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Nested... binomials coefficients? Can I have a proof that this number exists?
The number:
$$\binom{1}{\binom{2}{\binom{3}{\binom{4}{\vdots}}}}$$
If the number exists, then what is the closed form of that number?
|
Note that ${i\choose i+1} = 0$.
So $A_2 ={1 \choose 2} = 0$,
$A_3 ={1\choose{2\choose 3}} = {1 \choose 0} = 1$
Now $A_4={1\choose{2\choose {3\choose 4}}} = {1\choose{2\choose 0}} = {1\choose{1}} = 1$
Lastly
$$A_5={1\choose{2\choose {3\choose {4\choose 5}}}} = {1\choose{2\choose {3 \choose 0}}} = {1\choose{2\choose 1}} = {1\choose 2} = 0$$
If you consider a translation
$${1+1\choose{2+1\choose {3+1\choose {4+1\choose 5+1}}}} = {1+1\choose{2+1\choose {3+1 \choose 0}}} = {1+1\choose{2+1\choose 1}} = {1+1\choose 2+1} = 0$$
So you find a period for your expression.
Consider $$A_n = {1 \choose {2 \choose \underset{n}{\vdots} }}$$
Since $${k+1\choose{k+2\choose {k+3\choose {k+4\choose k+5}}}} = \ldots = 0 = {k+1\choose k+2}$$
We obtain $A_{10} = A_7 = A_4 = 1 $. To the general case take $n \mod 3 = r$ $A_n = A_{r + 3}$ or you could consider $A_0 = 1A_1 = 1$ then $$A_0 = A_3,A_1 = A_4, A_2= A_5 $$
That is $A_n = A_r$ (where $r = n \mod 3$)
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Showing that hyperbolic trigonometric functions parameterize the unit hyperbola I know that the same way circular trigonometry is defined over the circle $ x^2 + y^2 = 1 $, hyperbolic trigonometry is defined over the hyperbola $ x^2 - y^2 = 1 $.
What I don't know is how deduced the formulas
$$ \sinh x = \frac {e^x - e^{-x}} {2} \quad \text{and} \quad \cosh x = \frac {e^x + e^{-x}} {2} $$
are deduced.
My question is: How are the formulas for $ \sinh x $ and $ \cosh x $ deduced from the equation $ x^2 - y^2 = 1 $ of the unit hyperbola?
|
I'll prove $\sinh(x)$ = $\frac {e^x - e^{-x}} {2}$ and leave the proof of $\cosh(x)$ as an exercise for you. So, $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$
And we know that, $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
And we also know that $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...$$
Therefore, $$e^x - e^{-x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...)$$
Now, after we open the brackets, the negative terms will become positive and positive terms will become negative, so let's just do that
$$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - 1 + x - \frac{x^2}{2!} + \frac{x^3}{3!} - ...$$
And we are left with : $$2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!} + ...$$ and we'll take the $2$ common so we get,
$$2(x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...)$$
And, we saw earlier that $$\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$$.
So we get that $$e^x - e^{-x} = 2 \times \sinh(x)$$ and hence we get that $$\sinh(x) = \frac {e^x - e^{-x}}{2}$$
$Q.E.D$
Hope it helps
|
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|
Application of diagonalization of matrix - Markov chains Problem: Suppose the employment situation in a country evolves in the following manner: from all the people that are unemployed in some year, $1/16$ of them finds a job next year. Furthermore, from all the people that were employed in some year, $1/8$ of them loses their jobs the next year, while the rest keeps working. Suppose at this moment there are $4$ million people working and half a million people are unemployed. How will the employment situation look like over a year from now? In $2$ years? In $100$ years?
Attempt at solution: I know this is an application of diagonalization, where I have to use the fact that $A^k = P D^k P^{-1}$, with $D$ being some diagonal matrix.
I was thinking of setting up a difference equation. Let $x_t$ denote the number of working people in year $t$, and let $y_t$ denote the number of unemployed people in year $t$. Then should I set up a relation of the form \begin{align*} v_t = \begin{pmatrix} x_t \\ y_t \end{pmatrix} = A v_{t-1} ? \end{align*}
This would relate the employment situation at year $t$ to some previous year. I found the matrix $A$ as $A = \begin{pmatrix} 7/8 & 1/16 \\ 1/8 & 15/16 \end{pmatrix}$ by writing above the column the two options 'from a job' and 'from no job' and left of the rows I wrote 'to a job' and 'to no job'. The characteristic polynomial of this matrix is \begin{align*} \det(xI_2 - A) = x^2 - \frac{29}{16} x + \frac{13}{16} = \frac{1}{16} (16x - 13)(x-1) \end{align*} So the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = \frac{13}{16}$. An eigenvalue $v_1$ corresponding to $\lambda_1$ is a non-zero solution to \begin{align*} (\lambda_1 I_2 - A) v_1 = 0 \end{align*} or \begin{align*} \begin{pmatrix} 1/8 & 1/16 \\ 1/8 & 1/16 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{align*} This can be rowreduced. A possible eigenvector is $v_1 = \begin{pmatrix} -1/2 \\ 1 \end{pmatrix}$. Similarly I found an eigenvector $v_2$ corresponding to $\lambda_2$ as $v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. If I let $P = \begin{pmatrix} -1/2 & 1 \\ 1 & 1 \end{pmatrix}$, then $P^{-1} = \begin{pmatrix} -2/3 & 2/3 \\ 2/3 & 1/3 \end{pmatrix}$. But then if I want to write $D = P^{-1}A P$, it doesn't work ( I don't get a diagonal matrix).
Can someone point out where I went wrong please, and if this is the correct strategy to solve this problem?
|
The method is correct.
The eigenvector corresponding to $\lambda_1=1$ should be
$$v_1 = \begin{pmatrix} 1/2 \\ 1 \end{pmatrix}$$
The signs were wrong in the matrix $\lambda_1I_2-A$.
The signs were also wrong for the other eigenvector. You must have made the same mistake.
|
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|
proof by induction $2^n \leq 2^{n+1}-2^{n−1}-1$ My question is prove by induction for all $n\in\mathbb{N}$, $2^n \leq 2 ^{n+1}-2^{n−1}-1$
My proof
$1+2+3+4+....+2^n \leq 2^{n+1}-2^{n−1}-1$
Assume $n=1$,$1 ≤ 2$
Induction step
Assume statement is true for $n=k$, show true for $n=k + 1$
$1+2+3+4+....+2^k+2^k+1 ≤ 2 ^{k+1} −2^{k−1} - 1$
$2^{k+1} - 2^{k-1} -1 + 2^{k+1} ≤ 2^{k+1} - 2^{k-1} -1$
$4^{k+1} -2^{k-1} -1 ≤ 2^{k+1} - 2^{k-1} -1$
I do not know how to proceed from here, and i am confused because it seems to me this is not true.
|
Alternatively, we could prove it directly without induction:
\begin{align*}
2^{n + 1} - 2^{n - 1} - 1
&= (2^n + 2^n) - 2^{n - 1} - 1 \\
&= 2^n + (2^{n - 1} + 2^{n - 1}) - 2^{n - 1} - 1 \\
&= 2^n + 2^{n - 1} - 1 \\
&\geq 2^n + 2^{1 - 1} - 1 &\text{since } n \in \mathbb N \implies n \geq 1 \\
&= 2^n
\end{align*}
as desired. $~~\blacksquare$
|
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|
Solving a system of five polynomials I am trying to solve the following system of equations for tuple $\left(a,b,c,d,t\right) \in \mathbb{R}^{4} \times [0,1]$, with parameter $\ell\in\mathbb{R}$.
$$
\begin{eqnarray}
a\frac{t^{2}}{2} - bt + 1 = 0 \qquad (1)\\
a \frac{t^{3}}{6} - b \frac{t^{2}}{2} + t - \ell = 0 \qquad (2)\\
c \frac{t^{2}}{2} - dt + \left(d - \frac{c}{2} - 1\right) = 0 \qquad (3)\\
c \frac{t^{3}}{6} - d \frac{t^{2}}{2} + \left(d - \frac{c}{2} - 1\right)t + \left(\frac{c}{3} - \frac{d}{2} + 1 - \ell\right) = 0 \qquad (4)\\
at - b - ct + d = 0 \qquad (5)
\end{eqnarray}
$$
This arose while solving an optimal control problem, whose solution is known. That known solution tells that if $\frac{1}{6} \leq \ell \leq \frac{1}{4}$, then the above system of equations has unique solution given by $a = 24(1-4\ell)$, $b = 8(1-3\ell)$, $c = -24(1-4\ell)$, $d = 8(1-3\ell) - 24(1-4\ell)$, $t = \frac{1}{2}$. Otherwise, (when $\ell$ does not satisfy the stated inequality) the above system has no solution. While this seems like an easy exercise, I am not able to recover these results. I feel that these facts are not as difficult to establish as I am making it ... so any help or insight in proving them would be appreciated.
So far I have tried these (other than verifying the claimed solution):
*
*Brute force: from (5), substitute $t = \frac{b-d}{a-c}$ in the first four equations and try to solve the resulting four polynomials for $(a,b,c,d)$. As you can see, this soon leads to complications, specially in establishing the bound for $\ell$, and in using the fact that $0 \leq \frac{b-d}{a-c} \leq 1$.
*Let us call the LHS of (2), a cubic polynomial in $t$, as $p(t)$, and the LHS of (4), another cubic, as $q(t)$. Then (1) and (2) say that $p(t) = p^{\prime}(t) = 0$, meaning $p(t)$ has (at least) a double real root $t$. This, using (1), tells that the quadratic discriminant $= b^{2} - 2 a \geq 0$, and using (2), tells that the cubic discriminant $ = \frac{3a^{2}}{4}\ell^{2} + \frac{1}{2}\left(3ab + b^{3}\right)\ell + \left(\frac{2a}{3} - \frac{b^{2}}{4}\right) = 0$. Similarly, (3) and (4) tells that $q(t) = q^{\prime}(t) = 0$, meaning $q(t)$ has (at least) a double real root $t$, which in turn, yields two similar conditions on the quadratic discriminant of (3) and on the cubic discriminant of (4). It is not clear to me how to use the resulting relations together with (5).
*Consider the quadratic (in $t$) equation obtained by subtracting (1) from (3), the cubic (in $t$) equation obtained by subtracting (2) from (4), and the original equation (5), which is linear in $t$. If we call the cubic (in $t$) LHS obtained by subtracting (2) from (4), as $r(t)$, then the resulting three equations are $r(t) = r^{\prime}(t) = r^{\prime\prime}(t) = 0$, which tells that the cubic $r(t)$ has a real root with multiplicity 3, with the root being $\frac{b-d}{a-c}$. Again, not sure how to go from here to recover the desired results.
|
Puting $t=0$ in Eq(1) and $t=1$ in Eq(3) shows us that both $0$ and $1$ are not possible values for t. So, dividing by $t$ and $t-1$ should be fine.
Solving Eq(1) and Eq(2) for $a$ and $b$ in terms of $l$ and $t$ we get
\begin{eqnarray*}
a &=& \frac{6(t-2l)}{t^3} \\
b &=& \frac{2(2t-3l)}{t^2} \\
at - b &=& \frac{6t-12l-4t+6l}{t^2} = \frac{2(t-3l)}{t^2}
\end{eqnarray*}
Solving Eq(3) and Eq(4) for $c$ and $d$ in terms of $l$ and $t$ we get
\begin{eqnarray*}
c &=& \frac{-6\left(t-1+2l\right)}{(t-1)^3} \\
d &=& \frac{-2((2t+1)(t-1)+3l(t+1))}{(t-1)^3} \\
ct-d&=& -2\frac{\left[3t^2-3t+6lt-2t^2+t+1-3lt-3l)\right]}{(t-1)^3} \\
&=& -2\frac{t^2-2t+3lt-3l+1}{(t-1)^3} \\
&=& -2\frac{t-1+3l}{(t-1)^2}
\end{eqnarray*}
Using Eq(5) and above results
\begin{eqnarray*}
\frac{(t-3l)}{t^2}&=&-\frac{t-1+3l}{(t-1)^2} \\
3l(t^2-(t-1)^2) + t(t-1)(2t-1)&=& 0 \\
(3l-t(1-t))(2t-1) &=& 0
\end{eqnarray*}
If $t \ne \frac{1}{2}$ then we can see there are multiple solutions for the system but given $t\in(0,1)-{\frac{1}{2}}$, $l\in(0,\frac{1}{12})$ as $3l=t(1-t)$. But given the input condition that $l\in[\frac{1}{6},\frac{1}{4}]$, $3l$ cant be equal to $t(1-t)$ and hence $t=\frac{1}{2}$. The rest of the solution follows by substituting $t=\frac{1}{2}$ in the above equations.
|
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|
Linear transformation, transformation in base1 I have the following problem, I know how to solve it (knowing the answer) however I do not understand what I'm doing
We have the following base in $\mathbb{R}^4$, called $b_1$,$b_2$,$b_3$,$b_4$
$$B = \left\{
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
\smash{\underbrace{
\begin{array}{ccccc}
1 \\0 \\0 \\ 0 \\
\end{array}
}_{\text{ $b_1$}}}
,
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}} \smash{\underbrace{
\begin{array}{ccccc}
1 \\1 \\0 \\ 0 \\
\end{array}
}_{\text{ $b_2$}}},
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}} \smash{\underbrace{
\begin{array}{ccccc}
0 \\1 \\1 \\ 0 \\
\end{array}
}_{\text{ $b_3$}}},
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}} \smash{\underbrace{
\begin{array}{ccccc}
0 \\0 \\1 \\ 1 \\
\end{array}
}_{\text{ $b_4$}}}
\right\}
\
$$
And the following linear transformation from $\mathbb{R}^4 \rightarrow \mathbb{R}^4$:
$$
f\left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
\smash{\underbrace{
\begin{array}{ccccc}
1 \\0 \\0 \\ 0 \\
\end{array}
}} \right) = \left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
\smash{\underbrace{
\begin{array}{ccccc}
0 \\0 \\1 \\ 1 \\
\end{array}
}} \right) \text{ and } f\left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
\smash{\underbrace{
\begin{array}{ccccc}
0 \\1 \\1 \\ 0 \\
\end{array}
}} \right) = \left( \vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{array}}
\smash{\underbrace{
\begin{array}{ccccc}
1 \\1 \\0 \\ 0 \\
\end{array}
}} \right)
$$
Also, the rest of the transformations of the base remain the same, in other words: $f(b_2) = b_2$ and $f(b_4) = b_4$
What is the representing matrix of F in the base B?
What I did was the following:
I found the inverse of the matrix B, then multiplied this inverse by the left with the "transformed" matrix (using the values that f gives us). Doing this I get to the right answer, which is:
Answer = $\begin{bmatrix}0 & 0 &0&0\\0&1&1&0\\0&0&0&0\\1&0&0&1 \end{bmatrix}$
However I don't understand why the answer is this, I thought I was supposed to multiply the inverse by the right in order to get the transformation of the matrix in the other base. Am I missing something, or misunderstanding what's being asked?
Thanks in advance
|
You don't need to use inverse matrix (you need it if you change base). Let $f\colon V\to V$ be a linear transformation ($V$ for vector space). If $B=\{e_1, \ldots, e_n\}$ is a basis of $V$, and
$$
f(e_1) = a_{11}e_1 + a_{21}e_2 + \ldots + a_{n1}e_n\\
\cdots\\
f(e_n) = a_{n1}e_1 + a_{n1}e_2 + \ldots + a_{nn}e_n\\
$$
then matrix
$$
\begin{pmatrix}
a_{11} & \cdots & a_{1n}\\
\vdots & \vdots & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}
$$
called transformation matrix in the base $B$ (see wiki, for example). Just look:
$$
f(b_1) = b_4\\
f(b_2) = b_2\\
f(b_3) = b_2\\
f(b_4) = b_4\\
$$
And matrix is ...?
|
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|
Simplifying $ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$ who can simplify the following term in a simplest way?
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
(The answer is 1). Thanks for any suggestions.
|
HINT :
$$2ab+a^2+b^2=(a+b)^2$$
with
$$\cos^2 (x)+\sin^2(x)=1.$$
|
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|
Integral of square of variable in expression in square root in denominator I'm having trouble with following integral
$$
\int \frac{1}{\sqrt{1+4x^2+4y^2}} dy
$$
that according to WolframAlpha is
$$
\frac{1}{2} \ln( \sqrt{1+4x^2+4y^2} + 2y) + c
$$
which I can verify but I can't figure out what substitution to apply or what functions to use for integration by parts in original integral. What would be your first step to attack this integral please? Thank you.
|
Notice, the standard formula $$\color{red}{\int\frac{dx}{\sqrt{a^2+x^2}}=\ln(x+\sqrt{a^2+x^2})+c}$$ For derivation, let $x=a\tan \theta \implies dx=a\sec^2\theta d\theta $ $$\int\frac{dx}{\sqrt{a^2+x^2}}=\int\frac{a\sec^2 \theta d\theta}{\sqrt{a^2+(a\tan\theta)^2}}=$$ $$=\int\frac{\sec^2 \theta d\theta}{\sec\theta}=\int\sec\theta d\theta=\ln(\sec\theta+\tan\theta)+c$$ $$=\ln(\sqrt{1+\tan^2\theta}+\tan\theta)+c$$ Now, substituting the value of $\tan \theta=\frac{x}{a}$ $$=\ln(x+\sqrt{a^2+x^2})+c$$
In the given integration, $x $ is treated as a constant. Now we have $$\int \frac{1}{\sqrt{1+4x^2+4y^2}} dy $$ $$ =\int \frac{1}{\sqrt{(1+4x^2)+(2y)^2}} dy $$ Let, $2y=t \implies 2dy=dt\ or\ dy=\frac{dt}{2}$ $$ =\frac{1}{2}\int \frac{dt}{\sqrt{(1+4x^2)+t^2}}+c$$ $$ =\frac{1}{2}\ln(t+\sqrt{(1+4x^2)+t^2})+c $$ $$ =\frac{1}{2}\ln(2y+\sqrt{(1+4x^2)+(2y)^2})+c$$ $$ =\frac{1}{2}\ln(\sqrt{1+4x^2+4y^2}+2y)+c $$
|
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|
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
|
*
*$s=pr$ where $p=\frac{a+b+c}{2}$ and $r$ is the radius of the inscribed circle.
*$s=\sqrt{r\cdot r_a\cdot r_b\cdot r_c}$ where $r_a,r_b,r_c$ are the exradii of excircles.
|
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|
Given $\tan A + \tan B = 3x$ and $\tan A \tan B = 2x^2$, find $\tan A - \tan B$ Given $$\tan A + \tan B = 3x$$ and $$\tan A \tan B = 2x^{2},$$ how can one find $\tan A - \tan B$? I have tried substitution, but failed to find the answer.
Edit: Can this problem be solved using the formulas for sums and differences of tangents?
|
$tanA+tanB=3x$
Squaring both sides,we get
$tan^2A+tan^2B+2tanA tanB=9x^2$
Subtract $4tanA tanB$ from both sides,
$tan^2A+tan^2B-2tanA tanB=9x^2-4tanA tanB$
$(tan A-tan B)^2=9x^2-4tanA tanB$
Put $tanA tanB$ in above equation,
$(tan A-tan B)^2=9x^2-8x^2=x^2$
$tan A-tan B=\pm x$
|
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|
$6$ eigenvalues of a $4\times4$-matrix? I am struggling with determining the eigenvalues of the following (symmetric) matrix:
$$ A =\begin{pmatrix}
2 & 1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 2
\end{pmatrix} $$
What I did : $$ \operatorname{char}_X(A) =\det\begin{pmatrix}
2 -X & 1 & 0 & 0 \\
1 & 2 -X & 0 & 0 \\
0 & 0 & 2 -X & 1 \\
0 & 0 & 1 & 2 -X
\end{pmatrix} "=" \det\begin{pmatrix}
2 -X & 1 & 0 & 0 \\
0 & (2-X)^2-1 & 0 & 0 \\
0 & 0 & 2 -X & 1 \\
0 & 0 & 0 & (2-X)^2-1
\end{pmatrix} = $$ $$=(2-X)^2 ((2-X)^2-1)^2 =
(2-X)^2 (X-3)^2 (X-1)^2. $$
This yields $6$ eigenvalues. How could that happen? What is wrong here?
|
The asserted equality of two determinants is incorrect. Let $M$ be the matrix whose determinant appears first; let $N$ be the matrix whose determinant appears after the "equals" sign.
You multiplied the second row by $2-X$. That alters the determinant. Then you added $-1$ times the first row to the second, and that does not alter the determinant. Then you did the same in the third and fourth rows. So you would have
$$
\det M = \frac 1 {(2-X)^2} \det N = (X-3)^2(X-1)^2.
$$
Another thing to be aware of is this:
$$
\det\begin{bmatrix} A & 0 \\ 0 & B\end{bmatrix} = \det A\cdot\det B.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin^4x+\frac{1}{4}16\sin^4x\cos^4 x+\cdots+\frac{1}{4^n}\sin^4(2^{n-1}x)\cos^4(2^{n-1}x)$$
i could not solve further.Any hint will be useful.
|
Hint: Note that $\sin^2 \theta - \sin^4\theta = \sin^2\theta(1-\sin^2\theta) = \sin^2\theta\cos^2\theta = \dfrac{1}{4}\sin^2 2\theta$.
Hence, $\sin^4\theta = \sin^2\theta - \dfrac{1}{4}\sin^2 2\theta$.
Applying that here gives us $\dfrac{1}{4^k}\sin^4(2^k x) = \dfrac{1}{4^k}\sin^2(2^k x) - \dfrac{1}{4^{k+1}}\sin^2(2^{k+1}x)$.
So, we need to compute the limit of $\displaystyle\sum_{k = 0}^{n}\dfrac{1}{4^k}\sin^4(2^k x)$ $=\displaystyle\sum_{k = 0}^{n}\left[\dfrac{1}{4^k}\sin^2(2^k x) - \dfrac{1}{4^{k+1}}\sin^2(2^{k+1}x)\right]$.
This is a telescoping sum.
|
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|
Definite integral question Let $ f(x)$ be a quadratic equation with $f'(3)=3$. If $I=\int_{0}^{\frac{\pi}{3}}t \times \tan(t)dt $ and the value of integral$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan(\frac{x-3}{3})dx $ is equal to $kI$.Then find k.
My attempt:
Put $\frac{x-3}{3}=p$ in below integral
$$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan\left(\frac{x-3}{3}\right)dx=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\,dp $$
Now I cannot proceed further, $\tan p$ is an odd function but $f(3p+3)$ is not an even function, neither it is odd function. So their product $f(3p+3) \tan (p)$ cannot say even or odd function.
