Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Cyclic inequality How can we prove that:
$$a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10}+a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20}\geq 2(a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}), \ \forall\ a,b,c\geq 0.$$
I proved only that $32S_1+18S_2\geq 2 S$, using AM-GM inequality, where
$S_1=a^{60} c^{10} +b^{60}a^{10}+c^{... | Update 2019/10/05 (Simplify some proof.)
With computer, here is a solution:
Clearly, we only need to prove the case when $a, b, c > 0$.
Let $x=a^{10}, \ y = b^{10}, \ z=c^{10}$. The inequality is written as
$$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge
2x^5yz\sqrt[10]{\frac{x}{y}} + 2xy^5z\sqrt[10]{\frac{y}{z}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For positive integers $m,n$ if $\sqrt 7 - \frac{m}{n} > 0$ then prove that $\sqrt 7 - \frac{m}{n} > \frac{1}{{mn}}$ For positive integers $m,n$ if $\sqrt 7 - \frac{m}{n} > 0$ then prove that $\sqrt 7 - \frac{m}{n} > \frac{1}{{mn}}$.
| By condition, $7n^2 > m^2$. The square of an integer when divided by 7 moiety can be given in only 0, 1, 2 and 4. Therefore, none of the numbers $m^2+1$, $m^2+2$ is not divisible by 7, where $7n^2\ge m^2+3$. Then $n\sqrt7\ge\sqrt{m^2+3}\ge\sqrt{m^2+2+\frac1{m^2}} > m+\frac1m$ at $m > 1$, so $\sqrt7-\frac{m}n > \frac1{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $\cos(x)\cos(2x)\cos(4x)=1/8$ I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.
I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?
| You can use the double angle identity
$$ \sin 2x = 2\sin x \cos x
$$
By multiplying $\sin x$,
$$\begin{align*}
\cos(x)\cos(2x)\cos(4x) &= \frac{1}{8} \\
\frac{1}{2} \sin (2x) \cos(2x)\cos(4x) &= \frac{1}{8}\sin x \\
\frac{1}{4} \sin (4x)\cos(4x) &= \frac{1}{8}\sin x \\
\frac{1}{8} \sin (8x) &= \frac{1}{8}\sin x \\
\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3]... | Assuming $$\frac{p}{q}=2\sqrt{3}+3\sqrt[3]{2}$$
You will have :
$$54 = (\frac{p}{q} - 2\sqrt{3})^3$$
$$= \frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+18\frac{p}{q}-24\sqrt{3}$$
So finally :
$$54 -\frac{p^3}{q^3}-18\frac{p}{q}= -6\frac{p^2}{q^2}\sqrt{3}-24\sqrt{3} = -6 (\frac{p^2}{q^2}+4)\sqrt{3}$$
which is impossible becau... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$
Let $a_1 \in \mathbb{R}$. Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$.
This question seems hard to solve since we don't know what $a_1$ is. We want both $|x|,|x-a_1|$ to be as small as possible or as larg... | Set $b=a_1/2$ and the translation $x\mapsto x-b$, $y\mapsto y$. This leads to considering the function
$$
f(x)=\frac{1}{1+|x-b|}+\frac{1}{1+|x+b|}
$$
and it's not restrictive to assume $b\ge0$. Leave out the case $b=0$, for the moment. The function is even, so we just need to study it for $x\ge0$ and we can write it as... | {
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How do I show $\int_0^\infty \frac{x}{(x+1)^2}dx$ diverges I did $$\int_0^\infty \frac{x}{(x+1)^2}dx=\int_0^\infty \frac{1}{(x+2+\frac1x)}dx\le\int_0^\infty \frac{1}{x}dx$$ but $\int_0^\infty \frac{1}{x}dx$ diverges. So my logic fails.
| Let $c$ be the upper limit of the integral and take the limit as $c$ goes to infinity.
$$I = \int_0^\infty \frac{x}{(x+1)^2}dx = \lim_{c \rightarrow \infty} \int_0^c \frac{x}{(x+1)^2}dx.$$
Then compute the definite integral as normal;
$$I = \lim_{c \rightarrow \infty}\int_0^c \frac{x}{(x+1)^2} = \lim_{c\rightarrow\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum polynomial of $\sqrt{i+\sqrt{2}}$ over $ \mathbb{R}$ So here's what I know about the minimum polynomial $p(x)$ of $c=\sqrt{i+\sqrt{2}}$ over $\mathbb{R}$...
*
*$p(x)$ has to be degree 2
*$p(x)$ must be a multiple of $q(x)=x-\sqrt{i+\sqrt{2}}$ because $q(x)$ is the generator of the kernel of the sub... | Robert Israel's answer is excellent and complete, but I thought I might offer some supporting computations. Let $\alpha = i+\sqrt{2}$, and let $\beta = \sqrt{\alpha}$. Then $|\alpha| = \sqrt{3}$, so we can rewrite $\alpha$ as
$$\alpha = \sqrt{3}(\cos(\theta)+i\sin(\theta))$$
where $\cos(\theta) = \frac{\sqrt{2}}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702399",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find Nth formula of recursive formula $a_n=a_{n-1}+n(n-1)a_{n-2}$ $$a_n=a_{n-1}+n(n-1)a_{n-2}$$
$$a_0=1, a_1=-\frac{1}{2}$$
Is it possible to find explicit formula for $a_n$ just by using $a_0$ and $a_1$?
I know how to solve this problem if $a_n=Aa_{n-1}+Ba_{n-2}$ where $A$ and $B$ is some real number(constant), but i... | Let $\displaystyle\;b_n = \frac{a_n}{n!}$, the recurrence relation at hand can be transformed to
$$n ( b_n - b_{n-2} ) = b_{n-1}\quad\text{ with }\quad \begin{cases} b_0 = 1\\b_1 = -\frac12\end{cases}\tag{*1}$$
Let $f(z) = \sum_{n=0}^\infty b_n z^n$ be the OGF for $b_n$. If we multiple $(*1)$
by $z^n$ and start to sum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Sum of Consecutive Perfect Squares If $n\in\mathbb{Z}$ and $2n+1$ is a perfect square, then is it true that $n+1$ is a sum of two consecutive perfect squares?
| Since $2n + 1$ is odd, if it is a perfect square it is the square of an odd number. So write $2n+1 = (2k+1)^2$. Then we can solve to get $n+1 = \frac{(2k+1)^2 - 1}{2} + 1$, and further simplify this to $2k^2 + 2k + 1 = k^2 + k^2 + 2k + 1 = k^2 + (k+1)^2$, showing that $n+1$ is indeed the sum of two consecutive square... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerat... | This is exactly the kind of question which "looks very difficult" and at the same time is "extremely easy to answer". It appears that it is specially crafted to generate a complicated looking expression in order to intimidate a casual reader/student. The following evaluation shows that the difficulty is only superficia... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $ | 1 - \frac{x^2}{2} - \cos x | \leq \frac{1}{24}$ using Taylor Use Taylor's theorem to show that for $x \in [-1, 1] $ we have that $$ \left| 1 - \frac{x^2}{2} - \cos x \right| \leq \frac{1}{24}. $$
Attempt: Since $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \ldots $, I see that $1 - \frac{x^2}{2} = P_2 (x)... | Since the term in $x^3$ is $0$, your approximation $P_2(x)$ can also be thought of as $P_3(x)$.
By the Lagrange form of the remainder, the absolute value of the error in $P_3(x)$ is $\frac{1}{4!}|\cos(c_x)||x|^4$ for some $c_x$ between $0$ and $x$. We have $|\cos(c_x)|\le 1$, and $|x|\le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit calculation with Riemann integral Help me calculating the limit with Riemann integral:
$$a_n=\frac{1}{n\sqrt{n^2+1}}+\frac{2}{n\sqrt{n^2+4}}+\frac{3}{n\sqrt{n^2+9}}+...+\frac{n-1}{n\sqrt{n^2+(n-1)^2}}$$
| Hint: This can be written as
$$\lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^{n-1} \frac{r}{\sqrt{(n^2+r^2)}}$$
$$=\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n-1} \frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^2}}$$
$$=\int_0^1 \frac {xdx}{\sqrt{1+x^2}}$$
| {
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"url": "https://math.stackexchange.com/questions/1707742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Partial derivative of $f(x,y) = \frac{x^3+y^3}{x^2+y^2}$ at $f_1(0,0)$ To be more precise than the title, the function is actually piecewise
$$
f(x,y) = \begin{cases}
\frac{x^3+y^3}{x^2+y^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$
I checked that the function is continuous at $(0,0)$, so I then calculat... | Hint: Compute the limit (as $h$ goes to $0$) of $(f(0+h,0)-f(0,0))/h$ to get the value of the partial derivative at $(0,0)$.
