Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=... | I assume this means locus and not plane..
$$ \begin{align*}
|z-2i|&= 2|z+3|\\
|x+iy-2i|&= 2|x+iy+3| \\
|x+i(y-2)|&= 2|(x+3)+i(y)| \\
\sqrt{x^2+(y-2)^2}&=2\sqrt{(x+3)^2+y^2}\\
x^2+(y-2)^2&=4\left(x^2+6x+9+y^2\right)\\
x^2+y^2-4y+4&=4x^2+24x+36+4y^2 \\
3x^2+24x+32+3y^2+4y&=0\\
x^2+8x+\frac{32}{3}+y^2+\frac{4}{3}y&=0 \\
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proof without words: $1+8\times\text{triangular number}$ is an odd perfect square A recent question asked how to show that $8T_n+1$ is a perfect square if $T_n$ is a triangular number. This follows immediately from $T_n=\frac12 n(n+1)\implies 8T_n+1=4n^2+4n+1=(2n+1)^2$.
Can this be proven without words?
| $$
\begin{matrix}
\color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{green}{\bullet} \\
\color{orange}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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An example of evaluating an integral using partial fraction decomposition
How do you evaluate the integral
\begin{equation*}
\int \frac{3x^3+ 7x^2+ 5x+ 3}{(x+1)^2(x^2+1)}\,\mathrm{d}x
\,?
\end{equation*}
This is just an exercise that I wanted to have typed up for my class. Maybe it'll help other students t... | Notice that if we were to expand the denominator of the integrand, the highest power of $x$ would be $4$, which is less than the highest power in the numerator. So this expression is ripe to be rewritten with partial fraction decomposition. Since $x^2+1$ cannot be factored, the decomposition must look like
\begin{equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Multivariable limit using polar coordinates?
$$\lim_{(x,y)\to(0,0)} \frac {2\sin (x^2 + y^2) + y^3}{3x^2+3y^2}$$
I am actually very new to polar coordinates so I'm a little bit confused about this part. I substitute this into this:
$\frac {2sin (r^2) + (r cos \theta)^3}{3r^2}$
But I am not even sure if it's right and... | \begin{align}
\lim_{(x,y)\rightarrow (0,0)} \frac{2\sin(x^2+y^2)+y^3}{3x^2+3y^2} &= \lim_{r\rightarrow 0} \frac{2\sin(r^2)+r^3\sin^3\theta}{3r^2}\\
&=\lim_{r\rightarrow 0} \frac{2\sin(r^2)}{3r^2} + \lim_{r\rightarrow 0}\frac{r^3\sin^3\theta}{3r^2}\\
&= \frac{2}{3} + \lim_{r\rightarrow 0}\frac{r\sin^3\theta}{3}\\
&= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculate the sum $\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}$ I want to calculate the sum : $$\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1} $$
I differentiated it, so $$\sum_{n=1}^{\infty}x^{3n}$$
by putting $ x = \sqrt[3]{t}$ it becomes :
$$\sum_{n=0}^{\infty}t^n$$
so $$f'(x) = \frac{1}{1-x^3}$$
by integrating it I should g... | \begin{eqnarray*}
S=\sum_{n=0}^{\infty} \frac{x^{3n+1}}{3n+1} = \sum_{n=0}^{\infty} \int_{0}^x x^{3n} dx = \int_{0}^x \frac{dx}{1-x^3}
\end{eqnarray*}
Now do partial fractions
\begin{eqnarray*}
\frac{1}{1-x^3}= \frac{x+2}{3(1+x+x^2)}-\frac{1}{3(x-1)}
\end{eqnarray*}
Integrating this gives
\begin{eqnarray*}
S=\frac{1}{6... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the volume of the solid inside both $x^2+y^2+z^2=a^2$ and $(x-a/2)^2+y^2=(a/2)^2$ using a triple integral on cylindrical coordinates I am asked to find the volume of the solid inside both $x^2+y^2+z^2=a^2$ and $(x-a/2)^2+y^2=(a/2)^2$ using a triple integral on cylindrical coordinates.
I have this triple integral ... | In order to better understand the solid region $B$ enclosed by both surfaces in order to better evaluate our limits of integration, we can visualize the surfaces as follows:
In cylindrical coordinates, we can express the cylinder as:
$$ r = a \cos \theta \qquad (0 \leq \theta \leq \pi)$$
and the sphere as:
$$ r^2 + z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial sum relation Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and th... |
Is there a general solution for this type of problem
Consider for example $P(x)=(x^2+x+1)^n\,$.
$\sum_{k=1}^{2n} \frac{1}{x_k}$
Let $y_k = \frac{1}{x_k} \ne 0\,$ then $P(\frac{1}{y_k})=P(x_k)=0\,$, so $y_k$ are the roots of: $$Q(y)=y^{2n} P\left(\frac{1}{y}\right)=y^{2n}\left(\frac{1}{y^2}+\frac{1}{y}+1\right)^n=(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor expansion of composite functions Suppose one has a function of the form: $$f(x)=\sqrt{1+a\cos^{2}(x)}$$ where $a$ is some parameter.
Firstly, what is the correct method for Taylor expanding such a composite function? Should one just repeatedly use the chain rule $(f\circ g)'(x)=f'(g(x))g'(x)$ to determine the c... | Your method would work, but it'd involve computing a lot of unpleasant derivatives. A better method would be to expand $cos^2(x)$, then expand the square root:
$$\sqrt{1+a \cos^2(x)} = \sqrt{1 + a \left(1 - \frac{1}{2} x^2 + \mathcal{O}(x^4)\right)^2} \\
= \sqrt{1 + a \left(1 - x^2 + \mathcal{O}(x^4)\right)} \\
= \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Iterated sequence - general formula I have a question regarding writing a general formula for an iterated sequence of variables.
Say we have
$$a_0=1$$
$$a_1=\frac{3}{2}a_0$$
$$a_2=\frac{3}{2}a_1-\frac{1}{2}a_0$$
$$a_3=\frac{3}{2}a_2-\frac{1}{2}a_1$$
So in general we have $a_i=\frac{3}{2}a_{i-1}-\frac{1}{2}a_{i-2}$... | Starting from your general term
$$a_n=\frac{3}{2}a_{n-1}-\frac{1}{2}a_{n-2}$$
substituting value of $a_{n-1}$ in this , it will transform into $$a_{n}=\frac{7}{4}a_{n-2}-\frac{3}{4}a_{n-3}$$
which is also equals to $$a_n=\frac{2^{2+1}-1}{2^2}a_{n-2}-\frac{2^2-1}{2^2}a_{n-3}$$
again substituting the value of $a_{n-2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 : f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\fra... | Another way to process is to set $y=mx^\alpha$ with some well chosen $\alpha>0$ so that the various sums $x^a+y^b$ become homogeneous.
Of course $m=m(x,y)$ is not a constant, just the ratio of the rate of convergence of $x$ and $y$.
Part a:
With $y=mx^2$
$\displaystyle f(x,y)=\frac{x^2y}{x^4+y^2}=\frac{mx^4}{x^4+m^2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not re... | The inductive hypothesis is
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
=\frac{k}{k+1}.\tag1$$
You need to prove that (1) implies the statement got from (1) by replacing
$k$ by $k+1$. This is
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{(k+1)(k+2)}
=\frac{k+1}{k+2}\tag2$$
but instead of (2) you have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| $$\int_{0}^{1}\frac{x^2+1}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2+1}{x^4+1}\,dx = 2\int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx = 2\int_{0}^{1}\frac{1+x^2-x^4-x^6}{1-x^8}\,dx$$
hence the value of the integral is provided by the (Dirichlet) series
$$ 2\sum_{k\geq 0}\left(\frac{1}{8k+1}+\frac{1}{8k+3}-\frac{1}{8k+5}-\frac{1}{8k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
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Surface integral of a pyramid in a sphere Calculate surface integral: $$I = \iint \limits_{S} \frac{xy}{x^2 + y^2} d{S}$$ where S is surface determined by sides of pyramid $x + y + z = 1 \phantom\ (x, y, z \ge 0 ),$ inside sphere $(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 + z^2 \le \frac{1}{4}.$
How should I parameteri... | The parts of the surface that give a non-zero contribution to the integral are
1) the one over the face $x + y + z = 1$
2) the one over the face $z=0$.
(on the other two faces $x\cdot y=0$ and therefore $f(x,y)=0$).
