implies $x^2+y^2+z^2\equiv 3(x+y+z)\pmod{2}$, and so $f(x,y,z) \equiv 5 \equiv 1 \not\equiv 0 \pmod{2}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Integration with substitution I want to integrate (2x+1)/((x^2 - 6x + 14)^3)
I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14
I find that it makes turning the top of the fraction into terms of u very difficult
| $$\begin{align} \int \frac{2x + 1} {(x^2 - 6x + 14)^3} & = \int \frac{2x - 6 + 6 + 1}{(x^2 - 6x + 14)^3}\,dx \\ \\ & = \int\dfrac{2x - 6}{(x^2 - 6x + 14)^3}\,dx + \int \frac{7}{(x^2 - 6x + 14)^3}\,dx \\ \\ &= \int\dfrac{\overbrace{2x - 6}^{du}}{(\underbrace{x^2 - 6x + 14}_{u})^3}\,dx + \int \frac{7}{(\underbrace{x^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Integrate the following integral using partial fractions I want to integrate
$$\frac{1}{(1-u^2)^2}. $$
I have used the difference of two squares to get
$$\frac{1}{2(1-u^2)} + \frac{1}{2(1+u^2)} $$
and then integrated it to get
$$\frac{1}{2ln|1-u^2|} + \frac{1}{2ln|1+u^2|}.$$
Just wondering if this is correct? thanks
... | If you want to use partial fraction decomposition, then note that: $$\dfrac 1{(1-u^2)^2} = \dfrac 1{[(u-1)(u+1)]^2} = \dfrac 1{(u-1)^2(u+1)^2} $$
$$= \dfrac{A}{(u-1)} + \dfrac{B}{(u -1)^2} + \dfrac{C}{(u+1)} + \dfrac D{(u+1)^2}$$
Now try solving for the needed constant terms: $A,\, B, \,C,\, D$.
Spoiler I (to check you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Trouble with factoring polonomial to the 3rd degree I am having trouble factoring this problem:
$\displaystyle{-x^{3} + 6x^{2} - 11x + 6}$
I know the answer but i can't figure out how it is done with this.
I have tried by grouping and is doesn't seem to work. Can someone show me how to do this.
| $$\begin{align}
& -x^3+6x^2-11x+6 \\
=& -x^3+1+6x^2-11x+5 \\
=& (1-x^3)+6x^2-6x-5x+5 \\
=& (1-x)(1+x+x^2)-6x(1-x)+5(1-x) \\
=& (1-x)(1+x+x^2-6x+5) \\
=& (1-x)(x^2-5x+6) \\
=& (1-x)(x^2-2x-3x+6) \\
=& (1-x)(x(x-2)-3(x-2)) \\
=& (1-x)(x-2)(x-3)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/585075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How find this invertible matrix $C=\left[\begin{smallmatrix} A&B\\ B^T&0 \end{smallmatrix}\right]$ let matrix $A_{n\times n}$,and $\det(A)>0$, and the matrix $B_{n\times m}$,and such $rank(B)=m$,and let
$$C=\begin{bmatrix}
A&B\\
B^T&0
\end{bmatrix}$$
Find this Invertible matrix
$C^{-1}$
my try: I found this matrix Inv... | Let
$ C^{-1} = \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} $ with appropriate sizes (i.e. $X$ is $n \times n$, $Y$ is $n \times m$, $Z$ is $m \times n$ and $W$ is $m \times m$). Then,
$C C^{-1} = \begin{bmatrix} A & B \\ B^T & 0 \end{bmatrix} \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} = \begin{bmatrix} I_n & 0 \\ 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/585501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How to integrate the bump functions,i.e,$\int_a^{b}e^{-\frac{1}{x-a}+\frac{1}{x-b}}dx$,where $aSince $$\lim_{x\to{a}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=\lim_{x\to{b}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=0,$$ $e^{-\frac{1}{x-a}+\frac{1}{x-b}}$ is continuous on the interval $[a,b]$ (taking $0$ if $x=a$ or $b$).
So the integra... | $$\int_a^be^{-\frac{1}{x-a}+\frac{1}{x-b}}~dx$$
$$=\int_{a-\frac{a+b}{2}}^{b-\frac{a+b}{2}}e^{-\frac{1}{x+\frac{a+b}{2}-a}+\frac{1}{x+\frac{a+b}{2}-b}}~d\left(x+\dfrac{a+b}{2}\right)$$
$$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^{-\frac{1}{x+\frac{b-a}{2}}+\frac{1}{x-\frac{b-a}{2}}}~dx$$
$$=\int_{-\frac{b-a}{2}}^\frac{b-a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/585768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = ... | define
$$B_n = \sum_{k=1}^{3n} f_k$$
LEMMA $$B_n = f_{3n+2} - 1$$
PROOF by induction.