Can someone help me finding $k$.
|
Put $\frac{x-3}{3}=p$ in below integral
$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan\left(\frac{x-3}{3}\right)dx=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\ dp $
Let $f(x)=Ax^2+Bx+C\Rightarrow f'(x)=2Ax+B\Rightarrow f'(3)=6A+B=3$(given in the question)
$f(3p+3)=A(3p+3)^2+B(3p+3)+C=9Ap^2+18Ap+9A+3Bp+3B+C$
$=9Ap^2+(18A+3B)p+9A+3B+C=9Ap^2+3(6A+B)p+9A+3B+C$
$=9Ap^2+3(3)p+9A+3B+C=9Ap^2+9p+9A+3B+C$
$\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\ dp $
$=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9Ap^2+9p+9A+3B+C) 3 \tan(p) dp$
$=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9Ap^2+9A+3B+C) 3 \tan(p) dp+\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9p)\times 3 \tan(p) dp$
Function in first integral being an odd function will become zero.
$=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9p)\times 3 \tan(p) dp$
$=27\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}p\times tan(p) dp$
$=27\times 2\int_{0}^{\frac{\pi}{3}}p\times tan(p) dp$
$=54\int_{0}^{\frac{\pi}{3}}p\times tan(p) dp$
Therefore $k=54$
|
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|
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
|
We have that $\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial $(x^2-5)^2-24$, hence:
$$ x^4-10\,x^2+1 = \prod_{\xi_i\in Z}(x-\xi) $$
and $1\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial:
$$ (x-1)^4-10(x-1)^2+1 = x^4-4x^3-4x^2+16x-8. $$
By Viète's theorem, the sum of the roots of a polynomial $p(x)$ raised to the minus one power is given by $-\frac{[x^1]\,p(x)}{[x^0]\,p(x)}$, hence the answer is $-\frac{16}{-8} = \color{red}{2}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $dy/dx$ at $x=8$
Given that variables $xy=40$, find $dy/dx$ at $x=8$.
I used $40/8$ to get $y=5$.
So why is the answer $-5/8$ and not $5/8$?
|
We have that
$$xy=40 \implies y = \frac{40}{x}. $$
Remember that
$$\frac{d}{dx}\frac{1}{x} = - \frac{1}{x^2}$$
so
$$\frac{dy}{dx} = \frac{d}{dx}\frac{40}{x} = 40\frac{d}{dx}\frac{1}{x} =- \frac{40}{x^2}. $$
Therefore we have, when $x = 8$, that
$$\frac{dy}{dx} = -\frac{40}{8^2} = -\frac{40}{64} = -\frac58. $$
|
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|
Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
|
One of the easiest ways to prove $B$ is invertible is the following;
$$B=A^2-2A+2I$$
Using that $A^3=2I$, we can write
$$B=A^2-2A+2I= A^2-2A + A^3 = A(A^2+A-2I)=A(A+2I)(A-I)$$
Now let us prove $A$, $A+2I$ and $A-I$ are invertible.
*
*Supose there is $v$ such that $Av=0$, then we have $0=A^20=A^3v=2Iv=2v$. So we have $v=0$. So $A$ is invertible.
*Supose there is $v$ such that $(A+2I)v=0$, then we have $Av=-2v$ and we have $A^3v=-8v$. So we get $-8v=A^3v=2Iv=2v$. So we have $v=0$. So $(A+2I)$ is invertible.
*Supose there is $v$ such that $(A-I)v=0$, then we have $Av=v$ and we have $A^3v=v$. So we get $v=A^3v=2Iv=2v$. So we have $v=0$. So $(A-I)$ is invertible.
Since B is the product of three invertible matrices, B itself is invertible.
Remark: For item 1 above, we also simply present the inverse of $A$. From $A^3=2I$, we have $A^{-1}=\frac{1}{2}A^2$.
|
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|
Finding the equation of a circle. A circle of radius $2$ lies in the first quadrant touching both axis. Find the equation of the circle centered at $(6,5)$ and touching the above circle externally.
Let me share how I answered this question with your suggestions.
Since the radius of the first circle is $2$ and touches both axes. It follows that $h,k$ is $(2,2)$. The formula for this circle is
$$(x-2)^2 + (y-2)^2 = 4 .$$
Given that the center of the second circle is $(6,5)$, then we can get the distance to another circle. The distance between $(6,5)$ and $(2,2)$ is $5$.
We can subtract and get the radius of the second circle. $r=5-2$.
Then, the radius is $3$.
The equation of the second circle is
$$(x-6)^2 + (y-5)^5 = 9 .$$
Thank you everyone! :)
|
Equation of circle of radius 2 and touching the axes is $(x-2)^2+(y-2)^2=4$ and its center is $(2,2)$.
Center of required circle is (6,5).Radius of required circle is half of distance between (6,5) and (2,2) because these two circles touch externally.Therefore radius of required circle$=r=5-2$
Now use $(x-h)^2+(y-k)^2=r^2$ and get the equation.
|
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|
Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.
|
we will prove that $$\frac{x+y}{2}\le \frac{a+b}{2}$$ after squaring is this equivalent to $$2\sqrt{ab}\sqrt{\frac{a^2+b^2}{2}}\le \frac{(a+b)^2}{2}$$ squaring again we get $$ab\left(\frac{a^2+b^2}{2}\right)\le \frac{(a+b)^4}{16}$$ and this is equivalent to $$0\le (a-b)^4$$ and this is true.
|
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|
Regularity and invertibility of two parameterized matrices?
$$
C=
\begin{bmatrix}
1+a & 2 & 3 & 4 & 5 \\
1 & 2+a & 3 & 4 & 5 \\
1 & 2 & 3+a & 4 & 5 \\
1 & 2 & 3 & 4+a & 5 \\
1 & 2 & 3 & 4 & 5+a \\
\end{bmatrix}
$$
For which values of $a$ is the matrix $C$ regular?
$$
A=
\begin{bmatrix}
3a & a \\
-a & 1 \\
\end{bmatrix}
$$
Determine for which values of $a$ is there an inverse $A^{-1}$ and then solve it?
Please help....
|
Actually, there is an easier way to find the values for the matrix $C$ for which it is invertible:
Look at the following representation
$$
\begin{pmatrix}
1+a & 2 & 3 & 4 & 5 \\
1 & 2+a & 3 & 4 & 5 \\
1 & 2 & 3+a & 4 & 5 \\
1 & 2 & 3 & 4+a & 5 \\
1 & 2 & 3 & 4 & 5+a \\
\end{pmatrix}=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
\end{pmatrix}+
a \operatorname{Id}_{5\times5}=B+a \operatorname{Id}_{5\times5}
$$
which is the eigenvalue problem for matrix B in $\lambda=-a$.
That for $\lambda=0$ we have an eigenvalue is clear, the dimension of the eigenspace is $4$, which is clear because with Gauss we get
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5 \\
\end{pmatrix}\to
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0
\end{pmatrix}\
$$
so in fact we are looking for one other eigenvalue $\lambda_2$ with eigenspace dimension $1$, so that everything adds up to $5$ eventually.
If we look hard enough, we can easily figure out it has to be $\lambda=15\Leftrightarrow a=-15$. For this to be seen, observe that
$$
\begin{pmatrix}
1+a \\
1 \\
1 \\
1 \\
1
\end{pmatrix}+
\begin{pmatrix}
2 \\
2+a \\
2 \\
2 \\
2
\end{pmatrix}+
\begin{pmatrix}
3 \\
3 \\
3+a \\
3 \\
3
\end{pmatrix}+
\begin{pmatrix}
4 \\
4 \\
4 \\
4+a \\
4
\end{pmatrix}=
\begin{pmatrix}
10+a \\
10+a \\
10+a \\
10+a \\
10
\end{pmatrix}\equiv(-1)\begin{pmatrix}
5 \\
5 \\
5 \\
5 \\
5+a
\end{pmatrix}
$$
which is true for $a=-15$. Since now we have 5 dimensional eigenspace we are finished and conclude that for $a_1=0$ and $a_2=-15$ the matrix $C$ is not invertible.
For the second part just use Cramer's rule which will give you
$$
A^{-1}=\frac 1{a(3+a)}\begin{pmatrix} 1 & -a \\ a & 3a \end{pmatrix}
$$
which indeed is defined for all $a\in\mathbb{R}\backslash\{0,-3\}$
|
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|
Finding $P$ such that $P^TAP$ is a diagonal matrix
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_2(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
So here's the solution:
$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&1&-3/2\\ 0&-1/2&0&1 \end{array}\right)$$
Therefore, $$P = \left(\begin{array}{cc} 1&-3/2\\ 0&1 \end{array}\right) \\ P^TAP = \left(\begin{array}{cc} 2&0\\ 0&-1/2 \end{array}\right) $$
What was done here exactly? I'd be glad elaborate about the process.
Thanks.
|
Generally, in the process of diagonalization, it is easiest to approach via calculating the eigenvalues and corresponding eigenvectors to form an orthonormal eigenbasis.
Such an orthogonal matrix is guaranteed to exist by the Spectral Theorem since our matrix, $A$, is a real symmetric matrix.
step 1: calculate eigenvalues
Find the eigenvalues by finding the characteristic polynomial: $\det(A-\lambda I) = (2-\lambda)(4-\lambda) - 3\cdot 3 = 8-6\lambda + \lambda^2 - 9 = \lambda^2 - 6\lambda - 1$
Finding the roots of the characteristic polynomial will find our eigenvalues. Solving via the quadratic formula gives us $\frac{6\pm\sqrt{36+4}}{2}=3\pm \sqrt{10}$
step 2: find the eigenvectors
Now, we try to find the eigenvectors.
Eigenvector for $\lambda_1=3+\sqrt{10}$ would be a vector in the kernel of $A-\lambda_1 I$.
$rref\left(\begin{bmatrix} 2-3-\sqrt{10}&3\\3&4-3-\sqrt{10}\end{bmatrix}\right) = \begin{bmatrix}1&\frac{1-\sqrt{10}}{3}\\0&0\end{bmatrix}$, so the eigenvector $v_1$ is $\begin{bmatrix}\frac{-1+\sqrt{10}}{3}\\1\end{bmatrix}$.
Similarly, the eigenvector for $\lambda_2=3-\sqrt{10}$ would be a vector in the kernel of $A-\lambda_2 I$.
$rref\left(\begin{bmatrix} 2-3+\sqrt{10}&3\\3&4-3+\sqrt{10}\end{bmatrix}\right) = \begin{bmatrix}1&\frac{1+\sqrt{10}}{3}\\0&0\end{bmatrix}$, so the eigenvector $v_2$ is $\begin{bmatrix}\frac{-1-\sqrt{10}}{3}\\1\end{bmatrix}$
step 3: form an orthonormal basis for each eigenspace
A convenient thing about this situation is, thanks to the spectral theorem and the fact that our $A$ is real symmetric, vectors in different eigenspaces are already guaranteed to be orthogonal. Indeed $\langle v_1, v_2\rangle = (\frac{-1+\sqrt{10}}{3})(\frac{-1-\sqrt{10}}{3})+1\cdot 1 = 0$
If we had a repeated eigenvalue, then we would need to apply the gram-schmidt process to the basis vectors of its corresponding eigenspace. In our case, each eigenvalue is of multiplicity one, so we only need to normalize the vectors.
$u_1 = \frac{v_1}{\|v_1\|} = \begin{bmatrix} ((1+\sqrt{10})/(3 \sqrt{1+1/9 (1+\sqrt{10})^2)}\\ 1/\sqrt{1+1/9 (1+\sqrt{10})^2)}\end{bmatrix}$
These numbers were not very pretty to work with... oh well.
You have then $A = PDP^T$ where $P=[u_1,u_2]$ and $D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$. $P$ is an orthogonal matrix, so $P^T=P^{-1}$ and we have $P^T A P=D$
|
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|
How many pairs of $(x, y)$ satisfied this equation I need help to solve in $\mathbb{Z}$ the following equation
$$yx^{2}+xy^{2}=30$$ I tried to solve it by factor $30$ to $5\times 6$ and I get those two pairs $(2, 3) \& (3, 2) $... is their any other pairs of $(x, y) $? ..
|
Factor it as $xy(x+y)=30$. Write $30=2\times3\times5$. Therefore $y$ has to be a factor of $30$.
So $y=\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$.
However, if $y\geq6$ and $x\geq1$ then $xy(x+y)\geq42$. If $y\geq6$ and $x\leq-1$ then we must also have $x\leq-7$ since otherwise one factor is negative and two are positive. Then also $xy(x+y)\geq42$.
*
*$y=1$ gives $x(x+1)=30$, this gives $x=5 \vee x=-6$.
*$y=2$ gives $2x(x+2)=30$, this gives $x(x+2)=15$, thus $x=3$ or $x=-5$.
*$y=3$ gives $3x(x+3)=30$, this gives $x(x+3)=10$, thus $x=2$ or $x=-5$.
*$y=5$ gives $5x(x+5)=30$, this gives $x(x+5)=6$, thus $x=1$ or $x=-6$.
*$y=-1$ gives $-x(x-1)=30$, this gives $x(x-1)=-30$, which has no solutions.
*$y=-2$ gives $-2x(x-2)=30$, this gives $x(x-2)=-15$, which has no solutions.
*$y=-3$ gives $-3x(x-3)=30$, this gives $x(x-3)=-10$, which has no solutions.
*$y=-5$ gives $-5x(x-5)=30$, this gives $x(x-5)=-6$, thus $x=2$ or $x=3$.
*$y=-6$ gives $-6x(x-6)=30$, this gives $x(x-6)=-5$, thus $x=1$ or $x=5$.
*$y=-10$ gives $-10x(x-10)=30$, this gives $x(x-10)=-3$, which has no real solutions.
*$y=-15$ gives $-15x(x-15)=30$, which has no real solutions.
*$y=-30$ gives $-30x(x-30)=30$, which has no real solutions.
Conclusion: We get the pairs $(5,1), (1,5), (-6,1), (1,-6), (2,3), (3,2), \\ (2,-5), (-5,2), (5,-6), (-6,5), (3,-5), (-5,3)$
|
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|
How does this algebraic trick regarding partial fraction works? Suppose I have to evaluate the integral $$\int \frac{x}{(x-1)(2x+1)(x+3)} \, dx $$
I write it as $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$
where $a_1$, $a_2$, $a_3$ are constants. once I have foud relationaships between them I solve for the $a_{i}$ and subsequently substitute them into the integral.
It can be observed that for finding this constants we may use following trick.
Suppose we have to find $a_1$ then for time being remove the term beneath $a_1$ from the relation $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$ now put the value of $x$ that would appear if the removed term is equated to $0$ in $\frac{x}{(x-1)(2x+1)(x+3)}$. This the value of constant $a_1$.
How does this works ?
|
We have $$\frac{x}{(x-1)(2x+1)(x+3)}=\frac{a_1}{x-1}+\frac{a_2}{2x+1}+\frac{a_3}{x+3}.$$
Then, we get $$(x-1) \times \frac{x}{(x-1)(2x+1)(x+3)}=\frac{x}{(2x+1)(x+3)}=a_1+\frac{a_2(x-1)}{2x+1}+\frac{a_3(x-1)}{x+3}. $$ If we take $x=1,$ then we find
$$\frac{1}{3 \times 4}=\frac{1}{12}=a_1.$$ We proceed similarly, we get $$(2x+1) \times \frac{x}{(x-1)(2x+1)(x+3)}=\frac{x}{(x-1)(x+3)}=\frac{a_1(2x+1)}{x-1}+a_2+\frac{a_3(2x+1)}{x+3}.$$ If $x=-1/2$ then, $a_2=2/15.$ Finally,
$a_3=3/20.$ So, $$\frac{x}{(x-1)(2x+1)(x+3)}=\frac{1}{12(x-1)}+\frac{2}{15(2x+1)}+\frac{3}{20(x+3)}.$$
|
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|
Solving $x^2\equiv a\pmod{p}$ where $p\equiv5\pmod{8}$ and $a$ is a quadratic residue The exercise asks me to show that one of the values $x = a^{(p+3)/8}$ and $x = 2a \cdot (4a)^{(p-5)/8}$ is a solution to $x^2\equiv a\pmod{p}$, where $p\equiv5\pmod{8}$ is a prime and $a$ is a quadratic residue. My reasoning was that $a^{(p-1)/2} \equiv a^{(8k+5-1)/2} \equiv a^{4k+2} \equiv 1 \pmod{p}$ by Euler's criterion and therefore $a^{2k+1} \equiv \pm 1\pmod{p}$. Therefore $x = a^{(p+3)/8}$ has $x^2 = a^{2k+2} \equiv a \pmod{p}$ when $a^{2k+1} \equiv 1 \pmod{p}$. I am struggling to show that $x = 2a \cdot (4a)^{(p-5)/8}$ is a solution when $a^{2k+1} \equiv -1 \pmod{p}$.
|
Recall that $p \equiv 5 \mod 8$, and thus $p \not\equiv \pm 1 \mod 8$. Thus, the second supplement to the law of Quadratic Reciprocity shows that $2$ is a quadratic non-residue modulo $p$. In other words, the Legendre symbol $\left(\dfrac{2}{p}\right) = -1$. But Legendre's formula says that $\left(\dfrac{a}{p}\right) \equiv a^{\left(p-1\right)/2} \mod p$ for any integer $a$ coprime to $p$. Applying this to $a=2$, we obtain $\left(\dfrac{2}{p}\right) \equiv 2^{\left(p-1\right)/2} = 2^{4k+2} \mod p$ (since $\left(p-1\right)/2 = 4k+2$). Compared with $\left(\dfrac{2}{p}\right) = -1$, this becomes $2^{4k+2} \equiv -1 \mod p$.
Assume that $a^{2k+1} \equiv -1 \mod p$. Set $x = 2a \cdot \left(4a\right)^{\left(p-5\right)/8}$. Thus,
$x^2 = \left(2a \cdot \left(4a\right)^{\left(p-5\right)/8}\right)^2 = \left(2a \cdot \left(4a\right)^{k}\right)^2$ (since $\left(p-5\right)/8=k$)
$= 4a^2 \left(4a\right)^{2k} = a \left(4a\right)^{2k+1} = a \underbrace{4^{2k+1}}_{=2^{2\left(2k+1\right)} = 2^{4k+2} \equiv -1 \mod p} \underbrace{a^{2k+1}}_{\equiv -1 \mod p} \equiv a \left(-1\right)\left(-1\right) = a \mod p$,
qed.
|
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|
Evaluation of $ \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx$
Evaluation of $\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx$
$\bf{My\; Solution::}$ We can write it as $$\displaystyle \int \sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{\cos ^2x+2\cos x}}dx$$
$$\displaystyle = \int\sqrt{\frac{(1-\cos x)}{(1+\cos x)}\times \frac{(1+\cos x)}{(1+\cos x)}}\times \frac{1}{\sqrt{(1+\cos x)^2-1}}dx$$
$$\displaystyle = \int \frac{\sin x}{(1+\cos x)}\times \frac{1}{\sqrt{(1+\cos x)^2-1}}dx$$
Now Let $(1+\cos x) = t\;,$ Then $\sin xdx = -dt$
So Integral $$\displaystyle = -\int \frac{1}{t}\times \frac{1}{\sqrt{t^2-1}}dt =-\sec^{-1}(t)+\mathcal{C} = -\sec^{-1}\left(1+\cos x\right)+\mathcal{C}$$
So $$\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx = -\sec^{-1}\left(2\cos^2 \frac{x}{2}\right)+\mathcal{C}$$
But Answer Given as $\displaystyle = \sin^{-1}\left(\frac{1}{2}\sec^2 \frac{x}{2}\right)+\mathcal{C}$
Plz help me where i am wrong, Thanks
|
Everything is Ok, for matching the answer See here, $$\displaystyle -\int \frac{1}{t}\times \frac{1}{\sqrt{t^2-1}}dt =\csc^{-1}(t)+\mathcal{C} = \csc^{-1}\left(1+\cos x\right)+\mathcal{C}\\
= \csc^{-1}\left(2\cos^2 \frac{x}{2}\right)+\mathcal{C}$$
Now use, $$\csc^{-1}x=\sin^{-1}(1/x)$$ You will get answer as you want.
|
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|
Show that $\sqrt{6+4\sqrt{2}}-\sqrt{2}$ is rational using the rational zeros theorem Let $r=\sqrt{6+4\sqrt{2}}-\sqrt{2}$, then $r+\sqrt{2}=\sqrt{6+4\sqrt{2}}$.
Squaring both sides, we get $$r^2+2r\sqrt{2}+2=6+4\sqrt{2}$$ which is the same as $r^2-4=2\sqrt{2}$.
Squaring both sides again, we get $r^4-8r^2+16=8$ or $$r^4-8r^2+8=0.\tag{$\star$}$$
The rational zeros theorem tells us that the only possible rational solutions to ($\star$) are $±1$, $±2$, $±4$, $±8$.
I do not know where to go from here. Please help me complete this proof.
|
Check your second line, $r^2+2r\sqrt{2} + 2 = 6 + 4\sqrt{2} \ne r^2-4 -2\sqrt{2}$.
It does, however, equal $ (r-2) (r+2 \sqrt{2}+2) = 0$ which would imply that $r = 2$ or $r= -2(1+\sqrt{2})$.
Thus $2$ is the only rational root of your polynomial and $\sqrt{6+4\sqrt{2}} -\sqrt{2}= 2 $ (since $r =\sqrt{6+4\sqrt{2}} -\sqrt{2}$ and $(r-2)=\sqrt{6+4\sqrt{2}} -\sqrt{2} -2 = 0$).
|
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|
Reflection formula for Hurwitz Zeta function? In doing some calculus with Mathematica today, I found that
$$\zeta\left(3,\frac{1}{4}\right) - \zeta\left(3,\frac{3}{4}\right) = 2\pi^3$$
by numerically evaluating both sides. Here, $\zeta(x,y)$ denotes the Hurwitz Zeta function defined by
$$\zeta(x,y) = \sum_{n=0}^\infty \frac{1}{(y+n)^x},$$
called Zeta[.,.] in Mathematica. I would be very grateful if there's anyone who could provide some ideas for a proof of that or even a complete proof. I tried to find a kind of reflection theorem for this function but I didn't find anything, neither by hand nor by looking into the literature.
Thank you very much!
|
First define the $\beta(x)$ function by
\begin{align}
\beta(x) = \frac{1}{2} \, \left[ \psi\left(\frac{x+1}{2}\right) - \psi\left(\frac{x}{2}\right) \right] = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{x+k}
\end{align}
for which after two derivatives with respect to x and setting $x = 1/2$ the result becomes
\begin{align}
16 \, \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{3}} &= \frac{1}{8} \left[ \psi^{(2)}\left(\frac{x+1}{2}\right) - \psi^{(2)}\left(\frac{x}{2}\right) \right] \\
&= \frac{1}{8} \left[ 2(\pi^{3} - 28 \, \zeta(3)) - (-2)(\pi^{3} + 28 \, \zeta(3)) \right] = \frac{\pi^{3}}{2}
\end{align}
and finally
\begin{align}
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{3}} = \frac{\pi^{3}}{32}.
