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January 20

Finding limits using first principles

Dear Wikipedians:

I am working on proving the following statement using first principles in limits:

$\lim _{x\to -\infty }{\frac {4x^{3}+3}{5x^{3}-2}}={\frac {4}{5}}$

I first rewrote the above statement as follows:

For every number $\epsilon >0$ , there is a corresponding number $M<0$ such that $\left|{\frac {4x^{3}+3}{5x^{3}-2}}-{\frac {4}{5}}\right|<\epsilon$ whenever $x<M$ .

I then wrote the proof:

Given $\epsilon >0$ , let $M=-{\sqrt[{3}]{\frac {23}{5\epsilon }}}$ , then for all $x<M$ ,

$\left|{\frac {4x^{3}+3}{5x^{3}-2}}-{\frac {4}{5}}\right|$

$=\left|{\frac {23}{5(5x^{3}-2)}}\right|$

$<\left|{\frac {23}{5x^{3}-2}}\right|$

$<\left|{\frac {23}{5M^{3}-2}}\right|$

$=\left|{\frac {23}{5\left(-{\sqrt[{3}]{\frac {23}{5\epsilon }}}\right)^{3}-2}}\right|$

$=\left|{\frac {23}{-{\frac {23}{\epsilon }}-2}}\right|$

$<\left|{\frac {23}{-{\frac {23}{\epsilon }}}}\right|$

$=\left|-\epsilon \right|$

$=\epsilon$

My question is: is my proof above a valid first-principles limit proof?

Thanks for your help.

70.29.26.221 (talk) 03:31, 20 January 2011 (UTC)[reply]

No, you wrote:

\left|{\frac {23}{5x^{3}-2}}\right|

<\left|{\frac {23}{5M^{3}-2}}\right|

However;

If M<0 and x<M then |M|<|x|. Hence |1/M|>|1/x|. Taemyr (talk) 04:54, 20 January 2011 (UTC)[reply]

Which is not in conflict with what he wrote. The expression with M in the denominator is the larger. –Henning Makholm (talk) 05:01, 20 January 2011 (UTC)[reply]

(ec) It's the right general approach, and those of the details I have checked seem to hold up. However, it's cumbersome to convince oneself that your inequalities between absolute values are all valid (because in each case one must figure out the sign of the arguments in order to figure out whether to reverse the inequality). It would be a much more readable proof if you resolve the sign problem as soon as possible (e.g.

\textstyle \left|{\frac {23}{5(5x^{3}-2)}}\right|={\frac {23}{5(2-5x^{3})}}

because the denominator is always negative).

Afterwards, you could help yourself by

\textstyle {\frac {23}{10-25x^{3}}}<{\frac {25}{10-25x^{3}}}={\frac {1}{10/25-x^{3}}}

instead of carrying the 23 around everywhere. –Henning Makholm (talk) 05:01, 20 January 2011 (UTC)[reply]

Why not

\lim _{x\to -\infty }{\frac {4x^{3}+3}{5x^{3}-2}}=\lim _{x\to -\infty }{\frac {4+3x^{-3}}{5-2x^{-3}}}={\frac {4+3\lim _{x\to -\infty }x^{-3}}{5-2\lim _{x\to -\infty }x^{-3}}}={\frac {4+3\cdot 0}{5-2\cdot 0}}={\frac {4}{5}}

Bo Jacoby (talk) 08:34, 20 January 2011 (UTC).[reply]

Because that's not a proof from first principles, unless you prove the distributivity of 'lim' first. AndrewWTaylor (talk) 13:15, 20 January 2011 (UTC)[reply]

You would also need to prove that

\lim _{x\to -\infty }x^{-3}=0

. An easier task than proving distributivity, but still necessary. --Tango (talk) 14:11, 20 January 2011 (UTC)[reply]

Thank you all for your help. I'm glad that my proof worked out. 76.68.4.123 (talk) 02:06, 21 January 2011 (UTC)[reply]

Resolved

Integration of vector-valued functions

I've been reading the part of Lang's Real and Functional Analysis where he defines the Lebesgue integral of a function. His definition is supposed to apply to functions with values in a Banach space E, but I'm suspicious the definition may not be correct when E is infinite-dimensional.

