Quod Erat Demonstrandum

2014/05/29

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.

‘solution’

Let \int be “\int_0^x" and f be f(t).

Then

\int f=f-1

1=f-\int f

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

f=(1-\int)^{-1}1

Think about sum to infinity, we have

f=(1+\int+\int\int+\int\int\int+\dots)1

=1+\int_0^x1dx+\int_0^x\int_0^x1dx+\int_0^x\int_0^x\int_0^x1dx+\dots

=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots

=e^x

done…

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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