Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
... | You can actually continue from your third line by using that $\sqrt 2$ is irrational so if $$a^2+b^2+2\sqrt 2ab=3,$$ then actually $$a^2+b^2=3 \text{
and } 2\sqrt 2ab=0$$ so that either $a$ is zero or $b$ is $0$. Since we can't solve $a^2=3$ or $b^2=3$ in the rationals, you can conclude by using that $\sqrt 3$ is irr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find local maximum, minimum and saddle points of $f(x,y) = x^4 + y^4 - 4xy + 2$
Find maximum, minimum and saddle points of $f(x,y) = x^4 + y^4 - 4xy + 2$.
For critical points, $f_{x} = f_{y} = 0$, $f_{x} = 3x³ - 4y$ and $f_{y} = 3y³ - 4x$. Therefore $f_{xx} = 9x²$, $f_{yy} = 9y²$, and $f_{xy} = f_{yx} = -4$.
Hessian ... | By AM-GM $$x^4+y^4+2\geq4\sqrt[4]{x^4y^4}=4|xy|\geq4xy.$$
The equality occurs for $x=y=1$, which gives that $0$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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smallest possible sum of b and c
In the set of positive integers {$2, 6, 8, 9, 11, b, c$}, the mean is larger than the unique mode, which is larger than the median. What is the smallest possible sum of $b$ and $c$?
So I just started plugging in a few values to try to see what works, but values close to 6, 8 and 9 don... | First off, either $b = c$ or not. First, let's assume $b = c$. Then $b$ is the unique mode, so we must make sure that it is smaller than the mean and larger than the median. If $b \leq 9$, then the mode is not larger than the median. At the same time, the mean is equal to $\frac{2+6+8+9+11+2b}{7} = \frac{36 + 2b}{7}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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l'Hôpital vs Other Methods Consider the first example using repeated l'Hôpital:
$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} =... | $$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = \lim_{x\to0} \frac{12x^2}{12x^2+2} = \frac{0}{0+2} = 0$$
There. You can't apply l'Hospital there because the denominator doesn't go to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
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Modular arithmetic and probability $a$, $b$ and $c$ are chosen from $1$-$999$ with order and replacement, what is the probability that $a^2 + bc$ is divisible by $3$? (each choice is equally likely)
I split it up into 2 cases
$a^2$ has a remainder of $1$ while $bc$ has a remainder of $2$
$a^2$ has a remainder of $1$, $... | $b$ or $c$ have remainders of $0$ with probability $\frac{5}{9}$, rather than your $\frac13 \times 2$, since you have to take account of double counting the possibility they both do: $\frac13 \times 2 -\left(\frac13\right)^2 = \frac59$
That will turn your final calculation into $\frac4{27}+\frac5{27}=\frac13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve the equation $x^{x+y}=y^{y-x}$ in positive integers.
Let $x,y$ are positive integers. Solve the equation
$$x^{x+y}=y^{y-x}$$
smartplot(x^(x+y)=y^(y-x));
I used http://www.wolframalpha.com
| Here I will show that $y= x^{a/b}$ where a and be are two consecutive odd or even numbers:
Clearly x and y have common factors, lets suppose $y= k x$ we may write:
$y = x ^ {\frac{x+y}{x-y}}= x^{\frac{(k+1)x}{(k-1)x}}=x^{\frac{k+1}{k-1}}$
$k+1 $ and $k -1$ are two consecutive odd or even numbers:
Examples:
$k=3 ⇒ a=k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $a \sin\theta + b \cos\theta = c$ Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$
I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$
I just can't figure out the right steps to arrive at this sol... | Recall the sum formula for sine
$$A\sin(\theta+\alpha)=A\sin(\theta)\cos(\alpha)+A\cos(\theta)\sin(\alpha)$$
Equating this with the L.H.S gives
$$a\sin(\theta)+b\cos(\theta)=A\cos(\alpha)\sin(\theta)+A\sin(\alpha)\cos(\theta)$$
so we get the system
$$a=A\cos(\alpha)\\b=A\sin(\alpha)$$
and so $A=\sqrt{a^2+b^2}$.
Recall ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Basis for the space of 2 by 3 matrices whose nullspace contains $(2,1,1)$ To find a basis for the space of $2$ by $3$ matrices whose nullspace contains $(2,1,1)$
My attempt:
Let A= \begin{bmatrix}
a & b & c \\
d & e & f \\
\end{bmatrix}
Then $(2,1,1)$ is a solution to $Ax=0$
So I get,
$$2a+b+c... | You can use the Gram–Schmidt process to do this. For a certain $x_1$ and $x_2$ (such that $x$, $x_1$ and $x_2$ are linearly independent) you can get
\begin{align}
y &=\begin{bmatrix}a \\ b \\ c\end{bmatrix}=x_1-x\cdot \frac{x\cdot x_1}{x \cdot x}, \\
z &=\begin{bmatrix}d \\ e \\ f\end{bmatrix}=x_2-x\cdot \frac{x\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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find the value of ''a'' and ''b'' for which function is always increasing
Question Let f(x)= x$^{3}+ax^{2}+bx+5sin^{2}x$ be an increasing
function$\forall$ x $\in$$\mathbb{R}$$ Then$
(a) $a^2 -3b-15$<0
(b) $a^2 -3b-15$>0
(c)$a^2 -3b+15<$0
(d)a>0,b>0
My Approach i tried by making perfect square of the derivative
$\lef... | Choice $(a)$ is correct.
For $a,b \in \mathbb{R},\;$let $f:\mathbb{R} \to \mathbb{R}\;$be defined by
$$f(x)= x^3+ax^2+bx+5\sin^2(x)$$
Suppose $a,b$ are such that
$\qquad{\small{\bullet}}\;\;f\;$is an increasing function.
$\qquad{\small{\bullet}}\;\;a^2-3b -15 \ge 0$.
Our goal is to derive a contradiction.
Since $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at soluti... | Hint: Try maximizing and minimizing $ (a_k + \frac{2}{a_{k-1}})$ using
(1) If $a \le c$ and $b \le d$ then $a + b \le c + d$
(2) If $a \ge c$ and $b \ge d$ then $a + b \ge c + d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Limit of an Exponential Equation My problem is the $\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}$. I got $$\ln(B) = \ln\left(\lim_{x \rightarrow \infty} 3x \cdot \ln\left(1-\frac{4}{x}\right)\right)$$
which equates to $$\frac{\ln\left(1-\frac{4}{x}\right)}{\frac{1}{3x}} = \frac{0}{0}$$
so I can use L'... | Since $f(x)=x^{-12}$ is a continuous function in $x=e$, we obtain:
$$\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}=\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}\cdot\left(-12\right)}=\left(\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}}\right)^{-12}=e^{-12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$. How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?
I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be... | Induction on odd numbers is easier when expressed as $P(n_0)$ is true, $P(n)
\implies P(n+2)$.
Let $f(n)=n(n^2-1)$. Then $f(1)=24$ and $f(n+2)-f(n)=6 (n + 1)^2$.
If $n$ is odd, then $n+1$ is even and so $(n+1)^2$ is a multiple of $4$.
Therefore, $f(n+2)-f(n)$ is a multiple of $24$ and so is $f(n+2)$ by induction.
For... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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A curious way of generating series expansion for $\cos x$ If we take the approximation $\sin x \approx x$, then, using the trigonometric identity $1- \cos 2x = 2\sin^2 x$, and take $2\sin^2 x \approx 2x^2$, we get, after making the substitution $x \to x/2$, that
$$\cos x \approx 1 - \frac{x^2}{2}$$
Now, using the trig... | You get weird coefficients for terms beyond $x^2$ because you re using an inaccurate input for the double angle formula. If you have
$\cos x = 1-(x^2/2)+O(x^4)$,
you must accept
$\cos 2x = 1-(2x^2)+(x^4/2)+O(x^4)$;
and putting $x/2$ for $x$ then gives
$\cos x = 1-(x^2/2)+(x^4/32)+O(x^4)$
You did not make the erro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2509858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can we split a function while doing differentiation? I have a silly question here. Assume a function $$f(x)= \frac{x}{(x-1)^2} \cdot (x+2)$$
Can I write
$$\frac{d}{dx} f(x) = \frac{d}{dx} x \cdot \frac{d}{dx} (x-1)^{-2} \cdot \frac{d}{dx} (x+2)^{-1}$$
If not, then why?
| \begin{align}
f(x) &= \frac{x}{(x-1)^2} \cdot (x+2)\\
&= \frac{x^2+2x}{(x-1)^2}\\
&= \frac{x^2}{(x-1)^2} + \frac{2x}{(x-1)^2}\\
\end{align}
Now
\begin{align}
f^\prime(x) &= \frac{x^2}{(x-1)^2}\,\frac{d}{dx} + \frac{2x}{(x-1)^2}\,\frac{d}{dx}\\
&= -\frac{2x}{(x-1)^3}\ - \frac{2(x+1)}{(x-1)^3}\\
&= -2\left(\frac{x}{(x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$
I have to solve this system of equations \begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}
From the third equation I have done this to manipulate it:
$$ x^3+y^3-z^3=43-3xy \implies (x+y)(x^2-xy+y^2)-z^3=43... | $$ x^3+y^3-z^3=43-3xy \implies ...\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$
From the first equation: $\color{blue}{x+y}=\color{green}{z+1}$
$$(\color{blue}{x+y})^2=z^2+5$$
$$(\color{green}{z+1})^2=z^2+5$$
$$\vdots$$
It's easy from this point
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Consider the following table representing the distribution of the discrete random variable x: $$\begin{array}{|c|c|c|c|c|}\hline X & -4 & 0 & 1 & 4 \\ \hline f(x)=P(X=x) & 0.1& 0.3 & 0.2 & 0.4 \\\hline\end{array}$$
Compute:
a) \begin{array}E(x)&=\sum_x xf(x)\\&=(-4\cdot 0.1)+0+(1\cdot 0.2)+(4\cdot0.4) \\
&=-0.4+0.... | There is an error in part $c$, when you compute $\mathbb{E}[X^2]$.
$1.6+0+0.2+6.4=8.2 \neq 5$.
After obtaining $E(X^2)$ in part $c$, I will use it to check my answer in part $b$. Notice that $$E(X^2-X+1)=E(X^2)-E(X)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of the integral Taking $b,c \in \mathbb{R^+}$, the integral $$\int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{\sqrt{\left(a-x\right)^2+b}}$$
diverges, however the integral
$$\int_{-\infty}^{\infty} \left(\dfrac{1}{\sqrt{\left(a-x\right)^2+b}}-\dfrac{1}{\sqrt{\left(a-x\right)^2+c}}\right)\mathrm{d}x$$
converg... | As Jack D'Aurizio and Messney said in their comments, we can eliminate the constant $a$ by renaming our variable $x-a \rightarrow x$, leaving the evaluation of our integral as
$$\left. \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2517467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The number of integers $k$ for which the equation $x^3-27x+k$ has at least two distinct integer roots is? The number of integers $k$ for which the equation $x^3-27x+k$ has at least two distinct integer roots is?
My try:
Since imaginary numbers always form roots in pairs the question is the same as having two integer ro... | Substituting $c = -\frac{k}{ab}$ into the third equation gives
$$a+b-\frac{k}{ab}=0\quad\Rightarrow\quad k = ab(a+b).$$
Using that in the second equation and simplifying gives
$$(a+b)^2 = ab+27,\text{ or }a^2+ab+b^2=27.$$
Thus
$$b = \frac{1}{2}(-a\pm\sqrt{3(36-a^2)}\,).$$
Since $a$ and $b$ must both be integers, one qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges
Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.
I showed it using Abel's theorem and limit comparison test. Any other simpler method?
| Write
$$\frac{(n^3+1)^{\frac{1}{3}}-n}{\log n} = \frac{n^3+1-n^3}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)} = \frac{1}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)}$$
and observe that the denominator goes at $0$ faster than $\frac{1}{n^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519958",
"timestamp": "2023-03-29T00:00:00",
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Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2... | We first show for the $n=1$ case:
$2^3-3 = 8-3 = 5$
which is clearly divisible by 5. Next is the inductive step where we assume
$2^{3n} - 3^n=5m $ for some integer $m$.
We can rewrite this as
$2^{3n}=5m+3^n$.
Now we want to prove that $2^{3(n+1)}-3^{n+1}$ is divisible by 5. We first can simplify this a bit:
$2^{3(n+1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the probability of success in Hi-Lo? The way the game works is you are given natural numbers from 1 - 10, and one of these numbers is selected. You have to then guess whether the next randomly selected number will be higher or lower than the previous one, hence the name Hi-Lo.
For example, first number is 6. Ch... | Notice that we have independence here so the "success to fail ratio", if we carry on this experiment in infinitum, is the same as in an individual trial. Assuming that we lose in the case of a tie, the chances of winning given that the first number is $1$ is $\frac{9}{10}$, $2$ is $\frac{8}{10}$, $3$ is $\frac{7}{10}$,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}.$ >Then $\lim_{n\rightarrow \infty}a_{n}$
Let $a_{0}=1,a_{1}=2$ and for $n\geq 1\;, n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}.$
Then $\lim_{n\rightarrow \infty}a_{n}$
$\bf{Attempt:}$ $2a_{2} = a_{0} = 1$ so $\displaystyle a_{2} = \frac{1}{2}$
$\displaystyle 6a_{3}=2a_{2} = 1$ s... | Working through the first few $n$ values it can be seen that, for $n \geq 2$,
$$a_{n} = \frac{a_{0}}{n!}.$$
This yields the limit
$$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{a_{0}}{n!} = 0.$$
For verification:
\begin{align}
\frac{n(n+1) a_{n+1} - (n-1) a_{n}}{(n+1)(n+2)} &= \frac{a_{0} \, (n - (n-1))}{(n+2)... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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For which value(s) of a and b do we have, : $\lim_{x\to 0}{\frac{e^x+a x \sin(2x)-b \cos(2x)-x}{x^2}}=\frac{9}{2}?$ I have a limit
$$\lim_{x\to 0}{\frac{e^x+a x \sin(2x)-b \cos(2x)-x}{x^2}}=\frac{9}{2},$$
where $a,b\in\mathbb{R}$, and I want to find $a$ and $b$.
In other cases, I evaluate the limits of numerator and d... | From Taylor expansion at $x=0$ we have, $$\sin(2x)\sim 2x-\frac{(2x)^3}{6}$$
$$\cos(2x)\sim 1-\frac{(2x)^2}{4}$$
$$e^x \sim 1+x+\frac{x^2}{2}$$
Hence,
$$ \frac{e^x+ax\sin2x-b \cos2x-x}{x^2}\sim \frac{1+x+\frac{x^2}{2} +2ax^2 -ax\frac{(2x)^3}{6} -b-b\frac{(2x)^2}{4}-x}{x^2}\\= \frac{1-b + (\frac{1}{2} -b+2a)x^2 -ax\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show recursion in closed form I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$
where $ a_{0}=a_{1}=0$
I know what to do when I deal with sequence in form like this:
$ a_{n}=2a_{n-1}-a_{n-2}$
- when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this ... | One approach that works to get to the final form is to take the formal power series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
and try and rewrite it in terms of itself. Applying the initial conditions where necessary:
\begin{align}
f(x) &= \sum_{n=0}^{\infty} a_n x^n\\
&= a_0 + a_1 x +\sum_{n=2}^{\infty} \left(2a_{n-1}-a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms. The sum is t... | Hint:
If $f(x)=\dfrac1{1+x}+\dfrac2{1+x^2}+\dfrac4{1+x^4}+\cdots+\text{ up to } n+1\text{ terms}$
$$\dfrac1{1-x}+f(x)=\dfrac1{1-x}+\dfrac1{1+x}+\left(\dfrac2{1+x^2}+\dfrac4{1+x^4}+\cdots+\text{ up to } n+1\text{ terms}\right)$$
$$=\dfrac2{1-x^2}+\dfrac2{1+x^2}+\left(\dfrac4{1+x^4}+\cdots+\text{ up to } n\text{ terms}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and... Given $x, y, z\in \mathbb{R}$, such that $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of $x$.
| Let $xy=c$, $xz=b$ and $yz=a$.
