Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Is the vector space of R vector system sinx, sin(x-1), cos(x+2) linearly independent? Is the vector space of $R^R$, vector system $\sin{x}$, $\sin(x-1)$, $\cos(x+2)$ linearly independent?
| You can use the Wronskian to verify that the functions are linearly dependent.
Alternatively, using trigonometric formulas, note that:
$\quad\quad \sin(x-1) = \cos(1) \sin(x) - \sin(1) \cos(x)$
$\quad\quad \cos(x+2) = \cos(2) \cos(x) - \sin(2) \sin(x)$
So (multiply by $\cos(2)$ and $\sin(1)$ respectively):
$\quad\quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Corollary of Fermat's Little Theorem to $a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}$ I'm not sure what to call this proposition in relation to Fermat's Little Theorem (it is not the converse, though it seems related), but I am interested to know if the following holds:
$$
a^x \equiv 1 \pmod{p} \implies x \equ... | For each prime $p$, there exists a primitive root $r$ which has the property you want: $$r^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}.$$
For example, if we work modulo $7$, $3$ is a primitive root, because
\begin{align}
3^0 &\equiv 1 \pmod 7 \\
3^1 &\equiv 3 \pmod 7 \\
3^2 &\equiv 2 \pmod 7 \\
3^3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integration of rational functions How would you integrate a rational function like this:
$$\frac{2x+3}{x^2+2x+10}$$
| $$\int \frac{2x+3}{x^2+2x+10} dx$$
$$= \int \frac{2x+2+1}{x^2+2x+10} dx $$
$= \int \frac{2x+2}{x^2+2x+10} dx + \int \frac{1}{x^2+2x+10}dx$
Let solve them individually,
$$\int \frac{2x+2}{x^2+2x+10} dx$$
$$\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$$
$$= \log |x^2 + 2x + 10| + c$$
Other part -
$$\int \frac{1}{x^2+2x+10} dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the easiest way to solve the following differential equation? I am only just starting online courses on the subject of differential equations, and I passed the following differential equation
$$\frac{dy}{dx}= -2x + 3y - 5$$
I'm a bit confused by the use of two variables in the equation. What is the easiest way ... | We can solve equation by the Variation of parameters
$$\frac { dy }{ dx } =-2x+3y-5\\ \frac { dy }{ dx } -3y=0\\ \int { \frac { dy }{ y } } =3\int { dx } \\ \ln { y=3x+C } \\ y=C{ e }^{ 3x }\\ y=C\left( x \right) { e }^{ 3x }\\ { y }^{ \prime }={ C }^{ \prime }\left( x \right) { e }^{ 3x }+3{ C\left( x \right) e }^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Evaluate $\sum_{n=1}^{64}(-1)^n \left\lfloor \frac{64}{n} \right\rfloor \varphi(n)$
Evaluate $$\sum_{n=1}^{64}(-1)^n \left\lfloor \dfrac{64}{n} \right\rfloor \varphi(n).$$
We have $$S = \sum_{n=1}^{64}(-1)^n \left\lfloor \dfrac{64}{n} \right\rfloor \varphi(n) = \sum_{n=1}^{32}\left\lfloor\dfrac{64}{2n}\right\rfloor \... | The identity you mention is indeed very relevant; also relevant is its proof, which rewrites the sum as $$\sum_{d=1}^n \left\lfloor \frac nd\right\rfloor \phi(d) = \sum_{k=1}^n \sum_{d \mid k} \phi(d)$$ and then uses the fact that $$\sum_{d \mid k} \phi(d) = k. \tag{1}$$ In this problem, we can similarly rewrite $$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What is $\lim_{n\to\infty} \frac{x_n}{y_n}$ provided $x_n=2y_{n}-y_{n-1}$ and $y_n=3y_{n-2}-y_{n-2}$ with $x_0=y_0=1$? Suppose $x_n$ and $y_n$ satisfy:
$$
\begin{pmatrix}
x_n \\
y_n \\
\end{pmatrix}=\begin{pmatrix}
2&1 \\
1&1 \\
\end{pmatrix}\begin{pmatrix}
... | Hint: Let $z_{n}\equiv(x_{n},y_{n})^{\intercal}$. Then, we can rewrite
the recurrence as
$$
z_{n}=Mz_{n-1}\text{ (}n\geq1\text{)}
$$
where
$$
M\equiv\begin{pmatrix}2 & 1\\
1 & 1
\end{pmatrix}.
$$
Note that $M$ is diagonalizable. That is, we can write $M=SJS^{-1}$
where
$$
S\equiv\begin{pmatrix}\frac{1}{2}\left(1-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Trick to solving this Summation? I am currently following some slides I found to try and learn how to find the average case complexity for some algorithms. I got stuck when having to handle this summation though:
$$\sum_{i=1}^k i*2^i $$
According to the slides: it can be shown that the the summation is equal to:
$$(... | $\sum_\limits{i=1}^k i*2^i = 1\cdot2 + 2\cdot2^2 + 3\cdot 2^3 +\cdots +k2^k$
Multiply by $-(1-2)$.
Since $-(1-2) = 1$ it doesn't change the value of the sum, but look what happens as you apply the distributive property.
$-(1-2)(1\cdot2 + 2\cdot2^2 + 3\cdot 2^3 +\cdots +k2^k) = -(1\cdot 2 - 1\cdot 2^2 + 2\cdot 2^2 - 2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\lim_{x \to2 }x^{2}+4x=12$ Please check my prove
we begin by
$$x^{2}+4x-12<\epsilon $$
$$|x+6||x-2|<\epsilon $$
I will prove by use property
$\lim_{x->n}f(x)g(x)=\lim_{x->n}f(x)\lim_{x->n}g(x)$
then we divide into to case
$|x+6|<\sqrt{\epsilon }$ and $ |x-2|<\sqrt{\epsilon }$
and let$\epsilon ,\delta _{1}... | Let $\epsilon>0$.
We wish to find a $\delta>0$ such that if $\vert x-2\vert<\delta$, then $\vert x^2+4x-12\vert<\epsilon$.
We see that the $x^2-4x-12=(x-2)(x+6)$. Since $x\to2$ we can assume that eventually it will be true that $\vert x-2\vert<1$ which means that
\begin{eqnarray}
-1&<&x-2<1\\
1&<&x<3\\
7&<&x+6<9
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$ Show that
$$\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$$
How many ways are there to prove it ?
Is there a standard way ?
I was thinking about making it a Riemann sum.
Or telescoping.
What is the easiest way ?
What is the sho... | Observe
\begin{align}
\sum^n_{k=3}\frac{2k}{n^2+k^2+1} = \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+(k/n)^2}
\end{align}
then we have
\begin{align}
\frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How to do this without using trigonometric substitution? $$\int \frac {\rho^2}{(\rho^2+ h^2)^\frac 32} d\rho$$
| In addition to the pythagorean identities for circular trig functions, such as
$$ 1 + \tan^2 x = \sec^2 x $$
and for the hyperbolic trig functions, such as
$$ 1 + \sinh^2 x = \cosh^2 x $$
there is a similar identity involving rational functions:
$$ 1 + \left( \frac{y^2 - 1}{2y} \right)^2 = \left( \frac{y^2 + 1}{2y} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
| *
*Find one solution
Try to find it as $f_0(x) = ax^2 +bx + c$
$f_0(x+1) + f_0(x-1) = 2ax^2 + 2a + 2bx +2c$ => $f_0(x)=2x^2-x + 3$
*$f(x) = f_0(x) + g(x)$ => $g(x-1) + g(x+1) =0$
Find all $g(x)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the inverse element of $u=i+j+k$ with respect to the multiplication? What is the inverse element of $u=i+j+k$ with respect to the multiplication?
When I have $z = x + iy$ inverse is $\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}$ but what happens in my case?
| Try $\frac{-i-j-k}{3}$. This is a natural extension of the complex conjugate formulation to the context of having three imaginary units.
I think Don Antonio's answer gives a good indication of how to find the inverse. To give another example of this sort of thinking: in the hyperbolic numbers $z = x+jy$ where $j^2=1$ w... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$? Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$
I tried expanding using binomial theorem:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\... | Hint: First observe that $a_n -b_n\sqrt{3}=(2-\sqrt{3})^n$. Combine with the assumption, we get $a_n^2 - 3b_n^2 =1$ for all $n$. Then $\frac{a_n^2}{b_n^2}-3=\frac{1}{b_n^2}$. You should prove that $b_n \to \infty$. Hence, $\frac{a_n}{b_n} \to \sqrt{3}$ since $\frac{a_n}{b_n}>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.
