1. sumation of cos x

Write the summation notation of the maclaurin polynomial of order n for cos x
I worked out what the maclaurin of cos x by doing the derivations, plug in 0 and divide by n!. I get this

$1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} ...$

then for the sumation i got this

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

My question is what goes on top of the sigma? The top part is the stopping point when computing the sum, correct? I need to find the Nth so the top is not infinity.

Let's say I want the thrid maclaurin, n =3. would that be

$1 - \frac {x^2} {2!} + \frac {x^4} {4!}$

or

$1 + 0 * \frac {x^1} {1!} - \frac {x^2} {2!} + 0 * x^3$
counting on zero but the 0 coeffients remove those odd powered x's so the above is just the same as the previous.

With x^2i that will take into acount the 0 coeffients, for when sine at 0. It out puts 1, x^2, x^4 ..., so that forms the first example of the thrid maclaurin. Does that mean the top of sigma is just n?

The thing is my teacher wrote something like int[n/2 + 1.5] as the top of sigma but said he wasnt sure of that. and after runing different forms n/2 +.5, n/2 +1 it doesnt make sense to me why you would need something like that. the top part is just the stoping point so why the heck do you even need a little equation on the top ?!.

grrr

2. You left out the factorial (2i)! in the denomiator.
And what's wrong with the upper limit of your sum being n?

3. I fixed the factorial

I think im stuck on the fact that in class my instructor wrote int[n/2 + 1.5]. I just don't under stand why that would be needed. I do think that just n would work.

4. that integer value is ok if you use different terms, but unnecessary.
write exactly what your instructor placed on the board
and it's (2i)!, 2i! isn't clear at all

5. What the instructor wrote in class

$
\sum_{i = 0}^{int[\frac {n} {2} + 1.5]} \frac {(-1)^i x^{2i}} {2i!}
$

6. Originally Posted by diroga
What the instructor wrote in class

$
\sum_{i = 0}^{int[\frac {n} {2} + 1.5]} \frac {(-1)^i x^{2i}} {(2i)!}
$
I assume he/she meant the integer part....
Now, are you sure of the 1.5 and not .5?
Because if I let n=0, I get the first two term and not just the first term.
Most people say $S_1=1$, $S_2=1-{x^2\over 2}$,...

This claims that $S_0=1-{x^2\over 2}$ which seems odd.

7. Originally Posted by matheagle
I assume he/she meant the integer part....
do you mean the int[n/2 + 1.5] part?

Originally Posted by matheagle
Now, are you sure of the 1.5 and not .5?
Because if I let n=0, I get the first two term and not just the first term.
I forggot there was talk about this in the math center. I think .5 is correct. If i want the third term then it would be

$
\sum_{i = 0}^{int[\frac {n} {2} + 1.5]} \frac {(-1)^i x^{2i}} {(2i)!}
$

$int[\frac {3} {2} + .5] = 2$

$i = 0,1,2$

$

1 - \frac {x^2} {2!} + \frac {x^4} {4!}
$