Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Triangle and Maxium value Given any triangle ABC with $a \ge b \ge c$ such that $\frac{a^3+b^3+c^3}{\sin^3(A)+\sin^3(B)+\sin^3(C)}=7$, what is the maximum value of $a$?
| since
$$\sin^3{A}+\sin^3{B}+\sin^3{C}=(8R^3)^{-1}(a^3+b^3+c^3)$$
so
$$(8R^3)^{-}=\dfrac{1}{7}\Longrightarrow R^{-1}=\sqrt[3]{\dfrac{1}{56}}$$
so
$$a=2R\sin{A}\le 2R=\sqrt[3]{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{... | Or, after the first substitution in my other answer, substitute $u=\sinh \theta.$ Then the integral becomes:
$$2\int \frac{\sinh \theta \cosh \theta}{1+\sinh \theta +\cosh \theta} d \theta =
\int \frac{\sinh 2 \theta}{1 + e^\theta} d \theta.$$
The last integrand is a rational function of $e^\theta,$ so the integral o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596467",
"timestamp": "2023-03-29T00:00:00",
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Checking whether the number is composite Prove that $5^{125}-1$/ ($5^{25} - 1$) is composite
I have written $5^{125}-1$ as $(5^{25}-1)(5^{100}+5^{75}+5^{50}+5^{25}+1)$ but what should I do after this?
Sorry about earlier mistake in question ,
| Let $x = 5^{25}$.
$\begin{align}
5^{125}-1 &= x^5-1\\
&=(x^4 +x^3 +x^2 + x + 1)(x-1) \\
&= (x^4 + 9x^2 + 1 + 6x^3 + 6x + 2x^2 - 5x^3 - 10x^2 - 5x)(x - 1)\\
&= ((x^2 + 3x + 1)^2 - 5x(x + 1)^2)(x - 1)
\end{align}$
Put $x = 5^{25}$, just in the expression $5x$, you will get
$5^{125}-1=((x^2 + 3x + 1)^2 - (5^{13}(x + 1))^... | {
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Point on the graph of $y=\sqrt{4x+13}$ closest to $(5,0)$? Just did this question on an exam earlier today, I'm curious to see if I'm correct.
What point on the graph of $y=\sqrt{4x+13}$ is closest to $(5,0)$?
My answer: $(-1,3)$
| Let $P = (x,\sqrt{4x+13})$ be a generic point of your function, and pose $Q=(5,0)$.
We want to find the point $P$ such that the distance between $P$ and $Q$ is minimal. The distance is:
$$d(x) = \sqrt{(x-5)^2 + \left(\sqrt{4x+13} - 0 \right)^2} = $$
$$= \sqrt{(x-5)^2 + |4x+13|} $$
Minimize the distance is equivalent to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to derive this frequency response? Given this difference equation $y(k)$ ...
$$y(k) = \frac{1}{K^2} \sum_{m = k-K+1}^k \; \sum_{n = m-K+1}^m x(n) - \frac{1}{L^2} \sum_{m = k-L+1}^k \; \sum_{n = m-L+1}^m x(n)$$
... how does one derive this frequency response $H(f)$?
$$H(f) = \frac{1}{K^2} \left( \frac{\sin{\pi f K}}... | Note that $$\left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right]$$ is
$$\left[ z^{-K+1} + z^{-K+2} + \cdots + z^{-1} + 1 \right]^2$$
which is
$$\left[ \frac{(z^{-K}-1)}{(z^{-1}-1)}\right]^2$$
$$=\left[ \frac{(z^{-K/2}-z^{K/2})/z^{K/2}}{(z^{-1/2}-z^{1/2})/z^{1/2}}\right]^2$$
$$=\left[ \frac{(z^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fibonacci proof question: $f_{n+1}f_{n-1}-f_n^2=(-1)^n$ Show that
$$f_{n+1}f_{n-1}-f_n^2=(-1)^n$$
when $n$ is a positive integer and $f_n$ is the $n$th Fibonacci number.
| *
*For a basis $n=1$ the equality holds, as
$$f_{k+1}f_{k-1}-f_k^2=f_2f_0-f_1^2=1 \cdot 0 - 1^2=-1=(-1)^1.$$
*Assume the equality holds for $n=k$. Then we may assume that
$$f_{k+1}f_{k-1}-f_k^2=(-1)^k.$$
For the final inductive step, we wish to prove that $$f_{k+2}f_k-f_{k+1}^2=(-1)^{k+1}.$$
*We begin with the lef... | {
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Show by induction : $n^7-n$ is a multiple of 7 I have to prove this : "$n^7-n$ is a multiple of 7". This is what I have done this so far :
$P(n):n^7-n$
On putting $n=1,$
$P(1):1^7-1=0$, which is a multiple of 7.
So, $P(1)$ is true.
Let $P(k)$ be a multiple of 7.
So, $k^7-k$ is a multiple of 7.
So, $k^7-k=7m$, where $m... | Without induction:
$$n^7-n=n(n^6-1)=n(n^2-1)(n^4+n^2+1)$$
Now, $$n^4+n^2+1-(n^2-2^2)(n^2-3^2)=14n^2-35\equiv0\pmod7$$
So, $$n^7-n\equiv n(n-1)(n+1)(n-2)(n+2)(n-3)(n+3)\pmod7$$
The right hand side is the product of $7$ consecutive integers, hence divisible by $7$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Evaluate: $\int \frac{1}{x^7-x}\ \mathrm{d}x$ Evaluate:
$$\int \frac{1}{x^7-x}\ \mathrm{d}x$$
My approach to this question:
$$\int \frac{1}{x^7-x}\ \mathrm{d}x = \int \frac{1}{x(x^6-1)}\ \mathrm{d}x$$
$$\int \frac{1}{x(x^6-1)}\ \mathrm{d}x = \int \frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)}\ \mathrm{d}x$$
$$\frac{1}{x(x-1)(... | Yes there is. Note that:
$$\frac{1}{x^7-x} =\frac{x^5}{x^6-1}-\frac{1}{x}$$
Now use $t = x^6-1$ for the first integral, and...
Note how this trick works for any integral of the form $(x^n-x)^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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AP Calculus Derivative Slope $xy^2-x^3y = 2$
$y' = \frac{3x^2y-y^2}{2xy-x^3}$
find the x-coordinate of each point on the curve where the tangent line is horizontal.
I know that $3x^2y-y^2 = 0 \Rightarrow$ then $y=3x^2.$
If I plug y into the original I get
$x(3x^2)^2-x^3(3x^2)=2 \Rightarrow x(9x^4)-3x^5 = 2 \Rightarrow... | Yes, you've found the unique solution: $\quad x = \left(\frac 13\right)^{1/5}$.
Note: It also happens to be true that at $y = 0$, then $y' = 0$. But it turns out that $y = 0$ does not satisfy the original equation, and so does not lie on the curve. Hence, we have only the solution that you found.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor series of $(1+x)\ln(1+x)$ in $x=0$ How to determine the Taylor series of $(1+x)\ln(1+x)$ in $x=0$?
My idea is finding the second derivative of the expression, which is $\frac{1}{1+x}$.
The Taylor series of this expression is $1-x+x^2-x^3$ and so on. If I integrate then this series I get $\frac{x^2}{2}-\frac{x^3... | Another approach, if you already know the expansion $\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4}+\cdots$, is to literally multiply the series by $(1+x)$ and combine terms:
\begin{eqnarray*}
(1+x)\ln(1+x) & = & 1\cdot(x-\tfrac{x^2}{2}+\tfrac{x^3}{3}\cdots)+x\cdot(x-\tfrac{x^2}{2}+\tfrac{x^3}{3}\cdots)\\
& =... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{2 \sin x}{3}+\frac{\tan x}{3} > x$ for $x \in (0, \frac{\pi}{2})$ Please, help
Prove $\frac{2 \sin x}{3}+\frac{\tan x}{3} > x$
$x \in (0, \frac{\pi}{2})$
| Letting $f(x)=2\sin x+\tan x-3x$, then differentiate it. You'll get the answer.
$$f^{\prime}(x)=2\cos x+\frac{1}{\cos^2 x}-3=\frac{2\cos^3 x-3\cos^2 x+1}{\cos^2 x}=\frac{(\cos x-1)^2(2\cos x+1)}{\cos^2 x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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if $3^3 2^2 \ | a^2$ then $3^2 2 \ | a $ where a is integer if $3^3 2^2 \ | a^2$ then $3^2 2 \ |a $ where a is integer.
I just cannot see it. please explain this trivial remark.
| I answered a generalization
of this here:
If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?
Here is the question
and my answer:
Given $n$, what is the largest $m$
such that $m \mid a$ for all $a$ with $n \mid a^2$?
