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janhardo

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11 years, 39 days
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Transforming to 2d input from a derative...

janhardo 695 Maple
derivative 2dmath differentiation
January 15 2022
0 1

Thought always that the round d is reserved for function of two variables x,y , but  that seems to be not the case here ?

> 

restart;

Comparing Different Answers

 

Een antwoord ergens gegeven is

Int(sqrt(x^2+1), x) = (1/2)*x*sqrt(x^2+1)+(1/2)*ln(x+sqrt(x^2+1)) + C                                                             (vb)

 

Mple geeft

 

> 

Int(sqrt(x^2+1),x)=int(sqrt(x^2+1),x)+C[1];

Int((x^2+1)^(1/2), x) = (1/2)*x*(x^2+1)^(1/2)+(1/2)*arcsinh(x)+C[1]

 (1)
> 

 

De twee antwoorden lijken nog niet opelkaar !
In het gegeven antwoord staat er een ln en in Maple kan een expressie omgezet worden in ln termen
  

> 

convert(%,ln);

Int((x^2+1)^(1/2), x) = (1/2)*x*(x^2+1)^(1/2)+(1/2)*ln(x+(x^2+1)^(1/2))+C[1]

 (2)

(2)  is hetzelfde (vb)

Dezelfde integraal i sook gegeven als

> 

Int(sqrt(x^2+1),x)=((x+sqrt(x^2+1))^2+4*ln(x+sqrt(x^2+1))-(x+sqrt(x^2+1))^(-2))/8+C[2];

Int((x^2+1)^(1/2), x) = (1/8)*(x+(x^2+1)^(1/2))^2+(1/2)*ln(x+(x^2+1)^(1/2))-(1/8)/(x+(x^2+1)^(1/2))^2+C[2]

 (3)

Controle

een effectieve manier om twe antwoorden t evergelijken voor hetzelfde probleem is het verschil te berekenen van een vergelijking met de twee integralen

> 

#lhs(%);
> 

#rhs(%%);
> 

 

> 

#diff(lhs(%)-rhs(%)=0,x);
> 

NULL
> 

#diff(f,x);
> 

diff(lhs(%)-rhs(%)=0,x);

(x^2+1)^(1/2)-(1/4)*(x+(x^2+1)^(1/2))*(1+x/(x^2+1)^(1/2))-(1/2)*(1+x/(x^2+1)^(1/2))/(x+(x^2+1)^(1/2))-(1/4)*(1+x/(x^2+1)^(1/2))/(x+(x^2+1)^(1/2))^3 = 0

 (4)
> 

simplify(%);

0 = 0

 (5)

Strange that  diff(lhs(%)-rhs(%)=0,x);  is translated by 2 d input with round d notation for functions with two variables ?
The two integrals are functions of one variable
diff(f, x)

Download Controleren_dezelfde_antwoord_voo_expressies.mw

Is there a hotkey for toggling between m...

janhardo 695 Maple
2dmath editing hot-key
January 13 2022
2 8

Sometimes its easier when doing math in maple input mode to use first the 2d maple input mode and convert this to maple input
Is there a hotkey assigned in Maple to do this toggling from 1d input to 2d input ( also from 1d output to 2d output  )

Now it must be done by mouse

A maplet for complex integration ...

janhardo 695 Maple
complex-analysis interactive
January 12 2022
0 0

This is so useful to see geometrical mapping diagram to visualize Complex analysis

Something that also can be made for Maple 

Mapping Diagram for Cauchy Integral Formula – GeoGebra

Using GeoGebra for visualizing complex variable. (google.com)

I highly encourage everyone interested in complex variable to read Tristan Needham „Visual Complex Analysis” and try to solve problems with or without aid of GeoGebra. I hope that in this workshop we will manage to get a feeling of complex functions and as a final point understand how complex integration works. It is a common misconception that complex integration can't be visualized, and using Tristan Needham's ideas we will try to explore this idea. It's a pity that we don't have a lot of time, thus we will skip a lot of important information and construct only some graphs. 

There is so much experimenting with Geogebra software and doing too this in Maple ?

