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Calculus Question: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$ Can anyone help me to find $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$? Any help would be appreciated. Thanks in advance.
|
Rewrite
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\ dx.
\end{align}
$$
Let $\,u=\cos x$, then $\,du=-\sin x\,dx$. For $\,0 < x < \frac{\pi}{2}$, we have $\,0 < u < 1$. Now, the integral turns out to be
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\,dx&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sqrt{1-\cos^2 x})\,dx\\
&=-\frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du.\tag1\\
\end{align}
$$
Next, use Maclaurin series for natural logarithm:
$$
\ln(1-u^2)=-\sum_{n=1}^\infty \frac{u^{2n}}{n}.\tag2\\
$$
Substitute $\,(2)$ to $\,(1)$, yield
$$
\begin{align}
\frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du&=-\frac{1}{2}\int_0^1\sum_{n=1}^\infty \frac{u^{2n}}{un}\,du\\
&=-\frac{1}{2}\sum_{n=1}^\infty\int_0^1 \frac{u^{2n-1}}{n}\,du\\
&=-\frac{1}{4}\sum_{n=1}^\infty \left.\frac{u^{2n}}{n^2}\right|_{u=0}^1\\
&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}.\tag3
\end{align}
$$
The infinite series in $(3)$ is defined as Riemann zeta function $\,\zeta (2)=\dfrac{\pi^2}{6}$. Thus,
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}\\
&=-\frac{1}{4}\cdot \frac{\pi^2}{6}\\
&=\large\color{blue}{-\frac{\pi^2}{24}}.\\
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/792950",
"timestamp": "2023-03-29T00:00:00",
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|
Limit of term with roots How can I prove the following:
$$\lim_{k\rightarrow \infty} \frac{2}{3 (\sqrt{k+1} - \sqrt{k-1})(\sqrt{k+2} + \sqrt{k})} = \frac{1}{3}$$
|
We multiply by the conjugate of $\sqrt{k+1} - \sqrt{k-1}$ and we use the asymptotic limit
$$\sqrt{k+a}\sim_\infty \sqrt k$$
we find
$$\lim_{k\rightarrow \infty} \frac{2}{3 (\sqrt{k+1} - \sqrt{k-1})(\sqrt{k+2} + \sqrt{k})} =\lim_{k\rightarrow \infty}\frac23\frac{2\sqrt k}{4\sqrt k}=\frac13$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function - What is the coefficient of $x^{26}$? What is the coefficient of $x^{26}$ given the function:
$${\left[ {{{1 + {x^{10}}} \over {{{(1 - x)}^7}}}} \right]^2}$$
My work:
$${\left[ {{{1 + {x^{10}}} \over {{{(1 - x)}^7}}}} \right]^2} = \left( {1 + 2{x^{10}} + {x^{20}}} \right){1 \over {{{(1 - x)}^{14}}}} = \left( {1 + 2{x^{10}} + {x^{20}}} \right)\sum\limits_{n = 0}^\infty {\left( {\matrix{
{n + 13} \cr
{13} \cr
} } \right)} \cdot {x^n}$$
Recitation's notes work:
$$... = \left( {{x^{26}} + 2{x^{16}} + {x^6}} \right)\sum\limits_{n = 0}^\infty {\left( {\matrix{
{n + 13} \cr
{13} \cr
} } \right)} \cdot {x^n}$$
I don't understand why multiplying by $x^{26}$. What's the reason?
|
You are perfectly right. Using the notation $[x^n] f(x)$ for "the coefficient of $x^n$ in (the series expansion for) $f(x)$" we can write:
\begin{align}
[x^{26}] \left( \frac{1 + x^{10}}{(1 - x)^7} \right)^2
&= [x^{26}] (1 + x^{10})^2 \cdot (1 - x)^{-14} \\
&= [x^{26}] (1 + 2 x^{10} + x^{20})
\cdot \sum_{n \ge 0} (-1)^n \binom{-14}{n} x^n \\
&= [x^{26}] (1 + 2 x^{10} + x^{20})
\cdot \sum_{n \ge 0} \binom{13 + n}{13} x^n \\
&= ([x^{26}] + 2 [x^{16}] + [x^6]) \sum_{n \ge 0} \binom{13 + n}{13} x^n \\
&= \binom{39}{13} + 2 \binom{29}{13} + \binom{19}{13}
\end{align}
Looks like a typo of some sort to me (perhaps forgot the $[]$ for "coefficient of" in the above derivation?).
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the maximum of $\|Ax\|$ for all unit vectors $x \in\mathbb R^2$? $A = \left[\begin{matrix} 1 & 0 \\ 3 & -1 \\ \end{matrix} \right]$
How to find the maximum of $\|Ax\|$ for all unit vectors $x \in\mathbb R^2$?
My professor told me the answer is: $\sqrt{\frac{11+ \sqrt{117} }{2}}$
I know this has something to do with the characteristic polynomial of $A$, which is $X^2-11x+1$.
|
Let $x = (a,b)$, then $||Ax|| = ||(a, 3a - b)|| = \sqrt{a^2 + (3a - b)^2}$.
Consider: $f(a,b) = a^2 + (3a - b)^2 = a^2 + 9a^2 - 6ab + b^2 = 10a^2 - 6ab + b^2$ subject to: $g(a,b) = a^2 + b^2 = 1$. Using Lagrange Multiplier method:
$\nabla f = \lambda\cdot \nabla g$:
$(20a - 6b, 2b - 6a) = (2a\lambda, 2b\lambda)$. So:
$20a - 6b = 2a\lambda$
$2b - 6a = 2b\lambda$.
If $\lambda = 0$, then: $b = 3a$, and $10a = 3b \to a = b = 0$, this can't happen.
So: $\lambda \neq 0$, then if $a = 0$ or $b = 0$, then both $a = b = 0$, and this can't happen. So $a \neq 0$, and also $b \neq 0$. Then:
$\dfrac{10a - 3b}{b - 3a} = \dfrac{a}{b}$
So: $10ab - 3b^2 = ab - 3a^2 \to -a^2 + b^2 = 3ab \to 1 - t^2 = 3t$, with $t = \dfrac{a}{b}$. So:
$t^2 + 3t - 1 = 0 \to t = \dfrac{-3 \pm \sqrt{13}}{2}$. Together with $a^2 + b^2 = 1$, you can solve for $a$, and $b$, and then find $\sqrt{f_{max}}$
|
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|
Matrix equality has a certain solution I am wondering about the following matrix equality
$$ \begin{pmatrix} 0 \\ 1 & \lambda_{1} \\ & 1 & \lambda_{2} \\ && \ddots & \ddots \\
&&& 1 & \lambda_{k} \\ &&&& 1 & 0 \\
&&&&& \ddots & \ddots \\
&&&&&& 1 & 0 \end{pmatrix} \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \\ \vdots \\ c_{k+1} \\ c_{k+2} \\ \vdots \\ c_{k+1 + \ell}\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\0 \\ \vdots \\ 0 \\d_{1} \\ \vdots \\ d_{\ell}\\ \end{pmatrix}.
$$
with $\lambda_{i} \neq 0$ for $1 \leq i \leq k$. Is there a way to show that this equation has a solution for the $c_{i}$ and $d_{i}$ such that $c_{1} \neq 0$?
|
My own answer:
Let $c_{1} = 1$, then the equation can be rewritten to
$$
\left( \begin{array}{ccccc|cc} \lambda_{1} &&&&&& \\ 1&\lambda_{2}&&&&& \\ &1&\lambda_{2} &&&& \\ && \ddots& \ddots & & & & \\ &&&1 & \lambda_{k} \\ \hline &&&& 1 & 0 \\ &&&&& \ddots & \ddots \\ &&&&&&1 & 0\end{array} \right) \begin{pmatrix} c_{2} \\ c_{3} \\ c_{4} \\ \vdots \\ c_{k+1} \\ \hline c_{k+2} \\ \vdots \\ c_{k+1+\ell} \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ \hline d_{1} \\ \vdots \\ d_{\ell} \end{pmatrix}
$$
The matrix on the upper left is invertible on the first $k$ entries. Since in the second part everything is free to choose, there is always a solution. This completes the proof.
|
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|
How do I rewrite vectors in other basis' given change of coordinate matrices? $\displaystyle β= \begin{bmatrix}2\\2\\\end{bmatrix}$,$\displaystyle \begin{bmatrix}4\\-1\\\end{bmatrix}$
$\displaystyle C= \begin{bmatrix}1\\3\\\end{bmatrix}$,$\displaystyle \begin{bmatrix}-1\\-1\\\end{bmatrix}$
From B to C Change of coordinate matrix is:\begin{bmatrix}1 & 3/4 \\ -1 & -13/4\end{bmatrix}
From C to B Change of coordinate matrix is:\begin{bmatrix}13/10 & 3/10 \\ -2/5 & -2/5\end{bmatrix}
Find $\displaystyle\begin{bmatrix}2\\2\\\end{bmatrix}_C$and $\displaystyle \begin{bmatrix}1\\3\\\end{bmatrix}_β$
|
Let the basis vectors for $B$ be $\mathbf{b}_1$ and $\mathbf{b}_2$. Then one can write the $\beta$ vectors as
$$
\begin{align}
\boldsymbol{\beta}_1 &= 2\,\mathbf{b}_1 + 2\,\mathbf{b}_2 = \beta_{11}^b\mathbf{b}_1 + \beta_{12}^b\,\mathbf{b}_2 \\
\boldsymbol{\beta}_2 &= 4\,\mathbf{b}_1 - \,\mathbf{b}_2 = \beta_{21}^b\mathbf{b}_1 + \beta_{22}^b\,\mathbf{b}_2
\end{align}
$$
Similarly, if the basis vectors for $C$ are $\mathbf{c}_1$ and $\mathbf{c}_2$, we have
$$
\begin{align}
\mathbf{C}_1 &= \,\mathbf{c}_1 + 3\,\mathbf{c}_2 = C_{11}^c\mathbf{c}_1 + C_{12}^c\,\mathbf{c}_2\\
\mathbf{C}_2 &= -\mathbf{c}_1 -\,\mathbf{c}_2 = C_{21}^c\mathbf{c}_1 +C_{22}^c\,\mathbf{c}_2 \,.
\end{align}
$$
We can express $\boldsymbol{\beta}_1$ and $\boldsymbol{\beta}_2$ in the $\mathbf{c}$-basis as
$$
\boldsymbol{\beta}_1 = \beta_{11}^c\,\mathbf{c}_1 + \beta_{12}^c\,\mathbf{c}_2 ~,~~
\boldsymbol{\beta}_2 = \beta_{21}^c\,\mathbf{c}_1 + \beta_{22}^c \,\mathbf{c}_2 \,.
$$
Similarly, for $\mathbf{C}_1$ and $\mathbf{C}_2$,
$$
\mathbf{C}_1 = C_{11}^b\,\mathbf{b}_1 + C_{12}^c\,\mathbf{b}_2 ~,~~
\mathbf{C}_2 = C_{21}^b\,\mathbf{b}_1 + C_{22}^c \,\mathbf{b}_2 \,.
$$
The coordinate transformation matrices are
$$
\begin{bmatrix} \beta_{11}^c \\ \beta_{12}^c \end{bmatrix}
= \begin{bmatrix}
1 & 3/4 \\ -1 & -13/4
\end{bmatrix}
\begin{bmatrix} \beta_{11}^b \\ \beta_{12}^b\end{bmatrix}
$$
and
$$
\begin{bmatrix} C_{11}^b \\ C_{12}^b \end{bmatrix}
= \begin{bmatrix}
13/10 & 3/10 \\ -2/5 & -2/5
\end{bmatrix}
\begin{bmatrix} C_{11}^c \\ C_{12}^c \end{bmatrix}
$$
The required results follow directly from the transformation rules.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do you evaluate this limit? $\ln({x^3+2x^2+x})+ \frac{2}{x}$ How do you evaluate the following?
$$\lim_{x \to 0^+} \left [ \ln({x^3+2x^2+x})+ \frac{2}{x} \right ]$$
If I plug in $x$, I get $\infty-\infty$, which is undetermined, and I haven't been able to get the limit at a more manageable form. Can you please help me?
|
Lets put $x = 1/y$ so that $y \to \infty$ as $x \to 0^{+}$. Then we can see that
\begin{align}
f(x) &= \log(x^{3} + 2x^{2} + x) + \frac{2}{x}\notag\\
&= \log\left(e^{2y}\left(\frac{1}{y^{3}} + \frac{2}{y^{2}} + \frac{1}{y}\right)\right)\notag\\
&> \log\left(e^{2y}\left(\frac{1}{y^{3}} + \frac{2}{y^{3}} + \frac{1}{y^{3}}\right)\right)\notag\\
&= \log\left(\frac{4e^{2y}}{y^{3}}\right)\tag{1}
\end{align}
Next we can see that $e^{t} \geq 1 + t > t$ for all $t$. And therefore $(e^{t})^{4} > t^{4}$. Putting $t = y/2$ we can see that $$e^{2y} > \frac{y^{4}}{16}\tag{2}$$ It follows from $(1)$ and $(2)$ that $$f(x) > \log\left(\frac{y}{4}\right)$$ Since $y \to \infty$ it follows that $\log(y/4) \to \infty$ and hence $f(x) \to \infty$ as $x \to 0^{+}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial fraction integral Question:
$\int \dfrac{5 }{(x+1) (x^2 + 4) } dx $
Thought process:
I'm treating it as a partial fraction since it certainly looks like one.
I cannot seem to solve it besides looking at it in the "partial fraction" way.
My work:
1) Focus on the fraction part first ignoring the $\int $ and $dx$ for the moment.
Multiply $(x+1) (x^2+4)$ on both sides of the equation and get:
$5$ = $\dfrac{A }{(x+1) } $ + $\dfrac{Bx+C }{(x^2+4) } $
Note: the x^2 + 4 is irreducible which explains the Bx+c as the numerator.
$ 5 = A(x^2+ 4) + Bx + C(x+1) $
I tried x = -1 which knocks out C:
$5 = A5 + - B $
I also tried x = 0 which knocks out B
$5 = A4 + C$
A is a lot harder to knock out since the squared changes the picked value to be positive.
I decided to add the two found equation together and get
$10 = A9 - B + C $
Now I am officially stuck at this step.
|
Split the fraction.
$$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{P(x)}{(x+1)}+\dfrac{Q(x)}{(x^2+4)}}$$
The largest partial fraction's denominator is an order 2 polynomial, so the order of the partial fractions' numerators must become order 2 when cross multiplied with the other partial fractions' denominators. Ie: P(X) is order 0, Q(X) is order 1.
$$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{A}{(x+1)}+\dfrac{Bx+C}{(x^2+4)}}$$
Now that you know what the target looks like, perform the crossmultiplication.
$$\color{blue}{5 = A(x^2+4)+(Bx+C)(x+1)}$$
Expand the polynomial so we can compare the coefficients.
$$\color{blue}{5 = (A+B)x^2 + (B+C)x + (4A+C)}$$
Which allows us to conclude
$$\color{blue}{A=-B, C=-B, 5= 4A+C \implies B=-1, A=1, C=1}$$
So we have our partial fraction expansion.
$$\color{blue}{\therefore \dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{1}{(x+1)}+\dfrac{(x+1)}{(x^2+4)}}$$
Next use: $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x+a) = \dfrac{1}{x+a}}$, $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x^2+b^2)= 2\dfrac{x}{x^2+b^2}}$ and $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan \dfrac{x}{c} = \dfrac{c}{x^2+c^2}}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the two points where the shortest distance occurs on two lines Find the point P on $\vec{AB}$ and point Q on $\vec{CD}$ such that $\vec{PQ}$ is the shortest distance
between the lines AB and CD, given $\vec{AB} = \begin{pmatrix}
1\\
0\\
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}
,\vec{CD} = \begin{pmatrix}
0\\
1\\
1\\
\end{pmatrix}
+ v\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix}
$ the normal vector $n=\begin{pmatrix}
-3\\
-2\\
-2\\
\end{pmatrix}
$
and the shortest distance is $\frac{3}{\sqrt{17}}$. I figured all this out from 4 given points but don't know how to find points P and Q. Please help, I'm stuck...
|
Once you know that the normal vector is $n$, the vector equation
$$
\begin{pmatrix}
1\\
0\\
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}
+w\begin{pmatrix}
-3\\
-2\\
-2\\
\end{pmatrix}
= \begin{pmatrix}
0\\
1\\
1\\
\end{pmatrix}
+ v\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix}
$$
is equivalent to a system of three linear equations in three unknowns, which indeed has the unique solution
$$
u=\frac{19}{17},v=-\frac{15}{17},w=\frac{3}{17}.
$$
That tells you that the points on $\vec{AB}$ and $\vec{CD}$ are
$$
\begin{pmatrix}
1\\
0\\
2\\
\end{pmatrix}
+ \frac{19}{17}\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}
\quad\text{and}\quad\begin{pmatrix}
0\\
1\\
1\\
\end{pmatrix}
-\frac{15}{17}\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix},
$$
respectively. (It also tells you that the distance between the two lines is the norm of $\frac3{17}(-3\ {-2}\ {-2})$, or $\frac3{\sqrt{17}}$ as you indicated.)
|
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|
Zero of Polynomial If $(a-b)$, $a$, and $(a+b)$ are the zeroes of the polynomial $x^3-3x^2+x+1$ then what are the values of $a$ and $b$?
I have taken $f(x)=x^3-3x^2+x+1$ and equated $f(a-b)$, $f(a)$ and $f(a+b)$ to zero. But I could not reduce the equations.
|
So you got the equations
*
*$a^3 - 3a^2 b + 3ab^2 - b^3 - 3a^2 + 6ab - 6b^2 + a - b + 1 = 0$
*$a^3 - 3a^2 + a + 1 = 0$
*$a^3 + 3a^2 b + 3a b^2 + b^3 - 3a^2 - 6ab - 6b^2 + a + b + 1 = 0$
The challenge is how to systematically simplify these equations. It is too much to try and solve for one variable to plug into the other equations: instead, you want to combine the equations in ways that eliminate the more complicated terms.
I've written the monomials in order of their total degree, and arbitrarily broke ties by putting the highest powers in $a$ first. An approach is to try and eliminate the terms as far to the left when possible when we can.
For example, we can eliminate $a^3$ from the first and third equation by subtracting off the second:
*
*$-3a^2 b + 3ab^2 - b^3 + 6ab - 6b^2 - b = 0$
*$a^3 - 3a^2 + a + 1 = 0$
*$3a^2 b + 3a b^2 + b^3 - 6ab - 6b^2 + b = 0$
and so forth. Another thing we can do is to factor equations when we can:
*
*$b(-3a^2 + 3ab - b^2 + 6a - 6b - 1) = 0$
*$a^3 - 3a^2 + a + 1 = 0$
*$b(3a^2 + 3a b + b^2 - 6a - 6b + 1) = 0$
at which point we can make a significant simplification by splitting into two parts: in one of the two parts we add the equation $b=0$, and in the other part we insist on $b \neq 0$, and can divide $b$ out of the equations.
By repeating these steps, you can fairly quickly come upon something you can solve in the ordinary way.
|
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|
Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\frac{n^2}2) + (\frac{n}2) + n^2 + n$$
$$ =n^4 + 3n^3 + 8n^2 + 3n$$
Why does what I did above not work?
|
Required formulas:
$$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}2\right)^2$$
$$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$$
$$\sum_{i=1}^n i=\frac{n(n+1)}2$$
I will start off at the third step.
$$\sum_{i=1}^n i^3+3\sum_{i=1}^n i^2+2\sum_{i=1}^n i$$
$$=\left(\frac{n(n+1)}{2}\right)^2+3\cdot \frac{n(n+1)(2n+1)}{6}+2\cdot \frac{n(n+1)}{2}$$
$$=\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{2}+n(n+1)$$
$$=\frac{n^2(n^2+2n+1)}{4}+\frac{2n(2n^2+3n+1)}{4}+\frac{4n^2+4n}{4}$$
$$=\frac{n^4+2n^3+n^2+4n^3+6n^2+2n+4n^2+4n}{4}$$
$$=\frac{n^4+6n^3+11n^2+6n}{4}$$
$$\color{green}{\sum_{i=1}^n i(i+1)(i+2)=\frac{n^4+6n^3+11n^2+6n}{4}}$$
|
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|
How would you find $\tan(3\cdot\arctan(x))$? How would you find $\tan(3\cdot\tan^{-1}(x))$ as quickly as possible?
