Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the composition $f \circ g=?$ and $g\circ f=?$ a) $f(x) = \sqrt[3]{x}\ $ and $g(x) = x^3$
find $f\circ g=?$ and $g\circ f=?$
I have $f\circ g = f(g(x)) = f(x^3) = \sqrt[3]{x^3} = x$
So basically, first we replace $g(x)$ with its value, then we replace $x$ with the value of $f(x)$ inside $g(x)$?
$g\circ f = g(f... | so there is two different case and different result,for example let us suppose that
$f(x)=x^3$
and $g(x)=x^2$
then multiplication is directly $x^3*x^2=x^5$
for composition like $f\circ g=f(g(x))=(x^2)^3$
$g\circ f=g(f(x))=(x^3)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/607909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Finding the Sum of the square of two positive integers.
Write the following equation as the sum of the square of two integers, $a^2 +b^2$.
$$(8^2+5^2)(13^2+7^2)$$
I remember that you are supposed to do something with complex numbers or at least that is what my teacher did in class.
| From identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$, that is easy to check, we get:
$$(8^2+5^2)(13^2+7^2)=(8\cdot13+5\cdot7)^2+(8\cdot7-5\cdot13)^2=139^2+9^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
| Using the trick mentioned in a comment, observe that $0\leq x \leq 2$ for this to make sense, so substitute $x = 2\cos(t)$. Squaring and rearranging, $$4\cos^2(t) - 2 = \sqrt{2-\sqrt{2+2\cos(t)}}.$$ Apply the double angle formula and square again to get $$4\cos^2(2t) - 2 = -\sqrt{2+2\cos(t)}$$ whence $$4\cos(4t)^2-2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mixed number fractions vs regular fractions? $3\frac{1}{6}-1\frac{11}{12}$ I just passed Calculus 2 in college with an A and I'm rather embarrassed that I'm asking this question. My wife is taking an intermediate Algebra course in college and they gave her the below problem.
$$3\frac{1}{6}-1\frac{11}{12}$$
Well I gues... | $3 \dfrac16$ denoted $3+\dfrac16 = \dfrac{19}6$ and not $3 \cdot \dfrac16$.
Similarly, $1\dfrac{11}{12}$ denotes $1+\dfrac{11}{12} = \dfrac{23}{12}$ and not $1 \cdot \dfrac{11}{12}$.
Hence,
$$3 \dfrac16 - 1\dfrac{11}{12} = \dfrac{19}6 - \dfrac{23}{12} = \dfrac{38-23}{12} = \dfrac{15}{12} = \dfrac54$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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positive integer ordered pairs $(x,y,z)$ in $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$ Total no. of positive integer ordered pairs of $(x,y,z)$ that satisfy the equation $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$
$\bf{My\ Try:}$ Using Simple Guess $x=2\;,y=3\,z=6.$ satisfy $\displaystyle \frac{1}{x}+\frac{1... | The only solutions are $(2,3,6); (2,4,4); (3,3,3)$ (and of course its permutations). This can be argued as follows (in fact you are almost there):
As you have it, we need $\dfrac{x+y}{xy-x-y}$ to be an integer. Without loss of generality we can take $x \leq y$. We have $x+y \leq 2y$ and $xy-x-y \geq (x-2)y$. Hence, for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/611420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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max and min $x y - \ln(x^2 + y^2)$ Find max and min
$$x y - \ln(x^2 + y^2) , 1/4 \le x^2 + y^2 \le 4$$
Problem: With this hairy expression as my partial derivatives, I do not know what to do.
Attempt:
| $$y-\frac{x}{x^2+y^2}=0,x-\frac{y}{x^2+y^2}=0\Rightarrow x^2=y^2\Rightarrow x=\pm y$$
You have to minimize $$xy-\log (x^2+y^2)$$
with given conditions
$$1/4 \le x^2 + y^2 \le 4$$
If $1/4 \le x^2 + y^2 \le 4$ can you find what are the bounds for $x^2$??
If you know what are bounds for $x^2$,
take its minimum value and s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $|\sin n|+|\sin (n+1)| > 2\sin(1/2)$ for all $n\in \mathbb N$
Show that
$$|\sin{(n)}|+|\sin{(n+1)}|>2\sin{\dfrac{1}{2}},n \ge 1,n\in \mathbb N$$
My try: let
$$F(n)=|\sin{(n)}|+|\sin{(n+1)}|$$
then
$$F(n+\pi)=|\sin{(n+\pi)}|+|\sin{(n+\pi+1)}|=|\sin{(n)}|+|\sin{(n+1)}|=F(n)$$
and
$$F(\pi-n)=|\sin{(\pi-n)}|... | Using induction and the formula $$\sin(nx)=2^{n−1}\Pi_{k=0}^{n−1}\sin(x+\frac{\pi k}{n})$$
which is not very hard to prove by induction (again) for instance, $\sin(2x)=2\sin(x)\cos(x)=2\sin(x)\sin(x+\pi/2)$.
1). $n=1$ case is shown above $|\sin{(1)}|+|\sin{(1+1)}|=\sin(1)(1+2\sin(1+\frac{\pi}{2})>2\sin{\frac{1}{2}}$;
2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Trouble solving $\int\sqrt{1-x^2} \, dx$ I am trying to learn how to solve integrals and I've got the hang out of a lot of examples, but I haven't got the slightest idea how to solve this example, this is how far I've got:
$$
\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} - 2\int\frac{x^2}{\sqrt{1-x^2}} \, dx
$$
Can you plea... | subsititute $ x =sin\theta $, this gives,
$ dx = cos\theta d\theta$
hence, $ \int \sqrt{(1-x^2)} dx = \int (\sqrt{1 - sin^2\theta} ). cos\theta d\theta = \int cos^2 \theta d\theta $
Remember that, $cos(\alpha + \beta) = cos \alpha .cos \beta - sin \alpha . sin \beta $,
substituting, $\alpha = \beta = \theta$,
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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calculating mod 7 Problem: Calculate $10^{10^{10}} \pmod 7$
According to Fermat's little theorem: $a^{p-1}\equiv1 \pmod p$, so $10^6\equiv1 \pmod 7$ and $10^n\equiv4 \pmod 6$, $n$ being any integer, why can we write $10^{10^{10}}\equiv10^4 \pmod 7$?
Similarly, if it's $10^n\equiv1 \pmod 3$, n being any integer, then wo... | I believe your argument is correct. To compute $10^{10^{10}} \pmod{7}$ you can first compute $10^{10} \pmod{6}$. In turn, you can do this modulo $2$ and $3$, so you get that $10^{10} \equiv 4 \pmod{6}$, and $10^{10^{10}} \equiv 3^{4} \equiv 2 \cdot 2 \equiv 4 \pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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How prove this equation $A^2+B^2=C^2+D^2$ define:
plane $W:Ax+By+Cz+D=0$ and the
hyperboloid of one sheet $U:x^2+y^2-z^2=1$
if $$W\bigcap U=l_{1},W\bigcap U=l_{2}$$
where $l_{1},l_{2}$ be two straight lines
show that :$$A^2+B^2=C^2+D^2$$
My try: since
$$\begin{cases}
Ax+By+Cz+D=0\\
x^2+y^2-z^2=1
\end{cases}$$
t... | You need to make use of the fact that the intersection is a pair of straight lines.
Look at the last equation as a quadratic in $x$, i.e of the form $a x^2+b x +c=0$
Its discriminant is $b^2-4ac$ is given by
$$
\Delta(y) = 4\,{C}^{2}\,\left( {D}^{2}+2\,y\,B\,D-{y}^{2}\,{C}^{2}+{C}^{2}+{y}^{2}\,{B}^{2}+{y}^{2}\,{A}^{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^... | Both parts of the integral express the area under the curve given by $x^7+y^3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Prove that $\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots.$ Question. Let
$$
f(x)=\!\left\{\,\,\,
\begin{array}{ccc}
\displaystyle{\left\lfloor{1\over x}\right\rfloor}^{-1}_{\hphantom{|_|}}&\text{if} & 0\lt x\le 1, \\ & \\
0^{\hphantom{|^|}} &\text{if} & x=0.
