0
$\begingroup$

I have to solve or estimate the answer to an equation that is as follows:

$$P_\text{blocks} = \frac{398 \cdot 19^{65}}{\prod^{66}_{i=0} 78804 - i}$$

It doesn't take long to realize that this is an extremely small number. So small, in fact, that I can't find a calculator that doesn't return $\frac{1}{\infty}$.

I can estimate the numerator and denominator to make it a little more simple to understand:

$$P_\text{blocks} = \frac{5\cdot10^{85}}{8^{363}}$$

Is there a way to write this in scientific notation, or is it too large to deal with?

$\endgroup$
5
  • $\begingroup$ The denominator is between $8\text{^} {\mathbf{3}63}$ and $8\text{^} {\mathbf{3}64}$ $\endgroup$
    – Henry
    Jul 29 '14 at 20:40
  • $\begingroup$ What prompted you to ask this question? $\endgroup$ Jul 29 '14 at 20:44
  • $\begingroup$ If you're in the mood for reading... eeforumify.com/viewtopic.php?pid=496930#p496930 $\endgroup$
    – Tako M.
    Jul 29 '14 at 20:53
  • $\begingroup$ Essentially, someone asked for the minimum block density required for a player to jump from the bottom to the top of a 2-d grid-based platforming game (Everybody Edits). The equation signifies the % chance of my answer (0.000838%) to actually happen. $\endgroup$
    – Tako M.
    Jul 29 '14 at 21:02
  • $\begingroup$ A quick approximation to the denominator is $(78804-33)^{67}=78771^{67}$ while the numerator is about $1.1\times 19^{67}$. $\endgroup$
    – Henry
    Jul 29 '14 at 21:47
1
$\begingroup$

The traditional way of doing this would be with logarithms which (base $10$) would give a calculation like $$2.599883072+65\times 1.278753601 -(4.896548262+\cdots + 4.896184379)$$ which is about $-242.337678$ or $\overline{243}.662322$ depending on whether you are using a calculator or tables and taking the anti-logarithm gives about $4.5954 \times 10^{-243}$.

$\endgroup$
0
$\begingroup$

Wolfram Alpha gives your approximation as around $1.9196\times10^{-153}$.

http://bit.ly/Xbgkks

However, it gives the original expression to be around $4.5954\times10^{-243}$

http://bit.ly/XbgWqg

So I think something's gone wrong here because you seem to be off by factor of $4\times10^{89}$.

$\endgroup$
3
  • $\begingroup$ Completely forgot to check wolfram! $\endgroup$
    – Tako M.
    Jul 29 '14 at 20:27
  • $\begingroup$ @TakoM. I've updated my answer (I think you may have made a mistake with your approximation). $\endgroup$
    – Jam
    Jul 29 '14 at 20:28
  • $\begingroup$ Yes, when I added over 1000 to the exponent's base it really threw it off. It's not a problem, though. The purpose was to get a general idea that was more specific than "very small" $\endgroup$
    – Tako M.
    Jul 29 '14 at 20:48
0
$\begingroup$

This is not the simplest form technology can reach. It is trivial for an experienced programmer to solve this problem directly (by creating an unlimited-size data type). I'm not going to do it unless you really, really need it (just ask).

You could estimate $8^{264}$ fairly simply if you don't need an exact solution, though. The easiest way is to note that $2^{10}$ is roughly equal to $10^3$, so this suggests every $8^{10}$ adds a little more than nine digits. So you're probably looking at around $10^{79}$ as the denominator (I rounded up), but maybe a few factors larger due to the shift of $1024/1000$ (which you could further use to refine the estimate).

EDIT: Based on Wolfram Alpha there may be something wrong with your initial estimate in terms of factors of $10$ over factors of $8$. Your initial "simplification" estimate is certainly greater than unity even if my math powers estimate is way off.

EDIT EDIT: You might be surprised to learn that MS calculator (the one built into Windows) is remarkably accurate as an engineering calculator. The denominator is roughly $1.14\times10^{328}$, and the whole shebang is roughly $4.60\times10^{-243}$.

For any conceivable applied use, this number is utterly ridiculous. A better estimate: $0$.

$\endgroup$
7
  • $\begingroup$ The numerator is exact, but for the denominator I added 1196 to make it an even 80k. I also removed the negative progression. Not a good thing to add when you're dealing with exponents, but it's not like people will die if this number is wrong. I just want a vague idea. $\endgroup$
    – Tako M.
    Jul 29 '14 at 20:44
  • $\begingroup$ When you're multiplying together a huge set of numbers, a small change can make a huge difference. You added a factor of significantly more than 1196^66 to the denominator when you made that addition. You really can't estimate as loosely as this when dealing with iterated multiplication, unfortunately. $\endgroup$ Jul 29 '14 at 21:02
  • $\begingroup$ Also: I suggest rewriting the factor series as $78738 + i$ just for sanity sake. :P $\endgroup$ Jul 29 '14 at 21:04
  • $\begingroup$ "A better estimate: $0$." The purpose of this equation is to prove that it is not $0$, so I like $\frac{1}{\infty}$ better (even if it is hypothetically equal to $0$... well, let's not go there) $\endgroup$
    – Tako M.
    Jul 29 '14 at 22:10
  • $\begingroup$ I agree as far as the math is concerned: the fact that it is small but non-zero matters. But for the purpose of estimation, it seems silly to try to estimate a value that tiny. To estimate 0 is nearly as accurate as it is to estimate the exact value, and much faster. :) Let's just say it's a really freaking tiny value. $\endgroup$ Jul 29 '14 at 23:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.