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?

  • $\begingroup$ evaluate the expression for well chosen $x$ (e.g. $x=0,1,2$) $\endgroup$ – Surb Jan 27 '15 at 21:08
  • $\begingroup$ Would that prove uniqueness of the my solution? $\endgroup$ – Boby Jan 27 '15 at 21:11
  • $\begingroup$ 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 $a_1=a_2=a_3$. $\endgroup$ – Surb Jan 27 '15 at 21:12
  • 1
    $\begingroup$ Oh. I see. Thanks. I get. Thanks a lot $\endgroup$ – Boby Jan 27 '15 at 21:14

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