Factorial of big numbers

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katikeit

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Hi guys!
I really need to solve this expression without the help of the computer but only using a simple calculator...please help!!:wink:

4965! * 4950! / 4915! * 5000!

I've also tried to use the Stirling's approximation but the calculator can't afford it!

many thanks!!
 
katikeit said:
Hi guys!
I really need to solve this expression without the help of the computer but only using a simple calculator...please help!!:wink:

4965! * 4950! / 4915! * 5000!

I've also tried to use the Stirling's approximation but the calculator can't afford it!

many thanks!!
Click to expand...
I am assuming you mean

\(\displaystyle \dfrac{4965! * 4950!}{4915! * 5000!} = \dfrac{\displaystyle 4915! * \left(\prod_{i=4916}^{4965}i\right) * 4950!}{\displaystyle 4915! * 4950! * \left(\prod_{i=4951}^{5000}i\right)} = \dfrac{\displaystyle \prod_{i=4916}^{4965}i}{\displaystyle \prod_{i=4951}^{5000}i} = \dfrac{\displaystyle \left(\prod_{i=4916}^{4950}i\right) \left(\prod_{i=4951}^{4965}i\right)}{\displaystyle \left(\prod_{i=4951}^{4965}\right) \left(\prod_{i=4966}^{5000}i\right)} = \dfrac{\displaystyle \prod_{i=4916}^{4950}i}{\displaystyle \prod_{i=4966}^{5000}i} = \displaystyle \prod_{i = 4916}^{4950}\left(\dfrac{i}{i + 50}\right).\)

Now a hand calculator will handle multiplying all those fractions correctly, no overflow. But is easy to make a mistake, and it is slow. A spreadsheet is much quicker and much less error prone.

Edit: 4916 / 4966 = 0.9899. And 4950 / 5000 = 0.99. So

\(\displaystyle \dfrac{4965! * 4950!}{4915! * 5000!} = \displaystyle \prod_{i = 4916}^{4950}\left(\dfrac{i}{i + 50}\right) \approx 0.99^{35} \approx 0.70.\)
 
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JeffM said:
I am assuming you mean

\(\displaystyle \dfrac{4965! * 4950!}{4915! * 5000!} = \dfrac{\displaystyle 4915! * \left(\prod_{i=4916}^{4965}i\right) * 4950!}{\displaystyle 4915! * 4950! * \left(\prod_{i=4951}^{5000}i\right)} = \dfrac{\displaystyle \prod_{i=4916}^{4965}i}{\displaystyle \prod_{i=4951}^{5000}i} = \dfrac{\displaystyle \left(\prod_{i=4916}^{4950}i\right) \left(\prod_{i=4951}^{4965}i\right)}{\displaystyle \left(\prod_{i=4951}^{4965}\right) \left(\prod_{i=4966}^{5000}i\right)} = \dfrac{\displaystyle \prod_{i=4916}^{4950}i}{\displaystyle \prod_{i=4966}^{5000}i} = \displaystyle \prod_{i = 4916}^{4950}\left(\dfrac{i}{i + 50}\right).\)

Now a hand calculator will handle multiplying all those fractions correctly, no overflow. But is easy to make a mistake, and it is slow. A spreadsheet is much quicker and much less error prone.
Click to expand...

Elegant!!
 
JeffM said:
I am assuming you mean

\(\displaystyle \dfrac{4965! * 4950!}{4915! * 5000!} = \dfrac{\displaystyle 4915! * \left(\prod_{i=4916}^{4965}i\right) * 4950!}{\displaystyle 4915! * 4950! * \left(\prod_{i=4951}^{5000}i\right)} = \dfrac{\displaystyle \prod_{i=4916}^{4965}i}{\displaystyle \prod_{i=4951}^{5000}i} = \dfrac{\displaystyle \left(\prod_{i=4916}^{4950}i\right) \left(\prod_{i=4951}^{4965}i\right)}{\displaystyle \left(\prod_{i=4951}^{4965}\right) \left(\prod_{i=4966}^{5000}i\right)} = \dfrac{\displaystyle \prod_{i=4916}^{4950}i}{\displaystyle \prod_{i=4966}^{5000}i} = \displaystyle \prod_{i = 4916}^{4950}\left(\dfrac{i}{i + 50}\right).\)

Now a hand calculator will handle multiplying all those fractions correctly, no overflow. But is easy to make a mistake, and it is slow. A spreadsheet is much quicker and much less error prone.

Edit: 4916 / 4966 = 0.9899. And 4950 / 5000 = 0.99. So

\(\displaystyle \dfrac{4965! * 4950!}{4915! * 5000!} = \displaystyle \prod_{i = 4916}^{4950}\left(\dfrac{i}{i + 50}\right) \approx 0.99^{35} \approx 0.70.\)
Click to expand...

Wow! I can only say many many thanks JeffM!!:grin:
 
katikeit said:
Hi guys!
I really need to solve this expression without the help of the computer but only using a simple calculator...please help!!:wink:

4965! * 4950! / 4915! * 5000! \(\displaystyle \ \ \ \ \ \) Leaving a space in front of and after the "/" symbol isn't the same as putting needed grouping symbols around the fractions.

