How can the Power Series for Arc Tan be Proven for Homework?

In summary, the student attempted to find the taylor series for arctan(x) but was unsure how to get started. They found a "cheat" solution by looking at an integral and figured that the series looked like one. They then differentiated each term and found that the Taylor series (for arctan(x)) was 1/(1+x2). They integrated both sides to find that the sum of the terms was 1.
  • #1
whatlifeforme
219
0

Homework Statement


Prove.


Homework Equations


arctan(x) = x - [itex]\displaystyle \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9}[/itex]
for -1 < x < 1.

The Attempt at a Solution


[itex]\displaystyle \sum^{∞}_{n=1} \frac{x^{n+2}}{n+2}[/itex]
 
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  • #2
what methods have you been taught for finding the taylor series?
 
  • #3
we skipped taylor and mclaurin series as we did not have time.
 
  • #4
then why have you been set this question?? :confused:

ok, here's a "cheat" idea …

doesn't that series look a bit like an integral? :wink:
 
  • #5
Wait. is the question to prove that arctan is equal to its power series OR to find the series for arctan?

If it's to prove it, then you may want to consider what's happening the remainder term.
 
  • #6
whatlifeforme said:

Homework Statement


Prove.


Homework Equations


arctan(x) = x - [itex]\displaystyle \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9}[/itex]
for -1 < x < 1.

The Attempt at a Solution


[itex]\displaystyle \sum^{∞}_{n=1} \frac{x^{n+2}}{n+2}[/itex]

How did you get that answer?
 
  • #7
tiny-tim said:
then why have you been set this question?? :confused:

ok, here's a "cheat" idea …

doesn't that series look a bit like an integral? :wink:

yes, when our teach show used she used integrals in part of the problem, but I've forgotten now.
 
  • #8
What happens when you differentiate that series?
 
  • #9
I will ask again: How did you get the answer you presented in the OP?
 
  • #10
(just got up :zzz:)
whatlifeforme said:
yes, when our teach show used she used integrals in part of the problem, but I've forgotten now.

ok, call the series A(x),

then (as FeDeX_LaTeX :smile: said) find dA/dx …

and then integrate! :wink:
 
  • #11
tiny-tim: so differentiate each term, then go back and take the integral of each? where does that get me?
 
  • #12
whatlifeforme said:
tiny-tim: so differentiate each term, then go back and take the integral of each? where does that get me?

The derivative of arctan(x) is 1/(1+x2). Can you get that Taylor series? Then try integrating both sides (remember to add an arbitrary constant which you have to deal with!)
 

1. What is a power series?

A power series is a mathematical series that represents a function as a sum of infinitely many terms, each of which is a constant multiplied by a variable raised to a non-negative integer power.

2. How is the arc tangent function represented as a power series?

The arc tangent function, also known as arctan or tan-1, is represented as a power series by the formula: arctan(x) = x - x3/3 + x5/5 - x7/7 + ...

3. How is the convergence of the power series for arc tangent determined?

The power series for arc tangent converges for values of x between -1 and 1. The series may also converge for values outside of this range, depending on the value of x and the number of terms in the series. The convergence can be determined using various tests, such as the ratio test or the root test.

4. What is the interval of convergence for the power series of arc tangent?

The interval of convergence for the power series of arc tangent is -1 < x < 1. This means that the series will converge for all values of x within this interval and may or may not converge for values outside of this interval.

5. How is the power series for arc tangent used to approximate the value of the function?

The power series for arc tangent can be used to approximate the value of the function at a given point by adding up a finite number of terms in the series. The more terms that are added, the more accurate the approximation will be. This is useful in situations where the exact value of the function cannot be calculated, such as in complex mathematical equations or computer algorithms.

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