\end{a... | We observe that: if $x\in\big(\frac{1}{k+1},\frac{1}{k}\big]$, then $\frac{1}{x}\in[k,k+1)$, thus $\left\lfloor{1\over x}\right\rfloor=k$ and hence
$$
\left\lfloor{1\over x}\right\rfloor^{-1}=\frac{1}{k}, \quad \text{whenever}\,\, x\in\Big(\frac{1}{k+1},\frac{1}{k}\Big].
$$
Therefore
$$
\int_{1/n}^1{\left\lfloor{1\over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
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$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try ... | Every integer $n$
can be written in the form
$3m+k$,
where $m$ is a non-negative integer
and $k = 0, 1, $ or $2$.
(This is a particular case of
the result that
for any positive integer $j$,
every integer $n$
can be written in the form
$jm+k$, where $m$ is a non-negative integer
and $k$ is an integer such that
$0 \le k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 5
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How prove this $e=\frac{2}{1}\left(\frac{4}{3}\right)^{\frac{1}{2}}\left(\frac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\cdots$ I see this nice equation:
$$e=\dfrac{2}{1}\left(\dfrac{4}{3}\right)^{\frac{1}{2}}\left(\dfrac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\left(\dfrac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13... | This (PDF) has the proof of your equation.
New Wallis- and Catalan-Type Infinite Products for $\pi,e$, and $\sqrt{2+\sqrt 2}$
by Jonathan Sondow and Huang Yi
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$ let $a,b,c,d,e$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$
My try:
I have solved follow Four-variable inequality:
let $a... | I would have written this in a comment, but I don't have enough 'reputation' to do so...
@Macavity I'd just like to point out that $$\sum \frac{a_i}{b_i} \ge \frac{\big(\sum a_i \big)^2}{\sum a_i b_i}$$ holds when $a_i b_i$ is positive for all $i$. (To use Cauchy-Schwarz, you need to square-root and then square those t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x... | Mathematical induction and trigonometric function relations and derivatives should do it.
I'll assume $x/a\in (0,\pi).$
Step: $n = 0$
$$\frac{1}{x^2+a^2}=\frac{\sin{[\mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{1/2}}$$ as
$$\sin{[\mathrm{arccot}{(x/a)}]} = \frac{a}{(a^2+x^2)^{1/2}}. $$
Step: $n \Rightarrow n+1$
Assume
$$\left... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find minimum of $\sqrt{(a+7)^2+(a+10)^2}+\sqrt{(a-b)^2+(a+5)^2}+\sqrt{(c-a)^2+(c-3)^2}+\sqrt{(c-6)^2+(c-8)^2}$ let $a,b,c\in R$,then Find this mimimum of the value
$$\sqrt{(a+7)^2+(a+10)^2}+\sqrt{(a-b)^2+(a+5)^2}+\sqrt{(c-a)^2+(c-3)^2}+\sqrt{(c-6)^2+(c-8)^2}$$
My try: since
$$\sum_{i=1}^{n}\sqrt{a^2_{i}+b^2_{i}}=\s... | Hint:
An alternative to Minkowski - Consider a path through the points $(b, a), (a, -5), (-7, a+5), (c-10, c+5), (-4, 13)$. The LHS is the length of this path, which obviously gets minimised when all these points are on a straight line, in order. Can you find $a, b, c$ so that these are on a straight line?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of: $x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$
$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
$$x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$$
I put it in the form $x^6y +x^6z+y^6x+y^6z+z^6x +z^6y$. I tried AM-GM but it's not helping.
| Step:$1$ $$x^6y +x^6z+y^6x+y^6z+z^6x +z^6y \geq 6x^2y^2z^2(xyz)^{\frac{1}{3}}$$
Step:$2$$$\frac{\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}}{3} \geq \frac{1}{(xyz)^{\frac{2}{3}}}$$
$$(xyz)^{\frac{2}{3}} \geq 3$$
Next step : Substitute result of step:$2$ in step:$1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
... | Set $a=\sqrt{x-4}$ and $b=\sqrt{x-5}$ for $-2\leq x\leq +2$. Equivalently we have $a^2=x-4$, $b^2={x-5}$ for $-2\leq x\leq +2$. It's implies for $-2\leq x\leq +2$,
\begin{align}
a-b=-1 \implies & (a-b)(a+b)=-1\cdot(a+b)\\
\implies & a^2-b^2=-(a+b)\\
\implies & (x-4)-(x-5)=-(a+b)\\
\implies & -1=-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
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Compute $\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$ Compute the value of the following expression
$$\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )^2+\cdots+\left (\frac{1}{n-1}... | You can prove the claim by induction. For $n=1$ you have $1=1$ which is clearly true.
For $n+1$ you can compute
$$
\left(1+\frac{1}{2}+\ldots+\frac1n+\frac{1}{n+1}\right)^2 + \left(\frac{1}{2}+\ldots+\frac1n+\frac{1}{n+1}\right)^2 + \ldots + \left(\frac{1}{n}+\frac{1}{n+1}\right)^2 + \left(\frac{1}{n+1}\right)^2
$$
usi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Evaluating the primitive $\int \frac{\mathrm dx}{e^{2x} + e^x + 1} $ Could someone help me evaluate this?
$$\int \frac{\mathrm dx}{e^{2x} + e^x + 1} $$
I tried to solve it for hours with no success.
I tried Wolframalpha but it's giving a step by step solution that is too long that in an exam I won't even have the time ... | Let $u = e^x$. Then, $du = e^x\,dx$, or, equivalently, $dx = \frac{1}{u}du$. Thus, the integral becomes:
$$
\int\frac{dx}{e^{2x}+e^x+1} = \int\frac{du}{(u^2+u+1)u}$$
Now, hit it with partial fractions:
$$\begin{align}
\frac{1}{(u^2+u+1)u} &= \frac{Au+B}{u^2+u+1} + \frac{D}{u}\\
&=\frac{Au^2+Bu+Du^2+Du+D}{(u^2+u+1)u}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| Let $S = 1+2^1+2^2+...+2^{10}$
Multiply by 2
$2S = 2^1+2^2+2^3+...+2^{11}$
Substract the former from the latter:
$S = 2^{11}-1$
$1+2^1+2^2+...+2^{10} = 2^{11}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
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Prove that for every $x\in[0,1]$, this is true $\sqrt{1+2x}\geq x+1-(1/(2x^2))$ Prove that for every $x\in[0,1]$, this is true $\sqrt{1+2x}\geq x+1-(\frac{1}{2x^2})$
i proved that $x+1-\sqrt{1+2x}>0$ by: $(x+1)^2 -1-2x=x^2$ so $x+1>\sqrt{1+2x}$ but then don't know how to proceed for this question
Thank you in advance
| Each of the following are equivalent
$$
\begin{align}
\sqrt{1+2x}&\ge x+1-\frac{1}{2x^2}\tag{1}\\
\sqrt{1+2x}-(x+1)&\ge-\frac{1}{2x^2}\tag{2}\\
\frac{-x^2}{\sqrt{1+2x}+(x+1)}&\ge-\frac{1}{2x^2}\tag{3}\\[6pt]
\sqrt{1+2x}+(x+1)&\ge2x^4\tag{4}
\end{align}
$$
Reversible Operations:
$(2)$: subtract $x+1$ from both sides
$(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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convergence of series, exercise I must prove convergence or divergence of the following series:
$$
\sum_{n=1}^{\infty}
\left[\arctan\left(1 \over n\right) - {n \over n^{2} + 1}\right]
$$
I've tried the integral test:
$$
\int_{1}^{\infty}x\,\arctan\left(1 \over x\right)\,{\rm d}x$$ and this is $<1$ so I would conclude c... | Hint.
$$(\arctan t)'=t-\frac{t^3}{3}+\frac{t^5}{5}+\cdots$$
thus
$$
\left|\arctan\frac{1}{n}-\frac{1}{n}\right|\le \frac{1}{3n^3}+\frac{1}{5n^5}+\cdots\le
\frac{1}{3n^3}\left(1+\frac{1}{n^2}+\cdots\right)=\frac{1}{3n^3}\frac{n^2}{n^2-1}\le\frac{1}{n^3},
$$
for $n\ge 2$, and
$$
\left|\,\frac{n}{n^2+1}-\frac{1}{n}\,\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding order of pole I have some problems with the following excersice: Find the order of the pole of:
$$\frac{1}{(2\cos z -2 + z^2)^2}$$ at $z=0$. I thought it is maybe better to work here with $1/f$ and find the order of the zero at $z=0$. But I don't know how to continue.
