Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\s... | I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ \lim\limits_{x \to 0}\big( f(x) - g(x)\big)$ is not always equal to $ \lim\limits_{x \to 0} f(x) - \lim\limits_{x \to 0} g(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$
Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$
I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $(n)$ into th... | As you correctly have: $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$ Now, observe that $$\sum_{n=2}^{\infty}\frac{1}{n-1}=\sum_{n=1}^{\infty}\frac{1}{n} \qquad \text{and}\qquad\sum_{n=2}^{\infty}\frac{1}{n+1}=\sum_{n=3}^{\infty}\frac{1}{n}$$ Hence
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
roots of cubic equation complex
Given the cubic equation: $$x^3-2kx^2-4kx+k^2=0.$$
If one root of the equation is less than $1$, another root is in the interval $(1,4)$ and the third root is greater than $4$, then the value of $k$ lies in the interval $\left(a+\sqrt{b},b(a+\sqrt{6})\right)$ where $a,b\in \mathbb{N}$... | Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f \to \infty$ as $x \to \infty$ and $f \to -\infty$ as $x \to -\infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
\begin{align*}
k^2-6k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{... | It helps a lot that we know the answer in advance! Otherwise, some lucky guesswork (or knowledge of algebraic number theory) is required.
One can follow a quite similar line of reasoning to Rafael Bombelli (c.1526-1572). See Bombelli and the invention of complex numbers; or my answer to an earlier question here; or sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \... | HINT
We have that
$$\cot (2x)\cot\left(\frac{\pi }{2}-x\right)=\frac{\cos(2x)}{\sin(2x)}\frac{\sin x}{\cos x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Using a power series to show $\log(2) = \sum_{k = 1}^{\infty} \frac{1}{k \cdot 2^{k}}$ I want to show
$$\log(2) = \sum_{k = 1}^{\infty} \frac{1}{k \cdot 2^{k}}.$$
So, I start with the geometric series:
$$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} $$
Then take the integral to get
$$\log(1 - x) = -\sum_{n = 1}^{\inf... | Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
\begin{align}
\frac{1}{1 - x} &= \sum_{n = 0}^{\infty} x^{n} \\
\log(1 - x) &= -\sum_{n = 1}^{\infty} \frac{x^{n}}{n}
\end{align}
becoming
\begin{align}
2 &= \sum_{n=0}^{\infty} \frac{1}{2^{n}} \\
- \ln\left(1 - \frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof using AM-GM inequality The questions has two parts:
Prove
(i) $ xy^{3} \leq \frac{1}{4}x^{4} + \frac{3}{4}y^{4} $
and
(ii) $ xy^{3} + x^{3}y \leq x^{4} + y^{4}$.
Now then, I went about putting both sides of $\sqrt{xy} \leq \frac{1}{2}(x+y)$
to the power of 4 and it left me with
$$-x^{3}y \leq \frac{1}{4}x^{4}... | (i) follows from Young's inequality for $p = 4$ and $q = \frac43$:
$$xy^3 \le |x||y|^3 \le \frac{|x|^p}{p} + \frac{\left(|y|^3\right)^q}q = \frac{|x|^4}4 + \frac{\left(|y|^3\right)^{4/3}}{4/3} = \frac{|x|^4+3|y|^4}4 = \frac{x^4+3y^4}4$$
(ii) follows from (i):
$$xy^3 + x^3y \le \frac{x^4+3y^4}4 + \frac{3x^4+y^4}4 = x^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{x}{y^\frac{n-1}{n}}$ is constant, how do I prove $\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$? From:
$\frac{x}{y^\frac{n-1}{n}}=constant$
To:
$\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$
It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.
| We want
$\dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{y}{x}; \tag 0$
from
$\dfrac{x}{y^{\frac{n - 1}{n}}} = \alpha = \text{constant}, \tag 1$
assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:
$x = \alpha y^{\frac{n - 1}{n}}; \tag 2$
$x^n = \alpha^n y^{n - 1}; \tag 3$
$n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domai... | $$3(\sin^4x+\cos^4x)-1=3((\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x)-1=2-\frac32\sin^22x=\frac{5+3\cos4x}4$$
and your integral is also
$$\int_0^{n\pi}\frac{dx}{5+3\cos x}=n\int_0^{\pi}\frac{dx}{5+3\cos x}$$ as the cosine is an even function.
Using the Weierstrass substitution, the last integral is shown to be $\dfrac\pi4$.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$a(x-a)^2+b(x-b)^2=0$ has one solution; $a, b$ are not $0$. Prove $|a|=|b|$ $a(x-a)^2+b(x-b)^2=0$ has one solution; $a, b$ are not $0$. Prove $|a|=|b|$
simplifying the equation I got:
$$(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$$
solving for the Discriminant $D=0$, I got:
$$a(a^2b-2ab^2+b^3)=0$$
and since $a$ cannot equal $0$,... | Oooh boy! Incorporating hamam_Abdallah's answer into your efforts:
Note: you want $(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$ to have one solution.
So you naturally did the quadratic equation to get the solutions are
$x= \frac {2(a^2 +b^2) \pm \sqrt{D}}{2(a+b)}$ and for that to have one unique solution you need $D= 0$.
That's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Easier way to find eigenvalues of Matrices? I am trying to find eigenvalues for this matrix,
A =
$\begin{bmatrix}
3 & 2 & -3 \\
-3 & -4 & 9 \\
-1 & -2 & 5 \\
\end{bmatrix}$
I find the characteristic equation here:
$(\lambda I - A)
=
\begin{bmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \... | Here is a simple computation:
\begin{align}
\det(\lambda I - A)& = \begin{vmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}=\begin{vmatrix}
\lambda - 2 & 0 & \lambda - 2 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}\\& =(\lambda - 2 )\begin{vmatrix}
1 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to evaluate $\int\frac{1}{3+4x+4x^2}$?
How to evaluate
$$\int\frac{1}{3+4x+4x^2}\quad ?
$$
This is what I've done so far:
$$
\frac{1}{4} \int\frac{1}{x^2+x+\frac{3}{4}}
=\frac{1}{4} \int\frac{1}{(x+\frac{1}{2})^2 + \frac{1}{2}}
$$
$$
y = \frac{1}{a}\arctan\frac{u}{a},\quad \frac{dy}{du} = \frac{1}{a^2 + u^2}
$... | HINT:Complete the squares and use the results
$$\int\frac{1}{x^2 + a^2}dx= \frac{1}{a}\arctan \frac{x}{a}+C $$
and
$$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \frac{x}{a}+C$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:
Let the regular matrix $A$,$B$:
$$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right] \hspace{2cm} B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1... | Long story short, matrix multiplication is not commutative.
