# Taylor polynomials

1. Dec 14, 2011

### georg gill

http://bildr.no/view/1030479

The link above, it is my own and it is a bit disorderly, I think should explain taylor polynomials. In one assignent one had an assignment to derive taylor polynomials for

$$cost^2$$

If one use the derivation rules with chain one get 2t for first derivative and $$4t^4$$
for second and so on. If t=0 and we are looking at a maclaurin series then every term except the first cos(0) becomes zero. I can understand that one could say $$u=t^2$$ and derieve taylor series for cosx and just use afterwards $$u=t^2$$.

But what is even more confusing is that in one other assignment one were too find taylor polynomial for $$y=ln(x^2)$$. Here one uses chain rule and it does work:

$$y'=\frac{2}{x}$$ $$y''=\frac{-2}{x^2}$$ $$y''=\frac{4}{x^4}$$

What is the difference?

2. Dec 14, 2011

### vladb

Try to calculate the derivatives more carefully, because what you stated is false. The first term is not the only non-zero term (there are infinitely many non-zero terms). The chain rules gives a correct result. (perhaps you are forgetting about the product rule)

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook