Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving $\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}$ I've been going through some of my notes when I found the following inequality for $a,b,c>0$ and $abc=1$:
$$
\begin{equation*}
\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}
\end{equation*}
$$
This was what I attempted, but that ... | here is another way which is not use derivative:
first we need to know: $\sqrt{\dfrac{x+y+z}{3}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$ (1)
to prove it, we have $\sqrt{\dfrac{x+y}{2}} \ge \dfrac{\sqrt{x}+\sqrt{y}}{2} \implies \sqrt{\dfrac{x+y+z+t}{4}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}}{4}$
let $t= \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equation $\sec(3\theta/2) = -2$ - brain dead Find $\theta$ with $\sec(3\theta/2)=-2$ on the interval $[0, 2\pi]$. I started off with $\cos(3 \theta/2)=-1/2$, thus $3\theta/2 = 2\pi/3$, but I don't know what to do afterwards, the answer should be a huge list of $\theta$s, which I cannot seem to get.
| $$\sec\left(\frac{3\theta}{2}\right)=-2 \implies\cos\left(\frac{3\theta}{2}\right)=-\frac{1}{2} \implies \cos\left(\frac{3\theta}{2}\right)=\cos\left(\frac{2\pi}{3}\right)$$
Thus, we have a general solution as follows$$\frac{3\theta}{2}=2n\pi \pm \frac{2\pi}{3} \implies \theta=\frac{2}{3}\left(2n\pi\pm\frac{2\pi}{3}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A set of numbers Problem: Let $E(x)$ be the number defined by the following expression
\begin{equation*}
E(x)=\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}
\end{equation*}
where $x$ is a real number and
$\sqrt[3]{Z}$ denotes the real cubic root of the real number $Z$.
Det... | $$E(x)^3=(Y+Z)^3=Y^3+3YZE(x)+Z^3\\
=x^3-3x+3E(x)\sqrt[3]{\frac{x^6-6x^4+9x^2-x^6+6x^4-9x^2+4}4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using Eulers formula I am trying to figure out how
\begin{equation*}
e^{i(-1+i\sqrt{3})}=e^{-\sqrt{3}} (cos(1)-i sin(1))??
\end{equation*}
I know that Euler's formula states that
\begin{equation*}
e^{ix} = \cos(x) + i \sin(x)
\end{equation*}
but surely that would mean in this case $x=-1+i\sqrt{3}$? Can someone please... | As Shayan says,
$e^{i(-1 + i \sqrt{3})} = e^{-\sqrt{3} - i} = e^{-\sqrt{3}} e^{-i} = e^{-\sqrt{3}}(\cos(1) - i\sin(1)); \tag{1}$
it is also true that
$e^{i(-1 + i \sqrt{3})} = \cos(-1 + i \sqrt{3}) + i\sin(-1 + i \sqrt{3}), \tag{2}$
since for any $z \in \Bbb C$,
$e^{iz} = \cos z + i \sin z; \tag{3}$
in (3), $e^{iz}$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b... | Your initial approach was fine. Just use the expansion $\sqrt{1+\frac{a}{n}}=1+\frac{a}{2n}+O\left(\frac{1}{n^2}\right)$ to show that
$$\begin{align}
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}&=\left(1+\frac{a}{2n}+O\left(\frac{1}{n^2}\right)\right)\left(1+\frac{b}{2n}+O\left(\frac{1}{n^2}\right)\right)\\\\\
&=1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
Complex number $\frac{z}{z+1}=2+3i$ Given that $\frac{z}{z+1}=2+3i$, find the complex number $z$, giving your answer in the form of $x+yi$.
Can someone give me some hints for solving this question? Thanks
| $$\frac{z}{z+1}=2+3i\Longleftrightarrow$$
$$\left(\frac{z}{z+1}\right)(z+1)=(2+3i)(z+1)\Longleftrightarrow$$
$$\frac{z(z+1)}{z+1}=(2+3i)z+(2+3i)\Longleftrightarrow$$
$$z=(2+3i)z+(2+3i)\Longleftrightarrow$$
$$z-(2+3i)=((2+3i)z+(2+3i))-(2+3i)\Longleftrightarrow$$
$$(-1-3i)z=2+3i\Longleftrightarrow$$
$$\frac{(-1-3i)}{-1-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Determining the max area of a trapezoid with no known sides A trapezoid is drawn inside a semi circle cross section with the upper base length, being the length of the circle diameter, $d$, and a lower base, $x$, touching the lower sides of the semi circle.
What is the maximum length of the lower base, $x$, of the trap... | Answer:
I really do not know what the problem is when you got the steps absolutely right.
If differentiation is the problem:
$$\frac{dA}{dx} = \sqrt{d^2-x^2} + (x+d) \frac{-2x}{2\sqrt{d^2-x^2}}$$
$$\frac{dA}{dx} = 0$$
Assuming two functions $U = x+d$ and $V = \sqrt{d^2-x^2}$
$$\frac{dA}{dx} = V\frac{dU}{dx}+U\frac{dV}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\lim\limits_{n\to\infty}\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$ I don't know how to transform the expression $\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$
The solution, after transformation is
$\frac{n-1}{n}x^2+2\frac{1+...+(n-1)}{... | Hint:
$$1 + \cdots + (n - 1) = \frac{n(n-1)}{2} = \frac{n^2}{2} - \frac{n}{2}$$
and
$$
1^2 + 2^2 + \cdots + (n - 1)^2 = \frac{(n-1)n(2n-1)}{6} = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
integral $\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$ I want to compute this integral
$$\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
where $0<b \leq a$.
I have this results
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2sin^2(t)+b^2cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
But I don't know ... | Although there is already an accepted answer, I put this answer here, since someone might have use of the "tool" to work with the expression like this before integrating.
The case $a=b$ is trivial, so we assume $a>b$. We write the numerator $\cos 2t$ in the following way:
$$
\cos 2t=\cos^2t-\sin^2t=\alpha\bigl(b^2\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | Here is my favorite one. Suppose for the sake of contradiction that $\sqrt{2} = \frac{a}b$ for integers $a,b$. Then $2b^2 = a^2$. Let $p$ be an odd prime that is not congruent to $\pm 1 \pmod 8$. Then by quadratic reciprocity,
$$\left( \frac{2b^2}p \right) = \left( \frac{2}p \right) = -1 \ne \left( \frac{a^2}p \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 7
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Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
My thoughts
Let $A = \{\text {Exactly one 6 in... | I think you are over thinking this. We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't
care what number it is otherwise.
One of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How to find integrals of the form $\frac 1{(a^2+x^2)^n}$ How will we find the integrals of the form $\frac 1{(a^2+x^2)^n}$?
For example if we find the integral of $\frac{1}{(x^2+1)^3}$, I cannot see any substitution here. Moreover, if we do integration by parts, it becomes more and more messy. I have heard somewhere a... | Let your integral be $I_n$. Then we use integral by part:
Let $\frac{1}{(a^2+x^2)^{n}} = u$, $du = \frac{-2nx}{(a^2+x^2)^{n+1}}dx$
$dx = dv, v = x$
$I_n = \frac{x}{(a^2+x^2)^{n}} + \int \frac{2nx^2}{(a^2+x^2)^{n+1}}dx $
The integral on the RHS can be written as: $\int \frac{2n(x^2+a^2)}{(a^2+x^2)^{n+1}}dx - \int \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Integer partitioning Suppose we have an integer $n$. I we want to partition the integer in the form of $2$ and $3$ only; i.e., $10$ can be partitioned in the form $2+2+2+2+2$ and $2+2+3+3$.
