Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Demonstration of sum of powers of $2$ Theorem : For every natural number $p$:
$$\sum^p_{i=0} 2^i = 2^{p+1}-1$$
I trieed to demonstrate the theorem using induction
Demonstration :
$1)$ If we have $p=0$ then we get $2^0=2^{0+1}-1$ that is always true.
$2)$ Supposing that the first statement is true then we get that
$$\su... | Inductive hypothesis: $\quad\displaystyle\sum_{i=0}^p 2^i=2^{p+1}-1$ for some $p\ge 0$.
We must deduce that $\quad\displaystyle\sum_{i=0}^{p+1} 2^i=2^{p+2}-1$.
Decompose the sum: \begin{align*}\sum_{i=0}^{p+1} 2^i=2^{p+2}-1 &= \sum_{i=0}^p 2^i+2^{p+1}\\
&=2^{p+1}-1 +2^{p+1}\qquad\scriptstyle(\text{inductive hypothesis)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find $E(|X-Y|^a)$ where $X$ and $Y$ are independent uniform on $(0,1)$
Let $X,Y$ be independent $Uniform(0,1)$ random variables. Find $E(|X-Y|^a)$ where $a>0$.
My working:
Define $W=1$ if $X>Y$ and $W=0$ if $X<Y$. We seek $E(|X-Y|^a)=E[E(|X-Y|^a|W)]=E(|X-Y|^a|W=1)P(W=1)+E(|X-Y|^a|W=0)P(W=0)=E((X-Y)^a)P(X>Y)+E((Y-X)^... | You should have something like:
$$
E(|X-Y|^a)=\iint_{x>y}(x-y)^a\;dxdy+\iint_{y>x}(y-x)^a\;dxdy=2\iint_{x>y} (x-y)^a\;dxdy
$$
Now:
$$
\iint_{x>y}(x-y)^a\;dxdy=\int_{y=0}^1\int_{x=y}^1(x-y)^a\;dxdy=\frac{1}{(a+1)(a+2)}
$$
So:
$$
E(|X-Y|^a)=2\iint_{x>y}(x-y)^a\;dxdy=\frac{2}{(a+1)(a+2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How does $(x+3)^2 - 2^2$ become $(x+1)(x+5)$? I don't understand how $(x+3)^2 - 2^2$ can be transformed to equal $(x+1)(x+5)$. A short demonstration and/or reference to math rules would be very kind.
| Just multiply each part out:
$(x + 3)^{2}$
$= (x + 3)(x + 3)$
$= x^{2} + 3x + 3x + 9$
$= x^{2} + 6x + 9$
Also, $2^{2} = 4$.
So, putting the two above together, we get:
$(x + 3)^{2} - 2^{2}$
$= x^{2} + 6x + 9 - 4$
$= x^{2} + 6x + 5$
And, if you understand how to factor, we get $x^{2} + 6x + 5 = (x + 5)(x + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove that for any integer, $n^2 + 5$ is not divisible by $4$. So I got that there is two cases: odd or even.
If odd then say $n^2$ is $(2k+1)^2 = 4k^2 + 4k + 1.$
then $4k^2 + 4k + 1 + 5$ would need to be divisible by 4 and I don't know where to go from there.
| $n$ is even or odd.
$n$ is even: $n = 2m \Rightarrow n^2 + 5 = 4m^2 + 5 = 4(m^2 + 1) + 1$, which isn't divisible by $4$.
$n$ is odd: $n = 2m+1 \Rightarrow n^2 + 5 = 4m^2 + 4m + 1 + 5 = 4(m^2 + m + 1) + 2$, which isn't divisible by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 3
} |
Infinite Product computation How can we compute the infinite product:
Do we need Gamma function?
Edited: I forgot to add 1/e factor so the product converges.
The product becomes
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate $\int e^{\frac{x^2}{2}} \cdot x^{3} dx$? $\int e^{\frac{x^2}{2}} \cdot x^{3} dx=?$
I tried to do the substitution $du = x^3dx$, so that $u=\frac{x^4}{4}$. Then,
$\int e^{\frac{x^2}{2}} \cdot x^{3} dx= \int e^{\sqrt{u}} du$
However, I would have to do a second substitution ($v = \sqrt{u}$) which would g... | Let $t=\frac{x^2}{2}, dt=xdx$ to get
$\displaystyle \int e^{\frac{x^2}{2}}x^3\;dx=\int e^t(2t)dt=2\int te^t\;dt=2\left[te^t-e^t\right]+C=2\left[\frac{x^2}{2}e^{\frac{x^2}{2}}-e^{\frac{x^2}{2}}\right]+C$
$\;\;\;\;=\displaystyle e^{\frac{x^2}{2}}\left[x^2-2\right]+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to solve an Hyperbolic triangle when all is given except angle C and side c) Another Hyperbolic triangle problem (all given except angle $\angle C$, and side $c$)
I thought that after asking How to solve an hyperbolic Angle Side Angle triangle? I could solve all hyperbolic triangles, but I am still stumped with SS... | Somehow the following formula's work:
$$ \tanh\left(\frac{c}{2}\right) = \tanh\left(\frac{1}{2} (a-b)\right) \frac{\sin{\left(\frac{1}{2}(A+B)\right)}}{\sin{\left(\frac{1}{2}(A-B)\right)}} $$
and
$$ \tan\left(\frac{C}{2}\right) = \frac{1}{\tan\left(\frac{1}{2} (A-B)\right)} \frac{\sinh\left(\frac{1}{2}(a-b)\right)}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$ I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$
where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS:
$$
\begin {align}
a\cos A... | From the cosine rule:
$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$
So
$$\begin{align}a\cos(A)+b\cos(B)+c\cos(C)&=a\frac{b^2+c^2-a^2}{2bc}+b\frac{a^2+c^2-b^2}{2ac}+c\frac{a^2+b^2-c^2}{2ab}\\
&=a^2\frac{b^2+c^2-a^2}{2abc}+b^2\frac{a^2+c^2-b^2}{2abc}+c^2\frac{a^2+b^2-c^2}{2abc}\\
&=\frac{a^2b^2+a^2c^2-a^4+a^2b^2+b^2c^2-b^4+a^2c^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating $ \int \frac{1}{\sqrt{x^2+4}}\,dx $ using Trigonometric Substitution I'm reviewing integration by trigonometric substitution in anticipation of covering it in class next week. I seem to be a bit rusty and keep catching myself making various mistakes. On this particular problem I keep getting the same answ... | Your answer is also correct. Remember that antiderivates are only determined up to an additive constant.
$$\ln\left|{\sqrt{x^2+4}\over 2}+{x\over 2}\right|+C=\ln|\sqrt{x^2+4}+x|-\ln2+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Calculate the trace of all elements in $F_8$ I got the following exercise where you have to calc the trace of all elements in ${F_8}$ which is constructed as ${F_2}[x]$/(${x^3+x+1}$)${F_2}[x]$.
Up to now I did those steps:
1) Find all elements in ${F_8}$ which are in my opinion: $0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1$
2) T... | You can take advantage of the linearity of trace. To make the notation less cumbersome I denote the coset $x+\langle x^3+x+1\rangle$ by $\alpha$. In that case the elements of the field are
$$
\Bbb{F}_8=\{a_0+a_1\alpha+a_2\alpha^2\mid a_0,a_1,a_2\in\Bbb{F}_2\}.
$$
The minimal polynomial of $\alpha$ is $m(x)=x^3+x+1$. It... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\log_{12}54=a$ then $ \log_{6}12=?$ I am given $$\log_{12}54=a$$
So what will be value of $ \log_{6}12?$
I used base changing theorem and wrote expression as
$$\frac{\log_{6}54}{ \log_{6}12} =a$$
And then $$ \frac{1+\log_{6}9}{ a} = \log_{6}12$$ now what to do?
| Observe first that
$$\begin{align}a&=\log_{12}(54)=\log_{12}(2)+3\log_{12}(3)\\&=\frac{1}{\log_2(12)}+\frac{3}{\log_3(12)}=\frac{1}{2+\log_2(3)}+\frac{3}{1+\frac{2}{\log_2(3)}}\end{align}$$
From this, we can compute $\log_2(3)$ in terms of $a$. We just need to solve for $x$ in the equation $$a=\frac{1}{2+x}+\frac{3}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138949",
"timestamp": "2023-03-29T00:00:00",
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Derivative of an inverse Let $f(x)=2x^3+7x−1$, and let $g(x)$ be the inverse of $f(x)$. Then find $g′(191/4)$. I know only one way of doing this. Solving the cubic equation for x and then differentiating it. This is too too long. How to solve it more easily?
| In order to not confuse the variables of the inverse function with the variables of the function itself, let :
$y=f(x)=2x^3+7x−1=X$
$Y=g(X)=x$
One little picture says more than a long speech! : See below
$\frac{dg}{dX}=\frac{1}{\frac{df}{dx}}$
For $X=\frac{191}{4}$ we have $y=\frac{191}{4}=2x^3+7x-1$ which real root i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{... | Here are the steps
$$\lim\limits _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}
$$
$$= \lim\limits _{n\to \infty}\sqrt{\frac{n+\sqrt{n^2+1}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+\sqrt{n^2+1}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+\sqrt{n^2\left(1+\frac{1}{n^2}\right)}}{3n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a, b, c,x$ be elements of a group $G$. In each of the following, solve for $x$ in terms of $a, b, c$. Just checking my work.
