I'd like to test whether this expression is negative when a and b are constrained to lie within (0,1) interval. Any ideas how?
D[Integrate[((1 - (a*x))^12)*((1 + (a/b))/((1 + (a*x/b))^2)), {x, 0, 1}], a]
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1 Answer
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Clear["Global`*"]
expr1 = ((1 - (a*x))^12)*((1 + (a/b))/((1 + (a*x/b))^2)) // Simplify;
Evaluating the integral
int = Assuming[0 < a <= 1 && 0 < b <= 1,
Integrate[expr1, {x, 0, 1}] // Simplify];
Taking the derivative
expr2 = D[int, a] // Simplify[#, 0 < a <= 1 && 0 < b <= 1] &;
Numerically finding the minimum
NMinimize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b},
WorkingPrecision -> 20] // N
{-6., {a -> 8.10042*10^-9, b -> 0.91264}}
Limit[expr2, a -> 0]
(* -6 *)
The minimum occurs when a is near 0
Limit[expr2, a -> 0]
(* -6 *)
EDIT: numerically finding the maximum
NMaximize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b},
WorkingPrecision -> 20] // N
(* {-1.00973*10^-12, {a -> 0.981099, b -> 9.7192*10^-13}} *)
The maximum occurs for b near 0
Limit[expr2, b -> 0]
(* 0 *)
Consequently, expr2 is nonpositive everywhere in the region. Graphically,
Plot3D[expr2, {a, 0, 1}, {b, 0, 1},
PlotPoints -> 50,
Exclusions -> True,
WorkingPrecision -> 15,
ClippingStyle -> None]
answered Oct 28, 2019 at 13:59
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$\begingroup$ Thanks Bob. Shouldn't the last step be a maximization as I want to verify if the expression is always less than 0 e.g. that it's max value < 0? Also do you know why it isn't possible to do analytically rather than numerically? $\endgroup$ Commented Oct 28, 2019 at 14:25
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$\begingroup$ The expression is transcendental. Terms include product of high order polynomial times a log. Analytic solution is unlikely. $\endgroup$ Commented Oct 28, 2019 at 14:45