Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate
$\lim_{n \rightarrow \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know that $1 - \frac {2}{k(k+1)} = \fr... | A formal proof
with product notation.
$lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})
$
$\begin{array}\\
\prod_{k=2}^n (1-\dfrac{2}{k(k+1)})
&=\prod_{k=2}^n \dfrac{k(k+1)-2}{k(k+1)}\\
&=\prod_{k=2}^n \dfrac{k^2+k-2}{k(k+1)}\\
&=\prod_{k=2}^n \dfrac{(k+2)(k-1)}{k(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$ Evaluate:
$$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$$
We let
$$\alpha = \operatorname{arccos} \frac{2}{\sqrt5} \qquad \beta = \operatorname{arccos}\frac{3}{\sqrt{10}}$$
Then we h... | From $\sin(\alpha+\beta)=\frac1{\sqrt2}$, we should expect $|\cos(\alpha+\beta)|=\frac1{\sqrt2}$ from the formula $\sin^2 \theta + \cos^2 \theta = 1$.
\begin{align}
\cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)\color{blue}-\sin(\alpha)\sin(\beta)\\
&=\frac{2}{\sqrt5}\cdot \frac{3}{\sqrt{10}}-\frac1{\sqrt5}\cdot \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Finding the inverse of $A$
Find the inverse of $$A =\left[\begin{matrix}0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{matrix}\right]$$
My attempts:
$$A^{-1} = \frac {\operatorname{adj}A}{\det A}$$
As I can find the $\det A$ that is $\det A = -b$. Here, how can I find the inverse? Is there... | Let
$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} := \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right]$$
Let $\mathrm r^\top := \begin{bmatrix} b & c & d\end{bmatrix}$. From
$$\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Are $1+p^3+p^6$ and $1+p^4+p^8$ coprime? What are the primes $p$ for which $1+p^3+p^6$ and $1+p^4+p^8$ are coprime?
I know it is true for $p=2$ and $p=3$ and not true for any $p \equiv 1 \mod 6$. I conjecture that it true for all primes $p \equiv 5 \mod 6$.
Any counterexample $> 10^8$.
This is relevant to OEIS sequenc... | From comments, therefore CW:
$$ \left( x^{8} + x^{4} + 1 \right) $$
$$ \left( x^{6} + x^{3} + 1 \right) $$
$$ \left( x^{8} + x^{4} + 1 \right) = \left( x^{6} + x^{3} + 1 \right) \cdot \color{magenta}{ \left( x^{2} \right) } + \left( - x^{5} + x^{4} - x^{2} + 1 \right) $$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
$$ 2y + x = ±1 $$ can't be deduced.
Instead ano... | If the slant asymptote exists, we may denote
$$k=\lim_{x \to \infty}\dfrac{y}{x}.$$
Then $$0=\lim_{x \to \infty}\left(1+3\cdot \frac{y}{x}-4 \cdot \frac{y^3}{x^3}-\frac{1}{x^2}+\frac{y}{x^3}+\frac{3}{x^3}\right)=1+3k-4k^3=-(k-1)(2k+1)^2.$$
Thus,$$k=1,-\frac{1}{2}.$$
*
*If $k=1$, we denote $y=x+b$ and put it into th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Linear Algebra, Least squares I'm going through the Linear Algebra course on Brilliant.org, but can't figure out how I should know that $\frac{17}{3}$ is the best value for $c$ to fill in here.
My two questions are:
1) Why do we want to minimize the distance of
$$3c^2-34c+101$$
2) How can you derive where this distan... | You wish to minimize the distance of $c\pmatrix{1 \\ 1 \\ 1} -2 \pmatrix{1 \\ 2 \\ 3}$ to the vector $\pmatrix{5 \\ 0 \\ 0}$.
We have
$$\left\|c\pmatrix{1 \\ 1 \\ 1} -2 \pmatrix{1 \\ 2 \\ 3} - \pmatrix{5 \\ 0 \\ 0}\right\|^2 = \left\|\pmatrix{c-7 \\ c-4 \\ c-6}\right\|^2 = (c-7)^2+(c-4)^2+(c-6)^2 = 3c^2-34c+101$$
To mi... | {
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"url": "https://math.stackexchange.com/questions/2852388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here ... | HINT
We have that
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=$$
$$=\frac{1}{(x+1)}\color{red}{-\frac{1}{(x+2)}+\frac{1}{(x+2)}-\frac{1}{(x+3)}+\frac{1}{(x+3)}-\frac{1}{(x+4)}+\frac{1}{(x+4)}}-\frac{1}{(x+5)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Limit superior of a sequence of oscillating functions related to Chebyshev polynomials Let $n \in \mathbb N$ and consider the polynomial function $f_n \colon \mathbb R \to \mathbb R$ defined by
$$f_n(x) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} (1 - x^2)^{n-k} x^{2k}$$
for any $x \in \mathbb R$. (These functions are r... | This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 \sqrt{x^2 \left(x^2-1\right)} \qquad \text{and}\qquad v=1-2 x^2+2 \sqrt{x^2 \left(x^2-1\right)}$$ a CAS produced
$$f_n(x)=\frac{ \left(u^n+v^n\right)}{2 }+\frac{ \left(u^n-v^n\right)}{2 }\,\frac{\sqrt{x^2 \left(x^2-1\right)}
}{ x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$ Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$
I have done it until the the divisor is of second degree.
But here the degree of the remainder is $4$
This means the remainder will be of the form. : $ax^3+bx^2+cx+d$ which makes ... | Using Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd,
$$x^{10}+1\equiv0\pmod{x^2+1}$$
Again $x^3-1=(x-1)(x^2+x+1)$
$$x^{10}+1= x(x^9-1)+x+1\equiv x+1\pmod{x^2+x+1}$$
Now apply CRT
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Roots of $f$ and $f'$ for $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$ I have this question from an admission exams. Given that $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively ... | According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_{n-1}x^{n-1}+\cdots + a_0\\
p'_n(x) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2}+\cdots+a_1
$$
now
$$
S_{100}(x) = -\frac{a_{100}}{a_{101}}\\
S'_{100}(x) = -\frac{(101-1)a_{100}}{101 a_{101}}
$$
Here
$$
a_{101} = \frac{(-1)^{101}}{101!}\\
a_{100} = -a_{101}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to calculate the indefinite integral of $\frac{1}{x^{2/3}(1+x^{2/3})}$?
$$\int \frac{dx}{x^{2/3}(1+x^{2/3})}.$$
I substituted,
$$t=\frac{1}{x^{1/3}}$$
$$\frac{dt}{dx} = -\frac{1}{3x^{4/3}}$$
$$\frac{dt}{dx} = -\frac{t^4}{3}$$
Rewriting the question,
$$\int \frac{dx}{x^{2/3}+x^{4/3}}$$
$$-\frac{1}{3} \int \frac{dt... | $t=\sqrt[3]{x},\qquad dt=\frac{1}{x^{2/3}}dx$
$$\int \frac{1}{x^{\frac{2}{3}}+x^{\frac{4}{3}}}dx=3\int \frac{1}{t^2+1}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$.
Let $f(x)=x^3+ax^2+bx+c$ be a cubic polynomial with real coefficients and all real roots, also $|f(i)|=1$ where $i=\sqrt{-1}$. Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$.
As $f(i)=-i-a+ib+c=1$ and $f(i)=-i-a+i... | Alternatively, note that for $z=a+bi\in \mathbb Z, \bar{z}=a-bi \in \mathbb Z$, the norm is:
$$|z|=\sqrt{z\cdot \bar{z}}=\sqrt{a^2+b^2}.$$
So:
$$|f(i)|=|c-a+(b-1)i|=\sqrt{(c-a)^2+(b-1)^2}=1 \Rightarrow (c-a)^2+b^2-2b=0 \tag{1}$$
Let $x_1,x_2,x_3$ be the roots of $f(x)=0 \iff x^3+ax^2+bx+c=0.$
By the Vieta's formulas:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
T... | You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $\frac{1}{2}\sqrt{2}(1-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}-1}{2}$.
So the area of the polyg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Locus of a vertex of a triangle inside an equilateral one, under an integer constraint Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $F$ and $G$ on the side $AB$.
