Challenging Summations, Limits, and Derivatives

[Mandelbroth]

I'm going to talk with someone at a local university tomorrow to see if I can get out of AP Calculus. Essentially, I would like to be prepared for our meeting tomorrow. I'm good with integrals, so that shouldn't be a problem. However, I'm not quite as confident with derivatives, limits, and sums.

I'm aware of a similar thread for integrals, but upon cursory inspection there isn't a corresponding thread for these three. I would therefore like to ask the community here for some challenging problems for summations, limits, and derivatives, so I can prove to myself that I'm prepared.

Guidelines for problems:
I'm not partial to any specific "year" of calculus (and I don't make that distinction very well anyway), so any given difficulty level is fine. I'm looking mostly for calculation-type problems, but I like proofs too. I'm mainly looking for something to convince me I'll do fine on any problem he'll give me tomorrow.

 

[Office_Shredder]

Prove that |x| is not differentiable at x=0.
Calculate the derivative of f(x) = x² directly using the limit definition of the derivative
Find algebraically (not using a calculator)the intervals on which the following function is concave up and concave down:
$$f(x) = e^{-x^2}$$

Calculate the following limits
$$\lim_{x \to \infty} \frac{ \sqrt{1+ x + 3x^2}}{1-2x}$$
$$\lim_{x\to 0} \frac{\cos(x)-1}{x^2}$$

Do the following sums converge?
$$\sum_{n=1}^{\infty} \frac{1}{n!}$$
$$\sum_{n=1}^{\infty} \frac{ (2n)!}{(n!)^2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}$$

Find the radius of convergence of the following series, and decide whether it converges at the endpoints of the interval of convergence
$$\sum_{n=1}^{\infty} \frac{x^n}{ \sqrt{n}}$$

 

[lurflurf]

Ap calculus does not cover any difficult sums. Do you know

$$
\sum_{k=0}^n k(k+1)(k+2) \\
\sum_{k=0}^\infty x^k \\
\sum_{k=0}^\infty x^k/k! \\
\sum_{k=0}^\infty \sin(a+k \, \pi/2) x^k/k! \\
\sum_{k=0}^\infty x^{2k+1}/(2k+1)! \\
\sum_{k=1}^\infty x^k/k \\
\text{These while not hard are beyond any reasonable AP calculus expectation} \\$$$$\\
\sum_{k=0}^\infty (-)^k E_{2k} x^{2k}/(2k)! \\
\sum_{k=0}^\infty x^{3k}/(3k)! \\
\sum_{k=0}^\infty k^3 \, x^k $$

Also know some sums converge without needing to know the value like
$$\sum_{k=1}^\infty k^{-k} $$
I would say "It converges because bottom big one."

 

[Mandelbroth]

Prove that |x| is not differentiable at x=0.
Calculate the derivative of f(x) = x² directly using the limit definition of the derivative

Using limit definitions, the limit of the difference quotient for $|x|$ is not (singularly) defined.
The limit of the difference quotient for $x^2$ is somewhat messy, but obviously comes out to $2x$. I've done both of these before, so I won't go into detail here.

Find algebraically (not using a calculator)the intervals on which the following function is concave up and concave down:
$$f(x) = e^{-x^2}$$

$D_x^2[e^{-x^2}]=e^{-x^2}(4x^2-2)$. Noting that the roots of this equation are of odd multiplicity, we say that the roots of the equation, $x\in\{\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\}$, are inflection points.

I'll finish the rest later. Thank you very much.

 

[SteamKing]

I don't understand the OP. I thought one took AP calculus (at least the test) to earn college credit and skip the introductory course in the first semester of college. If the OP is not clear on some of the fundamentals, it seems to me that he would benefit from taking additional course work in the subject, not less. It does not bode well to skip the fundamentals when the OP encounters more advanced material.

 

[lurflurf]

^So every time you forget something you go register for a class on it? That would be a waste of time. The OP just wants to make sure nothing has been missed. Most people have a few gaps that need filling. Often the next class is a good time to do so.

 

[verty]

Find these derivatives:

$f(x) = ln( \sqrt{x - 2} + \sqrt{x} )$
$f(x) = e^{tanh(x)} (cosh^2(x) - 1)$
$f(x) = sinh^{-1}(cos(x^2))$

I'm not sure if hyperbolic ratios count.

 

[SteamKing]

^So every time you forget something you go register for a class on it? That would be a waste of time. The OP just wants to make sure nothing has been missed. Most people have a few gaps that need filling. Often the next class is a good time to do so.

No, that's not my point. As I said, AP calculus, AFAIK, is usually taken by folks in high school so they can get college credit and skip an introductory class when they first go to college. The OP was not entirely clear on his purpose in skipping this class when he clearly did not understand the basics. Is the fellow he is meeting going to give him a quiz? Are they just going to shoot the breeze about calculus? It's not clear to me.

