Power series without Taylor series

[Guest]

Find power series representations of sin(x) and cos(x) without using Taylor series methods. Hint: What is the relationship between f(x) and f"(x)?

I am having trouble solving this one without the use of Taylor series, and don't know how to start it. Any help with the first step or two would be greatly appreciated!

 

[pka]

Find power series representations of sin(x) and cos(x) without using Taylor series methods.

Isn’t that like asking to work a mathematics problem without using mathematics?
I sure that whoever wrote this problem has a very clear idea what is required. But it is not clear at all to me.
The representations
\(\displaystyle \L
\begin{array}{l}
\sin (x) = \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k x^{2k + 1} }}{{\left( {2k + 1} \right)!}}} \\
\cos (x) = \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k x^{2k} }}{{\left( {2k} \right)!}}} \\
\end{array}\)
are so very well known as to make the question strange in my view.

From the two one sees at once that sine is an odd function and cosine is even.
One also sees how their derivatives work.

 

[Guest]

Trig confusion

We haven't ever done anything with the Taylor series that much yet, let alone with sin and cos. I'm guessing he wants us to find out how to do it perhaps with the power series? I've never seen sin or cos used like this before.

 

[soroban]

Hello, ArcainineFalls531!

I had an promising idea . . . but I ran into a wall.
\(\displaystyle \;\;\)Maybe someone can see how it can be continued.

Find power series representations of \(\displaystyle \sin(x)\) and \(\displaystyle \cos(x)\) without using Taylor series methods.
Hint: What is the relationship between \(\displaystyle f(x)\) and \(\displaystyle f"(x)\)?

For both functions: \(\displaystyle \,f''(x)\:=\:-f(x)\)

Let the function be: \(\displaystyle \,f(x)\:=\:a_o\,+\,a_1x\,+\,a_2x^2\,+\,a_3x^3\,+\,a_4x^4\,+\,a_5x^5\,+\,\cdots\)

Then: \(\displaystyle \,f''(x)\:=\;2a_2\,+\,6a_3x\,+\,12a_4x^2\,+\,20a_5x^3\,+\,30a_6x^4\,+\,42a_7x^5\,+\,\cdots\)


Setting \(\displaystyle f(x)\,=\,-f''(x)\) and equating coefficients, we get:

\(\displaystyle \;\;\;a_o\,=\,-2a_2\;\;\Rightarrow\;\;a_2\,=\,-\frac{1}{2!}a_0\)

\(\displaystyle \;\;\;a_1\,=\,-6a_3\;\;\Rightarrow\;\;a_3\,=\,-\frac{1}{3!}a_2\)

\(\displaystyle \;\;\;a_2\,=\,-12a_4\;\;\Rightarrow\;\;a_4\,=\,-\frac{1}{12}a_2\,=\,\frac{1}{4!}a_o\)

\(\displaystyle \;\;\;a_3\,=\,-20a_5\;\;\Rightarrow\;\;a_5\,=\,-\frac{1}{20}a_3\,=\,\frac{1}{5!}a_1\)

\(\displaystyle \;\;\;a_4\,=\,-30a_6\;\;\Rightarrow\;\;a_6\,=\,-\frac{1}{30}a_4\,=\,-\frac{1}{6!}a_o\)

\(\displaystyle \;\;\;a_5\,=\,-42a_7\;\;\Rightarrow\;\;a_7\,=\,-\frac{1}{42}a_5\,=\,-\frac{1}{7!}a_1\)


We seem to have two sequences of coefficients:

\(\displaystyle \;\;\;a_o,\;\)-\(\displaystyle \frac{a_o}{2!},\;\frac{a_o}{4!}\,\;\)-\(\displaystyle \frac{a_o}{6!},\;\cdots\)

\(\displaystyle \;\;\;a_1,\;\)-\(\displaystyle \frac{a_1}{3!},\;\frac{a_1}{5!},\;\)-\(\displaystyle \frac{a_1}{7!},\;\cdots\)

which follow the patterns for \(\displaystyle \sin x\) and \(\displaystyle \cos x\) very nicely.


But now what?
\(\displaystyle \;\;\)How do I "separate" them and claim that the odd terms comprise the sine series?
\(\displaystyle \;\;\)How do I determine that \(\displaystyle a_o\,=\,a_1\,=\,1\) ?

 

[galactus]

If you have access to the American Mathematical Monthly, you can find it here:

More--and Moore--Power Series Without Taylor's Theorem
I. E. Leonard, James Duemmel
American Mathematical Monthly, Vol. 92, No. 8 (Oct., 1985) , pp. 588-589

You can access this if you subscribe to JSTOR, but my subscription ran out.

Some colleges have back issues of this magazine in the library.

 

[Guest]

I think this part for some reason has to do with the great old number e... We did something in class, wasn't too clear to me, but he somehow proved that a0=e, that a1=e0, and therefore, would equal 1... I know that has to relate somehow... I think?

But now what?
How do I "separate" them and claim that the odd terms comprise the sine series?

 

[galactus]

For what it's worth, I managed to dig this much up.