(A) for $n=1$ we have $f_1+f_2+f_3 = 1+1+2=4 = 5-1=f_5-1$
(B) suppose true for $n$, then:
$$B_{n+1} = B_n +f_{3n+1} +f_{3n+2}+f_{3n+3} \\
= f_{3n+2} - 1 +f_{3n+1} +f_{3n+2}+f_{3n+3} \\
= (f_{3n+1} +f_{3n+2}) +(f_{3n+2}+f_{3n+3})-1\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How do you calculate this sum $\sum_{n=1}^\infty nx^n$? I can not find the function from which I have to start to calculate this power series.
$$\sum_{n=1}^\infty nx^n$$
Any tips?.
Thanks.
| We can do it another way.
$S = x + 2x^2 + 3x^3 + \ldots $
It can be written as
$ \Rightarrow S = (x + x^2 + x^3 + \ldots)+(x^2 + x^3 + \ldots)+(x^3 + \ldots)+\ldots $
$\Rightarrow S = (x + x^2 + x^3 + ...)+x(x + x^2 + ...)+x^2(x + ...) + \ldots $
$\Rightarrow S = ( 1+x+x^2+ .. )\times( x+x^2+.. )$
$\Rightarrow S = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Change of Variables in differential equation Given the equation $zZ''(z) + Z'(z) + \alpha^2Z(z) = 0$ use the change of variables $x = \sqrt{\frac{z}{a}}$ where $a$ is a constant to map the problem to the differential equation $Z''(x) + \frac{1}{x}Z'(x) + \gamma^2 Z(x) = 0,$ where $\gamma = 2\alpha \sqrt{a}$
Attempt:
I ... | $$ \frac{dZ}{dz} = \frac{dZ}{dx}\frac{dx}{dz}=\frac{dZ}{dx}\frac{1}{2ax}$$
$$ \frac{d^{2}Z}{dz^{2}}=\frac{d}{dz}[\frac{dZ}{dx}\frac{1}{2ax}], $$
but $dz=2axdx$, so
$$ \frac{d}{dz}[\frac{dZ}{dx}\frac{1}{2ax}] = \frac{1}{2ax}\frac{d}{dx}[\frac{dZ}{dx}\frac{1}{2ax}] = \frac{1}{4a^{2}x^{2}} \frac{d^{2}Z}{dx^{2}} -\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$? I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question.
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the
nearest value of $y^2 - y... | Let's let
$$y=\sqrt{5+\sqrt{5-{\sqrt{5+{\sqrt5-\cdots}}}}}$$
and
$$z=\sqrt{5-\sqrt{5+{\sqrt{5-{\sqrt5+\cdots}}}}}$$
and assume that both limits exist. Note that
$$y^2=5+z\quad\text{and}\quad z^2=5-y$$
Hence
$$\begin{align}
y^2-y&=(5+z)-y\\
&=5-(y-z)\\
&=5-{y^2-z^2\over y+z}\\
&=5-{(5+z)-(5-y)\over y+z}\\
&=5-{z+y\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| We need to find the value of $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{\dots}}}}}$.
Step 1: Let $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y$
Step 2: Square both sides.
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y^2$$
Step 3: Recall that $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y$. So:
$$7\sqrt{7\sqrt{7\sqrt{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 7,
"answer_id": 5
} |
why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x ten... | Both of those expressions approach zero, so they have the same limit at 2. But they are far enough apart, when $x>2$, that dividing by $\sqrt{x^2-4}$ gives a different answer. The difference becomes 'something tiny' divided by 'something else tiny', which could be anything.
In this case, as B.S. says, Use the equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let $a;b;c>0$. Prove : $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$ Let $a;b;c>0$. Prove :
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$
P/s : Only use AM-GM and Cauchy-Schwarz
In th... | We need to prove that $$\sum\limits_{cyc}\frac{a^2-b^2}{b+c}\geq0$$ or $$\sum\limits_{cyc}(a^2-b^2)(a+b)(a+c)\geq0$$ or
$$\sum\limits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)\geq0$$ or
$$\sum\limits_{cyc}(a^2-b^2)a^2\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{a^4+b^4}{2}-a^2b^2\right)\geq0,$$
which is true by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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Fractions in Questions and Answers
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