diff --git a/content/Fourier Series (lec  28-29).md b/content/Fourier Series (lec  28-29).md
index 1660d8b..5556766 100644
--- a/content/Fourier Series (lec  28-29).md	
+++ b/content/Fourier Series (lec  28-29).md	
@@ -1,154 +1,177 @@
-
+#fourier
 Remember the heat flow equation? We obtained that it's solution could be expressed in the form: 
 $$\sum_{n=1}^\infty c_{n}\sin\left( \frac{n\pi x}{L} \right)\quad\text{for}\quad0\leq x\leq L$$
-But what is $c_{n}$? They are the coefficients of a fourier transform. We want to develop a way to compute them.
-Let's derive how to compute the coefficients of a fourier transform. (feel free to skip to the end)
-$f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ where L is length of the rod
+But what is $c_{n}$? They are the coefficients of a Fourier transform. We want to develop a way to compute them.
+Let's derive how to compute the coefficients of a Fourier transform. (feel free to skip to the end)
+$f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ where $L$ is length of the rod
 
-This is a Fourier series, it's a more general form of what we have above:
+This is a Fourier series: it's a more general form of what we have above.
 $f(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty\left( a_{n}\cos\left( \frac{n\pi x}{L} \right) + b_{n}\sin\left( \frac{n\pi x}{L}\right) \right)$
 $x \in [-L,L]$
-almost everywhere piecewise continuous (?)
-has a lot of benefits over taylor series. $f(x)$ doesn't have to be infinitely differentiable (analytic)
-f(x) can even have jump discontinuities
-lets assume the equation is true when $x \in [-L,L]$
+It converges to $f(x)$ almost everywhere (convergence will be discussed below)
+Has a lot of benefits over Taylor series. $f(x)$ doesn't have to be infinitely differentiable (analytic)
+$f(x)$ can even have jump discontinuities
+Let's assume the equation is true when $x \in [-L,L]$
+Integrate both sides, it will tell us the DC offset:
 $\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L (\text{put summation here}) \, dx$
 $\int _{-L}^L \cos\left( \frac{n\pi x}{L} \right) \, dx=\frac{L}{n\pi}\sin\left( \frac{n\pi x}{L} \right)|_{-L}^L=0$
 same for $\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\, dx=0$ (it equals 0)
 so
 $\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L0 \, dx$
-$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L0 \, dx$
-$a_{0}=\frac{1}{L}\int _{L}^{2?a_{0}L} f(x) \, dx$
-$\int _{-L}^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \cos\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx \right)+b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx$
+$\int _{-L} ^L f(x) \, dx=a_{0}L$
+$a_{0}=\frac{1}{L}\int _{-L}^{L} f(x) \, dx$
+Now let's multiply both sides by $\cos\left( \frac{m\pi x}{L} \right)$ and integrate both sides, this will tell us the $\cos$ components:
+$\int _{-L}^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \cos\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx +b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right)\right) \, dx$
 use trig identities (will be provided on exam):
 $\cos(\alpha)\cos(\beta)=\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))$
 $\sin(\alpha)\cos(\beta)=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta))$
 $\sin(\alpha)\sin(\beta)=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))$
-$\int _{-L}^L \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}\, dx=\frac{1}{2}(\int _{-L}^L \left( \cos \frac{(n-m)\pi x}{L} \right) \, dx+\cancelto{ 0 }{ \frac{\cos(n+m)\pi x}{L} }dx$
+$\int _{-L}^L \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}\, dx=\frac{1}{2}(\int _{-L}^L \left( \cos(\frac{(n-m)\pi x}{L} )+\cancelto{ 0 }{ \cos(\frac{(n+m)\pi x}{L} })\right)dx$
 $= \begin{cases}0 & n\ne m \\L & n=m\end{cases}$
-
-$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi}{L} \, dx=0;$
-$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi}{L} \, dx=\begin{cases}0 & n\ne m \\L & n=m\end{cases}$
-
-going back,
+$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L} \, dx=\int _{-L}^L \text{odd}\, dx=0$
+so:
+$\int _{-L} ^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=a_{m}L$
+Similarly can be done for when multiplying both sides by $\sin\left( \frac{m\pi x}{L} \right)$ and integrating both sides to find the $\sin$ coefficients:
+$\int _{-L}^L f(x)\sin\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \sin\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\cancelto{ \text{odd} }{ \int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\sin\left( \frac{m\pi x}{L} \right) } \, dx +b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\sin\left( \frac{m\pi x}{L} \right)\right) \, dx$
+$\int _{-L} ^L \sin\left( \frac{n\pi x}{L} \right)\sin\left( \frac{m\pi x}{L} \right) \, dx=\frac{1}{2}\int_{-L}^L \cos\left(\frac{(n-m)\pi x}{L} \right)-\cos\left( \frac{(n+m)\pi x}{L} \right)dx$
+$=\begin{cases}0, & n\ne m \\L, & n=m\end{cases}$
+so:
+$\int _{-L} ^L f(x)\sin\left( \frac{m\pi x}{L} \right)\, dx=b_{m}L$
+In conclusion:
 $$a_{m}=\frac{1}{L}\int _{-L}^L f(x)\cos \frac{m\pi x}{L} \, dx \quad\text{valid for all }m=0,1,2,\dots$$
 $$b_{m}=\frac{1}{L}\int _{-L}^L f(x)\sin \frac{m\pi x}{L} \, dx=b_{m} \quad m=1,2,\dots$$
-now we know how to compute the coefficients for Fourier series!
