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Simple-looking bound on root of unity I am trying to prove some bound and stuck with the following:
If $|n|\leq 3N/4$, then $\left|e^{2\pi in/N}-1\right|\geq\dfrac{n}{N}$
($n,N$ are integers)
How can I prove it?
|
$\sqrt{2-2\cos \frac{2\pi n}{N}}\geq\frac{n}{N}$
Let $\alpha=n/N$. We want to prove $\sqrt{2(1-\cos (2\pi \alpha))}\geq\alpha$ for $-\frac{3}{4}\leq\alpha\leq\frac{3}{4}$ This is plainly true for $\alpha<0$, so we will only conider positive $\alpha$.
Note $\sqrt{2(1-\cos (2\pi\alpha))}\geq\alpha\iff 1\geq \frac{1}{2}\alpha^2+\cos(2\pi\alpha)$. Let's consider the derivative of $f(\alpha)=\frac{1}{2}\alpha^2+\cos(2\pi\alpha)$: $$ f'(\alpha)=\alpha-2\pi\sin(2\pi\alpha) $$
There is only one root in $0<\alpha<3/4$, and we may note $f''(\alpha)=1-4\pi^2\cos(2\pi\alpha)$, so it is reasonable to deduce that there will be a local minimum for $f(\alpha)$ in the desired range. This means we only need to check $f(0)$ and $f(3/4)$.
$$1\ge \frac{1}{2}\cdot0^2+\cos(2\pi\cdot 0)=0+1=1$$
$$ 1\ge \frac{1}{2}\left(\frac{3}{4}\right)^2+\cos\left(2\pi\frac{3}{4}\right)=\frac{9}{32}-0=\frac{9}{32} $$
So the inequality is satisfied for $|\alpha|\le \frac{3}{4}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Matrix raised to 14th power Calculate $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)^{14}$
Whould I do it one by one, and then find a pattern? I sense $6^{14}$ on the diagonal, and zeroes in the "lower triangle", but the "upper triangle" I'm not sure. Was thinking $14 \cdot 6^{13} $ but that's not correct.
|
I didn't notice that this was suggested by Harald Hanche-Olsen until just now. Consider this an expansion on his answer.
Since the identity matrix commutes with any matrix, we can use the binomial theorem with
$$
\left(6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\right)^n
$$
while noting that
$$
\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
$$
To get
$$
\begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n
=6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}
$$
|
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|
Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now $D = (a-6)^2-4a = $ perfect square. So $a^2+36-12a-4a = k^2$. where $k\in \mathbb{Z}$
$a^2-16a+36=k^2\Rightarrow (a-8)^2-28=k^2\Rightarrow (a-8)^2-k^2 = 28$
Now How can i solve after that
Help Required
Thanks
|
You are on the right track. We need
$$(a+k-8)(a-k-8) = 28$$
Now note that $(a+k-8)$ and $(a-k-8)$ are of the same parity (Why?) and the product is even. Hence, both have to be even. Hence, the possible cases are:
$1$. $(a+k-8)(a-k-8) = 2 \cdot 14$
$2$. $(a+k-8)(a-k-8) = (-2) \cdot (-14)$
$3$. $(a+k-8)(a-k-8) = 14 \cdot 2$
$4$. $(a+k-8)(a-k-8) = (-14) \cdot (-2)$
Now solve each to get possible values of $a$.
|
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|
Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar{z} = 1$.
So $\left|z^2+z+4z\bar{z}\right| = |z|\cdot \left|z+4\bar{z}+1\right| = \left|z+4\bar{z}+1\right|$
Now Let $z = x+iy$. Then $\bar{z} = x-iy$ and $|z| =1\Rightarrow x^2+y^2 = 1$
So $\left|x+iy+4x-4iy+1\right| = \left|5x+1-3iy\right| = \sqrt{(5x+1)^2+9y^2}$
So Let $f(x) = \sqrt{(5x+1)^2+9(1-x^2)} = \sqrt{25x^2+1+10x+9-9x^2}$
So $\displaystyle f(x) = \sqrt{16x^2+10x+10}=4\sqrt{x^2+\frac{5}{8}x+\frac{5}{8}}$
So $\displaystyle f(x) = 4\sqrt{\left(x+\frac{5}{16}\right)^2+\left(\frac{5}{8}-\frac{25}{256}\right)}\geq 4\sqrt{\frac{27 \times 5}{256}} = 4\times \frac{3\sqrt{15}}{16} = \frac{3\sqrt{15}}{4}$
which is occur at $\displaystyle x = -\frac{5}{16}$
(2) Now I did not understand how can i solve (II) one
Help Required
Thanks
|
For part 2:
Let $a = z^2 + 1$ and $b = z$. We have $a - 1 = b^2$.
We have
\begin{align*}
&(|z^2 - z + 1| + |z^2 + z + 1|)^2\\
=\,& |a - b|^2 + |a + b|^2 + 2|a - b|\cdot |a + b| \\
=\,& 2(|a|^2 + |b|^2) + 2|a^2 - b^2|\\
=\,& 2 |a|^2 + 2|a - 1| + 2|a^2 - a + 1|\\
\ge\,& 2|a|^2 + 2\mathrm{Re}(1 - a) + 2\mathrm{Re}(a^2 - a + 1)\\
=\,& 2|a|^2 + 2\mathrm{Re}(a^2) + 4 - 4\mathrm{Re}(a)\\
=\,& 4[\mathrm{Re}(a)]^2 + 4 - 4\mathrm{Re}(a)\\
=\,& [2\mathrm{Re}(a) - 1]^2 + 3\\
\ge\,& 3
\end{align*}
where we have used $|u| \ge \mathrm{Re}(u)$,
and $|u|^2 + \mathrm{Re}(u^2) = 2[\mathrm{Re}(u)]^2$.
Also, when $z = \frac{1}{\sqrt 2}\mathrm{i}$,
we have $|z^2 - z + 1| + |z^2 + z + 1| = \sqrt 3$.
Thus, the minimum of $|z^2 - z + 1| + |z^2 + z + 1|$ is $\sqrt 3$.
|
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|
What's the algebraic property where you can flip the fractions in an equation? Earlier in algebra, we spent over 20 minutes trying to figure out
$$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e} \,\,\,\, \text{solve for }R_2 $$
when the teacher said "What you start out with is the same as what you learned in pre-algebra
$$
\frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e}
$$
subtract $\frac{1}{R_1}$ from both sides:
$$\frac{1}{R_2} = \frac{1}{R_e} - \frac{1}{R_1}$$
and then the math gods said 'you may flip as long as all are flipped'"
$$R_2 = R_e - R_1$$
What is the name of this algebraic property?
(Sorry, I couldn't find any good tags for use here.)
|
You can't just flip all willy-nilly.
If what you said was true, then the resistance of electrical circuits in series and parallel connections would be the same! That just isn't true.
$$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}\mathbf{\neq} \frac{1}{R_1 + R_2 + R_3 + \dots + R_n}$$
Fastest way to see this is by setting $R_1 = R_ 2 = \dots = R_n = R$ (say)
Does it feel like $\frac{n}{R} $ and $\frac{1}{nR}$ are the same?
Now, you see. You must either change your math gods or listen to them more carefully.
The flipping property is called the Invertendo:
$$\frac{a}{b} = \frac{c}{d} \iff \frac{b}{a} = \frac{d}{c} $$
There are many ways you can realize it and I leave it as an exercise to you to try.
The simplest way is by raising both sides of the equation to $-1$ :
$$\frac{a}{b} = \frac{c}{d}
\implies \left(\frac{a}{b}\right)^{-1} = \left(\frac{c}{d}\right)^{-1}
\implies \frac{a^{-1}}{b^{-1}} = \frac{c^{-1}}{d^{-1}}
\implies \frac{1}{a} \cdot \frac{b}{1} = \frac{1}{c} \cdot \frac{d}{1} \implies \frac{b}{a} = \frac{d}{c}$$
The following is the best approach to your question:
$$
\frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e}\\
\implies \frac{1}{R_2} = \frac{1}{R_e} - \frac{1}{R_1}
= \overbrace{\frac{R_1 - R_e}{R_1 R_e}}^{\text{Taking LCM}} \\
\implies \underbrace{R_2 = \frac{R_1 R_e}{R_1 - R_e}}_{\text{Applying Invertendo}}$$
|
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|
Solve $x^2+(-7-4i)x+9+15i=0.$ Solve $$x^2+(-7-4i)x+9+15i=0.$$
Using the quadratic formula, I get $$\frac12 (7+4i \pm \sqrt{-4i})$$ but that's not correct. How do you solve this? I get no help from looking at wolfram alpha.
|
Given:
$$x^2+(-7-4i)x+(9+15i) = 0$$
We use the well known formula to find
$$\begin{align}
x & = \frac{-b \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} \\
& = \frac{-(-7-4i)\pm \sqrt{(-7-4i)^2 - 4 \cdot 1 \cdot (9+15i)}}{2 \cdot 1} \\
& = \frac{7+4i \pm\sqrt{(49 + 56i -16) - (36+60i)}}{2} \\
& = \frac{7+4i \pm\sqrt{-3 - 4i}}{2} \\
\end{align}$$
Now to find $\sqrt{-3 - 4i}$ we note that this can expressed as $r (\cos(\theta)+i\sin(\theta))$ where $r = \sqrt{3^2 +4^2}$ and $\theta = \tan^{-1}\left(\frac{-3}{-4}\right)$.
$\sqrt{r (\cos(\theta)+i\sin(\theta)} = \pm \sqrt{r} \cdot \left( \cos \left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right) \right)$
Alternatively if you are not using a calculator we need to find $a + bi$ such that $(a+bi)^2 = -3 -4i$
Note: $(a + bi)^2 = (a^2 - b^2) + 2 \cdot a \cdot b \cdot i$
We therefore have two simultaneous equations to solve:
$a^2 - b^2 = -3$ and $2 \cdot a \cdot b = -4 \Rightarrow a \cdot b = -2$
Ignoring the signs we can see that $|b| \gt |a|$ as the the real part is negative and $a$ and $b$ must have opposite signs as the imaginary part is negative. A little trial and error should show you that $(1 - 2i)$ and $(-1 + 2i)$ are solutions to $\sqrt{-3-4i}$
I'm sure you can take it from there.
|
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|
Equal integral but only one of them converges absolutely . Consider the following integral
$$\int_0 ^\infty \frac{\sin x}{1+x} \, dx.$$
By integration by parts we get $$\int_0^\infty \frac{\cos x}{(1+x)^2}\,dx.$$
But according to Rudin , one of them is absolutely convergent and the other isn't. How do i prove it.
$$\int_0^\infty \left| \frac{\cos x}{ (1+x)^2}\right|\,dx \le \int_0^\infty \frac{1}{(1+x)^2} \,dx < \infty $$ This one is quite clear .
Another question is what is the necessary condition on a integral so that we can do INTEGRATION BY PARTS.
I find it amusing by the fact that even though both the integrals are same but one of them converges absolutely and the other doesn't .
Thanks
|
The integral $\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx$ diverges. In fact, we have
\begin{eqnarray}
\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx&=&\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{1+x}\right|\,dx=\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin(x+k\pi)}{1+k\pi+x}\right|\,dx\\
&=&\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin x}{1+k\pi+x}\right|\,dx
=\sum_{k=0}^\infty\int_0^\pi\frac{\sin x}{1+k\pi+x}\,dx\\
&\ge& \sum_{k=0}^\infty\int_0^\pi\frac{\sin x}{1+(k+1)\pi}\,dx=\sum_{k=0}^\infty\frac{2}{1+(k+1)\pi}=\infty.
\end{eqnarray}
Hence the integral $\int_0^\infty\frac{\sin x}{1+x}\,dx$ isn't absolutely convergent.In contrast we have
$$
\int_0^\infty\left|\frac{\cos x}{(1+x)^2}\right|\,dx\le\int_0^\infty\frac{1}{(1+x)^2}\,dx=1,
$$
i.e. the integral $\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx$ is absolutely convergent.
However, the integral $\int_0^\infty\frac{\sin x}{1+x}\,dx$ does converge, because we have
$$
\int_0^\infty\frac{\sin x}{1+x}\,dx=\sum_{k=0}^\infty(-1)^k\int_0^\pi\frac{\sin x}{1+x+k\pi}\,dx=:\sum_{k=0}^\infty(-1)^ka_k,
$$
where the sequence $(a_k)$ satisfies the following:
\begin{eqnarray}
a_{k+1}-a_k&=&\int_0^\pi\sin x\left(\frac{1}{1+x+k\pi+\pi}-\frac{1}{1+x+k\pi}\right)\,dx\\
&=&-\int_0^\pi\frac{\pi\sin x}{(1+x+k\pi)(1+x+k\pi+\pi)}\,dx\le 0,
\end{eqnarray}
and
$$
0\le a_k\le \int_0^\pi\frac{\sin x}{1+k\pi}\,dx=\frac{2}{1+k\pi}\to 0.
$$
Notice that for every $\theta>0$ we have
$$
\int_0^\theta\frac{\sin x}{1+x}\,dx=-\frac{\cos x}{1+x}\Big|_0^\theta-\int_0^\theta\frac{\cos x}{(1+x)^2}\,dx=1-\frac{\cos\theta}{1+\theta}-\int_0^\theta\frac{\cos x}{(1+x)^2}\,dx.
$$
It follows that
$$
\int_0^\infty\frac{\sin x}{1+x}\,dx=1-\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx.
$$
|
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|
Prove that Q($\sqrt{2}$, $\sqrt{3}$) is a field Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\ |\ a,b,c,d \in \mathbb{Q}\}$ is a field.
I am doing the subfield test, but having trouble in showing how to express the inverse in such a form. Anyone can help?
|
Consider doing this step-wise. Since $\mathbb{Q}(\sqrt{2})$ is obviously a ring, to show it is a field, we just need to find an inverse for $a + b \sqrt{2}$. Let $a, b \in \mathbb{Q}$:
$$(a + b \sqrt{2})(a - b \sqrt{2}) = a^2 - 2b^2 \implies (a + b \sqrt{2})(\frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2}) = 1$$
Now, given an element of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$, say $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$, we rearrange it into elements of $\mathbb{Q}(\sqrt 2)$:
$$
\begin{align*}
a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} &= (a + b \sqrt{2}) + (c + d \sqrt{2})\sqrt{3} \\
&= \alpha + \beta \sqrt{3}, \qquad \alpha, \beta \in \mathbb{Q}(\sqrt{2})
\end{align*}
$$
To show it has an inverse, apply the same process as before:
$$
\begin{align*}
(\alpha + \beta \sqrt{3})(\alpha - \beta \sqrt{3}) &= \alpha^2 - 3 \beta^2 \\
(\alpha + \beta \sqrt{3})(\frac{\alpha}{\alpha^2 - 3 \beta^2} - \frac{\beta}{\alpha^2 - 3 \beta^2} \sqrt{3}) &= 1
\end{align*}
$$
Since $\mathbb{Q}(\sqrt 2)$ is a field, $\frac{\alpha}{\alpha^2 - 3 \beta^2}$ and $\frac{\beta}{\alpha^2 - 3 \beta^2}$ are elements of $\mathbb{Q}(\sqrt 2)$, and so $\frac{\alpha}{\alpha^2 - 3 \beta^2} - \frac{\beta}{\alpha^2 - 3 \beta^2} \sqrt{3}$ is in $\mathbb{Q}(\sqrt 2, \sqrt 3)$.
All you need to do now is verify that you aren't dividing by $0$ anywhere.
|
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|
If $3^n+81$ is a perfect square, then positive integer value $n$ is If $3^n+81$ is a perfect square, Then calculation of a positive integer value of $n$.
$\bf{My\; Try}::$ When $n≤4,$ then easy to know that $3^n+81$ is not a perfect square.
Now let $n=k+4(k∈Z^{+}),$ then $3^{n}+81=81(3^{k}+1).$
So $3^{n}+81$ is a perfect square, and $81$ is square,
there must be a positive integer $x$, such that $3^{k}+1=x^2⇒3^k=(x−1)⋅(x+1)$
Means $(x+1)$ and $(x-1)$ must be a power of $3$ form
Now I did not understand how can i solve after that
Help Required
Thanks.
|
$(x-1)$ and $(x+1)$ must be power of 3. Then $x-1=3^a$ and $x+1=3^b$
$\Rightarrow 3^b-3^a = 2 \Rightarrow 3^a(3^{b-a}-1)=2$.
So $a=0$ and $b-a=1 \Rightarrow b=1$. Thus $x-1 = 3^a = 1$ and $x+1 = 3^b = 3$.
$\Rightarrow 3^k=1\cdot 3=3 \Rightarrow k = 1 \Rightarrow n=k+1=5$.
Therefore, the unique solution is $n=5$ with $3^5+81=324 = 18^2$.
|
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much
|
As you stated, $2b + a = ab$. We can get this even nicer: $a = \frac{2b}{b - 1}$.
Substitute into the square root expression:
$$
\begin{align*}
a + b + \sqrt{a^2 + b^2} &= \frac{2b}{b - 1} + b + \sqrt{\left( \frac{2b}{b - 1} \right)^2 + b^2} \\
&= \frac{b^2 + b}{b - 1} + \sqrt{ \frac{b^4 - 2b^3 + 5b^2}{(b - 1)^2} } \\
&= \frac{b^2 + b}{b - 1} + \frac{b}{b - 1} \sqrt{ b^2 - 2b + 5 } \\
&= \frac{b}{b - 1} (b + 1 + \sqrt{ b^2 - 2b + 5 }) \\
\end{align*}
$$
From here I don't see any other way to find the minimum but calculus. Take the derivative and set equal to zero. But it's really messy. On the other hand, Wolfram Alpha gives a minimum of $\frac{5}{2}$ for it.
|
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|
finding minimum of function Can you please give me some hints finding minimum of this function:
$ (r-1)^2 + (\frac{s}{r} -1)^2 + (\frac{t}{s}-1)^2 + (\frac{4}{t}-1)^2$
where $ 1 \le r \le s \le t \le 4 $,
$r,s,t \in \Bbb R $
|
Another way... As Daniel Fischer has done, define $a=r, b=\frac{s}r, c=\frac{t}s, d=\frac4t$. Then we have $a, b, c, d \ge 1$ and $abcd = 4$.
Now using the QM- AM-GM inequalities, we have
$$\sqrt{\frac{(a-1)^2 + (b-1)^2 + (c-1)^2+(d-1)^2}{4}}\ge \frac{a+b+c+d - 4}4 \ge \sqrt[4]{abcd}-1 = \sqrt{2}-1$$
with equality (i.e. minimum) iff $a=b=c=d=\sqrt2$.
|
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|
Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in N$ Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in \mathbb{N}$
Attempt:
For $n = 1,$ we have $3>2$ so the base case holds.
Assume $a_k > 2^k$ for some k
For $n = k+1$ we have:
$a_{k+1} = 2 a_{k} + 5$
|
You could infact obtain $a_n$ with reasonable ease; Note that
$$a_{n+1} + 5 = 2(a_n+5)$$
Hence, we have
$$a_{n+1} + 5 = 2(a_n+5) = 2(2(a_{n-1}+5)) = 2^k (a_{n-k+1}+5) = 2^n(a_1 + 5)$$
Hence, we get
$$a_{n+1} +5 = 2^n(3+5) = 2^{n+3} \implies a_n = 2^{n+2} - 5$$
Now show that for $n \geq 1$, we have
$$2^{n+2} - 5 > 2^n$$
|
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How would I show this problem through mathematical induction? I am trying to learn a lot of this on my own but I have never tried proving something through mathematical induction. Here is the problem below.
$$1+3+3^2 + \cdots + 3^n = \frac{3^{(n+1)}-1}{2}$$
for all $n\in\mathbb{N}_0$, using mathematical induction.
Note that here $\mathbb{N}_0$ means all integers $n \geq 0$.
I need to start with the basis step which would be:
If $n = 0$ then $3^0 = 1$, and $1 = 3^{0+1} = 3 - 1 = 2$, and then $2/2 = 1$. So I have proved the basis step but how do I do the inductive step for this?
