TRWBW | emes: yes emes: and x=e^log(x)=sum n=0..inf (log(x))^n/n! |

zazoru | TRWBW, hm yeah... but 332012 could be different ways... like if you chance the first ring with the second you still have 332012 as that... am function you defined |

TRWBW | zazoru: either i'm wrong, the book is wrong, or you are giving the book's question wrong. 3 possibilities. |

emes | so e^sin(0) = sum(n=0...oo) (sin(0))^n/n! ... except sin(0) = 0, and the first term of the series is undefined, 0^0 |

TRWBW | emes: well i kinda grimaced, because i don't like that, but i didn't say anything |

Olathe | 0^0 = 1 |

TRWBW | emes: it's a matter of disagreement emes: defining 0^0=1 is fundamentally a matter of choice, just like any other math definition emes: 0^0=1 works better for series expansion |

Olathe | Or, you can loosely use the limit style of thinking, and note that with any other value for the power, you always get a 1. |

TRWBW | emes: 0^0 undefined works better in terms of saying a^b is continous for real a emes: if i was writing, i would have written e^x=1+sum i=1..inf x^i/i!, but that's just me |

emes | well, the rest of the terms are zero when x = 0 |

TRWBW | emes: it seems your problem isn't a series question, it's a whether 0^0 is defined question emes: and like i said, you can do that either way |

emes | well, it is a series question, 0^0 just came up along the way |

TRWBW | emes: if you want, btw, you can define sqrt(9)=-3. i won't stop you. emes: if you want, you can define arcsin(0)=100*pi |

Olathe | Here, you can't define 0^0 as anything other than 1, though, since e^0 = 1. |

emes | how would I go about a second order approximation for the series when the terms are all zero after the first? TRWBW: the range of arcsin is defined |

TRWBW | Olathe: i can define 0^0=cuberoot("apple pie"). but i'll have an uphill climb proving other things. |

Olathe | TRWBW: Sure, but you can't in this problem without contradiction. |

TRWBW | emes: exactly. it is defined. by a person. not by the math king. |

Olathe | The math king will be furious at your dismissal ! |

emes | rex mathematicus |

TRWBW | emes: your teacher might disagree, but truth is he's already bitter because he's teaching high school math. |

emes | lol so to make a second order approximation of e^sin(x) around x=0, I need to take derivatives, correct, because the terms in the series don't cooperate (to bring us back on topic ;) |

TRWBW | emes: not quite. if you define 0^0=1, then the series is like you wrote it. if you don't it's like i wrote it, 1+sin(x)+sin(x)^2/2+... either way, you can make a second order approximation by taking the first terms |

emes | you're right, I was getting caught up by the zero though that couldn't be considered a polynomial approximation of course |

TRWBW | emes: eh, depends how you define polynomial. ;;) (=double ;) |

noway- | I am having trouble with a proof, http://www.mathbin.net/8714 |

Steve|Office | noway-: So S is both an element and a subset of A? |

noway- | Steve|Office: yes |

Steve|Office | Odd, well, Since S is a subset of A, it is an element of the power set of A. |

Olathe | What does the then mean ? |

Steve|Office | Heh, good question. I parsed it as "and." |