OpenLLM-France/wikimedia huggingface dataset:
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg5
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 5: Thur, 03 Sept 09

### Page 5-1

Note: Eq.(2)p.4-2 and Eq.(4)p.4-2
$y'(x)=p(x)\$
Integrate from a to x:
$\int _{s=a}^{s=x}y'(s)\,ds=\int _{s=a}^{s=x}p(s)\,ds$
$\left[y(s)\right]_{s=a}^{s=x}=y(x)-y(a)$, where $y(a)=constant\$,
$y(x)=\int _{s=a}^{s=x}p(s)\,ds+y(a)=\int _^{x}p(s)\,ds=\int _^p(x)\,dx+k$
another way: $y(x)=\int _^p(x)\,dx+k$
Where $\int _^p(x)\,dx=F(x)$ and $k=constant\$
$\Rightarrow \ y(a)=F(a)+k\$
$\Rightarrow \ k=y(a)-F(a)\$
$\Rightarrow \ y(x)=F(x)-F(a)+y(a)\$

### Page 5-2

But $F(x)-F(a)=\int _{s=a}^{s=x}p(s)\,ds\$
| $\displaystyle {\begin{aligned}\Rightarrow \ y(x)=\int _{s=a}^{s=x}p(s)\,ds+y(a)=\int _^{x}p(s)\,ds=\int _^p(x)\,dx+k\end{aligned}}$ | (1) |
Eq.(3)p.4-2 : Why this form of nonlinear 1st order ODE?
Most general form:
| $\displaystyle {\begin{aligned}F(x,y,y')=0\end{aligned}}$ | (2) |
Application:
| $\displaystyle {\begin{aligned}x^{2}y^{5}+6(y')^{2}=0\end{aligned}}$ | (3) |
Where $x^{2}y^{5}+6(y')^{2}\$ is defined as $F(x,y,y')\$
HW: Show that $F(x,y,y')=0\$ in Eq(3) is a nonlinear 1st order ODE.
Hint: Define the differential operator $D(.)\$ associated with Eq(3).

### Page 5-3

Form for exact nonlinear 1st order ODE:
$F(x,y,y')=0\$ is exact if $\exists \$ a function $\phi \ (x,y)$ such that
| $\displaystyle {\begin{aligned}F={\frac {d\phi \ }{dx}}={\frac {d}{dx}}\phi \ (x,y(x))={\frac {\partial \phi \ }{\partial x}}(x,y)+{\frac {\partial \phi \ }{\partial y}}(x,y){\frac {dy}{dx}}=0\end{aligned}}$ | (1) |
where:
$\exists \$ is defined as "there exists"
$\phi \ _{x}=M(x,y)$
$\phi \ _{y}=N(x,y)$
Multiply Eq1) thru by $dx\$ to get:
Eq.(3)p.4-2 : $M(x,y)dx+N(x,y)dy=0\$
NOTE: If $F(X,y,y')=0\$ does not have the form:
| $\displaystyle {\begin{aligned}M(x,y)dx+N(x,y)dy=0\end{aligned}}$ | (2) |
Then $F(.)\$ cannot be exact.
Application: Eq.(3)p.5-2 is not exact because of the nonlinear term $(y')^{2}\$

### Page 5-4

Exactness test (continued)
p5-3 Eq(1):
$M(x,y)=\phi \ _{x}(x,y)\$
$N(x,y)=\phi \ _{y}(x,y)\$
Since $\phi \ _{xy}=\phi \ _{yx}\$
and $\phi \ _{xy}={\frac {\partial ^{2}\phi \ }{\partial x\partial y}}\$
and $\phi \ _{yx}={\frac {\partial ^{2}\phi \ }{\partial y\partial x}}\$
and ${\frac {\partial ^{2}\phi \ }{{\partial x}{\partial y}}}=(\phi \ _{x})_{y}\$
and ${\frac {\partial ^{2}\phi \ }{{\partial y}{\partial x}}}=(\phi \ _{y})_{x}\$
| $\displaystyle {\begin{aligned}M_{y}=N_{x}\end{aligned}}$ | (1) |
Application: Eq.(1)p.4-3 Not Exact
$M(x,y)=2x^{2}+{\sqrt {y}}\Rightarrow \ M_{y}={\frac {1}{2{\sqrt {y}}}}\$
$N(x,y)=x^{5}y^{3}\Rightarrow \ N_{x}=5x^{4}y^{3}\$
$M_{y}\neq \ N_{x}\$
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg7

