Taylor Series = 0????

[katyt]

I'm supposed to find the Taylor series about 0 for tsin(t^2) - t^3 and I'm having some trouble.
Here are the derivatives I found and I think maybe I'm doing them wrong and that's what's messing me up:

1st deriv. = sin(t^2) + 2t^2(cos(t^2)) - 3t^2, fprime(0)=0
2nd = 6tcos(t^2) - 2t^3(sin(t^2)) - 6t, fdoubleprime(0)=0
3rd = 6cos(t^2) - 4t^4(cos(t^2)) - 18t^2(sin(t^2)) - 6, ftripleprime(0)=0

and you get the idea. I'm bad at differentiating, so does anybody spot anything I might've done wrong? Or does the series really just equal 0?????

Thanks for any help!
Katy

 

[tkhunny]

That's very flat!

You do need to be a little more careful with your derivatives (the 2nd is just a bit off). However, fixing that probably won't help you with your dilemma. Just keep chugging. You will not find a non-zero derivative until you get to #7. The next is #11. Like I said, that is VERY flat around zero (0).

Good luck.

 

[soroban]

Hello, katyt!

I'm supposed to find the Taylor series about 0 for \(\displaystyle t\cdot\sin(t^2)\,-\, t^3\)

How about this plan?

We know the Taylor series for \(\displaystyle \sin x\)

. . . . . . . . . . . . . . \(\displaystyle \L\sin(x)\;=\;x\,-\,\frac{x^3}{3!}\,+\,\frac{x^5}{5!}\,-\,\frac{x^7}{7!}\,+\,\cdots\)


Then: . . . . . . . . .\(\displaystyle \L\sin(t^2)\;=\;t^2\,-\,\frac{t^6}{3!}\,+\,\frac{t^{10}}{5!}\,-\,\frac{t^{14}}{7!}\,+\,\cdots\)

And: . . . . . . . .\(\displaystyle \L t\cdot\sin(t^2)\;=\;t^3\,-\,\frac{t^7}{3!}\,+\,\frac{t^{11}}{5!}\,-\,\frac{t^{15}}{7!}\,+\,\cdots\)

Hence: .\(\displaystyle \L t\cdot\sin(t^2)\,-\,t^3\;=\;-\frac{t^7}{3!}\,+\,\frac{t^{11}}{5!}\,-\,\frac{t^{15}}{7!}\,+\,\cdots\)


Or: . . . . .\(\displaystyle \L t\cdot\sin(t^2) - t^3\;=\;\sum^{\infty}_{n=1}\frac{(-1)^nt^{4t+3}}{(2n+1)!}\)

 

[katyt]

thank you!!! that's very helpful!