Proving Continuity & Finding MacLorin Polynom of f(x)

  • Thread starter Thread starter TheForumLord
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around two problems related to continuity and series expansion. The first problem involves proving the continuity of a function defined as a summation of terms involving powers of x, specifically in the interval [0, 1]. The second problem focuses on finding the Maclaurin polynomial of a function defined by its derivative in relation to its value at zero.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the definition of continuity and whether it can be shown using properties of polynomials. There is a suggestion to evaluate derivatives at specific points to find the Maclaurin polynomial.

Discussion Status

Some participants have provided guidance on evaluating derivatives and the implications of polynomial continuity. There is an ongoing clarification regarding the notation used in the first problem, with some confusion about whether a summation is intended.

Contextual Notes

There is a noted misunderstanding regarding the notation for summation, which has led to some confusion in the discussion. Additionally, participants are addressing the need for clarity in mathematical communication and notation.

TheForumLord
Messages
104
Reaction score
0

Homework Statement



A. Let g(x)= \sigma\frac{1}{sqrt(n)}(x^{2n}-x^{2n+1})
Prove g(x) is continuous 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).

Homework Equations


The Attempt at a Solution



Have no idea about it...

Thanks in advance
 
Physics news on Phys.org
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.
 
TheForumLord said:
[b

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).

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.
 
Well, B is completely understandable...
About A->I need to show it's continuous 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!
 
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 by a moderator:
That's excatly what I can't understand...how can I find the eadius of con. in this specific case?


tnx
 
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.
 
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 occur to anyone...

I didn't know how to write Upper case sigma in Latex so please don't judge me...
 
TheForumLord said:
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 occur to anyone...
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.
TheForumLord said:
I didn't know how to write Upper case sigma in Latex so please don't judge me...
[ 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
Thanks
 
  • #11
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.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K