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 7: Tues, 08Sept 09

### Page 7-1

Mediawiki Hints:
Alt+Shift+E = Edit
Alt+Shift+P = Preview
Alt+Shift+S = Save
p.6-3:
Case2: $h_{x}N=0\ \Rightarrow \ -h_{y}M+h(N_{x}-M-y)=0\Rightarrow \ {\frac {h_{y}}{h}}={\frac {1}{M}}\left(N_{x}-M_{y}\right)=-g(y)\$
HW: Complete details to obtain $h_{y}\$
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg35
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 35: Thurs, 12Nov09

### Page 35-1

Gauss Legendre quadrature (numerical integration}
Quadrature; QUAD-->quadrilateral-->Greek: measuring areas
Area $=\sum$ Quadrilaterals
Cubature; CUBE; Volume $=\sum$ cubes
$I(f):=\int _{-1}^{1}f(x)\,dx\$
$I_{n}(f):=\sum _{j=1}^{n}w_{j}f(x_{j})\,dx\$ with $\left\{x_{j}\right\}\$ the roots for $P_{n}(x)=0\$ , where n is the degree of $P_{n}(x)\$ and $w_{j}\$ being the weight
$-1<x_{1}<x_{2}<...<x_{n}<1\$

### Page 35-2

| $\displaystyle {\begin{aligned}I(f)=I_{n}(f)+E_{n}(f)\end{aligned}}$ | (1) |
| $\displaystyle {\begin{aligned}w_{j}={\frac {-2}{(n+1)P_{n}'(x_{j})P_{n+1}(x_{j})}}\end{aligned}}$ | (2) |
where j=1,2,...,n
| $\displaystyle {\begin{aligned}E_{n}(f)={\frac {2^{2n+1}(n!)^{4}}{(2n+1)\left[(2n)!\right]^{2}}}{\frac {f^{(2n)}(\eta \ )}{(2n)!}}\end{aligned}}$ | (3) |
for $\eta \ \in \left[-1,+1\right]\$
Ex: $n=2\$ (2 point interpolation)
Eq.(3) P.31-3 $P_{2}(x)={\frac {1}{2}}(3x^{2}-1)\$
$\Rightarrow \ x_{1,2}=\pm \ {\frac {1}{\sqrt {3}}}\$
Eq.(4) P.31-3 $P_{2}'(x)=3x,P_{3}(x)={\frac {1}{2}}(5x^{3}-3x)\$
$W_{1}={\frac {2}{(2+1)(3)({\frac {-1}{\sqrt {3}}}){\frac {1}{2}}\left[5({\frac {-1}{\sqrt {3}}})^{3}-3({\frac {-1}{\sqrt {3}}})\right]}}=1\$

### Page 35-3

$W_{2}=1\$
HW: verify table for Gauss Legendre quadrature in wikipedia, analytical expression of $\left\{x_{j}\right\}\$ and $\left\{w_{j}\right\},j=1,...,n\$ and $n=1,...,5\$ (n=integration points) after verifying the expression for $P_{n}(x)\$ with $n=1,...,6\$; (see HW p31-3 )
Evaluate numerically $\left\{x_{j}\right\}\$ and $\left\{w_{j}\right\}\$ and compute results with Abram & Stegum (see lecture plan)
Question: How does Gauss Legendre quadrature compare to other quadrature methods, e.g. trapezoidal rule?
Answer: Look at $E_{n}(f)\$, Eq.(3) P.35-2. Consider $f\in \mathrm {P} \ _{2n-1}\$...
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg39
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 39: Thurs, 19Nov09

