Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? $$\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$$
Do you have an idea about this serie? If it converges what is the sum?
| $\frac{2n+1}{n(n+1)}=\frac{1}{n}+\frac{1}{n+1}$
So,
$\frac{3}{1.2}-\frac{5}{2.3} +\frac{7}{3.4}...= \frac{1}{1}+\frac{1}{2}-\frac{1}{2} -\frac{1}{3}+\frac{1}{3}+\frac{1}{4}....$
That should give you the idea.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determining Complex Differentiability using Cauchy-Riemann Equations I need to find where $f(x+iy)=-6(\cos x+i\sin x)+(2-2i)y^3+15(y^2+2y)$ is complex differentiable.
I first rearranged the function into its real and imaginary parts:
$f(x+iy)=(-6\cos x+2y^3+15y^2+30y)+i(-6\sin x-6y^2)$
That means $u(x,y)=-6\cos x+2y^3... | We can simplify these equations into
$$ \sin{x} = -3 y^2, \quad y^2 + 5y + 5 = \cos{x} $$
Now, $x$ and $y$ are real, so the only way that the first equation can be satisfied is if $y \in [-1/\sqrt{3}, 1/\sqrt{3}]$, since otherwise $-3y^2$ will not be in the range of sine.
Now we turn to the second equation. The graph o... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrating a Rational Function I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will simplify it enough to solve in reasonably quick . I plugged it into wolfram and as usual it... | From partial fractions and two variable substitutions
$$\begin{align}
\int \frac{x^3+x+2}{x^4+2x^2+1}dx &= \int\frac{x}{x^2+1}dx + 2\int\frac{1}{x^2+1}\frac{1}{x^2+1}dx \\
&= \frac{1}{2}\int\frac{du}{u}+\int\frac{1}{(\tan v)^2+1}dv
\end{align}$$
where $u=x^2+1$ and $v=\arctan x$ (recall that $\frac{d}{dx}\arctan x = \f... | {
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Eigenvalues and Jordan form I have a $5\times 5$ matrix and I need to find the Jordan form and its inverse. I know how to find the inverse. But for the Jordan form I am screwed.
The matrix is $$\begin{bmatrix}3 & 0 & 0 & 0 & 0\\2 & 3 & 0 & 0 & 0\\1 & 0 & 2 & 2 & 0\\0 & 0 & 0 & 3 & 2\\0 & 0 & 0 & 0 & 2\end{bmatrix}.$$... | Note that
$$
M=\left[\begin{array}{cc|ccc}
3 & 0 & 0 & 0 & 0\\
2 & 3 & 0 & 0 & 0\\
\hline
1 & 0 & 2 & 2 & 0\\
0 & 0 & 0 & 3 & 2\\
0 & 0 & 0 & 0 & 2
\end{array}\right]
=\begin{pmatrix}A&0\\C&D\end{pmatrix}
$$
is lower block triangular. Therefore the eigenvalues of $M$ are precisely the eigenvalues of $A$ and $D$. Yet $A... | {
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Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\... | Hint: Multiply the original inequality by $a+b+c$ on the LHS and $3$ on the RHS, expand and eliminate common terms and you will arrive at something you have proved.
| {
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How to solve this irrational equation? How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$
| Note that, with $x=1$, $y=\sqrt{8x+1}$ takes the value $y=3$ and with $x=3$, $y=\sqrt{8x+1}$ takes the value $y=5$. Put the point $A(1, 3)$ and $B(3,5)$. The equation of the line passing two points $A$ and $B$ is $y = x + 2.$ And then we write $$\sqrt{8x + 1}-(x+2).$$ Similar to with $$(x+1- \sqrt{6x - 2}).$$
We writ... | {
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How do I show that a matrix is injective? I need to determine whether this matrix is injective
\begin{pmatrix}
2 & 0 & 4\\
0 & 3 & 0\\
1 & 7 & 2
\end{pmatrix}
Using gaussian elimination, this is what I have done:
\begin{pmatrix}
2 & 0 & 4 &|& 0\\
0 & 3 & 0 &|& 0\\
1 & 7 & 2 &|& 0
\end{pmatrix}
Divide row1 by 2, and th... | Look at vectors of the form $\left(\begin{array}{c}2x\\ 0\\ -x\end{array}\right)$. What does your matrix do to these vectors? What does injective mean?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Integral $\int_0^\pi \cot(x/2)\sin(nx)\,dx$ It seems that $$\int_0^\pi \cot(x/2)\sin(nx)\,dx=\pi$$ for all positive integers $n$.
But I have trouble proving it. Anyone?
| Note that your question is same as proving
$$\int_{0}^{\frac \pi 2} \sin (2nx)\cot (x)\;\text{d}x = \frac \pi 2$$
Let
$$f_n=\int_{0}^{\frac \pi 2} \sin (2nx)\cot (x)\;\text{d}x$$
We will first show that $f_{n+1}-f_n=0$ for any $n\in \mathbb N$.
\begin{align*}
f_{n+1}-f_n&=\int_{0}^{\frac \pi 2}(\sin(2(n+1)x)-\sin(2nx))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/321919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Combination of logarithms and exponents I am given this question and told to solve for $a,b,c$:
$$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} = \frac{y^{3/2}\ln(y)}{3\ln(x)}$$
I tried to convert all the logarithms to $\ln$ and remove the $\frac{\ln(x)}{\ln(y)}$ term from both sides of the equation, but eventually I am st... | $$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} =\frac{y^{8a}}{y^c}\times\frac{x^b}{x^2}\times\frac{1}2\times\log_x (y^{8a})=y^{8a-c}\times x^{b-2}\times\frac{1}2\times\frac{\ln(y^{8a})}{\ln x}\\=y^{8a-c}\times x^{b-2}\times\frac{1}2\times \frac{8a\ln(y)}{\ln x}=y^{8a-c}\times x^{b-2}\times\frac{8a}2\times \frac{\ln y}{\ln... | {
"language": "en",
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"question_score": "3",
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Further clarification needed on proof invovling generating functions and partitions (or alternative proof)
Show with generating functions that every positive integer can be written as a unique sum of distinct powers of $2$.
There are 2 parts to the proof that I don't understand. I will point them out as I outline the... | The answer is binary: 0,1,10,11,100,101,110,111,1000,...
to prove every number has a unique binary expansion is easy. More generally every number has a unique base $b$ expansion: just iterate the division algorithm which uniquely splits any number $n$ into $n = m b + c$ with $0 \le c < b$.
The reason for using generat... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest $x$ for $9x \equiv 3 \pmod {23}$ $9x \equiv 3 \pmod {23}$
How to derive the smallest $x$. I understand I can use the extended euclidean algorithm for eg $19x = 1 \pmod {35}$.
However, I not too sure how to work on it when it is $3 \pmod {23}$.
I am able to reach the step of $1 = -5(9) + 2(23$) after... | We employ an algorithm that was discussed here.
$\alpha-$Solve:
$\;9x \equiv 3 \pmod{23}, \text{ and } \; 23 = 9 \cdot 2 + 5,\quad -3 + 1 \cdot 9 = 6$
$\alpha-$Solve:
$\;5x \equiv 6 \pmod{9}, \text{ and } \; 9 = 5 \cdot 1 + 4,\quad -6 + 2\cdot 5 = 4$
$\alpha-$Solve:
$\;4x \equiv 4 \pmod{5}, \text{ ANS: } \; x = 1 \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326364",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Limits $\sqrt{2n^2-1}/(n+1)$ and $1 = 0.9999...$ $1)$ Find the limit (if it exists) of the following sequence:
$$\frac{\sqrt{2n^2-1}}{n+1} = x_n$$
Attempt: Rewrite as $$\frac{\sqrt{n^2(2 - \frac{1}{n^2})}}{n+1} = \frac{n\sqrt{(2-\frac{1}{n^2})}}{n+1} = \frac{\sqrt{2 - \frac{1}{n^2}}}{\frac{1}{n} + 1}$$ So as $n \right... | Another way for second question.
