7
We start by writing
\begin{align}
I &= \int \limits_0^1 \frac{1-2x}{1+x -x^2} \log(1+x) [\log(1+x)-4\log(x)] \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \left[\log^2\left(\frac{x^2}{1+x}\right) - 4 \log^2(x)\right] \mathrm{d} x \equiv J -4K \, .
\end{align}
Then
\begin{align}
J &= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \log^2\left(\...
5
That looks correct. It becomes a tiny bit simpler if you expand the fraction with $e^x$ instead of expanding with $e^{-x}$:
$$
\frac{1+e^x}{1+e^{-x}} = \frac{1+e^x}{1+e^{-x}} \frac{e^x}{e^x} = \frac{1+e^x} {e^x+1} e^x = e^x
$$
4
You have
$$\begin{aligned}
f(\epsilon)=\ln (1-\ln(\varepsilon)) &= \ln \left(\frac{-\ln(\varepsilon)}{-\ln(\varepsilon)}(1-\ln(\varepsilon))\right) = \ln(-\ln(\varepsilon)) + \ln \left(1 - \frac{1}{\ln \varepsilon} \right)\\
&= \ln(-\ln(\varepsilon)) - \frac{1}{\ln \varepsilon} - \frac{1}{2\ln^2 \varepsilon}- \frac{1}{3\ln^3 \varepsilon} -\dots
\end{...
4
Yes, it's correct, except that you need to ensure $x>0$ and $5x-2>0$. This is implicit in the question, since the logarithm only takes (strictly) positive arguments. So solving the quadratic equation you got, you will end up with two answers, but one of them will be negative and hence needs to be rejected.
4
It's wrong. Try $x=1$ and $y=0.05.$
In this case $RHS-LHS=-0.047...<0.$
4
Let $H_n=\sum_{k=1}^n{\frac{1}{k}}$, we have
$$ \sum_{k=0}^{n} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=(H_{3n+3}-1)-\sum_{k=1}^n{\frac{3}{3k+3}}=(H_{3n+3}-1)-(H_{n+1}-1) $$
Since $H_n=\ln n +\gamma+o(1)$ you have
$$ \sum_{k=0}^{n} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=\ln(3)+o(1) $$
and thus $$ \sum_{k=0}^{+\infty} \left(...
3
Most likely it is a typo, as using different notation $\ln$ and $\log$ in the same expression is bad style in my view.
However, it is not incorrect. The default base for the logarithm depends on the context, and very often it will be $2, 10$ or $e$.
2
Since $-1\le \cos x\le 1$ and $e^x>0$ with $e^x>1, \; x>0$, by IVT we have infinitely many solutions for $x\le 0$, one is the trivial $x=0$ the others can by found by numerical methods.
Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<\cos x<1$ that is for $n=0,1,2,\dots$ that is
$$-\frac{\pi}2 -2\...
2
We can use that $\log_AB=\frac{\log B}{\log A}$ therefore
$$x\log_bx=x\frac{\log x}{\log b}\to 0$$
2
It's at least inconsistent, yes. That being said, in many textbooks $\log x$ actually means the natural logarithm.
2
HINT
Let use $\log_AB=\frac{\log B}{\log A}$ to obtain
$$\frac{(\log x)^3}{\log2\cdot\log3\cdot\log5}=\dots$$
1
Hint: I would first put the ODE in standard form
$$\frac{dy}{dt} + P(t)y = Q(t)$$
which in your case is
$$\frac{dy}{dt} + \frac{\ln t}{t-6}y = \frac{4t}{t-6}$$
and then observe the intervals in which $P(t)$ and $Q(t)$ are continuous. Note that both $P(t)$ and $Q(t)$ are undefined at $t=6$.
After you find the intervals of continuity for $P(t)$ and $Q(t)$,...
1
Just find the intervals where the coefficients and right side of the normalized equation
$$
y'+\frac{\ln t}{t-6}y=\frac{4t}{t-6}
$$
exist and are all continuous. Then pick the one that contains $t=1$.
1
As suggested above, it is not technically incorrect. You can always check the Errata (Google 'textbook name': Errata) - if one exists to see if this was an error.