Let
$n = \prod p_i^{n_i}
$,
$m = \prod p_i^{m_i}
$,
and
$a
=\prod p_i^{a_i}
$.
$n | a^2$
me... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve patterns Let's take an example of the pattern
$T_1 = 1$
$T_2 = 6$ // $T_2 = 1 + 8 - 3 = 6$
$T_3 = 15$ // $T_3 = 6 + 12 - 3 = 15$
$T_4 = 28$ // $T_4 = 15 + 16 - 3 = 28$
If you noticed, the pattern is $T_n = T_{n-1} + 4n - 3$
So, is there a way to solve this without recursion?
| We solve:
$$T(n) = T(n-1) + 4n - 3$$
Thus
$$T(n) - T(n-1) = 4n - 3$$
since the difference of $$T(n) - T(n-1)$$ is linear ($4n-3$ is a linear term): the original function $T(n)$ must be quadratic.
Why?
Consider an arbitrary quadratic
$$H(n) = an^2 + bn + c $$
$$H(n) - H(n-1) = an^2 + bn + c - a(n-1)^2 -b(n-1) - c = $$
$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}... | Observe that
$(x - 1)f'(x) = (x - 1)(\sum_0^6 x^i) = x^7 - 1, \tag{1}$
and that the polynomial $x^7 - 1$ has exactly one real zero, $x = 1$. Thus the zeroes of $f'(x)$ must be the remaining zeroes of $x^7 - 1$, which are the six complex $7$-th roots of unity $e^{2 \pi i / 7}$ for $1 \le i \le 6$. This shows that $f'... | {
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"url": "https://math.stackexchange.com/questions/617030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}... | Since
$$
1+n^2 \le k+n^2\le n+n^2 \quad \forall k \in \{1,2,\ldots,n\},
$$
it follows that
$$
\frac{n(n+1)}{2(n+n^2)}=\sum_{k=1}^n\frac{k}{n+n^2}\le \sum_{k=1}^n\frac{k}{k+n^2}\le\sum_{k=1}^n\frac{k}{1+n^2}=\frac{n(n+1)}{2(1+n^2)} \quad \forall n \ge 1.
$$
Thus
$$
\lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I ca... | It looks like you are applying fraction addition rules to fraction multiplication:
$$\frac ab+\frac cd=\frac {a\cdot d+b\cdot c}{b\cdot d}$$
Instead you should use:
$$\frac ab\cdot \frac cd=\frac {a\cdot c}{b\cdot d}$$
So we have
$$\frac{x^2+16y^2}{x} \cdot \frac{x^2+4xy}{x-4y}=\frac {(x^2+16y^2)(x+4y)}{x-4y}=\frac {x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get the simplest form of this radical expression: $3\sqrt[3]{2a} - 6\sqrt[3]{2a}$. How to get the simplest form of this radical expression:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Here is my work:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Since the radicands are the same, we just add the coefficients.
$$-3\sqrt[3]{2a} \sqrt... | We can factor, as an alternative, to get the same result:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a} = 3\sqrt[3]{2a}(1 - 2) = -3\sqrt[3]{2a}$$
(I don't see any need to write: $\;-3\sqrt[3]{2a} = -3\sqrt[3]{2} \sqrt[3]{a})$
Note that $-3\sqrt[3]{2a}\sqrt[3]{2a} = -3\sqrt[3]{4a^2} \neq -3\sqrt[3]{2a} $
| {
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How find this $f^{(4)}(0)$ let
$$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... | According to the general Leibniz rule, $\displaystyle(f\cdot g)^{(n)}=\sum_{k=0}^n{n\choose k}f^{(k)}g^{(n-k)}$ . In this case, $n=4$ , $f(x)$ is $e^x$, and $g(x)=\dfrac1{1-\sin x}$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a,b$ are roots of $x^2-3cx-8d = 0$ and $c,d$ are roots of $x^2-3ax-8b = 0$. Then $a+b+c+d =$ (1) If $a,b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation
$x^2-10ax-11b=0$. Then the value of $\displaystyle \sqrt{\frac{a+b+c+d}{10}}=,$ where $a,b,c,d$ are distinct real numbers.
(2... | Is anything more given? If not there are 4 answes, and two are integers.
Here is one answer: a=b=c=d=0
The other answer: a=c =-11, b=d=-99
There are two other irrational solutioms for $a$,$b$, $c$,$d$ with $z=11$
Can you provide more info? I will edit this answer based on what you tell me
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with $\dfrac{dr}{d\theta}$ for $r=2\sec\theta\csc\theta$
Find $\dfrac{dr}{d\theta}$ for $r=2\sec\theta\csc\theta$.
My try:
$$r^{\prime}=2[\sec\theta(-\csc\theta\cot\theta)+\csc\theta(\sec\theta\tan\theta)]$$
$$r^{\prime}=2\left(\frac{1}{\sin\theta}(-\frac{1}{\cos\theta}\frac{\cos\theta}{\sin\theta})+\frac{1}{\... | Your initial derivative is correct: $$r^{\prime}=2[\sec\theta(-\csc\theta\cot\theta)+\csc\theta(\sec\theta\tan\theta)]$$
However, note that:
$$\csc x = \dfrac{1}{\sin x} \;\text{ and }\;\sec x = \dfrac 1{\cos x}$$
$$\begin{align} r' &= 2[\sec\theta(-\csc\theta\cot\theta)+\csc\theta(\sec\theta\tan\theta)] \\ \\& = 2\l... | {
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"timestamp": "2023-03-29T00:00:00",
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Max value of $n$ for which $3^n\mid(80!)!$
Calculate the max value of $n$ for which $(80!)!$ is divisible by $3^n$.
My Attempt:
The exponent of prime factor $p$ in $(n!)$ is given as
$$
v_p(n!) = \left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor \frac{n}{p^2}\right\rfloor+\left\lfloor \frac{n}{p^3}\right\rfloor+\l... | The exact answer will be $ v_3 ( (80!)!) = \lfloor \frac{80!}{3} \rfloor + \lfloor \frac{80!}{3^2} \rfloor + \ldots $.
If we ignore the floor function, and take the sum of the geometric progression to infinity, the answer would simply be
$$v_3 ((80!)!) \approx \frac{ 80!}{3} + \frac{80!}{3^2} + \ldots = 80! \times \fr... | {
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Is this epsilon-delta proof valid? Prove using $\epsilon- \delta$ that $ \displaystyle \lim_{x \to -4} \frac { x^2 + 6x + 8 }{x + 4} = - 2 $.
Here's a proposed proof:
For $\delta \leq 1$, i.e. $ | x + 4 | < 1 $ which guarantees $x < -1 $, one can argue:
$ \left| \dfrac { x^2 + 6x + 8 }{x + 4} + 2 \right| = \left|... | You have $x^2 + 6x + 8 = (x+2)(x + 4).$ Try factoring and canceling.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$ $$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$
where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
my try:since
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{... | Not a full answer, but hopefully helpful progress: rewriting
$$I_n=\sum_{k=1}^{n+1}\frac{H_{k}H_{n+2-k}}{k+1},$$
we recognize that
$$
\sum_{n=0}^\infty I_n x^{n+2} = \bigg( \sum_{k=1}^\infty \frac{H_k}{k+1} x^k \bigg) \bigg( \sum_{k=1}^\infty H_k x^k \bigg).
$$
Since $\sum_{j=1}^\infty \frac1j x^j = -\log(1-x)$ and $H_... | {
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"url": "https://math.stackexchange.com/questions/627936",
"timestamp": "2023-03-29T00:00:00",
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Solution of Trigonometric equations If we were to solve the trigonometric equation
$$\sqrt{13-18\tan x} = 6\tan x-3$$
by squaring both the sides, we would get two roots;
$\tan x = \frac{2}{3}$ and $\tan x=-\frac{1}{6}$
2/3 is okay, but when we substitute -1/6 in the equation and simp... | What you need to know is that
$$\sqrt{13-18\tan x}=6\tan x-3$$
is not the same as
$$13-18\tan x=(6\tan x-3)^2.$$
These are different equations.
Notice that
$$\sqrt{13-18\tan x}=6\tan x-3$$
$$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ 6\tan x-3\ge0$$
$$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ \tan x\ge\frac ... | {
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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 $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\m... | Fact I.
$$
\lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n^2}\right)^{\!n}=\mathrm{e}^a.
$$
Fact II. For every $a>0$, there exists a $b>0$, such that
$$
1+\frac{\ln a}{n}\le a^{1/n} \le 1+\frac{\ln a}{n}+\frac{b}{n^2}.
$$
Using the two facts:
$$
1+\frac{\ln a+\ln b+\ln c}{3n}\le\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\le
1+\... | {
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Jordan canonical forms determined by a minimal polynomial Find the Jordan canonical forms of all $9\times 9$ matrices over $\mathbb{C}$ with minimal polynomial $x^2(x-3)^3$.
