Maclaurin Series Homework: Is My Solution Correct?

[DevonZA]

Homework Statement

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.

Homework Equations

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
f^iv(t)=-8cos2t f^iv(0)=-8cos(0)=-8

... until four non-zero terms

Am I on the right track?

 

[Dr Transport]

Doing your differentials incorrectly...

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

 

[DevonZA]

Doing your differentials incorrectly...

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

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 f^vi have yielded non-zero terms thus far.

 

[Ray Vickson]

Thanks Doc.

The next derivative will then be:

View attachment 105906

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 f^vi have yielded non-zero terms thus far.

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
$$\left(\frac{d}{dt}\right)^k f(0), \; k=1, 2, \ldots, 14.$$

 

[DevonZA]

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
$$\left(\frac{d}{dt}\right)^k f(0), \; k=1, 2, \ldots, 14.$$

Hi Ray

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

 

[DevonZA]

Plugging in zero values for x I get 60480

 

[Ray Vickson]

View attachment 105907

Plugging in zero values for x I get 60480

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!

 

[DevonZA]

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!

(d/dt) = 2t⋅cos(t²) ?

 

[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.