What you are looking at with those path limits is the question of continuity of this partial derivative function at $(0,0)$.
| {
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"url": "https://math.stackexchange.com/questions/1707853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Confusion determining eigenvalues Matrix $A$:
$$A = \begin{bmatrix}
-5 &14& -8 \\
-9 &16& -6 \\
-9 &11& -1
\end{bmatrix} $$
Determining eigenvalue of Matrix $A - 1I$ hence $\lambda = 1$ (other eigenvalues are 4 and 5).
Matrix $A - 1I:$
$$\begin{bmatrix}
-6 &14& -8 \\
-9 &15& -6 \\
-9 &11& -2
\e... | Using gaussian elimination, your system is equivalent to the following, writing only the matrix, not the $x_i$ nor the right hand side since there are only zeros in the RHS anyway:
$$\left(\begin{matrix}
-6 &14& -8 \\
-9 &15& -6 \\
-9 &11& -2
\end{matrix}\right)$$
$$\left(\begin{matrix}
-3 &7& -4 \\
-9 &15& -6 \\
-9 &1... | {
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Existence of rational roots in a quadratic equation Consider the quadratic equation $(a+c-b)x^2 + 2cx + b+c-a = 0 $ , where a,b,c are distinct real numbers and a+c-b is not equal to 0. Suppose that both the roots of the equation are rational . Then
a) a,b, and c are rational
b)$c/(a-b)$ is rational
c)$b/(c-a)$ is ra... | The roots are
$$\frac{-2c\pm \sqrt{4c^2 - 4(c+(a-b))(c-(a-b))}}{2(a+c-b)}$$
$$=\frac{-c\pm (a-b)}{a+c-b}$$
$=\frac{a-b-c}{a-b+c}$ or $\frac{-a+b-c}{a-b+c}=-1$
For the first root,
$$=\frac{a-b+c-2c}{a-b+c}=1-\frac{2c}{a-b+c}$$
So $\frac{2c}{a-b+c}$ is rational, so its inverse is also rational
$\frac{a-b+c}{2c}=\frac{a-b... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let... | So for your solution:
$$I_2 = \int_{0}^{a} \frac{x}{\sqrt{a^2 - x^2}} dx$$
Let $u = a^2 - x^2, du = -2x dx, dx = -\frac{1}{2x} du$ then it is clear that
$$ I_2 = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du = - \sqrt{u} $$
Now evaluating back at $x$,
$$ = - \sqrt{a^2 - x^2} | _0^a = - \sqrt{0} + \sqrt{a^2} = a$$
I woul... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Residue of $f(z) = \frac{1}{z-\sin z}$ at $z=0$ My attempt:
$$ f(z) = \frac{1}{z-\sin z}$$
$$\frac{1}{z-(z-\frac{z^3}{6}+\frac{z^5}{120}-...)}$$
$$\frac{1}{z(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$Res(f(z),0) = \lim_{z \to 0} z \cdot \frac{1}{z(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$ = \lim_{z \to 0} \fra... | Two small hints in addition to the nice answer of @C.Dubussy
*
*If $f$ has a pole of order $n$ in $z_0$, the $Res$ of $f$ at $z_0$ is
\begin{align*}
Res(f,z_0)=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\left[(z-z_0)^nf(z)\right]
\end{align*}
*From your series expansion
\begin{align*... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the integral $\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$ $$\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$$$=2\int \frac{x^3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx+3\int \frac{x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
For these two integrals, I have tried Euler substitutions and vari... | Since $x^2 + 2x -3 = (x+1)^2 -4$, you can use the substitution $x+1 = 2 \cosh t$. This way you are going to get a rational function in $\cosh t$ and $\sinh t$.
| {
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"question_score": "2",
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Why is $x^3-x^2+x$ injective? The function $$ f(x) = x^3-x^2+x$$ is injective (as seen on the graph), but by doing f(a) = f(b) I can't get to the point where a = b.
Can this be indicated in an analytical way (without graph) ?
| Well, you should not give up hope on that $f(a)=f(b)$ approach :-):
$$f(a)=f(b)\\a^3-a^2+a=b^3-b^2+b\\a^3-b^3-(a^2-b^2)+(a-b)=0\\(a-b)(a^2+ab+b^2-a-b+1)=0$$
Which means that either $a=b$ or $a^2+ab+b^2-a-b+1=0$
Considering this as a quadratic with respect to a, we calculate it's discriminant as $$\Delta=(b-1)^2-4(b^2-b... | {
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Integral through $u$-substitution v. multiplying out I have the integral:
$ \int (x^2)(x^3-1)^2 \, dx $ Through $u$-substitution, I can write this is equal to $ \frac {1}{3} \int (3x^2)(x^3-1)^2 dx $ which equals $ \frac{1}{3}\cdot\frac{(x^3-1)^3}{3}$.
However, if rather than using $u$-substitution, I multiply wit... | $$
\underbrace{\frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} + \text{a constant}} = \underbrace{ \frac 1 3\cdot \frac{(x^3-1)^3} 3 + \text{a different constant}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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partial fractions
$$\frac{1}{1-x^2}$$
$$\frac{1}{1-x^2}=\frac{a}{1-x}+\frac{b}{1+x}$$
$$1=a+ax+b-bx$$
$$1=a+b+x(a-b)$$
$a+b=1$ and $x(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$
$$2a=1\Rightarrow a=\frac{1}{2}$$
$b=\frac{1}{2}$
$$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
Where I went wrong?
| You didn't do anything wrong.
$$\frac{1}{1-x^2} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)}$$
is equivalent to your answer of
$$\frac{1}{1-x^2} = \frac{1}{2(1-x)} + \frac{1}{2(1+x)}.$$
I think we can agree that both answers have a common term of $\dfrac{1}{2(1+x)}$. Now, notice:
$$\frac{1}{2(1-x)} = \frac{1}{2 \cdot [-1(x-1... | {
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Loop invariants in logic I am working on some questions about hoare calculus/logic. The given program $\pi$ is:
$ x:=0; y:=1; WHILE \; \lnot x=n \; DO (y:=2y+x + 1; \; x:= x + 1) $
The hoare-logic rules that I can use are in the list below)
H1 {[$t$\x]$\varphi$} x := $t$ {$\varphi$}, provided that $t$ is free for $x$... | Here's what we want to prove:
{y = 3*2^x - x - 2}
y := 2*y + x + 1;
x := x + 1;
{y = 3*2^x - x - 2}
Indeed, this follows easily after sticking another assertion between the two assignments:
{y = 3*2^x - x - 2}
y := 2*y + x + 1;
{y = 3*2^(x + 1) - x - 3}
x := x + 1;
{y = 3*2^x - x - 2}
To see why the first 3 lines hol... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
| Substituting $x=2\sin t$, we get:
$$ \int x^3\sqrt{2^2-x^2}dx=32\int \sin^3 t \cos^2 t dt=32\int(1-\cos^2 t)\cos ^2 t\sin t dt$$
$$=32\int (\cos^2 t-\cos ^4 t)\sin tdt$$
$$=-32\int z^2 dz + 32\int z^4 dz$$ Where $z=\cos t.$
| {
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"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$
| Let $z=e^{\frac{i\pi}{7}}$. Then $\cos (\frac{\pi}{7})=\frac{z+z^{-1}}{2}$, $\cos (\frac{2\pi}{7})=\frac{z^2+z^{-2}}{2}$, $\cos (\frac{3\pi}{7})=\frac{z^3+z^{-3}}{2}$. Also $0=\frac{z^7-1}{z-1}=1+z+z^2+z^3+z^4+z^5+z^6$. As $z\ne0,$ divide both sides by $z^3$ to get $$z^3+\frac1{z^3}+z^2+\frac1{z^2}+z+\frac1z+1=0.$$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Definite integral with trigononmetric functions I have arrived at definite integral with trigonometric functions
$I(a, b) = \int_0^{2 \pi} \frac{1 - a \sin(\theta) - b \cos(\theta)}{(1 +a^2 +b^2 - 2 a \sin(\theta) - 2 b \cos(\theta) )^{3/2}} \mathrm{d} \theta, \quad 0 \le a, b < 1$
I was expecting result with ellipti... | Using a CAS, $$\alpha\,I(a,b)=\left(a^2+b^2+1-2 \sqrt{a^2+b^2}\right) K(z(a,b))+\left(1-a^2-b^2\right)
E(z(a,b))$$ where $$z(a,b)=\frac{4 \sqrt{a^2+b^2}}{a^2+b^2+1+2 \sqrt{a^2+b^2}}$$ $$\alpha=\frac 12 \sqrt{a^2+b^2+1+2 \sqrt{a^2+b^2}}\,\left(a^2+b^2+1-2 \sqrt{a^2+b^2}\right)\,$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloo... | Since $n$ is a natural number we can divide $x$ (with remainder) by $n$ in order to express $x = nb +r_1$ for some $b \in \mathbb{N}$ and $r_1 < n$. Now $\lfloor \frac{x}{n} \rfloor = \lfloor \frac{nb+r}{n} \rfloor = \lfloor b + \frac{r}{n} \rfloor = b$ since $r<n$. On the other hand we have that $\lfloor x \rfloor = b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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In a right triangle, can $a+b=c?$ I understand that due to the Pythagorean Theorem, $a^2+b^2=c^2$, given that $a$ and $b$ are legs of a right triangle and $c$ is the hypotenuse of the same right triangle. However, most of the time, $a+b\neq c$. What I have been wondering is, is there any set of values for $a$, $b$, and... | Substituting $c = a + b$ into $a^2+b^2=c^2$ gives us $a^2+b^2=(a+b)^2$. Multiplying that out gives us $a^2+b^2=a^2+2ab+b^2$ which means $2ab=0$
So to satisfy both $c = a + b$ and $a^2+b^2=c^2$ either $a=0$ or $b=0$.
