I lost my calculator batteries in a flood .

AI Thread Summary
The discussion revolves around evaluating the expression \(\left(\frac{256^{16}-1}{256^{16}}\right)^{256^{16}}\). Participants eliminate possible answers based on mathematical reasoning, concluding that the expression cannot equal one or zero. The consensus emerges that the result approaches \(1/e\), which is approximately 0.3679, making option c) the correct choice. There is also a side conversation about using LaTeX for formatting equations in posts, with participants sharing tips on how to incorporate mathematical notation effectively. The dialogue highlights common miscalculations and clarifies the significance of limits in exponential expressions, reinforcing the importance of careful evaluation in calculus.
BobG
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I lost my calculator batteries in a flood...

What's this equal?

\left(\frac{256^{16}-1}{256^{16}}\right)^{256^{16}}

a) 2.178
b) 1.000
c) .3679
d) 0.000
 
Last edited:
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Well, it's obviously not one, because the number inside the parentheses is not one.

It's also obviously not zero for the same reason.

It's not 2.178, because the number inside the parentheses is less than one, and multiplying a fraction smaller than one by another fraction smaller than one must result in a fraction smaller than one.

By the process of elimination, it must be 0.3679.

- Warren
 
i get 1/e...i could be missing something in my work...or maybe even forgetting my calculus lol...

y = [(x-1)/x]^x
lny = x[ln(x-1) - lnx]
lny = ln(x-1 / x) / 1/x
using lhopitals...

lny = (x/(x-1))(1/x^2) / -1/x^2
lny = -x/ x-1 as x approaches infinity...lny = -1 so y = 1/e

But i may be wrong since i am assuming x goes to infinity while its 256^16.

Manu
 
As the number x in

\left(\frac{x-1}{x}\right)^{x}

approaches infinity, the result definitely does approach 1/e. Since 256^16 has plenty of significant digits, 1/e is close enough to an accuracy of only four decimals. :smile:

- Warren
 
And that's what 0.3679 is !
 
Hi Warren,

im such a dummy, lol, for 1/e in my windows calc. i kept using 2.178 which is answer a) for the constant e. Finally i looked e in my pocket handbook and found the right one, so it is answer c. And btw how did you use equation in the posts? Please do let me know tnx Warren.

Manu
 
Manu2380 said:
https://www.physicsforums.com/showthread.php?t=8997 And btw how did you use equation in the posts? Please do let me know tnx Warren.
The short answer is simply to click on the images to see their source code. You can just copy and paste the source into your own messages to include those equations. You'll figure out how it works in no time by example.

The long answer is to read this thread:

https://www.physicsforums.com/showthread.php?t=8997

- Warren
 
Manu, we have Chroot to thank for LaTex on PF.
 
No one silly enough to unthinkingly punch this in on their calculator, huh? :-p
 
  • #10
JFruit said:
:redface: No, i did, i readd tah title adn did the same but i got 0 in the end hmm!
:wink:

Strange :bugeye: I would have thought you'd get 1
 
  • #11
Would it not be d) 0.000 because in the parantheses would be a decimal and therefore as it grew exponentially it would lessen infinitely towards 0, and 0.000 just means that whatever number rounded to three decimal places would be 0? Or did y'all already establish that?
 
  • #12
BobG said:
No one silly enough to unthinkingly punch this in on their calculator, huh? :-p
>_< ..., whistles and looks over shoulder..
 
  • #13
The answer is something like .9999999999999999999999999999999 but has many more nines than i can fit. So this rounds to 1.000, since the answers go to 3 decimals.

That is the answer.
 
  • #14
My mistake, the real answer is more like 0.000000000000000000000001 to many more decimal places, so it rounds to 0.00.

That is the final answer.
 
  • #15
Tau:

It is neither 1 nor 0. Please read the earlier posts in the thread.

- Warren
 
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