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janhardo

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11 years, 38 days
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@Carl Love ThanksSeems that the seq...

janhardo 695
January 20 2022
0 0

@Carl Love 

Thanks

Seems that the seq command not symbolic sums can do , only numeric 
Wondering why it is not possible with the sum command to show some terms in de serie ( this is student functionality )

@janhardo   .....

janhardo 695
January 20 2022
0 0

@janhardo 
How to plot the graph of this function ?

NULL

> 

restart;
> 

with(plots):

Contour integration is a powerful technique in complex analysis that allows us to evaluate real integrals that we would not otherwise be able to do.
 The idea is to compute a corresponding complex integral over an appropriate closed path in the complex plane.  
Then we split the integral into the sum of several integrals and show that one or more of them is (or approaches) zero.  
Then we use other techniques from complex analysis to evaluate the other pieces, and finally we add everything together to obtain the value of the original integral.

I try to learn more on the fastest way witth the used concepts following a  example on youtube : https://www.youtube.com/watch?v=Iiom5V5faxc

 

Can i follow this step by step in Maple?

> 

int(cos(x)/(x^2+1), x = -infinity .. infinity)
> 

Int(cos(x)/(x^2+1),x=-infinity..+infinity)=int(cos(x)/(x^2+1),x=-infinity..+infinity) ;

Int(cos(x)/(x^2+1), x = -infinity .. infinity) = Pi*cosh(1)-Pi*sinh(1)

 (1)
> 

plot(int(cos(x)/(x^2+1),x=-infinity..+infinity)); # nothing to see?

 

This real integral seems to be difficult to solve by hand directly, but Maple has no trouble with it  
The first idea is to replace all x in  the real integral with a complex number z, cos( z) seems to be problematic  as it is unbounded and goes keeping away further from the orign if the modulus of z becomes really  large ( ok ,don't now exactly how it looks like )

for cos(z) can be taken  "e^(I z) *("?), ""then we get   "(∫)[GAMMA]^(e^(I z))/(z^(2)+1) ⅆz "now the real intergal is replaced by a complex integral  ( the big gamma sign stands for i think for integration path in complex plane)

Now with this complex integral  we can split it in two integrals  so "(∫)[GAMMA]^(e^(I z))/(z^(2)+1) ⅆz = (∫)[-R]^(R)(e^(I x))/(x^(2)+1) ⅆx *(1) "(along the real axis)  "+ (∫)[C]^()(e^(I z))/(z^(2)+1) ⅆz*(2) "(  C : along a arc  connecting points on the real axis )
The eidea is to evaluate both integrals (1),(2)and take the  limit from both as R goes to infinity
-  replacing cos(x) for
"e^(I z) "?" "
-  The idea of splitting a complex integral into a sum of a complex intergral on the real x-axis and one into a contour (arc) integral  ?

``

Download youtube_vb_uitleg_berekening_reel_integraal_vi_acontour_integraal.mw

@vv ThanksForget this edward's ...

janhardo 695
January 19 2022
0 0

@vv 

Thanks

Forget this edward's book you said for learning complex analysis, that's a good advise.
There is much study material to find for learning complex analysis

It's only that i do some investigations with Maple about complex analysis how define it.

There is another modern book about the Riemann Zeta from Alexsandar ivic

Don't worry, i  am learning for the most part  not complex analysis from the book of Edwards.  

@vv ThanksYes, i am real beginner I...

janhardo 695
January 19 2022
0 0

@vv 

Thanks

Yes, i am real beginner In this Complex Analysis.

Got interested in the Riemann hypothesis and wondered how he got his results.
Trying to track that with Maple seemed like a nice challenge.
The complex integral you asked about is in the book: riemann's zeta function : h.m .edwards  
P.S if you are interested in the book?

This integral is used to derive the functional formula of the Zeta(s) function as it seems.

