A personal mathematical discovery/dilemma
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A personal mathematical discovery/dilemma
2nd power numbers:
1, 4, 9, 16, 25, 36, 49=1×1, 2x2, 3×3, etc.
3, 5, 7, 9, 11, 13 = difference of the square roots above (4-1, 9-4, etc.)
2, 2, 2, 2, 2= 2 is the constant difference
3rd power numbers: let's do the same thing
1, 8, 27, 64, 125, 216, 1x1x1, 2x2x2, etc.
7, 19, 37, 61, 91 = differences again
12, 18, 24, 30
6, 6, 6, 6= constant of 6
4th power number:
1, 16, 81, 256, 625, 1296, 2401
15, 65, 175, 369, 671, 1105
40, 110, 194, 302, 434
70, 84, 108, 132
14, 24, 34,
10, 10 = 10 is the constant
EDIT: this should be 24, so pattern ends up being 2,6,24,120.
Constant pattern of 2, 6, 10 and differences of 4. Now when I did the 5th power the constant became 120 (I didn't type here cause too many calculations and lazy). Now multiply the first numbers and you get 2×6×10=120, why did 5th power change? Does 120 stay as a constant for awhile and then change if I continue this?
I'm obviously not the first to make this discovery of this pattern but I'm wondering if it has any deeper meaning that I don't understand (I'm thinking some significance in the difference of dimensions?) I tried to Google 2, 6, 10, 120 and it came up with nothing. Maybe I will do the tree for 6th power to see what I get. I think maybe milagros, jonsykkel, madness, jappe, or sunl can help for some reason but if anybody knows this pattern or significance please tell.
1, 4, 9, 16, 25, 36, 49=1×1, 2x2, 3×3, etc.
3, 5, 7, 9, 11, 13 = difference of the square roots above (4-1, 9-4, etc.)
2, 2, 2, 2, 2= 2 is the constant difference
3rd power numbers: let's do the same thing
1, 8, 27, 64, 125, 216, 1x1x1, 2x2x2, etc.
7, 19, 37, 61, 91 = differences again
12, 18, 24, 30
6, 6, 6, 6= constant of 6
4th power number:
1, 16, 81, 256, 625, 1296, 2401
15, 65, 175, 369, 671, 1105
40, 110, 194, 302, 434
70, 84, 108, 132
14, 24, 34,
10, 10 = 10 is the constant
EDIT: this should be 24, so pattern ends up being 2,6,24,120.
Constant pattern of 2, 6, 10 and differences of 4. Now when I did the 5th power the constant became 120 (I didn't type here cause too many calculations and lazy). Now multiply the first numbers and you get 2×6×10=120, why did 5th power change? Does 120 stay as a constant for awhile and then change if I continue this?
I'm obviously not the first to make this discovery of this pattern but I'm wondering if it has any deeper meaning that I don't understand (I'm thinking some significance in the difference of dimensions?) I tried to Google 2, 6, 10, 120 and it came up with nothing. Maybe I will do the tree for 6th power to see what I get. I think maybe milagros, jonsykkel, madness, jappe, or sunl can help for some reason but if anybody knows this pattern or significance please tell.
Last edited by gimp on 10 Jul 2017, 03:17, edited 1 time in total.
God Bless America
Re: A personal mathematical discovery/dilemma
I think you need way more data. My first thought is to check for many, many more numbers and plot them on a graph.
Maybe get some man to write a program that makes the calculations automagically.
Maybe get some man to write a program that makes the calculations automagically.
<veezay> antti also gonna get stabbed later this month
<nick-o-matic> niec
My fake plants died because I did not pretend to water them.
<nick-o-matic> niec
My fake plants died because I did not pretend to water them.
Re: A personal mathematical discovery/dilemma
Sorry but you can't count =D
The constant for 4th power is 24, not 10. In your calculations 65-15=40 and 132-108=34, and because of that you've got an extra line with 14 24 34 (these should be 24 24 24 already).
The powers go:
2 (2nd power)
6 (3rd power, prev*3)
24 (4th power, prev*4)
120 (5th power, prev*5)
720 (6th power, prev*6)
etc.
Why? Don't ask me :p if you came across this by yourself then pretty coal!