So, given an integer, how to calculate the total number of ways of doing such partitions and how many $2$'s and $3$'s are there i... | The number of partitions of $n$ into parts in the set $\{2,3\}$ is the coefficient of $x^n$ in
$$\left(1+x^2+x^4+x^6+\ldots\right)\left(1+x^3+x^6+x^9+\ldots\right)=\frac1{(1-x^2)(1-x^3)}\;.$$
For $n=10$, for instance, we can ignore powers of $x$ higher than $10$, so we need only consider
$$(1+x^2+x^4+x^6+x^8+x^{10})(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The value of $\int^{\pi/2}_0 \frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$ Problem :
Find the value of $\int^{\pi/2}_0 \frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$
Now how to use Leibnitz's rule : $\frac{d}{dx}(\int^{\psi(x)}_{\phi(x)} f(t)dt) = \frac{d}{dx}\{\psi(x)\} f(\psi(x)) -\frac{d}{dx}\{\phi(x)\} f... | Let $I(x)$ be the integral
$$I(x)=\int_0^{\pi/2}\frac{\log(1+x\sin^2 \theta)}{\sin^2\theta}d\theta$$
and assume that $x>0$ is real-valued.
Now, taking a derivative with respect to $x$ gives
$$I'(x)=\int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta=\frac{\pi}{2\sqrt{1+x}}$$
where this latter integral was evaluated using co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to integrate $\int_{-1}^1 \tan^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right )\,dx$? Evaluate
$$\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$
Could somebody please help integrate this without using Differentiation under the Integral Sign?
| First, integrate by parts to reduce the problem to calculating
$$\int \frac{x^2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$
now split into two more manageable terms
$$\int \frac{2}{\sqrt{1-x^2}(2-x^2)}\,dx + \int \frac{x^2-2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$
The left term is the only tricky one. Substitute $x= \sin u$ to get rid of the sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Limit of $\dfrac{(1+4^x)}{(1+3^x)}$? I don't remember how to find the limit in this case. I take $x$ towards $+\infty$.
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$
I do not know where to start. I would instinctively say that $1$ can't be right because $4^x$ goes faster than $3^x$ and thus one would move towards ... |
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$
$=\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}$+
$\lim\limits_{x\rightarrow \infty}\frac{4^x}{1+3^x}$
since $\lim\limits_{x\rightarrow \infty}\ (1+3^x)=\infty$
$\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}=0$
The leadind term in the denominator of $\frac{4^x}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Integrate $ \sin x /(1 + A \sin x)$ over the range $0$,$2 \pi$ for $A=0.2$ Wolfram Alpha indicates the following solution form:-
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + constant\right)^{2\pi}_0
$$
My thi... | $$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \int_0^{2\pi}\left(\frac 1 a-\frac{1}{a(1+a\sin x)}\right)\, dx=\frac{2\pi}a-\frac1a \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx =\frac{2\pi}a-\frac Ja $$
now consider
$$\begin{align}J &= \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx\\
&=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Derivative of Function with Rational Exponents $f(x)= \sqrt[3]{2x^3-5x^2+x}$ I have a question following:
$$f(x)=\sqrt[3]{2x^3-5x^2+x}$$
Here's what I did,
$$f(x)=\sqrt[3]{2x^3-5x^2+x}
\\ = (2x^3-5x^2+x)^{3\over2}
\\\\f'(x) = {3\over 2}(2x^3-5x^2+x)^{3\over2}(6x^2-10x+1)$$
Did I do this correctly?? Because I have diffe... | $$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})}
$$
let $f(x)$ be $2x^3-5x^2+x$ and $g(x)$ be $x^{(\frac{1}{3})}$
$$
d/dx \ g(f(x)) = g'(f(x))*f'(x)
$$
so
$$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})} = \frac{1}{3}(2x^3-5x^2+x)^{\frac{-2}{3}}*(6x^2-10x+1)
$$
which can be simplified further if need to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
| $$
x^2-y^2=(x-y)(x+y)=14^2
$$
Since $(x+y)-(x-y)=2y$ either both $x+y$ and $x-y$ need to be even or both need to be odd. Since their product is even, both need to be even. There are two possible factorizations: $14\cdot14$ and $2\cdot98$. $14\cdot14$ implies $y=0$, so we want $x-y=2$ and $x+y=98$. That is, $x=50$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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if three integer such diophantine equation How find $x+y+z$
following Diophantine equation
$$xy^2+yz^2+zx^2=x^2y+y^2z+z^2x+x+y+z$$
ie:$(x-y)(y-z)(z-x)=x+y+z$
where $x,y,z$ are integers.
can find $x+y+z$
I tried some values and got some near equalities when $x,y,z$ at least two are equal, $x+y+z=0$,for other ca... | Suppose $x\le y\le z$ and put $y=x+a=z-b$ where $a,b\ge 0$
Then you will find $$-a\cdot b\cdot-(a+b)=ab(a+b)=3y+a-b$$ so that $$3y= ab(a+b)+b-a$$
Then you can search for values of $a,b$ which give an integer multiple of $3$ on the right-hand side. So if $a=b=3$ you get $3y=54$ so that $y=18$.
Note that if $a$ and $b$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with sine in a right triangle Given a triangle $ABC$ with angles a,b & c, prove that if $\sin^2(a) + \sin^2(b) + \sin^2(c) = 2$ then the triangle is right angled (has an angle of $90^o$).
If I assume the triangle is right angled and have AB, AC and BC as sides, with BC being the base, then I can say that statem... | Notice, in $\Delta ABC$ , we have $a+b+c=180^o$
Given that
$$\sin^2 a+\sin^2 b+\sin^2 c=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (180^o-(a+b))=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (a+b)=2$$
$$\implies (\sin a\cos b+\cos a\sin b)^2=2-\sin^2 a-\sin^2 b$$ $$\implies \sin^2 a\cos^2 b+\cos^2 a\sin^2 b+2\sin a\sin b\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Please help to find term's coefficient in the following example I trying find the number of all solutions in the following:
$ x_1 + x_2 + x_3 + x_4 + x_5 = 24 $
where:
2 of variables are natural odd numbers
3 of variables are natural even numbers
none of variables are equal to $0$ or $1$
(all the variables are $>= 2$)... | Let's take it from the point where we ask for the coefficient of $x^{12}$ in:
$$ (1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $$
[Note, however, that a factor of $\binom{5}{2}=10$ is already left out of the calculation at this point, which would account for varying positions of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would ... | Let $r_1,r_2$ be the roots of $x^2+4x+6$. By Vieta's $r_1+r_2 = -4$ and $r_1r_2 = 6$.
Since $x^2+4x+6$ is a factor of $x^4+ax^2+b$, we have that $r_1,r_2$ are also roots of $x^4+ax^2+b$.
Then, since $x^4+ax^2+b$ is an even polynomial, $-r_1,-r_2$ are also roots of $x^4+ax^2+b$.
It is easy to see that $r_1,r_2,-r_1,-r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Is it a composite number? How do I prove $19\cdot8^n+17$ is a composite number?
Or is that number just a prime?
So I tried to find a divisor in the cases $ n = 2k $ and $ n = 2k + 1 $. But I had no success.
Do you have any ideas?
| Three cases:
*
*$n$ is even. Then $$19\cdot 8^n+17\equiv 1\cdot (-1)^n+17\equiv 1+17\equiv 0\pmod{\! 3}$$
*$n=4k+1$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+1}+17\equiv 6\cdot \left(8^2\right)^{2k}\cdot
8+4\equiv 6\cdot (-1)^{2k}\cdot 8+4$$
$$\equiv 48+4\equiv 52\equiv 0\pmod{\! 13}$$
*
*$n=4k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving a complex number inequality involving absolute values. Here is the relevant paragraph (from "Complex numbers from A to Z" by Titu Andreescu and Dorin Andrica) :
Original question : How does $\left | 1+z \right |=t$ imply $\left | 1-z+z^2 \right |=\sqrt{\left | 7-2t^2 \right |}$?
(I checked for $z=i$ , it seems... | I propose the following. As $|z|=1$ we have $z^{-1}=\bar{z}$. Then, defining $x=Re(z)$, you get :
$$|1-z+z^2|=|z| \times |z+z^{-1}-1| = |z+\bar{z}-1|=|2x-1|$$
Moreover, $|1+z|^2=(1+z)(1+\bar{z})=1+|z|^2+z+\bar{z})=2(1+x)$. Hence :
$$|1+z|+|1-z+z^2|=\sqrt{2(1+x)}+|2x-1|:=f(x)$$
For $x\in[-1,\frac{1}{2}]$, $f(x)=1+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Divisibility of a polynomial by another polynomial I have this question:
Find all numbers $n\geq 1$ for which the polynomial $x^{n+1}+x^n+1$ is divisible by $x^2-x+1$. How do I even begin?
