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Jordan form of the "multiplicative table" matrix I have to find the Jordan form of the $(10\times10)-$matrix $A$ with the $n$th row formed by $n(1,2,3,4,5,6,7,8,9,10), \ \ 1 \leq n \leq 10$
I have calculated the determinant of $(A-xI)$ using Gaussian elimination, and I have that the result is $x^9(x-1)$, so there are two eigenvalues $x=0$ with algebraic multiplicity $9$ and $x=1$ with algebraic multiplicity $1$ (and so also the geometric one is $1$). Now I have to understand the geometric multiplicity of $0$, and I think that it could be $9$ (again using Gaussian elimination), is this right?
|
I want to thank @Hagen von Eitzen to make me see where was the mistake.
Given $A-xI$ the matrix defined upon, with Gaussian elimination of the type: $n$-th column $-$ $n$ times the first column, we end up with a matrix of the form:
$\left(\begin{matrix}1-x & 2x & 3x &... \\ 2 & -x & 0 &... \\ 3 & 0 & -x &... \\ . & ... \\ . \\ . \end{matrix}\right)$
It's easy to prove with induction that the determinant of this matrix is $ \\x^9(x-(1+4+9+16+...+100))$
There are two eigenvalues:
$x=0$ which has algebraic multiplicity $9$ and geometric multiplicity $9$, this last because, given that $e_i:= (0,0,...,1,0,...,0)$ where $1$ is in the $i$th place, the vectors $e_2,...,e_{10}$ are a base for the eigenspace.
$x=\sum_{k=1}^{10} k^2 $ which has algebraic and geometric multiplicity $1$
So the matrix in Jordan form will be:
$\left(\begin{matrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \sum_{k=1}^{10} k^2=385 \\ \end{matrix}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1626282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Why can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits?
Why can we convert a base $9$ number to a base $3$ number by simply
converting each base $9$ digit into two base $3$ digits ?
For example $813_9$ can be converted directly to base $3$ by noting
\begin{array} \space 8_9&=22_3 \\ \space 1_9 &=01_3 \\ \space 3_9
&=10_3 \\ \end{array}
Putting the base digits together ,we get $$813_9=220110_3$$
I know it has to do with the fact that $9=3^2$ but I am not able to understand this all by this simple fact...
|
Let's look at what you base $9$ number actually means. $$813_9=8\cdot 9^2+1\cdot 9^1+3\cdot 9^0$$
If we wish to write this as powers of $3$ with coefficients between $0$ and $2$, we can simply do
\begin{align}
3\cdot 9^0&=1\cdot 3^1+0\cdot 3^0\\
1\cdot 9^1&=0\cdot 3^3+1\cdot 3^2\\
8\cdot 9^2&=2\cdot 3^5+2\cdot 3^4\\
\end{align}
Why can we do this?
Why can we write $$a_n\cdot 9^n=b_{2n+1}\cdot 3^{2n+1}+b_{2n}\cdot3^{2n}$$
First note that $9^n3^2n$, so when dividing the equation by that, we get
$$\frac{a_n\cdot 9^n}{3^{2n}}=a_n=3b_{2n+1}+b_{2n}=\frac{b_{2n+1}\cdot 3^{2n+1}+b_{2n}\cdot3^{2n}}{3^2n}$$
So actually, the problem comes down to writing a number $0\leq a_n<9$ as $3b_{2n+1}+b_{2n}$, where $0\leq b_{2n},b{2n+1}<3$. This is obviously possible.
Why does this work for binary and hexadecimal?
Actually, the answer is fairly similar. The problem can be reduced equivalently, resulting in the equation $$a_n=8a_{4n+3}+4a_{4n+2}+2a_{4n+1}+a_{4n}$$ Where $0\leq a_n<16$ and $0\leq a_{2n+3},a_{2n+2},a_{2n+1},a_{2n}<2$. The solvability of this equation is in my opinion a little less obvious, but still quite understandable; But, for the sake of a more thourough understanding, we could look at hexadecimal-octagonal conversion. This comes down to the easy equation $a_n=2b_{2n+1}+b_{2n}$, where $0\leq a_n<16$ and $0\leq b_{2n+1},b_{2n}<8$. This is cleary solvable. Doing this for octagonal-base 4 conversion and base 4-binary conversion shows with a recurrance-like approach that this indeed works for hexadecimal-binary conversion.
Hope this helped!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Use the partial fraction to evaluate $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$. $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$
My try：
$\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$ ＝
$\int \frac{\left(3y^2+3y+2\right)}{\left(y-1\right)(y+1)\left(y+1\right)}dy$
$\frac{A}{y-1}+\frac{B}{y+1}+\frac{C}{y+1}$ = $\frac{A(y+1)(y+1)+B(y-1)y+1)+C(y-1)(y+1)}{(y-1)(y+1)(y+1)}$ = $\frac{\left(3y^2+3y+2\right)}{\left(y-1\right)(y+1)\left(y+1\right)}$
so that
$A(y+1)(y+1)+B(y-1)y+1)+C(y-1)(y+1)=Ay^2+2Ay+A+By^2-B+Cy^2-C=3y^2+3y+2$
And then I got
A+B+C=3
2A=3
A-B-C=2
=>A=3/2 B=? C=?
And then I found that I cannot figure out the value of BandC. What is wrong with my steps? I am so confusing now.
|
The doubled factor $(y+1)^2$, does not contribute $\frac{B}{y+1}+\frac{C}{y+1}$. It contributes $\frac{B}{y+1}+\frac{C}{(y+1)^2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
On the divisors of $a^b + (\frac{a-b}2)^a + (\frac{b-a}2)^b+ b^a $ Let $a$ and $b$ be two odd natural numbers. Show that $\frac{a+b}{2}$ divides $$a^b + \left(\frac{a-b}2\right)^a + \left(\frac{b-a}2\right)^b+ b^a$$
|
Well, that's simple. You need just one fact (easy to prove by induction): $x+y$ divides $x^n+y^n$ if $n$ is odd.
Now, $a^b+\left({b-a\over2}\right)^b$ is divisible by $a+{b-a\over2}={a+b\over2}$, and $b^a+\left({a-b\over2}\right)^a$ is divisible by $b+{a-b\over2}$ which also happens to be ${a+b\over2}$, hence so must be the sum of the two.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating the integral $\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}}$ I successfully evaluated the following integral using partial fraction expansion, but am unsure of a few steps.
$$
\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}} = \int\left( \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1} \right) \\\ \\
\\
x^2-2x-1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2 \\\ \\
\\
x=1 \implies -2 = 2B \implies B = -1 \\\ \\
\begin{align}
x=i \implies & -2-2i = (D+Ci)(i-1^2) = -2i(D+Ci) \\
\implies & -C+Di = 1+i \\
\implies &\ C=-1,\ \ D=1
\end{align} \\\ \\
\\
x^2-2x-1 = A(x-1)(x^2+1) - (x^2+1) - (x-1)^3 \\\ \\
\\
x=0 \implies -1 = -A \implies A = 1 \\\ \\
\\
\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}} = \int\left(\frac{1}{x-1} - \frac{1}{(x-1)^2} + \frac{1-x}{x^2+1}\right) = \int \left( \frac{1}{x-1} - \frac{1}{(x-1)^2} - \frac{x}{x^2+1} + \frac{1}{x^2+1} \right) \\\ \\
\\
= \ln|x-1| + \frac{1}{x-1} - \frac{1}{2}\ln|x^2+1| + \arctan(x) + C\\\ \\
$$
My main points of confusion are:
*
*In general, how do you know which denominators to use in step 1?
*Is it okay to find the variables by plugging in values for $x$? Does this work in general? Is there anything I should be aware of, particularly where $i$ is involved?
|
In general you factor the denominator (this can be hard), then
every linear factor $(x-a)^r$ gives you $r$ terms $\frac {A_1}{x-a}$, $\frac {A_2}{(x-a)^2}$, $\ldots$, $\frac {A_r}{(x-a)^r}$
every quadratic factor $(x^2+ bx+c)^s$ (it has complex roots, otherwise factor it to linear terms and apply the above) gives you $s$ terms $\frac {B_1 x+C_1}{x^2+bx+c}$,$\frac {B_2x+C_2}{(x^2+bx+c)^2}$, $\ldots$, $\frac {B_sx+C_s}{(x^2+bx+c)^s}$
Next step is to solve for $A_i$, $B_j$ and $C_j$. It's a linear system that has unique solution. In the end integrate all the remaining integrals
|
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|
Is it possible to solve the following equation without using the Rational Root Theorem? Given $f(x)=x^4+2x^3+2x^2-2x-3$, where $x-1$ is a factor of $f(x)$, how is it possible to solve $f(x)$ without the Rational Root Theorem?
Here's my progress:
$$f(x)=x^4+2x^3+2x^2-2x-3$$
$$f(x)=(x-1)(x^3+3x^2-5x+3)$$
And there's where I got stuck.
I cheated, though, and tried solving this equation with the Rational Root Theorem. That's what I got:
$$f(x)=x^4+2x^3+2x^2-2x-3$$
$$f(x)=(x-1)(x^3+3x^2-5x+3)$$
$$f(x)=(x-1)(x+1)(x^2+2x+3)$$
$$f(x)=(x-1)(x+1)(x-(-1-\sqrt2))(x-(-1+\sqrt2))$$
Please, if you're answering this question, I'd really appreciate if you could explain your procedures and what technique you used.
Thank you very much.
|
$$\begin{aligned}
x^4+2x^3+2x^2-2x-3
&=(x^4+2x^2-3)+(2x^3-2x) \\
&=(x^2+3)(x^2-1)+2x(x^2-1) \\
&=(x^2-1)(x^2+3+2x) \\
&=(x^2-1)(x^2+2x+3)
\end{aligned}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$
How to start the proof by induction? Is there any way to show this?
|
Here is an inductive proof:
*
*For $n=2$ we get $$1-\frac{1}{2^{2}}=\frac{2+1}{2*2}=\frac{3}{4}$$
*Let $n=k>2$ and $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})=\frac{k+1}{2*k}$$
and now the induction step:
*Let $n=k+1$, then $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})(1-\frac{1}{(k+1)^{2}})=(\frac{k+1}{2*k})(1-\frac{1}{(k+1)^{2}})= \\
=(\frac{k+1}{2*k})\frac{(k+1)^{2}-1}{(k+1)^{2}}=(\frac{1}{2*k})\frac{k(k+2)}{(k+1)}=\frac{1}{2}\frac{(k+2)}{(k+1)}=\frac{(k+1)+1}{2*(k+1)}$$
thus the proof is complete.
|
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"timestamp": "2023-03-29T00:00:00",
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|
limit of $f(x) = \lim \limits_{x \to 0} (\frac{\sin x}{x})^{\frac 1x}$ Any ideas how to calculate this limit without using taylor?
$$f(x) = \lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac1x}$$
|
Here's a proof
that just uses
basic properties
of
$\sin, \cos$,
and
$\ln$.
Since
$-1 \le \cos(x) < 1$
and $\sin'(x) = \cos(x)$
and $\cos'(x) = -\sin(x)$,
$\sin(x)
=-\int_0^x \cos(t) dt
$
so
$|\sin(x)|
\le |x|
$.
Also,
since,
for $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
=\int_0^{x} \frac{dt}{1+t}
$,
$\ln(1+x)
\le x
$
and,
for $1 > x > 0$,
$\ln(1-x)
=\int_1^{1-x} \frac{dt}{t}
=-\int^1_{1-x} \frac{dt}{t}
=-\int^0_{-x} \frac{dt}{1+t}
$,
so,
if $\frac12 > x > 0$,
$-\ln(1-x)
=\int^0_{-x} \frac{dt}{1+t}
\ge \frac{x}{1-x}
\ge 2x
$.
Therefore,
for
$-\frac12 < x < 1$,
$|\ln(1+x)|
\le 2|x|
$.
$\begin{array}\\
\cos(x)
&=1-2\sin^2(x/2)\\
&=1+O(x^2)
\qquad\text{with the implied constant
being less than 1},\\
\text{so that}\\
\frac{\sin x}{x}
&=\frac1{x}\int_0^x \cos(t)dt\\
&=\frac1{x}\int_0^x (1+O(t^2))dt\\
&=\frac1{x} (t+O(t^3))|_0^x\\
&=\frac1{x} (x+O(x^3))\\
&=1+O(x^2))\\
\text{so}\\
\ln(\frac{\sin x}{x})
&=\ln(1+O(x^2))\\
&=O(x^2)\\
\text{so}\\
\frac1{x}\ln(\frac{\sin x}{x})
&=\frac1{x}(O(x^2))\\
&=O(x)\\
&\to 0 \text{ as }x \to 0\\
\text{so}\\
(\frac{\sin x}{x})^{1/x}
&\to 1\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)( x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$$
$$( x + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)( x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)$$
He then mltiplied both pairs,resulting in:
$$(x^2 - \sqrt{2}x + 1) and (x^2 + \sqrt{2}x + 1)$$
How do I get the first four imaginary points, or the two distinct because I realize that each has a conjugate.
|
$$x^4+1=0\Longleftrightarrow$$
$$x^4=-1\Longleftrightarrow$$
$$x^4=|-1|e^{\arg(-1)i}\Longleftrightarrow$$
$$x^4=e^{\pi i}\Longleftrightarrow$$
$$x=\left(e^{\left(2\pi k+\pi\right)i}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$x=e^{\frac{1}{4}\left(2\pi k+\pi\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-3$
So, the solutions are:
$$x_0=e^{\frac{1}{4}\left(2\pi\cdot0+\pi\right)i}=e^{\frac{\pi i}{4}}$$
$$x_1=e^{\frac{1}{4}\left(2\pi\cdot1+\pi\right)i}=e^{\frac{3\pi i}{4}}$$
$$x_2=e^{\frac{1}{4}\left(2\pi\cdot2+\pi\right)i}=e^{-\frac{3\pi i}{4}}$$
$$x_3=e^{\frac{1}{4}\left(2\pi\cdot3+\pi\right)i}=e^{-\frac{\pi i}{4}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\dim W_1$, $\dim W_2$, $\dim W_1+W_2$, $\dim W_1\cap W_2$ for the following spans? Let $W_1=\{(1,1,2,1), (3,1,0,0)\}$ and $W_2=\{(-1,-2,0,1), (-4,-2,-2,-1)\}$
Apparently $\dim W_1=\dim W_2=2$.
For $\dim W_1\cap W_2$, since $(-4,-2,-2,-1)$ can be expressed as $-(1,1,2,1)-(3,1,0,0)$, $\dim W_1\cap W_2=1$. But what if the spans are complicated, how do you find $\dim W_1\cap W_2$. Do you try matrix on each vector and see which ones has a nontrivial solution?
For $\dim W_1+W_2$, how do you know it without calculating $\dim W_1\cap W_2$?
|
Put the given vectors into the columns of matrices
\begin{align*}
w_1 &=
\left[\begin{array}{rr}
1 & 3 \\
1 & 1 \\
2 & 0 \\
1 & 0
\end{array}\right] &
w_2 &=
\left[\begin{array}{rr}
-1 & -4 \\
-2 & -2 \\
0 & -2 \\
1 & -1
\end{array}\right]
\end{align*}
The dimensions of $W_1$ and $W_2$ are the ranks of the matrices $w_1$ and $w_2$ respectively. Row reducing gives $\DeclareMathOperator{rref}{rref}$
\begin{align*}
\rref(w_1) &=
\left[\begin{array}{rr}
1 & 0 \\
0 & 1 \\
0 & 0 \\
0 & 0
\end{array}\right]
&
\rref(w_2) &=
\left[\begin{array}{rr}
1 & 0 \\
0 & 1 \\
0 & 0 \\
0 & 0
\end{array}\right]
\end{align*}
These reduced row-echelon forms show that both $w_1$ and $w_2$ have rank two. This proves that $\dim W_1=\dim W_2=2$, as you have observed.
Now, the dimension of $W_1+W_2$ is the rank of the matrix
$$
[w_1,w_2]
=
\left[\begin{array}{rrrr}
1 & 3 & -1 & -4 \\
1 & 1 & -2 & -2 \\
2 & 0 & 0 & -2 \\
1 & 0 & 1 & -1
\end{array}\right]
$$
Row reducing gives
$$
\rref[w_1,w_2]=
\left[\begin{array}{rrrr}
1 & 0 & 0 & -1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
This proves that $\dim (W_1+W_2)=3$. The dimension formula then implies that $\dim(W_1\cap W_2)=1$.
This method also provides bases for the spaces in question. The pivot columns of $\rref[w_1,w_2]$ correspond to the columns in $[w_1,w_2]$ that form a basis for $W_1+W_2$. The "free" columns of $\rref[w_1,w_2]$ correspond to the columns of $[w_1,w_2]$ that span $W_1\cap W_2$. Here, we have $\DeclareMathOperator{Span}{Span}$
\begin{align*}
W_1+W_2 &=
\Span
\left\{
\left[\begin{array}{r}
1 \\
1 \\
2 \\
1
\end{array}\right],
\left[\begin{array}{r}
3 \\
1 \\
0 \\
0
\end{array}\right],
\left[\begin{array}{r}
-1 \\
-2 \\
0 \\
1
\end{array}\right]
\right\}
&
W_1\cap W_1
&= \Span\left\{
\left[\begin{array}{r}
-4 \\
-2 \\
-2 \\
-1
\end{array}\right]
\right\}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is this identity about floor function true? When $a, b, c$ are positive integers, is this identity below is true for all $a, b, c$?
$$\left\lfloor \frac{\left\lfloor\frac ab \right\rfloor}c \right\rfloor =\left\lfloor \frac{\left\lfloor\frac ac \right\rfloor}b \right\rfloor $$
|
Using division with remainder write
$a=qbc+r$ with $0\le r<bc$ and then write $r=q'b+r'$ with $0\le r'<b$, as well as $r=q''c+r''$ with $0\le r''<c$. So $a=qbc+q'b+r'=qbc+q''c+r''$.
Because $0\le r<bc$ we conclude that $0\le q'<c$ and $0\le q''<b$.
Then
$$ \left\lfloor\frac{\left\lfloor\frac ab\right\rfloor}c\right\rfloor=
\left\lfloor\frac{qc+q'}c\right\rfloor=q$$
and
$$ \left\lfloor\frac{\left\lfloor\frac ac\right\rfloor}b\right\rfloor=
\left\lfloor\frac{qb+q''}c\right\rfloor=q.$$
So indeed
$$\left\lfloor\frac{\left\lfloor\frac ac\right\rfloor}b\right\rfloor=\left\lfloor\frac{\left\lfloor\frac ab\right\rfloor}c\right\rfloor =\left\lfloor\frac a{bc}\right\rfloor.$$
|
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|
Probability of an even number of sixes
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.
I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:
*
*The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.
*Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.
