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Prove that $\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2}J_2(x) +c$
Prove that $$\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac
{8}{x^2}J_2(x) +c$$ where $J_n (x)$ is the Bessel function of first kind
and order $n $.
My attempt:
I know the following recurrence relations:
$$\frac {d}{dx}\left[x^nJ_n (x)\right]=x^nJ_{n-1} (x)$$
$$\frac {d}{dx}\left[x^{-n}J_n (x)\right]=-x^{-n}J_{n+1} (x)$$
$$2J'_n(x)=J_{n-1} (x)-J_{n+1} (x)$$
$$2\frac {n}{x}J_n(x)=J_{n-1} (x)+J_{n+1} (x)$$
But I cannot figure out how to get the above expression. Please help.
|
I will use the second of your recurrence relations:
$$
\frac{d}{dx}[x^{-n}J_n(x)] = -x^{-n}J_{n + 1}(x)
$$
And consider the next cases
*
*$n = 2$
$$
\frac{d}{dx}[x^{-2}J_2(x)] = -x^{-2}J_{3}(x) ~~~\Rightarrow~~~ \int dx\; x^{-2}J_3(x) = -x^{-2}J_2(x) \tag{1}
$$
*$n = 3$
$$
\frac{d}{dx}[x^{-3}J_3(x)] = -x^{-3}J_{4}(x) ~~~\Rightarrow~~~ -x^{2}\frac{d}{dx}[x^{-3}J_3(x)] = x^{-1}J_{4}(x)
$$
Therefore
\begin{eqnarray}
\int dx\; x^{-1}J_{4}(x) &=& -\int dx\; x^{2}\frac{d}{dx}[x^{-3}J_3(x)] ~~~\mbox{make}~~~ u = x^2\mbox{ and } dv = d(x^{-3}J_3(x)) \\
&=& -x^{-1}J_3(x) + \int dx\; 2x^{-2}J_3(x) \\
&\stackrel{(1)}{=}& -x^{-1}J_3(x) + 2[-x^{-2}J_2(x)] \tag{2}
\end{eqnarray}
*$n = 4$
$$
\frac{d}{dx}[x^{-4}J_4(x)] = -x^{-4}J_{5}(x) ~~~\Rightarrow~~~ -x^4\frac{d}{dx}[x^{-4}J_4(x)] = J_{5}(x)
$$
or equivalently
\begin{eqnarray}
\int dx\;J_5(x) &=& -\int dx\; x^4\frac{d}{dx}[x^{-4}J_4(x)] ~~~\mbox{call}~~~ u = x^4\mbox{ and } dv = d(x^{-4}J_4(x)) \\
&=& - J_4(x) + \int dx\; 4 x^{-1}J_4(x) \\
&\stackrel{(2)}{=}& -J_4(x) + 4\left[-x^{-1}J_3(x) -2 x^{-2}J_2(x) \right] \\
&=& -J_4(x) -\frac{4}{x}J_3(x) - \frac{8}{x^2}J_2(x)
\end{eqnarray}
Add the integration constant to get
$$
\int dx\; J_5(x) = -J_4(x) -\frac{4}{x}J_3(x) - \frac{8}{x^2}J_2(x) + c
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2046887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding the determinant using row operations. So I have to find the determinant of $\begin{bmatrix}3&2&2\\2&2&1\\1&1&1\end{bmatrix}$ using row operations. From what I've learned, the row operations that change the determinate are things like swaping rows makes the determinant negative and dividing a row by a value means you have to multiply it by that value. Once you have an upper triangular matrix then you just multiply the diagonal and you should have the determinant. So how come I can't get the right answer? Here's what I did. I start by dividing the first row by 3. $3\begin{bmatrix}1&2/3&2/3\\2&2&1\\1&1&1\end{bmatrix}$. Then I subtract the second row by 2 times the first row $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\1&1&1\end{bmatrix}$. Then I subtract the third and first row. $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\0&1/3&1/3\end{bmatrix}$. Then I multiply the third row by 2 and subtract it from the second row. $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\0&0&1\end{bmatrix}$. So now I have an upper triangular matrix so now I just do $(3)(1)(2/3)(1)$ and I get 2. The answer to this question however is not 2 but 1. So where did I go wrong? If the things I read were true then I should've got the right answer but I didn't which tells me that there's another row operation rule that no one told me about.
|
Your error is when you multiply the third row by $2$ and subtract the second row; this introduces a factor $2$ that you have to remove.
Use a more systematic way:
\begin{align}
\begin{bmatrix}
3&2&2\\
2&2&1\\
1&1&1
\end{bmatrix}
&\to
\begin{bmatrix}
1&2/3&2/3\\
2&2&1\\
1&1&1
\end{bmatrix}
&&R_1\gets \tfrac{1}{3}R_1 &&\color{red}{3}
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&2/3&-1/3\\
0&1/3&1/3
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-2R_1\\R_3&\gets R_3-R_1\end{aligned}
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&1&-1/2\\
0&1/3&1/3
\end{bmatrix}
&& R_2\gets\tfrac{3}{2}R_2 && \color{red}{\tfrac{2}{3}}
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&1&-1/2\\
0&0&1/2
\end{bmatrix}
&&R_3\gets R_3-\tfrac{1}{3}R_2
\\[4px]&\to
\begin{bmatrix}
1&2/3&2/3\\
0&1&-1/2\\
0&0&1
\end{bmatrix}
&&R_3\gets 2R_3 &&\color{red}{\tfrac{1}{2}}
\end{align}
The determinant is
$$
3\cdot\frac{2}{3}\cdot\frac{1}{2}=1
$$
|
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"url": "https://math.stackexchange.com/questions/2046993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find $\int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx$
Find $$\int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx$$
What I have tried
First method was to try $u$ substitution
Let $u=\cos(x)$ then $-du=\sin(x)dx$ then $\sin(x)=\sqrt{1-u^2} $ which transforms our integral into
$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx=-\int \frac{1}{1+u\sqrt{1-u^2}}du=-\int \frac{1-u\sqrt{1-u^2}}{u^4-u^2+1}du$$
I think the might the denominator could be seperated since we get $u^2= \frac{1\pm \sqrt{3}i}{2}$ but I wouldn't know how to proceed after that.
Another method I have tried is this
Using the Weierstrass substitution
Let $\tan(x/2)=t$ , so $\sin(t) = \frac{2t}{1+t^2}$ , $\cos(t)=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2dt}{1+t^2}$
Which transforms our integral as such
$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)}dx= \frac{4t}{t^4-2t^3+2t^2+2t+1} dt$$
I can't seem to find any roots by rational root theorem for the denonminator so I don't know how to proceed once again..
|
\begin{align*} &~~~~~\int\frac{\sin x}{1+\sin x\cos x}{\rm d}x\\ &=\int\frac{2\sin x}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{(\sin x+\cos x)+(\sin x-\cos x)}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{\sin x+\cos x}{2+2\sin x\cos x}{\rm d}x+\int\frac{\sin x-\cos x}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{\sin x+\cos x}{3-(\sin x-\cos x)^2}{\rm d}x+\int\frac{\sin x-\cos x}{(\sin x+\cos x)^2+1}{\rm d}x\\ &=\int\frac{{\rm d}(\sin x-\cos x)}{3-(\sin x-\cos x)^2}-\int\frac{{\rm d}(\sin x+\cos x)}{(\sin x+\cos x)^2+1}\\ &=\frac{1}{2\sqrt{3}}\ln\left|\frac{\sin x-\cos x+\sqrt{3}}{\sin x-\cos x-\sqrt{3}}\right|-\arctan(\sin x+\cos x)+C \end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2048910",
"timestamp": "2023-03-29T00:00:00",
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|
Find the greatest value of $a^2\cdot b^3$ where $a+b=10$ Find the greatest value of $a^2\cdot b^3$ where $a,b$ are positive real numbers satisfying $a+b=10$. Determine the values of $a,b$ for which the greatest value is attained.
At first I tried to find the value by putting $a=10-b$, I found a expression, but I got not way to do anything.
If we apply A.M-G.M. inequality, I think we found the minimum value or something else. It is a pre-Olympiad level problem. So I need some answers or some good hints.
|
The Am-GM inequality says that $\{a_n\}$ are given then $\frac {\sum a_n}{n} \geq \sqrt[n]{\prod a_n}$.
Rewrite $a+b = 10$ as $2 \times \frac a2 + 3 \times \frac b3 = 10$. Now, apply AM-GM and see what happens, noting that equality occurs when $\frac a2 = \frac b3$.
Applying AM-GM with the set $\{\frac a2,\frac a2,\frac b3,\frac b3,\frac b3\}$, we get:
$$
\frac{\frac a2+\frac a2+\frac b3+\frac b3+\frac b3}{5} \geq \sqrt[5]{\frac a2\cdot\frac a2\cdot\frac b3\cdot\frac b3\cdot\frac b3}
$$
Simplifying, we get that $\frac{a+b}{5} \geq \sqrt[5]{\frac {a^2b^3}{2^23^3}}$.
Knowing that $a+b=10$ gives us:
$$
2 \geq \sqrt[5]{\frac {a^2b^3}{2^23^3}}
$$
Taking the fifth power of both sides:
$$32 \times 2^23^3 \geq a^2b^3$$
which equates to $a^2b^3 \leq 3456$. This is attained when $3a = 2b$ i.e. $a=4,b=6$.
You can also differentiate, but then I believe this is not allowed in Olympiad exams.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2049448",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Limit of $\left(1+\frac{2}{n^2}\right)^n $
Compute$$\lim_{n\to\infty}\left(1+\frac{2}{n^2}\right)^n $$
I dont know how to do it without using continuity of exponential function
I mean I cannot do this:
$$\lim_{x\to\infty}a_n =a \wedge \lim_{x\to\infty}b_n =b\Rightarrow \lim_{x\to\infty}{a_n}^{b_n} =a^b$$
|
Use Bernoulli:
$$\left(1+\frac{2}{n^2}\right)^n= \frac{1}{\left(\frac{n^2}{n^2+2}\right)^n}= \frac{1}{\left(1-\frac{2}{n^2+2}\right)^n}$$
And by Bernoulli
$$\left(1-\frac{2}{n^2+2}\right)^n \geq 1-\frac{2n}{n^2+2}=\frac{n^2-2n+2}{n^2+2}$$
Therefore
$$1 \leq \left(1+\frac{2}{n^2}\right)^n \leq \frac{n^2+2}{n^2-2n+2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2050128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$
How to evaluate this integral?
$$\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$
Maybe we can start by $$\int_0^\infty {\frac{{dx}}{{({x^2} + 1)\cosh ax}}} = \frac{1}{2}\left[ {{\psi _0}\left( {\frac{a}{{2\pi }} + \frac{3}{4}} \right) - {\psi _0}\left( {\frac{a}{{2\pi }} + \frac{1}{4}} \right)} \right]$$ in this. Then take the derivative with respect to $a$, but I'm failed to solve it!
|
The given integral equals
$$ I=\frac{1}{8}\int_{0}^{+\infty}\frac{2\cosh(7\pi x)}{(1+x^2)\cosh^3(3\pi x)}\,dx = \frac{1}{8}\int_{\mathbb{R}}\frac{\cosh(7\pi x)}{(1+x^2)\cosh^3(3\pi x)}\,dx $$
and by computing the residues of the integrand function at $x=i$ and at $(2k+1)\frac{i}{2}$ we get:
$$ I = \frac{\pi}{8}+\frac{2\pi i}{8}\sum_{k\geq 0}\lim_{z\to\frac{2k+1}{2}i}\frac{d}{dz}\frac{\cosh(7\pi z)(z-(2k+1)i/2)}{(1+z^2)\cosh^3(3\pi z)}+R$$
that simplifies to:
$$ I = \frac{\pi}{8}-\frac{28}{27\pi}\sum_{k\geq 0}\frac{(2k+1)}{ (4k^2+4k-3)^2 }+R = \color{blue}{\frac{\pi}{8}-\frac{7}{27\pi}+R}$$
where $R$ is the contribute given by the poles at $\frac{i}{6},\frac{5i}{6},\frac{7i}{6},\frac{11i}{6}$ and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2051160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
$f(0)=0$ ; $f(x):=\int_0^x \cos \frac 1t \cos \frac 3t \cos \frac 5t \cos \frac 7t\,dt$, $\forall x \ne 0$. Is $f$ differentiable at $0$? Let $f:\mathbb R \to \mathbb R$ be defined by $f(0)=0$ and
$$
f(x):=\int_0^x \cos \dfrac 1t \cos \dfrac 3t \cos \dfrac 5t \cos \dfrac 7t \,dt ,\quad\forall x \ne 0.
$$
Then is $f$ differentiable at $0$?
|
First write the integrand as
$$\begin{aligned}
\cos\frac{1}{t}&\cos\frac{3}{t}\cos\frac{5}{t}\cos\frac{7}{t}\\
&=\frac{1}{8}\Bigl(1+\cos\frac{2}{t}+\cos\frac{4}{t}+\cos\frac{6}{t}+\cos\frac{8}{t}+\cos\frac{10}{t}+\cos\frac{14}{t}+\cos\frac{16}{t}\Bigr).
\end{aligned}
$$
The, let $f_k(x)=\int_0^x\cos\frac{k}{t}\,dt$ and $f_k(0)=0$. We rewrite the difference quotient for $f_k$ (for positive $x$, say, since it is even) using a substitution and an integration by parts:
$$
\begin{aligned}
\frac{f_k(x)-f_k(0)}{x}&=\frac{1}{x}\int_0^x\cos\frac{k}{t}\,dt=\bigl[u=k/t\bigr]=\frac{k}{x}\int_{k/x}^{+\infty}\frac{1}{u^2}\cos u\,du\\
&=\frac{k}{x}\Bigl[\frac{1}{u^2}\sin u\Bigr]_{k/x}^{+\infty}+\frac{2k}{x}\int_{k/x}^{+\infty}\frac{1}{u^3}\sin u\,du
\end{aligned}
$$
Taking the absolute value and estimating (triangle inequality and $|\sin u|\leq 1$), we get
$$
\Bigl|\frac{f_k(x)-f_k(0)}{x}\Bigr|\leq\frac{x}{k}+\frac{2k}{x}\int_{k/x}^{+\infty}\frac{1}{u^3}\,du=\frac{2x}{k}
$$
We find that $f_k$ is differentiable at $0$ with $f_k'(0)=0$. Since your function $f$, by the first identity, can be written as
$$
f(x)=\frac{x}{8}+\frac{1}{8}\bigl(f_2(x)+f_4(x)+f_6(x)+f_8(x)+f_{10}(x)+f_{14}(x)+f_{16}(x)\bigr)
$$
we find that your function is differentiable at $0$, and
$$f'(0)=\frac{1}{8}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2051257",
"timestamp": "2023-03-29T00:00:00",
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|
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $\ln(2)=0.693$.
Now, let's do some rearrangement:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......$$
$$
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}.......$$
$$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}......$$
$$\frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......)$$
$$\frac{1}{2}\ln(2).$$
I know that mathematics can't be wrong, and I have done something wrong. But here is my question: where does the argument above go wrong?
|
Such regrouping of a series is just not guaranteed not to alter the limit or even preserve convergence.
It would be admissible if the series were absolutely convergent, that is the series of absolute values would converges. But, the current series does not converge absolutely.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2051691",
"timestamp": "2023-03-29T00:00:00",
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|
Functional equation: $f: \mathbb{R} \rightarrow \mathbb{R}$, $f((x + 1) f(y)) = y (f(x) + 1)$ Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$
Source: Vuong Lam Huy, on a Facebook group.
|
Let $f : \mathbb{R} \to \mathbb{R}$ satisfy
$$ f((x+1)f(y)) = (f(x)+1)y, \quad : \forall x,y \in \mathbb{R} \tag{1} $$
Lemma 1. $f$ is bijective, $f(0) = 0$ and $f(f(x)) = x$.
Proof. Plug $y = 0$ to $\text{(1)}$. Then $f((x+1)f(0)) = 0$ In particular, $f(f(0)) = 0$.
Now plugging $x = f(0)$ to $\text{(1)}$, we have $f((f(0)+1)f(y)) = y$ for all $y$. This proves that $f$ is bijective. Then by the previous step, $(x+1)f(0)$ is constant. This implies $f(0) = 0$.
Lemma 2. $f(-x) = -f(x)$ and $f(xy) = f(x)f(y)$.
Proof. From Lemma 2, replacing $y$ by $f(y)$ in $\text{(1)}$ gives
$$f((x+1)y) = (f(x)+1)f(y), \quad : \forall x,y \in \mathbb{R} \tag{2}$$
On the other hand, replacing $(x,y)$ by $(x-1,1)$ in $\text{(2)}$ gives $f(x) = (f(x-1) + 1)f(1)$. Since $f(1) \neq 0$, this gives
$$ f(xy) = (f(x-1)+1)f(y) = \frac{1}{f(1)}f(x)f(y). $$
So it suffices to show that $f(1) = 1$. To this end, we first show that $f$ is odd. Indeed,
$$ f(-x)^2 = f(1)f((-x)^2) = f(1)f(x^2) = f(x)^2$$
together with the injectivity of $f$ shows that $f(-x) = -f(x)$. Next, plugging $x = -1$ to $\text{(2)}$ gives $0 = (f(-1)+1)f(y)$ for any $y$, which implies $f(-1) = -1$. Therefore $f(1) = 1$ and the claim follows.
Proposition. $f(x) = x$.
Proof. From Lemma 2, for $x > 0$ we have $f(x) = f(\sqrt{x})^2 > 0$. Also, plugging $y = 1$ to $\text{(2)}$ shows that $f(x+1) = f(x) + 1$ for any $x \in \mathbb{R}$. So if $0 < x < y$, then
$$ f(y) = f(x(1 + \tfrac{y-x}{x})) = f(x)(1 + f(\tfrac{y-x}{x})) > f(x) $$
and $f$ is increasing on $[0,\infty)$.
On the other hand, for $n \in \Bbb{Z}_{>0}$ and $x \neq 0$ we have $f(x^n) = f(x)^n$ and this extends to all of $n \in \Bbb{Z}$ by $f(1/x)f(x) = f(1) = 1$. Then for any rational number $r = p/q$ with $p,q \in \Bbb{Z}$,
$$ f(2^r)^q = f(2^{rq}) = f(2^p) = f(2)^p = 2^p $$
and hence $f(2^r) = 2^r$. Since $f$ is increasing, for any $x> 0$ we can choose sequences of rational numbers $(l_n)$ and $(u_n)$ so that $l_n \uparrow \log_2 x$ and $u_n \downarrow \log_2 x$. Then
$$ 2^{l_n} = f(2^{l_n}) \leq f(x) \leq f(2^{u_n}) = 2^{u_n} $$
and by the squeezing lemma, $f(x) = x$. Since $f$ is odd, the same is true for all $x \in \mathbb{R}$. ////
|
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"timestamp": "2023-03-29T00:00:00",
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|
The Most general solution satisfying equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$ The most general value of $x$ satisfying the equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$, is found to be $x=2n\pi+\frac{7\pi}{4}$.
My approach:
$$
\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}=\sin (\frac{\pi}{4}+\frac{\pi}{2})=\sin\frac{3\pi}{4}\\\implies x=n\pi+(-1)^n\frac{3\pi}{4}
$$
If I consider the cosine function
$$
\cos x=\frac{\sin x}{\tan x}=-\sin x=-\sin\frac{3\pi}{4}=\cos(\frac{3\pi}{4}+\frac{\pi}{2})=\cos\frac{5\pi}{4} \implies x=2n\pi+\frac{5\pi}{4}
$$
Is their anything wrong with my approach ?How do I compare different forms of general solutions without inputting for $n$ ?
|
$\tan \theta = -1 =\tan (\frac{3\pi}{4}) =\tan(\frac{7\pi}{4})$.
