Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio We already know that, for any real number $t$ such that $t\geq-1$,
$$
\int_{0}^{\pi/2} \ln \left(1+t \sin^2 x\right) \mathrm{d}x = \pi \ln \left( \frac{1+\sqrt{1+t}}{2} \right).
$$
Prove that
$$
\int_{0}^{\pi/2} \ln \left(1+4\sin^4 x\right... | Let's put it with real numbers.
We have
$$
\int_{0}^{\pi/2} \ln \left(1+t\sin^4 x\right) \mathrm{d}x
= \pi \ln \left( \frac{1}{4} \sqrt{ 1 + \sqrt{1+t} }
\left( \sqrt{1 + \sqrt{1+t} } + \sqrt{2} \right) \right), \quad t \geq -1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Prove that: $\sum\limits_{cyc} \frac{a^2+2bc}{(b+c)^2}\geq \sum\limits_{cyc} \frac{3}{2}\frac{a}{b+c}$ Let $a, b, c > 0$.Prove that: $\sum\limits_{cyc} \frac{a^2+2bc}{(b+c)^2}\geq \sum \frac{3}{2}\frac{a}{b+c}$
p/s:
I tried to solve the problem by $S.O.S$. But I cannot solve it !!
I have:
The inequatily $\Leftrightarr... | There is a good reason why you couldn’t solve it : your inequality is false
as claimed. Let $\varepsilon\in[-\frac{1}{2},\frac{1}{2}]$ be a variable tending to zero. Putting $a=1-2\varepsilon,b=1+\varepsilon,c=1+\varepsilon$, we have (using Landau’s $O$-
notation)
$$
\begin{array}{lcl}
\sum\limits_{cyc}\frac{a^2+2bc}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Polynomial Division - "Define the largest natural number..." Would someone mind helping me with this question? The more detailed possible so I can have 100% of understanding. Thanks.
Question: Define the largest natural number m such that the polynomial
$$P(x) = x^5-3x^4+5x^3-7x^2+6x-2$$
be divisible by $(x-1)^m$.
| Using the Euclidean division of $x^5-3x^4+5x^3-7x^2+6x-2$ and $x-1$ we get:
$$x^5-3x^4+5x^3-7x^2+6x-2=(x^4-2x^3+3x^2-4x+2)(x-1)$$
Then apply the Euclidean division of $x^4-2x^3+3x^2-4x+2$ and $x-1$.
Then we get $x^4-2x^3+3x^2-4x+2=q(x-1)$.
Then apply the Euclidean division of $q$ and $x-1$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \i... | Well, here's another one. In fact, it is possible to prove that for any integer $n$ and any real number $\alpha$, we have
$$n=\lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil.$$
First, we have $$\alpha n\leq\lceil\alpha n\rceil,$$ therefore, $$n-\lceil\alpha n\rceil\leq (1-\alpha)n,$$ but then, since $n-\lceil\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Can you factor before finding derivative? Say the function is $y=\frac{x^2-1}{x-1}$
Can you factor functions before finding the derivative or does that not work?
| The functions $f(x) = \dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1}$ and $g(x) = x+1$ are equal at every real number $x$ except $x = 1$, where $f(x)$ is undefined, but $g(x)$ is defined.
Therefore, $f'(x) = g'(x) = 1$ at every real number $x$ except $x = 1$.
If you used the quotient rule, you get
$f'(x) = \dfrac{(x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding a mistake in the computation of a double integral in polar coordinates I have to find $P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right) $
$f(x)$ and $f(y)$ are given, which I will use in my solution below .
$$P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right) = \int\int_D f(x)\cdot... | Upper limit $-\color{red}{0.5}(\sqrt2)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trignometry-Prove that $(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$ Prove that $$(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$$
I tried solving the LHS and RHS seperately but they were not coming ... | $(\csc\theta-\sec\theta)(\cot\theta-\tan\theta)=\displaystyle\big(\frac{1}{\sin\theta}-\frac{1}{\cos\theta}\big)\big(\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\big)$
$\displaystyle=\big(\frac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{\cos^{2}\theta-\sin^2\theta}{\cos\theta\sin\theta}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the derivative of $y=x\sqrt{9-x}$ "Find the derivative of $y=x\sqrt{9-x}$."
So this is what I have and now I'm stuck.
\begin{align}
y' &= x \frac{d}{dx}\left[(9-x)^{1/2}\right] + (9-x)^{1/2} \frac{d}{dx}(x)\\
&= x \left[\frac{1}{2}(9-x)^{-1/2}\right] + (9-x)^{1/2} (1)
\end{align}
So I now that I need to multipl... | $$y=x\sqrt{9-x}$$
$$y'=x'\sqrt{9-x}+x(\sqrt{9-x})'=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(9-x)'=$$
$$=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(-1)=\sqrt{9-x}+\frac{-x}{2\sqrt{9-x}}=$$
$$=\frac{2(9-x)-x}{2\sqrt{9-x}}=\frac{18-3x}{2\sqrt{9-x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
| For $n = 2k$
$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k} + \frac{1}{2k+1} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k} \geq\\
\geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k+1 \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
show that there are $1\leq j_1< j_2< j_3\leq 13$ such that $\left | A_{j1}\cap A_{j2}\cap A_{j3} \right |\geq 3$. Given $A_i,A_2,...,A_{13}$ $\subset [10]$ when $\left | A_i \right |=5$ for all $i$.
Need to show that there are $1\leq j_1< j_2< j_3\leq 13$ such that $\left | A_{j_1}\cap A_{j_2}\cap A_{j_3} \right |\geq$... | Your claim is false as currently stated (though I think
it becomes true if $13$ is replaced by $14$). Here is a
counterexample :
$$
\begin{array}{lcllcllcl}
A_1 &=& \lbrace 1,2,3,4,5 \rbrace, &
A_2 &=& \lbrace 1,2,3,4,6 \rbrace, &
A_3 &=& \lbrace 1,2,5,6,7 \rbrace, \\
A_4 &=& \lbrace 1,2,7,8,9 \rbrace, &
A_5 &=& \lbra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find side BC of a triangle given AB, AC, and a relation between $\angle A$ and $\angle B$ A question from my class:
In triangle $ABC$, $3\angle A+2\angle B=180$ and $AB=10, AC=4$. So question is, what all can we comment on side $BC$. Can we find its exact length?
I have a crude solution involving trigonometry an... | Let $\frac{A}{2} = \beta$
$10(4\cos^3\beta - 3\cos \beta)= 4\cos \beta$
Since $\cos\beta$ is not equal to zero (if it were equal to zero, then A would be $180^\circ$ and the triangle would not exist),
$40\cos^2\beta=34$
$\cos^2\beta=\frac{17}{20}$, $\sin^2\beta=\frac{3}{20}$
$\cos A= \cos 2 \beta = \cos^2\beta - \sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear... | $\left( \frac{x^2 -3}{x^2 +1}\right) \Rightarrow \left(\frac {x^2}{x^2+1} - \frac {3}{x^2+1} \right) \Rightarrow \left(\frac{x^2+1}{x^2+1} - \frac {1}{x^2+1} -\frac{3}{x^2+1} \right)\Rightarrow \left(1 -\frac {4}{x^2+1}\right) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
$\sin x+\sqrt{5} \cos x$ in a form $c\cdot \sin (x+d)$ How can I rewrite $\sin x + \sqrt{5}\cdot \cos x$ in a form $c \cdot \sin (x+d)$??? How can I find the values for $c$ and $d$? I have no idea how to solve that algebraically.
Is there also a possibility to rewrite it in termes of $\cos$ instead of $\sin$? Or maybe ... | In general, if we have: $a\sin x + b\cos x$, you can write that as: $$\sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin x + \frac{b}{\sqrt{a^2 + b^2}}\cos x \right)$$
and call $y = \arccos\left(\frac{a}{\sqrt{a^2 + b^2}}\right)$, to finally use the formula for $\sin(x+y)$ on the other direction. Applying that to your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2... | $\frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2}$
$\frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2}$
$\frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
$\frac{2^{2x-2}}{2^{x-2}} 2^{2x-2} = 2^{3x-2}$
Then use the fact that : $\frac{a^{b}}{a^{c}} = a^{b-c}$ with $a\ne{0}$
So : $2^{2x-2-(x-2)} 2^{2x-2} = 2^{3x-2}$
$\Rightarrow$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written i... | EDIT: added second answer proving the main question.