For $n=k+1$, there are two ways the number of sixes are even:
a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n]$
b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
I have two questions
*
*How do I get from $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$
*Other than that, is my use of induction correct? Is it rigorous enough to prove the formula?
|
Let $E_n$ be the event that there are an even number of sixes rolled in $n$ rolls, and $O_n$ be the event that there are an odd number of sixes rolled in $n$ rolls. Let $X_i = 1$ if roll $i$ is a $6$, otherwise $X_i = 0$. Clearly, $$\Pr[E_n] + \Pr[O_n] = 1$$ for all $n$. Now, observe that $$\begin{align*} \Pr[E_n] &= \Pr[E_{n-1} \cap X_n = 0] + \Pr[O_{n-1} \cap X_n = 1] \\ &= \frac{5\Pr[E_{n-1}] + \Pr[O_{n-1}]}{6} \\ &= \frac{4 \Pr[E_{n-1}] + 1}{6}.\end{align*}$$ This establishes a recursion relationship with $a_n = \Pr[E_n]$, namely $$a_n = \frac{2}{3} a_{n-1} + \frac{1}{6}, \quad a_1 = \frac{5}{6},$$ since in $n = 1$ roll, the chance of an even number of sixes is equal to the chance that no six is rolled. Now consider a constant $c$ so that $$a_n - c = \frac{2}{3}(a_{n-1} - c).$$ If this equation is to be equivalent to the desired recursion, we must have $$c - \frac{2}{3}c = \frac{1}{6},$$ or $c = 1/2$. Therefore, letting $b_n = a_n - 1/2$, we obtain the auxiliary recursion $$b_n = \frac{2}{3} b_{n-1}, \quad b_1 = \frac{1}{3},$$ which has the obvious geometric sequence solution $$b_n = \left(\frac{2}{3}\right)^{n-1} b_1 = \left(\frac{2}{3}\right)^{n-1} \frac{1}{3} = \frac{2^{n-1}}{3^n}.$$ Consequently, $$\Pr[E_n] = a_n = b_n + \frac{1}{2} = \frac{2^{n-1}}{3^n} + \frac{1}{2} = \frac{1}{2}\left(1 + \left(\frac{2}{3}\right)^n\right),$$ as claimed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given M, can we find $2$ primes $a,b$ so that for all naturals $x,y$, $|a^x-b^y|>M$? For example, if $M = 2$, one can show that $3,17$ satisfy the above:
For any naturals $x,y$, $|3^x-17^y|>2$.
|
Choose $a$ to be a prime larger than $M + 1$, and let $b$ be the first prime congruent to $1 \pmod{2 \cdot a}$, guaranteed to exist by Dirichlet's theorem. Now $a^x \equiv a \pmod{2 \cdot a}$ and $b^y \equiv 1 \pmod{2 \cdot a}$, so $\vert a^x - b^y \vert \ge a-1 \gt M$. Using this method, for $M=3$ we have $(a,b)=(5,11)$, $5^x \equiv 5 \pmod{10}$, and $11^y \equiv 1 \pmod{10}$. In other words $5^x = \dots 5$ and $11^y = \dots 1$, so they differ by at least $4$.
|
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|
The sum of the following infinite series $\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
The sum of the following infinite series $\displaystyle \frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
$\bf{My\; Try::}$ We can write the given series as $$\left(1+\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots\right)-1$$
Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\cdots$$
So we get $\displaystyle nx=\frac{4}{20}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{4\cdot 7}{20\cdot 30}$
So we get $$\frac{nx\cdot (nx-x)}{2}=\frac{4\cdot 7}{20\cdot 30}\Rightarrow \frac{4}{20}\cdot \left(\frac{4-20}{20}\right)\cdot \frac{1}{2}x^2=\frac{4}{20}\cdot \frac{7}{30}$$
But here $x^2=\text{Negative.}$
I did not understand how can I solve it
Help me, Thanks
|
The numerators suggest that you could make use of a power series involving exponents that are rational numbers with denominator $3$.
$$\begin{align}
\sum_{n=1}^{\infty}\frac{4\cdot7\cdot\cdots\cdot(3n+1)}{(n+1)!10^n}
&=\sum_{n=1}^{\infty}\frac{\frac43\cdot\frac73\cdot\cdots\cdot\frac{3n+1}3}{(n+1)!\left(10/3\right)^n}\\
&=\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{\frac{3n+1}{3}}{n}\left(\frac{3}{10}\right)^n\\
&=\left[\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{\frac{3n+1}{3}}{n}x^n\right]_{x=3/10}\\
&=\left[\frac{1}{x}\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{\frac{3n+1}{3}}{n}x^{n+1}\right]_{x=3/10}\\
&=\left[\frac{1}{x}\int_0^x\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}t^{n}\,dt\right]_{x=3/10}\\
&=\left[\frac{1}{x}\int_0^x\sum_{n=1}^{\infty}\binom{-\frac{4}{3}}{n}(-t)^{n}\,dt\right]_{x=3/10}\\
&=\left[\frac{1}{x}\int_0^x\left(\left(1-t\right)^{-4/3}-1\right)\,dt\right]_{x=3/10}\\
&=\left[\frac{1}{x}\left[3\left(1-t\right)^{-1/3}-t\right]_{t=0}^{t=x}\right]_{x=3/10}\\
&=\left[\frac{1}{x}\left(3\left(1-x\right)^{-1/3}-x-3\right)\right]_{x=3/10}\\
&=\frac{10}{3}\left(3\left(1-\frac{3}{10}\right)^{-1/3}-\frac{3}{10}-3\right)\\
&=10\left(\frac{7}{10}\right)^{-1/3}-11\\
&=\sqrt[3]{\frac{10^4}{7}}-11\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.
Can you please explain?
|
$$(x^3+x^4+x^5+x^6+\cdots)^3=x^9(1+x+x^2+\cdots)^3=x^9\left(\dfrac1{1-x}\right)^3=x^9(1-x)^{-3}$$
Now, we need the coefficient of $x^3$ in $(1-x)^{-3}$
Now the $r+1,(r\ge0)$th term of $(1-x)^{-3}$ is $$\dfrac{-3(-4)(-5)\cdots(-r)(-r-1)(-r-2)}{1\cdot2\cdot3\cdot r}(-x)^r=\binom{r+2}2x^r$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the equation of tangent at origin to the curve $y^2=x^2(1+x+x^2)$ How do I find the equation of tangent at $(0,0)$ to the curve $y^2=x^2(1+x+x^2)$ ?
Differentiating and putting the value of $x$ and $y$ gives an indeterminate form.
Can we trace the curve and geometrically make tangents and find their equation ?
|
$$y^2=x^2+x^3+x^4$$
$$2y\frac{dy}{dx}=2x+3x^2+4x^3$$
$$\implies \frac{dy}{dx}=\frac{2x+3x^2+4x^3}{2y}$$
Now, equation of tangent line at (0,0) is (eqn 1):
$$y=\frac{dy}{dx}\big{|}_{(0,0)}(x)$$
Since $\dfrac{dy}{dx}$ yields 0/0 form, we use limits to evaluate the tangent:
$$\lim_{x,y\to0} \frac{2x+3x^2+4x^3}{2y} $$
$$ \equiv\lim_{x\to0} \frac{2x+3x^2+4x^3}{2|x|\sqrt{1+x+x^2}} $$
Here we find:
*
*$\lim_{x\to0^-} \dfrac{2x+3x^2+4x^3}{2|x|\sqrt{1+x+x^2}}=-1 $ and
*$\lim_{x\to0^+} \dfrac{2x+3x^2+4x^3}{2|x|\sqrt{1+x+x^2}}=+1 $
The limit is known to not exist, as LHL and RHL don't agree. but in my opinion it doesn't matter if this limit of slope function $\dfrac{dy}{dx}$ exists or not, since its a curve.
so from eqn 1
$$\implies y = \pm x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of $\int_{0}^{1}\frac{\arctan x}{1+x}dx$
Evaluation of $$\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x}dx = \int_{0}^{1}\frac{\arctan x}{1+x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}dx$$
Then $$\frac{dI}{da} = \frac{d}{da}\left[\int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}\right]dx = \int_{0}^{1}\frac{x}{(1+a^2x^2)(1+x)}dx$$
So we get $$I = \frac{1}{a^2}\int_{0}^{1}\frac{x}{(x+1)(x^2+k^2)}dx\;,$$ Where $\displaystyle k=\frac{1}{a^2}$
Now Using Partial fraction for $$\frac{x}{(x^2+k^2)(x+1)} = \frac{Ax+B}{x^2+k^2}+\frac{C}{x+1} = \frac{(A+C)x^2+(A+B)x+(B+Ck^2)}{(x^2+k^2)(x+1)}$$
So we get $A+C=0$ and $A+B=1$ and $B+Ck^2=0$
So we get $\displaystyle A=\frac{1}{1+k^2}$ and $\displaystyle B = \frac{k^2}{1+k^2}$ and $\displaystyle C=-\frac{1}{1+k^2}$
So $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\int_{0}^{1}\left[\frac{x+k^2}{x^2+k^2}-\frac{1}{x+1}\right]dx$$
So $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\left[\frac{1}{2}\ln(x^2+k^2)-k\tan^{-1}\left(\frac{x}{k}\right)-\ln(x+1)\right]_{0}^{1}$$
So we get $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\left[\frac{1}{2}\ln(1+k^2)-\ln(k)-k\tan^{-1}\left(\frac{1}{k}\right)-\ln 2\right]$$
So So we get $$\frac{dI}{da}=\frac{1}{1+a^2}\left[\frac{1}{2}\ln(1+a^2)-\frac{1}{a}\tan^{-1}(a)-\ln2\right]$$
Now integration of above expression is very lengthy, Is there is any other method
If yes Then plz explain here, Thanks
|
The substitution $x\mapsto \frac{1-x}{1+x}$ transforms the integral into
$$I=\int_0^1 \frac{\tan^{-1} x}{1+x}\,dx=\int_0^1 \frac{\frac{\pi}{4}-\tan^{-1} x}{1+x}\,dx.$$
Because $\displaystyle\,\,\,\, \tan^{-1} \frac{1-x}{1+x}=\tan^{-1}1-\tan^{-1} x=\frac{\pi}{4}-\tan^{-1} x.$
Taking the average of these two representations then gives
$$I=\frac12\int_0^1 \frac{\frac{\pi}{4}}{1+x}dx=\frac{\pi}{8}\ln2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Coordinates of a matrix So on the textbook, it gives an example:
If the basis of B matrix is{$\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \begin{bmatrix}0&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\ 1&1\end{bmatrix}, \begin{bmatrix}1&0\\ 1&0\end{bmatrix}$}
Then the B-coordinates of a matrix $\begin{bmatrix}a&b\\ c&d\end{bmatrix}$ are $\begin{bmatrix}a-c+d\\ b\\d\\c-d\end{bmatrix}$
But can I write the B-coordinates as $\begin{bmatrix}a-d\\ b\\c-d\\d\end{bmatrix}$? I found it much easier to write the column matrix in this way, but is it correct?
|
$$(a-c+d)\begin{bmatrix}1&0\\ 0&0\end{bmatrix}+b \begin{bmatrix}0&1\\ 0&0\end{bmatrix}+d\begin{bmatrix}0&0\\ 1&1\end{bmatrix}+(c-d)\begin{bmatrix}1&0\\ 1&0\end{bmatrix}=\begin{bmatrix}a&b\\ c&d\end{bmatrix},$$
while
$$(a-d)\begin{bmatrix}1&0\\ 0&0\end{bmatrix}+b \begin{bmatrix}0&1\\ 0&0\end{bmatrix}+(c-d)\begin{bmatrix}0&0\\ 1&1\end{bmatrix}+d\begin{bmatrix}1&0\\ 1&0\end{bmatrix}=\begin{bmatrix}a&b\\ c&{\color{red}{c-d}}\end{bmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
how we can calculate $ \frac {\sqrt {x^2} + \sqrt {y^2} }{2 \sqrt {xyz}}$? I teach math for Schools. How can Help me in the following past Olympiad question?
If $y,z$ be two negative distinct number and $x$ and $y$ be negate of each other, how we can calculate $ \displaystyle\frac {\sqrt {x^2} + \sqrt {y^2} }{2 \sqrt {xyz}}$?
1) $\frac {\sqrt{x}}{x}$
2)$\frac {\sqrt{-y}}{y}$
3) $\frac {\sqrt{z}}{z}$
4) $\frac {\sqrt{-z}}{-z}$
|
$$\frac{\sqrt{x^2}+\sqrt{y^2}}{2\sqrt{xyz}}=\frac{2\sqrt{x^2}}{2\sqrt{-x^2z}}$$ using $y=-x$. This yields: $$\frac{2\sqrt{x^2}}{2\sqrt{x^2}\sqrt{-z}}=\frac{1}{\sqrt{-z}}$$. Note that $z<0$, so $-z>0$, so $\sqrt{-z}$ is well-defined.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim_{x \to 0} \frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$ Evaluate $$\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$$
First I tried using L'Hopital's rule..but it's very lengthy
Next I have written the limits as
$$L=\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}}{\lim_{x \to 0}\frac{\tan x-\sin x}{x^3}}=\frac{L_1}{L_2}$$
Now by L'Hopital's Rule we get $L_2=0.5$
$$L_1=\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}$$
Now $L_1$ can also be evaluated using three applications of L'Hopital's Rule, but is there any other approach?
|
\begin{align}
\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&= \lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)\cos(\sin x)}{\tan x - \sin x}\\
&=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)+\tan(\sin x)-\tan(\sin x)\cos(\sin x)}{\tan x -\sin x}\\
&=\lim_{x\to 0}\left(\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}+\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\right)
\end{align}
Then, by mean value theorem, there is $c\in (-\frac{\pi}{2},\frac{\pi}{2})$ such that
$$
\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\sec^2 c
$$
and between $\sin x$ and $\tan x$. Then $\lim_{x\to 0} \frac{\tan(\tan x)-\tan(\sin x)}{\tan x - \sin x}=1$. Next, we will show that $\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}=1$.
\begin{align}
\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}&=\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))\cos x}{\sin x(1 - \cos x)}\\
&=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\cdot \frac{1-\cos(\sin x)}{\sin^2 x}\cdot \frac{\sin^2 x}{1-\cos x}\cdot \cos x\\
&=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\lim_{x\to 0}\frac{1-\cos(\sin x)}{\sin^2 x}\lim_{x\to 0}(1+\cos x)\lim_{x\to 0}\cos x\\
&=1\cdot \frac{1}{2}\cdot 2\cdot 1\\
&=1.
\end{align}
Therefore,
\begin{align}
\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)}{\tan x - \sin x}+\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\\
&=1+1\\
&=2
\end{align}
|
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|
Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$ Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
Let $I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
I took $\sqrt{2-x-x^2}$ as first function and $\frac{1}{x^2}$ as the second function and integrated it by parts,
$I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx=\sqrt{2-x-x^2}\int\frac{1}{x^2}dx-\int\frac{-1-2x}{2\sqrt{2-x-x^2}}\times\frac{-1}{x}dx$
$=\sqrt{2-x-x^2}\times\frac{-1}{x}-\frac{1}{2}\int\frac{1+2x}{x\sqrt{2-x-x^2}}dx$
Now i put $x=\frac{1}{t}$ in the integral $\int\frac{1+2x}{x\sqrt{2-x-x^2}}dx$ to get
$I=\sqrt{2-x-x^2}\times\frac{-1}{x}+\frac{1}{2}\int\frac{(t+2)dt}{t\sqrt{2t^2-t-1}}$
I do not know how to solve it further.
|
Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int\frac{\sqrt{(x+2)(1-x)}}{x^2}dx$$
Now Let $\displaystyle (x+2) = (1-x)t^2\;,$ Then $\displaystyle x = \frac{t^2-2}{t^2+1} = 1-\frac{3}{t^2+1}$
So $$\displaystyle dx = \frac{6t}{(t^2+1)^2}dt$$ and $$\displaystyle (1-x) = \frac{3}{t^2+1}$$
So Integral $$\displaystyle I = 18\int\frac{t^2}{(t^2+1)\cdot (t^2-2)^2}dt = 6\int\frac{(t^2+1)-(t^2-2)}{(t^2+1)\cdot (t^2-2)^2}dt$$
so we get $$\displaystyle I = 6\int\frac{1}{(t^2-2)^2}dt-6\int\frac{1}{(t^2+1)(t^2-2)}dt$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$
$$\int\frac{\sin(2x)}{2+\cos x} dx $$
Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 \int 1 - 2\frac{1}{u} du = -2 (u - 2\ln u) + C$$
After substituting for $u$, I get $4 \ln(2+ \cos x) - 2\cos x - 4 + C$. According to Wolfram Alpha, the solution is $4 \ln(2+ \cos x) - 2\cos x + C$. Not sure where my mistake is, but any help is greatly appreciated.
|
Consider integral
$$I=\int\frac{\sin(2x)}{2+\cos x}dx.$$ Through the linéarisation formula $\frac{\sin(2x)}{2}=\sin(x)\cos(x),$ we obtain
$$I=2\int\frac{\sin x\cos x}{2+\cos x}dx,$$ and by the change of variable $u=\cos x, \ x=\arccos x$ and $du=-\sin xdx,$ we arrive at
$$I=-2\int\frac{udu}{2+u}du=-2\int\frac{u+2-2}{2+u}du$$
$$=-2\left(\int du-2\int\frac{1}{2+u}du\right)=-2u+4\ln|u+2|+C$$ which gives the result by replacing $u$ by $\cos x$. Thanks
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the system of equation with answers as x=___+___s Solve the system:
\begin{cases}
-4x_1 + x_2=3 \\
8x_1-2x_2=-6
\end{cases}
I need the solution in the form of:
\begin{cases}
x_1=\underline{\hspace{1cm}}+\underline{\hspace{1cm}}\times s\\[0.3em]
x_2=\underline{\hspace{1cm}}+\underline{\hspace{1cm}}\times s\\[0.3em]
\end{cases}
I've attempted to due this multiple times and end with the same answer each time.
\begin{array}{ccc|c}
-4 & 1 & & \phantom13 \\
8 & -2 & & -6 \\
\end{array}
Multiplying row 2 by $\frac12$:
\begin{array}{ccc|c}
-4 & 1 & & \phantom13 \\
4 & -1 & & -3 \\
\end{array}
Adding row 1 to row 2:
\begin{array}{ccc|c}
-4 & 1 & & \phantom13 \\
0 & 0 & & 0 \\
\end{array}
thus $x_2$ is a free variable, and:
\begin{align*}
x_1 ={}& \frac34-\frac14s \\
x_2 ={}& 0 + 1s
\end{align*}
Is what I get for my answer.
|
Reading off from your last matrix equation, you have
$$-4x_1 + x_2 = 3 \Rightarrow x_1 = 1/4 x_2 - 3/4$$
Now choose $x_2 = s$. Then $x_2 = 0 + 1s$ is equivalent to making that choice, and substituting $x_2 = s$ gives $x_1 = 1/4s - 3/4$ in the first equation.
$$x_2 = 0 + 1s$$
$$x_1 = - 3/4 + 1/4s$$
Now if $x$ is a vector, $$x = (x_1, x_2) = (-3/4 + 1/4s, 0 + s) = (-3/4, 0) + (1/4, 1)s$$
or written more suggestively,
$$x =
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} =
\begin{bmatrix}
-3/4 + 1/4s \\
0 + s
\end{bmatrix} =
\begin{bmatrix}
-3/4 \\
0
\end{bmatrix} +
\begin{bmatrix}
1/4s \\
s
\end{bmatrix} =
\begin{bmatrix}
-3/4 \\
0
\end{bmatrix} +
\begin{bmatrix}
1/4 \\
1
\end{bmatrix}s $$
Which is in the form requested.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Does $f_n(x) = \dfrac{x^2}{\sqrt{x^2+1/n}}$ converge uniformly to $|x|$ and why? It is easy to see that $f_n(x) = \dfrac{x^2}{\sqrt{x^2+1/n}}$ converge pointwise to $|x|$
Question: is the convergence uniform?
Attempt:
Want to show $\forall \epsilon > 0$, ... , $|f_n(x) - f| < \epsilon$
$|f_n(x) - f| = |\dfrac{x^2}{\sqrt{x^2+1/n}} - \dfrac{x^2}{\sqrt{x^2}}|$
Take $x = \frac{1}{\sqrt{n}}$
$|f_n(x) - f| = |\dfrac{x^2}{\sqrt{x^2+1/n}} - \dfrac{x^2}{\sqrt{x^2}}| = (1-1/\sqrt{2})n$
So $(f_n)$ fails to converge uniformly, can someone confirm?
Also, can someone spot the place where uniform convergence fails for the above sequence in the accompanying figure below?
|
$$\left|\frac{x^2}{\sqrt{x^2+\frac{1}{n}}}-\frac{x^2}{\sqrt{x^2}}\right|=\left|\frac{x^2\left(\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right)}{\sqrt{x^2+\frac{1}{n}}\sqrt{x^2}}\right|\leq \left|\frac{x^2\left(\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right)}{\sqrt{x^2}\sqrt{x^2}}\right|$$
$$=\left|\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right|=\sqrt{x^2+\frac{1}{n}}-\sqrt{x^2}\leq \sqrt{x^2}+\sqrt{\frac{1}{n}}-\sqrt{x^2}=\sqrt{\frac{1}{n}},\,\forall x\in\mathbb R$$
In the last line we used the inequality $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b},\,\forall a,b>0$ for $a=x^2,\, b=\frac{1}{n}$
|
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|
How many ways can you collect six dollars from $8$ people if $6$ people give $0$ or $1$ dollars and $2$ people each give $0$, $1$, or $5$ dollars? I must use a generating function to solve this question:
In how many ways can you collect six dollars from eight people if six people give either $0$ or $1$ dollars and the other two people each give $0$, $1$, or $5$ dollars?
Here is what I have so far.
The generating function is $(1+x)^6(1+x+x^5)^2$. Then let $f(x)= (1+x)^6$, and $g(x) = (1+x+x^5)^2$. We want the coefficient of $x^6$. Then, we note f(x) has the expansion:
$(1+x)^6$ = $1$ + $6 \choose 1$$x$ + $6 \choose 2$$x^2$ + ... + $6 \choose 6$$x^6$
But here is where I get stuck. We can let $h(x) = f(x)g(x)$ where $a_0b_0 + (a_1b_0+a_0b_1)x + (a_2b_0+ a_1b_1 + a_0b_2)x^2 + ... + (a_rb_0+ a_{r-1}b_1 + a_{r-2}b_2 + ... + a_0b_r)x^r$,
But I can't figure out the expansion of $g(x) = $$(1+x+x^5)^2$.
Can someone help me find the expansion of $g(x)$, and help me solve the problem?
|
In general,
$$
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
$$
Thus
\begin{align}
(1+x+x^5)^2&=1+x^2+x^{10}+2(x+x^6+x^5)\\
&=1+2x+x^2+2x^5+2x^6+2x^{10}
\end{align}
and so the coefficient of $x^6$ is
\begin{align}
\binom{6}{6}+2\binom{6}{5}+\binom{6}{4}+2\binom{6}{1}+\binom{6}{0}&=1+2\cdot 6+15+ 2\cdot 6+1\\
&=41.
\end{align}
|
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|
Proving that $1^n+2^n+3^n+4^n$ $(n\in \Bbb N)$ is divisible by 10 when $n$ is not divisible by 4 I was solving some math problems to prepare for math contests and came across this one:
Prove that $1^n+2^n+3^n+4^n$ $(n\in N)$ is divisible by 10 if and only if $n$ is not divisible by 4.
So, from what I understand, we have to prove that:
$1+16^n+81^n+256^n$ $(n\in \Bbb N)$ is never divisible by 10;
$1+2 \cdot 16^n+3 \cdot 81^n+4 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10;
$1+4 \cdot 16^n+9 \cdot 81^n+16 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10;
$1+8 \cdot 16^n+27 \cdot 81^n+64 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10.
