Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way... | The answer is $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $
$\sum n^4 $ is not really a problem.
You can do it using this method here
https://www.coursera.org/learn/calculus1/lecture/uR7YK/what-is-the-sum-of-n-4-for-n-1-to-n-k
I don't know if this is helpful but this is how i got it... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Faster way to get a rational integral $\int_2^3 \frac{x^{100}+1}{x^3+1} \, dx$ So my problem is calculating this integral:
$$
\int_{2}^{3} \frac{x^{100}+1}{x^3+1} \, dx
$$
I know it can be done by polynomial division but that is really tedious I would have to divide $\approx 30$ times.
And i know that Mathematica would... | We have
$$ J=\int_{2}^{3}\frac{x-1}{x^3+1}\,dx = \frac{1}{3}\log\left(\frac{21}{16}\right)\tag{1}$$
by partial fraction decomposition, and that is the only annoying part, since
$$ I = \int_{2}^{3}\frac{x^{100}+1}{x^3+1}\,dx = -J+\int_{2}^{3}x\cdot\frac{x^{99}+1}{x^3+1}\,dx \tag{2}$$
and
$$ \frac{x^{99}+1}{x^3+1} = 1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if the sum of the digits of $a+b$, $b+c$ and $a+c$ is $<5$, then is the sum of the digits of $a+b+c$ less or equal than 50? This is a problem from a math olympiad:
Let $S(x)$ denote the sum of the digits of a natural number $x$ in its decimal representation. Do there exist three natural numbers $a$, $b$ and $c$ with $... | Counterexample (the sum is 51):
a = 555 444 555 5
b = 555 555 444 5
c = 444 555 555 5
a+b = 1111 000 000 0
a+c = 1000 000 111 0
b+c = 1000 111 000 0
a+b+c = 1555 555 555 5
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix information that can be extracted from the characteristic polynomial
Let $A\in M_{3}(\mathbb{R})$ with $f_{\lambda}(x)=x^3+4x^2+5x+2$
Find the eigenvalues, the trace and the determinant of $A$
$f_{\lambda}(x)=x^3+4x^2+5x+2$ the dividers of $2$ are $-1,-2$ we get that
$(-1)^3+4(-1)^2+5(-1)+2=0$ dividing the ... | One has for an $n\times n$ matrix:
$$ f_\lambda(x) = \sum_{k=0}^n x^{n-k} (-1)^{k} \ {\rm tr } \wedge^k A
= \sum_{k=0}^n x^{n-k} (-1)^k \sum_{i_1<...<i_k} \lambda_{i_1} ... \lambda_{i_k} $$
where $\lambda_1,...,\lambda_n$ are the $n$ eigenvalues of $A$.
For the traces one has in terms of elements of $A$:
$$ {\rm tr} \... | {
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Show that $9|2^{2n+1}+2^{4n+1}-4$
Show that $9|2^{2n+1}+2^{4n+1}-4$ ( n is a positive integer):
1-Using induction
2-Don't use induction
I have noticed that $2^m \equiv 2^{m+6} \pmod 9 $ but couldn't use it to solve the problem!
| $2^{2n+1}+2^{4n+1}-4 = f(2^{2n})$ for $f(x)=2x^2+2x-4=2(x-1)(x+2)$.
Now $2^{2n} \bmod 9$ is $1,4,7,1,4,7,\dots$ and
$f(1) \equiv 0 \bmod 9$
$f(4) = 36 \equiv 0 \bmod 9$
$f(7) \equiv 0 \bmod 9$
So $f(2^{2n}) \equiv 0 \bmod 9$ for all $n$.
| {
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Does this constitute a valid proof that $\frac{x^2}{1+x^4} \leq \frac{1}{2}$?
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$
First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expre... | Start from
$$x^4-2x^2+1=(x^2-1)^2\ge0$$
which is true for all $x\in\Bbb R$. Then rearrange:
$$2x^2\le1+x^4$$
$$\frac{x^2}{1+x^4}\le\frac12$$
where the last step is done because 2 and $1+x^4$ are always positive for $x\in\Bbb R$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| In order to prove that $5$ is a divisor of $P(n)$ for even $n$, consider the following cases:
*
*$n\equiv\color\red0\pmod{10} \implies n+5\equiv\color\red0+5\equiv5\pmod{10} \implies 5|P(n)$
*$n\equiv\color\red2\pmod{10} \implies n+3\equiv\color\red2+3\equiv5\pmod{10} \implies 5|P(n)$
*$n\equiv\color\red4\pmod{10}... | {
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"timestamp": "2023-03-29T00:00:00",
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If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the ... | Note that $a^2+b^2+a^2b^2 = (a^2+1)(b^2+1)-1$
Then this becomes a special case of the Brahmagupta-Fibonacci identity:
$
(x^2+y^2)(z^2+w^2) = (xz+yw)^2+(xw-yz)^2
$
Take $x = a, y = 1, z=b, w = 1$ and you get:
$
(a^2+1)(b^2+1) = (ab+1)^2+(a-b)^2 = (a^2+a+1)^2+1
$
since $b = a+1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge... | We have $1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ and thus $$3(ab+bc+ca)=3/2-3/2(a^2+b^2+c^2)$$.
Using this and the fact you found, we find that an equivalent inequality is
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}-\frac{3}{2}a^2-\frac{3}{2}b^2-\frac{3}{2}a^2\geq1$$
Now $\frac{a}{1-a}\geq a(9/4a+3/4) $ (since $1/(1-a)\g... | {
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"question_score": "4",
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Partial sum of $k>2$ consecutive positive integers has no solution in prime Progress:
Let $a_1,...,a_k$ a sequence of $k$ consecutive positive integers. If $k>2$, show that no partial sum of the sequence $a_1+...+a_k$ has a solution in prime.
I made a heuristic attempt and found a pattern in the progression below:
$$... | The sum $a+(a+1)+\dots + (a+k-1)$ is equal to $\frac{k(2a+k-1)}{2}$.
If $k>2$ then both $k$ and $(2a+k-1)>2$. So your number is the product of two numbers that are larger than $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If we know the GCD and LCM, how to find all unordered pairs? If we know that $\gcd(a,b)=12$ and $\operatorname{lcm}(a,b)=360$, how can we find all unordered pairs $a,b$?
The answer in the back of the book I got this from gives $(12,360), (24,180), (36,120), (60,72)$ as the solutions.
| When solving such questions, it is best to think in terms of prime factorizations instead of integers. So we start by prime-factorizing $360$ and $12$:
\begin{align*}
360 &= 2^3 \cdot 3^2 \cdot 5^1 \\
12 &= 2^2 \cdot 3^1
\end{align*}
$a$ and $b$ must divide $360$, since $360$ is the LCM, so they cannot have any prime f... | {
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"url": "https://math.stackexchange.com/questions/1922319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Proof of an inequality involving different exponents I was asked the following interesting question but I can not come up with a proof.
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a^3}{b^3 + c^3} + \frac{b^3}{a^3 + c^3} + \frac{c^3}{a^3 + b^3} \geq \frac{a^2}{b^2 + c^2} + \frac{b^2}{a^2 + c^2} + \fra... | I think the solution with derivative is very smooth:
$$f'(x)=\sum\limits_{cyc}\frac{a^x\ln a(b^x+c^x)-a^x(b^x\ln b+c^x\ln c)}{(b^x+c^x)^2}=$$
$$=\sum\limits_{cyc}\frac{a^x(b^x(\ln a-\ln b)-c^x(\ln c-\ln a))}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}(\ln a-\ln b)\left(\frac{a^xb^x}{(b^x+c^x)^2}-\frac{a^xb^x}{(a^x+c^x)^2}\right)=$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922546",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to find a trajectory $\sigma(t)$ to represent the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$? Given the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$, find a trajectory $\sigma(t)$ which represent it.
