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System of five equations I'd really appreciate help with this problem, because I'm stuck with it.
$$
\left\{
\begin{array}{c}
ab+ac+ad+ae=-1 \\
ba+bc+bd+be=-1 \\
ca+cb+cd+ce=-1 \\
da+db+dc+de=-1 \\
ea+eb+ec+ed=-1
\end{array}
\right.
$$
I've tried substituting $a+b+c+d+e$ with t but it didn't get me far. I've also come up with equations $(a-b)(c+d+e)=0$ or $(b-c)(a+d+e)=0$.
|
Let $t=a+b+c+d+e$. Then your equations can be transformed to $a(t-a)=b(t-b)=\ldots=e(t-e)=-1$.
From this $0=a(t-a)-b(t-b)=(a-b)(t-a-b)$ and similarly for all other pairs in place of $a,b$ as you found out yourself.
We conclude that $a=b$ or $a+b=t$ (and equally for other pairs).
If $a=b=c=d=e$, this leads to $4a^2 = -1$, hence one solution is
$$\tag1 a=b=c=d=e=\pm\frac12 i.$$
For the rest we may assume that not all variables are equal, wlog. assume $a\ne b$. Then $a+b=t$. Since we cannot have both $a=c$ and $b=c$ we conclude $a+c=t$ or $b+c=t$, hence $c=a$ or $c=b$. All in all, we see that $a,b,c,d,e$ take only two values. Up to symmetry, we have one of the following cases: $a\ne b=c=d=e$ or $a=b\ne c=d=e$.
In the first case, $t=a+4b$, i.e. $4ab=-1$ and $ab+3b^2=-1$, hence $3b^2=-\frac34$, i.e. $b=\pm\frac{1}2i$ and $a=-\frac1{4b}=b$, contrary to the assumption $b\ne a$.
From now on assume $a=b\ne c=d=e$.
Then $t=2a+3c$, i.e. $a^2+3ac=-1$ and $2ac+2c^2=-1$, hence $c=-\frac{1+a^2}{3a}$ (note that clearly $a\ne0$), and by inserting $\frac{-4 a^4
- 2 a^2
+ 2}{9 a^2}=-1 $.
From this we find $-4a^4+7a^2+2=0$, hence $a^2=2$ or $a^2=-\frac14$.
Hence we find solutions by letting (using $a^2=2$)
$$\tag 2 \mathrm{Two\ numbers\ are\ }\pm\sqrt 2\mathrm{,\ the\ other\ three\ are\ }\mp\frac{\sqrt2}{2}.$$
On the other hand $a^2=-\frac14$ leads to $c=a$, hence no additional solution.
In total we have thus two solutions from $(1)$ and twenty solutions from $(2)$.
Note that only the solutions of type $(2)$ are real.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A problem about multiples.
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a multiple of $p$. Find the largest possible value of $p$.
I have no idea how to solve this. Please help. Thank you.
|
Observe that $(ab,16)=1$
If $a=16A+1,b=16B-1\implies a+b=16(A+B)$
If $a=16A+3,b=16B+5\implies a+b=16(A+B)+8$
If $a=16A+7,b=16B+9\implies a+b=16(A+B+1)$
So, in all the cases, $8\mid (a+b)$
As to find the pairs, as $ab\equiv -1\pmod {16}$ ,so $a(-b)\equiv 1$ i.e.,$-b$ is inverse of $a\pmod {16}$.
we know by this, $\lambda(16)=4\implies a^4\equiv 1\pmod {16}\implies a^{-1}\equiv a^3$
For example, $a=3,a^3=27\equiv -5\pmod {16}\implies b\equiv 5$
But $16$ is small enough to allow mental arithmetic, we just need to multiply $m,n$ where $m\le n<16$ and $(mn,16)=1$
|
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|
Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $\leftarrow$
Let $n$ be odd and consider $n= x^2-y^2 s.t. y+1 < x$
We can factor the difference of two squares as: $(x+y)(x-y)$ and we note for $n$ to be composite (not prime) $(x-y) > 1$. Thus we have shown this direction.
$\rightarrow$
Let $n$ be odd an composite. By definition of composite we have $n=ab$ for some odd integers $a$ and $b$. Now, working backwards:
$\dfrac{4ab}{4} = \dfrac{(a^2 + 2b + b^2)}{4} - \dfrac{(a^2-2b +b^2)}{4} = (\dfrac{a+b}{2})^2 - (\dfrac{a-b}{2})^2$. Thus, we shall take $x = \dfrac{a+b}{2}$ and $y =\dfrac{a-b}{2}$. And we have $n = x^2 - y^2$.
How do I get $y+1 < x$ in this direction?
|
You can't. However, you can show that $y+1<x$, which is what you should have been asking. Here it is, with $y=(a-b)/2, x=(a+b)/2$:
$$
\begin{align}
y+1 = \frac{a-b}{2}+1&=\frac{a-b+2}{2}<\frac{a+b}{2}\\
&\Longleftrightarrow a-b+2<a+b\\
&\Longleftrightarrow 2<2b \\
&\Longleftrightarrow 1<b \\
\end{align}
$$
which you know is true, since if $n=ab$ is composite with $b\le a$ we must have $b>1$. (Note: in your statement we can't just have $a,b$ odd integers.) Other than that, your proof is just fine.
|
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|
Prove $\sqrt[3]{60}>2+\sqrt[3]{7}$ Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$
I try to both sides of the cubic equation, but it is quite complicated
|
When you cube it, you get:
$$
60 > \left(2+\sqrt[3]{7} \right)^3 =15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
Continue to get
$$
\frac{60-15}{6\sqrt[3]{7}} > \left(2+\sqrt[3]{7} \right).
$$
Now cube again to get:
$$
\left(\frac{60-15}{6\sqrt[3]{7}}\right)^3 > 15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
The rhs is now larger than at the beginning, because $\displaystyle\frac{\frac{60-15}{6\sqrt[3]{7}}}{\sqrt[3]{60}}=\frac{60-15}{6\sqrt[3]{7}\sqrt[3]{60}} >1$.
To show this we use:
$$
45>6\sqrt[3]{7\times 60} \Rightarrow 45^3=91125 >6^3420=216\cdot 420=90720
$$
|
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|
Finding the $n$-th derivative of $f(x) =e^x \sin x$, solving the recurrence relation I am trying to find a closed solution for the nth derivative of the function:
$f(x) = e^x \sin x$
So far I have been able to obtain the derivative as:
$f^{(n)}(x) = e^x S_n \sin x + e^x C_n \cos x$
The sequences S and C are defined as below:
$S_n = S_{n-1} - C_{n-1}$
$C_n = S_{n-1} + C_{n-1}$
$S_0 = 1$,
$C_0 = 0$
I have been able to further simplify this by combining the two equations and obtaining:
$C_n = 2S_{n-2}$
$S_n = S_{n-1} - 2 S_{n-3}$
However, I have no idea what to do now. Can anyone help me find the closed form solution?
|
\begin{align}
\left( \begin{array}{r} S_n\\C_n\\ \end{array} \right) &=
\left( \begin{array}{rr}
1 &-1\\
1 &1\\
\end{array}
\right)
\left(\begin{array}{c} S_{n-1}\\C_{n-1}\\ \end{array} \right)\\
&=\left( \begin{array}{rr}
1 &-1\\
1 &1\\
\end{array}
\right)^n
\left(\begin{array}{c} S_{0}\\C_{0}\\ \end{array} \right)
\end{align}
Approach 1: Compute eigen-decomposition of the matrix and continue.
Approach 2: Eigenvalues are $(1+i)$ and $(1-i)$. Therefore, $S_n$ must be of the form $A(1+i)^n+B(1-i)^n$. Now, use values of $S_0$ and $S_1$ to compute $A$ and $B$. $A=B=1/2$.
After further simplification, we get
$$S_n = \sqrt{2^n} \cos \frac{n\pi}{4}, C_n = \sqrt{2^n} \sin \frac{n\pi}{4}$$
|
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|
How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
|
HINT:
$$
\begin{align*}
x^3-12x-16&=x\underbrace{(x^2-4)}_{(x-2)(x+2)}-8x-16 \tag{1}\\
&=x(x-2)(x+2)-\underbrace{8x-16}_{-8(x+2)}\\
&=(x+2) \underbrace{(x^2-2x-8)}_{\text{quadratic equation}}=0
\\
\end{align*}$$
Now just solve the quadratic equation and you should get something like:
$$(x+2)(x+a)(x+b)=0\tag{2}$$
and find: $\,\,x=-2,x=-a \,\,\text{or}\,\,x=-b$
|
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|
Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ How to show the following equality?
$$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$
|
Related problems: (I), (II). This problem is a direct application of Fourier transform and Poisson summation formula. Recalling the definition of Fourier transform and the Poisson summation formula respectively
$$ F(w) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-ixw} dx \,,$$
$$ \sum_{-\infty}^{\infty} f(n) = \sqrt{2\pi}\sum_{-\infty}^{\infty} F(2n\pi)\,, $$
where $F$ is the Fourier transform of $f$. Advancing with our problem, first, we compute the Fourier transform of $ f(x)=\frac{1}{x^2+a^2} $ which is equal to
$$ F(w) = \sqrt{\frac{\pi}{2}}\frac{1}{a}e^{-a|w|}\,.$$
Applying Poisson formula, we have
$$ \sum_{n=0}^{\infty}\frac{1}{n^2+a^2} = \frac{\pi}{a}\sum_{n=0}^{\infty}e^{-2an\pi} = \frac{\pi}{a} \sum_{n=0}^{\infty}r^{n}=\frac{\pi}{a}\frac{1}{1-r}\,,\quad r = e^{-2 \pi a} \,,$$
$$\Rightarrow \sum_{n=0}^{\infty}\frac{1}{n^2+a^2} = \frac{\pi}{a} \frac{1}{1-e^{-2a\pi}}=\frac{\pi}{a} \frac{e^{2a\pi}}{e^{2a\pi}-1} \,. $$
Now, I leave it to you to manipulate the above expression to reach the form
$$ \sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2} $$
You can use the identity
$$ \coth x = \frac{\cosh x}{\sinh x} = \frac {e^x + e^{-x}} {e^x - e^{-x}} = \frac{e^{2x} + 1} {e^{2x} - 1} \,. $$
|
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|
Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.
|
$21a^2+10ab+b^2-10a-8b+12=0$
$(7a+b)(3a+b)-4(7a+b+3a+b)+12=0$
$(7a+b-4)(3a+b-4)=4--->(1)$
$5a^2-4ab-b^2=(5a+b)(a-b)=-9--->(2)$
If $7a+b-4=\frac 2 k, 3a+b-4=2k$,
Express $a,b$ in terms of $k,$ replace their values in (2).
|
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|
How to find limit of a sequence defined by recurrence formula I have the following problem:
Let $a_{n}$ be the recurrence
$$a_{n+1}=a_{n}+2a_{n-1}$$
with $a_{0}=0$ and $a_{1}=1$. Can you help me find
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}$$
for $n\geq 1$?
|
You can solve this recurrence fairly easily using the characteristic equation, to get $a_n = \frac{1}{3}(2^n - (-1)^n)$. Then, as $n \rightarrow \infty$, $a_{n+1}/a_n \rightarrow 2$.
Edit: Given a linear homogeneous recurrence relation with constant coefficients $a_n = \sum_i c_i a_{n-i}$, assume $a_n = \alpha^n$ satisfies the recurrence. Then we get the characteristic equation $\alpha^n = \sum_i c_i \alpha^{n-i}$. The characteristic equation is a polynomial whose roots (the possible values of $\alpha$ that satisfy the recurrence) define the solution to the recurrence. If the roots $\alpha_1,\alpha_2,\dots,\alpha_m$ are unique (like in this example), then the solution is a linear combination of the roots to the $n$th power, i.e. $a_n = k_1 \alpha_1^n + k_2\alpha_2^n + \dots + k_m \alpha_m^n$, where the values of $k_1,k_2,\dots,k_m$ are defined by the initial conditions. If there are repeated roots, the combination includes powers of $n$ as well, but in this example the roots are unique, so I'll skip that.
For us, we have $a_n = a_{n-1} + 2a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$. Then, the characteristic equation is $\alpha^n = \alpha^{n-1} + 2\alpha^{n-2}$, which simplifies to $\alpha^2 = \alpha + 2$. The quadratic solutions give $\alpha_1 = 2$ and $\alpha_2 = -1$. Then, the theorem states that $a_n = k_1 2^n + k_2 (-1)^n$. Using the initial conditions, we have $a_0 = k_1 2^0 + k_2 (-1)^0 = k_1 + k_2 = 0$ and $a_1 = k_12^1 + k_2(-1)^1 = 2k_1 - k_2 = 1$. That means that $k_1 = \frac{1}{3}$ and $k_2 = - \frac{1}{3}$, so $a_n = \frac{1}{3}(2^n - (-1)^n)$.
Now, $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1/3(2^{n+1}-(-1)^{n+1})}{1/3(2^n - (-1)^n)} = \lim_{n \to \infty}\frac{2^{n+1}-(-1)^{n+1}}{2^n - (-1)^n} = \lim_{n \to \infty} \frac{2-(-1^{n+1}/2^n)}{1 - (-1/2)^n} = \frac{2}{1} = 2$
|
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|
Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
|
Work from the outside in. Begin by differentiating it term by term: $$f\,'(x)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-\frac{d}{dx}(6)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-0\;.$$
Now you have to calculate the derivative of $x^2\sqrt{2x+5}$. This is a product, so you use the product rule:
$$\left[x^2\sqrt{2x+5}\right]'=x^2\left[\sqrt{2x+5}\right]'+\left[x^2\right]'\sqrt{2x+5}\;.$$
To complete the differentiation you’ll need the derivative of $x^2$, which is very easy, and the derivative of $\sqrt{2x+5}$. That one is also pretty easy once you rewrite the function as $(2x+5)^{1/2}$: the power rule and the chain rule will take care of it.
|
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|
Partial Fractions - Calculus Evaluate the integral: $$\int \dfrac{9x^2+13x-6}{(x-1)(x+1)^2} dx$$
For some reason I cannot get the right answer. I split up the equation into three partial fractions but I cannot seem to find A, B, or C from the three subsequent equations. Thanks!
|
From $(A+B) x^2 + (2A+C)x +(A-B-C) = 9x^2+13 x -6$, you need to solve for $A,B,C$. Since the coefficients of the polynomials in $x$ must match, this gives three equations $A+B=9$, $2A+C = 13$, and $A-B-C = -6$. If you solve these you will get your answer.
To check, move your mouse over the following:
$$\frac{5}{x+1}+\frac{5}{(x+1)^2}+\frac{4}{x-1} = \frac{9x^2+13x-6}{(x-1)(x+1)^2}$$
|
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|
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
|
By Holder
$$\sum_{cyc}\frac{a^3}{a+b}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a+b)}=\frac{(a+b+c)^2}{6}\geq\frac{ab+ac+bc}{2}.$$
Done!
The Holder for three sequences it's the following.
Let $a_i$,$b_i$, $c_i$, $\alpha$, $\beta$ and $\gamma$ be positive numbers. Prove that:
$$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}(c_1+c_2+...+c_n)^{\gamma}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}c_1^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+\left(a_2^{\alpha}b_2^{\beta}c_2^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+...+\left(a_n^{\alpha}b_n^{\beta}c_n^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}\right)^{\alpha+\beta+\gamma}.$$
In our case $\alpha=\beta=\gamma=1$, $a_1=\frac{a^3}{a+b},$ $a_2=\frac{b^3}{b+c}$, $a_3=\frac{c^3}{c+a},$
$b_1=a+b,$ $b_2=b+c$, $b_3=c+a$ and $c_1=c_2=c_3=1$.
|
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|
Evaluate the sum $\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$?
Evaluate the series
$$\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}=?$$
Can you help me ? This is a past contest problem.
|
Let
\begin{align}S_n & =\sum_{k=0}^{n}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\
& =\sum_{k=0}^{n}\left(\frac{1}{6}\frac{1}{4k+1}-\frac{1}{2}\frac{1}{4k+2}+\frac{1}{2}\frac{1}{4k+3}-\frac{1}{6}\frac{1}{4k+4}\right).
\end{align}
$$A_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right),$$
$$B_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right),$$ and
$$C_n=\sum_{k=0}^{n}\left(\frac{1}{4k+2}-\frac{1}{4k+4}\right).$$
It is easy check that
$$S_n=\frac{1}{3}B_n-\frac{1}{6}A_n-\frac{1}{6}C_n.$$
Therefore,
$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{2B_n-A_n-C_n}{6}=\frac{2\ln 2-\dfrac{\pi}{4}-\dfrac{1}{2}\ln 2}{6}=\frac{1}{4}\ln 2-\frac{\pi}{24}.$$
|
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|
How do you solve an equation like this? We have an equation:
$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$
How do you solve these types of equations? For example, if we have:
$\dfrac{1}{x} = \dfrac{3}{2} $, we use:
$1\times 2 = 3\times x$
$x = 1.5$
What is a similair approach to my equation?
|
$$ \frac{p-1}{q} = \frac{q-1}{2p+1} = \frac {3}{5}$$ so
$$\frac{p-1}{q}=\frac {3}{5}$$ and $$\frac {3}{5}=\frac{q-1}{2p+1}.$$
\begin{eqnarray}
5p-5&=3q&\\
6p+3&=&5q-5
\end{eqnarray}
$$5p-3q=5$$
$$6p-5q=-8$$
$$30p-18q=30$$
$$30p-25q=-40$$
$$-7q=-70$$
$$q=10$$
$$p=7.$$
|
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|
Joint PDF. Is my book wrong? Consider that X and Y have the joint pdf $f(x,y) = (2/3)(x+1)$ for $0 < x < 1$ and $0 < y < 1$ and 0 otherwise. What is $P(X < 2Y < 3X)$? My book say the answer is 73/162
But I keep doing $\int_0^1 \int_{x/2}^{3x/2} f(x,y) \, dy \, dx$ and I am not getting 73/162
|
The region you need to integrate is the blue region as shown in the figure below.
Hence, your integral should go as follows.
\begin{align}
\int_0^{2/3} \int_{x/2}^{3x/2} f(x,y) \, dy \, dx + \int_{2/3}^1 \int_{x/2}^{1} f(x,y) \, dy \, dx & = \int_0^{2/3} \int_{x/2}^{3x/2} \dfrac23(x+1) \, dy \, dx\\
& + \int_{2/3}^1 \int_{x/2}^{1} \dfrac23 (x+1) \, dy \, dx\\
& = \int_0^{2/3} \dfrac23x(x+1) \, dx\\
& + \int_{2/3}^1 \dfrac23 (1-x/2)(x+1) \, dx\\
& = \dfrac23 \left(x^3/3+x^2/2 \right)_0^{2/3}\\
& + \dfrac23 \left(x^2/2+x - x^3/6 - x^2/4 \right)_{2/3}^1\\
& = \dfrac23 \left((2/3)^3/3+(2/3)^2/2 \right)\\
& + \dfrac23 \left(1/2+1 - 1/6 - 1/4 \right)\\
& - \dfrac23 \left((2/3)^2/2 + (2/3) \right)\\
& + \dfrac23 \left((2/3)^3/6 + (2/3)^2/4 \right) \\
& = \dfrac{73}{162}
\end{align}
|
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|
Closed Formula Expression for Sum of Combinatorics I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:
$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the largest multiple of $3$ less than or equal to $n$.
I have tried the expansion of $(1+1+1)^n$ which got nowhere, and I dont know how to proceed. I figure the answer will be conditional on $n \pmod 6$.
Can someone lend help? Thank you.
|
Let $\omega$ be a third root of 1.
Then
$$(1+1)^n = \binom{n}{0} +\binom{n}{1}+ \binom{n}{2}+ \binom{n}{3}+ ...+\binom{n}{n} \,.$$
$$ (1+\omega)^n = \binom{n}{0} + \binom{n}{1}\omega+ \binom{n}{2} \omega^2+ \binom{n}{3}+ ...+ \binom{n}{n} \omega^n \,.$$
$$ (1+\omega^2)^n =\binom{n}{0} + \binom{n}{1} \omega^2+ \binom{n}{2} \omega+ \binom{n}{3}+ ...+ \binom{n}{n}\omega^{2n} \,.$$
Now, since $1+ \omega +\omega^2=0$, adding them only every third column remains.
