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**galoruce****Member**- Registered: 2018-04-24
- Posts: 2

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.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 25,862

Hi galoruce,

**Welcome to the forum!**

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,386

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

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**galoruce****Member**- Registered: 2018-04-24
- Posts: 2

Thanks,Bob. Lots of work. I'm not smart enough to use what you gave me. Can you substitute numbers for letters?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,386

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

**5**x^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 = **5**x^3+ **3*5**x^2 + 15x + 5 + 2x^2 + 4x + 2 - 7x - 7 + 9 .........(b)

Subtract (a) from (b)

**3*5**x^2 + 15x + 5 + 4x + 2 - 7 = **3x5**x^2 + 19x

Now calculate the next first difference by replacing x by x+1

**3x5**(x+1)^2 + 19(x+1) = **3*5**x^2 + **2*3*5**x + 15 + 19x + 19

Subtracting to get the second difference:

= **2*3*5**x + 16

Replacing x by x+1

**2*3*5**(x+1) + 16 = **2*3*5**x + **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

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