Since we've proven that the family of numbers in the form of $a^2+3b^2$ is closed unfer product, we just jave to prove that any prime divisor $d$ of $a^2 + 3b^2$ can be represented as $r^2 + 3s^2$ for some integers $r,s$.
Let $d\ \vert \ (a^2 + 3b^2)$ then, $dk = a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Introductory Induction proof that $n(n^2 +5)$ is divisible by $6$ I am in currently in a discrete mathematics class, and I've done well on every problem I've encountered. Unfortunately, I find myself weak at some of the seemingly straight forward induction problems. As is the case of the following corollary. I will att... | Method 1:
$$0\left(0^2+5\right)=0$$
$$1\left(1^2+5\right)=6$$
$$2\left(2^2+5\right)=18$$
$$3\left(3^2+5\right)=42$$
$$4\left(4^2+5\right)=84$$
$$5\left(5^2+5\right)=150$$
Since all of these are $0 \pmod 6$, every other number must be $0 \pmod 6$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/889232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Interesting combinatorial identities Let $n$ be a strictly positive integer and let $j=0,\dots,n-1$.
By using Mathematica I managed to guess the following identities:
\begin{eqnarray}
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m}{j} &=& \frac{1}{2} \binom{2 n}{2j + 1} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}... | We provide a closed form expression for our sum.
\begin{eqnarray}
&&\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+a}{j+a}=\\
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\binom{a+2 n}{a+2 j+1}-\binom{a+n}{a+j+1} \binom{a+2 n}{j} \, \cdot _3F_2(-j,a+j+1,a+n+1;a+j+2,a-j+2 n+1;1) =\\
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of irrational roots of the equation $3^x8^{\frac{x}{x+1}}=36$ Find the number of irrational solutions of the equation
$$3^x8^{\frac{x}{x+1}}=36.$$
| Write exponents in LHS of ${3}^{x} {8}^{\frac{x}{x+1}} = 36$ in terms of a common base:
$${3}^{x} {8}^{\frac{x}{x+1}} = {e}^{\ln{3^x}} e^{\ln{{8}^{\frac{x}{x+1}}}} = {e}^{x \ln{3}} {e}^{\frac{{x} \ln{8}}{x+1}} = {e}^{x \ln{3}+\frac{x \ln{8}}{x+1}} = 36$$
Taking the natural log of both sides gives $x\ln{3} + \frac{x\ln{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ As the title says, what would be the condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$? Would there be infinitely many tuples that satisfy the condition above?
| Let $i = \sqrt{-1}$. Then, $(a+bi)(c+di) = (ac-bd)+(ad+bc)i = 0+0i = 0$, so either $a+bi = 0$ or $c+di = 0$, i.e. either $a = b = 0$ or $c = d = 0$.
If $a = b = 0$, then we are left with $cd = c^2-d^2 = 0$, from which we get $c = d = 0$.
If $c = d = 0$, then we are left with $ab = a^2-b^2 = 0$, from which we get $a =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Generalization of a formula for 2x2-matrices It is well known that $$|det(v_1,...,v_n)|\le ||v_1||_2...||v_n||_2$$
with equality if and only if the vectors are pairwise orthogonal. For n = 2,
the following formula holds : $$det(\pmatrix {a&b\\c&d})^2=(a^2+b^2)(c^2+d^2)-(ac+bd)^2$$
How can this be generalized to mat... | Let me explain this formula in geometric language. Assume $a=(a_1,a_2),b=(b_1,b_2)$ be two vectors, then the determinant of $\det (a,b)$ is the area of the parallelogram $S$ with two sides $a,b$. Note that
$$S^2:=\left(\det (a,b)\right)^2=(ab_\perp)^2=a^2\left(b^2-\frac{(a \cdot b)^2}{a^2}\right)=a^2b^2-(a\cdot b)^2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerat... | To avoid absolute values
\begin{align}
\int \frac{\sqrt{1+x^2}}{x} dx
= &\int \frac x{\sqrt{1+x^2}} + \frac 1{x\sqrt{1+x^2}} \ dx\\
=&\ \sqrt{1+x^2} - \text{coth}^{-1}\sqrt{1+x^2}
\end{align}
which is valid for all domain $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to prove: $\left(\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}}}-1\right)^{4}=5$? Question:
show that: the beautiful ${\tt sqrt}$-identity:
$$
\left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\
+\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4}
=5
$$
Can you someone... | Let $x = 4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}$. Then, we have:
(1) $x = 4 - 3 \cdot 5^{1/4} + 2 \cdot 5^{2/4} - 5^{3/4}$
(2) $5^{1/4}x = 4\cdot 5^{1/4} - 3 \cdot 5^{2/4} + 2 \cdot 5^{3/4} - 5$ (Multiply (1) by $5^{1/4}$)
(3) $(5^{1/4}+1)x = -1 + 5^{1/4} - 5^{2/4} +5^{3/4}$ (Add (1) and (2))
(4) $5^{1/4}(5^{1/4}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
How to compute $\int \frac{x}{(x^2-4x+8)^2} \mathrm dx$? Can someone help me to compute:
$$\int \frac{x}{(x^2-4x+8)^2}\mathrm dx$$
And, in general, the type:
$$\int \frac{N(x)}{(x^2+px+q)^n}\mathrm dx$$
with the order of polynomial $N(x)<n$ and $n$ natural greater than 1?
| $$\int \frac{x}{(x^2-4x+8)^2} dx = \int\frac{x - 2 + 2}{(x^2 - 4x + 8)^2}\,dx $$
$$= \frac 12\int \frac{2x - 4}{(x^2 - 4x + 8)^2}\,dx + \int \frac 2{((x-2)^2 + 2^2)^2}\,dx$$
For the first integral, use $u = x^2 - 4x + 8 \implies du = (2x-4)\,dx$.
For the second integral, put $2\tan \theta = (x-2)\implies 2\sec^2 \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Asymptotic behaviour of $\prod_{p \leq x} (1 + 4/(3p) + C p^{-3/2})$ I'm reading Montgomery and Vaughan and in it they state quite simply
\begin{equation}
\prod_{p \leq x} \left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}} \right) \ll (\log x)^{4/3}
\end{equation}
as $x \rightarrow \infty$ and where $C$ is some constant. It st... | Basically you should just mimic the proof of Mertens' formula. We have that
\[\sum_{p \leq x} \log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) = \frac{4}{3} \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x}\left(\log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) - \frac{4}{3p}\right),\]
and the first term is asympt... | {
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"answer_count": 2,
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What is the non-trivial, general solution of these equal ratios? Provide non-trivial solution of the following:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$$
$a=?, b=?, c=?$
The solution should be general.
| Besides to the good answers posted for this question (this answer and this one), I think finding eigenvalues and eigenvectors of such system will help you to find the solution.
$$\begin{array}{l}\frac{a}{{b + c}} = \frac{b}{{c + a}} = \frac{c}{{a + b}} = t\\\left[ {\begin{array}{*{20}{c}}1&{ - t}&{ - t}\\{ - t}&1&{ - t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897118",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Determinant of a matrix with $t$ in all off-diagonal entries. It seems from playing around with small values of $n$ that
$$
\det \left( \begin{array}{ccccc}
-1 & t & t & \dots & t\\
t & -1 & t & \dots & t\\
t & t & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t & t & t & \dots& -1
\end{array}\right) =... | Using elementary operations instead of induction is key.
$$\begin{align}
&\begin{vmatrix}
-1 & t & t & \dots & t\\
t & -1 & t & \dots & t\\
t & t & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t & t & t & \dots& -1
\end{vmatrix}\\
&=
\begin{vmatrix}
-t-1 & 0 & 0 & \dots & t+1\\
0 & -t-1 & 0 & \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Verifying an antiderivative found in any integral table If $a > 0$, and $0 < b < c$.
\begin{equation*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
= \frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert .
\end{equation*}
(This is the antiderivati... | For the simplicity let your calculation be $f(x)$ and let the antiderivative given in calculus book be $g(x)$. Assuming that your calculation is correct, then the proper way to express the indefinite integral using your calculation is
$$
\int \frac{1}{b + c\sin(ax)}\ dx=f(x)+C_1,
$$
where $C_1$ is a constant of integra... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Im trying to find what I got wrong in simplifying The question states " Simplify: $$\frac{6\left ( 27^{2n+3} \right )}{9^{3n+6}}$$
What I did was factoring $9$ and and $27$ to make it $3$. In the end I got a $\frac{6}{27}$ answer, simplified: $\frac{2}{9}$. but my problem is that in the book, the answer is stated as $\... | $$\frac{6 \cdot27^{2n+3}}{9^{3n+6}}=\frac{2\cdot3\cdot[(3^3)]^{2n+3}}{(3^2)^{3n+6}}$$
$$=\frac{2\cdot3\cdot3^{6n+9}}{3^{6n+12}}=\frac{2\cdot3}{3^3}=\frac2{3^2}$$
| {
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Factoring a quadratic with number in front of $x^2$ I have not yet understood how to factor a quadratic that contains a number in front of $x^2$, without using the quadratic equation. I am used to just brute forcing numbers such that AB and A + B are solutions to it.
Example:
$7x^2 = 25x + 12$
....
$7x^2 - 25x - 12 = 0... | Method I: Completing the square:
$$7x^2 - 25x - 12
=7\left(x^2-\frac{25}7x-\frac{12}7\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{12}7+\frac{25^2}{14^2}\right)\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{961}{196}\right)\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{31^2}{... | {
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"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find out the no of digits in product between some prime. How many digits are there in?
$2^{17}*3^{2}*5^{14}*7$. help me.
| As André Nicolas has pointed out in the comments,
$$2^{17}\times 5^{14} = 2^3\times 2^{14}\times 5^{14} = 2^3\times 10^{14} = 8\times 10^{14}.$$
Therefore,
$$2^{17}\times 3^2\times 5^{14}\times 7 = 3^2\times 7\times 8\times 10^{14} = 504\times 10^{14} = 5.04\times 10^{16}.$$
So $2^{17}\times 3^2\times 5^{14}\times 7$... | {
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Sum of the series $\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$ How do I find the sum of the following infinite series:
$$\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$$
I think the sum can be convert... | The $n$-th term in the series is $\dfrac{2 \cdot 6 \cdots (2n-2)}{5 \cdot 10 \cdots 5n} = \dfrac{1}{n+1}\dbinom{2n}{n}\dfrac{1}{5^{n+1}} = \dfrac{C_n}{5^{n+1}}$, where $C_n = \dfrac{1}{n+1}\dbinom{2n}{n}$ is the $n$-th Catalan number. Thus, the sum is $\displaystyle\sum_{n = 1}^{\infty}\dfrac{C_n}{5^{n+1}}$.
The genera... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$
I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ ... | To conclude your induction proof, just multiply x both sides :
$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1) $
multiply $x$ both sides :
$\begin{align} \\ x^{n+1}-x &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\
x^{n+1}-1 -(x-1) &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x)+(x-1) \\ \e... | {
"language": "en",
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"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
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Problems with trigonometry getting the power of this complex expression I'm here because I can't finish this problem, that comes from a Russian book:
Calculate $z^{40}$ where $z = \dfrac{1+i\sqrt{3}}{1-i}$
Here $i=\sqrt{-1}$. All I know right now is I need to use the Moivre's formula
$$\rho^n \left( \cos \varphi + i ... | Oh my. We have:
$$1+i\sqrt{3} = 2\exp\left(i\arctan\sqrt{3}\right)=2\exp\frac{\pi i}{3}$$
$$\frac{1}{1-i} = \frac{1}{2}(1+i) = \frac{1}{\sqrt{2}}\exp\frac{\pi i}{4},$$
hence:
$$z=\frac{1+i\sqrt 3}{1-i} = \sqrt{2}\exp\frac{7\pi i}{12},$$
so:
$$ z^{40} = 2^{20}\exp\frac{70\pi i}{3}=2^{20}\exp\frac{4\pi i}{3}=-2^{20}\exp\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Families of Idempotent $3\times 3$ Matrices I did the following analysis for $2\times2$ real idempotent (i.e. $A^2=A$) matrices:
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}^2=\begin{bmatrix}a^2+bc&(a+d)b\\(a+d)c&bc+d^2\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}
$$
So in particular we have $(a+d)c=c$ and $(a+d)b=b$ s... | This is a partial answer, but something that you may find useful.
Let $A$ be a $3\times3$ matrix and denote $t=\text{tr}(A)$, $d=\det(A)$ and $a=a_1+a_2+a_3$, where $a_k=\det(A_{\hat k\hat k})$ is the subdeterminant corresponding to the $k$th diagonal element of $A$.
The characteristic polynomial of $A$ is
$$
p(x)
=
\d... | {
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"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
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Improper integral containing $\sqrt{\cos x-\frac1{\sqrt 2}}$ in the denominator How do I find the value of this integral--
$$I=\displaystyle\int_{0}^{\pi/4} \frac{\sec^2 x \ dx}{\sqrt {\cos x-\dfrac{1}{\sqrt 2}}}$$
I came across this integral too in physics.
| With $t=\cos(x)$, we see it is an elliptic integral,
$$
I = \int_{1/\sqrt{2}}^1 \frac{dt}{t^2\sqrt{(1-t^2)(t-1/\sqrt{2})}}
$$
added
Maple gets: if $0<a<1$, then
$$
\int _{a}^{1}\!{\frac {dt}{{t}^{2}\sqrt { \left( 1-t^2 \right)
\left( t-a \right) }}}{}
=-{\frac {\sqrt {2}{\bf K} \left( 1
/2\,\sqrt {-2\,a+2} \right)... | {
"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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Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \... | Let $z=\frac yx$. Then we have
$$\dfrac{d^nz}{dx^n}=\dfrac{d^n(yx^{-1})}{dx^n}=\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}\dfrac{d^{n-k}(x^{-1})}{dx^{n-k}}=$$
$$\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}(-1)^{n-k}(n-k)!x^{-1-n+k}=$$
$$\sum_{k=0}^n(-1)^{n-k}\dfrac{n!}{k!}x^{-1-n+k}\dfrac{d^ky}{dx^k}$$
So we have
$$(\pm1)n!x^{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Circles - point of intersection of tangents Question:
Let $A$ be the center of the cricle $x^2 + y^2 - 2x-4y-20=0$. Suppose that the tangents at the points $B(1,7)$ and $D(4,-2)$ on the cricle meet at point $C$. Find the area of the quadrilateral $ABCD$.
What I have done:
Well I have found the center of the circle and ... | The center of the circle is quickly found by derivatives of the equation
$$ \left. \begin{aligned}
\frac{{\rm d}}{{\rm d}x} (x^2+y^2-2x -4 y-20) &= 0 \\
\frac{{\rm d}}{{\rm d}y} (x^2+y^2-2x -4 y-20) &= 0
\end{aligned} \right\}
\begin{aligned} x &= 1 \\ y &= 2 \end{aligned} $$
I like to use homogeneous coordinate... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with d... | May be not the answer you wanted given that it is of degree 8. But it has integer coefficients, so may be of interest.
If $R_n(x)=T_n(\sqrt{1-x^2})$, where $T_n$ is the Chebyshev polynomial of degree $n$, then
$$
T_n(\sin t)=\cos nt
$$
for all $t$. Because $\cos \alpha=0$, iff $\alpha$ is an odd multiple of $\pi/2$, th... | {
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If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$ If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$.
| HINT:
We have $$1-\cos^2\theta+2\cos\theta-2=0\iff(\cos\theta-1)^2=0$$
Alternatively , $$1-\cos^2\theta+2\cos\theta-2=0\iff\cos^2\theta+1=2\cos\theta$$
Dividing by $\displaystyle\cos\theta$ (which is clearly $\ne0$) $$\cos\theta+\sec\theta=2$$
Can you calculate $$\cos^3\theta+\sec^3\theta=\cos^3\theta+\frac1{\cos^3\... | {
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"source": "stackexchange",
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The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer Prove by induction that this number is an integer:
$$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$
Progress
I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt... | Outline: For the (strong) induction step, we can use the fact that
$$(3+\sqrt{5})^{n+1}+(3-\sqrt{5})^{n+1}=\left[(3+\sqrt{5})^{n}+(3-\sqrt{5})^{n}\right]\left[(3+\sqrt{5})+(3-\sqrt{5})\right]-(3+\sqrt{5})(3-\sqrt{5})\left[(3+\sqrt{5})^{n-1}+(3-\sqrt{5})^{n-1}\right].$$
Note that $(3+\sqrt{5})+(3-\sqrt{5})$ and $(3+\sqr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
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Evaluation of $ \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
$(1)$ Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x}{\sin \left(x+\frac{\pi}{4}\right)}dx$
$(2)$ Evaluation of $\displaystyle \int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\cdot \frac{x^3}{\sqrt{1-x... | We will work on the third one.
