Thomas Dean

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16 years, 132 days

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These are questions asked by Thomas Dean

I have a problem with evaluation. The eval(s, eqns) does not evaluate s before evaluation with eqns.

How do I do this?

vars := indets(eqns);

lprint(vars)

{exp(-2*t), exp(1/4*(-11+73^(1/2))*t), exp(-1/4*(11+73^(1/2))*t), i[C1](t), i[R1](t), i[R2](t), i[R3](t), i[V1](t), v[1](t), v[2](t), v[3](t),    v[4](t), v[L1](t), v[R1](t), v[R2](t), v[R3](t), v[V1](t)}

for s in indets(others) do
    if has(s, i) or has(s, v) then print(s);eval(s, eqns); end if;
end do:

How do I print the body of a procedure.  In older versions, verboseproc controlled this.

restart

interface(verboseproc = 3)

interface(prettyprint = 1)

print(int)

                  _m140116321288416, [_m140116321288416]

print(type);
                 proc () option builtin = type; end proc,

                   [proc () option builtin = type; end proc]

 

 

restart;
with(Syrup);
ckt := "*\n\
V1 1 0 -10*exp(-2*t)\n\
R1 1 2 2\n\
R2 2 3 2\n\
C1 3 0 0.5F\n\
R3 2 4 4\n\
L1 4 3 1H\n\
.end";
volts,others:=Solve(ckt,'tran', 'returnall'):


How do I get an expression for diff(i1(t),t)?    I tried

eval(diff(i[R1](t),t), volts union others);
But, this looks very strange

diff(i2(t),t)?  I tried a similar expression, but, it is ver different than the one by OP.

How do I set the initial conditions?  I can use init=1, on L1 for i2(0)=1.  But, how about i1(0)=1?

 

# uname -a

Linux p9x79 5.4.0-48-generic #52-Ubuntu SMP Thu Sep 10 10:58:49 UTC 2020 x86_64 x86_64 x86_64 GNU/Linux

I use X with TWM.


When minized, the maple icon is about 1/4" diameter.  How do I change this to a larger icon?

int(1/(x-1)+sum((k+1)*x^k,k=0..2018)/sum(x^k,k=0..2019),x);

Direct integration produces a large, messy statement, including numbers like 2017, 2016,...,1

Looking at the parts, the numbers 2020, 2019, etc. are in each.

(**) sum((k+1)*x^k,k=0..2018);
x^2019*(2019*x-2020)/(x-1)^2+1/(x-1)^2

(**) sum(x^k,k=0..2019);
x^2020/(x-1)-1/(x-1)

Looks like this should reduct to some simple expession????

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