My method: each factor of the minimal polynomial corresponds to a type of Jordan blocks with their maximal orders equal to the multiplicity of the... | You are correct except that without a convention about how blocks are ordered, you need to count permutations of blocks for each of your sixteen matrices. I count 892. Following is my work where each row in the table represents one of your matrices, and the number under a block is the frequency of that block in the ma... | {
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Simplify the expression : $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\cdots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ How to simplify the expression:
$\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$
I am not getting any clue how ... | We have to find the value of $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ .
$2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15}\theta) $=$2^{14} \tan(2^{14}\theta) +2^{15} \dfrac{1}{\tan(2^{15}\theta)}$ = $2^{14} \tan(2^{14}\theta) +2^{15} \dfrac{1-\tan... | {
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Let L be the line of intersection of the planes $cx + y + z = c$ and $x - cy + cz = -1$, where c is a real number.
*
*Find symmetric equations for $L$
*As the number $c$ varies, the line $L$ sweeps out a surface $S$. Find an equation for the curve of intersection of $S$ with the horizontal plane $z = t$ (the trace o... | The equations for $L$ that you got are correct as can be easily verified.
From this we can write
$x = -1 - 2 c \lambda $
$y = c + (c^2 - 1) \lambda $
$ z = t = c + (c^2 + 1) \lambda$
From the third equation, we get
$ \lambda = \dfrac{t - c }{c^2 + 1 } $
Substitute this in the first two equations
$ x = -1 - 2 c \dfrac{t... | {
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Finding $ \lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}} $ without power series. I have to find: $\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}$
"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac {\ln(\frac{\arctan x}{x})} {x^2}
$
L'Hôpital's rule once is not enough, second L'Hôpital's ru... |
"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac
{\ln(\frac{\arctan x}{x})} {x^2}$
L'Hôpital's rule once is not enough, second L'Hôpital's rule seems worse. any help?
By applying L'Hôpital's rule trice and simplifying in each step we obtain:
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\frac{\ln \left( \fra... | {
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Spherical symmetry math For spherical symmetry how the last four equations calculations is done? ccan you explain please?
For reference see the equations 44
| We have $\rho(x) = \bigl(\sum_i x_i^2\bigr)^{1/2}$, hence
\begin{align*}
\partial_i \rho(x) &= \frac 1{2(\sum_j x_j^2)^{1/2}}\cdot 2x_i = \frac{x_i}{\rho(x)}\\
\partial_i^2\rho(x) &= \frac{\rho^2(x) - x_i^2}{\rho^3(x)}
\end{align*}
As $S$ is spherically symmetric, the values of $S$ only depend on $x$'s distance to... | {
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value of the value expression F (x)=[1+sinx]+[2+sin (x/2)]+[3+sin (x/3)]+.........+[n+sin (x/n)] for x belongs to 0 to pi. Where []denotes greatest integer function.
I'd converted f(x) as
f (x)=1+2+3+4+5+........+n+sin x+sin (x/2)+sin (x/3)+........+sin (x/n)-fractional part (1+sinx)-fractional part (2+sinx/2)-fraction... | If x = $\frac{\pi}{2}$ , then $sin(x) = 1$
If x = $\pi$ , then sin($\frac{x}{2}) = 1 $
In these cases, one of the sine-terms is 1.
For all other x, $0\le sin(\frac{x}{n}) < 1$ for all n.
So the great-integer function is zero for all sine-terms except
for the term $sin(\frac{\pi}{2})=1$, for which it is 1.
So, for x = ... | {
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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:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\... | This is a comment that’s too long for the usual format. Let $g(x)=\dfrac{1}{1+x^2}$. Let us generalize the original question by putting for any $y \geq x \geq 0$,
$$
\begin{array}{lcl}
u_n(x,y)&=&\frac{1}{n}\sum_{k=1}^{n} g\left(x+\frac{k}{n}(y-x)\right) \\
f_n(x,y)&=& u_{n+1}(x,y)-u_n(x,y) \\
\end{array}
$$
Conjectu... | {
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"answer_id": 6
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On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$ How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$
I guess it is only $x=2$.
Please help.
| Hint: raise both sides to the sixth power.
| {
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Prove : $\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$ Problem :
Let $a,b\in \mathbb{Z}$ such that $a\neq 0,b\neq 0$ ; $\left | a \right |\leq 100,\left | b \right |\leq 100$.
Prove that:
$$\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$$
Thanks :)
P/s : I have no ideas about this problem ! :(
| If $a$ and $b$ have the same sign ($0$ has the same sign as any integer for this purpose), then you have
$$\lvert a\sqrt{2} + b\sqrt{3}\rvert = \lvert a\rvert \sqrt{2} + \lvert b\rvert \sqrt{3} \geqslant \sqrt{2} > \frac{1}{350}.$$
So suppose $a$ and $b$ have opposite sign. Then
$$\lvert a\sqrt{2} + b\sqrt{3}\rvert = \... | {
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot... | The straight way is as given by @AndréNicolas, cleverly exploiting remarkable identities.
Alternatively, as we know the sum, let $s$, and the product, let $p$, by Vieta's formulas, we know that the two numbers are the roots of
$$t^2-st+p=0.$$
Then
$$t^3=t\cdot t^2=st^2-pt=s(st-p)-pt=(s^2-p)t-sp$$ and the sum of the cub... | {
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Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere.
Hints are also welcome.
| This is well-know inequality, actually this is a inequality from Vacs if you are familiar with Art of Problem Solving. Here's the proof:
Let $z\le y \le x$, so we have:
$$(y-x)(y-z) \le 0 \implies y^2 - xy - zy + xz \le 0 \implies y^2 + nz \le xy + zy$$
Now for the Right Hand Side:
$$27(x^2y + y^2z + z^2x + xyz) = 27( ... | {
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Sum $\sum_{k=0}^{2013}2^ka_{k}$ let real sequence $a_{0},a_{1},a_{2},\cdots,a_{n}$,such
$$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}+... | oh sorry, i omit some details !
by the symmetric property, $a_{0}=-a_{2013}$$,$$a_{1}=-a_{2012}$$,......$
the hint is :
$(a_{2}+4/3a_{2}+......+4/3a_{3}+......)/(ta_{0})=-a_{1}/a_{0}(a_{1}-4/3a_{1}-......-4/3a_{2}-......)$
we assume that :
$t_{1}=\frac{(a_{2}+4/3a_{2}+......+4/3a_{3}+......)}{(-a_{1}-4/3a_{1}-......-4/... | {
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if $G_{4n+1}=2G_{2n+1}-G_{n},G_{4n+3}=3G_{2n+1}-2G_{n}$,Find $G_{n}=n?$
let sequence $\{G_{n}\}$ such
$$G_{1}=1,G_{3}=3,G_{2n}=G_{n}$$
$$G_{4n+1}=2G_{2n+1}-G_{n},G_{4n+3}=3G_{2n+1}-2G_{n}$$
If such $G_{n}=n$, then we said $n$ is 'good'.
How many 'good' numbers $n$, such that $n<2^{100}?$
My try:
since
$$\begin{eq... | This is sequence A030101 in the OEIS. That is, $G(n)$ is the number obtained by reversing the digits of $n$ when written base $2$, e.g. $G(25)=G(11001_2)=10011_2=19$.
This is easy to check: If $n=d_0+2d_1+\cdots+2^rd_r$, then
$$
\begin{align}
G(n)&=d_r+\cdots+2^rd_0\\
G(2n)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot0\\
G(2n+1)&... | {
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Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\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^... | $\sqrt{3\cos^2(x)-\sin(2x)}=-\sin(x)$
You square the both sides
$3\cos^2(x)-\sin(2x)=\sin^2(x)$
By using double angle identity of $\sin$ and dividing all the terms by $\cos^2(x)$
$\tan^2(x)+2\tan(x)-3=0$
Quadratic equation
$\tan(x)=1\text{ or }-3$
Find all the solutions in the interval
$x=5.03\text{rad}$
don't forge... | {
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Solution for $4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$ I'm trying to get a solution for:
$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$
My main problem is that I don't know how to combine this potencys!
