Respuesta :
Answer:
(a)
[tex]\arctan(1) = \frac{\pi}{4}[/tex]
(b)
[tex]\pi = \sum\limits_{n=0}^{\infty} \frac{4(-1)^n}{2n+1}[/tex]
(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.
Step-by-step explanation:
From the information given we know that
[tex]\arctan(x) = \sum\limits_{n=0}^{\infty} \, (-1)^{n} \frac{x^{2n+1} }{2n+1}[/tex]
(a)
For that you need to find and angle [tex]\theta[/tex] such that [tex]\tan(\theta)[/tex]= 1, remember that
[tex]\tan(\frac{\pi}{4}) = 1[/tex] therefore
[tex]\arctan(1) = \frac{\pi}{4}[/tex]
(b)
[tex]\frac{\pi}{4} = \sum\limits_{n = 0}^{\infty} (-1)^{n} \frac{(1)^{2n+1}}{2n+1} = \sum\limits_{n = 0}^{\infty}\frac{ (-1)^{n} }{2n+1}[/tex]
Then you just multiply by 4 and get that
[tex]\pi = \sum\limits_{n=0}^{\infty} \frac{4(-1)^n}{2n+1}[/tex]
(c)
Using the alternating series test, since the sequence [tex]1/(2n+1)[/tex] 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 [tex]s_n[/tex] denotes the partial sum of the series then
[tex]\big| R_n \big| = \big| \pi - s_n \big| \leq \frac{4}{2(n+1)+1}[/tex]
Therefore we are looking for an [tex]n[/tex] such that
[tex]\frac{4}{2(n+1)+1} \leq 0.5[/tex]
we just have to solve that inequality, when you solve that inequality you get that
[tex]n \geq 2.5[/tex]
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
[tex]\frac{4}{2(n+1)+1} \leq 0.001[/tex]
When you solve that inequality you get that
[tex]n\geq 1998.5[/tex]
so, if you sum 1999 terms you get the desired accuracy.
