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For $2x+2y-3z\geq0$ and similar prove that $\sum\limits_{cyc}(7z-3x-3y)(x-y)^2\geq0$
Let $x$, $y$ and $z$ be non-negative numbers such that
$2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
$$(7z-3x-3y)(x-y)^2+(7y-3x-3z)(x-z)^2+(7x-3y-3z)(y-z)^2\geq0$$
I have a proof for the following weaker inequality.
Let $x$, $y$ and $z$ be non-negative numbers such that
$2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
$$(11z-4x-4y)(x-y)^2+(11y-4x-4z)(x-z)^2+(11x-4y-4z)(y-z)^2\geq0.$$
For the proof we can use the following lemma.
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$x+y+z\geq0$ and $xy+xz+yz\geq0$. Prove that:
$$(a-b)^2z+(a-c)^2y+(b-c)^2x\geq0.$$
Proof.
Since $x+y+z\geq0$, we see that $x+y$ or $x+z$ or $y+z$ is non-negative because if
$x+y<0$, $x+z<0$ and $y+z<0$ so $x+y+z<0$, which is contradiction.
Let $x+y\geq0$.
If $x+y=0$ so $xy+xz+yz=-x^2\geq0$, which gives $x=y=0$ and since $x+y+z\geq0$,
we obtain $z\geq0$, which gives $(a-b)^2z+(a-c)^2y+(b-c)^2x\geq0.$
Id est, we can assume $x+y\geq0$.
Now, $$(a-b)^2z+(a-c)^2y+(b-c)^2x=(a-b)^2z+(a-b+b-c)^2y+(b-c)^2x=$$
$$=(x+y)(b-c)^2+2(a-b)(b-c)y+(y+z)(a-b)^2$$
and since $x+y>0$, it's enough to prove that
$$y^2-(x+y)(y+z)\leq0,$$
which is
$xy+xz+yz\geq0,$
which ends a proof of the lemma.
Now we can prove a weaker problem.
From the condition we have:
$$\sum_{cyc}(2x+2y-3z)(2x+2z-3y)=\sum_{cyc}(9xy-8x^2)\geq0$$
and we see that $\sum\limits_{cyc}(11z-3x-3y)=5(x+y+z)\geq0$ and
$\sum\limits_{cyc}(11z-3x-3y)(11y-3x-3z)=9\sum\limits_{cyc}(9xy-8x^2)\geq0$
and by the lemma we are done!
This way does not help for the starting inequality.
Any hint please. Thank you!
|
We can use the substitutions $2x+2y - 3z = u, \ 2y+2z - 3x = v$ and $2z+2x - 3y = w$ for $u, v, w \ge 0$.
By solving the system of equations above, we get
$x = \frac{2u+v+2w}{5}, \ y = \frac{2u+2v+w}{5}$ and $z = \frac{u+2v+2w}{5}$.
The rest is the same as @yao4015's solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration of $\int_0^a \frac{\log\left|x^2 - 1\right|}{1 - x^2} dx$ I am trying to integrate something and after some changes of variable and an integration by parts I am stuck with this integral:
$$\int_0^a \frac{\log\left|x^2 - 1\right|}{1 - x^2} dx$$
I know that $a < 1$.
My question is twofold:
*
*Is it legal, as I know that $a < 1$ to say: Okay, $a < 1$ so $x^2 < 1$, so I can transform the integral into:
$$\int_0^a \frac{\log\left(1 - x^2\right)}{1 - x^2} dx$$
*How can I solve it?
Thanks in advance!
|
After some hours and a change of variable more I have been able to
'solve' the integral, although its result is in terms of the Polylogarithm,
as some of you have pointed out.
I was suggested in Twitter to try the expansion $log(1-x^2) = log(1-x) + log(1+x)$
and then the change $1-x = \exp(t)$ in the first case and $1+x = \exp(t)$ in
the second.
In the first case we have:
$$1-x = \exp(t); x = 1 - \exp(t); t = \log(1-x); dx = - \exp(t) dt$$
\begin{align}
\int_0^a \frac{\log(1-x)}{1-x^2} dx & = - \int_0^{\log(1-a)} \frac{t}{2 - \exp(t)} dt \\
& = -\frac{1}{4}\left[t\left(t - 2 \log\left(1 - \frac{\exp(t)}{2}\right)\right) - 2 Li_2\left(\frac{\exp(t)}{2}\right)\right]_0^{\log\left(1-a\right)}
\end{align}
In the second case we have:
$$1+x = \exp(t); x = \exp(t) - 1; t = \log(1+x); dx = \exp(t) dt$$
\begin{align}
\int_0^a \frac{\log(1+x)}{1-x^2} dx & = \int_0^{\log(1+a)} \frac{t}{2 - \exp(t)} dt \\
& = \frac{1}{4}\left[t\left(t - 2 \log\left(1 - \frac{\exp(t)}{2}\right)\right) - 2 Li_2\left(\frac{\exp(t)}{2}\right)\right]_0^{\log\left(1+a\right)}
\end{align}
Calling what's inside the brackets $f(y)$ we can see that our integral is:
\begin{align}
4I & = -\left[f(y)\right]_0^{\log(1-a)} - \left[f(y)\right]_{\log(1+a)}^0 \\
& = -\left[f(\log(1-a)) - f(0) + f(0) - f(\log(1+a))\right]\\
& = f(\log(1+a)) - f(\log(1-a))
\end{align}
Substituying back the logarithms inside $f$ we get:
\begin{align}
4I = & +\log(1+a)\left[\log(1+a) - 2 \log(1 - \frac{1+a}{2})\right] - 2 Li_2\left(\frac{1+a}{2}\right)\\
& -\log(1-a)\left[\log(1-a) - 2 \log(1 - \frac{1-a}{2})\right] + 2 Li_2\left(\frac{1-a}{2}\right)
\end{align}
We can now operate inside the brackets to get:
\begin{align}
\log(1+a) - 2 \log(1 - \frac{1+a}{2}) = & \log(1+a) - 2 \log(\frac{1-a}{2}) \\
= & \log(1+a) - \log(1-a) - \log(1-a) + 2 \log(2) \\
= & \log(\frac{1+a}{1-a}) - \log(1-a) + 2 \log(2) \\
= & 2 \tanh^{-1}(a) - \log(1-a) + 2 \log(2)
\end{align}
We can operate in the same way for $\log(1-a)$ term and get a similar
result but with the opposite sign for the $\tanh^{-1}$ and a $\log(1+a)$ term
instead of $\log(1-a)$.
Substituting back in the integral we get:
\begin{align}
4I = & + 2 \log(1+a) \tanh^{-1}(a) - \log(1+a) \log(1-a) + 2 \log(1+a) \log(2)\\
& + 2 \log(1-a) \tanh^{-1}(a) + \log(1-a) \log(1+a) - 2 \log(1-a) \log(2)\\
& + 2 \left[Li_2\left(\frac{1-a}{2}\right) - Li_2\left(\frac{1+a}{2}\right)\right] \\ \\
= & + 2 \tanh^{-1}(a) \left(\log(1+a) + \log(1-a)\right)\\
& + 2 \log(2) \left(\log(1+a) - \log(1-a)\right)\\
& + 2 \left[Li_2\left(\frac{1-a}{2}\right) - Li_2\left(\frac{1+a}{2}\right)\right] \\ \\
= & + 2 \tanh^{-1}(a) \log(1-a^2)\\
& + 4 \tanh^{-1}(a) \log(2)\\
& + 2 \left[Li_2\left(\frac{1-a}{2}\right) - Li_2\left(\frac{1+a}{2}\right)\right]
\end{align}
And that's mostly it... now you just calculate the Dilogarithm as you can and
have the answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
$
$det(B) = -det(C) -2det(D) = -(1-2) - 2(2-1) = -(-1)-2(1)=1-2=1, 1^4=1$
However, the correct answer is 16. I'm confused on where I made my wrong turn
|
Why did you do that? Just use determinant along the first row. This way is easier because there is a 0 in that row.
$$
\begin{vmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1 \\
\end{vmatrix}=1\begin{vmatrix}
1 & 2 \\
2 & 1 \\
\end{vmatrix}-0\begin{vmatrix}
1 & 2 \\
1 & 1 \\
\end{vmatrix}+1\begin{vmatrix}
1 & 1 \\
1 & 2 \\
\end{vmatrix}=(1*1-2*2)+(1*2-1*1)=\\
-3+1=-2
$$
and $(-2)^4=16$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating limits using taylor expansion $$\lim_{x\to 0^{+}} (\ln x)^3\left(\arctan\left(\ln\left(x+x^2\right)\right) + \frac{\pi}{2}\right) + (\ln x)^2$$
I have this limit in my sheet and the answer is $\frac 13$.
But I don't know how to approach any step using taylor expansion near zero (this is our lesson by the way). Please help me with the simplest way.
|
When $x\to 0^+$,
$$
\ln(x+x^2) = \ln x + \ln(1+x^2) = \ln x + o(x) \tag{1}
$$
and this goes to $-\infty$ as $x\to 0^+$. To continue, we would like to have $\arctan u$ with $u\to 0$, though. When $x<0$, we can use the fact that
$$
\arctan x + \arctan \frac{1}{x} = -\frac{\pi}{2} \tag{2}
$$
to rewrite (for $x$ small enough)
$$
\arctan \ln(x+x^2) = \arctan(\ln x + o(x))
= -\frac{\pi}{2} - \arctan\frac{1}{\ln x + o(x)}
$$
which also us to write, as $\frac{1}{\ln x + o(x)}\xrightarrow[x\to 0^+]{} 0^-$,
$$\begin{align}
\arctan \ln(x+x^2) + \frac{\pi}{2}
&= - \arctan\frac{1}{\ln x + o(x)}\\
&= - \arctan\frac{1}{\ln x}\left(\frac{1}{1 + o\left(\frac{x}{\ln x}\right)}\right)\\
&= - \arctan\left( \frac{1}{\ln x}+o\left(\frac{x}{\ln^2 x}\right)\right)\\
&= -\frac{1}{\ln x}+\frac{1}{3\ln^3 x}+o\left(\frac{1}{\ln^3 x}\right) .\tag{3}
\end{align}$$
From there, we get
$$\ln^3 x\left(\arctan \ln(x+x^2) + \frac{\pi}{2}\right) =
-\ln^2 x+\frac{1}{3}+o\left(1\right)
$$
and, finally,
$$\ln^3 x\left(\arctan \ln(x+x^2) + \frac{\pi}{2}\right) + \ln^2 x=
\frac{1}{3}+o\left(1\right) \xrightarrow[x\to 0^+]{} \boxed{\frac{1}{3}}.
$$
|
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|
Computing $\int \frac{7x}{(2x+1)} dx$ I'm trying to compute the following integral $$\int \frac{7x}{(2x+1)} dx$$
Unfortunately Wolfram Alpha gives me a different result, but other integral calculators say that my result is correct. So where is my error:
$$\int \frac{7x}{(2x+1)} dx = \frac{7}{2}\int \frac{2x}{(2x+1)} dx$$
$$\frac{7}{2}\int \frac{2x+1}{(2x+1)} - \frac{1}{(2x+1)} dx = \frac{7}{2}\int 1 dx - \int \frac{1}{(2x+1)} dx$$
Let $u=2x+1$
$$=\frac{7}{2}(x-\frac{1}{2}\int\frac{1}{u}du)=\frac{7x}{2}-\frac{7}{4}\ln(|2x+1|)+C$$
Wolfram Alpha tells me that it is $$\frac{7}{4} (2 x - \ln(2 x + 1) + \underbrace{1}_{different}) + C$$
Why is there an additional $1$?
|
If $y=F(x)$ be a primitive of the function $y=f(x)$ then for every real $k$ the function $y=F(x)+k$ is a primitive also.
The primitive of $\dfrac{7x}{2x+1}$ is $\dfrac{7}{4} (2x -\ln(2 x + 1))$, then for every real $k$ the function $y=\dfrac{7}{4} (2x -\ln(2 x + 1))+k$ is a primitive also.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Distance between ellipse and line What is the distance between the ellipse $$\frac{x^2}{9}+\frac{y^2}{16}=1$$ and the line $y=6-x$.
I think I need to be using Lagrange Multipliers but don't know how.
|
The minimum value it's the minimal distance between $x+y=6$ and $x+y=c$,
where $x+y=c$ is a tangent line to our ellipse
and since $\frac{xx_1}{9}+\frac{yy_1}{16}=1$ is a tangent line to the ellipse in $(x_1,y_1)$,
we obtain $\frac{x_1}{9}=\frac{1}{c}$ and $\frac{y_1}{16}=\frac{1}{c}$.
We can assume that $c\neq0$ because it's obvious that for $c=0$ we can non get a minimal value.
After using these substitutions to the equation of the ellipse we get for $(x_1,y_1)$:
$\left(\frac{9}{5},\frac{16}{5}\right)$ or $\left(-\frac{9}{5},-\frac{16}{5}\right)$.
The distance between $(x_1,y_1)$ and $x+y=6$ it's $\frac{|x_1+y_1-6|}{\sqrt2}$ and
we see that $\left(\frac{9}{5},\frac{16}{5}\right)$ gets a minimal distance and $\left(-\frac{9}{5},-\frac{16}{5}\right)$ gets a maximal distance.
Thus, the answer is $\frac{|\frac{9}{5}+\frac{16}{5}-6|}{\sqrt2}$ or $\frac{1}{\sqrt2}$
|
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|
Relations similar to $\sin(\pi/18) + \sin(5\pi/18) = \sin(7\pi/18)$ This I checked numerically (and can be proven analytically easily). I guess there are many similar relations between the numbers
$$\{ \sin\frac{m\pi}{q}, 1\leq m \leq q \} , $$
where $q$ is some integer.
Can anyone give some reference?
|
More generally, let $T_n(x) + 1$ have a factor $Q(x)$ (over the rationals) where $T_n$ is the $n$'th Chebyshev polynomial of the first kind. A root of $Q(x)$ will be $\cos(j \pi/n)$ for some integer $j$. Using trigonometric identities, we can then express $Q(\cos(\pi/n)) \sin(\pi/n)$ as a rational linear combination of sines of multiples of $j \pi/n$. In your case
$$ T_{18}(x)+1 = 2 x^2 (4 x^2-3)^2 (64 x^6 - 96 x^4 + 36 x^2 - 3)^2 $$
and $$(64 \cos^6(\theta) - 96 \cos^4(\theta) + 36 \cos^2(\theta) - 3) \sin(\theta) = \sin(7\theta) - \sin(5 \theta) - \sin(\theta)$$
leading to your identity with $\theta = \pi/18$.
Similarly, using $$T_{20}(x) + 1 = 2\, \left( 2\,{x}^{2}-1 \right) ^{2} \left( 256\,{x}^{8}-512\,{x}^{6}+
304\,{x}^{4}-48\,{x}^{2}+1 \right) ^{2}
$$
we get
$$ \sin \left( {\frac {9\,\pi}{20}} \right) -\sin \left( {\frac {7\,\pi}{
20}} \right) -\sin \left( \frac{\pi}4 \right) +\sin \left( {\frac {3\,\pi}{
20}} \right) +\sin \left( \frac{\pi}{20} \right) = 0
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that the spectrum of $K_n$ is $((n-1)^1, (-1)^{(n-1)})$ I am trying to prove that the spectrum of the complete graph $K_n$ is $((n-1)^1, (-1)^{(n-1)})$ (where superscripts denote multiplicities of eigenvalues, not exponents). I have part of the proof but having trouble completing it.
The adjacency matrix $A(K_n)$ is the $n \times n$ matrix:
$$ A(K_n) =\begin{pmatrix}
0 & 1 & 1 & \ldots & 1 & 1 \\
1 & 0 & 1 & \ldots& 1 & 1 \\
1 & 1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & 1 & 1 \\
1 & 1& \ldots & 1 & 0 & 1 \\
1 & 1 & \ldots & 1 & 1 & 0 \\
\end{pmatrix}
$$
The eigenvalues of A ($\lambda_1, \lambda_2, \ldots \lambda_n$) can be found by solving:
$$det(\lambda I - A) = 0$$
$$\begin{vmatrix}
\lambda & -1 & -1 & \ldots & -1 & -1 \\
-1 & \lambda & -1 & \ldots & -1 &- 1 \\
-1 & -1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & -1 & -1 \\
-1 & -1& \ldots & -1 & \lambda & -1 \\
-1 & -1 & \ldots & -1 & -1 & \lambda \\
\end{vmatrix} = 0
$$
I understand that one way to show $det(\lambda I - A) = 0$ for a given $\lambda$ is to show that the matrix $(\lambda I - A)$ is linearly dependent ie. one row is a linear combination of the others. From this it follows that $\lambda$ is an eigenvalue.
For $\lambda = (n-1)$ we have the matrix
$$\begin{pmatrix}
(n-1) & -1 & -1 & \ldots & -1 & -1 \\
-1 & (n-1) & -1 & \ldots & -1 &- 1 \\
-1 & -1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & -1 & -1 \\
-1 & -1& \ldots & -1 & (n-1) & -1 \\
-1 & -1 & \ldots & -1 & -1 & (n-1) \\
\end{pmatrix}
$$
It can be readily seen that any particular row is a linear combination of all of the other rows, specifically that $Row_i = \sum_{j \neq i} (-1)Row_j$, hence $\lambda = (n-1)$ is an eigenvalue.
Also for $\lambda = (-1)$ we have the matrix:
$$\begin{pmatrix}
-1 & -1 & -1 & \ldots & -1 & -1 \\
-1 & -1 & -1 & \ldots & -1 &- 1 \\
-1 & -1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & -1 & -1 \\
-1 & -1& \ldots & -1 & -1 & -1 \\
-1 & -1 & \ldots & -1 & -1 & -1 \\
\end{pmatrix}
$$
Clearly every row is identical hence each of the rows is a linear combination of any of the other rows, so the matrix is linearly dependent and $\lambda = (-1)$ is an eigenvalue also.
However I do not know how to show their multiplicities. (Can we appeal to the fact that the eigenvalue $(n-1)$ reduces the rank by 1 hence has multiplicity 1, while the eigenvalue $(-1)$ reduces the rank by $(n-1)$, hence has multiplicity $(n-1)$?)
Also I feel there may be a much simpler proof?
|
Hint: The $n - 1$ vectors
$$
\left\{
\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0 \end{bmatrix},
\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0 \end{bmatrix},
\ldots,
\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1 \end{bmatrix}
\right\}
$$
are all in the kernel of $\lambda I - A$ for $\lambda = -1$. Why are these vectors linearly independent? What does that say about the multiplicity of $-1$ as an eigenvalue?
|
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|
Limit without l'Hopital or Taylor series: $\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}$ find the limit without l'Hôpital and Taylor rule :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$
My Try :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}{x^3\sin x}=\\\lim\limits_{x \to 0}\frac{x\sin 2x- \sin^2 x}{2x^3\sin x}=$$?
what now ?
|
If you are allowed to use the well-known limit
$$\lim_{x \to 0}\frac{\sin x}{x}=1$$
then
$$\lim_{x \to 0}\frac{\tan x}{x}=1$$
follows easily and with a bit more effort (see here), you have:
$$\color{blue}{\lim_{x \to 0}\frac{\tan x-x}{x^3}=\frac{1}{3}\tag{1}}$$
Now for your limit and using $\color{blue}{(1)}$:
$$\lim_{x \to 0}\frac{x\cos x- \sin x}{x^3}=\lim_{x \to 0}\left(\cos x\frac{x- \tan x}{x^3}\right)=-\lim_{x \to 0} \cos x \color{blue}{\lim_{x \to 0}\frac{\tan x - x}{x^3}} = -\frac{1}{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $k$ such that $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}$ for $xyz=1$.
Find all $k\in\mathbb{R}^+$ such that for all $xyz=1$, $x,y,z\in\mathbb{R}^+$, we
have
$$\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}.$$
If we set $$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a},$$ where $a,b,c\in\mathbb{R}^+$, then the inequality is equivalent to $$\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\geq\frac{3}{2^k}.$$ For $k=1$ this is obviously false, as we take $b\to0$ and $c\to\infty$. In an olympiad test I saw that the case $k=\sqrt3$ is true (though I do not know how to prove that). I then thought that maybe this is true for all $k>1$. Am I correct and how do I prove it?
|
As shown below, an easy partial result is that $k_{\text{min}} \ge 1$.
Let $k$ be fixed with $0 < k < 1$.
WIth the substitutions
$$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a}$$
let $a = b = 1$, and let $c \to 0^{+}$. Then
\begin{align*}
\;&\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\\[6pt]
&\to \;\frac{1}{2^k} + 1 + 0\\[6pt]
&=\;\frac{1 + 2^k}{2^k}\\[6pt]
&<\;\frac{1 + 2}{2^k}\\[6pt]
&=\;\frac{3}{2^k}\\[6pt]
\end{align*}
It follows that $k_{\text{min}} \ge 1$, as claimed.
However, as shown below, the reasoning by which the OP (Yuxiao Xie) excluded the value $k=1$ is not correct.
