# MacLaurin series, inverse tan of x

1. Jun 9, 2007

### Joza

I was asked to find dy/dx of inverse tan of x , which is 1/(1+x^2)

Then its says, using that dy/dx ^^^^^ equal to a particular series, find the first 4 terms of inverse tan of x.

I'm confused??? What is it asking here??

2. Jun 9, 2007

### quasar987

It says find the 4 first terms of the maclaurin series of the inverse of tan(x). But it wants you to find those terms based only on the knowledge that the derivative of the inverse of tan(x) is 1/(1+x^2) (this means without any further calculations/differentiabtions)

Hint: the nth term of a Maclaurin series is of the form $$\frac{f^{(n)}(0)}{n!}x^n$$ and the series expansion of $$f(z)=\frac{1}{1-z}$$ you have somewhere in your notes.

3. Jun 9, 2007

### Joza

But it has given me a series there already, so what do I do with that?

4. Jun 9, 2007

### mathwonk

the point is that the maclaurin presription for finding the series is impractical since the derivatives are tedious to calculate. the obvious geometric series for 1/(1+x^2) can be integrated term by etrm far easier.

5. Jun 9, 2007

### Joza

Sorry I don't understand....I only have high school maths and I'm struggling a bit with it.

6. Jun 9, 2007

### Gib Z

Ok Well the series they gave you for 1/(1+x^2) can be done by polynomial long division, and is $1-x^2+x^4-x^6...$. And we also know that $dy/dx \arctan x = 1/(1+x^2)$

So It basically wanted you to substitute in the series for the derivative, ie $$\frac{dy}{dx} \arctan x = 1 - x^2 + x^4 - x^6 + x^8...$$

Then it wanted you to realise that the series for arctan x could be achieved by integrating both sides...high school math is all you needed :)