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johnmurray100 · Apr 9, 2009, 04:26 PM

The 1st term of the maclaurin series for Cos x is 1.
The Genreral Term is {(-1)^n}*{x(2*n)}/{(2*n)!}.

This will yield 1st term =1 only if n=0

I would expect Term 1 to be generated by setting n=1 not 0

Can you explain where my reasoning is wrong please

galactus · Apr 9, 2009, 04:57 PM

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

The first term is 1 because a 0 exponent yields 1 as well as 0! If you subbed in n=1, you would get

$\frac{-x^{2}}{2}$ . Just as the second term yields.

johnmurray100 · Apr 10, 2009, 03:04 AM

Hello Galactus
Thank very much for your reply.
The problem I have is that a similar expansion for the Sin Series requires setting n=1 to get the first term.
I am at a loss to know why n=0 for 1st term of Cos X series & n=1 for 1st term of Sin Series.

I would expect n=1 to yield 1st term of both Series.

The general term for Sin Series is {(-1)^(n+!)}*{(x)^(2*n-1)}/{(2*n-1)!}

Setting N=0 yields -1! In denominator.

I would appreciate any guidance on this please

galactus · Apr 10, 2009, 09:01 AM

Look at the series. It is rather obvious what you get when you plug in 0 or 1.

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

Now, plug in n=0:

$(-1)^{0}\cdot\frac{x^{2(0)+1}}{(2(0)+1)!}=(1)\frac{x ^{1}}{1!}=x$

Plug in n=1:

$(-1)^{1}\cdot\frac{x^{2(1)+1}}{(2(1)+1)!}=(-1)\cdot\frac{x^{3}}{6}$

See now?

Those are the first two terms of the expansion.

$x-\frac{x^{3}}{6}$

Let's look at the series for e:

$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

Plug in n=0:

$\frac{x^{0}}{0!}=\frac{1}{1}=1$

Plug in n=1:

$\frac{x^{1}}{1!}=x$

And so on and so forth

johnmurray100 · Apr 10, 2009, 02:48 PM

Hello galactus,
Thank you for reply.

The denominator you have used is (2*n+1)!

The denominator that I have is (2*n-1)!
Have I got the wrong denominator?

galactus · Apr 10, 2009, 03:26 PM

For the sine series? You have the wrong one. It is (2n+1)!