# Power series expansion proofs for the differential of sinx and cosx

• Feb 17th 2011, 04:58 AM
maximus101
Power series expansion proofs for the differential of sinx and cosx
Let sin x and cos x be defined by their power series expansions

how can we use term by term differentiation to show that

$d/dx$sin x = cos x and

$d/dx$cos x = − sin x.
• Feb 17th 2011, 05:35 AM
chisigma
Quote:

Originally Posted by maximus101
Let sin x and cos x be defined by their power series expansions

how can we use term by term differentiation to show that

$d/dx$sin x = cos x and

$d/dx$cos x = − sin x.

Is...

$\displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$ (1)

... so that...

$\displaystyle \frac{d}{dx} \sin x = \sum_{n=0}^{\infty} (-1)^{n} (2n+1)\ \frac{x^{2n}}{(2n+1)!}=$

$\displaystyle = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} = \cos x$ (2)

The same procedure is valid for $\cos x$...

Kind regards

$\chi$ $\sigma$
• Feb 17th 2011, 06:24 AM
Prove It
Quote:

Originally Posted by chisigma
Is...

$\displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$ (1)

... so that...

$\displaystyle \frac{d}{dx} \sin x = \sum_{n=0}^{\infty} (-1)^{n} (2n+1)\ \frac{x^{2n}}{(2n+1)!}=$

$\displaystyle = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} = \cos x$ (2)

The same procedure is valid for $\cos x$...

Kind regards

$\chi$ $\sigma$

I think this question is entirely circular reasoning.

You require knowing the derivatives of $\displaystyle \sin{x}$ and $\displaystyle \cos{x}$ in order to generate their series expansions in the first place...
• Feb 17th 2011, 08:21 AM
Krizalid
No it's not circular.

Note that he differentiate the power series (given hypothesis) and identified that series as cosine.
• Feb 17th 2011, 08:28 AM
Prove It
Quote:

Originally Posted by Krizalid
No it's not circular.

Note that he differentiate the power series (given hypothesis) and identified that series as cosine.

The first sentence of the question is "Let $\displaystyle \sin{x}$ and $\displaystyle \cos{x}$ be defined by their power series expansions..."

How is one supposed to generate the power series expansions without already knowing the derivatives of $\displaystyle \sin{x}$ and $\displaystyle \cos{x}$?
• Feb 17th 2011, 08:48 AM
SammyS
Quote:

Originally Posted by Prove It
The first sentence of the question is "Let $\displaystyle \sin{x}$ and $\displaystyle \cos{x}$ be defined by their power series expansions..."

How is one supposed to generate the power series expansions without already knowing the derivatives of $\displaystyle \sin{x}$ and
$\displaystyle \cos{x}$?

Exactly the point!

Let sin x and cos x be defined by their power series expansions …

So IF you were to define the sine & cosine functions by what we know their power series are. Guess what!! No surprise that $\displaystyle {{d}\over{dx}}\,\sin(x)=\cos(x)\quad \text{ and }\quad {{d}\over{dx}}\,\cos(x)=-\sin(x)$

It's not so much circular reasoning as it is an exercise to help students verify that the power series have the same properties as the functions they represent - indeed they are identical.