Ive also thought about another function that would bring me same difficulties:
$6^x=36*9.75^{x-2}$
What am I supposed to do?
| $$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$$
$$4^{2x+1}-4^{2x+3}=3^{3x+1}-3^{3x+2}$$
$$4\cdot4^{2x}-4^34^{2x}=3\cdot3^{3x}-3^23^{3x}$$
$$60\cdot4^{2x}=6\cdot3^{3x}$$
$$10\cdot4^{2x}=3^{3x}$$
$$10\cdot16^{x}=27^{x}$$
$$10=(27/16)^{x}$$
$$\log_{10} 10=\log_{10} (27/16)^{x}$$
$$1=x\log_{10}(27/16)$$
$$x=\frac{1}{\log_{10}27-\l... | {
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proof by maths induction not sure how to prove this:
for all positive intergers prove:
\begin{equation}
1+2(2)+3(2^2)+...+n(2^{n-1})=(n-1)(2^n)+1
\end{equation}
heres my try:
prove $n=1$ :
\begin{equation}
1=1
\end{equation}
assume true for $n=k$:
\begin{equation}
1+2(2)+3(2^2)+...+k(2^{k-1})=(k-1)(2^k)+1
\end{equati... | Let $S(n)=\sum_{i=1}^n i2^{i-1}$ Your hypothesis is that $S(n)=(n-1)2^n+1$ As you say, it works for $n=1$. Now assume it is true for $k$ and evaluate $$S(k+1)=S(k)+(k+1)2^k\\=(k-1)2^k+1+(k+1)2^k\\=2k2^k-2^k+1+2^k\\=[(k+1)-1]2^{k+1}+1$$
| {
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Find this limit without using L'Hospital's rule I have to find this limit without using l'Hôspital's rule:
$$\lim_{x\to 0} \frac{\alpha \sin \beta x - \beta \sin \alpha x}{x^2 \sin \alpha x}$$
Using L'Hôspital's rule gives:
$$\frac{\beta}{6(\alpha^2 - \beta^2)}$$
I am stuck where to begin without using the rule.
| Using the Taylor series
$$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$$
the numerator is
\begin{align*}
\alpha \left(\beta x - \frac{(\beta x)^3}{3!} + O(x^5)\right) - \beta \left(\alpha x - \frac{(\alpha x)^3}{3!} + O(x^5)\right) = \frac{\beta \alpha^3 - \alpha \beta^3}{6} x^3 + O(x^5)
\end{align*}
Then the f... | {
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Area of a triangle using vectors I have to find the area of a triangle whose vertices have coordinates
O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$
I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:
area $= \frac{1}{2}ab\sin... | The correct formula is $\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \cos{\theta} $
So what you really have is $\cos{\theta} = \cfrac{-1}{\sqrt{3}}$
Therefore $$\sin{\theta} = \sqrt{1 - \cos^2{\theta}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} ... | {
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How find this $a_{n},b_{n}$ let sequence $\{a_{n}\},\{b_{n}\}$ such
$$a_{1}=1,b_{1}=3$$
and
$$\begin{cases}
a_{n+1}=2+\dfrac{27a_{n}}{9a^2_{n}+4b^2_{n}}\\
b_{n+1}=\dfrac{27b_{n}}{9a^2_{n}+4b^2_{n}}
\end{cases}$$
Find $a_{n},b_{n}$
My idea: since
$$\dfrac{a_{n+1}-2}{b_{n+1}}=\dfrac{a_{n}}{b_{n}}$$
then I can't.Thank yo... | As stated in my previous comments, by setting $z_n=3a_n+2i b_n$ we get:
$$ z_{n+1} = 6+\frac{27}{\overline{z_n}}\tag{1} $$
with $z_1 = 3+6i$. By setting $z_n = 3\sqrt{3}\, w_n$ we get:
$$w_{n+1} = \frac{2}{\sqrt{3}}+\frac{1}{\overline{w_n}}\tag{2} $$
with $w_1=\frac{1}{\sqrt{3}}(1+2i)$. $(2)$ gives that the sequence $\... | {
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How can solve this differential equation (second equation)? How can I solve this differential equation?
$$
\frac{1}{h}\left(\frac{a}{\delta} - \frac{b}{\delta} - \frac{c}{E\delta^4}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left[A\,\frac{\mathrm{d}\delta}{\mathrm{d}x} \left(\frac{1}{\delta} + \frac{3B}{\delta^2}\right)\... | Let $y=\dfrac{d\delta}{dx}$ ,
Then $\dfrac{1}{h}\left(\dfrac{a}{\delta}-\dfrac{b}{\delta}-\dfrac{c}{E\delta^4}\right)=\dfrac{d}{d\delta}\left(Ay\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)\right)\dfrac{d\delta}{dx}$
$\dfrac{1}{h}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)=A\dfrac{d}{d\delta}\left(\left... | {
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Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s e... | Consider evaluation of sums of following form:
$$S_p\stackrel{def}{=}\sum_{k=1}^n\lfloor\sqrt[p]{k}\rfloor$$
where $p$ is a positive integer $\ge 2$. In the special case $p = 2$, this reduces to the sum we want to calculate.
Let $a = \lfloor \sqrt[p]{n} \rfloor$ and take a sufficiently small $\epsilon > 0$ such that
$$... | {
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"url": "https://math.stackexchange.com/questions/669460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 2
} |
Limit of $\sqrt[x]{\frac{\tan x}{x}}$ as $x \to 0$ I am trying to calculate the Limit
$$\lim_{x \to 0} \sqrt[x]{\frac{\tan x}{x}}$$
Wolfram Alpha says it's $1$. But I get
$$\lim_{x \to 0} \sqrt[x]{\frac{\tan x}{x}}$$
$$= \exp \lim_{x \to 0} \ln \left(\left(\frac{\tan x}{x}\right)^{1/x}\right)$$
$$= \exp \lim_{x \to 0} ... | If $L$ is the limit to be evaluated then we have $$\begin{aligned}\log L &= \log\left(\lim_{x \to 0}\sqrt[x]{\frac{\tan x}{x}}\right)\\
&= \lim_{x \to 0}\log\left(\sqrt[x]{\frac{\tan x}{x}}\right)\text{ (by continuity of }\log)\\
&= \lim_{x \to 0}\dfrac{\log\left(\dfrac{\tan x}{x}\right)}{x}\\
&= \lim_{x \to 0}\dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integral of complex questions? $$\int_0^{\pi/4} \frac {\sin x + \cos x}{\sin^4x+\cos^2x}dx$$
$$\int e^x\cot x(\csc x-1)dx$$
These two integrals are impossible to find. If anyone knows how to integrate them please help me.
I am not able to differentiate it so I tried Integrals on wolfram.com and I got this answer :
| As intimated above, split the first integral up as follows:
$$\begin{align}\int_0^{\pi/4} dx \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^2{x}} &= \underbrace{\int_0^{\pi/4} dx \frac{\sin{x}}{1-\cos^2{x}+\cos^4{x}}}_{u=\cos{x}} + \underbrace{\int_0^{\pi/4} dx \frac{\cos{x}}{1-\sin^2{x}+\sin^4{x}}}_{u=\sin{x}}\\ &= \int_{1/\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a... | You know their sum, their sum of products taken by two, and their product. Use Vieta's identities and the cubic formula. (Another approach would be by employing Newton's identities).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Trailing zeroes in factorials: are there any excluded values divisible by 5 other than $5$ and $30$? I've discovered that when this algorithm for counting zeroes on the end of $n!$ is applied to some $n\in\Bbb{N}$: $$f(n)=\sum_{k=1}^{k:n/5^k\le1}\left\lfloor\frac{n}{5^k}\right\rfloor\notin\{5,30\}$$ Are the any other n... | For $k \geqslant 2$, we have
$$f(5^k) = \frac{5^k-1}{5-1} \equiv 1 \pmod{5},$$
so the multiple
$$\frac{5^{k-1}-1}{4}\cdot 5$$
of $5$ is skipped. That produces the sequence $5,5+5^2 = 30, 5+5^2+5^3 = 155, 155+5^4 = 780,\dotsc$ of skipped multiples of $5$.
There are more, for example $f(350) = 70+14+2 = 86$, so $85$ is s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Hessian of Morse function on $S^{n}$ mistake I am trying to get that $f(x_{0},...,x_{n+1})=x_{n+1}$ has $index_{(0,...,0,1)}=n$
Can you find my mistake or post a partial solution?
My attempt
I evaluate df using inverse of stereographic proj. (omitting (0,...,0,-1))
$d f(\phi^{-1})=(\dfrac{4x_{1}^{2}\sum_{i=1}^{n}x_{i}... | In the first entry, you compute the derivative of a quotient
$$
\dfrac{f'g - f g'}{g^2}
$$
where $f = 4 x_1^2 \sum_i x_i^2$ and $g = (1 + \sum x_i^2)^2$.
The $f g'$ term should be
$$
\left(4 x_1^2 \sum_i x_i^2\right) \left( 2 (1 + \sum_i x_i^2) 2 x_1 \right) \\
16 x_1^3 (\sum_i x_i^2 )(1 + \sum_i x_i^2).
$$
Presuma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$ $a;b;c\in \mathbb{R}^+$ such that $a+b+c=6$.
Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$
Thanks :)
I have no ideas about this problem ! :(
| You can use Lagrange multipliers as @ Martín-Blas Pérez Pinilla said. This is an alternative solution.
Using AM-GM, we have $6=a+b+c\geq 3\sqrt[3]{abc}\iff \frac1{\sqrt[3]{abc}}\geq \frac12$.
On the other hands, we have
\begin{equation}
1+\frac1{a^3}=\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Infinite Sum with Combination I am trying to figure out what the following sum converges to:
$$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n),\qquad\qquad0<x<1$$
An answer would be great, but if you have an explanation, that'd be better!
| Since
$$
7\binom{7+n}{n}=(7+n)\binom{6+n}{n}
$$
and using negative binomial coefficients,
$$
\binom{k+n}{n}=(-1)^n\binom{-k-1}{n}
$$
we get
$$
\begin{align}
\sum_{n=0}^\infty\binom{6+n}{n}x^n(6+n)
&=\sum_{n=0}^\infty\binom{6+n}{n}x^n(7+n)-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{7+n}{n}x^n-\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve the equation $285x \equiv 177 \pmod{924}$ using continued fraction
Solve the equation $285x \equiv 177 \pmod{924}$ using continued fraction
My attempt(using Wikipedia notion):
Continued fraction form for $\frac{924}{285}$ is $[3;4,6,1,9]=[q_1;q_2,q_3,q_4,q_5]$
$\frac{924}{285}=\frac{h_n}{k_n}$
We know that $h_n... | $285x≡177\pmod{924}$ or $285x+924y=177$ can be solved using the extended euclidean algorithm for the gcd of $285$ and $924$.
The remainder sequence of the euclidean algorithm starts with $r_0=a=924$, $r_1=b=285$ and for the extended variant the Bezout factor sequence for the Bezout identity
$$v_k b≡r_k\;\pmod a\qquad (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$ Prove that $$\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$$
My try
*
*$n=0$ then ... | $\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{d}{dx}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{\sin x}{x}\right)$
$=\frac{d}{dx}\frac{1}{x^{n}}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy$ (inductive assumption)
$=\frac{-n}{x^{n+1}}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy+\frac{1}{x^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $x+y+z=xyz$, find $\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ I found this question in a maths worksheet of trigonometry (kinda odd, right?), but I dont know how to figure it out.
If $\displaystyle x+y+z=xyz$, find $\displaystyle\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2... | We can not make $A+B+C=\pi$
as $\displaystyle \tan(A+B+C)=\frac{\sum\tan A-\tan A\tan B\tan C}{1-\sum \tan A\tan B}$
$\displaystyle\sum\tan A=\tan A\tan B\tan C\implies \tan(A+B+C)=0\implies A+B+C=n\pi$ where $n$ is any integer [Clearly, you have taken a special value$(1)$ of $n$]
Now for any integer $m,$ $\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/691180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$
I could not approach the p... | Assuming
$$\frac a{b-c}+\frac b{c-a}+\frac c{a-b}=0$$
we also have
$$\frac a{(b-c)^2}+\frac b{(c-a)(b-c)}+\frac c{(a-b)(b-c)}=0$$
as well as
$$\frac a{(b-c)(a-b)}+\frac b{(c-a)(a-b)}+\frac c{(a-b)^2}=0$$
and
$$\frac a{(b-c)(c-a)}+\frac b{(c-a)^2}+\frac c{(a-b)(c-a)}=0$$
These three sum together as
$$\frac a{(b-c)^2}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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solving $2\cosh2x = 13\cosh x - 12$ I've been asked to solve:
$2\cosh2x = 13\cosh x - 12$
I showed earlier in the question that $\cosh2x = 2\cosh^2x -1$
So I can say that:
$2(2\cosh^2x -1) = 13\cosh x - 12$
$\therefore 4\cosh^2x -13\cosh x + 10 = 0$
$\therefore \cosh x = \frac{5}{4}$ or $\cosh x = 2$
From $\cosh x = \... | You are correct: the equation has four solutions. So either the official answer is wrong, or you have mis-copied the question. Perhaps it only asked for positive solutions? Or perhaps the official answer was $x=\pm\ln 2$ and $x=\pm\ln(2+\sqrt 3)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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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... | First, we transform the question $x+y+z=10$ for every positive integer to $a+b+c=7$ for every non-negative integer.
We look at a more general question of finding the number of solutions in non-negative integers to the equation $ a + b + c = n $. Since the value of $a$ can be any non-negative integer $0,1,2,3, \ldots, i... | {
"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": 7
} |
Compute limit of sequence Let $(x_n)$ be real sequences such that $x_{1}=\dfrac{1}{3}, x_{2n}=\dfrac{1}{3}x_{2n-1}, x_{2n+1}=\dfrac{1}{3}+x_{2n}, n=1,2,\cdots $.
Compute $$\lim_{x \to \infty} \sup x_{n} \text{ and } \lim_{x \to \infty} \inf x_{n}. $$
| $$
x_{2n}=x_{2n-1}/3=(1/3+x_{2n-2})/3=x_{2n-2}/3+1/9
$$
$$
x{2n+1}=1/3+x_{2n}=1/3+x_{2n-1}/3
$$
and we have $x_1=1/3$, we can add $x_0=0$, which is ok by the definition
then sovle the recursive sequence, we get
$$
x_{2n} = \frac{1}{6}(1-\frac{1}{3^n})
$$
$$
x_{2n+1} = \frac{1}{2} - \frac{1}{6}\frac{1}{3^n}
$$
so
$$
\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/700283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x). Problem :
If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).
My approach :
The given equation can be written as $$(x-y)f(x+y) -(x+y)f(x-y) =4xy(x-y)(x+y)$$
$$\Rightarrow \frac{(x-y)f(x+y)}{(x-y)(x+y)} -\frac{(x+y... | $\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy=(x+y)^2-(x-y)^2$
or, $\frac{f(x+y)}{x+y}-(x+y)^2=\frac{f(x-y)}{x-y}-(x-y)^2$
make the substitutions, $x\to \frac{x+y}{2}$ and $y \to \frac{x-y}{2}$
Then the above becomes $\frac{f(x)}{x}-(x)^2=\frac{f(y)}{y}-(y)^2=k(say)$
There you have $f(x)=x^3+kx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/702695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:
Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$
My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!
| $\dfrac{\sec \theta \cdot \sin\theta}{\tan\theta+\cot\theta}=\dfrac{\sec \theta \cdot \sin\theta}{\tan\theta+\frac{1}{\tan\theta}}$
$= \dfrac{\sec \theta \cdot \sin\theta\cdot \tan\theta}{\tan^2\theta+1}$
$=\dfrac{\sec\theta\cdot\sin\theta\cdot\frac{\sin\theta}{\cos\theta}}{\sec^2\theta}$
$=\dfrac{\sin\theta\cdot\sin\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How prove this inequality $\frac{a^2+bc}{3+a}+\frac{b^2+ca}{3+b}+\frac{c^2+ab}{3+c}\le \frac{7}{4}$ let $a,b,c\ge 0$,show that
$$\dfrac{a^2+bc}{2a+b+c}+\dfrac{b^2+ca}{2b+c+a}+\dfrac{c^2+ab}{2c+a+b}\le\dfrac{7}{12}(a+b+c)$$
My idea: Without loss of generality,we Assmue that
$$a+b+c=3$$
then we only prove this
$$\dfrac{... | After full expanding we need to prove that
$$\sum_{cyc}(2a^4+3a^3b+3a^3c-10a^2b^2+66a^2bc)\geq0$$ or
$$2\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+5\sum_{cyc}ab(a-b)^2+64abc(a+b+c)\geq0,$$
which is true by Schur.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/711943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How prove this inequality $3b+8c+abc\le 12$ if $a^2+4b^2+9c^2=14$ let $a,b,c>0$ and such $$a^2+4b^2+9c^2=14$$show that
$$3b+8c+abc\le 12$$
My idea: since
\begin{align*}3b+8c+abc&=3b+c(8+ab)=3b+\dfrac{1}{9}\cdot 9c(8+ab)\le 3b+\dfrac{1}{9}\cdot\dfrac{1}{4}[81c^2+(ab+8)^2]\\
&=3b+\dfrac{1}{36}[126-9a^2-36b^2+a^2b^2+16ab+... | $\dfrac{b^2+1}{2}\ge b \implies \dfrac32(b^2+1)\ge 3b$
$\dfrac{c^2+1}{2}\ge c \implies 4(c^2+1)\ge 8c$
Adding these two equations,
$4c^2+\dfrac32b^2+\dfrac{11}{2}\ge 3b+4c$
Now, $12-\dfrac{11}{2}=\dfrac{13}{2}$
We know, $7>\dfrac{13}{2}$ and $7=\dfrac{(a^2+4b^2+9c^2)}{2}$
So, if we can now prove that $\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How would you determine the transformation matrix? Suppose there exists a linear transformation $T$ where $T: \mathbb{R^3} \to \mathbb{R^5}$ and $T(\textbf{x}) = \text{A} \textbf{x}$. Given
$$ \text{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1\\ 2\\ 1 \\3 \end{pmatrix} ,
\quad
\text{A} \begin{... |
Isn't this the same as finding the change-of-coordinates matrix when
making a change of basis?