I don't really understand how to use $\tan(2x)$ or any other identity for that matter to solve this.
|
Begin with $$\tan(2.\arctan(x))=\frac{2\tan(\arctan(x))}{1-\tan^2(\arctan(x))}=\frac{2x}{1-x^2}.$$
Next,
$$\begin{align}\tan(3.\arctan(x))&=\tan(2\arctan(x)+\arctan(x))\\&=\frac{\tan(2\arctan(x))+\tan(\arctan(x))}{1-\tan(2\arctan(x))\tan(\arctan(x))}\\&=\frac{\frac{2x}{1-x^2}+x}{1-\frac{2x}{1-x^2}x}\\&=\frac{3x-x^3}{1-3x^2}\end{align}$$
A more general solution is obtained combining the binomial and Euler's formulas:
$$(1+i\tan(\theta))^3=(1+ix)^3=1+3ix-3x^2-ix^3=(\frac{\cos\theta+i\sin\theta}{cos\theta})^3=\frac{\cos3\theta+i\sin3\theta}{\cos^3\theta},$$
so that, by equating the ratios of imaginary parts over real parts,
$$\tan(3\arctan x)=\tan(3\theta)=\frac{3x-x^3}{1-3x^2}.$$
Similarly, $(1+ix)^4=1+4ix-6x^2-4ix^3+x^4$ gives
$$\tan(4\arctan(x))=\frac{4x-4x^3}{1-6x^2+x^4},$$
$(1+ix)^5=1+5ix-10x^2-10ix^3+5x^4+ix^5,$
$$\tan(5\arctan(x))=\frac{5x-10x^3+x^5}{1-10x^2+5x^4},$$
...
|
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"timestamp": "2023-03-29T00:00:00",
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|
Series Question: $\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$$
I tried to use partial fraction
$$\begin{align}\frac{n^2}{(4n^2-1)^3}&=\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}+\frac{1}{64(2n+1)^2}+\frac{1}{64(2n-1)^2}\\&-\frac{1}{32(2n+1)^3}+\frac{1}{32(2n-1)^3}\end{align}$$
I can compute
$$\sum_{n=1}^{\infty}\left[\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}\right]=-\frac{1}{64}$$
using telescoping series, but I cannot compute the rest. I believe there's a better way than this. Any help would be appreciated. Thanks in advance.
|
From where you left off...:
$\displaystyle \sum_{n=1}^\infty \dfrac{1}{32}\cdot \left(\dfrac{1}{(2n-1)^3} - \dfrac{1}{(2n+1)^3}\right) = \dfrac{1}{32}\cdot \left(1 - \dfrac{1}{3^3} + \dfrac{1}{3^3} - \dfrac{1}{5^3} + ... \right) = \dfrac{1}{32}$, and
$\displaystyle \sum_{n=1}^\infty \dfrac{1}{64}\cdot \left(\dfrac{1}{(2n-1)^2} + \dfrac{1}{(2n+1)^2}\right) = \dfrac{1}{32}\cdot \displaystyle \sum_{n=1}^\infty \dfrac{1}{(2n-1)^2} - \dfrac{1}{64} = \dfrac{1}{32}\cdot \dfrac{\pi^2}{8} - \dfrac{1}{64}$.
Thus adding these values and your partial answer to complete the calculation.
|
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|
Integral $\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\frac{17\pi^4}{360}$ Hi I am trying to integrate $$
\mathcal{I}:=\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\int_0^1\int_0^1 \frac{\log(1-x)\log(1-y)}{1-xy}dx \,dy
$$
A closed form does exist. I tried to write
\begin{align}
\mathcal{I} &=\int_0^1 \log(1-y)dy\int_0^1 \log(1-x)\frac{dx}{1-xy} \\
&= \int_0^1\log(1-y)dy \ \int_0^1 \sum_{n\geq0}(xy)^n\, \ln(1-x) \ dx \\
&= \sum_{n\geq 0}\frac{1}{n+1}\int_0^1 \log(1-y) y^n\, dy \\
&= \sum_{n\geq 0}\frac{1}{n+1}\int_0^1 \log(1-y)y^n\, dy = ?
\end{align}
I was able to realize that
$$
\mathcal{I}=\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2=\frac{17\zeta_4}{4}=\frac{17\pi^4}{360},\qquad \zeta_4=\sum_{n\geq 1} n^{-4}
$$
however this does not help me solve the problem. How can we calculate $\mathcal{I}$? Thanks.
|
Consider the double integral
\begin{align}
I = \int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy
\end{align}
which, upon expansion into series form, becomes
\begin{align}
I &= \sum_{n=0}^{\infty} \ \int_{0}^{1} x^{n} \ln(1-x) dx \ \int_{0}^{1} y^{n} \ln(1-y) dy \\
&= \sum_{n=0}^{\infty} \left( \int_{0}^{1} x^{n} \ \ln(1-x) \ dx \right)^{2}.
\end{align}
Without much difficulty it can be shown that
\begin{align}
\int_{0}^{1} x^{n} \ \ln(1-x) \ dx &= \left. \partial_{\mu} B(n+1, \mu+1) \right|_{\mu = 1} \\
&= B(n+1,1) \left( \psi(1) - \psi(n+2) \right) \\
&= \frac{H_{n+1}}{n+1}.
\end{align}
Now, returning to the primary integral, it is seen that
\begin{align}
I = \sum_{n=0}^{\infty} \left( \frac{H_{n+1}}{n+1}\right)^{2} = \sum_{n=1}^{\infty} \left( \frac{H_{n}}{n} \right)^{2} = \frac{17 \zeta(4)}{4}.
\end{align}
Hence
\begin{align}
\int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy = \frac{17 \zeta(4)}{4}.
\end{align}
|
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|
writing a number as a sum of odd integers How many ways are there of writing $n$ as a sum of odd integers, where the order doesn't matter?
For example, there are $2$ ways of writing $3$: $(1,1,1)$ and $(3)$.
|
A list of the first 10 cases to study:
1- 1 (1)
2- 1 (1,1)
3- 2 (1,1,1);(3)
4- 2 (1,1,1,1);(3,1)
5- 3 (1,1,1,1,1);(3,1,1);(5)
6- 4 (1,1,1,1,1,1);(3,1,1,1);(3,3);(5,1)
7- 5 (1,1,1,1,1,1,1);(3,1,1,1,1);(3,3,1);(5,1,1);(7)
8- 6 (1,1,1,1,1,1,1,1);(3,1,1,1,1,1);(3,3,1,1);(5,1,1,1);(5,3);(7,1)
9- 8 (1,1,1,1,1,1,1,1,1);(3,1,1,1,1,1,1);(3,3,1,1,1);(3,3,3);(5,1,1,1,1);(5,3,1);(7,1,1);(9)
10- 10 (1,1,1,1,1,1,1,1,1,1);(3,1,1,1,1,1,1,1);(3,3,1,1,1,1);(3,3,3,1);(5,1,1,1,1,1);(5,3,1,1);(5,5);(7,1,1,1);(7,3);(9,1)
One pattern I have noticed is that when n=${2^x}+1$, the first difference of the combination amounts goes up by 1. For example, starting when n=$5$, the combination number goes up steadily by ones. Then when n=$9$, it begins to go up steadily by twos.
|
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|
Showing that $3x^2+2x\sin(x) + x^2\cos(x) > 0$ for all $x\neq 0$ I got this question:
Show that for all $x\neq 0$, $3x^2+2x\sin(x) + x^2\cos(x) > 0$
I tried to show it but got stuck.
|
If $x > 0$, then: $f(x) = x(3x + 2\sin x + x\cos x)$. We need to prove:
$3x + 2\sin x + x\cos x > 0$ when $x > 0$.
$3x + 2\sin x + x\cos x \geq 3x + 2\sin x - x = 2(x + \sin x) > 0$ because $x + \sin x > 0$ as $(x + \sin x)' = 1 + \cos x \geq 0$. So: $x + \sin x > 0 + \sin0 = 0$.
If $x < 0$ we again need to show: $3x + 2\sin x + x\cos x < 0$. But:
$3x + 2\sin x + x\cos x \leq 3x + 2\sin x - x = 2(x + \sin x) < 0$ by the same argument as above. Done!
|
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|
How to find measurements of angle CBD?
I got $x + CBD = 110^{o}$. I think I should find measurements of angle CBD, so how to find it?
Do you have any ideas? Please help me!
|
*
*$\angle ACB = 180^\circ-(10^\circ+70^\circ)-(60^\circ+20^\circ) = 20^\circ$ and $\angle AEB = 180^\circ-60^\circ-(50^\circ+30^\circ) = 40^\circ$.
*Draw a line from point $E$ parallel to $AB$, labeling the intersection with $AC$ as a new point $F$ and conclude $\Delta CEF\sim\Delta ABC$,
\begin{align}
\angle CEF &= \angle CBA = 50^\circ+30^\circ = 80^\circ\\
\angle FEB &= 180^\circ-80^\circ = 100^\circ\\
\angle AEF &= 100^\circ-40^\circ = 60^\circ\\
\angle CFE &= CAB = 60^\circ+20^\circ = 80^\circ\\
\angle EFA &= 180^\circ-80^\circ = 100^\circ
\end{align}
*Draw a line $FB$ labeling the intersection with $AE$ as a new point $G$ and conclude $\Delta AFE\cong\Delta BEF$,
\begin{align}
\angle AFB &=\angle BEA = 40^\circ\\
\angle BFE &= \angle AEF = 60^\circ\\
\angle FGE &= 180^\circ-60^\circ-60^\circ = 60^\circ = \angle AGB\\
\angle ABG &= 180^\circ-60^\circ-60^\circ = 60^\circ\\
\end{align}
*Draw a line $DG$. Since $AD=AB$ (leg of isosceles) and $AG=AB$ (leg of equilateral), conclude $AD = AG$, $\Delta DAG$ is isosceles and
$$
\angle ADG =\angle AGD = \frac{180^\circ-20^\circ}2 = 80^\circ.
$$
*Since $\angle DGF = 180^\circ-80^\circ-60^\circ = 40^\circ$, conclude $\Delta FDG$ (with two $40^\circ$ angles) is isosceles, so $DF = DG$.
*With $EF = EG$ (legs of equilateral) and $DE = DE$ (same line segment) conclude $\Delta DEF\cong\Delta DEG$ by side-side-side rule, $\angle DEF =\angle DEG = x$, and
$$
\angle FEG = 60^\circ = x+x\quad\Rightarrow\quad \large\color{blue}{x=30^\circ}.
$$
|
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|
Small part of loop invariant i'm not getting Show that $y = \frac{1}{2} \cdot z \cdot (z + 1)$ is a loop invariant for the following WHILE loop:
y := 0;
z := 0
WHILE $\neg$(z = x) DO
z := z + 1;
y := y + z
In other words, proof that
$\{y = \frac{1}{2} \cdot z \cdot (z + 1)\}$
z := z + 1;
y := y + z
$ \{y = \frac{1}{2} \cdot z \cdot (z + 1)\}$
Used for this are the Hoare axioms:
H1 {[$t$\x]$\varphi$} x := $t$ {$\varphi$}, provided that $t$ is free for $x$ in $\varphi$
H2 from {$\varphi$} $\pi_1$ {$\psi$} and {$\psi$} $\pi_2$ {$\chi$} follows {$\varphi$} ($\pi_1;\pi_2$) {$\chi$}
H3 from {$\varphi \wedge \epsilon$} $\pi_1$ {$\psi$} and {$\varphi \wedge \neg \epsilon$} $\pi_2$ {$\psi$} follows {$\varphi$} IF $\epsilon$ THEN $\pi_1$ ELSE $\pi_2$ {$\psi$}
H4 from {$\varphi \wedge \epsilon$} $\pi$ {$\varphi$} follows {$\varphi$} WHILE $\epsilon$ DO $\pi$ {$\neg \epsilon \wedge \varphi$}
H5 if $\varphi \rightarrow \varphi`$ and $\psi \rightarrow \psi`$ are true in M = (D, I), then: from {$\varphi`$} $\pi$ {$\psi`$} follows {$\varphi$} $\pi$ {$\psi$}
1 {$y + z = \frac{1}{2} \cdot z \cdot (z + 1)$} y := y + z {$y = \frac{1}{2} \cdot z \cdot (z + 1)$} H1
2 {$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot ((z + 1) + 1)$} z := z + 1 {$y + z = \frac{1}{2} \cdot z \cdot (z + 1)$} H1
3 {$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$}
z := z + 1
{$y + z = \frac{1}{2} \cdot z \cdot (z + 1)$} H5, 2
4 {$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$}
z := z + 1; y := y + z
{$y = \frac{1}{2} \cdot z \cdot (z + 1)$} H2, 3, 1
5 {$y = \frac{1}{2} \cdot z \cdot (z + 1)$}
z := z + 1; y := y + z
{$y = \frac{1}{2} \cdot z \cdot (z + 1)$} H5, 4*
*This application of H5 is correct because:
$y + (z + 1)$
$= \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$= \frac{1}{2} \cdot z \cdot (z + 1) + \frac{1}{2} \cdot 2 \cdot (z + 1)$
$= \frac{1}{2} \cdot z \cdot (z + 1) + (z + 1)$
So the precondition in stap 3 is equivalent with $y = \frac{1}{2} \cdot z \cdot (z + 1)$.
It's this last part I'm not getting at all; how is the precondition {$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$} equivalent with $y = \frac{1}{2} \cdot z \cdot (z + 1)$? What happens in this here equation that shows us that this is correct?
$y + (z + 1)$
$= \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$= \frac{1}{2} \cdot z \cdot (z + 1) + \frac{1}{2} \cdot 2 \cdot (z + 1)$
$= \frac{1}{2} \cdot z \cdot (z + 1) + (z + 1)$
|
I'm not sure to understand your question ...
If we have :
$y = \frac{1}{2} \cdot z \cdot (z + 1)$
then, adding $(z+1)$ to both sides, we get :
$y +(z + 1) = \frac{1}{2} \cdot z \cdot (z + 1) + (z+1) = \frac{1}{2}[ z \cdot (z + 1) + 2 \cdot (z+1)] = \frac{1}{2}[ (z + 2) \cdot (z+1)] $;
it's all ...
In your calculation you have :
$y + (z + 1)$
$= \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$= \frac{1}{2} \cdot z \cdot (z + 1) + \frac{1}{2} \cdot 2 \cdot (z + 1)$
$= \frac{1}{2} \cdot z \cdot (z + 1) + (z + 1)$
and it's the same thing.
Having derived that :
$y + (z + 1) = \frac{1}{2} \cdot z \cdot (z + 1) + (z + 1)$
you "cancel" $(z+1)$ form both "sides" to get :
$y = \frac{1}{2} \cdot z \cdot (z + 1)$.
|
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|
How do I derive the formula for the sum of squares? I was going over the problem of finding the number of squares in a chessboard, and got the idea that it might be the sum of squares from $1$ to $n$. Then I searched on the internet on how to calculate the sum of squares easily and found the below equation:$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}.$$
And then I searched for an idea on how to come up with this equation and found this link, but what I would like to know is that if (hypothetically)I have to be the first person in the world to come up with this equation, then can someone please tell some ideas on how to approach a problem like this.
|
Set $S = \sum_{i=1}^{n}i^2$, and noting that
\begin{align}
S = 1^2 + 2^2 + \ldots + (n-1)^2 + n^2 \\
= 1 + 2 + 3 + \ldots + (n-1) + n \\
+2 + 3 + \ldots + (n-1) + n \\
+3 + \ldots + (n-1) + n \\
\ldots\\
(n-1) + n \\
+ n
\end{align}
Where there are n lines in that second sum. From this we get
\begin{align}
S &= \sum_{i=1}^n\sum_{j=i}^nj \\
&= \sum_{i=1}^n\left(\sum_{j=1}^nj - \sum_{j=1}^{i-1}j\right)\\
&= \sum_{i=1}^n\left(\frac{n(n+1)}{2} - \frac{i(i-1)}{2}\right)\\
&= \sum_{i=1}^n\frac{(-i + n+1)(i+n)}{2} \\
&= \sum_{i=1}^n (\frac i2 - \frac{i^2}{2} + \frac{n}{2} + \frac{n^2}{2}) \\
&= \frac{n^2}{2} + \frac{n^3}{2} + \frac{1}{2}\sum_{i=0}^{n}i - \frac{1}{2}\sum_{i=1}^{n}i^2 \\
&= \frac{n^2}{2} + \frac{n^3}{2} + \frac{n(1+n)}{4} - \frac{S}{2}
\end{align}
Moving the $S$ to the LHS and simplifying gives the equation you're looking for.
\begin{align}
S = \frac{(n^2 + n)(2n+1)}{6}
\end{align}
|
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|
Skanavi 2.003, Difference in answers, which is right? In Skanavi book i have a exercise to simplify an equation
$$\begin{align}
((\sqrt[4]{p} - \sqrt[4]{q})^{-2} + (\sqrt[4]{p} + \sqrt[4]{q})^{-2}) : \frac{\sqrt p + \sqrt q}{p-q}
\end{align}$$
Solving:
Replacing:
$$ a =\sqrt[4]{p} $$
$$ b=\sqrt[4]{q} $$
Equation is changed:
$$
((a - b)^{-2} + (a + b)^{-2}): \frac{a^2+b^2}{a^4-b^4}
$$
$$ A = ((a - b)^{-2} + (a + b)^{-2})$$
$$ B = \frac{a^2+b^2}{a^4-b^4}$$
$$ B = \frac{a^2+b^2}{a^4-b^4} => B = \frac{a^2+b^2}{(a^2-b^2)(a^2+b^2)} => B = \frac{1}{a^2 - b^2}$$
$$ A = ((a - b)^{-2} + (a + b)^{-2}) => A = \frac{1}{(a - b)^{2}} + \frac{1}{(a + b)^{2}} => A = \frac{(a + b)^2 + (a-b)^2}{(a^2 - 2ab + b^2)(a^2 + 2ab + b^2)} => A = \frac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2}{a^4 + 2a^2b^2 + b^4 - 2a^2b - 4a^2b^2 - 2ab^2 + a^2b^2+2ab^2+b^4} => A =\frac{2a^2 + 2b^2}{a^4 - 2a^2b^2 + b^4} => A = \frac{2(a^2+b^2)}{(a^2-b^2)^2}$$
$$ A : B $$
$$ \frac{2(a^2+b^2)}{(a^2-b^2)^2} : \frac{1}{a^2 - b^2} = \frac{2(a^2+b^2)(a^2 - b^2)}{(a^2-b^2)^2} = \frac{2(a^2+b^2)}{(a^2-b^2)}$$
My Answer is:
$$
\frac{2(\sqrt p +\sqrt q)}{(\sqrt p-\sqrt q)}
$$
Books answer is:
$$
\frac{2(\sqrt p +\sqrt q)^2}{(p- q)}
$$
Where is my mistake?
|
Since you have fourth roots you can do the following, to eliminate roots from the denominator.
$$\frac {2(a^2+b^2)}{a^2-b^2}=\frac {2(a^2+b^2)}{a^2-b^2}\cdot\frac {a^2+b^2}{a^2+b^2}=\frac {2(a^2+b^2)^2}{a^4-b^4}$$
Substitute your values into this and you will see that the two answers are equal to each other.
Alternatively, without any short cuts $$\frac{\sqrt p+\sqrt q}{\sqrt p-\sqrt q}-\frac {(\sqrt p+\sqrt q)^2}{p-q} =$$$$\frac {p\sqrt p+p\sqrt q-q\sqrt p-q\sqrt q-p\sqrt p-2p\sqrt q-q\sqrt p+p\sqrt q+2q\sqrt p+q\sqrt q}{(\sqrt p-\sqrt q)(p-q)}=0$$
|
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|
Adding combinations Show non-numerically that:
$${2\choose2} + {3\choose2} + {4\choose2} + {5\choose2} = {6\choose3}$$
The answer is as follows, but I have no idea how it was done:
$$
\begin{eqnarray}
&\phantom{=}& {2\choose2} + {3\choose2} + {4\choose2} + {5\choose2} \\
&=& {3\choose3} + {3\choose2} + {4\choose2} + {5\choose2} \\
&=& {4\choose3} + {4\choose2} + {5\choose2} \\
&=& {5\choose3} + {5\choose2} \\
&=& {6\choose3}
\end{eqnarray}
$$
Could anyone please explain this? Thank you!
|
If you want to choose $k$ items from $n$ items, you can do either way:
*
*choose $k$ items from the first $n-1$ items.
*choose the last item, and then choose $k-1$ items from the first $n-1$ items.
That means
${n\choose k} = {n-1\choose k} + {n-1\choose k-1}$
|
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|
Determinant of specially structured block matrix How do you compute the determinant of the following $nd \times nd$ block matrix?
$$M = \begin{bmatrix}A+B & A & A & \dots & A & A\\ A & A+B & A & \dots & A & A\\ A & A & A+B & \dots & A & A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ A & A & A & \dots & A+B & A\\ A & A & A & \dots & A & A+B\end{bmatrix}$$
where $A$ and $B$ are $d \times d$ matrices.
|
It has the same determinant as the matrix
$$
\begin{pmatrix}
B & 0 & \cdots & -B \\
0 & B & \cdots & -B \\
\vdots & \vdots & \ddots & \vdots\\
A & A & \cdots & A+B
\end{pmatrix},
$$
which has the same determinant as
$$
\begin{pmatrix}
B & 0 & \cdots & 0 \\
0 & B & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots\\
A & A & \cdots & nA+B
\end{pmatrix},
$$
which is easy to compute.
|
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|
Prove that $\lceil \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}\rceil = 2n^2+n+3$ First, the question: Prove that
$$\Bigg\lceil \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}\Bigg\rceil = 2n^2+n+3$$
The motive to this question is the following. I once saw the following "near-identity" in a facebook page
$$7(\sqrt{7+\sqrt{7}}-\sqrt{5+\sqrt{5}}) \approx \sqrt{6+\sqrt{6}}$$
So I thought of generating similar near-identities by directly calculating
$$E(k) = \frac{\sqrt{k+\sqrt{k}}}{\sqrt{k+1+\sqrt{k+1}}-\sqrt{k-1+\sqrt{k-1}}}$$
and then look for those entries which the expression is almost an integer (for example when k=6 we have $E(6) \approx 6.991418$)
Then I got the idea that we can also find near-identities of the form
$$N(\sqrt{k+3+\sqrt{k+2}} - \sqrt{k+2+\sqrt{k+1}}) \approx \sqrt{k+1+\sqrt{k}}$$
where $N, k$ is are positive integers, by calculating
$$F(k) = \frac{\sqrt{k+1+\sqrt{k}}}{\sqrt{k+3+\sqrt{k+2}}-\sqrt{k+2+\sqrt{k+1}}}$$
and then again look for almost-integer entries. Concretely in this case, my definition of "almost-integer" here is that the number is at most 0.001 away from the integer closest to it. Numerical evidence shows that the $k$'s for which $F(k)$ is an almost-integer are $25, 36, 49, 64, 81, \ldots$ which suggest something is at play. After some pattern-guessing, I ended up with the claim in the beginning.
Direct graph plotting seems to indicate that
$$f(x) = 2x+\sqrt{x}+3 - \Bigg(\frac{\sqrt{x+1+\sqrt{x}}}{\sqrt{x+3+\sqrt{x+2}}-\sqrt{x+2+\sqrt{x+1}}}\Bigg)$$
converges to 0 "from above" the x-axis, but that's as far as I can go.
