- #1
georg gill
- 153
- 6
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
[tex]cost^2[/tex]
If one use the derivation rules with chain one get 2t for first derivative and [tex]4t^4[/tex]
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 [tex]u=t^2[/tex] and derieve taylor series for cosx and just use afterwards [tex]u=t^2[/tex].
But what is even more confusing is that in one other assignment one were too find taylor polynomial for [tex]y=ln(x^2)[/tex]. Here one uses chain rule and it does work:
[tex]y'=\frac{2}{x}[/tex] [tex]y''=\frac{-2}{x^2}[/tex] [tex]y''=\frac{4}{x^4}[/tex]
What is the difference?
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
[tex]cost^2[/tex]
If one use the derivation rules with chain one get 2t for first derivative and [tex]4t^4[/tex]
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 [tex]u=t^2[/tex] and derieve taylor series for cosx and just use afterwards [tex]u=t^2[/tex].
But what is even more confusing is that in one other assignment one were too find taylor polynomial for [tex]y=ln(x^2)[/tex]. Here one uses chain rule and it does work:
[tex]y'=\frac{2}{x}[/tex] [tex]y''=\frac{-2}{x^2}[/tex] [tex]y''=\frac{4}{x^4}[/tex]
What is the difference?