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The sum of the coefficients of a polynomial is the sum of all the constants that multiply each term. One simple way to find this sum is to evaluate the polynomial at x = 1.
For instance, if we have the polynomial \(f_1(x) = x^2 + 4x + 2\), we can find the sum of its coefficients by computing \(f_1(1)\):

• \(f_1(1) = 1^2 + 4(1) + 2 = 1 + 4 + 2 = 7\)

This is a quick and handy trick because plugging in 1 removes all instances of x in a polynomial, leaving us with just the coefficients to add.
The sum of coefficients includes all terms, both even and odd degrees.

Terms of even degree in a polynomial are those where the exponent of x is an even number. For example, in \(f_1(x) = x^2 + 4x + 2\):

• The term \(x^2\) has an even degree of 2.
• The constant term 2 can be considered as \(x^0\), which has an even degree of 0.

To focus only on these even-degree terms, we need to ignore the odd exponent and coefficient terms. In the example of a computed polynomial \(f_2(x) = x^4 + 8x^3 + 24x^2 + 32x + 14\):

• The terms with even degrees are \(x^4\), \(24x^2\), and 14.

When finding the sum of coefficients for even-degree terms, we sum only the coefficients of these terms:

• \(1 + 24 + 14 = 39\)

This sum excludes any terms with odd degrees such as \(8x^3\) and \(32x\).

A recursive function is one that refers to itself within its definition to compute each term based on previous terms. Here, the polynomial \(f_n(x)\) is defined recursively based on \(f_1(x)\):

• \(f_2(x) = f_1(f_1(x))\)
• \(f_3(x) = f_1(f_2(x))\)
• And so forth...

This means each higher-order polynomial \(f_n(x)\) builds upon the previous polynomial by feeding it back through \(f_1(x)\).
Recursive functions are powerful for handling repetitive structures, allowing us to compute values for increasingly complex compositions based on simpler initial compositions.
However, it can be challenging to manage them, especially as n grows larger, so recognizing patterns and properties (such as sums of coefficients) becomes crucial.

Evaluating a polynomial involves replacing the variable, x, with a specific value and simplifying the expression to find the result. This is crucial in solving many algebraic problems, including finding the sum of coefficients.
To evaluate a polynomial like \(f_1(x) = x^2 + 4x + 2\) at specific x-values, follow these steps:

• For \(x = 1\), we get \(f_1(1) = 1^2 + 4(1) + 2 = 7\)
• For \(x = -1\), we get \(f_1(-1) = (-1)^2 + 4(-1) + 2 = -1\)

Evaluating the polynomial at x = 1 gives us the sum of all coefficients, while evaluating at x = -1 is useful for isolating even-degree terms. Specifically, for large compositions \(f_n(x)\), evaluations allow us to determine important properties such as sums and behaviors of the polynomial.
In conclusion, once we evaluate both \(f_n(1)\) and \(f_n(-1)\), we can easily find the sum of coefficients of even degree terms using:

• \(s_n = \frac{f_n(1) + f_n(-1)}{2}\)