The... | I don't think you will find a nice description of the curve without using coordinates.
Let $A = (-1, 0)$, $B = (1, 0)$ and $D = (x, y)$. Then it is easy to prove that
$$\alpha = 2 - \sqrt{(x - 1)^2 + y^2} \qquad \beta = 2 - \sqrt{(x + 1)^2 + y^2}$$
$$\gamma = \sqrt{(x + 1)^2 + y^2} + \sqrt{(x - 1)^2 + y^2} - 2$$
Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Reduction of Order Leads to Non-Elementary Integral
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
\begin{align}
\frac{dw}{dx}(x^2+x)-w(x^2+1)&=0 \\
... | Define the operators $D$ and $X$ by $(D\,h)(x):=h'(x)$ and $(X\,h)(x):=x\,h(x)$. For any function $\phi$, we also define the operator $\phi(X)$ to be $\big(\phi(X)\,h\big)(x):=\phi(x)\,h(x)$. Observe that
$$D^2-\left(\frac{X+1}{X}\right)\,D+\frac{1}{X}=\left(D-1-\frac{1}{X}+\frac{1}{X+1}\right)\,\left(D-\frac{1}{X+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| Another solution
\begin{align*}
\sqrt {49 - x^2} - \sqrt {25 - x^2} &= 3\\
\sqrt {49-x^2} &= 3+ \sqrt {25-x^2}\\
49-x^2 &= 25 -x^2 + 9 + 6\sqrt {25-x^2}\\
15 &= 6 \sqrt {25 - x^2}\\
25 - x^2 &= \frac {25} 4\\
x^2 &= \frac {75}4\\
x &= \pm \frac 5 2 \sqrt 3.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $p$ and $q$ such that the maximum and minimum values of $5 +6\cos\theta + 2 \cos2\theta$ satisfy $x^2-p x+q=2$
If the maximum and minimum values of $5 +6\cos\theta + 2 \cos2\theta$ satisfy $x^2-px + q = 2$, then what are $p$ and $q$?
My thinking that, maximum valus of $\cos\theta$ is $1$ and minimum is $-1$. No... | The minimum value is wrong. Let $f(t)=5+6t+2(2t^{2}-1)$. [Recall that $\cos (2x)=2\cos^{2}x-1$]. Therefore, $f'(t)=6+8t=0$ when $t=-3/4$ which is a possible value for $\cos x$. So the minimum value is $5+6(-3/4)+2(9/8-1)=3/4$. The quadratic with leading coefficient $1$ and roots $3/4$ and $13$ is given by $(x-\frac 3 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{... | By C-S
$$12=a^2+b^2+c^2=\frac{1}{2}(1^2+1^2)(a^2+b^2)+c^2\geq\frac{1}{2}(a+b)^2+c^2=\frac{1}{2}(2-c)^2+c^2,$$
which gives $$3c^2-4c-20\leq0$$ or
$$(3c-10)(c+2)\leq0$$ or
$$-2\leq c\leq\frac{10}{3}.$$
Now, we get $$\frac{10}{3}-(-2)=\frac{16}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculating the series $1/8+1/88+1/888+....$ I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=\dfrac18\displaystyle\sum_{k=0}^\infty\dfrac{1}{\sum_{n=0}^k10^n}$
where $$\displaystyle\sum_{n=0}^k10^n=\dfrac{10^{k+1}-1}{9}$$
then
$S=\dfrac98\displaystyle\sum_{k=0}^\infty\dfrac{1}{10^{k... | $$\frac{1}{8}+\frac{1}{88}+\frac{1}{888}+\dotsm = \frac{1}{8}\left( 1+\frac{1}{11}+\frac{1}{111}+\dotsm \right) =
\frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} \iff x= \frac{1}{10}$$
Where $\sum_{n=1}^{\infty} \frac{x^n}{1-x^n}$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Number of solution of $x^4-5x^3+(\lambda+2)x^2-5x+1=0$
Consider the bi-quadratic equation $E:x^4-5x^3+(\lambda+2)x^2-5x+1=0$ then, the real values of $\lambda$ so that $E$ has four different solutions is?
My attempts:
As $x=0$ is not a solution for any $\lambda$ hence we divide by $x^2$,\begin{align}x^2+\dfrac{1}{x^2... | Hint: Draw a graph $$f(x)= {-x^4+5x^3-2x^2+5x-1\over x^2}$$
and see for which $\lambda$ the line $y= \lambda$ cuts the graph of $f$ four times.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find no. of non negative integer solutions of $a+2b+3c = 200$ I want to find number of the non negative integer solutions of $a+2b+3c = 200$
(answer can contain $a,b,c = 0$)
Example one of the solutions is $a = 100, b = 50, c = 0$
I used stars and bars trick and here i get is
$\dfrac{\dbinom{200+2}{2}}{2 \times 3}$ but... | Consider $x:=a+b+c$, $y:=b+c$, and $z:=c$. Let $n:=200$. Then, $x\geq y\geq z\geq 0$ and $x+y+z=n$. Let $$T_n:=\big\{(x,y,z)\in\mathbb{Z}_{\geq 0}^3\,|\,x+y+z=n\big\}\,.$$ Then, the symmetric group $S_3$ of order $3!=6$ acts on $T_n$ by permuting the three entries of each element of $T_n$. The number of $S_3$-orbi... | {
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"url": "https://math.stackexchange.com/questions/2869279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Showing $\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$
I wish to show that if $z$ is real, then
$$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$
I have shown this result, although my inequality is the wrong way around.
I considered
\begin{align}
|z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(trian... | We have
$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{\left|e^{iz}\right|}{\left|z^2+1\right|}= \frac{1}{\left|z^2+1\right|}$$
and
$$0\le\left|{z^2+1}\right|= |z|^2+1$$
therefore the result follows.
| {
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"url": "https://math.stackexchange.com/questions/2870910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2... | Thanks everyone i had the answer for my stuck.
We have : $$LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$$
Or $$-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$$
Or $$\left(a-b\right)^2\left(\frac{1}{2\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Positive integers have the form $a^2+b^2+c^2+2^d$ For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that
$$n=a^2+b^2+c^2+2^d.$$
Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers
$$m=a^2+b^2+c^2$$
if and only if $m$ is n... | We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $x\in\{1,2,4\}$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $\equiv 0$ or $\equiv 3\bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $\bmod 4$. So, it only remain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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How many zeros does the polynomial have in the right half plane? The polynomial is $f(z) = z^4+\sqrt{2}z^3+2z^2-5z+2$
If you check the image of the imaginary axis, you see that there are no zeros, so we can use the right semicircle from $iR$ to $-iR$,and make $R$ arbitrarily big. The problem now is picking the right f... | We can use the following way.
For all real $k$ we have:
$$z^4+\sqrt2z^3+2z^2-5z+2=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\left(2k-\frac{3}{2}\right)z^2+(\sqrt2k+5)z+k^2-2\right).$$
Now, let $2k-\frac{3}{2}>0$ and $$(\sqrt2k+5)^2-4\left(2k-\frac{3}{2}\right)(k^2-2)=0.$$
We see that $k=2.65...$ is valid.
Id est, f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integral $\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$ I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then:
$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac... | Work in progress.
Since $\arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}$ for any $x\in[-1,1]$ and
$$ \int_{0}^{\pi/2}\left(\sin x\right)^{n+1/2}\,dx=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{n}{2}+\frac{3}{4}\right)}{\Gamma\left(\frac{n}{2}+\frac{5}{4}\right)} $$
we "just" need an explicit val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$
My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
| We have
$$(y^3+2y+1)-(y-1)^3=3y^2-y+2=\frac{(6y-1)^2+23}{12}>0$$
and
$$(y+2)^3-(y^3+2y+1)=6y^2+10y+7=\frac{(6y+5)^2+17}{6}>0\,.$$
That is,
$$(y-1)^3<y^3+2y+1<(y+2)^3$$
for all $y\in\mathbb{Z}$. If $y^3+2y+1$ is a cube, then either $y^3+2y+1=y^3$ or $y^3+2y+1=(y+1)^3$, which gives $y=-\dfrac12$, $y=-\dfrac13$, or $y=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 5
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A functional equation in a TST exam Find all functions $f:\Bbb {R} \rightarrow \Bbb {R} $ such that:
$$f(f(xy-x))+f(x+y)=yf(x)+f(y)$$
Source: 2018 Hong Kong TST 2 problem 3
I recently proved that $f(x+1)=f(x)+f(1).$
| To recap your result: setting $x=y=0$ shows us that
\begin{align*}
f(f(0))+f(0)&=0\cdot f(0)+f(0) \\
f(f(0))+f(0)&=f(0) \\
f(f(0))&=0.