\end{align}
Now,
\begin{align}
\zeta\left(3, \frac{1}{4}\right) - \zeta\left(3, \frac{3}{4}\right) &= 2^{6} \left[ \sum_{n=0}^{\infty} \frac{1}{(4n+1)^{3}} - \sum_{n=0}^{\infty} \frac{1}{(4n+3)^{3}} \right] \\
&= 2^{6} \, \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}} = 2^{6} \, \frac{\pi^{3}}{2^{5}} \\
&= 2 \, \pi^{3}.
\end{align}
|
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Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?
|
First, check the formula for $n=1$. So:
$$\frac1{1\cdot 3}=\frac34-\frac{2\cdot1+3}{2(1+1)(2+1)}$$
Since this is true, we have shown the so called base case.
Now substitute in the formula $n$ by $n+1$ to get the statement that you have to show:
$$\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{(n+1)(n+3)}=\frac34-\frac{2n+5}{2(n+2)(n+3)}\qquad (*)$$
The good news is that you can (and should) assume that the formula is valid for the $n$ first positive integers, so
$$\begin{align}&\left(\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{n(n+2)}\right)+\frac1{(n+1)(n+3)}\\
&=\frac34-\frac{2n+3}{2(n+1)(n+2)}+\frac1{(n+1)(n+3)}
\end{align}$$
and you should obtain $(*)$ with straightforward computings.
Perhaps my explanation is not very brief but I hope it to be useful.
|
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|
How do I integrate $\frac {\sin^3x}{\cos^2x}$ How do I integrate $\frac {\sin^{3}x}{\cos^{2}x}$. I have tried to convert to $\tan$, but I could not reach to conclusion. Any help will be appreciated.
Thanks.
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$$\int { \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } dx=\int { \frac { \sin { x } \cdot \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } dx=-\int { \frac { 1-\cos ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } d\cos { x } } } } =\\ =\int { d\left( \cos { x } \right) } -\int { \frac { 1 }{ \cos ^{ 2 }{ x } } d\left( \cos { x } \right) =\cos { x } +\frac { 1 }{ \cos { x } } +C } $$
|
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|
maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
|
By the arithmetic-quadratic mean
inequality,
$\frac{a+b+c}{3}
\le \sqrt{\frac{a^2+b^2+c^2}{3}}
= \sqrt{\frac{48}{3}}
=\sqrt{16}
=4
$
with equality iff
$a=b=c$.
Therefore
$a+b+c = 12$
and
$a=b=c = 4$.
|
{
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|
Why am I getting two answers for 8th root of continued fraction Find value of $x$:
$x=\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-....and\,so\, on}}}$
On solving ,we have $x^8=2207-\frac{1}{x^8}$
$x^8+\frac{1}{x^8}=2207$
$x^4+\frac{1}{x^4}=47$
$x^2+\frac{1}{x^2}=7$
$x+\frac{1}{x}=3$
$x=\frac{3+\sqrt{5}}{2}\,,\frac{3-\sqrt{5}}{2}$
But $x$ can take only one value.
The problem is which of these two values is to be accepted/rejected and why.
|
You want the value which is greater than $1$ - clearly by estimating.
From $x^8+\frac 1{x^8}=2207$ you know that if $x$ satisfies that equation so will $\frac 1x$. Put $\frac 1x$ in the original equation and modify the eighth root accordingly and you will see where the alternative answer comes from and that this is clearly less than $1$.
|
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|
Let $a^n = a^{n - 1} + a^{n -2}$. Show that for any $A, B$, $F(n) = Aa^n + Bb^n$ satisfies Fibonacci recurrence relation. $$\begin{align*}
F(n) &= Aa^n + Bb^n\\
&= A(a^{n-1}+a^{n-2}) + B(b^{n-1}+b^{n-2}) \\
&= Aa^{n -1} + Aa^{n-2} + Bb^{n -1} + Bb^{n-2}\\
&= a^{n -1} (A + A^{a-1}) + b^{n - 2} (B + bB)
\end{align*}$$
Use $$ \color{red}{ a > 1; \; \; b = \frac{-1}{a}.} $$
Would that work?
edit: Suppose recurrence relation given by $F(n) = F(n - 1) + F(n - 2)$ grows at the same rate as some function $a^n$. Then, $a^n = a^{n -1} + a^{n -2}$ by substitution. $a^n \to a^2 = a + 1$. One of its roots is negative and we call it $b$. Since $a^n, b^n$ satisfy Fibonacci recurrence relation, $F(n) = Aa^n + Bb^n$ also does. That's what I want to prove.
|
Let me try rewriting the question and giving an answer:
Restated question
Suppose that $a$ and $b$ are the two solutions to $x^2 - x - 1 = 0$, so that we have $a^2 = a + 1$ (and similarly for $b$) and hence also
\begin{align}
a^n &= a^{n-1} + a^{n-2}& \text{ (**) }
\end{align}
for $n \ge 2$, and similarly for $b$.
Show that for any $A, B$, the function
$$
F(n) = A a^n + B b^n
$$
has the Fibonacci property, i.e., that
$$
F(n) = F(n-1) + F(n-2)
$$
for all integers $n \ge 2$.
Solution to this rephrased problem
Suppose $n$ is any integer greater than 2. Then
\begin{align}
F(n)
&= Aa^n + B b^n & \text{ by definition of $F$}\\
&= A(a^{n-1} + a^{n-2}) + B (b^{n-1} + b^{n-2}) & \text{ by equation (**) }\\
&= A a^{n-1} + A a^{n-2} + Bb^{n-1} + B b^{n-2} & \text{ by distributive law, twice }\\
&= A a^{n-1} + B b^{n-1} + A a^{n-2} + B b^{n-2} & \text{ by commutative lawfor addition }\\
&= \left( A a^{n-1} + B b^{n-1} \right) + \left( A a^{n-2} + B b^{n-2} \right )& \text{parens added for clarity }\\
&= F(n-1)+ F(n-2) & \text{definition of $F$, twice }.
\end{align}
The second and last steps are justified because we have $n \ge 2$, in case you were wondering about that bit. So we've shown that $F$ satisfies the Fibonacci recurrence for all $n \ge 2$, as required.
|
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|
Convergence of improper integral $\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx$ I'm trying to determine whether the following improper integral is convergent or divergent.
$$
\int_{0}^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}dx
$$
At first, I substituted $t=\dfrac{\pi}{2} - x $
and then I used $1-\dfrac{1}{2}x^2 \le \cos x$.
But I couldn't determine. :-(
$$$$
Second attempt, I used $\sin x\le x $ on $[ 0, \frac{\pi}{6} ]$.
But I couldn't determine. :-[
Could you give me some advice?
Thanks in advance.
|
Since we are only interested in whether the improper integral converges or diverges, let us write out the integrand in terms of their power series and see what we get:
$$ 2 \sin x = 2 \cdot \sum_{n=0}^{\infty} (-1)^n \frac{x^{(2n+1)}}{(2n+1)!} = 2x - \frac{x^3}{3}+\frac{x^5}{60}+ O(x^7)$$
Then we have:
$$ 1 - 2 \sin x = 1 -2 \cdot \sum_{n=0}^{\infty} (-1)^n \frac{x^{(2n+1)}}{(2n+1)!} = 1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7) $$
So now we have (and we rationalize the integral):
$$\int_0^\frac{\pi}{6}\frac{x}{ \sqrt{1-2 \sin x}} dx= \int_0^\frac{\pi}{6} \frac{x}{\sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)}} \cdot \frac {\sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)}} {\sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)}} dx= \int_0^ \frac{\pi}{6} \frac{x}{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} \cdot \sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} dx$$
Then by long division:
$$\frac{x}{1-2 \sin x} = \frac{x}{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} = x + 2x^2 +4x^3+ \frac{23}{x}x^4+\frac{44}{3}x^5+ O(x^6) $$
So the integral simplifies to:
$$ \int_0^\frac{\pi}{6} \left( x + 2x^2 +4x^3+ \frac{23}{x}x^4+\frac{44}{3}x^5+ O(x^6) \right ) \cdot \sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} dx$$
Then by multiplication we get:
$$ \int_0^\frac{\pi}{6} \left (x+ x^2 + \frac{3x^3} {2} + \frac{7x^4}{3}+\frac{31x^5}{8}+O(x^6) \right )dx $$
Finally when we integrate piecewise and evaluate at the upper and lower endpoints:
$$ \int_0^\frac{\pi}{6} \left (x+ x^2 + \frac{3x^3} {2} + \frac{7x^4}{3}+\frac{31x^5}{8}+O(x^6) \right )dx = \left ( \frac{x^2}{2}+ \frac{x^3}{3} + \frac{3x^4}{8} + \frac{7x^5}{15} + O(x^6) \right)_0^{\frac{\pi}{6}} \approx 0.231477911 + 0 \approx 0.2315$$
Thus we conclude that the improper integral must converge.
|
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$U_n=\int_{0}^{1}x^n(2-x)^ndx,V_n=\int_{0}^{1}x^n(1-x)^ndx,n\in N$ For $U_n=\int_{0}^{1}x^n(2-x)^ndx,V_n=\int_{0}^{1}x^n(1-x)^ndx,n\in N$,which of the following statement is/are true
$(A)\ U_n=2^nV_n\hspace{1cm}(B)\ U_n=2^{-n}V_n\hspace{1cm}(C)\ U_n=2^{2n}V_n\hspace{1cm}(D)\ U_n=2^{-2n}V_n$
$V_n $does not change its form when $x$ is replaced by $1-x\Rightarrow$$V_n=\int_{0}^{1}x^n(1-x)^ndx$.
$U_n=\int_{0}^{1}(1-x)^n(1+x)^ndx$ .There is no way looking by which i can relate $U_n$ and $V_n$.
|
Given $$\displaystyle U_{n} = \int_{0}^{1}x^n\cdot (2-x)^ndx$$ and $$\displaystyle V_{n} = \int_{0}^{1}x^n\cdot (1-x)^ndx$$
Now Let $x=2y$ in $U_{n}\;,$ and $dx = 2dy$ and changing Limit, We get $\displaystyle $
$$\displaystyle U_{n}=2\int_{0}^{\frac{1}{2}}2^n\cdot y^n\cdot 2^n\cdot (1-y)^n dy $$
$$\displaystyle = 2^{n+n}\cdot 2\int_{0}^{\frac{1}{2}}y^n\cdot (1-y)^ndy = 2^{n+n}\cdot 2\int_{0}^{\frac{1}{2}}x^n\cdot (1-x)^ndx$$
Above we use the formula $$\displaystyle \int_{a}^{b}f(y)dy = \int_{a}^{b}f(t)dt$$
Now $$\displaystyle U_{n} = 2^{n+n}\cdot 2\int_{0}^{\frac{1}{2}}x^{n}\cdot (1-x)^ndx = 2^{n+n}\int_{0}^{1}x^n\cdot (1-x)^n = 2^{2n}\cdot V_{n}$$
So We get $$\displaystyle U_{n} = 2^{2n}\cdot V_{n}$$
above we use the formula $$2\displaystyle \int_{0}^{a}f(x)dx = \displaystyle \int_{0}^{2a}f(x)dx\;,$$ If $$f(2a-x) = f(x)$$
Like in above case $$\displaystyle 2\int_{0}^{\frac{1}{2}}x^n\cdot (1-x)^ndx = \int_{0}^{1}x^n\cdot (1-x)^ndx$$
Bcz here $$\displaystyle f(x)=x^n\cdot (1-x)^ndx.\;,$$ Then $$f(1-x) = (1-x)^n\cdot x^n = f(x)$$
|
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|
How to prove this identity of ceiling function? My book writes down this identity of least integer function:
$$\lceil x\rceil +\left\lceil x + \frac{1}{n}\right \rceil + \left\lceil x + \frac{2}{n}\right \rceil + \cdots +\left\lceil x + \frac{n -1}{n}\right \rceil = \lceil nx\rceil + n-1 $$.
It didn't deduce it, however. I googled a bit about ceiling function but couldn't find any deduction. It is more like Hermite's Identity of floor function. Can anyone show me how to deduce this?
|
Another way of doing it is $$f(x)=\lceil x\rceil+\left\lceil x+\frac1n\right\rceil+\dots +\left\lceil x+1-\frac1n\right\rceil - \left\lceil nx\right\rceil$$
We have $$
\begin{align}
f\left(x+\frac{1}{n}\right)=&\left\lceil x+\frac1n\right\rceil+\left\lceil x+\frac2n\right\rceil+\dots +\left\lceil x+1\right\rceil - \left\lceil nx+1\right\rceil \\
=&\left\lceil x+\frac{1}{n}\right\rceil+\left\lceil x+\frac2n\right\rceil+\dots +\left\lceil x\right\rceil - \left\lceil nx\right\rceil \\
=&\lceil x\rceil+\left\lceil x+\frac1n\right\rceil+\dots +\left\lceil x+1-\frac{1}{n}\right\rceil - \left\lceil nx\right\rceil\\
=&f(x)
\end{align}$$
So $f$ is $\frac1n$-periodic. And we only need to evaluate it over $\left(0,\frac1n\right]$, in which case
$$f(x) = n-1$$
This is because if $x\in(0,\frac{1}{n}]$, then $x,x+\frac{1}{n},\dots,x+\frac{n-1}{n}$ and $nx\in(0,1]$ and $$\left\lceil x\right\rceil = \left\lceil x+\frac{1}{n}\right\rceil=\dots=\left\lceil x+\frac{n-1}{n}\right\rceil=\left\lceil nx\right\rceil=1$$
|
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|
Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$ Find:
$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$
I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What method to use?
Result should be $\frac{1}{4}$
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L'Hospital's rule is not the α and ω of limits computation. First remove the square roots in the numerator:
$$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{(\sqrt{1+\tan x}+\sqrt{1+x})\sin^2x}$$
Now use equivalents:
*
*$\tan x-x=x+\dfrac{x^3}3+o(x^3)-x$, hence $\;\tan x-x\sim_0 \dfrac{x^3}3$
*$\sqrt{1+\tan x}+\sqrt{1+x}\xrightarrow[x\to 0]{}2$
*$\sin x\sim_0 x$
$$\text{So we have:}\hspace{8em}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}\sim_0\frac{\dfrac{x^3}3}{2x^2}=\frac x6\to 0.\hspace{8em}$$
|
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|
3Finding minimum $f(k)$ ($n$: fixed natural number and $k=0,1,2,\cdots,n-1$ I would appreciate if somebody could help me with the following problem
Q. Finding minimum $f(k)$ where $n \in \mathbb N$ and $k = 0,1,...,n-1$.
$$f(k)={2k+1 \choose k} \times {2n-2k-1\choose n-k}$$
I tried and find
$n=6$ then $f(2),f(3)$ has minimum
$n=7$ then $f(3)$ has minimum
|
It’s convenient to let $m=n-1$ and define
$$g_m(k)=\binom{2k+1}k\binom{2m+1-2k}{m+1-k}$$
for $k=0,\ldots,m$; we wish to choose $k$ so as to minimize $g_m(k)$.
Let $\ell=m-k$; then
$$g_m(k)=\binom{2k+1}k\binom{2m+1-2k}{m+1-k}=\binom{2m+1-2\ell}{m+1-\ell}\binom{2\ell+1}\ell=g_m(\ell)=g_m(m-k)\;.$$
That is, $g_m$ is symmetric over the range $k=0,\ldots,m$. Now
$$\begin{align*}
g_m(k+1)-g_m(k)&=\binom{2k+3}{k+1}\binom{2m-1-2k}{m-k}-\binom{2k+1}k\binom{2m+1-2k}{m+1-k}\\\\
&=\binom{2k+3}{k+1}\binom{2m-1-2k}{m-k}-\binom{2k+1}{k+1}\binom{2m+1-2k}{m-k}\\\\
&=\left(\frac{(2k+3)(2k+2)}{(k+2)(k+1)}-\frac{(2m-2k)(2m+1-2k)}{(m+1-k)(m-k)}\right)\binom{2k+1}{k+1}\binom{2m-1-2k}{m-k}\\\\
&=2\left(\frac{2k+3}{k+2}-\frac{2m+1-2k}{m+1-k}\right)\binom{2k+1}{k+1}\binom{2m-1-2k}{m-k}\\\\
&=\frac{2(2k+1-m)}{(k+2)(m+1-k)}\binom{2k+1}{k+1}\binom{2m-1-2k}{m-k}\;,
\end{align*}$$
which has the same algebraic sign as $2k+1-m$.
In particular, $g_m(k)>g_m(k+1)$ if $2k+1<m$, $g_m(k)=g_m(k+1)$ if $2k+1=m$, and $g_m(k)<g_m(k+1)$ if $2k+1>m$. Thus, $m=2r+1$, then $g_m(k)$ attains its minimum at $k=r$ and $k=r+1$, and if $m=2r$, then $g_m(k)$ attains its minimum at $k=r$.
Alternatively, $g_m$ attains its minimum at $k=\left\lfloor\frac{m}2\right\rfloor$ and at $k=\left\lceil\frac{m}2\right\rceil$.
|
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|
Evaluation of trignometric limit I want to find the following limit without L'Hospital.
$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$
Maybe I should try to get rid of the radical.
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We will use the fact that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This will help us rationalize the numerator. With $a=1$and $b=(\tan x)^{1/3}$ we can write
\begin{align*}
\frac{1+(\tan x)^{1/3}}{1-2\cos^2x}& = \frac{1+(\tan x)^{1/3}}{1-2\cos^2x} \cdot \left[\color{blue}{\frac{1-(\tan x)^{1/3}+(\tan x)^{2/3}}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}}\right]\\
& = \frac{1+\tan x}{1-2\cos^2x}\cdot \left[\frac{1}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}\right]
\end{align*}
Now use the following identity:
$$\cos 2x=2 \cos^2 x -1 =\frac{1-\tan^2x}{1+\tan^2x},$$
to get
\begin{align*}
\frac{1+(\tan x)^{1/3}}{1-2\cos^2x}&= \frac{1+\tan^2 x}{\tan x-1}\cdot \left[\frac{1}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}\right]
\end{align*}
Now use the fact that $x \to 3\pi/4$ to get the limit as $-1/3$.
|
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|
Investigate the limiting behavior at the origin on the three lines $x = 0 , y = 0$ and $y = x$.
Investigate the limiting behavior at the origin on the three lines $x = 0 , y = 0$ and $y = x$.
For the function $f :\mathbb R^2 \to \mathbb R$ given by
$$f(x,y)=\begin{Bmatrix} 0, \quad \{x,y\} = {0,0} \\ \frac{x^2}{x^2+y^2} \quad\{x,y\}, \neq {0,0} \end{Bmatrix}$$
$$\lim_{x \to y } \frac{x^2}{x^2+y^2} = \frac{x^2}{x^2+x^2} = \frac{x^2}{2x^2} =\frac{1}{2} $$
$$ \lim_{y \to x } \frac{x^2}{x^2+y^2}= \frac{y^2}{y^2+y^2} = \frac{1}{2} $$
$$\lim_{x \to 0 } \frac{x^2}{x^2+y^2} = \frac{0}{0+y^2} = 0$$
$$\lim_{y \to 0 } \frac{x^2}{x^2+y^2} = \frac{x^2}{x^2} = 1$$
How's that?
|
You have the right idea but note that approaching along the straight line $x=0$, it is not always true that $\lim_{x,y \to 0} f(0,y) = \lim_{x\to 0} f(x,y)$. The notation is incorrect but you did arrive at the right conclusion (the limit does not exist due to differing values as we change paths).
|
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|
Prove the Identity $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $ By considering the fact that $f(\pi/2)=1$, prove the identity
$\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $
This question was is a subsection in a chapter on Fourier series, can I use my understanding of Fourier series to prove the identity, or is there an easier way of making the proof?
EDIT Another part of this question asked to find the Fourier series of the periodic function $f(x)$ defined by $f(x)= |sin(x)|$, this is the function they are probably referring to in the proof.
|
I'll play around and see what happens.
We want
$\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}}
$
$\begin{array}\\
S
&=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}}\\
&=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\frac1{1-\frac{1}{4m^2}}\\
&=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^j}\\
&=4\sum_{m=1}^{\infty}(-1)^m\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^{j+1}}\\
&=4\sum_{m=1}^{\infty}(-1)^m\sum_{j=1}^{\infty}\frac{1}{(4m^{2})^{j}}\\
&=4\sum_{j=1}^{\infty}\sum_{m=1}^{\infty}(-1)^m\frac{1}{(4m^{2})^{j}}\\
&=4\sum_{j=1}^{\infty}\frac1{4^j}\sum_{m=1}^{\infty}(-1)^m\frac{1}{(m^{2})^{j}}\\
&=4\sum_{j=1}^{\infty}\frac1{4^j}\sum_{m=1}^{\infty}(-1)^m\frac{1}{m^{2j}}\\
&=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)(1-2^{1-2j})\\
&=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)+8\sum_{j=1}^{\infty}\frac1{4^{2j}}\zeta(2j))\\
\end{array}
$
According to
https://en.wikipedia.org/wiki/Riemann_zeta_function,
$\sum_{j=1}^{\infty}\frac{\zeta(2j)-1}{4^j}
=\frac16
$
and
$\sum_{j=1}^{\infty}\frac{\zeta(2j)-1}{16^j}
=\frac{13}{30}-\frac{\pi}{8}
$.
Therefore
$\begin{array}\\
\sum_{j=1}^{\infty}\frac{\zeta(2j)}{4^j}
&=\frac16+\sum_{j=1}^{\infty}\frac{1}{4^j}\\
&=\frac16+\frac{1/4}{1-1/4}\\
&=\frac16+\frac13\\
&=\frac12\\
\end{array}
$
and
$\begin{array}\\
\sum_{j=1}^{\infty}\frac{\zeta(2j)}{16^j}
&=\frac{13}{30}-\frac{\pi}{8}+\sum_{j=1}^{\infty}\frac{1}{16^j}\\
&=\frac{13}{30}-\frac{\pi}{8}+\frac{1/16}{1-1/16}\\
&=\frac{13}{30}-\frac{\pi}{8}+\frac{1}{15}\\
&=\frac12-\frac{\pi}{8}\\
\end{array}
$.
Putting this together,
$\begin{array}\\
S
&=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)+8\sum_{j=1}^{\infty}\frac1{4^{2j}}\zeta(2j)\\
&=-4(\frac12)+8(\frac12-\frac{\pi}{8})\\
&=-2+4-\pi\\
&=2-\pi\\
\end{array}
$.