### Page 39-1

Note: Other Application: Quantum Mechanics
discrete variables <-- coding theory
Ref: Nikiforov, et.al (1991)
Probability (Queing theory, birth and death)
Generating functions:
-Legendre Polynomial $P_{n}\$: Eq.(5) P.38-3
- ${r \choose k}\ =$ "r choose k" :
$(1+x)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{k}\$ for $\left|x\right\vert \leq 1\$
$A(\mu \ ,\rho \ ):=1-2\mu \ \rho \ +\rho \ ^{2})\$
From Eq.(6) P.38-3 and Eq.(7) P.38-3 : ${\frac {1}{\sqrt {A(\mu \ ,\rho \ )}}}=\alpha \ _{0}+\alpha \ _{1}(2\mu \ \rho \ -\rho \ ^{2})+\alpha \ _{2}(2\mu \ \rho \ -\rho \ ^{2})^{2}+...=\alpha \ _{0}+(2\mu \ \alpha \ _{1})\rho \ +(-\alpha \ _{1}+4\mu \ ^{2}\alpha \ _{2})\rho \ ^{2}+...\$
Where $(2\mu \ \rho \ -\rho \ ^{2})=4\mu \ ^{2}\rho \ ^{2}-4\mu \ \rho \ ^{3}+\rho \ ^{4}\$
and $\alpha \ _{0}=P_{0}(\mu \ )\$
and $2\mu \ \alpha \ _{1}=P_{1}(\mu \ )\$
and $(-\alpha \ _{1}+4\mu \ ^{2}\alpha \ _{2})=P+2(\mu \ )\$

### Page 39-2

| $\displaystyle {\begin{aligned}P_{0}(\mu \ )=\alpha \ _{0}=1\end{aligned}}$ | (1) |
| $\displaystyle {\begin{aligned}P_{1}(\mu \ )=2\mu \ \alpha \ _{1}=\mu \ \end{aligned}}$ | (2) |
| $\displaystyle {\begin{aligned}P_{2}(\mu \ )={\frac {1}{2}}(3\mu \ ^{2}-1)\end{aligned}}$ | (3) |
HW: Continue the power series development to find $P_{3},P_{4},P_{5}\$ and complete result to that obatined by Eq.(6) P.31-3 or Eq.(7) P.31-3
2 recurrence formulas
Plan: Find ${\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}\$ and ${\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}\$
1) ${\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}=-{\frac {1}{2}}A^{\frac {-3}{2}}{\frac {dA}{d\mu \ }}={\frac {\rho \ }{A^{\frac {-3}{2}}}}\$
Recall Eq.(5) P.38-3, now ${\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}={\frac {\rho \ }{A^{\frac {3}{2}}}}=\$
| $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }P_{n}'(\mu \ )\rho \ ^{n}\end{aligned}}$ | (4) |
Where $P_{0}(\mu \ )=1\Rightarrow \ P_{0}'=0\$ and $P_{n}'(\mu \ )={\frac {d}{d\mu \ }}P_{n}(\mu \ )\$

### Page 39-3

| 2) $\displaystyle {\begin{aligned}{\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}={\frac {-\rho \ +\mu \ }{A^{\frac {3}{2}}}}=\sum _{n=0}^{\infty }P_{n}(\mu \ )n\rho \ ^{n-1}\end{aligned}}$ | (5) |
Compare Eq(4) and Eq(5)
* $\displaystyle {\begin{aligned}\mu \ -\rho \ =\rho \ \end{aligned}}$
${\frac {\rho \ (\mu \ -\rho \ )}{A^{\frac {3}{2}}}}=(\mu \ -\rho \ )\sum _{n=1}^{\infty }P_{n}'(\mu \ )\rho \ ^{n}=\rho \ \sum _{n=0}^{\infty }P_{n}(\mu \ )n\rho \ ^{n-1}=\rho \ \sum _{n=1}^{\infty }P_{n}{\mu \ }n\rho \ ^{n-1}$
Where n=1 in summation due to factor u being introduced
$\Rightarrow \ \mu \ \sum _{n=1}P_{n}'\rho \ ''-\sum _{n=1}P_{n}'\rho \ ^{n-1}=\sum _{n=1}P_{n}\mu \ \rho \ ''\$
$\Rightarrow \ \mu \ P_{1}'\rho \ -P_{1}1\rho \ +\sum _{n=2}\left[-P_{n-1}'+\mu \ P_{n}'-nP_{n}\right]\rho \ ^{n}=0\$

