new course

This commit is contained in:
Zheyuan Wu
2025-01-13 10:46:58 -06:00
parent d471db49c4
commit 30ef4e9ca8
33 changed files with 184 additions and 12 deletions

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course_code=input('We will follow the naming pattern of {class}_L{lecture number}.md, enter the course code to start.\n')
start=input('enter the number of lecture that you are going to start.\n')
end=input('Enter the end of lecture (exclusive).\n')
start=int(start)
end=int(end)
while start<end:
# create a empty text file
fp = open(f'{course_code}_L{start}.md', 'w')
fp.write(f'# Lecture {start}')
fp.close()
start+=1
import os
from pathlib import Path
course_code=input('We will follow the naming pattern of {class}_L{lecture number}.md, enter the course code to start.\n')
start=input('enter the number of lecture that you are going to start.\n')
end=input('Enter the end of lecture (exclusive).\n')
start=int(start)
end=int(end)
cur_dir = os.path.dirname(os.path.abspath(__file__))
while start<end:
# create a empty text file
file_name = Path.joinpath(cur_dir, f'{course_code}_L{start}.md')
fp = open(file_name, 'w')
fp.write(f'# Lecture {start}')
fp.close()
start+=1
print("Complete")

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# Lecture 1
## Chapter 5: Differentiation
### The derivative of a real function
#### Definition 5.1
Let $f$ be a real-valued function on an interval $[a,b]$ ($f: [a,b] \to \mathbb{R}$).
We say that $f$ is _differentiable_ at a point $x\in [a,b]$ if the limit
$$
\lim_{t\to x} \frac{f(t)-f(x)}{t-x}
$$
exists.
Then we defined the derivative of $f$, $f'$, a function whose domain is the set of all $x\in [a,b]$ at which $f$ is differentiable, by
$$
f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}
$$
#### Theorem 5.2
Let $f:[a,b]\to \mathbb{R}$. If $f$ is differentiable at $x\in [a,b]$, then $f$ is continuous at $x$.
Proof:
We need to show that $\lim_{t\to x} f(t) = f(x)$.
Equivalently, we need to show that
$$
\lim_{t\to x} (f(t)-f(x)) = 0
$$
So for $t\ne x$, since $f$ is differentiable at $x$, we have
$$
\begin{aligned}
\lim_{t\to x} (f(t)-f(x)) &= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right)(t-x) \\
&= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right) \lim_{t\to x} (t-x) \\
&= f'(x) \cdot 0 \\
&= 0
\end{aligned}
$$
Therefore, differentiable is a stronger condition than continuous.
> There exists some function that is continuous but not differentiable.
>
> For example, $f(x) = |x|$ is continuous at $x=0$, but not differentiable at $x=0$.
>
> We can see that the left-hand limit and the right-hand limit are not the same.
>
> $$ \lim_{t\to 0^-} \frac{|t|-|0|}{t-0} = -1 \quad \text{and} \quad \lim_{t\to 0^+} \frac{|t|-|0|}{t-0} = 1 $$
>
> Therefore, the limit does not exist. for $f(x) = |x|$ at $x=0$.
#### Theorem 5.3
Suppose $f$ is differentiable at $x\in [a,b]$ and $g$ is differentiable at a point $x\in [a,b]$. Then $f+g$, $fg$ and $f/g$ are differentiable at $x$, and
(a) $(f+g)'(x) = f'(x) + g'(x)$
(b) $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
(c) $\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$, provided $g(x)\ne 0$
Proof:
Since the limit of product is the product of the limits, we can use the definition of the derivative to prove the theorem.
(a)
$$
\begin{aligned}
(f+g)'(x) &= \lim_{t\to x} \frac{(f+g)(t)-(f+g)(x)}{t-x} \\
&= \lim_{t\to x} \frac{f(t)-f(x)}{t-x} + \lim_{t\to x} \frac{g(t)-g(x)}{t-x} \\
&= f'(x) + g'(x)
\end{aligned}
$$
(b)
Since $f$ is differentiable at $x$, we have $\lim_{t\to x} f(t) = f(x)$.
$$
\begin{aligned}
(fg)'(x) &= \lim_{t\to x} \left(\frac{f(t)g(t)-f(x)g(x)}{t-x}\right) \\
&= \lim_{t\to x} \left(f(t)\frac{g(t)-g(x)}{t-x} + g(x)\frac{f(t)-f(x)}{t-x}\right) \\
&= f(t) \lim_{t\to x} \frac{g(t)-g(x)}{t-x} + g(x) \lim_{t\to x} \frac{f(t)-f(x)}{t-x} \\
&= f(x)g'(x) + g(x)f'(x)
\end{aligned}
$$
(c)
$$
\begin{aligned}
\left(\frac{f}{g}\right)'(x) &= \lim_{t\to x}\left(\frac{f(t)g(x)}{g(t)g(x)} - \frac{f(x)g(x)}{g(t)g(x)}\right) \\
&= \frac{1}{g(t)g(x)}\left(\lim_{t\to x} (f(t)g(x)-f(x)g(t))\right) \\
\end{aligned}
$$

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# Lecture 10

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# Lecture 11

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# Lecture 12

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# Lecture 13

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# Lecture 14

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# Lecture 15

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# Lecture 16

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# Lecture 17

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# Lecture 18

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# Lecture 19

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# Lecture 2

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# Lecture 20

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# Lecture 21

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# Lecture 22

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# Lecture 23

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# Lecture 24

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# Lecture 25

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# Lecture 26

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# Lecture 27

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# Lecture 28

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# Lecture 29

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# Lecture 3

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# Lecture 4

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# Lecture 5

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# Lecture 6

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# Lecture 7

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# Lecture 8

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# Lecture 9

6
pages/Math4121/_meta.js Normal file
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export default {
index: "Course Description",
"---":{
type: 'separator'
}
}

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pages/Math4121/index.md Normal file
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# Math 4121
Riemann integration; measurable functions; measures; the Lebesgue integral; integrable functions; $L^p$ spaces.
## Textbook
Principles of Mathematical Analysis by Walter Rudin
A radical Approach to Lebesgue's Theory of Integration by David
## Grade
| item | percentage |
| --- | --- |
| Homework | 40% |
| Midterm 1 | 15% |
| Midterm 2 | 15% |
| Final | 30% |
## Homework
Due every Monday.

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@@ -32,6 +32,9 @@ export default {
CSE347: {
type: 'page',
},
Math4121: {
type: 'page',
},
about: {
display: 'hidden'
},