+Now we know how to compute the coefficients for Fourier series!
 
 properties:
 for functions $f$, $g$, If $\int _{-L}^Lf(x)g(x) \, dx=\begin{cases}0 & f\ne g \\L & f=g \end{cases}$
 then $f, g$ are orthogonal
-the forier expantion is called an ortho normal expansion, taylor is not ortho normal.
+the Fourier expansion is called an ortho normal expansion, Taylor is not orthonormal.
 #end of lec 28
 #start of lec 29
-last lecture we derived how to find the coefficients in a fourier series.
+Last lecture we derived how to find the coefficients in a Fourier series.
 $f(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty\left( a_{n}\cos\left( \frac{n\pi x}{L} \right) + b_{n}\sin\left( \frac{n\pi x}{L}\right) \right)$
 $x \in [-L,L]$
-### Theorem:
-If $f$ and $f'$ are piecewise continuous on $[-L,L]$, then the fourier series converges to:
+### 1st convergence theorem:
+If $f$ and $f'$ are piecewise continuous on $[-L,L]$, then the Fourier series converges to:
 $\frac{1}{2}(f(x^-)+f(x^+))$ for all $x \in (-L,L)$
-Basically meaning, the fourier series converges.
-At $x=\pm L$ the fourier series converges to $\frac{1}{2}(f(-L^+)+f(L^-))$
+and on $x=\pm L$ the Fourier series converges to $\frac{1}{2}(f(-L^+)+f(L^-))$
 ![draw](drawings/Drawing-2023-11-22-13.15.26.excalidraw.png)
-### Theorem:
-If f(x) is continuous on $(-\infty,\infty)$ and $2L$ periodic and if $f'$ is continuous, then the taylor series converges to $f(x)$ everywhere
+Recall the definition of piecewise continuous: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a <u>finite</u> number of points of <u>jump</u> discontinuity (horizontal asymptotes not allowed).
+### 2nd Convergence theorem (uniform convergence):
+If $f(x)$ is continuous on $(-\infty,\infty)$ and $2L$ periodic and if $f'$ is piecewise continuous on $[-L,L]$, then its Fourier series converges to $f(x)$ everywhere (i.e., the Fourier series converges uniformly).
 ![draw](drawings/Drawing-2023-11-22-13.14.05.excalidraw.png)
-#ex lets compute the fourier transform of:
-$f(x)=\begin{cases}1, & -\pi\leq x\leq 0 \\x, & 0<x\leq \pi\end{cases}$
+#ex #fourier 
+Let's compute the Fourier transform of:
+$$f(x)=\begin{cases}1, & -\pi\leq x\leq 0 \\x, & 0<x\leq \pi\end{cases}$$
 $L$ here is $\pi$ clearly.
