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Why does it work only when h tends to zero?

  1. Jun 17, 2013 #1
    why does it work only when h tends to zero????

    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\\

  2. jcsd
  3. Jun 17, 2013 #2
    f(x+h)/f(x)=x only when h tends to zero
    however shouldnt it work for any value of h as it wasnt assumed that h was zero in the first step .....
    where is the problem????
  4. Jun 17, 2013 #3


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    Homework Helper

    What are you trying to do?
  5. Jun 17, 2013 #4
    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

    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)...
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