Thanks in advance!
| $$2\cos z-2+z^2=2\left(1-\frac{z^2}2+\frac{z^4}{24}-\ldots\right)-2+z^2=$$
$$=\frac{z^4}{12}-\frac{2z^6}{6!}+\ldots=z^4\left(\frac1{12}-\frac{2z^2}{6!}+\ldots\right)\implies$$
$$\left(2\cos x-2+z^2\right)^2=z^8\left(\frac1{144}+\frac{z^2}{3\cdot 6!}+\ldots\right)\implies$$
$$\frac1{(2\cos z-2+z^2)^2}=\;\;\ldots\ldots$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$ Let $x,y,z$ be real numbers such that $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.
| Set
$$u:=\cos x , v:=\cos y , w:=\cos z$$
We have
$$u + v + w = 0$$
and
$$\cos 3x + \cos 3y + \cos 3z = 4u^3 - 3u + 4v^3 - 3v + 4w^3 - 3w = 4u^3 + 4v^3 + 4w^3 = 0$$
So
$$u^3 + v^3 + w^3 = 0$$
Now consider
$$\cos 2x * \cos 2y * \cos 2z = \left(2u^2-1\right)\left(2v^2-1\right)\left(2w^2-1\right)$$
Further, we have
$$\le... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{... | Personally (assuming I wasn't supposed to use sharper tools), I would tackle it by first noting that the fraction should be about $1/n$ when $n$ is very large. So pull out a factor $1/n$, and simplify the rest, and you get
$$\frac{n+6}{n^2-6}=\frac1n\cdot\frac{1+6/n}{1-6/n^2}.$$
Now you just need to find a good bound f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solving a challenging differential equation How would one go about finding a closed form analytic solution to the following differential equation?
$$\frac{d^2y}{dx^2} +(x^4 +x^2+x+c)y(x) =0 $$
where $c\in\mathbb{R}$
| You can have a closed form solution in terms of HeunT (the Heun Triconfluent function)
$$ y( x ) ={C_1}\,{{\rm e}^{\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}}{\it HeunT} \left(-\frac{{3}^{2/3}\sqrt[3]{2} \left(4\,c-1\right)}{8}, \frac{3\,i}{2}, -\frac{{2}^{2/3}\sqrt[3]{3}}{2}, \frac{i\sqrt [3]{2}\,{3}^{2/3}}{3}x \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Determine the number of divisors in $K[x]$ of $1 + x^{15}$ and of $1+x^{120}$ where $K[x]$ is the set of all polynomials where coefficients are elements of $K$ $(0,1)$
Is this related to the problem of finding how many cyclic linear codes there are if $n = 15$ and $120$?
I've seen the following corollary in a book abou... | Notice that over the binary field $K$, $(x^{15}+1)^8 = x^{120} + 1$ by repeated squarings. So if we find the irreducible factorization of $x^{15} + 1$, we have that of $x^{120} + 1$ as an immediate consequence.
Since $15 = 3\cdot 5$, we already know a couple of easy factorizations of $x^{15} + 1$, as the sum of third ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$ How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$
| Starting with $F=\cos(a-b)+\cos(b-c)+\cos(c-a)$ expand the cosines via the addition formula for cosine, and get
$$\cos a \cos b + \cos b \cos c + \cos c \cos a \\
+\sin a \sin b + \sin b \sin c + \sin c \sin a.$$
Then after applying $(u+v+w)^2-u^2-v^2-w^2=2uv+2vw+2wu,$ the double $2F$ of our objective function may be ... | {
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"url": "https://math.stackexchange.com/questions/639890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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limit problem - can't get rid of $0$. I am trying to evaluate limit:
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}$$
I tried to use known limit in denominator to get:
$$\lim_{x\rightarrow 0}\frac{\frac{\arcsin}{x} - \frac{\arctan x}{x}}{x( \frac{e^x-1}{x}\cdot \frac{1}{x}+\frac{1-\cos x}{x^2}... | Just use L‘hospital multiple times:
$$ \lim_{x \rightarrow 0} \frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}
= \lim_{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}} - \frac{1}{x^2+1}}{e^x+\sin x -2x-1}
= \lim_{x \rightarrow 0} \frac{\frac{x}{\sqrt{(1-x^2)^3}} + \frac{2x}{(x^2+1)^2}}{e^x+\cos x -2}
= \lim_{x \rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Dice probability where sum of seven doesn't precede sum of eight. Given the question:
A pair of fair dice is rolled until the first sum of 8 appears. What is the probability that a sum of seven doesn't precede sum of 8?
My solution is:
P(sum of 7 doesn't precede 1st sum of 8) = 1 - P(sum of 7 precedes 1st sum of 8) $=... | Sums other than $7$ or $8$ are irrelevant: if we do not record anything that is not a $7$ or $8$, the game ends in one "step." The ordinary probability of a $7$ is $\frac{6}{38}$, and the probability of an $8$ is $\frac{5}{36}$. Thus if we record only a $7$ or $8$, the probability of $8$ is $\frac{5}{11}$.
More formall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\frac{dy}{dx}=\frac{y^2-1}{x^2-1}$ with initial condition $y(2)=2$
Solve the following differential equation: $\dfrac{dy}{dx}=\dfrac{y^2-1}{x^2-1}$, with the initial condition $y(2)=2$.
My attempt:
I notice that this is a separable differential equation, so I try to get it into the form $p(y) dy=f(x) dx$.
$\df... | Since you are considering the boundary condition $y(2)=2,$ it is only fair to restrict the domain of the function to $x\gt1.$ Therefore, we have that $$\frac{y'(x)}{y(x)^2-1}=\frac1{1-x^2}\implies\frac{2y'(x)}{y(x)^2-1}=\frac2{1-x^2}$$ $$\implies\left[\frac1{y(x)-1}-\frac1{y(x)+1}\right]y'(x)=\frac1{1+x}+\frac1{1-x}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Stuck with a tricky existence proof Show that there exists a continuous function $f: [-1, 1] \rightarrow \mathbb{R}$ such
$f(0) = 1$ and
$f(x) = \frac{2-x^2}{2} \cdot f(\frac{x^2}{2-x^2})$
$\forall x \in [-1, 1]$
I tried putting in $x = 1$ and $x = -1$ in the second condition to find that $f(1) = f(-1) = 0$.
I also... | I recall solving this problem before... was it on a Putnam? Sadly, I don't remember what leap of intuition led me to the answer.
One may take $f(x) = \sqrt{1-x^2}$. It is clear that $f$ is continuous on $[-1,1]$ and that $f(0) = 1$, and one sees that
$$\frac{2-x^2}{2} \cdot \sqrt{1-\frac{x^4}{(2-x^2)^2}} = \frac{2-x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a trick to evaluating a matrix multiplied by the sum of three vectors? I am looking at some old linear algebra test material along with some sample answers. I don't understand a bit of computation in the sample answer. Here's the exam question:
Examine the recurrence equation $x_n = Ax_{n-1} + u$, where
$$A =... | There appear to be some typos.
Yes the previous computations are relevant. As in the comment by Avitus,
$Ap = (a^2 + 1)p , Aq=q, Ar = 0$. Then
$$ \begin{align}x_n = \alpha_n p + \beta_n q + \gamma_n r
&= A(\alpha_{n-1}p + \beta_{n-1}q + \gamma_{n-1}r ) + u\\
&= (\alpha_{n-1}(a^2 + 1) + c)p + (\beta_{n-1} + b)q - acr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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ordered pair of unequal positive integer solution of $x+y+z+w = 20$ [1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$\bf{My\; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,z\in \mathbb{Z^{+}}$
$\bullet $ ... | For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) \le S
$$
so $3x+3 \le S$, i.e., $x\... | {
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"source": "stackexchange",
"question_score": "1",
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differentiate arctan (maclaurin?) I have this assignment:
Differentiate this expression:
$$ f(x) =\arctan \frac{x-1}{x+1} $$
There is also known that $-1 < x$ (Why is that important?). I do not know how to solve this problem... By using Maclaurin I can come up with this:
$$ f(x) = \arctan (g(x)) $$
$$ g(x) = \frac{x-... | The stating of $x \gt -1$ merely defines a valid domain for inner function so that differentiation is possible, since at $x=1$ we would have a singularity.