Thus
$$2^n+ (1+\omega)^n+(1+\omega^2)^n =3 \left( \binom{n}{0} +\binom{n}{3}+ \binom{n}{6}+ \binom{n}{9}+ ...+\binom{n}{3k} \right)$$
All you have left is to calculate $(1+\omega)^n$ and $(1+\omega^2)^n$ by writing them in polar/trig form.
P.S. Same trick with $\omega(1+\omega)^n$ and $\omega^2 (1+\omega^2)^n$ yields $\left( \binom{n}{2} +\binom{n}{5}+ \binom{n}{8}+ \binom{n}{11}+ ... \right)$ while $\omega^2(1+\omega)^n$ and $\omega (1+\omega^2)^n$ yields $\left( \binom{n}{1} +\binom{n}{4}+ \binom{n}{7}+ \binom{n}{10}+ ... \right)$.
|
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|
Area fractal pentagrams I When I saw this image I was a little curious.
How can I find the area of this fractal?
|
First,
$ Area\ of\ Star\ with\ 5\ petals\ = Area\ of\ pentagram\ ->\ step\ 1 $
please refer http://mathworld.wolfram.com/Pentagram.html for area formula of pentagram. because the tricky part is identifying the GP.
Idea is that from each of the 5 triangles (assuming the new triangle has side one third the length of base triangle on which it forms ,as fractals are regular figures) there are 2 triangles coming up .
after iteration 1 added area =area of 10 equilaterla triangles of side $ \frac{a}3 $
after iteration 2 added area =area of 20 equilaterla triangles of side $ \frac{a}{3^2} $
after iteration 3 added area =area of 40 equilaterla triangles of side $ \frac{a}{3^3} $
so as this is infinite as per your problem this goes on and on
Area = Area of STAR (as found in step 1) + area added after infinite iteration (let this be k)
$k\ = 10\times(\sqrt(3)/4)(\frac{a}{3})^2\ +\ 20\times(\sqrt(3)/4)(\frac{a}{3^2})^2\ +\ 40\times(\sqrt(3)/4)(\frac{a}{3^3})^2\ ...$
$k\ = 10\times(\sqrt(3)/4)a^2 [ \frac{1}{3^2} + \frac{2}{3^4} + \frac{2^2}{3^6} +\ ..... ]$
this within square brackets is a infinite GP with common ratio of $\frac{2}{3^2}\ $and first term is $\frac{1}{3^2} $
as summation of GP with infinite series is $\frac{firstterm}{1-commonratio} $
$k\ = 10\times(\sqrt(3)/4)a^2 \times [\frac{\frac{1}{3^2}}{1-\frac{2}{3^2}}]$
$k\ = 10\times(\sqrt(3)/4)a^2 \times [\frac{1}{7}]$
so
$Area = Area\ of\ PENTAGRAM\ + 10\times(\sqrt(3)/4)a^2 \times [\frac{1}{7}]$
|
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|
Induction: $\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}$ I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
|
With induction we prove:
*
*$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2^n$
*$0\binom{n}{0}+1\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}=n2^{n-1}$
*$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $
For the proofs we use Pascal's identity: $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$.
*
*If $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2^n$ then
$$\binom{n+1}{0}+\binom{n+1}{1}+\binom{n+1}{2}+\cdots+\binom{n+1}{n}+\binom{n+1}{n+1}=\\
\binom{n}{0}+\binom{n}{1}+\binom{n}{0}+\binom{n}{2}+\binom{n}{1}+\cdots+\binom{n}{n}+\binom{n}{n-1}+\binom{n}{n}=\\
\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}+\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n-1}+\binom{n}{n}=2^n+2^n=2^{n+1}.$$
*If $0\binom{n}{0}+1\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}=n2^{n-1}$ then
$$0\binom{n+1}{0}+1\binom{n+1}{1}+2\binom{n+1}{2}+\cdots+n\binom{n+1}{n}+(n+1)\binom{n+1}{n+1}=\\
\ldots = \\
2\left(0\binom{n}{0}+1\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}\right)+\\
\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2n2^{n-1}+2^n=(n+1)2^n.$$
*If $0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $ then
$$0^2\binom{n+1}{0}+1^2\binom{n+1}{1}+2^2\binom{n+1}{2}+\cdots+n^2\binom{n+1}{n}+(n+1)^2\binom{n+1}{n+1}=\\
\ldots =\\
2\left( 0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}\right)+\\
2\left(0\binom{n}{0}+1\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}\right)+\\
\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=\\
2n(1+n)2^{n-2}+2n2^{n-1}+2^n=(n+1)(2+n)2^{n-1}
$$
|
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|
Integral table contradiction? In a table of integrals, I see the following two formulas:
$\int \frac{dx}{(a+x)(b+x)} = \frac{1}{b-a}\ln\frac{a+x}{b+x}$, and
$\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}$.
How can these both be true? It seems like if we expand $(a+x)(b+x)$ out to $x^2+(a+b)x+ab$, we can apply the 2nd equation to get
$\int \frac{dx}{(a+x)(b+x)} = \frac{2}{\sqrt{4ab-(a+b)^2}}\tan^{-1}\frac{2x+a+b}{\sqrt{4ab-(a+b)^2}}$,
which is surely not equivalent to $\frac{1}{b-a}\ln\frac{a+x}{b+x}$ (one involves a logarithm and the other involves an arctan, so no amount of algebraic fussing can reconcile them, can it?!)
|
Take a look at the logarithmic form of the complex arctangent:
$$\tan^{-1}x=\frac12i\Big(\ln(1-ix)-\ln(1+ix)\Big)\;.\tag{1}$$
For instance, consider the integral
$$\int\frac{dx}{(1+x)(2+x)}=\int\frac{dx}{x^2+3x+2}\;.$$ The two integration formulas yield the antiderivatives
$$\ln\frac{1+x}{2+x}\tag{2}$$ and
$$\frac2i\tan^{-1}\frac{2x+3}i\tag{3}\;.$$
From $(1)$ we have
$$\begin{align*}
\frac2i\tan^{-1}\frac{2x+3}i&=\Big(\ln(1-(2x+3))-\ln(1+2x+3)\Big)\\
&=\ln\frac{-2-2x}{4+2x}\\
&=\ln\frac{-1-x}{2+x}\\
&=\ln\frac{1+x}{2+x}+\ln(-1)\\
&=\ln\frac{1+x}{2+x}+i\pi\;,
\end{align*}$$
using the principal value of the log. Thus, $(2)$ and $(3)$ really do differ only by a constant, and there is no problem.
|
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|
Sum of the sum of the sum of the first $n$ natural numbers I have here another problem of mine, which I couldn't manage to solve.
Given that: $$x_n = 1 + 2 + \dots + n \\ y_n = x_1 + x_2 + \dots + x_n
\\ z_n = y_1 + y_2 + \dots + y_n $$
Find $z_{20}$.
I know the answer but I'm having a hard time reaching it. I recognized that $x_n$ is obviously $\dfrac{n(n + 1)}{2}$, but how to express the other two in a closed form to allow the calculation? I even tried writing the relations in a recursive way, without success. Is there an easy way to solve this one? I was not allowed to use calculators.
Thanks,
rubik
|
Here is a slightly longer method than Brian M. Scott's, relying on knowing closed forms for $n$, $n^{2}$ and $n^{3}$.
You have: $$x_{n}=\frac{n(n+1)}{2}=\frac{1}{2}n^{2}+\frac{1}{2}n$$
Therefore: $$\begin{align}y_{n}&=\sum_{i=1}^{n}{\left(\frac{1}{2}i^{2}+\frac{1}{2}i\right)}=\frac{1}{2}\left(\sum_{i=1}^{n}i^{2}+\sum_{i=1}^{n}{i}\right)=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right) \\ &= \frac{1}{2}\left(\frac{1}{3}n^{3}+n^{2}+\frac{2}{3}n\right) \end{align}$$
Therefore:
$$\begin{align}z_{n}&=\sum_{i=1}^{n}{\left(\frac{1}{6}n^{3}+\frac{1}{2}n^{2}+\frac{1}{3}n\right)}=\frac{1}{6}\sum_{i=1}^{n}n^{3}+\frac{1}{2}\sum_{i=1}^{n}n^{2}+\frac{1}{3}\sum_{i=1}^{n}{n} \\ &= \frac{1}{6}\cdot\frac{n^{2}(n+1)^{2}}{4}+\frac{1}{2}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{1}{3}\cdot\frac{n(n+1)}{2} \\ &= \frac{1}{24}n(n+1)(n+2)(n+3)\end{align}$$
|
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|
Forward and Backward Euler. I want to consider this differential system:
$$
\ \frac{dx}{dt} = -y(t)\\
\frac{dy}{dt} = \ x(t)
$$
where $t>0$ with initial condition$ (x(0),y(0))=(1,0).$
First I want to show that this differential equation admits an invariant of $I = x(t)^2 + y(t)^2$ Also, Can someone help me to figure out if Forward Euler, backwards Euler, or implicit trapezoidal rule admit invariants similar to $I = x(t)^2 + y(t)^2$?
|
I would like to explain this, using explicit Euler, because it is the most elementary.
We have:
\begin{align}
\begin{pmatrix}
x \\ y
\end{pmatrix}'
=
\begin{pmatrix}
-y \\ x
\end{pmatrix}
= f(x,y)
\end{align}
Explicit euler looks like this: $z_{k+1} = z_k +h \cdot f(z_k)$, applied to our problem it becomes:
\begin{align}
\begin{pmatrix}
x_{k+1} \\ y_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k}
\end{pmatrix}
+h
\begin{pmatrix}
-y_{k} \\ x_{k}
\end{pmatrix}
=
\begin{pmatrix}
x_{k}-hy_k \\ y_{k}+h x_k
\end{pmatrix}
\end{align}
So if we want to know, whether $x(t)^2+y(t)^2$ are constant, we check
\begin{align}
x_{k+1}^2+y_{k+1}^2 = (x_k -h y_k)^2 +(y_k+ h x_k)^2 = (x_k^2+y_k^2)(1+h^2)
\end{align}
Give the initial values $x_0=1$ and $y_0=0$ and using recursion, we get
\begin{align}
x_{k+1}^2+y_{k+1}^2 &=(x_k^2+y_k^2)(1+h^2) =(x_{k-1}^2+y_{k-1}^2)(1+h^2)^2 =\ ... \ =(x_0^2+y_0^2)(1+h^2)^{k+1}\\ &= (1+h^2)^{k+1}
\end{align}
So it is no invariant.
Next implicit euler. The method reads $z_{k+1} = z_k +h \cdot f(z_{k+1})$
So we get
\begin{align}
& \begin{pmatrix}
x_{k+1} \\ y_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k}
\end{pmatrix}
+h
\begin{pmatrix}
-y_{k+1} \\ x_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k}-hy_{k+1} \\ y_{k}+h x_{k+1}
\end{pmatrix}
\\
& \Leftrightarrow
\begin{pmatrix}
x_{k+1}+ h y_{k+1} \\ y_{k+1} -h x_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k}
\end{pmatrix}
\\
& \Leftrightarrow
\begin{pmatrix}
1 & h \\ 1 & -h
\end{pmatrix}
\begin{pmatrix}
x_{k+1} \\ y_{k+1}
\end{pmatrix}
=
\begin{pmatrix}
x_{k} \\ y_{k}
\end{pmatrix}
\end{align}
So as you can see, you end up with a linear system you have to solve for $x_{k+1} $ and $y_{k+1}$. This is the price you pay for an implicit scheme. If you solve for $x_{k+1} $ and $y_{k+1}$ you can again check the invariant.
Good luck,
Thomas
|
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|
Why is $\sqrt{8}/2$ equal to $\sqrt{2}$? I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.
Thanks in advance
|
Other techniques to prove that result:
Technique $\mathbf 1$:
We may also prove that by solving this equation for $x$:
$$\begin{align}\frac{\sqrt x}{2}&=\sqrt2\\\sqrt x&=2\sqrt 2\\\left(\sqrt x\right)^2&=\left(2\sqrt 2\right)^2\\x&=4\cdot2=\color{green}{\boxed{\color{black}{8}}}\end{align}$$
Therefore: $$\frac{\sqrt8}{2}=\sqrt 2$$
I know, this is a terrible way. But hey, it works! ;-)
Technique $\mathbf 2$:
Just like in Technique $\mathbf 1$: We may also prove that result by solving this equation for $x$:
$$\begin{align}\frac{\sqrt 8}{2}&=\sqrt x\\
\left(\frac{\sqrt 8}{2}\right)^2&=\sqrt x^2\\
\frac84&=x
\implies x=\color{green}{\boxed{\color{black}2}}
\end{align}$$
Therefore: $$\frac{\sqrt8}{2}=\sqrt 2$$
Technique $\mathbf 3$:
If $a=b$ then this means $a-b=0$, and the opposite is true. So let's check for $\sqrt 2$ and $\sqrt 8/2$:$$\begin{align}\dfrac{\sqrt 8}{2}-\sqrt{2}&=\dfrac{\sqrt 8}{2}-\left(\sqrt{2} \cdot \dfrac{\sqrt{4}}{2}\right)\quad\to\quad\text{since $\sqrt 4=2$}\\
&=\dfrac{\sqrt 8}{2}-\dfrac{\sqrt 2\cdot\sqrt 4}{2}\\
&=\dfrac{\sqrt{8}-\sqrt{2}\cdot\sqrt{4}}{2}\\
&=\dfrac{\sqrt{8}-\sqrt{2\cdot4}}{2}\\
&=\dfrac{\sqrt 8-\sqrt 8}{2}=\dfrac{0}{2}=0
\end{align}$$
Therefore, we can easily conclude that: $$\dfrac{\sqrt 8}{2}=\sqrt 2$$
Technique $\mathbf 4$:
If $a=b$ then it follows that $a/b=1$, and the opposite is true. So let's check for $\sqrt 2$ and $\sqrt 8/2$: $$\begin{align}\dfrac{\sqrt2}{\dfrac{\sqrt8}{2}}&=\sqrt{2}\cdot\dfrac{2}{\sqrt{8}}=\dfrac{\sqrt2\cdot\sqrt 4}{\sqrt8}=\dfrac{\sqrt{2\cdot 4}}{\sqrt 8}=\dfrac{\sqrt8}{\sqrt8}=1
\end{align}$$ Therefore: $$\dfrac{\sqrt 8}{2}=\sqrt 2$$
Technique $\mathbf 5$:
We will do a proof by contradiction: We suppose that $\sqrt 8/2\ne\sqrt 2$, therefore there must be a difference between them which is not zero $\sqrt 8/2-\sqrt 2\ne0$, that we will call $x$. So let's check if $x\ne0$: $$\begin{align}x=\dfrac{\sqrt 8}{2}-\sqrt{2}&=\dfrac{\sqrt 8}{2}-\left(\sqrt{2} \cdot \dfrac{\sqrt{4}}{2}\right)\\
&=\dfrac{\sqrt 8}{2}-\dfrac{\sqrt 2\cdot\sqrt 4}{2}\\
&=\dfrac{\sqrt{8}-\sqrt{2}\cdot\sqrt{4}}{2}\\
&=\dfrac{\sqrt{8}-\sqrt{2\cdot4}}{2}\\
&=\dfrac{\sqrt 8-\sqrt 8}{2}=\dfrac{0}{2}=0
\end{align}$$But this is a contradiction, since $x$ must be non-zero but it turns out that it is equal to $0$. Therefore: $$\dfrac{\sqrt 8}{2}=\sqrt 2$$
Generalization: Let $x$ be a strictly positive number in $\Bbb R$, then: $$\dfrac{\sqrt{x^3}}{x}=\sqrt{x}$$
Proof: $$\dfrac{\sqrt{x^3}}{x}=\dfrac{\sqrt{x^3}}{\sqrt{x^2}}=\sqrt{\dfrac{x^3}{x^2}}=\sqrt x\qquad\blacksquare\\\,\\\textit{inspired by Paul Dirac's post :)}$$
|
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|
If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of these triples as $8^2 + 15^2 = 64 + 225= 289 = 17^2$. Other examples of this triple are $(3, 4, 5)$ and $(5, 12, 13)$.
Using Proof by Contradiction, show that: If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple.
|
Since I assume that you're new to some Mathematical topics, I'd use basic language.
First of all, the question itself teaches you that, if $(a,b,c)$ is a Pythagorean triplet, then it means that $a^2 + b^2 = c^2$. So if $\rm (foo_1, foo_2, foo_3)$ is a Pythagorean triplet, then $\rm (foo_1)^2 + (foo_2)^2 = (foo_3)^2$ (right?). So as you are given in the question, if $(a+1,b+1,c+1)$ is a triplet,then $(a + 1)^2 + (b + 1)^2 = (c + 1)^2$
You are left with two equations:$$\begin{align} a^2 + b^2 &= c^2 \\ (a + 1)^2 + (b + 1)^2 &= (c + 1)^2\end{align}$$Subtract, and see the magic. Hint: knowledge of "even," and "odd" numbers will help. Especially that odd numbers are in the form $2m + 1 $ and even ones in $2m$... $m$ is an integer by the way.
|
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|
Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
|
T(n)=2T(n-1)+n -----(1)
T(n-1)=2T(n-2)+n-1 ------(2)
T(n-2)=2T(n-3)+n-2 ------(3)
Substitute (3) in (2) and then (2) in (1)
T(n-1)=$2^2T$(n-3)+2(n-2)+(n-1)
T(n)= $2^3$T(N-3)+$2^2$(n-2)+2(n-1)+n
=> T(k)=$2^k$T(n-k) + $2^{k-1}$(n-(k-1))+........+$2^0$n
Now, since T(1)=1, let n-k=1 => k=n-1
T(n)=$2^{n-1}$T(1) + $2^{n-2}$(2) + $2^{n-3}$(3) + $2^{n-n}$(n) --------(4)
Multiply Equation (4) with 2
T(n)=$2^{n}$+$2^{n-1}$ + $2^{n-2}$ + $2^{n-3}$+...... + 2n ----------- (5)
Subtract Equation (5) with (4)
T(n)=$2^{n}$+$2^{n-1}$+$2^{n-2}$+$2^{n-3}$+........+2 -n
Using sum of GP terms, we get :
[ $2^{1}$($2^{n}$-1) - n ]/ (2-1)
= $2^{n+1}$ -2 -n
|
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|
How would you solve this recurrence equation: $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ How would you solve $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ ?
I don't understand the text in my textbook. I Would like somebody to explain it to me.
|
\begin{align}
a_{n+1} & = 6 \cdot 5^n + 2 a_n\\
& = 6 \cdot 5^n + 2 (6 \cdot 5^{n-1} + 2a_{n-1})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2a_{n-1}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 (6 \cdot 5^{n-2} + 2a_{n-2})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3a_{n-2}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3(6 \cdot 5^{n-3} + 2 a_{n-3})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3 \cdot 6 \cdot 5^{n-3} + 2^4 \cdot a_{n-3}
\end{align}
Hence, in general (you need to prove the claim below using induction)
$$a_{n+1} = 2^{j+1}a_{n-j} + 6 \left( \sum_{k=0}^{j} 2^k 5^{n-k} \right)$$
Setting $j=n$, we get that
\begin{align}
a_{n+1} & = 2^{n+1}a_0 + 6 \left( \sum_{k=0}^{n} 2^k 5^{n-k} \right) = 2^{n+1}a_0 + 6 \cdot 5^n \left( \sum_{k=0}^{n} \left(\dfrac25 \right)^k \right)\\
& = 2^{n+1}a_0 + 6 \cdot 5^n \cdot \dfrac{1-(2/5)^{n+1}}{1-2/5}\\
& = 2^{n+1}a_0 + 6 \cdot \dfrac{5^{n+1} -2^{n+1}}3\\
& = 2^{n+1}a_0 + 2 \cdot (5^{n+1} -2^{n+1})
\end{align}
EDIT
You can verify if it is true by plugging the value for $a_n$ and $a_{n+1}$ and computing $a_{n+1} - 2a_n$.
$$a_n = 2^{n}a_0 + 2 \cdot (5^{n} -2^{n})$$
Hence, $$2a_n = 2^{n+1}a_0 + 4 \cdot (5^{n} -2^{n})$$
Hence,
\begin{align}
a_{n+1} - 2a_n & = 2^{n+1}a_0 + 2 \cdot (5^{n+1} -2^{n+1}) - \left( 2^{n+1}a_0 + 4 \cdot (5^{n} -2^{n})\right)\\
& = 2 \cdot (5^{n+1} -2^{n+1}) - 4 \cdot (5^n - 2^n)\\
& = 2 \cdot 5^{n+1} -2^{n+2} - 4 \cdot 5^n + 4 \cdot 2^n\\
& = 10 \cdot 5^{n} - 4 \cdot 5^n\\
& = 6 \cdot 5^n
\end{align}
|
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|
Solve $x^2 + 10 = 15$ How do I solve the following equation?
$$x^2 + 10 = 15$$
Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.
