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

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Joined: 01 May 2006
Posts: 1798
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Kudos [?]: 102 [0], given: 0

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
SVP
Joined: 05 Jul 2006
Posts: 1516
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Kudos [?]: 124 [0], given: 39

This problem was sent (2003) by a famous legend on this forum stolyar
CEO
Joined: 20 Nov 2005
Posts: 2910
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 19

Kudos [?]: 132 [0], given: 0

Fig wrote:

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

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

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.
--------------------------------------------------------------------------------
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 124 [0], given: 39

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
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

ps_dahiya wrote:
Fig wrote:

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

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

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.
CEO
Joined: 20 Nov 2005
Posts: 2910
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 19

Kudos [?]: 132 [0], given: 0

Thank god. This ended constructively.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 124 [0], given: 39

I LOVE HAPPY ENDINGS DONT YOU GUYS??
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

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
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 124 [0], given: 39

Ride on mate glad u enjoyed it

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