The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
| $$ x^2 + y^2 = kxy $$
$$ x^2 - k xy + y^2 = 0. $$
Divide by $y^2,$ let $r = x/y.$
$$ r^2 - kr + 1 = 0 $$
Discriminant is
$$ k^2 - 4. $$
Thi is not a square unless $k = \pm 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A Trigonometry + complex number Equation the question is:
Show that:
$$
[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\= 2^{n+1} \sin n \frac{(\theta - \phi)}{2} \cos x \frac{\theta + \phi - \pi}{2}$$
I gave it a try following the pattern of the prev... | I find that your result in the question is valid only for even values of $n$ (as well containing typographical errors; it should be $\sin^n$ and $\cos(n...)$). My solution is as follows:
First note that
$$\cos\theta-\cos\phi=-2\sin\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right) \\
\sin\theta-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I prove that $\arcsin \left(\sqrt\frac{1 + \sin2x}{2} \right)= |x + \frac{\pi}{4}|$ the function $ f$ is defined as: $\arcsin \sqrt\frac{1 + \sin2x}{2}$
The function is defined on $\mathbb{R}$.
Prove that $f(x) = |x + \frac{\pi}{4}|$ for all $x$ in $[\frac{-3\pi}{4}, \frac{\pi}{4}]$
We have for all in $ x \in ... | Note that we can write
$$\begin{align}
\frac{1+\sin(2x)}{2}&=\frac{\cos^2(x)+2\sin(x)\cos(x)+\sin^2(x)}{2}\\\\
&=\frac{(\sin(x)+\cos(x))^2}{2}\\\\
&=\frac{2\sin^2(x+\pi/4)}{2}
\end{align}$$
Can you proceed now?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I co... | The above sequence is given by $a_i = \frac{i(i+5)}{2}$. The finite sum is given by
\begin{align}
S_n & = \sum_{i=1}^n a_i \\
& = \frac{1}{2} \sum_{i=1}^n i^2 + \frac{5}{2} \sum_{i=1}^n i \\
& = \frac{n(n+1)(2n+1)}{12} + \frac{5n(n+1)}{4} \\
& = \frac{n(n+1)(n+8)}{6}
\end{align}
EDIT: (In response to comments) To a cer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove $\int_{0}^{1}{ dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}?$ Variation of Ahmed's integral
$$\int_{0}^{1}{\mathrm dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}\tag1$$
M... | One may observe that, for $x \in [0,1]$,
$$
\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)=\arctan\left({1\over \sqrt{x^2+2}}\right)+\arctan\left({x\over \sqrt{x^2+2}}\right)
$$ then
$$
I:=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=I_1+I_2
$$with
$$
\begin{a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Two numbers $x,y$ are randomly chosen from the segment [0,5] find the probability that $x+y\ge 5$ and $\left|x-y\right|\le 1$ Two numbers $x,y$ are randomly chosen from the segment [0,5] find the probability that $x+y\ge 5$ and $\left|x-y\right|\le 1$.
So the total area is 25 and the yellow area that has the needed re... | You are correct. The area is easy to get with geometry: the area of the rectangle is $2\sqrt2\times\sqrt2=4$ and that of the triangle is $1\times1/2=1/2$ for a total of $9/2$. Divide by the total possible area ($5^2=25$) and you get a probability
$$
p=\frac{9/2}{25}=9/50=0,18.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Length of two sides in a quadrilateral with given angles I'm stuck finding the length of two sides in a quadrilateral for which I know all angles and the length of two sides.
All red objects are know ($a,b,\alpha,\beta,\gamma $ and $\delta$). I need to find the length of $c$ and $d$. I know it can't be too difficult ... | Suppose our quadrilateral has vertices $A,B,C,D$. We can deduce the length of $BD$ using the law of cosines (Ignore the labels):
We get that
$$BD^2 = a^2 + b^2 - 2ab \cos \alpha$$.
Next, we find the angles $\angle ABD$ and $\angle ADB$ using law of sines to get:
$$m\angle ADB = \sin^{-1}\left(\frac{a \sin\alpha}{\sqrt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. B... | By the Chinese remainder theorem, $a\equiv b\pmod{15}$ if and only if
\begin{cases}
a \equiv b\pmod{3}\\[4px]
a \equiv b\pmod{5}
\end{cases}
In the case $a=x^2$ and $b=1$, the first equation gives
$$
x=1+3k \qquad\text{or}\qquad x=2+3k
$$
From $x=1+3k$, we get
$$
1+6k+9k^2\equiv 1\pmod{5}
$$
that is, $k-k^2\equiv0\pmod... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know h... | $x^2 + y^2 = z^2$ and $y = x + 1 $ $\implies (3x+2z+1)^2 + (3x+2z+2)^2 = (4x+3z+2)^2$
And this series goes on forever by $\cases{x := 3x+2z+1 \\
y := 3x+2z+2 \\
z := 4x+3z+2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ (here $:=$ stands for assignment)
It is not very difficult to check that $\gcd(x,y,z) = 1$ at e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$. If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.
How does even even begin to pr... | When $n=2$ you have $\frac{a^n+b^n}{a+b}=a-b+\frac{2b}{a+b}$ which is not an integer in general.
So I assume $n=2k+1$ is an odd prime should be part of the statement.
In this case
$$\frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+...+b^{n-1}\\=
a^{n-1}+a^{n-2}b-2a^{n-2}b-2a^{n-3}b^2+3a^{n-3}b^2+3a^{n-4}b^3-...\\
=(a+b)(a^{n-2}-2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the interval of $c$ such that the rational function $\frac{x^2+2x+c}{x^2+4x+3c}$ takes all real values Find the interval of $c$ such that the rational function $$f(x)=\frac{x^2+2x+c}{x^2+4x+3c}$$ takes all real values $\forall$ $x\in D_f$
I tried in the following way:
Let $$y=\frac{x^2+2x+c}{x^2+4x+3c}$$ converti... | \begin{align}
f(x) &= \frac{x^2+2x+c}{x^2+4x+3c} \\
&= \frac{x^2+4x+3c - 2x - 2c}{x^2+4x+3c} \\
&= 1-\frac{2x + 2c}{x^2+4x+3c}
\end{align}
therefore $x^2+4x+3c$ cannot be zero.
$x^2+4x+3c = 0$ has no solution
from discrimant
\begin{align}
b^2 - 4ac &< 0 \\
16 & < 12c \\
3c & > 4 \\
c & > \frac{4}{3}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a $3\times 3$ matrix $A\not = I_3$ such that $A^3 = I_{3}$ Use the correspondence between matrices and linear transformation to find find a $3\times 3$ matrix $A$ such that $A^3 = I_{3}$ and find an $A$ matrix that is not $I_{3}$
Where $I_{3}$ is the identity matrix:
$$I_{3}=
\left[ {\begin{array}{ccc}
1 & 0 ... | Try
\begin{align}
A =
\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
1 & 0 & 0
\end{bmatrix}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
The integral of ${x^2 - 1 \over x^3 \sqrt{2x^4 - 2x^2 + 1}}$
$$\int {x^2 - 1 \over x^3 \sqrt{2x^4 - 2x^2 + 1}}dx$$
Substituting $u = x^2$, I get $${1\over 4}\int {u - 1 \over u^2 \sqrt{2u^2 - 2u + 1}}du$$
Then completing the square under the square root.
$${1\over 4}\int {u - 1 \over u^2 \sqrt{\left(\sqrt{2} u - \f... | Hint pull out $x^2$ common from numerator and $x^4$ in the root from denominator ie $x^2$ from numerator. Then put $1-\frac {1}{x^2}=u $ thus $\frac {2}{x^3}dx=du $ also write the new denominator as $1+(1-\frac {1}{x^2})^2$ hope you can handle from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Number of solutions of $x+2y+3z=2n+k$ where $1 \leq x,y,z,k \leq n$ is about $\frac 14 n^2$ Let $n \in \Bbb N$ and $k \in [n]$, where $[n]=\{1,...,n\}$. I want to show:
The number of solutions $(x,y,z)$ of the equation $x+2y+3z = 2n+k$ where $x,y,z \in [n]$ is at least $cn^2$, for some constant $c$ that is independent... | Since you are looking for "at least" then its obvious that $2n+k$ have less solutions when $k=1$ than when $k$ is very big or $k=n$ (because the partitions of a number grows exponentially), so we are left with $x+2y+3z = 2n+1$ in order to give the "at least" constant $c$.