I've also seen another equation like this:
\begin{align*}
x^2 & = 4 \\
x^2 + 4 & = 0 \\
(x - 2)(x + 2) & = 0 \\
x & = 2 \text{ or } -2
\end{align*}
So I guess I could near the end of my equation do the following:
$$x^2 + 5 = 0$$
and then go from there?
Is my first attempt at solving correct?
|
Your equation should be corrected at the second line to be $$x^2 -4 = 0$$ and similarly $$x^2-5=0$$.
Then $$(x-\sqrt{5})(x+\sqrt{5})=0$$ which implies $$x=\sqrt{5}$$ or $$x=-\sqrt{5}$$
The answer being irrational doesn't matter, as $x$ is in the real number set (or can even be complex number set, a quadratic equation always have a solution in the complex number set).
|
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|
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & 1 & 2 & 3 & 0 & 1 & 2 \\
0 & 0 & 1 & 2 & 0 & 0 & 1
\end{array}
$$
From this I get that:
$$
\det{A} = (1 \cdot 3 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 \cdot 0 + 3 \cdot 3 \cdot 0 \cdot 0 + 4 \cdot 2 \cdot 1 \cdot 1) - (3 \cdot 3 \cdot 0 \cdot 2 + 2 \cdot 2 \cdot 3 \cdot 1 + 1 \cdot 3 \cdot 2 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 0) = (12 + 0 + 0 + 8) - (0 + 12 + 0 + 0) = 8
$$
But WolframAlpha is saying that it is equal 0. So my question is where am I wrong?
|
The trick you are applying (Rule of Sarrus) only works for $ 3\times 3$ Matrices.
|
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|
How do we prove that two parametric equations are drawing the same thing? For example, if I have
$$\begin {align}
x(t) &= r\sin t\cos t\\
y(t) &= r\sin^2 t\\
\end {align}$$
and
$$\begin {align}
x(t) &= \frac r 2 \cos t\\
y(t) &= \frac r 2 (\sin t + 1)
\end {align}$$
How do we show that the two parametric equations draw the same line?
|
Some efforts to establish that two parametrizations represent the same curve are trickier than others. In your first set, you will need the "double-angle" trig identities, or their relatives (I am also going to write '$r$' as '$R$' , since it is intended to be a constant, rather than radius in polar coordinates):
$$\sin 2 \theta \ = \ 2 \sin\theta \cos\theta \ \rightarrow \ x(t) \ = \ R \ \sin t \cos t \ = \ R \cdot \frac{1}{2} \sin2t \ , $$
$$\sin^2 \theta \ = \ \frac{1}{2} (1 - \cos 2\theta) \ \rightarrow \ y(t) \ = \ R \ \sin^2 t \ = \ R \cdot \frac{1}{2} (1- \cos 2t) $$
$$ R \ \sin 2t \ = \ 2x \ , \ 2y \ = \ R - R \ \cos 2t \ \Rightarrow \ R \ \cos 2t \ = \ R - 2y $$
$$ R^2 \sin^2 2t \ + \ R^2 \cos^2 2t \ = \ R^2 \ = \ (2x)^2 \ + \ ( R - 2y)^2 \ = \ 4x^2 + \ R^2 \ - \ 4Ry \ + 4y^2 $$
$$ \Rightarrow \ 0 \ = \ 4x^2 \ - \ 4Ry \ + \ 4y^2 \ \Rightarrow \ x^2 \ + \ y^2 \ - \ Ry \ = \ 0$$
$$ \Rightarrow \ x^2 \ + \ (y^2 \ - \ Ry \ + \frac{R^2}{4}) \ = \ \frac{R^2}{4} \ \Rightarrow \ x^2 \ + \ (y - \frac{R}{2})^2 \ = \ (\frac{R}{2})^2 \ ,$$
that is, a circle centered at $(0, \frac{R}{2})$ with radius $ \frac{R}{2}$ . Since the trig functions both have a period of $\pi$, the parametrization starts on the circle with $t = 0$ at $(0,0)$, advances counter-clockwise to $(\frac{R}{2}, \frac{R}{2})$ at $t = \frac{\pi}{4}$, and continues around to return to the origin at $t = \pi$ .
With the second parametrization, we follow a similar process (so I won't show all the steps):
$$x(t) \ = \ \frac{R}{2} \cos t \ , \ y(t) \ = \ \frac{R}{2} ( \sin t \ + \ 1)$$
$$ R \ \cos t \ = \ 2x \ , \ 2y \ = R \ \sin t \ + \ R \ \Rightarrow \ R \ \sin t \ = \ R - 2y $$
$$ R^2 \sin^2 t \ + \ R^2 \cos^2 t \ = \ R^2 \ = \ (2x)^2 \ + \ ( R - 2y)^2 \ \Rightarrow \ x^2 \ + \ (y - \frac{R}{2})^2 \ = \ (\frac{R}{2})^2 \ .$$
So we have the same circle again. However, this parametrization has a period of $2 \pi$ , starts at $(\frac{R}{2}, \frac{R}{2})$ for $t = 0$ , advances counter-clockwise to $(0,R)$ at $t = \frac{\pi}{2}$ , and completes its cycle at $(\frac{R}{2}, \frac{R}{2})$ for $t = 2 \pi$ . (As mjqxxx has said, the circular trajectory is followed in the same direction, but at different "speeds" and "starting points".)
You may well imagine that an unlimited number of other parametrizations can be chosen, differing not only in "rate of travel", "direction of travel", and "initial point", but also using "speed functions" which are not constant (or even reverse themselves!). Because this is possible, there are many ways to "disguise" a parametric curve; consequently, there is no general method for checking whether two descriptions actually produce the same figure.
|
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|
Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
|
Case 1: At least one of $x,y,z < 0$. Let's say $z < 0$. If we let $f(x,y,z) = 2(x+y+z) - xyz$, it's then clear that $f(x,y,z) \leq f(t,t,z)$, where $t = \sqrt{\frac{x^2+y^2}{2}}$. It then suffices to check $f(t,t,z) \leq 10$. Let $z = -u$, so that $u$ is positive. Then we need to show
$$f(t,t,-u) = 2(2t - u) + t^2 u \leq 10$$
on the condition $2t^2 + u^2 = 9$. We can try to express everything in terms of $u$:
$$ 2(2t - u) + t^2 u \leq 10 \\
\Leftrightarrow 4(2t - u) + (9 - u^2)u \leq 20 \\
\Leftrightarrow u^3 - 5u + 20 \ge 8t = 8 \sqrt{\frac{9 - u^2}{2}} \\
\Leftrightarrow (u^3 - 5u + 20)^2 - 32 (9-u^2) \ge 0 \\
\Leftrightarrow (u-1)^2(u^4 + 2u^3 - 7u^2 + 24u + 112) \ge 0$$
The last inequality is clearly true, since $u^4 + 49 \ge 14u^2 > 7u^2$. So this case holds, with equality $z = -u = -1$, $x = y = t = 2$.
Case 2: $x,y,z \ge 0$. WLOG assume $x \ge y \ge z$.
Case 2a: $x^2 \leq \frac{9 + \sqrt{65}}{2}$. ($x$ is "small")
We check that $f(x,y,z) \leq f(x, \sqrt{y^2+z^2}, 0)$.
$$2(x+y+z) - xyz \leq 2(x+\sqrt{y^2+z^2}) \\
\Leftrightarrow 2(y+z - \sqrt{y^2+z^2}) \leq xyz \\
\Leftrightarrow \frac{4}{y+z+\sqrt{y^2+z^2}} \leq x \\
\Leftrightarrow 4 \leq x(y+z+\sqrt{y^2+z^2}
$$
The last inequality follows from
$$x(y+z+\sqrt{y^2+z^2}) \ge 2x\sqrt{y^2+z^2} = 2 x \sqrt{9-x^2} \ge 4$$
and the condition on $x$.
Thus we may assume $z = 0$. Then $f(x,y,z) = 2(x+y) \leq 2\sqrt{2(x^2+y^2)} = 2\sqrt{18} < 10$.
Case 2b: $x^2 \ge \frac{9 + \sqrt{65}}{2}$. ($x$ is "large")
We get a trivial bound on $y+z$ by Cauchy-Schwarz:
$$y + z \leq \sqrt{2(y^2+z^2)} = \sqrt{2(9-x^2)} \leq \sqrt{9 - \sqrt{65}} < 1$$
Then
$$f(x,y,z) = 2(x+y+z) - xyz < 2(x+y+z) \leq 2 (3+1) < 10.$$
All in all, it's possible to do this without calculus, although I can't understand why one would not use a tool like Lagrange multiplier that is especially well suited for this situation. It's exactly the power of those tools that render brute-force solutions like this one unnecessary.
|
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|
Evaluating$\int {1\over1-\sin2x}dx$ $$
\int {1\over1-\sin 2x}dx = \int {1\over \sin^2 x-2\sin x\cos x+\cos^2x}dx = \int {1\over (\sin x-\cos x)^2}dx
$$
From here I get two different answers, depending on whether I factor out $\sin x$ or $\cos x$.
Factoring out $\sin x$, this one is correct according to WolframAlpha:
$$=\int {1\over [\sin x(1-{\cos x\over \sin x})]^2}dx=\int {1\over \sin^2x(1-{\cos x\over \sin x})^2}dx = \int {1\over(1-\cot x)^2}d(1-\cot x)$$
$$={1\over \cot x-1}+C$$
But when I factor out $\cos x$:
$$=\int {1\over [\cos x({\sin x\over \cos x}-1)]^2}dx=\int {1\over \cos^2x({\sin x\over \cos x}-1)^2}dx = \int {1\over(\tan x-1)^2}d(\tan x-1)$$
$$={1\over 1-\tan x}+C$$
I bet it's just some stupid typo that I'm missing, I can't figure it out.
Thanks.
|
They are both correct. To see this, you can simply take the derivative of each antiderivative to get your original integral back. This shows that both primitives you found are equal, up to a constant.
$$\frac{d}{dx} \left(\frac{1}{\cot x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$
$$\frac{d}{dx} \left(\frac{1}{\tan x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$
\begin{eqnarray*}
\left(\cos x - \sin x\right)^2 &=& (\cos x - \sin x)(\cos x - \sin x)\\
&=& \cos^2 x - 2 \sin x \cos x + \sin^2 x\\
&=& 1 - \sin 2x
\end{eqnarray*}
In your first line, you have
\begin{eqnarray*}
(\sin x - \cos x)^2 &=& (\sin x - \cos x)(\sin x - \cos x)\\
&=& \sin^2 x - 2 \sin x \cos x + \cos^2 x\\
&=& 1-\sin 2x
\end{eqnarray*}
As you can see, these are equivalent.
|
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|
Choosing a 5 member team out of 12 girls and 10 boys We must choose a 5-member team from 12 girls and 10 boys. How many ways are there
to make the choice so that there are no more than 3 boys on the team?
The correct answer is $\binom{22}{5} - \binom{12}{1} \binom{10}{4} - \binom{10}{5}$.
I understand the $\binom{22}{5}$ part, but where I am confused at is the other two parts. I do not know how to get those parts. Can anyone help me understand how to get to the solution?
|
That solution proceeds by starting with $\binom{22}5$, the total number of possible $5$-person teams, and subtracting the $\binom{12}1\binom{10}4$ teams that have one girl and four boys and the $\binom{10}5$ teams that have five boys.
The problem could also be solved by noting that there are $\binom{12}2\binom{10}3$ teams with two girls and three boys, $\binom{12}3\binom{10}2$ teams with three girls and two boys, $\binom{12}4\binom{10}1$ teams with four girls and one boy, and $\binom{12}5$ teams with five girls, and calculating the sum
$$\binom{12}2\binom{10}3+\binom{12}3\binom{10}2+\binom{12}4\binom{10}1+\binom{12}5\;,$$
but it’s pretty clearly more efficient to calculate
$$\binom{22}5-\binom{12}1\binom{10}4-\binom{10}5\;.$$
|
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|
An application of the Mean Value Theorem I'm recalling this question from memory, so I may be messing it up a bit.
Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.
Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c$. Then $a+b-c=2a\xi+b \Rightarrow (a-c)/2a=\xi$.
I'm not sure if this is right or where to go from here.
|
First, if $a =0$, then we have $bx + c = 0 \implies x = - \frac{c}{b} = \frac{b/2}{b} = \frac{1}{2}$.
Now, suppose $a \neq 0$.
Note that $c = - \frac{a}{3} - \frac{b}{2}$, so you want to prove that the function $f(x) = ax^2 + bx - \frac{a}{3} - \frac{b}{2}$ has a root in $[0,1]$. We have $f(0) = - \frac{a}{3} - \frac{b}{2}$ and $f(1) = \frac{2}{3} a + \frac{1}{2} b$. Note that
$$f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{4}b^2 < 0$$ as $a \neq 0$. Thus, $f(0)$ and $f(1)$ have different signs, and by the Intermediate Value Theorem, there is a root in $[0,1]$.
|
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|
How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any?
|
Example 1: Find the points of intersection of the circles given by their equations as follows:
$(x - 2)^2 + (y - 3)^2 = 9$
$(x - 1)^2 + (y + 1)^2 = 16$
Solution to Example 1:
We first expand the two equations as follows:
$x^2 - 4x + 4 + y^2 - 6y + 9 = 9 $
$x^2 - 2x + 1 + y^2 + 2y + 1 = 16 $
Multiply all terms in the first equation by -1 to abtain an equivalent equation and keep the second equation unchanged
$-x^2 + 4x - 4 - y^2 + 6y - 9 = -9 $
$x^2 - 2x + 1 + y^2 + 2y + 1 = 16 $
We now add the same sides of the two equations to obtain a linear equation
$2x - 3 + 8y - 8 = 7 $
Which may written as
$x + 4y = 9 \textbf{ or } x = 9 - 4y $
We now substitute $x$ by $9 - 4y$ in the first equation to obatin
$(9 - 4y)^2 - 4(9 - 4y) + 4 + y^2 - 6y + 9 = 9 $
Which may be written as
$17y^2 -62y + 49 = 0 $
Solve the quadratic equation for y to obtain two solutions
$y = \frac{(31 + 8\sqrt{2})}{17} \approx 2.49 $
and $ y =\frac{31 - 8\sqrt{2}}{17} \approx 1.16 $
We now substitute the values of y already obtained into the equation $x = 9 - 4y $ to obtain the values for x as follows
$x = \frac{29 + 32\sqrt{2}}{ 17} \approx - 0.96 $
and $x = \frac{29 - 32\sqrt{2})}{17} \approx 4.37 $
The two points of intersection of the two circles are given by
$(- 0.96 , 2.49)$ and $(4.37 , 1.16)$
|
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|
How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$
|
Simply
$$\frac{x}{x+1}=\frac{x+1-1}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}$$
|
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|
Rationalize Below Equation How can I rationalize the following equation:
$$\frac{22}{4\sqrt[3]{9}+2\sqrt[3]{6}+\sqrt[3]{4}}$$
|
Let $a=2\sqrt[3]{3}, b=\sqrt[3]{2}$. Then this is equal to $\dfrac{22}{a^2+ab+b^2} = \dfrac{22(a-b)}{a^3-b^3}$.
Since $a^3-b^3 = 8(3)-2 =22$, we have that our desired answer is $\dfrac{22(a-b)}{a^3-b^3} = a-b = 2\sqrt[3]{3}-\sqrt[3]{2}$.
|
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|
Complicated limit calculation Find the limit of $$\left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}\right]x^{(mn-m)/n}$$ when $x\to\infty$ and $m,n$ are natural numbers.
Thanks in advance!
|
An even quicker way to the answer comes from observing that
$$\left ( x^m + 1 \right )^{\frac{1}{n}} - \left (x^m - 1 \right)^{\frac{1}{n}} = x^{\frac{m}{n}} \left [ \left ( 1 + \frac{1}{x^m} \right )^{\frac{1}{n}} - \left (1 - \frac{1}{x^m} \right)^{\frac{1}{n}} \right ]$$
$$ \approx x^{\frac{m}{n}} \left [ \left ( 1 + \frac{1}{n x^m} \right ) - \left (1 - \frac{1}{n x^m} \right) \right ]$$
$$ = \frac{2 x^{\frac{m}{n}}}{n x^m} $$
The value $2/n$ follows from the factor on the right of the original expression.
|
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|
For what values of $x$ does $\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$ converge? Please help me to find the values of $x$ in which the series $$\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$$ converges? I applied some tests for it but...:(
Thank you
|
We can use partial fractions to decompose this fraction. We need to find $A$ and $B$ such that
$$\frac{A}{x+n} + \frac{B}{x+n-1} \equiv \frac{1}{(x+n)(x+n-1)} \, . $$
If we cross multiply, we get $A(x+n-1)+B(x+n)\equiv1$. When $x=-n$ we get $-A\equiv1$ and when $x=1-n$ we get $B \equiv 1$. It now follows that:
$$\frac{1}{(x+n)(x+n-1)} \equiv \frac{1}{x+n-1} - \frac{1}{x+n} \, . $$
Let's examine these terms as $n$ advances from $1$. Substituting $n=1,2,3,\ldots$ gives:
$$\left( \frac{1}{x} - \frac{1}{x+1} \right) + \left( \frac{1}{x+1} - \frac{1}{x+2} \right) + \left( \frac{1}{x+2} - \frac{1}{x+3} \right) + \cdots = $$
$$\frac{1}{x} - \frac{1}{x+1} + \frac{1}{x+1} - \frac{1}{x+2} + \frac{1}{x+2} - \frac{1}{x+3} + \cdots $$
You should be able to see that each term is cancelled by the very next term. For this to be well-defined we need the denominators to be non-zero and for the terms to tend to zero. With that in mind, $x$ can be anything but zero or a negative integer. With that in mind we have:
$$\sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n-1)} = \frac{1}{x} \, . $$
|
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What is the value of D here? Number $S$ is obtained by squaring the sum of digits of a two digit number $D$. If the difference between $S$ and $D$ is $27$, then the two digit number $D$ is?
My thoughts:
Let the two digit number $D$ be $AB$.
And so $S=(A+B)^2$
If $\,S-D=27,\,$ then $\,(A+B)^2 -AB=27$
$$A^2 + 2AB + B^2 - AB=27$$
Now how to obtain the value of $D$ further?
|
To try and approach systematically:
We know $|S-D|=27$.
Let $D$ is comprised of the digits $A$ and $B$, with each of $A, B$ each a single digit integer. Let $A$ be the left-most digit (the "ten's digit of $D$), and let $B$ be the right-most digit (the "one's" digit of $D$).
