User:Egm6321.f10.team03.sigillo/Hwk7

=Problem 2: Summation of functions =

Given
Given is the definition of a function, which is the sum of two or more functions:
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$$f(x)=\sum\limits_{i}{{{g}_{i}}\left( x \right)}$$
 * $$\,(Eq. 2.1) $$
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Find
It has to be shown that if each gi(x) is an even function also f(x) is an even function and that if gi(x) is odd also f(x) is an odd function.

Solution
As first we need to look at how even and odd functions are defined.

Even functions: A function gi(x) is even if:
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$${{g}_{i}}\left( x \right)={{g}_{i}}\left( -x \right)$$
 * $$\,(Eq. 2.2) $$
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Odd functions: A function gi(x) is odd if:
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$${{g}_{i}}\left( x \right)={-{g}_{i}}\left( -x \right)$$
 * $$\,(Eq. 2.3) $$
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Now let's split the function g(x) into 3 parts: The part for all x<0, the part for x=0 and the part for all x>0, denoted by g-(x), g0(x) and g+(x) respectively. So that the function g(x) can now be written as:
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$$g\left( x \right)={{g}_{-}}\left( x \right)+{{g}_{0}}\left( x \right)+{{g}_{+}}\left( x \right)$$
 * $$\,(Eq. 2.4) $$
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The properties of the single functions are:
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$$\begin{align} & {{g}_{-}}\left( x \right)=\left\{ _{0\quad for\quad x\ge 0}^{\mathbb{Z}\quad for\quad x<0} \right. \\ & {{g}_{0}}\left( x \right)=\left\{ _{0\quad for\quad x\ne 0}^{\mathbb{Z}\quad for\quad x=0} \right. \\ & {{g}_{+}}\left( x \right)=\left\{ _{0\quad for\quad x\le 0}^{\mathbb{Z}\quad for\quad x>0} \right. \\ \end{align}$$ THE EVEN CASE:
 * $$\,(Eqs. 2.5) $$
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Now, accordingly to Eq. 2.2 and with the properties of Eqs. 2.5 for an even function gi(x) we can set g-(x) = g+(x) so that the function gi(x) can be rewritten as:
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$${{g}_{i}}\left( x \right)={{g}_{i}}_{+}\left( x \right)+{{g}_{_{i}0}}\left( x \right)+{{g}_{i}}_{+}\left( x \right)$$ Summing n even functions together to build the function f(x) yields to:
 * $$\,(Eq. 2.6) $$
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$$\begin{align} & {{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right) \\ & {{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right) \\ & \quad \vdots \quad \ \ +\quad \vdots \quad \ \ +\quad \vdots \quad \ \ \\ & \underline{{{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right)} \\ & {{f}_{+}}\left( x \right)+{{f}_{0}}\left( x \right)+{{f}_{+}}\left( x \right)\quad =f\left( x \right) \\ \end{align}$$ 'f(x)'' has the same form as Eq. 2.6 which we found to be even, therefore also the sum of all gi(x), which is f(x), has to be even!'''
 * $$\,(Eqs. 2.7) $$
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THE ODD CASE:

Accordingly to Eq. 2.3 and with the properties of Eqs. 2.5 for an odd function gi(x) we can set g-(x) = -g-(x) so that the function gi(x) can be rewritten as:
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$${{g}_{i}}\left( x \right)={{g}_{i}}_{-}\left( x \right)+{{g}_{_{i}0}}\left( x \right)-{{g}_{i}}_{-}\left( x \right)$$ Summing n odd functions together to build the function f(x) yields to:
 * $$\,(Eq. 2.8) $$
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$$\begin{align} & {{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right) \\ & {{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right) \\ & \quad \vdots \quad \ \ +\quad \vdots \quad \ \ +\quad \vdots \quad \ \ \\ & \underline{{{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right)} \\ & {{f}_{-}}\left( x \right)+{{f}_{0}}\left( x \right)-{{f}_{-}}\left( x \right)\quad =f\left( x \right) \\ \end{align}$$ 'f(x)'' has the same form as Eq. 2.8 which we found to be odd, therefore also the sum of all gi(x), which is f(x), has to be odd!'''
 * $$\,(Eqs. 2.9) $$
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Contribution

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The solution showed above was derived from scratch by the author without consulting last years Homework!
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Solved and posted by Egm6321.f10.team03.sigillo 21:02, 5 December 2010 (UTC)