So far I have that $x^{n+1}+x^n+1 = x^{n-1}(x^2-x+1)+2x^n-x^{n-1}+1,$ and so the problem is equivalent to finding $n$ such that ... | Let $\omega = e^{\frac{i\pi}{3}}$, then $\omega^2 - \omega + 1 = 0$.
If there is a $n$ such that $x^2 - x + 1 $ divides $P_n(x) = x^{n+1} + x^n + 1$, then
$$ 0 = P_n(\omega) = \omega^n(\omega+1)+1.$$
$$ \omega^n(\omega+1) = -1.$$
$\omega+1 = \sqrt{3}e^{\frac{i\pi}{6}}$ is a vector of length $\sqrt{3}$. After a rotati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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needs solution of the equation ${(2+{3}^{1/2}})^{x/2}$ + ${(2-{3}^{1/2}})^{x/2}$=$2^x$ $$\left(2+{3}^{1/2}\right)^{x/2} + \left(2-{3}^{1/2}\right)^{x/2} = 2^x.$$
Clearly $x = 2$ is a solution. i need others if there is any. Please help.
| square this equation we have
$$(2+\sqrt{3})^x+(2-\sqrt{3})^x+2=4^x$$
so
$$\left(\dfrac{2+\sqrt{3}}{4}\right)^x+\left(\dfrac{2-\sqrt{3}}{4}\right)^x+\dfrac{2}{4^x}=1$$
It is clear
$$f(x)=\left(\dfrac{2+\sqrt{3}}{4}\right)^x+\left(\dfrac{2-\sqrt{3}}{4}\right)^x+\dfrac{2}{4^x}$$ is decreasing,because use $y=a^x,0<a<1$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, unde... | To know which quadrant $x+y$ is in, you need to know the signs of both $\sin(x+y)$ and $\cos(x+y)$ In your case, $\sin(x+y) \gt 0$ and $\cos(x+y) \lt 0$. This puts $x+y$ in the second quadrant. So the solution is the one angle $x+y = \dfrac{3 \pi}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
?
I know the general method of how to find the partial fraction of functions but this seems a special case to me..
| For starters, write $x^6+1 = (x^2+1)(x^4-x^2+1)$. If you are allowing complex roots you can rewrite $x^2+1 = (x-i)(x+i)$. Then you can continue to factor $x^4-x^2+1$.
The way I came up with this initial factorization is by realizing that $\pm i$ is a root of $x^6+1$, which means $(x-i)$ and $(x+i)$ divide $x^6+1$, whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{... | Use Holder
$$\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)(a+b)(a+b)\ge (1+1)^3=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
| Letting $x=u^2, dx=2udu$ gives $\displaystyle2\int_0^1\frac{u^2\sqrt{1-u^2}}{u^2+2}du$; then letting $u=\sin\theta, du=\cos\theta d\theta$ gives
$\displaystyle2\int_0^{\pi/2}\frac{\sin^2\theta\cos^2\theta}{\sin^2\theta+2}d\theta=2\int_0^{\pi/2}\frac{\tan^2\theta\sec^2\theta}{(\sec^4\theta)(3\tan^2\theta+2)}d\theta=2\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$? I have to prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$.
I have proven using congruencies that $2^{15}-1$ is divided by $31$. However we have
$$2^5\equiv 10 \mod{11}$$
$$2^{15}\equiv 10^3=1000\equiv 10 \pmod {11}$$
Therefore
$$2^{15}-1\equiv 9 \pmod{11}.... | You are correct: the factorization you are given is wrong.
Since $N=2^{15}-1=(2^5)^3-1=(2^5-1)((2^5)^2+2^5+1)$ you get
$$
2^{15}-1=31\cdot(1024+32+1)=31\cdot 1057
$$
However, you can also use
$$
2^{15}-1=(2^3)^5-1=(2^3-1)((2^3)^4+(2^3)^2+2^3+1)
$$
so $N$ is also divisible by $7$. Divide $1057$ by $7$ to get
$$
N=31\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$
Let $a,b>0$ show that
$$\dfrac{1}{1+a}+\dfrac{2}{1+a+b}<\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$$
It suffices to show that
$$\dfrac{(3a+b+3)^2}{((1+a)(1+a+b))^2}<\dfrac{a+b}{ab}$$
or
$$(a+b)[(1+a)(1+a+b)]^2>ab(3a+b+3)^2$$
this idea can't ... | Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{1}{1+a}+\dfrac{2}{1+a+b}\right)^2\le\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}\right)$$
so suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}<1$$
since
$$\dfrac{4b}{(1+a+b)^2}<\dfrac{4b}{4(1+a)b}=\dfrac{1}{1+a}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke t... | Put $z = 0$, the LHS vanishes. So $z$ must be a factor. By symmetry $xyz$ must be a factor. As LHS is of third degree, the only other factor must be a constant, try $x=y=z=1$ to get that...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$\operatorname{lcm}(2x, 3y-x, 3y+x)$ Problem
Find $\textrm{lcm} (2x, 3y-x, 3y+x)$, where $y > x > 0$ and $\gcd(x,y) = 1$.
Attempt
I noticed after some numerical calculation that the answer seems to depend on the parity of $x$ and $y$. However, I am basically at a loss of how to approach this. Any help would be apprecia... | You need to distinguish a few cases.
First note that if $3 \nmid x$, then $\gcd(3y - x, 3y + x)$ is either $1$ or $2$. Indeed, otherwise either a prime $p \neq 3$ or $4$ would divide both $6y$ and $2x$, but $x$ and $y$ are coprime by hypothesis. Furthermore, that $\gcd$ is $2$ if and only if both $x$ and $y$ are odd. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Hints on solving $y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=0$
$$y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=0$$
Is there any simple method to solve this equation?
I need hints please $\color{red}{not}$ a full answer
| Applying the idea in my comment:
$$0 = y'' - \frac{x-1+1}{x-1}y' + \frac{1}{x-1}y = y'' -\frac{1}{x-1}y' - y' + \frac{1}{x-1}y.$$
Multiplying by $\dfrac{1}{x-1}$, this becomes
$$ 0 = \frac{1}{x-1}y'' - \frac{1}{(x-1)^2}y' - \frac{1}{x-1}y' + \frac{1}{(x-1)^2}y.$$
We can realize the first two terms together as $\dfrac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Is it true that $\sin x > \frac x{\sqrt {x^2+1}} , \forall x \in (0, \frac {\pi}2)$? Is it true that $$\sin x > \dfrac x{\sqrt {x^2+1}} , \forall x \in \left(0, \dfrac {\pi}2\right)$$ (I tried differentiating , but it's not coming , please help)
| Square both sides (assume $x\in(0,\pi/2)$):
$$\sin x>\frac{x}{\sqrt{x^2+1}}\iff 1-\cos^2 x>1-\frac{1}{x^2+1}$$
$$\iff \cos ^2 x<\frac{1}{x^2+1}=\frac{\sin^2 x+\cos^2 x}{x^2+1}$$
$$\iff x^2\cos^2 x+\cos^2 x<\sin^2 x + \cos^2 x$$
$$\iff \tan^2 x> x^2\iff \tan x>x,$$
which is true ($(\tan x-x)'>0$ for $x\in(0,\pi/2)$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$?
$\mathbb Z^+$ is the set of positive integers. Find all functions $f:\mathbb{Z}^+\rightarrow \mathbb{Z}^+$ such that
$$f(m^2+f(n))=f(m)^2+n\quad(\clubsuit)$$
Let $P(x,y)$ be the assertion: $f(x^2+f(y))=f(x)^2+y \; \forall x,y \in \mathbb{Z}^+.$
$... | Here is an incomplete idea to actually solve the quoted problem (as opposed to your specific question about $f(x+y)=y+f(x)$). This idea might go nowhere. Define $a$ to be $f(1)$. It's easy to show using induction that if $a=1$, then $f$ is the identity function.
