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
$$a=2R\sin{A}\le 2R=\sqrt[3]{7}$$
| {
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"url": "https://math.stackexchange.com/questions/595633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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
$$2\sqrt{x}=t-\dfrac{1}{t}\Longrightarrow x=\dfrac{1}{4}(t-\dfrac{1}{t})^2$$
so
$$dx=\dfrac{1}{2}(t-\dfrac{1}{t})(1+\dfrac{1}{t^2})dt=\dfrac{t^4-1}{2t^3}dt$$
$$I=\int\dfrac{1}{1+t}\cdot\dfrac{t^4-1}{2t^3}dt=\dfrac{1}{2}\int\left(1+\dfrac{1}{t}+\dfrac{1}{t^2}+\dfrac{1}{t^3}\right)dt=\dfrac{1}{2}\left(t+\ln{t}-\dfrac{1}{t}-\dfrac{1}{2t^2}+C\right)$$
so
$$I=\dfrac{1}{2}\left(\sqrt{x}+\sqrt{x+1}+\ln{(\sqrt{x}+\sqrt{x+1})}-\dfrac{1}{\sqrt{x}+\sqrt{x+1}}-\dfrac{1}{2(\sqrt{x}+\sqrt{x+1})^2}+C\right)$$
My question: have other methods? Thank you very much
| 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 obviously reduces to a rational function integral.
| {
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"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
$5^{125}-1=((x^2 + 3x + 1)^2 - (5^{13}(x + 1))^2)(x - 1)$
it is now of the form $(a^2-b^2)(x-1)$ where $a,b$ are integers and hence $(x^5-1)/(x-1)$ is a composite number.
| {
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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 minimize the square of the distance. That is:
$$f(x) = d^2(x) = (x-5)^2 + |4x+13| $$
Let's take the derivative of $f(x)$ (note that you have to take care about the absolute value!!!):
$$f'(x) = \left\{ \begin{array}{lr}2x - 10 + 4 & x \geq -\frac{13}{4}\\2x - 10 - 4 & x < -\frac{13}{4}\end{array}\right. = \left\{ \begin{array}{lr}2x - 6 & x \geq -\frac{13}{4}\\2x - 14 & x < -\frac{13}{4}\end{array}\right.$$
Now, we must pose $f'(x) = 0$, and then:
$$\left. \begin{array}{lr}2x-6 = 0 & x \geq -\frac{13}{4}\\2x-14 = 0 & x < -\frac{13}{4}\end{array}\right. \Rightarrow \left. \begin{array}{lr}x = 3 & x \geq -\frac{13}{4}\\x = 7 & x < -\frac{13}{4}\end{array}\right.$$
Cleary, the second solution is not admissible since $7 > -\frac{13}{4}$. Hence, the only solution is $x=3$ and then the point you are looking for is:
$$P=(3, \sqrt{12+13}) = (3, 5)$$
| {
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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 $H(f)$?
$$H(f) = \frac{1}{K^2} \left( \frac{\sin{\pi f K}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(K-1)} - \frac{1}{L^2} \left( \frac{\sin{\pi f L}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(L-1)}$$
My (attempt at) derivation, using only the first lowpass filter of $y(k)$:
$$\sum_{m = k-K+1}^k \; \sum_{n = m-K+1}^m x(n) = \\
\left[x(k-2K+2) + x(k-2K+3) + \dots + x(k-K+1) \right] +\\
\left[x(k-2K+3) + x(k-2K+4) + \dots + x(k-K+2) \right] +\\
\cdots +\\
\left[x(k-K+1) + x(k-K+2) + \dots + x(k) \right] =\\
\left[ x(k-2K+2) + 2x(k-2K+3) + \cdots + (K-1)x(k-K+1) + \cdots + 2x(k-1) + x(k) \right]$$
Applying $Z$-transform, we get
$$X(z) \left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 2z^{-1} + 1 \right]$$
Now, if the above is correct, then the question boils down to how can this
$$\left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right]$$
when evaluated on the unit circle $z=e^{j2\pi f}$, be equal to this?
$$\left( \frac{\sin{\pi f K}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(K-1)}$$
| 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
$$\left[ \frac{(z^{-K}-1)}{(z^{-1}-1)}\right]^2$$
$$=\left[ \frac{(z^{-K/2}-z^{K/2})/z^{K/2}}{(z^{-1/2}-z^{1/2})/z^{1/2}}\right]^2$$
$$=\left[ \frac{(z^{-K/2}-z^{K/2})}{(z^{-1/2}-z^{1/2})}\right]^2 z^{-(K-1)}$$
which for $z=e^{j2\pi f}$ is
$$\left( \frac{\sin{\pi f K}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(K-1)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 left side.
$$
\begin{align*}
f_{k+2}f_k-f_{k+1}^2&= \left( f_k-f_{k+1} \right)f_k-f_{k+1}^2 \\
&=f_k^2-f_{k+1}f_k-f_{k+1}^2 \\
&=-f_{k+1}\left( f_{k+1}+f_k \right)+f_k^2 \\
&=-f_{k+1}f_{k-1}+f_k^2 \\
&= \left( f_{k+1}f_{k-1}-f_k^2 \right)(-1)^1 \\
&=(-1)^k(-1)^1 \\
&=(-1)^{k+1}. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \blacksquare
\end{align*}
$$
Notes: On line 4 by definition $f_{k+1}+f_k=f_{k-1}$. On line 6 we substituted our assumption.
| {
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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 $m$ is any natural number.
Let $P(k+1)$ be true,
Hence, $(k+1)^7-(k+1)$ is a multiple of 7.
Now, how do I evaluate $(k+1)^7$ without using the binomial theorem ? Or is there some other way to do it ?
| 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",
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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:
$$\int \frac{1}{x^7-x}\ \mathrm{d}x = \int \frac{1}{x(x^6-1)}\ \mathrm{d}x$$
$$\int \frac{1}{x(x^6-1)}\ \mathrm{d}x = \int \frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)}\ \mathrm{d}x$$
$$\frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} + \frac{Dx+E}{x^2-x+1} + \frac{Fx+G}{x^2+x+1}$$
At this point I realized how brutal this question if going to be. Is there an easier way to solve the integral?
| 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",
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"timestamp": "2023-03-29T00:00:00",
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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
$x(3x^2)^2-x^3(3x^2)=2 \Rightarrow x(9x^4)-3x^5 = 2 \Rightarrow 9x^5 - 3x^5 = 2 -> x^5 = \frac{1}{3} \Rightarrow x = (\frac{1}{3})^{(\frac{1}{5})}$
Is this all I have to do?
| 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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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 $\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}$ and so on. But the solution is $x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}$. My question is, how does the $x$ at the beginning of the series appear?
| 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)\\
& = & x + (\tfrac11-\tfrac12)x^2 -(\tfrac12-\tfrac13)x^3+(\tfrac13-\tfrac14)x^4-\cdots \\
& = & x + \tfrac1{1\cdot2}x^2-\tfrac1{2\cdot 3}x^3 + \tfrac1{3\cdot 4}x^4\cdots \\
& = & x + \tfrac{x^2}2 - \tfrac{x^3}{6} +\tfrac{x^4}{12} - \cdots
\end{eqnarray*}
| {
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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",
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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}
$.
$n | a^2$
means that
$n_i \le 2a_i
$
or
$a_i
\ge \lceil \frac{n_i}{2} \rceil
$.
$m | a$
means that
$m_i \le a_i$.
If $m$ is as large as possible,
$m_i = a_i$,
so
$m_i
= \lceil \frac{n_i}{2} \rceil
$.
In this case,
$n = 3^32^2$,
so
$m
= 3^{\lceil \frac{3}{2} \rceil }2^{\lceil \frac{2}{2} \rceil }
= 3^{2}2^{1 }
$.
| {
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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
$$H(n) = an^2 + bn + c $$
$$H(n) - H(n-1) = an^2 + bn + c - a(n-1)^2 -b(n-1) - c = $$
$$an^2 + bn + c - a(n^2 - 2n + 1) - b(n-1) - c = $$
$$an^2 + bn + c - an^2 + 2an - a - bn + b - c = $$
$$bn + c + 2an - a - bn + b - c = $$
$$c + 2an - a + b - c = $$
$$2an - a + b$$
if $2a = u$ and $b - a = k$ then we get an arbitrary linear function $un + k$. Thus if the difference of two functions is linear the original recurrence function is quadratic.
Coming back to the problem at hand:
$$2an + b - a = 4n - 3$$
$$2a = 4 \rightarrow a = 2$$
$$b - a = -3 \rightarrow b - 2 = -3 \rightarrow b = -1$$
$$T(n) = 2n^2 - n + c$$
$$T(1) = 2 - 1 + c = 1 + c = 1 \rightarrow c = 0$$
$$T(n) = 2n^2 - n$$
| {
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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 $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}$
Now $\displaystyle f^{'}(x) = 1+x+x^2+x^3+x^4+x^5+x^6$
and $\displaystyle f^{''}(x) = 1+2x+3x^2+4x^3+5x^4+6x^5$
$\displaystyle f^{'''}(x) = 0+2+6x+12x^2+20x^3+30x^4 = 2\left(1+3x+6x^2+10x^3+15x^4\right)$
Now I did not understand how can i solve it
Help Required
Thanks
| 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 $f'(x)$ has no real zeroes; and since $f'(1) = 7 > 0$, $f'(x) > 0$ for all $x \in \Bbb R$. Thus
$f(x) = \sum_0^7 (x^r /.r) \tag{2}$
is monotonically increasing everywhere. Being of odd degree, $f(x)$ has at least one real zero; being monotonically increasing, it can have no more.
Hope this helps. Good Yule to One and All,
and as always,
Fiat Lux!!!
| {
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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 :
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
| 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
$$
\lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12.
$$
| {
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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 can't seem to figure out how to simplify it enough to get it right.
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
equals, $\frac{x^4 + 4x^3y - 16x^2y^2 - 64xy^3}{x^2-4xy}$. I then factored x out of the numerator and denominator to get $\frac{x(x^3 + 4x^2y - 16xy^2 - 64y^3)}{x(x-4y)}$ and cancelled out the factored x's to get $\frac{x^3 + 4x^2y - 16xy^2 - 64y^3}{x-4y}$. I don't know what to do from here though.
I've managed to get enough marks to be able to pass it but since it's a readiness test I want to understand all of the material going in.
| 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
$$\frac{x^2+16y^2}{x} \cdot \frac{x^2+4xy}{x-4y}=\frac {(x^2+16y^2)(x+4y)}{x-4y}=\frac {x^3+4x^2y+16xy^2+64y^3}{x-4y}$$
Of course, it could be that there was a negative sign mixup which lead to the result
$$\frac {x^3+4x^2y-16xy^2-64y^3}{x-4y}=\frac {(x^2-16y^2)(x+4y)}{x-4y}=(x+4y)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617594",
"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.
$$-3\sqrt[3]{2a} \sqrt[3]{2a}$$
Since everything is under the same index it becomes:
$$-3\sqrt[3]{2} \sqrt[3]{a}$$
Did I do this correctly, if not can anyone tell me what I should do?
Thanks :-).
| 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} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$
$$1-\sin{x}=1-x+\dfrac{1}{3!}x^3-\dfrac{1}{5!}x^5+\cdots$$
Follow I fell very ugly,maybe this problem have nice methods?
Thank you
| 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",
"source": "stackexchange",
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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) If $a,b,c,d$ are distinct real no. such that $a,b$ are the roots of the equation $x^2-3cx-8d = 0$
and $c,d$ are the roots of the equation $x^2-3ax-8b = 0$. Then $a+b+c+d = $
$\bf{My\; Try}::$(1) Using vieta formula
$a+b=10c......................(1)$ and $ab=-11d......................(2)$
$c+d=10a......................(3)$ and $cd=-11b......................(4)$
Now $a+b+c+d=10(a+c)..........................................(5)$
and $abcd=121bd\Rightarrow bd(ab-121)=0\Rightarrow bd=0$ or $ab=121$
Now I did not understand how can i calculate $a$ and $c$
Help Required
Thanks
| 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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"question_score": "2",
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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:
$$r^{\prime}=2[\sec\theta(-\csc\theta\cot\theta)+\csc\theta(\sec\theta\tan\theta)]$$
$$r^{\prime}=2\left(\frac{1}{\sin\theta}(-\frac{1}{\cos\theta}\frac{\cos\theta}{\sin\theta})+\frac{1}{\cos\theta}(\frac{1}{\sin\theta}\frac{\sin\theta}{\cos\theta})\right)$$
$$r^{\prime}=2(-\sec^2\theta+\csc^2\theta)$$
$$r^{\prime}=2\csc^2\theta-2\sec^2\theta$$
This is wrong though, the correct answer is: $$r^{\prime}=2\sec^2\theta-2\csc^2\theta$$
| Your initial derivative is correct: $$r^{\prime}=2[\sec\theta(-\csc\theta\cot\theta)+\csc\theta(\sec\theta\tan\theta)]$$
However, note that:
$$\csc x = \dfrac{1}{\sin x} \;\text{ and }\;\sec x = \dfrac 1{\cos x}$$
$$\begin{align} r' &= 2[\sec\theta(-\csc\theta\cot\theta)+\csc\theta(\sec\theta\tan\theta)] \\ \\& = 2\left(\frac{1}{\cos \theta}\left(-\frac{1}{\sin\theta}\frac{\cos\theta}{\sin\theta}\right)+\frac{1}{\sin\theta}\left(\frac{1}{\cos\theta}\frac{\sin\theta}{\cos\theta}\right)\right) \\ \\ \\& = -2 \csc \theta + 2\sec \theta \\ \\ &= 2 \sec \theta - 2\csc \theta\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/619579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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
$$
v_p(n!) = \left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor \frac{n}{p^2}\right\rfloor+\left\lfloor \frac{n}{p^3}\right\rfloor+\left\lfloor \frac{n}{p^4}\right\rfloor+\cdots
$$
So the exponent of prime factor $3$ in $(80!)$ is given as
$$
v_3(80!) = \left\lfloor \frac{80}{3}\right\rfloor +\left\lfloor \frac{80}{3^2}\right\rfloor+\left\lfloor \frac{80}{3^3}\right\rfloor+\left\lfloor \frac{80}{3^4}\right\rfloor+\cdots=36
$$
But I do not understand how to calculate exponent of $3$ in $(80!)!$.
| 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
$$v_3 ((80!)!) \approx \frac{ 80!}{3} + \frac{80!}{3^2} + \ldots = 80! \times \frac{ \frac{1}{3} } { \frac{2}{3} } = \frac{80!}{2}.$$
Since you already calculated that $ v_3 ( 80!) = 36 $, we know that the first 36 terms of the summation are integers, and the rest of the terms will thus contribute a very small error. In fact, we can hunt down the exact value of this error, by looking at the base 3 representation of $80!$.
Claim: $v_3 (n!) = \frac{n}{2} - R$ where $R$ is half the digit sum of $n$ in base 3.
Proof: This follows immediately by looking at the overestimation in each digit, which is $(0.1111111\ldots)_3 = \frac{1}{2}$. Hence the total overestimation is half the digit sum in base 3. $_\square$
As an explicit example, if we look at $v_3 (80!) = (222)_3 + (22)_3 + (2)_3 $, we have over approximated by $2 \times \frac{1}{2} + 2 \times \frac{1}{2} + 2 \times \frac{1}{2} + 2 \times \frac{1}{2} = 4$. A quick check shows that $ \frac{80}{2} - 4 = 36$, which agrees with your calculation.
