A ( r ) be the area of the region in the x y -plane with x > 0 and y > 0 given by the equation Γ ( x ) + Γ ( y ) < r Let m be the minimal value of Γ ( x ) for x > 0 , and evaluate the following limit: r → 2 m + lim r − 2 m A ( r ) The solution requires numerical computation.
Let
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This is merely confirmation of Joseph Newton's and must not be taken in any way as criticism.
Yes, Mathematica 12.1.1 gives that answer from that expression. Can be checked on a Raspberry Pi 0 running Raspbian as Wolfram Mathematica is included at no additional cost (Raspbian is free). It is interesting to note that evaluating the limit expression at smaller and smaller values of r breaks the Mathematica kernel at a value of r near 0 . 0 0 0 0 1 3 7 5 4 3 . This has been reported to Wolfram. At that value of r , the value of the limit expression is 7 . 3 3 1 7 8 , which is close enough to scored as a correct answer by Brilliant's 3 decimal digits standard. The Mathematica kernel issue may be due to issues in the division of the area to be integrated numerically.
Mathematica's confirmation of the second partial derivative:
∂ t 2 ∂ 2 Γ ( t ) = Γ ( t ) ψ ( 0 ) ( t ) 2 + Γ ( t ) ψ ( 1 ) ( t )
Sorry if this is off topic, but what is the piece of music in your profile picture? It seems like it's in either E major or C# minor. Is it classical?
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It's from the Rite of Spring, the bit with 6 solo violas
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This solution is not a complete proof, but shows roughly how the answer is obtained.
Let a be the value of x that minimises Γ ( x ) , that is Γ ( a ) = m . As r decreases towards 2 m (the minimum value of Γ ( x ) + Γ ( y ) ), the region Γ ( x ) + Γ ( y ) < r becomes a smaller and smaller loop around ( x , y ) = ( a , a ) , which as it turns out approaches the shape of a circle.
Consider the Taylor series expansion of Γ ( x ) around the point x = a : Γ ( x ) = Γ ( a ) + Γ ′ ( a ) ( x − a ) + 2 1 Γ ′ ′ ( a ) ( x − a ) 2 + … Now, Γ ( a ) = m , and since x = a is a minimum, Γ ′ ( a ) = 0 . We will ignore higher-order terms, and only use a second degree expansion. Intuitively we can do this because as x approaches a the terms ( x − a ) 3 , ( x − a ) 4 , … approach zero much faster than ( x − a ) 2 . Actually proving that the limit of the area remains the same if we remove these terms is more complex, and not shown here. This leaves Γ ( x ) = m + 2 1 Γ ′ ′ ( a ) ( x − a ) 2 , so the equation Γ ( x ) + Γ ( y ) < r becomes m + 2 1 Γ ′ ′ ( a ) ( x − a ) 2 + m + 2 1 Γ ′ ′ ( a ) ( y − a ) 2 < r ⟹ ( x − a ) 2 + ( y − a ) 2 < Γ ′ ′ ( a ) 2 ( r − 2 m ) This is a circle of radius Γ ′ ′ ( a ) 2 ( r − 2 m ) centred at ( a , a ) , which has area Γ ′ ′ ( a ) 2 ( r − 2 m ) π . Hence, the limit becomes r → 2 m + lim r − 2 m 1 Γ ′ ′ ( a ) 2 ( r − 2 m ) π = Γ ′ ′ ( a ) 2 π Γ ′ ′ ( a ) can be written in terms of the polygamma functions : ψ ( 0 ) ( x ) = d x d ln Γ ( x ) = Γ ( x ) Γ ′ ( x ) ψ ( 1 ) ( x ) = d x d ψ ( 0 ) ( x ) = ( Γ ( x ) ) 2 Γ ′ ′ ( x ) Γ ( x ) − ( Γ ′ ( x ) ) 2 = Γ ( x ) Γ ′ ′ ( x ) − ( ( ψ ( 0 ) ( x ) ) 2 ⟹ Γ ′ ′ ( a ) = Γ ( a ) ( ( ψ ( 0 ) ( a ) ) 2 + ψ ( 1 ) ( a ) ) The value of a can then be found numerically on a program such as Mathematica by looking for the positive root of ψ ( 0 ) ( x ) :
This gives an answer of 7 . 3 3 1 8 3 .