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

\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.

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

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

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?

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