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
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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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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008