As regards $I_1$, we have that $dS=\sqrt{3}dxdy$ and
$$I_1=\sqrt{3}\iint_{D_1}\frac{xy}{x^2 + y^2}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293646",
"timestamp": "2023-03-29T00:00:00",
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Prove $\int_{-\frac{1}{4}}^{\frac{1}{4}} \left(\left(H_{x+\frac{1}{4}}-H_{x-\frac{1}{4}}\right) \cos (2 \pi x)+\pi \right) \, dx=\text{Si}(\pi )$ I recently stumbled upon a curious relationship between harmonic numbers and the Wilbraham-Gibbs constant:
$$\int_{-\frac{1}{4}}^{\frac{1}{4}} \left(\left(H_{x+\frac{1}{4}}-H... | Using the relation between the Harmonic numbers and digamma function, namely $H_{x} = \gamma + \psi(x +1)$, the relation between the digamma function and the gamma function,
$$\psi(x) = \frac{d}{dx} \, \ln(\Gamma(x)),$$
and the integrals:
\begin{align}
\int_{0}^{1} \sin\left(\frac{\pi \, t}{2}\right) \, \ln\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Continuing "Pascal's triangle" for negative binomial exponents Does there exist a pattern for the coefficients in a negative binomial expansion? This question is not about the explicit formula, but rather if there exist a continuation for Pascal's triangle.
$$\begin{array}l
(a+b)^{-2} &=&&&& \color{red}?\\
(a+b)^{-1} &... | There is a continuation respecting the addition law
\begin{align*}
\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}
\end{align*}
This way we can write
\begin{array}{l|rrrrrrrrrr}
(1+x)^{-3}&\color{grey}{0}&\color{grey}{0}&1&-3x&6x^2&-10x^3&15x^4&-21x^5&38x^6&\ldots\\
(1+x)^{-2}&\color{grey}{0}&\color{grey}{0}&1&-2x&3x^2&-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Bound for the equation $a^2 = b^2 + 4n$ For a fixed natural number $n$, I want to study the integer solutions of $$a^2 = b^2 + 4n.$$ There is always at least one solution, namely $$(n+1)^2 = (n-1)^2 + 4n.$$ There can be others; for example, the solutions of $a^2 = b^2 + 24$ are $$(a,b)
= (\pm 7, \pm 5), \; \; (\pm 5, ... | If $|a|=n+2$ then $b^2=(n+2)^2-4n$, which is strictly greater than $n^2$ but less than $(n+2)^2$, hence it must be $(n+1)^2$, so $n^2+4=n^2+2n+1$ so $n=3/2$(not possible)
Now if $|a|=n+3$ then $(n+3)^2-4n$ is a perfect square, which is strictly greater than $(n+1)^2$ but less than $(n+3)^2$, hence it must be $(n+2)^2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calc... | $$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}}+\sqrt{\frac{1-\cos\frac{\pi}{6}}{2}}=$$
$$=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}+\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}}{2\sqrt2}=$$
$$=\frac{\sqrt3+1+\sqrt3-1}{2\sqrt2}=\sqrt{\frac{3}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit Relation Involving Second Moment Saw this limit relation in a paper, but I can't seem to prove it:
\begin{equation}
\mathbb{E}(X^2) = \lim_{t \rightarrow 0} \frac{2}{t^2} [\mathbb{E}_X(e^{tX})-1]
\end{equation}
I might be missing something basic, but does anyone know how? Not sure if we need to assume $\mathbb{E}... | $$
e^{tX} = 1 + tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots
$$
$$
e^{tX} - 1 = tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots
$$
\begin{align}
\operatorname{E}(e^{tX} - 1) & = \operatorname{E}\left( tX + \frac{t^2 X^2} 2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representation of $e$ as a descending series I saw on Wikipedia: List of representations of e that
$$e=3+\sum_{k=2}^{\infty}\frac{-1}{k!(k-1)k}$$
It was also mentioned that this identity come from consideration on ways to put upper bound of $e$.
Can anyone give me a hint how we can derive this identity? I have trie... | You can derive the following telescoping sum for $k\ge2$:
\begin{align*}
\frac{1}{k!}+\frac{1}{k!(k-1)k}
&=\frac{1}{k!}\left(1+\frac{1}{(k-1)k}\right)\\
&=\frac{1}{k!}\left(1+\frac{1}{k-1}-\frac{1}{k}\right)\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)k!}\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)k(k-1)!}\\
&=\frac{k-1}{kk!}+\frac{1}{(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
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Finding the square root $\sqrt{4-2\sqrt{3}}$ How do you solve this:
$$X^2=4-2\sqrt{3}$$
I don't really know how to solve for $X$. I initially got $X=\sqrt{4-2\sqrt{3}}$ but I don't know how to simplify it.
| Let $\,a=\sqrt{4-2\sqrt{3}}\,$, $\,b=\sqrt{4+2\sqrt{3}}\,$ and note that $b \gt a \gt 0\,$, then:
$$\require{cancel}
\begin{cases}
a^2+b^2 = 4-\cancel{2\sqrt{3}} + 4+\cancel{2\sqrt{3}} = 8 \\
ab = \sqrt{(4-2\sqrt{3})(4+2\sqrt{3})} = \sqrt{16 - 2 \cdot 3} = 2
\end{cases}
$$
It follows that:
$$
\begin{cases}
\begin{align... | {
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"url": "https://math.stackexchange.com/questions/2298543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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$f(x^2)=(f(x))^2=f(f(x))$, $\forall x\in\mathbb R$. Prove that, there are infinitely many polynomial $f$ of degree $\geq1$, such that
$f(x^2)=(f(x))^2=f(f(x)), \forall x\in\mathbb R$. can only find three of them namely, $f(x)=0,1,x^2$. ... | Clearly the degree of the polynomial must be $2$, and the leading coefficient must be $1$.
Let the polynomial be $x^2+bx+c$
we need:
$x^4+bx^2+c=(x^2+bx+c)^2=((x^2+bx+c)^2+b(x^2+bx+c)+c$.
Lets separate into the five coefficients to get this system:
$1=1^2=1^3$ (coefficients of $x^4$)
$0=2b=2b$ (coefficients of $x^3$)
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find all complex roots for $z^6 + (1 + i)z^3 + i = 0$ Find all complex roots of $z^6 + (1 + i)z^3 + i = 0$.
I've tried by first settings $w = z^3$, so then I get $$w^2 + (1 + i)w + i = 0$$
Now I rewrite it as $$(w+(\frac{1+i}{2}))^2 = -i + (\frac{1+i}{2})^2$$
Set $w+(\frac{1+i}{2}) = a+bi$
$$(a+bi)^2 = -i + 2i=i$$
Whic... | The first step is good. The roots of $w^2+(1+i)w+i=0$ can be computed with the usual formula:
$$
\frac{-(1+i)\pm\sqrt{(1+i)^2-4i}}{2}
$$
but it's easier finding two numbers having sum $-1-i$ and product $i$: they're $-1$ and $-i$.
Now just find the cube roots of $-1=e^{\pi i}$ and $-i=e^{3\pi i/2}$.
If you want to do ... | {
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"question_score": "2",
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"answer_id": 1
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Make a list the elements of the subgroup in S7 Consider the permutation $$a=\begin{pmatrix}1234567\\3276514\end{pmatrix}$$ in S7 .List the elements of the subgroup in S7.
Hello, I just graduated from high school and in my summer time I'm reading about abstract algebra I want to learn more about the subject, but I do ... | The notation
$$a = \begin{pmatrix} b_1 & b_2 & \cdots & b_n \\ c_1 & c_2 & \cdots & c_n \end{pmatrix}$$
denotes a function with $a(b_1) = c_1, a(b_2) = c_2,\dots,a(b_n) = c_n$. This is usually given with $a_1 = 1, a_2 = 2, \dots, a_n = n$ but if you think about it, the order that the columns are given in doesn't matter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove inequality $a^3b^3+2b^3c^3+3a^3c^3\le 0$ For $a^3+b^3+c^3=0$. Prove that $a^3b^3+2b^3c^3+3a^3c^3\le 0$
i think from $a^3+b^3+c^3=0$ we have one of the three numbers is zero, the other two are opposites.Suppose $a;b$ opposites and $c=0$
| For simplicity, let $a^3=x,b^3=y,c^3=z$. Then: $x+y+z=0\Rightarrow x=-(y+z).$
$$xy+2yz+3xz=$$
$$-y(y+z)+2yz-3z(y+z)=$$
$$-y^2-2yz-3z^2=$$
$$-(y+z)^2-2z^2\le0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Use Parseval's equation and the table of Fourier series to evaluate $\sum\frac{1}{(1+n^2)^2}$
Use Parseval's equation and the table of Fourier series to evaluate $$\sum\frac{1}{(1+n^2)^2}$$
So I have used this method before to show $$\sum\frac{1}{n^4} = \frac{\pi^4}{32}$$ however, for this question I am struggling.
B... | It is easy to show that the complex Fourier Series for $e^{bx}$, $x\in[0,2\pi]$ is given by
$$e^{bx}=\frac{e^{2\pi b}-1}{2\pi}\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}$$
whereupon isolating the series becomes
$$\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}=\frac{2\pi e^{bx}}{e^{2\pi b}-1} \tag1$$
Differentiating $(1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| Notice all odd numbers are:
$1 = 1^2 - 0^2$
$3 = 2^2 - 1^2$
$5 = 3^2 - 2^2$ etc..
so if odd numbers are $\{o_1,o_2,...\}$ then $o_n=n^2-(n-1)^2$.