The problem is, these equations seem even more complex than the starting one. How would you prove these? Am I on the right track or is there an easier and more elegant way to do this? Thanks.
|
To be slightly more concise:
Lemma
For any prime $p$, $x+x^2+\dots+x^{p-1} \equiv x(1+x+\dots+x^{p-2}) \equiv x(\frac{x^{p-1}-1}{x-1}) \equiv 0 \pmod p$ iff $x \not \equiv 1 \pmod p$
Note the following:
$$1^{n}+2^{n}+3^{n}+4^{n}\equiv 1+0+1+0 \equiv 0 \pmod 2$$
$$1^{n}+2^{n}+3^{n}+4^{n} \equiv 2^{n}+2^{2n}+2^{3n}+2^{4n} \equiv 0 \pmod 5$$
From the fact that $2$ is a primitive root of $5$, and from the Lemma since $n \not \equiv 0 \pmod 4$ .
Thus, we get that $1^n+2^n+3^n+4^n$ is divisble by $10$ if $n \not \equiv 0 \pmod 4$.
Note that a generalization is possible: $1^{n}+2^{n}+\dots+(p-1)^{n} \equiv 0 \pmod {2p} \Leftrightarrow n \not \equiv 0 \pmod {p-1}$
|
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|
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it
and also plz explain me where i have done wrong in above method
Thanks
|
\begin{align}
&\lim_{x\to\infty}\left(\left((x+1)(x+2)(x+3)(x+4)(x+5)\right)^{\frac15}-x\right)\\
&=\lim_{x\to\infty}\cfrac{\left((1+\frac1x)(1+\frac2x)(1+\frac3x)(1+\frac4x)(1+\frac5x)\right)^{\frac15}-1}{\frac1x}\\
&=\lim_{h\to0}\cfrac{\left((1+h)(1+2h)(1+3h)(1+4h)(1+5h)\right)^{\frac15}-1}{h}\\
&=f'(0)\\
f(x)&=\left((1+x)(1+2x)(1+3x)(1+4x)(1+5x)\right)^{\frac15}\\
f(x)^5&=(1+x)(1+2x)(1+3x)(1+4x)(1+5x)\\
\left(f(x)^5\right)'&=5f(x)^4f'(x)\\&=f(x)\left(\frac1{1+x}+\frac2{1+2x}+\frac3{1+3x}+\frac4{1+4x}+\frac5{1+5x}\right)\\
f'(0)&=\frac{\left.\left(f(x)^5\right)'\right|_{x=0}}{5f(0)^4}\\
&=\frac{1+2+3+4+5}{{5f(0)^3}}=\frac{1+2+3+4+5}{5}=3
\end{align}
|
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|
Constant length of segment of tangent
Prove that the segment of the tangent to the curve $y=\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}$ contained between the $y$-axis and the point of tangency has a constant length.
What I have done: First of all I found out slope $m$ of the tangent to the curve at any general point $(h,k)$, which after simplifying comes out to be
$m=\frac{1}{\sqrt{a^2-h^2}}[2h-\frac{a^2}{h}]$
After that I found the equation of the tangent at the general point which comes out to be
$y-k=\left(\frac{1}{\sqrt{a^2-h^2}}(2h-\frac{a^2}{h})\right)(x-h)$
Then I found the $y$-coordinate at $x=0$ which comes out to be $k+\frac{a^2-2h^2}{\sqrt{a^2-h^2}}$
Then I found distance between points $(h,k)$ and $(0,k+\frac{a^2-2h^2}{\sqrt{a^2-h^2}})$ which comes out to be $\sqrt{\frac{3h^4-3a^2h^2+a^4}{a^2-h^2}}$ which shows length of segment depends on $h$ but this was not we had to prove.
Please tell me where I have made the mistake.
|
First, note that the given function $$y = f(x;a) = \frac{a}{2} \log \frac{a + \sqrt{a^2-x^2}}{a - \sqrt{a^2-x^2}} - \sqrt{a^2-x^2}$$ satisfies the relationship $$f(ax;a) = a f(x;1)$$ for $a > 0$; thus, $a$ is a scaling factor for positive reals, and it suffices to consider only the special case $a = 1$. This simplifies the computation considerably. I leave it as an exercise to show that $$\frac{df(x;1)}{dx} = -\frac{\sqrt{1-x^2}}{x}.$$ Consequently, the equation of the tangent line at $(x,y) = (x_0, f(x_0;1))$ is given by $$y - f(x_0;1) = -\frac{\sqrt{1-x_0^2}}{x_0} (x - x_0).$$ The $y$-intercept of this line is therefore $$b = f(x_0;1) + \sqrt{1-x_0^2}.$$ The squared distance of the segment from the point of tangency to the $y$-intercept is $$D(x_0)^2 = x_0^2 + \left(b - f(x_0;1)\right)^2.$$ But $b - f(x_0;1)$ is simply $\sqrt{1-x_0^2}$, so $D(x_0)^2 = 1$ and the distance is constant. When we replace the scale factor $a$, we find that the distance is $a$ for $a > 0$; if $a < 0$, then the distance is $-a$, so we can say $D = |a|$ for all $a \ne 0$.
|
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|
Find the maximum value of $4x − 3y − 2z$ subject to $2x^2 + 3y^2 + 4z^2 = 1.$ Find the maximum value of $4x − 3y − 2z$ subject to $2x^2 + 3y^2 + 4z^2 = 1.$
My Attempt
let $S=4x − 3y − 2z$ and $ t=2x^2 + 3y^2 + 4z^2$. Then $t-s =2x^2 + 3y^2 + 4z^2 -(4x − 3y − 2z)= 2(x-1)^2 + 3(y+1/2)^2+4(z+1/4)^2 -3>=-3$
so $s<=t+3<=4$ as $t=1$ So minimum value of S is 4.
It my reasoning and answer correct? I am novice in this type of problems and not confident that I have arrived a right solution or not. Please help.
|
Hint: CS inequality $\implies$
$$(2x^2+3y^2+4z^2)(8+3+1) \ge (4x-3y-2z)^2$$
|
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|
approximate the root of perturbed polynomial
Approximate the root of $$f(x)=(x-1)(x-1)(x-3)(x-4)-10^{-6}x^6$$ near $r=4$.
Do I have to use iterative method of finding the root, such as Bisection, Secant, etc? Is there other way?
|
If
$f(x)
=(x-1)(x-2)(x-3)(x-4)-10^{-6}x^6
$,
if $c$ is small,
$\begin{array}\\
f(4+c)
&=(3+c)(2+c)(1+c)(c)-10^{-6}(4+c)^6\\
&=6c(1+c/2)(1+c/3)(1+c)-10^{-6}4^6(1+c/4)^6\\
&\approx 6c(1+c/2+c/3+c)-(2/5)^6(1+6c/4)\\
&=6c(1+11c/6)-(2/5)^6(1+3c/2)\\
&=c(6+(3/2)(2/5)^6) +11c^2-(2/5)^6\\
&\approx 6c -(2/5)^6\\
\end{array}
$
If this is zero,
$c = \frac{(2/5)^6}{6}
\approx 0.0006826
$
so the root is about
$4.0006826$,
which agrees very nicely
with Moo's much more accurate answer.
|
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|
Calculating the determinant by upper triangular reduction - can you check if it's correct? Exercise:
Calculate $$\begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
-x & x & 0 & \dots & 0 \\
0 & -x & x & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & x
\end{vmatrix}$$
My approach:
$$\begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
-x & x & 0 & \dots & 0 \\
0 & -x & x & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & x
\end{vmatrix} = x^n \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
-1 & 1 & 0 & \dots & 0 \\
0 & -1 & 1 & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & 1
\end{vmatrix} = x^n \frac {1} {a_0} \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
-a_0 & a_0 & 0 & \dots & 0 \\
0 & -1 & 1 & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & 1
\end{vmatrix} = x^n \frac {1} {a_0} \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
0 & a_0 + a_1 & a_2 & \dots & a_n \\
0 & -1 & 1 & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & 1
\end{vmatrix} = x^n \frac {1} {a_0} \frac {1} {a_0 + a_1} \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
0 & a_0 + a_1 & a_2 & \dots & a_n \\
0 & - a_0 + a_1 & a_0 + a_1 & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & 1
\end{vmatrix} = x^n \frac {1} {a_0} \frac {1} {a_0 + a_1} \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
0 & a_0 + a_1 & a_2 & \dots & a_n \\
0 & 0 & a_0 + a_1 + a_2& \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & 1
\end{vmatrix} = \dots = x^n \frac {1} {(a_0)(a_0 + a_1) \dots (a_0 + a_1 + \dots a_{n-1})} \begin{vmatrix}
a_0 & * & * & * & \dots & * \\
0 & a_0 + a_1 & *&*& \dots & * \\
0 & 0 & a_0 + a_1 + a_2 & * & \dots & * \\
\dots & \dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & 0 & \dots & a_0 + a_1 + \dots a_n
\end{vmatrix} = (a_0 + \dots a_n) x^n$$
|
To verify this by other means, we denote the above determinant as $D_n$ and expand by cofactors along the last column. This yields $D_{n}=xD_{n-1}+x^n a_n$ with base case $D_0 =a_0$. (The $n$-by-$n$ submatrix in the corner yields $(-x)^n$, cancelling the $(-1)^n$ factor from the cofactor expansion.) It can then be readily checked that your result satisfies both this recurrence relation and the base case, so we conclude by mathematical induction that it is correct for all $n$.
|
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|
How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?
My Attempt
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0), (-2,0) are solution of the given equation. Same way solving for $f(0,y)=0$ we get $ y=2 , -1$ are roots and hence (0,2), (0,-1) are solutions. I have tried to factorize $f(x,y)$ or writing it as sum of squares but could not succeed. Is there any other solutions? How do I find them?
|
Rewrite the equation as
$$\frac14\left((x+2y-4)^2 + 3(x-2y)^2-48\right)=0.$$
Setting $a:=x+2y-4$ and $b:=x-2y$, we must have that $a$ and $b$ are integers such that
$$a^2+3b^2=48.\tag1$$
The only integer solutions to (1) satisfy $(|a|,|b|)=(0,4)$ and $(|a|,|b|)=(6,2)$. This gives six possibilities for $a$ and $b$, which yields six solutions: $(x,y)$ = $(4,0)$, $(0,2)$, $(6,2)$, $(4,3)$, $(0,-1)$, and $(-2,0)$.
|
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|
$ p \in Q[x] $ has as a root a fifth primitive root of unity, then every fifth primitive root of unity is a root of $p$. I'm extremely stuck. Can't figure it.
The conjugate is easy: let $w$ be a primitive root of unity, then $w^{-1}$ will also be a root, that's easy. But I'm missing $w^2$ and $w^3$. Why would they be also roots of p?
|
The fifth roots of $1$ form a cyclic group of order $5$, so any of them, different from $1$, is primitive. If $w$ is one of the primitive roots, the others are $\omega^2$, $\omega^3=\omega^{-2}$ and $\omega^4=\omega^{-1}=\bar{\omega}$.
It is clear that $(x-\omega)(x-\omega^2)(x-\omega^3)(x-\omega^4)=x^4+x^3+x^2+x+1$. The minimal polynomial of $\omega$ is a factor of this degree $4$ polynomial, so it must have degree $2$ or $4$ (because a degree $3$ polynomial has a real root).
Thus we have to exclude that $\omega$ has degree $2$. If we assume this, then its minimal polynomial must be $(x-\omega)(x-\bar\omega)=x^2-(\omega+\bar\omega)x+1$, which implies $q=\omega+\bar\omega$ is rational.
Therefore also $(x-\omega^2)(x-\omega^3)$ would have rational coefficients and, indeed,
$$
\omega^2+\omega^3=\omega^2+\frac{1}{\omega^2}=
\left(\omega+\frac{1}{\omega}\right)^2-2=q^2-2
$$
Thus
$$
x^4+x^3+x^2+x+1=(x^2-qx+1)(x^2-(q^2-2)x+1)
$$
The coefficient of the term of degree $1$ in the product can be easily computed as
$$
-q-(q^2-2)
$$
so we have $-q^2-q+2=1$, so $q^2+q-1=0$. This is a contradiction, because $\sqrt{5}$ is not rational.
Therefore this polynomial divides $p$, as soon as $p(\omega)=0$.
|
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|
Area of bounded between $y=x-1$ and $(y-1)^2=4 (x+1)$ Find the area enclosed by curves $y=x-1$ and $(y-1)^2=4 (x+1)$.
I found point of intersection as $(0,-1)$ and $(8,7)$ and set up integral as:
$|\int_{-1}^{7} [(y+1)-\frac{(y-1)^2}{4}-1]dy|$. Am I heading in right direction?
|
Finding the points of intersection of the two curves: $(y-1)^2 = 4(y+2) \Rightarrow y^2-2y+1 - 4y-8 = 0\Rightarrow y^2-6y-7 = 0\Rightarrow (y+1)(y-7) = 0\Rightarrow y = -1, 7\Rightarrow A = \displaystyle \int_{-1}^7 \left(y+1- \left(\dfrac{(y-1)^2}{4}-1\right)\right)dy$. The difference between your answer and mine is the last $-1$ in your answer. I think it should be $+1$.
|
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|
Proof for $\forall x\in \mathbb{Z}^+, \exists t \in \mathbb{Z}, 5 \nmid x \to ((x^2= 5t + 1) \vee (x^2 = 5t – 1))$ I am trying to write a proof for the following: If $x$ is a positive integer that is not divisible by $5$, then $x^2$ can be written either as $x^2 = 5t+1$ or $x^2 = 5t−1$ for some integer $t$. So far, I have managed to get that if $x$ is not divisible by $5$, then it is one of the following cases, where $n$ represents a positive integer:
\begin{align*}
x & = 5n + 1\\
x & = 5n + 2\\
x & = 5n + 3\\
x & = 5n + 4
\end{align*}
I have tried substituting for $x$, but with two different variables $n$ and $t$ and everything, I'm not sure how I would prove they can be equal. Any help would be much appreciated. Thank you!
|
If $x = 5n-1$, then
$$x^2 = (5n-1)(5n-1) = 5n(5n-1) - (5n-1) = 5n(5n-1) - 5n+1 $$
When you divide $x^2$ by $5$, the first two terms above has no remainders, and therefore, $x^2$ modulo $5$ is $1$. (note: $t$ in this case would be equal to $n(5n-1)-n$)
If $x = 5n-2$, then
$$x^2 = (5n-2)(5n-2) = 5n(5n-2) - 2(5n-2) = 5n(5n-2) - 2\cdot 5n + 2\cdot 2 $$
When you divide $x^2$ by $5$, the first two terms above has no remainders, and therefore, $x^2$ modulo $5$ is $-1$.
Similar calculations would show the required result.
|
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|
Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Because $n^2 < 2^n$ by the hypothesis, $n^2 + 2n + 1$ < $2^n + 2n + 1$. As far as I know, the only way I can get $2^{n+1}$ on the right side is to multiply it by $2$, but then I get $2^{n+1} + 4n + 2$ on the right side and don't know how to get rid of the $4n + 2$. Am I on the right track, or should I have gone a different route?
|
There are, of course, many ways to force $2^{n + 1}$ to show up on the RHS. I like to bound the lower order terms as multiples of the leading term by repeatedly applying the given inequality $n \geq 5$ so that I can apply the inductive hypothesis in one shot. Indeed, observe that:
\begin{align*}
(n + 1)^2
&= n^2 + 2n + 1 \\
&< n^2 + 4n + 5 \\
&\leq n^2 + 4n + n &\text{since } n \geq 5 \\
&= n^2 + 5n \\
&\leq n^2 + (n)n &\text{since } n \geq 5 \\
&= 2(n^2) \\
&< 2(2^n) &\text{by the inductive hypothesis} \\
&= 2^{n + 1}
\end{align*}
as desired.
|
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|
Inverse of $4 \times 4 $ matrix Would you be so kind to provide me some examples of $4 \times 4 $ matrices that satisfy the relation $AB = I $
I am completely clueless to find randomly any two matrix $A$ & $B$. Do we have any rules or shortcuts to find them or do we have to do a trial and error method till we get the answer. I hope that's not the right way.
Please shed some light towards this!
|
The standard "beginner's" method for matrix inversion is Gauss-Jordan elimination.
Here are some $4 \times 4$ examples that work out reasonably simply.
$$A = \left[ \begin {array}{cccc} 2&-2&-1&-3\cr 2&-3&0&-3
\cr -1&-1&2&0\cr -1&3&-2&-2
\end {array} \right],\ B = A^{-1} =
\left[ \begin {array}{cccc} -12&11&-9/2&3/2\cr -10&9
&-7/2&3/2\cr -11&10&-7/2&3/2\cr 2&-2
&1/2&-1/2\end {array} \right]$$
$$A = \left[ \begin {array}{cccc} -1&0&1&0\\ 2&0&1&1
\\ -1&1&3&-1\\ 1&0&3&1\end {array}
\right],\ B = A^{-1} =
\left[ \begin {array}{cccc} -2&-1&0&1\\ 6&6&1&-5
\\ -1&-1&0&1\\ 5&4&0&-3
\end {array} \right]
$$
|
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|
Probability that the roots of a quadratic equation are real
Roots of the quadratic equation $x^2+5x+3=0$ are $4\sin^2\alpha+a$ and $4\cos^2\alpha+a$. Another quadratic equation is $x^2+px+q=0$ where $p,q\in\mathbb{N}$ and $p,q\in[1,10]$. Find the probability that the roots of second quadratic equation are real and that they are $4sin^4\alpha+b$ and $4\cos^4\alpha+b$.
$$p^2-4q\ge 0$$
If $p=1$, then no possibilities.
If $p=2$, then $q=1$.
If $p=7,8,9,10$, then $q\in[1,10]$.
But in this way there may be repetitions. I need to find the number quadratic equations first and then I can use the fact the difference in roots for both equations is same to reduce the total possibilities.
|
As roots of the quadratic equation $x^2+5x+3=0$ are $x_1=4\cos^2\alpha+a$ and $x_2=4\sin^2\alpha+a$, so we get the magnitude of difference b/w the roots as
$$\left|x_1-x_2\right|=\left|4\cos^2\alpha-4\sin^2\alpha\right|=\left|\pm\sqrt{5^2-4(3)}\right|=\sqrt{13}\tag{1}$$
Now consider the quadratic equation $x^2+px+q=0$, let its roots be $\delta_1$ and $\delta_2$.
For the roots to be real $D\ge0\implies p^2-4q\ge 0$. Since, $p,q\in \Bbb{N}$, so $p\ge2, q\ge 1$.
The difference of the roots is
$$\left|\delta_1-\delta_2\right|=|\pm\sqrt{p^2-4q}|=\sqrt{p^2-4q}\tag{2}$$
The roots also need be $\delta_1=4\cos^4\alpha+b$ and $\delta_2=4\sin^4\alpha+b$. So $(2)$ becomes
$$|\delta_1-\delta_2|=\sqrt{p^2-4q}=|4\cos^4\alpha-4\sin^4\alpha|=|4\cos^2\alpha-4\sin^2\alpha|=\sqrt{13}\qquad(\text{from (1)})\\
\implies p^2-4q=13$$
And on testing for different value of $p$ and $q$, we find that only the following ordered pair satisfy the required conditions.
$$(p,q)=\{(5,3),(7,9)\}$$
Total no. of ordered pairs $(p,q)$ are $10\times 10=100$ and only two of these satisfy the two given conditions, hence the probability is
$$P=\frac{2}{100}=\boxed{\frac{1}{50}}$$
|
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|
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requires going from:
Expression 1: $$-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2$$
to:
Expression 2: $$(a+b-c)(a-b+c)(a+b+c)(-a+b+c)$$
I can see that these two expressions are equal by working in reverse and multiplying out the second expression to get the first but how does one go from Exp.1 to Exp.2?
|
You can solve it in a simpler way: start from Al Kashi's formula:
$$a^2=b^2+c^2-2bc\cos A,$$
from which you deduce first $4b^2c^2\cos^2A= (b^2+c^2-a^2)^2$, and finally
\begin{align*}
4b^2c^2\sin^2A&=4b^2c^2- (b^2+c^2-a^2)^2\\
&=(2bc -b^2-c^2+a^2)(2bc +b^2+c^2-a^2)\\
&=[a^2-(b-c)^2][(b+c)^2-a^2]\\
&=[(a-b+c)(a+b-c)][(b+c-a)(b+c+a)].
\end{align*}
|
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|
What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I suppose that I can't do it in this way. Why?
EDIT:
This is my solution:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)
\Longleftrightarrow
\sin(2x) - \sin(x) = 0
\Longleftrightarrow
2\cos(\frac{3x}{2})\sin(\frac{x}{2}) = 0
\Longleftrightarrow
\cos(\frac{3x}{2}) = 0 \vee \sin(\frac{x}{2}) = 0$$
Proper solution is $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$
|
The solutions to $\sin(\frac{x}{2})=0$ or $\cos(\frac{3x}{2})=0$ are given by $$\frac{x}{2} = \pi k \text{ or } \frac{3x}{2} = \frac{\pi}{2}+\pi k, k \in \mathbb Z,$$ which can be rewritten as $$x = 2\pi k \text{ or } x = \frac{\pi}{3}+\frac{2\pi}{3}k, k \in \mathbb Z.$$
The solutions to $\cos(x) = \frac{1}{2}$ or $\sin(x)=0$ are given by $$x = \frac{\pi}{3}+2\pi k, \frac{5\pi}{3}+2\pi k, \pi k, k \in \mathbb Z.$$ These are the same sets of points.
|
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|
If $x$, $\{x\}$, $\lfloor x\rfloor$ are in G.P, find $x$.