If $a=2$, we can arrive at a contradiction:
$$
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Evaluating a function at a point where $x =$ matrix. Given $A=\left(
\begin{array} {lcr}
1 & -1\\
2 & 3
\end{array}
\right)$
and $f(x)=x^2-3x+3$ calculate $f(A)$.
I tried to consider the constant $3$ as $3$ times the identity matrix ($3I$) but the answer is wrong. Appreciate any ideas.
| We find that
$$
A^2 = \pmatrix{
-1 & -4\\
8 & 7}, \quad
-3A =
\pmatrix{
-3&3\\
-6&-9}, \quad
3I = \pmatrix{
3&0\\0&3
}
$$
Adding these together, we find
$$
f(A) = \pmatrix{
-1&-1\\
2&1
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Basis for a eigenspace (multiple choice problem) The following (multiple choice) problem is from a test review.
For the given matrix $A$, find a basis for the corresponding eigenspace for the given eigenvalue.
$$A = \begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix};\quad \lambda = 7.$$
The four given... | As @GTonyJacobs points out in the comments, you are correct but the key is also correct. Vector spaces can have many different bases!
Note that you have proven that $A-7I$ has rank one so the rank-nullity theorem implies that the nullspace of $A-7I$ has dimension two. This allows us to immediately eliminate choices (B)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the generating function from recurrence relation $t_n=(1+c~q^{n-1})~p~t_{n-1}+a +b+nbq, ~~n\ge 2$? How to find the generating function $T(z)=\sum_{n=0}^{\infty} t_n~z^n$ from the recurrence relation $$t_n=(1+c~q^{n-1})~p~t_{n-1}+a +b+nbq,\qquad n\ge 2.$$ given that $$t_1=b+(1+c~p)(a~q^{-1}+b),$$ and $$t_0=... | Consider $t_{n+1} = (1+c~q^{n})~p~t_{n} + a + b + (n+1)bq$ for which
\begin{align}
\sum_{n=0}^{\infty} t_{n+1} \, x^{n} &= p \, T(x) + c \, p \, T(q x) + \frac{a + b}{1-x} + b q \, \partial_{x} \left( \sum_{n=0}^{\infty} x^{n} \right) \\
\frac{T(x) - t_{0}}{x} &= p \, T(x) + c \, p \, T(q x) + \frac{a + b}{1-x} + b q \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding subgroups of $\mathbb{Z}_{13}^*$
I need to find all nontrivial subgroups of $G:=\mathbb{Z}_{13}^*$ (with multiplication without zero)
My attempt:
$G$ is cyclic so the order of subgroup of $G$ must be $2,3,4,6$
Now to look for $g\in G$ such that $g^2=e,g^3=e,g^4=e,g^6=e$
$$\begin{align}
&12^1=12\mod 13\\
&12... | Your solution is right. As, I am noting that(may be i'm wrong), you are applying that "A cyclic subgroup of order $4$ must contain $2$ elements of order $4$ and $1$ element of order $2$, and you searching those elements and listing them. You can reduce your calculation by searching one element of each order, and then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Can someone please check my work?: $\cos^2(x)=1-\sin(x)$ $$\begin{align}\cos^2(x)&=1-\sin(x)\\
1-\sin^2(x)&=1-\sin(x)\\
(1-\sin x)(1+\sin x)&= 1-\sin(x)
\end{align}$$
divide both sides by $1 - \sin(x)$
End up with $1 + \sin(x)$
The answer is supposed to be in radians between $0$ and $2 \pi$.
So I get $1+\sin(x)=0$
$$\s... | Hey there you are partially correct , one thing you missed in your solution
Follow my solution
$\cos^2x=1-\sin x$
$\Longrightarrow (1-\sin x)(1+\sin x)=1-\sin x$
From this equation we have
$1- \sin x=0$ and $1+ \sin x=0$
$\sin x = 1$ and $\sin x = -1$
$\sin x= \sin(\frac\pi2)$ and $\sin x= \sin(-\frac\pi2)$
We have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Determining $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$
$$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$$
Attempt:
Simplification of the root factor:
$$\sqrt{x^4+2x^3-x^2+2x+1}=\frac{1}{x}\sqrt{\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-1}.$$
Arranging the rest of the factors as:
$... | $\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x^2+2x+1)}dx$$
Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$
$$\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{x^2\cdot \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Inequality involving number of binary Lyndon words of length $n$ and $n+1$
Let $f(n)$ be the number of binary Lyndon words of length $n$. This sequence is given by OEIS entry A001037. Is it true that $2f(n) \ge f(n+1)$ for all positive $n$?
I have found a general formula to calculate $f(n)$:
$$
f(n) = \frac{1}{n} ... | Step 1: For $n \in \mathbb{N}_+$,$$
\frac{2^n}{n} - \frac{2^{\frac{n}{2}}}{2} \leqslant f(n) \leqslant \frac{2^n}{n} + \frac{2^{\frac{n}{6}}}{6}. \tag{1}
$$
Proof: For the upper bound,$$
n f(n) = \sum_{d \mid n} μ\left( \frac{n}{d} \right) 2^d \leqslant 2^n + \sum_{\substack{d \mid n,\ d < n\\μ(n / d) = 1}} 2^d. \tag{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integral calculation: $\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots) \, dx$ $$K=\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots) \, dx$$
I'm looking for an exact solution for this integral. Thank you very much.
| Given
$$
K = \int_0^1 ( x - \sqrt{x} + \sqrt[3]{x} - \sqrt[4]{x} + \cdots )
$$
Write this as
$$
K = - \int_0^1 \sum_{k=1}^\infty (-1)^k x^{1/k}
$$
Thus
$$
K = - \sum_{k=1}^\infty (-1)^k \int_0^1 x^{1/k} = \sum_{k=1}^\infty (-1)^k \left[ \frac{1}{1/k+1} x^{1/k+1} \right]_0^1
$$
So
$$
K = - \sum_{k=1}^\infty (-1)^k \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, al... | We have:
$\sin20^{\circ} = \sin(30^\circ - 10^\circ)= \frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ$
$\sin40^{\circ} = \sin(30^\circ + 10^\circ)= \frac{1}{2}\cos10^\circ + \frac{\sqrt{3}}{2}\sin10^\circ$
$\sin 60^\circ = \frac{\sqrt{3}}{2}$
$\sin80^\circ =\cos 10^\circ$
Therefore,
$A = \sin20^\circ \sin40^\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proving the integral series $\int _0^1\left(1-x^2\right)^n\,dx=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}$
We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$
Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$
I tried to int... | Integrate by parts to obtain:
$$I_n=\int_0^1 \left(1-x^2\right)^n\,dx=\left(x\left(1-x^2\right)^n\right|_0^1+2n\int_0^1 x^2\left(1-x^2\right)^{n-1}\,dx$$
$$\Rightarrow I_n=2n\int_0^1 \left(x^2-1+1\right)\left(1-x^2\right)^{n-1}\,dx$$
$$\Rightarrow I_n=2nI_{n-1}-2nI_n \Rightarrow I_n=\frac{2n}{2n+1}I_{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
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Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ... | $\cos^2(A-120^\circ)+\cos^2(A+120^\circ)=1+\cos^2(A-120^\circ)-\sin^2(A+120^\circ)$
and Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $\cos^2(A-120^\circ)-\sin^2(A+120^\circ)=1+\cos(2A)\cos(240^\circ)=?$
$\cos(240^\circ)=\cos(180^\circ+60^\circ)=-\cos60^\circ=?$
Now use $\cos2A=2\cos^2A-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
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Polynomial division remainder Given the polynomials $f,g,h \in \mathbb{R}\left[X\right]$ with $$f=(x-1)^n-x^n+1$$ $$g=x^2-3x+2$$ $$h=x^2-x$$ where $n\ge3$
Find the remainder of dividing the polynomial f to g. Prove that if $n$ is odd, $h$ divides $f$ with no remainder.