### Page 39-4

$\Rightarrow \ -P_{n-1}'+\mu \ P_{n}'-nP_{n}=0\$
| $\displaystyle {\begin{aligned}\Rightarrow \ \mu \ P_{n}'-P_{n-1}'=nP_{n}\end{aligned}}$ | (1) |
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg9
## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 9: Thur, 10Sept09

### Page 9-1

| $\displaystyle {\begin{aligned}c(x):=\int _^{x}b(s)\,ds=:{\bar {b}}\ (s)\end{aligned}}$ | (1) |
Note: Symbol notations:
$\Delta \$ defined as equal by definition
Non-symmetric notation $:=$ means equal by definition as well, better notation than $\Delta \$
Goal: Derive mathematical structure of a class of exact N1_ODEs
Exact L1_ODE_VC:
| $\displaystyle {\begin{aligned}{\bar {b}}\ (x)y'+\left[a(x)y+k\right]=0\end{aligned}}$ | (2) |
Application: Just invent any $a(x),b(x)\$
Let $a(x)=x^{4}\$
$b(x)=x\Rightarrow \ {\bar {b}}\ (x)={\frac {1}{2}}x^{2}\$
$k=10\$

### Page 9-2

| $\displaystyle {\begin{aligned}{\frac {1}{2}}x^{2}y'+\left[x^{4}y+10\right]=0\end{aligned}}$ | (3) |
HW: Show that Eq(3) L1_ODE_VC is "exact". See Note P.10-1
Question: How about N1_ODEs? Eq.(2)P.6-4
N1 means Nonlinear, 1st Order
| $\displaystyle {\begin{aligned}M_{y}(x,y)=a(x)c(y)\end{aligned}}$ | (4) |
| $\displaystyle {\begin{aligned}N_{x}(x,y)=b(x)c(y)\Rightarrow \ N(x,y)=\left(\int _^{x}b(s)\,ds\right)c(y)\end{aligned}}$ | (5) |
Where $\int _^{x}b(s)={\bar {b}}\ (x)\$
Eq.(2) P.6-4:
${\frac {1}{N}}(N_{x}-M_{y})={\frac {1}{{\bar {b}}\ (x)c(y)}}\left[b(x)c(y)-a(x)c(y)\right]=-f(x)\$

### Page 9-3

Eq.(4) P.9-2 $M(x,y)=a(x)\int _^{y}c(s)\$
Where $\int _^{y}c(s)={\bar {c}}\ (y)\$
| $\displaystyle {\begin{aligned}M(x,y)=a(x){\bar {c}}\ (y)\end{aligned}}$ | (6) |
From Eq.(6) P.9-3 , Eq.(4) P.9-2 and Eq.(3) P.4-2 obtain:
| $\displaystyle {\begin{aligned}{\bar {b}}\ (x)c(y)y'+a(x){\bar {c}}\ (y)=0\end{aligned}}$ | (7) |
Application: Consider the following
$a(x)=5x^{3}+2\$
$b(x)=x^{2}\Rightarrow \ {\bar {b}}\ (x)={\frac {1}{3}}x^{3}\$
$c(y)=y^{4}\Rightarrow \ {\bar {c}}\ (y)={\frac {1}{5}}y^{5}\$
| $\displaystyle {\begin{aligned}\left({\frac {1}{3}}s^{3}\right)\left(y^{4}\right)y'+\left(5x^{3}+2\right)\left({\frac {1}{5}}y^{5}\right)=0\end{aligned}}$ | (8) |
HW: Show Eq(8) is "exact" N1_ODE. Note P.10-1

