Question

how do you solve this alternating series?

Answer 1

@rational can you help?

Answer 2

@nincompoop @jim_thompson5910 @zepdrix @dan815 @perl @Loser66 @nincompoop

someone said you can help me.
please help :)

Answer 3

how do I know the series for arctan?

Answer 4

$$

\Large \arctan x = \sum^{\infty}_{n=0} \frac{(-1)^n}{2n+1} x^{2n+1}\quad\text{ for }|x| \le 1

$$

Answer 5

lets approximate arctan (1/2)

Answer 6

arctan (x) = x-(1/3)*x^3+(1/5)*x^5-(1/7)*x^7+(1/9)*x^9+...

Answer 7

\( \Large {
Suppose~ that~\\ S_k = \sum_{n=1}^k (-1)^{n-1} a_n\\
\left | S_k - L \right \vert \le \left | S_k - S_{k+1} \right \vert = a_{k+1}.}

\)

Answer 8

|S _k - L | is the error of approximating the sum in k terms, where L is the sum of the series

Answer 9

so all we need to solve is
a_k+1 <= 1/33

Answer 10

$$

\Large a_{k+1} = \frac{(\frac{1}{2})^{2(k+1)}}{2(k+1)+1} \leq \frac{1}{33}

$$

Answer 11

i mean it might be easier to answer this numerically, because the directions say plug in a few terms.
so i would compute:
arctan(1/2) - (1/2)
then
arctan(1/2) - ( (1/2) - 1/3*(1/2)^3 )

Answer 12

make sure you are in radian mode

Answer 13

1/33 = 0.03030303... by the way

Answer 14

|arctan(1/2) - (1/2)| = 0.03635.. too big since 1/33 = 0.030303...

then
|arctan(1/2) - ( (1/2) - 1/3*(1/2)^3 )| = 0.0053 this works,

Answer 15

I tried plugging in 1/2 in the series, and I got. 0.4634672
What do you think is the purpose of reminding that the derivative of arctan?

Answer 16

so
(1/2) - (1/3) * (1/2)^3 is a good approximation to arctan(1/2) , to within 1/33 = 0.0303...

Answer 17

(1/2) - 1/3 * (1/2)^3 = 0.458333333...

Answer 18

arctan(1/2) is exactly 0.46347609

Answer 19

the point of this exercise is to approximate arctan(1/2) to at least the first decimal place accuracy

Answer 20

using the power series, but truncating it (you stop after a few terms)

Answer 21

Thanks! :), I now got the answer.
I think that the derivative of arctan was placed in order to find the taylor series of arctan.

Answer 22

exactly

Answer 23

$$
\Large \sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}\\
\Large = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+ \cdots
$$