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For the inductive step, assume your claim holds for $n$ and then try to prove it for $n+1.$ That is,
Assume: $1+3+3^2+...+3^n=\frac{3^{n+1}-1}{2}$.
Then see if you can prove that $1+3+3^2+...+3^n+3^{n+1}=\frac{3^{(n+1)+1}-1}{2}$.
How to do this? Start by transforming the left hand side using what you already know:
$1+3+3^2+...+3^n+3^{n+1}=\frac{3^{n+1}-1}{2}+3^{n+1}.$
Now combine these fractions and see if you can massage them into $\frac{3^{(n+1)+1}-1}{2}$.
That is: $\frac{3^{n+1}-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2(3^{n+1})}{2}=\frac{3(3^{n+1})-1}{2}=\frac{3^1(3^{n+1})-1}{2}=\frac{3^{(n+1)+1}-1}{2}.$
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Stewart's Calculus (6 edn 2007) : p. 1097. §16.8, Exercise #4. Source: Stewart. Calculus: Early Transcendentals (6 edn 2007). p. 1097. §16.8, Exercise #4.
Use Stokes' Theorem to evaluate $\int \int_S curl \ \vec F \cdot d \vec S $
$4. \; F(x,y,z) = x^2 \ y^3\ z \ \vec i + \sin(x\ y\ z)\ \vec j + x\ y\ z\ \vec k$,
$S$ is the part of the cone $y^2 =x^2 + z^2$ that lies between the planes $y = 0$ and $y = 3$, oriented in the direction of the positive $y$-axis.
I calculated $ curl \ \vec F $ :
$$ curl \ \vec F = ( xz-xy \cos(xyz) \ , \ -yz+x^2y^3 \ , \ yz \cos(xyz)-3x^2y^2z ) $$
Now I have to evaluate the domain $S$, and I have no idea how. I have a lot of difficulties to find a domain, and when I should use $\int \int_S curl \ \vec F \cdot d \vec S $ or $\int_c \vec F \cdot d \vec r $, because I always have difficulties on how to evaluate the domain of integration.
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As Fantini points out, the boundary is a circle - we have a winner - of radius $3$.
Then, we fix $y=3$ and parameterise $(x,z) = (\sqrt{9-z^2},z)$ and then integrate from $z=3$ to $z=-3$ (going counterclockwise). One must also note that by the Implicit Function Theorem, there no unique function $x(z)$ in an open neighbourhood of $(x,y,z) = (0,3,3)$ or $(x,y,z)=(0,3,-3)$ and so we must also integrate from $z=-3$ to $z=3$ with the parameterisation $(x,z) = (-\sqrt{9-z^2},z)$. Now $dx = z/\sqrt{9-z^2}\; dz$ in the first case (opposite sign in the second).
Note that the $y$-component vanishes and so we are left with $\oint_{P} (27x^2z,3xz)\cdot (dx, dz) = - \int_{-3}^{3} (27z^2\sqrt{9-z^2}+3z\sqrt{9-z^2}) dz + \int_{-3}^{3} (-27z^2\sqrt{9-z^2}-3z\sqrt{9-z^2}) dz = -54 \int_{-3}^{3} z^2\sqrt{9-z^2} dz.$
So the problem reduces to (letting $u = \sqrt{9 - z^2}$):
$-54 \int_{-3}^{3} z^2\sqrt{9-z^2} dz = -108 \int_{0}^{3} u(9-u^2)\cdot \frac{-u}{\sqrt{9-u^2}} du = -108\cdot \frac{-81\pi}{16} = \frac{3^7 \pi}{4}.$
|
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Induction Proof: Sum of Products of $3$ Consecutive Numbers I'm still trying to learn induction. But stuck on the following question. It will be highly appreciated if someone can show me how to do it.
I need to prove the following using math induction.
$$1\cdot2\cdot3 + 2\cdot3\cdot4 + 3\cdot4\cdot5 + \cdots + n(n+1)(n+2)= \frac{n(n+1)(n+2)(n+3)}{4}$$
for $n = 1, 2, 3, \ldots$
|
The formula is true for the case $n=1$, since $1\times 2\times 3 = 6 = \frac{1\times 2 \times 3 \times 4}{4}$.
Now, suppose that the formula is true for $n$. We have that
$1\times 2 \times 3 + \cdots + n(n+1)(n+2) + (n+1)(n+2)(n+3) \\= \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) \\
= \frac{n(n+1)(n+2)(n+3)}{4} + \frac{4(n+1)(n+2)(n+3)}{4}\\
= \frac{(n+1)(n+2)(n+3)(n+4)}{4} $
So, since the $n+1$ case is true, the induction holds.
|
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finding the value of $f(\frac{1}{7})$ $f$ is a function mapping positive reals between $0$ and $1$ to reals. Let $f$ be given by, $f( \frac{x+y}{2} ) = (1-a)f(x)+af(y)$ where $y > x$ and $a$ being a constant. Also,$f(0) = 0$ and $f(1) = 1$. Find $f( \frac{1}{7} )$.
I have tried some things but it is not working out. How to find the $f( \frac{1}{7} )$ .
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Here are the calculations:
$$f(\frac{1}{7})=f(\frac{0+\frac{2}{7}}{2})=(1-a)f(0)+af(\frac{2}{7})=af(\frac{2}{7})$$
$$f(\frac{2}{7})=f(\frac{0+\frac{4}{7}}{2})=(1-a)f(0)+af(\frac{4}{7})=af(\frac{4}{7})$$
$$f(\frac{4}{7})=f(\frac{\frac{1}{7}+1}{2})=(1-a)f(\frac{1}{7})+af(1)$$
So we have
$$f(\frac{1}{7})=a^2f(\frac{4}{7})=a^2[(1-a)f(\frac{1}{7})+af(1)]$$
then
$$f(\frac{1}{7})=\frac{a^3f(1)}{1-a^2(1-a)}=\frac{a^3}{1-a^2(1-a)}$$
where $a$ satisfies $a^2(1-a)\ne1$.
|
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Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers. Let $m$ and $n$ be two integers. Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers.
I think I have to use the contrapositive to solve this. So I assume $\neg P\implies Q$ and I have to derive $\neg Q$? I know its easier to show odd numbers rather than even... but im still unsure how to do this practice problem.
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I choose to use a and b instead of m and n because I already answered this somewhere else using a and b. Let $a=2\cdot x+1$, $b=2\cdot y+1$, where x and y are integers. This way a and b are odd numbers by definition. For example if x is 0, $a=2\cdot 0+1=1$. If x is 1, $a=2\cdot 1+1=3$. Repeating the pattern x and y can be any integer and a and b will always be an odd number by definition.
Now lets substitute a and b in $a^2+b^2$.
$a^2+b^2=\left(2\cdot x+1\right)^2+\left(2\cdot \:y+1\right)^2$
Expand the right side
$a^2+b^2=4x^2+4\cdot \:x+1+4\cdot \:y^2+4\cdot \:y+1$
Factor out a four
$a^2+b^2=4\left(x^2+x+y^2+y\right)+2$
Now Lets Say we have a number $n=4\cdot k+2$ where k is any integer. If $k=0, n=4\cdot 0+2=2$.If $k=1, n=4\cdot 1+2=6$. If $k=2, n=4\cdot 2+2=10$. And so on. The point is the values of n are even number not divisible by four. You can test that out...$n=\left\{...-18-14,-10,-6,-2,2,6,10,14,18...\right\}$
So let say $x^2+x+y^2+y=k$ and since x and y are integers k must be an integer as well. Due to closure of integers under multiplication and addition.
So going back to $a^2+b^2=4\left(x^2+x+y^2+y\right)+2$ can also be written as $a^2+b^2=4\left(k\right)+2$. And we know that is the set of even numbers not divisible by four as we showed above.
So $a^2+b^2\in \left\{...-18-14,-10,-6,-2,2,6,10,14,18...\right\}$.
|
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Finding a Correlation between Bernoulli Variables? Let X and Y be Bernoulli random variables. We don't assume independence or identical distribution, but we do assume that all 4 of the following probabilities are nonzero.
Let a := P[X = 1, Y = 1], b := P[X = 1, Y = 0], c := P[X = 0, Y = 1],
and d := P[X = 0, Y = 0].
How do I obtain a formula for a correlation between random variables X and Y?
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Stefan Hansen's hint is a good one. Here is the complete derivation:
$${\rm E}[X]=a+b=p$$
$${\rm E}[Y]=a+c=q$$
\begin{align}
\mathrm{Var}(X) & ={\rm E}[(X - {\rm E}[X])^2] \\
& = {\rm E}[(X - p)^2] \\
&= p(1-p)^2 + (1-p)(-p)^2 \\
& = p (1-2p+p^2) + p^2 - p^3 \\
& = p - 2p^2 + p^3+p^2-p^3 \\
& = p - p^2 \\
& = p(1-p)
\end{align}
$$\sigma{_X} =\sqrt{\mathrm{Var}(X)} = \sqrt{p(1-p)} = \sqrt{(a+b)(1-(a+b))}$$
$$\sigma{_Y} =\sqrt{\mathrm{Var}(Y)} = \sqrt{q(1-q)} = \sqrt{(a+c)(1-(a+c))}$$
\begin{align}
\mathrm{Cov}(X, Y) &= \rm{E}[XY] - \rm{E}[X]\rm{E}[Y] \\
&=a - pq \\
&=a - (a+b)(a+c) \\
\end{align}
Finally, by substitution into the equation for $\rho_{XY}$:
\begin{align}
\rho_{XY}&=\frac{\mathrm{Cov}(X,Y)}{\sigma_{X}\sigma_{Y}} \\
&=\frac{a - (a+b)(a+c)}{\sqrt{(a+b)(1-(a+b))}\sqrt{(a+c)(1-(a+c))}} \\
&=\frac{a - (a+b)(a+c)}{\sqrt{(a+b)(1-(a+b))(a+c)(1-(a+c))}}
\end{align}
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Proving that a number is composite I have proved that the number $10^{5}2^{17}+1$ is composite by showing that it is divisible by 3 , using remainders. I want an alternative proof.I am looking for a very elementary proof that does not mention remainders.
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$10^5 \pmod 3 = (1)^5 \pmod 3 = 1 \pmod 3$.
$2^{17} \pmod 3 = (-1)^{17} \pmod 3 = -1 \pmod 3$.
Thus, $10^5 \times 2^{17} \pmod 3 = 1 \times (-1) \pmod 3 = -1 \pmod 3$.
Now, $10^5 \times 2^{17} + 1 \pmod 3 = -1 + 1 \pmod 3 = 0 \pmod 3$.
|
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$\inf$ and $\sup$ of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ Let $n\geq3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_{1},...,a_{n})$ of positive numbers. Find the supremum and the infimum of the set of numbers $$\sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}},$$ where we put $a_{n+1}=a_{1}$ and $a_{n+2}=a_{2}$.
Attempted solution:
$$n=3\rightarrow \sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}=1$$
$$n>3\wedge a_{i}>0\rightarrow \frac{a_{k}}{\sum_{i=1}^{n}a_{i}}< \frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}< \frac{\sum_{i=1}^{k}a_{i}+\sum_{i=k+3}^{n}a_{i}}{\sum_{i=1}^{n}a_{i}}$$
$$\rightarrow \sum_{k=1}^{n} \frac{a_{k}}{\sum_{i=1}^{n}a_{i}}< \sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}<\sum_{k=1}^{n} \frac{\sum_{i=1}^{n}a_{i}-a_{k+1}-a_{k+2}}{\sum_{i=1}^{n}a_{i}}$$
$$\rightarrow 1< \sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}< n-2$$
I don't see why these must be the tightest bounds. Can $\inf$ and $\sup$ be calculated from these inequalities?
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Your bounds are sharp (to my surprise).
Example: $a_k = x^k$. Then
\begin{align*}
\sum_{k=1}^n \frac{a_k}{a_k + a_{k+1} + a_{k+2}}
&= \sum_{k=1}^{n-2} \frac{x^k}{x^k + x^{k+1} + x^{k+2}}
+ \frac{x^{n-1}}{x^{n-1} + x^n + x}
+ \frac{x^n}{x^n + x + x^2} \\
&= \frac{n-2}{1 + x + x^2} + \frac{1}{1 + x + x^{-n+1}} + \frac{1}{1 + x^{-n+1} + x^{-n+2}}
\end{align*}
This tends to $1$ as $x\to\infty$ and to $n-2$ as $x\to 0^+$.
The only essential feature of the functions $x^k$ here is that each grows faster than the previous one (as $x\to\infty$) and shrinks faster than the previous one (as $x\to 0^+$).
|
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Find $\lim_{x\to 0} \left(\frac{x}{\arcsin x}\right)^{\frac{1}{x^2}}$. Find $\lim_{x\to 0} \left(\dfrac{x}{\arcsin x}\right)^{\frac{1}{x^2}}$.
$\lim_{x\to 0}f(x)=e^a$ with $a=\lim_{X\to 0}\dfrac{1}{x^2}\left(\dfrac{x}{\arcsin x}-1\right)=\lim_{X\to 0}\dfrac{x-\arcsin x}{x^3}$
Continue i used formular L'hospotal .
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My hint:
Put $$L=\lim_{x\to 0}\left(\frac{x}{\arcsin x}\right)^{\frac{1}{x^2}}=\exp\left\{\lim_{x\to 0}\frac{1}{x^2}\ln\frac{x}{\arcsin x}\right\}=\exp\left\{-\lim_{x\to 0}\frac{1}{x^2}\ln\frac{\arcsin x}{x}\right\}$$
We use Taylor' expansion, we get:
$$\arcsin x=x+\frac{x^3}{6}+o(x^3)\to \frac{\arcsin x}{x}=1+\frac{x^2}{6}+o(x^2)$$
Hence:
$$L=\exp\left\{-\lim_{x\to 0}\frac{1}{x^2}\ln\left(1+\frac{x^2}{6}+o(x^2)\right)\right\}=\exp\left\{-\frac{1}{6}\right\}=\frac{1}{\sqrt[6]{e}}$$
|
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Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}$
Now $\displaystyle f^{'}(x) = 1+x+x^2+x^3+x^4+x^5+x^6$
and $\displaystyle f^{''}(x) = 1+2x+3x^2+4x^3+5x^4+6x^5$
$\displaystyle f^{'''}(x) = 0+2+6x+12x^2+20x^3+30x^4 = 2\left(1+3x+6x^2+10x^3+15x^4\right)$
Now I did not understand how can i solve it
Help Required
Thanks
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Law of signs is covered in pre-calc/algebra. Using the standard law of signs, one has to show that there is exactly one real root between -1 and -2. Now I am rusty on how this is done (I am 60+ and I learnt it 40 years ago!) but goes something like this.
$$
\begin{align}
f(x) &= x^7/7 + x^6/x + x^5/5 + x^4/4 + x^3/3 + x^2/2 + x^1/1 + 1 && \hbox{No sign changes}\\
f(-x) &= -x^7/7 + x^6/x - x^5/5 + x^4/4 - x^3/3 + x^2/2 - x^1/1 + 1 && \hbox{7 sign changes}\\
\end{align}
$$
So we know that all the seven roots (if they exist) is to the left of $0$.
Similarly $f(x-1)$ has no sign changes but $f(x-2)$ has 7 sign changes. So all the roots have to be between $-2$ and $-1$.
Most of the calculations are done using synthetic division and all this is covered in the precalc/algebra course. Hope someone can expand on this.
I will also dust off my precalc books and check if no one answers this
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Mclaurin on $\arccos(\frac{n^2-1}{n^2+1})$ I have expanded $\lim_{n\to \infty} \arccos(\frac{n^2-1}{n^2+1})$ to $\arccos(1-\frac{2}{n^2})$ and now i dont know what to do. I wrote the function on walfram alpha and he told me that the result is $\frac{2}{n}+\frac{1}{3n^3}+O((\frac{1}{n})^6)$ you can explain to me how that result came out from where?
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Actually, for $x\in [0,+\infty[$,
$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$
To see why, write $x=\tan \theta$ for $\theta \in[0,\pi/2[$, then $\theta=\arctan x$ and
$$\frac{1-x^2}{1+x^2}=\frac{1-\tan^2 \theta}{1+\tan^2 \theta}=\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta +\sin^2\theta}=\cos 2\theta$$
And since $2\theta \in [0,\pi[$, $\arccos(\cos2\theta)=2\theta$
Now, for $|x|<1$
$$\arctan x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$$
And (don't forget to rename the sum variable)
$$\arccos \frac{n^2-1}{n^2+1}=\arccos \frac{1-1/n^2}{1+1/n^2}=2\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)n^{2k+1}}$$
By the way, since the LHS in the first equation is even, and the RHS is odd, you get, for any $x \in \Bbb R$,
$$\arccos \frac{1-x^2}{1+x^2}=2|\arctan x|$$
One way to "guess" the result in the first place, is to remember the half-angle formula: if $x=\tan (\theta/2)$, then $\cos \theta = \frac{1-x^2}{1+x^2}$.
Edit, for malloc
Here is a slightly different proof.
In the above, I apply the development of $\arctan$ on $]-1,1[$, and I let $x=1/n$.
But, still using
$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$
And with $x=n$, we can also use the development of $\arctan$ at $\infty$. For this, you will need, for $x>0$:
$$\arctan x + \arctan \frac 1 x = \frac {\pi}2$$
Thus if $n=x >1$ (so that the development of $\arctan \frac 1x$ holds),
$$\arctan n = \frac {\pi}2-\arctan \frac 1 n = \frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$
And finally (since $\arccos (-x)=\pi-\arccos(x)$),
$$\arccos \frac{n^2-1}{n^2+1}=\pi-\arccos \frac{1-n^2}{1+n^2}$$
$$=\pi-2\arctan n=\pi-2\left(\frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}\right)$$
$$\arccos \frac{n^2-1}{n^2+1}=2\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$
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Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
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For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\frac{1}{2}=\frac{n(n+1)}{2n(n+1)}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\frac{n(n+1)}{2n^{2}}=\frac{1}{2}\left(1+\frac{1}{n}\right),$$ and so it follows from the squeeze theorem that the limit is $\frac{1}{2}$.
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How to get the simplest form of this radical expression: $3\sqrt[3]{2a} - 6\sqrt[3]{2a}$. How to get the simplest form of this radical expression:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Here is my work:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Since the radicands are the same, we just add the coefficients.
$$-3\sqrt[3]{2a} \sqrt[3]{2a}$$
Since everything is under the same index it becomes:
$$-3\sqrt[3]{2} \sqrt[3]{a}$$
Did I do this correctly, if not can anyone tell me what I should do?
Thanks :-).
|
$$3\sqrt[3]{2a}-6\sqrt[3]{2a}=\sqrt[3]{2a}(3-6)=\sqrt[3]{2a}(-3)= -3\sqrt[3]{2a}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a,b$ are roots of $x^2-3cx-8d = 0$ and $c,d$ are roots of $x^2-3ax-8b = 0$. Then $a+b+c+d =$ (1) If $a,b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation
$x^2-10ax-11b=0$. Then the value of $\displaystyle \sqrt{\frac{a+b+c+d}{10}}=,$ where $a,b,c,d$ are distinct real numbers.
(2) If $a,b,c,d$ are distinct real no. such that $a,b$ are the roots of the equation $x^2-3cx-8d = 0$
and $c,d$ are the roots of the equation $x^2-3ax-8b = 0$. Then $a+b+c+d = $
$\bf{My\; Try}::$(1) Using vieta formula
$a+b=10c......................(1)$ and $ab=-11d......................(2)$
$c+d=10a......................(3)$ and $cd=-11b......................(4)$
Now $a+b+c+d=10(a+c)..........................................(5)$
and $abcd=121bd\Rightarrow bd(ab-121)=0\Rightarrow bd=0$ or $ab=121$
Now I did not understand how can i calculate $a$ and $c$
Help Required
Thanks
|
Thanks mathlove.
