You are not logged in.
Pages: 1
Please be kind. I just registered. You guys are Beethoven..Chopin..Mozart. I am a street-corner harmonica player. Is there a reason why the increases in the differences of consecutive cube numbers resemble a "6" times table? "Differences in differences" in consecutive numbers to the 4th power show a pattern involving a multiple of 6 also.
Offline
Hi galoruce,
Welcome to the forum!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
hi galoruce
Welcome to the forum.
Yes there is a reason and you can use some algebra to prove it. I'll give you an outline of what to do and perhaps you can do it for yourself. If not, post again and I'll fill in the gaps.
Start with a general expression for a cubic:
Then write out the expression if you increase x by 1:
If you expand those brackets and then create the difference expression by subtracting the first from the second you'll end up with a quadratic.
So first differences are always a quadratic.
Now repeat that by writing a second quadratic with x changed to (x+1) and subtract to get the second difference expression.
You should find it's a linear expression (ie. an x term and a constant)
The x term has a factor of 6.
And if you carry out one more difference calculation you'll end up with the third difference = 6a .
There's the proof.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Thanks,Bob. Lots of work. I'm not smart enough to use what you gave me. Can you substitute numbers for letters?
Offline
hi galoruce
I have chosen a 'random' cubic. The table above shows the values for f(x) = 5x^3 + 2x^2 - 7x + 9 for values of x from 1 to 8, with the first, second and third differences. What your post concerns is the final column. Yes, the values are all 30 and in the 6 times table. Notice also that 30 = 5 *6. This is no coincidence. The cubic coefficient will always show up in the third differences column as that number times 6. I'll try to show why:
The calculation for x is
5x^3 + 2x^2 - 7x + 9 ...............................(a)
For the next calculation x is replaced by x+1:
5(x+1)^3 + 2(x+1)^2 - 7(x+1) + 9 = 5x^3+ 3*5x^2 + 15x + 5 + 2x^2 + 4x + 2 - 7x - 7 + 9 .........(b)
Subtract (a) from (b)
3*5x^2 + 15x + 5 + 4x + 2 - 7 = 3x5x^2 + 19x
Now calculate the next first difference by replacing x by x+1
3x5(x+1)^2 + 19(x+1) = 3*5x^2 + 2*3*5x + 15 + 19x + 19
Subtracting to get the second difference:
= 2*3*5x + 16
Replacing x by x+1
2*3*5(x+1) + 16 = 2*3*5x + 2*3*5 + 16
Subtracting
2*3*5
I have tried to track the coefficient 5 throughout the algebra by highlighting it in red. Hopefully I got them all. You can see that the first difference 5 occurs as 3*5 in the calculation. After another difference it occurs as 2*3*5.
If you replace every 5 with 'a' you get the general calculation. So the third difference would then by 2*3*a.
Hope that helps,
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Pages: 1