-lets find the coefficients $a_{n}$ and $b_{n}$
-$a_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0 \cos\left( \frac{n\pi x}{\pi} \right)\, dx +\int _{0}^\pi x\cos(nx)\, dx\right)$
-using integration by parts for second term:
-$=\frac{1}{\pi}\left( \frac{1}{n}\cancelto{ 0 }{ \sin(nx) }|_{-\pi}^0 +\frac{1}{n}\left( x\cancelto{ 0 }{ \sin(nx) }|_{0}^\pi-\int _{0}^\pi \sin(nx) \, dx \right)\right)$
-$a_{n}=\frac{1}{n^2\pi}(\cos(nx)|_{0}^\pi)=\frac{1}{n^2\pi}((-1)^n-1) \quad n=0,1,2,\dots$
-now lets find $b_{n}$
+Let's find the coefficients $a_{n}$ and $b_{n}$
+Use the formula we derived earlier:
+$a_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0 1\cos\left( \frac{n\pi x}{\pi} \right)\, dx +\int _{0}^\pi x\cos(nx)\, dx\right)$
+Using integration by parts (for the second integral):
+$=\frac{1}{\pi}\left( \frac{1}{n}\cancelto{ 0 }{ \sin(nx) |_{-\pi}^0} + \frac{1}{n}x\cancelto{ 0 }{ \sin(nx) |_{0}^\pi}-\frac{1}{n}\int _{0}^\pi \sin(nx) \, dx \right)$
+$a_{n}=\frac{1}{n^2\pi}(\cos(nx)|_{0}^\pi)=\frac{1}{n^2\pi}(\underbrace{ \cos(n\pi) }_{ (-1)^n }-1)$
+$a_{n}=\frac{1}{n^2\pi}((-1)^n-1)\quad n=1,2,\dots$
+Now let's find $b_{n}$
 $b_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0\sin(nx) \, dx+\int _{0}^\pi x\sin(nx) \, dx \right)$
-$=\frac{1}{\pi}\left( \frac{-1}{n}\cos(nx)|_{-\pi}^0-\frac{1}{n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cos(nx)\, dx \right) \right)$
+$=\frac{1}{\pi}[  \frac{-1}{n}\underbrace{ \cos(nx)|_{-\pi}^0 }_{ 1-(-1)^n }-\frac{1}{n}(  \underbrace{ x\cos(nx)|_{0}^\pi }_{ \pi(-1)^n-0 }-\underbrace{ \int _{0}^\pi \cos(nx)\, dx  }_{ 0 } ) ]$
+$b_{n}=\frac{1}{n\pi}((-1)^n-1-\pi(-1)^n)$
 $b_{n}=\frac{1}{n\pi}((-1)^n(1-\pi)-1) \quad n=1,2,\dots$
-
-
-
-we find that 
-$a_{2n}=0$
-$a_{2k-1}=-\frac{2}{n^2\pi}$ for $k=1,2,\dots$
+we find that
+$$a_{2n}=0 \qquad n=1,2,3,\dots$$
+$$a_{2k-1}=-\frac{2}{n^2\pi} \qquad k=1,2,3\dots$$
 what about when $n=0$?
 $a_{0}=\frac{1}{\pi}\left( \int _{-\pi}^0 \, dx+\int _{0}^\pi x \, dx \right)$
-$=\frac{1}{\pi}\left( x|_{-\pi}^0+\frac{x^2}{2}|_{0}^\pi \right)$
-$=\frac{1}{\pi}\left( 0+\pi+\frac{\pi^2}{2} \right)=\frac{\pi}{2}+1$
+$a_{0}=\frac{1}{\pi}\left( x|_{-\pi}^0+\frac{x^2}{2}|_{0}^\pi \right)$
+$a_{0}=\frac{1}{\pi}\left( 0+\pi+\frac{\pi^2}{2} \right)$
+$$a_{0}=\frac{\pi}{2}+1$$
 
-#ex lets compute the fourier transform of $f(x)=x$ from $-\pi\leq x\leq \pi$
-we have to take a windowed form of $f$ to make this possible, $L=\pi$
-at the left and right edge of the interval, the fourier series is equal to 0. (from the previous theorem)
-find the coefficients:
+#ex Let's compute the Fourier transform of:
+$$f(x)=x \qquad -\pi\leq x\leq \pi$$
+We have to take a windowed form of $f$ to make this possible, $L=\pi$
+At the left and right edge of the interval, the Fourier series is equal to 0. (1st convergence theorem.)