My Solution for (2) one::
Given $a,b$ are the roots of $x^2-3cx-8d=0.$ So $a+b=3c.........(1)$ and $ab=-8d........(2)$
and $c,d$ are the roots of $x^2-3ax-8b=0.$ So $c+d=3a..........(3)$ and $cd=-8b..........(4)$
So $a+b+c+d = 3(a+c)......................(5)$
Now $\displaystyle \frac{a+b}{c+d} = \frac{3c}{3a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow a^2-8d=c^2-8b$
So $(a^2-c^2)=-8(b-d)\Rightarrow (a+c)\cdot(a-c)=-8(b-d).................(6)$
Now $(1)-(3)$, we get $a+b-c-d=3c-3a\Rightarrow (b-d) = 4(c-a)=-4(a-c)$
Now put $(b-d) = -4(a-c)$ in eqn... $(6)$, we get $(a+c)\cdot(a-c)=32(a-c)$
So $(a-c)\cdot(a+c-32) = 0$, Now $a\neq c$, bcz $a,b,c,d$ are distinct real no.
So $a+c=32$, Put into eqn...$(5)$
We get $a+b+c+d = 3(a+c) = 3\cdot 32 = 96\Rightarrow (a+b+c+d) = 96$
|
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|
How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$ Let $a,b,c,d,e$ are postive real numbers,show that
$$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$
My try: since
$$\Longleftrightarrow\sum_{sym}\left(\dfrac{a-b}{a+2b+c}+\dfrac{1}{2}\right)\ge\dfrac{5}{2}$$
$$\Longleftrightarrow \sum_{sym}\left(\dfrac{3a+c}{a+2b+c}\right)\ge 5$$
use Cauchy-Schwarz inequality,we have
$$\sum_{sym}\dfrac{3a+c}{a+2b+c}\sum_{sym}((3a+c)(a+2b+c))\ge\left(\sum_{sym}(3a+c)\right)^2$$
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$
becasue this is not hold
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$for $a,b,c,d,e>0$
and this method is from this simaler inequality :
How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$
then I can't,Thank you
|
You need to try a different fraction to add, so that you get better terms on applying Cauchy-Schwarz. For e.g. here we have the equivalent inequality:
$$\sum_{cyc} \left(\frac{a-b}{a+2b+c}+\frac15\right)\ge 1$$
$$\iff \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge 5$$
By Cauchy-Schwarz, we have:
$$ \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge \frac{\left(\sum_{cyc} (6a-3b+c)\right)^2}{\sum_{cyc}\left((a+2b+c)(6a-3b+c)\right)} = \frac{16(a+b+c+d+e)^2}{\sum_{cyc}(a^2+8ab+7ac)}$$
So it is sufficient if we can show that:
$$16(a+b+c+d+e)^2 \ge 5 \sum_{cyc}(a^2+8ab+7ac) $$
But we can express the above $LHS - RHS$ as:
$$4\sum_{cyc}(a-b)^2 + \frac32 \sum_{cyc}(a-c)^2 \ge 0$$
Added based on a comment in linked post - this application of Cauchy Schwarz requires the numerator of the fraction $6a-3b+c$ to be non-negative, and hence covers only the cases with a condition like $a,b,c,d,e \in [\frac1k,k]$ where $k^2=\frac73$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it is a correct identity, although I can't prove it.
I've tried to use the formula
$$ \cos(z) = \frac{e^ {iz} - e^ {iz}}{2} $$ but without any satisfying result.
This exercise is from chapter on series.
EDIT: I corrected a mistake in the formula I wanted to use.
|
If $z^{5}-1=0$, then $$(z^{4}+z^{3}+z^{2}+z+1)(z-1)=0,$$ because $z^{5}-1=(z^{4}+z^{3}+z^{2}+z+1)(z-1)$. This implies that $$
z^{4}+z^{3}+z^{2}+z+1=0$$ or $z-1=0$. Since
\begin{equation*}
z=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}=e^{i2\pi /5}\ne 1
\end{equation*}
is a root of $z^5-1=0 $ and
\begin{equation*}
z^{k}=\cos \frac{2k\pi }{5}+i\sin \frac{2k\pi }{5}=e^{i2k\pi /5},
\end{equation*}
for $k\in \left\{ 1,2,3,4\right\} $, we have
\begin{eqnarray*}
0 &=&z^{4}+z^{3}+z^{2}+z+1=\left( \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}%
\right) +\left( \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\right) +\cdots +1
\\
&=&\left( \cos \frac{8\pi }{5}+\cos \frac{6\pi }{5}+\cdots +1\right)
+i\left( \sin \frac{8\pi }{5}+\cos \frac{6\pi }{5}+\cdots +\sin \frac{2\pi }{%
5}\right) .
\end{eqnarray*}
So
\begin{equation*}
\cos \frac{8\pi }{5}+\cos \frac{6\pi }{5}+\cdots +1=0.
\end{equation*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum with Exponent I've got the following sum which I'm trying to figure out:
$$\sum_{x=0}^n 2^{-x}$$
Wolframalpha tells me that it's equal to $2 - 2^{-n}$ but I am interested in figuring out why, and how to get that result by hand.
Any help is greatly appreciated. Thanks.
Also, as you can tell, I could really use some help with MathJax...
|
This is a geometric series with first term $1$ and common ratio $1/2$.
There is a well-known formula for calculating such sums.
Your sum, $S$ is given by:
$$S = \frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2} + \cdots + \frac{1}{2^n}$$
This is a geometric series because to get from one term to the next we multiply the the common ratio $\frac{1}{2}$. The usual trick for geometric series is to multiply $S$ by the common ratio:
$$\frac{1}{2}S = \frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3} + \cdots + \frac{1}{2^{n+1}}$$
Taking the difference between $S$ and $\frac{1}{2}S$ gives us:
$$S-\frac{1}{2}S = \frac{1}{2^0}-\frac{1}{2^{n+1}}$$
Obviously $S-\frac{1}{2}S=\frac{1}{2}S$ and $\frac{1}{2^0}=\frac{1}{1}=1$. The last line the becomes
$$\frac{1}{2}S=1-\frac{1}{2^{n+1}}$$
Finally, we multiply both sides by $2$ to get $S$:
$$S=2-\frac{2}{2^{n+1}} = 2-\frac{1}{2^n} = 2-2^{-n} $$
|
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|
Integration of $\int \frac{dx}{a+f^2(x)}$ I want to solve a integral of the form:
$$
\int \frac{dx}{a+f^2(x)}
$$
in my particular case I got
$$
\int \frac{dx}{5+\cos^2(x)}
$$
in my case I followed this process:
$$
\int \frac{dx}{5+\cos^2(x)} \\
let \ t = tg(\frac{x}{2}) =>
dx = \frac{2dt}{1+t^2}\\
\int \frac{dx}{5+\cos^2(x)} = \int \frac{\frac{2\,dt}{1+t^2}}{5+(\frac{1-t^2}{1+t^2})^2}\\
$$
Expanding the denominator
$$
5+(\frac{1-t^2}{1+t^2})^2 = \frac{5(1+t^2)^2+(1-t^2)^2}{(1+t^2)^2}
$$
Rewriting all:
$$
\int \frac{2\,dt}{1+t^2} \frac{(1+t^2)^2}{5(1+t^2)^2+(1-t^2)^2} = \\
2 \int \frac{(1+t^2)}{5(1+t^2)^2+(1-t^2)^2}dt
$$
Now I am stuck, I can try to expand the denominator again, but then I cannot divide it for the numerator, something like this:
$$
2 \int \frac{1+t^2}{5(1+2t^2+t^4)+(1-2t^2+t^4)} \,dt = \\
2 \int \frac{1+t^2}{5+10t^2+5t^4+1-2t^2+t^4} \,dt = \\
2 \int \frac{1+t^2}{6t^4+8t^2+6} \,dt = \\
\int \frac{1+t^2}{3t^4+4t^2+3}\, dt
$$
But now I really don't know how to move next...
|
I you want a solution of $\displaystyle \int\frac{1}{5+\cos^2 x}dx$
$\bf{Solution::}$ Divide both $\bf{N_{r}}$ abd $\bf{D_{r}}$ by $\cos^2 x$, we get $\displaystyle \int\frac{\sec^2 x}{5\sec^2 x+1}dx = \int\frac{\sec^2 x}{6+5\tan^2 x}dx$
Now Let $\sqrt{5}\tan x= t$, Then $\displaystyle \sec^2 xdx = \frac{1}{\sqrt{5}}dt$
So $\displaystyle \int\frac{1}{\left(\sqrt{6}\right)^2+t^2}dt = \frac{1}{\sqrt{30}}\tan^{-1}\left(\frac{t}{\sqrt{6}}\right)+\mathbb{C}$
So $\displaystyle \int\frac{1}{5+\cos^2 x}dx = \frac{1}{\sqrt{30}}\tan^{-1}\left(\frac{\sqrt{5}\tan x}{\sqrt{6}}\right)+\mathbb{C}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$ $$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$
where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
my try:since
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{3}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}$$
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{n+1}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}$$
so
$$2I_n=\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)+\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)\left(\dfrac{1}{3}+\dfrac{1}{n+1}\right)+\cdots+\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)$$
Maybe this try is not usefull, so I think use other methods to solve it .
Thank you very much!
|
Continuing using both Greg Martin and Jack D'Aurizio's answers, I start with:
$$
\begin{align}
\sum_{k=1}^{m+1}\frac{(H_{k-1})^2}{k} &= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{(H_k+H_{k-1} )H_k}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ \left( (H_{k-1}+\frac{1}{k}) + H_{k-1} \right)(H_{k-1}+\frac{1}{k})}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ \left( 2H_{k-1}+\frac{1}{k}\right)(H_{k-1}+\frac{1}{k})}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ 2(H_{k-1})^2+\frac{3}{k}H_{k-1}+\frac{1}{k^2}}{k} } \\
3\sum_{k=1}^{m+1}\frac{(H_{k-1})^2}{k} &= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ \frac{3}{k}H_{k-1}+\frac{1}{k^2}}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \left( \frac{3}{k^2}H_{k-1}+\frac{1}{k^3} \right) } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \left( \frac{3}{k^2}H_{k-1} \right) } + H_m^{(3)}
\end{align}
$$
|
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|
How to construct change of basis matrix How do I construct a change of basis matrix?
For example in $\mathbb R^3$, how to construct matrix changing basis from $A$ to $B$?
$A=\begin{pmatrix} 1 \\ 0 \\5 \end{pmatrix}\begin{pmatrix} 4 \\ 5 \\5 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\4 \end{pmatrix}$
$B=\begin{pmatrix} 1 \\ 3 \\2 \end{pmatrix}\begin{pmatrix} -2 \\ -1 \\1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\3 \end{pmatrix}$
|
Let $M_{u,v}$ be the base change matrix from base $u$ to base $v$. Then the following identity holds for bases $u$,$v$, and $w$: $$ M_{u,v}=M_{w,v}M_{u,w}$$ In this case $A$ can be the role of $u$ and $B$ is $v$. Use the standard basis $e$ as the intermediate basis $w$. Then the expression is: $$M_{A,B}=M_{e,B}M_{A,e} = M_{B,e}^{-1}M_{A,e}$$ The basis change matrix from any basis to the standard basis is easy, just write the vectors as the columns of the matrix. So here have to calculate $$\left( \begin{array}{ccc}
1 & -2 & 1 \\
3 & -1 & 2 \\
2 & 1 & 3 \end{array} \right)^{-1}\left( \begin{array}{ccc}
1 & 4 & 1 \\
0 & 5 & 1 \\
5 & 5 & 4 \end{array} \right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the series identity
Prove an identity:
$$\sum_{n=2}^{ \infty } \frac{2}{(n^3-n)3^n}=- \frac{1}{2}+ \frac{4}{3} \cdot \sum_{n=1}^{ \infty } \frac{1}{n \cdot 3^n}$$
I've checked that the left-hand-side of this identity is convergent absolutely, hence I can write it as:
$$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+1)} \cdot \frac{1}{n3^n}$$
I've also calculated the sum $$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+1)}= \frac{3}{2}
$$
But now, I don't see what can I do with the right-hand-side and what to do with $$\sum_{n=2}^{ \infty }\frac{1}{n3^n}$$
|
Using Partial Fraction Decomposition,
$$\frac1{n^3-n}=\frac An+\frac B{n-1}+\frac C{n+1}$$
multiply either sides $n^3-n=n(n-1)(n+1)$ and compare the coefficients of the different powers of $n$ to find $A,B,C$
Alternatively,
$$\frac2{n(n-1)(n+1)}=\frac{n+1-(n-1)}{n(n-1)(n+1)}=\frac1{n(n-1)}-\frac1{(n+1)n}$$
Again, $\displaystyle \frac1{n(n-1)}=\frac{n-(n-1)}{n(n-1)}=\frac1{n-1}-\frac1n$
and $\displaystyle \frac1{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac1n-\frac1{n+1}$
So, the $n$th term $\displaystyle T_n=\left(\frac1{n-1}-\frac2n+\frac1{n+1}\right)\frac1{3^n}=\frac13\cdot\frac1{(n-1)3^{n-1}}-\frac2{n3^n}+\frac3{(n+1)3^{n+1}}$
$$\implies\sum_{n=2}^\infty T_n=\sum_{n=2}^\infty\left(\frac13\cdot\frac1{(n-1)3^{n-1}}-\frac2{n3^n}+\frac3{(n+1)3^{n+1}}\right)$$
$$=\frac13\sum_{n=2}^\infty\frac1{(n-1)3^{n-1}}-2\sum_{n=2}^\infty\frac1{n3^n}+3\sum_{n=2}^\infty\frac1{(n+1)3^{n+1}}$$
Adjust each summation so that $n$ ranges between $[1,\infty)$
|
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|
Joint To Marginal Density : Can't figure it out. Here goes the problem:
Problem:
Suppose $X$ and $Y$ have the joint density function:
$f(X,Y) = c \sqrt{1 - x^2 - y^2}, \,\,\,\,\, x^2 + y^2 \leq 1$
*
*Find $c$.
*Find the marginal densities of $X$ and $Y$
My Approach For Solution:
Part 1:
To obtain $c$, we need to solve:
$\int\int_A f(x,y) dy dx = \int \int_A c \sqrt{1 - x^2 - y^2} = 1$
Now, let:
$x = r\,cos(\theta)\, ,\, \textrm{and let}\, y = r\,sin(\theta)$, then we have:
$dy dx = r dr d\theta$
So the original integral becomes:
$c\int_0^{2\pi}\int_0^1 \sqrt{1 - r^2}\,r dr d\theta$
Substituting $1 - r^2 = u\, \textrm{and}\, -2rdr = du$, we get (note the change of limits of the inner integral):
$-\frac{c}{2}\int_0^{2\pi}\int_1^0 \sqrt{u} du d\theta$
which is now solved:
$$-\frac{c}{2}\int_0^{2\pi}\int_1^0 \sqrt{u} du d\theta = -\frac{c}{2}\int_0^{2\pi}[\frac{2}{3}u^{3/2}]_1^0 d\theta = \frac{c}{3}\int_0^{2\pi}[u^{3/2}]_0^1 d\theta = \frac{c}{3}\int_0^{2\pi} d\theta = \frac{c}{3}\cdot 2\pi$$
Equating this result to 1 gives us $\color{red}{c = \frac{3}{2\pi}}$
Part 2:
Part 2 is where I am stuck. Now, we have:
$$f_X(x) = \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} f(x,y) dy = \frac{3}{2\pi} \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}}\sqrt{1 - x^2 - y^2} \color{red}{dy}$$
A similar expression will come for $f_Y(y)$, i.e.
$$f_Y(y) = \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} f(x,y) dy = \frac{3}{2\pi} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}}\sqrt{1 - x^2 - y^2} \color{red}{dx}$$
My question is: $\color{red}{\textrm{Now, how do I evaluate these integrals in Part 2}}$?
|
defint $y=\sqrt{1-x^2}\sin(t)$, then $dy=\sqrt{1-x^2}\cos(t)dt$.
$$
f_X(x)=\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}c\sqrt{1-x^2-y^2}dy = c(1-x^2)\int_{-\pi/2}^{\pi/2}\cos^2(t)dt=c\frac{\pi}{2} (1-x^2)=\frac{3}{4}(1-x^2)
$$
p.s. your integral limit for $y$ is wrong since $y$ must be smaller than $\sqrt{1-x^2}$.
|
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|
Trigonometry or inequality problem Today, I saw this question:
If $x,y,z \in [0,\frac\pi 2]$, $x+y+z=\frac{3\pi}{4}$ and $\sec^2(x)\sec^2(y)\sec^2(z)=8$, calculate $E=\tan x\tan y+\tan y\tan z+\tan z\tan x$
My first thought was getting everything in term of $\tan$, and everything went as expected. Let $a=\tan x, b=\tan y, c= \tan z$. Then, after playing with the equations I got:
$$x+y+z=\frac{3\pi}{4}$$
$$\implies 1+a+b+c=ab+bc+ac+abc \tag1$$
And since $\tan^2x+1=\sec^2x$, by AM-GM we have
$$8=(1+a^2)(1+b^2)(1+c^2)\ge 8abc$$
$$\implies 1 \ge abc$$
So now I want to show that
$$(1+a^2)(1+b^2)(1+c^2)=8 \implies a+b+c\ge ab+bc+ac$$
in order to show that $LHS\ge RHS$ , so that they only achieve equality when $a=b=c=1$. Wolfram confirms that it is possible to do so, but I have not managed to prove it.
Edit: A solution to the problem
From $(1)$, we know:
$$a=\frac{(b+c)+(1-bc)}{(b+c)-(1-bc)}$$
Substituting $u=b+c$, $v=1-bc$, we have that $u^2+v^2=(1+b^2)(1+c^2)$ (A special case of the Brahmagupta-Fibonacci identity). Therefore
$$1+a^2=2\frac{u^2+v^2}{(u-v)^2}=2\frac{(1+b^2)(1+c^2)}{(u-v)^2}=\frac{8}{(1+b^2)(1+c^2)}$$
Since $x\le \frac{\pi}{2}$, we have that $y+z\ge \frac{\pi}{4}\implies \tan(y+z)=\frac{u}{v}\ge1$. Therefore, $u-v\ge 0$, and we can safely take the square root without losing solutions.
$$\implies (1+b^2)(1+c^2)=2(u-v)$$
$$\implies (1+b^2)(1+c^2)=2(b+c+bc-1)$$
$$\implies 1+b^2+c^2+b^2c^2=2b+2c+2bc-2$$
$$\implies (b-1)^2+(c-1)^2+(bc-1)^2=0$$
$\implies a=b=c=1, \implies E=3$
But that last inequality I pointed still intrigue me. Can we prove it with elementary methods? Also, is there a simple solution to the problem?
|
Note that each integer has a unique prime factorisation..
$8=2*2*2$
$$(1+a^2)(1+b^2)(1+c^2)=8=2*2*2=(1+1)*(1+1)*(1+1) \ \ \ \ \ \ (*)$$
Since all terms on the the $R.H.S$ are prime numbers ...
By Euclid's Lemma
$2$ must divide either $(1+a^2)$ or $(1+b^2)$ or $(1+c^2)$ or all of them
Let us assume $2$ divides $(1+a^2)$
Now we see only $a=1$ satisfies this
The other two can be determined using same way or by solving simultaneous equations (two variables and two equations)..
So $b=c=1$
$$x=y=z=\frac{\pi}{4}$$
$$E=1+1+1=3$$
(You can also do this by comparing numbers in $(*)$)
Hope you find this easier.
|
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|
determinant of the linear transformation $T(X) =\frac{1}{2} (AX+XA)$ Let $V$ vector space of all matrices $3\times3$, and let $A$ be the diagonal matrix :
$$
\begin{pmatrix}
1 & 0 & 0\\
0 & 2& 0 \\
0 & 0& 1\end{pmatrix}
$$
Compute thee determinant of the linear transformation $T(X) =\frac{1}{2} (AX+XA)$.
Any hints would be appreciated.
|
Note that your transformation is realized by a $9\times 9 $ matrix, as $X$ has $9$ elements.