Let $x=0.999999999999\dots $
$10x=9.999999999999\dots$
subtract both equations
$9x=9$
$x=1;$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
| $$ \frac1{x-y} + \frac1{x+y} = \frac{2x}{x^2-y^2}$$
Proceeding in this fashion we would be left with
$$ \text{The sum } = \frac{16x^{15}}{x^{16}-y^{16}} + \frac{16x^{15}}{x^{16}+y^{16}} = \frac{32x^{31}}{x^{32}-y^{32}} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\int^1_0\frac{x^b-x^a}{\ln x} d x=\ln{{\frac{1+b}{1+a}}}$ $$\int^1_0\frac{x^b-x^a}{\ln x} d x =\ln{{\frac{1+b}{1+a}}}$$ The part inside $\ln$ is absolute value. A solution including integration under the integral can be found here. Which used the identity $$\int^1_0x^{\alpha} dx =\frac{1}{1+\alpha}$$. M... | Define $$f(z) = \int^1_0 \frac{ x^{bz} - x^{az} }{\log x} dx.$$ Differentiating gives $$f'(z) = \int^1_0 \left(bx^{bz} - ax^{az} \right) dx = \frac{b}{bz+1} - \frac{a}{az+1}$$
and integrating this gives $$f(z) = \log(bz+1) - \log(az+1)+C$$ and setting $z=0$ indicates the constant is $0.$ So
$$\int^1_0 \frac{x^b-x^a}{\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How Can I Find The Value Of This Limit 10 find this limit:
$$\displaystyle\lim_{n\to+\infty}\left[\sum_{k=1}^{n}\left(\dfrac{1}{\sqrt{k}}- \int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt\right)-2\sqrt{n}\right]$$
| We have
$$\int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt=\frac{1}{\sqrt k}-\arctan\frac{1}{\sqrt k}$$
Now if we denote by
$$u_n=\sum_{k=1}^n\arctan\frac{1}{\sqrt k}-2\sqrt{n}$$
we have
$$u_n-u_{n-1}=\arctan\frac{1}{\sqrt n}-2\sqrt{n}+2\sqrt{n-1}\sim\frac{-7}{12n\sqrt{n}},$$
and since the series $\sum\frac{1}{n\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{n \to \infty}\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$ Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$
For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge t... | Note that, with the substitution $x \mapsto nx$, we have
$$ \frac{f(n)}{n} = \int_{0}^{\infty} \log^2 \left( \frac{x + 1}{x + n^{-2}} \right) \, dx. $$
Now, let us denote
$$ h_n (x) = \log^2 \left( \frac{x + 1}{x + n^{-2}} \right) \quad \text{and} \quad h(x) = \log^2 \left( \frac{x + 1}{x} \right). $$
Then we easily ob... | {
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Cauchy Integral Formula for Matrices How do I evaluate the Cauchy Integral Formula $f(A)=\frac{1}{2\pi i}\int\limits_Cf(z)(zI-A)^{-1}dz$ for a matrix $A=\left(\begin{array}{ccc}2&2&-5\\3&7&-15\\1&2&-4\end{array}\right)$ and a function $f(x)=3x^2+1$?
I have evaluated the function directly, using interpolation and using ... | Choose the contour $C$ so that $|z| > \|A\|$ for all $z$ in $C$. Then note that $(zI -A)^{-1} = \frac{1}{z}(I-\frac{A}{z})^{-1}$, and for $|z| > \|A\|$, we have $(zI -A)^{-1} = \frac{1}{z} \sum_{k=0}^\infty \frac{A^k}{z^k}$. Since $|z| > \|A\|$ for all $z$ in $C$, the convergence is uniform so we may interchange integr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding polynomial given the remainders Question: Find a polynomial $f(x) \in \mathbb{Q} (x)$ of minimal degree that has both the following properties:
When $f(x)$ is divided by $(x-1)^2$, the remainder is $2x$; and when $f(x)$ is divided by $(x-2)^3$, the remainder is $3x$.
Answer provided: $f(x)=(x-2)^3 \cdot (4x-3)+... | An alternative:
Note that if $f(x)$ satisfies the 2 properties, so does $f(x)-q(x-1)^2(x-2)^3, q \in \mathbb{Q}$. Thus the polynomial of minimal degree has degree 4. We have $f(x)=(x-2)^3(ax+b)+3x$. Note that $1$ is a repeated root of $f(x)-2x$, so $f(1)=f'(1)=2$.
$2=f(1)=(-1)^3(a+b)+3$ so $a+b=1$. $f'(x)=3(x-2)^2(ax+... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding the coefficient of $ x^n $ in the expansion of $ { ({\ log_e (1+x) })^2 } $ I've been trying to find the coefficient of $x^n$ in the expansion of $ { ({\log_e (1+x) })^2 } $.I wrote out the expansion of $ { ({\log_e (1+x) })^2 } $ explicitly and tried to generalize the terms involving $x^n$, but...so far no luc... | If $$y=\{\log(1+x)\}^2$$
So, $$\frac{dy}{dx}=2\frac{\log(1+x)}{(1+x)}=2 (1+x)^{-1} \log(1+x)$$
$$=2\left(1+\frac{x(-1)}{1!}+\frac{x^2(-1)(-2)}{2!}+\frac{x^3(-1)(-2)(-3)}{3!}+\cdots\right)$$
$$\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots\right)$$
$$=2 (1-x+x^2-x^3+\cdots )\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$x^4+y^4=2z^2$ has only solution, $x=y=z=1$ . How do I verify that the only solution in relatively prime positive integers of the equation $x^4+y^4=2z^2$ is $x=y=z=1$?
| Lemma I: If $(a,b,c)$ satisfies $a^2+b^2=c^2$, then $ab/2$ is not a square, nor twice a square.
Proof: By Euclid's formula, we can write $a=2pq$, $b=p^2-q^2$ $c=p^2+q^2$, so that $ab/2=pq(p+q)(p-q)$. If this is a square, then, as the four factors are pairwise coprime, there are $x,y,u,v$ such that $p=x^2$, $q=y^2$, $... | {
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Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$?
$Approach$:
$N$=$11^2$.$13^4$.$17^6$
$N^2$=$11^4$.$13^8$.$17^{12}$
This means $N... | If a divisor $d$ of $N^2$ is smaller than $N$ then $\frac{N^2}{d}$ is a divisor of $N^2$ larger than $N$.
Let's say that there are $n$ divisors of $N^2$ that are smaller than $N$. Then there are $n$ divisors of $N^2$ that are larger than $N$.
In total there are 585 divisors of $N^2$ and each one is either smaller, la... | {
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"url": "https://math.stackexchange.com/questions/340533",
"timestamp": "2023-03-29T00:00:00",
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Solving a system of congruences: CRT If $x \equiv 3 \pmod 6$ and $x \equiv 5\pmod 8$ then how is this $x \equiv 21\pmod {24}$?
I understand $24$ is $\operatorname{lcm}(6,8)$ but how to get from $3$ and $5$ to $21$?
This is is a Chinese Remainder Theorem problem.
| $x\equiv3\pmod 6\implies x\equiv0\pmod 3$ and $x\equiv 1\pmod 2$
Again, $x\equiv5\pmod 8\implies x\equiv1\pmod 2$
So, it suffices to use CRT with $x\equiv0\pmod 3$ and $x\equiv5\pmod 8$ which is legitimate as $(3,8)=1$
Alternatively,
So, $x$ can be written as $6a+3$ where $a$ is any integer
Similarly, $x=8b+5$ for ... | {
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Evaluate $\sum_{n=1}^{\infty}\frac{1}{n^3+3 n^2+2 n}$ Summing this series from $0$ to $\infty$, the result is $\frac{1}{4}$. I tried a lot, but I could not get this result. I think it´s wrong.
