V vicky877 New Member Nov 27, 2007 #1 hi what u think about that is it true or wrong comment.. (x)^0=1 justify yourself..
P Pommie Well-Known Member Most Helpful Member Nov 27, 2007 #2 1000, 100, 10, 1, 0.1, 0.01, 0.001 or, 16,8,4,2,1,½,¼ Notice a pattern. Mike.
jpanhalt Well-Known Member Most Helpful Member Nov 28, 2007 #3 x^n/x^n = 1 Which is to say, x^(n-n) = 1 X^0 = 1 QED John
L ljcox Well-Known Member Dec 1, 2007 #4 The infinite series for (1 + x)^n = 1 + nx + n(n-1) x/2 + n(n-1)(n-2) x^2/6 + ... Note that n is in all terms except the first. Therefore if n = 0, then (1 + x)^n = 1. QED
The infinite series for (1 + x)^n = 1 + nx + n(n-1) x/2 + n(n-1)(n-2) x^2/6 + ... Note that n is in all terms except the first. Therefore if n = 0, then (1 + x)^n = 1. QED
V vicky877 New Member Dec 4, 2007 #6 hi tkbits said: 0 ^ 0 is undefined. For n > 0, 0 ^ n = 0. For x not 0, x ^ 0 = 1. Click to expand... MAN U ARE REALLY VERY CONCEPTUALLY ..
hi tkbits said: 0 ^ 0 is undefined. For n > 0, 0 ^ n = 0. For x not 0, x ^ 0 = 1. Click to expand... MAN U ARE REALLY VERY CONCEPTUALLY ..
L ljcox Well-Known Member Dec 4, 2007 #7 tkbits said: 0 ^ 0 is undefined. For n > 0, 0 ^ n = 0. For x not 0, x ^ 0 = 1. Click to expand... I would have thought so, but if you set x = -1 in the series I posted above, 0^0 = 1. ???
tkbits said: 0 ^ 0 is undefined. For n > 0, 0 ^ n = 0. For x not 0, x ^ 0 = 1. Click to expand... I would have thought so, but if you set x = -1 in the series I posted above, 0^0 = 1. ???
W wschroeder New Member Dec 5, 2007 #8 Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1. Last edited: Dec 5, 2007
Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1.
L ljcox Well-Known Member Dec 6, 2007 #9 wschroeder said: Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1. Click to expand... I did not use the 0^1/0^1 = 0^(n-n) proof. I used the series posted above which does not have any division by zero. However, I can't see how 0^0 can = 1.
wschroeder said: Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1. Click to expand... I did not use the 0^1/0^1 = 0^(n-n) proof. I used the series posted above which does not have any division by zero. However, I can't see how 0^0 can = 1.