I noticed that the three polynomials have a common... | Use the remainder theorem: for any polynomial $p(x)$, the remainder when dividing it by $x - a$ is given by $p(a)$.
Notice that $f(1) = 0$, so $x - 1$ divides evenly into $f(x)$. Therefore, for some $p(x)$, we may write,
$$f(x) = (x - 1)p(x).$$
Further, since $f(2) = p(2)$ is the common remainder when dividing both $p(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrate $\int \frac{(x+1)}{(x^2+2)^2}dx$ This is the question I want to ask
Integrate $$\int \frac{(x+1)}{(x^2+2)^2}\,dx.$$
I tried it using algebric manipulation
Integrate $x/(x^2+2)^2+1/(x^2+2)^2$.
Then the latter part would not be solved.
| $\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{1}{(x^2+2)^2}dx$$
Let $x=\sqrt{2}\tan \theta\;,$ Then $dx = \sqrt{2}\sec^2 \theta d\theta$
So Integral $$\displaystyle I = \int\frac{\sqrt{2}\sec^2 \theta }{4\sec^4 \theta }d\theta = \frac{1}{2\sqrt{2}}\int \cos ^2\theta d\theta = \frac{1}{4\sqrt{2}}\int (1+\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Conditional expectation on Gaussian random variables If we suppose that the two independent random variables $X \sim \mathcal{N}(0,\sigma^2_x)$ and $N \sim \mathcal{N}(0,\sigma^2_n)$ and that $S = X + N$, how would I work out the conditional expectation $E[X\mid S=s]$?
| Since $X$ and $N$ are independent and normal, any combinations of $X$ and $N$ are normal. As any combination of $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ is also a combination of $X$ and $N$, that is, any combinations of $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are normal, then $X+N$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Definite Integration with Trigonometric Substitution I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$
My first step was to use $\sqrt{a^2+x^2}$ as $x=a\ta... | There was a typo in the current post. After enforcing the substitution $2x=3\tan \theta$, the integral ought to read
$$\begin{align}I&=\int_{\arctan(2/3)}^{\arctan(4/3)}\frac{\frac32 \sec^2\theta}{\frac94 \tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the r... | The required inequality is trivialized by the claim below. The equality case is when $a=b=c=1$.
Claim: If $a,b,c>0$ are such that $abc=1$, then $\displaystyle\sqrt{\frac{a}{a+8}}\geq \frac{a^{4/9}}{a^{4/9}+b^{4/9}+c^{4/9}}$. The equality holds if and only if $a=b=c=1$.
Proof: Note that the required inequality is eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 0
} |
Formulae for sequences Given that for $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$
deduce that $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3 = \frac{n^2(3n+1)(5n+3)}{4}$
So far:
the sequence $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3$ gives $2^3 + 3^3 + 4^3 +\cdots,$ when n=1.
The brackets in the formula for the second sequenc... | $$s_1=1^3+2^3+3^3+..n^3=(\frac{n(n+1)}{2})^2\\s_2=1^3+2^3+3^3+..n^3+(n+1)^3+...+(2n)^3=(\frac{2n(2n+1)}{2})^2\\s_2-s_1=(n+1)^3+...+(2n)^3=\\(\frac{2n(2n+1)}{2})^2-(\frac{n(n+1)}{2})^2=\\\frac{n^2}{4}(4(2n+1)^2-(n+1)^2)=\\\frac{n^2}{4}((4n+2+n+1)(4n+2-n-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding $4$ variables using $3$. if I have:
$ x=\dfrac{a-.5b-.5c+.25d}{a+b+c+d}$
$ y=\dfrac{\dfrac{b\sqrt{3}}{2}+\dfrac{c\sqrt{3}}{2}+\dfrac{d\sqrt{3}}{4}}{a+b+c+d}$
$ z=a+b+c+2d $
Then how do I get back to:
$ a= $ , $ b= $ , $ c= $ , and $ d= $ ?
When there is a divide by zero error, $ x=0 $ and $ y=0 $.
When
... | Hints: You can't get back $b$ and $c$ individually, but you can solve for $a$, $d$, and $e\equiv b+c$ as follows: transform your system into
\begin{align*}
(a+e+d)x&=a-0.5e+d\\
(a+e+d)y&=\frac{\sqrt{3}}{2}e+\frac{\sqrt{3}}{4}d\\
z&=a+e+2d
\end{align*}
which can be further written as
$$
\begin{pmatrix}
x-1& x+0.5&x-1\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Matrix Equation: Reduced Row Echelon Form Given a matrix $A$ and its reduced row echelon form $R$, how can one find an invertible matrix $P$ such that $PA=R$?
I was given a $4\times4$ matrix $A$, I found $R$, and I tried to find $P$ by inspection. Since it was too difficult, I tried finding $P^{-1}$ by inspection becau... | A matrix can be reduced with some sequence of three elementary row operations: swapping rows, multiplying a row by a constant, and adding one row to another. Luckily for us, each of these operations is linear, so each can be represented as a matrix multiplication. Then we just have to chain all of those matrix multipli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Mathematical Induction proof for a cubic equation. If $ x^3 = x +1$, prove by induction that $ x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and
$a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \fra... | Hint: Multiply $x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$ on the left by $x^3$ and on the right by $x+1$. Regroup and deduce a formula for $a_{n+1}$, $b_{n+1}$, $c_{n+1}$ in terms of $a_{n}$, $b_{n}$, $c_{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Number theory with binary quadratic I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
Given $$ \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} = 2$$ find $x-y$
I'm not sure if given choices is right... (A)2 (B)3 (C)4 (D)5 (E)6
I've tried to mo... | Recognize the squares of binomials in both the numerator and denominator to rewrite the equation as $$\begin{align}\frac{x^2-(y-1)^2}{y^2-(x-1)^2} &= 2, \end{align}$$ and thus, factoring, $$\require\cancel \begin{align}\frac{(x-y+1)\cancel{(x+y-1)}}{(y-x+1)\cancel{(y+x-1)}}&=2. \end{align}$$ Finally, let $t=x-y$ and mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4... |
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
Set $t=x^2,z=x^2+x+1$.
$\Longrightarrow$
$$\begin{align}10t^2-7tz+z^2&=(2t-z)(5t-z)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}$$
$$\boxed{\color{red}{x_{1,2}=\frac{1}{2}\pm\frac{\sqrt5}{2},\;x_{3,4}=\frac{1}{8}\pm \frac{\sqrt{17}}{8}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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creative method to obtain range of newton function ?! I am searching for more proof that the range of $y=\frac{x}{x^2+1}$ is $ \frac{-1}{2}\leq y \leq \frac{+1}{2}$
these are my tries :
domain is $\mathbb{R}$
first : $$\quad{y=\frac{x}{x^2+1}\\yx^2+y=x \rightarrow x^2y-x+y=0 \overset{\Delta \geq 0 }{\rightarrow} ... | By contradiction: assume $y > \frac{1}{2}$
$$\frac{x}{x^2 +1} > \frac{1}{2} \Rightarrow 0 > (x-1)^2$$
and $y<-\frac{1}{2}\Rightarrow 0 > (x+1)^2 $.
