Show that the nth divided difference of is 1
solution:
We have,
dash indicating that the term is to be excluded from denominator
I don't understand this step. Can any math helper,math expert explain me? Especially 1 on R. H. S.
Show that the nth divided difference of is 1
solution:
We have,
dash indicating that the term is to be excluded from denominator
I don't understand this step. Can any math helper,math expert explain me? Especially 1 on R. H. S.
Hey Chiro,
As per my understanding,L. H. S.-the term next to summation'(sigma') in R. H. S. is equal to 1 as stated in the problem.I think it is the nth divided difference of n arguments()
Next steps in the solution are as follows:Put x=0 on both sides, we have
0=1 - f{}
Hence f{ }= 1
If I am wrong, tell me.
Hey chiro,
Your reply suggests me that you are incognizant of divided difference properties. The term next to sigma' is based on the property which states that divided difference is a symmetrical function of all the arguments involved.
Try it when $n=0$.
It is easily verified that
$\dfrac{x}{x-x_0} = 1 + \dfrac{x_0}{1}\dfrac{1}{x-x_0}$?
So, let's check it out when $n=1$:
$\dfrac{x^2}{(x-x_0)(x-x_1)} = 1+\dfrac{x_0^2}{x_0-x_1}\dfrac{1}{x-x_0} + \dfrac{x_1^2}{x_1-x_0}\dfrac{1}{x-x_1}$
The common denominator on the RHS is
$(x_0-x_1)(x-x_0)(x-x_1)$
So, the question is, is the following equation true?
$(x_0-x_1)(x-x_0)(x-x_1) + x_0^2(x-x_1) - x_1^2(x-x_0) \stackrel{?}{=} (x_0-x_1)x^2$
Well, the LHS becomes
$(x_0-x_1)(x^2-x_0x-x_1x+x_0x_1) + x_0^2x - x_0^2x_1 - x_1^2x + x_1^2x_0$
$=(x_0-x_1)x^2 - x_0^2x-x_0x_1x+x_0^2x_1+x_0x_1x + x_1^2x - x_0x_1^2 + x_0^2x - x_0^2x_1 - x_1^2x + x_1^2x_0$
$=(x_0-x_1)x^2$
Hence, the LHS and the RHS are equal when $n=1$. To prove it is true for all $n$, you can attempt to use induction. You know it is true when $n=0$ (and when $n=1$), so assume it is true when $n=k$ and try to prove it for $n=k+1$.