Differentiate $f(x)$ using the chain rule, where $$f'(z)=(\tan^{-1} (z))'=\frac{1}{1+z^2}.$$
Thus, if $g(x) = \frac {x-1}{x+1}$, we have by the chain rule: $$\begin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer Consider the differential equation
$$x^2y''+3(x-x^2)y'-3y=0$$
$(a)$ Find the recurrence equation and first three nonzero terms of the series solution in powers of $$ corresponding to the larger root ... | Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}$
$\therefore x^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+3(x-x^2)\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}-3\sum\limits_{n=0}^\infty a_nx^{n+r}=0$
$\sum\limits_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The difference of two consecutive perfect squares is always odd I am working on another homework assignment about proofs. The question is:
Prove or find counterexample: the difference of two consecutive perfect squares is odd?
There is no counterexample correct? I am thinking this is always true. If I were to do 7^2-6^... | Well of course it is.
Consider $n^2$ and the next greater perfect square would be $(n+1)^2$, which factors to $n^2+2n+1$. Now what would the difference between them be?
$n^2+2n+1-n^2=2n+1$.
All odd integers conform to 2n+1 where n is an integer.
There is also this to consider.
$1=1^2$
$1+3=2^2$
$1+3+5=3^2$
$1+3+5+7=4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655142",
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"source": "stackexchange",
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How to prove $\left(|a+b|^p+|a-b|^p\right)^{1/p}\ge 2^{1/p}\left(a^2+(p-1)b^2\right)^{1/2}$ For real numbers $a, b$ and all $1\le p\le 2$, how to prove $$\left(|a+b|^p+|a-b|^p\right)^{1/p}\ge 2^{1/p}\left(a^2+(p-1)b^2\right)^{1/2}?$$
| let $m=|a+b|,n=|a-b|$
first we prove $\left (\dfrac{m^p+n^p}{2} \right)^2 \ge \left(p\left(\dfrac{m-n}{2}\right)^2+mn\right)^p$
WLOG, let $ m\ge n , x=\dfrac{m}{n}\ge 1 \iff \left (\dfrac{x^p+1}{2} \right)^2 \ge \left(p\left(\dfrac{x-1}{2}\right)^2+x\right)^p \iff f(x)=2\ln{\left (\dfrac{x^p+1}{2} \right)}-p\ln{\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find complex Fourier coefficients
*
*let $f(x) = \sum^{10}_{m=1}(-1)^m \sin(2^m x)$.
denote complex Fourier coefficients of $f(x)$ over $[-\pi, \pi]$ as $c_n = \frac{1}{2\pi} \int _{-\pi}^\pi f(x) e^{-inx}\,dx.$
Calculate $c_n$ for any $n$.
I don't have any idea how to even start.
Can you please expla... | The coefficient $c_{0}=0$, because $f(x)$ is odd
\begin{equation*}
f(-x)=\sum_{m=1}^{10}(-1)^{m}\sin (-2^{m}x)=-\sum_{m=1}^{10}(-1)^{m}\sin
(2^{m}x)=-f(x).
\end{equation*}
For $n<0$, $n=-\left\vert n\right\vert $. Then
\begin{eqnarray*}
c_{n} &=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)e^{-inx}\,dx=\frac{1}{2\pi }
\int_{... | {
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"url": "https://math.stackexchange.com/questions/656004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin... | For acute $\theta$, there's this trigonograph:
$$\sec\theta \cdot \csc\theta \;=\; 2\,|\triangle OPQ| \;=\; 1\cdot( \tan\theta +\cot\theta)$$
(Showing that this works in any quadrant is straightforward.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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How prove $3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$ let $a,b,c>0$ and such $abc=1$,show that
$$3(a^4+b^4+c^4)+2(a+b+c)\ge 5\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$$
my idea: maybe can use AM-GM inequality,
$$3(a^4+b^4+c^4)a^2b^2c^2+2(a+b+c)(a^2b^2c^2)\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
$$\Long... | Here is a straightforward (but not very elegant) solution : let us put
$$
D=3(a^4+b^4+c^4)+2(a+b+c)abc - 5(a^2b^2+b^2c^2+a^2c^2)
$$
We may assume without loss that $a\leq b\leq c$.
Then, if we put $u=b-a$ and $v=c-b$, we have
$$
D=u^4 + (8a + 2v)u^3 + (10a^2 + 12va + 13v^2)u^2 + (10va^2 + 28v^2a + 12v^3)u + (10v^2a^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/664190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Pseudo-pythagorean theorem Pythagoras' theorem is a special case of the Cosine theorem for a angle of $90°$. But also for an angle of 60° and 120°, "aesthetical" special cases derive:
$c^2=a^2+b^2\pm ab$
First question:
Are there further angle $x°$ with a rational number $x$, so that $\cos x$ is rational as well, thus ... | If $\cos\frac {2m\pi}n$ is rational with $\gcd(m,n)=1$, then the primitive $n$th root of unity $\zeta=\cos \frac{2m\pi}{n}+i\sin\frac{2m\pi}n$ is a root of the rational polynomial $X^2-2\cos \frac{2m\pi}{n}X+1$. On the other hand, we know that $\zeta$ is a root of $X^n-1$ and any factorization of this over the rational... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariab... | Eliminating $b,$ we have $$a^2+(2-a-c)^2+c^2=12$$
$$\implies 2a^2-2a(2-c)+2c^2-4c-8=0\iff a^2-a(2-c)+c^2-2c-4=0$$ which is a Quadratic Equation in $a$
As $a$ is real, the discriminant must be $\ge0$
$$\implies (2-c)^2-4(c^2-2c-4)\ge0$$
$$\implies -3c^2+4c+20\ge0\iff3c^2-4c-20\le0$$
Now, we know if $(x-\alpha)(x-\beta)... | {
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"url": "https://math.stackexchange.com/questions/669257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $(a+\sqrt{a^2+1})$ and $(b+\sqrt{b^2+1})$ are converse then prove that a and b are opposites $(a+\sqrt{a^2+1})\,(b+\sqrt{b^2+1})=1$ is supposed to equal: $b = -a$ but how do i get that? I've been trying to solve for like 2 days now.
| If
$$
(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$$
we have
$$
a+\sqrt{a^2+1}=\sqrt{b^2+1}-b \ \ \ \ \ (1)
$$
$$
b+\sqrt{b^2+1}=\sqrt{a^2+1}-a \ \ \ \ \ (2)
$$
$(1)+(2)$ then $a=-b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $a,b,c>0$, prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}$
For $a,b,c>0$, prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}.$$
(Source : Balkan Mathematical Olympiad 2005)
My work:
From the given inequality, we can have,
$$\frac{(a-b)^2}{... | from your last one:
LHS$\ge (|a-b|+|b-c|+|c-a|)^2 \ge (|a-b|+|b-c+c-a|)^2=4(a-b)^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proves of identities in inverse trigonometry Can someone please help me prove the following results from inverse trigonometry?
$$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x+y}{1-xy}( x>0, y>0, xy>1)$$
and
$$\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x+y}{1-xy} ( x<0, y< 0, xy>1)$$
I know the prove for $\tan... | Apply $\;\tan\;$ to both sides and use trigonometric identities:
$$\begin{align*}\bullet&\tan (\arctan x+\arctan y)=\frac{\tan\arctan x+\tan\arctan y}{1-\tan\arctan x\tan\arctan y}=\frac{x+y}{1-xy}\\{}\\\bullet&\tan\left(\pi+\arctan\frac{x+y}{1-xy}\right)=\tan\arctan\frac{x+y}{1-xy}=\frac{x+y}{1-xy}\end{align*}$$
| {
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Expressing a holomorphic function as an infinite sum I want to show that $\frac{1}{\sin z}=\frac{1}{z}+\sum_{n=1}^\infty\frac{(-1)^n2z}{z^2-n^2\pi^2}$. So it's easy to see that the L.H.S. minus the R.H.S. is an entire function, using the fact that the residue of the meromorphic function $\pi\csc\pi z$ at $n\in\mathbb{Z... | If you can't show that the difference is bounded, how about showing its derivative is $\equiv 0$? Let
$$g(z) = \frac{1}{\sin z} - \frac{1}{z} - \sum_{n=1}^\infty (-1)^n\left(\frac{1}{z - n\pi} + \frac{1}{z+n\pi}\right).\tag{1}$$
Then
$$g'(z) = -\frac{\cos z}{\sin^2 z} + \sum_{n\in\mathbb{Z}} \frac{(-1)^n}{(z-n\pi)^2}.\... | {
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Eliminate the parameter from the parametric equations $$x=\frac{3t}{1+t^3} , y=\frac{3t^2}{1+t^3} , t \neq -1,$$
and hence find an ordinary equation in x and y for this curve, The parameter t can be interpreted as the slope of the line joining the general point $(x,y)$ to the origin. Sketch the curve and show that th... | $$ x^3 + y^3 =\frac{27t^3}{(1+t^3)^3} +\frac{27t^6}{(1+t^3)^3}
=\frac{27t^3(1+t^3)}{(1+t^3)^3}=\frac{27t^3}{(1+t^3)^2}$$
$$ x^3 + y^3 = 3xy$$
| {
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Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$. Show that the sequence is upper bounded.
