gordonm888 Joined: Feb 18, 2015
• Posts: 3456
September 13th, 2021 at 1:42:20 PM permalink
I met with the mother of the student I will be teaching (but have not yet met the student.) I learned that the student is 14, and is a freshman in high school. He communicates well, and is usually highly focused on whatever he is doing, and especially on math.

I also got the textbook I will be teaching from. The first 356 pages is Advanced Algebra with these topics:

- Coordinate System in Three Dimensions (18 pages)
- Matrices (48 pages)
- Problem Solving and Variation (38)
- Sequences and Series (36)
- Complex Numbers (36)
- Conic Sections (42)
- Radicals, Exponents, Logarithms and Rational Expressions (44)
- Exponential and Logarithmic Function (24)
- Rational and Radical Functions (32)
- Writing Models for Functions (20)
- Analyzing Functions (41)

Then comes Calculus (301 pages)

- Limits and Continuity (48 pages)
- Derivatives (32)
- Applications of Derivatives (44)
- Integration (28)
- Applications of Integrals (32)
- Differential Equations (18)
- Logarithmic, Exponential and Transcendental Functions (18)

Interesting that when i was a student some of these topics were first introduced in college math courses (e.g., Matrices, Complex Variables, Series) after Calculus, but are now high school content before calculus. I guess I'm showing my age.

The textbook is excellent - much much better than anything I had as a student.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888 Joined: Feb 18, 2015
• Posts: 3456
January 14th, 2022 at 10:29:34 AM permalink
I NEED SOME HELP.

After a lesson on solving logarithmic equations, my autistic-gifted student went home and tried applying the techniques I had shown him to a very famous equation, and came up with an impossible/incorrect result. He recognized it as being incorrect and expressed that he was �troubled.�
He showed this (below) to me today, and I sat there puzzled until I had to admit that I had no clear answer as to what he had done wrong.

The very famous equation is:

eπ i = -1

First he squares both sides:
e2π i = 1

Then he writes:
e2π i = e0

Then he essentially takes the natural log of both sides:
2πi = 0

So, this result is obviously incorrect. He then squared both sides to get:
-4π2 = 0

Which troubled him because he recognized this could not be right.

I considered whether he went outside the domain of log(x), or had introduced an ambiguity in squaring functions, but I can�t seem to find the invalid operation.
Any help on this would be greatly appreciated.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888 Joined: Feb 18, 2015
• Posts: 3456
January 14th, 2022 at 10:29:37 AM permalink
I NEED SOME HELP.

After a lesson on solving exponential and logarithmic equations, my autistic-gifted student went home and tried applying the techniques I had shown him to a very famous equation, and came up with an impossible/incorrect result. He recognized it as being incorrect and expressed to me that he was �troubled.�
He showed this (below) to me today, and I sat there puzzled until I had to admit that I had no clear answer as to what he had done wrong.

The very famous equation is:

eπ i = -1

First he squares both sides:
e2π i = 1

Then he writes:
e2π i = e0

Then he essentially takes the natural log of both sides:
2πi = 0

So, this result is obviously incorrect. He then squared both sides to get:
-4π2 = 0

Which troubled him because he recognized this could not be right.

I considered whether he went outside the domain of ln(x), or had introduced an ambiguity by squaring functions, but I can�t seem to find the invalid operation.
Any help on this would be greatly appreciated.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
rsactuary Joined: Sep 6, 2014
• Posts: 1901
Thanks for this post from: January 14th, 2022 at 11:34:15 AM permalink
Yeah, I don't think you can SBS here, but can't really tell you why.

ETA: if you go to the original (famous) equation.. I don't think there is a way to write -1 in the form of e, to equate both sides of exponents (other than what is given).
unJon Joined: Jul 1, 2018
• Posts: 3146
Thanks for this post from: January 14th, 2022 at 11:42:14 AM permalink
Per Wolfram Alpha messing around the issue is the step where you take the natural log.

ln(e^2ipi) = 0 =/= 2ipi
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Ace2 Joined: Oct 2, 2017
• Posts: 1241
Thanks for this post from: January 14th, 2022 at 11:55:18 AM permalink
I don�t think algebra always applies to i the same way it does to �normal� numbers
It�s all about making that GTA
unJon Joined: Jul 1, 2018
• Posts: 3146
January 14th, 2022 at 12:52:28 PM permalink

Scroll down to the section � Problems with inverting the complex exponential function�. It even has a cool 3D graph to illustrate.

Heart of the problem is: �e^(w+2kπi) = e^w for any complex number w and integer k, since adding iθ to w has the effect of rotating ew counterclockwise θ radians.�

ETA: Gordon, in your problem it�s easy to see how above is your formula where w=0 and k=1.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
gordonm888 Joined: Feb 18, 2015
• Posts: 3456
January 14th, 2022 at 1:36:18 PM permalink
Quote: unJon

Scroll down to the section � Problems with inverting the complex exponential function�. It even has a cool 3D graph to illustrate.

Heart of the problem is: �e^(w+2kπi) = e^w for any complex number w and integer k, since adding iθ to w has the effect of rotating ew counterclockwise θ radians.�

ETA: Gordon, in your problem it�s easy to see how above is your formula where w=0 and k=1.

I think this is the answer I was looking for. Let me re-state it. A natural logarithm (ln) function is not the inverse of a complex exponential function of form e^(w+2kπi) because:
- in order for a function to have an inverse it must map distinct values to distinct values. Given e^(w+2kπi), every integer value of k maps e^(w+2kπi) to the same value, Therefore, one cannot take ln(e^(w+2kπi)) and assert that it is equal to (w+2kπi).

As unJon (and his reference) says, e^(w+2kπi) is a periodic or circular function. If you try to take the natural logarithm you destroy the periodical (or circular) nature of the function.

Many thanks, unJon!
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888 