Hi all,
You have probably seen these candy machines before. Tubes
that contain candies of different colors and drop candies into
a receiver once you inserted the coin.
I was watching these more than a quarter of an hour at lunch
time (waiting for someone to buy candy) and thought that...
This sounds neat! For example, if I expand f(x) as n terms
f(x) = a_0x^0+a_1x^1+a_2x^2+\cdots =0
then I can just differentiate n+1 times and that would leave me with
a_n=0 and in the same way to show that all other coefficients a_i
must also be zero. Is this reasonable?
Regards...
Sorry, I posted in the wrong section, this is not homework.
It is the standard way that perturbation theory is discussed in
most books. Since it has to do with series, these books just state
the results.
Michuco
PS. MOD, could you move this to the non homework section. Thanks
Hi all,
In order for the series
a_0 x^0 + a_1 x^1 + a_2 x^2 + a_3 x^3+\cdots
to be equal zero, where x can take any values,say,
from 0 to 1, the coefficients a_i must be all zero.
This sounds reasonable, but I am sure that there is some kind of
prove for this. Most books that...
My bad. Most of what I have read indicate that you use the boundary in the integration and I thought that also includes the evaluation of v. Thanks for clearing that out. Can you suggest any particular book that helps in evaluating tough(er) integrals. I am trying eventually integrate...
I am thinking of
\begin{equation*}
v = \int_{-\infty}^{\infty} x e^{-2 x^2} d x = 0
\end{equation*}
Don't I need the definite integral in this step also?
I take that by the parenthesis, you define u=x and dv=xe^{-2x^2}? But wouldn't this give v=0 with xe^{-2x^2} being an odd function?
Sorry, I am particularly densed today
u = x^2 -> du = 2xdx
dv = e^{2x^2} -> v = \sqrt{\pi/2}
\int ... = x^2\sqrt{\pi/2}|_{-\infty}^\infty -\int_{-\infty}^\infty \sqrt{\pi/2} 2 x dx
The first term looks like zero, and the second doesn't looks like it would converge.
I also tried u=e^{-2x^2} and dv = x^2dx...
So I did this three times and I got something like e^2x[x^3/2-3x^2/4+6x/8-16/6] + C
which sounds reasonable, but then when I tried
\int_{-\infty}^{\infty} x^2 e^{-2 x^2} dx
with integration by parts, I ended up with something like infinity ..?
Michuco
This is highly dependent on what you want to do. If
you want some simple 2d, contour, density plot
straight out of a data file with slight format modifications
such as legends, symbols, etc. then I would think
that any of the packages you mentioned should do
a decent job.
On the other...
I took calculus many years ago and my work now involves
a fair amount of math, like, I got to play around with
stuff like gamma function (which I found Nahin's An
Imaginary Tale an excellent read), numerical methods,
but nothing fancy.
I discovered Calculus made easy a while ago, and I...
Lugita15 and Frostfire,
Thanks for the suggestions. I am finishing Calculus
made easy and have downloaded a copy of Keisler
Elementary Calculus. Calculus by Larson, Hostetler
and Edwards doesn't look bad, but $170.00? I
paid $23. for my Thompson/Gardner copy.
Cheers,
Michuco
Hi all,
I am looking for calculus books that are fun to read. In
particular, I am looking for books that show, say, the
proof of
\int dx/x = log[x]
not just simply state the results. Something similar
to Thompson's Calculus made easy.
Thanks in advance,
Michuco
Thanks for the replies. I know of the polar coord conversion proof which I asume that wiki took from Weinsstein's MathWorld. I was wondering about the Feynman connection, if there
was one.
Regards,
Michuco