Now even numbers $\{e_1,e_2,...\}$ are simply $e_n=o_n+1^2$.
regarding the number 6, if any shorter representation exists, it must be one of the following:
1) $x^2 + y^2$ - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Can the nested radical expression $\sqrt{8 + \sqrt{48}}$ be simplified? Solving a pythagorean problem I came up with the following radical:
$$C = \sqrt{1^2+(\sqrt{3} + 2)^2}$$
I can simplify this to:
$$\sqrt{8+4\times\sqrt{3}} = \sqrt{8+\sqrt{48}}$$
Is there a way to simplify this further? I'm not satisfied with the ... | A nested radical $\sqrt{u + \sqrt{v}}$ can be simplified if $u^2 - v$ is a perfect square. Since $$8^2 - 48 = 64 - 48 = 16 = 4^2$$ this nested radical can be simplified.
Suppose that
$$\sqrt{8 + \sqrt{48}} = \sqrt{a} + \sqrt{b}$$
where $a$ and $b$ are rational numbers. Squaring both sides of the equation yields
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compare Fourier series $$g(x) = \sum_{n = 1}^{\infty} a_n \sin(nx)$$
$$h(x) = -\frac{a_1}{2 \pi } x $$
How to prove that $g(x) \ge h(x)$ on $ (0, \pi) $, where $a_n$ decreases to $0$ monotonically?
I tried to make a Fourier series from $R(x) = x$, I got $ \displaystyle R(x) = \sum_{n = 1}^{\infty} \frac{(-1)^n}{... | I don't see a way to use the Fourier series of $h$ (which you have slightly miscalculated, a factor of $-2$ is missing).
A summation by parts works, however. Note that
$$2\sin \alpha \sin \beta = \cos (\alpha - \beta) - \cos (\alpha + \beta),$$
so
\begin{align}
2 g(x) \sin (x/2)
&= \sum_{n = 1}^{\infty} a_n \bigl(2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dice roll: Probability of getting a given sum + a particular set of numbers The problem goes as: "An unbiased usual six-sided die is thrown three times. The sum of the numbers coming up is $10$. What is the probability that $2$ has appeared at least once?"
Options:
$A. 1/36$
$B. 5/36$
$C. 91/216$
$D. 1/18$
My approach:... | The ways to get a sum of $10$ are as follows:
$$1 + 3 + 6$$
$$1 + 4 + 5$$
$$1 + 5 + 4$$
$$1 + 6 + 3$$
$$\color{green}{2 + 2 + 6}$$
$$\color{green}{2 + 3 + 5}$$
$$\color{green}{2 + 4 + 4}$$
$$\color{green}{2 + 5 + 3}$$
$$\color{green}{2 + 6 + 2}$$
$$3 + 1 + 6$$
$$\color{green}{3 + 2 + 5}$$
$$3 + 3 + 4$$
$$3 + 4 + 3$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$ So I'm trying to come up with an answer to this question for hours now. I don't know what I'm doing wrong and none of the calculators on the internet couldn't help so I figured I should ask people.
What have I done so far:
$\frac{(\sin... | Expanding the square, using relationship $2 \sin a \cos a= \sin 2a$ and the definition of $\cot$, we have to show that
$$\frac{[(\sin20^\circ)^2 + (\cos20^\circ)^2] + \sin 40^\circ}{\cos40^\circ} = \dfrac{\cos25^\circ}{\sin25^\circ}$$
As the content of the square brackets is 1, it is equivalent to show that :
$$\sin25^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Manipulating an algebraic equation This came in a competition, I went back home and tried it but was unsuccessful.
If $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}$$ then -
$$1) ~ax =by
~~2)~ xy = ab
~~3)~ ay = bx
~~4) ~ax = b$$
It is clear that if the following given condition holds for the variables th... | $$0=\frac{a^3+3ab^2}{3a^2b+b^3} -\frac{x^3+3xy^2}{3x^2y+y^3}=$$
$$=\frac{(ay-bx)(3a^2x^2+3b^2y^2+a^2y^2+b^2x^2-8abxy)}{(3a^2b+b^3)(3x^2y+y^3)}=$$
$$\frac{(ay-bx)(3(ax-by)^2+(ay-bx)^2)}{(3a^2b+b^3)(3x^2y+y^3)},$$
which says that $ay=bx$ or
$$(ax-by)^2+(ay-bx)^2=0.$$
In the last case we obtain $ay=bx$ again.
Thus, the a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the locus of centres of circles I have to find the locus of centres of circles that pass through the point $(8,0)$ and tangent to circle $x^2+y^2=100$
$(x_0, y_0)$ - centres we are looking for
I decided to try something with distances, so I made a the equation $(x_0^2-8)^2+y_0^2=r^2$ It is the distance from center... | Note that $P = (8,0)$ is located inside the circle $O: x^2 + y^2 = 10^2$, which is centered at the origin and has radius $10$. A circle tangent to $O$ and passing through $P$ consequently will be internally tangent to $O$. Suppose such a circle, which we will label $Q$, has center $(x_Q, y_Q)$; then $(x_Q, y_Q)$ is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Compute $\sum\limits_n(-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}$ in terms of $\sum\limits_n\frac1{2^n-1}$ and $\sum\limits_n\frac1{2^n+1}$ Question:
let $$\sum_{n=1}^{+\infty}\dfrac{1}{2^n-1}=E,\sum_{n=1}^{+\infty}\dfrac{1}{2^n+1}=F$$ where $E,F$ are constant,(in fact,$E$ is Erdős-Borwein Constant ),Find the sum
$$f=\su... | It's not possible to express $f$ only with $E$ and $F$ because of
$\displaystyle\sum_{n=1}^\infty (-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}=$
$\displaystyle -\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n-1} +\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n+1}+\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n+i} -\frac{i}{2}\sum_{n=1}^\infty\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{... | So you got $\frac{1} {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}, a_1=1.$
Denote: $b_n=\frac{1}{a_n}$, then the equation becomes $b_{n+1}=-\frac{3}{4}b_n+\frac12$ with $b_1=\frac{1}{a_1}=1$.
Now apply the general formula: $$\bbox[lightgreen]
{
If \ \ y_{n+1}=ay_n+b,y_1=c, \ then: y_n=\left(c+\frac{b}{a-1}\right)a^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $x^4+(x-1)(x^2-2x+2)=0$ With $x \in \mathbb{R}$. Solve the equation $$x^4+(x-1)(x^2-2x+2)=0$$
Idea $1$ : $(x^2-x+1)(x^2+2x-2)=0$
Idea $2$ :$(2x^2+x-1)^2=(3x-3)^2$
Idea $3$ :$<=>x^4(x-1)x^2-2(x-1)^2=0$
Let $y=x-1 \dots$
I need another way because I am collecting methods of solving with quartic equati... |
Solve $x^4+(x-1)(x^2-2x+2)=0$
You can solve this using Newton-Raphson :
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$
Where \begin{align}&f(x)=x^4+x^3-3x^2+4x-2\\
&f'(x)=4x^3+3x^2-6x+4\\\end{align}
To guess $x_0$ :
\begin{align}&f(1)>0
\\&f(0)<0
\\\end{align}
So $x_0=1$ :
\begin{align}&x_1=0.5\\
&x_2=0.8\\
&x_3\approx0.6765\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as ... | There is a wonderful generalization of the solution for this problem:
$$A = \sqrt{(a + b + c)^2 – 4ac}.$$
Where $a$, $b$ and $c$ are the known areas, and $A$ is the unknown area.
Caveat: $a$ and $c$ must be the triangles whose long legs are the side of the square, in this case $7$ and $9$.
$$A = \sqrt{729 – 4 \times 9 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 8
} |
Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$. Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$.
$$6 \cdot \frac 1{2x\sqrt{(2x)^2-1}} \cdot 2$$
$$\frac{12}{2x\sqrt{4x^2-1}}$$
$$\frac 6 {x\sqrt{4x^2-1}}$$
Why is that wrong?
| You see to say that $\displaystyle \sqrt{x^2-1} = \frac 1 2 \sqrt{(2x)^2 - 1}.$ That is incorrect. In fact
$$
\sqrt{x^2-1} = \frac 1 2 \cdot 2\sqrt{x^2-1} = \frac 1 2 \cdot \sqrt{4}\cdot\sqrt{x^2-1} = \frac 1 2 \sqrt{4x^2-4} = \frac 1 2 \sqrt{(2x)^2-4}.
$$
Now let $y = 6\operatorname{arcsec}(2x).$ Then
\begin{align}
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Obtaining an explicit expression of $U_n$ given by $U_{n+1}=\frac{2}{3}U_n -1$ and $U_0=2$
Let $(U_n)_n\geq 0$ be the sequence defined by: $$U_0=2\qquad, U_{n+1}=\frac{2}{3}U_n -1\quad\text{for all } n \geq 0$$ .