It remains to show that $80!$ in base 3 has a digit sum of 220 (I don't know of an immediate way to do this), which would give you Daniel's result that $v_3 (80!) = \frac{80!}{2} - 110$.
Of course, this easily generalizes to other (prime) values.
| {
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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| \dfrac { x^2 + 8x + 16}{x + 4}\right| < \left| \dfrac { x^2 + 8x + 16}{x}\right| < |x^2 + 8x + 16| = |(x+4)^2| = (x+4)^2 \ . $
Let's require $(x+4)^2 < \epsilon$, which implies $ | x + 4 | < \sqrt \epsilon $. Therefore we have $\delta = \min \{1, \sqrt \epsilon \}$.
Is it a valid proof or are there any loopholes I'm unaware of? Side-note: I realize there are different -- and perhaps simpler -- ways to prove this, I just want to see if this very approach is valid.
| You have $x^2 + 6x + 8 = (x+2)(x + 4).$ Try factoring and canceling.
| {
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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
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{3}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}$$
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{n+1}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}$$
so
$$2I_n=\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)+\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)\left(\dfrac{1}{3}+\dfrac{1}{n+1}\right)+\cdots+\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)$$
Maybe this try is not usefull, so I think use other methods to solve it .
Thank you very much!
| 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 $H_k = \sum_{j=1}^k \frac1j$, we see that
$$
\sum_{k=1}^\infty H_k x^k = \frac{-\log(1-x)}{1-x}.
$$
Integrating (and forcing the constant term to equal $0$) then gives
$$
\sum_{k=1}^\infty \frac{H_k}{k+1} x^{k+1} = \frac{\log^2(1-x)}2.
$$
Therefore
$$
\sum_{n=0}^\infty I_n x^n = \frac1{x^2} \bigg( \frac1x\frac{\log^2(1-x)}2 \bigg) \bigg( \frac{-\log(1-x)}{1-x} \bigg) = \frac{-\log^3(1-x)}{2x^3(1-x)}.
$$
We can use this generating function to try to get information about $\{I_n\}$. For example, if we define $\{c_n\}$ by
$$
-\log^3(1-x) = \sum_{n=1}^\infty c_n x^n,
$$
then we conclude that
$$
I_n = \frac12 \sum_{k=1}^{n+3} c_k.
$$
Note that
$$
c_n = \sum_{i+j+k=n} \frac1{ijk},
$$
so that
$$
I_n = \frac12 \sum_{i+j+k\le n+3} \frac1{ijk}.
$$
| {
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"url": "https://math.stackexchange.com/questions/627936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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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 simplify it, it becomes $\sqrt{16}$ = $-4$
The solution to the question says that this must be rejected, as $\sqrt{16}$ = $|4|$, which cannot be equal to $-4$.
Why is this done? Why can't the square root of $16$ be equated to $-4$?
| 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
$$\sqrt{13-18\tan x}=6\tan x-3$$
$$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ 6\tan x-3\ge0$$
$$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ \tan x\ge\frac 12.$$
So, we know that $\tan x=-1/6$ is not a solution of the equation at the top.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$
As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$
But now when I approach that top limit this is what I get:
$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$
And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?
| 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:
$$
1+\frac{\ln a+\ln b+\ln c}{3n}\le\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\le
1+\frac{\ln a+\ln b+\ln c}{3n}+\frac{k}{n^2}
$$
and hence
$$
\left(1+\frac{\ln\sqrt[3]{abc}}{n}\right)^{\!n}\le
\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{\!n}
\le\left(1+\frac{\ln(\sqrt[3]{abc})}{n}+\frac{k}{n}\right)^{\!n}
$$
and as the both the left and right hand side tend to $\sqrt[3]{abc}$, so does the middle one.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 factor. Hence all possible Jordan blocks are (List A)
\begin{bmatrix} 0 \end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}
\begin{bmatrix} 3 \end{bmatrix}
\begin{bmatrix}3 & 1 \\ 0 & 3 \end{bmatrix}
\begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\0 & 0 & 3 \end{bmatrix}
Hence all possible matrices are
\begin{bmatrix} 0 & 1 & &&&&&& \\ 0 & 0 &&&&&&&\\&&3 & 1 & 0 &&&&\\ &&0 & 3 & 1 &&&&\\&&0 & 0 & 3 &&&&\\ &&&&&B&&&\end{bmatrix}
where $B$ are $4\times 4$ matrices chosen arbitrarily as a combination of all possible blocks listed in List A:
\begin{bmatrix} 0 &&& \\ &0&&\\&&0&\\&&&0\end{bmatrix}
\begin{bmatrix} 0 &&& \\ &0&&\\&&0&\\&&&3\end{bmatrix}
\begin{bmatrix} 0 &&& \\ &0&&\\&&3&\\&&&3\end{bmatrix}
\begin{bmatrix} 0 &&& \\ &3&&\\&&3&\\&&&3\end{bmatrix}
\begin{bmatrix} 3 &&& \\ &3&&\\&&3&\\&&&3\end{bmatrix}
\begin{bmatrix} 0 &&& \\ &3 & 1 & 0 \\ &0 & 3 & 1 \\&0 & 0 & 3 \end{bmatrix}
\begin{bmatrix} 3&&& \\ &3 & 1 & 0 \\ &0 & 3 & 1 \\&0 & 0 & 3 \end{bmatrix}
\begin{bmatrix} 0 &1&& \\ &0 & & \\ & &0 & \\& & &0 \end{bmatrix}
\begin{bmatrix} 0 &1&& \\ &0 & & \\ & &0 & \\& & &3 \end{bmatrix}
\begin{bmatrix} 0 &1&& \\ &0 & & \\ & &3 & \\& & &3 \end{bmatrix}
\begin{bmatrix} 0 &1&& \\ &0 & & \\ & &3 & 1 \\& & &3 \end{bmatrix}
\begin{bmatrix} 0 &&& \\ &0 & & \\ & &3 & 1 \\& & &3 \end{bmatrix}
\begin{bmatrix} 0 &&& \\ &3 & & \\ & &3 & 1 \\& & &3 \end{bmatrix}
\begin{bmatrix} 3 &&& \\ &3 & & \\ & &3 & 1 \\& & &3 \end{bmatrix}
\begin{bmatrix} 3 &1&& \\ &3 & & \\ & &3 & 1 \\& & &3 \end{bmatrix}
\begin{bmatrix} 0 &1&& \\ &0 & & \\ & &0 & 1 \\& & &0 \end{bmatrix}.
However, some classmates said I need to consider the primary factor and invariant factor thus cut off some of the possibilities. I am quite confused...
| 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 matrix.
\begin{array}{cccccrl}
\begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix} &
\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} &
\begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} &
\begin{bmatrix} 3 \end{bmatrix} &
\begin{bmatrix} 0 \end{bmatrix} & \mbox{permutations} \\ \hline
1 & 1 & & & 4 & 6!/4! = 30 \\
1 & 1 & & 1 & 3 & 6!/3! = 120 \\
1 & 1 & & 2 & 2 & 6!/2!/2! = 180 \\
1 & 1 & & 3 & 1 & 6!/3! = 120 \\
1 & 1 & & 4 & & 6!/4! = 30 \\
2 & 1 & & & 1 & 4!/2! = 12 \\
2 & 1 & & 1 & & 4!/2! = 12 \\
1 & 2 & & & 2 & 5!/2!/2! = 30 \\
1 & 2 & & 1 & 1 & 5!/2! = 60 \\
1 & 2 & & 2 & & 5!/2!/2! = 30 \\
1 & 2 & 1 & & & 4!/2! = 12 \\
1 & 1 & 1 & & 2 & 5!/2! = 60 \\
1 & 1 & 1 & 1 & 1 & 5! = 120 \\
1 & 1 & 1 & 2 & & 5!/2! = 60 \\
1 & 1 & 2 & & & 4!/2! = 12 \\
1 & 3 & & & & 4!/3! = 4 \\ \hline
& & & & & 892 & \mbox{total}
\end{array}
| {
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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 to proceed in such problem please suggest it will be of great help .. I got this problem from www.mathstudy.in
| 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) $=$2^{14} \tan(2^{14}\theta) +2^{15} \dfrac{1}{\tan(2^{15}\theta)}$ = $2^{14} \tan(2^{14}\theta) +2^{15} \dfrac{1-\tan^2(2^{14}\theta)}{2\tan(2^{14}\theta)}$ = $\dfrac{2^{15}}{2\tan(2^{14}\theta)}$ = $2^{14}cot(2^{14}\theta)$
By this you can prove that $2^{13} \tan(2^{13}\theta)+2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15}\theta) $=$2^{13}cot(2^{13}\theta)$
hence you can prove that
By this you can prove that $2^{12} \tan(2^{12}\theta)+2^{13} \tan(2^{13}\theta)+2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15}\theta) $=$2^{12}cot(2^{12}\theta)$
So the result of $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ is $2\cot\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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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 of S in the plane $z = t$)
I got the symmetric equations as $$\frac{x+1}{-2c}=\frac{y-c}{c^2-1}=\frac{z-c}{c^2+1}$$
I got part one.
| 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 - c}{c^2 + 1} $
$ y = c + (c^2 - 1) \dfrac{t - c}{c^2 + 1 } $
And this simplifies to
$ x = \dfrac{ c^2 - 1 }{c^2 + 1 } - \dfrac{2 c}{c^2 + 1} t $
$ y = \dfrac{2 c}{c^2 + 1} + \dfrac{c^2 - 1}{c^2 + 1} t $
Note that $ \bigg( \dfrac{c^2 - 1}{c^2 + 1} \bigg)^2 + \bigg( \dfrac{2c}{c^2 + 1} \bigg) ^2 = 1 $
Therefore, we can let $ \dfrac{c^2-1}{c^2 + 1} = \cos(\theta) $ and $ \dfrac{2c}{c^2 + 1} = \sin(\theta) $
Then
$ x = \cos(\theta) - \sin(\theta) t $
$ y = \sin(\theta) + \cos(\theta) t $
Hence $(x, y)$ is a rotation of $(1, t)$ about the origin by a variable angle $\theta$, or by simply squaring $x$ and $y$ and adding we get
$ x^2 + y^2 = 1 + t^2 $
Thus the trace of the surface at $z = t$ is a circle of radius $\sqrt{1 + t^2} $
And this also means the surface $S$ is the hyperboloid of one sheet
$ x^2 + y^2 - z^2 = 1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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 rule seems worse. any help?
Edit: I cannot use power series expansion. that is provided on the next semester.
|
"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( \frac{\arctan x}{x}\right) }{x^{2}}
&=&\lim_{x\rightarrow 0^{+}}\frac{\ln \left( \arctan x\right) -\ln \left(
x\right) }{x^{2}}\tag{1} \\
&=&\frac{1}{2}\lim_{x\rightarrow 0^{+}}\frac{x-\arctan x-x^{2}\arctan x}{
x^{2}\arctan x+x^{4}\arctan x}\tag{2} \\
&=&\lim_{x\rightarrow 0^{+}}\frac{-\arctan x}{4x^{2}\arctan x+x+2\arctan x}
\tag{3}\\
&=&\lim_{x\rightarrow 0^{+}}\frac{-1}{5x^{2}+8x\arctan x+8x^{3}\arctan x+3}
\tag{4}\\
&=&-\frac{1}{3}.\tag{5}
\end{eqnarray*}
ADDED. Explanation:
$\ \:(1)$: $\frac{d}{dx}\left( \ln \left( \arctan x\right) -\ln \left( x\right)
\right) =\dfrac{x-\arctan x-x^{2}\arctan x}{x\left( 1+x^{2}\right) \arctan x}$;
$\ \:(2)$: $\frac{d}{dx}\left( x-\arctan x-x^{2}\arctan x\right) =-2x\arctan x$;
$\ \:(2)$: $\frac{d}{dx}\left( x^{2}\arctan x+x^{4}\arctan x\right) =4x^{3}\arctan
x+x^{2}+2x\arctan x$;
$\ \:(3)$: $\frac{d}{dx}\left( -\arctan x\right) =-\dfrac{1}{1+x^{2}}$;
$\ \:(3)$: $\frac{d}{dx}\left( 4x^{2}\arctan x+x+2\arctan x\right) =\dfrac{5x^{2}+8x\arctan x+8x^{3}\arctan x+3}{1+x^{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/646297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 the origin, that is we have $S(x) = \tilde S(\rho(x))$ for some function $\tilde S$. $\tilde S$ is denoted by $S$ again in your paper, but I will write $\tilde S$ here. By the chain rule, we have
\begin{align*}
\Delta S &= \sum_i \partial_i^2(\tilde S \circ \rho)\\
&= \sum_i \partial_i(\partial_\rho\tilde S \circ \rho \cdot \partial_i S)\\
&= \sum_i \partial_\rho^2 \tilde S \circ \rho \cdot (\partial_i \rho)^2 + \partial_\rho \tilde S \circ \rho \cdot \partial^2_i \rho\\
&= \partial_\rho^2 \tilde S \circ \rho \cdot \sum_i (\partial_i\rho)^2 +
\partial_\rho\tilde S \circ \rho \cdot \sum_i \partial^2_i \rho
\end{align*}
Using the above, we have
\begin{align*}
\sum_i (\partial_i \rho)^2(x) &= \sum_i \frac{x_i^2}{\rho^2(x)}\\
&= 1\\
\sum_i \partial_i^2\rho(x) &= \sum_i \frac{\rho^2(x) - x_i^2}{\rho^3(x)}\\
&= \frac{d\rho^2(x) - \sum_i x_i^2}{\rho^3(x)}\\
&= \frac{d\rho^2(x) - \rho^2(x)}{\rho^3(x)}\\
&= \frac{d-1}{\rho(x)}
\end{align*}
Using this, we get
$$
\Delta S(x) = \partial_\rho^2 \tilde S(\rho(x)) + \frac{d-1}{\rho(x)} \cdot \partial_\rho \tilde S(\rho(x))
$$
or in the notation of the paper
$$ \Delta S = \partial_\rho^2 S + \frac{d-1}{\rho} \cdot \partial_\rho S
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/647276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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)-fractional part (3+sinx)+..
How to solve the fractional part ?
