# B Smallest non-zero derivative

1. Oct 13, 2016

### MiLara

What does a function's smallest non-zero derivative say about the function? For example, say we have a function that looks like 6x^3, if you keep taking the derivative of this function until you get the smallest non-zero derivative, in this case 6x^3 -> 18x^2 -> 36x -> 36, what is the significance of the number 36 to the function 6x^3? I know that each function has a specific smallest non-zero derivative, however, each non-zero derivative can be characteristic of an infinite amount of functions in you keep integrating it.

Is there anything to this thought, or am i just asking a pointless question?

also, i was playing with numbers and derived that for nx^l, when l is a whole number, the smallest non zero derivative is (l-1)!*n*l.

2. Oct 13, 2016

### Staff: Mentor

Smallest? If you're limited to polynomial functions, then perhaps you mean lowest degree.
For other functions, I don't think this is a meaningful term. For instance, with f(x) = cos(x), g(x) = ex, h(x) = 1/x, and many, many others.
The only significance I can think of is not the number you end up with (if you do indeed end up with a constant) is in the particular derivatives.
First derivative gives you the slope of a line tangent to a curve (or the velocity if the function represents displacement).
Second derivative indicates the concavity of the function (or acceleration if the function represents displacement).
Third derivative indicates the rate of change of the 2nd derivative. In a displacement, velocity, accleration context, the 3rd derivative indicates the "jerk."
There's one for the 4th derivative, but what it indicates escapes me at the moment.

3. Oct 13, 2016

### PeroK

If you were put in a room with a pen and paper and no calculator and asked to work out all the whole number cubes as high as you could, how would you do it?

Or, if you were asked to calculate the values of $6n^3$ for whole number $n$? Would the number $36$ be helpful in any way?

4. Oct 13, 2016

### MiLara

n 6n^3 6n^3/36
1 6 0.166666667
2 48 1.333333333 I am not seeing any relationships between 6n^3 and 36?
3 162 4.5
4 384 10.66666667
5 750 20.83333333
6 1296 36
7 2058 57.16666667
8 3072 85.33333333
9 4374 121.5
10 6000 166.6666667

5. Oct 13, 2016

### MiLara

I did perhaps stumble upon a solution for (6n)^3 using 36. I got that (6n)^3 = n^2*6n*36. I think that it is just coincidental since I could not generalize this for any function that looks like (xn)^l because not all (xn)^l = n^2*xn*smallest non zero derivative.

Last edited: Oct 13, 2016
6. Oct 14, 2016

### PeroK

It's easier to see the relationship between the number $6$ and $n^3$

---, ---, ---, 0
---, ---, 1, 1
---, 6, 7, 8
6, 12, 19, 27
6, 18, 37, 64
6, 24, 61, 125

You can keep generating the whole number cubes by using $6$ as the difference of differences of differences.

For $6n^3$ the difference of differences of differences is $36$.

7. Oct 14, 2016

### MiLara

I tried to follow what I thought was your algorithim but with replacing 6 with 36. This is what i got.
__,__,__,0
__,__,1,1
__,36,37,38
36,72,109,147
36,108,217,364
36,144,361,725

maybe I just need to expand this out further. Hmmm...I wonder if I can predict how many more columns I will need to get 6n^3.
Thank you for this little thought experiment.

8. Oct 14, 2016

### PeroK

You missed that for $n = 1$, $6n^3 = 6$. So, you were off on the wrong foot!

Also, for $n = 2$, $6n^3 = 48$. And not $38$.

9. Oct 14, 2016

### MiLara

Ok I got it for 6x^3

--,--,--,--0
--,--,6,6
--,36,42,48
36,72,114,162
36,108,222,384
36,144,336,750

However, when I try to do the same for another function, let's say 2x^4, it does not seem to work.
So, if I keep taking the derivative of 2x^4 -> 8x^3 -> 24x^2 -> 48x -> 48, then try to use 48 as the difference of differences of differences, i get the following

--,--,--,0
--,--,2,2
--,48,50,52

But when x=2, 2x^4 = 32, not 52.
Does this only work for cubes?

10. Oct 15, 2016

### PeroK

With $x^4$ it's the fourth derivative that is a constant! So, need an extra difference of differences.

It works for any polynomial. But, for a polynomial of order $n$ you must first calculate up to $n-1$ by hand before the $n^{th}$ term can be generated using the algorithm.