\end{align*}
Now from the original equation we set $y=1$ to obtain
\begin{align*}
f(f(0))+f(x+1)&=f(x)+f(1) \\
f(x+1)&=f(x)+f(1).
\end{align*}
Let us set $x=0$ in this equation to obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Linear Transformation of a Polynomial I have an operation that takes $ax^2+bx+c$ to $cx^2+bx+a$. I need to find if this corresponds to a linear transformation from $R^3$ to $R^3$, and if so, its matrix.
I know that
$$
ax^2+bx+c = \begin{bmatrix}a&b&c \end{bmatrix} \begin{bmatrix}x^2\\x\\1 \end{bmatrix}
$$
If I perform ... | $$T((ax^2+bx+c)+(a'x^2+b'x+c'))=T((a+a')x^2+(b+b')x+(c+c'))
\\=(c+c')x^2+(b+b')x+(a+a')
\\=(cx^2+bx+a)+(c'x^2+b'x+a')
\\=T(ax^2+bx+c)+T(a'x^2+b'x+c')$$ and linearity is confirmed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What method is most preferred to find nature of the series $\sum_{n=1}^{\infty} \frac{a^n}{a^n+x^n}$? I tried to use D'Alembert's ratio test to test the convergence of the given series $\sum\limits_{n=1}^{\infty} \frac{a^n}{a^n+x^n}$ but I could only solve for $\frac{a}{x} <1$ and it is coming out to be converging but... | $$\frac{a^n}{a^n + x^n} = \frac{1}{1 + (\frac{x}{a})^n}$$
Ratio test:
$$\frac{\frac{a^{n+1}}{a^{n+1} + x^{n+1}}}{\frac{a^n}{a^n + x^n}} = a \frac{a^n + x^n}{a^{n+1} + x^{n+1}} = \frac{1 + (\frac{x}{a})^n}{1 + (\frac{x}{a})^{n+1}}$$
*
*If $\vert \frac{x}{a} \vert \leq 1$, the above limits goes to $1$. And the ratio ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$ Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$
I tried:
$x=\sin\alpha$ and $y=\cos\beta$
$\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}$ but if I write $\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta... | \begin{align}
& \left| xy-\sqrt{(1-x^2)(1-y^2)}\,\right| \\[10pt]
= {} & \left| \sin\alpha\cos\beta - \left| \cos\alpha\sin\beta \right| \, \right| \\[10pt]
= {} & \begin{cases} \left| \sin\alpha\cos\beta - \cos\alpha \sin\beta\right| = \left|\sin(\alpha-\beta){} \right| & \text{if } \cos\alpha>0\ \&\ \sin\beta>0, \\[5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Taking $4$ shots or $6$?
Suppose your skill (the ability to make a shot) follows a binomial distribution (with some unknown $p$). You need to make at least half of the shots. Would you rather take $4$ or $6$? More precisely, what should be the $p$ which is the threshold of your decision?
My Try: assuming $0 < p < 1$.... | Probability of making AT LEAST half shots out of 6 is
$$
\binom{6}{3}p^3(1-p)^3 +\binom{6}{4}p^4(1-p)^2 +\binom{6}{5}p^5(1-p) +\binom{6}{6}p^3
$$
and for making at least 2 out of 4 is
$$
\binom{4}{2}p^2(1-p)^2 +\binom{4}{3}p^3(1-p) +\binom{4}{4}p^4
$$
For threshold $p$, equating both, we get:
$$
10p^4-36p^3+48p^2-28p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that t... | This is just a start, not a solution, but I can't fit it in a comment. Maybe it's enough for you to answer the question yourself. We can think of the problem this way. Partition $[m] = \{1,2,\dots\}$ into two sets $A$ and $B$ and let $\sum A$ and $\sum B$ be the sum of the elements of $A$ and $B$ respectively. Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Show that $8$ is an eigen value of $A$ Consider the following matrix:
$$A=\begin{bmatrix}B&&&C\\D&&&F\end{bmatrix}$$ where
$$B= \begin{bmatrix} 9 &1&1&1&1\\1&9&1&1&1\\1&1&9&1&1\\1&1&1&9&1\\1&1&1&1&9\end{bmatrix}$$ and $$C=\begin{bmatrix}1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\end{bmatrix}$$
and $$D=C^T$$... | Hint:
*
*Examine the first two rows of $A-8I$, they are identical, what can you conclude.
Same result holds if you replace $D$ and $F$ with other matrices of the same size.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?
I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it... | Yes, that question helps a lot! Note that by AM-GM inequality $a^2+b^2+c^2\ge 3(a^2b^2c^2)^{1/3}=3$. Since $x^2\geq x$ for all $x\geq 1$, it follows that
$$(a^2+b^2+c^2)^2\geq a^2+b^2+c^2\geq a+b+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do I eliminate $x$ and $y$ from the system $x^2 y= a$, $x(x+y)= b$, $2x+y=c$ to get a single equation in $a$, $b$, $c$? Alright, a homework problem.
I'm stuck at this question,
Eliminate $x$ and $y$ from the given equations to get a single
equation in terms of $a$ , $b$ and $c$
$$\begin{align}
x^2 y &= a \\
x(x... | Combine the first and second equations to eliminate $y$:
$x(x+\frac{a}{x^2})=b$
which tidies to
$x^2+\frac{a}{x}=b$ ....... (1)
combine the second and third equations to eliminate $y$:
$x(x+c-2x)=b$
which tidies to
$x^2-cx+b=0$
solving this quadratic gives
$x=\frac{c \pm \sqrt{c^2-4b}}{2}$
which you can substitute into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Show that $0$ has multiplicity $3$ in $M-4I$? Question from an exam:
Consider the matrix $M=$
\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3\end{bmatrix}
Show that $4$ is an eigen value of the above matrix with multiplicity $3$.
I considered $M-4I$ from which I got
\begi... | Since $M$ is symmetric, the geometric and algebraic multiplicity of its eigenvalues coincide.
If you finish your row reduction, you get
$$
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
\xrightarrow{\ \ \ \ \ \ \ \ \ \ }
\begin{bmatrix} 1&1&1&0&0&-1\\0&0&0&1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\left|\operatorname{Re}(3+i+2\bar z^2 -iz)\right| \le 6$, when $|z| \le 1$ So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y... | Note that $\operatorname{Re}(3+i+2\overline z^2-iz)=\operatorname{Re}(3+2\overline z^2-iz)$ and that therefore\begin{align}\bigl\lvert\operatorname{Re}(3+i+2\overline z^2-iz)\bigr\rvert&\leqslant\bigl\lvert3+2\overline z^2-iz\bigr\rvert\\&\leqslant3+2|z|^2+|z|\\&\leqslant6,\end{align}since $|z|\leqslant1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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In how many ways can $5$ teaching posts be filled if $2$ posts are reserved for American candidates? For the post of $5$ teachers, there are $23$ applicants, $2$ posts are reserved for American candidates and there are $7$ American candidates among the applicants. In how many ways can the selection be made ?
Answer in ... | this is because the no of ways you can select 2 American candidates out of 7 is 7C2-------M
and the number of ways you can select the rest is (23-7)C(5-2)-------------N
Thus, total ways = M*N
= 7C2 * 16C3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Closed form solution for this integral? I'm hitting a road block in finding an expression (closed form preferably) for the following integral:
\begin{equation}
\int^{+\infty}_0 x^b \left ( 1-\frac{x}{u} \right )^c \exp(-a x^3) dx
\end{equation}
where $a,b$ are positive constants; $b>1$ is an odd multiple of $0.5$, whil... | An integral of a product of two Meijer G-functions of rational powers of the argument gives a G-function:
$$\int_0^\infty x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx = \\
\int_0^u x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx +
(-1)^c \int_u^\infty x^b \left( \frac x u - 1 \right)^c e^{-a x^3} dx = \\
\Gamma(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } ... | In order to find an explicit value for the given series, we may recall that for any $|x|<\frac{1}{e}$ we have
$$ \sum_{n\geq 1}\frac{n^{n-1}}{n!} x^n = -W(-x) $$
with $W(x)$ being the inverse function of $x e^x$ (Lambert's function), by Lagrange inversion theorem.