Thus, $$\frac{x+y+z}{c+b}=\frac{1}{2},$$
$$\frac{x+y+z}{c+a}=\frac{1}{3}$$ and
$$\frac{x+y+z}{a+b}=\frac{1}{4},$$
which gives
$$\frac{c+a}{c+b}=\frac{3}{2}$$ and
$$\frac{a+b}{c+b}=2.$$
From here we obtain $$b=\frac{3}{5}a$$ and
$$c=\frac{1}{5}a$$ or
$$xz=\frac{3}{5}yz$$ and
$$xy=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the integral $\int_0^{\pi} \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}$ where $ab\neq 0$ I’ve got some problems evaluating the integral $\int_0^{\pi} \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}$. I’ve found a solution in my textbook as follows:
The integral
$\int \frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}=\frac{1}{ab}\arctan(\f... | The integrand is $\pi$-periodic and even around $0$. Therefore,
$$\begin{align}
\int_0^\pi \frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}\,dx&=2\int_0^{\pi/2}\frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}\,dx\\\\
&=2\int_0^{\pi/2}\frac{\sec^2(x)}{a^2\tan^2(x)+b^2}\,dx\\\\
&=2\int_0^\infty \frac{1}{a^2u^2+b^2}\,du
\end{align}$$
Can you fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplify the expression: $(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$ Simplify the expression:
$$(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$$
I used the general method:
$(2x+1)(2x^2+2x+1)(2x^4+4x^3+6x^2+4x+1)×...... | Write $2x + 1 = (x+1)^2 - x^2$ and use the factorization for $a^2 - b^2$ repeatedly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$T(n) = T(n-2) + log n$ Does anyone know how to solve this example?
I start with this :
$T(n) = T(n-2) + \log n$
$T(n-2) = T(n-4) + \log(n-2)$
$T(n-4) = T(n-6) + \log (n-4)$
Now i substitute back:
$T(n-2) = T(n-6) + \log(n-4) + \log(n-2)$
$T(n) = T(n-6) + \log(n-4) + \log(n-2) + \log n$
Now i see a pattern here so i su... | The correct recursion for even terms would be
$$T(2n) = T(2n-2) + \log(2n) = T(2n-2k) + \sum_{j=0}^{k-1}\log(2n-2j) $$
Substituting $k=n$ gives
\begin{align}
T(2n) &= T(0) + \sum_{j=0}^{n-1}\left[\log(2) + \log(n-j)\right]\\
&= T(0) + \log(2)\cdot n + \sum_{j=0}^{n-1} \log(n-j)\\
&= T(0) + \log(2)\cdot n + \log\left(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.
The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$
I ... | @Jaideep Khare already gave a nice answer to first question, namely he proved that,
$$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}=- \frac{n}{2}$$
where $x_i$'s satisfy
$$\color{red}{\frac{x^{n+1}-1}{x-1}= }x^n+x^{n-1}+...+x+1=0$$
It is easy to check that, $\color{blue}{x_k = e^{\frac{2r\pi}{n+1}}}$ with $0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Laplace transform $\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)$ How to find Laplace transform ?:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s) $$
I have been stuck on this problem for quite a bit. I only found in tables:
$$\mathcal{L}_x\left[\frac{1}{(x+8) \sqrt{x+8}}\right](s)=\frac{1}{\sqrt{2}... | The integral you need to compute is
$$
I(s)=\int_0^\infty dx\frac{e^{-sx}}{(x+9)\sqrt{x+8}}\ .
$$
Call $\sqrt{x+8}=t$. Then $x=t^2-8$ and $dx=2t dt$.
So
$$
I(s)=\int_{2\sqrt{2}}^\infty 2t dt\frac{e^{-s(t^2-8)}}{t(t^2+1)}=2e^{8s}\int_{2\sqrt{2}}^\infty dt\frac{e^{-s t^2}}{t^2+1}\ .
$$
This can be rewritten as
$$
I(s)=2e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
If $A+B+C=180^\circ, \cot A=2$ and $\cot B= \frac34$, what is $\sin C$?
$$\cot A = 2\\
\frac{\cos A }{ \sin A }= 2\\
\cot B = \frac34\\
\frac{\cos B }{ \sin B} = \frac34$$
What should I do with $A+B+C=180^\circ$?
| $\tan C = -\tan (A+B)= \dfrac{\tan A+\tan B}{\tan A\tan B- 1}=...=-\dfrac{11}{2}\implies \cot C = -\dfrac{2}{11}\implies \cot^2C= \dfrac{4}{121}\implies \csc^2C= 1+\cot^2C= \dfrac{125}{121}\implies \sin^2C = \dfrac{121}{125}\implies \sin C = \dfrac{11}{5\sqrt{5}}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove by induction with summation and factorials Sorry I'm not sure how to format the text for using sum. Could someone help me out with that as well. Much appreciated.
\begin{eqnarray*}
\sum_{r=1}^{n} (r^2+1)r! =n(n+1)!
\end{eqnarray*}
for all $n \geq 1 $.
| $$\sum_{r=1}^{n} (r^2+1)r! =n(n+1)!$$
$$\sum_{r=1}^{1} (r^2+1)r! = (1^2 + 1)1! = 2 * 1 = 2 = 1(1 + 1)$$
So it is true for $n=1$. Now let's see for $n=p$
$$\sum_{r=1}^p (r^2+1)r! =p(p+1)!$$
$$\sum_{r=1}^{p+1} (r^2+1)r! = (p+1)(p+2)!$$
$$\sum_{r=1}^{p} (r^2+1)r! + ((p+1)^2 + 1)(p+1)! = (p+1)(p+1)! *(p+2)$$
$$\sum_{r=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$
My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\... | Letting $u = x/(1 - x)^2$, we can write this as:
$$\lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u) + \frac{1}{2}(u^{2} - x^{2})}{u^{4}(1 - x^{2})^{4}}$$
$$= \lim_{u \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u)}{u^{4}} + \lim_{x \rightarrow 0}\frac{u^{2} - x^{2}}{2x^{4}}$$
The first limit is a bit mor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$ The question is to evaluate $$\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$$ if $x^2 \neq (n \pi-1) \forall n\in N$
I tried to rewrite the integral as $$\int \frac{x\sqrt{(2\sin(x^2+1))(1-\cos... | Hint:
Let $x^2+1=u$ then your integral simplifies to
$$\int\dfrac{\sqrt{8}}{8}\dfrac{du}{\left(\cos\frac{u}{2}\right)^\frac52}$$
now let $\sin\dfrac{u}{2}=w$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2542883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| $$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}=\lim\limits_{x \to 7} \frac{\sqrt{9+x-7} - \sqrt[3]{27+x-7}}{\sqrt[4]{16+x-7} - 2}=$$
$$=\lim\limits_{x \to 7} \frac{3\sqrt{1+\frac{x-7}{9}} - 3\sqrt[3]{1+\frac{x-7}{27}}}{2\sqrt[4]{1+\frac{x-7}{16}} - 2}=\frac{3}{2}\lim\limits_{x \to 7}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
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Find the sum of the series $S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $ $$ S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $$
I started by factorizing the denominator as $k^2+k+1$ and $k^2-k+1$
The numerator leaves a quadratic with $k$ and $k-1$ or a constant with $k+1$ and $k-1.$
I tried writing the individual ter... | So your term is equal to $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)$$
Now note $(k+1)^2-(k+1)+1=k^2+k+1$, so your term is: $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{(k+1)^2-(k+1)+1}\right)$$
and you can apply a telescoping series technique to establish the sum to $n$ is just half of $\frac{1}{1^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving the equivalence relationship within a relation $(a,b)R(c,d)$ To prove that for a relation $R$ on $\mathbb{R} \times \mathbb{R}$ such as $(a,b)R(c,d)$ exists if and only if $\exists k \in \mathbb{N}$ so that :
$k \le \sqrt{a^2+b^2} < k+1$
$k \le \sqrt{c^2+d^2} < k+1$
*
*I understand that to prove I need to sh... | Sometimes proving equivalence relations IS trivial.