@Kitonum  Thanks With this from y...

janhardo 695
January 19 2022
0 0

@Kitonum 

Thanks

With this from you , every  partiele sum can then be written

@janhardo There is new functionalit...

janhardo 695
January 19 2022
0 0

@janhardo 

There is new functionality for ODE solving
Is y(x) the same as P(t) ? see worksheeet.
And how to solve a differential equation with parameter t in the ODE (steps)

Question: is y(x) = P(t)    ?

 

Maple makes it easy to solve differential equations now in 2021 with the ODE(steps) .. great work.

-closed forms  solutions
- series forms  solutions

- transformation solutions 
As student you could see the steps and solving by hand in Maple 

 

Now only plotting the graph the solution or slopefield

The earlier question for a modern approach for solving differential equations is now be answered with this new ODE(step) functionality

No, its not because students still are doing the tasks by hand in Maple

> 

restart;
> 

with(Student:-ODEs);

[ChangeVariables, DifferentialOrder, Integrate, IntegratingFactor, IsolateHighestDerivative, LinearForm, ODEPlot, ODESteps, ReduceOrder, SeparateVariables, Solve, Test, Type]

 (1)
> 

 

> 

 

> 

verg:=diff(y(x), x) = k*y(x)^2 - y(x)/100:%;

diff(y(x), x) = k*y(x)^2-(1/100)*y(x)

 (2)
> 

ODESteps(verg);

"[[,,"Let's solve"],[,,(ⅆ)/(ⅆx) y(x)=k (y(x))^2-(y(x))/100],["•",,"Highest derivative means the order of the ODE is" 1],[,,(ⅆ)/(ⅆx) y(x)],["•",,"Separate variables"],[,,((ⅆ)/(ⅆx) y(x))/(k (y(x))^2-(y(x))/100)=1],["•",,"Integrate both sides with respect to" x],[,,∫((ⅆ)/(ⅆx) y(x))/(k (y(x))^2-(y(x))/100) ⅆx=∫1 ⅆx+_C1],["•",,"Evaluate integral"],[,,-100 ln(y(x))+100 ln(100 k y(x)-1)=x+_C1],["•",,"Solve for" y(x)],[,,y(x)=-1/((e)^(x/100+_C1/100)-100 k)]]"

 (3)

------------------------------------------------- end --------------------------------------------------------

> 

y(x) = -1/(exp(x/100 + _C1/100) - 100*k);

y(x) = -1/(exp((1/100)*x+(1/100)*_C1)-100*k)

 (4)
> 

 

====================Calculation made in Harald pleym  task==========================================

> 

restart:
> 

deq:=diff(P(t),t)=k*P(t)^2-1/100*P(t):%;

diff(P(t), t) = k*P(t)^2-(1/100)*P(t)

 (5)

The equation is separable.

> 

deq1:=deq/rhs(deq):%;

(diff(P(t), t))/(k*P(t)^2-(1/100)*P(t)) = 1

 (6)

Integration gives

> 

map(Int,deq1,t=0..T);

Int((diff(P(t), t))/(k*P(t)^2-(1/100)*P(t)), t = 0 .. T) = Int(1, t = 0 .. T)

 (7)
> 

eq:=map(factor,%):%;

Int(100*(diff(P(t), t))/(P(t)*(100*k*P(t)-1)), t = 0 .. T) = Int(1, t = 0 .. T)

 (8)

Partial decomposition of

> 

g:=100/(P(t)*(100*k*P(t)-1)):%;

100/(P(t)*(100*k*P(t)-1))

 (9)

gives

> 

g=convert(g,parfrac,P(t));

100/(P(t)*(100*k*P(t)-1)) = -100/P(t)+10000*k/(100*k*P(t)-1)

 (10)
> 

map(x->x*diff(P(t),t),(%));

100*(diff(P(t), t))/(P(t)*(100*k*P(t)-1)) = (-100/P(t)+10000*k/(100*k*P(t)-1))*(diff(P(t), t))