Now $z$ can run from $1$ up to $\lfloor \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$? How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$?
I tried show that difference between... | If you're trying to show the difference goes to $0$, here's a big hint:
$$
\sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}} = \frac{1}{n} \sum\limits_{k=1}^n \frac{2^{k/n}}{1 + \frac{1}{kn}}
$$
and hence
$$ 0 \leq
\sum\limits_{k=1}^n \frac{2^{k/n}}{n} - \sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}}= \frac{1}{n}\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the values of $k$ such that the equation $x^3+x^2-x+2=k$ has three distinct real solutions Find the values of $k$ such that the equation
$$x^3+x^2-x+2=k$$
has three distinct real solutions.
Please explain how to find the solution! Thanks
| We can sketch the graph $y=x^3+x^2-x+2$ as below (I used WolframAlpha):
Now we imagine drawing horizontal lines across this graph (setting $y=k$ in the original equation)
We can see that the only points where the line crosses the equation exactly three times are any points where $$y\text{ value of minimum point}<k<y\t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
| Or, with less variables, you can treat your fraction as reducible to some number $a$ (e.g. a decimal), which you can write as $\frac{a}{1}$. Then:
$\frac{a+a}{1+1} = \frac{2a}{2} = \frac{a}{1} = a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 4
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Calculate sum of coefficients of polynomial Let $$(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5) = \sum_{k=0}^{9} (A_k \cdot x^k)$$
Compute:
*
*$$\displaystyle \sum_{k=0}^{9} A_k$$
*$$\displaystyle \sum_{k=0}^{4} A_{2k}$$
I tried to figure out from Viete's Sums how to rewrite this but I can't find the coefficients for... | Let $P$ be your polynomial :
$$P(x)=(x+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)=\sum_{k=0}^9 A_kx^k$$
Then
$$\sum_{k=0}^9 A_k = \sum_{k=0}^9 A_k.1^k = P(1) = 2\times3\times4\times5\times6 = 720$$
And, as
$$\sum_{k=0}^9 (-1)^kA_k = P(-1) = 0$$
you can, by adding, find :
$$2\sum_{k=0}^4 A_{2k} = P(1)+P(-1)=720$$
so that
$$\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cd... | HINT
Use the fact that $$\cos^2(x) = \dfrac{1}{2}+\dfrac{\cos(2x)}{2}$$
Thus, the result is equal to
$$50 + \dfrac{1}{2}\sum\limits_{k=1}^{100} \cos\left(\dfrac{2\pi k}{101}\right)$$
And the last sum can be proven to be $-1$ using this result: How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? .
So th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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The sum of areas of triangles omitted infinitely many times within an equilateral triangle ABC
Triangle $ABC$ is an equilateral triangle. Four new equilateral triangles are formed by joining the midpoints of the sides $A, B, C$ in $\triangle ABC$ Triangle, with the white triangle in İmage 1, and it is taken out.
Ea... | $$\begin{align} S &=\sum_{n=1}^\infty a\cdot r^{n-1}\\ \\ & = \frac{1}4+\frac{3}{16}+\frac{9}{64}+\dots \\ \\
&= \frac{1}4\cdot\frac{3}{4}^{0}+\frac{1}4\cdot\frac{3}{4}^{1}+\frac{1}4\cdot\frac{3}{4}^{2}+\dots \\ \\ \end{align}$$
$$ a=\frac{1}4, r=\frac{3}4\implies S= 1 $$
| {
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"url": "https://math.stackexchange.com/questions/2255189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Have I correctly taken this function $f(x) = \frac{x^{2}}{1+2x}$ and $f ' (x) $and turned them into power series? $f(x) = \dfrac{x^{2}}{1+2x}$
To turn this into a power series I recall the similar looking geometric series,
$f(x) = \sum\limits_{n=1}^{\infty} x^{n} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^5 +\cdots$
And n... | Suggestion:
$$
\frac{1}{2x+1} \approx 1 - 2 x + 4 x^2 - 8 x^3 + 16 x^4 - 32 x^5 +\mathcal{O}\left( x^{6} \right)
$$
Now multiply by $x^{2}$.
$$
f(x) = \frac{x^2}{2 x+1} \approx x^2 -2 x^3+4 x^4-8 x^5+ \mathcal{O}\left( x^{6} \right)
$$
Suggestion:
$$
\frac{1}{(2 x+1)^2} \approx 1 - 4 x + 12 x^2 - 32 x^3 + 80 x^4 - 19... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Inequality with absolute value and variable on both sides I'm not able to understand how to get through this
$$\dfrac{|x+6|}{x+1} \leq x-2$$
$x$ can't, of course, be equal to $-1$.
I'm at a point where I'm not sure how to deal with the absolute value after reaching
$$|x+6| \leq (x-2)(x+1)$$
Thanks!
| $$\dfrac{|x+6|}{x+1} \leq x-2$$
Let's start with $\dfrac{|x+6|}{x+1} = x-2$.
\begin{array}{rclcrcl}
|x+6| &= &(x+1)(x-2) &| \\
x^2-x-2 &= &|x+6| &|\\
x^2-x-2 &= &x+6 &\text{or} & x^2-x-2 &= &-x-6 \\
x^2-2x-8 &= &0 &\text{or} & x^2+4 &= &0 \\
x &\in &\{-2,4\} &\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$ Is this proof correct?
Suppose $2^k$ is the largest power of $2$ in the sequence $n+1, n+2, ... 2n$
Then we can compute the power of 2 in the product as $n/2 + n/2^2 + ... n/2^k = n(1 + 2 + .... 2^{k-1})/2^k = n$.
| We can write as $$(n+1)(n+2)...(2n)=\frac{2n!}{n!}\tag1$$ Now let us calculate the power of $2$ contained in $2n!$ $$\text{Power of }2 \text{ in } 2!=\frac{2n}{2}+\frac{2n}{2^2}+\frac{2n}{2^3}+\frac{2n}{2^4}\ldots\\=2n\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots\right)\\=2n\left(\frac{\frac{1}{2}}{1-\frac12}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
show that $\sum_{i = 1}^{2k}\frac{ (-1)^{i+1}}{i} = \sum_{i = k+1}^{2k} \frac{1}{i}$ I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.
Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k... | We can use harmonic numbers
$H_k=\sum_{i=1}^k\frac{1}{i}$.
We obtain
\begin{align*}
\sum_{i=k+1}^{2k}\frac{1}{i}=\color{blue}{H_{2k}-H_k}
\end{align*}
on the other hand we get
\begin{align*}
\sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i}-H_{2k}
&=-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2k}\right)=-H_k\\
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258584",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Extremely ugly integral $\int_{-\pi}^{\pi} \frac{\operatorname{sign}(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$
Evaluate: $\DeclareMathOperator{\sign}{sign}$ $$\int_{-\pi}^{\pi} \dfrac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$$
My idea:
\begin{align*}I=\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)-|x|}{\... | I can't take full credit for this, Wolfram can be very helpful when you know how to coax it along.
For the integral
$$\int{dx\over\sin^2x+\cos x},$$
let
\begin{align*}
u &= \tan\left(\frac{x}{2}\right)\\
du &= \frac{1}{2}\sec^{2}\left(\frac{x}{2}\right)\,dx.
\end{align*}
Then we have
\begin{align*}
\sin x &= \frac{2\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Combinatorial summation problem: Find $ \sum_{k=1}^{n} k {n+k-1 \choose 3} $ Can we get a neat form for this summation:
$$ \sum_{k=1}^{n} k {n+k-1 \choose 3}? $$
General combinatorial tricks didn't work.
| Let's try:
$$\begin{aligned}
\sum_{k=1}^n k\binom{n+k-1}{3} &= \sum_{k=1}^n \sum_{i=1}^k \binom{n+k-1}{3} \\
&= \sum_{i=1}^n \sum_{k=i}^n \binom{n+k-1}{3}.
