Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $ Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $
Try:: Let $x = 2009$, Then expression convert into $(x+4)^4-4(x+2)^4+6x^4-4(x-2)^4+(x-4)^4$
$\left\{(x+4)^4+(x-4)^4\right\}-4\left\{(x+2)^4+(x-2)^4\right\}+6x^4$
But This is very Complicated... | Let's calculate the coefficients at powers of $x$.
First, it's clear the coefficient at $x^4$ is zero ($1+1-4-4+6=0$).
Next, from your grouping of terms it's obvious that coefficients at $x^3$ and $x$ are zero, too. Or from another point of view, our function is even.
Then again, the coefficient at $x^2$ is $2\cdot 6\... | {
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"url": "https://math.stackexchange.com/questions/494686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Proving a formula about binomial coefficients I found the following formula in a book without any proof:
$$\sum_{j=k}^{\lfloor\frac n2\rfloor}{\binom{n}{2j}}{\binom{j}{k}}=\frac{n}{n-k}\cdot2^{n-2k-1}{\binom{n-k}{k}}$$
where $n$ is a natural number and $k$ is an integer which satisfies $0\le k \le\frac n2$.
I've tried... | We will use induction to prove the identity (equivalent to the one given) that
1) $\displaystyle\sum_{j\ge0}\binom{n}{2j}\binom{j}{k}=2^{n-2k-1}\bigg[\binom{n-k}{k}+\binom{n-k-1}{k-1}\bigg]$ $\;\;$for $0\le k\le\frac{n}{2}$ and the identity
2) $\displaystyle\sum_{j\ge0}\binom{n}{2j+1}\binom{j}{k}=2^{n-2k-1}\binom{n-k-1... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $2^b-1$ does not divide $2^a + 1$ for $a,b>2$ I'm trying to prove $2^b-1$ does not divide $2^a + 1$ for $a,b>2$. Can someone give a hint in the right direction for this?
| Assume otherwise and let $a$ be the nonnegative integer such that $2^b-1\mid 2^a+1$ for some $b>2$.
If $2^a+1=k(2^b-1)$ with $k\ge 1$, then $$2^{a-b}+1 = 2^a+1-2^{a-b}(2^b-1)=(k-2^{a-b})(2^b-1).$$
If $a\ge b$, this shows that $2^b-1\mid 2^{a-b}+1$, contradicting minimality of $a$.
Therefore $a\le b-1$. But then we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the simultaneous equations $x + \frac{3x-y}{x^2+y^2} = 3 $, $ y – \frac{x+3y}{x^2+y^2} = 0$ Find all solution in $\mathbb{R}$ for the following system of equations:
\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y – \frac{x+3y}{x^2+y^2} = 0 \end{cases}
I've tried few method, but none bring a success.
First I tri... | Complexify. Let $z = x+iy$. Then for $x^2+y^2 \neq 0$:
$$\begin{align}
&&\left(x + \frac{3x-y}{x^2+y^2}\right) + i\left(y - \frac{x+3y}{x^2+y^2}\right) &= (3+i0)\\
&\iff& (x+iy) + \frac{(3-i)(x-iy)}{x^2+y^2} &= 3\\
&\iff& z + \frac{3-i}{z} &= 3\\
&\iff& z^2 - 3z + (3-i) &= 0\\
&\iff& \left(z-\frac32\right)^2 &= i - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495932",
"timestamp": "2023-03-29T00:00:00",
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Finding the equation of a circle given two points and the radius Can't seem to figure this out - the question is:
There are exactly two circles of radius $r = \sqrt{5}$ through the points $(6,3)$ and $(7,2)$. Find the equations of both circles.
I was thinking that I would find the equation of the line passing through... | Let A be (6,3) and B be (7,2).
Gradient of AB = $\frac { 2-3 }{ 7-6 }$ = -1
So gradient of perpendicular bisector of AB = $-\frac { 1 }{ 1 }$ = 1
Midpoint of AB is ($\frac { 6+7 }{ 2 }$, $\frac { 3+2 }{ 2 }$), which is ($\frac { 13 }{ 2 }$, $\frac { 5 }{ 2 }$).
Hence, the equation of the perpendicular bisector is $y-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove that $a_n=na/(1+n^2 b)$ is strictly decreasing given a, b >0
I have tried proving this directly, by setting
$\frac{na}{1+n^2b} > \frac{(n+1)a}{1+(n+1)^2b}$
and end up with
$\frac{na+n^3ab+2n^2ab+nab}{(1+n^b)(1+n^2b+2bn+b)} > \frac{na+a+n^3ab+n^2ab}{(1+n^b)(1+n^2b+2bn+b)}$
I simplify that down to...
$\frac{n^2b+nb... | Just rewrite it as
$$
a_n = \frac{a}{\frac{1}{n} + nb}.
$$
Since $\frac{1}{n} + nb$ is increasing for sufficiently large $n$ and for $b > 0$, it shows $a_n$ must be decreasing for sufficiently large $n$ and for $a > 0$.
To be rigorous we need to say precisely what sufficiently large means and prove it. By sufficiently ... | {
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"timestamp": "2023-03-29T00:00:00",
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Findinf volume enclosed using triple integrals I have to find the volume cut off from the paraboloid $4z=x^2+y^2$ by the plane $z=4$
I understand that the paraboloid will have its vertex at the orgin and its axis as the $z-axis$. now, if it is cut off by the plane $z=4$, i get the following limits for the triple intgra... | We can interpret this volume as $$\int_0^4 \text{area of disk at height } z\ dz.$$ More formally, we convert to a cylindrical coordinate system, $$\int_0^4 \underbrace{\int_0^{2\pi} \int_0^{\sqrt{4z}} r\ dr\ d\theta}_{\text{area of disk}}\ dz.$$
The area of a disk is $\pi r^2$ where $r$ is the radius, in this case $r=... | {
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"url": "https://math.stackexchange.com/questions/501139",
"timestamp": "2023-03-29T00:00:00",
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Which of the following are complete residue systems modulo $11$? Which of the following are complete residue systems modulo 11?
$(a)\quad 0,1,2,4,8,16,32,64,128,256,512$
$(b)\quad 1,3,5,7,9,11,13,15,17,19,21$
$(c)\quad 2,4,6,8,10,12,14,16,17,20,22$
$(d)\quad -5,-4,-3,-2,-1,0,1,2,3,4,5$
I have the answer which is th... | The set of integers {0, 1, 2, ..., n - 1} is called the least residue system modulo n. Any set of n integers, $\bf no\ two\ of\ which\ are\ congruent\ modulo\ n$, is called a $\bf complete\ residue\ system\ modulo\ n$.
http://en.wikipedia.org/wiki/Complete_residue_system_modulo_m#Residue_systems
{0,1,2,3,4,5,6,7,8,9,10... | {
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Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:
$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$
(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor
| This is a sequel to my previous work. The aim is to get the relation $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$ directly rather than just verifying it. In order to do so, I have to re-arrange the original question as
“If $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor $x^2+px+1$, then $(a^2-c^2)(a^2-c^2+bc)=(ab)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integrating $\int \frac{dt}{(t+2)^2(t+1)}$ I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \... | $$D_t\;\;\frac{1}{t+1}-\ln|t+2|+\ln|t+1|+c=\\=\frac{(t+2)\cdot0-1\cdot1}{(t+1)^2}-\frac{1}{t+2}+\frac{1}{t+1}+0=\\=\frac{-(t+1)-(t+1)(t+2)+(t+2)^2}{(t+2)^2(t+1)}=\\=\frac{-t-1-t^2-3t-2+t^2+4t+4}{(t+2)^2(t+1)}=\\=\frac{1}{(t+2)^2(t+1)}$$
Correct .. -
| {
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$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6 The problem is as follows: Prove that $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0 \, \text{mod n} \ \text{if and only if } n \equiv \pm 1 \, \text{mod} 6.$$
My idea is to of course rewrite the summation. We have $$1^2 + 2^2 + 3^2 + \cdots ... | For the proof, we use the following observations:
(i) Suppose that $n$ divides $\frac{(n-1)(n)(2n-1)}{6}$. Then
$6n$ divides $(n-1)(n)(2n-1)$, and therefore $6$ divides $(n-1)(2n-1)$.
