{NotebookFileName[],DateString[]}

style[
t_
]:=
Style[t,Bold,FontFamily->"Arial",FontSize->14]

$Assumptions={{x,y}∈Reals}

Mobius lesson

DSolve[{f'[x]==(1/2)Cos[(x/2)],f[π]==2},f[x],x]

DSolve[{y''[t]==-10,y'[0]==20,y[0]==2000},y[t],t]

y[t]/.DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t][[1]]

Solve[0==y[t]/.DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t][[1]],t]

Solve[0==y[t]/.DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t][[1]],t]//N

Sum[(i^3)-2(i^2),{i,1,20}]

Telescoping sum, sums of powers

Sum[n,{n,1,5}]

Sum[(1/(i(i+1))),{i,n}]

Sum[Log[((i+2)/(i+3))],{i,n}]

Sum[Log[((i+2)/(i+3))],{i,n}]==Log[3]-Log[n+3]//FullSimplify

Sum[Log[((i+2)/(i+3))],{i,n}]==Log[3]-Log[n+3]//FullSimplify

Gamma[x+1]==x!//FunctionExpand

Assuming[
{x>0},
x!==(x((x-1)!))//FullSimplify
]

log(3)-log(4)+log(4)-log(5)+...+log(n+2)-log(n+3)=log(3)-log(n+3)

Table[
{Sum[Log[((i+2)/(i+3))],{i,n}],Log[3]-Log[n+3],Sum[Log[((i+2)/(i+3))],{i,n}]==Log[3]-Log[n+3]},
{n,20}
]//N

Arithmetic series

Sum[m i+b,{i,n}]//Simplify

m((n(n+1))/2)+n b==Sum[m i+b,{i,n}]//Simplify

Future value calculation

You can pay P dollars now for a security that will give you $1 every year for N years. You can invest that cash in an account that will yield annual interest r. If you had just invested your P, you would have ((P(1+r))^N) after N years. But if you buy the security, you will have (((1+r))^(N-1)) from the dollar you got at the end of your first year, (((1+r))^(N-2)) from the dollar you got at the end of the second year, etc. Breakeven is when

p(((1+r))^n)-Sum[(((1+r))^(n-i)),{i,n}]

p(((1+r))^n)-Sum[(((1+r))^(n-i)),{i,n}]//Simplify

Solve[0==((1-(((1+r))^n)+p r (((1+r))^n))/r),n]

Solve[0==((1-(((1+r))^n)+p r (((1+r))^n))/r),n]

Solve[0==((1-(((1+r))^n)+p r (((1+r))^n))/r),n]/.{r->0.05,p->10}

Geometric series and Exponential

Sum[(E^(λ i)),{i,0,n}]

D[Sum[(E^(λ i)),{i,0,n}],λ]

D[Sum[(E^(λ i)),{i,0,n}],λ]//Together

Block[{frac=D[Sum[(E^(λ i)),{i,0,n}],λ]//Together,steps=3,num,den},
num=Numerator[frac];
den=Denominator[frac];
Table[
Simplify[{D[num,{λ,i}],D[den,{λ,i}]}->( {D[num,{λ,i}],D[den,{λ,i}]}/.λ->0)],
{i,0,steps}
]
]//Column

Block[{frac=D[Sum[(E^(λ i)),{i,0,n}],{λ,2}]//Together,steps=3,num,den},
num=Numerator[frac];
den=Denominator[frac];
Table[
Simplify[{D[num,{λ,i}],D[den,{λ,i}]}->( {D[num,{λ,i}],D[den,{λ,i}]}/.λ->0)],
{i,0,steps}
]
]//Column

Limit[D[Sum[(E^(λ i)),{i,0,n}],λ]//Simplify,λ->0]

Limit[D[Sum[(E^(λ i)),{i,0,n}],{λ,2}]//Simplify,λ->0]

Limit[D[Sum[(E^(λ i)),{i,0,n}],{λ,3}]//Simplify,λ->0]

Table[
Limit[D[Sum[(E^(λ i)),{i,0,n}],{λ,p}]//Simplify,λ->0],
{p,0,10}
]//Simplify//Column//TraditionalForm

Limit[D[Sum[(E^(λ i)),{i,0,n}],{λ,m}]//Simplify,λ->0]

Sum[(E^(λ i)),{i,1,∞}]

D[Sum[(E^(λ i)),{i,1,∞}],λ]

D[Sum[(E^(λ i)),{i,1,∞}],λ]//Together

Sum[(i^p),{i,0,n}]

ArcTan

Series[ArcTan[x],{x,0,5}]

SeriesCoefficient[ArcTan[x],{x,0,n}]

Sum of sines

Sin[ω i]//TrigToExp

Sum[Sin[ω i]//TrigToExp,{i,0,n}]//Simplify

Sum[Sin[ω i],{i,0,n}]//Simplify

Sum[Cos[ω i],{i,0,n}]//Simplify

How can I see this using the sum of angles identities?

S=sin(0)+sin(ω)+sin(2ω)+sin(3ω)+...+sin(nω)
C=cos(0)+cos(ω)+cos(2ω)+cos(3ω)+...+cos(nω)

sin((i+1)ω)=sin(i ω+ω)
=sin(i ω)cos(ω)+cos(i ω)sin(ω)

So, I can write

sin(ω)+sin(2ω)+...+sin((n+1)ω)=sin(0)cos(ω)+sin(ω)cos(ω)+...+sin(n ω)cos(ω)+
cos(0)sin(ω)+cos(ω)sin(ω)+...+cos(n ω)sin(ω)
=S cos(ω)+C sin(ω)
=S-sin(0)+sin((n+1)ω)

cos(ω)+cos(2ω)+...+cos((n+1)ω)=cos(0)cos(ω)+cos(ω)cos(ω)+...+cos(nω)cos(ω)
-sin(0)sin(ω)-sin(ω)sin(ω)-...-sin(n ω)sin(ω)
=C cos(ω)-S sin(ω)
=C-cos(0)+cos((n+1)ω)

S+sin((n+1)ω)=S cos(ω)+C sin(ω)
C-1+cos((n+1)ω)=C cos(ω)-S sin(ω)

These are two equations in the two unknowns C and S and obviously can be solved for C and S.

eqs1={
S+Sin[(n+1)ω]==S Cos[ω]+C Sin[ω],
C-1+Cos[(n+1)ω]==C Cos[ω]-S Sin[ω]
}

Solve[eqs1,{C,S}]

Solve[eqs1,{C,S}]//Simplify

Number of the beast

Block[{b=0},
bnums1=Table[
b=10b+6;
b,
{666}
]
];
{Length[bnums1], Short[bnums1]}

Total[bnums1]

Sum[(10^i),{i,666}]

(((10^667)-10)/9)==Sum[(10^i),{i,666}]

Solve[((9B)/6)+666==(((10^667)-10)/9),B]

((2((10^667)-10))/27)-444==B/.Solve[((9B)/6)+666==(((10^667)-10)/9),B][[1]]

((2((10^667)-10))/27)-444==Total[bnums1]

Divisible[(10^666)-1,27]