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# Week 1 Thursday 6/22 brief notes.
**Reading: Ch. 6.1, 6.2$\ast$, 6.3$\ast$, 6.4$\ast$, 6.7, 6.8.**
Last time we defined the natural logarithm function $\ln(x)=\int_1 ^x \frac{1}{t} dt$, and derived several of its properties. Let us continue.
## Definition of the number $e$.
Since $\ln(x)$ is continuous (by fundamental theorem of calculus), and that it increases from negative infinity to positive infinity, and that it is one-to-one, we must have $\ln(x)=1$ at a **unique** point $x$. We shall **define** this unique point $e$, that $\ln(e) = 1$. Numerically one can estimate it to be $e\approx 2.71828...$.
But wait a minute, is this $e$, the unique point where $\ln(x)=1$, the same as Bernoulli's limit $$
\lim_{N\to\infty} \left( 1+\frac{1}{N} \right)^N \stackrel{???}{=}e
$$
Let us pretend the limit $L= \lim_{N\to\infty} \left( 1+\frac{1}{N} \right)^N$, what we need to show is that indeed $\ln(L)=1$, so that it is precisely the unique value $L=e$.
Since $\ln(x)$ is continuous, we shall recall that **continuous functions preserves limit(!)**, so $$
\begin{align*}
\ln(L)&=\ln\left( \lim_{N\to\infty}\left( 1+\frac{1}{N} \right)^N \right) \\
&\stackrel{(!)}{=}\lim_{N\to\infty}\ln\left( 1+\frac{1}{N} \right)^N \\
& =\lim_{N\to\infty}\underbrace{N}_{\to\infty}\underbrace{\ln\left( 1+\frac{1}{N} \right)}_{\to 0}
\end{align*}
$$
But how do we compute this limit? We have some $\infty\cdot0$ situation. Notice we can re-expressed as an **indeterminant form** of the type $\frac{0}{0}$, which we can then apply **L'Hospital rule** (we will discuss L'Hospital rule with a bit more detail later) with respect to $N$: $$
\begin{align*}
\lim_{N\to\infty} \frac{\ln\left( 1+\frac{1}{N} \right)}{\frac{1}{N}} & \stackrel{L'H}{=}\lim_{N\to\infty} \frac{\frac{1}{1+\frac{1}{N}} \cdot \frac{-1}{N^2}}{\frac{-1}{N^2}}\\
&=\lim_{N\to\infty} \frac{1}{1+\frac{1}{N}}=1
\end{align*}
$$
Hence $\ln(L)=1$. And as $\ln(e)=1$, and that $\ln(x)$ is one-to-one, we must have $e=L$, or
> $$
e=\lim_{N\to\infty}\left( 1+\frac{1}{N} \right)^N
$$
Neat!
## Some integrals involving natural logarithm.
The natural logarithm $\ln(x)$ has domain only on positive $x > 0$. There is natural way to extend it by considering the composition $\ln(|x|)=\begin{cases}\ln(x) & \text{if } x > 0\\ \ln(-x) & \text{if } x < 0\end{cases}$ , which has derivative $$
\frac{d}{dx}\ln(|x|)=\begin{cases}
\frac{1}{x} & \text{if }x > 0\\ \\
\frac{-1}{x}(-1)=\frac{1}{x} & \text{if }x < 0
\end{cases}
$$
So $\frac{d}{dx}\ln|x|=\frac{1}{x}$ for all $x\neq 0$. Hence have the generic antiderivative
> $$
\int \frac{1}{x}dx=\ln|x|+C
$$
**Example.**
Find the antiderivative : $\int \frac{x}{x^2 + 7} dx$.
Using $u$-substitution, with $u=x^2 + 7$, we have $du=2xdx$, so $$
\begin{align*}
\int \frac{x}{x^2 + 7}dx &= \int \frac{du}{2u} \\
&= \frac{1}{2} \ln|u| + C \\
&= \frac{1}{2}\ln| x^2 + 7| + C \\
&= \ln\sqrt{x^2 + 7} + C. \quad\blacklozenge
\end{align*}
$$
**Example.**
Find the antiderivative : $\int \frac{\ln(x)}{x}dx$.
Let $u = \ln(x)$, then $du = \frac{dx}{x}$. So we have $$
\begin{align*}
\int \frac{\ln(x)}{x}dx&=\int udu\\
&= \frac{1}{2}u^2+C \\
&=\frac{1}{2}(\ln(x))^2 + C. \quad\blacklozenge
\end{align*}
$$
## The natural exponentiation function. (Ch 6.3$\ast$)
The natural exponentiation function $\exp(x)$ is **defined** to be the **inverse** of the natural logarithm function $\ln(x)$, as we recall $\ln(x)$ is a one-to-one strictly increasing function, so such an inverse exists. Now by the inverse relations, $$
\ln(a)=b \iff \exp(b)=a
$$But is this $\exp(x)$ the same as $e^x$ as we expect?
