Solving e^(a^2) x erfc(a) Equation”

[bhartish]

e^(a^2) x erfc(a) = e^(a^2 x erfc(a))

 

[tiny-tim]

hi bhartish!

(try using the X² icon just above the Reply box )

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[bhartish]

I have seen the application of this formula in one of the journal papers . I just want to know is there any such relation ( or even other such type ) between exponential and complimentary error function ?

 

[tiny-tim]

I have seen the application of this formula in one of the journal papers

which journal (and issue and page numner)?

 

[gb7nash]

e^(a^2) x erfc(a) = e^(a^2 x erfc(a))

So you think:

[tex]e^{a^{2}} \frac{2}{\sqrt{\pi}} \int_{a}^\infty e^{-t^{2}} dt = e^{a^{2} \frac{2}{\sqrt{\pi}} \int_{a}^\infty e^{-t^{2}} dt }[/tex] ?

Looks like nonsense to me. I would be very leery about this if there were no proof in this journal you're talking about.

 

[HallsofIvy]

It's easy to give a counterexample. If a= .5, then erf(a)= 0.52050 (http://www.geophysik.uni-muenchen.de/~malservisi/GlobaleGeophysik2/erf_tables.pdf)
so the left side is [itex]e^{.25}(0.52050)= 0.66833[/itex] while the right side is [itex]e^{.25*0.52050}= 1.13897[/itex]. Nuff said?

 

[bhartish]

Even I have tried this counter example but is there any substantiating answer through calculus ?