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The three integers X, Y, and Z. Is their product XYZ = zero

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 [#permalink] New post 12 Sep 2006, 10:46
ps_dahiya :) Thanks for your replies :)

Yes :) I agree... :) Theorems are the keys. In my eyes, 0^0=1 appears as 1+1=2 almost an evidence ;).

'Series', more a french word?, are pratical and concrete examples of the use of 0^0=1 :)

Why 0^0=1?

Since,
o e^0=1
o x^x=e^(x*ln(x))
o Lim x*ln(x) = 0 when x -> 0
Thus, 0^0=e^0=1
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 [#permalink] New post 12 Sep 2006, 11:10
This problem was sent (2003) by a famous legend on this forum stolyar
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 [#permalink] New post 12 Sep 2006, 11:32
Fig wrote:
ps_dahiya :) Thanks for your replies :)

Yes :) I agree... :) Theorems are the keys. In my eyes, 0^0=1 appears as 1+1=2 almost an evidence ;).

'Series', more a french word?, are pratical and concrete examples of the use of 0^0=1 :)

Why 0^0=1?

Since,
o e^0=1
o x^x=e^(x*ln(x))
o Lim x*ln(x) = 0 when x -> 0
Thus, 0^0=e^0=1

Could you please provide any authentic source that describes 0^0 = 1 ???

Before doing that see these:
This is from IBM not from Windows. The below link (of IBM) also cites IEEE 854 standard.
http://www2.hursley.ibm.com/decimal/daops.html#refpower
http://standards.ieee.org/reading/ieee/ ... _desc.html


OK, not yet convinced? How about this? This is from a trustred source:


-----------------------------------------------
Professor Keith Geddes
Symbolic Computation Group
School of Computer Science
University of Waterloo
200 University Avenue West
Waterloo ON N2L 3G1
CANADA

E-mail: kogeddes@uwaterloo.ca
URL: http://www.uwaterloo.ca/~kogeddes
-----------------------------------------------
--------------------------------------------------------------------------------
The algorithm for integration of elementary functions is known as the Risch algorithm. It will either compute the integral as an elementary function or else prove that the integral cannot be expressed as an elementary function.

For the integrand x^x, the Risch algorithm first converts it to exp-log form. Namely the integrand is:

exp(x*log(x)) .

We give names to the functions appearing here, namely

theta[1] = log(x), theta[2] = exp(x*theta[1]) .

We know that theta[1] is "transcendental over Q(x)" and we know that theta[2] is "transcendental over Q(x,theta[1])".

We now have the problem:

integral theta[2] dx .

Note that the "outermost" function in the integrand is theta[2], which is an exponential function.
Moreover, the integrand is a polynomial in theta[2].

For such a case, the integration theory tells us that the integral, if it exists as an elementary function, must be a similar polynomial in
theta[2],
namely

integral theta[2] dx = q(x)*theta[2] .

(Note that we need to know the theory behind this fact.)

We then differentiate both sides of the above equation, which yields

theta[2] = q'(x)*theta[2] + q(x)*(theta[2])' .

Now in this case, from the definition of theta[2] we have

(theta[2])' = (exp(x*log(x)))'
= exp(x*log(x)) * (log(x) + 1)
= theta[2] * (theta[1] + 1)

Therefore, we have

theta[2] = q'(x)*theta[2] + q(x)*(theta[1]+1)*(theta[2]) .

Since theta[2] is "transcendental over Q(x,theta[1])", we can equate coefficients of theta[2] on each side which gives

1 = q'(x) + (theta[1]+1)*q(x) .

Also, since theta[1] is "transcendental over Q(x)", equating
coefficients of this equation as polynomials in theta[1] gives

1 = q'(x) + q(x)
0 = q(x)

which has no solution.
(The second equation tells us that q(x) = 0 and therefore q'(x) = 0, which makes the first equation become 1 = 0, which is a contradiction.)

CONCLUSION: (integral x^x dx) does not exist as an elementary function.
--------------------------------------------------------------------------------
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 [#permalink] New post 12 Sep 2006, 11:38
I do not know whether this long dicussion is of a real benefit or not ,

but i would leave this question unanswered on the test day if it will take me to go through all of this allogarithms :lol:
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 [#permalink] New post 12 Sep 2006, 12:20
ps_dahiya wrote:
Fig wrote:
ps_dahiya :) Thanks for your replies :)

Yes :) I agree... :) Theorems are the keys. In my eyes, 0^0=1 appears as 1+1=2 almost an evidence ;).

