let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$

Question

Question: let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$

If I put $x=2$, we get $P(4)$ which is not required. I tried to find $P(x)$ to let it general polynomial to find its roots but unable to find

Please help!

If $x$ is a root of $P$, that is... if $P(x)=0$, then $x^2$ is also a root of $P$ as $P(x^2)=x^2(x^2+1)P(x)=x^2(x^2+1)0=0$. It follows then that the only possible roots are $0$ or roots of unity (e.g. $1,-1,i,-i,\frac{\sqrt{3}+i}{2},\dots$) as for any other values this would imply that $x,x^2,x^4,x^8,x^{16},\dots$ are all distinct roots and so $P$ would have infinitely many roots, an impossibility for a polynomial. — JMoravitz, May 06 '20 at 17:30
That step requires a bit more explanation I think... why do you suppose they are all roots? If building off of my earlier comment, they are possible roots but not proven to actually be roots yet from what I had written. — JMoravitz, May 06 '20 at 17:36
And you have $P(2)$ which gives you 4 points. How many points do you need to fix $P(x)$ completely? — gt6989b, May 06 '20 at 17:36
Can anybody help me in https://math.stackexchange.com/questions/3661973/can-we-integrate-int-0-frac-π2ln1cos2-2x-dx — Paras, May 06 '20 at 17:49

Travis Willse · Answer 1 · 2020-05-09T16:22:18.240

We're looking for a polynomial that satisfies: $$ \phantom{(\ast)} \qquad P(x^2) = x^2 (x^2 + 1) P(x) . \qquad (\ast) $$

Hint An outline of one possible solution:

As a few people have suggested in the comments, use $(\ast)$ to compute that $\deg P = 4$.
Use $(\ast)$ to deduce that $P$ is even.
Use $(\ast)$, again, to show that $0$ is a root of $P$; since $P$ is even, $0$ is at least a double root of $P$. If $r$ is one of the remaining $2$ roots, by evenness so is $-r$, so any polynomial $P$ satisfying $(\ast)$ has the form $$P(x) = a x^2 (x + r) (x - r)$$ for some $a, r$.
Use $(\ast)$ yet again, to show that $P(1) = 0$, so we may as well take $r = 1$.
Substitute our expression for $P$ into the condition $P(2) = 3$ to determine $a$.
Compute $P(3)$.

JMoravitz · Accepted Answer · 2020-05-06T18:00:10.743

Elaborating on some details:

$P(0)=P(0^2)=0^2(0^2+1)P(0)=0$

$P(1^1)=P(1)=1^2(1^2+1)P(1)$ so $P(1)=2P(1)$ implying $P(1)=0$

$0=P(1)=P((-1)^2)=(-1)^2((-1)^2+1)P(-1)=2P(-1)$ implying $P(-1)=0$

So, we have found three roots so far.

As for the degree of $P$, if we were to suppose $P(x)$ were a degree $n$ polynomial, it follows that $P(x^2)$ should be a degree $2n$ polynomial.

As $\underbrace{P(x^2)}_{\text{degree }2n}=\underbrace{x^2(x^2+1)}_{\text{degree }4}\times \underbrace{P(x)}_{\text{degree }n}$ it follows that $2n=4+ n$ and so $n=4$

By the fundamental theorem of algebra, it follows that $P(x)$ can be written in the form:

$$P(x)=x(x-1)(x+1)(ax+b)$$

for some values $a,b$

Let us try to find out those values of $a,b$ using $P(2)$ and $P(4)$.

From $P(2)$ we get that $3=P(2)=2(2-1)(2+1)(2a+b)$ and so $2a+b=\frac{1}{2}$. That is a good start but not quite enough to finish, so we use one more.

From $P(4)$ we get $P(4)=P(2^2)=2^2(2^2+1)P(2)=4\cdot 5\cdot 3 = 60$ and that $P(4)=4(4-1)(4+1)(4a+b)$ and so $4a+b=1$

Subtracting the first result away from the second, this gives $2a=\frac{1}{2}$ and $a=\frac{1}{4}$, thus $b=0$

As a result, we have that $$P(x)=x(x-1)(x+1)(\frac{1}{4}x+0)$$ or rewritten as you had arrived at in the comments $$P(x)=\frac{x^2(x^2-1)}{4}$$

From here, finding $P(3)$ is a matter of just plugging in the value and we get $P(3)=\frac{9\cdot 8}{4}=18$

You should have wriiten $2n =n+4$ instead of $n^2=4n$ – Paras May 06 '20 at 17:58 — Paras, May 06 '20 at 17:58
Good catch. Clearly wasn't paying attention. – JMoravitz May 06 '20 at 18:00 — JMoravitz, May 06 '20 at 18:00

Travis Willse · Answer 3 · 2020-05-06T20:15:12.873

Here's a solution which is not so efficient as some other methods in this case, but which illustrates

a more general concept, and
that you need not actually compute $P$ in order to determine $P(3)$.

Recall the (forward) finite difference operator $$\Delta[f](x) := \Delta_1 [f](x) = f(x + 1) - f(x) ,$$ which formally behaves like a derivative in some ways. The corresponding Taylor-like series is $$f(x) \sim \sum_{k = 0}^\infty \frac{1}{k!} \Delta^k[f](a) (x - a)_k ,$$ where $y_k$ denotes the falling factorial $y_k := y (y - 1) \cdots (y - k + 1)$.

If $f$ is a polynomial $\Delta f$ is a polynomial of lower degree (or the same degree, $-\infty$, if $f = 0$), so $\Delta^k[f] = 0$ for all $k > \deg f$, and in particular the above series terminates at order $k = \deg f$.

The iterated difference operators are given by $$\Delta_1^k [f](a) = \sum_{i = 0}^k {k \choose i} (-1)^{k - i} f(a + i) .$$ Now, since $P(x^2)$ and $x^2 (x^2 + 1)$ are both even in $x$, so is $P$. So we can compute, e.g., $P(3) = P(-3)$ by evaluating the series at $a = -2, x = -3$, and to do so we only need to know $P(-2), P(-1), \ldots, P(-2 + \deg P)$. Since $x - a = -1$, $(x - a)_k = (-1)_k = (-1)^k k!$, and so $$\phantom{(\ast)} \qquad P(3) = P(-3) = \sum_{k = 0}^{\deg P} \sum_{i = 0}^k {k \choose i} (-1)^i f(-2 + i) \qquad (\ast) .$$

We can do this efficiently for $P$: Comparing the degrees of both sides of $$\phantom{(\ast\ast)} \qquad P(x^2) = x^2 (x^2 + 1) P(x) \qquad (\ast)$$ gives that $\deg P = 4$, so to compute $P(3)$ we only need to know $P(0), P(-1) = P(1), P(-2) = P(2) = 3$. The constraint $(\ast)$ implies that $P(-1) = P(0) = P(1) = 0$, so all that remains to do is to substitute in $(\ast\ast)$. Now, reindexing the double summation gives $$P(3) = P(-3) = \sum_{i = 0}^k (-1)^i f(-2 + i) \sum_{k = 0}^4 {k \choose i} .$$ But the only nonzero values among $f(-2), \ldots, f(2)$ are $f(\pm 2) = 3$, so only the $i = 0, 4$ terms contribute, leaving $$\boxed{P(3) = f(-2) \sum_{k = 0}^4 {k \choose 0} + f(2) \sum_{k = 0}^4 {k \choose 4} = 5 \cdot f(2) + 1 \cdot f(2) = 6 f(2) = 6 \cdot 3 = 18} .$$

let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$

3 Answers3