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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
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. Sad Wink
Post 09 Aug 2011, 04:12
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Is mathematical talent inborn? Most probably yes, but I would be more careful about concluding mathematical abilities from the simple tests on ANS ("Approximate Number System") conducted in children. Arithmetic is an important part of Math but Math is also something more than that - it's an ability of abstractive thinking. I could imagine someone improbably skilled in geometry but inept in calculus. Wink
Post 09 Aug 2011, 07:02
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
Speaking of calculus, it took me almost ten minutes to evaluate the following definite integral. Embarassed

Am I a dummy? Rolling Eyes Confused


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Post 09 Aug 2011, 09:52
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typedef



Joined: 25 Jul 2010
Posts: 2913
Location: 0x77760000
typedef
Nope. You are not.

There's always a starting point.
Post 09 Aug 2011, 15:21
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
YONG wrote:
Speaking of calculus, it took me almost ten minutes to evaluate the following definite integral. Embarassed

Am I a dummy? Rolling Eyes Confused
Despite of the fact we have been corresponding with each other (here on this board) for over three years, I still don't know you. In fact, I know you even less than before. Wink

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. Rolling Eyes


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? Wink
Post 09 Aug 2011, 17:07
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typedef



Joined: 25 Jul 2010
Posts: 2913
Location: 0x77760000
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 Very Happy

Time is the only difference here.


Last edited by typedef on 09 Aug 2011, 18:55; edited 1 time in total
Post 09 Aug 2011, 18:43
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
typedef wrote:
^^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 Very Happy

Time is they only difference here.
Exactly, what I wrote above (about Wolfram's computational engine) was just a sarcastic comment (or should seem to be something like that). Smile
Post 09 Aug 2011, 18:46
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Xorpd!



Joined: 21 Dec 2006
Posts: 161
Xorpd!
YONG wrote:
Speaking of calculus, it took me almost ten minutes to evaluate the following definite integral. Embarassed

Am I a dummy? Rolling Eyes Confused

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?
Post 09 Aug 2011, 22:06
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
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.
Oh no! Mad

I am provocative, again! Rolling Eyes Mad

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? Rolling Eyes Wink

The integral problem was used as an example to show what a dummy I am. That is all.
Post 10 Aug 2011, 04:17
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
Xorpd! wrote:
Does Mathematica get π/√(27) for the original integral?
Yeah, that is the correct answer to my integral problem. But you have to show me your working! Otherwise, marks will be deducted! Wink
Post 10 Aug 2011, 04:23
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Xorpd!



Joined: 21 Dec 2006
Posts: 161
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/((x²-x+1)(exp(2x-1)+1)) = A
0

Reverse the order of integration by letting u = 1-x to get

1
∫ du/((u²-u+1)(exp(1-2u)+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/(x²-x+1) = A
0

Completing the square, we find that our next transformation should be

x-½ = 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 = 1-x

Then

1
∫ f(x)dx =
0

σ1
∫ f(σ⁻¹x)dσ⁻¹x =
σ0

0
∫ f(1-x)(-dx) =
1

1
∫ f(1-x)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/(x²-x+1)
and g(x) = 1/(exp(2x-1)+1) that transform according to the symmetric(A') and antisymmetric(A") representations of the group.

f(x) = 1/(x²-x+1) + 0
g(x) = ½ + (exp(1-2x)-exp(2x-1))/(2(exp(½-x)+exp(x-½))²)

Then

1
∫ dx/((x²-x+1)(exp(2x-1)+1)) =
0

1
∫ {(1/(x²-x+1))×½ + (1/(x²-x+1))×(exp(1-2x)-exp(2x-1))/(2(exp(½-x)+exp(x-½))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.
Post 10 Aug 2011, 06:32
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YONG



Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
Xorpd!, excellent working! Razz

Two questions:

(1) How did you know that you should put "u = 1 - x"?

(2) Could you finish everything within 10 minutes?

Rolling Eyes Wink
Post 10 Aug 2011, 08:24
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
YONG wrote:
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.
Oh no! Mad

I am provocative, again! Rolling Eyes Mad

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? Rolling Eyes Wink

The integral problem was used as an example to show what a dummy I am. That is all.
I'm not going to comment such emotional posts like that, sorry. If you normalize and moderate your language (lower the emotions level to an acceptable one) communication would be possible.
Post 10 Aug 2011, 12:12
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Xorpd! wrote:
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 = 1-x

Then

1
∫ f(x)dx =
0

σ1
∫ f(σ⁻¹x)dσ⁻¹x =
σ0

0
∫ f(1-x)(-dx) =
1

1
∫ f(1-x)dx
0
I like such approach with use of the group theory. This way we all can see a general and clear explanation of the first of substitutions. Smile
Post 10 Aug 2011, 12:16
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AsmGuru62



Joined: 28 Jan 2004
Posts: 1408
Location: Toronto, Canada
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.
Post 10 Aug 2011, 12:53
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
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.
Some general principles as an ability of logical reasoning will be the same, but you're right here - I haven't seen too many situations when higher mathematics would be crucial for effective programming. In everyday work programmers usually have to solve problems referring rather to combinatorics. Smile
Post 10 Aug 2011, 13:03
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