Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Convergence of Cauchy sequence $(x_n)$, where $x_n=1/n^2$. I'm just looking at an example where we're asked to prove that the sequence $(x_n)$ is Cauchy, where $x_n=1/n^2$. (the example is from here at time 7:15 minutes; this requires registration but is free) Anyway, the proof goes exactly like this:
Let $\varepsilon... | (i) Yes. This is a mistake.
(ii) The absolute value is necessary, OR they could have said $p > q$. However, they did define it wrong for their argument to work.
Here's how I believe that proof is supposed to go. (It is similar, but there appear to be typos in the above.
Let $\{x_n\}$ be a sequence defined by $x_n = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cauchy sequence $a_n = 1 + \frac12 + \frac14 + ... + \frac{1}{2^n}$ For the sequence $a_n = 1 + \frac12 + \frac14 + ... + \frac{1}{2^n}$, $n \ge 1$, find a formula $N = N(\epsilon)$ such that
for all $\epsilon > 0$ and for all $m,n \ge N(\epsilon)$, $|a_m - a_n| < \epsilon$.
I have tried many things but it's just not ... | Hint:
\begin{align*}
a_n
&= 1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} \\
&= \frac{1}{2^n} (2^n + 2^{n-1} + 2^{n-2} + \cdots + 1) \\
&= \frac{2^{n+1} - 1}{2^n} = 2 - \frac{1}{2^n}
\end{align*}
so
\begin{align*}
|a_n - a_m| &= \left| \left( 2 - \frac{1}{2^n} \right) - \left( 2 - \frac{1}{2^m} \right) \right| \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/670713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$
$=3(a^2+4a+7) $
where do I go from here.. the solution I ha... | Hint:
If $a$ is odd then by the Division algorithm $a = 2k + 1$ for some integer $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Ways of writing $n=2a+b$ with $a$ and $b$ are non-negative integers For a non-negative number $n$, let $r_n$ be the number of ways of writing
$n = a + 2b$, where $a$ and $b$ are non-negative integers.
For example,
$5 = 1 \cdot 5 + 2 \cdot 0 = 3 \cdot 1 + 1 \cdot 2 = 1 \cdot 1 + 2 \cdot 2$,
so $r_5 = 3$.
My quest... | Note that $a$ is determined by $n$ and $b$ via $a=n-2b$, so $r_n$ is the number of valid choices for $b$. Since $a$ and $b$ must be nonnegative, we get $b\geq 0$ and $n-2b\geq 0$, so $\frac n2\geq b \geq 0$. If $n=2k$ or $2k+1$, then the only integer values $b$ can take are $0, 1, \ldots, k$ so there are $k+1$ choices.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/671892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$?
Hmmm... So I didn't know what to do here so I tested some cases for a rule.
If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, ... | Your approach with $3^2$ and $2^1$ is very close. You only forgot to account for the factor $1 = 3^0\cdot2^0$. This brings us to $6$ distinct factors for this smaller example. You can calculate this by multiplying the number of all the different exponents used for each factor. $3$ can occur either $2$, $1$ or $0$ times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why aren't these two integration methods yielding the same answer? I'm trying to solve this (not homework, if it matters), and both u-substitution and integration by parts are both yielding two different answers. Where am I going wrong?
Equation: $$\int \frac{(4x^3)}{(x^4+7)}dx$$
u-substitution answer: $$=\ln\big|(x^4+... | The correct result is:
$$\int\frac{4x^3}{x^4+7}dx=\ln|x^4+7|+c$$.
With substitution: $u=x^4+7$ from which $du=4x^3dx$ hence the integral become:
$$\int\frac{du}{u}=\ln|u|+c$$.
Integration by parts it's not raccomanded in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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How do I put $\sqrt{x+1}$ into exponential notation? I think $\sqrt{x+1} = x^{1/2} + 1^{1/2}$. Is this incorrect? Why or why not?
| $$
9^{1/2} + 16^{1/2} = 3 + 4 = 7 \ne 5 = 25^{1/2} = (9+16)^{1/2}.
$$
So
$$
\left(\frac{9}{16}\right)^{1/2} + 1^{1/2} = \frac 3 4 + 1 = \frac 7 4 \ne \frac 5 4 = \left(\frac{25}{16}\right)^{1/2} = \left(\frac{9}{16}+1\right)^{1/2}
$$
In other words, if $x=\dfrac{9}{16}$ then $x^{1/2}+1^{1/2}$ is not at all the same as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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A good site documenting approximations of irrationals I'm thinking of Sloane here but I believe that only takes sequences/series into account.
Basically I've derived an interesting, appealing formula for e and want to know if it's already been discovered.
| We want to show equality
$$ e={2 \over 3} \left( {1 \over 1!} +{1 +2\over 2!} + {1 +2+3\over 3!} + {1 +2+3+4\over 4!} + ... \right) $$
Let's define a more general expression $f$ using the indeterminate $x$ at each summand inside the parenthese and derive the equality by finally inserting $x=1$:
$$ \begin{array}{}
f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
how to find this line integral and what is its answer evaluate the line integral $$\int_C (xy^2 dy-x^2y dx), $$ taken in the counter-clockwise sense along the cardioid $$r= a(1+\cos\theta)$$
here putting the parametric form of cardioid $x=a(2\cos t-\cos2t), y= a(2\sin t-\sin2t) $ and taking $\theta$ , $0 $ to $2\pi $ t... | If $x=a(2\cos t-\cos 2t)$ and $y=a(2\sin t-\sin 2t)$, then
$$
xy=a^2\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big),
$$
$$
x^2-y^2=a^2\big(4\cos 2t-4\cos 3t+\cos4t\big),
$$
and
$$
x\,dx-y\,dy=a^2\big(-4\sin 2t+6\sin 3t-2\sin4t\big)\,dt.
$$
Hence
$$
xy^2\,dy-x^2y\,dx=a^4\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big)
\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What does $\prod_{n\geq2}\frac{n^4-1}{n^4+1}$ converge to?
What does $\prod_{n\geq2}\frac{n^4-1}{n^4+1}$ converge to?
As far as I can tell, this has no closed-form solution (not saying much, I don't know much math), but a friend of mine swears he saw a closed-form solution to this in some text he doesn't remember.
Ru... | Start with the infinite product expansion of $\sin x$,
$$\sin x = x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2\pi^2}\right)$$
We get
$$\prod_{n=2}^\infty \left(1 - \frac{\alpha^2}{n^2}\right) = \frac{\sin(\alpha\pi)}{\alpha\pi(1-\alpha^2)}$$
In particular,
$$\begin{align}
\prod_{n=2}^\infty \left(1 - \frac{1}{n^2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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How to solve the following system of equations. $2x^2-3xy+2y^2=2\frac{3}{4}\\x^2-4xy+y^2+\frac{1}{2}=0$
I tried all the methods that I know, but I could't isolate $x$ or $y$ to form one equation.
| Multiply second equation by $2$,
$$2x^2-8xy+2y^2+1=0$$
Substract the first equation by the second,
$$2x^2-3xy+2y^2-\frac{11}{4}-(2x^2-8xy+2y^2+1)=0\\2x^2-3xy-2y^2-\frac{11}{4}-2x^2+8xy-2y^2-1=0\\5xy=\frac{15}{4}\\x=\frac{3}{4y}$$
Substitute $x=\frac{3}{4y}$ in the first equation,
$$2x^2-3xy+2y^2=\frac{11}{4}\\2(\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea:
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int... | Just like the comments given by @heropup and @Random Variable, this problem can be solved by using residue theorem and Jordan's lemma.
Note that the denominator of the integrand can be factorize by
\begin{equation}
z^4+z^2+1=(z^2-z+1)(z^2+z+1)
\end{equation}
It have four poles that
\begin{equation}
z_{1,2}=\pm\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Newbie: Find the intersection of a line and a circle and interpret geometrically Find the points of intersection of the line $x+y+k=0$ and the circle $x^2+y^2=2x$. Show that there are two points of intersection if: $-1-\sqrt2<k<-1+\sqrt2$, one point of intersection if: $k=-1\pm\sqrt2$, and none otherwise. Interpret the... | Everything is correct up until your last two lines. You seem to be assuming that $k^2 + 2k -1$ is equal to $(k-1)^2$, which is not true.
