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Taylor and Geometric Series questions

rmon

New member
Apr 22, 2012
3
I've spent all day on this problem and am wasting precious time needed for other work - please give any input you can! The question: given two wages, w1 and w2 where w2 > w1...

a. the difference between the wages as a proportion of the lower: a = (w2 - w1) / w2
b. the difference between the wages as a proportion of the higher: b = (w2 - w1)/w1
c. difference between the natural logs of the wages: c = lnw2 - lnw1

--- show that if b = y, than a = y + y^2 + y^3... and c = y + y^2/2 + y^3/3.....

***what I know (or think i know): the first is just a general geometric series, the second a taylor series. ive tried calculating the taylor series of c and seeing if it equals y + y^2/2... with "b" above plugged in for y. no discernible connection. is this how you would go about solving this problem? ive gotten embarrassingly little done for a days work on this problem, seem to be moving in circles. any help would be VERY much appreciated:)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
I've spent all day on this problem and am wasting precious time needed for other work - please give any input you can! The question: given two wages, w1 and w2 where w2 > w1...

a. the difference between the wages as a proportion of the lower: a = (w2 - w1) / w2
b. the difference between the wages as a proportion of the higher: b = (w2 - w1)/w1
c. difference between the natural logs of the wages: c = lnw2 - lnw1

--- show that if b = y, than a = y + y^2 + y^3... and c = y + y^2/2 + y^3/3.....

***what I know (or think i know): the first is just a general geometric series, the second a taylor series. ive tried calculating the taylor series of c and seeing if it equals y + y^2/2... with "b" above plugged in for y. no discernible connection. is this how you would go about solving this problem? ive gotten embarrassingly little done for a days work on this problem, seem to be moving in circles. any help would be VERY much appreciated
Hi rmon! :)

Your formulas are not quite right.
There are a few minus signs that seem to be missing.

To find out why, let's rewrite the expression for b = (w2 - w1)/w1:
y=(w2-w1)/w1
w2=w1(1+y)

Substitute into the expression for a = (w2 - w1) / w2.
You should find a=y/(1+y), which is almost the sum of the geometric series you mentioned, except for a number of minus signs.

Now substitute in the expression for c =
lnw2 - lnw1 = ln(w2/w1).
You should find that this is c=ln(1+y).
And the Taylor series of ln(1+y) is the series you mentioned, except for a number of minus signs.
 
Last edited:

rmon

New member
Apr 22, 2012
3
ILikeSerena:
Thank you SO much for your help, I really appreciate it:) From your reply I realized that I had completely forgot about substituting for w2 which seems silly in retrospect! I have one follow-up question: what do you mean by a number of minus signs? Oh, and I made things confusing by switching a and b when I wrote the question... Doing it your way with the correct substitutions I get the answer to the first question = y/(1-y) and the answer to the second as ln(l/l-y). You're right that the second is the sum of the taylor series expansion, the first does not seem to be.

Thank you again:)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
With your current results, there are no minus signs, but you'll get the series that you showed.

The first is the sum of a geometric series.
It can also be written as a Taylor expansion.
Writing it as a Taylor expansion means taking the derivative of y/(1-y) repeatedly.

Making a Taylor expansion becomes a bit easier if you rewrite y/(1-y)=1/(1-y) - 1.
Now you can take the derivative of 1/(1-y) repeatedly.
 

rmon

New member
Apr 22, 2012
3
Excellent, thanks!