$axb =c.$
$(axb)a^{-1}b^{-1} = ca^{-1}b^{-1}$
$x = ca^{-1}b^{-1}$
$x^2b = xa^{-1}c$
$x^{-1}(x^2b) = (xa^{-1}c)x^{-1}$
$xb = a^{-1}c$
$x = a^{-1}b^{-1}c$
$x^2a = bxc^{-1}$ and $acx = xac.... | You’re forgetting that the group operation need not be commutative: the order of the operands matters. In the first problem, for example, $(axb)a^{-1}b^{-1}$ does not simplify to $x$; in fact, it doesn’t simplify at all without further knowledge of the group in question. You need to multiply $axb$ on the left by $a^{-1... | {
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"url": "https://math.stackexchange.com/questions/1140930",
"timestamp": "2023-03-29T00:00:00",
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Integration using Algebraic Manipulation Trying to solve
$$\int^1_2\frac{x-4}{x^2}dx$$
All I have so far is re-writing it into two integrals which yields the following:
$$\int^1_2\frac{x}{x^2} \, dx - \int^1_2\frac{-4}{x^2} \,dx$$
Then $\frac{x}{x^2}$ can be re-written as $\frac{1}{x}$ which yields:
$$\int^1_2\frac{... | For one, you can rewrite $$\begin{align}\int^1_2\frac{x-4}{x^2}dx = \int^2_1\frac{4-x}{x^2}dx \\ =4\int^2_1\frac{1}{x^2}dx - \int^2_1\frac{x}{x^2}dx \\ =4\int^2_1\frac{1}{x^2}dx - \int^2_1\frac{1}{x}dx \end{align}$$ and you are correct that $\int \frac{1}{x}dx = \ln|x|+C$. As for the other term to be integrated, consid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144017",
"timestamp": "2023-03-29T00:00:00",
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Given $f(x)=3x+2$ for $x<0$ and $f(x)= 3x +3$ for $x\ge0$, what is $f(t^2 + 1)$? I need to find $f(t^2 + 1)$ given $$f(x) = \begin{cases} 3x+2 & x \lt 0\\ 3x+3 & x\ge 0\end{cases}$$
The textbook says the answer is $3t^2 + 6$, but how do you know which piecewise to plug $t^2 + 1$ into?
| $$t^2 + 1 \geq 1 \quad \forall t \in \mathbb R$$
(The square of any number is greater than or equal to $0$. So $t^2 + 1\geq 1$.)
So we use $f(x) = 3x + 3$. Then we have $$f(t^2 + 1) = 3(t^2+1) + 3 = 3t^2 + 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
| $\cot\dfrac{11\pi}{18}=\cot\left(\pi-\dfrac{7\pi}{18}\right)=-\cot\dfrac{7\pi}{18}$
Now $\cot\dfrac{2\pi}9-\cot\dfrac{7\pi}{18}=\dfrac{\sin\left(\dfrac{7\pi}{18}-\dfrac{2\pi}9\right)}{\sin\dfrac{2\pi}9\cdot\sin\dfrac{7\pi}{18}}=\dfrac1{2\sin\dfrac{2\pi}9\cos\dfrac\pi9}$
( as $\sin\dfrac{7\pi}{18}=\cos\left(\dfrac\pi2-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146390",
"timestamp": "2023-03-29T00:00:00",
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Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make ... | HINT:
If $f(n)=2^{4n+2}+3^{n+2}$
$f(m+1)-3f(m)=2^{4(m+1)+2}+3^{(m+1)+2}-3[2^{4m+2}+3^{m+2}]=2^{4m+2}(2^4-3)$ which is divisible by $13$
So, $13|f(m+1)\iff13|f(m)$
Alternatively try with $f(m+1)-2^4f(m)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Rationalize the denominator: $(\frac{3}{2x^2}) ^{1/4}$ My answer is $\frac{(6x^2)^{1/4}}{ 2x^2}$
However the book's answer is $\frac{(24x^2)^{1/4}}{2x}$
Where did the book get that from?
Here's my work:
$$\sqrt[4]{\frac{3}{2x^2}}= \frac{\sqrt[4]{3}}{\sqrt[4]{2x^2}}\cdot \frac{\sqrt[4]{2x^2}}{\sqrt[4]{2x^2}}$$
$$= \... | The simplest way is to multiply numerator and denominator by $(2x^2)^{\color{red}3/4}$, just because $\frac34+\frac14=1$:
\begin{align*}
\biggl(\frac3{2x^2}\biggr)^{1/4}&=\frac{3^{1/4}}{(2x^2)^{1/4}}=\frac{3^{1/4}(2x^2)^{3/4}}{2x^2}=\frac{(3\cdot8x^6)^{1/4}}{2x^2}\\
&=\frac{(24x^2)^{1/4}\cdot x}{2x^2}=\frac{\Bigl(\cfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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derivative of $\sec^2(x/12)$ Alright, so the derivative of $\sec^2(x/12)$ is $\frac{1}{6} \tan\left(\frac{x}{12}\right) \sec^2\left(\frac{x}{12}\right)$
But if you use chain rule, you get:
$$2 \sec\left(\frac{x}{12}\right) \left(\sec\left(\frac{x}{12}\right)\right)'$$
$$2 \sec\left(\frac{x}{12}\right) \sec\left(\fr... | You forgot to apply the chain rule again for the second equation on your post. You have $\sec\left(\frac{x}{12}\right)$, which is a composite. Applying the chain rule, we have
\begin{align}\frac{d}{dx} \left[sec^2\left(\frac{x}{12}\right)\right] &= 2* \sec\left(\frac{x}{12}\right) * \frac{d}{dx} \left[\sec\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$
Prove that the following inequality
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
holds for arbitrary real numbers $a$, $b$ and $c.$
Someone says, "It's very easy problem. It can also... | By C-S
$$\left(\sum_{cyc}\sqrt{a^2+(b-1)^2}\right)^2=\sum_{cyc}\left(a^2+(b-1)^2+2\sqrt{(a^2+(b-1)^2)(b^2+(c-1)^2}\right)\geq$$
$$\geq\sum_{cyc}(a^2+b^2-2b+1+2(ab+(b-1)(c-1)))=$$
$$=\sum_{cyc}(2a^2-2a+1+2ab+2ab-4a+2)=$$
$$=\sum_{cyc}(2a^2+4ab-6a+3)=2(a+b+c)^2-6(a+b+c)+9.$$
Thus, it remains to prove that
$$2(a+b+c)^2-6(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Closed form for $\int_0^\infty\arctan\Bigl(\frac{2\pi}{x-\ln\,x+\ln(\frac\pi2)}\Bigr)\frac{dx}{x+1}$ I'm trying to find a closed form for this integral:
$$I=\int_0^\infty\arctan\left(\frac{2\pi}{x-\ln\,x+\ln\left(\frac\pi2\right)}\right)\frac{dx}{x+1}$$
Its approximate numeric value is
$$I\approx3.380582528445346979395... | Computing a Related Contour Integral:
Define
$$f(z)=\frac{i}{2}\frac{z-1}{1+az}\left(\frac{1}{z-\ln{z}+\ln\left(\frac{\pi}{2}\right)}+\frac{1}{z-\ln{z}+2\pi i+\ln\left(\frac{\pi}{2}\right)}\right)$$
and let $\gamma$ denote a keyhole contour deformed around $[0,\infty]$. Restricting the argument between $0$ and $2\pi$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 3,
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If $n$ is an odd integer, does $x+y$ divide $x^n + y^n$? I believe that the answer is yes.
Here's my thinking:
*
*$x^n + y^n -(x+y)(x^{n-1} + y^{n-1}) = -x^{n-1}y -xy^{n-1}$
*$-x^{n-1}y -xy^{n-1} - (x+y)(-x^{n-2}y -xy^{n-2}) = x^{n-2}y^2 + x^2y^{n-2}$
*So, at each step, we get $(-1)^{c}(x^{n-c}y^{c} + x^{c}y^{n-c}... | Your reasoning is fine.
A more general result is that for any positive integer $n$, and integers $u,v$, $u^n-v^n$ is divisible by $u-v$. This theorem is easier to prove by induction since you are proving for all $n$.