The question then becomes a bit more philosophical. We have a result that satisfies the Pythagorean equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 11,
"answer_id": 4
} |
Derive identities for $\cos(4x)$ and $\sin(4x)$ using following fact So I need to use the fact that: $$\cos(4x) + i\sin(4x) = \left(\cos(x) + i\sin(x)\right)^4$$ to derive identities for $\cos(4x)$ and $\sin(4x)$ in terms of $\cos(x)$ and $\sin(x)$. I'm not sure how to go about this, could I please get some help.
| $$(\cos x +i \sin x)^4=\cos 4x+i \sin 4x$$
$$(\cos x +i \sin x)^4=\cos^4x+4\cos ^3x \cdot (i \sin x)+6\cos ^2x \cdot (i \sin x)^2+4\cos x \cdot (i \sin x)^3+(i \sin x)^4=$$
$$=\cos^4x-6 \cos ^2x \sin^2x+\sin^4x+$$
$$+i(4 \cos ^3 x \sin x-4\cos x\sin^3x)$$
Then $$\cos 4x=\cos^4x-6 \cos ^2x \sin^2x+\sin^4x$$
and $$\sin 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+\cdots$
Prove that
$$\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+-\cdots$$
If I am not mistaken, this identity was either proven by Newton or known t... | The series can be retrieved by expressing the Maclaurin expansion of $\sin(nx)$ considered as a function of $\sin x$, avoiding then the use of complex numbers. Moreover, it can be extended to the case of non-integer $n$.
Let $u$ a real number, we introduce the notations $y=\sin x$ et $f(y)=\sin ux$. Successive deriva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Find the number of solutions of the equation $\sin^22x-\cos^28x=\frac{1}{2}\cos10x$ Find the number of solutions of the equation $\sin^22x-\cos^28x=\frac{1}{2}\cos10x$ lying in the interval $(0,\frac{\pi}{2})$
I found the period of $\sin^22x-\cos^28x$ as $\pi$ and the period of $\frac{1}{2}\cos10x$ is $\frac{\pi}{5}$... | $$\sin^2 2x-\cos^2 8x=\frac{1}{2} \cos10x$$
Using the formulae, $$\sin^2\alpha=\frac{1-\cos 2 \alpha}{2}; \cos^2\alpha=\frac{1+\cos 2 \alpha}{2}$$
$$\frac{1-\cos 4x}{2}-\frac{1+\cos 16x}{2}=\frac{1}{2} \cos10x$$
$$-\cos 4x-\cos 16 x=\cos 10 x$$
$$2 \cos 10x \cos 6x + \cos 10x=0$$
$\cos 10x =0$ or $\cos 6x= -\frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$ How do we integrate
$$\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$$
Could someone give me some hint for this question?
| \begin{align}
I=\int\frac{x^2+20}{(x\sin x+5\cos x)^2}dx&=\int\frac{x^2+20}{(x^2+5^2)(\frac{x}{\sqrt{x^2+5^2}}\sin x+\frac{5}{\sqrt{x^2+5^2}}\cos x)^2}dx\\
&=\int\frac{\left(1-\frac{5}{x^2+5^2}\right)dx}{\cos ^2\left(x-\arctan (\frac{x}{5})\right)}\\
&=\int\frac{d\left(x-\arctan\left(\frac{x}{5}\right)\right)}{\cos ^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\... | As usual I prefer the elementary way. We have
\begin{align}
L &= \lim_{x \to \infty}\dfrac{\left(\sqrt{1 + \dfrac{x^{3}}{x + 1}} - x\right)\log x}{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x}\notag\\
&= \lim_{x \to \infty}\dfrac{\left(1 + \dfrac{x^{3}}{x + 1} - x^{2}\right)\log x}{\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}\left(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Eigenvalues of symmetric block matrices related How to find eigenvalues of following block matrices?
$M=\begin{bmatrix}
A & B & O & O & O & O & O & \cdots & O & B\\
B & A & B & O & O & O & O & \cdots & O & O\\
O & B & A & B & O & O & O & \cdots & O & O\\
O & O & B & A & B & O & O & \cdots & O & O\\
O & O & O & B &... | Let $n$ be the number of $3 \times 3$ blocks in each row and column of $M$.
Let $P$ be the $3n \times 3n$ permutation matrix such that $P_{k,3k-2} = P_{n+k,3k-1} = P_{2n+k,3k} = 1$ for $k = 1,2,\ldots,n$ and all other entries are $0$.
Then, we have $$PMP^{-1} = \begin{bmatrix}X&I_{n \times n}&I_{n \times n}\\I_{n \tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 -y^2 = 2000^2$ I found this answer here on AoPS. I agree with the answer till it multiplies $49$ by $2$. I think it should be multiplied by $4$ since there are $4$ possible cases:
1) $x+y, x-y$ is po... | Take the simpler case $x^2-y^2=(x-y)(x+y)=20^2=2^4\cdot 5^2$ We give a factor $2$ to each of $x,y$ then distribute the factors of $2^2 \cdot 5^2$ to $x+y,x-y$. There are $9$ of them: $1,2,4,5,10,20,25,50,100$ The case $x-y=8, x+y=50$ gives $x=29, y=21$. You cannot separately pick the signs of $x+y$ and $x-y$, beca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $ Show that if $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $
$\gcd(a,b)=d\Rightarrow\ d\mid a,b\Rightarrow\ \ d^2\mid a^2,b^2\Rightarrow\ d^2\mid\gcd(a^2,b^2)$.
But to complete the proof we must show that: $\gcd(a^2,b^2)\mid d^2$
How can I achieve this?
| Simply, write $a=da'$ and $b=db'$ such that $$\gcd(a', b') =1$$ Hence, $a^2= d^2(a')^2$ and $b^2=d^2(b')^2$.
Since, $\gcd(a', b') =1$, $\gcd((a')^2, (b')^2) =1$ and thus $$gcd(a^2, b^2) = d^2\gcd((a')^2, (b')^2) = d^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all continuous functions over reals such that $f(x)+f(y) = f(x+y)-xy-1$ for all $x,y \in \mathbb{R}$
Find all continuous functions over reals such that $f(x)+f(y) = f(x+y)-xy-1$ for all $x,y \in \mathbb{R}$.
I saw first that $f(0) = -1$ but then I am struggling to see how to get a formula for $f(x)$. If I do $x... | Let $g(x) = f(x) + 1-x^2/2$. Then $g$ is continuous and
$$g(x+y) = g(x) + g(y)$$
The only functions with those properties (continuous and additive) are those of the form $g(x) = ax$, for $a \in \mathbb{R}$. (See Continuous and additive implies linear )
Hence $f(x) = -1+ax + \frac{x^2}{2}$ for some constant $a$.