\end{aligned}$$
Using the known relation
$$
\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}
$$
you can obtain
$$\begin{aligned}
\binom{n+i-1}{4} + \sum_{k=i}^n \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got max... | Set $\sin^2(x) = 1 - \cos^2(x) = y^2$ Then you have
$$
v=\sqrt{a^2(1-y^2)+b^2 y^2}+\sqrt{b^2 (1-y^2)+a^2 y^2}
$$
One extremum is obtained (differentiate w.r.t. y) at
$$
\sqrt{a^2(1-y^2)+b^2 y^2}= \sqrt{b^2 (1-y^2)+a^2 y^2}
$$
or
$$
y^2 = 1/2
$$
The other extremum is obtained at $y=0$.
So the difference of $v^2$ betw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
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Find the exact value of $\frac{\cos( \beta)}{\cos( \beta) -1}$ with $\sin(\beta - \pi) = \frac{1}{3}$
Let $h$ be $$\frac{\cos( \beta)}{\cos( \beta) -1}$$
With $\sin(\beta-\pi)=\frac{1}{3}$ and $\beta \in
{]}\pi,\frac{3\pi}{2}{[}$ determine the exact value of $h(\beta)$.
I tried:
$$\sin(\beta-\pi) = \sin(\beta)\cos(... | You answer is very close.
Solve for$\beta$
Simplify
$$
\sin \left( \beta - \pi \right) = - \sin \beta = \frac{1}{3}
$$
The solutions are
$$
\beta = - \arcsin \frac{1}{3} + 2\pi k, \quad \beta = \pi + \arcsin \frac{1}{3} + 2 \pi k
$$
where $k$ is an integer.
The constraint given in the problem is satisfied when
$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\cos{y}=x+\frac{1}{x}$, possible for any values of $y$? If for real values of x, $\cos{y}=x+\frac{1}{x}$,
Then how do we prove that no value of $y$ is possible ?
| The value of $\cos(y)$ is between $-1$ and $1$. Therefore, for this equality to hold, $x+\frac{1}{x}$ must be between $-1$ and $1$.
*
*If $x>1$, then $x+\frac{1}{x}>x>1$.
*If $x=1$, then $x+\frac{1}{x}=2>1$.
*If $0<x<1$, then $\frac{1}{x}>1$, so $x+\frac{1}{x}>\frac{1}{x}>1$.
*If $x=0$, then $x+\frac{1}{x}$ is n... | {
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"url": "https://math.stackexchange.com/questions/2266973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Find the signal and zeros of $\cos x - \sin x$
Let $f$ be a function of domain $\mathbb{R}$ defined by:
$$f(x) = \cos x - \sin x$$
Find the signal and zeros of the function.
First I tried to find the zeros of $f(x)$:
$$0 = \cos x - \sin x \Leftrightarrow \\
0 = \sqrt{1-\sin^2x}-\sin x \Leftrightarrow \\
\sin x = \sqr... | $f(x)=\cos x−\sin x = \sqrt {2} \cos (x+\frac \pi 4)$
more generally
$a\cos x+b\sin x = \sqrt {a^2 + b^2} \cos (x - \tan^{-1}\frac ba)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Substitution Integral: $\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx$ $$\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx $$ I tried with substitution $t=3-x$ but I don't know what to do next.
Also how do I solve this limit: $$\lim_{x\to\infty}\frac{1}{x}\int_0^x\frac{dt}{2+\cos t} $$ I saw that most problems like... | Hint:
(1) $\displaystyle \frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}}=\frac{\sqrt{x}(\sqrt{x}+\sqrt{3-x})}{(\sqrt{x}+\sqrt{3-x})(\sqrt{x}+\sqrt{3-x})}$
(2) As the integrand is even and has a period of $2\pi$, $\displaystyle\int_{k\pi}^{(k+1)\pi}\frac{dt}{2+\cos t} =\int_{0}^{\pi}\frac{dt}{2+\cos t} $ for all integers $k$.
Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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maximum of combinatorial function Find the maximum of $\binom{30-k}{x} {(\frac{4}{5})}^{30-k}$ w.r.t. $k$, and find $k$ as a function of $x$ when the function maximises.
Note that as $k$ increases $\binom{30-k}{x}$ decreases and ${(\frac{4}{5})}^{30-k}$ increases.
| When $k=30-x$, the binomial coefficient equals $1$, and the power of $\frac45$ is simply $\left(\frac45\right)^x$. Each time we decrease $k$ by $1$, the binomial coefficient increases by the ratio $\frac{\binom{n+1}{x}}{\binom{n}{x}}$ for some $n$, while the power of $\frac45$ decreases by a ratio of $\frac45$. Thus, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Perhaps a Pell type equation Find all pairs of positive integers $(a,b)$ that satisfy $13^a+3=b^2$.
If $a$ is even then $3=(b-13^{a/2})(b+13^{a/2})$ which have no solutions. Now if the case $a=2k+1$ is odd then $b^2-13.(13^k)^{2}=3$ I cant proceed from here, please help. Any other methods for the latter case?
| A solution using elliptic curves, or Mordell equations, as Will Jagy suggests in the comments.
There are three cases:
$1)$ $\ $ $a=3k$, $k\in\mathbb Z^+$. Then $\left(13^k\right)^3+3=b^2$ is a Mordell equation.
See http://oeis.org/search?q=mordell&language=english&go=Search,
in particular (this link is inside the link)... | {
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"question_score": "2",
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Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$ Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$.
......
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
It takes $\infty - \infty $ form when $x=\infty $.
Now,
$$=\lim_{x\to \infty } (\sqrt {x-a} - \sqrt {bx})\times \dfrac {\sqrt {x-a}+\sqrt {bx}}{\s... | Take $\sqrt{x}$ out to get:
$$L=\lim_\limits{x\to\infty} \sqrt{x}\left(\sqrt{1-\frac{a}{x}}-\sqrt{b}\right)=\lim_\limits{x\to\infty} \sqrt{x}\left(1-\sqrt{b}\right).$$
If $b=1$, $L=0$.
If $b>1$, $L=-\infty$.
If $0\le b<1$, $L=+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Cauchy random variable question If X is a Cauchy random variable it has the density function as follows:
$f_X(x):= \frac{1}{\pi}\cdot\frac{1}{1+(x-\Theta)^2 }$ con $x \in R$ e $- \Theta<x<\Theta$
I have to demostrate that the random variable $Y=\frac{1}{X}$ is a Cauchy random variable too.
$$F_Y (a)=P(Y<a)=P(\frac{1}{X... | Let $h(x)=x^{-1}$. Then
\begin{align}
f_Y(y)&=f_X(x)(h^{-1}(y))\times\left\lvert \frac{dh^{-1}(y)}{dy}\right\rvert \\
&=\frac{1}{\pi}\times\frac{1}{1+(1/y-\theta)^2}\times\frac{1}{y^2} \\
&=\frac{1}{\pi}\times\frac{1}{\gamma\left(1+[\gamma^{-1}(y-\tau)]^2\right)},
\end{align}
where $\gamma\equiv (1+\theta^2)^{-1}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Are positive real numbers $x,y$ allowed to be taken out during this proof?
Prove $$\left(x^2 - y^2\right)\left(\frac1y - \frac1x\right) \ge 0$$ where $x$ and $y$ are positive real numbers.
Can I simplify $\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$
and then $(x^2 - y^2)(x-y) \ge 0$ cancelling out the $xy$?
Is this valid becau... | If $x,y$ are positive, then $xy>0$; so, when you have
$$\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$$
you're allowed to multiply by $xy$ (and the $\ge $ won't flip), to get
$$(x^2 - y^2)(x-y) \ge 0$$
And from that indeed you can get
$$(x^2 - y^2)(x-y)=(x+y)(x-y)(x-y)=(x+y)(x-y)^2 \ge 0$$
which is true, since $x+y>0$ and $(x-y)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the SUM of all the numbers of S. Let S be the set of all three digits numbers. Such that
*
*The digits in each number are all from the set
$\left\{1,2,3,\ldots,9\right\}$.
*Exactly one digit in each number is even.
Then, find the SUM of all the numbers of S.