But we also see that $y(-1) = -\frac{1}{2}$ and $y(1) = \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine if this series $ \sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ converges Determine if the following series converges:
$$
\sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}.
$$
(http://i.stack.imgur.com/qWiuy.png)
I don't know how to start.
| You want to use a comparison test. Remember with series that you can disregard a finite number of terms. When $n \geq 2$
$$n^7+13n^5+9n+2 \leq n^7 +13n^7 + 9n^7+n^2 = 24n^7,$$
which implies
$$ \frac{1}{n^7+13n^5+9n+2} \geq \frac{1}{n^7 +13n^7 + 9n^7+n^2} \geq \frac{1}{24n^7}.$$
For the numerator, you have
$$n^6 + 13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody plea... | Integrating by parts, we have
$$\begin{align}
I_n&=\int_0^1\frac{x^n}{(1+x^3)^{1/2}}dx\\\\
&=\left.\frac23 x^{n-2}(x^3+1)\right|_0^1-\frac23(n-2)\int_0^1x^{n-3}(x^3+1)^{1/2}dx \tag 1\\\\
&=\frac232^{1/2}-\frac23(n-2)\int_0^1\frac{x^{n-3}(x^3+1)}{(x^3+1)^{1/2}}\,dx \\\\
&=\frac232^{1/2}-\frac23 (n-2)I_n-\frac23(n-2)I_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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How to integrate $\int \frac{4}{x\sqrt{x^2-1}}dx$ In order to solve the following integral:
$$\int \frac{4}{x\sqrt{x^2-1}}dx$$
I tried different things such as getting $u = x^2 + 1$, $u=x^2$ but it seems that it does not work. I also tried moving $x$ to the radical and writing the denominator as $\sqrt{x^4-x^2}$ but it... | We have, taking $u=\sqrt{x^{2}-1}
$, $$\int\frac{4}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du.
$$ I think you can get from here. Expanding a little bit, note that $$ u=\sqrt{x^{2}-1}\Rightarrow du=\frac{x}{\sqrt{x^{2}-1}}dx
$$ and $$ x^{2}=u^{2}+1
$$ then $$4\int\frac{1}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{x^{2}}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Struggling with an inequality: $ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $ Prove that for every natural numbers, $m$ and $n$, this inequality holds:
$$
\frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1
$$
I tried to use Bernoulli's inequality, but I can't figure it out.
| If we change a little this inequality we will have:
$$ \sqrt[m \cdot n]{(1 + n )^n} + \sqrt[m \cdot n]{(1 + m)^m} \geq \sqrt[m \cdot n]{(1+n)^n \cdot (1+m)^m}$$
From the first part we will have:
$$ \sqrt[m \cdot n]{(1 + n )^n} + \sqrt[m \cdot n]{(1 + m)^m} \geq 1 + \frac{n}{m} + 1 + \frac{m}{n} = \frac{(m+n)^2}{m \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
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For which $x, y\in\mathbb{R ^+}$ do we have $|xy-\frac{1}{xy}|\le|x-\frac{1}{x}|+|y-\frac{1}{y}|$? I need to find all $x, y\in\mathbb{R^+}$ such that the following inequality holds.
$$\Big| xy-\dfrac{1}{xy}\Big|\le\Big|x-\dfrac{1}{x}\Big|+\Big|y-\dfrac{1}{y}\Big|$$
If I substitute $x=2$ and $y=3$ clearly that inequalit... | Since both members are positive, we can square and get this equivalent $$x^2y^2+\frac{1}{x^2y^2}-2\le x^2+\frac{1}{x^2}-2+y^2+\frac{1}{y^2}-2+2|x-\frac{1}{x}||y-\frac{1}{y}|$$
Then gettting all terms but the last one to the left, factorizing and multipying by $x^2y^2$ we get
$$ (x^2y^2+1)(x^2-1)(y^2-1) \le 2x^2y^2|x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
|
$$\int\frac{x^3}{(x^2+1)^{3/2}}dx$$
$u:=x^2,du=2xdx$
$$=\frac{1}{2}\int\frac{u}{(u+1)^{3/2}}du$$
$s:=u+1,ds=du$
$$=\frac{1}{2}\int\frac{s-1}{s^{3/2}}ds$$
$$=\sqrt s+\frac{1}{\sqrt s}$$
$s=u+1,u=x^2$
$$=\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}+C$$
$$\boxed{\color{blue}{=\frac{x^2+2}{\sqrt{x^2+1}}+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{... | HINT....The equation simplifies to $ a\sin x +a- \cos x=1$ can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the equation of the circle. Find the equation of the circle whose radius is $5$ which touches the circle $x^2 + y^2 - 2x -4y - 20 = 0$ externally at the point $(5,5)$
| The circle: $x^2+y^2-2x-4y-20=0$ has center $(1, 2)$ & a radius $=\sqrt{(-1)^2+(-2)^2-(-20)}=5$ & the unknown circle has a radius $5$ Hence the point $(5, 5)$ is the mid point of line joining their centers
Let the center of unknown circle be $(a, b)$ then the point $(5, 5)$ is mid point of lines joining the centers $(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x... | Using the substitution $\displaystyle \theta=\frac{1}{2}\arctan\left(\frac{x}{2}\right)$ so $\displaystyle \text{d}\theta=\frac{1}{x^2+4} \text{d}x$ and $\displaystyle x=2\tan(2\theta)$ we get $$\int_{0}^{\infty}\frac{\ln(x)}{x^2+4}\text{d}x=\int_{0}^{\frac{\pi}{4}}\ln\left(2\tan(2\theta)\right)\text{d}\theta= \\ =\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ to a large power Compute
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$
What is the easier way to do this other than multiplying the entire thing out?
Thanks
| Following the comment of Matt Samuel:
$$
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix} =
\begin{pmatrix}
1 & 2 \\
0 & 1
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}^3
=
\begin{pmatrix}
1 & 3 \\
0 & 1
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & 1 \\
0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
External bisectors of the angles of ABC triangle form a triangle $A_1B_1C_1$ and so on If the external bisectors of the angles of the triangle ABC form a triangle $A_1B_1C_1$,if the external bisectors of the angles of the triangle $A_1B_1C_1$ form a triangle $A_2B_2C_2$,and so on,show that the angle $A_n$ of the $n$th ... | You are on the right track, you just need to continue. You have a correct formula for $A_1$, so format it differently.
$$\begin{align}
A_1 &= \frac{\pi}2-\frac A2 \quad\text{(your formula)}\\
&= \frac{\pi}3+\frac{\pi}6+\left(-\frac 12\right)A \\
&= \frac{\pi}3+\left(-\frac 12\right)\left(A-\frac{\pi}3\right) \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many divisors of the combination of numbers?
Find the number of positive integers that are divisors of at least one of $A=10^{10}, B=15^7, C=18^{11}$
Instead of the PIE formula, I would like to use intuition.
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
Number of... |
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
$A,B \to $ there are $5^{0} \to 5^{7} = 8$ divisors. $B, C \to$ there are: $3^{0} \to 3^{7} = 8$ divisors. $A, C \to$ there are $2^{0} \to 2^{10} = 11$ divisors.
These are correct. Now $1$ is the only divisor of the three nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
| Let's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have
\begin{align*}
&& 2149 &= a_k + 2a_{2k+n+1} +2 \\
&\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Minus sign in logarithm of integral's solution I want to solve the following integral:
$
\int \frac{dp}{a(1-p)u-bp}
$
where $a$, $b$ and $u$ are some constants.
After integration I get:
$
p = -\frac{\log(-apu+au-bp)}{au+b} + C.
$
According to WolframAlpha it is equivalent to (citing precisely: "Which is equivalent for ... | \begin{align}
I &= \int \frac{dp}{a(1-p)u - b p} = \frac{1}{au} \, \int \frac{dp}{1 - \left( 1 + \frac{b}{au}\right) p } \\
&= \frac{1}{au+b} \, \int \frac{dx}{1-x} \hspace{10mm} \text{where} \quad x = \left(1+\frac{b}{au}\right) \, p \\
&= - \frac{1}{au + b} \, \int \frac{dx}{x-1} \\
&= - \frac{1}{au+b} \, \ln(x-1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$ Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx $ for $y\in[0,1].$
I tried to differentiate the given function by using DUIS leibnitz rule but the calculations are messy and I tried to solve directly by integrating it but th... | As observed by @MichaelGaluza, by Leibniz's rule, it suffices to show
$$
\sqrt{y^{4}+\left(y-y^{2}\right)^{2}}+\int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq0.
$$
Note that $y-y^{2}=y\left(1-y\right)>0$ for $y\in\left(0,1\right)$.
Now, for $y\leq\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that $f(x,y,z)$ is reducible if and only if $a,b,c,d$ is a geometric progression.