So
$$
\left(
\begin{matrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21} \\
x_{22} \\
x_{23} \\
x_{31} \\
x_{32} \\
x_{33}
\end{matrix}
\right) \quad\longmapsto\quad
\left(
\begin{matrix}
x_{11} \\
\tfrac{3}{2}x_{12} \\
x_{13} \\
\tfrac{3}{2}x_{21} \\
2x_{22} \\
\tfrac{3}{2}x_{23} \\
x_{31} \\
\tfrac{3}{2}x_{32} \\
x_{33}
\end{matrix}
\right)
$$
and this is realized by the $9\times 9$ diagonal matrix
$$
T=\mathrm{diag}\,(1,\tfrac{3}{2},1,\tfrac{3}{2},2,\tfrac{3}{2},1,\tfrac{3}{2},1).
$$
Clearly $\det T=1\times\tfrac{3}{2}\times 1\times\tfrac{3}{2}\times2\times\tfrac{3}{2}\times1\times\tfrac{3}{2}\times1$
|
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|
Prove that $ \frac12 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2} $ I can't figure out how to prove the following inequality:
$$
1/2 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2}
$$
Thanks
|
As suggested by Lucian, use Taylor series for each piece and expand around zero. Then
$$
4\sin^2\left(x\right) + \frac{1}{4\cos^2\left(2x\right)}
$$
gives 1/4 + 5 x^2 + (4 x^4)/3 + (56 x^6)/9 + (3964 x^8)/315 + ...
Replacing x by Pi/7 and taking into account that Pi^2 is almost 10, you notice that terms above x^4 are perfectly negligible. So, for x=Pi/7, you end with 1/4 + 5 Pi^2/196 + Pi^4/28812. Still assuming that Pi^2 is almost 10, the value you obtain is 14653/28812 which is 0.508573 which is in your bounds (0.5 , 0.585786). If you do not approximate Pi^2 by 10, you would obtain 0.505156
The exact value of your expression is 0.506041
|
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|
Help with limit calculation Can anyone help me with this limit please:
I have been trying to solve this for 2 hours with no success:
$$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$
|
You can simplify the fraction using
$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2.$$
The numerator is
$$\sum_{k=1}^{n}(3k-2)^3=\cdots=\frac 14 (27 n^4-18 n^3-9 n^2+4n).$$
The dinominator is
$$\left(\sum_{k=1}^{n}(3k-2)\right)^2=\cdots=\frac 14 n^2 (3 n-1)^2=\frac 14(9n^4-6n^3+n^2).$$
Hence, your limit will be
$$\lim_{n\to\infty}\frac{27 n^4-18 n^3-9 n^2+4n}{9n^4-6n^3+n^2}=\lim_{n\to\infty}\frac{27-18(1/n)-9(1/n^2)+4(1/n^3)}{9-6(1/n)+(1/n^2)}=3.$$
|
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|
Generating function and combinatorics I am studying now the concept of generating function, and I have a solved question in my book which I don't understand, completely. There it is:
What is the number of options to roll 10 different dice, so the sum
of the results will be 25?
Now for the solution:
f(x) = (x+x^2+x^3+x^4+x^5+x^6)^10
I believe this is because we have 6 options in each dice, and 10 different dice.
=x^10(1+x+x^2+x^3+x^4+x^5)^10
Here the professor took out the Common factor x^10 (I think by taking out x^6 and some more)
=x^10(1-x^6)^10 * (1+x+x^2....)^10
The "...." sign is a sign for an infinite arithmetic progression or until x^5? I don't know.
Now we have to find the "base" of x^25, which is the basic of a generatic function or so I believe. Now he makes so permutations here:
C(10+15-1,15) - 10*C(10+9-1,9) + C(10,2)*C(10+3-1,3) =
=C(24,9)-10*C(18,9)+36*C(6,3)=
=1,307,504 - 486,200+720 = 822,040
I believe it has to be related to this concept, that the book shows for Derivative.
For example, f'(x)=n/(1-x)^(n+1); f''(x)=n(n+1)/(1-x)^n+2
Now he shows for the k-th Derivative that:
1/k! * f'(x) in the k-th time, if we set x=0:
n(n+1)....(n+k-1) / k! = C(n+k-1,k)
I am sorry for the long, not so understandable question, but It's really hard for me to understand his methods.
Thank you!
|
Since you have 10 dice, and each die can be any integer from 1 to 6, your generating function is
$f(x)=(x+x^2+x^3+x^4+x^5+x^6)^{10}=(x(1+x+x^2+x^3+x^4+x^5))^{10}=x^{10}(1+x+x^2+x^3+x^4+x^5)^{10}$
$=\displaystyle x^{10}\big(\frac{1-x^6}{1-x}\big)^{10}=x^{10}(1-x^6)^{10}(1-x)^{-10}.$
Now we have to find the coefficient of $x^{25}$, and we can use the Binomial Formula in the second factor and the formula $(1-x)^{-n}=\sum_{k=0}^{\infty}\binom{n-1+k}{k}x^k$ in the third factor.
This gives $f(x)=x^{10}(1-10x^6+45x^{12}+\cdots)\big(\sum_{k=0}^{\infty}\binom{9+k}{k}x^k\big)$, so the coefficient of $x^{25}$ will be the coefficient of $x^{15}$ in $(1-10x^6+45x^{12}+\cdots)\big(\sum_{k=0}^{\infty}\binom{9+k}{k}x^k\big)$, which is
$\;\;\binom{24}{15}-10\binom{18}{9}+45\binom{12}{3}=831, 204.$
|
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|
Find a limit in an efficent way I'm trying to calculate the following limit:
$$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{x}}}$$
What I did is writing it as:
$${e^{\frac{1}{x}\ln \left( {\frac{{\sin x}}{x}} \right)}}$$
Therefore, we need to calculate:
$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x}$$
Now, we can apply L'Hopital rule, Which I did:
$$\Rightarrow cot(x) - {1 \over x}$$
But in order to reach the final limit two more application of LHR are needed. Is there a better way?
|
Elementary proof using well known limits and inequalities:
$$\frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x} = \frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\frac{{\sin x}}{x}-1 }{x} =\frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\sin x-x }{x^2}$$
but for $x >0$ we have: $x-\frac{x^3}{6} \le \sin x \le x \Rightarrow -\frac{x^3}{6} = x-\frac{x^3}{6}-x\le\sin x -x \le 0 \Rightarrow -\frac{x^2}{6} \le\frac{\sin x-x }{x^2} \le 0$ the rest is squeeze theorem.
|
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|
What is the limit of $\mathrm{e}^{At}$ when $t\to \infty$, for $A$ a square matrix? I was doing a matrix calculation and need to find
$$\lim_{t\to \infty} \mathrm{e}^{At}=?$$
What is the limit of $\mathrm{e}^{At}$ when $t\to \infty$, for $A$ a matrix?
|
Recall the definition of exponential function of one variable in terms of power series:
$$
e^x := \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots
$$
Similarly, we can define the matrix exponent of $n\times n$ matrix $A$:
$$
e^A : = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \dots = \sum_{n=0}^{\infty}\frac{A^n}{n!},
$$
where $I$ is the identity matrix of the size $n\times n$.
Assume now that $A$ is diagonalizable, i.e. it can be represented as $A = P^{-1}DP$, where
$$
D =
\begin{bmatrix}
d_{11} & 0 & \cdots& 0\\
0 & d_{22} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & d_{nn}
\end{bmatrix}
$$
is a diagonal matrix, i.e. it has non-zero entries only on its diagonal. Note that
positive integer $n$, the power $D^n$ is easy to compute:
$$
D^n = \underbrace{D \cdot D \cdot \dots \cdot D}_{n \text{ times }} = \left(\
\begin{bmatrix}
d_{11} & 0 & \cdots& 0\\
0 & d_{22} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & d_{nn}
\end{bmatrix}\ \right) ^n
=
\begin{bmatrix}
\big(d_{11}\big)^n & 0 & \cdots& 0\\
0 & \big(d_{22}\big)^n & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & \big(d_{nn}\big)^n
\end{bmatrix}
$$
Then for any positive integer $n$ we can compute $A^n$ as
$$
A^n = \left(P^{-1} D P \right) ^n = \underbrace{P^{-1} D P \cdot P^{-1} D P \cdot \dots \cdot P^{-1} D P }_{n\text{ times }}
=P^{-1} D^n P
$$
Therefore the exponent
$$
e^A = I + \left(P^{-1} D P\right) + \left(\frac{1}{2!} P^{-1} D^2 P\right) + \left(\frac{1}{3!} P^{-1} D^3 P\right) + \dots =
P^{-1}\left(\sum_{n=0}^{\infty}\frac{1}{n!}D^n\right)P=
\\
=
P^{-1}\sum_{n=0}^{\infty}\frac{1}{n!}\left(\ \begin{bmatrix}
\big(d_{11}\big)^n & 0 & \cdots& 0\\
0 & \big(d_{22}\big)^n & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & \big(d_{nn}\big)^n
\end{bmatrix}\ \right)P=\\
=
P^{-1}\left(
\begin{bmatrix}
\displaystyle\sum_{n=0}^{\infty}\dfrac{\left(d_{11}\right)^n}{n!} & 0 & \cdots& 0\\
0 & \displaystyle\sum_{n=0}^{\infty}\dfrac{\left(d_{22}\right)^n}{n!} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & \displaystyle\sum_{n=0}^{\infty}\dfrac{\left(d_{nn}\right)^n}{n!}
\end{bmatrix}\right)P
=
P^{-1}
\left(
\begin{bmatrix}
e^{d_{11}} & 0 & \cdots& 0\\
0 & e^{d_{22}} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & e^{d_{nn}}
\end{bmatrix}
\right)P
=
P^{-1} e^D P
$$
Finally, multiplying $A$ by $t$, you get
$$
e^{At} = P^{-1} e^{Dt} P =
P^{-1}
\left(
\begin{bmatrix}
e^{d_{11}t} & 0 & \cdots& 0\\
0 & e^{d_{22}t} & \cdots & 0\\
\vdots& \vdots & \ddots & 0\\
0 & 0& \cdots & e^{d_{nn}t}
\end{bmatrix}
\right)P,
$$
where $P^{-1}DtP = A$ is diagonalization of $At$.
|
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|
Finding $ \lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}} $ without power series. I have to find: $\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}$
"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac {\ln(\frac{\arctan x}{x})} {x^2}
$
L'Hôpital's rule once is not enough, second L'Hôpital's rule seems worse. any help?
Edit: I cannot use power series expansion. that is provided on the next semester.
|
\begin{align*}
\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}&=\lim_{x \to 0^+} (\frac{x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} +... }{x})^{\frac{1}{x^2}}\\
&=\lim_{x \to 0^+}(1-\frac {x^2}3+ \frac{x^4}{5} - \frac{x^6}{7} +...)^{1/x^2}\\
&=e^{-1/3}
\end{align*}
|
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|
Prove that the series is bounded Good afternoon everybody! I have the following problem:
" Prove that the series $s_n=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ is decreasing and bounded, therefore convergent".
In order to prove it is decreasing, I must prove that $s_n-s_{n+1} >0$. But $s_n- s_{n+1}=[1+\frac{1}{2}+...+\frac{1}{n}-\ln n]$-$[1+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}-\ln (n+1)]$= $-\ln n-\frac{1}{n+1}+\ln (n+1)$=$\ln (\frac{n+1}{n})-\frac{1}{n+1}$=$\ln (1+\frac{1}{n})-\frac{1}{n+1}$. If we take the interval $[1,1+\frac{1}{n}]$, the minimum of the function $f(x)=\frac{1}{x}$ is $\frac{n}{n+1}$, hence $\ln(1+\frac{1}{n})=\int_1^{1+\frac{1}{n}} \frac{dx}{x}>\frac{1}{n}\frac{n}{n+1}=\frac{1}{n+1}$. It implies that $s_n-s_{n+1}=\ln(1+\frac{1}{n})-\frac{1}{n+1}>\frac{1}{n+1}-\frac{1}{n+1}=0$.
It shows that the series is monotone. But how to prove the series is also bounded? Thank you!
|
Note that ${1 \over n+1} \le { 1 \over t } \le {1 \over n}$ for $t \in [n,n+1]$, and $\ln({n+1 \over n}) = \int_n^{n+1} {dt \over t}$, hence ${1 \over n+1} \le \ln({n+1 \over n}) \le {1 \over n}$.
This gives $s_n -s_{n+1} \ge 0$ as you showed above.
Using the other side of the bound and the fact that $\ln$ is non-decreasing we have $s_n = 1+{1 \over 2} +...+{1 \over n} - \ln n \ge \ln({2 \over 1}) + \ln({3 \over 2})+...+\ln({n+1 \over n}) -\ln n = \ln(n+1)-\ln n \ge 0$.
|
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|
Find all roots of $z^2=3-4i$. Find all roots of $z^2=3-4i$.
$z^2=3-4i$
$z^2+4i-3=0$
But how do I go on from here?
|
Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\frac{2}{x}$.
Substitute $y = -\frac 2x$ into $x^2 - y^2 = 3$, giving
$$x^2 - \frac{4}{x^2} = 3$$
Multiply by $x^2$:
$x^4 - 4 = 3x^2$
Factor
$$x^4 - 3x^2 - 4 = 0$$
$$(x^2 + 1) (x^2 - 4) = 0$$
$$(x^2+ 1) (x + 2) (x - 2) = 0$$
Given $xy = -2$
for $x = 2$ then $y = -1$ therefore $z = 2 - i$
for $x = -2$ then $y = 1$ therefore $z = -2 + i$
|
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|
Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$. Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$.
I'm not really sure how to start this one, but since I found it in a book on group theory, such an approach might also be possible.
|
As $\displaystyle 2^3\cdot3=24\equiv1\pmod{23},$
$\displaystyle 2^n+3^m\equiv0\pmod{23}\iff 2^{n+3m}\equiv-3^m\cdot2^{3m}\equiv-(3\cdot2^3)^m$
$\displaystyle\implies2^{n+3m}\equiv-1\pmod{23}$
Now as $23$ is prime, $\displaystyle a^2\equiv1,a\equiv\pm1\pmod{23}$ for $(a,23)=1$
and $\displaystyle2^5=32\equiv9,2^{10}\equiv9^2=81\equiv12,2^{11}\equiv2\cdot12\equiv1$
So, there is no integer exponent$(e)$ of $2$ such that $2^e\equiv-1\pmod{23}$
Hence, the system admits no solution in integers
Alternatively,
HINT:
As $3^3\equiv2^2\pmod{23},$
Case $1:$ If $\displaystyle m=3r+1,2^n+3^m=2^n+3^{3r+1}\equiv 2^n+3\cdot2^{2r}\pmod{23}$
So, we need $\displaystyle 2^{n-2r}\equiv-3\pmod{23}$
Observe that there is no integer exponent$(e)$ of $2$ such that $2^e\equiv-3\pmod{23}$
Case $2:$ If $\displaystyle m=3r+2$
Case $3:$ If $\displaystyle m=3r$
|
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|
Prove : $\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$ Problem :
Let $a,b\in \mathbb{Z}$ such that $a\neq 0,b\neq 0$ ; $\left | a \right |\leq 100,\left | b \right |\leq 100$.
Prove that:
$$\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$$
Thanks :)
P/s : I have no ideas about this problem ! :(
|
Let $\lambda = a\sqrt{2} + b\sqrt{3}$ where $a$, $b$ not both zero and $|a|,|b| \le 100$. We have
$$\begin{array}{rrcl}
& ( \lambda - a\sqrt{2})^2 - 3b^2 &=& 0\\
\implies & \lambda^2 + 2a^2 - 3b^2 &=& \sqrt{8} a \lambda\\
\implies & \lambda^4 - (4a^2+6b^2)\lambda^2 + (2a^2-3b^2)^2 &=& 0\\
\iff & \lambda^2 (4a^2 + 6b^2) &=& (2a^2 - 3b^2)^2 + \lambda^4\\
\implies & |\lambda| &\ge& \frac{|2a^2 - 3b^2|}{\sqrt{4a^2+6b^2}}
\end{array}$$
Since $\sqrt{3/2}$ is irrational and $2a^2 - 3b^2$ is an integer,
$|2a^2 - 3b^2|$ is at least $1$ and hence
$$|\lambda| \ge \frac{1}{\sqrt{4\times 100^2+6\times 100^2}} = \frac{1}{100\sqrt{10}} > \frac{1}{350}$$
|
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|
The factors of $5^n-3^n-2^n$ I have been assigned the following question.
Let $f(n):= 5^n-3^n-2^n$. Prove that
(a) $p$ divides $f(p)$ for each prime $p$;
(b) $p^{k+1}$ divides $f(n)$ for $n=p^k$, with $p=2,3,5$ and $k\geq 0$;
(c) $p^{k+2}$ divides $f(n)$ for $n=p^k$, with $p=19$ and $k\geq 0$.
Progress
I have been able to derive (a) from the Fermat's little theorem. I suspect that (b) and (c) can be reduce to (a), but I have been
unable to do it.
|
The cases $k=0$ are trivial as $f(p^k)=f(1)=0$.
Lemma. $p^{q^k}\equiv p\mod q$.
By simple induction on $k$, as for $k=0$ is trivially true, suppose $p^{q^k}\equiv p \mod q$. We have
$$
p^{q^{k+1}}=p^{q·q^k}=(p^{q^k})^q
$$
and, by induction, we can complete as
$$
p^{q^{k+1}}=p^{q·q^k}=(p^{q^k})^q \equiv p^q \equiv p
$$
(the latter equivalence is because of Fermat's Small Theorem.) $\qed$
To prove (b) in general, we will use a simple factorization of the generic binomial $x^p\pm y^p$ as $(x\pm y)(\text{another polynomial})$.
The three cases $p=2,3,5$ are similar but slightly different. First, you can remove the term with base $p$, as $p^{k+1}$ divides $p^{p^k}$ because $k+1≤p^k$.
As induction base we state that $5=5^1$ divides $3+2=3^{5^0}+2^{5^0}$ and similarly for the other combinations. Suppose now that $p^{k+1}$ divides $q^{p^k}\pm r^{p^k}$ (bases and signs are not arbitrary, either $p=5$ and sign is $+$, or $q=5$ and sign is $-$).
The trivial factorization of $x^5+y^5$ is $(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)$, the factorization of $x^3-y^3$ is $(x-y)(x^2+xy+x^2)$ and the factorization of $x^2-y^2$ is $(x-y)(x+y)$. If we set $x=q^{p^k}$ and $y=r^{p^k}$, we have that
$$
q^{p^{k+1}}\pm r^{p^{k+1}} = (q^{p^k})^p\pm(r^{p^k})^p
$$
and, applying the above factorization, we can writ this as a product
$$
(q^{p^k}\pm r^{p^k})(\dots)
$$
By induction hypothesis, the first term is divisible by $p^{k+1}$. The second term is a sum of products of the $x$ and $y$ as defined above, but now you can use again Fermat to show that the whole term is divisible by $p$. First of all, $x\equiv q$ and $y\equiv r$. Now we can calculate each case separately.
*
*$p=5$:
$$
x^4-x^3y+x^2y^2-xy^3+y^4 \equiv 3^4-3^3·2+3^2·2^2-3·2^3+2^4 = 55 \equiv 0 \mod 5
$$
*
*$p=3$:
$$
x^2+xy+x^2 \equiv 5^2+5·2+2^2 = 39 \equiv 0 \mod 3
$$
*
*$p=2$:
$$
x+y \equiv 5+3=8 \equiv 0 \mod 2
$$
So in each case the right term of the multiplication is divisible by $p$.
