Interesting Cal2 Questions

[homeskillet305]

My calculus2 professor gave us a list of problems and told us to get the answers by any means necessary (including online help) so if anyone would like to have a crack at any these I would greatly appreciate it!

 

[lookagain]

... (including online help) so if anyone would like to have a crack at any these I would greatly appreciate it!

*We* are not to have a crack at these. You are to have a crack at these, first,
by showing some work/attempts on them at this forum. Then, you can be critiqued.

 

[galactus]

Wow, these are rather challenging. Here is a hint on #2.

Note that \(\displaystyle tan(a/2)=cot(a/2)-2cot(a)\)

Thus, \(\displaystyle tan(a/2^{n})=cot(a/2^{n})-2cot(1/2^{n-1})\)

After hammering at it, you should see that the partial sum telescopes.

Look at the limit of the remaining term as \(\displaystyle n\to \infty\)

Try to get started and let me know how you proceed.

 

[homeskillet305]

thank you for the hint, i'll sit down with it later today and see where i get too.

 

[galactus]

For #6, think about the integral test for series.

Integrate \(\displaystyle \int_{1}^{1000000}\frac{1}{\sqrt{x}}dx\)

This should give you the integer part.


#7 is a classic. Do some searching around. There are various ways to approach it. Maybe try to find Euler's method of doing it.

 

[homeskillet305]

ok so
after the sums are done cancelling i'm left with (-2cot(2a))+(1/2^ncot(a/2^n)-1/2^(n-1)cot(a/2^(n-1))
correct?
so then the second term goes to zero when n->inf
and i'm left with -2cot(2a) correct so far or am i way off?

 

[galactus]

Well, here is how I would start it. Many times, the trick is finding the correct identity that will help.

Sometimes rather obscure identities can be implemented when finding the sum of an infinite series involving trig terms.

The nth partial sum is:

\(\displaystyle \frac{tan(a/2)}{2}+\frac{tan(a/4)}{4}+\cdot\cdot\cdot \frac{tan(a/2^{n})}{2^{n}}\)

Using the aforementioned identity, rewrite as:

\(\displaystyle \left(\frac{cot(a/2)}{2}-cot(a)\right)+\left(\frac{cot(a/4)}{4}-\frac{cot(a/2)}{2}\right)+\cdot\cdot\cdot +\left(\frac{cot(a/2^{n})}{2^{n}}-\frac{cot(a/2^{n-1})}{2^{n-1}}\right)\)

This series telescopes and all cancels out except for \(\displaystyle \frac{cot(a/2^{n})}{2^{n}}-cot(a)\)

Now, take the limit of the leftmost expression.

\(\displaystyle \lim_{n\to \infty}\frac{cot(a/2^{n})}{2^{n}}=\frac{1}{a}\)

So, we finally get \(\displaystyle \frac{1}{a}-cot(a)\)

But, one more little tidbit.

The given series starts at n=0, but the above approach started at n=1. So, we throw on a \(\displaystyle tan(a)\)

Thus, the sum of the series is \(\displaystyle \frac{1}{a}-cot(a)+tan(a)\)