Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Random walk (Exercise 3.11.39 from Grimmett and Stirzaker) A particle performs a random walk on the non-negative integers as follows. When at the point $n\ (> 0)$ its next position is uniformly distributed on the set $\{0, 1, 2, \ldots, n + 1\}$. When it hits $0$ for the first time, it is absorbed. Suppose it starts at... | For completeness, here is the solution to part(c):
Assuming that a transition from $i$ to $j$ is always a step of length 1 (regardless of the values of $i$ and $j$) we can write the expected number of steps, $\lambda_n$, given that the particle starts at $n \geq 1$ as,
\begin{align}
\lambda_n &= \frac{1}{n + 2}((\lambd... | {
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"url": "https://math.stackexchange.com/questions/1675227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Be... | First, we prove a simple lemma.
$$
\begin{align}
n & \gt 5\\
\implies n-1 &\gt 4\\
\implies (n-1)^2 &\gt 16 \gt 2 \\
\implies n^2 - 2n + 1 &\gt 2\\
\implies n^2 &\gt 2n +1\\
\end{align}
$$
Now, we start with our induction step.
$$
\begin{align}
n^2 &\lt 2^n \text{ From hypothesis}\\
\implies 2n^2 & \lt 2^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Proving correctness of a recursive function of multiplication by induction. Define multiplication of two numbers y and z as:
$$m_c(y,z)=\begin{cases}0&z=0\\m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)+y(z\mod c)&z\neq 0\end{cases}\tag{$\forall c\geq 2$}$$
Now I need to show that this is correct using induction... | You should do induction on $z$, so that you can assume in the inductive step that $m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)=cy\cdot\left\lfloor \frac zc\right\rfloor$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$ positive integers $\Rightarrow a+c \le n-1$
and $ad+bc \le bd-1.$
| From $\frac ab+\frac cd \lt 1$ and $a+c<n$ it is clear that $a<b;\space c<d$, $n\ge 3$ and $2\le a+c\le n-1$. Furthermore at least one of the two fractions should be less than $\frac 12$ therefore we can put $\frac {a}{2a+k}+\frac cd\lt 1$ where $k\ge 1$ so $$\frac ab+\frac cd\le \frac {a}{2a+1}+\frac cd\lt 1\iff \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$ What is the maximum of $\quad 2(a+b)-ab\ $ if we have: $a^2+b^2=8 \ (a,b\ real) $
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow $$
$$ab\le4 $$
On the other hand , from $\ a^2+b^2... | let $u=\sqrt{8+2ab} \implies ab=\dfrac{u^2-8}{2} \implies 2u-\dfrac{u^2-8}{2}=-\dfrac{(u-2)^2}{2}+6,|ab| \le \dfrac{a^2+b^2}{2}=4 \implies 0\le u \le 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requ... | $a+b-c=2(s-c)$ etc.
$$\implies\dfrac{(a+b-c)(b+c-a)(c+a-b)(a+b+c)}{b^2c^2}=\dfrac{16s(s-a)(s-b)(s-c)}{(bc)^2}$$
Now use $\triangle=\sqrt{s(s-a)(s-b)(s-c)}=\dfrac12bc\sin A$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I... | There is no problem. Your solution is correct, and the other solution is correct. But you need to do a little work to see that they give the same answers:
*
*$\cos \frac {3x}2 = 0$ gives $x = \frac{(2k+1)\pi}3:\ x \in \{\pm \frac{\pi}3, \pm\pi,\pm \frac{5\pi}3, \pm \frac{7\pi}3, \pm 3\pi, \ldots\}$
*$\sin \frac {x}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 4
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Integral $I=\int \frac{dx}{(x^2+1)\sqrt{x^2-4}} $ Frankly, i don't have a solution to this, not even incorrect one, but, this integral looks a lot like that standard type of integral $I=\int\frac{Mx+N}{(x-\alpha)^n\sqrt{ax^2+bx+c}}$ which can be solved using substitution $x-\alpha=\frac{1}{t}$ so i tried to find such s... | HINT:
Your integral:
$$\text{I}=\int\frac{1}{\left(x^2+1\right)\sqrt{x^2-4}}\space\text{d}x=$$
Subsitute $x=2\sec(u)$ and $\text{d}x=2\tan(u)\sec(u)\space\text{d}u$.
Then $\sqrt{x^2-4}=\sqrt{4\sec^2(u)-4}=2\tan(u)$ and $u=\text{arcsec}\left(\frac{x}{2}\right)$:
$$\int\frac{\sec(u)}{4\sec^2(u)+1}\space\text{d}u=\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Complicated Indefinite Integration $\displaystyle\int{\dfrac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\;\mathrm{d}x} =\frac{2}{3}\big(\sqrt{x}+2\big) $
I do not know how to get $\dfrac{2}{3}$. I assume $\sqrt{x}+2 = u$, then $\mathrm{d}u= \dfrac{1}{2}\,x^{1/2}$
But it is wrong answer....
So help me
Thank you
| Assuming you mean $\displaystyle \int \frac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\,dx$:
Let $\,u=\sqrt{x}+2\,$ and $\,du=\dfrac{dx}{2\sqrt{x}}$
Thus, we now have $$\dfrac{2}{3} \int u^2 \,du=\dfrac{2u^3}{9}+C=\dfrac{2\left(\sqrt{x}+2\right)^3}{9}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$
Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\... | By the basic properties of Chebyshev polynomials the numbers $x_k=\sin(k\pi/14)$, $k=-6,-5,\ldots,5,6, k\neq0$ are the twelve zeros of the polynomial
$$
\frac{U_{13}(x)}x=8192 x^{12}-24576 x^{10}+28160 x^8-15360 x^6+4032 x^4-448 x^2+14.
$$
By the usual relations (aka Vieta relations) of the zeros of a polynomial and it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many $4$ digit integer elements $X$ having no digit $0$ are in the set $C$ such that $X$ has exactly one $1$ or $X$ has exactly one $5$? Let $A = \{4~\text{digit integers}~X~\text{having no}~0~\text{such that}~X~\text{has exactly one}~1\}$.
Let $B = \{4~\text{digit integers}~X~\text{having no}~0~\text{such that}~X~... | You correctly calculated the number of four digit integers with non-zero digits that have exactly one $5$ and the number of four digit integers with non-zero digits having exactly one $1$. However, their intersection is the set of four digit integers with non-zero digits that have exactly one $1$ and exactly one $5$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Extracting the divergent part of an integral I want to evaluate the integral
$$ \int_0^1 \frac{2x(x-2)(1-x)}{(1-x)^2 + ax} \, \mathrm{d}x$$
in the limit of small $a$. For $a = 0$ this integral is divergent due to the $1/(1-x)$ pole. The exact expression for this integral is, according to Mathematica, a horrible combina... | Notice, $$\int_0^1\frac{2x(x-2)(1-x)}{(1-x)^2+ax}\ dx$$
using property of definite integral: $\int_a^bf(x)\ dx=\int_a^bf(1+b-x)\ dx$,
$$=\int_0^1\frac{2(1-x)(1-x-2)(1-1+x)}{(1-1+x)^2+a(1-x)}\ dx$$
$$=\int_0^1\frac{2x^3-2x}{x^2-ax+a}\ dx$$
$$=\int_0^1\left(2(x+a)+\frac{(a^2-a-1)x-a^2}{x^2-ax+a}\right)\ dx$$
$$=2\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$? How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$
where
$$f(x,y)=x^4+y^4-x^2-2xy-y^2$$
Answer:
$$f_x=4x^3-2x-2y$$
$$f_y=4y^3-2y-2x$$
Critical points are $(0,0), (... | Probably easier to write the boundary as
$$x=2\cos\theta\ ,\quad y=2\sin\theta\ .$$
Then
$$\eqalign{f(x,y)
&=16\cos^4\theta+16\sin^4\theta-4\cos^2\theta-8\cos\theta\sin\theta-4\sin^2\theta\cr
&=4(1+\cos2\theta)^2+4(1-\cos2\theta)^2-4\sin2\theta-4\cr
&=4+8\cos^22\theta-4\sin2\theta\cr
&=12-4\sin2\theta-8\sin^22\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$? The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$
First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n... | Let $f(z) = \sum_{n=0}^\infty a_n z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$ yields
$$\sum_{n=0}^\infty a_{n+1}z^n - \sum_{n=0}^\infty a_nz^n = \sum_{n=0}^\infty(2n+3)z^n $$
and hence
$$z^{-1}(f(z)-a_0) - f(z) = 2\sum_{n=0}^\infty (n+1)z^n + \sum_{n=0}^\infty z^n, $$ so that
$$f(z)(1-z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int_{0}^{\infty }\frac{\ln( 1+x^{4} )}{1+x^{2}}{d}x~,~\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x$ How to evaluate these two integrals below
$$\int_{0}^{\infty }\frac{\ln\left ( 1+x^{4} \right )}{1+x^{2}}\mathrm{d}x$$
$$\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}\mathrm{d}x$$... | For the first integral, we introduce a new parameter $a$. Let $I(a)$ be the integral
$$I(a)=\int_0^\infty \frac{\log(a+x^4)}{1+x^2}\,dx \tag 1$$
We note that it can be shown that $I(0)=0$. Our integral of interest is $I(1)$.