My idea: since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+... | Now,today I have solve this problem,I post my methods,I hope is not wrong?
since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$
then
$$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$
so
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\... | {
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Prove this trigonometry inequality I'm having difficulty proving that tan(26°) < 0.5 < tan(27°) . Any idea ? Thanks.
| OK, here is the solution. Feel free to ask if you don't understand any of the steps.
Let's first prove the left side of the inequality:
$\tan26 < \frac{1}{2} \Rightarrow 2sin26<cos26 \Rightarrow 4sin^2{26}<cos^2{26}$
$\Rightarrow 4sin^2{26} < 1-sin^2{26} \Rightarrow sin^2{26}<\frac{1}{5} \Rightarrow cos^2{52}>\frac{3}... | {
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Prove that the sequence $\{\int_0^{\pi/2} \sin(t^n)\, dt\}$ converges to 0. Prove that the sequence
$$
\left\{\int_0^{\pi/2} \sin(t^n)\, dt :n\in\mathbb N\right\}
$$
converges to 0.
| First split the integral at $t=1$. Estimating the first half is simple; since $0 \leq \sin x \leq x$ for $0 \leq x \leq 1$ we have
$$
0 \leq \int_0^1 \sin(t^n)\,dt \leq \int_0^1 t^n \,dt.
$$
For the second half, make the change of variables $t^n = (\pi/2)^n u$ to get
$$
\int_1^{\pi/2} \sin(t^n)\,dt = \frac{\pi}{2n} \i... | {
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What is the last digit of this expression: $4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$ The expression:
$4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$
I know that the last digit of each of the sumands is 6, but I have trouble proving that. I tried to prove it using induction, but then I realized that I don't h... | $4^2\equiv 6 \mod 10$ and $6^2\equiv 6\mod 10$, which is equivalent to $4(^2)^2=4^4$.
With the $(6 \mod 10)$ relation, we can continue doubling the exponent of $4$ with ease.
$(4^4)^2=4^8\equiv 6^2=6 \mod 10, \dots$
| {
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Probability Questions! Alex, Bret, and Chloe repeatedly take turns tossing a fair die. Alex begins; Bret always follows Alex; Chloe always follows Bret; Alex always follows Chloe, and so on. Find the probability that Chloe will be the first one to toss a six.
| The third player in line wins if the first player fails to roll a six ($5/6$) and then the second player in line wins. So $p_3 = \frac{5}{6}p_2$. Similarly, $p_2 = \frac{5}{6}p_1$. Finally, the first player in line wins if he rolls a six immediately ($1/6$) or if he fails to roll a six ($5/6$) and then the third pla... | {
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Find the possible values of $k$, if the equation has equal roots. The equation $x^2+5k=kx+x+19$ has equal roots. Find the possible values of $k$.
Um having problem in rearranging the equation;
$x^2+5k-kx-x-19=0$
$x^2+k(5-x)-x-19=0$
What is the next step?
| Try this
$$ x^2 - (k +1)x + (5k - 19) = (x - a)^2 = x^2 -2a + a^2 = 0$$
So now by inspection , $$5k - 19 = a^2 $$ $$(k+1) = 2a$$
$$ k = 2a -1 $$
$$5(2a -1) - 19 = a^2$$
$$a^2 - 10a + 24 = 0$$
$$(a -6)(a-4) = 0$$
$$a = 6 \ , \ a = 4$$
$$k = 2(6)-1 \ , \ k = 2(4)-1 $$
$$k= 11 \ , \ k = 7 $$
| {
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Have I done something wrong in solving the following pair of equations? Question given:
Solve,$3x^2-5y^2-7=0\\3xy-4y^2-2=0$
What I have done so far:
$$
3xy-4y^2-2=0\\\frac{3xy}{y}-\frac{4y2}{y}-\frac{2}{y}=0\\3x-4y-\frac{2}{y}=0\\3x=4y+\frac{2}{y}\\x=\frac{(4y+\frac{2}{y})}{3}
$$By substituting $x=\frac{(4y+\frac{2}{y}... | Hint: Heroic! Let us work more simply. Our two equations are $3x^2-5y^2=7$ and $3xy-4y^2=2$. Multiply the first through by $2$, the second by $7$, and subtract. We get $6x^2-21xy +18y^2=0$, or more simply $2x^2-7xy+6y^2=0$. Now we can solve for $y$ in terms of $x$. One could use the Quadratic Formula, but we get lucky,... | {
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Proof review: cauchy sequences Problem statement:
a) Let $(s_n)$ be a sequence such that $\mid s_{n+1} - s_{n} \mid < 2^{-n}$ for all $n \in \mathbb{N}$. Prove that $(s_n)$ is a Cauchy sequence and hence a convergent sequence.
b) Is the result in (a) treu if we only assume that $\mid s_{n+1} - s_{n} \mid < \frac{1}{n}... | For part (b), try this counter-example:
$$s_n = \sum_{k=1}^{n} \frac{1}{n}.$$
Then
$$|s_{n+1} - s_{n}| = \frac{1}{n+1} < \frac{1}{n},$$
but $s_n \to \infty$.
| {
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Problem with a simple exercise So I have to solve the equation $$y^2=4\tag{1.9.88 unit 3*}$$
I did this: $$y^2=4 \text{ means } \sqrt{y^2}=\sqrt{4}=>y=2$$
But I have a problem, $y$ can be either negative or positive so I need to do: $$\sqrt{y^2}=|y|=2=>y=2- or- y=-2$$
Is it right?
| For this type of problem, I always think, "How would it appear on a graph? What are the x-intercepts?" Of course, I do not have time to actually draw it out. But I can think, "Let $y^2$ be $x^2$. How can $x^2=4$ be turned into a function? I can just move $4$ to the left by subtracting $4$ on both sides, and replacing t... | {
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"timestamp": "2023-03-29T00:00:00",
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Please explain where the $2x$ came from in this cubic identity: The question is to factorize the difference of cube identity $(x + 1)^3 - y^3$
I obviously want to put it in the form of $(a - b)(a^2 + ab + b^2)$
My working out:
$(a - b) = (x + 1) - (y) = (x + 1 - y)$
$(a^2 + ab + b^2) = (x^2 + 1) + (x + 1)(y) + (y^2) = ... | You made an error while substituting into $(a^2 + ab + b^2)$ .
| {
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Increasing function on small interval given positive derivative Suppose that $f'(0)>0$. Does it imply that there exists a $\delta > 0 $ such that $f$ is increasing on $[0,\delta]$?
I think this is false and I've been trying to think of a counter example. I was thinking using the function
$f(x) = x^2sin(1/x)$ if $x \neq... | I came across this problem as well. The solution I found dealt with decreasing functions, but the problem is equivalent to the increasing case (just take $-f$ as defined below).
Consider the function $f$ on $[ 0 , 2/\pi ]$:
$$
f(x)=\begin{cases}
x^2\sin^2 \left( \frac 1x \right) -\frac 12 x&, \text{if } x\neq 0 \\
-\fr... | {
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Showing that $n$ is pseudoprime to the base $a$ Show that if $n=\frac{a^{2p}-1}{a^2-1},$ where $a$ is an integer, $a>1$, and $p$ is an odd prime not dividing $a(a^2-1)$, then $n$ is pseudoprime to the base $a$.
Let $n=\frac{a^{2p}-1}{a^2-1}=1+a^2+(a^2)^2+(a^2)^3+\cdots$. Then $n-1=\frac{a^{2p}-1}{a^2-1}-1=\frac{a^{2p}... | Write
$$\begin{align}
n-1 &= \frac{a^{2p}-a^2}{a^2-1}\\
&= \frac{(a^p-a)(a^p+a)}{a^2-1}\\
&= a\frac{a^{p-1}-1}{a^2-1}(a^p+a).
