Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to simplify this formula I have this equation: $\frac{\sqrt{(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(-a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(a-b+\sqrt{a^2+b^2-2ab\cos\gamma})(a+b-\sqrt{a^2+b^2-2ab\cos\gamma})}}{4}$
It calculates the area of a triangle using only 2 sides of a triangle $a, b$, and the angle of those sides, $\ga... | Let $c = \sqrt{a^2+b^2-2ab\cos\gamma}$ and $P_1,P_2,P_3,P_4$ be the 4 factors inside the square root. Notice
*
*$P_1 P_4 = (a+b+c)(a+b-c) = (a+b)^2 - c^2 = 2ab(1+\cos\gamma)$,
*$P_2 P_3 = (-a+b+c)(a-b+c) = c^2-(a-b)^2 = 2ab(1-\cos\gamma)$
We have
$$\frac{\sqrt{P_1P_2P_3P_4}}{4}
= \frac{\sqrt{(2ab)^2(1-\cos^2\gamma)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$? I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?
| $$\frac{1}{x^8(1+x^2)}=\frac{1+x^2-x^2}{x^8(1+x^2)}=\frac{1}{x^8}-\frac{1}{x^6(1+x^2)}=\cdots$$
and so on, finally you will get:
$$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892228",
"timestamp": "2023-03-29T00:00:00",
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2... | Well, If $a,b,c$ form an arithmetic progression then there is a $d$ so that $a=b-d$ and $c=b+d$.
And to show that $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ form an arithmetic progression we have to show there is a $k$ so that
$k = (a^2+ac+c^2) - (a^2+ab+b^2)$ and $k= (b^2+bc+c^2) -(a^2+ac+c^2)$ or in other words that
$(a^2+a... | {
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"url": "https://math.stackexchange.com/questions/3892856",
"timestamp": "2023-03-29T00:00:00",
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Triple Integral over a Superquadric How do I solve the following:
$\iiint _V (x^2+y^2+z^2) dxdydz $, where $V=\{x^{2n}+y^{2n}+z^{2n} \leq 1\}$ is a superquadric?
| By symmetry,
$$ \iiint _V (x^2+y^2+z^2) dxdydz=24\iiint_{V^+} x^2 dxdydz $$
where
$$ V^+=\{x^{2n}+y^{2n}+z^{2n} \leq 1, x\ge0,y\ge0,z\ge0\}. $$
Let
$$ x=r^{\frac1n}\sin^{\frac1n}(\theta)\cos^{\frac1n}(\phi), x=r^{\frac1n}\sin^{\frac1n}(\theta)\cos^{\frac1n}(\phi),z=r^{\frac1n}\cos^{\frac1n}(\theta). $$
Then $V^+$ becom... | {
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Proving $\binom{2n}{n}<4^{n-1}$ for all positive integers $n\geq 5$ For $n=5$, we get $\binom{10}{5}<4^4$. Asumme inductively that $\binom{2n}{n}<4^{n-1}$ for all positive integers $n\geq 5$. Then I need to prove that $\binom{2(n+1)}{n+1}<4^{n}$ for all positive integers $n\geq 5$. I've been looking around a bit, and I... | Using inductive argument, you can do something simpler.
$$\binom{2n+2}{n+1}=\frac{2(2n+1)}{n+1}\binom{2n}{n}$$
Now prove that
$$\frac{2(2n+1)}{n+1}<4.$$
Once the inequality is proven, we can multiply both sides by $\binom{2n}{n}$ and get,
$$\frac{2(2n+1)}{n+1}\binom{2n}{n}<4\binom{2n}{n}\to \binom{2n+2}{n+1}<4\binom{2n... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$ Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$
Case $S(1)$ is true:
$$2((1)+2)\leq((1)+2)^{2}$$
$$6\leq9$$
Case $S(n)$ is true for all $n=1,2,...$
$$2(n+2)\leq(n+2)^{2}(i)$$
Case $S\left(n+1\right)$
$$2(n+3)\leq(n+3)^{2}(ii)$$
From (i)
$$2(n... | Recurrence :
$2(n+1)\leq(n+2)^2 $
For n=1
$4\leq 9$ that is correct
Suppose : $ 2(n+1)\leq(n+2)^2 $
Let's show $ 2(n+2)\leq(n+3)^2 $
$2(n+2)\leq(n+3)^2 $
$\Leftrightarrow $ $2(n+2)\leq ((n+2)+1)^2 $
$\Leftrightarrow $$2(n+2)\leq (n+2)^2 +2(n+2)+1$
$\Leftrightarrow $$0\leq(n+2)^2 +1$ that is correct
So:
$ 2(n+2)\leq(n+3... | {
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Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ $$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$
$$x=?$$
I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that,... | Your way is fine with a solution for $x=-\frac12$, now we have that
$$3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1=x^2(3x ^ 2-x + 1)+(x-1)^2 > 0$$
since $3x ^ 2-x + 1>0$.
To find complex roots we can refer to Quartic equation.
| {
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Prove that $\prod_{i=1}^n(1+x_i)\leq \sum_{i=0}^n\frac{S^i}{i!}$, where $x_i\in\mathbb{R^+}$.
Let $x_1$, $x_2$, $\ldots$, $x_n$ be positive real numbers, and let $$S=x_1+x_2+\cdots+x_n.$$ Prove that $$(1+x_1)(1+x_2)\cdots(1+x_n)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots+\frac{S^n}{n!}.$$
My first thought is about ... | Induction, as you requested, works too. Just do it. Where are you stuck?
$$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$
Let $ S' = \sum_{i=1}^{k+1} x_i = S + x_{k+1}.$
$\prod_{i=1}^{k+1} (1 + x_i) \leq (1+x_{i+1}) (1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!})... | {
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Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$ Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$
The way I attempted to the question is the following:
I state that $a=\sqrt[3]{x+1}$ hence we have that $3a^3-3+a+a^2=-1$. Hence $3a^3-2+a+a^2=0$, so $a^2(3a+1)+a-3=0$. A... | $3a^3+a^2+a-2=0$
$(3a-2)(a^2+a+1)=0$
We can find this by using Rational Root theorem, or another way is by noting that $a^2+a+1$ is also a root because $a^2+a+1=0 \implies a^3=1$ (roots of unity) so
$$3a^3+a^2+a-2=3+a^2+a-2=a^2+a+1=0$$
| {
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Evaluate $ 2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx $ I've been trying to solve this this problem for so long but haven't had much luck, can someone please help me out?
The original question was:
Find the area of the part of the surface $z = \ln(x^2+y^2+2)$ that lies above the disk $x^2 + y^2 \le 1$ ... | This is not a solid of revolution. The proble asks the area of the surface $z\log \left(x^2+y^2+2\right)$ inside the cylinder $x^2+y^2=1$
$z=\log \left(x^2+y^2+2\right)$ and $D:x^2+y^2\le 1$
$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$
$$z_x=\frac{2 x}{x^2+y^2+2};\;z_y=\frac{2 y}{x^2+y^2+2}$$
$$S=\iint\limits_{D} \s... | {
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"timestamp": "2023-03-29T00:00:00",
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Sum of cube roots of two conjugate quadratic integers makes an integer. Consider the following expression:
$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$
This equals $4$, but how can I show this?