So far, I have this:
The standard equation for an ellipse is:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
So,
$4x^{2}+9y^{2}=36\}$... | For a circle $$x^2+y^2 = r^2$$
we can try
$$x=r\cos(t)$$
$$y=r\sin(t)$$
because
$$x^2+y^2 = r^2\cos^2(t)+r^2\sin^2(t) = r^2$$
So for an ellipse
$$a^2x^2+b^2y^2 = c^2$$
we can try
$$x=\frac{c}{a}\cos(t)$$
$$y=\frac{c}{b}\sin(t)$$
In our case we have
$$2^2x^2+3^2y^2 = 6^2$$
so let us try try
$$x=\frac{6}{2}\cos(t)$$
$$y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Posch Teller potential analytical classical solution I am trying to obtain the classical solution for a 1D Schroedinger equation with a symmetric Posch-Teller potential, the equation is given by,
$ \frac{d^2 x}{d t^2} - V \frac{\sinh(x)}{\cosh^3(x)} = 0 $
Where $V>0$. The boundary conditions are given by $x(t_0) = x_0... | I think there is an analytic solution, I am not sure how convenient its form is. We have the equation
$$\frac{d^2 x}{dt^2} = \frac{V\sinh(x)}{\cosh^3(x)}.$$
Multiply both sides of the equation by the first derivative to obtain
$$\frac{d^2 x}{dt^2} \, \frac{d x}{dt} = \left(\frac{V\sinh(x)}{\cosh^3(x)} \right)\frac{d x... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all $x$ so $\frac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square... Find all $x$ so $\dfrac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square...
Clearly 1 is solution, but I have to show that there is an infinity... I see that this happens when the numbers $1,2,...,x$ have a whole quadratic root. Am I on the right road?
| The main result is:
$$1^2+2^2+3^2+\dots + x^2=\frac{x(x+1)(2x+1)}{6}$$
So you need $\frac{(x+1)(2x+1)}{6}=y^2$ for some $y$.
Or:
$$2x^2+3x+1 = 6y^2$$
From here, we are essentially going to "complete the square" on the left side.
Multiply both sides by $8$ and you get:
$$16x^2+24x+8=(4x+3)^2-1=48y^2$$
So you need infin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the differential equation $(x+5)y'+5y=\frac{3\sin x}{(x+5)^4}, x>-5$ What does it mean to solve the differential equation? For example,
Solve the differential equation
$$
(x+5)y'+5y=\frac{3\sin x}{(x+5)^4}, x>-5
$$
I believe I am to put this in the form of $y' + P(x)y = Q(x)$. Then integrate using $$v(x)=e^{\in... | $$y'(x)(5+x)+5y(x)=\frac{3\sin(x)}{(5+x)^4}\Longleftrightarrow$$
Divide both sides by $5+x$:
$$y'(x)+\frac{5y(x)}{5+x}=\frac{3\sin(x)}{(5+x)^5}\Longleftrightarrow$$
Let $v(x)=\exp\left[\int\frac{5}{5+x}\space\text{d}x\right]=(5+x)^5$
Multiply both sides by $v(x)$:
$$y'(x)(5+x)^5+5y(x)(5+x)^4=3\sin(x)\Longleftrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
| This is a very intriguing problem. First of all, $x\geq 0$ and $a\geq\sqrt{a+x}$ must hold for real $x$.
Squaring both sides of the equation we get $x^2=a-\sqrt{a+x}$.
$(x^2-a)^2=a+x$
$\Rightarrow x^4-2ax^2-x+a^2-a=0$
$\Rightarrow (x^2-x-a)(x^2+x+(1-a))=0$
$\therefore x= \frac12\pm\sqrt{a+\frac14},-\frac12\pm\sqrt{a-\... | {
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"timestamp": "2023-03-29T00:00:00",
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Number of roots of $x^4 + x^3 \sin(x) + x^2 \cos(x) = 0$ How many roots do the following polynomial have?
$$x^4 + x^3 \sin(x) + x^2 \cos(x)$$
Obviously $0$ is a root. Dividing by $x^2$, I am left with
$$x^2 + x\sin(x) + \cos(x)$$
How do I know this polynomial has no root?
| Another way is as follows: the function $$f(x)= x^4 + x^3 \sin(x) + x^2 \cos(x)$$ is an even function (since the cosinus is even and the sinus is odd), hence it is enough to see at $x\ge 0$. The derivative $$f’(x)=(4+\cos(x))x^3+2x^2\sin(x) +2x\cos(x)$$ is clearly positive for $x$ positive. Consequently $f(x)$ is incr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Why $4a^3+27b^2<0 \iff x^3+ax+b=0$ has exactly three solutions? According to Exc. 8 Sec. 4.3 of the book Advanced Calculus by Fitzpatrick,
For numbers $a$ and $b$, prove that the following equation has exactly three solutions if and only $4a^3+27b^2<0$: $$ x^3+ax+b=0, \ \ \ \ \ x \ \text{in} \ \mathbb R.$$
Long tim... | Let $\alpha$, $\beta$ and $\gamma$ be the roots, then $\alpha+\beta+\gamma=0$.
The discriminant:
\begin{align*}
\Delta &= (\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2 \\
&=
-27\alpha^2 \beta^2 \gamma^2-
4(\beta \gamma+\gamma \alpha+\alpha \beta)^3 \\
&= -27b^2-4a^3
\end{align*}
For real coefficients $a$,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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From Complex number to polar I have to calculate $(3 + 3i)^{829}$
What I did:
$$|3 + 3i| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9}=\sqrt{18}$$
The argument of $3 + 3i$ is $45º$.
| The key is in the title: you need to write the number in polar form:
$$x+\mathrm i y = r(\cos\theta+\mathrm i \sin\theta)$$
where $r$ is the modulus of $x+\mathrm i y$ and $\theta$ is the argument.
In your case, you have $x + \mathrm i y = 3 + 3\mathrm i$.
The modulus is given by $\left|3+3\mathrm i\right|=\sqrt{3^2+3^... | {
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"timestamp": "2023-03-29T00:00:00",
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If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$ If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$
Should I manipulate the equation? I'm stuck on this problem. Answers are greatly appreciated.
| Trig-free solution: If $x^2+y^2=1$ then
$$
(x+y)^2 + (x-y)^2 = 2(x^2+y^2)=2
$$
so the largest value of $(x+y)^2$ is $2$, attained when $x=y$.
| {
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Find the smallest number $n > 1$ such that $n = a^2 = b^3 = c^3$ Find the smallest number $n > 1$ such that
\begin{equation}
n = a^2 = b^3 = c^5
\end{equation}
for three integers $a, b$ and $c$.
I thought about finding $\mathrm{lcm}(a^2,b^3)$ and $\mathrm{lcm}(b^3, c^5)$ by using the formula
\begin{equation}
\mathrm{... | Let $n=p_1^{q_1}p_2^{q_2}...p_n^{q_n}$ be the prime factorization of $n$. Then we see that for $n$ to be a square, a cube, and a fifth power, we must have $2|{q_1}, 2|{q_2}, ... 2|{q_n}$, $3|{q_1}, 3|{q_2}, ... 3|{q_n}$, and $5|{q_1}, 5|{q_2}, ... 5|{q_n}$. To find the smallest $n$, we can let $n$ be of the form $p_1^{... | {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\sum\limits_{k=0}^{n-1} \sin^t(\pi k/2n)$? How to evaluate $\displaystyle\sum_{k=0}^{n-1} \sin^t\left(\frac{\pi k}{2n}\right)$?
$t,n$ are constants $\in \Bbb{Z}$.
My try:
$$\begin{align}
\zeta:=e^{i\pi/2n} \implies & \sum_{k=0}^{n-1}\sin^t\left(\frac{\pi k}{2n}\right) \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1... | Here is an answer when $t = 2s$ is even.
Let $n, s$ be non-negative integers. Then with $\zeta = \mathrm{e}^{i\pi/n}$ and $\omega = \zeta^2 = \mathrm{e}^{2i\pi/n}$, we have
\begin{align*}
2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right)
&= \sum_{k=0}^{n-1} \left( \frac{\zeta^k + \zeta^{n-k}}{i} \right... | {
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"answer_id": 1
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Prove that G is laying on K and find the ratio of AH to HB. We are given $Δ$ $ABC$ with $∠C = 90°$ and $AC > BC$. The circle K has a diameter CH (where CH is a height in $Δ$ $ABC$) and intersects with $AC$ in point $P$ and with $BC$ in point $Q$. If $PQ$ bisects $CG$, where $G$ is the centroid of $Δ$ $ABC$, prove that ... |
We have that $PQ$ is another diameter of the circle having $CH$ as a diameter and $CM\perp PQ$.