Also, calculating $E(k)$ also yields the following two claims (which is also beyond my reach at this point)
$$\Bigg\lfloor \frac{\sqrt{2n(2n+1)+1+\sqrt{2n(2n+1)}}}{\sqrt{2n(2n+1)+1+\sqrt{2n(2n+1)+1}}-\sqrt{2n(2n+1)-1+\sqrt{2n(2n+1)-1}}}\Bigg\rfloor = 2n(2n+1)+n$$
and
$$\Bigg\lfloor \frac{\sqrt{2n(2n-1)+\sqrt{2n(2n-1)}}}{\sqrt{2n(2n-1)+1+\sqrt{2n(2n-1)+1}}-\sqrt{2n(2n-1)-1+\sqrt{2n(2n-1)-1}}}\Bigg\rfloor_{0.5} = 2n(2n-1)+n-0.5$$
where $\lfloor x \rfloor_{0.5}$ is the greatest integer or half-integer less than $x$, for example, $\lfloor 1.501 \rfloor_{0.5} = 1.5$
Any help would be highly appreciated.
(PS. this is not my homework, just a curiosity)
|
I do not know how much this could help you : for large values of $n$, the expansion of $$F(n)= \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}$$ is given by $$2 n^2+n+3-\frac{1}{4 n^2}+\frac{1}{16 n^3}+\frac{9}{16 n^4}-\frac{45}{64
n^5}+O\left(\left(\frac{1}{n}\right)^6\right)$$ and the difference between $F(n)$ and $G(n)=2 n^2+n+3$ is always negative and smaller than $1$. The largest difference is obtained for $n=-2$ for which $F(-2)=8.48432$ and $G(-2)=9$.
|
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|
Calculate $\int_{0}^{2\pi}{\sin^8x}\ {dx}$ I have started doing integration by parts:
$$\int_{0}^{2\pi}{\sin^8(x)}{dx}
= \int_{0}^{2\pi}{\sin^7(x)}\cdot{\sin(x)dx}
= \int_{0}^{2\pi}{\sin^7(x)}\cdot{d(-\cos(x))}
= \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi}
+ \int_{0}^{2\pi}{\cos(x)}{d(\sin^7(x))}
= \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi}
+ \int_{0}^{2\pi}{\cos^2(x) \cdot 7 \cdot \sin^6(x)}{dx} = \dots$$
Is there a better way to do it?
|
Consider Beta function
$$
\text{B}(x,y)=2\int_0^{\Large\frac\pi2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\ d\theta=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}.
$$
Rewrite
$$
\int_0^{\large2\pi}\sin^8x\ dx=4\int_0^{\Large\frac\pi2}\sin^8 x\ dx,
$$
then
$$
\int_0^{\large2\pi}\sin^8x\ dx=2\cdot\frac{\Gamma\left(\dfrac92\right)\cdot\Gamma\left(\dfrac12\right)}{\Gamma(5)}=2\cdot\frac{\dfrac72\cdot\dfrac52\cdot\dfrac32\cdot\dfrac12\cdot\Gamma^2\left(\dfrac12\right)}{4!}=\large\color{blue}{\frac{35}{64}\pi},
$$
where $\Gamma(n+1)=n\cdot\Gamma(n)$ and $\Gamma\left(\dfrac12\right)=\sqrt\pi$.
|
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|
inequality $\prod\limits_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$ $n$ is a positive integer, then
$$\prod_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$$
with mathematical induction, we can prove this.
But I would love to find a wonderful method without mathematical induction
Thank you!
|
Building on André Nicolas' idea, you may rewrite $\prod_{k=1}^n\frac{2k-1}{2k}$ as
$$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdots\frac{2n-1}{2n}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots (2n-1)2n}{(2\cdot 4 \cdot 6\cdot 8\cdots 2n)^2}=\frac{(2n)!}{(2^n\cdot 1\cdot 2\cdot 3\cdots n)^2}=\frac{1}{4^n}\binom{2n}{n}.$$
Now, $\binom{2n}{n}$ is the central binomial coefficient. Its asymptotic is both well-known and easily determined as
$$\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}},$$
so that asymptotically,
$$\prod_{k=1}^n \frac{2k-1}{2k}\sim \frac{1}{\sqrt{\pi n}}<\frac{1}{\sqrt{3n}}\quad(\pi=3.1416...).$$
Moreover, wikipedia also lists a property of central binomial coefficients which directly proves your result, namely,
$$\binom{2n}{n}\le \frac{4^n}{\sqrt{3n+1}}<\frac{4^n}{\sqrt{3n}}.$$
|
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|
Finding the explicit notation of $f(n)$, based on it's recursive description. I came across this problem on a HackerRank challenge.
The function $f(n)$ is
*
*$1$ if $n = 0$
*$2f(n - 1)$, if $n$ is odd
*$f(n -1) + 1$, if $n$ is even
I solved the problem using a recursive function and it worked just well. However, I am assuming that a program would work faster if instead of recursion we use an explicit function.
The site gave this solution:
$f(n) = Pow(2, (n+1)/2 + 1) - 1 - (2 \% n)$
Can someone help me arrive at this explicit formation?
|
To get a closed form, we will first rewrite the recurrence as:
\begin{align*}
f_n &= 2\, f_{n-2}+1+ \left(n\mod 2\right)
\end{align*}
Next, we will use generating functions, suppose $G(x)= \sum_{n\ge 0}f_n\, x^n$
\begin{align*}
\sum_{n\ge 2} f_n x^n &= 2\, \sum_{n\ge 2}f_{n-2}x^n+ \sum_{n\ge 2}x^n+ \sum_{n\ge 2}\left( n\mod 2\right) \\
G(x)-f_0x^0-f_1x^1 &= 2\,x^2\, G(x) + \frac{x^2}{1-x}+\frac{x^3}{1-x^2} \\
G(x) &= \frac{1+2\, x}{1-3\, x^2+2\, x^4} \\
G(x) &= (1+2\, x)\left(\frac{1}{1-\sqrt2 x}+\frac{1}{1+\sqrt2 x}\right)+\frac{1}{2(1+x)}-\frac{3}{2(1-x)}
\end{align*}
Extracting the coefficients $[x^n]$ from each of the partial terms:
\begin{align*}
\boxed{\displaystyle f_n = 2^{n/2}\left(1+(-1)^n\right)+2^{(n+1)/2}\left(1+(-1)^{n+1}\right)+\frac{(-1)^n-3}{2}}
\end{align*}
Not exactly that formula, but can be modified. And also, that formula needs to be written as:
\begin{align*}
f(n) = \operatorname{Pow}(2, \operatorname{floor}((n+1)/2) + 1) - 1 - (n\mod 2)
\end{align*}
|
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|
Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$ Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$
My approach :
$= \frac{1}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3} +\cdots \infty$
Now how to solve this I am not getting any clue on this please help thanks.
|
Starting with
\begin{align}
\frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}
\end{align}
then differentiate and multiply by $t$ to obtain
\begin{align}
\frac{t}{(1-t)^{2}} = \sum_{n=0}^{\infty} n t^{n}.
\end{align}
Repeating leads to
\begin{align}
\sum_{n=0}^{\infty} n^{2} \ t^{n} = \frac{t(1+t)}{(1-t)^{3}}.
\end{align}
Now let $t = 1/6$ for which
\begin{align}
\sum_{n=0}^{\infty} \frac{n^{2}}{6^{n}} = \frac{42}{125}.
\end{align}
|
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|
Which of the following functions on R are uniformly continous?
$a)\frac {1}{x^2+1} $
$b)\cos^3x$
$c)\frac {x^2}{x^2+2} $
$d)x\sin x$
$a)|\frac{1}{x^2+1}-\frac{1}{y^2+1}|\leq|\frac{|x-y|(|x|+|y|)}{(1+x^2)(1+y^2)}|\leq \frac{|x-y||x|}{(1+x^2)(1+y^2)} +\frac{|x-y||y|}{(1+x^2)(1+y^2)}<\frac{|x-y|}{1+y^2}+\frac{|x-y|}{1+x^2}$ ( since $\frac{|y|}{1+y^2}<1$ and $\frac{|x|}{1+x^2}<1)$
Can we take $L=\frac{2+x^2+y^2}{(1+x^2)(1+y^2)}$ for lipschitz?
b)from the definition of u.continuty $|\cos^3x-\cos^3y|=|\cos x-\cos y||1+\cos x\cos y|<|1+\cos x \cos y|$
$\leq |1+\cos(x+y)+\cos(x-y)||\leq|2+\cos(x-y)|<|2+\delta)| ($ since $\cos(x-y)<|x-y|<\delta )$ so if we choose $\delta=\epsilon-2$. f(x) is uniformly continous?
$c)|\frac{x^2}{x^2+2}-\frac{y^2}{y^2+2}|<|\frac{2|x-y||x+y|}{(x^2+2)(y^2+2)}$
how can I continue?
d)we must show there is a delta for all $\epsilon>0$ , for $|x-y|<\delta$, $|x\sin x-x\sin y|
<\epsilon$
but when we take $x=m+2k\pi $ and $y=n+2k\pi$ $|(m+2k\pi)(\sin(m+2k\pi))-(n+2k\pi) \sin(m+2k\pi)|$ so $|m\sin m-n\sin n+2k\pi(\sin n-\sin m)|$
if we choose k big enough this inequality can be greater than epsilon. so is not uniformly continous?
|
Hint If a function $f$ is lipschitzian then it's uniformly continuous and notice that if $f'$ is bounded then $f$ is lipschitzian. We can apply the above remark to a,b,c).
Now for d) let $x_n=2n\pi$ and $y_n=x_n+\frac\pi2$ then we have
$$y_n-x_n=\frac\pi2\xrightarrow{n\to\infty}\frac\pi2$$
and
$$f(y_n)-f(x_n)\xrightarrow{n\to\infty}\infty$$
so what we can conclude?
|
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|
Solving combinatorical problem using characteristic polynomial
How many $6$ length strings above $\left\{1,2,3,4\right\}$ are there such that $24$ and $42$ aren't allowed.
The suitable recurrence relation for this problem is: $a_{n+2} = 2a_{n-1} + 4a{n-2}$. Hence, the characteristic polynomial is: $x^2 -2x -4 = 0$.It's roots are: $1+\sqrt5,1-\sqrt5$.
So, the genral function is: $\alpha (1+\sqrt5)^n + \beta (1-\sqrt5)^n$
We know that: $a_0=1$ and $a_1=4$. Hence,
$\beta = \frac{1}{2} - \frac{3}{2\sqrt5}$,
$\alpha = \frac{1}{2} + \frac{3}{2\sqrt5}$
All in all,
$$(\frac{1}{2} + \frac{3}{2\sqrt5})(1+\sqrt5)^n + (\frac{1}{2} - \frac{3}{2\sqrt5})(1-\sqrt5)^n$$
Now, for $n=6$ the result is $1344$, but the book says $1216$.
Who's right?
|
We may solve such kind of problems using directed graph/finite automaton.
Consider the following graph:
\begin{align*}
A = \left[\begin{array}{rrrrrrr}
& \mathrm{I} & \mathrm{O} & \mathrm{2} & \mathrm{4} & \mathrm{24} & \mathrm{42}\\
\mathrm{I} & 0 & 2 & 1 & 1 & 0 & 0 \\
\mathrm{O} & 0 & 2 & 1 & 1 & 0 & 0 \\
\mathrm{2} & 0 & 2 & 1 & 0 & 1 & 0 \\
\mathrm{4} & 0 & 2 & 0 & 1 & 0 & 1 \\
\mathrm{24} & 0 & 0 & 0 & 0 & 0 & 0 \\
\mathrm{42} & 0 & 0 & 0 & 0 & 0 & 0
\end{array}\right]
\end{align*}
We can now obtain the generating functions for all entries by computing:
$$\left(I-x\, A\right)^{-1}$$
and the g.f. we require is the sum of first four columns in the first row of the inverted matrix, which turns out to be:
\begin{align*}
G(x) &= \dfrac{1+x}{1-2\, x-3\, x^2} \\
&= \dfrac{1}{4}\left(\dfrac{1+\dfrac{1}{\sqrt{17}}}{\dfrac{\sqrt{17}-3}{4}-x}-\dfrac{1-\dfrac{1}{\sqrt{17}}}{\dfrac{\sqrt{17}+3}{4}+x}\right)
\end{align*}
Extracting $[x^n]$ from the above gives us:
\begin{align*}
a_n &= \dfrac{1}{4}\left(\dfrac{1+\dfrac{1}{\sqrt{17}}}{\left(\dfrac{\sqrt{17}-3}{4}\right)^{n+1}}-(-1)^n\, \dfrac{1-\dfrac{1}{\sqrt{17}}}{\left(\dfrac{\sqrt{17}+3}{4}\right)^{n+1}}\right) \\
&\sim \dfrac{1}{4}\, \dfrac{1+\dfrac{1}{\sqrt{17}}}{\left(\dfrac{\sqrt{17}-3}{4}\right)^{n+1}}
\end{align*}
and the required answer is $a_6 = 2258$
|
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|
Confused about a linear equation So I am working through some notes on Linear Algebra and I cant seem to follow this one part. The question asks to
Solve:
$x+y-z+2w=-20$
$2x-y+z+w=11$
$3x-2y+z-2w+27$
I don't have a problem with putting the equation into matrix form and even reducing it. The way the notes explains it is what I don't understand at all. So first once its in echelon,
$$
\begin{bmatrix}
1 & 1 & -1 & 2 & -20\\
0 & 1 & -1 & 1 & -17\\
0 & 0 & 1 & 3 &-2\\
\end{bmatrix}
$$the notes states: "The solution is therefore
$$
X=\begin{pmatrix}
-w-3\\
-4w-19\\
-3w-2\\
2\\
\end{pmatrix}
$$
(This is the first place I am confused? Why are there $4$ rows now? And how did they get these numbers?).
Then, we can reduce and have
$$
\begin{bmatrix}
1&0&0&1&-3\\
0&1&0&4&-19\\
0&0&1&3&-2
\end{bmatrix}
$$
Now for the second confusing part, it then states: "Note, the solution is
$$
X=\begin{pmatrix}
-3\\
-19\\
-2\\
0\\
\end{pmatrix} +w\begin{pmatrix}
-1\\
-4\\
-3\\
1\\
\end{pmatrix}
$$
Note that the rank is $3$ and there is $1$ parameter.
Thanks a lot for the help guys. Hopefully I can understand it.
|
You have four unknowns, so a solution is a vector with four components.
The system, after the final reduction, can be written as
$$
\begin{cases}
x+w=-3\\
y+4w=-19\\
z+3w=-2
\end{cases}
$$
Since $w$ is a free variable, you can set it to any value you want; therefore the solutions are
$$
\begin{cases}
x=-3-w\\
y=-19-4w\\
z=-2-3w\\
w=w
\end{cases}
$$
that in vector form can be written as
$$
\begin{bmatrix}
-3-w\\
-19-4w\\
-2-3w\\
w
\end{bmatrix}=
\begin{bmatrix}
-3\\
-19\\
-2\\
0
\end{bmatrix}+
w\begin{bmatrix}
-1\\
-4\\
-3\\
1
\end{bmatrix}
$$
where $w$ is any number.
|
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|
How to prove $a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$ I was asked to prove an inequality:
For any $n$ positive numbers $\{a_i\}$ with $a_{1}a_{2}\cdots a_{n} = 1$ and $m \geq n-1$ be a non-negative integer,
$a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$
After several attempts, I think I had to use induction on both $m$ and $n$. In particular, I fix an $n$ and claim that the inequality holds for every $m$, and use induction on $n$.
The case where $n = 2$ could be done by rewriting the inequality as $a_1^2 + a_2^2 \geq a_1 + a_2$ and assume that $a_1 \leq 1 \leq a_2$.
In the inductive step, given $a_1^k + a_2^k \geq a_1 + a_2$ , by writing $a_1^{k+1} + a_2^{k+1} = (a_1^{k} + a_2^{k})(a_1 + a_2) - a_1^{k}a_2^{k} (a_1 + a_2)$, the result follows.
However, this method seems to fail for $n>2$, as the right-hand side of the inequality becomes more complicated. Should I try not to prove it by induction?
|
We will prove that
$$a_1^m + a_2^m + \cdots + a_n^m \ge a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1} \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}.$$
Left-hand side inequality:
Obviously we have $(x^{n-1}-1)(x^{m-n+1}-1) \ge 0$ for any $x>0$ (note that $m\ge n-1$). Expanding we get
$$x^m\ge x^{n-1} + (x^{m-n+1}-1).$$
Applying this inequality for $x=a_1,\ldots,a_n$ and taking the sum we get
$$\sum_{i=1}^n a_i^m \ge \sum_{i=1}^n a_i^{n-1} + \left(\sum_{i=1}^n a_i^{m-n+1}- n \right).$$
But by AM-GM inequality
$$\sum_{i=1}^n a_i^{m-n+1} \ge n\times (a_1a_2\cdots a_n)^{(m-n+1)/n} = n.$$
Thus we have
$$\sum_{i=1}^n a_i^m \ge \sum_{i=1}^n a_i^{n-1}.$$
Right-hand side inequality:
Denote $S=a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1}$. Using AM-GM inequality for $n-1$ numbers we have
$$S-a_1^{n-1} = a_2^{n-1} + \cdots + a_n^{n-1} \ge (n-1)a_2\cdots a_n = \frac{n-1}{a_1},$$
i.e.
$$S-a_1^{n-1} \ge \frac{n-1}{a_1}.$$
Similarly, we have the other inequalities for $a_2,\ldots,a_n$. Taking the sum of these $n$ inequalities we get
$$nS-(a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1}) \ge (n-1)\left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}\right).$$
The left-hand side of the last inequality is just $(n-1)S$. Divided both sides by $(n-1)$ we get the desired inequality.
|
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|
What is wrong in the following calculation for the inverse of a matrix? $\left[\begin{array}{ccc|ccc}
0 & 3 & 0 & 1 & 0 & 0\\
4 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 5 & 0 & 0 & 1
\end{array}\right]$
Exchange the first and the second columns:
$\left[\begin{array}{ccc|ccc}
3 & 0 & 0 & 0 & 1 & 0\\
0 & 4 & 0 & 1 & 0 & 0\\
0 & 0 & 5 & 0 & 0 & 1
\end{array}\right]$
Divide the rows by $3$, $4$, and $5$ respectively:
$\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & 0 & \frac{1}{3} & 0\\
0 & 1 & 0 & \frac{1}{4} & 0 & 0\\
0 & 0 & 1 & 0 & 0 & \frac{1}{5}
\end{array}\right]$
Obviously, the result is not the inverse of the initial matrix. Where is the mistake?
|
You shoul work on columns or on rows, never on both.
|
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|
Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
|
Multiply by the conjugate, then reduce:
\begin{align*}
\lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right)
&= \lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \cdot \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\
&= \lim_{x\to\infty} \frac{(x + \sqrt{x + \sqrt{x}}) - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\
&= \lim_{x\to\infty} \frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \cdot \frac{\dfrac{1}{\sqrt{x}}}{\dfrac{1}{\sqrt{x}}} \\
&= \frac{\sqrt{\lim\limits_{x\to\infty}\dfrac{1}{\sqrt{x}} + 1}}{\sqrt{1 + \lim\limits_{x\to\infty} \dfrac{\sqrt{x + \sqrt{x}}}{x}} + 1} \\
&= \frac{\sqrt{0 + 1}}{\sqrt{1 + 0} + 1} \\
&= \frac{1}{2}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Hyberbolic and Circular (Trig) Functions: Why no parabolic? There are circular (trig) functions which determine all the points on a unit circle:
and which relate to the area swept out by an angle subtended on the circle. -- These functions can of course be extended to relations to ellipses as well.
There are also hyperbolic functions which determine all the points on a hyperbola:
My question is why there are no analogs of these functions for parabolas (the other type of conic section):
Here I have defined $\mathrm {sinp}(\theta)$ and $\mathrm {cosp}(\theta)$ to be the x- and y-coordinates of points on a "unit parabola".
Is there any good reason why we should have these extremely useful transcendental functions (sin, cos, sinh, cosh, etc), but we can't (or don't) define analogous functions for parabolas?
NOTE: I recommended that this post get deleted because Henning's response in Do "Parabolic Trigonometric Functions" exist? explained theoretically why a "parabolic trigonometric function" is different than the circular and hyperbolic trigonometric functions. However johannesvalks' answer to this was very interesting, as well, and probably shouldn't be deleted.
|
Note that
(1) For $\cos(\phi)$ and $\sin(\phi)$ the argument is a 'surface'.
(2) For $\cosh(\zeta)$ and $\sinh(\zeta)$ the argument is a 'surface'.
So you should look for functions where the argument is a 'surface'.
Consider $y = x^2 - c$, where $c$ is positive.
For a given $x = \xi$ we have an arc formed by:
(3a) the curve $(0,-c) \rightarrow (\xi,\xi^2-c)$
(3b) the line $(\xi,\xi^2-c) \rightarrow (0,0)$
(3c) the line $(0,0) \rightarrow (0,-c)$
The surface of this arc is given by
(4) $\displaystyle A =
\int_0^\xi \left( \frac{\xi^2 - c}{\xi} x + c - x^2 \right) dx =
\frac{1}{6} \xi^3 + \frac{1}{2} c \xi$
So for a point $(x,y)$ on the curve $y = x^2 - c$ we obtain
(5) $A = \frac{1}{6} x \Big( x^2 + 3c \Big)$
And what we want is
(6a) $x = x(A)$
(6b) $y = y(A)$
To find the inverse for $A = \frac{1}{6} x \Big( x^2 + 3c \Big)$, we write $\displaystyle x = \frac{c}{\zeta} - \zeta$, then we get the equation
(7) $\Big[ \zeta^3 \Big]^2 + 6 A \Big[ \zeta^3 \Big] + c^3 = 0$,
whence
(8) $ \zeta^3 = \sqrt{9A^2 + c^3} - 3A $,
therefore
(9a) $\displaystyle x(A) = \frac{c}{\sqrt[3]{\sqrt{9A^2 + c^3} - 3A}} - \sqrt[3]{\sqrt{9A^2 + c^3} - 3A}$
The 'parabolic' functions are given by
(10a) $\displaystyle x(A) = \frac{c}{\sqrt[3]{\sqrt{9A^2 + c^3} - 3A}} - \sqrt[3]{\sqrt{9A^2 + c^3} - 3A}$,
(10b) $\displaystyle y(A) = \frac{c^2}{\sqrt[3]{\sqrt{9A^2 + c^3} - 3A}^2} + \sqrt[3]{\sqrt{9A^2 + c^3} - 3A}^2 - 3 c$,
but they can be expressed using radicals, as is not the case for goniometric and hyperbolic functions.
