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These are answers submitted by vv

ImportMatrix...

vv 13922
February 09 2018
0 2

Use

A:=ImportMatrix("yourpath\\-Validierung-Stahl-AI-.txt");

and you will have a 1214 x 1356  Matrix.

Numeric solution...

vv 13922
February 09 2018
1 2

It's a specially cooked system.

 

> 

restart;
> 

sys:={
21000 = ((10/1000)/60)*4181*983*(x-y),
H = -((10/1000)/60)*4181*ln(1-((x-y)/(x-40))),
H = 124+102*ln(x-40)+(7+2*ln(x-40)) };

{21000 = (4109923/6000)*x-(4109923/6000)*y, H = -(4181/6000)*ln(1-(x-y)/(x-40)), H = 131+104*ln(x-40)}

 (1)
> 

evalf[1000]( fsolve(sys, {x,y,H}, {x=50..100,y=30..100,H=400..500} )   );

{H = 486.9792717948501357867918018059386067855156547114174326869481418810967243566285398975787784552235204706669347559475195400290118640303716801287372634993661346148619910232495800569296330392394830627549743601915872902357242427148731055409553705001658423528217321565927918891190491572966479397271232198426513612569355038417041808873881442623841262343853418832298327693593790610240451083265130917877844785166083247626127110701898215838963336513184408835894194697247808253085787062731557015693465725077077852896999228843274976631083570662482036300138497124715790019203338925750433405104219573337388245857036034857797986214888125128213468426540193206945839888044251684815660805344557214322036920099275934107239459489909624991382336777779198068497280647366466732283885118035797277802575588920390167295368604982886417919721008912823848353938665076022764471765288194352037535498835268165004232995241166817494393305369865683989720565493276815812710671414695060596389471870171101177308498370799633961387014797870, x = 70.65750866865388962274962328977939489377294903091858411945917234945764190715981783600325358893585110961932863462405500054380580852731304211782069883061069513954397685796059926183531905585579097223962590053390294660021611110475792368859465250322207982971943756610525306678494949905387521858682023969792135750167009335860887007621303198544321368035359069175536312026362350448543477228656103669649753257153387272484737179648441014252898713320281206372610926492497110938583549273159942143012717266939216706865489265272371561718110362755576369288566379279748261854103496737662967801146998815897672787725783471077732933162054089997390996937068458728288379330756605842335927722582205633540658909087373621564984005454626854462329726540402400771083530825578130314424765827873541484037247233111699579267679369342370990364580403457147196305388562501611449040359242137916479015561482141032930179069021129023984141254333721654117596419374749676562545752857082962606395129368846731712926167576535994552285681007055, y = 40.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000958347308577476849155817266967208600479833803954882640800358625509636683984970528639967775046602065201922053597715478431016498682880079609787467682537323162308315760688601560985179585421913717270699604525982018646370272917837250840635796722371496634076989812851876433207535565380863444548970077618286474517739106107056439827996365038277140944772163420380808436265855437736016956601251559483196367236991702787708084161571078609077635832578435472524929603687671421461968192303179415316459681254027553729208024619289830444214278717989841247739063311010334060299714878719957897139070573801145583905093544680767085382367073228848646299934993482758918092665796726848329309766789651450780060432670622190}

 (2)
> 

evalf(%);

{H = 486.9792718, x = 70.65750867, y = 40.00000000}

 (3)
> 

 

approx...

vv 13922
February 04 2018
0 3

The exact answer should be Pr(infinity), where

Pr := proc(n::posint) local k; 
add((2/3)^k/3*`if`(is(sin(k)<=1/2),1,0),k=0..n)
end:

which of course cannot be computed by Maple.
Maple answers:
Pr(13); evalf(%);
     2839595/4782969
     0.5936887736
which is an approximation (not too good).

evalf[50](Pr(300));
    .59502775989667091140797025425657732088366035178640


 

 

Not periodic...

vv 13922
February 04 2018
0 0

You cannot because the function is not periodic (for any values of the parameters).