EDIT: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Doubt in solving $\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$ Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of
$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\f... | Use the definition of Principal Value,
$$\sec^{-1}\sqrt5=\tan^{-1}\sqrt{5-1}=\tan^{-1}2$$
If $\csc^{-1}\dfrac{\sqrt{10}}3=y,\csc y=\dfrac{\sqrt{10}}3$ and $0<y<\dfrac\pi2$
$\cot y=+\sqrt{\left(\dfrac{\sqrt{10}}3\right)^2-1}=\dfrac13\iff\tan y=3$
$\implies\csc^{-1}\dfrac{\sqrt{10}}3=\tan^{-1}3$
Like my answer in showing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding equation of a circle given three non - collinear points A circle is given which passes through three non collinear points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then prove that equation of this circle is given by
$$\begin{vmatrix}
x^2+y^2&x&y&1\\
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
x_3^2+y_3^2&x_3&y_3&1\... | The quadratic form $Ax^2+By^2+Cxy+Dx+Ey+F=0$ represents always a conic (eventualy degenerate in two right lines if it is a product of two linear equations) and it represents a circle if and only if $A=B\ne0$ and $C=0$. Then for a circle we have the equation
$$A(x^2+y^2)+Dx+Ey+F=0$$ When we want a circle passing throu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find maximum of $f(x)=\frac{x^m(1-x)^n}{x(1-x)} $ on $0< x< 1$ I would appreciate if somebody could help me with the following problem:
Q: Find maximum $f(x)$ on $0< x< 1$ ($n,m \in \mathbb{N}$)
$$f(x)=\frac{x^m(1-x)^n}{x(1-x)} $$
I try
$$f'(x)=(m-1) x^{m-2} (1-x)^{n-1}-(n-1) x^{m-1} (1-x)^{n-2}$$
...
| If $m=n=1$, then $f(x)=1$ is a constant function. Thus suppose $m>1$ or $n>1$.
Factoring $f'(x)$,
\begin{align}
f'(x)&=x^{m-2}(1-x)^{n-2}((m-1)(1-x)-(n-1)x)\\
&=x^{m-2}(1-x)^{n-2}((m-1)-(m+n-2)x)
\end{align}
and so $x=\frac{m-1}{m+n-2}$ satisfies $f'(x)=0$. Is $\frac{m-1}{m+n-2}$ between $0$ and $1$? If $m>1$ and $n=1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| $$\sqrt{24+8\sqrt{5}}=\sqrt{20+4+2\cdot 2\cdot 2\sqrt{5}}=\sqrt{(2\sqrt5)^2+2^2+2 \cdot 2 \cdot 2\sqrt{5}}=\sqrt{(2+2\sqrt5)^2}=2+2\sqrt5$$
or, to continue on your start,
$$\sqrt{6+2\sqrt5}=\sqrt{1^2+(\sqrt5)^2+2\sqrt5}=1+\sqrt5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
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How many elements are in the set $ \{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\}$ How many elements are in the set $$ \left\{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\right\}$$
My attempt:
$\left( \frac{2+i}{2-i} \right) ^n = \left( \frac{3}{5} + \frac{4}{5}i \right)^n=e^{\arctan(4/3)ni}$
So if fo... | I was recently given off line a solution to this problem which to me seems a bit down to earth but probably easier to grasp.
If we have $a+bi$ such that $a = 3 \mod 5$ and $b = 4 \mod 5 $, i.e. $a = 5n + 3, b = 5m + 4$ for $m,n \in \mathbb Z$, then $(a+bi)^2 = x + yi$ and $x = 3 \mod 5$ and $y = 4 \mod 5 $ (can be easi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int \frac{1-\sin x}{(1+\sin x)\cos x}dx$ Evaluate $$I=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}$$
I tried in the following way:
$$1-\sin x=1-\cos\left(\frac{\pi}{2}-x\right)=2 \sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
Similarly $$1+\sin x=2 \cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
So
$$I=... | One may write
$$
\begin{align}
I&=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}
\\\\&=\int \frac{(1-\sin x)\cos x dx}{(1+\sin x)\cos^2 x}
\\\\&=\int \frac{(1-\sin x)\cos x dx}{(1+\sin x)(1-\sin^2 x)}
\\\\&=\int \frac{(1-u)du}{(1+u)(1-u^2)}\quad (u=\sin x)
\\\\&=\int \frac{du}{(1+u)^2}
\\\\&=-\frac{1}{1+\sin x}+C.
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Rational Expression equivalent form EDIT: I know how to find the answer, but does anyone know why plugging in numbers for x does not work?
The Question: If the rational expression $\frac {3x^2}{3x-1}$ is rewritten in the equivalent form $\frac {\frac 13}{3x-1}+A$, what must expression A be in terms of x?
The four answe... | You can solve the equation for A and get
$A = \frac {3x^2-\frac 13}{3x-1}$.
In the numerator 3 can be factored out.
$A = \frac {3(x^2-\frac 19)}{3x-1}$
$x^2-\frac 19$ is equivalent to the third binomial formula $a^2-b^2=(a-b)\cdot (a+b)$.
Therfore $ x^2-\frac 19=(x-\frac 13)\cdot (x+\frac 13)$
$A= \frac {3\cdot (x-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
order $a$ = 5, $a^3b = ba^3$. show that that $ab = ba$. Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.
Here is what I think:
We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. N... | Since $a$ has order 5, we have $a = a^6$. Then:
$$ ab = a^6b = a^3(a^3b) = a^3(ba^3) = (a^3b)a^3 = (ba^3)a^3 = ba^6 = ba \;\;\;\;\square $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin ... | converting the given equation into $$\tan(x/2)$$ we get
$$8\,{\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}-12\,{\frac {\tan \left( x/2
\right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}-5
=0$$
setting $$\tan(x/2)=t$$ you have to solve this equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Dimension about space of matrices of order 3 over the field of the real. Consider the vector space of the matrices of order 3 over the field of the real $M_{3}\left(\Re\right)$ numbers. and let S be the subspace such that is spanned by the matrices of the form $AB-BA$. What is the dimension of S?
I was thinking in... t... | Let $X=AB-BA$ then $tr(X)=tr(AB-BA) = tr(AB) - tr(BA) = 0 $ we have the subspace of matrices with trace = 0, and I found the basis for the space:
$\left\{ \begin{pmatrix}0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
1 & 0 & 0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove $\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$ Today I found the identity : $$\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$$.
How to prove or disprove this?
Thank you.
| Hint : $$\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2^{2n+1}}(e^{\frac{ik\pi}{2n+1}}+e^{-\frac{ik\pi}{2n+1}})^{2n+1}.$$ Then use the binomial theorem and switch the sums to get $$\frac{1}{2^{2n+1}}\sum_{j=0}^{2n+1}\sum_{k=1}^n (-1)^{k+1} \binom{2n+1}{j}e^{\frac{ikj\pi}{2n+1}}e^{-\frac{ik\pi(2n+1-j)}{2n+1}}$$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
equation of an ellipse given its center and two tangent lines There exists an ellipse centered at (0,0) with two tangent lines given by $y=-\frac{1}{2}x + \frac{\sqrt{39}}{2}$ and $y=\frac{1}{3}x + \frac{7}{3}$. Find the ellipse.
So, I used the equation of one line and substituted it into the equation of an ellipse:
$... | You can prove that the tangent line to an ellipse of equation
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
at a point $(s,t)$ on the ellipse is
$$
\frac{sx}{a^2}+\frac{ty}{b^2}=1
$$
You can rewrite the equations of your tangent lines as
$$
\frac{x}{\sqrt{39}}+\frac{2y}{\sqrt{39}}=1,
\qquad
\frac{-x}{7}+\frac{3y}{7}=1
$$
Thu... | {
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"url": "https://math.stackexchange.com/questions/1757284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| Since the expressions you know the value of are homogeneous and symmetric in $a,b,c$, we know (or should know) they are expressible in terms of homogeneous polynomials in the elementary symmetric functions $E_1=a+b+c$, $E_2=ab+ac+bc$, $E_3=abc$ and in polynomial terms.
