Recent content by Aerospace93

Homework Statement Use the binomial series to expand the function as a power series. 1/(2+x)3 I have attached an image. I understand until the end of the second line. I don't see the reasoning used to follow through to the third line. the (-1)^n is because the sign alternates becoming...

1/3^2 * (3/5)^n. the geometric series will be convergent since |r|=3/5<1?

can it be simplified to 5^n-2 (3/5)^n+2

Homework Statement lim x-> infinity 3n+2/5n

Homework Statement ∫arctan(√x)dx. Using the substitution √x=t: ∫arctan(√x)dx = ∫arctan(t)dt2 This is what I've got written in a solution manual. I don't see why the dt would be squared. Could anyone care explaining me? thanks

By the ratio test I'm left with: 2 x limx->\infty |nn/(n+1)n+1)| which is the same as what i previously wrote: 2 x limx->\infty [n/(n+1)]n*1/(n+1). Dividing by the largest power of the polynomial in the denominator: 2 x limx->\infty [1/(1+1/n)]n*1/(n+1).- so it becomes: 2 x limx->\infty...

Homework Statement \sum ftom n=1 to \infty (-2)n/nn. The Attempt at a Solution limn->\infty | (-2)n+1/(n+1)n+1) x nn/(-2)n | = |-2|limn->\infty |(n/n+1)n*(1/n+1) | If it were only (n/n+1) then would the answer be 2e? Either way, how do you sole this the way it is?

(1/2)^x+(1/3)^x... which is the same as flipping them over and making the exponential -x. cool

i certainly did that

Homework Statement (2x+3x)/6x = 2-x+3-x I've tried moving the 6 above, splitting it up and so... but i can't figure how to do it. It must be pretty simple, but I am just not seeing it. all helps appreciated!

Without knowing about the form of the complementary solution at first, we'd figure that if G(x)=x2ekx then the particular solution should be something like this: (Ax2+Bx+c)ekx?? And say that you then worked out the complementary solution and it was: c1er1x+c2er1x, then the last term "c"...

that's wasnt my question. to put it in other words, why would you multiply it by x2...

Homework Statement y''+2y'+y=xe-x Homework Equations Yc=c1e-x+c2xe-x relevant info on textbook: "If any term of yp is a solution of the complementary equation, multiply yp by x (or by x2 if necessary)." >> i don't understand the part where it says "a solution of the complementary equation"...

could someone confirm this for me? thanks

4[cos(-∏/6)+isin(-∏/6)]*2[cos(3∏/4)+isin(3∏/4)] = 4[cos(11∏/6)+isin(11∏/6)]*2[cos(3∏/4)+isin(3∏/4)] ?