(ii) Conversely, if $6$ divides $(n-1)(2n-1)$, then $6n$ divides $(n-1)(n)(2n-1)$, and therefore $n$ divides $\frac{(n-1)(n)(2n-1)}{... | {
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Is this operation legal? Is this operation allowed?
Going from this:
$\left ( \frac{x^{2}+6}{x^{2}-4} \right )^{2}= \left ( \frac{5x}{4-x^{2}} \right )^{2}$
To this: $\left ( \frac{\left (x^{2}+6 \right )\left ( 4-x^{2} \right )}{\left (x^{2}-4 \right )5x} \right )^{2}= 1$
| Notice that the denominators are equal, save for a factor of $-1$: which has no impact since the fraction is squared. $$\; (4 - x^2)^2 = (-(x^2 - 4))^2 = (x^2 - 4)^2$$
So multiply both sides by $(x^2 - 4)^2$, and you'll cancel both denominators. Then there's no need to divide by $5x$.
After multiplying both sides by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\... | The matrix of that form is called the "companion matrix"
Let $\{e_1,e_2,e_3\}$ be the standard basis for $\mathbb{F}^3$.
Then we see that $Ae_1=e_2, Ae_2=e_3, Ae_3=ce_1+be_2+ae_3$
Give a $\mathbb{F}[x]$-module structure on $\mathbb{F}^3$ by $x$-action defined as $A$-multiplication.
Denote that $\mathbb{F}[x]$-module... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Fejer kernel has this $\sin$ closed form. Let $D_N$ be the $N$th Dirichlet kernel, $D_N = \sum_{k = -N}^N w^k$, where $w = e^{ix}$. Define the Fejer kernel to be $F_N = \frac{1}{N}\sum_{k = 0}^{N-1}D_k$. Then $$F_N = \frac{1}{N}\frac{\sin^2(N x/2)}{\sin^2(x/2)}$$.
So far I have $D_k = \frac{w^{k+1} - w^{-k}}{w-1... | Remember that $NF_N(x)=D_0(x)+\dots+D_{N-1}(x)$ where $D_n(x)$ is the Dirichlet kernel.
Therefore , if $\omega = e^{ix}$ we have
$$NF_N(x)=\sum_{n=0}^{N-1}\frac{\omega^{-n}-\omega^{n+1}}{1-\omega}$$
Then we have
$$\sum_{n=0}^{N-1}\omega^{-n}=1+1/\omega+\dots+1/\omega^{N-1}=\frac{1-\omega^{-N}}{1-\omega^{-1}}=\omega\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/514914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
| Just making sure what has to be done here, as mentioned above: we will multiply in the following way:
$$(11 \cdot 5 \cdot 6)\left(\frac{1}{5}+\frac{1}{6}\right)=\left((11 \cdot 5 \cdot 6)\frac{2}{11}\right)$$, in $\mathbb{Z_p}$
then we get $11\cdot6 + 11\cdot 5= 121 = 5 \cdot 6\cdot 2 = 60$, but this is only possible i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove by induction? For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
| Prove for the base case, n=1:
$$1^{2} = \frac{1(1+1)(2+1)}{6} = \frac{2\cdot3}{6}=1$$
The "sum" of just $1^2$ is indeed 1. Base case proven.
Now for the induction step, proving that it holds for n+1:
Observe that for $1^{2} + 2^{2} +... + n^{2}$, the sum including $n+1$ equals the original summation plus the $n+1$ term... | {
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Show that the sequence ${a_n}$ converges where $a_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}}$ for $n\geq 1$. The original question was to determine whether the sequence converges, but I have checked for extremely high values of $n$ and it seems as though it does converge. This lead me to wonder if there was an "eas... | Term by term, is less than
$$\sqrt{1+\sqrt{2^1+\sqrt{2^2+\sqrt{2^3...}}}}=\sqrt{1+\sqrt{2}\sqrt{1+2^{-1}\sqrt{2^2+\sqrt{2^3...}}}}$$
$$<\sqrt{1+\sqrt{2}\sqrt{1+2^{-0.5}\sqrt{2^2+\sqrt{2^3+...}}}}$$
$$=\sqrt{1+\sqrt{2}\sqrt{1+\sqrt{2+2^{-1}\sqrt{2^3...}}}}$$
$$\cdots$$
$$<\sqrt{1+\sqrt{2}\sqrt{1+\sqrt{2+\sqrt{2^2...}}}}... | {
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"timestamp": "2023-03-29T00:00:00",
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The identity $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \, \mathrm dx$ Let $p(x)$ be a polynomial, and assume that $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx $ converges.
How do you prove that $$ \int_{a}^{b} p(x) \cot \... | Basically, because of the Riemann-Lebesgue lemma. By summing a geometric sum, or by induction using trigonometric identities, one finds
$$\sum_{n=0}^N 2\sin (anx) = \cot \frac{ax}{2} - \frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}}.$$
So that yields
$$\int_a^b p(x) \cot \frac{ax}{2}\,dx = 2\sum_{n=0}^N \int_... | {
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"url": "https://math.stackexchange.com/questions/517080",
"timestamp": "2023-03-29T00:00:00",
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Find the equation of the plane that passes through the line of intersection of the planes... Find the equation of the plane that passes through the line of intersection of the planes $4x - 2y + z - 3 = 0$ and $2x - y + 3z + 1 = 0$, and that is perpendicular to the plane $3x + y - z + 7 = 0$.
This is what I got: $3x + 4... | Hint: The line of intersection of the 2 planes is parallel to
$$ \begin{pmatrix} 4 \\ -2 \\ 1 \\ \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 3 \\ \end{pmatrix} = \begin{pmatrix} -5 \\ -10 \\ 0 \end{pmatrix}.$$
Hint: The plane that you are interested in is parallel to $\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ an... | {
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Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x... | Given
$x=6a+5=6(a+1)-1$
$x=5b+4=5(b+1)-1$
$x=4c+3=4(c+1)-1$
$x=3d+2=3(d+1)-1$
therefore x will be of the form $(\text{L.C.M(3,4,5,6)}k-1)$ or,
$x=60k-1$ for some $k$.
Can you guess that $k$?
ANSWER:$k=1$, or $x=59$
| {
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$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| we can prove this follow
$$3^{14}x^4y^4z^4(x^3+y^3+z^3)\le (x+y+z)^{15}$$
then assuming $x+y+z=1$,and denoting
$t=3(xy+yz+xz),q=xyz$
$$\Longleftrightarrow 3^{14}q^4(1-t+3q)\le 1\Longleftrightarrow 1-3^{14}q^4(1-t)-3^{15}q^5\ge 0$$
since
$$3xyz(x+y+z)\le (xy+yz+xz)^2\Longleftrightarrow q\le\dfrac{t^2}{3^3}$$
so
$$1-3^{1... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that x is rational Let $x$ be a real number with the properties that $x^3+x$ and $x^5+x$ are rational.
