maths 1B last minute questions (1 Viewer)

InteGrand · Oct 31, 2016

Drsoccerball said:
Okay... I have questions about convergence and divergence.

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?

By the way, in a conditionally convergent series, the sum of positive terms in the series will diverge and the sum of negative terms will also diverge.

Drsoccerball · Oct 31, 2016

InteGrand · Oct 31, 2016

Drsoccerball said:
$Find whether each of the following series converge or diverge$
$$ a) $ \sum_{k = 1}^{\infty}{\sin^2{\frac{1}{k}}}$
$$ b) $ \sum_{k = 1}^{\infty}{\frac{\ln{k!}}{k^3}}$
$$ c) $ \sum_{k = 1}^{\infty}{\frac{(\ln{k})^3}{}{\sqrt{k^3-3k^2 +1}}}$

$\noindent a) Convergent. The summand is asymptotically equivalent to $\frac{1}{k^{2}}$ as $k \to \infty$, and $\sum \frac{1}{k^{2}}$ converges. Hence the given series converges.$

$\noindent b) Convergent. Let $a_{k} = \frac{\ln k!}{k^3}$. Note $a_{k}\ge 0$ for all $k\geq 1$. Also, $k! \leq k^k$, so $\ln k! \leq k\ln k$. So $a_{k}\leq \frac{k \ln k}{k^{3}} = \frac{\ln k}{k^2}$. But $\sum \frac{\ln k}{k^{2}}$ converges (hint to show this: compare it to a $p$-series, noting for example that $\ln k < k^{\frac{1}{2}}$ for all $k$ sufficiently large), so by the comparison test, $\sum a_{k}$ converges.$

$\noindent c) Again convergent. Let the summand be $a_{k}$ (note $a_k \geq 0$ for all $k$). Note in fact that $\ln k$ becomes smaller than \emph{any} fixed (positive) power of $k$ for all $k$ sufficiently large. So $\left(\ln k\right)^{3} \leq k^{\frac{1}{6}}$ (say) for all $k$ sufficiently large. So $0\leq a_k \leq b_k := \frac{k^{\frac{1}{6}}}{\sqrt{k^3 -3k^2 + 1}}$. But $\sum b_{k}$ converges, so by comparison, $\sum a_{k}$ converges. (Note $\sum b_{k}$ converges because the denominator is asymptotically equivalent to $k^{\frac{3}{2}}$ and $\sum \frac{k^{\frac{1}{6}}}{k^{\frac{3}{2}}}$ converges.)$

InteGrand · Oct 31, 2016

leehuan · Oct 31, 2016

InteGrand said:
Yeah, I wrote that it was trace.

Whoops, went blind there

Drsoccerball said:
$Find whether each of the following series converge or diverge$
$$ a) $ \sum_{k = 1}^{\infty}{\sin^2{\frac{1}{k}}}$
$$ b) $ \sum_{k = 1}^{\infty}{\frac{\ln{k!}}{k^3}}$
$$ c) $ \sum_{k = 1}^{\infty}{\frac{(\ln{k})^3}{}{\sqrt{k^3-3k^2 +1}}}$

Try to put a bit of intuition into it before you apply a test. Like IG said, for those ones involving logs since you have algebraic expressions there try to think about what they do. Because logs are 'dominated' by algebraic

Drsoccerball · Oct 31, 2016

So it didnt even matter what was in the numerator with ln's ? Youd still replace it wth something such that the bottom degree - top degree = something greater than 1? Isnt this fudging it?

Drsoccerball · Oct 31, 2016

Find the radius of convergence of $\sum{\frac{(\ln{(x)}x)^k}{k^k}}$

InteGrand · Oct 31, 2016

Drsoccerball said:
Find the radius of convergence of $\sum{\frac{(\ln{(x)}x)^k}{k^k}}$

Have you tried looking at the ratio test limit?

InteGrand · Oct 31, 2016

It'll converge (absolutely in fact) for any x > 0, because for any fixed x, the denominator k will eventually become bigger than |x*ln(x)| and the series can then be compared to a GP with common ratio smaller than 1. All the terms in the given series will eventually become less than GP terms with common ratio smaller than 1, so it'll converge.

So radius of convergence is infinite.

InteGrand · Oct 31, 2016

Drsoccerball said:
So it didnt even matter what was in the numerator with ln's ? Youd still replace it wth something such that the bottom degree - top degree = something greater than 1? Isnt this fudging it?

Which part is fudging? Note that for all k sufficiently large, (ln k)^3 < k^{1/6}, so the summand is smaller than something asymptotic to 1/(k^{3/2 – 1/6}), and the sum of 1/(k^{3/2 - 1/6}) converges by the p-test.

InteGrand · Oct 31, 2016

If you meant it didn't matter exactly what polynomial in ln(k) was in the numerator, then yes, that's right, it'd work with any polynomial in ln(k) (or any finite sum of powers of ln(k)), as I explained in the later post. It's not fudging because it's something we can prove.