Set $x=2a t$ then the third one becomes
$$I_3=\int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx=(2a)^2\int_{0}^{1}\sin^{-1}\left(\sqrt{\frac{1-t}{2}}\right)tdt \tag{1}$$
Set $\sin s=\sqrt{(1-t)/2}$, then $t=\cos(2s)$, $dt=-2\sin(2s)ds$. So (1) becomes
$$I_3=(2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907086",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve a limit with radicals I don't know how to solve this limit. What should I do?
$$
\lim_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}}
$$
Thank you!!
| Multiply the given expression by a special kind of $1$ :
$1 = \frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}} . \frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} }$
$\lim \limits_{x\to 0} \frac{\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} - \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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To find maximum value If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
| We can use the method of Lagrange Multipliers. Note that the constraint function is $$ g(A,B,C) = A+B+C $$ and the maximization function is $$ f(A,B,C) = \sin A + \sin B + \sin C. $$We can find $ \nabla f $ and $ \nabla g $ and we get $$ \begin {eqnarray*} \nabla g &=& \left< 1, 1, 1 \right>, \\ \nabla f &=& \left< \co... | {
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"url": "https://math.stackexchange.com/questions/907695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
The inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/( x^3+2x^2+2)$ What is the independent coefficient in the inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/(x^3+2x^2+2)$ ?
I have been calculating some combinations, but I don't know how I can calculate the inverse.
| $$
(ax^2 + bx + c)(2x^2 + 2) + (dx + e)(x^3 + 2x + 2) = 1\\
(2a + d)x^4 + (2b + e)x^3 + (2a + 2c + 2d)x^2 + (2b + 2d + 2e)x + (2c + 2e) = 1\\
$$
$$\left[ \begin{array}{ccccc}
2 & 0 & 0 & 1 & 0 \\
0 & 2 & 0 & 0 & 1 \\
2 & 0 & 2 & 2 & 0 \\
0 & 2 & 0 & 2 & 2 \\
0 & 0 & 2 & 0 & 2 \\
\end{array} \right]
\left[ \begin{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How do I simplify $\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$?
How do I simplify the following equation?
$$\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$$
I have no idea where to start. If I multiply either fraction by its denominator I will still end up with a square root. I know the end result should be $-5... | $$\begin{align} \frac{\sqrt{21} - 5}{2} + \frac{2}{\sqrt{21} - 5} &= \frac{\sqrt{21} - 5}{2} + \frac{1}{\frac{\sqrt{21} - 5}{2}} \\ &= \frac{\big(\frac{\sqrt{21} - 5}{2}\big)^2 + 1}{\frac{\sqrt{21} - 5}{2}} \\ &= \frac{2\big(\frac{(\sqrt{21} - 5)^2}{4} + 1\big)}{\sqrt{21} - 5} \\ &= \frac{\frac{21 + 25 - 10\sqrt{21}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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If $ab+bc+ca=0$ then the value of $1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is If $ab+bc+ca=0$
then the value of
$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...
| $ab+bc+ca=0$
$a(b+c)= -bc$
$b+c= -bc/a$
Adding both side $a$ then $b+c+a=a-bc/a$
$a(a+b+c)= a^2-bc$
$1/(a^2-bc)= 1/a(a+b+c)$
Similarly $1/(b^2-ca) = 1/b(a+b+c)$
and $1/(c^2-ab) = 1/c(a+b+c)$.
By adding all eqution we get
$1/a(a+b+c)+ 1/b(a+b+c)+ 1/c(a+b+c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Why a particular ring of integers is not generated by a single element It says here in the Sage documentation that the ring of integers in the number field obtained from
$$f(x) = x^3 + x^2 - 2x + 8$$
is not generated by a single element. How would one go about showing that this is the case?
| Let $\alpha$ be a root of $f(x) = x^{3} + x^{2} - 2x + 8$. We will show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha, \beta]$, where $\beta = (\alpha + \alpha^{2})/2$. Granting this, assume that $\mathbb{Z}[\vartheta]$ for some $\vartheta = a + b\alpha + c\beta$. Since $\{1, \vartheta, \vartheta^{2}\}$ is an integral b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Induction Proof: $\sum_{k=1}^{n}\frac{a-1}{a^k}=1-\frac{1}{a^n}, a \ne 0$ I'm having trouble showing equality for the $A(n+1)$ portion of the proof.
Prove by Induction: $$\sum_{k=1}^{n}\frac{a-1}{a^k}=1-\frac{1}{a^n}, a \ne 0$$
Base Case $(n=1)$: $$\frac{a-1}{a^1}=1-\frac{1}{a^1}$$
$$1-\frac{1}{a}=1-\frac{1}{a}$$
There... | Hint:
The idea behind induction is that once the case $A(1)$ is proved you assume $A(k)$ is true for all $1\leq k\leq n$ to conclude that it musy be true for $A(n+1)$.
$$A(n+1) = \sum_{k=1}^{n+1} \frac{a-1}{a^k} = \sum_{k=1}^n \frac{a-1}{a^k} + \frac{a-1}{a^{n+1}} = 1-\frac{1}{a^n} + \frac{a}{a^{n+1}} - \frac{1}{a^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^... | $$x^5+x^3+x-3=0\\(x^5-1)+(x^3-1)+(x-1)=0\\(x-1)(x^4+x^3+2 x^2+2x+3)=0$$
So the solution you wanted is $x=1$.
Sorry but it is just a solution ad hoc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
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How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4... | Let
$$
f(x,y,z)=\log(5y+2)+\log(2z+5)+\log(x+3y)+\log(3x+z)-\log(xyz).
$$
Then
$$
f_x=\frac{-1}{x}+\frac{1}{x+3y}+\frac{3}{3x+z};\\
f_y=\frac{-1}{y}+\frac{3}{x+3y}+\frac{5}{2+5y};\\
f_z=\frac{-1}{z}+\frac{1}{3x+z}+\frac{2}{5+2z}.
$$
So the FOC's yield
$$
x=1,\quad y=\sqrt{2/15}, \quad z = \sqrt{15/2}.
$$
With these val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Simplify rational expression How do I simplfy this expression?
$$\dfrac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
I tried to use the following rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$
But I did not get the right result.
Thanks!!
| $$\frac{\frac{x}{2}+\frac{y}{3}}{6x+4y}=\frac{6 \cdot \left ( \frac{x}{2}+\frac{y}{3}\right ) }{6 \cdot (6x+4y)}=\frac{3x+2y}{6 \cdot 2 \cdot (3x+2y)}=\frac{1}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equat... | We have $n=14a+10$ for some integer $a$. If we take this equation modulo $7$ we have that $n \equiv 10 \mod 7 \equiv 3 \mod 7$ (This because $14 \equiv 0 \mod 7$ and $10 \equiv 3 \mod 7$). You should look up modular arithmetic
where you will see how to deal with this sort of questions. Modular arithmetic is very import... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
If $\frac{1}{x^2+x+1}$, then find $y_n$ ($n^{th} $differention of the equation). The answer:
$$\frac{2(-1)^n \cdot n!}{\sqrt{3}r^{n+1}}\sin(n+1)\theta,$$
where $r=\sqrt{x^2+x+1}$ and $\theta=\cot^{-1}\frac{2x+1}{\sqrt{3}}$.