Whew!
|
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|
How could I find the sum of this infinite series by hand? $$\sum_{n=1}^{\infty}\frac{(7n+32)3^n}{n(n+2)4^n}$$
Thank you!
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Let the $n(\ge1)$th term $T_n=\dfrac{(7n+32)3^n}{n(n+2)4^n}=A\dfrac{\left(\dfrac34\right)^n}n+B\dfrac{\left(\dfrac34\right)^{n+2}}{n+2}$ where $A,B$ are arbitrary constants
$\implies\dfrac{(7n+32)}{n(n+2)}=\dfrac An+ \dfrac{B\left(\dfrac34\right)^2}{n+2}$
$\implies7n+32=A(n+2)+\dfrac{9Bn}{16}$
Comparing the constants, $32=2A\iff A=16$
Comparing the coefficients of $n, 7=A+\dfrac{9B}{16}\iff B=\dfrac{16(7-A)}9=\cdots=-16$
$\implies T_n=16\cdot\dfrac{\left(\dfrac34\right)^n}n-16\cdot\dfrac{\left(\dfrac34\right)^{n+2}}{n+2}=16\left(u_n-u_{n+2}\right)$
where $u_m=\dfrac{\left(\dfrac34\right)^m}m$
$\sum_{n=1}^\infty T_n$ clearly telescopes to $16(u_1+u_2)=?$
|
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|
Consecutive sets of consecutive numbers which add to the same total I'm looking at examples of numbers that can be written as the sum of integers from $j$ to $k$ and from $k+1$ to $l$. For example $15$ which can be written as $4+5+6$ or $7+8$. Or $27 = 2+3+4+5+6+7 = 8+9+10$. I have been able to find a few numbers which have two ways to satisfy the above equations. For example,
$$\begin{aligned}105
&= 1+2+\dots +14 = 15+16+\dots+20\\
&= 12+13+\dots+18 = 19+20+\dots+23
\end{aligned}$$
However, I have not been able to find any numbers that can be written as the sum in three ways of consecutive sums. That is, I have not been able to find an $X$ such that,
$$\begin{aligned}X
&= (a+1)+(a+2)+\dots +b = (b+1)+(b+2) +\dots +c\\
&= (d+1)+(d+2)+\dots +e = (e+1)+(e+2) +\dots +f\\
&= (g+1)+(g+2)+\dots +h = (h+1)+(h+2) +\dots +i\\
\end{aligned}$$
Does any such number $X$ exist? If so can you provide an example? If no such number exists can you provide a proof?
Thanks
|
Suppose,
$$X=(a+1)+(a+2)+\dots+b=(b+1)+(b+2)+\dots+c$$
which implies,
$$\Leftrightarrow \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2}$$
$$2b(b+1)=a(a+1)+c(c+1)$$
or the special Pythagorean triple,
$$(2a-2c)^2+(2a+2c+2)^2=(4b+2)^2$$
where $a,b,c\in\mathbb Z$. We need to find integer solutions of the system,
$$(2c-2a)=s^2-t^2,\quad (2a+2c+2)=2st,\quad (4b+2)=s^2+t^2$$
So,
$$a=\frac{|s^2-t^2-2st|-2}{4},\quad b=\frac{s^2+t^2-2}{4},\quad c=\frac{2st+(s^2-t^2)-2}{4}$$
and just let $s=(2m+1),t=(2n-1)$, where $m\geq n$. Since,
$$X=\bigg(c+\frac{1}{2}\bigg)^2-\bigg(b+\frac{1}{2}\bigg)^2=\bigg(\frac{2st+(s^2-t^2)}{4}\bigg)^2-\bigg(\frac{s^2+t^2}{4}\bigg)^2\\
=\frac{st(t+s)(s-t)}{4}=(2m+1)(2n-1)(m-n+1)mn$$
If the equation below has 3 or more integer solutions then the $z$ is exactly what you want:
$$z=(2m+1)(2n-1)(m-n+1)mn$$
where $m\geq n$. Alternatively,
$$\begin{aligned}
a &= \tfrac{1}{2}\Big(-1+\sqrt{(x^2-2xy-y^2)^2}\Big)\\
b &= \tfrac{1}{2}\big(-1+x^2+y^2\big)\\
c &= \tfrac{1}{2}\big(-1+x^2+2xy-y^2\big)
\end{aligned}$$
where $x>y$ and sign chosen so $a$ is positive. Then,
$$X = \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2} =\tfrac{1}{2}xy(x^2-y^2)$$
so it suffices to find three pairs of $x,y$ with the same $X$.
|
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|
Simplifying $\sum_{i=0}^n i^k\binom{n}{2i+1}$ What is the formula for
\begin{eqnarray}\sum_{i=0}^n i^k\binom{n}{2i+1}?\end{eqnarray}
I tried to use the identity
$$
\sum_{i=0}^ni(i-1)\cdots(i-p)\binom{n}{2i+1}=(n-p-2)(n-p-3)\cdots(n-2p-2)2^{n-2p-3}
$$
but got into a mess. Any idea?
|
Suppose we seek to evaluate
$$S(n, k) = \sum_{q=0}^{n}
q^k {n\choose 2q+1}.$$
Introduce
$$q^k =
\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(qz)}{z^{k+1}} \; dz.$$
Observe that with $k\ge 1$ we also get the correct value for $q=0.$
We obtain for the sum
$$\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}}
\sum_{q=0}^{n} {n\choose 2q+1} \exp(qz)
\; dz
\\ = \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}}
\sum_{q=0}^{n} {n\choose 2q+1} \exp(2q(z/2)) \; dz
\\ = \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp(-z/2)
\sum_{q=0}^{n} {n\choose 2q+1} \exp((2q+1)(z/2)) \; dz.$$
This is
$$\frac{1}{2} \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp(-z/2)
((1+\exp(z/2))^n-(1-\exp(z/2))^n) \; dz.$$
Substituting $z=2w$ we obtain
$$\frac{1}{2^{k+1}} \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w)
((1+\exp(w))^n-(1-\exp(w))^n) \; dw.$$
This has two pieces call them $A_1$ and $A_2.$ The piece $A_1$ is
$$\frac{1}{2^{k+1}} \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w)
(1+\exp(w))^n\; dw
\\ = \frac{1}{2^{k+1}} \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w)
\sum_{q=0}^n {n\choose q} 2^{n-q} (\exp(z)-1)^q
\; dw.$$
This is by convolution of generating functions
$$\frac{1}{2^{k+1}}
\sum_{q=0}^n {n\choose q} 2^{n-q} q!
\sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace q}.$$
The piece $A_2$ is
$$-\frac{1}{2^{k+1}} \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w)
(1-\exp(w))^n \; dw
\\ = (-1)^{n+1} \frac{1}{2^{k+1}} \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w)
(\exp(w)-1)^n \; dw.$$
This is
$$(-1)^{n+1} \frac{1}{2^{k+1}} n!
\sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace n}.$$
Observe that this vanishes if $k\lt n.$
Finally note concerning $A_1$ that if $k\lt n$ all the terms with
$q\gt k$ from the sum in $q$ cease to contribute owing to the Stirling
number being zero, so we may set the upper limit to $k$ in this case.
Similarly when $n\lt k$ the terms with $q\gt n$ cease to contribute
owing to the first binomial coefficient.
The end result is a polynomial in $n$ plus a correction term and we get
$$\frac{1}{2^{k+1}}
\sum_{q=0}^k {n\choose q} 2^{n-q} q!
\sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace q}
+ (-1)^{n+1} \frac{1}{2^{k+1}} n!
\sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace n}.$$
|
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|
Show that $ \int_0^2 e^{x^2-x} dx \in [2e^{-1/4},2e^2] $
Show that $ \int_0^2 e^{x^2-x} dx \in [2e^{-1/4},2e^2] $
If $f(x)\leq g(x)$ for $x\in[a,b]$ then $\int^b_af(x)dx\leq \int^b_ag(x)dx$
if $x\in [0,2]$ then $x^2-x\leq x$, so
$$0 \leq \int_0^2 e^{x^2-x} dx\leq \int_0^2 e^{x} dx = e^2-1 <2e^2$$
Which integral should I choose on the left side?
|
Since $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ we have, for $x\in[0,2]$,
$$-\frac{1}{4}\le x^2-x \le 2$$
Then
\begin{align*}
\int_0^2e^{-\frac{1}{4}}\,dx&\le\int_0^2e^{x^2-x}\,dx\le\int_0^2e^2\,dx\\
2e^{-\frac{1}{4}}&\le\int_0^2e^{x^2-x}\,dx\le 2e^2
\end{align*}
|
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|
Paramaterization of paraboloid and plane. Consider the paraboloid $z=x^2+y^2$. The plane $2x-4y+z-6=0$ cuts the paraboloid, its intersection being a curve. Find "the natural" parameterization of this curve.
I have set each equation equal to each other by solving for $z$, completed the square to reach $(x-1)^2+(y+2)^2=11$. This is where I am stuck. Any help would be appreciated!
|
Plugging $z=x^2+y^2$ into $2x-4y+z-6=0$ we get
\begin{align*}
2x-4y+x^2+y^2-6&=0\\
x^2+2x+\color{red}{1}+y^2-4y+\color{red}{4}&=6+\color{red}{1+4}\\
(x+1)^2+(y-2)^2&=11
\end{align*}
Then, a parameterization for the curve is
$$\begin{cases}x&=\sqrt{11}\cos t-1 \\ y&=\sqrt{11}\sin t+2\\z&=2\sqrt{11}(2\sin t -\cos t)+16\end{cases}\qquad 0\le t\le 2\pi$$
|
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|
If $a,b,c,d,e,f$ are non negative real numbers such that $a+b+c+d+e+f=1$, then find maximum value of $ab+bc+cd+de+ef$ $(a+b+c+d+e+f)^2=$ sum of square of each number (X)+ $2($ sum of product of two numbers (Y) $)$
$ab+bc+cd+de+ef \le Y$ since all are positive.
Therefore $1\ge X+(ab+bc+cd+de+ef)$
Edit: From AM GM inequality, $$X\ge 6(abcde)^{1/3}$$and for am-gm inequality for a,b,c,d,e,f: $$1\ge 6(abcdef)^{1/6}$$
Hence $X\ge 6.\frac{1}{36}$ and $X\ge \frac{1}{6}$.
Is this correct?
|
$$a+b+c+d+e+f=1\implies (a+c+e)(b+d+f)\le \frac14 \\ \implies (ab+bc+cd+de+ef)+(ad+af+cf+be)\le \frac14$$
We can show the first bracket can attain full value by say $a=b=\frac12$.
|
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|
integration by parts of $25\, (1-\sin^{2}x)$ I need help solving this integration of parts problem. I've tried a few different solutions and keep getting the wrong answer. This question is in regards to this problem take the integral by parts of:
$$\displaystyle \int (5-5\sin x)(5+5\sin x)dx$$
So first I multiply, and get $25-25\sin^2 x.$
Then i tried to use the formula, integral
$$\displaystyle (f(x)g'(x)) = f(x)g(x) - \int (f'(x)g(x)).$$
But it was to no avail. I know the answer, if you'd like it provided
but obviously more important is the how! Please help me and thanks.
|
Consider
\begin{align}
I &= \int (5 - 5 \sin x) \cdot (5 + 5 \sin x) \, dx \\
&= 25 \, \int (1-\sin x) \cdot (1 + \sin x) \, dx = 25 \, J
\end{align}
Let $dv = 1-\sin x$ then $v = x + \cos x$, $u=1+\sin x$, $du = \cos x$ and
\begin{align}
J &= (x+\cos x)(1 + \sin x) - \int (x + \cos x) \, \cos x \, dx \\
&= (x + \cos x)(1 + \sin x) - \int x \, \cos x \, dx - \int \cos^{2}x \, dx \\
&= (x + \cos x)(1 + \sin x) - x \, \sin x - \cos x - J \\
\end{align}
or
\begin{align}
J &= \frac{1}{2} \left[ (x + \cos x) (1 + \sin x) - x \, \sin x - \cos x \right] \\
&= \frac{1}{2} \left( x + \sin x \, \cos x \right) = \frac{2x + \sin(2x)}{4}
\end{align}
Now,
$$I = \frac{25}{4} \, \left[ 2 \, x + \sin(2 \, x) \right]$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$. I was curious if there are quadratic equations where $a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$.
So trivially if $c=0$, $a$ and $b$ can be arbitrary, and if either $a$ or $b$ is zero, this implies that $c=0$, and the other arbitrary.
Is there a way to find other solutions, if any exist?
So the equation to solve in integers is $$a^3b^2c^2+ab^2c+c=0.$$
Dividing out by $c$ gives $$a^3b^2c+ab^2+1=0.$$
We could factor to give $$ab^2(a^2c+1)=-1$$
So either
*
*$ab^2=1$ and $a^2c+1=-1$ or
*$ab^2=-1$ and $a^2c+1=1$
For 1. $a=1$ and $b=\pm1$ and $c=-2$
For 2. $a=-1$ and $b=\pm1$ and $c=-2$ and $c=0$ is forced.
Is this analysis correct?
Are the non-trivial solutions:
$x^2+x-2$ and $x^2-x-2$, with roots $-2$ and $2$ respectively?
|
Here's an alternate approach you might consider (+1 on yours, which works just fine).
Now, if $a,b,c$ satisfy the desired property and $c\ne 0,$ then by Rational Root Theorem we have that $abc$ is an integer factor of $c,$ from which we conclude that either
*
*$c=0,$ in which case any $a,b$ will do (as you observed), or
*$c\ne 0$ and $ab$ is an integer factor of $1,$ whence $a,b\in\{1,-1\}$ since $a,b\in\Bbb Z.$
If $c\ne0$ and $a=1,$ then we have by assumption that $$0=(bc)^2+b(bc)+c=b^2c^2+b^2c+c=c^2+2c,$$ whence $c=-2$ since $c\ne 0.$ Hence, $abc=-2b,$ so $$0=(-2b)^2+b(-2b)-2=4b^2-2b^2-2=2b^2-2=2-2,$$ and so either of $b=\pm 1$ works.
If $c\ne 0$ and $a=-1,$ then we have by assumption that $$0=-(bc)^2+b(bc)+c=-b^2c^2+b^2c+c=-c^2+2c,$$ whence $c=2,$ since $c\ne 0.$ Hence, again, $abc=-2b,$ so $$0=-(-2b)^2+b(-2b)+2=-4b^2-2b^2+2=-6b^2+2=-6+2=-4,$$ a contradiction. Hence, we cannot have $a=-1.$
Hence, the solutions are precisely those of the forms $x^2\pm x-2$ or $ax^2+bx$ for some $a,b\in\Bbb Z.$
|
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|
How to integrate following indefinite integal? The integral is
$$
\int\frac{x-\sin x}{1-\cos x} \,dx
$$
However, the only guess I have is that the denominator is the derivative of the numerator. Probably the integration by substitution will work here?
|
$$\begin{align*}
\int\frac{x-\sin x}{1-\cos x}dx
&= \int \frac{(x-\sin x)(1+\cos x)}{(1-\cos x)(1+\cos x)}\ dx \\
&= \int \frac{x + x\cos x - \sin x -\sin x \cos x}{1-\cos^2 x}\ dx\\
&= \int (x\csc ^2 x + x \cot x \csc x - \csc x - \cot x)\ dx\\
&= -\int x\ d(\cot x) - \int x\ d(\csc x) - \int \csc x\ dx - \int \cot x\ dx\\
&= -x \cot x + \int \cot x\ dx - x\csc x + \int \csc x\ dx - \int \csc x\ dx - \int \cot x\ dx\\
&= -x \cot x - x\csc x + C\\
&= -x(\cot x + \csc x) + C
\end{align*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$a,b$ be two positive integers , where $b>2 $ , then is it possible that $2^b-1 \mid 2^a+1$? If $a,b$ be two positive integers , where $b>2 $ , then is it possible that $2^b-1\mid2^a+1$ ? I have figured out that if $2^b-1\mid 2^a+1$, then $2^b-1\mid 2^{2a}-1$ , so $b\mid2a$ and also $a >b$ ; but nothing else. Please help. Thanks in advance
|
If $b$ is odd, then $b\mid 2a$ implies $b\mid a$, then $2^b-1\mid 2^a-1$ and hence $2^b-1\nmid 2^a+1$ (as $2^a+1$ is between $2^a-1$ and $2^a-1+(2^b-1)$).
If $b$ is even, $b=2c$ say, then $c\mid a$, hence $2^c-1\mid 2^a-1$ and $2^c-1\mid 2^b-1$. As $c>1$ this shows $d:=\gcd(2^a-1,2^b-1)>1$ (and of course odd) and so $\gcd(2^a+1,2^b-1)<\frac{2^b-1}{d}<2^b-1$.
|
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|
Evaluate $f(z)=\int_0^1 \frac{dt}{t-z}$ where $z\in \mathbb{C}-[0,1]$.
Evaluate $\displaystyle f(z)=\int_0^1 \frac{dt}{t-z}$ where $z\in \mathbb{C}-[0,1]$. Here, $0\leq t \leq 1$.
My Try:
Can we use the same integration rules here as in real integration? I mean does $$ f(z)=\int_0^1 \frac{dt}{t-z}=\left[\ln|t-z|\vphantom{\frac11}\right]_0^1\text{ ?}$$
|
One simple mistake is you have $\ln|t-z|$ where you must have intended $\ln|t|-\ln|z|$.
Trigger warning: If the phrase "multiple-valued function" causes you pain, then stop reading before this present sentence.
Notice that if $z=x+iy$ and $x$ and $y$ are real then
$$
e^z = e^x e^{iy} = e^x (\cos y+ i\sin y)
$$
and, since the function $y\mapsto \cos y + i\sin y$ is periodic (with period $2\pi$), the function $z\mapsto e^z$ is not one-to-one. Notice that $e^{x+iy} = e^{x+iy+2\pi in}$, since $e^{2\pi in}=1$. Consequently the inverse of $z\mapsto e^z$ is a multiple-valued function
$$
\operatorname{Log}(w) = \operatorname{Log}(e^{x+iy}) = x + iy +2\pi in
$$
for $n\in\mathbb Z$.
So
$$
\int\limits_{\begin{smallmatrix} \text{some path} \\ \text{from $a$ to $b$} \end{smallmatrix}} \frac{dz} z = (\text{one of the values of }\operatorname{Log}b) - (\text{one of the values of }\operatorname{Log}a).
$$
Which one of the values it is depends on which path from $a$ to $b$ it is. Specifically in the expression $x+iy + 2\pi in$, the value of $n$ increases by $1$ every time the path winds around $0$ in a counterclockwise direction. Picture $t$ moving along a straight line from $0$ to $1$, and let's suppose for the moment that $z=(1+i)/2$. Then $t-z$ starts as $-(1+i)/2$ and ends up as $(1-i)/2$, and the line from $z$ to $t$ changes directions by going from southwest to southeast, rotating $90^\circ$ counterclockwise, i.e. $\pi/2$ radians counterclockwise. Hence the imaginary part of the integral is $i \pi/2$. The two real parts begin subtracted are equal since $|-(1+i)/2| = |(1-i)/2|$, so the integral is $i\pi/2$.
If you want to do it by methods resembling the technical stuff you learn in first-year calculus, you can write:
\begin{align}
& \int_0^1 \frac{dt}{t-z} = \int_0^1 \frac{dt}{t-x-iy} = \int_0^1 \frac{(t-x+iy)\,dt}{(t-x)^2+y^2} \\[10pt]
= {} & \int_0^1 \frac{(t-x)\,dt}{(t-x)^2+y^2} + i \int_0^1 \frac{y\,dt}{(t-x)^2+y^2} \\[10pt]
= {} & \int_{x^2+y^2}^{(x-1)^2+y^2} \frac{du/2}{u} + i \int_0^1 \frac{dt/y}{\left( \frac{t-x} y \right)^2+1} = \int_{x^2+y^2}^{(x-1)^2+y^2} \frac{du/2}{u} + i \int_{-x/y}^{(1-x)/y} \frac {dv}{v^2+1} \\[10pt]
= {} & \frac 1 2 \log \frac{(x-1)^2+y^2} {x^2+y^2} + i \left( \arctan\frac{1-x} y - \arctan \frac{-x} y \right) \\[10pt]
= {} & \frac 1 2 \log \frac{(x-1)^2+y^2} {x^2+y^2} + i \arctan \frac{y}{y^2+x^2-x}.
\end{align}
This is a function of $x$ and $y$, and maybe after that we'd want to figure out whether we write it as a function of $z=x+iy$.
|
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|
A simple double inequality with roots Prove or disprove for integer $n$ greater than $1$:
$$\sqrt{1} + \sqrt{2} + ...+ \sqrt{n-1} < \frac{2n\sqrt{n}}{3} < \sqrt{1} + \sqrt{2} + ...+ \sqrt{n} $$
I think I know a solution, but I am looking for different approaches.
EDIT: Inspired by Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$.
|
$\sqrt{x}$ is a concave increasing function on $\mathbb{R}^+$, hence by the Hermite-Hadamard inequality:
$$\frac{1}{2}\cdot\frac{1}{1}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n-1}}+\frac{1}{2\sqrt{n}}\leq \int_{1}^{n}\sqrt{x}\,dx = \frac{2}{3}\left(n\sqrt{n}-1\right)\tag{1}$$
as well as:
$$ \frac{(2n+1)\sqrt{2n+1}-1}{3\sqrt{2}}=\int_{1/2}^{n+1/2}\sqrt{x}\,dx\leq\frac{1}{\sqrt{1}}+\ldots+\frac{1}{\sqrt{n}}\tag{2}$$
improving a bit your original inequality.
|
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|
Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible? Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible?
What is the best/simplest/elementary way to approach this? Any solutions or hints are greatly appreciated.
|
No, it is not. If we set $p(x)=6x^4+3x^3+6x^2+2x+5$, over $\mathbb{F}_7$ we have:
$$ p(x+2) = -\left(x^4-2x^3-2x+1\right) $$
that is a palyndromic polynomial, from which:
$$\frac{p(x+2)}{x^2} = -\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right) = -\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)+2$$
and:
$$ p(x+2) = -(x^2+2x+3)(x^2+3x-2),$$
since the previous line gives:
$$\begin{eqnarray*} p(x+2)=-(x^2+1)^2+2x(x^2+1)+2x^2&=&-(x^2+1-x)^2+3x^2\\&=&-(x^2+1-4x)^2+2(x+2)^2\end{eqnarray*} $$
and the RHS is now the difference of two squares.
