Quod Erat Demonstrandum

2018/03/22

a question about inequality with derivatives

Filed under: Fun,mathematics,NSS,Pure Mathematics — johnmayhk @ 3:46 下午
Tags: , ,

Question

Let p(x) be a polynomial with real coefficients. Prove that if p(x)-p'(x)-p''(x)+p'''(x)\ge 0 for any real x, then p(x) \ge 0 for any real x.

Solution (elementary)

p(x)-p'(x)-p''(x)+p'''(x)\ge 0

\Rightarrow (p(x)-p''(x))-(p(x)-p''(x))'\ge 0

\Rightarrow q(x)-q'(x)\ge 0 where q(x)=p(x)-p''(x)

\Rightarrow q(x)\ge q'(x) for any real x

Claim that q(x)\ge 0 for any real x.

Proof

Suppose q(x_0) < 0, by the condition q(x)\ge q'(x), we have q'(x_0) < 0.

That means q(x) is strictly decreasing at x_0. And it means that when there is an x_0 such that q(x_0) is negative, then the graph of y=q(x) is below the x-axis afterwards (x > x_0).

There are 3 cases about the graph of y=q(x).

Case 1: the graph cuts the x-axis

By the above argument, if the graph cuts the x-axis, it cuts once only.

Also, q(x) is a polynomial, we have

\displaystyle \lim_{x\rightarrow -\infty}q(x)=+\infty

\displaystyle \lim_{x\rightarrow +\infty}q(x)=-\infty

Thus, the degree of q(x), n, is odd; and the leading term, a_nx^n, is dominant for large x; and hence a_n < 0.

Now, compare a_nx^n with na_nx^{n-1}, the leading term of q'(x).

For x > n (say)

x^n > nx^{n-1}

a_nx^n < na_nx^{n-1} (a_n < 0

Also, leading terms will be dominant for large x, thus, there exist a large enough x_1 such that

q(x_1) < q'(x_1)

which contradicts that “ q(x)\ge q'(x) for any real x ".

Case 2: the graph is below the x-axis (may touch the x-axis at one point)

In this case, the degree of q(x) should be even with leading coefficient a_n < 0.

Similar to the argument above, obtaining that

there exist a large enough x_1 such that

q(x_1) < q'(x_1)

which contradicts that “ q(x)\ge q'(x) for any real x ".

Case 3: the graph is above the x-axis (may touch the x-axis at one point)

This is the remaining possible case and it must be true.

That is, q(x) \ge 0 for any real x, the claims is proved; and the degree of q(x) is even with positive leading coefficient.

Recall that q(x)=p(x)-p''(x); hence p(x) has even degree with positive leading coefficient.

Thus,

\displaystyle \lim_{x\rightarrow -\infty}p(x)=+\infty

\displaystyle \lim_{x\rightarrow +\infty}p(x)=+\infty

Hence, the polynomial p(x) attains its minimum value at certain value x=x_2.

Therefore, p''(x_2) \ge 0 ……………….. (*)

But, from above,

q(x) \ge 0 \Rightarrow p(x) \ge p''(x) for any real x.

If p(x_2) < 0, then p''(x_2) < 0 which contradicts to (*).

Therefore p(x) \ge 0 for any real x.

Q.E.D.

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