If $x$, $\{x\}$, $\lfloor x\rfloor$ are in Geometric Progression, find $x$; $x \neq 0$.
Here, $\{x\}=x-\lfloor x\rfloor$
Some properties are pretty evident:
$$0\leq \{x\} < 1 \tag{1}$$
$$\lfloor x \rfloor \leq x\tag{2}$$
But I can't seem to be able to use them to substitute and find an equation for $x$. Any help is appreciated.
|
Let $x=n+y$, $n=\lfloor x\rfloor$ and $y=\{x\}$ Since these numbers are in geometric progression, we have $x/y=y/n$, or
$$\frac {n+y}{y}=\frac yn$$
$$(n+y)n=y^2$$
$$y^2 - n^2 - ny = 0$$
$$y=\frac{n\pm n\sqrt{5}}2$$
$y$ is between 0 and 1, so $n(1\pm\sqrt{5})/2$ has to be between $0$ and $1$.
This is impossible for $(1+\sqrt{5})/2$ as the first positive multiple of 1.618 is more than 1. But for $(1-\sqrt{5})/2\approx-0.618$, we have have $n=-1$
So
$$n=-1\\y=n\frac{1-\sqrt{5}}2=\frac{\sqrt{5}-1}2$$
$$x=n+y=\frac{\sqrt{5}-3}2
$$
So $x=-\phi^{-2}$ and $\{x\}=\phi^{-1}$
|
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|
$\lim_{x \to -1} \frac{|x + 1|} { x^2 + 2x + 1}$? Is the solution: "no limit"
Because:
$$\frac{|x + 1| }{ (x^2 + 2x + 1)} = \frac{1}{(x + 1)}$$
Or is factoring different for $|x + 1|$?
|
You can see that
$$
\frac{|x+1|}{x^2+2x+1}\ne\frac{1}{x+1}
$$
by just plugging in $x=-2$; the left-hand side is
$$
\frac{|-2+1|}{4-4+1}=1
$$
whereas the right-hand side is
$$
\frac{1}{-2+1}=-1
$$
Better, $x^2+2x+1=(x+1)^2=|x+1|^2$, so the correct simplification is
$$
\frac{|x+1|}{x^2+2x+1}=\frac{|x+1|}{|x+1|^2}=\frac{1}{|x+1|}
$$
and so the limit is
$$
\lim_{x\to-1}\frac{|x+1|}{x^2+2x+1}=
\lim_{x\to-1}\frac{1}{|x+1|}=\dots
$$
|
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|
Integrating $\int^b_a(x-a)^3(b-x)^4 \,dx$ I came across a question today...
The value of $\int^b_a(x-a)^3(b-x)^4 \,dx$ is
First I tried the property $\int^b_af(x)=\int^b_af(a+b-x)$. I got $\int^b_a(x-a)^4(b-x)^3 \,dx$, which can be simplified to: $\dfrac{b-a}{2}\int^b_a(x-a)^3(b-x)^3 \,dx$. Well now what? Do I have to open these brackets and do the whole thing or is there a short way (as definite integrals always have)?
|
By integration by parts,
\begin{align}
\int_a^b(x-a)^3 (x-b)^4 dx &= \left[\frac{1}{5}(x-a)^3(x-b)^5\right]_a^b-\int_a^b \frac{3}{5}(x-a)^2(x-b)^5 dx\\
&=-\frac{3}{5}\left(\left[\frac{1}{6}(x-a)^2(x-b)^6\right]_a^b-\int_a^b \frac{1}{3}(x-a)(x-b)^6dx\right)\\
&=\frac{1}{5}\left(\left[\frac{1}{7}(x-a)(x-b)^7\right]_a^b-\int_a^b \frac{1}{7}(x-b)^7 dx\right)\\
&=-\frac{1}{35}\left[\frac{1}{8}(x-b)^8\right]_a^b\\
&=\frac{1}{280}(a-b)^8
\end{align}
|
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|
Find a mistake in integration So, the integral is:
$$I=\int \frac{3x+5}{x^2+4x+8}dx$$
and here is how I did it, but in the end, I got a wrong result:
$$x^2+4x+8=(x+2)^2+4=4\bigg[\bigg(\frac{x+2}{2}\bigg)^2+1\bigg]$$
$$I=\frac{1}{4} \int \frac{3x+5}{\big(\frac{x+2}{2}\big)^2+1}dx$$
substitution: $\frac{x+2}{2}=u$, $dx=2du$, $x=2u-2$
$$I=\frac{1}{2}\int\frac{3(2u-2)+5}{u^2+1}du=\frac{1}{2}\int\frac{6u-1}{u^2+1}du=$$
$$\frac{3}{2}\int\frac{2u-\frac{1}{3}}{u^2+1}du=\frac{3}{2}\int\frac{2u}{u^2+1}du-\frac{1}{2}\int\frac{du}{u^2+1}=$$
$$\frac{3}{2}\ln|u^2+1|-\frac{1}{2}\arctan(u)+C$$
$$I=\frac{3}{2}\ln\bigg|\frac{x^2+4x+8}{4}\bigg|-\frac{1}{2}\arctan\bigg(\frac{x+2}{2}\bigg)+C$$
Thank you for your time.
|
You're doing good. Maybe some passages can be done more easily (at least according to my tastes):
\begin{align}
\int\frac{3x+5}{x^2+4x+8}\,dx
&=\frac{1}{2}\int\frac{6x+10}{x^2+4x+8}\,dx\\[6px]
&=\frac{1}{2}\int\frac{6x+12-2}{x^2+4x+8}\,dx\\[6px]
&=\frac{3}{2}\int\frac{2x+4}{x^2+4x+8}\,dx-
\int\frac{1}{x^2+4x+8}\,dx
\end{align}
The first integral can be written directly as
$$
\frac{3}{2}\log(x^2+4x+8)
$$
and for the second one can do like you did, that is, $2t=x+2$, so $dx=2dt$ and the integral becomes
$$
\int\frac{2}{4t^2+4}=\frac{1}{2}\arctan t=\frac{1}{2}\arctan\frac{x+2}{2}
$$
By delaying the completion of the square we have to deal with less fractions.
Note that
$$
\log\frac{x^2+4x+8}{4}=-\log4+\log(x^2+4x+8),
$$
so the result is the same as yours, because a constant can be absorbed in the constant of integration.
|
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|
Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt:
$\displaystyle \binom{n}{7}=\binom{n}{8} $
$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$
$$ \frac{6}{7!} = \frac{n-7}{40320} $$
$$ n-7 = 48 $$
$$ n=55 $$
|
$\displaystyle \left(2+\frac{x}{3}\right)^{n}=\sum_{i=0} ^n {n \choose k}({\frac{x}{3}})^i 2^{n-i}$
Now for $i =7\implies{n \choose 7}({\frac{x}{3}})^7 2^{n-7}$
and for $i =8\implies{n \choose 8}({\frac{x}{3}})^8 2^{n-8}$
now to get the coefficients of $x^8$ & $x^7$ equal. $\implies$
${n \choose 7}({\frac{1}{3}})^7 2^{n-7}={n \choose 8}({\frac{1}{3}})^8 2^{n-8} \implies \frac{1}{n-7}=\frac{1}{8}* \frac{1}{3}*\frac{1}{2} \implies n=55$
|
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|
General solution expressed in $a_0$ and $a_1$ of a Fibonacci-like sequence?
What is the general solution expressed in $a_0$ and $a_1$ of a Fibonacci-like sequence ?
I mean if $a_0,a_1$ are given and $a_{n+1}:=a_n+a_{n-1}$
$(\begin{array}{cc}a_n&a_{n-1}\end{array})=(\begin{array}{cc}a_{n-1}&a_{n-2}\end{array})\left(\begin{array}{cc}1&1\\1&0\end{array}\right)=(\begin{array}{cc}a_{1}&a_{0}\end{array})\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^{n-1}$
or according to this formula; $a_n=a_1\sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n-k-1}{k}+a_0\sum\limits_{k=0}^{\left\lfloor\frac{n-2}{2}\right\rfloor}\binom{n-k-2}{k}$
|
The general solution will be as follows:
$a_n = \dfrac{\varphi_2a_0 - a_1}{\varphi_2 - \varphi_1}\varphi_1^{n} + \dfrac{a_1 - \varphi_1a_0}{\varphi_2 - \varphi_1}\varphi_2^{n}$, where $\varphi_{1,2} = \dfrac{1 \pm \sqrt{5}}{2}$ -- roots of equation $\varphi^2-\varphi-1 = 0$. You can check it by substituting this formula to the recurrence.
|
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|
proof - $(x,y) = (4,6)$ is the only solution for $x^3 + x^2 - 16 = 2^y$ I saw this question and on seeing the answers I believed it did not have to be so complicated. The pair $(x,y) = (4,6)$ only fits the equation.
Find all possible $(x, y)$ pairs for $x^3 + x^2 - 16 = 2^y$.
My efforts:
$$x^3 + x^2 - 16 = 2^y$$
$$x^3 + x^2 = 2^y + 2^4$$
$$x^2(x+1) = 2^4(2^{y-4} +1)$$
$$x^2(x+1) = 4^2(4^{0.5y-2} + 1)$$
Also note that $\gcd(x^2, x+1) = 1$. I think that this result is important.
Now the RHS and LHS look a lot alike. So I presume that there must be a reason to claim that $x$ or $y$ cannot take more than one value. Its adequate to prove that even one of them won't change. I am not able to find a reason why this must happen.
Any help would be appreciated.
|
$x^2(x+1) = 2^y + 16$
Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.
$$\begin{array}{n|c|c|}
\hline
n & n^2(n+1) & 2^n + 16 \\ \hline
0 & 0 & 17 \\
1 & 2 & 18 \\
2 & 12 & 20 \\
3 & 36 & 24 \\
4 & 80 & 32 \\
5 & 150 & 48 \\
6 & 252 & 80\\ \hline
\end{array}$$
Note that the table indicates that $(x,y) = (4,6)$ is a solution.
So any solution involving $y \ge 7$ will require $x \ge 5$.
We will show that there is no solution for $y \ge 7$.
So we can assume now that $x \ge 5$ and $y \ge 7$.
\begin{align}
x^2(x+1) &= 2^y + 16\\
x^2(x+1) &= 16(2^{y-4} + 1)\\
\end{align}
Note that $2^{y-4}+1$ must be an odd integer.
So if $x$ is an odd integer, $\gcd(x+1,2^{y-4}+1) = 1$.
$\quad$ Hence $x+1 | 16$
$\quad$ Remembering that $x \ge 5$, we must have $x = 7$ or $x = 15$.
$\quad$Case $1: x = 7$
\begin{align}
16(2^{y-4}+1) &= 392\\
2(2^{y-4}+1) &= 49 & \text{Has no solution.}\\
\end{align}
$\quad$Case $2: x = 15$
\begin{align}
16(2^{y-4}+1) &= 3600\\
2^{y-4}+1 &= 225 & \\
2^{y-4} &= 224 \\
2^{y-4} &= 32 \times 7 & \text{Has no solution.}\\
\end{align}
So if $x$ is an even integer, $\gcd(x^2,2^{y-4}+1) = 1$.
$\quad$ So $x^2 | 16$.
This can't happen since $x \ge 5$.
|
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"url": "https://math.stackexchange.com/questions/1689962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve the recurrence relation $a_n = 2a_{n-1} + 2^n$ with $a_0 = 1$ using generating functions Here is what I have so far, or what I know how to do, rather:
I am given this equation: $a_n = 2a_{n-1} + 2^n$ with $a_0 = 1$
So, with the $2a_{n-1}$, I know I can do the following.
We change the $a_n$ to $x^n$, so then we have $x^n = 2^{n-1}$. When we take away $n-1$ from both sides, we get $x = 2$, and the root is 2.
Then when have $a_n = A_12^n$, with $a_0 = 1$, to which then we do:
$1= a_0 = A_12^0 = A_1 = 1$, and $a_n = 1*2^n$.
But, that is not the correct answer, and I am not sure what to do with the $2^n$ part, which is necessary to get the answer. Could anybody help me? Any help is greatly appreciated.
|
Let
$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$$
We know for a fact that $\frac{1}{1-x}=1+x+x^2+\cdots$ so we can compute
\begin{align}
2xf(x)+\frac{1}{1-2x}&=2a_0x+2a_1x^2+2a_2x^3+\cdots+1+(2x)+(2x)^2+\cdots\\
&=1+(2a_0+2)x+(2a_1+2^2)x^2+(2a_2+2^3)x^3+\cdots\\
&=a_0+a_1x+a_2x^2+a_3x^3+\cdots\\
&=f(x)
\end{align}
Now we get a functional equation $2xf(x)+\frac{1}{1-2x}=f(x)$ which we can solve to get $$f(x)=\frac{-1}{(1-2x)(2x-1)}=\frac{1}{(1-2x)^2}=(1-2x)^{-2}$$
and now we make the Taylor series for that, and we see that
\begin{align}
f(x)&=(1-2x)^{-2}\\
f'(x)&=(-2)(-2)(1-2x)^{-3}=2\cdot2\cdot(1-2x)^{-3}\\
f''(x)&=(-2\cdot2)\cdot(-3\cdot2)\cdot(1-2x)^{-4}=2^2\cdot3!\cdot(1-2x)^{-4}\\
&\vdots\\
f^{(n)}&=2^n\cdot(n+1)!\cdot(1-2x)^{-n-2}
\end{align}
so that the coefficients in the Taylor series are $a_n=\frac{f^{(n)}(0)}{n!}=(n+1)2^n$.
So we get
$$a_n=(n+1)2^n$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1690386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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|
Use of residues to find I=$\int_0^\infty \frac{\sin^2(x)}{1+x^4} dx$ I'm working on the problem $$I=\int_0^\infty \frac{\sin^2(x)}{1+x^4} dx$$
I found 4 singularities and i would like to use the singularities in the 1st and 2nd quadrants to solve this integral; i.e. $z_0=e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$. My prof. said we should split this into two integrals using the trig identity $\sin^2(x)=\frac{1-\cos(2x)}{2}$ so we have
$$I=\frac{1}4 \int_{-\infty}^{\infty}\frac{dx}{1+x^4}- \frac{1}4\int_{-\infty}^{\infty}\frac{\cos(2x)}{1+x^4}$$
Or
$$I=\frac{2\pi i}{4}Res[\frac{1}{1+z^4}, z_0=e^{\frac{\pi i}{4}}, e^{\frac{3\pi i}{4}}]-\frac{2\pi i}{4}Res[\frac{e^{2ix}}{1+z^4},z_0=e^{\frac{\pi i}{4}},e^{\frac{3\pi i}{4}}]$$
I didn't have any trouble with the 1st residue but i'm a bit unsure how to handle the second residue. Should i let $e^{2ix}=e^{2iz}$ then substitute $z=z_0$? In that case I seems to get a really messy answer. Is there a cleaner way to deal with that second residue? I should add that I am a physics major and this is a problem for my Mathematical Methods class. I haven't taken any complex variable courses.
Thank you.
|
Let's start at the beginning because it appears that, although you have the correct formulation, you really do not seem to understand what is going on.(For example, why not use the singularities in the 3rd and 4th quadrants?)
Consider the contour integral:
$$\oint_C dz \frac{1-e^{i 2 z}}{1+z^4} $$
where $C$ is a semicircle of radius $R$ in the upper half plane. The contour integral is equal to
$$\int_{-R}^R dx \frac{1-e^{i 2 x}}{1+x^4} + i R \int_0^{\pi} d\theta \, e^{i \theta} \, \frac{1-e^{i 2 R e^{i \theta}}}{1+R^4 e^{i 4 \theta}}$$
The first integral is equal to
$$\int_{-R}^R dx \frac{1-\cos{2 x}}{1+x^4} $$
We now show that the magnitude the second integral vanishes as $R \to \infty$. We do this by using the inequality $\sin{\phi} \ge 2 \phi/\pi$ when $\phi \in [0,\pi/2]$. The magnitude of the second integral is then bounded by
$$\begin{align}\left |i R \int_0^{\pi} d\theta \, e^{i \theta} \, \frac{1-e^{i 2 R e^{i \theta}}}{1+R^4 e^{i 4 \theta}} \right | &\le R \int_0^{\pi} d\theta \, \left | \frac{1}{1+R^4 e^{i 4 \theta}} \right | + R \int_0^{\pi} d\theta \, \left | \frac{e^{i 2 R e^{i \theta}}}{1+R^4 e^{i 4 \theta}} \right |\\ &\le \frac{\pi R}{R^4-1} + \frac{2 R}{R^4-1} \int_0^{\pi/2} d\theta \, e^{-2 R \sin{\theta}} \\ &\le \frac{\pi R}{R^4-1} + \frac{2 R}{R^4-1} \int_0^{\pi/2} d\theta \, e^{-2 R (2 \theta/\pi)}\\ &\le \frac{\pi (R+1/2)}{R^4-1}\end{align} $$
which vanishes as $R \to \infty$.
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand inside $C$, or $z=e^{i \pi/4}$ and $z=e^{i 3 \pi/4}$. Thus,
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{1-\cos{2 x}}{1+x^4} &= i 2 \pi \left [\frac{1-e^{i 2 e^{i \pi/4}}}{4 e^{i 3 \pi/4}} + \frac{1-e^{i 2 e^{i 3 \pi/4}}}{4 e^{i 9 \pi/4}} \right ] \\ &= i \frac{\pi}{2} \left [-i \sqrt{2} - e^{-i 3 \pi/4} e^{i \sqrt{2}} e^{-\sqrt{2}} - e^{-i \pi/4} e^{-i \sqrt{2}} e^{-\sqrt{2}} \right ] \\ &= i \frac{\pi}{2} \left [-i \sqrt{2} + e^{i \pi/4} e^{i \sqrt{2}} e^{-\sqrt{2}} - e^{-i \pi/4} e^{-i \sqrt{2}} e^{-\sqrt{2}} \right ] \\ &= i \frac{\pi}{2} \left [-i \sqrt{2} + i 2 \operatorname{Im}{\left ( e^{i \pi/4} e^{i \sqrt{2}} e^{-\sqrt{2}}\right )} \right ] \\ &= \frac{\pi}{\sqrt{2}}-\pi e^{-\sqrt{2}} \sin{\left (\sqrt{2} + \frac{\pi}{4} \right )} \end{align}$$
Thus,
$$\int_0^{\infty} dx \frac{\sin^2{x}}{1+x^4} = \frac{\pi}{4 \sqrt{2}}-\frac{\pi}{4} e^{-\sqrt{2}} \sin{\left (\sqrt{2} + \frac{\pi}{4} \right )}$$
which can also be written as
$$\int_0^{\infty} dx \frac{\sin^2{x}}{1+x^4} = \frac{\pi}{4 \sqrt{2}} \left ( 1 -e^{-\sqrt{2}} \left [ \sin{\left (\sqrt{2} \right )} +\cos{\left (\sqrt{2} \right )} \right ] \right )$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1690986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?
It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
|
$\;\;\displaystyle\frac{1}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\sin x+\cos x}\right)$
where $1=A(\sin x+\cos x)^2+B(1-\sin x\cos x)=(A+B)+(2A-B)\sin x\cos x$.
Then $A=\frac{1}{3}$ and $B=\frac{2}{3}$,
so $\displaystyle\int\frac{1}{\sin^3 x+\cos^3 x}dx=\int\left(\frac{1}{3}\cdot\frac{\sin x+\cos x}{1-\sin x\cos x}+\frac{2}{3}\cdot\frac{1}{\sin x+\cos x}\right)dx$
$\displaystyle=\frac{1}{3}\int\frac{2\sin x+2\cos x}{2-2\sin x\cos x}dx+\frac{2}{3}\int\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx$
$\displaystyle=\frac{1}{3}\int\frac{\sin x+\cos x}{1+(\sin x-\cos x)^2}dx+\frac{2}{3\sqrt{2}}\int\csc\left(x+\frac{\pi}{4}\right)dx$
$\displaystyle=\frac{1}{3}\arctan(\sin x-\cos x)+\frac{2}{3\sqrt{2}}\ln\big|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\big|+C$
$\displaystyle=\frac{1}{3}\arctan(\sin x-\cos x)+\frac{2}{3\sqrt{2}}\ln\left\vert\frac{\sqrt{2}-\cos x+\sin x}{\sin x+\cos x}\right\vert+C$
|
{
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"url": "https://math.stackexchange.com/questions/1694424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Proof using strong induction a conjecture about $4^n$ Compute $4^1$, $4^2$, $4^3$, $4^4$, $4^5$, $4^6$, $4^7$, and $4^8$. Make a conjecture about the units digit of $4^n$ where $n$ is a positive integer. Use strong mathematical induction to prove your conjecture.
Solution:
$4^1$ = $4$
$4^2$ = $16$
$4^3$ = $64$
$4^4$ = $256$
$4^5$ = $1024$
$4^6$ = $4096$
$4^7$ = $16384$
$4^8$ = $65536$
Conjecture: For $n>0$, the unit digit of $4^n$ is $4$, if $n$ is odd and is $6$, if $n$ is even.
*
*Basis:
For $n = 1 \to 4^1 = 4$
For $n = 2 \to 4^2 = 16$
The conjecture is right for n = 1 and n = 2
*Inductive step:
$4^{k+1} = 4\cdot 4^k$
If $k$ is odd ($k+1$ is even), the unit digit of $4^k$ is $4$, but if we multiply a number that the unit digit is $4$ with a number that the unit digit is $6$ we get a number that the unit digit is $6$, because $4\times 4 = 16$.
