Question

Suppose that \(f(x) = \Sigma _{n = 0}^\infty {c_n}{x^n}\) for all X.

(a) If f is an odd function, show that

\({{\rm{c}}_{\rm{0}}}{\rm{ = }}{{\rm{c}}_{\rm{2}}}{\rm{ = }}{{\rm{c}}_{\rm{4}}}{\rm{ = L = 0}}\)

(b) If f is an even function, show that

\({{\rm{c}}_{\rm{1}}}{\rm{ = }}{{\rm{c}}_{\rm{3}}}{\rm{ = }}{{\rm{c}}_{\rm{5}}}{\rm{ = L = 0}}\)

Short answer

a) If f is an odd function we get\({c_0},{c_2},{c_4},{c_6} \ldots {c_{2n}} = 0\).

b) If f is an even function we get \({c_1},{c_3},{c_5},{c_7} \ldots {c_{2n + 1}} = 0\).

Step-by-step solution

Definition of Function

A function is a relationship between a group of inputs that each have one output. A function, in simple terms, is a relationship between inputs in which each input is associated to only one output.

Calculate f(x)  and show that it is odd function

Let us find f(x)

\(f(x) = \sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}} \to ({\bf{1}})\)

Remember that f(x) is an odd function, if and only if\({\rm{f}}({\rm{x}}) = - {\rm{f}}( - {\rm{x}})\)

\(\begin{array}{l}f( - x) = \sum\limits_{n = 0}^\infty {{c_n}} {( - x)^n}f( - x)\\ = \sum\limits_{n = 0}^\infty {{c_{2n}}} {( - x)^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {( - x)^{2n + 1}}f( - x)\\ = \sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} - \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}} \to ({\bf{2}})\end{array}\)

Since f(x) is odd, Using Eqn\(({\bf{1}})\)and Eqn(2), we can write

\(\begin{array}{c}f(x) = - f( - x)\sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}\\ = - \left[ {\sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} - \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}} \right]\sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}\\ = - \sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}\sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 1}^\infty {{c_{2n + 1}}} {x^{2n + 1}}\\ = - \sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}2\sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}}\\ = 0\end{array}\)

Therefore\({c_0},{c_2},{c_4},{c_6} \ldots {c_{2n}} = 0\)

Calculate f(x) and show that it is even function

Let us calculate f(x)

\(\left. {f(x) = \sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}} \to ({\bf{1}}} \right)\)

Remember that f(x) is an even function, if and only if \({\rm{f}}({\rm{x}}) = {\rm{f}}( - {\rm{x}})\)

\(\begin{array}{c}f( - x) = \sum\limits_{n = 0}^\infty {{c_n}} {( - x)^n}f( - x)\\ = \sum\limits_{n = 0}^\infty {{c_{2n}}} {( - x)^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {( - x)^{2n + 1}}f( - x)\\ = \sum\limits_{n = 0}^\infty {{c_{2n}}} {x^{2n}} - \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}} \to ({\bf{2}})\end{array}\)

Since f(x) is even, Using Eqn(1) and Eqn(2), we can write

\(\begin{array}{c}f(x) = f( - x)\sum\limits_{x = 0}^\infty {{c_{2n}}} {x^{2n}} + \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}\\ = \sum\limits_{x = 0}^\infty {{c_{2n}}} {x^{2n}} - \sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}2\sum\limits_{n = 0}^\infty {{c_{2n + 1}}} {x^{2n + 1}}\\ = 0\end{array}\)

Therefore, \({c_1},{c_3},{c_5},{c_7} \ldots {c_{2n + 1}} = 0\).