Suppose $k = 1$. Then as$\;\,b \to 0^{+}$ and $\,c \to \infty$,
\begin{align*}
\;&\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\\[6pt]
&\to\;1+ 0 + 1\\[6pt]
&=\;2\\[6pt]
&>\;\frac{3}{2}\\[6pt]
\end{align*}
which doesn't break the required inequality, hence the value $k=1$ is not automatically excluded.
In fact, I suspect that $k_{\text{min}} = 1$.
Update: It's not true, as I had previously thought, that $k_{\text{min}} = 1$.
As a simple counterexample, let $(x,y,z) = \left(\frac{1}{8},2,4\right)$. Then
$$\frac{1}{(1+x)}+\frac{1}{(1+y)}+\frac{1}{(1+z)} = \frac{64}{45} < \frac{3}{2}$$
|
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|
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...
$x_2 = 2 + \frac{1}{2} = 2.5$
$x_3 = 2.5 + \frac{1}{2.5} = 2.9$
$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$
$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$
$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$
$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$
$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$
$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$
$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$
but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.
|
HINT: Show that $x_{n + \lceil 2 x_n \rceil} > 2 x_n$.
|
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|
Question about specific volume integral. So I need to evaluate the following volume integral, and I'm not sure which bounds to use and that makes the question kind of vague for me at least.
I've tried to solve it myself, but each try leaves my with a different solution, so I was wondering if someone could show me how to do the following question:
Determine the value of $\iiint_Wz^2dV$, where $W$ is the solide region around the cone $z^2=3x^2+3y^2$ and inside the sphere $x^2+y^2+z^2 = 6z$.
|
How do you find limits of integration?
Inside the sphere... this is asking for spherical coordinates.
$x = \rho \cos\theta\sin\phi\\
y = \rho \sin\theta\sin\phi\\
z = \rho\cos\phi$
now plug these into the equations given.
$z^2 = 3x^2 + 3y^2\\
\rho^2 \cos^2\phi = 3\rho^2 \cos^2\theta\sin^2\phi + 3\rho^2 \sin^2\theta\sin^2\phi\\
\rho^2\cos^2\phi = 3\rho^2\sin^2\phi\\
\tan^2\phi = 3\\
\phi = \arctan\frac 1{\sqrt 3}
$
$x^2 + y^2 +z^2 =6z \\
\rho^2 \cos^2\theta\sin^2\phi + \rho^2 \sin^2\theta\sin^2\phi + \rho^2\cos^2\phi = 6\rho\cos\phi\\
\rho^2 = 6\rho\cos\phi\\
\rho(\rho-6\cos\phi) = 0$
|
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|
Prove this inequality $\sqrt[3]{(a+b+c)(b+c+d)}\ge\sqrt[3]{ab}+\sqrt[3]{cd}$
Let $a,b,c,d>0$ show that
$$\sqrt[3]{(a+b+c)(b+c+d)}\ge\sqrt[3]{ab}+\sqrt[3]{cd}$$
Idea: Hence, we need to prove that
$$(a+b+c)(b+c+d)\ge ab+cd+3\sqrt[3]{(ab)^2cd}+3\sqrt[3]{(cd)^2(ab)}$$
$$\Longleftrightarrow ac+bc+bd+ad+b^2+c^2\ge 3\sqrt[3]{(ab)^2cd}+3\sqrt[3]{(cd)^2(ab)}$$
I attempted a following proof by AM-GM inequality,But I am not able to solve the upper part, this inequality
|
It's just Holder:
$$(a+b+c)(b+c+d)\left(\frac{b}{b+c}+\frac{c}{b+c}\right)\geq$$
$$\geq\left(\sqrt[3]{a\cdot(b+c)\cdot\frac{b}{b+c}}+\sqrt[3]{(b+c)\cdot d\cdot\frac{c}{b+c}}\right)^3=\left(\sqrt[3]{ab}+\sqrt[3]{cd}\right)^3$$
|
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|
Find if $\sum\limits^{\infty}_{n=2} \frac{3}{n^2+3n}$ converges or not. If converges, find its sum. Find if $$\sum^{\infty}_{n=2} \dfrac{3}{n^2+3n}$$
converges or not. If converges, find the sum.
I used the integral test to determine whether the series is convergent or not.
I found $$3\int^{\infty}_{2} \dfrac{1}{x^2+3x}\text{ dx} = \ln(5) - \ln(2)$$
and thus I can conclude that the series converges. But my question asks if it converges, find the sum. How? I evaluated the series at some points, and here is what I get:
$$\dfrac{3}{10} + \dfrac{3}{18} + \dfrac{3}{28}+\dfrac{3}{40} \dots$$
I don't see a pattern or anything, so how can I find the sum?
|
Observe
\begin{align}
\frac{3}{n(n+3)} = \frac{1}{n}-\frac{1}{n+3}
\end{align}
which means
\begin{align}
\sum^\infty_{n=2}\left(\frac{1}{n}-\frac{1}{n+3}\right) =&\ \left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\ldots\\
=&\ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}.
\end{align}
|
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|
Using the substitution method for a simple integral I have been playing with the substitution rule in order to test some ideas with computational graphs. One of the things I'm doing is applying the substitution to well known, and easy, integrals. For example, let's use that method to find the indefinite integral for
$$f(x) = x^2$$
Using the rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, we get
$$F(x) = \frac{x^3}{3} + C$$
So, let's do the following substitution:
$$u = x^2$$
$$\frac{du}{dx} = 2x \Leftrightarrow dx = \frac{1}{2x} du$$
So, performing the substitution in the integral of $f(x)$ gives us
$$\int x^2 dx = \int u \frac{1}{2x} du = \frac{1}{2x} \frac{u^2}{2} + C = \frac{(x^2)^2}{4x} + C = \frac{x^3}{4} + C$$
Have I done anything wrong with the substitutions??
Thanks in advance!
|
You have one mistake. If you are changing $dx$ into $du$, then you need to convert all terms containing $x$ into $u$.
So we have,
$$\int x^2 dx = \int u \frac{1}{2\sqrt u} du$$
$$= \frac 12 \int \frac{1}{\sqrt{u}}\cdot u du = \frac 12 \int \sqrt u du$$
$$= \frac{1}{2} \cdot u^{\frac{3}{2}}\cdot\frac{2}{3}+ C$$
$$= \frac{u^{\frac 3 2}}{3} + C = \frac{x^3}{3} + C$$
|
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|
proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
& = 5
\end{align}
Assumption: $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n = k \in \mathbb{N}$.
$5 \mid (3^{2n-1} + 2^{2n-1}) \implies 3^{2n-1} + 2^{2n-1} = 5m$, $m \in \mathbb{Z}$
Proof: Let $n = k + 1$
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 3^{2k+1} + 2^{2k+1}\\
& = 3^{2k} \cdot 3^1 + 2^{2k} \cdot 2^1\\
& = 3^{2k} \cdot 3 + 2^{2k} \cdot 2
\end{align}
And here I got stuck. I don't know how to get from the last line to the Assumption. Either I am overlooking a remodeling rule or I have used a wrong approach.
Anyway, I am stuck and would be thankful for any help.
|
let $$T_n=3^{2n-1}+2^{2n-1}$$ and $$5|T_n$$ we have to Show that $$5|T_{n+1}$$ indeed we have $$T_{n+1}-T_n=3^{2n+1}+2^{2n+1}-3^{2n-1}-2^{2n-1}=3^{2n-1}\cdot 8+2^{2n-1}\cdot 3=3^{2n-1}(10-2)+2^{2n-1}(5-2)=5(2\cdot3^{2n-1}+2^{2n-1})-2(3^{2n-1}+2^{2n-1})$$
|
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|
Number of ways to put balls into bins How many ways can I put $12$ balls into $11$ bins, such that no bin contains more than $2$ balls?
The problem I am having is assigning balls to bins.
|
Here is my answer, going with the assumption that having identical bins means that having two in the first and one in each other is the same as having two in the last and one in each other.
Essentially, we are figuring out different ways to add up to $12$ using eleven numbers, were the numbers being added are various combinations of $0$'s, $1$'s, and $2$'s.
We have:
$$2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 $$
$$2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 $$
$$2 + 2+ 2 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 0 $$
$$2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 0 + 0 + 0 $$
$$2 + 2 + 2 + 2 + 2 + 1 + 1 + 0 + 0 + 0 + 0 $$
$$2 + 2 + 2 + 2 + 2 + 2 + 0 + 0 + 0 + 0 + 0 $$
So the total number of ways to arrange the balls in bins is $6$ ways, which was stated in the comments by Christian Blatter.
If the balls/bins are distinguishable, I would suggest looking at the answer supplied by Jack D'Aurizio
|
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|
Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$ Question: Compute this integral
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
My Approach:
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
$$=\int\frac{x+2}{\sqrt{x^2+5x}+6}\times \frac{{\sqrt{x^2+5x}-6}}{{\sqrt{x^2+5x}-6}}~dx$$
$$\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \underbrace {~\int\frac{(6x+12)}{x^2+5x-36}~dx~}_{\text{This one I know how to deal with} }$$
$$\text{Now:} ~\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5-1)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \frac{1}{2}\int\frac{(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$\Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)$$
$$ \underbrace{\frac{1}{2}\int \frac{\sqrt{t}}{t-36}~dt}_{\text{I can deal with this}} ~~- \frac{1}{2}\int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
Now I'm stuck. I am unable to calculate: $$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
P.S.: I am high school student so please try to use elementary integrals only; i.e. integration by parts and substitution. I don't know how to use complex numbers in integration, multiple integrals, error function, etc. (I don't know if it can be used here or not, just clarifying.)
As answered by @Kanwaljit Singh: Finally I have to compute:
$$\int \frac{1}{\sqrt{x^2+5x}-6}$$
But if I was able to compute it, I would have done it in the very first step, id est ;
$$\int \frac{x+2}{\sqrt{x^2+5x}+6}~dx =
\frac{1}{2}\int \frac{2x+5-1}{\sqrt{x^2+5x}+6}~dx
\\ \frac{1}{2}\int \frac{2x+5}{\sqrt{x^2+5x}+6}~dx ~- \frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)
\\ \underbrace{\frac{1}{2}\int \frac{1}{t+6}~dt}_{\text{Doable}} ~-~\frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \int \frac{1}{\sqrt{x^2+5x}+6}~dx $$
Reached to a similar step by a short path.
But how do I compute this one?
A screenshot of this question:
|
$$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
$$=\int \frac{\sqrt{x^2+5x}+6-6}{(\sqrt{x^2+5x})^2-6^2}~dx$$
$=\int \frac{\sqrt{x^2+5x}+6}{(\sqrt{x^2+5x}+6)(\sqrt{x^2+5x}-6)} - \int \frac{6}{x^2+5x-36} ~dx$
$$=\int \frac{1}{\sqrt{x^2+5x}-6} - \int \frac{6}{x^2+5x-36} ~dx$$
Hope you can proceed further.
|
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|
Find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ I wish to find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ as part of a larger problem. I can see that the solution will involve De Moivre's theorem somehow, but I cannot see how to apply it. I have looked at solutions of $z^7 - 1 = 0$ but to no avail. Can anyone suggest a method for solving this problem?
|
Note that $\tan^2 \frac\pi7$, $\tan^2 \frac{2\pi}7$, and $\tan^2 \frac{3\pi}{7}$ are the roots of the polynomial equation $$x^3 - 21x^2 + 35x - 7 = 0.$$
If we label these roots $r$, $s$, and $t$, then Vieta's formulae tell us that $$\begin{cases}r + s + t = 21,\\ rs + rt + st = 35,\\ rst = 7.\end{cases}$$
From these, we would like to calculate $r^3 + s^3 + t^3$. We can do this by taking $$(r + s + t)^3 - 3 (r+s+t)(rs + rt + st) + 3rst = 21^3 - 3\cdot 21 \cdot 35 + 3\cdot 7 = 7077.$$
We can get the polynomial equation that made this solution work by observing that if $\theta$ is a multiple of $\frac\pi7$, then $(\cos\theta + i\sin\theta)^7 = 1$. Expanding this polynomial and taking the imaginary part yields
$$7\cos^6\theta \sin \theta - 35 \cos^4 \theta \sin^3\theta + 21 \cos^2 \theta \sin^5\theta - \sin^7 \theta = 0$$
or
$$\tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7\tan\theta = 0.$$
If we set $x = \tan\theta$, then the solutions to this are $0$, $\pm \tan \frac\pi7$, $\pm \tan \frac{2\pi}7$, and $\pm \tan \frac{3\pi}7$. So we divide by $x$ and cut all exponents in half to get only the roots we're interested in.
|
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|
Number Theory proof involving Legendre symbol Suppose $a,b$ are integers s.t. $ab\equiv 1\pmod{p}$. Show $(\frac{a^2+a}{p})=(\frac{b+1}{p})$.
Not really sure on how to approach this proof.
If I take $a^2+a\equiv 0\pmod{p}$
$a\equiv -a^2\pmod{p}$
$b^2a\equiv -b^2a^2\pmod{p}$
$b\equiv -1\pmod{p}$ so $b+1\equiv 0\pmod{p}$
Not sure how to use this or if I'm able to use this to prove the goal.
OR I can say since $a^2+a\in Q_p$
$s^2\equiv a^2+a\pmod{p}$ for some $s\in U_p$ then multiply by $b^2$
|
Since $(b^2|p) = (b|p)^2 = 1$,
$$\left(\frac{a^2 + a}{p}\right) = \left(\frac{a^2 + a}{p}\right)\left(\frac{b^2}{p}\right)= \left(\frac{a^2b^2 + ab^2}{p}\right) =\cdots$$
Complete the argument using the condition $ab \equiv 1\pmod{p}$.
|
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|
Determine the acceleration given the position function of a particle
The position function of a particle in a test laboratory is $s(t) = \frac{10t}
{t^2+3}$.
Determine the acceleration of the particle of particle after 4 seconds.
The speed $v(t)$ of the particle is the derivative of the position $\frac{ds(t)}{dt}$.
I got $v(t) = \frac{10(-t^2+3)}{(t^2+3)^2}$
For the acceleration $a(t)$ I tried taking the derivative of $v(t)$ but got the wrong answer at $a(4)$.
$$a(t)=\frac{-2t\cdot (t^3+3)^2-10(-t^2+3)\cdot 2(t^2+3)\cdot(2t)}{(t^2+3)^4}$$
|
Let $S(t)$ be given by
$$s(t)=\frac{10t}{t^2+3}\tag1$$
Differentiating $(1)$ yields
$$\begin{align}
v(t)&=s'(t)\\\\
&=-10\frac{t^2-3}{(t^2+3)^2}\\\\
&=-10\frac{t^2+3-6}{(t^2+3)^2}\\\\
&=-10\frac{1}{t^2+3}+60\frac{1}{(t^2+3)^2}\tag 2
\end{align}$$
Differentiating $(2)$ yields
$$\begin{align}
a(t)&=s''(t)\\\\
&=20\frac{t}{(t^2+3)^2}-240\frac{t}{(t^2+3)^3}\\\\
&=\frac{20t(t^2-9)}{(t^2+3)^3}\tag 3
\end{align}$$
Evaluting $(3)$ at $t=4$, we find that
$$a(4)=\frac{560}{6859}$$
|
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|
If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How to prove further?
|
In a triangle,
$b^2=c^2+a^2−2ca \cos B$ .....(1)
From question,
$(a+2b+c)(a+b+c) = 3(a+b)(b+c)$
$a^2+ab+ac+2ab+2b^2+2bc+ca+cb+c^2=3ab+3ac+3b^2+3bc$
$a^2+c^2-ca=3b^2−2b^2$
$a^2+b^2-ca=b^2$ .....(2)
Comparing equation (1) and (2),
$a^2+b^2−2ca\cos B=a^2+b^2–ca$
$\cos B = \frac 12$
$B = 60°$
|
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|
Uniform convergence of $\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$
Where does the seriers $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$ converges ? On which interval does it converge uniformly ?
We have $$\lim_n \frac{x^n}{1+x^n}=\begin{cases}0 &\text{ , if } |x|<1\\\frac{1}{2} &\text{ , if } x=1\\\text{does not exists} &\text{ , if } x=-1\\1 &\text{ , if } |x|>1\end{cases}$$
So we have to test the convergence only on the interval $(-1,1)$.
Now let , $\displaystyle S_n=\sum_{k=1}^n \frac{x^k}{1+x^k}$. How I find the point wise sum ?
Edit: I'm confused for different answers and different comments..!!
|
Observe for all $|x| \leq R<1$ you have that
\begin{align}
\left|\sum^m_{n=k} \frac{x^n}{1+x^n}\right| \leq \sum^m_{n=k} \frac{|x|^n}{1-R^n} \leq \frac{1}{1-R}\sum^m_{n=k}R^n = \frac{R^k-R^{m+k+1}}{(1-R)^2}
\end{align}
which means the series is uniformly Cauchy on $[-R, R]$. Hence the series is uniformly convergent on any compact subset of $(-1, 1)$.
Edit: Suppose the series is uniformly convergent on all of $(-1, 1)$. Fix $\epsilon = \frac{1}{2}$, then there exists $N$ such that
\begin{align}
\left|\sum^\infty_{n=N}\frac{x^n}{1+x^n}\right| <\frac{1}{2}
\end{align}
for all $x \in (-1, 1)$. Let $x_\ell = \frac{\ell-1}{\ell}$, then observe that
\begin{align}
\sum^\infty_{n=N} \frac{x_\ell^n}{1+x_\ell^n} = \sum^\infty_{n=N} \frac{(\ell-1)^n}{\ell^n+(\ell-1)^n} \geq \frac{1}{2}\sum^\infty_{n=N}\left(\frac{\ell-1}{\ell}\right)^n = \frac{1}{2}\frac{\left(\frac{\ell-1}{\ell}\right)^N}{1-\frac{\ell-1}{\ell}} = \frac{\ell}{2}\left(1-\frac{1}{\ell} \right)^N \rightarrow \infty
\end{align}
as $\ell\rightarrow \infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2201652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Isometry and its inverse I got this affine map:
$$
f: R^3 \rightarrow R^3: \begin{pmatrix}x\\
y\\
z\\\end{pmatrix} \rightarrow A \cdot \begin{pmatrix}x\\
y\\
z\\\end{pmatrix} + \begin{pmatrix}0\\
-1\\
1\\\end{pmatrix}
$$
with $$ A = \begin{bmatrix}
1 & a_{12} & a_{22}\\
0 & 1 & a_{21}\\
0 & 0 & 1
\end{bmatrix}$$
Also given was this information about the inverse (which should also be an affine map?):
$$
g=f^{-1}: R^3 \rightarrow R^3: \begin{pmatrix}x\\
y\\
z\\\end{pmatrix} \rightarrow B \cdot \begin{pmatrix}x\\
y\\
z\\\end{pmatrix} + \bar b
$$
I have to find $\bar b$. Does anyone have an idea how to find it? I tried inputting some values but I can't seem to get there.
|
$\overline{b} = g(0) = f^{-1}(0)$, so solve $f(x) = 0$ so $Ax = (0,1,-1)$. The answer (which depends on the 3 constants in $A$) is the required $\overline{b}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2203667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$$
I guess one could say that:
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{n}{n+1}
$$
However, I'm not sure this is rigorous enough (for example, in $\sum_{i=1}^n\frac{1}{n+i}$ how do we really know that the index goes from $1$ to $n$) and I think this needs to be proven via induction.
So the base case is:
$$a_1 = \frac{1}{2} \le \frac{1}{2} = \frac{n}{n+1}$$
The step: suppose $a_n \le \frac{n}{n+1}$ then let's prove that $$a_{n+1} \le \frac{n+1}{n+2}$$
The above can be developed as:
$$
\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)} \le \frac{n}{n+1}+\frac{1}{n+2}
$$
This is where I get stuck. If I could somehow prove that the number of terms to the left $\le$ the terms to the left I would be golden. Or maybe there's another way.
|
$$\frac{1}{n+1} \leq \frac{1}{n+1}$$
$$\frac{1}{n+2}< \frac{1}{n+1}$$
$$\cdots$$
$$\frac{1}{2n} < \frac{1}{n+1}$$
Now add up!
Equality comes when $n=1$.
|
{
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"url": "https://math.stackexchange.com/questions/2204141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $y=\sin{(m\cos^{-1}\sqrt{x})}$ then prove that $\lim\limits_{x\to 0}\frac{y_{n+1}}{y_n}=\frac{4n^2-m^2}{4n+2}$ If $y=\sin{(m\cos^{-1}\sqrt{x})}$ then prove that $$\lim\limits_{x\to 0}\frac{y_{n+1}}{y_n}=\frac{4n^2-m^2}{4n+2}$$
Note: $y_n = D^ny(x)$. To use: Libnitz rule of successive Differentiation.