A viable alternative is to notice, through Stickelberger criterion, that since the discriminant of $p$, $\Delta=-1728=-12^3$, is a quadratic residue $\pmod{7}$, $p$ splits as the product of an even number of irreducible polynomials, hence $p$ cannot be irreducible.
|
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|
Calculate product of geometric progressions Let us $p_k(z)=\sum_{i=0}^{k-1}z^{i}.$ For any $m$ find coefficient of the term $z^m$ in the polynomial $\prod_{j=1}^n p_{2j}(z).$
Here are some attempts
$$p_2=1+z$$
$$p_4=1+z+z^2+z^3$$
$$p_6=1+z+z^2+z^3+z^4+z^5$$
$$\vdots$$
$$p_{2n}=1+z+z^2+z^3+\ldots+z^{2n-1}$$
Coefficient of the term $z^0:$ $1$
Coefficient of the term $z^1:$ $n-1$
Coefficient of the term $z^2:$ $(n-1)+(n-1)(n-2)+((n-1)+(n-2)+\cdots+1)=(n-1)(1+n-2+\frac{n}{2})=\frac12(n-1)(3n-2)=(n-1)^2+\frac{n(n-1)}{2}$
Coefficient of the term $z^3:$ $2(n-1)^2$
Coefficient of the term $z^4:$ $2(n-2)+(n-2)(n-3)+(n-1)^2+\frac{(n-2)(n-1)}{2}=(n-1)^2+(n-1)(n-2)+\frac{(n-2)(n-1)}{2}=\frac{(n-1)(5n-8)}{2}$
Coefficient of the term $z^5:4(n-2)+(n-1)(n-2)=(n-2)(n+3)$
Coefficient of the term $z^6:5(n-2)+(n-3)(n-2)+\frac{(n-2)(n-1)}{2}=(n-2)(n+2)+\frac{(n-2)(n-1)}{2}=\frac{(n-2)(3n+3)}{2}$
Coefficient of the term $z^7:2(n-1)(n-3)+2(n-3)^2=4(n-2)(n-3)$
$$\ldots$$
What is generalized formula?
|
$$\prod_{j=1}^{n}p_{2j}(z)=\prod_{j=1}^{n}\frac{1-z^{2j}}{1-z}=\sum_{r\geq 0}\binom{r+n-1}{n-1}z^r \prod_{j=1}^{n}(1-z^{2j})$$
so the coefficient of $z^m$ in the RHS is given by a weigthed sum (weigths are given by binomial coefficients) of partition numbers. Have a look at this related question.
|
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|
In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of ways to select $2$ similar and $3$ different letter = $\dbinom{4}{1}\times \dbinom{4}{3}=16$.
Number of ways of selecting $2$ similar + $2$ more similar letter and $1$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and $2$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and another $2$ other similar = $\dbinom{3}{1}\times \dbinom{3}{1}=9$
Number of ways to select $4$ similar and $1$ different letter = $\dbinom{2}{1}\times \dbinom{4}{1}=8$
Ways of selecting
$5$ similar letters = $1$
Total ways = $1+16+18+18+9+8+1= 71$
Well I have the solution But I am not able to fully understand it.
Or if their could be an $\color{red}{\text{alternate way}}$ than it would be great.
I have studied maths up to $12$th grade.
|
Ok so we are breaking this down by case:
Case 1: none of our letters are the same and we pick 5 distinct letters
Case 2a: exactly 2 of the letters are the same and the other 3 are distinct.
Case 2b: 2 of the letters are the same, 2 others are the same, and the fifth is distinct.
Case 3a: exactly 3 of the letters are the same and the other 2 are distinct.
Case 3b: exactly 3 of the letters are the same and the other 2 are also the same.
Case 4: exactly 4 of the letters are the same and the other 1 is distinct.
Case 5: all 5 letters are the same.
We see that any combination must be one of these $5$ cases, and there is no overlap in the $5$ cases. Thus we are going to sum up the number of total possibilities for each case.
Case 1: If we pick $5$ distinct letters, we pick one of each letter, so there is only $1$ way to do this.
Case 2a: First lets count the ways to pick the $2$ letters which are the same. It can't be E, so there are only $4$ ways to pick them. Second, we must pick $3$ more from the remaining $4$ letters (none of these can be the same letter as the first $2$ since that would be a different case). Thus we use $4$ choose $3$, which turns out to be $4$ options. We multiply these together to see that we have $4\times 4=16$ options for case 1.
Case 2b: We are now picking two pairs from four possible letters which have sufficient quantities. Thus the ways to pick the two pairs is $4$ choose $2$. The fifth letter is any of the remaining $3$ letters, so we have $6\times 3=18$ for this case.
Case 3a: Similarly, there are only $3$ ways to pick $3$ matching letters. Then we proceed to pick the other $2$ from $4$ choices, so we multiply $3$ by $4$ choose $2$ to get $3\times 6=18$ options for case 3.
Case 3b: There are $3$ ways to pick the triplet and the other pair only has $3$ options also, since it can't be the same letter as the triplet (that would be case 5). Thus there are $3\times 3=9$ possibilities.
Case 4: Hopefully you follow the pattern: there are only $2$ ways to pick $4$ of the same letter, and $4$ ways to pick the fifth letter, so we have $2\times 4=8$ for this case.
Case 5: Finally, there is only one way to pick $5$ of the same number.
Summing those together we get $1+16+18+18+9+8+1=71$ options. For me, this is the most straightforward approach once you understand it.
|
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|
Consider the following system and find the values of b for which the system has a solution So I have this system:
$$\left\{\begin{array}{c}
x_1 &− x_2 &+ 2x_3 &= 2 \\
x_1 &+ 2x_2 &− x_3 &= 2 \\
x_1 &+ x_2 & &= 2 \\
x_1 & & +x_3 &= α
\end{array}\right.$$
And we are asked to find the values of $\alpha$ for which the system has a solution. When I do the coefficient matrix and a Gauss-Jordan elimination i get two entirely zero rows with one of the augmented matrix solutions being $\alpha-2$ so am I correct in saying that the system will have an infinite number of solutions if $\alpha=2$?
The thing that concerns me is that we are then asked to find the value of $\alpha$ if the degree of freedom is $2$?
Any help would be greatly appreciated!
PS. Sorry for any formatting errors!
|
HINT: use Rouché–Capelli theorem
A system of linear equations with $\,n\,$ variables has a solution if and only if the rank of its coefficient matrix $\,A\,$ is equal to the rank of its augmented matrix $\,\left[\,A \mid b\,\right].\,$
Let us rewrite your system in matrix form $\,\mathbf A\,\vec{\boldsymbol{x}} = \vec{\boldsymbol{b}}$:
\begin{align}
\left\lbrace
\begin{array}{ccccccc}
x_1 & - & x_2 & +& 2\, x_3 & = & 2 \\
x_1 & + & 2\,x_2 & -& x_3 & = & 2 \\
x_1 & + & 2\,x_2 & & & = & 2 \\
x_1 & & &+& x_3 & = & \alpha \\
\end{array}
\right.
\iff
%\underbrace{
\begin{pmatrix}
1 & -1 & 2 \\
1 & 2 & -1 \\
1 & 1 & 0 \\
1 & 0 & 1 \\
\end{pmatrix}
%}_{\;\ \mathbf A}
%\overbrace{
\begin{matrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\\
\phantom{x_4}\end{matrix}
%}^{\vec{\boldsymbol{b}}}
=
%\underbrace{
\begin{pmatrix} 2 \\ 2 \\ 2 \\ \alpha \end{pmatrix}
%}_{\vec{\boldsymbol{b}}}
\end{align}
$$
\mathbf{A} = \begin{pmatrix}
1 & -1 & 2 \\
1 & 2 & -1 \\
1 & 1 & 0 \\
1 & 0 & 1 \\
\end{pmatrix}
, \quad b =\begin{pmatrix} 2 \\ 2 \\ 2 \\ \alpha \end{pmatrix}
\implies
\mathbf{B} := \left[ \, \mathbf A \,\left\lvert \;\, \vec{\boldsymbol{b}} \right.\, \right] = \left[
\begin{array}{ccc|c}
1 & -1 & 2 & 2 \\
1 & 2 & -1 & 2 \\
1 & 1 & 0 & 2 \\
1 & 0 & 1 & \alpha \\
\end{array}
\right]
$$
It is easy to see that rank $\,\mathbf A = 2\,$ because its first row is linear combination of the second and the third rows: $\,\mathbf A\left[1\right] = 3\mathbf A\left[2\right] - 2\mathbf A\left[2\right]\,$
$$
\mathbf A\left[1\right] = 3\mathbf A\left[2\right] - 2\mathbf A\left[2\right]
\quad\iff\qquad
\begin{array}{ccc}
& \begin{pmatrix} 1 & \phantom{-}1 & \phantom{-}0 \end{pmatrix} & \times \,3 \\
- & \begin{pmatrix} 1 & \phantom{-}2 & -1 \end{pmatrix} & \times \,2 \\
\hline
& \begin{pmatrix} 1 & -1 & \phantom{-}2 \end{pmatrix} & \\
\end{array}
$$
Therefore we need to find values of $\,\alpha\,$ for which rank $\,\mathbf B = 2$.
I hope you can pick it from here.
|
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Permutation of multiset
How many 8-permutation are there of the letters of the word
'ADDRESSES'?
My textbook suggests that we should divide the situation into cases where a different letter is removed. In other words, for the multiset $\{1 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$, we count the number of permutation of the follow set:
*
*$\{0 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$
*$\{1 \cdot A, 1 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$
*$\{1 \cdot A, 2 \cdot D, 0 \cdot R, 2 \cdot E, 3 \cdot S\}$
*$\{1 \cdot A, 2 \cdot D, 1 \cdot R, 1 \cdot E, 3 \cdot S\}$
*$\{1 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 2 \cdot S\}$
It is easy to show that the total number of 8-permutation is $15120$. I am not happy with this case-dividing computation and I want to have a direct computation of the result. Accidentally, I find that $$C_8^9\frac{8!}{2!2!3!}=15120.$$ I further test this formula, e.g. 3-permutation of a multiset with 4 elements, and it actually works. So I think there should be a nice explanation why the formula works. Can anyone explain that for me?
|
There are nine letters in ADDRESSES, but we have eight positions to fill. Thus, we must omit one of the letters. If we omit the A, we can fill two of the eight positions in $\binom{8}{2}$ ways. We can then fill one of the reamining six positions with an R in $\binom{6}{1}$ ways. We can fill two of the remaining positions with E's in $\binom{5}{2}$ ways. Finally, we fill the remaining three positions with S's in $\binom{3}{3}$ ways. Hence, there are $$\binom{8}{2}\binom{6}{1}\binom{5}{2}\binom{3}{3} = \frac{8!}{2!6!} \cdot \frac{6!}{1!5!} \cdot \frac{5!}{2!3!} \cdot \frac{3!}{0!3!} = \frac{8!}{2!1!2!0!3!} = \frac{8!}{2!2!3!}$$
$8$ permutations of the letters of ADDRESSES in which A is omitted. Since there is only one A and one R in ADDRESSES, there are the same number of $8$ permutations in which an R is omitted (interchange the roles of A and R in the above argument).
By similar argument, the number of $8$ permutations in which a D is omitted is
$$\binom{8}{1}\binom{7}{1}\binom{6}{1}\binom{5}{2}\binom{3}{3} = \frac{8!}{2!3!}$$
Since there are two D's and two E's in addresses, there are the same number of permutations in which an E is omitted.
The number of $8$ permutations in which an S is omitted is
$$\binom{8}{1}\binom{7}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2} = \frac{8!}{2!2!2!}$$
Adding the cases above yields
\begin{align*}
\frac{8!}{2!2!3!} + \frac{8!}{2!2!3!} + \frac{8!}{2!3!} + \frac{8!}{2!3!} + \frac{8!}{2!2!2!} & = \frac{8!}{2!2!3!}(1 + 1 + 2 + 2 + 3)\\
& = 9 \cdot \frac{8!}{2!2!3!}\\
& = \frac{9!}{2!2!3!}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
check if large number $(9^{81}+6)$ is divisible by $11$ I would like to know if there is a mathematical way to check whether number $9^{81}+6$ is divisible by $11$, without actually calculating the whole number.
|
$$9^1 \equiv -2 \bmod{11}$$ $$9^2 \equiv 4 \bmod{11}$$ $$9^6 \equiv 64 \equiv -2 \bmod{11}$$ $$9^{30} \equiv -32 \equiv 1 \bmod{11}$$ $$9^{81} \equiv 9^{30} 9^{30} 9^{6} 9^{6} 9^{6} 9^2 9^1 \equiv (1)(1)(-2)(-2)(-2)(4)(-2) \equiv 64 \equiv 9 \bmod{11}$$ $$9^{81} + 6 \equiv 15 \equiv 4 \bmod{11}.$$
EDIT: I accidentally wrote $-32 \equiv -1 \bmod{11}$ instead of $-32 \equiv 1 \bmod{11}$. The minus signs fortuitously cancelled!
|
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|
Taylor expansion $\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$ I would like to use Taylor expansion around $y=0$ for that expression but i got stuck
$$\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$$
Taylor expansion around $y=0$
$$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+o\left(y^3\right) $$
$$\displaystyle {1 \over y^2 + y^3}= {1 \over y^2 (1+y)} = {1 \over y^2}{1 \over 1+y} = {1 \over y^2}(1-y+\frac{y^2}{2} + o(y^2)).$$
$$\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}=\left(-\frac{y^2}{2}+\frac{y^3}{6}+o\left(y^3\right) \right) \cdot \left( {1 \over y^2}(1-y+\frac{y^2}{2} + o(y^2))\right)
= \left(-\frac{y^2}{2}+\frac{y^3}{6}+o\left(y^3\right) \right) \cdot \left( (\dfrac{1}{y^2}-\dfrac{1}{y}+\frac{1}{2} + o(1))\right)\\
= \left(-\frac{y^2}{2}+o\left(y^2\right) \right) \cdot \left( (\dfrac{1}{y^2}-\dfrac{1}{y}+\frac{1}{2} + o(1))\right)\\
=
$$
i'm stuck here
am i right
|
No, your expansion is not right.
Although you wrote that $\frac{1}{y^2 + y^3} = y^{-2}(1 - y + y^2/2 + \cdots)$, you seem to have put all of that into the denominator. It shouldn't be there.
Instead, you should have
$$ (-y^2/2 + y^3/6 + \cdots ) y^{-2} (1 - y + y^2/2 + \cdots),$$
from which it is relatively easy to collect coefficients and describe the first coefficients of the Taylor expansion.
|
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|
Polynomial Divisibilty Test I recently came across a question in a book and I was wondering how to go about solving this. I just need a hint about how I could approach it.
I have to show that $x^{6n+2} - x^{6n+1} + 1$ is always divisible by $x^2 - x + 1$ where $n = \{1,2,3,4,\cdots\}$
|
Let $\alpha,\beta$ be two (complex) roots of $x^2-x+1$. Then $x^2-x+1=(x-\alpha)(x-\beta)$. Note that $x^2-x+1$ divides $x^3+1=(x+1)(x^2-x+1)$, and hence also $x^6-1=(x^3-1)(x^3+1)$. So both $\alpha,\beta$ satisfy $x^6=1$. Thus we have $\alpha^{6n+2}-\alpha^{6n+1}+1=(\alpha^6)^n\cdot\alpha^2-(\alpha^6)^n\cdot\alpha+1=\alpha^2-\alpha+1=0$, and same for $\beta$. So both $\alpha,\beta$ are roots of $x^{6n+2}-x^{6n+1}+1$. So $x-\alpha,x-\beta$ divide $x^{6n+2}-x^{6n+1}+1$, so also $(x-\alpha)(x-\beta)=x^2-x+1$ divides $x^{6n+2}-x^{6n+1}+1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
figuring out an integer function $f(1) = 1\\
f(2) = 2\\
f(3) = 6\\
f(4) = 20\\
f(5) = 70\\
f(6) = 252\\
f(7) = 924\\
f(8) = 3432\\
f(9) = 12870$
Then what is $f(n)$ (where $n > 0$)?
I though about many many possibilities but still cannot figure out the expression.
|
You could approach this kind of question in this way. You'd like to find what is the relation between $f_n$ and $f_{n+1}$ (you believe that there is some) so that's why most of the time it's a good idea to analyze expressions like $\frac{f_{n+1}}{f_n}$ or $f_{n+1}-f_n$. Let's look at $\frac{f_{n+1}}{f_n}$. You may notice that $\frac{f_{n+1}}{f_{n}}\sim 4$ as the sequence grows, more precisely:
$$
\begin{array}{ccl}
\frac{f_2}{f_1}& = & 2 \\
\frac{f_3}{f_2} & = & 3 \\
\frac{f_4}{f_3} & = &3.(3) \\
\frac{f_5}{f_4} & = & 3.5 \\
\frac{f_6}{f_5} & = & 3.6 \\
\frac{f_7}{f_6} & = & 3.(6)\\
\frac{f_8}{f_7} & = & 3.(714285) \\
\frac{f_9}{f_8} & = & 3.75 \\
\end{array}
$$
So you might expect that $\frac{f_{n+1}}{f_{n}} = 4 - g_n$ where $g_n \to 0$ as $n\to +\infty$. Upon more precise calculation we get:
$$
\begin{array}{ccl}
\frac{f_2}{f_1}& = & 4-\frac{2}{1} \\
\frac{f_3}{f_2} & = & 4-\frac{2}{2} \\
\frac{f_4}{f_3} & = &4-\frac{2}{3} \\
\frac{f_5}{f_4} & = &4-\frac{2}{4} \\
\frac{f_6}{f_5} & = &4-\frac{2}{5} \\
\frac{f_7}{f_6} & = & 4-\frac{2}{6}\\
\frac{f_8}{f_7} & = & 4-\frac{2}{7} \\
\frac{f_9}{f_8} & = & 4-\frac{2}{8} \\
\end{array}
$$
So we could see that $\frac{f_{n+1}}{f_n} = 4-\frac{2}{n} = \frac{2(2n-1)}{n}$. From here
$$\frac{f_{n+1}}{1} =\frac{f_{n+1}}{ f_n } \times \frac{f_{n}}{f_{n-1}} \times \dots \times \frac{f_2}{f_1} = \frac{2(2n-1)}{n}\times \frac{2(2n-3)}{n-1} \dots = \frac{2^n(2n-1)!!}{n!}$$
which is $\binom{2n}{n}$ as already mentioned
|
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|
Inequality related AM-GM It looks easy but I cannot find the solution.
Let $x,y,z$ be positive reals and satisfy $x^2+y^2+z^2=2$, prove that
$$ \left(x^3+\sqrt{3}xyz\right)\left( y^3-\sqrt{3}xyz\right)\leqslant 1. $$
I invented this inequality about a month ago. Essentially it can be proved by brutal force ( homogenize the inequality and then massive expanding), by SOS ( the usual method of express the polynomial into sum of square), or the Lagrange Multiplier Method (very painful task).
I am looking for a smart solution where the use of computer is minimum.
|
If $y^2-\sqrt3xz<0$ then our inequality is obviously true.
But for $y^2-\sqrt3xz\geq0$ by AM-GM we obtain:
$$(x^3+\sqrt3xyz)(y^3-\sqrt3xyz)=xy(x^2+\sqrt3yz)(y^2-\sqrt3xz)\leq$$
$$\leq\frac{1}{27}(x^2+y^2+xy+\sqrt3yz-\sqrt3xz)^3=$$
$$=\frac{1}{216}(2x^2+2y^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$
$$=\frac{1}{216}(6-x^2-y^2-3z^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$
$$=\frac{1}{216}\left(6-(x-y+\sqrt3z)^2\right)^3\leq\frac{6^3}{216}=1.$$
Done!
|
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|
Solving the simultaneous equations $x + y - z = 12$ and $x^2 + y^2 - z^2 = 12$ I found the following question online (it was a past question on the BMO):
find all positive integer solutions $x,y,z$ solving the simultaneous equations
$ x + y - z = 12 \ $ and $\ x^2 + y^2 - z^2 = 12 $ .
Normally these questions have a simple and elegant solution and although I've managed to solve it, the answer I found was neither! I wondered if anyone can see a simpler way to tackle the problem.
The way I solved this was to use the first equation to eliminate $z$ in the second question. I then factorised the resulting equation to obtain:
$(x-12)(y-12) = 66$ .
Then, using the fact that $66=2\times 3\times 11$ and also the fact that the equations are symmetric in $x$ and $y$, I worked out the possible factorisations of $66$ into two factors. From this I worked out possible solutions and then checked to see that they indeed were. (In total I found $8$ solutions.) Although this method worked, it seemed a bit cumbersome and I hope someone can improve on it! Please note that this question is aimed at someone with at most A-level standard maths abilities meaning it shouldn't need complex mathematics to solve it!
|
I'll try a generalization.
From
$x+y-z = n
$
and
$x^2+y^2-z^2 = m
$,
$z = x+y-n$
so
$z^2
=(x+y)^2-2n(x+y)+n^2
=x^2+2xy+y^2-2n(x+y)+n^2
$.
Therefore
$m
=x^2+y^2-(x^2+2xy+y^2-2n(x+y)+n^2)
=-2xy+2n(x+y)-n^2
$
so
$xy-n(x+y)
=-(m+n^2)/2
$
or
$(n^2-m)/2
=xy-n(x+y)+n^2
=(x-n)(y-n)
$.
The integer solutions
thus depend on the
factors of
$N=(n^2-m)/2$.
Note that if
$n$ and $m$ have different parity,
there are no integer solutions.
For each
$N
=ab
$
either
$x=n+a,
y=n+b
$
or
$x=n-a,
y=n-b
$.
The largest solution
is gotten by setting
$a=N$ and $b = 1$
so
$x = n+N$
and
$y = n+1$.
|
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|
Non-associative: set with a binary operation, but has inverses and identity I've been thinking about an example of some set with a binary operation which would satisfy all axioms of groups except for associativity. I'm new to Group Theory, so I would appreciate your knowledgeable insight.
My example:
$S={\mathbb{C} - \{0\}}$; $H = (S, \text^)$, where ^ is exponentiation. Can we consider this as a correct example? The identity under ^ appears to be 1. The inverse appears to be $\frac{2i\pi}{\ln(x)}$.
|
The object you are describing is called a loop. If you want a high powered example that fits in one mouthful, try the multiplicative group loop of non-zero octonions. As a more elementary alternative, I thought I'd give you a finite loop to ponder on (although you may need to work a bit to verify all the properties). I present Loop 8.1.4.0:
$$\begin{array}{c|c|c|c|c|c|c|c|}
\circ& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline
1 & 1 & 7 & 5 & 0 & 6 & 2 & 4 & 3\\\hline
2 & 2 & 6 & 7 & 5 & 0 & 3 & 1 & 4\\\hline
3 & 3 & 0 & 6 & 7 & 5 & 4 & 2 & 1\\\hline
4 & 4 & 5 & 0 & 6 & 7 & 1 & 3 & 2\\\hline
5 & 5 & 2 & 3 & 4 & 1 & 7 & 0 & 6\\\hline
6 & 6 & 4 & 1 & 2 & 3 & 0 & 7 & 5\\\hline
7 & 7 & 3 & 4 & 1 & 2 & 6 & 5 & 0\\\hline
\end{array}$$
Here $0$ is the identity, and the inverses are $0,3,4,1,2,6,5,7$ for $0\dots7$ respectively. Consider $1\circ 1\circ 2$ to show non-associativity and $1\circ 2$ for non-commutativity.