'Series', more a french word?, are pratical and concrete examples of the use of 0^0=1 :)

Why 0^0=1?

Since,
o e^0=1
o x^x=e^(x*ln(x))
o Lim x*ln(x) = 0 when x -> 0
Thus, 0^0=e^0=1

Could you please provide any authentic source that describes 0^0 = 1 ???

Before doing that see these:
This is from IBM not from Windows. The below link (of IBM) also cites IEEE 854 standard.
http://www2.hursley.ibm.com/decimal/daops.html#refpower
http://standards.ieee.org/reading/ieee/ ... _desc.html


OK, not yet convinced? How about this? This is from a trustred source:


-----------------------------------------------
Professor Keith Geddes
Symbolic Computation Group
School of Computer Science
University of Waterloo
200 University Avenue West
Waterloo ON N2L 3G1
CANADA

E-mail: kogeddes@uwaterloo.ca
URL: http://www.uwaterloo.ca/~kogeddes
-----------------------------------------------
--------------------------------------------------------------------------------
The algorithm for integration of elementary functions is known as the Risch algorithm. It will either compute the integral as an elementary function or else prove that the integral cannot be expressed as an elementary function.

For the integrand x^x, the Risch algorithm first converts it to exp-log form. Namely the integrand is:

exp(x*log(x)) .

We give names to the functions appearing here, namely

theta[1] = log(x), theta[2] = exp(x*theta[1]) .

We know that theta[1] is "transcendental over Q(x)" and we know that theta[2] is "transcendental over Q(x,theta[1])".

We now have the problem:

integral theta[2] dx .

Note that the "outermost" function in the integrand is theta[2], which is an exponential function.
Moreover, the integrand is a polynomial in theta[2].

For such a case, the integration theory tells us that the integral, if it exists as an elementary function, must be a similar polynomial in
theta[2],
namely

integral theta[2] dx = q(x)*theta[2] .

(Note that we need to know the theory behind this fact.)

We then differentiate both sides of the above equation, which yields

theta[2] = q'(x)*theta[2] + q(x)*(theta[2])' .

Now in this case, from the definition of theta[2] we have

(theta[2])' = (exp(x*log(x)))'
= exp(x*log(x)) * (log(x) + 1)
= theta[2] * (theta[1] + 1)

Therefore, we have

theta[2] = q'(x)*theta[2] + q(x)*(theta[1]+1)*(theta[2]) .

Since theta[2] is "transcendental over Q(x,theta[1])", we can equate coefficients of theta[2] on each side which gives

1 = q'(x) + (theta[1]+1)*q(x) .

Also, since theta[1] is "transcendental over Q(x)", equating
coefficients of this equation as polynomials in theta[1] gives

1 = q'(x) + q(x)
0 = q(x)

which has no solution.
(The second equation tells us that q(x) = 0 and therefore q'(x) = 0, which makes the first equation become 1 = 0, which is a contradiction.)

CONCLUSION: (integral x^x dx) does not exist as an elementary function.
--------------------------------------------------------------------------------


Thanks a lot for your help :)

I understand the reasoning of the professor above. We cannot calculate the integral of x^x. But, I must say that i don't know what is the link to 0^0 on it?

The function x^x still exists. If u plot it, u will see that, when x tends to 0, x^x tends to 1.

I'm also sorry, I searched on Internet (google, etc...). And I have only found the impact of 0^0 through the example of 'series'.

Perhaps, I should send an email to one of my former teachers ;)

I like also the point of view of Yezz. :)

About the useness of this discussion, at least and a bit selfish ;), I simply say me... On GDAY 0^0 is not definied as 1/x ! Got it Fig ;) :)

Last edited by Fig on 12 Sep 2006, 12:35, edited 1 time in total.
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 [#permalink] New post 12 Sep 2006, 12:28
Thank god. This ended constructively. :wink: :wink:
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 [#permalink] New post 12 Sep 2006, 12:37
I LOVE HAPPY ENDINGS DONT YOU GUYS?? :lol: :lol: :lol: :wink:
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 [#permalink] New post 12 Sep 2006, 12:41
Yes ;)

Finally, all can fall on u Yezz ;) What a question do u bring us here ? ;) Subject of a such long discussion here ;) Loosing time and energy ;)

More seriously, It's a good one :) Thx :)
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 [#permalink] New post 12 Sep 2006, 12:53
Ride on mate :lol: glad u enjoyed it :lol:
  [#permalink] 12 Sep 2006, 12:53
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