Prove that $x$ is rational. Denote $a=x^3+x$; $b=x^5+x$. We can multiply and add them together until we get the desired result. I also know some non-elementary proofs of this, but have you some nice elementary p... | $$x^5+x^3=ax^2$$
$$x^5+x=b$$
$$x^3-ax^2-x-b=0$$
If $a=0$ then the only option is $b=0$ and $x=0$. Assume $a\neq0$.
$$x^3+x=a$$
$$a-x-ax^2-x-b=0$$
$$ax^2+2x+b-a=0$$
So we get that $$x=\frac{-2\pm2\sqrt{1-a(b-a)}}{2a}$$
If $\sqrt{1-a(b-a)}$ is rational we are done. Assume it is not. Plug in $x^3+x=a$.
We get $$\frac{-4}{... | {
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"timestamp": "2023-03-29T00:00:00",
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Multiplication by $x^2$ linear maps $T:P(\mathbb{R}) \mapsto P(\mathbb{R})$ defined by
$(Tp)(x) = x^2p(x)$
Verify that multiplication by $x^2$ is a linear map.
Additivity: $x^2(p+q) = x^2p+x^2q$
Homogeneity: $x^2(ap) = a(x^2p)$
Is this a correct verification?
| It seems that you've just written down the definition of linearity, but I expect that the problem is looking for a more thorough proof using the definition of $P(\mathbb{R})$ - so it's really asking why the two equalities you wrote are true.
Let's start with polynomials $f = a_0 + a_1 x + a_2 x^2 + \cdots+ a_n x^n$, an... | {
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Finding the limit of $\frac{3^{n+1}}{4^{n-1}}$ as $n\rightarrow\infty$ How would one find the limit of the following.
as $n\rightarrow\infty$
$$\frac{3^{n+1}}{4^{n-1}}$$
I did the following
$$\frac{3\cdot 3^m}{1/4 \cdot 4^m}$$
$$\frac{3}{\frac{1}{4}}\ \frac{3^m}{4^m}$$
The limit if zero.
Another one I did is
$$\frac{2^... | $\lim_{n\to\infty}\frac{3^{n+1}}{4^{n-1}}=\lim_{n\to\infty}\frac{3^{n}\cdot 3}{4^{n}\cdot 4^{-1}}=\lim_{n\to\infty}\frac{3^{n}\cdot 3\cdot 4}{4^{n}}=12\lim_{n\to\infty}(\frac{3}{4})^{n}$=$12\cdot 0$=$0$, because $\lim_{n\to\infty}(\frac{3}{4})^{n}=0$.
Your answer is correct.
and the another
$\lim_{n\to\infty}\frac{2^{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence of a sequence $\frac{1}{1+n^3}$ How can I prove by integral test that the sequence $1+ \dfrac{1}{1+2^3} + \dfrac{1}{1+3^3} + \dots + \dfrac{1}{1+n^3}$ is convergent?
Thank you. Is there a way that I can integrate $\dfrac{1}{1+n^3}$ ?
| Just in case somebody doesn't feel like doing calculus today:
$$\sum_{n=2}^\infty \frac{1}{1+n^3} < \frac{1}{2} \sum_{n=2}^\infty \frac{2}{n^3-n} = \frac{1}{2} \sum_{n=2}^\infty \left(\frac{1}{(n)(n-1)} - \frac{1}{(n+1)(n)}\right) = \frac{1}{2} \cdot \frac{1}{2(2-1)} = \frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$. The $50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$ after the decimal point.
$\underline{\bf{My\; Try}}::$ Let $\left(\sqrt{50}+7\right)^{50} = I+f$, where $I = $Integer part and $f = $ fractional part. and $0\leq f<1$
Now Le... | Consider the number $N=(\sqrt{50}+7)^{50}+(\sqrt{50}-7)^{50}$.
Using the Binomial Theorem, or otherwise, we can show that $N$ is an integer.
Our number $(\sqrt{50}+7)^{50}$ is a little below $N$. How much below? Courtesy of the calculator, $(\sqrt{50}-7)^{50}\approx 4\times 10^{-58}$. So the $50$-th digit after the dec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Another Basic Integration Question, Possibly By Substitution What's the integral of $f(x)=(1-x^2)^{1/2}$? I tried making $x=\sin(t)$ and doing integration by substitution but I don't think I arrived to the correct answer. All responses are appreciated...
| Recall the three most common trigonometric substitutions and how they proceed.
Let $x=\sin\theta$. Then $\,dx = \cos\theta \,d\theta$. And $\theta = \arcsin x$.
$$\begin{align} \int\sqrt{1-x^2}\,dx
& =\int\sqrt{1 - \sin^2\theta}\,\cos\theta \,d\theta \\ \\
& =\int\sqrt{\cos^2\theta}\,\cos\theta \,d\theta \\ \\
& = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem :
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach :
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get :
$ y = \sqrt{x + 2005}$
Now putting this value of y in (ii) we get :
$ \Rightar... | Hint: $$2005 + x- (2005+y)=y^2 - x^2 \quad \Longrightarrow \quad (x-y)(x+y+1) = 0.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Prove that $\log^25 + \log^27 > \log12$. Prove that $\log^25 + \log^27 > \log12$.
What I tried so far:
$\log^25 + \log^27 > \log3 + \log4$
$(\log5 + \log7)^2 - 2 \cdot \log5 \cdot\log7 > \log3 + \log4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
Thanks in advance.
EDI... | From $5^3=125$ and $7^6=117649$, we deduce that ${\sf log}(5) \geq \frac{2}{3}$
and ${\sf log}(7) \geq \frac{5}{6}$.
From $3(6^7)=839808$ and $5^9=1953125$, we deduce that $3(6^7) \leq 5^9$ and hence
$12^8 \leq 10^9$. So ${\sf log}(12) \leq \frac{9}{8}$.
Finally, we have
$$
{\sf log}(5)^2+{\sf log}(7)^2 \geq \big(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended u... | The LDU-decomposition gives
$$ L=\small \begin{bmatrix}
1 & . & . & . \\
1/a & 1 & . & . \\
1/a & 1/(a+1) & 1 & . \\
1/a & 1/(a+1) & 1/(a+2) & 1 \\
\end{bmatrix} \\ D= \small \begin{bmatrix}
a & . & . & . \\
. & (a^2-1)/a & . & . \\
. & . & (a^2+a-2)/(a+1) & . \\
. & . & . & (a^2+2a-3)/(a+2) \\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form,... | I first would eliminate the negative exponents of $x$ by multiplying $f(x)$ by $x^3$ to get
$$g(x)=x^3f(x)=x^6-12\,x^4+48\,x^2-64$$
If you are right in assuming that $f(x)=h(x)^3$ then $g(x)=p(x)^3$ and $g'(x)=3p'(x)p(x)^2$
Now we assume that $p$ is a polynomial then
$$p(x)^2 \mid \text{gcd}(g(x),g'(x))$$
and
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
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Partial Fractions and power of a factor with $x^2$ I just started working with partial fractions and hit a wall with splitting this one:
$$ \frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} $$
I get here:
$$ \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}$$
Then on to:
$$ (Ax + B)(x + 2) + (Cx ... | You already have
\begin{align*}
&\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}\\
&= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}\\
&= \frac{(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2}{(x+2)(x^2+x+1)^2}.
\end{align*}
Multiplying both sides by $(x+2)(x^2+x+1)^2$ we... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the functi... | We want to show that $$ f(y) = \dfrac {\left( 1 - y^{-1} \right)^2}{\left( 1 - y^{-1-c} \right) \cdot \left( 1 - y^{1+c} \right)} $$ is decreasing in $ y > 1 $. Indeed, we can differentiate. It suffices to show that $ f'(y) < 0 $, for $ y > 1 $, if $ c \in (0, 1) $.