Assuming $BC=a<b=AC$, we have $CH=PQ=\frac{ab}{\sqrt{a^2+b^2}}$ and $$CG=\frac{2}{3}CM=\frac{2}{3}AM=\frac{\sqrt{a^2+b^2}}{3}.$$
Moreover, since $CPQ$ is similar to $CBA$,
$$ CP = \frac{a}{\sqrt{a^2+b^2}}PQ = \frac{a^2 b}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof of Euler's theorem in complex numbers I am an engineer and not a mathematician, but I have used the formula stated below many number of times.$$e^{i\theta} = \cos{\theta} + i\sin{\theta}$$ Just out of curiosity, what is the proof to this theorem? I know that adding the Taylor series of the sines and the cosines a... | It can also be seen with series. Robert mentioned this but here are the details. It needs to be noted that the series expansions for $e^z$, $\sin x$, and $\cos x$, with $z \in \Bbb C$ and $x \in \Bbb R$, each have an infinite radius of convergence. Also, these series all converge absolutely. Absolute convergence ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
What is the Taylor series expansion of $\frac{2z+1}{(z+1)^2}$ at $z=i$
What is the Taylor series expansion of $\frac{2z+1}{(z+1)^2}$ at $z=i$?
I try to use that $a_n=\frac{f^{(n)}(i)}{n!}$, but the formula is too tedious. How should I proceed?
| We are looking for a representation
\begin{align*}
\frac{2z+1}{(z+1)^2}=\sum_{n=0}^\infty a_n(z-i)^n\qquad\qquad a_n\in\mathbb{C}
\end{align*}
We obtain
\begin{align*}
\frac{2z+1}{(z+1)^2}&=\frac{2(z-i)+(1+2i)}{((z-i)+(1+i))^2}\\
&=\frac{1}{(1+i)^2}\cdot\frac{2(z-i)+(1+2i)}{\left(1+\frac{z-i}{1+i}\right)^2}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method I came across the following integral:
$$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$
I know it could be solved with resuide method, and I want to know if there are some real methods can sov... | Here is a variant of @tired's solution. In this solution, you are only required to know:
*
*How to expand a rational function into partial fraction decomposition,
*How to compute complex logarithm, and
*The following claim:
Claim. For any $z \in \Bbb{C}\setminus(-\infty, 0]$ we have
$$ I(z) := \int_{0}^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
If $20x=\pi$, what is $\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}$? If $20x=\pi$,
what is $$\frac{\cos 4x - \cos 8x}{\cos 4x\cdot \cos 8x}?$$
I've tried using the factor formula on the numerator but I haven't managed to get anywhere with it...
This is a multiple choice question with options $4$, $2$, $1$, $-1$, and... | $$\\ 20x=\pi \\ 10x=\frac { \pi }{ 2 } \\ \frac { \cos { \left( 10x-6x \right) -\cos { \left( 10x-2x \right) } } }{ \cos { \left( 10x-6x \right) \cos { \left( 10x-2x \right) } } } =\frac { \sin { \left( 6x \right) -\sin { \left( 2x \right) } } }{ \sin { \left( 6x \right) \sin { \left( 2x \right) } } } =\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Write the following as a fraction $1+\dfrac{1}{6x}+\dfrac{1}{x^2+3x}$
I keep getting answer like $\dfrac{6x^2 + 19x + 9}{6x(x+3)}$ but I think
it's wrong because I get different answers. Please help me
Greetings from Russia!
| \begin{align}
1 + \frac{1}{6x} + \frac{1}{x^2+3x} &= 1 + \frac{1}{6 x} + \frac{1}{x(x+3)} \\
&= 1 + \frac{x + 3}{6x(x+3)} + \frac{6}{6x(x+3)} = 1 + \frac{x + 9}{6x(x+3)} \\
&= \frac{6x^2 + 19x + 9}{6x(x+3)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopita... | Dumping my answer here to a different but similar question which was labeled as a duplicate and would no longer accept the answer that I had already typed.
$\lim\limits_{t\to1}\frac{5}{1-t^5}-\frac{2}{1-t^2}=$
$\lim\limits_{t\to1}\frac{5}{(1-t)(1+t+t^2+t^3+t^4)}-\frac{2}{(1-t)(1+t)}=$
$\lim\limits_{t\to1}\frac{5+5t-2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 10
} |
rational solutions of $x^4+x^2y^2+y^4=x^2$ I am trying to find rational solutions of $x^4+x^2y^2+y^4=x^2$, except for $(\pm 1,0), (0,0) $. I guess these three solutions are the only ones, but I failed to prove it. It seems that $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2) $ may be useful, but I do not know how to proceed.
| Let $a=x^2,b=y^2$. Then the equation simplifies to $a^2 + ab+b^2 = a$, which on grouping becomes $a^2 + a(b-1) + b^2=0$. Solving for $a$,
$$
a= \frac{1-b + \sqrt{1+b^2-2b-4b^2}}{2} = \frac{1-b + \sqrt{1-2b-3b^2}}{2}
$$
(The other root would be negative, but a is positive).
Therefore, we would like $1-2b-3b^2$ to be a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Let $a, b \in \mathbb{R}^+$. Prove that $\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$.
Let $$a, b \in \mathbb{R}^+$$ Prove that $$\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$$
We can make the denominators common on the LHS by AM-GM but the problem is... | Consider
$$\frac{2a^2+3b^2}{2a^3+3b^3}(a+b) = 1 + a b \frac{2a+3b}{2a^3+3b^3}$$
and from Holder, $(2a^3+3b^3)(2+3)^2 \geqslant (2a+3b)^3$, so we may write
$$LHS \times(a+b) \leqslant 2 + 25ab\left(\frac1{(2a+3b)^2} + \frac1{(2b+3a)^2}\right)$$
so it is enough to show that
$$25ab[(2b+3a)^2+(2a+3b)^2] \leqslant 2(2a+3b)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $m$ and $n$ so that the given function has the range $[-3, 5]$ Find $m$ and $n$ real numbers so that $f(x) = \frac{3x^2 + mx + n}{x^2 + 1}$ takes all and only the values from the interval $[-3, 5]$.
I started by solving the following double inequality:
$$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$$
From the left... | $\frac{3x^2 + mx + n}{x^2 + 1}$ reaching a maximum value of $5$ implies that $5(x^2+1)-3x^2+mx+n = 2x^2-mx-(n-5)$ has a double root. Similarly, $\frac{3x^2 + mx + n}{x^2 + 1}$ reaching a minimum value of $3$ implies that $3x^2+mx+n-(-3)(x^2+1) = 6x^2+mx+(n+3)$ has a double root. Quadratic expressions have a double ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How can the equation of a hyperbola be $xy=1$? We know that the standard form of a hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which is not coherent with $xy=1$. So, how can we say that $xy=1$ is a hyperbola?
| Actually equations in the form of $Ax^2+2Bxy+C^2+Dx+Ey=F$ can be simplified to an equation with variables $u,v$ depends on $x,y$ such that $\{u,v\}$ is an orthogonal basis for $\Bbb R^2$, i.e. one can rotate the graph in $uv$ coordinate into $xy$ plane.
Let $Ax^2+2Bxy+C^2+Dx+Ey=F$.
Then this equation can be written as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .
| Variational Approach
Without loss of generality, assume that $a+b+c=1$. Then
$$
\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\tag{1}
$$
has an interior extreme when
$$
\frac{\delta a}{(1-a)^2}+\frac{\delta b}{(1-b)^2}+\frac{\delta c}{(1-c)^2}=0\tag{2}
$$
for any variation so that
$$
\delta a +\delta b+\delta c=0\tag{3}
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $A+B+C=\pi$, prove: $\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$
If $A+B+C=\pi$, prove that:
$$\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$$
My Attempt:
Here, $A+B+C=\pi$
Now,
$$\begin{align}
LHS &=\cos(B+2C)+\... | For the RHS, you can use a known formula for the product of three cosines (see eg. here):
$\cos x \; \cos y \; \cos z = \frac{1}{4} \Big(\cos (x+y-z) + \cos (y+z-x) + \cos (z+x-y) + \cos (x+y+z)\Big)$
Considering your RHS, you set $x = \frac{B-C}{2}$, $y = \frac{C-A}{2}$, $z = \frac{A-B}{2}$, and plug it into the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\int \frac{\sqrt{9-x^2}}{x}$ I don't understand how the integral
$$\int \frac{\sqrt{9-x^2}}{x}=\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$$
I keep getting $-3/x +c$ as the answer.