The case $c=1$ gives
(11a) $\displaystyle x(A) = \frac{1}{\sqrt[3]{\sqrt{9A^2 + 1} - 3A}} - \sqrt[3]{\sqrt{9A^2 + 1} - 3A}$,
(11b) $\displaystyle y(A) = \frac{1}{\sqrt[3]{\sqrt{9A^2 + 1} - 3A}^2} + \sqrt[3]{\sqrt{9A^2 + 1} - 3A}^2 - 3$.
|
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|
How to prove inequality $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ Question:
Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$
when I solved this problem, I have see following three variables inequality:
Assumming that $a,b,c>0,a+b+c=3$, show that :$$f(a,b,c)=\dfrac{a}{a+bc}+\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\ge\dfrac{3}{2}$$
solution can see:inequality
I found this three answer all is not true,
1、such as dear @Macavity, in fact
$$\sum_{cyc}\dfrac{a^2}{a^2+abc}\ge\dfrac{16}{\sum_{cyc}(a^2+abc)}$$
the Right not
$\dfrac{16}{4+\sum_{cyc}abc}$
2、and the @ante.ceperic is also not true.in fact
$$a^2+b^2+c^2+d^2+abc+bcd+cda+dab\le 8$$ is not true with $a+b+c+d=4$
such let $a=3$
|
Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$P=\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$
Now using Cauchy Schwarz:
$$\left(\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\right)((a+bc)+(b+cd)+(c+da)+(d+ab))\ge (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$$
Now:
$$P\ge\frac{4+2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})}{4+ab+bc+cd+da}$$
Let $S_1=\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd},S_2=ab+bc+cd+da$
Now we need to prove:
$$\frac{4+2S_1}{4+S_2}\ge2\implies S_1-S_2\ge2$$
If we prove $S_1\ge6$ and $S_2\le4$, then $S_1-S_2\ge6-4=2$. Both inequalities can be proved easily and both hold when $a=b=c=d=1$:
$$(a+b+c+d)(b+c+d+a)\ge(ab+bc+cd+da)^2\implies S_2^2\le16\\\text{ since }a,b,c,d>0\implies S_2\le4$$
|
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|
Markov chains by hand If I have a starting point: $A_T=[0,1]$ at $T=1$ and a one step transition matrix of:
$B=\left[ \begin{align} &\frac34 & \frac14& \\& \frac1{20}& \frac {19}{20} &\end{align} \right]$
I can compute $A_nB=A_{n+1}$
And keep reiterating this, but it is slow by hand, I assume there is a faster way? Would diagonalizing the matrix help somehow?
|
$B$ can be diagonalized as $B = PDP^{-1}$, where
$P = \begin{bmatrix}-5 & 1\\1 & 1\end{bmatrix}$, $P = \begin{bmatrix}7/10 & 0\\0 & 1\end{bmatrix}$, $P^{-1} = \begin{bmatrix}-1/6 & 1/6\\1/6 & 5/6\end{bmatrix}$.
Then $B^n = PD^nP^{-1} = \begin{bmatrix}-5 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}(7/10)^n & 0\\0 & 1\end{bmatrix}\begin{bmatrix}-1/6 & 1/6\\1/6 & 5/6\end{bmatrix}$.
$$\boxed{\therefore B^n = \dfrac{1}{6}\begin{bmatrix}
5\left(\dfrac{7}{10}\right)^n + 1 & 5 - 5\left(\dfrac{7}{10}\right)^n\\
1 - \left(\dfrac{7}{10}\right)^n & \left(\dfrac{7}{10}\right)^n + 5
\end{bmatrix}}$$
As $B$ is a right stochastic matrix (because the row sums are $1$, not the column sums), we need to right multiply by $B$:
$A_n = A_0 B^n = \begin{matrix}[0 \quad 1]\\\mbox{} \end{matrix} \times
\dfrac{1}{6}
\begin{bmatrix}
5\left(\dfrac{7}{10}\right)^n + 1 & 5 - 5\left(\dfrac{7}{10}\right)^n\\
1 - \left(\dfrac{7}{10}\right)^n & \left(\dfrac{7}{10}\right)^n + 5
\end{bmatrix}\\
\mbox{}\\
\boxed{A_n = \dfrac{1}{6}\left[1 - \left(\dfrac{7}{10}\right)^n \qquad \left(\dfrac{7}{10}\right)^n + 5\right]}$
|
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|
Using calculus of residues, prove that $\int z^2log[(z+1)/(z-1)]dz$ Using calculus of residues, how can it be proven that
$$
\int z^2\log\left[\frac{z+1}{z-1}\right]\;dz
$$ taken round the circle $\left\vert z\right\vert=2$ has the value $\frac{4\pi i}{3}$?
|
Note: I'm assuming that $\log \frac{z+1}{z-1}$ is defined on $\mathbb{C}\setminus [-1,1]$, or at least that a branch-cut that lies entirely inside the disk $\{ z : \lvert z\rvert < 2\}$ is used. For arbitrary branch-cuts intersecting the contour, the integral depends on the exact choice of the branch-cut.
One way is to substitute
$$w = \frac{z+1}{z-1}.$$
The circle $\lvert z\rvert = 2$ becomes $\left\lvert w - \frac{5}{3}\right\rvert = \frac{4}{3}$, and since
$$z = \frac{w+1}{w-1},$$
we have $dz = -\frac{2dw}{(w-1)^2}$, but the circle $\left\lvert w - \frac{5}{3}\right\rvert = \frac{4}{3}$ is traversed clockwise, so overall we get the integral
$$2\int_{\lvert w-5/3\rvert = 4/3} \frac{(w+1)^2\log w}{(w-1)^4}\,dw,$$
which has a pole of order $3$ (the numerator has a simple zero in $1$ from the $\log$ factor) in $1$ as the only singularity enclosed by the contour. That integral can be evaluated by the residue theorem in the usual manner.
Taking the principal branch of the logarithm, we can expand the numerator
$$\bigl((w-1)^2 + 4(w-1) + 4\bigr)\left((w-1) - \frac{(w-1)^2}{2} + \frac{(w-1)^3}{3} + O\left((w-1)^4\right)\right)\\
= \dotsc + (1 - 4\frac{1}{2} + 4\frac{1}{3})(w-1)^3+ \dotsc,$$
so the residue is $\frac{1}{3}$. Choosing a different branch of the logarithm doesn't affect the residue, since only the coefficients of $(w-1)^k$ for $k\in \{-4,-3,-2\}$ change.
Another way to evaluate the integral is to rewrite and expand the logarithm
$$\log \frac{z+1}{z-1} = \log \frac{1+z^{-1}}{1-z^{-1}} = 2m\pi i + 2 \sum_{k=0}^\infty \frac{1}{(2k+1)z^{2k+1}}.$$
Since the Laurent series converges uniformly on the contour, we can interchange summation and integration, and obtain
$$\int_{\lvert z\rvert = 2} z^2\log \frac{z+1}{z-1}\,dz = 2m\pi i \int_{\lvert z\rvert = 2} z^2\,dz + 2\sum_{k=0}^\infty \frac{1}{2k+1}\int_{\lvert z\rvert = 2} z^{1-2k}\,dz = \frac{2}{3}\int_{\lvert z\rvert = 2} \frac{dz}{z}.$$
|
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|
Generating Function for the number of ways representing positive integer with odd numbers I had an exam and this question struck me out of nowhere, making me sad :)
Let $f(n)$ denote the number of possibilities representing $n$ using odd positive single-digit numbers $[1,3,...,9]$
For example, $f(6)=4$ because:
$6=1+1+1+1+1+1$, $6=3+3$, $6=5+1$, $6=1+1+1+3$
Note that there is no difference between $1+5$ and $5+1$ and both of the possibilities count as one.
Find the generating function representing $f(n)$
|
Since you are not making a difference based on the order of the numbers, the only important thing here is how many of each numbers is present in a partition. That means we can model 1's with this generating function:
$$1+x+x^2+x^3+\cdots=\frac{1}{1-x}$$
$1=x^0$ is the case when we don't take any ones, $x$ is the case when we take 1 one, $x^2$ when we take 2 ones and so on... The same can be done for other odd integers, for example:
for 3: $$1+x^3+x^6+x^9+\cdots=\frac{1}{1-x^3}$$
for 5: $$1+x^5+x^{10}+x^{15}+\cdots=\frac{1}{1-x^5}$$
and so on...
That means that the whole generating function will be:
$$\frac{1}{(1-x)\left(1-x^3\right)\left(1-x^5\right)\left(1-x^7\right)\left(1-x^9\right)}$$
In general, if you have a set $S$ of numbers you want to partition into, you can use the generating function
$$\frac{1}{\prod_{s\in S}\left(1-x^s\right)}$$
|
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|
Which are the two last digits of $a_n$? We have the sequence $(a_n)$ with $a_1=3$ and the recursive formula:
$$a_{n+1}=3^{a_n} , \forall n$$
Which are the two last digits of $a_n$ ?
How can I find them? I have to find $a_n \pmod {100}$ ,right?So,do I have to find the formula of $a_n$ ?
Could I use the Eulers'Theorem?
We would have:
$$(3,100)=1$$
$$3^{\phi(100)} \equiv 1 \pmod{100} \Rightarrow 3^{40} \equiv 1 \pmod {100}$$
But...how can I use this?
|
As for integer $\displaystyle n>0, a_n\equiv-1\pmod4\equiv3,a_n-1\equiv2\pmod4\implies\frac{a_n-1}2\equiv1\pmod2$
$$a_{n+1}=3^{a_n}=3\left(10-1\right)^{\frac{a_n-1}2}\equiv3\left(-1+\binom{\frac{a_n-1}2}110\right)\pmod{100}$$
$$\implies a_{n+1}\equiv15a_n-18\pmod{100}$$
Now let us find the period of $\displaystyle15\pmod{100}$
which will be same as that of $\displaystyle\frac{15}5\pmod{\frac{100}5}$ i.e., the period of $\displaystyle3\pmod{20}$
$\displaystyle3^1\equiv3,3^2\equiv9,3^3=27\equiv7,3^4=81\equiv1\pmod{20}$ or usingCarmichael Function $\displaystyle\lambda(20)=4$
$\displaystyle\implies a_n$ will have a period $4\pmod{100}$ or a divisor of $4$
Finally,
$$a_1=3$$
$a_2=3^3=27$ also $\equiv15a_1-18\pmod{100}\equiv15\cdot3-18\equiv27$
$$a_3\equiv15a_2-18\pmod{100}\equiv?$$
$$a_4\equiv15a_3-18\pmod{100}\equiv?$$
|
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|
Reducing a matrix of the form J-n*In, where J is the all-ones matrix. How can I reduce a matrix like this? The goal is to prove that the dimension of the generated space is 1. It's the matrix of the homogeneous system.
$\left( \begin{array}{cccccc}1-n&1&\dots &\dots&\dots&1 \\ 1 & 1-n & 1 & \dots & \dots & 1 \\ \vdots & 1 & \ddots & \ddots & & \vdots \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \vdots && 1 & 1-n & 1 \\ 1 &1 & \dots & \dots & 1 & 1-n\end{array} \right)$.
|
The matrix diagonalizes to $$\begin{pmatrix}
0&0&0&0\\
0&-n&0&0\\
0&0&-n&0\\
0&0&0&-n\\
\end{pmatrix}$$
Let $A=\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
\end{pmatrix}$
And if $X=\begin{pmatrix}
x_1\\
\vdots \\
x_n
\end{pmatrix}$ let $\sum X=x_1+\cdots x_n$.
Then we have $$AX=\sum X \begin{pmatrix}
1\\
\vdots \\
1
\end{pmatrix}$$
Now consider the eigenvalue problem
$$(A-nI)X=\lambda X$$
or equivalently,
$$AX=(n+\lambda) X=\sum X \begin{pmatrix}
1\\
\vdots \\
1
\end{pmatrix}$$
Now if $\lambda +n \neq 0$ then $X$ is a multiple of $\begin{pmatrix}
1\\
\vdots \\
1
\end{pmatrix}$ and we have that $\lambda +n=n$ so $\lambda$ is zero with an eigenspace of dimension one.
Otherwise $\lambda=-n$ and this has an eigenspace of dimension $n-1$
as can be easily checked the vectors $$\begin{pmatrix}
1\\
0\\
\vdots \\
0\\-1\\
0\\
\vdots\\
0\\
\end{pmatrix}$$ are eigen vectors.
Finally it would be nice to check the minimal polynomial which should be $x(x+n)$
|
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|
Polynomial multiplication modulo polynomial Suppose we are working on finite field $F_{16}$ and have pritimive polynomial $z^4+z+1$. I stuck at how to compute polynomial modulo. For example, we have $z^5+z+1$ mod $z^4+z+1$. I use the usual division, I obtain the remainder is $-z^2+1=z^2+1$ because each coefficient is over $F_2$. But I don't know whether this is the correct way to compute remainder. Can anyone help me?
|
I am assuming that $z^4+z+1$ is the irreducible polynomial used to construct the field $\mathbb{F}_{16}$ (this is different from the concept of primitive element). In that case $z^4+z+1 \equiv 0$.
\begin{align*}
z^4+z+1 & \equiv 0 \pmod{z^4+z+1}\\
z^4 & \equiv z+1 \pmod{z^4+z+1}\\
z^5 & \equiv z^2+z \pmod{z^4+z+1}\\
z^5+z+1 & \equiv z^2+1 \pmod{z^4+z+1}
\end{align*}
So you have the right answer but using the above congruences can help you reduce your work.
|
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|
Differentiation of a function $f:\mathbb{Q}\to \mathbb{Q}$(Rational Calculus) Assume that $f:\mathbb{Q}\to \mathbb{Q}$ is given such that $\forall a\in \mathbb{Q}$ the following limit, exists
\begin{equation}
\lim_{x\to a} \frac{f(x)-f(a)}{x-a}\in \mathbb{R}
\end{equation}
Is it true to say that the above limit is a rational number?
|
It is false: I would define something piecewise-constant and discontinuous at irrational $x$ such as $f : \mathbb{Q} \to \mathbb{Q}$ by:
$f(x) = 1$ for $x > \frac{1}{\sqrt{2}}$
$f(x) = \frac{1}{2}$ for $\frac{1}{3\sqrt{2}}<x<\frac{1}{2\sqrt{2}}$
$f(x) = \frac{1}{3}$ for $\frac{1}{4\sqrt{2}}<x<\frac{1}{3\sqrt{2}}$, and so on for positive $x$, and $f(0) = 0$.
Similarly for negative $x$ define:
$f(x) = -1$ for $x < -\frac{1}{\sqrt{2}}$
$f(x) = -\frac{1}{2}$ for $-\frac{1}{2\sqrt{2}}<x<-\frac{1}{3\sqrt{2}}$
$f(x) = -\frac{1}{3}$ for $-\frac{1}{3\sqrt{2}}<x<-\frac{1}{4\sqrt{2}}$
and so on. Then $f$ is differentiable (since constant) at all rational $x$ other than 0, and $f'(0) = \sqrt{2}$.
|
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|
Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.
My gut feeling is telling me to multiply the above series by 8, then subtract it from the original, but it doesn't quite get me there because I'm going to have a whole lot of extra terms that I do not want.
|
If
$\color{blue}{n = 1}$, $\sum$ = 1
$\color{blue}{n=2}$, $\sum = 1^3 + 3^3 = 28 = 2^2 \cdot 7 = \color{blue}{2}^2(2\cdot \color{blue}{2}^2-1) $
$\color{blue}{n = 3}, \sum = 1^3 + 3^3 + 5^3 = 153 = 3^2 \cdot 17 = \color{blue}{3}^2(2\cdot \color{blue}{3}^2 - 1) $
$\color{blue}{n = 4}, \sum = 1^3 + 3^3 + 5^3 + 7^3 = 496 = 2^4\cdot 31 = \color{blue}{4}^2(2\cdot \color{blue}{4}^2 - 1)$
So, we can suppose that $$\sum_{k=1}^{n} (2k-1)^3 = n^2(2n^2 - 1)$$ This formula can be proved by method of mathematical induction.
|
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|
Express $\log_5 288$ in terms of decimal logarithms $\log 2$ and $\log 3$ Assuming $a=\log 2$ and $b=\log 3$ (log is the base 10 logarithm). I have to find $\log_5 288$. How can I do this?
I've tried transforming $\log2$ to $\frac{\log_5 2}{\log_5 10}$ and same for $b$. Then it's $$\frac{5\log_5 2+2\log_5 3}{\log_5 10}=\frac{\log_5 288}{\log_5 10}=5\log2+2\log5=5a+3b$$ Is that correct?
|
$$\log 2=\frac{\log_5 2}{\log_5 10}=\frac{\log_5 2}{\log_5 2+\log_5 5}=\frac{\log_5 2}{\log_5 2+1}\qquad \log 3=\frac{\log_5 3}{\log_5 10}=\frac{\log_5 3}{\log_5 2+1} $$
Hence
$$\log_5 2=\frac{\log 2}{1-\log2}=\frac{a}{1-a}\qquad \log_5 3= \log 3 (\log_5 2+1)=b\left(\frac{a}{1-a}+1\right)=\frac{b}{1-a}\ .$$
Hence
$$\log_5 288=\log_5 2^5\cdot3^2=5\log_5 2+2\log_5 3=5\frac{a}{1-a}+2\frac{b}{1-a}=\frac{5a+2b}{1-a}\ .$$
|
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|
Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x<2\\
\frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3
\end{cases}
$$
My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.
If someone could shed light on this topic, that would be much appreciated.
|
The cdf $F(x)$ is given by
$$F(x)
=\begin{cases}
0 & \text{ for } x <-3\\
\int_{-3}^{x}(t+3) \, dt & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x <2\\
\frac{1}{2}+\int_{-2}^{x}(3-t) \, dt & \text{ for } 2 \leq x <3\\
1 & \text{ otherwsie}
\end{cases}
$$
My suggestion is to plot the pdf $f(x)$ and then think in terms of the area of triangle(s) being swept as you move along the $x-$axis.
|
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|
Picture/intuitive proof of $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$? Is there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$ is true?
Note I'm not looking for a computational proof like the one linked to, more a proof without words or intuitive style proof, thanks.
|
Enhancing my diagram for the angle-sum formula (currently featured in Wikipedia) to use three angles will get you pretty close ...
Thus,
$$\begin{align}
\cos(\alpha+\beta+\gamma) &= \cos\alpha \cos\beta \cos\gamma - \sin\alpha \sin\beta \cos\gamma - \sin\alpha \cos\beta\sin\gamma - \cos\alpha \sin\beta\sin\gamma \\
\sin(\alpha+\beta+\gamma) &= \sin\alpha \cos\beta \cos\gamma + \cos\alpha \sin\beta \cos\gamma + \cos\alpha \cos\beta \sin\gamma - \sin\alpha \sin\beta \sin\gamma
\end{align}$$
With $\alpha = \beta = \gamma = \theta$, these become ...
$$\begin{align}
\cos 3\theta &= \cos^3\theta - 3 \sin^2\theta \cos\theta \\
\sin 3\theta &= 3\cos^2\theta \sin\theta - \sin^3\theta
\end{align}$$
... which the Pythagorean identity helps us rewrite as ...
$$\begin{align}
\cos 3\theta &= \cos^3\theta - 3 (1-\cos^2\theta) \cos\theta = 4\cos^3\theta - 3 \cos\theta \\
\sin 3\theta &= 3(1-\sin^2\theta) \sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta
\end{align}$$
Off-hand, I don't know of a diagram that goes directly from $\cos 3\theta$ to $4\cos^3\theta-3\cos\theta$.
|
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|
Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
|
Use AM > HM on $\frac{a+b+c}{a+b}, \frac{a+b+c}{b+c}$ and $\frac{a+b+c}{c+a}$.
|
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|
How to determine the eigenvectors for this matrix I have the matrix $$\left( \begin{array}{ccc}
-\alpha & \beta \\
\beta/K & -\alpha/K
\end{array} \right)$$ for which the eigenvalues are $$\lambda_{1,2}=-\dfrac{\alpha}{2}-\dfrac{\beta}{2K}\pm\dfrac{\sqrt{(K\alpha-\beta)^2+4K\alpha\beta }}{2K}$$ I have a problem with finding the eigenvectors. It seems that the eigenvectors are zero vectors but then eigenvectors cannot be zero. What am I missing?
|
To find the eigenvalues of a matrix $A$, we solve $\mathrm{det}(A-\lambda I)=0$ for $\lambda$. In this case:
$\begin{align}
\left|\begin{array}{cc} -\alpha-\lambda & \beta \\ \frac{\beta}{K} & \frac{-\alpha}{K}-\lambda \end{array}\right| &= (-\alpha-\lambda)\left(\frac{-\alpha}{K}-\lambda\right)-\frac{\beta^2}{K} \\
&=\frac{\alpha^2}{K}+\left(\alpha+\frac{\alpha}{K}\right)\lambda + \lambda^2 - \frac{\beta^2}{K} \\
&=\lambda^2 + \left(\alpha+\frac{\alpha}{K}\right)\lambda + \frac{\alpha^2-\beta^2}{K} = 0,
\end{align}$
implies that our eigenvalues are:
$\begin{align}
\lambda_{1,2} &= -\frac{\alpha+\frac{\alpha}{K}}{2} \pm \frac{\sqrt{\left(\alpha+\frac{\alpha}{K}\right)^2 - 4\frac{\alpha^2-\beta^2}{K}}}{2} \\
&=-\frac{K\alpha+\alpha}{2K} \pm \frac{\sqrt{(K\alpha+\alpha)^2 - 4K(\alpha^2-\beta^2)}}{2K} \\
&=-\frac{(K+1)\alpha}{2K} \pm \frac{\sqrt{\left((K+1)^2-4K\right)\alpha^2 + 4K\beta^2}}{2K} \\
&=-\frac{(K+1)\alpha}{2K} \pm \frac{\sqrt{(K-1)^2\alpha^2 + 4K\beta^2}}{2K}.
