Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How find this $f^{(4)}(0)$ let
$$f(x)=\dfrac{e^x}{1-\sin{x}}$$
Find the value of
$$f^{(4)}(0)=?$$
My try: let
$$\dfrac{e^x}{1-\sin{x}}=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots$$
so
$$e^x=(1-\sin{x})(a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots)$$
since
$$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cd... | $$
\begin{array}{lcl}
\frac{1}{1-\sin(x)} &=& 1+\sin(x)+\sin^2(x)+\sin^3(x)+\sin^4(x)+O(x^5) \\
&=& 1+\bigg(x-\frac{x^3}{6}\bigg)+\bigg(x-\frac{x^3}{6}\bigg)^2+
\bigg(x-\frac{x^3}{6}\bigg)^3+\bigg(x-\frac{x^3}{6}\bigg)^4+O(x^5) \\
&=& 1+\bigg(x-\frac{x^3}{6}\bigg)+\bigg(x^2-\frac{x^4}{3}\bigg)+
\bigg(x^3\bigg)+\bigg(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the minimum of a function of two variables
Find the smallest value of $\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}\quad \quad$ for $a,b$ real and positive.
What I've done so far:
Let $F(a,b)=\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab... | Comment to the earlier hint: Consider a triangle XOY where $|OX|=7$ and $|OY|=\sqrt{50}$. Then that expression is equal to $|XY|$. The minimum for $|XY|$ is achieved when $∠XOY=135^∘$.
Then the law of cosine give you the minimum for $|XY|$ namely the minimum value for the original expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Explain the origin of the number of divisors and sum of divisors formulas. I know the basic formulas which are:
For a number $n = p_1^{a_1} p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}$, we have
$d(n) = ( a_1 + 1 )( a_2 + 1 ) \cdot \ldots \cdot (a_k+1)$ and
$S(n) = \frac{p_1^{a_1+1} - 1}{p_1-1} \cdot \frac{p_2^{a_2+1} - 1}{p... | Note that $\frac{x^{n+1}-1}{ x - 1} = x^n + x^{n-1} + \dots + x + 1$. Also note that $(a_1 + \dots + a_n) \cdot (b_1 + \dots + b_m)$ is the sum of all possible combinations of $a_ib_j$. From this you should be able to figure out $S(n)$.
Note that, in the expression of $S(n)$, if instead of the divisors we could have $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $p$ is prime and $p>3$ and $k,l,m,n,p\in\mathbb N$ and $p^k+p^l+p^m=n^2$, prove that $8\mid p+1$. If $k,l,m,n,p\in\mathbb N$ and a prime number $p>3$ that satisfies $$p^k+p^l+p^m=n^2$$ is chosen, prove that $8\mid p+1$.
$n^2$, when divided by $8$, gives a remainder $1$ (it can't give the remainders $0$ and $4$, beca... | When $p\equiv 1\mod 8$, we now that $P=p^k+p^l+p^m\equiv 1+1+1\equiv 3\not \equiv 1\mod 8$. When $p\equiv 5\mod 8$, we get $P\equiv 1+1+1\equiv 3\not\equiv 1\mod4$, so this won't work either. When $p\equiv 3\mod 8$, we get $3+3+3\equiv 1\mod 8$ (and the other possibilities won't work), so the only case we still have to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I prove $F_{n+1}^2 - F_nF_{n+2} = (-1)^n$ using induction? $F_n$ refers to the $n$ term of the Fibonacci Sequence.
I think I'm supposed to prove this by induction. I already have the base case. I am at:
$\text{F}_\text{k+1}^2 - F_k\text{F}_\text{k+2} + \text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k... | Hint
\begin{align}
F_{n+1}^2-F_nF_{n+2} &= F_{n+1}^2-(F_{n+2}-F_{n+1})F_{n+2}=F_{n+1}^2-F_{n+2}^2+F_{n+1}F_{n+2} \\ &=F_{n+1}(F_{n+1}+F_{n+2})-F_{n+2}^2=F_{n+1}F_{n+3}-F_{n+2}^2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Largest value of the function $f(x) = \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$ Largest value of the function $f(x) = \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}\;\;,$ where $x\in \mathbb{R}$
My Try:: Let $y=f(x)=\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$
$\displaystyle \Rightarrow y^2=(x^4-3x^2-6x+13)+(x^4-x^2+1)-2\sqrt{(... | Hint
$$f(x)=\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let
$$A(x,x^2),B(3,2),C(0,1)$$
so
$$f(x)=|AB|-|AC|\le|BC|=\sqrt{10}$$
then we have
$$AB:y-1=\dfrac{1}{3}(x-0)$$
so
$$\begin{cases}
y=x^2\\
y=\dfrac{1}{3}x+1
\end{cases}$$
$$\Longrightarrow x=\dfrac{1-\sqrt{37}}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $m^{2} + n^{2} = \csc^{2}\left(\theta\right)$ $$
\mbox{If}\quad\left\lbrace%
\begin{array}{rcl}
m^{2} + m'^{2} + 2mm'\cos\left(\theta\right) & = & 1
\\
n^{2} + n'^{2} + 2nn'\cos\left(\theta\right) & = & 1
\\
mn + m'n' + (mn' + m'n)\cos\left(\theta\right) & = & 0
\end{array}\right\rbrace
\quad\mbox{then pr... | Using the same set of notations used by user84413,
$\displaystyle ax+by+(ay+bx)\cos\theta=0\ \ \ \ (1)$
$\displaystyle a^2+b^2+2ab\cos\theta=1\iff (b+a\cos\theta)^2=1-a^2\sin^2\theta\ \ \ \ \ (2)$
Similarly, $\displaystyle x^2+y^2+2xy\cos\theta=1\iff(y+x\cos\theta)^2=1-x^2\sin^2\theta\ \ \ \ \ (3)$
Multiplying $(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
| Remark: Perhaps there are alternative proofs of $(ab)^{3/2} + (bc)^{3/2} + (ca)^{3/2} \ge ab + bc + ca$ without using Chebyshev sum inequality.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$M = \sum_{\mathrm{cyc}} \frac{a(b + c + a^2)}{(b^2 + c^2 + a)(b + c + a^2)}
\le \sum_{\mathrm{cyc}} \frac{a(b + c + a^2)}{(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Complex roots of the polynomial $bz^{2}+2az+b$ are on the unit circle I want to show that the roots of the polynomial $bz^{2}+2az+b$ ($a,b$ are real) when $\left|\frac{a}{b}\right|\leq1$ (which is equivalent to the discriminant not being positive) are on the unit circle. The roots are given by $$z_{\pm}=\frac{-a\pm\sqr... | Let $z = \rho (\cos \theta + i \sin \theta)$ be a root of polynomial $z^2+cz+1$, where $c\triangleq 2a/b$. Then:
$$\rho^2\sin(2\theta)+c\rho\sin\theta = 0$$
and
$$\rho^2\cos(2\theta)+c\rho\cos\theta + 1 = 0.$$
Multiplying the first relation by $\cos\theta$ and the second by $\sin \theta$, and then subtracting, we get:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
integration by trig substitution I solved the following integral by using trig substitution $(u = 3 sec \theta)$
$$
\int \frac{dx}{x^{2}+10x+16}=\int \frac{du}{u^2-9}
$$
and got the same result as in the textbook which looked for $\int \frac{du}{u^{2}-a^2}$. Below is the process and result given in the textbook.
$$
\i... | This may not be what you want, but do you notice the following?
$$\begin{align}\int\frac{1}{x^2+10x+16}&=\int\frac{1}{(x+5)^2-9}dx\\&=\int\frac{1}{u^2-9}du\\&=\frac 16\int\left(\frac{1}{u-3}-\frac{1}{u+3}\right)du\\&=\frac 16(\ln|u-3|-\ln|u+3|)+C\\&=\frac 16\ln\left|\frac{u-3}{u+3}\right|+C\end{align}$$
Here, I set $x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor's series when x goes to infinity Let $f(x) = \frac {x^3}{(x+1)^2}$. Find constants a, b, c, so that $f(x) = ax + b + \frac cx + o(\frac 1x)$ as $x$ goes to $\pm \infty$. So i know that i can't take Taylor series as $x$ goes to infinity. So i am assuming i have to make some kind of substitution. I tried making $... | Here is how you approach the problem. Put $x=\frac{1}{t}$ and then find the Laurent series at the point $t=0$
$$ f(x) = \frac {(1/t)^3}{(1/t+1)^2}=\frac{1}{t(1+t)^2} ={t}^{-1}-2+3\,t-4\,{t}^{2}+5\,{t}^{3}+ ( {t}^{4}) .$$
Now, substitute $t=1/x$ in the above expansion. See related problem.