$\tan (\pi -\frac{3\pi}{4}) = \tan(\pi +\frac{3\pi}{4})$.
$\cos \theta =\frac{1}{\sqrt{2}}$.
$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
$\cos (2\pi -\frac{7\pi}{4}) =\frac{1}{\sqrt{2}} = \cos \frac{7\pi}{4}$.
Hence, the principal value for both $\tan \theta$ and $\cos \theta$ is $\frac{7\pi}{4}$.
Thus, the general value of $\theta$ is $2n\pi + \frac{7\pi}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2052250",
"timestamp": "2023-03-29T00:00:00",
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|
General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\Big)=\sqrt{a^2+b^2}\big(\sin y.\cos x+\cos y.\sin x\big)=\sqrt{a^2+b^2}.\sin(y+x)=2
$$
$\frac{a}{\sqrt{a^2+b^2}}=\sin y$ and $\frac{b}{\sqrt{a^2+b^2}}=\cos y$.
$$
{\sqrt{a^2+b^2}}=\sqrt{8}=2\sqrt{2}\\\tan y=a/b=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}-\frac{1}{2}.\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}}=\frac{\sin(\pi/3-\pi/4)}{\sin(\pi/3+\pi/4)}=\frac{\sin(\pi/3-\pi/4)}{\cos(\pi/3-\pi/4)}=\tan(\pi/3-\pi/4)\implies y=\pi/3-\pi/4=\pi/12
$$
Substituting for $y$,
$$
2\sqrt{2}.\sin(\frac{\pi}{12}+x)=2\implies \sin(\frac{\pi}{12}+x)=\frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\\\implies \frac{\pi}{12}+x=n\pi+(-1)^n\frac{\pi}{4}\implies x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{12}
$$
What's going wrong with the approach ?
|
$$(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$$
divide by $2\sqrt{2}$:
$$\frac{\sqrt{3}-1}{2\sqrt{2}}\cos x+\frac{\sqrt{3}+1}{2\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$$
now use
$$\cos a \cos b+\sin a \sin b=\cos (a-b)$$
$$
\cos^2 a+\sin^2 a=\bigg(\frac{\sqrt{3}-1}{2\sqrt{2}}\bigg)^2+\bigg(\frac{\sqrt{3}+1}{2\sqrt{2}}\bigg)^2=1
$$
and the fact that $$a=\arctan(\frac{\sqrt{3}+1}{\sqrt{3}-1})=\frac{5\pi}{12}$$
we have $$\cos(\frac{5\pi}{12}-x)=\frac{1}{\sqrt{2}}=\cos\frac{\pi}{4}$$
which gives
$$\frac{5\pi}{12}-x=\pm\frac{\pi}{4}+2k \pi$$
$$\boxed{x=2k\pi+\frac{2\pi}{3},\quad x=2k\pi+\frac{\pi}{6}}$$
|
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|
Use Mathematical Induction to prove equation? Use mathematical induction to prove each of the following statements.
let $$g(n) = 1^3 + 2^3 + 3^3 + ... + n^3$$
Show that the function $$g(n)= \frac{n^2(n+1)^2}{4}$$ for all n in N
so the base case is just g(1) right? so the answer for the base case is 1, because 4/4 = 1
then for g(2) is it replace all of the n's with n + 1 and see if there is a concrete answer?
|
Show that: $$1^3 + 2^3 + 3^3 + ... + n^3= \frac{n^2(n+1)^2}{4}$$
$(1)$ First step is to evaluate the expression for $n=1$, $$1^3=\frac{1\times(1+1)^2}{4}$$
Which is $1=1$, which means we can proceed.
$(2)$ Now assume it is true for $n$, $$1^3 + 2^3 + 3^3 + ... + n^3= \frac{n^2(n+1)^2}{4}$$
$(3)$ Lastly prove for $n+1$ and you have proved it for all $n\in\mathbb N$,
$$1^3 + 2^3 + 3^3 + ... +n^3+ (n+1)^3= \frac{(n+1)^2(n+2)^2}{4}$$
After "inserting" $n+1$, replace the left side as it follows:
$$ \frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$
Simply "tidy up" the left side to get the final form;
$$\frac{(n+1)^2(n+2)^2}{4}=\frac{(n+1)^2(n+2)^2}{4}$$
And that's an example of a proof with induction.
|
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|
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
|
I. The method described by the OP is a bit round-about. To continue, solve for $y$ in $3y-2=k^2$ then substitute into $y^2-x^2(3y-2)=0$. We get,
$$(2 + k^2 - 3 k x) (2 + k^2 + 3 k x)=0$$
Solve for $k$,
$$k =\frac{3x\pm\color{blue}{\sqrt{-8+9x^2}}}{2}$$
Thus one needs to solve $-8=z^2-9x^2$ in the integers. It is a Pell-like equation with a very small number of solutions and easily done.
II. Or, more quickly, just solve for $y$ in the original equation,
$$2x^2 + y^2 = 3x^2y$$
to get,
$$y =\frac{3x^2\pm x\color{blue}{\sqrt{-8+9x^2}}}{2}$$
which gives the same easy condition to solve.
|
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|
Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to.
Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
using only real-analytic techniques.
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am unable to do by hand.
|
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am unable to do by hand.
Because you seem to be looking for an approach without complex numbers; here is a not so neat, 'brute force' but real-valued approach.
Since $x^4 - x^3 + x^2 - x+ 1$ has no real roots, it has a factorization of the form:
$$\begin{align}
x^4 \color{blue}{-1} x^3 \color{red}{+1} x^2 \color{green}{-1} x\color{purple}{+ 1} & =\left( x^2+ax+b \right)\left( x^2+cx+d \right) \\
& = x^4 + \left( \color{blue}{a+c} \right)x^3 + \left(\color{red}{ac+b+d} \right)x^2 + \left( \color{green}{ad+bc}\right)x + \color{purple}{bd}
\end{align}$$
So you need to solve the system:
$$\left\{\begin{array}{rcr}
\color{blue}{a+c} & = & \color{blue}{-1} \\
\color{red}{ac+b+d} & = & \color{red}{1} \\
\color{green}{ad+bc} & = & \color{green}{-1} \\
\color{purple}{bd} & = & \color{purple}{1}
\end{array}\right.$$
Clearly $d=\tfrac{1}{b}$. If you suspect that a factorization exists (or if you'd simply try it) of the form $\left( x^2+ax+1 \right)\left( x^2+cx+1 \right)$, the system reduces to the far simpler:
$$\left\{\begin{array}{rcr}
a+c & = & -1 \\
ac+2 & = & 1 \\
\end{array}\right. \iff a = \frac{-1\pm\sqrt{5}}{2} \;,\; c = \frac{-1\mp\sqrt{5}}{2}$$
|
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|
How to factorize $a^2-b^2-a+b+(a+b-1)^2$? The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.
|
Expand it and treat it as a quadratic expression in $a.$ That is if $C\ne 0$ then $Ca^2+Da+E=\;[a-(D- F)/2C]\;[a-(D+F)/2C)]\;$ where $F=\sqrt {D^2-4CE}.$
In this Q, we get $D^2-4CE=4b^2-4b+1=(2b-1)^2$ so it will simplify.
That is $$a^2-b^2-a+b+(a+b-1)^2=$$ $$= a^2-b^2-a+b +(a^2+2ab+b^2)-2(a+b)+1=$$ $$=2a^2+2ab -3a+b-1=$$ $$=2a^2+a(2b-3)+(b-1).$$ Let $C=2, D=2b-3,$ and $E=b-1.$
|
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|
$\sin{x}+\sin{2x}+\sin{3x}=1+\cos{x}+\cos{2x}$ - solving a trigonometric equation I'm looking for hints regarding such equation:
$$ \sin{x}+\sin{2x}+\sin{3x}=1+\cos{x}+\cos{2x} $$
I'd be particularly interested in any clever method for dealing with such problem.
|
$$\sin x +\sin 3x=2\cdot\sin 2x \cdot\cos x$$
$$\cos 2x=2\cos^2x-1$$
Then
$$2\cdot\sin{2x}\cdot\cos x+\sin{2x}=1+\cos{x}+2\cos^2x-1$$
$$\sin{2x}\cdot(2\cos x+1)=\cos{x}\cdot(2\cos x+1) $$
$$(2\cos x +1)\cdot(\sin 2x -\cos x)=0$$
$$(2\cos x +1)\cdot(\cos x)\cdot(2\sin x -1)=0$$
$$\cos x=-\frac{1}{2} \Rightarrow x=\pm\frac{2\pi}{3}+2k\pi$$
$$\cos x =0 \Rightarrow x=\frac{\pi}{2}+k\pi$$
$$\sin x =\frac{1}{2} \Rightarrow x=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}+2k\pi$$
|
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|
Find the last Digit of $237^{1002}$? I looked at alot of examples online and alot of videos on how to find the last digit But the thing with their videos/examples was that the base wasn't a huge number. What I mean by that is you can actually do the calculations in your head. But let's say we are dealing with a $3$ digit base Number... then how would I find the last digit.
Q: $237^{1002}$
EDIT: UNIVERSITY LEVEL QUESTION.
It would be more appreciated if you can help answer in different ways.
Since the Last digit is 7 -->
*
*$7^1 = 7$
*$7^2 = 49 = 9$
*$7^3 = 343 = 3$
*$7^4 = 2401 = 1$
$.......$
$........$
*$7^9 = 40353607 = 7$
*$7^{10} = 282475249 = 9$
Notice the Pattern of the last digit. $7,9,3,1,7,9,3,1...$The last digit repeats in pattern that is 4 digits long.
*
*Remainder is 1 --> 7
*Remainder is 2 --> 9
*Remainder is 3 --> 3
*Remainder is 0 --> 1
So, $237/4 = 59$ with the remainder of $1$ which refers to $7$. So the last digit has to be $7$.
|
Simple version without the notation:
$7 \times 1 = 7$
$7 \times 7 = 49$
$7 \times 9 = 63$
$7 \times 3 = 21$
Just look at the last digit in each case.
So the last digit of $7^1$ is $7$. The last digit of $7^2$ is $9$. The last digit of $7^3$ is $3$. And, the last digit of $7^4$ is $1$.
Thus the last digit of $7^5$ is also $7$. And the last digit of $7^9$ is $7$ (because $7^4 \times 7^4 \times 7 = 7^9$, and the last digits thereof are $1 \times 1 \times 7$.)
And the last digit of $7^{51}$ is the same as the last digit of $7^{47}$, which is the same as the last digit of $7^{43}$, which is the same as the last digit of $7^{39}$ (see the pattern?)...which is the same as the last digit of $7^{7}$, which is the same as the last digit of $7^3$, which is $3$.
By the same logic, the last digit of $7^{1002}$ is the same as the last digit of $7^2$, which is $9$.
|
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|
How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$
$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$
Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?
|
One way is to use Simpson's rule.
Without it, one could argue what is faster, but:
If $A=(a+b)/2$ and $D=(b-a)/2$, then your first integrand is
$$
(x-A+D)(x-A-D)=(x-A)^2-D^2.
$$
Thus
$$
\begin{aligned}
\int_a^b (x-a)(x-b)\,dx &=\frac{1}{3}\bigl((b-A)^3-(a-A)^3\bigr)-2D^3\\
&=\frac{1}{3}\bigl(D^3+D^3)-2D^3=-\frac{1}{6}(b-a)^3.
\end{aligned}
$$
For the second one you write the integrand as
$$
\bigl((x-A)^2-D^2\bigr)(x-A)=(x-A)^3-D^2(x-A)
$$
and do a similar calculation. I leave that to you.
|
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|
Help verify the proof that $\min \left({\lfloor{N/2}\rfloor, \lfloor{(N + 2)/p}\rfloor}\right) = \lfloor{(N + 2)/p}\rfloor$ for $N \ge p$ Well, I think that I have this correct, however my proof seams cumbersome, so can someone check these results or suggest a simplified proof.
Prove that ($N \ge p$)
\begin{equation*}
\min \left({\left\lfloor{\frac{N}{2}}\right\rfloor, \left\lfloor{\frac{N + 2}{p}}\right\rfloor}\right) = \left\lfloor{\frac{N + 2}{p}}\right\rfloor
\end{equation*}
for prime $p \ge 3$ with $N \ge p$ and $N \in \mathbb{Z}^{+}$. To see this, let $N = k\, p + m$ where $k \in \mathbb{Z}^{+}$ and $m \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{p - 1}\right\}$ by the Quotient Remainder Theorem, then show that this equation holds for all values of $k$. Starting with
\begin{equation*}
\left\lfloor{\frac{N + 2}{p}}\right\rfloor = \left\lfloor{\frac{k\, p + m + 2}{p}}\right\rfloor = k + {\delta}_{2}
\end{equation*}
where
\begin{equation*}
{\delta}_{2} =
\left\lfloor{\frac{m + 2}{p}}\right\rfloor =
\begin{cases}
0, & \text{for } m \le p - 3, \\
1, & \text{for } m \in \left\{{p - 2, p - 1}\right\}.
\end{cases}
\end{equation*}
Also
\begin{equation*}
\left\lfloor{\frac{N}{2}}\right\rfloor =
\left\lfloor{\frac{k\, p + m}{2}}\right\rfloor =
\left\lfloor{\frac{k\, p}{2} + \frac{m}{2}}\right\rfloor.
\end{equation*}
Therefore
\begin{equation*}
k + {\delta}_{2} \le \left\lfloor{\frac{k\, p}{2} + \frac{m}{2}}\right\rfloor.
\end{equation*}
If $k$ is even then we have
\begin{equation*}
k + {\delta}_{2} \le \frac{k\, p}{2} + \left\lfloor{\frac{m}{2}}\right\rfloor.
\end{equation*}
Solving for $k$ gives
\begin{equation*}
k \ge \frac{2 \left({{\delta}_{2} - \left\lfloor{m/2}\right\rfloor}\right)}{p - 2}.
\end{equation*}
If ${\delta}_{2} = 0$ or ${\delta}_{2} = 1$ then $k \ge 2$ since $k$ is even and ${\delta}_{2} - \left\lfloor{m/2}\right\rfloor \le 1$. Now if $k$ is odd then we have three cases to consider starting with
\begin{equation*}
k + {\delta}_{2} \le \frac{\left({k + 1}\right) p}{2} + \left\lfloor{\frac{m - p}{2}}\right\rfloor.
\end{equation*}
Solving for $k$ gives
\begin{equation*}
k \ge \frac{2 \left({{\delta}_{2} - p/2 - \left\lfloor{\left({m - p}\right)/2}\right\rfloor}\right)}{p - 2}.
\end{equation*}
With $m = 0$, then ${\delta}_{2} = 0$ and
\begin{equation*}
k \ge \frac{2}{p - 2} \left({- \frac{p}{2} + \left\lceil{\frac{p}{2}}\right\rceil}\right) =
\frac{1}{p - 2} \ge 1
\end{equation*}
since $\left\lceil{p/2}\right\rceil = \left({p - 1}\right)/2 + \left\lceil{1/2}\right\rceil = \left({p + 1}\right)/2$ with $p \ge 3$. For the last two cases $m = p - 2$ or $m = p - 1$, $\delta = 1$, then
\begin{equation*}
k \ge \frac{4 - p}{p - 2} \ge 1
\end{equation*}
for $p \ge 3$. Thus the primary equation holds for $p \ge 3$.
|
You can prove this a bit easier using for example using proof by contradiction:
Let's assume $\left\lfloor{\frac{N}{2}}\right\rfloor < \left\lfloor{\frac{N + 2}{p}}\right\rfloor$. This implies $\frac{N}{2} < \frac{N+2}{p}$ or after simplification $p < 2+\frac{4}{N}$.
Now cases where $N\leq 4$ can be easily checked by hand, so let's assume $N>4$, but then $p<3$ and we have a contradiction with $p\geq 3$.
Notice that the proof did not use the assumption that $p$ is a prime, so it whole applies for any natural number $p$.
|
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|
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it:
Basecase: substitute $1$ for $n$, everything works out.
Inductive step: assume that $$\prod _{i=1}^{n}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x}$$
then $$\prod _{i=1}^{n+1}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1})$$
Let $a$, $b$ and $c$ be positive real numbers, then $\dfrac {c}{a}=b\Leftrightarrow ab=c$, thus
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=1-x^{2^{n+1}} \Leftrightarrow \dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$$
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}}$$
Applying polynomial division, we see that indeed $\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}} = 1+x^{2^{n+1}}$.
Thus $\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$.
However, the exercise was in a chapter on binomial coefficients and pascal's triangle, furthermore we didn't mention polynomial division in class. Which makes me think that there was another solution that I was "supposed" to see. How was I supposed to prove it?
|
On the left-hand side, the ratio of $\Pi_n$ over $\Pi_{n-1}$ is
$$1+x^{2^n},$$
and on the right-hand side,
$$\frac{\dfrac{1-x^{2^{n+1}}}{1-x}}{\dfrac{1-x^{2^n}}{1-x}}=\frac{1-(x^{2^n})^2}{1-x^{2^n}}.$$
Alternatively,
$$\frac{1-x^{2^{n+1}}}{1-x}=\sum_{k=0}^{2^n}x^k.$$
Then
$$(1+x^{2^n})\sum_{k=0}^{2^n}x^k=\sum_{k=0}^{2^n}x^k+\sum_{k=0}^{2^n}x^{k+2^n}=\sum_{k=0}^{2^{n+1}}x^k.$$
|
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|
Why is $x^x-(\sin x)^x\sim\frac{1}{6}x^3 $ when $x\to 0$? I'm learning Taylor's expansion.
The given solution of the problem is:
when $x\to 0$, $x^x-(\sin x)^x=x^x(1-(\frac{\sin x}{x})^x)\sim\frac{1}{6}x^3 $
But I don't know how to use Taylor's formula to get:
$x^x(1-(\frac{\sin x}{x})^x)\sim\frac{1}{6}x^3 $
|
Just use
$$\sin(x) \approx x - \frac{x^3}{6}$$
$$\frac{\sin x}{x} \approx 1 - \frac{x^2}{6}$$
$$x^x\left(1 - \left(1 - \frac{x^2}{6}\right)^x\right)$$
Now the bracket factor
$$\left(1 - \frac{x^2}{6}\right)^x$$
is small, so you can use Binomial expansion: $(1+X)^a = 1 + aX + \ldots$
That is
$$x^x\left(1 - \left( 1 - x\cdot \frac{x^2}{6}\right)\right) = x^x\left(1 - 1 + \frac{x^3}{6}\right) = x^x\left(\frac{x^3}{6}\right)$$
For $x\to 0$ $x^x \to 1$ hence you remain with
$$\frac{x^3}{6}$$
|
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|
Prove that if {a;b} $\in \mathbb R^+$ then $a^2+b^2>ab$ I have tried factoring it already, but it doesn't seem to evolve much:
First I multiply each side by $2$:
$ 2(a^2+b^2)>2ab$
Then I substitute using the relation $(a+b)^2=a^2+2ab+b^2$ and it becomes:
$2(a^2+b^2)>(a+b)^2 - (a^2+b^2)$
and then:
$3(a^2+b^2)>(a+b)^2$
And that's pretty much it, I'm stuck.
|
$$
(a-b)^2 \geq 0 \\
(a-b)^2 = a^2 - 2ab + b^2 \\
a^2 + b^2 \geq 2ab \geq ab
$$
|
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|
Is there a matrix $A$ such that: $A^4=\begin{pmatrix}0 & 2 & -1 & 1\\ 0 & 0 & 3 &1\\ 0 & 0& 0 & 4\\ 0 & 0 & 0 & 0 \\ \end{pmatrix} ~?$
Is there a matrix $A\in \mathbb{C}^{4\times 4}$ such that: $$A^4=\left(\begin{array}{cccc}
0 & 2 & -1 & 1\\
0 & 0 & 3 &1\\
0 & 0& 0 & 4\\
0 & 0 & 0 & 0 \\
\end{array} \right) ~?$$
Any hint is appreciated. Thanks a lot!
|
If an $n\times n$ matrix satisfies $A^m=0$ for some $m$ then it satisfies $A^n=0$.
notice that if $A^4$ was equal to that matrix then $(A^4)^4=0$, but then $A^4=0$.
|
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|
Calculate limit $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule How to calculate limit: $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule?
