Question

We can use this power series to approximate the constant  . a) First, evaluate arctan(1) . (You do not need the series to evaluate it.) b) Use your answer from part (a) and the power series above to find a series representation for  . (The answer will be just a series – not a power series.) c) Verify that the series you found in part (b) converges. d) Use your convergent series from part (b) to approximate  with |error| < 0.5. e) How many terms would you need to approximate  with |error| < 0.001?

Answer 1

(a)

`\arctan(1) = (\pi)/(4)`

(b)

`\pi = \sum\limits_(n=0)^(\infty) (4(-1)^n)/(2n+1)`

(c)

Therefore if you sum any three terms of it you get the desired accuracy.

(d)

If you sum 1999 terms you get the desired accuracy.

Explanation:

From the information given we know that

`\arctan(x) = \sum\limits_(n=0)^(\infty) \, (-1)^(n) (x^(2n+1) )/(2n+1)`

(a)

For that you need to find and angle
`\theta` such that
`\tan(\theta)`= 1, remember that

`\tan((\pi)/(4)) = 1` therefore

`\arctan(1) = (\pi)/(4)`

(b)

`(\pi)/(4) = \sum\limits_(n = 0)^(\infty) (-1)^(n) ((1)^(2n+1))/(2n+1) = \sum\limits_(n = 0)^(\infty)( (-1)^(n) )/(2n+1)`

Then you just multiply by 4 and get that

`\pi = \sum\limits_(n=0)^(\infty) (4(-1)^n)/(2n+1)`

(c)

Using the alternating series test, since the sequence
`1/(2n+1)` is decreasing and its limit tends to 0 when n tends to infinity the series is convergent.

(d)

Using the estimation theorem of alternating series we know that if
`s_n` denotes the partial sum of the series then

`\big| R_n \big| = \big| \pi - s_n \big| \leq (4)/(2(n+1)+1)`

Therefore we are looking for an
`n` such that

`(4)/(2(n+1)+1) \leq 0.5`

we just have to solve that inequality, when you solve that inequality you get that

`n \geq 2.5`

Therefore if you sum any three terms of it you get the desired accuracy.

(e) For this part you need to solve the following inequality

`(4)/(2(n+1)+1) \leq 0.001`

When you solve that inequality you get that

`n\geq 1998.5`

so, if you sum 1999 terms you get the desired accuracy.