\end{align}$
Now, an eigenvector corresponding to the eigenvalue $\lambda_i$ is in the nullspace of $\left(\begin{array}{cc} -\alpha-\lambda_i & \beta \\ \frac{\beta}{K} & \frac{-\alpha}{K}-\lambda_i \end{array}\right).$
It appears that the vector $\bf{v}_i=\left(\begin{array}{c} \beta \\ \alpha + \lambda_i \end{array}\right)$ would work.
|
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|
Can you show that the LHS equals the RHS in this equation, by showing how I can get the expression on the RHS? $$
\frac{1^2+2^2+...+(n-1)^2}{n^3} = \frac{(n-1)n(2n-1)}{6n^3}
$$
Can someone show me step by step how I can transform the LHS to the RHS? If possible, using high school-level math.
I have now edited the title, because I didn't formulate myself right. I would like to know how I can derive the expression on the RHS, without knowing it.
|
Consider the geometric progression
$$
\sum_{k=1}^{n-1} x^k=\frac{x^{n}-x}{x-1}.\tag1
$$
Differentiating $(1)$ with respect to $x$ yields
$$
\sum_{k=1}^{n-1} kx^{k-1}=\frac{(n-1)x^{n}-nx^{n-1}+1}{(x-1)^2}.\tag2
$$
Multiplying $(2)$ by $x$ yields
$$
\sum_{k=1}^{n-1} kx^{k}=\frac{(n-1)x^{n+1}-nx^{n}+x}{(x-1)^2}.\tag3
$$
Differentiating $(3)$ with respect to $x$ yields
$$
\sum_{k=1}^{n-1} k^2x^{k-1}=\frac{(-2n^2+2n+1)x^{n}+n^2x^{n-1}+(n-1)^2x^{n+1}-x-1}{(x-1)^3}.\tag4
$$
Setting $x=1$ to $(4)$ and evaluating the RHS at $x\to1$ by L'Hospital's rule yields
\begin{align}
\sum_{k=1}^{n-1} k^2&=\lim_{x\to1}\frac{(-2n^2+2n+1)x^{n}+n^2x^{n-1}+(n-1)^2x^{n+1}-x-1}{(x-1)^3}\\
&=\large\color{darkgreen}{\frac{n(n-1)(2n-1)}{6}}.\tag5
\end{align}
Hence, using $(5)$ we obtain
$$
\large\frac{1}{n^3}\sum_{k=1}^{n-1} k^2=\color{blue}{\frac{1^2+2^2+\cdots+(n-1)^2}{n^3}=\frac{n(n-1)(2n-1)}{6n^3}}.
$$
|
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|
How find this integral $I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$ Find this integral
$$I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$$
since
$$(ax+by)^2=a^2x^2+b^2y^2+2abxy$$
so
$$I=I_{1}+I_{2}+I_{3}$$
where
$$I_{1}=\iiint_{x^2+y^2+z^2\le 1}ax^2\,dV=a\iiint_{x^2+y^2+z^2\le 1}x^2\,dV$$
since
$$\iiint_{x^2+y^2+z^2\le 1}x^2\,dV=\iiint_{x^2+y^2+z^2\le 1}y^2\,dV$$
maybe have other methods? Thank you
|
With (almost) no computations:
The domain $x^2+y^2+z^2\leqslant1$ is invariant by the rotations of the plane $(x,y)$ hence $$I(a,b)=(a^2+b^2)\,J,$$ where
$$
J=\int\!\!\!\iint_{x^2+y^2+z^2\leqslant1}x^2\mathrm dx\mathrm dy\mathrm dz.
$$
By symmetry with respect to the $(x,y,z)$ axes, $3J=K(1)$, where
$$
K(t)=\int\!\!\!\iint_{x^2+y^2+z^2\leqslant t^2}(x^2+y^2+z^2)\mathrm dx\mathrm dy\mathrm dz.
$$
By differentiation, using the fact that the surface of the sphere of radius $t$ is $4\pi\,t^2$,
$$
K'(t)=4\pi\,t^2\cdot t^2=4\pi\,t^4.
$$
Since $K(0)=0$, for every nonnegative $t$,
$$
K(t)=\frac45\pi\,t^5,
$$
and, finally,
$$
I(a,b)=\frac13\,\frac45\pi\,(a^2+b^2).
$$
|
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|
How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?
|
the equation
$$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$
is equivalent to
$$2^{n-2}=\frac{n+1}{5-n}$$
because $2^{n-2}>0$ follow that $\frac{n+1}{5-n}>0,0<n<5$
we need to check for $n=1,2,3,4$
for $n=1$ we get $$2^{-1}=\frac{2}{4}=\frac{1}{2}$$
for $n=2$ we get $$2^0=\frac{3}{3}=1$$
for $n=3$ we get $$2^1=\frac{4}{2}=2$$
for $n=4$ we get $$2^2\neq\frac{5}{1}=5$$
so the integer solutions are $n=1,n=2,n=3$
|
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|
Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{align} x(x+1)=y+1 \\
y(y+1)=z+1\\
z(z+1)=x+1 \end{align}$$
Multiplying all we get $$xyz=1 \tag{1}$$ and adding all we get
$$x^2+y^2+z^2=3 \tag{2}$$
Now from Original Equations
$$\begin{align} x^2=y+1-x\\
y^2=z+1-y\\
z^2=x+1-z \end{align}$$ Multiplying all and Using $(1)$ we get
$$(y+1-x)(z+1-y)(x+1-z)=1 $$ $\implies$
$$xy+yz+zx-3=(x-y)(y-z)(z-x) \tag{3}$$ I am unable to proceed further..
|
using the RMS GM inequality
$$\sqrt{\frac{x^2+y^2+z^2}{3}}\geq\sqrt[3]{xyz}$$
with equality if and only if $x=y=z$, plugging in your values for $xyz$ and$x^2+y^2+z^2$ we get that
$$\sqrt{\frac{3}{3}}\geq\sqrt[3]{1}$$
$$1=1$$
thus the only possible solutons are those you already stated.
|
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|
Upper bound of $\sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}$ I am looking for an upper bound of the following sum
$$
S_p:= \sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}.
$$
The upper bound should be independent of $p$, of course. Numerical experiments indicate that
$$
S_p \le \frac53
$$
with the maximum attained for $p=3,4$. However, I am not able to prove it.
I could only prove $S_p \le 3$. Here is, what I did:
First, we see
$$
\frac{p+1}{p-j+1} = 1 + \frac j{p-j+1}\le 1+j,
$$
hence
$$
S_p \le \sum_{j=1}^p (1+j) \frac1{2^j} .
$$
Using the power series
$$
(1-x)^{-2} = \sum_{j=0}^\infty (1+j) x^j , \ |x|<1,
$$
I can estimate
$$
S_p \le (1-\frac12)^{-2} -1 = 3.
$$
How can the bound be improved?
|
First, note that that the change of summation index $j\leftarrow p+1-j$ shows that
$$
S_p=\frac{p+1}{2^{p+1}}\sum_{j=1}^p\frac{2^j}{j}
$$
Thus
$$
\frac{2^{p+1}}{p+1}S_p=\sum_{j=1}^p\frac{2^j}{j}=\frac{2^p}{p}+\frac{2^{p}}{p}S_{p-1}
$$
Similarly$$\eqalign{
\frac{2^{p+2}}{p+2}S_{p+1} &=\frac{2^{p+1}}{p+1}+\frac{2^{p+1}}{p+1}S_{p}\cr
&=\frac{2p}{p+1}\left(\frac{2^{p+1}}{p+1}S_p-\frac{2^{p}}{p}S_{p-1}\right)+\frac{2^{p+1}}{p+1}S_{p}
}$$
This can be rearranged as follows
$$\eqalign{
\frac{S_{p+1}}{S_p} &= \frac{(p+2)(3p+1)}{2(p+1)^2}-\frac{p+2}{2(p+1)}\cdot\frac{S_{p-1}}{S_p}
}$$
Now, If for some $p\ge2$ we have $S_p<S_{p-1}$ then
$$
\frac{S_{p+1}}{S_p}<\frac{(p+2)(3p+1)}{2(p+1)^2}-\frac{p+2}{2(p+1)}=\frac{p(p+2)}{(p+1)^2}<1.
$$
That is $S_{p+1}<S_p$, and we have proved that
$$
S_p<S_{p-1}\quad \Longrightarrow\quad S_{p+1}<S_p
$$
Now it is easy to check that $S_5=\frac{8}{5}<\frac{5}{3}=S_4$, hence the sequence
$(S_p)_p$ is decreasing starting from $p=4$. Since
$$S_1=1~<S_2=\frac{3}{2}<S_3=\frac{5}{3}=S_4=\frac{5}{3}$$
we conclude that
$$\max\{S_p:p\geq 1\}=\frac{5}{3}$$
and it is exactly attained at $p=3,4$.$\qquad\square$
Remark. as Did suggested, we have also the simple lower bound
$$
S_p\geq\frac{p+1}{2^{p+1}}\cdot\frac{2^{p+1}-2}{p}
$$
So, $\ell~{\buildrel{\rm def}\over =}~\lim\limits_{p\to\infty}S_p\geq 1$. This implies that
$$\inf\{S_p:p\geq 1\}=1=S_1.$$
Finally, with a little more work, it can be proved that $\ell=1$.
|
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|
Solve x in logarithm equation I am trying to solve $x$ for $2log_{10} (x-4) - log_{10}4(x-1) = 0$
I have the key with the answer 10 and have confirmed this is correct using Wolfram Alpha but which steps should I take to reach that answer?
|
Start with original problem:
$$2 \log_{10} (x-4) - \log_{10}4(x-1) = 0$$
A rule of logarithm: $2 \log_{10} (x-4)=\log_{10}(x-4)^2$:
$$ \log_{10} (x-4)^2 - \log_{10} (4x-4)= 0$$
Another rule of logarithm: Express as the left side of the equation as a single product:
$$ \log_{10} \frac{(x-4)^2}{4x-4}= 0$$
Convert to exponential form:
$$ 10^{\log_{10} \frac{(x-4)^2}{4x-4}}= 10^0$$
or
$$\frac{(x-4)^2}{4x-4}=1$$
and when algebraically rearranging
$$x^2-8x+16=4x-4$$
or
$$x^2-12x+20=0$$
which is now a quadratic equation that you can solve.
|
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|
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
|
$$n^4-1 = (n-1)(n+1)(n^2+1)$$ The factors $n-1$ and $n+1$ take care of $n \equiv \pm 1 \mod 5$, while if $n \equiv \pm 2 \mod 5$, $n^2 + 1 \equiv 2^2 + 1 \equiv 0 \mod 5$.
|
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|
Is this graph connected Define the following graph on the vertex set ${\mathbb N}_{\geq1}\>$:
Two numbers $a$, $b\in {\mathbb N}_{\geq1}$ are connected by an edge (written $a \ \mathcal{R} \ b)$ if and only if $a+b \ | \ ab-1$.
Clearly $1$ is isolated. Can we connect all integers greater than $2$ to $2$?
For example: $$2014 \ \mathcal{R} \ 147 \ \mathcal{R} \ 4175 \ \mathcal{R} \ 3891 \ \mathcal{R} \ 142 \ \mathcal{R} \ 43 \ \mathcal{R} \ 7 \ \mathcal{R} \ 3 \ \mathcal{R} \ 2.$$ Therefore $2014$ can be connected to $2$ (written $2014\sim2$).
Question: Is this graph connected?
Motivation :
I worked on the Machin formula and I wondered if we had $$\arctan \frac{1}{a} + \arctan \frac{1}{b} = \arctan \frac{1}{c}$$ where $a,b,c$ are integers and this happens if $c=\frac{ab-1}{a+b}$ is an integer.
EDIT : My apologie I forgot to mention that the graph is restricted to positive integer.
|
A partial answer :
We have $a \sim b$ if and only if there exist a sequence of integers $a_1, \ldots, a_n$ such that $a \ \mathcal{R} \ a_1 \ \mathcal{R} \ \cdots \ \mathcal{R} \ a_n \ \mathcal{R} \ b$. The relation $ ab-1 = c (a + b) $ can be written as $(a-c)(b-c)=c^2+1$ and can be solves $a=c+d$ and $b=c+ \dfrac{c ^ 2 + 1} d$ where d is a divisor of $c^ 2 +1$.
If $c$ is even: All divisors of $c^2+1$ are congruent to $1$ modulo $4$ then $a$ is congruent to $b$ modulo $4$.
If $c$ is odd: $d$ is congruent to $1$ modulo $4$ and $\dfrac{c^2}d+1$ is congruent to $2$ modulo 4 or vice versa.
If $c=4k+1$ then $a$ is congruent to $2$ modulo $4$ and $b$ is congruent to $3$ modulo $4$ or vice versa.
If $c = 4k +3$ then $a$ is congruent to $0$ modulo $4$ and $b$ to $1$ modulo $4$ or vice versa.
Therefore there is probably three components in the graph (but I did not prove this):
The first formed only by $1$, the second one formed by integer congruent to $0$ modulo $4$ or $1$, and the last by integer congruent to $2$ or $3$ modulo $4$.
|
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|
For which $a$ does this system have one solution, infinite solutions, or no solution? Consider the system of linear equations:
\begin{cases} x+y+az=1 \\ x+ay+z=1 \\ ax+y+z=1 \end{cases}
For which $a$ does the system have:
*
*no solution
*one solution
*an infinite number of solutions?
So first I reduced the matrix to echelon form
\begin{pmatrix}{cc|cc}
1&1&a&1\\1&a&1&1\\a&1&1&1\end{pmatrix}
and after raw operations got to
\begin{pmatrix}{cc|c}
1&1&a&1\\0&a-1&1-a&0\\0&0&-a^2-a+2&1-a\end{pmatrix}
Therefore $-a^2-a+2=1-a$ so when $a=1, a=-1$ there are infinitely many solutions, but when is there one solution?
|
If $a = 1$, the system reduces to $x + y + z = 1$, and it has more unknowns than equation, hence infinite solutions. If $a \neq 1$, then subtract the second from the first equation:
$(a-1)(z-y) = 0$. So: $z-y = 0$, and $y = z$. Similarly subtract the third from the second equation: $(a-1)(y-x) = 0$. So $y - x = 0$, and $x = y$. Thus: $x = y = z$, and $x + x + ax = 1$. So: $(a+2)x = 1$. So if $a = -2$, system has no solutions, and if $a \neq -2$, it has unique solution: $x = y = z = \dfrac{1}{a+2}$
|
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|
How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$
Find the value:
$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$
I use computer have this reslut
$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{2}\dfrac{\pi\ln^2{2}}{\sqrt{2}}-\dfrac{3}{2}\dfrac{\pi^2\ln{2}}{\sqrt{2}}$$
where $C$ is Catalan constant.
My idea: let $$\tan{x}=t,\sin{2x}=\dfrac{2t}{1+t^2},dx=\dfrac{1}{1+x^2}$$
then
$$I=\int_{0}^{\infty}\ln^2{(1+t^4)}\dfrac{\dfrac{2}{1+t^2}}{2-\left(\dfrac{2t}{1+t^2}\right)^2}\cdot\dfrac{dt}{1+t^2}=\int_{0}^{\infty}\dfrac{\ln^2{(1+t^4)}}{(t^2-1)^2}dt$$
Then I can't.Thank you
|
Here is an approach. I'll continue from the step you have already reached with the correction suggested in Fabien comment which is considering the integral
$$ I = \int_{0}^{\infty} \frac{\ln^2(1+x^4)}{1+x^4} dx. $$
To evaluate the above integral we consider the integral
$$ F = \int_{0}^{\infty} (1+x^4)^{\alpha} dx = \frac{\pi}{2\sqrt{2}\Gamma( 3/4 )}{\frac {\Gamma ( -1/4-\alpha ) }{
\Gamma \left( -\alpha \right) }},$$
which can be evaluated using the beta function techniques (use the substitution $(1+x^4)=\frac{1}{t}$). Now $I$ follows from $F$ as (see related techniques )
$$ I = \lim_{\alpha \to -1}\frac{d^2F(\alpha)}{d\alpha^2}=\frac{\pi \,\sqrt {2}}{48}\left({\pi}^{2}-36\,\pi \,\ln \left( 2 \right) +108 \ln^2( 2 )+12\,\psi'\left( 3/4 \right) \right) .$$
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|
If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c^2(a^{n-2})=a^{n-2}(b^2+c^2)=a^{n-2}\times a^2=a^n \implies b^n+c^n<a^n$
But I stuck in part (b). Can I prove it in a way I proved part (a)?
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We need to show: $\left(\dfrac{b^2}{a^2}\right)^{x} + \left(\dfrac{c^2}{a^2}\right)^{x} > 1$ for $0< x = \dfrac{n}{2}< 1$, and $a^2 = b^2 + c^2$. Let $u = \dfrac{b^2}{a^2}$, and $v = \dfrac{c^2}{a^2}$, then $0 < u,v < 1$, and $u+v = 1$, and we now prove: $u^x + v^x > 1$.
Consider $f(x) = u^x + v^x$ on $(0,1)$. We have: $f'(x) = u^x\cdot \ln u + v^x\cdot \ln v < 0$ because $\ln u < 0$, and $\ln v < 0$ for $0 < u,v < 1$. Thus $f(x)$ decreases, and $f(x) > f(1) = u+v = 1$.
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|
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Then write an explicit expression for (a) $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$, and (b) an expression for $J_1+2J_3$ in terms of $J_0$.
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$.
Remember that
$J_k(z)=c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz$ where $c_k=\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}$
$J_{k-1}(z)=c_{k-1} z^{k-1} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{3}{2}}dz$ where $c_{k-1}=\frac{1}{2^{k-1} \Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}$
$$\frac{d}{dz}(z^kJ_k(z)) =\frac{d}{dz}(z^k c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k \frac{d}{dz}(z^{2k} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k [2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)+ z^{2k} (e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1] $$
But $e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1=0$. So
$$\frac{d}{dz}(z^kJ_k(z)) =c_k 2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz) =\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =\frac{1}{2^k-1 (k-\frac{1}{2})\Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}kz^{k-1}z^k\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =z^k c_{k-1}kz^{k-1}\int_{-1}^1 \frac{e^{izs}(1-z^2)^{k-\frac{1}{2}}}{k-\frac{1}{2}}dz$$
How do I contunie from here?
Write an explicit expression for $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$
We know that $J_\frac{1}{2}(x)=\sqrt{\frac{2}{\pi}}\frac{\sin(x)}{\sqrt{x}}$ when $c_\frac{1}{2}=\frac{1}{\sqrt{2}\Gamma(1)\Gamma(\frac{1}{2})}$. By the formula in the previous part, we now that
$$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x) \iff x^kJ_k(x)=\int_\textbf{R} x^kJ_{k-1}(x)dx \iff J_k(x)=\int_\textbf{R} J_{k-1}(x)dz$$
$$J_\frac{3}{2}(z) =\int_\textbf{R}J_\frac{1}{2}(x) \,dx= \sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx =\sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx=(1+i)\frac{\pi}{2}$$
$$J_\frac{5}{2}(x) =\int_\textbf{R}J_\frac{3}{2}(x) \,dx =\int_\textbf{R}(1+i){\frac{\pi}{2}} \,dx $$
Does $J_\frac{5}{2}(x)=\infty$?
Write an explicit expression for $J_1+2J_3$ in terms of $J_0$
By the formula in the previous part,
$$J_1(x) =\int_\textbf{R} J_0(x)\,dx$$
$$J_3(x) =\int_\textbf{R}J_2(x)=\int_{\textbf{R}^2} J_1(x)\,dx^2=\int_{\textbf{R}^3} J_0(x)\,dx^3 $$
So $J_1+2J_3=\int_\textbf{R} J_0(x)\,dx+2\int_{\textbf{R}^3} J_0(x)\,dx^3$
Please check
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This is only a partial answer and would be far too long for a comment, but I must quote which is limited in comments.
How are x and z related? Are the $J_k$ the Bessel functions of the first kind?
For them it is correct that $$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x)$$
see http://dlmf.nist.gov/10.6#ii or Abramowitz and Stegun 9.1.30.
So I guess you just mix-up $z$ and $x.\;$ Even with these assumptions your last statements
By the formula in the first part,
$$xJ_1(x)=x^2J_0(x) \iff J_1(x)=xJ_0(x)$$
$$x^3J_3(x)=x^6J_2(x)=x^9J_1(x)=x^{12}J_0(x) \iff J_3(x)=x^9J_0(x)$$
are blatantly wrong because you obviously omit the derivatives or product rule! For example with $k=0:$
$$xJ_0(x)= \frac{d}{dx}(x^1J_1(x))=J_1(x) + xJ_1'(x)$$
And if you know a bit about Bessel functions you would immediately see your relations cannot be correct (all higher order functions would vanish at the zeroes of $J_0$!)