Your equation $k^2 + 2k -1 = 0$ is correct, and it has roots $k = -1 \pm \sqrt2$, which is probably what you were expecting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/686099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.
| Since $$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}}},$$ you have $$x^2=4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}},$$ so $$(x^2-4)^2=4-\sqrt{4+\sqrt{4-\cdots}}.$$ Hence, $$(x^2-4)^2=4-x,$$ which you can try to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these sta... | $x^3 - y^3 = (x - y)(x^2 + x y + y^2)$. Now
$x^2 + x y + y^2 \equiv (x + 3 y)^2 - 3 y^2 \mod 5$, and since $3$ is not a quadratic residue mod $5$ we find that there are no solutions to
$x^2 + x y + y^2 \equiv 0 \mod 5$ other than the trivial $(0,0)$. Thus
$x^3 \equiv y^3 \mod 5$ only when $x \equiv y \mod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Differentiation wrt parameter $\int_0^\infty \sin^2(x)\cdot(x^2(x^2+1))^{-1}dx$ Use differentiation with respect to parameter obtaining a differential equation to solve
$$
\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}dx
$$
No complex variables, only this approach. Interesting integral and it should have a nice ODE. I ha... | Well, as an alternative (like Mr. Ron Gordon did).
\begin{align}
\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\
&=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/691798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Solving $\arcsin(1-x)-2\arcsin(x)=\pi/2$ \begin{eqnarray*}
\arcsin(1-x)-2\arcsin(x) & = & \frac{\pi}{2}\\
1-x & = & \sin\left(\frac{\pi}{2}+2\arcsin(x)\right)\\
& = & \cos\left(2\arcsin(x)\right)\\
& = & 1-2\left(\sin\left(\arcsin(x)\right)\right)^{2}\\
& = & 1-2x^{2}\\
x & = & 2x^{2}\\
x\left(x-\frac{1}{2}\right) &... | As $\displaystyle \sin\left(\frac\pi2\pm A\right)=\cos A,$
$\displaystyle\sin\left(\frac\pi2\pm2\arcsin x\right)=\cos(2\arcsin x)=1-2\left[\sin(\arcsin x)\right]^2=1-2x^2$
So, $\displaystyle x=\frac12$ corresponds to $\displaystyle\arcsin(1-x)+2\arcsin x=\frac\pi2$ as $\displaystyle\arcsin\frac12=\frac\pi6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed?
| $$ \begin{align}\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}&=
\dfrac{(x-3)(3(x-3)(x+2)^2-2(x+2)^3)}
{(x-3)^4}\\
&= \dfrac{3(x-3)(x+2)^2-2(x+2)^3}{(x-3)^3}\\
&= \dfrac{(x+2)^2(2(x-3)-2(x+2))}{(x-3)^3} \\
&= \dfrac{(x+2)^2(3x-9-2x-4)}{(x-3)^3}\\
&= \dfrac{(x+2)^2(x-13)}{(x-3)^3}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Interpretation of a homogeneous transformation matrix of the plane I have the transformation matrix $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}$. This $3\times 3$ matrix is a homogeneous transformation matrix in $2-D$ space.
My book says that this matrix translates the line $y=x+1$ to $y=x$. I don't see how. L... | It appears you've either put the translation elements in the wrong position or you're using column vectors instead of row vectors:
$$
\begin{pmatrix}a&b&1\end{pmatrix}
\begin{pmatrix}1&0&\color{#00A000}{0}\\0&1&\color{#00A000}{0}\\\color{#C00000}{0}&\color{#C00000}{-1}&1\end{pmatrix}
=\begin{pmatrix}a&b-1&1\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometric Substitution
I am having trouble with this problem even though everything I did seemed right to me since we went over a similar one in my class. I used the method of setting up a triangle, my hypotenuse is $\sqrt{54+9x^2}$ and my sides are $\sqrt{54}$ and $3x$. I got $\tan(t)=3x/\sqrt{54}$ so
$$x=\sqrt{... | Sorry, I misunderstood your question the first time round.
First, I'm getting
$$9 \sec \theta \tan \theta + 9 \ln | \sec \theta + \tan \theta | + C $$
as my intermediate answer. From there, you need to take out all the thetas and put in $x$. From what you wrote, I'm assuming you see how we get $\tan \theta = x / \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Fourier Series for $|\cos(x)|$ I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.
I understand its an even function, so all the $b_n$s are $0$
$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$
$$a_n = \frac 2 \pi \int _0^\pi |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2... | Although $ \int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx. $$
We can evaluate it as follows, as can be seen in the plot below
$$a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Proof: $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ iff R is commutative We want to show that for some ring $R$, the equality $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ if and only if $R$ is commutative.
Here's my proof --- I'm not sure if the first part stands up to examination. I'd be grateful if someo... | It looks good! The only holes are the following typos: $x(a+b)=xa+b$ and $-ba=ab=0$
In fact, you don't even need the substitution $x=a-b$, since you can simply multiply out the right-hand side normally assuming distributivity for the ring!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Fourier transform of $\operatorname{erfc}^2\left|x\right|$ Could you please help me to find the Fourier transform of
$$f(x)=\operatorname{erfc}^2\left|x\right|,$$
where $\operatorname{erfc}z$ denotes the the complementary error function.
| We have
$$ \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|x|) e^{-i\xi x} \, dx = \frac{4}{\xi}e^{-\xi^{2}/4} \left\{ \operatorname{erfi}\left( \frac{\xi}{2} \right) - \operatorname{erfi}\left( \frac{\xi}{2\sqrt{2}} \right) \right\}, \tag{1} $$
where $\operatorname{erfi}$ is the imaginary error function defined by
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the greatest common divisor (gcd) of $n^2 - 3n - 1$ and $2$ Find the greatest common divisor (gcd) of $n^2 - 3n - 1$ and $2$ considering that $n$ is an integer. Thanks.
| $2$ only has two divisors, 0 and 1. So $gcd(a,2)=1$ or $2, \forall a \in \mathbb{Z}$.
Case 1: $n$ is even. Then $n^2 \equiv 0 \pmod{2}, 3n\equiv 0 \pmod{2}.$ So then $gcd(n^2-3n-1,2)=1$ because $n^2-3n-1 \equiv 0+0-1 \equiv 1 \pmod{2}$.
Case 2: $n$ is odd. Then $n^2 \equiv 1 \pmod{2}, 3n \equiv 1 \pmod{2}.$ So then $gc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/705892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \left |\sin(x) - x + \frac{x^3}{3!} \right | < \frac{4}{15}$ Prove $ \left |\sin(x) - x + \dfrac{x^3}{3!} \right | < \dfrac{4}{15}$ $\forall x \in [-2,2]$
By Maclaurin's formula and Lagrange's remainder we have $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{\sin(\xi)}{5!}x^5$ for some $0<\xi<2$
subbing this in we get... | From the Leibniz rule it is known that, if the sequence of $\frac{x^4}{5!},\frac{x^6}{7!},\frac{x^8}{9!},...$ is decreasing, which is the case for $x^2<6\cdot7=42$, $|x|\le6$ to get a round number, then
$$
0\le\frac{x^4}{5!}-\frac{x^6}{7!}
\le \frac{\sin x}{x}-1+\frac{x^2}{3!}
\le\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Matrices and determinants question. Establish that if A is the matrix
\begin{bmatrix}
b+c & a^2 & a \\
c+a & b^2 & b \\
a+b & c^2 & c \\
\end{bmatrix}
then $|A| = -(a-b)(b-c)(c-a)(a+b+c)$.