Your theorem then follows when $n$ is odd by setting $u=x,v=-y$ so $u-v=x+y$ and $u^n-v^n=x^n+y^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the domain of this function in the x,y,z graph Find the domain of this function in set notation:
$f(x,y)=\dfrac{\sqrt{x^2+y^2+4}}v$
$v=y-x$
Could I say that $x^2+y^2+4 \geq 0$ and $y-x \geq 0$ and based off those statements put that into set notation? and get:
$D= \{(x,y)|x^2+y^2+4 \neq 0$ and $y-x \neq 0\}$
| Since $v \neq 0$ by definition, then $y-x \neq 0$ ---> $y \neq x$. Therefore, the line y= x is excluded from the domain of f. Also, since $\sqrt{x^2+y^2+4}\geq 0$ and $y \neq x$, $\sqrt{x^2+y^2+4}$. So the domain is $ D = \{(x,y)| \sqrt{x^2+y^2+4}\geq 0$ where $y\neq x$} Surb gives an excellent sketch of the domain a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the quadratic congruences, or show them to be unsolvable.
Solve the following congruences, or show them to be unsolvable:
$(a) 3x^2 - 5x +7 \equiv 0 \mod 13$
Since $9 \cdot 3 \equiv 1 \mod 13$, $9 \cdot 5 \equiv 6 \mod 13$ and $9 \cdot 7 \equiv 11 \mod 13$ we can multiply the above congruence by $9$ to obtain $... | About b). Yes multiplication by 8 is correct. The polynomial you get is the same as $ x^2 +4 x + 3 \equiv 0 \pmod {13} $. Clearly you get $(x+1)(x+3) \equiv 0 \pmod {13} $. So $ x=10$ or $ x=12 \pmod {13} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have
Base case let $p(2)$
\begin{align*}
\frac{1}{2}+ \frac{1}{2+2} + \... | what would be p(n+1)? think about where that falls in your solution. p(n+1) is 1/(2n+2)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? How can I prove that
$$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$
I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
| $$
\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}
$$
So
$$
\begin{align}\sum_n^\infty \frac{1}{n(n+1)} &= \left( \frac{1}{1}-\frac{1}{2} \right)
+ \left( \frac{1}{2}-\frac{1}{3} \right) + \cdots\\ &= \frac{1}{1} + \left( -\frac{1}{2}+\frac{1}{2} \right) + \left( -\frac{1}{3}+\frac{1}{3} \right) + \cdots \\
&=1+ 0 + 0 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Composition of two rotations of the same angle $\alpha$ fixing points $a,b \in{S^2}$ Let $g,h$ be rotations of the same angle $\alpha$ around fixed points $a,b \in{S^2}$. Show $gh$ fixes $c \in{S^2}$ on the great circle that forms the perpendicular bisector of the segment $ab$, such that $ac$ and $bc$ make an angle $\a... | Let
$$
\mathbf{K_a}=
\left[\begin{array}{ccc}
0& -a_3 & a_2 \\
a_3 & 0 & -a_1 \\
-a_2 & a_1 & 0
\end{array}\right]
$$
where $a = [a_1, a_2, a_3]$
Then
$$
\mathbf{R_a} = \mathbf{I} + (\sin\alpha) \mathbf{K_a} + (1-\cos\alpha)\mathbf{K_a}^2
$$
is the rotation by the vector $a$. (This is called Rodrigues' Formula.)
Y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1164722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the sum of $n$ terms of an A.P. is $2n+3n^2$, find the $r^{th}$ term. If the sum of $n$ terms of an Arithmetic Progression is $2n+3n^2$, find the $r^{th}$ term.
Note: This question is from the book Higher Algebra by H.S. Hall & S.R. Knight and its answer is $6r-1$. (As listed at the end of the book).
What I've alre... | We know that the sum of the first $n$ terms is $S_n = 2n + 3n^2$, so the sum of the first $(n - 1)$ terms is $S_{n-1} = 2(n - 1) + 3(n - 1)^2$. But if the $n$th term is $t_n$, then recall that $S_{n-1} + t_n = S_n$. Hence, it follows that:
\begin{align*}
t_n
&= S_n - S_{n-1} \\
&= (2n + 3n^2) - (2(n - 1) + 3(n - 1)^2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
| For the equation.
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$
The easiest way is to factor the square.
$$p^2=ks$$
Then decisions can be recorded.
$$x=p$$
$$y=s-p$$
$$z=p-k$$
For the more General case formula there. Number of solution for $xy +yz + zx = N$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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Technique to calculate rank of this matrix. 1) How can we calculate the rank of this matrix for $n > 3$ ?
\begin{matrix}
1^2 & 2^2 & 3^2 & \cdots & n^2 \\
2^2& 3^2 & 4^2 &\cdots & (n+1)^2 \\
3^2& 4^2 & 5^2 &\cdots & (n+2)^2 \\
\vdots&\vdots &\vdots &\ddots & \vdots\\
n^2& (n+1)^2 & (n+2)^2 &\cdot... | We claim that the determinant of matrix $M$ given by $M_{i,j}=(a_i+b_j)^m$ where $i,j\in\mathbb{N}_n$ is $0$ if $n>m+1$. We have
$$
\begin{aligned}
\det(M)
&=\begin{vmatrix}
(a_1+b_1)^m & (a_1+b_2)^m & \ldots & (a_1+b_n)^m\\
(a_2+b_1)^m & (a_2+b_2)^m & \ldots & (a_2+b_n)^m\\
\ldots & \ldots & \ldot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Maximum of a trigonometric expression Maximize $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$ I broke it in 2 parts but then realized they don't have their maxima at the same points. So I am stuck. This is what I did. I wrote it as $$3\cos x+4\sin x+2\sin x(24\cos x+7\sin x)$$ Now we can maximize the first two terms a... | Note that if $ \tan (x) = \frac {3}{4} $, then $ \tan (2x) = \frac {24}{7} $
With $ 48 \sin x \cos x = 24 \sin (2x) $ and $ 14 \sin^2 x = 7 - 7 \cos (2x) $
And consider the Pythagoras Triplets $ (3,4,5), (7,24,25) $
We have
$ f(x) = ( 4 \sin x + 3 \cos x ) + (24 \sin (2x) - 7 \cos (2x) ) + 7 $
Express them in terms of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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AIME 2013 Solutions (divisiblity) Problem 2
Find the number of five-digit positive integers, $n$, that satisfy the following conditions:
(a) the number $n$ is divisible by $5,$
(b) the first and last digits of $n$ are equal, and
(c) the sum of the digits of $n$ is divisible by $5.$
Obviously, if $n$ satisfies divisibli... | Well you need to find the number of solution to the following equations:
$$b + c + d = 0$$
$$b + c + d = 5$$
$$b + c + d = 10$$
$$b + c + d = 15$$
$$b + c + d = 20$$
$$b + c + d = 25$$
I would recommend you using generating funtions. So since $0\ge b,c,d \ge 9$, consider the following generating function:
$$(1 + x + x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Sum of the roots of an equation.
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.
I used Vieta's to get the wrong answer.
If $$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3$$
And $p, b, c$ are... | We have:
$$ p(x)=8x^3-3x^2-3x-1 = 9x^3-(x+1)^3 $$
hence $p(x)=0$ iff
$$ \left(1+\frac{1}{x}\right)^3 = 9 $$
and the only real root of the above equation is given by
$$ x = \frac{1+\sqrt[3]{81}+\sqrt[3]{9}}{8} $$
leading to the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $f(x) = x^3$ is continuous at $ x=2.$ I need to give an $\epsilon, \delta$ proof that $x^3$ is continuous at $x=2$. I got that $|x^3 -8| = |x-2| |x^2 +2x +4|$, and I need to assume $|x-2| < \delta$. How can I choose $\delta$ in terms of $\epsilon$ so that $|x-2| |x^2 +2x +4| < \epsilon$ ?
We learned the "adding ... | Let $\epsilon > 0$ and $\delta < \min{1, \epsilon/19}$. Let $x$ be within $\delta$ of $2$. Observe that
$$
|x^3 - 8 | = |x-2||x^2+2x+4| = |x-2| |(x-2)^2 + 6x| = |x-2| |(x-2)^2 + 6(x-2) + 12|.
$$
Applying the triangle inequality to the final absolute value, we find that
$$
|x^3 - 8 | \leq |x-2|^3 + 6 |x-2|^2 + 12 |x-2|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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3 Gents, 3 Ladies, and one round table The question is how many ways the aforesaid ladies and gentlemen be seated around the round table, such that two and only two ladies sit together.
The answer says 72.
My understanding:
(#ways to arrange 3 gentlemen at the table) // (3-1)!
*(#ways to select 2 ou... | There are $\binom{3}{2}$ to select the two women who sit together and $2!$ ways to arrange them within that block. That leaves four open seats, two of which are adjacent to the two women who sit together, so there are $2$ ways to sit the third woman in a seat that is not adjacent to the other two women and $3!$ ways t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Finding the Laurent series of a given function How do I find the Laurent series expansion for the function
$\frac{z}{(z+1)(z+2)}$
which converges for $1 < |z| < 2$ and diverges elsewhere.