Proceeding, we differentiate $(1)$ to obtain
$$\begin{align}
I'(a)&=\int_0^\infty\frac{1}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688024",
"timestamp": "2023-03-29T00:00:00",
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Induction Proof $k^2 \times 2^k$ I need help on this proof. I am not able to do it after setting m=m+1.
Prove by induction on n that sum of $k^2 \times 2^k$ from $k=1$ to $n$ is equal to $(n^2-2n+3) \times 2^{n+1}-6$
Base case:
Let $k=1$ so L.H.S side is $2$
Let $n=1$ so R.H.S side is $2$
Inductive hypothesis:
Let $n=m... | The induction hypothesis is that
$$\tag1\sum_{k=1}^m k^22^k=(m^2-2m+3)2^{m+1}-6$$
Now,
$$\sum_{k=1}^{m+1}k^22^k=\sum_{k=1}^mk^22^k+(m+1)^22^{m+1}$$
Inductive Step:
From $(1)$,
$$\sum_{k=1}^{m+1}k^22^k=\sum_{k=1}^mk^22^k+(m+1)^22^{m+1}=(m^2-2m+3)2^{m+1}-6+(m+1)^22^{m+1}=2^{m+1}(m^2-2m+3+m^2+2m+1)-6=2^{m+1}(2m^2+4)-6=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derive a polynomials formula by setting $n = 1$, $n = 2$, $n = 3$ to determine the coefficients. Derive a formula for the sum of squares $1^2 + 2^2 + 3^2 + … + n^2$.
Hint: assume the formula is a polynomial of degree 3, i.e. $an^3 + bn^2 + cn + d$, and use the cases of $n=0, n=1, n=2$, and $n=3$ to determine its coeff... | Another derivation is using proof by induction; these are the $\color{green}{\mathrm{three}}$ steps to carry out:
Step 1: Basis Case: For $i=1$: $$\sum^{i=k}_{i=1} i^2=\frac{1(1+1)(2\times 1+1)}{6}= \frac{2\times 3}{6}=1$$ So statement holds for $i=1$.
Step 2: Inductive Assumption: Assume statement is true for $i=k$: ... | {
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"timestamp": "2023-03-29T00:00:00",
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Geometry - Pentagon This is a tough program I have hard time to find the answer. Can anyone help me? Thank you very much in advance.
Problem - In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answ... | Let $F$ be the intersection of $AC$ and $BE$. By angle chasing, we obtain the following picture:
Now $\triangle BAF$ and $\triangle BEA$ are similar, so
$$BF:BA = BA:BE,$$
or
$$\frac yx = \frac{x}{x+y}=\frac{1}{1+y/x}.\tag{1}$$
Solving that for $y$ in terms of $x$ we have
$$\frac yx = \frac{\sqrt{5}-1}2.$$
Now appl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$
I found that:
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$
and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{... | Notice that the first few terms are:
$(1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)\cdots$
which gives: $1/2+1/4=3/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An approximation for $e^{ix}$ By the Taylor expansion of $\log(1+z)$ one can show that
$$e^{ix} = (1+ix)e^{-x^2/2+r(x)}$$
where $\lvert r(x)\rvert \leq \lvert x\rvert^3$ for all $x$ such that $\lvert x\rvert \leq 1/2$.
In a paper I saw the same claim except for the part where the author claims the the bound on $r(x)$ i... | If we define \begin{align}
r(x)=ix+\frac{x^2}{2}-\log (1+ix)\tag{1}\end{align}
then $e^{ix} = (1+ix)e^{-x^2/2+r(x)}$.
Using the Taylor expansion of $\log(1+z)$ in $(1)$ we have \begin{align}
r(x)&= ix+\frac{x^2}{2}-\left(ix+\frac{x^2}{2}-i\frac{x^3}{3}-\frac{x^4}{4}-i\frac{x^5}{5}+\cdots\right)\\
&=i\frac{x^3}{3}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute all possible values of $\cot{\theta} - \frac{6}{z}$. Note that $ 0 < \theta < \frac{\pi}{2}$.
Let $\theta = \arg{z}$ and suppose $z$ satisfies $|z - 3i| = 3$.
Compute all possible values of $\displaystyle \cot{\theta} - \frac{6}{z}$. Note that $\displaystyle 0 < \theta < \frac{\pi}{2}$.
$\bf{My\; Try::}$ Let ... | Since $\theta=\arg z$, note that $z=3\cos\theta+3i(1+\sin\theta)$ is not correct.
Let $z=x+yi$ where $x,y\in\mathbb R$. Then, since we have
$$x^2+(y-3)^2=3^2$$we can write
$$x=3\cos\alpha,\qquad y=3+3\sin\alpha$$
where $-\pi/2\lt\alpha\lt \pi/2$.
So, we can have
$$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}=\frac{3\cos\alpha}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Recurrence Relation problem I am trying to solve the recurrence relation:
$$G_n = \frac{1-G_{n-1}}{4}$$
$$G_0 = 0$$
$$G_1 = \frac{1}{4}$$
I am told that the answer is
$$G_n = \frac{1}{5}\left(1+\left(\frac{-1}{4}\right)^{n+1}\right)$$
I have found the characteristic equation to be
$$G_n = B\left(\frac{-1}{4}\r... | You have $G_n=\frac{1}{4}-\frac{1}{4}G_{n-1}$. With the ansatz $G_n=r^n$ we have for the homogeneous
$$r^n=-\frac{1}{4}r^{n-1}$$
$$r=-\frac{1}{4}$$
and for the particular
$$K=\frac{1}{4}-\frac{1}{4}K$$
$$K=\frac{1}{5}$$
so the general solution is $G_n=\frac{1}{5}+B\left(-\frac{1}{4}\right)^n$ and we plug in the initial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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determinant of 3x3 matrix algebra not matching CAS Find
$$\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}$$
we will expand along first row
$$ \begin{aligned}
\begin{vmatrix}
... | Why not first add twice the first row to the third one to make things slightly simpler?
$$\begin{vmatrix}i&2&\!\!-1\\3&1+i&2\\0&5&2-i\end{vmatrix}=-5\begin{vmatrix}i&\!\!-1\\3&2\end{vmatrix}+(2-i)\begin{vmatrix}i&2\\3&1+i\end{vmatrix}\stackrel{\text{develop by third row}}=$$
$$=-5(2i+3)+(2-i)(-1+i-6)=-10i-15+(2-i)(-7+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707810",
"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(... | Another approach is to note that $$\int_0^1\left(1-x^a\right)^bdx=\int_0^1\left(1-y\right)^b\tfrac{1}{a}y^{1/a-1}dy=\frac{1}{a}\text{B}\left(\frac{1}{a},\,b+1\right)=\frac{b!\Gamma\left(\tfrac{1}{a}+1\right)}{\Gamma\left(b+\tfrac{1}{a}+1\right)},$$ so $$\frac{\int_0^1\left(1-x^{50}\right)^{100}dx}{\int_0^1\left(1-x^{50... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Using Lagrangian multipliers to find maximum and minimum values of $f(x)=\frac{1}{\sqrt{(x+1/2)^2+y^2+z^2}}$ for $x^2+y^2+z^2=1$ We define $$f(x)=\frac{1}{\sqrt{(x+1/2)^2+y^2+z^2}}$$ and find the maximum and minimum values of $f$ such that $$x^2+y^2+z^2=1$$
Here is what I have so far:
$f(x)$ approaches a maximum as $1/... | You have to solve the four equations simultaneously for $\lambda, x, y, z$. The second equation gives either $\lambda=-1$ or $y=0$. But if $\lambda=-1$ then the first equation becomes $1=0$. Therefore $y=0$, and similarly $z=0$ from the third equation. The fourth equation then gives $x^2=1$, so either $x=1$ or $x=-1$.