Can anybody help me?
| $\frac{1}{n(n+1)(n+2)} = \frac{1}{2(n+1)}(\frac{1}{n}-\frac{1}{n+2}) = $
$ = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}$
So the sum equals: $\displaystyle{\frac{1}{2\times1\times(1+1)} - \lim_{n\rightarrow +\infty}{\frac{1}{2n(n+1)}}} = \frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/341650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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Help with evaluating a sum I am trying to evaluate the following sum:
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n}$$
So far I have written the sum as
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n} = \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \frac{1}{5^n} = \sum_{n=1}^{\infty} \frac{1}{n5^n} - \sum_{n=1... | $$\begin{align}
\sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=1}^\infty \frac{1}{(n+1)5^n} &= \sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=2}^\infty \frac{1}{n5^{n-1}} \\
&=\sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=1}^\infty \frac{1}{n5^{n-1}}+1\\
&=1+\sum_{n=1}^\infty \left(1-5\right)\frac{1}{n5^n}\\
&=1-4\sum_{n=1}^\inft... | {
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"timestamp": "2023-03-29T00:00:00",
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How can prove this inequality(8) Let $a,b,c,x,y,z >0$ and $A=a^2+b^2+c^2,\ B=x^2+y^2+z^2,\ C=ax+by+cz$. By Cauchy-Schwarz inequality, we always have $C^2\le AB$. If $C^2<AB$, prove that
$$
\frac{A}{C+\sqrt{2(AB-C^2)}}<\frac{a+b+c}{x+y+z}<\frac{C+\sqrt{2(AB-C^2)}}{B}.
$$
I created this inequality. Are there any nice pro... | It seems the following.
The inequality is not strict. For instance, put $a=b=c=x=1$, $y=z=0$. Then
$A=3$, $B=C=1$, and $C^2<AB$. But
$$\frac{a+b+c}{x+y+z}=3=\frac{C+\sqrt{2(AB-C^2)}}{B}.$$
The non-strict inequality can be easily proved by means of spherical trigonometry.
Consider the following vectors of unit length in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Laurent series expansions, complex variables Given $f(z)=\frac{1}{z^2(1-z)}$ I am to find two Laurent series expansions. There are two singularities, $z=0$ and $z=1$. So for the first expansion, I used the region $0<|z|<1$ and I got $\sum_{n=0}^\infty z^n+\frac{1}{z}+\frac{1}{z^2}$. The second expansion is for the regi... | The problem is that for $|z|>1$ the absolute value of $z$ is greater than $1$ (ok that sound ridicolus but it is really the problem) as the geometric sequence wont work in the given form.
We just change it a bit.
$$\frac{1}{z^2(1-z)} =\frac{1}{z^2} \cdot \frac{1}{1-z} = \frac{1}{z^2} \cdot \frac{1}{\frac{z}{z}(1-z)}=\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $1Let $a,b>0$, $a\neq b$, $n\in\mathbb N$, $n>1$, and $a^n-b^n=a^{n+1}-b^{n+1}$.
How can we prove that
$1<a+b<\dfrac{2n}{n+1}$
Thank you everyone. I have proven it. My method:
$a+b-1=a+b-\dfrac{a^{n+1}-b^{n+1}}{a^n-b^n}=\dfrac{ab(a^{n-1}-b^{n-1})}{a^n-b^n}>0$
on the other hand
$$a+b-\dfrac{2n}{n+1}=a+b-\d... | Firstly, a point of caution. In your solution, you had a step which went:
\begin{align}
& \frac{(n+1)ab(a^{n-1}-b^{n-1})-(n-1)(a^{n+1}-b^{n+1})}{(n+1)(a^n-b^n)}<0 \\
& \Longleftrightarrow ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1})
\end{align}
Note that in order to do this, you must first WLOG assume $a>b$, so that... | {
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Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$ How to simplify the expression:
$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$
If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.
Thank you.
| Put
$$x_0:=0,\quad x_1:=\sqrt{5},\quad x_2:=\sqrt{5+2\sqrt{5}},\quad x_3:=\sqrt{5+2\sqrt{5+2\sqrt{5}}}\ ,$$
and so on, which amounts to
$$x_0:=0,\qquad x_{n+1}:=\sqrt{5+2x_n}\quad(n\geq0)\ .$$
Then
$$x_{n+1}-x_n=\sqrt{5+2x_n}-\sqrt{5+2x_{n-1}}={2(x_n-x_{n-1}) \over \sqrt{5+2x_n}+\sqrt{5+2x_{n-1}}}\ .$$
As $x_1-x_0>0$ t... | {
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Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.
I know that for proving the $\gcd(a,b) = c$ you need to prove
*
*$c|a$ and $c|b$
*$c$ is the greatest number that divides $a$ and $b$
Number 2 is what I'm struggling with. Does anybody have any ideas?
| Are you allowed to use Fermat's Little theorem?
$5^{\phi(14)} = 5^6 \equiv 1 \mod 14$
$5^{98} \equiv 5^{96}5^2 \equiv 25 \equiv -3 \mod 14$ so $14|5^{98} + 3$
and $5^{99} \equiv -3*5 \equiv -15 \equiv -1 \mod 14$ so $14|5^{99} + 1$
So $14$ is a common divisor.
Hmm, but how to show it is the greatest common divisor?
Wel... | {
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What is the remainder when $25^{889}$ is divided by 99? What is the remainder when $25^{889}$ is divided by 99 ?
$25^3$ divided by $99$ gives $26$ as a remainder.
$25*(25^3)$ divided by $99$ gives (remainder when $25*26$ is divided by $99$) as a remainder.
i.e. $25*(25^3)$ divided by $99$ gives $56$ as a remainder.
$(2... | Euler’s Number of $99$
= $99.\frac{2}{3}.\frac{10}{11}$
= $60$
From Fermat’s Theorem we know
$25^{60 × K} \mod {99} = 1$ (where K is any natural number)
Note that $25$ and $99$ are co-primes (∵ they don’t have any common factors other than $1$)
Putting $K =15$, we have, $25^{900} \mod {99} = 1$
Let’s assume $25^{899} \... | {
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How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3}... | Yes, This problem in 1963 IMO.http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908
| {
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"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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Series $\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z} \right)$ If $z$ is an integer, the sum of the series
$$\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)$$ is easy since it is a telescoping series. But if $z$ is a fraction, say $z=3/2$, I don't see why the series sums to $$\frac{8}{3}-\ln 4$$
Is there... | For $\vert x \vert < 1$, if we define $f(x)$ as$$f(x)=\sum_{n=0}^{\infty} \left(x^n - x^{n+z}\right)$$ then $$\int_0^1 f(x) dx = \sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right)$$
But we have $$f(x) = (1-x^z) \sum_{n=0}^{\infty} x^n = \dfrac{1-x^z}{1-x}$$
Hence,
$$\sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Interpret the equation $17+28y+4x^2+4y^2=8x$ geometrically
Interpret the equation geometrically:
$$17+28y+4x^2+4y^2=8x$$
I have drawn the bend and now I got the expression $(y-2)=(1-x)^2$ but that is the wrong expression. What should it be?
| Bring all terms onto one side. Club all the 'x' terms together and all the 'y' terms together.
You get:
$(4x^2 - 8x) + (4y^2 + 28y) + 17 = 0$
Simplify the above equation as :
$(4x^2 - 8x + 4) + (4y^2 + 28y + 49) + 13-49 = 0$
$4{(x-1)}^2 + 4{(y+7/2)}^2 = 36$
${(x-1)}^2 + {(y+7/2)}^2 = 3^2$
Hence it is a circle with cent... | {
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Evaluate $\int_{0}^{1000} \frac{e^{-10x}\sin x}{x} \text{d}x$ to within $\pm 10^{-5}.$ Evaluate $$\displaystyle \int_{0}^{1000} \frac{e^{-10x}\sin x}{x} \text{d}x$$ to within $\pm 10^{-5}$.
| Thinking more about this, one don't need to approximate the integral by sending the upper limit to $\infty$ nor know how to evaluate the integral over $[0,\infty)$.
For $x > 0$, $|\frac{\sin x}{x}| < 1$ and $e^{-x}$ drops off to $0$ very quickly.
If we cutoff the integral at $1$ instead of $1000$. The error:
$$\left|\i... | {
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Laplace inverse transform of the complex expression I have an expression below which I need to do the laplace transform
Any help is highly appreicated.
The expression is :
$$
\frac{\exp\left(\frac{x}{2}\sqrt{(U/D)^2+4s/D}\right)}{s\sqrt{(U/D)^2+4s/D}}
$$
The integral of Ron Gordon's expression:
The first expression ... | I will reduce this to an integral over the real line. Consider the integral in the complex plane:
$$\displaystyle \oint_C \frac{dz}{z} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}} e^{t z}$$
where $C$ is the following contou... | {
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Residue Formula application Using the Residue formula, I've been trying to prove $$\int_0^{2\pi}\frac{1}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta=\frac{2\pi}{ab},\quad\quad a,b\in\Bbb R.$$First, it seems like the formula should be wrong (unless perhaps we assume $a,b\in\Bbb R^+$) since the right-hand side can be negat... | You can use trigonometric substitution to solve this problem.
