# Math Help - question on series

1. ## question on series

hi
well looks like i found a good site to help me with my calc hw.
have a question on power series

the first one i wanna know if i did it right

1) find power series for f(x)=ln(6+x)
ln(6+x) = 1/(6+x) = (1/6) /(1-(-x/6)) = a/1-r
ln(6+x) = sum ar^n = sum (1/6)(-x/6)^n
(-x/6)<1 = -6<x<1

2) Use power series to find the integral of 3/(4-x^2) and then approximate the integral from x=0 to x=.9

kinda stuck on this one there are no examples like this in our book im guessing just do the same thing i did in question 1 right? could someone work this one out.

thanks

oh how do you enter in math formulas in on here?

2. You let out the integral in (1)

Originally Posted by amoochio46
hi
well looks like i found a good site to help me with my calc hw.
have a question on power series

the first one i wanna know if i did it right

1) find power series for f(x)=ln(6+x)
ln(6+x) = INTEGRAL 1/(6+x) = (1/6)INTEGRAL /(1-(-x/6)) = a/1-r
ln(6+x) = sum ar^n = sum (1/6)(-x/6)^n
(-x/6)<1 = -6<x<1

2) Use power series to find the integral of 3/(4-x^2) and then approximate the integral from x=0 to x=.9

kinda stuck on this one there are no examples like this in our book im guessing just do the same thing i did in question 1 right? could someone work this one out.

thanks

oh how do you enter in math formulas in on here?

3. (2) You could use partial fractions and that may work out better, but I'll do it directly...

$\int_0^{.9} {3dx\over 4-x^2} = (3/4) \int_0^{.9} {dx\over 1-(x^2/4)}=(3/4) \int_0^{.9}\sum_{k=0}^{\infty} (x^2/4)^k$.

Now use the geometric series with $r=x^2/4$

${1\over 1-r} =\sum_{k=0}^{\infty} r^k$ since $-1.

4. Originally Posted by amoochio46
[snip]
oh how do you enter in math formulas in on here?

5. Originally Posted by matheagle
You let out the integral in (1)
not sure i get what your saying. what integral did i leave out?

6. ln(6+x) = INTEGRAL 1/(6+x) = (1/6)INTEGRAL /(1-(-x/6)) = a/1-r

You left out the integral where I wrote INTEGRAL

ln(6+x) does NOT equal 1/(6+x)

ln(6+x) = INTEGRAL 1/(6+x)