So, we differentiate using the Quotient Rule: $$ f'(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A Non-recursive Fibonacci Sequence How can I determine the general term of a Fibonacci Sequence? Is it possible for any one to calculate F2013 and large numbers like this?
Is there a general formula for the nth term of the Fibonacci Sequence?
If there is a combinatorial formula, that is more preferable; some formula li... | When
(1) $F_{n+2} = F_{n+1} + F_{n}$
one can try $F_n = \phi^n$, then we obtain
(2) $\phi^{n+2} - \phi^{n+1} - \phi^n = 0$,
thus
(3) $\phi^n \Big( \phi^2 - \phi - 1 \Big) = 0$,
ignoring the trivial case $F_n = 0$ we obtain
(4) $\phi = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}$
So we get
(5) $F_n = a \left( \frac{1}{2} + \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$? All is in the title:
If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?
I can't conclude from the fact that $k^2 - 1$ is divisible by $8$, that then $k^4-1$ is divisible by $16$.
| Hint: $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = (k^2 - 1)\Big((k^2 - 1) + 2\Big)$$
ADDED per comment: So yes, we have that if $(k^2 - 1)$ is divisible by $8$, then $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = 8b(k^2 + 1)$$ for some integer $b$.
And now, if $k^2 - 1$ is divisible by 8, it is even, then so is $k^2 + 1$.
That is, $k^2 + 1 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How ... | let the nos be
n-1, n, n+1, n+2
we need to prove that (n-1)(n)(n+1)(n+2)+1= square no.
rearranging it : (n-1)(n+2)(n)(n+1)+1= square no.
LHS= (n-1)(n+2)(n)(n+1)+1
=(n^2+n-2)(n^2 +n)+1
= (k-2)(k) + 1 (by replacing (n^2 +n) by k)
= k^2 -2k +1
= (k-1)^2
hence proved , LHS is a square no.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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How find the value of $\cos{x}+\cos{y}$ let
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$$
Find the value $$cos{x}+\cos{y}=?$$
this following My ugly solution: let
$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$$
then
$$\cos{x}=\dfrac{1-a^... | Here is a slight simplification of your method : one has
$$
\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\frac{1}{1-\alpha},
\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\frac{1}{1-\beta} \tag{1}
$$
with $\alpha=\tan(x)\tan(\frac{y}{2})$ and $\beta=\tan(y)\tan(\frac{x}{2})$.
Your equation then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solve the following non-homogeneous recurrence relation: Find the solution to the following non-homogenous recurrence relation:
$a_{n+2} - 4a_{n+1} + 4a_{n} = 2^n$ for $a_0=1, a_1 = 2$.
I have found from the characteristic polynomial the general homogenous solution is:
$a_{n} = c_{1}2^n + c_{2}n2^n$ where $c_1, c_2$ ar... | Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and add over $n \ge 0$. Recognize e.g.
\begin{align}
\sum_{n \ge 0} a_{n + k} z^k
&= \frac{A(z) - a_0 - a_1 z - \ldots - a_{k - 1} z^{k - 1}}{z^k} \\
\sum_{n \ge 0} 2^n z^n
&= \frac{1}{1 - 2 z}
\end{align}
to get
$$
\frac{A(z) - 1 - 2 z}{z^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/535946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$123^{561}$ find last $2$ digits Modular Exponentiation mod $100$? $123^{561}$
Find the last $2$ digits
Can't I work this out using modular exponentiation working mod $100$?
$123^2 = 29\pmod{100}$
$123^4 = (123^2)^2 = 41\pmod{100}$
$123^8 = (123^4)^2 = 81\pmod{100}$
$123^{16} = (123^8)^2 = 61\pmod{100}$
$123^{32} = (12... | $123^1$ is $23$, not $29$ mod $100$. Your solution is otherwise correct, although Euler's theorem would be more efficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/536497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice... | The continued fraction method works like this: Suppose $x = a^2 + b$, where $a = \lfloor \sqrt x \rfloor$. Then
$$
\begin{align}
x &= \sqrt{a^2 + b}\\
x-a &= \sqrt{a^2 + b} - a\\
\frac{1}{x-a} &= \frac{1}{\sqrt{a^2 + b} - a}\\
&= \frac{1}{\sqrt{a^2 + b} - a}\frac{\sqrt{a^2 + b} + a}{\sqrt{a^2 + b} + a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 13,
"answer_id": 1
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When to stop doubling down? My question is similar to this one but very specificly different
When to stop in this coin toss game?
Imagine a game where you would start with $100. Every time you can roll a die (d6), if it is 1-5 you double the winnings, but if it is a 6 you lose everything.
How would you calculate the i... | After first roll:
$P_1(\$200) = \frac{5}{6}$, $P_1(0) = \frac{1}{6}$.
Expected value (average winning) $E_1[X] = P_1(\$200)\cdot 200 + P_1(0) \cdot 0 = \frac{5}{6}\cdot 200 = \frac{5}{3}\cdot 100$.
After $2$ rolls:
$P_2(\$400) = \frac{5^2}{6^2}$, $P_2(0) = 1 - P_2(\$400) = \frac{6^2-5^2}{6^2}$.
$E_2[X] = P_2(\$400)\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Denoting the set of initial segments of a binary sequence The index is an infinite, innumerable binary sequence in $\{0,1\}$. $ I= \{f \mid f: \Bbb {N} \to \{0,1\}\} $
Is there a way to get a set $X_i$ from the infinite index number $10110\ldots$
$ X_i = \{1,10,101,1011,10110,\ldots\}$ where $i \in I$ and $i=10110\ldot... | $${ X }_{ a }={ { \left\{ \left\lfloor \frac { a }{ { 10 }^{ n } } \right\rfloor ,\left\lfloor \frac { a }{ { 10 }^{ n-1} } \right\rfloor ,...,\left\lfloor \frac { a }{ { 10 }^{ 0 } } \right\rfloor \right\} } }\\n=\left\lfloor \log _{ 10 }{ a } \right\rfloor $$ This works for only finite sequences but you can di... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution verification: $\lim\limits_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$ I am trying to find the following limit
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2}$$
and I did the following steps:
\begin{align}
\require{cancel}
&\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} \\
&\lim_... | Excellent work. You've made a correct and tight case for the limit being $-1$.
Just replace
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only $(+1)$ on top.
With
Now here, the top portion goes to $1$ because in the first term of the numerator there is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/546922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx... | from the equation of the ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
$$ y = \sqrt{\frac{a^2b^2 - x^2b^2}{a^2}} = \frac{b}{a}\sqrt{a^2-x^2}$$
since proceding with change of variable in integration is, using x = a*cos(Θ)::
$$
\\
\sqrt{a^2 - x^2} = \sqrt{a^2-a^2cos^2\theta} =asin\theta
\\
\int_{}^{}\sqrt{a^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
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How do I find the total area of total area of the red circles C_n,n=1,2,3 (Calculus II)
I am very confused with this problem. I have been working on it for about an hour and still nothing. Help would be greatly appreciated!
| One possible solution for this uses Decartes' Circle Theorem, which states that for four mutually externally tangent circles with radii $a,b,c,d$, then the following equality holds:
$$
2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Convergence of $x_{n+1} = \frac12(x_n + \frac2{x_n}).$
Let $x_1=1$ and $$x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).$$ Prove or disprove $(x_n)$ is convergent and show the limit.
