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YONG
Check this out.
I used to think that either double consumption of Lugol's solution or serious illness during infancy was the cause of the high IQ of one of the forum members. It seems that I was wrong. 

09 Aug 2011, 04:12 

YONG
Speaking of calculus, it took me almost ten minutes to evaluate the following definite integral.
Am I a dummy?


09 Aug 2011, 09:52 

typedef
Nope. You are not.
There's always a starting point. 

09 Aug 2011, 15:21 

MHajduk
YONG wrote: Speaking of calculus, it took me almost ten minutes to evaluate the following definite integral. I don't know your real mathematical skills and can't judge them relying only on your words. Besides, if you think that mastery in such kind of intellectual activity as mathematics encloses only in speed then you're wrong, of course. My previous comment was by no means directed to you neither was allusion to you, so I don't understand what can cause your provocative response. BTW, Wolfram computational engine calculates even more complicated problems in a milliseconds, faster than any human being  does it mean that this mathematical application is a real mathematician? 

09 Aug 2011, 17:07 

typedef
^^Nope, it's just a computer. The only thing that computers have that beat the human mind is time..
Otherwise, humans were, are, and will always be smarter than computers. That's a fact. Because we make them. If you could process 2 Billion thoughts a second then you'd get 500 questions on a test.. LOL Time is the only difference here. Last edited by typedef on 09 Aug 2011, 18:55; edited 1 time in total 

09 Aug 2011, 18:43 

MHajduk
typedef wrote: ^^Nope, it's just a computer. The only thing that computers have that beat the human mind is time.. 

09 Aug 2011, 18:46 

Xorpd!
YONG wrote: Speaking of calculus, it took me almost ten minutes to evaluate the following definite integral. Ten minutes is really good for this integral, IMO. Another integral that uses the same trick is from the CRC Handbook of Chemistry and Physics: π/2 ∫ dθ/(1+(tan(θ))^m) = π/4 0 This last integral also appeared in the Putnam exam with m = √(2) and so was free points for those who read the CRC book. Does Mathematica get π/√(27) for the original integral? 

09 Aug 2011, 22:06 

YONG
MHajduk wrote: My previous comment was by no means directed to you neither was allusion to you, so I don't understand what can cause your provocative response. I am provocative, again! This is a Math thread, meaning that you are not allowed to jump steps. Please follow the "traditional" approach: Step 1. Accuse me of being irritating. Step 2. Accuse me of being provocative. Step 3. Accuse me of blackmailing you. Remember? The integral problem was used as an example to show what a dummy I am. That is all. 

10 Aug 2011, 04:17 

YONG
Xorpd! wrote: Does Mathematica get π/√(27) for the original integral? 

10 Aug 2011, 04:23 

Xorpd!
Quote: Yeah, that is the correct answer to my integral problem. But you have to show me your working! Otherwise, marks will be deducted! Well, if you let 1 ∫ dx/((x²x+1)(exp(2x1)+1)) = A 0 Reverse the order of integration by letting u = 1x to get 1 ∫ du/((u²u+1)(exp(12u)+1)) = B = A 0 If we recall that if A = B, then also A = ½(A+B) and also that 1/(z+1)+1/(1/z+1) = 1 then 1 ∫ ½dx/(x²x+1) = A 0 Completing the square, we find that our next transformation should be x½ = ½√(3)tan(θ) And we get π/6 ∫ dθ/√(3) = A π/6 and the result follows. Another way to see this is via group theory. Let C_s be the group consisting of the identity: Ex = x and the reflection through the point x = ½: σx = 1x Then 1 ∫ f(x)dx = 0 σ1 ∫ f(σ⁻¹x)dσ⁻¹x = σ0 0 ∫ f(1x)(dx) = 1 1 ∫ f(1x)dx 0 Using the character table for C_s: Code: Oh darn, I can't seem to get the σ to show correctly above... We can project out the parts of the functions f(x) = 1/(x²x+1) and g(x) = 1/(exp(2x1)+1) that transform according to the symmetric(A') and antisymmetric(A") representations of the group. f(x) = 1/(x²x+1) + 0 g(x) = ½ + (exp(12x)exp(2x1))/(2(exp(½x)+exp(x½))²) Then 1 ∫ dx/((x²x+1)(exp(2x1)+1)) = 0 1 ∫ {(1/(x²x+1))×½ + (1/(x²x+1))×(exp(12x)exp(2x1))/(2(exp(½x)+exp(x½))²)}dx 0 Since the second term in the curly braces above transforms as the antisymmetric representation of C_s, it integrates to zero and we get the result arrived at previously. 

10 Aug 2011, 06:32 

YONG
Xorpd!, excellent working!
Two questions: (1) How did you know that you should put "u = 1  x"? (2) Could you finish everything within 10 minutes? 

10 Aug 2011, 08:24 

MHajduk
YONG wrote:


10 Aug 2011, 12:12 

MHajduk
Xorpd! wrote: Another way to see this is via group theory. Let C_s be the group consisting of the identity: 

10 Aug 2011, 12:16 

AsmGuru62
IMO, abstract principles in math/calculus are not the same as abstract principles of programming. Example: I can produce pretty cool abstract GUI framework in both C++ and Asm (with virtuals, inheritance and stuff), but I have no idea about the integrals  something from school ... , but that is faded memory.


10 Aug 2011, 12:53 

MHajduk
AsmGuru62 wrote: IMO, abstract principles in math/calculus are not the same as abstract principles of programming. Example: I can produce pretty cool abstract GUI framework in both C++ and Asm (with virtuals, inheritance and stuff), but I have no idea about the integrals  something from school ... , but that is faded memory. 

10 Aug 2011, 13:03 

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