# Thread: Theorem involving delta function: proof by induction

1. ## Theorem involving delta function: proof by induction

Hi Folks,

I am working through a theorem involving the delta function. I have some queries in the early stages of derivation. See attached.

Thanks

2. 1. There is k -th derivative of delta function and differencial of k-th derivative
gives k+1 derivative.
2. The k-th derivative of delta function at $\pm \; \infty$ =0.
You may think delta function being proportional to $e^{-x^2}$.

3. Originally Posted by zzzoak
1. There is k -th derivative of delta function and differencial of k-th derivative
gives k+1 derivative.
2. The k-th derivative of delta function at $\pm \; \infty$ =0.
You may think delta function being proportional to $e^{-x^2}$.
I am confused now because in the first line of attachment
$d x^{k+1} = x^k$ but then for the delta function you state that

$d \delta ^k (x) = \delta^{k+1}$
Are both treated differently? Thanks

4. Yes, they are treated differently.

Delta Function -- from Wolfram MathWorld
(17) equation.

For k-th derivative we write
$
\delta^{(k)}
$

and k-th degree
$
x^{k}.
$

5. Originally Posted by zzzoak
Yes, they are treated differently.

Delta Function -- from Wolfram MathWorld
(17) equation.

For k-th derivative we write
$
\delta^{(k)}
$

and k-th degree
$
x^{k}.
$
Thanks,

I have written out the problem fully, just have 2 queries highlighted in red. Hopefully it can be read!!

1) At the Start, integration by parts query
2) how we got from (k+1)(-1)(-1)^kK!=(-1)^(k+1)!

6. I am not sure you have some problems with this line

$
(-1)(-1)^k \; (k+1)k! \; = \; (-1)^{k+1} \; (k+1)! \; \; ?
$

7. Originally Posted by zzzoak
I am not sure you have some problems with this line

$
(-1)(-1)^k \; (k+1)k! \; = \; (-1)^{k+1} \; (k+1)! \; \; ?
$
Yes, Im wondering how the LHS equates to the RHS? Ie the k -th derivative goes to k+1 etc

Thanks

8. (k+1)!=(k+1)k! and (-1)(-1)^k=(-1)^(k+1)

Cheers