Solution for (c) follows later.
|
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"question_score": "2",
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|
Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot y=1$, then $x^3+y^3=?$ [key says 110]
$x+y=-1$ and $x\cdot y=2$, then $x^3+y^3=?$ [key says 5]
$x-y=-1$ and $x\cdot y=2$, then $x^3 -y^3=?$ [key says -7]
$x+y=\frac{1}{3}$ and $x\cdot y=\frac{1}{9}$, then $x^3+y^3=?$ $\left[-\dfrac{2}{27}\right]$
|
If the problem cannot be reduced to the symmetric case then Groebner basis can be used to solve the problem. But the calculation by hand is very cumbersome so I use Maxima and got the following output. poly_reduced_grobnercalculates a Groebner base and poly_pseudo_dividedoes the division. The first element of the result list is the list of coefficients the second and the third are the numerator and the denomiantor of the remainder. This is the number we are looking for.
(%i4) load(grobner)
Loading maxima-grobner $Revision: 1.6 $ $Date: 2009-06-02 07:49:49 $
(%i5) poly_reduced_grobner([-5+y+x,x*y-1],[x,y])
(%o5) [y+x-5,y^2-5*y+1]
(%i6) poly_pseudo_divide(y^3+x^3,%,[x,y])
(%o6) [[y^2-x*y-10*y+x^2+5*x+25,15],110,1,7]
(%i7) poly_reduced_grobner([1+y+x,x*y-2],[x,y])
(%o7) [y+x+1,y^2+y+2]
(%i8) poly_pseudo_divide(y^3+x^3,%,[x,y])
(%o8) [[y^2-x*y+2*y+x^2-x+1,-3],5,1,7]
(%i9) poly_reduced_grobner([1+y+x,2+x*y],[x,y])
(%o9) [y+x+1,y^2+y-2]
(%i10) poly_pseudo_divide(y^3+x^3,%,[x,y])
(%o10) [[y^2-x*y+2*y+x^2-x+1,-3],-7,1,7]
(%i11) poly_reduced_grobner([(-1)/3+y+x,x*y-1/9],[x,y])
(%o11) [3*y+3*x-1,9*y^2-3*y+1]
(%i12) poly_pseudo_divide(y^3+x^3,%,[x,y])
(%o12) [[9*y^2-9*x*y-6*y+9*x^2+3*x+1,3],-2,27,7]
|
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|
Double Integral Confusion A buddy was asking me for help with one of his MV Calc problems, and I ended up getting the same answer as him so I figured I'd ask it here...
Question Find $$\iint_{R} (x-1) \, dA$$ where $R$ is the region enclosed by $y=x$ and $y=x^3$ in the first quadrant. So naturally I told him to compute $$\int_{0}^{1} \int_{x^3}^{x} (x-1) \, dy \, dx$$ as $x$ varies from $0$ to $1$ and $y$ from $x^3$ to $x$. Using this integral, we got $\frac{-7}{60}$. His textbook gives an answer of $\frac{-1}{2}$.
|
As mentioned in the comments, your answer is correct. I'll provide a solution here for future readers.
First, draw a picture. I'm going to use Wolfram Alpha to show one online--it should be simple enough to graph by hand. We can see the intersections are at $(0, 0)$ and a $(1, 1)$. Thus, we have our limits of integration. Evaluation is as follows:
$$\begin{align}
\int_0^1\int_{x^3}^x(x-1)\,dy\,dx &= \int_0^1y(x-1)\Bigg|_{x^3}^x\,dx\\
&= \int_0^1(x-x^3)(x-1)\,dx\\
&= \int_0^1(-x^4+x^3 + x^2-x)\,dx\\
&= -\frac{x^5}{5} + \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2}\Bigg|_0^1 \\
&= \frac{-1}{5} + \frac{1}{4} + \frac{1}{3} -\frac{1}{2}\\
&= \frac{-7}{60}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola.
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also for $\sqrt {x}-\sqrt {y}=1 $)
|
Squaring, we get $x + y + 2\sqrt{xy} = 1$; moving the $2\sqrt{xy}$ term to one side, and squaring again, we obtain $$
x^2 -2xy + y^2 -x -y + 1 = 0$$
The value of the determinant $$\begin{vmatrix}A & B \\ B & C\\\end{vmatrix}
$$ will tell us which type of curve is represented by a quadratic relation $$Ax^2 + 2Bxy + Cy^2 + Dx + Ey+F = 0;$$ we have a parabola if and only if the determinant is 0. (Details) Here $A=1, B=-1, C=1$, so the determinant is $\begin{vmatrix}1 & -1 \\ -1 & 1\\\end{vmatrix} = 0$, and this is indeed a parabola, truncated because it omits the points where $x<0$ or $y<0$.
|
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|
How to solver this equation $\sum_{i=1}^{6}x_{i}x_j=-3,j=1,2,3,4,5,6,j\neq i$ Let $x_{i}\in R,i=1,2,3,4,5,6$ such that
$$\begin{cases}
x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{1}x_{5}+x_{1}x_{6}=-3\\
x_{2}x_{1}+x_{2}x_{3}+x_{2}x_{4}+x_{2}x_{5}+x_{2}x_{6}=-3\\
x_{3}x_{1}+x_{3}x_{2}+x_{3}x_{4}+x_{3}x_{5}+x_{3}x_{6}=-3\\
x_{4}x_{1}+x_{4}x_{2}+x_{4}x_{3}+x_{4}x_{5}+x_{4}x_{6}=-3\\
x_{5}x_{1}+x_{5}x_{2}+x_{5}x_{3}+x_{5}x_{4}+x_{5}x_{6}=-3\\
x_{6}x_{1}+x_{6}x_{2}+x_{6}x_{3}+x_{6}x_{4}+x_{6}x_{5}=-3
\end{cases}$$
Find the $x_{i}$.
My idea: let
$$S=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}$$
then
$$x_{i}(S-x_{i})=-3,i=1,2,3,4,5,6$$
then follow I fell ugly,Thank you
|
Following on David Peterson's hint: This shows there are at most two different values among the $x_i$ (solve the quadratic). Call them $a,b$. We go through the cases:
1)all the same. Each equation becomes $5a^2=-3$, which gives $x_i=\pm\sqrt{\frac 35}i$, which is acceptable over $\Bbb C$ but not $\Bbb R$
2)five the same, one different. Let the odd one be $a$. We then get $5ab=-3, b(a+4b)=ab+4b^2-3, 4b^2=4ab$, one of $a,b$ is zero, which fails
3)four the same, two the other. Let the four be $a$. We get $b(4a+b)=4ab+b^2=-3=a(3a+2b)=3a^2+2ab, (b+3a)(b-a)=0$ We reject $a=b$ as we covered that in case 1. $-3a^2=-3$ so $a=\pm1, b=\mp3$
4)three of one, three of the other. Summing $x_i=\frac 12(S \pm \sqrt{12+S^2})$ and noting three of the signs are plus and three minus gives $S=3S, S=0$ and $a=-b$. This gives $-a^2=-3$ and we have three $x$'s are $\sqrt 3$ and three are $-\sqrt 3$
|
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|
Solution for $4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$ I'm trying to get a solution for:
$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$
My main problem is that I don't know how to combine this potencys!
Ive also thought about another function that would bring me same difficulties:
$6^x=36*9.75^{x-2}$
What am I supposed to do?
|
For the first one, we have
$$
4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2} \implies\\
3^{3x+2}-3^{3x+1} = 4^{2x+3}-4^{2x+1} \implies\\
(3-1)3^{3x+1} = (4^2 - 1)4^{2x+1} \implies\\
2\cdot 3^{3x+1} = 5\cdot 3\cdot4^{2x+1} \implies\\
3^{3x} = 5 \cdot 2^{4x+1}
$$
I think that's the simplest we can get it. From there, I suppose we'd have to solve using logs.
That is, let $\log(x)$ be the logarithm of your choosing. We have
$$
\log(3^{3x}) = \log(5 \cdot 2^{4x+1}) \implies\\
3x \log(3) = \log(5) + (4x+1)\log(2) \implies\\
(3 \log 3 - 4 \log 2)x = \log 5 + \log 2 \implies\\
x = \frac{\log 5 + \log 2}{3 \log 3 - 4 \log 2} = \frac{\log(10)}{\log(3^3/2^4)} = \frac{1}{\log_{10}27 - \log_{10}16}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
$n$th derivative of $e^x \sin x$ Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:
Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.
Taking the derivatives: $f^{(n)}(x) = \dfrac{1}{2i} \cdot ((1+i)^n e^{(1+i)x} - (1-i)^n e^{(1-i)x})$
Now, I use that: $$ (1+i)^n = {\sqrt{2}}^n \cdot \left(\cos\dfrac{n \pi}{4} + i \sin\dfrac{n \pi}{4}\right) \\(1 - i)^n = \sqrt{2}^n \cdot \left( \cos \dfrac{-n \pi}{4} + i \sin \dfrac{-n \pi}{4} \right)$$
Plugging that mess, I get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\left(\cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4}\right) e^{ix} - \left(\cos \dfrac{-n \pi}{4} + i \sin \dfrac{- n \pi}{4}\right) e^{-ix} \right)$$
But, $e^{ix} = \cos x + i \sin x$, and using Moivre's theorem, that makes: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\cos \left( x + \frac{n \pi}{4}\right) + i \sin \left( x + \frac{n \pi}{4}\right) - \left(\cos \left( - x - \frac{n \pi}{4}\right) + i \sin \left( -x -\frac{ n \pi}{4}\right)\right)\right)$$
and since $\cos$ is an even function, and $\sin$ is odd, we get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot 2i \sin \left(x + \dfrac{n \pi}{4}\right)$$
Simplifying, the answer would be $f^{(n)}(x) = e^x \cdot \sqrt{2}^n \cdot \sin\left(x + \dfrac{n \pi}{4}\right)$. I'm almost positive that this is it, but I just want to be sure. Thank you in advance!
|
Yes, your work is correct. But there is an easier way. Put $g(x)= e^{(1+i)x}$ and note that $f(x)$ is the imaginary part of $g(x)$ for real $x$. Now $g^{(n)}(x) =(1+i)^n e^{(1+i)x}$. Since $1+i=2^{1/2} e^{\pi i/4}$, we have $g^{(n)}(x) =2^{n/2}e^{n\pi i/4}e^{(1+i)x}=2^{n/2}e^{x + i(x +n\pi /4)}$. Thus $f^{(n)}(x)$ is the imaginary part of $g^{(n)}(x)$, which is $2^{n/2}e^x \sin (x +n\pi /4)$ as desired.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
find the limits find the limit
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{i = 0}^{n + 1} {{{\left( {\begin{array}{*{20}{c}}
{n + 1}\\
i
\end{array}} \right)}^3}} }}{{\sum\limits_{i = 0}^n {{{\left( {\begin{array}{*{20}{c}}
n\\
i
\end{array}} \right)}^3}} }}$$
I use Maple to get the
$$\sum\limits_{i = 0}^n {{{\left( {\begin{array}{*{20}{c}}
n\\
i
\end{array}} \right)}^3} = hypergeom\left( {\left[ { - n, - n, - n} \right],[1,1], - 1} \right)} $$
but I don't know how to compute it,Can anyone help me?
thank you very much!
|
Let $\;\displaystyle\text{Fr}_n = \sum\limits_{k=0}^n \binom{n}{k}^3\;$ be the $n^{th}$
Franel number (OEIS
A000172 ).
Method 1 (Asymptotic expansion)
By my answer to a related question, $\text{Fr}_n$ has following asymptotic expansion:
$$\text{Fr}_n = 8^n \frac{2}{\pi\sqrt{3}n}\left[
1
- \frac{1}{3n}
+ \frac{1}{27n^2}
+ \frac{1}{81n^3}
+ \frac{1}{243n^4} + \ldots
\right]$$
This immediately implies
$\displaystyle\quad \lim_{n\to\infty} \frac{\text{Fr}_{n+1}}{\text{Fr}_n}
= \lim_{n\to\infty} \frac{8^{n+1} n}{8^n (n+1)} = 8.
$
Method 2 (Poincare-Perron theorem)
In $1890s$, Franel has shown $\text{Fr}_n$ satisfy a recurrence relation of the form:
$$\text{Fr}_{n+2} = \alpha_n \text{Fr}_{n+1} + \beta_n \text{Fr}_{n}
\quad\text{ where }\quad
\begin{cases}
\alpha_n &= \frac{7 (n+1)(n+2) + 2}{(n+2)^2},\\
\beta_n &= \frac{8(n+1)^2}{(n+2)^2}
\end{cases}
$$
The key is as $n$ becomes large, both coefficients $\alpha_n$ and $\beta_n$ converge to some finite limit. The "limit" of the recurrence relation has the form
$$x_{n+2} = 7 x_{n+1} + 8 x_n$$
and its characteristic polynomial $\lambda^2 - 7 \lambda - 8 = (\lambda + 1)(\lambda - 8)$ has roots $-1$ and $8$.
Since distinct roots of the the characteristic polynomial has distinct modulus, Poincare-Perron theorem$\color{blue}{^{[1]}}$ tell us
*
*either $\text{Fr}_n$ vanishes for all sufficient large $n$
*or $\lim\limits_{n\to\infty} \frac{\text{Fr}_{n+1}}{\text{Fr}_n}$ converges to one of the characteristic root. i.e converges to $-1$ or $8$.
Since $\text{Fr}_n$ is positive for all $n$, it doesn't vanish for large $n$
and $\frac{\text{Fr}_{n+1}}{\text{Fr}_n}$ cannot converge to a negative number. This leaves us $\lim\limits_{n\to\infty}\frac{\text{Fr}_{n+1}}{\text{Fr}_n} = 8$ as the only possibility.
Method 3 (Binomial identities + Stirling approximations)
Let $\;\;\displaystyle C_n = \binom{n}{\lfloor\frac{n}{2}\rfloor}.\;\;$ For fixed $n > 1$ and integer $k$ such that $0 \le k \le n$, define
$$z_k = \binom{n}{k},
\quad
x_k = \begin{cases}\displaystyle \binom{n}{k-1},& k > 0\\ \\ 0,& k = 0\end{cases}
\quad\text{ and }\quad
y_k = \begin{cases}\displaystyle \binom{n}{k}, & k < n\\ \\ 0, & k = n\end{cases}
$$
Pascal identity tell us $z_k = x_k + y_k$.
Notice
$$\text{Fr}_n = \sum_{k=0}^n z_k^3\quad\text{ and }\quad
\text{Fr}_{n-1} = \sum_{k=0}^n x_k^3 = \sum_{k=0}^n y_k^3$$
We have
$$\text{Fr}_n - 8\text{Fr}_{n-1} = \sum_{k=0}^n \left( (x_k+y_k)^3 - 4 (x_k^3 + y_k^3) \right) = -3 \sum_{k=0}^n z_k (x_k - y_k)^2$$
Notice all $z_k \le C_n$, we obtain
$$|\text{Fr}_n - 8\text{Fr}_{n-1}|
\le 3 C_n \sum_{k=0}^n (x_k - y_k)^2
= 3 C_n \sum_{k=0}^n \left(2(x_k^2 + y_k^2) - z_k^2\right)
$$
It is well known
$$
\sum_{k=0}^n z_k^2 = \binom{2n}{n} = C_{2n}
\quad\text{ and }\quad
\sum_{k=0}^n x_k^2 = \sum_{k=0}^n y_k^2 = \binom{2n-2}{n-1} = \frac{n}{2(2n-1)}C_{2n}
$$
Together with a lower bound of $\text{Fr}_{n-1} \ge C_{n-1}^3$, this leads to
$$
|\text{Fr}_n - 8\text{Fr}_{n-1}| \le \frac{3}{2n-1}C_n C_{2n}
\quad\implies\quad
\left|\frac{\text{Fr}_n}{\text{Fr}_{n-1}}-8\right| \le \frac{3}{2n-1}\frac{C_n C_{2n}}{C_{n-1}^3}
$$
Using Stiring approximation, we have
$$C_n \sim 2^n \sqrt{\frac{2}{\pi n}}
\quad\implies\quad
\left|\frac{\text{Fr}_n}{\text{Fr}_{n-1}}-8\right|
\stackrel{<}{\sim} 6 \sqrt{\frac{\pi}{n}}
$$
With this, we can conclude $\;\;\lim\limits_{n\to\infty}\frac{\text{Fr}_n}{\text{Fr}_{n-1}} = 8$.
Notes
*
*$\color{blue}{[1]}$ I can't find any good reference for Poincare-Perron theorem online.
Please look at Chapter 8 of Saber Elaydi's book An Introduction to Difference Equations for more details.
|
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|
Identity for this recursive relation with multiple terms I have a recursive relation algorithm which is defined as follows:
$$F_n = 3(F_{n-1} - F_{n-2}) + F_{n-3}$$
$$F_0 = 0$$
$$F_1 = 1$$
$$F_2 = 4$$
From calculating the first few values, I know this is equivalent to $F_n = n^2$, however I don't quite know how to derive that. I tried substituting and telescoping, but I couldn't find an appropriate pattern.
How would I do this?
|
Proof
We want to show that $F_n = n^2$ for all $n \geq 0$. This is clearly true for $n=0,1,2$ since $F_0=0, F_1=1, F_2=4$. For $F_3$, we get that $F_3 = 3(F_2-F_1)+F_0 = 3(4-1)+0 = 9 = 3^2$. So it is true for $n=3$.
Assume for some $k\geq 3$ that $F_i = i^2$ for all $i \leq k$. We want to show that $F_{k+1} = (k+1)^2$. By the recurrence relation, $F_{k+1} = 3(F_k-F_{k-1})+F_{k-2}$. By the inductive hypothesis, we know that $F_k = k^2, F_{k-1} = (k-1)^2, F_{k-2}=(k-2)^2$. So
$\begin{align}
F_{k+1} &= 3(k^2-(k-1)^2)+(k-2)^2 \\
&= 3[k^2-(k^2-2k+1)]+(k^2-4k+4) \\
&= 3(2k-1)+k^2-4k+4 \\
&= 6k-3+k^2-4k+4 \\
&= k^2+2k+1 = (k+1)^2
\end{align}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding a real number c for polynomial (proof) The question is to find a real number c for which $ x\ge c%+$ implies $$x^4-4x^3+7x-9 \ge1000$$.
I was given the hint that $x>10$, then $4x^3<0.4x^4$, so $x^4-4x^3>0.6x^4$.
Problem is, I'm not understanding this line of reasoning, particularly how x>10 goes to the next step involving the $0.4x^4$.
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If $x > 10$ then $0.4x > 4$ and so $4x^3 < (0.4x)x^3 = 0.4x^4$. Therefore $x^4 - 4x^3 > x^4 - 0.4x^4 = 0.6x^4$.
Furthermore, if $x > 10$, then $7x - 9 \geq 7(10) -9 = 61$. Therefore, if $x > 10$, $x^4 - 4x^3 + 7x - 9 > 0.6x^4 + 61$. So how big should $x$ be so that $0.6x^4 \geq 1000?$
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|
Area of a triangle using vectors I have to find the area of a triangle whose vertices have coordinates
O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$
I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:
area $= \frac{1}{2}ab\sin{C}$
These are my steps for doing this:
$\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \sin{\theta} $
Let $\mathbf{a} = \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix}$ and let $\mathbf{b} = \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix}$
$\therefore \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix} = (5\sqrt{3})(15)\sin{\theta} $
$\therefore \sin{\theta} = -\dfrac{1}{\sqrt{3}}$
If I substitute these values into the general formula:
area $= \frac{1}{2}ab\sin{C}$
I get:
area $= \frac{1}{2}(5\sqrt{3})(15)(-\dfrac{1}{\sqrt{3}})$
$\therefore$ area $= -\dfrac{75}{2}$
However this isn't right, the area should be $\dfrac{75}{\sqrt{2}}$
I feel I'm missing something really obvious but I can't spot it, can anyone help?
Thank you.
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Since your vectors are in $\mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
$Area=(1/2) | a \times b |$.
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|
How do I prove $\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$? How do I prove $$\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$$?
This could be proven if $$\lim_{n\to\infty} \frac{\prod_{k=1}^n (1+\frac{1}{k})}{n} = 1$$ but I can neither prove this one.
Please help!