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve the equation $x^5 - 5x^4 - 5x^3 +25x^2 +4x -20 = 0$ given that its roots are of the form $+a, -a, +b, -b, c$? I understand that it is an easy problem, but I am not able to solve it at all! Any clue on how to approach this problem will be amazing! Thank you. (PS: I am just beginning to learn math).
| Let $P(x)=x^5 - 5x^4 - 5x^3 +25x^2 +4x -20$.
Note a equation that would have $-a,a,-b,b,c$ as it's roots would be of the form $(x-a)(x+a)(x-b)(x+b)(x-c)=(x^2-a^2)(x^2-b^2)(x-c)$.
So we get
$$P(x)=x^5 - 5x^4 - 5x^3 +25x^2 +4x -20 = (x^2-a^2)(x^2-b^2)(x-c)=(x^4-(a^2+b^2)x^2+a^2b^2)(x-c)=x^5-cx^4+\dots$$
This tells us t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Evaluation of $\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx$ How can we evaluate following integral:
$$I=\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx$$
I tried using the property $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$ but it not helping because of the denominator. How should I proceed?
| Assume $a>0$. One may write
$$
\begin{align}
\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx &=a^2\int_{0}^{3} \frac{t\sqrt{3-t}}{\sqrt{1+t}} dt \qquad (x=at)
\\\\&=2a^2\int_{1}^{2} (u^2-1)\sqrt{4-u^2}\: du \qquad \left(u=\sqrt{1+t}\:\right)
\\\\&=8a^2\int_{\pi/6}^{\pi/2} (3-4\cos^2 v)\cos^2 v\: dv \qquad \left(u=2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
(Easier) way to determine determinant of a almost-circulant matrix Consider the matrix:
$$\begin{pmatrix} \color{red}{-x+y} &-1&0&-1 \\ -1&\color{blue}{x+y}&-1&0
\\ 0&-1&\color{red}{-x+y}&-1 \\ -1&0&-1&\color{blue}{x+y} \end{pmatrix}$$
Is there any easy way to calculate the determinant of this matrix?
I have read up o... | Substract line two from line four and develop by first column (factoring out $\;x+y\;$ from last row):
$$\begin{vmatrix}-x+y&-1&0&-1\\-1&x+y&-1&0\\0&-1&-x+y&-1\\0&-x-y&0&x+y\end{vmatrix}=(-x+y)(x+y)\begin{vmatrix}x+y&-1&0\\-1&-x+y&-1\\-1&0&1\end{vmatrix}+$$
$$+(x+y)\begin{vmatrix}-1&0&-1\\-1&-x+y&-1\\-1&0&1\end{vmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that a positive integer is composite Let $ n $ be a positive integer that can be written as a sum of two relatively prime squares in two distinct ways, that is $ n = a^2 + b^2 = c^2 + d^2 $ so that $ gcd(a, b) = gcd(c, d) = 1, $ then $ n $ is composite.
So I have successfully factored $ n $ into $ \displaystyle \... | I carefully read Brillhart's first paper and realized I was unsure about the final step. So, naive or not, here is a lemma.
LEMMA Suppose, with positive integers $n,H,E,F,$ such that $$ nH = EF, $$
while
$$ 1 < E < F < n. $$
Define
$$ g = \gcd(n,E). $$
THEN both $g > 1$ and $g < n.$
PROOF First, $g \leq E < n.$ Next, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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recurrence relation unable to solve I am trying to solve recurrence relation : $$z_n = 2z_{n-1} + z_{n-2}
\;\;\;\;\;z_0=1\;\;\;z_1=3$$
Could you please help to provide a solution. I got stuck with Lamdas..
Are there some simple methods to solve any kind of these problems perhaps ? Like a good cookbook ?
Thanks
| Here is an answer based upon generating functions.
With
\begin{align*}
A(t)=\sum_{n=0}^\infty{z_nt^n}
\end{align*}
we obtain
\begin{align*}
\sum_{n=2}^\infty z_nt^n&=2\sum_{n=2}^\infty z_{n-1}t^n+\sum_{n=2}^\infty z_{n-2}t^n\tag{1}\\
&=2t\sum_{n=1}^\infty z_{n}t^{n}+t^2\sum_{n=0}^\infty z_{n}t^{n}\tag{2}\\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Why do we not include $c$ in the computation of the definite integral? Why is it when evaluating the definite integral we commonly opt to omit the constant $c$
$$\int_1^2x^2 \, dx= \left.\frac{x^3}{3} \right|_1^2 =\frac{2^3}{3}-\frac{1^3}{3}=\frac{7}{3}$$
But when evaluating the indefinite integral we include it?
$$\in... | $$
\text{Becuase } \left(\frac{2^3}{3}+c\right) - \left(\frac{1^3}{3}+c\right) = \frac{7}{3} \ne \left(\frac{7}{3}+c\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$ Integrate $$\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$$
I have tried to do Tangent substitution, but there is huge power.
What method is better to use here?
| I suspect the OP did not intend for this, but let's go with the residue theorem. Even if we make no simplification to the denominator, things end up pretty simple. Note that, when $z=e^{i x}$,
$$\cos^6{x}+\sin^6{x} = \left (\frac{z+z^{-1}}{2} \right )^6 - \left (\frac{z-z^{-1}}{2} \right )^6 = \frac1{16} \left (3 z^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Grouping students into groups of 3 I have 6 students.
I wanted to divide the students into discussion groups of size 3.
I wished to do this such that all possible groups of 3 students get to discuss together for exactly 1 round.
I could do this trivially in 10 rounds so that the assignment of students into groups (A an... | $$\frac1{3!}\binom9{3,3,3}=\frac1{3!}\frac{9!}{3!3!3!}=280$$
The second factor of LHS equals the number of arrangments when $9$ students are divided in $3$ distinghuishable groups of $3$. Since the groups are not distinghuishable we must repair the multiple counting. For that the first factor of LHS.
Applying it on $6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $a_{214}$ nonintegral where $a_0 = 1, a_{n+1}=\frac{1 + a_0^5 + \cdots + a_n^5}{n + 1}?$ Define the sequence $a_n$ by the following.
$$a_0 = 1,$$
$$a_{n+1}=\frac{1 + a_0^5 + \cdots + a_n^5}{n + 1}$$
$a_{214}$ is nonintegral.
（https://oeis.org/A108394）
Please tell me the proof.
(By computing the sequence modul... | EDIT: This isn't fully correct and contains some imprecise calculations. The core is using modulo 215. To misquote the OP: "By computing the sequence modulo 215 it is easy to show that $a_{215}$ is not integral."
ORIGINAL:
Doing the calculations modulo 215 gives the following:
$$a_0=1$$
$$a_1=\frac{1+1^5}{1}=2$$
$$a_2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
| We can write the integral as follows
\begin{align*}
\int x^3\sqrt{4-x^2}dx&=\int x(x^2-4)\sqrt{4-x^2}dx+\int 4x\sqrt{4-x^2}dx\\
&=\int(4-x^2)\sqrt{4-x^2}(-xdx)-2\int (4-x^2)^{1/2}(-2xdx)\\
&=\int(4-x^2)^{3/2}(-xdx)-2\int (4-x^2)^{1/2}(-2xdx)\\
&=\frac{1}{2}\frac{(4-x^2)^{5/2}}{5/2}-2\frac{(4-x^2)^{3/2}}{3/2}+C\\
&=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
for $f(x)=\ln({1-\frac{1}{x^2})}$ find $\lim_{n\to\infty}{f(1)+f(2)+...+f(n)}$
Let $f:(1, \infty)\to \Bbb R$ with $f(x)=\ln\left({1-\frac{1}{x^2}}\right)$
Find $$\lim\limits_{n\to\infty}\left[f(1)+f(2)+...+f(n)\right]$$
What I have done so far is
$$\begin{align}
S_n&=\sum_{k=2}^n{f(k)} \\
&= \ln\left(\prod_{k=2}^n \l... | Since $f(n)=\ln \frac{(n-1)(n+1)}{n^2}$ for all $n\ge 2$,
$$
\sum_{k=2}^n f(n)=\ln \frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdot \cdots \cdot \frac{(n-1)(n+1)}{n\cdot n} =\ln \frac{n+1}{2n}.
$$
Therefore,
$$
\lim_{n\to\infty}\sum_{k=2}^n f(n)= -\ln 2.
$$
(I excluded $f(1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Understanding Leibniz formula: $D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}f D^kg$ I would like some clarification for better understanding of Leibniz formula: $$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}f D^kg$$
If I use the formula with the following expression: $f(x)= x^3e^x$
$$D^n (f g) = \su... | $$D^n (gf) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}e^x D^k x^3 = \binom{n}{0} e^x x^3 + \binom{n}{1} e^x 3x^2 + \binom{n}{2} e^x 6x + \binom{n}{3} e^x6$$
Second expression you wrote is correct but the first formula you wrote you made mistake.
