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...
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
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...
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...
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...
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...
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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...