\end{align}$$
Since $p$ is odd, $a^2-1$ divides $a^{p-1}-1$, so we have written $n-1$ as a product of three integers, and the last one, $a^p+a$ is even, whether $a$ is even or odd. Hence $n-1$ i... | {
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Express in terms of $x$ and $y$ when the values of $x$ and $y$ are given. Given,
$x=1+3a+6a^2+10a^3+\ldots$
$y=1+4b+10b^2+20b^3+\ldots$
$s=1+3ab+5(ab)^2+7(ab)^3+\ldots$
Express $s$ in terms of $x$ and $y$.
My work:
I could see how the first sequence works, but could not find how the second sequence works, until I wro... | Assume, $|a|,|b|<1$
You can find s by using formula for infinite GP in the last step.
$s=\frac{1+\frac{2ab}{1-ab}}{1-ab}=\frac{1+ab}{(1-ab)^2}$
Also note that taylor series for $\frac 1 {(1-x)^n}$ is :
$ 1+{n\choose 1} x +{n+1\choose2}x^2 + {n+2\choose3}x^3... $
Hence $x$ stands for $n=3$ and $y$ for $n=4$
Now find $x$... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that any palindrome with an even number of digits is divisible by 11 Confusing myself here, need some clarification..
First, we consider the palindrome $abccba$. We can see this can be written as
$$a(10^5 + 10^0) + b(10^4 + 10^1) + c(10^3 + 10^2) = a(10^5 + 1) + 10b(10^3 + 1) + 100c(10 + 1)$$ So essentially we se... | Hint: Notice that $10^k=(-1)^k \pmod{11}$. So, for $k$ odd and $k'$ even, $10^{k} + 10^{k'}=0\pmod{11}$.
| {
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If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
| since
$$\sum_{cyc}\dfrac{a}{b+c^2}=\sum_{cyc}\dfrac{a^2}{ab+ac^2}$$
Use Cauchy-Schwarz inequality,we have
$$\left(\sum_{cyc}\dfrac{a^2}{ab+ac^2}\right)\cdot\sum_{cyc}(ab+ac^2)\ge\left(\sum_{cyc}a\right)^2=1$$
so we only prove this following inequality
$$\dfrac{1}{\sum_{cyc}(ab+ac^2)}\ge\dfrac{9}{4}$$
$$\Longleftrightar... | {
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How do I prove this trigonmetric identity? I need to prove that the following identity is true:
$$
\frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x
$$
This isn't homework; just a practice exercise. But I keep getting stuck! Thanks much.
| Let's start with our original expression:
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}$$
We need to prove that:
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x$$
Step 1: Recall that $\tan x = \dfrac{\sin x}{\cos x}$. This means that $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\dfrac... | {
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Can this be converted into a polynomial equation for $x$? I came upon this monster equation while fiddling with the area of pentagons:
$$\sqrt{(a+b+c-x)(a+b-c+x)(a+c-b+x)(b+c-a+x)}+\sqrt{(c+d+x)(c+d-x)(c-d+x)(d-c+x)}=4T$$
Where $a,b,c,d,e,T$ are known quantities. So my question is:
*
*Can this be converted into a p... | If you have an equation of the form $\sqrt a + \sqrt b = c$ then by taking squares we get $2\sqrt {ab} = c^2-a-b$, and then $4ab = (c^2-a-b)^2$.
You can also develop the product in the equation $(\sqrt a + \sqrt b - c)(\sqrt a - \sqrt b - c)(- \sqrt a + \sqrt b - c)(- \sqrt a - \sqrt b - c) = 0$, which gives you (again... | {
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Integrate $(x^2+1)^\frac{1}{3}$ I tried many times to integrate this function using integration by parts, substitution as $x^2+1=t$ without any conclusion... is it integrable? If it is integrable, how can I integrate it?
Thank you in advance
| Let $u=(x^2+1)^\frac{1}{3}$ ,
Then $x=(u^3-1)^\frac{1}{2}$
$dx=\dfrac{3u^2}{2(u^3-1)^\frac{1}{2}}du$
$\therefore\int(x^2+1)^\frac{1}{3}~dx=\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
Case $1$: $|u|\leq1$ , i.e. $\left|(x^2+1)^\frac{1}{3}\right|\leq1$
Then $\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
$=\int\dfrac{3u^3}{2i(1... | {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\r... | First Approach - Using Euler sums
We begin by finding the Maclaurin series expansion for $\arctan^2 x$.
Since
$$\arctan x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1}, \qquad |x| < 1,$$
applying the Cauchy product to the product between two inverse tangent functions leads to:
\begin{align}
\arctan^2 x &= \ar... | {
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"timestamp": "2023-03-29T00:00:00",
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Cramer's Rule Question Use Cramer's rule to solve this system for z:
$$2x+y+z=1$$
$$3x+z=4$$
$$x-y-z=2$$
so my work is:
$$\frac{\left|\begin{matrix}
2 & 1 & 1\\
3 & 0 & 4\\
1 & -1 & 2
\end{matrix}\right|}{\left|\begin{matrix}
2 & 1 & 1\\
3 & 0 & 1\\
1 & -1 & 1
\end{matrix}\right|}$$
which gives $\frac{3}{-3}$ or $-1$. ... | The determinant in the denominator is incorrect. It should be
$$\left|\begin{matrix}
2 & 1 & 1\\
3 & 0 & 1\\
1 & -1 & {\color{red} {-1}}
\end{matrix}\right|$$
which evaluates to (according to Sarrus' Rule):
$$(2\cdot0\cdot-1) + (1\cdot1\cdot1) + (1 \cdot 3 \cdot -1) - (1\cdot0\cdot1) - (1\cdot3\cdot-1) - (1\cdot2\cdot... | {
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Solving an equality in 2 variables I need to prove that
$$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$
if $a+b = 1$ and $a b \le 1/4$
I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get
$a b \le 4$ in the inequality ?
| $$
\begin{align}
\hspace{-1cm}\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2
&\ge2\left(a+\frac1a\right)\left(b+\frac1b\right)\tag{1}\\
&=\frac2{ab}\left(a^2+1\right)\left(b^2+1\right)\\
&\ge8\left(a^2+1\right)\left(b^2+1\right)\tag{2}\\
&=8\small\left(\left(a-\tfrac12\right)^2+\left(a-\tfrac12\right)+\tfrac54\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the integral : $\int\frac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$ Find the integral : $\int\dfrac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$
Please guide which substitution fits in this I am not getting any clue on this .. thanks..
| After using the substitution posted you'll have to do a bit of work but should get this at your final answer
$$2\sqrt{x}+\ln(\sqrt{x}-1)-\ln(\sqrt{x}+1)-\ln(x-1)-3x^\frac{1}{3}+\ln(x^\frac{2}{3}+x^\frac{1}{3}+1)-2\ln(x^\frac{1}{3}-1)-\ln(x^\frac{1}{3}+x^\frac{1}{6}+1)+2\ln(x^\frac{1}{6}-1)+\ln(x^\frac{1}{3}-x^\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/717902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\fr... | Substitute $t= \sin2x$
\begin{align}
&\int \cos 2x\ln \frac{\cos x+\sin x}{\cos x-\sin x}dx\\
=&\>\frac12\int \tanh^{-1}t\>dt\overset{ibp}=\frac12
t \tanh^{-1}t+\frac14\ln(1-t^2)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 4
} |
Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}... | Applying AM-GM on $2ab$ and $a^2+b^2$ we find $2ab(a^2+b^2) \leq \left(\frac{(a+b)^2}{2} \right)^2$ or $a^2+b^2 \leq \frac{(a+b)^4}{8ab}$. Plugging this in, we find
$$
\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{(a+b)^3}{\sqrt[3]{2(a+b)\frac{(a+b)^4}{8ab}}} = \frac{(a+b)^3\sqrt[3]{4ab}}{\sqrt[3]{(a+b)^5}} = 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find Max value of this expression: $P=(x-2yz)(y-2zx)(z-2xy)$ Let $x$, $y$ and $z$ be positives and satisfying $x^2+y^2+z^2=2xyz+1$ .
Find a maximum of this expression:
$$P=(x-2yz)(y-2zx)(z-2xy).$$
| Set $y=x$ and $z=1$. Then $x^2+y^2+z^2=2x^2+1=2xyz+1$ and
$$(x-2yz)(y-2zx)(z-2xy)=(-x)(-x)(1-2x^2)=x^2(1-2x^2)$$
which clearly has no minimum...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
standard Taylor series using substitution Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?