But yeah, do your Excel homework :p
The constant for 4th power is 24, not 10. In your calculations 65-15=40 and 132-108=34, and because of that you've got an extra line with 14 24 34 (these should be 24 24 24 already).
The powers go:
2 (2nd power)
6 (3rd power, prev*3)
24 (4th power, prev*4)
120 (5th power, prev*5)
720 (6th power, prev*6)
etc.
Why? Don't ask me :p if you came across this by yourself then pretty coal!
But yeah, do your Excel homework :p
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Re: A personal mathematical discovery/dilemma
Oh, in that case it's ez. It's the factorial sequence.
https://oeis.org/search?q=2%2C6%2C24%2C ... &go=Search
https://oeis.org/search?q=2%2C6%2C24%2C ... &go=Search
<veezay> antti also gonna get stabbed later this month
<nick-o-matic> niec
My fake plants died because I did not pretend to water them.
<nick-o-matic> niec
My fake plants died because I did not pretend to water them.
Re: A personal mathematical discovery/dilemma
Oh damn! yeah well that's embarassing. I guess the mistake came down to 132-108=24 not 34... Well it still has a pattern though like you have pointed out. I found it myself kind of, I was taking a test where the question was 1, 8, 27, ?, 125, 216. I eventually figured out the answer two ways, one with the constant 6 as the difference of the differences and the other with them being powers of 3. I wondered if there was a pattern if I tried different exponents and obviously made an error when doing the 4th power.
Will have to look into Ruben's post now.
Will have to look into Ruben's post now.
God Bless America
Re: A personal mathematical discovery/dilemma
yeah, but how? I mean, we're talking about differences of differences of differences of powers... That always amount to constant factorial numbers if you take a sufficient amount of differences. That's interesting imo :DRuben wrote:Oh, in that case it's ez. It's the factorial sequence.
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Re: A personal mathematical discovery/dilemma
I don't understand all the math lingo in Rubens link, but I do see the pattern of 1,2,6,24,120 is differences of differences of powers and can also be found out with logic like this:
1×1=1
1×2=2
2×3=6
6×4=24
24×5=120
So the first number is the solution (they are the constants in mine) of the last equation and the second number goes up by one (just like the exponents in mine).
1×1=1
1×2=2
2×3=6
6×4=24
24×5=120
So the first number is the solution (they are the constants in mine) of the last equation and the second number goes up by one (just like the exponents in mine).
God Bless America
Re: A personal mathematical discovery/dilemma
A factorial of a given integer is a product of all the numbers up to that integer (integer = "whole" number, e.g. 3 but not 3.2).
1! = 1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
5! = 1*2*3*4*5 = 120
etc.
so those numbers from the pattern represent the factorial sequence.
Funfact: the factorial sequence is famous for its extremely rapid increase, e.g. 20! ~ 2.4*10^18, or 2432902008176640000 :p
1! = 1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
5! = 1*2*3*4*5 = 120
etc.
so those numbers from the pattern represent the factorial sequence.
Funfact: the factorial sequence is famous for its extremely rapid increase, e.g. 20! ~ 2.4*10^18, or 2432902008176640000 :p
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Re: A personal mathematical discovery/dilemma
Sooo I spent some time trying to derive a formula proving this, and I haven't managed to derive a formula as such, but I noticed a tight connection with the Pascal's triangle, which also corresponds to the subsequent coefficients of expansions of (a+b)^n and (a-b)^n.
Namely, for each power n, the subsequent coefficients of the expansion are numbers from the nth row of Pascal's triangle (the very top row with a single 1 is taken as the 0th row), e.g.:
(a+b)^2 = a^2 + 2ab + b^2 (coefficients 1, 2, 1 -> 2nd row)
(a+b)^3 = a^2 + 3(a^2)b + 3a(b^2) + b^3 (coefficients 1, 3, 3, 1 -> 3rd row)
etc.
(For (a-b)^n every second coefficient is negative, starting from positive, so e.g. (a-b)^3 = a^2 - 3(a^2)b + 3a(b^2) - b^3).)