Suppose $a<b$. Let $k=\frac{b}{a},t=\tan\frac{\theta}{2}$. Then $k>1$ and
$$ \sin\theta=\frac{2t}{t^2+1}, \cos\theta=\frac{t^2-1}{t^2+1},d\theta=\frac{2t}{t^2+1}dt $$
and hence
\begin{eqnarray*}
\int_0^{2\pi}\frac{1}{a^2\cos^2\theta+b^2\sin^2\theta}d\theta&... | {
"language": "en",
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"source": "stackexchange",
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Determine the definite limit The following limit
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
evaluates to 1/12.
This is my progress so far:
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
$$\lim_{x\to 1}\frac{1 + \sqrt{x}}{2(1 - x)} - \frac{1 + \sqrt[3]{x} + \sq... | OK, without l'Hôpital... do $x \mapsto u^6$:
$$
\begin{align*}
\lim_{x \to 1} \frac{1}{2 (1 - \sqrt{x})} - \frac{1}{3 (1 - \sqrt[3]{x})}
&= \lim_{u \to 1} \frac{1}{2 (1 - u^3)} - \frac{1}{3 (1 - u^2)} \\
&= \lim_{u \to 1}
\frac{3 (1 - u^2) - 2 (1 - u^3)}{6 (1 - u^2)(1 - u^3)} \\
&= \lim_{u \to 1}
\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you simplify $\tan 10A$ in terms of $5A$? How do you simplify $\tan 10A$ in terms of $5A$?
I just need a few steps to get me going. All help is appreciated.
Thanks!
| Using de Moivre's formula,
$$\cos 10x+i\sin 10x=(\cos x+i\sin x)^{10}$$
$$=\sum_{0\le r\le 10}\binom {10}r(\cos x)^{10-r}(i\sin x)^r$$
$$=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(i\sin x)^{2s}+\binom {10}{2s+1}(\cos x)^{2n-2s-1}(i\sin x)^{2s+1}$$
$$=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(\si... | {
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Find all values x, y and z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes. Find all positive integers x, y, z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes.
It seems trivial that the only... | Suppose without loss of generality $x \ge y$. Then $z^2+1 \le (x^2+1)^2$, so $z < x^2+1$.
Note that $x^2+1 \mid (z^2+1)-(x^2+1)=(z-x)(z+x)$.
Since $(x^2+1)$ is a prime number and
$$0 < z-x < z+x < x^2+x+1<2(x^2+1),$$
we must have $x+z = x^2+1$ and $z-x=1$, from which we obtain $x=2$, and then $z=3, y=1$.
| {
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Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Hint: Follow the example in lectures to show that $2^n -1$ is not divisible by 3.
In lectures, the example showed the $3 \mid 2^n -1 \iff n$ is even. So wha... | In fact we can prove more general result.
This holds true any number of the form $3k\pm1$ where $k$ is an integer
because $2^3=8\equiv1\pmod 7\implies 2^{3k}\equiv1$
$\implies 2^{3k+1}\equiv2\pmod 7\not\equiv1$
and $2^{3k-1}=2^{3(k-1)}\cdot2^2\equiv4\pmod 7\not\equiv1$
We know any prime $>3$ can be written as $6r\pm1$... | {
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"timestamp": "2023-03-29T00:00:00",
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Factorize $a^2-ab-bc\pm c^2$ I got this question in a test but it did not specify the variable with respect to which I was supposed to factorize
$$a^2-ab-bc\pm c^2$$
where it could be just $a(a-b)-c(b\pm c)$ but no common factor over all terms. I feel I may be missing something. The $\pm$ is there because I cannot reme... | We observe that
\begin{eqnarray}
a^2-ab-bc+c^2&=&a^2+c^2-b(a+c)\\
&=&(a+c)^2-b(a+c)-2ac\\
&=&(a+c)^2-b(a+c)+\frac{b^2}{4}-\frac{b^2+8ac}{4}\\
&=&\left(a+c-\frac{b}{2}\right)^2-\frac{b^2+8ac}{4}.
\end{eqnarray}
Hence, if $b^2+8ac\geq 0$ then
$$
a^2-ab-bc+c^2=\left(a+c-\frac{b}{2}+\sqrt{\frac{b^2+8ac}{4}}\right)\left(a+c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$
| $\dfrac{S^{2}}{s}=(s-a)(s-b)(s-c)\leq \left(\dfrac{(s-a)+(s-b)+(s-c)}{3}\right)^{3}=\dfrac{s^{3}}{27}$
$\therefore$ $S\leq\dfrac{s^{2}}{3\sqrt{3}}=\dfrac{(a+b+c)^{2}}{12\sqrt{3}}\leq\dfrac{1}{12\sqrt{3}}\cdot 3(a^{2}+b^{2}+c^{2})$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Summation of sequence $a_n - a_{n-1} = 2n$ $(a_1,a_2,a_3,..)$ be a sequence such that $a_1$ =2 and $a_n - a_{n-1} = 2n$ for $n \geq 2$. Then $a_1 + a_2 + .. + a_{20}$ is equal to?
$a_1$ = 2
$a_2$ = 2 + 2x2
$a_3$ = 6 + 2x3
$a_4$ = 12 + 2x4
$a_5$ = 20 + 2x5
$a_n$ = $b_n$ + $g_n$ ,here $g_n$=2xn
$b_3$-$b_2$=4
$b_4$-$b_... | As $a_n-a_{n-1}=2n, a_n$ can be at most quadratic.
Let $a_n=An^2+Bn+C$ where $A,B,C$ are arbitrary constants
So, $2n=a_n-a_{n-1}=A(2n-1)+B=2An+B-A$
Comparing the coefficients of $n,2A=2, A=1$
Comparing the constants $B-A=0\implies B=A=1$
So, $a_n= n^2+n +C$
$2=a_1=2+C\implies C=0\implies a_n= n^2+n $
So, $$\sum_{1\le r... | {
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Elementary Diophantine equation Solve $(x+y)(xy+1)=2^z$ in positive integres. My attempts is to use $x+y=2^a$, $xy=2^b-1$ and therefore $x,y$ are the roots of the quadratic equation $w^2-2^aw+2^b-1=0$. I try to analyze its dicriminant but it seems to be a dead end...
| ETA: Jyrki's answer and comments already says this it turns out, and so he really should get the bounty.
Let $x$ and $y$ be positive integers satisfying $x+y\ge 16$. Then $x+y$ must be a power of $2$. Write then, $x+y=2^a$ for some $a \ge 4$. Then $xy+1$ must also be a power of $2$, and as the strict inequality $xy+1 \... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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"answer_id": 2
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In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
| You are looking for number of partitions of 12 in parts greater than 1. Because their number is not so big we can present all of them in a list bellow
$$\begin{array}{c|c}
2+2+2+2+2+2 & 1 \\
2+2+2+2+4 & 5 \\
2+2+2+3+3 & 10 \\
2+2+2+6 & 4 \\
2+2+3+5 & 12 \\
2+2+4+4 & 6 \\
2+3+3+4 & 12 \\
3+3+3+3 & 1 \\
2+2+8 & 3 \\
2... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$ The question is basically in the title: Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$
I get how to do it from $7\mid $ and $7\mid y$ to $7\mid x^2+y^2$, but not the other way around.
Help is appreciated! Thanks.
| $x^2,y^2$ can be $0^2\equiv0, (\pm1)^2\equiv1,(\pm2)^2\equiv4, (\pm3)^2\equiv2\pmod 7$
Observe that for no combination except $0,0$ of $x^2+y^2 \equiv0\pmod 7$
Alternatively,
If $(7,xy)=1, x^2+y^2\equiv0\pmod 7\implies \left(\frac xy\right)^2\equiv-1\pmod 7$
But we know $-1$ is a Quadratic residue $\pmod p$ iff pri... | {
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solution set for congruence $x^2 \equiv 1 \mod m$ if $m$ is an integer greater than 2, and a primitive root modulo $m$ exists, prove that the only incongruent solutions of $x^2 \equiv 1 \mod m$ are $x \equiv \pm 1 \mod m$.
