How to Derive the Taylor Series for log(x)?

In summary, you were trying to solve the following problem yourself but couldn't figure out how the given Taylor series for log(x) is found.
  • #1
PainterGuy
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69
Hi,

I was trying to solve the following problem myself but couldn't figure out how the given Taylor series for log(x) is found.

?hash=bd0a599335ca11df42aea03e1955cda2.jpg


Taylor series for a function f(x) is given as follows.

?hash=bd0a599335ca11df42aea03e1955cda2.jpg


Question 1:
I was trying to find the derivative of log(x).
?hash=fec28be32c6920fd38d0c29f7bb45c83.jpg


My calculator gives it as d{log(x)}/dx=1/In(x)*x.

How do I convert log(e)/x to 1/In(x)*x? Logarithm base change formula isn't working for me. Thank you for your help!
 

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Last edited:
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  • #2
The log function in the example is the natural log, so log(e) = 1.
 
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  • #3
Thank you!

Question 1:
vela said:
The log function in the example is the natural log, so log(e) = 1.

I believe that you are referring to Example 6 shown below.

taylor_logx-jpg.jpg


You can see that the Taylor series in Example 6 doesn't have "x" and "x²" in denominator for the first two terms. But the Taylor series found by me has "x" and "x²" in denominator as shown below. Where am I going wrong?

?hash=9dbb0151b0b8b6721b0e6e6da4430946.jpg
Question 2:

?hash=9dbb0151b0b8b6721b0e6e6da4430946.jpg


Thank you for your help.
 

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  • #4
PainterGuy said:
Thank you!

Question 1:I believe that you are referring to Example 6 shown below.

View attachment 245372

You can see that the Taylor series in Example 6 doesn't have "x" and "x²" in denominator for the first two terms. But the Taylor series found by me has "x" and "x²" in denominator as shown below. Where am I going wrong?

View attachment 245373

I think you've calculated f'(x) = 1/x and f''(x) = -1/x^2, but forgot that these derivatives are evaluated at x = a, and should therefore be 1/a = 1 and -1/a^2 = -1.
 
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  • #5
On your calculator, "log" means ##\log_{10}## whereas in the example "log" means ##\log_e## or what on your calculator is called ##\ln##. It's pretty common in mathematics to dispense with the ##\ln## notation and use ##\log## to refer to the natural logarithm.
 
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  • #6
Thanks a lot!

I understand it now.

The following is a related example with a=2.

?hash=606c3838b1a7170a296b83086b9a1955.jpg


Also:

?hash=606c3838b1a7170a296b83086b9a1955.jpg
 

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1. What is a Taylor series for log(x)?

A Taylor series for log(x) is a mathematical representation of the natural logarithm function using a finite sum of terms. It is used to approximate the value of log(x) for any given input value of x.

2. How is a Taylor series for log(x) calculated?

A Taylor series for log(x) is calculated using the formula: log(x) = (x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 + ... + (-1)^(n+1) * (x-1)^n/n, where n is the number of terms used in the approximation.

3. What is the purpose of using a Taylor series for log(x)?

The purpose of using a Taylor series for log(x) is to approximate the value of log(x) for any given input value of x, especially when the input value is not a whole number or is a very large or small number. It is also used in calculus to simplify complex functions and make them easier to work with.

4. What is the difference between a Taylor series for log(x) and a Maclaurin series for log(x)?

A Taylor series for log(x) is centered around any given value of x, while a Maclaurin series for log(x) is centered around x = 0. This means that a Maclaurin series for log(x) is a special case of a Taylor series for log(x) where the input value of x is 0.

5. How accurate is a Taylor series for log(x)?

The accuracy of a Taylor series for log(x) depends on the number of terms used in the approximation. The more terms that are used, the more accurate the approximation will be. However, as the number of terms increases, the complexity of the calculation also increases.

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