A rapid sequence of formal manipulations:
$$ \sum_{n\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Does $x^2+3x^4+o(x^4)=x^2+o(x^3)$ hold true? My textbook asks if the following holds true or false for $x\rightarrow0$:
$$x^2+3x^4+o(x^4)=x^2+o(x^3)$$
I think I understand what the little-o notation means. For $x\rightarrow x_o$ we have:
$$\lim_{x\rightarrow x_0}{\frac{f(x)}{g(x)}=0}$$
and we say $f(x)=o(g(x))$.
So to ... | Recall that by definition
$$f(x)=o(g(x)) \iff f(x)=g(x)\cdot \omega(x) \quad \omega(x) \to 0$$
therefore as $x \to 0$ we have
$$x^2+3x^4+o(x^4)=x^2+3x^4+x^4\cdot \omega(x)=$$$$=x^2+x^3(3x+x\cdot \omega(x)))=x^2+x^3\cdot \omega_1(x)=x^2+o(x^3)$$
but for the limit in your example the key point is that
$$o(x^{n+1})=x\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Value of $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$ Is there a closed form expression of
$$\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$$
, where $\{x\}=x-\lfloor x \rfloor$ denotes the fractional part of $x$? $\text{I think this limit will converge to a point between $0$ and $... | By periodicity, $\log_2 (k)$ and $\log_2 (k) - n$ has the same fractional part. Hence
\begin{align*}
\lim_n \frac 1{2^n} \sum_1^{2^n} \{\log_2(k)\} &= \lim_n \frac 1{2^n} \sum_1^{2^n} \left\{\log_2 \left(\frac k{2^n}\right) \right\} \\&= \lim_n \frac 1{2^n} \sum_1^{2^n} \log_2 \left( \frac k{2^n }\right) - \left\lfloo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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An optimization problem in Euclidean Geometry - finding "the smallest" inscribed triangle What is the smallest perimeter of a triangle that can be inscribed in a triangle with sides of lengths $5$, $9$, and $10$?
I have read that such a triangle has its vertices at the feet of altitudes from the three vertices of the g... | Suppose we have a triangle $\Delta ABC$ with $|AB| = c$, $|BC| = a$ and $|AC| = b$. Let $E$ be the foot lying on $AB$, so that $AB \perp EC$. Without loss of generality, assume $b \ge a$. Then
\begin{align*}
|AE|^2 + |EC|^2 &= |AC|^2 = b^2 \\
|BE|^2 + |EC|^2 &= |BC|^2 = a^2 \\
|AE| + |BE| &= |AB| = c
\end{align*}
Subtr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} ... | You incorrectly arrive at
$$
\frac{4(x^2-x-2)}{(x^2+2)(x+4)}\ge0
$$
which should be instead
$$
\frac{4(x^2-x-2)}{(x^2+2)(x+4)}\le0
$$
Moreover you do the “illegal” step of removing the denominator, which changes sign at $-4$.
The factors $4$ and $x^2+2$ can be removed because they're positive. Factoring the numerator ... | {
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"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Proving $-\cos\alpha = \sin(\alpha-90^\circ)$ from $\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$ $$-\cos\alpha = \sin(\alpha-90^\circ)$$ but
$$\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$$
How does this work out?
$$\begin{align}
\sin\alpha\cdot\cos\beta &=\phantom... | Write
$$\sin(\alpha - 90) = \sin\alpha\cos 90 - \sin 90\cos\alpha $$ $$= \sin \alpha \cdot 0 - 1\cdot\cos \alpha = -\cos \alpha.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\int_0^{\infty} \frac{x}{\sinh(\sqrt{3}x)} dx$ I was asked, by a high school student in the UK, how to calculate the following integral:
$$\int_0^{\infty} \frac{x}{\sinh(\sqrt{3}x)} dx.$$
It has been a long time since I have done any calculus and most of my immediate thoughts used mathematics that he is unl... | Smooth change of variables just to write it in a more "elegant" way.
$$\sqrt{3}x = y$$
The measure becomes $\text{d}y = \sqrt{3}\text{d}x$ and remember that $y = \sqrt{3}x$ which then combines with the other square root. Hence the integral becomes
$$\frac{1}{3}\int_0^{+\infty} \frac{y\ \text{d}y}{\sinh(y)}$$
Now:
$$\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Given that $m$ is a real number not less than $-1$ The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.
Here's what I've tried:.
Using Vieta's formulas:
$rs = m^2-3m+3... | It is correct:
$$(r+s)^2 - 2rs = 6 \iff \\
(-2(m-2))^2-2(m^2-3m+3)=6 \iff \\
(4m^2-16m+16)-2m^2+6m-6=6 \iff \\
2m^2-10m+4=0.$$
And the roots are:
$$m_{1,2}=\frac{5\pm \sqrt{17}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the Maximum of the Chi-Square density function Below is a problem which I did and I believe my answer is correct. I would like
somebody to confirm that (if true) and provide some additional comments about my style.
Thanks,
Bob
Problem:
Show that $\chi^2_v$ pdf has a maximum at $v - 2$ if $v > 2$.
Answer:
The $\... | What you did makes a lot of sense, but what if it $v-2$ was a saddle point?
A better way to see it is using the double derivative test, i.e.
\begin{equation}
f'(x) =
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving Homogenous First Order ODE Using Substitution Problem
Find the general solution of the following homogeneous equation.
$$ ty' = y + \sqrt { t^2 - y^2} $$
Attempt
I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, whi... | Your substitution is fine.
$$ty' = y + \sqrt { t^2 - y^2}$$
$$ty' - y = \sqrt { t^2 - y^2}$$
$$\frac {ty' - y}{t^2} = \frac 1t \sqrt { 1 - (\frac yt)^2}$$
$$ \left (\frac yt \right )'= \frac 1t \sqrt { 1 - (\frac yt)^2}$$
$$ \int \frac {d\left (\frac yt \right )}{\sqrt { 1 - (\frac yt)^2}}= \int \frac {dt}t $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the relation between coefficients of quadratic equation
My attempt:
I could prove $|c| < 1$.
As it given that $f\lt 1$, so $f(0)\lt 1$.
I have solved using triangular inequality. Is there any other way?
ans is ABCD
| On the boundaries $x=0$ and $x=1$ the inequality corresponds to
*
*$x=0 \implies |c|\le 1$
*$x=1 \implies |a+b+c|\le 1 \implies 0\le|a+b|\le 2$
Let wlog $a \ge 0$, since the extrema values are also reached at $x=-\frac{b}{2a}$ we need to consider $3$ cases:
$1) \quad 0\le-\frac{b}{2a}\le1 \iff -2a\le b\le 0$
*
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using Gram-Schmidt to find the QR decomposition I'm having problems doing a QR decomposition of a matrix...
Let $A=\begin{bmatrix}
{1} & {1} & {0} \\
{0} &{1} &{1} \\
{1} & {0} &{1}
\end{bmatrix}$
Find the QR decomposition for A
Here's what I've been doing:
I choose this basis, $B=\left \{(1,0,1), (1,1,0), ... | You're computing $u_2$ wrongly.
I find it useful to set up a systematic way, where the information can be picked up easily.
Let $v_1$, $v_2$ and $v_3$ be the three columns of $A$.