Note that I do not want to make use of the following line of reasoning: 4 is a solution to $x^3-6x-40=0$, that this cubic ... | First note that $392=2^3\times7^2$ and so $\sqrt{392}=14\sqrt{2}$, and hence
$$\sqrt[3]{20+\sqrt{392}}=\sqrt[3]{20+14\sqrt{2}}.$$
Next, in the hope of finding a simple expression for the cube root, we compute
$$(20+14\sqrt{2})(20-14\sqrt{2})=20^2-2\times14^2=8,$$
which shows that
$$\sqrt[3]{20+14\sqrt{2}}\sqrt[3]{20-14... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902338",
"timestamp": "2023-03-29T00:00:00",
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sum of binomial coefficients expansion to prove equation I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$
I already expanded
$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}... | $$\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}=\frac{1\times 2}{2}+\frac{2\times 3}{2}+\frac{3\times 4}{2}+...+\frac{(n-1)\times n}{2}=\frac{1}{2}(1\times 2+2\times 3+3\times 4+...+(n-1)\times n)=\frac {1}{2}\times \frac{(n-1)n(n+1)}{3}=\frac{(n-1)n(n+1)}{6}$$
| {
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Uniform convergence of $\frac{n^3x}{1+n^4x^2}$ I have been trying to prove the uniform convergence to f = 0 on [0,1] for the sequence of functions $\frac{n^3x}{1+n^4x^2}$.
So far i have attempted to prove pointwise converge as such;
Let $ N = 1/\epsilon x $
For every $n \ge N$ we have $ \vert\frac{n^3x}{1+n^4x^2} - ... | One strategy to check for uniform convergence is to find the maximum $M_n$ of $f_n(x) = \frac{n^3x}{1 + n^4 x^2}$ over $[0,1]$ using calculus, then check whether $M_n \to 0$ (keeping in mind that $0 < f_n(x) \leq M_n$ for all $x$).
That said, the same idea works with any upper bound for $f_n(x)$ that approaches $0$ as ... | {
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Is my $\epsilon$-$\delta$ calculation correct? I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta... | You want
$$
|x^2+1||x+1||x-1|<\varepsilon
$$
near $x=1$.
First step is to control the quantity $|x^2+1||x+1|$ near $x=1$. So first
restrict $x$ so that $|x-1|<1=\delta_1$. This means
$0<x<2$, and hence $|x+1|<3$ while $x^2+1<5$. Hence
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^2+1||x+1|<15,
$$
and thus
$$
|x-1|<1... | {
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Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$ I have tried to wrap my head around this for some time now, and quite frankly I am stuck.
Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$
Then what is the va... | Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$,
$$\tan(2n+1)x=\dfrac{\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}}{\binom{2n+1}0-\binom{2n+1}2t^2+\cdots+(-1)^n\binom{2n+1}{2n}t^{2n}}$$
where $t=\tan x$
So, the roots of $$\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots... | {
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can I finish this solution by this method, if I can please tell me how $m^2+m=n^3+n^2+n$, find solutions in natural numbers. so this was the problem and answer is that it doesn't have solutions in natural numbers.
It was pretty hard to think about this problem, I wanted to analyze the ends
of the numbers, for example, ... | Another approach:
$$m^2+m=\frac{m^3-1}{m-1}-1$$
$$n^3+n^2+n=\frac{n^4-1}{n-1}-1$$
Therefore:
$$\frac{m^3-1}{m-1}=\frac{n^4-1}{n-1}$$
Due to Fermat's little theorem:
$n^4-1 \equiv 0 \mod (5)$,$\rightarrow n^4-1=5t$
$m^3-1$ can be a multiple of 5 with some condition:
$m\equiv (0, 1, 2, 3, 4)\ mod (5)$
If $m \equiv 1 \ mo... | {
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Minimizing $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ over positive reals with $a+b=1$. Why is the minimum not $18$?
If $a,b \in R^+$ such that $a+b=1$, then find the minimum value of $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$
We can write $$\left(a+\frac{1}{a}\right)^{2}+\... | Another variation on the root-mean-square inequality (or Cauchy-Schwarz with $c = d = 1$) is:
$$\left(x + \frac{1}{x} \right)^2 + \left(1-x+\frac{1}{1-x} \right)^2 \ge \frac{\left(x + \frac{1}{x} +1-x+\frac{1}{1-x} \right)^2}{2} \tag{$0 \le x \le 1$}$$
$$= \frac{1}{2} \left(1 + \frac{1}{x} + \frac{1}{1-x} \right)^2 = \... | {
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find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem I'm trying to prove with squeeze theorem that the limit of the following series equals 1:
$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$
For the left side of the inequality I did:
$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{... | First of all we have, for any $ n\in\mathbb{N}^{*} $, the following : $$ \sqrt[n]{n}=1+\frac{\ln{n}}{n}\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x} $$
Since : \begin{aligned}0\leq\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x}&\leq n^{\frac{1}{n}}\\ &\leq 2\end{aligned}
We have : \begin{aligned} 1\leq \sqrt[n]{n}\leq 1+\frac{... | {
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find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$ I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$
Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-... | Everything is fine ! A little bit more can be said:
$f$ is strictly decreasing on $(-\infty,\frac{1}{2}]$
and
$f$ is strictly increasing on $[\frac{1}{2}, \infty).$
| {
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Series expansion of $x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$ at $x=1$ In various texts it is stated that a good approximation of,
$x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$,
about $x=1$ is given by,
$\frac{4\sqrt{2}}{3} (x-1)^{3/2}$.
Plotting the graphs this is indeed the case, however, I am not sure how this approximati... | The Approximation in the Question
I recognized this as the result of an integral that uses the trigonometric substitution $x=\sec(\theta)$, so I took the derivative of ${\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}$ and got $2{\textstyle\sqrt{t^2-1}}$. Thus,
$$
{\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x... | {
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Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
| Using Newton’s identities,
$$x_1^2+x_2^2+x_3^2= (x_1+x_2+x_3)^2- 2\sum x_ix_j=3^2-k=6$$
and therefore $k=3$.
| {
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Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.
Wh... | Let radius of circle be $r$. So side of square $a=2r$.
Let the circle touch $KR$ at $X$. So $KX=a/2=r$
Let $KQ=x$. So $PF=a-2x=2(r-x)$
Use power of point of $Q$, $$PQ\cdot QF=QX^2$$
| {
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Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$ I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:
$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?
Thank you very much!
| Using algebra
Rewriting
$$a_{n+1}=\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)$$ as
$$a_{n+1}=a_n-\left(a_n-\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)\right)=a_n-\frac{a_n^2-3}{2 a_n}$$ we can recognize the itegrative scheme for solving $a^2=3$; so the limit is $\sqrt 3$.
Considering the function $f(a)=a^2-3$, starting w... | {
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A question of Roots of unity
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
\cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.
I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.
| The roots of unity for $z^9=1$ are $e^{i \frac{2\pi k}9},\>k=0, \pm1,\pm2,\pm3,\pm4$ and their sum is equal to zero, i.e.
$$0=1+ e^{i\frac{2\pi k}9} + e^{-i\frac{2\pi k}9}
+ e^{i\frac{4\pi k}9} + e^{-i\frac{4\pi k}9}
+ e^{i\frac{6\pi k}9} + e^{-i\frac{6\pi k}9}
+ e^{i\frac{8\pi k}9} + e^{-i\frac{8\pi k}9}
$$
which l... | {
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Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$... | When you know $a+b$, you can express $a^n+b^n$ in terms of powers of $c=ab$. A few cases
*
*$a^2+b^2=(a+b)^2-2ab$
*$a^3+b^3=(a+b)^3-3ab(a+b)$
What about $a^5+b^5$? The relations are obtained from the expansion of $(a+b)^5$:
\begin{align}
(a+b)^5&=a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)\\
&=a^5+b^5+5ab((a+b)^3-3ab(a+b))+1... | {
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"answer_count": 5,
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How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$. How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$ . Justify your answer.