In order to find these expressions I prefer a more... | {
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"url": "https://math.stackexchange.com/questions/1759141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to show that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is strictly increasing for $x \geq 1$? I am trying to prove that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is a strictly increasing function for $x \geq 1$.
I try to do this by showing that $f'(x)>0$ for all ... | Simple, use derivative, not approximations. You get:
$(x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2}))$
so all are positive for $x \geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Hard summation involving binomial and quadratic What is $$\sum \frac{2r^2-98r+1}{(100-r)({100\choose r})}$$ Where $r\in [1,99]$I have reduced it to $$\frac{(2r^2-98r+1)}{(100){99\choose r}}$$ what to do further? Partial fractions don't seem to help.
| Using that
$$2r^2-98r+1=(r+1)^2-r(100-r)$$
yields a telescopic sum :
$$\begin{align}\sum_{r=1}^{99}\frac{2r^2-98r+1}{(100-r)\binom{100}{r}}&=\sum_{r=1}^{99}\frac{(2r^2-98r+1)\cdot r!\cdot (100-r)!}{(100-r)\cdot 100!}\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(2r^2-98r+1)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x^2+y^2=2z^2$, positive integer solutions Determine all positive integer solutions of the equation $x^2+y^2=2z^2$.
First I assume $x \geq y$, and I have $x^2-z^2=z^2-y^2$. Then I have $(x-z)(x+z)=(z-y)(z+y)$, but from here, I don't know how it can help me to describe solutions (I know that there are infinitely many). ... | Just a start:
Note that $$\left(\frac{x-y}{2}\right)^2 + \left(\frac{x+y}{2}\right)^2=\frac{x^2+y^2}{2}=z^2.$$
And $\frac{x-y}{2}$ and $\frac{x+y}{2}$ are integers (why?)
So you need to find solutions to $u^2+v^2=z^2$.
| {
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"url": "https://math.stackexchange.com/questions/1767109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that this limit yields $\gamma=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k= n}^{(n-1)^2}\frac{1}{k}$ $\gamma =0.5772156...$ is Euler's constant
Show that this limit yields
$$\gamma=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=n}^{(n-1)^2}\frac{1}{k}$$
| Since
$$\frac{1}{k} \leqslant \int_{k-1}^k \frac{dx}{x} = \log k - \log(k-1) \leqslant \frac{1}{k-1},$$
it follows that
$$\sum_{k = n}^{(n-1)^2}\frac{1}{k} \leqslant \log(n-1)^2 - \log(n-1) = \log(n-1) \leqslant \sum_{k = n}^{(n-1)^2}\frac{1}{k} + \frac{1}{n-1} - \frac{1}{(n-1)^2},$$
and
$$ \sum_{k = 1}^{n-1}\frac{1}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Game theory probability I have a variant of two-finger morra game, where the winner is determined by the parity of the sum of the two numbers thrown, but the amount won or lost is the product of the two numbers. There are two player - Alice and Bob
If Alice plays:
1 finger with probability $p$
2 fingers with probabili... | I suppose that Alice wins if the sum of the numbers is even. If $a \{1,2\}$ and $b \in \{1,2\}$ are the strategies played by Alice and Bob, respectively. The payoff of Bob is:
$$B(a,b) = \begin{cases}
-1 & \text{if}~a = b = 1 \\
2 & \text{if}~a = 1, b = 2 ~\text{or}~ a=2, b=1\\
-4 & \text{if}~a = b = 2 \\
\end{cases}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $a, b$ so that $\lim_{n \to \infty}((1-n^3)^{\frac{1}{3}}-an-b)=0$ If we go by definition, then we have to find $a,b$, so that for every $\varepsilon >0, \exists N(\varepsilon), $ so that $\forall n>N(\varepsilon)$
$$|(1-n^3)^{\frac{1}{3}}-an-b| < \varepsilon$$
My attempt,
$$ \left| \frac{1-n^3-an\left(\sqrt[3]{1... | You want $(1-n^3)^{\frac 13} \rightarrow an+b$
If $(1-n^3)^{\frac 13} = an+b + \delta$
Then $(1-n^3) = (an+b + \delta)^3$
$1-n^3 = (an+b + \delta)(a^2n^2+b^2 + \delta^2+2abn+2an \delta + 2b \delta)$
$1-n^3 = a^3n^3+ab^2n + an\delta^2+2a^2bn^2+2a^2n^2 \delta + 2abn \delta$
$+a^2bn^2+b^3 + b\delta^2+2ab^2n+2abn \delta + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conditional Probabilities multi coloured balls from a bag I have two bags
Bag1 : Contains 4 red and 5 black balls
Bag2 : Contains 3 red and 7 black balls
One ball is drawn from bag1 and 2 balls are drawn form bag2. Find the probability of 2black balls and 1 red ball.
How do I go about solving this problem?.
This is wha... |
This is what I have come up with so far:
$$P(B,RR) + P(R,BB) = \frac{5}{9} \cdot \frac{7}{10} \cdot \frac{3}{9} + \frac{4}{9} \cdot \frac{7}{10} \cdot \frac{6}{9}$$
That's not quite okay. The probability of drawing the red from bag-2 is twice that. It is the old "two ways to draw the same thing" problem: red then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Sum of three consecutive cubes equals a perfect square I have found this problem in an old German textbook:
Find all sets of three consecutive integers such that the sum of their cubes is a perfect square.
We can write
$$S = (x-1)^3 + x^3 + (x+1)^3 = (x-1+x+1)((x-1)^2 - (x-1)(x-1) + (x+1)^2) + x^3$$
which reduces to
$... | What you can do (but not ask from seventh graders) is to turn the equation into an elliptic curve in Weierstrass normal form, by multiplying both sides by $9$ and putting $9S=Y^2$ and $X=3x$. Then we get
$$Y^2=X^3+18X$$
which indeed denotes an elliptic curve and we can use for instance SageMath to return its integral p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 0
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Is $\frac{1010103010101}9$ prime or composite? $1010103010101$ obviously divisible by $9$.
Is $\frac{1010103010101}9$ prime or composite?
The answer would be obtained without using WolframAlpha
| The polynomial $ x^6 + x^5 + x^4 + 3x^3 + x^2 + x + 1 $ can be factored through the following manipulations:
$$x^3 (x^3 + x^2 + 1) + x^4 + 2x^3 + x^2 + x + 1 = (x^3 + 1)(x^3 + x^2 + 1) + x^4 + x^3 + x = (x^3 + x + 1)(x^3 + x^2 + 1) $$
Now, plug in $ x = 10^2 $ to get a factorization of your number, which easily confir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a ... | Too long for a comment.
The Engel form of Cauchy-Schwarz is not the right way:
$$\frac{(x^2)^2}{8x^3+5y^3}+\frac{(y^2)^2}{8y^3+5z^3}+\frac{(z^2)^2}{8z^3+5x^3} \geq \frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$
So we should prove that
$$\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\geq\frac{x+y+z}{13}$$
which is equivalent to
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
"answer_count": 8,
"answer_id": 7
} |
Cubic Spline Interpolation My problem is to find a interpolating cubic spline to the points
$$\left\{(0,0), \left(\frac{\pi}{2}, 1\right), \left(\pi,0\right), \left(\frac{3\pi}{2}, -1\right),(2\pi,0)\right\}$$
I did as follows:
I set up the system
$$ \begin{cases}
& hg_0+2(h+h)g_1+hg_{2} = 6\left(\dfrac{y_{2}-y_1}{h... | Find the Cubic Spline for the points:
$$\left\{(0,0), \left(\frac{\pi}{2}, 1\right), \left(\pi,0\right), \left(\frac{3\pi}{2}, -1\right),(2\pi,0)\right\}$$
We will make use of this algorithm, which is certainly equivalent to the one you are using.
Since we have equal spacing, all of the $h_i = \dfrac{\pi}{2}$.
