Welcome to NRICH.

 
Log Power Series
By Zhidong Leong on Sunday, July 29, 2001 - 04:48 am:

How do you express log in a power series (something like a0 + a1x + a2x2+...).

By Arun Iyer on Sunday, July 29, 2001 - 08:27 am:

The function I am writing is slightly different than the one you asked for.

log(1+x)=x-x2/2+x3/3-.....
valid only if -1<x<1

love arun

By Zhidong Leong on Sunday, July 29, 2001 - 12:35 pm:

Hmm....
But how do you get it?
Also, why are there a lot of these kinds of strange functions?

By Arun Iyer on Sunday, July 29, 2001 - 07:11 pm:

1/(1-y) = 1 + y + y2 + y3 + ......
for |y| < 1

Put y = -x
1/(1+x) = 1 - x + x2 - x3 + ......
Integrating both sides we get,
log(1+x)=x - x2/2 + x3/3 -.....upto + C (constant of integration)

We can easily see that when x = 0, C = 0.
Hence,
log(1+x)=x - x2/2 + x3/3 -....

What do you actually mean by strange functions?
love arun

By Zhidong Leong on Tuesday, July 31, 2001 - 01:12 pm:

I mean strange functions such as cos, sin, tan, log, etc
All these equal strange infinite series.
But why are these functions created?

By Arun Iyer on Tuesday, July 31, 2001 - 06:41 pm:

Are you familiar with calculus?

If you are then you would find it interesting to note that such infinite series are quite helpful in solving many tough integration problems.

You might come across them if you see the Onwards and Upwards section.
click here for one such discussion.

love arun

By Brad Rodgers on Tuesday, July 31, 2001 - 08:44 pm:

There is a way to expand nearly any given function into a series that converges for certain values. It's called Taylors expansion and rests upon the following assumptions:

(the following passage is copied straight from teach yourself calculus by P. Abbott)

"1. Any function which will be considered is capable of being expanded in this form

2. Subject in some cases to certain conditions, the series is convergent.

3. The successive derivatives, f'(x), f''(x),f'''(x),...,fn(x) all exist.

In accordance with assumption 1 assume that f(x+h) can be expanded in ascending powers of h as follows:

f(x+h)=A0+A1h+A2h2+A 3h3+... Equation 1

where the coeffecients A0, A1, A2, are functions of x but do not contain h.

Since this is to be true for all values of h, let h=0.

Then on substitution, you find A0=f(x)

Since the series in Equation is an identity, you can assume that if you differentiate both sides with respect to h, keeping x constant, the result in each case will be another identity.

Repeating the process, you find:

(i) f'(x+h)=A1+2A2h+3A3h2+4A 4h3+...

(since f'(x)=0 when you differentiate with respect to h, as x is kept constant.

Similarly:

(ii) f''(x+h)=2A2+3*2A3h+4*3A4h2+...

(iii) f'''(x+h)=3*2A3+4*3*2A4h+...

and so on for higher derivatives.

In all these results put h=0. Then:

(from (i)) f'(x)=A1
(from (ii)) f''(x)=2A2
(from (iii)) f'''(x)=3*2A3
.
.
.

Which leads to

A1=f'(x)
A2=f''(x)/2!
A3=f'''(x)/3!
.
.
.
An=fn(x)/n!

and so on.

Substituting for these in Equation 1 you obtain Taylor's expansion:

f(x+h)=f(x)+h*f'(x)/1!+h2f''(x)/2!+h3f'''(x)/3!+...+h nfn(x)/n!

"

This equation will hold so long as the assumptions made hold, which (for a certain range of values) is almost always.

[Editor: For a Taylor expansion of the form a0 + a1x + a2x2+.. to exist for a particular function it is necessary that the function be defined for all small enough real numbers. In this particular case, log0 is not defined so no Taylor expansion exists at the origin. If it did we would have a0=log0, which would be nonsense!]