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|
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
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Divide through by $\cos x$ to get
\begin{align}
\tan x + 1 & = \frac 12 \sec x \tag{1}\\
\text{Square both sides: } \qquad\qquad\tan^2 x + 2 \tan x + 1 & = \frac 14 \sec^2 x = \frac{1}{4} (\tan^2 x + 1) \\
3 \tan^2 x + 8 \tan x + 3 & = 0 \\
\tan x & = \frac{-8 \pm \sqrt{28}}{6} = \frac{-4 \pm \sqrt{7}}{3}.
\end{align}
Check the two solutions with the original equation before squaring (equation $(1)$). Because $x \in [0, \pi]$ and $\tan x < 0$ in both cases, we know that $\sin x \ge 0$ and $\cos x \le 0$, so $\sec x \le 0$.
If $\tan x = \frac{-4 + \sqrt{7}}3$, then $\tan x + 1 = \frac{-1 + \sqrt{7}}3 > 0$, contradicting $(1)$ and $\sec x \le 0$.
If $\tan x = \frac{-4 - \sqrt{7}}3$, then $\tan x + 1 = \frac{-1 - \sqrt{7}}3 < 0$. This looks good.
If you want to be more sure, compute $\sin x$ and $\cos x$ from the chosen $\tan x$:
\begin{align}
\sin x & = \frac{4 + \sqrt 7}{\sqrt{3^2 + (-4 - \sqrt{7})^2}} \\
& = \frac{4 + \sqrt 7}{\sqrt{32 + 8\sqrt 7}} = \frac{1}{4}\sqrt{8 + 2\sqrt{7}}
= \frac{1}{4}(1 + \sqrt{7}) \\
\cos x & = -\sqrt{1 - \sin^2 x} = -\frac{1}{4}\sqrt{8 - 2\sqrt 7} = \frac{1}{4}(1 - \sqrt{7}).
\end{align}
It's obvious that $\sin x + \cos x = \frac 12$, $\sin x \ge 0$ and $\cos x \le 0$.
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|
How do you calculate this limit $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{n^2+k^2}$? How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$
Wolfram alpha is unable to calculate it.
This is a question from a questions booklet, and the options for the answer are--
$\begin{align}
&A) \dfrac{\pi}{2} \\
&B) \log 2 \\
&C) \dfrac{\pi}{4} \\
&D) \dfrac{1}{2} \log 2
\end{align}$
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Clearly,
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k^2} &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\frac{k^2}{n^2}}
\stackrel{\text{Riemann sum}}\longrightarrow \int_0^1 \frac{x\,dx}{1+x^2}=\left.\frac{1}{2}\log (1+x^2)\right|_0^1\\
&=\frac{1}{2}\log 2.
\end{align}
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|
Evaluation of $ \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx$ Evaluation of $\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx$
$\bf{My\; Solution:}$ Using $\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$
$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$ and $\displaystyle \cos \phi = \frac{n}{\sqrt{x^2+n^2}}$ and $\displaystyle \tan \phi = \frac{x}{n}\Rightarrow \phi = \tan^{-1}\left(\frac{x}{n}\right)$
So Integral is $$\displaystyle = \int \sec^2(x-\phi)\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx$$
Now Let $$\displaystyle (x-\phi) = y\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)=y$$. Then $$\displaystyle \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx = dy$$
So Integral is $$\displaystyle \int \sec^2(y)dy = \tan y +\mathbb{C} = \tan\left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)+\mathbb{C}$$
So $$\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx = \left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathbb{C}$$
My Question is , Is there is any other solution other then that, because It is very Complex and Lengthy.
If Yes, The please write here
Thanks
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This is a shorter, but perhaps uglier, solution:
Using the identities $n^2+x^2=(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2$ and
$n=\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)$, we can rewrite the integral as
$\displaystyle \int\frac{(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2-[\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)]}{(n\cos x+x\sin x)^2} dx$
$\displaystyle=\int\frac{(n\cos x+x\sin x)[n\cos x+x\sin x-\cos x]-(n\sin x-x\cos x)[-n\sin x+x\cos x+\sin x]}{(n\cos x+x\sin x)^2} dx$
$\displaystyle=\int\bigg(\frac{n\sin x-x\cos x}{n\cos x+x\sin x}\bigg)^{\prime} dx$
$\;\;\;\;\displaystyle=\frac{n\sin x-x\cos x}{n\cos x+x\sin x}.$
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|
Finding a limit to negative infinity with square roots: $\lim\limits_{x\to -\infty}(x+\sqrt{x^2+2x})$ Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$
I start by multiplying with the conjugate:
$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\right)\right]$$
$$\lim_{x\to-\infty} {x^2 - (x^2 + 2x)\over x - \sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2x\over x - \sqrt{x^2+2x}}$$
divide by highest power of denominator
$$\lim_{x\to-\infty} {(\frac1x)(-2x)\over (\frac1x) x - ({1\over \sqrt{x^2}})\sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2\over 1 - \sqrt{1+\frac2x}} = {-2 \over 1-\sqrt{1 + 0}} = {-2 \over 0}$$
but I know this is wrong as the answer is $-1$. Where did I mess up? Thanks.
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For better clarity set $\dfrac1x=-y\implies y\to0^+$
$$\lim_{x\to-\infty}(x+\sqrt{x^2+2x})=\lim_{y\to0^+}\frac{\sqrt{1-2y}-1}y$$
$$=\lim_{y\to0^+}\frac{1-2y-1}{y(\sqrt{1-2y}+1)}$$
Hope you can take it home from here
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|
Upper bound on $ \binom{a}{m+1}\sum ^m_{j=0} \binom{a-m-1}{j}/\binom{b}{j+m+1}$ Given $a,b,m$ such that $0<2m<a<b$.
I would like to find out upper bound of $$S = \binom{a}{m+1}\sum ^m_{j=0} \frac{\binom{a-m-1}{j}}{\binom{b}{j+m+1}}$$
Anyone can help me please?
Thank you so much.
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I begin by using this identity: $\int_0^1 t^{\alpha} (1-t)^{\beta}=\frac{!}{\alpha+\beta+1}\frac{1}{\binom{\alpha+\beta}{\alpha}}$, for some positive $\alpha$ and $\beta$. This can be alternatively expressed as follows: for some positive $n$ and $k$, we have $\frac{1}{\binom{n}{k}}= (n+1) \int_0^1 t^{k}(1-t)^{n-k} dt$. Applying this to the right hand side, we have:
$$S=(b+1) \binom{a}{m+1} \sum_{j=0}^m \binom{a-m-1}{j} \int_0^1 t^{j+m+1}(1-t)^{b-j-m-1} dt\\
= (b+1) \binom{a}{m+1} \int_0^1 (1-t)^{b-m-1} t^{m+1} \left[\sum_{j=0}^m \binom{a-m-1}{j} \left( \frac{t}{1-t}\right)^j \right] dt \\
\leq (b+1) \binom{a}{m+1} \int_0^1 (1-t)^{b-m-1} t^{m+1} \left[\sum_{j=0}^{a-m-1} \binom{a-m-1}{j} \left( \frac{t}{1-t}\right)^j \right] dt
\\
= (b+1) \binom{a}{m+1} \int_0^1 (1-t)^{b-m-1} t^{m+1} \left[ \left(1+ \frac{t}{1-t}\right)^{a-m-1} \right] dt
\\
= (b+1) \binom{a}{m+1} \int_0^1 t^{m+1} (1-t)^{b-a} \\
= \frac{b+1}{b-a+m+2} \frac{ \binom{a}{m+1}}{\binom{b-a+m+1}{m+1}}
\\
=(b+1)\frac{a!(b-a)!}{(b-a+m+2)!(a-m-1)!}
$$
In case I have made arithmetic mistakes, I am sure you can pick it up and mend it, since I hope the main logistics are clear
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A problem on Mean Value Theorem If $f''(x)$ exists on $[a,b]$ and $f'(a)=f'(b)$, then :
$$f(\frac{a+b}{2})=\frac 1 2[f(a)+f(b)]+\frac{(b-a)^2}{8}f''(c)$$
for some $c\in(a,b)$.
I tried but was unable to think of a function and was unable to use the given condition except for Rolle's Theorem which does not yield anything useful(yet).
Any hints or help will be appreciated.
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there exists $x_0$ and $y_0$ in $[a,b]$ such that
\begin{align}
f(a) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(a - \frac{a+b}{2}) + \frac{1}{2}f''(x_0)(a-\frac{a+b}{2})^2 \\
f(b) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(b - \frac{a+b}{2}) + \frac{1}{2}f''(y_0)(b-\frac{a+b}{2})^2
\end{align}
Multiplying by $\frac{1}{2}$ and summing together these two equations gives
\begin{align}
f(\frac{a+b}{2}) = \frac{1}{2}[f(a) + f(b)] - \frac{f''(x_0) + f''(y_0)}{2}\frac{(a-b)^2}{8}
\end{align}
Use Darboux's theorem to find a $c \in [a,b]$, such that $f''(c) = \frac{f''(x_0) + f''(y_0)}{2}$
The above part $\textbf{didn't}$ use the assumption $f'(a) = f'(b)$, thus the conclusion is different, instead of + , I have a -. I correct in the following:
there exists $x_1$ and $y_1$ in $[a,b]$ such that
\begin{align}
f(\frac{a+b}{2}) = f(a) + f'(a)(\frac{a+b}{2} - a) + \frac{1}{2}f''(x_1)(a-\frac{a+b}{2})^2 \\
f(\frac{a+b}{2}) = f(b) + f'(b)(\frac{a+b}{2} - b) + \frac{1}{2}f''(y_1)(b-\frac{a+b}{2})^2
\end{align}
Since $f'(a) = f'(b)$, multiplying by $\frac{1}{2}$ and summing together these two equations gives
\begin{align}
f(\frac{a+b}{2}) = \frac{1}{2}[f(a) + f(b)] + \frac{f''(x_1) + f''(y_1)}{2}\frac{(a-b)^2}{8}
\end{align}
The same theorem allows to get the conclusion.
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Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
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Add both:
$$\frac{2x}{x^2+4x+3}=\frac{Ax+A+Bx+3B}{x^2+4x+3}$$
$$\frac{2x}{x^2+4x+3}=\frac{(A+B)x+(A+3B)}{x^2+4x+3}$$
Then how this must be valid for all $x$ then $A+B=2$ and $A+3B=0$
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|
Evaluate the limit of $\ln(\cos 2x)/\ln (\cos 3x)$ as $x\to 0$ Evaluate Limits
$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{\ln(\cos(3x))}$$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
\begin{align*}
\lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\stackrel{LHR}{=}\\
&= \lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\\
&=\lim _{x\to \:0}\frac{\left(\ln \left(\cos \left(2x\right)\right)\right)'}{\left(\ln \left(\cos \left(3x\right)\right)\right)'}\\
&=\lim _{x\to \:0}\left(\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}}\right)\\
&=\lim _{x\to \:0}\left(\frac{2\sin \left(2x\right)\cos \left(3x\right)}{3\cos \left(2x\right)\sin \left(3x\right)}\right)\\
&\stackrel{LHR}{=}\lim _{x\to \:0}\frac{\left(2\sin \left(2x\right)\cos \left(3x\right)\right)'}{\left(3\cos \left(2x\right)\sin \left(3x\right)\right)'}\\
&=\lim _{x\to \:0}\left(\frac{4\cos \left(2x\right)\cos \left(3x\right)-6\sin \left(3x\right)\sin \left(2x\right)}{9\cos \left(3x\right)\cos \left(2x\right)-6\sin \left(2x\right)\sin \left(3x\right)}\right)\\
&=\lim _{x\to \:0}\left(\frac{2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)}{3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)}\right)\\
&=\frac{\lim _{x\to \:0}\left(2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)\right)}{\lim _{x\to \:0}\left(3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)\right)}=\dfrac{4}{9}
\end{align*}
*
*Could we do it in others ways?
|
As told in answers, if you are just concerned by $$\lim_{x\to 0}\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}$$ the shortest way is L'Hopital's rule using the hints given by RecklessReckoner and JimmyK4542. You will then arrive to $$\lim_{x\to 0}\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}=\frac{a^2}{b^2}$$
But, if you want to know the path to this limit, Taylor expansion is really simple. Start with the series expansion $$\cos(t)= 1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^5\right)$$ and the series expansion of $$\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$$ in which you replace now $y$ by $(-\frac{t^2}{2}+\frac{t^4}{24})$ to arrive to $$\log\Big(cos(t)\Big)=-\frac{t^2}{2}-\frac{t^4}{12}+O\left(t^6\right)$$ Now, replace $t$ by $ax$ for getting the numerator and $t$ by $bx$ for getting the denominator and perform the polynomial division. You should then get $$\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}= \frac{a^2}{b^2}+\frac{1}{6} a^2 \left(\frac{a^2}{b^2}-1\right)x^2+O\left(x^3\right)$$ which shows you that the limit is reached from above if $a \gt b$ and from below if $a \lt b$.
|
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|
Trigonometric problem: $2^{\sin{x}} + 2^{\cos{x}} \ge 2^{(1-1/{\sqrt2})}$ Show that:
$$\large2^{\sin{x}} + 2^{\cos{x}} \ge 2^\left({1-\frac{1}{\sqrt{2}}}\right)$$
This looks like an am gm problem to me where we should be using the fact that am is more that or equal to gm but I am having problem solving this equation after this:
$\large\left(2^{\sin{x}} + 2^{\cos{x}}\right)/2 \ge \sqrt{2^{\sin{x}+\cos{x}}}$
Please solve the sum step wise. Thanks
|
By the am-gm inequality,
\begin{align}
\frac{2^{\sin x}+2^{\cos x}}{2}
&\ge 2^{\frac{1}{2}(\sin x+\cos x)}\\
\end{align}
Multiplying both sides of the inequality by $2$ yields
$$2^{\sin x}+2^{\cos x}\ge2\cdot 2^{\frac{1}{2}(\sin x+\cos x)}$$
Since $\color\red{\sin{x}+\cos{x}=\sqrt{2}\sin(x+\pi/4)}$, we have
\begin{align}
RHS
&=2\cdot2^{\frac{1}{2}(\sin{x}+\cos{x})}\\
&=2^{1+\frac{1}{\sqrt{2}}\sin(x+\pi/4)}\\
&\ge2^{1-\frac{1}{\sqrt{2}}}
\end{align}
as $-1\le \sin(x+\pi/4)\le 1.$
Hence
$$\color\red{2^{\sin x}+2^{\cos x}}\ge 2^{1+\frac{1}{\sqrt{2}}\sin(x+\pi/4)}\ge \color\red{2^{1-\frac{1}{\sqrt2}}}$$
|
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|
Prove $\sum \limits_{cyc} \frac{1+b^2+c^4}{a^+b^2+c^3}\geq 3$
If $a,b,c$ are positive real numbers, prove $$\sum \limits_{cyc} \frac{1+b^2+c^4}{a+b^2+c^3}\geq 3$$
Additional info: We should only use Cauchy (preferred to used at least once and more than AM-GM) and AM-GM. We are not allowed to use induction.
Things I have done so far: My idea is about separating question into $3$ inequality and proving them one by one and then sum all of them to prove question inequality.So$$\sum \limits_{cyc} \frac{1}{a+b^2+c^3} \geq 1$$$$\sum \limits_{cyc} \frac{b^2}{a+b^2+c^3} \geq 1$$
$$\sum \limits_{cyc} \frac{c^4}{a+b^2+c^3} \geq 1$$
UPDATE
As Macavity pointed in comments, the separating idea is not useful because first inequality does not hold.So any hint for starting step is appreciated.
|
Using Cauchy-Schwarz, note
$$LHS = \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a+b^2+c^3)^2} \ge \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a^2+b^2+c^2)(1+b^2+c^4)} = \sum_{cyc} \frac{a+b^2+c^3}{a^2+b^2+c^2}$$
As the denominators of the cyclic sum are now the same, it only remains to show that
$$\sum_{cyc} (a + b^2+c^3) \ge 3(a^2+b^2+c^2) \iff \sum_{cyc} a + \sum_{cyc} a^3 \ge 2\sum_{cyc} a^2$$
which follows from AM-GM as $a+a^3 \ge 2a^2$.
|
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|
If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$. If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$.
I tried the following,
As we know, $(a+b)^2=a^2+b^2+2ab$
Therefore, $(4x^2+y^2)^2=(4x^2)^2+(y^2)+2(4x^2*y^2)$
$=16^4+y^4+8x^2y^2$
What should I do know? Please help. I am stuck.
|
Hint:
$(2x+y)^2-4xy=4x^2+y^2$
You can take it from here?
|
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|
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written in the same form: $49$ is already shown, and $7$ is $2^2+3\cdot 1^2$.
I'm looking for a general proof of this using infinite descent. I tried to let $a^2+3b^2=xy$, where $x,y$ are positive integer factors of the LHS, but couldn't continue from here I'm afraid. However, I have managed to prove that integers of the form $a^2+3b^2$ are closed under multiplication, i.e. the product of two of them is of the same form. I was hoping this would help.
I also had another question if you have time: Is the representation in the form $a^2+3b^2$ unique? For example, is it possible for an integer, say $N$, to be $N=a^2+3b^2=x^2+3y^2$ with $a\ne x$ and $b\ne y$?
EDIT: Sorry, here is the "closed under multiplication" proof I found:
Using Diophantus' Identity we have $(a^2+3b^2)(x^2+3y^2)=(a^2+(\sqrt{3}b)^2)(x^2+(\sqrt{3}y)^2=(ax+\sqrt{3}\cdot \sqrt{3} y)^2)+(a\cdot \sqrt{3} y - \sqrt{3} b\cdot x)^2=(ax+3by)^2+3(ay-bx)^2$ as required.
EDIT (2): I am very sorry. The question should be:
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.
|
For your second question. Not unique. $$28=4^2+3\cdot 2^2=5^2+3\cdot 1^2=1^2+3\cdot 3^2.$$
$$84=6^2+3\cdot 4^2=3^2+3\cdot 5^2.$$
|
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|
How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? I am able to solve equation of the type $ |5x+1|=|11-2x|$. I square both the side and my equation becomes $ (5x+1)^2=(11-2x)^2 $ further simplification gives me $ (5x+1)=\pm (11-2x)$.
I get have difficulties in solving equation of the type
$|x+1| - |1-x|=2~~$ and
$~~|x-1|=|x|+a$
Any steps and or links that could be of help will be greatly appreciated.
|
In this case, $|x + 1| - |1 - x| = 2$, we have,
$$
|x + 1| = \begin{cases} x+1 \ \text{if} \ x \geq -1\\ -x-1 \ \text{if} \ x < -1 \end{cases}
\quad \text{and} \quad
|1 - x| = \begin{cases} 1 - x \ \text{if} \ x \leq 1\\ x-1 \ \text{if} \ x \geq 1 \end{cases}
$$
If $x < -1, \ |x + 1| =-x-1$ and $|1 - x| = 1 - x$, then
$$
|x + 1| - |1 - x| = 2 \quad \Rightarrow \quad -x-1-(1 - x) = 2 \ \Rightarrow \ -2 = 2 \quad \text{Absurd!}
$$
If $-1 \leq x < 1, \ |x + 1| =x + 1$ and $|1 - x| = 1 - x$, then
$$
|x + 1| - |1 - x| = 2 \quad \Rightarrow \quad x + 1 - (1 - x) = 2 \ \Rightarrow \ 2x = 2 \quad \Rightarrow \quad x = 1
$$
If $x > 1, \ |x + 1| = x + 1$ and $|1 - x| = x - 1$, then
$$
|x + 1| - |1 - x| = 2 \quad \Rightarrow \quad x + 1 - (x - 1) = 2 \ \Rightarrow \ 2 = 2 \quad \text{valid for all} \quad x > 1
$$
Thus, $S = \{x \in \mathbb{R} : x \geq 1\}$. The other issue resolves also analyzing cases.
|
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|
Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
$$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$
I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that:
$$\frac{3v^2}{w^3}\geq\frac{9u^3-9uv^2+w^3}{u^4}+\frac{7}{4u},$$
which is a linear inequality of $v^2$,
which says that it's enough to prove our inequality for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$.
Hence, $c=3-2a$, where $0<a<\frac{3}{2}$ and we get
$$(2a-1)^2(24-45a+26a^2-4a^3)\geq0,$$
which is true for $a\in\left(0,\frac{3}{2}\right)$.
Done!
|
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|
How to calculate limit of a function having factorial in denominator For $n$ tending to infinity find the following limit
$$\frac{2^n}{n!}.$$
I have a feeling that it is multiplication of many numbers with the last one turning to $0$ but the first one is finite so limit should be $0$. But I am not sure and neither am I able to put it in mathematical form.