I know that if a primitive root mod $m$ exists, then $m = 1, 2, 4, p^m,$ or $2p^m$, where p is an... | $1,2,4$ can be dealt easily.
For $p^m$ divides $(x^2-1)=(x-1)(x+1)$
Now, $x+1-(x-1)=2\implies (x-1,x+1)$ divides $2$
So, either $p^m$ divides $(x-1)$ or $(x+1)$ resulting in exactly $2$ in-congruent solutions
If $2\cdot p^m$ divides $(x^2-1)=(x-1)(x+1)$
Clearly, $x$ is odd
So, $p^m$ divides $\frac{x-1}2\cdot\frac{x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sums that are pythagorean and normal I noticed that
$3^2+4^2+15^2=9^2+13^2$
and also
$3+4+15=9+13$
Is there an easy way to find all pairs of disjoint sets of positive integers whose sum are the same and whose sum of squares are the same? How common are they?
| It is easy to generate two such sets of arbitrary length.
Let us choose $s_1,s_2,\ldots,s_{n-1}$ and $t_1,t_2,\ldots,t_{m-1}$ fairly freely and write
$$
\begin{array}{cc}
\sigma_1 = \sum_{i=1}^{n-1} s_i &~& \sigma_2 = \sum_{i=1}^{n-1} s_i^2 \\
\tau_1 = \sum_{i=1}^{m-1} t_i &~& \tau_2 = \sum_{i=1}^{m-1} t_i^2 \\
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding the Matrix Power of a matrix and limit Find the matrix power, $A^k$, of
$$A=\begin{pmatrix}a & 1-a \\ b & 1-b\end{pmatrix}$$
$$D=P^{-1}AP$$
$$A^k=PD^kP^{-1}$$
I think that
$$P=\begin{pmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{pmatrix} \ \ \ \text{and}\ \ \ \ P^{-1}=\frac{b}{1+b-a}\begin{pmatrix}1&\frac{1-a}{b}\\... | A = $\begin{bmatrix}1-a \\ b & 1-b\end{bmatrix}$
Finding the eigenvalues and eigenvectors and writing the matrix in Jordan Normal Form yields:
$\displaystyle A = \begin{bmatrix}a & 1-a \\ b & 1-b\end{bmatrix} = P.D.P^{-1} = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & a-b\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integer solutions of $n^3 = p^2 - p - 1$ Find all integer solutions of the equation, $n^3 = p^2 - p - 1$, where p is prime.
| Rewrite the expression as $p(p-1)=(n+1)(n^2-n+1)$. Case 1: $p|n+1$. It follows that $p\leq n+1$. But then also $p-1 \geq n^2-n+1$, and from these we get $n \geq n^2-n+1$ which can only be true for $n=1$ which gives $p=2$.
Case 2: $p|n^2-n+1$. We write $\frac{n^2-n+1}{p}=k$. Plugging this into the equation we get $p=kn+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/370425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$
Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$
No idea if I'm on the right track. May... | This may be solved by complex variables. Put $z = e^{i\theta}$ to obtain
$$\int_0^{2\pi} \frac{d\theta}{1-a\cos\theta+a^2} =
\int_{|z|=1} \frac{1}{iz} \frac{dz}{1-a/2(z+1/z)+a^2} =
\int_{|z|=1} \frac{1}{iz} \frac{z dz}{z-a/2(z^2+1)+za^2} \\=
-i \int_{|z|=1} \frac{dz}{z-a/2(z^2+1)+za^2}$$
The two poles (use e.g. the qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/370711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum value of $T=\frac{2}{3}(\cos 2A-\cos 2B)-\tan\frac{C}{2}$ Let $ABC$ be a triangle. Find the maximum value of
$$T=\frac{2}{3}(\cos 2A-\cos 2B)-\tan\frac{C}{2}$$
Please give me some hints. I don't know where to start
Thanks
| Hint.
$$T=-\frac43\sin(A+B)\sin(A-B)-\frac{1}{\tan\frac{A+B}{2}}.$$
If $A+B\ge \pi/2$,
$$T\le \frac43\sin(A+B)-\frac{1}{\tan\frac{A+B}{2}},$$
where equality holds for $A-B=-\pi/2$.
If $A+B<\pi/2$,
$$T\le \frac43\sin^2(A+B)-\frac{1}{\tan\frac{A+B}{2}},$$
where equality holds for $A=0$.
Note that if you set $t=\tan\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show abelian groups of order 3240? Show how to get all abelian groups of order $2^3 \cdot 3^4 \cdot 5$.
I just started learning this and was wondering how you would do this?
Is this correct?
$2^3 \cdot 3^4 \cdot 5 = 3240$. Therefore the number of abelian groups of order $3240$ is $3 \cdot 4 = 12$.
Is this the entire pr... | You are implicitly using the Fundamental Theorem of Finite Abelian Groups. You should cite this by name.
Of order $8=2^3$, you can have $\mathbb{Z}_8$ or $\mathbb{Z}_2\times \mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$. Similar options exist for the $3^4$.
To fully explain, I would write ou... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Differential Equation: Complex Eigenvalue For the following system
$$ x'=\left( \begin{array}{ccc}
\frac{-1}{2} & 1 \\
-1 & \frac{-1}{2} \end{array} \right)x $$
To find a fundamental set of solutions, we assume that $$ x = Ee^{rt}$$
According to the solution, this set of linear algebraic equations is obtained [1]:
$$ ... | We are given:
$$ x'=\left( \begin{array}{ccc}
\frac{-1}{2} & 1 \\
-1 & \frac{-1}{2} \end{array} \right)x $$
To find the characteristic polynomial, we solve:
$$|A - \lambda I| = 0$$
So, for your matrix, we would write:
$$ \left| \begin{array}{ccc}
\frac{-1}{2} - \lambda & 1 \\
-1 & \frac{-1}{2} - \lambda \end{array} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this determinant bounded? Let $D_n$ be the determinant of the $n-1$ by $n-1$ matrix such that the main diagonal entries are $3,4,5,\cdots,n+1$ and other entries being $1$. i.e.
$$D_n= \det \begin{pmatrix}
3&1&1&\cdots&1\\
1&4&1&\cdots&1\\
1&1&5&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots &n+1
\end{p... | Since adding a multiple of one row to another doesn't change a matrix's determinant, we can subtract the first row from each of the other rows to see that
$$D_n= \det \begin{pmatrix}
3&1&1&\cdots&1\\
-2&3&0&\cdots&0\\
-2&0&4&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-2&0&0&\cdots &n
\end{pmatrix}.$$
Then we subtr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\... | $y=xa^x$
$$y'=x'a^x+x(a^x)'=a^x+xa^x\ln a=a^x(1+x\ln a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/376642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Is $\frac{\lfloor{x}\rfloor+1}{2} \le \lfloor\frac{x}{2}\rfloor + 1$ The answer seems to be yes.
Here's my reasoning:
For $x < 1$, $\frac{1}{2} < 1$
For $1 < x < 2$, $1 = 1$
For $2 \mid x$, $\frac{x}{2} + \frac{1}{2} < \frac{x}{2} + 1$
For $2 \mid x-1$, $\frac{x+1}{2} = \frac{x-1}{2} + 1$
For all other values, the valu... | Set $n = \lfloor \frac{x}{2} \rfloor$. Then $x< 2n + 2$ and thus $\lfloor x \rfloor < 2n+2$ which implies $\lfloor x \rfloor \leq 2n+1$ as $\lfloor x \rfloor$ is an integer. This gives
$$\frac{\lfloor x \rfloor + 1}{2} \leq \frac{2n+2}{2} = n+1 = \lfloor \frac{x}{2} \rfloor + 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding normalised eigenvectors... I'm trying to find the eigenvector/eigenvalues of the $2\times2$ matrix:
\begin{pmatrix}4 & 2 \\ 2 & 3 \end{pmatrix}
This is my work:
$$\det(A-\lambda I) = \lambda^2-7 \lambda+8=0 \iff \lambda=\frac{7+\sqrt{17}}{2} \ \lor \ \lambda= \frac{7-\sqrt{17}}{2}$$
$x_1$ (eigenvector)=\begin{p... | If $\mathbf{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\mathbf{x}=\lambda\mathbf{x}$ and $(A-\lambda I)\mathbf{x}=\mathbf{0}$.