If $x = 2$, I get uncertainty $\frac{0}{0}$
|
$\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}
$
Putting $x = y+2$,
$\begin{array}\\
\dfrac{(2^x)-4}{\sin(\pi x)}
&=\dfrac{(2^{y+2})-4}{\sin(\pi (y+2))}\\
&=4\dfrac{(2^{y})-1}{\sin(\pi y)}
\qquad\text{since } 2^{y+2} = 4\cdot 2^y
\text{ and }\sin(\pi (y+2))=\sin(\pi y + 2\pi)=\sin(\pi y)\\
&=4\dfrac{e^{y\ln 2}-1}{\sin(\pi y)}\\
&\approx 4\dfrac{y\ln 2}{\pi y}
\qquad\text{since } e^z \approx 1+z \text{ and } \sin(z) \approx z
\text{ for small }z\\
&= \dfrac{4\ln 2}{\pi }
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$ Find coefficient of $x^n$ in
$(1+x+2x^2+3x^3+.....+nx^n)^2$
My attempt:Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2}{(1-x)^2}$. (Ignoring terms which have powers of x greater than $x^n$)
So one can say that
coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$
=coefficient of $x^n$ in $(1-x+x^2)^2(1-x)^{-4}$
Is there a shorter way.
|
Hint: use
$$ (f(x)g(x))^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}(x)g^k(x).$$
The coefficient of $x^n$ is $(f(x)g(x))^{(n)}(0)/n!$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
|
The root function $f(x) = \sqrt{x}$ is defined to be nonnegative.
So $\sqrt{2} = 1.4...$, however the equation $x^2 - 4=0$ has to solutions:
If $x$ is such a solution, then so is $-x$ since $(-x)^2-4 = (-1)^2x^2-4 = x^4-4 = 0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have $A={1\over3}$ and $B={-1\over 3}$
Finding C and D is a bit tedious
I wonder what is the closed form for
$$\sum_{n=1}^{\infty}{1\over an+b}=F(a,b)?$$
This way is not a good approach. Can anyone help me with a better approach to tackle this problem? Thank you.
|
Note that
$$S = \sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}= \sum_{n=1}^\infty \frac{1}{12n}\left\{\frac{1}{4n^2-1}-\frac{1}{16n^2-1} \right\}$$
Hence
$$ \frac{1}{12n}\left\{\frac{1}{4n^2-1}-\frac{1}{16n^2-1} \right\}= \frac{1}{24n}\left\{\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{1}{4n+1}-\frac{1}{4n-1} \right\}$$
Using the digamma function we have
$$S = \frac{1}{24}\left\{- \psi \left(\frac{3}{2}\right)-\psi \left(\frac{1}{2}\right)+\psi \left(\frac{5}{4}\right)+\psi \left(\frac{3}{4}\right) \right \} = \frac{1}{12}(1-\log 2)$$
Since
$$\psi(x+1) = -\gamma + \sum \frac{x}{n(n+x)}$$
Note that
$$\psi\left(\frac{1}{2} \right) = -\gamma -2\log(2)$$
$$\psi\left(\frac{1}{4} \right) = -\gamma -\frac{\pi}{2}-3\log(2)$$
$$\psi(1+x) = \psi(x)+\frac{1}{x}$$
$$\psi(1-x) = \psi(x)+\pi \cot(\pi x)$$
|
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|
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help would be greatly appreciated.
|
Personally I like this method which is the same thing as the other answers.$\sqrt{4+2\sqrt{3}}=\sqrt{a}+\sqrt b$ squaring both sides $4+2\sqrt 3=a+2\sqrt{ab}+b$ for this to be true we must equate the parts with and without the radicals so $a+b=4$ and $2\sqrt{ab}=2\sqrt{3}$ so $ab=3$ now we have a system of equations that can be solved through substitution $a(4-a)=3\to a^2-4a+3=0$ which has roots 3, 1. So if a=1 then b=3 and vice versa if a=3. So our answer is $\sqrt{4+2\sqrt{3}}-\sqrt 3=1+\sqrt{3}-\sqrt3$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
proving $ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{n-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}$ proving $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{\color{red}{m}-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \Rightarrow 1-n+\frac{n(n-1)}{2}+\cdots \cdots (-1)^{n-1}\frac{n.(n-1)\cdot (n-2)\cdots(n-m+2)}{(m-1)!}$
Added
writting LHS as $\displaystyle \binom{n}{0}-\left(\binom{n-1}{0}+\binom{n-1}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\cdots \cdots +(-1)^{n-1}\left(\binom{n-1}{m-2}+\binom{n-1}{m-1}\right)=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \binom{n}{0}-\binom{n-1}{0}+\binom{n-1}{1}-\cdots +(-1)^{n-1}\binom{n-1}{m-2}-\left(\binom{n-1}{1}-\binom{n-1}{2}+\cdots +(-1)^n\binom{m-1}{m-1}\right)$
wan,t be able to solve after that, help me to solve it
|
HINT: Fix a random $n$ and do induction on $m$.
|
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|
Can I express $x^7y+xy+x+1$ by a repeated use of operation $xy+x+y+1$? A certain calculator can only give the result of $xy+x+y+1$ for any two real numbers $x$ and $y$.
How to use this calculator to calculate $x^7y+xy+x+1$ for any given $x$ and $y$?
When $x$ and $y$ are equal, it will give $(x+1)^2$. But I cannot proceed beyond that.
|
Give $x^7$ and $y$ to the calculator, to get $R_1 = x^7y + x^7+y +1$.
Give $x$ and $y$ to the calculator, to get $R_2 = xy+x+y+1$.
Add $R_1$ and $R_2$ to get, $R = R_1+R_2 = x^7y + x^7 + x+xy + 2y+1$.
Subtract $x^7$ and $2y$ from $R$ to get the required.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $AC: CB$ in $\triangle XYZ$ Problem:
In $\triangle XYZ$, $XY = 4$, $YZ = 7$, and $XZ = 9$. Let $M$ be the midpoint of $\overline{XZ}$, and let $A$ be the point on $\overline{XZ}$ such that $\overline{YA}$ bisects angle $XYZ$. Let $B$ be the point on $\overline{YZ}$ such that $\overline{YA} \perp \overline{AB}$. Let $\overline{AB}$ meet $\overline{YM}$ at $C$. Find $AC: CB$.
Attempt:
We know that $XM$ = $MZ$ = $\dfrac{9}{2}$, and by the Angle-Bisector theorem,
$$\dfrac{4}{x} = \dfrac{7}{9-x}$$
$$\implies 36-4x = 7x$$
$$\implies x= \dfrac{36}{11}=XA$$
Therefore, $AZ$ = $\dfrac{63}{11}$, and $AM$ = $\dfrac{9}{2} - \dfrac{36}{11}= \dfrac{27}{22}.$
Also, by the Menelaus theorem,
$$\dfrac{ZM}{MA} \times \dfrac{AC}{CB} \times \dfrac{YB}{YZ} = 1$$
$$\implies \dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1$$
From here, I got stuck. I'm wondering if one could use mass points, since the problem wants to find the ratio of lengths, and not specific side lengths. Any help is appreciated!
|
You have already got
$$\dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1\tag1$$
Also, we have
$$YB=\frac{AY}{\cos \frac Y2}\tag2$$
So, we want to find $\cos\frac Y2$ and $AY$.
By the law of cosines,
$$9^2=4^2+7^2-2\cdot 4\cdot 7\cos Y\implies \cos Y=-\frac 27$$
from which we have
$$\cos\frac Y2=\sqrt{\frac{1-\frac 27}{2}}=\sqrt{\frac{5}{14}}$$
Also, the length of $AY$ is given by $$AY=\sqrt{YX\cdot YZ\left(1-\frac{YZ^2}{(YX+YZ)^2}\right)}=\frac{4\sqrt{70}}{11}$$
(see here or here for this formula)
So, from $(1)$ and $(2)$, we get
$$\color{red}{AC:CB=3:8}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the triple integral problem Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$
, where $Q=3x^2+2y^2+z^2+2xy$.
|
Let $\Sigma$ be the symmetric $3 \times 3$ matrix which is specified by $\frac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1} \mathrm{x} = Q(\mathrm{x})$ for $\mathrm{x}\in \Bbb{R}^3$. Then $\Sigma$ satisfies
$$ \Sigma^{-1} = \begin{pmatrix} 6 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \qquad
\Sigma = \frac{1}{10} \begin{pmatrix} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}, \qquad
\det \Sigma = \frac{1}{40}. $$
Now introduce a random vector $X = (X_1, X_2, X_3)$ which has multivariate normal distribution $\mathcal{N}(0, \Sigma)$. Then its density is given by
$$ f_{X}(\mathrm{x}) = \frac{1}{\sqrt{(2\pi)^3\det \Sigma}} \exp\{ -\tfrac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1}\mathrm{x} \}. $$
From this, we can compute the integral as follows:
\begin{align*}
\int_{\Bbb{R}^3} (x^2 - 2xy)e^{-Q(\mathrm{x})} \, d\mathrm{x}
&= \sqrt{(2\pi)^3\det \Sigma} \int_{\Bbb{R}^3} (x^2 - 2xy) f_{X}(\mathrm{x}) \, d\mathrm{x} \\
&= \sqrt{(2\pi)^3\det \Sigma} \cdot \Bbb{E}[X_1^2 - 2X_1 X_2] \\
&= \sqrt{(2\pi)^3\det \Sigma} \cdot (\Sigma_{11} - 2\Sigma_{12}),
\end{align*}
where $\Sigma_{ij} = \operatorname{Cov}(X_i, X_j)$ is the $(i,j)$-entry of the covariance matrix $\Sigma$. Computing everything, we obtain
$$ \int_{\Bbb{R}^3} (x^2 - 2xy)e^{-Q(\mathrm{x})} \, d\mathrm{x} = \frac{2\pi^{3/2}}{5\sqrt{5}}. $$
|
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|
Help find closed form for:$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)$ What is the closed form for
$$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)?$$
My try:
I have found a few values of $F(k)$, but was unable to find a closed form for it.
$F(0)=0$
$F(1)={2\over \pi}$
$F(2)=\left({2\over \pi}\right)^2$
$F(3)=\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^3$
$F(4)={1\over 2}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4$
$F(5)={1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$
|
Note that
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)! } = \cos(x)$$
You can reach the result by integrating $k$-times.
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+k)! } = \frac{1}{x^{k}}\int^{x}_0 \mathrm{d}t_{k-1}\int^{t_{k-1}}_0 \mathrm{d}t_{k-2} \cdots\int^{t_1}_0\mathrm{d}t_0\cos(t_0) $$
For example when $k=1$
$$\sum_{n=0}^\infty \frac{(-1)^n(\pi/2)^{2n}}{(2n+1)! } = \frac{2}{\pi}\int^{\pi/2}_0 \mathrm{d}t_0 \cos(t_0)\,= \frac{2}{\pi}$$
For $k=2$
\begin{align}
\sum_{n=0}^\infty \frac{(-1)^n(\pi/2)^{2n}}{(2n+2)! } &= \left( \frac{2}{\pi}\right)^2\int^{\pi/2}_0 \mathrm{d}t_{1}\int^{t_1}_0\mathrm{d}t_0\cos(t_0) \\ &= \left( \frac{2}{\pi}\right)^2\int^{\pi/2}_0 \mathrm{d}t_{1}\sin(t_1)\\
& = \left( \frac{2}{\pi}\right)^2
\end{align}
For
$$f_k(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+k}}{(2n+k)!}$$
We can define recursively
$$f_0(x) = \cos(x)$$
$$f_k(x) =\int^x_0 f_{k-1}(t)\,dt$$
By solving the recursive formula
$$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+k}}{(2n+k)!} = \begin{cases} \sum_{n=0}^{\lceil k/2 \rceil-2}\frac{x^{2n+1}(-1)^{\lceil k/2 \rceil+n}}{(2n+1)!}-(-1)^{\lceil k/2 \rceil}\sin(x) & \text{If $k$ is odd} \\ \sum_{n=0}^{k/2-1}\frac{x^{2n}(-1)^{n+k/2}}{(2n)!}-(-1)^{k/2}\cos(x) & \text{If $k$ is even}\end{cases}$$
Finally we have the closed form
$$F(k) = \begin{cases}\left( \frac{2}{\pi}\right)^k \sum_{n=0}^{\lceil
k/2 \rceil-2}\frac{(\pi/2)^{2n+1}(-1)^{\lceil k/2
\rceil+n}}{(2n+1)!}-\left( \frac{2}{\pi}\right)^k(-1)^{\lceil k/2
\rceil} & \text{If $k$ is odd} \\ \left(
\frac{2}{\pi}\right)^k\sum_{n=0}^{k/2-1}\frac{(\pi/2)^{2n}(-1)^{n+k/2}}{(2n)!}&
\text{If $k$ is even}\end{cases}$$
To check the correctness of the formula
For $k=5$ we have
$$F(5)=\left(\frac{2}{\pi}\right)^5 \sum_{n=0}^{1}\frac{(\pi/2)^{2n+1}(-1)^{3+n}}{(2n+1)!}-\left( \frac{2}{\pi}\right)^5(-1)^{3} = {1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$$
|
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|
If $a+b+c+d+e=0$ so $180(a^6+b^6+c^6+d^6+e^6)\geq11(a^2+b^2+c^2+d^2+e^2)^3$ Let $a$, $b$, $c$, $d$ and $e$ be real numbers such that $a+b+c+d+e=0$. Prove that:
$$180(a^6+b^6+c^6+d^6+e^6)\geq11(a^2+b^2+c^2+d^2+e^2)^3$$
I tried Holder, uvw and more, but without success.
|
Let $x = (a,b,c,d,e)$.
By homogeneity, let $g(x) = a^2+b^2+c^2+d^2+e^2 = 1$. Further, as stated in the question, $h(x) =a+b+c+d+e=0$.
Let $$f(x) = a^6+b^6+c^6+d^6+e^6 $$ and we have to show, under the conditions above, that $f(x) \geq 11/180$. For $x_0 = \frac{1}{\sqrt{30}}(-3,-3,2,2,2)$ we have equality, $f(x_0)=11/180$, and also for all permutations in $x_0$, and the inversion to $-x_0$. These are $2 \cdot \binom{5}{2}$ = 20 points with $f=11/180$.
The proof is geometric. Consider the body $f(x) = $const. . This is the hypersurface of what has been called an N(5)-dimensional ball in p(6)-norm, see here. A nice visualization is given in here. Notice that an N-dimensional ball in p($\infty$)-norm is a hypercube. These balls are convex bodies, hence they have extremals where for given $f(x) = $const. , the Euclidean distance from the origin $g(x)$ has a maximum. Clearly, without further restrictions, these extrema occur at $x_1 = \frac{1}{\sqrt{5}}(\pm 1,\pm 1,\pm 1,\pm 1,\pm 1)$.
Due to the second condition $h(x)$, we need to consider the normal projection of these extrema $x_1$ to the space normal to $\frac{1}{\sqrt{5}}(1,1,1,1,1)$. An illustration can be given for the three-dimensional case, i.e. $f_3(x) = a^6+b^6+c^6 =$const. under the condition $h_3(x) =a+b+c=0$. We have $g_3(x) = a^2+b^2+c^2$. The normal space here is a plane, and the projections of $g_3$ (unit circle, green) and $f_3$ (red) onto this plane are given in the following figure, with a suitably chosen constant:
With a suitably chosen constant, $g_3$ and $f_3$ will touch at 6 positions, since there are 6 equivalent extrema (out of the 8 extrema $x_1$, 2 are parallel to $(1,1,1)$ and the other 6 are equivalent in value).
Back to the problem, for $x_{10} = \frac{1}{\sqrt{5}}(1,1,1,1,1)$, the normal projection is the nullvector. Considering one minus-sign, we have e.g. $x_{11} =\frac{1}{\sqrt{5}}(-1,1,1,1,1)$ and the projection is the vector $n_{11} = \frac{1}{\sqrt{20}}(-4,1,1,1,1)$. Considering two minus-signs, we have e.g. $x_{12} =\frac{1}{\sqrt{5}}(-1,-1,1,1,1)$ and the projection is the vector $n_{12} = \frac{1}{\sqrt{30}}(-3,-3,2,2,2)$. Further cases are given by permutations and inversions. The other cases of 3,4, and 5 minus-signs follow by inversion to $-x_{1n}$.
It is easy to see that indeed, for $f(x)=$const. , the value for the euclidian distance $g(x)$ for $n_{12}$ is the largest one for the $n$-vectors and therefore, by convexity, the largest value that $g(x)$ attains at all. Since we know already that $f(x) = 11/180$ holds with equality at that point (rather at the 20 equivalent points), and that the Hessian at this point is positive definite, this proves the claim. $\qquad \Box$
|
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|
Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large number of steps I achieved the correct answer:
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx=\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}+C$$
I was able to check my answer using Mathematica.
expr = D[1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(2 x^2), x];
Assuming[x >= 3, FullSimplify[expr]]
Which returned the correct response:
Sqrt[-9 + x^2]/x^3
Mathematica returns the following answer:
Integrate[Sqrt[x^2 - 9]/x^3, x, Assumptions -> x >= 3]
-(Sqrt[-9 + x^2]/(2 x^2)) - 1/6 ArcTan[3/Sqrt[-9 + x^2]]
Which I can write to make more clear.
$$-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}+D$$
Now, you can see that part of my answer is there, but here is my question. How can I show that
$$\frac16\sec^{-1}\frac{x}{3}\qquad\text{is equal to}\qquad -\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}$$
plus some arbitrary constant? What identities can I use? Also, can anyone share the best web page for inverse trig identities?
Update: I'd like to thank everyone for their help. The Trivial Solution's suggestion gave me:
$$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$
Then the following identity came to mind:
$$\tan^{-1}x+\tan^{-1}\frac1x=\frac{\pi}{2}$$
So I could write:
\begin{align*}
\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}
&=\frac16\tan^{-1}\frac{\sqrt{x^2-9}}{3}-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac16\left(\frac{\pi}{2}-\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac{\pi}{12}-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}
\end{align*}
Using Olivier's and Miko's thoughts, I produced this plot in Mathematica.
Plot[{1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(
2 x^2), -(1/6) ArcTan[3/Sqrt[x^2 - 9]] - Sqrt[x^2 - 9]/(
2 x^2)}, {x, -6, 6},
Ticks -> {Automatic, {-\[Pi]/12, \[Pi]/12}}]
Which shows that the two answers differ by $\pi/12$, but only for $x>3$.
|
Pulled straight off wikipedia.
You seem to have used ** intersection of tan($\theta$) and arcsec(x) **
Also, that "constant" you were referring to is probably $\frac{\pi}{12}$.
|
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|
Quadratic equation system $A^2 + B^2 = 5$ and $AB = 2$ Given a system of equations
$A^2 + B^2 = 5$
$AB = 2$
what is the correct way to solve it?
