Problem 126 - Calculus of a Single Variable

Question

Determine all polynomials \(P(x)\) such that $$ P\left(x^{2}+1\right)=(P(x))^{2}+1 \text { and } P(0)=0 $$

Step-by-Step Solution

Answer

The required polynomial is \(P(x) = 1\) for \(x \geq 1\) and \(P(0) = 0\).

Step 1: Plug Point P(0) = 0 into the Equation

Plugging \(P(0) = 0\) to \(P((x^{2}+1) = (P(x))^{2} + 1\) we get \(P(1) = 1\).

Step 2: Identify Recursive Nature of Function

We notice that the polynomial has recursive character. Each value \(P(x)\) depends on \(P(x^2 + 1)\), and each \(P(x)\) depends on another result, \(P(y)\), where \(y > x\). If we can find \(P(1)\), we can use the function to find \(P(x)\) for all \(x > 1\). Since \(P(1) = 1\), we can use the function to find the values for all \(x > 1\).

Step 3: Find P(x) for x > 1

Using the recursive character of the function, we find that \(P(x)\) for \(x > 1\) is equal to \(1, \forall x \geq 1\). This is because, given the recursive nature of the function, we can trace back each \(P(x)\) for \(x > 1\) to \(P(1) = 1\).

Step 4: Interpret Information

This implies that the required polynomial is a constant polynomial, \(P(x) = 1\) for \(x \geq 1\). And \(P(0) = 0\).

Key Concepts

Recursive Nature

When solving polynomial functions, understanding their recursive nature can be quite enlightening. Recursive simply means that certain values in the function depend on other values in a systematic pattern. In this particular exercise, we see that each value of the polynomial, denoted as \( P(x) \), is linked to the value \( P(x^2 + 1) \).
This relationship tells us that knowing the value of \( P(x) \) gives us the ability to compute the next value in the sequence, namely \( P(x^2 + 1) \). It forms a loop or cycle where once you have a starting point, you can deduce the subsequent values.

• This characteristic is powerful because it helps in tracing back the values from \( P(x) \).
• Initially focusing on the known value such as \( P(1) = 1 \) sets the stage for finding all other values with \( x > 1 \).

By trusting this recursive system, we efficiently solve polynomial problems without the need to test each value individually.

Constant Polynomials

Constant polynomials occur when a polynomial function simplifies to a single constant value for most or all inputs. In some cases, like in this exercise, specific values of \( x \) result in the polynomial yielding the same constant number.
The exercise concludes that for \( x \geq 1 \), \( P(x) = 1 \), making it a constant polynomial in that domain. Constant polynomials hold a special place due to their simplicity and ease of use.

• This happens because once the base condition \( P(1) = 1 \) is established, the recursive nature forces all subsequent values for \( x \geq 1 \) to also be 1.
• The equations around \( P(x) \) align to support only one possible result for each \( x \).

Such insights help us realize that not all polynomials need to exhibt varying outputs; sometimes they are perfectly constant.

Function Values

In polynomial functions, values are the results you get by plugging inputs into the function. Generally, function values denote how the polynomial behaves over different inputs.
For \( P(x) \), the given problem indicates a shifting pattern where only specific values are specifically defined: \( P(0) = 0 \) and \( P(1) = 1 \). As mentioned, using \( P(x^2+1) = (P(x))^2 + 1 \), we found that \( P(x) \) for \( x \geq 1 \) equals 1.

• This idea highlights that certain function values are pivotal to deducing results; such anchor points were critical in unraveling the overall function behavior in this example.
• The pattern for values quickly reduced into a clean, constant path, simplifying the function greatly.

Knowing such key values streamlines understanding and solving polynomial equations efficiently.

Polynomial Equations

A polynomial equation is an algebraic expression consisting of variables and coefficients, involving only operations of addition, subtraction, multiplication, and non-negative integer exponents.
In this problem, the polynomial equation \( P(x^2 + 1) = (P(x))^2 + 1 \) connects \( P(x) \) at different stages, initially intriguing due to its complexity.

• Solutions often require understanding patterns and drawing insights, exactly as seen when realizing \( P(x) = 1 \) for all \( x \geq 1 \).
• When facing such equations, breaking them into smaller components like solving for \( P(0) = 0 \) aids in approachability.

The beauty of polynomial equations lies in leveraging such basic specifics to unlock generalized results. Everything fits together like puzzle pieces, each contributing to simplifying the overall function.