The power of odd and even function?

Logic Level 2

Let f : R → R f:ℝ \rightarrow ℝ be a real-valued function

Is it possible to separate f f into the sum of 2 functions, where one of them is odd and the other one is even?

i.e. Do there exist 2 functions P P and Q Q , where P P is even and Q Q is odd, such that f ≡ P + Q f\equiv P+Q ?

Furthermore: If this answer is always, then are P P and Q Q unique?

Little Hint: you may assume the fact that zero function is the only function which is both even and odd

Never Sometimes Always, unique Always, not unique

Raymond Chan by Raymond Chan , 20, USA

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2 solutions

Raymond Chan
Dec 23, 2017

Define P ( x ) = 1 2 [ f ( x ) + f ( − x ) ] P(x)=\frac { 1 }{ 2 } \left[ f\left( x \right) +f\left( -x \right) \right] and Q ( x ) = 1 2 [ f ( x ) − f ( − x ) ] Q(x)=\frac { 1 }{ 2 } \left[ f\left( x \right) -f\left( -x \right) \right]

Then, { P ( − x ) = 1 2 [ f ( − x ) + f ( x ) ] = 1 2 [ f ( x ) + f ( − x ) ] = P ( x ) Q ( − x ) = 1 2 [ f ( − x ) − f ( x ) ] = − 1 2 [ f ( x ) − f ( − x ) ] = − Q ( x ) \begin{cases} P(-x)=\frac { 1 }{ 2 } \left[ f\left( -x \right) +f\left( x \right) \right] =\frac { 1 }{ 2 } \left[ f\left( x \right) +f\left( -x \right) \right] =P(x) \\ Q(-x)=\frac { 1 }{ 2 } \left[ f\left( -x \right) -f\left( x \right) \right] =-\frac { 1 }{ 2 } \left[ f\left( x \right) -f\left( -x \right) \right] =-Q(x) \end{cases}

Therefore, P P is even and Q Q is odd

Note that P ( x ) + Q ( x ) = 1 2 [ f ( x ) + f ( − x ) ] + 1 2 [ f ( x ) − f ( − x ) ] = f ( x ) P(x)+Q(x)=\frac { 1 }{ 2 } \left[ f\left( x \right) +f\left( -x \right) \right] +\frac { 1 }{ 2 } \left[ f\left( x \right) -f\left( -x \right) \right] =f\left( x \right)

Since zero function is the only function which is both even and odd

So, f f is definitely possible to separate into the sum of an even function and an odd function

Uniqueness

Let f ( x ) = M ( x ) + N ( x ) f\left( x \right) =M(x)+N(x) for some even function M M and odd function N N

By definitions, { M ( − x ) = M ( x ) N ( − x ) = − N ( x ) \begin{cases} M(-x)=M(x) \\ N(-x)=-N(x) \end{cases}

Sub f f back to P P : P ( x ) = 1 2 [ f ( x ) + f ( − x ) ] = 1 2 [ M ( x ) + N ( x ) + M ( − x ) + N ( − x ) ] = 1 2 [ M ( x ) + N ( x ) + M ( x ) − N ( − x ) ] = M ( x ) P(x)=\frac { 1 }{ 2 } \left[ f\left( x \right) +f\left( -x \right) \right] =\frac { 1 }{ 2 } \left[ M(x)+N(x)+M(-x)+N(-x) \right] =\frac { 1 }{ 2 } \left[ M(x)+N(x)+M(x)-N(-x) \right] =M(x)

Similarly, Q ( x ) = 1 2 [ f ( x ) − f ( − x ) ] = 1 2 [ M ( x ) + N ( x ) − M ( − x ) − N ( − x ) ] = 1 2 [ M ( x ) + N ( x ) − M ( x ) + N ( − x ) ] = N ( x ) Q(x)=\frac { 1 }{ 2 } \left[ f\left( x \right) -f\left( -x \right) \right] =\frac { 1 }{ 2 } \left[ M(x)+N(x)-M(-x)-N(-x) \right] =\frac { 1 }{ 2 } \left[ M(x)+N(x)-M(x)+N(-x) \right] =N(x)

Therefore, the even part and odd part are unique

1 Helpful 1 Interesting 0 Brilliant 0 Confused
Marta Reece
Dec 23, 2017

Let the function be represented by a McLaurin series.

The series contains odd and even terms.

Sum of the odd terms is odd. Sum of the even terms is even.

The two sums are unique.

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