I see immediately that the answers are
*
*$A=1, B=2$
*$A=2, B=1$
*$A=-1, B=-2$
*$A=-2, B=-1$
but I don't understand the correct way of getting there. I have tried to isolate one of the variables and put the resulting expression into one of the equations, but this didn't get me anywhere. What is the correct way of solving this problem?
|
If
$a^2+b^2 = c$
and
$ab = d$
then
$(a+b)^2
=a^2+2ab+b^2
= c+2d
$
and
$(a-b)^2
=a^2-2ab+b^2
= c-2d
$.
Therefore
$a+b = \sqrt{c+2d}$
and
$a-b = \sqrt{c-2d}$
so
$a = \frac12(\sqrt{c+2d}+\sqrt{c-2d})
$
and
$b = \frac12(\sqrt{c+2d}-\sqrt{c-2d})
$.
If $c=5, d=2$,
then
$\sqrt{c+2d} = 3$
and
$\sqrt{c-2d} = 1$
so
$a= 2$
and
$b = 1$.
If you want to get
all the solutions,
you can write
$a = \frac12(\pm\sqrt{c+2d}\pm\sqrt{c-2d})
$
and
$b = \frac12(\pm\sqrt{c+2d}\mp\sqrt{c-2d})
$
for all possible signs.
|
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|
Question on completing the square obtaining the form $a(x+p)^2+q$ From the last term in completing the square from the quadratic $ax^2+bx+c$,
I was just wondering how $$-\left(\frac{b}{2a}\right)^2+c = \left(c-\frac{b^2}{4a}\right)$$
I would have gotten $$-\left(\frac{b}{2a}\right)^2+c=\left(c-\frac{b^2}{(2a)^2}\right)=\left(c-\frac{b^2}{4a^2}\right)$$
Books final answer was the following when completing the square $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$
|
You forgot to multiply the $\left(\frac{b}{2a}\right)^2$ by $a$.
|
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|
show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$
Let the sequence $\{a_{n}\}$ be such that $a_{1}=1, a_{2}=100$, and $$a_{n+2}=\left\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$$
Prove that the sequence $\{a_{n}\}$ is periodic.
I have used a computer and found the periodic is $T=874$, but how to prove it?
|
This is not an answer, but rather an observation.
The following graph shows the set of points $P = \{(a_n, a_{n+1}) : n \geq 1\}$.
$\hspace{8em}$
Notice that they are confined in a very narrow region and are clustered near an ellipse. This ellipse is not hard to identify. Indeed, if a sequence $(b_n)$ satisfies
$$ b_{n+2} = \frac{4}{3}b_{n+1} - b_n, $$
then it follows that
$$ \det \begin{pmatrix} b_{n+1} & b_n \\ b_{n+2} & b_{n+1} \end{pmatrix} = b_{n+1}^2 - \frac{4}{3}b_{n+1}b_n + b_n^2 $$
is constant, since
$$ \begin{pmatrix} b_{n+2} & b_{n+1} \\ b_{n+3} & b_{n+2} \end{pmatrix}
= \begin{pmatrix} 0 & 1 \\ -1 & \frac{4}{3} \end{pmatrix}
\begin{pmatrix} b_{n+1} & b_n \\ b_{n+2} & b_{n+1} \end{pmatrix}
\quad \text{and} \quad
\det \begin{pmatrix} 0 & 1 \\ -1 & \frac{4}{3} \end{pmatrix} = 1. $$
Thus the points $(b_n, b_{n+1})$ stays forever on the ellipse
$$f(x, y) := x^2 - \frac{4}{3}xy + y^2 = \text{const}.$$
If we can somehow show that $f(a_n, a_{n+1})$ is also bounded by some perturbation argument, then since the region $f(x, y) \leq c$ is bounded and $P$ has only integer points, we can argue that $(a_n)$ is eventually periodic. But at this point, I am not sure if this observation will be really useful.
|
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|
Finding a basis for this subspace of $n\times n$ matrices $M(n,\mathbb{F})$ denotes the set of all $n\times n$ matrices with entries in $\mathbb{F}$.
Assume that $M(n,\mathbb{F})$ is a vector space over $\mathbb{F}$. Suppose $(\mu_{ij}) \in M(n,\mathbb{F})$ where $\mu_{ij} \mu_{ji}=1$ for all $i,j$. Let $U = \{(m_{ij})\in M(n,\mathbb{F}): m_{ij} = \mu_{ij}m_{ij} \text{ for all } i,j\}$.
Suppose $\mu_{ii}=1$ for all $i$; find a basis for $U$ and find the dimension of $U$.
I wrote out few cases (when $n = \{2, 3, 4\})$ and these help me see that the dimension of $U$ should be given by the expression $\frac12 n(n+1)$.
Case $n=2$:
$\begin{pmatrix}
m_{11} & m_{12} \\
m_{21} & m_{22}
\end{pmatrix} =
\begin{pmatrix}
m_{11} & m_{12} \\
\mu_{21}m_{12} & m_{22}
\end{pmatrix}$
Case $n=3$:
$\begin{pmatrix}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33}
\end{pmatrix} =
\begin{pmatrix}
m_{11} & m_{12} & m_{13} \\
\mu_{21}m_{12} & m_{22} & m_{23} \\
\mu_{31}m_{13} & \mu_{32}m_{23} & m_{33}
\end{pmatrix}$
Case $n=4$:
$\begin{pmatrix}
m_{11} & m_{12} & m_{13} & m_{14} \\
m_{21} & m_{22} & m_{23} & m_{24} \\
m_{31} & m_{32} & m_{33} & m_{34} \\
m_{41} & m_{42} & m_{43} & m_{44}
\end{pmatrix} =
\begin{pmatrix}
m_{11} & m_{12} & m_{13} & m_{14} \\
\mu_{21}m_{12} & m_{22} & m_{23} & m_{24} \\
\mu_{31}m_{13} & \mu_{32}m_{23} & m_{33} & m_{34} \\
\mu_{41}m_{14} & \mu_{42}m_{24} & \mu_{43}m_{34} & m_{44}
\end{pmatrix}$
Doing the following
$$\begin{pmatrix}
m_{11} & m_{12} \\
m_{21} & m_{22}
\end{pmatrix} =
\begin{pmatrix}
m_{11} & m_{12} \\
\mu_{21}m_{12} & m_{22}
\end{pmatrix}=
\begin{pmatrix}
m_{11} & 0 \\
0 & 0
\end{pmatrix} +
\begin{pmatrix}
0 & m_{12} \\
\mu_{21}m_{12} & 0
\end{pmatrix}+
\begin{pmatrix}
0 & 0 \\
0 & m_{22}
\end{pmatrix}$$
we can see that the $2\times 2$ matrices in $U$ are spanned by 3 linearly independent matrices, which agrees with the dimension listed above. Repeat this process for greater dimensions. Essentially my solution depends greatly upon the reader picking up on the pattern.
QUESTION: Is there a better way to present the solution to this problem?
|
Hint:
If $D$ is the diagonal and $U$ is the upper triangular part you can write
$$M = D+U+ K_1 U^T K_2$$
Now can you find $K_1,K_2$ matrices in terms of the $\mu$s which accomplish what you want?
|
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|
Inequality with a+b+c=1 Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{1}{4}.$$
Progress: This is equivalent to show that
$$\dfrac{1}{a^2+5a}+\dfrac{1}{b^2+5b}+\dfrac{1}{c^2+5c}\le \dfrac{1}{4abc}.$$
I'm not sure how to proceed further. Also equality doesn't occur when $a=b=c$, which makes me more confused.
Edit: The original inequality is as follows:
Let $a_1,a_2,\cdots{},a_n$ be $n>2$ positive reals such that $a_1+a_2+\cdots{}+a_n=1$. Prove that,
$$\sum_{k=1}^n \dfrac{\prod\limits_{j\ne k}a_j}{a_k+n+2}\le \dfrac{1}{(n-1)^2}.$$
|
We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{bc}{5+a}\leq\frac{1}{16}$.
Indeed, we need to prove that $$\sum\limits_{cyc}\frac{bc}{6a+5b+5c}\leq\frac{a+b+c}{16}$$ or
$$\sum\limits_{cyc}\left(\frac{a}{16}-\frac{bc}{6a+5b+5c}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{6a^2+5ab+5ac-16bc}{6a+5b+5c}\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)(3a+8c)-(c-a)(3a+8b)}{6a+5b+5c}\geq0$$ or
$$\sum\limits_{cyc}(a-b)\left(\frac{3a+8c}{6a+5b+5c}-\frac{3b+8c}{6b+5a+5c}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2(15a+15b+7c)}{(6a+5b+5c)(6b+5a+5c)}\geq0.$$
Done!
|
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|
How to integrate $\int\sqrt{\frac{4-x}{4+x}}$? Let
$$g(x)=\sqrt{\dfrac{4-x}{4+x}}.$$
I would like to find the primitive of $g(x)$, say $G(x)$.
I did the following: first the domain of $g(x)$ is $D_g=(-4, 4]$. Second, we have
\begin{align}
G(x)=\int g(x)dx &=\int\sqrt{\dfrac{4-x}{4+x}}dx\\
&=\int\sqrt{\dfrac{(4-x)(4-x)}{(4+x)(4-x)}}dx\\
&=\int\sqrt{\dfrac{(4-x)^2}{16-x^2}}dx\\
&=\int\dfrac{4-x}{\sqrt{16-x^2}}dx\\
&=\int\dfrac{4}{\sqrt{16-x^2}}dx-\int\dfrac{x}{\sqrt{16-x^2}}dx\\
&=\int\dfrac{4}{\sqrt{16(1-x^2/16)}}dx+\int\dfrac{-2x}{2\sqrt{16-x^2}}dx\\
&=\underbrace{\int\dfrac{1}{\sqrt{1-(x/4)^2}}dx}_{\text{set $t=x/4$}}+\sqrt{16-x^2}+C\\
&=\underbrace{\int\dfrac{4}{\sqrt{1-t^2}}dt}_{\text{set $t=\sin \theta$}}+\sqrt{16-x^2}+C\\
&=\int\dfrac{4\cos\theta}{\sqrt{\cos^2\theta}}d\theta+\sqrt{16-x^2}+C\\
\end{align}
So finally, I get
$$G(x)=\pm\theta+\sqrt{16-x^2}+C'.$$
With wolframalpha I found some different answer. Could you provide any suggestions?
Also, multiplying by $4-x$ is it correct at the beginning? because I should say then that $x\neq 4$.
|
Solving a more general problem, using integration by parts:
$$\mathcal{I}_\text{n}\left(x\right)=\int\sqrt{\frac{\text{n}-x}{\text{n}+x}}\space\text{d}x=x\sqrt{\frac{\text{n}-x}{\text{n}+x}}+\text{n}\int\frac{x}{\left(\text{n}+x\right)^2\sqrt{\frac{2\text{n}}{\text{n}+x}-1}}\space\text{d}x$$
Now, substitute $\text{u}=\text{n}+x$ and $\text{d}\text{u}=\text{d}x$:
$$\int\frac{x}{\left(\text{n}+x\right)^2\sqrt{\frac{2\text{n}}{\text{n}+x}-1}}\space\text{d}x=\int\frac{\text{u}-\text{n}}{\text{u}^2\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}$$
Seperate the integral:
$$\int\frac{\text{u}-\text{n}}{\text{u}^2\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}=\int\frac{\text{u}}{\text{u}^2\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}-\int\frac{\text{n}}{\text{u}^2\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}=$$
$$\int\frac{1}{\text{u}\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}-\text{n}\int\frac{1}{\text{u}^2\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}$$
So:
*
*$$\int\frac{1}{\text{u}\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}=\int\frac{1}{\sqrt{\text{u}}\sqrt{2\text{n}-\text{u}}}\space\text{d}\text{u}=2\arcsin\left\{\frac{\sqrt{\text{u}}}{\sqrt{2}\sqrt{\text{n}}}\right\}+\text{C}_1$$
*Substitute $\text{s}=\frac{2\text{n}}{\text{u}}-1$ and $\text{d}\text{s}=-\frac{2\text{n}}{\text{u}^2}\space\text{d}\text{u}$:
$$\int\frac{1}{\text{u}^2\sqrt{\frac{2\text{n}}{\text{u}}-1}}\space\text{d}\text{u}=-\frac{1}{2\text{n}}\int\frac{1}{\sqrt{\text{s}}}\space\text{d}\text{s}=\text{C}_2-\frac{\sqrt{\frac{2\text{n}}{\text{u}}-1}}{\text{n}}$$
|
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|
Solving proportions with $3$ ratios,$ x:3:y = -2:3:-4$ Proportion seems simple enough for me. Example is $4:x = 2:5,$ and the answer is $x = 10.$
My problem is how do I solve for proportions with $3$ ratios like $x:3:y = -2:3:-4$ ?
Do I write it like
$$\frac{\frac{x}{3}}{y} = \frac{\frac{-2}{3}}{-4} $$
?
|
We have: $x:3:y=-2:3:-4$
$\Rightarrow x:3=-2:3$
$\Rightarrow \dfrac{x}{3}=-\dfrac{2}{3}$
$\Rightarrow x=-2$
and
$\Rightarrow 3:y=3:-4$
$\Rightarrow \dfrac{3}{y}=-\dfrac{3}{4}$
$\Rightarrow \dfrac{y}{3}=-\dfrac{4}{3}$
$\Rightarrow y=-4$
$\therefore x=-2,y=-4$
|
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|
Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$
could some help me with this, thanks
|
As @dxiv says in the comments, we can write $$f (x) = (x-\frac {1}{2})^3 +\frac {1}{4}x +\frac {3}{8} $$ Thus, substituting $u=x- \frac {1}{2}$ gives us $f (u)=u^3+\frac {1}{4}(u+\frac {1}{2}) +\frac {3}{8} = u^3+\frac {1}{4}u +\frac {1}{2} $. Thus, $$f (f (u)) =(u^3+\frac {1}{4}u +\frac {1}{2})^3 +\frac {1}{4}[u^3 +\frac {1}{4}u +\frac {1}{2}] +\frac {1}{2}$$ As $x $ goes from $\frac {1}{4} $ to $\frac {3}{4} $,. $u $ goes from $-\frac {1}{4} $ to $\frac {1}{4} $.
Hope you can take it from here.
|
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|
Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$ Show that
$$1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024} <2$$
I know that the denominators are perfect squares starting from that of $1$ till $32$. Also I know about this identity
$$\frac{1}{n(n+1)} > \frac{1}{(n+1)^2} > \frac{1}{(n+1)(n+2)}.$$
But I am not able to implement it
Please help me.
|
For $n>0$, we have
$$\frac{1}{(n+1)^2}\leq \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
and by sum,
$$1+\sum_{n=1}^{31}\frac{1}{(n+1)^2}\leq 2-\frac{1}{32}<2$$
|
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|
discrete mathematics combinatorics - counting
I have $52$ cards from $4$ series ($13 \times 4=52 \text{ cards}$). Each series are
numbered from $1$ to $13$. In how many possible ways you can draw from the
deck $6$ cards, so that all your six cards has two numbers exactly?
My answer:
First card - $\binom{13}{1} \cdot \binom{8}{4}$ (choosing from the $8$ possibilities, $4$ places I can put my cards. choosing $4$ because I can't have more than $4$ cards of the same number)
Second card - $\binom{13}{1} \cdot \binom{7}{4}$ (same, but $7$)
Total - First card + Second card
Not quite sure about my answer. I am pretty sure I am counting more than there is. Help will be much appreciated.
|
The number of different ways to obtain $2$ out of $13$ different numbers is
\begin{align*}
\binom{13}{2}
\end{align*}
Next we consider all valid configurations which can be obtained with two numbers.
Since there are always four cards with the same number and we take six cards from the deck there are three types of valid configurations
*
*$(4,2)$: Four cards with the first number and two cards with the second number. There are
\begin{align*}
\frac{6!}{4!2!}=\binom{6}{2}
\end{align*}
different configurations of this type.
*$(3,3)$: Three cards with the first number and three cards with the second number , giving
\begin{align*}
\frac{6!}{3!3!}=\binom{6}{3}
\end{align*}
different configurations.
*$(2,4)$: Two cards with the first number and four cards with the second number, giving $$\frac{6!}{2!4!}=\binom{6}{2}$$ different configuration as in the first case.
We conclude:
There are
\begin{align*}
\binom{13}{2}\left(2\binom{6}{2}+\binom{6}{3}\right)=3900
\end{align*}
different configurations to draw six cards with two numbers from the deck.
|
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|
Solve the system of equations ... Solve the system of equations :
(EDIT : The problem does not say anything about the nature of $x$ and $y$ (integer, natural number ,..etc.) )
$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3} $
$2x + {\frac {1} {x+y}} = \frac {13} {3}$
I do not know how to approach these types of problems. I tried finding value of $\frac {1} {x+y}$ in terms of $x$ and $y$, but it complicates the problem even more.
Can anyone provide a pointer to what should be done ?
|
Perhaps this can help: Let $x+y=v$ and $x-y=w$, then the equations can be written as
\begin{align*}
3\left(v^2+\frac{1}{v^2}\right)+w^2 & = \frac{85}{3}\\
\left(v+\frac{1}{v}\right)+w & = \frac{13}{3}.
\end{align*}
Now let $v+\frac{1}{v}=t$, then the system can be rewritten as
\begin{align*}
3t^2+w^2 & = \frac{85}{3}+6\\
t+w & = \frac{13}{3}.
\end{align*}
Now solve for $t$ and $w$...
|
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|
Positive pairs of integral values satisfying $2xy − 4x^2 +12x − 5y = 11$ The number of positive pairs of integral values of $(x, y)$ that solves
$2xy − 4x^2 +12x − 5y = 11$ is?
I rearranged it to $(2x-5)(y+1-2x)=6$, which took quite a bit of time.
So it can be $2*3$ , $3*2$, $6*1$ or $1*6$ which gives us 2 possible positive integral pairs. Answer: 2.
Is there a faster way to do similar problems?
|
First note that:
$$y=\frac{11 - 12 x + 4 x^2}{2 x-5}$$
Applying division:
$$y=\frac{(2x-1)(2x-5)+6}{2x-5}=(2x-1)+\frac{6}
{2x-5}$$
To be an integer $2x-5$ must divide $6$ so $2x-5$ must be $-1,-3,1,2$ or $3$ or $6$. Check then by the positive integral solutions.
|
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|
Any solution for $\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$ I tried to solve this triple integral but couldn't integrate the result.
$$\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$$ and the surface to integrate in is $$x^2+y^2+z^2\le1$$
Is there any way to transform the integral into polar coordinates?
|
Trivial by symmetry: the unit ball $B\subset\mathbb{R}^3$ is invariant under the transformation $(x,y,z)\mapsto(z,y,x)$, hence:
$$\iiint_B \frac{x^2+2y^2}{x^2+4y^2+z^2}\,d\mu = \frac{1}{2}\left(\iiint_B \frac{x^2+2y^2}{x^2+4y^2+z^2}\,d\mu+\iiint_B \frac{z^2+2y^2}{x^2+4y^2+z^2}\,d\mu\right) = \frac{1}{2}\mu(B) = \color{red}{\frac{2\pi}{3}}.$$
|
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|
Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^2+b^2+2ab = (a+b)^2$
Hence $\sqrt{2}|z| \geq a + b$ but $a = |\operatorname{Re}z|~,~ b = |\operatorname{Im}z|$
Did I make a mistake somewhere, if yes I would appreciate it if it could be pointed out and perhaps provide some guideline on how to prove this.
|
Let $z=a+bi$. Then $|z|^2=a^2+b^2$ and Re$z=a$ and Im$z=b$. Observe that
$$(\sqrt{2}|z|)^2=2(a^2+b^2)$$ and $$\Big(|\text{Re}z|+|\text{Im}z|\Big)^2=a^2+2|ab|+b^2.$$
Now,
$$
\begin{align}
(|a|-|b|)^2\geq 0&\iff a^2+b^2-2|ab|\geq 0\\
&\iff2(a^2+b^2)\geq a^2+2|ab|+b^2\\
&\iff (\sqrt{2}|z|)^2\geq\Big(|\text{Re}z|+|\text{Im}z|\Big)^2\end{align}
$$
Since $\sqrt{2}|z|$ and $|\text{Re}z|+|\text{Im}z|$ are nonnegative, the result follows.
|
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|
Sum of reciprocals of the triangle numbers
Consider the sum of $n$ terms :
$S_n = 1 + \frac{1}{1+2} + \frac {1}{1+2+3} + ... + \frac {1}{1+2+3+...+n}$
for $n \in N$.