+Find the coefficients:
 $a_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\cos(nx) \, dx=0$
-Why is it zero? because the integrand is an odd function. (odd times even is odd.) and because we are integrating from $-\pi$ to $\pi$ (symmetric interval)
+$$a_{n}=0$$
+Why is it zero? because the integrand is an odd function. (odd times even is odd.) and because we are integrating from $-\pi$ to $\pi$ (a symmetric interval)
 definition of odd: $f(x)=-f(-x)$
 definition of even: $f(x)=f(-x)$
 odd times even is odd.
 odd times odd is even.
 even times even is even.
 huge exam time saving technique.
-find $b_{n}$
-$b_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\sin(nx) \, dx=\frac{2}{\pi}\int _{0}^\pi x\sin(nx) \, dx$
-using integration by parts:
-$b_{n}=\frac{2}{n}(-1)^{n+1}$
-another take away: if $f$ is odd, the cos terms are 0
-if $f$ is even, the sin terms are 0.
+Find $b_{n}$:
+$b_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\sin(nx) \, dx=\frac{2}{\pi}\int _{0}^\pi x\sin(nx) \, dx$ <- that's even, don't be a silly goose and say it's $0$
+Using integration by parts:
+$b_{n}=\frac{2}{\pi}\left( x\left( -\frac{1}{n}\cos(nx)|_{0}^\pi \right)-\int_{0}^\pi -\frac{1}{n}\cos(nx) \, dx \right)$
+$b_{n}=\frac{2}{\pi}\left( -\frac{\pi}{n}(-1)^n+\frac{1}{n^2}\cancelto{ 0 }{ \sin(nx)|_{0}^\pi } \right)$
+$$b_{n}=\frac{2}{n}(-1)^{n+1}$$
+another take away: if $f$ is odd, the $\cos$ terms are $0$
+if $f$ is even, the $\sin$ terms are $0$.
 
-if f is only defined between 0 and L:
-you can create an odd extension: $\bar{f}=\begin{cases}f(x)  & 0\leq x\leq L \\-f(-x), & -L\leq x<0 &  & \end{cases}$
-this will contain only sin terms
-you also have a choise to extend it as an even function, symmetrically across the y axis.
-$\bar{f}=\begin{cases}f(x)  & 0\leq x\leq L \\f(-x), & -L\leq x<0 &  & \end{cases}$
-this will contain only cos terms.
-#end of lec 29
+if $f$ is only defined between $0$ and $L$:
+you can create an odd extension: $\bar{f}(x)=\begin{cases}f(x)  & 0\leq x\leq L \\-f(-x), & -L\leq x<0 &  & \end{cases}$
+this will contain only $\sin$ terms.
+You also have a choice to extend it as an even function, symmetrically across the $y$ axis.
+$\bar{f}(x)=\begin{cases}f(x)  & 0\leq x\leq L \\f(-x), & -L\leq x<0 &  & \end{cases}$
+This will contain only $\cos$ terms.
+#end of lec 29 
 #start of lec 30
-from last lecture: 
+From last lecture: 
 $f(x)$ is defined on $[0,L]$
-odd extention:
+odd extension:
 $\bar{f}(x)=\begin{cases}f(x), & 0\leq x\leq L \\-f(-x,) & -L\leq x<0\end{cases}$
-and the $a$ coeffecients (cos terms) are zero.
-not only that, but the b terms are:
-$\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right); \qquad b_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$
+and the $a$ coefficients ($\cos$ terms) are zero.
+not only that, but the $b$ coefficients are:
+$b_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$
 $$b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$$
-"How about that, this is called a foureir sin series."
+and $\bar{f}(x)$ is:
+$$\bar{f}(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$$
+"How about that, this is called a Fourier sine series."
 For even extension:
 $\bar{f}(x)=\begin{cases}f(x), & 0\leq x\leq L \\f(-x,) & -L\leq x<0\end{cases}$
-and the $b$ coeffecients (sin terms) are zero.
-not only that but the a terms are:
-$\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$
-$a_{n}=\frac{2}{L}\int _{0}^L f(x)\cos\left( \frac{n\pi x}{L} \right)\, dx$
+and the $b$ coefficients ($\sin$ terms) are zero.
+not only that but the $a$ coefficients are:
+$a_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \cos\left( \frac{n\pi x}{L} \right) \, dx$
+$$a_{n}=\frac{2}{L}\int _{0}^L f(x)\cos\left( \frac{n\pi x}{L} \right)\, dx$$
+and $\bar{f}$ is:
+$$\bar{f}(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$$
+Remember that $\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ was the expansion of the eigen value function from the heat equation?
