Calculating the Taylor Series for Arctan(x): Explained and Illustrated

In summary: I would take x=-1 for the first term and x=1 for the second term.In summary, the series is:- (33/5) - (34/7) + (35/9) - (36/11)+...
  • #1
vucollegeguy
29
0
The series is:

(33/5) - (34/7) + (35/9) - (36/11)+...

Looking at this, I'm guessing I can use the Taylor Series for arctan(x) but I don't know how to apply it or where to begin.

Any help is greatly appreciated.
 
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  • #2
You have 33/5 - 34/7 + 35/9 - 36/11 +...
= 33(1/5 - 3/7 + 32/9 - 33/11 +...)
[tex]=~3^3~\sum_{n = 0}^{\infty}(-1)^n~\frac{3^n}{2n + 5}[/tex]

The series above is an alternating series. Do you know a test for determining whether such a series converges?
 
  • #3
I can use the alternating series test where I let bn=(3n)/(2n+5), right?
 
  • #5
The title of your post is "sum the series". I doubt that is what you mean. You probably mean test it for convergence.

Anyway, remember the alternating series test will only tell you a series is convergent. If the test for convergence fails, that does not tell you the series diverges. So the alternating series might not (hint, hint) be the end of the story for this problem.
 
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  • #6
This is what I'm confused about. This is a practice exam for my final. And question specifically says "Sum the following series."

I did what mark44 did and factored out the 3^(3) but I didn't come up with the sum formula. When I factored it out, it looked as if it were from the Taylor Series of arctan(x).

Now, I don't know where to go from here.
 
  • #7
If you forget about convergence issues and want to "sum" the series in the sense of taking the Taylor expansion of some function and insert some value for x so that the expansion matches your series, then this is an example of a "resummation" method.

The idea is then that the series represents a finite number that was derived formally correctly, but it is an expansion around some point yielding a divergent series. But the terms of the expansion will contain all the information about the number which you have to "decode".

You are on the right track with the arctan function. If you look at two successive terms and forget about the missing terms at the start, what should you take for x?
 

FAQ: Calculating the Taylor Series for Arctan(x): Explained and Illustrated

What is the Taylor Series?

The Taylor Series is a mathematical concept that represents a function as an infinite sum of terms. It is used to approximate functions and calculate their values at specific points.

What is the formula for summing the Taylor Series?

The general formula for summing the Taylor Series is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... + f^n(a)(x-a)^n/n!, where f(a) is the value of the function at the point a and f'(a), f''(a), etc. are the derivatives of the function at the point a.

How accurate is the Taylor Series approximation?

The accuracy of the Taylor Series approximation depends on the number of terms used in the sum. The more terms you include, the closer the approximation will be to the actual function value. However, the series may not converge for all values of x, so it is important to choose a point a where the series converges.

What are some real-world applications of the Taylor Series?

The Taylor Series has many applications in fields such as physics, engineering, and economics. It is used to approximate functions in order to make predictions and solve complex problems, such as calculating the trajectory of a projectile or predicting stock market trends.

Are there any limitations to using the Taylor Series?

One limitation of using the Taylor Series is that it can only approximate functions that are smooth and have continuous derivatives. It may not work for functions with discontinuities or sharp corners. Additionally, the accuracy of the approximation may decrease as the distance from the chosen point a increases.

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