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Hint:
\begin{align}
\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)} &= \frac{1}{z} \lim_{n \to \infty} \frac{ \left( \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n}{n-1} \right)^z n!}{(1+\frac{z}{1})(1 + \frac{z}{2}) (1 + \frac{z}{3}) \cdots (1 + \frac{z}{n}) \cdot n! } \\
&= \frac{1}{z} \lim_{n \to \infty} \frac{ \left\{ \left( 1 + \frac{1}{1} \right) \cdot \left( 1 + \frac{1}{2} \right) \cdot \left( 1 + \frac{1}{3} \right) \cdots \left( 1 + \frac{1}{n} \right) \right\}^z}{(1+\frac{z}{1})(1 + \frac{z}{2}) (1 + \frac{z}{3}) \cdots (1 + \frac{z}{n}) }
\end{align}
using that as $n \to \infty$, $\frac{n}{n-1}, \frac{n+1}{n} \to 1$.
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|
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.
Still I have absolutely no idea how to generalize it for $n$ first terms...
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Hint in the form of an example: If you start the sum at $0$, you can write
$$\begin{align}
\lfloor\sqrt0 \rfloor+\lfloor\sqrt1 \rfloor+\lfloor\sqrt2 \rfloor+\cdots+\lfloor\sqrt{11} \rfloor&=0+1+1+1+2+2+2+2+2+3+3+3\\
&=3+3+3+3+3+3+3+3+3+3+3+3\\
&\quad-1-1-1-1-1-1-1-1-1\\
&\quad-1-1-1-1\\
&\quad-1\\
&=12\cdot3-(1+4+9)\\
&=(11+1)\lfloor\sqrt{11}\rfloor-(1^2+2^2+\cdots+\lfloor\sqrt{11}\rfloor^2)
\end{align}$$
Remark: The main reason for including the unnecessary term $0$ is that it makes it easier to describe the pattern of what's being subtracted. (Note, there are not $11$ terms in $1^2+2^2+\cdots+\lfloor\sqrt{11}\rfloor^2$, but only $\lfloor\sqrt{11}\rfloor=3$ terms.) Properly understood, the pattern explains the generalization to $\lfloor\sqrt[p]n\rfloor$ in Claude Leibovici's answer.
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Integration by parts. How can I integrate $ \int_{3}^8 \ln \sqrt{x+1}\ dx$ by parts ?
Is this step right ?
$ \int_{3}^8 \frac{1}{2}\ln(x+1)\ dx $ = $ \frac{1}{2} \int_{3}^8\ln(x+1)\ dx$
$f^{'}(x) = 1 , f(x) = x , g(x)= ln(x+1) $
$ \left[ x\ln(x+1) \right]_{x=3}^{x=8} - \int_{3}^8 x\frac{1}{x+1}\ dx $
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Everything you wrote is right.
Now take the derivative of $g(x)=\ln (x+1)$. So $g'(x)=\frac{1}{x+1}$. Then by the integration by parts formula you have
$$\frac{1}{2}\int_3^8 \ln(x+1)dx= \frac{1}{2}\left( x\ln(x+1) \bigg|_{3}^8 - \int_3^8 \frac{x}{x+1}dx\right).$$
Altogether
$$\frac{1}{2}\int_3^8 \ln(x+1)dx= 4\ln (9)-\frac{3}{2} \ln 4 - \frac{1}{2}\int_3^8 \frac{x}{x+1}dx$$
Finally the last integral can be solved by writing $\frac{x}{x+1} = 1-\frac{1}{x+1}.$
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Find the least nonnegative residue of a number Find the least nonnegative residue of $4^{47} \bmod 12$. I began by having $4^2$ congruent to $4 \bmod 12$ but I am not sure where to go from there.
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Since $4^2\equiv 4$ we\ have
$$
4^4\equiv 4 \mod 12 \\
4^8 \equiv 4 \mod 12 \\
4^{16}\equiv 4 \mod 12 \\
4^{32}\equiv 4 \mod 12 \\
4^{40}= 4^{32}\cdot 4^8 \equiv 4^2 \equiv 4 \mod 12 \\
4^{44}= 4^{40}\cdot 4^4 \equiv 4^2 \equiv 4 \mod 12 \\
4^{46}= 4^{44}\cdot 4^2 \equiv 4^2 \equiv 4 \mod 12 \\
4^{47}= 4^{46}\cdot 4^1 \equiv 4^2 \equiv 4 \mod 12 \\
$$
So the answer is $4$.
I used the binary expansion of $47$
$$
47=32+8+4+2+1$$
You can also use induction to show that $4^k \equiv 4 \mod 12$ for $k\ge 1$
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Find the surface area obtained by rotating $y=1+3x^2$ from $x=0$ to $x=2$ about the y-axis. Find the surface area obtained by rotating $y= 1+3 x^2$ from $x=0$ to $x = 2$ about the $y$-axis.
Having trouble evaluating the integral:
Solved for $x$:
*
*$x=0, y=1$
*$x=2, y=13$
$$\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\sqrt\frac{y-1}3'}^2\,dy$$
I got stuck at 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+(1/12)+(1/(y-1))}
any help would be great thanks
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This can be done using Pappus's Theorem and integrating in $x$:
$$
\begin{align}
\int_0^22\pi x\,\overbrace{\sqrt{y'^2+1}\,\mathrm{d}x}^{\mathrm{d}s}
&=\int_0^22\pi x\sqrt{36x^2+1}\,\mathrm{d}x\\
&=\frac\pi{36}\int_0^2\sqrt{36x^2+1}\,\mathrm{d}(36x^2+1)\\
&=\frac\pi{36}\int_1^{145}\sqrt{u}\,\mathrm{d}u\\
&=\frac\pi{36}\left[\frac23u^{3/2}\right]_1^{145}\\[4pt]
&=\frac\pi{54}\left(145^{3/2}-1\right)
\end{align}
$$
where we used the substitution $u=36x^2+1$.
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|
Answer Says $\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6} = 0$. I say DNE. What did I do wrong? I was asked to find
$$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6}$$
Observe that setting y=mx results in $$\lim_{(x,mx)\rightarrow(0,0)}\frac{x^3(mx)^2}{x^4+(mx)^6} = 0$$
The textbook solution then proved that the limit is 0 using the squeeze theorem.
However, I tried to set y=x^(4/6) and I got:
$$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3(x^{\frac{4}{6}})^2}{x^4+(x^{\frac{4}{6}})^6} = \lim_{x\rightarrow0}\frac{x^4}{x^4+x^4} = \frac{1}{2}$$
So I concluded that the limit does not exist.
I am not convinced that my solution is correct, I would really appreciate to know the reason why I am wrong.
Thank you
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Another approach:
$$\frac{x^3y^2}{x^4+y^6}\stackrel{\text{Polar Coord.}}=\frac{r^5\cos^3\theta\sin^2\theta}{r^4(\cos^4\theta+r^2\sin^6\theta)}=r\frac{\cos^3\theta\sin^2\theta}{\cos^4\theta+r^2\sin^6\theta}\xrightarrow[r\to0]{}0$$
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|
If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a^2+b^2+c^2)+2(ab+bc+ca) = 49$
So $23+2(ab+bc+ca) = 49\Rightarrow (ab+bc+ca) = 13$
Now from $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31\Rightarrow \frac{(a+1)\cdot (b+1)+(b+1)\cdot (c+1)+(c+a)\cdot (a+1)}{(a+1)(b+1)(c+1)} = 31$
So $\displaystyle \frac{(ab+bc+ca)+2(a+b+c)+3}{1+(a+b+c)+(ab+bc+ca)+abc} = 31\Rightarrow \frac{13+2\cdot 7+3}{1+7+13+abc} = 31$
So $\displaystyle \frac{30}{21+abc} = 31\Rightarrow 21\times 31+31(abc) = 30\Rightarrow (abc) = \frac{30-21\times 31}{31}=-\frac{621}{31}$
Now How can I calculate $a^3+b^3+c^3$
Is there is any better method by which we can calculate $abc$
Help me
Thanks
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HINT:
Where you have left of,
we can derive $$a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+b+ca)]$$
Now, $(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$
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|
How can this trig equation be simplified? We have $9+40\sin^2x=-42\sin x\cos x$.
I know this simplifies to $7\sin x+3\cos x=0$, but how?
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Along with Trafalgar Law's hint, we can use the fact that $\sin^2(x) + \cos^2(x) = 1$. Multiply both sides by $9$ to get $9 = 9\sin^2(x) + 9\cos^2(x)$. We then have
$$\begin{aligned}
9\sin^2(x) + 9\cos^2(x) + 40\sin^2(x) &= -42\sin(x)\cos(x)\\
49\sin^2(x) + 42\sin(x)\cos(x) + 9\cos^2(x) &= 0\\
(7\sin(x) + 3\cos(x))^2 &= 0
\end{aligned}$$
which gives the desirable equation you want to have.
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|
Calculate $\iint_D 3dxdy$ where $D = \{(x, y) : (x+y)^2 + (2x - y)^2 \le 4 \}$ Calculate $\displaystyle\iint_D 3dxdy$ where $D = \{(x, y) : (x+y)^2 + (2x - y)^2 \le 4 \}$.
I tried to solve this can I failed.
Can you please give me some hints?
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Assume that $D:5x^2+2y^2-2xy=4$. Since $4-10<0$, $D$ is an ellipse and it is sufficient that we calculate area of it. Now, consider $A=\left[ \begin{array}{cc} 5 & -1 \\ -1 & 2 \end{array} \right]$. Let us call $\lambda_1$ and $\lambda_2$ the eigenvalues of $A$. One can see, $D$ in new coordinate is $\lambda_1X^2+\lambda_2Y^2=4$. So, the area of $D$ is $\pi\sqrt{\frac4{\lambda_1}}\sqrt{\frac{4}{\lambda_2}}=\frac{4\pi}{\sqrt{\lambda_1\lambda_2}}$. The product of the eigenvalues is the determinant of the matrix $A$. Therefore the solution is $$3\times\frac{4\pi}{\sqrt{9}}=4\pi.$$
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|
Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
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I have come up with an answer with myself. Using CS inequality
$$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$
$$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$
Hence we have
$$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$
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Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$ $a;b;c\in \mathbb{R}^+$ such that $a+b+c=6$.
Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$
Thanks :)
I have no ideas about this problem ! :(
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Use Holder inequality and we have:
$$\left(1 + \frac 1{a^3}\right)\left(1 + \frac 1{b^3}\right)\left(1 + \frac 1{c^3}\right) \ge \left(1 + \frac 1{abc}\right)^3$$
From $AM-GM$ we have:
$$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$
$$\frac 63 \ge \sqrt[3]{abc}$$
$$2^3 \ge {abc}$$
$$8 \ge abc$$
So using this we minimize the right hand side:
$$\left(1 + \frac 1{a^3}\right)\left(1 + \frac 1{b^3}\right)\left(1 + \frac 1{c^3}\right) \ge \left(1 + \frac 1{abc}\right)^3 \ge \left(1 + \frac 1{8}\right)^3 = \left(\frac 9{8}\right)^3 = \frac{729}{512}$$
Now to find when the equality happens. Note that we have equality in the first inequality when $\frac{1}{a^3} = k\cdot 1$, $\frac{1}{b^3} = k\cdot 1$ and $\frac{1}{c^3} = k\cdot 1$.
From this we conclude that equality holds when $a=b=c$. For the second equality we have inequality when $8=abc \implies 8=a^3 \implies a=b=c=2$
So the minimum value occurs at $a=b=c=2$ and it's $\frac{729}{512}$
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|
Infinite Sum with Combination I am trying to figure out what the following sum converges to:
$$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n),\qquad\qquad0<x<1$$
An answer would be great, but if you have an explanation, that'd be better!
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Related techniques: (I), (II). Follow the steps:
1) simplify $(n+6){ n+6\choose n} $ as
$$ (n+6){ n+6\choose n} = \frac{1}{6!}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2. $$
2) use the series identity
$$ \sum_{n=0}^{\infty} x^{n+1}=\frac{x}{1-x} \longrightarrow (*) $$
3) Applying the operators $D(xD)(x^2D)^5 $ to both sides of $(*)$ , where $D=\frac{d}{dx}$, gives
$$ \sum_{n=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n+5}=D(xD)(x^2D)^5 \frac{x}{1-x} $$
$$ \implies \frac{1}{6!}\sum_{n=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n}\\
=\frac{1}{6!\,x^5}D(xD)(x^2D)^5 \frac{x}{1-x} . $$
Note: The operator $(x^2 D)^5$ means
$$ (x^2 D)^5 = (x^2D)(x^2D)(x^2D)(x^2D)(x^2D). $$
|
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|
An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we take $f(x)=\sqrt{(x+1)} -\sqrt{(x-1)} -\sqrt{(4x-1)}$ then we can see that
$f(x)$ is defined when $x \geq 1$ and $f(1) \geq 0\mbox{ and }f'(x) \geq 0$ so, the function is monotone increasing and it will never appear zero.
so, my question is , In this type of equation where the solution actually does not exist, then why should we get this type of solution?
my solution procedure is,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$
|
$-\frac{x}{x}= \frac{x}{x}$ has no solutions at all, since $-\frac{x}{x}\neq \frac{x}{x}$ no matter what $x$ is.
But we can square both sides, and then what happens?
$\frac{x^2}{x^2}= \frac{x^2}{x^2}$ is an equation that is true for all nonzero numbers.
By applying a non-invertible operation to both sides, we can turn an equation with no solutions into one with uncountably infinitely many solutions.
|
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|
Find the equation of a circle containing three given points Problem:
Determine the equation of the circle that passes through three points, $J(-3, 2)$, $K(4, 1)$, and $L(6, 5)$.
I thought of using systems like so:
$$\left\{ \begin{array}{rcl} (x+3)^2 + (y-2)^2 = r^2 \\ (x-4)^2 + (y-1)^2 = r^2 \\ (x-6)^2 + (y-5)^2 = r^2 \\ \end{array} \right.$$
After equating the expressions on the left hand side of each equation, expanding and simplifying, I found out that $y = 7x -2$. I decided to substitute the $y$ into two of the expressions to solve for $x$. But it only gives me $50x^2 - 50x + 25 = 50x^2 - 50x + 25$, which does not help. Can someone please help me out? Thanks!
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$\begin{vmatrix}
x^2+y^2&x&y&1\\
(-3)^2+2^2&-3&2&1\\
4^2+1^2&4&1&1\\
6^2+5^2&6&5&1\\
\end{vmatrix}=0$
$30(x^2+y^2)-60x-300y+30=0$
$x^2+y^2-2x-10y+1=0$
|
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Solve $2^x+7=y^2$ for integer $(x,y)$ How many ordered solutions $(x,y)$ are there to the equation $2^x+7=y^2$ , where $x$ and $y$ are integers?
I tried taking cases for $x$ and $y$ like they are even or odd but I couldn't solve further.
|
Given: $2^x+7=y^2$
Since $L.H.S.$ is odd, $R.H.S$. must be odd.
Putting $y=2m+1$,
$2^x+6=4m^2+4m$,
or $2^{x-1}+3=2m^2+2m$
this forces $x=1$
putting it in original equation, we get $y=3$ and $y=-3$
Thus we have solutions $(1,3)$ and $(1,-3)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A definite integral involving logarithmic functions: $\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x$ I'm trying to show $$
\int_0^1 \frac{\ln x \cdot \ln(1+x)}{1-x}dx=-\frac{1}{4}\pi^2 \ln(2)+\zeta(3).
$$
I am unsure how to approach this integral as I do not know how to use a power series representation for the integrand. I cannot use the generating function
$$
\frac{\ln(1-x)}{1-x}=-\sum_{n=1}^\infty H_n x^n
$$
since I do not have this in my integrand so it is not so easy to approach. Thanks for the help!
|
\begin{align}J&=\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x\\&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln t}{1-t}dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x \frac{\ln t}{1-t}dt\right)dx\\&=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1+x)(1-tx)}dtdx\\
&=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{\ln(tx)}{(1+t)(1-tx)}dtdx+\int_0^1\int_0^1 \frac{\ln(tx)}{(1+x)(1+t)}dtdx\\
&\overset{u(x)=tx}=-\zeta(2)\ln 2-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt+2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\
&\overset{\text{IBP}}=-\zeta(2)\ln 2-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln u}{1-u}du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln t}{1-t}dt+\\&2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-2\zeta(2)\ln 2+\int_0^1 \frac{\ln^2 t}{1-t}dt-J+2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-2\zeta(2)\ln 2+2\zeta(3)-J+2\ln 2 \times -\frac{1}{2}\zeta(2)\\
&=-3\zeta(2)\ln 2+2\zeta(3)-J\\
&=-3\times \frac{\pi^2}{6}\ln^ 2+2\zeta(3)-J\\
J&=\boxed{\zeta(3)-\frac{1}{4}\pi^2\ln 2}
\end{align}
NB:
\begin{align}\int_0^1 \frac{\ln x}{1+x}dx&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2t\ln t}{1-t^2}dt\\
&\overset{x=t^2}=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\
&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\
&=-\frac{1}{2}\zeta(2)\\
\end{align}
Moreover, i assume that:
\begin{align}\int_0^1 \frac{\ln x}{1-x}dx&=-\zeta(2)\\
\zeta(2)&=\frac{1}{6}\pi^2\\
\int_0^1 \frac{\ln x}{1-x}dx&=2\zeta(3)\\
\end{align}
|
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|
Prove that $\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$ Prove that $$\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$$
My try
*
*$n=0$ then $\frac{\sin x}{x}=\frac{1}{x} \int_0^x \cos y\,dy$ (true)
*Assuming $n=k$ (true)
*Prove $n=k+1$
|
Let $f_n(x)=\int_0^x y^n \cos(y+n\pi/2) dy$. Then, using integration by parts,
$$\int_0^z \frac{1}{x^{n+1}} f_n(x) dx =
-\frac{1}{nz^n} f_n(z)
+ \frac{1}{n} \sin(z+n\pi/2).$$
Now, again using integration by parts, this time on $f_n(z)$, it follows that the right hand side equals
$$\frac{1}{z^n} \int_0^z y^{n-1} \sin(y+n\pi/2)dy.$$
Note that $\sin(y+n\pi/2) = \cos(y+(n-1)\pi/2)$, which follows from $\sin(a+b)=\sin a \cos b + \sin b \cos a$.
Hence,
$$\int_0^z \frac{1}{x^{n+1}} f_n(x) dx = \frac{1}{z^n} f_{n-1}(z).$$
Now use your induction trick.
|
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|
If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$
I could not approach the problem at all though I think I could have done something by using Cauchy-Schwarz inequality, but could not pull this together. Please help.
|
You only need to prove that
$$
\begin{multline}
\left(\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}\right)\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) \\
= \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2}
\end{multline}$$
Where did that come from?
Well, consider this simple equality
$$
\frac{a}{b - c}\left(\color{green}{\frac{1}{c - a} + \frac{1}{a - b}} + \frac{1}{b - c}\right) = \color{green}{\frac{ac - ab}{(a - b)(b - c)(c - a)}} + \frac{a}{(b - c)^2}
$$
When you take cyclic sum of left and right hand sides, the first fraction in RHS dissapears.
In other words, we can write 2 similar ones:
$$
\begin{align}
\frac{b}{c - a}\left(\frac{1}{b - c} + \frac{1}{a - b} + \frac{1}{c - a}\right) = \frac{ab - bc}{(a - b)(b - c)(c - a)} + \frac{b}{(c - a)^2} \\
\frac{c}{a - b}\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) = \frac{bc - ac}{(a - b)(b - c)(c - a)} + \frac{c}{(a - b)^2}
\end{align}
$$
Three last equations add up to the first equation since those big fractions right after the equal sign cancel out.
|
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|
Integral $\int_0^\infty \frac{1}{(1+x^m)(1+x^2)}\,dx$ I saw somewhere that the above integral is equal to $\pi/4$ for all real number $m$.
This seems to be surprising. Does anyone have a nice proof?
|
Let
$$\begin{align}
I(m) &= \int_0^\infty \frac{dx}{(1+x^m)(1+x^2)}\tag{$x = t^{-1}$}\\
&= \int_0^\infty \frac{t^{-2}\,dt}{(1+t^{-m})(1+t^{-2})}\\
&= \int_0^\infty \frac{dt}{(1+t^{-m})(1+t^2)}\\
&= I(-m).