Hello, could someone explain to me how to find the invertible elements of a set as large as $\mathbb {Z_ {990}}$ and $... | For $k$ to be invertible in $\mathbb Z_n$ means there is an integer $a$ so that $ak \equiv 1 \pmod n$. And $ak\equiv 1\pmod n$ means there is an integer $b$ so that $ak = 1+bn$ which means $ak -bn = 1$.
By Bezouts Lemma that is possible if and only if $k,n$ are relatively prime.
So find all the numbers $k$ so that $k$... | {
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Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$ There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent seri... | *
*When the function is rational, the standard procedure is to start with finding its expansion into partial fractions. In the present case, you get $$f(z)=\dfrac{1}{z(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z}.$$ Hence, making use of your substitution $w=z-1$ (or $z=1+w$), we obtain
\begin{align}
\dfrac{1}{z-1}-\dfrac{1}{z}&=... | {
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Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$ I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$
So far I have:
*
*Base case: true
*Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$... | Notice that
$$3^{n+1}(n+1) = 3 \cdot 3^n (n+1) = \frac{3}{4} \cdot 3^n \cdot 4(n+1)$$
Because of the 3/4, you can move this term into the parentheses and combine it with the $3^n(2n-1)+1$ part. And because of the $3^n$, you can move the $4(n+1)$ into the parentheses and combine with the $2n-1$ part, to get
$$ \frac{3}{... | {
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Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$ Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$
My attempt: multiply both side... | I gave a solution years ago.
Denote $A = \sum_{k=1}^n \frac{1}{\frac{1}{k} + a_k}$,
$B = \sum_{k=1}^n \frac{1}{a_k}$.
The function $f(x) = \frac{1}{1 + \frac{1}{x}}, \ x > 0$ is concave. By Jensen's inequality, we have
\begin{align}
A &= f(\tfrac{1}{a_1}) + 2f(\tfrac{1}{2a_2})
+ 3(\tfrac{1}{3a_3}) + \cdots + n f(\tfrac... | {
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$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum. I need to find supremum of:
$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$
I found out that:
$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2... | We pick a sub-sequence of $A$ in the following fashion: $n=1, 2, \cdots, $, and $m$ depends on $n$ via $m=\lfloor \sqrt{a_n+1} \rfloor$.
Define $a_n = n^3+3n^2, b_m=m^2+2m=(m+1)^2-1=\lfloor\sqrt{a_n+1}+1 \rfloor^2-1$. Then $\frac{a_n b_m}{a_n^2+b_m^2} \in A$ and we will prove that $\lim_{n\to \infty} \frac{a_n b_m}{a_n... | {
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How can I find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ I need to find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ so that I may find the relationship between $I_4$ and $I_3$, I have done the following:
Setting up for Integration by parts
$$u=(1+x^3)^n \implies u^{'}=3nx^2(1+x^3)^{n-1... | Notice that $$
I_n - I_{n - 1} = \int_0^1 {x^9 (1 + x^3 )^{n - 1} dx} .
$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Stochastic Integration: Riemann Sums I am reading this book:
Introduction to Stochastic Differential Equations
On page 63, in the lemma in the middle of the page the author has the following
$\displaystyle \sum_{k=0}^{m_n - 1}W(\tau_k^n)(W(t_{k+1}^n) - W(t_k^n)) = \frac{W^2(T)}{2} - \frac{1}{2}\sum_{k=0}^{m_n - 1}(W(t... | For ease of notation we replace $m_n,t_k^n,\tau_k^n$ with $m,t_k,\tau_k$ respectively. Starting on the right, we have the following:
\begin{align}
\text{RHS } & = \frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 + W(t_{k+1})W(t_k) - \frac{1}{2}W(t_k)^2 + W(\tau_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1}... | {
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"answer_count": 1,
"answer_id": 0
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How to prove that every number of the form $245\cdot 22..11-239$ is congruent to $10^m \mod 41$? Consider Numbers:
\begin{align}
245&\cdot 211-239\\
245&\cdot 2211-239\\
245&\cdot 22211-239\\
245&\cdot 222211-239{}{}{}\\
245&\cdot 2222\ldots11-239
\end{align}
I am stuck, how to prove that every number of this form is c... | Since $1..1$ with $n$ ones is equal to $\frac{10^n-1}{9}$, we can write the general term of the sequence as
$$
a_n = 245 \cdot \Big(200 \cdot \frac{10^n-1}{9} + 11\Big) - 239.
$$
Looking modulo 41 we have $245 \equiv -1$, $200 \equiv -5$, $-239 \equiv 7$ and $9^{-1} \equiv -9$ (since $9 \times (-9) = -81 \equiv 1$). We... | {
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Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, u... | Great job so far! Notice that all terms of the form $\binom{n}{i}a^{n-i}$ are divisible by $a$ if and only if $i<n$. This means that the remainder is simply $(2x)^n$.
| {
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Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $ The limit can be rewritten as
$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$
Recall the inequality,
$$ \cos x < \frac{\sin x}{x} < 1$$
holds for $ x\in (-\pi/2, \pi/2) $.
This provides... | It is a fine and nice argument with trig functions.
Another way to proceed is to use asymptitics (see them as Taylor expansions at $x=0$.
Then you can write $$\lim_{x \rightarrow 0} \frac{-(\sin(x) -x)}{x^2} = \lim_{x \rightarrow 0} -(\frac{-x^3/3!}{x^2} )= \lim_{x \rightarrow 0} x/6 = 0$$
| {
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"answer_id": 1
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Calculating conditional variance using two different methods
The stock prices of two companies at the end of any given year are modeled with random variables $X$ and $Y$ that follow a distribution with joint density function $f(x, y) = 2x$ for $0<x<1, x<y<x+1$ and $0$ otherwise. What is the conditional variance of $Y$... | The main mistake is that the conditional density is actually $$f(y \mid X=x) = \begin{cases} 1 & x < y < x+1 \\ 0 & \text{otherwise}\end{cases}$$
It is very important to consider the support of the density.
In the first approach, you have
$$E[Y \mid X=x] = \int_x^{x+1} y \, dy = \frac{(x+1)^2 - x^2}{2} = x + \frac{1}{... | {
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Let $A=(-1,0),B=(1,0), C$ be points in $\mathbb{R}^2.$ What is the locus of points $\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$}? Let $A=(-1,0),\ B=(1,0),\ C$ be points in $\mathbb{R}^2.$ What (shape(s)) is the locus of points $L(C)=\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$ } ?
Obviously t... | Let $C(a,b),s(x,y)$ where $y\not=0$.
Also, let $U(u,0)$ be the intersection point of the angle bisector with $x$-axis.
Then, we have
$$\begin{align}&sA:sB=AU:BU
\\\\&\iff (1-u)\sqrt{(x+1)^2+y^2}=(1+u)\sqrt{(x-1)^2+y^2}
\\\\&\iff u=\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}\end{a... | {
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What's the most elementary way to solve this trigonometric problem?
