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bhobba

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There is an excellent math teacher out here in Aus (he won young Australian of the year) who makes good videos. Here is one about the old conundrum 0^0:

As you can see in my comment, I'm afraid I have to disagree with Eddie here. 0^0 is undefined because you get different answers depending on how you take the double limit. It shows why one must be really careful with what it means by powers. You really need calculus to define them properly. You will find it in real analysis or honours calculus textbooks, but sketch it define log to base e, Ln(x) as integral 1 to x 1/y dy. We see a problem immediately at x =0. Differentiating Ln(xy), you get 1/x. Thus Ln(xy) = ln (x) + C. If x = 1 we see C = Ln(y) or Ln(xy) = Ln(x) + Ln(y). Let e^x be the inverse of Ln(x). Let a = e^x, b = e^y, and we have e^(x+y) = e^(Ln(a) +Ln(b)) = e^Ln(a*b) = a*b = e^x*e^y. We can, from these relations, work out all the other fundamental relations of logs and powers. As I explained in the comments, this greatly helps clarify 0^0. Basically, Eddie was a bit sloppy and an example of why you need real analysis.

Thanks

Bill

As you can see in my comment, I'm afraid I have to disagree with Eddie here. 0^0 is undefined because you get different answers depending on how you take the double limit. It shows why one must be really careful with what it means by powers. You really need calculus to define them properly. You will find it in real analysis or honours calculus textbooks, but sketch it define log to base e, Ln(x) as integral 1 to x 1/y dy. We see a problem immediately at x =0. Differentiating Ln(xy), you get 1/x. Thus Ln(xy) = ln (x) + C. If x = 1 we see C = Ln(y) or Ln(xy) = Ln(x) + Ln(y). Let e^x be the inverse of Ln(x). Let a = e^x, b = e^y, and we have e^(x+y) = e^(Ln(a) +Ln(b)) = e^Ln(a*b) = a*b = e^x*e^y. We can, from these relations, work out all the other fundamental relations of logs and powers. As I explained in the comments, this greatly helps clarify 0^0. Basically, Eddie was a bit sloppy and an example of why you need real analysis.

Thanks

Bill

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