GS1
$u_1=v_1$
$\langle u_1,u_1\rangle=2$
GS2
$\alpha_{12}=\dfrac{\langle u_1,v_2\rangle}{\langle u_1,u_1\rangle}=\dfrac{1}{2}$
$u_2=v_2-\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ ... | Your mistake is to believe that $\theta$ is the polar argument. It is not, because
$$\tan\phi=\frac yx=\frac ba\tan \theta\ne\tan\theta.$$
You can fix by taking the differential
$$(\tan^2\phi+1)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$
and substituting
$$\left(\frac{b^2}{a^2}\tan^2\theta+1\right)\,d\phi=\frac ba(\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$ Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$
My working:
Let $f(x)=\frac{ax^2+x-2}{a+x-2x^2}=y$
therefore,
$$(a+2y)x^2+(1-y)x-(2+ay)=0$$
Now, for $x$ to have real values, the discr... | So, we need $$(14+4a^2)^2\ge4(1+8a)^2$$
$$\iff(4a^2+14-2-8a)(4a^2+14+2+8a)\ge0$$
$4a^2+14+2+16a=4(a^2+4a+4)=4(a+2)^2\ge0$
So,the sufficient condition is $4a^2+14-2-16a=4(a^2-4a+3)=4(a-1)(a-3)\ge0$
$\implies$ either $a\ge$max$(1,3)$ or $x\le$min$(1,3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$ To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is
\begin{align}
\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T
\end{align}
in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In ... | Assume $a\ne 0$, We know that
$$\det(A+uv^T)=(1+v^TA^{-1}u)\det(A)$$
Here $A=aI, u = be, v=e$
\begin{align}\det(aI+bee^T)&=(1+be^T(aI)^{-1}e)\det(aI) \\
&=\left(1+\frac{bu}a\right)a^u\\
&=a^{u}+bua^{u-1}\end{align}
If $a=0$ and $u>1$, then the determinant is $0$.
If $a=0$ and $u=1$, then the determinant is $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
$ \frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2}$. Where can I find some more series of this class? After reading Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ and noodling around on wolfram alpha, I discovered
$$ \begin{align}
&\coth(x \pi)=\frac{x}{\pi... | There is some nice things to see here:
$$\tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}}$$
$$\Big(\tanh(x \pi) \Big)'= \Big( \frac{1}{\pi}\sum_{n=-\infty}^\infty{\frac{4x}{(2n+1)^2+4x^2}} \Big)'$$
$$ \pi\operatorname{sech}^2(\pi x)=\frac{1}{\pi}\sum_{n=-\infty}^{\infty}\frac{4\left(4n^2+4n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Coefficient of a generating function I have the following generating function,
$$f(x)=(1+x+x^2+x^3)^4$$
I want to find the $[x^5]$ coefficient, to do so I wrote,
$$[x^5]f(x)=[x^5](1+x+x^2+x^3)^4=[x^5]\left(\frac{1-x^4}{1-x}\right)^4=
[x^5](1-x^4)^4(1-x)^{-4}
$$
I am wondering how to proceed from here?
|
We obtain
\begin{align*}
\color{blue}{[x^5]}&\color{blue}{(1-x^4)^4(1-x)^{-4}}\\
&=[x^5]\sum_{j=0}^\infty\binom{-4}{j}(-x)^{j}(1-x^4)^4\tag{1}\\
&=\sum_{j=0}^5\binom{3+j}{3}[x^{5-j}](1-x^4)^4\tag{2}\\
&=\sum_{j=0}^5\binom{8-j}{3}[x^{j}]\sum_{k=0}^4\binom{4}{k}(-1)^kx^{4k}\tag{3}\\
&=\binom{8}{3}\binom{4}{0}-\binom{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Consider the equation $a_n +b_n\sqrt{2} = (1+\sqrt{2})^n$ where $a_n, b_n \in \mathbb{Z} \ge 1$. Prove that $\gcd(a_n, b_n) = 1$. Consider the equation $a_n +b_n\sqrt{2} = (1+\sqrt{2})^n$ where $a_n, b_n \in \mathbb{Z} \ge 1$. Prove that $\gcd(a_n, b_n) = 1$.
I know that $(1+2\sqrt{3}) = 1^3 + 3(1)^2(\sqrt{2})^2 + 3(1... | We have
$$
\pmatrix{a_{n+1} \\ b_{n+1}}
=
\pmatrix{1 & 2 \\ 1 & 1}
\pmatrix{a_n \\ b_n}
$$
and so
$$
\pmatrix{a_n \\ b_n}
=
\pmatrix{1 & 2 \\ 1 & 1}^n
\pmatrix{a_0 \\ b_0}
=
\pmatrix{1 & 2 \\ 1 & 1}^n
\pmatrix{1 \\ 0}
$$
Therefore, $\pmatrix{a_n \\ b_n}$ is the first column of $\pmatrix{1 & 2 \\ 1 & 1}^n$. Write
$$
\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16... | A similar formula for $x^3$ is $$x^3=(x-1)(x^2+x+1)+1$$
You can extend this to higher powers
$$x^n= (x-1)(x^{n-1}+x^{n-2}+...+1)+1$$
Which is the basis for the sum of geometric series formula.
$$1+x+x^2+...+x^{n} = \frac {x^{n+1}-1}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
A specific homogeneous polar differential equation In an assignment of our school, we are asked to solve
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y - 3}{x - y - 1}$$
by turning it into a homogeneous polar differential equation (equation of the form $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = F\left(\frac{y}{x... | Attached a plot showing the solutions
$$
S_1\to 2 \tan ^{-1}\left(\frac{x+y-3}{x-y-1}\right)=\log \left(\frac{x^2-4 x+y^2-2 y+5}{2 (x-2)^2}\right)+2 \log (x-2)+C_0\\
S_2\to \tan ^{-1}\left(\frac{y-1}{x-2}\right)-\frac{1}{2} \log \left(\frac{(y-1)^2}{(x-2)^2}+1\right)=\log (x-2)+C_1
$$
with $C_0=0, C_1 = -1.13$
I hope t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$ I was solving questions on beta and gamma functions and then I came across this question.
$$ \int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx $$
Generally in questions of beta functions integrals of the type $\int x^n(1−x^m) \, dx$ can... | Apply the substitution $x^4 = \frac{1-u}{1+u}$. Then
$$ I := \int_{0}^{1} \frac{(1-x^4)^{3/4}}{(1+x^4)^2} \, dx
= \frac{1}{2^{9/4}} \int_{0}^{1} u^{3/4}(1-u)^{-3/4} \, du
= \frac{1}{2^{9/4}} B(\tfrac{7}{4},\tfrac{1}{4})
= \frac{3\pi}{2^{15/4}}.$$
Explanations. First substitute $x^4 = t$ to obtain $ I = \frac{1}{4} \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by... | First $x^2 \le 8$. Second, continuing from what you have written, i.e.,
$$\frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x,$$
we have
$$\frac{-17}{\sqrt{8-0^2}+\sqrt{25-0^2}} \ge \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}} \ge x \implies x \le \frac{-17}{\sqrt{8}+5} \implies x^2 > 8.$$
Contradiction! So no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $n^3+(n+1)^3>(n+2)^3$ Here is the question: for which natural numbers do the following inequalities hold true? State a claim and prove
it:
$n^3+(n+1)^3>(n+2)^3$.
I think this statement holds when $n\ge6$, where $n^3+(n+1)^3=559$, and $(n+2)^3=512$.
I thought about using induction, but I couldn't express $(n+1)^3+... | Need to show:
$n^3\gt \dfrac{(n+2)^3-(n+1)3}{1}=$
$=3(t^2)$, $t \in (n+1,n+2)$
(MVT)
Hence:
$3(n+2)^2 \gt (n+2)^3-(n+1)^3$.
For $n \ge 6:$
$n^3 > 3(n+2)^2$.
Since:
The inequality is true for $n=6$.
$f(x):= x^3 -3(x+2)^2$.
$f'(x) = 3x^2 -6(x+2) =$
$3(x^2-2x -4) = $
$3[( x-1)^2 -5] >0$, for $x \ge 6$,
hence $f(x)$ is str... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What are the strict local maximisers of $x^T A x$ in the simplex? I am dealing with a matrix in this form.
