- #1
smoothman
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Hi there guys. My first post here. I heard this forum was really helpful so I've signed up lol.I'm trying to get to grips with using Taylor's/McLaurin's formula for series expansions...My main problem lies with expansions of Logarithmic functions..
I want to work out how to expand Logs when numbers are used instead of "x".
for example i want to expand ln 2 instead of ln x
or ln 12 instead of ln (1 + x). i can expand the function when there is an x involved inside the ln function.. but if its just a number such as (ln 2) then for some reason my expansion doesn't seem to work...
for example i can work out ln (1 + x) using mclaurins:
f(x) = ln (1 + x)
[tex]f'(x) = 1/(1+x) [/tex]
[tex]f''(x) = -1/(1+x)^2 [/tex]
[tex]f'''(x) = 2/[(1+x)^3[/tex]
so now we plug that into mclaurins we have:
f(0) = ln 1 = 0
[tex] f'(0)x = \frac{1}{1+0}x = x[/tex]
[tex] \frac{f''(0)}{2!}x^2 = \frac{-1}{2!(1+0)^2}x^2 = -\frac{x^2}{2}[/tex]
[tex] \frac{f'''(0)}{3!}x^3 = \frac{2}{3!(1+0)^3}x^3 = \frac{x^3}{3}[/tex]
so we deduce that
[tex] ln (1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}...[/tex]so I am fine with expanding logs that have an x involved...
NOW HERES THE REAL PROBLEM
how would i prove something like: ln 2 (which has no x involved in the ln function)..
ln 2 = 1 - 1/2 + 1/3 - ... = [tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]
so far i can work out ..
f(x) = ln 2
f'(x) = 1/2
f''(x) = 0...
so i plug it into mclaurins i get:
i can neglect f(0) since the sum is from 1 to infinity..
[tex] f'(0)x = \frac{1}{2}x = \frac{1}{2}x[/tex]
[tex] \frac{f''(0)}{2!}x^2 = \frac{0}{2!}x^2 = 0[/tex]
as you can see here I am immediately going wrong since the first term should just have been 1, then -1/2 ... as shown above on the BLUE text.
please guide me on here. and how do you acutally deduce that the terms of ln 2 is equivalent to
[tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]
thnk you very much guys =)
I want to work out how to expand Logs when numbers are used instead of "x".
for example i want to expand ln 2 instead of ln x
or ln 12 instead of ln (1 + x). i can expand the function when there is an x involved inside the ln function.. but if its just a number such as (ln 2) then for some reason my expansion doesn't seem to work...
for example i can work out ln (1 + x) using mclaurins:
f(x) = ln (1 + x)
[tex]f'(x) = 1/(1+x) [/tex]
[tex]f''(x) = -1/(1+x)^2 [/tex]
[tex]f'''(x) = 2/[(1+x)^3[/tex]
so now we plug that into mclaurins we have:
f(0) = ln 1 = 0
[tex] f'(0)x = \frac{1}{1+0}x = x[/tex]
[tex] \frac{f''(0)}{2!}x^2 = \frac{-1}{2!(1+0)^2}x^2 = -\frac{x^2}{2}[/tex]
[tex] \frac{f'''(0)}{3!}x^3 = \frac{2}{3!(1+0)^3}x^3 = \frac{x^3}{3}[/tex]
so we deduce that
[tex] ln (1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}...[/tex]so I am fine with expanding logs that have an x involved...
NOW HERES THE REAL PROBLEM
how would i prove something like: ln 2 (which has no x involved in the ln function)..
ln 2 = 1 - 1/2 + 1/3 - ... = [tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]
so far i can work out ..
f(x) = ln 2
f'(x) = 1/2
f''(x) = 0...
so i plug it into mclaurins i get:
i can neglect f(0) since the sum is from 1 to infinity..
[tex] f'(0)x = \frac{1}{2}x = \frac{1}{2}x[/tex]
[tex] \frac{f''(0)}{2!}x^2 = \frac{0}{2!}x^2 = 0[/tex]
as you can see here I am immediately going wrong since the first term should just have been 1, then -1/2 ... as shown above on the BLUE text.
please guide me on here. and how do you acutally deduce that the terms of ln 2 is equivalent to
[tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]
thnk you very much guys =)
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