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Proving ln(x) using infinite series

  • Thread starter darewinder
  • Start date
14
0
1. Homework Statement

Well we are given a series of steps done with the number "x" and in the end the end value is ln(x). Basically we are asked to prove why it isnt a coincedience

2. Homework Equations

I put the steps into an equation, but i cant prove it.

ln(x) =[tex]^{lim }_{n->inf}[/tex] [tex](x^\frac{1} {2^n} -1)*2^n[/tex]


3. The Attempt at a Solution

Well plugging in gives me inf times 0 so i thought of solving it using Hopital's rule but i cant get to a form where it is 0/0. I tried factoring, rationalizing but i couldnt get anywhere. I would appreciate if you guys can help me do this little bit.

and I would also appreciate if anyone can show me how to come up with a sequence for this. Second part of the question askes me to come up with my own limit of a sequence to get ln(x). I just flipped changed the sine of - inside the brackets to make it + and added the minus sign to the -2n at the end, but thats like the same thing. So if you guys can give me some ideas it would be great!! :)

Thank you
 
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Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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Since n only occurs in 2n, you can simplify by letting a= 2n. Then the problem becomes showing that
[tex]ln(x)= \lim_{a\rightarrow \infty}(x^(1/a)-1)(a)[/tex]

Now I would be inclined to "reverse" the function: If y= (x1/a-1)(a), then x= (y/a+ 1)a. Do you recognize the limit of that as a common formula for ey? And if x= ey, then y= ln(x).
 
14
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tanks a bunch ill get back on you on the reverse function let me work it out. Im eating right now :) But the limit makes sense (argg why didn't i think of that)
 
14
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hmm i have a question about this, x= (y/a+ 1)a.

Wouldn't x=1 when we take the limit as a>>inf?

Thanks for your help

never mind i see how x wouldn't equal to 1 because y has some "a" terms in it so we don't know the ratio. But i dont see how x = e^y. Is the x formula like an equation for the e function?
 
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