My question Here is : How do I write $U_n$ with a function of $n$ ?
| You can use the method of generating functions. Let $U(x)=\sum_{n=0}^\infty U_n x^n$. We wish to solve the recurrence
$$
U_{n+1}=\frac{2}{3}U_n -1;\quad (n\geq0)\tag{1}
$$
where $U_0=2$. Multiply both sides of (1) by $x^n$ and sum on $n$ to get that
$$
\frac{U(x)-2}{x}=\frac{2}{3}U(x)-\frac{1}{1-x}.\tag{2}
$$
Solve for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find $(x + 1)^{−3}$ in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$ In a previous question, I established that $x^5 + x^4 + 1$ is irreducible in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$, so it will not be divisible by $x+1$. Hence $x^5+x^4+1$ will be coprime to $(x+1)^3 = x^3+x^2+x+1$
So far I have used the Euclidean algorithm to get:
... | Here's my approach to this one, which does not really depend too much on the standard theorems on polynomial rings--except for the Euclidean Algorithm--but does rely on a few simple observations about the polynomial
$p(x) = x^5 + x^4 + 1 = 0 \in \Bbb Z_2[x]/(x^5 + x^4 + 1) \tag{1}$
and a lot of grinding away at synthet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| HINT: use the substituion
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
where $$\tan\left(\frac{x}{2}\right)=t$$
then you will have
$$\frac{16t(1-t^2)}{(1+t^2)^2}-\frac{2\sqrt{6}t}{1+t^2}-\sqrt{6}\frac{(1-t^2)}{1+t^2}+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find when $\frac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$ is an integer
Find all natural numbers $n$ such that $$\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$$ is an integer.
For $n = 1$ we have $\dfrac{18}{3} = 6$.
For $n = 2$ we have $\dfrac{126}{18} = 7$.
For $n = 3$ we have $\dfrac{972}{126} = \dfrac{54}{7}$
For $n... | Hint: It looks like considering the top and bottom $\mod 4$ narrows it down quite a bit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof by induction, factorials and exponents I've spent many hours trying to find proof by induction to this problem:
$$(n!)^2<\left(\frac{(n+1)(2n+1)}{6}\right)^n$$
for n > 1.
Does anyone have any idea how to solve it? Any help would be appreciated.
| Using AM-GM inequality (which can be proved by induction on the number of terms),
$$\sqrt[n]{1^2\cdot 2^2\cdot3^2 \cdots n^2} \le \frac{1^2 + 2^2+3^2+\cdots + n^2}{n}$$
Equality holds iff $1^2 = 2^2 = \ldots = n^2$, which means equality does not hold for $n>1$.
On the left hand side,
$$(n!)^2 = 1^2 \cdot2^2 \cdot 3^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Infinitely many positive integers $n$ such that $n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1)$
Prove that there exist infinitely many positive integers $n$ such that $$n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1).$$
We need to prove that there are infinitely many $n$ such that $\prod_{m=0}^{n-... | Since $(m-i)(m-1+i)=\left(m^2-m+1\right)+i$, multiplying both sides by their conjugates gives
$$
(m^2-m+1)^2+1=\left(m^2+1\right)\left((m-1)^2+1\right)\tag{1}
$$
For $m\ge2$, we have that $n=m^2-m+1\gt m$. Therefore, $(1)$ shows that
$$
\left.n^2+1\,\,\middle|\,\,\prod_{k=0}^{n-1}\left(k^2+1\right)\right.\tag{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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The dimension of the class of matrices that suffice a diagonalization property. Let $A$ be a $2\times 2$ real matrix with distinct real eigenvalues. For a $2\times 2$ real matrix $B$, let $C(B)$ denote the $4\times 4$ matrix whose block form is
\begin{equation*}
\begin{bmatrix}
A & B\\
0 & A
\end{bmatrix}_.
\e... |
Disclaimer: I tried to keep this as low level as possible and it took a fair bit of work. There are probably more slick ways to do this with more advanced theory.
Let $x$ and $y$ denote the eigenvalues of $A$. Then the eigenvalues of $C(B)$ are also $x$ and $y$ (its characteristic polynomial is $(\lambda - x)^2(\lamb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Planes and Augmented Matrices I seem to be having trouble with solving the following system of equations so that an infinite number of solutions would arise. The question is "Find the values of $p$ and $q$ for which the following system of equations has an infinite number of solutions and clearly explain my reasoning:"... | Begin with the augmented matrix
$$
\left(
\begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 1 & -1 & 1 & 3 \\ -2 & p & 2 & q\end{array}
\right)
$$
and row reduce this by
*
*multiplying the first row by $-1/2$ and adding it to the second row, and
*adding the first row to the last row
to get:
$$
\left(
\begin{array}{ccc|c} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Orthonormalbasis of a isometry $2 \times 2$ matrix I am a bit stuck in this easy question.
Let $A\in \mathbb R^{2 \times2}$ be a isometry so $A^{-1} = \bar{A}^T$. Show that $\left\{\begin{pmatrix} a \\ c \end{pmatrix},\begin{pmatrix} b \\ d \end{pmatrix}\right\}$ is a orthonormal basis.
$A = \begin{pmatrix} a&b \\ c&d... | Hint: Write
$$
v_1 = \pmatrix{a\\c} \qquad v_2 = \pmatrix{b\\d}
$$
Verify that for $A = \pmatrix{a&b\\c&d}$, we have
$$
A^TA = \pmatrix{
\langle v_1,v_1 \rangle & \langle v_2,v_1 \rangle\\
\langle v_1,v_2 \rangle & \langle v_2,v_2 \rangle}
$$
Since $A^{-1} = {A^T}$, we have ${A^T}A = I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equivalent quadratic form I want to find an integral quadratic form $a^2+2bxy+z^2$ with $b\equiv 0 \mod 81$ which is properly integrally equivalent to $3x^2+2xy+12y^2$.
In matrix worm that means that there exists a $2\times 2$-matrix $T$ over $\mathbb{Z}$ with $\det(T)=1$ and with $$T^T \begin{pmatrix} 3 & 1 \\ 1 & 12... | There is an easy step-by-step solution:
we want $81|12n^2-10n+2$, which is equivalent to $81|6n^2-5n+1$.
$81|6n^2-5n+1$ implies $3|-5n+1$, so $n\equiv 2\pmod{3}$, so $n=3k+2$ for some $k$ ($k\in\{0,1,2,\dots,26\}$ if you're looking for a solution $n\in\mathbb{Z}_{81}$).
put $n=3k+2$ into $81|6n^2-5n+1$.
you will get $8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
sum of integer parts of roots
If $\alpha,\beta,\gamma$ are real roots of the equation $\displaystyle \frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-2} = \frac{11}{54}x^2.$ Then $\lfloor \alpha \rfloor + \lfloor \beta \rfloor +\lfloor \gamma \rfloor $
$\bf{Attempt}$ From equation $$\frac{x^2-3x+2+x^2-2x+x^2-x}{x(x-1)(x-2)} = ... | HINT: use that $$\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x-2}-\frac{11}{54}x^2=-\frac{(x-3) \left(11 x^4+22 x^2-96 x+36\right)}{54 (x-2) (x-1) x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T$?
Find the basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T:M^{\mathbb R}_{2 \times 2} \to M^{\mathbb R}_{2 \times 2}$ defined as $T(A)=A-A^t$ for all $A \in M^{\mathbb R}_{2 \times 2}$.
Let $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$. ... | You can use the fact that every square matrix is, in a unique way, the sum of a symmetric and an antisymmetric matrices.
Any symmetric matrix belongs to the kernel of $T$, whereas the antisymmetric matrix
$$
A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$
does not belong to $\ker T$. Since the space of symmetric mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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For which $c$ there exists $t\in\mathbb{R}^+$ such that $x_1(t)=0$ and $x_2(t)>0$ for the system $x_1'(t)=-200kx_2(t)-c\ ,\ x_2'(t)=-100kx_1(t)$? For which $c\in\mathbb{R}$ there exists $t\in\mathbb{R}^+$ such that $x_1(t)=0$ and $x_2(t)>0$ for the system $x_1'(t)=-200kx_2(t)-c\ ,\ x_2'(t)=-100kx_1(t)\ ,\ x_1(0)=10^4\ ... | You already expressed $t$ (such that $x_1(t)=0$ in terms of $c$). By the way, there should be
$$t=\frac{ln\left(\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}\right)}{200\sqrt2k}.$$
So it remains to make the respective substitution in the formula for $x_2(t)$ and then to find when $x_2(t(c))>0$.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find inverse Laplace transform of $\frac{\sqrt{s} \sqrt{a+s}}{b+s}$ How to find inverse Laplace transform of
$$\frac{\sqrt{s} \sqrt{a+s}}{b+s}$$
where $a$ and $b$ constants, ${a,b}>0$
I tried to solve it, but I couldn't.
