Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Apostol's method of exhaustion to find area under x^2 I'm a high school student currently going through Apostol's calculus. I'm not that familiar with proofs, but I learned Calculus in school up to partial fractions, but we focused more on problems instead of the concept/proof, so please bear with me. I'm stuck in the ... | You should use the left-hand inequality
$$
\frac{b^3}3-\frac{b^3}n<A\qquad\text{for all $n\ge1$}\tag1
$$
to prove that $A\ge\frac{b^3}3$. The argument by contradiction follows the same lines as the proof you've given: If $A<\frac{b^3}3$, then you can use algebra to rearrange (1) into the statement
$$
n<\frac{b^3}{\fra... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for every $n\in\mathbb{N}$, $n^2$ is divisible by 3 or has a form $3k+1$? I tried to do this by induction, but it doesn't make any sense: $n^2=3k$ or $n^2=3k+1$
*
*option: $(n+1)^2= 3k+2n+1$
*option: $(n+1)^2= 3k+2n+2$
Is there any other way on proving this problem?
| For all $n$ natural number, $n = 3k$ or $n=3k+1$ or $n=3k+2$
When $n=3k$, $n^2=9k^2=3(3k^2)$, thus is divisible by 3
When $n=3k+1$, $n^2=9k^2+6k+1=3(3k^2+2k)+1$, thus is in the form of $3n+1$
When $n=3k+2$, $n^2=9k^2+12k+4=3(3k^2+4k+1)+1$, thus is in the form of $3n+1$
Thus, covering all the cases and finishing the pro... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that the line $x+y=q$ intersects the ellipse $x^2-2x+2y^2=3$ at two different points if $q^2<2q+5$ What should be the proper way to answer this question?
My working is as follows, but is it the right way to answer the question, since it seems like I am only showing that $q^2<2q+5$?
$x+y=q \tag{1}$
$x^2-2x+2y^2=3 \... | If $y=x-q$ intersects the ellipse $$x^2-2x-2y^2=3 \implies x^2-2x+2(x-q)^2=3$$
Then the roots of this quadratic have to be real. So we demand $B^2 >4AC$ in the simplified quadratic,
$$3x^2-(4q+2)x+2q^2-3=9 \implies q^2 <2q+5$$
| {
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Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}... | After your first line you may set
$$y= \log_2(x^2+7)$$
for better readability and continue to calculate with $\log_2$.
You get
$$\log_{\frac 34}(\frac 13 y) + \log_{\frac 12}\frac 12 y=-2$$
Now, using $\log_b a =\frac{\log_2 a}{\log_2 b}$ you can isolate $\log_2 y$ (I leave the intermediate steps to you):
$$\log_2 y... | {
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"url": "https://math.stackexchange.com/questions/3602742",
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$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y)dy$. Find $f(x)$ $f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$. Find $f(x)$
I've tried taking $\int_{0}^{1} (xy^2 + x^2y) f(y) dy$ to be $k(x)$ since it comes out to be a function of $x$. That transforms our equation to $f(x) = x + k(x)$.
$f(y) = y + k(y) \implies (xy^2 + x... | Let $A=\int_0^{1} y^{2}f(y)dy$ and $B=\int_0^{1} yf(y)dy$. Then $f(x)=x+xA+x^{2}B$. Multiply by $x$ and integrate to get $B=\frac 1 3 +\frac A 3 +\frac 1 4 B$. Similarly multiply by $x^{2}$ and integrate to get $A=\frac 1 4 +\frac A 4 +\frac 1 5 B$. Solve these two equations for $A$ and $B$.
| {
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Is it possible to apply L'Hospital to $ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8}))) $? I have a question about how to calculate this limit; can I apply L'Hospital?
$$
\lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8})))
$$
Is it possible to ma... | $$\lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8})))$$
$$ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )
= \lim_{x \to\infty}\left(\dfrac{1}{e^{x^2}}+\dfrac{2}{x} \right)
= 0$$
$$ \lim_{x \to\infty}\sin^2\left(\frac{2}{x^4}\right)x^8
= \left[ 2\lim_{x \to\infty}
\... | {
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Question regarding floor function subtraction Let:
*
*$a,b,c$ be integers with $a \ge b$ and $c > 0$
There exists integers $s,t$ such that:
$$\left\lfloor\frac{a}{c}\right\rfloor = \frac{a - s}{c}\text{ },\text{ }\left\lfloor\frac{b}{c}\right\rfloor = \frac{b-t}{c}$$
Does it follow that:
$$\left\lfloor\frac{a}{c}\... | What you've done looks correct. Here's a somewhat different way to show it, although not really much simpler, but it does include a few points you don't show explicitly. Note you can use a similar proof in the case $c \lt 0$ to confirm your proposition, but you would need to change it to use $s \le t$ instead.
You have... | {
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Complex numbers algebra problem involving cyclic summation
Let $a_1$, $a_2$, $a_3\in \mathbb{C}$ and $|a_1|=|a_2|=|a_3|=1$.
If $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, find $|a_1 + a_2 + a_3|$
What I have done till now:
First, I tried to attack the required sum directly.
Let $\alpha=|a_1 + a_2 + a_3|$ , then squaring both ... | Note that each term in the sum has length $1$. If we add $1$ as a fourth complex number to this sum, we get $0$. In this way, we get a (possibly degenerate) quadrilateral with four sides of equal length, producing a rhombus. In particular this means that pairs of these terms (including $1$) must be negatives of each ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing the binomial sum $\sum_{0\le i
Find the sum
$$\sum_{0\le i<j\le n}j\binom ni.$$
My 1st attempt: replacing $j$ with $n-j$. So the expression turns out to be
$$S=\sum n\binom ni-\sum j\binom ni$$
So adding the original and the final expression we get simply $S=n2^{n-1}$.
My second attempt: considering 3 part... | \begin{align}
\sum_{0 \le i<j\le n} j \binom{n}{i}
&= \sum_{i=0}^n \sum_{j=i+1}^n j \binom{n}{i}\\
&= \sum_{i=0}^n \binom{n}{i} \sum_{j=i+1}^n j \\
&= \sum_{i=0}^n \binom{n}{i} \frac{(n+i+1)(n-i)}{2} \\
&= \frac{1}{2}\sum_i \binom{n}{i} (n^2+n-i(i-1)-2i) \\
&= \frac{1}{2}(n^2+n)\sum_i \binom{n}{i} -\frac{1}{2}\sum_i \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given: $\frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y},$ prove: a) $x^x\times y^y\times z^z =1,$ b) $x\times y \times z =1.$ Given: $ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y} $
Prove: $ a)\ x^x\times y^y\times z^z =1 \\ b) \ x\times y \times z =1 $
Here is... | Here the common base is 10, then let all tharatios b equal to $l$, then
$$\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=k \implies x=10^{k(y-z)}, y=10^{k(z-x)}, z=10^{k(x-y)}.$$ $$F=x^x. y^y. z^z=10^{k(xy-xz)}.10^{k(yz-yx)}. 10^{(zx-zy)}=10^0=1$$
$$G=x.y.z=10^{k(y-z)}.10^{k(z-x)}.10^{k(x-y)}=10^0=1.$$
| {
"language": "en",
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On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators - Part II This post is inspired by this earlier question:
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators.
I quote:
A... | There are lots of counterexamples . . .
A counterexample with $y=2$:
$$\frac{1}{2}=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$$
A counterexample with all odd denominators:
$$\frac{1}{9}=\frac{1}{11}+\frac{1}{63}+\frac{1}{231}$$
A counterexample such that none of $y,x_1,x_2,x_3$ is a multiple of one of the others:
$$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of solution in cubic residue I need to find the number of non zero solutions in $\mathbf Z_p $ to the equation $x+y=1,$ where $x\in \Gamma$ and $y\in g\Gamma$ where $ g $ is primitive root modulo an odd prime $p$ and $\Gamma=\langle g^3\rangle$. I could do it when both $x$ and $y$ are in $\Gamma$.
Clearly, it is... | Let's count the solutions with $x \in \Gamma$ and $y \in g^n \cdot \Gamma$ as $n$ varies; it should only depend on $n \mod 3$.
Naturally $p \equiv 1 \mod 3$, so write $p-1=3m$ with $m$ even, so $(-1)^m = 1$.
We have:
$$\frac{1}{|\Gamma|} \sum_{x \in \Gamma} x^k = \begin{cases} 1, & k \equiv 0 \mod m, \\
0, & \text{othe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620169",
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Where Is my wrong The random variables given for the PDF distribution function:
$$
p_X(x) =
\begin{cases}
\frac{a}{7}x, & 2 <x\leq 3 \\
0, & x\notin(2,3] \\
\end{cases}
$$
$a)$ Find value of parametric $a$, and $F_X(x)$ (CDF).