Let $a,b,c,d$ be real numbers not all $0$, and let $f(x,y,z)$ be the polynomial in three variables defined by
$$f(x,y,z) = axyz + b(xy + yz + zx) + c(x + y + z) + d.$$
Prove that $f(x,y,z)$ is reducible if and only if $a,b,c,d$ i... | We have:
$$
\begin{align*}
f(x,y,z) & = axyz + b(xy + yz + zx) + c(x + y + z) + d.\\
& = axyz + b(xy) + b(yz + zx) + c(x+y)+ cz + d\\
& = xy(az+b) + (x+y)(bz+c) + (cz+d)\\
\end {align*}
$$
The only way we can make it reduce further is if $(az+b), (bz+c)$ and $(cz+d)$ are related geometrically, i.e., $(bz+c)$ and $(cz+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $\alpha$ and $\beta$ are the zeroes of $p(x) =x^2- px +q = 0$ Find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.
| For convenience, we reverse the sign of $q$, without loss of generality.
$$x^2-px-q$$ is the characteristic equation of the recurrence
$$x_{n+2}=px_{n+1}+qx_n,$$
that has the general solution
$$a\alpha^n+b\beta^n.$$
With $a=b=1$,
$$x_0=2,\\
x_1=\alpha+\beta=p,\\
x_2=\alpha^2+\beta^2=px_1+qx_0=p^2+2q,\\
x_3=\alpha^3+\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar exam... | From $r = \sqrt{4+2\sqrt{3}}-\sqrt{3}$ we get $r+\sqrt{3}=\sqrt{4+2\sqrt{3}}$ and, squaring both sides,
$$
r^2+2r\sqrt{3}+3=4+2\sqrt{3}
$$
and so
$$
r^2-1=2(1-r)\sqrt{3}
$$
Square again:
$$
r^4-2r^2+1=12-24r+12r^2
$$
so
$$
r^4-14r^2+24r-11=0
$$
The rational root test only allows $1$, $-1$, $11$ and $-11$ as roots. Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
... | HINT :
Using$$\binom{n}{i}=\binom{n}{n-i}$$
just add the two sums you write to get $$2S=100\binom{100}{0}+100\binom{100}{1}+\cdots+100\binom{100}{99}+100\binom{100}{100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Method of partial fractions when denumerator cannot be factorized? Suppose I'm given an expression: \begin{align*} P(x) = \frac{1+x}{1-2x-x^2}. \end{align*} The denumerator cannot be readily factorized, so I found the zeros, which are \begin{align*} \lambda_1 = -1 - \sqrt{2} \qquad \lambda_2 = -1 + \sqrt{2}. \end{align... | In general when you have some factors in denumerator which is not factor-able you have to get its numerator as $ax+b$ and then equate the fractions and compute $a$ and $b$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the general solution of $\cos(x)-\cos(2x)=\sin(3x)$ Problem:
Find the general solution of $$\cos(x)-\cos(2x)=\sin(3x)$$
I tried attempting this by using the formula$$\cos C-\cos D=-2\sin(\dfrac{C+D}{2})\sin(\dfrac{C-D}{2})$$
Thus, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Right... | *
*Expand the trig functions $$ \cos(x)+1-2 \cos^2(x) = 4 \sin(x) \cos^2(x)-\sin(x)$$
*Use the tangent half angle substitution $t=\tan(x/2)$, $\cos(x)=\frac{1-t^2}{1+t^2}$ and $\sin(x) = \frac{2 t}{1+t^2}$ $$ \frac{2 t^2 (3-t^2)}{(1+t^2)^2} = \frac{2 t (3 t^4-10 t^2+3)}{(1+t^2)^3}$$
*Collect terms $$\frac{2 t (t+1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac... | $\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \sin \left(x-\frac{1}{x}\right)dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt\Rightarrow \left(x+\frac{1}{x}\right)dx = xdt$
and Changing Limits
Now Using $$\disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Let $(a, b, c)$ be a Pythagorean triple. Prove that $\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2$ is greater than 8 and never an integer.
Let $(a, b, c)$ be a Pythagorean triple, i.e. a triplet of positive integers with $a^2 + b^2 = c^2$.
a) Prove that $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > 8$$
b) Prove that there ... | If $(a,b,c) \in \mathbb{N}^3$ is a Pythagorean triple, then without loss of generality, we may assume that $a=\left(m^2-n^2\right)d$, $b=2mnd$, and $c=\left(m^2+n^2\right)d$ for some positive integers $m,n,d$ with $m>n$, $\gcd(m,n)=1$, and $m\not\equiv n\pmod{2}$.
We now have
$$t:=\frac{c}{a}+\frac{c}{b}=\frac{m^2+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Help with a dominated convergence theorem problem The problem is to find an integrable function that bounds $f(x)={\dfrac{{{n^{\frac{3}{2}}}x}}{{1 + {n^2}{x^2}}}}$ on $[0,1]$ so we can calculate $\displaystyle\int_0^1 {\frac{{{n^{\frac{3}{2}}}x}}{{1 + {n^2}{x^2}}}dx} $ using dominated convergence theorem. I know by tak... | By AM-GM: $\dfrac{1+n^2x^2}{n^{3/2}x^{3/2}} = \dfrac{1}{(nx)^{3/2}} + (nx)^{1/2} = \dfrac{1}{(nx)^{3/2}} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3}$
$\ge 4 \sqrt[4]{\dfrac{1}{(nx)^{3/2}}\cdot \dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}}$ $= \dfrac{4}{3^{3/4}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why I am getting different answer? I have just started learning single variable calculus. I'm confused in a problem from sometime. I didn't get why my answer is different from the book.
$$
\require{cancel}
\begin{align}
&\int\sin x \sin 2x \sin 3x\,dx\\
&=\int\sin x\;\,2\sin x\cos x \left(3\sin x - 4\sin^3 x\right)\,dx... | Both solutions are correct, they only differ by a constant. See also Wolfram Alpha.
This is similar to the following situation: Both $f(x) = \sin^2(x)$ and $g(x) = -\cos^2(x)$ are antiderivatives of $2\sin(x)\cos(x)$. Even if they look quite different, they only differ by a constant: $f(x) = 1 + g(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to factor $4x^2 + 2x + 1$? I want to know how to factor $4x^2 + 2x + 1$? I found the roots using quadratic equation and got $-1 + \sqrt{-3}$ and $-1 - \sqrt{-3}$, so I thought the factors would be $(x - (-1 + \sqrt{-3}))$ and $(x - (-1 - \sqrt{-3}))$
However, according to MIT's course notes, the factors are $(1 - (... | $$4x^2+2x+1$$
Finding roots by quadratic rule as follows
$$x=\frac{-2\pm\sqrt{2^2-4(4)(1)}}{2(4)}$$ $$=\frac{-2\pm2i\sqrt{3}}{8}$$
$$=-\frac{1}{4}\pm\frac{i\sqrt 3}{4}$$
$$x=-\frac{1}{4}+\frac{i\sqrt 3}{4}\ \vee\ x=-\frac{1}{4}-\frac{i\sqrt 3}{4}$$
Edit:
Now, we have the factors as follows $$4x^2+2x+1=4\left(x+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
nice classical nonhomogeneous inequality Let $a,b,c$ be positive reals and $abc=1$. Prove that $$10(a^4+b^4+c^4)+21\ge 17(a^3+b^3+c^3).$$
I have found a solution using MV and I'm wondering if there is a nice solution.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$10(81u^4-108u^2v^2+18v^4+12uw^3)+21w^4-17(27u^3-27uv^2+3w^3)\geq0$$
or $f(u)\geq0$, where
$$f(u)=270u^4-360u^2v^2+60v^4+40uw^3-153u^3w+153uv^2w-10w^4.$$
But by Schur $w^3\geq4uv^2-3u^3$.