$A=\begin{pmatrix}
0&a & a & a & c & c &c & c\\
a& 0& a &a & c& c& c& c\\
a& a &0 &a & c& c & c & c\\
a& a& a &0 & c &c &c &c \\
c& c & c & c & 0& b& b &b \\
c& c & c & c& b&0 & b & b\... | Consider the gradient of $x^{T}Ax$. Since $A$ is symmetric, the gradient is $2Ax$. Let $x = [x_1\ x_2\ x_3\ldots x_7\ x_8]^{T}$, $l = x_1 + x_2 + x_3 + x_4$ and $r = x_5 + x_6 +x_7 + x_8$, then the gradient is actually (stupid notation..
$$[2(a(l - x_i) + cr)_{i=1, 2, 3, 4} \ 2(cl + b(r - x_i))_{i = 5, 6, 7, 8}]^T$$
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate the integral with and without Green's theorem Evaluate the line integral $\oint_C y^2dx + xdy$ when $C$ has the vector equation $\alpha(t)=(2\cos^3t)i+(2\sin^3t)j$, $0 \leq t \leq 2\pi$.
My attempt
BY USING GREEN's THEOREM,
i.e., $P=y^2$, $Q=x$
we get $\iint_c 1-2y dxdy$
Now putting $x=r\cos^3t$ and $y=r\sin^... | Your second method
Use $$\int_0^{2\pi}\sin^n x\ dx=\dfrac{2\pi}{2^{2k}}{2k\choose k}$$
for even $n=2k$ and for odd $n$ it is zero, then
$$-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t (1-\sin^2t) dt + 12 \int_0^{2 \pi} (1-\sin^2t)^2\sin^2t dt$$
$$-24\int_0^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution for $\frac{3}{x-9} \gt \frac{2}{x+2}$ What I did:
$\frac{3}{x-9} \gt \frac{2}{x+2}$
$3(x+2) \gt 2(x-9)$
$3x+6 \gt 2x-18$
$x \gt -24$
When typing this in in symbolab, it showed me that the solution is $-24 \lt x \lt -2$ or $x \gt 9$
What did i do wrong ?
How come i didnt get the correct solution ?
| HINT
The first step
$$\frac{3}{x-9} \gt \frac{2}{x+2}\iff 3(x+2) \gt 2(x-9)$$
is wrong since $x-9$ and $x+2$ are not both positive (or both negative) in general.
To solve properly use that
$$\frac{3}{x-9} \gt \frac{2}{x+2}\iff \frac{3}{x-9} - \frac{2}{x+2}>0\iff\frac{3(x+2)-2(x-9)}{(x-9)(x+2)}>0$$$$\iff\frac{x+24}{(x-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}$ Greetings I saw here (among the last integrals) that: $$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi C$$
Where $C$ is Catalan's constant. Did this integral appear here befor... | Since the integrand is an even function, we can write
$$ I
= \frac{1}{2}\int_{-1}^{1} \log\left(\frac{1-x}{1+x}\right)\log\left(\frac{1-x^2}{1+x^2}\right)\,\frac{dx}{x}. $$
Now deforming the line contour $[-1, 1]$ to the semicircular contour from $-1$ to $1$ and substituting $x = e^{i\theta}$,
$$ I = -\frac{i}{2} \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Pre-calculus Complex coordinate law of cosine help
Let $\theta = \angle BAC$. Then we can write $\cos \theta = \dfrac{x}{\sqrt{2}}$
Find x.
Currently, that is all I have. I think I should use Law of cosine. Where should I progress?
| $$a = BC = \sqrt{2^2+3^2}=\sqrt{13}$$
$$b = AC = \sqrt{3^2+1^2}=\sqrt{10}$$
$$c = AB = \sqrt{2^2+1^2}=\sqrt{5}$$
so $$\cos \theta = {b^2+c^2-a^2\over 2bc} = {1\over 5\sqrt{2}}\implies x=1/5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to do $\lvert x - 1 \rvert < \lvert x-3 \rvert$
Solve for $x$ such that $$\lvert x - 1 \rvert < \lvert x-3 \rvert$$
I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work?
| Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $\ge 0$ and $x-3$ can be $<$ or $\ge 0$.
Or in other words $x$ can be $<$ or $\ge 1$ or $x$ can be $< $ or $\ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x \ge 3$ both) or redundant (if $x\ge 3$ then $x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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How to find the slant asymptote for $y=\frac{x^2\arctan\left(x\right)}{3x+3}$ I got stuck at the end and wonder if you can give some guide on how to proceed to find the slant asymptote?
$$y=\frac{x^2\arctan\left(x\right)}{3x+3}$$
$$y= kx+m\\
\frac{f(x)}{x} \rightarrow k\\
f(x) - kx \rightarrow m $$
Solution:
$$\lim ... | $\arctan(x)=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)=\frac{\pi}{2}-\frac{1}{x}+O\left(\frac{1}{x^3}\right)$ for $x\to \infty$, hence
$$ \lim_{x\to +\infty}\left[\frac{x^2\arctan(x)}{3x+3}-\frac{\pi}{6}x\right]=\lim_{x\to +\infty}-\frac{(2+\pi ) x}{6 (1+x)}=-\frac{\pi+2}{6} $$
and the equation of the slant asymptot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$? I am trying to evaluate $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$. I have found that $15^{15} \equiv 1 \pmod{7}$ and that $16^{16} \equiv 2 \pmod{7}$.
To evaluate $15^{15} \pmod{17}$, I did the following:
$$15 = 2 \... | Euler's Theorem states that
$$ a^{\phi(n)}\equiv1 \pmod{n}$$
For all $n$, where $\phi(n)$ is the Euler totient function, denoting the number of positive integers less than $n$ relatively prime to $n$. For primes, $\phi(n)=n-1$. Thus
$$ a^{n-1}\equiv1\pmod n. $$
This result is also known as Fermat's little theorem. Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How To Evaluate $ \int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx $? $$
\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx
$$
What I tried was
*
*To convert $N^r$ into $sin2x$
*To Use Identity $(a+b)^3 = (a^3 + b^3)(a^2 + b^2 - ab) $
But none of them proved useful ?
How do I evaluate this Integr... | Hint
Start with: $\sin x=\frac{\tan x}{\sec x}$ and $\sec^2 x =\tan^2 x +1$. Then,
$$\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx = \int \sec^2x\,\frac{\tan^2x}{\left(\tan x+1\right)^2\left(\tan^2x-\tan x +1\right)^2}\,\mathrm{d}x.$$
Now use $u=\tan x$. Then use $t=\left(u+1\right)\left(u^2-u+1\right)$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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what is the characteristic polynomial and minimal polynomial of A and B? What is the characteristic polynomial and minimal polynomial of $A$ and $B$ ?
$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},B=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 1 \... | $A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},$ first i break them into two jordan nlock forms that $P=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$ and $Q=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$
now , the characteristic and minimal polynomial of $P = (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Counting number of occurences Let $1\leq k\leq n$. We find all finite sequences of positive integers with sum $n$. Suppose that the term $k$ occurs a total of $T(n, k)$ times in all these sequences.
Find $T(n, k)$.
As an example to make things more clear: $T(3,1)=5$, $T(3,2)=2$ and $T(3,3)=1$
This on base of the sum... | With $q$ the length of the sequence and $k$ marked with $u$ we get the
generating function
$$G(z, u) = \sum_{q\ge 0} (z+z^2+\cdots+uz^k+\cdots)^q
\\ = \sum_{q\ge 0} z^q (1+z+\cdots+uz^{k-1}+\cdots)^q
\\ = \sum_{q\ge 0} z^q \left(\frac{1}{1-z} + (u-1)z^{k-1}\right)^q
\\ = \frac{1}{1-z(1/(1-z)+(u-1)z^{k-1})}
\\ = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using Laplace transforms to evaluate$\int_{0}^{\infty}\frac{\sin^2(x)}{x^2(x^2 + 1)} dx$ Recently I've been playing around with Feynman's Trick to evaluate integrals. Obviously, one of it's many great features is that it allows derivatives to make expressions simpler. I was wondering whether Laplace Transforms could eq... | Noting that the iterated integral
$$I_1=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dt\right)\,dx$$
is finite, the Fubini-Tonelli Theorem guarantees that the iterated integral formed by interchanging the order of integration,
$$I_2=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Find the coefficient of $x^{10}$ We have been given the following function.
$f(x)$= $x$ +$x^2$ + $x^4$ + $x^8$ + $x^{16}$ + $x^{32}$ + ...upto infinite terms
The question is as follows:
What is the coefficient of $x^{10}$ in $f(f(x))$?