| Differentiate:
$$\left(\sqrt{9-x^2}-3\log(\sqrt{9-x^2}+3)+3\log x+C\right)'=$$
$$=-\frac x{\sqrt{9-x^2}}+\frac{3x}{\sqrt{9-x^2}}\cdot\frac1{\sqrt{9-x^2}+3}+\frac3x=$$
$$=\frac x{\sqrt{9-x^2}}\left(-1+\frac3{\sqrt{9-x^2}+3}\right)+\frac3x=$$
$$=\require{cancel}\frac x{\cancel{\sqrt{9-x^2}}}\frac{-\cancel{\sqrt{9-x^2}}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the sum $\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \cdots$ Suppose $|x| < 1$. Can you give any ideas on how to find the following sum?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$
| Prove first that for any $x\in(-1,1)$ the identity
$$ \frac{1}{1-x}=\prod_{k\geq 0}\left(1+x^{2^k}\right) \tag{1}$$
follows from a telescoping product or the fact that every $n\in\mathbb{N}^*$ has a unique representation in base-$2$. By considering $\frac{d}{dx}\log(\cdot)$ of both sides of $(1)$, we get:
$$ \frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Using elementary row or column operations to compute a determinant How do you use elementary row or column operations to find the determinant of the following matrix?
$$\begin{bmatrix} 1 & 7 & -3\\
1 &3 & 1\\
4&8&1\end{bmatrix}$$
| The key idea in using row operations to evaluate the determinant of a matrix is the fact that a triangular matrix (one with all zeros below the main diagonal) has a determinant equal to the product of the numbers on the main diagonal. Therefore one would like to use row operations to 'reduce' the matrix to triangular f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
To find the locus of a point
A point moves such that the sum of its distances from the coordinate axes is equal to its distance from the circle $x^2+y^2=4$.
I tried and got this equation for the locus:
$$x+y=\sqrt{x^2+y^2}-2$$
Should I replace $x+y$ with $|x|+|y|$ because distances are absolute? In that case, how wil... | Not only should you replace $x$ and $y$ with $|x|$ and $|y|$, you should also take the absolute value of the right-hand side because it represents a distance too:
$$|x|+|y|=\left\vert\sqrt{x^2+y^2}-2\right\vert$$
To solve this for $y$, we can first restrict ourselves to non-negative $x$ and $y$, since substituting $x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_... | hint:
factor the denominator as follows
$x^2-1=(x-1)(x+1)$.
the numerator $x$ goes to $1$.
in the denominator, $(x+1)$ goes to $2$ but $(x-1)$ tends to $0$.
so, the denominator goes to $0$.
but $x$ is less than $1$, so the denominator goes to $0$ by negatives values, let's say it tends to $0^-$.
thus the final limit is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find the $n$th derivative of $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2$ In this case $n=16$ and the point is $x=1$.
I know $\sqrt{x}$ is not differentiable in zero, but this function actually has left derivatives.
Also, I know that $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2=x^2+x-2+2\sqrt{(x^2-1)(x-1)}$, so it is enogth to compute the ... | HINT:
Since $x\ge 1$, we may write $\sqrt{(x^2-1)(x-1)}=(x-1)\sqrt{x+1}$.
Then, use the General Leibniz Rule for product differentiation
$$\frac{d^n}{dx^n}\left((x-1)\sqrt{x+1}\right)=\sum_{k=0}^n\binom{n}{k}\frac{d^k}{dx^k}(x-1)\frac{d^{n-k}}{dx^{n-k}}(x+1)^{1/2}$$
Finish by recognizing that the only surviving terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$ Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$
given that $a_0 = 0$, and $a_1 = 3$
My Attempt:
Get the characteristic equation and solve it.
For homogeneous equation
$x^2 -4x + 4 = 0$
$x = 2 $ or $ x = 2$
Hence, $a_n^h = (A+Bn)\cdot2^n $
Guess a particular solution: $n^22^nC$
$n^22^nC... | $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Incorrect method to find a tilted asymptote Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.
I would like to find out, though, what is wrong with the following incorre... | The problem with your second approach is that you've kept more precision than your approximation actually has — you can say $y \approx x$, but you don't have enough precision to clarify that more specifically to $y \approx x+1$ (or any other translate).
In more detail,
$$ \frac{1}{1 + \frac{2}{x}} \approx 1 - \frac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
What is $\sin(nx)$ iteration in terms of $\sin A$ and $\cos A$? I want to use sum angle formulas, $\sin(A+B)=\sin A\cos B+\cos A\sin B$
to get for any angles, $\sin(nA)$ in terms of powers of $\sin A$ and $\cos A$.
I know there are other ways, but I want to use trigonometry and iteration on that. The goal is to carry ... | The angle addition formulas for sine and cosine are nicely captured by this matrix formula
$$
\left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) =
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right)
\left( \begin{array}{c} s_{n-1} \\ c_{n-1}\end{array} \right)
$$
where $s_k = \sin(kx)$, $c_k = \cos(kx)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove $4p-3$ is a square knowing that $n\mid p-1$ and $p\mid n^3-1$, $p$ prime I really need some help at this problem:
Let $p$ be a prime number and $n$ a natural number, $n\ge2$ such that $n \mid p-1$ and $p \mid n^3-1$. Prove that $4p-3$ is a square.
So $p \mid (n-1)(n^2+n+1)$
What if $p \mid n-1$?
Treating t... | Given that $p\mid(n^3-1)$, thus we have $p\mid(n-1)$ or $p\mid(n^2+n+1)$. We also have $n\mid(p-1)$. Let us make cases.
Case 1: $p\mid(n-1)$ and $n\mid(p-1)$. Since $p-1\ge 1$ and $n>1$. Thus we can conclude that $p\mid(n-1)\implies p\le n-1$ and $n\mid(p-1)\implies n\le p-1$. Thus we have $n\le p-1\le n-2\implies n\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the following Definite Integral $\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$, where $\{.\}$ is the fractional part of $x$. Integrate: $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$$
My attempt: We know that within the interval $[0,1)$, the following is true:
$$\{x+1\}=x$$
$$\{x^2+2\}=x^2$$... | When you say $\{x\}=x+1$ for $x\in [-1,0) $ you are saying that the fractional part of $-0.3$ is $0.7$. Was that your intended meaning?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove that the sequence $\frac{\sin(k)+k^2}{\root\of{1+3k^4}}$ converges using the $\epsilon$ criteria. Propose that the limit $L = \frac{1}{\root\of3}$.
Given $\epsilon \gt 0 \exists k_\epsilon$ such that for all $k \gt k_\epsilon$ we have $\lvert\frac{\sin(k)+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert \lt ... | First, noting that $|\sin(k)|\le 1$, then given $\epsilon>0$ we have
$$\left|\frac{\sin(k)}{\sqrt{1+3k^4}}\right|\le \frac{1}{k^2}<\frac{\epsilon}{2}$$
whenever, $k>K_1=\sqrt{\frac{2}{\epsilon}}$.
Next, we have
$$\left|\frac{k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|=\left|\frac{1}{\left(\sqrt3k^2+\sqrt{1+3k^4}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$
So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\... | The substitution is incorrect : the tangent is not bijective on the interval $[0,2\pi]$. First, you need to restrict yourself to an interval on which the tangent behaves better. Using the $\pi$-periodicity of the function you want to integrate, you can show that:
$$\int_0^{2 \pi} \frac{1}{a \sin^2 (\theta)+b \cos^2 (\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Why $(x-2)^2$ cannot be divided by $(x-2)(x-3)$ I know the question may seem ridiculous. My question is:
Why can't $(x-2)^2$ be divided by $(x-2)(x-3)$?
I know the answer is obvious if we do the division by hand. However, we also know that if remainder of $\frac{P(x)}{Q(x)}$ is zero, then $Q(x)$ is a divisor of $P(x)... | If you divide $(x-2)^2$ with $(x-2)(x-3)$ you get $\frac{(x-2)}{(x-3)}$. The polynomial $q(x)$ divides the polynomial $p(x)$ if there exist a polynomial $g(x)$ such that $p(x) = g(x)\times q(x)$. You have that $\frac{(x-2)}{(x-3)} = \frac{1}{x-3} + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Number of polynomials of degree less than 4 satisfying 5 points Let polynomial $P(x)$ have the property that $P(1),$ $P(2),$ $P(3),$ $P(4)$ and $P(5)$ are equal to $1$, $2$, $3$, $4$, $5$ in some order. How many possibilities are there for the polynomial $P,$ given that the degree of $P$ is strictly less than $4$?
| By Lagrange interpolation, given the values of $P(1)$, $P(2)$, $P(3)$, $P(4)$, and $P(5)$, there is a unique polynomial of degree $\leq 4$ taking those values. So the question is, for which permutations of the numbers $1$ through $5$ will the resulting $P(x)$ actually have degree $<4$? To determine this, we can look ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
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Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k... | First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}(k+2)(k+4)=\frac{2\cdot1^3+21\cdot1^2+67\cdot1}{6}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}(k+2)(k+4)=\frac{2n^3+21n^2+67n}{6}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}(k+2)(k+4)=$
$\color\red{\sum\limits_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Laplace transform of a gaussian function? How do I find the Laplace transform of this Gaussian function $f(t)=\sqrt{t}e^{-\frac{1}{t}}$.