We have:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying the result formula for depressed Cubic After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt... | With the benefit of hindsight we notice that $10+\sqrt{108}=10+6\sqrt{3}$ and
$$10+6\sqrt{3}=(1+\sqrt{3})^3.$$
Similarly,
$$-10+6\sqrt{3}=(-1+\sqrt{3})^3.$$
Take the (real) cube roots and subtract.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Let's try
$$\frac{x^2+3}{x^6(x^2+1)}=\frac ax+\frac b{x^2}+\frac c{x^3}+\frac d{x^4}+\frac e{x^5}+\frac f{x^6}+\frac{gx+h}{x^2+1}\implies$$
$$x^2+3=ax^5(x^2+1)+bx^4(x^2+1)+cx^3(x^2+1)+dx^2(x^2+1)+ex(x^2+1)+f(x^2+1)+(gx+h)x^6$$
and now:
$$x=0\implies 3=f\;,\;\;coef. \;x^7\implies 0=a+g\;,\;\;coef.\;x^6\implies 0=b+h$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{... | there is a high-school method.
since $\frac{a^2+1}{a^2+2}$ is an even function, and it's decreasing when $x \lt 0$, and increasing when $x \gt 0$, all we need to do is to find the min value of $\mid a \mid$.
since $x^2 + (a - 3) x - (a - 2) = 0$ has root(s) as real number(s), it follows that
$$
\Delta = (a - 3)^2 + 4 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 20... | Let $f,g,p,u,k,d,x,y,z>0$ such that $u\geq k$:
$$y=\left(\frac{x\left(g+u\right)p}{g+k}\right),x=\left(\frac{z\left(d+u\right)p}{d+k}\right),z=\left(\frac{y\left(f+u\right)p}{f+k}\right)$$
Then we have the functions :
$$h\left(x\right)=\frac{x}{13+\frac{5p^{2}\left(x+u\right)^{2}}{\left(x+k\right)^{2}}}$$
Wich is conve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 5,
"answer_id": 3
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How to solve $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } }$ for $\cos(\alpha)$ using half-angle formula. I have $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } } $
How can I get $\cos(\alpha ) $ from this?
I know this identitiy.
$\cos(\frac{\alpha }{2} )=\sqrt{\frac{1+\cos(\alpha ) }{2} } $
But j... | $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } } $
$\cos^2(\frac{\alpha }{2} )=\frac{a^2}{a^2+b^2}$
$\frac{1}{2}(1 + \cos(\alpha))=\frac{a^2}{a^2+b^2}$
$\cos(\alpha)=\frac{2a^2}{a^2+b^2} - 1$
half-angle formula
$\cos^2(\frac{\alpha }{2} ) = \frac{1}{2}(1 + \cos(\alpha))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
How can I prove this sum?
$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$
My try:... | \begin{align*}
\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi}{(5n+1)^{2}}-\frac{\phi}{(5n+4)^{2}}\right] & =\frac{2}{125}\left(5+2\sqrt{5}\right)\pi^{2}\\
\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi^{-1}}{(5n+2)^{2}}-\frac{\phi^{-1}}{(5n+3)^{2}}\right] & =\frac{2}{125}\left(5-2\sqrt{5}\right)\pi^{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
I want to know if my attempt is correct.
First $a^{13} \equiv (a^3)^4 \cdot a \equiv a^4 \cdot a \equiv a^3 \cdot a^2 \equiv a \cdot a^2 \equiv a^3 \equiv a \pmod 3$.
Second, $(a^7)^2 \cdot a^{-1} \equiv a^2 ... | By Euler's Theorem (or you may even use Fermat's theorem),
$a^2 \equiv 1 \pmod {3} \implies a^{12} \equiv 1 \pmod 3 \implies a^{13} \equiv a \pmod 3$
$a^6 \equiv 1 \pmod {7} \implies a^{12} \equiv 1 \pmod 7 \implies a^{13} \equiv a \pmod 7$
$a^{12} \equiv 1 \pmod {13} \implies a^{13} \equiv a \pmod {13} $
$\therefore a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$ The question
Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$
What I've tried
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac... | I'll reproduce the answer @C.Dubussy have just deleted:
$$ \prod_{n=2} \left( 1 - \frac{1}{n^4} \right) = \prod_{n=2}^∞ \left( 1 + \frac{1}{n^2}\right) \prod_{n=2}^∞ \left( 1 - \frac{1}{n^2}\right) = \frac{\sin{iπ}}{iπ} \prod_{n=2}^∞ \frac{n-1}{n} \prod_{n=2}^∞ \frac{n+1}{n} $$
And because the last product gives $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
Here is what I got so far (using Cauchy's integral formula) :
$$\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n} ~=~ \sum\limits_{n=0}^\infty \left(\frac{1}{2\pi i}\... | An ODE-based proof. If we set
$$ A(t) = \sum_{n\geq 0}\binom{2n}{n}t^n $$
we have an analytic function in a neighbourhood of the origin with radius of convergence $\rho\geq\frac{1}{4}$.
We may notice that:
$$ A'(t) = \sum_{n\geq 1}\binom{2n}{n} n t^{n-1} = \sum_{n\geq 0}\binom{2n}{n}(4n+2) t^n $$
while:
$$ \frac{d}{dt... | {
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Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$
Evaluation of $$\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
Put $x=\cos 2 \theta\;,$ Then $dx = -2\sin 2 \theta d\theta$ and Changing Limit, We get
$$I = \int_{0}^{\frac{... | multiply the integrand by
$$
1=\frac{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}
$$
doing the alegbra correctly this indeed eliminates the roots in the denominator and we end up with
$$
I=\frac{1}{2}\int_{0}^{1}\frac{-\sqrt{1-x}... | {
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Using $\epsilon-\delta$ proof to prove continuity
Use an $\epsilon-\delta$ proof to show that
$f : R \setminus \left \{ \frac{-3}{2} \right \} \rightarrow R$ , $$f(x) = \frac{3x^2-2x-5}{2x+3}$$
is continuous at $x = -1$
Hello there. Can anyone here help me with this? I know i need to show that $|x-l|<\delta$ imp... | Fix $\varepsilon>0$ and let $x\in\mathbb{R}\setminus\{-\frac{3}{2}\}$ such that $|x-(-1)|=|x+1|<\delta$ for $\delta>0$ to be chosen later. As $f(-1)=0$, we have :
\begin{align}
\left|f(x)-f(-1)\right|&=\left|\frac{3x^2-2x-5}{2x+3}-0\right|\\
&=\left|\frac{3x^2-2x-5}{2x+3}\right|.
\end{align}
Now observe that
\begin{ali... | {
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Modulo Equations I am trying to solve a problem involving modulo arithmetic but I am not sure what method to use as I have never done this style of question before nor do I have any examples to work from.
The question is:
Solve for $x$ where $x^5 = 11 \mod (35)$
So I though Eculid's GCD algorithm would help but I jus... | Let $x\in\mathbb{Z}$ be such that $x^5\equiv 11\pmod{35}$. Then, $x^{15}=\left(x^5\right)^3\equiv 11^3\equiv 1\pmod{35}$. Since $\gcd(x,35)=1$, $x^{12}=x^{\text{lcm}(5-1,7-1)}=x^{\lambda(35)}\equiv 1\pmod{35}$, where $\lambda$ is the Carmichael function. Thus, $x^3=x^{\gcd(15,12)}=1\pmod{35}$, whence $1\equiv x^6\eq... | {
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with t... | Compare each group, term by term, with the next group.
$$\frac14-\frac17+\frac15-\frac18+\frac16-\frac19=\frac3{4\cdot7}+\frac3{5\cdot8}+\frac3{6\cdot9}>\frac{3^2}{6\cdot9}>\frac19>\frac1{10}$$
Each group is greater than the next group, so the alternating series test applies.
| {
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Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equ... | $100 = 2^2 \times 5^2$. It suffices to show that the two sums are equal mod $4$ and mod $25$.
If $x$ is even, $x^4 \equiv x^8 \equiv 0 \mod 2^2$, while if $x$ is odd,
$x^4 \equiv x^8 \equiv 1 \mod 2^2$. So $\sum_{i=1}^{99} x^4 \equiv \sum_{i=1}^{99} x^8 \mod 2^2$.
Similarly, if $x$ is divisible by $5$, $x^4 \equiv x^8... | {
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Is there any integer $n>1$ such that $3^n - 1$ is divisible by $2^n - 1$? Is there any integer $n>1$ such that $3^n - 1$ is divisible by $2^n - 1$?
I guess not. For every even integer $n$, we can show that $3^n - 1$ is not divisible by $2^n - 1$ because $2^n -1$ is a multiple of $3$, but $3^n-1$ is not. Would anyone ... | Here is a proof by contradiction. Assume there exists such $n$.
