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 | 回應

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

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

• 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 | 回應