Question

On the first problem set, question 1A3-e, the answer is given as being an even function. However, if one were to use another squared function, such as (1+x)^2, and then square it, they would get a result similar to that in 1A3-d, which is noted to be neither even nor odd. So, in my estimation, it's not fair to say that for all Jsub0(x) that Jsub0(x^2) is an even function. Am I missing something here?

Answer 1

The answer here depends on the definition of an even function. We have an even function if it produces the same value for -x as it produces for x. Any function that takes x^2 as its input will have this quality, because (-x)^2 = x^2.

The function presented in part d of the question does not have this property. We can see this by plugging in a pair of test values such as 1 and -1, which produce different outputs. If we changed it to (x + 1)^2, we would have a function that's superficially similar but it is not a function of x^2. This function is centered around x = -1, but any function of x^2 would be centered around x = 0.

Answer 2

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However, if one were to use another squared function, such as (1+x)^2
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Let \( J_0(x)= (1+x)^2 \)
then \( J_0(x^2)= (1+x^2)^2 \)
which is different from \( K_0(x)= (1+x)^4 \)
see attached graph.