I need your help

Jai Ganesh · 2005-07-10 22:56:17

I'll give my opinion tomorrow on Mathsy's proof, when I start the day afresh;
Regarding including n!, it would be of immense help as combinatorics works mostly on
factorials;
nPr= n!/(n-r)!
and
nCr = n!/(n-r)!*r!
nPr gives the number of Permutations
and
nCr gives the number of combinations.
There is an approximation for factorials of higher numbers;
it is called the James-Stirling formula
n! is approximately equal to
√(2*Pi*n) multiplied by (n/e)^n
This is a brilliant approximation as the value of n goes up.

mathsyperson · 2005-07-10 23:00:55

You probably know more about this kind of thing than me, but wouldn't the algorithm for n! take (n-1) multiples, meaning that it would be just as complicated as the 2^n one, if not more so because it doesn't have the prime factor shortcut?

Also, I was messing around with the calculator putting in multiple decimal points and it gave me an answer! It didn't make sense and I know it doesn't matter at all, but it's still quite interesting.

...57423423230896109004106619977392256259918212890625
+ ...42576576769103890995893380022607743740081787109376
----------------------------------------------------------------------------
...(1)00000000000000000000000000000000000000000000000001

Jai Ganesh · 2005-07-10 23:23:21

42576576769103890995893380022607743740081787109376

Mathsy, did you get this number too by exhaustion, just like you got the 30 digits????:)

mathsyperson · 2005-07-10 23:27:44

Sorry, but no.
I extended the 5 one, because that's easy, then took all of its digits away from 9 to come up with that.
Does that mean you found something wrong with my proof?

Jai Ganesh · 2005-07-10 23:29:27

This has got nothing to do with the proof,
I was just a bit inquisitive!:D

mathsyperson · 2005-07-10 23:34:32

Phew! I haven't been proven wrong...yet.
By the way, I love the factorial approximation.

MathsIsFun · 2005-07-11 01:03:04

OK, a newer version is available (here, as usual).

Caution: x^y only uses the integer portion of y and will be very slow, because I am using the simplest "brute force" method of multiplying "y" times. I will work on an improved version as time permits.

Jai Ganesh · 2005-07-11 16:37:25

Both x! and x^y work, but for larger values, as you rightly pointed out, it becomes very slow.
However, a useful tool.

MathsIsFun · 2005-07-11 18:47:15

Added X/Y, so now it has the basic functions of a calculator.

I don't have the time right now to convert my prime factorization tool to javascript to speed up the X^Y function, so if you want to calculate 2^2500 then do 2^100, then copy-paste this to X and raise it another 25.

Jai Ganesh · 2005-07-11 19:14:12

Consider giving a link to the calculator on the Index page,
since not many may be aware, when the link is in this forum.

Jai Ganesh · 2005-07-11 22:12:21

mathsyperson wrote:

We are only interested in the 10^6 coefficient of the square of this. That will be equal to 2(y*(6*10^0))+a constant made up of the expansion of the other powers of 10, but that does not involve y.
So, the 10^6 coefficient is 12y+c, which should only be equal to y for one value of y.
We are only interested in the last digit of 12y+c, so it can be called 2y+c without affecting our purposes. However big c is, it can also have all of its digits except the last removed.

Mathsy, how did you get '2(y*(6*10^0))+a constant made up of the expansion of the other powers of 10, but that does not involve y'?

Last edited by Jai Ganesh (2005-07-11 22:13:59)

mathsyperson · 2005-07-11 23:08:41

I had 'y*10^6+1*10^5+0*10^4+9*10^3+3*10^2+7*10^1+6*10^0' and I wanted to square this.
I would need to multiply each term by every other term to do this, which would involve multiplying the 2 coefficients and then writing *10^(the 2 powers of 10 added up). This means that in this case, the term involving 10^6 would be:
(y*10^6)(6*10^0)+(1*10^5)(7*10)+(0*10^4)(3*10^2)+(9*10^3)²+(3*10^2)(0*10^4)+(7*10)(1*10^5)+(6*10^0)(y*10^6).
The only terms in this expansion which involve y are the first and last, meaning that it could be written as 2(y*10^6)(6*10^0)+c, because all the other terms are independant of y and so will remain the same. Also, if the 10^5 coefficient is >9 or the 10^4 coefficient is >99 etc, then they will affect the result, but the entire expansion of any of these does not involve y, so it will just add to the constant.

Does that help?

mathsyperson · 2005-07-12 01:48:51

I'm bored, so:

...8509890062166509580863811000557423423230896109004106619977392256259918212890625
+ ...1490109937833490419136188999442576576769103890995893380022607743740081787109376
---------------------------------------------------------------------------------------------------------------------
...(1)0000000000000000000000000000000000000000000000000000000000000000000000000000001

Sorry, no exhaustion.

Jai Ganesh · 2005-07-12 16:43:08

mathsyperson wrote:

I'm not good at proofs, but this is my attempt:
c is a constant, so there is only one value for y.
As shown in the expansion of y109376, the important coefficient will always take the form 2(y*6), because 6 is always the last digit. This means that this proof can be carried forward for any point on the chain of the magic number.[/proof]
There's probably a flaw in there somewhere, could someone check it please?