Determine a range of validity for this series.
| Hint
Start with Taylor expansion $$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+O\left(y^6\right)$$ Rise to the third power to obtain $$\frac{1}{(1+y)^3}=1-3 y+6 y^2-10 y^3+15 y^4-21 y^5+O\left(y^6\right)$$ Now, replace $y$ by $\frac{4 x}{5}$ to get your result.
I am sure that you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/723199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Analysis problem from Romanian Contest - 2 sequences which forms another one Let $a,b$ be 2 real numbers, and the sequences $(a_n)_{n \geq 1}, (b_n)_{n \geq 1}$ defined by $a_{1}=a$, $b_{1}=b$, $a^2+b^2 <1$ and
\begin{cases}
a_{n+1}=\frac{1}{2}\left(a_{n}^{2}-\frac{b_{n}^{2}}{n^{2}}\right), \mbox{ }(\forall) n \geq 1\... | i try to find the regular of this sequence :
$a_{2}=\frac{1}{2}$$(a^2-b^2)$$,$$b_{2}=-2ab$
$a_{3}=\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4})$$,$$b_{3}=(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{3}{2})$
$a_{4}=\frac{1}{2}((\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4}))^{2}$$-(\frac{1}{2}$$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The MacLaurin series of $e^{\frac{x^2}{2}}$ Could someone please explain to me how you derive the MacLaurin series for $e^{ \frac{x^2}{2}}$?
I understand how it is derived from the MacLaurin series for $e^x$ where it is
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum_{n=1}^\infty\frac{x^{2n}}{2^nn!}$
and y... | $$e^x=\sum^\infty_{k=0}\dfrac{x^k}{k!}\\
\implies e^{x^2/2}=\sum^\infty_{k=0}\dfrac{(x^2/2)^{2}}{k!}=\sum^\infty_{k=0}\dfrac{x^{2k}}{2^kk!}$$
Differentiate:
$$e^{x^2/2}|_{x=0}=1\\
e^{x^2/2}x|_{x=0}=0\\
(e^{x^2/2}+x^2e^{x^2/2})|_{x=0}=1$$
E.t.c. Every differential $\dfrac{d^{2n}}{dx^{2n}}e^{x^2/2}=1$, and every differen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find constants of function I have this equality :
$$f(x)=\frac{9}{(x-1)(x+2)^2}$$
I am required to find the constants A, B and C so that,
$$f(x) = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^{2}} $$
How do we go about solving such a question?
I am not sure on how to solve such questions. Approach and Hints to ... | From
\begin{equation}
\frac{9}{(x-1)(x+2)^2} = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^2}
\end{equation}
it follows, after multiplying each side with $(x-1)(x+2)^2$:
\begin{equation}
9=A(x+2)^2+B(x+2)(x-1)+C(x-1)
\end{equation}
which, after regrouping, gives
\begin{equation}
9=(A+B)x^2 + (4A+B+C)x + (4A-2B-C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/729350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
How to prove this trig identity? If $A+B+C=\pi$ then prove:$$\sin^2A+\sin^2B+\sin^2C=2-2\cos A\cos B\cos C$$
I am completely lost on this, please help.
| $A+B+C = \pi \implies A + B = \pi - C \implies \sin(A+B)=\sin(C) \land \cos(A+B)=-\cos(C)$
$\sin^2(A) + \sin^2(B) + \sin^2(C) \\ = \sin^2(A) + 1-\cos^2(B) + 1 − \cos^2(A+B) \\ =2+(\sin^2(A)-\cos^2(B))-\cos^2(A+B) \\ = 2+\cos(B-A)\cos(A+B)-\cos(A+B)\cos(A+B) \\ = 2+(\cos(B-A)+\cos(A+B))\cos(A+B) \\ = 2 - 2\cos(A)\cos(B)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$. Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.
I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.
| [I'm assuming by function you mean cubic polynomial, otherwise this is simple.]
Observe that the first three equations are satisfied by the function $2x$.
Hence, the polynomial $f(x) - 2x$ has roots $x=1, 2, 3$, or that $f(x) = 2x + A(x-1)(x-2)(x-3)$.
Finally, set $x=4$, we get $\pi = 8 + A\times 3\times 2 \times 1$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Using EigenValues to form a diagonal matrix After going through my Linear algebra note, I know if for any matrix $A$, we find the eigenvalues and eigenvectors , we can construct a matrix P, such that $P^{-1}AP$ is a diagonal matrix.
Now for this matrix for example
$\left( \begin{array}{ccc}
1 & -2 \\
1 & 4 \end{array} ... | It was noted in the comments that the eigenvector for $2$ is wrong and the OP needs further help in finding an eigenvector for $2$.
Let $\begin{pmatrix} v_1\\ v_2\end{pmatrix}$ be an eigenvector for $2$. It holds that $\begin{pmatrix} 1 & -2\\ 1 & 4\end{pmatrix}\begin{pmatrix} v_1\\ v_2\end{pmatrix}=2\begin{pmatrix} v_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/732552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How find this $\frac{3x^3+125y^3}{x-y}$ minimum value
let $x>y>0$,and such $xy=1$, find follow minimum of the value
$$\dfrac{3x^3+125y^3}{x-y}$$
My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequalit... | continuation:
let $x-y=t$, $xy=1$, we get:
$$x=\frac{1}{2}t+\frac{1}{2}\sqrt{t^2+4},\\y=-\frac{1}{2}t+\frac{1}{2}\sqrt{t^2+4}$$
then we have
$$f=\frac{3x^3+125y^3}{x-y}\\=\frac{-61t^3+64t^2\sqrt{t^2+4}-183t+64\sqrt{t^2+4}}{t}$$
we calculate the differential:
$$g=\frac{df}{dt}=-\frac{2(61t^3\sqrt{t^2+4}-64t^4-128t^2+128... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proof that $y^2=x^3+x$ has a unique integer solution Prove that the equation $y^2=x^3+x$ has only one integer solution, namely $x=y=0$.
| Write $y^2=x(x^2+1)$. If $x=0$ then $y=0$ and if $y=0$ then $x=0$. We can then assume that $xy\not=0$, and replace $x$ with $-x$ or $y$ with $-y$ so that $x>0$, $y>0$. Observe that $x$ and $x^2+1$ are coprime, so since $x(x^2+1)$ is a square, so is $x$ and so is $x^2+1$. This is impossible because $x^2+1=a^2$ is imposs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to factor $x^4+2x^3+4x^2+3x+2$ I'm trying my hand on these types of expressions.
How to factorize $x^4+2x^3+4x^2+3x+2$ into two (or more) polynomials with rational coefficients. please write step by step solution.
| I think the best way is by intuition.
$x^4+2x^3+4x^2+3x+2$
$=(x^4+x^3+x^2)+(x^3+x^2+x)+(2x^2+2x+2)$
$=x^2(x^2+x+1)+x(x^2+x+1)+2(x^2+x+1)$
$=(x^2+x+1)(x^2+x+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/736479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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have trouble with this limit question
a) By considering the areas of the triangle OAD, the sector OAC and the triangle OBC,
show that
$(\cos \theta)(\sin \theta) < \theta < \frac{\sin\theta}{\cos\theta}$
I find out:
Area of OAD=$\frac{1}{2}OD\cdot AD \cdot \sin \theta$
Area of OAC=$\frac{1}{2}OC^2 \theta$
Area of OBC=... | The geometry has been discussed in comments. So we have
$$\frac{1}{2}\cos\theta\lt \frac{1}{2}\theta.$$
Multiply through by $2$. We get $\cos\theta\sin\theta\lt \theta$.
Divide both sides by $\theta\cos\theta$. We get
$$\frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{1}$$
We also got from the geometry that
$$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/738858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Can we solve this using stars and bars? The number of ways of distributing 12 identical oranges among 4 children so that every child gets at least one and no child more than 4 is 31.
My try: First of all give each child 1 orange and we are left with 8 oranges.
Then
3 3 2 0
3 3 1 1
3 2 2 1
2 2 2 2
and permuting each, w... | As SandeepThilakan said in the comments above, the coefficient of $x^{12}$ in the expansion of the polynomial $(x^1+x^2+x^3+x^4)^4$ counts the way that between 1 to 4 objects distributed to 4 groups gives a total of 12 objects.
