Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove by complete induction a floor formula I cant prove by complete induction this formula:
$$(-1)^{\lfloor\frac{n+1}2\rfloor+1}=(-1)^{\lfloor\frac{n}2\rfloor+1+n}$$
I keep failing when I apply $(-1)^{\lfloor\frac{n+1+1}2\rfloor+1}=(-1)^{\lfloor\frac{n+1}2\rfloor+1+n+1}$. It does not give me the same values. For exam... | Alt. hint (without induction): use Hermite's identity $\displaystyle\,\lfloor x \rfloor + \left\lfloor x + \frac{1}{2} \right\rfloor = \lfloor 2x \rfloor\,$ for $\displaystyle\,x=\frac{n}{2}\,$:
$$\left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n+1}2\right\rfloor = \left\lfloor 2 \cdot \frac{n}{2}\right\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $x=8+3\sqrt{7},$ then what is value of $\sqrt{x} -\frac{1}{\sqrt{x}}$?
If $x=8+3\sqrt{7},$ then what is value of $\sqrt{x} -\frac{1}{\sqrt{x}}$?
This question is somewhat different than I thought. I only know how to find the value when root is not given . Please help me.
| Note that $\left ( \sqrt{x}- \frac{1}{\sqrt{x}} \right )^2 = x + \frac{1}{x} - 2$. Further note that $8+3 \sqrt{7} > 1$, so $\sqrt{x}-\frac{1}{\sqrt{x}} > 0$.
$$x=8+3 \sqrt{7} \implies \frac{1}{x} = \frac{8-3\sqrt{7}}{8^2 - (3\sqrt{7})^2} = \frac{8-3\sqrt{7}}{64-63} = 8-3\sqrt{7}.$$
Thus, $x+\frac{1}{x} = 16$, which im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3}$ for $ax^3 + bx^2 + cx + d$ Using Vieta's formulas, I can get $$\begin{align} \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\&= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\... | Let $y = 1/x$. So you ask for the value of $f(y) = y_1^3 + y_2^3 + y_3^3$ where the $y_i$ are solutions of $dy^3 + cy^2 + by + a = 0$. With this equation, write
$$- d f(y) = c (y_1^2 + y_2^2 + y_3^2) + b (y_1 + y_2 + y_3) + 3 a.$$
Note the identity $$y_1^2 + y_2^2 + y_3^2=(y_1+y_2+y_3)^2-2(y_1y_2+y_1y_3+y_2y_3).$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$a= \frac{b^2-2b+16}{b^3} b= \frac{8a^3}{a^2-2a+16} a=? b=?$ it's part of my bigger homework, i got to state like that and i need some nice way to calculate a and b. I was trying to do this by setting b value from second equations to first equations, but it feels like so much work, can i do it smarter way?
If it's impo... | HINT:
$8 a^3 b^3= 8 a^2(b^2-2b+16) = b^4(a^2-2a+16)$.
setting $a = x b$ you have
$8 x^3 b^6= 8 x^2 b^2(b^2-2b+16) = b^4(x^2 b^2-2 x b +16)$
Equating the second and third expression gives a quadratic equation in $x$ which you can solve for $x = x (b)$.
Equating the first and second expression gives
$ x = \frac{b^2-... | {
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"url": "https://math.stackexchange.com/questions/2420671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving Inequality involving exponential function How can I solve the following inequality algebraicly for $x$?
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot (\frac{x}{50})^\frac{5}{2}\cdot e^\sqrt{\frac{x}{50t}}<x\ \ \ \ \ \ \ (t>0)$$
I suppose it can be solved with using Lambert W function but I don't know how I can introduce Lam... | $$\frac{\sqrt{5}}{\sqrt{t}}\cdot (\frac{x}{50})^\frac{5}{2}\cdot e^\sqrt{\frac{x}{50t}}<x\ \ \ \left(t,x\in\mathbb{R}\right)$$
Reshape your inequality into an inequality of only one function. Bring for that all $x$ to the left-hand side of your inequality:
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{5}{2}}}{50^\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the determinant of this n by n matrix? Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new... | If you denote:
$$D_k=\begin{vmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{vmatrix}$$
and expand it by the last column (or row), you get
\begin{align}D_k=\begin{vmatrix}
0 & x_1 & x_2 & \cdots& x_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Definite integral $\int_0^1\sqrt{x^2+1}\, dx$. Use trig substitution? I have an integral:
$$\int_0^1\sqrt{x^2+1}\, dx$$
but I have gotten stuck. Here's the work I have done already:
I'm not sure where to go from here. Using a trig identity doesn't seem like it would work. Integration by parts doesn't work (at least... | You don't need a trigonometry substitution. Integrate by parts. Let $a\in\mathbb R$. I'm using $a^2$ instead of $a$ here because $x=a\tan t$, $x=a\cot t$ seem like possible substitutions. You've already tried $x=\tan \theta$ for your problem. At least one answer has shown how it can solve the problem.
$$I:=\int \sqrt{x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\... | squaring both sides of your inequality (the last one)
$$\frac{1}{3k+1}\cdot \frac{(2k+1)^2}{(2k+2)^2}\le \frac{1}{3k+4}$$
this is equivalent to
$$(3k+4)(2k+1)^2\le (3k+1)(2k+2)^2$$
simplifying this we obtain: $$k>0$$ and this is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\lim\limits_{x\to 4}\frac{2x\sqrt{x}+x-8\sqrt{x}-4}{x+\sqrt{x}-6}$ by just factoring Calculate $$\lim_{x\to 4}\dfrac{2x\sqrt{x}+x-8\sqrt{x}-4}{x+\sqrt{x}-6}$$ by just factoring
Factoring $2x\sqrt{x}+x-8\sqrt{x}-4$ gives us $(x-4)(2\sqrt{x}+1)$
$\color{red}{2x\sqrt{x}+x-8\sqrt{x}-4=x(2\sqrt{x}+1)-4(2\sqrt{x}+... | Let $x=t^2$
If
$$x=4 $$
Then
$$t=2$$
We have to find
$$\lim_{t\to 2} \frac{2t^3+t^2-8t-4}{t^2+t-6} $$
$$\lim_{t\to 2} \frac{\color{blue}{(t-2)}(t+2)(2t+1)}{\color{blue}{(t-2)}(t+3)} $$
I think you can handle further.
$\color{red}{OR}$
In your equation factories as follows
$$(x-4)=(\sqrt x +2)\color{blue}{(\sqrt x-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$5$ kids toss one die each. Find the number of ways that the sum of dice points is $22$. I have already done these following steps and I'm already lost. Can someone please help me?
$x_1 + x_2 + x_3 + x_4 + x_5 = 22$
when $1 \leq x_n \leq 6$ and $n = 1,2,3,4,5$.
Then,
\begin{align*}
f(x) & = (x+x^2+x^3+x^4+x^5+x... | As with the other answers, here is an expansion of the generating function.
$$
\begin{align}
x^5\left(\frac{1-x^6}{1-x}\right)^5
&=x^5\sum_{j=0}^5\binom{5}{j}\left(-x^6\right)^j\sum_{k=0}^\infty\binom{-5}{k}(-x)^k\tag{1}\\
&=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^{j+k}\binom{5}{j}\binom{-5}{k}x^{k+6j+5}\tag{2}\\
&=\sum_{j=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Algorithm to get the number of sorted combinations? Lets have a set of number S = {A,B,C,...,n} where A < B < C .... < n
how many ways can you sort their combinations?
Take this small example:
If the set is {1,2,8} (so n = 3) i can sort their combination in ascending order like this: 1,2,1+2,8,1+8,2+8,1+2+8 which is eq... | This is not a complete answer, but in order to get you started:
At least for small $N$ you can read off the cases from the Hasse diagram. E.g. for your example $N=3$ we have
$$\begin{matrix}
& & & & A+B+C \\
& & & \huge\diagup & \\
& & B+C \\
& & \huge| & & \\
& & A+C \\
& \huge\diagup & & \huge\diagdown \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Distribute $N$ indistinguishable balls, where $N$ is even, into $n$ distinguishable boxes if each box can contain at most $N/2$ balls How many ways are there to distribute $N$ ($N$ is an even number) indistinguishable balls into $n$ distinguishable boxes if each box can contain at most $\frac{N}{2}$ balls.
