Quod Erat Demonstrandum


just a so-called solution

Filed under: Fun — johnmayhk @ 3:09 下午

From an old file in hard disk…

Find f(x) such that

\int_0^x f(t)dt=f(x)-1.


Let \int be “\int_0^x" and f be f(t).


\int f=f-1

1=f-\int f

1=f(1-\int)  (factorization)


Think about sum to infinity, we have






3 則迴響 »

  1. Obviously, the solution is absolutely wrong! How can we treat the integral sign like this! There is no such thing.

    迴響 由 Simon YAU YAU — 2014/06/01 @ 10:53 上午 | 回覆

  2. can we use operator theory to explain such intuitive idea?

    迴響 由 Justin — 2014/06/02 @ 1:27 上午 | 回覆

    • One can. In fact this is a classical way to solve integral equations I believe. For example, whenever the operator norm for the integration operator is less than 1, and we work in a Banach space, then the solution above makes perfect sense.

      迴響 由 Soarer — 2014/06/05 @ 5:45 下午 | 回覆

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