# Maclaurin series

1. Sep 14, 2016

### DevonZA

1. The problem statement, all variables and given/known data

Note - I do not know why there is a .5 after the ampere. I think it is an error and I have asked my lecturer to clarify.
2. Relevant equations

3. The attempt at a solution
f(t)=sint2 f(0)=sin(0)2=0
f'(t)=2sintcost f'(0)=sin2(0)=0
f''(t)=2cos2t f''(0)=2cos2(0)=2
f'''(t)=-4sin2t f'''(0)=-4sin2(0)=0
fiv(t)=-8cos2t fiv(0)=-8cos(0)=-8

.... until four non-zero terms

Am I on the right track?

2. Sep 14, 2016

### Dr Transport

Doing your differentials incorrectly....

$i = \sin(t^2)$, the first derivative is $2t\cdot\cos(t^2)$

3. Sep 14, 2016

### DevonZA

Thanks Doc.

The next derivative will then be:

right?

I have put in the 0 value up until the eighth term and I am still getting zero as the answer.
Only f''(t), f'''(t) and fvi have yielded non-zero terms thus far.

4. Sep 14, 2016

### Ray Vickson

If you want the first four nonzero terms in the Maclauren expansion of $f(t) = \sin(t^2)$ you will need the first 14 derivatives of $f$; that is, you will need
$$\left(\frac{d}{dt}\right)^k f(0), \; k=1, 2, \ldots, 14.$$

5. Sep 14, 2016

### DevonZA

Hi Ray

I got another non-zero term at the fourth non zero term at the 10th derivative.

6. Sep 14, 2016

### DevonZA

Plugging in zero values for x I get 60480

7. Sep 14, 2016

### Ray Vickson

You will get another one at the 14th derivative.

It must be obvious by now that the way I did it was using some other approach!

8. Sep 14, 2016

### DevonZA

(d/dt) = 2t⋅cos(t2) ?

9. Sep 14, 2016

### pasmith

You can obtain the first four terms of the series for $\sin(t^2)$ simply by obtaining the first four terms in the series for $\sin x$ and substituting $x = t^2$.

Only if you have a more complicated function than $at^n$ as the argument, such as $\sin(t + 3)$ or $\sin(e^t)$, will calculating derivatives compete with "write down the series for $\sin(x)$ and substitute for $x$" in efficiency.