1
Here is an entirely different way to do it. We have $$abc=ab+bc+ca\neq 0$$ so divide by $abc$ to obtain $$\frac 1a+\frac 1b+\frac 1c=1$$
We also have $2^a=3^b=5^c=x$ by the properties of logarithms so that $x^{\frac 1a}=2$ etc
Now consider $x^{\left(\frac 1a+\frac 1b+\frac 1c\right)}$
1
Using
$$\log_{b}(x) = \frac{\ln(x)}{\ln(b)}$$
then
$$\log_{2}(x) \, \log_{3}(x) \, \log_{5}(x) = \log_{2}(x) \, \log_{3}(x) + \log_{2}(x) \, \log_{5}(x) + \log_{3}(x) \, \log_{5}(x)$$
becomes
$$\frac{\ln^{3}(x)}{\ln2 \, \ln3 \, \ln5} = \frac{\ln^{2}(x)}{\ln2 \, \ln3} + \frac{\ln^{2}(x)}{\ln2 \, \ln5} + \frac{\ln^{2}(x)}{\ln3 \, \ln5}$$
and reduces to
$$ \ln(...
1
We can use the fact of logarithms that as $\log_BA = \frac{log A}{log B}$
then expression $\lim_{b \to \infty} x\log_b(x)$ can be rewritten as
$\lim_{b \to \infty} \frac{x\log x}{log b}$ and as $b \to \infty$ then $\log b \to \infty$ and hence making $\frac{1}{\log b} \to 0$ and in
$\lim_{b \to \infty} \frac{x\log x}{log b}$ b is the variable which is ...
1
Since you can choose any base for log, use base 4. Then $\log_4(1/4)=-1$. Then we know that $1/4<1/3<1/2$, so $\log_4(1/4)<\log_4(1/3)<\log_4(1/2)$ or $-1<\log_4(1/3)<-0.5$. Can you take it from here?
1
Note that\begin{align}\log(xy)-x=C&\iff\log(xy)=x+C\\&\iff xy=e^xe^C\\&\iff y=e^C\frac{e^x}x.\end{align}So, your level curve is just the graph of the map $x\mapsto e^C\dfrac{e^x}x$.
1
You could use numerical methods
You could use maclaurin's expansion
$e^x-\cos x = x+x^2+\frac{x^3}{6}$
Upto 3 degree would be good
Now this becomes a cubic equation
$x=0$
Now this becomes a quadratic equation
$1+x+\frac{x^2}{6}$
Roots are -4.725,-1.281
For large negative x $e^{-x} \approx 0$ so $\cos x = 0$ which gives $x \approx -(2n+1)\frac{\pi}{2}...
1
Let us go with brute force:
$$ \frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3} = \int_{0}^{1}x^{3k}(1+x-2x^2)\,dx $$
leads to:
$$ \sum_{k\geq 0}\left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=\int_{0}^{1}\frac{1+x-2x^2}{1-x^3}\,dx=\int_{0}^{1}\frac{2x+1}{x^2+x+1}\,dx $$
where the RHS equals
$$ \left[\log(x^2+x+1)\right]_{0}^{1} = \color{red}{\log 3} $$...
1
In general, we have
$$\sum_{k=0}^\infty(\frac{1}{nk+1}+\frac{1}{nk+2}+\frac{1}{nk+3}+\dots+\frac{1}{nk+n-1}-\frac{n-1}{nk+n})=\ln(n)$$ for any integer $n>1$.
Check for $n=3$, you will get the result.
Check my related post.
1
Define $f:\mathbb{R}\setminus\{\pm1,\pm3\}\to\mathbb{R}$ by $$f(x):=\frac{1}{\log_2\big((x-2)^2\big)}+\frac{1}{\log_2\big((x+2)^2\big)}$$ for all real numbers $x\neq \pm1,\pm2,\pm3$, and $f(\pm 2)$ is defined to be $\dfrac14$. Note that $f$ is a continuous even function (i.e., $f(-x)=f(x)$). Therefore, it suffices to solve $f(x)=y$ for $x\geq 0$.
...
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