Ans is (n^2+n-2)/2, (n^2+n)/2, (n^2+n+2)/2
| 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 = $\frac{\pi}{2},$ we have the sum
2 + 2 + 3 + ... + n = $\frac{n(n+1)}{2}+1 = \frac{n^2+n+2}{2}$
For x = $\pi,$ we have the sum
1 + 3 + 3 + ... + n =$\frac{n(n+1)}{2}+1 = \frac{n^2+n+2}{2}$
and for all other x we have simply the sum
1 + 2 + 3 + ... + n = $\frac{n(n+1)}{2} = \frac{n^2+n}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/647767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$
But I can't prove
$$f(n)\le f(n+1)$$
since
$$f(n+1)=\dfrac{n+1}{(n+1)^2+1}+\dfrac{n+1}{(n+1)^2+2^2}+\cdots+\dfrac{n+1}{(n+1)^2+n^2}+\dfrac{n+1}{(n+1)^2+(n+1)^2}$$
$$f(n)=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
so
$$f(n+1)-f(n)=\left(\dfrac{n+1}{(n+1)^2+1}-\dfrac{n}{n^2+1}\right)+\left(\dfrac{n+1}{(n+1)^2+2^2}-\dfrac{n}{n^2+2^2}\right)+\cdots+\left(\dfrac{n+1}{(n+1)^2+n^2}-\dfrac{n}{n^2+n^2}\right)+\dfrac{1}{2(n+1)}$$
so
$$f(n+1)-f(n)=\sum_{k=1}^{n}\dfrac{k^2-n^2-n}{(k^2+n^2)((n+1)^2+k^2)}+\dfrac{1}{2(n+1)}$$
This problem is my found it,can you help to solve this problem?
| 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 $y \geq x \geq 0$,
$$
\begin{array}{lcl}
u_n(x,y)&=&\frac{1}{n}\sum_{k=1}^{n} g\left(x+\frac{k}{n}(y-x)\right) \\
f_n(x,y)&=& u_{n+1}(x,y)-u_n(x,y) \\
\end{array}
$$
Conjecture $f_n(x,y)$ can be written
$$
f_n(x,y)=(y-x)h_n(x,y)\prod_{k=1}^n g\left(x+\frac{k}{n}(y-x)\right)g\left(x+\frac{k}{n+1}(y-x)\right)
$$
where $h_n$ is a polynomial with nonnegative coefficients in $x,y$.
I have checked this conjecture up to $n=10$, and it clearly is stronger
than the original claim.
| {
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"url": "https://math.stackexchange.com/questions/648367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 6
} |
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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/649570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 2
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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
$$\lvert a\sqrt{2} + b\sqrt{3}\rvert = \frac{\lvert a\sqrt{2} + b\sqrt{3}\rvert\cdot\lvert a\sqrt{2} - b\sqrt{3}\rvert}{\lvert a\sqrt{2}-b\sqrt{3}\rvert} = \frac{\lvert 2a^2 - 3b^2\rvert}{\lvert a\rvert\sqrt{2} + \lvert b\rvert\sqrt{3}}.$$
Since $2a^2-3b^2$ is a nonzero integer, and $\lvert a\rvert, \lvert b\rvert \leqslant 100$, we have
$$\frac{\lvert 2a^2 - 3b^2\rvert}{\lvert a\rvert\sqrt{2} + \lvert b\rvert\sqrt{3}} \geqslant \frac{1}{100(\sqrt{2}+\sqrt{3})} >\frac{1}{315}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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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 $x\cdot y=1$, then $x^3+y^3=?$ [key says 110]
$x+y=-1$ and $x\cdot y=2$, then $x^3+y^3=?$ [key says 5]
$x-y=-1$ and $x\cdot y=2$, then $x^3 -y^3=?$ [key says -7]
$x+y=\frac{1}{3}$ and $x\cdot y=\frac{1}{9}$, then $x^3+y^3=?$ $\left[-\dfrac{2}{27}\right]$
| 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 cubes of the roots must be
$$(s^2-p)s-2sp=s^2-3sp.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
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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:
$$27(x^2y + y^2z + z^2x + xyz) = 27( z(y^2 + zx) + x^2y + xyz) \le 27 (z(xy + zy) + x^2y + xyz) = 27 ( 2xyz + z^2y + x^2y) = 27y(2xz + x^2 + z^2) = 27y(x+z)^2 = 4 \cdot 3^3y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)$$
Now from AM-GM we have:
$$y + \frac{x+z}{2} + \frac{x+z}{2} \ge 3 \sqrt{y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)}$$
$$(y + \frac{x+z}{2} + \frac{x+z}{2})^3 \ge 3^3 y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)$$
So we have:
$$27(x^2y + y^2z + z^x + xyz) \le 4 \cdot 3^3y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right) \le (y + \frac{x+z}{2} + \frac{x+z}{2})^3 = 4(x+ y+z)^3$$
Hence the proof. As you can note this inequality is even stronger, because $xyz > 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/653574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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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
$$-(n+1)a_{n+1}=2013(a_{0}+a_{1}+\cdots+a_{n})\cdots\cdots (2)$$
then
$(2)-(1)$,we have
$$na_{n}-(n+1)a_{n+1}=2013a_{n}$$
then
$$(n+1)a_{n+1}=(2013-n)a_{n}$$
then
$$\dfrac{a_{n+1}}{a_{n}}=\dfrac{2013-n}{n+1}$$
so
$$\dfrac{a_{n}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n-2}}\cdots\dfrac{a_{1}}{a_{0}}=\cdots$$
so
$$\dfrac{a_{2013}}{a_{0}}=\dfrac{1}{2013}\cdot\dfrac{2}{2012}\cdots\dfrac{2013}{1}=1?$$
then How can find this sum?
| oh sorry, i omit some details !
by the symmetric property, $a_{0}=-a_{2013}$$,$$a_{1}=-a_{2012}$$,......$
the hint is :
$(a_{2}+4/3a_{2}+......+4/3a_{3}+......)/(ta_{0})=-a_{1}/a_{0}(a_{1}-4/3a_{1}-......-4/3a_{2}-......)$
we assume that :
$t_{1}=\frac{(a_{2}+4/3a_{2}+......+4/3a_{3}+......)}{(-a_{1}-4/3a_{1}-......-4/3a_{2}-......)}$
$t_{2}=\frac{(a_{3}+4/3a_{3}+......+4/3a_{4}+......)}{(-a_{2}-4/3a_{2}-......-4/3a_{3}-......)}$
so,
$a_{0}(1+1+4/3+......-a_{0}-4/3a_{0}-......-4/3\cdot1/2(a_{0}+a_{1})-......)$$=$$(\frac{a_{0}}{a_{1}}\cdot{\frac{a_{1}}{a_{2}}}$$\cdot......\cdot{\frac{a_{2012}}{a_{2013}}})\cdot{\frac{a_{0}}{a_{2013}}}\cdot$$t_{1}\cdot{t_{2}}......\cdot{t_{n}}$$=t_{1}\cdot{t_{2}}......\cdot{t_{n}}$
$=\frac{(a_{n-1}+4/3a_{n-1}+......+4/3a_{n-2}+......)}{(-a_{1}-4/3a_{1}-......-4/3a_{2}-......)}=1$
therefore, your solution holds !
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$\begin{eqnarray}G_{1}&=&1,\\
G_{2}&=&1,\\
G_{3}&=&3,\\
G_{4}&=&G_{2}=1\\
G_{5}&=&2G_{3}-G_{1}=5,\\
G_{6}&=&G_{3}=3,\\
G_{7}&=&3G_{3}-2G_{1}=9-2=7\\
G_{8}&=&G_{4}=1,\\
G_{9}&=&2G_{5}-G_{2}=10-1=9,\\
G_{10}&=&G_{5}=5,\\
G_{11}&=&3G_{5}-2G_{2}=15-2=13,
G_{12}&=&G_{6}=3,\\
G_{13}&=&2G_{7}-G_{3}=14-3=11,\\
G_{14}&=&G_{7}=7,\\
G_{15}&=&3G_{7}-2G_{3}=21-6=15,\end{eqnarray}$$
so when $n=1,3,5,7,9,,15,\cdots$ is 'good"
But How find numbers? when $n<2^{100}?$
Thank you
| 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
$$
\begin{align}
G(n)&=d_r+\cdots+2^rd_0\\
G(2n)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot0\\
G(2n+1)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot1\\
G(4n+1)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot0+2^{r+2}\cdot1\\
G(4n+3)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot1+2^{r+2}\cdot1
\end{align}
$$
from which the identities $G(2n)=G(n)$, $G(4n+1)=2G(2n+1)-G(n)$, and $G(4n+3)=3G(2n+1)-2G(n)$ are easily verified.
The upshot is that the OP's "good" numbers are those that are palindromes when written in binary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 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:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^2 x \\
\Longrightarrow &3\left(1-\sin^2 x\right)-\sin 2x &=& \sin^2 x \\
\Longrightarrow & 4\sin^2 x + \sin 2x - 3 &=& 0 \\
\Longrightarrow &2\left(1-\cos 2x\right)+\sin 2x - 3 &=& 0\\
\Longrightarrow &-2\cos 2x + \sin 2x &=& 1\end{array}$$
So, what's next?! Thank you in advance!
| $\sqrt{3\cos^2(x)-\sin(2x)}=-\sin(x)$
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 forget to check this value in the original equation.
Here when you substitute this value in the equation it equals zero but if the interval is $[0,2\pi]$, one of the solutions is $\frac{\pi}{4}$ but it doesn't work, it works if the right side is $\sin(x)$ not $-\sin(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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?
| $$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$$
$$4^{2x+1}-4^{2x+3}=3^{3x+1}-3^{3x+2}$$
$$4\cdot4^{2x}-4^34^{2x}=3\cdot3^{3x}-3^23^{3x}$$
$$60\cdot4^{2x}=6\cdot3^{3x}$$
$$10\cdot4^{2x}=3^{3x}$$
$$10\cdot16^{x}=27^{x}$$
$$10=(27/16)^{x}$$
$$\log_{10} 10=\log_{10} (27/16)^{x}$$
$$1=x\log_{10}(27/16)$$
$$x=\frac{1}{\log_{10}27-\log_{10}16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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{equation}
prove for
\begin{equation}
1+2(2)+3(2^2)+...+k(2^{k-1})+(k+1)(2^{k+1-1})=(k+1-1)(2^k+1)+1
\end{equation}
therefore:
\begin{equation}
(k-1)(2^k)+1+(k+1)(2^{k+1-1})=(k+1-1)(2^{k+1})+1
\end{equation}stuck from there onwards maybe understanding this wrongly.
| 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 $$S(k+1)=S(k)+(k+1)2^k\\=(k-1)2^k+1+(k+1)2^k\\=2k2^k-2^k+1+2^k\\=[(k+1)-1]2^{k+1}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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 fraction can be written as
\begin{align*}
\frac{\dfrac{\beta \alpha^3 - \alpha \beta^3}{6} x^3 + O(x^5)}{\alpha x^3 + O(x^5)} \to \frac{\beta \alpha^3 - \alpha \beta^3}{6\alpha}
\end{align*}
as $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/660046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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 $= \frac{1}{2}ab\sin{C}$
These are my steps for doing this:
$\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \sin{\theta} $
Let $\mathbf{a} = \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix}$ and let $\mathbf{b} = \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix}$
$\therefore \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix} = (5\sqrt{3})(15)\sin{\theta} $
$\therefore \sin{\theta} = -\dfrac{1}{\sqrt{3}}$
If I substitute these values into the general formula:
area $= \frac{1}{2}ab\sin{C}$
I get:
area $= \frac{1}{2}(5\sqrt{3})(15)(-\dfrac{1}{\sqrt{3}})$
$\therefore$ area $= -\dfrac{75}{2}$
However this isn't right, the area should be $\dfrac{75}{\sqrt{2}}$
I feel I'm missing something really obvious but I can't spot it, can anyone help?
Thank you.
| 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 $$\sin{\theta} = \sqrt{1 - \cos^2{\theta}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$$
Finally, the area of the triangle is:
$$
Area = \frac{1}{2} (5 \sqrt{3}) (15) \frac{\sqrt{2}}{\sqrt{3}} = \frac{75 \sqrt{2}}{2}
$$
We can just multiply $\frac{\sqrt{2}}{\sqrt{2}}$ to the area, and then we get the answer you posted:
$$
Area = \frac{75 \sqrt{2}}{2} \left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{75}{\sqrt{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
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 you
| 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 $\{w_n\}_{n\in\mathbb{N}}$ is the orbit of an element in $PSL(2,\mathbb{C})$. Iterating twice we can remove the complex conjugation and get:
$$w_{n+2}=\frac{2}{\sqrt{3}}+\frac{1}{\frac{2}{\sqrt{3}}+\frac{1}{w_n}}.\tag{3}$$
In order to have $a_n$ and $b_n$ we only need to find a closed form for the sequence given by:
$$t_{n+1} = \frac{2}{\sqrt{3}}+\frac{1}{t_n}.\tag{4}$$
Since the roots of the polynomial $x^2-\frac{2}{\sqrt{3}}x-1$ are $-\frac{1}{\sqrt{3}}$ and $\frac{3}{\sqrt{3}}$, with the choice
$$ t_n = \frac{A\left(-\frac{1}{\sqrt{3}}\right)^{n+1}+B\left(\frac{3}{\sqrt{3}}\right)^{n+1}}{A\left(-\frac{1}{\sqrt{3}}\right)^{n}+B\left(\frac{3}{\sqrt{3}}\right)^{n}}$$
equation $(4)$ is fulfilled, and we only need to choice $A$ and $B$ such that $t_1$ and $t_2$ fit with our starting values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/668098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How can solve this differential equation (second equation)? How can I solve this differential equation?
$$
\frac{1}{h}\left(\frac{a}{\delta} - \frac{b}{\delta} - \frac{c}{E\delta^4}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left[A\,\frac{\mathrm{d}\delta}{\mathrm{d}x} \left(\frac{1}{\delta} + \frac{3B}{\delta^2}\right)\right],
$$
where $a,b,c,E,h,A,B$ are constants.
| Let $y=\dfrac{d\delta}{dx}$ ,
Then $\dfrac{1}{h}\left(\dfrac{a}{\delta}-\dfrac{b}{\delta}-\dfrac{c}{E\delta^4}\right)=\dfrac{d}{d\delta}\left(Ay\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)\right)\dfrac{d\delta}{dx}$
$\dfrac{1}{h}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)=A\dfrac{d}{d\delta}\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)y$
$\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y~d\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)=\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)d\delta$
$\int\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y~d\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)=\int\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta^2}+\dfrac{3(a-b)B}{\delta^3}-\dfrac{c}{E\delta^5}-\dfrac{3Bc}{E\delta^6}\right)d\delta$
$\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)^2\dfrac{y^2}{2}=-\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}+\dfrac{3(a-b)B}{2\delta^2}-\dfrac{c}{4E\delta^4}-\dfrac{3Bc}{5E\delta^5}\right)+c_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 etc.
Still I have absolutely no idea how to generalize it for $n$ first terms...
| 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
$$\lfloor \sqrt[p]{k} + \epsilon \rfloor = \lfloor \sqrt[p]{k} \rfloor\quad\text{ for all } k = 1, 2, \ldots, n$$
The introduction of the $\epsilon$ piece make the function $\lfloor \sqrt[p]{x} + \epsilon\rfloor$ continuous at the jumps of $\lfloor x \rfloor$. This allows us to rewrite $\mathcal{S}_p$ as a Riemann-Stieljes integral.
$$
\mathcal{S}_p = \sum_{k=1}^n \lfloor \sqrt[p]{k} \rfloor
= \sum_{k=1}^n \lfloor \sqrt[p]{k} + \epsilon \rfloor
= \int_{1^-}^{n^+} \lfloor \sqrt[p]{x} + \epsilon \rfloor d \lfloor x \rfloor
$$
Integrate by part, we get
$$
\mathcal{S}_p
= \bigg[ \lfloor \sqrt[p]{x} + \epsilon \rfloor \lfloor x \rfloor \bigg]_{x=1-}^{n^+}
- \int_{1^-}^{n^+} \lfloor x \rfloor d \lfloor \sqrt[p]{x} + \epsilon \rfloor
= na
- \int_{1+\epsilon}^{\sqrt[p]{n}+\epsilon} \lfloor (s-\epsilon)^p \rfloor d \lfloor s \rfloor
$$
Convert the last integral back into a sum, we find
$$
\mathcal{S}_p
= na - \sum_{s=2}^a (s^p - 1)
= (n+1)a - \sum_{s=1}^a s^p
= (n+1)a - \frac{B_{p+1}(a+1)-B_{p+1}(0)}{p+1}
$$
where $B_k(x)$ is the Bernoulli polynomial of order $k$
For the special case $p = 2$, the sum we want is
$$
\bbox[8pt,border: 1px solid blue;]{
\sum_{k=1}^n \lfloor\sqrt{k}\rfloor = \mathcal{S}_2 =
(n+1) a - \frac16 a(a+1)(2a+1),\quad a = \lfloor\sqrt{n}\rfloor
}
$$
This is equivalent to the formula in Markus Scheuer's answer.