 

[Mandelbroth]

$$\sum_{n=1}^{\infty} \frac{1}{n!}$$

Gee. This sum is pretty e-mazing, huh?

$$\sum_{n=1}^{\infty} \frac{ (2n)!}{(n!)^2}$$

Using Stirling's asymptotic formula for the Gamma function, we set up the limit $$\lim_{n\rightarrow +\infty}[\frac{2(\frac{2n}{e})^{2n}\sqrt{\pi n}}{2\pi n (\frac{n}{e})^{2n}}]=\lim_{n\rightarrow +\infty}[\frac{4^n}{\sqrt{\pi n}}]\neq 0$$
The series does not converge.

$$\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}$$

This is just $\zeta(\frac{3}{2})$.

Find the radius of convergence of the following series, and decide whether it converges at the endpoints of the interval of convergence
$$\sum_{n=1}^{\infty} \frac{x^n}{ \sqrt{n}}$$

This is helpful. I forgot these kinds of problems existed. Thankfully, I remember that it kind of goes like this:
$$r=\frac{1}{\limsup_{n\rightarrow +\infty}\sqrt[n]{\frac{1}{\sqrt{n}}}}=\frac{1}{\limsup_{n\rightarrow +\infty}\frac{1}{\sqrt[2n]{n}}}=1$$
The sum doesn't converge at the endpoints (id est, the series converges on the open interval $(-1,1)$). This is shown by the comparison test (I used the harmonic series).

Find these derivatives:

$f(x) = ln( \sqrt{x - 2} + \sqrt{x} )$
$f(x) = e^{tanh(x)} (cosh^2(x) - 1)$
$f(x) = sinh^{-1}(cos(x^2))$

I'm not sure if hyperbolic ratios count.

Again, I'm not partial. I think I'll just do one of these though.

I hope you don't mind, but I prefer writing hyperbolic trig functions in terms of the normal trig functions, so I'll be ready for the words "Suppose we have an angle $h$..."

So, we have $f(x)=e^{\frac{\tan(ix)}{i}}(\cos^2(ix)-1)$, where the $\cos^2(\cdot)$ denotes squaring and not $\cos(\cos(\cdot))$. Then, we have $f'(x)= e^{\frac{\tan(ix)}{i}}D_x[\cos^2(ix)]+\cos^2(ix) D_x[e^{\frac{\tan(ix)}{i}}]-D_x[e^{\frac{\tan(ix)}{i}}]$, and we also can verify $D_x[e^{\tan(ix)/i}]=\sec^2(ix)e^{\tan(ix)/i}$ by the chain rule. For $\cos^2(ix)$, we get $D_x[\cos^2(ix)]=\frac{\sin(2ix)}{i}$. Substituting these in, we obtain the result: $$f'(x)= e^{\frac{\tan(ix)}{i}}\frac{\sin(2ix)}{i}+\cos^2(ix) \sec^2(ix)e^{\tan(ix)/i}-\sec^2(ix)e^{\tan(ix)/i}=e^{\frac{\tan(ix)}{i}}(\frac{\sin(2ix)}{i}-\sec^2(ix)+1).$$

The OP was not entirely clear on his purpose in skipping this class when he clearly did not understand the basics.

I'm not sure if this is meant as an insult or not. I'll assume it isn't meant that way, but I'd like to point out that I'm offended that you would say it. If you've seen some of the other work I do around here (misconceptions about uniform probability distributions over natural numbers aside ), I think you'd agree that I understand "the basics." If I didn't "clearly" understand the basics, this option wouldn't be open to me in the first place. The university official in question has decided that I probably don't need the introductory course, and thus is offering me a way out. I'm not a very confident person, though. I just want to be sure of my abilities. This kind of comment does not help.

The concept of the meeting later today is to talk about math (because math is fun) and then, to my understanding, take a final exam for the university's calculus course to prove I know the material. Then, by some hocus pocus, I'll get college credit for the class and be allowed to take something more entertaining (my intention is to take Calculus 3 or Differential Equations).

 

[WannabeNewton]

You don't have to take the class you know. You can just take the BC exam, no need for a year long BC class.

 

[verty]

So, we have $f(x)=e^{\frac{\tan(ix)}{i}}(\cos^2(ix)-1)$, where the $\cos^2(\cdot)$ denotes squaring and not $\cos(\cos(\cdot))$. Then, we have $f'(x)= e^{\frac{\tan(ix)}{i}}D_x[\cos^2(ix)]+\cos^2(ix) D_x[e^{\frac{\tan(ix)}{i}}]-D_x[e^{\frac{\tan(ix)}{i}}]$, and we also can verify $D_x[e^{\tan(ix)/i}]=\sec^2(ix)e^{\tan(ix)/i}$ by the chain rule. For $\cos^2(ix)$, we get $D_x[\cos^2(ix)]=\frac{\sin(2ix)}{i}$. Substituting these in, we obtain the result: $$f'(x)= e^{\frac{\tan(ix)}{i}}\frac{\sin(2ix)}{i}+\cos^2(ix) \sec^2(ix)e^{\tan(ix)/i}-\sec^2(ix)e^{\tan(ix)/i}=e^{\frac{\tan(ix)}{i}}(\frac{\sin(2ix)}{i}-\sec^2(ix)+1).$$

This is a very nice method, I haven't seen it before. I have learned something today. This avoids having to learn things like sinh(2x) = 2 sinh(x) cosh(x), etc.