Te equation $A=A^2B$ is not correct. From the equation $$(A^2)^{-1} = A^{-1}B$$
you have to multiply it by $A$ from the left to get
$$A(A^2)^{-1} = B$$
and the multiply it by $A^2$ from the right and you get
$$A=B\cdot A^2$$
which is not the same as $A=A^2B$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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How to solve a recurrence relation with generating functions? I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n \geq 2$
This is what I was thinking,
\begin{align*}
g(x) &= a_0x^0+a_1x^1+a_2x^2+...+... | You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,\ldots,N$ to get that $a_n$ depends on $\sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us ass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why do we automatically assume that when we divide a polynomial by a second degree polynomial the remainder is linear? Given question:
If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
what is the remainder when the polynomial is divided by $x^2 − 2x −... | While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)
If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
Evaluating $ \int \frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {\rm d}x$, where $S_n = \sin^n(x) + \cos^n(x)$
Prove that
$$ \int \frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {\rm d}x = 2x - \arctan \left( \frac{\tan2x}{2 + \tan^22x} \right) + C$$
where $S_n = \sin^n(x) + \cos^n(x)$.
Even differentiating the right doe... | Let $c=\cos2x$. Then applying the double angle identities,
$$\sin^2x=\frac{1-\cos2x}2,\quad\cos^2x=\frac{1+\cos2x}2,$$
to each occurrence of $\sin^nx$ and $\cos^nx$, we get
$$\begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\\
S_{10}=2^{-4}(1+10c^2+5c^4)\\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
\end{ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$
Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of
$$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$
Source: IME entrance exam (... | We solve it using complex residues by way of enrichment. With the
function
$$f(z) = \frac{1}{z-1} \frac{(n+1)/z}{z^{n+1}-1}$$
we have for $\zeta_k = \exp(2\pi i k/(n+1))$ with $1\le k\le n$ that
$$\mathrm{Res}_{z=\zeta_k} f(z)
= \frac{1}{z-1}
\left.\frac{(n+1)/z}{(n+1)z^n}\right|_{z=\zeta_k}
= \frac{1}{\zeta_k-1}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3038472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 6
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Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$ Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
parity.
According to the previous theorem,My try is the followin... | Since $A$ may be any odd number $\ge3$ and we can find one or more triples for any odd leg $A\ge 3$ using a function of $(m,A).$
We know that $A=m^2-n^2\implies n=\sqrt{m^2-A}.$
$$\text{We can let }n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le \frac{A+1}{2}$$
$\text{Note: }n\in \mathbb{R}\implies \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why is my proof for “if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$” is false?
$$\text{if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$
$$x(4-x^2)>-1$$
$$x>\dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 \leqslant x \leq... | You made two mistakes and recovered:
$$\dfrac{1}{x^2-4}<2 \not\Rightarrow 1<2(x^2-4) \ \ (\text{it must be} \ 1\color{red}>2(x^2-4), \text{because} \ x^2-4\le 0)\\
4.5<x^2 \not\Rightarrow -2.121<x<2.121 \ (\text{it must be} \ 4.5\color{red}>x^2 \Rightarrow -2.121<x<2.121)$$
Alternatively, you can prove it as follows:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $a_n > 0$ prove that $\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}$ converges I have an interesting task: If $a_n > 0$, prove that $$\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}$$ converges.
I thought that it will be simple because ratio test gives me:
$$\frac{u_{n+1}}{u_n}= \frac{a_{n+... | Hint:
$$
\eqalign{
& \sum\limits_{1\, \le \,n\,} {{{a_n } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right) \cdots \left( {a_n + 1} \right)}}} = \cr
& = \sum\limits_{1\, \le \,n\,} {{{\left( {a_n + 1} \right) - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right) \cdots \left( {a_n + 1} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof verification that $\{x_n\} = 0,\underbrace{77\dots 7}_{\text{n times}}$ is a Cauchy sequence.
Given a sequence $\{x_n\}$:
$$
x_n = 0,\underbrace{77\dots 7}_{\text n\ times}
$$
Prove that $\{x_n\}$ is a Cauchy sequence.
Recall the definition of a fundamental sequence:
$$
x_n\ \text{is fundamental} \ \iff \f... | Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + \cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = \frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = \frac{1}{1+r},\ r \in \Bbb R_{>0}
$$
Then for $m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
My Attempt
\begin{align}
\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{... | Hint.
$$
\tan(\mbox{arccot}(a)) = \frac 1a
$$
and
$$
\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}
$$
so
$$
\tan(\mbox{arccot}(a)+\mbox{arccot}(b)) = \frac{a+b}{a b -1}
$$
etc.
NOTE
You can try now
$$
\tan(\mbox{arccot}(a)+\mbox{arccot}(b)+\mbox{arccot}(c)) = d = \frac{a (b+c)+b c-1}{abc-(a+b+c)}
$$
and then
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Integer Solutions of the Equation $u^3 = r^2-s^2$ The question says the following:
Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.
The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $\gcd(r,s) = 1$and $r+s \equiv 1 \pmod ... | $1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$
...
The difference between two consecutive squares on the left will give you a cube:
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...
Which means the solutions are pairs of this form: $(\frac{n(n-1)}{2}, \frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12... | Above equation shown below:
$n^4+2n^3+6n^2+12n+25=m^2$
As pointed out by Will Jagy the only positive
integer solution to above equation is
$(n,m)=(8,75)$
Also $m$ is a multiple of five as mentioned by Rhyes hughes
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, ... | You correctly derived:
$$\frac{2a}{c+b}=\frac{-a+b+c}{a}$$
Your mistake is here:
Then I took the original equations and did $\frac{a+b-c}{c}=1$
Instead, denote: $\frac{b+c}{a}=t$, then the above equation becomes:
$$\frac2{t}=-1+t \Rightarrow t^2-t-2=0 \Rightarrow t_1=-1 \ \ \text{and} \ \ t_2=2.$$
Also note that:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see ... | Use the formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$.
We get:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=\\
\frac{(\sqrt{x-a}+\sqrt{x-b})((x-a)-\sqrt{(x-a)(x-b)}+(x-b))}{\sqrt{x-a}+\sqrt{x-b}}=\\
2x-a-b-\sqrt{(x-a)(x-b)}=a-b \Rightarrow \\
(x-a)(x-b)=(2x-2a)^2 \Rightarrow \\
3x^2+(b-7a)x+4a^2-ab=0 \Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal ... | My favorite way is to consider
$$
f(x)=\left(\frac{1}{x^3}+\frac{1}{x^2}\right)^{1/3}-\left(\frac{1}{x^3}+1\right)^{1/3}
=\frac{\sqrt[3]{1+x}-\sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
\lim_{x\to0^+}f(x)=\lim_{x\to0}\frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=\frac{1}{3}
$$
Without Ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be... | We have that $(2x+a)(x+b)=2x^2+5x-3$, where $x=-\frac a2$ and $x=-b$ are the solutions you require.
Then $$(2x^2+(a+2b)x+ab=2x^2+5x-3$$
leads to $$a+2b=5; ab=-3$$
We set $b=\frac12(5-a)\to\frac12a(5-a)=-3 \to a^2-5a-6=0$
$$\to a=-1, 6$$
$$\to b= 3, -\frac 12$$
So $$2x^2+5x-3=(2x-1)(x+3)=(2x+6)(x-\frac12)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
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Evaluating $\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$
How can we find the value of $$\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$$ using elementary methods?
With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.