So $S = (A + B)^2$ and $D = 10 \cdot A + B.\;\;\;\;$(*)
Then
$$|S - D| = 27 \iff |(A + B)^2 - (10A + B)| = 27, \quad (0 < A < 10, \;0 \le B < 10)$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\iff (A+B)^2 - (10A + B) = 27,\quad\text{OR}\quad (10 A + B) - (A + B)^2 = 27$$
$$\text{with}\quad(0 < A < 10, \;0 \le B < 10)$$
(*) For example, if $D = 54$, then $D = 5 \cdot 10 + 4$, $S = (5 + 4)^2 = 81,\; S - D = 27$; $\quad\;$ and if $D = 73 = 7\cdot 10 + 3\;\;, S = (7 + 3)^2 = 10^2 = 100; \;S - D = 27.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A nasty integral of a rational function I'm having a hard time proving the following $$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}.$$
Mathematica has no problem evaluating it while I haven't the slightest idea how to approach it. Of course, I would like to prove it without the use of a computer. Is this an explicit form of a special function I fail to recognize?
|
Some progess: The integrand actually decomposes as
$$\frac{1}{2} \left( \frac{x^2 + 2x + 1}{x^6 + 4x^5 + 3x^4 - 4x^3 - 2x^2 + 2x + 1} + \frac{x^2 - 2x + 1}{x^6 - 4x^5 + 3x^4 + 4x^3 - 2x^2 - 2x + 1} \right).$$
Note that the second term is the same as the first term, except with $-x$ instead of $x$. Thus, with some substitutions, the integral becomes
$$\frac{1}{2} \int_{-\infty}^\infty \frac{y^2}{y^6 - 2y^5 - 2y^4 + 4y^3 + 3y^2 - 4y + 1} \; dy.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/266181",
"timestamp": "2023-03-29T00:00:00",
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|
What is a good bound for $\int^{2\pi}_{0}\cos^k(x)dx$? The integral $$\int^{2\pi}_{0}\cos^k(x)dx$$comes up in one of Rudin's problems. He asked me to do a detailed approximation of it. I am wondering how to do it. This feels really elementary, but I need a lower bound that works for any $k$. Integrating by parts would not really help though the function is explicitly integrable.
|
If $k$ is odd, then the integral is zero. If $k$ is even, note that $$\int_0^{2\pi} \cos^{2m}(x) dx = 4 \int_0^{\pi/2} \cos^{2m}(x) dx$$
Let $I_m = \displaystyle \int_0^{\pi/2} \cos^{2m}(x) dx$
Then
\begin{align}
I & = \int_0^{\pi/2} \cos^{2m-1}(x) d(\sin(x))\\
& = \left. \cos^{2m-1}(x) \sin(x)\right \vert_0^{\pi/2} + \int_0^{\pi/2}\sin(x) (2m-1) \cos^{2m-2}(x) \sin(x) dx\\
& = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) \sin^2(x) dx\\
& = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) (1-\cos^2(x)) dx\\
& = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) dx - (2m-1) \int_0^{\pi/2} \cos^{2m}(x) dx\\
2mI_{2m} & = (2m-1) I_{2m-2}
\end{align}
Hence, we get that
$$I_{2m} = \dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 I_0$$
Now $I_0 = \int_0^{\pi/2} 1 dx = \dfrac{\pi}2$. Hence, we get that
$$I_{2m} = \dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 \dfrac{\pi}2$$
Hence, $$\int_0^{2\pi} \cos^{2m}(x) dx = 2 \pi \left(\dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 \right) = 2\pi \dfrac{(2m)!}{4^m \cdot m! \cdot m!} = \dfrac{2 \pi}{4^m} \dbinom{2m}m$$
By Stirling, we have $$\dbinom{2m}m \sim \dfrac{2^{2m+1}}{\sqrt{4 \pi m}} = \dfrac{4^m}{\sqrt{\pi m}}$$
Hence, we have that $$\int_0^{2\pi} \cos^{2m}(x) dx \sim \dfrac{2 \pi}{\sqrt{\pi m}} = \dfrac{2 \sqrt{\pi}}{\sqrt{m}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the valuation of $\lim_{x\to0}\ x^2\left(1+2+3+...+\left[\frac1{|x|}\right]\right)$? Compute the following limitation:
$\lim_{x\to0}\ x^2\left(1+2+3+...+\left[\frac1{|x|}\right]\right)$.
|
Recall that $$1 + 2 + 3 + \cdots + \cdots + n = \dfrac{n(n+1)}2$$
If $\vert x \vert \in \left(\dfrac1{n+1}, \dfrac1n\right]$, we have $\lfloor 1/\vert x \vert \rfloor = n$ and hence we get that
$$1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor = \dfrac{\lfloor 1/\vert x \vert \rfloor(\lfloor 1/\vert x \vert \rfloor+1)}2 = \dfrac{n(n+1)}2$$
Then $$x^2 \left(1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor \right) = \dfrac{x^2n(n+1)}2 \in \left(\dfrac{n}{2(n+1)}, \dfrac{n+1}{2n} \right]$$
Hence, as $n \to \infty$ i.e. as $\vert x \vert \to 0$, we get the limit as $1/2$.
|
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How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\
0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\
0 & \cdots & \cdots & \cdots & \cdots & 2 & 5
\end{pmatrix}$$
is equal to $\frac13(4^{n+1}-1)$?
More generally:
How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)?
$$M_n(a,b,c) = \begin{pmatrix}
a & b & 0 & 0 & 0 & \cdots & 0 \\
c & a & b & 0 & 0 & \cdots & 0 \\
0 & c & a & b & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\
0 & \cdots & \cdots & 0 & c & a & b \\
0 & \cdots & \cdots & \cdots & \cdots & c & a
\end{pmatrix}$$
Here $a,b,c$ can be taken to be real numbers, or complex numbers.
In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that
$$m_{ij} = \begin{cases}
a & i = j, \\
b & i = j - 1, \\
c & i = j + 1, \\
0 & \text{otherwise.}
\end{cases}$$
There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type $M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly.
Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:
$$\begin{align}
\det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\
\det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\
\det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc
\end{align}$$
But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.
|
For the more general case of a tridiagonal Toeplitz matrix mentioned at the end on the question, the direct method (not using induction) I used in this answer can be adopted as follows. First note that the answer will depend only on $a$ and the product $p=bc$, since conjugation by a diagonal matrix with as diagonal entries a geometric progression $1,t,t^2,\ldots,t^{n-1}$ can be used to move any nonzero factor$~t$ from $b$ to $c$. If $p=0$ then we have a triangular matrix, and the determinant will be $a^n$; with this trivial case put aside, I can therefore assume (for reasons to become clear) without loss of generality that $c=-1$. So we want to know
$$
D=
\begin{vmatrix} a & -p & 0 & 0 & \cdots & 0\\ -1 & a & -p & 0 & \cdots & 0\\ 0 & -1 & a & -p & \ddots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & 0\\ \vdots & \vdots & \ddots & -1 & a & -p\\ 0 & 0 & \cdots &0 & -1 & a \end{vmatrix}.
$$
The quadratic polynomial $X^2-aX+p$ will be of importance; let $r$ be a root of this polynomial. Multiplying our matrix from the left by
$$
\begin{pmatrix}1&r&r^2&\ldots&\ldots&r^{n-1}\\
0&1&r&r^2&\ddots&r^{n-2}\\
0&0&1&r&\ddots&r^{n-3}\\
\vdots & \ddots & \ddots & 1& \ddots & \vdots \\
\vdots & \ddots & \ddots & \ddots& \ddots & r \\
0&0& \ldots& \ldots &0 & 1
\end{pmatrix},
$$
which has determinant$~1$, gives due to the fact that $-r^2+ar-p=0$ (and so $ar-p=r^2$), the matrix
$$
R=\begin{pmatrix}a-r&0&0&\ldots&0&r^n\\
-1&a-r&0&\ddots&0&r^{n-1}\\
0&-1&a-r&\ddots&0&r^{n-2}\\
\vdots & \ddots & \ddots & \ddots & \vdots&\vdots \\
\vdots & \ddots & \ddots & -1 & a-r& r^2 \\
0&0& \ldots &0&-1 & a
\end{pmatrix}.
$$
The diagonal entries are equal to the other root $s=a-r$ of $X^2-aX+p$ (possibly equal to$~r$), except for the last entry which is $s+r$.
Now one can recognise that $C=sI-C_P$ where $C_P$ is the companion matrix of
$$
P=r^n+r^{n-1}X+r^{n-2}X^2+\cdots+r^2X^{n-2}+rX^{n-1}+X^n.
$$
Since $C_P$ has as characteristic polynomial $\det(IX-C_P)=P$, we get as our determinant
$$
\det(R)=P[X:=s]=r^n+r^{n-1}s+r^{n-2}s^2+\cdots+r^2s^{n-2}+rs^{n-1}+s^n.
$$
So our answer can be written as
$$
D=\det(R) = \begin{cases}
\displaystyle\frac{r^{n+1}-s^{n+1}}{r-s}&\text{if }r\neq s,\\ \\
\displaystyle (n+1)r^n&\text{if }r=s.\end{cases}
$$
Of course one can take by the quadratic formula concretely $r=\frac{a+\sqrt{a^2-4p}}2$ and $s=\frac{a-\sqrt{a^2-4p}}2$, but this does not make the answer more transparent.
|
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|
Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$.
Then, $dx = 3\cos\theta\,d\theta$
and, $$\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta$$
So, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = \int \cot^2 \theta \ d\theta = -\cot\theta - \theta + C$$
Returning to the original variable, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = -\frac {\sqrt{9 - x^2}}{x} - \sin^{-1}\left(\frac{x}{3}\right) + C$$
I don't understand why $\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta \,$ instead of $\sqrt{9-x^2} = |3||\cos\theta| = |3|\cos\theta$. I feel like I have problems understanding this because I am not sure what is the purpose of the absolute value signs in this case, are they to indicate that, for example, $|\cos\theta| = \pm\cos\theta$? If that's the case, why do we choose $3$ to be positive instead of negative?
|
Think about it: How would using $\;|\,3\,|\;$ change the result?$\quad$ After all, $\;|\,3\,| \;= \quad?$
Also note: $\,|\cos\theta| \,= \,\cos\theta \,\ge \,0\;$ since the range of $\,\theta\,$ is $\;\;-\dfrac{\pi}{2} \,\le\, \theta \,\le \dfrac{\pi}{2}$.
You seem to be confused about the what "$|\;\cdot\;|$" means.
$\;| \,a \,|$ is not the same as $\,\pm a\,$.
Rather, we define
*
*$|\, a \,| = -a\,$ if $\,a \lt 0$.
*$|\, a\, | = \;\,\;a\,$ if $\,a \geq 0$.
This ALWAYS returns a non-negative result.
By definition, $\,\sqrt{\;\;}\,\;$ returns only the non-negative root of a square or quadratic. So $\,\sqrt{a^2}=|a|,\,$ while solving for, say, $\,x^2-a = 0\,$ we obtain two roots: $\,x\, =\,\pm \sqrt{a}\,$.
|
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|
Rotation around slant line What is the volume when $f(x) = x^2$ is rotated around the line $y = x$?
For each individual $x$, I was considering the difference between $\begin{pmatrix} x\\x \end{pmatrix}$ and the projection of $\begin{pmatrix} x^2\\x \end{pmatrix}$ onto $\begin{pmatrix} x\\x \end{pmatrix}$ (which would give vectors perpendicular to the line $y = x$ and of the correct length, going from a point on $y=x$ to the corresponding point on $y=x^2$).
I would get $\begin{pmatrix} x\\x \end{pmatrix} \cdot \left(1 - \frac{x^3+x^2}{2x^2}\right)$, and I could integrate the length of that squared times $\pi$, going from $0$ to $1$ to find the volume.
I would get $\pi \cdot \int_0^1 2*\left(x - \frac{x^3+x^2}{2x}\right)^2 = \frac{\pi}{60}$, but the correct answer is $\frac{\sqrt{2}\pi}{60}$.
Incidentally, $\frac{d\begin{vmatrix} x\\x \end{vmatrix}}{dx} = \sqrt{2}$.
I'm missing something, and can't figure out exactly how this (rigorously) fits together.
|
[Sorry, previous answer was wrong]
You are right in saying that "Incidentally, $\frac { d {\begin{matrix}x\\x\\ \end{matrix}} }{dx} = \sqrt{2}$. In changing the coordinates, from integrating on the x-axis, to integrating on the line y=x, you also need to account for the difference in volume, which is precisely this amount.
For example, if you consider the length of the line on $y=x$ from $x=0$ to $1$, you do not say that it has length $\int_{x=0}^1 1 dx = 1$, but it has length $\int_{x=0}^1 \sqrt{2} dx = \sqrt{2}$. The Jacobian takes care of this.
|
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|
Why does $(2/p)=\prod_{k=1}^{(p-1)/2}2\cos\left(\frac{2\pi k}{p}\right)$? Browsing one of my favorite video game sites, there was a post asking about the derivation of the formula
$$
\left(\frac{2}{p}\right)=\prod_{k=1}^{(p-1)/2} 2\cos(2\pi k/p).
$$
I know that $\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$, but I'd never seen a formula in terms of cosine, it seems computational more involved. Anyway, out of curiosity is there a derivation or reference to a derivation? Many thanks.
|
For odd prime $p,$
$$\cos px+i\sin px=(\cos x+i\sin x)^p=\sum_{0\le r\le p\cos^{p-r}x(i\sin x)^r}$$
Equating the Real parts, $$\cos px=\cos^px-\binom p2 \cos^{n-2}x\sin^2x+\binom p4\cos^4x\sin^4x+\cdots$$
$$=\cos^px(1+\binom p2+\binom p4+\cdots)+\cdots+(-1)^{\frac{p-1}2}\cos x$$
$$=2^{p-1}\cos^px+\cdots+(-1)^{\frac{p-1}2}\cos x$$
If we put $x=\frac{2k\pi}p,\cos px=1$ where $k$ is any integer.
So, the roots of $$2^{p-1}\cos^px+\cdots+(-1)^{\frac{p-1}2}\cos x-1=0$$ are $\cos \frac{2k\pi}p$ where $0\le k\le p-1$
So, $$\prod_{0\le k\le p-1}\cos \frac{2k\pi}p=\frac1{2^{p-1}}\implies \prod_{1\le k\le p-1}\cos \frac{2k\pi}p=\frac1{2^{p-1}}$$ as $\cos\frac{2k\pi}p=1$ for $k=0$
Now, $\cos\frac{2(p-k)\pi}p=\cos(2\pi-\frac{2k\pi}p)=\cos\frac{2k\pi}p\implies \cos\frac{2k\pi}p\cos\frac{2(p-k)\pi}p=\cos^2\frac{2k\pi}p$
$$\implies \prod_{1\le k\le p-1}\cos \frac{2k\pi}p=\prod_{1\le k\le \frac{p-1}2}\cos\frac{2k\pi}p\prod_{\frac{p-1}2< k\le p-1}\cos\frac{2k\pi}p=\prod_{1\le k\le \frac{p-1}2}\cos^2\frac{2k\pi}p$$
So, $$\prod_{1\le k\le \frac{p-1}2}\cos^2\frac{2k\pi}p=\frac1{2^{p-1}}\implies \prod_{1\le k\le \frac{p-1}2}\left(2\cos\frac{2k\pi}p\right)^2=1$$
Now, $\cos\frac{2k\pi}p<0$ if $\frac\pi2<\frac{2k\pi}p<3\frac\pi2\implies \frac p4< k<\frac{3p}4$ which in our case reduces to $\frac p4< k\le \frac{p-1}2$
So $k$ has $\mu=\frac{p-1}2-\lfloor \frac p4\rfloor$ values for which $\cos\frac{2k\pi}p<0$,
$$(-1)^\mu\prod_{1\le k\le \frac{p-1}2}2\cos\frac{2k\pi}p=1\implies \prod_{1\le k\le \frac{p-1}2}2\cos\frac{2k\pi}p=(-1)^\mu$$
Observe that $\mu,\frac{p^2-1}8$ have same parities.
For example, for $p=8n-1, \frac{p^2-1}8=\frac{(8n-1)^2-1}8=8n^2-2n$ and $\mu=4n-1-(2n-1)=2n$
|
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|
Proving:$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$ How to prove that :
$$(1+x)(1+x^2)(1+x^3)\cdots (1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$$
|
$$(1+x^r)(1+x^{n+1-r})=1+x^r+x^{n+1-r}+x^{n+1}\ge 1+x^{n+1}+2\sqrt{x^r\cdot x^{n+1-r}}$$ applying A.M.$\ge$ G.M. with $x^r,x^{n+1-r}$ assuming $x>0$
So, $$(1+x^r)(1+x^{n+1-r})\ge(1+x^{\frac{n+1}2})^2 $$
Now, $1\le r<n+1-r\implies r<\frac{n+1}2$
If $n$ is odd $=2m+1$ (say), $r<\frac{2m+1+1}2\implies 1\le r\le m=\frac{n-1}2$,
so $n+1-r=2m+1+1-r$ will vary in $[m+2=\frac {n+3}2,n]$
so $1+x^{\frac{n+1}2}$ will remain unpaired.
So, $$\prod_{1\le r\le n}(1+x^r)=(1+x^{\frac{n+1}2})\prod_{1\le r\le\frac{n-1}2}(1+x^r)(1+x^{n+1-r})\ge (1+x^{\frac{n+1}2})(1+x^{\frac{n+1}2})^{\left(2\cdot\frac{n-1}2\right)}=(1+x^{\frac{n+1}2})^n$$
Similarly if $n$ is even $=2m$ (say), $1\le r<\frac{2m+1}2\implies 1\le r\le m=\frac n2,$
so $n+1-r=2m+1-r$ will vary in $[m+1=\frac n2+1,n]$
$$\prod_{1\le r\le n}(1+x^r)=\prod_{1\le r\le\frac n2}(1+x^r)(1+x^{n+1-r})\ge (1+x^{\frac{n+1}2})^{\left(2\cdot\frac n2\right)}=(1+x^{\frac{n+1}2})^n$$
|
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Find the equation of the sphere $ x^2+y^2+z^2-2x-4y+8z=15$ I'm not sure how you get from this:
$x^2+y^2+z^2-2x-4y+8z=15$
To:
$(x^2-2x+1) + (y^2-4y+4) + (z^2+8z+16)-1-4-16=15$
How do you get the $1,4,16$?
|
It's grouping like terms and completing the square. For example, if you collect the terms in $x$ you have $x^2 - 2x$, so you want a constant term that will make that a perfect square. Since $(x-1)^2 = x^2 - 2x + 1$, you need a $1$. Similarly for the other two.
In general, if you have $ax^2 + bx$, to complete the square you need to add a constant term of $b^2/4a$, which makes the entire thing equal to $(\sqrt(a) x + \frac{b}{2\sqrt{a}})^2$. You can easily verify this by multiplying it out.
Here, in the first case, you have $a=1, b=-2$, so your constant term is $(-2)^2/(4*1) = 1$.
Similarly, in the second case $a=1,b=-4$, so your constant term is $(-4)^2/(4*1) = 4$.
In the third case, $a=1,b=8$, so your constant term is $8^2/(4*1) = 16$.
Once you've completed the squares, you can simplify the equation to
$$(x-1)^2 + (y-2)^2 + (z+4)^2 = 36,$$
which is much neater and more informative than the original form--you can immediately read off the coordinates of the center of the sphere and its radius.
|
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find the sum of the alternating series How to find the sum of the infinite series
$$\frac{1}{12}-\frac{1\cdot 4}{12 \cdot 18 } + \frac{1\cdot 4\cdot 7}{12\cdot 18\cdot 24} - \frac{1 \cdot 4 \cdot 7\cdot 10}{12 \cdot 18 \cdot 24 \cdot 30}+...$$
I understood the answer posted in Yahoo Answer till the last but one step:
That is how did he get: $ \lim_{n \to \infty} S_n = 0 $
Other steps I understood.
Thanks in advance
|
Using the fact that
$$
\prod_{k=0}^{n}(12+6k) = 6^{n+1}(n+2)! = 3^{n+1} 2^{n+2} \frac{(n+2)!}{2}
$$
we see that the $n^{\text{th}}$ term is
$$
\begin{align*}
(-1)^n \frac{\prod_{k=0}^{n}(1+3k)}{\prod_{k=0}^{n}(12+6k)} &= \frac{\prod_{k=0}^{n}\left(\frac{1}{3}-k\right)}{\frac{(n+2)!}{2}} \cdot \frac{1}{2^{n+2}} \\
&= \frac{3}{(n+2)!} \cdot \frac{2}{3} \prod_{k=0}^{n}\left(\frac{1}{3}-k\right) \cdot \frac{1}{2^{n+2}} \\
&= -\frac{3}{(n+2)!} \cdot \prod_{k=0}^{n+1}\left(\frac{2}{3}-k\right) \cdot \frac{1}{2^{n+2}} \\
&= -3 \cdot \binom{2/3}{n+2} \cdot \frac{1}{2^{n+2}}.