For the case $p = 3, 4, 5$. this leads to
$$\begin{align}
\sum_{k=1}^n \lfloor\sqrt[3]{k}\rfloor = \mathcal{S}_3 & =
(n+1) a - \frac14\left((a+1)^4 - 2(a+1)^3 + (a+1)^2\right)\\
\sum_{k=1}^n \lfloor\sqrt[4]{k}\rfloor = \mathcal{S}_4 & =
(n+1) a - \frac15\left(
(a+1)^5 - \frac52(a+1)^4 + \frac53 (a+1)^3 - \frac16(a+1)
\right)\\
\sum_{k=1}^n \lfloor\sqrt[5]{k}\rfloor = \mathcal{S}_5 & =
(n+1) a - \frac16\left(
(a+1)^6-3(a+1)^5 + \frac52 (a+1)^4 - \frac12 (a+1)^2
\right)\\
\end{align}
$$
reproducing the formulas in Claude Leibovici's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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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
$$\lim_{x \to 0} \sqrt[x]{\frac{\tan x}{x}}$$
$$= \exp \lim_{x \to 0} \ln \left(\left(\frac{\tan x}{x}\right)^{1/x}\right)$$
$$= \exp \lim_{x \to 0} \frac{\ln(\tan(x)) - \ln(x)}{x}$$
Using L'Hospital:
$$= \exp \lim_{x \to 0} \frac{\frac{1}{\tan(x)\cos^2(x)} - \frac{1}{x}}{1}$$
$$= \exp \lim_{x \to 0} \frac{1}{\sin(x)\cos(x)} - \frac{1}{x}$$
$$= \exp \lim_{x \to 0} \frac{1}{\sin(2x)} - \frac{1}{x}$$
$$= \exp \lim_{x \to 0} \frac{x - \sin(2x)}{\sin(2x) x}$$
But when I calculate
$$\lim_{x \to 0} \frac{x - \sin(2x)}{\sin(2x) x}$$ with Wolfram Alpha I get $\pm \infty$ . So the limit of $\lim_{x \to 0} \sqrt[x]{\frac{\tan x}{x}}$ should be $e^{\pm \infty} = 0 \text{ or } \infty \neq 1$. Which is both wrong.
Where is my mistake?
| If $L$ is the limit to be evaluated then we have $$\begin{aligned}\log L &= \log\left(\lim_{x \to 0}\sqrt[x]{\frac{\tan x}{x}}\right)\\
&= \lim_{x \to 0}\log\left(\sqrt[x]{\frac{\tan x}{x}}\right)\text{ (by continuity of }\log)\\
&= \lim_{x \to 0}\dfrac{\log\left(\dfrac{\tan x}{x}\right)}{x}\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{\tan x}{x} - 1\right)}{\dfrac{\tan x}{x} - 1}\cdot\dfrac{\dfrac{\tan x}{x} - 1}{x}\\
&= \lim_{x \to 0}1\cdot\frac{\tan x - x}{x^{2}}\text{ (because }y = \frac{\tan x}{x} - 1 \to 0\text{ and }\lim_{y \to 0}\frac{\log(1 + y)}{y} = 1)\\
&= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{2}\cos x}\\
&= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{2}\cdot 1}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{2}} + x\cdot\frac{1 - \cos x}{x^{2}}\\
&= 0 + 0\cdot \frac{1}{2} = 0\end{aligned}$$ The first limit is calculated here without any series expansion or LHR and the second limit is pretty standard based on $1 - \cos x = 2\sin^{2}(x/2)$ and using $\lim_{x \to 0}\dfrac{\sin x }{x} = 1$. It follows that $L = e^{0} = 1$.
Update: It is also possible to use the approach of the linked answer above to calculate the limit $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}}$$ Clearly if $0 < x < \dfrac{\pi}{2}$ then we have $\sin x < x < \tan x$ so that $$0 < \frac{\tan x - x}{x^{2}} < \frac{\tan x - \sin x}{x^{2}} = \tan x\cdot\frac{1 - \cos x}{x^{2}}$$ Taking limits as $x \to 0^{+}$ and using Squeeze Theorem we get $$\lim_{x \to 0^{+}}\frac{\tan x - x}{x^{2}} = 0$$ And to handle $x \to 0^{-}$ we can put $ x = -y$ and get the left hand limit also as $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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:
$$\begin{align}\int_0^{\pi/4} dx \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^2{x}} &= \underbrace{\int_0^{\pi/4} dx \frac{\sin{x}}{1-\cos^2{x}+\cos^4{x}}}_{u=\cos{x}} + \underbrace{\int_0^{\pi/4} dx \frac{\cos{x}}{1-\sin^2{x}+\sin^4{x}}}_{u=\sin{x}}\\ &= \int_{1/\sqrt{2}}^1 \frac{du}{1-u^2+u^4} + \int_0^{1/\sqrt{2}}\frac{du}{1-u^2+u^4}\\ &= \int_0^{1}\frac{du}{1-u^2+u^4}\end{align}$$
Now, the roots of the denominator are at $u^2=e^{\pm i \pi/3}$. Split this integral up yet again by partial fractions:
$$\begin{align}\frac1{1-u^2+u^4} &= \frac1{(u^2-e^{i \pi/3})(u^2-e^{-i \pi/3})}\\ &= \frac1{i 2 \sin{(\pi/3)}} \left [\frac1{u^2-e^{i \pi/3}} - \frac1{u^2-e^{-i \pi/3}} \right ]\\ &= \frac{2}{\sqrt{3}} \Im{\left [\frac1{u^2-e^{i \pi/3}} \right ]}\end{align}$$
Now use the fact that
$$\int \frac{du}{u^2-a^2} = \frac1{2 a} \log{\left ( \frac{u-a}{u+a}\right )}+C$$
which may be derived using (what else?) partial fractions. Note that in our case, $a=e^{i \pi/6}$ is complex so we do not have to worry about poles along the interval of integration. So, we thus have:
$$\begin{align}\int_0^1 \frac{du}{1-u^2+u^4} &= \frac{2}{\sqrt{3}} \Im{\left [\int_0^1 \frac{du}{u^2-e^{i \pi/3}} \right ]} \\ &= \frac{2}{\sqrt{3}} \Im{\left [\frac{1}{2 e^{i \pi/6}} \left [\log{\left ( \frac{u-e^{i \pi/6}}{u+e^{i \pi/6}}\right )} \right ]_0^1 \right ]}\\&= \frac{2}{\sqrt{3}} \Im{\left [\frac{1}{2 e^{i \pi/6}} \left [\log{\left ( \frac{1-e^{i \pi/6}}{1+e^{i \pi/6}}\right )} - \log{(-1)} \right ]\right ]}\\&= \frac{2}{\sqrt{3}} \Im{\left [\frac{1}{2 e^{i \pi/6}} \left [\log{\left ( (-i)\tan{\left ( \frac{\pi}{12}\right)}\right )} - \log{(-1)} \right ]\right ]}\\ &= \frac{1}{\sqrt{3}} \Im{\left [e^{-i \pi/6} \left (\log{\tan{\frac{\pi}{12}}} + i \frac{\pi}{2} \right )\right ] }\\ &= \frac{\pi}{4} + \frac{\log{(2+\sqrt{3})}}{2 \sqrt{3}} \end{align}$$
Therefore
$$\int_0^{\pi/4} dx \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^2{x}} = \frac{\pi}{4} + \frac{\log{(2+\sqrt{3})}}{2 \sqrt{3}}$$
A couple of notes about the result:
1) Mathematica returns $\frac{1}{12} \left(3 \pi +\sqrt{3} \log \left(7+4 \sqrt{3}\right)\right)$. This is equivalent to the derived result. I leave it to the reader to prove that to him/herself.
2) I took $\log{(-i)} = \log{(-1)} + i \pi/2$, so that I got cancellation and didn't need to worry about the how to treat the argument of $-1$. Of course, I did take the argument of $i$ to be $\pi/2$, but that just means I used the principal value of the argument.
| {
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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
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a^2+b^2+c^2)+2(ab+bc+ca) = 49$
So $23+2(ab+bc+ca) = 49\Rightarrow (ab+bc+ca) = 13$
Now from $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31\Rightarrow \frac{(a+1)\cdot (b+1)+(b+1)\cdot (c+1)+(c+a)\cdot (a+1)}{(a+1)(b+1)(c+1)} = 31$
So $\displaystyle \frac{(ab+bc+ca)+2(a+b+c)+3}{1+(a+b+c)+(ab+bc+ca)+abc} = 31\Rightarrow \frac{13+2\cdot 7+3}{1+7+13+abc} = 31$
So $\displaystyle \frac{30}{21+abc} = 31\Rightarrow 21\times 31+31(abc) = 30\Rightarrow (abc) = \frac{30-21\times 31}{31}=-\frac{621}{31}$
Now How can I calculate $a^3+b^3+c^3$
Is there is any better method by which we can calculate $abc$
Help me
Thanks
| 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).
| {
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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 numbers $n:5|n$ that I could add to the set?
Now, a companion question would be: What are the multiples of $2$ that get skipped?
Just because a number gets skipped, doesn't mean it's divisible by $5$; I found $40!$ to $49!$ have $10$ trailing zeroes, but $50!$ has $12$, skipping $11$. $$f(n)\notin\{5,11,17,18,23,29,30,...\}$$
Am I doing something wrong, or are these numbers supposed to be skipped?
| 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 skipped. Generally, if we call $s_5(n)$ the sum of the digits in the base-$5$ representation of $n$, then we have
$$f(n) = \frac{n-s_5(n)}{4}.$$
$f(n)$ skips $k-1$ values when $n$ is a multiple of $5^k$ but not of $5^{k+1}$, and among the skipped values is a multiple of $5$ whenever $k-1 \geqslant 5$, or the remainder of $f(n)$ modulo $5$ is between $1$ and $k-1$ (inclusive). The latter condition translates to $1 \leqslant s_5(n) \bmod{5} \leqslant k-1$ for $2 \leqslant k \leqslant 5$, and I used that to find the example $350 = 2\cdot 5^3 + 4\cdot 5^2$ as the smallest number that is not a power of $5$ and leads to a skipped multiple of $5$ in the $f(n)$ sequence.
| {
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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))
$d f(\phi^{-1})=(\dfrac{4x_{1}^{2}\sum_{i=1}^{n}x_{i}^{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{2}},...,\dfrac{4x_{n}^{2}\sum_{i=1}^{n}x_{i}^{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{2}})$
and so $(0,...,0,1)$ is a critical point.
Then the first two rows for Hessian:
$Hessian(a_{1i})=(\dfrac{(8x_{1}\sum_{i=1}^{n}x_{i}^{2}+8x_{1}^{3})(1+\sum_{i=1}^{n}x_{i}^{2})^{2}-16x_{1}^{3}\sum_{i=1}^{n}x_{i}^{2}(1+\sum_{i=1}^{n}x_{i}^{2})}{(1+\sum_{i=1}^{n}x_{i}^{2})^{4}},...,\dfrac{-16x_{n}^{2}x_{1}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{3}})=(0,...,0)$
$Hessian(a_{2i})=(\dfrac{-16x_{1}^{2}x_{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{3}},\dfrac{(8x_{2}\sum_{i=1}^{n}x_{i}^{2}+8x_{2}^{3})(1+\sum_{i=1}^{n}x_{i}^{2})^{2}-16x_{2}^{3}\sum_{i=1}^{n}x_{i}^{2}(1+\sum_{i=1}^{n}x_{i}^{2})}{(1+\sum_{i=1}^{n}x_{i}^{2})^{4}},...,\dfrac{-16x_{n}^{2}x_{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{3}})=(0,...,0)$
So the Hessian is just zeroes.
But the $index_{(0,...,0,1)}=n$. I am supposed to get n-negative values in the diagonal.
thanks
| 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
$$
\left(4 x_1^2 \sum_i x_i^2\right) \left( 2 (1 + \sum_i x_i^2) 2 x_1 \right) \\
16 x_1^3 (\sum_i x_i^2 )(1 + \sum_i x_i^2).
$$
Presumably by plugging in $n = 1$ into your formula above, you can see where you went off the rails.
You seem to have $8$ instead of $16$. (This is the only term I checked).
I don't see, however, why computing only the first row of the Hessian suffices to show you that the eigenvalues are not all negative -- can you explain your reasoning? Or am I supposed to see, from the first row, the pattern of the others? If so, I still don't see why the e-values can't all be negative. (Except that maybe your swapping 8 for 16 would ruin it...)
Later comment, now that you've fixed the $8$ vs $16$ thing:
Looking more closely at your work, let's consider the unit circle in 2-space. Stereographic projection from the south pole takes the unit disk to the upper hemisphere (in this case, the interval $[-1, 1]$ to the upper circle) by
$$
(x, 0) \mapsto (\frac{2x}{1 + x^2} , \frac{1-x^2}{1 + x^2})
$$
The derivative you want to compute is
\begin{align}
d(f \circ \phi^{-1}) &= d(\frac{1-x^2}{1 + x^2})/dx \\
&= \frac{(1-x^2)' (1 + x^2) - (1-x^2)(1 + x^2)'}{(1 + x^2)^2}\\
&= \frac{(-2x) (1 + x^2) - (1-x^2)(2x)}{(1 + x^2)^2}\\
&= \frac{(-2x - 2x^3) - (2x - 2x^3)}{(1 + x^2)^2}\\
&= \frac{(-2x - 2x^3) + (-2x + 2x^3)}{(1 + x^2)^2}\\
&= \frac{- 4x}{(1 + x^2)^2}
\end{align}
Since, near $x = 0$, this is just $-4x$ to first order, the second derivative at $0$ will be $-4$, and not $0$. Presumably by plugging in $n = 1$ to your formula above you can discover where you went off the rails.
