Finite Differences & Leading Coefficients Relation

[dvmckay23]

Homework Statement

Determine the relation that exists between the nth finite difference and the leading coefficient.

Homework Equations

... I'm not too sure how to html it up properly, but the numbers/"n"s following the "a"s are meant to be sub-script:
f(x) = anx^n + an-1x^n-1 + ... + a2x^2 + a1x + a0

The Attempt at a Solution

If someone could even tell me the proper direction for the line of thought here. All I can figure out thus far is that the difference is the same sign (positive or negative) of the coefficient. Any hints would be GREATLY appreciated!

Thanks!
~D

 

[HallsofIvy]

Are you assuming an nth degree polynomial?

Try some easy cases first. What is the first difference of ax+ b?
What is the second difference of ax^2+ bx+ c?
What is the third difference of ax^3+ bx^2+ cx+ d?

 

[Architeuthis Dux]

The relation between the nth finite difference and the leading coefficient can be described using the general equation for a polynomial function:

f(x) = anxn + an-1xn-1 + ... + a2x^2 + a1x + a0

The nth finite difference is calculated by taking the difference between the values of f(x) at x = n and x = n-1. This can be represented as:

Δnf(x) = f(n) - f(n-1)

Substituting the general polynomial equation, we get:

Δnf(x) = (anxn + an-1xn-1 + ... + a2x^2 + a1x + a0) - (an-1xn-1 + ... + a2x^2 + a1x + a0)

Simplifying, we get:

Δnf(x) = anxn - an-1xn-1

We can see that the nth finite difference is directly related to the leading coefficient, an. This relationship can be generalized to:

Δnf(x) = anxn

This means that the nth finite difference is equal to the leading coefficient multiplied by xn. This relationship holds true for all polynomial functions.

In conclusion, the relationship between the nth finite difference and the leading coefficient is that the nth finite difference is equal to the leading coefficient multiplied by xn.