Find the least rational number $r$ such that $S_n < r$, for all $n \in N$.
My attempt :
$S_n = 2(1-\frac{1}{2} + \frac {1}{2} - \frac{1}{3} + .... + \frac {1}{n} - \frac {1}{n+1}) = 2(1 - \frac {1}{n+1}) $
Now what to do with that '$r$' thing ?
How to proceed ?
|
Let $r=2$, and we can see that
$$\frac{2n-2}n=2-\frac2n<2$$
Similarly, as $n\to\infty$, the limit is $2$, so this is the least rational number satisfying the inequality.
|
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|
Prove that if $a$ and $b$ are integers with $a\not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$, then $a|b$. Prove that if $a$ and $b$ are integers with $a\not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$, then $a|b$.
I will use the backwards and forwards method to prove the proposition. A represents the hypothesis, and B represents the conclusion.
|
$a, b$ are integers with $a \not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$.
B: $a|b$ ($b$ is divisible by $a$)
B1:
How can I show that one integer (namely, $b$) is divisible by another (namely, $a$)?
Show that there is an integer $k$ such that $\dfrac{b}{a} = k$.
A1: $ax^2 + bx + b − a = 0$ (A)
$\therefore x = \dfrac{ -b \pm \sqrt{b^2 - 4a(b-a)} }{ 2a }$
$ \implies x = \dfrac{ -b \pm \sqrt{b^2 - 4ab + 4a^2} }{ 2a }$
$b^2 - 4ab + 4a^2 = (2a - b)^2$
$ \therefore x = \dfrac{ -b \pm \sqrt{(2a - b)^2} }{ 2a }$
$ \implies x = \dfrac{ -b \pm (2a - b) }{ 2a }$
$ \implies x = \dfrac{ 2a - 2b }{ 2a }$
$ = \dfrac{ a - b }{ a }$
$ = 1 - \dfrac{ b }{ a }$
$\implies \dfrac{b}{a} = 1 - x$ where $x$ is a positive integer (A)
$\therefore \dfrac{b}{a} = k$ where $k$ is an integer (B1).
$Q.E.D.$
I would greatly appreciate it if the members of MSE could please take the time to review my proof.
|
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|
Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
|
Cauchy-Schwarz actually works pretty well.
We have with Cauchy-Schwarz if we put $x_1=\sqrt{\frac{a}{a+3b+3c}}$, $x_2=\sqrt{\frac{b}{b+3c+3a}}$ and $x_3=\sqrt{\frac{c}{c+3a+3b}}$ as well as $y_1=\sqrt{a(a+3b+3c)}$, $y_2=\sqrt{b(b+3c+3a)}$ and $y_3=\sqrt{c(c+3a+3b)}$
$$
\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)≥\left(x_1y_1+x_2y_2+x_3y_3\right)^2\iff\\
\left(\frac{a}{a+3b+3c}+\frac{b}{b+3c+3a}+\frac{c}{c+3a+3b}\right)\left(a(a+3b+3c)+b(b+3c+3a)+c(c+3a+3b)\right)≥(a+b+c)^2
$$
so its enough to prove
$$
\frac{(a+b+c)^2}{\sum_{cyc}a(a+3b+3c)}≥\frac37\iff\\
7(a^2+b^2+c^2+2ab+2bc+2ca)≥3(a^2+b^2+c^2+6ab+6bc+6ca)\iff\\
a^2+b^2+c^2≥ab+bc+ca
$$
which is true.
|
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|
For which $a$, does $p^2+p+(a-1)$ have only rational solutions?
For which $a$, does $p^2+p+(a-1)$ have only rational solutions?
The roots of $p^2+p+(a-1)$ are $\frac{-1+\sqrt{(5-4a)}}{2}$ and $\frac{-1-\sqrt{(5-4a)}}{2}$, and they are rational iff $\sqrt{5-4a}$ is rational, then iff $5-4a$ is a square of some rational. Then what?
|
It has only rational solutions if $x^2+x+a-1=(x-r)(x-s)=x^2-(r+s)x+rs$ with rational $r,s$. This says that $a=rs+1=r(-1-r)+1=1-r-r^2$. Hence the quadratic equation has only rational solutions if $a$ is representable by $1-r-r^2$ for some rational $r$. For example, $a=\sqrt{2}$ is not of this form.
|
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|
If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$? Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$. Denote the abundancy index of $y \in \mathbb{N}$ by $I(y)=\sigma(y)/y$.
If $\sigma(N)=2N$, then $N$ is said to be perfect.
Euler proved that every odd perfect number has the form $q^k n^2$ where $q$ is prime (called the Euler prime) with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
If $k=1$, then since $q \geq 5$, then we have
$$I(q)I(n^2)=2 \Longleftrightarrow I(n^2)=\dfrac{2q}{q+1} \geq \dfrac{5}{3}.$$
If $k>1$, then we can use the bound
$$I(q^k) < \dfrac{q}{q-1} \leq \dfrac{5}{4}$$
so that
$$I(n^2)=\dfrac{2}{I(q^k)} > \dfrac{2(q-1)}{q} \geq \dfrac{8}{5}.$$
Here is my question:
If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$?
|
In an answer to a related MSE question, we get that
$$I(n^2) = \dfrac{2{q^k}(q - 1)}{q^{k+1} - 1} = 2 - 2\dfrac{q^k - 1}{q^{k+1} - 1}.$$
Then $I(n^2) \geq 5/3$ implies that
$$2 - 2\dfrac{q^k - 1}{q^{k+1} - 1} \geq \dfrac{5}{3}$$
$$\dfrac{1}{6} \geq \dfrac{q^k - 1}{q^{k+1} - 1}$$
$$q^{k+1} - 6q^k + 5 \geq 0,$$
which does not force $k=1$.
$k=1$ does follow if it is known a priori that $q=5$. In this case, $I(n^2) \geq 5/3$ and $q=5$ imply that $I(n^2)=5/3$.
UPDATE (September 03 2017 - Manila time)
Suppose that $k > 1$, so that we have $k \geq 5$ (since $k \equiv 1 \pmod 4$). We are given that
$$\frac{2}{I(q^k)}=I(n^2) \geq \frac{5}{3}.$$
It follows that
$$I(q^k) \leq \frac{6}{5},$$
whereupon we obtain
$$I(q^5) \leq I(q^k) \leq \frac{6}{5}.$$
It follows that
$$\frac{q^6 - 1}{q^6 - q^5} = \frac{q^6 - 1}{{q^5}(q - 1)} \leq \frac{6}{5}.$$
Consequently, since $q \geq 5 > 1$, we have
$$5q^6 - 5 = 5(q^6 - 1) \leq 6(q^6 - q^5) = 6q^6 - 6q^5$$
which implies that
$$q^6 - 6q^5 + 5 \geq 0.$$
When $q=5$, we obtain
$$q^6 - 6q^5 + 5 = {q^5}(q - 6) + 5 = {5^5}\cdot(-1) + 5 < 0.$$
|
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How do you compute the $\gcd(1+n+n^2,1+n+n^2+s+2ns+s^2)$ I would like to prove the following claim which I think is true:
Claim: Let $n,$ $m$ and $s$ be positive numbers. Fix $s$, then for every positive number $n$ the $\gcd(1+n+n^2,1+n+s+n^2+2ns+s^2)$ will be equal to a divisor of $1+5s^2+s^4.$
For example for every positive number $n$ if we set $s=8$ the $\gcd(1+n+n^2,73+n+16n+n^2)=1,7,631$ or $4417$. We can see that $1+5*8^2+8^4=4417$ and $4417=7*631$.
I came to the claim by moving numbers around in GAP. If it is wrong a counter example would be awesome.
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No way a statement like this can be right.
$$\gcd(1+n+n^2, 1+n+s+n^2+2ns+s^2) = \gcd(1+n+n^2, s(1+2n+n^2)) = \gcd(1+n+n^2, sn) = \gcd(1+n+n^2, s)$$. So simply take $s = n^2+n+1$, then it can never be $s |1+5s^2+s^4$, unless $s=1$.
|
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Prove $n^5+n^4+1$ is not a prime I have to prove that for any $n>1$, the number $n^5+n^4+1$ is not a prime.With induction I have been able to show that it is true for base case $n=2$, since $n>1$.However, I cannot break down the given expression involving fifth and fourth power into simpler terms. Any help?
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$n^5 + n^4 + 1 = n^5 + n^4 + n^3 – n^3 – n^2 − n + n^2 + n + 1$
$\implies$$ n^3(n^2 + n + 1) − n(n^2 + n + 1) + (n^2 + n + 1)$
=$ (n^2 + n + 1)(n^3 − n + 1)$
Hence, for $n>1$, $n^5 + n^4 + 1$ is not a prime number.
|
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If $f(1)$ and $f(i)$ real, then find minimum value of $|a|+|b|$ A function $f$ is defined by $$f(z)=(4+i)z^2+a z+b$$ $(i=\sqrt{-1})$for all complex number $z$ where $a$ and $b$ are complex numbers. If $f(1)$ and $f(i)$ are both purely real, then find minimum value of $|a|+|b|$
Now
$f(1)=4+i+a+b$ which means imaginary part of $a+b=-1$ and $f(i)=-4-i+a \cdot i+b$ which gives imaginary part of $a \cdot i+b=1$ but how to proceed further to find desired value?
|
Setting $a = a_1 + i a_2, b = b_1 + i b_2$ for four real unknowns $a_k, b_k$.
Real $f(1) = 4 + i + a + b$ and real $f(i) = -4 - i + a i + b$ give the equations
$$
\text{Im}(f(1)) = 1 + a_2 + b_2 = 0 \\
\text{Im}(f(i)) = -1 + a_1 + b_2 = 0
$$
Then we have to minimize
$$
g(a_1, a_2, b_1, b_2)
= \lvert a \rvert + \lvert b \rvert
= \sqrt{a_1^2 + a_2^2} + \sqrt{b_1^2 + b_2^2}
$$
which reduces to a function of two unknowns if we apply the two equations above:
$$
b_2 - 1 = - a_1 \\
b_2 + 1 = - a_2 \\
h(b_1, b_2)
= \sqrt{(b_2 - 1)^2 + (b_2 + 1)^2} + \sqrt{b_1^2 + b_2^2} \\
= \sqrt{2b_2^2 + 2} + \sqrt{b_1^2 + b_2^2} \\
\ge \sqrt{2}
$$
Here is a graph:
$h$ is shown in red colour, $\partial_{b_1} h$ in blue and $\partial_{b_2} h$ in green.
So we have a minimum at $b = 0$ which implies $a = 1-i$ via the constraints.
The minimal value for $\lvert a \rvert + \lvert b \rvert $ is $\sqrt{2}$.
|
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|
Optimization and maximum geometry What is the side length of the largest square that will fit inside an equilateral triangle with sides of length 1.
I created two equations:
Square: Area=$x^2$
Triangle: Area= $\sqrt{3 /4}$
However, how can I find the maximum??
I did $\sqrt{3 /4}=x^2$ ang got fourth root $3 / 2$ which is wrong
|
$1)$ Suppose that $DEFG$ is a square. Let's call $x$ the side of the square.
We also have that $\Delta ADE$ is an equilateral triangle, so $AD=x$.
Now, for the triangle $BDG$ we have:
$$\sin 60°=\frac{DG}{BD}=\frac{x}{BD} \rightarrow BD=\frac{2\sqrt{3}x}{3}$$
$$AB=1=AD+BD=x+\frac{2\sqrt{3}x}{3} \rightarrow x=\frac{3}{3+2\sqrt{3}}=2\sqrt{3}-3$$
It means that doesn't make sense talk about largest square because we have just one and its side is $x=2\sqrt{3}-3$.
$2)$ What could make sense is talk about the biggest rectangle inside the triangle. In order to do that we can call $DE=x$ and then $AD=x$. Let's also call $DG=y$, so
$$\sin 60°=\frac{DG}{BD}=\frac{y}{BD} \rightarrow BD=\frac{2\sqrt{3}y}{3}$$
and then
$$AB=1=AD+BD=x+\frac{2\sqrt{3}y}{3}$$
and we have to maximize:
$$A=xy=\left(1-\frac{2\sqrt{3}y}{3}\right)y$$
and the maximum happens for (using the quadratic standard approach)
$$y=\frac{\sqrt{3}}{4} \rightarrow x=\frac{1}{2}$$
what give us
$$A_{max}=\frac{\sqrt{3}}{8}$$
|
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|
How do I divide a square root of a fraction by another square root of another fraction? My teacher gave this challenger in class today and I can't figure how to solve it:
$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$
|
Given,
$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$
On rationalizing,
$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}} \cdot \frac{\sqrt{\frac13} + \sqrt{\frac12}}{\sqrt{\frac13} + \sqrt{\frac12}}$
= $\frac{\sqrt{\frac29} + \sqrt{\frac13} - \sqrt{\frac12} - \sqrt{\frac34}}{\frac{-1}{6}}$
= $-6 \left( \frac{\sqrt2}{3} + \frac{1}{\sqrt 3} - \frac{1}{\sqrt 2} - \frac{\sqrt3}{2} \right)$
= $-6 \left( \frac{2\sqrt2 + 2\sqrt3 - 3\sqrt2 - 3\sqrt3}{6} \right)$
= $- \left( - \sqrt2 - \sqrt3 \right)$
= $\sqrt2 + \sqrt3$
|
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|
Solve $y'=\frac{x+y-2}{x-y}$ ODE
$$y'=\frac{x+y-2}{x-y}$$
$$\begin{vmatrix}
1 & 1\\
1 & -1
\end{vmatrix}\neq0 $$
Substation: $x=u+\alpha$ , $y=v+\beta$
$\frac{dy}{dx}=\frac{dv}{du}$
$$\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}\begin{pmatrix}
\alpha \\
\beta
\end{pmatrix} =\begin{pmatrix}
2 \\
0
\end{pmatrix}\Rightarrow \alpha=1\text{ , }\beta=1 $$
So:
$x=u+1$
$y=v+1$
$$y'=\frac{dv}{du}=\frac{u+1+v+1-2}{u+1-(v+1)}$$
$$\frac{dv}{du}=\frac{u+v}{u-v}$$
$$\frac{dv}{du}=\frac{u+v}{u-v}$$
$$\frac{dv}{du}=\frac{1+\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}\Rightarrow v=uz$
$$\frac{dv}{du}=z+u\frac{dz}{du}=\frac{1+z}{1-z}$$
How should I continue?
|
$u\frac{dz}{du}=\frac{1+z}{1-z}-z=\frac{z^2+1}{1-z}$ thus
$$\frac{1-z}{1+z^2}dz=\frac{1}{u}du$$
or
$$\left(\frac{1}{1+z^2}-\frac{z}{1+z^2}\right)dz=\frac{1}{u}du$$
as a result
$$\tan^{-1}(z)-\frac12\ln(1+z^2)=C+\ln(u)$$
|
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|
Derivative of arcsin In my assignment I need to analyze the function
$f(x)=\arcsin \frac{1-x^2}{1+x^2}$
And so I need to do the first derivative and my result is:
$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
But in the solution of this assignment it says
$f'(x)=-\frac{2x}{|x|(1+x^2)}$
I don't understand how they get this. I checked my answer on online calculator and it is the same.
|
Divide et impera.
Under the square root you have
$$
1-\frac{(1-x^2)^2}{(1+x^2)^2}=
\frac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}=\frac{4x^2}{(1+x^2)^2}
$$
so the big square root is actually
$$
\frac{2|x|}{1+x^2}
$$
Thus your formula becomes
$$
-\frac{4x}{(1+x^2)^2}\frac{1+x^2}{2|x|}=-\frac{2x}{|x|(1+x^2)}
$$
|
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|
Find all possible positive integer $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $
Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
I don't know how to start with. Any hint or full solution will be helpful.
|
Let's suppose that $3^{n-1}+5^{n-1}\mid 3^n+5^n$, so there is some positive integer $k$ such that $3^n+5^n=k(3^{n-1}+5^{n-1})$. Now, if $k\ge 5$ we have $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$$
This means that $k\le 4$. On the other hand, $3^n+5^n=3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$, then $k\ge 4$ and thus we deduce that $k=4$. In this case we have $3^n+5^n=4(3^{n-1}+5^{n-1})$, which gives us the equation $5^{n-1}=3^{n-1}$, but if $n>1$ this equation is impossible.
Hence $n=1$, and we can easily check that $2\mid 8$.
|
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|
How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$
|
Letting $u=\sqrt[4]{x}$, you have that: $\frac{\sqrt[4]x -1}{x-1}=\frac{u-1}{u^4-1}=\frac{1}{1+u+u^2+u^3}\to \frac{1}{4}$ as $u\to 1$ and hence also as $x\to 1$.
Letting $v=\sqrt[3]{x+7}$ we get that $\frac{\sqrt[3]{x+7}-2}{x-1}=\frac{v-2}{v^3-8}=\frac{1}{v^2+2v+4}\to \frac{1}{12}$ as $v\to 2$ and hence as $x\to 1$.
Finally, let $w=\sqrt{3x+1}$ we get that $\frac{2\sqrt{3x+1}-4}{x-1}=2\frac{w-2}{\frac{1}{3}(w^2-4)}=6\frac{1}{w+2}\to \frac{3}{2}$ as $w\to 2$ and hence as $x\to 1$.
So the limit is $$\frac{\frac{1}{12}+\frac{3}{2}}{\frac{1}{4}}=\frac{19}{3}.$$
This is, of course, just hiding L'Hopital.
|
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|
Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a−b)^2 = (a-b)(a-b) = a^2 -2ab +b^2$. To Prove that $(a−b)^2 = a^2 −b^2 $ if b = 0 I substituted b with zero both in the expanded expression and the original simplified and I got $(a−b)^2 = (a-0)^2 = (a-0)(a-0) = a^2 - a(0)-a(0)+0^2 = a^2$ and the same with $a^2 -2ab +b^2$ which resulted in $a^2 - 2a(0) + 0^2 = 2a$ or if I do not substite the $b^2$ I end up with $a^2 + b^2$. That's what I got when I try to prove the expression true for $b=0$.
As for the part where $b=a$, $(a−b)^2 = (a-b)(a-b) = a^2-2ab+b^2$, if a and b are equal, let $a=b=x$ and I substite $a^2-2ab+b^2 = x^2-2(x)(x) + x^2 = x^2-2x^2+x^2 = 1-2+1=0$ I do not see where any of this can be reduced to $a^2-b^2$ unless that equals zero......I do see where it holds but I do not see how would a solution writting out look.After typing this it seems a lot clearer but I just can't see how to phrase a "solution".