If $n$ is even, notice $2^n-1 \equiv 0 \pmod 3$. Thus, as $3^n-1$ is not dividible by $3$, $n$ is odd. $$2^n-1 \equiv 7 \pmod {12}$$So there exists such a $p$ that $$p \equiv \pm 7 \pmod {12}, 2^n-1 \equiv 0 \pmod p \Rightarrow 3^n-1 \equiv 0 \pmod p$$ ... | {
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Find the remainder of $\sum_{i=0}^{99} 2^{i^2}$ when dividing by 7 and determine if the quotient is even or odd I've recently had this problem in an exam and couldn't solve it.
Find the remainder of the following sum when dividing by 7 and determine if the quotient is even or odd:
$$\sum_{i=0}^{99} 2^{i^2}$$
I know the... | By Fermat's little theorem
$$ 2^{i^2}\!\!\!\pmod{7}=\left\{\begin{array}{ll}\color{green}{1}&\text{if } i\equiv 0\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 1\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 2\pmod{6}\\\color{green}{1}&\text{if } i\equiv 3\pmod{6}\\\color{blue}{2}&\text{if } i\equiv 4\pmod{6}\\\color{blue}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve $276 x\equiv 90\pmod {666}$
Solve $276 x\equiv 90\pmod {666}$
I found using Euclidean algorithm that $\gcd (276,666)=6$
then I divided by $6$ and I got:
$$46x\equiv 15\pmod {111}$$
and I found that $\gcd(46,111)=1$ using Euclidean algorithm
I am stuck here and don't know what to do
| You correctly reduced the problem to $46x \equiv 15 \pmod{111}$. Applying the Euclidean algorithm to find $\gcd(46, 111)$ yields
\begin{align*}
111 & = 2 \cdot 46 + 19\\
46 & = 2 \cdot 19 + 8\\
19 & = 2 \cdot 8 + 3\\
8 & = 2 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
so $\gcd(46, 111) = 1$ as you... | {
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Simplify $\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$ Simplify $$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
I know there is another easier method except the one I answered. I cannot find it. Can you please help? Thanks in advance.
| For a general approach related to lab's hint: We want to (hopefully) simplify $\sqrt{4 \pm 2\sqrt{3}}$. Let's look at $\sqrt{4 + 2\sqrt{3}}.$
$$
\sqrt{4 + 2\sqrt{3}} = a + b\sqrt{3}
$$
Let's find $a$ and $b$. First square both sides.
$$
4 + 2\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}.
$$
This gives us two equations:
$$
\... | {
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"source": "stackexchange",
"question_score": "4",
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Prove $\frac{1^5}{1^5}+\frac{1^5-2^5}{1^5+2^5}+\frac{1^5-2^5+3^5}{1^5+2^5+3^5}+\frac{1^5-2^5+3^5-4^5}{1^5+2^5+3^5+4^5}+\cdots?$ On my previous page I proposed,
$$\frac{1^3}{1^3}+\frac{1^3-2^3}{1^3+2^3}+\frac{1^3-2^3+3^3}{1^3+2^3+3^3}+\frac{1^3-2^3+3^3-4^3}{1^3+2^3+3^3+4^3}+\cdots=3-\frac{\pi^2}{4}$$
Jack D'Aurizio prov... | We need some preliminary lemma.
$$ S_N=\sum_{n=1}^{N}n^5 = \frac{1}{12}N^2(N+1)^2 (2N^2+2N-1) \tag{1} $$
$$ P_N=\sum_{n=1}^{2N}(-1)^{n+1} n^5 = N^2(5-20 N^2-16 N^3) \tag{2}$$
$$ D_N=\sum_{n=1}^{2N-1}(-1)^{n+1} n^5 = N^2(5-20 N^2+16 N^3) \tag{3}$$
They give:
$$ S = \sum_{N\geq 1}\left(\frac{D_N}{S_{2N-1}}+\frac{P_N}{S... | {
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"source": "stackexchange",
"question_score": "2",
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Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods and without gamma functions? I know that
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
but all of the methods I've found seem to be too complicated for an early calculus student. Is there any method of calculating this with real methods and without gam... | We first transform the integral by putting $u=x^2$, $$
\int_0^{\infty} \sin \left(x^2\right) d x=\frac{1}{2} \int_0^{\infty} \frac{\sin u}{\sqrt{u}} d u
$$
Using the Gaussian integral: $\int_0^{\infty} e^{-u v^2} d v=\frac{\sqrt{\pi}}{2 \sqrt{u}} $, where $u$ is a constant, we have
$$
\begin{aligned}
I &=\frac{1}{2} \i... | {
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Finding centre of sphere inscribed in tetrahedron Given the tetrahedron with vertices defined by vectors $a=(-4, -3, 1)$, $b=(8,3,1)$, $c= (2, 6, 1)$, $d=(4,3,3)$, find the centre of the sphere inscribed in the tetrahedron.
My train of thought: consider the intersection of the four bisectors of the vertices of the tetr... | The four vertices are $A=(4,3,3),B=(2,6,1),C=(-4,-3,1),D=(8,3,1)$.
The four faces of the tetrahedron are $BCD:z-1=0$; $ACD:x-2y+2z-4=0$; $ABD:x+2y+2z-16=0$; and $ABC:3x-2y-6z+12=0$. So the distance of a point $(a,b,c)$ from the three faces is $|c-1|,\ |\frac{1}{3}|a-2b+2c-4|,\ \frac{1}{3}|a+2b+2c-16|,\ \frac{1}{7}|3a-2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Eigenvalue of block matrix of order $2n$ How to find eigenvalues of following block matrix?
$$P=\begin{bmatrix} A & B \\ B & A \end{bmatrix}$$
Where,
$A=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\... | Note that the eigenvalues of $P$ are real and in $[-3,3]$.
Let $U_n$ be the matrix that is derived from $A_n$ by putting the entries $[1,n],[n,1]$ equal to $0$.
Then $\det(A_n\pm B_n-\lambda I_n)=\det(A_n-\lambda I_n)\pm \det(U_{n-1}-\lambda I_{n-1})=p_n(\lambda)\pm q_{n-1}(\lambda)$.
The roots of $p_n$ are $2\cos(\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Need help with Differential equation concept Having trouble with this differential equation.
Find the general solution to the differential equation:
$$ \frac {dy}{dx}= \frac {√y(x^2 + x − 4)} {(x^2 + 1)(x − 1)} $$
I don't know where to start.
| $$\text{y}'\left(x\right)=\frac{\sqrt{\text{y}(x)}(x^2+x-4)}{(x^2+1)(x-1)}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x\tag1$$
Substitute $\text{u}=\text{y}\left(x\right)$ and $\text{d}\text{u}=\text{y}'\le... | {
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"source": "stackexchange",
"question_score": "1",
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Need help solving $x^4-3x^3-11x^2+3x+10=0$ Solve $x^4-3x^3-11x^2+3x+10=0$
I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function.
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x_{1,2}=-\frac b{4a}-S \pm 0.5\sqrt{-4S^2-2P+ \frac q S}$$
$$x_{3,4}=-\frac b{4a} + S \pm 0.5\... | One can check that $1$ and $-1$ are both roots of this polynomial. After factoring out $x-1$ and $x+1$, you're left with a quadratic polynomial.
As a general rule, the rational root theorem is a good place to start for questions like these.
| {
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"question_score": "3",
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Prove $\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43$ for $a^2 + b^2 + c^2 + d^2 = 4$
Let $a,b,c,d \geqslant 0$ and $a^2+b^2+c^2+d^2=4$. Prove that
$$\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43.$$
I try some reverse AM-GM techniques but fail. I don'... | Alternative proof:
We have
$$\frac{5}{18} + \frac{1}{18}d^2
- \frac{1}{4 - d} = \frac{(2 - d)(d - 1)^2}{18(4 - d)} \ge 0.$$
Thus, we have
\begin{align*}
&\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c}\\
\le\,& \frac{5}{18}\sum_{\mathrm{cyc}} ab + \frac{1}{18}\sum_{\mathrm{cyc}}abd^2 \\
\le\,& \frac{5}{... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$
| Note that if $x^2=y^3$, then $y=x^{2/3}$. Therefore, we find that
$$x^{x^{2/3}}=\left(x^{2/3}\right)^x \tag 1$$
Taking the logarithm of both sides of $(1)$ yields
$$x^{2/3}\log(x)=\frac23 x\log(x) \tag2$$
Solutions to $(2)$ are $x=1$ and $x=27/8$. For $x=1$, $y=1$ and for $x=27/8$, $y=9/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I compute $\tan(.5\arctan(x))$? The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.
| Recall that $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan a \tan b}$. Also let $c = \tan({\frac{\arctan x}{2}})$.
\begin{align}
c &= \tan(\arctan x - \frac{\arctan x}{2})\\
&= \frac{\tan(\arctan x) - \tan(\frac{\arctan x}{2})}{1 + \tan(\arctan x)\tan(\frac{\arctan}{2})}\\
&= \frac{x - c}{1 + xc}\\
c^2x + c &= x- c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such th... | One solution is to rewrite $\gamma\alpha\gamma^{-1}=\beta$ as $\gamma\alpha=\beta\gamma$.