I have done the partial fractions for $f(z) = \frac{2}{z+2} - \frac{1}{z+1}$.
| As you found already, $f(z) = \frac{2}{2+z} - \frac{1}{1+z}$
Let's look at each fraction separately:
$-\frac{1}{1+z}$ is analytic in $|z| \gt 1$ and since $|\frac{1}{z}| \lt 1$ we have:
$$
-\frac{1}{1+z} = -\frac{1}{z}\cdot\frac{1}{1-(-\frac{1}{z})} = -\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}-\dots) = -\frac{1}{z}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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The number of ways of writing an integer as a sum of two squares Given an integer $m=pq$, where $p,q$ are both primes such that $p\equiv 1 \pmod{4}, q\equiv 1 \pmod{4}$.
It is known that $p$ can be written as a sum of two squares (of positive integers) in a unique way, and the same for $q$. Prove that $m$ can be writte... | @Elaqqad
This is my solution based on your idea and notations. Please point out wherever there is a mistake.
There exist coprime integers $x,y,z,t$ such that
\begin{align*}
&a-c=xy\\&
a+c=zt\\&
d-b=xz\\&
d+b=yt
\end{align*}
Therefore we have that
\begin{align*}
&a=\frac{1}{2}(xy+zt)\\&
c=\frac{1}{2}(zt-xy)\\&
d=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimise $x+y$ under the condition $2x+32y \leq 9xy$ where $x,y > 0$ Let $x$ and $y$ be positive real numbers such that $2x+32y\leq 9xy$. Then find the smallest possible value of $x+y$. I tried using AM-GM but it is difficult to find the case of equality. Is Cauchy-Schwarz also helpful? Thanks.
| here is how i approached. i will look at the boundary of $$2x + 32 y = 9xy $$ which is given by the hyperbola $$ y = \frac{2x}{9x - 32}, \frac{dy}{dx} = -\frac{64}{(9x-32)^2} $$ this hyperbola has slope of $-1$ at $$ -\frac{64}{(9x-32)^2} = -1 \to \left(x = \frac{40}9,y = \frac{10}9, x + y = \frac{50} 9\right) \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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A cubic factor of a nine degree polynomial The polynomial $x^3−3x^2+4x−1$ is a factor of $x^9+px^6+qx^3+r$. Find $p+q+r$. How do I start with the problem?
| As David wrote, the division must be done very carefully. To avoid it, we could use a real brute force approach writing $$\Delta=(x^9+px^6+qx^3+r)-(x^3−3x^2+4x−1)\sum_{i=0}^6 a_ix^i=0$$ This should give $$\Delta=(a_0+r)+(a_1-4 a_0) x+(3 a_0-4 a_1+a_2) x^2+ (-a_0+3 a_1-4
a_2+a_3+q)x^3+(-a_1+3 a_2-4 a_3+a_4) x^4+(-a_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How to show $\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$ How to show
$$\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$$
| hint: $\displaystyle \sum_{n=0}^\infty\dfrac{1}{(24n+5)(24n+19)} = \dfrac{1}{14}\displaystyle \sum_{n=0}^\infty\left(\dfrac{1}{24n+5} - \dfrac{1}{24n+19}\right)= \dfrac{1}{14}\displaystyle \sum_{n=0}^\infty\left(\displaystyle \int_{0}^1\left(x^{24n+4} - x^{24n+18}\right)dx\right)= \dfrac{1}{14}\displaystyle \int_{0}^1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Algebra of trigonometry and some limits I am singlehandedly trying to get through a book on divergent series, however one sequence is really strange and I cant seem to understand what the authors use to come to these conclusions!
Imagine the following is our sum. ($0 < x < 2{\pi}$)
As the authors attempt to determine ... | let us pick a small $n$ and try what they are doing. i will pick $n = 10.$ we have
$$S_1 = \frac{\sin(3x/2) - \sin (x/2)}{2\sin x/2}, S_2 = \frac{\sin(5x/2) - \sin (x/2)}{2\sin x/2}, \cdots, S_{10}=\frac{\sin(21x/2) - \sin (x/2)}{2\sin x/2} $$
now we can compute $$\begin{align} S &= 2\sin^2(x/2)\left(S_1 + S_2 + S_3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show $f$ is differentiable at $x=4$ $f$ is continuous. $f$ is defined on $(0,1) \cup (1, \infty)$.
$$f(x) = \begin{cases} \frac{4\sqrt x}{(\ln x)^2} & \text{if } x\in (0,1) \cup (1,e^4),\\
\frac{e^2}{64} (ln x) (8-lnx) & \text{ if } x\in[e^4, \infty)
\end{cases}$$
Need to show that $f$ is differentiable at $x = e^4$
I... | i will try this. make a change of variable $$x = e^4(1+h).$$ we can now write
$$\begin{align} \ln x &= \ln \left(e^4(1+h) \right) = 4 + \ln(1+h) = 4 + h + \cdots\\
(\ln x)^{-2} &=(4+h \cdots)^{-2} = 4^{-2} - 24^{-3}h+ \cdots=\frac 1{16}-\frac h{32}+\cdots\\
\sqrt x &= e^2(1 + \frac 12 h + \cdots)= e^2 + \frac{e^2h}2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1184709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$ for $a,b,c>0$ If $a$, $b$, and $c$ are positive real numbers, show that
$$
\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \geq \sqrt{b^2 + bc + c^2}.
$$
What values of $a$, $b$, and $c$ ( or relation among them) make both sides equal ( makes... | Consider the points $A\equiv (0,0),B \equiv (a,0), C,D$ on the plane such that $AC=b$ and $\angle CAB=60^{\circ}$ and $AD=c$ and $\angle DAB=60^{\circ}$. Then $BC=\sqrt{a^2-ab+b^2}$ and $BD=\sqrt{a^2-ac+c^2}$. Now we also have $CD=\sqrt{b^2+bc+c^2}$. Consider the triangle $BCD$ and apply triangle inequality to get $BC+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve trigonometric equations with sin and cos I have two equations for two unknowns $u_k$ and $v_k$:
$\tan{u_1} \tan{u_k} + \cos{(v_1 - v_k)} = 0$
$\tan{u_2} \tan{u_k} + \cos{(v_2 - v_k)} = 0$
where $u_1$, $v_1$, $u_2$ and $v_2$ are parameters.
I managed to find a general solution for $v_k$ by expanding $\cos{(v_i - v... | $$\tan{u_1} \tan{u_k} + \cos{(v_1 - v_k)} = 0\\
\tan{u_2} \tan{u_k} + \cos{(v_2 - v_k)} = 0$$
You reached:
$$\tan{u_k} = \frac{\sin{(v_2 - v_1)}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}} \cos{v_k},\tan{u_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\sin{(v_2 - v_1)}} \sin{v_k}$$
Try calculating: $\sin^2v_k+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate. $y = (1 − \sec x) / \tan x$ I have no idea what I have done wrong. Please criticise.
Differentiate $y = \frac{1 - \sec x}{\tan x}$.
\begin{align*}
\frac{d}{dx} \left[ \frac{1 - \sec x}{\tan x} \right]
&= \frac{\tan x \frac{d}{dx} [ 1 - \sec x ] - (1 - \sec x) \frac{d}{dx} \tan x }{(\tan x)^2} \\
... | Hint:
$$
\dfrac{d}{dx} \tan x= \sec^2 x
$$
Here you was wrong.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
How do I calculate the number of permutations of the list $(6, 6 ,5, 4)$? I have the list $l = (6, 6, 5, 4)$ and want to how to calculate the possible number of permutations.
By using brute force I know that there are 12 possible permutations:
$$\{(6, 5, 6, 4),
(6, 6, 5, 4),
(5, 6, 6, 4),
(6, 4, 5, 6),
(6, 5, 4, 6)... | For a list of $n$ distinct elements the number of permutations if $n!$. If you have $4$ distinct elements, the number of permutations is $24$.
However in this case we have the number $6$ appearing twice. For every ordering of the four numbers $6,6',5,4$ there is another ordering switching the $6$ and the $6'$, so you n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\s... | From the classic for $a+bi$ can be seen in polar form as
\begin{align}
a + b i = \sqrt{a^{2} + b^{2}} \, e^{i \tan^{-1}(b/a)}
\end{align}
for which
\begin{align}
1 + i = \sqrt{2} \, e^{i \tan^{-1}(1)} = \sqrt{2} \, e^{\pi i/4} = 2^{1/2} e^{\pi i/4 + 2n \pi i}
\end{align}
for $n$ being integer. Now
\begin{align}
(1+i)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Problem with definite integral $\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$ $$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
The question says 'evaluate the integral using the suggested substitution. It gives $u=\cos x$. But I think Let $u=1-2\sin x$ is better.
