Talk:Bernoulli's inequality

asdf

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

This has been fixed. -- Toby Bartels

a little bit more

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

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

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]

The reason why it is not mentioned that the inequality applies for all x≥−2 cannot be because for non-integer n values - aⁿ is meaningless for negative a values (−2≤x<−1), as far as this argument should end by requiring x≥0. The condition was modified in https://en.wikipedia.org/w/index.php?title=Bernoulli%27s_inequality&diff=955387483&oldid=935247787 --Polsko40 (talk) 15:23, 9 January 2022 (UTC)[reply]
The proof in Proof for integer exponent does not accord with the restriction given, but with the older x ≥ −2. --Polsko40 (talk) 15:23, 9 January 2022 (UTC)[reply]

As written right now, case 1 for integer r seems unnecessary and the limit could be wider. Case 2 states that for r >= 0 we need x >= -2. If this is so, then for r >= 1 (case 1) we could have x >= -2 as well, because if r >= 1 then we also have r >= 0, so case 2 applies. There is no reason to restrict x >= -2 to x >= -1. It would be more meaningful if in the case when r >= 1 we could have a WIDER range for x, but not a smaller range. — Preceding unsigned comment added by 2600:6C54:7CF0:86B0:D1E0:BC73:4512:E9D6 (talk) 07:26, 31 August 2024 (UTC)[reply]

Please add History

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

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)

I think it is the opposite. If h(t) = x(t)^(y(t)) and lim x(t) as t-> t0 = 0 and lim y(t) as t -> t0 = 0, then lim h(t) as t -> t0 will, at least in most situations, have a well-defined value. That value will be different for different functions x(t) and y(t), but the limit typically does have a value (intuitively, if x gets to zero faster, then the value is zero and if y gets to zero faster, then the value is 1). It is because the limit can have different values (which are well defined for each choice of functions), that we cannot say what the value of 0^0 is.

Another way to put it, if you plot the graph of the function z(x,y) = x^y in the right half plane (I would stay away from x<0), then you will see that the function approaches different values when (x, y) approaches (0, 0) from different directions (along a slanted straight line, along a parabola, etc.). The value of z in general does approach a definite value when you approach (0, 0) along each direction, so, the limits do exist. Because the limits can take different values, there is no way to assign a value to z(0,0) = 0^0 so that the function is continuous at zero. So, we say that 0^0 is not well defined.

In the case of the Bernoulli identity, I would tend to think that we first fix r = 0 and then look at it as a function of x. So, in this case, when x gets to -1, and r 'was already at zero' I would say that (1 + x)^r = 1. — Preceding unsigned comment added by 2600:6C54:7CF0:86B0:D1E0:BC73:4512:E9D6 (talk) 07:19, 31 August 2024 (UTC)[reply]

Convexity proof

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]