For example if the equivalence is based adding or multiplication of two numbers symmetry and reflexivity are automatic as addition and multiplication is commutative.
However this one, although very easy, is not as trivial as you think. Note $(a,b) R (c,d)$ if there e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving trigonometric identity: $ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $
Prove that
$$ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $$
This is my working -
$$\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos... | $$\frac{\tan x+\tan y}{\tan x-\tan y} =\frac{\frac{\sin y\cos x+\cos y\sin x}{\cos x\cos y}}{\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}}= \frac{\sin y\cos x+\cos y\sin x}{\sin y\cos x-\cos y\sin x}=\frac{\sin (y+x)}{\sin (y-x)} $$
which gives the results since we know that
$$ \sin(x-y) =\sin y\cos x-\cos y\sin x $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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What is $\operatorname{ord}_{22}(5^6)$?
Find $\operatorname{ord}_{22}(5^6)$.
So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I co... | By the Chinese remainder theorem, it is enough to find the order of $5^6$ $\!\!\pmod{2}$ (which is obviously $1$) and the order of $5^6\equiv5$ $\!\!\pmod{11}$. Since $5$ is a quadratic residue $\!\!\pmod{11}$, due to
$$\left(\frac{5}{11}\right)=\left(\frac{11}{5}\right)=\left(\frac{1}{5}\right)=1$$
the order of $5\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Show that $\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$ for $n\ge 5.$ Question is to show that
$$\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$$
when $n\geq5$, $x \in \mathbb R $ and $i$ is an imaginary unit.
My effort:
$$|\int_{-n}^{n}e^{iy^2}dy|\leq \int_{-n}^{n}|e^{iy^2}|dy=\int_{-n}^{n}|\cos(y^2)+i\sin(y^2)|dy$$
$$ \l... | $$\left|\int_{-n}^{n}e^{iy^2}dy\right|^2 =\left|\int_{-n}^{n}\cos{y^2}dy +i\int_{-n}^{n}\sin{y^2}dy\right|^2 \\= 4\left(\int_{0}^{n}\cos{y^2}dy \right)^2 + 4\left(\int_{0}^{n}\sin{y^2}dy \right)^2\\= 4[I^2_n + J^2_n]$$
This is somehow related to this: Prove only by transformation that: $ \int_0^\infty \cos(x^2) dx = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Calculate $\lim\limits_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$ In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.
Here is my approach/what I have done so fa... | $$\int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x=\int\limits_0^\infty \frac{n^3 \sin(y)}{n^4y + n y(1 +n^3 y^3)} \, d y=\frac{1}{n} \int\limits_0^\infty \frac{\sin(y)}{y + y^3 +y/n^3} \, d y$$
Here we make the change of variables $y=n x$.
The integral $$ \int\limits_0^\infty \frac{\sin(y)}{y + y^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| It is quite practical to exploit the identity
$$ (1+x)^2 (1-2x+3x^2-4x^3) = 1-5x^4-4x^5 \tag{A}$$
from which
$$\begin{eqnarray*} \sqrt{1-2x+3x^2-4x^3} &=& \frac{\sqrt{1-5x^4-4x^3}}{1+x} \\&=&\left(1+O(x^4)\right)(1-x+x^2-x^3+O(x^4))\tag{B}\end{eqnarray*}$$
and the coefficient of $x^3$ in the RHS of $(B)$ is trivially $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac... | $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}...=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots=\dfrac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Limit of $\lim_{n\to\infty}n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdots\sin\left(\frac{1}{n}\right)}$ $$\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$$
I have noticed ... | $$\begin{align}\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)&=\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions a... | I like the Taylor expansion approach, but wondered if there was a more "Cal I" approach.
Using
$$
A-B = (A^{1/6} - B^{1/6})(A^{5/6} + A^{4/6}B^{1/6} + A^{3/6}B^{2/6} + \cdots + B^{5/6})
$$
with $A = (x^2+1)^3$ and $B = (1+x^3)^2$ gives
$$
\begin{multline*}
(x^2+1)^{1/2} - (x^3+1)^{1/3}\\
=\frac{(x^2+1)^3 - (x^3+1)^2}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$\iiint (y^2+z) dxdydz$ under region $x+y+z=2, x=2, y=1, z=y$ I'm having trouble trying to find the interval.
I tried mapping it to $u=x+y+z$, $v=x$, $w=y$ to get
$$u=2$$
$$v=2$$
$$w=1$$
$$u-v-w=w$$
Is this in the right direction?
| Your region is the union of the two regions $\{(x,y,z)\in\mathbb{R}^3:2-y-z\leq x\leq 2, z\leq y\leq 1, 0\leq z\leq 1\}$ and $\{(x,y,z)\in\mathbb{R}^3:2-y-z\leq x\leq 2, -z\leq y\leq 1, -1\leq z\leq 0\}$.
This can be determined by sketching the region in the $(y,z)$ plane.
Your integral is
$$\int_0^1\int_z^1\int_{2-y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solution for this inequality $a^2 - 4b + 8 > 0$ I dont know where I messed up, this is what I have done:
This is the isolated x of a function
$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)$$
The root above must be positive and greater than $0$ so a function could have inflection points.
So: $a^2 - 4 b ... | To guarantee that exist
$$\sqrt{a^2 - 4 b + 8} $$
you have to set
$$a^2 - 4 b + 8 \geq 0\implies a^2\geq4b-8$$
and consider 2 cases
*
*$$4b-8<0 \implies a^2\geq0>4b-8$$
*$$4b-8\geq0 \implies -\sqrt{4b-8}\leq a^2\leq \sqrt{4b-8}$$
To solve
$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)>0 \iff\pm\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Splitting the square root of complex function into real and imaginary parts I have these functions below: $$\sqrt{(x+iy)^2-a^2}$$
$$\frac{b(x+iy)}{\sqrt{(x+iy)^2-a^2}}$$
How do I split these to get the real and imaginary parts of these functions?
If anyone could help me out, that really would be helpful!!!!
It will hel... | Let $\sqrt{(x+iy)^2-a^2}=c+id$
$\implies (x+iy)^2-a^2=(c+id)^2$
Comparing the real & the imaginary parts
$\implies c^2-d^2=x^2-y^2-a^2\ \ \ \ (1)$
$2cd=2xy$
$(c^2+d^2)^2=(c^2-d^2)^2+(2cd)^2=(x^2-y^2-a^2)^2+(2xy)^2=(x^2+y^2)^2+a^4-2a^2(x^2-y^2)$
$c^2+d^2=\sqrt{(x^2+y^2)^2+a^4-2a^2(x^2-y^2)}\ \ \ \ (2)$
Use $(1),(2)$
O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& =... | $6\equiv1\pmod7, ~8\equiv-1\pmod7$, hence $$6^n+8^n\equiv1^n+(-1)^n\pmod7$$ $1^n+(-1)^n=0~\iff~n$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the probability of getting at least one pair in Poker? Problem:
A 5-card hand is dealt from a perfectly shuffled deck of playing cards. What is the probability of each
of the hand has at least two cards with the same rank.