MultiplicativeOrder...

vv 13922
February 03 2018
0 0 
findOrderOf:= proc(g::posint, p::prime)
  local G,i;
  G:=1;
  for i from 1 to p-1 do
    G:=g*G mod p;
    if(G = 1) then return i end if;
  end do;  
end proc;

Maple has it builtin:  NumberTheory:-MultiplicativeOrder

You should read showstat(NumberTheory:-MultiplicativeOrder)
and try to understand the algorithm for an efficient implementation.

 

typo...

vv 13922
February 03 2018
1 2 
f := n -> sum(i^3, i=1..n);

But note that f(...) will fail if i is assigned, so a better definition is:

f := n -> sum('i'^3, 'i'=1..n);

 

solve...

vv 13922
February 02 2018
2 2

solve(y>0);

or

Y:=numer(y)*denom(y);
SolveTools:-SemiAlgebraic([Y>0]);

obtain the answer in a few seconds (Maple 2017).

But there are 56 regions and it depends on your final aim to use them.

Edit. You may try to change the order of the variables and hope for simpler regions.

simple...

vv 13922
February 01 2018
0 0

The Maple implementation NumberTheory:-Totient is of course faster (for large numbers), but the simplest one is probably:

eulerphi:=(n::posint) -> add(`if`(igcd(k,n)=1,1,0),k=1..n);

 

typo...

vv 13922
January 30 2018
0 0

Should be

solve(SysEqE,{diff(phi[1](t),t),diff(phi[3](t),t)});

op...

vv 13922
January 29 2018
0 0

op(0,a);

But you should try to see how this appeared, because it is a nonsense (probably from a programming error).

I don't think it has a special name;...

vv 13922
January 26 2018
0 2

I don't think it has a special name; it is a multiplication by a diagonal matrix.

But are you sure it is useful? AFAIK the eigenvector algorithms always do a row (or column) normalization.

evalc...

vv 13922
January 25 2018
2 1

Use
evalc(Re(expr));

 

Solution...

vv 13922
January 25 2018
3 3
> 

restart;
> 

eq1 := diff(u1(x), x, x)+diff(u2(x), x)+int(2*x*s*(u1(s)-3*u2(s)), s = 0 .. 1) = 6*x^2+3*x*(1/10)+8;
eq2 := diff(u1(x), x)+diff(u2(x), x, x)+int((3*(s^2+2*x))*(u1(s)-2*u2(s)), s = 0 .. 1) = 21*x+4/5;
bcs := u1(0)+(D(u1))(0) = 1, u2(0)+(D(u2))(0) = 1, u1(1)+(D(u1))(1) = 10, u2(1)+(D(u2))(1) = 7;

EQ1 := diff(u1(x), x, x)+diff(u2(x), x)+a*x = 6*x^2+3*x*(1/10)+8;
EQ2 := diff(u1(x), x)+diff(u2(x), x, x)+b*x+c = 21*x+4/5;

diff(diff(u1(x), x), x)+diff(u2(x), x)+int(2*x*s*(u1(s)-3*u2(s)), s = 0 .. 1) = 6*x^2+(3/10)*x+8

  

diff(u1(x), x)+diff(diff(u2(x), x), x)+int(3*(s^2+2*x)*(u1(s)-2*u2(s)), s = 0 .. 1) = 21*x+4/5

  

u1(0)+(D(u1))(0) = 1, u2(0)+(D(u2))(0) = 1, u1(1)+(D(u1))(1) = 10, u2(1)+(D(u2))(1) = 7