\end{align*}
$$
Hence the value of the sum is
$$
\begin{align*}
-3 \sum_{n=2}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} &= 4-3-3\frac{2}{3}\cdot\frac{1}{2}-3 \sum_{n=2}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} \\
&= 4 - 3 \sum_{n=0}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} \\
&= 4 - 3 \left(1+\frac{1}{2}\right)^{2/3}.
\end{align*}
$$
|
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|
Rational function partial fraction help? So I have to integrate
$$
\frac{3x+2}{(x)(x+1)^3}
$$ so I have the terms
$$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$
... and then I have
$$
3x+2=A(x+1)^3 + B(x)(x+1)^2 +C(x)(x+1)^2+ D(x)(x+1)^2
$$
... here I tried to equal the quotients and I have $A=2$..how do I find the others?
|
From $$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$ you have: $$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}$$ Now let $x=0$ then $$A(0+1)^3=3\times0+2\to A=2$$ Let $x=-1$ then $$D(-1)=3\times(-1)+2=-1\to D=-1$$ Set $x=1$ then $$8A+4B+2C+D=3\times1+2=5\to 16+4B+2C-1=5\to4B+2C=-10$$
Set $x=2$ then $$27A+18B+6C+2D=3\times 2+2=8\to 54+18B+6C-2=8\to18B+6C=-44$$ Now solve the two final equation respect to $B$ and $C$.
|
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|
How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
|
\begin{eqnarray}
2^{n-2}\frac{(2n-5)(2n-7)\ldots3\cdot1}{(n-1)(n-2)\ldots3\cdot2\cdot1}&=&
2^{n-2}\frac{(2n-4)(2n-5)\ldots\cdot3\cdot2\cdot1}{[(n-1)(n-2)\ldots2\cdot1][(2n-4)(2n-6)\ldots4\cdot2]}\\
&=&2^{n-2}\frac{(2n-4)!}{[(n-1)(n-2)\ldots2\cdot1][2^{n-2}(n-2)(n-3)\ldots2\cdot1]}\\
&=&\frac{(2n-4)!}{(n-1)!(n-2)!}=\frac{1}{n-1}\cdot\frac{(2n-4)!}{[(n-2)!]^2}\\
&=&\frac{1}{n-1}{2n-4\choose n-2}.
\end{eqnarray}
|
{
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|
Evaluating $\lim_{n \to \infty} \oint_{ |z| = 1/4} \frac{1}{(4 z(1-z))^n} \frac{\mathrm{d}z}{z (1-2 z)} = \frac{1}{2}$ While working on an earlier question involving $\sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}}$ I rewrote the sum as a contour integral, using generating functions:
$$
\sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}} = \sum_{j=0}^n \left( \frac{1}{4^n} \binom{n+j-1}{j} \right) 2^{n-j} = [t^n] \left( \sum_{j=0}^\infty \frac{t^j}{4^n}\binom{n+j-1}{j} \cdot \sum_{j=0}^\infty t^j 2^j \right)
$$
Now, using well known generating functions, for $|t|<1/2$:
$$
\sum_{j=0}^\infty t^j \binom{n+j-1}{j} = \frac{1}{(1-t)^n} \quad \sum_{j=0}^\infty (2t)^j = \frac{1}{1-2t}
$$
We get
$$
\sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}} = [t^n] \left( \frac{1}{1-2t} \frac{1}{\left(4(1-t)\right)^n} \right) = \frac{1}{2 \pi i} \oint \frac{1}{1-2t} \left(\frac{1}{4 t(1-t)} \right)^n \frac{\mathrm{d} t}{t}
$$
where the Cauchy integral formula was used along with $n! [t^n] g(t) = g^{(n)}(0)$.
Now, Byron Schmuland showed that the large $n$ limit of the left-hand-side equals $\frac{1}{2}$.
Question: Can one demonstrate $$ \lim_{n \to \infty} \frac{1}{2 \pi i} \oint_{ |t| = \rho} \frac{1}{1-2t} \left(\frac{1}{4 t(1-t)} \right)^n \frac{\mathrm{d} t}{t} = \frac{1}{2}$$ using asymptotic methods? Here $0<\rho<\frac{1}{2}$.
|
As $t$ traces around the origin, so does $u=4 t (1-t)$. Solving for $t$ we get two solutions, one, $2t=1-\sqrt{1-u}$ maps a $u$-path around the origin into the $t$-path around the origin, and another, $2t=1+\sqrt{1-u}$, maps a $u$-path around the origin into the $t$-path around $t=1$. This suggests a change of variables, $t = \frac{1}{2} \left(1-\sqrt{1-u}\right)$
$$
\frac{1}{2 \pi i} \oint \frac{1}{\left(4 t(1-t) \right)^n} \frac{\mathrm{d}t}{t(1-2t)} = \frac{1}{2 \pi i} \oint \frac{1}{2} \frac{1+\sqrt{1-u}}{1-u} \frac{\mathrm{d}u}{u^{n+1}} = [u^n] \frac{1}{2} \frac{1+\sqrt{1-u}}{1-u}
$$
where the Cauchy formula was used in the last equality. This gives
$$\begin{eqnarray}
\sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}} &=& [u^n] \frac{1}{2} \frac{1+\sqrt{1-u}}{1-u} = [u^n] \frac{1}{2} \left( \frac{1}{1-u} + \frac{1}{\sqrt{1-u}} \right) \\ &=& \frac{1}{2} + \frac{1}{2} \binom{-1/2}{n} = \frac{1}{2} + \frac{1}{2^{2n+1}} \binom{2n}{n}
\end{eqnarray}
$$
Although this is not quite what you are asking for, this reproduces Byron's closed form result leading to the evaluation of the limit.
Alternatively, you could infer the large $n$ asymptotioc from the near-singularity-behavior of the generating function $g(u) = \frac{1}{2} \frac{1+ \sqrt{1-u}}{1-u}$:
$$
g(u) = \frac{1}{2} \frac{1}{1-u} + \mathcal{o}\left(\frac{1}{1-u}\right)
$$
implying the limit equals $[u^n] g(u) \sim \frac{1}{2}$.
|
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|
closest point to on $y=1/x$ to a given point I feel like I'm missing something basic - given a point $(a,b)$ how do I find the closest point to it on the curve $y=1/x$? I tried the direct approach of pluggin in $y=1/x$ into the distance formula but it leads to an order-4 polynomial...
|
Well. We have by the distance formula that the point on $xy = 1$ closest to $(a,b)$ will be the solution to:
$$\frac{d}{dx}\sqrt{(a-x)^2+(b-\frac{1}{x})^2} = 0$$
Which has the same soutions as
$$\frac{d}{dx}\left( (a-x)^2+(b-\frac{1}{x})^2\right)= 0\\
-2(a-x)+\frac{2(b-\frac{1}{x})}{x^2} = 0\\
x^4-ax^3+bx-1=0$$
Which has as the first real solution:
$$x==\frac{a}{4}-\frac{1}{2} \surd \left(\frac{a^2}{4}+\frac{2^{1/3} (-4+a b)}{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}+\frac{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}{3\ 2^{1/3}}\right)-\frac{1}{2} \surd \left(\frac{a^2}{2}-\frac{2^{1/3} (-4+a b)}{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}-\frac{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}{3\ 2^{1/3}}-\frac{a^3-8 b}{4 \sqrt{\frac{a^2}{4}+\frac{2^{1/3} (-4+a b)}{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}+\frac{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}{3\ 2^{1/3}}}}\right)$$
And the others are just as complicated. This was produced with Mathematica and simplifying under the assumption that the root was real ,$b>-a$ and, $x>0$.
This is such a simply posed problem, but it seems as if the solution is incredibly complicated.
|
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|
Proving $ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $ by induction I have the Following Proof By Induction Question:
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $$
Can Anybody Tell Me What I'm Missing.
This is where I've Gone So Far.
Show Truth for N = 1
LHS = (1) (2) = 2
RHS = $$ \frac{(1)(1+1)(1+2)}{3} $$
Which is Equal to 2
Assume N = K
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) = \frac{(k)(k+1)(k+2)}{3} $$
Proof that the equation is true for N = K + 1
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) + (k+1) (k + 2)$$
Which is Equal To:
$$ \frac{(k)(k+1)(k+2)}{3} + (k+1) (k + 2)$$
This is where I've went so far
If I did the calculation right the Answer should be
$$\frac{(k+1)(k+2)(k+3)}{3}$$
|
It might also be helpful to go backwards to see that it makes sense.
Specifically,
$$S_{n+1} - S_n = \frac{(n+2)(n+1)(n)}{3} - \frac{(n+1)(n)(n-2)}{3} = \frac{1}{3}\left(n^3 + 3n^2 + 2n - n^3 + n\right) = n(n+1)$$
Going backwards, you can more easily see that $$S_{n+1} = S_n + n(n+1)$$
|
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|
Is there a counterexample to the claim that if $\mathbf y\cdot\mathbf y=1$ and $\mathbf x\cdot\mathbf y=c$ then $\mathbf x=c\mathbf y$? Let $x,y$ be arbitrary vectors where $\mathbf{y} \cdot \mathbf{y} = 1$ and $c$ be a real valued scalar. If
$\mathbf{x} \cdot \mathbf{y} = c = c (\mathbf{y} \cdot \mathbf{y} ) = (c \mathbf{y} ) \cdot \mathbf{y} $, then
$\mathbf{x} = c \mathbf{y}$
Is this true? Feels like a sloppy conclusion no?
|
$\mathbf{x} = c \mathbf{y} + $ some vector perpendicular to $\mathbf{y}$
|
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|
Deriving $e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+...$ I know how to prove equation:
$$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$
How can I now derive the series:
$$e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$$
Those two seem very similar to me...
|
If we are allowed to use calculus,
let $$z=1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots$$
So, $$\frac{dz}{dy}=0+1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots=z$$
$$\implies \frac{dz}z=dy$$
Integrating both sides, $\log z=y+c$
For $y=0,z=1\implies c=\log1-0=0\implies \log z=y\iff z=e^y$
|
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|
$a+b+c+d+e=abcde$ What is $\max(a)?$ The question from my nephew is this:
If $a, b, c, d, e \in N^+$ and $a+b+c+d+e=a\times b\times c\times d\times e$, then what's the the maximum possible value of $a$? Thanks ahead:)
|
$5$. This is clearly possible with $b=2$, $c=d=e=1$. It remains to show that $b+c+d+e< 6(bcde-1)$ or $6bcde>b+c+d+e+6$. However, we cannot have $b=c=d=e=1$, so $bcde\ge 2$ and, thereby, $3bcde\ge 6$. Thus we are left with $3bcde>b+c+d+e$. Assuming that $b$ is the largest number, we have $b\ge 2$, so $\frac 32bcde\ge 3cde\ge c+d+e$. Finally, $\frac32bcde>b$.
|
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|
Find $a, b, c, d \in \mathbb{Z}$ such that $2^a=3^b5^c+7^d$ Solve $2^a=3^b5^c+7^d$ over the positive integer.
I know $a$ is even because: $(-1)^a \equiv2^a = 3^b5^c+7^d \equiv1 \ (mod\ 3)$
|
One solution to your equation is $a = 2$, $b = 0$, $c = 0$ and $d = 0$. Another one would be $a = 3$, $b = 0$, $c = 0$, $d = 1$. Yet another one would be $a = 4$, $b =1$, $c = 1$, $d = 0$. And $a = 5$, $b = 0$, $c = 2$, $d =1$.
One is clever to recognize that $a + b + c+d = 2(a - 1)$ for all $a,b,c,d \in \mathbb{N}$ that satisfy your equation. Since we want positive integer solutions, it turns out that there is only one tuple which satisfies the above: $(6,1,1,2)$.
|
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|
Polynomial inequalities Just wanted to check my maths:
a) $\displaystyle \frac{(x^3-4)}{x-2} \leq x+2$ I got that $x\leq 1$.
b) $\displaystyle \frac{(2x-3)(x+1)}{x^2} \leq -2$
I got that $\displaystyle -\frac{3}{4} \leq x \leq 1$
Right/wrong? Thanks :). I just worked on it as if it were an equals sign... so I'm not sure I'm correct.
|
In case of the 2nd one,
$\displaystyle \frac{(2x-3)(x+1)}{x^2} \leq -2$
Multiplying both sides by $x^2$ (Considering $x\ne0)$as it is +ve so no problem of sign changing,
$\Rightarrow(2x-3)(x+1)\leq -2x^2$
$\Rightarrow 2x^2-x-3\leq -2x^2$
$\Rightarrow 4x^2-x-3\leq 0$
$\Rightarrow(2x)^2-2.2x.\frac{1}{4}+1/16-1/16-3\leq0$
$\Rightarrow(2x-1/4)^2-49/16\le0$
$\Rightarrow-\frac{7}{4}\le2x-1/4\le7/4$
$\Rightarrow-3/4\le x\le 1$
This solution must be without 0.
$x\in(-3/4,1)-{0}$
|
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|
Suppose that today is tuesday. what day of the week will it be in $3^{100000} + 3^{10000} + 3^{1000}$ days? Suppose that today is tuesday. what day of the week will it be in $3^{100000} + 3^{10000} + 3^{1000}$ days ?
I figured that, eventually, I will add $3^{100000} + 3^{10000} + 3^{1000}$ and $\mod\ 7$ that answer...
I tryed with the Fermat little theorem but cant quit figure how to apply it here.
$m = n(\mod p-1)$
then , $a^m = a^n (\mod p)$
doing that I realised that $3^{x} \mod 7 = 4$ for all $x \mod 10 = 0$
when I try to apply Fermat, I end up with :
$m = 100000 \mod 6 = 4 \\ 3^4 = 3^{100000} \mod 7 $
which make no sense...
Am I using the right theorem for the problem ? If so, how do I apply it...
Thanks
|
$3^6 = 1 \mod 7$. Hence $3^k \mod 7 = 3^{k-6n} \mod 7$.
$100000 = 6\cdot 16666 + 4$. $10000 = 6\cdot 1666 + 4$. $1000 = 6\cdot 166 + 4$. Hence $3^{100000} + 3^{10000} + 3^{1000} \mod 7 = 4+4+4 \mod 7 = 5 \mod 7$.
Hence it will be Sunday.
|
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|
Tangent line to solution set of equations Determine the parametric equation for the tangent line at the point $P = (1,1,1)$ to a curve which is described by the solution set of the following equations:
$x^2 + y^2 + z^2 = 3$,
$3x + 4y + 5z = 12$
I dont have a clue where to start. How can I differentiate functions which are determined by a solution set of equations?
|
Hints: to find the solution set, you can set the equations equal to one another, or add them, to find the solution set:
$x^2 + y^2 + z^2 = 3$,
$+$
$3x + 4y + 5z = 12$
$$x^2 + 3x + y^2 + 4y + z^2 + 5z \color{blue}{\bf{-15}} = 0\tag{1}$$
$$x^2 + 3x \color{blue}{\bf -4} + y^2 + 4y \color{blue}{\bf -5} + z^2 + 5z \color{blue}{\bf - 6} = 0$$ $$\implies (x +4)(x - 1) + (y+5)(y-1) + (z+6)(z-1) = 0\tag{2}$$
The curve of interest and the solution set of the two given equations are the solutions to $(1)$ = $(2)$. (Note that $P$ is among the solutions.) This curve is what you want to differentiate.
|
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|
Algebraically Solve Limit $$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}$$
I can solve it using l'Hôpital but just cannot find a way to do it algebraically.
|
$$\begin{align}
\lim_{x\to 0}\frac{2\sqrt{x+1}-x-2}{x^2}
&= \lim_{x\to 0}\frac{2\sqrt{x+1}-(x+2)}{x^2} \frac{2\sqrt{x+1}+(x+2)}{2\sqrt{x+1}+x+2}\\
&= \lim_{x\to 0}\frac{4(x+1)-(x+2)^2}{x^2} \frac{1}{2\sqrt{x+1}+x+2}\\
&= \lim_{x\to 0}\frac{4(x+1)-(x^2+4x+4)}{x^2} \frac{1}{2\sqrt{x+1}+x+2}\\
&= \lim_{x\to 0}\frac{-x^2}{x^2}\frac{1}{2\sqrt{x+1}+x+2}\\
&=\lim_{x\to 0}\frac{-1}{2\sqrt{x+1}+x+2}\\
&=-\frac{1}{4}
\end{align}$$
|
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Hensel Lifting and solving with mods I need to use the Hensel-Newton method (aka Hensel Lifting) to find all solutions of:
$$x^2 + x + 47 \equiv 0\:(\text{mod } 343)$$
Note: $$ 343 = 7^3 $$
I don't really understand Hensel Lifting so I am not sure where to begin. Can someone help give me a better understanding of Hensel Lifting and how to start this problem?
|
Let's start modulo $7$.
By trial and error, we find $x^2+x+5\equiv 0\pmod 7$ has the solutions $x\equiv 1$, $x\equiv 5$.
Now if $x^2+x+47\equiv 0\pmod{49}$ then $x=x_0+7y$ with $x_0\in\{1,5\}$.
Hence $x^2+x+47=(x_0^2+14x_0y+49y^2)+(x_0+7y)+47\equiv (x_0^2+x_0+5)+7\cdot(2x_0y+y+6)\pmod{49}$, where we already know that the left hand side is a multiple of $7$. Hence we have to solve
$$ \frac{x_0^2+x_0+5}7+(2x_0+1)y+6\equiv0\pmod 7.$$
If $x_0=1$, this leads to $3y+7\equiv 0$, i.e. $y\equiv 0$. If $x_0=5$, we obtain $11y+11\equiv 0$, i.e. $y\equiv 6$.
Thus the only solutions modulo $49$ are $x\equiv 1\pmod{49}$, $x\equiv 47\equiv -2\pmod{49}$.
Now if $x^2+x+47\equiv 0\pmod{343}$, then $x=x_1+49z$ with $x_1\in \{1,-2\}$.
Hence
$$ x^2+x+47 = x_1^2+2\cdot 7^2x_1z+7^4z^2+x_1+7^2z+47\\\equiv(x_1^2+x_1+47)+7^2(2x_1+1)z\pmod{343}.$$
Again, this is trivially a multiple of $49$ and we are lead to solve
$$\frac{x_1^2+x_1+47}{49}+(2x_1+1)z\equiv 0\pmod{7}.$$
With $x_1=1$, this is $3z+1\equiv 0$, i.e. $z=2$ and $x\equiv99\pmod{343}$.
With $x_1=-2$, this is $-3z+1\equiv0$, i.e. $z=-2$ and $x=-100\pmod{343}$.
|
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|
Derivative of $x^2\sqrt{1+x}$
Given that $f(x)=x^2\sqrt{1+x}$, show that $f'(x)=\dfrac{x(ax+b)}{2\sqrt{1+x}}$ where $a$ and $b$ are constants to be found.
I first tried using the product rule: $f'(x)=2x\sqrt{1+x}+\dfrac{x^2}{2\sqrt{1+x}}$, but now I'm stuck.