EDITED.
Numerical solution for $a=1,b=1$.
Plot:
| One method of consideration may be applied to the following. By using the series expansion
$$\sqrt{1-x} = 1 - \sum_{k=1}^{\infty} \frac{(2k-2)! \, x^{k}}{2^{2k-1} \, k! \, (k-1)!}.$$
Now the fraction in question becomes:
\begin{align}
\frac{\sqrt{s \, (s+a)}}{s+b} &= \frac{\sqrt{\left(s+ \frac{a}{2}\right)^{2} - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\lim _{x\to \pi/2}\left(\tan x-1\right)\left(1-\tan\left(\frac{x}{2}\right)\right) =\infty\cdot 0$? $\lim _{x\to \pi/2}\left(\tan x-1\right)\left(1-\tan\left(\frac{x}{2}\right)\right) $ Is there a way to solve this limit without any fancy trigonometric identity?
| Assuming that you can use Taylor series.
First, let $x=y+\frac \pi 2$ to make $$(\tan (x)-1)\left(1-\tan \left(\frac{x}{2}\right)\right) =(\cot (y)+1)\left(\tan \left(\frac{y}{2}+\frac \pi 4 \right)-1\right) $$ Now, using $$\cot(y)=\frac{1}{y}-\frac{y}{3}+O\left(y^3\right)$$ $$\tan \left(\frac{y}{2}+\frac \pi 4 \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the measure of angle $MAN$. We construct two points $M,N$ such that perimeter of triangle $MCN$ is equal to half of the perimeter of the square.Find the measure of angle $MAN$.
The case $CM=CN$ leads to a simple equation so we can find the measure of $MAN$ using trigonometry.But in general case I don't no How to u... | I think I understood about which square you say!
I think your problem is the following.
Let $ABCD$ is a square with $AB=a$. $M\in BC$ and $N\in CD$ such that $CN+CM+MN=2a$. Find $\measuredangle MAN$.
Let $MC=x$ and $CN=y$.
Hence, $$x+y+\sqrt{x^2+y^2}=2a,$$
$$AM=\sqrt{(a-x)^2+a^2}=\sqrt{2a^2-2ax+x^2}=\sqrt{\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2350264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| $$1+\dfrac a{x-a}=\dfrac x{x-a}$$
$$1+\dfrac a{x-a}+\dfrac{bx}{(x-a)(x-b)} =x\left[\dfrac1{x-a}+\dfrac b{(x-a)(x-b)}\right]=\dfrac{x^2}{(x-a)(x-b)}$$
$$1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}=\cdots=\dfrac{x^3}{(x-a)(x-b)(x-c)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to use mittag-leffer expansion to compute $\texttt{sech} z$? A question concerning a paper from Kesten (1986) In Kesten's 1986 paper (Limit distribution of Sinai's Random Walk) we read:
The reference is E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, 4th ed. (Cambridge Univ. Press, Cambridge, 1952). ... | Note first that $\texttt{cosh} (0) = 1$ so
$$ 1 = \frac{1}{\cosh(0)} = \texttt{sech} (0) = \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 }$$
Therefore
$$ \Bbb{E}\{ e^{-\theta L}; L>0\} =\frac{1}{2\theta} \bigg[1 - \texttt{sech}(\sqrt{2\theta})\bigg]\\
= \frac{1}{2\theta} \bigg[ \sum_{k=0}^\infty \frac{4\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to calculate multiplicative inverses in $GF(2^3)$ without the Euclidean algorithm The problem I have concerns finite field arithmetic in $GF(p^k)$.
I know how to find multiplicative inverses using the extended Euclidean algorithm, but for my exams I need to calculate multiplicative inverses in $GF(2^3)$ without it.... | As an alternative, build the table of the so called discrete logarithm. Let $a$ be a root of $f = x^{3} + x + 1$. Then
\begin{align}
&a^{0} = 1\\
&a^{1} = a\\
&a^{2} = a^{2}\\
&a^{3} = a + 1\\
&a^{4} = a^{2} + a\\
&a^{5} = a^{2} + a + 1\\
&a^{6} = a^{2} + 1\\
&a^{7} = 1\\
\end{align}
So to compute the inverse of $a^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
| Let $3x^2-2xy=k$.
Since $x^2=4y^2+4$, we obtain:
$$k=3x^2-2xy\geq3x^2-(x^2+y^2)=2(4+4y^2)-y^2=8+7y^2\geq8.$$
Hence,
$$3x^2-2xy=k\left(\frac{x^2}{4}-1\right)$$ or
$$\left(3-\frac{k}{4}\right)x^2-2xy+ky^2=0,$$
which gives
$$1-k\left(3-\frac{k}{4}\right)\geq0$$ or
$$\frac{k^2}{4}-3k+1\geq0,$$ which is
$k\geq2(3+2\sqrt2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
N... | the trick is here to write
$$3(a^2+b^2+c^2)\geq (a^2+b^2+c^2)(a+b+c)$$ this is equivalent to
$$(a-b)(a^2-b^2)+(b-c)(b^2-c^2)+(c-a)(c^2-a^2)\geq 0$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ I have the following problem:
How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$
My attempt:
\begin{align*}
\exp(ix)=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}
\end{align*}
Now split the sum in 4 su... |
We obtain
\begin{align*}
\cos(x)&=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\
&=\sum_{{n=0}\atop{n \text{ even}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}+\sum_{{n=0}\atop{n \text{ odd}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\
&=\sum_{{n=0}}^{\infty} \frac{x^{4n}}{(4n)!}-\sum_{{n=0}}^{\infty} \frac{x^{4n+2}}{(4n+2)!}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solving for $x$ in the absolute value equation $$|x-1|+|2x-3|=|3x-4|$$
I tried to solve the cases separately for the range of values of $x$, but it did not work. Can anyone please show me the way.
| Twisted/Confusing way to solve this problem:
$|x-1|+|2x-3|=|3x-4|$...eq(I). Let us call the three expressions as $A, B, C$
, so we have $A+B=C$
$\begin{matrix} \text{when} & \text{expression}\\ x\geq1&A=x-1 & i\\x<1& A=-(x-1) & ii\\ x\geq\frac{3}{2} & B=2x-3 & iii\\ x<\frac{3}{2} & B=-(2x-3) & iv\\ x\geq \frac{4}{3} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to calculate per unit cost? I need help understanding this math problem from my TEAS Mometrix Practice exam. It gives an explanation but I still don't understand. The question is:
Mandy can buy $4$ containers of yogurt and $3$ boxes of crackers for $\$9.55$. She can buy $2$ containers of yogurt and $2$ boxes of cra... | There are several ways to solve this or similar problems we have
$$4 \cdot x + 3 \cdot y = 9.55 \\
2 \cdot x + 2 \cdot y = 5.90$$
One approch as given in your answer is to multiply the second equation by 2 or -2 you then have $4 \cdot x$ in both equations so can cancel for $x$ by adding the two equations or subtracti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any bod... | Since $1= 2-1$, you can multiply by $2-1$ and it won't change the value.
$$(2-1)(2^0+2^1+2^2+\cdots + 2^n) $$
$$=2(2^0+2^1+2^2+\cdots + 2^n) -1(2^0+2^1+2^2+\cdots + 2^n)$$
$$= (2^1+2^2+2^3+\cdots + 2^{n+1}) - (2^0+2^1+2^2+\cdots +2^n)$$
$$= 2^{n+1}-2^0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 1
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Possibilities for the side and angle of the triangle We have the triangle ABC with $c=18$, $\alpha=\frac{\pi}{6}$ and the median through $C$ has the length $5$. I want to determine all the possibilities for $a$ and $\gamma$.
I have done the following:
From the median we have the following:
$$m_c=\sqrt{\frac{2a^2+2b^2... | I think it is easier to find $C$ by straightedge and compass, then study $\gamma$.
Let $M$ be the midpoint of $AB$: $C$ must lie on a circle centered at $M$ with radius $5$, and on a fixed line from $A$. It follows that there are two solutions, namely $C_1$ and $C_2$ in the diagram below.
By the secant-tangent theorem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Application Of De Moivre & Euler's Formulae Can anyone show me how I can prove that $\sin x \cos (3x) = \frac{1}{4} \sin (7x) - \frac{1}{4}\sin (5x) + \frac{1}{2}\sin x$?
I tried using Euler's formulae
$$\sin x= \frac{e^{ix}-e^{-ix}}{2i}$$
and
$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$
but the simplification didn't help at... | It should be $\sin { x } \cos ^{ 2 }{ 3x } $
$$\frac { 1 }{ 4 } \sin { 7x } -\frac { 1 }{ 4 } \sin { 5x } +\frac { 1 }{ 2 } \sin { x } =\frac { 1 }{ 4 } \left( \sin { 7x } -\sin { 5x } +2\sin { x } \right) =\\ =\frac { 1 }{ 4 } \left( 2\sin { \frac { 7x-5x }{ 2 } \cos { \frac { 7x+5x }{ 2 } } } +2\sin { x } \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving the inequality $\frac{x}{x+1} - \frac{1}{x-3} - 2 \leq 0$ I tried to solve
$$ \dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq 0$$
and I got
$$S = \{x \in \mathbb{R} \mid x \leq -\sqrt{5} \vee x \geq \sqrt{5}\}$$
but I have plotted and here is the function graph. It doesn't seem to be a quadratic inequality.