A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is ... | Here's an alternative trigonometric approach that doesn't assume knowledge of the divided triangle formulae, but at the expense of additional work algebraically.
Half the battle with trigonometry is getting started: which formulae capture the key details of your set-up? If we want to do trigonometry separately on the ... | {
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"answer_id": 5
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The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$ The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____
My approach is as follow
Already this question is solved Maximum value of argument but I... | $$\left| {\arg\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$$ $$= \left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right) - \arg\left(2\sin \frac{\t... | {
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"url": "https://math.stackexchange.com/questions/3972107",
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"source": "stackexchange",
"question_score": "1",
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How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both ... | The solution makes use of the Pythagorean identities.
I. $\arcsin(3/5)=\theta\implies \sin(\theta)=3/5$. Using the identity: $\cos^2(\theta)=1-\sin^2(\theta)\iff \cos(\theta)=\sqrt{1-\sin^2(\theta)}$. Hence, $\cos(\arcsin(3/5))=\sqrt{1-(\frac{3}{5})^2}=\sqrt{\frac{16}{25}}=4/5$.
II. $\arctan(7/24)=\theta\implies \tan(\... | {
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Prove $H$ is a normal subgroup of $K$ $G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1 & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i & ... | as for $H$ in $K$ you could notice that $H$ is of index 2 in K so it is normal.
| {
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Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$ Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).
My Work :-
Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\fra... | Hint: $$\frac{1}{105}=\frac{1}{4}(\frac{1}{3\cdot 5}-\frac{1}{5\cdot 7})$$ $$\frac{1}{315}=\frac{1}{4}(\frac{1}{5\cdot 7}-\frac{1}{ 7\cdot 9})$$ do you
observe a pattern........
| {
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Integral residue $\int_{0}^{\infty} {\frac{\cos\left(x\right)}{x^{2} + 2x + 4}}\,{\rm d}x. $ $$
\mbox{I have the integral:}\quad
\int_{0}^{\infty}
{\frac{\cos\left(x\right)}{x^{2} + 2x + 4}}\,{\rm d}x.
$$
I tried to use the residue theorem and semicircle and calculate the integral form minus infinity to infinity, but m... | I suppose you are calculating
$$\int_{-\infty }^{\infty } \frac{\cos (z)}{z^2+2 z+4} \, dz=\frac{e^{-\sqrt{3}} \pi \cos (1)}{\sqrt{3}}$$
Which can be done by integrating $\frac{e^{i z}}{z^2+2 z+4}$ along large upper semicircular contour, using Jordan's lemma, calculating residue at $-1+\sqrt{3} i$ and taking real part... | {
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$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$ I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$
My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$... | The minimal polynomial of your matrix is $x^3$. To find $f(A)$ for a holomorphic function $f$ defined on some neighbourhood of $\sigma(A) = \{0\}$, you need to find a polynomial $p \in \Bbb{C}[x]$ of degree $< 3$ such that
$$p(0) = f(0), \quad p'(0) = f'(0), \quad p''(0)=f''(0)$$
and then $f(A)=p(A)$.
Trying out $p(x) ... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How do i solve this integration question using the washer and shell method? What is the volume of a solid enclosed by $y = (x-1)^2$ and $y = 4$ revolved around $x = - 3$?
I tried the washer method and the shell method and got different answers each time and I'm really confused please help!
My set up for the washer meth... | Washer Method
The larger radius comes from the right side of the parabola $y = (x - 1)^2$, while the smaller radius comes from the left side of that parabola. Rewriting that parabola in terms of $x$, we have:
$$ y = (x - 1)^2 \Rightarrow x = 1 \pm \sqrt{y} . $$
Then, $ R(y) = (1 + \sqrt{y}) - (-3) $ and $ r(y) = (1 - \... | {
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Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $ Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $
I did these steps:
*
*Basis step
$$P(10): \ \ 1024 \geq 1000 \ (True)$$
*Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1... | You already proved the base case
Now suppose that $n^3\le 2^n$ is true and
let's prove it for $n+1$
$(n+1)^3=n^3+\left(3n^2+3n+1\right)\tag{1}<\ldots$
for $n\ge 4$ we have $3n^2+3n+1 < n^3$ indeed, adding $n^3$ to both sides we get
$n^3+3 n^2+3 n+1<n^3+n^3\to (n+1)^3<2n^3\to n+1<n\sqrt[3]{2}\to n(\sqrt[3]{2}-1)>1\to n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions in a field Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution.
My attempt:
Suppose we have $x = b^{1/3} \in F$. We have that
$$x^3 = (b^{1/3})^3) = b^{\frac{1}{3} \cdot 3} = ... | You offer no justification for your claim that $b^{1/3} = c^{1/3} \implies b = c$.
In essence, that's what the problem requires you to prove.
As an indication that your proof is not valid, note that you never used the hypothesis that $F$ is an ordered field.
Thus, let $F$ be an ordered field, and suppose $x,y\in F$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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The formulas of prostapheresis: memorization technique This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,
\begin{cases}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\
\sin\alpha-\sin\be... | You can memorize the pattern
$$f(a)+\varepsilon f(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$
where $f$, $g$ and $h$ are either $\sin$ or $\cos$, and $\varepsilon$ and $\delta$ are either $1$ or $-1$. Given $f$ and $\varepsilon$, you then need a strategy to find $g$, $h$ and $\delta$.
Step 1: Fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 1
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Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.
I wanted to calculate $2 \cdot Im(x) = ... | $$x=\frac{a+b+c+ab+bc+ca}{1+abc}$$
$$=\frac{abc}{abc}\cdot\frac{a+b+c+ab+bc+ca}{1+abc}$$
$$=\frac{\frac{a+b+c+ab+bc+ca}{abc}}{\frac{1+abc}{abc}}$$
$$=\frac{\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}{\frac{1}{abc}+1}$$
$$=\frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Link between thetrahedral numbers and combinatorics The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1... | Your last link says
The numbers you're seeing are binomial coefficients. To go from $(0,0)$ to $(3,n)$ you have to take $3+n$ steps: $3$ right and $n$ up. You have to choose which $3$ of the $3+n$ steps that are going to go to the right, and the number of ways to choose is $\binom{3+n}{3}$
Your question is why $\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to evaluate $\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$ How can I approach:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$$
I tried the usual differentiation under the integral sign but it didn't work so well.
I also tried to rewrite it the following 2 ways:
$$\int _0^1\frac{\ln \left(1+\sqr... | Substitute $t=\sqrt x$ in $I= \displaystyle\int _0^1\frac{\ln (1+\sqrt{x})}{1+x^2}\:dx$
\begin{align}
I &
=2\int _0^1\frac{t \ln \left(1+t\right)}{1+t^4}dt
\overset{t\to\frac1t}= \int _0^\infty\frac{t \ln \left(1+t\right)}{1+t^4}dt - \int _1^\infty\overset{t^2\to t}{\frac{t \ln t}{1+t^4}}dt \\
&= -\frac12\int _0^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$ let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$
maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x... | Hint:
It suffices to prove that
$$\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \ge \left(\frac{x^{2n+1} + y^{2n+1}}{2}\right)^{\frac{1}{2n+1}}.$$
Since this inequality is symmetric and homogeneous, assume $y = 1$ and $x \ge 1$. It suffices to prove that
$$\ln (x^{n+1} + 1) - \ln (x^n + 1) \ge \frac{1}{2n+1}\ln \frac{x^{2n+1} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Using matrix algebra to solve a differential vector equation So I am solving a simple problem under the influence of gravity, with no resistance or extra forces.