My solution... | Since $N$ is even, let $N = 2k$. Then the question can be rephrased as follows: In how many ways can $2k$ indistinguishable balls be placed in $n$ boxes if at most $k$ balls can be placed in one box?
Clearly, $n \geq 2$, for otherwise we would not be able to distribute all the balls to the boxes given the restriction... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$ by using Mean Value Theorem? Given Question is Prove that
$$\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$$
when $0<a<b<1$ using Mean Value Thoerem.
I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and... | Using $f (x)=arcsin (x) $ in the interval $[a,b] $ we have $\frac {\arcsin (b)-\arcsin (a)}{b-a}=\frac {1}{\sqrt {1-c^2} }$ now we have $a <c <b $ hence $-a^2>-c^2>-b^2$ adding 1 and recipeocating we have $\frac {1}{1-a^2}<\frac {1}{1-c^2}<\frac {1}{1-b^2} $ hence taking root completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Guessing the value of $n$ $A$ and $B$ play game, $A$ choose $n$ where $n \in \{1, 2,\ldots 1001\}=S$.
$B$ has to guess the value of $n$ by choosing a number of subsets of $S$, then $A$ will tell $B$ the number of subsets $B$ choose that contain $n$.
Do the same operation for $3$ times, let $k_1, k_2, k_3$ be the numb... | On the first round ask the six questions.
\begin{eqnarray*}
\{ i \mid i \equiv 1 \pmod 7 \} \\
\{ i \mid i \equiv 1 \pmod 7 \text{ or } i \equiv 2 \pmod 7 \} \\
\vdots \\
\{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \text{ or } 5 \text{ or } 6\pmod 7 \} \\
\end{eqnarray*}
On the second round a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^... | from your second equation we get
$$t^2=\frac{1}{1+y}$$ for $$y> -1$$ and taking the square root we have
$$t_{1,2}=\pm\frac{1}{\sqrt{1+y}}$$ thus we get
$$x=\pm\sqrt{1+y}-\pm\frac{1}{\sqrt{1+y}}$$
from the first equation we get
$$0=t^2+tx-1$$ solving this we get
$$t_{1,2}-\frac{1}{2}x\pm\sqrt{\frac{x^2}{4}-1}$$
from her... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$ Question:
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $... | multiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get
$$2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$ and we have
$$2^n\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\geq 2$ and $$3^n\cdot 3<3^n\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have
$$2^{n+1}+3^{n+1}<2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find minimum value of sum of areas of squares
I tried using some trigonometry but couldn't arrive at any helpful inequality
| From trivial applications of similitude we get $y=\dfrac{15}{7}-\dfrac{3 x}{4}$
The sum of the area of the two squares is
$f(x)=x^2+\left(\dfrac{15}{7}-\dfrac{3 x}{4}\right)^2$
Derivative is
$f'(x)=2x+2\left(-\dfrac{3}{4}\right)\left(\dfrac{15}{7}-\dfrac{3 x}{4}\right)$
$f'(x)=0$ if $x=\dfrac{36}{35}$
This value is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into v... | Let $I = (a+b)^7-a^7-b^7$.
$$a^7+b^7 = (a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$$
and
$$(a+b)^6 = a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$
Then
\begin{eqnarray*}
I &=& (a+b)(7a^5b+14a^4b^2+21a^3b^3+14a^2b^4+7ab^5)\\
&=&7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\
& =& 7ab(a+b)(a^2+ab+b^2)^2
\end{eqnarray*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| Hint $ $ Use $\ ax^{n+1}\!+by^{n+1} = (x\!+\!y)(ax^n\!+by^n) - xy(ax^{n-1}\!+by^{n-1})$ to solve for $\,xy,x\!+\!y$
Remark $ $ The recurrence is $\,(S-x)(S-y)f_n = (S^2 - (x+y)S+xy)f_n = 0\,$ where $\,Sf_n = f_{n+1}.\,$ It has solutions $\,f_n = x^n,\,y^n\,$ so also $\,f_n = ax^n + bx^n$ by linearity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Determining derivatives of trigonometric functions For what function $f$ and number $a$ is the limit
$$\lim_{x \to \pi/4} \frac {\tan x - 1}{4x-\pi}$$
the value of $f’(a)$?
All I’m asking is how I would begin solving this problem.
|
HINT, for the limit use:
$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}\tag1$$
So, we get:
$$\lim_{x\to\frac{\pi}{4}}\space\frac{\tan\left(x\right)-1}{4x-\pi}=\lim_{x\to\frac{\pi}{4}}\space\frac{\frac{\sin\left(x\right)}{\cos\left(x\right)}-1}{4x-\pi}=\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Is the limit $\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$ zero? I have this limit:$$\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$$
At first sight seems that limit equals 0. But WolframAlpha says that there is no limit. I tried to prove it. I considered cases $y = kx$, and so on. I never got to find ... | It seems to me like the limit is zero. Write the function as
$$
\frac{(x-y)+y}{(x-y)^2+y^2}+\frac{2y}{(x-y)^2+y^2}
$$
The second fraction goes to zero since
$$
\left|\frac{2y}{(x-y)^2+y^2}\right|=\frac{|2y|}{(x-y)^2+y^2}\le \frac{2|y|}{y^2}=\frac2{|y|}\stackrel{y\to\infty}\to 0
$$
For the first fraction, use the Cauchy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given this matrix $A$, find $A^{144}$ I know this is a very common question to ask when one is making their way into Linear Algebra (i.e. given a matrix, find the result of that matrix to the nth-power).
I'm given this matrix: $$
A=
\left[ {\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} } \right]$$
I'm a... | Note that $A^4 = I$, and $A^{144} = (A^4)^{36}$, and you're done. If you really want to go with that $A^6 = A^2$ instead, then we have that
$$A^{144} = (A^6)^{24} \\
= (A^2)^{24} = A^{48} \\
= (A^6)^8 = (A^2)^8 \\
= A^{16}= A^4\times A^{12}\\
= A^4 \times(A^6)^2 = A^4\times(A^2)^2\\
= A^8 = A^2\times A^6\\
= A^2\times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
| Hint:
$$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}=\frac{1}{\frac{x^2-x+1}{x}}=\frac{1}{\frac{x^2+x+1-2x}{x}}=\frac{1}{\frac{x^2+x+1}{x}-2}$$
Note, that we have to assume $x\neq 0$ in this process. You will have to check if that is problematic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-... | If $u=\alpha$ and $v$ are the roots
$$
\begin{align}
x^2+ax+b+1
&=(x-u)(x-v)\\
&=x^2-(u+v)x+uv
\end{align}
$$
Then $a=-(u+v)$ and $b=uv-1$. Since $u=\alpha\in\mathbb{Z}$, and $a\in\mathbb{Z}$, we know that $v\in\mathbb{Z}$. Therefore,
$$
\begin{align}
a^2+b^2
&=(u+v)^2+(uv-1)^2\\
&=u^2+v^2+u^2v^2+1\\
&=\left(u^2+1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How would I solve the following question about matrices? $A= \begin{pmatrix}-1&3\\ -2&6\end{pmatrix}$
Find two $2\times 2$ matrices $B$ and $C$ such that $AB=AC$ but $B\ne C$.
I have tried to do some row operations along with multiplication but I keep getting the wrong answer.
Any help?
| $B=\left(
\begin{array}{ll}
3 & 0 \\
2 & 1 \\
\end{array}
\right);\;C=\left(
\begin{array}{ll}
-6 & 9 \\
-1 & 4 \\
\end{array}
\right)$
$AB=AC=\left(
\begin{array}{ll}
3 & 3 \\
6 & 6 \\
\end{array}
\right)$
I did in this way
Called $B=\left(
\begin{array}{ll}
a & b \\
c & d \\
\end{array}
\right);\;C=\left(
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Question regarding the proof of $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^{2}$ Let $n$ be a positive integer. Prove that
$$\sum_{k=0}^{n} (-1)^k \binom{n}{k}^{2} = \begin{cases}
0 & \text{if $n$ is odd} \\
(-1)^m \binom{2m}{m}, & \text{if $n$ = 2m}
\end{cases}$$
So I wrote out a first few terms to get a feel of the proble... | If you have polynomials $f(x)=\sum_k a_kx^k$ and $g(x)=\sum_k b_kx^k$
then the $x^n$-coefficient of $f(x)g(x)$ is
$$c_n=a_0b_n+a_1b_{n-1}+\cdots+a_nb_0=\sum_{k=0}^n a_{k}b_{n-k}.$$
Here $a_k=\binom nk$ and $b_k=(-1)^k\binom nk$, and here
$$c_n=\sum_{k=0}^n (-1)^{n-k}\binom nk\binom{n}{n-k}.$$
Can you relate this to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the inequality $a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$ for positive $a, b, c$ I faced problem proving this inequality for positive $a$, $b$, $c$:
$a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$
I tried to simplify it and I got that:
$bc^3 + a^3 c + a b^3 \geq a b^2 ... | An AM-GM proof
By AM-GM inequality $$ab^3+abc^2\geq2\sqrt{a^2b^4c^2}=2ab^2c\tag{1}$$ Similarly we get $$bc^3+bca^2\overset{\text{AM-GM}}\geq2abc^2\tag{2}$$ and
$$ca^3+cab^2\overset{\text{AM-GM}}\geq2a^2bc\tag{3}$$
Hence adding $(1),(2),(3)$ we get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Magic squares in combinatorics Let $P_{3}(r)$ be the number of 3 x 3 magic squares that are symmetric to their main diagonal. Prove that $P_{3}(r) \leq (r+1)^3$.