Attempt:
$y_1=-1/2\,{\dfrac {m\cos \left( m\arccos \left( \sqrt {x} \right) \right)
}{\sqrt {x}\sqrt {1-x}}}$ then
$4y_1^2x(1-x)=m^2(1-y^2)$
differentiating n times
$$8y_1y_{n+1}x(1-x)+n4(1-2x)2y_1y_n+\frac{n(n-1)}{2}(-8)2y_1y_{n-1}=-2m^22yy_n$$
Problem is here $y_{n+1}, y_n, y_{n-1}$ are all present. What to do to prove the result.
|
We have $y= \sin(m \cos ^{-1} \sqrt{x})$ & using the notation $y_n=\frac{d^n y}{dx^n}$. You start fine ... differentiating (multiple use of the chain rule) squaring & using Pythagorus to get to
\begin{eqnarray*}
4x(1-x)y_1^{2}=m^2(1-y^2)
\end{eqnarray*}
Now differentiate this
\begin{eqnarray*}
8x(1-x)y_1 y_2 +(4-8x)y_1^{2}+2 m^2y_1 y=0
\end{eqnarray*}
Now cancel $2y_1$ and keep differentiating
\begin{eqnarray*}
4x(1-x) y_2 +2(1-2x)y_1+ m^2 y=0 \\
4x(1-x) y_3 +6(1-2x)y_2+ (m^2-4) y_1 =0 \\
4x(1-x) y_4 +10(1-2x)y_3+ (m^2-16) y_2=0 \\
\end{eqnarray*}
It is easy to show by induction that
\begin{eqnarray*}
4x(1-x) y_{n+2} +2(2n+1)(1-2x)y_{n+1}+ (m^2-4n^2) y_n=0. \\
\end{eqnarray*}
Now substiute $x=0$ and the required result follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2205411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\displaystyle\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$ using squeeze theorem
Find $$\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$$ using squeeze theorem
Tried using binomial expansion, but have no idea on how to continue.
|
$$\begin{align*}&x-\sqrt{x+1}\sqrt{x+2}=\frac{x^2-(x+1)(x+2)}{x+\sqrt{x+1}\sqrt{x+2}}=\frac{-3x+2}{x+\sqrt{x^2+3x+2}}=\\{}\\
&\frac{-3+\frac2x}{1+\sqrt{1+\frac3x+\frac2{x^2}}}\xrightarrow[x\to\infty]{}\frac{-3+0}{1+\sqrt{1+0+0}}=-\frac32\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2206605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Evaluation of the series $\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}$. Find the following sum
$$\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}$$
Could someone give some hint to proceed in this question?
|
Consider that $$ \frac{(n+2)^2}{2^n} - \frac{(n+3)^2}{2^{n+1}} = \frac{n^2+2n-1}{2^n} = \frac{n^2+2n+3}{2^n} - \frac{4}{2^n}.$$
So, if the required sum is denoted by $S$, we have
$$ \sum_{n=1}^{\infty} \frac{(n+2)^2}{2^n} - \frac{(n+3)^2}{2^{n+1}} = S - 4 \sum_{n=1}^{\infty} \frac{1}{2^n}.$$
The left hand side is a telescoping sum, and on the right hand side we have an infinite geometric series summation.
So we easily get $\dfrac{9}{2} = S - 4$ so that $S = \dfrac{17}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2209638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\beta=\frac{c}{a}$$
Therefore the roots of $9x^2-2x+7=0$ are $(\alpha+2)$ and $(\beta+2)$. So, $$\alpha+\beta+4=\frac{2}{9}\implies4-\frac{b}{a}=\frac{2}{9}\implies\frac{b}{a}=\frac{34}{9}$$ and
\begin{align*}
(\alpha+2)(\beta+2)=\frac{7}{9}\\
\Rightarrow\alpha\beta+2(\alpha+\beta)+4=\frac{7}{9}\\
\Rightarrow\frac{c}{a}-2\frac{b}{a}+4=\frac{7}{9}\\
\Rightarrow\frac{4a-2b+c}{a}=\frac{7}{9}\\
\end{align*}
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.
There is another similar question: If the roots of $px^2+qx+r=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then what will be the expression of $r$ in terms of $a$, $b$, and $c$?
|
Since $ax^2+bx+c=0$ has the same roots as $x^2+\frac{b}{a}x+\frac{c}{a}=0$ we can notice that if $(a_1,b_1,c_1)$ satisfies the requirements then also does $(ta_1,tb_1,tb_2)$ since $4a-2b+c\neq 0$ we can get $4a-2b+c=\frac{7t}{9}$ for any $t$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2209958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt,
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
|
Move everything to the left hand side and substitute $u=-3\tan x$ to get
\begin{align*}0&=\frac{2}{\sin x\cos x}-1-3\tan x\\ &=2\left(\tan x+\frac 1{\tan x}\right)-1-3\tan x\\ &=-1-\frac{6}{-3\tan x}-\frac{3\tan x}{3}\\ &=\frac u3 -1-\frac 6u \\ &= \frac{(u-6)(u+3)}{3u}\end{align*}
Thus we get that $u=-3$ or $u=6$. The first gives $x=\pi n+\frac \pi 4$ for all $n$, and the second gives $x=\pi n - \arctan 2$ for all $n$.
The solutions that satisfy your conditions are for $n=0$ in the first case and for $n=1$ in the second case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2211823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Calculate the determinant of $A-2A^{-1}$ given the characteristic and minimal polynomials of $A$ Given the characteristic polynomial and minimal polynomial of $A$ being $(x+2)^{6}(x-1)^{3}$ and $(x+2)^{2}(x-1)^{3}$, respectively. How do I determine the characteristic polynomial and minimal polynomial of $A-2A^{-1}$.
I tried to algebraically rearrange this in order to have two separate determinant calculations but haven't had success so far.
Also I have the Jordan Canonical Form of $A$ and must determine that of the $A-2A^{-1}$ matrix.
|
$A$ is similar to a triangular matrix with a diagonal of $-2$ repeated $6$ times and $1$ repeated $3$ times. Therefore, $\det A= (-2)^6 \cdot 1^3 = 2^6$. For simplicity, let's assume $A$ is triangular.
We have $\det (A-2A^{-1}I) \det A = \det (A^2-2I)$.
The diagonal of $A^2$ is $(-2)^2$ repeated $6$ times and $1^2$ repeated $3$ times.
Therefore, the diagonal of $A^2-2I$ is $(-2)^2-2$ repeated $6$ times and $1-2$ repeated $3$ times and so $\det (A^2-2I)=-2^6$.
Therefore, $\det (A-2A^{-1}I)=-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2212508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $s(N(10^{n}-1)) = s(10^{n}-1)$
Let $N$ be a positive integer, and let $n$ be the number of digits in $N$ in decimal representation. Also let $s(n)$ denote the sum of the digits of $n$. Prove that $s(N(10^{n}-1)) = s(10^{n}-1)$.
For example, we have $43 \cdot 99 = 4257$ and $s(4257) = 18 = s(99)$. How do we prove this result in general, and does it hold for all $m \geq n$?
|
Here I am assuming that the last digit $a_0 \neq 0$, if that's not the case then go for the last non-zero digit in your number $N$ and modify the proof accordingly.
Let
$$N=a_{k-1}10^{k-1}+\dotsb +a_0, \qquad \text{ where } a_i \in \{0,1,\ldots ,9\} \text{ and } a_0 \neq 0$$
Then
\begin{align*}
N(10^k-1) & =[a_{k-1}10^{k-1}+\dotsb +a_0](10^k-1)\\
& =a_{k-1}10^{2k-1}+\dotsb +a_010^{k}-(a_{k-1}10^{k-1}+\dotsb +a_0)\\
& =a_{k-1}10^{2k-1}+\dotsb +(a_0-1)10^{k}+((9-a_{k-1})10^{k-1}+(9-a_{k-2})10^{k-2}+\dotsb +(10-a_0)).\\
\end{align*}
Thus
\begin{align*}
s(N(10^k-1)) &= \overbrace{9+9+\dotsb+9}^{k}\\
&=s(10^k-1).
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2213480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Can the definite integral $ I(n) := \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2}dx$ be computed? For $n \in \mathbb N$, $n \geq 1$, consider the following integral expression:
\begin{equation} I(n) := \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2} dx \end{equation}
My attempts to use any of the obvious (elementary) few methods that I am accustomed with (i.e partial integration, substitution, series expansion of the integrand) in order to find a general antiderivative were all futile. There might still be a way to compute an antiderivative using complex analysis, but this is currently beyond my capabilites. However, although Wolfram Alpha does also not seem to be able to either to compute a general antiderivate or compute the above expression for general $n \in \mathbb N$, it will yield (after possibly some refreshing) an explicit value for any $n$ I've tested so far (all $n$ between $1$ and $15$). Namely, one gets the following result for $1 \leq n \leq 15:$
\begin{equation} I(n)= (n-\frac{1}{2}) \ln(2n-1) - (n-1) \ln(2n). \end{equation}
This suggest that there is indeed a way to compute the above expression explicitly. Any help is highly appreciated.
|
Hint. By making the change of variable,
$$
x=2\sqrt{2n-1}\:u ,\quad dx=2\sqrt{2n-1}\:du,\quad u=\frac{x}{2\sqrt{2n-1}},
$$ one has
$$
\frac{\pi}{2n}\cdot I(n)=\ln(2\sqrt{2n-1})\int_0^1\frac{\sqrt{1-u^2}}{a^2-u^2}\:du+\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du \tag1
$$ with
$$
a:=\frac{n}{\sqrt{2n-1}}>1, \quad (n>1).\tag2
$$ The first integral on the right hand side of $(1)$ may be evaluated by two changes of variable, $u=\sin \theta$ and $t=\tan\theta$, obtaining, for $a>1$,
$$
\int_0^1\frac{\sqrt{1-u^2}}{a^2-u^2}\:du=\frac12\int_0^\infty\frac{1}{\left((a^2-1)t^2+a^2\right)\cdot\left(t^2+1\right)}\:dt=\frac{\pi \left(a-\sqrt{a^2-1}\right)}{2 a}.\tag3
$$
Similarly, the second integral on the right hand side of $(1)$ is such that
$$
\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du= \frac{1}{2}\int_0^\infty\frac{2\ln t-\ln(1+t^2)}{\left((a^2-1)t^2+a^2\right)\cdot\left(t^2+1\right)}\:dt,\quad (a>1),\tag4
$$ transforming the integrand by a partial fraction decomposition, using the standard results
$$
\begin{align}
\int_0^\infty\frac{\ln t}{t^2+\alpha^2}\:dt&=\frac{\pi}{2}\cdot\frac{\ln\alpha}{\alpha},\quad \alpha>0, \qquad (\star)
\\\int_0^\infty\frac{\ln (1+t^2)}{t^2+\alpha^2}\:dt&=\pi\cdot\frac{\ln(\alpha+1)}{\alpha},\quad \alpha>0,\qquad (\star \star)
\end{align}
$$ one gets
$$
\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du=\frac{\pi\sqrt{a^2-1}}{2 a} \cdot\ln\left(\frac{\sqrt{a^2-1}+a}{a}\right)-\frac{\pi}2 \ln 2,\quad (a>1).\tag5
$$
Combining $(3)$ and $(5)$ with $(2)$ gives
$$
I(n) := \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2} dx= \left(n-\frac{1}{2}\right) \ln(2n-1) - (n-1) \ln(2n)
$$
for all real numbers $n$ such that $n>1$.
Edit. To prove $(\star)$, one may perform the change of variable $t=\alpha u$, $\alpha>0$, getting
$$
\int_0^\infty\frac{\ln t}{t^2+\alpha^2}\:dt=\frac1\alpha \cdot\int_0^\infty\frac{\ln \alpha+ \ln u}{u^2+1}\:du=\frac{\pi}{2}\cdot\frac{\ln\alpha}{\alpha}
$$ since
$$
\int_0^\infty\frac{1}{u^2+1}\:du=\left[\frac{}{}\arctan u \frac{}{}\right]_0^\infty=\frac \pi2, \quad \int_0^\infty\frac{\ln u}{u^2+1}\:du=0
$$ (the latter is seen by making $u \to 1/u$).
To prove $(\star\star)$, one may perform the change of variable $t=\alpha u$, $\alpha>0$, getting
$$
\int_0^\infty\frac{\ln (1+t^2)}{t^2+\alpha^2}\:dt=\frac1\alpha \cdot\int_0^\infty\frac{\ln (1+\alpha^2u^2)}{u^2+1}\:du
$$ differentiating the latter integral with respect to $\alpha$ gives a classic evaluation
$$
\frac{d}{d\alpha}\int_0^\infty\frac{\ln (1+\alpha^2u^2)}{u^2+1}\:du=\int_0^\infty\frac{2\alpha u^2}{(u^2+1)(1+\alpha^2u^2)}\:du=\frac{\pi}{\alpha+1}
$$ then integrating one gets $(\star\star)$.
|
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"url": "https://math.stackexchange.com/questions/2215805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Finding Trigonometric Fourier Series of a piecewise function Find the Fourier Trigonometric series for:
$$f(x)=
\begin{cases}
\sin(x) & 0\leq x \leq \pi
\\ 0 & \pi\leq x \leq 2\pi,
\\
\end{cases}\quad
f(x+2\pi)=f(x).$$
I tried to find the series of this function, but when I plot up to 50 terms with Wolfram, it doesn't resemble the function so I guess I made a mistake finding the Fourier series.
This is what I did:
The length of the interval is $\boxed{L= 2\pi}$.
I calculated the coefficients as follows
$$
\begin{align*}
a_0&=\displaystyle \dfrac 1 L \int_{0}^ {2\pi} f(x) \, dx=\dfrac 1 L \left(\int_{0}^ {\pi} \sin(x) \, dx +\int_{\pi}^ {2\pi} 0 \, dx\right)\\
\\
a_n&= \dfrac 2 L \int_{0}^ {2\pi} f(x) \cos\left(\dfrac {2n\pi x} {L}\right) \, dx\\
&=\dfrac 2 L \left(\int_{0}^ {\pi} \sin(x) \cos\left(\dfrac {2n\pi x} {L}\right) \, dx +\int_{\pi}^ {2\pi} 0 \, dx\right)\\
&=\dfrac 1 \pi \cdot \dfrac {\cos(n \pi)+1} {(1-n^2)}\\
\\
b_n&= \frac 2 L \int_{0}^ {2\pi} f(x) \sin\left(\frac {2n\pi x} {L}\right) \, dx\\
&=\frac{2}{L} \left(\int_{0}^ {\pi} \sin(x) \sin\left(\frac {2n\pi x} {L}\right) \, dx +\int_{\pi}^ {2\pi} 0 \, dx\right)\\
&=\dfrac 1 \pi \cdot \dfrac {- \sin(n \pi)} {(n^2-1)}\\
&=0
\end{align*}
$$
I computed the series using
$$ \displaystyle a_0+\sum_{n=1}^\infty \Big[a_n\cdot \cos\left(\dfrac {2n\pi x} {L}\right)+b_n \cdot \sin\left(\dfrac {2n\pi x} {L}\right)\Big]$$
Finally, the wrong Fourier series of $f(x)$ that I found is:
$$ \displaystyle \dfrac {1} {\pi} +\sum_{n=2}^\infty \Big[\displaystyle \dfrac 1 \pi \cdot \dfrac {\cos(n \pi)+1} {(1-n^2)}\cdot \cos\left( n x\right)+0 \Big]$$
*I took initial $n=2$ to avoid an undetermined series at $n=1$
Any ideas on where my mistakes are?
|
\begin{align}
a_0 &= \frac{1}{2\pi}\int_{0}^{2\pi} \sin x \cdot \mathbf{I}_{[0,\pi]}\, dx = \frac{1}{2\pi}\int_{0}^{\pi} \sin x \, dx
= \frac{1}{\pi}\\
a_n &= \frac{2}{2\pi}\int_{0}^{2\pi} \sin x \cdot \mathbf{I}_{[0,\pi]} \cdot \cos\tfrac{2\pi n x}{2\pi}\, dx
= \frac{1}{\pi} \int_{0}^{\pi} \sin x \cdot \cos nx \, dx
= \frac{1}{\pi} \frac{\cos (\pi n) + 1}{1-n^2}\\
&= \frac{1}{\pi}\frac{1 + (-1)^n}{1-n^2}
= \begin{cases} \frac{2}{\pi}\frac{1}{1-n^2} & \text{if $n$ even}\\ 0 & \text{if $n$ odd}\end{cases}\\
b_n &= \frac{2}{2\pi}\int_{0}^{2\pi} \sin x \cdot \mathbf{I}_{[0,\pi]} \cdot \sin\tfrac{2\pi n x}{2\pi}\, dx
= \frac{1}{\pi} \int_{0}^{\pi}\sin x \cdot \sin (nx)\,dx\\
&= \frac{1}{\pi}\cdot\begin{cases}\tfrac{\pi}{2} & \text{if $n=1$}\\0 & \text{if $n>1$}\\\end{cases}
= \begin{cases}\tfrac{1}{2} & \text{if $n=1$}\\0 & \text{if $n>1$}\\\end{cases}
\end{align}
Hence the Fourier series of $f(x)$ over $(0,2\pi)$ is given by
\begin{align}
f(x) &\sim
a_0+\sum_{n=1}^{\infty}\left [a_n \cos (nx) + b_n\sin(nx)\right]\\
&= \frac{1}{\pi} + \sum_{n=1}^{\infty}\frac{2}{\pi}\frac{1}{1-(2n)^2}\cos((2n)x) + \frac{1}{2}\sin((1)x)\\
&= \frac{1}{\pi} + \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2nx)}{1-4n^2} + \frac{1}{2}\sin(x)
\end{align}
Plotted with 20 terms using Wolfram Alpha:
Note
I used Wolfram Alpha to compute these integrals, but it skipped the special case of $b_1$. This is likely to do with a symbolic division that implicitly assumed $n\neq 1$. Either way, this special case follows near-directly from the principle of orthogonality, i.e.,
$$
\frac{1}{\pi}\int_0^{2\pi} \sin(nx)\sin(mx)\,dx = \begin{cases}1 &\text{if $n=m$}\\
0 & \text{if $n\neq m$}\end{cases}
$$
So in our case, $$\frac{1}{\pi} \int_0^{2\pi} \sin(x)\mathbf{I}_{(0,\pi)}\sin(nx)\,dx
= \frac{1}{\pi} \begin{cases} \int_0^{\pi} \sin(x)^2\,dx & \text{for $n=1$}\\
\int_0^{\pi} \sin(x)\sin(nx)\,dx & \text{for $n>1$}\end{cases}$$
This latter point highlights why I knew that Wolfram Alpha hadn't given me everything,
$$\int_{0}^a \sin(x)^2 \,dx > 0,\quad\text{for $a>0$}$$
so we wouldn't expect the corresponding coefficient to be zero. If, on the other hand, the function were something like $f(x)=\sin(5x)\cdot \mathbf{I}_{(0,\pi)}$ or $f(x)=\cos(2x)\cdot \mathbf{I}_{(0,\pi)}$, then we'd know to check $b_5$ or $a_2$ respectively.