Thank you for your help.
|
$$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n-2}$$
As $n \rightarrow \infty$, $\left ( \frac{2}{3} \right )^{n-2} \rightarrow 0$
Therefore, from the Squeeze Theorem $$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$
|
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|
If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$
$a,b,c,d>0$ satisfying $a^3+b^3+c^3+d^3=1$. Prove
$$\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$$
I tried to go the normal way, by Cauchy-Schwarz, but that doesn't work. So I tried to incorporate this newly learned idea, since $a,b,c,d<1$ we can write the left as a power series:
$$\sum_{n=0}^{\infty}(bcd)^n+(cda)^n+(dab)^n+(abc)^n$$
If we can show, $(bcd)^n+(cda)^n+(dab)^n+(abc)^n\ge (K(a^3+b^3+c^3+d^3))^n$ for some suitable constant $K$ we can finish. But I can't really do it. Can someone help me?
|
The AM-GM inequality yields
$$ abcd \le \frac{1}{4 \sqrt[3]{4}} \iff abc \le \frac{1}{4 \sqrt[3]{4}d} \iff \frac{1}{1-abc} \le \frac{1}{1 - \frac{1}{4 \sqrt[3]{4} d}}. $$
It follows that
$$ \sum_{cyclic} \frac{1}{1-abc} \le \sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1}.$$
Therefore it is sufficient to show that
$$ \sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1} \le \frac{16}{3} \iff 4 - \sum_{cyclic} \frac{-1}{4 \sqrt[3]{4} d - 1} \le \frac{16}{3} \iff \sum_{cyclic} \frac{-1}{4 \sqrt[3]{4} d - 1} \ge \frac{-4}{3} \iff \sum_{cyclic} \frac{1}{1 - 4 \sqrt[3]{4}d} \ge \frac{-4}{3}. $$
Now using a form of the Cauchy-Schwarz inequality usually referred to as Titu's Lemma (after Titu Andreescu), we find that
$$
\sum_{cyclic} \frac{1}{1 - 4 \sqrt[3]{4}d} \ge \frac{(1 + 1 +1 + 1)^2}{\sum_{cyclic} (1 - 4 \sqrt[3]{4}d)} = \frac{16}{4 - 4 \sqrt[3]{4}} (a + b + c + d).
$$
Hence it suffices to show that
$$ \frac{16}{4 - 4 \sqrt[3]{4}(a + b + c + d)} \ge \frac{-4}{3} \iff \frac{1}{1 - \sqrt[3]{4}(a + b + c + d)} \ge \frac{-1}{3} \iff 3 \ge -1 + \sqrt[3]{4} (a + b + c + d) \iff 4 \ge \sqrt[3]{4} (a + b + c + d) \iff \sqrt[3]{16} \ge a + b + c + d. $$
The last inequality follows from Hölder's inequality since
$$ (a^3 + b^3 + c^3 + d^3)(1 + 1 + 1 + 1)(1 + 1 + 1 + 1) \ge (a + b + c + d)^3 \iff \sqrt[3]{16} \ge a + b + c + d. $$
Therefore the conclusion follows and equality holds for $ a = b = c = d = \frac{1}{\sqrt[3]{4}}.$
|
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|
Apostol (6.25.40): Find $\int {x^{-2}}\sqrt{2 - x - x^2} dx$ I've been struggling with this exercise from Apostol for some time (Section 6.25, Question 40). The integral is
$$
\int\frac{\sqrt{2-x-x^2}}{x^2}\, dx
$$
with a Hint of "multiply numerator and denominator by $\sqrt{2-x-x^2}$".
NB: the answer supplied in the book is
$$
-\frac{\sqrt{2-x-x^2}}{x} +\frac{\sqrt{2}}{4}\log(\frac{\sqrt{2-x-x^2}}{x} - \frac{\sqrt{2}}{4}) - \arcsin\frac{2x+1}{3} + C
$$
I performed the hint and I obtained the following:
$$
\int\frac{\sqrt{2-x-x^2}}{x^2}\, dx = \int\frac{\sqrt{2-x-x^2}}{x^2}\cdot \frac{\sqrt{2-x-x^2}}{\sqrt{2-x-x^2}}\cdot dx
$$
$$
= \int\frac{2-x-x^2}{x^2\sqrt{2-x-x^2}}\cdot dx
$$
$$
= \int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx - \int\frac{x}{x^2\sqrt{2-x-x^2}}\cdot dx- \int\frac{x^2}{x^2\sqrt{2-x-x^2}}\cdot dx
$$
$$
= \int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx - \int\frac{dx}{x\sqrt{2-x-x^2}} - \int\frac{dx}{\sqrt{2-x-x^2}}
$$
Now I can easily show that
$$
\int\frac{dx}{\sqrt{2-x-x^2}} = \arcsin{\frac{2x+1}{3}}
$$
I have not tackled the integral $\int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx $ though I suspect a substitution of $x+0.5 = 1.5 \sin{t}$ would help since I already know that $2-x-x^2 = (\frac{3}{2})^2 - (x+\frac{1}{2})^2$. The integral I am struggling with (at the moment) is $\int\frac{dx}{x\sqrt{2-x-x^2}} $.
So far I have tried the following:
*
*An initial substitution of $x+\frac{1}{2} = \frac{3}{2}\sin(t)$ which resulted in the following:
$$
\int\frac{dx}{x\sqrt{2-x-x^2}} = -2\int\frac{dt}{1-3\sin(t)}
$$
*A second substitution of $u = \tan\frac{t}{2}$ gave the following:
$$
-2\int\frac{dt}{1-3\sin(t)} = -4\int\frac{du}{u^2-6u+1}
$$
*Using integration by partial fractions, I can show that
$$
-4\int\frac{du}{u^2-6u+1} = \frac{-1}{\sqrt{2}}\log|\frac{u-3-2\sqrt{2}}{u-3+2\sqrt{2}}|
$$
*This would mean that
$$
-2\int\frac{dt}{1-3\sin(t)} = \frac{-1}{\sqrt{2}}\log|\frac{\tan\frac{t}{2}-3-2\sqrt{2}}{\tan\frac{t}{2}-3+2\sqrt{2}}|
$$
*Since $x+\frac{1}{2} = \frac{3}{2}\sin(t)$ implies $\cos(t) = \frac{2}{3}\sqrt{2-x-x^2}$ and $\tan\frac{t}{2} = \frac{1-\cos(t)}{\sin(t)}$ we start to get something that is clearly not looking like anything close to the answer
So my questions that I am looking for answers are:
*
*Is this the right approach for this integral or am I overlooking something more obvious?; and
*Will a similar approach be required for the integral $\int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx$ ?
|
For the integral $\int\dfrac{dx}{x\sqrt{2-x-x^2}}$, we can substitute $x=\dfrac{1}{t}$, so that $dx=-\dfrac{1}{t^2}dt$ and the integral becomes $\int\dfrac{-\frac{1}{t^2}dt}{\frac{1}{t}\sqrt{2-\frac{1}{t}-\frac{1}{t^2}}}=-\int\dfrac{dt}{\sqrt{2t^2-t-1}}$ which can be evaluated easily. I think this should work out.
|
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|
Evaluation of $\int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
Evaluation of $\displaystyle \int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
My Try:: Let $\displaystyle I = \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx = \int \ln(\cos x+\sqrt{\cos 2x})\cdot \csc^2 xdx$
So $\displaystyle I = -\ln\left(\cos x+\sqrt{\cos 2x}\right)\cdot \cot x+\int \frac{1}{\left(\cos x+\sqrt{\cos 2x}\right)}\cdot \left(-\sin x-\frac{1}{2\sqrt{\cos 2x}}\cdot 2\cdot \sin 2x\right)\cdot \cot xdx$
Now How Can I solve after that
Help me
Thanks
|
Hint: After integrating by parts, try substituting $u=\sin{x}$. Also, it will be helpful to recall the trigonometric identity, $\cos{(2x)}=\cos^2{x}-\sin^2{x}=2\cos^2{x}-1=1-2\sin^2{x}$.
Let $I$ be your integral, then:
$$\begin{align}
I
&=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cot{x}\left(\sin{x}+\frac{\sin{(2x)}}{\sqrt{\cos{(2x)}}}\right)}{\cos{x}+\sqrt{\cos{(2x)}}}\mathrm{d}x\\
&=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cos{x}\left(1+\frac{2\cos{x}}{\sqrt{\cos{(2x)}}}\right)}{\cos{x}+\sqrt{\cos{(2x)}}}\mathrm{d}x\\
&=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cos{x}\left(1+\frac{2\sqrt{1-\sin^2{x}}}{\sqrt{1-2\sin^2{x}}}\right)}{\sqrt{1-\sin^2{x}}+\sqrt{1-2\sin^2{x}}}\mathrm{d}x\\
&=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\left(1+\frac{2\sqrt{1-u^2}}{\sqrt{1-2u^2}}\right)}{\sqrt{1-u^2}+\sqrt{1-2u^2}}\mathrm{d}u\\
&=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\frac{1}{\sqrt{1-2u^2}}-\sqrt{1-u^2}}{u^2}\mathrm{d}u.
\end{align}$$
This last integral may then be split up into two fairly easy integrals.
|
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|
How to solve $(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$
Solve $$(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$$
I have a very similar problem like this on my homework, and I have no clue how to set it up or even start. How could I set this up?
|
$$(x-3)(\frac{dy}{dx})+y=6e^x \Rightarrow \frac{dy}{dx}+\frac{1}{x-3}y=\frac{6 e^x}{x-3} \Rightarrow y'(x)+\frac{1}{x-3}y(x)=\frac{6 e^x}{x-3} $$
The solution of the homogeneous problem is the following:
$$y'_h(x)+\frac{1}{x-3}y_h(x)=0 \Rightarrow \frac{dy_h}{dx}=-\frac{1}{x-3}y_h \Rightarrow \frac{dy_h}{y_h}=-\frac{dx}{x-3} \Rightarrow \int \frac{dy_h}{y_h}=-\int \frac{dx}{x-3} \\ \Rightarrow \ln y_h=-\ln{|x-3|} +c \Rightarrow \ln {|y_h|}=\ln{(|x-3|)^{-1}} +c \Rightarrow e^{\ln {|y_h|}}=e^{\ln{(|x-3|)^{-1}} +c} \\ \Rightarrow |y_h|=e^c \frac{1}{|x-3|} \Rightarrow y_h=\pm e^c \frac{1}{x-3} \overset{\pm e^c=C}{\Rightarrow} y_h(x)=\frac{C}{x-3}$$
The solution of the non-homogengeneous problem is the following:
We suppose that the solution is of the form: $$y_p(x)=\frac{C(x)}{x-3}$$
Replacing this at the problem we get:
$$y'_p(x)+\frac{1}{x-3}y_p(x)=\frac{6 e^x}{x-3} \Rightarrow \frac{C'(x)(x-3)-C(x)}{(x-3)^2}+\frac{1}{x-3}\frac{C(x)}{x-3}=\frac{6 e^x}{x-3} \Rightarrow \frac{C'(x)(x-3)-C(x)+C(x)}{(x-3)^2}=\frac{6 e^x}{x-3} \Rightarrow C'(x)=6 e^x \Rightarrow C(x)=6 e^x+c_2$$
So, $$y_p(x)=\frac{6e^x+c_2}{x-3}$$
The solution of the initial problem is equal to the sum of the homogeneous and the non-homogeneous problem $$y=y_h+y_p$$
|
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|
What is the probability that a person has both of these attributes? The probability that a blue-eyed person is left-handed is $\frac{1}{7}$.
The probability that a left-handed person is blue-eyed is $\frac{1}{3}$ .
The probability that a person has neither of the attributes is $\frac{4}{5}$.
What is the probability that a person has both attributes?
I just don't know how they got the answer
Answer: $\frac{1}{45}$
|
Draw a Venn diagram!
Let probabilities be as follows:
$a$=Blue only
$b$=Blue and Left
$c$=Left only
$d$=None
From probabilities given,
$$\begin{align}
\frac b{a+b}=\frac 17 \quad \Rightarrow a&=6b\\
\frac b{b+c}=\frac 13 \quad \Rightarrow c&=2b\\
d&=\frac 45\\
\end{align}$$
As the probabilites sum to 1,
$$\begin{align}
a+b+c+d&=1\\
6b+b+2b+\frac 45&=1\\
9b&=\frac 15\\
b&=\frac 1{45}\end{align}$$
|
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|
Integral of $R(R^2+y^2)^{-3/2}$ with respect to $y$ $$\int_0^\infty \frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}dy$$
The indefinite integral seems to be $$\frac{-R}{\sqrt{R^2+y^2}}+C$$
$R$ is a constant
|
Using $y=Rx$, then $x=\tan(\theta)$ we get
$$
\begin{align}
\int_0^\infty\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y
&=\frac1R\int_0^\infty\frac{\mathrm{d}x}{\sqrt{1+x^2}\left(1+x^2\right)}\\
&=\frac1R\int_0^{\pi/2}\cos(\theta)\,\mathrm{d}\theta\\
&=\frac1R
\end{align}
$$
It seems that you have miscomputed the primitive.
Computing the Primitive
Combining the previous substitutions with $y=R\tan(\theta)$, we get
$$
\begin{align}
\int\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y
&=\frac1R\int\cos(\theta)\,\mathrm{d}\theta\\
&=\frac1R\sin(\theta)+C\\
&=\frac1R\frac{y}{\sqrt{R^2+y^2}}+C
\end{align}
$$
|
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|
Proving that a number is non-negative? The numbers $a$,$b$ and $c$ are real.
Prove that at least one of the three numbers
$$(a+b+c)^2 -9bc \hspace{1cm} (a+b+c)^2 -9ca \hspace{1cm} (a+b+c)^2-9ab$$
is non-negative.
Any hints would be appreciated too.
|
$$\sum [(a+b+c)^2-9bc]=3\sum[a^2+b^2+c^2-ab-bc-ca]=\frac32\sum (a-b)^2\ge0$$
If each $(a+b+c)^2-3bc<0,$ $$\sum [(a+b+c)^2-3bc]<0$$
|
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|
Finding the sum of $3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3$ I see this:
$$A=3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3=3\cdot ([4^{\log n}-1]/3)=n^2-1$$
The base of logarithm is $2$, and $n$ is $2,4,8,\dots$
Anyone could describe me how this sum was calculated? Some hints or some tutorial for this?
|
Maybe we need to take a few steps back ( though I imagine this has all been covered in your course). If $a$ is any real number other than $1$ and $r$ is a positive integer, then $1 + a + a^{2} +\ldots +a^{r-1} = \frac{a^{r}-1}{a-1}.$ If you need to verify that, multiply the left side by $a-1$ and notice that there is a lot of cancellation. If $b$ is another real number, then we see that $b + b.(a) + b.(a^{2}) +\ldots +b.(a^{r-1}) = b.\frac{a^{r}-1}{a-1}.$
In this problem, $b =3, a= 4$ and $r = \log_{2}(n)$ (where, although it not explcitly said, $n$ must be a (positive integer) power of $2)$. Then $a^{r} = 2^{2\log_{2}(n)} = n^{2}.$ Also, $b = 3,$ and $a-1 = 3,$ so the answer does indeed simplify to $n^{2}-1.$
|
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|
Can one use logarithms to solve the equations $2=3^x + x$ and $2=3^x x$? Could someone explain how would you solve:
$$2=3^x + x$$
and
$$2=3^x \cdot x$$
I can only solve halfway through.
And why is
$$10^{\log (x)}= x$$
Thanks
|
By definition, a log is a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.
Below is a simple example,
\[ 10^{2}=100 \]
So
\[ \log_{10} 100=\log_{10} 10^{2}=2\log_{10} 10= 2 \]
And
\[
10^{\log_{10} 100}=10^{2} = 100
\]
Generally
\[
b^{\log_{b}(x)}=x
\]
Also, the solutions to both of those problems cannot be found using elementary functions.
To solve these equations, we must use the Lambert W function. This function will provide the value of $x$ in equations that take the form $z=xe^{x}$.
\[
z=xe^{x}\Longleftrightarrow W(z)=x
\]
Since you're curious, here are the solutions
\[ 2=3^{x}x=x3^{x}=xe^{\ln 3^{x}}=xe^{x\ln 3} \]
\[ 2\ln 3=(x\ln 3)e^{x\ln 3} \]
\[ W(2\ln 3) = x\ln 3 \]
\[ x = \frac{W(2\ln 3)}{\ln 3} \]
And after some trial and error,
\[ 2=3^{x}+x \]
\[ 2-x=3^{x} \]
\[ (2-x)3^{2}=3^{x}3^{2} \]
\[ (2-x)\frac{3^{2}}{3^{x}}=9 \]
\[ 9=(2-x)\frac{3^{2}}{3^{x}}=(2-x)3^{2-x}=(2-x)e^{\ln 3^{2-x}}=(2-x)e^{(2-x)\ln 3} \]
\[ 9\ln 3=((2-x)\ln 3)e^{(2-x)\ln 3} \]
\[ W(9\ln 3)=(2-x)\ln 3 = 2\ln 3-x\ln 3 \]
\[ W(9\ln 3) - 2\ln 3= -x\ln 3 \]
\[ x=\frac{2\ln 3 - W(9\ln 3)}{\ln 3} \]
|
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|
Denest $\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$ Is it possible to denest following radical to sum of terms with smaller root count inside?
$\sqrt{20+10 \sqrt{2}-4 \sqrt{5}-2 \sqrt{10}}=\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$
I've found that it cannot be denested into $a+b\sqrt{2}+c\sqrt{5}+d\sqrt{10}$ by squaring and equating coefficients. Resulting system didn't have rational solutions.
Maybe it can be denested to something as sum of $\sqrt{a+b\sqrt{c}}$ or $\sqrt{a+b\sqrt{c}+d\sqrt{e}}$ or even $\sqrt[4]{a+b\sqrt{c}}$?
For example, following can be denested:
$$\sqrt{8+2 \sqrt{2}-2 \sqrt{6}}=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}$$
$$\sqrt{110-60 \sqrt{3}+46 \sqrt{5}-28 \sqrt{15}}=2\sqrt{15+6\sqrt{5}}-\sqrt{50+22\sqrt{5}}$$
$$\sqrt{48+12 \sqrt{2}+16 \sqrt{5}-12 \sqrt{6}+4\sqrt{10}-4\sqrt{30}}=$$
$$=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}+\sqrt{30+15\sqrt{2}}-\sqrt{10-5\sqrt{2}}$$
|
Well if you're still looking for something:
$$\sqrt{2+\sqrt{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}$$
and
$$\sqrt{10-2\sqrt{5}}=2 \cdot \sqrt{\frac{5-\sqrt{5}}{2}}$$
so
$$\sqrt{\left(2+\sqrt{2}\right) \cdot \left(10-2\sqrt{5}\right)}=2\cdot \sqrt{\frac{5-\sqrt{5}}{2}} \cdot \left(\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with degree 3 such that $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots
Note that $\dfrac{\pi}{12}$, $\dfrac{9\pi}{12}$, $\dfrac{17\pi}{12}$ are solution of equation $\cos3\theta=\dfrac{1}{\sqrt{2}}$ and $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are distinct number.
We have $\cos3\theta=4\cos^3\theta-3\cos\theta$. Let $x=\cos\theta$, therefore
$\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots of $4x^3-3x=\dfrac{1}{\sqrt{2}}.$
I want method similar to this to find $P(x)$.
Thank you.
|
You may try below :
Say $r_1, r_2, r_3, r_4$ are roots of polynomial,
$P(x) = \large x^4-\left(\sum r_1\right)x^3 + \left(\sum r_1r_2\right)x^2 - \left(\sum r_1r_2r_3\right)x + r_1r_2r_3r_4 $
Note that the roots define the polynomial only upto a constant factor.
|
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"url": "https://math.stackexchange.com/questions/905303",
"timestamp": "2023-03-29T00:00:00",
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|
How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$? Please help me to find a closed form for this integral:
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx
$$
Routine textbook methods for this complicated integral fail.
|
Let's give a (sketch of the) proof of the following closed form evaluation.
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G,
$$
where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.
Observe that, by the change of variable $\displaystyle u=\frac 1x$, we have
$$
I=\int_0^1\frac{u-1-u\ln u}{\left(1+u^2\right) \ln^2 u} \mathrm du=\int_0^1\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right) \ln^2 u} \mathrm du.
$$
We set
$$
I(s):=\int_0^1u^s\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right) \ln^2 u} \mathrm du, \quad s\geq0.
$$
We are allowed to differentiate $I(s)$ twice to obtain
$$
I''(s)=\int_0^1u^s\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right)} \mathrm du.
$$
Using the standard expansion $\displaystyle \frac{1}{1-u^4}=\sum_{k=0}^{\infty}u^{4k}, \, |u|<1,$ and performing the termwise integration,
$$
\begin{align}
I''(s)=\sum_{k=0}^{\infty}\int_0^1u^{s+4k}(1-u^2)(u-1-u\ln u) \mathrm du
\end{align}
$$
gives
$$
\begin{align}
I''(s)=\sum_{k=0}^{\infty}\left(\frac{1}{(4k+2+s)^2}-\frac{1}{(4k+4+s)^2}+\frac{1}{(4k+2+s)}-\frac{1}{(4k+1+s)}\right).