The above is actually an example of a Bol loop, which satisfies the more complicated weak associativity property $a(b(ac))=(a(ba))c$. For general loops, there are smaller examples; the smallest non-associative loop has order $5$ - here is one of them:
$$\begin{array}{c|c|c|c|c|}
\circ & 0 & 1 & 2 & 3 & 4\\\hline
0 & 0 & 1 & 2 & 3 & 4\\\hline
1 & 1 & 4 & 0 & 2 & 3\\\hline
2 & 2 & 3 & 4 & 1 & 0\\\hline
3 & 3 & 0 & 1 & 4 & 2\\\hline
4 & 4 & 2 & 3 & 0 & 1\\\hline
\end{array}$$
Note that this does not have two-sided inverses, since we have identities like $1\circ 2=0$ and $3\circ 1=0$, so that the left inverse of $1$ is $2$ and the right inverse is $3$. For non-associativity just consider $1\circ 1\circ 1$ (this loop is not even power-associative).
|
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|
Errors in partial fractions decomposition I decompose the fraction $\frac{x^2}{(x^2+1)^3}$ into $$\frac{x^2}{(x^2+1)^3}=\frac{Ax+B}{(x^2+1)^2}+\frac{Cx+D}{(x^2+1)^3}.$$
From here I found: $A=C=0,$ and $B=-D=1$
Now $$\frac{x^2}{(x^2+1)^3}=\frac{1}{(x^2+1)^2}-\frac{1}{(x^2+1)^3}$$
but the right side is not equal to the left side. How is this possible.
And, and similarly another example How is this possible
$$\frac{x^5+1}{x^4(x^2+1)}=\frac{Ax^3+Bx^2+Cx+D}{x^4}+\frac{Ex+F}{x^2+1}.$$
Find, $A=B=C=F=0,$ and $D=E=1$
Now write, $$\frac{x^5+1}{x^4(x^2+1)}=\frac{1}{x^4}+\frac{1}{x^2+1}.$$
but the right side is not equal to the left side. How is this possible.
Please help me. Thanks.
|
The first expression you got it right, I do the second for you:
$\dfrac{x^5+1}{x^4(x^2+1)}=\dfrac{x^5}{x^4(x^2+1)}+\dfrac{1}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1+x^2-x^2}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1+x^2}{x^4(x^2+1)}-\dfrac{x^2}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1}{x^4}-\dfrac{1}{x^2(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1}{x^4}-\dfrac{1}{x^2}+\dfrac{1}{x^2+1}$
|
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|
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$, then $n=17$ How can one show that :
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$ so $n=17$
I tried using the fact that any number $a\in\mathbb Q$ so $a=\frac{x}{y}$ such that $\gcd(x,y)=1$
So $\frac{n+15}{n+1}=\frac{x^2}{y^2}$
But here I'm stuck.
|
If we let $m=n+1$, the equation
$$
\frac{m+14}{m}=\frac{x^2}{y^2}
$$
is equivalent to
$$
m=\frac{14y^2}{(x-y)(x+y)}
$$
we can assume that $x\ge0$ and $y\gt0$ and $(x,y)=1$. Thus,
$$
(x-y,x+y)\mid2\quad\text{and}\quad(x-y,y)=1\quad\text{and}\quad(x+y,y)=1
$$
If $(x-y,x+y)=2$, then, since $(x-y,y)=1$, $y$ must be odd and the numerator only has one factor of $2$. Since the denominator has two factors of $2$, $m\not\in\mathbb{Z}$. Therefore,
$$
(x-y,x+y)=1
$$
and since either both $x-y$ and $x+y$ are even or both are odd, both must be odd. Because $y\gt0$, we cannot have $x-y=x+y$.
Therefore, since $(x-y)(x+y)\mid7$, we have either
$$
x-y=1\quad\text{and}\quad x+y=7\quad\implies\quad x=4,y=3,\color{#C00000}{m=18}
$$
or
$$
x-y=-1\quad\text{and}\quad x+y=7\quad\implies\quad x=3,y=4,\color{#00A000}{m=-32}
$$
or
$$
x-y=-1\quad\text{and}\quad x+y=1\quad\implies\quad x=0,y=1,\color{#0000F0}{m=-14}
$$
Thus,
$$
\begin{array}{c}
\color{#C00000}{n=17}&\text{or}&\color{#00A000}{n=-33}&\text{or}&\color{#0000F0}{n=-15}\\
\color{#C00000}{\Downarrow}&&\color{#00A000}{\Downarrow}&&\color{#0000F0}{\Downarrow}\\
\color{#C00000}{\sqrt{\frac{n+15}{n+1}}=\frac43}&&\color{#00A000}{\sqrt{\frac{n+15}{n+1}}=\frac34}&&\color{#0000F0}{\sqrt{\frac{n+15}{n+1}}=0}
\end{array}
$$
|
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|
Nilpotent matrix and relation between its powers and dimension of kernels Given a 4x4 matrix $T$ over $\mathbb{R}$ such that $T^4 = 0 $, $k_i = \textsf{dim} Ker(T^i)$, I need to check which of the following sequences, $$k_1\leq k_2 \leq k_3 \leq k_4,$$ is NOT possible :
$ 1)\; 1\leq 3 \leq 4 \leq 4$
$2) \; 2\leq3\leq4\leq4$
$3) \; 3 \leq 4 \leq 4\leq 4$
$4)\; 2 \leq 4 \leq 4 \leq 4$
The only relevant thing I could recall relating to this is the fact that for nilpotent operators $$ \{0 \} \subset Ker(T) \subset Ker(T^2) \subset \ldots \subset Ker(T^{n-1})$$
But the equality in the choices is putting me off. A hint would be welcome. Thanks in advance.
|
Consider the possible lists invariant factors
of $T$. Since $T$ is $4 \times 4$, then its characteristic polynomial is $T^4$. The invariant factors must divide the characteristic polynomial, hence are all of the form $T^i$. The characteristic polynomial is the product of all invariant factors, so the list of invariant factors must be $T^{i_1}, T^{i_2}, \ldots, T^{i_t}$, where $i_1, \ldots, i_t$ are positive integers, and $i_1 + \cdots + i_t = 4$. Moreover, each invariant factor divides the following invariant factor, so we have $i_1 \leq \cdots \leq i_t$. The possible partitions of $4$ are:
\begin{align*}
4 &= 4\\
4 &= 3+1\\
4 &= 2 + 2\\
4 &= 2 + 1 + 1\\
4 &= 1 + 1 + 1+ 1
\end{align*}
Let $k$ be our base field. We consider $k^4$ as a $k[x]$-module where $x$ acts as $T$. By the structure theorem for finitely generated modules over a PID
, then $k^4$ is isomorphic to one of the following $k[x]$-modules.
\begin{align*}
\frac{k[x]}{(x^4)} &\longleftrightarrow 1, 2, 3, 4\\
\frac{k[x]}{(x^3)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 2, 3, 4, 4\\
\frac{k[x]}{(x^2)} \oplus \frac{k[x]}{(x^2)} &\longleftrightarrow 2, 4, 4, 4\\
\frac{k[x]}{(x^2)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 3, 4, 4, 4\\
\frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 4, 4, 4, 4
\end{align*}
The numbers on the right are the corresponding dimensions $\dim(\ker(T^j))$ for $j = 1, \ldots, 4$. These numbers were obtained by considering the action of $x$ on the standard basis. For instance, $\frac{k[x]}{(x^3)} \cong k \oplus kx \oplus kx^2$ is a $k$-vector space with basis $1,x,x^2$. Multiplying by $x$ annihilates $x^2$, but not $1$ or $x$ (because we are modding out by $x^3$). Similarly, multiplying by $x^2$ annihilates $x$ and $x^2$ but not $1$.
EDIT: Alternatively, consider the possible Jordan canonical forms for $T$. Since $T$ is nilpotent, then its only eigenvalue is $0$, so its Jordan canonical form is
$$
J =
\begin{pmatrix}
0 & * & 0 & 0\\
0 & 0 & * & 0\\
0 & 0 & 0 & *\\
0 & 0 & 0 & 0
\end{pmatrix}
$$
where the $*$ are either $0$ or $1$. This yields $8$ possibilities for $J$ and with a bit of calculation, one can compute the dimensions of the kernels of $J, J^2, J^3, J^4$ for each $J$. Actually, it's not so hard to find which list is impossible. If $\dim(\ker(T)) = 1$, then $T$ has rank $3$, so we must have that its Jordan canonical form is
$$
J =
\begin{pmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{pmatrix} \, .
$$
As you noted in the comments,
$$
J^2 =
\begin{pmatrix}
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix} \, .
$$
and which has a $2$-dimensional kernel. This rules out the first choice.
|
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|
Prove by induction $\sum \frac {1}{2^n} < 1$ Prove by induction $\sum \frac {1}{2^n} < 1$
Well supposing the base case has been shown to be true, I start with the induction step:
Suppose true for n = k:
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} < 1$$
Want to show this is true for:
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} < 1$$
Now this is where i am getting stuck, should i be tryig to show that
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} < 1 +\frac{1}{2^{k+1}}$$ or should i be attempting to show that
$1 + \frac{1}{2^{k+1}} < 1$ which is utter nonsense. So i am stuck on what exactly to prove here.
Note i also thought of maybe using the geometric series of $\frac {1}{2} $ as some upper bound but then that would provide me a value that was greater than 1 so it didn't work out
|
For $n=k+1$:
$$\frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k}\right) < \frac{1}{2} + \frac{1}{2}\times 1 =1.$$
|
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|
What are the methods of dividing numbers to get weird values like $16\over 17$ without a calculator? I tried estimating it to somewhere near $16\over 20$, but it's a far stretch from getting the actual $16\over 17$. How can one do so? Conventionally, I think for numbers such as $50\over 17$, or for any large numbers, we have methods to do division and we can get $2+{16\over 17}$, but we're still left with $16\over 17$, which I have no idea how to find (or at least estimate till a good accuracy).
More interestingly, how would computers or calculators even find these values?
|
Please don't upvote me or downvote me as I have obviously copied and edited Mr. Hardy's answer. I agree with Mr Hardy that long division is the answer, but there is a small trick that makes subtraction a bit easier
Long division:
$$
\begin{array}{cccccccccc}
& & & 0 & . & 9 & 4 & 1 & 1 & 7 & 6 \\ \\
17 & ) & 1 & 5 & . & 9 & 9 & 9 & 9 & 9 & 9 \\
& & 1 & 5 & & 3 \\ \\
& & & & & 6 & 9 \\
& & & & & 6 & 8 \\ \\
& & & & & & 1 & 9 \\
& & & & & & 1 & 7 \\ \\
& & & & & & & 2 & 9 \\
& & & & & & & 1 & 7 \\ \\
& & & & & & & 1 & 2 & 9 \\
& & & & & & & 1 & 1 & 9 \\ \\
& & & & & & & & 1 & 0 & 9 \\
& & & & & & & & 1 & 0 & 2 \\ \\
& & & & & & & & \text{etc.}
\end{array}
$$
|
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|
Prove that $2^n>n^4$ for all $n\geq 17$ I'm always a bit fuzzy on how to solve induction problems involving inequalities. I've managed to get somewhere though, but it looks like I have to go down four levels of induction to prove.
This is what I have so far:
Base ($n=17$):
$$2^{17}>17^4\Rightarrow131072>83521$$
Step:
Assumption: $2^k>k^4$
\begin{align}
2^{k+1}&>(k+1)^4\\
2\cdot 2^{k}&>(k+1)^4\\
2^k+2^k&>k^4+4k^3+6k^2+4k+1
\end{align}
From here, it looks like I need to claim that $2^k+2^k>k^4+2^k$, but in order to do this, I would then need to prove that $2^k>4k^3+6k^2+4k+1$, and then again probably one or two more times.
Is there a different approach I can take other than going down this long and tedious route?
|
all that is required for the induction step is $$ (n+1)^4 < 2 n^4 $$
for large enough $n,$ so
$$ \frac{(n+1)^4}{ 2 \, n^4} < 1. $$
This is the same as
$$ \left( \frac{n+1}{n} \right)^4 = \left( 1 + \frac{1}{n} \right)^4 < 2 $$
This does not work for $n \leq 5,$ indeed
$(6/5)^4 = 1296/ 625,$ but it does work for $n=6$ as $(7/6)^4 = 2401/1296$ and $2 \cdot 1296 = 2592.$
Meanwhile $$ = \left( 1 + \frac{1}{n} \right)^4 $$
is strictly decreasing in $n,$ so it is less than $2$ for all $n \geq 6.$ It may help to consider $n=10,$ as $(11/10)^4 = 14641/10000 = 1.4641 < 2.$
If $n \geq 17$ and $$ \frac{n^4}{2^n} < 1, $$ then also $n \geq 6,$ so when $n \geq 17,$
$$ \color{red}{\frac{(n+1)^4}{2^{n+1}} = \frac{n^4}{2^n} \cdot \frac{(n+1)^4}{2 n^4} < \frac{n^4}{2^n} < 1} $$
No real tricks here: if you have positive things $A,B$ that depend on some $n,$ and you want to show $A < B,$ you can try for $A-B < 0$ or for $\frac{A}{B} < 1.$ Either way might give you something simpler than separately considering $A,B.$ The induction step then becomes showing that the left hand side decreases as $n$ is increased to $n+1.$
|
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|
How many digits will $5^{4^{3^2}}$ produce? According to Wolfram Alpha, $5^{4^{3^2}}$ evaluates to an integer with $183\,231$ digits. How does one find out how many digits such a calculation will produce?
|
A number $n \in \mathbf N^+$ has $\lfloor\log_{10}n\rfloor+1$ digits. For $n = 5^{4^{3^2}}$, we have
\begin{align*}
\log_{10} 5^{4^{3^2}} &= 4^9 \cdot \log_{10} 5\\
&= 262\,144 \cdot \log_{10} 5\\
&\approx 183\,230.8
\end{align*}
So $n$ has $183\,231$ digits.
|
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|
Integrating $x^5 \arcsin x$ Integrating $$\int_0^1 x^5 (\sin^{-1}x) \, dx$$
the answer is $\dfrac{11 \pi}{192}$.
I did substitute $x=\sin^2 u$ and obtained $$2\int_0^{\pi/2}( \sin^{12}x)( \cos x) \, dx$$ but got the incorrect answer. Why isn't this substitution working?
$$\int_0^{\pi/2}( \sin^m x)( \cos^n x)=\frac{(m-1)(m-3)\cdots(1\text{ or } 2)(n-1)(n-3) \cdots (1\text{ or } 2)k}{(m+n)(m+n-2)(m+n-4)\cdots(1\text{ or } 2)}$$ $k=\frac{\pi}{2}$ if $m$ and $n$ are even else $1$.
|
We may use the substitution $x=\sin(\theta)$ followed by integration by parts to get:
$$\begin{eqnarray*} I = \int_{0}^{\pi/2}\theta\cos(\theta)\sin(\theta)^5\,d\theta &=& \frac{\pi}{12}-\frac{1}{6}\int_{0}^{\pi/2}\sin(\theta)^6\,d\theta\\&=&\frac{\pi}{12}-\frac{1}{6}\int_{0}^{\pi/2}\cos(\theta)^6\,d\theta\tag{1}\end{eqnarray*}$$
By exploiting the formula $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and the binomial theorem we have:
$$ \cos^6(\theta) = \frac{5}{16}+\frac{15}{32}\cos(2\theta)+\frac{3}{16}\cos(4\theta)+\frac{1}{32}\cos(6\theta)\tag{2} $$
hence:
$$ I = \frac{\pi}{12}-\frac{1}{6}\cdot\frac{\pi}{2}\cdot \frac{5}{16}=\color{red}{\frac{11\pi}{192}}\tag{3}$$
follows.
|
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|
Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot x \cdot (1+\sqrt{\cos x})}$$
Using the rule $\frac{1-\cos x}{x} = 0$ for $x\to0$ is useless because we would end up with
$$\frac{0}{0\cdot x \cdot \sqrt{\cos x}} = \frac{0}{0}$$
But hey, perhaps we can rationalize again?
$$\frac{1-\cos x}{(x^2+x^2\sqrt{\cos x})}\cdot \frac{(x^2-x^2\sqrt{\cos x})}{(x^2-x^2\sqrt{\cos x})}$$
Resulting in
$$\frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 - x^4\cdot\cos x} = \frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 \cdot (1-\cos x)}$$
Cancelling
$$\frac{(x^2-x^2\sqrt{\cos x})}{x^4} = \frac{1-\sqrt{\cos x}}{x^2}$$
Well that was hilarious. I ended up at the beginning! Dammit.
My second attempt was to use the definition $\cos x = 1 - 2\sin \frac{x}{2}$:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2}$$
And then rationalize
$$\frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2} \cdot \frac{1+\sqrt{1 - 2\sin \frac{x}{2}}}{1+\sqrt{1 - 2\sin \frac{x}{2}}}$$
$$\frac{1-(1 - 2\sin \frac{x}{2})}{x^2+x^2\sqrt{1 - \sin \frac{x}{2}}} = \frac{- 2\sin \frac{x}{2}}{x^2\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
I want to make use of the fact that $\frac{\sin x}{x} = 1$ for $x\to0$, so I will multiply both the numerator and denominator with $\frac{1}{2}$:
$$\frac{-\sin \frac{x}{2}}{\frac{x}{2}\cdot x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Then
$$\frac{-1}{x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Well clearly that's not gonna work either. I will still get $0$ in the denominator.
The correct answer is $\frac{1}{4}$. I can kind of see why is the numerator $1$, but no idea where is that $4$ going to come out of.
I don't know how am I supposed to solve this without L'Hopital.
|
May be you can use the fact that $$\sqrt{1+x}=1+\frac{x}{2}+o(x)$$ and $$\cos(x)=1-\frac{x^2}{2}+o(x^2).$$ Therefore
$$...=\lim_{x\to 0}\frac{x^2}{4x^2}=\frac{1}{4}$$
|
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|
Definite integral of cosine times a quotient of two quartic polynomials If $f(x)=\frac{\left(x^2+x+1\right)^2+\left(x^4+x^2+1\right)}{\left(x^2+x+1\right)^2-\left(x^4+x^2+1\right)}$,then find $\int_{-\pi}^{\pi}\cos x\cdot f(x)dx$
Let $I=\int_{-\pi}^{\pi}\cos x\cdot f(x)dx=\int_{-\pi}^{\pi}\cos x\cdot\frac{\left(x^2+x+1\right)^2+\left(x^4+x^2+1\right)}{\left(x^2+x+1\right)^2-\left(x^4+x^2+1\right)}dx$
$I=\int_{-\pi}^{\pi}\cos x\cdot\frac{\left(x^2-x+1\right)^2+\left(x^4+x^2+1\right)}{\left(x^2-x+1\right)^2-\left(x^4+x^2+1\right)}dx$
But there seems no way for simplification,Please help me in solving this integral.
|
The function $f$ simplifies to $f(x)=x+1/x$, so
$$
\int_{-\pi}^\pi \cos(x)f(x)\,dx
$$
diverges.
Addition
This addition was asked for in a comment. The integrals
$$
\int_{-\pi}^0 (x+1/x)\cos x\,dx\quad\text{and}\quad \int_0^{\pi}(x+1/x)\cos x\,dx
$$
both diverges. To see this, we note that $x+1/x\approx 1/x$ and $\cos x\approx 1$ as $x\approx 0$ (you should make this more precise using a comparison theorem as it is stated in your book), so your function behaves like $1/x$ when $x\approx 0$. Since, for any $a>0$, the integrals
$$
\int_{-a}^0\frac{1}{x}\,dx\quad\text{and}\quad\int_0^a\frac{1}{x}\,dx
$$
both diverges, the integrals under inspection diverge.
|
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|
How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$
Thanks in advance
Update: I actually solved it, and this is the way that I wanted to:
$\lim \:_{x\to \:\infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)\:=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x+\sqrt{1+\sqrt{x}}}+\sqrt{x}}\right)=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x^2\left(1+\frac{\sqrt{1+\sqrt{x}}}{x}\right)+\sqrt{x}}}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x\left(\frac{1}{x}+\frac{1}{\sqrt{x}}\right)}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x}\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{x}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1}\right)=\frac{0}{2}=0$
Thanks for your two answers guys. I really appreciate it ! This community is awesome !
|
$$\begin{align}\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x&=\frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x}\\
&\le \frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{1+\sqrt x}\\
&=\frac{\sqrt{1+\sqrt x}}{1+\sqrt x}\\
&=\frac1{\sqrt{1+\sqrt x}}\\
&\to 0\end{align}$$
|
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Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.
Attempt:
I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.
|
Let $$\displaystyle \tan^{-1}\left(\frac{1}{2}\right)=\alpha\Rightarrow \tan \alpha = \frac{1}{2}$$ and $$\displaystyle \tan^{-1}\left(\frac{1}{3}\right)=\beta\Rightarrow \tan \beta = \frac{1}{3}$$
So $$\displaystyle \tan(\alpha+\beta) = \frac{\tan \alpha+\tan \beta}{1-\tan \alpha\cdot \tan \beta} = \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}=1$$
So we get $$\displaystyle \tan(\alpha+\beta)=1=\tan \frac{\pi}{4}\Rightarrow \alpha+\beta = n\pi+\frac{\pi}{4}\;,n\in \mathbb{N}$$
But above $\displaystyle 0<\tan^{-1}\left(\frac{1}{2}\right)<\frac{\pi}{6}$ and $\displaystyle 0<\tan^{-1}\left(\frac{1}{3}\right)<\frac{\pi}{6}$
So we get $\displaystyle \alpha+\beta = \frac{\pi}{4}$
|
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|
How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \cos^2x)(1) \longrightarrow\,
= \sin^2x - \cos^2x
\end{align}
Is that true?
|
Note that $a^2 + b^2 = (a+b)^2 - 2ab$
$$(\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x =(\sin^2 x + \cos^2 x)^2 - 2(\sin x\cos x)^2 = \\ 1 -\frac{ \sin^2 2x}{2}$$
Note the following results:
$$ \sin^2 x + \cos^2 x = 1$$
$$ \sin x \cos x = \frac{\sin 2x}{2}$$
|
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|
Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation This is how I tried so far...
