Recent content by Refraction

Thanks, that worked perfectly!

Would there be a reasonably simple way to get to the two results given without substituting them into the first equation? I have a feeling that I'm supposed to do it that way, and not just by replacing x with the answers given. It seems like I could get to that form with some rearranging and...

what did you mean by putting the expression given as x into the equation? Putting it into the second one there/the other way around?

Sorry, I left out a part of the equation above. It was actually x = [what I already gave], I've edited that in now. It just seems like that way it would be easier to rearrange into the form needed, I'm not 100% sure about that though.

Homework Statement I've been given the following quartic equation in a question: x = (1 + 2a + a^2)x - (2a + 3a^2 + a^3)x^2 + (2a^2 + 2a^3)x^3 - a^3x^4 and need to show that two of the solutions (that aren't 0 and 1) can be given by: x= \frac{(2+a) \pm \sqrt{a^2 - 4}}{2a} 2. The attempt...

By the coefficient t for ln(1+t), do you mean when it would be t ln(1+t)? (making it (x-1)ln(x) after the substitution)

t=x-1 and when t=0, x=1 ? I just realized that for a function like xln(1+x) you can find the series for ln(1+x) and multiply the compact form by x to give the expansion for xln(1+x). It looks like I need to do something similar here, but I'm not sure what that last step would be now. I...

I'm still not having much luck figuring this out, any chance you could be a bit more specific about what you meant?

f'(x) is ln(x) + 1, and in this question it's at x = 1 not x = 0 (which wouldn't be defined for f'(x) anyway) so it's f'(1) = ln(1) + 1 = 1 in that case.

Sorry, it was supposed to be t in there, anyway I'll see if I can figure out what you meant.

I guess it would be t - (t^2)/2 + (t^3)/3 - (t^4)/4 + (t^5)/5 etc. if it was about t = 0? Not sure what you mean by the relationship between x and t though.

Homework Statement For f(x) = xln(x), find the taylor series expansion of f(x) about x = 1, and write the infinite series in compact form. 2. The attempt at a solution I can find the expansion itself fine, these are the first few terms: 0 + (x-1) + \frac{(x-1)^{2}}{2!} -...

The numerator kind of looks like a_{n+2}=2a_{n}+a_{n+1} to me, so it would be like this: c4 = (2*2) + 3 = 7 c5 = (2*3) + 7 = 13 c6 = (2*7) + 13 = 27 etc, it only works when n > 2 which seems to fit into what you mentioned as well. Hope that helps a bit!

Well the area bounded by the lines looks something like this: So with the reversed order of integration (dy dx) for the first double integral, R1, the inner integral is from y = 0 to y = x, and the outer integral is from x = 0 to x = 3.

The line was x = y in the original question, I just used it as y = x for when the order is reversed (so it's in the first half of the reversed order integral now).