P.S: This is my first time asking a question here so whatever I did wrong I am sorry in advance and appreciate the feedback.
|
This is a geometrical approach.
Let $a-b=d$. Then $d^2 + b^2 = a^2$ and the using Converse of Pythagoras theorem there must be a triangle having $a, b, a-b$ as sides, which is impossible because $b + (a - b) = a \not \gt a$.
|
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|
How can I find $\sum_{cyc}\sin x\sin y$ $x$, $y$ & $z$ are real number such that
$$\frac{\sin{x}+\sin{y}+\sin{z}}{\sin{(x+y+z)}}=\frac{\cos{x}+\cos{y}+\cos{z}}{\cos{(x+y+z)}}=2$$
find the value
$$\sum_{cyc}\sin x\sin y$$
All help would be appreciated.
|
Since
\begin{align*}
2 \cos(x+y+z) &= \cos x + \cos y + \cos z \\
2 \sin(x+y+z) &= \sin x + \sin y +\sin z
\end{align*}
\begin{align*}
2 e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz}
\end{align*}
Multiplying throughout by $e^{-ix}$, we get
\begin{align*}
2e^{i(y+z)} = 1 + e^{i(y-x)} + e^{i(z-x)}
\end{align*}
Equating the real parts, we get
\begin{align*}
2\cos(y+z) &= 1 + \cos(y-x) +\cos(z-x)\\
2(\cos y \cos z - \sin y \sin z) & = 1 + \cos x \cos y + \sin x \sin y + \cos z \cos x + \sin z \sin x
\end{align*}
Similarly, we get
\begin{align*}
2(\cos x \cos z - \sin x \sin z) &= 1 + \cos x \cos y + \sin x \sin y + \cos z \cos y + \sin z \sin y\\
2(\cos x \cos y - \sin x \sin y) &= 1 + \cos x \cos z + \sin x \sin z+ \cos z \cos y + \sin z \sin y\\
\end{align*}
Adding, and canceling $2 \sum \cos x \cos y$ from the sides, we get
$$\sum \sin x \sin y = -\frac{3}{4}$$
|
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|
Prove $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square
Prove the number $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square.
I tried studying $a \mod b$ for different integers $b$, without succes.
|
Let $$k^2 = 2\cdot 10^{2016} + 16$$ Then we have $(k-4)(k+4)=2\cdot 10^{2016}$. Thus we have two factors of $2\cdot 10^{2016}$ with a difference of $8$.
Now if these factors, say $a$ and $b$, both contain $5$ in their prime factorization, then we have that $|a-b|$ is a multiple of $5$ which is not $8$. Thus let $a=2^k$ and $b= 2^m 5^n$, where $k,m,n$ are non-negative integers. We know that $ab = 2^{m+k}5^{n}= 2^{2017}5^{2016}$. Thus $m+k=2017$ and $n=2016$. Now:
$$a-b = 2^k - 2^m5^n = \pm 8$$
$$2^{k-3} - 2^{m-3}5^n = \pm 1$$
Its easy to see that $k\geq 3$ and $n\geq 3$, since both can't be smaller than $3$. Since $\pm 1$ is not divisible by $2$ we have that either $m=3$ or $k=3$.
Thus our possible solutions for $(m,k,n)$ is $(3,2014,2016)$ and $(2014,3,2016)$, neither of which satisfies the last equation. Hence, we have a contradiction.
|
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|
Show that $a^3+a^2b+ab^2+b^3$ is not a prime, if a and b are positive integers I approached it by attempting to factor it and then show that one factor can't be one and the other can't be prime. This gets nowhere, as you can't factor this expression. Is there another way to do it?
|
$a^3+b^3+a^2b+ab^2\\=(a+b)(a^2-ab+b^2)+a^2b+ab^2\\=(a+b)(a^2-ab+b^2)+ab(a+b)\\=(a+b)(a^2-ab+b^2+ab)\\=(a+b)(a^2+b^2)$
|
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|
Show that determinant of $\small\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$ is divisible by $19$ Using that the numbers
$228,
323$
and
$456$
are
divisible
by
$19$.
Show
that
the
determinant of matrix
$\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$
is
divisible
by
$19$.
|
$$\det\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}22 & 2 & 8\\ 32& 2 & 3 \\ 45 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}\color{red}{228} & 2 & 8\\ \color{red}{323}& 2 & 3 \\ \color{red}{456} & 5 & 6\end{pmatrix}=\color{red}{19}\cdot\det\begin{pmatrix}12 & 2 & 8\\ 17& 2 & 3 \\ 24 & 5 & 6\end{pmatrix}.$$
|
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|
Prove that $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0\Rightarrow a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$ If $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0$, prove that $a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$.
I guess that the given expression is true if and only if $a=b=c=0$. Is it true? Or,is there any other alternatives?
|
Maybe you mean that
$$2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})-a-b-c=$$
$$=(\sqrt[4]a+\sqrt[4]b+\sqrt[4]c)(\sqrt[4]a+\sqrt[4]b-\sqrt[4]c)(\sqrt[4]a-\sqrt[4]b+\sqrt[4]c)(-\sqrt[4]a+\sqrt[4]b+\sqrt[4]c),$$
which is not necessary.
|
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|
Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$
|
If there is a solution it has more than 4000 digits, which makes me think there is no solution beyond the one two already mentioned.
In Mathematica,
i=1;
ans=Solve[5 x ^2-7x+1==y^2&&x>0&&y>0,{x,y},Integers];
ans2=ans/.C[1]->i//RootReduce
Dynamic@i
Dynamic[IntegerLength@ans2[[All,All,2]]]
and then run
While[FreeQ[PrimeQ[ans2[[All,All,2]]],{True,True}],i++;ans2=ans/.C[1]->i//RootReduce]
Setting i=0 will stop the loop with the two known solutions:
{{x->3,y->5},{x->11,y->23}}
but there appear to be no easily found solutions beyond i=1
|
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|
Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt,
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
|
Obviously $a \not = 0$. Then we have $a^4 = 2a^3 - 2a^2$ by multiplying the condition by $a^2$, so the equation becomes: $a^3 - a^2 + 2$. Similarly $a^3 = 2a^2 - 2a$, so the equation becomes: $a^2 - 2a + 2 = 0$
|
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|
Coefficient of $x^2$ in $(x+\frac 2x)^6$ I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
|
We have $$\displaystyle\left(x+\dfrac{2}{x}\right)^6=\sum_{k=0}^{6}\binom{6}{k}x^{6-k}\left(\frac{2}{x}\right)^k=\sum_{k=0}^{6}2^k\binom{6}{k}x^{6-2k}.$$ Then, $6-2k=2\Leftrightarrow k=2,$ so $$\text{coef }x^2=2^2\binom{6}{2}=60.$$
|
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|
smallest positive integer $r$ so that $5^{33333}≡ r (mod 11)$ Title is pretty self explanatory. I tried using the division algorithm to get some a hint about the $5^{33333}$. This was not helpful.
|
$$5\equiv 5\mod 11\\
5^2=25\equiv 3\mod 11\\
5^3= 5^2\cdot 5\equiv 3\cdot 5=15\equiv 4\mod 11\\
5^4= 5^3\cdot 5\equiv 4\cdot 5=20\equiv 9\mod 11\\
5^5= 5^4\cdot 5\equiv 9\cdot 5=45\equiv 1\mod 11\\
$$
So, you know that $5^5\equiv 1\mod 11$.
You also know that $33333=5\cdot 6666+3$
|
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|
Does flipping the negative in a fraction flip all terms both sides? This is an expression that contains a negative in the denominator. If I were to take the negative and place it in the numerator, would this change all positive terms to negative and vice-versa?
Negative in denominator:
$\frac{a+3-y^2}{-2}$
After flipping negative to the numerator: $\frac{-a-3+y^2}{2}$ (Is this correct?)
|
Yes, but your language is a little vague and misleading. Things to notice:
$1/1 = 1$ (Duh)
$1/(-1) = -1$ (Not so duh; but pretty easy)
$1*1 = 1; (-1)*1= 1*(-1) = -1; (-1)(-1) = 1$ ("A negative times a negative is a positive". To some this is obvious; to some this makes no sense. But most know this as a rule [$*$])
$a(b + c) = ac + ac$ (this is the distributive law)
So if we know all those thing we can conclude:
$\frac {a + 3 - y^2}{-2} = $
$\frac{a+3 - y^2}{(-1)2}=$
$\frac 1{-1}\frac{a+3 -y^2}{2} =$
$- \frac{a+3 -y^2}{2} =$
$\frac{(-1)(a+3-y^2)}2 =$
$\frac{(-1)a + (-1)3 -(-1)y^2}2=$
$\frac{-a -3 +y^2}2$
which can be simplified as "you flip the sign everywhere".
Post-script: Or, d'oh, we can do what everyone else suggests:
$\frac {a + 3 - y^2}{-2} = 1*\frac {a + 3 - y^2}{-2}=\frac{-1}{-1}\frac {a + 3 - y^2}{-2}= \frac{-(a+3-y^2}{-(-2)} = \frac {-a - 3 + y^2}{2}$
=====
[$*$] why is "a negative times a negative a positive"
Welp.... if $x = -5$ means $5 + x = 0$ and $3*5$ means $5+5+5$. So $(-3)*5= x$ means $x + 5+5+5 = 0$ so $(-3)*5 = -5 -5 -5$. So $(-3)*(-5) = -(5) -(-5) - (-5) = 5+5+5 = 3*5$.
|
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|
Max and min of a function with absolute values I have this function :
$f(x) =
\begin{cases}
\left\lvert\frac{x-3}{x+3}\right\rvert\log \left\lvert\frac{x+3}{x-3}\right\rvert -3, & \text{if $x\neq \pm3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
How do you find the derivative?
I usually split the functions when there are absolute values but this time I've found some irregularities about the min and max points.
I would split in this way:
$f(x)_1 =
\begin{cases}
\left(\frac{x-3}{x+3}\right)\log \left(\frac{x+3}{x-3}\right)-3, & \text{if $x\lt -3 \land x \gt +3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
$f(x)_2 =
\begin{cases}
\left(\frac{3-x}{x+3}\right)\log \left(\frac{-x-3}{x-3}\right)-3, & \text{if $ -3 \lt x \lt +3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
And so the derivatives are:
$f(x)_1' =
\begin{cases}
\left(\frac{6}{(x+3)^2}\right)\left(\log \left(\frac{x+3}{x-3}\right) -1\right), & \text{if $x\lt -3 \land x \gt +3$} \\[2ex]
0, & \text{if $x= \pm3$}
\end{cases}$
$f(x)_2' =
\begin{cases}
\left(-\frac{6}{(x+3)^2}\right)\left(\log \left(\frac{-x-3}{x-3}\right) -1\right), & \text{if $ -3 \lt x \lt +3$} \\[2ex]
0, & \text{if $x= \pm3$}
\end{cases}$
I don't know what I got wrong.. I know for sure that it's something related the absolute value.I should get two max and a minimum.
Can you help me please?
Thanks in advance!
|
I think what you have is correct. Namely, you can prove using your derivatives on different intervals that the function is strictly decreasing of $x<-3$. On $(-3,3)$ and on $(3,\infty)$ the function is probably concave and hence has one local maximum on each interval. Comparing values at those local maxima with $-3$, values at $x=\pm3$, you get your solution.Plotting the function usually helps.
|
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|
$y=x^2-x^3$ Area under curve I have to use the limit process to fund the area of the region between the graph of the function, and the x-axis over the given interval.
$f(x)=x^2-x^3$ in the interval of $[-1,0]$
I started the problem by solving for $\Delta x$, and got that $\Delta x=\frac{1}{n}$. Then I got $c_i=a+i\Delta x$, and got $c_i=-1+\frac{i}{n}$
Then I plugged it into the limit process:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(f(c_i)\Delta x)$
Long Story short I got $1$ to be my answer is it correct? If not then how do I get the correct answer?
|
To find the area under the curve, evaluate:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(f(c_i)\Delta x)$
$\Delta x = \frac{b-a}{n} = \frac{0 - (-1)}{n} = \frac{1}{n}$
$c_i = a + i\Delta x = -1 + \frac{i}{n}$
$f(c_i) = (c_i)^2 - (c_i)^3 = \frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2$
So we have,
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n((\frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2)\Delta x)$
The $\Delta x$ is constant, so we can take it out of the summation:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(\frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2)\Delta x$
Distribute the summation.
$$\lim_\limits{n \to \infty}(\sum_\limits{i=1}^n\frac{4i^2}{n^2} - \sum_\limits{i=1}^n\frac{5i}{n} - \sum_\limits{i=1}^n\frac{i^3}{n^3} + \sum_\limits{i=1}^n2)\Delta x$$
$$\lim_\limits{n \to \infty}(\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6} - \frac{5}{n}\frac{n(n+1)}{2} - \frac{1}{n^3}\frac{n^2(n+1)^2)}{4} + 2n)(\frac{1}{n})$$
$$\lim_\limits{n \to \infty}(\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6n} - \frac{5}{n}\frac{n(n+1)}{2n} - \frac{1}{n^3}\frac{n^2(n+1)^2)}{4n} + \frac{2n}{n})$$
Evaluate the limit:
$$= \frac{8}{6} - \frac{5}{2} - \frac{1}{4} + \frac{2}{1} = \frac{7}{12}$$
I wrote alot, so I might have made a mistake, but the final answer is correct so I probabbly did not
|
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|
Derive $\cos(3\theta)=(4\cos\theta)^3 − 3\cos\theta$ I'm having trouble with the following derivation:
Q:
We can use Euler's Theorem ($e^{i\theta} = \cos\theta + i\sin\theta$), where $e$ is the base of the natural logarithms, and $i = \sqrt{-1}$, together with the binomial theorem as above, to derive a number of trigonometric identities. E.g., if we consider $(e^{i\theta})^2$, we can evaluate it two different ways. First, we can multiply exponents, obtaining $e^{i \, 2\theta}$ and then applying Euler's formula to get $\cos(2\theta) + i \sin(2\theta)$, or we can apply Euler's formula to the inside, obtaining $(\cos\theta + i \sin\theta)^2$, which we then evaluate via the binomial theorem:
\begin{align}
\cos(2\theta) + i \sin(2\theta) &= (\cos\theta + i \sin\theta)^2 \\
&= (\cos\theta)^2 + 2 i \cos\theta \sin\theta + i^2 (\sin\theta)^2 \\
&=(\cos\theta)^2 + 2 i \cos\theta \sin\theta − (\sin\theta)^2
\end{align}
Equating real and imaginary parts gives us
\begin{align}
\cos(2\theta) &= (\cos\theta)^2 − (\sin\theta)^2 \\
\sin(2\theta) &= 2 \cos\theta \sin\theta
\end{align}
We can then rewrite the first of these identities, using $1=(\sin\theta)^2+(\cos\theta)^2$ to get $(\cos\theta)^2=1−(\sin\theta)^2$, whence the familiar
$$\cos(2\theta)=1−2(\sin\theta)^2$$
Use this same approach to show $\cos(3\theta)=(4\cos\theta)^3−3\cos\theta$.
A:
This is my work so far:
\begin{align}
e^{i \, 3\theta} &= \cos(3\theta) + i \sin(3\theta) = (\cos\theta)^3 + 3 i (\cos\theta)^2 \sin\theta - 3\cos\theta (\sin\theta)^2 - i(\sin\theta)^3 \\
\cos(3\theta) &= (\cos\theta)^3 - 3(\sin\theta)^2 \cos\theta \\
\sin(3\theta) &= 3\sin\theta(\cos\theta)^2 - (\sin\theta)^3
\end{align}
But now I'm unsure how to get $\cos(3\theta) = (4 \cos\theta)^3 − 3\cos\theta$ from what I've derived.
|
Continue from your answer,
$\cos(3\theta) = (\cos \theta)^3 - 3(\sin \theta)^2\cos \theta$
$= (\cos \theta)^3 - 3(1-\cos^2 \theta)\cos \theta$
$= (\cos \theta)^3 - 3\cos \theta(1-\cos^2 \theta)$
$= \cos^3 \theta - 3\cos \theta + 3\cos^3\theta$
$= 4\cos^3 \theta - 3\cos \theta$
Similarly you can change $\cos^2\theta = 1 - \sin^2\theta$ to get the formula of $\sin(3\theta)$.
|
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|
What is the greatest area difference between two "nice triangles"? We call a triangle "nice if all angles are between $45$ and $90$ degrees (including $90$ and $45$ itself) and all sides are between $1$ and $2$ (including $1$ and $2$ itself). What is the greatest area difference between two "nice triangles"?
My attempt: Because we have side and angle limits the best way to finding area is using the formula $S=bc\cos{A}$. We should find the greatest and lowest area. But here I got stuck and I don't know how to have both limits with each other. First I thought that the maximum area is $A=90^\circ $ and $b=c=2$. But then I saw that then we have$a=2\sqrt{2}>2$. Could you please give a way?
|
The Maximal triangle has two sides of length $2.$
for any triangle that had only 2 sides of lenght 2, we could extend the second longest side until it was length 2, and create a larger triangle.
$Area = \frac 12 BC\sin a\\
B,C = 2$
maximize $2\sin a$
constrained by:
$4 \sin \frac {a}{2} \le 2$
$\sin \frac {a}{2}\le \frac 12\\
\frac {a}{2}\le 30\\
a\le 60$
Since $\sin a$ is strictly increasing between 0 and 90.
$a = 60\\
Area = 2\sin 60 = \sqrt 3$
then
minimize $\frac 12 \sin a$
constrained by:
$4 \sin \frac {a}{2} \ge 1$
and again $a = 60$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
What is $x$, if $3^x+3^{-x}=1$? I came across a really brain-racking problem.
Determine $x$, such that $3^x+3^{-x}=1$.
This is how I tried solving it:
$$3^x+\frac{1}{3^x}=1$$
$$3^{2x}+1=3^x$$
$$3^{2x}-3^x=-1$$
Let $A=3^x$.
$$A^2-A+1=0$$
$$\frac{-1±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$
$$\frac{-1±\sqrt{-3}}{2}=0$$
I end up with
$$\frac{-1±i\sqrt{3}}{2}=0$$
which yields no real solution. And this is not the expected answer.
I'm a 7th grader, by the way. So, I've very limited knowledge on mathematics.
EDIT
I made one interesting observation.
$3^x+3^{-x}$ can be the middle term of a quadratic equation:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$
|
Just building upon previous comments, it doesn't have as you pointed out a Real result but the Imaginary solution can be analytically found as:
$3^x = \dfrac{1\pm\sqrt{3} i}{2}$
Reexpressing rhs in polar notation:
$3^x = e^{i \dfrac{\pi}{3}}$
And changing lhs basis to $e$
$3^x = e^{\ln{3^x}} = e^{x\ln{3}} $
Then:
$\boxed{x = i \dfrac{\pi}{3\ln{(3)}}}$
Note: this is the principal value solution. Due to the periodicity of the function, any $x = i \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$, for $n\in \mathbb{Z}$ will also be a solution. Also the 2nd quadrant values need to be considered $x = - i \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
}
|
Continuity and integration using Dominated Convergence theorem I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.
