Simplifying Equations with Exponential Terms

riordo
Messages
16
Reaction score
0

Homework Statement


Can you help me simplify the following equation: -e^((-1/2)*x^2)*(x^2-1)+2*e^((-1/2)*x^2)*x.

Homework Equations





The Attempt at a Solution



I've been guessing that you can combine the e^((-1/2)*x^2) components and thus end up with (x^2-1)x+e^((-1/2)*x^2)*x. Please let me know. Thanks.
 
Physics news on Phys.org
The most I can simplify it is just an exp term and a polynomial...

e^{-\frac{x^2}{2}}(-x^2+2x+1)
 
riordo said:

Homework Statement


Can you help me simplify the following equation: -e^((-1/2)*x^2)*(x^2-1)+2*e^((-1/2)*x^2)*x.

Homework Equations





The Attempt at a Solution



I've been guessing that you can combine the e^((-1/2)*x^2) components and thus end up with (x^2-1)x+e^((-1/2)*x^2)*x. Please let me know. Thanks.
Yes, you can factor out the e(-1/2)x^2x2 but you don't appear to have done it correctly! removing e(-1/2)x^2x2 from the first term leaves -(x2-1) and from the second leaves x. Removing it from both gives e(-1/2)x^2x2(-(x2-1+x)= -e-(1/2)x^2x2(x2- x+ 1).
 
Thank you. It appears that either you can combine the negative and positive e^((-1/2)*X^2) terms or eliminate the term by adding a negative value on both sides of the = sign. However that seems to me to make it more complicated. When I plug this into Mathematica software it simplifies to -e^((-1/2)*x^2)*x(x^2-3). I don't understand how it comes to this solution especially the (x^2-3). Let me know if you have any insight to what I am missing.
Thank you very much.
 
Solved. Thanks for the help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top