Let me show where you made mistake
$$D^n (f g) = \sum\limits_{k=0}^{n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$
Find all integer solutions $(x, y)$ of the equation
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$
What have done is that:
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so,
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we... | If $0<x\le y$ then $\frac23 = \frac1x+\frac1y \le \frac1x+\frac1x$ and so $x\le3$.
If $x<0$ then $\frac23 < \frac1{-x}+\frac23=\frac1y$ and so $y=1$.
Therefore $(x,y) \in \{ (3,3) , (-3,1) , (1,-3), (2,6),(6,2)\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:
*
*$f(x)-1$ is divisible by $(x-1)^3$
*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f ... | When a repeated factor $(x-a)^3$ is present in polynomial $P(x)$, then $P(a) = P'(a) = P''(a) = 0$
So set $f(1) - 1 = 0, f'(1) = 0, f''(1) = 0$
$f(x) - 1 = ax^5 + bx^4 + cx^3 - 1$
$f'(x) = 5ax^4 + 4bx^3 + 3cx^2$
$f''(x) = 20ax^3 + 12bx^2 + 6cx$
Hence your three equations are:
$a + b + c - 1 = 0 $
$5a + 4b + 3c = 0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$
Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$.
We have $f(0) = 0$ and $f(x) = 2f(2x) - x$, but I am not sure how to convert this functional equation into something that is easier to solve. Maybe using ind... | We prove that there is only one function $f:\mathbb R\to\mathbb R$ that satisfies $2f(2x)=f(x)+x$ and
\begin{align*}
\lim_{t\to0}tf(xt)=0
\end{align*}
for all $x\in\mathbb R$, namely $f(x)=x/3$. Note that if $f$ is continuous, it satisfies this limit condition (why?)!
By iterating $f(x)=\frac{1}{2}f(x/2)+x/4$ one finds... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Cubic modular equation
If I found the sol of (a), I could find sol of (b) and (c) by using Hensel's lemma.
I want to know the way of solving (a) without testing all the cases (ex testing 1,2,3,4,...10)
I think that there will be other way not testing all residues. It takes such a long time...
Or is there short cut for... | Note the polynomial in a) can be written as
$$x^3-3x^2-x-1=(x-1)^3-3x+1-x-1=(x-1)^3-4(x-1)-4.$$
Hence, setting $x-1=y$, it is enough to solve $y^3=4y+4 $.
We'll use Cardano's method: set $y=u+v$. The last equation becomes
$$u^3+v^3+3uv(u+v)=4(u+v)+4,$$
and adding the condition $3uv=4$, it comes down to
$$\begin{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Series $1^3-1.5^3+2^3-2.5^3+...$ I came across a question today...
Sum to $n$ term of the series $$1^3-1.5^3+2^3-2.5^3+...$$ is?
I divided this series in two different series $$\left(1^3+2^3+3^3+...\right)-\left(1.5^3+2.5^3+3.5^3+...\right)$$
$$\Rightarrow\sum^{\frac{n+1}{2}}_{r=1}r^3-\sum^{\frac{n-1}{2}}_{r=1}\left(... | Hints: You can represent the series as $$(\frac{-n}2)^3$$
Assume even number of terms
$$n^3 - (n+1)^3 = -(3n^2+3n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?... | The approach shown by Elliot G is perfectly fine, but if you like a viable alternative, you may consider that:
$$ \frac{1}{x^2+1} = \frac{1}{(x+i)(x-i)} = \frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right) $$
hence by squaring both sides:
$$ \frac{1}{(x^2+1)^2} = -\frac{1}{4}\left(\frac{1}{(x-i)^2}+\frac{1}{(x+i)^2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.
$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equati... | We have
$$
\begin{align}
&4 \cos x − 3 \sin x \\
&= 5 \left( \frac{4}{5} \cos x - \frac{3}{5} \sin x \right) \\
&= 5 \left( \sin (\tan^{-1} (\tfrac{4}{3})) \cos x
- \cos (\tan^{-1} (\tfrac{4}{3})) \sin x \right) \\
&= 5 \sin\left(\tan^{-1} (\tfrac{4}{3}) - x \right) \\
&= 5 \sin\left(x - (\tan^{-1} (\tfrac{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $a+b+..=a^7+b^7+..=0$ show that $a(a+b)..=0$ Question:
Suppose $a,b,c,d$ are real numbers such that
$a+b+c+d=a^7+b^7+c^7+d^7=0$
Show that
$a(a+b)(a+c)(a+d)=0$
My attempt: Using $a+b+c+d=0$, I get
$a(a+b)(a+c)(a+d)= 0
\implies a=0, \text{or}$ $ a(bc+cd+db)+bcd=0$
How can I use $a^7+b^7+c^7+d^7=0$ to prove $a(bc+c... | Since $d=-(a+b+c)$ and $d^7=-(a^7+b^7+c^7)$, we obtain
$$(a+b+c)^7-a^7-b^7-c^7=0$$ or
$$7(a+b)(a+c)(b+c)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$$ or
$$(a+b)(a+c)(a+d)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0.$$
Thus, it remains to prove that if $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Easy partial fraction decomposition with complex numbers There is an easy method to perform a partial fraction decomposition - described here, under the "Repeated Real Roots" title, for the coefficient A2.
The problem is - this method doesn't work in some cases, for example:
$$
\frac{1}{(2 + j \omega)^2 \, (4 + j \omeg... | The partial fraction expansion is
$$\frac{1}{(2+j\omega)^2(4+j\omega)}=\frac{A}{2+j\omega}+\frac{B}{(2+j\omega)^2}+\frac{C}{4+j\omega}\tag 1$$
To find $C$, we multiply both sides of $(1)$ by $4+j\omega$ and take the limit as $j\omega \to -4$. Proceeding, we find
$$\begin{align}
\lim_{j\omega \to -4}\frac{(4+j\omega)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Considering the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Consider the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Find a closed form expression for all x which converge and hence evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$.
Attempt at the solu... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is there a difference of $\frac{-1}{6}$ when integrating $(1-2x)^2$ in two different ways? I integrated $(1-2x)^2$ by expanding the expression first, that is $1 - 4 x + 4 x^2$ and I get $$x - 2 x^2 + \frac{4 x^3}{3} +C$$
When I integrate by substitution I get $$\frac{-(1-2x)^3}{6} + C$$
Expanding $(-(1-2x)^3)/6 +C$... | Indefinite integral can differ by a constant (you denote it by $C$). In this case you just have $C=-1/6$.
This is simply given by the fact that indefinite integral (or for this purpose better name would be antiderivative) of $f(x)$ is a function (not the function) $F(x)$ such that $F'(x) = f(x)$. And since derivative o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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The function $\lim_{n\to\infty}({4^n+x^{2n}+\frac{1}{x^{2n}})}^{1/n}$ is non-derivable at
The function $$f(x)=\lim_{n\to\infty}\left(4^n+x^{2n}+\frac{1}{x^{2n}}\right)^{\frac{1}{n}}$$ is non-derivable at how many points?
The limit is of $\infty^0$ form. Is it an indeterminate form or is it simply equal to $1$?
$$\li... | If $\frac14 < x^2 < 4$, then we can write
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}4\left(1+\left(\frac{x^2}{4}\right)^n+\left(\frac{x^{-2}}{4}\right)^n\right)^{1/n}\\\\
&=4
\end{align}$$
If $x^2>4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I go about solving this? I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$
| Let
$$
I_1=\int_0^\pi \frac{dx}{\sqrt{n^2+1}+\sin(x)+n\cos(x)}, I_2=\int_0^\pi \frac{n dx}{\sqrt{n^2+1}+n\sin(x)+\cos(x)}.
$$
Let $u=\tan\frac{x}{2}$. Then $x=2\arctan u$, $\sin x=\frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}$. Thus
\begin{eqnarray}
I_1&=&\int_0^\infty \frac{1}{\sqrt{n^2+1}+\frac{2u}{1+u^2}+n\frac{1-u^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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How is the last "=" true? How can the last equality be true?
$$
G(u)=\frac{g}{(1+u)^g-1}-\frac1u=\frac{g}{gu + \cdots + u^g} - \frac{1}{u}=\sum_{k=0}^\infty c_ku^k
$$
| If you look at the left-hadn side of the equals sign, you have
$$
\begin{align}
\frac{g}{gu + \cdots + u^g} - \frac{1}{u}
&= \frac{1}{u}\frac{g}{g + {g \choose 2} u + \cdots + u^{g-1}} - \frac{1}{u} \\
&= \frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
\end{align}
$$
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$ If we have
$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$
Then, what will be the set of $x$ for which this equation is true?
I tried to solve it by putting $x = \sin a$ or $\cos a$ but got no result.
I am totally stuck on how to do it.
| Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$
2\arcsin x =\begin{cases}
\arcsin(2x\sqrt{1-x^2}) \;\;;2x^2\le 1 \\
\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1, 0< x\le 1\\
-\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2> 1, -1< x \le 0
\end{cases}
$$
So the given equation clearly holds... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
$$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$
My thoughts were turning the right hand side to $abc$ as $abc=1$ ho... | If $a,b,c> 0$ and $abc=1$, then by Cauchy-Schwarz: $$\sum_{\text{cyc}}\frac{a}{2+bc}=\sum_{\text{cyc}}\frac{a^2}{2a+abc}=\sum_{\text{cyc}}\frac{a^2}{2a+1}\ge \frac{(a+b+c)^2}{2(a+b+c)+3}$$
And by AM-GM $a+b+c\ge 3\sqrt[3]{abc}=3$, so: $$(a+b+c)^2-2(a+b+c)-3=((a+b+c)-3)((a+b+c)+1)\ge 0$$
Equality holds if and only if $a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. ... | Hint:
$$ 1\pm 2\pm 3\pm 4\ldots \pm n = \frac{n(n+1)}{2}-a_2-a_3-\ldots-a_n, \qquad a_k\in\{0,2k\}.$$
Consider $(1+x^4)(1+x^6)\cdot\ldots\cdot(1+x^{2n})$. How many monomials appear in the expansion?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps:
*
*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
*${\displays... | You are good except for your last step; $\tan(\theta)$ is not $x^2$, and in particular $\csc$ is not $1/\sec$.
One way to identify what $\tan(\theta)$ is in terms of $x$ is to consider the triangle that justifies the substitution $x=\sec(\theta)$ in the first place. This has a leg of $\sqrt{x^2-1}$, which means the hyp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve by factoring, then perform synthetic division. $${x^4 - 3x^3 + 3x^2 - x = 0}$$
$${x(x^3 - 3x^2 + 3x - 1)}$$
Now, we have to perform synthetic division, if I'm correct we get the 1 to divide by from the X outside of the parentheses from the second equation, which is above this paragraph, and is also the factored f... | $$x^3-3x^2+3x-1=x^3-x^2-2x^2+2x+x-1=x^2(x-1)-2x(x-1)+(x-1)=(x-1)(x^2-2x+1)=(x-1)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge? Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge?
I used the integral test and found that it does, but it was a bit cumbersome. Is there an easier way?
Thanks
| By partial fraction decomposition,
$$\begin{eqnarray*}\sum_{n\geq 3}\frac{n^2-2n-3}{(n-2)^2 (n+2)^2}&=&\frac{1}{32}\sum_{n\geq 3}\left(\color{blue}{\frac{7}{x-2}-\frac{7}{x+2}}+\color{red}{\frac{10}{(x+2)^2}-\frac{6}{(x-2)^2}}\right)\\&=&\frac{1}{32}\left[\color{blue}{\frac{175}{12}}\color{red}{-6\,\zeta(2)+10\left(\ze... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a ... | Let's reform this inequality in a way such that we can comprehend it better. Define $a=\dfrac{y}{x}$ and $b=\dfrac{z}{y}$, therefore $\dfrac{x}{z}={1\over ab}$. We can suume without lose of generality that $a,b\le1$ We need to prove that $$\dfrac{x}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{y}{8+5\left(\dfrac{z}{y}\right)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
"answer_count": 8,
"answer_id": 3
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Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that,
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how can I simplify th... | From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\dfrac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\dfrac{x+y}{1-xy} & \mbox{if } xy>1\end{cases} $$
$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{ if }x^2<1$$
Finally use $$\arcsin x=\arctan\dfrac x{\sqrt{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area inside the circle $r = 10 \sin \theta$ and above the line $r = 2 \csc \theta$. Function Plotter graph:
I think the formula is
$$A = \frac 1 2 \int_{\alpha}^{\beta} (\text{outer})^2 - (\text{inner})^2 d\theta$$
where $\alpha, \beta$ are where they intersect in $[0, 2\pi]$.
This is what I got based on th... | We can do this all with classical geometry.
The region in question is a circle or radius 5 less a section of a circle with angle $2\cos^{-1} \frac 35$ plus 2 $3-4-5$ right triangles.
$25\pi - 25\cos^{-1} \frac 35 + 12$
Lets see if calculus gives us the same answer.
limits of integration.
$2\csc \theta = 10\sin \theta\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Calculating cosine of dihedral angle
Let $O,A,B,C$ be points in space such that $\angle AOB=60^{\circ},\angle BOC=90^{\circ},\angle COA=120^{\circ}$ Let $\theta$ be the acute angle between the planes $AOB$ and $AOC$. Find $\cos\theta.$
Here is what I did:
I let $AO=OB=BA=1, OC=1$. Then we can compute $BC=\sqrt{2},AC=... | You can also use the Spherical Law of Cosines.
We imagine a sphere centered at $O$, and we take $A$, $B$, $C$ the points where the rays from $O$ meet that sphere as vertices of a curvilinear triangle on the surface of that sphere. The "sides" of that triangle are arcs of great circles, and the lengths of those arcs are... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square.
Since $n^2+n+43$ is odd,if it's a perfect square it can be written as: $8k+1$,then:
$$n^2+n+43=8k+1\Rightarrow\ n^2+n+42=8k\ \Rightarrow\ n(n+1)\equiv6\pmod8$$
So $... | $n^2+n+43=m^2$ iff $4n^2+4n+172=4m^2$ iff $(2n+1)^2+171=(2m)^2$.
Let $u=2n+1$ and $v=2m$. Then $171 = v^2-u^2 =(v+u)(v-u)$.
Since $171 = 3 \cdot 3 \cdot 19$, we can try all possibilities for $v+u$ and $v-u$ and then find $n$ and $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Trying to compute a Riemann-sum as an integral where the $\lim_{n\to\infty} a\cdot n +\sum_{n^2}^{4n^2} \dfrac{1}{k+n}$
I have the following problem: I want to figure out , for what value of a does this limit converge and what value does it then converge to.
I'm assuming that I'm supposed to rewrite the sum as an integ... | Unless there is some miraculous cancellation -- and there will not be -- the term $a \cdot n$ with $a \neq 0$ will cause divergence. So let's take $a = 0$ and focus on the behavior of the sum. This would be a typical Riemann sum for $f(x) = (1 + x)^{-1}$ if not for the limits that are given.
Nevertheless, we can find ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{... | Here's another approach, using recurrence relations.
Note that
$$\begin{align}
a_1&=\frac 17\qquad\text{and}\\
a_n&=a_{n-1}\left(\frac{2n-1}{2n+5}\right)\\
(2n+1)a_n+4a_n&=(2n-1)a_{n-1}\\
a_n&=\frac 14\big(b_{n-1}-b_n\big)\\
\end{align}$$
where $b_m=(2m+1)a_m$.