$b)$ find the probability that the random variables is part of interval $(-\frac{... | You're wrong at the last line by substracting $\frac{28}{35}$ twice, since you forgot to flip the sign the second time.
$$F_X(\frac{5}{2}) - F_X(2) = \frac{1}{5}\frac{25}{4} - \frac{4}{5} - \big(\frac{1}{5}4 - \frac{4}{5}\big) = \frac{1}{5}\big(\frac{25}{4} - 4\big)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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AIME I 2007 Problem $10$. Spot the overcounting?
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
My... | You took into account that some rows already have two shaded squares, but you didn’t take into account that some rows don’t have any shaded squares at all yet, and you need to shade them in both of the last two columns. So in case $2$ you can only choose $2$ of $4$ squares, with the third one fixed, and in case $3$ you... | {
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Point out my fallacy in the three-of-kind problem, please How many hands of three-of-kind are there from a standard 52-card deck?
$$
13 \binom{4}{3} \binom{12}{2}\binom{4}{1}^2
$$
is the correct answer:
*
*$13$ : #ways to choose the denomination of the three-of-a-a-kind
*$\binom{4}{3}$ : #ways to choose their sui... | The $52 {3 \choose 2}$ counts each three of a kind three times. It counts each three of a kind once for having each of the three cards first. And in fact $52 {3 \choose 2}=3 \cdot \left(13 {4 \choose 3}\right)$
| {
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Prove $\lim_{t\rightarrow 0}\left[-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t\right] = \frac{4}{45}$ The author shows the following limit being taken
$\lim_{t\rightarrow 0}\left[-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t\right] = \frac{4}{45}$
I don't see how you could get anything but $\infty$..... | This is an approach which mirrors the use of Taylor series.
Let's observe that $$\arctan t =\int_{0}^{t}\frac{dx}{1+x^2}\tag{1}$$ and the expression under limit can be written as $$\left(1+\frac{t^2}{3}\right)\cdot\frac{1}{t^4}\left(\frac{\arctan t} {t} - \frac{1}{1+t^2/3}\right)$$ and hence the limit in question equal... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$ Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$
My Attempt:
$$p^3x-p^2y-1=0$$
$$p^2y=p^3x-1$$
$$y=px-\frac {1}{p^2}$$
This is solvable for $y$ so differentiating both sides w.r.t $x$
$$\frac {dy}{dx} = p + x \cdot \frac {dp}{dx} - (-2) p^{-3} \cdot... | The initial equation is
$$y'x-y-\frac1{y'^2}=0$$
and by differentiating,
$$y''x+2\frac{y''}{y'^3}=0$$
We split in two cases:
$$y''=0\to y=ax+b$$
and
$$y'=-\sqrt[3]{\frac2x}\to y=-\frac32\sqrt[3]{2x^2}+c.$$
Plugging back in the equation
$$p=a\to a^3x-a^2(ax+b)-1=0$$ implies $b=-\dfrac1{a^2}$, and
$$p=-\sqrt[3]{\frac2x... | {
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Understanding why a binary operator is associative. ( On a property of "fractional part of a sum" operation) Define the binary operation which gives the fractional part of a real number $x$ like so:
$$x +_1 y = \{x + y\}$$
where
$$\{x\}=x-\lfloor x \rfloor$$
Now we want to show that this operator is associative on the... | We have that $(x +_1 y) +_1 z = (x+y+z) - (p+q)$ and that $x +_1 (y +_1 z) = (x+y+z) - (r+s)$. If we take the floor of both of these expressions both of the left hand sides will be equal to zero so we get.
$$0 = \left \lfloor (x+y+z) - (p+q) \right \rfloor = \left \lfloor (x+y+z) - (r+s) \right \rfloor$$
Because $p+q$ ... | {
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Show that there is a $C$ such that $\frac{1}{n^{1+\alpha}} \leq C(\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha})$
1.(a) Show that if $\alpha>0$, then there is a constant $C$ such that for any $n\in \mathbb N$,
$$\frac{1}{n^{1+\alpha}} \leq C(\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha})$$
[Suggestion: Write
$$\frac {1}{n^\... | We choose $\displaystyle C=1+\dfrac{1}{\alpha}$ if $\alpha\ge 1$ and $C=1+\dfrac{2}{\alpha(\alpha+1)}$ otherwise. The inequality $\displaystyle \dfrac{1}{n^{1+\alpha}}\le C\left(\dfrac{1}{n^\alpha}-\dfrac{1}{(n+1)^\alpha}\right)$
is equivalent to $$\left(1+\dfrac{1}{n}\right)^\alpha\ge\dfrac{C}{C-\frac{1}{n}}=1+\dfrac{... | {
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If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x... | Hint.
As $x^2-3x+2 = (x-2)^2+(x-2)$ calling $F(x) = f(x^2+x)$ we have
$$
F(x)+2F(x-2)=3x(3x-5)
$$
| {
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"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $y_1(x)$ and $y_2(x)$ from the recurrence relation of $r_2$ only, by using the Frobenius method The given original equation is:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
The series I have is:
$$x^r\left[\left(r^2-r+r+x^2-\frac{1}{4}\right)C_0+\left(r^2+r+1+r+x^2-\frac{1}{4} \right)C_1x\right]+\sum^\infty_{n=2}\lef... | $$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
1) This is Bessel's equation of order $\dfrac 12$
2) Indicial equation:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
Substitute $y=x^a$
$$a(a-1)x^a+ax^a+x^{a+2}-\frac{1}{4})x^a=0$$
Take the coefficient of the lowest power of $x$:
$$P(a)=a^2-\dfrac 14$$
The indicial equation:
$$(a... | {
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"url": "https://math.stackexchange.com/questions/3646395",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evalua... | We have that $q$ is a root of
$$
X^6 +X^5+\dots +X+1=X^3(X^3+X^{-3} +X^{2}+X^{-2}+ X+X^{-1}+1).
$$
Hence $q+q^{-1}$ is a root of
$$
Y^3 - 3Y +Y^2 -2 +Y +1= Y^3+Y^2-2Y-1.
$$
Hence $\frac{1}{q+q^{-1}}$ is a root of
$$
Z^3+2Z-Z-1.
$$
Exactly the same is true for the roots $q^2, q^4$, so we get that the sum of the three r... | {
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If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$ Question -
Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that
$$
a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}
$$
My try -
i tried putting $a+2 = x, b+2=y , c+2=z$
then we get... | Let $c=\min\{a,b,c\}$.
Thus, we need to prove that:
$$a^2+b^2+c^2-\frac{(a+b+c)^2}{3}\geq\frac{2+a}{2+b}+\frac{2+b}{2+a}-2+\frac{2+b}{2+c}-\frac{2+b}{2+a}+\frac{2+c}{2+a}-1$$ or
$$\frac{2}{3}((a-b)^2+(c-a)(c-b))\geq\frac{(a-b)^2}{(2+a)(2+b)}+\frac{(c-a)(c-b)}{(2+a)(2+c)},$$ which is true because
$$\frac{2}{3}>\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $f(x)$ is a common factor of $g(x)$ and $h(x)$ find $f(x)$ Given that $f(x)$ is a common factor of $g(x)=x^4-3x^3+2x^2-3x+1$ and $h(x)=3x^4-9x^3+2x^2+3x-1$ find $f(x)$
I tried to factorised $g(x)$ but it doesn't have any rational roots as I've already tried 1 and -1.
So how do I solve this?
| $$x^4-3x^3+2x^2-3x+1=x^4-3x^3+x^2+x^2-3x+1=$$
$$=(x^2+1)(x^2-3x+1).$$
$$3x^4-9x^3+2x^2+3x-1=3x^4-9x^3+3x^2-x^2+3x-1=$$
$$=(x^2-3x+1)(3x^2-1).$$
Can you end it now?
The first factorization we can get by the following way:
$$x^4-3x^3+2x^2-3x+1=x^2\left(x^2+\frac{1}{x^2}+2-3\left(x+\frac{1}{x}\right)\right)=$$
$$=x^2\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $x+y+z=1$ prove $ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $ Question -
Let $x, y, z$ be non-negative real numbers which satisfies $x+y+z=1$. Prove that
$$
\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sq... | Since $\frac{1}{12}<\frac{1}{8},$ it's enough to prove that:
$$\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{8}}\leq\sqrt3.$$
Now, by C-S
$$\left(\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{8}}\right)^2\leq\sum_{cyc}\frac{x+\frac{(y-z)^2}{8}}{2x+y+z}\sum_{cyc}(2x+y+z)=\sum_{cyc}\frac{4x+\frac{(y-z)^2}{2}}{2x+y+z}$$ and it's enough to prove th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).
