⋅ (x - cⱼ) is a noden function so if we multiply that by an odd function, the product will be a noden function. we already know that (x - c₁) ⋅ (x - c₂) ⋅. x is an odd function so P(x) is an odd function. the product of noden functions of the form (x - c) is a noden function if there exists a noden factor which is not part of a conjugate pair-all instances of conjugate pairs have already been removed of course. Another way interpret this is that 1 is an even function, so P(x) is an even function. we removed all the even functions, and we are left with 1, that means P(x) was an even function all along. I'm going to denote that operation as a method, RemoveConjugatePairs. Remember that the product of any conjugate pair is always an even function. Now, let's remove all the conjugate pairs. Remember, factors of the form xₖ are odd functions, so when there are an even number of them, their product is an even function, otherwise (if the n is odd), the product is an odd function. Now, let's determine what n is, which tells us how many factors of the form xₖ there are. Let's assign p as P(x) without a, we can ignore it for the sake of the overall parity of P(x). I'm going to use p as a variable in the context of algorithms and := as an assignment operator. When we only care about parity, we can ignore the even functions/factors in a factored polynomial. The product of any conjugate pair is always an even function.įrom those properties, we can see that multiplying by an even function doesn't affect the parity of the product. Their product is x² - a², which is an even function. The simplest example is (x a) and (x - a), they are conjugates of each other (I'll refer to them as a conjugate pair). Noden ⋅ noden = even if both nodens are linear of the form (x - c) and are conjugates of each other. When I use any of those terms multiple times, they do not necessarily refer to the same function. For this section, I'm going to use even to denote an even function, odd to denote an odd function, and noden to denote a function that is neither odd nor even-all of which are polynomials. Here are some properties of odd, even, and noden functions (each function is strictly of that parity). And factors of the form (x - cⱼ) are neither odd nor even (let's call them noden for short) since x - cⱼ = x¹ - cⱼ ⋅ x⁰. Factors of the form xₖ are odd functions because xₖ = xₖ¹. ⋅ (x - cₘ) where P(x) is a polynomial in factored form and each factor is linear-this is possible for any polynomial according to The Linear Factorization Theorem.Ī is an even function because a = a ⋅ x⁰. Another scenario would be is what if there were literally so many factors that replacing x's with -x's, negating, and then simplifying the whole expression was too cumbersome? So, I've devised a method that comes directly from the basic arithmetic properties of odd and even functions. I've thought about your question for a while. You could just get the explicit definition of f(-x) and/or -f(-x) and see if they are equal to the definition of f(x) to determine if f(x) is odd, even, or neither. H ( x ) = − 2 x 5 h(x)=-2x^\greenD5 h ( x ) = − 2 x 5 h, left parenthesis, x, right parenthesis, equals, minus, 2, x, start superscript, start color #1fab54, 5, end color #1fab54, end superscriptį ( x ) = a x n f(x)=ax^ odd start color #1fab54, start text, o, d, d, end text, end color #1fab54 G ( x ) = 3 x 2 g(x)=3x^\purpleC2 g ( x ) = 3 x 2 g, left parenthesis, x, right parenthesis, equals, 3, x, start superscript, start color #aa87ff, 2, end color #aa87ff, end superscript
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