really simple question about MacLaurin series? anyone?

Since you said that you haven't learned this topic, you can do some research to help you. For example, take a look at this Taylor series Wikipedia article:

http://en.wikipedia.org/wiki/Taylor_series

A Maclaurin series is just a Taylor series centered at 0.

a)

The Maclaurin series of e^x is:

e^x = 1 + x + x^2/2! + x^3/3! + ... .

Replacing x with -x^2 gives:

e^(-x^2) = 1 - x^2 + (-x^2)^2/2! + (-x^2)^3/3! + ... .

==> e^(-x^2) = 1 - x^2 + x^4/2! - x^6/3! + ... .

Cutting this off to the first four terms gives:

e^(-x^2) ≈ 1 - x^2 + x^4/2! - x^6/3!.

b)

Cutting the result from a) to two terms gives:

e^(-x^2) ≈ 1 - x^2.

Multiplying both sides by x gives:

x*e^(-x^2) ≈ x(1 - x^2)

==> x*e^(-x^2) ≈ x - x^3.

I hope this helps!

If you've not covered this yet in class this is not going to be very helpful.

Maclaurin series of a function is the Taylor series of the function about zero. In general Maclaurin series for f(x) is

f(0) + x f'(0) + x² f''(0)/2 + x³ f'''(0)/6 + … + x^n f^(n)(0)/n! + …

where f^(n) is the nth differential d^n f/dx^n

So a) exp(-x²)

1 - x² + x^4/2 - x^6/6 + …

for b) multiply through by x

x - x³

y not jst use e^x=1+x+x^2/2!+x^3/3!+x^4/4!+ ..etc

an sub x=-x^2?

>e^(-x^2)=1-x^2+x^4/2-x^6/6+..etc..just pretend u hv worked out dn/dx^n(e^(-x^2)?!..saves time & effort!

>x*e^(-x^2)=x-x^3

it could be that straightforward. the sequence for ln(a million+x^4) could be something like ((-a million)^n-a million)((x^4n)/n) so in case you combine which you will get the sequence for the vital of ln(a million+x^4)