Recent content by Kuno

I got it with u = -2/x, thanks. I'm not sure why I didn't do that in the first place.

Homework Statement \int \frac{dx}{x^{2} e^{\frac{-2}{x}}} The Attempt at a Solution I'm not sure where to begin.

The switch would be the best method if you were looking at a single character. I would try looping through an array testing each character.

I'm sure the indicator is lying, how long have you had the phone? Lithium Ion batteries tend to go bad after a certain amount of cycles. My old phone used to be like that too after having it for a while.

I have both Western Digital (100gig) and Seagate (120gig) and they've been running fine for over several years or so. However, I still suggest backing up everything.

Okay I see it now, thanks.

Homework Statement find a_{n} for f(x)\ =\ \frac{1}{1+3x} The Attempt at a Solution I got: f(0) = 1 f'(0) = -3 f''(0) = 9 The answer I ended up with was: a_{n} \ = \ {(-1)}^n\frac{3^n}{n!} However, the answer in the back of my book has the same answer except it's not divided by n!.

You could do the anti-snowflake. http://math.rice.edu/~lanius/frac/anti.html

It does not exist because the values oscillate between 1 and -1.

/Applications/Utitilities/Grapher.app Not sure if that's exactly what you're looking for. But it allows you enter equations and nice way. For example, if you type (5x^2)/2 it will format it for you.

Although I used a different book, that looks like exactly the same order my first year calculus course went. I would suggest trying to understand the concepts rather than just spitting out problems.

Do they give you how long it took or the distance at all? I'm thinking it may be possible to use the mean value theorem. But then you would need a position function and the time it took to get there.

I'm not sure if you mean 2 times sin x squared or 2 times sin 2x? But it if is former, then... 2(sin x)^2 - cos x - 1 = 0 2(1 - [cos x]^2) - cos x - 1 = 0 2 - 2[cos x]^2 - cos x - 1 = 0 -2[cos x]^2 - cos x + 1 = 0 from here you just treat it as you would a quadratic equation.

http://www.thinkgeek.com/tshirts/generic/8a5b/" It may no be very practical, but hey, it's a start.

let a_{n} = \frac{k^2 - 1}{3k^4 + 1} and b_{n} = \frac{1}{k^2} then show that a_{n} > b_{n} for every n Okay, I finally got it...