Power series expansion

[kilroymcb]

I'm asked to find the power series expansion for f(0)=sin^2(x) out to y=t sub 8

I've done that. I found the derivatives out to the 8th, then I plugged in zero into each.
I'm getting this:
0th=sin^2(0)=0
1st=2sin(0)cos(0)=0
2nd=2cos^2 (0) - 2sin^2(0) = 2
3rd = -8cos(0)sin(0)=0
4th=(-16(cos^2(0)) - sin^2(0) = -16
5th = 128(cos(0)sin(0)) = 0
6th = 256(cos^2(0) - sin^2(0)) = 256
7th = 2048(cos(x)sin(x)) = 0
8th = -2048sin^2(0) + 2048cos^2(0) = 2048

This gives me the expansion 0 + 2x^2/2! + (-16x^4)/4! + 256x^6/6! + 2048x^8/8!
Now, I need to come up with the summation formula... but that -16 is throwing me for a loop, sitting there all by itself. I can't see where I made a mistake... if not, what could the formula be?

 

[galactus]

Actually, you have a few errors.

\(\displaystyle \L\\f^{4}(0)=-8\\f^{6}(0)=32\\f^{8}(0)=-128\\f^{10}(0)=512\\f^{12}(0)=-2048\)


See the pattern?.

 

[soroban]

Hello, kilroymcb!

Your derivatives are off . . .

Find the power series expansion for: \(\displaystyle \:f(x)\:=\:\sin^2(x)\) about \(\displaystyle x = 0\) out to \(\displaystyle y\,=\,t_8\)

We have the general form:
\(\displaystyle f(x)\;=\;f(0)\,+\,\frac{f'(0)}{1!}x\,+\,\frac{f''(0)}{2!}x^2\,+\,\frac{f'''(0)}{3!}x^3\,+\,\frac{f^{(4)}(0)}{4!}x^4\.+\,\frac{f^{(5)}(0)}{5!}x^5\,+\,\frac{f^{(6)}(0)}{6!}x^6\,+\,\frac{f^{(7)}(0)}{7!}x^7\,+\,\frac{f^{(8)}(0)}{8!}x^8\)

Differentiate:
\(\displaystyle \;f(x) \:=\:\sin^2(x)\)
\(\displaystyle \;f'(x)\:=\:2\cdot\sin(x)\cdot\cos(x) \:=\:\sin(2x)\)
\(\displaystyle \:f''(x)\:=\:2\cdot\cos(2x)\)
\(\displaystyle \,f'''(x)\:=\:-4\cdot\sin(2x)\)
\(\displaystyle f^{(4)}(x)\:=\:-8\cdot\cos(2x)\)
\(\displaystyle f^{(5)}(x)\:=\:16\cdot\sin(2x)\)
\(\displaystyle f^{(6)}(x)\:=\:32\cdot\cos(2x)\)
\(\displaystyle f^{(7)}(x)\:=\:-64\cdot\sin(2x)\)
\(\displaystyle f^{(8)}(x)\:=\:-128\cdot\cos(2x)\)


Let \(\displaystyle x\,=\,0:\)
\(\displaystyle \begin{array}{ccccc}f(0) & = & sin^2(0) & = & 0 \\
f'(0) & = & sin(2x) & = & 0 \\
f''(0) & = & 2\cdot\cos(0) & = & 2 \\
f'''(0) & = & -4\cdot\sin(0) & = & 0 \\
f^{(4)}(0) & = & -8\cdot\cos(0) & = & -8\\
f^{(5)}(0) & = & 16\cdot\sin(0) & = & 0 \\
f^{(6)}(0) & = & 32\cdot\cos(0) & = & 32\\
f^{(7)}(0) & = & -64\cdot\sin(0) & = & 0 \\
f^{(8)}(0) & = & -128\cdot\cos(0) & = & -128\end{array}\)


And we have:

\(\displaystyle \L f(x) \:=\:0\,+\,\frac{0}{1!}x\,+\,\frac{2}{2!}x^2\.+\.\frac{0}{3!}x^3\,-\,\frac{8}{4!}x^4\,+\,\frac{0}{5!}x^5\,+\,\frac{32}{6!}x^6\,+\,\frac{0}{7!}x^7 \,-\,\frac{128}{8!}x^8\)

. . . . . \(\displaystyle \L=\:\frac{2}{2!}x^2\,-\,\frac{2^3}{4!}x^4\,+\,\frac{2^5}{6!}x^6\,-\,\frac{2^7}{8!}x^8\)


I found the pattern . . . You can simplify the fractions if you like.