Summing from $2$ to $n$ by telescoping and adding $a_1$:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\frac{1}{7\sqrt{2}} \leq \int_{0}^1 \frac{x^6}{\sqrt{1+x^2}}dx \leq \frac{1}{7}$
Prove that $\displaystyle \dfrac{1}{7\sqrt{2}} \leq \int_{0}^1 \dfrac{x^6}{\sqrt{1+x^2}}dx \leq \dfrac{1}{7}$.
My book says let $f(x) = \dfrac{1}{\sqrt{1+x^2}}$ and $g(x) = x^6$. Then $\displaystyle \int_0^1 \dfrac{x^6}{\sqrt... | Achtung: overkill. $g(x)=\frac{1}{\sqrt{1+x}}$ is a convex function on $[0,1]$ (since its logarithm is convex) bounded between $1$ and $\frac{1}{\sqrt{2}}$. Since $g'(1)=-\frac{1}{4\sqrt{2}}$,
$$\forall x\in(0,1),\qquad \frac{(1-x)}{4\sqrt{2}}+\frac{1}{\sqrt{2}}\leq g(x)\leq (1-x)+\frac{x}{\sqrt{2}}\tag{1}$$
holds by ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Approximation of $\sqrt{2}$ I got the following problem in a chapter of approximations:
If $\frac{m}{n}$ is an approximation to $\sqrt{2}$ then prove that $\frac{m}{2n}+\frac{n}{m}$ is a better approximation to $\sqrt{2}.$(where $\frac{m}{n}$ is a rational number)
I am stuck in this problem. Any help will be appreciate... | $\frac{m}{n}$ approximately equals $\sqrt 2$
Suppose $\epsilon$ is our the error in our estimate. i.e. $\frac{m}{n} + \epsilon = \sqrt 2$ or
$(\frac{m}{n} + \epsilon)^2 = 2$
$\frac{m^2}{n^2} + 2\frac{m}{n}\epsilon + \epsilon^2 = 2$
Now we need to solve for $\epsilon.$
If $\epsilon$ is small then then $\epsilon^2$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{2\cos x}{\cos 2x + 1 }= \sec x$ Prove that $\dfrac{2\cos x}{\cos 2x + 1 }= \sec x$.
So far I have:
$\dfrac{2\cos x}{\cos 2x + 1 }= \dfrac 1 {\cos x}$
Where do I go from here?
| $$\frac{2\cos(x)}{1+\cos(2x)}=\sec(x)\Longleftrightarrow$$
Use $\sec(x):=\frac{1}{\cos(x)}$
$$\frac{2\cos(x)}{1+\cos(2x)}=\frac{1}{\cos(x)}\Longleftrightarrow$$
$$2\cos^2(x)=1+\cos(2x)\Longleftrightarrow$$
Use $\cos(2x)=2\cos^2(x)-1$
$$2\cos^2(x)=1+2\cos^2(x)-1\Longleftrightarrow$$
Use $1-1+2\cos^2(x)=0+2\cos^2(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Sum of the series $\sum \frac{n}{2^{n}}$ I know that the series converges by d'Alembert ratio test, where $\lim\left ( \frac{A_{n+1}}{A_{n}} \right )= \frac{1}{2}$, but I don't know how to calculate the sum of the serie. Thanks for the help.
| \begin{align*}
S&= \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+.....\\
\frac{1}{2}S&=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\\
S-\frac{1}{2}S&=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......\\
S-\frac{1}{2}S&=1
\end{align*}
Thus $S=2$
| {
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"url": "https://math.stackexchange.com/questions/1788459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $ $$ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $$
I thought of substituting $ x-\frac{1}{x} $ as $t$ but it gets stuck midway. I am close but I think I need to sustitute something else here.
| Let $$I = \int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}}dx = -\int\frac{(1+x^2)}{x^2(x-\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}dx$$
So $$I = -\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})\sqrt{(x-\frac{1}{x})^2+2}}dx$$
Now Put $\displaystyle x-\frac{1}{x} = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt$
So $$I = -\i... | {
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"url": "https://math.stackexchange.com/questions/1788945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $... | Let $x = \ln t$. Then $dx = dt/t$ and we have
$$\int \sqrt{1 + e^x} \, dx = \int \frac{\sqrt{1 + t}}t \, dt.$$
Now let $u = t + 1$. Then $du = dt$ and we have
$$\int \frac{\sqrt{1+t}}t \, dt = \int\frac{\sqrt{u}}{u-1} \, du.$$
Now let $v = \sqrt{u}$. Then $u = v^2$ and so $du = 2v \, dv$, and we have
$$\int\frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the equation $ x^{13} = 1$ by radicals? (And $x^{19}=1$) Is there any elementary way to solve the equation $ x^{13}= 1 $ by means of radicals? If not, how to get all the solutions?
Remark: The transcendental form of the solution by means of sines and cosines is not allowed, but only radicals, since this eq... | First, we establish some ground rules. All cubics and quartics can be solved by radicals, agreed? Thus, given,
$$\frac{x^{13}-1}{x-1}=x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0\tag1$$
to solve for $x$ in radicals all we have to do is to "break-up" the $12$-deg into quadratics, cubics, or quartics.
I. M... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive integers $a,b$ satisfying $a^3+a+1=3^b$ How to prove that $a=b=1$ is the only positive integer solution to the following Diophantine equation?$$a^3+a+1=3^b$$
| Partial solution, Check the equation mod 9 $\big(\forall b>1,3^b\equiv 0 \pmod 9\big)$. Nine cases to check:
$a=0 \pmod 9, a^3+a+1\equiv 1 \pmod 9\\
a=1 \pmod 9, a^3+a+1\equiv 3 \pmod 9 \\
a=2 \pmod 9, a^3+a+1\equiv 2 \pmod 9\\
a=3 \pmod 9, a^3+a+1\equiv 4 \pmod 9\\
a=4 \pmod 9, a^3+a+1\equiv 6 \pmod 9\\
a=5 \pmod 9, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find $(x+y)$.
We know that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find the expression $(x+y)$.
My work so far:
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$
$$\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}\left(x+\sqrt{x^2+... | Notice that we have
$$-\left(y+\sqrt{y^2+1}\right)=\left(x-\sqrt{x^2+1}\right)$$ From your third line. Now $$x+y=\sqrt{x^2+1}-\sqrt{y^2+1}\tag{1}$$ In the same fashion, we can get$$x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\tag{2}$$ Now adding $\text{(1)}$ and $\text{(2)}$ together gives $x+y=0$. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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problem with Combination and Permutation Four married couples have bought 8 seats in a row for a concert.
In how many ways can they be seated:
a)if each couple is to sit together?
(8)(1)(6)(1)(4)(1)(2)(1)
b)if all men sit together?
n-k+1=8-4+1=5
so, (5)(4! 4!)=5! 4!
c)If no man sits next to his wife ?(This is a non-tri... | a) and b) are correct.
For c), we will use Inclusion/Exclusion. There are $8!$ arrangements without restriction. From this we need to subtract the number of bad arrangements, where at least one couple are next to each other.
First we count the number of arrangements where Couple A are together. Tie them together with r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$ for $a,b,c >0$ and $a+b+c=3$
Let $a,b,c >0$ with $a+b+c=3$. Prove that
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52.$$
What I did:
It is cyclic inequality so I assume $c= \min\{ a,b,c \}$.
I consider the first case where $a\ge b\ge c$ then
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac... | Perhaps it would help to take $a=1+x$, $b=1+y$, $c=1+z$, so that $x+y+z=0$
Now the inequality looks like:
$$(1+x)^{|y-x|} + (1+y)^{|z-y|} + (1+z)^{|x-z|} \geq \frac{5}{2}$$
Using bernoulli's inequality, the expression is more or less
$$1 + x|y-x| + 1 + y|z-y| + 1 + z|x-z| \geq \frac{5}{2}$$
$$x|y-x| + y|z-y| + z|x-z| \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Discriminant formula issue Could you please explain to me, how to get the formula of discriminant ? How can I visualize it, any articles, lectures?
I can memorize it $b^2 - 4ac$ But, want to understand it.
Thanks.
| You must have seen a method called "completing the square" before you saw the quadratic formula in school right? This is where it comes from, in fact, if you know how to complete the square, you can derive the whole quadratic formula on your own.
