These are chat archives for ellisonbg/phys202-2015

May 2015
Randy Sterbentz
May 11 2015 08:15
How do you incorporate that extra r into the function? The f input is already a defined function, whether through lambda or separately defined. So how do you define a new function that modifies an existing function? And just f = f*r doesn't work (for obvious reasons...)
Noah Jan Miller
May 11 2015 17:19
Thank you, @JAmarel, it works now. I used lambda r, t to make a new function that calls f(r,t) and multiplies by r.
Brian E. Granger
May 11 2015 17:26
Yep you have to define a new function that computes and returns r*f(r,t)