Totally.
Let $P = \begin{pmatrix} 1 & 1& 0 \\ 1 &2 & 1 \\ 1 & 3& 1\end{pmatrix}$
It's easy to see that $P\in GL_3(\mathbb R)$.
Your equalities can be rewrite:
$$AP= \begin{pmatrix} 1&1&2 \\ 1&2&3\\ 2&2&4\\ 1&2&3 \\3 &-1&... | {
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"url": "https://math.stackexchange.com/questions/713329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation of a series with certain property The sequence $\{a_n\}$ has the property that $a_1 + a_2+\cdots+a_n = n ^3~~\forall n$. Compute the value of $\frac{1}{a_2} + \frac{1}{a_3}+\ldots+ \frac{1}{a_{2014-1}}$.
I know that I have to somehow a arrange the denominators in such a order that I could take advantage of t... | First observe that
$$
a_n=(a_1+\cdots+a_{n-1}+a_n)-(a_1+\cdots+a_{n-1})=n^3-(n-1)^3=3n^2-3n+1.
$$
Thus
\begin{align}
\frac{1}{a_2-1}+\cdots+\frac{1}{a_{2014}-1}&=\frac{1}{3 (2^2-2)}+\cdots+\frac{1}{3 (2014^2-2014)}=\frac{1}{3}\sum_{k=2}^{2014}\frac{1}{k(k-1)}\\ &=\frac{1}{3}
\sum_{k=2}^{2014}\left(\frac{1}{k-1}-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$a+b\le 1$$ Find the minimum of
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$
My try:
I can find this minimum，use Holder inequality
$$(\dfrac{1}{a^2}+\dfrac{1}{b^2})(a+b)^2\ge (1+1)^3=8$$
But
$$ab\le \dfrac{(a+b)^2}{4}\le\dfrac{1}{4}$$
so for
$$a... | Here care must be taken while using AM-GM so that the constraint and equality condition is not violated. Hence rewrite the objective as:
$$\left(\frac12 ab+ \frac12 ab+\frac{1}{32a^2}+\frac{1}{32b^2}\right)+ \frac{31}{32}\left( \frac1{a^2}+\frac1{b^2}\right)$$
The first part is a sum of four terms with constant produc... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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solving complex numbers with powers algebraically Find algebraically the value of :$\left(2^{0.5} + 6^{0.5} - \left( 2^{0.5} - 6^{0.5} \right)i \right)^4$
Below are my works
I try to simplify inside. but i found that i can't add $2^{0.5}$ and $6^{0.5}$ together.
| Simplify by manipulating the inner expression:
$$\begin{align}\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{6})i &= (\sqrt{2} + \sqrt6i) + (\sqrt{6} - \sqrt2i)\\
&= (\sqrt{2} + \sqrt6i) - (\sqrt{2} + \sqrt6i)i
\\&= (\sqrt{2} + \sqrt6i)(1 - i)\end{align}$$
Now, let $$\begin{align}z &= (\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find constants $a$ and $b$ such that $ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$
Find constants $a$ and $b$ such that $$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$$
First,$a$ should be positive to make sure the limit is meaningful as $x \to 0^-$ .
Then I ch... | Let's substitute $t\mapsto tx$:
$$
\begin{align}
\lim_{x\to0}\frac1{bx-\sin(x)}\int_0^x\frac{t^2\,\mathrm{d}t}{\sqrt{a+t}}
&=\lim_{x\to0}\frac{x^3}{bx-\sin(x)}\color{#00A000}{\int_0^1\frac{t^2\,\mathrm{d}t}{\sqrt{a+tx}}}\\
&=\color{#00A000}{\frac1{3\sqrt{a}}}\lim_{x\to0}\frac{x^3}{bx-\sin(x)}
\end{align}
$$
If $b\ne1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I... | Since $24=3\cdot 8$, work modulo $3$ and modulo $8$ and then put the answer back together using the Chinese remainder theorem.
*
*A safe prime $p>7$ is always $p\equiv 2 \bmod 3$. This can be shown by considering the options for $q\bmod 3$. Since $p>7$, we have $q>3$, and so $q\not\equiv 0\bmod 3$. And if $q\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$ Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$.
I tried some elementary ways while all failed.
| From Sum of tangent functions where arguments are in specific arithmetic series and Prove the trigonometric identity $(35)$,
$$\tan13x=\frac{\binom{13}1t-\binom{13}3t^3+\binom{13}5t^5-\binom{13}7t^7+\binom{13}9t^9-\binom{13}{11}t^{11}+t^{13}}{\cdots}$$
where $\displaystyle t=\tan x$
Now if $\displaystyle\tan13x=0,13x=n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How to compute the integral $(1+t^2) / (1-t^2) $ compute $\displaystyle\int \frac{1+t^2}{1-t^2}dt$
I tried splittting them to two parts, and computed $\displaystyle\int \frac{1}{1-t^2}dt$ using trig sub, but i don't know how to compute the second part.
| Use the trig substitution as you did before, but don't stop there! use $$t=\tan{\frac{\theta}{2}}$$
This will give you the following identities:
$$1+t^2=\sec^2{\frac{\theta}{2}}$$
Thus
$$\frac{1}{1+t^2}=\cos^2{\frac{\theta}{2}}$$
and
$$1-\frac{1}{1+t^2}=1-\cos^2{\frac{\theta}{2}}=\sin^2{\frac{\theta}{2}}$$
Giving
$$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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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... | The indetermination is of the form $1^{+\infty}$, suggesting the application of fundamental limit:
$$\lim_{w\rightarrow 0^+}(1+w)^{\frac{1}{w}}=e. $$
Remembering that
$$\lim_{x\rightarrow 0}\frac{a^x-1}{x}=\ln a, $$
we have
$$\lim_{x\rightarrow 0^+} \left(\frac{3^x+5^x}{2} \right)^{\frac{1}{x}}=\lim_{x\rightarrow 0^+} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplifying a radical with fraction and sum What are the steps to simplify
$(1+(\frac{1}{2}(x^3-\frac{1}{x^3})^2))^ \frac{1}{2}$
to
$\frac{1}{2}(x^3+\frac{1}{x^3})$ ?
| Hint: Try using the identities $(a-b)^2=a^2-2ab+b^2$ and $(a+b)^2=a^2+2ab+b^2$ (which are actually the same identity).
Be careful: check whether the result holds for $x\lt0$.
I am assuming you meant
$$
\left(1+\left(\frac12\left(x^3-\frac1{x^3}\right)\right)^2\right)^{1/2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inequality: $x^2+y^2+z^2 \geq \sqrt{2}x(z+y)$ How can I prove the following inequality:
$$x^2+y^2+z^2 \geq \sqrt{2}x(z+y)?$$
Thanks!
| We use here that for all $(a,b)\in \mathbb{R}^2$ :
$$a^2+b^2\ge 2ab.$$
Using this with $a=\frac{x}{\sqrt{2}}$ and $y$ gives us
$$\frac{x^2}{2}+y^2 \ge \sqrt{2}xy.$$
Then we use again the first inequality with $a=\frac{x}{\sqrt{2}}$ and $z$ that gives us $$\frac{x^2}{2}+z^2 \ge \sqrt{2}xz.$$
Then sum this two inequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$ I am trying to evaluate this integral.
$$
I=\int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2
$$ Note $$
\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}, \ |x| < 1.
$$
I was trying to do use this series expansion but wasn't sure how to go ab... | Using Richard Feynman's favorite method, the method of differentiation under the integral sign.
$$
\begin{align}
I(\alpha)&=\int_0^\infty\frac{\ln^2(1+\alpha x)}{x^{\frac{3}{2}}}dx\\
\frac{dI(\alpha)}{d\alpha}&=\int_0^\infty\frac{2x\ln(1+\alpha x)}{x^{\frac{3}{2}}(1+\alpha x)}dx\\
I'(\alpha)&=2\int_0^\infty\frac{\ln(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
How to find this angle? $P$ is a point in $\triangle ABC$. If $\angle PAB = 10^\circ$, $\angle PBA = 20^\circ$, $\angle PAC = 40^\circ$ and $\angle PCA = 30^\circ$, find $\angle B$.
| Let $\angle PBC=x$. so for the condition $\angle PCB=80^\circ-x$.
