why does it work only when h tends to zero????

$\displaystyle f(x + h) = xf(x)\\

Ef(x) = xf(x)\\

E = x\\

\ln E = hD\\

\ln x = hD\\

f(x) = y\\

y\ln x = hDy\\

y\ln x = h\frac{{dy}}{{dx}}\\

\int {} \ln xdx = h\int {} dy/y\\

x\log x/e = h\ln y + \ln c\\

x\log x/e = h\ln y/C\\

(x/h)\ln x/e = \ln y/C\\

C(x/e)\frac{{x/h}}{1} = y = f(x)\\

f(x + h) = C((x + h)/e)\frac{{(x + h)/h}}{1}\\

f(x + h)/f(x) = (((x + h)/e)\frac{{(x + h)/h}}{1})/(x/e)\frac{{x/h}}{1}\\

f(x + h)/f(x) = ((x + h)/x)\frac{{x/h}}{1})*((x + h)/e)\\

f(x + h)/f(x) = (1 + h/x)\frac{{x/h}}{1}*((x + h)/e)\\

\mathop {\lim }\limits_{h \to 0} f(x + h)/f(x) = \mathop {\lim }\limits_{h \to 0} e*((x + h)/e)\\

\mathop {\lim }\limits_{h \to 0} f(x + h)/f(x) = \mathop {\lim }\limits_{h \to 0} (x + h)\\

\mathop {\lim }\limits_{h \to 0} f(x + h)/f(x) = x\\

\\$

Re: why does it work only when h tends to zero????

why is it true only when h tends to zero ...

but we havent assumed h to be zero in our initial assumption....???

Re: why does it work only when h tends to zero????

What is this series of equalities?

Re: why does it work only when h tends to zero????

some latex error...ignore <br/> everwhere

Re: why does it work only when h tends to zero????

I understand that <br/> is spurious. (On this forum, LaTeX does not like newlines in formulas and requires spaces at least every 50 characters. In this case, it is better to have one formula per line.) I am asking about the status of these equations. Are they a proof of some fact? Does every subsequent formula follow from the previous one or is it the other way around? What are the new symbols E and D?

Contrary to a possible stereotype, mathematicians would not parse a text consisting of mathematical formulas only (unless it is written by a genius). They require a plain text explanation of the purpose, method and so on.

Re: why does it work only when h tends to zero????

PURPOSE

to find a function such that when it is increased by h(Constant) at x it becomes x times it value at x

E and D

E is shift operator such that Ef(x)=f(x+h) , E^2 f(x)=f(x+2h) and so on s.t. E^nf(x)=f(x+nh)

D is differtial operator that is d/dx...

STEP 4

using taylors series

e^hD= E

Re: why does it work only when h tends to zero????

Quote:

Originally Posted by

**mpx86** PURPOSE

to find a function such that when it is increased by h(Constant) at x it becomes x times it value at x

E and D

E is shift operator such that Ef(x)=f(x+h) , E^2 f(x)=f(x+2h) and so on s.t. E^nf(x)=f(x+nh)

D is differtial operator that is d/dx...х.

Thanks, this helps. Another thing that is unclear is whether this argument is taken from a textbook and you need help understanding it or if it is something you came up with yourself and it needs checking. Also, it is not clear whether the required f(x) has been found: the last line says $\displaystyle \lim_{h \to 0} f(x + h)/f(x) = x$, which is an immediate corollary of the requirement in the first line, and not only in the limit, but for every h.

More seriously, the line E = x and the following lines don't make sense to me: the left-hand side is an operator (i.e., a mapping from functions to functions), and the right-hand side is either a number x or an identity function. These entities have different types. Similarly, it is not clear what a logarithm of an operator means.

Re: why does it work only when h tends to zero????

i am asking why does it yields correct answer only when h tends to zero ....we havent assumed it to be tending to zero in our initial assumption... why does it yields h tending to zero at the end...

i thing it is due to defination of derivatives (which we have used in our derivation) which has been defined for h tending to zero.... besides it just doesnt look like the complete answer.... this is a question done by me and is not written in any textbook (yielding more chances for error)...

moreover it always yield correct answers when h tends to zero....(try different functions ) .....

however the process yields correct answer (exactly) when E is not a dependent quantity or a variable that is when E=constant.....

Re: why does it work only when h tends to zero????

so no one knows ..... no problem..... i will figure it out

myself..........