TRWBW | mingkus: um, nope mindwarp: it's f(x)=log(1+x), f'(x)=1/(1+x) |

futurist | 0 and ln2 |

TRWBW | futurist: f'(x) not f(x) |

futurist | oh, 1/2 and 1 |

mingkus | you are trying to prove this right x/2 < log(1+x) < x ?? |

TRWBW | futurist: and if using the mean value you get x*f'(t) for some t in [0,x] which means t also in [0,1], what can you say about x*f'(t)? futurist: if f'(t) is in [1/2,1], what can you say x*f'(t) is in? |

futurist | lemme think about that for a second |

TRWBW | futurist: if a is in [1/2,1], what can you say a*b is in? |

futurist | oh [b/2,b] -- i was starting at the x*f'(t) thing |

mingkus | futurist: x/2 < log(1+x) < x for x in [0,1], is that what you want to prove? |

TRWBW | futurist: okay, so back to the last one futurist: if f'(t) is in [1/2,1], what can you say x*f'(t) is in? |

futurist | mingkus yeah oh, [x/2,x] hmm weird |

TRWBW | futurist: look helpful? |

mingkus | futurist: is it log(1 + x) or ln(1 + x) |

Safrole | How about using the convexity of the log function? Is that applicable to this problem? |

futurist | so x/(1+x) is in the desired intervale |

TRWBW | ? |

futurist | f'(x) = 1/(1+x) |

mingkus | not if it log |

futurist | i mean, x/(1+x) is in the interval we want to show log(1+x) is in ln |

TRWBW | futurist: you want to show that f(x) is in [x/2,x]. MVT tells you f(x)=f'(t)*x for some t in [0,x]. you just showed f'(t)*x is in [x/2,x] |

futurist | log = ln in this book |

mingkus | log at base e is equal to ln k i know how to do it |

futurist | mingkus yes |

action | TRWBW shaves his head |

TRWBW | shaves his head oops shakes |

mingkus | devide in 2... |

Safrole | lol |

mingkus | x/2 < log(1+x) and log(1+x) < x |

TRWBW | futurist: you want to show that f(x) is in [x/2,x]. MVT tells you f(x)=f'(t)*x for some t in [0,x]. you just showed f'(t)*x is in [x/2,x] |

mingkus | or ln sorry |

futurist | ah |