Note: Note that, you only nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of $\left(\dfrac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n$ I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\m... | While $\sqrt[n]a\to 1$, it is not correct to say $(n\sqrt[n]a-n)\to 0$. Actually, $\sqrt[n]{1+\epsilon}\approx 1+\frac1n\epsilon$ so $n\sqrt[n]a-n\approx a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Identify $\sum(-1)^{n-k}2^{2k}\binom{n+k}{2k}$ Does anybody know what the following sum evaluates to?
$$
\sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k}
$$
| Generating functions to the rescue again! Defining
$$
c_n = \sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k},
$$
we have
\begin{align*}
\sum_{n=0}^\infty c_n x^n &= \sum_{n=0}^\infty x^n \sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k} \\
&= \sum_{k=0}^\infty {(-1)^{k}}2^{2k} \sum_{n=k}^\infty (-x)^n \binom{n+k}{2k} \\
&= \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\sin^5 x$ I know $\sin x$ integrates to $-\cos x$ but ive never seen $\sin^5(x)$ integrated.
would I need to expand it to $\sin x\sin x\sin x\sin x\sin x$ and then how would I complete the integration from here?
| Here's another answer:
We know from using integration by parts that
$$
\int\sin^5x dx = -\frac{1}{5}\sin^4x\cos x + \frac{4}{5}\int\sin^3x dx
$$
Applying intgration by parts to $\int\sin^3xdx$ means
$$
\int\sin^5x dx = - \frac{1}{5}\sin^4x\cos x + \frac{4}{5}(-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x)
$$
$$
= - \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Does this equation have positive integer solutions? The only solution I can find for $a^2 + b^2 + c^2 = d^2 + e^2 + f^2$ is $a=0$, $b=0$, $c=0$, $d=0$, $e=0$, $f=0$. Are there any positive integer solutions? Any where none of $a,b,c,d,e,f$ are equal?
| Solving $a^2 + b^2 + c^2 = d^2 + e^2 + f^2$ has many many (boring) positive integer solutions.
Choose any bijection $\theta: \{a,b,c\} \to \{d,e,f\}$. Then, choose any positive integer values for $\{a,b,c\}$.
It must be true that $a^2 + b^2 +c^2 = \theta(a)^2 + \theta(b)^2 + \theta(c)^2 = d^2 + e^2 + f^2.$ As stated ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving convergence of a sequence Let the following recursively defined sequence:
$a_{n+1}=\frac{1}{2} a_n +2,$
$a_1=\dfrac{1}{2}$.
Prove that $a_n$ converges to 4 by subtracting 4 from both sides.
When I do that, I get:
$2(\frac{1}{2} a_{n+1} -2)=(\frac{1}{2} a_n -2)$, so $y=2y$,
which is true only for $0$. But I'm n... | Just a generalization of the approach for you. Note that
$$
a_n=\frac{1}{2}a_{n-1}+2
$$
So if you subtract this expression from the $a_{n+1}$ you have above, you get rid of the constant. Also denote $\Delta a_{n+1}=a_{n+1}-a_{n}$ and you get:
$$
\Delta a_{n+1}=\frac{1}{2} \Delta a_{n}=\frac{1}{2^2} \Delta a_{n-1}=\ldo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/644049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
If $x,y\in\mathbb Z$, solve $x^2+xy+y^2=x^2y^2$. If $x,y\in\mathbb Z$, solve $$x^2+xy+y^2=x^2y^2$$
We could try some factorizations. $x^2+y^2=xy(xy-1)$. We may as well add $2xy$ to both sides: $(x+y)^2=xy(xy+1)$. Then we could subtract $x^2y^2$ from both sides: $(x+y+xy)(x+y-xy)=xy$
Or we could somehow use the fact tha... | $$ x^2+xy+y^2=x^2y^2 \quad \Longleftrightarrow \quad (1-y^2)x^2+yx+y^2=0$$
$$ \delta=y^2-4(1-y^2)y^2=y^2(4y^2-3)$$
$$4y^2-3=p^2 \quad \implies \quad y^2=1\cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to factorize $2x^2+5x+3$? I'm doing pre-calculus course at coursera.org and I'm in trouble with this solution $$2x^2 +5x +3 = (2x+3)(x+1)$$
By trial, using ac-method I got stuck:
$$
ac = (2)(3) = 6\\
6 + ? = 5 \Rightarrow~ ? = 5 - 6 = -1
$$
Then,
$$2x^2+6x-x+3 = 2x(x+3)-x+3$$
At this point I could not get the answe... | Complete squares:
$$2x^2+5x+3 = 2(x^2+2\alpha x + \alpha ^2) + (3- 2 \alpha ^2)$$
where $\alpha = \frac{5}{4}$. Then
$$(x+\frac{5}{4})^ 2 = \frac{1}{2}\left(\frac{25}{8}-3\right) $$
$$x+\frac{5}{4} = \pm \sqrt{\frac{1}{16}} $$
$$x= \frac{-5}{4} \pm \frac{1}{4} = \frac{-5 \pm 1}{4}$$
$$ x\in \left\{-1, \frac{-3}{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/644975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ Show that the triangle which satisfy the inequality $\dfrac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ is right angled.
My work:
$\sin^2 A+\sin^2 B+\sin^2 C=2(\cos^2 A+\cos^2 B+\cos^2 ... | From where you have left off,
$$\sin^2A+\sin^2B+\sin^2C=2\implies1-\cos^2A+1-(\cos^2B-\sin^2C)=2$$
$$\implies\cos^2A+\cos^2B-\sin^2C=0$$
Using $\displaystyle\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$
$$\implies\cos^2A+\cos^2B-\sin^2C=\cos^2A+\cos(B+C)\cos(B-C)$$
As $\displaystyle A+B+C=\pi, B+C=\pi-A,\cos(B+C)=\cos(\pi-A)=-\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove
$$f(n)\le f(n+1)$$
this inequality is found when I deal this follow limit:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\... | Take $g(x):=\frac{1}{n^2+x^2}$ and apply the Euler-MacLaurin's summation formula.
,
Then we obtain
$$
f(n)=\frac{\pi}{4}-\frac{1}{4n}+h(n),\qquad |h(n)|\leq\frac{3^{3/2}-2}{32n^2}.
$$
From this
$$
f(n+1)-f(n)=\frac{1}{4n(n+1)}+H(n),\qquad |H(n)|\leq\frac{3^{3/2}-2}{16n^2}.
$$
It yields
$$
f(n+1)-f(n)\geq\frac{(6-3^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 4
} |
On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$ How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$
I guess it is only $x=2$.
Please help.
| A proof by induction.
Let:
$$f_n(x) = \sqrt[3]{6+\sqrt[3]{6+\ldots+\sqrt[3]{6+x}}},\ g_n(x) = \sqrt{2+\sqrt{2+\ldots+\sqrt{2+x}}}$$
With $n$ terms. Then for $n=1$ you can easily solve the cubic equation to show that $f_1=g_1 $ only at $x=2$ (over the reals).
Now assume our claim is true for $n$, i.e. that $f_n(x)=g_n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Taylor Series Expansion of $\frac{1}{1+x^2}$ about $x=a$ Let $$f(x)=\frac{1}{1+x^2}$$ Consider its Taylor series expansion about a point $a\in \mathbb{R}$. What is the radius of convergence of this series??
About $x=0$ we could expand it like
$$(1+x^2)^{-1}= 1-x^2+(-1)(-2) \frac{x^2}{2!}+(-1)(-2)(-3)\frac{x^3}{3!}+\do... | With $u=x-a$:
$$
\begin{align}
\frac1{1+(u+a)^2}
&=\frac1{1+a^2+2au+u^2}\\
&=\frac1{1+a^2}\frac1{1+\frac{2au+u^2}{1+a^2}}\\
&=\frac1{1+a^2}\left(1-\frac{2au+u^2}{1+a^2}+\left(\frac{2au+u^2}{1+a^2}\right)^2-\left(\frac{2au+u^2}{1+a^2}\right)^3+\dots\right)\\[4pt]
&=\frac1{1+a^2}-\frac{2a}{(1+a^2)^2}u+\frac{3a^2-1}{(1+a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show quartic polynomial has no real solutions To show a lower bound for the runtime of an algorithm, I need to show that
$$
3 x^4 - \frac{64}{5} x^3 + \frac{192}{5} x^2 - \frac{192}{5} x+ 12 > 0
$$
for all real numbers $x\in \mathbb{R}$.