I've also tried to use the Stirling's approximation but the calculator can't afford it!

many thanks!!
Click to expand...

katikeit,

make sure in the future you type in needed grouping symbols to indicate what you want, such as:

(4965!*4950!)/(4915!*5000!)

JeffM said:
Edit: 4916 / 4966 = 0.9899. \(\displaystyle \ \ \ \ \) *\(\displaystyle \ 4916/4966 \approx 0.9899.\)*
And 4950 / 5000 = 0.99. So
Click to expand...


JeffM said:
Now a hand calculator will handle multiplying all those fractions correctly, no overflow. But is easy to make a mistake, and it is slow. A spreadsheet is much quicker and much less error prone.
Click to expand...
But the spreadsheet isn't allowed to be used, because that violates the OP's requirement that a computer (other than hand-held calculator) can't be used.
 
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See Jeff,

You even "bowled" Denis over!!!!
 
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katikeit said:
Hi guys!
I really need to solve this expression without the help of the computer but only using a simple calculator...please help!!:wink:

(4965!*4950!)/(4915!*5000!)

I've also tried to use the > > > Stirling's approximation < < < but ...
Click to expand...

Here is a method using Stirling's approximation:


\(\displaystyle ln(x) \ \ stands \ \ for \ \ log_e(x), \ \ \) the natural logarithm of x


\(\displaystyle ln(n!) \ \ is \ \ asymptotic \ \ to \ \ \ n*ln(n) \ - \ n.\)


Let K = (4965!*4950!)/(4915!*5000!)

Let A = 4965

Let B = 4950

Let C = 4915

Let D = 5000


\(\displaystyle Then \ \ K \ \ = \ \dfrac{(A!)*(B!)}{(C!)*(D!)}\)


\(\displaystyle ln(K) \ = \ ln(A!) \ + \ ln(B!) \ - \ ln(C!) \ - ln(D!)\)


\(\displaystyle ln(K) \ \approx \ [A*ln(A) \ - \ A] \ + \ [B*ln(B) \ - \ B] \ - \ [C*ln(C) \ - C ] \ - \ [D*ln(D) \ - \ D]\)


\(\displaystyle ln(K) \ \approx \ A*ln(A) \ - \ A \ + \ B*ln(B) \ - \ B \ - \ C*ln(C) \ + C \ - \ D*ln(D) \ + \ D\)


\(\displaystyle ln(K) \ \approx \ A*ln(A) \ + \ B*ln(B) \ - \ C*ln(C) \ - \ D*ln(D) \ - \ A \ - \ B \ + \ C \ + \ D\)


\(\displaystyle - A \ - \ B \ + \ C \ + \ D \ = \ -4965 \ - \ 4950 \ + \ 4915 \ + \ 5000 \ = \ 0\)


\(\displaystyle ln(K) \ \approx \ 4965*ln(4965) \ + \ 4950*ln(4950) \ - \ 4915*ln(4915) \ - \ 5000*ln(5000)\)


\(\displaystyle ln(K) \ \approx \ -0.353\)


\(\displaystyle e^{ln(K)} \ \approx \ e^{-0.353}\)


\(\displaystyle K \ \approx \ 0.70\)
 
Denis said:
> > > BUT the < < < OP asked for a method that can be handled by a simple calculator! Here:
"...only using a simple calculator...please help!!:wink:
Click to expand...

But nothing. I used Stirling's approximation that the OP mentioned, and I did my work on a scientific calculator only.

And all of that relatively congested, use of symbolism (repeated use of capital pi) used for the multiplying, which would

not have been expected to be known to the OP used by JeffM, could have been looked up on the Internet,

just as the formula for Stirling's approximation was looked up.

Now run along and don't forget to take your emoticon with you, Denis.
 
Denis said:
the stuff you showed cannot be handled by a simple calculator,
and "simple calculator" is a condition as per OP.
Click to expand...


Hello, Denis. I don't understand. As an example, none of the calculations I enter in my scientific calculator causes an overflow/error.


The largest magnitude (ignoring the sign) of the individual number at any one time for one of my calculations appears to be

5,000*ln(5,000), which is about 42,586.
 
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lookagain said:
And all of that relatively congested, use of symbolism (repeated use of capital pi) used for the multiplying, which would

not have been expected to be known to the OP used by JeffM, could have been looked up on the Internet,
Click to expand...
I of course recognize that my contribution was nowhere near as helpful to the student as was your contribution reminding her about the proper use of grouping symbols. Undoubtedly, that was what she really was looking for.

Nevertheless, I expected someone who knew Stirling's approximation would also know the symbol for repeated multiplication.

And guess what? She did know it.

katikeit said:
I can only say many thanks JeffM!!
Click to expand...

Furthermore, demonstrating that expressions like the one she mentioned can often be greatly simplified by a little bit of canceling may help her on other problems. Sometimes doing it even permits an exact answer, which seldom happens using Stirling's approximation.
 
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