I have tried that,
The only even digit can have a... | Note that if you permute $3$ distinct digits $x,y,z$, then the sum of the six 3-digit number is $222\cdot(x+y+z)$. If you permute $3$ digits $x,x,z$, with $x\not=z$ then the sum of the three 3-digit number is $111\cdot(2x+z)$. Therefore your sum is
$$222\sum_{x<y\ x,y\in\{1,3,5,7,9\}}\sum_{z\in\{2,4,6,8\}}(x+y+z)+111\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to prove the following in general: $\lim\limits_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$? To prove
$$\lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$$
The proof is easy when we take $n$ as positive integer and $a$ any positive real number.
In my book it is given that the result is true even when $n$ is any... | $$\begin{array}{rcl}
\displaystyle \lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p/q}-a^{p/q}}{x-a} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x^{p/q}-a^{p/q}\right)\left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)}{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find integer part of the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$ for $x+y=3$
If $M$ is the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$. Subject to $x+y=3$. Find the value of $\lfloor M \rfloor$.
I think the maximum of $M$ occurred when $x=y=3/2$. And I just put the value of that in M and found 34.... | $x^{4}y+x^{3}y+x^{2}y+xy+xy^{2}+xy^{3}+xy^{4}=xy+xy(x+y)+xy(x^{2}+y^{2})+xy(x^{3}+y^{3})=xy+3xy+xy((x+y)^{2}-2xy)+xy((x+y)^{3}-3x^{2}y-3xy^{2})=4xy+xy(9-2xy)+xy(27-3xy(x+y))=13xy-2(xy)^{2}+xy(27-9xy)=40xy-11(xy)^{2}.$
Now take $xy=k$ and you have $k\leq \frac{9}{4}$. Now our expression is $40k-11k^{2}$. Using derivativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Easy way to find out limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$? What's an easy way to find out the limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$?
I don't think binomial expansion like with $\left (1-\frac{1}{n^2} \right )^n = \left (1+\frac{1}{n} \rig... | Using the binomial theorem and the fact that $\binom{n}{k}\leq\frac{n^k}{k!}$, we have: $$1\leq\left(1+\frac1{n^2}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac1{n^{2k}}\leq\sum_{k=0}^n\frac{n^k}{k!n^{2k}}=\sum_{k=0}^n\frac{1}{k!n^k}\leq1+\frac1n+\sum_{k=2}^n\frac1{n^2}\leq1+\frac2n.$$ Therefore, by the squeeze theorem, the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
What is the limit of this sequence? How does it relate to the exponential function?
Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$
I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.
Edit:
I'm trying to prove a result that wou... | $$A_n=\left(1+\frac{x^2}{n^2}\right)^n\implies \log(A_n)=n \log\left(1+\frac{x^2}{n^2}\right)$$ Now, using Taylor series for large values of $n$ $$\log\left(1+\frac{x^2}{n^2}\right)=\frac{x^2}{n^2}-\frac{x^4}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ $$ \log(A_n)=\frac{x^2}{n}-\frac{x^4}{2 n^3}+O\left(\frac{1}{n^6}\right)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
How to solve integral arctan I solved this integral: $$ \int_0^{\pi/2}\frac{dx}{5+4\cos x} $$ and I obtained $\frac{2}{3} \arctan \frac{1}{3}$ is it correct? how do I find the answer in $\pi$?
|
and I obtained $\frac{2}{3} \arctan \frac{1}{3}$ is it correct?
Yes, that is correct!
How do I find the answer in $\pi$?
What do you mean, "in $\pi$"?
Perhaps the model answer looks different (containing $\pi$)? You can use the property:
$$\boxed{\arctan a + \arctan \frac{1}{a} = \frac{\pi}{2}} \implies \color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Put fraction in "arctan-friendly" form I would like to put $\int\frac{1}{(2x^2+x+1)}dx$ into something like $\int\frac{1}{(u^2+1)}dx$. What is the quickest way to proceed? I know that previous fraction can be rewritten as $2t^2+t+1 = \frac{7}{8}\left( \left( \frac{4t+1}{\sqrt{7}} \right)^2 +1 \right)$, but I don't have... | Note that
$$(ax+b)^2=a^2x^2+2abx+b^2$$
Now we want to complete the square on $2x^2+x+1$. We then have $a^2=2, 2ab=1 $ $\implies 4a^2b^2=1 \implies b^2=\frac 1 8$. Thus, we write
$$2x^2+x+1 = \left(2x^2+x+\frac 1 8\right)+\frac 7 8 = \left(\sqrt 2 x+\frac{1}{2\sqrt 2}\right)^2+\frac 7 8$$
Thus, letting $\sqrt2 x+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the probability that "A" wins the game. In a game, A and B play against each other. They throw a die alternatively until they get the face of 6. Once one gets 6, he wins. i.e., there is only one winner and they keep playing until one of them wins.
Consider that B plays first.
What is the probability of A winning?... | For $A$ to win on their first turn we need $B$ to not get a 6 and $A$ to get a 6, for probability $\frac{5}{6}\cdot\frac{1}{6}$.
For $A$ to win on their second turn, we need both to miss on their first try, $B$ to miss on their second try, and $A$ to get 6 on their second try, for probability $\left(\frac{5}{6}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac... | To assume that a series is convergent to compute the value is not valid Ixion. i.e. $S=1+2+4+6+8+...$ is obviously not convergent although $S=1+2(1+2+3+4+...)=1+2S$ would imply $S=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Calculate the sum of $\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$ I would like to calculate the sum for $x\in(-1,1)$ of this: $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$$
So far I managed that
$$\int \frac{x^n}{n}dx = \frac{x^{n+1}}{n(n+1)}, $$
and
$$\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x), $$
and
$$ \int -\log(... | Using partial fraction decomposition $$\frac 1{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ So
$$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty} \frac{x^{n+1}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x\sum_{n=1}^{\infty} \frac{x^{n}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x \log(1-x)-\sum_{n=1}^{\infty} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left... | $$\lim _{ x\to a } \left( 2-\frac { x }{ a } \right) ^{ \tan \left( \frac { \pi x }{ 2a } \right) }=\lim _{ x\to a }{ \left[ { \left( 1+\left( 1-\frac { x }{ a } \right) \right) }^{ \frac { a }{ a-x } } \right] } ^{ \frac { a-x }{ a } \tan \left( \frac { \pi x }{ 2a } \right) }=\\ ={ e }^{ \lim _{ x\rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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From $a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$ to $a_n$ to be proven by induction
Find and Prove by induction an explicit formula for $a_n$ if $a_1=1$ and for $n \geq 1$
$$a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$$
Checking the pattern:
$$a_1=1$$
$$a_2= 2 \cdot 1$$
$$a_3= (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_4= (\frac{4}... | Rewrite:
$$\frac{a_{n+1}}{(n+1)^{n+1}}=\frac{a_n}{n^n} . \frac{1}{n+1}$$
Thus:
$$\frac{a_{n+1}}{(n+1)^{n+1}}=\frac{a_n}{n^n}.\frac{1}{n+1}
=\frac{a_{n-1}}{n^{n-1}}.\frac{1}{n}.\frac{1}{n+1}
=\frac{a_{n-2}}{n^{n-2}}.\frac{1}{n-1}.\frac{1}{n}.\frac{1}{n+1}$$
$$=...$$
$$=\frac{a_1}{1}.\frac{1}{(n+1)!}$$
Therefore:
$$a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the following are real numbers
*
*$$\frac{1}{z}+\frac{1}{\overline{z}}$$
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
1.$$\frac{1}{z}+\frac{1}{\overline{z}}$$
$$\frac{\overline{z}}{z\cdot \overline{z}}+\frac{z}{z\cdot\overline{z}}$$
$$\frac{\overline{z}+z}{z\cdot \overline{z}}$$
$$\frac{2Re(z... | we have $$\frac{z+\overline{z}}{z\overline{z}}=\frac{2x}{x^2+y^2}$$ if $$z=x+iy$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Compute $\sum\limits_{n=1}^\infty\frac{b_{n-1}}{n 2^n}$ where $b_n=\sum\limits_{k=0}^n\frac{2^k}{k+1}$ Define the sequence $a_n := \frac{2^n}{n+1}$. Then, define the sequence $b_n$, the partial sums of $a_n$; i.e.:
$$b_n=\sum_{k=0}^{n} {a_k}$$
The problem is to compute:
$$\sum_{n=1}^{\infty} {\left(\frac{b_{n-1}}{n 2^n... | Consider the following.