\end{align}
$$
Now, recall the following series representation of the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$,
$$
\psi(u+1) = -\gamma + \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{u+k}
\right), \quad u >-1,
$$ where $\gamma$ is the Euler-Mascheroni constant and, by differentiation, giving$$
\psi'(u+1) = \sum_{k=1}^{\infty} \frac{1}{(u+k)^2}, \quad u>-1.
$$
Hence
$$
I''(s)=\frac{1}{16}\psi'\left(\frac{s+2}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+4}{4}\right)+\frac{1}{4}\psi\left(\frac{s+4}{4}\right)+\frac{1}{4}\psi\left(\frac{s+1}{4}\right)-\frac{1}{4}\psi\left(\frac{s+3}{4}\right)-\frac{1}{4}\psi\left(\frac{s+2}{4}\right).
$$
We have, as $s \rightarrow +\infty$, $I(s) \rightarrow 0$ and $I'(s) \rightarrow 0$, leading to
$$
I(s)=\log \Gamma \left(\frac{s+2}{4}\right)-\log \Gamma \left(\frac{s+4}{4}\right) +
4 \left(\psi\left(-2, \frac{s+4}{4}\right) + \psi\left(-2, \frac{s+1}{4}\right) -
\psi\left(-2, \frac{s+2}{4}\right) - \psi\left(-2, \frac{s+3}{4}\right)\right),
$$
then, with the use of Wolfram Alpha, as $s \rightarrow 0$ we get
$$
I=I(0)=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G.
$$
|
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|
How to find $\int \frac{x^4-4}{x^2\sqrt{4+x^2+x^4}} \,\mathrm dx$ Integrate $$\int \frac{(x^4-4)}{(x^2\sqrt{4+x^2+x^4})}\mathrm dx$$
My try:
$$\int \frac{(x^2-4/x^2)}{(\sqrt{4+x^2+x^4})}\mathrm dx\\
=\int \frac{ (x^2-4/x^2)}{(\sqrt{(x^2+1/2)^2+15/4})}\mathrm dx\\$$
Let $t=x^2$
$$=\int \frac{(t^2-4)}{2t\sqrt t\sqrt{(t+1/2)^2+15/4}}\mathrm dt=?$$
Answer:Spoiler
$$\frac{\sqrt{4+x^2+x^4}}{x}$$
|
$$\int \frac{(x^4-4)dx}{(x^2\sqrt{4+x^2+x^4})}$$
$$ \int \frac{(x^2-4x^{-2})dx}{(\sqrt{4+x^2+x^4})} $$
$$ \int \frac{(x-4x^{-3})dx}{(\sqrt{4x^{-2}+1+x^2})} $$
Take, $$ u = 1+x^2+4x^{-2} \implies du = (2x-8x^{-3})dx \implies du/2 = (x-4x^{-3})dx$$
So integral =
$$ \int \frac{du/2}{\sqrt{u}} = \sqrt{u} + C \implies \frac{\sqrt{4+x^4+x^2}}{x} +C $$
|
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|
How to evaluate $\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$ How to evaluate: $$\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$$
I have done a substantial work on it:
Let $x^5z^3=x^5-1$. So
$$x^5(z^3-1)=1\implies 5x^4(z^3-1)\mathrm{d}x+x^5(-3z^2\mathrm{d}z)=0\implies \mathrm{d}x=\frac{3xz^2\mathrm{d}z}{5(z^3-1)}$$
So:
$$\int \frac{\mathrm{d}x}{x^4[x(x^5-1)]^{1/3}}\text{ or }\int x^{-13/3}(x^5-1)^{-1/3}\mathrm{d}x\\
=\int x^{-13/3}(x^5z^3)^{-1/3}.\frac{3xz^2\mathrm{d}z}{5(z^3-1)}=\frac35\int \frac{x^{-13/3}x^{-5/3}z^{-1}xz^2\mathrm{d}z}{x^{-5}}\\
=\frac35\int z\;\mathrm{d}z=\frac{3}{10}\left(\frac{x^5-1}{x^5}\right)^{2/3}+\mathcal{C}$$
Is my working correct? Is there an easier way?
Actually(thanks to @Jean-ClaudeArbaut):
*
*$x^5(\color{red}{1-z^3})=1$
*$5x^4(z^3-1)\mathrm{d}x+x^5(\color{red}{3}z^2\mathrm{d}z)=0$
|
$$
\int \dfrac{1}{x^4\left[x\left(x^5-1\right)\right]^{1/3}}dx\tag{1}
$$
use the sub $u = 1/x^3\implies du = -\dfrac{3}{x^4} dx$.
thus Eq. (1) becomes
$$
\int \dfrac{1}{\left(\dfrac{1}{u^2}-\dfrac{1}{u^{1/3}}\right)^{1/3}}\dfrac{du}{-3} = \dfrac{1}{-3}\int \dfrac{u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}du
$$
then $v = \left(1-u^{5/3}\right)^{2/3}$
$$
dv = \dfrac{2}{3}\dfrac{-\dfrac{5}{3}u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}\rightarrow -\dfrac{9}{10}dv = \dfrac{u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}du
$$
I think this correct, but then again I have made mistakes in the past!
|
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|
How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three variable inequality
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.
Prove that
$$\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$
also see:can see
：http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=243
from this equality,I have see a nice methods:
I think this Four varible inequality is also true
First,Thank you Aditya answer,But I read it your solution,it's not true
|
Let's use Lagrange Multipliers:
Let use take $f$ to be:
$$f(a,b,c,d)=\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)$$
subject to constraint $abcd=1$:
$$g(a,b,c,d)=abcd=1$$
Here are the equations we need to solve:
$$\frac{-1}{(a+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda bcd\\
\frac{-1}{(b+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda acd\\
\frac{-1}{(c+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda abd\\
\frac{-1}{(d+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda abc\\
abcd=1
$$
Let $\displaystyle Z:=\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}$
$$Z=\frac{\lambda}a+\frac1{(a+3)^2}=\frac{\lambda}b+\frac1{(b+3)^2}=\frac{\lambda}c+\frac1{(c+3)^2}=\frac{\lambda}d+\frac1{(d+3)^2}$$
Since $a=b=c=d$ seems to satisfy the equations from intuition. If you don't believe see this page on W|A, which says that when $ab(a+3)(b+3)\ne0$(which is true as $a,b,c,d>0$) the solutions(solved for two variables) are:
$$a=b,\lambda=-\frac{(a b (a+b+6))}{((a+3)^2 (b+3)^2)}$$
Since, now $a=b=c=d(=1)$, the value of $f$ seems to be:
$$f(1,1,1,1)=4\times\frac14-4\times\frac14=0$$
Which apparently is its minimum value.
Now for the general case of n-variables the equations would be:
$$\frac{-1}{(x_1+(n-1))^2}+\sum_{cyc}\dfrac{1}{(\sum_{cyc}x_i+1)^2}=\lambda \frac{\prod x_i}{x_1}=\frac{\lambda}x_1\tag{cycle equation 'n' times}\\\prod x_i=1$$
Again let $\displaystyle Z:=\sum_{cyc}\dfrac{1}{(\sum_{cyc}x_i+1)^2} $
Now $$Z=\frac{\lambda}x_i+\frac{1}{(x_1+(n-1))^2}$$
Again extending that logic we would get $$x_1=x_2=\cdots=x_n(=1)$$
So the value of $f$ here would be:
$$f(1,1,\cdots,1)=\sum_{i=1}^n\frac{1}{1+\sum_{j=1,j\ne i}^n1}-\sum_{i=1}^n\frac{1}{(n-1)+1}\\=n\times\frac1{1+(n-1)}-n\times\frac1{(n-1)+1}=0$$
Which again is ...
|
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|
Finding substitution in the integral $\int{\frac{2+3x}{3-2x}}dx$ In a problem sheet I found the integral $$\int{\frac{2+3x}{3-2x}}dx.$$
In the solution the substitution $z=3-2x$ is given which yields $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2}dz$. We have
$$\int{\frac{2+3x}{3-2x}}dx = \int{\frac{2+3\left(\frac{3-z}{2}\right)}{z}}\left(-\frac{1}{2}dz\right)=\cdots$$
How do I recognise these kind of substitutions and how do I choose them?
After reading the line $z=3-2x \Rightarrow x=\frac{3-z}{2} \Rightarrow dx=-\frac{1}{2}dz$. I was able to solve the integral. My main problem is, how do I recognise from the integrand that I have to substitute and solve for the free variable $x$. Is there a trick to remember?
Dmoreno's hint which I was at first not able to follow but after some thinking made me realise that equations like $\frac{a+bx}{c-dx}$ can be alternated like:
$$\frac{a+bx}{c-dx} = \left(\frac{a+bx}{c-dx} + \frac{b}{d}\right) - \frac{b}{d} = \frac{ad+bc}{d(c-dx)} - \frac{b}{d},$$
therefore $$\int{\frac{a+bx}{c-dx}}dx = \frac{ad+bc}{d}\cdot\int{\frac{1}{c-dx}}dx - \frac{b}{d}\cdot \int dx,$$
which made my problem much easier.
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Hint: note that
$$ \frac{2+3x}{3-2x} = \frac{13}{2 (3-2x)}-\frac{3}{2} \tag{1}$$
Can you take it from here?
Addendum:
\begin{array}{cccccc}
\color{blue}{2} & \color{blue}{+} & \color{blue}{3 x }& | & \color{red}{3} & \color{red}{-} & \color{red}{2x} \\ \hline
-9/2 & + & 3 x & | & \color{purple}{-\frac{3}{2} } & & \\
\color{green}{13/2} & + & 0 & |
\end{array}
Since $\color{blue}{D} = \color{red}{d} \, \color{purple}{c} + \color{green}{r}, $ it follows that:
$$2+ 3x = -(3-2x)\frac{3}{2} +\frac{13}{2},$$
which readily yields eq. $(1)$.
|
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|
Is it true that $\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1$? This question is inspired by the formula $$\displaystyle\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots = \zeta(2)-1,$$
see for instance this question.
It appears that for any integer $n\geq 0$, we have
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1.$$
I believe that Yiorgos' answer could be generalized to prove the above formula. Can anyone pull it off?
Side question: If $n$ is replaced by a complex number $s$, for which values of $s$ does the integral converge? (We interpret $x^s$ as $\exp(x\log s)$, where $\log s$ is the principal branch of the complex log.) If the above formula is true, then for any integer $n\geq 0$, have $$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((n+1)x^n-nx^{n-1}) dx = \zeta(n+2) - 1.$$
Is it true that for complex value of $s$ for which the integral converges, we have
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1?$$
|
To answer the side question.
Assume that $ \Re s >1$. Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ the initial integral can be rewritten as:
$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(s+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(s+2)}dt \\&= \frac{1}{s+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-(k+1)^{-(s+1)}\right)\\&= \frac{1}{s+1}\left(\sum_{k=1}^\infty \frac{1}{k^{s+2}}-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}\right)
\end{align}$$
Thus
$$
\int_0^1 \lfloor x^{-1} \rfloor^{-1} (s+1)\:x^s \:\mathrm dx = \zeta(s+2)-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}, \quad \Re s >1 \tag1
$$ and we have
$$
\int_0^1 \lfloor x^{-1} \rfloor^{-1} s\:x^{s-1} \:\mathrm dx = \zeta(s+1)-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}, \quad \Re s >2. \tag2
$$
Then we may substract $(2)$ from $(1)$, using
$$
\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}=\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}\left(1-\frac{1}{(k+1)}\right)=\zeta(s+1)-1
$$
to get, by analytic continuation,
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1, \quad \Re s >0.$$
|
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|
How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either.
|
Let $u=k\sin x$, so $du=k\cos x dx$. Then
$\displaystyle\int \frac{\sqrt{1-k^2\sin^2 x}}{\sin x} dx=k\int\frac{\sqrt{1-u^2}}{u}\cdot\frac{1}{\sqrt{k^2-u^2}}du=k\int\sqrt{\frac{1-u^2}{k^2-u^2}}\cdot\frac{1}{u}du$.
Now let $z=\sqrt{\frac{1-u^2}{k^2-u^2}}$, so $u^2=\frac{1-k^2z^2}{1-z^2}$ and $2udu=\frac{2z(1-k^2)}{(1-z^2)^2}dz$.
Then $\displaystyle\int k\sqrt{\frac{1-u^2}{k^2-u^2}}\cdot\frac{1}{u}du=k\int z\frac{1-z^2}{1-k^2z^2}\frac{z(1-k^2)}{(1-z^2)^2}dz=k(1-k^2)\int\frac{z^2}{(1-z^2)(1-k^2z^2)}dz.$
Then $\frac{z^2}{(1-z^2)(1-k^2z^2)}=\frac{A}{1-z}+\frac{B}{1+z}+\frac{C}{1-kz}+\frac{D}{1+kz}\implies A=B=\frac{-1}{2(k^2-1)}\text{ and } C=D=\frac{1}{2(k^2-1)},$
so $\displaystyle\frac{k}{2}\int\left(\frac{1}{1-z}+\frac{1}{1+z}-\frac{1}{1-kz}-\frac{1}{1+kz}\right) dz$
$=\displaystyle\frac{k}{2}\left[\ln\left\vert\frac{1+z}{1-z}\right\vert-\frac{1}{k}\ln\left\vert\frac{1+kz}{1-kz}\right\vert\right]+C=k\tanh^{-1}t-\tanh^{-1}\frac{t}{k}+C,$ where $\displaystyle z=\sqrt{\frac{1-k^2\sin^{2}x}{k^2-k^2\sin^{2}x}}=\frac{\sqrt{1-k^2\sin^{2}x}}{k\cos x}$ for $k>0$ and $\cos x>0$
and $\displaystyle t=\frac{1}{z}=\frac{k\cos x}{\sqrt{1-k^2\sin^{2}x}}$.
|
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|
How to evaluate the following two integral combined with anti-trigonometric function and trigonometric function？ \begin{align*}
&\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} \\
&\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx}.
\end{align*}
A few days ago,my e-friend asked me two integral,but he also doesn't know how to solve it,can somebody show me how to find their closed values?
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In fact, they are similar to Coxeter’s integrals, we use Sangchul Lee's result can get
\begin{align*}
\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} & = \frac{{{11\pi ^2}}}{72},\\
\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} & = \frac{{{\pi ^2}}}{6}.
\end{align*}
We can also try other methods!
$$\int_0^{\frac{\pi }{2}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{5 \pi ^2}{24}$$
$$\int_0^{\frac{\pi }{3}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{2 \pi ^2}{15}$$
$$\int_0^{\frac{\pi }{5}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{71 \pi ^2}{900}$$
|
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|
If the equation $2x^2-7x+12=0$ has two roots alpha and beta ,then the value of alpha/beta+beta/alpha is If the equation $2x^2-7x+12 =0$ has two roots $\alpha$ and $\beta$ ,
then the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is
note $x=\frac{-7+\sqrt{47}}{4},\frac{-7-\sqrt{47}}{4}$
then
$$\frac{\frac{-7+\sqrt{47}}{4}}{\frac{-7-\sqrt{47}}{4}}+\frac{\frac{-7-\sqrt{47}}{4}}{\frac{-7+\sqrt{47}}{4}}$$
so 96/2
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First, note that your quantity can be rewritten as
$\frac{\alpha^2 + \beta^2}{\alpha \beta}$,
and the numerator and denominator here are both symmetric polynomials in $\alpha$ and $\beta$.
On the other hand, for any polynomial, the coefficients themselves are in fact symmetric polynomials in the roots. By dividing by the leading coefficient (which doesn't change the roots), we may as well assume that a polynomial is monic (which in our case gives us $x^2 - \frac{7}{2} x + 6$).
Now, if $\alpha$ and $\beta$ are the roots of a monic polynomial $x^2 + bx + c$, we must have
$x^2 + b x + c = (x - \alpha)(x - \beta) = x^2 + (- (\alpha + \beta)) x + \alpha \beta$, and comparing like coefficients gives
$-b = \alpha + \beta$ and $c = \alpha \beta$.
In fact, these elements generate the ring of symmetric polynomials in $\alpha$ and $\beta$, so we can express all such polynomials as polynomial combinations of these. In particular, we find that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = (-b)^2 - 2 c = b^2 - 2c$. So, for a monic polynomial, the ratio is
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{b^2 - 2c}{c} = \frac{b^2}{c} - 2$.
Evaluating for $b = -\frac{7}{2}$, $c = 6$ gives the result for your polynomial.
|
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|
How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$? How to get $f(x)$, if we know that $f(f(x))=x^2+x$?
Is there an elementary function $f(x)$ that satisfies the equation?
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Here’s a technique for finding the first few terms of a formal power series representing the fractional iterate of a given function like $f(x)=x+x^2$. I repeat that this is a formal solution to the problem, and leaves unaddressed all considerations of convergence of the series answer.
I’m going to find the first six terms of $f^{\circ1/2}(x)$, the “half-th” iterate of $f$, out to the $x^5$-term. Let’s write down the iterates of $f$, starting with the zero-th.
\begin{align}
f^{\circ0}(x)&=x\\
f^{\circ1}=f&=x&+x^2\\
f^{\circ2}&=x&+2x^2&+2x^3&+x^4\\
f^{\circ3}&\equiv x&+3x^3&+6x^3& + 9x^4& + 10x^5& + 8x^6\\
f^{\circ4}&\equiv x &+ 4x^2& + 12x^3& + 30x^4& + 64x^5& + 118x^6\\
f^{\circ5}&\equiv x& + 5x^2& + 20x^3& + 70x^4& + 220x^5& + 630x^6\\
f^{\circ6}&\equiv x& + 6x^2& + 30x^3& + 135x^4& + 560x^5& + 2170x^6\\
f^{\circ7}&\equiv x& + 7x^2& + 42x^3& + 231x^4& + 1190x^5& + 5810x^6\,,
\end{align}
where the congruences are modulo all terms of degree $7$ and more.
Now look at the coefficients of the $x$-term: always $1$. Of the $x^2$-term? In $f^{\circ n}$, it’s $C_2(n)=n$. The coefficient of $x^3$ in $f^{\circ n}$ is $C_3(n)=n(n-1)=n^2-n$, as one can see by inspection. Now, a moment’s thought (well, maybe several moments’) tells you that $C_j(n)$, the coefficient of $x^j$ in $f^{\circ n}$, is a polynomial in $n$ of degree $j-1$. And a familiar technique of finite differences shows you that
\begin{align}
C_4(n)&=\frac{2n^3-5n^2+3n}2\\
C_5(n)&=\frac{3n^4-13n^3+18n^2-8n}3\,,
\end{align}
I won’t go into the details of that technique. The upshot is that, modulo terms of degree $6$ and higher, you have $f^{\circ n}(x)\equiv x+nx^2+(n^2-n)x^3+\frac12(2n^3-5n^2+3n)x^4+\frac13(3n^4-13n^3+18n^2-8n)x^5$.
Now, you just plug in $n=\frac12$ in this formula to get your desired series. And I’ll leave it to you to go one degree higher, using the iterates I’ve given you.
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|
Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational How can I prove that the sum $\sqrt 2+\sqrt[3] 3$
is an irrational number ??
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If $x=\sqrt2+\sqrt[3]3$, then
$$
\begin{align}
3
&=(x-\sqrt2)^3\\
&=x^3-3\sqrt2x^2+6x-2\sqrt2\\
(x^3+6x-3)^2&=2(3x^2+2)^2\\
0&=x^6-6x^4-6x^3+12x^2-36x+1
\end{align}
$$
Thus, $x$ is an algebraic integer. Since $2\lt x\lt3$, $x\not\in\mathbb{Z}$, so $x\not\in\mathbb{Q}$. In this answer, it is shown that a rational algebraic integer is an integer.
Alternative approach
If $x=\sqrt2$ and $y=\sqrt[3]3$, then $x^2-2=0$ and $y^3-3=0$. Thus, both $x$ and $y$ are algebraic integers. Therefore, $x+y=\sqrt2+\sqrt[3]3$ is also an algebraic integer.
Note that this approach, while seeming simpler, requires that one knows that the sum of two algebraic integers is again an algebraic integer. This fact can be shown using some linear algebra.
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|
Sum of a series problem involving cubes For any odd integer $n$, evaluate $n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3$
How would you go about solving such a problem?
Any help would be appreciated.
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I would use the following ideas:
(1): Standard formula: $1^3+2^3+\cdots+a^3=\left(\frac{a(a+1)}{2}\right)^2$
(2): Use (1) to get an expression for $2^3+4^3+6^3+\cdots+(2a)^3=2^3\cdot(1^3+2^3+\cdots+a^3)$
(3): Use (1) and (2) to get an expression for $1^3+3^3+5^3+\cdots+(2a-1)^3$ by noting that
$$1^3+3^3+5^3+\cdots+(2a-1)^3=\left(1^3+2^3+\cdots+(2a)^3\right)-\left(2^3+4^3+\cdots+(2a)^3 \right)$$
(4) Your expression breaks down to an expression from (3) minus an expression from (2) (since your $n$ is supposed to be odd--if your $n$ were even you could still use these ideas to get your expression for the alternating sum.)
|
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|
prove that $ a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$ I have:
let $a$, $b$ and $c$ be non-negative real numbers with sum $2$. Prove that
$$a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$$
I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.
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This inequality is come from? This inequality can use this
$$\Longleftrightarrow \frac{(a^2+b^2+c^2)(a+b+c)^4}{16}-2(\sum{a^3b^3}+4a^2b^2c^2)=\frac{3abc(a+b+c)}{4}\sum{(a-b)^2}+\sum\frac{c(2a+2b+c)(a^2+b^2-c^2)^2}{16}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2}+\frac{a^2b^2c^2}{4}$$
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|
How do you find the minimum percentage who have studied all four subjects? At college, 70% students studied Maths, 75% students studied English, 85% studied french and 80% studied german. What percentage at least must have studied all 4?
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Using the Generalized Principle of Inclusion-Exclusion, we compute:
$N(0)=100$
$N(1)=70+75+80+85=310$
$N(2)=x_1+x_2+x_3+x_4+x_5+x_6=X$
$N(3)=y_1+y_2+y_3+y_4= Y$
$N(4)=Z$
$100\binom{0}{0}-310\binom{1}{0}+X\binom{2}{0}-Y\binom{3}{0}+Z\binom{4}{0}= 0 $ Study Nothing
$\hphantom{100\binom{0}{1}-}310\binom{1}{1}-X\binom{2}{1}+Y\binom{3}{1}-Z\binom{4}{1}=D$ Study One subject
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+}X\binom{2}{2}-Y\binom{3}{2}+Z\binom{4}{2}=C$ Study two subjects
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+X\binom{2}{2}-}Y\binom{3}{3}-Z\binom{4}{3}=B$ Study three subjects
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+X\binom{2}{2}-Y\binom{3}{3}+}Z\binom{4}{4}=A$ Study four subjects.
$x_i$ = study any two subjects
$y_i$ = study any three subjects
$100 - 310 + X - Y + Z = 0 => 100 - 310 + C + 3B + 6A - B - 4A + A = 0 =>C + 2B + 3A = 210$ ... (1)
$310 - 2X + 3Y - 4Y = D => D = 310 - 2(C+3B+6A) + 3(B+4A) - 4A = 310 - 2C - 3B - 4A$
$X - 3Y + 6Z = C => X = C +3Y - 6Z = C + 3(B+4A) - 6A = C + 3B +6A$
$Y - 4Z = B => Y = B + 4A$
$Z = A$
After working backwards, we have gotten to a point we have reduced it to three variables,namely, percentage of students who study two subjects (C), percentage of students who study three subjects (B) and finally percentage of students who study four subjects (A) with following equation $ C + 2B + 4A = 210$. If we have C and B we could get A. But I still wonder how you could get the minimum A with just four numbers in the question.
Thanks
Satish
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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.