MHB Taylor and Geometric Series questions

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The discussion revolves around solving a problem involving the differences between two wages, w1 and w2, expressed as proportions and natural logarithms. The user initially struggles with the relationships between these differences and their representation as geometric and Taylor series. A key insight is that substituting w2 in terms of w1 and y reveals the correct forms of the series, correcting earlier mistakes in the user's calculations. The conversation highlights the importance of proper substitutions and the connection between geometric series and Taylor expansions. Ultimately, the user gains clarity on how to approach the problem with the right formulas and methods.
rmon
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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. I've 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? I've gotten embarrassingly little done for a days work on this problem, seem to be moving in circles. any help would be VERY much appreciated:)
 
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rmon said:
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. I've 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? I've 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:
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:)
 
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.
 
Excellent, thanks!
 
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