Note we have defined the number $e$ as the unique point where $\ln(e)=1$, so for any number $x$ we have $$
x= x \cdot 1 = x \cdot \ln(e) =\ln(e^x)
$$which by the property of inverse, $$
\exp(x)=e^x
$$
In other words $e^x$ is the inverse to $\ln(x)$, and we have $$
e^x = y \iff \ln(y)=x
$$as well as cancellation laws $$
\begin{align*}
e^{\ln(x)} = x \text{ , for $x > 0$}\\
\ln(e^x)=x \text{ , for all $x$}
\end{align*}
$$
**Example.**
Solve for all values of $x$ if $3e^{4x^2-7} = 5$.
$\blacktriangleright$ Note $$
\begin{align*}
3 e^{4x^2 - 7 } = 5 &\implies e^{4x^2 - 7} = \frac{5}{3}\\
&\implies 4 x^2 - 7 = \ln\left( \frac{5}{3} \right) \\
&\implies 4x^2 = 7 + \ln(5 / 3)\\
&\implies x = \pm \sqrt{\frac{7+\ln(5 / 3)}{4}}
\end{align*}
$$
## Properties of natural exponentiation.
Just as the properties of the natural logarithm, we have analogously the following:
> (1) $e^x e^y = e^{x+y}$
> (2) $\frac{e^x}{e^y}=e^{x-y}$
> (3) $(e^x)^r = e^{xr}$
Proof of (1). $\blacktriangleright$ Using the properties of log, note that $\ln(e^x e^y) = \ln(e^x)+\ln(e^y) = x+y$, so by inverse relation, we have $e^x e^y = e^{x+y}$. $\blacksquare$
Proof of (2). $\blacktriangleright$ Note $\ln( \frac{e^x}{e^y})=\ln(e^x) - \ln(e^y)=x-y$, so $\frac{e^x}{e^y}=e^{x-y}$. $\blacksquare$
Proof of (3). $\blacktriangleright$ Note $\ln((e^x)^r)=r\ln(e^x)=rx$, so $(e^x)^r = e^{rx}$. $\blacksquare$
## Derivative of natural exponentiation.
To derive the derivative of $\exp(x) =e^x$, we use the fact that it is the inverse function to $\ln(x)$, and exploit chain rule. Since $\ln(e^x)=x$, differentiating both sides give $$
\frac{d}{dx}\ln(e^x)= \frac{d}{dx}(x)=1
$$
But by chain rule, $$
\frac{d}{dx}\ln(e^x)= \frac{1}{e^x}\cdot \frac{d}{dx}e^x
$$So $$
\frac{1}{e^x} \frac{d}{dx}e^x = 1
$$
In other words,
> $$
\frac{d}{dx}e^x = e^x
$$
The natural exponentiation function is a wonderful function whose **derivative is itself!**
What about derivatives of exponential functions with a different base?
**Example.**
Consider $f(x)=3^x$, find $f'(x)$.
$\blacktriangleright$ First, take the natural log of both sides, so we get $\ln(f(x))=\ln(3^x)=x\ln(3)$. Now differentiate both sides and use chain rule, we get $$
\frac{f'(x)}{f(x)}=\ln(3)\implies f'(x)=f(x)\ln(3)=3^x\ln(x). \quad\blacklozenge
$$
So in general, for any positive $a > 0$, we have
> $$
\frac{d}{dx}a^x = a^x\ln(a).
$$
## Limits and graphs of natural exponentiation.
Using the graph of $\ln(x)$ and the fact that $\exp(x)=e^x$ is its inverse, we can obtain the graph of $\exp(x)$ as an increasing, concave up graph with limits
> $$
\begin{align*}
\lim_{x\to +\infty} e^x & =+\infty \\
\lim_{x\to -\infty} e^x & = 0
\end{align*}
$$
Here are the graphs of $e^x$ and $\ln(x)$:
![[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 14.40.01.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 14.40.01.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 14.40.01.excalidraw.dark.svg|dark exported image]]%%
**Example.**
Determine the following limit $$
\lim_{x\to\infty} \frac{e^{3x}}{4e^{3x} +e^{2x}}
$$ $\blacktriangleright$ Note by dividing top and bottom by $e^{3x}$, we get $$
\lim_{x\to\infty} \frac{1}{4 + e^{-x}}= \frac{1}{4+0}=\frac{1}{4}. \quad \blacklozenge