Thoughts of this integra... | Here is an elementary approach, although it turned into a crossover with FDP's answer.
First note that from here we have:
$$\color{blue}{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}=\frac{3}{2}\int_0^1 \frac{\arctan x\ln(1+x^2)}{x}dx$$
$$\overset{IBP}=\frac32 \underbrace{\ln x\arctan x\ln(1+x^2)\bigg|_0^1}_{=0}-\frac32 \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Solving : $2n^{\frac{1}{2}}-1.5n^{\frac{1}{3}}+n^{\frac{1}{6}}=y$ for $n$ in terms of $y$ I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{\frac{1}{2}}-bn^{\frac{1}{3}}+cn^{\frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no a... | Let $x=n^{1/6}$ (i.e. $n=x^6$):
$$2x^3-\frac 3 2x^2+x=y$$
Subtract $y$ and divide both sides by $2$:
$$x^3-\frac 3 4x^2+\frac 1 2x-\frac 1 2y=0$$
In order to reduce this to a depressed cubic, let $x=z+\frac 1 4$:
$$z^3+\frac{5z}{16}+\frac{3}{32}-\frac 1 2y=0$$
In order to turn this into a quadratic, let $z=w+\frac{5}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of Infinity of Trigo to Pi I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular$ \to 8$-sided r... | \begin{align}
\sum_{n=0}^m 2^n \left(2 \sin \frac{90^\circ}{2^n} - \sin \frac{180^\circ}{2^n} \right) &=\sum_{n=0}^m 2^n \left(2 \sin \frac{\pi}{2^{n+1}} - \sin \frac{\pi}{2^n} \right)\\
&=\sum_{n=0}^m \left(2^{n+1} \sin \frac{\pi}{2^{n+1}} - 2^n\sin \frac{\pi}{2^n} \right)\\
&=\left(2^1 \cdot \sin \frac{\pi}{2} -2^0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving that a triangle is isosceles. In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the c... | Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=\frac{a+b-c}{2}$$ and $$CD=\frac{a+c-b}{2}.$$
Thus, $$CG_2=\frac{1}{3}\sqrt{2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2}$$ and
$$BG_1=\frac{1}{3}\sqrt{2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2},$$ which gives
$$2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2=2c^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all functions satisfying $f(x+1)=\frac{f(x)-5}{f(x)-3}$ Find all functions satisfying
$$f(x+1)=\frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-\frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-\frac{2}{g(x)}$$
Any clue here?
| Let $f(x)=g(x)+3$ ,
Then $g(x+1)+3=\dfrac{g(x)+3-5}{g(x)+3-3}$
$g(x+1)+3=\dfrac{g(x)-2}{g(x)}$
$g(x+1)+3=1-\dfrac{2}{g(x)}$
$g(x+1)=-2-\dfrac{2}{g(x)}$
Let $g(x)=\dfrac{h(x+1)}{h(x)}$ ,
Then $\dfrac{h(x+2)}{h(x+1)}=-2-\dfrac{2h(x)}{h(x+1)}$
$\dfrac{h(x+2)}{h(x+1)}=-\dfrac{2h(x+1)+2h(x)}{h(x+1)}$
$h(x+2)+2h(x+1)+2h(x)=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do I find out that the following two matrices are similar? How do I find out that the following two matrices are similar?
$N =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$
and $M=
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
... | Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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On calculating the limit of the infinite product $\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$
Let $S_n=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$. What is the value of $\lim_{n \to \infty} S_n$ ?
What I attempted:-
$\log S_n=\sum_{k=3}^n \log (1-\tan^4\frac{\pi}{2^k})$.
Since $\lim_{x \to 0} \frac{\tan x}{x}=1$, $\tan^4... | What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$\begin{split}
S_n&=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})\\
&=\prod_{k=3}^n (1-\tan^2\frac{\pi}{2^k})(1+\tan^2\frac{\pi}{2^k})\\
&=\prod_{k=3}^n (\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$ Let x and a be real numbers > 0. Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$
My idea is that I'm going to use $a>b \iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$\frac{1}{2}(x+\frac{a}{x}) \ge \sq... | Because by AM-GM $$x+\frac{a}{x}\geq2\sqrt{x\cdot\frac{a}{x}}=2\sqrt{a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Show that $CD\perp AB $. Let $\triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.
If $\angle BAA_1=30°$ and $D\in [AB] $ s.t. $CD=AB $ show that $CD\perp AB $.
My idea : I draw $A_1T\perp AB $, $T\in [AB]$.
I have to show that $A_1T $ is the middle line in $\triangle CBD $ $\Leftrightarrow$ $A_... | Let $A = (1,0),$ $B=(0,0),$ and $C = \left(2 - \frac2{\sqrt3},\frac23\right).$
Since $2 - \frac2{\sqrt3} \approx 0.845299,$
angles $\angle CAB$ and $\angle CBA$ are acute,
and since the distance from $C$ to $\left(\frac12,0\right)$ is greater than
$\frac12,$ the angle $\angle ABC$ also is acute.
Then $A_1 = \left(1 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the given limit as $x$ approaches $\infty$
If $f(x) = 8x^3+3x$ then, $$\lim_{x \to \infty} \frac{f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$$ is?
My attempt:
It is clear that the function cannot be easily inverted. So, there must be something in the limit given itself that may simplify the problem.
Honestly, I have no cl... | Consider the equation
$$y=8x^3+3x$$ As you say, there is only one real root which is given by
$$x(y)=\frac{1}{2} \left(\frac{\sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2
y}}{2^{2/3}}-\frac{1}{\sqrt[3]{2} \sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2 y}}\right)$$
Now, tedious but doable,for each piece, use Taylor expansion for infinitely... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$ I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify ... | $$\begin{align}
\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-\sqrt{2}&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\sqrt{2}\\
&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\frac{4\sqrt{2}-2}{4-\sqrt{2}}\\
&=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}\\
&=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}~\cdot~\frac{4+\sqrt{2}}{4+\sqrt{2}}\\
&=\frac{-10\sqrt{2}-12}{14}\\
&=\frac{-5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$) I need to simplify \begin{equation}
\frac{x^{5}}{x^{2}+1}
\end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
\begin{equation}
x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+...
\end{equation}
| When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
*
*First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
*Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
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Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$.
I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} \dots \dot... | Our OP SunShine has correctly deduced that $\deg p(x) = 1$, so that
$p(x) = ax + b; \tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; \tag 2$
$p(x + 1) = ax + a + b; \tag 3$
$xp(x + 1) = ax^2 + ax + bx; \tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; \tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; \tag 6$
$(a^2 - a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we... | With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=\frac4{4-t}+\frac{9t}{9t-1},t\in\left(\frac14,4\right)$$
Then
$$h(t)=\frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$\log h(t)=\log(-9t^2+72t-4)-\log(4-t)-\log(9t-1)$$
$$(\log h(t))'=\frac{-18t+72}{-9t^2+72... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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A verification request upon a combinatorial problem An urn contains $100$ balls numbered from $1$ to $100$. Four are removed at random without being replaced. Find the probability that the number on the last ball is smaller than the number on the first
ball.