I have from a table that $L\Big[{\frac{1}{\sqrt{t}} e^{-\frac{a^2}{4t}}}\Big ]=\sqrt{\frac{\pi}{s}}e^{-a\sqrt{s}}$ for a Gaussian Normal Distribution.
If I set $a=2$ then the formula ... | The LT you seek may be written as
$$\begin{align} \int_0^{\infty} dt \, \sqrt{t} \, e^{-1/t} e^{-s t} &= 2 \int_{0}^{\infty} du \, u^2 \, e^{-(s u^2+1/u^2)} \\ &= 2 e^{2 \sqrt{s}} \int_0^{\infty} du \, u^2 \, e^{-\left (\sqrt{s} u + 1/u \right )^2} \\ \end{align}$$
Sub $v=\sqrt{s} u+1/u$, which implies two branches of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Why doesn't this limit exist? I'm trying to compute this limit:
$\lim_{x\to +\infty}x^2-x^2\sqrt{1+\frac{(-1)^x}{x^2}}$
which, according to the results, is not defined; I initially thought about a $\infty - \infty$ form but it looks like I'm missing something...
| Note first that there's a potential issue with $(-1)^x$ being ambiguous or undefined for non-integer values of $x$. That's not a fundamental problem, however; we can just consider the question as asking about a sequence.
With that in mind, consider we can multiply by the conjugate root:
\begin{align*}
x^2 - x^2\sqrt{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a real number $t$ such that $\alpha$ is represented with respect to the basis Doing some self study as this topic really stumps me, every time I see it I cringe. I believe this question requires knowledge of change of basis and using $\phi_{BD}(\alpha)$.
Question: Let $\alpha\in End(\Bbb R^3)$ be represented with... | Because the matrix is diagonal, the form is very simple. If $B=\{b_1,b_2,b_3\}$, the matrix $$\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$$ means that $$\tag{*} \alpha b_1=0,\ \ \alpha b_2=b_2,\ \ \alpha b_3=4b_3.$$ So
$$
\begin{bmatrix}
0 \\
0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Probability of taking at least one ball of specific color, no reposition, order doesn't matter
A bag contains 9 balls. 4 are red, 3 are white and 2 are yellow. Three
balls are extracted in a succession, without reposition.
What is the probability of extracting at least 1 yellow ball?
I did:
*
*probability of 1 y... | There are $\binom{7}{3}$ ways of extracting 3 balls that are not yellow, and $\binom{9}{3}$ possibile extractions. So the probability of extracting 3 non-yellow balls is
$$
p=\frac{\binom{7}{3}}{\binom{9}{3}}.
$$
Then the probability of extracting at least one yellow ball is
$$
\overline{p}=1-p=1-\frac{\binom{7}{3}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $a, b$ and $c$ from this system of linear equations? Find $a$, $b$ and $c$ from:
$$
\begin{align}
a+c &= 3 \\
b+a &= 2 \\
c+b &= -1 \\
\end{align}
$$
I tried the following way, the answer comes wrong:
$$(a+c) + (b+a) = 3+2$$
| $$a+c=3.....(1)$$
$$b+a=2.....(2)$$
$$c+b=-1.....(3)$$
Try your way and solve like this-
$$(a+c)+(b+a)=3+2$$
$$2a+c+b=5.......(4)$$
Now, subtract equation(3) from (4)
$$ \begin{array}\ 2a+c+b=5 \\\;\;\;\;\;-c-b=1\\ \hline2a\;\;\;\;\;\;\;\;\;\;\;=6
\end{array}$$
$$2a=6\;\implies a=3$$
Now, Put the value value of $a$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx$? Suppose $g>0$, $g$ decreasing on $[a,b]$ where $0<a<b<1$. Is it true or false that
$$\displaystyle\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx ?$$
| The inequality does not hold. Counterexample: $g(x)=1$ and $0 \lt a=\frac{1}{4} \lt b=\frac{3}{4} \lt 1$ then:
$$\int_{1/4}^{1/2}\frac{dx}{(1-x)^2} = \frac{1}{1-x} \bigg|_{1/4}^{1/2} = \;\frac{2}{3} \quad \lt \quad 2\; = \frac{1}{1-x} \bigg|_{1/2}^{3/4} = \int_{1/2}^{3/4}\frac{dx}{(1-x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Arithmetic sequence common difference involving cube or square!
Evaluate $$\sum_{k = -3}^2 k^3$$
If I evaluate it using the expression above by putting the lower and upper limits it produces $-27$ for an answer, but when evaluated using the closed form expression of the geometric sequence where I put first term $a = ... | \begin{align}
S
&= \sum_{k=-3}^2 k^3 \\
&= (-3)^3 + (-2)^3 + (-1)^3+ 0^3 + 1^3 + 2^3 \\
&= (-3)^3 - (2)^3 + 2^3 - 1 + 1 \\
&= -27
\end{align}
In general you could transform into a combination of sums of $j^k$, which is the topic of Faulhaber's formula.
\begin{align}
S
&= \sum_{k=-3}^2 k^3 \\
&= \sum_{j=1}^6 (j-4)^3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute limit if it exist of $a_1 = 2$, $a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$ $$a_1 = 2$$
$$a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$$
I tried to use induction but it looks as if the sequence isn't monotonic.
| Let $\displaystyle\;b_n = \frac{a_n + 1}{a_n - 1}$, we have
$$
b_{n+1} = \frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n + \frac{1}{a_n} + 2}{a_n + \frac{1}{a_n} - 2} = \left(\frac{a_n + 1}{a_n - 1}\right)^2 = b_n^2$$
Together with
$\displaystyle\;b_1 = \frac{a_1-1}{a_1+1} = \frac{2+1}{2-1} = 3$, we get
$$b_n = b_1^{2^{n-1}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $n^2 \leq 2^n + 1$ by induction Suppose $n \in \mathbb{N}$, prove the given inequality.
So far I have this:
Show that the above is true for $n=1$.
$$ 1 \leq 3 $$
Suppose that it holds for all $n$. Show that it holds for $n+1$
$$ (n+1)^2 \leq 2^{n+1} + 1 $$
$$ \color{red}{n^2} + 2n + 1 \leq 2^n + \color{red}{2^n... | You can first look for what you need.
Assume $n^2\le 2^n+1$. Then
$$
(2^{n+1}+1)-(n+1)^2=2\cdot2^n-n^2-2n=2\bigl((2^n+1)-n^2\bigr)+(n^2-2n-2)
$$
which is surely $\ge0$ so long as $n^2-2n-2\ge0$.
This happens when $n\ge1+\sqrt{3}$, so for $n\ge3$.
Now you can give the proof.
*
*Case $n=0$: $0^2\le 2^0+1$, true
*Cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Surface area of a cone intersecting a horizontal cylinder I'm trying to find the surface area of the cone $x^2 + y^2 = z^2$ above the $xy$-plane and under the cylinder $y^2 + z^2 = 16$. I thought of using spherical coordinates $x = \rho \sin \phi \cos \theta,\quad y = \rho \sin \phi \sin \theta,\quad z= \rho \cos \th... | $x = \rho \cos\theta \sin\phi\\
y = \rho \sin\theta \sin\phi\\
z = \rho \cos\phi$
On the surface of the cone.
$\phi = \frac {\pi}{4}$
$x = \frac {\sqrt 2}{2}\rho \cos\theta\\
y = \frac {\sqrt 2}{2}\rho \sin\theta \\
z = \frac {\sqrt 2}2\rho$
We can drop the $\frac {\sqrt 2}2$ across terms and still have a good perameta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel to (a) X axis (b) parallel to Y axis.
Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel
to (a) X axis (b) parallel to Y axis.