| using linearity with respect to the first column, those matrix have the same determinant:
\begin{bmatrix}
b+c & a^2 & a \\
c+a & b^2 & b \\
a+b & c^2 & c \\
\end{bmatrix}
\begin{bmatrix}
b+c + a& a^2 & a \\
c+a + b& b^2 & b \\
a+b +c & c^2 & c \\
\end{bmatrix} whose determinant is $a+b +c$ m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to Prove This Trigonometry Identity? I have to prove that:
$$\tan^2\theta \sin^2\theta = \tan^2\theta - \sin^2 \theta$$
Here is what I have tried
$$\tan^2\theta \sin^2\theta$$
$$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$
$$=\frac{\sin^4\theta}{\cos^2\theta}$$
Not much of an attempt, ... | $$\tan^2\theta \sin^2\theta+\sin^2 \theta=\sin^2\theta(\tan^2\theta +1) = \sin^2\theta\cdot\sec^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$
Alternatively, $$\frac1{ \sin^2\theta}-\frac1{\tan^2\theta}=\csc^2\theta-\cot^2\theta=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/711580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\r... | If one know that $\displaystyle \int_0^{\dfrac{\pi}{2}}\log (\sin (x))dx=-\dfrac{\pi}{2}\log(2)$
so then :
$\displaystyle \int_0^\dfrac{\pi}{2}\dfrac{x}{\tan x }dx=\dfrac{\pi}{2}\log(2)$ (by integration by parts)
Let $I=\displaystyle \int_0^\dfrac{\pi}{2}\dfrac{x}{\tan x }dx$
Change of variable $u=\tan x$ :
$I=\disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
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Product of two number $a$ and $b$ How to prove that if $a, b$ are two positive integers so that $a^2+b^2=4100$ and $a<b<2a$ then $ab=2000$.
| As pointed by another user, $a,b$ must be positive integers, otherwise it's easy to find a counterexample.
Let's find some limitation for $a$. From the condition we have:
$$4100 = a^2 + b^2 > 2a^2 \iff a^2 < 2050 \iff a < \sqrt{2050} \approx 45.27$$
Since $a$ is integer we can conclude $a\le 45$
Now for the lower bound... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Do either of the following series converge: $\sum_{n = 1}^\infty \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots{2n}}$ Does $$\sum_{n = 1}^\infty \frac{1\cdot3\cdot5\cdots{2n-1}}{2\cdot4\cdot6\cdots{2n}}$$ converge and additionally does the following converge $$\sum_{n = 1}^\infty \frac{1^2\cdot3^2\cdot5^2\cdots{... | Hint:
$$ \frac{1\cdot3\cdot5\dots{2n-1}}{2\cdot4\cdot6\dots{2n-2}}= \frac{3}{2} \frac{5}{4}... \frac{2n-1}{2n-2} >1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/715033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another combinatorics problem: $\sum\limits_{k = 0}^n (-1)^k \binom{2n-k}k2^{2n-2k}=2n+1$ I am dealing with problem 10F in the textbook A Course in Combinatorics by J. H. van Lint and R. M. Wilson.
Problem 10F: Prove directly the equality
$\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{... | Finally got it. First, note that we can take the inner sum to be $\sum_{k=0}^\infty$ since $\binom{2n-k}{k} = 0$ when $k > n$. Next, observe that
$$
\binom{2n-k}{k} = \binom{2(n-k) + k}{k} = \binom{2j + k}{k}
$$
where $j = n - k$. Then
\begin{align*}
\sum_{n \geq 0} \sum_{k \geq 0} (-1)^k \binom{2n - k}{k}2^{2(n-k)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\fr... | Integrate by parts: $\int udv=uv-\int v du$, where
$$u=\ln\frac{\cos x+\sin x}{\cos x-\sin x}\Rightarrow du=\frac{\frac{(\cos x-\sin x)(-\sin x+\cos x)-(\cos x+\sin x)(-\sin x +\cos x) }{(\cos x-\sin x)^2}}{\frac{\cos x+\sin x}{\cos x-\sin x}}=...=\frac{2}{\cos 2x}dx$$
and
$$ dv=\cos 2x dx \Rightarrow v=\frac{1}{2}\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Integrate $\int\frac{5x-7}{x^2-3x+2}$ I want to integrate $\int\frac{5x-7}{x^2-3x+2}$ but my result differs from the one on Wolframalpha http://www.wolframalpha.com/input/?i=integrate+%285x-7%29%2F%28x%5E2-3x%2B2%29
I did the following steps:
$$\frac{5x-7}{(x-2)(x-1)} = \frac{A}{x-2}+\frac{B}{x-1}$$
$$5x-7 = A(x-1)+B(x... | They are both valid. $$\frac{d}{dx}\ln(a-x)=\frac{-1}{a-x}=\frac{1}{x-a}=\frac{d}{dx}\ln(x-a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/727894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$. $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.
(a) -1 (b) 1 (c) 2 (d) -2
I am getting the value of k: $-17/29$ after eq... | By the polynomial remainder theorem, the remainder of $f(x)$ when divided by $x - a$ is equal to $f(a)$. So we have
$$p(3) = q(3)$$
$$27k + 41 = 15 + k$$
$$k = -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/730163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $\sum x_k= \sqrt{n}$, then $\sum\frac{x_k^2}{(x_k^2+1)^2}\leq\frac{n^2}{(n+1)^2}$ A new question has emerged after this one was successfully answered by r9m: If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$. I thought of this generaliza... | Alternative way to reach the result, that is already proposed in the other answers:
Consider the non-linear maximization problem $$\max_{x_i}\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}$$ subject to $x_1+x_2+\ldots+x_n=\sqrt{n}$ and $x_i\ge0$. In that case the Lagrange function is defined as $$Λ(x, λ)=\sum_{i=1}^{n}\frac{x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integrate with square root in square $\int \left(1 + \sqrt{\frac{x-1}{x+1}}\right)^2 dx$
How would you attack this? I've tried variable substitution with $t = \sqrt{\frac{x-1}{x+1}}$
| Expand the square
$$\int {\left( {1 + \frac{{x - 1}}{{x + 1}}} \right)dx + 2\int {\sqrt {\frac{{x - 1}}{{x + 1}}} } } dx$$
Rearrange the first integral and collect $2$
$$2\left[ {\int {\left( {1 - \frac{1}{{x + 1}}} \right)} \;dx + \int {\sqrt {\frac{{x - 1}}{{x + 1}}} dx} } \right]$$
First integral is easy; for the se... | {
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"url": "https://math.stackexchange.com/questions/733844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b,c > 0 $? Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b> 0$ and $c\geq 0$ ?
Is there a general proof for that relation ?
| Suppose $a,b,c \in \mathbb N$, $a < b$.
$$\color{red}{\dfrac ab}
= \dfrac{a(b+c)}{b(b+c)}
= \dfrac{ab+ac}{b(b+c)}
\color{red}{<} \dfrac{ab+bc}{b(b+c)}
= \dfrac{b(a+c)}{b(b+c)}
= \color{red}{\dfrac{a+c}{b+c}}
\color{red}{<} \dfrac{b+c}{b+c} = \color{red}1$$
Suppose $a,b,c \in \mathbb N$, $a > b$.
$$\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove it? Let $y_0\geqslant 2$, $y_n=y_{n-1}^2-2$, $n\in\mathbb{N}_+$, set $\displaystyle S_n=\sum_{k=0}^{n}\frac{1}{y_0\cdots y_k}$, how to prove
$$\lim_{n\to\infty}S_n=\frac{y_0-\sqrt{y_0^2-4}}{2}.$$
Do you have some idea?
I can only get that
$$y_n=\begin{cases}y_0, & n=0 \\ \left(\frac{y_0^2-2+\sqrt{y_0^4-4y... | Since, $y_n=y_{n-1}^2-2$, we have $y_n^2=(y_{n-1}^2-2)^2=y_{n-1}^2(y_{n-1}^2-4)+4$.
or, $y_n^2-4=y_{n-1}^2(y_{n-1}^2-4)=y_{n-1}^2y_{n-2}^2(y_{n-2}^2-4)=\ldots=(\prod\limits_{i=0}^{n-1}y_i^2)(y_0^2-4)$.
Thus, $\dfrac{y_n^2-4}{\prod\limits_{i=1}^{n-1}y_i^2}=y_0^2-4$ for $n\ge1$.
$\implies \dfrac{\sqrt{y_n^2-4}}{\prod\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\sum_{n=1}^\infty \frac{1}{n! \sin(n)}$ diverge or converge? Does the series $$ \sum_{n=1}^\infty \frac 1 {n!\sin(n)}$$ converge or diverge? Even the necessary condition of the convergence is difficult to verify.
| Since at least one person seemed uncomfortable with the details of Hagen von Eitzen's answer, I thought it would be a nice exercise to flesh it out.