+then $\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$ is also an expansion of some related eigen value function. It's interesting to note.
 
-remember that $\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ was the expansion of the eigen value function from the heat equation?
-then $\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$ is also an expansion of some related eigen value function.
-
-#ex Fourier sine series for $f(x)=x^2$ from $0\leq x\leq \pi$
-well that means we want the odd extension:
+#ex #fourier Fourier sine series for: 
+$$f(x)=x^2 \qquad 0\leq x\leq \pi$$
+Well that means we want the odd extension:
 ![draw](drawings/Drawing-2023-11-24-13.15.17.excalidraw.png)
-the $\cos()$ ($a_{n}$) terms are zero.
-the b terms are:
+the $a_{n}$ (cosine) terms are zero.
+the $b_{n}$ terms are:
 $b_{n}=\frac{2}{\pi}\int _{0}^\pi x^2\sin(nx) \, dx$
-$=-\frac{2}{\pi n}\left[ x^2\cos(nx)|_{0}^\pi-2\int _{0}^\pi x\cos(nx)\, dx \right]$
+$=-\frac{2}{n\pi}\left[ x^2\cos(nx)|_{0}^\pi-2\int _{0}^\pi x\cos(nx)\, dx \right]$
 $=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n}\left( x\cancelto{ 0 }{ \sin(nx) }|_{0}^\pi-\int _{0}^\pi \sin(nx)\, dx \right) \right]$
-$b_{n}=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n^2}\cos(nx)|_{0}^\pi \right]=\frac{2\pi}{n}(-1)^{n+1}+\frac{4}{n^3\pi}((-1)^n-1)$
-for $n=1,2,3,\dots$ note no $n=0$ so no divison by zero problems here.
+$b_{n}=-\frac{2}{n\pi}[ \pi^2(-1)^n-\frac{2}{n^2}\underbrace{ \cos(nx)|_{0}^\pi }_{ (-1)^n-1 }]=\frac{2\pi}{n}(-1)^{n+1}+\frac{4}{n^3\pi}((-1)^n-1)$ for $n=1,2,3,\dots$
+Note no $n=0$ so no division by zero problems here.
 
-#ex fourier cosine series of $f(x)=\sin(x)$ for $0\leq x\leq \pi$
+#ex #fourier Fourier cosine series of $f(x)=\sin(x)$ for $0\leq x\leq \pi$
 ![draw](drawings/Drawing-2023-11-24-13.23.08.excalidraw.png)
-$a_{n}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(nx)\, dx$ for $n=0,1,2,\dots$
-use trig identity: (by the way the identites will be provided in the exam.)
+$b_{n}$ (sine) terms are all zero obviously, as it's asking for a Fourier cosine series, i.e., we are doing an even extension.
+$a_{n}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(nx)\, dx$ for $n=0,1,2,\dots$ <-Odd, but not symmetrical bounds. We can't rule out that it's zero.
+>Don't be a silly goose and try changing the bounds by removing that 2 in the front. If you did, you'd also have to change $\sin(x)$ to $\bar{f}$ which is $abs(\sin(x))$ and then you're integrating $a_{n}=\frac{1}{\pi}\int _{-\pi}^\pi |\sin(x)|\cos(nx)\, dx$ which is even$\times$even.
 
-$=\frac{2}{\pi} \frac{1}{2}\int _{0}^\pi (\sin(1-n)x+\sin(n+1)x)\, dx$
-integrating gives you:
+Use trig identity: (by the way the identities will be provided in the final exam.)
+$=\frac{2}{\pi} \frac{1}{2}\int _{0}^\pi \left[\sin((1-n)x)+\sin((n+1)x)\right]\, dx$
+Integrating gives you:
+$\frac{1}{\pi}( \frac{-1}{1-n}\underbrace{ \cos((1-n)x)|_{0}^\pi }_{ (-1)^{n+1}-1 } +\frac{-1}{n+1}\underbrace{ \cos((n+1)x)|_{0}^\pi }_{ (-1)^{n+1}-1 })$
+$a_{n}=-\frac{1}{\pi} \frac{1}{n+1}(-1)^{n+1}+\frac{1}{\pi} \frac{1}{n+1}+\frac{1}{\pi} \frac{1}{-(1-n)}(-1)^{n+1}+\frac{1}{\pi} \frac{1}{1-n}$
 $a_{n}=-\frac{1}{\pi} \frac{1}{n+1} (-1)^{n+1}+\frac{1}{\pi} \frac{1}{n+1}+\frac{1}{\pi} \frac{1}{n-1}(-1)^{n-1}-\frac{1}{\pi} \frac{1}{n-1}$
-what about when n=0 or n=1?