For part (a) if we differentiate w.r.t $x$ we get $y'=\large-\frac{2x+y}{x+2y}$, this when equated t... | If $x_1,x_2$ with $x_1=x_2$ are the roots of the equation $$x^2+xy+y^2-7=0$$
we need the discriminant$=0$ $$\implies y^2-4(y^2-7)=0\iff y^2=?$$ and consequently $$x_1=x_2=-\dfrac y2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Factorize expression $(4a+3)^2-a^4$ The question is to factorize $(4a+3)^2-a^4.$
The answer I ended up with was $(4a+3-a^2)(4a+3-a^2).$
I got this by difference of two squares $\cdots.$
I would like to know if this is correct.
| SIMPLE
The question was to factorize $(4a+3)^2-a^4.$
IF you use difference of two squares we will get
LET $4a+3$ be $X$ and $a^2$ be $Y$
the $X^2-Y^2$
WE GET
$(4a+3-a^2)(4a+3+a^2).$
UNDERSTOOD
Further each of the quadratic equation can be factorised
$(x-1)(x-3)(x-/7+2)(x+/7+2)$
SO THE FINAL AWSER IS HERE
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding the transformation I am trying to find the transformation of $x = 2$ under $T(z) = \frac{z+1}{z-1}$
My approach was as follows,
I put $u +iv = \frac{z+1}{z-1} $ and then rationalized the RHS.
I get $u = \frac{3-y^2}{1+y^2}$ and $v = \frac{4y}{1+y^2}$
I was trying to substitute $y = tan(\theta)$ and prove its a... | Note that if $\Re(z)=2$,
then $\forall y\in \mathbb{R}$,
$z=2+iy$
$$T(z)=\frac{2+iy+1}{2+iy-1}=\frac{3+iy}{1+iy}=\frac{(1-iy)(3+iy)}{1+y^2}=\frac{3+y^2-2iy}{1+y^2}=2+\frac{1-y^2-2iy}{1+y^2}$$
Now notice that:
$$|T(z)-2|^2=\frac {1}{(1+y^2)^2} ((1-y^2)^2+2^2y^2)=\frac{1+2y^2+y^4}{(1+y^2)^2}=1$$
Which means that the dist... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I want to find a pair of integers $X$, $Y$ which satisfy $X^2 - 2Y^2=1$ such that $X > Y > 50$
Find a pair of integers $X, Y$, which satisfy $X^2 - 2Y^2 = 1$, such that $X > Y > 50$.
I have started by finding a pair of much smaller integers that work: $X(1) = 3$ and $Y(1) = 2$.
When I looked up a solution it was as... | Another (more naive) way of looking at this: we want all possible $k$ for which $(Y+k)^2-2Y^2=1$. Solving this for $Y$ (in terms of $k$) we get
$$
Y=k\pm\sqrt{2k^2-1}
$$
We will want to ignore the minus, since our solutions must be positive and we only want the solutions for when $\sqrt{2k^2-1}$ is an integer.
By inspe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Method for finding sum of series $\sum_{r = 1}^{10} \binom{10}{10 - r}\binom{20}{r}$? I came across this in a problem:
$$\sum_{r = 1}^{10} \binom{10}{10 - r}\binom{20}{r}$$
Please help me solve this and give a general method to solve such sums.
| Using $$(1+x)^{20} = \binom{20}{0}+\binom{20}{1}x+\binom{20}{2}x^2+\cdots\cdots +\binom{20}{20}x^{20}.........(1)$$
Similarly $$(1+x)^{10} = \binom{10}{0}+\binom{10}{1}x^{10}+\binom{10}{2}x^{2}+\cdots \cdots \cdots+\binom{10}{10}x^{10}.....(2)$$
Now coeficients of $x^{10}$ in multiplication of $(1)$ and $(2)$
$$(1+x)^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Maximum value of a vector If a,b,c are unit vectors , then we have to know $|a-b|^2$$+|b-c|^2$$+|c-a|^2$ does not exceed .
I tried it a lot , but not able to get how to start it .
| Using $||\mathbf{a}||^2=\langle\mathbf{a},\mathbf{a}\rangle$ and the linearity properties of the inner product (if you don't want this you may just expand on components) you may prove:
$||\mathbf{a}+\mathbf{b}+\mathbf{c}||^2+||\mathbf{a}-\mathbf{b}||^2+||\mathbf{b}-\mathbf{c}||^2+||\mathbf{c}-\mathbf{a}||^2=3(||\mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Counting the number of ways three dice can sum to $n$ I want to count in how many ways three dice can sum to a given number without brute forcing it. In fact, I would like to do it using generating functions and without having to expand out the product.
To do this, I have thought of making a differential equation out ... | Here is an alternate approach. We could at first transform $y(x)$ so that an expansion after that becomes less cumbersome. Maybe this variant is also useful for your needs.
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We also use Iverson brackets
\begin{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum of the value $\max\left(\frac{1}{ac}+b,\frac{1}{a}+bc,\frac{a}{b}+c\right)$ Let $a,b,c>0$ find the following minimum of the value
$$\max\left(\dfrac{1}{ac}+b,\dfrac{1}{a}+bc,\dfrac{a}{b}+c\right)$$
the book is hint:when $a=b=c=1$,then the minimum of the value is $2$
| Let $M =\max \left[ \frac{1}{ac}+b, \frac{1}{a}+bc, \frac{a}{b}+c \right]$
Case I: $c\ge 1$
\begin{align*}
M &=\max \left[ \frac{1}{a}+bc, \frac{a}{b}+c \right]
\tag{$\frac{1}{a}+bc \ge \frac{1}{ac}+b$} \\
& \ge \max \left[ 2\sqrt{\frac{bc}{a}}, 2\sqrt{\frac{ac}{b}} \right]
\tag{$AM \ge GM$} \\
&... | {
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Prove that the function f(x,y) is not differentiable I have the following problem:
Prove that the function:
$f(x,y)=
\ \begin{cases}
\frac{x^3-x\cdot y^2}{x^2+y^2} & (x,y)\neq (0,0) \\
\\0 & (x,y)=(0,0)
\end{cases}
\\$
is continuous on $R^2$ and has its first order partial derivatives.
everywhere on $... | We have $${ f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ x } } =1\\ \\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Decide whether or not a function is continuous Is the piecewise function
$$f(x)=\begin{cases}
\frac{x^2-x-6}{x-3} & x>3 \\
\frac{\sqrt{x}-3}{x-9} & x \leq 3
\end{cases}
$$
continuous?
I've done the following:
$\lim_{x\rightarrow3^+}\frac{\sqrt{x}-3}{x-9}=\frac{\sqrt{3}-3}{3^+ -9}=\frac{\sqrt{3}-3}{6}$
$\l... | Let us find the value of $f(3)$ first (i.e. $x=3$).
$$f(3)=\frac{\sqrt{3}-3}{3-9}=-\frac{\sqrt{3}-3}{6}$$
Now, let us find the left-side limit as $x\to3$ (i.e. $x\lt3$).
$$\lim\limits_{x\to3^-}(f(x))=\frac{\sqrt{3}-3}{3-9}=-\frac{\sqrt{3}-3}{6}$$
The left-side limit as $x\to3$ is equal to $f(3)$. Let us find the right-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find the value of $\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$ How to find the exact value of the following limit?
$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$
| HINT:
$$\dfrac{e^{x\cos x^2}-e^x}{x^5}=e^x\cdot\dfrac{e^{x(\cos x^2-1)}-1}{x^5}$$
$$\dfrac{e^{x(\cos x^2-1)}-1}{x^5}=\dfrac{e^{x(\cos x^2-1)}-1}{x(\cos x^2-1)}\cdot\dfrac{x(\cos x^2-1)}{x^5}$$
$$\dfrac{\cos x^2-1}{x^4}=-\left(\dfrac{\sin x^2}{x^2}\right)^2\cdot\dfrac1{(1+\cos x^2)}$$
Now use $\lim_{h\to0}\dfrac{e^h-1}h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to disprove $8q^2+15=7m^2$ for any integral value of $q$ and $m$
Question Statement:-
Show that the quadratic equation $x^2+7x-14(q^2+1)=0$, where $q$ is an integer, has no integral roots.