I followed the Wikipedia instructions and checked the paper they are based on.... | this equal to:
$15x^4-6x^3+192x^2-192x+60>0$
$15x^4-6x^3+192x^2-192x+60=(3x^2-8.702x+25.0905)(5x^2-6.83x+2.3709)+.5112$
$3x^2-8.702x+25.0905>0,5x^2-6.83x+2.3709>0 \implies \\15x^4-6x^3+192x^2-192x+60>0.5 >0$
$25.0905=\dfrac{14439589717}{57550000},2.3709=\dfrac{272893639}{11510000}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding all positive integers $x,y,z$ that satisfy $3^x - 5^y = z^2$
Find all positive integers $x,y,z$ that satisfy:
$$3^x - 5^y = z^2.$$
I think that $(x,y,z)= (2,1,2)$ will be the only solution. But how to prove that?
| Notice that:
$$\begin{align}3^x - 5^y &\equiv (-1)^{x} - 1 \mod 4\\
z^2 &\equiv (-1)^x - 1 \mod{4}\\
&\equiv \begin{cases} 0 & \text{$x$ even,}\\ 2 & \text{$x$ odd}\end{cases} \end{align}$$
Since $2$ is a quadratic non-residue modulo $4$, $x$ must be even $\implies x = 2a$ for positive integer $a$. Furthermore, $4| z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the volume inside an elyptical cylinder and a sphere I'm trying to find the volume bounded by a sphere and an elyptical cylinder. The sphere is given by $x^2+y^2+z^2=1$ and the elyptical cylinder by $2x^2+y^2-2x=0$.
My first attempt with spherical coordinates didn't go well, now I did some work with cylindric... |
The equation for the ellipse states $x^2+y^2-2x=0$ (...) from which I have $\rho(\theta)= \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.
The equation of the ellipse is
\begin{equation*}
2x^{2}+y^{2}-2x=0
\end{equation*}
from which I've obtained
\begin{equation*}
\rho (\theta)=\frac{2\cos \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Center of Mass, Multivariable Calculus I have a solid with the bounds $z=2x^2+2y^2$ where $z=c$ and this solid has a uniform density of B. I need to find the mass and the center of mass of this solid. I know how to find a normal center of mass, but I do not know how to set up an integral for this problem, but I think i... | The mass of the solid is defined as
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV,$$
that is, the integral of body density at each point over the volume of the body. In this case we have $\rho \equiv B$ which is constant, therefore the mass will be a multiple of the body's volume:
$$M = \iiint\limits_{\mathcal{B}} \rho ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the value of A, B and C in the identities. $6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Find the value of A, B and C.
I started it like this:
$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Solving the right hand side:
$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$
$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx ... | Equating x coefficients you will get A+B=5.You already have AB=6.Solving you will get A=2 or 3.This will be the easier approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/654198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum $\sum_{k=0}^{2013}2^ka_{k}$ let real sequence $a_{0},a_{1},a_{2},\cdots,a_{n}$,such
$$a_{0}=2013,a_{n}=-\dfrac{2013}{n}\sum_{k=0}^{n-1}a_{k},n\ge 1$$
How find this sum
$$\sum_{k=0}^{2013}2^ka_{k}$$
My idea: since
$$-na_{n}=2013(a_{0}+a_{1}+a_{2}+\cdots+a_{n-1})\cdots\cdots(1)$$
so
$$-(n+1)a_{n+1}=2013(a_{0}+a_{1}+... | Method 1 (generating function)
Let $a(z) = \sum\limits_{k=0}^\infty a_k z^k$, notice
$$a(z)\left(\frac{z}{1-z}\right) = \left(\sum_{k=0}^\infty a_k\right)\left( \sum_{\ell=1}^\infty z^\ell\right) = \sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}a_k\right)z^n$$
and
$$\left(z\frac{d}{dz}\right)a(z) = \left(z\frac{d}{dz}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/654374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why is it trivial that $\left(1+\frac{2\ln3}{3}\right)^{-3/2}\leq\frac{2}{3}$? Can someone tell me why $$\left(1+\dfrac{2\ln3}{3}\right)^{-3/2}\leq\dfrac{2}{3}$$ is trivial because for me its not and I will need to do the calculation to see it.
| You have that
\begin{align*}
\left(1+\frac{2\ln3}{3}\right)^{-\frac{3}{2}} &= e^{ -\frac{3}{2}\ln\left(1+\frac{2}{3}\ln3\right) } \\
&=e^{ \frac{3}{2}\ln\frac{1}{1+\frac{2}{3}\ln3} } \\
&\leq e^{ \frac{3}{2}\ln\left( 1-\frac{1}{2}\cdot\frac{2}{3}\ln3 \right) } \\
&\leq e^{ -\frac{3}{2}\frac{1}{3}\ln 3 } = e^{ -\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/654662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$
This is what I did:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^... | HINT:
We have $$1-\sin2x+2\cos2x=0$$
Using Double-Angle Formulas, $$1-\frac{2t}{1+t^2}+2\frac{1-t^2}{1+t^2}=0$$ where $t=\tan x$
Solve the Quadratic Equation in $\tan x$
As $x\in\left[\frac{3\pi}2,2\pi\right], t=\tan x<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Converting rectangular equation to polar equation How do I convert the following rectangular equation to a polar equation:
$$
(x- 3)^2 + (y+2)^2 = 1
$$
I was able to simplify it to the following:
$$
r^2 - 6r\cos(\theta) + 4r\sin(\theta) + 12 = 0
$$
but I am at a loss as to how to simplify further...
I'm trying to get r... | Given that you're trying to solve for $r$, threat the equation $r^2-6r\cos\theta+4r\sin\theta+12=0$ as a quadratic equation, where $r$ is the only variable and everything else is treated as constants.
But before we begin solving for $r$, it would be best to take the terms $-6r\cos\theta$ and $4r\sin\theta$ and factor o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with Polar coordinates and the length of the curve. I have a test coming up today and I was going over our past midterms and this question came up. I tried it but its not working, please any hints or solution in how to do it will be really helpful.
Question:
Consider the polar curve $r = min(1 - cos(\theta), 1 + s... | Both of the polar curves in this problem are cardioids: $ \ r = 1 - \cos \theta \ $ has its symmetry axis along the $ x-$ axis with the cusp opening in the positive $ x-$ direction and $ \ r = 1 + \sin \theta \ $ has its symmetry axis along the $ y-$ axis, its cusp opening in the negative $ y-$ direction.
The points ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/657341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof for maths induction prove: $$ \sum_{r=1}^{n}[r^{2}+1](r!)=n[(n+1)!]$$ for all $n \in N$
prove $n=1$, $(1^2+1)(1!)$ = $1[(1+1)!]$
assume true for $n=k$, $(k^2+1)$$(k!)$= $k$$[(k+1)!]$
I got to here : $[k^{2}+1](k!)+[(k+1)^2+1](k+1)! = (k+1)[(k+2)!]$
| If
$$ \sum_{r=0}^{n}(r^{2}+1)r!=n(n+1)!$$
then
$$ \sum_{r=0}^{n+1}(r^{2}+1)r!=\sum_{r=0}^{n}(r^{2}+1)r!+((n+1)^2+1)(n+1)!=$$
$$=n(n+1)!+((n+1)^2+1)(n+1)!=$$
$$=(n+1)!(n+((n+1)^2+1))=$$
$$=(n+1)!(n+n^2+2n+1+1)=$$
$$=(n+1)!(n^2+3n+2)=$$
$$=(n+1)!(n+2)(n+1)=$$
$$=(n+1)(n+2)!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find $\frac{a^3}{a^6 + 1}$ given $a$ is a root of a quadratic equation
If $a$ is a root of the equation $x^2 - 3x + 1 = 0$, then find the value of $\frac{a^3}{a^6 + 1}$.
So, I figured we can use the quadratic formula, and formed the following equation:
$$a=\frac{-(-3)+\sqrt{9-4}}{2(1)}\implies a=\frac{3+\sqrt5}2$$
Bu... | $a^2=3a-1$ then
$$\frac{a^3}{a^6 + 1}
\\=\frac{a^3}{(a^2+1)(a^4-a^2 + 1)}
\\=\frac{a^3}{(3a)(9a^2-6a+1-3a+1+1)}
\\=\frac{a^2}{(3)(9a^2-9a+3)}
\\=\frac{a^2}{(9)(3a^2-3a+1)}
\\=\frac{a^2}{(9)(9a-3-3a+1)}
\\=\frac{a^2}{(9)(6a-2)}
\\=\frac{3a-1}{(18)(3a-1)}
\\=\frac{1}{18}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/660488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Quadratic residue of $-1$ in composite modulus It is true for each odd prime number p that if $x^2\equiv-1 \pmod p$ then $p\equiv1\pmod 4$
I've observed that it should be true for all composite integers, whose prime factors are congruent to $1$ modulo $4$. However I couldn't find any remark on the internet whether it's... | For an odd $n$, there are $x$ with $x^2 \equiv -1 \pmod{n}$ if and only if all prime divisors of $n$ are of the form $p = 4m+1$.
The necessity follows from $x^2 \equiv -1 \pmod{n} \Rightarrow x^2 \equiv -1 \pmod{d}$ for all divisors $d$ of $n$, in particular its prime divisors.
The sufficiency follows from the Chinese ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/660878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The equation $3^n+4^m=5^k$ in positive integer numbers Please help me to prove that the equation $3^n + 4^m = 5^k$
where $n$, $m$, $k$ are positive integer numbers has only the solution $n=m=k=2$.