MY ATTEMPT
If the first removed ball is $2$, then there is $1... | Here is a simple argument:
Either the number on the last ball is greater than the number on the last ball or less than the first ball's number. Hence the required probability is $\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $x \geq y>x/2$ then is it true that $x \pmod y < x/2$? I'm not sure how to prove this statement, which I believe is true:
given $x,y \in Z$ such that $x \geq y > \frac {x}{2}$ then $x$ (mod $y$) < $\frac {x}{2}.$
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x \geq y \geq \frac{x}{2};$ consi... | This is true. Suppose that $0\leq \frac{x}{2} < y \leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 \leq r <y$. Here, $r$ is your $x \bmod y$. We claim that $r < \frac{x}{2}$.
Suppose not, so that $r \geq \frac{x}{2}$. Then $x=yq+r \geq yq+\frac{x}{2}$. Subtracting $\frac{x}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to find the range of $y=\frac{x^2+1}{x+1}$ without using derivative? The only thing I know with this equation is $y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
| Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=\frac{x^2+1}{x+1}$
$$\Rightarrow yx + y = x^2 +1 \Rightarrow -x^2 + yx +(y-1)=0 \Rightarrow x = \frac{-y \pm\sqrt{y^2 + 4(y-1)}}{-2} \Rightarrow \frac{-y \pm \sqrt{(y-(-2-\sqrt5))(y - (\sqrt5-2))}}{-2}$$
The Lowest Value so wi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a Cauchy problem, differential equation I have the following Cauchy problem
\begin{cases} y'(x) + \frac{1}{x^2-1}y(x) = \sqrt{x+1} \\ y(0) = 0 \end{cases}
I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= \frac{1}{x^2-1}$ :
$$\int A(x)dx=\int \frac{1}{x^2-1}dx= \frac{1}{2} \log\Big(\fra... | Note that since you have on the right side $\sqrt{x+1}$, we may assume that $x\geq -1$. Moreover the coefficient $\frac{1}{x^2-1}$ implies that $x\not=\pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the argumen... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem on characteristic polynomial of a matrix Without computing $\det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 \times 4$ matrix given by:
$$A = \begin{bmatrix} 0 & 0 & 0 & -4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$
| To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm
In your peculiar case however, your matrix has the form of a companion matrix:
$$
A=\begin{bmatrix}0&0&\dots &0&-c_{0}\\1&0&\dots &0&-c_{1}\\0&1&\dots &0&-c_{2}\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\dots &1&-c_{{n-1}}\e... | {
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"source": "stackexchange",
"question_score": "2",
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If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$, then find $\alpha$, $\beta$ in $(0,\frac\pi2)$
If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$,
where $\displaystyle \alpha, \beta \in \bigg(0,\frac{\pi}{2}\bigg)$. Then value of $\alpha,\beta$ are
Try: I am trying to conver... | Hint:
For real $\cos^2\alpha,\sin^2\beta$
$$\dfrac{\cos^4\alpha+4\sin^4\beta+1+1}4\ge\sqrt[4]{\cos^4\alpha\cdot4\sin^4\beta}$$
The equality will occur if $\cos^4\alpha=4\sin^4\beta=1$
| {
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"source": "stackexchange",
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How to see $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$? I was reading an example where the purpose was to compute a certain Galois group. Along the way, the writer says : note $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$. But how do I note this? I understand you can factorize by $x^2-1$, since when I draw on the unit circle I see that $-1$... | Also, by using your idea we obtain:
$$x^6-1=(x^2-1)(x^4+x^2+1)=(x^2-1)(x^4+2x^2+1-x^2)=$$
$$=(x^2-1)((x^2+1)^2-x^2)=(x^2-1)(x^2-x+1)(x^2+x+1).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers. Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers.
WLOG, let a$\leqq b\leqq c$,
| Hint: given $a \le b$ and $\frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
| {
"language": "en",
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"source": "stackexchange",
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Verifying $ \sum_{k=0}^{\infty}{3k \choose k}\frac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\frac{2}{27}\right)^k=\frac{1}{4}$ $$
\sum_{k=0}^{\infty}{3k \choose k}\dfrac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\dfrac{2}{27}\right)^k=\dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
\sum_{k=0}^{\infty}\left\{{3k \cho... | By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{4}{27}\right)^k x^{2k} = \frac{\cos\left(\frac{1}{3}\arcsin x\right)}{\sqrt{1-x^2}} \tag{1}$$
and by partial fraction decomposition
$$ \frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+\frac{1}{3k-1}+\frac{1}{3k-2}\... | {
"language": "en",
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Creating an integral to represent the volume of the intersection of two balls in cartesian coordinates The question states:
Let $A$ be the intersection of the balls
$x^2+y^2+z^2\leq 9$ and $x^2+y^2+(z-8)^2\leq 49$
I am asked to just set up the iterated triple integral that represents the volume of $A$ in cartesian coor... | \begin{align}
V &= \int_{8-7}^{\frac32}\left(\int\int_{x^2+y^2 \le 49-(z-8)^2}\mathrm dx\mathrm dy\right)\mathrm dz + \int_{\frac32}^{3}\left(\int\int_{x^2+y^2 \le 9-z^2}\mathrm dx\mathrm dy\right)\mathrm dz\\
&=\int_{1}^{\frac32}\int_{-\sqrt{49-(z-8)^2}}^{\sqrt{49-(z-8)^2}}\int_{-\sqrt{49-(z-8)^2-y^2}}^{\sqrt{49-(z-8)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\righ... | Let $\displaystyle I =\frac{1}{2}\int\frac{2\cos^2 x\cdot \sin x}{\sin x-\cos x}dx=\frac{1}{2}\int\frac{(1+\cos 2x)\cdot \sin x}{\sin x-\cos x}dx$
So $\displaystyle I =\frac{1}{4}\int \frac{2\sin x}{\sin x-\cos x}dx+\frac{1}{2}\int\frac{\cos 2x\cdot \sin x}{\sin x-\cos x}dx$
Now writting
$2\sin x=(\sin x+\cos x)+(\sin... | {
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"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Write a Limit to calculate $f'(0)$ Let $f(x) = \frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+... | $f'(0) = \displaystyle \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \left (\dfrac{1}{h} \right ) \left ( \dfrac{2}{1 + h^2} - 2 \right )$
$= \lim_{h \to 0} \left (\dfrac{1}{h} \right ) \left ( -\dfrac{2h^2}{1 + h^2} \right ) = \lim_{h \to 0} -\dfrac{2h}{1 + h^2} = 0. \tag 1$
| {
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"source": "stackexchange",
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Fourier transform of the Klein group Obviously we have the elements of $G = \lbrace e, a, b, c\rbrace\;$ where $\;c=ab$
Define:
$$\chi : G \rightarrow \mathbb{Z}$$
How would I construct the transform for each of the elements of $G$?