How I have tried:
$$y=\frac{1}{x^2+x+1+\frac{1}{4}-\frac{1}{4}}$$
$$y=\frac{1}{(x^2+x+\frac{1}{4... | HINT:
Let $x^2+x+1=(x-a)(x-b)$
So, we can write $\displaystyle a,b=\frac{-1\pm\sqrt3i}2$
$$\implies\frac1{x^2+x+1}=\frac1{a-b}\left(\frac1{x-b}-\frac1{x-a}\right)$$
Now from this and many others, $$\frac{d^n(1/(x-a))}{dx}=(-1)^n\frac{n!}{(x-a)^{n+1}}$$
For $\displaystyle x-a=x-\left(\dfrac12+i\dfrac{\sqrt3}2\right)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show that two expressions are equivalent I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$
This expression is supposed to be equivalent to
$$\sqrt{x^2+1} \quad.$$
I tried to algebrai... | $$\\an-inovative-solution\\ \\x=tan(a) \Rightarrow \sqrt{x^2+1}=sec(a) \\\frac{(\sqrt{x^2+1}+x)^2+1}{2(\sqrt{x^2+1}+x)}=\\\frac{(sec(a)+tan(a))^2+1}{2(sec(a)+tan(a))}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a)}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a))}=\\as-we-know-(sec^2(a)=1+tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the limit of $\frac{1}{n^2}\sum_{k=1}^n k^2\sin\left(\frac{\theta}{k}\right)$ I am trying to find the limit of
$$\frac{1}{n^2}\sum_{k=1}^n k^2\sin\left(\frac{\theta}{k}\right)\qquad \theta>0$$
as $n\to \infty$. I know that since $x-\frac{x^3}{6}\leq \sin x \leq x$ then
$$\frac{1}{n^2}\sum_{k=1}^n k^2\sin \left( \f... | We have:
$$\frac{\theta}{n^2}\sum_{k=1}^{n} k = \theta\frac{n+1}{2n}= \frac{\theta}{2}+O\left(\frac{1}{n}\right) $$
while:
$$\frac{\theta^3}{6n^2}\sum_{k=1}^{n}\frac{1}{k}=\frac{\theta^3}{6}\cdot\frac{H_n}{n^2}=O\left(\frac{1}{n}\right),$$
since, obviously:
$$H_n = \sum_{k=1}^{n}\frac{1}{k}\leq\sum_{k=1}^{n} 1 = n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$ Could anyone give me hints on how to solve this limit without L'Hospital rule?
$$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$$
I tried using standard formualaes but got nowhere.
| If $L$ is the desired limit then $$\begin{aligned}\log L &= \log\left\{\lim_{x \to 0}\left(\frac{e^{x} + e^{-x} - 2}{x^{2}}\right)^{1/x^{2}}\right\}\\
&= \lim_{x \to 0}\log\left(\frac{e^{x} + e^{-x} - 2}{x^{2}}\right)^{1/x^{2}}\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{2\cosh x - 2}{x^{2}}\right)\\
&= \lim_{x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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The remainder of $1^1+2^2+3^3+\dots+98^{98}$ mod $4$ How can I solve this problem:
If the sum $S=(1^1+2^2+3^3+4^4+5^5+6^6...+98^{98})$ is divided by $4$ then what is the remainder?
I know that all the even terms I can ignore since $(2n)^{2n}=4^nn^{2n})$ which is divisible by $4$,but i dont know what to do next it. ... | $$S=1^1+2^2+3^3+4^4+5^5+6^6+7^7+...97^{97}+98^{98}\equiv \\S\equiv1^1+0+3^3+0+5^5+0+7^7+...+97^{97}+0\\\equiv1^1+3^3+5^5+7^7+...+97^{97}\\\equiv1^1+0+3^3+0+5^5+0+7^7+...+97^{97}+0\\\equiv1^1+3^3+5^5+7^7+...+97^{97}\equiv\\S\equiv1+(4k-1)^3+(4k+1)^5+(4k-1)^7+(4k+1)^9+(4k-1)^{11}...+(4k+1)^{97}\\\equiv1+(-1)^3+1^5+(-1)^7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve $x+3y=4y^3,y+3z=4z^3 ,z+3x=4x^3$ in reals
Find answers of this system of equations in reals$$
\left\{
\begin{array}{c}
x+3y=4y^3 \\
y+3z=4z^3 \\
z+3x=4x^3
\end{array}
\right.
$$
Things O have done: summing these 3 equations give $$4(x+y+z)=4(x^3+y^3+z^3)$$
$$x+y+z=x^3+y^3+z^3$$
I've also tried to show tha... | Suppose that we had $x>1$. Then, since $4x^3-3x > x$, we have $z>x$, and similarly $y>z$, $x>y$, contradiction. By symmetry, we conclude that $x,y,z\in [-1,1]$.
So there exist $\alpha,\beta,\gamma \in [0,\pi]$ with $x=\cos \alpha$, $y=\cos \beta$, $z=\cos \gamma$. By the formula for $\cos 3\alpha$, we can rewrite th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/924596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Solve equation with two unknowns I have these equations.
$2\pi r_1+2\pi r_2=24$
and $\pi r_1^2+\pi r_2^2=20$
and to solve them I did the following steps
Step 1 : $\frac {2\pi r_1+2\pi r_2}{2}= \frac {24}{2}$
Step 2 : $\pi r_1+\pi r_2= 12$
Step 3 : $\pi r_1= 12 - \pi r_2$
Step 4 : $\pi r_1^2+(12 - \pi r_2)^2=20$
Step 5 ... | $\begin{cases}2\pi r_1+2\pi r_2=24 & (1) \\ \pi r_1^2+\pi r_2^2=20 & (2)\end{cases}$
Equation $(1)$ gives $r_1=\frac{12-\pi r_2}{\pi}$
Plug that into $(2)$, we have $\pi\big(\frac{12-\pi r_2}{\pi}\big)^2+\pi r_2^2=20 \Rightarrow 2\pi r_2^2-24r_2+\frac{144}{\pi}-20=0$.
This quadratic equation has no solutions, and there... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a $2\times 2$ block matrix I would like to know the proof for:
The determinant of the block matrix
$$\begin{pmatrix} A & B\\ C& D\end{pmatrix}$$
equals
$$(D-1) \det(A) + \det(A-BC) = (D+1) \det(A) - \det(A+BC)$$
when $A$ is a square matrix, $D$ is a scalar, $C$ is a row vector and $B$ is a column vector... | We add an extra column and row to this matrix:
$$\left(\begin{array}{cc} A & B & 0 \\ C & D & 0 \\ 0 & 0& 1\end{array}\right).$$
This new matrix has the same determinant as the original.
Now we perform some row and column operations that don't change the determinant. First add $D-1$ times the last row to the second-to-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Trignometric problem (using De Movier's Theorem)
Ok so this question, I started out writing tan as sin and cos in the right side of the equation, simplified as much as possible and ended up with a very (sort of) fascinating equation which is
Where s = sin thetha and c= sin thetha
As you can see the denominator and nu... | Using DeMoivre's Theorem and the Binomial Theorem, you get:
$\cos 5\theta + i\sin 5\theta = (\cos \theta + i\sin \theta)^5$
$= \cos^5\theta + 5i\cos^4\theta\sin\theta + 10i^2\cos^3 \theta\sin^2\theta + 10i^3\cos^2 \theta\sin^3\theta + 5i^4\cos \theta \sin^4 \theta + i^5\sin^5\theta$
$= \cos^5\theta + 5i\cos^4\theta\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor ... | Another way to solve it....
First:
Note that
$$\frac{d^2}{dx^2}\Bigl(nx^{n+1}\Bigr)=n^2(n+1)x^{n-1}=(n^2+n^3)x^{n-1}\qquad(1)$$
we will use this later.
Second:
We know that
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\qquad|x|<1$$
Taking the derivative respect to $x$, we have:
$$\begin{array}{rcl}
\frac{d}{dx}\Bigg(\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Calculus Limits Problem L'Hopital's Rule is not allowed.