$r$ in this problem seems to be the sum of each row and column
This is the first time I've dealt with magic squares, didn't even know of their existence befo... | Suppose that $a$ and $b$ have been selected:
$$\begin{pmatrix}
a & ? & ?\\
? & b & ? \\
? & ? & ?\\
\end{pmatrix}$$
We can immediately deduce the following value:
$$\begin{pmatrix}
a & ? & ?\\
? & b & ? \\
? & ? & (r-a-b)\\
\end{pmatrix}$$
Using symmetry we obtain these two values:
$$\begin{pmatrix}
a & ? & \frac{r-b}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}... | Alternatively, you can always double check the result with generating functions (like here and here for more examples):
$$f(t)=\sum\limits_{n=0}x_n\cdot t^{n}=2 + \sum\limits_{n=1}x_n\cdot t^{n}= 2 + \sum\limits_{n=1}(3x_{n-1}-8)\cdot t^{n}=\\
2 + 3\sum\limits_{n=1}x_{n-1}\cdot t^{n}-8\sum\limits_{n=1} t^{n}=2+3t\sum\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Minimum of $\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $. $\forall n \in \mathbf{N}$, prove that the function $f(x)=\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $ attains its minimum at $x=\frac{1}{2}$.
Now it suffices to prove that
\begin{equation*}
\sum_{l=0}^{n} \binom{n}{l}^2... | \begin{align*}
n &> 0 \text{.} \\
f'(1/2) &= \left. \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)x^{2l-1}(1-x)^{2(n-l)} - 2(n-l)x^{2l}(1-x)^{2(n-l)-1} \right) \right|_{x = 1/2} \\
&= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)(1/2)^{2l-1}(1/2)^{2(n-l)} - 2(n-l)(1/2)^{2l}(1/2)^{2(n-l)-1} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limit using taylor expansions How can I evaluate this limit? Trying with Taylor doesn't seem to give me the right result, why ?
$$\lim_{x\to 0} \frac{\sqrt{1+4x} -1 -\sin(2x)}{\log(1+x^2)}$$
with Taylor I can approximate $\sin(2x)$ to $ 2x $
and $\log(1+x^2)$ to $ x^2 $. If I plug in these in the limits ... | binomial theorem:
$(a+b)^n) = a^n + na^{n-1}b + \frac {n(n-1)}{2} a^{n-1}b^2 \cdots$
works for fractional exponents
$(1+4x)^\frac 12 = 1 + \frac 12 (4x) - \frac 18 (4x)^2+ \cdots%$
$\sin x = x - \frac 16 x^2+\cdots\\
\sin 2x = (2x) - \frac16 (2x)^3+ \cdots$
numerator: $1 + 2x - 2x^2 - 1 - 2x + \frac 86 x^3 = -2x^2- \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Summation of $1\cdot 3\cdot 5\cdot 7 + 3\cdot 5\cdot 7\cdot 9 ...$
Find the sum of:
$1 \cdot 3\cdot 5\cdot 7 + 3\cdot 5\cdot 7\cdot 9+...$ till $n$ terms.
My attempt:
I got the $i^{th}$ term to be $(2i-1)(2i+1)(2i+3)(2i+5)$
Expansion gives: $16i^4 +64i^3+56i^2+-16i-15$
Required: $$\sum\limits_{i=1}^n (16i^4 +64i^3+56... | Let
$$
f(k) = (2k-1) \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}
.
$$
Then
$$
f(k+1) = \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}(2k+9)
,
$$
so
$$
f(k+1)-f(k) = 10 \, \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}
,
$$
and your sum can be rewritten as the telescoping sum
$$
\frac{1}{10}\sum_{k=0}^{n-1} \Bigl( f(k+1)-f(k) \Bigr)
= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Arithmetic of complex numbers
If $z = \cos x + i\sin x$ , show that
$$\frac{2}{1+z} =1-i\tan\left(\frac{x}{2}\right),$$
I get up to:
$$\frac{1+\cos x -i\sin x}{1+\cos x}.$$
| Showing that the two numbers are equal can be done by showing that the magnitude and the argument for both numbers are the same.
Magnitude:
$$1+z = 1+\cos x + i\sin x$$
$$|1+z| = \sqrt{1 + 2 \cos x + \cos^2 x + \sin^2 x} = \sqrt{2+2\cos x}$$
$$| \frac{2}{1+z}| = \frac{\sqrt{2}}{\sqrt{1 + \cos x}} $$
$$|1-it| = \sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving Recurrence Relation of $T(n)=4T(n-2)+2$ Question
Solve Recurrence Relation of $T(n)=4T(n-2)+2$
Base case-: $T(1)=1,T(2)=2$
My Approach/solution
$$T(n)=4T(n-2)+2$$
$$T(n-2)=4T(n-4)+2 \tag{1}$$
$$T(n-4)=4T(n-6)+2 \tag{2}$$
Using $(1)$ and $(2)$ in my equation
$$\begin{align*}
T(n)&=4\cdot (4T(n-4)+2)+2\\
&=4^{2... | i think the right solution is given by $$T(n)=\frac{1}{12} \left(9\ 2^n+(-1)^{n+1} 2^n-8\right)$$
at first solve the equation $$T(n)=4T(n-2)$$ with $$T(n)=q^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forg... | For $x=y=\frac{1}{\sqrt2}$ we get a value $-1$.
We'll prove that it's a minimal value.
We need to prove that
$$x^4-6x^2y^2+y^4\geq-1.$$
Indeed,
$$x^4-6x^2y^2+y^4+1\geq x^4-6x^2y^2+y^4+(x^2+y^2)^2=2(x^2-y^2)^2\geq0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a... | $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)$$
$$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = 3abc$$
$$\begin{align*} a^5 + b^5 + c^5 & = (a + b + c)(a^4 + b^4 + c^4) - (ab + bc + ca)(a^3 + b^2 + c^3) + abc(a^2 + b^2 + c^2) = \\ & = -3abc(ab + bc + ca) - 2abc(ab + bc ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"? How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"?
My attempt: Let $F(n, k)$ be the number of bit strings of length $n$ that contain exactly $k$ blocks of $10.$ Note that for $k \neq0, $ $F(0, k) = F(1, k)= 0.$
Consider a bit... | Thanks for the insight. I used a similar technique for a similar problem. I wrote this up for my assignment which is due in a week. Although I believe that I found the recurrence relation correctly, I am unable to go about extracting the coefficients from a quadratic raised to a negative k-th power. Could you please be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ab+bc+ca+abc=4$ then $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$, via AM-GM
Suppose, for positive reals $a$, $b$, $c$, that
$$ab+bc+ca+abc=4$$
Prove that
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$$
I applied AM-GM on the first equality ie, $a$, $b$, $c$, and $abc$ to get $$ab+bc+ca \geq 3\qquad\text{and}\qquad abc \le... | Let $a=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z}$, where $x$, $y$ and $z$ be positives.