Answer:
By a rank, I mean a card like a $2$ or a king. The set of all poker hands is ${52... | As Eric Fisher pointed out in his answer, the probability of obtaining at least one pair can be found by subtracting the probability of obtaining no pairs from $1$. If no pairs are obtained, we must select five different ranks. There are $\binom{13}{5}$ ways to select the five ranks. For each rank, we must select on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.... | Here are some interesting observations, which are too long to be included as a comment.
A useful reference can be found here.
Denoting $\displaystyle\sum_{r=1}^n r^m=\sigma_m$, we note that
$$\begin{align}\sigma_3&=\sigma_1^2\tag{1}\\
\frac{\sigma_5}{\sigma_3}&=\frac {4\sigma_1-1}{3}\tag{2}\\
\frac {\sigma_7}{\sigma_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| $$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}\sqrt[3] {\cos 3x}}{x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^6x-\cos^32x\cos^23x}{6x^2}=$$
$$=\lim_{x\rightarrow0}\frac{\cos^2x(1-\cos^2x)(128\cos^8x-256\cos^6x+200\cos^4x-69\cos^2x+9)}{6x^2}=$$
$$=\frac{128-256+200-69+9}{6}=2.$$
I used $$x^6-y^6=(x-y)(x^5+x^4y+x^3y^2+x^2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Evaluate integral : How to evaluate integral
$$\int \frac{x^2}{\sqrt{x^2-25}}dx=?$$
My Try :
$$\int \frac{x^2+25-25}{\sqrt{x^2-25}}dx=\int \frac{x^2-25}{\sqrt{x^2-25}}+\frac{25}{\sqrt{x^2-25}}dx$$
$$\int \frac{x^2-25}{\sqrt{x^2-25}}+\int\frac{25}{\sqrt{x^2-25}}dx$$
Now what ?
| Continuing with your method, $$\int \frac{x^2}{\sqrt{x^2-25}}\,dx=\int \left(\sqrt{x^2-25} + \frac{25}{\sqrt{x^2-25}}\right)\,dx$$ Now $$\int \sqrt{x^2-25}\, dx=\frac{x\sqrt{x^2-25}}2-\frac{25}2\ln(x+\sqrt{x^2-25})+C_1 \tag{1}$$ using integration by parts, with $f(x)=x^2-25$ and $g'(x)=\dfrac1{\sqrt{x^2-25}}$.
Also, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find this binomial summation How to calculate this?$$\sum_{x = 0}^n \sum_{y = 0}^n a^x b^y\binom{x + y}{y}$$
I tried the following approach by replacing $x + y = t$ and then solving $\sum_{t\ =\ 0}^{2n}\sum_{r=max(0,\ t - n)}^{min(n,\ t)} a^r b^{t - r}\binom{t}{r}$ but this got me nowhere. Maybe we can reduce it... | We can generate a recurrence like this.
Let's denote $S(t) = \sum_{r\ =\ max(0, t - n)}^{min(n, t)}a^rb^{t - r}\binom{t}{r}$. Now $\forall 1 \leq t \leq n,\ S(t) = (a+b)S(t - 1)$
So, for calculating $S(t),\ \forall t > n$ notice \begin{align*}
S(n + 1) = (a+b)^{n+1} - \binom{n + 1}{0}a^{n + 1} - \binom{n + 1}{0}b^{n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Feasibility and boundedness of non-linear programming In linear programming, we have the following results:
*
*primal problem infeasible --> dual problem infeasible or unbounded
*dual problem infeasible --> primal problem infeasible or unbounded
*primal problem unbounded --> dual problem infeasible
*dual problem ... | Here's an example involving a primal-dual pair of semidefinite programming problems in which the primal has multiple optimal solutions but the dual is infeasible.
Our SDP primal-dual pair has the primal problem:
$\min \mbox{tr}(CX)$
subject to
$\mbox{tr}(A_{i}X)=b_{i}$, $i=1, 2, \ldots, m$
$X \succeq 0$
with the du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Aczel's inequality Can anyone here please explain how to notice Aczel's identity in a certain question. I mean in questions involving inequalities, on understanding the inequality one gets a hint to apply a certain inequality e. g. Cauchy Schwarz, Jensen's inequality, Chebyshev's inequality, etc. So how does a que... | Suppose $$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2 \tag{A}$$
The quantity on the RHS is nonnegative so for signs to work out there are just two cases:
$$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1) > 0 \text{ and } (b_1 ^ 2 + b_2 ^ 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that this polynomial has three different real roots There is polynomial $f(x)=x^3 +ax^2+bx+c$ and $ab=9c$ and $b<0$. Like in the title. I've only come up so far with $9x^3 + 9ax^2 + 9bx+ab=0$ and factoring here does nothing.
| I have expanded Eric Towers' comments into an answer. Hope this can help OP. The second method is more efficient and error-prone doesn't work well, unlike the first one.
Method 1: Discriminant
Given that $ab=9c$ and $b<0$, the discriminant is
\begin{align}
\Delta &= 18(1)abc-4a^3c+a^2b^2-4(1)b^3-27(1)^2c^2 \\
&= 162c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Solving the quadratic congruence (using some lemma)
Find all $x$'s such that $x^2 \equiv 17 \pmod{ 128}$
Now, I've read one solution here (fixed URL). What I don't understand is how in the first place one can find a particular solution, $a$? (in order to find the all four).
| $$x^2 \equiv 17 \pmod{128}$$
$$x^2 \equiv 1 \pmod 4 \implies x \in \{1,3\}$$
\begin{align}
(1+4n)^2 &\equiv 1 \pmod{16} \\
1 + 8n &\equiv 1 \pmod{16} \\
8n &\equiv 15 \pmod{16} \\
n &\in \{ \ \}
\end{align}
\begin{align}
(3+4n)^2 &\equiv 1 \pmod{16} \\
9 + 8n &\equiv 1 \pmod{16} \\
8n &\equiv 8 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \f... | This one is a cakewalk. Just use the definition $$K(1/2)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-(1/2)\sin^{2}x}}=\frac{\sqrt{2}}{2}\int_{0}^{\pi}\frac{dx}{\sqrt{2-\sin^{2}x}}$$ and then put $t=\tan (x/2)$ so that $dx=2/(1+t^2)\,dt$ and $$2-\sin^{2}x=2-\frac{4t^2}{(1+t^2)^{2}}=2\cdot\frac{1+t^{4}}{(1+t^{2})^{2}}$$ and hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Probability for a Poisson Process I have a problem with a solution. I find it hard to understand some of the steps in the solving process. Can someone help me understand whats going on here?
"Calculate the probability P(X(1)=1, X(4)=4|X(2)=2, X(3)=3, X(5)=5) for a Poisson process X(t) with arrival rate λ > 0."
Solutio... | No, it's not using the independent increments property, but it's kind of a preparation to use it… let's consider a simpler example.
You want to calculate: $$P(X(1) = 1, X(2) = 3)$$ unfortunately $X(1)$ and $X(2)$ aren't independent so we cannot split this into two probabilities… but we can rewrite this a bit to get inc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary Inequality from a local olympiad test The problem I'm stuck with is the following :
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$a+x \ge b+y \ge c+z \ge 0 $ and $ a+ b+c = x+y+z$
Prove that $ay+bx \ge ac+xz$
As easy as it sounds, I didn't achieve any significant progress after several tri... | \begin{align}
(ay+bx) - (ac+xz) &= a(\color{red}{y-c}) + x(b-z) \\
&= a(\color{red}{a+b - x -z}) + x(b-z)\\
&= a(a-x) + (a+x)(b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a^2-x^2) + (a+x)(b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(a-x+2b-2z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(\color{red}{a+b-x-z}+b-z)\\
&= \tfrac12(a-x)^2 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How many tries on average before I see the same value $N$ times in a row? If I repeatedly roll a fair, $X$-sided die, on average how many rolls should I expect to make before I happen to roll the same value $N$ times in a row?