I know how to prove it for $n=m=k$.
If $3^x + 4^x = 5^x$ then $(3/4)^x + 1 = (5/4)^x$,
and this equation has at most one so... | Considering the equation modulo 3 and 4 separately, we see that $2 | n \implies n = 2a$ and $2 | k \implies k = 2c$ for some positive integers $a, c$. Rewriting, we get
$$\begin{align}
3^{2a} + 4^m &= 5^{2c}\\
2^{2m} &= (5^c)^2 - (3^a)^2\\
2^{2m} &= (5^c + 3^a)(5^c - 3^a)\end{align}$$
We thus allow $5^c + 3^a = 2^y$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/662121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How prove $\measuredangle CDE=2\measuredangle ABE$ In rectangular $ABCD$,and $E\in AC$,such
$$BE=\sqrt{2}\cdot AE$$
show that
$$\measuredangle CDE=2\measuredangle ABE$$
My try: let $$AB=a,AD=b,\dfrac{AE}{AC}=k,$$
then
$$AE=k\sqrt{a^2+b^2},BE=k\sqrt{2(a^2+b^2)}$$
I know have this nice relsut
$$AE^2+EC^2=BE^2+ED^2$$
th... | Probably not the best way to do it, but maybe you can simplify it.
Let $F \in AB$ such that $AB\perp EF$ and let $G \in CD$ such that $CD \perp EG$.
Let $|AE|=x\Rightarrow |BE|=x\sqrt2$
Let $|EG|=y$
Finally let $\measuredangle ABE=\alpha$, $\measuredangle BAC = \measuredangle ACD = \beta$ and $\measuredangle ADC=\gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/663182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example.
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|... | Divide the domain to
(1) $x \le -1$
(2) $-1 \le x \le 3$
(3) $x \ge 3$
solve the equation in each domain
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Linear algebra : eigenvalues of an integral operator on polynomials Consider the linear transformation
$$
T : \left\{
\begin{array}{ccc}
\mathbb{R}_n[X] & \to & \mathbb{R}_n[X] \\
P & \mapsto & \int_0^1 (X + t)^n\,P(t)\,dt
\end{array}\right.
$$
where $\mathbb{R}_n[X]$ denotes the vector space of polynomials with real ... | Using the adhoc convention $\frac{x}{0}=x$, Newton’s binomial formula yields
$$
T(X^j)=\sum_{i=0}^{n} \frac{\binom{n}{i}}{n+1+j-i}X^i \tag{1}
$$
If we denote by ${\cal B}_n=(1,X,X^2, \ldots, X^{n})$ the canonical
basis of ${\mathbb R}_n[X]$, the matrix $A$ of $T$ relatively to ${\cal B}_n$ is
$A=(a_{ij})_{1\leq i,j \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to solve $i^2+i+1\equiv 0\pmod{2^p-1}$ in general? While looking at the Mersenne numbers (for prime $p$, the number $2^p-1$), I noticed that only certain of them had any solution to the modular equation $i^2+i+1\equiv 0\pmod{2^p-1}$, e.g., $p=3,5,7,13,17,19,31,37,...$ but not $p=11,23,29,...$.
I have got... | Just as an alternate view of the particular solutions to this modular equation, if $$k^2+k+1\equiv 0\pmod q$$ then $$k^2\equiv -k-1\pmod q$$ so that solution is a quadratic residue $\mod q.$ Taking note of the comments, we have $$(k^2+k+1)(k-1)=k^3-1\implies k^3\equiv 1\pmod q$$ Since we also have $$k(k+1)\equiv -1\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/666269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s e... |
The following is valid for $n\geq 1$
\begin{align*}
\sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor
=n\lfloor\sqrt{n}\rfloor
-\frac{1}{3}\lfloor\sqrt{n}\rfloor^3-\frac{1}{2}\lfloor\sqrt{n}\rfloor^2+\frac{5}{6}\lfloor\sqrt{n}\rfloor
\end{align*}
For convenient calculations we consider two aspects:
*
*We introduce a variable ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 0
} |
Is the following scheme for generating $p_n=(1/3)^n$ stable or not. $p_n=(5/6)p_{n-1}-(1/6)p_{n-2}$. Is the scheme for generating $p_n=(1/3)^n$ stable or not?
$$p_n= \frac{5}{6} p_{n-1} - \frac{1}{6}p_{n-2}.$$
| From your recursion, we have
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
=\left[\begin{array}{c}5/6&-1/6\\1&0\end{array}\right]
\left[\begin{array}{c}p_{n-1}\\p_{n-2}\end{array}\right]
$$
The eigenvalues of the matrix are
$$
\frac12\quad\text{for }\left[\begin{array}{c}1/2\\1\end{array}\right]
$$
and
$$
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrals involving reciprocal square root of a quartic For two integration below, what is the ratio of them?
$$I_1 = \int_0^1\frac{dt}{\sqrt{1-t^4}}$$ $$I_2 = \int_0^1 \frac{dt}{\sqrt{1+t^4}}$$
What is the ration $\frac{I_1}{I_2}$?
I do not have any thoughts for solving this question so could anyone give me some hin... | Hint:
$$I_1(x)=\int_x^1\frac{dt}{\sqrt{1-t^4}}=\frac{1}{\sqrt{2}}\int_{0}^{\arccos x}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$
$$I_2(x)=\int_0^x \frac{dt}{\sqrt{1+t^4}}=\frac{1}{2}\int_{0}^{\arccos\frac{1-x^2}{1+x^2}}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$
and since $$\arccos 0 = \arccos \frac{1-1^2}{1+1^2}$$
$$\frac{I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
If a and b are odd integers, then $8\mid (a^2-b^2)$ Prove: If $a$ and $b$ are odd integers, then $8\mid (a^2-b^2)$
So far I have: $$8\mid \left((2n+1)^2-(2m+1)^2\right)\Longleftrightarrow 8\mid \left(4n^2+4n+1-4m^2-4m-1\right) $$
Is this right so far? Where do I go from here? Step by step explanation please!
| $$4n^2+4n+1-4m^2-4m-1=4n^2+4n-4m^2-4m=4(n^2-m^2+n-m) \\
4[(n-m)(n+m)+(n-m)]=4(n-m)(n+m+1)$$
Now
$$(n-m)+(n+m+1)=2n+1$$
is odd, thus one of $(n-m)$ or $(n+m+1)$ must be even.
P.S. A simpler solution is the following
$$a^2-1=(a-1)(a+1)$$
is the product of two consecutive even numbers. Thus one is divisible by $4$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a... | Note that $a,b,c$ are the roots of the equation $$x^3-(a+b+c)x^2+(bc+ac+ab)x-abc=0$$ which we know to be $$x^3-7x^2+13x-\frac {621}{31}=0 \dots (A)$$
Add the three equations for $a, b, c$ to obtain
$$(a^3+b^3+c^3)-7(a^2+b^2+c^2)+13(a+b+c)-\frac {3\cdot 621}{31}=0$$
Note that if we define $P_n=a^n+b^n+c^n$ we can multip... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find all complex values for $a$ where there's no solution for the non homogeneous system
Find all complex values for $a$ where there's no solution for the non homogeneous system
$$\left\{ {\begin{array}{*{20}{c}}
{x + (1 - a)y + z = 1} \\
{2x - y + z = 3} \\
{3x - ay + (a - 1)z = 4} \\
\end{array}} \right... | If you rref you get this
$ [1 \ 0 \ 0 \ \frac{3a - 4}{2a - 3} \\ \ 0 \ 1 \ 0 \ \frac{1}{2a - 3} \\ \ 0 \ 0 \ 1 \ \ \ \ \ \ 0 \ \ ] $
So $$ 2a - 3 \ne 0 $$
$$ a \ne \frac{3}{2}$$
Now you still need to check a = 1 because if your calculations are correct , (I didn't check them) to get a 1 in the last row you have to di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying long fractions How would I go about simplifying long fractions, such as the likes of this:
$((8+\frac{3}{4}) + (3\frac{2}{3}))$ / $((4+\frac{2}{5}) - (1\frac{7}{8}))$
The correct answer is ($4 + \frac{278}{303}$)
I'm not really sure how to approach this problem,
regards.
| Combine the numerator into an improper fraction:
$$\begin{align}8 + \frac{3}{4} + 3\frac{2}{3} &= 8 + 3 + \frac{3}{4} + \frac{2}{3} \\
&= \frac{132}{12} + \frac{9}{12} + \frac{8}{12} \\
&= \frac{149}{12}\end{align}$$
Do the same for the denominator:
$$\begin{align}4 + \frac{2}{5} - 1\frac{7}{8} &= 4 - 1 + \frac{2}{5} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/678286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to use $x + \frac{1}{x} = 7$ to compute $x^2 + \frac{1}{x^2}$. I am not sure how to approach this question:
You know that $x + \frac{1}{x} = 7$. Compute $x^2 + \frac{1}{x^2}$.
