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?

  • $\begingroup$ 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

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 $4.5954 \times 10^{-243}$.


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


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


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

  • $\begingroup$ 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

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: $0$.

  • $\begingroup$ 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

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