I am unable to find any serious flaw, yet,
I am not fully convinced.
I shall try with a much simpler number, viz. y76.
y76 is (y*10^2)+(7*10^1)+(6*10^0).
When this number is squared,
[This is of the form (a+b+c)²
which is equal to a² + b² + c² + 2ab + 2bc + 2ac]
we get
10000y² + 4900 + 36 + 14000y + 840 + 1200y
which is the same as
10000y² + 15200y + 5776
From this, it is clear, for any y belonging to Natural Numbers,
the last two digits are not affected. They continue to be 76.
But, does this conclusively prove there exists a value y such that
the last three digits of y76² are y76?

Mathsy, I am confused.

Last edited by Jai Ganesh (2005-07-12 16:45:23)

mathsyperson · 2005-07-12 21:19:19

In your example, you have 10000y²+15200y+5776.
As y is the hundreds digit in the original number, y76, then we are only interested in the hundreds digit of the square.
This will be the units digit of 2y+7.
We want to prove that there is one value of y betwwen 0 and 9 such that y76²=...y76.
We now know that the hundreds digit of the square is 2y+7, so we need to solve y=2y+7.
However, as it is only the units digit of 2y+7 that we are interested in, then we are allowed to add multiples of 10 to it without affecting the result. To make things easier, we can change the equation to y=2y-3.
This is a linear equation with one answer, in this case being 3, which is right because the 376²=...376.

The proof is basically the same for whichever number you want to do it on, y76, y9376, y2607743740081787109376, whatever.
This is because the important digit will always end up as 2y+c, as I explained originally.

MathsIsFun wrote:

Proofs are just way too strict.

I agree!

Jai Ganesh · 2005-07-12 21:32:51

Your reply seems convincing!
I shall take hard copies of your posts,
study them all again, then again,
before finally saying,
"You are right, Mathsy!"
Give me just a few days!

MathsIsFun · 2005-07-12 21:41:00

mathsyperson wrote:

MathsIsFun wrote:
Proofs are just way too strict.

Where did I say that?

Anyway, I agree with me.

mathsyperson · 2005-07-12 21:48:21

Woo! I'm convincing!

To MathsIsFun, judging by the address, it seems like the very first topic made here!
http://www.mathsisfun.com/forum/viewtopic.php?id=1

Jai Ganesh · 2005-07-12 23:46:26

MathsIsFun wrote:

mathsyperson wrote:
MathsIsFun wrote:
Proofs are just way too strict.
Where did I say that?
Anyway, I agree with me.

Mathematical proof is based axioms and proven theorems, and proofs are viewed throough a microscope.
Euler had claimed x^4+y^4+z^4=w^4 had no solutions;
In 1988, Naom Elkies of Harvard University discovered that
2,682,440^4 + 15,365,639^4 + 18,796,760^4 = 20,615,673^4
Merely testing the first million numbers cannot constitute a mathematical proof!

mathsyperson · 2005-07-12 23:52:55

Good old disproof by counter-example.
By far the easiest kind of proof.

MathsIsFun · 2005-07-13 00:18:32

You see, I am right, they are too strict! You can try your best, and still they can come undone.

Maybe they should be called "beyond reasonable doubt"s

And you're right mathsy, the counter-example is by far the easiest. All cats are black! But how about this white one here?

Jai Ganesh · 2005-07-13 17:01:10

Mathematical Induction is a brilliant way of proving.
For example, the sum of the first n Natural Numbers is
n(n+1)/2.
For n=1, we get 1.
For n=2, we get 3.
Therefore, the summation formula works for 1 and 2.

Let's assume it is true for an arbitrary Natural Number k.
Therefore, the sum of the first k Natural Numbers would be k(k+1)/2.

For k+1, the sum would be k(k+1)/2 + (k+1),
that is [k(k+1) + 2(k+1)]/2, or [(k+1)(k+2)]/2

And, according to the formula we had at the beginning of the proof,
the sum of the first k+1 Natural Numbers would be [(k+1)(k+2)]/2

Since the two are equal, this can be said to be true for sum up to any natural number.

Similarly, it can be proved that:-
the sum 1² + 2² + 3² + 4² +... n² = [n(n+1)(2n+1)]/6
and also
the sum 1³ + 2³ + 3³ + 4 ³ + ......... n³ = [n(n+1)/2]²

Last edited by MathsIsFun (2005-07-13 18:21:38)

MathsIsFun · 2005-07-13 18:22:41

(just neatened up your excellent post a little for clarity, hope you don't mind)

In this type of proof you:

a) assume it works for n, then
b) check to see if it remains true for (n+1)

Is it also necessary to prove it for n=1 ?

Roraborealis · 2005-07-13 18:52:20

MathsIsFun wrote:

Where did I say that?

Here. Fifth post down.

Jai Ganesh · 2005-07-13 19:04:11

MathsIsFun wrote:

Is it also necessary to prove it for n=1 ?

First, the formula or equation is checked for values like 1,2,3 etc.Only if this test is pased, we asume it is true for an arbitrary number k. Thereafter, if it is proved true for k+1, it becomes a 'proof beyond doubt'.

Math Is Fun Forum

#51 2005-07-10 22:56:17

Re: I need your help

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