If you have learned about vector spaces, then you can think of the integral as an dot (inner) product over vectors $1, \sin(nx),\cos(nx)$ for $n=1,\dotsc,\infty$. The inner product measures similarity. In this case we are asking about amount of "$sin(x)$"-ness if we take a $\sin(x)$ function and lop off the latter half. The answer, perhaps unsurprisingly, is a half.
|
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|
Implicit Differentiation Quesrion $$x^y = y^x$$ is the equation
My solution is :$$\frac { dy }{ dx } =\left( \ln { y-\frac { y }{ x } } \right) /\left( \ln { x-\frac { x }{ y } } \right) $$
Just wondering if this is correct!
|
Your solution does not match mine:
$$
x^y = y^x \implies x\ln y = y\ln x \implies \frac{\ln y}{y} = \frac{\ln x}{x} \\
\frac{\frac1y\cdot y-1\cdot\ln y }{y^2}dy = \frac{\frac1x\cdot x-1\cdot\ln x }{x^2}dx\\
\frac{1-\ln y}{y^2} dy= \frac{1-\ln x}{x^2}dx\\
\frac{dy}{dx} = \frac{1-\ln x}{1-\ln y}\cdot\frac{y^2}{x^2}
$$
As I said, that does not seem to reduce to your answer.
|
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|
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} dt + B\int {1 \over t+w_2}dt \\= {w - 1\over w - w_2} \ln|t + w| + {1- w_2\over w - w_2}\ln|t + w_2| + C $$
After finding the value of $w, w_2$ final answer I got is
$${\sqrt{5} + 1\over 2\sqrt{5}}\ln|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\ln|t + 1/2 + \sqrt{5}/2| + C$$
But the given answer is :
$$\bbox[7px,Border:2px solid black]{ \frac{\ln\left(\left|t^2+t-1\right|\right)}{2}+\frac{\ln\left(\left|2t-\sqrt{5}+1\right|\right)-\ln\left(\left|2t+\sqrt{5}+1\right|\right)}{2\cdot\sqrt{5}}+C}$$
Where did I go wrong ? especially that first term of the answer is a mystery to me.
|
Here's an alternate method: we have $$I = \int \frac{t+1}{t^2+t-1}dt$$
Note that $\frac{d}{dt}(t^2+t-1) = 2t+1$
Then write $$2I = \int \frac{2t+1}{t^2+t-1} +\frac{1}{t^2+t-1}dt$$
$$2I = \ln|t^2+t-1|+\underbrace{\int \frac{1}{(t+\frac{1}{2})^2-\frac{5}{4}} dt}_{J}$$
Note that $$\frac{1}{(t+\frac{1}{2})^2-\frac{5}{4}}= \frac{1}{(t+\frac{1}{2}+\frac{\sqrt{5}}{2})(t+\frac{1}{2}-\frac{\sqrt{5}}{2})} = \frac{A}{t+\frac{1}{2}+\frac{\sqrt{5}}{2}}+\frac{B}{t+\frac{1}{2}-\frac{\sqrt{5}}{2}}$$
Then $A(t+\frac{1}{2}-\frac{\sqrt{5}}{2})+B(t+\frac{1}{2}+\frac{\sqrt{5}}{2})=1$
$t=-\frac{1}{2}+\frac{\sqrt{5}}{2}\implies B=\frac{1}{\sqrt{5}},\quad t=-\frac{1}{2}-\frac{\sqrt{5}}{2}\implies A=-\frac{1}{\sqrt{5}}$
$$J = \frac{1}{\sqrt{5}}\int -\frac{1}{t+\frac{1}{2}+\frac{\sqrt{5}}{2}}+\frac{1}{t+\frac{1}{2}-\frac{\sqrt{5}}{2}} dt = \frac{1}{\sqrt{5}}\bigg(\ln|t+\frac{1}{2}-\frac{\sqrt{5}}{2}|-\ln|t+\frac{1}{2}+\frac{\sqrt{5}}{2}|\bigg)+C$$
Then $$I = \frac{1}{2}\ln|t^2+t-1|+\frac{1}{2\sqrt{5}}\bigg(\ln\bigg|t+\frac{1}{2}-\frac{\sqrt{5}}{2}\bigg|-\ln\bigg|t+\frac{1}{2}+\frac{\sqrt{5}}{2}\bigg|\bigg)+C$$
|
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|
Direct proof that $\frac{1}{\sqrt{1-x}} = \frac{1}{2\pi}\int_0^{2\pi} \frac{d\theta}{1-x\cos^2\theta}$? In a totally indirect way, I've proven to myself that
$$\frac{1}{\sqrt{1-x}} = \frac{1}{2\pi}\int_0^{2\pi} \frac{d\theta}{1-x\cos^2\theta}.$$
The proof was by expanding the integrand as a power series in $x\cos^2\theta$, integrating term-by-term, and comparing to the binomial expansion of the left side.
But I would like a direct proof. Do you know one? I would be happy either with a method based on antidifferentiation, or a contour integral, or a double integral resulting from squaring, or some other method I haven't thought of. I am just trying to avoid the power series in $x$.
|
Here's the contour integral solution:
$$
\cos\theta=\frac{z+1/z}{2},
$$
where $z$ is belongs to the unit circle $C_1=\{z\in\mathbb C:\ z=e^{i\theta}\}$
and $d\theta=-i\,dz/z$ so
$$
\int_0^{2\pi}\frac{d\theta}{1-x\cos^2\theta}=\oint_{C_1}\frac{i4zdz}{xz^4+2(x-2)z^2+x}.
$$
If $x=0$, we get $-i\int_{C_1}dz/z=2\pi$. Otherwise, the roots of the denominator are at
$$
z^2=\frac{2-x\pm2\sqrt{1-x}}{x}=\frac{1}{x}\left(1\pm\sqrt{1-x}\right)^2,
$$
thus, $$z_1^\pm=\frac{1\pm\sqrt{1-x}}{\sqrt{x}},\qquad z_2^\pm=-\frac{1\pm\sqrt{1-x}}{\sqrt{x}}.$$
For $0<x<1$, only $z_1^-$ and $z_2^-$ fall within the unit circle and the residue formula gives
$$
-8\pi\left[\frac{1}{4x(z_1^-)^2+4(x-2)}+\frac{1}{4x(z_2^-)^2+4(x-2)}\right]=-2\pi\frac{2}{-2\sqrt{1-x}}=\frac{2\pi}{\sqrt{1-x}}.
$$
|
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|
mathematical induction with exponent $$\frac 13 + \frac 1{3^2} + \frac 1{3^3} + \dots + \frac 1{3^n} + = \frac 12 \times \left( 1 - \frac{1}{3^n} \right)$$
Step 1 - $n=1$
$$\begin{align}
\frac 1 {3^1} & = \frac 1 2 \times \left( 1 - \frac 1 {3^1} \right) \\
\frac 1 3 & = \frac 1 2 \times \left( 1 - \frac 1 3 \right) \\
\frac 1 3 & = \frac 1 2 \times \frac 2 3 \\
\frac 1 3 & = \frac 1 3 \\
\end{align}$$
Step 2 - n$=k$
$$ \frac 13 + \frac 1{3^2} + \frac1 {3^3} + \dots + \frac 1 {3^k} = \frac 12 \times \left( 1 - \frac 1 {3^k} \right)$$
Step 3 - $n=k+1$
Having problems solving step 3.
|
By hypothesis, we know that
$$
\frac{1}{3}+\frac{1}{3^2}+\dots+\frac{1}{3^k}=\frac{1}{2}\bigg[1-\frac{1}{3^k}\bigg]
$$
Hence, by adding to both sides by $\frac{1}{3^{k+1}}$ we get
$$\begin{align}
\frac{1}{3}+\frac{1}{3^2}+\dots+\frac{1}{3^k}+\frac{1}{3^{k+1}}&=\frac{1}{2}\bigg[1-\frac{1}{3^k}\bigg]+\frac{1}{3^{k+1}}\\
&=\frac{3^k-1}{2\cdot 3^k}+\frac{1}{3^{k+1}}\\
&=\frac{3^{k+1}-3+2}{2\cdot 3^{k+1}}\\
&=\frac{3^{k+1}-1}{2\cdot 3^{k+1}}\\
&=\frac{1}{2}\bigg[1-\frac{1}{3^{k+1}}\bigg]
\end{align}$$
|
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|
Is it true that $f$ must be constant? Let $f: \Bbb R \to \Bbb R$ be a function which satisfies $f(x) = f(2x-1)$ for all $x \in \Bbb R$ and $f$ is continuous at $x=1$. Is it true that $f$ must be constant?
|
From your assumption we have $f(x)=f(\frac{x}{2}+\frac{1}{2})$ for all $x$. For $x\in \mathbb R$ and substituting successively $x$ by $\frac{x}{2}+\frac{1}{2}$, we get
$$f(x)=f\Big(\frac{x}{2}+\frac{1}{2}\Big)=f\Big(\frac{1}{2}\Big(\frac{x}{2}+\frac{1}{2}\Big)+\frac{1}{2}\Big)=f\Big(\frac{x}{2^2}+\frac{1}{2}+\frac{1}{2^2}\Big)=\cdots =f(\frac{x}{2^n}+\frac{1}{2}+\cdots \frac{1}{2^n}\Big).$$
Note that $\frac{x}{2^n}+\frac{1}{2}+\cdots +\frac{1}{2^n} \to 1$ as $n\to \infty$ with fixed $x$. Then by the continuity of $f$ at $1$ we get $f(x)=f(1)$ for all $x$.
|
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|
Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.
|
Notice that
$$
\log_uv\log_vw=\dfrac{\log v}{\log u}\cdot\dfrac{\log w}{\log v}=\dfrac{\log w}{\log u}=\log_uw \quad \forall u,v,w>0
$$
therefore
\begin{eqnarray}
(\log_ab+1)(\log_bc+1)(\log_ca+1)&=&(\log_ab\log_bc+\log_ab+\log_bc+1)(\log_ca+1)\\
&=&(\log_ac+\log_ab+\log_bc+1)(\log_ca+1)\\
&=&\log_ac\log_ca+\log_ac+\log_ab\log_ca+\log_ab+\log_bc\log_ca+\log_bc+\log_ca+1\\
&=&\log_aa+\log_ac+\log_cb+\log_ab+\log_ba+\log_bc+\log_ca+1\\
&=&(1+\log_aa)+(\log_ab+\log_bc+\log_ca)+(\log_ba+\log_cb+\log_ac)\\
&=&1+1+8+13=23.
\end{eqnarray}
|
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|
How to do junior math Olympiad question Let $a$ and $b$ be such that $0<a<b$. Suppose that $a^3 = 3a-1$ and $b^3 = 3b-1$. Find $b^2 -a$.
|
Since $0<a<b$ and $a^3=3a-1$ and $b^3=3b-1$, then we need to find the two positive real roots of $x^3-3x+1=0$ with $a$ being the smaller root and $b$ being the bigger root.
Following Cardano's solution, if we let $x=u+v$ then
$$
\begin{align*}
x^3-3x+1&=0\\
(u+v)^3-3(u+v)+1&=0\\
u^3+3u^2v+3uv^2+v^3-3u-3v+1&=0\\
u^3+v^3+1+3u^2v-3u+3uv^2-3v&=0\\
(u^3+v^3+1)+3(u+v)(uv-1)&=0\\
\end{align*}
$$
So we want to find the simultaneous solution to the following system:
$$
\begin{align*}
u^3+v^3&=-1\\
uv&=1.
\end{align*}
$$
By substituting $u=\frac1v$ into the first equation we have
$$
\frac1{v^3}+v^3=-1\iff v^6+v^3+1=0.
$$
Then the quadratic formula tells us
$$v^3=\frac{-1+i\sqrt{3}}{2}=e^{2\pi i/3}\iff v=e^{2\pi i/9}\implies u=e^{-2\pi i/9}$$
or
$$v^3=\frac{-1-i\sqrt{3}}{2}=e^{4\pi i/3}\iff v=e^{4\pi i/9}\implies u=e^{-4\pi i/9}.$$
So the two real positive solutions are
$$
\begin{align*}
a&=e^{-4\pi i/9}+e^{4\pi i/9}=2\cos\left(\frac{4\pi}{9}\right)\\
b&=e^{-2\pi i/9}+e^{2\pi i/9}=2\cos\left(\frac{2\pi}{9}\right)\\
\end{align*}
$$
and we see that
$$
\begin{align*}
b^2-a&=4\cos^2\left(\frac{2\pi}{9}\right)-2\cos\left(\frac{4\pi}{9}\right)\\
&=4\cos^2\left(\frac{\frac{4\pi}{9}}{2}\right)-2\cos\left(\frac{4\pi}{9}\right)\\
&=4\left(\frac{1+\cos\left(\frac{4\pi}{9}\right)}{2}\right)-2\cos\left(\frac{4\pi}{9}\right)\\
&=2+2\cos\left(\frac{4\pi}{9}\right)-2\cos\left(\frac{4\pi}{9}\right)\\
&=2.
\end{align*}
$$
|
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|
Finding the exact value of a radical How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
|
While the accepted solution is fine if you're willing to solve a quadratic equation, it's worth developing some theory to avoid this for two reasons:
*
*To save effort spent solving quadratic equations, replacing it by effort spent doing these different and exciting calculations instead :)
*So that we can also figure out something like $\sqrt[3]{1351 + 780 \sqrt{3}}$, where we'd have to solve a cubic equation instead, and who knows how to do that?
The theory is that if $(a + b \sqrt 3)^2 = 97 + 56 \sqrt 3$, then we also have $(a - b \sqrt{3})^2 = 97 - 56 \sqrt 3$. (You can check this by comparing the expansions. It ultimately comes from the fact that $\sqrt3$ and $-\sqrt3$ are "symmetric" as far as the rational numbers are concerned: you can't tell them apart by their algebraic properties.)
Multiplying these two equations together and applying difference of squares, we get
$$(a + b\sqrt 3)^2 (a - b\sqrt 3)^2 = (97 + 56 \sqrt3)(97 - 56 \sqrt 3) \implies (a^2 - 3b^2)^2 = 97^2 - 3 \cdot 56^2 = 1.$$
So now we have three equations for $a$ and $b$, not two:
\begin{align}
a^2 + 3b^2 &= 97 \\
2ab &= 56 \\
a^2 - 3b^2 &= \pm1.
\end{align}
The first two come from comparing coefficients in $(a + b \sqrt 3)^2 = 97 + 56 \sqrt 3$, as before. The third comes from the fact that if $(a^2 - 3b^2)^2 = 1$, then either $a^2 - 3b^2 = 1$ or $a^2 - 3b^2 = -1$.
If we consider $a^2 - 3b^2 = 1$ as the first possibility, then together with $a^2 + 3b^2 = 97$ we have two linear equations in $a^2$ and $b^2$. Adding them together, we get $2a^2 = 98$, so $a^2 = 49$, and $a = \pm 7$. Subtracting them, we get $6b^2 = 96$, so $b^2 = 16$, and $b = \pm 4$. Since $2ab = 56$, $a$ and $b$ are both positive or both negative, so we get $7 + 4\sqrt3$ and $-7 - 4\sqrt3$ as our two solutions (both valid).
The second case, when $a^2 - 3b^2 = -1$, tells us $2a^2 = 96$ and $6b^2 = 98$, which does not give us a rational solution. (Actually, it tells us that $a = 4 \sqrt 3$ and $b = \frac{7}{\sqrt 3}$, giving us the same answer in a "backwards" way.) But we can ignore it: we already got an answer.
|
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|
Find the value of $\sin (x)$ if $\cos(x)=\frac 1 3$ $$\cos (x)=\left (\frac{1}{3}\right)$$for $0°\lt\ x\lt90°$
Find the exact value of $\sin(x)$.
Give your answer as a surd.
I found the value with my calculator
$$\sin (x)=\left(\frac{\sqrt 8}{3}\right)$$
Is there a way of doing this by hand?
|
$\sin^2 x + \cos^2 x = 1$
$\sin^2 x = 1 - \cos^2 x$
$\sin x = \pm \sqrt{1 - \cos^2 x}$
$\sin x = \pm \sqrt{1 - \left( \frac 13\right)^2}$
$\sin x = \pm \sqrt{1 - \frac 19}$
$\sin x = \pm \sqrt{\frac 89}$
$\sin x = \pm \frac {\sqrt 8}3$
|
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|
Determinant doubt How do we prove, without actually expanding, that
$$\begin{vmatrix}
\sin^2 {A}& \cot {A}& 1\\
\sin^2 {B}& \cot{B}& 1\\
\sin^2 {C}& \cot{C}& 1
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried applying cosine double angle formula but couldn't get anywhere.
|
It is enough to show that
$$ \det\begin{pmatrix}\sin^3 A & \cos A & \sin A \\ \sin^3 B & \cos B & \sin B \\ \sin^3 C & \cos C & \sin C \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^3 & \frac{b^2+c^2-a^2}{bc} & a \\ b^3 & \frac{a^2+c^2-b^2}{ac} & b \\ c^3 & \frac{a^2+b^2-c^2}{ab} & c \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^3 b c & b^2+c^2-a^2 & abc \\ b^3 a c & a^2+c^2-b^2& abc \\ c^3ab & a^2+b^2-c^2 & abc \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^2 & b^2+c^2-a^2 & 1 \\ b^2 & a^2+c^2-b^2& 1 \\ c^2 & a^2+b^2-c^2 & 1 \end{pmatrix}=0 $$
or that
$$ \det\begin{pmatrix}a^2 & b^2+c^2 & 1 \\ b^2 & a^2+c^2& 1 \\ c^2 & a^2+b^2 & 1 \end{pmatrix}=0 $$
that is pretty simple.
By barycentric coordinates, this is equivalent to the collinearity of the centroid, symmedian point and symmedian point of the anticomplementary triangle (collinearity of $X(2),X(6)$ and $X(69)$ according to ETC). Pretty trivial since a triangle and its anticomplementary triangle have the same centroid.
|
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|
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
|
Multiplying by $x-2$ changes sign of the inequality when $x-2<0$ though you could multiply by something which is always $>0$ for example multiplying by $(x-2)^2$(assuming $x\neq2$) we get
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}\\(1-x)(x-2)<3x(x-2)<(x-2)(x-3)$$
Which is equivalent with
$$(4x-1)(x-2)>0\land (x-2)(2x+3)<0$$
|
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|
Cyclicity:To find the units digit of a number
Can you please help me to find the units digit of this number:
33^34^35^36^37^38 .. I got this question in CSR 2017
|
$33^4\equiv 3^4\equiv 81\equiv 1\pmod{10}$.
$34^{35^{36^{37^{38}}}}=4k$ for some $k\in\mathbb Z^+$.
$33^{4k}\equiv \left(33^4\right)^k\equiv 1^k\equiv 1\pmod{10}$.
Euler's theorem is relevant here, but I've been able to explain this simply without it.
Edit: The multiplicative order of $33$ modulo $10$ is $4$.
You could use Euler's theorem to notice that $\phi(10)=4$ and continue.
To explain it more simply, notice the pattern:
$33^0\equiv 1\pmod{10}$
$33^1\equiv 3\pmod{10}$
$33^2\equiv 3^2\equiv 9\pmod{10}$
$33^3\equiv 3^3\equiv 27\equiv 7\pmod{10}$
$33^4\equiv 3^4\equiv 81\equiv 1\pmod{10}$
$33^5\equiv 33\cdot 33^4\equiv 33\cdot 1\equiv 3\pmod{10}$
...
Since $34^{35^{36^{37^{38}}}}$ is divisible by $4$, we get the answer $1$.
|
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|
Simple inequality $\frac{4x+1}{2x-3}>2$
$$\frac{4x+1}{2x-3}>2$$
I have started with looking at positive/negative situations and there are 2 or that both expressions are positive or both are negative.
If both a positive we can solve for $$\frac{4x+1}{2x-3}>2$$
$${4x+1}>2(2x-3)$$
$${4x+1}>4x-6$$
$${0x}>-7$$
Or both are negative and then $${4x+1}<2(2x-3)$$
And again $$0x<-7$$
So there is not answer?
|
For $x > \frac{3}{2}$, we rearrange to $$4x+1 > 4x -6 \iff 1 > -6$$ and so it holds for all $x > \frac{3}{2}$.
If $x < \frac{3}{2}$, we rearrange to $$4x+1 < 4x-6 \iff 1 < -6$$ and hence it does not hold for $x < \frac{3}{2}$.
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Which point of the sphere $x^2+y^2+z^2=19$ maximize $2x+3y+5z$? Which point of the sphere $x^2+y^2+z^2=19$ maximize $2x+3y+5z$?
So I assume that there is a point maximizing $2x+3y+5z$. How can I calculate the exact value of this point?
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By the Cauchy-Schwarz inequality
$$ 2x+3y+5z \leq \sqrt{x^2+y^2+z^2} \sqrt{2^2+3^2+5^2} = 19 \sqrt{38} $$
and equality is achieved if $(x,y,z)=\lambda(2,3,5)$, i.e. for $\lambda=\frac{1}{\sqrt{2}}$.
|
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
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$x^2+y^2+z^2$ only depends on the squared distance of $(x,y,z)$ from the origin and the constraint $x+y+z=1$ tells us that $(x,y,z)$ lies in a affine plane. The problem is solved by finding the distance between such plane and the origin: since the plane is orthogonal to the line $x=y=z$,
$$ \min_{x+y+z=1}x^2+y^2+z^2 = \left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2 = \frac{1}{3}$$
and we are done.
|
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In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$? In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$?
$1$. $18°$
$2$. $54°$
$3$. $60°$
$4$. $30°$
My Attempt:
$$A=2B=3C$$
Let $\angle A=x$ then $\angle B=\dfrac {x}{2}$ and $\angle C=\dfrac {x}{3}$
Now,
$$x+\dfrac {x}{2} +\dfrac {x}{3}=180°$$
$$\dfrac {6x+3x+2x}{6}=180°$$
$$\dfrac {11x}{6}=180°$$
$$x=\dfrac {180\times 6}{11}$$
So, which is the correct option?
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As A=2B=3C,
B=1.5C
Therefore, summing the angles, A+B+C= 3C+1.5C+C=180 degrees.
5.5C=180
C=36/1.1
=360/11
= 32.77 degrees.
|
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How do I prove using the definition of limit that $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$? I want to use the definition of limit in order to prove $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$, but I almost crack my head open trying to do this, it seems to be easy way but y cant find a $\delta$ such that $|\frac {x^2+y^3}{y^2 - x + xy}| \leq \epsilon$, any idea of how I might be able to find an inequality to transform the denominator in a nicer thing like $\frac{1}{\left\lVert (x,y)\right\rVert}$?