I've found questions on here with answers when $X=2$, or where $N=2$, or where you're looking... | Let $E_k$ denote the expected number of rolls that are still needed if $k$ rolls have passed with equal result. We write recurrence relations for $E_k,0\leq k\leq N-1$ and calculate the wanted expectation $E=E_0$ from them.
We obtain
\begin{align*}
E_0&=E_1+1\tag{1}\\
E_1&=\frac{1}{X}(E_2+1)+\frac{X-1}{X}(E_1+1)\\
E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find orthonormal vectors for given vectors using Gram-Schmidt Doing an exercise after completing the G.S. Lecture 17 and stuck.
Suppose we have three vectors $\begin{pmatrix}1 \\ 1 \\0 \end{pmatrix}$ , $\begin{pmatrix}1 \\ 0 \\1 \end{pmatrix}$, $\begin{pmatrix}0 \\ 1 \\1 \end{pmatrix}$ and the aim is to transform ... | HINT
$u_2-u_3$ is orthogonal to $u_1$
in this way you can simplify the calculation assuming $u_1*=u_1$, $u_2*=u_2-u_3$
NOTE
When you calculate c you should use $u_3$, $u_1$ and $b$.
Remember the meaning of the procedure: you are subtracting to $u_3$ the components of $u_3$ on orthogonal vectors $u_1$ and $b$, to do thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Calculate the determinant of $A-5I$ Question
Let $ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lam... | Row-reducing the matrix $A-5I$ is quicker.
$$
\begin{gather}
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
6 & 2 & 8 & 9 & 10 \\
11 & 12 & 8 & 14 & 15 \\
16 & 17 & 18 & 14 & 20 \\
21 & 22 & 23 & 24 & 20
\end{pmatrix} \longrightarrow
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
23 & 6 & -1 & 2 & 0 \\
32 & 9 & 6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Find the value of $n$. Write the value of $n$ if the sum of n terms of the series $1+3+5+7...n =n^2$.
I'm not getting the right value if I proceed with the general formula for finding sum of n terms of a arithmetic series. The general summation formula for arithmetic series is $\frac{n(2a+(n-1)d)}{2}$, where $a$ is the... | If we add $2+4+...+(n-1) = 2(1+2+3+...+\frac{n-1}{2})$ to both sides of the equation, we get
$$1+2+3+...+n = n^2+2\bigg(1+2+3+...+\frac{n-1}{2}\bigg)$$
which can be written as:
$$\frac{n(n+1)}{2} = n^2+2\bigg[\frac{\big(\frac{n-1}{2}\big)\big(\frac{n+1}{2}\big)}{2} \bigg]$$
that is
$$\frac{n^2+n}{2} = n^2+\frac{n^2-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find max value of $a$ Let $a, x, y, z \in \mathbb{R}_{>0}$ such that $xyz=1$.
Find maximum value of $a$ such that is satisfied the inequality
$$ \sum_{cyc} \frac{x}{(x+1)(y+a)} \ge \frac{3}{2(1+a)} $$
After doing some calculation the max value of $a$ seems to me to be $2$ but I can’t prove it.
| Let $y=z$ and $x\rightarrow+\infty.$
Thus, $$\frac{1}{a}\geq\frac{3}{2(1+a)},$$
which gives $a\leq2$.
We'll prove that $a=2$ it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{x}{(x+1)(y+2)}\geq\frac{1}{2}$$ or
$$\sum_{cyc}(2x^2y+xy-3x)\geq0.$$
Now, let $x=\frac{p}{q}$ and $y=\frac{q}{r}$, where $p$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given a fair coin, on average what is the number of tosses needed to get 3 heads in a row? My attempt:
Let $H$ be the outcome heads, and $T$ be tails. Let $X$ be the random variable "number of tosses required to get 3 consecutive heads". The minimum number of trials (tosses) is 3. Thus we have infinite sample space:
$$... | The more general case of asking for the average number $E$ of rolls of a fair $X$-sided die ending with a run of $N$ consecutive heads is
answered in this MSE post as
\begin{align*}
E=\frac{X(X^N-1)}{X-1}\tag{1}
\end{align*}
Here we apply (1) by considering a coin as two-sided die ($X=2$) and tosses ending in heads wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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For a primitive pythagorean triple $(a, b, c)$, prove $(c+b), (c-b)$ are both squares. Problem
For a primitive pythagorean triple $(a, b, c)$ (with $a$ odd, $b$ even), prove $(c+b), (c-b)$ are both squares.
My thoughts
Having $a$ be odd, gives $a^2$ as odd also.
From $a^2 + b^2 = c^2$ we have $a^2 = (c-b)(c+b)$.
I have... | Given $A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$
$$C-B=m^2+n^2-2mn=(m-n)^2$$
$$C+B=m^2+n^2+2mn=(m+n)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Error while integrating reciprocal of irreducible quadratic $\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$ I am trying to derive a formula for indefinite integral of reciprocal of irreducible real quadratic.
$$\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$$, $b^2 - 4ac \lt 0$. According to WolframAlpa it should come out as: $$\frac{... | In my opinion you can save many hassles with a slightly different method. Set $\sqrt{4ac-b^2}=D$ and do
\begin{align}
\int\frac{1}{ax^2+bx+c}\,dx
&=\int\frac{4a}{4a^2x^2+4abx+4ac}\,dx \\[6px]
&=\int\frac{4a}{(2ax+b)^2+(4ac-b^2)}\,dx \\[6px]
&=\int\frac{4a}{D^2t^2+D^2}\frac{D}{2a}\,dt
&& 2ax+b=Dt,\quad dx=\frac{D}{2a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove :$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$ In fact initially I wanted to prove that
$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$$
Which by the accepted answer here fails to be true.
@Barry Cipra advised me to ask this question in a different post:
How can I now prove that:
$$\sqrt... | For $n\geq 2$,
\begin{align*}
\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} & = 2^{1/4+1/8+1/16+\ldots+1/2^n} \cdot 3^{1/8+1/16+1/32+\ldots+1/2^n}\cdot \ldots \cdot n^{1/2^n} \\
& < 2^{1/2}\cdot 3^{1/4} \cdot 4^{1/8} \cdot \ldots \cdot n^{1/2^{n-1}} \\
& = \sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}} \\
&<3\end{align*}
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Righ... | The left hand side of $$\frac{1}{|x+1|} < \frac{1}{2x}$$ is positive so the right hand side must be positive. Therefore $x>0$ which implies $x+1>0$ and $|x+1|=x+1.$
Upon substitution, the inequality becomes $$\frac{1}{x+1} < \frac{1}{2x}$$ with the condition $x>0.$ The solution is $0<x<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cro... | it is $$(x^2-1)(3(x^2-1)^2+7(x^2-1)+4)=(x^2-1)x^2(3x^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How do I integrate these functions? I deleted my last question, since some of you wanted me to rewrite the question properly. I feel sorry for inconvenience, but please understand that this is the first time I use $\texttt{MathJax}$.
Up to now, I tried every method I know in integration, like substituition, partial fra... | Here is an alternative evaluation for the second integral.
\begin{align*}
\int_0^1 \frac{dx}{(1 + x^2) [x^2 (1 - x)]^{1/3}} \, dx
&= \int_0^1 \sum_{n = 0}^\infty (-1)^n x^{2n} \frac{1}{x^{2/3} (1 - x)^{1/3}} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 x^{2n - 2/3} (1 - x)^{-1/3} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
} |
Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
| Assume we know $a_1$ and $b_1$.
