Shannan Johnson

2017-08-11 11:50:53 UTC

Discussion:

(too old to reply)

Shannan Johnson

2017-08-11 11:50:53 UTC

James Waldby

2017-08-11 16:04:37 UTC

To be more newsgroup-friendly, instead of just posting a link

to a picture of an equation, include an ASCII-form copy of the

equation in the post. That is, explicitly say in the post that

X = { p in P_2(R) : 0 = p'(-1) - 2 (integral(0 to 1) p(x)dx) }.

Also, to avoid ambiguity, spell out whether P_2(R) refers to

second-degree polynomials over the reals with real coefficients.

There may be a more-elegant approach to solving this problem, but

one approach is to compute p'(-1) - 2 (integral(0 to 1) p(x)dx)

symbolically, supposing p(x) = a x^2 + b x + c. That is, compute

p'(x) = 2ax + b so p'(-1) = b-2a. Similarly, evaluate the integral

to a/3 + b/2 + c or some such, and work through the algebra to

obtain the implication that 3c = -4a, if I didn't make a mistake.

to a picture of an equation, include an ASCII-form copy of the

equation in the post. That is, explicitly say in the post that

X = { p in P_2(R) : 0 = p'(-1) - 2 (integral(0 to 1) p(x)dx) }.

Also, to avoid ambiguity, spell out whether P_2(R) refers to

second-degree polynomials over the reals with real coefficients.

There may be a more-elegant approach to solving this problem, but

one approach is to compute p'(-1) - 2 (integral(0 to 1) p(x)dx)

symbolically, supposing p(x) = a x^2 + b x + c. That is, compute

p'(x) = 2ax + b so p'(-1) = b-2a. Similarly, evaluate the integral

to a/3 + b/2 + c or some such, and work through the algebra to

obtain the implication that 3c = -4a, if I didn't make a mistake.

--

jiw

jiw

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