I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}}{n}\leq x^{-\frac{3}{2}} $$
But $$\int_0^1x^{-\frac{3}{2}}dx=\frac{-2}{\sqrt{x}}\biggr|_0^1$$ which doesn't seem right. Any help will be appreciated.
|
Let $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$. Then, differentiating with respect to $n$ reveals
$$\begin{align}
\frac{df_n(x)}{dn}&=\frac{\sqrt{x}}{1+n^2x^2}-\frac{2n^2x^{5/2}}{(1+n^2x^2)^2}\\\\
&=\frac{\sqrt{x}(1-n^2x^2)}{(1+n^2x^2)^2}\tag1
\end{align}$$
From $(1)$, it is easy to see that $\frac{df_n(x)}{dn}=0$ when $n=1/x$. Moreover, we can see from $(1)$ that $f_n(x)$ is a maximum when $n=1/x$. Therefore, we have
$$\sup_{n} \left(\frac{n\sqrt{x}}{1+n^2x^2}\right)=\frac{1}{2\sqrt{x}}$$
whereby we have a dominating function that is integrable on $[0,1]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2130472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $\mathcal{S}=\{ax+by\mid x,y\ge0 \text{ and } x,y\in \mathbb{Z}\}$.
Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $$\mathcal{S}=\{ax+by \mid x,y\geq0 \hspace{0.25cm} \text{ and } \hspace{0.25cm} x,y\in \mathbb{Z}\}.$$
Since $a,b\in \mathbb{N}$ and $\gcd(a,b)=1$, then there exists $x',y'\in\mathbb{Z}$ such that $$ax'+by'=1.$$ Let $\alpha\in\mathcal{S}$. Then $$\alpha=ax+by$$ where $a,b\in \mathbb{N}$ with $\gcd(a,b)=1$ and $x,y\in \mathbb{Z}$ with $x,y\geq 0$.
If we were to multiply the first equation by $\alpha$ we will get $$\alpha ax'+\alpha by'=\alpha.$$ But $\alpha=ax+by$. So, $$\alpha ax'+\alpha by'=\alpha=ax+by.$$ We can rewrite this equation as follows:
$$a(\alpha x'-x)=b(y-\alpha y').$$ Than $a(\alpha x'-x)| b$ or $a(\alpha x'-x)| (y-\alpha y')$ or $a(\alpha x'-x)| b(y-\alpha y')$.
From here I am unsure how to proceed. Any tips
|
$S \subset \mathbb N$ (let's include $0$ in our natural numbers) clearly. $a\mathbb N \subset S$ and $b\mathbb N \subset S$ so if $a = 1$ or $b =1$, $S = \mathbb N$.
So it suffices to assume $a > 1$ and $b > 1$.
Okay some bold claims:
1) $n = a(b-1) + b(-1) = a(-1) + b(a-1)=ab-a-b \not \in S$.
2) For any $n > ab-a-b$, $n \in S$.
And as $S \subset \mathbb N$, follows that $S = \{ak + bj| 0\le k <b, 0 \le j < a\} \cup \{n > ab - a - b| n \in \mathbb N\}$.
...
Clain 3: If $n = ax + by$ then $n = aw + bv \iff a = x + kb; b = x-ka$ for some $k \in \mathbb Z$.
Pf: $a(x+kb) + b(x - ka) = ax + by +akb - akb = ax + by$
Let $aw + bv = ax + by$ and let $m = w-x$. Then $ax + by = aw + bv = a(x+m) + bv =ax + bv+am = ax + b(v+\frac{am}b)$
So $y = v + \frac{am}b$ is an integer. So $b|am$ but $\gcd(a,b) = 1$ so $b|m$. So $m = kb$ and $w = x + kb$ and $v = y - \frac{akb}b =y-ak$.
So now the first claim:
Let $n = a(b-1) +b(-1) = a(w) + bv$. Then $w = b-1 + kb$ and $v= -1 - ka$. If $k \ge 0$ then $-1-ka \le -1 < 0$. If $k > 0$ then $k \ge 1$ and $w = b-1 + kb \le -1 < 0$.. So there are no $x,y \ge 0$ so that $ax +by = n$
Now the second claim.
Well, first claim 4: There exist $0 < W < b$ and $0 < V < a$ so that $aW - bV = 1$
By euclids algorithm we know there exist $x, y$ so that $ax + by = 1$. As $a,b > 0$ it's it must be that one of $x,y$ is positive and the other is negative. Now by archimedian principle we can find an integer $k$ so that $0 \le x + kb < b$. Let $W = x + kb$ and let $V = ka - y$ so $aW - bV = a(x + kb) + b(y - ka) = 1$
If $W = 0$ then $1=aW - bV = -bV$ which is impossible as $b \not \mid 1$.
So $0 < W= x+kb < b$ so
$\frac{-x}b < k < \frac {b-x}b = 1 - \frac xb$
$- \frac{xa}b - y < ka - y = V < a - \frac {xa}b$
So $V < a$ and if $V \le 0$ then $aW - bV > aW > 1$. So $0 < V$.
Okay, now the second claim.
Let $n = ab -a -b + 1= ab -a -b +aW -bV$
$=a(W -1) + b(a-1-V)$
$W \ge 1$ so $W-1\ge 0$. $a > V$ so $a-V > 0$ so $a-V \ge 1$ so $a-1-V \ge 0$. So $a-a-b+1 \in S$.
Suppose $n \in S$ and $n > ab -b -a$.
Then $n = ax + by; x,y \ge 0$ wolog $0 \le x < b$. (Else substitute with $x' = x + kb$ so that $0 \le x < b$). $ax + by > a(b-1) + b(-1)$ but $x \le b-1$ so $y > -1$ i.e $y \ge 0$.
$n+1 = ax +by + aW -bV=a(x+W) + b(y-V) = a(x+W -b) + (y-V +a)$. $x+(W-b) > x \ge 0$. $y-V + a = y +(a-V) > y \ge 0$. So $n_1 \in S$.
So Claim 2 follows by induction.
===
Example $a = 5$ and $b=7$. All integers $n > 35- 5 - y = 23$ are in $S$.
$3*5 - 2*7 = 1$ so $W =5; V = 2$ $24 ==a(W -1) + b(a-1-V)=5*2 + 7*2$. $25 = 5(2+3) + 7(2-2) = 5*5; 26 = 5(5+3) + 7(0-2)= 5(5+3 -7) + 7(0-2+5) = 5+3*7; $ etc.
So $S = \{0,5,10,15,20,7,12,17,22,14,19,21, n\ge 24\}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2131050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$
I suspect some clever use of the geometric series will do, but I don't know how.
|
Take a geometric series with term $r$ and take derivative with respect to $r$ both sides:
$\frac{d}{dr}\sum_{n=0}^{\infty}r^n=\frac{d}{dr}\frac{1}{1-r}.$
Derivate,
$\sum_{n=0}^{\infty}nr^{n-1}=\frac{1}{(1-r)^2}.$
At the left hand side, for $n=0$ the term is zero, so we can start the sum at $n=1$, and also multiply $r$ both sides to balance the exponent:
$r\sum_{n=1}^{\infty}nr^{n-1}=r\frac{1}{(1-r)^2}.$
Now you have
$\sum_{n=1}^{\infty}nr^{n}=\frac{r}{(1-r)^2}.$
For $r=\frac{1}{3}$,
$\sum_{n=1}^{\infty}n\left(\frac{1}{3}\right)^{n}=\frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2}.$
Operate,
$\sum_{n=1}^{\infty}n\frac{1}{3^n}=\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{3}{4}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2131479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Antiderivative of a product of a rational function and a square root. Let $0<x<1$ and $p$ be a non-negative integer.
Consider a following definite integral:
\begin{equation}
{\mathbb I}_p(x):=\int\limits_1^x \frac{(1-v^2)^p}{v^{2p+2}} \cdot \sqrt{\frac{1-v}{1+v}} dv
\end{equation}
We have computed it in a following way. Firstly we substituted for the square root and reduced the problem to integrating a rational function. Then we decomposed the resulting rational function into simple fractions and by integrating terms by term we completed the computation. The result reads:
\begin{equation}
{\mathbb I}_p(x) = (-1)^p 2 \binom{p-\frac{1}{2}}{-\frac{1}{2}} \text{arctanh}(\sqrt{\frac{1-x}{1+x}})+\frac{\sqrt{1-x^2}}{x^{2 p+1}} \cdot\sum\limits_{l=0}^{2 p} {\mathcal C}_{l,p} x^l
\end{equation}
where the coefficients read:
\begin{eqnarray}
&&{\mathcal C}_{l,p} := \sum _{q=0}^{\left\lfloor \frac{l}{2}\right\rfloor }
\frac{\sqrt{\pi } 2^{-l+2 q+1} (-l+2 p+q)! \binom{4 p+2}{l-2 q} \binom{-l+2 p+2 q+1}{2 q+1} }
{\left(-q-\frac{3}{2}\right)! (-l+2 p+2 q+1) (-l+2 p+2 q+1)!}
\, _3F_2\left(-p-1,l-4 p-2 q-2,2 q-l;-2 p-1,-2 p-\frac{1}{2};1\right)\\
&&=
\left(
\begin{array}{c}
\{-1\} \\
\left\{-\frac{1}{3},\frac{1}{2},\frac{1}{3}\right\} \\
\left\{-\frac{1}{5},\frac{1}{4},\frac{2}{5},-\frac{5}{8},-\frac{1}{5}\right\} \\
\left\{-\frac{1}{7},\frac{1}{6},\frac{3}{7},-\frac{13}{24},-\frac{3}{7},\frac{11}{16},\frac{1}{7}\right\} \\
\left\{-\frac{1}{9},\frac{1}{8},\frac{4}{9},-\frac{25}{48},-\frac{2}{3},\frac{163}{192},\frac{4}{9},-\frac{93}{128},-\frac{1}{9}\right\} \\
\left\{-\frac{1}{11},\frac{1}{10},\frac{5}{11},-\frac{41}{80},-\frac{10}{11},\frac{171}{160},\frac{10}{11},-\frac{149}{128},-\frac{5}{11},\frac{193}{256},\frac{1}{11}\right\} \\
\end{array}
\right)
\end{eqnarray}
Above we have given numerical values of those coefficients for $p=0,\dots,5$ (rows) and $l=0,\dots,2 p$ (columns). Now the question is can we come up with some closed form solution for the coefficients. Note that even at the first glance we see that in case $l=0$ and $l=p$ there is a closed form. What about other values of $l$?
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With
$\ds{t \equiv {1 - v \over 1 + v} \implies
v = {1 - t \over 1 + t}}$:
\begin{align}
{\mathbb I}_{p}\pars{x} & = \int_{1}^{x}{\pars{1 -v^{2}}^p \over v^{2p + 2}} \,\root{1 - v \over 1 + v}\,\dd v =
\stackrel{t^{2}\ \mapsto\ t}{=}\,\,\,
-\,2^{\,2p + 1}\int_{0}^{\pars{1 - x}/\pars{1 + x}}
t^{p + 1/2}\pars{1 - t}^{-2p - 2}\,\dd t
\\[5mm] & =
\bbx{\ds{-2^{2p + 1}\,
\mrm{B}\pars{{1 - x \over 1 + x},p + {3 \over 2},-2p - 1}}}
\end{align}
where $\ds{\,\mrm{B}}$ is the Incomplete Beta Function. Note that
$\ds{\Re\pars{p + {1 \over 2}} > - 1 \implies \Re\pars{p} > -\,{3 \over 2}}$.
|
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"url": "https://math.stackexchange.com/questions/2133670",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.
Substituting dx in the integral:
$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$
I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.
Thanks
|
I would let $u=x-1$ which makes the integrand $(u+1)^2(u-1)/u^2.$ After you multiply out the top you have a sum of powers of $u$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 4
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|
$\lim_{z \to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}$ Find $$\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16}$$
Note that
$$z=\exp(\pi i/3)=\cos(\pi/3)+i\sin(\pi/3)=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^2=\exp(2\pi i/3)=\cos(2\pi/3)+i\sin(2\pi/3)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^3=\exp(3\pi i/3)=\cos(\pi)+i\sin(\pi)=1$$
$$z^4=\exp(4\pi i/3)=\cos(4\pi/3)+i\sin(4\pi/3)=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$$
So,
\begin{equation*}
\begin{aligned}
\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16} & = \dfrac{1+8}{-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}+4\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)+16} \\
& = \dfrac{9}{\dfrac{27}{2}+i\frac{3\sqrt{3}}{2}} \\
& = \dfrac{6}{9+i\sqrt{3}} \\
& = \dfrac{9}{14}-i\dfrac{\sqrt{3}}{2} \\
\end{aligned}
\end{equation*}
But, when I check my answer on wolframalpha, their answer is $$\dfrac{245}{626}-i\dfrac{21\sqrt{3}}{626}.$$
Can someone tell me what I am doing wrong?
|
You can make your life very easy by noticing that $(\mathrm e^{\mathrm i \pi/3})^3=\mathrm e^{\mathrm i \pi} = -1$.
This means that if $z=\mathrm e^{\mathrm i \pi/3}$, then $z^3=-1$ and hence $z^4=-z$.
$$\frac{z^3+8}{z^4+4z+16} = \frac{7}{3z+16}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2136937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
Find the $x$ such $|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$ Find $x$, such there exsit four point $A,B,C,D$ on the plane,such
$$|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$$
I think it's nice problem. I'm looking for an arbitrary quadrilateral six relations，and this six trigonmetry have some relations.such
$$\sin^2{x}+\cos^2{x}=1.\tan{x}\cot{x}=1,$$$$\dfrac{1}{\sin{x}}=\csc{x},\dfrac{1}{\cos{x}}=\sec{x},1+\cot^2{x}=\csc^2{x},1+\tan^2{x}=\sec^2{x}$$
|
There are no such $x\in\mathbb R$.
We may suppose that
$$0\lt x\lt \frac{\pi}{2},\quad A(0,0),\quad B(\sin x,0),\quad C(c_1,c_2),\quad D(d_1,d_2)$$
with $c_2\ge 0$.
Since $C(c_1,c_2)$ where $c_2\ge 0$ is both on the circle $X^2+Y^2=|AC|^2$ and on the circle $(X-|AB|)^2+Y^2=|BC|^2$, we get
$$c_1=\frac{\sec^2x+\sin^2x-\cos^2x}{2\sin x}=\frac{\frac{(1-t)(t+2)}{t+1}}{2\sin x},\quad c_2=\frac{\sqrt{X_1}}{2\sin x}$$
where
$$t=\cos(2x)$$
and
$$X_1=(2\sin x\sec x)^2-(\sec^2x+\sin^2x-\cos^2x)^2=\frac{t(1-t)(t^2+3t+4)}{(t+1)^2}$$
Also, since $D(d_1,d_2)$ is both on the circle $X^2+Y^2=|AD|^2$ and on the circle $(X-|AB|)^2+Y^2=|BD|^2$, we get
$$d_1=\frac{\cot^2x+\sin^2x-\csc^2x}{2\sin x}=\frac{-\frac{t+1}{2}}{2\sin x},\quad d_2=\pm\frac{\sqrt{X_2}}{2\sin x}$$
where
$$X_2=(2\sin x\cot x)^2-(\cot^2x+\sin^2x-\csc^2x)^2=\frac{(t+1)(7-t)}{4}$$
Now multiplying the both sides of
$$\tan^2x=(c_1-d_1)^2+(c_2-d_2)^2$$
by $4\sin^2x$ gives
$$\begin{align}&4\sin^2x\tan^2x=\left(\frac{(1-t)(t+2)}{t+1}+\frac{t+1}{2}\right)^2+\left(\sqrt{X_1}\mp \sqrt{X_2}\right)^2\\\\&\implies 4\cdot \frac{1-t}{2}\cdot \frac{1-t}{t+1}-\left(\frac{(1-t)(t+2)}{t+1}+\frac{t+1}{2}\right)^2-X_1-X_2=\mp 2\sqrt{X_1X_2}\\\\&\implies (t-2)(t+3)=\mp 2\sqrt{X_1X_2}\end{align}$$
Squaring the both sides gives
$$(t-2)^2(t+3)^2=4\cdot\frac{t(1-t)(t^2+3t+4)}{(t+1)^2}\cdot \frac{(t+1)(7-t)}{4}$$
Multiplying the both sides by $t+1$ gives
$$2t^4+t^3-3t^2-t+9=0$$
which can be written as
$$2\left(t^2+\frac t4-1\right)^2+\frac{7}{8}t^2+7=0$$
This has no real solutions since the LHS is always positive.
Hence, there are no $x\in\mathbb R$ satisfying the conditions given in the question.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2137636",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Can the quadratic formula be used with variable coefficients? Could we use the quadratic formula on an expression such as $z^2xy - zx^2y+y = 0$ to find $z$ in terms of $x$ and $y$?
|
Write $$z^2xy - zx^2y + y = 0$$
as $$(xy)z^2 - (x^2y)z + (y) = 0$$
so that you can see $a = xy$, $b=-x^2y$ and $c=y$.
Then
\begin{align}
z &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\
z &= \dfrac{x^2y \pm \sqrt{x^4y^2-4xy^2}}{2xy} \\
z &= \dfrac{x^2y \pm |y|\sqrt{x^4-4x}}{2xy} \\
z &= \dfrac{x^2 \pm \sqrt{x^4-4x}}{2x} \\
\end{align}
Notice that $y$ has disappeared. That is because
$(xy)z^2 - (x^2y)z + (y) = 0 \iff y(xz^2 - x^2z + 1) = 0$.
So we need to notice that $y=0$ is a solution. If $y\ne 0$ then we get
$z = \dfrac{x^2 \pm \sqrt{x^4-4x}}{2x}$; where we are going to need to require that $x \ne 0$ and, if we only want real-valued solutions, $x^4-4x \ge 0$.
|
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|
Pseudoinverse of $2\times 2$ matrix How can I find the Moore-Penrose pseudoinverse of the $2 \times 2$ complex matrix
$$A=\begin{pmatrix}0&a\\0&b\end{pmatrix}$$
for $a \neq 0$ and $b \neq 0$?
Here I want to use the limit formula
$$A^+=\lim_{\epsilon \to 0} (\epsilon I+A^*A)^{-1}A^*$$
since $\mbox{rank}(A)=1$, which is not full rank. Any help, please?
|
Computing eigendecompositions using SymPy:
>>> from sympy import *
>>> a, b = symbols('a b')
>>> M = Matrix([[0,a],[0,b]])
>>> (M.T * M).eigenvects()
[(0, 1, [Matrix([
[1],
[0]])]), (a**2 + b**2, 1, [Matrix([
[0],
[1]])])]
>>> (M * M.T).eigenvects()
[(0, 1, [Matrix([
[-b/a],
[ 1]])]), (a**2 + b**2, 1, [Matrix([
[a/b],
[ 1]])])]
We now build the matrices in the SVD:
>>> U = (1/sqrt(a**2 + b**2)) * Matrix([[a,-b],[b,a]])
>>> S = diag(sqrt(a**2 + b**2),0)
>>> V = Matrix([[0,1],[1,0]])
>>> U * S * V.T
Matrix([
[0, a],
[0, b]])
The SVD is
$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} = \begin{bmatrix} \frac{a}{\sqrt{a^{2} + b^{2}}} & - \frac{b}{\sqrt{a^{2} + b^{2}}}\\ \frac{b}{\sqrt{a^{2} + b^{2}}} & \frac{a}{\sqrt{a^{2} + b^{2}}} \end{bmatrix} \begin{bmatrix} \sqrt{a^{2} + b^{2}} & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}^{\top}$$
Hence, the pseudoinverse is
$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix}^{+} = \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{a^{2} + b^{2}}} & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} \frac{a}{\sqrt{a^{2} + b^{2}}} & - \frac{b}{\sqrt{a^{2} + b^{2}}}\\ \frac{b}{\sqrt{a^{2} + b^{2}}} & \frac{a}{\sqrt{a^{2} + b^{2}}} \end{bmatrix}^{\top} = \color{blue}{\begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix}}$$
Verifying,
$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix} \begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} = \begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix}$$
$$\begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix} \begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix} = \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix}$$
Also, both products of the given matrix and its pseudoinverse are symmetric, as required.