$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin... | Consider the integral
\begin{align}
\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx
\end{align}
with the substitution $u = \cos(x)$. Making the desired substitution the integral becomes
\begin{align}
I = \int_{\sqrt{3}/2}^{1} u \sqrt{ \frac{1 - 2 \sqrt{1-u^{2}}}{1 - u^{2} } } du
\end{align}
Now make the substitutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: i... | Let $t = x - \tan^{-1}\frac15$ to integrate
\begin{align}
& \int\frac{1}{5\cos x+\sin x+7}~dx
= \int \frac1{\sqrt{26}\cos t +7}dt\\
=& \int \frac{2d(\tan\frac t2)}{(7+\sqrt{26})+(7-\sqrt{26})\tan^2\frac t2}\\
=& \frac2{\sqrt{23}}\tan^{-1} \left( \sqrt{\frac{ 7-\sqrt{26}}{7+\sqrt{26}}}\tan\frac t2 \right)+C
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$? I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$
I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.
| Setting $u=\sqrt{x-1}$ gives you
$$ \int \frac{2u \, du}{(1+u^2)^2u}= 2\int \frac{du}{(1+u^2)^2} \tag{1}$$
There are two ways to do this: the first is to change variables again to $u=\tan{\theta}$, so $du=\sec^2{\theta} \, d\theta$:
$$ \begin{align*}
2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} &= 2\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding x in matrix algebra The question asked me to calculate $|A|$ for the given matrix below and state whether or not it is invertible.
$$
\left(
\begin{matrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
x & x & 3x & 1\\
0 & x & x & 0\\
\end{matrix}
\right)
$$
I g... | Recall that performing row replacement operations does not affect the determinant. Hence, if we replace row $3$ with (row $3$ + $(-x) \cdot $ row $1$), then replace row $3$ with (row $3$ - row$2$) and replace row $4$ with (row $4$ - row $2$), we obtain:
$$
\begin{vmatrix}
1 & 0 & 2 & 1\\
0 & x &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is this a theorem or a conjecture? A few months ago, I read The Irrationals by Julian Havil. I remember reading that if you order the rationals between 0 and 1 in this way: $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\cdots$ and then construct intervals, the first one being $[\f... | First few intervals : $(1/3,1/2),(2/5,3/7),(7/17,5/12),(12/29,17/41),...$
The general construction is related to the Farey sequence.
Define freshman's sum $\frac{a}{b}\oplus \frac{c}{d}=\frac{a+c}{b+d}$.
This sum satisfies the median property : $\frac{a}{b}<\frac{c}{d}\Rightarrow \frac{a}{b}<\frac{a}{b}\oplus \frac{c}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Proving divergence of a sequence by proving the sequence is increasing. We define a sequence recursively by $$a_{n+1}=\frac{1}{4}({a_n}^2+a_n+2)~~~~~~~(a_1=3)$$
By showing $a_n$ is increasing prove that $a_n$ does not converge.
Not sure how to do this one. I tried showing that $a_{n+1}-a_n>0$ but couldn't get it to wor... | $\begin{array}\\
a_{n+1}
&=\frac{1}{4}({a_n}^2+a_n+2)\\
&=\frac{1}{16}(4{a_n}^2+4a_n+8)\\
&=\frac{1}{16}(4{a_n}^2+4a_n+1+7)\\
&=\frac{1}{16}((2{a_n}+1)^2+7)\\
\end{array}
$
Therefore
$2a_{n+1}+1
=\frac{1}{8}((2{a_n}+1)^2+7)+1
=\frac{1}{8}((2{a_n}+1)^2+15)
$.
Letting
$b_n
=2a_n+1
$,
$b_{n+1}
=\frac{1}{8}(b_n^2+15)
$.
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Converting a rotated ellipse in parametric form to cartesian form I have a rotated ellipse in parametric form:
$$\begin{pmatrix}y \\ z\end{pmatrix} = \begin{pmatrix}a\cos t + b\sin t \\ c\cos t + d\sin t\end{pmatrix} \tag{1} $$ or,
$$(y,z) = (a\cos t + b\sin t , c\cos t + d\sin t) \tag{2} $$
By using $$\cos^2 t + \sin^... | You can also find the eigenvalues of the matrix
$\begin{pmatrix}\frac{c^2+d^2}{(ad-bc)^2}&\frac{-ac-bd}{(ad-bc)^2}\\\frac{-ac-bd}{(ad-bc)^2}&\frac{a^2+b^2}{(ad-bc)^2}\end{pmatrix}$
in order to diagonalise it.
The eigenvalues are $\lambda=\frac{s^2\pm\sqrt{(s^2-2\det{})(s^2+2\det{})}}{2(\det{})^2}$, where $\det{} =ad-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How is $x^\frac{9}{4} \cdot x^\frac{1}{2} = x^\frac{11}{4}$ How is $x^\frac{9}{4} \cdot x^\frac{1}{2} = x^\frac{11}{4}$?
I don't really understand it.
| $$x^{9/4}=(x^{1/4})^9.$$
$$x^{1/2}=(x^{1/4})^2.$$ Letting $y=x^{1/4}$, we get:
$$x^{9/4}\cdot x^{1/2} = y^9\cdot y^2=y^{11} = (x^{1/4})^{11}=x^{11/4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Parametrization of an intersection Hello I'm trying to parametrize the intersection of the sphere of centre $(0,0,0)$ and radius $R$ and the cylinder of centre $(R/2,0,0)$ and radius $R/2$. (the intersection is the surface)
Any hints of how to proceed? My guess is to switch into cylinder coordinates but I can't figure ... | Given your equations:
$${x^2} + {y^2} + {z^2} = {R^2}$$
$${\left( {x - \frac{R}{2}} \right)^2} + {y^2} = {\left( {\frac{R}{2}} \right)^2}$$
Indroducing:
$$\begin{gathered}
x = \frac{R}{2}(\cos (\varphi ) + 1) \hfill \\
y = \frac{R}{2}\sin (\varphi ) \hfill \\
\end{gathered} $$
the second equation is satisfied.
Fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help evaluating $ \int \sqrt{{x}^{2} + 3} \; dx $ Can you help me evaluating the following indefinite integral?
$$
\int \sqrt{{x}^{2} + 3} \; dx
$$
Please, don't give a full solution, just some hint on which method to use...
** UPDATE **
Thank you very much to everybody for the useful comments and suggestions. I'm sorr... | Substitution $\sqrt{x^2+3}=u-x$ will work
$\sqrt{x^2+3}=u-x\\ x^2 +3 = u^2 - 2ux + x^2\\3=u^2-2ux \\ 2ux = u^2-3 \\ x = \frac{u^2-3}{2u} \\ u - x = u - \frac{u^2-3}{2u}\\u-x=\frac{u^2+3}{2u} \\
dx = \frac{2u\cdot2u-2(u^2-3)}{4u^2}du \\ dx = \frac{2u^2+6}{4u^2}du \\ dx = \frac{u^2+3}{2u^2}du \\ \int{\frac{u^2+3}{2u^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatri... | Hint: It is related to the Fibonacci numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\} \sin \left\{\theta+m\cdot \frac{\pi}{4}\right\}} = 4\sqrt{2}$ If $\displaystyle 0 < \theta < \frac{\pi}{2}$ and $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\}\cdot \sin \left\{\t... | HINT:
Can you try tis way?
$$\cot\theta-\cot\left(\frac{2\pi}3+\theta\right)=\frac{\sin\dfrac{2\pi}3}{\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}=\frac{\sqrt3}{2\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}$$
$$=\frac{\sqrt3}{\cos\dfrac{2\pi}3+\cos\left(\dfrac{2\pi}3+2\theta\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sq... | Useful inequality chain
Given $a \ge 0$ and $b > -\frac{1}{2}a$:
$a+b-2\frac{b^2}{a} \le a+b-\frac{b^2}{a+b} = \frac{a^2+2ab}{a+b} \le \sqrt{a^2+2ab} \le a+b-\frac{1}{2}\frac{b^2}{a+b} \le a+b$
Solution
Given $x \ge 1$:
$\sqrt{x}+\frac{1}{2}-\frac{1}{4\sqrt{x}+2} \le \sqrt{x+\sqrt{x}} \le \sqrt{x}+\frac{1}{2}$.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| Use this trigonometric identity:
$$
\frac{3\sin x+2\cos x}{2\sin x+3\cos x} = \overbrace{\frac{12}{13} + \frac5{13}\tan\left( x - \varphi \right)}^{\text{So integrate this function.}} \text{ where }\varphi = \arctan\frac 2 3.
$$
NOTE: I initially mislaid the denominator in $5/13$ and just had $5$ as the coefficient. @... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
What will be $E(X)$ and $Var(X)$? Given a discrete random variable
$$P(X = k) = \frac{1}{2^k}, \ \ \ k=1,2,...$$
How to calculate the expected value and Variance of $X$?
| Let $E(x)=\sum_{k=1}^\infty kx^k$ and $D(x)=\sum_k x^k$. Then
$$
E(x)+D(x)=\sum_k(k+1)x^k=\frac{d}{dx}\left(\sum_{k=1}^\infty x^{k+1}\right)=\frac{d}{dx}\frac{x^2}{1-x}=\frac{x(2-x)}{(1-x)^2}\cdot
$$
In particular,
$$
E(0.5)=\frac{0.5(2-0.5)}{(1-0.5)^2}-D(0.5)=3-1=2.
$$
This takes care of the expectation. Similarly, le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Mean of the difference between uniform random variables. I have two uniform random variables $B$ and $C$ distributed between $(2,3)$ and $(0,1)$ respectively. I need to find the mean of $\sqrt{B^2-4C}$. Could I plug in the means for $B$ and $C$ and then solve or is it more complicated than that?