I have tried adding $x + \frac{1}{x}$ to get $\frac{x^2 +1}{x}$ but can't see if this was useful or not.
I need help in getting started.
| Thank you to Yiyuan Lee and awllower who really answered the question.
I am only adding this for completeness, and to remind myself of the solution.
We are given $x + \frac{1}{x} =7$ and want to use this to find $x^2 + \frac{1}{x^2}$.
Using the fact $$\left(x +\frac{1}{x}\right)^2 = x^2 +\frac{1}{x^2} +2$$
and rewrit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
How to compute the value of $1 \cdot \frac{2}3 \cdot \frac 45 \cdots$, as $n$ goes to infinity? It's supposed to be $\dfrac 12\sqrt{\dfrac \pi n}$.
| From the Wallis's product in the following form ( Advanced Calculus by Angus Taylor, formula (20.7-9))
\begin{equation*}
\frac{\pi }{2}=\lim_{n\rightarrow \infty }\left( \frac{2\cdot 4\cdots (2n)}{
1\cdot 3\cdots (2n-1)}\right) ^{2}\frac{1}{2n+1}\tag{1}
\end{equation*}
follows that the limit in the title is zero.
ADD... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give ... | The answers actually
$$
(x^2 + 4)(x^3 + 1)\\
x^5 + 4x^3 + x^2 + 4\\
(x^5 + 4x^3)(x^2 + 4)\\
x^3(x^2 + 4) 1(x^2 + 4)\\
(x^2 + 4)(x^3 + 1)
$$
(Sorry, too lazy to actually change them to exponents)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a trigonometric limit What is the limit as $x$ approaches $0$ of: $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}?$$
We cannot use L'Hôpital's rule or anything advanced like Taylor series. I reduced it to, by considering the numerator's conjugate:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{1+x... | $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}\sim_0\frac{\sqrt{1+x+\frac{x^3}3} - \sqrt{1+x-\frac{x^3}6}}{x+\frac{x^2}2}\sim_0\frac{\frac{1}{2}\left(\frac{x^3}3+\frac{x^3}6\right)}{x}\sim_0\frac14 x^2$$
so the desired limit is $0$.
Edit Without Taylor series the calculus is tedious: let
$$f(x)=\sqrt{1+\tan x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Dimension and its consequences Find a basis for $S=[\mathrm{(1,2,3),(3,4,7),(5,-2,3)}]\subseteq \Bbb R^3$ and give the dimension.
Then, putting all the vectors as the columns of a new matrix:
$A=\begin{bmatrix}
1 & 3 & 5 \\
2 & 4 & -2 \\
3 & 7 & 3
\end{bmatrix}$
By row reducing:
$A_R=\begin{bmatrix}
1 & 0 & -1... | By putting the vectors as the columns of the matrix, the leading entries in the row echelon form indicate columns of the original matrix that form a basis of its column space. In this case, we have:
$$
\overbrace{\begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & -2 \\ 3 & 7 & 3 \\ \end{bmatrix}}^A
\xrightarrow{\text{row operations... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove continuity for cubic root using epsilon-delta I am trying to prove that a function is continuous at a point a using the $\epsilon$-$\delta$ theorem. I managed to find a $\delta$ in this case $|2x^2+1 - (2a^2+1)| < \epsilon$. But I have a hard time when the function under consideration is $f(x) = \sqrt[3]{x}$. Tha... | $$|\sqrt[3]{x}-\sqrt[3]{a}| = |\sqrt[3]{x}-\sqrt[3]{a}| \times \frac {| x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} = \frac {|x - a|}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} \le \frac {| x - a |}{| {ax} |^{\frac 1 3}}$$
The fina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
N-th power of matrix Find the formula for the n-th power of this matrix.
$$
\pmatrix{1&1\\1&0}
$$
Well $f^2 = \pmatrix{2&1\\1&1}$ and $f^3 = \pmatrix{3&2\\2&1}$ and $f^4 = \pmatrix{5&3\\3&2}$
and $f^5 = \pmatrix{8&5\\5&3}$...
I can't seem to find the pattern here. Would someone mind helping me
| Hint: For $F = \pmatrix{1&1\\1&0}$, show (perhaps inductively) that
$$
F^n = F^{n-1} + F^{n-2}
$$
Inductive step:
$$
F^{n+1} = F \cdot F^n = F(F^{n-1} + F^{n-2}) = F^n + F^{n-1}
$$
For the base case, note that $F^0 = I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many "Two pair" poker hands are there in a standard deck? From the solutions to our midterm, I know the answer of this question to be:
$$\binom{13}{2} \times \binom{4}{2} \times \binom{4}{2} \times \binom{11}{1} \times \binom{4}{1}$$
However, I initially thought it would be:
$$\binom{13}{1} \times \binom{4}{2} \tim... | The difference between yours and the solution is $\binom{13}{2}$ versus $\binom{13}{1} \times \binom{12}{1}$. If you write them out you will see that $2 \binom{13}{2} = \binom{13}{1} \times \binom{12}{1}$, the 2 double counting as you have noticed (that KKQQ5 and QQKK5 are essentially the same).
(Note that $\binom{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^... | Your answer is correct, but there is a somewhat simpler method. You are looking for a polynomial $P=a+bx+cx^2+dx^3+ex^4$ such that $P\times(1+x-\frac{x^3}6)\equiv 1\pmod{x^5}$. This gives (by comparing coefficients of $x^0,x^1,\ldots,x^4$) the equations
$$
\begin{align}a&=1\\a+b&=0\\b+c&=0\\-\tfrac16a+c+d&=0\\-\tfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/692406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solving $6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$, $a,b,p,q \neq 0$ Is there an efficient way to find solutions to the equation:
$6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$ and $a,b,p,q \neq 0$
If the equation has no solutions, how could you prove that, that is the case?
Edit: The equation can... | Partial answer:
You can factor:
$$\frac{1}{6} (6a + 1)(6b + 1) = 36(p + \frac{13}{36})(q + \frac{19}{36}) + \frac{11}{36}$$
$$6(6a+1)(6b+1) = (36p + 13)(36q + 19) + 11$$
Note that $RHS \equiv LHS \equiv 6 \pmod {36}$. Some trial and error with $p, q$ yields $(5, 7, 1, 4), (2, 32, 1, 8), (3, 27, 1, 10), (9, 11, 1, 12)$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/694494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove, that two equations are equivalent EDIT: Missed something very important! Sorry! We have $x^4+1=2(2x-1)^{1/4}$ not $x^4+1=2\sqrt{2x-1}$. One friend of mine told me that the equation $x^4+1=2(2x-1)^{1/4}$, where $x\geq \frac{1}{2}$
is equivalent to $$x^4+1=2x$$. How did he obtain this? The equation $x^4+1=2(2x-1)^... | What you can say is that $$x^4 +1 = 2\sqrt{2x - 1}\tag{1}$$ $$x^4 + 1 = 2x\tag{2}$$
share one solution: when $x = 1$, both equations are satisfied.
But each equation has a second solution not shared by the other.
*
*Real solutions to $(1)$ are $x = 1,\;x\approx 0.68682$.
*Real solutions to $(2)$ are $\;x = 1,\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$? I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if ther... | $$\sum {n\choose k}x^n$$ is easy to sum, using the binomial theorem. Then if you can express $n^k$ in terms of ${n\choose0},{n\choose1},\dots,{n\choose k}$, you can get a formula for $\sum n^kx^n$. Expressing powers of $n$ in terms of those binomial coefficients can be done using Stirling numbers, which I invite you to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Different solutions of $x+y+z=10$ where $x$, $y$, $z$ are all positive integers and $x, y, z \leq 10$ The number of solutions to the equation $x+y+z=10$ where $x,y,z$ are positive integers, is given by ${k−1 \choose n−1}$, where in this case $k=10,n=3$, giving us ${9 \choose 2} = 36$
Now we have
$x + y + z = 10$ wit... | One way is to consider this as a multiset of 3 types of elements, 7 in total (as there is at least 1 of each type, substract that). So the number of solutions is:
$$
\left(\!\!\binom{3}{7} \!\! \right) = \binom{3 + 7 - 1}{7} = \binom{9}{7} = 36
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/700216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Is this fraction even possible to put into partial fractions? I'm to integrate $\int\frac{x}{x^2+6x+13}dx$
But I'm finding it impossible to do anything with it.
Since $x^2+6x+13$ is an irreducible quadratic factor, and the only factor it means that the partial fraction should be in the form...
$\frac{Ax+B}{x^2+6x+13}$ ... | Hints:
$$\frac x{x^2+6x+13}=\frac12\frac{2x+6}{x^2+6x+13}-\frac3{(x+3)^2+4}=$$
$$\frac12\frac{(x^2+6x+13)'}{x^2+6x+13}-\frac32\frac{\left(\frac{x+3}2\right)'}{1+\left(\frac{x+3}2\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
a quadratic equation for two unknown number find values of $p$ such that the equation $4x^2 + 3px - 2p = 0$ has? below are a few choices of the value p:
a) 2 real roots
b) 1 real roots
c) no roots or complex roots
so far i did for a) 2 real roots
$$(3p)^2 - 4(4)(-2p) > 0\\
9p^2 +32p > 0\\
9\left( p^2+\frac{32p}{9} \ri... | $D=(3p)^2+4\cdot4\cdot2p=9p^2+32p=p(9p+32)$
if $D>0$ then there is 2(different) real roots
$D=9p^2+32p=p(9p+32)>0$
$p\in(-\infty;-32/9)\cup(0;+\infty)$
if $D=0$ then there is 1(or 2 equal) real root
$p=0$ and $p=-32/9$
if $D<0$ then there is no roots or 2 complex roots
$p\in(-32/9;0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/706694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\,x>1$, then $\lim\limits_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x$. How can I prove that
$$
\lim_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x,
$$
whenever $x>1$. Here $\left\lfloor \cdot\right\rfloor$ denotes ... | Since $y-1< \lfloor y\rfloor\le y$, for every $y\in\mathbb R$, then
$$
\frac{x^{n+1}-1}{x^{n}}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}< \frac{x^{n+1}}{x^n-1},
$$
and hence
$$
x-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}<x+\frac{x}{x^n-1},
$$
or
$$
-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/707294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How does $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$? Good afternoon my wonderful friends!
Whenever I do this equation I set it up using the difference of two cubes, which is as follows:
$(a+b)^3 = (a+b)(a^2-ab+b^2) = a^3 + b^3$
Whenever I try to use this formula I always get:
$(x+h) x^2 - xh + h^2$
Simplify:
$x^3 - x^2h... | $$(a+b)^3 \neq a^3 + b^3$$
for example for $a=1$ and $b=1$ you get $(a+b)^3=(1+1)^3=2^3=8 \neq 2 = 1^3 + 1^3 = a^3 + b^3$
To evaluate $(a+b)^n$ one can use the property of the Pascal's triangle
in your case:
$$(a+b)^3 = (a+b)*(a+b)*(a+b)=(a^2 + ab + ab + b^2) * (a+b)=$$
$$(a^2 + 2ab + b^2)* (a+b) = a^3 + a^2b + 2a^2b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to algebraically prove $\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$? Need help trying to prove this problem algebraically.
$$\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$$
The farthest I've got is simplifying the RHS to $$nm + \frac{n(n-1)}{2!} + \frac{m(m-1)}{2!}$$
but not sure what to do after that.
| $${n+m\choose2}=mn+{n\choose2}+{m\choose2}$$
Without loss of generality we will prove this identity by induction on $m$.
Notice that for $m=2$
\begin{align}{n+2\choose2}&=\frac{(n+2)!}{n!2!}=\frac{(n+2)(n+1)n!}{n(n-1)(n-2)!2!}=\frac{(n+2)(n+1)}{n(n-1)}\cdot{n\choose2}\\&=\frac{n(n-1)+4n+2}{n(n-1)}\cdot{n\choose2}={n\ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/711311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
} |
Rationalizing mixed denominators? How would I rationalize the following Fraction?
$$ \frac {2}{5-\sqrt2+\sqrt3}$$
I have considered the idea of multiplying by the same radicals, but the 5 prevents that.
| $$\begin{align}\frac{2}{5-\sqrt2+\sqrt3}\left(\frac{5+\sqrt2-\sqrt3}{5+\sqrt2-\sqrt3}\right)&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\\
&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\left(\frac{20-2\sqrt6}{20-2\sqrt6}\right)\\
&=\frac{200-20\sqrt6+40\sqrt2-4\sqrt{12}-40\sqrt3+4\sqrt{18}}{400-24}\\
&=\frac{200-20\sqrt6+52\sqrt2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/712854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Derivatives of Logarithmic function Determine $f'(x)$ for $f(x) = ln(x + \sqrt{x^2 + 1})$
My handbook has the answer as $\displaystyle\frac{1}{\sqrt{x^2 + 1}}$ with no steps on how they got there.
I tried to get there, but somewhere I am getting things wrong. This is what I have
\begin{align}
f'(x) & = \displaystyle\fr... | Why, in your second line, did you divide the derivative by $$x + \sqrt{x^2 + 1}\;\;?$$
Your first step is correct:
$$f'(x)=\displaystyle\frac{1}{x + \sqrt{x^2 + 1}}\cdot\tfrac{d}{dx}(x + \sqrt{x^2 + 1})$$
But when we take the derivative of $x + \sqrt{x^2 + 1} = x + (x^2 + 1)^{1/2},$ we get
$$f'(x) = \dfrac 1{x+\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$ bounded?
$$f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$$Is this function bounded?
So obviously this converges because $|\frac1k\sin\frac x{2^k}|<|\frac x{2^k}|$ and $\sum\frac x{2^k}$ converges by the integral test.
Now I need to show that there exists a $N$ ... | The function is unbounded. In fact, take $x_m = \frac{2^{3m}}7 2\pi$; then the sequence $f(x_m)$ tends to infinity. To see this, note that
$$
\sin\Big( \frac n72\pi \Big) \approx \begin{cases}
0.781831, &\text{if }n\equiv1\pmod 7, \\
0.974928, &\text{if }n\equiv2\pmod 7, \\
-0.433884, &\text{if }n\equiv4\pmod 7.
\end{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim\limits_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\frac1x}$ Evaluate
$$
\displaystyle\lim_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
$$
And actually I have my answer and just need someone to verify this for me since I haven't done something like this for a long time.
First... | Everything is good until you try to simplify $e^{\frac{1}{2}\ln 3 +\frac{1}{2}\ln 5}$. By the rules of logarithms, $\frac{1}{2}\ln 3 + \frac{1}{2}\ln 5 = \ln \sqrt{3} + \ln \sqrt{5} = \ln (\sqrt{3}\cdot \sqrt{5}) = \ln \sqrt{15}$. So the final answer is $e^{\ln \sqrt{15}} = \sqrt{15}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $n^7+7$ can never be a perfect square. Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n \equiv 5 \pmod{8}$ or $n \equiv 9 \pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
| probably easier than previous solutions:
Suppose that $n^7+7=x^2, x \in \mathbb{N}$ has a solution. Then we must have that $x^2+11^2=x^2+121=n^7+128=n^7+2^7$. Now $n+2|n^7+2^7=x^2+11^2$. If $11|n+2$, then $\nu_{11}(n^7+2^7)=\nu_{11}(n+2)$. This also means that $11|x^2+11^2 \implies 11|x$. Let $x=11y$, then we get $\nu_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$ Given six numbers $x,y,z,a,b,c$ which satisfy the following relations
$y^2+yz+z^2=a^2$
$z^2+zx+x^2=b^2$
$x^2+xy+y^2=c^2$
Express $x+y+z$ in terms of $a,b,c$
My attempt:
$\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-... | Let $A=a^2,B=b^2,C=c^2$. We have
$$
\begin{array}{lcl}
B-A &=& (x-y)(x+y+z) \\
C-A &=& (x-z)(x+y+z) \\
C-B &=& (y-z)(x+y+z) \\
\end{array}\tag{1}
$$
Then, if we put $D=(B-A)^2+(C-A)^2+(C-B)^2$, we deduce
$$
\begin{array}
D
&=& (x+y+z)^2\Bigg((x-y)^2+(x-z)^2+(y-z)^2\Bigg) \\
&=& 2(x+y+z)^2\Bigg(x^2+y^2+z^2-xy-xz-yz\Big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Bessel function integral How to solve the integral for $J_1{(2x\sin{\frac{\theta}{2}})}$ at $[0,\pi]$? If solving by Matlab, please provide me the source. Thank you!
| A simple analytical result is derived from the series representation of $J_1$:
$$J_1(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (k+1)!} \left ( \frac{z}{2}\right )^{2 k+1} $$
as well as the integral
$$\int_0^{\pi/2} du \, \sin^{2 k+1}{u} = \frac{2^{2 k}}{\displaystyle (2 k+1) \binom{2 k}{k}}$$
Thus
$$\begin{align}\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help with discrete math proof? I am having trouble proving the following:
If $x\in R$ and $x > 0$, then $x^4+1 \geq x^3+x$.
Work: I tried to rearrange the equation as $x^4-x^3-x+1 \geq 1$, but that does not really help. I also tried proof by cases where case 1 would be that x is irrational and case 2 would be that x i... | As $x\in \mathbb{R}$ and $x>0$, then we will analyze the sign of the expression
$$x^4-x^3-x+1. $$
Indeed,
$$x^4-x^3-x+1=x^3(x-1)-1(x-1)=(x-1)(x^3-1) $$
Now, as $x^3-1=(x-1)(x^2+x+1)$, then
$$x^4-x^3-x+1=(x-1)(x-1)(x^2+x+1) =(x-1)^2(x^2+x+1).$$
Note that $(x-1)^2(x^2+x+1)\geq 0$ because $x>0$ (it can be zero if $x=1>0$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Factorising a complex polynomial over $\mathbb{C}$ I'm given $f(z)=z^6-1$ to factorise over $\mathbb{C}$.