I don't know if it is my algebra or derivation that stops me. Could anyone give a hint?
|
You can avoid the product rule by putting everything under one square root:
$$x^{2} \sqrt{1 + x} \; = \; \sqrt{x^4(1 + x)} \; = \; \sqrt{x^4 + x^5} \; = \; \left(x^4 + x^5 \right)^{\frac{1}{2}}$$
Now apply the power rule for derivatives:
$$f'(x) \; = \; \frac{1}{2} \left(x^4 + x^5 \right)^{-\frac{1}{2}} \cdot \left(4x^3 + 5x^4\right)$$
$$ = \;\; \frac{4x^3 + 5x^4}{2\sqrt{x^4 + x^5}}$$
$$ = \;\; \frac{4x^3 + 5x^4}{2x^2\sqrt{1 + x}}$$
$$ = \;\; \frac{4x + 5x^2}{2\sqrt{1 + x}}$$
$$ = \;\; \frac{x\left(4 + 5x\right)}{2\sqrt{1 + x}}$$
|
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|
Finding minimum and maximum value of complex number Let a be a positive real number and let $M_a =\{z \in C^*: |z+\frac{1}{z}|=a\}$ Find the minimum and maximum value of $|z|$ when z$\in M_a$
|
Let $z=r(\cos \theta+i\sin \theta)$
$z^{-1}=r^{-1}(\cos \theta-i\sin \theta)$
We have ,
$|z+\frac{1}{z}|=|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|$
So we have,
$\displaystyle|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|=\sqrt{(r^2+1/r^2)+2\cos 2\theta}$
Thus we have ,
$(r^2+1/r^2)+2\cos 2\theta=a^2$
Let $r^2=x$,
$\Rightarrow x^2+1=x(a^2-2\cos2\theta)$
$\Rightarrow x^2-x(a^2-2\cos2\theta)+1=0$
Solving the quadratic we have,
$\displaystyle x=\frac{(a^2-2\cos 2\theta)+\sqrt{(a^2-2\cos 2\theta)^2-4}}{2}$ and $\displaystyle x=\frac{(a^2-2\cos 2\theta)-\sqrt{(a^2-2\cos 2\theta)^2-4}}{2}$
Clearly x attains its maximum value at $2\theta=0$ using the first expression,
The value is $\displaystyle \frac{(a^2+2)+\sqrt{(a^2+2)^2-4}}{2}$
So the maximum value of $r=\sqrt{\displaystyle \frac{(a^2+2)+\sqrt{(a^2+2)^2-4}}{2}}$
This on simplification turns out to be, $\displaystyle r=\frac{a+\sqrt{a^2+4}}{2}$
Minimum value can also be obtained in that way.
|
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|
Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0}^{k} (-\frac{1}{2})^k + (-\frac{1}{2})^{k + 1} = \frac{2^{k+1} + (-1)^k}{3 \times 2^k}+(-\frac{1}{2})^{k + 1}$, by adding $(-\frac{1}{2})^{k + 1}$ to both sides.
I want to show that the right side is equal to:
$\frac{2^{k+1+1} + (-1)^{k+1}}{3 \times 2^{k+1}}$
Thank you!
|
We start from $\frac{2^{k+1}+(-1)^{k}}{3\cdot 2^k} +\frac{(-1)^{k+1}}{2^{k+1}}$. Multiply numerator and denominator of the first term by $2$, and numerator and denominator of the second term by $3$. Now we can add safely and get
$$\frac{2^{k+2}+2(-1)^{k}+3(-1)^{k+1}}{3\cdot 2^{k+1}}.$$
We need to verify that $2(-1)^k+3(-1)^{k+1}=(-1)^{k+1}$. This is clear, since $2(-1)^k+2(-1)^{k+1}=0$.
|
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|
Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $(b-c)$ and $(c-a)$. The given expression may have a coefficient a constant factor which is nonzero. Let it be $m$.
$$\therefore a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
Please guide further on how to find this coefficient.
|
From your last line $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)=m(a-b)(b-c)(c-a);$$
putting $a=0$, $b=1$, $c=2$ we get
$$
0(1^2-2^2)+1(2^2-0^2)+2(0^2-1^2)=m(0-1)(1-2)(2-0)\\
\implies 0+4-2=2m \implies 2m=2 \implies m=1.
$$
|
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|
Does the Rational Root Theorem ever guarantee that a polynomial is irreducible? I know that the Eisenstein's criterion can guarantee it (when applicable), but what about this?
Also, is there another test that you can use to test irreducibility besides these two? For example, $x^4+2x^2+49$, over the rationals?
|
With respect to your specific question (and quartics in general), the RRT doesn't solve the problem but it can help. Once you verify that your quartic has no roots, the only way it can factor is as a product of quadratic polynomials. Additionally, you can guarantee that the quadratic equations are monic and have integer coefficients. Observing that the constant terms multiply to $49$, we have four options:
$$x^4 + 2x^2 + 49 = (x^2 + ax + 7)(x^2 + bx + 7)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax - 7)(x^2 + bx - 7)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax + 1)(x^2 + bx + 49)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax -1)(x^2 + bx - 49)$$
The vanishing of the cubic term gives $a = -b$ in each scenario, and the vanishing of the $x$ term eliminates scenarios 3 and 4. Comparing the $x^2$ terms in the first two scenarios gives:
$$14 - a^2 = 2$$
$$-14 - a^2 = 2$$
And it is easily seen that neither of these have solutions in the integers.
|
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|
Ellipse in polar coordinates I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants $a$ and $b$ in this format -
Where $2a$ is the total height of the ellipse and $2b$ being the total width.
You can then find the radial length, $r$, at any angle $\theta$ to major axis as...
$$r(\theta) = \sqrt{(b \sin(\theta))^2 + (a \cos(\theta))^2}$$
...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:
$$r(\theta) = \frac{ab}{\sqrt{(b \cos(\theta))^2 + (a \sin(\theta))^2}}$$
Here is the link to the Wikipedia page:
Can someone explain this, please? Why divide by the hypotenuse? Why the $ab$? Thank you!
|
Polar Equation from the Center of the Ellipse
The equation of an ellipse is
$$
\left(\frac{x}{a}\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag1
$$
Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(1)$, we get
$$
r^2\cos^2(\theta)+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag2
$$
and we can solve $(2)$ for $r^2$ to get the polar equation
$$
r^2=\frac{\overbrace{a^2\!\left(1-e^2\right)}^{b^2}}{1-e^2\cos^2(\theta)}\tag3
$$
Polar Equation from a Focus of the Ellipse
Centered at the right focus
$$
\left(\frac{x+ae}a\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag4
$$
Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(4)$, we get
$$
r^2\cos^2(\theta)+2aer\cos(\theta)+a^2e^2+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag5
$$
which gives the quadratic equation in $r$:
$$
\frac{r^2\left(1-e^2\cos^2(\theta)\right)}{1-e^2}+2aer\cos(\theta)-a^2\!\left(1-e^2\right)=0\tag6
$$
whose solution is
$$
r=\frac{a\!\left(1-e^2\right)}{1+e\cos(\theta)}\tag7
$$
|
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|
How many solutions to the diophantine equation: $a + b + c + d = 22$
where
$\{a,b,c,d\}$ are distinct integers,
and
for each $x \in \{a,b,c,d\}, 1 \le x \le 9$.
Is there an elegant solution?
|
Count only the cases with $a<b<c<d$ and multiply with $4!=24$ afterwards.
Necessarily, $d\ge 7$, as otherwise $a+b+c+d\le 3+4+5+6=18<22$.
If $d=7$, then necessarily $a=4, b=5, c=6$, which gives us one solution $(4,5,6,7)$.
If $d=8$, we have the solution $(3,5,6,8)$, by increasing $c$ $(2,5,7,8)$ and $(3,4,7,8)$.
If $d=9$, there are again few enough solutions to count "by hand": $(1,4,8,9)$, $(2,3,8,9)$, $(1,5,7,9)$, $(2,4,7,9)$, $(2,5,6,9)$, $(3,4,6,9)$.
So the grand total is $240$.
|
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|
Is there a step by step checklist to check if a multivariable limit exists and find its value? Do we rely on certain intuition or is there an unofficial general crude checklist I should follow?
I had a friend telling me that if the sum of the powers on the numerator is smaller then the denominator, there is a higher chance that it may not exists.
And if the sum of powers on the numerator is higher then the denominator, most likely, it exists.
Also if there is a sin or cos or e-constant, it most likely exists.
How can I know what to do at the first glance given so little time exists for me in exam to ponder? If I spend all my time on figuring out a two-path test when the limit exists, that would be a huge disaster.
Is this one of those cases where practice makes perfect?
Example: $$\lim_{(x,y)\to(0,0)}\frac{(\sin^2x)(e^y-1)}{x^2+3y^2}$$
Please give me a hint and where do you get the hint.
Example: $$\lim_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{4x^4+y^6}\right)$$
I need a hint for this too.
Common methods I have learnt for reference:
Two-Path test, Polar Coordinates, Spherical Coordinates, Mean Value Theorem using inequalities.
|
I think there is no common method for all types of limits. You need significantly decrease the range of possible functions to get at least some kind of a road map.
For this two particular limits I suggest you the following two "brand new" approaches:
*
*The first one is usage of equivalences (or more general use of Taylor series expansion). Since $\sin x\sim x$ and $e^y-1\sim y$ we get
$$
\begin{align}
\lim\limits_{(x,y)\to(0,0)}\frac{\sin^2 x(e^y-1)}{x^2+3y^2}&=
\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}\frac{\sin^2 x}{x^2}\frac{e^y-1}{y}\\
&=\left(\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}\right)\left(\lim\limits_{(x,y)\to(0,0)}\frac{\sin^2 x}{x^2}\right)\left(\lim\limits_{(x,y)\to(0,0)}\frac{e^y-1}{y}\right)\\
&=\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}\\
\end{align}
$$
The last limit does exists and equals $0$. Indeed
$$
0\leq \left|\frac{x^2y}{x^2+3y^2}\right|=\frac{x^3|y|}{x^2+3y^2}\leq\frac{x^2|y|}{2\sqrt{3}xy}=\frac{1}{2\sqrt{3}}x\;\mathrm{sign}\; y
$$
$$
\lim\limits_{(x,y)\to(0,0)}\frac{1}{2\sqrt{3}}x\;\mathrm{sign}\; y=0
$$
Hence by mean value theorem
$$
\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}=0
$$
so
$$
\lim\limits_{(x,y)\to(0,0)}\frac{\sin^2 x(e^y-1)}{x^2+3y^2}=0
$$
*The second one is usage of continuiuty of fucntions. In your particular case this is $\exp$, then
$$
\lim\limits_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{x^4+3y^6}\right)=\exp\left(-\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{4x^4+y^6}\right)
$$
Now we introduce polar coordinates to evaluate the internal limit
$$
\begin{align}
\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{4x^4+y^6}&=
\lim\limits_{r\to 0}\frac{r^2}{4r^4\cos^4\phi+r^6\sin^6\phi}\\
&=\lim\limits_{r\to 0}\frac{1}{r^2(4\cos^4\phi+r^2\sin^6\phi)}
\end{align}
$$
Again we use inequalities
$$
r^2(4\cos^4\phi+r^2\sin^6\phi)\leq r^2(4\cos^2\phi+r^2\sin^2\phi)\\\leq r^2\max(4,r^2)(\cos^\phi+\sin^2\phi)=r^2\max(4,r^2)
$$
and even more for $r<2$ we can say that
$$
r^2(4\cos^4\phi+r^2\sin^6\phi)\leq r^2\max(4,r^2)\leq 4r^2
$$
hence
$$
\begin{align}
\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{4x^4+y^6}&=
\lim\limits_{r\to 0}\frac{1}{r^2(4\cos^4\phi+r^2\sin^6\phi)}\\
&\geq\lim\limits_{r\to 0}\frac{1}{4r^2}=+\infty
\end{align}
$$
and
$$
\lim\limits_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{x^4+3y^6}\right)=\exp(-\infty)=0
$$
|
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|
separable equation3 Is this solution correct?
Equation:
$x \sin y~dx+(x^2+1)\cos y~dy=0$
solve:
$$x \sin y\,dx=-(x^2+1)\cos y\,dy
\\ \frac{x}{-(x^2+1)}dx=\frac{\cos y}{\sin y}\,dy
\\ \frac{x}{-(x^2+1)}dx=\cot y\,dy
\\ \implies -\frac{1}{2}\ln(|x^2+1|)=-\cot^{-1} y \,dy$$.
|
You almost had it, but not quite.
From your second-to-last line:
$$\frac{-x}{(x^2+1}\,dx = \cot(y)\,dy$$
$$\int\frac{-x}{(x^2+1)}\,dx = \int\cot(y)\,dy$$
$$-\frac{1}{2}\ln(x^2+1) + C= \int\frac{\cos(y)}{\sin(y)}\,dy$$
For the left integral, let $u = \sin(y) \implies du=\cos(y) \,dy$.
$$-\frac{1}{2}\ln(x^2+1) + C= \int\frac{du}{u}\,dy$$
$$-\frac{1}{2}\ln(x^2+1) + C= \ln|\sin y|$$
This can be solved for $y$, but it's not pretty (and may introduce some domain error):
$$C\exp\left(-\frac{1}{2}\ln(x^2+1)\right)= \sin y$$
$$y = \arcsin\left(C\exp\left(-\frac{1}{2}\ln(x^2+1)\right)\right)$$
|
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|
Complex Analysis Gauss Mean Value Theorem Have I done this correctly?
Evaluate
$\displaystyle\int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})d\theta$
Gauss MVT:
$$f(z_0)=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(z_0+re^{i\theta})d\theta$$
So we have the following:
$$\frac{1}{2\pi}f(z_0)=\frac{1}{2\pi}\displaystyle\int_0^{2\pi} \sin^3(3e^{i\theta} + \frac{\pi}{4})d\theta $$
With,
$$z_0=\frac{\pi}{4}$$ and
\begin{align*}
f(z_0) &= sin^3(z_0) \\
f(\frac{\pi}{4})&=\frac{1}{2\sqrt2} \\
\frac{1}{2\pi}f(z_0)&=\frac{1}{2\pi}\cdot \frac{1}{2\sqrt2} \\
&= \frac{1}{4\sqrt{2}\pi}\\
\end{align*}
Thanks.
|
Since $\sin^3 z$ is a holomorphic function, its average over a circle centered at $\pi/4$ is equal to $\sin^3(\pi/4)=2^{-3/2}$. The average is $\frac{1}{2\pi}\int_0^{2\pi}\dots$, therefore
$$ \int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})\,d\theta = 2\pi \cdot 2^{-3/2}$$
|
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|
Please help me to find the value of $ABCDE$ (step by step) $ABCD\times E = DCBA$ with $A,B,C,D$, and $E$ distinct decimal digits (and $ABCD$ representing the concatenation of those digits). How can I find the value each of them?
|
First of all, note that $ABCD$ and $DCBA$ have the same digit-sum; this means in particular that they both have the same value $\mod 9$ (call it $r$). But working $\mod 9$, this means that $r\cdot E = r$, meaning that either $E\equiv 1$ (in which case we must have $E=1$ and then trivially $A=D, B=C$ and the solution is invalid) or $r\equiv0$ — in other words, $ABCD$ (and likewise $DCBA$) is a multiple of 9.
Next, we can use the 'alternating digit sum', or in other words work$\mod 11$ : $ABCD\equiv D-C+B-A\equiv q$, say, and $DCBA\equiv A-B+C-D\equiv -q$. So $E\cdot q \equiv -q\bmod 11$, leaving the possibilities $E\equiv-1$ (impossible, since the first positive number $\equiv -1 \bmod 11$ is $10$) or $q\equiv0$ — in other words, $ABCD$ (and likewise $DCBA$) is also a multiple of $11$, and so in fact it's a multiple of $99$.
With this in hand, we can look at $DCBA$ : since we know it's a multiple of $99$, and in fact a multiple of $99$ by a 2-digit number, we can write $DCBA=99\cdot XY$. But $99\cdot XY = (100-1)\cdot XY = XY00-XY = XZ\bar{X}W$ where $Z=Y-1$, $W=10-Y$, and $\bar{X}=9-X$ (note that $Y$ can't be $0$ since otherwise that would leave $A=0$, but $A$ is also the first digit of a number), so $D=X, C=Y-1, B=9-X, A=10-Y$. Likewise, we have $ABCD=99\cdot ST$ (with $ST\cdot E = XY$) yielding $A=S, B=T-1, C=9-S, D=10-T$. In other words, $E\cdot ST = XY$ with $X=10-T, Y=10-S$.
Now work mod 11 again: $XY\pmod{11}\equiv Y-X\equiv (10-S)-(10-T)\equiv T-S\equiv ST$. But since $XY=E\cdot ST$, then either $E\equiv 1\bmod 11$ (which we can rule out) or $XY\equiv ST\equiv 0$ - in other words, $XY$ and $ST$ are both multiples of $11$, i.e. $X=Y$ and $S=T$. Now it's a simple trial-and-error of finding $S$ and $E$ such that $E\cdot S = X = 10-S$; since $E\cdot S$ must be a single digit, there are only a couple of cases to consider and it's easy to find $S=2, E=4, X=8$ (the only alternative, $S=1, E=9, X=9$ gives $D=X=9=E$, so it can be ruled out), leading to $A=2, B=2-1=1, C=9-2 = 7, D=10-2 = 8$ and the final result:
$2178 \cdot 4 = 8712$.
|
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Range of $\frac{1}{2\cos x-1}$ How can we find the range of $$f(x) =\frac{1}{2\cos x-1}$$
Since range of $\cos x$ can be given as : $-1 \leq \cos x \leq 1$
therefore we can proceed as :$$\begin{array}{rcl}
-2 \leq & 2\cos x & \leq 2 \\
-2-1 \leq & 2\cos x -1 & \leq 2-1\\
-3 \leq & 2\cos x -1 & \leq 1 \\
\frac{-1}{3} \leq & \frac{1}{2\cos x-1} & \leq 1
\end{array}$$
Is it the range? Please suggest and guide.
|
I think your calculation is right
|
{
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|
How can prove this equation. if $a+b=c+d=e+f=\dfrac{\pi}{3}$,
$\dfrac{\sin{a}}{\sin{b}}\cdot\dfrac{\sin{c}}{\sin{d}}\cdot\dfrac{\sin{e}}{\sin{f}}=1$,
Prove that:
$\dfrac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\dfrac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\dfrac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$
|
Consider an equilateral triangle $ABC$, and let $D$ be on $BC$ so that $\angle{BAD}=a$, so $\angle{DAC}=\frac{\pi}{3}-a=b$. Let $E$ be on $AC$ so that $\angle{CBE}=c$, so $\angle{EBA}=\frac{\pi}{3}-c=d$. Let $F$ be on $AB$ so that $\angle{ACF}=e$, so $\angle{FCB}=\frac{\pi}{3}-e=f$.By the sine version of Ceva's theorem and the given condition $\frac{\sin{a}}{\sin{b}}\cdot \frac{\sin{c}}{\sin{d}}\cdot \frac{\sin{e}}{\sin{f}}=1$, $AD, BE, CF$ are concurrent at a point, which we shall call $P$.
Extend $AD$ to points $A_1, A_2$ s.t. $\angle{A_1CB}=a+f, \angle{A_2BC}=b+c$. We have $\angle{CA_1A}=\pi-\angle{A_1CA}-\angle{A_1AC}=\pi-b-(e+f+a+f)=e$. Similarly $\angle{BA_2A}=d$. Thus $APC$ is similar to $ACA_1$ and $APB$ is similar to triangle $ABA_2$. Therefore $\frac{AA_1}{AC}=\frac{AC}{AP}=\frac{AB}{AP}=\frac{AA_2}{AB}$, so $AA_1=AA_2$, so $A_1=A_2$. Now
$$\frac{\sin{(b+2c)}}{\sin{(a+2f)}}=\frac{\frac{A_1P}{\sin{(a+2f)}}}{\frac{A_2P}{\sin{(b+2c)}}}=\frac{\frac{CP}{\sin{e}}}{\frac{BP}{\sin{d}}}=\frac{CP\sin{d}}{BP\sin{e}}$$
Similarly, we get $$\frac{\sin{(d+2e)}}{\sin{(c+2b)}}=\frac{AP\sin{f}}{CP\sin{a}}$$ $$\frac{\sin{(f+2a)}}{\sin{(e+2d)}}=\frac{BP\sin{b}}{AP\sin{c}}$$ so multiplying gives the desired equality $$\frac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\frac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\frac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$$
|
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|
Solve for $X$ in a simple $2\times2$ equation system. I posted a similar question recently but I still have problem with this problem and would appreciate any help!
$$\left[ \begin{array}{cc} 9 & -3\\ 5 & -5\end{array} \right] - X \left[ \begin{array}{cc} -9 & -2\\ 8 & 5\end{array} \right] = E$$ With $E$ i pressume they mean the identity matrix $\left[ \begin{array}{cc} 1 & 0\\ 0 & 1\end{array} \right]$.