Does s... | Consider $x-3> 0$
$$\dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq 0$$
$$\Leftrightarrow x-\dfrac{x+1}{x-3}-2(x+1)\leq 0 \space\space ||x+1 > 0$$
$$\Leftrightarrow \dfrac{x(x-3)}{x-3}-\dfrac{x+1}{x-3}-\dfrac{2(x+1)(x-3)}{x-3}\leq 0$$
$$\Leftrightarrow x^2-3x-x-1-2(x^2-2x-3) \leq 0 \space \space ||x -3>0$$
$$\Leftrightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Rational integral $$\int\dfrac{x^2}{1+x^4}dx$$
I tried many standard approaches, but I didn't get too far! Here's the most promising of them:
$$\int\dfrac{dx}{\frac{1}{x^2}+x^2}$$
knowing that $\left(1/x+x\right)^2=\frac{1}{x^2}+x^2+2$ we can change variables $1/x+x=t$. Unluckily this doesn't work either.
| Hint: Express
$$\frac{x^2}{1+x^4}=\frac{ax+b}{x^2-\sqrt{2}x+1}+\frac{cx+d}{x^2+\sqrt{2}x+1}$$
and find $a,b,c,d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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For $0I only prove that for $0<x<y\leq 1/2$ the result is right as following:
the inequality is equivalent to
$$x^{2x+1}+2x^{x+1}y^y+x y^{2y}<y x^{2y}+2y^{x+1}x^y+y^{2x+1}.$$
First, $x^{2x+1}<y^{2x+1}$ because of $$\left(\frac{y}{x}\right)^{2x+1}>1;$$
Second, $2x^{x+1}y^y<2y^{x+1}x^y$ because of
$$\frac{2y^{x+1}x^y}{2x... | $$\frac{x^x+y^y}{x^y+y^x}<\frac{x^x+y^y}{2\sqrt{x^yy^x}}\ \ (\because a+b>2\sqrt{ab})$$
$$=\frac{1}{2}\left(\sqrt{\frac{x^{2x}}{x^yy^x}}+\sqrt{\frac{y^{2y}}{x^yy^x}}\right)$$
$$<\frac{1}{2}\sqrt{2}\sqrt{\frac{x^{2x}+y^{2y}}{x^yy^x}}\ \ \left(\because a+b<\sqrt{2(a^2+b^2)}\right)$$
$$<\frac{1}{2}\sqrt{2}\sqrt{\frac{2y^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Need help showing $\cos(x + y) / \sin(x - y) = -2$ if $\tan(\pi/4 - x) = 3 \tan(\pi/4 - y)$. I've tried manipulating the given relations using basic trig identities (which is all I know) but I wasn't able to prove anything.
| $\tan(\pi/4 - x) = 3 \tan(\pi/4 - y)$
$\iff \frac{\sin(\pi/4 - x)}{\cos(\pi/4-x)}=\frac{3 \sin(\pi/4 - y)}{\cos(\pi/4-y)}$
$\iff \frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}=\frac{3 \cos(y)- 3\sin(y)}{\cos(y)+\sin(y)}$
$\iff \cos(x)\cos(y)+\sin(y)\cos(x)-\sin(x)\cos(y)-\sin(x)\sin(y)= 3\cos(x) \cos(y)+3\sin(x)\cos(y)-3\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Let $F = Z_3$. Is $f(x) = x^4 + x^3 + x + 2$ irreducible in $F[x]$? I'm posting to check if I did this right.
First approach was to check roots:
$f(0) \neq 0$,
$f(1) \neq 0$,
$f(2) \neq 0$.
Second approach: Brute Force
My options for degree 1 were:
$x$, $x+1$, $x+2$
My options for degree 2 were:
$x^2$, $x^2 + 1$, $x^... | All you have to do to check that your factorization is correct is multiply it out.
$\begin{align}
(x^2 + 1)(x^2 + x + 2) &= x^2(x^2 + x + 2) + (x^2 + x + 2) \\
&= x^4 + x^3 + 2x^2 + x^2 + x + 2 \\&= x^4 + x^3 + 3x^2 + x + 2\\& = x^4 + x^3 + x + 2\end{align} $
Which is indeed what you started with.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Special palindromic pattern of $(\frac{10^n-1}{9})^2$ I noticed a repetitive, palindromic pattern when I've performed exponentiation with $10$. That is: $$(\frac{10^n-1}{9})^2 \text{ has a palindromic pattern, }\forall\:n\in [1;9]\:(n\in \mathbb{N})$$
We define $f(n)=(\frac{10^n-1}{9})^2$.
*
*$f(1) = \color{red}{1}$
... | \begin{align*}
11 \times 11 &= \begin{matrix} 1 & 1 & \\ & 1 & 1 & + \\ 1 & 2 & 1 \end{matrix} \\ \\
111 \times 111 &= \begin{matrix} 1 & 1 & 1 & & \\ & 1 & 1 & 1 & \\ & & 1 & 1 & 1 & + \\ 1 & 2 & 3 & 2 & 1 \end{matrix} \\ \\
1111 \times 1111 &= \begin{matrix} 1 & 1 & 1 & 1 & & & \\ & 1 & 1 & 1 & 1 & & \\ & & 1 & 1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Given 2 roots and a remainder of a cubic polynomial, find the remainder of $f(3)$
$f(x)$ is a cubic polynomial and the coefficient of $x^3$ is $k$. If the polynomial has exactly $2$ roots and these roots are $2$ and $-5$ and $f(1) = -48$, find the possible values of $f(3)$.
I've gotten part of the cubic equation to b... | We have also $2k+a=0$ or $-5k+a=0$.
In the first case $f(x)=k(x-2)^2(x+5)$, which gives $-48=6k$ or $k=-8$ and
$f(3)=-8\cdot1^2\cdot8=-64$.
In the second case $f(x)=k(x+5)^2(x-2)$, which gives $-48=-36k$ or $k=\frac{4}{3}$ and
$f(3)=\frac{4}{3}\cdot8^2\cdot1=\frac{256}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $|\cos z\ |^2 + |\sin z\ |^2 \geq 1$ My attempt:
For $z :=x+iy$,
$$\cos z =\cos x \cos iy - \sin x \sin iy \\ = \cos x \cosh y - i \sin x \sinh y\\ \sin z = \sin x \cosh y + i \cos x \sinh y$$
So
$$|\cos z\ |^2 + |\sin z\ |^2 = \cos ^2 x \cosh ^2 y + \sin ^2 x \sinh ^2 y + \cosh^2 y \sin^2x + \cos ^2x \sinh ^... | One of my favourite trig. identities is
\begin{align}\lvert \sin{(x+iy)} \rvert^2 &= \lvert \sin{x}\cosh{y} + i\cos{x}\sinh{y} \rvert^2 \\
&= \sin^2{x}\cosh^2{y}+\cos^2{x}\sinh^2{y} \\
&= \sin^2{x}(1+\sinh^2{y})+(1-\sin^2{x})\sinh^2{y} \\
&= \sin^2{x}+\sinh^2{y}.\end{align}
To get the equivalent for $\lvert \sin{(x+iy)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| I assume your formula is $a+b=\dfrac ab+\dfrac ba,$ then $a^2+b^2=ab(a+b)$ and since $a^2+b^2=(a+b)^2-2ab,$ we have $$x^2-2y=xy,$$ where $x=a+b$ and $y=ab.$ From here you can find $x$ in-terms of $y$ as $$x=\dfrac{y\pm\sqrt{y^2+8y}}{2}.$$ Then $$\color{Green}{a^2+b^2=\left(\dfrac{y\pm\sqrt{y^2+8y}}{2}\right)y},$$ where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$... Find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$ and contain at least one each of $\alpha$, $\beta$, and $\gamma$.
My attempt: using inclusion/... | Since there are five letters available and each word has seven positions to fill, there are $5^7$ words. From these, we must exclude those cases in which one or more of the letters $\alpha$, $\beta$, or $\gamma$ is missing.