My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\im... | Let $\arctan x=y;x=\tan y, -\dfrac\pi2 <y<\dfrac\pi2$
As $\cos y>0,$
$$\cos y=+\dfrac1{\sqrt{1+x^2}};\sin y=\tan y\cdot\cos y=\dfrac x{\sqrt{1+x^2}}$$
$$\cos(\arctan x)+x\sin(\arctan x)=\sqrt{1+x^2}$$
$\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}=\sqrt{{(\sqrt{1+x^2}})^2-1}=\sqrt{x^2}=|x|$ for real $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Simplify the algebraic fraction $\frac{4x^2-8x+3}{4x^2+2x-12}$. Find the quotients as algebraic fractions of the following division:
$$\frac{4x^2-8x+3}{4x^2+2x-12}$$
The answer is
$$\frac{2x-1}{2(x+2)}$$
The method that I am applying to solve:
Write what $2$ numbers multiplied give me the far most right number in the n... | $4x^2 - 8x + 3$
When you have a coefficent on the $x^2$ term, it gets a little bit more complicated to factor. There are two main techniques. Since 3 is prime, we can start with the guess
$(Ax - 3)(Bx - 1)$
Now we look for $A,B$ such that $AB = 4$ and $-A - 3B = -8$
Our candidates are $\{(1,4),(2,2), (4,1)\}$ and we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Cartesian equation for the curvature of a superellipse? I have an application where I need to determine the curvature of a superellipse at various points along the curve, where the Cartesian form of the superellipse is defined as:
$$\frac{x^n}{a^n}+\frac{y^n}{b^n}=1$$
For an ellipse, the Cartesian equation for the curv... | Using Hessian matrix for implicit function $F(x,y)=0$,
\begin{align}
F(x,y) &= \frac{x^n}{a^n}+\frac{y^n}{b^n}-1 \\
F_x &= \frac{nx^{n-1}}{a^n} \\
F_y &= \frac{ny^{n-1}}{b^n} \\
\nabla F &=
\begin{pmatrix}
F_{x} \\
F_{y}
\end{pmatrix} \\
F_{xx} &= \frac{n(n-1)x^{n-2}}{a^n} \\
F_{yy} &= \frac{n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. When $ab/(a+b)$ is an integer, where $a,b$ are positive integers?
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
print a,b,q1 div q2;
end if;
end for;
end for;
the Magma code given above outputs t... | The conjecture is true - just cancel the gcd $(A,B) =:c\,$ to reduce to the simpler coprime case.
$\begin{align}{\bf Theorem}\ \ \ A\!+\!B\mid AB &\iff\, a+b\mid c,\ \ \ \ a = A/c,\, b = B/c \\[.4em]
&\ \smash[t]{\overset{\times\ c}\iff}\ \:\! A\!+\!B\mid (A,B)^2\end{align}$
Proof $\,\ \dfrac{AB}{A+B} = \dfrac{acb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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The tangent line to the curve of intersection of the surface $x^2+y^2=z$ and the plane $x+z=3$ at the point $(1,1,2)$ passes through The tangent line to the curve of intersection of the surface $x^2+y^2=z$ and the plane $x+z=3$ at the point $(1,1,2)$ passes through
(A)$(-1,-2,4)$
(B)$(-1,4,4)$
(C)$(3,4,4)$
(D)$(-1,4,0)... | Two surfaces, paraboloid $z=x^2+y^2$ and plane $z=3-x$, intersect on ellipse:
$$x^2+y^2=3-x \iff x^2+x+y^2=3 \iff (x+\frac{1}{2})^2-\frac{1}{4}+y^2 = 3 \iff (x+\frac{1}{2})^2+y^2=\frac{13}{4} \\$$
$$\iff \frac{(x+\frac{1}{2})^2}{\frac{13}{4}}+\frac{y^2}{\frac{13}{4}}=1$$
Now the tangent on the intersection curve at the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Verify Stokes' Theorem for Hemisphere I am trying to answer the following question, but am having difficulty getting the same result for each side of Stokes' theorem.
Question:
Verify Stokes' Theorem for the hemisphere $D: x^2 + y^2 + z^2 = 9, z\geq0$ its bounding circle $C: x^2 + y^2 = 9, z=0$ and the vector field $\o... | Let's clarify our setting first. Note that a normal vector to the sphere of radius $3$ at $\vec{x}$ is $\vec{n}=\frac{\vec{x}}{\|\vec{x}\|}=\frac{\vec{x}}{3}$. Our hemisphere is given by
$$
S=\{(x,y,\varphi(x,y)): x^2+y^2\le 9\},
$$
with $\varphi(x,y)=\sqrt{9-x^2-y^2}$. Therefore the surface measure is
$$
dS(\vec{x})=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $
Evaluate $$\sum_{n,k} \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.
Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation ... | Generally similar sums can be evaluated using the Beta function:
$$
B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}=
\frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}.
$$
Applying this in your case ($x=n-k,y=k$) one has:
$$
\binom{n}{k}^{-1}=(n+1)\int_0^1 t^{n-k}(1-t)^{k}dt
$$
or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Does $a^2 + b^2 = 2 c^2$ have any integer solution?
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?
I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I... | It is equivalent to solving the circle equation $x^2 + y^2 = 2$ in rational numbers.
Take one point $(1,1)$ on this curve and consider the line with slope $t$ passing through that line: $y = t x - t + 1$, substitute this in $x^2 + (t x - t + 1)^2 - 2 = 0$ and divide out $(x-1)$ to find the second point of intersection ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Proving $\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B)$. Could you give me some guidance on proving the following trig identity?
$$\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B).$$
| Consider $\cos(x\mp y)=\cos x\cos y\pm\sin x\sin y$, add them together, you'll get $$\cos(x+y)+\cos(x-y)=2\cos x\cos y.$$
Now let $x=A+B,\,y=A-B$, so $2\cos(A+B)\cos(A-B)=\cos(2A)+\cos(2B)=2\cos^2A-1+1-2\sin^2B$,
hence we get the desired result as $\cos^2A-\sin^2B=(\cos A+\sin B)(\cos A-\sin B)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that for all $n\in\mathbb N, (\sqrt3+i)^n+(\sqrt3-i)^n=2^{n+1}\cos(\frac{n\pi}6)$ I want to prove by induction that for all $n \in \mathbb{N}$, $$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\left(\frac{n\pi}{6} \right)$$
I can prove the identity using direct complex number manipulation and de M... | My try is not that 'pure' in the sense of using the trig. equation.
Let $ A = \sqrt 3 + i$, $B = \sqrt 3 - i$. With some initial step verified, we can write
\begin{align*}
A^{n+1} + B^{n+1}&= (A^n + B^n)( A+ B) - AB (A^{n-1} + B^{n-1}) \\
& = 2^{n+2}\left( 2 \cos(\frac{n\pi}{6})\cos(\frac{\pi}{6}) - \cos \frac{(n-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$f(m^2 + f(n)) = f(m)^2 + n$ for all natural numbers Find all functions such that $$ f(m^2 + f(n)) = f(m)^2 + n$$ for all natural numbers (1,2,3,4,..)
I have been stuck on this problem.
I have tried to set f(1) = k, however, did not make any progress.
| Let $P(m, n)$ denote $f(m^2+f(n)) = f(m)^2 + n$. Note that
\begin{align*}
P(1, y) \implies f(1+f(y)) &= f(1)^2 + y\\
\implies P(f(x), 1+f(y)) \implies f(f(x)^2 + f(1)^2 + y) &= f(f(x))^2 + 1 + f(y) \qquad (1)\\
P(x, y) \implies f(x^2+f(y)) &= f(x)^2 + y\\
P(f(1), x^2+f(y)) \implies f(f(x)^2 + f(1)^2 + y) &= f(f(1))^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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On $\sum_{x=0}^\infty x^n r^x$ I am looking for either a closed form or recursive expression for $$\sum_\limits{x=0}^\infty x^n r^x\ $$ that does not include differential operators, where $n \in \mathbb N_0$. It is clear to me that $$\sum_\limits{x=0}^\infty x^n r^x = r \cfrac{\text d}{\text dr}\sum_\limits{x=0}^\infty... | Terminology ... polylogarithm
$$
\operatorname{Li}_s(z) = \sum_{k=0}^\infty \frac{z^k}{k^s}
$$
So your question asks about the polylogarithm $\operatorname{Li}_{-n}(r)$.