As he says, the irrationality measure says that there are only finitely many positive integers $p,q$ such that $|\pi - \frac{p}{q}| < q^{-8}$. Since the left side is neve... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Algebra Question ( possible values) If
$$k = \dfrac{a}{b+c} = \dfrac{b}{a+c} = \dfrac{c}{b+a}$$
How many possible values of $k$ are there?
| Add like this:
$$(b+c)k+(a+c)k+(b+a)k=a+b+c$$
$$k=\frac{a+b+c}{b+c+a+c+b+a}=\frac12$$
if $a+b+c\ne0$, else $a=-(b+c)\Longrightarrow\dfrac a{b+c}=-1$.
So there are $2$ possible values for $k$, because first is e.g. for $a=b=c=1$ and the second e.g. for $a=b=1, c=-2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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relation between sides of a triangle and circumradious prove that $a.b.c=4R. S_{\bigtriangleup ABC}$, ($S$ stands for surface) and ($R$ stands for circumradious)
any response would be appreciated
| A good starting point is to use the formula for area $S$ of $\bigtriangleup ABC$:
$S = \dfrac{1}{2} \cdot ab \cdot sinC$. We now prove this formula:Let $H$ be the point on side $BC$ such that $AH$ is perpendicular to $BC$, and let $h = AH$. So $S = \dfrac{1}{2}\cdot AH \cdot BC = \dfrac{1}{2} \cdot h \cdot BC = \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding an asymptotic expansion for $\sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}}$ It is well known that an asymptotic expansion of the n-th harmonic number is $$H_{n}= \sum_{k=1}^{n} \frac{1}{k} \sim \ln(n) + \gamma + \frac{1}{2n} -\frac{1}{12n^{2}} + O(n^{-4}).$$
How could we find an asymptotic expansion for the sum $ \di... | $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Convergence of series $A_n = \sqrt{\sum_{k=n}^{\infty} a_k} - \sqrt{\sum_{k=n+1}^{\infty}a_k} $ if series $a_n$ converges I must show that if a series $\sum_{n=1}^{\infty} a_n$ with positive terms converges, then the series $\sum_{n=1}^{\infty} A_n$, where $A_n = \sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{... | Fix some sequence $(b_k)_{k\geqslant1}$ and define $c_k=b_k-b_{k+1}$ for every $k\geqslant1$. Then, for every $n\geqslant0$,
$$
\sum_{k=1}^nc_k=b_1-b_{n+1}.
$$
If the sequence $(b_k)$ converges, this implies that the series $\sum\limits_kc_k$ converges and that its sum is $b_1-b_\infty$, where $b_\infty=\lim\limits_{k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Integrate $\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx$. HW Since this is a homework problem, a hint would be appreciated to help me get this started, since I have no idea how to start this. Thanks
Here's the problem:
Compute the improper integral:
$$\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx$$
| Here is the full solution
Using the substitution $x=\sin \theta$, the limit of the integrand changes from $x=0$ to $\theta =0$ and from $x=1$ to $\theta = \frac{\pi}{2}$ since $\theta=\arcsin(x)$.
Also, $dx=\cos \theta d\theta\Leftrightarrow dx=\sqrt {1 - {{\sin }^2}\theta } d\theta \Leftrightarrow dx=\sqrt {1 - {x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Trig Reduction with Pythagoras If $\sin 10 = p$, then determine $$\tan^2 30^\circ \times \tan^2190^\circ$$ in terms of p.
| $\tan^2 30^\circ$ is easy because it's a common angle; we know that it is equal to $\frac{1}{3}$
Note that $\tan^2 190^\circ = \tan^2 10^\circ = \frac{\sin^2 10^\circ}{\cos^2 10^\circ}$
Since $\cos^2 \theta = 1 - \sin^2 \theta$, we find that $\tan^2 190^\circ = \frac{p^2}{1 - p^2}$
Thus, the final answer is $\frac{p^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the last two digits of the number $9^{9^9}$
Find the last two digits of the number $9^{9^9}$ .
[Hint: $9^9 \equiv 9 \pmod {10} $; hence, $9^{9^9}$ = $9^9+10k$ ;now use the fact that $9^9 \equiv 89 \pmod {100}$]
| $$9^{9^9}=(10-1)^{9^9}\equiv(-1)^{9^9}+9^9\cdot10^1\cdot(-1)^{9^9-1}\equiv-1+10\cdot9^9\pmod{100}$$
Now, $\displaystyle9^9=(10-1)^9\equiv-1\pmod{10}\implies10\cdot9^9\equiv-10\pmod{10\cdot10}$
$$\implies9^{9^9}\equiv-1-10\pmod{100}\equiv100-11$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this $\cos{x}+\cos{y}+\cos{z}=1$ Question:
let $x,y,z\in R$ and such $x+y+z=\pi$,and such
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$
then
$$a+b+c=\pi$$
and
$$\tan{\dfrac{a}{4}}+\tan{\dfrac{... |
Looking down at the positive octant , ( arrow tips are coordinate axes).
$x+y+z=\pi$ ( the cyan colored plane. )
$\cos{x}+\cos{y}+\cos{z}=1$ , ( the pink colored area. )
$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$ , (the light gray colored area. )
I can see three solutions where the gray ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/749758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How to find $\lim_{x\to\ln 2} \frac{2e^{3x}-16}{3e^{2x}-12}$? Need to find without using L'Hopital's rule or derivatives.
I know the answer is 2, but how can I find this analytically, without using limit tables?
Thanks!
| Let $u = e^x$. Then
\begin{align}
\lim_{x \rightarrow \ln 2} \frac{2e^{3x}-16}{3e^{2x}-12} &= \lim_{u \rightarrow 2} \frac{2u^3-16}{3u^2-12} &\text{substitute } u = e^{2x} \\
&= \lim_{u \rightarrow 2} \frac{2(u^3-8)}{3(u^2-4)} & \text{factor a 2 from top and a 3 from bottom} \\
&= \lim_{u \rightarrow 2} \frac{2(u-2)(u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/750312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Multiplying Adjacent Matrices? My teacher hasn't explained it too well, so i'm looking for an explanation:
$$A =
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}$$
$$A^2 =
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2
\end{pmatrix}$$
I would have thought the first number in $A^2$ would be... | $A^2$ is simply $A\cdot A$, where the $\cdot$ denotes matrix multiplication.
Matrices are not multiplied elementwise (other than for addition), but as follows:
The matrix element of the resulting matrix which sits on row $i$ and column $j$ is calculated by multiplying each element of $i$-th row of the first matrix with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How find this system $a^2+b^2=3,a^2+c^2+ac=4,b^2+c^2+\sqrt{3}bc=7$ Find the this system real solution
$$\begin{cases}
a^2+b^2=3\\
a^2+c^2+ac=4\\
b^2+c^2+\sqrt{3}bc=7
\end{cases}$$
I think that one can use Geometry to solve this system. Maybe there exist an algebraic method.
$$a^2+b^2=\sqrt{3}^2$$
$$a^2+c^2-2ac\cos{(1... | Assuming $a,b,c>0$, then as you noted the equalities are just cosine laws for $3$ triangles which form a larger triangle with sides $2,\sqrt3,\sqrt7$, because the angles add up to $2\pi$. That triangle is right, because $4+3=7$, so we can find the lengths analytically if we draw it like this:
Here we set $A=(0,0)$, $B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Finding the limit of a sequence by diagonalising a matrix Consider the sequence described by:
$\frac11 , \frac32 , \frac75 , ... ,\frac {a_{n}}{b_{n}}$
where $ a_{n+1} = a_n +2b_n $ and $b_{n+1} = a_n+b_n$
Find a matrix $A$ such that
$$\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_... | Hint: From $\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_{n} \end{bmatrix}$ find $\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A^k \begin{bmatrix} a_{0} \\b_{0} \end{bmatrix}$, for a suitable $k\in \mathbb N$.
