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unsolvable math problem

Anonymous Sun 24 Mar 22:43:38 2019 No.10494448 View Reply Original Report

Quoted By: >>10494627 >>10495111 >>10495648

*blocks you're path*

Anonymous

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Anonymous Sun 24 Mar 2019 22:46:37 No.10494455 Report

Quoted By: >>10494607 >>10495208 >>10495209 >>10495214 >>10496947

tfw 1^infinity doesn't equal 1

Anonymous

Anonymous Sun 24 Mar 2019 23:42:41 No.10494561 Report

Quoted By:

minus a whole page

Anonymous

Anonymous Mon 25 Mar 2019 00:02:24 No.10494607 Report

Quoted By: >>10494683

>>10494455
Infinity is not something for which exponentiation is usually defined. We have definitions for integers/reals/complex/complex matrices, and others I'm sure. But infinity is none of those things, so I'm not sure why you think the answer would be intuitive.

Anonymous

Anonymous Mon 25 Mar 2019 00:08:29 No.10494627 Report

Quoted By: >>10494640 >>10494641 >>10494651

>>10494448
I assume that $p$ is a prime number and that $m$ is a whole number. What's the problem? For every prime number $p$ , there's a whole number $m$ that satisfies the equations. There are as many solutions as there are primes.

Anonymous

Anonymous Mon 25 Mar 2019 00:12:29 No.10494640 Report

Quoted By:

>>10494627
The problem is "are all perfect numbers of that form where p is prime?". Come back when you finish high school, bud

Anonymous

Anonymous Mon 25 Mar 2019 00:12:37 No.10494641 Report

Quoted By: >>10494681 >>10495860

>>10494627
https://en.wikipedia.org/wiki/Perfect_number#Even_perfect_numbers

Anonymous

Anonymous Mon 25 Mar 2019 00:17:14 No.10494651 Report

Quoted By: >>10494681 >>10495860

>>10494627
The problem is "are all perfect numbers of that form where 2^p - 1 is prime?". Come back when you finish high school, bud.

Anonymous

Anonymous Mon 25 Mar 2019 00:29:58 No.10494681 Report

Quoted By: >>10494694 >>10494696

>>10494651
>>10494641
I see. Well I'm not a mathematician. But Now I understand that there an addition condition that $(2^p - 1)$ must also be prime....i.e. a Mersenne prime.

So you say the question is: "are all perfect numbers of that form where 2^p - 1 is prime?"

And according to the article, Euler showed that all even perfect numbers take that form for which $2^p - 1$ is also prime.

Anonymous

Anonymous Mon 25 Mar 2019 00:30:28 No.10494683 Report

Quoted By: >>10494698 >>10495208 >>10495209 >>10495214

>>10494607
1^1 = 1, 1^100 = 1, 1^10000000 = 1, so it shouldn't matter what the exponent is if the base is one, however, it is apparently indeterminate.

Anonymous

Anonymous Mon 25 Mar 2019 00:34:38 No.10494694 Report

Quoted By: >>10494715

>>10494681
>And according to the article, Euler showed that all even perfect numbers take that form for which $2^p - 1$ is also prime.
What's your point? No one has proven that all perfect numbers are of that form

Anonymous

Anonymous Mon 25 Mar 2019 00:35:56 No.10494696 Report

Quoted By:

>>10494681
odd perfect numbers are still unsolved
and the number of mersenne primes is still unsolved

Anonymous

Anonymous Mon 25 Mar 2019 00:35:59 No.10494698 Report

Quoted By:

>>10494683
(9/10)^10 ~ 0.349
(99/100)^100 ~ 0.366
(999/1000)^1000 ~ 0.368
...
Limit = 1/e

Anonymous

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Anonymous Mon 25 Mar 2019 00:44:45 No.10494715 Report

Quoted By: >>10494811

>>10494694
I was just restating what I got out of the article for clarity. If my statement was wrong, then I would know what I thought was the problem would be wrong, and I would have to re-read the article until I understood what the problem was. It was confusing to me, because OP's picture did not mention that m was any perfect number and that 2^p -1 also had to be a prime number. I guess you mathematicians like reading things in context....nothing wrong with that, It just confused outsiders like me.

Anonymous

Anonymous Mon 25 Mar 2019 01:37:15 No.10494811 Report

Quoted By: >>10496452

>>10494715
>OP's picture did not mention that m was any perfect number and that 2^p -1 also had to be a prime number
Because it's a well known problem. It's like writing $\zeta(z) = 0$ and everyone will know you're talking about the riemann hypothesis

Anonymous

Anonymous Mon 25 Mar 2019 04:45:03 No.10495111 Report

Quoted By: >>10495119 >>10496444 >>10497132 >>10497529

>>10494448
p = 0 and m = 0.

Is this the famous p = mp problem from computer science? Because p = 0 and m = 0 also solve that fucking problem.