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Note that
$$
x=y^2\implies\frac{x^2+y^3}{y^2-x+xy}=\frac{y^4+y^3}{y^2-y^2+y^3}=y+1\overset{y\to0}\to1
$$
and that
$$
x=0\implies\frac{x^2+y^3}{y^2-x+xy}=\frac{y^3}{y^2}=y\overset{y\to0}{\to}0
$$
These show that the limit does not exist.
|
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Given: $\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$, find $f(x)$.
$$\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$$
On differentiating, I get,
$$f(x)\sin x\cos x = {f^\prime(x)\over f(x)}{1\over 2( b^2 - a^2)}$$
$$\sin 2x (b^2 - a^2) = {f^\prime( x)\over (f(x))^2} $$
On integrating,
$${-1\over f(x)} = {-(b^2 - a^2)\cos 2x\over 2} \implies f(x) = { 2\over(b^2 - a^2)\cos 2x}$$
The answer given is $\displaystyle f(x) = {1\over a^2 \sin^2 x + b^2 \cos^2 x}$.
I am unable to get the given result, the closest I got is, $$f(x) = {2\over b^2 \cos^2x -b^2\sin^2 x- a^2\cos^2x+ a^2\sin^2 x}$$.
How to simplify further to get the given answer ?
Related but not duplicate.
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The solutions are OK (contrary to what I first wrote). After integrating, you can check that you have
$$
\frac{1}{f}=\frac{(b^2-a^2)\cos 2x}{2}
$$
and, according to the solution in the book,
$$
\frac{1}{f}=a^2\sin^2x+b^2\cos^2x.
$$
The difference should be a constant (the integrating constant). And indeed,
$$
\frac{(b^2-a^2)\cos 2x}{2}-\bigl(a^2\sin^2x+b^2\cos^2x\bigr)=-\frac{a^2+b^2}{2}.
$$
Thus, everything is in order.
|
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|
show that $a^2-1$ is a quadratic nonresidue mod $p$ if and only if $p\equiv 6\pmod 7$ Question: Let prime number $p>7$, for positive integer $a,b,c$ such $1<a<b<c<p$, and
$$a+b+c\equiv a^3+b^3+c^3\equiv a^5+b^5+c^5\equiv\frac{p-1}{2}\pmod p$$
show that:
$a^2-1$ is a quadratic nonresidue mod $p$ if and only if $p\equiv 6\pmod 7$
I think following identity can work?
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)$$
and
$$a^5+b^5+c^5=(a+b+c)^5-5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ac)$$
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Since $ab+bc+ac$ and $abc$ are rational fractions in $a+b+c,a^3+b^3+c^3,a^5+b^5+c^5$,
you can show that for $p \neq 2,3$, then $a,b,c$ satisfy these relations mod $p$ if and only if they are roots of a certain cubic equation.
(in fact, you get the equation $8a^3+4a^2-4a-1 = 0$)
It turns out that over $\Bbb C$, its roots are $\cos(2\pi/7), \cos(4\pi/7), \cos(6\pi/7)$, so the Galois group of the cubic is cyclic, and we know how it factors mod $p$ according to $p$ mod $7$ :
If $p \neq 7$, the cubic has $3$ roots mod $p$ iff $p \equiv \pm 1 \pmod 7$, and it has no root otherwise.
Moreover, $a^2-1 = \cos(2\pi/7)^2-1 = - \sin(2\pi/7)^2 = (i\sin(2\pi/7))^2$, so this has a square root $a'$ if and only if there is a $7$th root of unity $a+a' = \exp(2i\pi/7)$ if and only if $p \equiv 1 \pmod 7$.
Thus, the cubic splits and $a^2-1$ is a not square mod $p$ if and only if $p \equiv 6 \pmod 7$.
|
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|
for what values of N element N is $3^{n+1}>n^4$ So I'm doing homework for my logic class...
For what values of n element of natural numbers is $3^{n+1}>n^4$ and we are supposed to use induction. This is what I have so far...
Proof: The inequality holds for n=1. Since $3^2>1^4= 9>1$. Assume $3^{k+1}>k$, where $k\geq 1$. We show $3^{k+2}>(k+1)^4$
$3^{k+2}=3^2+3^k>3k^4=k^4+k^4+k^4...$ I'm not sure what to do from here.
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The inequality $3^{n+1}>n^4$ is true for $n=1,\ 2$ and all $n\ge5$. (The inequality is false for $n=3$ and $n=4$ because $81\not>81$ and $243\not>256$.)
We can prove the inequality for $n\ge5$ by induction.
The base case ($n=5$) is true: $3^6>5^4, \ \ 729>625$.
Now suppose the inequality is true for some $k\ge5$:
$$
3^{k+1} > k^4. \tag{1}
$$
But
$$
3^{k+2}=3^{k+1}+2\cdot3^{k+1}, \quad\mbox{ while }\quad
(k+1)^4 = k^4 + 4k^3+ 6k^2+ 4k+ 1,
$$
and for every $k\ge5$ we have
$$
2\cdot3^{k+1} > 2\cdot k^4 \ge 2\cdot5k^3> 4k^3+ 6k^2+ 4k+ 1
$$
by hypothesis $(1)$. Thus we have $3^{k+2}>(k+1)^4$; our induction step is completed.
Therefore, by the principle of mathematical induction, our inequality $$3^{n+1}>n^4$$ is true for all $n\ge5$.
|
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|
Probability of drawing a spade on the first draw, a heart on the second draw, and an ace on the third draw My answer is
$(\frac{1}{52})(\frac{1}{51})(\frac{2}{50})$ + $2(\frac{1}{52})(\frac{12}{51})(\frac{3}{50})$ + $(\frac{12}{52})(\frac{12}{51})(\frac{4}{50})$
= $\frac{1}{204}$
But the answer in the textbook is
$(\frac{13}{52})(\frac{13}{51})(\frac{2}{50})$ + $2(\frac{13}{52})(\frac{13}{51})(\frac{3}{50})$ + $(\frac{13}{52})(\frac{13}{51})(\frac{4}{50})$
= $\frac{13}{850}$
Which answer is correct?
|
The probability of drawing a spade followed by a heart followed by the ace of spades or hearts is
$$\frac{12\times13}{51\times 50}\times\frac2{52}.$$
The probability of drawing a spade followed by a heart followed by the ace of diamonds or clubs is is
$$\frac{13^2}{51\times 50}\times\frac2{52}.$$
These add to
$$\frac{13\times25}{51\times50}\times\frac2{52}=\frac1{204}.$$
|
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|
Which is greater $x_1$ or $x_2$? $$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$
$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ We have to find which is greater among $x_1$ and $x_2$
If we add both we get $$x_1+x_2=\pi$$ If we use formulas we get
$$x_1=\arccos\left(\frac{6\sqrt{2}-4}{15}\right)$$ and
$$x_2=\arcsin\left(\frac{3+8\sqrt{2}}{15}\right)$$
but how to compare now?
|
We have
$$\frac{3}{5} > \frac{1}{2} \implies \arcsin \frac{3}{5} > \arcsin \frac{1}{2} = \frac{\pi}{6}$$
and
$$\frac{2\sqrt{2}}{3} > \frac{\sqrt{3}}{2} \implies \arcsin \frac{2\sqrt{2}}{3} > \arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3},$$
hence
$$x_2 > \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}.$$
|
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|
$10^n+1$ as a product of consecutive primes I am interested to know the examples or the systematic rule whether $10^n+1$ is a product of consecutive primes.
For example,
when $n=1$,
we have $10^1+1=11$ is a product of a single prime 11. So, a True example.
When $n=2$,
we have $10^2+1=101$ is a product of a single prime 101. So, a True example.
When $n=3$,
we have $10^3+1=1001$ is a product of a product of consecutive primes $7*11*13$. So, a True example.
When $n=4$,
we have $10^4+1=10001$ is NOT a product of a product of consecutive primes $73*137$. So, a False example.
Are there more examples or the systematic rule whether $10^n+1$ is a product of consecutive primes?
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I'm finding that $11$ divides $10^n+1$ for all odd $n$. $7*11*13$ divides $10^n+1$ for odd $n$ divisible by $3$. But $17$ is a divisor only if $n$ is an odd multiple of $8$. Thus it seems $10^n+1$ is not a product of consecutive primes for odd $n>3$. If even $n$ is an odd multiple of $2$, $101$ divides $10^n+1$, but $103$ seems to be a divisor only for odd $n$ in sequence $17,119,221...$. If even $n$ is an odd multiple of $4$, $73$ and $137$ are consecutive divisors of $10^n+1$, but they are not consecutive primes. If even $n$ is an odd multiple of $8$, as already noted, $17$ is a divisor, but not $13$, and $19$ is a divisor only if $n$ is an odd multiple of $9$. Thus it seems, for $n>3$, $10^n+1$ could be the product of consecutive primes only if $16$ divides $n$. Is there a way to rule out these $n$ as well? For $n=16$ or any odd multiple of $16$, $10^n+1$ contains $353*449*641*1409*69857$, which are not consecutive primes. For $n=32$ and its odd multiples, $10^n+1$ contains $19841*976193*6187457*834427406578561$, again not consecutive primes. Thus if $10^n+1$ is not the product of consecutive primes when $n$ is a given power of $2$, then it is not the product of consecutive primes for any odd multiple of that power of $2$. This shortens the search among the even $n$ remaining.
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Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my soliution:
$$S(x)=1+\frac{x^{4}}{4\cdot2^{4}}+\frac{x^{7}}{7\cdot2^{7}}+\frac{x^{10}}{10\cdot2^{10}}+\cdots=1+\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}=1+S_1(x)$$
$$(S_1(x))_x'=\left(\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}\right)_x'=\sum_{n=1}^\infty \frac{x^{3n}}{2^{3n+1}}=\frac{1}{x}\sum_{n=1}^\infty \left(\frac{x}{2}\right)^{3n+1}$$
Now let's take $\frac{x}{2}=y$, then
$$S_2(y)=\sum_{n=1}^\infty y^{3n+1}=y^4+y^7+y^{10}+\cdots=\frac{y^4}{1+y^3},|y|\le1$$
$$\left(S_1(y)\right)'=\frac{1}{2y}\cdot\frac{y^4}{1-y^3}=\frac{1}{2}\cdot\frac{y^3}{1-y^3}$$
$$S_1(y)=\frac{1}{2}\int\frac{y^3}{1-y^3}dy=\frac{1}{2}\int\left(-1+\frac{1}{1-y^3}\right)dy=-\frac{1}{2}y+\frac{\sqrt{3}}{6}\arctan\left(\frac{2\left(y+\frac{1}{2}\right)}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\lvert y-1\rvert+C$$
$$S_1(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|+C$$
$$S_1(0)=0, C=-\frac{\sqrt{3}\pi}{36}$$
$$S(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|-\frac{\sqrt{3}\pi}{36}+1$$
$$S(1)=\frac{3}{4}+\frac{\sqrt{3}}{3}\arctan\left(\frac{2\sqrt{3}}{3}\right)+\frac{1}{6}\ln(7)-\frac{\sqrt{3}\pi}{36}$$
By writing this for about 2 hours I deserve extra 50 points or at least good answers... Ha ha, thanks!
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Sometimes when faced by one of these "sections" of a well-known series it helps to think "roots of unity". The other solutions given here are surely the way to go in this case, but it's worth knowing the generic technique, so here goes.
Let $G(x)=\sum_0^\infty \frac{x^{n+1}}{n+1}$ we see that we want "one-third" of this series. So with $\omega:=\exp{\frac{2\pi}{3}}$ we have that the given series is the value at $x=\frac{1}{2}$ of
$$\frac{1}{3}\left[G(x)+\omega^2\ G(\omega x)+ \omega\ G(\omega^2 x)\right].$$
We'd get a couple of other series by taking the other obvious multipliers, $(1,1,1)$ and $(1,\omega,\omega^2)$.
As $G(x)=\log (1-x)$ there's then nothing left to do apart from some tedious arithmetic evaluating the modulus and argument of things like $1-\frac{1}{2}\omega$.
|
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|
Vectors representing the principal axes of an ellipse The principal axes of the ellipse given by the equation $ 9x^2+6y^2-4xy=4$ are along the directions of the vectors
a) $2i+3j$ and $3i-2j$
b) $2i+j$ and $i-2j$
c) $3i+2j$ and $2i-3j$
d) $i+2j$ and $2i-j$
e) $i+j$ and $i-j$
Seems trivial however I do not have much idea on how to find those vectors.
|
Notice that
$$ 9x^2+6y^2-4xy = \begin{pmatrix} x& y \end{pmatrix} \begin{pmatrix} 9 & \frac{-4}{2} \\ \frac{-4}{2} & 6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$
We'll define $\vec{x} = \begin{pmatrix} x \\ y \end{pmatrix} $. Now:
$$ 9x^2+6y^2-4xy = \vec{x}^T \begin{pmatrix} 9&-2\\ -2&6 \end{pmatrix} \vec{x} $$
The matrix $\begin{pmatrix} 9&-2\\ -2&6 \end{pmatrix}$ can be diagonalized, which we prefer because then in the new basis there will be no xy term, meaning the new basis will give us the principal axes of the ellipse.
Let's find the eigenvalues:
$$ (\lambda-9)(\lambda - 6) - 4 = \lambda^2 - 15\lambda + 50 = 0$$
$$ \lambda = 5 ,10 $$
For $\lambda = 5$ we'll find the eigen vector:
$$\begin{pmatrix} 4&-2 \\ -2 & 1 \end{pmatrix} v = 0$$
It is easy to see that $v=\begin{pmatrix} 1\\2 \end{pmatrix} $ solves the equation. The original matrix is a symmetric matrix meaning the other eigen vector will be orthogonal to the one we just found meaning it will be $\begin{pmatrix} 2\\-1 \end{pmatrix} $ So the answer is option (d).
|
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If $a,b,c$ are positive, prove that $\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \geq \frac{9}{a+b+c}$ If $a,b,c$ are positive real numbers, prove that
$$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$
|
your inequality is equivalent to
$$2\,{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}-a{c}^{2}+2\,{b}^{3}-{b}^{2}c-b{c
}^{2}+2\,{c}^{3}>0
>0$$ after Clearing the denominators and this is equivalent to
$$(a-b)(a^2-b^2)+(a-c)(a^2-c^2)+(b-c)(b^2-c^2)\geq 0$$ which is true.
the equal sign holds if $$a=b=c$$
|
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If $a, b, c$ are positive and $a+b+c=1$, prove that $8abc\le\ (1-a)(1-b)(1-c)\le\frac{8}{27}$ If $a, b, c$ are positive and $a+b+c=1$, prove that $$8abc\le\ (1-a)(1-b)(1-c)\le\dfrac{8}{27}$$
I have solved $8abc\le\ (1-a)(1-b)(1-c)$ (by expanding $(1-a)(1-b)(1-c)$)
but do not get how to show that $(1-a)(1-b)(1-c)\le\frac{8}{27}$
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\begin{eqnarray*}
a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
9abc \leq (ab+bc+ca)(a+b+c) \\
8abc \leq 1-(a+b+c)+(ab+bc+ca)-abc
\end{eqnarray*}
Thus the first inequality is shown.
\begin{eqnarray*}
4((a+b)(a-b)^2+(b+c)(b-c)^2+(c+a)(c-a)^2)+a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
8(a^3+b^3+c^3)-3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)-6abc \leq 0 \\
27(a+b+c)(ab+bc+ca)-27abc \leq 8(a+b+c)^3
\end{eqnarray*}
Now add $27 -27(a+b+c)$ to the left hand side & use $a+b+c=1$.
Thus the second inequality is shown.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Finding $\lim_{n \to \infty} x_n$ Let $x$ be a positive real number. A sequence $x_n$ of real numbers is defined as follows:
$$x_1=\frac 12\left(x+\frac5x\right)$$ and $$x_{n+1}=\frac 12\left(x_n+\frac 5{x_n}\right) \quad\forall n\geq 1$$
The question is to show that $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^{2^{n}}$$ and hence or otherwise evaluating $$\lim_{n \to \infty} x_n$$
To show first part I tried using induction. It is clear that $n=1$ is true. Let it be true for $n=k$. So $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^2=\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^{2^{n+1}}$$
I tried manipulating this but could not proceed. Any ideas? Thanks.
|
If it true for $n=k$ then we have $\frac{x_k - \sqrt{5}}{x_k + \sqrt{5}} = \left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)^{2^k}$.
Then $$\frac{x_{k+1} - \sqrt{5}}{x_{k+1} + \sqrt{5}} = \frac{x_k + \frac{5}{x_k} - 2\sqrt{5}}{x_k + \frac{5}{x_k} + 2\sqrt{5}} = \frac{x_k^2 - 2x_k\sqrt{5} +5}{x_k^2 + 2x_k\sqrt{5}+5}$$
which is a square $$\frac{(x_k-\sqrt{5})^2}{(x_k+\sqrt{5})^2} = \left(\left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)^{2^k}\right)^2 = \left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)^{2^{k+1}}$$
as required.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find maximum and minimum value of inverse function .
We have to find maximum and minimum value of $x^2 +y^2$
My try
how can I proceed
|
Let $\lfloor xy \rfloor=n$, then
$$\frac{x^2+y^2}{1-x^2y^2} = \tan \frac{n\pi}{4}$$
You've checked that no solutions for $0\le xy \le 1$, hence the tangent value must be negative which is achievable when $n=3,7,\ldots, 4k-1, \ldots$
Now
\begin{align}
\frac{x^2+y^2}{1-x^2y^2} &= \tan \left( k\pi-\frac{\pi}{4} \right) \\
x^2+y^2 &= x^2y^2-1
\end{align}
The required minimum value occurs when $xy=3$, therefore
$$\fbox{$\min \{ x^2+y^2 \}=8$}$$
and there're no maxima.
|
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|
$a+b+c+d+e=79$ with constraints How many non-negative integer solutions are there to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$?
I get that for $a\ge7$ you do $79-7=72$, $\binom{72+5-1}{5-1}=\binom{76}4$. For $b\ge35$ I think it's $\binom{47}4$ and I'm not too sure what it is for $3\le c\le41$ and I also have no clue as to how to do them all at the same time.
|
We may solve this problem by finding the number of solutions with the following constraints
(1) $a\ge7 $ and $c\ge3$
(2) $a\ge7 $ and $c\ge42$
(3) $a\ge7 $, $b\ge35$ and $c\ge3$
(4) $a\ge7 $, $b\ge35$ and $c\ge42$
The answer to the problem is the answer of (1) - the answer of (2) - the answer of (3) + the answer of (4)
(1) Let $a'=a-7$ and $c'=c-3$. The equation is equivalent to $a'+b+c'+d+e=79-7-3 $, where the variables are nonnegative integers. The number of solutions is $\binom{79-7-3+4} {4 }=\binom {73}{4}$.
(2) Similar to (1), the number of solutions is $\binom{79-7-42+4} {4 }=\binom {34}{4}$.
(3) The number of solutions is $\binom{79-7-3-35+4} {4 }=\binom {38}{4}$.
(4) is impossible as $a+b+c>79$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to work out value of $a$ and $b$ in exponents? I am a student and I need a step by step solution in working let the value of $a$ and $b$ in this question.
$$1800 = 2^3 × 3^a × 5^b$$
I don't know what steps to take in order to work out what the powers are.
Thank you and help is appreciated.
|
Sweep through the prime numbers $2, 3, 5, 7, 9, 13, \dots $ as indicated by @projectilemotion.
First prime: $p=\boxed{2}$
$$
\frac{1800}{2^{3}} = 255
$$
Decomposition is now $1800 = 2^{3} \times ?$
Next prime: $p=\boxed{3}$
$$
\underbrace{\frac{225}{3} = 75}_{1}
\qquad \Rightarrow \qquad
\underbrace{\frac{75}{3} = 25}_{2}
\qquad \Rightarrow \qquad
\frac{25}{3} \notin \mathbb{Z}
$$
The prime was used $\color{blue}{2}$ times.
Decomposition is now $1800 = 2^{3} 3^{\color{blue}{2}} \times ?$
Next prime: $p=\boxed{5}$
$$
\underbrace{\frac{25}{5} = 5}_{1}
\qquad \Rightarrow \qquad
\underbrace{\frac{5}{5} = 1}_{2}
\qquad \Rightarrow \qquad
\frac{1}{5} \notin \mathbb{Z}
$$
The prime was used $\color{red}{2}$ times.
Final decomposition
$$
\boxed{1800 = 2^{3} \, 3^{\color{blue}{2}} \, 5^{\color{red}{2}}}
$$
Start with a list of primes:
$$
p =\left\{
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,
67, 71
\right\}
$$
If you were not given the $2^{3}$ at the beginning, you would start with $p=2$
$$
\underbrace{\frac{1800}{2} = 900}_{1}
\qquad \Rightarrow \qquad
\underbrace{\frac{900}{2} = 450}_{2}
\qquad \Rightarrow \qquad
\underbrace{\frac{450}{2} = 225}_{\color{green}{3}}
\qquad \Rightarrow \qquad
\frac{225}{2} \notin \mathbb{Z}
$$
The $\color{green}{3}$ successful divisions reveal $p^{\color{green}{3}}=2^{\color{green}{3}}$ is a factor.
|
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|
Question on a tricky Arithmo-Geometric Progression::
$$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty$$
This summation was irritating me from the start,I don't know how to attempt this ,tried unsuccessful attempts though.
|
Apply $(1-x)^{-2}=1+2x+3x^2+4x^3+\cdots\infty$.