Edit: Mark has already answered 1. and 2. As for 3., your solution i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5 \begin{pmatrix}n+1\\2\end{pmatrix}+1$ $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5
\begin{pmatrix}n+1\\2\end{pmatrix}+1$
How would you go on proving this equation? Doesn't have to be induction..
| $$(k+1)^3=k^3+3k^2+3k+1\quad=>\quad(k+1)^3-k^3=3k^2+3k+1\quad=>$$
$$=>\quad\sum_0^n(3k^2+2k+1)=\sum_0^n\Big[(3k^2+3k+1)-k\Big]=\sum_0^n\underbrace{(3k^2+2k+1)}_{\large(k+1)^3-k^3}-\sum_0^nk=$$
$$=(n+1)^3-\frac{n(n+1)}2=n^3+\underbrace{3n^2+3n}_{6\tfrac{n(n+1)}2}+1-\frac{n(n+1)}2=n^3+5\cdot\underbrace{\frac{n(n+1)}2}_{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Problem with trigonometric equation I am having trouble solving this equation
$$4\cdot \sin \theta + 2 \cdot \sin 2\theta =5$$
Thank you for your help.
| \begin{align}
4\sin \theta+2\sin 2\theta&=5\\
4\sin \theta+2\cdot2\sin \theta\cos\theta&=5\\
4\sin \theta+4\sin \theta\cos\theta&=5\\
4\sin \theta(1+\cos\theta)&=5
\end{align}
Take square both sides.
\begin{align}
4^2\sin^2 \theta(1+\cos\theta)^2&=5^2\\
16(1-\cos^2\theta)(1+\cos\theta)^2&=25
\end{align}
Let $x=\cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/754542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the System of Equations in $x$ and $y$ \begin{equation}
x+\frac{3\,x-y}{x^2+y^2}=3 \tag{1}
\end{equation}
\begin{equation}
y=\frac{x+3\,y}{x^2+y^2} \tag{2}
\end{equation}
| I am assuming that you are solving this equation over the reals only. This approach is motivated because you have $x^2 + y^2$, and 'anti-symmetric' coefficients.
Set $z = x+iy $.
Take the first equation, add it to $i$ times the second equation, we get
$$ (x+iy) + \frac{ (3-i) ( x-iy) } { x^2 + y^2} = 3.$$
Converting th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Showing $\lim_{n\to\infty} \left( \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4}$ How could I go about proving the following limit:
$$
\lim_{n\to\infty} \left( \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4}
$$
| Hint: Write each term individually as
$$\frac{n}{n^2 + k^2} = \frac{1}{n} \frac{1}{1 + (k/n)^2}$$
and try to recognize a Riemann sum.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
... | I am going to prove it by modulo. In modulo $$
x^{2}+x+1 \equiv 0\left(\bmod x^{2}+x+1\right) \Rightarrow x+1\equiv -x^2\textrm{ and } x^3 \equiv 1 \left(\bmod x^{2}+x+1\right).
$$
In modulo $x^{2}+x+1$, we have
$$
\begin{aligned}
(x+1)^{2 n+1}+x^{n+2} \equiv &\left(-x^{2}\right)^{2 n+1}+x^{n+2} \\
\equiv &-x^{4 n+2}+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Does the series $\sum_{n=1}^\infty \left|\frac{\cos2^n}{n}\right|$ converge or not? $$\sum_{n=1}^\infty~\left|\frac{\cos2^n}{n}\right|$$
I just confused what to do.
I am sorry, i don't understand. But this task is important for me. Can you give full solution?
| I guess the homework tag has now become irrelevant and it's about understanding the argument. So let's spell out what user2566092 hinted.
The addition theorem for the cosine is
$$\cos (x+y) = \cos x \cos y - \sin x \sin y.$$
Specialising $y$ to $x$, we obtain the double-angle formula
$$\cos (2x) = \cos^2 x - \sin^2 x.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Interesting association between tangent lines of slope one and ellipses Why is it that a tangent line with slope $1$ to an ellipse centered at the origin will have a transformation of $\pm \sqrt{a^2 +b^2}$ where $a$ and $b$ are the major and minor axis of the ellipse?
For example: The tangent line of slope one to the e... | It's actually quite simple. An ellipse centered at the origin with semimajor and semiminor axes $a$ and $b$ can be transformed to a circle by a suitable scaling of the horizontal axis by a factor of $b/a$. Since such a linear transformation preserves tangency, the resulting line has a slope that is also scaled by the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
eigenvectors of a matrix Good Day,
I have a matrix of
\begin{bmatrix}
28 & 10\\
10 & 19
\end{bmatrix}
I have found the eigenvalues...
first eigenvalue (v1) : 24 + 5sqrt5 = rd off to 35.18
second eigenvalue (v2) : 24 - 5sqrt5 = rd off to -12.81
Following the page above,
if eigenvalue = 35.18, -6.18x+10y = 0 and 10x-16... | You shouldn't round, if you can avoid it. Your matrix is
\begin{bmatrix}
29 & 10 \\
10 & 19
\end{bmatrix}
whose characteristic polynomial is
$$
X^2 - 48X + 451
$$
The roots are given by the formula
$$
\frac{48\pm\sqrt{48^2-4\cdot 451}}{2}
$$
so they are $24+5\sqrt{5}$ and $25-5\sqrt{5}$, so you computed correctly. An e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/759646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integration by reduction I have learnt how to integrate by reduction formula but this one seems to give me hell someone to lift me by telling me what to do or simply to solve it.
\begin{equation}
I_n=\int\sec^n x\,dx
\end{equation}
| Write the integral as
\begin{align}
\int\sec^n x\,dx=\int\sec^{n-2}x\sec^2 x\,dx
\end{align}
Then use integration by parts. Let
\begin{align}
u&=\sec^{n-2}x\\
du&=(n-2)\sec^{n-3}x\sec x\tan x\,dx=(n-2)\sec^{n-2}x\tan x\,dx\\
\end{align}
and
\begin{align}
dv&=\sec^{2}x\\
v&=\int\sec^{2}x\,dx=\int\,d(\tan x)=\tan x\\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
is $324+455^n$ ever prime Another question that I can only solve in part.
Is there an $n$ such that $324+455^n$ is prime?
When $n$ is odd, this is false since
$$ 324+455^n = (2\cdot 3^2)^2+(5\cdot 91)^n \equiv (-1)^2+(5\cdot 15)^n \equiv 1+(6\cdot 3)^n \equiv 1+(-1)^n\pmod{19}$$
so that $19\mid 324+455^n$ whenever $n... | Let's analyse three possible cases:
If $n\equiv 1\pmod 2$, we have $455^n+324 \equiv 0\pmod {19}$ because
$$\begin{eqnarray}
(455^1+324) & = & 19\times 41 \\
(455^2-1) & = & 19\times 10896\\
\end{eqnarray}$$
If $n\equiv 2\pmod 4$, we have $455^n+324 \equiv 0\pmod {17}$ because
$$\begin{eqnarray}
(455^2+324) & = & 17\ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this expression: $∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$?
How to evaluate: $$∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$$ If anyone knows the answer, please help.
| $$
\int \frac{\sin 2x dx}{(1 - \cos 2x)^4} = \frac{1}{2}\int\frac{\sin 2x d(2x)}{(1 - \cos 2x)^4} = \frac{1}{2}\int \frac{d(1 - \cos 2x)}{(1 - \cos 2x)^4} = \frac{1}{2}\int u^{-4}du
$$
$$
= -\frac{1}{6}(\cos 2x)^{-3} = -\frac{1}{6\cos^3(2x)} + C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/761695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Trigonometric problem: Elevation angle The elevation of the top of a tower $KT$ from a point $A$ is $27^\circ$. At another point $B$, $50$ meters nearer to the foot of the tower where $ABK$ is a straight line, the angle of elevation is $40^\circ$. Find the height of the tower $KT$.
| Let $AK=d$, $BK=d-50$, and $KT=h$. Then
$$
\tan 27^\circ=\frac{h}{d}\qquad\rightarrow\qquad d=\frac{h}{\tan 27^\circ}\tag1
$$
and
$$
\tan 40^\circ=\frac{h}{d-50}\qquad\rightarrow\qquad d-50=\frac{h}{\tan 40^\circ}\tag2
$$
Substitute $(1)$ to $(2)$, yield
\begin{align}
\frac{h}{\tan 27^\circ}-50&=\frac{h}{\tan 40^\circ}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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When is the sum of consecutive squares a prime? For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?