### Page 9-4

L2_ODE_VC with missing dependant variable
$P(x)y''+Q(x)y'+R(x)y=S(x)\$
Where $R(x)\rightarrow \ 0\$ due to missing dependant variable y.
Eq.(1)P.2-3 $p:=y'\rightarrow \ P(x)p'+Q(x)p=S(x)\$ is a L1_ODE_VC
Solution: Eq.(4)P.8-2
Exact N2_ODEs:
General N2_ODEs: $F(x,y,y',y'')=0\$
Application: $(x^{3}+2x^{5}y^{2})y''+(x^{\frac {3}{2}}+10){\sqrt {y}}y'+y^{100}=0\$
$F=0\$ is exact means $\exists \phi \ (x,y,y')\$ such that $F(x,y,y',y'')={\frac {d}{dx}}\phi \ (x,y,y')\$
# University of Florida/Egm6321/f09.team1.gzc/Mtg10

Mtg 10: Sat, 22 Jan 11
    page10-1 HW2.9:Use (1) p.8-1 to generate P$_5$(x), and matlabl command"roots"to comp. the roots of P$_5$(x)to check values in table on p. 7-5. Plot the roots on [-1,+1] using matlab "plot" command (plot dots "." with coordinator (x$_i$,y$_i$), i = 1,...,5 : use "markersinge" 15) x$_i$: roots of P$_5$(x) y$_{i:}$ Plot (x, y, '.' , 'markersige' , 15) Repeat the above for P$_{10}$(x) observe the location of the roots near end points -1 and +1 (prepare for Runge phenomenon) NOTE: Lagrange interp. cont'd p.9-4   page10-2 Simpson's rule (simple) $[a,b]\ x_{0}=a,\ x_{1}={\frac {a+b}{2}},\ x_{2}=b\ \color {red}(1)$ ${\color {blue}{\underline {Method1:}}}\ f_{2}(x)=P_{2}(x){\overset {\color {red}(2)}{=}}c_{2}x^{2}+c_{1}x+c_{0}$ c$_0$, c$_1$, c$_2$ unknowns p$_2$=(x$_i$) = f(x$_i$) i= 0, 1, 2 (3) 3 equations for 3 unknowns {c$_i$} Method 2:Use lagrangeinterp.(2) p.8-3 (1) p.9-2   ${\color {red}{(4)}}=\sum _{i=0}^{\color {blue}{n=2}}{\color {blue}{\underset {l_{i}(x)}{\underbrace {\color {black}l_{i,2}(x)} }}}f(x_{i})$ Equiv. of meth1 and meth 2: (3)   $p_{2}(x_{j})=\sum _{i=0}^{2}\color {blue}{\underset {\delta _{ij}}{\underbrace {\color {black}l_{i}(x_{j})} }}f(x_{i})=f{x_{j}}\ \color {red}(5)$ page10-3 $l_{0{\color {blue},2}}=l_{0{\color {blue},2}}=\prod _{j=0{\color {red}j\neq i}}^{n=2}{\frac {x-x_{j}}{x_{0}-x_{j}}}={\frac {(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}}\ {\color {blue}\in P_{2}}$ It can be verified that l$_0$(x$_0$)=1 , l$_0$(x$_1$) = l$_0$(x$_2$) = 0 l$_i$(x$_j$) = δ$_{ij}$i,j = 0. 1. 2 ${\color {blue}{\underset {\color {blue}l_{1}(x)}{\underbrace {\color {black}l_{1{\color {blue}(i=1),2(n=2)}}(x)} }}}=\prod _{j=0,{\color {red}{j\neq 1}}}^{2{\color {blue}(n=2)}}{\frac {x-x_{j}}{x_{i{\color {blue}(i=1)}}-x_{j}}}={\frac {(x-x_{0})(x-x_{2})}{{\color {blue}{\underset {>0}{\underbrace {\color {black}(x_{1}-x_{0})} }}}{\color {blue}{\underset {<0}{\underbrace {\color {black}(x_{1}-x_{2})} }}}}}\ {\color {blue}\in P_{2}}$ $l_{1}(x_{1})=1\ ,\ l_{1}(x_{0})=l_{1}(x_{2})=0$ page10-4 ${l}_{1}^{''}(x_{1})<0$ HW$^*$2.10: Use(2) & (3)p.10-2 to find expression for {c$_i$} in terms (x$_i$, f(x$_i$)) i=0,1,2. HW$^*$2.11: Use (4) p.10-2 to derive simple Simpson's rule HW$^*$2.11:f(x)-e$^x$1/x on [o,1] S10and on [-1,1]S11 Consider n=1(Trap), 2(Simp), 4, 8, 16   page10-5 Constrast f$_n$(x) as in (2) p.8-3. Plot f , f$_n$ , n= 1, 2, 4, 8, 16 Compare ${I}_{n}=\int _{a}^{b}f_{n}(x)dx$ n=1, 2, 4, 8 and compare to I (use WA with more digits) For n=5 plot l$_0$, l$_1$, l$_2$ How would l$_3$, l$_4$, l$_5$ look like? HW$^*$2.13:show ${\underset {Simple\ Trap.}{\underbrace {{\color {red}(1)}{\color {blue}p.7-3}} }}\Rightarrow {\underset {Compare\ Trap.}{\underbrace {{\color {red}(1)}{\color {blue}p.7-4}} }}$ ${\underset {Simple\ Trap.}{\underbrace {{\color {red}(2)}{\color {blue}p.7-4}} }}\Rightarrow {\underset {Compare\ Trap.}{\underbrace {{\color {red}(4)}{\color {blue}p.7-4}} }}$
# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg37

