1. ## maclaurin series

or this question, i found that my coefficient of x^4 is wrong... after applying the maclaurin series formula, i would get the coefficient of X^4 is -5/96.... but the exact ans is -1/96... can anyone check which part is wrong?

2. ## Re: maclaurin series

the first vector is the derivatives of $\sqrt{\cos(x)}$ unevaluated.

the second vector is the above evaluated at $x=0$

the third vector is the second further scaled by $k!$ and are thus the coefficients of the Maclaurin series.

you can figure out where you went wrong.

3. ## Re: maclaurin series

hi, can you do it in the form of y^2 = cos x ? it would be easier for me to check where's my mistake. thanks!

4. ## Re: maclaurin series

Originally Posted by delso
hi, can you do it in the form of y^2 = cos x ? it would be easier for me to check where's my mistake. thanks!
why would you do it that way?

There are many places to use implicit differentiation. This isn't one of them.

5. ## Re: maclaurin series

maybe what you said is correct! if i use your method, there'll be less differentation will be performed

6. ## Re: maclaurin series

for this, my coefficient of x^4 which is 8/4! = 1/3 .. but the ans should be 13/24... can you tell me which part contain mistake? or any shortcut to do this question rather than perform so many times of implicit differentiation?

7. ## Re: maclaurin series

Originally Posted by romsek
why would you do it that way?

There are many places to use implicit differentiation. This isn't one of them.
sorry, i couldnt understand your working after -1/2 (sin x )/(cos x )^-0.5 ... can you check my working that i've post earlier please..

8. ## Re: maclaurin series

same format as last time.

letting $u=e^x-1$ and expanding about $u=0$ might save you some writing.

9. ## Re: maclaurin series

for the question y= cos x )^0.5 , to find the coeffeicient of x^3 and X^4 , must using the equation of proven part to proceed... can you do it from that part please?