Now,
$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty\\
=\dfrac{1}{4}\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+\cdots\infty\right)\\
=\dfrac{1}{4}\left(1-\dfrac{1}{2}\right)^{-2}\hspace{25pt}\text{ here }x=\dfrac{1}{2}.\\
=\dfrac{1}{4}\times4=1.$
|
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|
Infinite Sum Calculation: $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$ Problem
Show the following equivalence: $$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Good afternoon, dear StackExchange community. I'm studying real analysis (the topic right now is exchanging limits) and I can't wrap my head around this problem. It might be a duplicate, but I couldn't find a thread that helped me to solve this problem.
Also, I know that both terms equals $\frac{\pi^2}{8}$ (courtesy of Wolframalpha), but I think I should show the original problems via exchanging limits of double series or with analyising the summands, because we didn't introduce the identity $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi}{6}$ yet.
My Attempts
If we consider the first summands of both sums, we have:
\begin{array}{|c|c|c|c|}
\hline
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{(2k+1)^2} & 1 & \frac{1}{9} & \frac{1}{25} & \frac{1}{49} & \frac{1}{81} & \frac{1}{121} & \frac{1}{169} \\ \hline
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{n^2} & / & 1 & \frac{1}{4} & \frac{1}{9} & \frac{1}{16} & \frac{1}{25} & \frac{1}{36} \\ \hline
\end{array}
We immediately see that in $\frac{1}{n^2}$ is every summand of $\frac{1}{(2k+1)^2}$ is included. Additionally, we see that the extra summands in $\frac{1}{n^2}$ have the form $\frac{1}{(2x)^2}$.
Therefore, we could write,
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$$
and the original equation would be
$$\Rightarrow 1 + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \left( 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\right)$$.
But I don't know how this could help me.
Anyways, series and infinite sums give me headaches and I would appreciate any help or hints. Thank you in advance.
Wow! Thank you for the really quick replies!
|
Whoa, back up a bit!
$$\begin{align}\sum_{n=1}^{\infty}\frac{1}{n^2} &= 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\\&= \frac1{2(0)+1} + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\\&=\sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}\end{align}$$
Notice that $\frac1{(2k)^2}=\frac1{4k}$ so that
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac14\sum_{k=1}^{\infty} \frac{1}{k^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
Subtract $\frac14\sum_{k=1}^{\infty} \frac{1}{k^2}$ from both sides to get
$$\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=\frac34\sum_{k=1}^{\infty}\frac{1}{k^2}$$
|
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|
Differentiation - Power rule Find derivative of this function
$$g(t)=-3t(6t^4-1)^{4/3}$$
I have tried it till the answer:
$ -3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) $
$-3t x \frac{4}{3} (6t^4-1)^\frac{1}{3} ( 6X4 t^3 -0 ) $
$ -4t (6t^4 -1)^\frac{4}{3} (24t^3) $
$-96t(6t^4-1)^\frac{1}{3} $
However , I checked and saw that the answer is
$$-3(6t^4-1)^{1/3}(38t^4-1)$$
I just started learning differentiation, and I don't understand my mistake on why I can't achieve the answer. Thanks!
|
We can separate the $-3$ first:
$$g'(t)=-3(t(6t^4-1)^{4/3})'$$
Then apply the product rule and chain rule:
$$=-3(t'(6t^4-1)^{4/3}+t((6t^4-1)^{4/3})')$$
$$=-3((6t^4-1)^{4/3}+4/3\cdot t(6t^4-1)^{1/3}(6t^4-1)')$$
$$=-3((6t^4-1)^{4/3}+4/3\cdot t(6t^4-1)^{1/3}(24t^3))$$
Simplify:
$$=-3((6t^4-1)(6t^4-1)^{1/3}+4/3\cdot t(6t^4-1)^{1/3}(24t^3))$$
$$=-3((6t^4-1)^{1/3}(4/3\cdot t(24t^3)+6t^4-1))$$
$$=-3(6t^4-1)^{1/3}(32t^4+6t^4-1)$$
$$=-3(6t^4-1)^{1/3}(38t^4-1)$$
|
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|
Approximating arc length I have the integral expression for the arc length:
$$ \int \sqrt{1 + \frac{1}{4x}\,}\,\mathrm{d}x $$
and need to approximate the arc length of the curve: $y = $$ \sqrt{x}$
in between x = 0 and x = 4.
Should I be using the trapezoidal rule or Simpson's rule for this? And how can I begin to apply it?
|
The formula for arc length of $y = \sqrt{x}$ from $x = 0$ to $x=4$ is
$$ \int_{0}^{4} \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx$$
where here $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$, squaring this results in $\left( \frac{dy}{dx} \right)^2 = \frac{1}{4x}$. Hence our task is to integrate
$$\int_{0}^{4} \sqrt{1 + \frac{1}{4x}} dx$$
Let $$u = \sqrt{ 1 + \frac{1}{4x} } \Rightarrow u^2 = 1 + \frac{1}{4x} \Rightarrow 2udu = \frac{-1}{4x^2}dx$$
From that we also obtain that $x = \frac{1}{4(u^2 - 1)}$ so that $x^2 = \frac{1}{16(u^2 - 1)^2}$, then $\frac{-1}{4x^2} = -4(u^2-1)^2$, so that we get
$$dx = \frac{-u}{2(u^2 -1)^2}du$$
Hence, with our sub, we get the integral
$$\int_{x=0}^{x=4} \frac{-u^2}{2(u^2 - 1)^2} du$$
Now I would do another sub, letting $u = \sec(\theta)$ so that $du = \sec(\theta)\tan(\theta)d\theta$, and we get with Phythagoras' Theorem that
$$\int_{x=0}^{x=4} \dfrac{-\sec^2(\theta)}{2\tan^4(\theta)} \sec(\theta)\tan(\theta)d\theta = -\dfrac{1}{2} \int_{x=0}^{x=4} \dfrac{\sec^3(\theta)}{\tan^3(\theta)} d\theta = -\dfrac{1}{2} \int_{x=0}^{x=4} \csc^3(\theta)d\theta,$$
This can be solved via integration by parts and is definitely done in detail somewhere on the internet. You should arrive at the answer
$$= \dfrac{-1}{2} \left( \dfrac{-\csc(\theta)\cot(\theta) + ln|\csc(\theta) - \cot(\theta)|}{2} \right)$$
I will leave it to you to do the back substitutions and plug in the bounds. Also, I feel like I pulled out a sledge hammer to solve this, I'm sure someone will post a more elegant solution.
|
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|
If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$ THE PROBLEM: If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$
MY THOUGHT PROCESS: We have to prove that $ab+bc+ca=0$.
One method using which we can do this is, if we can somehow obtain the equation $k(ab+bc+ca)=0$ we can deduce that $ab+bc+ca=0$ using the zero product rule.
Another method to do this would be to obtain an expression which has $(ab+bc+ca)$ in it. The identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ has $2(ab+bc+ca)$ in it. Therefore if we can somehow show that $(a+b+c)^2=a^2+b^2+c^2$ our task would be over.
MY ATTEMPT: I have proved the result using the first approach. I tried to do it using the second one but could not proceed far. I was facing problems showing that $(a+b+c)^2=a^2+b^2+c^2$. If i get any help i shall be very grateful.
|
Using the addition formulas:
$$2\cos\left(\theta+\frac{2\pi}{3}\right)=-\cos\theta-\sqrt3\sin\theta$$
$$2\cos\left(\theta+\frac{4\pi}{3}\right)=-\cos\theta+\sqrt3\sin\theta$$
Then on the one hand
$$\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)=$$
$$\frac{\cos^2\theta-3\sin^2\theta}{4}+\frac{-\cos^2\theta+\sqrt3\sin\theta\cos\theta}{2}+\frac{-\cos^2\theta-\sqrt3\sin\theta\cos\theta}{2}=$$
$$\frac{\cos^2\theta-3\sin^2\theta}{4}-\cos^2\theta=\frac{\cos^2\theta-3\sin^2\theta-4\cos^2\theta}{4}-\cos^2\theta=\frac{-3\cos^2\theta-3\sin^2\theta}{4}=-\frac{3}{4}$$
On the other hand
$$\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)\right)^2=$$
$$\left(\frac{\cos^2\theta-3\sin^2\theta}{4}\right)^2+\left(\frac{-\cos^2\theta+\sqrt3\sin\theta\cos\theta}{2}\right)^2+\left(\frac{-\cos^2\theta-\sqrt3\sin\theta\cos\theta}{2}\right)^2=$$
$$\frac{\cos^4\theta-6\sin^2\cos^2+9\sin^4}{16}+\frac{\cos^4\theta-2\sqrt3\sin\theta\cos^3\theta+3\sin^2\theta\cos^2\theta}{4}+\frac{\cos^4\theta+2\sqrt3\sin\theta\cos^3\theta+3\sin^2\theta\cos^2\theta}{4}=$$
$$\frac{\cos^4\theta-6\sin^2\cos^2+9\sin^4}{16}+\frac{2\cos^4\theta+6\sin^2\theta\cos^2\theta}{4}=$$
$$\frac{\cos^4\theta-6\sin^2\cos^2+9\sin^4+8\cos^4\theta+24\sin^2\theta\cos^2\theta}{16}=$$
$$\frac{9}{16}(\cos^4\theta+2\sin^2\theta\cos^2\theta+\sin^4\theta)=\frac{9}{16}\left(\cos^2\theta+\sin^2\theta\right)^2=\frac{9}{16}.$$
If we square the first equation and subtract it from the second we get $0$.
Since
$$\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)(a+b+c)=$$
$$a\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)\right)$$
and
$$\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2(a^2+b^2+c^2)=$$
$$a^2\left(\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)\right)^2\right)$$
we conclude
$$(a+b+c)^2-(a^2+b^2+c^2)=0$$
and $$ab+bc+ca=0$$
|
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|
Prove: $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}$ Let $a,b,c$ be the lengths of the sides of triangle $ABC$ opposite $A,B,C$, respectively, and let $m_a,m_b,m_c$ be the lengths of the corresponding angle medians. Prove:
$$\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}.$$
Source: I thought about it while solving the American Mathematical Monthly's problem 11945, which asked to prove $\frac{a}{w_a}+\frac{b}{w_b}+\frac{c}{w_c}\ge 2\sqrt{3}$, where $w_a,w_b,w_c$ are respective angle bisectors.
I think it can be solved by making up (inequality-) constrained optimization problem, but I am interested in solution within elementary geometry.
|
Another way.
By the Ptolemy and C-S twice we obtain:
$$\sum_{cyc}\frac{a}{m_a}=\sqrt{\sum_{cyc}\left(\frac{a^2}{m_a^2}+\frac{2ab}{m_am_b}\right)}\geq\sqrt{\sum_{cyc}\left(\frac{a^2}{m_a^2}+\frac{2ab}{\frac{c^2}{2}+\frac{ab}{4}}\right)}=$$
$$=2\sqrt{\sum_{cyc}\left(\frac{a^2}{2b^2+2c^2-a^2}+\frac{2ab}{ab+2c^2}\right)}=2\sqrt{\sum_{cyc}\left(\frac{a^4}{2a^2b^2+2a^2c^2-a^4}+\frac{2a^2b^2}{a^2b^2+2c^2ab}\right)}\geq$$
$$\geq2\sqrt{\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^2b^2-a^4)}+\frac{2(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+2c^2ab)}}\geq2\sqrt{1+2}=2\sqrt3.$$
|
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|
Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \text{if} \ x<0 \\
\end{array}
\right.$$
For the first equation, we have
$$|x^2-2x|\Rightarrow\left\{\begin{array}{rcr}
x^2-2x & = & 1, \ \text{if} \ x^2\geq 2x \\
x^2-2x & = & -1, \ \text{if} \ x^2<2x \\
\end{array}
\right.$$
and for the second equation, we have
$$|x^2+2x|\Rightarrow\left\{\begin{array}{rcr}
x^2+2x & = & 1, \ \text{if} \ x^2+2x\geq 0 \\
x^2+2x & = & -1, \ \text{if} \ x^2+2x<0 \\
\end{array}
\right.$$
Solving for all of these equations, we get
$$\left\{\begin{array}{rcr}
x^2-2x & = 1 \Rightarrow& x_1=1+\sqrt{2} \ \ \text{and} \ \ x_2=1-\sqrt{2}\\
x^2-2x & =-1 \Rightarrow& x_3=1 \ \ \text{and} \ \ x_4=1\\
x^2+2x & = 1 \Rightarrow& x_5=-1-\sqrt{2} \ \ \text{and} \ \ x_6=-1+\sqrt{2}\\
x^2+2x & =-1 \Rightarrow& x_7=-1 \ \ \text{and} \ \ x_8=-1
\end{array}
\right.$$
So we have the roots $$\begin{array}{lcl}
x_1 = & 1+\sqrt{2} \\
x_2 = & 1-\sqrt{2} \\
x_3 = & -1+\sqrt{2} \\
x_4 = & -1-\sqrt{2} \\
x_5 = & 1 \\
x_6 = & -1
\end{array}$$
But according to the book, the answer is
\begin{array}{lcl}
x_1 & = & 1+\sqrt{2} \\
x_4 & = & -1-\sqrt{2} \\
x_5 & = & 1 \\
x_6 & = & -1
\end{array}
What happened to $x_2$ and $x_3$? Any other way to solve this equation quicker?
|
$x^{2}-2x = 1 $ ,if $x^{2}\geq 2x$ and $x\geq 0$ (because$\left |x^{2}-2x \right | = 1 $ if $x\geq 0$)
but $1-\sqrt{2} \leq 0$
$x_{3} =-1+\sqrt{2}$ is not correct solution because of same reason in second equation
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2280212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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|
find the value $\frac{49b^2+39bc+9c^2}{a^2}=52$ Let $a,b.c$ be real numbers such that
$$\begin{cases}a^2+ab+b^2=9\\
b^2+bc+c^2=52\\
c^2+ca+a^2=49
\end{cases}$$
show that
$$\dfrac{49b^2+39bc+9c^2}{a^2}=52$$
I have found this problem solution by geomtry methods.solution 1,can you someone have Algebra methods?
|
Hint:
$$\begin{cases}a^2+ab+b^2=x\\
b^2+bc+c^2=y\\
c^2+ca+a^2=z
\end{cases}$$
Then
$$\begin{cases}
(a-c)(a+b+c) = x-y\\
(b-a)(a+b+c) = y-z\\
(c-b)(a+b+c) = z-x\\
\end{cases}$$
Let $(a+b+c)^{-1}=k$, then
$$\begin{cases}
(a-c) = (x-y)k\\
(b-a) = (y-z)k\\
(c-b) = (z-x)k\\
\end{cases}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2281982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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|
Find $A^{50}$ for a $3\times 3$ matrix
If $$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$$ then $A^{50}$ is:
*
*$\begin{bmatrix}1&0&0\\25&1&0\\25&0&1\end{bmatrix}$
*$\begin{bmatrix}1&0&0\\50&1&0\\50&0&1\end{bmatrix}$
*$\begin{bmatrix}1&0&0\\48&1&0\\48&0&1\end{bmatrix}$
*$\begin{bmatrix}1&0&0\\24&1&0\\24&0&1\end{bmatrix}$
For $A$, the eigenvalues are $1, 1, -1$, but I don't know the procedure any further.
|
Sometimes a good way to start a problem like this is just to try a few multiplications and look for a pattern.
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$$=$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 0 & 1 \end{array} \right)$ $=A^2$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 0 & 1 \end{array} \right)$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$$=$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
2 & 0 & 1 \\
1 & 1 & 0 \end{array} \right)$ $=A^3$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
2 & 0 & 1 \\
1 & 1 & 0 \end{array} \right)$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$$=$
$\left( \begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
2 & 0 & 1 \end{array} \right)$ $=A^4$
Now the pattern should be clear enough to be able to solve the multiple-choice question. If not, do two more.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
My attempt,
As I know from the original equation,
$\alpha+\beta+\gamma=-a$
$\alpha\beta+\beta\gamma+\alpha\gamma=b$
$\alpha\beta\gamma=-c$
I've tried to expand $(\alpha+\beta+\gamma)^3$ which is equal to $\alpha^3+\beta^3+\gamma^3+3\alpha^2\beta+3\alpha\beta^2+3\alpha^2\gamma+6\alpha\beta\gamma+3\beta^2\gamma+3\gamma^2\alpha+3\gamma\beta$
Basically, I know I've to find what's the value of $\alpha^3+\beta^3+\gamma^3$, $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ and $\alpha^3\beta^3\gamma^3$. But I
m stuck at it. I would appreciate can someone explain and guide me to it. Thanks a lot.
By the way, I would appreciate if someone provides another tactics to solve this kind of routine question. Thanks a lot.
|
Let $p_k = \alpha^k + \beta^k + \gamma^k$. By Newton identities, we have
$$\begin{align}
p_1 + a = 0 &\implies p_1 = -a\\
p_2 + ap_1 + 2b = 0&\implies p_2 = a^2 - 2b\\
p_3 + ap_2 + bp_1 + 3c = 0&\implies p_3 = -a^3 + 3ab - 3c
\end{align}
$$
In particular, we have
$$\alpha^3 + \beta^3 + \gamma^3 = p_3 = -a^3 + 3ab - 3c$$
Apply Newton identities to the polynomial
$$x^3 + \frac{b}{c} x^2 + \frac{a}{c} x + \frac{1}{c}$$
which have roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$, we get
$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} =
-\frac{b^3}{c^3} + 3\frac{ba}{c^2} - \frac{3}{c}\tag{*1}$$
By Vieta's formula, we have $\alpha\beta\gamma = - c$. Multiply LHS of $(*1)$ by $\alpha^3\beta^3\gamma^3$ and RHS by the same number $-c^3$, we get
$$\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = 3c^2 - 3abc + b^3$$
As a result, the polynomial with roots $\alpha^3, \beta^3, \gamma^3$ is
( by Vieta's formula again)
$$x^3 - (\alpha^3+\beta^3+\gamma^3) x^2 + (\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3) - \alpha^3\beta^3\gamma^3\\
= x^3 + (a^3 -3ab + 3c)x^2 + (3c^2 - 3abc + b^3) x + c^3
$$
How to remember the Newton identifies
In general, given a polynomial of the form
$$P(x) = x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n$$
with roots $\lambda_1, \ldots, \lambda_n$. The sequence of numbers $p_k = \sum_{i=1}^n\lambda_i^n$ satisfies a bunch of identities.
$$\begin{cases}
p_k + a_1 p_{k-1} + a_2 p_{k-2} + \cdots + a_{k-1} p_1 + \color{red}{k} a_k = 0, & k \le n\\
p_k + a_1 p_{k-1} + a_2 p_{k-2} + \cdots + a_{n-1} p_{k-n+1} + a_n p_{k-n} = 0,
& k > n
\end{cases}$$
For any particular $k$, you can obtain the corresponding identity by multiplying $p_{k-\ell}$ term with $a_\ell$ and sum over all available $0 \le \ell \le n$.
When $k \le n$, you will have terms like $p_0$, $p_{-1}$,... Just replace all appearance of $p_0$ by $\color{red}{k}$ and forget all $p_\ell$ with negative $\ell$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Probability problem gives different answers with different methods
In a box there are $4$ red ball, $5$ blue balls and $6$ green balls.$4$ balls are picked at random. What is the probability the at least one ball is green and at least one ball red $?$
Method $1$:
Now,in how many ways can this be done $?$
$1.$ $1$ red,$1$ green,$2$ blue.
$2.$ $2$ red,$1$ green,$1$ blue.
$3.$ $1$ red,$2$ green,$1$ blue.
$4.$ $2$ red,$2$ green,$0$ blue.
$5.$ $3$ red,$1$ green,$0$ blue.
$6.$ $1$ red,$3$ green,$0$ blue.
So the all possible cases are obtained by adding these cases,i.e. ,
$${^6C_1} \times {^5C_2} \times {^4C_1} + {^6C_2} \times {^5C_1} \times {^4C_1} + {^6C_1} \times {^5C_1} \times {^4C_2} + {^6C_2} \times {^4C_2} + {^6C_3} \times {^4C_1} + {^6C_1} \times {^4C_3} \\= 914.$$
So the probability is ${914\over {({_{15} C_4})}}={914\over 1365}$
Method $2$:
Probability of at least $1$ red ball and at least $1$ green ball = $1$- Probability of no red or green ball, i.e. all balls are blue =$1-{5\over 1365} = {1360\over 1365}.$
Why these two methods give different answers? Are one or both of them wrong $?$ How?
|
Let $R$ be the number of reds and $G$ be the number of greens.
