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# taylor`s or maclaurin`s series

Posted on 2005-04-05
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You are given the value of log 4=0.6021.Find the approximate value of log 404 using taylor`s or maclaurin`s series.
Hint:Use the formula for log (1+e^x) base e
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Question by:aniltalks
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ozo earned 252 total points
ID: 13715482
Is this homework?
Can you build a Taylor or Maclaurin series for  log (1+e^x) base e?
You know that log(404) = log(4)+log(10²)+log(404/400)
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Expert Comment

ID: 13743565
Hi aniltalks,
McLaurin's series is given by
________________________________________________________________________
f(x) = f(0)/0! + f'(0)/1! + f''(0)/2! + f'''(0)/3! + ....
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Given f(x) = ln(1 + e^x)
f'(x) = e^x/(1 + e^x)
f''(x) = [(1+e^x)e^x - e^x] / (1 + e^x)² = (1 - e^x + e^2x / (1 + e^x)²
...
.....

Bye
---
Harish
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LVL 37

Expert Comment

ID: 13743637
Sorry..
f''(x) = [(1+e^x)e^x - e^2x] / (1 + e^x)² = e^x / (1 + e^x)²

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LVL 37

Expert Comment

ID: 13743647
Ozo...
log(404) = log(4)+log(10²)+log(404/400)
means
log(404) = log(4)+log(10²)+log(404)-log(400)

log(404) gets cancelled and you are left with
0 = log(4)+log(10²)-log(400)

What does it have to do with the given problem?
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LVL 37

Expert Comment

ID: 13743675
Again a mistake :(
I have left the x coeffecients themselves !

Mclaurin's series is given by
________________________________________________________________________
f(x) = 1 f(0)/0! + x f'(0)/1! + x² f''(0)/2! + ... +  x&#8319; f&#8319;(0)/n! + ....
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
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Assisted Solution

Harisha M G earned 248 total points
ID: 13743684
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