Question

What are the first 3 non-zero terms in the power series of arcsin(2x)?

Answer 1

In general, do you know how to find the terms of a power series?

Answer 2

sort of

Answer 3

Go on :)

Answer 4

I was getting 1+(1/2)arcsin(4)+(1/2)arcsin(192) but you can't take the arcsin of 4

Answer 5

That's because you're applying the formula wrong... the power series is a function of x. Yet your sum is not a function of x.

Answer 6

So then what do I do?

Answer 7

Given f(x), the power series of f(x) is (assuming we are doing the power series around 0):
\[f(x)=\sum_{n=0}^{\infty}\frac{ 1 }{ n! } f^{(n)}(0)x^n\]

Answer 8

In this case f(x)=arcsin(2x). Then you apply the power series formula above. Understanding me? :P

Answer 9

so arcsin(2x) would go where the f^n is?

Answer 10

Oh sorry, the notation \[f^{(n)}(x)\]
is another notation for:
\[\frac{ d^ny }{ dx^n }(x)\]

Answer 11

Let me give you an example.

Answer 12

If f(x)=sin(x). Then the power series of f(x) i.e. the power series of sin(x) is:
\[f(x)=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{d^ny}{dx^n}(0)x^n=\frac{1}{0!}y(0)x^0+\frac{1}{1!}\frac{dy}{dx}(0)x^1+\frac{1}{2!}\frac{d^2y}{dx^2}(0)x^2+\frac{1}{3!}\frac{d^3y}{dx^3}(0)x^3...\]
=\[\sin(0)x^0+(\frac{d}{dx}(\sin(0)))x^1+\frac{1}{2}(\frac{d^2}{dx^2}(\sin(0)))x^2+\frac{1}{6}(\frac{d^3}{dx^3}(\sin(0)))x^3+...\]

Answer 13

\[=\sin(0)x^0+\cos(0)x^1-\sin(0)x^2-\frac{1}{6}\cos(0)x^3+...\]

Answer 14

Lol, it looks really messy... but do you get what to do now?

Answer 15

Do you follow what i've done?

Answer 16

I think I do. Thanks

Answer 17

np :)