Anonymous

Anonymous Mon 25 Mar 2019 04:50:51 No.10495119 Report

Quoted By:

>>10495111

Anonymous

Anonymous Mon 25 Mar 2019 06:04:33 No.10495208 Report

Quoted By:

>>10494455
>>10494683
Well, $\infty$ isn't a number, so $1^{\infty}$ isn't a number either. They're undefined.
Despite this, $\infty$ can still be used to describe the behavior of numbers, so it's not wrong to use it in a limit. (That said, if you see $\infty$ outside the context of a limit, you can safely disregard the equation as invalid.)
So let's take a look at some limits of the form $1^{\infty}$ !
$ \begin{align*} \lim_{x\to\infty} 1^{x} &= 1 \\ \lim_{x\to\infty} \left(1+\frac{z}{x}\right)^{x} &= e^{z} \\ \lim_{x\to\infty} \left(1-\frac{z}{x}\right)^{x} &= \frac{1}{e^{z}} \end{align*} $
So, you see, $1^{\infty}$ can mean anything. We call these expressions "indeterminate forms" when they show up in a limit (again, if they're not in a limit, they're just undefined).
Other such indeterminate forms include:
$ \frac{0}{0} \begin{align*} \lim_{x\to0} \frac{x}{x} &= 1 \\ \lim_{x\to0^{+}} \frac{x}{x^2} &= \infty \\ \lim_{x\to0} \frac{x^2}{x} &= 0 \\ \lim_{h\to0} \frac{f{\left(x+h\right)} - f{\left(x\right)}}{h} = f'{\left(x\right)} \end{align*} \frac{\infty}{\infty} \begin{align*} \lim_{x\to\infty} \frac{x}{x} &= 1 \\ \lim_{x\to\infty} \frac{x}{x^{2}} &= 0 \\ \lim_{x\to\infty} \frac{x^{2}}{x} &= \infty \end{align*} \infty - \infty \begin{align*} \lim_{x\to\infty} \left(x - x\right) &= 0 \\ \lim_{x\to\infty} \left(x^2 - x\right) &= \infty \\ \lim_{x\to\infty} \left(\sqrt(x^2 + 4x + 1) - x\right) &= 2 \end{align*} 0\times\infty \begin{align*} \lim_{x\to\infty} e^{-x}x &= 0 \\ \lim_{x\to\infty} x\sin{\left(\frac{1}{x}\right) &= 1 \\ \lim_{x\to\infty} \left(x^{2}\times\frac{1}{x}\right) &= \infty \end{align*} 0^{0} \begin{align*} \lim_{x\to0} x^{0} &= 1 \\ \lim_{x\to0} 0^{x} &= 0 \end{align*} \infty^{0} \begin{align*} \lim_{x\to\infty} x^{0} &= 1 \\ \textrm{Ran out of space. Just take my word for this one.} \end{align*} [\eqn] You can turn any indeterminate form into another, by the way.$

Anonymous

Anonymous Mon 25 Mar 2019 06:06:08 No.10495209 Report

Quoted By:

>>10494455
>>10494683
Well, $\infty$ isn't a number, so $1^{\infty}$ isn't a number either. They're undefined.
Despite this, $\infty$ can still be used to describe the behavior of numbers, so it's not wrong to use it in a limit. (That said, if you see $\infty$ outside the context of a limit, you can safely disregard the equation as invalid.)
So let's take a look at some limits of the form $1^{\infty}$ !
$ \begin{align*} \lim_{x\to\infty} 1^{x} &= 1 \\ \lim_{x\to\infty} \left(1+\frac{z}{x}\right)^{x} &= e^{z} \\ \lim_{x\to\infty} \left(1-\frac{z}{x}\right)^{x} &= \frac{1}{e^{z}} \end{align*} $
So, you see, $1^{\infty}$ can mean anything. We call these expressions "indeterminate forms" when they show up in a limit (again, if they're not in a limit, they're just undefined).
Other such indeterminate forms include:
$ \frac{0}{0} \begin{align*} \lim_{x\to0} \frac{x}{x} &= 1 \\ \lim_{x\to0^{+}} \frac{x}{x^2} &= \infty \\ \lim_{x\to0} \frac{x^2}{x} &= 0 \\ \lim_{h\to0} \frac{f{\left(x+h\right)} - f{\left(x\right)}}{h} = f'{\left(x\right)} \end{align*} \frac{\infty}{\infty} \begin{align*} \lim_{x\to\infty} \frac{x}{x} &= 1 \\ \lim_{x\to\infty} \frac{x}{x^{2}} &= 0 \\ \lim_{x\to\infty} \frac{x^{2}}{x} &= \infty \end{align*} \infty - \infty \begin{align*} \lim_{x\to\infty} \left(x - x\right) &= 0 \\ \lim_{x\to\infty} \left(x^2 - x\right) &= \infty \\ \lim_{x\to\infty} \left(\sqrt(x^2 + 4x + 1) - x\right) &= 2 \end{align*} 0\times\infty \begin{align*} \lim_{x\to\infty} e^{-x}x &= 0 \\ \lim_{x\to\infty} x\sin{\left(\frac{1}{x}\right) &= 1 \\ \lim_{x\to\infty} \left(x^{2}\times\frac{1}{x}\right) &= \infty \end{align*} 0^{0} \begin{align*} \lim_{x\to0} x^{0} &= 1 \\ \lim_{x\to0} 0^{x} &= 0 \end{align*} \infty^{0} \begin{align*} \lim_{x\to\infty} x^{0} &= 1 \\ \textrm{Ran out of space. Just take my word for this one.} \end{align*} $
You can turn any indeterminate form into another, by the way.

