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tech / sci.math / Re: Complex Logarithm, Taylor Series Bound

SubjectAuthor
* Complex Logarithm, Taylor Series BoundMidi Midi
`* Re: Complex Logarithm, Taylor Series BoundDavid Petry
 `- Re: Complex Logarithm, Taylor Series BoundDavid Petry

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Complex Logarithm, Taylor Series Bound

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Subject: Complex Logarithm, Taylor Series Bound
From: midispam...@gmail.com (Midi Midi)
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 by: Midi Midi - Wed, 11 May 2022 18:32 UTC

Hi all,

working my way through Shiryaev's book on probability theory, I have encountered an identity for the principal value of the logarithm:

for any complex number z with abs(z)<1/2, we have

log(z+1) = z + theta*abs(z)^2,

where theta is some number depending on z with abs(theta)<=1.

My question is: Can someone give me a hint why abs(theta) should be less than 1? Using standard bounds for the real-valued Taylor expansion, I can only see that abs(theta)<=2.

Thanks a lot!

Re: Complex Logarithm, Taylor Series Bound

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Date: Wed, 11 May 2022 19:16:21 -0700 (PDT)
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Subject: Re: Complex Logarithm, Taylor Series Bound
From: davidlpe...@gmail.com (David Petry)
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 by: David Petry - Thu, 12 May 2022 02:16 UTC

On Wednesday, May 11, 2022 at 11:32:53 AM UTC-7, Midi Midi wrote:
> Hi all,
>
> working my way through Shiryaev's book on probability theory, I have encountered an identity for the principal value of the logarithm:
>
> for any complex number z with abs(z)<1/2, we have
>
> log(z+1) = z + theta*abs(z)^2,
>
> where theta is some number depending on z with abs(theta)<=1.
>
> My question is: Can someone give me a hint why abs(theta) should be less than 1? Using standard bounds for the real-valued Taylor expansion, I can only see that abs(theta)<=2.

Using the Taylor series for log(1+x) gives

(log(1+x)-x)/x^2 = -1/2 - x/3 + x^2/4 ...

which gives

|(log(1+x)-x)/x^2| = | -1/2 - x/3 + x^2/4 ...| <= 1/2 + |x|/3 + |x|^2/4 ... (if |x| < 1)

and if |x| <= 1/2, this gives

|(log(1+x)-x)/x^2| <= 1/2 + 1/2*3 + 12^2*/4 ... = 4*(log(2) - 1/2) < 1

Re: Complex Logarithm, Taylor Series Bound

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Subject: Re: Complex Logarithm, Taylor Series Bound
From: davidlpe...@gmail.com (David Petry)
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 by: David Petry - Tue, 17 May 2022 02:40 UTC

On Wednesday, May 11, 2022 at 7:16:30 PM UTC-7, David Petry wrote:
> On Wednesday, May 11, 2022 at 11:32:53 AM UTC-7, Midi Midi wrote:
> > Hi all,
> >
> > working my way through Shiryaev's book on probability theory, I have encountered an identity for the principal value of the logarithm:
> >
> > for any complex number z with abs(z)<1/2, we have
> >
> > log(z+1) = z + theta*abs(z)^2,
> >
> > where theta is some number depending on z with abs(theta)<=1.
> >
> > My question is: Can someone give me a hint why abs(theta) should be less than 1? Using standard bounds for the real-valued Taylor expansion, I can only see that abs(theta)<=2.
> Using the Taylor series for log(1+x) gives
>
> (log(1+x)-x)/x^2 = -1/2 - x/3 + x^2/4 ...

I guess I meant to write (log(1+x)-x)/x^2 = -(1/2 - x/3 + x^2/4 ... )

>
> which gives
>
> |(log(1+x)-x)/x^2| = | -1/2 - x/3 + x^2/4 ...| <= 1/2 + |x|/3 + |x|^2/4 ... (if |x| < 1)
>
> and if |x| <= 1/2, this gives
>
> |(log(1+x)-x)/x^2| <= 1/2 + 1/2*3 + 12^2*/4 ... = 4*(log(2) - 1/2) < 1

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