That page shows $\operatorname{Li}_s(z)$ for $s=-1,-2,-3,-4$, the first four
examples you computed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ . What is the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ ?
I have been stuck on this problem with no direction. I have tried multiplying the sequence with $x$ and trying out $S-Sx$ but have gotten nowhere. Any help?
Thanks.
| Hint:
$F'(x)=S=1\cdot2x+2\cdot3x^2+3\cdot4x^3+\cdots$
$F(x)=x^2+2x^3+3x^4+\cdots=x^2(1+2x+3x^2+\cdots)=\dfrac{x^2}{(1-x)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sum $\sum \frac{1}{(4k-3)(4k-2)(4k-1)(4k)}$ I am stuck on this problem for quite a while now, and I don't seem any closer to the solution. So, here it is:
$S = 1/4! + 4!/8! + 8!/12! + 12!/16! + ......$
I crossed out the factorials first, and it could be easily represented by the general term,
$T = \frac{1}{(4n-3)(4n-... | \begin{align*}
& \frac{1}{(4n-3)(4n-2)(4n-1)(4n)} \\
&= \frac{1}{3}\left(\frac{1}{(4n-1)(4n-2)(4n-3)} - \frac{1}{4n(4n-1)(4n-2)}\right) \\
& = \frac{1}{3\cdot 2}\left( \frac{1}{(4n-2)(4n-3)} - \frac{2}{(4n-1)(4n-2)} + \frac{1}{4n(4n-1)}\right) \\
& = \frac{1}{3\cdot 2\cdot 1}\left( \frac{1}{4n-3} - \frac{3}{4n-2} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find the intergral $I_{A}=\int_{0}^{2\pi}\frac{\sin^2{x}}{(1+A\cos{x})^2}dx$ Let $A\in (0,1)$be give real number ,find the closed form intergral
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx$$
This integral comes from a physical problem，following is my try:
since
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin... | Use integration by parts
$$\int \frac{\sin^2 x\ dx}{(1+A\cos x)^2}$$
$$=\int \sin x\cdot \frac{\sin x}{(1+A\cos x)^2}\ dx$$
$$=\sin x\cdot \frac{1}{A(1+A\cos x)}-\int \frac{\cos x}{A(1+A\cos x)}\ dx$$
$$=\frac{\sin x}{A(1+A\cos x)}-\frac{1}{A^2}\int \frac{(1+A\cos x)-1}{1+A\cos x}\ dx$$
$$=\frac{\sin x}{A(1+A\cos x)}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3694900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Angular integrals used in QED I am reading a research paper
and am stuck at a point where the author uses angular integrals. I don't have any idea about it and would like help.
The angular integral is:
$$I_k (y)=\int_0^\pi {\sin^2(\theta)\cos^{2k}(\theta) \over \beta^2(y)-\cos^2 (\theta)} d\theta$$
The author dire... | An interesting approach to these integrals is the use of complex integration. In this case, we can use the fact that
with
\begin{equation}
f(\sin\theta,\cos\theta) = \frac{\sin^2\theta \cos^{2k} \theta}{\beta^2-\cos^2\theta}
\end{equation}
we have
\begin{eqnarray}
\int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What are the steps to factor $x^2 - 1$ into $(x+1)(x-1)$? Does $(x+1)(x-1) = x^2+1x-1x-1$? If so where are the $+1x$ and the $-1x$ when it is being factored from $x^2-1$ into $(x+1)(x-1)$?
What exactly are we dividing $x^2-1$ by to get $(x+1)(x-1)$ and how did you know what to divide it by?
| Method 0:
$+1x - 1x = 0$
So $(x+1)(x-1) = x^2 + 1x - 1x -1 =$
$x^2 +(1x-1x) -1 = $
$x^2 + 0 -1 =$
$x^2 -1 $.
Method 1:
$x^2 - 1 = $
$x^2 + x - x - 1 = $
$(x^2 +x) - (x+1) = $
$(x\cdot x + x\cdot 1) + (-1)\cdot(x+1) =$
$x\cdot(x + 1) + (-1)\cdot (x+1) =$
$\color{blue}x\cdot\color{red}{(x + 1)} + \color{blue}{(-1)}\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What mistake am I making while deriving the expansion of $\cos(\alpha + \beta)$ I was trying to derive the formula for expansion of $\cos (\alpha + \beta)$ by equating the ratio of lengths of two specific chords to the ratio of angles opposite to them but I'm not getting the correct results. Here's how I'm doing it :
... | You make a mistake exactly where you suspected. Note the ratio of the lengths of the chords is not equal to the ratio of the angles subtended by them. You could see this by applying the law of cosines on $\triangle AOB$ and $\triangle AOC$: $$AB^2=OB^2 + OA^2 -2OA\cdot OB\cos\alpha \\ =1+1-2\cos\alpha\\=2(1-\cos\alpha)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the saddle point of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2+(-x_1+x_2)y_3$ Finding the saddle points of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2$+$(-x_1+x_2)y_3$ subject to the constraints $x_1+x_2+x_3=1, y_1+y_2+y_3=1$. Show that the saddle point is $x=(\frac{1/3}{1/3},\fr... | At a saddle point, all of the partial derivatives are 0. This leads to a system of equations:
\begin{align}
x_1 -2x_2 +x_3=0,\tag{1}\\
2x_1 -2x_3=0,\tag{2}\\
x_2-x_1=0,\tag{3}\\
y_1+2y_2-y_3=0, \tag{4}\\
-2y_1 +y_3=0,\tag{5}\\
y_1-2y_2=0. \tag{6}
\end{align}
Adding 1 to both sides of equation (4) leads to
\begin{equati... | {
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"url": "https://math.stackexchange.com/questions/3704400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(... | Note that $x^2-6x-9 = (x-3)^2 - 18$. So after pulling out the factor of $\frac 12$, it suffices to compute
$$\frac{d}{dx} \left(\frac{x-3}{x(x+3)}\right)^2$$
and
$$\frac{d}{dx} \left(\frac{1}{x(x+3)}\right)^2.$$
These obviously only require finding the derivative of what's inside, since the derivative of $(f(x))^2$ is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
| Observe that $(x+1)$ divides all quadratics: the original equation is
$$\sqrt{(x+1)(x+7)} + \sqrt{(x+1)(x+2)} = \sqrt{(x+1)(6x+13)},$$
and rearranging we obtain the following:
$$(\sqrt{x+1})(\sqrt{x+7}+\sqrt{x+2}-\sqrt{6x+13})=0.$$
We are doing some trickery with allowing square roots to venture into $\mathbb C$ here, ... | {
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"url": "https://math.stackexchange.com/questions/3707727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$. (a) Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ such that $V = W_{1}\oplus W_{2}$. If $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are bases for $W_{1}$ and $W_{2}$, respectively, show that $\mathcal{B}_{1}\ca... | Another approach without appealing dimensionality.
*
*Suppose that $\mathcal B_1$ is a basis for $W_1$ and that $\mathcal B_2$ is a basis for $W_2$. Since $V = W_1 \oplus W_2$, any vector in $V$ can be uniquely written as a sum of a vector in $W_1$ and a vector in $W_2$; but also, at the same time, every vector in $... | {
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"source": "stackexchange",
"question_score": "4",
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Sum $\sum_{n = 1}^{\infty}\left[\frac1n\sin nx + \frac{1}{n^2}\cos nx\right]$ I want to find the following sum by using the complex methods for series ($z = \cos nx + i \sin nx$).
$$
\sum_{n = 1}^{\infty}\left[\frac1n\sin nx + \frac{1}{n^2}\cos nx\right]
$$
Here is my attempt:
$$
S_N = \sum_{n = 1}^{N}\frac1n\sin nx +... | Continue with
\begin{align}
\sum_{n = 1}^{\infty}\frac{\sin nx }n
= -\Im \ln(1 - z)
= \frac i2 [\ln( 1-z) -\ln (1-\bar z)]
=\frac i2\ln(-z)= \frac{\pi-x}2
\end{align}
and use the result to evaluate
$$
\sum_{n = 1}^{\infty}\frac{1- \cos nx }{n^2}=
\sum_{n = 1}^{\infty}\int_0^x \frac{\sin nt }{n}dt
= \int_0^x \frac{\p... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$ I was training for upcoming Olympiads, working on inequalities, and the following inequality came up:
$$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$$ with the obvious delimitations $3b\geq a;\: 3a\geq b.$
I've been pondering the question for quite a while, and... | As hinted by Quasi's solution, repeated squaring works for this problem. This approach should definitely be in your bag, especially since it's so easy to get rid of square roots.