Anonymous

Anonymous Mon 25 Mar 2019 06:09:45 No.10495214 Report

Quoted By:

>>10494455
>>10494683
Well, $\infty$ isn't a number, so $1^{\infty}$ isn't a number either. They're undefined.
Despite this, $\infty$ can still be used to describe the behavior of numbers, so it's not wrong to use it in a limit. (That said, if you see $\infty$ outside the context of a limit, you can safely disregard the equation as invalid.)
So let's take a look at some limits of the form $1^{\infty}$ !
$ \begin{align*} \lim_{x\to\infty} 1^{x} &= 1 \\ \lim_{x\to\infty} \left(1+\frac{z}{x}\right)^{x} &= e^{z} \\ \lim_{x\to\infty} \left(1-\frac{z}{x}\right)^{x} &= \frac{1}{e^{z}} \end{align*} $
So, you see, $1^{\infty}$ can mean anything. We call these expressions "indeterminate forms" when they show up in a limit (again, if they're not in a limit, they're just undefined).
Other such indeterminate forms include:
$ \frac{0}{0} \begin{align*} \lim_{x\to0} \frac{x}{x} &= 1 \\ \lim_{x\to0^{+}} \frac{x}{x^2} &= \infty \\ \lim_{x\to0} \frac{x^2}{x} &= 0 \\ \lim_{h\to0} \frac{f{\left(x+h\right)} - f{\left(x\right)}}{h} = f'{\left(x\right)} \end{align*} \frac{\infty}{\infty} \begin{align*} \lim_{x\to\infty} \frac{x}{x} &= 1 \\ \lim_{x\to\infty} \frac{x}{x^{2}} &= 0 \\ \lim_{x\to\infty} \frac{x^{2}}{x} &= \infty \end{align*} \infty - \infty \begin{align*} \lim_{x\to\infty} \left(x - x\right) &= 0 \\ \lim_{x\to\infty} \left(x^2 - x\right) &= \infty \\ \lim_{x\to\infty} \left(\sqrt(x^2 + 4x + 1) - x\right) &= 2 \end{align*} 0\times\infty \begin{align*} \lim_{x\to\infty} e^{-x}x &= 0 \\ \lim_{x\to\infty} x\sin{\left(\frac{1}{x}\right)} &= 1 \\ \lim_{x\to\infty} \left(x^{2}\times\frac{1}{x}\right) &= \infty \end{align*} 0^{0} \begin{align*} \lim_{x\to0} x^{0} &= 1 \\ \lim_{x\to0} 0^{x} &= 0 \end{align*} \infty^{0} \begin{align*} \lim_{x\to\infty} x^{0} &= 1 \\ \textrm{Ran out of space. Just take my word for this one.} \end{align*} $
You can turn any indeterminate form into another, by the way.

Anonymous

Anonymous Mon 25 Mar 2019 12:07:01 No.10495648 Report

Quoted By: >>10495860

>>10494448
So, what is the problem? Dead batteries in your calculator?

Anonymous

Anonymous Mon 25 Mar 2019 14:15:46 No.10495860 Report

Quoted By:

>>10495648
Read
>>10494641
>>10494651

Anonymous

Anonymous Mon 25 Mar 2019 18:50:36 No.10496444 Report

Quoted By: >>10497132 >>10497529

>>10495111
If the solution was obvious apparently only to you, you should consider the possibility that you don't actually understand the problem, i.e. that the problem as it is in your mind isn't the actual problem, you absolute brainlet.

Anonymous

Anonymous Mon 25 Mar 2019 18:55:24 No.10496452 Report

Quoted By:

>>10494811
No it's not asspie

Anonymous

Anonymous Mon 25 Mar 2019 19:01:51 No.10496462 Report

Quoted By:

Can someone please explain to me what is unsolvable about this math problem? What is the goal?

Anonymous

Anonymous Mon 25 Mar 2019 23:02:42 No.10496947 Report

Quoted By:

>>10494455
You mean like e? Woa :O

Anonymous

Anonymous Tue 26 Mar 2019 00:40:24 No.10497132 Report

Quoted By:

>>10495111
>>10496444
these two trips cancelled each other

Anonymous

Anonymous Tue 26 Mar 2019 04:13:49 No.10497529 Report

Quoted By:

>>10495111
Based
>>10496444
Cringe

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