My working is as follows up to the point I don't understand:
$f(-1)=0$ and $f(1)=0$
So $(z+1)$ and $(z-1)$ are factors
$(z+1)(z-1)=z^2-1$
$(z^2-1)(z^4+pz^3+qz^2+rz+s)=z^6-1$
$z^6+pz^5+qz^4-z^4+rz^3-pz^3+... | This solution refers to the original post: $z^6 + 1$. We have: $z^3 + i$ and $z^3 - i$ are factors. And $f(z) = z^3 + i$ has $f(i) = i^3 + i = -i + i = 0$. So $z - i$ is a factor, and you can easily find that: $z^3 + i = (z - i)(z^2 + iz - 1)$. Let $g(z) = z^3 - i$. Then $g(-i) = 0$. So $z + i$ is a factor of $g(z)$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$ To prove
$$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$
My approach :
$$
\begin{align}
\text{L.H.S.} & = \sin\frac{A}{2}+\sin \frac{B}{... | I will prove,
$$\sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} - \sin \frac{\pi}{2}= 4\sin \frac{\pi -A}{4}\sin\frac{\pi -B}{4} \sin\frac{\pi-C}{4}$$
Let $z_1=e^{iA/2}$, $z_2=e^{iB/2}$, $z_3 = e^{iC/2}$, $z_4=e^{i\pi/2}=i$
If $z_0=e^{i\theta}$, $\sin(\theta)= \frac1{2i}\left(z_0 - \frac1{z_0}\right)$
Therefore, LHS, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/727852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Two people A and B throwing dice A begins by throwing a dice until he gets $6$, then B does the same thing.
What is the probability that A throws more times than B?
I try to solve it, but I got 2 different answers:
1. We mark: $X=$How many times A throws, $Y$=How many times B throws.
$$P(X>n|Y=n)=\frac{P(X>n)\cdot P(Y=... | To find the probability that A throws more times than B, we need to sum the probabilities that A throws two or more times given that B throws once, A throws three or more times given that B throws twice and so on.
So we need to evaluate
$\sum_{k=1}^{\infty}P(A_{throws} > k,B_{throws}=k)$
Now for B throwing once, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Harder Trigonometry Identity ($\sec A+\csc A$) How do I prove:
$\sin A (1 + \tan A) + \cos A (1 + \cot A) = \sec A + \csc A$
I've tried expanding the brackets by multiplying sin A and cos A to the left hand side but to no avail. Where should I start from?
| Another possibility, just continuing to get common denominators:
$$\sin A \left(1 + {\sin A\over \cos A}\right) + \cos A\left(1 + {\cos A\over \sin A}\right) \\ = \sin A \left({\cos A + \sin A\over \cos A}\right) + \cos A \left({\sin A + \cos A\over \sin A}\right) \\ = {\sin A(\cos A + \sin A)\over \cos A} + {\cos A (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Prove that $a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2$ In thinking about a base case in this problem, I came up with the following question.
Given real numbers $a \ge b \ge c \ge d \ge 0$, prove that the following holds:
$a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2 \tag{A}$
My attempt:
After simplification, this redu... | Here's a method without expanding everything:
$$\begin{align}(a^2-b^2)+(c^2-d^2)&\ge((a-b)+(c-d))^2\\
(a-b)(a+b)+(c-d)(c+d)&\ge(a-b)^2+(c-d)^2+2(a-b)(c-d)\\
(a-b)(a+b-(a-b))+(c-d)(c+d-(c-d)&\ge2(a-b)(c-d)\\
(a-b)b+(c-d)d&\ge(a-b)(c-d)\\
(a-b)(b-c+d)+(c-d)d&\ge0\end{align}$$
And that's obvious. If you want to see when e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Trignometry: Find the value in terms of $ \ l \ $ If $ \ \tan x + \sec x \ = \ l \ , $
find the value of secant x in terms of $ \ l \ $ .
I do not know how to solve it but i do know that the basic trig identities are involved.
Pls help me!!!
| Simply use the definitions:
$$\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{1 + \sin x}{\cos x}.$$
So we seek $x$ such that
$$1 + \sin x = L \cos x \quad \text{or} \quad L \cos x - \sin x = 1.$$
Let $t = \tan(x/2)$. Then
$$\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = 2\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/736035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Limits - solution needed: $\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$ how to solve this problem? (without using l'Hopital rule)
$$\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$$
thanks for helping.
| It is much simpler to put $x = \dfrac{\pi}{2} - t$ so that we have $\sin x = \cos t, \cos x = \sin t$ and $\sin 2x = \sin (\pi - 2t) = \sin 2t$. Also as $x \to \pi/2$ we have $t \to 0$. Clearly we then see that $$\begin{aligned}L &= \lim_{x \to \pi/2}\frac{1 - \sin x + \cos x}{\sin 2x - \cos x}\\
&= \lim_{t \to 0}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).$$
I ended up doing it in Mathematica, is there any other way?
Thanks in advance!
| Recall that
$$\sin(x) = x-x^3/6+x^5/120-x^7/5040+O(x^9)$$
$$\tan(x)=x+x^3/3+2x^5/15+17x^7/315+O(x^9)$$
thus
$$\begin{align}
\tan(\sin(x)) &= x-x^3/6+x^5/120+17x^7/315+O(x^9)\\
&\;+\;(x-x^3/6+x^5/120+17x^7/315+O(x^9))^3/3\\
&\;+\;2(x-x^3/6+x^5/120+17x^7/315+O(x^9))^5/15\\
&\;-\;(x-x^3/6+x^5/120+17x^7/315+O(x^9))^7/5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Discrete Math Induction: $\sum^n_{i=1} \frac1{i(i+1)}$ For $\sum^n_{i=1} \frac1{i(i+1)}$ Find a formula and proofs that it holds for all n ≥ 1.
How would I find the formula for this one that can hold for all n ≥ 1?
| Notice that:
$$\frac{1}{i} - \frac{1}{i + 1} = \frac{1}{i(i+1)}$$
Afterwards, observe the cancellations that occur in:
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \dots - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1}\\
= 1 - \frac{1}{n + 1}$$
Then, we have to prove this by induction (although ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value range of parameter inequality. Assume $a,b,c>0$ satisfies :$$5c-3a \leq b \leq 4c-a$$ and $$c\ln b \geq a + c\ln c$$. Find the value range of $\frac{b}{a}$.
My approach:
because $a>0$ and I want to construct the target expression : $\frac{b}{a}$. I divide the first inequality by $a$ in the three part. and I get ... | Here what I did:
We have the following inequality:
$$5c-3a\leq b\leq 4c-a.\quad\quad (E_0)$$
Since $b>0$, then I can divide by $b$:
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{4c-a}{b}.\quad (1)$$
Further $(1)$ can be seen as:
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{5c-3a+2a-c}{b},$$
or
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{5c-3a}{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{-2nk} \binom{n}{2k}\left(\binom{2k}{k}^n\right)$ I can see numerically that $$\lim_{n \to \infty} \sum_{k=0}^{\lfloor n/2 \rfloor} 2^{-2nk} \binom{n}{2k}\left(\binom{2k}{k}^n\right) = 1$$ but how can you prove this? Using Stirling's approximation doesn't seem to be enough.... | $$
\sum_{k=1}^{\lfloor n/2\rfloor}2^{-2nk}\binom{n}{2k}\binom{2k}{k}^{\large n}
=\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}\left[4^{-k}\binom{2k}{k}\right]^{\large n}\tag{1}
$$
The $k=0$ term is $1$ so we need to determine how the rest of the terms go to $0$. Let's look at the ratio between consecutive terms in $(1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Taylor series and Maclaurin Series expansion
Hi! I am currently working on some calc2 online homework problems on Taylor series and Maclaurin series. I have tried a few different answers to this question, but I am really not sure how to go about solving this problem. If anyone can help me I would greatly appreciate it... | Observe that
$$\frac{1}{1 - 3x} = \frac{1}{-2 - 3(x - 1)} = -\frac{1}{2 + 3(x - 1)} = -\frac{1/2}{1 + (3/2)(x - 1)}.$$
Therefore
$$\frac{1}{1 - 3x} = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3(x - 1)}{2}\right)^n = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3}{2}\right)^n (x - 1)^n, \quad \left|\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
question on surds i already asked this question but the answer I got did not match the one in the book $$\sqrt{ 3x }= x + \sqrt {3}$$
Give x in the form
$$A \sqrt {B} + C $$
Can you show me how this is done step by step.