*
*Apply "trigonometric expression of ceva's theorem"-$ \frac{\sin 10^\circ}{\sin 40^\circ}\times \frac{\sin x}{\sin 20^\circ}\times \frac{\sin 30^\circ}{\sin(80^\circ-x)} =1$
*From that we have $\frac{\sin x}{\sin(80^\circ-x)}=4\cos 10^\circ \sin 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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To Find $A^{50}$ $$A=\begin{bmatrix}1 & 0&0\\1 & 0&1\\0&1&0\end{bmatrix}$$
Find $A^{50}$ ?
Now from Cayley–Hamilton theorem, I get $A^3-A^2-A+I=0$ and $A^{50}=(A^4)^{12}A^2$ so I found $A^4$ which is $-2A-I$, then we have $A^{50}=B^{12}A^2$ where $B =A^4$ was calculated, now should I again use Cayley–Hamilton theorem ... | If you do a few calculations: \begin{equation} A^2= \begin{bmatrix} 1&0&0\\1&1&0\\1&0&1 \end{bmatrix} \end{equation}\begin{equation} A^4= \begin{bmatrix} 1&0&0\\2&1&0\\2&0&1 \end{bmatrix} \end{equation}\begin{equation} A^6= \begin{bmatrix} 1&0&0\\3&1&0\\3&0&1 \end{bmatrix} \end{equation}
...so then from here you can de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$\overline{x} \times \overline{a} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ I'm trying to solve this exercise:
Let $\overline{a} \neq \overline{0}$, $\overline{b}$ be two vectors of the Euclidean vector space $V_{3}$. Prove the equation $\overline{x} \times \overline{a} = \over... | We know that $\vec x \times \vec a = \vec b$ means that $\vec a \perp \vec b$
If the two vectors are not perpendicular then there are no solutions. The following system has no solutions:
$$\left\{ \begin{gathered}
{a_3}{x_2} - {a_2}{x_3} = {b_1} \\
{a_1}{x_3} - {a_3}{x_1} = {b_2} \\
{a_2}{x_1} - {a_1}{x_3} = {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/732753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to prove that $\frac{r}{R}+1=\cos A+\cos B+\cos C$? How do we prove that for any triangle this holds: $$\frac{r}{R}+1=\cos A+\cos B+\cos C$$ I can use this beautiful identity to prove several geometric inequalities, but I have no idea how to prove the identity itself. Can anyone give me hints?
| here is mechanical solution:
$\cos A+\cos B+\cos C-1=\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}=\dfrac{(a+b-c)(b+c-a)(c+a-b)}{2abc}=\dfrac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{2abc(a+b+c)}=\dfrac{8S^2}{abc(a+b+c)}=\dfrac{\dfrac{S}{s}}{\dfrac{abc}{4S}}=\dfrac{r}{R}$
$S=\sqrt{s(s-a)(s-b)(s-c)},s=\dfrac{a+b+c}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).$$
I ended up doing it in Mathematica, is there any other way?
Thanks in advance!
| Claim: Suppose for some integer $n\gt1$,
$$
f(x)=x+a_nx^n+a_{2n-1}x^{2n-1}+a_{3n-2}x^{3n-2}+o\left(x^{3n-2}\right)
$$
and
$$
g(x)=x+b_nx^n+b_{2n-1}x^{2n-1}+b_{3n-2}x^{3n-2}+o\left(x^{3n-2}\right)
$$
Then
$$
\begin{align}
&f(g(x))-g(f(x))\\
&=\left((n-1)(a_{2n-1}b_n-a_nb_{2n-1})+\frac{n(n-1)}{2}a_nb_n(b_n-a_n)\right)x^{... | {
"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": 2
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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... | Hint
Squaring both sides gives $3x = (x + \sqrt{3})^2 = x^2 + 2\sqrt{3}x + 3$.
Rewrite this to get $x^2 + (2\sqrt{3} - 3)x + 3 = 0$. Solve this second degree equation to get
$$x = 1.5 -\sqrt{3} \pm \sqrt{(1.5-\sqrt{3})^2 - 3}$$
Now try to rewrite this on the form $A\sqrt{B}+C$. Be sure to check which root actually is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$
When $n$ is $1$:
$$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$
$$ 4 + 9 + 25 = 38 $$
When $n$ is $k$:
$$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$
Finally, when $n$ is $k+1$:
$$ 2^{2^{k+1}} + 3^{2^{... | Hint:
$$2^{n+6}+3^{n+6}+5^{n+6}=2^6\cdot2^n+3^6\cdot 3^n+5^6\cdot5^n\equiv7\cdot2^n+7\cdot3^n+7\cdot5^n\ \ (\text{mod }19)$$
What can $2^k\text{ mod }6$ be?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/744328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculate for $(1+\tan 20^\circ)(1+\tan 25^\circ)$. Help me with my works I have no idea what I am doing here,
I started with $\tan 20^\circ=\tan(45^\circ-25^\circ)=(1-\tan 25^\circ)/(1+\tan 25^\circ)$
I am sure the work I have shown so far are ok, but how do you get $1+\tan 20^\circ=2/(1+\tan 25^\circ)$ from that?
| $1 = \tan(45^\circ) = \tan(20^\circ+25^\circ) = \frac{ \tan 20^\circ + \tan 25^\circ}{1- \tan 25^\circ \tan 20^\circ} $
so
$2 = 1+ \tan 20^\circ + \tan 25^\circ + \tan 20^\circ \tan 25^\circ = (1+ \tan 20^\circ)(1+ \tan 25^\circ)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How find this value of $x$ such $\log_{\frac{1}{12}}{(x^2+2x-3)}>x^2+2x-16$ if $$\log_{\frac{1}{12}}{(x^2+2x-3)}>x^2+2x-16$$
Find the value of $x$
My idea: since $$x^2+2x-3>0\Longrightarrow x>1 ,or, x<-3$$
$$x^2+2x-3<\left(\dfrac{1}{12}\right)^{x^2+2x-16}$$
let
$$(x+1)^2=y\ge 4$$
so
$$\left(\dfrac{1}{12}\right)^{y-... | Let, $x^2+2x-16=y$. So, $x^2+2x-3=y+13$.
Now, we have, $y+13<\bigg(\dfrac1{12}\bigg)^{y} \implies 12^y(y+13)<1$
From this we can conclude, when $y<-1$ or $x^2+2x-15<0$ or $x\in[-5,3]$ there exists solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/747917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.
| On the LHS you have two factors of $2^{n+1}$ so it is simply a case of factoring out a common factor on the left hand side:
$$2^{n+1} + 2^{n+1} = 2^{n+1} \cdot (1 + 1) = 2^{n+1}\cdot 2 = 2^{n+2}$$
For a more combinatorial proof we could also have used double counting. Imagine you want to figure out the number of ways t... | {
"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": 3
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How to differentiate $\frac{2x^5}{\tan x}$ $$\frac{2x^5}{\tan x}$$
I can differentiate $2x^5$ ($10x^4$) and $\tan x$ ($\sec^2 x$) but can't do that one
Is there a rule I can apply?
| How to differentiate $2x^5/\tan x$
$$\begin{align}
\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac uv\right) & = \dfrac{v\dfrac{\mathrm du}{\mathrm dx} - u\dfrac{\mathrm dv}{\mathrm dx}}{v^2} \\
\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{2x^5}{\tan x}\right)& = \tan x\cdot10x^4 - 2x^5\cdot(\sec x)^2)/(\tan x)^2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How find this $x^3-5x+10=2^y$ let $x,y$ is positive integer,and such
$$x^3-5x+10=2^y$$
find all $x,y$.
since $$x=1\Longrightarrow 1^3-5+10=6$$ can't
$$x=2,2^3-5\cdot 2+10=8=2^3$$
so $x=2,y=3$
$$x=3,LHS=27-15+10=22$$
$$x=4,LHS=64-20+10=54$$
$$x=5,LHS=125-25+10=110$$
$$x=6,LHS=216-30+10=236$$
$$\cdots$$
I find $$(x,y)=(... | First $(10,4)$ and $(10,6)$ are not solutions, neither is $(10,y)$ for any integer $y$, since $(10)^3-5(10)+10=2^y$ would imply $5\mid2^y$, a contradiction.