Question 1:
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \ ?$$
I tried to cross multiply $\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}$with $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ and I got $x+2$ on both LHS and RHS thus I conclude
$\frac{\sqrt{3+x}+1}{\sqrt{6+x}+... | For real $x, \left|\sin\dfrac{x+\pi}{x-\pi}\right|\le1$
$$\lim_{x\to\pi}\sin\frac{x-\pi}{x+\pi}\sin\dfrac{x+\pi}{x-\pi}$$
$$=0\cdot\text{ a finite real number}=0$$
$$\lim_{x\to-2}\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}=\lim_{x\to-2}\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}\cdot\lim_{x\to-2}\frac{6+x-4}{3+x-1}$$
$$=\frac{\sqrt{3+(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the $\Theta$ notation? I know that to prove that f(n) = $\Theta$(g(n)) we have to find c1, c2 > 0 and n0 such that
$$0 \le c_1 g(n) \le f(n) \le c_2 g(n)$$
I'm quite new with the proofs in general.
Let assume that we want to prove that $$an^2+bn+c=\Theta(n^2)$$
where a,b,c are constants and a > 0
I'll sta... | The intuition is to make $\frac{\left|b\right|}{n} \le \frac{a}{2}$ and to make $\frac{|c|}{n^2} \le \frac{a}{4}$. That way, you can show the following inequality:
$$ \frac{a}{4} = a - \frac{a}{2} - \frac{a}{4} \le a+\frac{b}{n}+\frac{c}{n^2} \le a + \frac{a}{2} + \frac{a}{4} = \frac{7a}{4}$$
Therefore, let $c_1 = \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ a... | $$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2 \implies 2 < x^2 < 6 \implies \sqrt{2} < |x| < \sqrt{6}$$
So, the solutions are $x$ such that $-\sqrt{6} < x < -\sqrt{2}$ or $\sqrt{2} < x < \sqrt{6}$. Drawing the number line is very helpful for these exercises.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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More Generating Functions problems (a) For this problem, define a nonstandard die as a 6-sided die that is equally likely to come up on each side, but has a different set of numbers than the usual 1,2,3,4,5,6 on its sides. A standard die will be the usual fair die bearing the numbers 1,2,3,4,5,6.
Is it possible to des... | Use a generating function for the dice. Normal dice have faces with 1 to 6 dots, so they are represented by the polynomial:
$\begin{align}
D(z)
&= z + z^2 + z^3 + z^4 + z^5 + z^6 \\
&= z \frac{1 - z^6}{1 - z}
\end{align}$
Throwing two dice gives the distribution:
$\begin{align}
D^2(z)
&= z^2 + 2 z^3 + 3 z^4 + 4 z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove by induction that $n(n+1)(n+5)$ is multiple of $3.$ $n(n+1)(n+5) = 3d$
I cannot figure out how to solve this homework question. A friend gave me a solution I couldn't make sense of, and I hope there's something easier out there. Also, what would be the general approach towards questions of this form?
| The proof can be done without induction, just see that $n$ is of one of the three forms
$n=3k$, $n=3k+1$ or $n=3k+2$. and check each of these cases
Suppose $n=3k$ and $n(n+1)(n+5)=3k(3k+1)(3k+5)=3(k(3k+1)(3k+5))=3d$ where $d=k(3k+1)(3k+5)$,
Next suppose that $n=3k+1$, hence
$n(n+1)(n+5)=(3k+1)(3k+2)(3k+6)=3((3k+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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How to represent Fermat number $F_n$ as a sum of three squares? Let $F_n=2^{2^n}+1$ be the Fermat number. How to represent the Fermat number $F_n$ for $n \geq 3$ as a sum of three squares of different natural numbers? For example for $n=3$ we have
$$
F_3=257=5^2+6^2+14^2.
$$
Is there any simple procedure to w... | Let $X=2^{2^{n-1}}$. Then $X$ is congruent to $1$ modulo $3$, so
$$
F_n=X^2+1=\bigg(\frac{2X+1}{3}\bigg)^2+
\bigg(\frac{2X-2}{3}\bigg)^2+
\bigg(\frac{X+2}{3}\bigg)^2
$$
The values are all distinct, except when $X=1$ or $4$ (which are excluded since
$n\geq 3$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{... | The second answer is correct. The mistake in the first method is that your computation of the integral (after substitution) is wrong.
It holds:
$\int\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}u=\frac{2}{4} \sqrt{1+u} +C_2=\frac{1}{2}\sqrt{1+2x^2}+C_2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to show this equals 1 without "calculations" We have $$ \sqrt[3]{2 +\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = 1 $$
Is there any way we can get this results through algebraic manipulations rather than just plugging it into a calculator?
Of course, $(2 +\sqrt{5}) + (2-\sqrt{5}) = 4 $, maybe this can in some way help?
| Let $a=\sqrt[3]{2+\sqrt{5}}$ $b=\sqrt[3]{2-\sqrt{5}}$. Obviously $a b =-1$.
Now let's forget what values of $a$ and $b$ and solve the system:
$$
ab=-1\\
a+b=1
$$
They imply
$$
a^2-a-1=0
$$
namely $a=\frac {1\pm \sqrt{5}}{2}$. So $a=\frac {1+\sqrt{5}}{2}, b=\frac {1-\sqrt{5}}{2}$. Now taking $a^3,b^3$ we can easily ge... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the series converge or diverge? I want to check, whether $$\sum\limits_{n=0}^{\infty }{\frac{n!}{(a+1)(a+2)...(a+n)}}$$
converges or diverges.
$a$ is a constant number
Ratio test
$$\begin{align}
& \frac{a_{n}}{a_{n-1}}=\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}\cdot \frac{(a+1)(a+2)...(a+(n-1))}{(n-1)!}=\frac{n}{a+n... | Since:
$$(a+1)(a+2)\cdot\ldots\cdot(a+n)=\frac{\Gamma(a+n+1)}{\Gamma(a+1)}$$
assuming $\Re(a)>1$ we can write the original series as:
$$ S = \sum_{n\geq 1}\frac{\Gamma(n+1)\Gamma(a+1)}{\Gamma(a+n+1)}=\sum_{n\geq 1}n\cdot B(n,a+1)=\sum_{n\geq 1}n\int_{0}^{1}x^{n-1}(1-x)^a\,dx
\tag{1}$$
but $\sum_{n\geq 1}n x^{n-1}=\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal. Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal.
Part A:
$$T(x,y,z)=\begin{pmatrix} 1 & 1 & 0... | Eigenvectors solve the equation $(A-\lambda I) v = 0$; eigenvectors are vectors, not matrices.
For $\lambda = 1$, you should have
$$\begin{align*}
(A-I )v &= 0 \\
\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find this integral $I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$ Question:
Find this integral
$$I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$$
let $x\to -x$,so
$$I=\int_{-\pi}^{\pi}\dfrac{x\sin(x) \cot^{-1}{(2014^{-x})}}{1+\cos^4(x)}dx$$
and note... | We have
\begin{equation}
I=\frac{\pi^2}{2}\int_0^1\frac{1}{1+x^4}dx\tag{1}
\end{equation}
Substituting $t=x^4$ yields
\begin{equation}
I=\frac{\pi^2}{8}\int_0^1\frac{t^{-\frac{3}{4}}}{1+t}dt
\end{equation}
According to Part 11. Scientia 18, 2009, 61-75 by Khristo N. Boyadzhiev, Luis A. Medina, and Victor H. Moll, the a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer wo... | Hint:
$$\lim_{x \to \infty} \frac{(\sqrt{x^4 + ax^2 +1} - \sqrt{x^4 + bx^2 +1})(\sqrt{x^4 + ax^2 +1} + \sqrt{x^4 + bx^2 +1})}{\sqrt{x^4 + ax^2 +1} + \sqrt{x^4 + bx^2 +1}}$$
The correct answer is $\frac{1}{2}(a-b)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Verify Demorgan's Law Algebraically If $\overline X \equiv \text { not }X$,
De Morgan's Laws are stated as:
*
*$ \overline{(A + B)}= \overline A\cdot \overline B$
*$ \overline{(A\cdot B)} = \overline A + \overline B$
Verify the above laws algebraically.
I can prove this using truth tables and logic gates but alge... | Lemma: $ A + B = 1 \land AB = 0 \implies A = \overline B $
Proof:
Given $ A + B = 1 $, then:
$ (A + B) \overline B = 1 \cdot \overline B = \overline B $
$\implies A \overline B + B \overline B = A \overline B + 0 = A \overline B = \overline B $ (1)
Given $ AB = 0 $, we combine it with (1) to obtain:
$ A \overline B +... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum of $x^2+y^2+z^2$ subject to $ax+by+cz=1$
If $ax+by+cz=1$, what is the minimum of $x^2+y^2+z^2$?
