Taylor and Geometric Series questions

In summary: I was getting a bit lost trying to do it that way.In summary, the difference between the wages as a proportion of the lower is a = (w2 - w1) / w2. The difference between the wages as a proportion of the higher is b = (w2 - w1)/w1. The difference between the natural logs of the wages is c = lnw2 - lnw1.
  • #1
rmon
3
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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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  • #2
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.
 
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  • #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:)
 
  • #4
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.
 
  • #5
Excellent, thanks!
 

FAQ: Taylor and Geometric Series questions

What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It is used to approximate a function with a polynomial, which can then be evaluated at different points to approximate the original function.

How is a Taylor series calculated?

A Taylor series is calculated by taking derivatives of a function at a given point and evaluating them at that point. The coefficients of the resulting polynomial are then used to construct the Taylor series.

What is the purpose of a Taylor series?

The purpose of a Taylor series is to approximate a function with a simpler polynomial function that can be evaluated more easily. This is useful in various fields such as physics, engineering, and economics where complex functions need to be approximated.

How is a geometric series related to a Taylor series?

A geometric series is a special case of a Taylor series where the function being approximated is a geometric sequence. This means that the function can be expressed as a constant multiplied by a variable raised to some power. In this case, the Taylor series only has a finite number of terms.

What is the convergence of a Taylor series?

The convergence of a Taylor series refers to how well the series approximates the original function. A Taylor series may converge to the original function for some values of x, but not for others. This is why it is important to consider the convergence of a Taylor series when using it to approximate a function.

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