# Taylor series finding sin(x^2)+cos(x) from sin(x^2) and cos(x) alone

1. Apr 10, 2012

### IntegrateMe

If I want to find the taylor series at x = 0 for sin(x^2)+cos(x)...

sin(x^2) = x^2 - x^6/3! + x^10/5! - x^14/7! ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ...

So why does sin(x^2) + cos(x) = 1 + x^2/2! + x^4/4! + 121x^6/6! ...?

Thanks!

2. Apr 10, 2012

### Poopsilon

I get -121x^6/6!, other then that just give the fractions a common denominator and add.

3. Apr 10, 2012

### IntegrateMe

yes, sorry, it is a minus sign...can you explain that more elaborately? I think I'm starting to understand, but I'm not quite there yet.

4. Apr 10, 2012

### Poopsilon

Well the 1 + x^2/2! + x^4/4! is hopefully obvious. As for the -121x^6/6!, you need to add those terms in your two taylor series which have an x^6 in them, so that is - x^6/3! and - x^6/6!. But before you can add these fractions you need to give them a common denominator. Remember 6! = 6*5*4*3*2*1 and 3! = 3*2*1.

5. Apr 10, 2012

### IntegrateMe

Nope :\

6. Apr 10, 2012

### Poopsilon

Ok forgetting these series are infinite for a minute just consider the numbers you wrote without the ... . Then you are basically adding polynomials correct? So what do you do when you add polynomials well you add the x^2 terms together and you add the x terms together etc. Maybe you could be more specific about what you aren't understanding since I'm not sure how weak/strong your foundation is.