de-CH
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Taylor polynomial of 2nd degree
dgl-05-02
multiple
840
randRangeExclude(-8, 8, [-1,0,1] )randRangeExclude(-8, 8, [-1,0,1, A, -A] )ArandRange(2, 5)12*F*X0 - (A+B)*F4*F*F*X0*X0 - 4*F * (A+B)*F *X0 + 2* F + (A+B)*(A+B)*F*F

Let us be given the function f with f(x) = e^{Fx^2 - (A+B)*F x + A*B*F}.

Determine the second-degree Taylor polynomial at x_0 = X0.

x\displaystyle \color{teal}T_2 (x) =A0 + " + " + A1 + " (x- " + X0 + ") + " + fractionReduce(A2,2) + " (x- " + X0 + ")^2"

We are looking for \displaystyle \color{teal}T_2 (x) = f(X0) + f'(X0)(x- X0) + \frac 12 f''(X0)(x- X0)^2.

A direct computation shows that \displaystyle f(X0) = 1, and we calculate the two missing values \displaystyle f'(X0) and \displaystyle f''(X0) as follows:

We first get \displaystyle f'(x) = (2*Fx- (A+B)*F)\cdot f(x) and therefore \displaystyle f'(X0) = A1.

By differentiating this again, we obtain \displaystyle f''(x) = (4*F*Fx^2 - 4*F * (A+B)*F x + 2* F + (A+B)*(A+B)*F*F )\cdot f(x), so \displaystyle f''(X0) = A2.

Substituting now yeilds \displaystyle T_2(x) = A0+ A1 (x- X0) + fractionReduce(A2,2) (x- X0) ^2.