If you want to simplify it some, then consider a change of variables: $ x = 3a -b , y = 3b-a$.
This gives us $ a = \frac{3x+y}{8}, b = \f... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed... | Let $S$ be the sum in question.
Let $f(x)=(1+x)^n+(1-x)^n$.
Then $f'(1)=2S$.
Now $f'(x)=n(1+x)^{n-1}-n(1-x)^{n-1}$ and so $f'(1)=n 2^{n-1}$.
Therefore, $S=n 2^{n-2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find t... | $$\text{You wrote: } x=\frac{y+2}{y^2+2y+1} $$
$$x(y^2+2y+1)=y+2$$
$$
xy^2 + ( 2x-1 )y +(x-2)=0 \tag 1
$$
$$
ay^2 + by + c = 0, \quad\text{where } a=x,\, b=(2x-1), \, c = x-2 \tag 2
$$
$$
y = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{1-2x\pm \sqrt{(2x-1)^2 -4x(x-2)}}{2x} \tag 3
$$
except that $(2)$ can be a valid solution... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \si... | Well, in a case very unusual for these type of questions the equation can be factored.
$\sin^2 x + 2\cos^2x + 3\sin x \cos x = (\sin x + \cos x)(\sin x + 2\cos x)=0$.
so
we have solutions when $\sin x = -\cos x$ or $\sin x=-2\cos x$. Looking at the unit circle it is clear that these can only occur in the 2nd and 4th q... | {
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"source": "stackexchange",
"question_score": "1",
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Find the range of $f:x \mapsto a+b\cos x$ The function $f:x \mapsto a+b\cos x$, is defined for $0 \le x\le 2\pi$. Given that $f(0) = 10$ and $f\left(\frac{2}{3}\pi\right) = 1$, find the values of $a$ and $b$, the range of $f$, and the exact value of $f\left(\frac{5}{6}\pi\right)$.
I was able to get the value of a and b... | Your calculation for $a,b$ is correct, and so is the calculation of $f\left(\frac56\pi\right)$.
For calculating the range of $f$, consider the following hint:
*
*For all $x$, you $-1\leq \cos x \leq 1$.
*The bounds above are strict, i.e., there exist values of $x$ for which $\cos x$ reaches values of $-1$ and $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all possible values $m$ such that $f(x)=x^2-5mx+10m-4$ has 2 roots and one of them is twice as the other Consider the polynomial $f(x)=x^2-5mx+10m-4$, find $m$ such that there exist a number $a$ that satisfies $f(a)=f(2a)=0$. This was my attempt:
$f(a)=f(2a)$
$a^2-5ma+10m-4=4a^2-10ma+10m-4$
$a^2-5ma=4a^2-10ma$
... | The given equation can be written as
$\begin{align}
f(x)& =x^2-4-5mx+10m\\
&=(x-2) (x+2) -5m(x-2) \\
&=(x-2) (x+2-5m)
\end{align}$
Now, we have two possibilities: $\alpha=2\beta$ or $\beta=2\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Value of $\alpha$ for which $x^5+5\lambda x^4-x^3+(\lambda\alpha-4)x^2-(8\lambda+3)x+\lambda\alpha-2=0$ has roots independent of $\lambda$
Consider the equation $$x^5 + 5\lambda x^4 -x^3 + (\lambda \alpha -4)x^2 - (8\lambda +3)x + \lambda\alpha - 2 = 0$$ The value of $\alpha$ for which the roots of the equation are in... | The question may be phrased incorrectly, as it is not possible to make the set of all roots independent of $\lambda$. The question I will answer is: For what value of $\alpha$ does the equation have some roots which are independent of $\lambda$?
As demonstrated in the question, $\alpha=-\frac{64}{5}$ is one possibilit... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why is $|x|'=\frac{x}{|x|}$? Why is $|x|'=\frac{x}{|x|}$ ?
May someone explain this clearly? If $x>0$ then it should be 1, and if $x<0$ then it's $-1$...
| You can derive this equation by writing the absolute value as $|x| =\sqrt{x^2}$. From here we see that
$$
\frac{d}{dx} |x| = \frac{d}{dx} \sqrt{x^2} = \frac{1}{2 \sqrt{x^2} } \left( \frac{d}{dx} x^2 \right) = \frac{2x}{2 \sqrt{x^2} } = \frac{x}{\sqrt{x^2} } = \frac{x}{|x|}
$$
where we use the chain rule. Notice ... | {
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"url": "https://math.stackexchange.com/questions/3726935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that this function all over the positive integer gives us this sequence? Firstly, we have this sequence : $1,1,2,1,2,3,1,2,3,4,...$ which is the sequence of integers $1$ to $k$ followed by integers $1$ to $k+1$. We could say a fractal sequence.
Secondly, we have this formula : $$a_n=\frac{1}{2}(2n+\lfloor\sqrt{... | $\newcommand{\bb}[1]{\left( #1 \right)}$
$\newcommand{\f}[1]{\left\lfloor #1 \right\rfloor}$
Key observation: Given $m \in \Bbb{Z}^+$, if:
$$
\sum_{k=1}^m k = \frac{m(m+1)}{2} < n \leq \frac{(m+1)(m+2)}{2} = \sum_{k=1}^{m+1} k
$$
then:
$$
a_n = n - \frac{m(m+1)}{2}
$$
Now rewrite your formula in the following manner:
\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $d^k + (a-d)^k = a[d^{k-1}-d^{k-2}(a-d)+\dots+(a-d)^{k-1}]$ It is stated to me as part of a larger problem that for odd $a > 2$, positive integer $n$, $k=a^n$, and positive integer $d$, we have
\begin{equation}
d^k + (a-d)^k = a[d^{k-1}-d^{k-2}(a-d)+\dots+(a-d)^{k-1}].
\end{equation}
I have tried expanding but... | This is just a different form of the factorization for positive odd $n$ $$x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \cdots + y^{n-1}) = (x+y) \sum_{k=0}^{n-1} (-1)^k x^{n-k-1}y^{k}.$$ This is because
$$\begin{align*}
(x+y) \sum_{k=0}^{n-1} (-1)^k x^{n-k-1} y^k
&= \sum_{k=0}^{n-1} (-1)^k x^{n-k} y^k + \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometry Proof to Find Maximum area of $\triangle PIE$
Circle $\omega$ is inscribed in unit square $PLUM,$ and points $I$ and $E$ lie on $\omega$ such that
$U, I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE.$
I'm not sure if there is a solution possible without trigonometry... | I would provide a solution without trigonometry, as asked in the OP. Consider the circle as centered in point $(1/2,1/2)$ of a Cartesian plane, so that the square corners are $L(0,0)$, $U(0,1)$, $M(1,1)$, $P(1,0)$. The circle equation is $(x-1/2)^2+(y-1/2)^2=1/4$. Let us call $k$ the slope of a line passing through $U... | {
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"url": "https://math.stackexchange.com/questions/3730695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:... | if $a=b=c$ is not true one can multiply three expressions of $x$ in from OP's thisrd step of $x$ to get $x^3=-1$. So $a+b+c \ne 0$ from OP's second equatiom. Hence either $x^3=-1$ or $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Second system of equations I've solved that system and recieved $y=\frac{1}{4}$, $x = -\frac{4}{5}$ but there are one more pair of $(x, y)$ in book and I don't know what I have to do to find its.
\begin{cases} 2x - 3xy + 4y=0 \\ x + 3xy -3x = 1 \end{cases}
In book there are two answers and the second answer is $(1, -2)... | In case you made a typo and the 2nd equation is actually $x+3xy-3y=1$, that equation can be factored:
$$
x-1 + 3y(x-1) = 0
$$
$$(x-1)(3y+1)=0$$
Which means that either $x=1$ or $y=-\frac{1}{3}$
Substitute that into the first equation to get the 2 pairs.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the cardinality of $\big\{(x,y,z)\mid x^2+y^2+z^2= 2^{2018}, xyz\in\mathbb{Z} \big\}$.
What is the cardinality of set $\big\{(x,y,z)\mid x^2+y^2+z^2= 2^{2018}, xyz\in\mathbb{Z} \big\}$?
