# Taylor and Geometric Series questions

#### rmon

##### New member
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
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.

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#### rmon

##### New member
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
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
Excellent, thanks!