This is a great summary of the conversation and provides a clear explanation of the different methods for defining the number e. In summary, the conversation discussed the two main methods for defining e: using Taylor's series or using the function log(x). The first method involves using the limit of difference quotients to find the value of a for which C_a = 1, which is then defined as e. The second method involves using the inverse function of log(x), exp(x), which is defined as the number e. The conversation also mentioned the hyperreals
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  • #2
Do you have a question?
 
  • #3
There is no question. This is just a link to his web page.

By the way, you don't have to use Taylor's series to get the derivative of the exponential. With [itex]f(x)= a^x[/itex], [itex]f(x+h)= a^{x+h}= a^xa^h[/itex] so the "difference quotient" is [itex]/frac{a^xa^h- a^x}{h}= a^x\left(\frac{a^h- 1}{h}\right)[/itex]. The derivative is given by [itex]\frac{da^x}{dx}= a^x\left(\lim_{h\to 0} \frac{a^{h}- 1}{h}[/itex]. The limit depends only on a, not on x- it is a constant with respect to x. Writing that constant as [itex]C_a[/itex], the derivative is [itex]C_a a^x[/itex].

Of course, [itex]C_a[/itex] depends on what a is. If, for example, a= 2, the difference quotient is [itex]\frac{2^h- 1}{h}[/itex]. Taking h= 0.001, we get the approximation [itex]\frac{2^{0.001}- 1}{0.001}= \frac{1.0006934- 1}{0.001}= 0.6934[/itex] for [itex]C_2[/itex]. Similarly for a= 3, and taking h= 0.001, [itex]\frac{3^{0.001}- 1}{0.001}= \frac{1.001099- 1}{0.001}= 1.0099[/tex] as an approximation for [itex]C_3[/itex].

Notice that [itex]C_2[/itex] is less than 1 while [itex]C_3[/itex] is larger than 1. So there exist a value of a, between 0 and 1 such that [itex]C_a= 1[/tex]. We define "e" to be equal to that value of a. That is, [tex]frac{d e^x}{dx}= e^x[/itex].
 
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  • #4
I don't know why I was not allowed to edit the previous post. Hopefully, this will come out correctly!

By the way, you don't have to use Taylor's series to get the derivative of the exponential. With [itex]f(x)= a^x[/itex], [itex]f(x+h)= a^{x+h}= a^xa^h[/itex] so the "difference quotient" is [itex]\frac{a^xa^h- a^x}{h}= a^x\left(\frac{a^h- 1}{h}\right)[/itex]. The derivative is given by [itex]\frac{da^x}{dx}= a^x\left(\lim_{h\to 0} \frac{a^{h}- 1}{h}\right)[/itex]. The limit depends only on a, not on x- it is a constant with respect to x. Writing that constant as [itex]C_a[/itex], the derivative is [itex]C_a a^x[/itex].

Of course, [itex]C_a[/itex] depends on what a is. If, for example, a= 2, the difference quotient is [itex]\frac{2^h- 1}{h}[/itex]. Taking h= 0.001, we get the approximation [itex]\frac{2^{0.001}- 1}{0.001}= \frac{1.0006934- 1}{0.001}= 0.6934[/itex] for [itex]C_2[/itex]. Similarly for a= 3, and taking h= 0.001, [itex]\frac{3^{0.001}- 1}{0.001}= \frac{1.001099- 1}{0.001}= 1.0099[/itex] as an approximation for [itex]C_3[/itex].

Notice that [itex]C_2[/itex] is less than 1 while [itex]C_3[/itex] is larger than 1. So there exist a value of a, between 0 and 1 such that [itex]C_a= 1[/itex]. We define "e" to be equal to that value of a. That is, [itex]\frac{d e^x}{dx}= e^x[/itex].
 
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  • #5
Another way to handle the exponential is to first define [itex]log(x)= \int_1^x\frac{1}{t}dt[/itex]. We can immediately derive several properties. First, since 1/t is not defined for t= 0, this define log(x) only for x> 0. And then the derivative, 1/x, is positive so log(x) is always increasing. Clearly log(1) is 0 so, since the function is increasing, log(x) is positive for x> 1 and negative for x< 1. It goes to positive infinity for x going to positive infinity and goes to negative infinity for x going to 0.