It is obvious that we can do Lagrangian multiplier
$$W=x^2+y^2+z^2-\lambda (ax+by+cz-1)$$
| Using the Cauchy-Schwartz Inequality:
$(x^2+y^2+z^2)\cdot (a^2+b^2+c^2)\geq (ax+by+cz)^2$ , Now Given $(ax+by+cz) = 1$
So $\displaystyle (x^2+y^2+z^2)\geq \frac{1}{(a^2+b^2+c^2)}$ and equality hold when $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Proving that $x-\left\lfloor \frac{x}{2} \right\rfloor\; =\; \left\lfloor \frac{x+1}{2} \right\rfloor$ How would you prove that if $x$ is an integer, then
$$x-\left\lfloor \frac{x}{2} \right\rfloor\; =\; \left\lfloor \frac{x+1}{2} \right\rfloor$$
I tried to start by saying that if $x$ is an even integer, then:
$$\left... | If $x$ is even, then $x=2k$ where $k \in \mathbb{Z}$.
We then have: $$x-\left\lfloor \frac{x}{2} \right\rfloor=2k-\lfloor k \rfloor=k=\left\lfloor k+\frac{1}{2} \right\rfloor=\left\lfloor \frac{2k+1}{2} \right\rfloor=\left\lfloor \frac{x+1}{2} \right\rfloor$$
If $x$ is odd, then $x=2k+1$ where $k \in \mathbb{Z}$.
We ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Estimating $\int_0^{\sqrt 2 / 2} \sin (x^2) dx$ with Taylor Series I seem to be having trouble with part of this question (Reference: Apostol Volume 1, Section 7.8, Question 8). The full question states:
(a) If $0 \leq x\leq \frac{1}{2}$, show that $\sin x = x - x^3/3! + r(x)$, where $|r(x)| \leq (\frac{1}{2})^5/5!$
(b... | Write $\sin x = x - x^3/3! + E(x)$, where $E(x)$ is the error and you have shown that $\lvert E(x) \rvert \leq 1/2^5 5!$ for $0 \leq x \leq \frac{1}{2}$. Equivalently, $\sin x^2 = x^2 - x^6/3! + E(x^2)$, where $\lvert E(x^2) \rvert \leq 1/2^5 5!$ for $0 \leq x \leq 1/\sqrt 2$.
Then
$$\int_0^{1/\sqrt{2}} \sin x^2 dx = \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small... | NTS $\cos A\cos B\cos C \leq 1/8$
Now, since $ABC$ is a triangle, only one of the angles can possibly be equal or greater than $\pi/2$. So, without loss of generality, let $B, C<\pi/2$. Then $\cos B>0,\cos C>0.$
Case 1: $A\ge\pi/2$
$\Rightarrow \cos A<0 \Rightarrow \cos A\cos B\cos C<0<1/8$
Case 2: $A<\pi/2$ $\Righta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integra... | If by $\sin(x)^2$ you mean $(\sin(x))^2$, we could follow
$$\cos(2x)=\cos(x)^2-\sin(x)^2=1-2\sin(x)^2$$
to have
$$\int \sin(x)^2 = \int \frac{1}{2} - \frac{1}{2}\cos(2x) = \frac{1}{2}x-\frac{1}{4}\sin(2x)$$
Then go ahead and use integration by parts we done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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What am I doing wrong? Finding a limit as $x$ approaches $0$ $$\lim_{x\to 0} {a-\sqrt{a^2-x^2}\over x^2} =$$
$${a-\sqrt{a^2-x^2}\over x^2}\cdot{a+\sqrt{a^2-x^2}\over a+\sqrt{a^2-x^2}} = $$
$$a^{2} - a^{2} - x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}} $$
$$-x^{2}\over ax^{2}+x^{2}\sqrt{a^{2}-x^{2}}$$
$$-1\over a+\sqrt{a... | Should be
$${a^{2} - (a^{2} - x^{2})\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}}}=
{a^{2} - a^{2} +x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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the sum of $1-\frac{1}{5}+\frac{1}{9}-\frac{1}{13}+.....$ I thought this was the real part of the series: $\sum_{n=0}^\infty \frac{i^n}{1+2n}$, with $i=\sqrt{-1}$. When taking the real part I am left with: $\sum_{n=0}^\infty \frac{\cos(n\pi/2)}{1+2n}$. I know this sum is around 0.866973, but I have no idea how to come ... | We can write $$\displaystyle I = 1-\frac{1}{5}+\frac{1}{9}-\frac{1}{13}+......\infty = \int_{0}^{1}\left(1-x^{4}+x^{8}-x^{12}+.......\infty\right)$$
So we get $$\displaystyle I =\int_{0}^{1}\frac{1}{1+x^4}dx=\frac{1}{2}\int_{0}^{1}\frac{(x^2+1)-(x^2-1)}{x^4+1}dx =\frac{1}{2}\int_{0}^{1}\frac{x^2+1}{x^4+1}dx-\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How many (decimal) digits does $2^{3021 377}$ have? I was wondering, how many (decimal) digits does $2^{3021377}$ have?
We have $2^4=16,\, 2^5=32,\, 2^6=64$ and $2^7=128,\, 2^8=256, \, 2^9=512$ but $2^{10}=1024,\, 2^{11}=2048, \, 2^{12}=4096, \, 2^{13}=8192$, so there is no periodicity in the power.
On the other hand,
... | $$\left\lfloor 3021377\log_{10}2\right\rfloor +1 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve an equation with three terms, with the unknown inside a square root, inside a third root, in two of them? The equation is $$\\ \sqrt[3]{\sqrt{a}+b}+\sqrt[3]{-\sqrt{a}+b}=k.$$
How do I find$\ a$?
| $$k=(b+c)^{1/3}+(b-c)^{1/3}$$ where $c=\sqrt a$. We use $(x+y)^3=x^3+y^3+3xy(x+y)$.
$$k^3=b+c+b-c+3(b^2-c^2)^{1/3}k=2b+3k(b^2-c^2)^{1/3}$$
Now subtract $2b$, divide by $3k$, cube, subtract from $b^2$, and you have $c^2$, which is $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Stereographic projection proof that is geometrical. Given tangents $VP$ and $ZP$ on circle intersecting at $P$.
Prove: $YX=XW$
Heres what I do know. Obviously segments $ON, OV, OS$ are all congruent since they are just radii. I also know that $VP$ and $ZP$ are congruent since they are tangent lines intersecting. There... | This is a coordinate-based proof. Choose (w.l.o.g.) the following coordinates for your points:
\begin{align*}
N&=\begin{pmatrix}0\\1\end{pmatrix} &
V&=\frac{1}{a^2+1}\begin{pmatrix}2a\\a^2-1\end{pmatrix}\\
S&=\begin{pmatrix}0\\-1\end{pmatrix}&
Z&=\frac{1}{b^2+1}\begin{pmatrix}2b\\b^2-1\end{pmatrix}
\end{align*}
This is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The integral of $x^3/(x^2+4x+3)$ I'm stumped in solving this problem. Every time I integrate by first dividing the $x^3$ by $x^2+4x+3$ and then integrating $x- \frac{4x^2-3}{x+3)(x+1)}$ using partial fractions, I keep getting the wrong answer.
| Hint
If you start doing the long division, you should arrive to $$\frac{x^3}{x^2+4 x+3}=x-4+\frac{13 x+12}{x^2+4 x+3}$$ For the last term, decompose in simple fractions and get $$\frac{x^3}{x^2+4 x+3}=x-4-\frac{1}{2 (x+1)}+\frac{27}{2 (x+3)}$$
I am sure that you can take from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 ... | Your second step goes wrong
$2^{2x}=2^x+2^x$ but not $2^2*2^x$
simpler way
$3^x=(2^2)^{3x+1}/2^{2x}$
$3^x=2^{6x+2-2x}$
$3^x=2^{4x+2}$
Then apply log to the base e on both sides and use your calculator.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Simplification of geometric series. Can someone please help simplify this series?
$$\sum_{k=1}^\infty k\left(\frac 12\right)^k$$
In general, $$\sum_{k=1}^\infty\left(\frac 12\right)^k = \frac{1}{1-\frac{1}{2}} =2.$$
However, I am confused with the $k$ in front of the term $k\big(\frac 12\big)^k$.