-$a_{0}=\frac{4}{\pi}=\frac{2}{\pi}\int _{0}^\pi \sin(x) \, dx$
+Assuming that $n\ne0,1$. (note: $n=-1$ is a non-issue since negative coefficients are never considered when taking a Fourier transform.)
+So what is $a_{0}, a_{1}$?
+$a_{0}=\frac{2}{\pi}\int _{0}^\pi \sin(x) \, dx=\frac{4}{\pi}$
 $a_{1}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(x) \, dx=\frac{1}{\pi}\int _{0}^\pi \sin(2x)\, dx=0$
-"0 is a very very special number it took humanity many numbers of years to invent 0" referring to when dividing by 0.
-additionally we know that the terms cancel when:
+<i>"zero is a very very special number it took humanity many numbers of years to invent zero"</i> referring to when dividing by 0.
+Additionally we know that the terms cancel when:
 $a_{2k-1}=0$ for $k=1,2,\dots$
-$a_{2k}=\frac{2}{\pi} \frac{1}{2k+1}+\frac{2}{\pi} \frac{1}{2k-1}$ for $k=1,2,\dots$
+$a_{2k}=\frac{2}{\pi} \frac{1}{2k+1}-\frac{2}{\pi} \frac{1}{2k-1}$ for $k=1,2,\dots$
 then:
-$$\bar{f}(x)=\frac{2}{\pi}+\frac{2}{\pi}\sum_{k=1}^\infty\left( \frac{1}{2k+1}+\frac{1}{2k-1} \right)\cos(2kx)$$
-
+$$\bar{f}(x)=\frac{2}{\pi}+\frac{2}{\pi}\sum_{k=1}^\infty\left( \frac{1}{2k+1}-\frac{1}{2k-1} \right)\cos(2k\pi x)$$
+Even with 10 terms, we get a pretty good approximation:
+![fouriercosineofsin.png](drawings/fouriercosineofsin.png)
 We have prepared ourselves now, now we start solving PDE's. He's encouraging us to attend the lectures in these last two weeks. He's making it sound like PDE's are hard.
\ No newline at end of file
diff --git a/content/Partial differential equations (lec 30-33).md b/content/Partial differential equations (lec 30-34).md
similarity index 76%
rename from content/Partial differential equations (lec 30-33).md
rename to content/Partial differential equations (lec 30-34).md
index af66de4..ca4056e 100644
--- a/content/Partial differential equations (lec 30-33).md	
+++ b/content/Partial differential equations (lec 30-34).md	
@@ -231,9 +231,74 @@ you complaining that your exams are hard theyre not hard. I'm talking about 40 y
 now we consider a guitar string:
 ![draw](drawings/Drawing-2023-12-01-13.49.58.excalidraw.png)
 assuming the thickness of the string is much smaller than the length of the string, which is true.
-$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L, t>0$
-^ Reminds me of the wave equation from phys 130.
+$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L,\quad t>0$
+^ Reminds me of the wave equation from Phys 130.
+Since the string is tied down on the ends we have the following initial conditions:
 $u(t,0)=u(t,L)=0 \qquad t>0$
 $u(0,x)=f(x)$  $0\leq x\leq L$
 $\frac{ \partial u }{ \partial t }(0,x)=g(x)$ $0\leq x\leq L$
-#end of lec 33
\ No newline at end of file
+^IBVP of the system.
+#end of lec 33
+#start of lec 34
+The wave equation follows many phenomena in electrical engineering.
+separation of variables:
+$u(t,x)=X(x)T(t)$
+plug in to equation:
+$T''X=\alpha^2TX''$
+$\frac{T''}{\alpha^2T}=\frac{X''}{X}=-\lambda$
+^ #evp !
+consider the $X$ side:
+$X''+\lambda X=0, \quad X(0)=X(L)=0$
+We've solved this before.
