Taylor Series for ln(x+1) at x=1

In summary, the 4th term of the Taylor series centered at x=1 for f(x)=ln(x+1) is -0.375. To find this term, it is easier to start with the series for 1/(1+x), integrate it to get the series for ln(1+x), and then select the 4th term (the x^3 term). It is recommended to avoid using decimals and to write the series in its entirety.
  • #1
Thallen
1
0
Find the 4th term of the Taylor series centerd at x=1 for f(x)=ln(x+1)

f(x)=ln(1=x)
f'(x)=(1+x)^-1
f"(x)=(-1)[(1+x)^-2]
f"'(x)=(2)[(1+x)^-3]
f""(x)=-6[(1+x)^-4)]

Plug in 1:
.6931
.5
-.25
.25
-.375

What do I do next? (Also, is the 4th term the 4th term starting with f(x)? or the 4th derivative?)
 
Physics news on Phys.org
  • #2
it would be easier to start with the series for 1/(1+x) which is close to the geometric series. then integrate that to get the series for ln(1+x). then just look at the correct coefficient.
 
  • #3
Don't use decimals, please, your teacher should be happier that way. And the 4'th (non-zero) term is exactly that, write down the taylor series, and pick out the 4th term in it (the x^3 term here).
 

What is the Taylor Series for ln(x+1) at x=1?

The Taylor Series for ln(x+1) at x=1 is ln(x+1) = (x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 + .... It is an infinite series that approximates the natural logarithm function at x=1 by using derivatives of ln(x+1) at x=1.

How is the Taylor Series for ln(x+1) at x=1 derived?

The Taylor Series for ln(x+1) at x=1 is derived using the Taylor Series formula, which is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + .... In this case, a=1 and f(x) = ln(x+1). By taking the derivatives of ln(x+1) and plugging in a=1, we can simplify the formula to get the Taylor Series for ln(x+1) at x=1.

Why is the Taylor Series for ln(x+1) at x=1 important?

The Taylor Series for ln(x+1) at x=1 is important because it allows us to approximate the value of ln(x+1) at x=1 without actually calculating the natural logarithm function. This can be useful in situations where calculating the natural logarithm is difficult or time-consuming.

How accurate is the Taylor Series for ln(x+1) at x=1?

The Taylor Series for ln(x+1) at x=1 is accurate as long as the value of x is close to 1. The more terms we include in the series, the more accurate the approximation will be. However, if x is far from 1, the series will not be a good approximation for ln(x+1).

What is the purpose of using a Taylor Series for ln(x+1) at x=1 instead of just calculating the natural logarithm?

The purpose of using a Taylor Series for ln(x+1) at x=1 is to simplify the calculation of ln(x+1) at x=1. The Taylor Series formula allows us to use derivatives to find the value of ln(x+1) at x=1, which can be easier and more efficient than calculating the natural logarithm directly.

Similar threads

B Mean-Value Theorem, Taylor's formula, and error estimation
    • Calculus
Replies
3
Views
1K
I Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?
    • Calculus
Replies
9
Views
822
B Why do I not understand what seems to be basic Calculus?
    • Calculus
Replies
11
Views
2K
I Taylor expansion of f(x)=arctan(x) at infinity
    • Calculus
Replies
2
Views
1K
B Why is Big-O about how rapidly the Taylor graph approaches that of f(x)?
    • Calculus
Replies
19
Views
2K
MHB Using Standard Taylor Series to build other Taylor Series
    • Calculus
Replies
2
Views
1K
I Taylor series and variable substitutions
    • Calculus
Replies
3
Views
1K
MHB How can I develop ln(x) into a series for x >= 1 in fluid dynamics?
    • Calculus
Replies
1
Views
875
A Summing over continuum and uncountable numerocities
    • Calculus
Replies
1
Views
907
B Calculating shaded area: I'm getting discrepancies between methods
    • Calculus
Replies
3
Views
304

Hot Threads

Recent Insights

Back
Top