First, $a_{n+1}+b_{n+1}=3(a_n+b_n) = \cdots = 3^n(a_1+b_1)$.
And secondly, $a_{n+1}-b_{n+1}=-(a_n - b_n) = \cdots = (-1)^n (a_1 - b_1)$.
And
$$ a_{n+1} = \frac{(a_{n+1}+b_{n+1})+(a_{n+1}-b_{n+1})}{2}$$
$$ b_{n+1} = \frac{(a_{n+1}+b_{n+1})-(a_{n+1}-b_{n+1})}{2}$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.
In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.
I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to bui... | Using the identity
\begin{align}
\cos3\alpha+\cos3\beta+\cos3\gamma
&=
1-\frac{r\,(3\,\rho^2-(r+3\,R)^2)}{R^3}
\tag{1}\label{1}
,
\end{align}
where $r$ is the inradius, $R$ is the circumradius
and $\rho$ is the semiperimeter
of the triangle,
condition
\begin{align}
\cos3\alpha+\cos3\beta+\cos3\gamma&=1
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting f... | "$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8"$
"From here, we get"
"$2^x + 15 = 2^8$"
Uh....What? No, you don't.
You get
$2^x . 2^{-3}*2^3 + 15. 2^{-3}2^3 . 2^x = 2^8*2^3"$
$2^x + 15*2^x = 2^{8+3}$
So $16 *2^x = 2^{11}$
$2^{x+4} = 2^{11}$
$x+4 = 11$
$x = 7$.
====
I'd find it easier to do. Let $y = x-3$ so
$2^y + \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
limit with summation and product
Given $L=\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\bigg(\frac{1}{r!}\prod^{r}_{i=1}\left(\frac{i}{2}+\frac{1}{3}\right)\bigg)$. then $\lfloor L \rfloor$ is
Try: $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1}{r!}\bigg[\left(\frac{1}{2}+\frac{1}{3}\right)\cdot \left(\... | Comment too long,
so entered as an answer:
Wolfy also says that
$\sum_{n=1}^∞ \prod_{j=1}^n(1/2 + 1/(b j))
= 2^{1+2/b} - 1
$.
Wolfy also says that
(the first line is what it returns,
the remaining lines
are my attempt
to properly simplify)
$\begin{array}\\
\sum_{n=1}^∞ \prod_{j=1}^n(\frac1{a} + \frac1{b j})
&= -\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+... | Because if $$1^2-2^2+...+(-1)^n(n-1)^2=\frac{(-1)^nn(n-1)}{2}$$ then
$$1^2-2^2+...+(-1)^n(n-1)^2+(-1)^{n+1}n^2=$$
$$=(-1)^{n+1}\left(n^2-\frac{n(n-1)}{2}\right)=(-1)^{n+1}\frac{n(n+1)}{2}.$$
The base is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $n \in \mathbb{N}\setminus\{1\}$ then $\gcd(n^2-1,3n+1)=1$ if and only if $n$ is even If $n\in\mathbb N\setminus\{1\}$ then $\gcd(n^2-1,3n+1)=1$ if and only if $n$ is even.
Can somebody help me prove this problem?
| Let $a=n^2-1$, and let $b = 3n+1$.
Let $d = \text{gcd}(a,b)$.
First, suppose $n$ is odd.
Then $a,b$ are both even, so $d > 1$.
Next, suppose $n$ is even.
Then $a,b$ are both odd, so $d$ must be odd.
Now simply note that
\begin{align*}
9a-(3n-1)b &= 9(n^2-1)-(3n-1)(3n+1)\\[4pt]
&=(9n^2-9)-(9n^2-1)\\[4pt]
&=-8\\[4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$
I tried homogenization and the BW (https://artofproblemsolving.com/comm... | In principle it is, if not always, almost very often that such a problem can be solved using techniques from optimization. For instance one can consider the following maximization problem:
\begin{equation}
\max_{a,b,c\in C} f(a,b,c)
\end{equation}
where the constraint set $C:=\{a,b,c\in\mathbb{R}_+:abc=1\}$ and $f(a,b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Solve the initial value problem $x^2y'+y(x-y)=0$ $x^2y'+y(x-y)=0$
$y(1)=-1$
Can someone point me in the right direction on this? I started doing this:
$y'+p(x)y=q(x)y^n$
So
$p(x)=x-y, q(x)=1/y^2, n=2$
$x^2y'+y(x-y)=(1/y^2)y^2=1$
Then ran out of ideas...
(I don't want you give give me the answer, just steer me the righ... | $$y'+\frac{y}{x}-\frac{y^2}{x^2}=0$$
$$xy'+y=\frac{y^2}{x}$$
$$(xy)'=\frac{(xy)^2}{x^3}$$
Substitute $z=xy$:
$$z'=\frac{z^2}{x^3}$$
integrate:
$$\int \frac {dz}{z^2}=\int\frac{dx}{x^3}=\frac{-1}{2x^2}+K$$
$$ \frac {1}{z}=\frac{1}{2x^2}+K$$
$$ \frac {1}{xy}=\frac{1}{2x^2}+K$$
$$ x=y(\frac 1 2 +Kx^2)$$
$$ y=\frac x {(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fourier series expansion of $\frac{\pi^4}{96}$ and $\frac{\pi^4}{90}$ I have to prove that:
\begin{equation}
\frac{\pi^4}{96}=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^4} \qquad \mbox{and} \qquad \frac{\pi^4}{90}=\sum_{n=1}^{+\infty}\frac{1}{n^4}
\end{equation}
My idea:
I've expanded $f(x)=|x|$ in a real Fourier series $\fo... | From what you've already proved:
$$\sum_{n=1}^\infty\frac 1{n^4}=\sum_{n=0}^\infty\frac 1{(2n+1)^4}=\sum_{n=1}^\infty\frac 1{(2n)^4},$$
whence
$$\sum_{n=1}^\infty\frac 1{n^4}-\sum_{n=1}^\infty\frac 1{(2n)^4}=\frac{15}{16}\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{96}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the integral $\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$ The question is $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$$
I have tried to multiply both numerator and denominator by $1-\sqrt{x}$ but can't proceed any further, help!
| We have
$\mathrm{\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\ dx}$
Put $\mathrm{\sqrt{x} = \cos t}$
Differentiate Both Sides w.r.t. $\mathrm{x}$
$\mathrm{\frac{dx}{2\sqrt{x}}= - sint \ dt}$
$\mathrm{dx =- 2 \ cost\ sint \ dt}$
So, we get now:
$\mathrm{-\int \sqrt{\frac{1- cost}{1+ cost}}\ \times 2\ cost \ sint \ dt}$
= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
limit of: $\begin{align} \lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n} \end{align}$ How to get the limit of:
$\begin{align}
\lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n}
\end{align}$
I have formed it to:
$\begin{align}
\lim_{n\to \infty}\frac{n(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{n}
\end{align}$
And tried to use L'Hopital b... | $$\lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n}=\lim_{n\to \infty}\frac{-2\sqrt{n}}{\sqrt{n-2\sqrt{n}}+\sqrt{n}}=\lim_{n\to \infty}\frac{-2\sqrt{n}}{\sqrt{n}-1+\sqrt{n}}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the next cyclic inequality Let $a,b,c>0$ such that $$a+b+c=1$$
prove that
$$cyclic\sum \frac {ab}{\sqrt {c+ab}}\le\frac {1}{2} $$
By symetrie, i proved it by assuming $a=b=c$ but i cannot justify this hypothesis.
| By C-S $$\left(\sum_{cyc}\frac{ab}{\sqrt{c+ab}}\right)^2=\left(\sum_{cyc}\frac{ab}{\sqrt{c(a+b+c)+ab}}\right)^2=\left(\sum_{cyc}\frac{ab}{\sqrt{(a+c)(b+c)}}\right)^2\leq$$
$$\leq\sum_{cyc}ab\sum_{cyc}\frac{ab}{(a+c)(b+c)}.$$
Thus, it's enough to prove that
$$\sum_{cyc}ab\sum_{cyc}\frac{ab}{(a+c)(b+c)}\leq\frac{(a+b+c)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{... | There's a small mistake. You seem to think that to study the convergence of the sequence$$\left(\frac{\sqrt{n-2\sqrt n}}{\sqrt n}\right)_{n\in\mathbb N}$$is equivalent to the convergence of its square. It is true in this case because it is a sequence of real numbers greater than $0$; but it is false in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$... | Here is yet another answer, which does not require any of the full, explicit formulae for sums of powers.