Thus, the condition gives
$$\frac{4xy}{(x+z)(y+z)}+2c\left(\frac{x}{y+z}+\frac{y}{x+z}\right)+\frac{4xyc}{(x+z)(y+z)}=4$$ or
$$\frac{2c(x^2+y^2+xz+yz+2xy)}{(x+z)(y+z)}=4-\frac{4xy}{(x+z)(y+z)}$$ or
$$\frac{2c(x+y)(x+y+z)}{(x+z)(y+z)}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Exercises EG.3 and EG.4 in Probability with martingales, by David Williams
Let $G$ be the free group with two generators $a$, $b$. Start at time 0 with unit element 1, the empty word. At each second multiple the current word on the right by one of the four elements $a$, $a^{-1}$, $b$, $b^{-1}$ choosing each with proba... | I have only been able to solve this through guidance of Did.
$\def\P{\mathbb{P}}$
$\def\Ex{\mathbb{E}}$
$\def\Z{\text{\ensuremath{\mathbb{Z}}}}$
$\def\I{\textrm{I}}$
$\def\as{\textrm{a.s.}}$
EG.3 Use the Markov Chain theory developed in Markov chains, by Norris,
to calculate the hitting time to reach sate $0$ from stat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Write the parametric equation of the revolution surface generated by the line when it rotates around the axis $Oz$. Write the parametric equation of the revolution surface generated by the line whose equation is $2x - y + z = 1, x + y - 3z = 2$ when it rotates around the axis $Oz$.
If we let $z=t$ and we add the two e... | The line has equation given by the system
$
\left\{
\begin{array}{l}
2x - y + z = 1\\
x + y - 3z = 2\\
\end{array}
\right.
$
The planes have normal vectors $(2,-1,1)$ and $(1,1,-3)$
The intersecting line is parallel to their cross product
$(2,-1,1)\times(1,1,-3)=(2,7,3)$
and passes through a point common to the two p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$ Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$
My attempt led me to have $\quad n=2k+1:\quad k\in\mathbb N$
The expression under square root is odd, so the square root's value is also odd.
I assumed $\sqrt{1+5^n+6^n+11^n}=2a+1:\quad a\... | It is never an integer. If $n\in\mathbb N$, then$$1+5^n+6^n+11^n=1+5^n+(5+1)^n+(10+1)^n\equiv3\pmod5,$$but every perfect square is congruent to $0$, $1$, or $4\pmod5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How could we express the space in an other way? I want to calculate a triple integral over the space $$D=\{(x,y,z)\mid |x|\leq 1, |y|\leq 1, z\geq 0, x^2+y^2+z^2\leq 1\}$$
$$$$
We have the following:
*
*$|x|\leq 1\Rightarrow -1\leq x\leq 1$
*$|y|\leq 1\Rightarrow -1\leq y\leq 1$
*$x^2+y^2+z^2\leq 1 \Rightarrow z^... | Note that $x^2+y^2+z^2\leq 1$ implies $|x|\leq 1, |y|\leq 1$:
$$|x|^2=x^2\leq x^2+y^2+z^2\leq 1 \implies |x|\leq 1.$$
Hence, $D$ is the upper-half unit ball centered at the origin
$$D=\{(x,y,z)\in\mathbb{R}^3\;:\; z\geq 0, x^2+y^2+z^2\leq 1\}.$$
By using the cartesian coordinates, we have (see your last line)
$$\iiin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is the polynomial $x^4+2x^3 y+4x^2 y^2+2 x y^3+y^4$ Schur positive? I tried to check that whether the symmetric polynomial $p=x^4+2x^3 y+4x^2 y^2+2 x y^3+y^4$ is Schur positive.
We have Schur polynomials $s_0 = 1$, $s_1 = x+y$, $s_2=x^2 + xy + y^2$, $s_3 = x^3 + x^2y + xy^2 + y^3$, $s_4 = x^4 + x^3y + x^2y^2 + xy^3 + y... | The Jacobi-Trudi identity gives
$$
s_{(3,1)}=\left|\begin{array}\mbox{s}_3&s_4\\s_0&s_1\end{array}\right|=x^3y+x^2y^2+xy^3
$$
and
$$
s_{(2,2)}=\left|\begin{array}\mbox{s}_2&s_3\\s_1&s_2\end{array}\right|=x^2y^2
$$
So $p= s_4+s_{(3,1)}+2s_{(2,2)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On the convergence of $\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$ Does the following series absolutely converge, conditionally converge or diverge?
$$\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$$
My answer:
$$ 0<\frac{1}{\sqrt{n}}<\left|\frac{\cos(\pi n/2)}{n}\right| $$
an... | This series is conditionally convergent. Its terms are:
$$\frac{1}{n}\cos\left(\frac{n\pi}{2}\right) =
\left\{
\begin{array}{llll}
\frac{1}{n} & \mbox{if } n=4k , k>0\\
0 & \mbox{if } n=4k+1 , k\ge0\\
-\frac{1}{n} & \mbox{if } n=4k+2 , k\ge0\\
0& \mbox{if } n=4k+3, k\ge0\\
\end{array}
\right. $$
By remov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the limit $\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n$ NOTE: L'Hopital's and Taylor series not allowed!
Taking the log and exponenting the entire thing I get $$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}e^{n\ln\left(\frac{1+\sqrt[n]{a}}{2}\right... | $$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{\sqrt[n]{a}-1}{2}\right)^{\frac{2}{\sqrt[n]{a}-1}\cdot\frac{1}{2}\cdot\frac{\sqrt[n]a-1}{\frac{1}{n}}}=e^{\frac{1}{2}\ln{a}}=\sqrt{a}$$
I used the following.
We know that $\frac{\sqrt[n]{a}-1}{2}\rightarrow0$ for $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Probability Problem: A box contains 100 balls A box contains 100 balls, numbered from 1 to 100. If two balls are selected at random and at the same time from the box, what is the probability that the numbers on the balls will be:
a) consecutives
b) 2 multiples of 6
c) odd and even
d) 2 divisors of 60
a) $P=\left(\dfr... |
A box contains $100$ balls, numbered from $1$ to $100$. If two balls are selected at random, what is the probability that the numbers on the balls will be consecutive?
There are $\binom{100}{2}$ ways to select two of the $100$ balls.
A pair of consecutive numbers is determined by the smaller of the numbers. There... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{xy}\ge4$ given that $x+y=1$ and conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge2$
Let $x,y\in\mathbb R^+$ and $x+y=1$
1- Prove that $\frac{1}{xy}\ge4$
2- Conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge25$
I have tried to start from $x+y=1$ or $x\ge0\land y\ge0$ and reach $\frac{1}{xy}\ge4$ but ... | given $X + Y = 1$, proof that $\frac{1}{XY} \ge 4$, where $X$ and $Y$ are real positive integers.
Since $X$ and $Y$ are positive integers then, $X \lt 1$ and $Y \lt 1$ for they to sum to $1$.
$X$ and $Y$ can be both equal
So that $X + Y = 1$
this then becomes $X$ or $Y$ $= \frac{1}{2} = 0.5$
But if $X$ and $Y$ are dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solve system $\cos^2x+\cos^2y+\cos^2z=1,$ $\cos x+\cos y+\cos z=1,$ $x+y+z=\pi$ I want to solve the system:
$$\cos^2(x)+\cos^2(y)+\cos^2(z)=1,$$
$$\cos(x)+\cos(y)+\cos(z)=1,$$
$$x+y+z=\pi.$$
I tried to prove that only one of cosines can be not a zero, but I just prove that one or three cosines can not be zero.
I get,... | The hint.
Since $z=\pi-x-y$, we obtain
$$\cos{x}+\cos{y}-\cos(x+y)=1$$ or
$$2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=2\cos^2\frac{x+y}{2}$$ or
$$\cos\frac{x+y}{2}\left(\cos\frac{x-y}{2}-\cos\frac{x+y}{2}\right)=0$$ or
$$\sin\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}=0.$$
I think the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Algebraic proof of a simple inequality Prove $\frac{a+b+c}{2} ≥ \frac{bc}{b+c} + \frac{ca}{c+a} + \frac{ab}{a+b}$
I feel like this problem would have a much simpler solution than just expaning out and simplifying. Any pointers are appreciated
| By C-S
$$\sum_{cyc}\frac{ab}{a+b}=\sum_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}}\leq\sum_{cyc}\frac{1}{\frac{(1+1)^2}{a+b}}=\sum_{cyc}\frac{a+b}{4}=\frac{a+b+c}{2}.$$
Another way:
It's $$\sum_{cyc}\frac{ab}{a+b}\leq\frac{a+b+c}{2}$$ or
$$\sum_{cyc}\left(\frac{a+b}{4}-\frac{ab}{a+b}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proof of an inequality by induction Prove using induction that
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{n^2} \le 2-\frac{1}{n}$$
for all positive whole numbers $n$.