Lemme go ahead and compute the second derivative:
\begin{align}
H &= d(\frac{- 4x}{(1 + x^2)^2} )/dx \\
&= \frac{(- 4x)'(1 + x^2)^2 + 4x( (1+x^2)^2 )'}{ (1 + x^2)^4} \\
&= \frac{- 4(1 + x^2)^2 + 4x( 2(1+x^2)(2x) )}{ (1 + x^2)^4} \\
&= \frac{- 4(1 + x^2)^2 + 16x^2(1+x^2)}{ (1 + x^2)^4} \\
&= \frac{ (-4(1 + x^2) + 16x^2)(1+x^2)}{ (1 + x^2)^4} \\
&= \frac{ (-4 + 12x^2)(1+x^2)}{ (1 + x^2)^4}
\end{align}
Now plug in $x = 0$ to get
\begin{align}
H_0 &= \left. \frac{ (-4 + 12x^2)(1+x^2)}{ (1 + x^2)^4} \right|_{x=0} \\
&= \frac{ (-4 )(1)}{ (1)^4} \\
&= -4.
\end{align}
| {
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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{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{a^3}\geq 9\frac1{\sqrt[3]{2^8a}}.
\end{equation}
I am sure, from this point you can get continue it by yourself to get minimum $P={\frac{9^3}{8^3}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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
$$
\begin{align}
\sum_{n=0}^\infty\binom{6+n}{n}x^n(6+n)
&=\sum_{n=0}^\infty\binom{6+n}{n}x^n(7+n)-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{7+n}{n}x^n-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{-8}{n}(-x)^n-\sum_{n=0}^\infty\binom{-7}{n}(-x)^n\\[3pt]
&=\frac7{(1-x)^8}-\frac1{(1-x)^7}\\[6pt]
&=\frac{6+x}{(1-x)^8}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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_nk_{n-1}-h_{n-1}k_n=(-1)^n \Rightarrow$$924k_{5-1}-285h_{5-1}=(-1)^5$.
Thus, when we find $h_{4}$ we'll get the equation:
$-285h_{4} \equiv (-1)^5 \pmod{924} \Rightarrow$
$ 285h_{4} \equiv (-1)^{4}\pmod{924} \Rightarrow$
$ 285h_{4}(-1)^{4} \equiv 1\pmod{924} \Rightarrow$
$x=h_{4}(-1)^{4}177$ is a solution, because $h_{4}(-1)^{4}$ is $285^{-1}$ modulo 924, thus $285x \equiv 285*285^{-1}b \equiv b \pmod{924}$
Let's find $h_{4}$:
$\frac{h_1}{k_1}=\frac31 \Rightarrow h_1=3$
$\frac{h_2}{k_2}=3+\frac14=\frac{13}4 \Rightarrow h_2=13$
Using $h_n=q_nh_{n-1}+h_{n-2}$:
$h_3=6*13+3=81$
$h_4=1*81+13=94$ Bingo.
Thus $x=94*(-1)^4*175$.
This question is from an exam. The correct answer is $x \equiv 153,461,769 \pmod{924}$
Did I do something wrong? how can $x$ have multiple options(edit: how can I find all the options)?
Any help would be highly appreciated!
| $285x≡177\pmod{924}$ or $285x+924y=177$ can be solved using the extended euclidean algorithm for the gcd of $285$ and $924$.
The remainder sequence of the euclidean algorithm starts with $r_0=a=924$, $r_1=b=285$ and for the extended variant the Bezout factor sequence for the Bezout identity
$$v_k b≡r_k\;\pmod a\qquad (\text{short version of }u_ka+v_kb=r_k)$$
starts with $v_0=0$, $v_1=1$
Then the algorithm proceeds iterating over $k=1,2,...$
\begin{align}
q_k&=r_{k-1}\,\text{div}\,r_k,\\
r_{k+1}&=r_{k-1}-q_kr_k, \\
v_{k+1}&=v_{k+1}-q_kv_k
\end{align}
resulting in the table
$$\begin{array}{c|c|c|c|}
k& q_k&r_k&v_k\\\hline
0 & & 924 & 0 \\
1 & 3 & 285 & 1 \\
2 & 4 & 69 & -3 \\
3 & 7 & 9 & 13 \\
4 & 1 & 6 & -94 \\
5 & 2 & 3 & 107 \\
6 & & 0 & -308 \\
\end{array}$$
that is, in the end we get the $gcd(285,\,924)=3$ and
$$107\cdot 285 ≡ 3 \pmod{924}.$$
Fortunately, $177=3⋅59$ is divisible by $3$, so $3x≡3⋅59⋅107\pmod{3⋅308}$ has the 3 solutions
\begin{align}
x&=59⋅107\mod 308&&=153, \\
x&=153+308&&=461&&\text{ and } \\
x&=461+308&&=769.\end{align}
| {
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Prove that $\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$ Prove that $$\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{\sin x}{x}=\frac{1}{x} \int_0^x \cos y\,dy$ (true)
*Assuming $n=k$ (true)
*Prove $n=k+1$
| $\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}\frac{d}{dx}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy$ (Product Rule)
$=\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}x^{n-1}\cos\left(x+\frac{(n-1)\pi}{2}\right)$ (Fundamental Theorem of Calculus)
$=\frac{-n}{x^{n+1}}\left(|\frac{1}{n}y^n\cos\left(y+\frac{(n-1)\pi}{2}\right)|_0^x+\int_{0}^{x} \frac{1}{n}y^{n}\sin\left(y+\frac{(n-1)\pi}{2}\right)dy\right)+\frac{1}{x^n}x^{n-1}\cos\left(x+\frac{(n-1)\pi}{2}\right)$ (integration by parts)
$=\frac{-1}{x^{n+1}}\left(x^n\sin\left(x+\frac{n\pi}{2}\right)-\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right)dy\right)+\frac{1}{x^n}x^{n-1}\sin\left(x+\frac{n\pi}{2}\right)$ (trigonometric identities)
$=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right)dy$ (first and last terms cancel)
| {
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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}$
First I thought of taking x,y and z as $\displaystyle \tan A, \tan B,$ and $\tan C,$ making $A+B+C=\pi$, but couldnt solve ahead. Any solution not involving trigonometry would do as well.
I also think that this question does not even relate to trigo.....or does it?
| 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,$ $\displaystyle m(A+B+C)=mn\pi\implies \tan(mA+mB+mC)=\tan mn\pi=0$
$\displaystyle\implies\sum\tan mA=\tan mA\tan mB\tan mC$
Now set $m=3$ and apply $\tan3x$ formula
| {
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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 problem at all though I think I could have done something by using Cauchy-Schwarz inequality, but could not pull this together. Please help.
| 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}+\frac b{(c-a)^2}+\frac c{(a-b)^2}\\+\frac {a+c}{(a-b)(b-c)}+\frac {a+b}{(c-a)(b-c)}+\frac{b+c}{(c-a)(a-b)}=0\tag 1$$
With the three later fractions, we can multiply by $1$ as follows:
$$\frac {(a+c)(c-a)}{(a-b)(b-c)(c-a)}=\frac {c^2-a^2}{(a-b)(b-c)(c-a)}$$
Doing this to each yields
$$\frac {c^2-a^2}{(a-b)(b-c)(c-a)}+\frac {b^2-c^2}{(a-b)(b-c)(c-a)}+\frac {a^2-b^2}{(a-b)(b-c)(c-a)}=0$$
This equivalence should be clear by inspection, which transforms $(1)$ into
$$\frac a{(b-c)^2}+\frac b{(c-a)^2}+\frac c{(a-b)^2} = 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 = \frac{5}{4}$ I can say:
$e^x + e^{-x} = \frac{5}{2}$
$\therefore e^{2x} - \frac{5}{2}e^x + 1 = 0$
Solving this quadratic gives:
$e^x = 2$ and $e^x = \frac{1}{2}$
$\therefore x = \ln2$ and $x = \ln \frac{1}{2}$
If we do the same thing for $\cosh x=2$
$e^x + e^{-x} = 4$
$\therefore e^{2x} - 4e^x + 1 = 0$
Solving this quadratic gives:
$e^x = 2 \pm \sqrt{3}$
$\therefore x = \ln(2 + \sqrt{3})$ and $x = \ln(2 - \sqrt{3})$
When I put these four values on $x$ (separately) into my original equation they all work. However, the mark scheme to this question accepts only:
$x= \ln 2$ and $x = \ln(2 + \sqrt{3})$
How come they're not taking the negative square root when solving the quadratic in terms of $e^x$ into account?
Thank you :)
EDIT:
Here is a photo of the original question:
| 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)$?
| {
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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$ with $x, y, z \leq 10$ (where $x,y,z$ are positive integers and can be the same)
What are the different methods by which we can solve this?
| 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 , \ldots $ (see note below), we can represent this as the generating function $$ A(x) = 1 + x + x^2 + \cdots + x^i + \cdots . $$ We have the same generating function for the possible values of $b$ and $c$. So our generating function for the number of solutions is $A(x) \times B(x) \times C(x) = [A(x)]^3$. However, finding this product could be extremely tedious.
We instead transform $ A(x) $ into the rational function $ \frac{1}{1-x} $, which we recognize from the sum of a geometric progression. Thus, we are interested in $ [A(x)]^3 = \frac{1}{(1-x)^3} $. This can be expanded using the negative binomial theorem, which gives
$$ \frac{1}{(1-x)^3} = {2 \choose 2} + {3 \choose 2} x + { 4 \choose 2 } x^2 + \cdots + { i+2 \choose 2 } x^ i + \cdots. $$
Therefore, the answer when $n=7 $ is given by $ { 9 \choose 2 } = 36$. This agrees with what we know from the stars and bars method. $_\square$
Note: It might be confusing why we allow $a$ to be any non-negative integer, even those which are larger than $n$, which clearly would not lead to a solution. Consider what would happen if we let $a = n+1$ or $a = n+2$ or any larger integer: In the final product of polynomials, the exponents of these terms would be larger than $n,$ so they will not contribute to the term we want, which has exponent $n.$
| {
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"url": "https://math.stackexchange.com/questions/700216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Compute limit of sequence Let $(x_n)$ be real sequences such that $x_{1}=\dfrac{1}{3}, x_{2n}=\dfrac{1}{3}x_{2n-1}, x_{2n+1}=\dfrac{1}{3}+x_{2n}, n=1,2,\cdots $.
Compute $$\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}
$$
so
$$
\lim_{n \to \infty} \sup x_{n} = 1/2
$$
$$
\lim_{n \to \infty} \inf x_{n} = 1/6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/700283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x). Problem :
If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).
My approach :
The given equation can be written as $$(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)f(x-y)}{(x-y)(x+y)} =4xy$$
$$\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy$$
Now we know that $$(x+y)^2 -(x-y)^2 = 4xy$$
$\Rightarrow \frac{f(x+y)}{x+y} =(x+y)^2$....(i)
& $\frac{f(x-y)}{x-y} = (x-y)^2$...(ii)
Now putting y =0 in (i) and (ii) we get :
$\frac{f(x)}{x} =x^2$ $\Rightarrow f(x) =x^3$
But the answer is $f(x) =x^3 +kx$ ( where k is any constant ) please clarify this part thanks...
| $\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy=(x+y)^2-(x-y)^2$
or, $\frac{f(x+y)}{x+y}-(x+y)^2=\frac{f(x-y)}{x-y}-(x-y)^2$
make the substitutions, $x\to \frac{x+y}{2}$ and $y \to \frac{x-y}{2}$
Then the above becomes $\frac{f(x)}{x}-(x)^2=\frac{f(y)}{y}-(y)^2=k(say)$
There you have $f(x)=x^3+kx$
| {
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"url": "https://math.stackexchange.com/questions/702695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:
Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$
My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!
| $\dfrac{\sec \theta \cdot \sin\theta}{\tan\theta+\cot\theta}=\dfrac{\sec \theta \cdot \sin\theta}{\tan\theta+\frac{1}{\tan\theta}}$
$= \dfrac{\sec \theta \cdot \sin\theta\cdot \tan\theta}{\tan^2\theta+1}$
$=\dfrac{\sec\theta\cdot\sin\theta\cdot\frac{\sin\theta}{\cos\theta}}{\sec^2\theta}$
$=\dfrac{\sin\theta\cdot\sin\theta}{\sec\theta\cdot\cos\theta}$
$=\dfrac{\sin^2\theta}{1}=\sin^2\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/709134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this inequality $\frac{a^2+bc}{3+a}+\frac{b^2+ca}{3+b}+\frac{c^2+ab}{3+c}\le \frac{7}{4}$ let $a,b,c\ge 0$,show that
$$\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{a^2+bc}{3+a}+\dfrac{b^2+ca}{3+b}+\dfrac{c^2+ab}{3+c}\le \dfrac{7}{4}$$
then I can't,Thank you very much
| After full expanding we need to prove that
$$\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,$$
which is true by Schur.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How prove this inequality $3b+8c+abc\le 12$ if $a^2+4b^2+9c^2=14$ let $a,b,c>0$ and such $$a^2+4b^2+9c^2=14$$show that
$$3b+8c+abc\le 12$$
My idea: since
\begin{align*}3b+8c+abc&=3b+c(8+ab)=3b+\dfrac{1}{9}\cdot 9c(8+ab)\le 3b+\dfrac{1}{9}\cdot\dfrac{1}{4}[81c^2+(ab+8)^2]\\
&=3b+\dfrac{1}{36}[126-9a^2-36b^2+a^2b^2+16ab+64]\\
&=\dfrac{1}{36}(a^2b^2-9a^2-36b^2+16ab+108b+190)
\end{align*}
then we only prove this inequality
$$a^2b^2-9a^2-36b^2+16ab+108b+190\le 36\times 12=432$$
$$\Longleftrightarrow a^2b^2-9a^2-36b^2+16ab+108b\le 242$$
then I can't prove it,Thank you for you help
| $\dfrac{b^2+1}{2}\ge b \implies \dfrac32(b^2+1)\ge 3b$
$\dfrac{c^2+1}{2}\ge c \implies 4(c^2+1)\ge 8c$
Adding these two equations,
$4c^2+\dfrac32b^2+\dfrac{11}{2}\ge 3b+4c$
Now, $12-\dfrac{11}{2}=\dfrac{13}{2}$
We know, $7>\dfrac{13}{2}$ and $7=\dfrac{(a^2+4b^2+9c^2)}{2}$
So, if we can now prove that $\dfrac12a^2+\dfrac12b^2+\dfrac12c^2\ge abc+\dfrac12$, Then we are done.
Can you do it from here?
I think weighted AM-GM should help.
EDIT: As contributed by Barto, here is the complete solution of the question.
To prove, $a^2+b^2+c^2\ge 2abc+1$, we can re-write this equation as,
$14(a^2+b^2+c^2)\ge 28abc+a^2+4b^2+9c^2 \implies 13a^2+10b^2+5c^2\ge 28abc$
By weighted AM-GM, $13a^2+10b^2+5c^2\ge28\sqrt[28]{a^{26}b^{20}c^{10}}\ge 28abc$ from @MarkBennett's comment.
Hence, proved.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How would you determine the transformation matrix? Suppose there exists a linear transformation $T$ where $T: \mathbb{R^3} \to \mathbb{R^5}$ and $T(\textbf{x}) = \text{A} \textbf{x}$. Given
$$ \text{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1\\ 2\\ 1 \\3 \end{pmatrix} ,
\quad
\text{A} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} \hspace{6pt}1 \\ \hspace{6pt}2\\ \hspace{6pt}2\\ \hspace{6pt}2 \\-1 \end{pmatrix} ,
\quad
\text{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3\\ 4\\ 3 \\2 \end{pmatrix} ,
$$
how can we then determine the transformation matrix $\text{A}$? What are some "standard methods" to do it? Isn't this the same as finding the change-of-coordinates matrix when making a change of basis?
|
Isn't this the same as finding the change-of-coordinates matrix when
making a change of basis?
