M. next Messages 380 Reaction score 0 Thread starter Oct 19, 2012 #1 How is it that: See figure: Given: See figure too In details, I don't get the maths and simplification that took place! Thanks! Attachments Nor.PNG 3.4 KB · Views: 577 Eqqqq.PNG 2.1 KB · Views: 532
How is it that: See figure: Given: See figure too In details, I don't get the maths and simplification that took place! Thanks!
dextercioby Science Advisor Insights Author Messages 13,419 Reaction score 4,227 Oct 19, 2012 #2 Well, what is [exp(-x^2)]^2 equal to ?
jtbell Staff Emeritus Science Advisor Homework Helper 2025 Award Messages 16,110 Reaction score 8,392 Oct 19, 2012 #4 Remember, exp(a) = e^a. So exp(a)exp(a) = (e^a)(e^a) = ... ?
M. next Messages 380 Reaction score 0 Oct 20, 2012 #5 it is supposed to be e^2a. Correct me if am wrong.
jtbell Staff Emeritus Science Advisor Homework Helper 2025 Award Messages 16,110 Reaction score 8,392 Oct 20, 2012 #6 Right. So that should tell you what [e^(-x^2)][e^(-x^2)] is.
TheDragon Messages 10 Reaction score 0 Oct 20, 2012 #7 woah get your math straight. that's not true: exp(a^2) =(e^a)(e^a) the integral of exp[-x^2] is defined only from -infinity to infinity.
woah get your math straight. that's not true: exp(a^2) =(e^a)(e^a) the integral of exp[-x^2] is defined only from -infinity to infinity.