+the only non-trivial solutions for that Eigen value problem is:
+$\lambda_{n}=(\frac{n\pi}{L})^2$
+$X_{n}(x)=\sin\left( \frac{n\pi x}{L} \right)$ for $n=1,2,3,\dots$
+$\frac{T''}{\alpha^2T}=-\left( \frac{n\pi}{L} \right)^2$
+$T_{n}''+\left( \frac{\alpha n\pi}{L} \right)^2T_{n}=0$
+characteristic equation:
+$r^2+\left( \frac{\alpha n\pi}{L} \right)^2=0$
+$r_{1,2}=\pm i \frac{\alpha n\pi}{L}$
+"Don't memorize the steps. If you try to memorize you will mess up the final for sure. Ask yourself, why am I doing this here?"
+$T_{n}(t)=b_{n}\cos\left( \frac{\alpha n\pi}{L}t \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right)$
+$u_{n}(t,x)=\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$
+if you sum all these <u>modes</u>, you get the solution:
+$$u_{n}(t,x)=\sum_{n=1}^\infty\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$$
+
+"mathematics and reality do align very well, if the speed of the string is different the solution differs aswell."
+$u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$
+^ that's starting to look familiar.
+$\implies b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right) \, dx$
+where does it converge? well $f(x)$ and $f'(x)$ are both continuous, so it converges everywhere.
+$\frac{ \partial u }{ \partial t }(0,x)=g(x)=\sum_{n=1}^\infty\underbrace{  a_{n} \frac{\alpha n\pi}{L} }_{  }\sin\left( \frac{\alpha n\pi}{L} \right)$
+$a_{n} \frac{\alpha n\pi}{L}$ are the Fourier $\sin$ coefficients of $g(x)$
+$a_{n}=\frac{2}{\alpha n\pi}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$
+$\alpha^2$ is the Hooke modulus of the string btw.
+remember the heat equation, the amplitude is exponentionally decreasing,
+here the amplitude is oscillatory and doesnt increae in time. boi-oi-oi-oing
+to make it more releasitic we have to add a term for resistance, and we end up with a b in the characteristic equation for T.
+btw this equation models the electromagnetic feild, to some approximation.
+the lowest mode is called the fundemental mode, the following terms after are called harmonics.
+If two instruments play the same note (same fundemental frequency), they still sound different! and that's because of the difference in harmonics.
+The modes are standing waves in the string.
+"my claim is that any object, including social objects ,can be described by waves. Everything is a wave."
+you can model elementary particle behaviours with solitons (non linear waves.)
+in life in the real world, all waves have finite speed.
+So thats why its important to learn the wave equation. its the prototype to waves.
+"waves are the fundamental object. [...]. So that's why it's important, these are the fundamental objects of nature here."
+$f(x)=\begin{cases}x, & 0\leq x\leq \frac{\pi}{2}\\ \ \pi-x, & \frac{\pi}{2}<x\leq \pi\end{cases}$
+$\alpha^2=1$
+$g(x)=\sin(x)$
+$b_{n}=\frac{2}{\pi}\left( \int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx +\int _{\frac{\pi}{2}} ^\pi (\pi-x)\sin(nx) \, dx\right)$
+$b_{n}=\frac{4}{n^2\pi}\sin\left( \frac{n\pi}{2} \right)$
+$b_{2k}=0$
+half of the b coefficients are 0.
+$b_{2k-1}=\frac{4}{(2k-1)^2\pi}(-1)^{k+1}$
+plug in g(x) to get a_n terms
+however theres a short cut here. from definition of g(x) earlier:
+$g(x)=a_{1}\sin(x)+2a_{2}\sin(2x)+3a_{3}\sin(3x)+\dots$
+but g(x) is sin x.