I began by showing that it is true for $n=1$
I then assumed that it is true for $n=p$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{... | It can be much simpler.
You have equality for $n = 1$.
Adding a new $n+1$ term adds $\frac{1}{(n+1)^2}$ to LHS and $\frac{1}{n(n+1)}$ to the RHS. Guess which one is smaller for all $n>1$?!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can I quickly determine the eigenvalues of this matrix? I am working on observability and detectability in controls and I ran across this example that I didn't understand. The author deliberately sought the form of this matrix, because of its "block-form" in order to quickly find the eigenvalues
\begin{bmatrix}
l_{... | Note that the characteristic polynomial is easy to evaluate by using the Laplace expansion of a determinant (along the blue column or row):
\begin{align}&\det\begin{bmatrix}
\color{Blue}{l_{11}-x} & -1 & -1 & 0 & 0 \\
\color{Blue}{0} & -1-x & 0 & 0 & 0 \\
\color{Blue}{0} & -1 & -1-x & 0 & 0 \\
\colo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Ordinary generating function of $n\cdot 2^{n-1}$ demonstration How I could demonstrate that the ordinary generating function of $n\cdot 2^{n-1}$ is
$$\frac{x}{(1-2x)^2}?$$
| From $a_n=n2^{n-1}$ and the way generating functions are defined, we have
$$f(x)=\sum\limits_{n=0} a_n\cdot x^n=\sum\limits_{n=0}n2^{n-1}\cdot x^n=\sum\limits_{n=1}n2^{n-1}\cdot x^n=x\left(\sum\limits_{n=1}n2^{n-1}\cdot x^{n-1}\right)=\\
x\left(\sum\limits_{n=1}n(2x)^{n-1}\right)=\frac{x}{2}\left(\sum\limits_{n=1}2n(2x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve.
Thank you people, but I need the thinking process, not the answer.
| In the original version, let $y=x^2+2$. Then, we have $y^2-6xy+8x^2=0$, which gives us $(y-2x)(y-4x)=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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On the summation $\sum \limits_{n=1}^{\infty} \arctan \left ( \frac{1}{n^3+n^2+n+1} \right )$ Here is a problem that I ran into. I seriously doubt if there is a closed form but you never know.
Evaluate the series
$$\mathcal{S} = \sum_{n=1}^\infty \arctan \left ( \frac 1 {n^3+n^2+n+1} \right) $$
I searched in vain to at... | Something tells me this might be difficult to do by hand. Plugging in your product formula into Mathematica and then Simplifying, this is what comes out:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Expand using binomial theorem Yesterday, my maths lecturer did this:
$[(\cos x+i\sin x)+(\cos x-i\sin x)]^4$
Expanded to:
$8\cos2x+2\cos4x+6$
I've tried using binomial theorem to get to this answer but I can never seem to get there. How can I do it?
When I do it I get to:
$2\cos4x+4(\cos3x+i\sin3x)(\cos x-i\sin x)+6(\c... | If we combine like terms, we get $(2\cos x)^4$. The Double-angle formula for $\cos$ is: $$\cos 2x=2\cos^2x-1$$ We start with $(2\cos x)^4=16\cos^4x$ and get $4(2\cos^2x)^2=4(\cos2x+1)^2=4\cos^22x+8\cos 2x+4,$ and the first term gives $2(2\cos^22x)=2\cos4x+2$, for a grand total of $2\cos 4x+8\cos 2x+6$.
From where you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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derivation of the cubic formula In my research I came across the following derivation of the cubic formula that was almost complete. However at the end it says to make the substitution $u=z^3$ which would determine the roots. I am not quite sure how to do this and then how to determine the roots and I am looking for s... | The cubic formula was derived from a series of substitutions. It's probably better to memorize the process of deriving the formula, rather than memorizing the actual formula.
Starting with the general cubic, make a substitution such that the squared term is removed. Let's denote the depressed cubic as$$x^3+qx+r=0$$Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\the... | $$\begin{align}
%
\cos(3\theta) - \cos(2\theta) &=
%
\mathcal R \left\{e^{3i\theta}\right\} - \left\{e^{2i\theta}\right\} \\ \\ &=
%
\mathcal R \left\{\left(e^{i\theta}\right)^3 - \left(e^{i\theta}\right)^2\right\} \\ \\ &=
%
\mathcal R \left\{\left(\cos(\theta) + \rm i \sin(\theta) \right)^3 - \left(\cos(\theta) + \rm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Use the method of generating series to solve the recurrence: $b_n=2b_{n-1}-b_{n-2}$ with $b_0=0, b_1=5$ This is my attempt:
Let
\begin{equation}
B(x):=\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Multiplying our recurrence through by $x^n$ and then summing from $2$ to $\infty$
\begin{equation}
\sum^{\infty}_{n=2}... | Following your steps, we have that
\begin{align}b_n&=[x^n]\frac{5x}{(x-1)^2}=5[x^{n-1}](1-x)^{-2}=5[x^{n-1}]\sum_{k=0}^{\infty}\binom{-2}{k}(-x)^k\\
&=5\binom{-2}{n-1}(-1)^{n-1}=5(-1)^{n-1}\frac{(-2)(-3)\cdot (-2-(n-1)+1)}{(n-1)!}\\&=\frac{5n!}{(n-1)!}=5n.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Area of a square inside a square created by connecting point-opposite midpoint
Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each.
$H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively.
What will the area of the square formed in the middle be?
I know that this problem can be solved by trigonometry ... | I hope that the following makes sense. I would have been better if you had labelled points on the figure.
The side length of the square is $1$.
The area of one of the large triangles is $A_{\text{large}} = 1\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$.
The hypotenuse of the large triangle is $\sqrt{1^2 + \frac{1}{4}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 15,
"answer_id": 11
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Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
| Observe that the coefficient of $x^1$ in $$(x-1)(x-2)(x-3)$$ is $$2\cdot3+1\cdot2+1\cdot3$$
or the coefficient of $x^{4-2}$ in $$(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)=(x^2-(1+4)x+1\cdot4)(x^2-(2+3)x+2\cdot3)$$
is $$2\cdot3+(1+4)(2+3)+1\cdot4$$
So, the required sum
$$=\sum_{1\le r_1<r_2\le n}r_1r_2=\dfrac{(\sum_{r=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Derivative of $\sin^{-1}(x)$ I can find this using the fact that $\sin(\sin^{-1}(x)) = x$, for all $x\in[-1,1].$
Now, differentiate.
$$\frac{d}{d\sin^{-1}(x)}\sin(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x)= \frac{d}{dx} x= 1$$
$$\cos(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x) = 1$$
$$\frac{d}{dx} \sin^{-1}(x) = \f... | You could do it like this:
first, define $f$ by $f(x) = \sin^{-1}x,$
where $-1 \leq x \leq 1$ and
$-\frac\pi2 \leq \sin^{-1}x \leq \frac\pi2.$
(Recall that $\sin^{-1}$ is not a real function for $|x| > 1$
and that $\sin$ is not a one-to-one function, so we have to choose
an appropriate domain and range of $\sin^{-1}$;... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Strange functional equation: $f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left( \frac{1}{6} -4 \ f \left(\frac{-4\pi^2}{z}\right)\right)$ I would like to get some information about the following functional equation:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\fra... | That formula puts no restrictions at all on your function on the half-plane $\Re(z) > 0$.