I understand if the pr... | We can try to solve like a geometric series:
$$
S_n = \frac{1}{2} + 2\frac{1}{2^2} + 3\frac{1}{2^3} + \dots + (n - 1)\frac{1}{2^{n - 1}} + n\frac{1}{2^n} \\
\frac{1}{2}S_n = \frac{1}{2^2} + 2\frac{1}{2^3} + 3\frac{1}{2^4} + \dots + (n - 1)\frac{1}{2^{n}} + n\frac{1}{2^{n+1}} \\
S_n - \frac{1}{2}S_n = \frac{1}{2} + (2 -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Use row reduction to show that the determinant is equal to this variable. Show
determinant of:
\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}
is equal to $(b - a)(c - a)(c - b)$
I'm not sure if you can use squares or square roots hmmm.. please help me. I'm sure it's a simple question. Much appreciated.
| We see immmediatley that the determinant is a polynomial in $a,b,c$, homgenous of degree $3$. And if $a=b$ or $b=c$ or $c=a$, two columns are identical and make the determiniant zero.
On the other hand, if $a,b,c$ are pairwise distinct, the determinant must be nonzero as any vanishing linear combination of rows would r... | {
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Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\s... | $$x(5+2\sqrt{6})=289\\
\Rightarrow x=\frac{289}{(5+2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}\\
\Rightarrow x=-289(2\sqrt{6}-5)\\$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\... | Hint: The question asks: If $z^3=8$, show that $|z|^2=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and... | Let S = $\frac{a}{b+c}$ + $\frac{b}{c+a}$ + $\frac{c}{a+b}$
M = $\frac{b}{b+c}$ + $\frac{c}{c+a}$ + $\frac{a}{a+b}$
N = $\frac{c}{b+c}$ + $\frac{a}{c+a}$ + $\frac{b}{a+b}$
Now M + N = 3.
M + S >= 3 (AM-GM Property)
N + S >= 3 (AM-GM Property)
M + N + 2S >= 6 (Adding above inequalities)
2S >= 3
S >= $\frac{3}{2}$
| {
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"answer_count": 6,
"answer_id": 3
} |
Solve differential equation, $x'=x^2-2t^{-2}$ Solve differential equation:
$x'=x^2- \frac{2}{t^2}$
Maybe is it sth connected with homogeneous equation?
I have no idea how to solve it.
| Let $x=\dfrac{1}{u}$ ,
Then $x'=-\dfrac{u'}{u^2}$
$\therefore-\dfrac{u'}{u^2}=\dfrac{1}{u^2}-\dfrac{2}{t^2}$
$u'=\dfrac{2u^2}{t^2}-1$
Let $v=\dfrac{u}{t}$ ,
Then $u=tv$
$u'=tv'+v$
$\therefore tv'+v=2v^2-1$
$tv'=2v^2-v-1$
$t\dfrac{dv}{dt}=(2v+1)(v-1)$
$\dfrac{dv}{(2v+1)(v-1)}=\dfrac{dt}{t}$
$\int\dfrac{dv}{(2v+1)(v-1)}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integral of rational function with higher degree in numerator How do I integrate this fraction:
$$\int\frac{x^3+2x^2+x-7}{x^2+x-2} dx$$
I did try the partial fraction decomposition:
$$\frac{x^3+2x^2+x-7}{x^2+x-2} = \frac{x^3+2x^2+x-7}{(x-1)(x+2)}$$
And:
$$\frac{A}{(x-1)}+\frac{B}{(x+2)}=\frac{A(x+2)}{(x-1)(x+2)}+\fra... | You need to use polynomial long division, first, so the degree in the numerator is less than that of the denominator to get
$$I = \int \left(x+1 + \dfrac{2x-5}{x^2 + x-2}\right)\,dx$$
THEN you can use partial fraction decomposition given the factors you found for the denominator.
$$I = \frac {x^2}{2} + x + \left( I_2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Base conversion between base r^3 to r^2 The problem asks the following:
If $(\alpha2)_{r^3} = (r3\beta)_{r^2},$ find $\alpha, \beta,$ and r.
First off, I assume $r^2$ > 3 and $r^3 > 2$.
I know that $(r3\beta)_{r^2}$ = (10 _ _ _ _ )$_{r}$
I've attempted to convert to base 10 and set them equal to each other:
$(\alph... | Let:
\begin{align*}
\alpha &= Ar^2 + Br + C \\
2 &= Dr^2 + Er + F \\
3 &= Gr + H \\
\beta &= Ir + J
\end{align*}
where each uppercase coefficient is some integer in $\{0, 1, \ldots, r - 1\}$. Then by substituting into our expressions in base $10$, we have that:
$$
(Ar^2 + Br + C)r^3 + (Dr^2 + Er + F) = r^5 + (Gr + H)r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a$ cannot be a prime The sides of a triangle are of length $a,b,c$ where $a,b,c$ are integers and $a>b$, angle opposite to $c$ is $60$ degrees. Prove that $a$ Cannot be a prime
| By the Law of Cosines, we have
$$
c^2=a^2+b^2-2ab\cos(\angle ACB)=a^2+b^2-ab\implies(c+b)(c-b)=a(a-b).
$$
Now, suppose $a$ is prime. Then, either $a|c+b$ or $a|c-b$. But by the triangle inequality, $a>c-b$ so we must have
$$
a|c+b.
$$
Next, because the angle opposite $c$ is $60$ degrees, either of the remaining angle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $A^{10}$ is equal to linear combination of $A^k, k = 1,...,9$ and identity matrix. Let $A=\begin{bmatrix}2&0&1\\0&1&0\\1&0&1\end{bmatrix}$.
I did this with brute force and it was messy, is there a more theoretical way?
| The characteristic polynomial of $A$ is
$$P_A(t)= \det (t I_3 - A) = t^3- 4 t^2 + 4 t -1$$
The matrix $A$ satisfies the polynomial equality
$$P_A(A) = A^3 - 4A^2 + 4A - 1 =0$$
and therefore $A^3 = 4A^2 - 4A + 1$ and so $A^{10} = A^7(2A-1)^2$, a polynomial of degree $9$.
We can find a unique polynomial $R(t)$ of degr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why is $x^5 \sin x$ an odd function? Why is $x^5 \sin x$ an odd function?
Is the result just wrong? Because $f(-x)= (-x)(-x)(-x) \sin(-x) = (-x)(-x)(-x)(-x)(-x) (-\sin x) = (-x^5)(-\sin x) = x^5 \sin x$
| If $f(x) = x^5 \sin x$, then
$$f(-x) = (-x)^5 \sin (-x) = -x^5 \left( - \sin x\right) = x^5 \sin x = f(x),$$
and so $f$ is actually even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}}$
Evaluate the limit
$$
y=\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}}
$$
My Attempt:
When $n\rightarrow \i... | As already commented by Lucian, $$\prod_{k=0}^n k^k=H(n)$$ which is the hyperfactorial function which also has a Stirling-like series $$H(n)=A e^{-\frac{n^2}{4}} n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}} \left(1+\frac{1}{720
n^2}-\frac{1433 }{7257600 n^4}+O\left(\left(\frac{1}{n}\right)^6\right)\right)$$ where $A$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $\sum \frac{1}{n^2+7n+9}$ How do we prove that $$\sum_{n=0}^{\infty} \dfrac{1}{n^2+7n+9}=1+\dfrac{\pi}{\sqrt {13}}\tan\left(\dfrac{\sqrt{13}\pi}{2}\right)$$
I tried partial fraction decomposition, but it didn't work out after that. Please help me out. Hints and answers appreciated. Thank you.
| Start with the infinite product expansion of $\cos x$
$$\cos x = \prod_{k=0}^\infty \left(1 - \frac{x^2}{(k+\frac12)^2\pi^2}\right)$$
Taking logarithm of $\cos(\pi x)$ and differentiate, we have
$$-\pi\tan(\pi x) = \sum_{k=0}^\infty \frac{2x}{x^2-(k+\frac12)^2}
\quad\implies\quad
\sum_{k=0}^\infty\frac{1}{(k+\frac12)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
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.