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Series[log[x], {x, 0, 3}]

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  Series[log[x], {x, 0, 3}] Hongyi Zhao <hszhao.cn@gmail.com> - 2011-05-21 10:44 +0000
    Re: Series[log[x], {x, 0, 3}] Erik Max Francis <max@alcyone.com> - 2011-05-22 10:56 +0000
    Re: Series[log[x], {x, 0, 3}] Oliver Jennrich <oliver.jennrich@gmx.net> - 2011-05-22 10:56 +0000
    Re: Series[log[x], {x, 0, 3}] Peter Pein <petsie@dordos.net> - 2011-05-22 10:58 +0000

#2546 — Series[log[x], {x, 0, 3}]

Hi all,

I do the following computation within Mathematica:

Series[log[x], {x, 0, 3}]

Then I get:

log[0]+(log^\[Prime])[0] x+1/2 (log^\[Prime]\[Prime])[0] x^2+1/6 (log^(3))
[0] x^3+O[x]^4

Any hints on this result?  The log[0] is illegal in my mind.

Regards.
--
.: Hongyi Zhao [ hongyi.zhao AT gmail.com ] Free as in Freedom :.

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#2593

Hongyi Zhao wrote:
> I do the following computation within Mathematica:
> 
> Series[log[x], {x, 0, 3}]
> 
> Then I get:
> 
> log[0]+(log^\[Prime])[0] x+1/2 (log^\[Prime]\[Prime])[0] x^2+1/6 (log^(3))
> [0] x^3+O[x]^4
> 
> Any hints on this result?  The log[0] is illegal in my mind.

?Series

First, note that `log` has nothing to do with the logarithm function 
`Log`.  As far as Mathematica is concerned, you're referencing an 
unknown function that you just happened to name `log`.  So it should not 
and make any connection between any real-world function.

Second, this is just the standard power series for an arbitrary 
function.  I presume your complaint is that you really meant the 
function to be `Log`, and the logarithm of 0 is undefined, but _you're_ 
the one who specified a Taylor series expansion around x = 0 (a 
Maclaurin series).  So if the function you're thinking of is not defined 
at x = 0, then you shouldn't request the expansion around that point.

-- 
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
  San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis
   An undevout astronomer is mad.
    -- Edward Young

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#2594

Hongyi Zhao <hszhao.cn@gmail.com> writes:

> Hi all,
>
> I do the following computation within Mathematica:
>
> Series[log[x], {x, 0, 3}]
>
> Then I get:
>
> log[0]+(log^\[Prime])[0] x+1/2 (log^\[Prime]\[Prime])[0] x^2+1/6 (log^(3))
> [0] x^3+O[x]^4

As you should.

>
> Any hints on this result?  The log[0] is illegal in my mind.

Why? Mathematica doesn't know what 'log' is, so it treats it (correctly)
as a general function.

If you want to calculate the series of the logarithm, use

Series[Log[x],{x,0.3}]

(Note: Log, not log)

-- 
Space - The final frontier

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#2606

Am 21.05.2011 12:44, schrieb Hongyi Zhao:
> Hi all,
> 
> I do the following computation within Mathematica:
> 
> Series[log[x], {x, 0, 3}]
> 
> Then I get:
> 
> log[0]+(log^\[Prime])[0] x+1/2 (log^\[Prime]\[Prime])[0] x^2+1/6 (log^(3))
> [0] x^3+O[x]^4
> 
> Any hints on this result?  The log[0] is illegal in my mind.
> 
> Regards.
> --
> .: Hongyi Zhao [ hongyi.zhao AT gmail.com ] Free as in Freedom :.
> 

Hi,

try to use Mathematica Syntax for Log; the outcome is not what you might
expect:

In[1]:= Series[Log[x],{x,0,3}]
Out[1]= Log[x]+O[x]^4

but shifting x by one gives a nice taylor series:

In[2]:= Series[Log[1+x],{x,0,3}]
Out[2]= x-x^2/2+x^3/3+O[x]^4

Peter

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