Labeling the axes, let us use the downward vertical as our positive $y%-axis, with $x$ being standard.
Of course, the easiest way to actually define the posit... | The problem is that you started with a wrong assumption, that the gravity is the only force. If you just have a mass suspended at some point, and you release it, it will not oscillate, but instead it will go straight down. You need to add the tension in the suspending wire.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$ How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as
$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfra... | Hint:
$f(x)=[\tan^{-1}(x)][\tanh^{-1}(x)]$
$f'(x)=\dfrac{ \tan^{-1}(x)}{1-x^2}+\dfrac{ \tanh^{-1}(x)}{1+x^2} $
If $|x|<1$:
$\displaystyle f'(x)=\left(\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{2n+1} x^{2n+1} \right) \left(\sum_{n=0}^{+\infty} x^{2n} \right) + \left(\sum_{n=0}^{+\infty} \dfrac{1}{2n+1} x^{2n+1} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\fra... | You can make the problem more general since you have
$$S_p=\frac{\prod_{n=1}^p (2n) }{\prod_{n=1}^p (2n-1) }=\frac{2^p \Gamma (p+1) } {\frac{2^p \Gamma \left(p+\frac{1}{2}\right)}{\sqrt{\pi }} }=\sqrt{\pi }\,\,\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}$$ Take logarithms, use Stirling approximation and cont... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
find the matrix exponential Let
\begin{equation*}
A =
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}\end{equation*}
\begin{equation*}
B =
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix}\end{equation*}
\begin{equation*}
C =
\begin{pmatrix}
4 & 1 & 0 & -4 & 0 & 0\\
0 & 4 & -1 & 0 & -4 ... | Rye, since we discussed this a little bit in your previous question: Note that for the case $f=\exp$ there is a standard way of defining $f(a)$ for non-normal elements of a unital Banach algebra:
$$f(a)=\sum_{k=0}^\infty\frac{a^k}{k!}$$
You can find more details about this in the first pages of Murphy's book. This can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$ This is a math Olympiad problem.
my attempt :
By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100... | hint
$$x^2+y^2-8x=9\iff$$
$$(x-4)^2+y^2=25$$
$$=(\pm 3)^2+(\pm 4)^2$$
$$=0^2+(\pm 5)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find a lower bound on a smallest solution of $\ln( \frac{x}{a} ) +\frac{1}{x}=0$. The equaiton
\begin{align}
\ln( \frac{x}{a} ) +\frac{1}{x}=0
\end{align}
has two solutions for a sufficiently large $a$ (e.g., $a>3$).
I want a lower bound on the smaller of the two solutions.
I have found an upper bound on the la... | The function$$f(x)=\log \left(\frac{x}{a}\right)+\frac{1}{x}$$ has two solutions as soon as $a >e$.
As said in comments and answers, the smallest solution is given by
$$x=-\frac{1}{W_{-1}\left(-\frac{1}{a}\right)}$$
For the secondary branch, there exist nice bounds
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-(u+1)}\right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$ I was solving a question and saw a pattern. Can someone prove it, please?
We know
$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$
So we... | This answer gives a proof by induction. Here instead I will give a proof using Hockey-stick identity:
$$\sum_{i=r}^n \binom ir = \binom{n+1}{r+1}$$
Note that
$$k(k+1)\cdots(k+r) = \frac {(k+r)!}{(k-1)!} = (r+1)!\binom{k+r}{r+1}$$
Hence your sum
$$=(r+1)!\sum_{k = 1}^{n} \binom {k+r}{r+1} =(r+1)!\sum_{k = {r+1}}^{n+r} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find a matrix $A$ with no zero entries such that $A^3=A$ I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?
| It is useful to think about a problem like that in geometric terms.
We want to have $ A^3=A $ which could be thought as some kind of rotation of the plane.
Along these lines what we want is if we rotate a vector by $\theta $ under $A$ to be equal to a rotation by $3 \theta$ corresponding to $A^3$.
A rotation by $\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
} |
Domain of a function quick question $$f(x)=\arcsin\left(\frac{x-1}{2x}\right)$$
We need to find the domain of this function.
My try:
$$-1\leq \frac{x-1}{2x} \leq 1 $$
We can split this into
$$-1\leq \frac{x-1}{2x} \quad\text{and} \quad\frac{x-1}{2x} \leq 1$$
My idea is to solve for this two inequalities and then take t... | Here it is another approach for the sake of curiosity.
I would start with squaring both sides as it is done next:
\begin{align*}
\left|\frac{x-1}{2x}\right| \leq 1 & \Longleftrightarrow \left(\frac{x-1}{2x}\right)^{2} \leq 1\\\\
& \Longleftrightarrow \frac{x^{2} - 2x + 1 - 4x^{2}}{4x^{2}} \leq 0\\\\
& \Longleftrightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding $\frac{DK}{DI}$ In triangle $ABC,$ where $AB = 8, AC = 7,$ and $BC = 10,$ $I$ is the incenter. If $AI$ intersects $BC$ at $K$ and the circumcircle of $\triangle ABC$ at $D,$ find $\frac{DK}{DI}.$
I first drew a diagram, but I was unsure where to go from here.
| I have doubts that you correctly compute the result, since the value $IK$ given in comments is incorrect. The correct result is:
$$\frac{DK}{DI}=\frac23.
$$
The details are given below.
Let $x,y,z$ being the distances from the vertices $A,B,C$ to the tangent points of the incircle. From the equations $x+y=c, y+z=a, z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2... | First of all, since $t= \sqrt{1-r^2}$, we should have $dt/dr = \frac {-r}{\sqrt{1-r^2}}$ and $t^2 + r^2 = 1$.
Now
$$\int_0^1 \frac {r^3}{\sqrt{1-r^2}}dr = \int_{\color{red}1}^{\color{red}0} \frac {r^3}{\sqrt{1-r^2}}\frac {\sqrt{1-r^2}}{-r}dt = \int_0^1r^2dt = \int_0^1(\color{red}{1-t^2})dt = [t-\frac{t^3}3]_0^1 = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$
$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?
