Calculues Questions

1. Feb 12, 2010

TheForumLord

1. The problem statement, all variables and given/known data

A. Let $$g(x)= \sigma\frac{1}{sqrt(n)}(x^{2n}-x^{2n+1})$$
Prove g(x) is continous in [0,1].

B. Let f be a function such as f(0)=1 and there's a neighouhood of x=0 in which :
$$f ' (x)= 1+(f(x))^{10}$$ .
Find the MacLorin Polynom of degree 3 of f(x).

2. Relevant equations
3. The attempt at a solution

2. Feb 12, 2010

Staff: Mentor

For A you need to show that g is continuous at each point in [0, 1]. What's the definition of continuity of a function at a point? Do you have to do this by using the definition of continuity or can you use the fact that this is a polynomial and all polynomials are everywhere continuous?

For B you need f(0), f'(0), f''(0), and f'''(0). You already are given that f(0) = 1, and you have f'(x), which you can evaluate at x = 0.

To find f''(0), you need to find f''(x), which you can do by differentiating f'(x), and then evaluate f''(x) at x = 0.
To find f'''(0), differentiate f''(x), and then evaluate at x = 0.

3. Feb 12, 2010

LCKurtz

That's "Maclaurin Polynomial".

Well, don't you just need f(0), f'(0), f''(0) and f''(0) to calculate that series? You are given formulas for f(0) and f'(x). You must need a couple more derivatives. You might need the chain rule. Show us your derivatives.

4. Feb 13, 2010

TheForumLord

Well, B is completely understandable...
About A->I need to show it's continous using power-series theorems...If I'll prove that the given power-series is convergeing uniformly to g - I'll be done...I've no idea about it...
I'll be delighted to get some help

Thanks!

5. Feb 13, 2010

HallsofIvy

Staff Emeritus
Power series? g(x) is a polynomial! You don't need to worry about any power series or convergence! A is not asking about a limit as n goes to infinity is it? It is just about a single polynomial for a fixed value of n.

Or was that $\sigma$ supposed to be $\Sigma$? That is, is this a sum over all n? In that case, because it is a power series, it converges uniformly inside its radius of convergence. You need only show that its radius of convergence includes [0, 1].

Last edited: Feb 13, 2010
6. Feb 13, 2010

TheForumLord

That's excatly what I can't understand....how can I find the eadius of con. in this specific case?

tnx

7. Feb 13, 2010

Staff: Mentor

TheForumLord,
A fair amount of time has been wasted because we didn't understand what you were trying to communicate. It has now come to light that your first problem problem is a summation. The Greek alphabet has upper case letters and lower case letters. In particular, upper case sigma, $\Sigma$, is used to represent summations. Lower case sigma, $\sigma$, is used in statistics to represent the population standard deviation. I interpreted $\sigma$ in this problem as a constant. It didn't occur to me that you really meant a summation.

Also, at this stage of your mathematical education, you really ought to learn how to spell "calculus." It's clear to me that you're not likely to be in the finals of a math spelling bee, but at least get calculus right.

8. Feb 13, 2010

TheForumLord

Dear Mark44... My english is pretty lame indeed but in this particular case, writing calculus in a wrong way was just a typing mistake - which can occure to anyone....

I didn't know how to write Upper case sigma in Latex so plz don't judge me...

9. Feb 13, 2010

Staff: Mentor

Sure, anyone can make a typo, but you can eliminate at least some of them by looking over what you've written before you hit the submit button.
[ tex] \sigma[/tex] or [ itex] \sigma[/itex] (without the leading space) produces $\sigma$.
[ tex] \Sigma[/tex] or [ itex] \Sigma[/itex] (without the leading space) produces $\Sigma$.

Same for all the rest of the Greek letters.

10. Feb 13, 2010

TheForumLord

Thanks

11. Feb 14, 2010

HallsofIvy

Staff Emeritus
Better is \sum :
$$\sum$$.

By the way, just clicking on a formula in any post will show you the LaTex code used for it.

Your series can be written by separating the even and odd powers- it is $\sum a_mx^m$ with
$$a_m= \sqrt{\frac{2}{m}}$$
if m is even and
$$a_m= \sqrt{\frac{2}{m-1}}$$
if m is odd.
As for finding the radius of convergence, using the ratio test gives
$$|x|< \sqrt{\frac{m+1}{m}}$$
if n= 2m and
$$|x|< \sqrt{\frac{m}{m-1}}$$
if n= 2m+1

What is the limit of those as n goes to infinity?

Of course, you will need to check if the sum converges at x= 1 but that is easy.