Now, if we look at the subsequent differences in gimp's pattern, taking "n" as the power and "k" as the number in question (1,2,3,...), the first row is the powers:
k^n (0th row)
Then there's the difference of the above minus the above for (k-1):
k^n - (k-1)^n (1st row)
Then there's the difference of the above minus the above for (k-1):
(k^n - (k-1)^n) - ((k-1)^n - (k-2)^n) = k^n - 2(k-1)^n + (k-2)^n (2nd row)
Then there's the difference of the above minus the above for (k-1):
(k^n - 2(k-1)^n + (k-2)^n) - ((k-1)^n - 2(k-2)^n + (k-3)^n) = k^n - 3(k-1)^n + 3(k-2)^n - (k-3)^n (3rd row)
then
k^n - 4(k-1)^n + 6(k-2)^n - 4(k-3)^n + (k-4)^n (4th row)
k^n - 5(k-1)^n + 10(k-2)^n - 10(k-3)^n + 5(k-4)^n - (k-5)^n (5th row)
...
And as you can see, the coefficients of each difference are numbers from the corresponding row of Pascal's triangle =)
Then, as we take some n, say n=3, we expect the final constant to be n! = 1*2*3 = 6. It should sit in the nth difference - 3rd row above, which is described as
k^n - 3(k-1)^n + 3(k-2)^n - (k-3)^n
Funnily enough, if we expand the above equation using n=3, we find that all the terms with k cancel out, and we're left with
-(1*0^3)+(3*1^3)-(3*2^3)+(1*3^3) = - 0 + 3 - 24 + 27 = 6 = 3! :)
This is true for all powers, e.g. for 5th power:
-(1*0^5)+(5*1^5)-(10*2^5)+(10*3^5)-(5*4^5)+(1*5^5) = - 0 + 5 - 320 + 2430 - 5120 + 3125 = 120 = 5!
(The above formulas start with negative coefficients as they are odd powers. Even powers would start from +. This comes from expansion of the (a-b)^n formula.)
Actually realised that the conclusion can be written as a neat formula :D
Where c0, ..., ck are coefficients of the nth row of Pascal's triangle :>
Dunno if anybody will try to follow it but was fun regardless :> If someone goes "hoh nice" then even better ;)
Namely, for each power n, the subsequent coefficients of the expansion are numbers from the nth row of Pascal's triangle (the very top row with a single 1 is taken as the 0th row), e.g.:
(a+b)^2 = a^2 + 2ab + b^2 (coefficients 1, 2, 1 -> 2nd row)
(a+b)^3 = a^2 + 3(a^2)b + 3a(b^2) + b^3 (coefficients 1, 3, 3, 1 -> 3rd row)
etc.
(For (a-b)^n every second coefficient is negative, starting from positive, so e.g. (a-b)^3 = a^2 - 3(a^2)b + 3a(b^2) - b^3).)
Now, if we look at the subsequent differences in gimp's pattern, taking "n" as the power and "k" as the number in question (1,2,3,...), the first row is the powers:
k^n (0th row)
Then there's the difference of the above minus the above for (k-1):
k^n - (k-1)^n (1st row)
Then there's the difference of the above minus the above for (k-1):
(k^n - (k-1)^n) - ((k-1)^n - (k-2)^n) = k^n - 2(k-1)^n + (k-2)^n (2nd row)
Then there's the difference of the above minus the above for (k-1):
(k^n - 2(k-1)^n + (k-2)^n) - ((k-1)^n - 2(k-2)^n + (k-3)^n) = k^n - 3(k-1)^n + 3(k-2)^n - (k-3)^n (3rd row)
then
k^n - 4(k-1)^n + 6(k-2)^n - 4(k-3)^n + (k-4)^n (4th row)
k^n - 5(k-1)^n + 10(k-2)^n - 10(k-3)^n + 5(k-4)^n - (k-5)^n (5th row)
...
And as you can see, the coefficients of each difference are numbers from the corresponding row of Pascal's triangle =)
Then, as we take some n, say n=3, we expect the final constant to be n! = 1*2*3 = 6. It should sit in the nth difference - 3rd row above, which is described as
k^n - 3(k-1)^n + 3(k-2)^n - (k-3)^n
Funnily enough, if we expand the above equation using n=3, we find that all the terms with k cancel out, and we're left with
-(1*0^3)+(3*1^3)-(3*2^3)+(1*3^3) = - 0 + 3 - 24 + 27 = 6 = 3! :)
This is true for all powers, e.g. for 5th power:
-(1*0^5)+(5*1^5)-(10*2^5)+(10*3^5)-(5*4^5)+(1*5^5) = - 0 + 5 - 320 + 2430 - 5120 + 3125 = 120 = 5!