If $k$ is even ($k+1$ is odd), the unit digit of $4^k$ is $6$, but if we multiply a number that the unit digit is $6$ with a number that the unit digit is $4$ we get a number that the unit digit is $4$, because $ 6\times 4 = 24$.
So, the unit digit of $4^{k+1}$ is $4$, if $n$ is odd and is $6$, if $n$ is even.
|
To be proven: If $n$ is a positive integer, the units digit of $4^n$ is $4$ if $n$ is odd and $6$ if $n$ is even.
Proof by strong induction: (Base cases $n=1$ and $n=2$.)
Now assume that $k\ge 3$ and the result is true for all smaller positive values of $k$. The goal is to prove the theorem for $n=k$.
Let $\ell =k-2$; note that $\ell\ge 1$.
If $k$ is odd, then $\ell$ is odd, and strong induction implies that the units digit of $4^\ell$ is $4$. Then the units digit of $4^k$ is the units digit of $4^\ell \cdot 4^2$, which is the units digit of $4 \cdot 16$, which is $4$.
If $k$ is even, then $\ell$ is even, and strong induction implies that the units digit of $4^\ell$ is $6$. Then the units digit of $4^k$ is the units digit of $4^\ell \cdot 4^2$, which is the units digit of $6 \cdot 16$, which is $6$.
Thus, we have proven that the units digit of $4^k$ is $4$ if $k$ is odd and $6$ if $k$ is even. This proves the result.
|
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"url": "https://math.stackexchange.com/questions/1696365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
the system of diophantine equations: $x+y=a^3$; $xy=\dfrac{a^6-b^3}{3}$ has only trivial solutions. Without using Fermat's Last Theorem, how can one prove that the following system of diophantine equations has only trivial solutions:
$$x+y=a^3$$
$$xy=\dfrac{a^6-b^3}{3}$$
We suppose of course that $\gcd(x,y)=\gcd(a,x)=\gcd(a,y)=\gcd(a,b)=1$
|
$
\begin{cases}
1)&x = a^3 - y \\
2)&xy = \frac{a^6 - b^3}{3} \\
\end {cases}
$
First, subsitute $1)$ into $2)$
$(a^3 - y)y = \frac{a^6 - b^3}{3}$
$3y\left(a^3-y\right)=a^6-b^3$
$3a^3y-3y^2=a^6-b^3$
$-3y^2+3a^3y-a^6+b^3=0$
$y = \frac{-3a^3\pm \sqrt{-3a^6+12b^3}}{2\left(-3\right)}$
$y = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$
Substitute $y = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$ into $1)$
$x = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$
Subsitute $x = \frac{3a^3+\sqrt{12b^3-3a^6}}{6}$
and $y = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$ into $1)$
$\frac{3a^3\pm\sqrt{12b^3-3a^6}}{6} + \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6} = a^3$
$\:\frac{3a^3\pm \sqrt{12b^3-3a^6}}{3}\:\:=\:a^3$
$12b^3 = 3a^6 \implies 4b^3 = a^6$
$GCD(a,b) = 1$, Thus there are only trivial solutions $(a = 0, b = 0)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
it suffices to
$$4\sqrt{abcd}+64abcd\ge 1$$
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Here is a way using Schur and smoothing. Let $a$ be the minimum among $a, b, c, d$ and say $3p = b+c+d, \,q = bc+cd+db, \,r = bcd$. We define
$$F(a, b, c, d) = 3\sum_{cyc} a^2+64abcd $$
We will first show $F(a, b, c, d) \ge F(a, p, p, p)$.
$$\iff 3(b^2+c^2+d^2-3p^2) \ge 64a(p^3-r)$$
But $b^2+c^2+d^2-3p^2 = 2(3p^2-q)$, so we have to show
$$\iff 3(3p^2-q) \ge 32a(p^3-r) \tag{1}$$
By Schur inequality
$$\sum_{cyc} b^3+3bcd \ge \sum_{cyc} bc (b+c) \implies \frac43p(3p^2-q) \ge p^3-r$$
so it is enough for $(1)$ to show
$$3(3p^2-q) \ge 32a\cdot \frac43p(3p^2-q) \iff (3p^2-q)(9-128ap) \ge 0 $$
Now $3p^2 \ge q$ is easily shown and $9-128ap = 9-128(1-3p)p > 0 $ as $p \ge \frac14$ for $a$ to be the minimum. Hence we have shown $(1)$ holds true.
Now all that is needed is to show $F(1-3p, p, p, p) \ge 1$.
$$\iff 3((1-3p)^2+3p^2)+64(1-3p)p^3 \ge 1$$
$$\iff 2(1+2p)(1-3p)(1-4p)^2 \ge 0$$
which is now obvious. Equality is iff $3p=1$, i.e. $a=0, b=c=d=\frac13$ or when $4p=1$, i.e. $a=b=c=d=\frac14$.
|
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|
The best way of integrating irrational functions Ok, so, here is the example integral:
$$I=\int\frac{x-2-\sqrt{-x^2-4x+4}}{x^2-\sqrt{-x^2-4x+4}}dx$$
I always solve these types of integrals using Euler's substitutions, but, recently, I came across some more difficult integrals like the one above which, after using Euler's substitution, become the integral of a rational function which is too complex to calculate, or at least to me.
So, what I did with the integral above is use second Euler's substitution:
$$\sqrt{ax^2+bx+c}=x\cdot t \pm \sqrt{c}$$
which gives:
$$\sqrt{-x^2-4x+4}=x\cdot t-2$$
Then, we have:
$$x=\frac{4-2t}{1-t^2},\ \ \ dx=\frac{2(t^2-1)+2t(4-2t)}{(1-t^2)^2}dt$$
After the substitution, the integral becomes:
$$I=\int \frac{\frac{4-2t}{1-t^2}-2-\frac{4-2t}{1-t^2}\cdot t+2}{\frac{(4-2t)^2}{(1-t^2)^2}-\frac{4-2t}{1-t^2}\cdot t+2}\cdot \frac{2(t^2-1)+2t(4-2t)}{(1-t^2)^2}dt$$
Which, after some calculation is:
$$I=\int\frac{8-40t+24t^3-8t^4}{2t^3-6t^2+14t-14}dt$$
Then, I can divide them and get two integrals, one from the table and the other one of a rational function with numerator's degree lower than denominator's.
Then, I should integrate that remaining rational function, but I'm not sure how, since I can't factor the polynomial in denominator.
My question is, is there a simpler way to calculate integrals like this one and if there isn't, what do I do with the remaining integral of rational function? What if I get a polynomial of degree $5$, $6$, $7$ etc. in denominator?
Thank you for your time.
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There is a mistake in your computation. I added my comments here since there is a limited number of characters I can put into the comment window.
\begin{equation}
x=\frac{4-2t}{1-t^2},\ \ \ dx=\frac{-2(t^2-4 t+1)}{(1-t^2)^2}dt
\end{equation}
Then after the substitution, the integral becomes:
\begin{equation}
I=\int \frac{\frac{4-2t}{1-t^2}-2-\frac{4-2t}{1-t^2}\cdot t+2}{\frac{(4-2t)^2}{(1-t^2)^2}-\frac{4-2t}{1-t^2}\cdot t+2}\cdot \frac{-2(t^2-4 t+1)}{(1-t^2)^2}dt
\end{equation}
Which, after some calculation is:
\begin{equation}
I=\int \frac{2 (t-2) ((t-4) t+1)}{(t+1) (t (t-2) (2 t+5)+9)} dt
\end{equation}
Then, I can divide them and get two integrals:
\begin{equation}
I=\int \frac{-2}{(t+1)} dt + \int \frac{6 t^2-16 t+14}{2 t^3+t^2-10 t+9} dt =-2 \log (t+1) +\int \frac{6 t^2+2 t -10}{2 t^3+t^2-10 t+9}+\int \frac{24- 18 t}{2 t^3+t^2-10 t+9}=-2 \log (t+1)+\log (2 t^3+t^2-10 t+9)+2 \int \frac{12- 9 t}{2 t^3+t^2-10 t+9}
\end{equation}
|
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|
calculate $\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}$? prove the following equation :
$$\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}={s+2r-1\choose r} - {s+2r-1\choose r-1}$$
|
Suppose we seek to verify that
$$\frac{1}{r+1} \sum_{q=0}^r (q+1) {2r-q\choose r-q}
{s-2+q\choose q} = {s+2r-1\choose r} - {s+2r-1\choose r-1}.$$
Introduce the integral representation
$${2r-q\choose r-q}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{2r-q}}{w^{r-q+1}}
\; dw$$
Note that this is zero when $q\gt r$ so we may extend $q$ to infinity,
getting for the sum
$$\frac{1}{r+1} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{2r}}{w^{r+1}}
\sum_{q\ge 0} (q+1)
{s-2+q\choose q} \frac{w^q}{(1+w)^q}
\; dw.$$
Now observe that
$$z \times \frac{1}{(1-z)^{s-1}}
= \sum_{q\ge 0} {s-2+q\choose q} z^{q+1}$$
and hence
$$\frac{1}{(1-z)^{s-1}} + (s-1) \frac{z}{(1-z)^s}
= \sum_{q\ge 0} (q+1) {s-2+q\choose q} z^q.$$
Evaluating this at $z=w/(1+w)$ so that $1-z = 1/(1+w)$ yields
$$(1+w)^{s-1} + (s-1) w (1+w)^{s-1}.$$
Substitute this into the integral to obtain
$$\frac{1}{r+1} {s+2r-1\choose r}
+ \frac{s-1}{r+1} {s+2r-1\choose r-1}.$$
This is
$$\frac{1}{r+1} {s+2r-1\choose r}
+ \frac{s+r}{r+1} {s+2r-1\choose r-1}
- {s+2r-1\choose r-1}.$$
Now
$$\frac{s+r}{r+1} {s+2r-1\choose r-1}
= \frac{1}{r+1} \frac{(s+2r-1)!}{(s+r-1)! (r-1)!}
\\ = \frac{r}{r+1} \frac{(s+2r-1)!}{(s+r-1)! r!}
= \frac{r}{r+1} {s+2r-1\choose r}.$$
Collecting everything we obtain
$$\frac{1}{r+1} {s+2r-1\choose r}
+ \frac{r}{r+1} {s+2r-1\choose r}
- {s+2r-1\choose r-1}
\\ = {s+2r-1\choose r}
- {s+2r-1\choose r-1}$$
as claimed.
|
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|
find $\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$ find $\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$
I had $(1-x)^{-\frac{p}{q}}$ in mind.
$$S=\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$
$$S+1=1+\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$
$$S+1=1+\frac{3}{2!\cdot3}+\frac{3\cdot(3+2)}{3!\cdot9}+\frac{3\cdot(3+2)\cdot(3+4)}{4!\cdot27}+\cdots$$
$$S+1=1+\frac{3}{2!}\left(\frac{\frac{2}{3}}{2}\right)+\frac{3\cdot(3+2)}{3!}\left(\frac{\frac{2}{3}}{2}\right)^2+\frac{3\cdot(3+2)\cdot(3+4)}{4!}\left(\frac{\frac{2}{3}}{2}\right)^3+\cdots$$
$$S+1=\left(1-\frac{2}{3}\right)^\frac{-3}{2}$$
I got $S=3\sqrt{3}-1$
But answer given is $S=3\sqrt{3}-4$
|
In the last step, you miss some multiple of $3$, and you miss one term of the expansion.
$$
S=\sum_{n\geq 1} \frac{(2n+1)!!}{(n+1)!3^n}=\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\frac{(-2)^{n+1}}{3^n}=3\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\left(\frac{-2}{3}\right)^{n+1}=3\left(\left(1-\frac{2}{3}\right)^{-\frac{1}{2}}-1-\frac{1}{3}\right)=3\sqrt{3}-4.
$$
|
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|
How was this factoring of $1-2x-x^2$ achieved? If I factor $1-2x-x^2$ using the quadratic formula I get
$$x=\frac{2\pm \sqrt{4-4(-1)(1)}}{2(-1)}$$
$$x=\frac{2\pm \sqrt{8}}{-2}$$
$$x=-1 \pm \sqrt{2}$$
Let $\alpha = -1 +\sqrt{2}$ and $\beta=-1-\sqrt{2}$.
So $1-2x-x^2=-(x-\alpha)(x-\beta)$.
In the image below, where did $1-\alpha x$ and $1-\beta x$ come from?
|
Your $\alpha$ and $\beta$ are not the same as the $\alpha$ and $\beta$
in the image you copied.
In the image, they are factoring
$$1−2x−x^2 = -\left(\frac 1\alpha - x\right)\left(\frac 1\beta - x\right).$$
You can confirm that using their values
$\alpha = 1 + \sqrt2$ and $\beta = 1-\sqrt2$,
we have
\begin{align}
\frac 1\alpha &= \frac 1{1 + \sqrt2} = -1+\sqrt2, \\
\frac 1\beta &= \frac 1{1 - \sqrt2} = -1-\sqrt2, \\
\end{align}
which matches the factorization you found
(roots at $x = -1+\sqrt2$ and $x = -1-\sqrt2$).
|
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|
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
|
I'm not sure why you're using this complicated method. If you have an equation of the form
$$
a\sin^2x+b\sin x\cos x+c\cos^2x=d
$$
you can just observe that this is equivalent to
$$
a\sin^2x+b\sin x\cos x+c\cos^2x=d\sin^2x+d\cos^2x
$$
so it becomes
$$
(a-d)\sin^2x+b\sin x\cos x+(c-d)\cos^2x=0
$$
This is Roman83's solution, but I'd like to discuss it in general, then comparing it with your method.
Note that $\cos x=0$ gives a solution if and only if $a-d=0$, in which case you get the equation
$$
\cos x(b\sin x+(c-d)\cos x)=0
$$
which is easy. If $a-d\ne0$, you can divide both sides by $\cos^2x$ and get
$$
(a-d)\tan^2x+b\tan x+(c-d)=0
$$
which is a quadratic in $\tan x$.
Your method leads to the same quadratic, but doing a lot of work. First you observe that
$$
\sin^2x=\frac{1-\cos2x}{2},\qquad
\cos^2x=\frac{1+\cos2x}{2}
$$
so you get, after multiplying by $2$ both sides,
$$
a(1-\cos2x)+2b\sin x\cos x+c(1+\cos2x)=2d
$$
hence
$$
(c-a)\cos2x+b\sin2x+a+c-2d=0
$$
Now the Weierstrass substitution gives, setting $t=\tan x$,
$$
(c-a)\frac{1-t^2}{1+t^2}+b\frac{2t}{1+t^2}+a+c-2d=0
$$
that becomes
$$
(c-a)-(c-a)t^2+2bt+(a+c-2d)+(a+c-2d)t^2=0
$$
and, reordering terms,
$$
2(a-d)t^2+2bt+2(c-d)=0
$$
|
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|
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$
I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two binomials,but i could not.And i do not have any other method to solve it.
|
Let $p(x)=1+x+2x^2+3x^3$ and $q(x)=1+x+2x^2+3x^3+4x^4$. Then
$$(q(x))^4-(p(x))^4=(q(x)-p(x))A(x)$$
for some polynomial $A(x)$. But $q(x)-p(x)=4x^4$, so $(q(x))^4-(p(x))^4$ is divisible by $x^4$. It follows that all coefficients of $x^k$ for $k\le 3$ are $0$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$. If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$.Find $k$
Let $\frac{\pi}{32}=\theta$
$\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\frac{1}{\sin\theta}+\frac{1}{\sin2\theta}+\frac{1}{\sin4\theta}+\frac{1}{\sin8\theta}+\frac{1}{\sin16\theta}$
Now i am stuck.
|
We have the formula ie $\sin(2x)=2sinxcosx$ so we know value of $sin(45)$ substitute $45=2x$ so youll get $sin(22.5)=\pi/8$ . thus solving double angle identity many times you get all sines . then the work is almost done. To convert cos to sin use $sin^2(x)+cos^2(x)=1$
|
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|
If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$
If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$.
What I did is to subtract I from both sides of the equation:
$$A^{27}-I=A^{64}-I=0$$
then:
\begin{align*}
A^{27}-I &= (A-I)(A+A^2+A^3+\dots+A^{26})=0\\
A^{64}-I &= (A-I)(A+A^2+A^3+\dots+A^{63})=0.
\end{align*}
So from what I understand, either $A=I$ (as needed) or $A+A^2+A^3+\dots+A^{26}=0$ or $A+A^2+A^3+\dots+A^{63}=0$.
At this point I got stuck. By the way, I found out that $A$ is an invertible matrix because if $A^{27}=I$ then also $A^{26}A=AA^{26}=I$ then $A^{26}=A^{-1}$.
Also I thought to use the contradiction proving by assuming that $A+A^2+A^3+\dots+A^{63}=0$, but because $A^{27}=I$, then:
$$A+A^2+A^3+\dots+A^{26}+I+A^{28}+\dots+A^{53}+I+A^{55}+\dots+A^{63}=0$$
but yet nothing.
Would appreciate your guidance, thanks!
|
\begin{align*}
& I = A^{64} = A^{2(27) + 10} = (A^{27})^2A^{10} = A^{10}\\
\implies & I = A^{27} = A^{2(10) + 7} = (A^{10})^2A^7 = A^7\\
\implies & I = A^{10} = A^{1(7)+3} = (A^7)^1A^3 = A^3\\
\implies & I = A^7 = A^{2(3) + 1} = (A^3)^2A = A.
\end{align*}
This is nothing more than the Euclidean algorithm applied to the exponents. The same procedure can be used to show that if $A^p = I$ and $A^q = I$ with $p$ and $q$ coprime, then $A = I$.
|
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|
How many solutions in integers to the following equation
What is the number of positive integer solutions $(a, b)$ to $2016 + a^2 = b^2$?
We have,
$2016 = (b-a)(b+a) = 2^5 \cdot 3^2 \cdot 7$
$b - a = 2^{t_1} 3^{t_2} 7^{t_4}$ and
$a - b = 2^{t_5} 3^{t_6} 7^{t_7}$
Thus, we must have $t_1 + t_5 = 5$ and $t_2 + t_6 = 2$ and $t_4 + t_7 = 1$.
Total, $\binom{6}{1} \cdot\binom{3}{1} \cdot\binom{2}{1} = 36$.
But the actual answer is $12$, where am I going wrong?
|
Just to close the problem out, since $t_1$ can not be either $0$ or $5$ (by parity), there are only $24$ possible choices of the $t_i$. Inspection shows that $b$ is always positive, but that we switch the sign of $a$ when we exchange the choices $\{t_1,t_2,t_4\},\;\{t_5,t_6,t_7\}$. Thus we can reject exactly half of the sextuples, getting the desired answer of $\fbox {12}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
|
Let $t=\sin^2(x)$.
$$2(t^3+(1-t)^3)-3(t^2+(1-t)^2)+1=6t^2-6t+2-6t^2+6t-3+1=0.$$
|
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|
Is the intersection of $x^2 + y^2 + z^2 = 1$ and $x = \frac{1}{2}$ a manifold in $\mathbb{R}^3$? Is the intersection of $x^2 + y^2 + z^2 = 1$ and $x = \frac{1}{2}$ a manifold in $\mathbb{R}^3$?
I think that it is, because it can be parameterized by $f(x) = (\frac{1}{2},\sqrt{\frac{3}{4}} \cos x, \sqrt{\frac{3}{4}} \sin x)$ for $0 < x < 2\pi$.
Is this correct?
|
Let
$$f:\Bbb R^3\to\Bbb R^2,\quad f(x,y,z)=(x^2+y^2+z^2-1,x-1/2).$$
Then, your space is $f^{-1}(0)$. Now,
$$df=\begin{pmatrix}
2x & 2y & 2z \\
1 & 0 & 0
\end{pmatrix}
$$
which has full rank for all $(x,y,z)\in f^{-1}(0)$, so yes it is a manifold.
|
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|
The Diophantine Equation: $x^3-3=k(x-3)$ I wish to know how to resolve the diophantine equation: $x^3-3=k(x-3)$ ? The problem is:
Find all integers $x\ne3$ such that $x-3\mid x^3-3$.
- From 250 Problem's in Elementary Number Theory, by W. Sierpinski
I have found the solutions to this problem by using rules in divisibility of numbers, as below:
$$x-3\mid x-3 \Rightarrow x-3\mid x^3 - 3x^2$$
$$x-3\mid x^3-3 \quad and \quad x-3 \mid x^3-3x^2$$
Thus, $x-3$ divides their difference:
$$\Rightarrow x-3\mid 3x^2-3 $$
On the other hand:
$$x-3\mid x-3 \Rightarrow x-3 \mid 3x^2-9x$$
Hence their difference is divisible by $x-3$:
$$\Rightarrow x-3\mid 9x-3 \quad and \quad x-3\mid 9x-27$$
$$\Rightarrow x-3\mid 24 \Rightarrow 24=k(x-3)\Rightarrow x=\frac{24}{k}+3$$
So we have the general solution to find the integer values of $x$. But I did not manage to solve the diophantine equation related to the problem: $$x^3-3=k(x-3)$$
Any hints and helps about this equation?