I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$
| The sum of the squares of the first $x$ consecutive integers, starting from $n+1$, equals
$$(n+1)^2+(n+2)^2+\ldots+(n+x)^2=\frac{(n+x)(n+x+1)(2(n+x)+1)}{6}-\frac{n(n+1)(2n+1)}{6}$$ $$=\frac{1}{6}\cdot x\cdot(6n^2+6nx+2x^2+6n+3x+1).$$
For this to be prime we must have $x\mid 6$ or
$$6n^2+6nx+2x^2+6n+3x+1\mid6.$$
Solvin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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system of linear congruences How can I solve this system of linear congruences? $$2x\equiv 0\, \text{mod 3}\\ 3x\equiv 2\, \text{mod 5}\\ 5x\equiv 4\, \text{mod 7}$$
I don´t know where to start, I am having a lot of troubles, can you lend me a hand please?
| Let's solve this equation without using the CRT. so $2x = 3k$, and this requires that $ 2 | k$. So $k = 2t$. So $x = 3t$, and $3(3t) = 5m + 2$. So $9t = 5m + 2 = 9m + 2 - 4m$, and $t = m + \dfrac{2 - 4m}{9}$. So $2 - 4m = 9p$, and $m = \dfrac{2 -9p}{4} = -2p + \dfrac{2 - p}{4}$. So $p = 2 - 4q$. So $t = -2(2 - 4q) + q ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/764014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Odd series convergence Prove that we have following inequality:
$1+ \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{397} > \frac{9}{4}$
Anybody can help me to figure it out?
| Note that $\frac{1}{2k-1}+\frac{1}{2k+1}=\frac{4k}{4k^2-1}>\frac{4k}{4k^2}=\frac{1}{k}$.
Denote $1+\frac{1}{3}+\ldots+\frac{1}{2k-1}$ by $S_k$, then $S_{199}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{397}$, so
\begin{align}
S_{199}&>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\frac{1}{198}\\
&>1+\frac{1}{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there complex solutions for $z^3=\bar z$ I'm asked to solve $z^3=\bar z$. I got $z=0, 1, -1$. Are there any complex solutions $a+bi$ to this though?
| $(a+bi)^3 = (a^3 - 3ab^2) + (3a^2b - b^3)i = a - bi$
$a^3 - 3ab^2 = a, 3a^2b - b^3 = -b$
If $a = 0$, then $-b^3 = -b$, and the only reals which statisfy that is $b = 0, 1, -1$
If $b = 0$, then similarly $a^3 = a$ so $a = 0, 1, -1$
So far this gives $0, -1, 1, i, -i$
If $a, b \neq 0$,
$a^2 - 3b^2 - 1 = 0, b^2 - 3a^2 - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/766599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove the inequality $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$
If $a,b,c\in\mathbb R^+$ prove that:
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$
| By Cauchy Schwarz Lemma:
$$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geqslant \frac{(a+b+c)^2}{2(a+b+c)}$$
Also By $HM\leqslant AM$
$$\frac 3 {\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}}\leqslant \frac{2(a+b+c)}{3}$$
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\geqslant\frac9 {2(a+b+c)}$$
Add both of them. Then app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$
| Applying the inequality $AM \ge GM $ on {$ab,bc,ca$}, we get:
$$\frac{ab+bc+ca}{3} \ge (a^2.b^2.c^2)^{\frac13}$$
$\Rightarrow$ $(abc)^{\frac23} \le \frac{12}{3} = 4$
i.e. the maximum value of abc is $4^{\frac32}$ = $8$.
Also, maximum value of $8$ is attained for $a=b=c=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any a... | Note that these are actually the same answer, since:
$$ \sinh^{-1} x = \ln \left(x + \sqrt{x^2+1} \right) $$
and
$$ \frac{2x-1}{\sqrt{3}} + \sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2 + 1} = \frac{2x - 1 + 2\sqrt{x^2 - x + 1}}{\sqrt{3}}$$
So the answer you got is also correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\frac{2^{4n + 2} + 1}{5}$ is composed for all $n \ge 2$ The problem is to prove that $$\frac{2^{4n + 2} +1 }{5}$$ is composed for every $n \ge 2$.
I've tried this way: write $2^{4n + 2} + 1$ as $1 - (-4)^{2n + 1}$ and $5 = 1 - (-4)$, then $$\frac{2^{4n + 2} + 1}{5} = \frac{1 - (-4)^{2n + 1}}{1 - (-4)} = 1 -... | One approach. First observe that we must have
$$2^{4n + 2} = - 1 \pmod 5$$
This is true because it is equivalent to
$$4 * 16^n = -1 \pmod 5$$
And there are various ways of showing this holds for all $n$.
So we know the expression is an integer. As for compositeness, a hint is as follows. Note that
$$4x^4 + 1 = (2x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding primes $p$ such that $ \dfrac {p+1}2$ and $\dfrac{p-1}4$ are primes How many odd primes $p$ are there such that both $ \dfrac {p+1}2$ and $\dfrac{p-1}4$ are primes ?
| Think modulo 12. Either $p = 3$ or $p \cong \{1,5,7,11\} \mod 12$. $3$ doesn't work because $(3-1)/4$ isn't an integer. $\frac{\{1,5,7,11\}+1}{2} \cong \{1,3,4,6\} \mod 12$ so your $\frac{p+1}{2}$ prime requirement eliminates any $p \not \cong 1 \mod 12$. If $p \cong 1 \mod 12$ and $p>1$, then $12$ divides $p-1$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The ... | You could've used Heron's formula straight away, actually.
$$\begin{align}
T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\
& = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\
& = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\
& = \tfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How to solve this limit without using L'Hôspital? This limit wildly appeared in a test and, well, I could not solve it without using L'Hôspital rule - which were not allowed in the test. Can anyone help me?
$$\lim_{x \to +\infty}\sqrt{x-\sqrt{x}}-\sqrt{x-1}$$
Answer, thanks to @user71352.
$$
\lim_{x\to\infty}\frac{\sq... | $$
\sqrt{x-\sqrt{x}}-\sqrt{x-1}=\frac{x-\sqrt{x}-x+1}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}=\frac{1-\sqrt{x}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}=\frac{\frac{1}{\sqrt{x}}-1}{\sqrt{1-\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{x}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use logarithmic differentiation to find the derivative of $y = (1 +\frac1x)^{2x}$ Can someone guide me through solving this problem?
$$\dfrac{\mathrm d}{\mathrm dx}\left(1 +\dfrac1x\right)^{2x}$$
| Take logs on both sides and differentiate,
\begin{align}
y &=\left(1+\frac{1}{x}\right)^{2x}\\
\ln y &=2x \ln \left(1+\frac{1}{x}\right)\\
\frac{\frac{dy}{dx}}{ y} &=2 \ln \left(1+\frac{1}{x}\right)-2 x\frac{\frac{1}{x^2}}{1+\frac{1}{x}} \\
\frac{dy}{dx} &=2 y\ln \left(1+\frac{1}{x}\right)-2 yx\frac{\frac{1}{x^2}}{1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Riemann-Stieltjes integral Compute the following Riemann-Stieltjes integral:
The integral from 0 to 3 of $dg(x)/(sqrt(1 + x^2))$ where $g(x) = |2x - 2| + |3x - 6| - |x - 3|$
What I have tried:
Using formula: sum of $f(c_i)*(g(x_(i+1)) - g(x-i))$.
