September 13th, 2021 at 1:42:20 PM
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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.

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.

January 14th, 2022 at 10:29:34 AM
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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:

First he squares both sides:

Then he writes:

Then he essentially takes the natural log of both sides:

So, this result is obviously incorrect. He then squared both sides to get:

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.

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:

e

^{π i}= -1First he squares both sides:

e

^{2π i}= 1Then he writes:

e

^{2π i}= e^{0}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:

-4π

^{2}= 0Which 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.

January 14th, 2022 at 10:29:37 AM
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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:

First he squares both sides:

Then he writes:

Then he essentially takes the natural log of both sides:

So, this result is obviously incorrect. He then squared both sides to get:

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.

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:

e

^{π i}= -1First he squares both sides:

e

^{2π i}= 1Then he writes:

e

^{2π i}= e^{0}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:

-4π

^{2}= 0Which 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.

January 14th, 2022 at 11:34:15 AM
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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).

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).

January 14th, 2022 at 11:42:14 AM
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Per Wolfram Alpha messing around the issue is the step where you take the natural log.

ln(e^2ipi) = 0 =/= 2ipi

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.

January 14th, 2022 at 11:55:18 AM
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I don’t think algebra always applies to i the same way it does to “normal” numbers

It’s all about making that GTA

January 14th, 2022 at 12:52:28 PM
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This Wikipedia page has an explanation: https://en.m.wikipedia.org/wiki/Complex_logarithm

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.

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.

January 14th, 2022 at 1:36:18 PM
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Quote:unJonThis Wikipedia page has an explanation: https://en.m.wikipedia.org/wiki/Complex_logarithm

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.

link to original post

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!

January 14th, 2022 at 2:07:02 PM
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I think a related concept is that x

Clearly you cannot take the square of x = -1 and calculate the result as 1 and then take the square root of 1, and calculate that result as +/-1 and somehow claim that -1 = +/-1.So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

^{1/2}is not single-valued, therefore x^{2}and x^{1/2}are only inverse functions if you restrict the domain of x^{1/2}to "all non-negative numbers" or "all negative numbers."Clearly you cannot take the square of x = -1 and calculate the result as 1 and then take the square root of 1, and calculate that result as +/-1 and somehow claim that -1 = +/-1.