Is the periodic function a vector space?

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SUMMARY

The set of all periodic functions of period 1 does not form a vector space. This conclusion is drawn from the examination of the function f(x) = 2*x mod 2, which fails to satisfy the scalar multiplication property of vector spaces. Specifically, when multiplying f(x) by a scalar, such as 1/2, the resulting function does not maintain periodicity, as demonstrated by the calculations involving specific values like x=17. Thus, the periodic functions of period 1 do not meet the criteria for closure under scalar multiplication.

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golmschenk
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Homework Statement


Is the following a vector space:
The set of all periodic functions of period 1? ( i.e. f(x+1)=f(x) )

Homework Equations


If v1 and v2 are in V then v1 + v2 is in V

If v1 is in V then c*v1 is in V where c is a scalar

The Attempt at a Solution


I'm thinking no because a function like:
f(x) = 2*x mod 2
If you multiply by 1/2 you have
f(x) = x mod 2

What I'm wondering is if the mod 2 is effected by the scalar multiplication of 1/2. Is it or is this not a subspace? Thanks for your time.
 
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golmschenk said:
f(x) = 2*x mod 2
If you multiply by 1/2 you have
f(x) = x mod 2
Let's check this using x=17:
\frac{(2\cdot5)\mod 2}{2} = \frac{\mathop{\mathrm{remainder}}(10,2)}{2} = \frac{0}{2} = 0,​
while
5\mod 2 = \mathop{\mathrm{remainder}}(5,2) = 1.​
Evidently, the two expressions are not the same. Can you see why this is?
 
Right. It's a scalar multiple of the entire function. Got it. Thanks!
 

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