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 37: Thurs, 17Nov09

### Page 37-1

P.36-4 continued
$P_{n}(x)\$
$P_{1}(x)=x\$
$Q_{n}(x)\$
From P.18-1 :
| $\displaystyle {\begin{aligned}Q_{1}(x)={\frac {1}{2}}xlog\left({\frac {1+x}{1-x}}\right)-1=x\tanh ^{-1}x-1\end{aligned}}$ | (1) |
HW show $Q_{1}(x)={\frac {1}{2}}xlog\left({\frac {1+x}{1-x}}\right)-1=x\tanh ^{-1}x-1\$
ref K p33 for $Q_{2},Q_{3}...\$
| $\displaystyle {\begin{aligned}Q_{n}(x)=P_{n}(x)\tanh ^{-1}x-2\sum _{j=1,3,5}^{J}{\frac {2n-2j+1}{(2n-j+1)j}}P_{n-j}(x)\end{aligned}}$ | (2) |
$Q_{0}(x)=\tanh ^{-1}(x)\ \$ is odd
HW: Use Eq(2) to show when $Q_{n}\$ is even or odd, depending on "n"
HW: Plot $\left\{P_{0},P_{1},...,P_{4}\right\}\$ and $\left\{Q_{0},Q_{1},...,Q_{4}\right\}\$
Legendre function $L_{n}(x)=P_{n}(x)\$ or $Q_{n}(x)\$ solution of Legendre solution

### Page 37-2

| $\displaystyle {\begin{aligned}\left\langle L_{n},L_{m}\right\rangle =0\end{aligned}}$ | (1) |
for $n\neq m\$
for $L_{n}=P_{n},\left\langle L_{n},L_{n}\right\rangle =\left\langle P_{n},P_{n}\right\rangle ={\frac {2}{2n+1}}\$
for $\left\langle L_{n},L_{n}\right\rangle =\left\langle P_{n},Q_{n}\right\rangle \ =$ HW
$\left\langle L_{n},L_{m}\right\rangle =\left\langle P_{n},P_{m}\right\rangle \ =$ Eq.(3) P.33-1
$\left\langle L_{n},L_{m}\right\rangle =\left\langle P_{n},Q_{m}\right\rangle \ =0$ for $n\neq m\$
Proof: Legendre equation, Eq.(1) P.14-2 : 2) $\left[(1-x^{2})y'\right]'+n(n+1)y=0\$
Where
| $\displaystyle {\begin{aligned}\left[(1-x^{2})y'\right]'=(1-x^{2})y''-2xy'\end{aligned}}$ | (2) |
$\Rightarrow \ \left[(1-x^{2})L_{n}'\right]'+n(n+1)L_{n}=0\$
Multiply by $L_{m}\$ and integrate from -1 to +1:
$\int _{-1}^{1}L_{m}\left[(1-x^{2})L_{n}'\right]'\,dx+n(n+1)\int _{-1}^{1}L_{m}L_{n}\,dx=0\$
Where $L_{m}\left[(1-x^{2})L_{n}'\right]'=\alpha \ \$