If $z^2=x^2+y^2$ then how ca... | The relationship between $x,y,z$ and the side of the triangle, $a$ is interesting:
$$3(x^4+y^4 + z^4 + a^4) = (x^2+y^2+z^2 + a^2)^2 \tag{1}$$
Please see here, near reference $19$
If you put $z^2 = x^2 + y^2$ in $(1)$, you'll get
$$3(x^4 + y^4 + (x^2+y^2)^2 + a^4) = (2x^2 + 2y^2 + a^2)^2$$
Solving for $a$, you get
$$a =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Help finding Eigenvectors The matrix is \begin{equation*}
A =
\begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & -2 \\
3 & 2 & 1
\end{pmatrix}
\end{equation*}
I got the eigenvalues $\lambda_1 = 1, \lambda_2 = 1 + 2i$, and $\lambda_3 = 1-2i$. I am only concerned with the complex valued eigenvectors. For $\lambda_2$, I got the eigenv... | You're correct. The eigenvalues are $\lambda_1=1,\lambda_2=1-2i,\lambda_3=1+2i$
$$v_1=\left[\begin{array}{rrr|r}
0 & 0 & 0 & 0 \\
2 & 0 & -2 & 0 \\
3 & 2 & 0 & 0 \\
\end{array}\right]\sim\left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 & 1 & \frac{3}{2} & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right]\longrightarrow\left[\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$ Let $ABC$ be acute triangle. Prove that
$$\sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$$
My attempt:
$$\sqrt{\tan A} + \sqrt{\tan B}\geq 2\sqrt[4]{\tan A\cdot\tan B}$$
At here I think I need to prove $$2\sqrt[4]{\tan A\cdot\tan B}\ge2\sqrt{\cot\frac{A}... | Remark: $\tan A,\ \tan B,\ \tan C,\ \cot\frac{A}{2},\ \cot\frac{B}{2},$ and $\cot\frac{C}{2}$ are all positive because $0<A,\ B,\ C<\frac{\pi}{2}$. This is used in this proof, for example in cancellings or taking square roots of both sides of inequalities.
The function $f\left(x\right)=\tan x$ is convex on the interval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the fami... | Let $f(x)=x^3-x^2+2x-1, g(x)=x^3+x^2-1$. We have
$$-g(x)\cdot g(-x)=-(x^3+x^2-1)((-x)^3+(-x)^2-1)\\
=(x^3+x^2-1)(x^3-x^2+1) = x^6 - (x^2-1)^2\\
=x^6-x^4+2x^2-1=f(x^2)$$
Now if $g(x)=(x-r)(x-s)(x-t)$, then
$$f(x^2)= - (x-r)(x-s)(x-t) (-x-r)(-x-s)(-x-t)\\=(x^2-r^2)(x^2-s^2)(x^2-t^2)
\\\implies f(x)=(x-r^2)(x-s^2)(x-t^2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.
The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
x &= 1\\
x &= -\sqr... | Note that
$x^9=1=\cos 2n\pi +\sin 2n\pi=e^{i2n\pi}$
$\implies x=e^{i2n\pi/9}; n=0,1,2,...,8$ are roots.
Assume $e^{i2\pi/9}=a$. Then the powers of it coprime to $9$ viz. $a,a^2,a^4,a^5,a^7 $ and $a^8$ are generators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled
Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).
I am looking for a quick/efficient way to resolve t... | A word of length $n$ is either a word of length $n - 1$ and a $b$ or a $c$, or a word of length $n - 2$ and $aa$. Call the number of words of length $n$ $x_n$, you see $x_0 = 1$, $x_1 = 2$, and:
$\begin{align*}
x_{n + 2}
&= 2 x_n + x_{n - 2}
\end{align*}$
Solving this using generating functions is routine. Defi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is the function differentiable in the point $(0,0)$? I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.
Considering my function: $f(x, y) = x^3 - 3xy^2$
The partial derivative of $x$ at the position $x,y$ is:
$... | Besides from the solution proposed by @SonGohan, you can prove it by the definition.
The candidate to be the derivative is given by $L(x,y) = (3x^{2} - 3y^{2},-6xy)$.
Now you can prove that
\begin{align*}
\lim_{(x,y)\to(0,0)}\frac{|f(x,y) - f(0,0) - L(x,y)((x,y) - (0,0))^{T}|}{\|(x,y)\|} = 0
\end{align*}
Indeed, this i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$
If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$
I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ug... | Using the denesting formula
$$\sqrt{a\pm\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have
$$f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}} \\= 2\sqrt{\frac{x+\sqrt{x^2-16(x-4)}}{2}}=2\sqrt{\frac{x+\sqrt{(8-x)^2}}{2}}$$
Now $\sqrt7<8$, therefore
$$f(\sqrt 7)=2\sqrt{\frac{\sqrt 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Calculation in Routh's theorem The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it... | Note that (just observe to know it, no need to expand it!)
$\mathrm{LHS}(zx + x + 1)(xy + y + 1)(yz + z + 1) - (xyz - 1)^2$ is a polynomial in $z$ of degree at most $2$.
Thus, if we can find three distinct constants $a, b, c$ such that $\mathrm{LHS} = \mathrm{RHS}$ for $z = a, b, c$ respectively,
then we have $\mathrm{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000... | Just to give a different approach, calculating mod $2000$ and using the fact that
$$3^{2000}=(1-10)^{1000}=1-1000\cdot10+\cdots\equiv1\mod2000$$
we have
$$\begin{align}
7^{2005}-1&=-1-(3-10)^{2005}\\
&\equiv-1-3^{2005}+2005\cdot3^{2004}\cdot10-{2005\choose2}3^{2003}\cdot10^2+{2005\choose3}3^{2002}\cdot10^3\\
&\equiv-1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Use the $\epsilon$-$\delta$ definition of a limit to prove $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$ $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$
You should start by writing $\frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6} + 1$ in the way $\left ( x-3 \right )g\left ( x \right )$
(a) Determine $... | This is a bit complicated question. You need to find a suitable bound for the rational function $g(x) $ for values of $x$ in a certain neighborhood of $3$. In formal terms we need two positive numbers $h, K$ such that $|g(x) |<K$ whenever $|x-3|<h$.
This clearly requires an analysis of numerator and denominator of $g$.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verif... | The problem is not too hard if you consider
$$f(x)=\tan (A(x))\,\tan (B(x))$$ Using logarithmic differentiation
$$\frac{f'(x)}{f(x)}=A'(x) \csc (A(x)) \sec (A(x))+B'(x) \csc (B(x)) \sec (B(x))$$
This gives $f'(0)=0$.
Repeating the process knowing that $f(0)=\frac 13$ and $f'(0)=0$ (this simplifies a lot the calculation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
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Can someone help me finish the solution to this differential equation? Find the general solution of $(x-4y-3)dx-(x-6y-5)dy=0.$
The rest of my solution:
$x-4y=3$
$x-6y=5$
The point of intersection is $(-1, -1)$.
Let
$x=u-1$; $dx=du$
$y=v-1$; $dy=dv$
$(u-4v)du-(u-6v)dv=0$
Let $u=vz; du=vdz+zdv$
$(vz-4v)(vdz+zdv)-(vz-6v)d... | You find :
$$\ln((z-2)^2)=\ln(cv(z-3))$$
$$(z-2)^2=cv(z-3)$$
$z=\frac{u}{v}$
$$(\frac{u}{v}-2)^2=cv(\frac{u}{v}-3)$$
$$(u-2v)^2=cv^2(u-3v)$$
$u=x+1$ and $v=y+1$
$$((x+1)-2(y+1))^2=c(y+1)^2((x+1)-3(y+1))$$
$$(x-2y-1)^2=c(y+1)^2(x-3y-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $f'(x)=\sqrt{x+1}$ and $h(x)=f(x^2+x)$, what is $h'(x)$? I was thinking it could be plugging in $x^2+x$ to the $f'(x)$, then using the Chain Rule to solve it, but I'm not sure if it is right. Please help!