What I mean is, for example take another group $G$ as $G = \mathbb{Z_5} = \lbrace0,1,2... | Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $\chi(g)= \pm 1$. This implies the character table will take the form:
\begin{align*}
\begin{array}{c | c c c c }
& e & a & b & c\\
... | {
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"source": "stackexchange",
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Finding real $a$, $b$, $c$ such that $x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$ has $1+i$ as a zero, and one negative integer as a zero with multiplicity $2$
Find $a, b, c \in \mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x... | If $ 1 + i $ is the root of the polynomial
$$
p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
$$
and $a,b,c\in\mathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)\cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
\begin{align}
x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
=&
(x^2-2x+2)\big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2s\big)
\\
=&x^5... | {
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"source": "stackexchange",
"question_score": "1",
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Finding out the value of $\angle DQC$ in a trapezium $ABQD$ where $\angle DCB$ = 30$^\circ$ In this below diagram, $\angle ABC=60^\circ, \angle DCB=30^\circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?
At... |
\begin{align}
\operatorname{perimeter}(APQD) &= \operatorname{perimeter}(ABCD) \\
(3+\sqrt 3)x + 2y + CQ+DQ &= (6+2\sqrt 3)x + 2y \\
CQ +DQ &= (3+\sqrt 3)x
\end{align}
\begin{align}
\operatorname{area}(APQD) &= \operatorname{area}(ABCD) \\
\frac 12(3x+2y+CQ)(\sqrt 3x) &= \frac 12(4x+2y)((\sqrt 3x)) \\
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| Hint If we rewrite $u = \cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $\cos^6 x \leq \cos^2 x$ and $\sin^4 x \leq \sin^2 x$, and in both cases equality holds (for $x \in [0, \frac{\pi}{2}]$) only for $x = 0, \frac{\pi}{2}$.
| {
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Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then ... | By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=\frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There ... | {
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Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from her... | The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=\int_0^1\frac{1-x^3}{... | {
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"question_score": "10",
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Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=... | Set $f_n(x)=\cos x\cos2x\cdots\cos nx$ for simplicity. We want to prove by induction that
$$
\lim_{x\to0}\frac{1-f_n(x)}{x^2}=\frac{1}{12}n(n+1)(2n+1)
$$
The statement is true for $n=1$. Assume it is true for $n$. Then
$$
f_{n+1}(x)=f_n(x)(\cos nx\cos x-\sin nx\sin x)
$$
Thus
\begin{align}
\frac{1-f_{n+1}(x)}{x^2}
&=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all possible values of $x$ if $\frac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real
If the expression $\dfrac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real, find the set of all possible values of $x$.
My Attempt
$$
-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\f... | You are right: there are more solutions. Actually, there is exactly one more solution in $(0,2\pi)$. Note that\begin{align}-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}=0&\iff-\tan x-1+1-2\sin^2\left(\frac x2\right)-2\sin x=0\\&\iff-\tan x-1+\cos x-\sin x.\end{align}Now, let $c=\cos x$ and let $s=\sin x$. ... | {
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Prove that $\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
Prove that $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...
$\displaystyle|f(x,y)-L|=\left |\frac... | To begin with, notice that
\begin{align*}
\begin{cases}
|x| = \sqrt{x^{2}} \leq \sqrt{x^{2}+y^{2}}\\\\
|y| = \sqrt{y^{2}} \leq \sqrt{x^{2}+y^{2}}
\end{cases}\Longrightarrow
\begin{cases}
\displaystyle\frac{|x|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
\displaystyle\frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1
\end{cases}
\end{align*}
Fu... | {
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"source": "stackexchange",
"question_score": "2",
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Integrate using trigonometric substitution. Am I on the right path? I have been trying to solve:
$$\int \frac{\sqrt{x^2-9}}{x^3} dx$$
I am letting $ x = 3\sec \theta$ and so $dx = 3 \sec \theta \tan \theta$
So then I have:
$$\int \frac{\sqrt{9\sec^2 \theta - 9}}{27 \sec^3 \theta} dx$$
$$\int \frac{\sqrt{9(\sec^2 \thet... | Yes, your solution so far is correct. Now, to integrate $\int \sin^2{\theta}\,d\theta$, use the half-angle formula for the sine function:
$$
\sin^2{\theta}=\frac{1-\cos(2\theta)}{2}.
$$
Also, a bit later, you're going to need this formula:
$$
\sin{(2\theta)}=2\sin{\theta}\cos{\theta}.
$$
| {
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"url": "https://math.stackexchange.com/questions/3098752",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$? Since
$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$
then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,
$$\dfrac{1}{2} \ar... | It's even nicer if you use simple trigonometry.
Consider the following Figure, for positive $x$ (for negative $x$ see edit, thanks to Steven for pointing this out in comment).
$\hskip1.in$
The triangle $OAB$ is right-angled, and such that $\overline{OA} = 1$, and $\overline{AB} = x$. We have
\begin{equation}\angle BO... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$ I try to prove the following equation:
for $x \geq 2$
$$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$$
To my opinion I have to us... | From Euler's summation formula (also page 50 here)
$$\sum\limits_{y< n\leq x}f(n)= \int\limits_{y}^{x}f(t)dt + \int\limits_{y}^{x}\{t\}f'(t)dt-\{x\}f(x)+\{y\}f(y) \tag{1}$$
and
$$\int \frac{\ln{x}}{x^3} dx = -\frac{\ln{x}}{2x^2}- \frac{1}{4 x^2} + C \tag{2}$$
we have
$$\sum\limits_{2\leq n \leq x}\frac{\ln{n}}{n^3}... | {
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"question_score": "3",
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"answer_id": 1
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Conflicting situations arising while solving for $x$ Given $$x^{1/4}+\frac{1}{x^{1/4}}=1$$
find $$x^{1024}+\frac{1}{x^{1024}}=?$$
My approach let $a=x^{1/4}$
$$a+\frac{1}{a}=1$$
$$a^2+1-a=0 $$
multiplying by $(a+1)$
$$(a+1)(a^2+1-a)=0 $$
$$a^3+1=0, a^3=-1$$
by solving we get value of $x$
$$x^3=1$$
Now the answer to the... | Since $a^3=-1$, one has $a^6=1$ and
$$ x=(x^\frac14)^4=a^4=-a. $$
So
$$ x^{1024}+\frac{1}{x^{1024}}=a^{1024}+\frac{1}{a^{1024}}=a^{6\cdot17+4}+\frac{1}{a^{6\cdot17+4}}=a^4+\frac{1}{a^4}=-a-\frac{1}{a}=-1$$
| {
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"url": "https://math.stackexchange.com/questions/3101546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $21^{1234}\pmod{100}\equiv \ ?$ The I'm having trouble to do this only by hand (no software or calculator). I tried the following:
\begin{align}21^{1234}(\text{mod} \ 100) &= 21^{1000}21^{200}21^{20}21^4(\text{mod} \ 100)\equiv41^{500}41^{100}41^{15}41^2(\text{mod} \ 100)\\
\end{align}
It's not reasonable to conti... | $3^4< 100 < 3^5 = 243 \equiv 43 \pmod{100}$
So $3^{5k}\equiv 43^{k}\pmod {100}$.