Most importantly, for our purposes, [itex]log(x^y)= \int_1^{x^y}\frac{1}{t}dt[/itex]. Let [itex]u= t^{1/y}[/itex] so that [itex]t= u^y[/itex] and [itex]dt= yu^{y-1}du[/itex]. When t= 1, u= 1 and when [itex]t= x^y[/itex] [itex]u= x[/tex]. So the integral becomes [itex]\int_1^x\frac{1}{x^y}\left(yu^{y-1}\right)dx= y\int_1^y \frac{1}{u} du= y log(x)[/itex].

Since log(x) is an increasing function, from the positive real numbers to all real numbers, it has an inverse function from all real numbers to the positive real numbers. Call that inverse function "exp(x)". If y= log(x) then x= exp(y) and if y= exp(x) then x= log(y). Since the derivative of log(x) is 1/x, the derivative of y= Exp(x) is 1/(1/y)= y= Exp(x). If y= Exp(x) then x= log(y). If x is not 0, [itex]1= \frac{1}{x}log(y)= log(y^{1/x})[/itex]. Going back the other way, [itex]y^{1/x}= Exp(1)[/tex] so [tex]y= (Exp(1)^x[/itex]. That is, the function, Exp(x), is this number, [itex]Exp(1)[/itex] to the x power. We define "e" to be Exp(1), the number, x, such that log(x)= 1.
 
  • #6
These may be mathematically equivalent ways. However, put yourself in the shoes of a department curriculum committee that has to decide what is taught when and make recommendations that make pedagogical sense. Would you choose one of your two suggested methods or would you recommend waiting until after students have seen series expansions?
 
  • #8
jedishrfu said:
I liked the hyperreals approach allowing you to skip the notion of limits initially in Calculus. Keisler's Calculus book goes that route and its arguably more intuitive to students initially than limits are.

https://www.math.wisc.edu/~keisler/calc.html
Agreed. Learning is a drill-down process where you first get an overview and gradually break it down until you fully get it. It makes no sense to me to start with a very detailed account.
 
  • #9
HallsofIvy said:
Notice that C2C2C_2 is less than 1 while C3C3C_3 is larger than 1. So there exist a value of a, between 0 and 1 such that Ca=1Ca=1C_a= 1. We define "e" to be equal to that value of a. That is, dexdx=exdexdx=ex\frac{d e^x}{dx}= e^x.
If we substitute ##n=1/h## and look for the ##a## for which
$$\lim_{n\to\infty} \left(\frac{a^{1/n}-1}{1/n}\right) = 1$$
with a little rearrangement, we get:
$$a=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n$$
which is another definition of ##e## that many people are familiar with.
 
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  • #10
There's a similar presentation in John Roe’s “Elementary Geometry” sections 4.1 – 4.4.
https://books.google.com.sg/books?id=Xmb44wyy7gEC
4.1: Dot product
4.2: Pythagorean theorem
4.3: Series definitions of sine and cosine, then defining angle through cosine and the dot product
4.4: SOHCAHTOA
 

What is SOHCAHTOA?

SOHCAHTOA is a mnemonic device used to remember the trigonometric ratios of sine, cosine, and tangent. It stands for "Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, and Tangent equals Opposite over Adjacent."

How do you use SOHCAHTOA?

To use SOHCAHTOA, you must first identify which side of a right triangle is the opposite, adjacent, and hypotenuse. Then, you can use the mnemonic device to remember which trigonometric ratio to use to solve for the missing side or angle.

Why is SOHCAHTOA important?

SOHCAHTOA is important because it helps us solve problems involving right triangles and trigonometry. It is a useful tool in fields such as engineering, physics, and navigation.

What are some common mistakes when using SOHCAHTOA?

Some common mistakes when using SOHCAHTOA include using the wrong trigonometric ratio, not labeling the sides correctly, and forgetting to convert angles to the correct unit (degrees or radians).

Are there any limitations to SOHCAHTOA?

SOHCAHTOA is limited to right triangles and cannot be used for other types of triangles. It also assumes that the triangle is in a two-dimensional plane and does not take into account three-dimensional objects.

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