Consider the quadratic equation that you're used to $$ax^2 + bx + c = 0.$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
| $(1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+(abc)$
Applying AM≥GM on $(a+b+c)+(ab+bc+ca)+(abc)$
$$\frac{(a+b+c)+(ab+bc+ca)+(abc)}{7}≥(a^{4}b^{4}c^{4})^\frac{1}{7}$$
So $(a+b+c)+(ab+bc+ca)+(abc)≥7(a^{4}b^{4}c^{4})^\frac{1}{7}$
If we add 1 on LHS so now LHS>RHS
So $(1+a)^{7}(1+b)^{7}(1+c)^{7}>7^{7}(a^{4}b^{4}c^{4})$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In how many ways can $2m$ objects be paired and assigned to $m$ boxes? Been refreshing my statistics knowledge and came across this problem while going through a book.
How many ways can $2m$ objects be paired and assigned to $m$ boxes?
The book claims the answer is
$$ \frac{(2m)!}{2^m} $$
Which makes sense, however, ... | If $m = 1$, we have a set $\{1,2\}$ and there's only $1$ unordered pair we can form.
If $m = 2$, we have a set $\{1,2,3,4\}$ and there are $3$ pairings we can form, namely,
$$\{\{1,2\}, \{3,4\}\}$$
$$\{\{1,3\}, \{2,4\}\}$$
$$\{\{1,4\}, \{2,3\}\}$$
For a general $m$, we have a set $\{1,2,\dots,2m\}$ and the number of pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803926",
"timestamp": "2023-03-29T00:00:00",
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Integrating trig function I'm stuck at this problem:
$$ \int{\sqrt{(\sin^2 x)^2 + (2\sin x \cos x)^2}dx} = \int{\sqrt{\sin^2 x \sin^2 x + 4\sin^2 x \cos^2 x} dx}$$
I tried a few trig identities: $\sin^2 x = \frac{1-\cos 2x}{2} $ and $ \cos^2 x = \frac{1+\cos 2x}{2} $ and $\sin^2 x + \cos^2 x = 1$. Keep hitting dead en... | Try a substitution with the last expression that you have
$$\int\sqrt{\sin^4(x)+4\sin^2(x)\cos^2(x)}dx\quad u=\cos(x)$$.
This leads us to the following integral where we do another substitution:
$$\int\sqrt{3u^2+1}du \quad\!\! \text{substitute}\quad\!\! u=\sqrt{\dfrac{a}{b}}\tan (v) \quad\!\!\text{where}\quad\!\! a=... | {
"language": "en",
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Solutions to equation
Let $\alpha(n)$ be the number of pairs $(x, y)$ of integers such that $x+y = n, 0 \le y \le x$, and let $\beta(n)$ be the number of triples $(x, y, z)$ such that $x + y + z = n$ and $0 \le z \le y \le x.$ Find a simple relation between $\alpha(n)$ and the integer part of the number $\frac{n+2}{2}... | The triples that satisfy your condition are of two types. Type 1 are the triples with smallest element $0$, and Type 2 are the ones with smallest element $\ge 1$.
There are just as many triples of Type 1 with sum $n$ as there are pairs $(y,x)$ with $0\le y\le x$ such that $y+x=n$.
There are just as many triples of Typ... | {
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Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy... | $$y=-x, z=1$$
$${y = \frac{-1 + x - \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x + \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$
$${y = \frac{-1 + x + \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x - \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$
All integer solutions:
$x=-3, y=-2, z=6$
$x=-2, y=-3, z=6$
$x=-2, y=0, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$
When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The e... | For the maximum values, we could also show it using derivatives.
Consider the function $$F=\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ Compute derivatives; after simplifications, we get $$F'_x=\frac{y z \left(y z-x^2\right)}{(x+y)^2 (x+z)^2 (y+z)}$$ $$F'_y=\frac{x z \left(x z-y^2\right)}{(x+y)^2 (x+z) (y+z)^2}$$ $$F'_z... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $|2x-1| -|x+5| = 3$
Problem : Solve the equation $|2x-1| - |x+5| = 3$
In my attempt to solve the problem, I only manage to get one of the solutions.
Attempted Solution
$$\begin{equation}
\begin{split}
|x|-|y| & \leq |x-y| \\
\implies |2x-1| - |x+5| & \leq |(2x-1)-(x+5)| \\
\implies 3 &\leq |x-6| \\... | Because $|2x-1|-|x+5|$ is also $\leq |(x+5) - (2x-1)|$. (Multiplying through by $-1$ doesn't change the absolute value...)
The generic method is this:
*
*$x+5$ is nonnegative for $x \geq -5$ and negative for $x < -5$.
*$2x-1$ is nonnegative for $x \geq 1/2$ and negative for $x < 1/2$.
*On the interval $(-\infty,-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t: $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^... | Consider:
$$ \frac{1}{(x^2-1)^2} = \frac{1}{(x+1)^2(x-1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2}$$
Try to decompose the above
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{... | Yet another method —Taylor expansion (to first order):
We use that $\ln(1+u) = u + o(u)$ when $u\to 0$.
Then $\ln(x+4) = \ln\left(1+\frac{x}{4}\right) + \ln 4 = \frac{x}{4} + o(x) + \ln 4$, and
$$
\frac{\ln(x+4)- \ln 4}{x} = \frac{\frac{x}{4} + o(x)}{x} = \frac{1}{4} + o(1) \xrightarrow[x\to0]{} \frac{1}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Computing an infinite product $\prod_{n=1}^{\infty} \frac{1}{2}(1+\cos\frac{x}{2^n})$ I would like to compute the infinite product
$\displaystyle f(x)=\prod_{n=1}^{N\rightarrow\infty} \frac{1}{2}\left({1+\cos\frac{x}{2^n}}\right)$ for a given real $x$.
Since the terms in the product are positive and smaller than $1$, t... | Take the well known product and square:
$$\prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$$
$$\prod_{n=1}^{\infty} \cos^2\frac{x}{2^n}=\frac{\sin^2(x)}{x^2}$$
Now use the formula:
$$\cos^2 \frac{t}{2}=\frac{1}{2} (1+\cos t)$$
To get:
$$\prod_{n=0}^\infty \frac{1}{2} \left( 1+\cos \frac{x}{2^n} \right)=\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Recursive Sequence $a_n = \frac{1}{2} (a_{n-1} + 5) $ I got this question in which they ask me to explain why it is convergent and evaluate its limit.
$$a_1=3\;and\;a_n = \frac{1}{2} (a_{n-1} + 5) \\ n=2,3,4,... $$
To prove it's convergent, I show that it is increasing and bounded above by 5. Also, I find its limit by... | It might help to solve the recurrence relation
Define $b_n\equiv a_n+c$ and find the value of $c$ that makes $b_n$ a geometric sequence
$$a_n = \frac{1}{2} (a_{n-1} + 5)\implies b_n-c = \frac{1}{2} (b_{n-1}-c + 5)$$
$$\implies b_n = \frac{1}{2} b_{n-1}-\frac c2 + \frac 52 +c$$
$$\implies b_n = \frac{1}{2} b_{n-1} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating the following $\int \sqrt{\tan x+1}\,dx$
Question: Integrate the following, $$\int\sqrt{\tan x+1}\;dx.$$
Wolfram Alpha returns a non-elementary answer. Can someone please spot the mistake I have made here:
First consider this integral:
$$\int \frac{1}{(x+1)\sqrt{x+3}} \, dx = -\sqrt{2}\tanh^{-1}\frac{\sq... | Let $I=\int \sqrt{\tan x+1} d x$ and $u=\sqrt{\tan x+1}$.
\begin{aligned}
I &=\int \frac{2 u^{2} d u}{\left(u^{2}-1\right)^{2}+1} \\
&=2 \int \frac{u^{2}}{u^{4}-2 u^{2}+2} d u \\
&=2 \int \frac{d u}{u^{2}+\frac{2}{u^{2}}-2} \\
&= 2 \int \frac{\left(1+\frac{\sqrt{2}}{u}\right)+\left(1-\frac{\sqrt{2}}{u}\right)}{u^{2}+\f... | {
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"url": "https://math.stackexchange.com/questions/1812219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $|a_{2n}-a_{n}|<\frac{10}{27}$ Let sequence $$a_{1}=1,a_{n+1}=\dfrac{2}{2a_{n}+1}$$
show that
$$\left|a_{2n}-a_{n}\right|<\dfrac{10}{27}$$
and the constant $\frac{10}{27}$ A smaller number instead
$$a_{n+1}+\dfrac{1+\sqrt{17}}{4}=2(1+\sqrt{17})\cdot\dfrac{a_{n}+\frac{1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{1}$$... | For convenience, let $p=\frac{1+\sqrt{17}}{4}$ and $q=\frac{-1+\sqrt{17}}{4}$
Let $b_{n}=\frac{a_{n}+p}{a_{n}-q}$, then your last equation can be rewritten as follows:
$b_{n+1}=\frac{p}{q}b_{n}$
Since the above is a geometric sequence, we have $b_{n}=b_{1}r^{n-1}$, where $r=\frac{p}{q}$.