Totally.
Let $P = \begin{pmatrix} 1 & 1& 0 \\ 1 &2 & 1 \\ 1 & 3& 1\end{pmatrix}$
It's easy to see that $P\in GL_3(\mathbb R)$.
Your equalities can be rewrite:
$$AP= \begin{pmatrix} 1&1&2 \\ 1&2&3\\ 2&2&4\\ 1&2&3 \\3 &-1&2\end{pmatrix}$$
Finally:
$$ A = \begin{pmatrix} 1&1&2 \\ 1&2&3\\ 2&2&4\\ 1&2&3 \\3 &-1&2\end{pmatrix} P^{-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Summation of a series with certain property The sequence $\{a_n\}$ has the property that $a_1 + a_2+\cdots+a_n = n ^3~~\forall n$. Compute the value of $\frac{1}{a_2} + \frac{1}{a_3}+\ldots+ \frac{1}{a_{2014-1}}$.
I know that I have to somehow a arrange the denominators in such a order that I could take advantage of the property. To add them, I have to multiply the denominators to obtain a common denominators.
| First observe that
$$
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.
$$
Thus
\begin{align}
\frac{1}{a_2-1}+\cdots+\frac{1}{a_{2014}-1}&=\frac{1}{3 (2^2-2)}+\cdots+\frac{1}{3 (2014^2-2014)}=\frac{1}{3}\sum_{k=2}^{2014}\frac{1}{k(k-1)}\\ &=\frac{1}{3}
\sum_{k=2}^{2014}\left(\frac{1}{k-1}-\frac{1}{k}\right)
=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2014}\right)=\frac{671}{2014},
\end{align}
since
$$
\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$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}$$
so for
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$ minimum I can't find it,thank you
| Here care must be taken while using AM-GM so that the constraint and equality condition is not violated. Hence rewrite the objective as:
$$\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)$$
The first part is a sum of four terms with constant product. Hence this achieves minimum when each term is the same. Hence for this part,
$$\implies 16ab^3=16a^3b=1 \implies a=b=\tfrac12$$
So for the first part, the minimum is $\frac12$. For the second part, we have
$$\frac1{a^2}+\frac1{b^2} = \left(\frac1a-\frac1b\right)^2+\frac2{a^2b^2} \ge 0+ \frac8{(a+b)^2} \ge 8$$
which is also achieved when $a=b=\frac12$.
Hence the minimum of the objective within the constraints is achieved when $a=b=\frac12$, and the value is $\frac{33}4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solving complex numbers with powers algebraically Find algebraically the value of :$\left(2^{0.5} + 6^{0.5} - \left( 2^{0.5} - 6^{0.5} \right)i \right)^4$
Below are my works
I try to simplify inside. but i found that i can't add $2^{0.5}$ and $6^{0.5}$ together.
| Simplify by manipulating the inner expression:
$$\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{6})i)^4 \\
&= (\sqrt{2} + \sqrt6i)^4(1 - i)^4\end{align}$$
Then,
$$\begin{align}|z| &= |\sqrt{2} + \sqrt6i|^4\cdot|1 - i|^4 \\
&= (\sqrt{2 + 6})^4 \cdot (\sqrt{1 +1})^4 \\
&= 256\end{align}$$
On the other hand, $$\begin{align}\arg{z} &= 4\arg(\sqrt{2} + \sqrt6i) + 4 \arg(1 - i) \\
&= 4 \tan^{-1}\frac{\sqrt{6}}{\sqrt{2}} -4 \tan^{-1}1 \\
&= \frac{\pi}{3}\end{align}$$
Hence
$$z = 256e^{i\frac{\pi}{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find constants $a$ and $b$ such that $ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$
Find constants $a$ and $b$ such that $$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$$
First,$a$ should be positive to make sure the limit is meaningful as $x \to 0^-$ .
Then I check the limit of the numerator,say $- \frac{a}{2}<x<0$. For t in the interval(x,0), there's $\frac{1}{\sqrt{a+t}}<M$, where $M>0$,so $$| \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} | <M |\int^x_0t^2dt| =M \frac{-x^3}{3},$$ by using the sandwich theorem I get $\lim_{x \to 0^-} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} =0$,thing should be the same when $x>0$ .
So the limit is the form $0/0$ . Apply L'Hopital's rule twice I get
$$\begin{align} \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} &= \lim_{x \to 0} \frac{x^2}{ \sqrt{a+x}(bx-\sin x)} \\ &= \lim_{x \to 0} \frac{2x} { \frac{bx-\sin x}{2 \sqrt{a+x}}+ \sqrt{a+x} (b- \cos x)} \\ &= \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3bx+2ab-2(a+x)\cos x -\sin x} \end{align}$$Now the limit becomes $ \frac{0}{2ab-2a}$,so $2a(b-1)=0$ and $b=1$ cause $a$ is positive.
Apply L'Hopital's rule again $$\begin{align} \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3x+2a-2(a+x)\cos x -\sin x} &= \lim_{x \to 0} \frac{4( \sqrt{a+x} + \frac{x}{2 \sqrt{a+x}})}{3+2a \sin x- 2 \cos x + 2 x \sin x - \cos x} \\ & = \lim_{x \to 0} \frac{2(2a+3x)}{\sqrt{a+x} (3-3 \cos x + 2a \sin x + 2x \sin x )} \\ &= \frac{4a}{\sqrt{a} 0} \quad ?!\end{align}$$
and I fail to solve it.
Thanks in advance.
| Let's substitute $t\mapsto tx$:
$$
\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}
$$
If $b\ne1$, the limit on right would be $0$. Thus, we need $b=1$.
In this answer, it is shown that $\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$, therefore, we want
$$
\frac1{3\sqrt{a}}\cdot6=1\implies a=4
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 can see that $p \equiv 11 \pmod {24}$ and $p \equiv 23 \pmod {24}$ are consistent with that, but can anyone explain why the other possible congruences for a prime $p$ (such as $p \equiv 1 \pmod {24}$) are excluded by $p$ being a safe prime?
My reasoning so far is:
For a prime $p \equiv 1 \pmod 6$ and $p \equiv (1,7,13,19) \pmod {24}$, or $p \equiv 5 \pmod 6$ and $p \equiv (5,11,17,23) \pmod {24}$.
For a safe prime $p = 2q + 1$ it cannot be true that $p \equiv 1 \pmod 6$, otherwise $2q$ would be divisible by 6 and $q$ would not be prime. This eliminates 1, 7, 13 and 19.
Likewise $p = 2q+1 \equiv 5 \pmod {24}$ and $p = 2q+1 \equiv 17 \pmod {24}$ cannot hold, otherwise $q$ would have to be even: $q \equiv 2 \pmod {24}$ or $q\equiv 8\pmod {24}$ respectively.
This leaves $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ as possible congruences.
Is this correct and sufficient, and/or is there a better way of demonstrating that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ can and must hold?
| 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 1 \bmod 3$, then $p=2q+1$ is divisible by $3$, so it is not a prime. Hence $p\equiv q\equiv 2 \bmod 3$.
*Now consider the options for $q\bmod 8$. Since $q$ is prime, $q\equiv 1,3,5,7\bmod 8$ and, respectively, $p$ would be $p\equiv 2q+1\equiv 3,7,3,7\bmod 8$, so there are only two possibilities $p\equiv 3$ or $7\bmod 8$.
Now we can use the Chinese remainder theorem. Solving:
$$\begin{cases} x\equiv 2\bmod 3,\\
x\equiv 3 \text{ or } 7 \bmod 8, \end{cases}$$
leads to only two solutions modulo $24$, namely $11$ or $23\bmod 24$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$ Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$.
I tried some elementary ways while all failed.
| From Sum of tangent functions where arguments are in specific arithmetic series and Prove the trigonometric identity $(35)$,
$$\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}$$
where $\displaystyle t=\tan x$
Now if $\displaystyle\tan13x=0,13x=n\pi$ where $n$ is any integer
$\displaystyle\implies x=\frac{n\pi}{13}$ where $0\le n\le13-1$
As $\displaystyle\tan0=0,\tan\frac{n\pi}{13},1\le n\le12$ are the roots of
$\displaystyle\binom{13}1-\binom{13}3t^2+\binom{13}5t^4-\binom{13}7t^6+\binom{13}9t^8-\binom{13}{11}t^{10}+t^{12}=0\ \ \ \ (1)$
As $\displaystyle\tan(r\pi-y)=-\tan y,$ for any integer $r$
the given relation $\displaystyle(i)\sum\limits_{k=1}^{12}\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}=2\sum\limits_{k=1}^6\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}$
$\displaystyle(ii)u_k=\tan^2\frac{k\pi}{13},1\le k\le6$
shall be the roots of
$\displaystyle\binom{13}1-\binom{13}3u+\binom{13}5u^2-\binom{13}7u^3+\binom{13}9u^4-\binom{13}{11}u^5+u^6=0\ \ \ \ (2)$
Now, $\displaystyle y_k=\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}=\frac{3u_k-u_k^2}{1-3u_k}$ (using $\tan3A$ formula )
$\displaystyle\implies u_k^2=3u_k-y_k(1-3u_k)=3u_k(1+y_k)-y_k$
So, we need to transform Equation $\#(2)$ in terms of $y$
Then apply Vieta's Formulas.
But my Question is why $\displaystyle\tan\frac{k\pi}{13}$ is paired with $\displaystyle\tan\frac{3k\pi}{13},$ but not with $\displaystyle\tan\frac{2k\pi}{13}$ or $\displaystyle\tan\frac{5k\pi}{13}$ etc.?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/718346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to compute the integral $(1+t^2) / (1-t^2) $ compute $\displaystyle\int \frac{1+t^2}{1-t^2}dt$
I tried splittting them to two parts, and computed $\displaystyle\int \frac{1}{1-t^2}dt$ using trig sub, but i don't know how to compute the second part.
| Use the trig substitution as you did before, but don't stop there! use $$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}}$$
Giving
$$\frac{1+t^2-1}{1+t^2}=\frac{t^2}{1+t^2}=\sin^2{\frac{\theta}{2}}$$
So overall (use line 3 and 5!)
$$\frac{1-t^2}{1+t^2}=\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}$$
Using Double Angle Formula this gives
$$\frac{1-t^2}{1+t^2}=\cos{\theta}$$
and
$$dt = \frac{1}{2}\sec^2{\frac{\theta}{2}}$$
You may not need it all but have a go now :D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/718982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluate the limit $\lim\limits_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\frac1x}$ Evaluate
$$
\displaystyle\lim_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
$$
And actually I have my answer and just need someone to verify this for me since I haven't done something like this for a long time.
First, to deal with the pesky $1/x$, I take the natural log inside the limit:
\begin{align}
\lim_{x\to0+}\ln\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
&= \lim_{x\to0+}\frac{1}{x}\ln\left(\frac{3^x+5^x}{2}\right)\\
&= \lim_{x\to0+}\frac{\ln(3^x+5^x)-\ln2}{x}\\
&= \lim_{x\to0+}\frac{3^x\ln3+5^x\ln5}{3^x+5^x}......L'Hopital's \;Rule\\
&=\frac{\ln3+\ln5}{2}\\
&=\frac{1}{2}\ln3+\frac{1}{2}\ln5
\end{align}
And since what we calculated was the limit the of the natural log, the final answer would be $\displaystyle e^{\frac{1}{2}ln3+\frac{1}{2}ln5}=e^{\sqrt{3}+\sqrt{5}}$. Please tell me if I did this correctly, thanks.
| The indetermination is of the form $1^{+\infty}$, suggesting the application of fundamental limit:
$$\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^+} \left(1+\frac{3^x+2^x}{2}-1 \right)^{\frac{1}{x}}=\lim_{x\rightarrow 0^+} \left( 1+\frac{3^x+5^x-2}{2}\right)^{\frac{2}{3^x+5^x-2}\cdot \frac{3^x+5^x-2}{2x}}=$$
$$=e^{\lim_{x\rightarrow 0^+} \frac{1}{2}\left(\frac{3^x-1}{x}+\frac{5^x-1}{x} \right)} =e^{\frac{1}{2}(\ln 3+\ln 5)}=e^{(\ln 3\cdot 5)^{\frac{1}{2}}}=\sqrt{15}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplifying a radical with fraction and sum What are the steps to simplify
$(1+(\frac{1}{2}(x^3-\frac{1}{x^3})^2))^ \frac{1}{2}$
to
$\frac{1}{2}(x^3+\frac{1}{x^3})$ ?
| Hint: Try using the identities $(a-b)^2=a^2-2ab+b^2$ and $(a+b)^2=a^2+2ab+b^2$ (which are actually the same identity).
Be careful: check whether the result holds for $x\lt0$.
I am assuming you meant
$$
\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.$$
Then sum this two inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/724617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$ I am trying to evaluate this integral.
$$
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.
$$
I was trying to do use this series expansion but wasn't sure how to go about it because of the square of the logarithm. And also it seems than we then will have $I\propto \int_0^\infty x^{n-3/2}dx$ which will diverge. Thanks
| Using Richard Feynman's favorite method, the method of differentiation under the integral sign.
$$
\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+\alpha x)}{\sqrt{x}(1+\alpha x)}dx.
\end{align}
$$
Let $\,x=t^2\;\Rightarrow\;dx=2t\,dt$, then
$$
\begin{align}
I'(\alpha)&=2\int_0^\infty\frac{\ln(1+\alpha t^2)}{t(1+\alpha t^2)}\cdot2t\,dt\\
&=4\int_0^\infty\frac{\ln(1+\alpha t^2)}{(1+\alpha t^2)}dt.
\end{align}
$$
To solve the integral part, again we use the method of differentiation under the integral sign.
$$
\begin{align}
I(\beta)&=\int_0^\infty\frac{\ln(1+\alpha\beta t^2)}{(1+\alpha t^2)}dt\\
\frac{dI(\beta)}{d\beta}&=\int_0^\infty\frac{\alpha t^2}{(1+\alpha\beta t^2)(1+\alpha t^2)}dt\\
I'(\beta)&=\int_0^\infty\left[\frac{1}{(\beta-1)(1+\alpha t^2)}-\frac{1}{(\beta-1)(1+\alpha\beta t^2)}\right]dt\\
&=\frac{1}{\beta-1}\int_0^\infty\left[\frac{1}{1+\alpha t^2}-\frac{1}{1+\alpha\beta t^2}\right]dt.