The original question i... | Use change of variables:
$$X = \sqrt{B^2-4C}
$$
$$Y=C$$
$$B = \sqrt{X^2+4Y}$$
$$C=Y$$
Jacobian:
$$
J = det\left( \begin{array}{ccc}
\frac{\partial B}{\partial X} &\frac{\partial C}{\partial X} \\
\frac{\partial B}{\partial Y} & \frac{\partial C}{\partial Y} \\
\end{array} \right) = det \left( \begin{array}{ccc}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why can $5^{2n+3} + 3^{n+3} \cdot 2^{n} \quad \forall n \in \mathbb{N}$ never be a prime number? It seems to be true for the first thousand $n$ but I really can't think of a way to prove this statement.
Any kind of help will be appreciated!
| Oops...I think I found a proof.
We can notice that for every $n$ our sum seems to be divisible by $19$.
So we have to prove that
$$5^{2n+3} + 3^{n+3} \cdot 2^{n} \equiv 0 \pmod{19}\quad \forall n \in \mathbb{N}$$
We can prove this using induction.
Base-case:
\begin{align*}
&19|152&
\end{align*}
Inductive step:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$
| If we consider the equation of a certain type.
$$(n-4)^2+(n-3)^2+(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2+$$
$$+(n+4)^2=(k-3)^2+(k-2)^2+(k-1)^2+k^2+$$
$$+(k+1)^2+(k+2)^2+(k+3)^2+(k+4)^2$$
The solutions can be written simply using the sequence. The next element which is obtained from the previous one.
$$p_2=3p_1+4s_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove this form $2^{n+2}-7a^2_{n}$ always is square numbers. Let sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=-1$,and
$$a_{n}+a_{n-1}+2a_{n-2}=0,n\ge 3$$
show that
$2^{n+2}-7a^2_{n}$ is square numbers.
such as
$$2^3-7=1,n=1$$
$$2^4-7=9=3^2,n=2$$
$a_{3}=-1$
$$2^{5}-7=25=5^2,n=3$$
$a_{4}=3$
$$2^6-7\cdot 9=1^2,n=4$$
| We have always $2^{n+2}-7a^2_{n}=b_n^2$, where $b_1=-1$, $b_2=-3$ and $b_n+b_{n-1}+2b_{n-2}=0$ (induction).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find all primitive triples (a,b,c)? (Pythagorean Triples) how to find all primitive triples when one value of $(a,b,c)$ is given? For example in this case $a = 45$.
What is the procedure to find the primitive triples ?
Conditions for primitive triples are:
*
*$a^2 + b^2 = c^2$
*$(a,b) = 1$
*$a$ and $c$ are... | We can find Pythagorean triples for a given side $A$ by solving Euclid's formula for $n$ and then testing a finite number of $m$-values to see which generates an integer for $n$.
$$A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil$$
The lower limit ens... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find $11^{644} \mod 645$ Can someone just explain to me the basic process of what is going on here? I understand everything until we start adding 1's then after that it all goes to hell. I just need some guidance. The Problem with the solution is attached.
Thanks in advance..
Find $11^{644} \mod 645$
| The rather intimidating notation used in the textbook is effectively just finding the binary representation of 644.
$644 = 512+128+4 = 1010000100_{2}$
It does this by extracting factors of two and noting when the result is odd, corresponding to successive digits (from the right) of the binary number above.
so putting t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving by strong induction for a sequence of integers, $2^n$ divides term $n$ Provided the following sequence of integers $t_1, t_2, t_3$,... is defined as:
$t_1 =4, t_2 =8$ and $t_n= $ $ 6t_n$$_-$$_1$ - $4t_n$$_-$$_2$ for all integers $n \geq 3$
How do we prove that $2^n$ divides $t_n$ for all integers $n \geq 1$ ... | Let
$u_n = t_n/2^n$,
or
$t_n = 2^n u_n$.
Since
$t_n
=6 t_{n-1}-4t_{n-2}
$,
$2^nu_n
=6 \cdot 2^{n-1}u_{n-1}-4\cdot 2^{n-2}u_{n-2}
=3 \cdot 2^{n}u_{n-1}- 2^{n}u_{n-2}
$,
or
$u_n
=3 u_{n-1}- u_{n-2}
$.
Since $u_1$ and $u_2$
are integers
(
$u_1 = t_1/2 = 2$
and
$u_2 = t_2/4 = 2$
),
all following
$u_n$ are integers.
Even mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Lagrange's multiplier not working
Given the function $f(x,y):=xy+x-y$. Let $D:=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq25\wedge x \geq 0\}$. Find the absolute maximum and minimum of $f$ on $D$.
My working is as follows:
$\begin{array}
& f_x(x,y)=y+1=0 & \qquad \qquad f_y(x,y)=x-1=0 \\
\Rightarrow y=-1 & \qquad \qquad \Righ... | Let us focus our attention on the boundary of $D$. From your system we deduce
$$
(4\lambda^2-1)y=1-2\lambda,
$$
and then $x=-y$. This gives us a single admissible point $P_0$ whose coordinates are $x=\frac{5}{\sqrt{2}}$ and $y=-\frac{5}{\sqrt{2}}$. At this point, $f(P_0)=-\frac{25}{2}+5 \sqrt{2} \approx -5.42893$.
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Question about a solution of a system of three non linear equations in three unknowns Let $a$, $b$ and $c$ be positive real numbers such that $$ a + \frac{1}{b} = 3$$ $$b + \frac{1}{c} = 4$$ $$ c + \frac{1}{a} = \frac{9}{11} $$ then $$ a \times b \times c =?$$
I tried doing this problem but I was unsuccessful. Tried... |
Let $a$, $b$ and $c$ be positive real numbers such that
\begin{align}
a + \frac{1}{b} &= 3 \tag{1}\label{1}
\\
b + \frac{1}{c} &= 4 \tag{2}\label{2}
\\
c + \frac{1}{a} &= \frac{9}{11} \tag{3}\label{3}
\end{align}
\eqref{1}$\times$\eqref{2}$\times$\eqref{3}
$-$\eqref{1}$-$\eqref{2}$-$\eqref{3} gives:
\begin{align}
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\frac{t^2}{t+1}$ to the form $\frac{1}{t+1} +t -1$ One of my attempts would look like below.
$\frac{t^2}{t+1}$ = $\frac{t \times t+1-1}{t+1}$ = 1+ $\frac{t-1}{t+1}$ = $\frac{t-1+1-1}{t+1} + 1$ = $\frac{-2}{t+1} +2$
Also, I put into t an arbitrary number in both equations and have gotten the same answer.... | Several responses have you make the following factorization:
$$\frac{t^2}{t+1} = \frac{(t+1)(t-1) + 1}{t+1}$$
But in general, you can take a slightly more iterative approach:
$$\frac{t^2}{t+1} = \frac{t^2 + t - t}{t+1} = \frac{t\cdot (t+1)}{t+1} - \frac{t}{t+1} = t - \big(\frac{t+1-1}{t+1}\big) = t - 1 + \frac1{t+1}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rew... | Because there is alot of solutions, let me present you mine: the equation you stated is equivalent to $$(p+10)(p+q+265)=4658=2\cdot 17\cdot 137$$
and because $p+10>2$ then $p+10\geq 10$ but also $p+q+265\geq 2\cdot 137$ hence $p+2=17$ and $p+q+265=2\cdot 137$
In negative solutions (without (negative) prime assumptions... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\lim\limits_{n \to \infty} \frac{2n^2+2}{3n^3+1}=0$ directly from the definition of limit. One of my homework questions is:
Prove $\lim\limits_{n \to \infty} \frac{2n^2+2}{3n^3+1}=0$ directly
from the definition of limit.