Drsoccerball · Oct 31, 2016

InteGrand said:
It'll converge (absolutely in fact) for any x > 0, because for any fixed x, the denominator k will eventually become bigger than |x*ln(x)| and the series can then be compared to a GP with common ratio smaller than 1. All the terms in the given series will eventually become less than GP terms with common ratio smaller than 1, so it'll converge.

So radius of convergence is infinite.

This makes sense but when we do it by force we get |xlnx| < e and this isnt true for all x? ( what i did was ratio test is less than 1)

Edit: nvm got it was missing a k+1 in the denominator

InteGrand · Oct 31, 2016

Drsoccerball said:
This makes sense but when we do it by force we get |xlnx| < e and this isnt true for all x? ( what i did was ratio test is less than 1)

$\noindent If $x \ln x = 0$, we obviously have convergence (as the series is just summing $0$ then), so assume now $x \ln x \neq 0$. Let $a_{k} = \left( \frac{x \ln x}{k}\right)^{k}$, then$

$\begin{align*}\left|\frac{a_{k+1}}{a_{k}}\right| &= |x \ln x| \frac{k^k}{(k+1)^{k+1}} \\ &= |x \ln x| \frac{1}{k+1}\times \left(\frac{k}{k+1}\right)^{k}\\ &\to 0 \quad \text{as } k \to \infty,\end{align*}$

$\noindent because of the $\frac{1}{k+1}$ factor.$

Edit: OK.

Drsoccerball · Oct 31, 2016

InteGrand · Oct 31, 2016

Drsoccerball said:
$$ Find the length of the cardioid $ r = 1 + \cos{\theta}$

See here for an example etc. for how to find arc length in polar coordinates: http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx .

$\noindent The length is given by $L = \int \, \mathrm{d}{s} = \int_{0}^{2\pi} \sqrt{r^{2} + \left( \frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}}\, \mathrm{d}\theta$, where $r \equiv r(\theta) = 1+\cos \theta$.$

Drsoccerball · Oct 31, 2016

InteGrand said:
See here for an example etc. for how to find arc length in polar coordinates: http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx .

$\noindent The length is given by $L = \int \, \mathrm{d}{s} = \int_{0}^{2\pi} \sqrt{r^{2} + \left( \frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}}\, \mathrm{d}\theta$, where $r \equiv r(\theta) = 1+\cos \theta$.$

When you evaluate the integral you get 0 ?

InteGrand · Oct 31, 2016

Drsoccerball said:
When you evaluate the integral you get 0 ?

No, you did same thing as Mr_Kap haha. Need to use absolute values.

Drsoccerball · Oct 31, 2016

InteGrand said:
No, you did same thing as Mr_Kap haha. Need to use absolute values.

ahahahaha

Drsoccerball · Oct 31, 2016

InteGrand said:
No, you did same thing as Mr_Kap haha. Need to use absolute values.

$r^2 = 1 + 2\cos{\theta} + \cos^2{\theta}, \frac{dr}{d\theta} = -\sin{\theta}$

$I = \int_0^{2 \pi}{\sqrt{2 + 2\cos{x}}}dx$

$I = \sqrt{2} \int_0^{2 \pi}{\sqrt{1 + \cos{x}}}dx$

$I = \sqrt{2} \int_0^{2 \pi}{\frac{\sin{x}}{\sqrt{1 - \cos{x}}}}dx$

$I = 2\sqrt{2} [-\sqrt{1- \cos{x}}]_0^{2 \pi}$
?

Edit: Imm guessing you mean that |-cosx| but why ????????/

InteGrand · Oct 31, 2016

Drsoccerball said:
$r^2 = 1 + 2\cos{\theta} + \cos^2{\theta}, \frac{dr}{d\theta} = -\sin{\theta}$

$I = \int_0^{2 \pi}{\sqrt{2 + 2\cos{x}}}dx$

$I = \sqrt{2} \int_0^{2 \pi}{\sqrt{1 + \cos{x}}}dx$

$I = \sqrt{2} \int_0^{2 \pi}{\frac{\sin{x}}{\sqrt{1 - \cos{x}}}}dx$

$I = 2\sqrt{2} [-\sqrt{1- \cos{x}}]_0^{2 \pi}$
?

Edit: Imm guessing you mean that |-cosx| but why ????????/

$\noindent We need to integrate something like $\sqrt{1+\cos x}$. To do this, we can recall that $1+\cos x = 2\cos^{2}\frac{x}{2}$. Taking square roots of both sides yields $\sqrt{1+\cos x} = \sqrt{2}\left| \cos \frac{x}{2}\right|$. So the function we'll end up integrating is (a multiple of) $\left|\cos \frac{x}{2}\right|$. Why absolute values? Because $\sqrt{a^2} = |a|$.$

maths 1B last minute questions (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)