\end{align}
$$
Note that
$$
\int_0^\infty\frac{1}{1+k y^2}dy=\frac{\pi}{2\sqrt{k}}.
$$
Therefore
$$
\begin{align}
I'(\beta)&=\frac{1}{\beta-1}\left(\frac{\pi}{2\sqrt{\alpha}}-\frac{\pi}{2\sqrt{\alpha\beta}}\right)\\
&=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{\sqrt{\beta}-1}{\sqrt{\beta}(\beta-1)}\right)\\
&=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{\left(\sqrt{\beta}-1\right)}{\sqrt{\beta}(\beta-1)}\cdot\frac{\left(\sqrt{\beta}+1\right)}{\left(\sqrt{\beta}+1\right)}\right)\\
&=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{1}{\sqrt{\beta}\left(\sqrt{\beta}+1\right)}\right)\\
\frac{dI(\beta)}{d\beta}&=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{1}{\sqrt{\beta}}-\frac{1}{\sqrt{\beta}+1}\right)\\
I(\beta)&=\frac{\pi}{2\sqrt{\alpha}}\int\left(\frac{1}{\sqrt{\beta}}-\frac{1}{\sqrt{\beta}+1}\right)d\beta\\
&=\frac{\pi}{2\sqrt{\alpha}}\left(2\sqrt{\beta}-2\sqrt{\beta}+2\ln\left(\sqrt{\beta}+1\right)+\text{C}_1\right)\\
&=\frac{\pi}{2\sqrt{\alpha}}\left(2\ln\left(\sqrt{\beta}+1\right)+\text{C}_1\right).\\
\end{align}
$$
For $\beta=0$ implying $I_\beta(0)=0$, then $\text{C}_1=0$ and
$$
I_\beta(1)=\int_0^\infty\frac{\ln(1+\alpha t^2)}{(1+\alpha t^2)}dt=\frac{\pi\ln 2}{\sqrt{\alpha}}.
$$
Now, plug in $I_\beta(1)$ to $I'(\alpha)$.
$$
\begin{align}
I'(\alpha)&=4\cdot\frac{\pi\ln 2}{\sqrt{\alpha}}\\
\frac{dI(\alpha)}{d\alpha}&=4\pi\ln 2\cdot\alpha^{-\frac{1}{2}}\\
I(\alpha)&=4\pi\ln 2\int\alpha^{-\frac{1}{2}}\,d\alpha\\
&=(4\pi\ln 2)\left(2\alpha^{\frac{1}{2}}+\text{C}_2\right)
\end{align}
$$
For $\alpha=0$ implying $I_\alpha(0)=0$, then $\text{C}_2=0$. Thus
$$
I_\alpha(1)=\int_0^\infty\frac{\ln^2(1+x)}{x^{\frac{3}{2}}}dx=\boxed{\color{blue}{\large8\pi\ln 2}}
$$
$$$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/730869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
How to find this angle? $P$ is a point in $\triangle ABC$. If $\angle PAB = 10^\circ$, $\angle PBA = 20^\circ$, $\angle PAC = 40^\circ$ and $\angle PCA = 30^\circ$, find $\angle B$.
| Let $\angle PBC=x$. so for the condition $\angle PCB=80^\circ-x$.
*
*Apply "trigonometric expression of ceva's theorem"-$ \frac{\sin 10^\circ}{\sin 40^\circ}\times \frac{\sin x}{\sin 20^\circ}\times \frac{\sin 30^\circ}{\sin(80^\circ-x)} =1$
*From that we have $\frac{\sin x}{\sin(80^\circ-x)}=4\cos 10^\circ \sin 40^\circ=4\cos 10^\circ\cos 50^\circ=2(\cos 60^\circ+\cos 40^\circ)=1+2\cos 40^\circ$
*Now see .........$\sin 60^\circ-\sin 20^\circ=2\cos 40^\circ\sin 20^\circ$. from this equation we have $1+2\cos 40^\circ=\frac{\sin 60^\circ}{\sin 20^\circ}=\frac{\sin x}{\sin(80^\circ-x)}$. so comparing we can say $x=60^\circ$
*so $\angle B=20^\circ+60^\circ=80^\circ$.
-
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/731078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To Find $A^{50}$ $$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 to find $B^{12}$ or is there a better possibility?
| 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 deduce the answer...
On a more rigorous note: note that $A^2=\begin{bmatrix} 1&0&0\\1&1&0\\1&0&1 \end{bmatrix} =(\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}+I)$.
Now $\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}^2 =0$.
So that $(\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}+I)^{n}=\binom{n}{n-1} \begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}^1I^{n-1}+I^{n}=n\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}I+I$.
(This last step is just using the binomial theorem combined with the result directly above it. You can write out the full binomial expansion and then just reduce to zero all higher powers of the nilpotent matrix.)
So then $A^{50}=(A^2)^{25}$, so that $n=25$ and the result follows.
| {
"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
} |
$\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} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ and find if it's the only one.
It does not seem that hard, my approach is the following, but I seem to be missing something:
Let $B = \left \{ \overline{\epsilon_{1}}, \overline{\epsilon_{2}}, \overline{\epsilon_{3}} \right \}$ be a base of $V_{3} $
Then the vectors $\overline{a}$. $\overline{b}$ can be represented as:
$\overline{a} = a_{1}\overline{\epsilon_{1}} + a_{2}\overline{\epsilon_{2}} + a_{3}\overline{\epsilon_{3}} = \sum_{i=1}^{3}a_{i}\overline{\epsilon_{i}}$
$\overline{b} = \sum_{j=1}^{3}b_{j}\overline{\epsilon_{j}}$
I also analyze the vector $\overline{x}$ as follows:
$\overline{x} = x_{1}\overline\epsilon_{1} + x_{2}\overline\epsilon_{2} +x_{3}\overline\epsilon_{3}$
$\overline{x} \times \overline{a} = \overline{b}$
$(x_{2}a_{3} - x_{3}a_{2})\overline\epsilon_{1} + (x_{3}a_{1} -x_{1}a_{3})\overline\epsilon_{2} + (x_{1}a_{2} - x_{2}a_{1})\overline\epsilon_{3} = \sum_{j=1}^{3}b_{j}\overline\epsilon_{j}$
And I get:
$x_{2}a_{3} - x_{3}a_{2} = b_{1}$
$x_{3}a_{1} - x_{3}a_{3} = b_{2}$
$x_{1}a_{2} - x_{3}a_{1} = b_{3} $
I am aware that $\langle\overline{a},\overline{b} \rangle =0$ means that the two vectors at vertical with each other, but how to I proceed from here?
| 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} = {b_3} \\
\end{gathered} \right.$$
We know that $\vec a\perp\vec b$ which means $a_1b_1+a_2b_2+a_3b_3=0$
we can solve for $b_3$ getting $b_3=-\dfrac{a_1b_1+a_2b_2}{a_3}$ and substitute into the previous system
$$\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} = -\dfrac{a_1b_1+a_2b_2}{a_3} \\
\end{gathered} \right.$$
Solving the system we get:
$$\left\{ \begin{gathered}
{x_1} = \frac{{{b_2}{b_3} - {a_1}{b_3}{x_3}}}{{{a_1}{b_1} + {a_2}{b_2}}} \\
{x_2} = - \frac{{{b_1}{b_3} - {a_2}{b_3}{x_3}}}{{{a_1}{b_1} + {a_2}{b_2}}} \\
{x_3} = {x_3} \\
\end{gathered} \right.$$
Which shows that the solution is not unique
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/732753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\frac{r}{R}+1=\cos A+\cos B+\cos C$? How do we prove that for any triangle this holds: $$\frac{r}{R}+1=\cos A+\cos B+\cos C$$ I can use this beautiful identity to prove several geometric inequalities, but I have no idea how to prove the identity itself. Can anyone give me hints?
| here is mechanical solution:
$\cos A+\cos B+\cos C-1=\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}=\dfrac{(a+b-c)(b+c-a)(c+a-b)}{2abc}=\dfrac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{2abc(a+b+c)}=\dfrac{8S^2}{abc(a+b+c)}=\dfrac{\dfrac{S}{s}}{\dfrac{abc}{4S}}=\dfrac{r}{R}$
$S=\sqrt{s(s-a)(s-b)(s-c)},s=\dfrac{a+b+c}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/734395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\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^{3n-2}+o\left(x^{3n-2}\right)
\end{align}
$$
Applying the claim to
$$
\tan(x)=x+\frac13x^3+\frac2{15}x^5+\frac{17}{315}x^7+o\left(x^7\right)
$$
and
$$
\sin(x)=x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7+o\left(x^7\right)
$$
yields
$$
\begin{align}
&\tan(\sin(x))-\sin(\tan(x))\\[9pt]
&=\left(2\left(-\frac2{15}\cdot\frac16-\frac13\cdot\frac1{120}\right)+3\cdot\frac13\cdot\frac16\left(\frac16+\frac13\right)\right)x^7+o\left(x^7\right)\\
&=\frac1{30}x^7+o\left(x^7\right)
\end{align}
$$
Proof of Claim: Simply, compose the series:
$$
\begin{align}
f(g(x))
&=\left(x+b_nx^n+b_{2n-1}x^{2n-1}+b_{3n-2}x^{3n-2}+o\left(x^{3n-2}\right)\right)\\
&+a_n\left(x+b_nx^n+b_{2n-1}x^{2n-1}+b_{3n-2}x^{3n-2}+o\left(x^{3n-2}\right)\right)^n\\
&+a_{2n-1}\left(x+b_nx^n+b_{2n-1}x^{2n-1}+b_{3n-2}x^{3n-2}+o\left(x^{3n-2}\right)\right)^{2n-1}\\
&+a_{3n-2}\left(x+b_nx^n+b_{2n-1}x^{2n-1}+b_{3n-2}x^{3n-2}+o\left(x^{3n-2}\right)\right)^{3n-2}\\
&+o\left(x^{3n-2}\right)\\
&=x+(a_n+b_n)x^n+(a_{2n-1}+b_{2n-1}+na_nb_n)x^{2n-1}\\
&+\left(a_{3n-2}+b_{3n-2}+na_nb_{2n-1}+\frac{n(n-1)}{2}a_nb_n^2+(2n-1)a_{2n-1}b_n\right)x^{3n-2}\\
&+o\left(x^{3n-2}\right)
\end{align}
$$
Note that the first three terms are symmetric in $a$ and $b$, so they will vanish in the difference.
Similarly,
$$
\begin{align}
g(f(x))
&=x+(a_n+b_n)x^n+(a_{2n-1}+b_{2n-1}+na_nb_n)x^{2n-1}\\
&+\left(a_{3n-2}+b_{3n-2}+na_{2n-1}b_n+\frac{n(n-1)}{2}a_n^2b_n+(2n-1)a_nb_{2n-1}\right)x^{3n-2}\\
&+o\left(x^{3n-2}\right)
\end{align}
$$
Subtract.
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
question on surds i already asked this question but the answer I got did not match the one in the book $$\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} $$
this is where I got stuck:
$$ \frac {x^2 +2x \sqrt{3} +3}{3x} $$
| Hint
Squaring both sides gives $3x = (x + \sqrt{3})^2 = x^2 + 2\sqrt{3}x + 3$.
Rewrite this to get $x^2 + (2\sqrt{3} - 3)x + 3 = 0$. Solve this second degree equation to get
$$x = 1.5 -\sqrt{3} \pm \sqrt{(1.5-\sqrt{3})^2 - 3}$$
Now try to rewrite this on the form $A\sqrt{B}+C$. Be sure to check which root actually is a solution to the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/743508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 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^{k+1}} + 5^{2^{k+1}} $$
I try by expanding, by subtraction, but no solution /:
| Hint:
$$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)$$
What can $2^k\text{ mod }6$ be?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate for $(1+\tan 20^\circ)(1+\tan 25^\circ)$. Help me with my works I have no idea what I am doing here,
I started with $\tan 20^\circ=\tan(45^\circ-25^\circ)=(1-\tan 25^\circ)/(1+\tan 25^\circ)$
I am sure the work I have shown so far are ok, but how do you get $1+\tan 20^\circ=2/(1+\tan 25^\circ)$ from that?
| $1 = \tan(45^\circ) = \tan(20^\circ+25^\circ) = \frac{ \tan 20^\circ + \tan 25^\circ}{1- \tan 25^\circ \tan 20^\circ} $
so
$2 = 1+ \tan 20^\circ + \tan 25^\circ + \tan 20^\circ \tan 25^\circ = (1+ \tan 20^\circ)(1+ \tan 25^\circ)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this value of $x$ such $\log_{\frac{1}{12}}{(x^2+2x-3)}>x^2+2x-16$ if $$\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-17}>y-4$$
Then I can't.Thank you
| Let, $x^2+2x-16=y$. So, $x^2+2x-3=y+13$.
Now, we have, $y+13<\bigg(\dfrac1{12}\bigg)^{y} \implies 12^y(y+13)<1$
From this we can conclude, when $y<-1$ or $x^2+2x-15<0$ or $x\in[-5,3]$ there exists solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does $2^{n+1} + 2^{n+1} = 2^{n+2}$? Simple question, why does:
$2^{n+1} + 2^{n+1} = 2^{n+2}$ ?
Furthermore, why does this only work for powers of 2?
Thanks.
| On the LHS you have two factors of $2^{n+1}$ so it is simply a case of factoring out a common factor on the left hand side:
$$2^{n+1} + 2^{n+1} = 2^{n+1} \cdot (1 + 1) = 2^{n+1}\cdot 2 = 2^{n+2}$$
For a more combinatorial proof we could also have used double counting. Imagine you want to figure out the number of ways that $n+2$ can form a group of people including zero people, so that each person can choose whether or not to join a group, in essence each person has two choices, and since we have $n+2$ people we have $2\times 2\times 2 \times \cdots \times 2 = 2^{n+2}$. On the other hand if we take those $n+2$ people and split them in two groups where the two groups share $\lfloor\frac{n+2}{2}\rfloor$ among themselves (each group then has a total of $n+1$ people) and write out the set of subsets of each group and then take the union of both set of subsets we see that is equal to the set of subsets of $n+2$ people (if we do not allow for repetitions). Since those individual groups of people of $n+1$ could form $2^{n+1}$ different groups then we have $$2^{n+1} + 2^{n+1} = 2^{n+2}$$
The algebraic proof is a little easier to understand though.
| {
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How to differentiate $\frac{2x^5}{\tan x}$ $$\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 \\
& = \tan x\cdot10x^4-2x^5.(1+(\tan x)^2))/(\tan x)^2\\
& = 10x^4/\tan x - 2x^5.\sec2x \\
& = 10x^4*\cot x - 2x^5\sec2x \\
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How find this $x^3-5x+10=2^y$ let $x,y$ is positive integer,and such
$$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)=(2,3)$$
I only find $x\le 7$ this solution.
maybe this have other solution.and This problem is from Mathematical olympiad problems
Thank you
| First $(10,4)$ and $(10,6)$ are not solutions, neither is $(10,y)$ for any integer $y$, since $(10)^3-5(10)+10=2^y$ would imply $5\mid2^y$, a contradiction.
Now, rearrange the given equation to get
$$
(x^2-5)x=2(2^{y-1}-5)\tag{1}
$$
Now either $x=2k+1$ or $x=2k$ for some integer $k>1$;
let $x=2k+1$, direct substitution gives
$$(4k^2+4k+1-5)(2k+1)=2(2^{y-1}-5)$$
or
$4(k^2+k-1)(2k+1)=2(2^{y-1}-5)$, a contradiction, since it would imply $2\mid5$.
Hence, $x$ is even and $x^2-5$ is odd (since $d=\gcd(x^2-5,x)\mid5$ so that $d=1$ or $5$, if $d=5$ we get a contradiction from the above equation, hence $d=1$.