$$
## Integral of $e^x$.
Since $\frac{d}{dx} e^x = e^x$, we then have the indefinite integral $$
\int e^x dx = e^x + C
$$
**Example.**
Compute $$
\int e^{-5x} dx
$$
$\blacktriangleright$ Note if $u= -5x$, then $du = -5dx$, so we get $$
\int e^{u} \frac{du}{-5}= \frac{-1}{5} e^u + C = \frac{-1}{5} e^{-5x} + C. \quad\blacklozenge
$$
**Example.**
Compute the indefinite integral: $$
\int e^x \cos(e^x - 23)dx
$$
$\blacktriangleright$ We can apply $u$-substitution. Let $u=e^x -23$, then $du=e^x dx$. Hence the integral becomes $$
\int \cos(u) du = \sin(u) + C = \sin(e^x -23) + C. \quad\blacklozenge
$$
**Example.**
Compute the indefinite integral: $$
\int x^2 e^{x^3-2}dx
$$
$\blacktriangleright$ Let us try $u$-substitution again, where $u=x^3-2$, then $du=3x^2 dx$, so we get $$
\int e^u \frac{du}{3}=\frac{1}{3}e^u+C=\frac{1}{3}e^{x^3-2}+C. \quad\blacklozenge
$$
## Hyperbolic functions. (6.7)
We are in a position to build two new wonderful functions out of what we have so far: The hyperbolic cosine function $\cosh(x)$ and the hyperbolic sine function $\sinh(x)$. They are defined as follows:
> $$
\begin{align*}
\cosh(x) = \frac{e^x + e^{-x}}{2} \\
\sinh(x) = \frac{e^x - e^{-x}}{2}
\end{align*}
$$
Their graphs are as follows:
![[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 16.04.04.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 16.04.04.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 16.04.04.excalidraw.dark.svg|dark exported image]]%%
Here, $\cosh(x)$ is an even function, while $\sinh(x)$ is an odd function.
## Derivatives of hyperbolic cosine and hyperbolic sine.
By directly calculating them, you will find that
> $$
\begin{align*}
\frac{d}{dx}\cosh(x) & =\sinh(x) \\
\frac{d}{dx}\sinh(x) & =\cosh(x)
\end{align*}
$$
They are kind of like the trigonometric functions, except without the pesky negative sign business!
## Hyperbolic relations.
The two hyperbolic functions are also algebraically related by the following relation:
> $$
\cosh^2(x) - \sinh^2(x)=1
$$
You can derive this by directly expanding it from their definitions.
Notice the difference with the Pythagorean relation $\cos^2(x)+\sin^2(x)=1$.
Now why are they called hyperbolic functions? To explain this I'll tell you that the trigonometric functions are sometimes called **circular functions**, for the following reason: If we set $X=\cos(t)$ and $Y=\sin(t)$, then by Pythagorean relation we have $X^2 + Y^2 = 1$, which is a unit circle.
If we do the same for the hyperbolic functions, let $X = \cosh(t)$ and $Y=\sinh(t)$, then by the hyperbolic relation we have $X^2 - Y^2 =1$, which is actually the graph of a unit hyperbola!
But what does this $t$ mean in $\cosh(t)$ or $\sinh(t)$?
In the case of the circular functions, $\cos(t)$ and $\sin(t)$ are the $X$ and $Y$ coordinates of the unit circle $X^2+Y^2=1$ when the **circular sector** has area $\frac{t}{2}$.
In the case of the hyperbolic functions, $\cosh(t)$ and $\sinh(t)$, they are the $X$ and $Y$ coordinates of the unit hyperbola $X^2-Y^2=1$ when the **hyperbolic sector** has area $\frac{t}{2}$:
![[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 22.31.08.excalidraw.svg]]
%%[[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 22.31.08.excalidraw.md|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 22.31.08.excalidraw.dark.svg|dark exported image]]%%
These hyperbolic functions (and their inverses) will be helpful later for us to compute various integrals. We will revisit this.
## L'Hospital rule. (6.8)
As we saw earlier, L'Hospital's rule was helpful in finding limits that has an indeterminant form $\frac{0}{0}$, it will also work for $\frac{\infty}{\infty}$ as well.
Let me formally state what it says.
> **L'Hospital rule.**
> Suppose $f,g$ both differentiable, and $g'(x)\neq0$ near the point $x=a$, and the limit $$
\lim_{x\to a} \frac{f(x)}{g(x)}
$$ is of the **standard** indeterminant forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Then $$
\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}
$$ **provided that the right hand side, the limit of the ratio of derivatives, exist.**
> Here, $a$ can be any limit point, one-sided, or infinity.