  

diff(diff(u1(x), x), x)+diff(u2(x), x)+a*x = 6*x^2+(3/10)*x+8

  

diff(u1(x), x)+diff(diff(u2(x), x), x)+b*x+c = 21*x+4/5

 (1)
> 

sol:=dsolve({EQ1,EQ2},{u1(x),u2(x)}):
> 

U1:=unapply( eval(u1(x),sol),x):
U2:=unapply( eval(u2(x),sol),x):
> 

sys:=eval([(lhs-rhs)(eq1),(lhs-rhs)(eq2),bcs],[u1=U1,u2=U2]):
> 

map(coeffs,sys,x):
> 

consts:=solve(%):
> 

SOL:=simplify(eval(sol,consts));

{u1(x) = (-55897*exp(1-x)+27844*exp(2-x)+35538*exp(1+x)+(229194*x^2+1583820*x-1422696)*exp(1)+(-45192*x^2-385320*x+335340)*exp(2)-300822*x^2-1458780*x-55263*exp(x)+1319226)/(-49980*exp(2)+232200*exp(1)-250080), u2(x) = (-55897*exp(1-x)+27844*exp(2-x)-35538*exp(1+x)+(464400*x^3-681738*x^2+1399212*x-1095936)*exp(1)+(-99960*x^3+160164*x^2-309456*x+259476)*exp(2)-500160*x^3+669894*x^2-1398996*x+55263*exp(x)+1038390)/(-49980*exp(2)+232200*exp(1)-250080)}

 (2)
> 

U1:=unapply( eval(u1(x),SOL),x):             # Check
U2:=unapply( eval(u2(x),SOL),x):             #
simplify(eval({eq1,eq2,bcs},[u1=U1,u2=U2])); #

{1 = 1, 7 = 7, 10 = 10, 21*x+4/5 = 21*x+4/5, 6*x^2+(3/10)*x+8 = 6*x^2+(3/10)*x+8}

 (3)
> 

 


Download sysode-int.mw

By recurrence...

vv 13922
January 24 2018
1 1

 

n is supposed to be a positive integer.

We compute the indefinite integral.

 

 

> 

J:=n -> Int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x);

proc (n) options operator, arrow; Int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x) end proc

 (1)
> 

value(combine(J(n+1)-J(n)));

(1/2)*T*sin(2*Pi*x*(2*n+1)/T)/(Pi*(2*n+1))

 (2)
> 

K:=unapply(%,n);

proc (n) options operator, arrow; (1/2)*T*sin(2*Pi*x*(2*n+1)/T)/(Pi*(2*n+1)) end proc

 (3)
> 

J(n)=value(J(1))+Sum(K(k),k=1..n-1);  # Answer

Int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x) = -x+2*T*((1/2)*cos(Pi*x/T)*sin(Pi*x/T)+(1/2)*Pi*x/T)/Pi+Sum((1/2)*T*sin(2*Pi*x*(2*k+1)/T)/(Pi*(2*k+1)), k = 1 .. n-1)

 (4)
> 

value(%);  # Version of the answer

int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x) = -x+2*T*((1/2)*cos(Pi*x/T)*sin(Pi*x/T)+(1/2)*Pi*x/T)/Pi+((1/8)*I)*T*(exp((2*I)*Pi*x*(2*n+1)/T)*LerchPhi(exp((4*I)*Pi*x/T), 1, n+1/2)-exp((6*I)*Pi*x/T)*LerchPhi(exp((4*I)*Pi*x/T), 1, 3/2)+exp(-(6*I)*Pi*x/T)*LerchPhi(exp(-(4*I)*Pi*x/T), 1, 3/2)-LerchPhi(exp(-(4*I)*Pi*x/T), 1, n+1/2)*exp(-(2*I)*Pi*x*(2*n+1)/T))/Pi

 (5)
> 

simplify(diff( (lhs-rhs)(%), x)); # Check

0

 (6)

Maths...

vv 13922
January 23 2018
0 0

This is a maths problem, not a Maple one. See https://en.wikipedia.org/wiki/Cycles_and_fixed_points

Of course it is easy to implement any of the formulae.

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