If you know $f(z_1)$ for some $z_1$ with $\Re(z_1) > 0$, then it tells you the value of $f(z_2)$ where $z_2 = \frac{-4\pi^2}{z_1}$. But $\Re(z_2) < 0$. And if you try to use the formula again, the new value is for $\frac{-4\pi^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2534373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\displaystyle 1 <\cos A+\cos B+\cos C$ Prove that $\displaystyle 1 < \cos A+\cos B+\cos C \leq \frac{3}{2}$ where $A+B+C = \pi$
Attempt: $\cos A+\cos B+\cos C= 1+4\cos A\cdot \cos B\cdot \cos C$
Now $\displaystyle \cos A\cdot \cos B\cdot \cos C=\frac{1}{2}\cos A\left[\cos (B-C)-\cos A\right]\leq \frac{1}{2}... | use that $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=1+\frac{r}{R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\... | We can simplify the problem by finding the minimun of
$$ g(t) = \frac{\sqrt{3}}{4}t + \frac{\sqrt{(9-t)(t-1)}}{4} $$
with $t = x^2, \ t \in [1,9]$
A straightforward approach is just to take the first derivative
$$ g'(t) = \frac{\sqrt{3}}{4} + \frac{5-t}{4\sqrt{-9+10t-t^2}} $$
Solving $g'(t)=0$ leads to
$$ \sqrt{16 - (t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Show $\frac 1 3 C_0+\left(\frac 2 3 \right)\left(\frac 1 3 \right)^2 C_1+\left(\frac 2 3\right)^2 \left(\frac 1 3 \right)^3 C_2 +\dots =\frac 1 2$ From a puzzle involving Markov chains I have the following expression, and calculating $50$ terms of the sum I strongly believe $$\frac 1 3 C_0 + \left(\frac 2 3 \right) \le... | Use the generating function for the Catalan numbers:
$$c(x) = \sum_{n=0}^\infty C_n x^n = \frac{2}{1+\sqrt{1-4x}}.$$
So, $$\begin{align}\frac 1 3 C_0 + \left(\frac 2 3 \right) \left(\frac 1 3 \right)^2 C_1
+ \left(\frac 2 3 \right)^2 \left(\frac 1 3 \right)^3 C_2 + \dots &= \frac13 \sum_{n=0}^\infty C_n \left(\frac29\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equations with the floor function I am trying to solve the following problem:
For what real numbers $x$ is: $\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=0$?
I have no idea how to do this, please help me
| Hint:
We know:
$x,y\in [z,z+1)$ then $\lfloor x\rfloor=\lfloor y\rfloor$
$\lfloor a+b\rfloor=\lfloor a\rfloor+b$ for integer $b$
With those 2:
$\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=\lfloor2x\rfloor-3-3(\lfloor x\rfloor+2)=\lfloor2x\rfloor-3\lfloor x\rfloor-9=0\\\implies\lfloor2x\rfloor-3\lfloor x\rfloor=9$
Now divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Generalization of Dirichlet integral Is it correct that
$$\lim_{L\to\infty} \int_0^L d x \int_0^L dy \int_0^L d z \frac{\sin(x+y)}{x+y}\frac{\sin(y+z)}{y+z}\frac{\sin(z+x)}{z+x} =\frac {\pi^3}{16}. $$
or does anybody has a reference for this? The value $\frac{\pi^3}{16}$ comes from numerics.
The identity
$$ \lim_{L\to... | A lengthy method, but it works. First, we rewrite:
$$\sin(x+y) \sin(y+z) \sin(z+x)=\frac{1}{4} \left(\sin(2x)+\sin(2y)+\sin(2z)-\sin(2(x+y+z)) \right)$$
Now let's change the variables (keeping the same letters for simplicity):
$$x \to \frac{1}{2}x^2, \qquad y \to \frac{1}{2}y^2, \qquad z \to \frac{1}{2}z^2$$
$$I= 2 \in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_0^{\sqrt 3} \frac{1}{1+x^2} \sin^{-1} (\frac{2x}{1+x^2}) dx$ The question is to evaluate $$\int_0^{\sqrt 3} \frac{1}{1+x^2} \sin^{-1} (\frac{2x}{1+x^2}) dx$$
I used the substitution $x=\tan a$ and it led me to the answer $\pi^2 /9$.However the correct answer is $\frac{7 \pi^2}{72}$.I couldn't get why ... | $x=\tan y\implies0\le y\le\dfrac\pi3$
Using Principal values
$$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=\sin^{-1}(\sin2y)=\begin{cases}2y&\mbox{if }0\le2y\le\dfrac\pi2 \\
\pi-2y & \mbox{if } \dfrac\pi2 \le2y\le\pi \end{cases}$$
| {
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The sum of all integers $n$ satisfying $\frac{1}{4} < \sin\frac{\pi}{n} < \frac{1}{3}$ Asked in China Junior Math Olympiad. (No calculators or tables allowed.)
Find the sum of all integers $n$ satisfying the following inequality:
$$\frac{1}{4} < \sin\frac{\pi}{n} < \frac{1}{3}$$
| $\dfrac{1}{4} < \sin\dfrac{\pi}{n}
\implies
\sin\dfrac{\pi}{n} > 0
$
$
-\pi < \dfrac{\pi}{n} < \pi
\implies
n > 0
$
$
n=1
$ does not work
$
\implies n \ge 2
$
$
-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}
\implies
\sin(x)
$ increasing
$
\sin\dfrac{\pi}{6} = \dfrac{1}{2} > \dfrac{1}{3}
\implies
n \ge 7
$
$\dfrac{1}{4} < \sin ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:
Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.
This problem is seems to be ... | You have already written down $\frac{c}{a} = \alpha \cdot \beta$, so there you have it.
| {
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"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all prime numbers $p$ such that $16p+1$ is a perfect cube What I have attempted:
Suppose $16p+1=k^3$ where $k \in Z$ then $16p=k^3-1=(k-1)(k^2+k+1)$ so we can say that $k=17$ and thus $p=17^3+17+1=4931$ which is prime.
How would I find the remaining numbers?
| You had $$16p=k^3-1=(k-1)(k^2+k+1)$$
Because $k$ is odd, $k-1$ is even and $k^2+k+1$ is odd. If $k^2+k+1$ is odd, then $k-1$ must be a multiple of $16$. But for $k-1$ to be a multiple of $16$ other than $16$, $p$ would have to not be a prime. Therefore, $k-1 = 16$ and $k = 17$.
That means that $k^2+k+1$ must be our pri... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Integral asked on MIT bee qualifier 2012 $\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx$ I have been trying to solve this one but I have no clue, I put it into wolfram and the result is absurd, considering this one is taken from MIT bee qualifier 2012, this is the integral:
$$\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \, dx$$
| You can rewrite the integrand as
$$ \frac{(x-1)(x+1)}{(x+1)^2\sqrt{x^2\left(x + \dfrac{1}{x} + 1\right)}} = \frac{x^2 - 1}{(x^3 + 2x^2 + x)\sqrt{x + \dfrac{1}{x} + 1}} \\
= \frac{1 - \dfrac{1}{x^2}}{\left( x + \dfrac{1}{x} + 2 \right)\sqrt{x + \dfrac{1}{x} + 1}} $$
Then make the substitution $u^2 = x + \dfrac{1}{x} + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Question about series involving zeta function I was observing the series $$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)-1}{2n+1}$$
And Wolfram alpha says that it does not converge. But I'm convinced that this is wrong since $$\sum_{n=1}^{\infty} (\zeta(2n+1)-1) = \frac{1}{4}$$
I would imagine then, by comparison test, that th... | Perhaps I can fill in the missing details for you to help show how
$$I = \int^\infty_0 \frac{\sinh x - x}{x} \cdot \frac{dx}{e^x (e^x - 1)} = 1 - \gamma - \frac{1}{2} \ln (2).$$
To do this we will follow Jack's suggestion and use Frullani's theorem together with the following integral representation for the Euler-Masch... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$'s General Solution I was looking for the general solution of the differential equation $$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$$
My work
We need to get the auxillary equation of the given differential equation above...
The auxillary equation would be:
$$r^5 + r^4 - 7r^3 - 1... | Hint: Since you know that two roots are $-2$ and $3$, you can exploit factoring by grouping or polynomial long division:
\begin{align}\begin{aligned}
&r^5+r^4-7r^3-11r^2-8r-12 \\&\quad= r^5+2r^4-r^4-2r^3-5r^3-10r^2-r^2-2r-6r-12\\
&\quad= (r+2)(r^4-r^3-5r^2-r-6)\\
&\quad= (r+2)(r^4-3r^3+2r^3-6r^2+r^2-3r+2r-6)\\
&\quad= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
expected value, conflicting answers I am having a problem trying to understand why the two methods for finding the expected are resulting in different solutions. I am not sure if I am doing a mistake somewhere. Your help is appreciated
And I was also wondering how I could use the given ( a third method maybe?) $E(Y)$ ... | Method 1 is correct.