Thus,
$$f'(u)=1080u^3-720uv^2+40w^3-459u^2w+153v^2w\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $x,y,z>0$ and $xyz=32,$ Then the minimum of $x^2+4xy+4y^2+4z^2$ is If $x,y,z$ are positive real no. and $xyz= 32\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is
$\bf{My\; Try::}$ Here I have Used $\bf{A.M\geq G.M}$ Inequality
So $$\displaystyle \frac{x^2+4xy+4y^2+4z^2}{4}\geq \left(x^3\cdot y^3\cdot z^2\right)^{... | Applying the AM-GM is the right strategy, but you need to do it a bit differently.
$$x^2+4xy+4y^2+4z^2$$
$$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$
$$\ge 6\sqrt[6]{x^2 \cdot 2xy \cdot 2xy \cdot 4y^2 \cdot 2z^2 \cdot 2z^2}$$
$$= 6\sqrt[6]{64x^4y^4z^4}$$
$$= 12(xyz)^{2/3}$$
$$= 12 \cdot 32^{2/3}$$
Equality holds when $x^2 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Inverse Trigonometric Function: Find the Exact Value of $\sin^{-1}\left(\sin\left(\frac{7\pi}{3}\right)\right)$ $$\arcsin\left(\sin\left(\frac{7\pi}{3}\right)\right)$$
I cannot use this formula, correct? $f(f^{-1}(x))=x$
The answer in the book is $\frac{\pi}{3}$
How do I approach solving a problem such as this?
The in... | Let $\sin^{-1}\sin\dfrac{7\pi}3=x$ where $-\dfrac\pi2\le x\le\dfrac\pi2$
$$\implies\sin x=\sin\dfrac{7\pi}3$$
$\implies x=n\pi+(-1)^n\dfrac{7\pi}3$ where $n$ is any integer
If $n$ is even $=2m$(say), $x=2m\pi+\dfrac{7\pi}3=\dfrac{(6m+7)\pi}3\implies -\dfrac\pi2\le\dfrac{(6m+7)\pi}3\le\dfrac\pi2$
$\iff-3\le12m+14\le3\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no... | We can compute the number of real roots using Sturm's Theorem.
$$
\begin{array}{rll}
\text{Sturm Chain}&+\infty&-\infty\\\hline
x^6+6x^5+30x^4+120x^3+360x^2+720x+720&+\infty&+\infty\\
6x^5+30x^4+120x^3+360x^2+720x+720&+\infty&-\infty\\
-5x^4-40x^3-180x^2-480x-600&-\infty&-\infty\\
-48x^3-432x^2-1728x-2880&-\infty&+\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 0
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Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$... | Let the numerator and the denominator
$$N= \sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\ldots+\sqrt{10+\sqrt{99}}$$
$$D =\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\ldots+\sqrt{10-\sqrt{99}}$$
Apply the denesting formulas
$$\sqrt{a\pm\sqrt c} = \frac{1}{\sqrt2} \left( \sqrt{a+\sqrt{a^2-c}} \pm
\sqrt{a-\sqrt{a^2-c}} \right)$$
t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 5
} |
Need help with an arithmetic sequence proving question It is given that $a_1, a_2, a_3, \ldots ,a_n$ are consecutive terms of an Arithmetic progression. I have to prove that
$$\sum_{k=2}^n (\sqrt{a_{(k-1)}} + \sqrt{a_k} )^{-1} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n} }$$
Using Mathematical induction I showed that it is tru... | Actually the induction is not such a mess:
$$
\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{a_m - a_1 } + \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{a_{m+1} - a_m}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{d(m - 1)}+ \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{d}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is the least number of (fixed) parameters I can ask for, when calculating area of a triangle of unknown type? I need to calculate the area of a triangle, but I don't know, whether it is right angled, isoscele or equilateral.
What parameters do I need to calculate the area of a triangle of unknown type?
| All you need is the lengths of each side of the triangle.
By Heron's Formula, we know for a triangle with sides $a,b,c$, we have
$$A=\sqrt{s(s-a)(s-b)(s-c)}\text{ ,where }s=\frac{a+b+c}2$$
Reference:
https://en.wikipedia.org/wiki/Heron%27s_formula
EDIT:
In response to suggestion by @Hurkyl , I now add the case of ASA ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Angle bisector between two vectors, which are expressed by unit non-orthogonal vectors Given vectors $\,a=2m-2n\,$ and $\,b=3m+6n\,$, where $\ \left\lvert m \right \rvert =\left\lvert n \right \rvert =1\,$ and $\,\angle\left(m,n\right)=\dfrac{2\pi}{3},\,$ find vector of angle bisector of the angle $\angle\left(a,b\righ... | First, compute angle $\theta := \angle (a,b)$ between $a$ and $b$ using cosine law.
$$
\left\langle a, b \right\rangle = \left\| a\right\| \left\| b\right\| \cos \theta
\implies
\cos \theta = \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| },
$$
where $\left\langle \,\cdot\,, \cdot \,\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse Laplacetransform of rational function with multiple pole I have to calculate the inverse Laplacetransorm of this function using Residue calculus
$$
\frac{s^4 + 6s^3 - 10s^2 + 1}{s^5}
$$
but I can't find any Residue rule that would solve this. Can you show me how to solve this without using partial fractions(I d... | If $F(s)=\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)$ is the Laplace Transform of $f(t)$, then the Inverse Laplace Transform is given by
$$f(t)=\frac{1}{2\pi i}\int_{\sigma -i\infty}^{\sigma +i\infty}e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)\,ds \tag 1$$
where he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding values of $a$ with which two equations are equivalent; getting rid of radical sign Two equations are given:
$$x^2+(a^2-5a+6)x=0$$
$$x^2+2(a-3)x+a^2-7a+12=0$$
We need to find the values of $a$ that will render them equivalent.
From the first equation,
$$x=-a^2+5a-6$$
From the second,
$$x=\frac{-2a+6\pm2\sqrt{a... | Notice, we have $$x^2+(a^2-5a+6)x=0\tag 1$$
$$x^2+2(a-3)x+a^2-7a+12=0\tag 2$$
Now, since the coefficients of $x^2$ in both the equations is $1$ hence both the above equations will be equivalent to each other if their corresponding coefficients of $x$ & constant terms are equal hence, by comparison, we have the followin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove the root is less than $2^n$
A polynomial $f(x)$ of degree $n$ such that coefficient of $x^k$ is $a_k$. Another constructed polynomial $g(x)$ of degree $n$ is present such that the coefficeint of $x^k$ is $\frac{a_k}{2^k-1}$. If $1$ and $2^{n+1}$ are roots of $g(x)$, show that $f(x)$ has a positive root less than... | Assume that you meant that the coefficient of $x^k$ in $g(x)$ is $\frac{a_k}{2^k-1}$ for each $k=1,2,\ldots,n$ and that the constant term of $f(x)$ is $0$. If $f$ has no root in the interval $\left(1,2^{n}\right)$, then either it is strictly positive or strictly negative on this interval. Without loss of generality, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve for $v$ - simplify as much as possible Solve for $v$. Simplify the answer.
$$-3 = -\frac{8}{v-1}$$
Here is what I tried:
$$-3 = \frac{-8}{v-1} $$
$$(-8) \cdot (-3) = \frac{-8}{v-1} \cdot (-8) $$
$$24 = v-1$$
$$25 = v$$
| I'd probably start by taking a reciprocal of both sides:
$$
\begin{align}
-3&=-\frac{8}{v-1}\\
-\frac{1}{3}&=-\frac{v-1}{8}\\
(-8)-\frac{1}{3}&=-\frac{v-1}{8}(-8)\\
\frac{8}{3}&=v-1\\
\frac{8}{3}+1&=v\\
\frac{11}{3}&=v
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper ... | Notice, $$\frac{x^2+1}{x^2-5x+6}=\frac{(x^2-5x+6)+5(x-1)}{x^2-5x+6}$$
$$=1+5\frac{x-1}{x^2-5x+6}$$
$$\implies \frac{x-1}{x^2-5x+6}
=\frac{x-1}{(x-2)(x-3)}$$ $$=\frac{A}{x-2}+\frac{B}{x-3}$$
By comparing the corresponding coefficients, we get $A=-1$, $B=2$, hence
$$\frac{x-1}{x^2-5x+6}=-\frac{1}{x-2}+\frac{2}{x-3}$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"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.