How should I go on and solve this for the $2\times2$ matrix $X$? a full development so I can follow your solution would be very much appreciated!
Thank you kindly for you help!
|
I will express it as equations, as I think that is easier (at least for beginners).
With
$$X=\begin{pmatrix} x_{11} & x_{12}\\ x_{21} & x_{22} \\ \end{pmatrix}$$
At first we make the multiplication
\begin{align*}
X \cdot \begin{pmatrix} -9 & -2\\ 8 & 3 \end{pmatrix}&=
\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \cdot
\begin{pmatrix} -9 & -2 \\ 8 & 3 \end{pmatrix}\\
&= \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\
-9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\
\end{pmatrix}
\end{align*}
So our equation is
$$\begin{pmatrix} 9 & -3 \\ 5 & -5 \end{pmatrix} - \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\
-9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\
\end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$
Now we write it as a system of equations:
\begin{align*}
1&=9- (-9 x_{11} +8x_{12})\\
0&= -3-(-2x_{11} +3x_{12}) \\
0&=5-(-9x_{21} +8 x_{21}) \\
1&=-5 - (-2x_{21}+3x_{22}) \\
\end{align*}
If you need help solving this system tell me.
|
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|
Showing two graphs isomorphic using their adjacency matrices I don't know how to solve the following problem:
Show that two simple graphs $G$ and $H$ are isomorphic if and only if there exists a permutation matrix $P$ such that $A_G=PA_HP^t$.
Here $A$ is the adjacency matrix. I have a feeling this shouldn't be very difficult, but my linear algebra is not very good, am I missing something obvious?
|
Here’s an example that may get you thinking in the right direction. Consider $PAP^t$, where $P$ is the permutation matrix
$$\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}$$
and, as an illustrative example,
$$A=\begin{bmatrix}
1&2&3&4\\
2&3&4&5\\
0&1&2&3\\
3&2&1&0
\end{bmatrix}\;.$$
We have
$$PA=\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}
\begin{bmatrix}
1&2&3&4\\
2&3&4&5\\
0&1&2&3\\
3&2&1&0
\end{bmatrix}=
\begin{bmatrix}
3&2&1&0\\
1&2&3&4\\
0&1&2&3\\
2&3&4&5
\end{bmatrix}\;,
$$
and then
$$
PAP^t=\begin{bmatrix}
3&2&1&0\\
1&2&3&4\\
0&1&2&3\\
2&3&4&5
\end{bmatrix}
\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}=
\begin{bmatrix}
0&3&1&2\\
4&1&3&2\\
3&0&2&1\\
5&2&4&3
\end{bmatrix}\;.
$$
The first row of $A$ is the second row of $PAP^t$, and the first column of $A$ is the second column of $PAP^t$. The second row and second column of $A$ are the fourth row and column of $PAP^t$. The fourth row and column of $A$ are the first row and column of $PAP^t$. And the third row and column of $A$ are still the third row and column of $PAP^t$. In other words, both the rows and columns have been permuted by the permutation $(1,2,4)$ in cycle notation or
$$\pmatrix{1&2&3&4\\2&4&3&1}\tag{1}$$
in two-line notation. If $A$ is the adjacency matrix of a graph, $PAP^t$ is just the adjacency matrix of the same graph after the vertices have been renumbered according to the permutation $(1)$.
|
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|
Number of solutions to $x^3 - c^2x^2 + p^2 \equiv 0 \mod p^3$ I would like to find the number of solutions ($\mod p^3$) to $x^3 - c^2x^2 + p^2 \equiv 0 \mod p^3$, where $p$ is prime and $c$ is an integer not divisible by $p$. I think Hensel's Lemma will be useful, but I'm not entirely sure how to apply it.
Help would be greatly appreciated.
|
If $x$ satisfies $x^3 - c^2x^2 + p^2 \equiv 0 \pmod {p^3}$, $x$ also satisfies
$$x^3 - c^2x^2 + p^2 \equiv 0 \pmod {p^2}$$
So $x^2(x-c^2)\equiv 0 \pmod {p^2}$. So $x$ satisfies $p\mid x$ or $p^2\mid(x-c^2)$ and $x$ satisfies only one formula. (Because $c$ is not disible by $p$.)
If $p\mid x$, then there exists $y\in\mathbb{Z}$ s.t. $x=py$. Substitute $x$ with $py$ then we get
$$(1-c^2y^2)p^2\equiv 0\pmod{p^3}$$
This formula is equivalent to
$$1-c^2y^2\equiv 0\pmod{p}$$
So the number of $y$ is 2, namely $y\equiv c^{-1} \pmod{p}$ and $y\equiv-c^{-1} \pmod{p}$. So $x\equiv \pm c^{-1} p + kp^2 \pmod{p^3}$ ($k$ is arbitary integer) is solution of this equation, therefore the number of this equation is $2p$.
If $p^2\mid(x-c^2)$, then there exists $z$ s.t. $x=c^2+zp^2$. Substitute $x$ with $c^2+zp^2$ and rearrange it then we get
$$c^4 zp^2 +p^2\equiv 0 \pmod{p^3}$$
This equation is equivalent to
$$c^4 z + 1\equiv 0 \pmod{p}$$
and it has only one soultion $z\equiv 1 \pmod p$, so the number of solutions of this equation is $2p+1$.
|
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|
A finite summation involving $2013$ Can you help me compute the summation below?
$$1+\frac{1}{2}+\cdots+\frac{1}{2013}+\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\cdots+\frac{1}{2012\cdot 2013}+\cdots+\frac{1}{1\cdot2\cdots2013}$$
|
Here's how it works for $n=3$.
Find a common denominator:
$$\begin{align}
1 + S &= 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{1\cdot2} + \frac{1}{1\cdot3} + \frac{1}{2\cdot3} + \frac{1}{1\cdot2\cdot3}\\
&=\frac{1\cdot2\cdot3 + 2\cdot3 + 1\cdot3 + 1\cdot2 + 3 + 2 + 1 + 1}{1\cdot2\cdot3}\\
&= \frac{(1+1)(1+2)(1+3)}{1\cdot2\cdot3}\\
&= \frac{2\cdot3\cdot4}{1\cdot2\cdot3}\\
&= 4
\end{align}$$
So, $S = 3$.
Generally, $(1 + S)n! = (x + 1)(x + 2) \cdots (x + n)\big|_{x=1} = (n + 1)!$, which gives $S = n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$(1-a^2)(1-b^2)(1-c^2)=8abc\; (a,b,c\in \mathbb{Q}^{+})$ has infinitely many solutions Prove that:
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
has an infinite number of solutions $(a,b,c)$.
Now we found this infinite
$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$
My Question: Have other form solution?
|
The equation,
$$(1-a^2)(1-b^2)(1-c^2)-8abc=0$$
is just a quadratic in any of the variables. Hence,
$$c=\frac{8ab\pm y}{2(-1+a^2+b^2-a^2b^2)}$$
where,
$$64a^2b^2-4(-1+a^2+b^2-a^2b^2)(1-a^2-b^2+a^2b^2) = y^2$$
This quartic polynomial to be made a square is easily reducible to an elliptic curve. From initial rational point {$a,b$} = {$5/6,\; 2/11$}, one can find an infinite more.
|
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|
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
|
From the original inequality $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
Let’s set $$x= \sqrt{3a + b^3}$$ $$y= \sqrt{3b + c^3}$$ $$z= \sqrt{3c + a^3}$$
By AM-GM we know that $$x^2-2xy+y^2 \ge 0$$ $$y^2-2yz+z^2 \ge0$$ $$z^2-2xz+x^2 \ge 0$$ So we then can set up the following inequalities
$$x^2+y^2\ge2xy$$
$$y^2+z^2\ge2yz$$
$$z^2+x^2\ge2xz$$
Multiply each inequality above by $z$, $x$, $y$ in that order which yields
$$zx^2+zy^2\ge2xyz$$
$$xy^2+xz^2\ge2xyz$$
$$yz^2+yx^2\ge2xyz$$
Add these inequalities and factor to get
$$xy(x+y)+xz(x+z)+zy(y+z) \ge 6xyz$$
Now divide both sides by $xyz$ for
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\ge6$$
Now replace $x$, $y$ and $z$ with the terms on the LHS of the original inequality above
$$\frac{\sqrt{3a + b^3}+\sqrt{3b + c^3}}{\sqrt{3c + a^3}}+\frac{\sqrt{3a + b^3}+\sqrt{3c + a^3}}{\sqrt{3b+ c^3}}+ \frac{\sqrt{3b + c^3}+\sqrt{3c + a^3}}{\sqrt{3a+ b^3}} \ge 6$$
with equality if and only if $a=b=c=1$.
|
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|
Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis? I don't understand where that formula came from. Could someone explain? For example any point $(x,y)$ on the ellipse from the two foci $(-c,0)$ and $(c,0)$ is equal to $2a$ where $2a$ is the distance of the major axis. Where did this idea come from?
|
OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We draw an ellipse by holding a pen taut against the string. The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable. First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic. Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$. We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
|
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|
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
|
Let $s = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + \dots$
Then $2s = 1 + 2/2 + 3/4 + 4/8 + 5/16 + \dots$
And then subtracting terms with similar denominators gives: $2s - s = 1 +1/2+1/4+1/8+1/16+\dots = 2$
|
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|
How to effectively calculate $(1/\sqrt1 + \sqrt2) + (1/\sqrt2 + \sqrt3) +\cdots + (1/\sqrt{99} + \sqrt{100})$ I have this series:
$$\frac{1}{\sqrt1 + \sqrt2} +\frac{1}{\sqrt2 + \sqrt3} +\frac{1}{\sqrt3 + \sqrt4} +\cdots+\frac{1}{\sqrt{99} + \sqrt{100}} $$
My question is, what approach would you use to calculate this problem effectively?
|
Hint :$\displaystyle \frac{1}{\sqrt{(n+1)}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$(By multiplying the numerator and the denominator by multiplying $(\sqrt{n+1}-\sqrt{n})$ to both numerator and denominator.)
So we have,
$(1/(\sqrt1 + \sqrt2)) + (1/(\sqrt2 + \sqrt3)) + .. + (1/(\sqrt{99} + \sqrt{100}))=\displaystyle \sum_{n=1}^{99}\frac{1}{\sqrt{(n+1)}+\sqrt{n}}=\sum_{n=1}^{99}\sqrt{n+1}-\sqrt{n}=\sqrt{100}-\sqrt{1}=10-1=9$
|
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|
Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum or minimum and determine the same.
|
$${\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\frac{\sum_{i=1}^{n}i^3}{n^4}=\frac{(n(n+1))^2}{4n^4}=\frac{1}{4}(1+\frac{1}{n})^2$$
$$\Rightarrow \displaystyle \lim_{n\to\infty}{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\lim_{n\to\infty}\frac{1}{4}(1+\frac{1}{n})^2=1/4$$(Using the fact that $\lim_{n\to\infty}1/n=0)$
|
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|
Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated. A deck of 52 cards is shuffled, and we deal a bridge of 13 cards. Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated.
Here is what I did:
$Cov(X,Y) = E[XY]-E[X]E[Y]$
uncorrelated means $Cov(X,Y) = 0$, hence $E[XY]=E[X]E[Y]$
$E[X] = \sum_{k=0}^{k=4} k \frac{\dbinom{4}{k} \dbinom{48}{13-k}}{\dbinom{52}{13}} $
$E[Y] = \sum_{k=0}^{k=13} k \frac{\dbinom{13}{k} \dbinom{39}{13-k}}{\dbinom{52}{13}} $
Are the summations above correct? and how do I calculate $E[XY]$?
|
We show that although $X$ and $Y$ are not independent, the conditional expectation of $Y$, given $X=x$, is equal to the plain expectation of $Y$.
Given that $x=0$ (no Aces), we are choosing $13$ cards from the $48$ non-Aces. The expected number of spades is then $13\cdot \frac{12}{48}=\frac{13}{4}$.
Given that $x=1$ (one Ace), there are two possibilities: (i) the Ace is a spade or (ii) it is not.
(i) If the Ace is a spade (probability $\frac{1}{4}$), then we have $1$ assured spade. In addition, we are choosing $12$ cards from the $48$ non-spades, so the expected number of additional spades is $12\cdot\frac{12}{48}$. Thus (i) makes a contribution of $\frac{1}{4}\cdot\left(1+12\cdot\frac{12}{48}\right)$ to the conditional expectation.
(ii) If the Ace is a non-spade (probability $\frac{3}{4}$), the expected number of spades is $12\cdot \frac{12}{48}$. Thus
$$E(Y|X=2)= \frac{1}{4}\cdot\left(1+12\cdot\frac{12}{48}\right)+\frac{3}{4}\left(12\cdot \frac{12}{48} \right).$$
This simplifies to $\frac{13}{4}$.
A similar analysis works for $X=2$, $3$, and $4$. For instance, if $x=2$, then with probability $\frac{1}{2}$ the Ace of spades is included among the two Aces, and with probability $\frac{1}{2}$ it is not. The analogue of the calculation we made for $x=1$ yields
$$E(Y|X=2)= \frac{1}{2}\cdot\left(1+11\cdot\frac{12}{48}\right)+\frac{1}{2}\left(11\cdot \frac{12}{48} \right),$$
which again simplifies to $\frac{13}{4}$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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|
Finding the locus of a complex number Find the locus of $\arg\left(\frac{z-3}{z}\right) = \frac{\pi}{4}$ where $z$ represent complex number.
Working: $\arg\left(\frac{z-3}{z}\right) $ can be written as $\arg(z-3)-\arg(z) = \frac{\pi}{4}$, or $\arg\left((x-3)+iy\right) - \arg(x+iy)=\frac{\pi}{4}$.
If we take tangent to both side we get :
$$
\begin{align*}
\tan \left[\arg\left((x-3)+iy\right) -\arg (x+iy)\right] = \tan \frac{\pi}{4},\\
\tan\left[\frac{\arg\left((x-3)+iy\right) - \arg(x+iy)}{1+ \arg\left((x-3)+iy\right) \arg(x+iy)}\right] = 1.
\end{align*}
$$
Please suggest further...
|
So, $$\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$$
$$\implies \arctan\left(\frac{ \frac y{(x-3)}-\frac yx}{1+ \frac y{(x-3)}\frac yx}\right)=\frac\pi4$$
$$\implies \frac{3y}{(x-3)x+y^2}=\tan\left(\frac\pi4\right)=1$$ assuming $x(x-3)\ne0$
$$\implies x^2-3x+y^2=3y$$
If $x=3,\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$ reduces to $$\arctan \left( \frac y0\right)-\arctan \left( \frac y3\right)=\frac\pi4$$
$$\text{Now, }\arctan \left( \frac y0\right)= \begin{cases}
\ \frac{\pi}2 & \text{if } y>0 \\
\ -\frac{\pi}2 & \text{if } y<0\\
\text{ undefined} & \text{if } y= 0
\end{cases}$$
$$\text{ If }y>0, \arctan \left( \frac y3\right)=\frac\pi2-\frac\pi4=\frac\pi4$$
$$\implies \frac y3=\tan\left(\frac\pi4\right)=1\implies y=3\text{ at }x=3$$
$$\text{ If }y<0, \arctan \left( \frac y3\right)=-\frac\pi2-\frac\pi4=-\frac{3\pi}4$$
$$\implies \frac y3=\tan\left(-\frac{3\pi}4\right)=\tan\left(\pi-\frac{\pi}4\right)=-\tan\left(\frac\pi4\right)=-1\implies y=-3\text{ at }x=3$$
Similarly, for $x=0$
|
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|
If $x > 1$, prove that $f(x) = \dfrac{1}{\sqrt{x^2+1}}\log(x+\sqrt{x^2-1})$ $$f(x) = 2\int_{0}^{1}\dfrac{du}{u^2(1-x)+1+x}$$
I have used partial fractions but solves nothing.
|
Since $x>1$ the denominator better reads as $(1+x)-u^2(x-1)$. Now, let's change variables $u = \sqrt{\frac{x+1}{x-1}} w$, giving:
$$
f(x) = \frac{2}{1+x} \sqrt{\frac{x+1}{x-1}} \int_0^\sqrt{\frac{x-1}{x+1}} \frac{\mathrm{d}w}{1-w^2} = \frac{1}{\sqrt{x^2-1}} \int_0^\sqrt{\frac{x-1}{x+1}} \left(\frac{1}{1+w} + \frac{1}{1-w}\right) \mathrm{d}w
$$
Giving
$$
f(x) = \frac{1}{\sqrt{x^2-1}} \log \frac{1+\sqrt{\frac{x-1}{x+1}} }{1-\sqrt{\frac{x-1}{x+1}}} = \frac{1}{\sqrt{x^2-1}} \log \left( \frac{\sqrt{x-1} + \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x-1}} \cdot \color\green{\frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} + \sqrt{x-1}}} \right) = \frac{1}{\sqrt{x^2-1}} \log \left( \frac{ (x+1)+(x-1)+2\sqrt{x^2-1} }{(x+1)-(x-1)} \right) = \frac{1}{\sqrt{x^2-1}} \log \left( x+\sqrt{x^2-1} \right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Please help with this boundary value problem I came across the following problem and I am stuck on it :
Let $u$ be a solution of the boundary value problem $u''+\dfrac{1}{t}u'=f(t),t \in (0,1)$ and $u'(0)=a,u(1)=b.$ Define for $x^2+y^2 \leq 1,v(x,y)=u(\sqrt {x^2+y^2})$ and $g(x,y)=f(\sqrt {x^2+y^2})$.Then $v$ is a solution of the PDE $v_{xx}+v_{yy}=g$ in $\{(x,y):x^2+y^2 < 1\}$ and $v(x,y)=0$ on $\{(x,y):x^2+y^2 = 1\}$ if
1.$\space a>0 ,b>0$
2.$\space a>0,b=0$
3.$\space a=0,b=0$
4.$\space a<0,b=0$.
I have to determine which of the above four options is/are true? Can someone point me in the right direction? Thanks in advance for your time.
|
Attacking the problem backwards from Poisson equation to its radial symmetric equation is more intuitive in my opinion.
$\newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}}$
$\newcommand{\pb}[2]{\frac{\partial }{\partial #2}\left(#1\right)}$
$\newcommand{\ppt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$
Consider the Poisson's equation $\Delta v = v_{xx} + v_{yy} = g$ in polar coordinates $(r,\theta)$:
$$
r = \sqrt{x^2 + y^2} \text{ and } \theta = \arctan(y/x) + C
$$
We have:
$$\pp{r}{x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r}, \text{ and } \pp{r}{y} = \frac{y}{r}
$$
$$\pp{\theta}{x} = \frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}} = -\frac{y}{r^2}, \text{ and } \pp{\theta}{y} = \frac{x}{r^2}
$$
Now
$$
\pp{v}{x} = \pp{v}{r} \pp{r}{x} + \pp{v}{\theta} \pp{\theta}{x} = \pp{v}{r} \frac{x}{r} - \pp{v}{\theta} \frac{y}{r^2}
$$
and
$$
\begin{aligned}
v_{xx}= \pb{\pp{v}{x}}{x} &= \ppt{v}{r}\pp{r}{x}\frac{x}{r} + \pp{v}{r}\pb{\frac{x}{r} }{x} - \ppt{v}{\theta}\pp{\theta}{x}\frac{y}{r^2} - \pp{v}{\theta}\pb{\frac{y}{r^2}}{x}
\\
&= \ppt{v}{r}\frac{x^2}{r} + \pp{v}{r}\left(\frac{1}{r} - \frac{x^2}{r^3}\right)
+ \ppt{v}{\theta}\frac{y^2}{r^4} + \pp{v}{\theta}\frac{2xy}{r^4}
\end{aligned}\tag{1}
$$
Similarly we have:
$$
v_{yy} = \ppt{v}{r}\frac{y^2}{r} + \pp{v}{r}\left(\frac{1}{r} - \frac{y^2}{r^3}\right)
+ \ppt{v}{\theta}\frac{x^2}{r^4} - \pp{v}{\theta}\frac{2xy}{r^4}\tag{2}
$$
Exploiting $x^2+y^2 = r^2$, (1) plus (2) gives:
$$
\Delta v = \ppt{v}{r} + \frac{1}{r}\pp{v}{r}+\frac{1}{r^2}\ppt{v}{\theta}.
$$
If a function is radial symmetric, meaning it can be represented using only $r$ for it is a constant for fixed $r$, i.e., it is a constant on any fixed circle, and rotating does not change its value. Hence $$\displaystyle \ppt{v}{\theta} = 0.$$ And the Laplacian of a radial symmetric function is again radial symmetric for no involvement of $\theta$. Here I use $r$ for the $t$ in your question.
