Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$d > \sigma$ ... (1)

$\exp^{-(\frac{d^2}{2\sigma^2})} < 10^{-0.5}$ ... (2)

Is (1) <==> (2) true ?

EDIT: > replaced by < in the 2nd expression

share|cite|improve this question
What? Could you please clearify your question... – draks ... Jun 4 '12 at 12:50
@draks I've edited it – shn Jun 4 '12 at 14:13
If $r>0$, this is true. What makes you worry? There is just one step missing... – draks ... Jun 4 '12 at 17:18
@draks The program code that I made does not return the same result with the two equations. I've edited my post, can you check again ? – shn Jun 5 '12 at 12:49
If $d=100$ and $\sigma=1$ then (1) is certainly true and (2) is certainly false. – Gerry Myerson Jun 5 '12 at 12:56

[EDITED replacing > with < in the 2nd equation]

This is the best I could come up with:

$(1) \iff (2)$ means that $(1) \implies (2) \wedge (1) \impliedby (2) \quad \forall d, \sigma $

This can be proven to be false since from $(2)$ you have that:

$log_e{\exp^{-{d^2\over{2\sigma^2}}}} \lt log_e{10^{-0,5}} \rightarrow -{d^2\over{2\sigma^2}} \lt log_e{10^{-0,5}} \rightarrow -{d^2\over{2\sigma^2}} \lt -0,5\cdot log_e{10}$

Assuming $\sigma \ne 0$, I can multiply both terms by $2\sigma^2$

$(2)$ $-d^2 \lt -0,5\cdot 2\sigma^2\cdot \log_e{10} \rightarrow -d^2 \lt -\sigma^2\cdot \log_e{10} \rightarrow d^2 \gt \sigma^2\cdot \log_e{10}$

By applying the square root operator, I get

$(2)$ $d \gt \sqrt{\log_e{10}}\cdot \sigma$

I assumed that $d$ and $\sigma$ are real numbers. To prove that $(1) \iff (2)$ does not hold you just need to find a single counterexample:

$\sigma = 1 \quad d = 1.1 \quad \rightarrow \sqrt{\log_e{10}}\cdot \sigma = 1.51 $

Does it makes sense to you? I tried. :)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.