asa12

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4 years, 229 days
i would also not like to ask, but if not ask, what should i do?

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These are replies submitted by asa12

@Kitonum 

A := Matrix([[-2,1,1],[0,2,0],[-4,1,3]]);

A:=[<-2,1,1>, <0,2,0>, <-4,1,3>];
S1 := LinearAlgebra:-Basis(A);
 

A:=[<-2,0,-4>, <1,2,1>, <1,0,3>];
S1 := LinearAlgebra:-Basis(A);

and

A:=`<|>`(`<,>`(-2, 0, -4), `<,>`(1, 2, 1), `<,>`(1, 0, 3));
S1 := LinearAlgebra:-NullSpace(A);

what is wrong with maple 12?


 

but output are not [1,4,0],[1,0,4],[1,0,1]
 

@AmusingYeti 

moreover, how to translate this condition statement?

 

if op(Expr)==indets(Expr3,{string,name})[countvar2-ii+1]:

 

i tried

if Expr.args==list(Expr3.free_symbols)[countvar2-ii]:

but not correct

@AmusingYeti 

 i use this expression in maple is easy

test1 := OO(AA(A,B),C)

but in python and sympy

need

from sympy.abc import x,y,z,a,b,c

but OO and AA these kind of expression can not be used

how to do?

@Kitonum 

actually I am finding geometry axioms of newton equations 

is this correct?

or is it possible?

as I have a sense that there are many possible 

systems that can embedded in newton equation

 

@vv 

slope is tangent of angle

a right triangle with total 180 degree of 3 interior angles 

@Ramakrishnan 

actually i would like to find a system which after solve this system will return v=u+at  and  s=u*t+1/2*a*t^2

solution := solve(unknownsys, [...]);

solution := [ v=u+at , s=u*t+1/2*a*t^2 ];

but there is only 2 equations for 5 variables, 

is there some equations missing?

and

is there any function input these solution set that can return an invariant?

@tomleslie 

it really works 

this prove that there is no need to use the most 

simplified logic , this intermediate result can also 

work

And consider wildcard bit as 1 in summation

Is correct method

Thank you very much

@eilers 

following the doing of exponential, it should have a sum, 

i do not understand why remove sum

@eilers 

if specify limit on integral, which range is correct?

i choose from 0 to 1, then does it mean that it can remove sum(..., x=0..infinity)?

hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x=0..1)), x=0..infinity);

@Axel Vogt 

when using maple 12

Warning, unable to evaluate the functions to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

no graph

then i try

Digits:=15:
f:= x ->
  evalf( Int('t -> Re(exp(-t*LambertW(-1/t)))', 1 .. x, method=_d01ajc) )
 +evalf( Int('t -> Im(exp(-t*LambertW(-1/t)))', 1 .. x, method=_d01ajc) )*I:
g:= (-1)/f+1;
g:= f;
[Re@g, Im@g]:
plot(%, 0 .. 2, -10 .. 10, color=[red,blue, grey]);
 
still no graph
 
which version of maple can plot?

@Mariusz Iwaniuk 

thank you

 

@tomleslie 

sorry i use a wrong formula from googled result

after refer from

https://en.wikipedia.org/wiki/List_of_representations_of_e

exponential1 := sum((1/n!), n=0..infinity);
exponential1 is not a decimal number, it is exp(1)
 
hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x)), x=0..infinity);
 
hoyeung2:= sum((Int(exp(LambertW(1/(-1+x!))*(-1+x!)), x)), x=0..infinity);
 
how to evalute hoyeung1 or hoyeung2 as a decimal number?
 
how to evalute hoyeung^x as a decimal number function is func1 := proc(x) return hoyeung^x end proc:
 
but i do not know whether
sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x))*m^x, x=0..infinity) = hoyeung^x
or
 
sum((Int(exp(LambertW(1/(-1+x!))*(-1+x!)), x))*m^x, x=0..infinity) = hoyeung^x
 
if not, using hoyeung^1 , then enough to calculate hoyeung^x just using power
 
can limit(1+(Int(exp(LambertW(1/(-1+x))*(-1+x)), x)))^x, x=infinity) = hoyeung^1 ?

 

@eilers 

power of something or differential of power such diff(y(x), x) ^ n

i do not know so want to find it

@Markiyan Hirnyk 

i use reasoning to guess two equations and dsolve, this is what i got. hope you are clear.

sol := dsolve(ln(diff(y(x),x)) = diff(y(x),x)^(1/(1-x)), y(x));
ma := subs(_C1=1, rhs(sol));
Int(exp(LambertW(1/(-1+x))*(-1+x)), x)+1;
with(plots):
complexplot(evalf(subs(x=m, ma)), m = -Pi .. Pi);

sol := dsolve(ln(diff(y(x),x)) = diff(y(x),x)^(1/(1-y(x))), y(x));
ma := subs(_C1=1, rhs(sol));
x-Intat(1/exp((-1+_a)*LambertW(1/(-1+_a))), _a = y(x))-_C1 = 0;
with(plots):
complexplot(evalf(subs(x=m, ma)), m = -Pi .. Pi);

@Markiyan Hirnyk 

How about the second one?

why dsolve not return Y(x) =

but return y(x) in a function = 0?

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