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maths

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1000, 100, 10, 1, 0.1, 0.01, 0.001

or,

16,8,4,2,1,½,¼

Notice a pattern.

Mike.
 
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
 
tkbits said:
0 ^ 0 is undefined.

For n > 0, 0 ^ n = 0.
For x not 0, x ^ 0 = 1.
I would have thought so, but if you set x = -1 in the series I posted above, 0^0 = 1.

???
 
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.
 
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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.
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.
 
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