In Method 2, it is not true that $P((Y-2/3)^2 \leq w) = P(Y \leq \sqrt{w}+2/3)$ because $Y \leq \sqrt{w}+2/3$ does not imply $(Y-2/3)^2 \leq w$. You have to be careful when taking the square root. What is true is that
\begin{align*}
(Y-2/3)^2 \leq w &\Longleftrightarrow Y-2/3 \leq \sqrt{w} \text{ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Laurent Series for different domains I would like to determine the Laurent series for the function
$$f(z) = \frac{1}{z(z-2)^3}$$
for two different domains
$$\vert{z}\vert<2$$
and
$$\vert{z}\vert>2$$
But, I am unsure when I am computing the Laurent series when I am to take into consideration the different domains. I u... | For $|z|>2$ :
$$f(z) = \frac{1}{z(z-2)^3} = \frac{1}{z[z^3(1-\frac{2}{z})^3]}=\frac{1}{z^4(1-\frac{2}{z})^3}$$
A known geometric series is :
$$\frac{1}{1-w} = \sum_{n=0}^\infty w^n ,\quad |w| < 1$$
From that, we can derive :
$$\frac{1}{(1-w)^3} = \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)w^n, \quad |w|<1$$
So, applying ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{a_{n}} \right)^{a_{n}}=e$ if $\lim\limits_{n\rightarrow\infty} a_{n}=\infty$ What would be the nicest proof of the following theorem:
If $\lim\limits_{n \rightarrow \infty} a_{n} = \infty$, then $\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} ... | For any $a_n$ exists $x\in \mathbb N$ such that $x\le a_n\le x+1$ and
$$\left(1 + \frac{1}{x +1} \right)^{x} \leq \left(1 + \frac{1}{a_n} \right)^{a_n} \leq \left(1 + \frac{1}{x} \right)^{x+1}$$
and
$$\left(1 + \frac{1}{x +1} \right)^{x} =\frac{\left(1 + \frac{1}{x +1} \right)^{x+1}}{1 + \frac{1}{x +1} } \to \frac e 1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2554111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Calculate $\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x}$ Beside using l'Hospital 10 times to get
$$\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} = 3$$ and lots of headaches, what are some elegant ways to calculate the limit?
I've tried to write the functions as powers of $e$ or as power seri... | $$\begin{array}{cl}
& \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2554448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to solve a system of linear equations modulo n? For example,
$4x - 10y \equiv 8\pmod {20}$
$7x + 2y \equiv 5\pmod {20}$
It resembles linear diophantine equations and the Chinese Remainder Theorem, but I don't know how to actually solve it..
| First you can multiply the system by any number that has an inverse, that is $\gcd(x,20)=1$.
So in particular you cannot multiply or divide by $2,4,5,10$ as you would not multiply or divide by $0$ in a normal non-modular system. Well you can do it, but you'll loose equivalence in the way.
For instance here we would lik... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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The order of a modulo p.
If $p$is prime and $\operatorname{ord}_p(a)=4$, then $1+a+a^2+a^3≡ 0\bmod p$, where $\operatorname{ord}_p(a)$ is the order of $a$ modulo $p$.
I think it is true statement
$\operatorname{ord}_p(a)=4$, then $a^4≡1\bmod p$ so $a^4-1≡0\bmod p$
since $a^4-1=1+a+a^2+a^3$, then $a^4-1=1+a+a^2+a^3\e... | $a^4-1 \neq 1+a+a^2+a^3$, but it is true that $a^4-1=(a-1)(a^3+a^2+a+1)=0 \mod p $, and since there are no zero divisiors, either $a=1$, in which case its order cannot be $4$, or, as you conclude: $(a^3+a^2+a+1)=0$
| {
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"question_score": "5",
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Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$,
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$
Also we know that all the expressions are different from zero and defined. Including the expressions we received during the soluti... | Applying sum to product rule on LHS and sine of the sum of 2 angles on RHS
$$2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$$
$$4\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)=6\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha+\beta}{2}\right)$$
$$2\cos \left(\frac{\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Simplify expression $\frac{2\cos(x)+1}{4\cos(x/2+π/6)}$ How to simplify the following expression: $$\frac{2\cos(x)+1}{4\cos\left(\frac x2+\fracπ6\right)}$$
I got to: $ \dfrac{2\cos(x)+1}{4\cos\left(\dfrac x2\right)\cdot \dfrac{\sqrt3}2-\sin(x) \cdot \frac 12}$
| Let $\dfrac{x}{2}+\dfrac{\pi}{6}=y$. Then $x=2y-\dfrac{\pi}{3}$.
\begin{align*}
\frac{2\cos x+1}{4\cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)}&=\frac{2\cos\left(2y-\dfrac{\pi}{3}\right)+1}{4\cos y}\\
&=\frac{2\cos2y\cos\dfrac{\pi}{3}+2\sin2y\sin\dfrac{\pi}{3}+1}{4\cos y}\\
&=\frac{2\cos^2y-1+2\sqrt{3}\sin y\cos y+1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Evaluate the limit $\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$ Can someone please check my conclusion to the evaluation of the following limit?
$$\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$$
I got that the limit is equal to infinity. If limit is equal to infinity does this mean that limit does not exist?
| The limit is not infinity. If you need to do it the hard way, consider
$$
x(\sqrt[4]{16x^4+x^2+1}-2x)
=x\frac{\sqrt{16x^4+x^2+1}-4x^2}{\sqrt[4]{16x^4+x^2+1}+2x}
=\frac{x(x^2+1)}{(\sqrt[4]{16x^4+x^2+1}+2x)(\sqrt{16x^4+x^2+1}+4x^2)}
=\frac{x^3(1+\frac{1}{x^2})}
{x^3
\Bigl(\sqrt[4]{16+\frac{1}{x^2}+\frac{1}{x^4}}+2\Big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Finding $x$ values of $\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $
Solve $$\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $$ for $0^\circ<x<120^\circ$
I simplify it to
$$2 \cos 6x + 2 = 3 + \cos 3x $$
There is $\cos 6x $ and $\cos 3x$ how do I merge them together? To solve the equation from $0^\circ$ to $12... | Hint: Use the identity
$$\cos 2y=2\cos^{2}y-1$$
with $y=3x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I evaluate $\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$
$$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$$
I'm not really comfortable with more than 1 sigma's and that's why this question is confusing me. I don't think it's possible to reduce the number of variables to 1 here.
The answer ... | Since all terms in the sum are non-negative, one can arbitrary change the order of summation without changing the value of the sum. In particular, we can group terms with same value of $k = a + b$ and sum over them first. Notice
$$\sum_{a+b=k, a, b \ge 1}ab
= \sum_{a=1}^{k-1} a(k-a)
= k \frac{(k-1)k}{2} - \frac{(k-1)k(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Taylor series expansion of sin(x) I understand that Taylor series expansion for $\sin(x)$ is derived as follow:
$$
\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-...
$$
Now, what exactly is the first, second, and third term?
Is the first term just $\sin(x) = x$?
Is the second term $\sin(x) = x-\frac{x^3}{3!}$?
| Terms are the members of a summation (whatever the formula) and they are enumerated in the order of their appearance (left to right, starting from first)
*
*$x$ is the first term,
*$-\dfrac{x^3}{3!}$ is the second term,
*$\dfrac{x^5}{5!}$ is the third term.
When speaking of a polynomial, a term is said of the $n^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
I think that the only solutions are $n = 0,1,2,4$, but I have no idea on how to prove it.
I tried to write $3^n-1$ as $1+3+3^2+...+3^{n-1}$ and manipulate the sum but found my self at the equall... | Well $3^n - 1 = (3-1)(1 + 3 + 3^2 + ... + 3^{n-1}) = 2(1 + 3 + 3^2 + ... + 3^{n-1})$
So $2^n|3^n - 1$ if and only if $2^{n-1}|(1 + 3 + 3^2 + ... + 3^{n-1})$.
If $n$ is odd and greater than one $(1+3 + 3^2 + .... + 3^{n-1})$ is odd so we can assume $n$ is even.
Let $n = 2m$ then $2^{2m}|3^{2m} - 1=(3^m -1)(3^m+1)$. So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Find the indefinite integral $\int\frac{dx}{(x^2+2x+5)^2}$ I need help with the indefinite integral
\begin{align}
& \int\frac{dx}{(x^2+2x+5)^2} \\[10pt]
= {} & \int\frac{dx}{((x+1)^2+4)^2} = \int\frac{du}{(u^2+4)^2} & & x+1=u,\quad du=dx \\[10pt]
= {} & \frac{1}{16} \int\frac{du}{(\frac{u^2}{4}+1)^2} \\[10pt]
= {} & ... | If you write $s = \tan(t)$ then you obtain $\sec^2(t)\,\mathrm{d}t = \mathrm{d} s$ and
$$
\int \frac{1}{(1+s^2)^2}\,\mathrm{d}{s}=\int \cos^2(t)\,\mathrm{d}{t}
$$
you can then solve the second integral by using the double-angle formula for cosine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Binomial to the power of five, equality proof I want to find out when the equality $(x+y)^5=x^5+y^5$ for real numbers $y$ and $x$ holds.