$43^2 = (40 + 3)^2 = 1600 + 2*3*40 + 9 \equiv 49 \pmod {100}$
So $3^{10k} \equiv 49^k \pmod {100}$.
$49^2 = (50-1)^2 = 2500 - 100 + 1 \equiv 1 \pmod {100}$.
So $3^{20k}\equiv 1\pmod{100}$ so
$3^{1234} \equiv 3^{14} \equiv 49*3^4 = (50-1)(8... | {
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"url": "https://math.stackexchange.com/questions/3101714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $f(x)$ where $f(x)(A-\frac{B}{x+B/A})+Cf(x+\frac{B}{A})=0$. $A, B, C > 0$, $x$ is complex and $Re(x)>0$. My guess is that $f(x)=0$ but I don't know how to prove it.
| For $f(x)\left(A-\dfrac{B}{x+C}\right)+Df(x+C)=0$ ,
$Df(x+C)=\dfrac{B-A(x+C)}{x+C}f(x)$
$f(Cx+C)=\dfrac{B-A(Cx+C)}{D(Cx+C)}f(Cx)$
$f(C(x+1))=\dfrac{B-AC(x+1)}{CD(x+1)}f(Cx)$
$f(C(x+1))=\dfrac{-AC\left(x-\dfrac{B}{AC}+1\right)}{CD(x+1)}f(Cx)$
$f(C(x+1))=-\dfrac{A}{D}\dfrac{x-\dfrac{B}{AC}+1}{x+1}f(Cx)$
With reference to... | {
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probability of $3$ girls sitting together in back row of adjacent seat in vans In how many ways $9$ boys and $3$ girls can be selected in two vans,each having numbered seats,$3$ in the front and $4$ at the back? How many sitting arrangements are possible $3$ girls should sit together in a back row of adjacent seat? Now... | We have $9$ boys $+$ $3$ girls $=12$ people
Total seats in one van $=3+4=7$
Total seats in $2$ vans $=7\times2=14$
So, we have to arrange $12$ people in $14$ seats.
Number of ways $= \ ^{14}P_{12}=\dfrac{14!}{(14-12)!}=7(13!)$
Now the number of ways to choose back seats is $2$ and we can arrange girls in adjacent seats... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many solutions has the following equation? Determine how many real solutions has the following equation:
$\sqrt[5]{2x+1+2\sqrt{{x^2+1}}}+\sqrt[5]{2x+1-2\sqrt{{x^2+1}}} = 2$
My first idea was to use the graph for the square root function: f:C->D, f(x)= $\sqrt[a]{b}$, if a is odd then f(x) is increasing on R so it sh... | Let $f(x)$ be a function defined by the expression.
Notice that the second term has a value of $0$ when $x=\frac{3}{4}$ and the first term has a value of $\sqrt[5]{5}$ which is less than $2$.
The derivative of $f(x)$ is
$$ \frac{1}{5}(2x+1+2\sqrt{x^2+1})^{-4/5}\left(2+\frac{2x}{\sqrt{x^2+1}}\right)+ \frac{1}{5}(2x+1-2\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $a < b$ and $c < d$ then $a + c < b + d$ I'm having a tough time with this questions. Could someone perhaps give me a hint? I'm only allowed to use the axioms of real numbers:
A1. $a + b = b + a$
A2. $a + (b + c) = (a+b) + c$
A3. $a + 0 = a$
A4. $a + (-a) = 0$
M1. $a \cdot b = b \cdot a$
M2. $a \cdot (b \... | Use $x < y\iff x + d < y+d$, to prove that $a < b \iff a+c < b+ c$ and to prove that $c< d \iff b + c < b + d$.
and finish it off with $x < y; y< z \implies x < z$.
to prove $a + c < b+c$ and $b+c < b+d\implies a+c < b+d$.
| {
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"source": "stackexchange",
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How to compute remainder of division of $P(x)$ by $x^2 -3x+2$?
The remainder of division of $P(x)$ by $x^2−1$ is $2x+1$, and the remainder of division of the same polynomial by $x^2−4$ is $x+4$. Compute the remainder of division of $P(x)$ by $x^2−3x+2$.
I will translate these into math equations
$$P(x) = (x^2-1)Q(x)+... | $P(x)=(x^2-1)Q(x)+2x+1;$
$P(x)=(x^2-4)R(x)+x+4;$
$P(x)=(x-1)(x-2)S(x)+ax+b.$
1)$P(1)=2(1)+1=a+b;$
2)$P(2)=2+4=2a+b;$
$3=a+b$; and $6=2a+b;$
$a=3$; $b=0;$
Remainder: $ax+b=3x.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Decide all solutions to the following trigonometric equation: $\sin \left( 4\,x \right) =\cos \left( 3\,x \right).$ in the interval $[0, 2π[ $. Decide all solutions to the following trigonometric equation:
$$
\sin \left( 4\,x \right) =\cos \left( 3\,x \right).
$$
in the interval $[0, 2π[$.
I start by expanding $\sin \... | Hint: Note that
$$
\sin(4x)=\cos(3x) = \sin\left(\frac{\pi}2-3x\right),
$$ and $\sin x = \sin y$ if and only if $x-y = 2n\pi$ or $x+y= (2n+1)\pi$ for some $n\in \Bbb Z$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In a geometric sequence, the sum of the first six terms is 9 times the sum of the first three terms. If the first term is f5, what is the third term? I solved this question and my answer was 20, but apparently, it's supposed to be -25/2 and I don't understand how that works out.
$S_6=9(S_3)$, $t_1=5$
$\frac{1-r^6}{1-r}... | Your answer is correct. Here it is worked out again in a different manner:
Remembering a geometric sequence is in the form $a,ax,ax^2,ax^3,\cdots$ we are told the sum of the first six terms is equal to 9 times the sum of the first three terms.
$a+ax+ax^2+ax^3+ax^4+ax^5 = 9(a+ax+ax^2)$
Letting $a=5$ and moving to one s... | {
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a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc =... | If $a=b$ then $b=c$ and $a=b=c$, which says that $abc$ is not defined.
Let $(a-b)(a-c)(b-c)\neq0.$
Thus, since
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc},$$
$$b-c=\frac{c-a}{ac}$$ and
$$c-a=\frac{a-b}{ab},$$ we obtain
$$(a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2},$$ which gives
$$abc=1$$ or $$abc=-1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\... | Hint: Note that this is a periodic sequence, meaning: $x_{n+6}=x_n$.
Therefore, this sequence diverges.
| {
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"source": "stackexchange",
"question_score": "10",
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Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve.
$\sqrt{x+1} = \sqrt{x+2}$
$x... | Yes, you can square both sides. But you do actually have to square both sides.
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$(\sqrt{x+1} - \sqrt{x+2})^2 = (\sqrt{x+3})^2$
$x+1 - 2\sqrt{x+1}\sqrt{x+2} + x + 2 = x + 3$.