Given:
\begin{align}
b_{n} &= \sum_{k=0}^{n} \frac{2^{k}}{k+1} \\
S &= \sum_{n=1}^{\infty} \frac{b_{n-1}}{2^{n} \, n}.
\end{align}
Now:
$$b_{n} = \frac{1}{2} \, \sum_{k=1}^{n+1} \frac{2^{k}}{k} $$
which leads to
\begin{align}
S &= \frac{1}{2} \, \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{2^{k}}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Compute $\int_0^\frac {\pi} {2} \frac {\cos x}{(1+\sqrt{\sin (2x)})^2}\,dx$
Compute$$\int_0^\frac {\pi} {2} \frac {\cos x}{(1+\sqrt{\sin (2x)})^2}\,dx$$
I think there's no closed form antiderivative of it. I tried WolframAlpha but it didn't help much.
| By setting $x=\frac{t}{2}$, then exploiting symmetry:
$$ I=\int_{0}^{\pi/2}\frac{\cos(x)}{\left(1+\sqrt{\sin(2x)}\right)^2} = \frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1+\cos(x)}+\sqrt{1-\cos(x)}}{\left(1+\sqrt{\sin(x)}\right)^2}\,dx $$
and the last integral can be written in the following form:
$$ I = \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ wil... | $$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$
$$=\sum_{k=1}^n 4!\binom{k+3}{4}$$
$$=4!\sum_{k=1}^n \binom{k+3}{4}$$
Using the hockey-stick identity:
$$=4!\binom{k+4}{5}$$
$$=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
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Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmat... | Just variation of the loup blanc's answer...
Interesting it would be also to check whether it is possible to use
$B^2-\text{trace}(B)B+\det(B)I=0$ ?
Indeed
$B^2+\det(B)I= A\pm\sqrt{\det(A)}I=\text{trace}(B)B=tB $
If we name $ C=A\pm\sqrt{\det(A)}I$ ( $C$ can be directly calculated) then $\text{trace}(C/t)=t$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\int \sqrt{2x + 3} dx$ Integrate: $\int \sqrt{2x + 3} \ dx$
Doing some guesswork gives me: $\frac{1}{2}(2x + 3)^{3/2}$ to try. Differentiating this gives: $2\frac{3}{2}\frac{1}{2}(2x + 3)^{1/2}$ = $\frac{6}{4}(2x + 3)^{1/2}$
Too bad the answer should be: $\frac{1}{3}(2x + 3)^{1/2}$, so clearly I'm missing so... | Here is how you could approach this:
Your answer is within a constant factor of the real answer, so you have
$$\int \sqrt{2x+3}dx=c*(2x+3)^{\frac{3}{2}}$$
When we differentiate $c*(2x+3)^{\frac{3}{2}}$ we want to get $\sqrt{2x+3}$, so differentiate and solve for $c$:
$$\sqrt{2x+3}=\frac{d}{dx}\big(c*(2x+3)^{\frac{3}{2}... | {
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"url": "https://math.stackexchange.com/questions/2304558",
"timestamp": "2023-03-29T00:00:00",
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How to calculate product $\prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right)$? How can I calculate the following product of series? $$ \prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right) $$....Can I take a geometric series and compare it with that?
| Multiply through by $\frac12=1-\frac1{2^{2^0}}$ to get
$$\begin{align}\left(1-\frac1{2^{2^0}}\right)\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)&=\left(1-\frac1{2^{2^0}}\right)\left(1+\frac1{2^{2^0}}\right)\prod_{k=1}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^1}}\right)\left(1+\frac1{2^{2^1}}\right)\prod_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Find an upperbound for the rational function I know
$$\lim_{(x, y)\to (0,0)} \frac{x^3 + y^4}{x^2 + y^2} = 0$$ so
$$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert \le f(x, y)$$ for some simpler $f(x, y)$ whose limit is also $0$.
How do I find the function $f(x, y)$? In other words, how do I get the upper bound... | Let $x=r\cos{t}, y=r\sin{t}$, then $$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert=|r(\cos^3{t}+r\sin^4{t})|\leq |r||1+r| = \sqrt{x^2+y^2}+x^2+y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integration $\int \frac{\sec(x)}{\tan^2(x) + \tan(x)}$ How to evaluate the following integral $$\int \frac{\sec(x)}{\tan^2(x) + \tan(x)} \ dx$$
Thank you.
| $$\int\frac{\sec{x}}{\tan^2x+\tan{x}}dx=\int\frac{\cos{x}}{\sin{x}(\sin{x}+\cos{x})}dx=$$
$$\int\left(\frac{1}{\sin{x}}-\frac{1}{\sin{x}+\cos{x}}\right)dx=-\int\left(\frac{d(\cos{x})}{\sin^2{x}}-\frac{d\left(\cos\left(\frac{\pi}{4}+x\right)\right)}{\sqrt2\sin^2\left(\frac{\pi}{4}+x\right)}\right)=$$
$$=-\int\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
Evaluate the series $\sum _{n=1}^{\infty} \frac{n}{5^n}$ $$\sum _{n=1}^{\infty}\frac{n}{5^n}$$
I tried to plug in $n=1,2,3,4,...$ but I can't use common ratio to solve problem.
I think there is another way like using differentiation or integral but I don't no exactly what to do.
| I am not using calculus let the sum be $S$ thus $S=\frac{1}{5}+\frac{2}{5^2}$...now $\frac{S}{5}=\frac{1}{5^2}+\frac{2}{5^3}+...$ subtracting two we have $\frac{4S}{5}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...$ which is GP with common ratio less than 1 using formula that sum of such gp is $\frac{a}{1-r}$ we have $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers. Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers.
My attempt:
For $n=1$, trivially there are integer solutions.
For $n=2$, we have the solutions
$$a=b... | If $n$ is such that $2n+1$ is prime, then the only $n$th powers mod $2n+1$ are $0$ and $\pm 1$ (since the group of units mod $2n+1$ is cyclic of order $2n$). It follows easily that $a^n+b^n=c^n+n$ has no solutions mod $2n+1$ if $n>3$. In particular, this handles the cases $n=5$ and $n=6$ (as well as infinitely many o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The val... | \begin{align}
k&=\frac{4\cos^2\frac\pi7-1}{\cos\frac{\pi}{7}(2\cos^2\frac\pi7-1)}\\
&=\frac{2\cos\frac{2\pi}{7}+1}{\cos\frac{2\pi}{7}\cos\frac{\pi}{7}}\\
&=\frac{2\cos\frac{2\pi}{7}+1}{\frac{1}{2}(\cos\frac{3\pi}{7}+\cos\frac{\pi}{7})}\\
&=\frac{4\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}}{\sin\frac{\pi}{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$? $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
in terms of summation i can write it as
$$S_{n}=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$$
How to continue from this point?
used partial frac... | $$
\begin{align}
&\sum_{n=1}^\infty(-1)^{n-1}\frac{n^2}{n^3+1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n+e^{-2\pi i/3}}\right)\tag{1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n-1-e^{2\pi i/3}}\right)\tag{2}\\
&=\frac13\sum_{n=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Tom Apostol - Calculus Vol. 1: Method of Exhaustion in Introduction I'm new to proof based math, and I really want to get better from pursuing the discipline by myself. I actually had a fear, last year, of proof-based math after my first semester in college. It was pretty brutal - I had never seen or experienced proofs... | Remember, he states before it that, for all $n\ge1$, $A < \frac{b^3}{3} + \frac{b^3}{n}$. This is actually the same with the statement:
$$∀n\ge1,\Biggl(A < \frac{b^3}{3} + \frac{b^3}{n}\Biggr)$$
Thus since we assumed that $A > \frac{b^3}{3}$, this yields to the fact that $A - \frac{b^3}{3} > 0$, thus we can change som... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Need to prove formula for $\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$ I need help showing that for $|z|<1$ we have
$\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$ = $\frac{z}{1-z}$
I tried using $\sum_{n=0}^{\infty} (-z^{2^k})^{n} = \frac{1}{1+z^{2^k}}$ but that hasn't gotten me very far. Any help ... | Let
$$\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}} = \sum_{k=0}^{\infty} a_n z^n$$
and examine contributions to the $a_n$ from the various terms in the $k$ sum.
