Math conundrum

[S2000magician]

Technically, 0/0 is known is an indeterminate form; it’s undefined. Other indeterminate forms are:

• ∞/∞
• 0×∞
• ∞ – ∞
• 0^0
• 0^∞
• 1^∞

As bchad mentions, if these arise as limits of functions of a single variable (and those functions are differentiable), you can sometimes use l’Hôpital’s Rule to calculate the limit. However, if you’re merely computing a ratio of two static numbers, it’s not a limit, and it’s simply undefined.

[Turd_Fergeson]

nerd fight!

[Mobius_Strip]

@comp\_sci\_kid

lim x->inf x^0 = 1, lets x = z |z^0 - 1| lt e z^0 when z<>0 by definition is 1 so 0

I can’t decide if you generated some deep insight by extending the analysis into the complex plane, or just ended up scattering a few random symbols into your post to make your elusive thought process even less comprehensible?

[rawraw]

This thread is awesome ha ha. Wish I would’ve had a better math education growing up

[comp_sci_kid]

@Mobius Striptease

@comp\_sci\_kid

lim x->inf x^0 = 1, lets x = z |z^0 - 1| lt e z^0 when z<>0 by definition is 1 so 0

I can’t decide if you generated some deep insight by extending the analysis into the complex plane, or just ended up scattering a few random symbols into your post to make your elusive thought process even less comprehensible?

looks straightforward limit proof to me,

[Mobius_Strip]

@comp\_sci\_kid

@Mobius Striptease

@comp\_sci\_kid

lim x->inf x^0 = 1, lets x = z |z^0 - 1| lt e z^0 when z<>0 by definition is 1 so 0

I can’t decide if you generated some deep insight by extending the analysis into the complex plane, or just ended up scattering a few random symbols into your post to make your elusive thought process even less comprehensible?

looks straightforward limit proof to me,

indeed it is, it also appears that the confirmation of the Riemann hypothesis is a direct simple corollary from the lemma in step 2 of your reasoning

[Greenman72]

@Turd_Fergeson

nerd fight!

Word. I feel like we all belong on an episode of The Big Bang Theory.

[comp_sci_kid]

@Mobius Striptease

@comp\_sci\_kid

@Mobius Striptease

@comp\_sci\_kid

lim x->inf x^0 = 1, lets x = z |z^0 - 1| lt e z^0 when z<>0 by definition is 1 so 0

I can’t decide if you generated some deep insight by extending the analysis into the complex plane, or just ended up scattering a few random symbols into your post to make your elusive thought process even less comprehensible?

looks straightforward limit proof to me,

indeed it is, it also appears that the confirmation of the Riemann hypothesis is a direct simple corollary from the lemma in step 2 of your reasoning

Not sure if i should be impressed because you know Riemann hypothesis?

[Mobius_Strip]

As impressive as that is indeed, I wouldn’t want to steal your thunder with the limit proof you sketched above. Beautiful work.

[comp_sci_kid]

@Mobius Striptease

As impressive as that is indeed, I wouldn’t want to steal your thunder with the limit proof you sketched above. Beautiful work.

Why thank you, it is always nice to hear a word of encouragement. Let me know how your middle school algebra class did this fall. I heard there are a lot of bright students and you are an outstanding teacher

[rawraw]

@comp\_sci\_kid

@Mobius Striptease

As impressive as that is indeed, I wouldn’t want to steal your thunder with the limit proof you sketched above. Beautiful work.

Why thank you, it is always nice to hear a word of encouragement. Let me know how your middle school algebra class did this fall. I heard there are a lot of bright students and you are an outstanding teacher

[JohnyMac]

Ha ha I loved this exchange…it brought all the quants out to play! I too am of a more “fundamental” persuasion and would probably simply leap to the conclusion that the subject ratio is simply not useful for evaluating this particular financial circumstance. No doubt a different analytical angle would prove more useful!

But then again I would miss out on this scintilating Math exploration.

[Greenman72]

Just FYI–I asked Siri, “What is zero divided by zero?”

She gave me a real smartass answer.

[CFABLACKBELT]

@bchad

Technically it’s undefined, because the zero in the denominator is simply undefined.

However, you can often use l’Hôpital’s Rule to find the limit of something where the denominator approaches zero, so if you have a formula or function that happens to be zero at one point, you can take the limit of the numerator and the limit of the denominator to see whether f(0+delta) / g(0 + delta) is infinite, or finite or zero. f(x) is simply the numerator of the expression, expressed as a function, and g(z) is the denominator. That’s often a useful substitute.

You can look up l’Hôpital’s rule on wikipedia, it’s not that tricky if you’ve ever taken calculus, athough you may have forgotten it from high school.

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As an example, you could have the expression y=(x^2 - 1)/ (x-1), which technically is undefined at x=1. However, if you rearrange the expression as y=(x+1)(x-1)/(x-1), it’s clear that the plot looks like a line y=x+1, but there’s an undefined point (a hole, if you will) at x=1. But you know if you set x to 1+delta and make delta closer and closer to zero, the function will get closer and closer to y=2, and that’s often more useful than just throwing up your hands and saying it’s undefined.

L’Hopital’s rule is a little more complex than that, but the example gives you the basic idea of how it works.

A few years ago I took it upon myself to learn calculus on my own. Was probably one of the best things I’ve ever studied. I def want to take some higher level maths and physics courses once I get through year-one of my current career pivoting.