 (11)
> 

map(Int,%,t=0..T,continuous);# via Carl na foutmelding

Int(100*(diff(P(t), t))/(P(t)*(100*k*P(t)-1)), t = 0 .. T, continuous) = Int((-100/P(t)+10000*k/(100*k*P(t)-1))*(diff(P(t), t)), t = 0 .. T, continuous)

 (12)
> 

eq1:=rhs(%)=rhs(eq):%;

Int((-100/P(t)+10000*k/(100*k*P(t)-1))*(diff(P(t), t)), t = 0 .. T, continuous) = Int(1, t = 0 .. T)

 (13)
> 

value(%);

100*ln(P(0))-100*ln(100*k*P(0)-1)-100*ln(P(T))+100*ln(100*k*P(T)-1) = T

 (14)

Substitution of the initial data yields

> 

subs(P(0)=200,T=t,%);

100*ln(200)-100*ln(20000*k-1)-100*ln(P(t))+100*ln(100*k*P(t)-1) = t

 (15)
> 

P(t)=solve(%,P(t));

P(t) = -200/(20000*k*exp((1/100)*t)-exp((1/100)*t)-20000*k)

 (16)
> 

 

Download nieuw_in_maple_2021-ODE_steps_-vergeleken_met_populatie_dv_van_Harald_pleym.mw

@Kitonum I like this presentation o...

janhardo 695
January 17 2022
0 0

@Kitonum 

I like this presentation of the graphs too, with this plot description, although its more work to achieve.

But it should be also not complicated to let the borders disappear as a option for Maplesoft developers ?

@Carl LoveThanksThe function As sol...

janhardo 695
January 17 2022
0 0

@Carl Love

Thanks

The function 
As solution from the differential equation is continuous 
At what stage we can see that P(t)(population notation) should be continuous from a given differential equation  ? 

Was there not a advisor function for classifying  differential equations 

@acer "It seems to be some kin...

janhardo 695
January 17 2022
0 0

@acer 

Thanks

"It seems to be some kind of duplication of your querying here"
In that information is nothing to find how to get a prime notation in maple plaintext input as i earlier determined.
As you mentioned before in your earlier post: it doesn't exist the prime notation for maple plaintext input.

@acer  Thanks I can remember diff...

janhardo 695
January 17 2022
0 0

@acer 

Thanks

I can remember differential equations in Maple  worksheet  who are showned the prime notation.

Could this be done via a alias ? 

@acer "3) A single apostrophe ...

janhardo 695
January 17 2022
0 0

@acer 

"3) A single apostrophe ' which comed from a key on a standard keyboard. This is used as a so-called unevaluation-quote in Maple. In 2D Input this can be used as the prime."

Yes, this is working in 2D 

y'=x -> solve DE  

But not for maple input yet?

How to plot the real part of the zeta fu...

janhardo 695
January 17 2022
0 0

How to plot the real part of the zeta function on the critical line in 2D /3D ?

This is a video explaining all Riemann zeta function - YouTube

Drawing a vertical plane at  x =1/2  on the real axis .

Can be helpful too : LMFDB - The L-functions and modular forms database
note: for people who can possible solve the riemann hypothese ?

@acer Ok, i see Its first chec...

janhardo 695
January 16 2022
0 0

@acer 

Thanks

Ah, i see 

Its first checking if the typesetprime setting has a value prime: yes or no 
Its not a default setting that prime=true

That you than can come up with this Differential equation  in a default worksheet.

y`= x ; 

@Carl Love  Should also help to si...

janhardo 695
January 16 2022
0 0

@Carl Love 

Should also help to simplify  v more ..or doesn't matter ?

@janhardo What a hassle to get this...

janhardo 695
January 16 2022
0 0

@janhardo 

Takes some time to get it working?

Typesetting:-Settings(typesetprime =true);
                             false

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