$r^2= 9 \cos 2 ( \theta)$
$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and
$r^2= x^2 + y^2$
so, it will become
$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$
|
Hint Polar and Cartesian coordinates are related by the equations
$$x = r \cos \theta, \quad y = r \sin \theta,$$ so a Pythagorean trigonometric identity gives $$x^2 + y^2 = (r \cos \theta)^2 + (r \sin \theta)^2 = r^2 (\cos^2 \theta + \sin^2 \theta) = r^2 .$$ Now, simply rearrange and substitute.
|
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|
What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$ What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$
The lines $x+y=2$ and $x-y=2$ are perpendicular to each other and the circle is touching both the lines,these lines are tangents to the circle.Let points of tangency be $P$ and $Q$,let the center of the circle be $O$ and let the point where the lines $x\pm y=2$ meet be $R$.$OP$ is perpendicular to $PR$ and $OQ$ is perpendicular to $QR$,therefore $OPRQ$ is a square and the point $(-1,1)$ does not lie on any of the lines $x\pm y=2$.
But now i am stuck,how to solve further.Please help me.
|
Let $(h,k)$ be the center of the required circle.
The perpendicular distance from $(h,k)$ to $x+y-2=0$
=The perpendicular distance from $(h,k)$ to $x-y-2=0$
=The distance between $(h,k)$ to $(-1,1)$.
The first two equality gives $|\frac{h+k-2}{\sqrt 2}|=|\frac{h-k-2}{\sqrt 2}|$, which will imply either $k=0$ or $h=2$
So, you have two cases,
Case,I $k=0$
From the seond and third equality gives,
$\frac{|h-2|}{\sqrt 2}=\sqrt{(h+1)^2+(1)^2}$
squaring on both sides,
or $ \frac{(h-2)^2}{2}=(h+1)^2+1$
or $h^2-4h+4=2h^2+4h+2+2$
or $h^2+8h=0$
0r $h=0,-8$
Radius$=\sqrt{(-8+1)^2+(1)^2}=5\sqrt 2 or \sqrt{(0+1)^2+(1)^2}=\sqrt 2$
Case-II, $h=2$
$\frac{|2+k-2|}{\sqrt 2}=\sqrt{(2+1)^2+(k-1)^2}$
squaring on both sides,
or, $k^2= 18+2k^2-4k+2$
or, $k^2-4k+30$
This doesn't have real solutions. So case-II is not valid.
|
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|
An Inequality for sides and diagonal of convex quadrilateral from AMM Let $\square ABCD$ be a convex quadrilateral. If the diagonals $AC$ and $BD$ have mid-points $E$ and $F$ respectively, show that:
$$\overline{AB} + \overline{BC} +\overline{CD} + \overline{DA} \ge \overline{AC}+\overline{BD}+2\overline{EF}$$
where, $\overline{XY}$ denotes the length of the line segment $XY$.
The problem was 11841 from May 2015 issue of AMM magazine. While writing complex numbers/vectors for vertices reduces the problem to the well known Hlawka's Inequality for Inner-product spaces, I am interested in purely geometrical solutions.
(It's way past the last date of submission, so I believe it's safe to ask for alternative solutions here.)
Edit: We have the following reformulation, (not that it makes the job any easier though!)
If $H,I,J$ and $K$ are the midpoints of $AD,DC,CB$ and $BA$ respectively, then $\square HIJK$ is a parallelogram and it's easy to see that the diagonals $\overline{HJ}$ and $\overline{IK}$ intersect at $G$, which is the midpoint of $\overline{EF}$ as well.
Then we have the equivalent reformulation of the question:
In a parallelogram $\square HIJK$, whith diagonals intersecting at $G$ and $F$ be an interior point, we need to show:
$$\overline{FH}+\overline{FI}+\overline{FJ}+\overline{FK} > \overline{IJ}+\overline{KJ}+2\overline{FG}$$
|
The simplest special case is when $\overline{ABCD}$ is a square with side length $a$. In this case, the diagonals are of equal length and given by $a\sqrt{2}$.
$4a \ge 2a\sqrt{2},\quad 4 \ge 2\sqrt{2},\quad$the conjecture holds.
The next simplest case is when $\overline{ABCD}$ is a rectangle of side lengths $a$ and $b$. In this case the diagonals are still equal and given by $\sqrt{a^2 +b^2}$, so
$2(a+b)\ge 2\sqrt{a^2 + b^2}, \quad a+b\ge \sqrt{a^2 + b^2}, \quad a^2 + 2ab + b^2 \ge a^2 + b^2, \quad 2ab \ge 0, \quad$ and the conjecture holds.
The next case is when $\overline{ABCD}$ is a parallelogram, with sides $a, b$ and interior angles $h, k$. The diagonals can be given by the law of cosines, $d_1 = \sqrt{a^2 + b^2 + 2abcos(h)}$, and $d_2 = \sqrt{a^2 + b^2 + 2abcos(k)}$. These can both be written in terms of one interior angle, since ${h, k} \in [0, \pi]$ and $h=\pi-k$, and $cos(\pi-h) = -cos(h)$. Ie: $d_{1,2} = \sqrt{a^2 + b^2 \pm 2abcos(h)}$.
$2(a+b) \ge \sqrt{a^2 + b^2 + 2abcos(h)} + \sqrt{a^2 +b^2 -2abcos(h)}$
$4(a^2 + 2ab + b^2) \ge a^2 + b^2 + 2abcos(h) + 2\sqrt{(a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))} + a^2 +b^2 -2abcos(h)$
$4a^2 + 8ab + 4b^2 \ge 2a^2 + 2b^2 + 2\sqrt{(a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))}$
$a^2 + 4ab + b^2 \ge \sqrt{(a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))}$
$(a^2 + 4ab + b^2)^2 \ge (a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))$
(in the interest of space, I’m gonna skip the foiling and ask you to trust me that this simplifies down to the following:)
$2a^3b+4a^2b^2+2ab^3\ge-a^2b^2cos(h)$
That $cos(h)$ term is bounded in $[-1, 1]$ so worst case scenario, the right hand side of that equality is negative and the inequality is obviously true. Best case scenario, $2a^3b+2a^2b^2+2ab^3\ge 0$ and the conjecture holds.
In all of the above cases, $\overline{EF} = 0$, because for all parallelograms, the diagonals intersect at their midpoints.
The next case is a right trapezoid, with sides $b_1, b_2, h,$ and the last side can be given by $\sqrt{h^2 + (b_2-b_1)^2}$. Assume $b_1 \le b_2$. This is the first case in which the line $\overline{EF}$ will come into play. In a right trapezoid, the line $\overline{EF}$ lies on the perpendicular bisector of $h$, and its length is given by $\frac{b_2-b_1}{2}$. And the lengths of the diagonals $d_1, d_2$ can be given by $d_1 = \sqrt{h^2 +b_1^2},$ and $d_2=\sqrt{h^2+b_2^2}$. Finally we can write out our inequality as:
$h + b_1 + b_2 + \sqrt{h^2 + (b_2-b_1)^2} \ge \sqrt{h^2 + b_1^2} + \sqrt{h^2 + b_2^2} +2\frac{(b_2-b_1)}{2}$
But alas, in attempting to solve this inequality I realized I would have to square a four term polynomial. I have already shown geometrically that the conjecture holds for all parallelograms. The next cases to consider are this case of the right trapezoid, then the general trapezoid, and finally the general quadrilateral. If you still want a purely geometric solution, perhaps I’ll come back and give it another shot.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1466077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
If line through point $P(a,2)$ meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at A and D and meets the coordinate axis at B and C If line through point $P(a,2)$ meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $A$ and $D$ and meets the coordinate axes at $B$ and $C$ so that $PA$, $PB$, $PC$, $PD$ are in geometric progression, then the possible values of $a$ can be
$(A)5\hspace{1cm}(B)8\hspace{1cm}(C)10\hspace{1cm}(D)-7$
I could not solve this question, I inferred from question that $PA\cdot PD=PB\cdot PC$ and $PA\cdot PD=PT^2$, where $T$ is the point of tangency.
But I could not solve further. This is a multiple correct choice type question. Please help me.
Thanks.
|
WLOG we can take $a>0$, $A$ nearer to $P$ than $D$ and notice that the only case we must consider is when $B$ is the intersection with $x$-axis and $C$ is the intersection with $y$-axis, for otherwise those four segments cannot form a geometric progression. If $b$ is the $x$ coordinate of $B$, the equation of line $PB$ is $y=2(x-a)/(a-b)+2$ so that the $y$ coordinate of $C$ is $y_C=-2b/(a-b)$.
Combining this equation with that of the ellipse, we can readily find the $y$ coordinate of $A$ and $D$:
$$
y_A= \frac{2 \left(3 \sqrt{a^2-2 a b+9}-a b+b^2\right)}{a^2-2 a b+b^2+9},
\quad
y_D= \frac{2 \left(-3 \sqrt{a^2-2 a b+9}-a b+b^2\right)}{a^2-2 a b+b^2+9}.
$$
We know that $PA:PB=PB:PC=PC:PD$ and this relation also holds for the $y$ components of the segments, that is:
$$
(y_P - y_A):(y_P - y_B) = (y_P - y_B):(y_P - y_C) = (y_P - y_C):(y_P - y_D).
$$
Inserting here the expressions given above for $y_C$, $y_A$, $y_D$, as well as $y_P=2$ and $y_B=0$, we can solve for $a$ and $b$. The only acceptable positive solution is:
$$
a=3 \sqrt{2+\sqrt{13}}\approx 7.10281,
$$
but of course the opposite value, by symmetry, is also a valid solution. As you can see, this is not far from your $(D)$ choice but it is not the same. So the exercise is wrong.
|
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"timestamp": "2023-03-29T00:00:00",
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|
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$. I have a homework as follow:
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$.
Please help to prove it.
EDIT: MY ATTEMPT
Suppose that $5\mid a^2-2b^2$, then $a^2-2b^2=5n$,where $n\in Z$,
then $a^2-2b^2=(a+\sqrt2b)(a-\sqrt2b)=5n$,
Since 5 is a prime number, we get that $5\mid (a+\sqrt2b)$or $5\mid (a-\sqrt2b)$,
If $5\mid (a+\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
If $5\mid (a-\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
Thus $5\nmid a^2-2b^2$.
|
This is probably what Bernard intended.
The non-zero quadratic residues mod $5$ are $1$ and $4$. Plugging these into $a^2-2b^2$ gives
$$
\begin{array}{c|ccc}
&1&4&a^2\\\hline
1&4&2\\
4&3&1\\
b^2
\end{array}
$$
Since none of the entries are $0$, we get that $a^2-2b^2\not\equiv0\pmod5$ if neither $a$ nor $b$ are $0\pmod5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1471722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be equal to
$(A)15^\circ\hspace{1cm}(B)75^\circ\hspace{1cm}(C)115^\circ\hspace{1cm}(D)135^\circ$
Let AD be the median.Angle $ADC=45^\circ,ACD=30^\circ,CAD=105^\circ$.and $BD=CD$Then i am stuck,how to find angle $B?$Please help me.Thanks.
|
If you have access to Geogebra you can easily find the answer yourself:
Now we you only need to prove it. :)
Edit: It seems that the inclination of the median is not clear, but making the two choices on the same figure brings an immediate solution:
In triangle $AED$ we have $\frac{EC}{DC} = \frac{\sin 45^\circ}{\sin 105^\circ}=\sqrt{3}-1$. In triangle $ADC$ we have $\frac{DC}{AC} = \frac{\sin 15^\circ}{\sin 135^\circ}=\frac{\sqrt{3}-1}{2}$
This means that $\frac{2DC}{AC} = \frac{EC}{DC}$ so $DC \cdot BC = AC \cdot EC$. Thus $A,B,D,E$ are on the same circle and $\angle ABC = 105^\circ$ and $\angle EBC = \angle EAD = 15^\circ$.
So the right position of $A$ is in fact the position of $E$ in the second figure, and $\angle B = 15^\circ$.
I used freely the well known facts: $\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\sin 105^\circ = \frac{\sqrt{6}+\sqrt{2}}{4}$. These can be deduced immediately using formulas for $\sin(a\pm b)$.
|
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"url": "https://math.stackexchange.com/questions/1471974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Approximation of non-analytic function I have a function which is of the form
\begin{equation}
f(x) = \frac{1 - x^{1/2} + x - x^{3/2} + \ldots}{1+x^{1/2} - x + x^{3/2} - \ldots}.
\end{equation}
Intuitively, I would assume that for small $x$, it holds
\begin{equation}
f(x) \approx \frac{1-x^{1/2}}{1+x^{1/2}}
\end{equation}
and then, furthermore,
\begin{equation}
f(x) \approx 1 - a x^{1/2} + \ldots
\end{equation}
where $a$ is some factor. My question is: How can I determine $a$ and the range of $x$ for which this approximation is valid? Obviously, I cannot use a Taylor approximation since $f$ is not analytic and the derivative diverges in the origin.
Let me point out that I am not so much interested in the specific example above, which I have just invented. Much rather, I would like to know what is the general theory and methods behind this type of fractional functions.
|
Note, that the numerator, is
$$ \sum_{k=0}^\infty (-x^{1/2})^k = \frac{1}{1 + x^{1/2}}, \quad |x| < 1 $$
and the denominator equals
$$ 2 -\sum_{k=0}^\infty (-x^{1/2})^k = 2 - \frac 1{1 + x^{1/2}} $$
Hence
\begin{align*}
f(x) &= \frac{\frac 1{1 + x^{1/2}}}{\frac{2(1 + x^{1/2}) - 1}{1 + x^{1/2}}}\\
&= \frac 1{2(1 + x^{1/2}) - 1}\\
&= \frac 1{1 + 2x^{1/2}}\\
&= \sum_{k=0}^\infty (-2x^{1/2})^k\\
&= 1 - 2x^{1/2} + 4x \mp \cdots
\end{align*}
That is, $a = 2$ seems like a good choice.
Addendum: If we write $f$ as $g(\sqrt x)$ with
$$ g(y) = \frac{1 - y + y^2 \mp}{1 + y - y^2 \pm } $$
then we can Taylor expand $g$ nicely, giving
$$ g(y) = 1 - 2y + 4y^2 \mp $$
and hence
$$ f(x) = 1 - 2x^{1/2} + 4x \mp $$
and the range of convergence can be obtained by that of $g$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Linear autonomous systems in the plane: When do phase curves rotate clockwise? For a linear autonomous system in the plane
$$ \mathbf{\dot{x}} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mathbf{x} \qquad (a,b,c,d \in \mathbb{R})$$
with determinant $D = ad - bc$ and trace $T = a + d$ we have the characteristic polynomial
$$ \chi(\lambda) = \lambda^2 - T\lambda + D$$
and the eigenvalues
$$ \lambda_{1, 2} = \frac{T \pm \sqrt{\Delta}}{2}, \qquad \Delta = T^2 - 4D $$
We know that the phase curves will rotate around the origin iff $\Delta < 0$.
Question: How can one determine using $a, b, c, d$, whether the phase curves turn clockwise or counter-clockwise?
Counterexample: The System
$$ \mathbf{\dot{x}} = \begin{pmatrix} 1 & s\\ -s & 1 \end{pmatrix}\mathbf{x} \qquad (s \in \mathbb{R} \setminus \{ 0 \})$$
will rotate clockwise for $s > 0$ and counter-clockwise for $s<0$, but in both cases we have
$$ D = 1 + s^2 > 0, \qquad T = 2, \qquad \Delta = -4s^2 < 0 $$
so these quantities will not suffice to determine the orientation.
|
Inspired by amd's help I did the math:
The original system
$$ \mathbf{\dot{x}} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mathbf{x}, \qquad D = ad - bc, \qquad T = a + d $$
has the eigenvalues
$$ \lambda_{\pm} = \frac{T \pm i\sqrt{|\Delta|}}{2}, \qquad |\Delta| = 4D - T^2 > 0 $$
with corresponding eigenvectors
$$ v_{\pm} = \begin{pmatrix} -a+d \mp i\sqrt{|\Delta|}\\ -2c \end{pmatrix} $$
We want to relate it to the system
$$ \mathbf{\dot{w}} = \begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix}\mathbf{w}, \qquad \alpha = \Re\left(\lambda_{+}\right) = \frac{T}{2}, \qquad \beta = \Im\left(\lambda_{+}\right) = \frac{\sqrt{|\Delta|}}{2} > 0 $$
which is spinning counter-clockwise because of $\beta > 0$ (as pointed out by amd) and has the same eigenvalues, but now with the corresponding eigenvectors
$$ w_{\pm} = \begin{pmatrix} \pm i\\ 1 \end{pmatrix} $$
We are looking for a Matrix $B$ such that
$$ \begin{pmatrix} a & b\\ c & d \end{pmatrix} = B\begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix}B^{-1}, \qquad \mathbf{x} = B\mathbf{w} $$
Such a matrix is uniquely defined by the condition
$$ (v_{+} \mid v_{-} ) = B(w_{+} \mid w_{-} ) = B\begin{pmatrix} i & -i\\ 1 & 1\end{pmatrix} $$
and hence computes to
$$ B = \begin{pmatrix} -a+d - i\sqrt{|\Delta|} & -a+d + i\sqrt{|\Delta|}\\ -2c & -2c \end{pmatrix} \frac{1}{2i} \begin{pmatrix} 1 & i\\ -1 & i \end{pmatrix} = \begin{pmatrix} -\sqrt{|\Delta|} & d - a\\ 0 & -2c\end{pmatrix} $$
with determinant $\det B = 2c\sqrt{|\Delta|}$ yielding the result
*
*$c > 0$: The matrix $B$ preserves orientation as $\det B > 0$, hence $\mathbf{x} = B\mathbf{w}$ rotates counter-clockwise just like $\mathbb{w}$.
*$c < 0$: The matrix $B$ inverts orientation as $\det B < 0$, hence $\mathbf{x} = B\mathbf{w}$ rotates clockwise.
*$c = 0$: Leads to real eigenvalues $a$ and $d$, so the system does not rotate.
Should you be surprised that the orientation only depends on $c$, then please note: If
$$ B = \begin{pmatrix} e & f\\ g & h \end{pmatrix} \in GL(2,\mathbb{R}) $$
is an invertible real matrix, then we have $g \neq 0$ or $h \neq 0$ and
$$ B\begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix}B^{-1} = \begin{pmatrix} ? & -\beta\frac{e^2 + f^2}{\det B}\\ \beta\frac{g^2 + h^2}{\det B} & ? \end{pmatrix} $$
and since $g^2 + h^2 >0$ the sign of the "lower left coefficient" only changes if $\det B < 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series?
I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:
\begin{equation}
y(x) = \tan x
\end{equation}
\begin{equation}
y^{\prime} = \frac{1}{\cos^{2} x}
\end{equation}
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x}
\end{equation}
This can be rewritten as:
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime}
\end{equation}
Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$:
\begin{equation}
y^{\prime} = y^{2}
\end{equation}
And therefore
\begin{equation}
\frac{1}{\cos^{2} x} = \tan^{2} x
\end{equation}
Now, evalueting this equation at $x = \pi / 4$
\begin{equation}
\frac{1}{(\sqrt{2}/2)^{2}} = 1^{2}
\end{equation}
\begin{equation}
2 = 1
\end{equation}
|
Here is a simple one:
$$
x=y\\
x^2=xy\\
2x^2=x^2+xy\\
2x^2-2xy=x^2-xy\\
2(x^2-xy)=1(x^2-xy)\\
2=1
$$
The error is quite obviously division by zero (from the 5th to 6th step).
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 2
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|
If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$? If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$ ?
I can find by trial and error, that $x=\sqrt 2$. But, what is the general process to answer questions like this?
|
Taking the logarithm base 2 of each side, we have
$\log_2 (x^{x^4}) = x^4\log_2 (x)= \log_2 4 = 2$
Multiplying each side by four we have
$4x^4\log_2 (x) = x^4\log_2 (x^4) = 8$
Relabeling $x^4$ as $u$, we have
$u\log_2(u) = 8$
Utilizing the Lambert W Function, we have that
$u = e^{W(8\ln 2)}=e^{W(4\ln 4)}=4$
Replacing back, $x^4=4$ and the result follows.
For the conclusion to the problem, we found earlier that $x^4=4$ and so either $x^2=2$ or $x^2=-2$. Assuming that we require $x$ to be real, it must be the first.
$\begin{array}{rl}
x^{x^2} + x^{x^8} &=x^2 + x^{(x^4)^2}\\
&=2+x^{16}\\
&=2+(x^4)^4\\
&=2+4^4\\
&=2+256\\
&=258\end{array}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1480675",
"timestamp": "2023-03-29T00:00:00",
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|
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
|
Use the linearisation formulae (inverse of the duplication formulae):
\begin{align*}
&=3\frac{1-\cos2\theta}2+\frac52\sin2\theta-2\frac{1+\cos2\theta}2\\
&=\frac12-\frac52(\cos2\theta-\sin2\theta)=\frac12-\frac{5\sqrt 2}2\cos\Bigl(2\theta-\frac\pi4\Bigr),
\end{align*}
whence
\begin{align*}
3\sin^2\theta+&5\sin\theta\cos\theta-2\cos^2\theta=0\iff\cos\Bigl(2\theta-\frac\pi4\Bigr)=\frac1{5\sqrt2}\\
&\iff 2\theta-\frac\pi4\equiv\pm\arccos\frac1{5\sqrt2}\mod2\pi\\
&\iff \theta\equiv\frac\pi8\pm\frac12\arccos\frac1{5\sqrt2}\mod\pi.
\end{align*}
A shorter method:
First observe we cannot have $\cos\theta=0$, for it would imply $\sin\theta=0$ and we cannot have both. So we can divide the equation by $ \cos\theta$, obtaining:
$$3\tan^2\theta+5\tan\theta-2=0$$
Now the quadratic equation $t^2+5t-2=0$ has discriminant equal to $49$ and roots $\;\Bigl\{-2,\dfrac13\Bigr\}$. Thus we have to solve
$$\begin{cases}\tan\theta=-2\\\tan\theta=\dfrac13\end{cases}\iff\begin{cases}\theta\equiv-\arctan2\mod\pi\\\theta\equiv\arctan\dfrac13\mod\pi=\dfrac\pi2-\arctan 3\mod\pi\end{cases}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1481232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integral of $x\arctan x/(1+x^2)^{3/2}$ Find
$$\int \frac{x\arctan x}{(1+x^2)^{3/2}}\operatorname{dx} $$
I tried many ways to do it but I can't resolve it. What are the steps to solve this exercise?
|
One way is to just integrate by parts,
$$
\begin{split}
\int \frac{x}{(1+x^2)^{3/2}}\arctan x&=-\frac{1}{\sqrt{1+x^2}}\arctan x+\int\frac{1}{(1+x^2)^{3/2}}\,dx\\
&=-\frac{1}{\sqrt{1+x^2}}\arctan x+\frac{x}{\sqrt{1+x^2}}.
