Group of polynomials with real coefficients under adition - antiderivative that passes through (0,0)?Let G be the group of all polynomials with real coefficients under addition. For each f in G, let ∫f denote the antiderivative of f that passes through the point (0,0). Show that the mapping f → ∫ f from G to G is a homomorphism. What is the kernel of this mapping? Is this mapping a homomorphism if ∫f denotes the antiderivative of f that passes through (0,1)?My question here is what does the bolded sentence mean? Does it mean the n-th antiderivative of f such that ( x , y ) = ( 0 , 0 ) is a solution after n consecutive antiderivatives?

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Group of polynomials with real coefficients under adition - antiderivative that passes through (0,0)?Let G be the group of all polynomials with real coefficients under addition. For each f in G, let ∫f denote the antiderivative of f that passes through the point (0,0). Show that the mapping    f  →  ∫  f from G to G is a homomorphism. What is the kernel of this mapping? Is this mapping a homomorphism if ∫f denotes the antiderivative of f that passes through (0,1)?My question here is what does the bolded sentence mean? Does it mean the n-th antiderivative of f such that   (  x  ,  y  )  =  (  0  ,  0  ) is a solution after n consecutive antiderivatives?

Kylan Simon · Accepted Answer

Step 1The antiderivative of a polynomial is a family of polynomials one degree higher than the original, the family being parametrized by the additive constant of integration. The given &quot;initial condition&quot;, that the polynomial should pass through (0,a), fixes the constant to be a.Step 2Example: the antiderivative of   x  +  2 is       x    2    +  2  x  +  c. We must have   c  =  0 in order for 0 to be a root of this polynomial, so the homomorphism maps   x  +  2 to       x    2    +  2  x. It would map to       x    2    +  2  x  +  1 if you required the curve to pass through (0, 1) instead.

Micah Haynes · Accepted Answer

Explanation:The text you&#039;ve highlighted isn&#039;t very clearly phrased. It just means that   ∫  f is the unique polynomial h whose derivative h′ is equal to f and is such that   h  (  0  )  =  0 (so the graph of h &quot;passes through the point&quot; (0,0)). The final part of the question is asking what happens if you change your mind and require   h  (  0  )  =  1 instead. (The requirement that h be an antiderivative of f, i.e., that       h    ′    =  f, does not determine the constant term of h.)

Group of polynomials with real coefficients under adition - antiderivative that passes through (0,0)

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