\end{align}$$
But
$$\begin{align}
I(m) + I(-m) &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{1}{1+x^{-m}}\right)\frac{dx}{1+x^2}\\
&= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{x^m}{1+x^{m}}\right)\frac{dx}{1+x^2}\\
&= \int_0^\infty \frac{dx}{1+x^2}\\
&= \frac{\pi}{2}.
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x). Problem :
If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).
My approach :
The given equation can be written as $$(x-y)f(x+y) -(x+y)f(x-y) =4xy(x-y)(x+y)$$
$$\Rightarrow \frac{(x-y)f(x+y)}{(x-y)(x+y)} -\frac{(x+y)f(x-y)}{(x-y)(x+y)} =4xy$$
$$\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy$$
Now we know that $$(x+y)^2 -(x-y)^2 = 4xy$$
$\Rightarrow \frac{f(x+y)}{x+y} =(x+y)^2$....(i)
& $\frac{f(x-y)}{x-y} = (x-y)^2$...(ii)
Now putting y =0 in (i) and (ii) we get :
$\frac{f(x)}{x} =x^2$ $\Rightarrow f(x) =x^3$
But the answer is $f(x) =x^3 +kx$ ( where k is any constant ) please clarify this part thanks...
|
The additional term comes from the relation $(x-y)k(x+y)-(x+y)k(x-y)=0$ for any $k$
|
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|
Polynomials with Integer Coefficients and irrational roots Is there a polynomial with integer coefficients which has √2 +√7  as a root?
|
$$
x = \sqrt{2} + \sqrt{7} \\ (x-\sqrt{2})^2=7 \\ x^2-2\sqrt{2}x+2=7 \\ x^2-5=2\sqrt{2}x \\ (x^2-5)^2=8x^2 \\ x^4-18x^2+25=0
$$
|
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|
Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:
Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$
My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!
|
"I know! Let's stand him on his head!"
$$ \frac{\tan \theta \ + \ \cot \theta}{\sec \theta \ \cdot \ \sin \theta} \ = \ \frac{\tan \theta \ + \ \cot \theta}{\tan \theta } \ = \ 1 \ + \ \frac{\cos^2 \theta }{\sin^2 \theta } \ = \ \frac{\sin^2 \theta \ + \ \cos^2 \theta }{\sin^2 \theta } \ = \ \frac{1}{\sin^2 \theta } $$
$$ \Rightarrow \ \ \frac{\sec \theta \ \cdot \ \sin \theta}{\tan \theta \ + \ \cot \theta} \ = \ \sin^2 \theta \ \ . $$
As I look through the thread again, this is rather like dani_s's proof, without involving cotangent...
|
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|
Express y in terms of x Question:
$$
\text{It is given that } y= \frac{3a+2}{2a-4} \text{and }x= \frac{a+3}{a+8} \\
$$
$$
\text{Express } y \text{ in terms of } x.
$$
From using $x$ to solve for $a$, I discovered that
$$
a = \frac{8x-3}{1-x}
$$
Then I proceeded to substitute $a$ into $y$. I did this twice to ensure no mistakes are made, and my final answer for both was
$$
\frac{22x-7}{20x-10}
$$
There's a problem, the correct answer is
$$
\frac{7-22x}{10-20x}
$$
This makes me want to cry, more so because I checked it twice and I was very careful about my working out, here it is:
$$
\frac{2+ 3(\frac{8x-3}{1-x})}{ 2(\frac{8x-3}{1-x}) -4 }
$$
$$\rightarrow{}$$
$$
\frac{(\frac{2(1-x) + 3(8x-3)}{1-x})}{(\frac{-4(1-x)+2(8x-3)}{1-x})}
$$
$$\rightarrow{}$$
$$
\frac{2(1-x) + 3(8x-3)}{1-x} * \frac{1-x}{-4(1-x)+2(8x-3)}
$$
$$\rightarrow{}$$
$$
\frac{22x-7}{1-x} * \frac{1-x}{20x-10}
$$
The $(1-x)$'s cancel out
$$\rightarrow{}$$
$$
\frac{22x-7}{20x-10}
$$
Can someone please tell me as to what I did incorrectly in the process? Thank you in advance!
|
You got the correct answer. Just multiply both numerator and denominator by -1
$$\frac{22x-7}{20x-10} = \frac{-1}{-1} \times \frac{7-22x}{10-20x}$$
|
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|
How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$a+b\le 1$$ Find the minimum of
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$
My try:
I can find this minimum，use Holder inequality
$$(\dfrac{1}{a^2}+\dfrac{1}{b^2})(a+b)^2\ge (1+1)^3=8$$
But
$$ab\le \dfrac{(a+b)^2}{4}\le\dfrac{1}{4}$$
so for
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$ minimum I can't find it,thank you
|
By AM-GM
$$16ab+16ab+a^{-2}+b^{-2}\ge16$$
But
$$2\sqrt{ab}\le a+b\le 1\implies ab\le\frac14\implies -31ab\ge-\frac{31}4$$
Summing we get, with equality iff $a=b=\frac12$
$$S\ge\frac{33}4$$
|
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|
Find the minimum value of $\frac{1}{2}(x^{2}+2bxy+9y^{2})-y$ Find the minimum value of $f=\frac{1}{2}(x^{2}+2bxy+9y^{2})-y$
I know that there's a theorem which states $P(x)=\frac{1}{2}x^{T}Ax-x^{T}b$ reaches min at the point $Ax=b$ at $P_{min}=-\frac{1}{2}b^{T}A^{-1}b$.
I first tried to write the function $f$ in the form of $P(x)=\frac{1}{2}x^{T}Ax-x^{T}b$, so
$\frac{1}{2}(\begin{bmatrix}x &y \end{bmatrix}
\begin{bmatrix}1&2b\\2b&9\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix})
-\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}$
And $P_{min}=-\frac{1}{2}b^{T}A^{-1}b$
$=-\frac{1}{2}\begin{bmatrix}0&1\end{bmatrix}\frac{1}{9-4b^{2}}\begin{bmatrix}9&-2b\\-2b&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}$
$=-\frac{1}{2(9-4b^{2})}$
However the answer key says that the min is $-\frac{1}{2(9-b^{2})}$, where did I go wrong?
|
Hint: Your matrix A is not right. It should be:
1 b
b 9.
In other words, a(11) = 1, a(12) = b = a(21), and a(22) = 9.
|
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|
Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I can see that $p \equiv 11 \pmod {24}$ and $p \equiv 23 \pmod {24}$ are consistent with that, but can anyone explain why the other possible congruences for a prime $p$ (such as $p \equiv 1 \pmod {24}$) are excluded by $p$ being a safe prime?
My reasoning so far is:
For a prime $p \equiv 1 \pmod 6$ and $p \equiv (1,7,13,19) \pmod {24}$, or $p \equiv 5 \pmod 6$ and $p \equiv (5,11,17,23) \pmod {24}$.
For a safe prime $p = 2q + 1$ it cannot be true that $p \equiv 1 \pmod 6$, otherwise $2q$ would be divisible by 6 and $q$ would not be prime. This eliminates 1, 7, 13 and 19.
Likewise $p = 2q+1 \equiv 5 \pmod {24}$ and $p = 2q+1 \equiv 17 \pmod {24}$ cannot hold, otherwise $q$ would have to be even: $q \equiv 2 \pmod {24}$ or $q\equiv 8\pmod {24}$ respectively.
This leaves $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ as possible congruences.
Is this correct and sufficient, and/or is there a better way of demonstrating that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ can and must hold?
|
I suggest you change to "otherwise $q$ would have to be even: $q \equiv 2$ or $14 \pmod{24}$, or $q \equiv 8$ or $20\pmod{24}$ respectively."
I would approach it from the other direction, checking candidates for $q$, because multiplication is safer than division in modular atithmetic. There are $12$ odd congruency classes $\pmod{24}$. The four classes $3, 9, 15, 21$ can be discarded as candidates for $q$ because of divisibility by $3$ (apart from the number $3$ itself). Any odd number $\equiv 1 \pmod{6}$ may also be safely discarded, because if $q\equiv 1 \pmod{6}$, then $p = 2q + 1 \equiv 3\pmod 6$. So we discard the four classes $1, 7, 13, 19$ as candidates for $q$. Thus we're left with four candidates for $q$: $5, 11, 17, 23$. In each case $2q + 1 \equiv 11$ or $23\pmod{24}$.
That being said, I would also find it faster, easier and better to work out modulo $12$. $p \equiv 11$ or $23 \pmod{24}$ is exactly the same as saying $p\equiv 11 \pmod{12}$. In this case we only have six odd congruency classes, and four of them are removed by observing that $q \equiv 5 \pmod6$. The remaining two possibilities ($q \equiv 5$ and $q\equiv 11$) both produce $p \equiv 11$.
|
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|
Simplifying a radical with fraction and sum What are the steps to simplify
$(1+(\frac{1}{2}(x^3-\frac{1}{x^3})^2))^ \frac{1}{2}$
to
$\frac{1}{2}(x^3+\frac{1}{x^3})$ ?
|
If what you have is $$ \sqrt{ 1 + \frac{1}{2}(x^3 - \frac{1}{x^3})^2} $$
$$ \sqrt{ 1 + \frac{(x^6 - 1)^2}{2x^6}} = \sqrt{ \frac{x^{12 } + 1}{2x^6}} = \sqrt{ \frac{1}{2}(x^6 + \frac{1}{x^6})}$$
So they are not equal.
|
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|
Average Value Half-Disk Find the average value of the x-coordinate of a point in the half-disk $R = {(x,y) : x^2 + y^2 \leq 25, x\geq 0}$ Also, out of curiosity what would be the value of the y coordinate?
I'm not really sure how to approach this problem, since we're not given a function f over R.
|
$$x_c = \dfrac{\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} x dy dx}{\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} dydx}$$
First the denominator:
$$\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} dydx=\int_{x=0}^{5} 2\sqrt{25-x^2}dx$$
Using $x=5\sin(\theta)$ then $dx=5\cos{\theta}d\theta$
$$\int_{x=0}^{5} 2\sqrt{25-x^2}dx=\int_{\theta=0}^{\pi/2} 2\sqrt{25-25\sin^2{\theta}}5\cos{\theta}d\theta=\int_{\theta=0}^{\pi/2} 50\cos^2{\theta}d\theta=\int_{\theta=0}^{\pi/2} 50\frac{1+cos{2\theta}}{2}d\theta=\frac{25\pi}{2}$$
Second the numerator:
$$\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} x dy dx=\int_{x=0}^{5} 2x \sqrt{25-x^2} dx$$
Same variable change:
$$\int_{x=0}^{5} 2x \sqrt{25-x^2} dx=\int_{\theta=0}^{\pi/2} 10\sin{\theta}\sqrt{25-25\sin^2{\theta}}5\cos{\theta}d\theta=\int_{\theta=0}^{\pi/2} 250\sin{\theta}\cos^2{\theta}d\theta=-\frac{250\cos^3(\pi/2)}{3}+\frac{250\cos^3(0)}{3}=\frac{250}{3}$$
Dividing:
$$x_c = \dfrac{\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} x dy dx}{\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} dydx}=\frac{\frac{250}{3}}{\frac{25\pi}{2}}=\frac{20}{3\pi}$$
|
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|
Compute the square root of a complex number This is a follow up to a previous question. I solved the equation $z^4 - 6z^2 + 25 = 0$ and I found four answer to be $z = \pm\sqrt{3 \pm 4i}$.
However someone in the comment said that the answer is going to be $2+i$, $2-i$, $-2+i$, $-2-i$. I cannot understand how we can find these roots from the answer that I found. How are we supposed to compute the square root of a complex number?
|
Hint:
Let $x + yi = \sqrt{a + bi}$. Then $(x+ yi)^2 = a + bi$. Then solve for $x$ and $y$ and you will generally have two sets of values for the square root $ \sqrt{a + bi}$
Example:
Say you want to compute $\sqrt{3 + 4i}$. Then assume the square root is $a + bi$. That is $a + bi = \sqrt{3 + 4i} \implies (a + bi)^2 = (a^2 - b^2) + 2abi = 3 + 4i$. Now solve the equations $ (a^2 - b^2) = 3$ and $2ab = 4$ to find $a$ and $b$.
|
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Is there some way to simplify $\sum_{i=1}^n \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) $ To obtain a closed form. Is there some way to simplify $\sum_{i=1}^n \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) $?
Does it have a closed form? It's the last piece of a puzzle I need to solve a similar question Differentiate $P_{x_n}(z) = \prod_{i=1}^n\frac{1+z+z^2+...+z^{i-1}}{i}$ twice to calculate the variance of involutions.
Taking the example
$\sum_{i=1}^4 \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) = (\frac{2-1}{2} + \frac{3-1}{2} + \frac{4-1}{2})(\frac{1-1}{2}) + (\frac{1-1}{2} + \frac{3-1}{2} + \frac{4-1}{2})(\frac{2-1}{2}) + (\frac{1-1}{2} + \frac{2-1}{2} + \frac{4-1}{2})(\frac{3-1}{2}) + (\frac{1-1}{2} + \frac{2-1}{2} + \frac{3-1}{2})(\frac{4-1}{2}) = 2(\frac{1-1}{2}\frac{2-1}{2})+ 2(\frac{1-1}{2}\frac{3-1}{2}) + 2(\frac{1-1}{2}\frac{4-1}{2}) + 2(\frac{2-1}{2}\frac{3-1}{2}) + 2(\frac{2-1}{2}\frac{4-1}{2}) + 2(\frac{3-1}{2}\frac{4-1}{2})$
I feel that since the sums are decreasing in number by one with regard to $\frac{i-1}{2}$ there should be a factorial term in the closed form, I'm just not sure how to obtain it. Any help would be appreciated.
|
To answer my own question,
$\sum_{i=1}^n \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) = \sum_{i=1}^n \sum_{j=1}^n(\frac{j-1}{2})(\frac{i-1}{2}) - \sum_{i=1}^n\frac{(i-1)^2}{4} $
Which simplifies to
$\frac{n(n-1)}{4}\frac{n(n-1)}{4} - \frac{1}{4}(\frac{(n-1)n(2(n-1)+1)}{6})$
or just $\frac{(n - 2) (n - 1) n (3 n - 1)}{48}$
|
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|
Inequality: $x^2+y^2+z^2 \geq \sqrt{2}x(z+y)$ How can I prove the following inequality:
$$x^2+y^2+z^2 \geq \sqrt{2}x(z+y)?$$
Thanks!
|
Obviously, is enough to prove for $x\ge0$, $y\ge0$, $z\ge0$. By homogeneity, can be supposed wlog that $x^2+y^2+z^2=1$ Using Lagrange multipliers with the problem
$$f(x,y,z)=\sqrt{2}x(z+y),\qquad x^2+y^2+z^2=1,$$
we have:
$$\sqrt{2}(z+y)=2\lambda x,$$
$$\sqrt{2}x=2\lambda y,$$
$$\sqrt{2}x=2\lambda z,$$
with nontrivial solutions for $\lambda=0,\pm1$...
|
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|
To Find $A^{50}$ $$A=\begin{bmatrix}1 & 0&0\\1 & 0&1\\0&1&0\end{bmatrix}$$
Find $A^{50}$ ?
Now from Cayley–Hamilton theorem, I get $A^3-A^2-A+I=0$ and $A^{50}=(A^4)^{12}A^2$ so I found $A^4$ which is $-2A-I$, then we have $A^{50}=B^{12}A^2$ where $B =A^4$ was calculated, now should I again use Cayley–Hamilton theorem to find $B^{12}$ or is there a better possibility?
|
Let $f(x) = x^3-x^2-x-1 = (x^2-1)(x-1) = (x-1)^2(x+1)$.
Since $f(A) = 0$, we want to find polynomial $q(x), r(x)$ such that
$$x^{50} = q(x)f(x) + r(x)\tag{*1}$$
with $\deg r(x) \le \deg f(x) - 1 = 2$. If we can figure out what is $r(x)$, then
$$A^{50} = q(A)f(A) + r(A) = r(A)$$
Write $r(x)$ as $a x^2 + b x + c$. To fix the coefficients, evaluate both side of $(*1)$ at
$1$ and $-1$ and the derivative at the double root $1$, we get:
$$
\begin{cases}
1 &= a + b + c\\
1 &= a - b + c\\
50 &= 2a + b
\end{cases}
\quad\implies\quad
\begin{cases}
a &= 25\\
b &= 0\\
c &= 24\\
\end{cases}
\quad\implies\quad
r(x) = 25 x^2 - 24.
$$
As a result,
$$A^{50} = 25 A^2 - 24 I
= 25 \begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
1 & 0 & 1
\end{pmatrix}
- 24 \begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0\\
25 & 1 & 0\\
25 & 0 & 1
\end{pmatrix}
$$
|
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|
$\overline{x} \times \overline{a} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ I'm trying to solve this exercise:
Let $\overline{a} \neq \overline{0}$, $\overline{b}$ be two vectors of the Euclidean vector space $V_{3}$. Prove the equation $\overline{x} \times \overline{a} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ and find if it's the only one.
It does not seem that hard, my approach is the following, but I seem to be missing something:
Let $B = \left \{ \overline{\epsilon_{1}}, \overline{\epsilon_{2}}, \overline{\epsilon_{3}} \right \}$ be a base of $V_{3} $
Then the vectors $\overline{a}$. $\overline{b}$ can be represented as:
$\overline{a} = a_{1}\overline{\epsilon_{1}} + a_{2}\overline{\epsilon_{2}} + a_{3}\overline{\epsilon_{3}} = \sum_{i=1}^{3}a_{i}\overline{\epsilon_{i}}$
$\overline{b} = \sum_{j=1}^{3}b_{j}\overline{\epsilon_{j}}$
I also analyze the vector $\overline{x}$ as follows:
$\overline{x} = x_{1}\overline\epsilon_{1} + x_{2}\overline\epsilon_{2} +x_{3}\overline\epsilon_{3}$
$\overline{x} \times \overline{a} = \overline{b}$
$(x_{2}a_{3} - x_{3}a_{2})\overline\epsilon_{1} + (x_{3}a_{1} -x_{1}a_{3})\overline\epsilon_{2} + (x_{1}a_{2} - x_{2}a_{1})\overline\epsilon_{3} = \sum_{j=1}^{3}b_{j}\overline\epsilon_{j}$
And I get:
$x_{2}a_{3} - x_{3}a_{2} = b_{1}$
$x_{3}a_{1} - x_{3}a_{3} = b_{2}$
$x_{1}a_{2} - x_{3}a_{1} = b_{3} $
I am aware that $\langle\overline{a},\overline{b} \rangle =0$ means that the two vectors at vertical with each other, but how to I proceed from here?
|
$$
0 \cdot x_1+ a_3\cdot x_2 -a_2\cdot x_3=b_1\\
-a_3\cdot x_1+0\cdot x_2+a_1\cdot x_3=b_2\\
a_2 \cdot x_1 -a_1 \cdot x_2 + 0 \cdot x_3=b_3\\
$$
This is a system of linear equation. We are given that the vector a is nonzero. Without loss of generosity, let us assume that a_3 is nonzero.