Unsubstitting, we have $\frac{... | {
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"url": "https://math.stackexchange.com/questions/1814625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that $ \left\lfloor \frac{n-1}{2}\right\rfloor +\left\lfloor \frac{n+2}{4}\right\rfloor + \left\lfloor\frac{n+4}{4} \right\rfloor =n$
Show that :$$ \forall \in \mathbb{Z}, \left\lfloor \dfrac{n-1}{2}\right\rfloor +\left\lfloor \dfrac{n+2}{4}\right\rfloor + \left\lfloor\dfrac{n+4}{4} \right\rfloor =n$$
Solu... | Because the least common multiple of the denominators is $4$, the effect of the floor functions will repeat with period $4$. You can convince yourself of this by trying $n$ from $0$ to $12$ or so. This means there are four cases depending on $n \bmod 4$ and they try all four cases. Each column in the middle of the t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort ... | Just another way to compute the antiderivative.
Change variable $$x^2-16=t^2\implies x=\sqrt{t^2+16}\implies dx=\frac{t}{\sqrt{t^2+16}}\,dt$$
So, $$I=\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx=\int \sqrt{t^2+16}\,dt$$ One integration by parts $$u=\sqrt{t^2+16}\implies du=\frac{t}{\sqrt{t^2+16}}\,dt$$ $dv=dt \implies v=t$.... | {
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"timestamp": "2023-03-29T00:00:00",
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Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
| The important binomial theorem states that
\begin{equation}
\sum_{k=0}^n\binom{n}{k} x^ky^{n-k}=(x+y)^n
\end{equation}
Setting $x=1$ and $y=1$, we have the following relation
\begin{equation}
\sum_{k=0}^n\binom{n}{k} =2^n
\end{equation}
and setting $x=-1$ and $y=1$, we have the following relation
\begin{equation}
\sum_... | {
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Series expansion of $\int x^xdx$ The indefinite integral:
$$J=\int x^xdx$$
has no known closed form solution.
Expanding in series the function $f=x^x$ we get:
$$f\simeq\sum_{k=0}^N \dfrac{x^k\ln(x)^k}{k!}$$
So we can write:
$$J\simeq\sum_{k=0}^N\int\dfrac{x^k\ln(x)^k}{k!}dx$$
and we get:
$$J\simeq x+\sum_{k=1}^N\dfrac... | $\int x^x~dx=\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\dfrac{x^n(\ln x)^n}{n!}dx$
For $\int x^n(\ln x)^n~dx$ , where $n$ is any non-negative integers,
$\int x^n(\ln x)^n~dx$
$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$
$=\dfrac{x^{n+1}(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How would you find the exact roots of $y=x^3+x^2-2x-1$? My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.
I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{... | Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$
The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$
( coming from the $t^7 - 1$ factor in numerator )
for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i} + e^{-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
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Complex Numbers and Binomial Expansion
I was able to show the above by equating the solution of $cos \ 5\theta=0$ (which gives $\pi/10$) and the solution of ${16cos}^5\theta - {20cos}^3\theta+5cos\theta = 0$ (which is what you get when you use De Moivre's Theorem with binomial expansion). However, I do not understand ... | What you have to realize about the equation $\cos5\theta=0$ is that $\frac{\pi}{10}$ is a solution because $\cos\frac{\pi}{2}=0$, but also that $\frac{7\pi}{10}$ is a solution because $\cos\frac{35\pi}{10}=\cos\frac{5\pi}2=\cos\left(2\pi+\frac{\pi}2\right)=\cos\frac\pi 2=0$. Therefore, we can use this equation to find ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute covariance matrix random walk
Consider a random walk on the square lattice $\mathbb{Z}^2$ with diagonal jumps of size $2$, i.e. the jump probabilities are
$$P(X_1 = x) =
\begin{cases}
\frac{1}{4} & \quad \text{if } x \in \{(2,2), (-2,2), (2,-2), (-2,-2) \} \\
0 & \quad \text{otherwise } \\
\end{ca... | Suppose that $i=1$ and $j=2$ and $d=2$. Then we have
\begin{align}
& \sum_{x \in \mathbb{Z}^d} x^{(i)} x^{(j)} P(X_1 = x) = \sum_{x \in \mathbb{Z}^2} x^{(1)} x^{(2)} P(X_1 = x) \\[10pt]
= {} & \underbrace{2\cdot 2\cdot \frac 1 4}_{x \,=\, (2,2)} + \underbrace{2 \cdot(-2)\cdot \frac 1 4}_{x \,=\, (2,-2)} + \underbrace{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Recursive to non recursive function $$
f(x) =
\begin{cases}
0 & x=1 \\
f(x-1)+1 & \frac{f(x-1)}{x-1} < p \\
f(x-1) & \text{otherwise} \\
\end{cases}
$$
Where $p$ is a constant less that or equal to 1.
And x is a whole number greater than 0.
How do I get a non recursive solution for $f(x)$?
| To make Nithin's answer more precise, we have
$$f(x) = \begin{cases}
\left\lceil (x-1) p \right\rceil & \text{ if } p \leq 1 \\
x - 1 & \text{ if } p > 1
\end{cases}$$
You can easily prove that by induction on $x$:
*
*If $x = 1$, then $f(x) = 0 = x-1 = \left\lceil (x-1)p \right\rceil$.
*If $p > 1$ and $f(x-1) = x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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In $\triangle ABC$, if $\tan A$, $\tan B$, $\tan C$ are in harmonic progression, then what is the minimum value of $\cot \frac{B}{2}$?
In a $\triangle ABC$, if $\tan A$, $\tan B$, $\tan C$ are in harmonic progression, then what is the minimum value of $\cot(B/2)$?
$\bf{My\; Try::}$ Here $A+B+C=\pi\;,$ Then $\tan A+\t... | We have $\cot A+\cot C=2\cot B\ \ \ \ (1)$
and Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$ says $$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$
Eliminating $\cot C,$
we get $$\cot^2A-2\cot A\cot B+1-2\cot^2B=0$$ which a Quadratic Equation in $\cot A$
As $\cot A$ is real, the discrim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that for all odd integer $n, n^{4}=1\pmod {16}$ My answer: $n=2k+1$
$n^{4}=(2k+1)^{4}$=$16k^{4}+32k^{3}+8k^{2}+24k+1$.
I do not know how to conclude; really needed help here.
| $$n^4-1=(n^2+1)(n+1)(n-1)$$
If $n$ is odd, then all the factors are even, and one of the factors $(n+1)$ or $(n-1)$ is divisible by $4$. Hence $2^4|n^4-1$, or in other words $n^4\equiv1\pmod{16}$.
This can be generalized to see that $n^{2^k}\equiv1\pmod{2^{k+2}}$ for odd $n$:
$$n^{2^k}-1=\left(\left(n^{2^{k-1}}+1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 13,
"answer_id": 3
} |
Prove using induction that $C(n) = C(n/5) + C(3n/4) + n$ is $O(n)$ I was hoping someone could take a look at my answer to this question and check if it's correct and offer some advice/help on how to correct if it's wrong.
Question:
Consider the function $C: \mathbb N \rightarrow \mathbb N$ defined by:
$C(n) \begin{case... |
I have a feeling I'm doing something wrong
So, anyway, you have to be partially right and partially wrong. :-) But your feeling is right, because if we want to prove inductively the claim with $n_0$ we have to have also $\lfloor \frac{n}{5} \rfloor\ge n_0$ and $\lfloor \frac{3n}{4} \rfloor\ge n_0$, which makes the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.