Then
\begin{align*}
P(R\geq 1, G\geq 1) &= 1-P(R=0 \cup G = 0) \\
&= 1-[P(R=0)+P(G=0)-P(R=0, G=0)]\\
&= 1 -\left[\frac{\binom{4}{0}\binom{11}{4}}{\binom{15}{4}}+\frac{\binom{6}{0}\binom{9}{4}}{\binom{15}{4}}-\frac{\binom{5}{4}\binom{10}{0}}{\binom{15}{4}}\right] \\&= \frac{914}{1365}\end{align*}
where the second line is true by inclusion-exclusion. This agrees with your first answer.
|
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|
Evaluating the integral $\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}}dx$ The integral to be evaluated is $$\int \frac{(x-1)}{(x+1) \sqrt{x^3+x^2+x}}dx$$
I split the integral to obtain
$$\int \frac{\sqrt x}{(x+1) \sqrt{x^2+x+1}}dx - \int \frac{1}{(x+1)\sqrt{x^3+x^2+x}}dx$$
But I could not proceed any further, as I am not able to find any substitution, nor could I find any further simplification.
I tried by multiplying the numerator and denominator with $\sqrt {x-1}$ so as to obtain $ x^3-1$ inside the radical, but that didn't help either.
How should I proceed to evaluate this integral?
|
Notice
$$\begin{align}
\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}dx
= &\int \frac{x-1}{(x+1)\sqrt{x+1+x^{-1}}}\frac{dx}{x}\\
= &\int \frac{x-1}{(x+1)\sqrt{x+1+x^{-1}}}\frac{d(x+1+x^{-1})}{x-x^{-1}}\\
= &\int \frac{d(x+1+x^{-1})}{(x+2+x^{-1})\sqrt{x+1+x^{-1}}}
\end{align}
$$
Let $u = \sqrt{x + 1 +x^{-1}}$, the indefinite integral evaluates to
$$\int \frac{du^2}{(1+u^2)u} = 2\int \frac{du}{1+u^2}
= 2\tan^{-1}(u) + \text{ const.}
= 2\tan^{-1}\sqrt{x+1+x^{-1}} + \text{ const.}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2286670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Matrix representation problem Consider the next linear operator:
$$T:M_{2\times2}(\mathbb{R})\longrightarrow M_{2\times2}(\mathbb{R})$$
such that $T\left(\begin{array}{cc}a&b\\c&d\end{array}\right)=\left(\begin{array}{cc}0&a+b\\0&0\end{array}\right)$. I need find the $Im(T)$ and the $Ker(T)$ but I don't how is the matrix representation of these linear operator, I've already calculate the images of the canonical vector basis:
$$T\left(\begin{array}{cc}1&0\\0&0\end{array}\right)=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$
$$T\left(\begin{array}{cc}0&1\\0&0\end{array}\right)=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$
$$T\left(\begin{array}{cc}0&0\\1&0\end{array}\right)=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)$$
$$T\left(\begin{array}{cc}0&0\\0&1\end{array}\right)=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)$$
should I use a block matrix for the representation of $T$? I need help please.
An example of my work, if use the netx representation (I don't know if is it right):
$$[T]_{\beta}=\left(\begin{array}{cc|cc}0&1&0&0\\0&0&0&0\\\hline0&1&0&0\\ 0&0&0&0\end{array}\right)$$
making elemental operations by rows I have:
$$[T]_{\beta}=\left(\begin{array}{cc|cc}0&1&0&0\\0&0&0&0\\\hline0&0&0&0\\ 0&0&0&0\end{array}\right)$$
then we obtain $(a,b,c,d)=(a,0,c,d)$, in matrix form:
$$\left(\begin{array}{cc}a&0\\c&d\end{array}\right)=a\left(\begin{array}{cc}1&0\\0&0\end{array}\right)+c\left(\begin{array}{cc}0&0\\1&0\end{array}\right)+d\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$$
|
We note that the vector space $M_{2\times2}(\mathbb{R})$ has dimension $4$, since we have the following basis for the vector space:
$$
\mathbf{v}_1 = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \qquad
\mathbf{v}_2 = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} \qquad
\mathbf{v}_3 = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} \qquad
\mathbf{v}_4 = \begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} $$
We have that the transformation $T$ works as follows:
\begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} &\xrightarrow[T]{}
\begin{pmatrix} 0 & a+b \\ 0 & 0 \end{pmatrix} \\
T(a \mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 + d\mathbf{v}_4) &= (a+b)\mathbf{v}_2 \\
A \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} &= \begin{bmatrix}0 \\ a+b
\\ 0 \\ 0\end{bmatrix} \\[8pt]
A &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
\end{align*}
This is the matrix representation of the linear operator $T$.
The kernel of $T$ is the set of all elements that get mapped to the identity element $\mathbf{0} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.
\begin{align*}
\ker T &= \left\{ \begin{pmatrix} a & -a \\ c & d \end{pmatrix} \bigg\rvert \,\,\, a,c,d \in \mathbb{R} \right\} \\
&= \left\{A = [a_{ij}] \in M_{2\times 2}(\mathbb{R}) \mid a_{11} + a_{12} = 0 \right\}
\end{align*}
The image of $T$ is the set of all elements that can result from the transformation $T$, which is:
\begin{align*}
\text{im } T &= \left\{ \begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix} \bigg\rvert \,\,\, b \in \mathbb{R} \right\} \\
&= \left\{A = [a_{ij}] \in M_{2\times 2}(\mathbb{R}) \mid a_{11} = a_{21} = a_{22} = 0 \right\}
\end{align*}
Another way to find the kernel and image of $T$ is to find the nullspace (kernel) and column space (image) of $A$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f,g,h,i\in \{0,1,2\}$
$$M^{T}M= \begin{pmatrix}
a & d & g\\
b & e & h\\
c & f & i
\end{pmatrix}\begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix} $$.
sum of diagonal entries $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2 = 5$$
How can I form different cases?
|
The numbers $a^2$ etc. are either $0$, $1$or $4$. To get them adding to $5$ you need five $1$s, or a $4$ and a $1$ and all the rest $0$s.
|
{
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"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
|
With algebraic identities, this is actually rather natural:
Remember if we had all powers of $x$ down to $x^0=1$, we could easily factor:
$$x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}=\frac{(x^3-1)(x^3+1)}{x-1}=(x^2+x+1)(x^3+1),$$ so we can write:
\begin{align}x^5+x+1&=(x^2+x+1)(x^3+1)-(x^4+x^3+x^2)\\
&=(x^2+x+1)(x^3+1)-x^2(x^2+x+1)\\
&=(x^2+x+1)(x^3+1-x^2).
\end{align}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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|
In Fig : what is $\angle DAC= ?$ In Fig : what is $$\angle DAC= ? $$
in triangle $\triangle ABC $ And $|AB|=|AC|$And Point $D$ in the triangle $\triangle ABC $
such that : $\angle BCD =40,\angle CBD =20$
My Try:
$$\frac{\sin 30}{AD}=\frac{\sin \angle BAD}{BD}$$
and :
$$\frac{\sin 10}{AD}=\frac{\sin \angle DAC}{DC}$$
So :
$$\sin \angle DAC =\frac{\sin 10}{\frac{\sin 30 \times BD}{\sin \angle BAD}}\times DC$$
$$\sin \angle DAC =\frac{\sin 10 \times \sin \angle BAD}{\sin 30 \times BD}\times DC$$
Now what?
|
Let $x = \angle DAB, y = \angle DAC$. Since sum of all angles of the outer triangle is $360^\circ$, so $x + y = 80$. You can get another equation using the sine rule for the sides $BD, AD, CD$;
$$ 1 = \frac{BD}{AD} . \frac{AD}{CD} . \frac{CD}{BD} = \frac{\sin x}{\sin 30} . \frac{\sin 10}{\sin y} . \frac{\sin 20}{\sin 40}; $$
so solving the two equations $$\frac{\sin x}{\sin y} = \frac{\sin 30 \sin 40}{\sin 10 \sin 20}$$ and $$x+y = 80$$ will give you the angle $y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2293913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find natural (arc length) parametrization of a curve. If a curve is defined like this:
$$
\rho = 1 - \cos(x)
$$
find natural parametrization of it.
So this is how the curve looks like when when I use polar coordinates $x = \rho \cos(t)$, $y = \rho\sin(t)$:
$$
\alpha(t) = \bigl((1 - \cos(t)) \cos(t), (1 - \cos(t)) \sin(t)\bigr).
$$
Now I found natural parameter to be $s = 4 \arcsin(\sqrt{s/8})$.
Can anyone tell me if this is good?
How does my curve look alike when I plug natural parameter in it?
Am I doing something wrong?
Sorry for the bad post. :/
|
The polar graph $\rho = f(\theta)$ parametrized by
$$
(x, y) = \bigl(f(t)\cos t, f(t) \sin t\bigr)
\tag{1}
$$
has speed
$$
\frac{ds}{dt} = \sqrt{f(t)^{2} + f'(t)^{2}}.
$$
For $f(t) = 1 - \cos t$, the speed is
$$
\sqrt{(1 - \cos t)^{2} + (\sin t)^{2}}
= \sqrt{2(1 - \cos t)}
= 2|\sin \tfrac{t}{2}|.
$$
If $0 \leq t \leq 2\pi$, then $\sin \frac{t}{2}$ is non-negative, so the arclength function is
$$
s = \int_{0}^{t} 2\sin \tfrac{u}{2}\, du
= 4(1 - \cos \tfrac{t}{2}),
$$
or $t = 2\arccos(1 - \frac{s}{4})$.
The double-angle formulas for the circular functions read
\begin{align*}
\cos(2\theta) &= \cos^{2}\theta - \sin^{2}\theta, \\
\sin(2\theta) &= 2\sin\theta \cos\theta,
\end{align*}
and $\cos(\arccos u) = u$ while $\sin(\arccos u) = \sqrt{1 - u^{2}}$.
Substituting $\theta = \arccos(1 - \frac{s}{4})$ gives
$$
\left.
\begin{aligned}
\cos t &= (1 - \tfrac{s}{4})^{2} - 1 + (1 - \tfrac{s}{4})^{2}
= 2(1 - \tfrac{s}{4})^{2} - 1, \\
\sin t &= 2(1 - \tfrac{s}{4}) \sqrt{1 - (1 - \tfrac{s}{4})^{2}}
= 2(1 - \tfrac{s}{4})\sqrt{\tfrac{s}{4}(2 - \tfrac{s}{4})}.
\end{aligned}
\right\}
\tag{2}
$$
The arc length parametrization is found by substituting (2) into (1). As a numerical check, the diagram shows the trig parametrization (red) overlaid by the putative arc length parametrization (blue).
|
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|
Manipulating a Summation Series
Summation: How does$$\begin{align*} & \frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\tag1\\ & =\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k\tag2\\ & =\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k=\sum\limits_{k=1}^n\frac 1{k+n}\tag3\end{align*}$$
Given that$$\sum\limits_{k=1}^n\frac 1{(2k)^3-2k}=\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k$$
I'm not sure how they got from the first step to the second. The $\tfrac 12\sum\limits_{k=1}^n\tfrac 1k$ didn't change, so that must mean that$$\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac n{2n+1}=\sum\limits_{k=1}^n\frac 1{2k-1}$$But I don't see how. I tried expanding the LHS, and rearranging to get the RHS, but that didn't go far.$$\begin{align*}\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}\right)+\frac n{2n+1} & =\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 1{2k+1}\right)+\frac n{2n+1}\\ & =\frac 12\sum\limits_{k=1}^n\frac {4k}{4k^2-1}+\frac n{2n+1}\\ & =2\sum\limits_{k=1}^n\frac k{(2k)^2-1}+\frac n{2n+1}\end{align*}$$I feel like I'm close, but I just can't seem to finish it. I believe that you'll need to use a summation identity.
Questions:
*
*How do you get from step $(1)$ to step $(2)$?
*How do you get from $(2)$ to $(3)$
*Is there a PDF that lists all the summation rules and identities?
|
*
*Note that
$$\sum_{k=1}^n\frac{1}{2k+1}=\sum_{k=2}^{n+1}\frac{1}{2k-1}=\sum_{k=1}^{n}\frac{1}{2k-1}-1+\frac{1}{2n+1}$$
Therefore,
\begin{align*}
\frac{1}{2}\sum_{k=1}^n\frac{1}{2k-1}+\frac{1}{2}\sum_{k=1}^n\frac{1}{2k+1}+\frac{n}{2n+1}&=\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}+\frac{1}{2(2n+1)}+\frac{n}{2n+1}\\
&=\sum_{k=1}^n\frac{1}{2k-1}
\end{align*}
*Note that
$$\sum_{k=1}^n\frac{1}{2k-1}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{2k}=\sum_{k=1}^{2n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}$$
Therefore,
\begin{align*}
\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\\
&=\sum_{k=n+1}^{2n}\frac{1}{k}\\
&=\sum_{k=1}^{n}\frac{1}{k+n}\\
\end{align*}
|
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|
Calculate $\sum_{n=1}^{\infty}{c_n\over n}$ when $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$ Let $(c_n)$ be defined as $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$. I want to get $\sum_{n=1}^{\infty}{c_n\over n}$. I would appreciate any suggestions how to start on this. It is my suspicion that I would have to resort to integration.
|
Break the sum into three pieces
\begin{eqnarray*}
S=\sum_{n=0} ^{\infty} \left( \frac{2}{3n+1} + \frac{1}{3n+2} +\frac{-3}{3n+3} \right) \\
\int_0^1 x^n dx = \frac{1}{n+1}
\end{eqnarray*}
Now use the above formula ... interchange the sum & integral & one has ...
\begin{eqnarray*}
S= \int_0^{1} \frac{2+x-3x^2}{1-x^3} dx =\int_0^{1} \frac{2+3x}{1+x+x^2} dx =\frac{\pi}{6 \sqrt{3}} + \frac{3}{2} \ln 3.
\end{eqnarray*}
|
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|
Fraction issue, I don't know why I'm having this result. I don't know how to get this result: $\frac{38}{17}$
This is the equation, can anyone explain why?
$(\frac{x}{4}) - (x - \frac{5}{6}) = (1 + \frac{2(x-5)}{3})$
I did this : LCM $= 12$.
Then: $$3(x) - 2(x-5) = 4(1 + 2(x-5)$$
$$3x - 2x + 10 = 4(1 + 2x - 10)$$
$$3x - 2x + 10 = 4 + 8x - 40$$
$$-7x = -46$$
$$\frac{(-46)}{-7}$$
|
$$ \frac{x}{4} - \left(x - \frac{5}{6}\right) = 1 + \frac{2(x-5)}{3} $$
First add the $x/4$ and $-x$ on the LHS and distribute the $2/3$ on the RHS,
$$ \frac{-3}{4}x + \frac{5}{6} = 1 + \frac{2}{3}x - \frac{10}{3} $$
Move all the constants to the LHS and $x$'s to the RHS,
$$ \frac{5}{6} - \frac{6}{6} + \frac{20}{6} = \frac{8}{12}x + \frac{9}{12}x$$
Simplify,
$$ \frac{19}{6} = \frac{17}{12} x $$
Multiply both sides by $\frac{12}{17}$,
$$ \frac{38}{17} = x $$
|
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|
Volume of Rotation Let $W_n$ be the region in an $xy$-plane that’s bounded by the $y$-axis and the parabola $x=y^2-n^2$.
Find the value of $n$ for which the volume obtained by rotating $W_n$ about the $y$-axis is precisely 16 times the volume obtained by rotating $W_n$ about the $x$-axis.
|
Lets start with the y-axis revolution.
We are integrating from $-n$ to $n$.
We get $2π\displaystyle \int_0^{n}(n^2-y^2)^2\,dy = \frac{16}{15}πn^5$
I put a $2$ in front for symmetry and ease in calculation.
Next, we do x-axis revolution. This one is a little bit tricky, we have to rewrite in terms of $x$.
$x=y^2-n^2$ turns into $y=\sqrt{x+n^2}$
No plus or minus because either one would produce the needed solid. Using shells would cover the volume of the solid TWICE.
We get $π\displaystyle \int_{-n^2}^{0}(x+n^2)\,dx = \frac{1}{2}πn^4$.
So we get $\displaystyle \frac{16}{15}πn^5=16 \times \frac{1}{2}πn^4$.
That means $\frac{16}{15}πn^5=8πn^4$.
We get $\frac{16}{15}n=8$.
Therefore, $\boxed{n=\displaystyle \frac{15}{2}}$.
|
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|
Determine the minimal polynomial of $\alpha$ over $\Bbb Q(\sqrt{-2})$ Let $\alpha$ be a root of the polynomial $x^4-4x^2+2.$
How should I determine the minimal polynomial of $\alpha$ over $\Bbb Q(\sqrt{-2})$?
|
Let $\alpha$ be a root of the polynomial $x^4-4x^2+2$
$$x^4-4x^2+2$$
$$\left(x^2 - \sqrt{2} -2\right) \left( x^2 + \sqrt{2} -2\right)$$
$$\left(\sqrt{2 - \sqrt2} - x\right) \left(\sqrt{2 + \sqrt2} - x\right) \left(\sqrt{2 - \sqrt2} + x\right) \left(\sqrt{2 + \sqrt2} + x\right)$$
Then
$$\alpha = \left(\sqrt{2 \pm \sqrt2} \pm x\right) $$
Where the polynomial $\alpha$ has no solutions in $\Bbb{Q}(i\sqrt2)$
|
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|
Understanding proof of period of sin(ax/b) + cos(cx/d) I've been trying to understand this proof for determining that the period of $\sin(\frac{ax}{b}) + \cos(\frac{cx}{d})$ where $a, b, c, d$, are integers and $a \over b$, $c \over d$ are in lowest terms is $\frac{2 \pi *lcm(b,d)}{gcf(a, c)}$. The proof goes like this:
*
*Since $\sin (\frac {ax}{b})$ repeats every $\frac{2b\pi}{a}$ and $\cos (\frac {cx}{d})$ repeats every $\frac{2d\pi}{c}$, both functions complete an integral number of periods from $x = 0$ to $x = z$ if $\frac{z}{\frac{2b\pi}{a}}$ and $\frac{z}{\frac{2d\pi}{c}}$ are integers.
*Since $\frac{az}{2b\pi}$ and $\frac {cz}{2d\pi}$ are both integers, the numerator of $z$ must be divisible by $2\pi$, $b$, and $d$. Thus the numerator of $z$ is at least $2\pi$: times the least common multiple of $b$ and $d$.
*Since $az$ and $cz$ must both be integers, the denominator of z can be no greater than the greatest common factor of a and c. Putting this together, we find the minimum $z$ when the numerator is minimized and the denominator is maximized, or $z = \frac{2 \pi *lcm(b,d)}{gcf(a, c)}$.
I completely understand point 1. For point 2, I understand why $\frac{az}{2b\pi}$ and $\frac {cz}{2d\pi}$ are both integers, but what I don't understand is how it reasons about the numerator of z. I see that $z$ as a whole should be divisible by $2 \pi$, $b$, and $d$ but how does one know that that means $2\pi \times lcm(b,d)$ and not $\pi \times lcm(2, b, d)$? Point 3 is the most confusing; is the conclusion that $az$ and $cz$ must be integers because they are numerators of a fraction in lowest terms? More confusing is the conclusion that the denominator must be the greatest common factor of $a$ and $c$. Does anyone know of another way to explain the steps in this proof?
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$\DeclareMathOperator{\lcm}{lcm}$It may help to work instead with the $1$-periodic functions
$$
C(x) = \cos(2\pi x),\qquad
S(x) = \sin(2\pi x),
$$
so that $S(\frac{a}{b}x) + C(\frac{c}{d}x)$ is $\ell$-periodic if and only if $\sin(\frac{a}{b}x) + \cos(\frac{c}{d}x)$ is $2\pi\ell$-periodic.
The question is, what is the (smallest positive) period $\ell$ of $S(\frac{a}{b}x) + C(\frac{c}{d}x)$? The steps of your proof become:
*
*Both $S(\frac{a}{b}x)$ and $C(\frac{c}{d}x)$ complete an integer number of periods for $0 \leq x \leq \ell$ if $m = \frac{a}{b}\ell$ and $n = \frac{c}{d}\ell$ are integers. Particularly, $\ell = \frac{mb}{a} = \frac{nd}{c}$ is rational. Write $\ell = \frac{p}{q}$.
*The numerator $p$ must be divisible by $b$ and by $d$, hence by $\lcm(b, d)$.
*Since $a\ell = mb$ and $c\ell = nd$ are integers, the denominator $q$ is no larger than $\gcd(a, c)$.
The smallest positive rational number satisfying these conditions is found by taking the smallest numerator divisible by $b$ and $d$ and the largest denominator dividing both $a$ and $c$, i.e., taking $p = \lcm(b, d)$, and $q = \gcd(a, c)$, so that $\ell = \frac{\lcm(b, d)}{\gcd(a, c)}$.
|
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|
Finding the orthogonal projection of $(2,1,3)$. $V=\mathbb R^3,u=(2,1,3) $ and $W$={$(x,y,z):x+3y-2z=0$}.
Vectors ortogonal to $W$=Span{$(1,3,-2)$},Vectors on $W$=Span{$(-1,1,1)$}.
So,orthogonal basis is {$(1,3,-2),(-1,1,1)$}.Then,orthonormal basis is
{$\frac{1}{\sqrt {14}}(1,3,-2),\frac{1}{\sqrt {3}}(-1,1,1)$}.