--> $1/\sqrt(2)*(1 - 5) + 1/\sqrt(5)*(1 - 1) = -2\sqrt(2)$
I'm not sur... | Redefining the function g as follows
\begin{align}
g\left( x \right) &= \left| {2x - 2} \right| + \left| {3x - 6} \right| - \left| {x - 3} \right|
\\
&= \left\{ \begin{array}{l}
2x - 2,\,\,\,\,\,x \ge 1 \\
2 - 2x,\,\,\,\,\,0 < x < 1 \\
\end{array} \right. + \left\{ \begin{array}{l}
3x - 6,\,\,\,\,\,x \ge 2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find the equation $\cot x=\frac{\sin 20^\circ - \sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$ let $x\in R$, and such
$$\cot x =\frac{\sin 20^\circ -\sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$$
Find $x$
my idea:
$$\cot x=\csc 80^\circ - \cot 20^\circ$$
then I can't
| Let $\theta=20^\circ$, and defin
$\Delta=\sin\theta-\sin4\theta\cos\theta+\sqrt{3}\sin4\theta\sin\theta
$.
Then
$$\eqalign{2\Delta&=2\sin\theta-(\sin5\theta+\sin3\theta)+\sqrt{3}(\cos3\theta-\cos5\theta)\cr
&=2\sin\theta-\sin5\theta-\sqrt{3}\cos5\theta\cr
&=2\sin\theta-2(\sin5\theta\cos60^\circ+\cos5\theta\sin60^\circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had s... | Once it is noticed that
$$
x^4-4x^3+2x^2+4x+4=(x-1)^4-4(x-1)^2+7
$$
the square can be completed to get
$$
\left((x-1)^2-2\right)^2+3=\color{#00A000}{\left((x-1)^2-2-i\sqrt3\right)}\color{#0000FF}{\left((x-1)^2-2+i\sqrt3\right)}
$$
Solving for $(\alpha+i\beta)^2=2+i\sqrt3$, we get $\alpha^2-\beta^2=2$ and $2\alpha\beta=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 2
} |
To prove or disprove $abc=1$ $\implies$ $ \sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4} > \dfrac92$ Let $a,b,c$ be positive real numbers such that $abc=1$ , then is it true that
$ \sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4} > \dfrac92$ ?
| With the hope that the problem remains unchanged;
First see that:
$$
\sqrt{a^4+a^2b^2+b^4}\geq \sqrt 3ab
$$
Hence:
$$
\sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4}\\\geq \sqrt 3(ab+bc+ac)\geq 3\sqrt 3> \dfrac92
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
About a combinatorics problem I'm dealing with this problem. Show that $\sum\limits_{j = 1}^n {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}i\\j\end{array}} \right)} \left( {\begin{array}{*{20}{c}}j\\
k\end{array}} \right) = 0$ for $1 \le i,k \le n$ and $i \ne k$.
Attempt: Since
$\left( {\begin{arra... | The solution for odd values of $i + k$ is here.
\begin{equation}
\sum\limits_{j = k}^i {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}
i\\
j
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right) = \sum\limits_{a = 0}^{i - k} {{{\left( { - 1} \right)}^a}\left( {\begin{array}{*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/786172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\rightarrow 1}\frac{\sin{(x^2-1)}}{x-1}$ My problem is to calculate
$$\lim_{x\rightarrow 1}\frac{\sin{(x^2-1)}}{x-1}$$
I evaluated to $\frac{\sin(x+1)(x-1)}{x-1}$ and then to $\sin(x+1)$ but I wonder about the result of limit as x approach 1 of $\sin{(x+1)}$. I cannot think of any result I cannot connect t... | Notice
$$ \frac{\sin ( x^2 - 1 )}{x-1} = \frac{\sin ( x^2 - 1 )}{x-1} \cdot \frac{ x+1}{x+1} = \frac{\sin ( x^2 - 1 )}{x^2-1} \cdot (x+1) \to_{x \to 1} 2 $$
since
$$ \frac{ \sin y}{y} \to_{y \to 0} 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
calc the length of $y = x^2-1$ I want to know the lenght of the curce $y = x^2-1$. To do this I use the integral method and get the integral $$\int \sqrt{(2x)^2+1)}dx$$. Substitute $2x = u$ and then $u+\sqrt{u^2+1} = s$ gives $$\frac{s^4-1}{8s^2}+\frac{1}{2}\ln(s)$$
The answer should be $\sqrt{5}+\ln(2+\sqrt{5})$. So t... | I wrote about this topic together with two colleagues some years ago. We also derived almost the same formula as you gave which can be found as $L_{kurve}$ at the beginning of page 2 in the following (Danish) article:
LMFK-article on parabola length
As you can see, your answer is off by a factor $2$. This accounts for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Verify a vector is an eigenvector of a matrix I have been asked to verify whether $v = \begin{bmatrix}1\\4\end{bmatrix}$ is an eigenvector of $A = \begin{bmatrix}-3&1\\-3&8\end{bmatrix}$? If yes, find the eigenvalue.
The way that I approached this question is to find eigenvalues, then use eigenvalues to verify whether ... | This is the long way!
Recall that $v$ is an eigenvector of $A$ if $Av=\lambda v$ for some $\lambda$.
Here we have
$$
\begin{bmatrix}
-3 & 1 \\ -3 & 8
\end{bmatrix}
\begin{bmatrix}
1\\ 4
\end{bmatrix}=
\begin{bmatrix}
1\\ 29
\end{bmatrix}
$$
But is $\begin{bmatrix}
1\\ 29
\end{bmatrix}$ a scalar multiple of $\begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/794396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Finding eigenvectors of a matrix I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$.
Here is how I find eigenvalues:
$$\begin{align*}
\det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda... | Well, the polynomial $\lambda^3+1$ has two more (complex) roots, which means a rotation in a $2$ dimensional subspace.
In your last equation substitute $\lambda=-1$ and, say, $x=1$ to find one eigenvector.
(You are right: there are infinitely many eigenvectors if there is one as they always form a subspace.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a -... | All the other answers are fine from a mathematical point of view, but they ignore the fact that using the quadratic formula is a very bad way to solve quadratic equations in computer code (using floating point arithmetic).
The problem arises when one of the roots is near zero. In this case, either the "$+$" or the "$-$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$ Question:
let $a,b,c>0$,show that:
$$\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$$
maybe th... | By Hölder's inequality we have $$\left(\sum_{cyc}\sqrt{\frac{a}{2bc + ca + abc}}\right)^2 \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right) \geq 27$$ Hence it suffices to prove that $$27 \geq \frac{81}{2} \cdot \frac{abc}{(a + b + c)^3} \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right)$$ which is obvious aft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ ... | Note
$$ \frac{1}{x^2-x+1}=\frac{1+x}{1+x^3}. $$
So
\begin{eqnarray*}
I&=&\int_0^1 \frac{\ln^2x}{x^2-x+1}dx=\int_0^1 \frac{(1+x)\ln^2x}{1+x^3}dx\\
&=&\sum_{n=0}^{\infty}(-1)^n \int_0^1(1+x)x^{3n}\ln^2xdx\\
&=&\sum_{n=0}^{\infty}(-1)^n \int_0^1(x^{3n}+x^{3n+1})\ln^2xdx\\
&=&2\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{(3n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 4
} |
Calculus Question: $\int_2^\infty\frac{\log^3(x-1)}{x^2}dx$ I have just taken calculus quiz but I could not find $\displaystyle \int_2^\infty\frac{\log^3(x-1)}{x^2}dx$? Any help would be appreciated. Thanks in advance.
EDIT:
Forgot to mention, my tutor gave us hints about this question.
*
*Use Taylor series
*$\dis... | By substitutions, the following integrals are equivalent:
\begin{align*}
\int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= \int_{1}^{\infty} \, \frac{\log^3(x)}{(1+x)^2} \, dx\\
&= -\int_{0}^{1} \, \frac{\log^3(x)}{(1+x)^2}\, dx \tag 1
\end{align*}
$(1)$ can be written as a sum, consider:
\begin{align*}
\int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Summation with Ceilinged Logarithmic Function According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$$
\sum_{i = 1}^{n}\left \lceil \log_2{(i)} \right \rceil = n\left \lceil \log_2{(n)} \right \rceil - 2^{\left \lceil \log_2{(n)} \right \rceil} + 1
$$
Now, I was wondering if someo... | Based on @vonbrand's answer, using integration by parts is a good approximation:
Concretely,
$$
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n.n(\left \lceil \log_a{(n)} \right \rceil - 1) - 1(\left \lceil \log_a{(1)} \right \rceil - 1) - \int_{1}^{n}i(\left \lceil \log_a{(i)} \right \rceil - 1) di
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are ... | Let $-1\le a \le 1$ and:
\begin{align*}
I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx \tag 1\\
\frac{\partial}{\partial a}I(a) &= \int_{0}^{1} \, \frac{1}{(1+a\, x)\sqrt{1-x^2}} + \frac{1}{(1-a\, x)\sqrt{1-x^2}} \, dx\\
&= \frac{1}{\sqrt{1-a^2}}\, \left(\arcsin\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 7
} |
Linear Algebra determinant reduction Prove, without expanding, that
\begin{vmatrix}
1 &a &a^2-bc \\
1 &b &b^2-ca \\
1 &c &c^2-ab
\end{vmatrix} vanishes.