My attempt at a solution:-
For the given quadratic equation having rational coefficients to have integral roots the discrim... | If $x^2+7x-14(q^2+1)=0$, where $x,q\in\mathbb Z$, then $$7\mid 7\left(2\left(q^2+1\right)-x\right)=x^2$$ so by Euclid's Lemma $7\mid x$, so $7^2\mid x^2$ and $7^2\mid 7x$, so $$7^2\mid x^2+7x=7\left(2\left(q^2+1\right)\right),$$
so $7\mid 2\left(q^2+1\right)$. Since $\gcd(7,2)=1$, by Euclid's Lemma $7\mid q^2+1$, i.e. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Solve the integral $ \int{\frac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx $ My attempt:
\begin{align} \int{\frac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx &\\ &=\int{\frac{x+\sqrt{x^2+5x+6}}{x+\sqrt{x^2+5x+6}}}dx+\int{\frac{-\sqrt{x^2+5x+6}-\sqrt{x^2+5x+6}}{x+\sqrt{x^2-5x+6}}}dx \\
&=x-\int{\frac{\sqrt{x^2+5x+6}}{x+\sqrt... | Hint:
$\int\dfrac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}~dx$
$=\int\dfrac{x-\sqrt{(x-3)(x-2)}}{\sqrt{(x+3)(x+2)}+x}~dx$
$=\int\dfrac{\left(x-\sqrt{(x-3)(x-2)}\right)\left(\sqrt{(x+3)(x+2)}-x\right)}{\left(\sqrt{(x+3)(x+2)}+x\right)\left(\sqrt{(x+3)(x+2)}-x\right)}~dx$
$=\int\dfrac{x\sqrt{(x+3)(x+2)}}{5x+6}~dx+\int\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Limit with sum of cube roots I am trying to evaluate
$$\lim \limits_{n \to \infty} \left(\sqrt[3]{{1 \over n^4}} + \sqrt[3]{{2 \over n^4}} + \sqrt[3]{{3 \over n^4}} + \dots + \sqrt[3]{{n \over n^4}} \right)$$
I simplify the expression to
$${\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n} \over \sqrt[3]{n^... | Let $f(x)=x^\frac 13$.
$$S_n=\sum_{k=1}^n (\frac{k}{n^4})^{\frac 13}\;\;$$
is a Riemann sum and
$$\lim_{n\to+\infty}S_n=$$
$$\lim_{n\to+\infty}\frac{1-0}{n} \sum_{k=1}^n f(0+k\frac{1-0}{n})$$
$$=\int_0^1 x^\frac{1}{3}dx$$
$$=\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Diophantine equation $7b^2+7b+7=a^4$.
Solve in positive integers $(a,b)$ the equation
$$7b^2+7b+7=a^4.$$
As the left side is divisible by $7$ we must have $7|a^4\implies 7|a$. Let $a=7c$, then the given equation reduces to $$b^2+b+1=343c^4.$$
I'm not sure how to proceed. Any hints or solutions are welcome.
| If
$$
b^2+b+1=343c^4
$$
then $b^2+b+1-343c^4=0$
$$
b=\frac{-1-\sqrt{1-4(1-343c^4)}}{2}=\frac{-1-\sqrt{1372c^4-3}}{2}
$$
So there is an integral solution iff there exist $d$ such as $1372c^4-3=d^2$, that is
$$
1372c^4-d^2=3
$$
A solution for the last equation is $c=1$, $d=37$. Hence $b=18$ or $b=-19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot ... | $18^{n+1}=2^{n+1}⋅27$
$\implies (3^{2}⋅2)^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}⋅2^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}=3^{3}$
$\implies 2(n+1) = 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Using the Runge Kutta's Method to solve a 2nd derivative question Given that the motion of two bodies subject to a gravitational force of $\frac{d^2x}{dt^2}$ = $\frac{-x}{(x^2+y^2)^{3/2}}$, $x(0) = 0$, $\frac{dx}{dt}$$(0) = -0.5$ and $\frac{d^2y}{dt^2}$ = $\frac{-y}{(x^2+y^2)^{3/2}}$, $y(0) = 1$, $\frac{dy}{dt}(0) = -... | There is an alternative method to my previous answer when the acceleration function is defined as a 2nd order diff. equation $$\ddot{\mathbf{Y}} = f(t,\mathbf{Y},\dot{\mathbf{Y}})$$
with the state vector $\mathbf{Y} = (x,y)$ in your case, holding the positions in the two coordinates, and $\mathbf{V} = \dot{\mathbf{Y}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
maximum value of $xy+yz+zx.$ given $x+2y+z=4$ if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
| $$xy+z(x+y)=xy+(4-x-2y)(x+y)=4x+4y-x^2-2y^2-2xy=k\text{(say)}$$
$$x^2+2x(y-2)+2y^2-4y-k=0$$
As $x$ is real, the discriminant $$4(y-2)^2-4(2y^2-4y-k)\ge0$$
The equality occurs if $x=-\dfrac{2(y-2)}2$
$$\iff k\le4-y^2\le4$$
The equality occurs if $y^2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
A limit without invoking L'Hopital: $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^2}$ The following limit
$$\ell=\lim_{x \rightarrow 0} \frac{x \cos x - \sin x}{x^2}$$
is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic l... | We have,
$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} = \lim_{x \to 0} \dfrac{\cos x -1}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2} $$
$$ = -2\lim_{x \to 0} \dfrac{\sin^2 \left(\frac{x}{2}\right)}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2} $$
The first limit is zero since $\displaystyle \lim_{x \to 0} \dfrac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$$
I have no idea on how to face this limit. Its value at $0$ seems to be $0$, but its limit equals $\frac{23}{24}$. I tried l'Hôpital but the new expressions were ... | Preliminary remark. I turned the previous hint into a full answer because it has been downvoted two times. It seems that the little-o notation is not so popular :-(
Use Taylor'expansion. Note that
$$\frac{x}{1-x^2}=x(1+x^2+o(x^2))=x+x^3+o(x^3),$$
and
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4).$$
Then
$$\cos\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)$ I've managed to solve it by rewriting the expression as
$$\frac{1 - \frac{\sqrt{x^2 +x + 1}}{x e^{\frac{1}{x}}} }{ \frac{1}{x e^{\frac{1}{x}}} }$$
then applying L'Hospital's rule.
This took up one whole page and was very hairy, even after s... | Multiplying by $\displaystyle\frac{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}$ we have
$$\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)=\lim_{x\to\infty}\frac{x^2 (e^{\frac{2}{x}} -1)-x-1 }{x e^{\frac{1}{x}} + \sqrt{x^2+x+1} }=\lim_{x\to\infty}\frac{x (e^{\frac{2}{x}} -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
New Question: Find $x,y,z$ s.t. $x+y+z=1$ and $x^2+y^2+z^2=1$, and $x,y,z \in ( - 1,0) \cup (0,1)$ If in my earlier question [1], we relieve the constraints such that $x,y,z$ can also get values between $(−1,0)$, it seems that the system has a solution. Now, the question is how we can find this solution?
$$\begin{case... | The set of solutions become a circle which intersect of next two spheres.
$$(x-1/3)^2+(y-1/3)^2+(z-1/3)^2=2/3$$
$$x^2+y^2+z^2=1$$ except $x=y,y=z,z=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
maximum value of the expression : $2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1$ I have to find out the maximum value of the expression :
$2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1.$
| Let $(x,y,z)=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) \in S^2$
\begin{align*}
2\cos \phi+3\sin \phi & \le \sqrt{2^2+3^2} \\
2x+3y &= \sin \theta (2\cos \phi+3\sin \phi) \\
& \le \sqrt{13} \, |\sin \theta| \\
2x+3y+z & \le \sqrt{13} \, |\sin \theta|+\cos \theta \\
& \le \sqrt{13+1} \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
| As noted by @Daniel, it equals $1/(1-x)(1-x^2)(1-x^3)$. Another way to put that is $$\frac{(1+x+...+x^5)(1+x^2+x^4)(1+x^3)}{(1-x^6)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Note that $a = \frac{1+\sqrt{5}}{2}$ satisfies the equation $a^2 - a - 1 = 0$
Substituting $a=x+\frac{1}{x}$ gives:
$$0 = \left(x+\frac{1}{x}\right)^2 - \left(x+\frac{1}{x}\right) - 1 = x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}$$
$$\iff \quad x^4 - x^3 + x^2 - x + 1 = 0$$
Multiplying by $x+1$ results in $x^5+1=0$, so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
What is the value of $\prod_{n=1}^\infty (1+\frac{1}{n^2})$? How would I go about finding $\prod_{n=1}^\infty (1+\frac{1}{n^2})$ ? Setting $a_k = \prod_{n=1}^k (1+\frac{1}{n^2})$, the first few partial products are $a_1 = 2$, $a_1 = 2.5$, $a_3 \approx 2.78$, $a_4 \approx 2.591$, $a_{20} \approx 3.50$, and they eventual... | Six years too late
Similar to @Felix Marin's answer to this question but focusing on the partial product
$$P_k=\prod_{n=2}^{k} \left( 1+\frac{1}{n^2}\right)=\prod_{n=2}^{k} \left( \frac{n^2+1}{n^2}\right)=\frac{\prod_{n=2}^{k}(n+i)\,\prod_{n=2}^{k}(n-i) } {\prod_{n=2}^{k}n\,\prod_{n=2}^{k}n }$$ that is to say
$$P_k=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the lowest value possible of $f(x) = e^{1-0.5x}+e^{0.5x-1}$
Let $f$ be the function defined by: $$f(x) = e^{1-0.5x}+e^{0.5x-1}$$
Find the values of $k$ for which the equation $f(x) = k$ has two non
negative solutions, using the graphical calculator.
