Tool to check the parity of a function (even or odd functions): it defines the ability of the function (its curve) to verify symmetrical relations.

Even or Odd Function - dCode

Tag(s) : Functions

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Definition: A function is even if the equality $$ f(x) = f(-x) $$ is true for all $ x $ from the domain of definition.

An even function will provide an identical image for opposite values.

__Example:__ Determine whether the function is even or odd: $ f(x) = x^2 $ (square function) in $ \mathbb{R} $, the calculation is $ f(-x) = (-x)^2 = x^2 = f(x) $, so the square function $ f(x) $ is even.

**Graphically**, this involves that opposed abscissae have the same ordinates, this means that the ordinate y-axis is an axis of symmetry of the curve representing $ f $.

Having proved this equality for a single value like $ f(1) = f(-1) $ does not allow to conclude that there is parity, only to say that 1 and -1 have the same image by the function $ f $.

Definition: A function is odd if the equality $$ f(x) = -f(-x) $$ is true for all $ x $ from the domain of definition.

An odd function will provide an opposite image for opposite values.

__Example:__ Determine whether the function is even or odd: $ f(x) = x^3 $ (cube function) in $ \mathbb{R} $, the calculation is $ -f(-x) = -(-x)^3 = x^3 = f(x) $, so the cube function $ f(x) $ is odd.

**Graphically**, this involves that opposed abscissae have opposed ordinates, this means that the origin (central point) (0,0) is a symmetry center of the curve representing $ f $.

NB: if an odd function is defined in 0, then the curve passes at the origin: $ f(0) = 0 $

Having proved equality for a single value like $ f(2) = -f(-2) $ does not allow us to conclude that there is imparity, only to say that 2 and -2 have opposite images by the function $ f $.

A function is neither odd nor even if neither of the above two equalities are true, that is to say: $$ f(x) \neq f(-x) $$ and $$ f(x) \neq -f(-x) $$

__Example:__ Determine the parity of $ f(x) = x/(x+1) $, first calculation: $ f(-x) = -x/(-x+1) = x/(x-1) \neq f(x) $ and second calculation: $ -f(-x) = -(-x/(-x+1)) = -x/(x-1) = x/(-x+1) \neq f(x) $ therefore the function $ f $ is neither even nor odd.

In trigonometry, the functions are often symmetrical:

The cosine function $ \cos (x) $ is even.

The sine function $ \sin (x) $ is odd.

The tangent function $ \tan (x) $ is odd.

Developments in convergent power series or polynomials of even (respectively odd) functions have even degrees (respectively odd).

Yes, the function $ f(x) = 0 $ (constant zero function) is both even and odd because it respects the 2 equalities $ f(x) = f(-x) = 0 $ and $ f(x) = -f(-x) = 0 $

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