Expanding this binomial yields $(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4$
Factoring the right side of the above equality gives $x^5+y^5+5xy(x^3+2x^2y+2xy^2+y^3)$
If $(x+y)^5=x^5+... | Since $$x^3+2x^2y+2xy^2+y^3=(x+y)(x^2-xy+y^2)+2xy(x+y)=$$
$$=(x+y)(x^2+xy+y^2)=(x+y)\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right),$$
we see that $x^3+2x^2y+2xy^2+y^3=0$ for $x+y=0$ only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to proof: $\cos(10°)*\cos(30°)*\cos(50°)*\cos(70°)=\frac{3}{16}$ How to proof: $\cos(10°)*\cos(30°)*\cos(50°)*\cos(70°)=\frac{3}{16}$
I tried to: $\frac{8*3^{1/2}\sin(10°)*\cos(10°)*\sin(40°)*\sin(20°)}{16*\sin(10°)} $ and got nowhere
| $$\begin{align}\cos(10°)\cos(30°)\cos(50°)\cos(70°) &= \sin(80°)\cos(30°)\sin(40°)\sin(20°) &\\=& \cos(30°)\sin(80°)\cdot\dfrac{-1}{2}(\cos(60°) - \cos(20°))&\\=& \cos(30°)\cdot\dfrac{-1}{2}(\sin(80°)\cos(60°) - \cos(20°)\sin(80°))&\\=& \cos(30°)\cdot\dfrac{-1}{2}\left(\dfrac{1}{2}(\sin(140°)+\sin(20°)) - \\\dfrac12(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x)... | Multipying both side by $e^{-2\ln x}=1/x^2$ we get $$y'=\frac{2}{x}y+x^3\Longleftrightarrow (\frac{y}{x^2})'= x\Longleftrightarrow \frac{y}{x^2}=x^2/2+c\\\Longleftrightarrow y(x)= x^4/2+cx^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
H... | Let $a=\sqrt{2+\sqrt3}\,$, $b = \sqrt{2-\sqrt3}\,$, then:
$$\require{cancel}
a^2+b^2 = 2+\bcancel{\sqrt{3}}+2 - \bcancel{\sqrt{3}} = 4 \\
ab = \sqrt{(2+\sqrt3)(2-\sqrt3)} = \sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4-3} = \sqrt{1} = 1
$$
It follows that:
$$(a-b)^2 = a^2+b^2-2ab = 4 - 2 \cdot 1 = 2$$
Since $\sqrt{2+\sqrt3} \gt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 12,
"answer_id": 7
} |
Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then h... | To do this more cleanly but also rigorously, note that given $\epsilon > 0$, it is possible to find $N$ such that $N > \dfrac{1}{\epsilon}$. Then for all $n > N$, we have that $\dfrac{1}{n} < \epsilon$. From here, we can achieve the desired result:
$$
\epsilon > |{\dfrac{1}{n}}| = |\dfrac{2n + 1 - 2n}{n}| = |\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Let $x=(1+i\sqrt 7\;)/(2\sqrt 2\;).$ Since $x\bar x=x^2\bar x^2=1$ and $x+\bar x=1/\sqrt 2\;$ we have $$((1+i\sqrt 7\;)/2)^4+(1-i\sqrt 7\;)/2)^4=4(x^4+ \bar x^4)=$$ $$=4((x^2+\bar x^2)^2-2x^2\bar x^2)=4((x^2+\bar x^2)^2-2)=$$ $$=4( ((x+\bar x)^2-2x\bar x)^2-2)=$$ $$=4((1/\sqrt 2\;)^2-2)^2-2)=4((1/2-2)^2-2)=4(9/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Expand $a^5 + b^5 + c^5$ in terms of Schur polynomials How to expand certain sums of powers in terms of Schur polyomials. I have been gaining proficiency with symmetric polynomials, today I would like to expand:
$$ a^5 + b^5 + c^5 = \sum_{\lambda_1 + \lambda_2 + \lambda_3 = 5} s_\lambda (a,b,c) \tag{$*$} $$
My first t... | The power sum symmetric function $p_n$ is the alternating sum of hooks of size $n$, that is:
$$p_n=\sum_{i=0}^{n-1}(-1)^{i}s_{(n-i,1^{i})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine all p $\in \mathbb{R}$ such that the following integral converges The integral is:
$$\int_0^{+\infty} \frac{(\cos(x) - 1)x^2}{x^p + (x + 1)^6} dx$$
Now, what I did was this:
$$\frac{(\cos(x) - 1)x^2}{x^p + (x + 1)^6} \leq \frac{\cos(x)x^2}{x^p + (x + 1)^6} \leq \frac{\cos(x)x^2}{x^p + x^6} \leq \frac{x^2}{x... | Since
$$ \left|\int_0^{\infty}f(x)\ dx\right| \le \int_0^{\infty} |f(x)|\ dx $$
and $|\cos x-1| = 1-\cos x$, we consider
$$ \int_0^\infty \frac{(1-\cos x)x^2}{x^p + (x+1)^6}\ dx \le \int_0^\infty \frac{2x^2}{x^p + (1+x)^6}\ dx \le \int_0^1 2x^2 + \int_1^\infty \frac{2x^2}{x^p + x^6} $$
The first term clearly converges,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding PDF of $Y = 1 - X^2$, piecewise monotonic Have been banging my head against this question from Casella and Berger for some time, not understanding where my gap in knowledge is here:
Given:
$f_X(x)=\frac38 (x+1)^2 \qquad Y = 1-X^2 $
Find the pdf of $Y$, seems simple enough...
Method: Given that $g(x)$ is not mo... | $$\frac{2-y}{\sqrt{1-y}} = \frac{1}{\sqrt{1-y}} + \frac{1-y}{\sqrt{1-y}} = \frac{1}{\sqrt{1-y}} + \sqrt{1-y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An interesting integral with log: $\int_{0}^{\pi/2}\ln^2\left(\tan^2\left({x\over 2}\right)\right)\mathrm dx=\frac{\pi^3}2$
$$\int_{0}^{\pi/2}\ln^2\left(\tan^2\left({x\over 2}\right)\right)\mathrm dx=\frac{\pi^3}2\tag1$$
This is an interesting integral, I came aross it while I was solving a circuit board problem. My ... | $\newcommand{\Log}{\operatorname{Log}}$First set $x=2\arctan(t)$ so that:
\begin{align}
I:=\int^{\pi/4}_0 \ln^2\left(\tan^2\left(\frac x 2\right)\right)\,dx=8\int^1_0 \frac{\ln^2(t)}{1+t^2}\, dt
\end{align}
Now we set $t=\frac{1}{u}$ so that:
\begin{align}
I=-8\int^1_\infty \frac{\ln^2(u)}{1+u^2}\,du = 8\int^\infty_1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this fo... | Well done! Your answer is correct. To make it a bit more presentable, let us simplify this further to get: $$f’(x) = \frac{\frac{\sqrt{\frac{1-x}{1+x}}}{2}\times \frac{-2x}{x^2-2x+1}}{2}(1-x)$$ $$=-\frac{(1-x)^{\frac32}}{2\sqrt{1+x}(x^2-2x+1)}$$ $$=-\frac{1}{2\sqrt{1-x^2}}$$
EDIT:
Well, you have your derivative calcul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Using the Left-Inverse to "Solve" an Impossible System of Equations I was working with the following system of equations:
$$\begin{split}
\begin{bmatrix}
4 & 0\\
0 & 5\\
0 & 0\\
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\end{bmatrix}
& =
\begin{bmatrix}
1\\
1\\
... | Here is a simpler example of what you are doing. Consider the argument
\begin{align}
2 \times 3 &= 8 \\
\frac 12 \times 2 \times 3 &= \frac 12 \times 8 \\
3 &= 4
\end{align}
If you put $``="$ between two quantities that aren't equal, then you cannot believe any conclusions that may follow.
\begin{equation}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 7,
"answer_id": 3
} |
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.