For $n=0$ none of the terms in the sum contributes (the $k=0$ contribution is
$z-z^2+z^3-z^4+\ldots$).
For $n$ odd only the $k=0$ term contributes,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation of product of three terms in AP Here is an interesting relationship:
$$\sum_{r=1}^n r(r+m)(r+2m)=\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m} r\right)\tag{1}$$
$$\text{i.e.} \;\; 1\cdot(1+m)(1+2m)+2\cdot (2+m)(2+2m)+3\cdot (3+m)(3+2m)+\cdots +n(n+m)(n+2m)=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+2m))$$
which gi... | I will use the identity
$$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2$$
It can be proved without direct expansion. (https://math.stackexchange.com/a/1215805/268334)
\begin{align}
&\;\sum_{r=1}^n r(r+m)(r+2m)\\
=&\;\sum_{r=1}^n [(r+m)^3-m^2(r+m)]\\
=&\;\sum_{r=1}^{n+m}r^3-\sum_{r=1}^mr^3-m^2\sum_{r=1}^n (r+m)\\
=&\;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a \equiv b \pmod{1008}$
Suppose that the positive integers $a$ and $b$ satisfy the equation $$a^b-b^a = 1008.$$ Prove that $a \equiv b \pmod{1008}$.
Taking the equation modulo $1008$ we get $a^b \equiv b^a \pmod{1008}$, but I didn't see how to use this to get that $a \equiv b \pmod{1008}$.
| Suppose
$a^b-b^a = n$
with all variables
positive integers.
If $n=1$,
this is Catalan's problem
for which the only solution is
$a=3, b=2$.
Assume
$n \ge 2$.
I will show that there
at most a finite number of solutions
and give bounds for
$a$ and $b$
in terms of $n$.
For $n=1008$,
the only solution is
$a=n+1, b=1$.
If $b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
A long trigonometric/hyperbolic integral Evaluate:
$$ \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}=\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx=-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right) $$
In the first integral let $x=\sin\theta$, $dx=\cos\theta d\theta$
In the second integ... | $$I = \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}$$
$$\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx$$
$$-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right) $$
Apply integration By Parts:
$$I_1 = \int \frac{\sqrt{1-x^2}}{x^2}dx$$
$$\sqrt{1 - x^2}\left( -\frac{1}{x}\right) - \int -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove by induction that for $n$ $ \in \mathbb N $ , $ 2 n - 18 < n^2-8n +8 $? Question:
Prove for $n$ $ \in \mathbb N $ , $ 2 n\ -\ 18\ <\ n^2-8n\ +8 $
My attempt:
$ Base\ Case:\ n\ =\ 1,\ it\ holds. $
$I.H:\ Suppose\ 2k-18\ <\ k^2-8k+8,\ where\ k\ is\ a\ natural\ number.$
$ Then,\ \left(k+1\right)^2-8\left(k+... | Assuming $\quad2n-18 < n^2-8n+8\quad $ then
$\begin{array}{ll}
2(n+1)-18 = (2n-18)+2 < & (n^2-8n+8)+2\\
&=[n^2-6n+1]-2n+9\\
&=[(n+1)^2-8(n+1)+8]-(2n-9)\\
\end{array}$
So the induction works when $2n-9\ge 0$ that is when $n\ge 5$
You'll get to verify the initial step for $n=5\ :\ \begin{cases}2n-18=-8\\
n^2-8n+8=25-40+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$... | Let's rewrite $y=ux^2$ in order to render the denominator homogeneous.
Note: $u$ is variable, write it $u=\frac y{x^2}$ if you prefer.
$\displaystyle f(x,y)=\frac{x^3u^4x^8}{(x^4+u^2x^4)^2}=\underbrace{\frac{u^4}{(1+u^2)^2}}_\text{bounded}\ \underbrace{x^3}_{\to 0}\to 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Does this function have a closed form? $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$
$$k_b(x)=\sum^\infty_{n=0}{\frac{(n+x)^2}{n!}}$$
I noticed these functions closely relate to $e$. By looking at them I was able to determain a closed from for $k_b$.
$$k_b(x)=e(x^2+2x+2)$$
Though I still have not found a short form ... | This is too long for a comment.
Considering $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$ and using the result from Wolfram Alpha, it seems that the expression could simplify to
$$k_a(x)=\frac{(2+x-x^2)+e\,(2-2x+x^2)\left(\Gamma (x+2)-(x+1)\, \Gamma (x+1,1) \right) }{\Gamma (x+2)}$$ which, as user404188 noticed, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove: If n is a perfect square, then n+3 is not a perfect square Use either direct proof, proof by contrapositive, or proof by contradiction.
Using proof by contradiction method
Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)
There exists integers x and y such that $n = x^2$ and $n+3... | $$
3=y^2-x^2 \ge (x+1)^2-x^2= 2x+1
$$
which is false if $x>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+... | Using production rule and rule for derivative of complicated function:
$$(f\cdot g)' = f'\cdot g + f \cdot g' \\ f(\varphi(x))' = f' \cdot (\varphi(x))'$$
We have:
$$(2x-1)'\cdot(x+3)^{\frac{1}{2}}+(2x-1)\cdot((x+3)^{\frac{1}{2}})'\cdot(x+3)'$$
$$(2x-1)' = 2 \\ ((x+3)^{\frac{1}{2}})' = \frac{1}{2}\cdot(x+3)^{\frac{-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find General Solution of First Order DE.
Find general solution of
$$y'+\frac{2y}{x}=\frac{3}{x^2}$$
and find solution for which
$$y(2)=-1$$
I know
$$xy'+\frac{r(x)y}{x}=3r(x)x$$
$$(ry')=ry'+r'y$$$$r'=\frac{r}{x},\frac{r'}{r}=\frac{1}{x},ln(r)=ln(x),r=x$$$$\frac{xy'}{2}+y=\frac{3}{2x}$$
$$\frac{(xy)'}{2}=... | It is a non-homogeneous linear ODE and IVP. Find the general solution in two steps:
1) Solve the homogeneous equation:
$$y'+\frac{2y}{x}=0 \Rightarrow\frac{1}{y}dy=-\frac{2}{x}dx \Rightarrow y=\frac{C}{x^2}.$$
2) Assume $y=\frac{C(x)}{x^2}:$
$$y'+\frac{2y}{x}=\frac{3}{x^2} \Rightarrow \frac{C'(x)x^2-2xC(x)}{x^4}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve this Differential Equation in y and x.
Solve the following DE:
$$y'=\frac{y+y^2}{x+x^2}$$
in particular
$$y(2)=1$$$$\frac{y'}{y+y^2}=\frac{1}{x+x^2}$$$$\int \frac{dy}{y+y^2}=\int \frac{dx}{x+x^2}$$$$y'dx=dy$$$$ln(y)-ln(y+1)=ln(x)-ln(x+1)+C$$
but where to from here?
| This gives
\begin{align*}
\frac{y}{y+1} &=e^{C}\frac{x}{x+1}\\
2y&=\frac{e^Cx+x+1}{e^Cx-x-1}.
\end{align*}
Now use $y(2)=1$, to get $e^C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I find the term of a recursive sequence? I have $\{a_n\}$ the following sequence:
$a_1 =-1$
$a_k a_{k+1} = - a_k - \frac{1}{4}$
How can I find $a_1 a_2 \cdots a_n$?
| $a_{k+1} = -\frac{4a_k+1}{4a_{k}}$
$a_2 = -3/4$
$a_3 = -2/3=-4/6$
$a_4 = -5/8$
$a_5 = -3/5 = -6/10$
$a_6 = -7/12$
So, we prove that $a_n = -(n+1)/(2n)$
Base case: $a_1 = -1$
IH: $a_{k} = -(k+1)/(2k)$
Then, $a_{k+1} = -\frac{4a_k+1}{4a_{k}} = -\frac{-2(k+1)/k + 1}{-2(k+1)/k} = -\frac{-2k-2+k}{-2(k+1)} = -((k+1)+1)/(2(k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
For how many integers $(3n)^4-(n-10)^4$ is a perfect square?