Using the binomial expansion theorem:
$$(x+x^2+x^3+x^4)^4 \\ = ((x+x^2)+x^2(x+x^2))^4 \\ = x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How do I find the $x$ intercepts for $-x^2-3x+3$ How do I find the $x$ intercepts? $-x^2-3x+3$?
I converted the function into vertex form but I am stuck at $3/4= -(x^2+1.5)$.
Can someone give me and idea what I can do?
| Alternatively, you can complete the square.
$x^2 + 3x - 3 = (x + \frac{3}{2})^2 - 3 - \frac{9}{4} = 0$. So, $(x + \frac{3}{2})^2 = \frac{21}{4} \Rightarrow x = -\frac{3}{2} \pm \frac {\sqrt{21}}2.$
Thus the x-intercepts are $(\frac{-3+\sqrt{21}}{2},0)$ and $(\frac{-3-\sqrt{21}}{2},0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/740161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
| Looking at $(x+1)^4 + (x-1)^4=16$, I recall that $16=2^4$. So I look at the equation trying to see if I could get a $2^4$ in there. This sugguests that we guess $x=1$ to make the left term a power of $2$.
Plugging in $x=1$,
$$ (1+1)^4 + (1-1)^4 = 2^4 + 0^4 = 16 $$
So $x=1$ is a solution.
This problem has a symmetry pro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ As the title states, trying to solve $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$
| Hints: Expand the denominator as a geometric series in $-x^4$:
$$\frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}=1-x^4+x^8-x^{12}+...$$
Multiply by $1+x^2$ to obtain a series form for the integrand.
$$\frac{1+x^2}{1+x^4}=\frac{1}{1+x^4}+\frac{x^2}{1+x^4}=1+x^2-x^4-x^6+x^8+x^{10}-x^{12}-x^{14}+...$$
All that's left is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Finding the lengths of the sides of a triangle given 3 angles only. If a right $\triangle ABC$ has $\angle A= 90^\circ$, $\angle B=45^\circ$, $\angle C=45^\circ$. Is there a way of finding the lengths of the sides $a$, $b$ & $c$ without knowing any of their lengths ? Normally we use $(\cos{x})^2+(\sin{x})^2=1$ as the h... | Proof:if x,y and z belong to real numbers and$x<y<z$ then
$(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times((1-\frac{x}{z})\times\sqrt\frac{(x+z)}{(z-x)})=\sin A \tag{1}$
$(\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})-(((\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Indeterminate limit that is supposed to be solved with De L'Hospital's rule Last week my Maths teacher gave the class this exercise taken from our text book. We are working on De L'Hospital's rule at the moment and this exercise is from that part of the book so everybody assumed that was the right procedure to solve it... | \begin{eqnarray*}
\frac{\sin ^{2}x-x^{2}\cos ^{2}x}{x^{2}\sin ^{2}x} &=&\frac{\left( \sin
x-x\cos x\right) \left( \sin x+x\cos x\right) }{x^{2}\sin ^{2}x} \\
&=&\frac{\frac{\left( \sin x-x+x-x\cos x\right) }{x^{3}}\frac{\left( \sin
x+x\cos x\right) }{x}}{\frac{x^{2}\sin ^{2}x}{x^{4}}} \\
&=&\frac{\left( \frac{\sin x-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/744172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Integrating a function with substitution Totally forgot how to integrate.
$$ \int \frac{1}{x^2 \sqrt{x^2+4}}dx$$
Just need a tip, for this what would I use to substitute?
| $$\int \frac{1}{x^2 \sqrt{x^2+4}}dx = \int \frac{1}{8(\frac{x}{2})^2 \sqrt{(\frac{x}{2})^2+1}}dx= \int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}dx$$
$$=\int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}dx\frac{\frac{d(tan^{-1}(\frac{x}{2}))}{dx}}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/746175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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What's wrong with this recursion of counting codes of length $n$ formed by $a$, $b$, and $c$ such that no three consecutive letters are distinct I found the following problem in a combinatorics book and gave it a try.
Let $B_n$ denote the set of codes of length $n$ formed by using the letters $a$, $b$, and $c$, none o... | You have two cases to consider: strings that end in $x x$ and strings that end in $x y$ ($x \ne y$). Call the numbers of each of length $n$ respectively $u_n$ and $v_n$, set up recurrences for both. You are interested in $u_n + v_n$.
*
*$\ldots xx$: You can add a new $x$ (1 possibility) to get a new $\ldots xx$,
or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find sequence function and general rule the function $$a_{n+2}=3a_{n+1}-2a_n+2$$
is given, and $$a_0=a_1=1, (a_n)_{n\ge0}$$
multiplying everything by $$/\sum_{n=0}^\infty x^{n+2}$$
also adding $$\sum_{n=0}^\infty (a_{n+2}x^{n+2}+a_1x+a_0)-a_1x-a_0=\sum_{n=0}^\infty (3a_{n+1}x^{n+2}+a_0)-a_0-\sum_{n=0}^\infty 2a_nx^{n+2... | The generating function I'm getting is:
$
\begin{align}
R(x) &= \dfrac{1-3\, x+4\, x^2}{1-4\, x+5\, x^2-2\, x^3}\\
&= \dfrac{2}{1-2 \, x} + \dfrac{1}{1-x} - \dfrac{2}{{\left(1-x\right)}^{2}}\\
&= \sum_{n\ge 0} \left(2^{n+1}-2\, n-1\right)\, x^n
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/753074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find two triangles of longest side length 25? I'm using the quadratic Diophantine equations to solve for two integer triangles of longest side $25$. It's been shown that for $a^2+b^2=c^2$, which goes to $x^2+y^2=1$ where $x=\frac ac, y=\frac bc, a=t^2−1, b=2t, c=t^2+1$.
If I want to solve for $c=25$, how will I go abo... | All Pythagorean triples, i.e. triplets of positive integers such that $a^2+b^2=c^2$, can be expressed as $a=k(m^2-n^2)$, $b=2kmn$, $c=k(m^2+n^2)$; where $m$ and $n$ are coprime, $m>n$ and $m$ and $n$ have opposite parity. (Up to interchanging of $a$ and $b$.)
So in your case you want $k(m^2+n^2)=25$. This gives you the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/753294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the inequality $({1+\frac{a}b})^n$ + $(1+\frac{b}a)^n$ $\geq$ $2^{n+1}$ Let $a$ and $b$ be positive real numbers and let $n$ be a natural number
prove that
$$\left({1+\frac ab}\right)^n+\left(1+\frac ba\right)^n\ge2^{n+1}.$$
| $a,b$ are positive, so
$$\dfrac{a^k}{b^k}+\dfrac{b^k}{a^k}\ge2\Longleftrightarrow a^{2k}+b^{2k}\ge2a^kb^k\Longleftrightarrow\big(a^k-b^k\big)^2\ge0,$$
and using the binomial theorem (three times total) we get
$$\left(1+\frac ab\right)^n+\left(1+\frac ba\right)^n=\sum_{k=0}^n\binom nk\left(\dfrac{a^k}{b^k}+\dfrac{b^k}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/753593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help... | When numbers are so small you can proceed by attempt.
However $a^2+b^2=6$ doesn't have no integer solution: wlog you can take $a,b\geq0$.
So try: $a=0\Rightarrow b=\sqrt6$ (not valid in $\mathbb Z$).
$a=1\Rightarrow b=\sqrt5$ (not valid as above)
$a=2\Rightarrow b=\sqrt2$ (idem).
$a\geq3\Rightarrow b^2<0$.
But maybe yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from sol... | You have
$$(2^x+2^y)^2=4^x+4^y+2^{x+y+1}=2^{x+1}+2^{y+1}+2^{x+y+1}$$
If $a=2^x$, $b=2^y$, then
$$(a+b)^2=2(a+b+ab)>2(a+b)$$
so $a+b>2$.
I'm looking for an upper bound. If I find it, I'll edit this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$ I would like to show that
$$ \sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64} $$
I've been working o... | For positive integer $n$
If $\sin(2n+1)x=0, (2n+1)x=m\pi\iff x=\frac{m\pi}{2n+1} $ where $m$ is any integer
From $(3)$ of this,
$\displaystyle \sin(2n+1)x=2^{2n}s^{2n+1}+\cdots+(2n+1)s=0$ where $s=\sin\frac{m\pi}{2n+1}$
So the roots of $\displaystyle 2^{2n}s^{2n+1}+\cdots+(2n+1)s=0 $ are $\sin\frac{m\pi}{2n+1}; 0\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
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