When I tried working on it I found the sequence was bounded by
square root of 2 and it is was monotone. But apparently the sequence is... | Let $y_n = x_{n}^2$, then
\begin{align*}
\\y_{n+1} = x_{n+1}^2 &= \frac{1}{4}\left(x_n + \frac2{x_n}\right)^2
\\ &=\left(\frac{x_n}{2}\right)^2 +\left(\frac{1}{x_n}\right)^2 + 1 \ge 2\sqrt{\frac{x_n}{2}\cdot\frac{1}{x_n}} +1 = \sqrt{2}+1 \gt 2 \tag{1}
\end{align*}
Thus by (1), we have $x_n \gt \sqrt{2}~~(n\ge 2)$. Als... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Show that prime $p=4n+1$ is a divisor of $n^{n}-1$
Show that the prime number $p=4n+1$ is a divisor of $n^{n}-1$
Ok, the question itself is simple as hell, but I couldn't think of a simple way to solve this question. I tried to solve the question by using $p\equiv 1 \pmod n$ but only to fail miserably... I couldn't ... | We'll use this lemma which I'll leave for you to prove:
Ic $p=4n+1$ is prime, then $a^n\equiv 1\pmod {p}$ if and only if $a\equiv x^4\pmod p$ for some $x$ not divisible by $p.$
We'll show that $-4$ is always a fourth power, $\pmod p$.
If $n$ is even, then $p\equiv 1\pmod 8,$ so $2$ is a square, and hence $4$ is a fou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/553332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
A problem on indefinite integration $$\int\frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx$$
I tried some substitutions, but none succeeded in simplifying the expression. Please help.
| Disclaimer I'm cheating, I get the final answer from WA and reverse engineering out the steps. People have any intuition how to get the steps without cheating, please update this answer.
$$\begin{align}
\int \frac{x^4 - 2}{x^2 \sqrt{x^4+x^2+2}} dx
= & \int \frac{2x^4 + x^2 - (x^4 + x^2 + 2)}{x^2 \sqrt{x^4+x^2+2}} dx... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\d... | For the proof in case $n\geq 2$, I found Muirhead's inequality very useful.
*
*Expand the inequality. You have $$3xyz+\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^3yz}-3x^2y^2z^2 \geq 0$$, or stated another way, $$xyz(3-x^2+y^2+z^2)+(\sum_{cyc}{x^2y^2}-\sum_{cyc}{x^2y^2z^2})\geq0.$$
Therefore it is enough to prove that $$x^2+y^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$\sum\sin(\frac{k\pi}{4})$ absolute convergent, conditional convergent,divergent? Would the following be abs convergent, conditional convergent or divergent.
$\sum\sin(\frac{k\pi}{4})$
I know sin(x) is between $-1<x<1$
y=sin(x) is oscillating would it be $(-1)^n$
| Sorry about the old answer, that was just garbage.
The real answer is that it's not convergent. You can try grouping them in terms of $8$ like I had done, but the problem is that, because it ends at $\infty$, you don't have a point to end at. That means the answer could be the same as any of the following eight:
$\disp... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $x_n = (\prod_{k=0}^n \binom{n}{k})^\frac{2}{n(n+1)}$ then $\lim_{n \to \infty} x_n = e$ I try to prove the following:
$$x_n = \left(\prod_{k=0}^n \binom{n}{k}\right)^\frac{2}{n(n+1)}$$
$$\lim_{n \to \infty} x_n = e$$
I want to use double sided theorem, so I've proven that
$$x_n \ge \frac{n}{\sqrt[n]{n!}}$$
As it kn... | Too late, more than likely too advanced but done for the fun of deriving asymptotics.
$$\prod_{k=0}^n \binom{n}{k}=\frac{\big[\Gamma (n+1)\big]^n}{G(n+1)\, G(n+2)}$$ where $G(.)$ is the Barnes G-function.
Using Stirling-like approximations
$$\log(G(p))=\frac{1}{4} p^2 (2 \log (p)-3)+\frac{1}{2} p (-2 \log (p)+2+\log (... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Asymptotic expansion of an integral I came up with a simpler example which illustrates the technical difficulty I have encountered in my work.
Consider an integral depending on parameter $\epsilon$:
\begin{equation}
\int\limits^{\infty}_{1 + \frac{2 \epsilon^{2}}{R^{2} - \epsilon^{2}}}
\frac{1}{t^{2}-1} \frac{1}{\sqrt{... | First make the change of variables $x = \epsilon^2\frac{t+1}{t-1}$ to get
$$
I(\epsilon) = \int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = \frac{1}{2} \int_{\epsilon^2}^{R^2} \frac{dx}{x\sqrt{R^2-x}}.
$$
Fix $0 < \delta < R^2$ and split the integral into t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Functional equation: $f\left(\frac{x-1}{x}\right)+ f\left(\frac{1}{1-x}\right)= 2- 2x$ There is a function given $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x
,f\colon \Bbb R\setminus\{0,1\}\to \Bbb R$ How many fuction exist? I have no idea how to start
| $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x\tag{1}$
Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(1)$
$\displaystyle f\left(\frac{1}{1-x}\right)+f\left(x\right) = \frac{2}{x}\tag{2}$
Similarly Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(2)$
$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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convergence of a function serie to norm 2 need to show if the following function serie converge on $||.||_2$ on [1,5]:
$$\sum _1^\infty {sin^2(nx) \over n^2} $$
I have no idea how to approach that one, would like
for some directions...
| We can in fact compute what the series converges to. Note that
$$\sum_{n\ge 1}\frac{\sin^2(nx)}{n^2}
= \sum_{n\ge 1}\frac{1}{2} \frac{1-\cos(2nx)}{n^2}
= \frac{\pi^2}{12} - \frac{1}{2} \sum_{n\ge 1} \frac{\cos(2nx)}{n^2}.$$
Re-write this as follows:
$$\frac{\pi^2}{12} - 2 x^2 \sum_{n\ge 1} \frac{\cos(n2x)}{(n2x)^2}.$$
... | {
"language": "en",
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"source": "stackexchange",
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How to solve this system of equations. Solve the system of equations: $$\begin{cases}\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy} \\\left(\sqrt{3}+xy\right)^{\log_2x}+\dfrac{x}{\left(\sqrt{3}-xy\right)^{\log_2y}}= 1+\dfrac{x}{y}\end{cases}$$
My try:
$\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy}\\\Leftrightarrow xy\left(... | Subsitute $xy=1 , y=1/x,and \, log y=-logx$ :-
$log(3^{1/2}+1)^{logx}+ x*log(3^{1/2}-1)^{logx}=1+x^2$
then take Log at both sides of your equation then your 2nd equation becomes
$logx*log(3^{1/2}+1)+logx+ logx*log(3^{1/2}-1)=2logx$
$log(3^{1/2}+1)+ log(3^{1/2}-1)=1$
$log((3^{1/2}+1)(3^{1/2}-1))=1$
which is always t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/559171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Which step is wrong in this proof
Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give
$$x=−1−\frac{1}{x}$$
Substitute this back into the $x$ term in the middle of the original eq... | What you have proved is that there is no real number $x$ such that $x^2+x+1=0$.
On the other hand, the two complex solutions of $x^2+x+1=0$ do indeed satisfy $x^3=1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
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Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x +... | A short method :
Since $$\sin(x+\alpha)=(\cos \alpha) \sin x + (\sin \alpha) \cos x$$
we have $\sin(x+\alpha) \in \text{Span}\{\sin x, \cos x\}$
This implies that the rank of the family $(\sin(x+\alpha),\sin(x+\beta),\sin(x+\gamma))$ is $2$. That means that there are linearly dependants.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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which one is larger $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$? Which is larger? $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$?
Algebraic justilation does not help.
| $\sqrt[n]{x+\delta}-\sqrt[n]{x}\ \boxed{\phantom{A} }\sqrt[n]{x}-\sqrt[n]{x-\delta}$
$\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}\ \boxed{\phantom{A} }2\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Cubic Poynomial : In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the......... Question:
In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the equation are
(a) all real and equal
(b) all real and distinct
(c) one real and two imaginary
(d) all... | The answer is (c): The equation has one real root and two nonreal complex conjugate roots.