Here it is. He first defines step functions (each "step" being a set with finite measure) and their integrals in the obvious way. He then defines the L¹-norm on the space of step functions by ||f||₁ = ∫|f|. Finally, he defines a function f to be integrable if it is the pointwise limit almost everywhere of an L¹-Cauchy sequence of step functions f_n, and defines its integral as the limit of ∫f_n (that this is well-defined requires proof of course).

My problem is that this definition would appear to exclude certain functions which you'd obviously want to consider integrable, and is therefore underinclusive. Here is an example. Take the measure space to be N with the counting measure, and E to be a Hilbert space with countable orthonormal basis e_m. Now define f by f(n) = (1/n)e_n. Unless I'm mistaken, f is not the limit almost everywhere of a Cauchy sequence of step maps (Cauchy for Lang's norm), and therefore not integrable. However, the sequence f(n) is a summable family (see fr:famille sommable), though not absolutely summable.

If E is finite-dimensional, a family of its elements is summable if and only if it is absolutely summable, so my objection does not apply in that case.

My questions are as follows. First, am I correct? Second, if so, is this the generally accepted definition of integration of Banach-valued functions? If not, what is? Shouldn't a summable family always be integrable for the counting measure, not just absolutely summable ones? 82.120.62.252 (talk) 10:06, 20 January 2011 (UTC)[reply]

The definition you summarize seems to be equivalent to what our article calls a Bochner integral. It appears to be true that your function is not Bochner integrable under this definition even though it is summable. I don't think it is surprising a priori that moving from Rⁿ to a more complicated codomain would reveal differences between what is possible with a general theory of integration (which must work for all measures) to what a specialized notion of summability can achieve. –Henning Makholm (talk) 16:41, 20 January 2011 (UTC)[reply]

Another possibility is the Pettis integral which (I think) does reduce to ordinary summing for the counting measure. Our article about it is stubby, but presumably it has less nice convergence properties than the Bochner integral over arbitrary measures. –Henning Makholm (talk) 17:21, 20 January 2011 (UTC)[reply]

Thanks a lot. The Bochner integral does indeed appear to be equivalent. I guess since this thing has a name, there must be good reasons for the restriction. Since continuous functions on a rectangle are integrable, at least it's doing some kind of job. If only Lang had mentioned the name! 82.120.62.252 (talk) 21:39, 20 January 2011 (UTC)[reply]

I found it by googling lebesgue integral banach space, as the first result returned. –Henning Makholm (talk) 02:03, 21 January 2011 (UTC)[reply]

does this work?

is the following question a working way to obfuscate the options "install" and "dont install" (not necessarily in that order?)

“

If you had twin servants you couldnt tell apart, and one always did what you asked and the other always did the opposite, would you tell the one you happen to be talking with to:

"Do the opposite of what your twin would do if I asked him not to install the Yahoo Toolbar"
"Do the same thing your twin would do if I asked him not to install the Yahoo Toolbar."

”

I think this indeed works. The first option is ~~install~~ don't install. -- Meni Rosenfeld (talk) 14:01, 20 January 2011 (UTC)[reply]

Nice, but isn't the first option is "don't install" because it resolves to "the opposite of the opposite of don't install" ? Gandalf61 (talk) 14:11, 20 January 2011 (UTC)[reply]

I agree, the 1st is "don't install". Perhaps Meni missed the word "not"? --Tango (talk) 14:15, 20 January 2011 (UTC)[reply]

That one I did notice. It's the "opposite" in the beginning that I didn't integrate despite noticing and despite double-checking. Pretty embarassing. -- Meni Rosenfeld (talk) 14:25, 20 January 2011 (UTC)[reply]

I'd ask them to wear the same clothes so I could tell them apart ;-) (yes sometimes I have these evil urges) Dmcq (talk) 14:23, 20 January 2011 (UTC)[reply]

well, one of them will follow your directions and wear the same clothes as the other one, and the other won't and will wear something different. so whichever one is wearing the same clothes as the ... wait a minute... you ba---rd!!

Is this a Raymond Smullyan puzzle? Michael Hardy (talk) 18:47, 20 January 2011 (UTC)[reply]

Hi, OP here. It's an artistic statement - a work of Performance art. I asked the reference desk this question almost in a way that a devious programmer could who was thinking of the most sneaky possible way of giving the user an "option" of whether to install the Toolbar. It's art. 91.183.62.45 (talk) 20:16, 20 January 2011 (UTC)[reply]