In trying to follow: Prove that $\lim \limits_{n\to\infty}\frac{n}{n^2+1} = 0$ from th... | Let $\epsilon>0$ be given and let $N>\frac{4}{3\epsilon}$. If $n>N$, then $n>\frac{4}{3\epsilon}$ or $$\epsilon>\left|\frac{4}{3n}\right|$$
Thus :
$$\epsilon>\left|\frac{4n^2}{3n^3}\right|$$
$$\epsilon>\left|\frac{2n^2+2n^2}{3n^3}\right|\geq\left|\frac{2n^2+2}{3n^3}\right|>\left|\frac{2n^2+2}{3n^3+1}\right|=\left|\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof by Induction $3^n > n^3$ I am trying to prove the following, however I'm stuck at the Induction hypothesis
Prove by induction that, for all integers $n$, if $n\geq 5$, then $3^n>n^3$
What I have Done:
Base Case:
$n = 5$
$3^5 > 5^3$
$243 > 125$ so TRUE
Assume True for $n = k$
$3^k > k^3$ must be TRUE
Should be T... | Supposing that $3^k\gt k^3$, you need to prove $3^{k+1}\gt (k+1)^3$.
For inductive step : $$\begin{align}3^{k+1}&\gt 3k^3\\&=k^3+k^3+k^3\\&\gt k^3+3k^2+3^2k\\&=k^3+3k^2+3k+6k\\&\gt k^3+3k^2+3k+1\\&=(k+1)^3.\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Working out $\tan x$ using sin and cos expansion
Using only the series expansions $\sin x = x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...$ and $\cos x = 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...$
Find the series expansions of the $\tan x$ function up to the $x^5$ term.
So it is:
$$
\frac {x- \dfrac{x^3} {3!} + \d... | The following way may be easier.
Since $\tan x$ is an odd function, it can be represented as $\tan x=ax+bx^3+cx^5+\cdots.$
Since $\sin x=\cos x\tan x$, one has
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\left(ax+bx^3+cx^5+\cdots\right).$$
Comparing the coefficients gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Which is larger, $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$? The clue given by the text is to "use the fact that $\sqrt{x}$ is increasing."
I was able to get the correct answer here by squaring both expressions. But I don't think I made use of the text-prided clue, so am wondering if perhaps I'm missing the "deeper... | Since $x\mapsto\sqrt{x}$ is an increasing function we have the following
\begin{align*}
2\sqrt{15}&>2\sqrt{12}\\
8+2\sqrt{15}&>8+2\sqrt{12}\\
3+2\sqrt{15}+5&>2+2\sqrt{12}+6\\
(\sqrt{3}+\sqrt{5})^2&>(\sqrt{2}+\sqrt{6})^2
\end{align*}
Therefore, $\sqrt{3}+\sqrt{5}>\sqrt{2}+\sqrt{6}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
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Integral by using substitution (How to proceed?) Using the substitution $x=a\sin\theta$, or otherwise, find $\int\frac{1}{x^2\sqrt{a^2-x^2}}dx$.
My attempt,
$x=a\sin\theta$
$dx=a\cos (\theta)d\theta$. Then $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin ^2(\theta)}$
The given answer is $-\frac{\sqrt{a^2-x^2}}{a^2x}+c$
How to proce... | Using your substition:
$$\int\frac1{x^2\sqrt{a^2-x^2}}\;\rightarrow\;\int\frac{a\cos\theta\;d\theta}{a^2\sin^2\theta\cdot a\cos\theta}=-\frac1{a^2}\int\frac{d\theta}{-\sin^2\theta}=-\frac1{a^2}\cot\theta+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculating the value of $\cos \left(\frac{1}{2} \arccos \frac{3}{5}\right)$ I need help finding the value of $$\cos \left(\frac{1}{2} \arccos \frac{3}{5}\right)$$
My try:
let $\theta = \frac{1}{2} \arccos \frac{3}{5}$, then $2\theta = \arccos \frac{3}{5}$ and $\theta \in [0,\frac{\pi}{2}]$
Now let $A = 2\theta$, and $... | Hint with a trigonometric identity:
$$\cos x=2\cos^2\frac x2-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How can I solve this integral with complex number? $n$ here is a complex number such that $n=n_r+in_i$
How can I solve this integral?
$$\int_{0}^{\infty}\frac{x^4}{|x^2-n^2|^2} d x=? $$
| Let's change the notation to a more friendly one: $\;n:= a+bi\;$ , so:
$$x^2-n^2=x^2-(a^2-b^2+(2abi))=(x^2-a^2+b^2)-2abi\implies$$
$$|x^2-n^2|^2=(x^2-a^2+b^2)^2+4a^2b^2$$
Thus your integral is
$$\int_0^\infty\frac{x^4}{(x^2-a^2+b^2)^2+4a^2b^2}dx\stackrel{M=\max\{a,b\}}\ge\frac14\int_M^\infty\frac{x^4}{x^4+M^4}dx=$$
$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $n \in \Bbb Z$, determine $\gcd(3n^2 + 7n + 4, n + 2)$. I factored $3n^2+7n+4$ to $(3n+4)(n+1)$ and because there isn't a common factor of those and $n+2$ I said that the gcd is $1$, but is there any othere way to go about it that would come up with a gcd that isn't $1$?
| We have
$$3n^2+7n+4 = (n+2)(3n+1) + 2$$
Hence, if $d$ is a common divisor of $3n^2+7n+4$ and $n+2$, then $d$ must divide $2$. Hence, $\gcd(3n^2+7n+4,n+2) = 1 \text{ or }2$. $3n^2+7n+4$ is always even. Further, $n+2$ has the same parity as $n$.
Hence, if $n$ is odd, $$\gcd(3n^2+7n+4,n+2) = 1$$
while if $n$ is even $$\gc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\lim_{n\to \infty}n^2({1\over{n^3+1^3}}+{1\over{n^3+2^3}}+\cdots+{1\over{n^3+n^3}}).$ Find the limit of $$\lim_{n\to \infty}n^2({1\over{n^3+1^3}}+{1\over{n^3+2^3}}+\cdots+{1\over{n^3+n^3}}).$$
I'm not sure how to evaluate this limit. Any hints or solutions are greatly appreciated. I turned this into ... | Your integral is correct, since the terms of your sum can be rewritten as $$\frac{n^2}{n^3+i^3} = \frac{1}{n} \cdot \frac{1}{1+(i/n)^3}$$, which means that your sum is a Riemann sum for the corresponding integral.
If you are having a hard time getting started on evaluating the integral, you can factor $x^3+1 = (x+1)(x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Intriguing Indefinite Integral: $\int ( \frac{x^2-3x+1/3 }{x^3-x+1})^2 \mathrm{d}x$
Evaluate
$$\int \left( \frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 \mathrm{d}x$$
I tried using partial fractions but the denominator doesn't factor out nicely. I also substituted $x=\dfrac{1}{t}$ to get
$$\frac{-1}{9} \int \left(\frac... | Partial fractions also works, this may be a bit more work than trying to find a clever ad hoc solution, but the advantage is that it is a straightforward method.
If $\alpha$ is a root of the denominator, then putting $x = \alpha + t$ and performing a so-called Laurent expansion around $t=0$ (which is analogous to a Ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
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Compute the values of two infinite products whose factors are the same I have the following question:
How to prove that
$(1-\frac{1}{2})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{6})\cdot (1+\frac{1}{7})\cdot (1-\frac{1}{8})\cdot\ ...\ = \frac{1}{2}\ $, but
$(1-\frac{1}{2})\cdot (... | The first is straightforward as already answered by @MarcoCantarini.
For the second
$$P = (1-\tfrac{1}{2})(1-\tfrac{1}{4}) (1+\tfrac{1}{3})(1-\tfrac{1}{6})(1-\tfrac{1}{8}) (1+\tfrac{1}{5}) (1-\tfrac{1}{10}) (1-\tfrac{1}{12}) (1+\tfrac{1}{7}) \cdots
$$
here's an alternate approach. Add a factor $(1+1/1)$ to get
$$
\begi... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^2+b^2+c^2\geq [2(a-b)^2(b-c)^2(a-c)^2]^{1/3}$ Mathematica seems to know that this statement is true, yet I am struggling to prove it. Possible useful inequalities are Minkowski and the geometric mean. Using the geometric mean inequality I can prove a less strict bound:
$$(a-b)^2+(b-c)^2+(a-c)^2 \geq 0 \im... | Let $x=a-b$ and $y=b-c$. Then $a-c=x+y$. Thus, we are left with
$$
(x+y+c)^2+(y+c)^2+c^2\ge(2x^2y^2(x+y)^2)^{1/3}\tag{1}
$$
To find the minimum of the left-hand side of $(1)$, we set the derivative with respect to $c$ to $0$:
$$
2(x+y+c)+2(y+c)+2c=0\tag{2}
$$
$(2)$ says that $c=-\frac{x+2y}3$. Plugging this into the le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\frac {2n}{(a+b)^n} \le \frac {1}{a^n} + \frac {1}{b^n}$ for $a,b>0, n\in\mathbb{N}$ by induction prove using induction:
$$ \frac {2n}{(a+b)^n} \le \frac {1}{a^n} + \frac {1}{b^n} $$
$$a,b \gt 0 , n \in N$$
my attempt:
*
*base $n=1$:
$$ \frac {2}{(a+b)} \le \frac {1}{a} + \frac {1}{b}$$
$$2ab \le b(a+b) + ... | Direct Proof
First, I would re-write the inequality so that you want to show $2n\leq \frac{(a+b)^{n}}{a^n}+\frac{(a+b)^{n}}{b^n}$. Then, we will use two facts which I will prove before finishing.