First, find the eigenvector corresponding to the eigenvalue $λ=\frac{7+\sqrt{17}}{2}$:
$$\begin{align*} &\quad\quad\quad\quad\left(\begin{array}{c|c}
A-\lambda I & 0
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve a cubic polynomial? I've been having trouble with this question:
Solve the equation, $$5x^3 - 24x^2 + 9x + 54 = 0$$ given that two of its roots are equal.
I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: $a$, $a$, $b$, but I am still unsuccessful.
Any help would be g... | By Vieta's formula, $$\begin{cases}2a+b=\dfrac{24}5\\a^2+2ab=a(a+2b)=\dfrac95\\a^2b=-\dfrac{54}5 \end{cases}$$
Notice that $0$ is not a root of the original equation, so both $a,b$ are nonzero. Divide the second equation by the third equation to get $\dfrac{a+2b}{ab}=\dfrac2a
+\dfrac1b=-\dfrac1{6}$. Multiplying by the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$, $a_n\in \mathbb Q$ and not $a_n$ not monotonic How can I construct a sequence $\{a_n\}$ of positive rational numbers, which is not monotonic such that $$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$$
I thought a lot on this question but I always come up with a monotonic sequence. Any help ... | From Madhava formula, we have
$$\pi = 4 - \dfrac43 + \dfrac45 - \dfrac47 \pm = \sum_{n=0}^{\infty}\dfrac{(-1)^n 4}{(2n+1)}$$
We can write $\dfrac1{2n+1} = \dfrac{k\sqrt{n}+1}{2n+1} - \dfrac{k\sqrt{n}}{2n+1}$. Hence, we have
$$\pi = \sum_{n=0}^{\infty} \left( (-1)^n\dfrac{\sqrt{n}+4}{2n+1} + (-1)^{n+1}\dfrac{\sqrt{n}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$ Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$
| This uses the hints by Raymond Manzoni and Cameron Williams. I am not sure if the answer is correct, but this should give an idea of how to proceed. I believe the answer by Raymond Manzoni is correct.
$$ \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)} = \dfrac{2^n(2^{n+1})!\sqrt{\pi}}{4^{2^n}(2^n)!^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 5,
"answer_id": 2
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Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$
This should be fairly straightforward but the proof seems to be alluding me.
I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perh... | You can use the Taylor series of $\sin(x)$ about $x=0$:
$$\sin(x) = \sum_{n=0}^\infty {\frac {(-1)^n x^{2n+1}} {(2n+1)!}}$$
The first few terms are:
$$\sin(x) = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots$$
Thus, set:
$$x - \frac {x^3}{3!} < \sin x\\
x - \frac {x^3}{3!} < x - \frac {x^3} {3!} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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The unique root on (1,2) For any given $n\geq 2$, let $x^n=\sum\limits_{k=0}^{n-1}x^{k}$ be the equation, prove: there is only one real root which in (1,2).
| We have $x^n=\frac{1-x^n}{1-x}$, which rearranges to $0=\frac{1-x^n}{1-x}-x^n=\frac{1-x^n-x^n+x^{n+1}}{1-x}$. For this to be zero (on $x>1$), $f(x)=x^{n+1}-2x^n+1$ must be zero. We now have the following facts:
*
*$f(2)=2^{n+1}-2^{n+1}+1=1>0$.
*$f(1)=1-2+1=0$.
*$f'(x)=(n+1)x^{n}-2nx^{n-1}=0$ has two solutions, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find max of $x^7+y^7+z^7$
Find max of $x^7+y^7+z^7$ where $x+y+z=0$ and $x^2+y^2+z^2=1$
I tried to use the inequality:$$\sqrt[8]{\frac {x^8+y^8+z^8} 3}\ge\sqrt[7]{\frac {x^7+y^7+z^7} 3}$$ but stuck
| The condition gives $x^2+xy+y^2=\frac{1}{2}$.
Also $$x^7+y^7+z^7=x^7+y^7-(x+y)^7=-7xy(x+y)(x^2+xy+y^2)^2=$$
$$=-\frac{7}{4}xy(x+y)\leq\frac{7}{4}\sqrt{\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}}=\frac{7}{8\sqrt2}\sqrt{\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}}\leq$$
$$\leq\frac{7}{8\sqrt2}\sqrt{\frac{4}{27}}=\frac{7}{12\sqrt6}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculation of ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$ (1) Total no. of integer ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$
(2) Total no. of integer ordered pair $(x,y,z)$ in $x+yz = 1\;\;,y+zx = 1\;\;,z+xy = 1$
My Try:: (1) Clearly $ x = 0,y = 0,z = 0$ are the solution of given eq... | Hint to (2): subtract 2nd equation out 1st:
$$(x-y)(1-z)=0.\ \ \ (1)$$
Addition:
Similarly
$$(x-z)(1-y)=0,\ \ \ (2)$$
$$(z-y)(1-x)=0.\ \ \ (3)$$
If $z=1$, then $x+y=1, xy=0$ (from your $1$st and $3$rd equations) whence we obtain two solutions
$$z=1, x=1, y=0,$$
$$z=1, x=0, y=1.$$
Similarly for $y=1$ we get one more sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$
I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.
Any help is app... | Let $ t_n := \frac{s_n}{3^n} $. Hence, $ t_n = t_{n-1} + 3^{-n}\left(2^{n-2} - 1\right) $. Also, $ t_1 = 0 $. Hence, $$ t_n = \sum_{k = 2}^n \frac{1}{4}\left(\frac{2}{3}\right)^n - 3^{-n} $$ This is a geometric series and easily evaluated to arrive at $$ t_n = \frac{1}{2}\cdot 3^{-n} \cdot \left(1 - 2^{n}\right) + \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$ I need the closed-form for the $n$-th derivative ($n\geq0 $):
$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)$$
Thanks!
By following the suggestion of Hermite polynomials:
$$H_n(x)=(-1)^ne^{x^2}\frac{\partial^... | Here we show that OPs approach is regrettably not correct and provide an alternative to answer the problem.
This answer is based upon the $n$-th derivative of the composite of two functions. It is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:
Hoppe Form of Generalized... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$
However, the answer is 3
| Given: $\boxed{3\sqrt{x+13} = x+9}$
Isolate the radical, x's cancel:
$\dfrac{3\sqrt{x+13}}{3} = \dfrac{x+9}{3}$
$\sqrt{x+13} = \dfrac{x+9}{3}$
Square both sides:
$\left(\sqrt{x+13}\right)^2 =\left(\dfrac{x+9}{3}\right)^2$
$x+13 = \left(\dfrac{x+9}{3}\right)\left(\dfrac{x+9}{3}\right)$
Move the 9 to the other side:
$9(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n I'm reading elementary number theory and trying to understand the following problem: If $x^2\equiv 1 \pmod{n}$, $n=pq$, $p$ and $q$ are odd primes and $x \not\equiv \pm 1 \pmod{n}$, then e... | If $n=p\cdot q$ divides $x^2-1=(x+1)(x-1)$
If odd prime $r$ divides $x-1$ and $x+1,$ it will divide $(x+1)-(x-1)=2$ which is impossible $\implies r$ can divide exactly one of $x-1,x+1$
Now let us check the possibilities
Case $1:$ $p$ divides $x-1$ and $q$ divides $x-1\implies (x-1)$ is divisible by lcm$(p,q)=pq=n\i... | {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education... | In this case, the following trick also works: Dividing both the numerator and the denominator by $\cos^4 x$, we can use the substitution $ t = \tan x$ to obtain
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2}
&= \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x)^2} \sec... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$
Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c}
+ \frac{b}{c+a} + \frac{c} {a+b}$$
Using Nesbitt's inequality: $I \ge \frac 32$
I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$... | This not a complete solution, just a 'case' of a problem.