-Thank you in advance.
|
You made a mistake at the beginning. $(x^3-3)-(x^3-3x^2)=3x^2-3$, not just $3x^2$. Thus, you have $(x-3) \mid (9x-3)$ by subtracting from $3x^2-9x$ and then $(x-3) \mid 24$ by subtracting from $9x-27$.
Here is the set of all of the integer divisors of 24:
$$\{-24, -12, -8, -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6, 8, 12, 24\}$$
This is the set of all possible $x-3$. Therefore, we can add all of the numbers by $3$ to get the set of all possible $x$. Thus, the following is the solution set:
$$\{-21, -9, -5, -3, -1, 0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 15, 27\}$$
|
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|
Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac{2}{x\cos 2x}-\frac{1}{x^2}\log(\frac{1+\tan x}{1-\tan x}))}{2x}$
I am stuck here.
|
You may observe that, by the Taylor expansion, you get, as $x \to 0$,
$$
\begin{align}
\tan x&=x+\frac{x^3}3+O(x^5) \tag1
\\ \frac{1+\tan x}{1-\tan x}&=1+2 x+2 x^2+\frac{8 x^3}{3}+O(x^5) \tag2
\end{align}
$$ giving
$$
\begin{align}
\log\left(\frac{1+\tan x}{1-\tan x}\right)&=2 x+\frac{4 x^3}{3}+O(x^5) \tag3
\\\\\frac1x\log\left(\frac{1+\tan x}{1-\tan x}\right)&=2 +\frac{4 x^2}{3}+O(x^4) \tag4.
\end{align}
$$ Then
$$
\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\large \frac1x}-e^2=e^{\large \frac1x\log\left(\frac{1+\tan x}{1-\tan x}\right)}-e^2 \tag5
$$ rewrites
$$
\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\large \frac1x}-e^2=e^2 \left(e^{\large \frac{4 x^2}{3}+O(x^4)}-1\right)=e^2 \left(\frac{4 x^2}{3}+O(x^4)\right)\tag6
$$
yielding, as $x \to 0$,
$$
\lim_{x\to 0}\frac{\left(\tan\left(\frac{\pi}4+x\right)\right)^{\large \frac1x}-e^2}{x^2}
= \frac{4e^2}3. \tag7
$$
|
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|
Question Regarding Integration Within Summation
It expresses an integral within a summation procedure.
|
$$\sum _{1}^{\infty}\log (\frac{k+a+1}{k+a})-\log(\frac{1+k+b}{k+b})$$
$$=\sum_1^\infty \log\frac{k+b(k+a+1)}{k+a(k+b+1)}$$
now sum it over
and expanding the expression gives
$$=\lim _{k \to \infty}\log \frac{(1+b)(2+a)}{(1+a)(2+b)}\frac {(2+b)(3+a)}{(2+a)(3+b)}.... \frac{(k+a+1)}{(k+b+1)}$$
as the consecutive expressions will cancel out in each iteration
$$=\lim _{k \to \infty} \log\frac{(1+b)(k+a+1)}{(1+a)(k+b+1)}$$
$$=\log\frac{1+b}{1+a}$$
|
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|
Convergence of binomial series I know that $f(x)=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$ converges for $|x|<1$
What I then have to show is that $(1+x)f'(x)=\alpha f(x)$ for $|x|<1$ and that any such $f$ is of the form $c(1+x)^\alpha$ for some constant c, and to use that fact to establish the binomial series.
I tried taking $$(1+x)f'(x)=(1+x)\left ( \sum_{n=0}^{\infty}\binom{\alpha}{n} x^n \right )'=(1+x)\left ( \sum_{n=0}^{\infty}\frac{\alpha !}{n!(\alpha-n)!} x^n \right )' = (1+x) \sum_{n=0}^{\infty}\frac{n \alpha !}{n!(\alpha-n)!} x^{n-1}=\alpha\sum_{n=0}^{\infty}\frac{ (\alpha-1) !}{(n-1)!((\alpha-1)-(n-1))!} (x^{n-1}+x^n)$$
But I'm not sure if my approach was right, and I'm not sure how to deal with the 2 $x$ terms if it was right.
I have no idea where to start for the second part.
|
$$\begin{align*}(1+x)f'(x)
&= (1+x) \sum_{n=1}^\infty \binom a n n x^{n-1}
= (1+x) \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n \\
&= \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n + \sum_{n=1}^\infty \binom a {n} n x^n \\
&= a + \sum_{n=1}^\infty \binom a {n+1} (n+1) x^n + \binom a {n} n x^n
= a + \sum_{n=1}^\infty a_n x^n
\end{align*}$$
where
$$\begin{align*}
a_n = \binom a {n+1} (n+1) + \binom a {n} n
&= \frac{a!}{(n+1)!(a-n-1)!}(n+1) + \frac{a!}{n!(a-n)!}n \\
&= \frac{a!a}{n!(a-n)!}
\end{align*}$$
Hence
$$(1+x)f'(x) = a + \sum_{n=1}^\infty a_n x^n = a + \sum_{n=1}^\infty \frac{a!a}{n!(a-n)!} x^n = af(x)
$$
The second part can be solved using logarithms, as RRL pointed out.
|
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|
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
|
$a_{n+1} = a_n \frac{n+1}{2n+1}$ Each number in the sequence is slightly more that (1/2) as the one before it.
for $n>3, \dfrac{a_{n+1}}{a_n} < 0.6$
$a_n < 0.4 (0.6)^{n-2}$
The sequence converges to 0.
|
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|
Integral of quotients of $\sin$ function I am trying to calculate the definite integral
$$\int_0^\pi \frac{\sin(\frac{21}{2}x)}{\sin(\frac{1}{2}x)} dx.$$
Wolfram Alpha says here that the answer is $\pi$. I replaced 21 by other constants and think that in general, $\int_0^\pi \frac{\sin(\frac{n}{2}x)}{\sin(\frac{1}{2}x)} dx = \pi$ for all odd $n \in \mathbb Z$.
However I have no idea how to approach this problem. I tried substituting $u = x/2$ to simplify the integral a bit to
$$2\int_0^{\pi/2} \frac{\sin(nu)}{\sin(u)}.$$
Then I thought that I could maybe use the identity $\sin(nx) = \sin(x)\cos( (n-1)x) + \sin((n-1)x)\cos(x)$. For instance, since
\begin{align*}
\sin(3x) &= \sin(2x)\cos(x) + \cos(2x)\sin(x) \\
&= 2\sin(x) \cos^2(x) + \cos^2(x)\sin(x) - \sin^3(x) \\
&= \sin(x)(3\cos^2(x) - \sin^2(x))
\end{align*}
the integral would become
$$ 2\int_0^{\pi/2} 3\cos^2(x) -\sin^2(x) dx $$
which I am able to solve:
\begin{align*}
2\int_0^{\pi/2} 3\cos^2(x) -\sin^2(x) dx &= 2[\frac{3}{2}(x+\sin(x)\cos(x)) - \frac{1}{2}(x - \sin(x)\cos(x))]^{\pi/2}_0 \\
&= [2x + 4\sin(x)\cos(x)]^{\pi/2}_0 \\
&= \pi.
\end{align*}
However I don't know how to generalize this approach because the expansion for $\sin(21x)$ would have lots of unwieldy terms. Is there another way to do this problem that I am missing?
|
This is exactly the closed form of the Dirichlet's Kernel when N = 10.
$\sum _{n=-N}^{N} $ $e^{-inx}$ = $\frac{sin(\frac{(2N+1)x}{2})}{sin(\frac{x}{2})}$
Integration the series term by term, the terms with positive values of n will cancel the terms with negative ones, and the value of the integral when $n=0$ is $ \int_0 ^\pi e^{0ix} = \pi $, thus $ \int _0 ^{\pi} \frac{sin(\frac{(2N+1)x}{2})}{sin(\frac{x}{2})} dx = \pi $ for any integer value of N.
|
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"timestamp": "2023-03-29T00:00:00",
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|
For any real number $x$, if $ x^3+2x+33\neq 0$, then $x+3 \neq0$ How to solve this type of sum using indirect proof. Appreciate if anyone can explain it step by step.
|
Your statement is true. But we can even establish that the converse is true.
Note that $$x^3+2x+33=(x+3)(x^2-3x+11)$$
But, since the polynomial discriminant is negative, for any real number, $x^2-3x+11 \neq 0$.
So we have that if $x^3+2x+33=0 \Rightarrow x+3=0$.
Also, since $(-3)^3+2(-3)+33=0$, we have that $x+3=0 \Rightarrow x^3+2x+33$.
Thus, we have that $x^3+2x+33=0 \Leftrightarrow x+3=0$.
This implies that $$x^3+2x+33 \neq 0 \Leftrightarrow x+3 \neq 0$$
|
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|
All 3 digit numbers the sum of whose digits is not greater than 16 My working: Let the digits of the numbers be $x,y,z$ where
\begin{align}1\leq x&\leq 9\\0\leq y,z&\leq 9\\x+y+z &\leq 16 \end{align}
I tried to solve it by making different cases here and using counting but it got really complicated and i couldnt get very far.
So how do i solve such questions?
|
Let $h$ denote the hundreds digit, $t$ denote the tens digit, and $u$ denote the units digit. Since the digit sum is at most $16$,
$$h + t + u \leq 16 \tag{1}$$
where $1 \leq h \leq 9$, $0 \leq t \leq 9$, and $0 \leq u \leq 9$.
Let $h' = h - 1$. Then $h'$ is a non-negative integer satisfying the inequalities $0 \leq h' \leq 8$. Substituting $h' + 1$ for $h$ in inequality 1 yields
\begin{align*}
h' + 1 + t + u \leq 16\\
h' + t + u \leq 15 \tag{2}
\end{align*}
Inequality 2 is an inequality in the non-negative integers.
If we let $s = 15 - (h' + t + u)$, then we obtain the equation
$$h' + t + u + s = 15 \tag{3}$$
which is an equation in the non-negative integers. If we ignore, for the moment, the restrictions on $h'$, $t$, and $u$, a particular solution of equation 3 corresponds to the placement of three addition signs in a row of fifteen ones. For instance,
$$1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 1 1 +$$
corresponds to the solution $h' = 4$, $t = 5$, $u = 6$, and $s = 0$, while
$$1 1 1 + 1 1 1 1 + 1 1 + 1 1 1 1 1 1$$
corresponds to the solution $h' = 3$, $t = 4$, $u = 2$, and $s = 6$. Hence, the number of solutions of equation 3 in the non-negative integers is
$$\binom{15 + 3}{3} = \binom{18}{3}$$
since we must select which three of the eighteen symbols (fifteen ones and three addition signs) will be addition signs.
However, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$. We must exclude those solutions.
Suppose $h' > 8$. Then $h' \geq 9$. Let $h'' = h' - 9$. Then $h''$ is a non-negative integer. Substituting $h'' + 9$ for $h'$ in equation 3 yields
\begin{align*}
h'' + 9 + t + u + s & = 15\\
h'' + t + u + s & = 6 \tag{4}
\end{align*}
Equation 4 is an equation in the non-negative integers with
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions.
Suppose $t > 9$. Then $t \geq 10$. Let $t' = t - 10$. Then $t'$ is a non-negative integer. Substituting $t' + 10$ for $t$ in equation 3 yields
\begin{align*}
h' + t' + 10 + u + s & = 15\\
h' + t' + u + s & = 5 \tag{5}
\end{align*}
Equation 5 is an equation in the non-negative integers with
$$\binom{5 + 3}{3} = \binom{8}{3}$$
solutions.
By symmetry, there are also $\binom{8}{3}$ solutions in which $u > 9$.
Notice that since $9 + 10 = 19 > 15$, at most one of the conditions $h' \leq 8$, $t \leq 9$, and $u \leq 9$ can be violated simultaneously in equation 3. Hence, the number of solutions of equation 3 subject to those restrictions is
$$\binom{18}{3} - \binom{9}{3} - 2\binom{8}{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Question about an inequality which seems right but not easy to prove The origin problem is as follows:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$
prove:$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq 4+\frac{1}{4}[(a−b)^2+(b−c)^2+(c−d)^2+(d−a)^2]$$
The solution is easy enough, which is to plug $4 = a+b+c+d $ into to $RHS$ and then move it to $LHS$, notice:
$$[\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-d)^2}{d}+\frac{(d-a)^2}{a}](a+b+c+d) \geq [(a−b)^2+(b−c)^2+(c−d)^2+(d−a)^2]$$
which would finally lead to the proof of the problem.
However, when I tried to solve the problem, I applied the Lagrange identical equation to the $RHS$ of the inequality, then I left out some quadratic term which finally leads to the inequality:
$$\color{Blue}{\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq a^2+b^2+c^2+d^2}$$
Which seems quite right, but I'm not really sure if it is. However, numerical tests imply that it is right. But I'm still not sure.
That stuff is kind of like Chebyshev inequality but it can't be directly used here.
From another point of view, my question is whether the following strengthening of the origin inequality is right or not:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$
then:$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq 4+\frac{1}{4}[(a−b)^2+(b−c)^2+(c−d)^2+(d−a)^2+(a-c)^2+(b-d)^2]$$
|
$\sum\limits_{cyc}\frac{a^2}{b}-4=\sum\limits_{cyc}\left(\frac{a^2}{b}-2a+b\right)=\sum\limits_{cyc}\frac{(a-b)^2}{b}\geq\frac{\left(\sum\limits_{cyc}|a-b|\right)^2}{4}\geq\frac{1}{4}\sum\limits_{cyc}(a-b)^2$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
I am not able to find an idea on how to proceed with the above questions. I have found only the obvious solution $(1,1,1)$.
Could you please provide some hints and ideas on how to proceed with the above question? Also, can we find the solutions?
Thanks.
|
Let $y=1+a, z=1-a$. Then
$$x^3+2(1+3a^2)=2(1+a^2)+x^2$$
$$x^2-x^3=4a^2.$$
Let $1-x=4p^2$, then
$$x^2(1-x)=(4p^2-1)^24p^2=(2a)^2.$$
Let $a=p(4p^2-1)$. Then
$$(x,y,z)=(1-4p^2, 1+p(4p^2-1), 1-p(4p^2-1))$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
evaluate $\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$ I just have a quick question about limits like this one: $$\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$$
leaving it as is i get $$\lim_{x\rightarrow 0}\frac{x}{\left ( x-1 \right )-\left( x+1\right)}$$
$$\lim_{x\rightarrow 0}\frac{x}{-2}$$
$$= 0$$
which is wrong, so by saying $\left | x-1 \right |=\left | 1-x \right |$ I get $$\lim_{x\rightarrow 0}\frac{x}{\left ( 1-x \right )-\left( x+1\right)}$$
$$\lim_{x\rightarrow 0}\frac{x}{-2x}$$
$$\lim_{x\rightarrow 0}\frac{1}{-2}$$
$$=\frac{1}{-2}$$
which, I'm pretty sure, is correct. I guess my question is, if this were an exam and I had no access to a calculator or a graphing program or something, is there anyway to intuitively see that the first, obvious, answer is incorrect?
or in a situation like this do you just have to evaluate through trial and error?
|
another way: $\frac{x}{\sqrt{(x-1)^2}-\sqrt{(x+1)^2}} \\ =\frac{x}{\sqrt{(x-1)^2}-\sqrt{(x+1)^2}} \cdot \frac{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}}{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}} \\ =\frac{x [ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{(x-1)^2-(x+1)^2} \\ =\frac{x[ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{([x-1]-[x+1])([x-1]+[x+1])} \text{ note I used difference of squares formula here } \\ =\frac{ x[ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{(-2)(2x)} \\ =\frac{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}}{-4} \text{ note this function is continuous at } x=0 $
|
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|
Bounding a sum involving a $\Re((z\zeta)^N)$ term This is a follow up to this question. Any help would be very much appreciated.
Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ or some other $N>ak^2$.
Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$.
Here there are five questions of varying intricacy. An answer to four is what I am hoping to achieve myself but an answer to an earlier part should go a long way towards helping and obviously an answer to part 5. would be amazing.
I have given a fairly trivial bound below which is good for my needs. If I don't get a better answer by the end of the bounty period I will accept my own (CW) answer and grant charMD the bounty.
*
*Simplify, where $v\in\{1,2,\dots,(k-1)/2\}$, $$1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}.$$
*Upper bound, where $v\in\{1,2,\dots,(k-1)/2\}$, $$\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\leq
f_2(v,k,N).$$
*Simplify, where $v\in\{1,2,\dots,(k-1)/2\}$, $$\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right).$$
*Upper bound $$\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\leq f_4(k,N).$$
*Sum $$\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2
\left(\frac{2\pi\,v}{k}\right)\left(1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right).$$
|
Consider first
$$\begin{align}
1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}&\leq 1+\sin^2\left(\frac{2\pi
v}{k}\right)\left|\Re\left((\alpha_v\zeta)^N\right)\right|+|\alpha_v|^{2N}
\\&\leq 1+\sin^2\left(\frac{2\pi
v}{k}\right)\left|(\alpha_v\zeta)^N\right|+|\alpha_v|^{2N}
\\&\leq 1+\sin^2\left(\frac{2\pi
v}{k}\right)\left|\alpha_v\right|^N+|\alpha_v|^{2N}
\\&\leq 1+\sin^2\left(\frac{2\pi
v}{k}\right)3^N+3^{2N}
\end{align}$$
In terms of efficiency, I am interested in $k$ large but $N=\mathcal{O}(k^2)$ and for $N\approx \frac{k}{2}\mod k$,
$$-\Re\left((\alpha_v\zeta)^N\right)\approx +\Re\left((\alpha_v)^N\right).$$
The largest problem is that
$$\alpha_v=2\cos\left(\frac{2\pi v}{k}\right)+\zeta^{-1}$$
has a large real part for $k$ large and $v$ small but as $v\rightarrow \frac{k-1}{2}$
$$\alpha_v\approx -1,$$
rather than $\alpha_v\approx 3$ as is the case for $v$ small. Hopefully this doesn't make the bound unusable (should find out soon).
Therefore
$$
\begin{align} &\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\\ &\leq \frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1+\sin^2\left(\frac{2\pi
v}{k}\right)3^N+3^{2N}\right)
\\&=\frac{1}{4^{2N-1}}\left((1+3^{2N})\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)+3^{N}\sum_{v=1}^{\frac{k-1}{2}}\tan^2\left(\frac{2\pi v}{k}\right)\right).
\end{align}$$
Now using $\sec^2A=1+\tan^2A$ and this answer of Joriki this is equal to
\begin{align*}
\frac{1}{4^{2N-1}}\left((1+3^{2N})\left(\frac{k-1}{2}+\frac{k(k-1)}{2}+3^{N}\frac{k(k-1)}{2}\right)\right)&=2\frac{k-1}{4^{2N}}\left[(k+1)3^{2N}+k\cdot 3^N+k+1\right]
\end{align*}
Any sharpening would be most welcome in an answer. An ideal answer would be a good bound of the form:
$$\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\leq f(N,k)e^{-\pi^2(2N-1)/k^2},$$
with the 'smaller' $f(N,k)$ the better. This answer here leads to an
$$f(N,k)=6(k^2-1)\left(\frac{3}{4}e^{\pi^2/k^2}\right)^{2N}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the extrema on sphere Let $f(x,y,z) = xe^{yz}$ . Find the extrema on the sphere $x^2+y^2+z^2=1$ . We have $e^{yz}=2tx$ , $xze^{yz}= 2ty$ , $xye^{yz}=2tz$ solving by subtitutions i found that $z(x^4-2)=0$ implies that if $z=0$ then $y=0$ and $x= \pm1$ . if the other part is zero then i found that the solution does not exist . Hence $( 1,0,0)$ is absolute maximum and $(-1,0,0)$ is absolute minimum . Is there any problem here ? Then i was asked to find the extrema for the unit sphere and for the sphere of radius $3$ to solve simultenously and i got stucked cqn somebody help me for that . ?
|
The unit sphere is a compact subset of $\mathbb{R}^3$ and $f$ is continuous, so the extrema exist and are reached.
Let $g=0$ be the constraint, with
\begin{align}
g(x,y,z)=x^2+y^2+z^2-1.
\end{align}
We are going to use the Lagrange multipliers theorem ; to do this, we first compute the differentials
\begin{align}
\mathrm{d}g(x,y,z)&=2x\mathrm{d}x+2y\mathrm{d}y+2z\mathrm{d}z,
\end{align}
\begin{align}
\mathrm{d}f(x,y,z)&=\mathrm{e}^{yz}\mathrm{d}x+xz\mathrm{e}^{yz}\mathrm{d}y+xy\mathrm{e}^{yz}\mathrm{d}z.
\end{align}
Thus, the critical points can be found by cancelling every sub-determinant of the matrix
\begin{align}
\begin{pmatrix}
2x & 2y & 2z \\
\mathrm{e}^{yz} & xz\mathrm{e}^{yz} & xy\mathrm{e}^{yz}
\end{pmatrix}.
\end{align}
This gives us the following conditions :
$$x^2z-y=0\quad\quad(1)$$
and
$$x^2y-z=0\quad\quad(2)$$
and
$$y^2-z=0\quad \text{or}\quad x=0.\quad\quad(3)$$
$\bullet$ If $x^2z-y=0$, then plugging $y=x^2z$ in $(2)$ gives $x^4=1$ or $z=0$. If $x^4=1$, then $y^2=z^2$ and we have
$$1=x^2+y^2+z^2=1+2z^2$$
that is $z=0$ and then $y=0$. So we get a first couple of critical point $(x,y,z)=(\pm1,0,0)$ for which $f(\pm1,0,0)=\pm1$. If $z=0$, then we get the same conclusion.