Proof: This Wiki article classifies the nature of the roots.
The discriminant of a cubic equation $ax^3+bx^2+cx+d=0\;$ is
$$\Delta = 18 a b c d - 4b^3d + b^2 c^2 -4ac^3 - 27 a^2 d^2.$$
*
*If $\Delta > 0,$ then the equation h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$\displaystyle f(x,y,z)=3log(x^2+y^2+z^2)-2x^2-2y^3-2z^3$ $(x,y,z)\neq(0,0,0)$ has only one extreme value $\displaystyle log(\frac{3}{e^2})$ Show that the function defined by $\displaystyle f(x,y,z)=3log(x^2+y^2+z^2)-2x^2-2y^3-2z^3$
$(x,y,z)\neq(0,0,0)$ has only one extreme value $\displaystyle log(\frac{3}{e^2})$
Atte... | Stationary points of $f$ can be found from the system
$$\nabla{f}=\vec{0}$$
which is equivalent to
$$
\begin{cases}
\dfrac{\partial{f}}{\partial{x}}=0, \\
\dfrac{\partial{f}}{\partial{y}}=0, \\
\dfrac{\partial{f}}{\partial{z}}=0,
\end{cases}
\quad\Leftrightarrow\quad
\left \lbrace {\matrix{
x\left(\dfrac{3}{x^2 + y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Getting rid of the square roots in the expression $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ I need to find an alternate form of $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ without square roots for a problem that I'm working on, but it's rather complicated to do. What we can do is
$\sqrt{a} + \sqrt{b} + \sqrt{c} = d \iff \sqrt{a} =... | You are asking a special case of the following problem in abstract algebra: Suppose $x$ is a solution to $p(x) = 0$ and $y$ solves $q(y)=0$, for polynomials $p,q$ (with, say, integer coefficients); find a polynomial (with, again, integer or whatever coefficients) that has $x+y$ as a solution. In the "$\sqrt{a} + \sqrt... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding indefinite integral $\int{ \mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$ Could anyone help me how to solve this indefinite integral?
$$\int{\mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$$
| Firstly let us find the anti-derivative in case $\alpha=0$. This is pretty straightforward:
\begin{eqnarray}
\int\frac{dx}{\sqrt{\sin(x)^3+\sin(x)}}&\underbrace{=}_{y=\sin(x)}&\int\frac{dy}{\sqrt{1-y^2} \sqrt{y} \sqrt{1+y^2}}\\
&\underbrace{=}_{z=y^4}& \frac{1}{4} \int z^{-7/8} (1-z)^{-1/2} dz\\
&=& \frac{1}{4} B_{[\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Solve for z(t) from the simultaneous equation using Laplace transform Solve for $z(t)$ from the simultaneous equation using Laplace transform
$$
y' + 2y + 6 \int\limits_0^t z \mathrm{d}t = -2 u(t) \\
y' + z' + z = 0$$
subject to $y(0) = -5$ and $z(0) = 6$.
| First note that if $u(t)$ is the Heaveside step function, then
$$\int_0^t z(s)\,dt = u(t)\ast z(t)$$
where $\ast$ denotes convolution. So we can rewrite the system as
$$\left\{\begin{aligned} y^{\prime}+2y + 6u(t)\ast z(t) &= -2u(t)\\ y^{\prime} + z^{\prime} + z &= 0\end{aligned}\right.$$
Noting that $y(0) = -5$ and ... | {
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"question_score": "1",
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How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers. Prove that
$$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}\in N$$
This problem from this
My try: let
$$x=\sqrt[3]{35+15\sqrt{6}}+\sqrt[3]{35-15\sqrt{6}}$$
t... | Denotes $t=\sqrt[3]{15 \sqrt{6}+35}+\sqrt[3]{35-15 \sqrt{6}},$ then $t^3+15 t-70=0.$
$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}$
$=\sqrt[5]{1782+405t}-t$
Since $(t+2)^5-(1782+405t)=(t+5)^2 \left(t^3+15 t-70\right)=0$, we get $\sqrt[5]{1782+405t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Find a power series solution centered at 0 (Differential equations Here's the problem: $$(x-1)y''+y'=0$$ This is the work that I've already done: $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n$$
$$y''=\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n$$
I then plug those into the original equation:... | You have
$$
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1}-\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n = 0.
$$
Rewrite the first sum as
$$
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1} = \sum_{m=1}^{\infty}(a_{m+1})(m+1)(m)x^m.
$$
Rewrite the second sum as
$$
\begin{align}
\sum_{n=0}^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Indefinite Integral $\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$ How can I evaluate this integral?
$$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$$
| Write $$3\sin x+2\cos x=A(2\sin x+3\cos x)+B\frac{d(2\sin x+3\cos x)}{dx}$$
$$\implies 3\sin x+2\cos x=A(2\sin x+3\cos x)+B(2\cos x-3\sin x)$$
$$\implies 3\sin x+2\cos x=(2A-3B)\sin x+(3A+2B)\cos x$$
Solve for $A,B$ equating the coefficients of $\cos x,\sin x$
So, $$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx=A+B... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substi... | Calculating this integral can be done by parts:
$$I=\int x^2\sqrt{a^2-x^2} dx= -\int x(\frac{1}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}} )' dx=$$
$$=-\frac{x}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}}+\frac{1}{3}\int(a^2 -x^2)\sqrt{a^2-x^2}dx.
$$
$$\frac{4}{3}I=-\frac{x}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}}+a^2\int\sqrt{a^2-x^2} dx. (1)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Solving $(z+1)^5 = z^5$ The question says to solve this equation: $(z+1)^5 = z^5$
I did. Just want to find out if I did it properly and if my run-around logic makes sense.
First I begin my writing the equations as:
$$ (z+1)^5 = z^5$$
$$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$
So $$ \mathbf... | If you divide both sides by $z^5$ (note that $z\ne0$, since for $z=0$ we get $0=1$), you get
$$
\left(1+\frac1z\right)^5=1.
$$
The expression in brackets cannot be $1$, so we are left with the four non-trivial fifth roots of unity:
$$
1+\frac1z=e^{2\pi i k/5},\ \ k=1,2,3,4.
$$
So we get four solutions, namely
$$
z=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/577847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
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question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$ Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have
$$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$
therefore,
$$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{... | Here is one way forward. Note that $\sin (\arcsin (x))=x$ and $\cos (\arcsin(x))=\sqrt{1-x^2}$ for $|x|\le 1$. Therefore, we have for $|x|\le 1$
$$\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h} \tag 1$$
Now, in THIS ANSWER, I showed using only basic inequalities th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Evaluate $\tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)$ without using a calculator Evaluate without using a calculator:
$\displaystyle{\tan\left(2\sin^{-1}\left(\sqrt{5} \over 5\right)\right).}$
So I built my triangle hyp=$5$, adj=$2\sqrt{5}$, opp=$\sqrt{5}$.
$$
\tan\left(2\theta\right) = 2\sin\left(\theta\right)\cos\... |
Using the identity $\tan{2A}=\dfrac{2\tan A}{1-\tan^2 A}$, $\tan \theta=\dfrac{\sqrt{5}}{\sqrt {20}} =\dfrac{1}{2}. \text{Hence} \tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)=\dfrac{2\times \dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How can I prove this statment? I have this problem to solve:
Prove that when $$x\rightarrow 0$$ $$\sqrt{1+x}-1 \sim \frac{x}{2}$$
Can someone give me a tip? or show me the way?