$-2\sqrt{x+1}\sqrt{x+2} = -x$
$(-2\sqrt{x+1}\sqrt{x+2})^2 = (-x)^2$
$4(x+1)(x+2) = x^2$
$4x^2 +12x + 8 = x^... | {
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Show that the pseudoinverse of $T$ is $1/4T$. If $T$ is a selfadjoint matrix then, by wiki, the pseudo inverse of $T$ is given by
$$T^{+}= \lim_{t \to 0}(T^2+tI)^{-1} T. \tag{1} $$
*
*I don't understand under which norms we have
$$\lim_{t \to 0}\|T^{+}-(T^2+tI)^{-1} T \|=0?$$
*Also if $T=\left(\begin{array... | The norm is irrelevant, as all norms on $M_2(\mathbb{C})$ are equivalent, being a finite dimensional vector space.
The determinant of $T^2+tI$ is $(2+t)^2\color{red}{-4}=t^2+4t$. So you need to compute the limit of
$$
\frac{1}{t^2+4t}\begin{pmatrix}t & t \\ t & t \end{pmatrix}=
\frac{1}{t+4}\begin{pmatrix}1 & 1 \\ 1 & ... | {
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Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \fra... | It is faster using equivalents. Recall that a polynomial is asymptotically equivalent to its leading term, so
$$\frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}\sim_{-\infty}\frac{\sqrt{4x^2}}{\sqrt[3]{x^3}}=\frac{2|x|}{x}=-2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Having trouble computing $\int_3^5\frac{t}{1+0.1t} dt $ $$\int_3^5\frac{t}{1+0.1t} dt $$
For some reason this is equal to:
1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))
I have no idea how to reduce to that.
| $$
\frac{x}{1+0.1x}=\frac{x}{1+0.1x}\cdot\frac{10}{10}=
\frac{10x}{10+x}=10\left(\frac{x}{10+x}\right)=\\
10\left(\frac{-10+10+x}{10+x}\right)=
10\left(\frac{-10}{10+x}+\frac{10+x}{10+x}\right)=
10\left(-\frac{10}{10+x}+1\right)=\\
10\left(1-\frac{10}{10+x}\right)=10-\frac{100}{10+x}.
$$
$$
\int\left(10-\frac{100}{10+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a$, $b$ and $c$ are sides of a triangle, then prove that $a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c)$ $\leqslant$ $3abc$
Let $a$, $b$ and $c$ be the sides of a triangle. Prove that $$a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c) \leqslant 3abc$$
SOURCE: BANGLADESH MATH OLYMPIAD (Prepara... | Using the Ravi substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that
$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz\geq 0$$ But this is AM-GM:
$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2\geq 6\sqrt[6]{x^6y^6z^6}=6xyz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to express $z +\frac{1}{z}$ in polar form If we write $z=re^{i\theta}$ then we can express $z + \frac{1}{z}= \frac{r^{2}e^{i2\theta} +1}{r}. e^{-i\theta}$ Can we simplify it again and express it as $Re^{i \phi}$ ?
| $$z+\frac{1}{z}=\left(r+\frac{1}{r}\right)\cos\theta+i\left(r-\frac{1}{r}\right)\sin\theta=$$
$$=\sqrt{\left(r+\frac{1}{r}\right)^2\cos^2\theta+\left(r-\frac{1}{r}\right)^2\sin^2\theta}e^{\left(\arctan\left(\frac{r^2-1}{r^2+1}\tan\theta\right)\right)i}=$$
$$=\sqrt{r^2+\frac{1}{r^2}+2\cos2\theta}e^{\left(\arctan\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Confusion with Summation notation I need to compute the value of this:
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)$$
the $N$ is a gaussian noise with mean=2 and standard deviation=2.
The question:
is this equivalent to:
$$
\frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z... |
The scope of the sigma operator $\Sigma$ is solely defined via arithmetic precedence rules. The scope is given by the expression that follows immediately the $\Sigma$ and is valid respecting the arithmetic precedence rules up to an operator with precedence level equal to '$+$' or up to the end if no such operator foll... | {
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"timestamp": "2023-03-29T00:00:00",
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Does this pattern of summing polygonal numbers to get a square repeat indefinitely? I am using the table of polygonal numbers on this site:
http://oeis.org/wiki/Polygonal_numbers
The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3... | This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)
$$\sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$
$$\sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$
$$\sum_{k=0}^{4+2}{[T... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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evaluate the surface integral $f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = \sqrt{x^2 + y^2}$ with $z \leq 2$ Evaluate the surface integral of scalar function $\int_{S} f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = \sqrt{x^2 + y^2}$ with $z \leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z ... | Your approach is almost correct but for one erroneous formula:
$\mathrm{d}S = \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A $
$$
\begin{align*}
\mathrm{d}\vec{S} = \langle -z_x,-z_y, \ 1 \rangle \ \mathrm{d}A
\end{align*}
$$
therefore, $$|\mathrm{d}\vec{S}| = \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A $$
$z_x$ and $z_y$ values what ... | {
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Solving $\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$
$$\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$$
I tried much of elementary methods to solve above integral but is not advancing.
Any methods from elementary to advanced are appreciated.
| Hint. One may write
$$
\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}=\frac14\sum_{n=0}^\infty\frac1{4^n}\left(\sin x\right)^{2n+\frac13}
$$ then one is allowed to perform a termwise integration
$$
\int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\sum_{n=0}^\infty\frac1{4^{n+1}}\int_0^{\Large \frac{\pi}2}\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \f... | From $\left\lfloor x+\frac{1}{2}\right\rfloor=\lfloor 2x\rfloor-\lfloor x\rfloor$ we have
$$\left\lfloor \frac{n}{2^k}+\frac{1}{2}\right\rfloor=\left\lfloor \frac{n}{2^{k-1}}\right\rfloor-\left\lfloor \frac{n}{2^k}\right\rfloor$$
therefore
$$\sum_{k=1}^\infty \left\lfloor \frac{n}{2^k}+\frac{1}{2}\right\rfloor=\sum_{k=... | {
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"timestamp": "2023-03-29T00:00:00",
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Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is co... | Since $\frac{4}{x}>\frac{3}{x}$, we have three cases:
*
*$\left [ \frac{3}{x} \right ]=0$ and $\left [ \frac{4}{x} \right ]=5.$ Easy to show that it's impossible.