Any hints ?
| \begin{align}
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab
\end{vmatrix}
&=\begin{vmatrix}
1 & a & a^2-bc \\
0 & b-a & b^2-a^2+bc-ca \\
0 & c-b & c^2-b^2+ca-ab
\end{vmatrix} \\
&=\begin{vmatrix}
b-a & b^2-a^2+bc-ca \\
c-b & c^2-b^2+ca-ab
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$ Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq
2xy(x+y+xy)$ for $x,y \in \mathbb{R}^+$.
I started by multiplying everything out on the RHS to get the equivalent statement
\begin{align*}
x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \ge... | By AM-GM, we have $$x^3+y^3+x^2y^3+x^3y^2 \geq 4 \sqrt[4]{x^3y^3x^2y^3x^3y^2} = 4x^2y^2.$$
Similarly, we have $$x^3y^2 + y^2 + x^2 + x^3 \geq 4 \sqrt[4]{x^3y^2y^2x^2x^3} = 4x^2y$$ and $$x^2y^3+ x^2 + y^2 + y^3 \geq 4 \sqrt[4]{x^2y^3x^2y^2y^3} = 4xy^2.$$ Adding these inequalities yields the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$p(x)=0$ with real coefficient has purely Imaginary roots.Then the equation $p(p(x)) = 0$ has If the Quadratic equation $p(x)=0$ with real coefficient has purely Imaginary roots.Then the
equation $p(p(x)) = 0$ has
$\bf{OPTIONS::}$
$(a)\;\; $ Only purely Inaginary Roots. $\;\;\;\;\;\;(b)$ all real roots.
$(c)$ Two re... | If a quadratic equation has purely imaginary roots, then it can be written in the form $p(x)=k(x-ai)(x+ai)=k(x^2+a^2)$ for some $k,a\ne0$.
Hence $p(p(x))=k[k^2(x^2+a^2)^2+a^2]=k^3x^4+2k^3a^2x^2+k^3a^4+ka^2$. By the quadratic equation, this gives us
$$x^2=\frac{-2k^3a^2\pm\sqrt{4k^6a^4-4k^3(k^3a^4+ka^2)}}{2k^3}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Find a formula for $1 + 3 + 5 + ... +(2n - 1)$, for $n \ge 1$, and prove that your formula is correct. I think the formula is $n^2$.
Define $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$
Then $p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n = (n + 1)^2$
So $p(n + 1): n^2 + 2n = (n + 1)^2$
The equality above is incorrect, so e... | $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$
$p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n+1 =n^2+2n+1= (n + 1)^2$
next term after $2n-1$ is $2n+1$ is not $2n$ as you mean
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$ Question:
Prove that
$$\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk=\frac{1}{16\pi}\Gamma^4\left(
\frac{1}{4}\right)$$
My attempt
Start by the transformation
$$k \to \frac{2\sqrt{k}}{1+k}$$
Hence we have
$$\int^{1}_0 K\left(\frac{2\sqrt{k}}{1+k}\right)\,\frac{1... | Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx
\end{align}
where $K(x)$ is the complete elliptic integral of the first kind. It can be shown that the hypergeometric form is
\begin{align}
K(x) = \frac{\pi}{2} \ {}_{2}F_{1}(\frac{1}{2}, \frac{1}{2}; 1; x^{2}).
\end{align}
By using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
solving a system of equations dealing with Lorentz transformations Can anyone help me to find the solutions of this system of equations:
$$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
I know the answer:
$$x= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$y= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $... | The second equality says that
$$y^2=c^2z^2+1$$
Replace $y^2$ in the first and third equalities. You get:
$$c^2x^2-v^2(1+c^2z^2)=c^2,$$
$$v(1+c^2z^2)+c^2zx=0,$$
Therefore:
$$c^2x^2-v^2c^2z^2=c^2+v^2,\;\;(E_1)$$
$$vc^2z^2+c^2zx=-v,\;\;(E_2)$$
Now, you get from the first equality $(E_1)$:
$$x^2=\dfrac{c^2+v^2+c^2v^2z^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculate a sum $1+4e+9e^2+...+n^2e^{n-1}$, where e is nth root of unity Its similar to Calculate a sum involving nth root of unity
I dont get how to do it when $e$ is not $1$.
Please help!
| Start with: $1 + x + x^2 +...+ x^n = \dfrac{1 - x^{n+1}}{1-x}$, differentiate both sides with respect to $x$:
$1 + 2x + 3x^2 + ...+ nx^{n-1} = \dfrac{nx^{n+1} - (n+1)x^n + 1}{(1 - x)^2}$, multiply $x$ both sides again:
$x + 2x^2 + 3x^3 + ...+ nx^n = \dfrac{nx^{n+2} - (n+1)x^{n+1} + x}{(1 - x)^2}$, and differentiate aga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/811729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the jordan form the matrix let characteristic polynomial $P_A(x)=(x+2)^4(x-3)^2$ and
minimal polynomial $m_A=(x+2)^3(x-3)$ find the jordan form that possible.
we know $q_6=\frac{f_6}{f_5}$ ($f_i$ is gcd{det of i x i submatrices which isnt equal to 0})
$q_6=\frac{f_6}{f_5}=\frac{(x+2)^4(x-3)^2}{f_5}=(x+2)^3(x-3)^2... | We are given that the characteristic and minimal polynomial of a matrix $A$ are
\begin{align*}
p_A(x) &= (x+2)^4(x-3)^2 & m_A(x) &= (x+2)^3(x-3)
\end{align*}
This tells us that $A$ is a $6\times 6$ matrix whose eigenvalues are $\lambda_1=-2$ and $\lambda_2=3$. The Jordan form of $A$ is thus of the form
$$
\begin{bmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^\pi\frac{\cos nx}{a^2-2ab\cos x+b^2}\, dx$
How to compute the following integral
\begin{equation}
\int_0^\pi\frac{\cos nx}{a^2-2ab\cos x+b^2}\, dx
\end{equation}
I have been given two integral questions by my teacher. I cannot answer this one. I have also searched the similar question here but it lo... | Let $I_n(a,b)$ be the desired integral. Note that $I_n(a,b)=I_n(b,a)$, and $I_n(a,b)=I_n(-a,-b)$. So, we may suppose that $|b|< a$ Note that
$$\eqalign{
\frac{a^2-b^2}{a^2-2ab\cos x+b^2}&=\frac{a}{a-e^{ix}b}+\frac{be^{-ix}}{a-e^{-ix}b}\cr
&=\sum_{n=0}^\infty \left(\frac{b}{a}\right)^ne^{inx}+\frac{be^{-ix}}{a}\sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/815855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 1
} |
Kernel of linear transformation in $\Bbb R^3$ Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be a linear transformation satisfying
\begin{align*}
T(0,1,1) =& (-1,1,1) \\
T(1,0,1) =& (1,-1,1) \\
T(1,1,0) =& (1,-1,0) .
\end{align*}
Is it necessary true that $\ker(T) = \operatorname{Sp}\{(1,-1,1)\}$ ?
Well, I tried to say that w... | We can simply put our vectors in the matrix and do row operations in the following way - we are trying to get a basis in the right part:
$$\left(
\begin{array}{ccc|ccc}
0 & 1 & 1 &-1 & 1 & 1\\
1 & 0 & 1 & 1 &-1 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
0 & 1 & 1 &-1 & 1 & 1\\
0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
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