How to get a prime and subscript notatio...

janhardo 695 Maple
derivative 2dmath typesetting
January 10 2022
1 4

I am studying the help pages for declare() statement in the PDEtool package. 
It can be used also for calculus(prime notation) for 1 variable and two variables (subscript notation) as i understand it now.

There is no input possible in Maple by using a prime notation and indexed functions notation?

Maybe with a alias ? ..remember that I can be chanced I (default) into the i for complex numbers 

How to plot the approximation graph for ...

janhardo 695 Maple
plotting riemann-zeta
January 10 2022
0 1

I must feed the not trivial zeros numbers into this aproximation formula ?

 

Riemann hypothese and staircase of primes

 

> 

restart;
> 

with(NumberTheory)
> 

PrimeCounting(1)

0

 (1)
> 

pi(Pi)

2

 (2)
> 

PrimeCounting(10000)

1229

 (3)
> 

numelems(select(isprime, [seq(1 .. 10000)]))

1229

 (4)

The prime counting function is approximated by Li(x) and x/ln(x).

> 

plot([PrimeCounting(x), Li(x), x/ln(x)], x = 1 .. 500, legend = [pi(x), Li(x), x/ln(x)])

The staircase of primes approximated by two functions
Interesting is the video: How i learned to love and fear the Riemann Hypothesis

https://www.quantamagazine.org/how-i-learned-to-love-and-fear-the-riemann-hypothesis-20210104/

> 

NULL
> 

ps:=Array(1..30):
y:=0:
for n from 1 to 30 do
 if is(n,prime)
     then ps[n]:=plot([[n,y],[n,y+1],[n+1,y+1]]):
     y:=y+1;
     else ps[n]:=plot([[n,y],[n+1,y]]):
 end if ;
od;
with(plots):
display({seq(ps[n],n=1..30)}):  
> 

plot([PrimeCounting(x)] ,x = 1 .. 35, legend = [pi(x)]):
> 

plot([PrimeCounting(x), Li(x), x/ln(x)], x = 1 .. 35, legend = [pi(x), Li(x), x/ln(x)])

 

 

 

 

Prime counting function
What found RIEMANN for the prime counting function in relation to the zeta function after he defined the zeta function?

 

He found further a function what follows exactly the shape of the prime counting function

Final discovery v. Riemann.  

- step in the omhoog in de priemtelfunctie = log(p) (zie video)

 

Using the logarithmic primecount function( from Chebyshev) (approximation)
Further  analyse with this Chebyshev approximation formula in relation to the not trivial zero points from Riemann zeta function ( zeros) gives another real function for approximating the primecounting function what uses the non trivial zeros from Riemanns zeta function  in this function:

 

"(not trivial zeros ) u[k ] = "i*w[k]+v[k]   
Number now all nottrivial zeros in the upperhalfplane from down to bottom,  as u[1], u[2], u[3, () .. ()]

> 

"`ϕ`(x) := x-ln(2Pi)-1/(2 )ln(1-1/(x^(2))) - (∑)2/(|u[k]|) x^(v[k]) cos(w[k] ln(x)-alpha[k])"
> 

NULL

Its only alpha[k] that must be calculated out of the not trivial zeros and i must have a list of  serie of not trivial zeroes from the zeta function. => see Hardy's Z(t) ? from this ..... alpha[k]  can be calculated ?
All not trivial zeros are complex numbers laying on a line ,but  orginating from (0,0) in the complex plane as  vectors to the points    

> 

varphi(x):= x - ln(2*Pi) - 1/2*ln(1 - 1/x^2) - sum(2*x^v[k]*cos(w[k]*ln(x) - alpha[k])/abs(u[k]), k = 1 .. infinity);

x-ln(2*Pi)-(1/2)*ln(1-1/x^2)-(sum(2*x^v[k]*cos(-w[k]*ln(x)+alpha[k])/abs(u[k]), k = 1 .. infinity))

 (5)

This formula seems to be correct .
Now how to make a plot ?
Hardy's Z(t) function shows the not trivial zeros in the upperhalfplane of the critical strip of the Riemann zeta function  as zeros in this Z(t) real function : derived from a alternating serie ?

Download priem_staircase_en_riemann_functie.mw

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