Hot answers tagged homework
3
I find it useful to plot the function after solving for the constant from the ICs.
We have:
$$y(x) = -\dfrac{8}{e^{4 x}-2}$$
A plot of $y(x)$ shows:
Of course, analytically, we care about where that denominator is $0$, so we have:
$$e^{4 x}-2 = 0 \rightarrow x = \dfrac{\ln 2}{4} \approx 0.173286795139986$$
From the plot, you can also see what happens ...
3
Take the $\ln$ of the bottom. We get $x\ln x$. Do it again. We get $\ln x+\ln\ln x$.
Take the $\ln$ of the top. We get $(\ln x)^{\ln x}\ln\ln x$. Do it again. We get $\ln x\ln\ln x+\ln\ln\ln x$.
The ratio of the $\ln\ln$'s is
$$\frac{\ln x\ln\ln x+\ln\ln\ln x}{\ln x+\ln\ln x}.\tag{1}$$
This $\to\infty$ as $x\to\infty$. To see that, divide top and bottom ...
3
First note that, we can evaluate the integral using the beta function
$$ \int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^m}= 2\int_{0}^{\infty}\frac{dx}{(1+x^2)^m}={\frac {\sqrt {\pi }\,\Gamma \left( m-\frac{1}{2} \right) }{\Gamma \left( m
\right) }},
$$
where $\Gamma(n)$ is the gamma function. Now, you can see that
$$ C = {\frac {\,\Gamma \left( m
...
2
Suppose there is $s \in (0,1)$ with $f'(s) > A/2$ and $f(s) \ge 0$. Then for
$t \in [s,1]$, $f'(t) = f'(s) + \int_s^t f''(x)\ dx > A/2 - A(t-s)$,
and $$f(1) = f(s) + \int_s^1 f'(x)\ dx > \int_s^1 (A/2 - A(x-s))\ dx =
A s (1-s)/2 > 0$$
which is impossible. Similarly in three other cases (in two of them you consider $f(0)$ rather than $f(1)$).
2
You have reduced it to row echelon form. You can go further to reduced row echelon form as follows:
$$
\begin{pmatrix}
1 & 0 & 1 & -2 &|&0 \\
0 & 1 & -2 &2 &|&0 \\
0 & 0 & 0 & 0 &|&0 \\
0 & 0 & 0 & 0 &|&0 \\ \end{pmatrix}
$$
You can read this matrix as:
...
2
The answer to the second question is "yes".
Consider the affine curves given by the equations $y^n-x$ over some field $k$. Their function field $F$ is rational: $F=k(x,y)=k(y)$. Hence the projective closure of all of these curves is birational to $\mathbb{P}^1_k$.
However these projective closures are pairwise non-isomorphic: for $n>1$ the curves possess ...
2
HINT: At this point you’re expected to know the binomial expansion:
$$(k+x)^n=\sum_{i=0}^n\binom{n}ik^{n-i}x^i\;.$$
The first four terms are the terms for $i=0,1,2,3$:
$$\binom{n}0k^n,\quad\binom{n}1k^{n-1}x,\quad\binom{n}2k^{n-2}x^2,\quad\text{and}\quad\binom{n}3k^{n-3}x^3\;.$$
Since the coefficients of $x^2$ and $x^3$ are equal, you know that
...
1
The series $$1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$$
is indeed divergent, because it has infinitely many partial sums equal to $1$, and infinitely many partial sums equal to $0$. Inserting parentheses as
...
1
I agree with you nearly up to this point:
$$ds \ = \ 3a \sin t \cos t \ , $$
but because we are taking the square-root $ \ \sqrt{\cos^2 t \ \sin^2 t} \ $ to obtain the infinitesimal arclength element , it would be better that we write the result as $ \ ds \ = \ 3a \cdot | \sin t \cos t \ | \ = \frac{3}{2} a \ | \sin 2t \ | \ . $
The curve itself is ...
1
For part a) you should check that the symmetry of $P(x)$ with respect to $x$ gives
$$\int_{-\infty}^{\infty}Ne^{-|x|/a} dx=2\int_{0}^{\infty}Ne^{-x/a}dx$$
and then do a change of variable $\displaystyle u=\frac{x}{a}$
For part b) note that by definition $\displaystyle\langle f(x) \rangle=\int_{-\infty}^{\infty}f(x)Ne^{-|x|/a}dx$ and that both $x$, $x^2$ ...
1
The notation in Dummit and Foote is actually pretty iffy. In the hint, they are referencing an earlier problem in which the setup is the same except that the $K_i$ are called $G_i$. Thus, you should interpret $g_i$ as an element of $K_i$.
If you would like a further hint as to the actual solution, feel free to ask.
1
For part i)...
i) For the maximum rate of change, try taking the gradient.
The gradient vector is $<2y^{1/2}, xy^{-1/2}>$.
The maximum rate of change will occur in the direction of $<2*(4)^{1/2}, 3*(4)^{-1/2}> = <4, 3/2>$.
The maximum rate of change is then $\sqrt {4^2 + (3/2)^{2}} = \sqrt {73}/2$
As for part ii)...
To find the ...
1
If $Y = -X$, $f_{-X}(-x) = f_X(x)$ is the same as $f_Y(t) = f_X(-t)$ where $t = -x$. Saying $X$ and $Y$ have the same distribution says that $f_Y(t) = f_X(t)$.
So this is exactly the same.
And how could you possibly argue with
$ {\mathbb P}(X(\omega)\le -x) = {\mathbb P}(X(\omega)\le -x)$?
1
Note that
$-2^2 = - (2 \times 2) = -4$ and is not $(-2) \times (-2) = 4$. Hence,
$$(-2^2)^3 = (-4)^3 = (-4) \times (-4) \times (-4) = - 64$$
In general, when $m$ is a positive integer, we have
$$-a^m = - (\underbrace{a \times a \times \cdots \times a}_{m \text{times}})$$
and is not
$$(-a)^m = (\underbrace{(-a) \times (-a) \times \cdots \times (-a)}_{m ...
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