\end{split}
$$
Here, the last step might seem mystery. I know it by heart, but one way to conclude it is:
$$
\begin{split}
\frac{1}{(1+x^2)^{3/2}}
&=\frac{(1+x^2)-x^2}{(1+x^2)^{3/2}}\\
&=\frac{1}{\sqrt{1+x^2}}-\frac{x^2}{(1+x^2)^{3/2}}
\\
&=D(x)\frac{1}{\sqrt{1+x^2}}+xD\frac{1}{\sqrt{1+x^2}}\\
&=D\frac{x}{\sqrt{1+x^2}}.
\end{split}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the partial sums of $4+44+444+\cdots$
Find the sum to $n$ terms of $4+44+444+\cdots $
My attempts:
*
*Used successive difference method
*Used $4+(4+40)+(4+40+400)\dots$ method
Failed to get a formula for partial sum in both ways. What am I missing?
|
We know that
$$ \sum_{k=0}^n {\frac{1}{x^k}} =
\frac{\frac{1}{x^{n+1}}-1}{\frac 1x - 1} =
\frac{x^{n+1}-1}{x^n(x-1)}$$.
Taking the derivative of both sides results in
$$\displaystyle \sum_{k=1}^n \frac{k}{x^{k+1}} =
\frac{x^{n+1}-(n+1)x^n+n}{x^{n+1}(x - 1)^2}$$.
Multiplying both sides by $x^{n+1}$, we get
$$\sum_{k=1}^n kx^{n-k} = \dfrac{x^{n+1} - (n+1)x + n}{(x-1)^2}$$
Summing the first $n$ terms and letting $x = 10$ we get
\begin{align}
4 + 44 + 444 + \dots + 444\ldots4
&= 4(1 + 11 + 111 + \dots + 111\ldots1)\\
&= 4[1+ (1+x) + (1+x+x^2) + \cdots + (1+x+x^2 + \cdots + x^{n-1})]\\
&= 4[nx^0 + (n-1)x^1 + (n-2)x^2 + \cdots + 1x^{n-1}]\\
&= 4[1x^{n-1}+ 2x^{n-2} + 3x^{n-3} + \cdots + nx^0]\\
&= 4\sum_{k=1}^{n} kx^{n-k}\\
&= 4\dfrac{x^{n+1} - (n+1)x + n}{(x-1)^2}\\
&= \frac{4}{81}(10^{n+1} - 10(n+1) + n)
\end{align}
Checking the first few terms.
$4 = \frac{4}{81}(100 - 20 + 1) = 4$
$4 + 44 = \frac{4}{81}(1000 - 30 + 2) = 48$
$4 + 44 + 444 = \frac{4}{81}(10000 - 40 + 3) = 492$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding all positive integer solutions for $x+y=xyz-1$ How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2, z=1$
Any hints would be appreciated.
|
We have
$$z=\frac{x+y+1}{xy}=\frac 1y+\frac 1x+\frac{1}{xy}\le 1+1+1$$$$\Rightarrow z=1,2,3$$
Also,
$$zxy-x-y=1\iff z^2xy-zx-zy+1=z+1\iff (zx-1)(zy-1)=z+1$$
So, for $z=1$, we have $(x-1)(y-1)=2$.
For $z=2$, we have $(2x-1)(2y-1)=3$.
For $z=3$, we have $(3x-1)(3y-1)=4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1484330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
When is $2^n +3^n + 6^n$ a perfect square? Find all $n$ for which $2^n+3^n+6^n$ is a perfect square.
I do not have a specific idea how to solve this one
|
If $2^n+3^n+6^n=x^2$ for some positive integer $x$, then we know from a comment by user236182 that $n=2k$ for some positive integer $k$. Hence,
$$\left(x-2^k\right)\left(x+2^k\right)=3^{2k}\left(1+2^{2k}\right)\,.$$
Since $x$ is obviously odd, $\gcd\left(x-2^k,x+2^k\right)=1$, so $3^{2k}$ divides either $x-2^k$ or $x+2^k$. However, $3^{2k}>1+2^{2k}$ implies that $x+2^k=m\cdot 3^{2k}$ for some positive integer $m$. If $m\geq 2$, then $$2x=\left(x+2^k\right)+\left(x-2^k\right)>x+2^k=m\cdot 3^{2k}\geq 2\cdot 3^{2k}$$ leads to $x>3^{2k}$. Therefore, $$9^{2k}<x^2=2^{2k}+3^{2k}+6^{2k}<9^{2k}\,,$$
which is absurd. Hence, $m=1$. That is, $x+2^k=3^{2k}$ and $x-2^k=1+2^{2k}$. Therefore, $$3^{2k}=x+2^k=\left(1+2^k+2^{2k}\right)+2^{k}=1+2^{k+1}+2^{2k}\,.$$ If $k>1$, then $1+2^{k+1}+2^{2k}<3^{2k}$, which is a contradiction. Hence, $k=1$, or $n=2$, is the only possible solution. As $n=2$ yields $x=7$, we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I evaluate this without using taylor expansion :$\lim_{x \to \infty}x^2\log(\frac {x+1}{x})-x\ $?
How do I evaluate this without using Taylor expansion?
$$\lim_{x \to \infty}x^2\log\left(\frac {x+1}{x}\right)-x$$
Note: I used Taylor expansion at $z=0$ and I have got $\frac{-1}{2}$
Thank you for any help
|
Another option is to use the following (very convenient) inequality:
$$
\frac{2x}{2+x} \leq \ln(1+x) \leq \frac{x}{\sqrt{1+x}}
$$
which holds for $x \geq 0$. In your case, you have $\ln(1+\frac{1}{x})$, leading to
$$
x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x \leq x^2\ln\!\left(1+\frac{1}{x}\right) -x \leq x^2\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x}}} - x.
$$
*
*For the LHS:
$$
x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x = x\left(\frac{2}{2+\frac{1}{x}} - 1\right) = x\left(\frac{-\frac{1}{x}}{2+\frac{1}{x}}\right) = \frac{-1}{2+\frac{1}{x}} \xrightarrow[x\to\infty]{} -\frac{1}{2}.
$$
*And the RHS:
$$\begin{align}
x^2\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x}}} - x
&= x\left(\frac{1}{\sqrt{1+\frac{1}{x}}} - 1\right)
= x\left(\frac{1-\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}}}\right)\\
&= x\left(\frac{-\frac{1}{x}}{\sqrt{1+\frac{1}{x}}\left(1+\sqrt{1+\frac{1}{x}}\right)}\right)\\
&= \frac{-1}{\sqrt{1+\frac{1}{x}}\left(1+\sqrt{1+\frac{1}{x}}\right)}
\xrightarrow[x\to\infty]{} -\frac{1}{2}.
\end{align}$$
It only remains to invoke the squeeze theorem.
(I also strongly suggest bookmarking the "useful inequalities sheet" linked at the beginning.)
|
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|
Finding the roots of $x^3+ax+a=0$ If the roots $x_1,x_2,x_3$ of the equation $x^3+ax+a=0$ satisfy $$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8$$ then the roots of the equation are?
I wrote the Vieta formulas down but then I'm stuck.Please give your suggestions.
|
$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8 \iff \frac{x_1^3}{x_1x_2}+\frac{x_2^3}{x_2x_3}+\frac{x_3^3}{x_1x_2}=-8 \iff \frac{-ax_1-a}{x_1x_2}+\frac{-ax_2-a}{x_2x_3}+\frac{-ax_3-a}{x_1x_3}=-8 \iff \frac{a}{x_1x_2}+\frac{a}{x_2x_3} + \frac{a}{x_1x_3} + \frac{a}{x_1}+ \frac{a}{x_2}+ \frac{a}{x_3}=8 (1)$
Because $x_1x_2x_3=-a$ (Vieta), from (1) we have:
$-(x_1 + x_2 +x_3) - (x_1x_2 + x_1x_3+x_2x_3) = 8$
Should be easy from here.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Limits without L'Hopitals Rule ( as I calculate it?) Prove that:
$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$
|
Let $\displaystyle z = \frac{1}{y}\;,$ So when $z\rightarrow \infty\;,$ Then $y\rightarrow 0$
So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1+\sqrt{1+2y}-2\sqrt{1+y}}{y^2} = \lim_{y\rightarrow 0}\frac{1+(1+2y)^{\frac{1}{2}}-2(1+y)^{\frac{1}{2}}}{y^2}$$
Now Using $$\displaystyle \bullet (1+t)^n = 1+nt+\frac{n(n-1)t^2}{2}+\frac{n(n-1)(n-2)}{6}t^3+.....$$
So we get $$\displaystyle \lim_{y\rightarrow 0}\frac{1+(1+ y-\frac{y^2}{2}-\frac{y^3}{6}+...)-2(1+\frac{y}{2}-\frac{y^2}{8}-\frac{z^3}{16}....)}{y^2}$$
So we get limit $$\displaystyle = -\frac{1}{4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all factored pairs of (a,b) such that...
Determine all possible ordered pairs (a, b) such that
*
*$a − b = 1$
*$2a^2 + ab − 3b^2 = 22$
I've gone as far as factoring the left side of the second equation:
*
*$(a-b)(2a + 3b) = 22$
*$(1)(2a + 3b) = 22$
*$2a + 3b = 22$
I understand that a different (and arguably easier) way of solving this question is to directly substitute the first equation into the second, but I'd like to learn/use this approach.
|
$$
\begin{cases}
a − b = 1 \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2(1 + b)^2 + (1 + b)b − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2b^2+4b+2 + b+b^2 − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
4b+2 + b = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
5b+2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
5b = 20
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
b = \frac{20}{5}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
b = 4
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + 4 \\
b = 4
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 5 \\
b = 4
\end{cases}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1491415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
find the area of the region lying inside the circle $r=6$ and inside the cardioid $r=4-3\sin \theta$. Well, I drew a graph to visualise it and I found the interceptions $\theta=\arcsin \left(-\frac{2}{3}\right)$. From the graph, by symmetry, I found that the area of region from $\theta$ to $\pi/2$ and from $\pi/2$ to $\pi-\theta$.
So How would I apply the formula now?
|
You want to compute the area integral
$$\int_A {1} dA$$
where $A$ is the intersection of the interior of the circle and the cardioid i.e.
\begin{align*}
A &= \{(x, y) : r = \sqrt{x^2 + y^2} \leq 6\} \cap \{(x, y) : r = \sqrt{x^2 + y^2} \leq 4 - 3 \sin \theta \}\\
&= \{(x, y) : r \leq \min(6, 4 - 3 \sin \theta) \}
\end{align*}
So let
$$f(\theta) = \min(6, 4 - 3 \sin \theta)$$
and you can rewrite
$$A = \{(x, y) : r \leq f(\theta)\}$$
and the problem is equivalent to computing area of the region bounded by the closed curve $r = f(\theta)$ and so is the same as computing
$$A = \int_0^{2\pi} \frac{1}{2} f(\theta)^2 d\theta$$
[See this link for details.]
Now you have to split the integration range to the region where $f$ takes value 6 and where $f$ take value $4 - 3 \sin \theta$. It is easily solved: for $\theta \in [0,2\pi)$
$$6 \leq 4 - 3 \sin \theta \iff 2 \leq -3 \sin \theta \iff -\frac{2}{3} \geq \sin\theta \iff \theta \in [\theta_1, \theta_2]$$
where
$$\theta_1 = \pi - \arcsin\left(-\frac{2}{3}\right); \theta_2 = 2 \pi + \arcsin\left(-\frac{2}{3}\right); 0 < \theta_1 < \theta_2 < 2\pi.$$
This means
$$f(\theta) = \begin{cases}
6 &\text{if }\theta_1 \leq \theta \leq \theta_2\\
4 - 3 \sin \theta &\text{otherwise}
\end{cases}$$
Going back to the integral
\begin{align*}
A &= \int_0^{2\pi} \frac{1}{2} f(\theta)^2 d\theta\\
&= \int_0^{\theta_1} \frac{1}{2} (4 - 3 \sin)^2 d\theta + \int_{\theta_1}^{\theta_2} \frac{1}{2} 6^2 d\theta + \int_{\theta_2}^{2\pi} \frac{1}{2} (4 - 3 \sin)^2 d\theta
\end{align*}
(You should be able to complete the computation.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
multiplication over gf(16) Can some one show me how to do multiplication over gf(16) step by step
I found this example online, http://userpages.umbc.edu/~rcampbel/Math413Spr05/Notes/12-13_Finite_Fields.html#An_Example.
An Example:
$x^4 = x+1$
$x^5 = x^2+x$
$(x^2+x+1) (x^3+x^2+1)$
$= x^5 + 2x^4 + 2x^3 + 2x^2 + x + 1$
$= x^2 + 1$
I know $x^4 = x+1$, because $x^4 mod (x^4 + x + 1)$, but how does $x^5 = x^2+x$?
|
By distributivity, all you actually In the general casehave to know is how to multiply $1, x,x^2,x^3$ with each other, i.e. express $x^4,x^5,x^6$. In the general case, here is how it goes:
\begin{align*}
x^4&=\color{red}{x+1} &x^5&=x\cdot x^4=x(x+1)=\color{red}{x^2+x}\\
x^6&=x\cdot x^5=\color{red}{x^3+x^2}&x^7&=x\cdot x^6=x^4+x^3=\color{red}{x^3+x+1} \\
x^8&=(x^4)^2=(x+1)^2=\color{red}{x^2+1}&&\&c.
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Write the expression in Euler's formula
$a=2+2i$ , $b=5e^{i\frac{\pi}{3}}$
such that $$\frac{b^5}{a^3}=Re^{\theta i}$$
find: $R$ and $\theta$
$$R=\sqrt{2^2+2^2}=\sqrt{8}$$,
$$tan^{-1}=\frac{2}{2}\rightarrow \theta=\frac{\pi}{4}$$
$$a=\sqrt{8}e^{\frac{\pi}{4}i}, a^3=({\sqrt{8}e^{\frac{\pi}{4}i}})^3=(\sqrt{8})^3e^{\frac{3\pi}{4}i}$$
$$b^5=5^5e^{\frac{5\pi}{3} i}$$
$$\frac{b^5}{a^3}=\frac{5^5e^{\frac{5\pi}{3} i}}{(\sqrt{8})^3e^{\frac{3\pi}{4}i}}$$
$$\frac{b^5}{a^3}=\color{red}{5^5-(\sqrt{8})^3}*e^{(\frac{5\pi}{3}-\frac{3\pi}{4})i}=(5^5-(\sqrt{8})^3)*e^{\frac{11}{12}\pi i}$$
$$R=5^5-(\sqrt{8})^3, \theta=\frac{11}{12}\pi$$
I can not find my mistake
|
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(2+2i\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(|2+2i|e^{\arg(2+2i)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(\sqrt{2^2+2^2}e^{\tan^{-1}\left(\frac{2}{2}\right)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(\sqrt{8}e^{\tan^{-1}\left(1\right)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(2\sqrt{2}e^{\frac{\pi}{4}i}\right)^3}=$$
$$\frac{5^5e^{5\cdot\frac{\pi}{3}i}}{2^3\left(\sqrt{2}\right)^3e^{3\cdot\frac{\pi}{4}i}}=$$
$$\frac{3125e^{\frac{5\pi}{3}i}}{8\left(2\sqrt{2}\right)e^{\frac{3\pi}{4}i}}=$$
$$\frac{3125e^{\frac{5\pi}{3}i}}{16\sqrt{2}e^{\frac{3\pi}{4}i}}=$$
$$\frac{3125}{16\sqrt{2}}e^{\left(\frac{5\pi}{3}-\frac{3\pi}{4}\right)i}$$
$$\frac{3125}{16\sqrt{2}}e^{\frac{11\pi}{12}i}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots$. Show that: $$(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots = \dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}.$$
By the duplication theorem and $\Gamma(z+1)=z\,\Gamma(z)$, we have: $$\dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}=\dfrac{\Gamma\left(-\dfrac{z}{2}\right)}{z\,\Gamma\left(-z\right)\Gamma\left(\dfrac{z}{2}\right)} \cdot 2^{-z}.$$
Using the definition $\Gamma\left(z\right) = \dfrac{1}{z}\prod_{k=1}^\infty \left(1+\dfrac{1}{k}\right)^z \left(1+\dfrac{z}{k}\right)^{-1}$, and simplifying terms, we obtain:$$\dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}=2^{-z}(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots .$$
How can I get rid of the $2^{-z}$, and am I going about this the right way?
Cheers!
|
We can make use of Euler's product for the gamma function, Whittaker and Watson, page 237, §12.11, also (31) of MathWorld:
$$
\frac{1}{\Gamma(z)} =
\lim_{n\rightarrow\infty}
\frac{z \, (1+ z) \cdots (n-1 + z)}{1 \cdot 2 \cdots (n - 1)} n^{-z}.
\qquad (1)
$$
According to (1), we have
$$
\begin{align}
\frac{1}{\Gamma(1 + \frac z 2)}
&=
\lim_{n\rightarrow\infty}
\frac{(1 + \frac{z}{2}) \, (2+\frac{z}{2}) \cdots (n + \frac{z}{2})}{1 \cdot 2 \cdots (n - 1)} n^{-1 - z/2} \\
&=
\lim_{n\rightarrow\infty}
\left(1 + \frac{z}{2}\right) \,
\left(1 + \frac{z}{4}\right) \cdots \left(1 + \frac{z}{2n}\right) n^{-z/2}.
\qquad (2)
\end{align}
$$
Similarly,
$$
\begin{align}
\frac{1}{\Gamma(\frac 1 2 - \frac z 2)}
&=
\lim_{n\rightarrow\infty}
\frac{(\frac{1}2 - \frac{z}{2}) \, (\frac{3}{2} -\frac{z}{2}) \cdots (\frac{2n - 1}{2} -\frac{z}{2})}{1 \cdot 2 \cdots (n - 1)} n^{-1/2 + z/2}
\\
&=
\lim_{n\rightarrow\infty}
\frac{1}{2} \frac{1 \cdot 3 \cdots (2n - 1)}{2 \cdot 4 \cdots (2n-2) \sqrt{n} }
\left(1 - z\right) \,
\left(1 - \frac{z}{3}\right) \cdots \left(1 - \frac{z}{2n-1}\right) n^{z/2},
\\
&=
\lim_{n\rightarrow\infty}
\frac{1}{\sqrt{\pi}} \,
\left(1 - z\right) \,
\left(1 - \frac{z}{3}\right) \cdots \left(1 - \frac{z}{2n-1}\right) n^{z/2},
\qquad (3)
\end{align}
$$
where in the last step we have used Wallis's product formula:
$$
\lim_{n\rightarrow\infty}
\left(
\frac{1}{2} \frac{1 \cdot 3 \cdots (2n - 1)}{2 \cdot 4 \cdots (2n-2) \sqrt{n} }
\right)^2
=
\lim_{n\rightarrow\infty}
\frac{1}{2} \frac{1 \cdot 3 \cdot 3 \cdot 5 \cdots \cdot (2n - 1) \cdot (2n - 1)}
{2 \cdot 2 \cdot 4 \cdot 4 \cdots \cdot (2n - 2) \cdot (2n) }
= \frac{1}{2} \times \frac{2}{\pi} = \frac{1}{\pi}.
$$
Multiplying the two yields
$$
\begin{align}
\frac{1}{\Gamma(1 + \frac z 2)\,\Gamma(\frac 1 2 - \frac z 2)}
&=
\frac{1}{\sqrt\pi}
\left(1 - z\right) \,
\left(1 + \frac{z}{2}\right) \,
\left(1 - \frac{z}{3}\right) \,
\left(1 + \frac{z}{4}\right) \,
\cdots,
\end{align}
$$
which is the desired result.
|
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|
Transforming the quadratic form $x_1x_2 + x_2x_3 + x_3x_1$ to its normal form. I need to transform $x_1x_2 + x_2x_3 + x_3x_1$ to its normal form described by the inertia theorem. How can I find the transformation? I am allowed to complete squares and perform rotations to find the linear change of coordinates. I know how to do a rotation for quadratic forms over $\mathbb{R}^2$ but not $\mathbb{R}^3$. How would I go about manipulating this expression?
|
The matrix of the quadratic form for the given basis $x_1, x_2, x_3$ is
$$
A = \frac 12\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$
For simplicity, I will diagonalize $2A$ first. Eigenvalues of $2A$ can be found by solving
\begin{align}
\lambda^3 - 3 \lambda - 2 & = 0 \\
(\lambda + 1)^2(\lambda - 2) & = 0.
\end{align}
For $\lambda = -1$, the corresponding eigenvector must satisfy
$$
\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
u_1\\
u_2\\
u_3
\end{pmatrix}=
\begin{pmatrix}
-u_1\\
-u_2\\
-u_3
\end{pmatrix} \Longleftrightarrow
\begin{aligned}
u_1 + u_2 + u_3 = 0.
\end{aligned}
$$
We can pick two arbitrary independent orthogonal solutions. I will pick
$$
u = \begin{pmatrix}
1/\sqrt 6 \\
1/\sqrt 6 \\
-2/\sqrt 6
\end{pmatrix} \quad \text{ and } \quad
v = \begin{pmatrix}
1/\sqrt 2 \\
-1/\sqrt 2 \\
0
\end{pmatrix}.
$$
For $\lambda = 2$, we solve for $w$ from
$$
\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
w_1\\
w_2\\
w_3\end{pmatrix}
=
\begin{pmatrix}
2w_1\\
2w_2\\
2w_3
\end{pmatrix} \Longleftrightarrow
\begin{aligned}
-2w_1 + w_2 + w_3 & = 0\\
w_1 - 2w_2 + w_3 & = 0\\
\end{aligned}.
$$
It is easy to see that $w = \begin{pmatrix}1/\sqrt 3\\1/\sqrt 3\\1/\sqrt 3\end{pmatrix}$ is a solution.
Therefore,
$$
2A = \begin{pmatrix}
1/\sqrt 6 & 1/\sqrt 2 & 1/\sqrt 3\\
1/\sqrt 6 & -1/\sqrt 2 & 1/\sqrt 3\\
-2/\sqrt 6 & 0 & 1/\sqrt 3
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 2
\end{pmatrix}
\begin{pmatrix}
1/\sqrt 6 & 1/\sqrt 2 & 1/\sqrt 3\\
1/\sqrt 6 & -1/\sqrt 2 & 1/\sqrt 3\\
-2/\sqrt 6 & 0 & 1/\sqrt 3
\end{pmatrix}^T.
$$
The factor $2$ (or its reciprocal) can be pushed to the middle matrix of eigenvalues. This does not change the eigenvectors. The matrix of eigenvectors is the change of coordinates that you want.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1501408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.