(The above formulas start with negative coefficients as they are odd powers. Even powers would start from +. This comes from expansion of the (a-b)^n formula.)
Actually realised that the conclusion can be written as a neat formula :D
Where c0, ..., ck are coefficients of the nth row of Pascal's triangle :>
Dunno if anybody will try to follow it but was fun regardless :> If someone goes "hoh nice" then even better ;)
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Re: A personal mathematical discovery/dilemma
Some people have fun with counting n!-s, some people have fun with counting !bear-s. I don't drink bear, so counting n!-s is still more fun imo :)Madness wrote:Wtf dude, you have a weird idea of fun!
Re: A personal mathematical discovery/dilemma
I've always loved maths, especially working out stuff with pen and paper =) Especially things like this where there's no obvious solution are especially interesting for me, and the satisfaction of finally working it out is massive for my :> teh joy of maths!
also
also
iCS wrote:Some people have fun with counting n!-s, some people have fun with counting !bear-s.
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Re: A personal mathematical discovery/dilemma
It should be
Maybe later I'll try to figure out the proof.
Maybe later I'll try to figure out the proof.
Re: A personal mathematical discovery/dilemma
Heh, didn't realise that combinations corresponded to teh pascal triangle rows :p nice one! And the rest is definitely a more rigoristic way of describing teh even/odd alterations. GG :>
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Re: A personal mathematical discovery/dilemma
pretty sure i would lose math battle to a 3rd grader so next time leave me out of your fantasiesgimp wrote:I think ....can help.
Last edited by Sunshine on 11 Nov 2017, 04:49, edited 1 time in total.
Re: A personal mathematical discovery/dilemma
You still helped me smile jappe
Anyways wow pawq nice find using Pascal's triangle for coefficients to create a formula, pretty creative idea.
Anyways wow pawq nice find using Pascal's triangle for coefficients to create a formula, pretty creative idea.
God Bless America
Re: A personal mathematical discovery/dilemma
for a general polynomial of the n-th order p(x) = a_n * x^n + ... a_0 the n-th difference (aka discrete derivative) is clearly a_n * n!
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Re: A personal mathematical discovery/dilemma
Not gonna lie, I don't know what any of that means.
God Bless America
Re: A personal mathematical discovery/dilemma
Just going add that what you're essentially doing in the first post is you're getting the nth derivative of x^n. (except instead of looking at the slope over an infinitesimally small point, you are looking at the slope over a range of 1 (i.e. x=0 to x=1))
For example, n=3
f(x)=x^3=
1st derivative: f'(x)=3*(x^2)
2nd derivative: f''(x)=3*2*x
3rd derivative f'''(x)=3*2*1=6
In general terms,
(d^n/(dx)^n) (x^n) = n*(n-1)*(n-2)* ... * 1 = !n
Edit: I think mila just said exactly what I said in more technically correct terms (https://calculus.subwiki.org/wiki/Discrete_derivative)
I haven't done math in 6 years so I'm probably not the best reference
For example, n=3
f(x)=x^3=
1st derivative: f'(x)=3*(x^2)
2nd derivative: f''(x)=3*2*x
3rd derivative f'''(x)=3*2*1=6
In general terms,
(d^n/(dx)^n) (x^n) = n*(n-1)*(n-2)* ... * 1 = !n
Edit: I think mila just said exactly what I said in more technically correct terms (https://calculus.subwiki.org/wiki/Discrete_derivative)
I haven't done math in 6 years so I'm probably not the best reference
Re: A personal mathematical discovery/dilemma
to be more precise for the discrete case, f(x)=x^3, f'(x)=3x^2+O(x), f''(x)=6x+O(1), f'''(x) = 6sunl wrote:For example, n=3
https://en.wikipedia.org/wiki/Big_O_notation
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