# need help with limit proof

• Nov 2nd 2007, 07:55 AM
MKLyon
need help with limit proof
I'm having trouble thinking this out.
I have this problem:
Suppose that for all x in (−c, c) the identity
a0 + a1x + .... + an-1x^(n-1) + (an + A(x))x^n = b0 + b1x + .... + bn-1x^(n-1) + (bn + B(x))x^n, where the limit as x goes to 0 of A(x) = lim as x goes to 0 of B(x) = 0. How would I show that a0 = b0, a1 = b1, ...... , an = bn. It seems to make sense, but I'm not quite show sure how to show it. How would you do it? Thanks for any help on this.
• Nov 2nd 2007, 07:59 AM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
I'm having trouble thinking this out.
I have this problem:
Suppose that for all x in (−c, c) the identity
a0 + a1x + .... + an-1x^(n-1) + (an + A(x))x^n = b0 + b1x + .... + bn-1x^(n-1) + (bn + B(x))x^n, where the limit as x goes to 0 of A(x) = lim as x goes to 0 of B(x) = 0. How would I show that a0 = b0, a1 = b1, ...... , an = bn. It seems to make sense, but I'm not quite show sure how to show it. How would you do it? Thanks for any help on this.

$\displaystyle a_0 + a_1x+...+ (a_n +A(x))x^n = b_0+b_1x+...+(b_n+B(x))x^n$
Take the limit $\displaystyle x\to 0$ of both sides, and we get,
$\displaystyle a_0=b_0$.
Subtract these from both sides to get,
$\displaystyle a_1x+...+(a_n+A(x))x^n = b_1x+...+(b_n+B(x))x^n$
Divide by $\displaystyle x\not = 0$ to get,
$\displaystyle a_1+...+(a_n+A(x))x^{n-1}=b_1+...+(b_n+B(x))x^{n-1}$.
Take the limit again,
$\displaystyle a_1 = b_1$
Keep on repeating this argument.
• Nov 2nd 2007, 08:13 AM
MKLyon
That's really clever. Thank you. One other question:
If the polynomial p(x) = the sum from k = 0 to n of akx^k for all x in (-c,c), what would be the coefficients ak?

Thanks again for the help.
• Nov 2nd 2007, 10:36 AM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
That's really clever. Thank you. One other question:
If the polynomial p(x) = the sum from k = 0 to n of akx^k for all x in (-c,c), what would be the coefficients ak?

Thanks again for the help.

For any polynomial $\displaystyle p(x)$ we can write $\displaystyle p(x) = \sum_{n=0}^{\infty} \frac{p^{(n)}(0)}{n!}x^n$*.

*)This is actually a finite sum because eventually the derivative is zero. Take for example $\displaystyle p(x) = 1+x+x^2$ then $\displaystyle p'''(x) = 0$ so $\displaystyle p^{(n)}(x) = 0$ for all $\displaystyle n\geq 3$.
• Nov 2nd 2007, 11:24 AM
MKLyon
If the polynomial (I'll write out the sum in long form) is:
a0 + a1x^1 + .... + anx^n and it equals zero, doesn't it mean all the coeffcients ak must be zero? Is this the answer to my second question?
• Nov 2nd 2007, 11:32 AM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
If the polynomial (I'll write out the sum in long form) is:
a0 + a1x^1 + .... + anx^n and it equals zero, doesn't it mean all the coeffcients ak must be zero? Is this the answer to my second question?

A polynomial $\displaystyle f(x)$ which is non-zero has at most $\displaystyle \deg f(x)$ zeros. So if $\displaystyle f(x)$ is always zero on an interval then it must be the zero polynomial because it can only have finite number of zeros.