# Solve $b_1 e^{-a_1x^2}-b_2 e^{-a_2x^2}-b_3 e^{-a_3x^2}=0, \forall x$

Suppose, \begin{align*} b_1 e^{-a_1x^2}-b_2 e^{-a_2x^2}-b_3 e^{-a_3x^2}=0, \forall x \end{align*} Assume $a_1,a_2,a_3, b_1,b_2, b_3>0$

What are the possible values of $a_1,a_2,a_3, b_1,b_2, b_3>0$ that satisfy this equation?

One solution is: \begin{align*} b_1=b_2+b_3\\ a_1=a_2=a_3 \end{align*}

Is this a unique solution? Or is there other solutions?

• evaluate the expression for well chosen $x$ (e.g. $x=0,1,2$) – Surb Jan 27 '15 at 21:08
• Would that prove uniqueness of the my solution? – Boby Jan 27 '15 at 21:11
• for example $x=0$ implies $b_1=b_2+b_3$, then you can play around with the equation $b_2(e^{-a_1x^2}-e^{-a_2x^2})+b_3(e^{-a_1x^2}-e^{-a_3x^2})=0$ to show that $a_1=a_2=a_3$. – Surb Jan 27 '15 at 21:12
• Oh. I see. Thanks. I get. Thanks a lot – Boby Jan 27 '15 at 21:14