| If $f'(x) = \sqrt{x+1}$, then $f(x) = \int \sqrt{x+1} dx = \frac{2}{3}(x+1)^{\frac{3}{2}} + c$. You can solve that integral using u-substitution. Then we also know that $h(x) = f(x^2 + x)$. So then using function notation, we know that $h(x) = f(x^2 + x) = \frac{2}{3}(x^2 + x+1)^{\frac{3}{2}} + c$. We can now different... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$? What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)
I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect e... | Here is another answer I came up with:
Notice that $\lfloor x\rfloor$ and $3$ are integers, so $x^3$ is an integer, say $x=\sqrt[3]{y^3+z}$, where $0\leq z<(y+1)^3-y^3$, and y and z are integers. Now we have $y^3+z-y=3$, or $(y-1)y(y+1)=3-z.$ Since $z\geq0$, $(y-1)y(y+1)\leq 3$ and $y\leq1$. However, notice that if $y<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding $\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$
Find $$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$$
I solved the simpler versions of these:
$$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r}\right)\;\;,\;\;\sum_{r=2}^{\infty}\ln\left(1-\frac{1}
{r^2}\right)$$
by factorizing which yields a telescoping su... | Hint:
$$1-\dfrac1{r^3}=\dfrac{(r-1)(r^2+r+1)}{r^3}=\dfrac{f(r-1)g(r-1)}{f^3(r)}$$
where $g(r-1)=r^2+r+1\implies g(r)=(r+1)^2+(r+1)+1=r^2+3r+3$
and $f(r)=r$
Similarly, $$1-\dfrac1{(r+1)^3}=\dfrac{r(r^2+3r+3)}{(r+1)^3}=\dfrac{f(r)g(r)}{f^3(r+1)}$$
$$\implies\prod_{r=2}^n\left(1-\dfrac1{r^3}\right)=\prod_{r=2}^ng(r)\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this solution is genuine I cannot prove that the solution to :$$y'=-\frac{x\sqrt{1-y^2}}{y\sqrt{1-x^2}}$$
is:
$$y=\sqrt{x^2-2C\sqrt{-x^2+1}-C^2},\:y=-\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}$$
By substituting the solution(s) back into the original equation. I have taken derivative and attempted to equate LHS with RHS but I... | Square the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$
Taking the derivative:
$$2yy'= 2x+2C\frac{x}{\sqrt{1-x^2}}$$
So
$$yy' = x + C\frac{x}{\sqrt{1-x^2}}=\frac{x\sqrt{1-y^2}}{\sqrt{1-x^2}}$$
Assuming $x$ is not $0$, then
$$\sqrt{1-x^2} + C=\sqrt{1-y^2}$$
Squaring both sides:
$$1-x^2+2C\sqrt{1-x^2}+C^2=1-y^2$$
Then,
$$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$ It's an inequality based on two found on the website MSE (see the reference):
Let $a,b,c>0$ then we have:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(... | As username says it's not symmetric so here there is a refinement and a symmetric formula :
$a,b,c\in[0.95,1]$:
$$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(0.75(abc)^{\frac{1}{3}}+0.75(a+b+c))(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$ I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$?
I tried to solve it as follows:
I state $a\in\mathbb{N}$, so that $\sqrt{x^2+2x+4}<a<\sqrt{4x^2+2x+1}$
Hence $x^2+2x+4<a^... | With the lower limit,
$$\begin{equation}\begin{aligned}
& (x + 1)^2 \lt x^2 + 2x + 4 \lt (x + 2)^2 \\
& x + 1 \lt \sqrt{x^2 + 2x + 4} \lt x + 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This means, with $a \in \mathbb{N}$ as you stated,
$$a \ge x + 2 \tag{2}\label{eq2A}$$
With the upper limit,
$$\begin{equation}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$b=a^p+1$ is a perfect square. Show that $p|(b-9)$ $p$ is a prime, and $a$ is a positive integer, $b=a^p+1$ is a perfect square. Show that $p|(b-9)$
It seem very interesting problem.if let $a^p+1=x^2$,it is clear $p\neq 2$
I have prove : $x$ is odd
proof:if $x$ is even number,then $(x+1,x-1)=1$,and note $a^p=(x+1)(x-... | I am in 8th grade, so please if I make mistakes in the calculation then please tell me, I would delete the answer.
You proved that $x$ is odd. That would mean that $x^2$ is odd too.
So the equation $x^2=a^p+1$
If we subtract $1$ from both sides then $a^p$ would be even. Which implies that $a$ is even.
Since $x$ is odd,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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$63^{63^{63}} \mod 100$ I need to find $63^{63^{63}} \bmod 100$.
This is what I've got so far:
Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$
Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$
Using this I got that $63^{63} \equiv ... | $63^7 = (60 +3)^7= ..... + {7\choose 2}60^2\cdot 3^5 + 7\cdot 60\cdot 3^6 + 3^7 \equiv 20\cdot 3^6 +3^7\pmod{100}$.
And as $2\cdot 3^6 \equiv 2\cdot 3^2 \equiv 8 \pmod {10}$ and as $3^7 = 9^3\cdot 3 = (10-1)^3 \cdot 3 = 3(100^3 - 3\cdot 100 + 3\cdot 10 - 1)\equiv 90-3 \equiv 87 \pmod {100}$
So have $63^7\equiv 20\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Prove that there is no positive rational number a such that $a^2 = 3$. I solved a similar question $a^4 = 2$ but I didn't find it hard. This one is taking a lot of my time and I still don't understand how to solve this.
Here's what I have tried so far.
edit:
\begin{align}
a^2 &= 3 \\
a &= \left(\frac mn\right) \\
a^2 ... | Alright, make the straightforward assumption that $a= \frac mn$ with $m,n$ in lowest terms and they are integers. Thus, we get $(\frac mn)^2 = 3$, which soon becomes $m^2 = 3n^2$. Thus, $m^2$ is a multiple of $3$.
However, for $m^2$ to be a multiple of $3$, it needs to have an even number of factors of $3$. If it had a... | {
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"url": "https://math.stackexchange.com/questions/4068869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In p... | The left direction is correct. An alternative way to prove this is using the Cauchy condensation test.
The right direction is not correct. You say: for $0 < p \le 1$ the series $\sum_n 1/n^p$ is divergent but you did not prove this!
Alternatively, you can also use the integral test or the Cauchy condensation test for t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve one system of equations Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$
I tried solving this using cross multiplication method but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in ... | As @Aderinsola Joshua commented, use successive eliminations
From $(1)$, you have $y=-\frac{a x+c z}{b}$.
Plug in $(2)$ and then $z=\frac{c x \left(a^2-b^2\right)}{a \left(b^2-c^2\right)}$
Plug in $(3)$ and simplify to get
$$\frac{b c (a-b) (a+b) (a-c) (a+c) \left(a^2
\left(b^2-c^2\right)^2-x^2\right)x}{a^2 \left(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$ I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.
Do you have any ideas? :)
EDIT:
Do you please have step-by-step ... | For $n=4$, you can make it shorter since
$$A=\frac 1{1+\sin^4(x)}=\frac 1{(\sin^2(x)+i)(\sin^2(x)-i)}$$
Using partial fractions and double angle formula
$$A=\frac{i}{\cos (2 x)-(1-2 i)}-\frac{i}{\cos (2 x)-(1+2 i)}$$ Using now the tangent half-angle
$$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i... | {
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"url": "https://math.stackexchange.com/questions/4070996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When i... | Using induction,
If $10^r=1+3^sk$ where $3\nmid k$
$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$
for $s+1\le2s+1\iff s\ge0,3s\ge s+1\iff 2s\ge1$
Now for the base case $r=1=3^0, s=2$
So, by using weak induction, $$10^{3^a}-1$$ is divisible by $3^{a+2}$ for $a\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here... |
Let set: $\begin{cases}f(x,y)=x^2+y^2\\g(x,y)=6x^2+2xy+6y^2\end{cases}$
The curve $(\mathcal E): g(x,y)=9$ is an ellipse centered at the origin with major axis supported by $y=-x$ and minor axis supported by $y=x$.
