Why does it work only when h tends to zero?

  • Thread starter mpx86
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  • #1
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why does it work only when h tends to zero????

[itex]\hat{}\[\begin{array}{l}
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\\
\\

\end{array}\][/itex]
 

Answers and Replies

  • #2
10
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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????
 
  • #3
statdad
Homework Helper
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What are you trying to do?
 
  • #4
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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

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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