Thanks in advance
| $$\sqrt { 1+x } -1=\sqrt { 1+x+\frac { { x }^{ 2 } }{ 4 } -\frac { { x }^{ 2 } }{ 4 } } -1=\sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 }-\frac { { x }^{ 2 } }{ 4 } } -1\\$$For $x$ very small$$ \frac { { x }^{ 2 } }{ 4 } <<{ \left( \frac { x }{ 2 } +1 \right) }^{ 2 },$$therefore you can ignore the term $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuc... | I like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} > L$ and lets assume that it converges to some constant $c$.
Now, we can write ,
$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 4,
"answer_id": 1
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Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
| As $\displaystyle\binom n2=\frac{n(n-1)}2=\frac12\cdot n^2-\frac12\cdot n$
$$\sum_{2\le n\le m}\binom n2=\frac12 \sum_{2\le n\le m}n^2-\frac12\sum_{2\le n\le m} n$$
$$=\frac12\left( \sum_{1\le n\le m}n^2-1\right)-\frac12\left(\sum_{1\le n\le m} n-1\right)$$
$$=\frac12 \frac{m(m+1)(2m+1)}6-\frac12\frac{m(m+1)}2$$
$$=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 2
} |
evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+... | $$\frac{x+1}{(x^2+x+1)(x^3-x^2+1)}=\frac{2x+3}{7(x^2+x+1)}-\frac{2x^2-x-4}{7(x^3-x^2+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
let sequence $\{a_{n}\}$ such $a_{1}=1$,and
$$a_{n+1}a_{n}=n,n\ge 1$$
show that
$$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$
(2): I consider we... | Hint : $a_n=\displaystyle\frac{(n-1)!!}{(n-2)!!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root
$a,b$ are real numbers. Roots may or may not be real.
In this question, the aim is to find values of and b ,eg. 2,4.
| Is this what you want? If an equation $x^2+ax+b=0$ has roots $v,v^2-2$, then by Vieta's formulas, we have
$$v+(v^2-2)=-\frac{a}{1}=-a, v(v^2-2)=\frac{b}{1}=b.$$
Then, we have $(a,b)=(-v^2-v+2,v^3-2v)$.
So, you can get $(a,b)$ if you know the value of $v$.
Edit : OK. The answers for the "new" question is the followings ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Find all the positive integer Find all the positive integers (x,y), such that
a) $1!+2!+3!+\cdots+ x!=y^2$
b)$1!+2!+3!+\cdots+x!=y^z$
| For part b, assuming $z$ is also a positive integer,for $x\ge 27$, first note that the last digit of summation is 3, so the last digit of $y$ must be $3$ or $7$ and $z=4k+1$ or $z=4k+3$
$1!+2!+\cdots +x!\equiv 9 \pmod{27}$
But $y^{4k+3}$ or $y^{4k+1}$ cannot be $27t+9$. So all answers, if any, must be less than $27$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But ... | We can recognize both expressions as squares:
*
*$3+2\sqrt2=2+2\sqrt2+1=(\sqrt2)^2+2\cdot \sqrt2\cdot 1+1^2=(\sqrt2+1)^2$, and
*$4-2\sqrt3=3-2\sqrt3+1=(\sqrt3-1)^2$.
This means that $\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=2+\sqrt2-\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Establish $\int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1}}{2 \cos(\pi a /2)}$ when $-1 < a < 1$ My attempt at a solution: (this is homework, btw)
Let $f(z) = \frac{z^a}{z^2 + b^2}dz$ then the singularities of $f$ occur at $\pm ib$.
$$
Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib}
$$
$... | My corrected answer (thanks to Daniel Fischer and robjohn):
We use a circle contour centered at the origin with the origin cut out (since $a$ can be negative).
(Also we make the branch cut for $x^a$ along the positive real axis.)
$\gamma = [r, R]$
$\mu_R = Re^{i \theta}$, where $0 < \theta < 2\pi$
$\kappa = [-R, -r]$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\c... | $\displaystyle \begin{aligned}\int \frac{1}{\sin ^3 x+\cos ^3 x} d x & =\frac{1}{3} \int \left(\frac{2}{\sin x+\cos x}+\frac{\sin x+\cos x}{\sin ^2 x-\sin x \cos x+\cos ^2 x}\right) d x \\& =\frac{2}{3} J+\frac{1}{3} K \\\\J&=\int \frac{1}{\sin x+\cos x} d x \\& =\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}... | I am not sure about Leibitz, but the terms are decreasing in absolute value, and the series is alternating, so you are good.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why does the graph of $e^{1/z}$ look like a dipole? I was looking at the color wheel graph of $e^{1/z}$, and my girlfriend commented that it looked just like a dipole. Does anyone have an explanation for that, why the geometry would be so similar? I guess as we follow e.g. the red color from the left side as it goes up... | Try $e^{1/z^2}$, you will get an even better match. This is due to the following:
The potential due to the dipole you have in the figure is given by $\phi(x,y) = -\dfrac{x}{x^2+y^2}$. Hence, the electric field is given as
$$\vec{E} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, \dfrac{2xy}{(x^2+y^2)^2}\right) \tag{$\star$}$$
If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 3,
"answer_id": 1
} |
Find the domain of the function $f(x)=\dfrac{64+x^2+x}{9x^2+27x+9}$.
Find the domain of the function $f(x)=\dfrac{64+x^2+x}{9x^2+27x+9}$.
I don't know how to start nor how to do it. Please help me!
| The domain of a function is the set of argument values for which the function is defined.
We need to make $\displaystyle\frac{64+x^2+x}{9x^2+27x+9}$ defined
$\displaystyle\implies 9x^2+27x+9\ne0\iff x^2+3x+1\ne0$
Now, the roots of $\displaystyle x^2+3x+1=0$ are $x=\frac{-3\pm\sqrt5}2$ which will make the given functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Residues to solve an improper integral I'm asked to solve the following improper integral:
$$\int_0^\infty \frac{\rm {Log}^2(t)}{1+t^2}dt. $$
Do I consider the function $f(z) = \frac{\rm{Log}^2(z)}{1+z^2}$ or some variant?
Is the correct contour the so called "keyhole" that skips the $(0,+\infty)$ interval?
Any hints... | To use a keyhole contour $C$, consider
$$\oint_C dz \frac{\log^3{z}}{1+z^2}$$
You can show that the integrals about the circular arcs about the origin, both large and small, vanish as theire respective radii go to $\infty$ and $0$. The contour integral is therefore equal to
$$\int_0^{\infty} dx \frac{\log^3{x} - (\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that the polynomial is irreducible Let $a \in 3 + 4\mathbb{Z}$ and $f\left( x \right) = \left( {x + 1} \right)\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 2} \right) + a$ a polynomial. Using the homomorphism $\pi :\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$, see if $f$ is irreducible over $\mathbb{Q}$.
My... | Modulo $2$, we have
$$(x^2+x+1)^2 = x^4 + 2x^3+3x^2+2x+3$$
so any monic factor of $ x^4 + 2x^3+3x^2+2x+3$ in $\mathbf Z/4\mathbf Z$ reduces to $x^2+x+1$ modulo $2$. There are only $4$ possible factors to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove using mathematical induction pt 2 Assumed that i asked a question like 30 min ago thinking i got the hang of this, seems not.