Now, rearrange the given equation to get
$$
(x^2-5)x=2(2^{y-1}-5)\tag{1}
$$
Now either $x=2k+1$ or $x=2k$ for some integer $k>1$;
let $x=2k+1$, direct substituti... | {
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"url": "https://math.stackexchange.com/questions/756355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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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)
| A very fast solution
Maximum value of $\displaystyle4\sin\theta+3\cos\theta$ is $\displaystyle{5}$. This value is achieved when $\displaystyle\tan\theta=\dfrac{4}{3}$ by differentiating.So $\cos\theta=\dfrac{3}{5}$ and $\sin\theta=\dfrac{4}{5}$.
Hence $4\cos\theta-3\sin\theta=0$.
| {
"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": 13
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Showing that $\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$ I was reading an article in which it was stated that, with a change of variable, one could show that:
$$\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$$
I tried with $t = 1 + \frac{1}{x}$ but that doesn't work out,... | With the substitution $u=\frac{1}{x}$ we get
$$\int_1^\infty \frac{dx}{1+x^2} = \int_0^1 \frac{u^{-2}}{1+u^{-2}}du=\int_0^1 \frac{du}{1+u^2}$$
which implies
$$\int_0^\infty \frac{dx}{1+x^2} = \int_0^1 \frac{dx}{1+x^2} + \int_1^\infty \frac{dx}{1+x^2} = 2\int_0^1 \frac{dx}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$
3) $a+b+c=0$
4) $c=0$
Update
I just tried to su... | We know that $c^2+bc+a = 0$ and $c^2+ac+b = 0$. Subtract to get $c(b-a)-(b-a) = 0$ or $(b-a)(c-1) = 0$. This means that either $b = a$ or $c = 1$. If $c = 1$, then $1+a+b = c+a+b = 0$.
| {
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"url": "https://math.stackexchange.com/questions/759780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all primes $(p,q)$ for perfect squares. Find all prime pairs $(p,q)$ such that $2p-1, 2q-1, 2pq-1$ are all perfect squares.
Source: St.Petersburg Olympiad 2011
I have only found the pair $(5,5)$ so I am thinking that maybe a modulo $5$ approach could work.
| Although Ivan Loh's answer is correct, I believe I may have found a much simpler and elementary proof that $(5,5)$ is the only pair.
Assume without loss of generality that $q\geq p$ and write $2pq-1 = x^2$,
$2q-1 = y^2$, with $x$ and $y$ positive integers which are necessarily both odd.
We have the following inequali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
For which n does the inequality $2 \uparrow^{n+1}n > 3\uparrow^n 3 +2$ hold? For which n does the following inequality hold ?
$$2 \uparrow^{n+1}n > 3\uparrow^n 3 + 2$$
where $\uparrow$ stands for knuth's up-arrow notation.
I need this inequality to prove that
$$f_{\omega+1}(n) > G(n)$$
for $n\ge 8$
where $f_{\omega+1}(... | The inequality holds precisely when $n \ge 4$.
When $n=3$, we have
$$2 \uparrow^4 3 = 2 \uparrow^3 4 = 2\uparrow^2 (2 \uparrow^2 4) = 2 \uparrow^2 65536 < 3\uparrow^3 3 + 2 = 3\uparrow^2(3\uparrow^2 3) +2 = 3\uparrow^2 7625597484987 + 2$$
Using the fact that $2 \uparrow^m (n+2) > 3\uparrow^m n + 2$,(proven here) we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/761745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How should I try to evaluate this integral $\int_0^\pi \sqrt{1+4\sin^2\frac x2 - 4\sin\frac x2}\;dx$? Suppose that we are given the following integral:
$$\int_0^\pi \sqrt{1+4\sin^2\frac x2 - 4\sin\frac x2}\;dx.$$
(Original screenshot)
And the answer is one of these :-
*
*$4\sqrt3-4-\frac\pi3$
*$\pi-4$
*$... | \begin{align}
\int_0^\pi\sqrt{4\sin^2\frac{x}{2}-4\sin\frac{x}{2}+1}\,dx&=\int_0^\pi\sqrt{\left(2\sin\frac{x}{2}-1\right)^2}\,dx\\
&=\int_0^\frac{\pi}{3}\left(1-2\sin\frac{x}{2}\right)\,dx+\int_\frac{\pi}{3}^\pi\left(2\sin\frac{x}{2}-1\right)\,dx\\
&=\left[x+4\cos\frac{x}{2}\right]_0^\frac{\pi}{3}+\left[-4\cos\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/761947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A different type binomial expansion problem Suppose we have $$(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}.$$
What will be the value of $a_0^2 - a_1^2 + a_2^2 - \cdots + a_{2n}^2$?
The answer is $a_n$, but I can't solve it.
See, what I've done is substitute $x$ as $-\frac{1}{x}$ and I've got:
${\frac{(x... | Since
$$
(1+x+x^2)^n=\sum_{k=0}^{2n}a_kx^k\tag{1}
$$
we can look at the following in two ways
$$
\begin{align}
\left(1+\frac1x+\frac1{x^2}\right)^n
&=\sum_{k=0}^{2n}a_k\frac1{x^k}\\
&=\sum_{k=0}^{2n}a_kx^{-k}\tag{2}
\end{align}
$$
or as
$$
\begin{align}
\left(\frac1{x^2}+\frac1x+1\right)^n
&=\left(\frac{1+x+x^2}{x^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Non-Homogenous System Find the general solution of $\vec{x^{'}}=\begin{pmatrix} 1&1\\4&1 \end{pmatrix}x+\begin{pmatrix}2\\-1\end{pmatrix}e^{t}$
I found the eigenvalues to be $\lambda_1=3 \;\; \lambda_2=-1$
Next I calculated the eigenvectors to be $e_1=\begin{pmatrix}1\\2 \end{pmatrix}$ and $e_2=\begin{pmatrix}1\\-2 \en... | It looks like your $\phi^{-1}(t)$ went astray.
I got:
$$\phi^{-1}(t) = \begin{pmatrix} \dfrac{1}{2 e^{3 t}} & \dfrac{1}{4 e^{3 t}} \\ \dfrac{e^{t}}{2} & -\dfrac{e^{t}}{4} \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Probability of first actor winning a "first to roll seven with two dice" contest? Two players P and Q take turns, in which they each roll two fair and independent dice. P rolls the dice first.
The first player who gets a sum of seven wins the game. What is the probability that player
P wins the game?
| The probability that $ P $ eventually wins the game can be represented as the sum of an infinite geometric series.
For example,
The probability of $ P$ winning in the first round itself is $ \dfrac{1}{6} $. (Since out of the $ 36 $ possible outcomes, there are $ 6 $ outcomes in which the sum of the dice is equal t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
| So let's track each term in $4x^2 + 2x + 1$.
Since $x \equiv 2 \pmod{3}$, we get the following:
$$ 4x^2 \equiv 16\equiv 1 \pmod 3$$
$$2x \equiv 4 \equiv 1 \pmod 3$$
$$1 \equiv 1 \pmod 3$$
Adding these together, we get:
$$4x^2 + 2x + 1 \equiv 0 \pmod 3$$
We conclude that $3\mid (4x^2+2x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to solve this equation or system of equations? I want to solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2.$$
I tried. Put $a = \sqrt{2 x-3}\geqslant 0$ and $b =\sqrt{3 x-2}\geqslant 0 $.
Suppose
$$5x-4=m(2x-3)+n(3x-2)$$
then $m=\dfrac{2}{5}$ and $n=\dfrac{7}{5}$.
Therefore
$$5x-4=\dfrac{2}... | A "classical" solution:
1) Conditions for the existence of radicals give us $x\geq \frac{3}{2}.$
2) The condition for the existence of equality $(5X-4)\sqrt{2x-3} >(4X-5)\sqrt{3x-2}$ give us: $x>2.$
3)We write the equation in the form $(5X-4)\sqrt{2x-3}-2=(4X-5)\sqrt{3x-2}$and raise a squared:
$$2x^3-3x^2-3x+6=4(5x-4)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Algebra Iranian Olympiad Problem If:
$x^2+y^2+z^2=2(xy+xz+zy)$
and $x,y,z \in R^+$
Prove:
$\frac{x+y+z}{3} \ge \sqrt[3]{2xyz}$
I tried my best to solve this thing but no use.
Hope you guys can help me.Thanks in advance.
| If we put $a=\sqrt{x}$ and $b=\sqrt{y}$, the degree two equation (in $z$)
$x^2+y^2+z^2-2(xy+xz+yz)=0$ has two solutions, $(a-b)^2$ and $(a+b)^2$. By
cyclically permuting $x,y,z$, we may assume $z=(a+b)^2$. The inequality
to be shown is then equivalent to $(x+y+z)^3 \geq 54xyz$, or
$(a^2+b^2+(a+b)^2)^3 \geq 54(a^2b^2(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.