Recent content by Sabricd

Hi, I'm not exactly sure how to solve the following non-homogeneous ODE by variation of parameters. Solve the given non-homogeneous ODE by the variation of parameters: x^2y" + xy' -1/4y = 3/x + 3x Can someone please point me in the right direction? Help will be much appreciated...

Thank you so much Mark! :)

Hello, I need help with this problem. I need to find the first three terms of the Taylor series for the function f(x)= (1 + x)^(1/3) to get an estimate for 1.06^(1/3). Hence I did: f(x)= (1 + x)^(1/3) f'(x)= (1/3)(1 + x)^(-2/3) f''(x)= (-2/9)(1 + x)^(-5/3) f(a) + f'(x)/1! * (x - a) +...

Hi, OK. So would it be correct if I do: f(x)=(e^x-1)/x f(x)=e^x/x - 1/x \Sigma(e^x)/x -\Sigma(1/x) (1/x)\Sigma(e^x) -\Sigma(1/x) (1/x)\Sigma(x^n/n!) - \Sigma(1/x) ...is this correct? that's what I have so far and I'm not really sure if it's right. Thank you, -Sabrina

Would it be pointless to treat it as f(x)=(e^x -1)/x and then take its fourth derivative and use Taylor series then?

I'm sorry I'm not sure I understood that :( -Sabrina

Hello, I'm kind of stuck in this problem. I have to express the integral as a power series. the integral of (e^x -1)/x I thought about evaluating it as f(x)=(e^x -1)/x and treating it as a Taylor series is that correct? Could I have any other hints? I would really appreciate it...

So when you have alternating series you cannot use the p-series test?

Hello, I have to determine whether the given series diverges or converges. \sum cos(n*pi)/(n^(3/4)) where n= 1 and goes to infinity. I tried a couple numbers for n and got: -1 + 1/(2)^(3/4) - 1/(3)^(3/4) Hence I came up with the series: \sum((-1)^n)/(n)^(3/4) where n=1 and goes to...

Nevermind! I figured out why. I guess I confused the alternating series test with some other one. :) Thanks!

Hello, I have to determine whether the series converges or diverges. It is \Sigma (-1)^n * cos(Pi/n) where n=1 and goes to infinity. First I took the absolute value of the function and got the limit from n to infinity of cos(pi/n) and as a result I got 1 because cos(0)=1. However my...

Ok thanks! :)

Yes! Sorry I'm still kind of lost. I'm teaching myself Cal2..but I cannot understand how that is derived. Would you mind explaining to me how you simplified it? Thanks!

Hello! Yes! This is probably silly but I am not sure how to simplify... =1 + 2/(1!)*x + 6/(2!)*x^2 + 24/(3!)*x^3 + 120/(4!)*x^4 ... I am trying to solve a Maclaurin series but I don't have that much experience with factorials. I would really REALLY appreciate if you could give me some...

what happens if your comparison test fails and you limit comparison test is negative or zero? Does that ever happen?