Indeed if you calculate its equation in the $45^\circ$ rotated basis then you get it in its reduced form... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a closed formula for $\sum_{n\geq 0} f(n)x^n.$ Given that $f(n) =f(n-1) + 2f(n-2)$ for $n \geq 2$, with $f(1) = 1$ and $f(2) = 3$ find a closed formula for $$\sum_{n\geq 0} f(n)x^n.$$
Here I try to do the following $ \sum_{n=0}^{\infty}f(n)x^{n} = \sum_{n=1}^{\infty}f(n-1)x^{n} + 2\sum_{n=2}^{\infty}f(n-2)x^{n}$ b... | Let $F(x)$ denote the generating function for $f(n)$, i.e.
$$F(x)=\sum_{n\ge1}f(n)x^n$$
Multiply both sides of the recurrence by $x^n$ and sum both sides starting from $n=3$:
$$\sum_{n\ge3}f(n)x^n=\sum_{n\ge3}f(n-1)x^n+2\sum_{n\ge3}f(n-2)x^n$$
Now manipulate this as need to get everything in terms of $F(x)$:
$$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Standard normal distribution - Calculating the probability of an electron Figure $1$ shows $Bohr$'s model for the hydrogen atom. We let $(X, Y)$ denote the position of the electron in relation to the proton-core. In quantum mechanics one can never determine $(X, Y)$ exactly, but we are simplifying the model by using p... | You cannot integrate the normal density by substitution over a finite interval. You need tables or a computer.
Your answer should have been $\Phi(2.15)-\Phi(-2.15) \approx 0.968$
Among other errors, you have the false steps $$\int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \to \int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx$$ $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question related to internal angle bisector Internal angle bisector of $\angle A$ of triangle $\Delta ABC$, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of ∆ABC then.
(a) $AD = \frac{{2bc}}{{b + c}}\cos \frac{A}{2}$
(b) $... | You know that $\displaystyle \small \frac{BD}{DC} = \frac{c}{b}$. Add $1$ to both sides and you get,
$\displaystyle \small \frac{a}{DC} = \frac{b+c}{b}$
Now we know that $\displaystyle \small \frac{a}{\sin A} = \frac{c}{\sin C} \implies \sin C = \frac{c}{a} \sin A$
In $\triangle ADC, \displaystyle \small \frac{\sin C}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Induction with divisibility QUESTION: Prove that $16 \mid 19^{4n+1}+17^{3n+1}-4$ for all $n \in \mathbb{N}$
This is what I have so far but I'm not sure where to go from here.
PROOF:
Let $16 \mid 19^{4n+1}+17^{3n+1}-4$ equal $S(n)$.
Base case, let $n=0$.
\begin{align*}
16 &\mid 191+171-4\\
= 16 & \mid 32\\
= 2
\end{... | To show $16 \mid 19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4$ given $16 \mid 19^{4k+1}+17^{3k+1}-4$, note that
$19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4=$
$17^3\color{blue}{(19^{4k+1}+17^{3n+1}-4)} + \color{blue}{(19^4-17^3)}\cdot19^{4n+1}+
\color{blue}{(17^3\cdot4-4)},$
and all of the blue terms are divisible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate $\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$? $$\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$$
I've tried combining the sum, telescoping series and even trying to make an Nth partial sum but nothing seems to budge. I'm not sure where to go...
| We have \begin{align}
&\sum _{n=1}^m \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right)=\sum _{n=1}^m \frac{n}{2 n-1}-\sum _{n=1}^m \frac{n+2}{2 (n+2)-1}\\
&=\sum_{n=1}^m \frac{n}{2 n-1}-\sum _{n=3}^{m+2} \frac{n}{2 n-1}\\
&=\frac{1}{2\cdot 1-1}+\frac{2}{2\cdot 2-1}+\sum _{n=3}^m \frac{n}{2 n-1}-\sum
_{n=3}^m \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end wit... | We must select the $5$ digits which must appear, although we know the $0$ must appear. So in reality we must only select $4$ digits out of $6$ possibilities.
Notice the sum of all those numbers is a multiple of $3$ ($1+2+6+7+8+9 = 33)$. Therefore the condition is that the two numbers we do not pick add up to a multiple... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find value of $x^3+y^3+z^3$ if $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$ If $x,y$ and $z$ are positive real numbers such that $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$, then what is the value of $x^3+y^3+z^3$?
I first thought of making them all equal because in that case the product is maximum, but obviously that... | Let's find the minimal value of $x+y+z$ when $x^3y^4z^5 = C$.
At point of tangency between $x^3y^4z^5 = C$ and $x+y+z=A$ the normal is proportional to:
$$
\left(\frac{3}{x},\frac{4}{y},\frac{5}z\right)
$$
On the other hand, it should be proportional to $(1,1,1)$. It is possible when $x/3 = y/4= z/5 = k$. From
$$
(3k)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do you find the least value when an inverse trigonometric function is in the denominator?
The function:
$f(x)=\frac{4\pi^2}{3\arccos{(x^4-2x^2)}}+\frac{5\pi}{3}$
If $B=\frac{m}{\pi}$
where $m$ is the minimum value which $f(x)$ can take. Find the value of $B$.
The choices given in my book are as follows:
$\begin{ar... | $x^4-2x^2=x^2(x^2-2) \ge -1$, and it takes value the value $-1$ when $x^2=1$.
As $x^2=1$, $\arccos(x^4-2x^2)$ has the largest value of $\pi$. Hence $$m=\frac{4\pi^2}{3\pi}+\frac{5\pi}{3}=\frac{9\pi}{3}=3\pi$$
Hence you are right, $$B=\frac{m}{\pi}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
\infty} (a_n-3)6^n$.
First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the for... | In this answer we consider the sequence $b_n=|3-a_n|6^n$, where $a_{n+1}=\sqrt{6+a_n}$ with initiial condition $a_0\geq0$. We show that $b_n$ is
*
*bounded and monotone increasing if $0\leq a_0<3$ (see \eqref{three}).
*bounded and monotone decreasing when $a_0>3$ (see \eqref{twop}).
As a consequence,
*$b_n(a_0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 1
} |
Computing or approximating $\sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ I'm trying to compute $\sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ when $N$ is large. Is there an asymptotic formula that tells me $\lim_{n \to \infty} \frac{1}{f(N)} \sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ for... | By Stirling's formula
$$
\log \binom{N}{n} \!= n\log \left( {\frac{N}{n}} \right) - (N - n)\log \left( {1 \!-\! \frac{n}{N}} \right) - \frac{1}{2}\log \left( { n\left( {1 \!-\! \frac{n}{N}} \right)} \right) + \mathcal{O}(1),
$$
provided $1\leq n \leq N-1$.
Now
$$
\sum\limits_{n = 1}^{N-1} {\log \left( {n\left( {1 - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.