Then by using Gaussian elimination, we get:
$$
\begin{array}{ccc|c}
1 & 0 & -a_1/a_3 & -b_2/a_3 \\
0 & 1 & -a_2/a_3 & b_1/a_3 \\
0 & 0 & 0 & (a_1b_1+a_2b_2+a_3b_3)/a_3
\end{array}
$$
From linear algebra, we know that this system only has solution iff
$$
a_1b_1+a_2b_2+a_3b_3=0
$$
which is equivalent to the dot product is zero.
|
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|
How to integrate $\int_{1}^{3} {\frac{x^2 - 1}{x^4 + 1}}\, dx$ $$\int_{1}^{3} {\frac{x^2 - 1}{x^4 + 1}}\, dx$$
Well, I can simplify the numerator:
$$\int_{1}^{3} {\frac{(x - 1)(x+1)}{x^4 + 1}}\, dx$$
But I have no idea how to simplify the denumenator:
$$\ x^4+1=? \ $$
How to solve this integral?
|
We will first compute the integral without bounds. Start off by rewriting the integral using partial fractions:
$$\int \frac{-2\sqrt{2}x - 2}{4(x^2 + \sqrt{2}x + 1)} + \frac{2 - 2\sqrt{2}x}{4(-x^2 + \sqrt{2}x - 1)} \ \mathrm{d}x.$$
Note that we can split this into two integrals to get
$$\int \frac{-2\sqrt{2}x - 2}{4(x^2 + \sqrt{2}x + 1)}\ \mathrm{d}x + \int \frac{2 - 2\sqrt{2}x}{4(-x^2 + \sqrt{2}x - 1)} \ \mathrm{d}x.$$
Factoring out the constants on both sides, we obtain
$$\frac{1}{2\sqrt{2}} \int \frac{\sqrt{2} - 2x}{-x^2 + \sqrt{2}x - 1}\ \mathrm{d}x +
\frac{1}{4} \int \frac{-2\sqrt{2}x - 2}{x^2 + \sqrt{2}x +1} \ \mathrm{d}x.$$
Now, make the change of variable
$$ u = -x^2 + \sqrt{2}x - 1 \qquad \therefore \qquad \mathrm{d}u = (\sqrt{2} - 2x)\mathrm{d}x.$$
Hence, rewriting the integrand, we have
$$\frac{1}{2\sqrt{2}} \int \frac{1}{u}\ \mathrm{d}u +
\frac{1}{4} \int -\frac{\sqrt{2}(2x + \sqrt{2})}{x^2 + \sqrt{2}x +1} \ \mathrm{d}x,$$
which is the same as writing
$$\frac{1}{2\sqrt{2}} \int \frac{1}{u}\ \mathrm{d}u -
\frac{1}{2\sqrt{2}} \int \frac{2x + \sqrt{2}}{x^2 + \sqrt{2}x +1} \ \mathrm{d}x.$$
We can do a similar change of variable here by letting
$$ s = x^2 + \sqrt{2}x + 1 \qquad \therefore \qquad \mathrm{d}s = (\sqrt{2} + 2x)\mathrm{d}x.$$
This gives
$$\frac{1}{2\sqrt{2}} \int \frac{1}{u}\ \mathrm{d}u -
\frac{1}{2\sqrt{2}} \int \frac{1}{s} \ \mathrm{d}s.$$
It is now easy to integrate both sides and substitute $u$ and $s$ back after integration. You should obtain
$$ \int \frac{x^2 - 1}{x^4+1} \ \mathrm{d}x =
\frac{1}{2\sqrt{2}} \left[\log(-x^2 + \sqrt{2}x - 1) - \log(x^2 + \sqrt{2}x + 1)\right] + C,$$
for some constant $C$. It now remains to integrate from $1$ to $3$, which should be a straightforward task from here.
|
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|
How to evaluate Ahmed's integral? How to show that:
$$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$$
I saw this on Wolfram.
|
Define $f\left(t\right):=\int_{0}^{1}\frac{\arctan\left(t\sqrt{x^{2}+2}\right)}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}dx$ so$$\begin{align}f\left(\infty\right)&=\frac{\pi}{2}\int_{0}^{1}\frac{dx}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}\\&=\frac{\pi}{2}\left[\arctan\frac{x}{\sqrt{x^{2}+2}}\right]_{0}^{1}\\&=\frac{\pi^{2}}{12},\\f^\prime\left(t\right)&=\frac{1}{1+t^{2}}\int_{0}^{1}\left(\frac{1}{x^{2}+1}-\frac{t^{2}}{1+t^{2}\left(x^{2}+2\right)}\right)dx\\&=\frac{1}{1+t^{2}}\left(\frac{\pi}{4}-\frac{t}{\sqrt{1+2t^{2}}}\arctan\frac{t}{\sqrt{1+2t^{2}}}\right).\end{align}$$Substituting $u=t^{-1}$ and using $\arctan\frac{1}{\theta}=\frac{\pi}{2}-\arctan\theta$, Ahmed's integral $A:=f\left(1\right)$ satisfies$$\frac{\pi^{2}}{12}-A=f\left(\infty\right)-f\left(1\right)=\frac{\pi^{2}}{16}-\left(\frac{\pi^{2}}{12}-A\right)\implies A=\frac{5\pi^{2}}{96}.$$
|
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|
$f$ is an even function defined on $(-5,5)$
If $f$ is an even function defined on $(-5,5)$ on the interval, then find four real values of x satisfying $f(x)=f(\frac{x+1}{x+2})$.
My book gives the answer as $\frac{\pm3 \pm\sqrt{5}}{2}$
And the solution is given as:
Since, f is an even function,
then, $f(x)=f(-x)$,$\forall x\in (-5,5)$
$$\implies f(x)=f(\frac{1-x}{2-x}) \implies x=\frac{1-x}{2-x} \implies x=\frac{3\pm\sqrt{5}}{2}$$
Again, $$f(-x)=f(\frac{x+1}{x+2}) \implies -x=\frac{x+1}{x+2} \implies x=\frac{-3\pm\sqrt{5}}{2}$$
What I can't understand is that how can we say $\implies f(x)=f(\frac{1-x}{2-x}) \implies x=\frac{1-x}{2-x}$? It is nowhere said that $f(x)$ is a one-one function then how can we say so? Also if we do use this logic then $f(x)=f(\frac{x+1}{x+2})$
which gives $x=\frac{-1\pm\sqrt{5}}{2}$ which is not given as a solution.
|
You are correct; we can't conclude that $x=\frac{1-x}{2-x}$ just from $f(x) = f\left(\frac{1-x}{2-x}\right)$. But we don't need to; we need to make the opposite inference: If $x = \frac{1-x}{2-x}$, then $f(x) = f\left(\frac{1-x}{2-x}\right)$, which is what we are looking for.
There might of course be many other $x$ for which $f(x) = f\left(\frac{1-x}{2-x}\right)$ but $x\ne\frac{1-x}{2-x}$. For example, $f$ might be a constant function so that $f(x) = f\left(\frac{1-x}{2-x}\right)$ for all $x\ne 2$. But the question only asks us to find four such values, and this we can certainly do.
|
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|
Exponential and Logarithmic Functions - $\log_2(x^2-4x-28)= 2$ I got $x= \frac{45}{x-4} $
I know its wrong. Can someone show me how to solve.
Thanks
|
Problem: We need to solve for $x$ in the given equation
$$\log_2(x^2-4x-28)=2$$
Remember that if $\log_a(b)=y$, then $a^y=b$. In our equation, $a=2$, $b=x^2-4x-28$, and $y=2$. Therefore $x^2-4x-28=2^2=4$. This is easy to solve for.
$$x^2-4x-28=4$$
$$x^2-4x-32=0$$
Factor it
$$(x-8)(x+4)=0$$
$$\color{green}{\boxed{x=8, \ -4}}$$
Hope I helped!
|
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|
$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$
When $n$ is $1$:
$$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$
$$ 4 + 9 + 25 = 38 $$
When $n$ is $k$:
$$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$
Finally, when $n$ is $k+1$:
$$ 2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}} $$
I try by expanding, by subtraction, but no solution /:
|
Not with induction:
The residues of each term cycle since $a^{2^{n+1}} = (a^{2^n})^2$
$2^{2^n} \pmod {19}$ with $4, 16, 9, 5, 6, 17, 4, \dots$
$3^{2^n} \pmod {19}$ with $9, 5, 6, 17, 4, 16, 9, \dots$
$5^{2^n} \pmod {19}$ with $6, 17, 4, 16, 9, 5, 6, \dots$
By adding terms pairwise, you can see that the statement is true. This also provides some motivation for induction using $k+6$ rather than $k+1$.
|
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|
Improper integrals in curve length I am supposed to find the length of curve of the following:
*
*$ y = \sqrt{2-x^2}$ ; $0\le x\le 1$
*$y =\ln(\cos x) $; $0\le x\le \frac{\pi}{3}$
I followed the directions found from this question : Length of a curve y = 1 - √x to solve till the integral. So currently i have this for the 2 questions:
*
*$\ell = \int\limits_{0}^{1} \sqrt{1+\frac{x^2}{2-x^2}} \ dx $
*$\ell = \int\limits_{0}^{\pi/3} \sqrt{1+(\frac{-\sin x}{\cos x})^2} \ dx $
However, i am having difficulties integrating the integrals and evaluating due to the square root. Can someone guide me in integrating the integrals above?
All help and suggestions are appreciated. Thank you!
|
In comments you've said that you've already managed to calculate the first integral, so let's talk about the second.
We have $$ \sqrt{1+\left(\frac{-\sin x}{\cos x}\right)^2} = \sqrt{1+\frac{\sin^2 x}{\cos^2 x}} =\sqrt{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}} = \frac{1}{|\cos x|}$$
For $x\in(0,\frac\pi 3)$ we have $\cos x > 0$, so the integral to be calculated is $$ \ell = \int_0^{\frac\pi 3}\frac{dx}{\cos x}$$
The standard substitution for similar integrals is $t=\tan\frac{x}{2}$, $x = 2\arctan t$, which gives $$ dx = \frac{2\,dt}{1+ t^2}$$ $$\cos x = \cos^2\frac x 2 - \sin^2\frac x 2 = \frac{\cos^2\frac x 2 - \sin^2\frac x 2 }{\cos^2\frac x 2 + \sin^2\frac x 2 } = \frac{1 - \tan^2\frac x 2 }{1 + \tan^2\frac x 2 } = \frac{1-t^2}{1+t^2}$$
$$t_1 = \tan 0 = 0, \qquad t_2 = \tan\frac\pi 6 = \frac{1}{\sqrt{3}}$$
so the integral to calculate is
$$ \ell = \int_0^{1/\sqrt{3}} \frac{2 dt}{1-t^2}$$
Can you continue?
|
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|
Evaluate the integral $\int\frac{1}{2+3\sin x}\,\text{d}x$. Please evaluate the integral,
$$\int \frac{1}{2+3\sin x}\,\text{d}x.$$
What I have tried is to substitute $\sin x = \sqrt{1-x^x}$ but I was stuck in a maze. Also, I did look a the wolfram solution. Can anyone propose a different solution from Wolfram, perhaps simpler with a bit of explanation?
|
use Weierstrass substitution: $\sin\left(x\right)=\frac{2t}{1+t^2}$, where $t=\tan\left(\frac{x}{2}\right)$
$⇔2\arctan\left(t\right)=x$,$dx=\frac{2dt}{1+t^2}$
$$\int_{ }^{ }\frac{dx}{2+3\sin\left(x\right)}
$$$$=\int_{ }^{ }\frac{\frac{2dt}{1+t^2}}{2+3\left(\frac{2t}{1+t^2}\right)}
$$$$=\int_{ }^{ }\frac{2dt}{2t^2+6t+2}
$$$$=\int_{ }^{ }\frac{dt}{t^2+3t+1}$$
$$=\int_{ }^{ }\frac{dt}{\left(t+\frac{3}{2}\right)^2-\frac{5}{4}}$$
substitute $t+\frac{3}{2}=u$
$$\int_{ }^{ }\frac{du}{u^2-\frac{5}{4}}$$
now we need another substitution:$\frac{2}{\sqrt{5}}u=v,\frac{dv}{du}=\frac{2}{\sqrt{5}}⇔du=\frac{\sqrt{5}}{2}dv$
$$\frac{4}{5}\int_{ }^{ }\frac{\frac{\sqrt{5}}{2}dv}{v^2-1}$$
$$\frac{2}{\sqrt{5}}\arctan\left(v\right)+C$$
undo substitution $\frac{2}{\sqrt{5}}u=v$, we have:$$\frac{2}{\sqrt{5}}\arctan\left(\frac{2}{\sqrt{5}}u\right)+C$$
undo substitution $t+\frac{3}{2}=u$, we have:$$\frac{2}{\sqrt{5}}\arctan\left(\ \frac{2\tan\left(\frac{x}{2}\right)+3}{\sqrt{5}}\right)+C$$
we're done.
|
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|
LogSine Integral $I=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta$ These are known as LogSine integrals at $2\pi/3$, so I will call the integral Ls as this is common in the literature. I am trying to prove
$$
Ls=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta=-\frac{13\pi^3}{162}-2Gl_{2,1}\big(\frac{2\pi}{3}\big)
$$
where $Gl_{2,1}$ can be reduced to one-dimensional polylogarithmic constants. I know we can write
$$
\ln^2\big(2\cos \frac{\theta}{2}\big) =\big(\ln 2+\ln \cos\frac{\theta}{2}\big)^2=\ln^2 2+\ln^2 \cos \frac{\theta}{2} +2\ln 2 \ln \cos \frac{\theta}{2},
$$
but am totally stuck at this point. Thanks
|
Using the principal brach of $\log z$,
$$\log(1+e^{2ix}) = \log(e^{i x}(e^{-ix}+ e^{i x})) = \log(e^{ix})+ \log(2 \cos x) = ix + \log(2 \cos x) .$$
Squaring both sides,
$$ \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx = \int_{0}^{\pi /6} \Big( ix + \log(2 \cos x) \Big)^2 \ dx .$$
Then equating the real parts on both sides of the equation and rearranging,
$$ \begin{align} \int_{0}^{\pi/3} \log^{2} \left( 2 \cos \frac{x}{2}\right) \ dx &= 2 \int_{0}^{\pi /6} \log^{2}(2 \cos x) \ dx \\ &= 2 \int_{0}^{\pi /6} x^{2} \ dx + 2 \ \text{Re} \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx \\ &= \frac{\pi^{3}}{324} + 2 \ \text{Re} \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx . \end{align}$$
Now make the substitution $z = e^{2ix}$.
Then
$$\int_{0}^{\pi/3} \log^{2} \left( 2 \cos \frac{x}{2} \right) \ dx = \frac{\pi^{3}}{324} + \text{Re} \frac{1}{i} \int_{C} \frac{\log^{2}(1+z)}{z} \ dz$$
where $C$ is a portion of the unit circle in the first quadrant of the complex plane.
But since we're using the principal branch of $\log z$, $\log(1+z)$ is analytic on the complex plane for $\text{Re}(z) > -1$.
So the path doesn't matter.
And therefore
$$ \int_{0}^{\pi /3} \log^{2}\left( 2 \cos \frac{x}{2} \right) \ dx = \frac{\pi^{3}}{324} + \text{Re} \frac{1}{i} \int_{0}^{e^{i \pi/3}} \frac{\log^{2}(1+z)}{z} \ dz .$$
You can find an antiderivative of the integrand in terms of polylogarithms by integrating by parts twice.
$$ \begin{align} \int \frac{\log^{2}(1+z)}{z} \ dz &= \log^{2}(1+z)\log(-z) - 2 \int \frac{\log(1+z) \log(-z)}{z} \ dz \\ &= \log^{2}(1+z) \log(-z) + 2 \text{Li}_{2}(1+z) \log(1+z) - 2 \int \frac{\text{Li}_{2}(1+z)}{1+z} \ dz \\ &= \log^{2}(1+z) \log(-z) + 2 \text{Li}_{2}(1+z) \log(1+z) - 2 \text{Li}_{3}(1+z) + C \end{align} $$
Evaluating the integral at the limits and then simplifying a bit,
$$ \int_{0}^{\pi /3} \log^{2}\left( 2 \cos \frac{x}{2} \right) \ dx = \frac{7 \pi^{3}}{324} - \frac{\pi}{6} \log^{2}(3) + \log(3) \text{Im} \ \text{Li}_{2}(1+e^{i \pi /3}) + \frac{\pi}{3} \text{Re} \ \text{Li}_{2}(1+e^{i \pi /3}) $$
$$ - 2 \ \text{Im} \ \text{Li}_{3}(1+e^{i \pi /3}) \approx 0.439089177455491 .$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/751227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
If $a_{n+1}=1+\frac{1}{a_{1}+a_{2}+\cdots+a_{n}-1}$ then $0define the sequence $\{a_{n}\}$ and such
$$a_{1}=a,a_{n+1}=1+\dfrac{1}{a_{1}+a_{2}+\cdots+a_{n}-1},n\ge 1$$
Find the all real number $a>0$,such
$$0<a_{n}<1,n\ge 2$$
My try: since $$a_{2}=1+\dfrac{1}{a_{1}-1}=1+\dfrac{1}{a-1}$$
then
$$0<1+\dfrac{1}{a-1}<1$$
so
$$0<a<1$$
and $$a_{3}=1+\dfrac{1}{a_{1}+a_{2}-1}=1+\dfrac{1}{a+\dfrac{a}{a-1}-1}=\dfrac{a^2}{a^2-a+1}$$
if $$a_{3}\in(0,1)\Longrightarrow 0<a<1$$
so maybe I guess if $0<a<1$,then $a_{n}\in(0,1),n\ge 2$
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First, we try to simplify the given sequence:
$$a_{n+1} - 1 = \frac{1}{a_1 + \dots + a_n - 1}$$
$$a_1 + \dots + a_n - 1 = \frac{1}{a_{n+1} - 1}$$
$$a_1 + \dots + a_n = 1 + \frac{1}{a_{n+1} - 1}$$
Now, we do some subtraction for $n\ge2$:
$$(a_1 + \dots + a_{n-1} + a_n) - (a_1 + \dots + a_{n-1}) = \left(1 + \frac{1}{a_{n+1} - 1}\right) - \left(1 + \frac{1}{a_{n} - 1}\right)$$
$$a_n = \frac{1}{a_{n+1} - 1} - \frac{1}{a_n - 1}$$
$$\frac{1}{a_{n+1} - 1} = \frac{a_n^2 - a_n + 1}{a_n - 1}$$
$$a_{n+1} - 1 = \frac{a_n - 1}{a_n^2 - a_n + 1}$$
$$a_{n+1} = \frac{a_n^2}{a_n^2 - a_n + 1}$$
Now,
$$0 < a_{n+1} < 1$$
$$\iff 0 < \frac{a_n^2}{a_n^2 - a_n + 1} < 1$$
But $a_n^2 - a_n + 1$ is always positive for all real $a_n$ (consider its discriminant). Hence, we can multiply throughout to get:
$$0 < a_n^2 < a_n^2 - a_n + 1$$
From here, we can deduce that $a_n \not = 0, a_n < 1 \implies 0 < a_{n+1} < 1 \implies 0 < a_{n+2} < 1 \implies \dots$. As such, $a_2 \not = 0, a_2 < 1 \implies 0 < a_n < 1$ for all positive integers $n \ge 2$.
But $0<a_2<1 \implies a_2 \not = 0, a_2 < 1$. So it suffices to find the values of $a_2$ for which $0< a_2 < 1$. But $ a_2 = 1 + \frac{1}{a - 1}$. Hence,
$$0< 1 + \frac{1}{a - 1} < 1$$
$$-1 < \frac{1}{a-1} < 0$$
From $\frac{1}{a-1} < 0$, we have $$a - 1 < 0 \implies a < 1$$
We also have (from $-1 < \frac{1}{a-1}$):
$$-a + 1 > 1 \implies a < 0$$
To conclude, the valid range of $a$ is given by $a < 0$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/753010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}$ For positive real numbers with $a+b+c=abc$ prove that
$$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}$$
I made the substitution $a=\tan(\alpha), b = \tan(\beta), c= \tan(\gamma)$ with the constraint $\alpha+\beta+\gamma=\pi$
My inequality reduces to proving,
$$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma) \le\frac32$$
But I am stuck on it. Any help with either inequality would be appreciated.
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By AM-GM
$$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}\frac{1}{\sqrt{\frac{abc}{a+b+c}+a^2}}=\sum_{cyc}\sqrt{\frac{a+b+c}{a(a+b)(a+c)}}=$$
$$=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{a+b}\right)=\frac{3}{2}.$$
Done!
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/759592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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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.