If 2Inf(x) = -x^2, does sq.rt [2Inf(x)] = ix?

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If 2Inf(x) = -x^2, does sq.rt [2Inf(x)] = ix?

2006-10-26 07:24:09 · 4 answers · asked by Martin F 6 in Science & Mathematics ➔ Mathematics

4 answers

Yup, I did wonder what you are trying to do..............

If you want to find f(x) then

lnf(x) = -(x^2)/2

and so f(x) = e^[-(x^2)/2]

which looks much nicer if you write it down as

f(x) = sq.rt[1/e^(x^2)]

2006-10-26 08:16:26 · answer #1 · answered by Stewart H 4 · 0⤊ 0⤋

Yes thats correct
The RHS can be expressed as ix because -x^2 = sqrt^-1*sqrt(x^2) = ix

2006-10-26 08:24:46 · answer #2 · answered by quark_sa 2 · 0⤊ 0⤋

yes.it cannot be +/-ix as log cannot me negative
so i think it should be ix

2006-10-26 07:32:48 · answer #3 · answered by raj 7 · 0⤊ 0⤋

That's a tough one.

2006-10-26 07:26:07 · answer #4 · answered by Anonymous · 0⤊ 0⤋