So $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac12n(6n^2-3n-1) \text{ for all } n\in\mathbb N$$
This time it seems way harder with the squares.
so i did the steps and got stuck on the 3rd step(Again)... | Hypothesis:
$$
1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1)
$$
Thesis:
$$
1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)
$$
By the induction hypothesis
$$
1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=
\frac{1}{2}n(6n^2-3n-1)+(3(n+1)-2)^2
$$
Write this as a polynomial in $n$; write
$$
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$... | It's more simple to prove that, let $\sqrt{2\sqrt{2\sqrt{2}...}}=t$. Then, $t^2=2t \rightarrow t=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/601045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?
| multiple the numbers before parantheses then it's formula for the general will be obtained
$$\sum_{n=1}^\infty(\frac{1}{2\cdot 2^{n-1}}+\frac{1}{3\cdot 2^{n-1}})$$
and it's equal to following geometric serie:
$$\frac{5}{6}\sum_{n=1}^\infty\frac{1}{2^{n-1}}=\frac{10}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Trigonometry - parameter For which real parameter values $n$ equation $\frac{4n+3}{6}-\sin4x\cos4x-(n+\frac{2}{3})\sin(4x-\frac{\pi}{4})=0$ have exactly three solutions for $x \in [\frac{\pi}{16}, \frac{5\pi}{16}]$
Well I rewrite this equation as $4n+3-6\sin4x\cos4x-(3n+2)\sqrt2(\sin4x-\cos4x)=0 $ , but I don't know wh... | First, I apologize for the length of the answer and my incapability to minimize answer as HINT only.
Squaring should be avoided as it immediately introduces Extraneous roots
Using $\sin2A=2\sin A\cos A$
$$4n+3-3\sin8x-(6n+4)\sin\left(4x-\frac\pi4\right)=0$$
Let $\displaystyle 4x-\frac\pi4=y$
$\displaystyle\implies(i)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/606000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluate limit of a sequence... NBHM $2013$ Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$
All I could do was to see that $$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$
Just because $\sin(2n\pi+\theta)=\sin(\theta)$..
So, we now have $$\li... | Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$
As $\sin(2n\pi+\theta)=\sin(\theta)$ we would have :
$$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$
So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/606559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $a,b,c \in \mathbb N$ and $\gcd(a^2-1,b^2-1,c^2-1)=1$, prove that $\gcd(ab+c,bc+a,ca+b)=\gcd(a,b,c)$. The whole problem is in the title. I'm out of ideas and can't think of a way to solve this, so I'd like to get some help. Thanks.
| *
*if $n$ divides $a$, $b$ and $c$ then it divides $(ab+c,bc+a,ca+b)$. Hence
$$gcd(a,b,c)\le gcd(ab+c,bc+a,ca+b) \quad [1]$$
*if $n$ divides $ab+c$, $bc+a$ and $ca+b$, then it divides
$$c(ca+b)-(bc+a)=a(c^2-1)$$
$$b(bc+a)-(ab+c)=c(b^2-1)$$
$$a(ab+c)-(ca+b)=b(a^2-1)$$
Hence $$gcd(ab+c,bc+a,ca+b)\le gcd(a(c^2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/606631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
| You can use an “unknown relation”: denote by ? either $<$ or $>$; so long as you do operations that don't change the direction of inequalities, the symbol will always mean the same; so, when you're done you'll know which one it is.
\begin{align}
1+3\sqrt{2}&\mathrel{?}3\sqrt{3}\\
(1+3\sqrt{2})^2&\mathrel{?}(3\sqrt{3})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How to Plot $\sqrt{\frac{a^2+(b-1)^2}{a^2+(b+1)^2}}=2$ How to plot this complex division?
$$
\sqrt{\frac{a^2+(b-1)^2}{a^2+(b+1)^2}}=2
$$
| The plot of $\sqrt{\frac{a^2+(b-1)^2}{a^2+(b+1)^2}}=2$ is the following:
Work out $a^2+(b-1)^2$ and $a^2+(b+1)^2$. This will give:
$$\frac{a^2+(b-1)^2}{a^2+(b+1)^2} = \frac{1-4b}{a^2+b^2+2 b+1}$$
This would make it easier to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find equation of a tangent on $y= \sin2x$ Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$
What I calculated:
Intersections: $$\sin2x= \frac{1}{2}$$
...
$$0=tg^2x-4tgx-1$$
$$tgx_{1}=2+\sqrt{3}$$
$$x_{1}=75+k\pi$$
$$tgx_{2}=2-\sqrt{3}$$
$$x_{2}=15+ k\pi$$
$$T_{1}(75+k\pi,\frac{1}{2})$$ a... | You calculated your equations using the given $x$ (angles) in terms of degrees, not in radians: specifically, you need to multiply each angle in your equations of the lines $\,(75^\circ, 15^\circ\,)$ by the conversion factor $\frac \pi{180^\circ}$ to obtain those values in radians, which is how the solution equations a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 is If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 is
$A. 1/4$
$B. 1/7$
$C. 1/8$
$... | Shouldn't the answer be $24/100$? This is because there are 40 numbers which satisfy $7^x = 2^x = 1 \mod 5$, and $20$ each for $7^x = 2^x = 2,3,4 \mod 5$, so we can't just assign equal probabilities to each event.
( Saying equal likeliness of the events $1+4,2+3,3+2,4+1$ as the sum of the remainders out of $4\cdot 4=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Prove Trig Identity
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$
This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\righ... | You are missing a factor of $2$ in the last line of your answer. Anyway, you were on the right track. You just need to use the same type of identity (as you used in the beginning) for the first term and the double angle formula for the second.
$$
2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{2A - B - C}{2}\right) + 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/611324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$ After numerical analysis it seems that
$$
\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
$$
Could someone prove the validity of such identity?
| Yes, we can prove it. We can change the order of summation in
$$\begin{align}
\sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
&= \sum_{k=1}^\infty \frac{2k(2k+1)}{4^{2k+2}}\sum_{n=1}^\infty \frac{1}{n^{2k+2}}\\
&= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{2k(2k+1)}{(4n)^{2k+2}}\\
&= \sum_{n=1}^\infty r''(4n),
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.
I thought of it this way: $$f(x)=\begin{ca... | Let f(x)=|x|.We know $$f(x)=\begin{cases}
-x \text{ if x<0}\\
x\text{ if x>0}
\end{cases}$$
What if f(x)=|x-1|. We must have,
$$f(x)=\begin{cases}
-(x-1) \text{ if x<1}\\
x-1\text{ if x>1}
\end{cases}$$
Note that there are 3 terms in the given function.
$|2x-1|$, $3|2x+4|$ and 7.
The first term is postive if x$\gt \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must
be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$
My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using... | First Answer:
\begin{align*}
2\alpha +\beta = -\frac{3b}{a} &\Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
&\Rightarrow {\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\\
&\Rightarrow -3\alpha^2-\frac{6b}{a}\alpha=\frac{3c}{a} \\
&\Rightarrow a{\alpha}^2+2b\alpha+c=0\tag{0}\\
&\Rightarrow 2a{\alpha}^3+4b{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How prove this equality $(x^2+2)(y^2+2)(z^2+2)\ge (\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2$ let $x,y,z>0$,and such
$$x+y+z=3$$
prove or disprove this
$$(x^2+2)(y^2+2)(z^2+2)\ge (\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2\tag{1}$$
I know this well know inequality 1
$$(a^2+2)(b^2+2)(c^2... | $(\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2 \le 3(y^2+yz+z^2+z^2+zx+x^2+z^2+zx+x^2) \iff $
$(x^2+2)(y^2+2)(z^2+2)- 3(y^2+yz+z^2+z^2+zx+x^2+z^2+zx+x^2) \ge0 \iff x^2y^2z^2+2y^2z^2+2x^2z^2-2z^2-3yz-3xz+2x^2y^2-2y^2-3xy-2x^2+8\ge 0 $$ \tag 2$
let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz,\to u=1\ge v \ge w$,
$(2)$ be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.