$\bullet$ If $x^2y-z=0$, then we can mimic what we've done just above with $y$ and $z$ exchanged.
$\bullet$ If $y^2-z=0$, then $(1)$ gives $x^2y^2=y$ while $(2)$ gives $x^2y=y^2$. If $y=0$, then $z=0$ and $x=\pm1$ and we get the above critical point anew. If $y\neq0$, we can divide by $y$ in the preceding relations to get $x^2=y$ and $x^2=1/y$, so that $\pm x=y=1$ is the only possibility. But this is impossible under the constraint $g=0$ (i.e. on the unit sphere, only one coordinate can be equal to $\pm1$).
$\bullet$ Finally, if $x=0$, then $(1)$ and $(2)$ immediately imply $y=z=0$ which is impossible under the constraint $g=0$.
Conclusion : There is only two critical points, and they are given with the corresponding value of $f$ by :
\begin{align}
(x,y,z)=(-1,0,0)\quad,\quad f(-1,0,0)=-1,
\end{align}
\begin{align}
(x,y,z)=(1,0,0)\quad,\quad f(1,0,0)=1.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a,b,c$ are the sides and $A,B,C$ are the angles of a triangle. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then,
$a,b,c$ are the sides of a $\triangle ABC$ and $A,B,C$ are the respective angles. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then $\sin^2 \bigl(\frac{A}{2}\bigr), \sin^2 \bigl(\frac{B}{2}\bigr)$ and $\sin^2\bigl(\frac{C}{2}\bigr)$ are in which type of progression?(Arithmetic,geometric,harmonic)
My try,
I immediately noticed that $1$ was a root and since the product of the roots was $1$ I got that $a,b,c$ were in harmonic progression. I replaced the reciprocals of $a,b,c$ by $p,q,r$. Wrote $\sin^2\bigl({\theta\over 2}\bigr)$ as $\frac{1-cos(\theta)}{2}$ and replaced $\cos$ by the sides but did not succeed. I also tried by replacing it with $\frac{(s-b)(s-c)}{bc}$ but still could not get the required answer.
If I assume the result I am getting the answer by simplifying the expression.
Please give an elegant solution.
|
As you noticed, the roots are,
$$x = 1\hspace{1 cm}\mathrm{ or }\hspace{1 cm}x = \frac{(a-b)c}{a(b-c)}$$
The roots are equal,$$1 = \frac{(a-b)c}{a(b-c)}.$$
Therefore, we can find
$$b = \frac{2ac}{a+c}.$$
Now let's talk about trigonometry. The half angle formula for sine is
$$\sin^2{\frac{W}{2}} = \frac{1 - \cos{W}}{2}$$
and the cosine law is
$$w^2 = u^2 + v^2 - 2 u v \cos{W}$$
Therefore,
$$-\frac{\cos{W}}{2} = \frac{w^2 - u^2 - v^2}{4 u v}$$
Therefore,
$$\sin^2{\frac{W}{2}} = \frac{1}{2} + \frac{w^2 - u^2 - v^2}{4 u v} = \frac{w^2 - (u - v)^2}{4uv}.$$
Now we can find that
$$\sin^2{\frac{A}{2}}=\frac{\left(a^2+2 a c-c^2\right) \left(a^2+c^2\right)}{8 a c^2 (a+c)}$$
$$\sin^2{\frac{B}{2}}=\frac{1}{4} \left(-\frac{a}{c}+\frac{4 a c}{(a+c)^2}-\frac{c}{a}+2\right)$$
$$\sin^2{\frac{C}{2}}=\frac{-a^4+2 a^3 c+2 a c^3+c^4}{8 a^2 c (a+c)}$$
We then just need to check the progression types.
|
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|
what is the set of points in the complex plane which satisfies |z| = Re(z) + 2? what is the set of points in the complex plane which satisfies |z| = Re(z) + 2?
so $ \sqrt{x^2+y^2} = x + 2 $ this is not a circle or anything and it asks me to sketch it what should I do
|
$$
\sqrt{x^2+y^2} = x = 2 \iff x^2+y^2 =(x+2)^2 \\
\iff x^2+y^2=x^2+4x+4 \\
\iff y^2=4x+4
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}.......(1)$
By Heron's formula,$\Delta=\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}}$
$\Delta^2=\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}$
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=16\Delta^2$
Putting in $(1)$
$16\Delta^2=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$
By formula $R=\frac{abc}{4\Delta}$,we get
$a^2+b^2+c^2=8R^2$
By sine rule,$a=2R\sin A,b=2R\sin B,c=2R\sin C$
$\sin^2A+\sin^2B+\sin^2C=2$
I am stuck here.
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Without loss of generality suppose $\cos C\ne 0 $.
Let the area of the triangle be T. The LHS is $16 T^2. $
The RHS is $8(\frac {1}{2}(a b \sin C)^2) (4\sin^2 C)c^2/(a^2+b^2+c^2)=32 T^2c^2/(a^2+b^2+c^2).$
And the denominator above is $a^2+b^2+c^2=(c^2+2 a b \cos C)+c^2=2(c^2+a b \cos C).$ So we have $$1=c^2/(\sin^2 c)(c^2+a b \cos C)\iff$$ $$\iff c^2=c^2\sin^2 C+a b \cos C \sin^2 C$$ $$\iff c^2\cos^2 C=a b \cos C\sin^2 C\iff$$ $$ \iff c^2\cos C=a b \sin^2 C=a b (1-\cos^2 C)\iff$$ $$\iff a b\cos^2C +c^2\cos C-a b=0.$$ Solving this quadratic in $\cos C$ we have $$\cos C=(-c^2\pm \sqrt {c^4+4 a^2b^2}\;)/2 a b.$$ $$\text { But } \cos C=(a^2+b^2-c^2)/2 a b.$$ $$\text {Therefore } \pm \sqrt {c^4+4 a^2b^2} =a^2+b^2$$ from which $c^4=(a^2+b^2)^2-4 a^2 b^2=(a^2-b^2)^2.$ So $$c^2=a^2-b^2 \lor c^2=b^2-a^2$$ and the triangle is right-angled.
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-(4n-2)^2)$$
$$=2(4)+2(6)+2(12)+2(14)+2(20)+2(22)+\cdots+2(8n-4)+2(8n-2)$$
$$=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]$$
How should I proceed further?
|
We know that $$\begin{align*}1+2+3+...+(4n+1)&=\frac{(4n+1) \cdot (4n+2)}{2}\\&=(2n+1)(4n+1)\end{align*}$$
So, $$\begin{align*}&2+3+(2+2)+(3+2)+6+7+...+(4n-2)+(4n-1)+(4n-2+2)+(4n-1+2)=(2n+1)(4n+1)-1 \\&\implies 2(2+3+...(4n-1))+2n \cdot 2=(2n+1)(4n+1)-1 \\&\implies 2(4+6+6+7+...+(8n-2))=2((2n+1)(4n+1)-4n-1)\\&=16n^2+4n\\&=\fbox{4n(4n+1)}\end{align*}$$
This is what you need.
|
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How to solve this question in more time efficient way?
Q) if$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}$$ then find $4z^2(x^2+y^2)$a)$(x^2+y^2)^{3}$b)$(x^2-y^2)^3$c)$(x^2-y^2)^2$d)$(x^2+y^2)^2$
Ans:c
i solved this in a very long way:
$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}=z\tan 2a$$$$\implies x= \frac{z\tan 2a}{\sin a} , y=\frac{z\tan 2a}{\cos a}$$
$$x^2+y^2=\frac{4z^2\tan ^22a}{\sin ^2 2a}=\frac{4z^2}{\cos^2 2a}$$
$$\implies z^2=\cos^2 2a(x^2+y^2)/4\ldots (1)$$
now$$x\sin a= y\cos a\implies -2x^2\sin^2 a=-2y^2\cos^2a$$
adding $x^2$ both sides
$$x^2(1-2\sin^2 a)= x^2-2y^2\cos^2 a$$$$=x^2-2y^2+2y^2\sin^2a$$ $$=x^2-y^2-y^2+2y^2\sin^2 a$$$$=x^2-y^2-y^2(1-2\sin^2 a)=x^2-y^2-y^2\cos 2a$$$$=(x^2+y^2)\cos 2a= x^2-y^2\implies \cos 2a= \frac{x^2-y^2}{x^2+y^2}\ldots (2)$$
hence from (1) and (2)
$$4z^2(x^2+y^2)= \left( \frac{x^2-y^2}{x^2+y^2} \right)^2 (x^2+y^2)^2= (x^2-y^2)^2$$
now you can see that i got the answer but there was a log way to find this. This question has been asked in an exam so there is noway its solution could be soo long there got to be some shorter way.
So how could i solve this question is more time efficient way?
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$x\sin\alpha=y\cos\alpha\iff\tan\alpha=\dfrac yx$
$\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2xy}{x^2-y^2}$
$$\implies\sin\alpha=\dfrac zx\cdot\dfrac{2xy}{x^2-y^2}=\dfrac{2yz}{x^2-y^2}$$
Use $$\dfrac1{(\sin\alpha)^2}-\dfrac1{(\tan\alpha)^2}=1$$
OR
similarly find $\cos\alpha=?$
Now find $\sin^2\alpha+\cos^2\alpha$
|
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How to solve the integral $\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$? I have seen this integral:
$$\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$$
In this integral: $a$ and $b$ are constants.
I have try with two ways, but failed:
*
*$u = \sqrt {{a^2} - {x^2}}$
*$x = a\sin t$
It seems that they are not true-way to solve this integral.
Any suggestion for solving this integral?
Now, I am not a student; so, this is not my exercise.
Sorry about my English. If my question is not clear, please comment below this question.
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First off I would let $x=-a\cos\theta$. Then
$$\begin{align}\int_0^a\frac{\sqrt{a^2-x^2}}{b-x}dx&=\int_{\frac{\pi}2}^{\pi}\frac{a^2\sin^2\theta}{b+a\cos\theta}d\theta\\
&=\int_{\frac{\pi}2}^{\pi}\frac{a^2(1-\cos^2\theta)}{b+a\cos\theta}d\theta\\
&=\int_{\frac{\pi}2}^{\pi}\left(-a\cos\theta+b+\frac{a^2-b^2}{b+a\cos\theta}\right)d\theta\end{align}$$
Now that last term may look a little intimidating, but I just did it this morning, so we get$$\begin{align}\int_0^a\frac{\sqrt{a^2-x^2}}{b-x}dx&=\left[-a\sin\theta+b\theta+\frac{a^2-b^2}{\sqrt{b^2-a^2}}\cos^{-1}\left(\frac{b\cos\theta+a}{b+a\cos\theta}\right)\right]_{\frac{\pi}2}^{\pi}\\
&=a+b\frac{\pi}2+\sqrt{b^2-a^2}\left(\cos^{-1}\left(\frac ab\right)-\pi\right)\end{align}$$
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Check: Radius of Convergence of the Sum of these Complex Taylor Series I just found the following Taylor series expansions around $z=0$ for the following functions:
*
*$\displaystyle \frac{1}{z^{2}-5z+6} = \frac{1}{(z-2)(z-3)} = \frac{-1}{(z-2)} + \frac{1}{(z-3)} = \sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}} - \sum_{n=0}^{\infty}\frac{z^{n}}{3^{n+1}} = \sum_{n=0}^{\infty}\left( \frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}\right)z^{n}$
*$\displaystyle \frac{1}{1-z-z^{2}} = \frac{1}{\left[z-\left(\frac{1-\sqrt{5}}{-2} \right) \right]\left[z-\left(\frac{1+\sqrt{5}}{-2} \right)\right]} = \frac{2\sqrt{5}}{5(1-\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1-\sqrt{5}} \right)z} - \frac{2\sqrt{5}}{5(1+\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1+\sqrt{5}} \right)z} = \frac{\sqrt{5}}{5} \sum_{n=0}^{\infty}\left(\frac{(-1)^{n}2^{n+1}}{(1-\sqrt{5})^{n+1}} - \frac{(-1)^{n}2^{n+1}}{(1+\sqrt{5})^{n+1}} \right)z^{n}$
I wanted to confirm that the radius of convergence for the first series is $|z|<2$ and that the radius of convergence for the second series is $|z|<\frac{-1+\sqrt{5}}{2}$.
I know that for a series $\sum_{n=1}^{\infty}a_{n}z^{n}$ with radius of convergence $R_{1}$ and $\sum_{n=1}^{\infty}b_{n}z^{n}$ with radius of convergence $R_{2}$, the radius of convergence of the series $\sum_{n=1}^{\infty}(a_{n}+b_{n})z^{n}$ is some $R$ where $R \geq \min(R_{1}, R_{2})$. And it can even be the case that if both of the series have finite $R_{1}$ and $R_{2}$, that $R$ can be infinite!
So, I wanted to make sure this was not the case here, and that I have the correct radii of convergence for both. If not, how do I go about finding them (preferably in the least icky way possible)?
Thank you.
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From the representation
\begin{align*}
\frac{1}{z^2-5z+6}&=\frac{1}{(z-2)(z-3)}\tag{1}\\
\frac{1}{1-z-z^2}&=\frac{1}{\left(z-\left(\frac{1-\sqrt{5}}{-2}\right)\right)\left(z-\left(\frac{1+\sqrt{5}}{-2}\right)\right)}\tag{2}\\
\end{align*}
we can already deduce the radius of convergence.
Both expressions have two simple poles. Recall the radius of convergence is the distance from the center of the series expansion to the nearest singularity.
In case of (1) we have two simple poles at $z=2$ and $z=3$. So, the radius of convergence $R$ is
$$R=\min\{2,3\}=2$$.
In case of (2) we have two simple poles at $z=\frac{1\pm\sqrt{5}}{2}$. So, the radius of convergence $R$ is
$$R=\min\left\{\left|\frac{1\pm\sqrt{5}}{2}\right|\right\}=-\frac{1+\sqrt{5}}{2}$$
|
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Trigonometric limit $\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$ The limit is
$$\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ I was able to solve it using L'hopital and the answer that I got was $2$.
Can you please confirm if the answer is right and suggest some other way to evaluate the limit without using L'hopital?
|
Multiply both deminator and numerator to the $\left( 1+\sqrt { 2 } \sin { x } \right) $ and $\left( 1+\tan { x } \right) $ respectively
\begin{align*}
\lim_{x\to\frac{\pi}{4}}{\frac{1-\tan x}{1-\sqrt{2}\sin x}}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\left(1-\tan^2\!x\right)\left(1+\sqrt2\sin x\right)}{\left(1-2\sin^2\!x\right)\left(1+\tan x\right)}}={} \\
{}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\frac{\cos(2x)}{\cos^2\!x}\left(1+\sqrt2\sin x\right)}{\cos(2x)\left(1+\tan x\right)}}={} \\
{}={}&\lim_{x\to\frac{\pi}{4}}\frac{\left(1+\sqrt2\sin x\right)}{\cos^2\!x\left(1+\tan x\right)}=2.
\end{align*}
|
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Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ I am being asked to find the sum of the following convergent series :
$$\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}$$
Attempting to generalize from partial sums yields nothing of interest:
$s_1 = \frac{5}{6}$
$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43}{36}$
$s_3 = \frac{43}{36} + \frac{35}{216} = \frac{293}{216}$
$s_4 = \frac{293}{216} + \frac{97}{1296} = \frac{1855}{1296} $
I do not see a pattern here...
How must I proceed to find the sum of this series?
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Note that the sum $\sum_{n=0}^{\infty}r^{n}=\frac{1}{1-r}$ for $|r|<1$. According to this fact $\sum_{n=0}^{\infty}\frac{2^{n}+3^{n}}{6^{n}}=\sum_{n=0}^{\infty}\frac{2^{n}}{6^{n}}+\sum_{n=0}^{\infty}\frac{3^{n}}{6^{n}}=\sum_{n=0}^{\infty}(\frac{2}{6})^{n}+\sum_{n=0}^{\infty}(\frac{3}{6})^{n}=\frac{1}{1-1/3}+\frac{1}{1-1/2}=\frac{7}{2}$.
|
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|
Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$ $$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$
What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin
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Using Taylor series for $\cos u$, $\sqrt{1+u}$, $\ln(1+u)$, and $\sin u$ when $u\to 0$. When $x\to \infty$, $\frac{1}{x} \to 0$, so
$$\cos\frac{\pi}{x} = 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right)$$
and
$$\sqrt{ \cos(\frac{\pi}{x}) } = \sqrt{ 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right) } = 1- \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right)$$
so that
$$\ln \sqrt{ \cos(\frac{\pi}{x}) } = \ln\left (1- \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right) \right) = - \frac{\pi^2}{4x^2} + o\left(\frac{1}{x^2}\right)$$
and finally
$$
x^2 \sin \ln \sqrt{ \cos(\frac{\pi}{x}) } = x^2 \sin\!\left( - \frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right)
= -x^2 \left(\frac{\pi^2}{4x^2} + o\!\left(\frac{1}{x^2}\right)\right) = - \frac{\pi^2}{4} + o(1)$$
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Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this is the part I don't understand
$\rightarrow \frac{6}{32}$
How did the numerator simplify to $6n^3$?
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It is worth generalizing the given limit: for positive integers $m$,
$$\begin{align*} a_n(m) &= \frac{n^m}{\binom{mn}{m}} \\ &= \frac{n^m m! \, (m(n-1))!}{(mn)!} \\ &= \frac{n^m m!}{(mn)(mn-1)\cdots(m(n-1)+1)} \\ &= \prod_{k=1}^m \frac{nk}{m(n-1) + k} \\ &= \prod_{k=1}^m \frac{k}{m + (k-m)/n}\end{align*}.$$ Consequently, $$\lim_{n \to \infty} a_n(m) = \prod_{k=1}^m \lim_{n \to \infty} \frac{k}{m+(k-m)/n} = \prod_{k=1}^m \frac{k}{m} = \frac{m!}{m^m}.$$
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Is the number of sequences with equal 0's and 1's small? I think that is n choose n/2. Though when I try to get a feel of what that number should be I get a much larger number than I expect:
i.e. we have
$$ n C \frac{n}{2} = \frac{n!}{\frac{n}{2}! \frac{n}{2}! }$$
using Stirling's?
Recall:
$$n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right) ^n$$
What I tried:
$$ \frac{n!}{\frac{n}{2}! \frac{n}{2}! }
\sim
\frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right) ^n}{ \sqrt{2 \pi \frac{n}{2}} \left( \frac{\frac{n}{2}}{e} \right) ^\frac{n}{2} \sqrt{2 \pi \frac{n}{2}} \left( \frac{\frac{n}{2}}{e} \right) ^\frac{n}{2}}
\sim
\frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right) ^n}{ \sqrt{2 \pi \frac{n}{2} 2 \pi \frac{n}{2} } \frac{\frac{n}{2} }{e ^\frac{n}{2}} ^\frac{n}{2} \frac{\frac{n}{2} }{e ^\frac{n}{2}} ^\frac{n}{2}}
\sim
\frac{\sqrt{2 \pi n} \frac{n^n}{e^n} }{ \sqrt{ \pi^2 n^2 } \frac{n^{n}}{e^n} \frac{1}{2^{n}}} \sim \sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}
$$
which I feel has to be wrong.
Why do I feel that? Because n choose n/2 is the number of sequences that have half zero and the other half 1. This number should be much less than the total number of binary sequences (because it you have half 1's and half 0's then to change it to a sequence that is unbalanced should be easy...just change any bit! which means intuitively that half 1's and half zeros should be really small).
Basically, shouldn't most sequences not be balanced? i.e isn't the number of balanced sequences small?
Say the number of not balanced sequences should be
$$2^n - nC\frac{n}{2}$$
shouldn't the above quantity be super large, like $2^n$?
How do I show that the majority of the sequences are unbalanced?
|
Your derivation is correct, and does tell you that most sequences are unbalanced. The expression $\sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}$ is much smaller than $2^n$ when $n$ is large, by a factor of about $\sqrt{n}$. For instance, if $n=1000000$, your estimate says that only about $1/1000$ of all the binary sequences are balanced. The number of unbalanced sequences is about $$2^n-\sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}=2^n\left(1-\frac{\sqrt{2}}{\sqrt{\pi}\sqrt{n}}\right),$$
and when $n$ is large, the second factor on the right-hand side is almost $1$.
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What is the probability of getting 2 same colour sweets and 1 different colour sweet? A little box contains $40$ smarties: $16$ yellow, $14$ red and $10$ orange.
You draw $3$ smarties at random (without replacement) from the box.
What is the probability (in percentage) that you get $2$ smarties of one color and another smarties of a different color?
Round your answer to the nearest integer.
Answer given is $67$. I don't get it. Is it not:
$$\left(\frac{16}{40} \times \frac{15}{39} \times\frac{24}{38}\right) + \left(\frac{14}{40} \times\frac{13}{39} \times\frac{26}{38}\right) +\left(\frac{10}{40} \times\frac{9}{39} \times\frac{30}{38}\right)= 22?$$
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Your options are "Exactly two yellow smarties, exactly two red smarties, or exactly two orange smarties."
If $P$ represents your final probability, you need to add up the following probabilities:
$P($exactly two yellows$) \, + \, P($exactly two reds$) \, + \,P($exactly two orange$)$
The probability of getting exactly two yellow smarties is $3\cdot \frac{16}{40}\cdot \frac{15}{39}\cdot \frac{24}{38}$. The reason we multiply by $3$ is that there are three different ways to choose two yellow smarties, i.e. $\binom{3}{2}=3$. The first two draws can be yellow, the first and the last can be yellow, or the last two draws can be yellow.