Fact 1: Since $a,b>0$ it follows that $(a+b)^{n}=\sum_{i=0}^{n}\binom{n}{i}a^{i}b^{i}\geq \binom{n}{1}ab^{n-1}+\binom{n}{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$4$-element subsets of the set $\{1,2,3,\ldots,10\}$ that do not contain any pair of consecutive numbers Find the number of subsets of the set $\{1,2,3,\ldots,10\}$ that contain $4$ elements and do not include any pair of consecutive numbers.
For example $\{1,2,5,7\}$ is not an example of such a subset since $1,2$ are ... | Let's apply your strategy of using the Inclusion-Exclusion Principle.
There are $\binom{10}{4}$ four-element subsets of a ten-element set. From these, we must exclude those subsets containing at least two consecutive numbers.
A string of at least two consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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40th derivative of a function I would like to have some verification to see if my answer is correct. The given function is $f(x)=ln(1+x^2)$ and I need the 40th derivative at $x=0$. Here is my work: Using series one can manipulate $\frac{1}{1+x}=1-x+x^2-x^3+x^4...=SUMx^n(-1)^n$ into $\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8...... | Taking the $40$th derivative at 0 will yield you the coefficient for the $x^{40}$ term multiplied by $40!$
The general series for natural log is:
$$
ln(x+1) = \frac {x^1}1 - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4...
$$
replacing $x^2$ makes:
$$
ln(x^2+1) = \frac {x^2}1 - \frac {x^4}2 + \frac {x^6}3 - \frac {x^8}4..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
| Define $f(\alpha)=\int\dfrac{dx}{\alpha+x^2}=\frac{1}{\sqrt\alpha}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)$
You can recover the original integral by putting $\alpha=1$ later,
Now, differentiating both sides.
$f'(\alpha)=\int\dfrac{-dx}{(\alpha+x^2)^2}=\frac{-1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, with $\mathbf{a+b+c=0}$. The angle between any two of these vectors is $120^\circ$. Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, such that $\mathbf{a}+\mathbf{b}+\mathbf{c} = \mathbf{0}$. Show that the angle between any two of thes... | Geometrically, $\mathbf a + \mathbf b + \mathbf c = \mathbf 0$ means we can form a triangle $\triangle XYZ$ whose oriented sides $\overrightarrow{XY}, \overrightarrow{YZ}, \overrightarrow{ZX}$ are parallel copies of $\mathbf a$, $\mathbf b$, $\mathbf c$.
Since $\mathbf a$, $\mathbf b$, $\mathbf c$ are unit vectors, $\t... | {
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"url": "https://math.stackexchange.com/questions/1264990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help understanding how to factor completely $x^3-x^2-x+1$ I need someone to help explain the steps to completely factor the problem $x^3-x^2-x+1$.
Here is what I have done so far:
$x^3-x^2-x+1$
to
$x^3-x^2+-1(x+1)$
Since there is a negative I changed $-x+1$ to $-1(x+1)$
to
$x^2(x-1)+-1(x+1)$
to
$(x^2-1)(x-1)(x+1)... | $$x^3 - x^2 - x + 1 \to x^2(x - 1) + (-1)(x-1)$$
$$\to (x-1)(x^2 - 1) \to (x-1)^2(x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Domain of the function $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$ will be? I tried solving this question by
$1.$ $-1$ and $4$ will not be in domain because denominator can not be zero .
$2.$ Either both denominator and numerator will be positive or negative so that whole term in root becomes positive.
But I am not able... | The zeroes of the denominator are $\frac{4 \pm \sqrt{32}}{2} = 2(1 \pm \sqrt{2})$.
Therefore the denominator is negative iff $2(1 - \sqrt{2}) < x < 2(1 + \sqrt{2})$.
The upper part is positive if $x<0$, zero if $x=0$, and negative is $x>0$
So they ae both positive if $x < 2(1 - \sqrt{2})$.
They are both negative if $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question regarding proving by induction I am struggling with a math problem I have been assigned. The problem is as follows:
Let $X_1 = -3$ and $X_2 = 0$. Given that for every natural number $n \geq 2, X_{n+1} = 7X_n - 10X_{n-1}$, prove by induction that for every $n$ belonging to $\mathbb{N}$, $X_n = 2 \cdot 5^{n-1} -... | Assume true for $n = k$, Which means
${X_k} = 2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}$
${X_{k + 1}} = 7{X_k} - 10{X_{k - 1}}$
Required to prove for $n = k + 1$ That is required to prove
${X_{k + 1}} = 2 \cdot {5^k} - 5 \cdot {2^k} = 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) - 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Describe the set of odd primes such that $\left(\frac{-5}{p}\right) = 1$ (Legendre Symbol) Okay, so $\left(\frac{-5}{p}\right) = 1$.
I am assuming that I can start this by saying
$\left(\frac{-5}{p}\right) = \left(\frac{5}{p}\right) \times \left(\frac{-1}{p}\right)$.
There are well defined rules for $\left(\frac{-1}{p... | $\left(\frac{-5}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{5}{p}\right)=(-1)^{\frac{p-1}{2}}\left(\frac{5}{p}\right)\stackrel{\text{QR}}=(-1)^{\frac{p-1}{2}}\left(\frac{p}{5}\right)$
$(-1)^{\frac{p-1}{2}}=\begin{cases}1, &p\equiv 1\pmod{\!4}\\-1, &p\equiv -1\pmod{\!4}\end{cases}$
$0^2\equiv \color{#0bc}{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $\int_{-1}^{1} (x^{4/3} + 4x^{1/3}) dx$? I started by integrating it this way:
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
What is wrong with it?
| You are asking "What is wrong with it?"
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
Check the derivative
$$(\frac{3}{7}\frac{7}{3} x^{4/3} + \color{red}{\frac{16}{4}\frac{4}{3}}x^{1/3}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$... | Testing divisibility by 9 gives
$$
48 + x + y \equiv 0 \pmod {9}.
$$
Testing divisibility by 11 gives
$$
12 - x + y \equiv 0 \pmod {11}
$$
So $x - y \equiv 1 \pmod {11}$ and $x + y \equiv 6 \pmod{9}$.
Since $x, y$ are both digits, $x = 1 + y$, and $x + y$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that if $\sum \limits_{k=1}^{n} x_n=1$ then $\sum \limits_{k=1}^{n} x_n^2 \geq {1 \over n}$ Prove that if $\sum \limits_{k=1}^{n} x_n=1$ then $\sum \limits_{k=1}^{n} x_n^2 \geq {1 \over n}$ where $\{x_k\}_1^n$ are real numbers which are not all the same.
I tried to prove it by induction. The base was trivial. Now... | This is false. For instance, $x_1 = 1$ and rest of $x_i$'s zero violates the inequality. Your inequality should be the other way around, which follows from Cauchy Schwarz inequality.
By Cauchy Schwarz, we have
$$\left \vert\vec{a} \cdot \vec{b} \right \vert \leq \left \Vert \vec{a}\right \Vert \left \Vert \vec{b}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proof without using induction How to prove that
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
without using induction.
If we don't know the right side of this expression, how to get right expression. I tried with partial sums and binomial formula but can't get it.
So the problem is:
$$1^2+2^2+...+n^2=?$$
Thanks for repli... | Assume we know $\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$. Compute the following cubes
$$\begin{align}
1^3&=1\\
(1+1)^3&=1^3+3\cdot 1^2+3\cdot 1+1^3\\
\vdots&=\vdots\\
n^3&=(n-1)^3+3(n-1)^2+3(n-1)+1^3\\
(n+1)^3&=n^3+3n^2+3n+1^3\end {align}$$
Add these equations together and cancel the cubes you have on both sides you get
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
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Finding roots of 5 th degree taylor expansion of $e^x$ I need to find the roots [actually their count and sign] of :
$$
{\mathcal T}(e^x,5)=\frac{x^5}{5!}+\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$$
*
*It[Root] is clearly negative.
*It should have odd number of roots.
*Either One or Three or Five.
*It is e... | A similar way to @Daniel Fischer's answer (if not quite as clean) to show that the derivative is positive:
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{24}\big(x^4+4x^3+12x^2+24x+24\big)\\
&= \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+23)\big)\\
&> \frac{1}{24}\big((x^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
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Fractions in Questions and Answers
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