There are three ways to exclude one of the letters $\alpha$, $\beta$, or $\gamma$ and four ways... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(... | It is:
$$L=\lim_{x\to 0} \frac {x+ax\cos x-b\sin x}{x^3}\stackrel{LR}=\lim_{x\to 0} \frac {1+a\cos x-ax\sin x-b\cos x}{3x^2}\stackrel{1+a-b=0; LR}=\lim_{x\to 0} \frac {-a\sin x-a\sin x-ax\cos x+b\sin x}{6x}\stackrel{LR}=\lim_{x\to 0} \frac {-a\cos x-a\cos x-a\cos x+ax\sin x+b\cos x}{6}=1 \Rightarrow -3a+b=6.$$
Hence:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-... | $$\sum_{cyc}\frac{b-c}{a}=\frac{\sum\limits_{cyc}(a^2b-a^2c)}{abc}=\frac{(a-b)(a-c)(b-c)}{abc}$$ and
$$\sum_{cyc}\frac{a}{b-c}=\frac{\sum\limits_{cyc}a(a-b)(c-a)}{(a-b)(b-c)(c-a)}=$$
$$=\frac{\sum\limits_{cyc}(a^2c-a^3-abc+a^2b)}{-(a-b)(a-c)(b-c)}=\frac{\sum\limits_{cyc}(a^2c+a^2b+abc-a^3+abc-3abc)}{-(a-b)(a-c)(b-c)}=$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Inequality $\sum\limits_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$ I recently came across the following exercise:
Prove that $$\sum_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$$
for every natural number $n \ge 1$.
I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arr... | By C-S
$$\sum_{k=1}^n\frac{1}{n+k}=1-\sum_{k=1}^n\left(\frac{1}{n}-\frac{1}{n+k}\right)=1-\frac{1}{n}\sum_{k=1}^n\frac{k}{n+k}=$$
$$=1-\frac{1}{n}\sum_{k=1}^n\frac{k^2}{nk+k^2}\leq1-\frac{\left(\sum\limits_{k=1}^nk\right)^2}{n\sum\limits_{k=1}^n(nk+k^2)}=1-\frac{\frac{n(n+1)^2}{4}}{n\cdot\frac{n(n+1)}{2}+\frac{n(n+1)(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ I have the following question.
Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ for all positive integer $n$.
My attempt:
Firstly, notice that $$n^{30}-n^{14}-n^{18}+n^2$$
$$=n^2(n-1)^2(n+1)^2(n^2+n+1)(n^2-n+1)(n^2+1)^2(n^4-n^2+1)(n^4+1)(n^8+1... | Hint:
Consider the factors of $n^{30}-n^{18}-n^{14}+n^2$
$$(n-1)^2,n^2,(n+1)^2,\left(n^2+1\right)^2,n^2-n+1,n^2+n+1,n^4+1,n^4-n^2+1,n^8+1$$
for any $n>1$ we have an even number and a multiple of $3$ in the first three factors.
With similar arguments it can be shown that at least one of the factor is a multiple of $5,7,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$ I recently found a simple slowly converging infinite series approximation to $\sqrt{x}$ which may be well known, but I am struggling to prove it and fully understand its properties. The formula is
$$\sqrt{x} = \sum_{k=0}... | Let $t=1-\frac1x$. Then, we have $\sqrt x =\left(1-t\right)^{-1/2}$.
Applying the generalized binomial theorem reveals
$$\left(1-t\right)^{-1/2}=\sum_{k=0}^\infty (-1)^k\binom{-1/2}{k}t^k\tag1$$
for $|t|<1$ ($x>1$).
We can write the term $(-1)^k\binom{-1/2}{k}$ as
$$\begin{align}
(-1)^k\binom{-1/2}{k}&=\frac{(1/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Indices/Exponents: Simplify the following expression I've been asked to simplify this expressing it with positive indices/exponents. Could someone please show me in steps how to do this? The answer is shown next to it. Thanks a lot.
$$\dfrac{{(\large a^\frac{-5}{2})}^2 b^2}{\large {a}^3(b^\frac{1}{4})^6} = \dfrac{b^\f... | $$ \begin{eqnarray*}
\left( a^{-\frac 52} \right)^2 = a^{-5}
\\ \frac 1{a^3}=a^{-3}
\\ \dfrac 1 { \left( b^{\frac 14} \right)^6}=b^{-\frac 32}
\end{eqnarray*} $$
so
$$ \begin{eqnarray*}
\dfrac {\left( a^{-\frac 52} \right)^2b^2 }{a^3\left( b^{\frac 14} \right)^6}
&=a^{-5}a^{-3}b^2 b^{-\frac 32}
\\&=a^{-8}b^\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence or divergence of $ \sum \frac{n(n+1)}{4^n} $
Convergence or divergence of
$$ \sum_{n=1}^{\infty} \frac{n(n+1)}{4^n} $$
Considering
$$ a_n= \frac{n(n+1)}{4^n} \leq \frac{n(n+2)}{4^n} =b_n$$
As the integral test is conditioned on having a function positive and decreasing. $a_n$ is first increasing then d... | The series is convergent (by the Root test: $\lim_\limits{n\to\infty} \frac{\sqrt[n]{n(n+1)}}{4}=\frac14<1)$ and its sum can be calculated. Consider the series (for $|x|<1$):
$$\sum_{n=0}^{\infty} x^{n+1}=x+x^2+x^3+x^4+\cdots=\frac{x}{1-x}.$$
Take its derivative:
$$\sum_{n=0}^{\infty} (n+1)x^n=1+2x+3x^2+4x^3+\cdots=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
If $a_n = \sum\limits_{r=0}^n \binom{n}{r} b_r$, prove $(-1)^n b_n = \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s$ Suppose that sequences of real numbers satisfy:
\begin{align*}
a_n &= \sum\limits_{r=0}^n \binom{n}{r} b_r \\
\end{align*}
Prove that:
\begin{align*}
(-1)^n b_n &= \sum\limits_{s=0}^n \binom{n}{s} (-1)^... | See binomial transform for more details. Here's a proof using (exponential) generating series.
Let
$$ B(x) = \sum_{n = 0}^\infty b_n \frac{x^n}{n!} $$
be the exponential generating series for $(b_n)$.
Then
\begin{align}
e^x B(x) &= \left( \sum_{i = 0}^\infty \frac{x^i}{i!} \right)\left( \sum_{j = 0}^\infty b_j \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
If $ab+1$ is perfect square there is a $k$ such that $ak+1$ and $bk+1$ are perfect squares
Show that if $ab+1$ is a perfect square for positive integers $a$ and $b$, then there is a positive integer $k$ such that $ak+1$ and $bk+1$ are both perfect squares.
I tried to prove that there is a $k$ such that $\gcd(ak+1, bk... | Just to give a path to the answer given by Ghartal there...
It's reasonably clear that $k$ is going to be dependent on both $a$ and $b$. So, taking $n$ defined by $n^2=ab+1$, we can try $k=a+b$ and see what happens:
$ak+1 = a(a+b)+1 = a^2+ab+1 = a^2+n^2$
$bk+1 = b(a+b)+1 = b^2+ab+1 = b^2+n^2$
So then it's easy to fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
solve $\sin x+\cos x=1$
How do we solve $\sin x+\cos x=1$?
I have solved it easily by making the substitution,
$\sin x+\cos x=1=R\sin(x+a)$ which gives the solutions $x=2n\pi, 2n\pi+\pi/2$. But when I do as follows
$$
(\sin x+\cos x)^2=1\implies 2\sin x\cos x=0\implies \sin 2x=0\implies2x=n\pi
$$
For even and odd
$$
... | One way is this: if $\sin x + \cos x = 1$, then both $\sin x$ and $\cos x$ are non negative, otherwise the other one would be greater than $1$. If one of them is different from $0$ or $1$, one has
$$1 = \sin^2 x + \cos^2 x \lt \sin x + \cos x = 1$$
This is impossible. The only remaining possibilities are $(\sin x, \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Proving the product expansion of $\sin\theta$. Where did I go wrong? You can prove the expansion$$\frac {\sin\theta}{\theta}=\left\{1-\left(\frac {\theta}{\pi}\right)^2\right\}\left\{1-\left(\frac {\theta}{2\pi}\right)^2\right\}\ldots$$By taking the expansion$$\sin n\phi=2^{n-1}\sin\phi\cos\phi\left(\sin^2\frac {\pi}n-... | We have
\begin{align}
\sin\theta&=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\
&=2\sin\frac{\theta}{2}\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)..........(1).
\end{align}
Similarly in (1) changing $\theta$ into $\frac{\theta}{2}$ and $\frac{\pi}{2}+\frac{\theta}{2}$ successively, we have
$$\sin\frac{\theta}{2}=2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Construct $4 \times 4$ magic square with fixed "1" The method I have found to generate $4\times 4$ magic squares gives me a result in which the number "1" is at of the corners of the square. How can we extend this to a method to generate a magic square, for a fixed location of number "$1$"?
The number "$1$" can be in $... | At first let's define a more powerful magic square,
which we will call $\color{Red}{\text{super-magic square}}$.
By a $\color{Red}{\text{super-magic square}}$ we mean a magic square such the the sum of any arbitrary row is equal to the sum of any arbitrary column is equal to sum of any arbitrary diagonal.