Similarly we can find $P($exactly two reds$)$ and $P($exactly two orange$)$.
Thus $P=3\left( \frac{16}{40}\cdot \frac{15}{39}\cdot \frac{24}{38}\right)+ 3\left( \frac{14}{40}\cdot \frac{13}{39}\cdot \frac{26}{38}\right)+3\left( \frac{10}{40}\cdot \frac{9}{39}\cdot \frac{30}{38}\right)\approx 67$%.
|
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Prove limit of a sequence in Newton's method Given the $ f(x)=x^3+x-1 $, I have shown so far that $ f$ has a unique root $r\in(0,1)$ and that for the sequence $(x_{n}), n>=0$ produced by Newton's method we have $$\lim_{n\to\infty} x_{n}=r$$ for every $x_{0}\in\mathbb{R}$.
How do I prove that $$\lim_{n\to\infty} \frac{x_{n+1}-r}{(x_{n}-r)^2}=\frac{3r}{3r^2+1}$$ ?
|
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^3+x_n -1}{3x_n^2+1} = \frac{2x_n^3+1}{3x_n^2+1}$$
hence
$$x_{n+1}-r =\frac{2x_n^3+1}{3x_n^2+1} - r = \frac{2x_n^3 - 3x_n^2r+1-r}{3x_n^2+1} = \frac{2x_n^3 - 3x_n^2r+r^3}{3x_n^2+1} = \frac{(x_n-r)^2(2x_n+r)}{3x_n^2+1}$$
Therefore
$$\frac{x_{n+1}-r}{(x_n-r)^2} = \frac{2x_n+r}{3x_n^2+1} \to \frac{3r}{3r^2+1}$$
as $n \to \infty$.
|
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|
Sum of the series $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$ If $|x|<1$, find the sum of infinite terms of following series:
$$\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$$
Could someone give me hint to solve this problem. I wrote $r_{th}$ term of the series but can't find sum of $n$ terms. If somehow I can get sum of $n$ term then I can put $n \to \infty $ to get sum of infinite terms.
|
Hint: $$\frac{x^{2n}}{(1-x^{2n+1})(1-x^{2n+3})}=\frac{1}{x(1-x^2)}\left(\frac{1}{1-x^{2n+1}}-\frac{1}{1-x^{2n+3}}\right)$$
|
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|
Proof of an inequality in $\mathbb{C}$ Let $z\in \mathbb{C}, n \geq 2$. Show this complex inequality
$$|z^n-1|^2\le |z-1|^2\left(1+|z|^2+\dfrac{2}{n-1}\Re{(z)}\right)^{n-1}$$
For $n=2$ the inequality is easy to prove:
$$|z^2-1|^2\le |z-1|^2\left(1+|z|^2+2\Re{(z)}\right)$$
$$\Longleftrightarrow |z+1|^2\le 1+|z|^2+2\Re{(z)}$$
$$\Longleftrightarrow z+\overline{z}\le 2\Re{(z)}$$
which is in fact an equality, thus is exact.
for $n=3$.
$$\Longleftrightarrow |z^2+z+1|^2\le (1+|z|^2+\Re{(z)})^2$$
$$\Longleftrightarrow(z^2+z+1)(\overline {z^2}+\overline{z}+1)\le |z|^4+2|z|^2+1+\Re^2{z}+2\Re{(z)}+2|z|^2\cdot\Re{(z)}$$
$$\Longleftrightarrow |z|^4+|z|^2+|z|^2(z+\overline{z})+1+(z+\overline{z})+(z^2+(\overline{z})^2)\le |z|^4+2|z|^2+1+\Re^2{z}+2\Re{(z)}+2|z|^2\cdot\Re{(z)}$$
$$\Longleftrightarrow |z|^2+(\Re{(z)})^2\ge z^2+(\overline{z})^2=(z+\overline{z})^2-2|z|^2$$
$$\Longleftrightarrow 3|z|^2\ge 3(\Re{(z)})^2$$
It is clear
Is it true for a general $n$?
|
Firstly: $$\frac{z^n-1}{z-1}=z^{n-1}+z^{n-2}+\cdots1 \hspace{2cm}(1)$$
Let $z_k, \; k=0,\cdots,n-1$ be the roots of unity, i.e. the roots of $z^n-1=0$, and take $z_0=1$. Using Vieta's relations for the sum of the roots (https://en.wikipedia.org/wiki/Vieta's_formulas), one has $\sum\limits_{i=0}^{n-1} z_i=0$. Using that $z_0=1$, we get $\sum\limits_1^{n-1}z_i=-1$. With $z_k = \cos\theta_k+I\sin\theta_k$, and separating the real and imaginary parts, we get: $$\sum\limits_{k=1}^{n-1}\cos\theta_k=-1$$ $$\sum\limits_{k=1}^{n-1}\sin\theta_k=0$$
Equation $(1)$ can be written now as: $$\frac{z^n-1}{z-1}=(z-z_{n-1})(z-z_{n-2}\,)\cdots(z-z_1)$$, hence: $$\vert\frac{z^n-1}{z-1}\vert^2=\prod\limits_{k=1}^{n-1}\vert z-z_k\vert^2$$
We use now the inequality between the geometric and arithmetic means (and $z=x+ I y$):
$$\left( \prod\limits_{k=1}^{n-1}\vert z-z_k\vert^2 \right)^{\frac{1}{n-1}}\le\frac{\sum\limits_{k=1}^{n-1}\vert z-z_k\vert^2}{n-1}=\frac{\sum\limits_{k=1}^{n-1}(1+x^2+y^2-2x\cos\theta_k-2y\sin\theta_k)}{n-1}=1+x^2+y^2+2\frac{x}{n-1},$$
where we used the above-derived equalities for the sum of $\sin$ and $\cos$.
Putting it all together:
$$\vert\frac{z^n-1}{z-1}\vert^2 \le (1+\vert z\vert^2+2\frac{\Re(z)}{n-1})^{n-1}$$.
We can re-state the problem also as:
"For an arbitrary point in a plane, the squared product of distances to the vertices of any regular polygon inscribed in the unit circle (with $n$ vertices, that contains the point $(1,0)$, which is not to be included in the product), is at most equal to the power $n-1$ of the sum of the squared distance to the point $(-\frac{1}{n-1},0)$ and the constant $1-(\frac{1}{n-1})^2$ ".
The minimum of the right-hand side term reads: $\left[ 1-(\frac{1}{n-1})^2\right]^{n-1}$. Notably is the limit of it, for $n \to \infty$.
|
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|
You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life. This is from the lecture notes in this course of discrete mathematics I am following.
The professor is writing about how fast binomial coefficients grow.
So, suppose you had 2 minutes to save your life and had to estimate, up to a factor of $100$, the value of, say, $\binom{63}{19}$. How would you do it? I will leave this (hopefully intriguing!) question hanging and maybe come back to the topic of efficiently estimating binomial coefficients later.
Any ideas/hints on how to do it?
|
With pen and paper, Stirling's approximation:
$$
\begin{align}
{63 \choose 19}
&= \frac{63!}{19! 44!} \\
&\doteq \frac{\sqrt{2\pi 63}}{\sqrt{2\pi 19}\sqrt{2\pi 44}}
\left( \frac{63}{e}\right)^{63}
\left( \frac{e}{19}\right)^{19}
\left( \frac{e}{44}\right)^{44} \\
&\doteq \sqrt{\frac{60}{2 \cdot 3 \cdot 20 \cdot 50}} \cdot
20^{63} \cdot 5^{-19} \cdot 15^{-44} \\
&\doteq \frac{1}{10} \cdot 2^{63} \cdot 10^{63} \cdot 2^{19} \cdot 10^{-19}
\cdot 3^{-44} \cdot 2^{44} \cdot 10^{-44} \\
&= 10^{-1} \cdot 2^{126} \cdot 3^{-44} \\
&= 10^{-1} \cdot (2^{10})^{12} \cdot 2^6 \cdot (3^2)^{-22} \\
&\doteq 50 \cdot 10^{-1+36-22} \\
&= 5 \cdot 10^{14}
\end{align}
$$
using various estimates including $2^{10} \doteq 10^3$.
Crude, but if instead you had to estimate ${630 \choose 19}$ then you might not want to do cancelling by writing out integer factors.
|
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|
How do I determine if matrix A is diagonalizable? I am trying to figure out how to determine the diagonalizability of the following matrix:
$A=\begin{pmatrix}
1 &0 &0 &0 \\
2&1 & -3 & -2\\
3& 0 & 0 &-9 \\
-1& 0& -1& 0
\end{pmatrix}$
There are two distinct eigenvalues: $λ_1=λ_4=1$,$λ_2=3$ and $λ_3=-3$.
Can someone help to define geometric multiplicity?
And I found that matrix is diagonalizable geometric multiplicity is equal to the algebraic multiplicity.
|
The characteristic polynomial is
$$
\det
\begin{pmatrix}
1-X &0 &0 &0 \\
2&1-X & -3 & -2\\
3& 0 & 0-X &-9 \\
-1& 0& -1& 0-X
\end{pmatrix}
=(1-X)^2(3-X)(-3-X)
$$
You want to determine the geometric multiplicity of the eigenvalue $1$, that is the dimension of the eigenspace, which is the null space of $A-I$. Let's do an elimination:
\begin{align}
A-I=\begin{pmatrix}
0 &0 &0 &0 \\
2&0 & -3 & -2\\
3& 0 & -1 &-9 \\
-1& 0& -1& -1
\end{pmatrix}
&\to
\begin{pmatrix}
-1& 0& -1& -1 \\
2&0 & -3 & -2\\
3& 0 & -1 &-9 \\
0 &0 &0 &0
\end{pmatrix}
\\[6px]&\to
\begin{pmatrix}
1& 0& 1& 1 \\
0& 0 & -5 & -4\\
0& 0 & -4 &-12 \\
0 &0 &0 &0
\end{pmatrix}
\\[6px]&\to
\begin{pmatrix}
1& 0& 1& 1 \\
0& 0 & 1 & 4/5\\
0& 0 & 0 &-44/5 \\
0 &0 &0 &0
\end{pmatrix}
\\[6px]&\to
\begin{pmatrix}
1& 0& 1& 1 \\
0& 0 & 1 & 4/5\\
0& 0 & 0 &1 \\
0 &0 &0 &0
\end{pmatrix}
\end{align}
So the matrix $A-I$ has rank $3$ and its null space has dimension $1$.
The matrix is not diagonalizable.
|
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|
I do not understand how to solve this The two sequences of numbers { 1, 4, 16, 64, . . .} and { 3, 12, 48, 192, . . .} are mixed as follows: { 1, 3, 4, 12, 16, 48, 64, 192, . . .}. One of the numbers in the mixed series is 1048576. Then the number immediately preceding it is
|
The second sequence is just 3 times the previous sequence. Note that no number in the first sequence is divisible by 3 while all numbers in the second are. The mixed sequence is in the form $\{4^0, 3\cdot 4^0, 4^1, 3\cdot 4^1, 4^2, 3 \cdot 4^2, 4^3, 3 \cdot 4^3, \ldots\}$. Seeing that 1048576 is not divisible by 3 it must be from the first sequence. Looking at the mixed set you can see that each $4^k$ is preceded by $3 \cdot 4^{k - 1}$. Since $4^k = 1048576$, $3 \cdot 4^{k - 1} = \frac{3}{4} \cdot 4^k= \frac{3}{4} \cdot 1048576 = 786432$ so the preceding term is 786432 .
|
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|
Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}}
+ \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \\
&= \lim_{n \to \infty} \frac{\sqrt{n}}{n} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots
+ \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr]
\end{align}
$$
Now using Cauchy first thm of limits
$$
a_n = \frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n+1}}
$$
The answer should be $\frac{1}{2\sqrt{2}}$.
But the answer is $1/\sqrt{2}$.
|
If one attempts to use Cauchy's First Limit Theorem, then we start with
$$\begin{align}
\lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\sqrt{n}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}} \tag 1\\\\
\end{align}$$
The Theorem reveals that since $\lim_{n\to \infty}\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=0$, then $\lim_{n\to \infty}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=0$
Therefore, the limit on the right-hand side of $(1)$ is of indeterminate form (i.e., $\infty \times 0$) and we need to pursue its evaluation judiciously.
We revert to the left-hand side of $(1)$ and note that both $\sqrt{n}$ and $\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$ are monotonically increasing, divergent sequences. We can invoke, therefore, the Stolz-Cesaro Theorem and write
$$\begin{align}
\lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\frac{\sum_{k=1}^{n+1}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}-\sum_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}}{\sqrt{n+1}-\sqrt{n}} \\\\
&=\lim_{n\to \infty}\frac{\frac{1}{\sqrt{2n+2}+\sqrt{2n+4}}}{\sqrt{n+1}-\sqrt{n}}\\\\
&=\lim_{n\to \infty}\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{2n+2}+\sqrt{2n+4}}\\\\
&=\frac{1}{\sqrt{2}}
\end{align}$$
as was to be shown!
|
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|
Prove using factorials that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$ Prove using factorials that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$
I think I'm having a bit of algebra problem with this proof. Here is my work thus far:$$\frac{n!}{k!(n-k)!}+2\frac{n!}{(k+1)!(n-k-1)!}+\frac{n!}{(k+2)!(n-k-2)!}\stackrel{?}{=}\frac{(n+2)!}{(k+2)!(n-k)!}$$$$\frac{(k+1)(k+2)n!}{(k+2)!(n-k)!}+\frac{2(k+2)(n-k)n!}{(k+2)!(n-k)!}+\frac{(n-k-1)(n-k)n!}{(k+2)!(n-k)!}\stackrel{?}{=}\frac{(n+2)!}{(k+2)!(n-k)!}$$$$\frac{n!({k^2}+{3k+2}+{2kn}-2k^2+4n-4k+n^2-kn-n+k^2+k)}{(k+2)!(n-k)!}\stackrel{?}{=}\frac{(n+2)!}{(k+2)!(n-k)!}$$$$\frac{n!(n^2+3n+2)}{(k+2)!(n-k)!}\stackrel{?}{=}\frac{(n+2)!}{(k+2)!(n-k)!}$$But from here I am stuck. How would I algebraically make these two sides match?
|
This result, and its generalizations, are made intuitively obvious by the relationship between Pascal's triangle and binomial coefficients; e.g., $$\begin{array}{ccccccc} \ldots & \binom{n}{k} & & \binom{n}{k+1} & & \binom{n}{k+2} & \ldots \\ & \ldots & \binom{n+1}{k+1} & & \binom{n+1}{k+2} & \ldots & \\ & & \ldots & \binom{n+2}{k+2} & \ldots & & \end{array}$$ and recalling that each coefficient is the sum of the two coefficients diagonally above it.
|
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|
Only valid for Pythagoraean triples $\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$? $$\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$$
Where (a,b,c) are the Pythagoraean Triples and are satisfy by the Pythagoras theorem $a^2+b^2=c^2$
An example of Pythagoraean triple (3,4,5)
It is true that this continued fraction is only valid for Pythagoraean Triple only?
I try other numbers and it seem that the only numbers that are valid are the Pythagoraean Triples. Can anyone verify this? Or show some examples where is also work for other numbers.
|
Solving the Recurrence
Let
$$
x = \sqrt 2 + \frac{b}{x}
$$
Hence,
$$
x = \frac{\sqrt 2 x + b}{x} \implies x^2 = \sqrt 2 x + b \\
x^2 - x\sqrt2 - b = 0
$$
The roots of this equation are
$$
x_1, x_2 = \frac{\sqrt2 \pm \sqrt{ \sqrt{2}^2 + 4b}}{2} = \frac{\sqrt{2} \pm \sqrt{2 +4b}}{2}
$$
Proving that this works for all Pythogaren triplets
Let us consider the always-positive root
$$x_1 = \frac{ \sqrt{2} + \sqrt{2 + 4b} }{2}$$
Let us first compute $x_1^2$ and produce Pythogorean triplets:
$$
x_1^2 = \left (\frac{ \sqrt{2} + \sqrt{2 - 4b} }{2} \right)^2= \\
\frac{2 + (2 + 4b) + 2 \cdot \sqrt 2 \cdot \sqrt{2 + 4b}}{4} = \\
\frac{4 + 4b + 2 \cdot \sqrt 2 \cdot \sqrt 2 \sqrt{1 + 2b}}{4}
\\
\\
x_1^2 = 1 + b + \sqrt{1 + 2b}
$$
We want $x_1 = \sqrt{a + c}$, so we pick $a = \sqrt{1 + 2b}$, $c = 1 + b$
that gives us
$$
a^2 + b^2 = \sqrt{1 + 2b}^2 + b^2 = \\
1 + 2b + b^2 \\
= (1 + b)^2 \\
= c^2
$$
which verifies that the Pythogarean triplets $(a, b, c)$ satisfy this equation.
Non Pythogaren triplet integer solutions
Since we know that
$$
x_1^2 = 1 + b + \sqrt{1 + 2b}
$$
We can pick $a = 1 + \sqrt{1 + 2b}$, $c = b$ Giving us non-triplet numbers. It can be seen that this can never be a pythogarean triplet by seeing that
$$
b^2 = b^2 \\
c^2 = c^2 \\
a^2 = 1 + (1 + 2b) + 2 . 1 . \sqrt{1 + 2b}
$$
Which cannot be re-arranged into a triplet.
However, they are still integer solutions such that
$x1 = \sqrt{a + c}$ for a given $b$.
|
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|
Find all real numbers $a,b$ such that $|a|+|b|\geq\frac{2}{\sqrt{3}}$ and $|a\sin x+b\sin{2x}|\leq 1$ for all real $x$. Find all real numbers $a,b$ such that $|a|+|b|\geqslant\frac{2}{\sqrt{3}}$ and $|a\sin(x)+b\sin(2x)|\leqslant 1$ for all real $x$.
We could write the inequality as
$$
\left|\frac{a}{\sqrt{a^2+b^2}}\sin (x)+\frac{b}{\sqrt{a^2+b^2}}\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}}
$$
and let
$$
\cos (y)=\frac{a}{\sqrt{a^2+b^2}}
$$
and
$$
\sin (y) = \frac{b}{\sqrt{a^2+b^2}}.
$$
The inequality is equivalent to
$$
\left|\cos (y)\cos\left(\frac{\pi}{4}-x\right)-\sin (y)\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}},
$$
but the terms are different, so we cannot use the cosine addition formula.
|
Let $f(x)=a\sin x+b\sin 2x$. Note that $f(-x)=-f(x)$, so we can assume $a\ge 0$. Also $f(\pi-x)=a\sin x-b\sin 2x$, so we can also assume $b\ge 0$.
We look first at the maximum value of $k\sin x+\sin 2x$. Differentiating and putting $c=\cos x$ we get $kc+2(2c^2-1)=0$, so $4c^2+kc-2=0$, so $c=\frac{1}{8}(\sqrt{32+k^2}-k)$. [It is obvious that the maximum is for positive $c$.]. Hence setting $s=\sin x$, we have $s=\sqrt{1-c^2}=\sqrt{1-\frac{1}{64}(\sqrt{32+k^2}-k)^2}$. So the maximum value of the expression is $ks+2sc=\frac{1}{32}(3k+\sqrt{32+k^2})\sqrt{32+2k(\sqrt{32+k^2}-k)}$.
We want to find the right value of $k$ to minimise. But first we must divide by $1+k$, so that we are keeping $a+b$ constant. From this point on a mess becomes a truly horrible mess.
Minimising a positive quantity is the same as minimising its square. So we start by squaring and dividing by $(1+k)^2$ to get: $$\frac{128+80k^2-k^4+32k\sqrt{32+k^2}+k^3\sqrt{32+k^2}}{128(1+k)^2}$$
We now differentiate and find that the result factorises as $(k-2)g(k)$ where $$g(k)=\frac{k^4+4k^3+24k^2+128k-256-\sqrt{32+k^2}(k^3+4k^2+8k-64)}{64(k+1)^3\sqrt{32+k^2}}$$
So we need to show that $g(k)>0$ for $k>0$. Put $r(k)=k^4+4k^3+24k^2+128k-256$ and $s(k)=\sqrt{32+k^2}(k^3+4k^2+8k-64)$, so that $g(k)=r(k)-s(k)$. We have $r(k)=(k^2+32)(k^2+4k-8)$. With a little work we also find $r(k)^2-s(k)^2=256(k-2)(k+2)^2(k^2+32)$, so $|r(k)|>|s(k)|$ for $k>2$. Since $r(k)$ is obviously positive for $k>2$ we have $g(k)>0$ for $k>2$. But $|r(k)|<|s(k)|$ for $k<2$ and $s(k)$ is obviously negative for $k\le 2$, so again $g(k)>0$ for $k\le 2$.
Thus we have established that we minimise the maximum value of $f(x)$ by taking $k=2$ or $a=2b$. Looking back we find that the maximum is then achieved for $c=\frac{1}{8}(6-2)=\frac{1}{2}$ and hence taking $a=\frac{4}{3\sqrt3},b=\frac{2}{3\sqrt3}$ we get a maximum value of 1.
So the only possible values of $a,b$ are $\pm\frac{4}{3\sqrt3},\pm\frac{2}{3\sqrt3}$.
All I can say is that there has to be a better way, but at the moment I cannot see it!
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.