This is the real case. The complex case should be easy to tackle.
|
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|
rational numbers of the form $\frac{a^3+b^3}{c^3+d^3}$ Show that all positive rational numbers can be written in the form
$$\frac{a^3+b^3}{c^3+d^3}$$
where $a,b,c,d$ are positive integers.
|
Xam commented on 2017 Feb 13 that an answer was given here by zabelman on 2005 Dec 17. Also on 2017 Feb 13, timon92 mentioned the following answer found in a post by mathlove on 2015 Mar 11:
For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that
$$r=\frac{a^\color{red}{3}+b^\color{red}{3}}{c^\color{red}{3}+d^\color{red}{3}}.$$
For $r=p/q$ where $p,q$ are positive integers, we can take
$$(a,b,c,d)=(3ps^3t+9qt^4,\ 3ps^3t-9qt^4,\ 9qst^3+ps^4,\ 9qst^3-ps^4)$$
where $s,t$ are positive integers such that $3\lt r\cdot(s/t)^3\lt 9$.
For $r=2014/89$, for example, since we have $(2014/89)\cdot(2/3)^3\approx 6.7$, taking $(p,q,s,t)=(2014,89,2,3)$ gives us $$\frac{2014}{89}=\frac{209889^3+80127^3}{75478^3+11030^3}.$$
|
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|
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sqrt{x+2}}}{(x-2)\sqrt{1+\sqrt{x+2}+\sqrt3}}$
I am stuck here.
|
Following my comment, the question would make "more sense" (seeing that you know the method of multiplying with a conjugate expression) if the numerator was $\sqrt{1+\sqrt{x+2}}-\sqrt3$.
If you did copy the question correctly, then it might have been a bit of a trick question since you don't need that method; it is not an indeterminate form. Other answers (by S.C.B. and Jaroslaw Matlak) explain the way to go.
If the numerator is as I suggested, then your idea of multiplying with the conjugate expression is fine and you can apply that trick twice in order to cancel out a common factor $\color{red}{x-2}$:
$$\begin{align}\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}
& = \lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}\color{blue}{\frac{\sqrt{1+\sqrt{x+2}}+\sqrt3}{\sqrt{1+\sqrt{x+2}}+\sqrt3}} \\[7pt]
& = \lim_{x\to2}\frac{1+\sqrt{x+2}-3}{\left(x-2\right)\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)} \\[7pt]
& = \lim_{x\to2}\frac{\sqrt{x+2}-2}{\left(x-2\right)\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)}\color{blue}{\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}} \\[7pt]
& = \lim_{x\to2}\frac{\color{red}{x-2}}{\color{red}{\left(x-2\right)}\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)\left(\sqrt{x+2}+2\right)} \\[7pt]
& = \lim_{x\to2}\frac{1}{\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)\left(\sqrt{x+2}+2\right)}= \frac{1}{8\sqrt{3}} \\[7pt]
\end{align}$$
Perhaps this still helps.
|
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|
Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ Given the positive numbers $a, b, c$ satisfy $a+b+c\le \sqrt{3}$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$
My Try (Edited from Comments):
By Cauchy Schwarz, we have that $$ (a^2+1)(1+3) \ge \left(a+\sqrt {3}\right)^2 \rightarrow \frac{a}{\sqrt{a^2+1}} \le \frac{2a}{a+\sqrt{3}} $$ I need a new method
|
By AM-GM and P-M we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum_{cyc}\frac{a}{\sqrt{a^2+3\cdot\frac{1}{3}}}\leq\sum_{cyc}\frac{a}{\sqrt{4\sqrt[4]{\frac{a^2}{27}}}}=
\frac{3\sqrt[8]{27}}{2}\frac{\sum\limits_{cyc}\sqrt[4]{a^3}}{3}\leq$$
$$\leq\frac{3\sqrt[8]{27}}{2}\left(\frac{a+b+c}{3}\right)^{\frac{3}{4}}\leq\frac{3\sqrt[8]{27}}{2}\left(\frac{\sqrt3}{3}\right)^{\frac{3}{4}}=\frac{3}{2}$$
|
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|
A problem with two dimensional vectors and scalar multiples
Let $\vec{x}$ and $\vec{y}$ be non-zero vectors. Show that any two-dimensional vector can be expressed in the form
$s \vec{x} + t \vec{y}$, where $a$ and $b$ are real numbers, if and only if of the vectors $\vec{x}$ and $\vec{y}$, one vector is not a scalar multiple of the other vector.
I am not quite sure how to approach this problem. Must the vectors $\vec{x}$ and $\vec{y}$ parallel for this to be true? I just have a suspicion, but I am actually struggling to find a good place to begin.
|
Suppose $\vec{x} = k \vec{y}$. Then, $s\vec{x} + t\vec{y} = (s+kt)\vec{x}$. The locus of such points as $s$ and $t$ range over the reals is a line. So, not every two-dimensional vector can be expressed as $$s\vec{x} + t\vec{y}.$$For the other direction, let $\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ and $\vec{y} = \begin{pmatrix} c \\ d \end{pmatrix}$. Let $\begin{pmatrix} w \\ z \end{pmatrix}$ be an arbitrary two-dimensional vector.
We want to find real numbers $s$ and $t$ such that
$$s \begin{pmatrix} a \\ b \end{pmatrix} + t \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} w \\ z \end{pmatrix}.$$This gives us the system of equations
\begin{align*}
as + ct &= w, \\
bs + dt &= z.
\end{align*}Solving this system in $s$ and $t$, we find
$$s = \frac{dw - cz}{ad - bc} \quad \text{and} \quad t = \frac{az - bw}{ad - bc}.$$
Both of these quantities are well-defined as long as $ad \neq bc$. Thus, we can conclude if we show that $ad = bc$ implies that of the two non-zero vectors $\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ and $\vec{y} = \begin{pmatrix} c \\ d \end{pmatrix}$, one vector is a scalar multiple of the other vector.
So, suppose $ad = bc$, If $a \ne 0$, then $d = \frac{c}{a} \cdot b$, so $$\frac{c}{a} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} c \\d \end{pmatrix}.$$ The case where $c \ne 0$ is similar.
If $a= c =0$, then both vectors are of the form $\begin{pmatrix} 0 \\ z \end{pmatrix}$, so one must be a multiple of the other.
|
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|
Solve the following system of homogeneous linear equation. $$2x-y+z=0, 3x+2y-z=0,x+4y+3z=0$$
$$
\begin{vmatrix}
2 & -1 & 1 \\
3 & 2 & -1 \\
1 & 4 & 3 \\
\end{vmatrix}
$$
By reducing row $$R_1=R_1-2R_3\\R_2=R_3-3R_3$$
we get
\begin{vmatrix}
0 & -9 & -5 \\
0 & -10 & -10 \\
1 & 4 & 3 \\
\end{vmatrix}
Expanding along C3 and solving the determinant we get 40
But how to solve further ?
|
Once all equations are equal to zero then:
$$\begin{vmatrix}
0 & -9 & -5 \\
0 & -10 & -10 \\
1 & 4 & 3 \\
\end{vmatrix}=\begin{vmatrix}
0 & -9 & -5 \\
0 & 1 & 1 \\
1 & 4 & 3 \\
\end{vmatrix}$$
And now $R_1=R_1+9R_2$ and get
$$\begin{vmatrix}
0 & 0 & 4 \\
0 & 1 & 1 \\
1 & 4 & 3 \\
\end{vmatrix}$$
The above system is equivalent to
$$4z=0\to z=0\\
y+z=0\to y=0\\
x+4y+3z=0\to x=0$$
|
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|
How to validate the following inequality? $\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}.$ I want to find the least values of $a$ and $b$ for which the above inequality holds good for all nonnegative real values of $x, y, z.$
|
$$\tag{1}\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}$$
Let us proceed by successive equivalent propositions, reaching at the end two unique values for $a$ and $b$.
First of all, you do not need three variables: you can amalgamate $x$ and $y$ into a single variable $t:=x+y$.
$$\tag{2}\sqrt{2(t+z)}\leq a\sqrt{t}+b\sqrt{z}$$
Set now $T=\sqrt{t}$ and $Z=\sqrt{z}.$
$$\tag{3}\sqrt{2(T^2+Z^2)}\leq aT+bZ.$$
As all the quantities in (3) are positive, we have the following equivalent inequality:
$$\tag{4}2(T^2+Z^2)\leq a^2T^2+b^2Z^2+2abTZ.$$
$$\tag{5} (a^2-2)T^2+(b^2-2)Z^2+2abTZ \geq 0. \ \ \text{for any positive} \ T,Z $$
Dividing by $Z^2$, and setting $V=T/Z$, one gets the quadratic inequality:
$$\tag{6}(a^2-2)V^2+(2ab)V+(b^2-2) \geq 0 \ \ \text{for any positive} \ V$$
Taking into account
*
*the sign of discriminant $\Delta=4a^2b^2-4(a^2-2)(b^2-2)=8(a^2+b^2-2)$ and
*the sign of dominant coefficient $a^2-2,$
the only possible case is when $a^2-2\geq0$ and $a^2+b^2-2\leq 0$.
The only (positive) values of $a$ and $b$ fulfilling these two conditions are
$a=\sqrt{2}$ and $b=\sqrt{2}$
|
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|
To find value of $\sin 4x$ Given $\tan x = \frac { 1+ \sqrt{1+x}}{1+ \sqrt{1-x}}$. i have to find value of $\sin 4x$. i write $\sin 4x=4 \frac{ (1-\tan^2 x)(2 \tan x)}{(1+\tan^2 x)^2}$ but it seems very complicated to do this? Any other methods?
Thanks
|
$a=\sqrt{1+y}\qquad b=\sqrt{1-y}\qquad t=\tan(x)=\frac{1+a}{1+b}\qquad \sin(4x)=\frac{4t(1-t^2)}{(1+t^2)^2}$
$a^2+b^2=2$
$a^2-b^2=2y$
Let's have fun:
$\displaystyle{\sin(4x)=4\left(\frac{1+a}{1+b}\right)\frac{(1+b)^2-(1+a)^2}{(1+b)^2}\frac{(1+b)^4}{((1+a)^2+(1+b)^2)^2}}$
$((1+a)^2+(1+b)^2)^2=(1+2a+a^2+1+2b+b^2)^2=4(2+a+b)^2$
$(1+b)^2-(1+a)^2=(2+a+b)(b-a)$
We have a good simplification already : $\displaystyle{\sin(4x)=\frac{(1+a)(1+b)(b-a)}{2+a+b}}$
$(2+a+b)(2-a-b)=(4+2a+2b-2a-a^2-ab-2b-ab-b^2)=2-2ab=2(1-ab)$
$(1-ab)(1+ab)=1-a^2b^2=1-(1-y^2)=y^2$
$2y^2\sin(4x)=(1+a)(1+b)(b-a)(2-a-b)(1+ab)$
$(b-a)(1+ab)=b-a+a(1-y)-b(1+y)=-(a+b)y$
$(1+a)(1+b)(2-a-b)=2+a+b-b^2-a^2-a^2b-ab^2=a+b-b(1+y)-a(1-y)=(a-b)y$
$2y^2\sin(4x)=-(a+b)y(a-b)y=-2y^3$
$$\bbox[10px,border:2px solid]{\sin(4x)=-y}$$
|
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|
Computing $7^{13} \mod 40$ I wanted to compute $7^{13} \mod 40$. I showed that
$$7^{13} \equiv 2^{13} \equiv 2 \mod 5$$
and
$$7^{13} \equiv (-1)^{13} \equiv -1 \mod 8$$.
Therefore, I have that $7^{13} - 2$ is a multiple of $5$, whereas $7^{13} +1$ is a multiple of $8$. I wanted to make both equal, so I solved $-2 + 5k = + 8n$ for natural numbers $n,k$ and found that $n = 9, k = 15$ gave a solution (just tried to make $3 + 8n$ a multiple of $5$. Therefore, I have that
$$7^{13} \equiv -73 \equiv 7 \mod 40.$$
Is this correct? Moreover, is there an easier way? (I also tried to used the Euler totient function, but $\phi(40) = 16$, so $13 \equiv -3 \mod 16$, but I did not know how to proceed with this.)
|
You could note that $\varphi(5) = 5-1 = 4$ and $\varphi(8) = 8- 4 = 4$. Hence $7^4 \equiv 1 \pmod 5$ and $7^4 \equiv 1 \pmod 8$, in which case $7^4 \equiv 1 \pmod{40}$.
Hence $7^{13} \equiv (7^4)^3 \cdot 7 \equiv 7 \pmod{40}$.
|
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|
Solve for the general solution of the equation $(2y^2+3xy-2y+6x)dx + x(x+2y-1)dy=0$ I know that this equation is neither exact nor homogeneous but after I put it in the form \[\frac{dy}{dx} + \frac{2y^2+3xy-2y+6x}{x(x+2y-1)}=0\]
I don't know how to proceed with it. Maybe some substitutions I overlooked?
Thanks!
|
Or you can calculate directly an integrating factor.
Integrating factor. If you write your equation as the line-field represented by the one-form $$(2y^2+3xy-2y+6x)dx + x(x+2y-1)dy=0$$ set $A(x,y) = 2y^2+3xy-2y+6x$ and $B(x,y) = x(x+2y-1)$ and look for a function $\mu(x,y)$ such that the form $\mu \, A \, dx + \mu \, B \, dy$ is exact, i.e. you need to find $\mu$ such that
$$\frac{\partial (\mu \, A)}{\partial y} - \frac{\partial (\mu \, B)}{\partial x} = 0$$ which expanded gives you
$$A \, \frac{\partial \mu}{\partial y} - B \, \frac{\partial \mu}{\partial x} + \left(\frac{\partial A}{\partial y} - \frac{\partial B}{\partial x}\right) \, \mu = 0$$ First calculate the divergence $$\frac{\partial A}{\partial y} - \frac{\partial B}{\partial x} = 4y+3x-2 - 2x-2y +1 = x + 2y - 1 = \frac{1}{x}\, B$$ Therefore, if you take $\mu = \mu(x)$ to depend only on $x$, then your equation for $\mu$ reduces to
$$- B \, \frac{\partial \mu}{\partial x} + \frac{1}{x}\, B \, \mu = 0$$ which allows you to cencel-out $B$ on both side yielding
$$- \frac{\partial \mu}{\partial x} + \frac{1}{x} \mu = 0$$ or if you prefer rewrite it as
$$- x\, \frac{d\mu}{d x} + \mu = 0$$ Well, one very simple non-tirival solution of the latter equation is $\mu(x) = x$ and this is your integrating factor.
A change of variables. Alternatively, because of the factor $(x+2y-1)$ in term $B$ you can consider the new variable $$w = \frac{1}{4} \, (x+2y - 1)^2$$ whose differential is
$$dw = \frac{1}{2} \, (x+2y - 1) \, d(x+2y) = \frac{1}{2}\, (x+2y - 1) \, dx + (x+2y - 1) \, dy $$ If we multiply both sides by $x$ we get
$$x\, dw = \frac{1}{2}\, x \, (x+2y - 1) \, dx + x\, (x+2y - 1) \, dy =
\frac{1}{2}\, (x^2+2xy - x) \, dx + B \, dy$$ i.e.
$$B \, dy = x \, dw - \frac{1}{2}\, (x^2+2xy - x) \, dx$$
so the equations becomes
$$0 = A\, dx + B \, dy = (2y^2+3xy-2y+6x)\, dx - \frac{1}{2}\, (x^2+2xy - x) \, dx + x \, dw$$
$$0 = \left(2y^2+3xy-2y+6x - \frac{x^2}{2} - xy + \frac{x}{2}\right) \, dx + x \, dw$$
$$0 = \left((2y^2+2xy-2y) +6x - \frac{x^2}{2} + \frac{x}{2}\right) \, dx + x \, dw$$
The change of variable $w = \frac{1}{4} \, (x+2y - 1)^2$ can be expanded as
$$4 \, w = x^2+4y^2 + 1 + 4xy - 2x - 4y = (4y^2 + 4xy - 4y )+ x^2 - 2x + 1$$ and thus
$$2y^2 + 2xy - 2y = 2 \, w - \frac{x^2}{2} - x + \frac{1}{2}$$
which I can plug back into the equation, replacing $2y^2 + 2xy - 2y$ and obtain the new equation
$$0 = \left(2 \, w - \frac{x^2}{2} - x + \frac{1}{2} +6x - \frac{x^2}{2} + \frac{x}{2}\right) \, dx + x \, dw$$
$$0 = \left(2 \, w - x^2 + \frac{11 \, x}{2} + \frac{1}{2}\right) \, dx + x \, dw$$ in other words, the equation looks like
$$\big(2 \, w + b(x) \big) \, dx + x \, dw = 0$$ which after representing $w = w(x)$ as a function of $x$ turns into the linear equation
$$x\, \frac{dw}{dx} + 2\, w + b(x) = 0$$
|
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|
Divide $1^2,2^2..,81^2$ numbers into 3 groups. Given $1^2,2^2,3^2,.....,81^2$ numbers. How can I divide them into $3$ groups with $27$ numbers in each so that they have the same sum.
Is there any algorithm to solve this task?
Thanks in Advance.
|
We take a sequence of 9 consecutive squares, $n^2, (n+1)^2,...,(n+8)^2$. Then,
$$(n+0)^2 + (n+4)^2 + (n+8)^2 = 3n^2 + 24n + 80$$
$$(n+1)^2 + (n+5)^2 + (n+6)^2 = 3n^2 + 24n + 62$$
$$(n+2)^2 + (n+3)^2 + (n+7)^2 = 3n^2 + 24n + 62$$
So, we can divide 27 squares into 3 groups of equal sum, by rotating the dominant group out of these 9. We can then divide total 81 into 3 groups of 27 each, which in turn can be divided into 3 groups of 9 with equal sum each. All that is left is to take one 9-group each from these 3 sets.
|
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}
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What is the remainder when $15^{40}$ divided by $1309$? I know that $$ 15 \equiv 1\pmod{7}, N \equiv 1\pmod{7},$$ but cannot proceed further.
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Note that $1309=7 \cdot 11 \cdot 17$. Let $N=15^{40}$.
Since $15 \equiv 1\pmod{7}$, $N \equiv 1\pmod{7}$.
Since $15^{10} \equiv 1\pmod{11}$, $N=\big(15^{10}\big)^4 \equiv 1\pmod{11}$.
Since $15 \equiv -2\pmod{17}$, $15^4 \equiv (-2)^4 \equiv -1\pmod{17}$ and $N=\big(15^4\big)^{10} \equiv (-1)^{10}=1\pmod{17}$.
Therefore $N-1$ is divisible by $7$, $11$, and $17$, and hence by $1309$.
The remainder is $1$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2151562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Limit as $x \to \infty$ of $\frac{x^5+x^3+4}{x^4-x^3+1}$ Suppose we have to find the following limit
$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}$
Now, if we work with the De L'Hopital rule with successive differentiations we get $L=+\infty$
But if we work like this instead: $$L=\lim_{x\to\infty}\frac{x^5(1+\frac{1}{x^2}+\frac{4}{x^5})}{x^5(\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^5})}$$ then $L$ does not exist.
What is correct and what is false here? I'm a little confused.
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First off, division by $0$ is undefined. Try this, $$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}=\lim_{x\to\infty}\frac{x+\frac{1}{x}+\frac{4}{x^4}}{1-\frac{1}{x}+\frac{1}{x^4}}=\infty$$
This shows more clearly the limit without resulting in $\frac{1}{0}$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2152654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
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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.