Since I have very limited knowledge in number theory, I tried using logarithms and then manipulating the equation so that we... | For $n \in \mathbb N$, consider the equation
$$ x^2 + y^2 + z^2 = 2^n $$
where $x,y,z$ are integers. Since $x \mapsto -x$, $y \mapsto -y$, $z \mapsto -z$ does not change the equation, we may assume $x,y,z \ge 0$. We may henceforth suppose $x \ge y \ge z$.
Note that there is no solution when $n=1$.
Suppose $n \ge 2$. Si... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove by mathematical induction, that $81\times 3^{2n} - 2^{2n}$ is divisible by $5$ where $n \in \mathbb{Z}^+$
For all $k$, the equation I came up with is $3^{4+2k} = 5m + 2^{2k}$ where $m$ is a positive integer.
For all $k+1$, the expression is $3^{6+2k} - 2^{2k+2}$.
I tried to plug in the first equation to reach a... | $$81\cdot3^{2n}-2^{2n}=5k\implies
\\\begin{align}81\cdot3^{2(n+1)}-2^{2(n+1)}=81\cdot9\cdot3^{2n}-4\cdot2^{2n}&=9\cdot(81\cdot3^{2n}-2^{2n})+5\cdot2^{2n}
\\&=9\cdot5k+5\cdot2^{2n}
\\&=5k'.\end{align}$$
Shorter:
$$81\cdot3^{2n}-2^{2n}=9^n-4^n\mod 5\implies 9^{n+1}-4^{n+1}=9\cdot9^n-4\cdot4^n=4(9^n-4^n)\mod5.$$
Yet sho... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$
My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Ri... | You can use the method of Lagrange multipliers. If $f(x,y)=x+y$, then, for every $(x,y)\in\Bbb R^2$, $\nabla f(x,y)=(1,1)\ne(0,0)$. Therefore, the maximum of $f$ can only be attained at the boundary of the disk, that is, when $2x^2+3y^2=1$. So, solve the system$$\left\{\begin{array}{l}1=4\lambda x\\1=6\lambda y\\2x^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof for the general formula for $a^n+b^n$. Based on the following observations. That is
$$a+b = (a+b)^1 \\ a^2+b^2 = (a+b)^2-2ab \\ a^3+b^3 = (a+b)^3-3ab(a+b) \\ a^4+b^4= (a+b)^4-4ab(a+b)^2+2(ab)^2\\ a^5+b^5
= (a+b)^5 -5ab(a+b)^3+5(ab)^2(a+b)\\\vdots$$
I came to make the following conjecture as general formula... | $a$ and $b$ are roots of $x^2=(a+b)x-ab$. Therefore, $a^{n+2}=(a+b)a^{n+1}-(ab)a^n$ and analogously for $b$.
Let $p_n=a^n+b^n$. Then $p_{n+2}=(a+b)p_{n+1}-(ab)p_n$ is a simple recurrence. The initial values are of course $p_0=2$ and $p_1=a+b$.
This recurrence is a special case of Newton's identities.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Definite integral $\int_{-\infty}^\infty \frac{\log(x^2+a^2)}{(x-ib)^2} dx$ How can I evaluate the integral
$$\int_{-\infty}^\infty \frac{\log(x^2+a^2)}{(x-ib)^2} dx?$$
Here $a, b$ are positive real constants. When I plug this expression in MATLAB, I obtain the answer as
$$ - \frac{\mathrm{log}\!\left(x - a\, \mathrm{i... | Let $f \colon (0,\infty)^2 \to \mathbb{C}, \, f(a,b) = \int_{-\infty}^\infty \frac{\log(a^2+x^2)}{(x-\mathrm{i} b)^2} \, \mathrm{d} x$. We can get rid of $a$ by letting $x = a t$ and using the result $\int_{-\infty}^\infty \frac{\mathrm{d} t}{(t-\mathrm{i} c)^2} = 0$ for $c > 0$:
$$ f(a,b) = \frac{1}{a} \int \limits_{-... | {
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"timestamp": "2023-03-29T00:00:00",
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Locus of the circumcenter of triangle formed by the axes and tangent to a given circle.
A circle centered at $(2,2)$ touches the coordinate axes and a straight variable line $AB$ in the first quadrant, such that $A$ lies the $Y-$ axis, $B$ lies of the $X-$ axis and the circle lies between the origin and the line $AB$.... | The first thing to note is that $\triangle OAB$ is always a right triangle, and its incenter is always $(2,2)$. Thus, hypotenuse $AB$ is always a diameter of the circumcircle, and the circumcenter is therefore the midpoint of $AB$. If the line passing through $AB$ has equation $$\frac{x}{a} + \frac{y}{b} = 1,$$ then ... | {
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} |
Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$
My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\fr... | Due to homogeneity, assume that $abc = 1$.
Let $p = a+b+c, q = ab+bc+ca, r=abc=1$.
We need to prove that $p\cdot \frac{q}{r} + 3 \ge 4 \cdot \frac{p}{\sqrt[3]{r}}$
or $pq + 3 \ge 4 p$.
Since $q^2 \ge 3pr$, it suffices to prove that
$p \cdot \sqrt{3p} + 3 \ge 4p$ or $\frac{1}{3}(\sqrt{3p} - 3)(3p - \sqrt{3p} - 3)\ge 0$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Ambiguity with $\sin 2x$ and $\cos 2x$ squaring So basically I was learning about trigonometric functions of multiples of $x$, and they were defined as
$$\sin 2x= 2\sin x\cos x$$
and
$$\cos 2x= \cos^2 x- \sin^2 x$$
Now, I wanted to work from $\sin 2x= 2\sin x\cos x$ and, using $\sin^2 2x+ \cos^2 2x=1$, get the formula ... | Unnecessarily squaring produces ambiguities. For example, if $x=1$, then starting with the squared equation as follows produces ambiguity
\begin{align*}
x^2&=1^2=1\Rightarrow x=\pm 1
\end{align*}
Unnecessarily squaring $\cos 2x=\cos(x+x)=\cos^2x-\sin^2x=1-\sin^2x-\sin^2x=1-2\sin^2x$ results similarly.
As for $\cos x=\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
total differentiability of $\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$ if the denominator equals $0$ Let $f:\mathbb{R}^3\to\mathbb{R}$ be given by $f(x,y,z)=\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$ if the denominator is not equal to $0$ and otherwise by $f(x,y,z)=0$.
Find all the points where $f$ is differentiable.
$f$ is surely di... | Choose $(x,y,z) \neq 0$ and look at $f((tx,ty,tz)) = t^3\frac{x^3z^4}{(x^2+y^2)(y^2+z^2)}$. From this, if $f$ is differentiable, we would expect that
$Df(0) = 0$.
Note that ${x^2 \over x^2+y^2} \le 1$ and ${z^2 \over z^2+y^2} \le 1$ so
$|f((x,y,z))| \le |xz^2|$.
Note that $|xz| \le {1 \over 2} (x^2+z^2)$.
Suppose $|z| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
find the coefficient of $x^5$ in $(1+2x-3x^2)^6$ I know that I should first factor the expression within the brackets to $-(3x+1)^6(x-1)^6$. However, after that I do not know what to do.
| We can also apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ of an expression.
We obtain
\begin{align*}
\color{blue}{[x^5]}&\color{blue}{(1+2x-3x^2)^{6}}\\
&=[x^5]\sum_{j=0}^{6}\binom{6}{j}(-3x^2)^j(1+2x)^{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Continuity of $a^x+b$ with $a, b \in \mathbb R$ Let $a,b \in \mathbb{R}$ with $a > 0$. find $a$, $b$ so the function would be continuous
$$
f(x) = \begin{cases} a^x + b, & |x|<1 \\
x, & |x| \geq 1 \end{cases}
$$
I got $b = -a^x+x$ as my answer, but I'm unsure.