The answer I have in the book is:
$$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
this is where I got s... | I figured out the problem: The solution to $\color{blue}{\sqrt 3}(x) = x + \sqrt 3$ is indeed $$\frac {1}{2} \sqrt{3} + \frac {3}{2}$$
But that's not the problem you posted. In the above, only $3$ is under the radical sign. In your post, you have $\sqrt{3x}$
In the event that the problem should read: $$\sqrt 3(x) = x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Several (advanced) L'Hospital problems Problems :
$$
\begin{align}
&\text{A}.\ \lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{B}.\ \lim_{x\rightarrow 0}\left(\cot x-\frac{1}{x}\right)\\
&\text{C}.\ \lim_{x\rightarrow0}\frac{\sin^{-1}x-x}{\tan^{-1}x-x}\\
&\text{D}.\ \lim_{x\rightarrow 1}(2-x)^{\t... | For example A,D and E
*
*put $y=\lim_{x\to a} f(x)$
*Apply ln as $\ln y=\ln\lim_{x\to a} f(x)=\lim_{x\to a} \ln f(x)$
*use L'Hopital rule
*$\ln y=l\implies y=e^l$
For A
$$y=\lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}
\\ \ln y=\lim_{x\rightarrow1}\ln (2-x)^{\tan\left(\frac{\pi x}{2}\right)}
\\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/745427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Expressing a Polynomial as a sum of cube roots of integers How do you prove $x^3-3x^2-6x-4$ has a zero of the form $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$, for distinct positive integers a,b,c
| Use a classic method for solving a cubic (not having to remember the formula).
First eliminate the squared term by setting $y=x-1$ to obtain $$y^3-9y-12=0$$
Now we look for solutions of the form $y=v+w$, where $y^3=(v+w)^3=(v^3+w^3)+3vw(v+w)$ which we can rewrite as $$y^3-3vw\cdot y-(v^3+w^3)=0$$
Comparing coefficients... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/745609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the easiest way to integrate $y=\frac {x+4}{\sqrt{-x^2-2x+3}}$? What is the easiest way to integrate $y=\frac{x+4}{\sqrt{-x^2-2x+3}}$ ? I tried to integrate it by making numerator in form: $-2x-2$ and then pulling it under differential, but the result drastically differed from Mathematica output.
| Since $-x^2-2x+3=-(x+3)(x-1)$, both roots $x_1=-3,x_2=1$ are real. As such it is known that e.g. the Euler substitution
\begin{align}
\sqrt{-x^{2}-2x+3}&=t(x-x_2)=t(x-1)\\[2ex] &\Leftrightarrow x=\frac{t^{2}-3}{t^{2}+1}=1-\frac{4
}{1+t^{2}},\quad dx=\frac{8t}{\left( 1+t^{2}\right) ^{2}}dt, \\[2ex]
& \Leftrightarrow\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/748472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does $2^{n+1} + 2^{n+1} = 2^{n+2}$? Simple question, why does:
$2^{n+1} + 2^{n+1} = 2^{n+2}$ ?
Furthermore, why does this only work for powers of 2?
Thanks.
| Note that $2^{n+1}=2\cdot 2^n=2^n+2^n$
If you try it with $3$ you get $3^{n+1}=3\cdot 3^n=3^n+3^n+3^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/750710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solutions for quartic Suppose I have an equation in the form $(x-a)^4 + (x-b)^4 = c$. What is a clever way to find all four solutions? I have tried expanding and then used long division. However, I believe a better way is to reduce to cubic, then to quadratic. I don't know how to do this.
| The solutions of $$(x-a)^4 + (x-b)^4 = c$$ are less complex that we could think. They are, after some tedious simplifications given by $$x_1=\frac{1}{2} \left(a+b-\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_2=\frac{1}{2} \left(a+b+\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_3=\frac{1}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
the geometric explain of $t = x-\frac{a}{3}$ in the simplify of cubic equation $x^3+ax^2+bx+c=0$ Assume $$f(x) = x^3+ax^2+bx+c$$
we have $$f''(x)=2a+6x$$. we get $x = -\frac{a}{3}$
Magically, If we take the transformation: $$t = x -\left(-\frac{a}{3}\right)$$. we can transform the above equation into $$f(x)=g(t) = t^3+... | Let's look at the quadratic term of $f(t - \dfrac{a}{3})$. It came from $2$ sources. One is the $2$nd term of the cubic expansion $(t - \dfrac{a}{3})^3$: $-3t^2\cdot \dfrac{a}{3}$ which is: $-at^2$, and the other term comes from the $1$st term of $a(t - \dfrac{a}{3})^2$ which is: $at^2$. So the coefficient of $t^2$ is:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra
| Let $a = \sin x, b = \cos x$. Then we need to find the minimum of the function $\cos x + \sin x + \sec(x) \csc(x)$ which is $2+ \sqrt{2}$ at $x = \frac{\pi}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How find this limit $\lim_{n\to \infty} \left(\frac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\cdot \cdots$ Find this limit
$$\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)$$
I know we must use this
$$n!\approx\left(\df... | $$L=\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)=\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot (n!)^2}\right)^{\frac{1}{n}}\cdot\lim_{n\to\infty}\sqrt[n]{n!}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| Very important fact of physics and harmonics and digital signal processing:
The sum of two sinusoids of equal frequency is another sinusoid of that same frequency, regardless of amplitude or phase.
Or stated mathematically:
If
$$f(x) = A_1\cos\left(\omega\,x + p_1\right) + A_2\cos\left(\omega\,x + p_2\right)$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 14,
"answer_id": 11
} |
How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ According to wolfram alpha this is true: $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$
But how do you show this? I know of no rules that works with addition inside square roots.
I noticed I could do this:
$\sqrt{24} = 2\sqrt{3}\sqrt{2}$
But I still don't see how I ... | On can easily discover the denesting using my simple radical denesting algorithm.
$\ w = 5+\sqrt{24}\,$ has norm $\,n = ww' = 5^2-24 = 1.\,$ Subtracting out $\,\sqrt{n}=1\,$ yields $\,4+\sqrt{24}.$
This has trace $\,t = 8,\,$ so dividing $\,\sqrt{t} = 2\sqrt{2}\,$ out of $\,4+\sqrt{24}=4+2\sqrt{6}\,$ yields
$$ \frac{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\int_0^{\infty}\int_0^{\infty}e^{-(x^2 + 2xy\cos(\alpha)+y^2)}\mathrm dx\mathrm dy=\frac{\alpha}{2\sin(\alpha)}$ I'm having difficulty with a question. It says
By putting $x=r\cos(\theta), y=r\sin(\theta)$, prove that $$\int_0^{\infty}\int_0^{\infty}e^{-(x^2 + 2xy\cos(\alpha)+y^2)}\mathrm dx\mathrm dy=\fra... | \begin{align}
\frac{1}{1+\sin{(2\, \theta)}\, \cos{\alpha}} &=\frac{1}{1+2\sin{\theta}\cos{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{\sec{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{1+\tan{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{\sin{(\alpha)}^2+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How have they done the algebra here? Proof by induction
\begin{align}&4-\frac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^k\\
=&4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}
\\
=&4-\frac{(k+1)+2}{2^{(k+1)-1}}
\end{align}
Original image
Can someone explain these steps to me please? Did the $2^{k-1}$ change to $2^k$ by multiplying nu... | When you look at the second line notice that they have an extra $2$ in the numerator they multiplied by $1$, which can also be written as $1 = \frac{2}{2}$. So just for completeness:
\begin{align*} &4-\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k \\
& = 4-1\cdot\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k \\
& = 4-\frac{2}{2}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Bombelli's wild thought of cubic equations In many books, like Visual Complex Analysis. talk about the real original of complex number. the author begin with this equation:
$$x^3=15x+4$$
Then the author use the formula $$x=\sqrt[3]{q+\sqrt{q^2-p^3}}+\sqrt[3]{q-\sqrt{q^2-p^3}}$$
to say that the equation has a root $$x =... | He (Bombelli) probably used the Cardano formula to obtain the $\textbf{one}$ root solution you just described $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$
He then had to find the cube roots of each of the radicals above. I am not sure exactly how he found them but he may have used the following sometimes used polynomial to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How find the $AP+\frac{1}{2}BP$ minmum value An equilateral triangle $ABC$ such $$AB=BC=AC=2a>0$$
A circle $O$ is inscribed in triangle $ABC$,and the point $P$ on the circle $O$.
Find the minimum $$AP+\dfrac{1}{2}BP$$
My idea: let $$A(-a,0),B(a,0),O(0,\dfrac{\sqrt{3}}{3}a)$$
then the circle equation is $$
x^2+(y-\dfra... | The solution of Omri Nissan Solan.
Let $K$ and $L$ be common points of the circle with $AB$ and $BC$ respectively
and let $BO\cap KL=\{M\}$.
Thus, our circle it's the Apollonian circle of $\Delta KLB$.
Id est, $AP+\frac{1}{2}PB=AP+PM\geq AM=\frac{a\sqrt7}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
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.