Then $\gamma\alpha(1)=\beta\gamma(1)$, and from $\alpha(1)=2$ we get $\gamma(2)=\beta\gamma(1)$.
Similarly, $\gamma(5)=\beta\gamma(2)$ and $\gamma(1)=\beta\gamma(5)$.
Combining these, we get $\gamma(2)=\beta\gamma(1)=\beta\beta\be... | {
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"url": "https://math.stackexchange.com/questions/1802679",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$ Is there a simple intuitive way to show that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
I sense it could be done more simple than this:
1 - take the derivative
$f'(x)=1-\frac{x}{\sqrt{x^2-1}}<0$ if $x>1$ so the slope of $f(x)$ in $(1,\infty)$ is negat... | Notice that $(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})=x^2-(x^2-1)=1$, and that $x+\sqrt{x^2-1}>x>1$, and therefore, we have that $x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}<\frac{1}{1}=1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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What are all the uses of the determinant? I've learned how to calculate the determinant but what is the determinant used for? So far, I only know that there is no inverse if the determinant is 0.
| *
*It allows you to evaluate cross products and find the general equation of the plane if given $3$ points. For instance: $A(1,1,0),\, B =(1,0,1),\,C=(0,1,2)$ $$B-A=(1,0,1)-(1,1,0)=(0,-1,1)$$ and $$C-A=(0,1,2)-(1,1,0)=(-1,0,2)$$ You now use the cross-product of $$(B-A)\times(C-A)=\begin{bmatrix}i & j & k \\0 & -1 & 1... | {
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Strong Pseudoprime to base square? Question:
Let n>1 be an odd composite integer and let a be an integer with (a,n)=1.
Show that, if n is a strong pseudoprime to base a, then n is also a strong pseudoprime to base $a^2$
What I have done is that : let $n-1=2^{s}t$ then assumption implies that
$1. a^t \equiv 1 \pmod n$... | Yes, the strong pseudoprime test doesn't require that you test $a^{n-1} \equiv 1 \bmod n$, but in any case that also works. Your proof basically works OK but gets a little unclear, perhaps because of the confusion between $a^{2k}$ and $(a^2)^k$.
So to restate the strong pseudoprime rule:
Define $t,d$ with $t\cdot 2^d ... | {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic expansion of ratio function I want to expand the following function:
$$
f(x)=\frac{1}{(1-e^{-x})}
$$
$f(x)$ can be rewritten as
$$
f(x) \sim \frac{1}{x-x^2/2 + x^3/2/3}
$$
But I want to express big-oh notation such that
$$
f(x) = \frac{1}{x} + .... +O(x^2)
$$
up to $x^2$ order.
How to do it?
| Around $0$:
$$
f(x) = \frac{1}{x-\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)}
= \frac{1}{x}\frac{1}{1-\frac{x}{2}+\frac{x^2}{6}+o(x^2)}
$$
(Advice: when doing Taylor expansions, avoid the equivalents $\sim$, which are made to capture the first term only. Confusions and mistake may arise very quickly otherwise.)
Now, use the fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding the hypotenuse ( Trigonometric Identities!) If the sides of a right-angled triangle are $\cos 2a + \cos 2b + 2\cos(a+b)$ and $\sin 2a + \sin 2b + 2\sin(a+b)$, find the hypotenuse-
I can simplify this but I end up having a lot of terms in the end :/
| $$\cos 2a+\cos 2b+2\cos(a+b)=\cos^2 a-\sin^2 a + \cos^2b-\sin^2b+2\cos a\cos b-2\sin a\sin b=(\cos a+\cos b)^2-(\sin a+\sin b)^2$$
$$\sin2a+\sin2b+2\sin(a+b)=2\sin a\cos a+2\sin b\cos b+2\sin a\cos b+2\cos a\sin b=2(\sin a+\sin b)(\cos a+\cos b)$$
Let $\cos a+\cos b=u$, $\sin a+\sin b=v$.
$$\text{Hypotenuse}=\sqrt{(u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is divisible by $7$.
Here it seems Fermat's theorem has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's ... | You want to prove the statement:
$[a^3+b^3+c^3\equiv0\pmod7]\implies[a\equiv0\pmod7]\vee[b\equiv0\pmod7]\vee[c\equiv0\pmod7]$
Instead, prove the equivalent statement:
$[a\not\equiv0\pmod7]\wedge[b\not\equiv0\pmod7]\wedge[c\not\equiv0\pmod7]\implies[a^3+b^3+c^3\not\equiv0\pmod7]$
Now, observe the following:
*
*$n\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Bilinear transformation which maps $z=(\infty, i, 0)$ and $w= (-1, -i, 1)$ I have three equations after simplifying this a bit
$a+c=0$
$ai+b-c=0$
$b-d=0$
How do I proceed further?
If you care to know this is from the chapter Complex Variables
| We want $w(z)=\frac{az+b}{cz+d}$, such as $w(\infty)=-1,w(0)=1,w(i)=-i$.
This lead to the system :
$$\begin{align}\begin{cases} \frac{a}{c}=-1\\ \frac{b}{d}=1 \\ \frac{ai+b}{ci+d}=-i \end{cases} &\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{ai+b}{-ai+b}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2 \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + ... | We have x= 2 (mod 3) and x= 4 (mod 5).
From the first x= 2+ 3i for some integer i.
From the second x= 4+ 5j for some integer j.
2+ 3i= 4+ 5j so 3i- 5j= 2.
Now use the "Chinese remainder theorem to solve that 'Diophantine equation'.
3 divides into 5 once with remainder 2: 5- 3= 2
2 divides into 3 once with remainder 1: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function. Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function.
|
Hint: In order to answer such questions the systematic approach provided by @coffeemath is very helpful, since it provides you a method to always find the odd and even part of a function $f$ regardless of its shape.
You want to split the function $f$ into the sum of an even function $g$ and and odd function $h$
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$ How many answers do this equation have?
$3^{2x}-34(15^{x-1})+5^{2x}=0$
My Attempt:$3^{2x}+5^{2x}=34(15^{x-1})$.Now what to do?
| $$3^{2x}-34(15^{x-1})+5^{2x}=0$$
$$3^{2x}-\frac{34}{15}(15^{x})+5^{2x}=0$$
$$15\cdot3^{2x}-34\cdot 3^x \cdot 5^x+15\cdot5^{2x}=0$$
$$15\cdot\left(\frac{3}{5}\right)^{2x}-34\cdot \left(\frac{3}{5}\right)^{x}+15=0$$
$\left(\frac{3}{5}\right)^{x}=t$
$$15\cdot t^{2}-34\cdot t+15=0$$
$t=\frac35$ or $t=\frac53$
$x=1$ or $x=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How can I use Bessel's equation to solve the Lengthening Pendulum differential equation? Taking a small extract of this previous bounty question of mine:
It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2... | Hint: In the pendulum example the variable $l$ plays the role of $x$ and $\theta$ the role of $y$.
Keeping this in mind and comparing
\begin{align*}
\frac{d^2\theta}{dl^2}+\frac{2}{l}\frac{d\theta}{dl}+\frac{g}{v^2}\frac{\theta}{l}=0
\end{align*}
with
\begin{align*}
\frac{d^2\theta}{dl^2}+\left(\frac{1-2a}{l}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$ for $x,y,z \geq 0$ This inequality is wrong - see the accepted answer (it appears there is no general inequality for these two expressions).
On the left we have harmonic mean of pairwise ge... | Wrong Hint:
$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)^{-1} \leq \frac{2}{3} $
$\to 2.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \leq \frac{2}{3}.\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3\sqrt{xyz}}=\frac{2}{9}.\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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.