*$\left [ \frac{3}{x} \right ]=1$ and $\left [ \frac{4}{x} \right ]=4,$ which is impossible again and
*$\left [ \frac{3}{x} \right ]=2$ and $\left [ \fr... | {
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Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tr... | Remember that $$ \sqrt{a}+\sqrt{b}= {a-b\over \sqrt{a}-\sqrt{b}}$$
so $$\sqrt{x+y}+\sqrt{x+3}={y-3\over \sqrt{x+y}-\sqrt{x+3}} $$
and now we have $${y-3\over \sqrt{x+y}-\sqrt{x+3}}=\frac{1}{x}\left(y-3\right)$$
Obviuosly $y\neq 3$ so we have $$\sqrt{x+y}-\sqrt{x+3}=x$$
Since $$\sqrt{x+y}+\sqrt{x}=x+3$$
we have now $$... | {
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Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic... | Given $$\int\dfrac{x+4}{x^2+2x+5}dx=\int\dfrac{x+4}{(x+1)^2+4}dx$$
Use u-substitution: $u=x+1$ and we get
\begin{align}
\int\dfrac{u+3}{u^2+4}du &= \int\dfrac{u}{u^2+4}du+\int\dfrac{3}{u^2+4}du \\
&= \dfrac12\ln|u^2+4|+\dfrac32\arctan\left(\dfrac u2\right) +c \\
&= \dfrac12\ln|(x+1)^2+4|+\dfrac32\arctan\left(\dfrac{x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$ without expansions in limits
Evaluate $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$
One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{\log_e (1+x)^{1/x}} = e^... | If I may suggest, the problem of
$$y=\frac{(1+x)^{\frac1 x} - e + \frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{\frac1 x}\implies \log(a)= {\frac1 x}\log(1+x)$$
$$ \log(a)={\frac1 x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\sum \sqrt{\frac{a^2}{6a^2+5ab+b^2}}\le \frac{\sqrt{3}}{2}$ Let $a,b,c\in R^+$ prove that the inequality $$\sqrt{\frac{a^2}{6a^2+5ab+b^2}}+\sqrt{\frac{b^2}{6b^2+5bc+c^2}}+\sqrt{\frac{c^2}{6c^2+5ca+a^2}}\le \frac{\sqrt{3}}{2}$$
My try:$$\sum\limits_{cyc} \sqrt{\frac{a^2}{6a^2+5ab+b^2}}=\sum\limits_{cyc} \sqrt{\f... | SOS works for symmetric inequalities. For cyclic inequalities SOS helps sometimes,
but for your inequality it does not help or at least, does not help immediately.
Also, your second way gives a wrong inequality.
Try $c=0.0001$, $b=0.01$ and $a=1$.
Your first idea gives a proof.
Indeed, by AM-GM
$$\sum_{cyc}\frac{a}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the recurrence $b_1 = 2$ for $b_n = 3b_{n-1} + 5$
Solve the recurrence $b_1 = 2$ for $$b_n = 3b_{n-1} + 5$$
I've tried solving this problem using iteration, but the formula I get in the end is wrong. It is not a closed formula since there's still recurrence. I think my error is starts from the last line below. ... | Let $a_{n} = b_n+c$ for some $c$ so that $a_{n+1}=3a_n$ (i.e. $a_n$ is geometric sequence). Then $$ a_{n+1}-c= 3a_n-3c+5\implies c = 5/2$$
and $$a_n= a_03^n$$ so $$ b_n = a_03^n-5/2$$
Since $b_1 = 2$ we get $3a_0=9/2$ so $a_0 =3/2$ so we finaly have $$ b_n = {3^{n+1}-5\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$? This is a problem that I stumbled upon in one of my books.
Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):
$
\rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\
\rightarrow R\cos x... | \begin{eqnarray}
5\cos x - \sin x &=& 4\\
\frac{5}{\sqrt{26}}\cos x-\frac{1}{\sqrt{26}}\sin x&=&\frac{4}{\sqrt{26}}\\
\cos\alpha\cos x-\sin\alpha\sin x&=&\frac{4}{\sqrt{26}}\\
\end{eqnarray}
Giving
$$\cos(x+\alpha)=\frac{4}{\sqrt{26}}$$
So either
$$ x+\alpha=\arccos\left(\frac{4}{\sqrt{26}}\right)=38.33^\circ $$
or
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{... | For $x>1$, $\frac{1}{x-1}=\frac{1}{x}\frac{1}{1-1/x}$ is a convergent geometric series. Let's return to your title inequality. The left-hand side is $\sum_{n\ge 0}\frac{1}{n!x^n}$; the right-hand side is $\sum_{n\ge 0}\frac{1}{x^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to find the integration factor and solve $(x^3+xy^2-y) dx$ $+$ $(y^3+x^2y+x) dy$ $=$ $0$? $(x^3+xy^2-y)dx+(y^3+x^2y+x)dy=0$
I tried to find on Wolfram Alpha but it showed only the solution not in step-by-step, I know that the factor is $\frac{1}{x^2+y^2}$ but how to find it?
| $$(x^3+xy^2-y)dx+(y^3+x^2y+x)dy=0 \tag 1$$
If we a-priori know that the integrating factor $\mu$ is a function of $(x^2+y^2)$ it is easy to find that $\mu=\frac{1}{x^2+y^2}$ : For example see the Aleksas Domarkas's answer.
But what to do if we don't know that $\mu$ is a function of $(x^2+y^2)$ ?
Suppose that we did tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find all polynomials such that $p(x^2-2x)=p(x-2)^2$
Find all polynomials $p\in \mathbb{C}[x]$ such that $$p(x^2-2x)=p(x-2)^2$$
We can not say anything specific about the degree since both sides are of the degree $2n$.
Also by copering the coeficients we see that the leading coefficent must be $1$.
Setting $$x^2-2x = ... | Substitute $y = x-1$, we see that $p(y^2-1) = (p(y-1))^2$. Take $q(x) = p(x-1)$ we get $q(x^2) = (q(x))^2$. Take $r(x) = q(e^x))$. we get $r(2x) = r(x)^2$. Take $s(x) = \log(r(x))$ we get $s(2x) = 2s(x)$. As $s(x)$ is defined at $(0,\infty)$ and continuous we know that the only possibility is $s(x) = cx$ for some const... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$ I am looking for other methods to find the general integral
$$C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2x}$$
To do so, I first preformed $u=x-\pi$:
$$C(a,b)=\int_{-\pi}^{\pi}\frac{xdx}{a+b\cos^2x}+\pi\int_{-\pi}^{\pi}\frac{dx}{a+b\cos^2x}$$
The sub $x\mapsto -x$ prov... | By making change of variable $u=2\pi -x$, we have
$$\begin{align*}
C(a,b)&=\int_0^{2\pi}\frac{2\pi -u}{a+b\cos^2 u}du=\int_0^{2\pi}\frac{2\pi}{a+b\cos^2 u}du-C(a,b),
\end{align*}$$ hence $\displaystyle C(a,b)=\pi\int_0^{2\pi}\frac{1}{a+b\cos^2 u}du$. On the other hand, we find that
$$\begin{align*}
\int_0^{2\pi}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Equality of $2^x + 2^{4-x} \geq 8$
Want to find the value(s) of $x$ for which equality holds in $$2^x + 2^{4-x} \geq 8$$
I've found it by solving $2^x + 2^{4-x} = 8$:
$$2^x + 2^{4-x} = 8 \Rightarrow 2^{2x} - 8 \cdot 2^x + 16 = 0 \Rightarrow (2^x - 4)^2 = 0$$
so clearly $x = 2$.
However, the notes I've been reading ... | $(2^{x/2}-2^{(2-x/2)})^2 \ge 0;$
$2^x +2^{4-x} -2(2^{x/2}2^{(2-x/2)}) \ge 0;$
$2^x +2^{4-x} \ge 8;$
Equality for $2^{x/2}= 2^{(2-x/2)}$ (Why?);
$2^x=2^2$, or $x=2$.
| {
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"url": "https://math.stackexchange.com/questions/3155460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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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.