CASE 2. $x=2k$, in this case, $k$ is either odd or even,
i) $k$ is even, hence $x=4t$, for some $t>1$, substitution into the rearranged equation gives $[(4t)^2-5]\cdot4t=2(2^{y-1}-5)$, a contradiction, since it implies $2\mid5$.
ii) $k$ is odd, $x=4t+2$, substitute:
$[(4t+2)^2-5]*(4t+2)=2(2^{y-1}-5)$ or
$(64t^3+96t^2+28t-2)=2(2^{y-1}-5)$ or
$(64t^2+96t+28)t=2^y-8$ so that $2\mid t$, hence $t=2r$ and $x=8r+2$;
substitution gives
$[(8r+2)^2-5]\cdot(8r+2)=2(2^{y-1}-5)$ or
$[(8r+2)^2-5]\cdot(4r+1)=(2^{y-1}-5$ or
$64r^3+48r^2+7r+1=2^{y-3}$
From here we have $r \equiv 9 \pmod{16}$ or $r=16s+9$ and hence $x=128s+74$, choosing integer values for $s$ can help narrow the search along with $x=8r+2$ where $r$ is odd.
| {
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Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| A very fast solution
Maximum value of $\displaystyle4\sin\theta+3\cos\theta$ is $\displaystyle{5}$. This value is achieved when $\displaystyle\tan\theta=\dfrac{4}{3}$ by differentiating.So $\cos\theta=\dfrac{3}{5}$ and $\sin\theta=\dfrac{4}{5}$.
Hence $4\cos\theta-3\sin\theta=0$.
| {
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Showing that $\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$ I was reading an article in which it was stated that, with a change of variable, one could show that:
$$\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$$
I tried with $t = 1 + \frac{1}{x}$ but that doesn't work out, especially because the lower bound doesn't become $0$.
| With the substitution $u=\frac{1}{x}$ we get
$$\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}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$
3) $a+b+c=0$
4) $c=0$
Update
I just tried to sub $c$ into both of the equations:
$c^2+cb+a=0$ and $c^2+ac+b=0$ which then gives us the equality
$c^2+cb+a=c^2+ac+b$
$ => cb+a=ac+b$
$=> b(c-1)=a(c-1)$
which gives me then a=b which is contradictory because the integers are supposed to be distinct.
Update #2
Ok it looks like 1) is true, 2) is true, 3) is true, and 4) is false .. right?
| We know that $c^2+bc+a = 0$ and $c^2+ac+b = 0$. Subtract to get $c(b-a)-(b-a) = 0$ or $(b-a)(c-1) = 0$. This means that either $b = a$ or $c = 1$. If $c = 1$, then $1+a+b = c+a+b = 0$.
| {
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Finding all primes $(p,q)$ for perfect squares. Find all prime pairs $(p,q)$ such that $2p-1, 2q-1, 2pq-1$ are all perfect squares.
Source: St.Petersburg Olympiad 2011
I have only found the pair $(5,5)$ so I am thinking that maybe a modulo $5$ approach could work.
| Although Ivan Loh's answer is correct, I believe I may have found a much simpler and elementary proof that $(5,5)$ is the only pair.
Assume without loss of generality that $q\geq p$ and write $2pq-1 = x^2$,
$2q-1 = y^2$, with $x$ and $y$ positive integers which are necessarily both odd.
We have the following inequalities.
$$x = \sqrt{2pq-1} < \sqrt{2pq} \leq q\sqrt{2}$$
$$y = \sqrt{2q-1} < \sqrt{2q}$$
For the first inequality we have used the assumption that $p\leq q$. Now we have that
$$x^2 - y^2 = (2pq-1) - (2q-1) = 2q(p-1)$$
So $2q(p-1) | (x+y)(x-y)$. Since $q$ is prime, we must now have either
$q|(x-y)$ or $q|(x+y)$. Additionally, $x$ and $y$ are both odd, so
$x-y$ and $x+y$ are both even. Combining this with the preceding statement we see that either $2q|(x-y)$ or $2q|(x+y)$.
The former cannot occur since $x-y < x < q\sqrt{2} < 2q$, so if
$2q|(x-y)$ then $x = y$ which would imply that $p = 1$.
So $2q|(x+y)$. Now $x + y > 0$ so we must have
$$2q \leq x + y < q\sqrt{2} + \sqrt{2q}$$
Rearranging this we get
$$(2-\sqrt{2})q < \sqrt{2q}$$
so
$$(2-\sqrt{2})^2q < 2$$
and so $q < \frac{2}{(2-\sqrt{2})^2} < 5.8$. Since both $p$ and $q$ are $1$ mod $4$, this leaves $p = q = 5$ as the only possible pair.
| {
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For which n does the inequality $2 \uparrow^{n+1}n > 3\uparrow^n 3 +2$ hold? For which n does the following inequality hold ?
$$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)$$
for $n\ge 8$
where $f_{\omega+1}(n)$ is a function from the
fast growing hierarchy and G(n) is Graham's
sequence
$$G(1) = 3\uparrow^4 3$$
$$G(n+1) = 3\uparrow^{G(n)} 3$$
for all n > 0.
| The inequality holds precisely when $n \ge 4$.
When $n=3$, we have
$$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$$
Using the fact that $2 \uparrow^m (n+2) > 3\uparrow^m n + 2$,(proven here) we have for $n \ge 4$:
$$2 \uparrow^{n+1} n \ge 2\uparrow^{n+1} 4 = 2\uparrow^n(2\uparrow^n 4) > 2 \uparrow^n 5 > 3 \uparrow^n 3 + 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How should I try to evaluate this integral $\int_0^\pi \sqrt{1+4\sin^2\frac x2 - 4\sin\frac x2}\;dx$? Suppose that we are given the following integral:
$$\int_0^\pi \sqrt{1+4\sin^2\frac x2 - 4\sin\frac x2}\;dx.$$
(Original screenshot)
And the answer is one of these :-
*
*$4\sqrt3-4-\frac\pi3$
*$\pi-4$
*$\frac{2\pi}3 - 4 - 4\sqrt3$
*$4\sqrt3-4$
(original screenshot )
Is there a way to check each answer and reach to the question (like in case of indefinite integral I can differentiate the options and match it with question and tick the correct answer) or maybe confirm that this value is correct for the question without solving.
| \begin{align}
\int_0^\pi\sqrt{4\sin^2\frac{x}{2}-4\sin\frac{x}{2}+1}\,dx&=\int_0^\pi\sqrt{\left(2\sin\frac{x}{2}-1\right)^2}\,dx\\
&=\int_0^\frac{\pi}{3}\left(1-2\sin\frac{x}{2}\right)\,dx+\int_\frac{\pi}{3}^\pi\left(2\sin\frac{x}{2}-1\right)\,dx\\
&=\left[x+4\cos\frac{x}{2}\right]_0^\frac{\pi}{3}+\left[-4\cos\frac{x}{2}-x\right]_\frac{\pi}{3}^\pi\\
&=\frac{\pi}{3}+4\cos\frac{\pi}{6}-4-\pi+4\cos\frac{\pi}{6}+\frac{\pi}{3}\\
&=4\sqrt{3}-4-\frac{\pi}{3}
\end{align}
If this is multiple choices, I'd immediately answer 1 because its value greater than $0$ and we can easily notice that the answer must have term of $\pi$ because there is term integral of $1$ that yield $x$.
| {
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A different type binomial expansion problem Suppose we have $$(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}.$$
What will be the value of $a_0^2 - a_1^2 + a_2^2 - \cdots + a_{2n}^2$?
The answer is $a_n$, but I can't solve it.
See, what I've done is substitute $x$ as $-\frac{1}{x}$ and I've got:
${\frac{(x^2-x+1)}{x^2}}^n = a_0 - \frac{a_1}{x} + \frac{a_2}{x^2}+...$
I've got the alternating signs but I can't get the squares of the numbers.
| Since
$$
(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}\right)^n\\
&=\sum_{k=0}^{2n}a_kx^{k-2n}\\
&=\sum_{k=0}^{2n}a_{2n-k}x^{-k}\tag{3}
\end{align}
$$
Therefore, $(2)$ and $(3)$ show that $a_k$ is palindromic; that is,
$$
a_k=a_{2n-k}\tag{4}
$$
Furthermore, using $(1)$ and substituting $x\mapsto-x$, we get
$$
(1-x+x^2)^n=\sum_{k=0}^{2n}(-1)^ka_kx^k\tag{5}
$$
Using $(1)$, $(4)$, $(5)$, and the formula to multiply power series, we get the coefficient of $x^{2n}$ in $(1+x+x^2)^n(1-x+x^2)^n$ is
$$
\sum_{k=0}^{2n}(-1)^ka_ka_{2n-k}=\sum_{k=0}^{2n}(-1)^ka_k^2\tag{6}
$$
We can use $(1)$ to get that the coefficient of $x^{2n}$ in $(1+x^2+x^4)^n$ is
$$
a_n\tag{7}
$$
Noting that $(1+x+x^2)^n(1-x+x^2)^n=(1+x^2+x^4)^n$, $(6)$ and $(7)$ show that
$$
\sum_{k=0}^{2n}(-1)^ka_k^2=a_n\tag{8}
$$
| {
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Non-Homogenous System Find the general solution of $\vec{x^{'}}=\begin{pmatrix} 1&1\\4&1 \end{pmatrix}x+\begin{pmatrix}2\\-1\end{pmatrix}e^{t}$
I found the eigenvalues to be $\lambda_1=3 \;\; \lambda_2=-1$
Next I calculated the eigenvectors to be $e_1=\begin{pmatrix}1\\2 \end{pmatrix}$ and $e_2=\begin{pmatrix}1\\-2 \end{pmatrix}$ Next I form the fundamental matrix $$\phi(t)= \begin{pmatrix} e^{3t}&e^{-t}\\2e^{3t}&-2e^{-t} \end{pmatrix}$$
Finding the inverse yields $\phi^{-1}(t)=\dfrac{1}{4} \begin{pmatrix}2e^{-t}&e^{-t}\\2e^{3t}&-e^{3t} \end{pmatrix}$
I proceeded to multiply the matrix by $g(t)$ to obtain $$\begin{pmatrix} 4-1\\4e^{4t}+e^{4t} \end{pmatrix} \implies \begin{pmatrix} 3\\5e^{4t} \end{pmatrix}$$
I integrated that and multiplied by $\phi(t)$ to get $\dfrac{1}{4} \begin{pmatrix} 3te^{3t}+\frac{5}{4}e^{3t}\\6te^{3t}-\frac{5}{2}e^{3t} \end{pmatrix}$
I left the $\dfrac{1}{4}$ out from the inverse to avoid working with fractions. Now the book gets $x_h+\dfrac{1}{4}\begin{pmatrix}1\\-8 \end{pmatrix}e^{t}$ where $x_h$ is the solution to the homogenous. Is the book wrong or am I wrong somewhere?
| It looks like your $\phi^{-1}(t)$ went astray.
I got:
$$\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}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability of first actor winning a "first to roll seven with two dice" contest? Two players P and Q take turns, in which they each roll two fair and independent dice. P rolls the dice first.
The first player who gets a sum of seven wins the game. What is the probability that player
P wins the game?
| The probability that $ P $ eventually wins the game can be represented as the sum of an infinite geometric series.
For example,
The probability of $ P$ winning in the first round itself is $ \dfrac{1}{6} $. (Since out of the $ 36 $ possible outcomes, there are $ 6 $ outcomes in which the sum of the dice is equal to $ 7 $.)
The probability of $ P $ winning in the second round is $ \dfrac{5}{6} \times \dfrac{5}{6}\times \dfrac{1}{6} $. This means that for $ P $ to win th second round, he has to lose the first round, then $ Q $ has to lose his turn and finally $ P $ wins. This goes on.
The final answer is
$$ \large\dfrac{1}{6} + \left( \dfrac{5}{6}^{2} \times \dfrac{1}{6} \right) + \left( \dfrac{5}{6}^{4} \times \dfrac{1}{6} \right) + \dots = \displaystyle\sum\limits_{i=0}^{\infty} \dfrac{1}{6} \times \dfrac{5}{6}^{2n}$$
This is an infinite geometric series and its sum is equal to
$$ \large\dfrac{\dfrac{1}{6}}{\left( 1 - \dfrac{5}{6}^2 \right)} = \boxed{\dfrac{6}{11}} $$
| {
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If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
| So let's track each term in $4x^2 + 2x + 1$.
Since $x \equiv 2 \pmod{3}$, we get the following:
$$ 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$$
We conclude that $3\mid (4x^2+2x+1)$.
| {
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How to solve this equation or system of equations? I want to solve the equation
$$(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}{5}(2x-3)+\dfrac{7}{5}(3x-2)=\dfrac{2}{5}a^2 + \dfrac{7}{5}b^2.$$
Similarly, we have
$$4x-5=\dfrac{7}{5}a^2+\dfrac{2}{5}b^2.$$
The given system of equation are written
$$\begin{cases}
\left (\dfrac{2}{5}a^2 + \dfrac{7}{5}b^2\right )a -\left (\dfrac{7}{5}a^2 + \dfrac{2}{5}b^2\right )b=2,\\
3a^2 -2b^2=-5.
\end{cases}$$
Equavalent to
$$\begin{cases}
2(a^3-b^3)-7ab(a-b)=10,\\
3a^2 -2b^2=-5.
\end{cases}$$
Now, I can not solve the last system of equations. How can I solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2$$
or solve the system of equations
$$\begin{cases}
2(a^3-b^3)-7ab(a-b)=10,\\
3a^2 -2b^2=-5.
\end{cases}$$
| A "classical" solution:
1) Conditions for the existence of radicals give us $x\geq \frac{3}{2}.$
2) The condition for the existence of equality $(5X-4)\sqrt{2x-3} >(4X-5)\sqrt{3x-2}$ give us: $x>2.$
3)We write the equation in the form $(5X-4)\sqrt{2x-3}-2=(4X-5)\sqrt{3x-2}$and raise a squared:
$$2x^3-3x^2-3x+6=4(5x-4)\sqrt{2x-3}$$
4) Note $\sqrt{2x-3}=t>1 $ and consequently $x=\frac{t^2+3}{2}.$
5) We obtain the equation in $t$:$$t^6+6t^4-40t^3+3t^2-56t+6=0<=>(t-3)(t^5+3t^4+15t^3+5t^2+3t-2)=0$$
with the solution $t=3$ and then $x=6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Algebra Iranian Olympiad Problem If:
$x^2+y^2+z^2=2(xy+xz+zy)$
and $x,y,z \in R^+$
Prove:
$\frac{x+y+z}{3} \ge \sqrt[3]{2xyz}$
I tried my best to solve this thing but no use.
Hope you guys can help me.Thanks in advance.
| If we put $a=\sqrt{x}$ and $b=\sqrt{y}$, the degree two equation (in $z$)
$x^2+y^2+z^2-2(xy+xz+yz)=0$ has two solutions, $(a-b)^2$ and $(a+b)^2$. By
cyclically permuting $x,y,z$, we may assume $z=(a+b)^2$. The inequality
to be shown is then equivalent to $(x+y+z)^3 \geq 54xyz$, or
$(a^2+b^2+(a+b)^2)^3 \geq 54(a^2b^2(a+b)^2)$. We are then done because
$$
(a^2+b^2+(a+b)^2)^3 -54(a^2b^2(a+b)^2)=2\Bigg((b-a)(2a+b)(a+2b)\Bigg)^2
$$
As guessed by CalvinLin, equality is reached exactly when
$(x,y,z)=(1,1,4)$ up to permutation.
| {
"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.