This problem could be solved very easily using the calcula... | For the second way:
\begin{equation}
\frac{1}{y} \cdot e + y \cdot \frac{1}{e} < e + e^{-1}\\\frac{1}{y} \cdot e + y \cdot \frac{1}{e} - e - e^{-1}<0\\\frac{e^2+y^2-e^2y-y}{ey}<0
\end{equation}
Now we know that the denominator is always positive, so we have to solve $e^2+y^2-e^2y-y<0$. This is a parabola thus it s ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Sum of an infinite series of fractions I am taking a fun walk into Number Theory land, and I am conducting an investigation about an infinite sum of fractions. These fractions have to do with the amount of composites within an even number. All the fractions have to do with the reciprocal of a prime.
This is my sum:
$\f... | The partial sums can be rearranged as
$$\sum_{k=2}^n a_k = \sum_{k=2}^n \frac{1}{p_k}\left(1 - \sum_{j=2}^{n-1} \frac{1}{p_j}\right)
= \sum_{k=2}^n \frac{1}{p_k} - \sum_{2\le j < k \le n}\frac{1}{p_jp_k}\\
= \sum_{k=2}^n \frac{1}{p_k} - \frac12\left[ \left(\sum_{k=2}^n \frac{1}{p_k}\right)^2 - \left(\sum_{k=2}^n \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
dividing a group 0f 10 men and women into couples of same sex
There're 10 men and 10 women. We need to divide them into couples of same sex.
The answer is: $$\left( \frac{10!}{5!*2^5} \right)^2$$ I don't understand why.
I know that we can divide into couples where there's exactly 1 man and 1 woman in each couple in... | You are right about the total possible number of couples regardless of sex. The number $\frac{20!}{10!*2^{10}}$ includes every possibility: out of the 10 couples, they can be of the same gender, or some of them can be of 1 man and 1 woman. The number $10!$ represents the case in which none of the couples are of the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
| Considering the telescoping sum
$$\sum_{n=2}^{\infty} \left[ \frac{1}{n^2(n-1)^2}-\frac{1}{n^2(n+1)^2} \right]=\frac{1}{4}$$
Also $$\frac{1}{n^2(n-1)^2}-\frac{1}{n^2(n+1)^2}=\frac{4}{n^{5}}+O\left( \frac{1}{n^7} \right)$$
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^5} &=
1+\frac{1}{16}+\sum_{n=2}^{\infty}
\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
find the maximum of sum of roots without calculus A question asks me to evaluate the maximum and minimum value of the following function:
$$y=\sqrt{x}+\sqrt{27+x}+\sqrt{13-x}, 0\le x\le13$$
I have tried to compute y using $$x=0,1,2,...,13$$ I believe such maximum occurs at x=9, and the minimum value lies on boundary po... | Shifting the OP's function $9$ to the left we get a function of the form
$$g(x)=\sqrt{x+(n+1)^2}+\sqrt{x+\bigl(n(n+1)\bigr)^2}+\sqrt{n^2-x}.$$
Now it's easy to prove without calculus simply by squaring that $\sqrt{x+a^2}\leq \frac{1}{2a}x+a$ and $\sqrt{a^2-x}\leq-\frac{1}{2a}x+a$ for positive $a$ and $-a^2< x<a^2$. Eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to find $\sum_{i=1}^n\left\lfloor i\sqrt{2}\right\rfloor$ A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)). A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).
If $n = 5$ then
$$\left\lfloor1\sqrt{2}\right\rfloor+ \left\lfloor2\sqrt{2}\right\rfloor + \left\lfloor3\sqrt{2}\right\rfloor +\left\lfloor4 \sqrt{2}\righ... | It is clear that, because the sequence $\{<n\sqrt{2}>\}$(the fractional part) is equidistributed over the interval $[0,1)$, we have
$$\tag{1}\sum_{n=1}^N \lfloor n\sqrt{2} \rfloor = \frac{N(N+1)\sqrt{2}}{2}-\sum_{n=1}^N <n\sqrt{2}>\label{1}$$
and for the latter sum,
$$\tag{2}\frac{1}{N}\sum_{n=1}^N <n\sqrt{2}> \to \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 0
} |
The limit of a function with sum of two roots. I need to find the limit of the following function:
$$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$
I derived it to the form of:
$$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqr... | For $x<0$,
$$f(x)=-x(1+\frac{2}{x})^{\frac{1}{2}}+x(1+\frac{1}{x})^\frac{1}{3}$$
$$=-x-1+x+\frac{1}{3}+\frac{1}{x}\epsilon(x).$$
the limit is $$-\frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\int y \ln y \; \mathrm{d}y$ Can you please show me how to integrate this?
$$\int y \ln y \; \mathrm{d}y$$
I attempted integration by parts but it is getting more complicated.
| Do you know integration by parts?
$$\int fg' = fg - \int f' g.$$
Let's pose $f = \log(y)$ and $g' = y$. Then:
$$f' = \frac{1}{y}, g = \frac{1}{2}y^2.$$
Joining all:
$$\int y\log(y) dy = \frac{1}{2}y^2\log(y) - \int \frac{1}{y}\frac{1}{2}y^2 dy = \\
\frac{1}{2}y^2\log(y) - \frac{1}{4}y^2 + c. $$
If $h(y) = y\log(y)$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Definition of the binomial coefficient $a_{n}=\binom{-\frac{1}{2}}{n}$ Applying the definition of the binomial coefficient I can't figure out how to simplify the following expression:
$$a_{n}=\binom{-\frac{1}{2}}{n}$$
I want to find:
$$a_{n}=(-1)^n\frac{(2n-1)!}{(n-1)!2^{2n}n!}$$
I'm stuck at the beginning.
$$a_{n}=\f... | Here we use the following definition of the binomial coefficient
\begin{align*}
\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}
\qquad\qquad\alpha\in\mathbb{C},\quad n\in \mathbb{N}\tag{1}
\end{align*}
It is also convenient to use double factorials for a more compact notation
\begin{align*}
(2n)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \fra... | Let $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=t\ $ implies $a=e^{(b-c)t}$, $b=e^{(c-a)t}$ & $c=e^{(a-b)t}$
so now, using values of a, b, c: $$a^ab^bc^c=e^{a(b-c)t}e^{b(c-a)t}e^{c(a-b)t}=e^{(ab-ac+bc-ab+ac-bc)t}=e^0=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What is the contrast used in calculating the $2^p$ ANOVA table? ${{}}$ According to my statistics textbook, you need the contrast of each effect and interaction in order to compute the sum of squares (for ANOVA) of that effect/interaction. Unfortunately, it never explains what contrast is.
Here are how the contrasts ar... | What I think is the most frequently seen definition (but I haven't checked all the books, even among those I have at hand) is this:
A contrast is a linear combination in which the sum of the coefficients is $0$, but not all coefficients are $0$.
A bit more precisely, if you multiply all the coefficients by the same s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the integral part of the product $\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdots \frac{2016}{2015}.$
Find the integral part of the following number $$T = \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dfrac{8}{7} \cdots \dfrac{2016}{2015}.$$
We can show that $T = 2017\int_{0}... | Although @winther and @achillehui did this the smart way and used the standard approximations I took the "battering ram" approach by first rewriting the product in the form
\begin{equation}
P=\dfrac{2^{2016}(1008!)^2}{2016!}
\end{equation}
This gave
\begin{equation}
\ln P=2016\ln(2)+\sum_{k=1}^{1008}\ln\left(\dfrac{k}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
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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.