**Example.**
Compute the limit $$
\lim_{x\to 1} \frac{\ln(x)}{x-1}=?
$$
$\blacktriangleright$ Note this is an indeterminant form of type $\frac{0}{0}$, let us apply L'Hospital rule to see if the limit of the ratio of the derivatives exist:
$$
\lim_{x\to 1} \frac{\ln(x)}{x-1} \stackrel{L'H}{=}\lim_{x\to 1} \frac{\frac{1}{x}}{1}=1.
$$
Since the limit of the ratio of derivatives exist, and is $1$, we conclude the original limit is $1$ as well. $\blacklozenge$
**Example.**
Compute the limit $$
\lim_{x\to\infty} \frac{e^{3x}+ 3x}{2e^{3x}+x^2+1}=?
$$
$\blacktriangleright$ Note this is an indeterminant form of type $\frac{\infty}{\infty}$. So we attempt L'Hospital rule: $$
\begin{align*}
\lim_{x\to\infty} \frac{e^{3x}+ 3x}{2e^{3x}+x+1} & \stackrel{L'H}{=}\lim_{x\to\infty} \frac{3e^{3x}+ 3}{6e^{3x}+1} ...\left( \frac{\infty}{\infty} \right) \\
& \stackrel{L'H}{=} \lim_{x\to\infty} \frac{9e^{3x}}{18e^{3x}}=\frac{1}{2}. \quad\blacklozenge
\end{align*}
$$
Why does L'Hospital rule work? First, we only really need to consider the case of $\frac{0}{0}$ as the type $\frac{\infty}{\infty}= \frac{1 / 0}{1 / 0} = \frac{0}{0}$. Here I give you the geometric intuition:
Suppose $f(x)\to0$ and $g(x)\to0$ as $x\to a$, and that $f,g$ both differentiable, with $g'(x)\neq0$ near $x$. Since $f,g$ differentiable, this means locally it is a line (as approximated by its tangent line). So we can zoom in at $x=a$, which both $f,g$ approaches 0. So locally near $x=a$ we have the following picture:
![[1 teaching/smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 17.16.56.excalidraw.svg]]
%%[[1 teaching/smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 17.16.56.excalidraw|🖋 Edit in Excalidraw]], and the [[smc-summer-2023-math-8/notes/week-1/---files/week-1-thursday-notes 2023-06-22 17.16.56.excalidraw.dark.svg|dark exported image]]%%
So the ratio of the heights of the function $$
\begin{align*}
\frac{f(x)}{g(x)} & = \frac{f(a+h)}{g(a+h)} \\
& =\frac{f(a+h)-0}{g(a+h)-0} \\
& =\frac{\frac{f(a+h)-f(a)}{h}}{\frac{g(a+h)-g(a)}{h}}\to \frac{f'(a)}{g'(a)} \text{ as } h\to0\\
\end{align*}
$$
Geometrically, **if you have two right triangles with the same base, the ratio of the heights is the ratio of the slopes of the hypotenuse** !
There are other types indeterminant forms, such as $0\cdot\infty$, $\infty-\infty$, $0^0$, $\infty^0$, and $1^\infty$. To deal with these exotic indeterminant forms, convert them to the standard types $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and then you can apply L'Hospital rule. (See the text 6.8 for now.)
**Example.** Here is a limit of $\infty - \infty$ type, albeit a contrived example: $$
\begin{align*}
&\lim_{x\to\infty} x-\sqrt{x^2+1} \quad....(\infty - \infty) \\
&= \lim_{x\to\infty}x\left( 1-\frac{\sqrt{x^2+1}}{x} \right) \\
&= \lim_{x\to\infty}x\left( 1-\sqrt{1+\frac{1}{x^2}} \right)...(\infty\cdot 0) \\
&= \lim_{x\to\infty} \frac{\left( 1-\sqrt{1+\frac{1}{x^2}} \right)}{\left( \frac{1}{x} \right)} ...\left( \frac{0}{0} \right) \\ \\
&\stackrel{L'H}{=}\lim_{x\to\infty} \frac{ \frac{1}{\sqrt{1+ \frac{1}{x^2}} \cdot x^3} }{-\frac{1}{x^2}} \\
&=\lim_{x\to\infty} \frac{-1}{x\sqrt{1+\frac{1}{x^2}}} = 0. \quad\blacklozenge
\end{align*}
$$(By the way, this limit can be done more "easily" using the conjugates. So, L'Hospital rule need not always be the "shortest" way of evaluating limits! More practically, L'Hospital is useful when taking the derivatives actually simplifies the expression. Also, one has to watch out when can we even apply L'Hospital rule!)
///