So orthogonal projection of
(2,1,3)=$<\frac{1}{\sqrt {14}}(1,3,-2),(2,1,0)>\frac{1}{\sqrt {14}}(1,3,-2)$+$<\frac{1}{\sqrt {3}}(-1,1,1),(2,1,0)>\frac{1}{\sqrt {3}}(-1,1,1)>$
=$\frac{-1}{14}(1,3,-2)+\frac{2}{3}(-1,1,1)=(\frac{-25}{42},\frac{34}{42},\frac{22}{42})$
But it's correct answer is = $\frac{1}{
14}
(29,17,40)$
Where i'm missing?
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If $W = \{(x,y,z): x+3y-2z=0\}$, then $x=2z-3y$, and every element of $W$ looks like
$$\left(\begin{array}{c} x \\ y \\ z \end{array}\right)
\ \ = \ \ \left(\begin{array}{c} 2z-3y \\ y \\ z \end{array}\right)
\ \ = \ \ y\left(\begin{array}{c} -3 \\ 1 \\ 0 \end{array}\right)
+z\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right)$$
This means that $W$ is spanned by $(-3,1,0)^{\top}$ and $(2,0,1)^{\top}$.
The vector $(-1,1,1)^{\top}$ lies in $W$ - you can check this:
$$\det\left(\begin{array}{ccc}
-1 & -3 & 2 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) = 0$$
but it's not enough on its own to span $W$ since $\dim W =2$.
|
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|
Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, taking into account the time zone). Here's one question that I've been stuck with:
With $a, b, c$ $\in \mathbb{R}^{+}$ such that $ab + bc + ca + abc =2$, find the max of
$ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$
I'm just curious what kind(s) of techniques these junior high students can use, since as far as I remember when I was at that age, calculus and derivatives aren't taught until high school.
|
If $a=b=c=\sqrt3-1$ then we get a value $\frac{3\sqrt3}{4}$.
We'll prove that it's a maximal value.
Indeed, let $x=\frac{a+1}{\sqrt3}$, $y=\frac{b+1}{\sqrt3}$ and $z=\frac{c+1}{\sqrt3}$.
Hence, the condition gives $x+y+z=3xyz$ and we need to prove that
$$\sum_{cyc}\frac{x}{3x^2+1}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\frac{x}{3x^2+\frac{3xyz}{x+y+z}}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\frac{x+y+z}{3(x+y)(x+z)}\leq\frac{3}{4}$$ or
$$9(x+y)(x+z)(y+z)\geq8(x+y+z)^2$$ and since
$$9(x+y)(x+z)(y+z)\geq8(x+y+z)(xy+xz+yz)$$ it's just
$$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that
$$xy+xz+yz\geq x+y+z$$ or
$$xy+xz+yz\geq(x+y+z)\sqrt{\frac{3xyz}{x+y+z}}$$ or
$$(xy+xz+yz)^2\geq3xyz(x+y+z)$$ or
$$\sum_{cyc}z^2(x-y)^2\geq0.$$
Done!
|
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|
Conversion of $x^2-2xy-y^2-x+y=0$ to $Ax^2-BY^2=C$ to solve diophantine equation. Given $x^2-2xy-y^2-x+y$ we see that $\Delta=(-2)^2-4(1)(-1)=4+4=8>0$, therefore it is a hyperbola, I was required to convert it to $AX^2-BY^2=C$, so first I removed the $xy$ term by noticing that rotation of $-\pi/8$ converted it to
$$\sqrt2x^2-\left(\frac{\sqrt{2+\sqrt2}+\sqrt{2-\sqrt2}}{2}\right)x-\sqrt2y^2+\left(\frac{\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2}}{2}\right)y=0$$
Also now I found the center of the conic by using $\partial F/\partial x=\partial F/\partial y =0$ which gave me
$$(x_0,y_0)=\left(\frac{\sqrt{2+\sqrt2}+\sqrt{2-\sqrt2}}{4\sqrt2},\frac{\sqrt{2+\sqrt2}+\sqrt{2-\sqrt2}}{4\sqrt2}\right)$$
Now I get by translation to origin by using $(x_0,y_0)$
$$x^2-y^2=\frac{\sqrt{2+\sqrt2}\sqrt{2-\sqrt2}}{8}=\frac{\sqrt2}{8}$$
But the RHS is not an integer, therefore how should we solve the diophantine equation $x^2-2xy-y^2-x+y$, since I saw that we needed to convert it to Pell type equation $X^2-DY^2=N$. Should I square both sides?
|
Discriminant of the quadratic (in terms of $x$ for example) must be a perfect square, thus $$(2y+1)^2-4(-y^2+y)=z^2$$or $$8y^2+1=z^2$$Its a Pell equation now.
|
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|
Infinite zeros in infinite series The problem:
Given that
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$
My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} -\frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
\frac{\pi}{12} & = \frac{1}{3} - \frac{1}{9} + \frac{1}{15} - \frac{1}{21} + \frac{1}{27} -\frac{1}{33} + \frac{1}{39} - \frac{1}{45} + \ldots\\
\\
& = 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + 0 + 0 - \frac{1}{21}
\end{align}
$$
now add them together:
$$
\begin{align}
\frac{\pi}{4} + \frac{\pi}{12} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
& + 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + \ldots \\
\end{align}
$$
and we will get:
$$
\begin{align}
\frac{\pi}{3} & = 1 + 0 + \frac{1}{5} - \frac{1}{7} + 0 -\frac{1}{11} + \frac{1}{13} + 0 + \ldots \\
& = 1 + \frac{1}{5} - \frac{1}{7} -\frac{1}{11} + \frac{1}{13} + \ldots
\end{align}
$$
My questions:
*
*I inserted/removed infinite zeros into/from the series, is that OK?
*My solution relies on the fact that $\Sigma a_n + \Sigma b_n = \Sigma (a_n + b_n)$ and $k \Sigma a_n = \Sigma k a_n$. Is this always true for convergent infinite series? If so, why is it? (yeah I know this is a stupid question, but since I'm adding infinite terms up, I'd better pay some attention.)
*Bouns question: Can I arbitrarily (arbitrariness isn't infinity, you know) insert/remove zeros into/from a convergent infinite series, without changing its convergence value?
|
It is a rewarding exercise to write, or read, detailed rigorous proofs of some simple, "obvious" results.
The definition of $x=\sum_{j=1}^{\infty}y_j$ is that $x=\lim_{n\to \infty}S_n$ where $S_n=\sum_{j=1}^ny_j.$ A useful way to state (or define) that the sequence $S=(S_n)_n$ converges to $x$ is that for any $r>0 $ the set $$F(S, r)=\{n: S_n\not \in [-r+x,r+x]\}$$ is a finite set.
Let $x=\sum_{j=1}^{\infty}y_j.$ Insert some $0$'s into the sequence $(y_j)_j$ to produce a new sequence $(z_i)_i.$ For each $i$ we have either $z_i=0$ or $z_i=y_{g(i)}$ where $g(i)\leq i.$
Let $T=(T_m)_m$ where $T_m=\sum_{i=1}^mz_i.$
Let $i_0$ be the least (or any) $i$ such that $z_i=y_1.$
Then for every $m\geq i_0$ we have $T_m=\sum_{ \{g(i):i\leq m\}}y_{g(i)}= S_{f(m)}$ where $f(m)=\max \{g(i):i\leq m\}.$
Note that for any $n$ there exists $m_0$ such that $z_{m_0}=y_{n+1}$ so there exists $m_0$ such that $f(m_0)=n+1.$
Now for any $r>0$ we have $$F(T,r)=\{m:T_m\not \in [-r+x,r+x]\}\subset \{m\geq i_0:S_{f(m)} \not \in [-r+x.r+x]\}\cup \{m:m<i_0\}$$ $$=\{m\geq i_0: f(m)\in F(S,r)\}\cup \{m:m<i_0\}.$$ Observe that for any $n$ the set $\{m\geq i_0: f(m)=n\}$ is a finite set, because there exists $m_0$ such that $f(m_0)=n+1,$ and hence $m\geq m_0\implies f(m)>n.$ And recall that $F(S,r)$ is a finite set because the sequence $S$ converges to $x.$
So for any $r>0$ the set $F(T,r)$ is a subset of a union of a finite collection of finite sets: $$F(T,r)\subset (\cup_{n\in F(S,r)}\{m\geq i_0:f(m)=n\})\cup \{m:m<i_0\}.$$ Therefore $F(T,r)$ is finite for every $r>0$. Therefore $T$ converges to $x.$ Therefore $x=\sum_{i=1}^{\infty}z_i.$
|
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|
If $p \equiv 1 \pmod{4}$, then $p \nmid a^2+3b^2$? Suppose that $a, b$ are co-prime numbers and $p$ is a prime number in the form $4k+1$. Can $3a^2+b^2$ have a prime divisor like $p$?
I really don't know how to think about this! Please help!
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Suppose we have $$a^2\equiv -3b^2\pmod p$$
Now, since $\gcd(a,b)=1$ we can exclude the case where one, hence both, of $a,b$ are divisible by $p$. Thus we can divide by $b^2$ to see that $-3$ must be a square $\pmod p$.
Assume that $p\equiv 1 \pmod 4$. Quadratic reciprocity then tells us that $-3$ is a square $\pmod p$ if and only if $p\equiv 1\pmod 3$. Thus we require $p\equiv 1 \pmod {12}$.
It is easy to work backwards. That is, given $a$ such that $a^2\equiv -3 \pmod p$ then $a^2+3\times 1^2$ is divisible by $p$.
Conclusion: if $p\equiv 1 \pmod {12}$ then we can find coprime $a,b$ with $a^2+3b^2\equiv 1 \pmod {p}$.
Worth noting: the same technique can applied to the case $p\equiv -1 \pmod 4$. Again we require that $=3$ be a square $\pmod p$ but again reciprocity forces us to require that $p\equiv 1 \pmod 3$ Thus we are lead to considering $p\equiv 7 \pmod {12}$
Conclusion: if an odd prime $p$ divides $a^2+3b^2$ with coprime $a,b$ then $p\equiv 1,7\pmod {12}$. Conversely, any such prime divides an expression of the desired form.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2321157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
need help for proof of $\sum_{i=1}^{n} \left(3 \cdot (\frac{1}{2})^{i-1} +2 \, i \right)$ So I'm asked to start with:
$$\sum_{i=1}^{n} \left(3 \cdot (\frac{1}{2})^{i-1} +2 \, i \right)$$
My notes say to start by taking out the 3, use i-1 and make it multiply by $\frac{1}{2}$, and then transforming the last part by taking out the 2:
$$3 \cdot \sum_{i=1}^{n} (\frac{1}{2})^{i-1} \cdot \frac{1}{2} +2\sum_{i=1}^{n} i$$
then I got:
$$\frac{3}{2}\sum_{i=1}^{n} (\frac{1}{2})^{i-1} + 2 \times \frac{n(n+1)}2$$
then:
$$\frac{3}{2} \times \frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} + n \times ( n + 1 ) $$
evaluates to:
$$\frac{3}{2} \times 2 \times \left(1-(\frac{1}{2})^n\right)$$
finally:
$$3[1-(\frac{1}{2})^n] + n (n+1)$$
Can someone step me through this? I don't understand going from the initial problem to the 1st step, pulling out the 3 and i-1.
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hint
$$S (x)=\sum_{i=1}^n3x^{i-1}=3\frac {x^n-1}{x-1} $$
$$\sum_{i=1}^n 2i=2 (1+2+3+... n)=2\frac {n (n+1)}{2} $$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2321815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integrate the expression $\int\frac{x}{1+\sin x}\ \text dx$
Integrate the expression $\int\frac{x}{1+\sin x}\ \text dx$
I really do not see any identity working here. I tried rewriting in terms of $\cos x$ and got $$\int\frac{\frac{x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}\ \text dx$$
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$$I = \int\frac{x}{1+\sin x}\, dx$$
Multiply by the conjugate of $1 + \sin x$
$$\int\frac{x}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}\, dx = $$
$$\int\frac{x}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}\, dx = \int\frac{x-x\sin x}{1-\sin^2 x}\, dx$$
$$\int\frac{x-x\sin x}{\cos^2 x}\, dx$$
$$\int\frac{x}{\cos^2 x}\, dx- \int\frac{x\sin x}{\cos^2 x}\, dx$$
Apply Integration by Parts $$\int f(x)g(x)'\,dx = f(x)g(x) - \int f'(x)g(x)\, dx$$
where
$$\int\frac{x}{\cos^2 x}\, dx$$
$f(x) = x,\implies f'(x) = 1$
$g(x) = \tan x \implies g'(x) = \frac{1}{\cos^2x}$
$$x\tan(x) - \int\tan(x)$$
$$x\tan(x) + \ln(\cos x) +C\tag1$$
And
$$\int\frac{x\sin x}{\cos^2 x}\, dx$$
$f(x) = x\sin x,\implies f'(x) = \sin x + x\cos x$
$g(x) = \tan x \implies g'(x) = \frac{1}{\cos^2x}\tan x$
$$x\sin x \tan x - \int(\sin x + x\cos x)\tan x\, dx$$
$$x\sin x \tan x - \int \sin x\tan x + x\cos x\tan x\, dx$$
$$x\sin x \tan x - \int x\sin x + \sin x\tan x\, dx$$
$$x\sin x \tan x - \left(\int x\sin x \, dx + \int \sin x\tan x\, dx\right)$$
$$x\sin x \tan x - \left(-x\cos x + \sin x + \int \sin x\tan x\, dx\right)$$
$\int\sin x\tan x\, dx = \int \frac{\sin^2 x}{\cos x}\, dx = \int \frac{1 - \cos^2 x}{\cos x}\, dx = \int \frac{1}{\cos x} - \cos x \, dx = \int \frac{1}{\cos x}\, dx - \int \cos x \, dx =$
$ \ln\left(\tan(x) + \sec(x)\right) - \sin(x) +C$
$$x\sin x \tan x - \left( -x\cos x + \sin x + \ln\left(\tan(x) + \sec(x)\right) - \sin(x)\right) + C$$
$$x\sin x \tan x + x\cos x - \ln\left(\tan(x) + \sec(x)\right) + C\tag2$$
Also...
$$I = \big(x\tan(x) + \ln(\cos x)\big) + \big(x\sin x \tan x + x\cos x - \ln\left(\tan(x) + \sec(x)\right)\big) + C$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2322303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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|
Limits problem $ \lim_{x\to \pi^+/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$ $$ \lim_{x\to \pi^+/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$
I tried to solve this by L'hostipal's rule..but that doesn't give a solution..appreciate if you can give a clue..
$ \lim_{x\to \pi^+/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$
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\begin{align}
\lim_{x\to \frac{\pi^+}{2}} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}&= \lim_{x\to \frac{\pi^+}{2}} \frac{\sqrt{2\cos^2x}}{\sqrt{\pi}-\sqrt{2x}}\\
&= \lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}\cos x}{\sqrt{\pi}-\sqrt{2x}}\\
&= \lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}\cos x(\sqrt{\pi}+\sqrt{2x})}{\pi-2x}\\
&= \lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}\sin (\frac{\pi}{2}-x)(\sqrt{\pi}+\sqrt{2x})}{2(\frac{\pi}{2}-x)}\\
&= \lim_{x\to \frac{\pi^+}{2}} \frac{\sin (\frac{\pi}{2}-x)}{(\frac{\pi}{2}-x)}\lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}(\sqrt{\pi}+\sqrt{2x})}{2}\\
&= (1)\left[\frac{-\sqrt{2}(\sqrt{\pi}+\sqrt{\pi})}{2}\right]\\
&=-\sqrt{2\pi}
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2323198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find extremas of $f(x,y) = xy \ln(x^2+y^2), x>0, y>0$ As the title says I need to find extreme values(maximum and minimum) of
$$f(x,y) = xy \ln(x^2+y^2), x>0, y>0$$
I don't understand how to find critical points of this problem.
I start with finding partial derivative and set derivatives equal to zero. And that is where I am stuck currently. So any help would be appreciated.
So:
$
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy* ln(x^2+y^2)) = y*ln(x^2+y^2) + \frac{2xy^2}{x^2+y^2}
$
$
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy* ln(x^2+y^2)) = x*ln(x^2+y^2) + \frac{2x^2y}{x^2+y^2}
$
So we have now:
$
\nabla f(x,y) = (0,0)
$
$
y*ln(x^2+y^2) + \frac{2xy^2}{(x^2+y^2)^2} = 0
$
And
$
x*ln(x^2+y^2) + \frac{2x^2y}{(x^2+y^2)^2} = 0
$
After trying to solve these equations I get that $x=y$ Is that correct?. So I don't understand what are then critical points as It can't be (0,0)?
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It's easier to work with polar coordinates $x = r \cos \theta, y = r \sin \theta$ where (in your case) $r > 0$ and $0 < \theta < \frac{\pi}{2}$. Then we have
$$ f(x,y) = xy \ln(x^2 + y^2) = r \cos \theta r \sin \theta \ln(r^2) = r^2 \ln r \sin(2 \theta) = g(r,\theta). $$
The partial derivatives of $g$ are given by
$$ \frac{\partial g}{\partial r} = r (2 \ln r + 1) \sin(2 \theta), \\
\frac{\partial g}{\partial \theta} = 2 r^2 \ln r \cos(2 \theta). $$
Since $r > 0$ and $\sin 2\theta > 0$ in our range, the equation $\frac{\partial g}{\partial r} = 0$ implies that $2 \ln r + 1 = 0$ or $r = e^{-\frac{1}{2}}$. The second equation then implies that $\cos 2 \theta = 0$ so $\theta = \frac{\pi}{4}$. Hence, we have one critical point at $r = e^{-\frac{1}{2}}$ and $\theta = \frac{\pi}{4}$. The second partial derivatives of $g$ are given by
$$ \frac{\partial^2 g}{\partial r^2} = (2 \ln r + 3) \sin(2 \theta), \\
\frac{\partial^2 g}{\partial r \partial \theta} = 2 r(2 \ln r + 1) \cos(2 \theta), \\
\frac{\partial^2 g}{\partial \theta} = -4 r^2 \ln r \sin(2 \theta). $$
Hence, the Hessian at $r = e^{-\frac{1}{2}}, \theta = \frac{\pi}{4}$ is given by
$$ \begin{pmatrix} 2 & 0 \\ 0 & 2e^{-1} \end{pmatrix}. $$
Since the Hessian is positive definite, $(e^{-\frac{1}{2}}, \frac{\pi}{4})$ a local minimum point for $g$ (and so $\frac{e^{-\frac{1}{2}}}{\sqrt{2}} \left( 1, 1 \right)$ is a local minimum for $f$).
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2324079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$ Here's the questions
There are two equations:
$a\cos x+b\cos y-c=0 $ and $ a\sin x+b\sin y-d=0$ .
For instance
What is the value of $x$ and $y$ in following question?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
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just a hint
From the first equation
$$2\cos (x)=2-3\cos (y) $$
and from the second
$$2\sin (x)=2-3\sin (y) $$
thus
$$(2\sin (x))^2+(2\cos (x))^2=$$
$$4=8-12\cos(y )-12\sin(y)+9$$
$$\cos(y)+\sin (y)=\frac {13}{12}$$
$$\sqrt {2}\cos (y-\frac {\pi}{4})=\frac {13}{12}$$
You can finish.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2324467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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|
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.
LHS $= 2\cos^{2}3x-1$
$=2{(4\cos^{3}x-3\cos{x})}^2-1$
$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$
$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$
Still thank you for the answers!
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Note that you can use the identity $\cos^2 x+\sin^2 x=1$(or alternatively, $\sin^2 x=1-\cos^2 x$).
Take each side of the equation to the power of $2,3$ to get $$9 \sin^2 x=9-9\cos^2 x$$
$$24\sin^4 x=24-48 \cos x^2+24 \cos^4 x$$
$$16\sin^6 x=16-48 \cos^2 x+48 \cos^4 x-16\cos^6 x$$
Tedious, but it probably will work.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2324685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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|
If $b>a,$ find the minimum value of $|(x-a)^3|+|(x-b)^3|,x\in R$ If $b>a,$ find the minimum value of $|(x-a)^3|+|(x-b)^3|,x\in R$
Let $f(x)=|(x-a)^3|+|(x-b)^3|$
When $x>b,f(x)=(x-a)^3+(x-b)^3$
When $a<x<b,f(x)=(x-a)^3-(x-b)^3$
When $x<a,f(x)=-(x-a)^3-(x-b)^3$
I am stuck here.The answer given in my book is $\frac{(b-a)^3}{4}.$
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Let $a<x<b$.
Hence, since $f(x)=x^3$ is a convex function, by Jensen we obtain:
$$|(x-a)^3|+|(x-b)^3|=(x-a)^3+(b-x)^3\geq2\left(\frac{x-a+b-x}{2}\right)^3=\frac{(b-a)^3}{4}$$
Let $x\geq b$. Hence,
$$|(x-a)^3|+|(x-b)^3=(x-a)^3+(x-b)^3\geq(b-a)^3>\frac{(b-a)^3}{4}.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2326375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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.