Question

(c) Since π/4 = arctan, you can find the value of π/4 using your expansion. How many terms of the expansion would you have to add in order to obtain π correct to four decimal places?
edit: forgot d and e...
(d) In an earlier assignment you found the Euler-Machin identity
\[\pi/4 = 4\arctan \frac{ 1 }{ 5 } - \arctan \frac{ 1 }{ 239 }\]
Using your series for arctan x to evaluate arctan 1/5 and arctan 1/239, how many terms would you have to find pi correct to 4 decimal places?
(e) find pi to 4 decimal places using the Maclaurin series for arctan x

done parts a and b
(a) Find the Maclaurin series for arctan(x)
(b) Show that the Maclaurin series for arctan(x) converges on x=1

Answer 1

for a
\[\huge \sum_{n=0}^{\infty} \frac{ (-1)^nx^{2n+1} }{ 2n+1 }\]

Answer 2

google is great :D

Answer 3

lol, I did part a by successive derivation

Answer 4

Not sure how to approach b.

For part c, since it's alternating, does approximation go by the next term you don't count in the series?

Answer 5

for b, i think you just need to use the summation series , and input x = 1

Answer 6

im prob wrong, i'd ask the other wizards of math on here >.<

Answer 7

Wizards of OS lend me your energy :* @Hero @ranga @nincompoop @shamil98 @bibby

Answer 8

d/dx arctan = 1/(1+x^2)
meh ik dat much o.o

Answer 9

Ask @math&ing001 ?? Idk :<

Answer 10

\[\pi/4 = 4\arctan \frac{ 1 }{ 5 } - \arctan \frac{ 1 }{ 239 }\] I meant

Answer 11

For (a), there's a rather nice way of getting the Maclaurin series for \(\large \arctan x\). Recall that
\[\large \frac{d}{dx}\arctan x= \frac{1}{1+x^2}\]
So, if we can come up with a Maclaurin series for \(\large \dfrac{1}{1+x^2}\) and then integrate termwise, we've got our series for \(\large \arctan x\).

Now, what's the series representation for \(\large\dfrac{1}{1+x^2}\)? Well, we know for \(\large |x|<1\) that
\[1+x+x^2+x^3+\ldots = \sum_{n=0}^{\infty}x^n=\large\frac{1}{1-x}\](this is a geometric series, by the way).

So, we now see that if \(\large |x|<1\), then clearly \(\large |-x^2|=|x^2|<1\) and thus
\[\large\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty}(-1)^nx^{2n}\]
Therefore, we see that
\[\large \begin{aligned}\arctan x &= \int\frac{1}{1+x^2}\,dx \\ &= \int\left(\sum_{n=0}^{\infty}(-1)^nx^{2n}\right)\,dx \\ &= C + \sum_{n=0}^{\infty}\left((-1)^ n\int x^{2n}\,dx\right)\\ &= C+\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}\end{aligned}\]
Now, we know that \(\large \arctan 0=0\); thus
\[\large 0=\arctan 0 = C+\sum_{n=0}^{\infty} \frac{0^{2n+1}}{2n+1}=C \implies C=0\]
Therefore, the Maclaurin series for arctan x is
\[\large \arctan x=\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}.\]

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For (b), when x=1, the series we found in (a) turns into
\[\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\]You can then show that this series converges using the alternating series test.

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I will have to think about the other parts a little more before answering those, but I hope this other way of computing the Maclaurin series makes sense! :-)

Answer 12

Gosh. I'd write a verbose paragraph but all I can really say is thank you for putting in this much work. You're an amazing person. Now I must read and digest this :p

Answer 13

Woops; the \((-1)^n\) term went missing in my post for some reason... sigh.

Answer 14

If only there was a nice way to edit your post... >_>. The \(\large(−1)^n\) term should appear in the following lines:

Where I say \(\large \displaystyle =C+\sum_{n=0}^{\infty}\frac{\color{red}{(−1)^ n}x^{2n+1}}{2n+1}\) in the integration steps.

Where I say \(\large\displaystyle 0=\arctan 0=C+\sum_{n=0}^{\infty}\frac{\color{red}{(−1)^n}0^{2n+1}}{2n+1}=C\implies C=0\)

And then where I officially claimed \(\large \displaystyle \arctan x=\sum_{n=0}^{\infty}\frac{\color{red}{(−1)^n} x^{2n+1}}{2n+1}\).

Sorry about that! :-/

Answer 15

lol

Answer 16

FYI: I have parts a and b.

Answer 17

I can explain this to you bibby :>

Answer 18

lmao I'll repost it. leave me alone

Answer 19

NO. I GOT THIS. SIT.

Answer 20

I'll be back in a bit bro :*

Answer 21

kay :[