Let $v(x,y) =u(\sqrt{x^2 + y^2}) =u(r)$, $g(x,y) = f(r)$. Then
$$
u'' + \frac{1}{r}u' = f(r)
$$
Using your favorite method of solving ODE, here I cook up using product rule reverse direction:
$$
(ru')' = rf(r)
$$
Integrating:
$$ u'(r) = \frac{1}{r}(\int^r_0 sf(s) ds+ r u'(r)|_{r=0}) \tag{3}
$$
Since the limit exists when $r\to 0$:
$$
a = u'(0) = \lim_{r\to 0}\frac{\int^r_0 sf(s) ds}{r} = \lim_{r\to 0} rf(r) =0
$$
Now for $b$, since $v = 0$ when $r= 1$, hence $b = u(1) = 0$.
Hence $a = b =0$.
If you would like to get a form of the solution, integrating (3) again yields:
$$ u(r) - u(0) = \int^{r}_0 u'(\tau)d\tau = \int^{r}_0 \frac{1}{\tau}\int^{\tau}_0 sf(s) ds d\tau.
$$
|
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|
The least possible value How to find the least possible value for :$$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$$
For every real $x$
|
$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$
$=(y+2)^2+(y+1)^2+y^2+(y-1)^2+(y-2)^2$.Let $(y=(x-3))$
$=5y^2+2(1^2+2^2)\ge2(1^2+2^2)$
Equality occurs at $y=0\Rightarrow x=3$
|
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|
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
|
No. You should have $2(x+4)^2$ on the right after multiplying. Then $$0\le2(x+4)^2-4(x+4)\\0\le(x+4)^2-2(x+4)\\0\le(x+4)\bigl((x+4)-2\bigr)\\0\le(x+4)(x+2).$$ Now, that last inequality is true whenever $x+4$ and $x+2$ have the same sign or one of them is $0$--that is, whenever $x+4\le0$ ($x\le-4$) or $x+2\ge0$ ($x\ge -2$)--but we can't allow $x=-4$ in our original inequality. Hence, our given inequality is true whenever $x<-4$ or $x\ge -2$.
|
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|
Find remainder of $F_n$ when divided by $5$
Let $\{ F_n\}$ be the sequence of numbers defined by $F_1=1=F_2;\, F_{n+1}=F_n+F_{n-1}$ for $n \geq 2$. Let $f_n$ be the remainder left when $F_n$ is divided by $5$. Then $f_{2000}$ equals
(A) $0$ $~~~~~~~~~~~~~~~~$ (B) $1$ $~~~~~~~~~~~~~~~~$ (C) $2$$~~~~~~~~~~~~~~~~$ (D) $3$
I found that $F(1)=1$, $F(2)=1$, $F(3)=2$, $F(4)=3$, $F(5)=5$, $F(6)=8$, $F(7)=13$, $F(8)=21$, $F(9)=34$, and $F(10)=55$. But I need a systematic pattern to find $F_n$.
|
$$F_n=F_{n-1}+F_{n-2}=F_{n-2}+F_{n-3}+F_{n-2}=2F_{n-2}+F_{n-1}$$
$$=2(F_{n-3}+F_{n-4})+F_{n-3}=3F_{n-3}+2F_{n-4}$$
$$=3(F_{n-4}+F_{n-5})+2F_{n-4}=5F_{n-4}+3F_{n-5}$$
$$\text{ So,} F_n\equiv 3F_{n-5}\pmod 5 $$
As $F_1=F_2=1\implies F_0=F_2-F_1=0\equiv0\pmod 5$
$\implies 5\mid F_n$ if $5\mid n$
Alternatively, from Lemma#$5$ of this , $F_n\mid F_m \iff n\mid m\text{ or } n=2$ for $m\ge n\ge 1$
We find $F_2=F_1=1,F_3=2,F_4=3, F_5=5$
So, $F_5\mid F_m \iff 5\mid m$ for $m\ge1$
|
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|
Solving $x+2y+5z=100$ in nonnegative integers I have not done combinatorics since high school, so this is an embarrassingly simple question.
We can solve the diophantine equation $x+y+z=100$ in nonnegative integers using the "bars and boxes" combinatorial method. We have $100$ dots, and we want place 2 partition markers among them, so the answer is ${ 102 \choose 2}$. Is there a way to generalize this (by a change of variable, perhaps) to equations like $x+2y+5z=100$? I know we can handle the case where we need to solve in positive integers by making the substitutions $x\rightarrow x+1$ and so forth, but I can't think of a way to apply a similar technique when the coefficients aren't $1$.
If not, is there a slick way to handle these more general equations?
|
We can also solve it without generating functions as follows.
First, as a lemma, what is the number of nonnegative solutions to $x + 2y = n$? For any choice of $y$ such that $0 \le 2y \le n$, we have a unique solution (take $x = n - 2y$), so the number of solutions is the number of such choices, i.e. $\lfloor \frac{n}{2} \rfloor + 1$ (or, if you like, $\lceil \frac{n+1}{2} \rceil$).
Now the number of solutions to $x + 2y + 5z = n$: for each choice of $z$ such that $0 \le 5z \le n$, the number of solutions is that of $x + 2y = n - 5z$. The total number of solutions is got by adding them up.
For $n = 100$, the number $n - 5z$ takes values (enumerating the odd and even ones separately) $0, 10, \dots, 100$ and $5, 15, \dots, 95$, so the total number of solutions is
$$\begin{align}
&\phantom{=} \left(\left\lceil \frac{0+1}{2} \right\rceil + \left\lceil \frac{10+1}{2} \right\rceil + \dots + \left\lceil \frac{100+1}{2} \right\rceil\right) + \left(\left\lceil \frac{5+1}{2} \right\rceil + \left\lceil \frac{15+1}{2} \right\rceil + \dots + \left\lceil \frac{95+1}{2} \right\rceil \right)\\
&= \left(1 + 6 + \dots + 51\right) + \left(3 + 8 + \dots + 48\right) \\
&= 286 + 255 = 541.
\end{align}$$
|
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|
How to find the value of $\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$
Determine
$$
\sin\left(\pi \over 14\right) + 6\sin^{2}\left(\pi \over 14\right)
-8\sin^{4}\left(\pi \over 14\right)
$$
My idea: Let $\displaystyle{\sin\left(\pi \over 14\right)} = x$.
|
Try using the Identities
$$\sin^4\theta = \dfrac{3-4\cos2\theta +\cos 4\theta}{8}$$
$$\sin^2\theta = \dfrac{1-\cos2\theta}{2}$$
$$\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$$
Let $\theta = \dfrac{\pi}{14}$
$$\sin{\theta}+6\dfrac{1-\cos2\theta}{2}-8\dfrac{3-4\cos2\theta +\cos 4\theta}{8}$$
$$\sin{\theta}+3 - 3\cos 2\theta-3+4\cos2\theta-\cos 4\theta$$
$$=\sin{\theta}+\cos 2\theta-\cos4\theta\tag1$$
Replacing $\theta$ with $\dfrac{\pi}{14}$ in $(1)$ we get
$$= \cos\left(\dfrac{\pi}{7}\right) - \cos\left(\dfrac{2\pi}{7}\right) + \cos\left(\dfrac{\pi}{2} - \dfrac{\pi}{14}\right)$$
$$= \cos\left(\dfrac{\pi}{7}\right) - \cos(\dfrac{2\pi}{7}) + \cos\left(\dfrac{3\pi}{7}\right)$$
$$= \cos\left(\dfrac{\pi}{7}\right) + \cos\left(\dfrac{3\pi}{7}\right)+ \cos\left(\dfrac{5\pi}{7}\right)=\dfrac{1}{2}$$
\begin{cases}
Known\\
\sum_{i=0}^{n-1} \cos((2i+1)x) = \dfrac{\sin2nx}{2\sin x}\\
\text{for n = 3} \\
\sum_{i=0}^{2} \cos((2i+1)x) = \dfrac{\sin2\cdot 3 \cdot x}{2\sin x}\\
= \dfrac{\sin6x}{2\sin x} \\
\text{Let } x = \dfrac{\pi}{7}\\
= \dfrac{\sin \dfrac{6\pi}{7}}{2\sin \dfrac{\pi}{7}} = \dfrac{\sin (\pi -\dfrac{6\pi}{7})}{2\sin \dfrac{\pi}{7}} = \dfrac{\sin (\dfrac{\pi}{7})}{2\sin \dfrac{\pi}{7}}\\
=\dfrac{1}{2}
\end{cases}
|
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|
Please, I need a more detailed explanation of the particular solution of the problem with vectors Here is the problem and its solution (link to the source if you are interested):
Two different points $A$ and $B$ are given. Find a set of such points $M$, that $\overrightarrow{MA}\cdot\overrightarrow{MB}=k^2$, where $k$ is a given non-zero number.
The solution:
Let coordinates of points be: $A(x_a, y_a), B(x_b, y_b), M(x, y)$.
Then:
$$
\begin{align}
& \overrightarrow{MA}=(x_a - x, y_a - y), \overrightarrow{MB}=(x_b - x, y_b - y), \\
& \overrightarrow{MA}\cdot\overrightarrow{MB}= (x_a - x)(x_b - x)+(y_a - y)(y_b - y)=k^2
\end{align}
$$
Transform the last expression, opening brackets:
$$
x_a x_b - (x_a+x_b)x + x^2 + y_a y_b - (y_a+y_b)y + y^2 = k^2
$$
At this point I ask someone to provide a more detailed explanation, please, cause I failed to understand the next steps. What a transformation has been made here?
$$
\left( x - \frac{x_a + x_b}{2} \right)^2 - \frac{(x_a - x_b)^2}{4} +
\left( y - \frac{y_a + y_b}{2} \right)^2 - \frac{(y_a - y_b)^2}{4} = k^2 \\
\left( x - \frac{x_a + x_b}{2} \right)^2 + \left( y - \frac{y_a + y_b}{2} \right)^2
= k^2 + \frac{(x_a - x_b)^2}{4} + \frac{(y_a - y_b)^2}{4}
$$
Thus, this is a circumference with a center at the middle of segment $AB$ and its radius. (This I also failed to understand)
$$
r = \sqrt{k^2 + \frac{|AB|^2}{4}}
$$
|
Hints:
$$x^2\pm a=\left(x\pm\frac{a}{2}\right)^2-\frac{a^2}{4}$$
and this is what they do in the first step, with both $\,x\,,\,y\,$ . For example,
$$y^2-(y_a+y_b)y=\left(y-\frac{y_a+y_b}{2}\right)^2-\frac{(y_a+y_b)^2}{4}\;,\;\;etc.$$
$$|AB|^2=(x_a-x_b)^2+(y_a-y_b)^2\;\ldots$$
|
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|
Weird calculus limit How to find the following limit?
$$
\lim_{n \to \infty} \dfrac{ 5^\frac{1}{n!} - 4^\frac{1}{n!} }{ 3^\frac{1}{n!} - 2^\frac{1}{n!} }
$$
Edit done to the question.
Thank you!
|
Let $x = \dfrac1{n!}$. We then have $$\lim_{n \to \infty} \dfrac{5^{1/n!} - 4^{1/n!}}{3^{1/n!} - 2^{1/n!}} = \lim_{x \to 0} \dfrac{5^x-4^x}{3^x-2^x}$$
Now $$\lim_{x \to 0}\dfrac{a^x - 1}x = \log(a) \tag{$\star$}$$
\begin{align}
\lim_{x \to 0} \dfrac{5^x-4^x}{3^x-2^x} & = \lim_{x \to 0} \dfrac{\dfrac{5^x-1}x-\dfrac{4^x-1}x}{\dfrac{3^x-1}x-\dfrac{2^x-1}x}\\
& = \dfrac{\lim_{x \to 0} \dfrac{5^x-1}x-\lim_{x \to 0} \dfrac{4^x-1}x}{\lim_{x \to 0} \dfrac{3^x-1}x-\lim_{x \to 0} \dfrac{2^x-1}x}\\
& = \dfrac{\log(5) - \log(4)}{\log(3) - \log(2)}\\
& = \dfrac{\log(5/4)}{\log(3/2)}
\end{align}
|
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|
Convergence or divergence of $\sum\limits^{\infty}_{n=0} \frac{(2n+1)^{2}} {3^{n}(2n)!}$
Show if the series converges or diverges.
$$\sum^{\infty}_{n=0} \frac{(2n+1)^{2}} {3^{n}(2n)!}$$
Can someone please help with proving this? (I think it converges)
|
$$\dfrac{(2n+1)^2}{(2n)!} = \dfrac{4n^2+4n+1}{(2n)!} = \dfrac{(2n)(2n-1)+6n+1}{(2n)!} = \dfrac1{(2n-2)!}+ 3 \cdot \dfrac1{(2n-1)!} + \dfrac1{(2n)!}$$
Now $$\sum_{n=0}^{\infty}\dfrac{(2n+1)^2}{(2n)!}x^n = \sum_{n=0}^{\infty}\dfrac{x^n}{(2n)!} + 3 \cdot\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-1)!} + \sum_{n=1}^{\infty}\dfrac{x^n}{(2n-2)!}$$
\begin{align}
\sum_{n=0}^{\infty}\dfrac{x^n}{(2n)!} & = \dfrac{\exp(\sqrt{x})+\exp(-\sqrt{x})}2\\
\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-1)!} & = \sqrt{x} \cdot \dfrac{\exp(\sqrt{x})-\exp(-\sqrt{x})}2\\
\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-2)!} & = x \cdot \dfrac{\exp(\sqrt{x})+\exp(-\sqrt{x})}2
\end{align}
Hence,
$$\sum_{n=0}^{\infty} \dfrac{(2n+1)^2}{(2n)!} x^n = \dfrac{(x+\sqrt{x}+1) \exp(\sqrt{x})+(x-\sqrt{x}+1) \exp(-\sqrt{x})}2$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/347702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Increasing and decreasing annuity Going in circles.....
Find an expression for the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of $n$, and then decrease by 1 each period up to a final payment of 1.
Okay... We know that for the first $n$ payments, we have an increasing annuity $(Ia)_n$ which equals $\frac{\ddot{a}_n-n\nu^n}{i}$ and we have a decreasing annuity $(Da)_{n-1}$ which equals $\frac{(n-1)-a_{n-1}}{i}$, but because this annuity starts decreasing at period n to get the present value we have to multiply $(Da)_{n-1}$ by $\nu^n$.
Thus, our compound annuity is equal to:
$\frac{\ddot{a}_n-n\nu^n}{i}$+$\nu^n\frac{(n-1)-a_{n-1}}{i}$ = $\frac{(1+i)a_n-n\nu^n}{i}$+$\frac{\nu^n(n-1)-\nu^na_{n-1}}{i}$ = $\frac{(1+i)a_n-\nu^n-\nu^na_{n-1}}{i}$ = $\frac{(1+i)a_n-\nu^n(1+a_{n-1})}{i}$
and now I'm stuck... Finan gives the answer as $a_n\cdot{\ddot{a}_n}$
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The present values of the payments can be written as
$$\begin{align}
PV &= a_n + a_nv + a_n v^2 + a_n v^3 +a_n v^4 + \ldots +a_n v^n\\
& = a_n(1+v+v^2+v^3+v^4+\ldots + v^n)\\
& = a_n \cdot{\ddot{a}_n}
\end{align}$$
Alternatively, you could write the present value as $$
Ia_n + Da_{n-1} v^n$$ and exploit the relationships between $a_n$ and $\ddot{a}_n$ . i.e .
$$\begin{align}
Ia_n + Da_{n-1} v^n & = \frac{\ddot{a}_n - nv^n}{i} + \frac{n-1-a_{n-1}}{i}v^n\\
& = \frac{\ddot{a}_n-nv^n+nv^n-v^n-a_{n-1}v^n}{i}\\
& = \frac{\ddot{a}-v^n-a_{n-1}v^n}{i}\\
& = \frac{1+a_{n-1}-v^n-a_{n-1}v^n}{i}\\
& = \frac{1-v^n}{i} + \frac{a_{n-1}-a_{n-1}v^n}{i}\\
& = a_n + a_{n-1}\frac{1-v^n}{i}\\
& = a_n + a_{n-1}\cdot a_n\\
& = a_n\left(1+a_{n-1}\right)\\
& = a_n\cdot \ddot{a}_n
\end{align}$$
where I have used the fact that $\ddot{a}_n = 1+ a_{n-1}$ and $a_n = \frac{1-v^n}{i}$.
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"url": "https://math.stackexchange.com/questions/348175",
"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $\alpha $ given $2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha )$ Given:
$$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$
where $$0 <\alpha < 90^\circ, $$
find $α.$
The issue I have with this question is the $-3$ on the right hand side, it really complicates things and I don't know how to deal with it.
When I usually come across equations of the form $$a\sin \theta + b\cos \theta $$ it's a relatively straight forward process of converting them into another equation of the form
$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right)$$
where $$\cos \alpha = {a \over {\sqrt {{a^2} + {b^2}} }}$$
and $$\sin \alpha = {b \over {\sqrt {{a^2} + {b^2}} }}$$
(or some other variant of this).
I cant wrap my head around this, specifically. If
$$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$
this must mean the square root expression prefixing the cosine addition identity must be $-3$, as
$$\sqrt {{a^2} + {b^2}} \cos(\theta + \alpha ) \equiv - 3\cos(\theta + \alpha ).$$
I am aware the square root operation outputs negative values I just dont know how to deal with that in this instance.
I hope this doesn't read too convoluted, thank you.
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You can use the so-called double angle formula. Notice that
$$-3\cos(\theta+\alpha) \equiv -3\cos\theta\cos\alpha + 3 \sin\theta\sin\alpha $$
Now you can compare coefficients. You want $-3\cos(\theta+\alpha) \equiv 2\sin\theta - \sqrt{5}\cos\theta$, and so you need to find an $\alpha$ for which $3\sin\alpha = 2$ and $3\cos\alpha = \sqrt{5}$. It follows that
$$\frac{3\sin\alpha}{3\cos\alpha} \equiv \tan\alpha = \frac{2}{\sqrt{5}}$$
You $\alpha$ is then $\arctan(2/\sqrt{5}) \approx 41.8^{\circ}$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/349185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Bessel functions of the first kind How would I show that
$$J_1(x)+J_3(x)=\frac 4x J_2(x)$$
Using the series definition of the Bessel Function, which is
$$J_p(x)=\sum ^\infty _{n=0} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac x2\right)^{2n+p}$$
I know that it can be achieved using some reindexing, but I am not sure how. I would greatly appreciate any help. Thank you.
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Yes, there is an index shift. Rewrite
$$J_1(x) = \frac{x}{2} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+1)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$
$$J_3(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n-1)!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$
Then
$$\begin{align}J_1(x)+J_3(x) &= \frac{x}{2} + \sum_{n=1}^{\infty} (-1)^n \left [ \frac{1}{n! (n+1)!} -\frac{1}{(n-1)! (n+2)!} \right ]\left ( \frac{x}{2} \right ) ^{2 n+1}\\ &= \frac{x}{2} + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1} \\ &= \frac{4}{x} \left[\frac{1}{2} \left ( \frac{x}{2} \right )^2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+2} \right ]\\ &= \frac{4}{x}\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+2} \\\therefore J_1(x)+J_3(x) &= \frac{4}{x} J_2(x) \end{align}$$
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{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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.