For how many integers $n$ the expression
$$(3n)^4-(n-10)^4$$
becomes a perfect square?
$$$$One way is: Let $x=3n$ and $y=n-10$ to get $$x^4-y^4=z^2.$$ This equation does not have non-trivial solutions in integers (I do not want to discuss about this).... | $$(3n)^4-(n-10)^4=m^2\Rightarrow\begin{cases} 9n^2=x^2+y^2\\n^2-20n+100=x^2-y^2\text{ or}\space 2xy\end{cases}$$ WLG we can consider $n,x,y$ without common factors then $n$ cannot be even because the well known parametrization of Pythagorean triples we used above.
$$n\text{ odd}\Rightarrow 10n^2-20n+100=2x^2\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to pr... | This solution builds on the idea mentioned by @Bernard in one of the comments above.
Let $u=x+\dfrac 1x$.
$$y=\frac {x^2+1}{x^2+x+1}=\frac {x+\frac 1x}{x+1+\frac 1x}=\frac u{u+1}=\frac 1{\frac 1u+1}$$
For $x>0$, we have $u>0$, and $u_{\text{min}}=2$*.
*
*Hence $y_\text{min}=\dfrac 1{\frac 1{u_{\text{min}}}+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
how to enumerate and index partial permutations with repeats I know how to count permutations of, say, p red balls and q white balls and r blue balls. I also know how to count permutations of part of a set of distinct objects. But I don't know an efficient way to count partial permutations with repetition.
To illustr... | We can count the number of different sequences with the help of exponential generating functions.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we can calculate the wanted number of sequences as
\begin{align*}
12!&[z^{12}](1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\frac{z^8}{8!})(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Weird differential equation How to solve the given differential equation(given $ f(1)=e$ and $f(0)=0$, $f'(0)=1$,$f''(0)=0.$)
$$f''(x)=2xf'(x)+4f(x).$$
Where $f'(x)$ is first derivative and $f''(x)$ is second derivative.
I was trying to guess that the function is exponential(of form of $e^{x^2}$) but couldn't get the ... | Frobenius Method:
To solve $y''-2xy'-4y=0$ consider Frobenius solution $y = \sum_{n=0}^{\infty} a_n x^{n+r}$ hence $y' = \sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}$ and $y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}$ hence:
$$ \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}-2x\sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}-4\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find coefficient of $x^2$ in a complicated expansion
Find the coefficient of $x^2$ in the expansion of $(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$
I noticed that since we only care about the coefficient of $x^2$, only the coefficients of $x^2$ inside the square brackets, as well as the constant terms, will matter. From there... | Note that$(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$ is equal to
$$4[(1+2x)^6+6\cdot 3x^2+1+o(x^2)]-x^2\cdot (1+1+o(x))\\
=4[(1+6\cdot 2x+\binom{6}{2}(2x)^2+18x^2+1+o(x^2)]-2x^2+o(x^2)\\
=8+48x+310x^2+o(x^2)$$
where we used the Binomial Theorem and the little-o notation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Baby Rudin: Theorem 3.20 (c) and (d) In theorem 3.20, Rudin offers a proof to the limits of the following sequences:
$$u_n = n^{\frac{1}{n}}$$
$$v_n = \frac{n^{\alpha }}{(1+p)^n}$$ with $\alpha$ a real number and $p > 0$.
In both the proofs, Rudin uses inequalities which I am not familiar with. For $u_n$, the following... | For the first one, assuming $x\geq0$ :
$$
\begin{align}
(1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\
&=1+nx+\frac{n(n-1)}{2}x^2+\cdots+x^n \\
&\geq \frac{n(n-1)}{2}x^2
\end{align}
$$
For the second one :
$$
\begin{align}
\binom{n}{k}&= \frac{n(n-1)\cdots(n-k+1)}{k!} \\
&=\frac{1\left(1-\frac{1}{n}\right)\cdots\left(1-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$ I was trying to calculate the following limit:
$$\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$$
and, feeding it into WolframAlpha, I found the limit is $0$.
But; the function isn't defined from the left of $3$
My knowledge is that the function must be defined in an open interva... | First, note $x$ must be not less than $3$ for a root to be defined, and greater than $3$ for a denominator to be nonzero (hence a function defined). So any 'open interval' you may consider must be a subset of the domain, which is, at most, $(3,\infty)$.
Then:
$$\begin{align}
\frac{x^3-27}{\sqrt{x-3}} & = \frac{(x-3)(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find a basis for the subspace of $\mathbb R^4$ spanned by the 4 vectors. Find a basis for the subspace of $\mathbb R^4$ spanned by the 4 vectors.
$\alpha_1=(1,1,2,4),\alpha_2=(2,-1,-5,2),\alpha_3=(1,-1,-4,0),\alpha_4=(2,1,1,6)$
Check for linear dependence : $ c_1\alpha_1+c_2\alpha_2+c_3\alpha_3+c_4\alpha_4=0$
$\left[\b... | Operate row reduction on the transposed matrix, i.e. write the vectors as row vectors:
$$\begin{bmatrix}
1&1&2&4\\
2&-1&-5&2\\
1&-1&-4&0\\
2&1&1&6
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&1&2&4\\
0&-3&-9&-6\\
0&-2&-6&-4\\
0&-1&-3&-2
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&1&2&4\\
0& 1&3&2\\
0&-2&-6&-4\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two unit vectors $x,y \in \mathbb{R}^n$ satisfying $x.y = \frac{\sqrt{3}}{2}$ Two unit vectors $x$ and $y$ in $\mathbb{R}^n$ satisfy x.y= $\frac{\sqrt{3}}{2}$
Evaluate
A) angle between $x$ and $y$ in Radian.
I got $\frac{\pi}{6}$
B) $y.(3x-5y)$
Looked at solving for $x$ and $y$, but think am going down completely the... | Well, for the first question, we have - where $\phi$ is the angle between $x$ and $y$:
$$\cos\phi=\frac{x\cdot y}{|x||y|}=\frac{\sqrt{3}}{2}\Rightarrow\phi=\frac{\pi}{6}$$
since $\phi\in[0,\pi)$
For the second question, we have:
$$y\cdot(3x-5y)=y\cdot(3x)-5y\cdot y=3x\cdot y-5y^2=2\frac{\sqrt{3}}{2}-5|y|^2=3\frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Squares and Rationalization I was solving this question, and I'm hitting a wall.
If ${1\over{\sqrt{2011+\sqrt{2011^2-1}}}}=\sqrt{m}-\sqrt{n}$, where $m$ and $n$ are positive integers, what is the value of $m+n$?
I tried to solve this question with two approaches:
${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2... | Hint
$\dfrac1{\sqrt{2011+\sqrt{2011^2-1}}}$
$=\dfrac1{\sqrt{2011+2\sqrt{1005*1006}}}$
$=\dfrac1{\sqrt{\sqrt{1005}+\sqrt{1006}}^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not . Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not .
Answer:
Since $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})]=0 , \ \ and \ \ - 1\leq \sin(2x) \leq 1 $,... | Let us consider the following two sequences:
$$a_n=n\pi+\frac{\pi}{4}$$
$$b_n=n\pi+\frac{3\pi}{4}$$
and let $f(x)=\ln\left(1+\frac{1}{x}\right)+\sin(2x)$
So, we see that:
$$\begin{align*}f(a_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a cubic polynomial. If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting i... | Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Divisibility of $n\cdot2^n+1$ by 3. I want to examine and hopefully deduce a formula that generates all $n\geq0$ for which $n\cdot2^n+1$ is divisible by $3$. Let's assume that it is true for all $n$ and that there exist a natural number $k\geq0$ such that $$n\cdot2^n+1=3k,$$
Now I want to proceed with induction, but cl... | For any natural number $ n $ we have that $ n \equiv i $ mod $ 3 $ where $ i \in \{0,1,2\} $. So let us consider each case.
If $ n \equiv 0 $ mod $ 3 $ then clearly $ n2^{n}+1 \equiv 1 $ mod $ 3 $ so this is not a valid case.
Suppose now that $ n \equiv 1 $ mod $ 3 $ and write $ n=3k+1 $. Then $$ n2^{n}+1=(3k+1)2^{3k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
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.