| Since $f(x) = a^x + b$ will be continuous on $|x| < 1$ for $a > 0$, we only need to match up this portion of $f$ with it's definition on $|x| \geq 1$ at the endpoints $x = \pm 1$. Evidently, $f(-1) = -1$ and $f(1) = 1$. So we need $a^{-1} + b = -1$ and $a + b = 1$. Using the latter gives $b = 1 -a$, so substitution yie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=... | $$\frac{1}{\sin^3\alpha} -\frac{1}{\cos^3 \alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha \cos^3 \alpha}= -\frac{(\sin\alpha-\cos\alpha)(1+\sin\alpha\cos \alpha)}{(\sin \alpha \cos\alpha)^3}$$ Now,$$ (\sin \alpha -\cos \alpha )^2 =\frac 14 \implies 1-2\sin\alpha\cos\alpha =\frac 14 \\\implies \sin\alpha \cos\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Compute $ \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $ Find $\displaystyle \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $
I tried to do like this :
Let $A=\displaystyle \lim_{x \to\frac{\pi}{2}} \{1^{\sec^2 x} + 2^{\sec^... | Let $f_n(x)$ be the sequence given by
$$f_n(x)=\left(\sum_{k=1}^n k^{\sec^2(x)}\right)^{\cos^2(x)}$$
Then, we have
$$\begin{align}
f_n(x)&=\left(n^{\sec^2(x)}\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
&=n\left(\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3752948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | In fact, substitution makes it very easy. Starting with
$$\int \frac{u^3}{(u^2 + 1)^3}\ du$$
take $v = u^2 + 1$, then $dv = 2u\ du$ so $u\ du = \frac{1}{2} dv$. Then we get
$$\begin{align}
\int \frac{u^3}{(u^2 + 1)^3}\ du &= \int \frac{u^2 \cdot u}{(u^2 + 1)^3}\ du\\
&= \int \frac{u^2 \cdot \overbrace{(u\ du)}^{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 5
} |
Prove that a polynomial ring is integrally closed
Let $V \subseteq {\mathbb{A}}^2_{\mathbb{C}}$ be the curve defined by $x^2-y^2+x^3=0$, and let $\mathbb{C}\left [ V \right ]$ the coordinate ring of $V$. Let $\Theta :=\bar{y}/\bar{x} \in \mathbb{C}\left ( V \right )$. I must show that the ring $B:=\mathbb{C}\left [ V ... | Let us show that the $\Bbb{C}[V][\Theta]$ is in fact generated by $\Theta$ as a $\Bbb{C}$-subalgebra of $\Bbb{C}(V).$
Define a morphism $\phi$ as follows:
\begin{align*}
\phi : \Bbb{C}[x,y]&\to\Bbb{C}[t]\\
x&\mapsto t^2 - 1,\\
y&\mapsto t^3 - t.
\end{align*}
It is not difficult to check that this factors through the qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-... | For the second part of your question,
For $y \neq x$,
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l = \left(\frac{y}{y-x} - \frac{x}{y-x}\right) - \frac{y-x}{y} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l= -(y-x)\left(\frac{1}{y} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Extreme points of a function at domain ends Consider the following function: $f(x) = x\sqrt{9-x^2}$
$\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad $
$f'(x) = \frac{-2x^2+9}{\sqrt{9-x^2}}$ and $D(f) = [-3,3]$ therefore the critical points of the function are $x_{c_i} = \left\{ -3, -\frac{3\sqrt2}{2}... | If $D(f)=[-3,3]$, then $f( \pm 3)=0.$
Let $x \in (-3,0)$, then $f(x)<0=f(-3)$, henc $f$ has a local maximum at $x=-3.$
Let $x \in (0,3)$, then $f(x)>0=f(3)$, henc $f$ has a local minimum at $x=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | $$\int \frac{x^3}{\left(4x^2+9\right)^{\frac{3}{2}}}dx$$
if $4x^2+9=t^2$ as in the answer of the user @Anton Vrdoljak you have:
$$=\int \frac{t-9}{32t^{\frac{3}{2}}}\ dt=\frac{1}{32}\cdot \int \left(\frac{1}{t^{\frac{1}{2}}}-\frac{9}{t^{\frac{3}{2}}}\right)dt=\frac{1}{32}\left(2t^{\frac{1}{2}}-\left(-\frac{18}{t^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos ... | After multiplying the numerator and denominator of the expression in the limit by $$\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}$$, I get $$\lim_{x \to 0} \frac{1+x\sin(x)-\cos(x)}{x\tan(x) \left(\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}\right)}$$
It is clear that $$\lim_{x \to 0}\left( \sqrt{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Sum of squares and linear sum
For which positive integer $n$ can we write $n=a_1+a_2+\dots+a_k$ (for some unfixed $k$ and positive integers $a_1,a_2,\ldots,a_k$) such that $\sum_{i=1}^k a_i^2 = \sum_{i=1}^k a_i + 2\sum_{i<j}a_ia_j$?
When $k=1$, the equation is $a_1^2=a_1$, so $a_1=1$ and $n=1$ is the only possibility... | Partial answer.
If we write $m = \sum_i a_i^2$, then the identity $$\left(\sum_i a_i\right)^2 = \sum_i a_i^2 + 2\sum_{i < j} a_i a_j$$ allows us to translate our condition to $$ 2m = n^2 + n.$$ We know that $m \equiv n^2\mod 2$, hence a necessary condition is $4\mid n^2 - n$, which means $n\equiv 0, 1\mod 4$.
Now we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | Let us ise Euler's transformation
$$I=\int \frac{dx}{(x^2-4x+13)^2}=\int \frac{dx}{(x-a)^2 (x-b)^2},~~a,b=2\pm 3i.$$
Let $$t=\frac{x-a}{x-b} \implies x=\frac{bt-a}{t-1} \implies dx=\frac{a-b}{(t-1)^2}.$$
Then $$I=(b-a)^{-3} \int \frac{dt}{t^2(t-1)^2}=(a-b)^{-3}\int \frac{u^2 du}{(u-1)^2}, u=1/t .$$
Next use $u=v+1$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
} |
Find the value of $\lim _{a \to \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x $ Find the value of : $$
\lim _{a \rightarrow \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x
$$ I have tried ... | We can rewrite the integral as
$$\lim_{a\to\infty} \frac{1}{a}\int_0^\infty \frac{x^2+1}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx + \int_0^\infty \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx$$
which we are allowed to split up since both pieces are absolutely convergent. Then notice that
$$ \frac{x}{1+x^4}\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$a = \log_{40}100, b = \log_{10}20$.How can I express $b$ depending only on $a$? Let $a = \log_{40}{100}, b = \log_{10}{20}$. How can I express $b$ depending only on $a$? I tried using the formula to change the base from $40$ to $10$, but couldn't get it just depending on $a$.
I used the base change formula $\log_a b =... |
Find $a=\log_{40}100$ only in terms of $b=\log_{10}20$.
\begin{align*}
2b&=2\log_{10}20=\log_{10}(20)^2=\log_{10}(40\cdot10)=\log_{10}40+1=2\log_{10^2}40+1=\frac2a+1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is my variance negative? Let $Y_1 \sim Gamma(2,1)$, and $Y_2|Y_1 \sim f_{Y_2}(Y_2)$, where $y_1\geq y_2\geq 0$.
Where
$$f_{Y_2}(y_2) = \frac{1}{y_1}$$
$$f_{Y_1}(y_1) = y_1 \exp( -y_1)$$
Goal: Find standard deviation of $Y_2 - Y_1$
Start with
$$Var(Y_1) = 1, E(Y_1) = 1$$
$$E(Y_2|Y_1) = \int^{y_1}_0 \frac{y_2}{y_1}dy... | I found my mistake.
$$E(Y_1) = 2$$
$$Var(Y_1) = 2$$
Therefore
\begin{align*}
Var(Y_2) = & E\left (Var(Y_2|Y_1)\right) + Var\left ( E(Y_2|Y_1)\right ) \\
= & E \left ( \frac{y^2_1}{12}\right ) + Var \left( \frac{y_1}{2} \right ) \\
= & \frac{6}{12} + \frac{2}{4}\\
= &1
\end{align*}
In addition... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that the following sequence $(\sin\frac{\pi k}{3})_{k=1}^\infty$ diverges $$x_k := \sin \left(\frac{\pi k}{3} \right),\qquad x_{\infty} := \lim_{k \to \infty}{x_k}$$
I tried to prove by contradiction. Assume limit exists and is $L$. Fix $\epsilon=\frac{\sqrt{3}}{4}$. There exists $K\in\mathbb{Z}^{+}$ such that $... | Let's consider 3 sub sequences
$$k=6n \Rightarrow \sin\frac{\pi k}{3}=\sin 2\pi n=0 $$
$$k=6n+1 \Rightarrow \sin\frac{\pi k}{3}=\sin \left(2\pi n +\frac{\pi }{3} \right) = \frac{\sqrt{3} }{2}$$
$$k=9n+1 \Rightarrow \sin\frac{\pi k}{3}=\sin \left(2\pi n +\pi+\frac{\pi }{3} \right) = -\frac{\sqrt{3} }{2}$$
So we have 3 l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Existence of limit for sequence $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right)$ with initial values $x_0=5,x_1=10$ Let $x_0=5,x_1=10,$ and for all integers $n\ge2$ let $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right).$ By induction, we have $\forall m\in\mathbb Z_{\ge0}\enspace x_m>0,$ so we can avoid division by $0$... | Let $$x_{n+1}=\tfrac{1}{2}(x_n+\frac{a}{x_{n-1}})$$
Then for $d_n=x_n-\sqrt{a}$, \begin{align}
x_{n+1}-\sqrt{a}&=\tfrac{1}{2}(x_n-\sqrt{a})+\frac{a}{2}\left(\frac{1}{x_{n-1}}-\frac{1}{\sqrt{a}}\right)\\
d_{n+1}&=\tfrac{1}{2}d_n-\frac{\sqrt{a}}{2}\frac{d_{n-1}}{d_{n-1}+\sqrt{a}}=\tfrac{1}{2}d_n-\frac{1}{2}\frac{d_{n-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game st... | Working under the assumption that the intended interpretation of the question was merely asking the probability that $B$ wins (i.e. distinguishing between the term "rounds" as iterating whenever A has a turn and "turns" iterating whenever either A or B has a turn) two other approaches have already been written. Here I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^... | \begin{align}
\text{Let } & y = x\tan \theta, \\[8pt]
\text{so that }& dy = x\sec^2\theta\,d\theta \\[8pt]
\text{and } & x^2 + y^2= x^2\sec^2\theta,
\end{align}
and as $y$ goes from $0$ to $1$ then $\theta$ goes from $0$ to $\arctan(1/x)$.