### Page 37-3

Integrate $\alpha \ \$ by parts:
| $\displaystyle {\begin{aligned}-\int _{-1}^{1}(1-x^{2})L_{n}'L_{m}'\,dx+(n)(n+1)\left\langle L_{n},L_{m}\right\rangle =0\end{aligned}}$ | (1) |
Interchange n and m:
| $\displaystyle {\begin{aligned}-\int _{-1}^{1}(1-x^{2})L_{m}'L_{n}'\,dx+(m)(m+1)\left\langle L_{m},L_{n}\right\rangle =0\end{aligned}}$ | (2) |
Eq(1)-Eq(2):
$\left[n(n+1)-m(m+1)\right]\left\langle L_{n},L_{m}\right\rangle =0$
Where $\left[n(n+1)-m(m+1)\right]\neq 0$ since $n\neq m\$
$\Rightarrow \ \left\langle L_{n},L_{m}\right\rangle =0\$ when $n\neq m\$
cf.K.p41
# PlanetPhysics/Bernoulli Equation and its Physical Applications

The Bernoulli equation has the form
   ${\begin{matrix}{\frac {dy}{dx}}+f(x)y=g(x)y^{k}\end{matrix}}$
where $f$ and $g$ are continuous real functions and $k$ is a constant ($\neq 0$, \,$\neq 1$).\, Such an equation is got e.g. in examining the motion of a body when the resistance of medium depends on the velocity $v$ as $F=\lambda _{1}v+\lambda _{2}v^{k}.$ The real function $y$ can be solved from (1) explicitly.\, To do this, divide first both sides by $y^{k}$.\, It yields
   ${\begin{matrix}y^{-k}{\frac {dy}{dx}}+f(x)y^{-k+1}=g(x).\end{matrix}}$
The substitution
   ${\begin{matrix}z:=y^{-k+1}\end{matrix}}$
transforms (2) into ${\frac {dz}{dx}}+(-k+1)f(x)z=(-k+1)g(x)$ which is a linear differential equation of first order.\, When one has obtained its general solution and made in this the substitution (3), then one has solved the Bernoulli equation (1).
# PlanetPhysics/Bibliography Doctoral Theses Supervised by Grothendieck
### Doctoral theses supervised by Alexander Grothendieck

(original list in French)