When $\gcd(a+b,a+c,b+c)=1 \implies \text { } a,b,c \text { are not all odd or not all even}$
Your expression is
$\dfrac{a(c+a)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)}{(a+b)(b+c)(c+a)}$
Denoting $a+b=k, b+c=l$ and $c+a=m$
$$klm |akm+blk+cml \implies a=p_1l, b=p_2m, c=p_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 1,
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Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$ Find the range of:
$$y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$$
What I tried:
Let:$$\log_e\frac{x^2+e}{x^2+1}=X,$$
then $$y=\sqrt {\sin X}+\sqrt{\cos X}$$
$$y_{max}at X=\pi/4$$
The... | A related problem. First, we study the expression
$$ \frac{x^2+e}{x^2+1}=1+\frac{e-1}{x^2+1} \longrightarrow_{|x|\to\infty} 1, $$
which implies $ y(x)\longrightarrow_{|x|\to \infty} 0 $. To find the maximum of the function $y(x)$, Let's study the function
$$ h(t)=\sqrt{\sin(t)}+\sqrt{\cos(t)}. $$
The maximum of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Equation with high exponents I would appreciate any help with this problem:
$ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x^1+1x^0=0 $
I know that when $x$ isn't zero $x^0=1$ so the equation could be re-written as $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x+1=0 $. I am not sure what to do from here. I have tried using wolframalpha t... | As the equation is Reciprocal Equation of the First type,
Divide either sides by $x^4,$ to get $$x^4+\frac1{x^4}+2\left(x^3+\frac1{x^3}\right)+2\left(x^2+\frac1{x^2}\right)+5\left(x+\frac1x\right)+3=0$$
Put $x+\frac1x=y$ to reduce the equation the degree $\frac82=4$
Can you take it form here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Extracting the coefficient from a generating function Consider (for fixed $r$) the following function:
$$f(z) = \frac{1}{1-z-z^2-\cdots-z^r} = \frac{1}{1-z\frac{1-z^r}{1-z}}=\sum_{j=0}^\infty\left(z\frac{1-z^r}{1-z}\right)^j$$
(Assume everything is ok with regards to convergence.)
The text I am reading claims that if w... | $$f(z) = \sum_{j=0}^\infty \left( z \frac{1 - z^r}{1-z} \right)^j = \sum_{j=0}^{\infty}z^j \left( \sum_{k=0}^j {j\choose k} (-z^r)^k \right) \left( \sum_{m=0}^{\infty} {m+j-1 \choose j-1}z^m \right)$$
Change the order of summation. Note that in order to get a term of $z^n$, we need $j + rk + m = n$, or that $m= n - rk ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/403277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving summation $2n+2^2(n-1)+2^3(n-2)+....+2^n$ Can anyone help me with this summation? I tried to use the geometric series on this, but I can't use that.
$$2n+2^2(n-1)+2^3(n-2)+....+2^n$$
I am trying to do this for studying algorithms. Can we get a closed form for this ?
| Notice that
$$S_n=2n+2^2(n-1)+2^3(n-2)+\cdots+2^{n-1}(2)+2^n(1)$$
$$S_{n-1}=2(n-1)+2^2(n-2)+2^3(n-3)+\cdots+2^{n-1}(1)+2^n(0)$$
differ by
$$S_n-S_{n-1}=2+2^2+2^3+\cdots+2^n=2(2^n-1)$$
Therefore, $$S_n=2(2^n-1) + S_{n-1} =2(2^n-1) + 2(2^{n-1}-1) + S_{n-2}$$and so on until
$$S_n=2(2^n-1 +2^{n-1}-1+2^{n-2}-1+\cdots+2^{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/403384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$ I was recently asked this question which stumped me.
How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?
| $$\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1}- n =\left(\dfrac{\sqrt{x_1}}{\sqrt{x_n}}-\dfrac{\sqrt{x_n}}{\sqrt{x_1}} \right)^2+\left(\dfrac{\sqrt{x_2}}{\sqrt{x_{n-1}}}-\dfrac{\sqrt{x_{n-1}}}{\sqrt{x_2}} \right)^2+...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 7,
"answer_id": 4
} |
$n^5-n$ is divisible by $10$? I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to ... | $$n^5-n=n(n^2+1)(n+1)(n-1)= n(n^2-4)(n+1)(n-1)+5n(n-1)(n+1)=(n-2)(n-1)n(n+1)(n+2)+5n(n-1)(n+1)$$
$(n-2)(n-1)n(n+1)(n+2)$ is even and divisible by 5, since it is the product of 5 consecutive integers.
$5(n-1)n(n+1)$ is also even and divisible by $5$.
Note: Both expressions are also divisible by $3$, so $n^5-n$ is actual... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 6
} |
Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\di... | HINT: Pull out everything that you can: $6x^3-2x^2=2x^2(3x-1)$, and $9x^3-x=x(9x^2-1)$. Then notice that $9x^2=(3x)^2$, so that $9x^2-1=(3x)^2-1^2=(3x-1)(3x+1)$. Finally, do the cancellations that are now apparent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Why does $(k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor = 1,\;k\ge2\;\implies\;\text{isPrime}(k)$ Let $k$ be a integer such that $k\ge2$
Why does
$$(k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor = 1$$
only when $k$ is prime?
Example:
$$\pi(n) = \sum _{k=4}^n \left((k-2)!-k \left\lfloor \frac{k!}{(k-1... | If $k$ is not a prime, then the equation is $0$ modulo $k$. (We have $(k-2)! \equiv 0 \pmod{k}$.)
Conversely if $k$ is a prime, then by Wilson's theorem $(k-2)! \equiv 1 \pmod{k}$.
Edit: (To complete the argument.) If $k$ is a prime, then $(k-2)! = 1 + l k$ for some $l \ge 0$. Equivalently $l = \frac{(k-2)! - 1}{k}$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to find the Laurent expansion for $1/\cos(z)$ How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$
| Note $\frac{1}{\cos z}=-\frac{1}{\sin (z-\frac{\pi}{2})}$.
Let $t:=z-\frac{\pi}{2}$. Then $0<|t|<1$, $\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots$, so
$$
\begin{align}
\frac{1}{\sin t}&=\frac{1}{ t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots } \\
&=\frac{1}{t}\frac{1}{1-\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up ge... | You're on the right track so far; you've used the definition of $R_n$ to express the right side of the equation. Now just do this for the left side as well. You want to show that for any $n\geq 0$,
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
This can be done directly:
$$\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| \begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1\times0) + 1\\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\\
= &12
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$ Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test... | I'm not sure about your result; the ratio test is inconclusive anyway. The way I look at this, I express each term as
$$b_k = \frac{a_k}{2 k+1}$$
where
$$a_k = \frac{1}{2^{2 k}} \binom{2 k}{k}$$
Using, e.g., Stirling's formula, you may show that
$$a_k \sim \frac{1}{\sqrt{\pi k}} \quad (k \to \infty)$$
so that the rati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Did I derive this correctly? I derived this $$(2x+1)^2 \sqrt{4x+1}$$
and got $(8x+4)(\sqrt{4x+1})$+$\frac{2}{\sqrt{4x+1}}(2x+1)^2$
Is this correct?
I ask because Wofram Alpha gave me a different answer.
Thanks in advance.
| Your answer:
$$\begin{aligned}
&(8x+4)(\sqrt{4x+1})+\frac{2}{\sqrt{4x+1}}(2x+1)^2
\\
=& \frac{(8x+4)(4x+1)}{\sqrt{4x+1}}+\frac{2(2x+1)^2}{\sqrt{4x+1}}
\\
=& \frac{32x^2 + 24x + 4}{\sqrt{4x+1}} + \frac{8x^2 + 8x + 2}{\sqrt{4x+1}}
\\
=& \frac{40x^2 + 32x + 6}{\sqrt{4x+1}}
\\
=& \frac{2(10x + 3)(2x+1)}{\sqrt{4x+1}}.
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Is This A Derivative? I am in a little over my head. This all began with my reading how each level of pascals triangle adds to $2^n$, where n=row# starting with n=0. I then though, "wouldn't it be clever if the rows added to something else--like say $3^n$ instead?" Or even better generalize it for any constant, $a^n$.