Consider the sum
$$
S(n, p) = \sum_{m = 1}^{n} m^p\, .
$$
I can rewrite this as
$$
S(n, p) = n^{p+1}\sum_{m = 1}^{n} \frac{1}{n}\;\left(\frac{m}{n}\right)^p\, .
$$
The sum in the above expression is a Riemann sum w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
The complex equation $x^3 = 9 + 46i$ has a solution of the form $a + bi$ where $a,b\in \mathbb Z$. Find the value of $a^3 + b^3$ The complex equation $x^3 = 9 + 46i$ has a solution of the form $a + bi$ where $a,b\in \mathbb Z$. Find the value of $a^3 + b^3$ .
| Let $x=a+bi$, where $\{a,b\}\subset\mathbb R$.
Thus, $$(a+bi)^3=9+46i$$ or
$$a^3-3ab^2=9$$ and
$$3a^2b-b^3=46,$$
which gives
$$46(a^3-3ab^2)=9(3a^2b-b^3)$$ or
$$(2a+3b)(23a^2-48ab+3b^2)=0,$$ which gives
$$2a+3b=0.$$
Thus, $$3\cdot\frac{9}{4}b^3-b^3=46,$$ which gives $b=2$, $a=-3$ and $a^3+b^3=-19.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Geometry Problem: Find the area of $ABCD$ square
My Problem is:
The $E-$ point on the $AC$ diagonal is marked on the $ABCD$ square.If $BE= 13$, $CE=17$ find the square area.
My way:
$$a=\frac{k+17}{\sqrt2}$$
$$13^2=k^2+\frac {(k+17)^2}{2}-\frac {2k(k+17)}{\sqrt2}×\frac{\sqrt2}{2} \Rightarrow k^2+289=338 \Rightarrow... | Let $ABE_1E$ be a parallelogram.
Thus, $BECE_1$ is isosceles trapezoid ($BE=E_1C=13$) and since $EE_1\perp BC$,
we obtain:
$$BE_1^2+EC^2=BE^2+E_1C^2$$ or
$$BE_1^2+17^2=13^2+13^2,$$ which gives $$BE_1=7.$$
Id est, $$AC=AE+EC=7+17=24$$ and
$$S_{ABCD}=\frac{1}{2}\cdot24^2=288.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ equal subintervals and taking limit.
Answer:
The length of the i... | $$\begin{align} \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x& = \lim_{n \to \infty} \sum_{i=1}^{n} [4-3 (2+i \frac{2}{n} )] \frac{2}{n}
\\&= -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n}\\ &= -\large \lim_{n \to \infty} \left(\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
B... | No there isnt. This is a well known phenomenon with a latin name, Casus Irreducibilus.
If a third degree equation with real coefficients is irreducible, and has all real solutions then these solutions cannot be expressed as roots of real numbers.
| {
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"url": "https://math.stackexchange.com/questions/2623575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is... | By definition of integral without substitutions:
$$(x^2+1)^7x^3=(x^2+1)^7(x^2+1-1)x=(x^2+1)^8x-(x^2+1)^7x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Convergence of $(x_n)_n$ if $2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdots x_n}$ $\forall~n$ I became interested in solving a question in a recently deleted post concerning the relation $$2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_n}$$ for $n\in\mathbb{N}$.
The question was whether it converges and if so w... |
Your answer is correct in fact we have the following relations
$$\color{blue}{x_{n+2} =2018^{\frac{2}{n+2}}x_{n+1}^{\frac{n}{n+2}}}$$and
$$\color{red}{x_{n+1}=2018^{1+\frac{2}{n(n+1)}}x_1^{-\frac{1}{n(n+1)}}.} \to 2018$$
See the prove below
Let us establish a recursive formula3. we have $$2018^2=x_{n+1}\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Searchingthe polar form of $z = 2+i(1+\sqrt{3})$ Find the polar form of:
$$z = 2+i(1+\sqrt{3})$$
Is there any other way to solve that in different way than by finding $|z|$ that is really "ugly" ($|z| = \sqrt{8+2\sqrt{3}}$)?
| you are right $z=\rho \cdot e^{i\cdot \theta}$ with $\rho=|z| = \sqrt{8+2\sqrt{3}}$. But Note that $$ \frac{1 + \sqrt3}{2} = \cos\frac{π}{3}+\sin\frac{π}{3}=\sqrt2\cos(\frac{π}{3}-\frac{π}{4} )=\sqrt2\cos\frac{π}{12}$$
and then ,
$$\theta=\arg (z)=\arctan\left(\frac{1 + \sqrt3}{2}\right)=\color{red}{\arctan\left(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the... | Hint: at the point $x_0=\sqrt{c}$ the slope of the tangent is $2x_0=2\sqrt{c}$, the slope of the normal is $k=-\frac{1}{2x_0}=-\frac{1}{2\sqrt{c}}$. Now use the fact that you know the point coordinates to get the normal equation as $y=kx+m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Flip a coin 5 times. What is the probability that heads never occurs twice in a row? Suppose I flip a coin $5$ times in a row. There are $2^5$ possible outcomes, i.e: HHHTH, HTTTT, HTHTH, etc. I want to know the probability that heads never occurs twice in a row.
I drew out $32$ events that can occur, and I found out t... | Let head be represented by 0 and tail by 1. Then a sequence of heads and tails can be represented by a binary number.
For 2 tosses ($n = 2$), the possible sequences are:
00
01
10
11
There are $2^2$ total possible sequences and among them the number of sequences where heads never occurs twice in a row is 3. So for $n =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n... | You were going right till the very last step. You need to apply the limit after computing the sum. The limit converts to Riemann integral:
$$\lim_{n\to \infty}\sum_{k=1}^{n} \frac{k^4}{n^5} = \int_{0}^{1} x^4 dx = \frac{1}{5}$$
Since you are not familiar with integration, another method is to find $\sum_{k=1}^{n} k^4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Regular polygon areas in ratio 3:2 Two regular polygons are inscribed in the same circle of radius $r$. First one has $k$ sides and second has $p$ sides. We are given that their areas have a ratio of $1.5$.
Calculate the area of a regular polygon inscribed in the same circle, having number of sides the sum of the other... | There are only limited possibilities for $n$ because we require $n\ge3$ to get any area and also $\frac32\frac n2\sin\frac{2\pi}n<\pi$ because the bigger polygon has to be smaller than the circumcircle. So we enumerate the possibilities:
$$\begin{array}{r|l}n&\text{Area}\\\hline
3&1.299\\
4&2.000\\
5&2.378\\
11&2.974\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of the given matrix in $a, b, c$?
If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of
$$
\begin{pmatrix}
a & b & c \\
b & c & a \\
c & a & b \\
\end{pmatrix} \,\,?
$$
(A) $p^2+6q \quad$
(B) $1 \quad$... | $$\det\begin{pmatrix}a&b&c\\ b&c&a \\ c&a&b\end{pmatrix}=(a+b+c)\det\begin{pmatrix}1&1&1\\ b&c&a \\ c&a&b\end{pmatrix}=0$$
since $a+b+c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.