Finding the nth+1 derivative via Leibniz

[Gallagher]

Homework Statement

I'm told to find the nth derivative of a function via http://www.math.osu.edu/~nevai.1/H16x/DOCUMENTS/leibniz_product_formula_H6.pdf.

Homework Equations

Then I'm asked to show that f⁽ⁿ⁺¹⁾=f⁽ⁿ⁾+3n(n-1)f^(n-2) evaluated at x=0 when n>1

The Attempt at a Solution

I can solve this just fine when plugging in a constant for n. For example, when n=2, f '" = f " + 6 f, no problem. My question though is whether this is enough to prove that the above equation balances for all n>1?

If not, how would I show this? Would I need to use the general form of f⁽ⁿ⁾ I found via the Leibniz formula? How would this work? I ask this because it is of the form f⁽ⁿ⁾ = Ʃ_k=0ⁿ (n!/k!(n-k!) * (d^k/dx^k*f) * (d^n-k/dx^n-k*g)) and am not sure how I would transform that into f⁽ⁿ⁺¹⁾ or f^(n-2) with the summation and combination.

Thank you in advance!

 

[vela]

If not, how would I show this? Would I need to use the general form of f⁽ⁿ⁾ I found via the Leibniz formula? How would this work? I ask this because it is of the form f⁽ⁿ⁾ = Ʃ_k=0ⁿ (n!/k!(n-k!) * (d^k/dx^k*f) * (d^n-k/dx^n-k*g)) and am not sure how I would transform that into f⁽ⁿ⁺¹⁾ or f^(n-2) with the summation and combination.

Are you using f to stand for more than one function here? It appears on both the left and in the summation. How does g fit into all of this?

 

[SammyS]

Homework Statement

I'm told to find the nth derivative of a function via http://www.math.osu.edu/~nevai.1/H16x/DOCUMENTS/leibniz_product_formula_H6.pdf.

Homework Equations

Then I'm asked to show that f⁽ⁿ⁺¹⁾=f⁽ⁿ⁾+3n(n-1)f^(n-2) evaluated at x=0 when n>1

The Attempt at a Solution

I can solve this just fine when plugging in a constant for n. For example, when n=2, f '" = f " + 6 f, no problem. My question though is whether this is enough to prove that the above equation balances for all n>1?

If not, how would I show this? Would I need to use the general form of f⁽ⁿ⁾ I found via the Leibniz formula? How would this work? I ask this because it is of the form f⁽ⁿ⁾ = Ʃ_k=0ⁿ (n!/k!(n-k!) * (d^k/dx^k*f) * (d^n-k/dx^n-k*g)) and am not sure how I would transform that into f⁽ⁿ⁺¹⁾ or f^(n-2) with the summation and combination.

Thank you in advance!

It looks like a straight up proof by induction. You claim to now how to do the base step: that's showing it's true when n=2 .

For the inductive step:

Assume the statement is true when n = k .

From that assumption (plus your mathematical skill) show that the statement is true for n = k+1.​

In other words:

Assume that the following is true for k > 1 :

f^(k+1) = f^(k)+3k(k-1)f^(k-2) evaluated at x=0 .​

Show that the following follows from this assumption :

f^((k+1)+1) = f^(k+1)+3(k+1)((k+1)-1)f^((k+1)-2)​

which simplifies to:

f^(k+2) = f^(k+1)+(3k+3)(k)f^(k-1)​

All evaluated at x=0 .