... | You may rewrite your $$\frac{a^n}{2^n}(1+1)^n$$ as $$\left(\frac{a}{2} + \frac{a}{2}\right)^n$$
Unfortunately neither of the two examples you mention appear to be specific examples of general patterns.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/409803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| We use the identity: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
let $x=a-b ,y=b-c, z=c-a$ we see $x+y+z=0$ so $(a-b)^3+(b-c)^3+(c-a)^3-3(a-b)(b-c)(c-a)=0 $
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 5
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Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac... | Here is another heuristic argument along probabilistic lines that there are finitely many solutions.
The distance between perfect squares near $n$ is approximately $2\sqrt{n}$. Thus, the probability of a given integer $n$ to be a perfect square is approximately $\frac1{2\sqrt{n}}$. Summing the probability that $2^n-23$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$
Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$.
How many nice methods do you know for solving this problem? Thank you everyone.
My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so $$x^{2000}+\dfrac{1}{x^{2000}}=... | $$x+x^{-1}=\frac{\sqrt5+1}2$$
$$\implies 2x^2-x+2=\sqrt5x$$
$$\text{On squaring,} (2x^2-x+2)^2=5x^2$$
$$\implies 1-x+x^2-x^3+x^4=0$$
$$\text{ or, }x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2\cdot x\cdot\frac1x=\left(\frac{\sqrt5+1}2\right)^2-2=\frac{\sqrt5-1}2$$
$$\implies x+\frac1x-\left(x^2+\frac1{x^2}\right)=1\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
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Factor $x^5-1$ into irreducibles in $\mathbb{F}_p[x]$ I have to factor the polynomial $f(x)=x^5-1$ in $\mathbb{F}_p[x]$, where $p \neq 5$ is a generic prime number.
I showhed that, if $5 \mid p-1$, then $f(x)$ splits into linear irreducible.
Now I believe that, if $5 \nmid p-1$ but $5 \mid p+1$, then $f(x)$ splits into... | Hints (for you to understand, complete/prove)
Over any field
$$x^5-1=(x-1)(x^4+x^3+x^2+ x +1)$$
Now, the roots of the second factor above are the roots of unity of order $\,5\,$ different from 1 itself. Since these roots (including $\;1\;$) form a cyclic group of order$\; 5\;$, we want to now when $\,5\mid p-1\,$ . In ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to determine the rank and determinant of $A$? let $A$ be $$A_{a} = \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$$
How can I calculate the rank of $A$ by the Gauss' methode and $\det A$?
| Notice that $A_a= J+aI_4$ with $$J= \left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \right)$$ Let $P_J(X)$ be the characteristic polynomial of $J$, then $\det(A_a)=P_J(1-a)$. By noticing:
$$\left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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proving $\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+$ Without Induction i proved that:
$$
\begin{align}
& {} \quad \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)\cdot 2n} \\[10pt]
& =\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\text{ for }n\in \math... | Let it be true for n
Then for $n+1$ we have,
$\displaystyle \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =(\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n-1)\cdot 2n})+\frac{1}{(2n+1)\cdot 2(n+1)}$(Using induction hypothesis for $n$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Determine whether a multi-variable limit exists $\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$ I need to determine whether the next limit exists:
$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$$
Looking at the numerator $(-1-\frac{x^2}2)$ it immediately reminds me of ... | HINT
Find the limit as $x \to 0$ when $y=x$ and show that it is not equal the limit as $x\to 0$ when $y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$4\times4$ matrices minimal polynomial Anyone can tell me how to find the minimal polynomial of $4\times4$ matrices belong to $M_4(\mathbb{R})$, because I know how to find $2\times2$ and $3\times3$ matrices, but never try to find $4\times4$. I think this should be linear algebra problem. Can anyone show me an example? ... | $A=\begin{pmatrix}
3&-4&0&2\\
-4&-5&-2&4\\
0&0&3&-2\\
0&0&2&-1\\
\end{pmatrix}$
Minimal polynomial of A will be the minimal polynomial of
$X=\begin{pmatrix}
3&-4\\
-4&-5\\
\end{pmatrix}$ multiplied by the minimal polynomial of
$Y=\begin{pmatrix}
3&-2\\
2&-1\\
\end{pmatrix}$
Reason:We have $A^k=\begin{pmatrix}
X^k&C_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $ax - a^2 = bx - b^2$ for $x$ Method 1
Solve for x
$$ax - a^2 = bx - b^2$$
Collect all terms with x on one side of the equation
$$ax - bx = a^2 -b^2$$
Factor both sides of the equation
$$(a -b)x = (a+b)(a - b)$$
Divide both sides of the equation by the coefficient of $x$ (which is $a-b$)
$$x = a + b$$ (whe... | If $a=b$, the left side of the equation $(a-b)(x - (a+b)) = 0$ is always equal to the right side, so $x$ can be anything.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/419786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Compute this limit $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ using L'Hôpital's rule I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\le... | Hint:its easy to prove $$\sin(x+y)+\sin(x-y)=2\sin(x)\cos(y)$$
then put $y=\frac{A+B}{2}$,$x=\frac{A-B}{2}$
$$ \sin(x)\sim x$$ $$\ cosx\sim1-\frac{x^2}{2}$$ because $$\lim_{x\to0}\frac{sinx}{x}=\lim_{x\to0}\frac{cosx}{1-\frac{x^2}{2}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| A Lagrange multiplier proof:
Suppose $a^2b^2c^2$ is maximized. Then since $a + b + c = 0$, two of $a,b$, and $c$ are of one sign, and the third is of the other sign. (Clearly none are zero). Replacing $a,b,$ and $c$ by their negatives if needed, one can assume two are positive, and permuting the variables if necessary ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
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How to find solutions to this equality $\; \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$ We have the following equality:
$$ \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$
Some of the solutions I found:
*
*$\mathrm{x} = 0$
*Also for $\mathrm{a}=0$, every $\mathrm{x}$ is a solution I believe
I tried get... | I've solve upto some extent
$$ x = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$
$$ {a^2x(1-x)(1-ax(1-x))-x=0}\implies x=0$$
$$ {a^2(1-x)(1-ax(1-x))-1=0}$$
$$ {a^2(x-1)(1+ax(x-1))+1=0}$$
put x-1 =t
$$ {a^2t(1+at(t+1))+1=0}$$
$$ {a^2t(1+at^2+at)+1=0}$$
$$a^3t^3+a^3t^2+a^2t+1=0$$
for this equation if we take a=1
$$t^3+t^2+t+1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove if $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} $ is a real number If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $
i found one link that had a similar problem.
Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$
| HINT:
Let $z=\cos A+i\sin A, w=\cos B+i\sin B$
$\implies z\cdot w=\cos(A+B)+i\sin(A+B)$
Using $\cos C-\cos D=-2\sin\frac{C-D}2\sin\frac{C+D}2$
and $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2$
$$\frac{z-w}{1-zw}=\frac{\cos A-\cos B+i(\sin A-\sin B)}{1-\{\cos(A+B)+i\sin(A+B)\}}$$
$$=\frac{-2\sin\frac{A-B}2\sin\frac{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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Question about modular arithmetic and divisibility If $$a^3+b^3+c^3=0\pmod 7$$
Calculate the residue after dividing $abc$ with $7$
My first ideas here was trying the 7 possible remainders and then elevate to the third power
$$a+b+c=x \pmod 7$$
$$a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc=x^3\pmod 7$$
$$3(a+b+c)(ab+bc+ac)-3abc... | If one of $a,b,c$ is divisible by $7,abc\equiv0\pmod 7$
Else
$n^3\equiv \begin{cases} 1 &\mbox{if } n \equiv 1,2,4\pmod 7 \\
-1 & \mbox{if } n \equiv 3,5,6\pmod 7 \end{cases} \pmod 7$
Observe that for no combination of $a,b,c$ $$a^3+b^3+c^3\equiv0\pmod 7$$
$$\implies a^3+b^3+c^3\equiv0\pmod 7\implies 7\text{ must divid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/428038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Given a matrix $A$ and what it maps two vectors to, is $0$ an eigenvalue of it? Studying for my Algebra exam, and this question popped out with no solution in a previous exam:
Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 ... | The answer for (I) looks good.
Good for (II) might be to note that
$$
A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}2\\4\\-6\end{bmatrix}
$$
and
$$
A\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}
$$
which gives the first two columns of $A$:
$$
A=\begin{bmatrix}1&2&x\\2&4&y\\-3&-6&z\end{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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