I tried solving it myself and I found the answer too but the method of solving was... | $f(x)$ contains no tenth power.
$f^2(x)$ has $x^2\cdot x^8$, with a coefficient $2$.
$f^4(x)$ has $x\cdot x\cdot x^4\cdot x^4$ with a coefficient $\dfrac{4!}{2!2!}=6$ and $x^2\cdot x^2\cdot x^2\cdot x^4$ with a coefficient $\dfrac{4!}{3!1!}=4$.
$f^8(x)$ has $x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x^2\cdot x^2$ appe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$4ac-b^2\leq 3a(a+b+c)$ in quadratic Suppose that the polynomial $(b+c)x^2+(a+c)x+(a+b)$ doesn't have real roots, where $a,b,c\in\mathbb{R}$. Prove that $4ac-b^2\leq 3a(a+b+c)$.
The quadratic not having real roots means that $$(a+c)^2-4(b+c)(a+b)<0$$ which translates to $(a^2+2ac+c^2)-4(b^2+ab+ac+bc)<0$, or
$$a^2+c^2-... | We have $$(a+c)^2<4(a+b)(b+c)$$ or
$$a^2+2ac+c^2<4b^2+4ab+4ac+4bc$$ or
$$c^2+a^2+4b^2-2ac-4bc+4ab<8ab+8b^2$$ or
$$(c-a-2b)^2<8b(a+b),$$ which gives
$$a^2(c-a-2b)^2\leq8a^2b(a+b).$$
In another hand, we need to prove that
$$4ac-b^2\leq 3a^2+3ab+3ac$$ or $$a(c-a-2b)\leq2a^2+ab+b^2,$$ which is obvious for $a(c-a-2b)<0.$
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$
The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos... | This is an impossible situation. If you use the identity
$$\tan(\theta) = {\sin(\theta)\over\cos(\theta)},$$
you get $\cos(\theta) = 1/6$.
However, $\cos(\theta) = \sqrt{7}/4$ by the Pythagorean identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ (both arbitrary functions)
To make more clear what I'm talking about I will give a example.
Rewrite $f(x,y) = 1+x-y-xy$ as $g(x)h(y)$
If we choose $g(x) = (1+x)$ and $h(y) = (1-y)$ we have
$f(x,y) = g(x... | Without loss of generality let $f(x) = ax^2 + b$ and $g(y) = cy^2 + d$. Now
$$
1-x^2y^2 = f(x)g(y) = acx^2y^2 +adx^2 = cby^2 +bd
$$
By comparing terms we obtain:
\begin{align}
ad&=0\\
cb&=0\\
ac&=-1\\
bd&=1
\end{align}
By $bd=1$ we obtain $a=c=0$. Hence this problem has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do I evaluate the integral $\int_0^1\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$. stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was def... | If your purpose is to evaluate
$$ S=\sum_{k\geq 0}\left[\frac{1}{10k+1}-\frac{1}{10k+4}+\frac{1}{10k+6}-\frac{1}{10k+9} \right]$$
you do not need an indefinite integral, just an integral over $(0,1)$: that's a huge difference, in some cases.
Actually, since $1+9=4+6=10$, you may just invoke the identities
$$ \sum_{k\g... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How do I use the definition of partial derivative to get $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2 + y^2)^2}$? $\\f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2}, & \text{if $(x,y) \ne (0,0)$} \\
0, & \text{if $(x,y) = (0,0)$}
\end{cases}$
Using the definition of partial derivatives:
\begin{align*}
f_x(x,y) &= \lim_{h\to 0}... | Notice that
\begin{align}&\phantom{ = }\frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&= \frac{\color{red}{xy(x^2 +y^2)} + hy(x^2 +y^2) \color{red}{-xy(x^2 + y^2)} - xy(2hx+ h^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&=\frac{hx^2y+hy^3 -2hx^2y-h^2xy}{h(x^2 + 2hx + h^2 + y^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the residue with a Laurent series expansion. I have a question about the following problem:
"Detect the error in the following argument. The function $f(z)=\frac{1}{z(z-1)^2}$ has an isolated singularity at $z=0$. The Laurent series is $f(z)=\frac{1}{(z-1)^3}-\frac{1}{(z-1)^4}+\frac{1}{(z-1)^5}-...$
for $|z-1|... | We consider the expansion of $f$ at $z_0=1$ since the center $1$ is indicated by the terms $\frac{1}{(z-1)^k}$.
The function
\begin{align*}
f(z)=\frac{1}{z(z-1)^2}=\frac{1}{z}-\frac{1}{z-1}+\frac{1}{(z-1)^2}\tag{1}
\end{align*}
is to expand around the center $z_0=1$. Since there are isolated singularities, nam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Help find the mistake in this problem of finding limit (using L'Hopital)
Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$
Attempt
\begin{align*}
&\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\
= &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\
= &\lim_{x \to 0} \l... | $$\lim_{x\to0}\left(\dfrac1{x^2}-\dfrac1{\tan^2x}\right)=\lim_{x\to0}\dfrac{\tan x-x}{x^3}\cdot\lim_{x\to0}\dfrac{\tan x+ x}x\cdot\left(\lim_{x\to0}\dfrac x{\tan x}\right)^2$$
Now the limit for the last two are too simple for L'hospital
For $\lim_{x\to0}\dfrac{\tan x-x}{x^3}$ either use L'hospital or Are all limits so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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What is the coefficient of the term $x^3y^5$, as a result of the binomial expansion of the following term? We have the term
$(1+xy+y^2)^n$
If we expand it using the binomial theorem, why is the factor of the term $x^3y^5$ the following: $4{n\choose 4}$? (The binomial coefficient n choose 4 multiplied by 4)
| It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^3y^5]}&\color{blue}{(1+xy+y^2)}\\
&=[x^3y^5]\sum_{j=0}^n\binom{n}{j}(xy)^j(1+y^2)^{n-j}\tag{1}\\
&=[y^5]\binom{n}{3}y^3(1+y^2)^{n-3}\tag{2}\\
&=\binom{n}{3}[y^2]\sum_{j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n$ Given the following series:
$$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n,$$ where $$a_n = \frac{2^2.4^2\cdots(2n)^2}{3.4.5.6\cdots (2n+2)} x^{2n+2}$$
Now we try Ratio test first: $$\frac{a_{n+1}}{a_n} = \frac{2n^2+4n+2}{n+2}x^2$... | Apply Ratio Test
$$
\begin{align}
\frac{2^2}{3\cdot4}x^4 + \frac{2^2\cdot4^2}{3\cdot4\cdot5\cdot6}x^6+\cdots
&=\sum_{k=2}^\infty\frac{(k-1)!^22^{2k-2}}{(2k)!/2!}x^{2k}\\
&=\frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}}\\
\end{align}
$$
The Ratio Test says
$$
\begin{align}
\lim_{k\to\infty}\frac{a_{k+1}}{a_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Help with induction proof I need help with the following induction proof which I am not sure if I am doing correctly.
$$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$$
I check for $n=1$ (Base case)
$$\frac{1}{1\cdot3}=\frac13$$
$$\frac{1}{2\cdot1+1}=\frac13$$
Now, is this the correct nex... | Your idea is correct but $$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1))(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
Should have been $$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
Good luck finishing it up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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What is $26^{32}\bmod 12$? What is the correct answer to this expression:
$26^{32} \pmod {12}$
When I tried in Wolfram Alpha the answer is $4$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $0.$
| First, note that $26 \equiv 2 \pmod {12}$, so $26^{32} \equiv 2^{32} \pmod {12}$.
Next, note that $2^4 \equiv 16 \equiv 4 \pmod {12}$, so $2^{32} \equiv \left(2^4\right)^8 \equiv 4 ^8 \pmod {12}$, and $4^2 \equiv 4 \pmod {12}$.
Finally, $4^8 \equiv \left(4^2\right)^4 \equiv 4^4 \equiv \left(4^2\right)^2 \equiv 4^2 \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
If $A=\sum_{n=1}^{3027}\sin(n\pi/2018)$ and $B=\sum_{n=1}^{3027}\cos(n\pi/2018)$, evaluate $A(1-\cos(\pi/2018))+B\sin(\pi/2018)$
Let
$$A=\sum_{n=1}^{3027}\sin\frac{n\pi}{2018} \qquad B=\sum_{n=1}^{3027}\cos\frac{n\pi}{2018}$$
Evaluate
$$A\left(1-\cos\frac{\pi}{2018}\right)+B\sin\frac{\pi}{2018}$$
I don't know... | I'll generalize the problem a bit, writing $m$ for $3027$ and $2\theta$ for $\pi/2018$.
