n^n

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  1. james22's Avatar
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    Re: n^n
    (Original post by dugdugdug)
    But I thought the beauty of maths is that there is no ambiguity, with there ever only being one universal answer to a problem or are we delving into the higher echelons?
    We only say that 0^0=1 because thats how we have defined it, the definition of exponents doesn't cover the cases of 0^0 so we have to extend its definition, however depending on what sort of maths we are doing it makes sense to use a more appropriate definition.

    It's a bit like when we extend multiplication to vectors, we have to change the definition to make it fit our new maths.
  2. dugdugdug's Avatar
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    Re: n^n
    (Original post by james22)
    We only say that 0^0=1 because thats how we have defined it, the definition of exponents doesn't cover the cases of 0^0 so we have to extend its definition, however depending on what sort of maths we are doing it makes sense to use a more appropriate definition.

    It's a bit like when we extend multiplication to vectors, we have to change the definition to make it fit our new maths.
    So I suppose it's a bit like defining n! as nx(n-1)x(n-2)x ..... x 3x2x1 for convenience because we know n! exists for non-integers.
  3. around's Avatar
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    Re: n^n
    (Original post by dugdugdug)
    But I thought the beauty of maths is that there is no ambiguity, with there ever only being one universal answer to a problem or are we delving into the higher echelons?
    maths is a human activity, isn't it?
  4. nuodai's Avatar
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    Re: n^n
    (Original post by dugdugdug)
    But I thought the beauty of maths is that there is no ambiguity, with there ever only being one universal answer to a problem or are we delving into the higher echelons?
    Sorry to disappoint. Maths is about exploring consequences of assumptions; if two strands of maths use different assumptions then their consequences need not agree.

    Edit: To make my point clearer, it makes no sense to ask what something is unless you say (or it's already understood) how it's defined. If I asked what ``+-\times-+" was equal to then it wouldn't make any sense unless I defined what I meant by that string of symbols, and 0^0 is no different. It's sometimes convenient to define something which is written in the same way to mean two different things in mathematics. If \alpha, \beta are ordinal numbers (or cardinal numbers, or sets) and we're working in the context of set theory then it's useful to define \alpha^{\beta} in terms of sets of functions. But if \alpha, \beta are (real or) complex numbers and we're working in the context of (real or) complex analysis then it's usual to define \alpha^{\beta} = \exp( \beta \log \alpha ) where \exp z = \sum_0^{\infty} \frac{z^n}{n!} and \log z = \log \left| z \right| + i\arg z (and \log is defined as usual for real numbers). And if \alpha, \beta are natural numbers and we're teaching exponentiation to schoolchildren then we'd want to ignore all the technicalities above and say that \alpha^{\beta} is what you get when you multiply \alpha by itself \beta times. What do you get when you multiply 0 by itself 0 times? (The answer is: it really depends what you mean by "multiplying zero", and by "multiplying [something] zero times".)

    So 0^0 can be what you want it to be, as long as you define it appropriately and your definition is consistent with whatever you choose to do with it.
    Last edited by nuodai; 06-04-2012 at 09:28.
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