#### An incomplete list of "theses directed by A. Grothendieck":

* Schemes of reductive groups, by M. Demazure
[1] Sch\'emas en groupes r\'eductifs; short version of the thesis by M. Demazure, published in Bull. SMF. M\'em {\mathbf 93} (1965), 369-413. (Demazure's thesis is published "in totto" in SGA 3.
* Derived categories in $l$-adic cohomology.
[2] Cat\'egories d\'eriv\'ees en cohomologie l'-adique, (1969), thesis of J.-P. Jouanolou.
* Ample sheaves over group schemes and homogeneous spaces, thesis of Michel
Raynaud, published in Springer Lecture Notes {\mathbf119} (1970). [3] Faisceaux amples sur les sch\'emas en groupes et espaces homog\'enes; thesis of Michel Raynaud.
* Lefschetz theorems in coherent cohomology nad etale cohomology;
[4] Th\`eor\'emes de Lefschetz en cohomologie coh\'erente et en cohomologie \'etale; thesis of Mich\'ele Raynaud, Bull. SMF M\'em. 41 (1976).
* Complexe cotangent et d\'eformations, Springer Lecture Notes {\mathbf 239} (1971); thesis of Luc
Illusie.
* (On) Tannakian Categories , Thesis by Saavedra Rivano, published in Springer
Lecture Notes , 265 (1972). [6] Cat\'egories Tannakiennes, Springer Lecture Notes 265 (1972); thesis of Saavedra Rivano.
* \htmladdnormallink{non-Abelian {http://planetphysics.us/encyclopedia/AbelianCategory3.html} Cohomology}, Thesis by J. Giraud.
[7] Cohomologie non ab\'elienne, Grundlehren Math. Wiss. 179 (1972), Springer; thesis of J. Giraud.
* Ring topos and relative schemes, Thesis by Monique Hakim.
[8] Topos annel\'es et sch\'emas relatifs, Ergebnisse Math. \htmladdnormallink{GR {http://planetphysics.us/encyclopedia/SR.html}.} {\mathbf 64} (1972), Springer; thesis of Monique Hakim.
* Cohomologie cristalline, Springer Lecture Notes 407 (1974); thesis of Pierre Berthelot.
* Derived Categories and Abelian Categories, Thesis by Jean-Louis Verdier.,
[10] Des cat\'egories d\'eriv\'ees des cat\'egories ab\'eliennes; thesis of Jean-Louis Verdier, later published in Ast\'erisque {\mathbf 239} (1996), SMF (G. Maltsiniotis, ed.)
* Grothendieck Categories , Thesis of Hong Suan Xsin.
[11] Gr-Categories; thesis of Hong Suan Xsin.
* Isotopic study of embeddings of a topological $1$-complex into a complex surface; thesis
of Yves Ladegaillerie, short version published in Topology {\mathbf 23} no. 3(1984), 303-311.
* Jacobiennes g\'en\'eralis\'ees globales relatives; thesis of Carlos Contou-Carr\'ere, short version
published in the Grothendieck Festschrift , vol. II (1990).
# PlanetPhysics/Bibliography for Axiomatics and Mathematical Physics Foundations in Categories
### A Bibliography for Axiomatic Theories and Categorical Foundations of Mathematical Physics and Mathematics

\subsubsection{a. Foundations of Mathematics, Logics and Formal Logics: Axiomatics, Categories, Topoi and Higher Dimensional Algebra}
# PlanetPhysics/Bibliography for Experimental Physics

This is a contributed topic containing a current bibliography for experimental physics of both recently published papers and archived preprints in Experimental Physics.

## Experimental Physics: Recent Literature

Entry in progress
# Visual Basic for Applications/Loops

This lesson introduces loops.

## Objectives and Skills

Objectives and skills for loops include:
* Performing operations repetitively by using a Visual Basic Do...Loop structure
* Performing operations repetitively by using Visual Basic For...Next and For Each...Next structures

## Readings

* Wikipedia: Control flow
* Microsoft: Do...Loop Statement
* Microsoft: For...Next Statement

## Activities

In these activities you will create macros that use input boxes to obtain user input, variables, conditions, and loops to perform calculations, and message boxes to display results. Your macros should include Option Explicit, Dim, InputBox, MsgBox, titles on the dialog boxes, an icon on the message box, and use appropriate data types for your variables.
* Average Calculation
** Create a macro that asks the user to enter grade scores. Start by asking the user how many scores they would like to enter. Then use a loop to request each score and add it to a total. Finally, calculate and display the average for the entered scores.
** Revise the script several times to compare the code required when the loop control structure is based on While, Do While, Do Until, and For Next statements, respectively.
* Multiplication Table
** Review MathsIsFun: 10x Printable Multiplication Table.
** Create a macro that uses nested for loops to generate a multiplication table. Rather than simply creating a 10 by 10 table, ask the user to enter the starting and ending values.
* Multiplication Table with Conditions
** Review MathsIsFun: 10x Printable Multiplication Table.
** Create a script that uses nested for loops to generate a multiplication table. Use a condition with (<variable> Mod 2) to test for odd or even values. Generate the multiplication table only for even values. For odd values, skip display and continue to the next iteration of the loop.
* Multiplication Table with User Input
** Review MathsIsFun: 10x Printable Multiplication Table.
** Create a script that uses nested for loops to generate a multiplication table. Rather than simply creating a 10 by 10 table, ask the user to enter the starting and ending values. Use an Exit Do / Exit For statement to terminate the loop if the count exceeds 10.

## See Also

* EasyExcelVBA: Repeat Statements
* EasyExcelVBA: Combining Condition and Repeat Statements