Then
\begin{align*}
& \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \\[8pt]
= {} & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $a,b,c$ if ${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$ If $${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$$
and $$(a+b+c)=21, a\gt6$$
We have to find $a,b,c$
My attempt
I use a little fact that the sum of first $n$ odd numbers is $n^2$
From that i get $$a^2+b^2=c^2$$
Which means that the solutions are ... | If $a=2n-1$ so $$1^2+3^2+...+a^2=n^2=\left(\frac{a+1}{2}\right)^2.$$
Id est, $$(a+1)^2+(b+1)^2=(c+1)^2$$ or
$$a^2+b^2+2a+2b+2=(22-a-b)^2$$ or
$$ab-23(a+b)=-241$$ or
$$(a-23)(b-23)=288.$$
Also we have $a\geq7$, $b+c\leq14$, $c>b$ and $c>a$.
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
System of congurences and the Chinese Remainder Theorem I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $... | From the Chinese remainder theorem:
$$x = x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3$$
$$= 2 \cdot 35 \cdot \color{red}{1} + 1 \cdot 21 \cdot \color{red}{4} + 1 \cdot 15 \cdot \color{red}{6} = 244$$
The general solution is when $x \equiv 244 \pmod {\text{lcm}(3,5,7)} \Rightarrow x \equiv 244 \pmod {105} \equiv 34 \pmod {105}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[... | The limit can also be shown using HM-GM-AM.
Setting $u = x^2$ and considering $u\to +\infty$ we have
$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u \leq \sqrt[n]{\prod_{k=1}^n (k+u)} - u \leq \frac{\sum_{k=1}^n(k+u)}n - u = \frac{n+1}{2}$$
For the LHS we have
$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u = \frac{n - \sum_{k=1}^n\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
A problem involving the roots of quartic polynomial $x^4+px^3+qx^2+rx+1$ Let $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ be the roots of the following polynomial
$$P(x)=x^4+px^3+qx^2+rx+1$$
Show that
$$(1+{\alpha_1}^4)(1+{\alpha_2}^4)(1+{\alpha_3}^4)(1+{\alpha_4}^4)=(p^2+r^2)^2+q^4-4pq^2r.$$
I came across this prob... | Squaring both sides
$$(x^4+qx^2+1)^2=x^2(px^2+r)^2$$
Let $x^2=y$
$$(y^2+qy+1)^2=y(py+r)^2$$
$$\iff y^4+y^2A+1=y^3B+yC$$
Let $z=1+y^2\implies y=\sqrt{z-1}$
$$\implies(z-1)^2+(z-1)A+1=\pm\sqrt{z-1}((z-1)B+C)$$
$$ z^2+zE+F=\pm\sqrt{z-1}(zG+H)$$
Squaring both sides
$$z^4+\cdots+F^2=(z-1)(\cdots+H^2)$$
$$z^4+\cdots+F^2+H^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in t... | Write $ax^9+bx^8+1=(x^2-x-1)\sum\limits_{k=0}^7c_kx^{7-k}$ which gives the initial conditions $c_7=-1$ and $-c_7-c_6=0\implies c_6=1$. Notice that $c_0=a$ and $c_1-c_0=b$ on equating coefficients.
Further, we have $c_i=c_{i-1}+c_{i-2}$ for $i>1$ which is the negative Fibonacci sequence shifted by one. In particular, $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\the... | If you multiply and divide by $3$, you get
$$ \int (x^2 -1)(x^3 - 3x)^{4/3}dx = \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx $$
changing variable to $u = x^3 - 3x$ you have $du = (3x^2 - 3x)dx$ so
$$
\begin{split}
\int (x^2 -1)(x^3 - 3x)^{4/3}dx &= \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx\cr
&= \frac{1}{3} \int u^{4/3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Indefinite integral of $\sin^8(x)$ Suppose we have the following function:
$$\sin^8(x)$$
We have to find its anti-derivative
To find the indefinite integral of $\sin^4(x)$, I converted everything to $\cos(2x)$ and $\cos(4x)$ and then integrated. However this method wont be suitable to find the indefinite integral $\si... | By expanding
\begin{align}
\sin^8x
&=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)=\\
&=\frac{1}{128}\left(
\frac{e^{8ix}-e^{-8ix}}{2}
-8\frac{e^{6ix}-e^{-6ix}}{2}
+28\frac{e^{4ix}-e^{-4ix}}{2}
-56\frac{e^{2ix}-e^{-2ix}}{2}+35\right)=\\
&=\frac{1}{128}\left[\cos8x-8\cos6x+28\cos4x-56\cos2x+35\right]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $ By Wolfram Alpha I know that the limit
$$
\lim_{(x,y)\to (0,0)} \dfrac{8 x^2 y^3 }{x^9+y^3}=0.
$$
I have tried to prove that this limit is $0$, by using polar coordinate, the AM–GM inequality and the change of variable $ x^9= r^2 \cos^2(t) $ and $y^3= ... | Actually, that function isn't even bounded near $(0,0)$ and therefore, the limit at that point doesn't exist. You can check that$$\frac{8x^2(-x^3+x^6)^3}{x^9+(-x^3+x^6)^3}=\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}$$and that$$\lim_{x\to0}\left|\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding a Mistake for a Particular Form of Inequality My book depicts that the following problem uses ${x^3\over (1+y)(1+z)}+{(1+y)\over 8}+{(1+z)\over 8} \ge {3x\over 4} $.
Let $x, y, z$ be positive real numbers such that $xyz = 1$. Prove that $$ {x^3\over (1+y)(1+z)}+{y^3\over (1+z)(1+x)}+{z^3\over (1+x)(1+y)}\geq{3... | The inequality $$ {x^3\over (1+y)(1+z)}+{(1+y)}+{(1+z)} \ge {3x} $$ is indeed true by AM-GM with the equality for
$$ {x^3\over (1+y)(1+z)}=1+y=1+z,$$ which gives $y=z$ and $x^3=(1+y)^3$ or $x=1+y,$ which with the condition $xyz=1$ gives $$(1+y)y^2=1.$$
But in the original inequality the equality occurs for $x=y=z=1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum ... | $$\log(1-2x)=-\sum_{n=1}^\infty\frac{(2x)^n}n\;,\;\;|2x|<1\iff |x|<\frac12\implies$$
$$(x-1)\log(1-2x)-2x=\sum_{n=1}^\infty\frac{2^nx^{n+1}-2^nx^n}n-2x$$
and
$$\sum_{n=1}^\infty\frac{2^n(n-1)x^{n+1}}{n^2+n}=\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n+1}}^{:=f(x)}-\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n^2+n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How would I simplify this function $\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{\dots}}}}$ How do I simplify $\rho(x)$ into simple terms?
$$\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x+\sqrt{x-\sqrt{\dots}}}}}}}}$$
where the subtracting and the adding follows the Thue–Morse sequence $$+,-,-,+,-,+,+,-,-,+,+,-,+... | I wouldn't call this an exact closed form, but a 'close' one indeed. I remember a result in the paper (Page $28$); A chronology of continued square roots and other continued compositions by Dixon J. Jones; he refers another problem reffered in $1899$ by Karl Bochow (Problem 1740. Zeitschrift f¨ur mathematischen und
nat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
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.