Talk:Bernoulli's inequality

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asdf[edit]

In generalized inequality: if x=0, both inequalities are valid. -- anon

This has been fixed. -- Toby Bartels

a little bit more[edit]

1-x<e^(-x) 

 Jackzhp 20:00, 6 November 2006 (UTC)[reply]

This entry should be for real values of r, not just for integer values[edit]

Bernoulli's inequality true for all real values of r ≥ 1. This is important for many applications, so it is less useful to just describe it for integer values of r. And I do not think that it is worthwhile including an induction proof, which only gives the result for integer values of r.

Here's a quick proof for all real values of r ≥ 1. Unfortunately, I don't have the Wikipedia skills to directly edit this entry, so maybe someone else can do it.

Let f(x) = (1+x)^r - 1 - rx. The extended mean value theorem says that

f(x) = f(0) + f'(0)x + (1/2)f(y)x^2

for some y between 0 and x. But f(0) = 0 and f'(0) = 0, so

f(x) = r(r-1)(1+y)^{r-2}x^2.

for some y between 0 and x. It is clear that this last expression is nonnegative if x > -1 and r ≥ 1. —Preceding unsigned comment added by 76.118.41.44 (talk) 14:59, 4 November 2007 (UTC)[reply]

For n integer values the inequality also applies for x≥−2[edit]

For some reason this is not mentioned anywhere, but upon trying to prove the inequality I stumbled upon that fact. It can be easily shown that for even n values the inequality applies for all real x values, and for odd ns, for all x≥−2.

  • Maybe it is not mentioned because for non-integer n values - an is meaningless for negative a values (−2≤x<−1).

[Shir Peled] —Preceding unsigned comment added by 128.139.226.37 (talk) 18:09, 16 December 2007 (UTC)[reply]

I agree, it should definitely be mentioned in the article that for whole number exponents Bernoulli's inequality holds for x≥−2 - in fact this follows from the inductive proof already given in the article, because what is used there is that x+2 ≥ 0, i.e. x ≥ −2 Joel Brennan (talk) 15:55, 5 November 2021 (UTC)[reply]

Please add History[edit]

Does anyone know which Bernoulli came up with the inequality, or when? Will someone please add a history section, links to the proper Bernoulli, and a link to this article from the proper Bernoulli's page? Thank you. Yoda of Borg (talk) 04:12, 10 October 2008 (UTC)[reply]


More info at http://web01.shu.edu/projects/reals/history/bernoull.html. I can fill out the article later, but I'm not an expert on whether what I would add would be considered plagiarizing this source. Yoda of Borg (talk) 04:19, 10 October 2008 (UTC)[reply]

Question[edit]

Shouln't it be specified that for x=-1 and r=0 the first written Bernoulli's equation doesn't work? (Because then there is 0^0). — Preceding unsigned comment added by 178.235.177.80 (talk) 00:52, 14 December 2020 (UTC)[reply]

0^0 is only undefined in the context of limits (which is not the context here) so it is fine in the case x = −1, r = 0 - if an expression approaching 0 is raised to the power of an expression approaching 0 then the result is indeterminate, but if the whole number 0 is raised to the power of the whole number 0 then the answer is simply 1 (because any real number a to the power of the whole number 0 is defined to be 1 as it is the product of zero copies of a)

Convexity proof[edit]

I don't see how the inequality derived for 0 < \alpha < 1 can be "reversed" outside that domain, because I think convexity is only proved for alpha between 0 and 1.

It's more straightforward to prove f(x) = (1 + x)^n is convex over x >= -1, n >= 1, then by convexity f lies above its tangent line at x=0, 1 + nx. Wqwt (talk) 20:49, 21 August 2023 (UTC)[reply]

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