For exam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The "building blocks" of the constraints and... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $u=1$ and we need to prove that
$$(27-15\sqrt3)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq a^2+b^2+c^2$$ or
$$\frac{(27-15\sqrt3)v^2}{w^3}\geq3u^2-2v^2$$ or $f(w^3)\geq0,$ where
$$f(w^3)=(27-15\sqrt3)u^3v^2-(3u^2-2v^2)w^3.$$
We see that $f$ decreases, whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Permutations of seven numbers such that none of the sums are divisible by 3 How many permutations $(a_1,\dots, a_7)$ of $(2,3,4,5,6,7,8)$ exist such that none of the sums
$a_1$,
$a_1+a_2$,
...
$a_1+a_2+\dots a_7$
are divisible by $3$?
I tried to study the cases of the possible $3$-remainders for members of such permuta... | We can't start with a number that is congruent $0 \mod 3$. so we must start with one that is congruent $\pm 1 \mod 3$. If we start $1 \mod 3$ we can't have a $-1 \mod 0$ before we have the second $1\mod 3$. If we start with a $1 \mod 3$ then have some zeros modulo three, another $1 \mod 3$ we will have no more $1\mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simple question - How $x^4 +y^4 -2(x-y)^2$ having a minimum at $(0,0)$ I know this is a silly question, but I need to prove that
$$ f(x,y) = x^4 + y^4 -2 (x-y)^2 $$
is having a minima at $(0,0)$.
I tried doing maxima and minima method to it but the determinant i.e $AC- B^2= 0 $
So, we need to prove it algebraically on... | If $x=\sqrt2$ and $y=-\sqrt2$ then we get a value $-8$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$x^4+y^4-2(x-y)^2\geq-8$$ or
$$x^4+y^4+8-2(x-y)^2\geq0.$$
Now, by AM-GM $$x^4+y^4+8\geq2\sqrt{8(x^4+y^4)}.$$
Thus, it remains to prove that
$$2\sqrt{8(x^4+y^4)}\geq2(x-y)^2$$ or
$$8(x^4+y^4)\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Evaluate $\lim_{x \to \infty}{\frac{1}{2^x-5^x+3^x}}$. Why is my solution wrong?
Evaluate $$\lim_{x \to\infty}{\frac{1}{2^x-5^x+3^x}}. $$
My attempt:
$$\lim\limits_{x \rightarrow \infty}{\frac{1}{2^x-5^x+3^x}}
= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{1-\frac{5^x}{2^x}+\frac{3^x}{2^x}}} \\= \li... | Because it is not true that $0\times\infty\times0=0$. In fact, $0\times\infty\times0$ is indeterminate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\... | $${1+\cos\theta\over\tan^2\theta}=\cos^2\theta{1+\cos\theta\over1-\cos^2\theta}=\cos^2\theta{1\over1-\cos\theta}$$
so
$$\lim_{x\to1}{1+\cos\pi x\over\tan^2\pi x}=\lim_{x\to1}{\cos^2\pi x\over1-\cos\pi x}={1\over1+1}={1\over2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Modular Arithmetic Combination I'm trying to find the remainder of $\frac{2^{2010}}{35}$. The first two parts of the question asked for the remainder when dividing $2^{2010}$ by 5 and then by 7.
To solve for the first two problems, I used Euler's Theorem:
$$2^{2010}=(2^{3})^{670}$$
Let $a=2$ and $n=5$, then
$$2^{3}\equ... | $$2^{2010} \pmod{35}$$
Note that $2^2 \equiv 4 \equiv -1 \pmod 5$
So $2^{2010} \equiv (2^2)^{1005} \equiv (-1)^{1005} \equiv -1 \pmod 5$
Note that $2^3 \equiv 8 \equiv 1 \pmod 7$
So $2^{2010} \equiv (2^3)^{670} \equiv 1 \pmod 7$
The Chinese remainder theorem states that there is an isomorphism
$$f:\mathbb Z_{35} \to \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$ I have no idea how hard this conjecture is to prove:
Any even number $n\ge 36$ can be written as $n=a+b+c+d$ where $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb Z^+$.
Small exceptions are $n=2, 4, 6, 10, 12, 14, 20, 26, 34.\,$ Tested for $n\le 10,000$.
A counter-e... | $$a^2+b^2+c^2=d^2$$
$$a=2ps$$
$$b=2ks$$
$$c=s^2-p^2-k^2$$
$$d=s^2+p^2+k^2$$
$$2n=2ps+2ks+s^2-p^2-k^2+s^2+p^2+k^2$$
$$qt=n=s(p+k+s)$$
Lay on multipliers and pick the right.
Or other item.
$$a=2s(p-k)$$
$$b=2s(p+k)$$
$$c=p^2+k^2-2s^2$$
$$d=p^2+k^2+2s^2$$
$$2n=2p^2+2k^2+4ps$$
$$p(p+2s)=n-k^2$$
It remains to try all po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7... | Assuming both nested square roots are well-defined, we have $a=\sqrt{7b}$ and $b=\sqrt{2a}$, from which
$ab=\sqrt{14 ab}$ and $ab=\color{blue}{14}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
How do I show that the following series converges in $\mathbb{C}$? Problem Statement
Does the series
$$S := \sum_{n=0}^{\infty}{\frac{1}{2n^2 - i^n}}$$
converge in $\mathbb{C}$?
Attempt
Since
$$i^n = \begin{cases}
1, n \equiv 0 \pmod 4 \\
i, n \equiv 1 \pmod 4 \\
-1, n \equiv 2 \pmod 4 \\
-i, n \equiv 3 \pmod 4 \... | Your series is absolutely convergent since $\sum_{n\geq 1}\frac{1}{2n^2-1}$ is convergent.
What about computing a closed form for it? We are dealing with
$$ \sum_{m\geq 0}\left(\frac{1}{2(4m+1)^2-i}+\frac{1}{2(4m+2)^2+1}+\frac{1}{2(4m+3)^2+i}+\frac{1}{2(4m+4)^2-1}\right)$$
and in general
$$ \sum_{n\geq 0}\frac{1}{(n+a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Range of $|x+3|+|x-3|=6$ I can solve this equation by som tedious algebra, I got $x_1=3$ and $x_2=-3$. But according to the book the solutions are given by $x\in[-3,3]$, which means that for example $x=1$ and $x=2$ are solutions as well. How can I algebraically show this? Or can I interpret the absolutes as distances a... | $$|x+3|=\begin{cases}
x+3, & \text{ if } x\geq -3\\
-x-3,& \text{ if } x<-3
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & \text{ if } x> 3\\
-x+3,& \text{ if } x\leq3
\end{cases}$$
Thus,
$$|x+3|+|x-3|=\begin{cases}
2x, & \text{ if } x> 3\\
6, &\text{ if }-3 \leq x \leq3\\
-2x,& \text{ if } x<-3
\end{cases}$$
When $x>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ We have
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$
I would like to know if there exist a closed form for
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\le... | This is not an answer to this question but an attempt to answer a generalized question. Our generalization consists in replacing the power two in the last term in the parentheses by a power three.Now, by starting from the (corrected--the term $(1-x^2)$ in the denominator in front of the first power of the arc sine shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not su... | It is always true that
$$0\le (|a|-|b|)^2=a^2+b^2-2|a||b|$$
$$|a||b|\le \frac{a^2+b^2}{2}.$$
$$-ab\le|a||b|\le \frac{a^2+b^2}{2}.$$
As $a\not =0$ and $b\not =0$ in this case, it follows
$$-ab\le|a||b|\le \frac{a^2+b^2}{2}<a^2+b^2,$$
and it holds that
$$-ab<a^2+b^2.$$
This result leads to a contradiction considering you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$s(n^x)$ is not a perfect square for all $x$
Does there exist an $n$ such that $s(n^x)$ is not a perfect square for all positive integers $x$ where $s(m)$ denotes the sum of the digits of a positive integer $m$?
If $n = 5$, for example, then take $x = 8$ to get $5^8 = 390625$, which has digit sum $25 = 5^2$.
If $n = ... | Here is a very handwavy argument that none exists. There are $9 \cdot 10^{k-1}$ numbers with $k$ digits, of which $\sqrt{10^k}-\sqrt{10^{k-1}}=10^{\frac k2}-10^{\frac{k-1}2}=(\sqrt{10}-1)10^{\frac{k-1}2}$ are squares. We imagine that the numbers of interest are randomly squares or not, so a $k$ digit number has proba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.
I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{... | $$P_1(x)=\frac{x^{24}-1}{x^2-1}$$
$$P_2(x)=\frac{x^{12}-1}{x-1}$$
$$\frac{P_1(x)}{P_2(x)}=\frac{x^{24}-1}{x^{12}-1}\cdot\frac{x-1}{x^2-1}=\frac{x^{12}+1}{x+1}$$
Then Ruffini's rule tells us that the remainder of this reduced division is the polynomial $x^{12}+1$ evaluated at $-1$, i.e. 2. When the top and bottom of $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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Fractions in Questions and Answers
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