Adapting the identities proven here, we can write
$$\begin{align}
A &:= \sum_{n=1}^{m}\sin 2n\theta = \frac{\sin(m+1)\theta}{\sin\theta}\;\sin m\theta \\[4pt]
B &:= \sum_{n=1}^{m}\cos 2n\theta = \frac{\sin(m+1)\theta}{\sin\theta}\;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that for each $n \geq 1$ it holds that $2^{n-1} F_n \equiv n \pmod{5}$ I want to show that for each $n \geq 1$ it holds that:
$$2^{n-1} F_n \equiv n \pmod{5}$$
where $F_n$ is the $n$-th Fibonacci number.
Could you give a hint how we can show this?
The sequence $F_n$ of Fibonacci numbers is defined by the recurrenc... | Use the general term formula:
$$2^{n-1} F_n \equiv 2^{n-1}\cdot \frac{\phi^n-\psi^n}{\sqrt{5}} \equiv \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2\sqrt{5}} \equiv\\
\frac{2{n\choose 1}\sqrt{5}+2{n\choose 3}(\sqrt{5})^3+2{n\choose 5}(\sqrt{5})^5+\cdots +2{n\choose 2\lfloor{\frac{n-1}{2}\rfloor}+1}(\sqrt{5})^{2\lfloor{\frac{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Formal proof of the sequence involving double factorials $x_n = \frac{(2n)!!}{(2n-1)!!}$ is not bounded. I'm trying to prove the following:
Let $n\in \mathbb N$ and
$$
x_n = \frac{(2n)!!}{(2n-1)!!}
$$
Where:
$$
(2n)!! = 2\cdot 4 \cdot 6 \cdot \dots \cdot 2n \\
(2n - 1)!! = 1\cdot 3 \cdot 5 \cdot \dots \cdot (... | As simple as it gets:
LEMMA: For $k\in N$ and $k>1$
$$\frac{2k}{2k-1}\gt\frac{\sqrt{3k+1}}{\sqrt{3k-2}}\tag{1}$$
Proof: It's trivial. Both sides are positive, square them:
$$\frac{4k^2}{4k^2-4k+1}\gt\frac{3k+1}{3k-2}$$
..which is equivalent to:
$$4k^2(3k-2)\gt(4k^2-4k+1)(3k+1)$$
$$12k^3-8k^2\gt12k^3+4k^2-12k^2-4k+3k+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$
Given a sequence
$$
\begin{cases}
x_{n+1} = \frac{x_n^4 + 1}{5x_n} \\
x_1 = 2 \\
n \in \mathbb N
\end{cases}
$$
Prove it has lower bound at $1\over 5$ and upper bound at $2$
I've tried to find a closed form for the ... | Hint:
This is not a detailed answer, but this plot says all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that we can't find effective bounds on the point guaranteed by the Mean Value Theorem. I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:
For each real number $M$ and each real n... | Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $\xi \in (0, 1)$ for which $f'(\xi) = 0$, then $\frac{1}{3} \leq \xi \leq \frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$ If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$
$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$
Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$
I am stuck here.
| Hint.- Just to give another way. Developing the product and simplify you have the equivalent inequality
$$7xyz\le xy+xz+yz$$ and you have too $$\frac13\ge\sqrt[3]{xyz}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How many ordered pairs satisfy $\log(x^3+\frac{1}{3}y^3+\frac{1}{9})=\log x+\log y$? How many ordered pairs satisfy $\log(x^3+\frac{1}{3}y^3+\frac{1}{9})=\log x+\log y$?
It simplifies to $x^3+\frac{1}{3}y^3+\frac{1}{9}=xy$
I dont know how to proceed further.
| $$\log\left(x^3+\frac{1}{3}y^3+\frac{1}{9}\right)=\log x+\log y\tag1$$
First of all, we have to have
$$x^3+\frac{1}{3}y^3+\frac{1}{9}\gt 0\quad\text{and}\quad x\gt 0\quad\text{and}\quad y\gt 0\tag2$$
Under $(2)$, we have
$$(1)\implies x^3+\frac{1}{3}y^3+\frac{1}{9}-xy=0\tag3$$
As lab bhattacharjee suggests in the comme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=$? Consider the convergent integral (I looked up difficult indefinite integrals on google images and then I saw this integrand and I was like hey let's see if it converges)
$$I=\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
We have the numerical approximatio... | Since $$x^4 + 1 = x^2 (x^2 + x^{-2}) = x^2 ((x + x^{-1})^2 - 2),$$ we have for $x > 0$
$$\frac{x^2-1}{(x^2+1)\sqrt{x^4 + 1}} = \frac{x^2 (1 - x^{-2})}{x^2 (x + x^{-1}) \sqrt{(x + x^{-1})^2 - 2}} = \frac{1-x^{-2}}{(x+x^{-1})\sqrt{(x+x^{-1})^2 - 2}}.$$ Now with the substitution $$u = x + x^{-1}, \quad du = (1 - x^{-2}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Knowing when a pattern is a linear homogeneous recurrence relation I'm quite stuck on finding a pattern here:
Suppose we have:
*
*1 inch letters: f, i, t
*2 inch letters: a, c, d, e, g, n, o, p, s, u
And we want to find a recurrence relation for making a banner of length $n$ inches. We can repeat letters and orde... | To complete @weee's solution, use generating functions. Note that $A_0 = 1$ (just one empty banner) and $A_1 = 3$ (number of one-inch banners). Define $g(z) = \sum_{n \ge 0} A_n z^n$, write your recurrence as:
$\begin{equation*}
A_{n + 2} = A_{n + 1} + 10 A_n
\end{equation*}$
Multiply the recurrence by $z^n$, sum over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Concerning the product of all unique positive divisors
If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
(A) $n^3$
(B) $n^4$
(C) $n^6$
(D) $n^8$
(E) $n^9$
I tri... | Since the product of positive divisors of $n$ is $n^{\frac{d(n)}{2}}$. We want
$$n^{\frac{d(n)}{2}}=n^2.$$
This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding a quadratic equation using roots
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_{1/2}$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_{1/2}=\large{-b\pm{\sqrt{4ac}}\over2a}$
and from here:
$$\begin{align}x_1... | Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convert Polynomial to Sum I have to convert the polynomial: $P_4=x^4+7x^3-13x^2-103x-84$ into the form: $$P_4(x)=\sum_{i=0}^4 a_i(x-1)^i$$ using Horner's Method.
I can evaluate both of them using Horner's Method, but I can't figure out how to convert them into one another.
| Note that:
$$a_0+a_1(x-1)+a_2(x-1)^2+a_3(x-1)^3+a_4(x-1)^4=x^4+7x^3-13x^2-103x-84 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_0}=1+7-13-103-84=\color{red}{-192};\\
(x-1)[a_1+a_2(x-1)+a_3(x-1)^2+a_4(x-1)^3]=x^4+7x^3-13x^2-103x+108 \stackrel{\div (x-1)}{\Rightarrow} \\
a_1+a_2(x-1)+a_3(x-1)^2+a_4(x-1)^3=x^3+8x^2-5x-108 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove if $f$ is convex on $(a,b)$, then $f$ is bounded on every closed subinterval of $(a,b)$ Prove :
$f : (a,b) \to \mathbb{R} $ is convex, then $f$ is bounded on every closed subinterval of $(a,b)$
where $f$ is convex if $f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda)f(y), \forall x,y \in (a,b), \forall... | Here is an analytic proof that a convex function $f:[\alpha,\beta] \to \mathbb{R}$ is bounded on the closed interval.
Take $M = \max (f(\alpha),f(\beta))$. Note that any $x \in [\alpha,\beta]$ is of the form $x = \lambda\alpha + (1-\lambda)\beta$ where $0 \leqslant \lambda \leqslant 1$.
Hence, we have for all $x \in [\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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