+so $a_{1}=1$ and every other term is 0
+plug this in to get solution:
+$$u(t,x)=\left( \frac{4}{\pi}\cos t+\sin t \right)\sin(x)+\sum_{k=1}^\infty \frac{4(-1)^{k+1}}{\pi(2k+1)^2}\cos((2k+1)t)\sin((2k+1)x)$$
+typo in his notes, not 2k-1 its 2k+1
+#end of lec 34
diff --git a/content/Separation of variables & Eigen value problems (lec 26-28).md b/content/Separation of variables & Eigen value problems (lec 26-28).md
index 279e61b..9e2fa01 100644
--- a/content/Separation of variables & Eigen value problems (lec 26-28).md	
+++ b/content/Separation of variables & Eigen value problems (lec 26-28).md	
@@ -21,7 +21,8 @@ $\Delta u=u(t+\Delta t,x)-u(t,x)$
 Amount of heat to change temperature over $\Delta t$ with $\Delta u$ is $\underbrace{ C(x) }_{ \text{specific heat capacity } }\underbrace{ \rho(x) }_{ \text{density} }\Delta u\underbrace{ a\Delta x }_{ \text{volume} }$
 
 $C(x)\rho(x)a\Delta x(u(t+\Delta t,x)-u(t,x))=a\Delta t(k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x))+Q(t,x)\Delta xa\Delta t$
->I'm gonna be honest, I'm lost in this derivation already. What is $Q(t,x)$?
+$Q(t,x)$ is a source term, a term to model internally produced heat.
+>I'm gonna be honest, I'm lost in this derivation already.
 
 divide by $a\Delta x\Delta t$
 $C(x)\rho(x)\frac{(u(t+\Delta t,x)-u(t,x))}{\Delta t}=\frac{ k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x)}{\nabla x}+Q(t,x)$
@@ -32,7 +33,7 @@ thermodynamics can be very important for electrical engineers, for instance the
 ## Separation of variables & Eigen value problems
 #ex #SoV #evp
 (This is more of a case study than an example.)
-Rewrite the equation we derived by grouping the constant terms into one constant $D$
+Rewrite the equation we derived by grouping the constant terms into one constant $D$ and assume for this problem that the internally generated heat is $0$ (i.e., $Q(x,t)=0$)
 $\frac{ \partial u }{ \partial t }=D \frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0$
 boundary conditions:
 $u(t,0)=u(t,L)=0 , \quad t>0$ (simple case)
@@ -116,7 +117,7 @@ lets go back to the problem and focus on $T$:
 $\frac{T'}{DT}=\frac{X''}{X}=-\lambda$
 $\frac{T'}{T}=-\left( \frac{n\pi}{L} \right)^2D$
 this is a separable equation.
-We can treat the function T as a variable:
+We can treat the function $T$ as a variable:
 $\frac{dT}{dt} \frac{1}{T}=-\left( \frac{n\pi}{L} \right)^2D$
 $\int{dT} \frac{1}{T}=\int-\left( \frac{n\pi}{L} \right)^2Ddt$
 $\ln(T)=-\left( \frac{n\pi}{L} \right)Dt+c_{n}$
diff --git a/content/_index.md b/content/_index.md
index e516061..080b50d 100644
--- a/content/_index.md
+++ b/content/_index.md
@@ -24,8 +24,8 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Systems of linear equations (lec 21-22)](systems-of-linear-equations-lec-21-22.html)
 [Power series (lec 22-25)](power-series-lec-22-25.html)
 [Separation of variables & Eigen value problems (lec 26-28)](separation-of-variables-eigen-value-problems-lec-26-28.html)
-[Fourier series (lec 28-29)](fourier-series-lec-28-29.html) (raw notes, not reviewed or revised yet.)
-[Partial differential equations (lec 30-33)](partial-differential-equations-lec-30-33.html) (raw notes, not reviewed or revised yet.)
+[Fourier series (lec 28-29)](fourier-series-lec-28-29.html)
+[Partial differential equations (lec 30-34)](partial-differential-equations-lec-30-34.html) (raw notes, not reviewed or revised yet.)
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
 [Big LT table (.png)](drawings/bigLTtable.png)
diff --git a/content/drawings/Drawing-2023-11-22-13.15.26.excalidraw.md b/content/drawings/Drawing-2023-11-22-13.15.26.excalidraw.md
index 5fc5fcc..28da041 100644
--- a/content/drawings/Drawing-2023-11-22-13.15.26.excalidraw.md
+++ b/content/drawings/Drawing-2023-11-22-13.15.26.excalidraw.md
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 ```
 %%
\ No newline at end of file
diff --git a/content/drawings/Drawing-2023-11-22-13.15.26.excalidraw.png b/content/drawings/Drawing-2023-11-22-13.15.26.excalidraw.png
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diff --git a/content/drawings/fouriercosineofsin.png b/content/drawings/fouriercosineofsin.png
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