mersenneforum.org  

Go Back   mersenneforum.org > New To GIMPS? Start Here! > Homework Help

Reply
 
Thread Tools
Old 2019-09-29, 12:34   #12
wildrabbitt
 
Jul 2014

6778 Posts
Default

Quote:
For roots of positive real numbers a, and real values of k, everything is fine -- you've got a real-valued logarithm, defined by an integral.
Can you explain how in



\( 64^3 = e^{3log(64)} \)


the real-valued logarithm is defined by an integral and what the integral is?


or rather


\( 64^{1/2} = e^{(1/2)log(64)} \) ? ( as you said roots).

Last fiddled with by wildrabbitt on 2019-09-29 at 12:39
wildrabbitt is offline   Reply With Quote
Old 2019-09-29, 14:22   #13
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

10011100011002 Posts
Default

Quote:
Originally Posted by wildrabbitt View Post
Can you explain how in



\( 64^3 = e^{3log(64)} \)


the real-valued logarithm is defined by an integral and what the integral is?


or rather


\( 64^{1/2} = e^{(1/2)log(64)} \) ? ( as you said roots).
Your application to the square root of 64 is perfectly correct.

The (natural) log of a positive real number x is defined as

\log(x)\;=\;\int_{1}^{x}dt/t

which is perfectly well-defined. (Here, "t" is what is called a "dummy variable," used simply to avoid using the same symbol to denote two different things in the same statement. It doesn't matter what symbol you use, as long as it isn't already being used for something else.) So, in particular,

\log(64) \; = \; \int_{1}^{64}dt/t\text{.}

Making obvious substitutions, you can even demonstrate the usual "laws of logarithms" directly from this definition. For example, assuming a and b are positive real numbers,

\log(a*b) \; = \; \int_{1}^{a*b}dt/t\;=

\int_{1}^{a}dt/t\;+\;\int_{a}^{a*b}dt/t\text{. Substitute }t \; = \; a\tau \text{, }dt \; = \; ad\tau\text{ in the second integral; }\tau \; = \; 1\text{ to }b

\log(a*b} \; = \; \int_{1}^{a}dt/t \; + \; \int_{1}^{b}d\tau/\tau \; =

\log(a) \; + \; \log(b)

In the complex plane, this comes to grief. You can of course still write

\log(z) \; = \; \int_{1}^{z}dt/t\text{,}

but now, unlike with the positive real numbers, there are myriad paths from 1 to z, and the answer you get depends on the "path of integration."

Suppose, for example, you take the path

t \; = \; exp^{i\theta}\text{, }\theta \; = \; 0\text{ to }2\pi

which winds counterclockwise around the unit circle centered at the origin once, and takes you back to where you started, at z = 1. You get

\log(1) = \int_{0}^{2\pi}id\theta \; = \; 2\pi i

which isn't 0, the answer you would get by integrating over a "path" consisting of the single point t = 1.

You can wind around the circle any number of times, counterclockwise or clockwise, and get any integer multiple of 2*pi*i, as a value of log(1).

This problem can be avoided by making a "branch cut" emanating from 0 (say a ray), and defining a logarithm in the complement of the branch cut. This is essentially declaring by fiat that Thy path of integration shall not intersect the branch cut!. However, this can result in the usual "laws of exponents" giving wrong results.

A classic example of the sort of trouble that can arise is misapplying the rule, valid for positive real a and b, that

\sqrt{\frac{a}{b}} \; = \; \frac{\sqrt{a}}{\sqrt{b}}

Using, on the one hand, a = 1 and b = -1; and, on the other, a = -1 and b = 1. This leads to

\frac{\sqrt{1}}{\sqrt{-1}}  \; = \;  \frac{\sqrt{-1}}{\sqrt{1}}

which, upon "multiplying up" gives

\sqrt{1}\times\sqrt{1} \;= \; \sqrt{-1} \times\sqrt{-1}\text{, so that}

1 \; = \; -1

Misapplication of the rule has resulted in equating the two equal and opposite square roots of -1.

If you allow complex exponents, things become truly bizarre. For example,

i^{i} \; = \; \exp(i\log(i)) \; =

\exp(i(\frac{\pi i}{2} \; + \; 2n\pi i)) \; =

\exp(-\frac{\pi}{2} \; - \; 2n\pi)

which gives an infinity of real values, one for each integer value of n.

Last fiddled with by Dr Sardonicus on 2019-09-29 at 14:27 Reason: xiginf topsy
Dr Sardonicus is offline   Reply With Quote
Old 2019-09-29, 16:23   #14
wildrabbitt
 
Jul 2014

44710 Posts
Default

Thanks.



Is the case of \(i^i\)



....which gives an infinity of real values, one for each integer value of n.


Is the above also due to misapplying the roots rule you mentioned?



Quote:
A classic example of the sort of trouble that can arise is misapplying the rule, valid for positive real a and b, that

https://www.mersenneforum.org/cgi-bi...{a}}{\sqrt{b}}

?


/* damn */

Last fiddled with by wildrabbitt on 2019-09-29 at 16:26 Reason: pasted copied latex which came out as links, and forgot to say thanks
wildrabbitt is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
FATAL ERROR, can somebody explain? Nils Hardware 11 2012-07-21 18:06
Can anyone explain/prove this equality? davieddy Math 9 2009-11-07 07:42
Can anyone explain 'iterations' for factoring? petrw1 PrimeNet 8 2007-08-11 18:28
Could someone please explain my blurb? jasong jasong 5 2007-07-19 00:43
Could someone explain how the Fermat factoring programs work? jasong Information & Answers 3 2006-09-12 02:25

All times are UTC. The time now is 05:48.


Mon Oct 25 05:48:58 UTC 2021 up 94 days, 17 mins, 0 users, load averages: 0.81, 0.99, 0.99

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.