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Question Number 1988 by Rasheed Soomro last updated on 28/Oct/15

x^{2} = \frac{f (x) + f (- x)}{2}

f (x) = ?

Answered by 123456 last updated on 28/Oct/15

supossing that f is poly

f (x) = \underset{i = 0}{\sum^{n}} a_{i} x^{i}

f (- x) = \underset{i = 0}{\sum^{n}} a_{i} {(- 1)}^{i} x^{i}

x^{2} = \frac{f (x) + f (- x)}{2}

f (x) + f (- x) = 2 x^{2}

\underset{i = 0}{\sum^{n}} a_{i} x^{i} + \underset{i = 0}{\sum^{n}} a_{i} {(- 1)}^{i} x^{i} = 2 x^{2} (n ⩾ 2)

\underset{i = 0}{\sum^{n}} a_{i} [1 + {(- 1)}^{i}] x^{i} = 2 x^{2}

i \neq 2 \Rightarrow a_{i} [1 + {(- 1)}^{i}] = 0 \Rightarrow a_{i} = 0 \lor i \equiv 1 (\mod 2)

i = 2 \Rightarrow 2 a_{2} = 2 \Rightarrow a_{2} = 1

so

f (x) = \underset{i = 0}{\sum^{n}} a_{i} x^{i}, a_{i} = {\begin{cases} \forall a_{i}, i \equiv 1 (\mod 2) \\ a_{i} = 0, i \equiv 0 (\mod 2) \lor i \neq 2 \\ a_{2} = 1 \end{cases}

- - - - - - - - -

f (x) = a_{1} x + x^{2} + a_{3} x^{3} + a_{5} x^{5} + \cdot \cdot \cdot

f (- x) = - a_{1} x + x^{2} - a_{3} x^{3} - a_{5} x^{5} - \cdot \cdot \cdot

f (x) + f (- x) = 2 x^{2}

\frac{f (x) + f (- x)}{2} = x^{2}

Answered by prakash jain last updated on 28/Oct/15

Let g (x) be any odd function .

g (x) = - g (- x)

f (x) = x^{2} + g (x)

f (- x) = x^{2} + g (- x) = x^{2} - g (x)

\frac{f (x) + f (- x)}{2} = x^{2}

e x a m p l e s

f (x) = x^{2} + \sin x

f (x) = x^{2} + \underset{i = 0}{\sum^{n}} a_{i} x^{2 i + 1}