13613 Reputation

19 years, 253 days

Differentiable = > Defined...

@pagan If a function is not defined at a point, the question of its possible continuity or differentiability at that point cannot come up.

Notice also that the help page for isdifferentiable specifically refers to piecewise defined functions.
?isdifferentiable
f:=piecewise(x=3,7,1/(x-3)):

isdifferentiable(f,x,1);
false
isdifferentiable(1/(x-3),x,1);
true

Differentiable = > Defined...

@pagan If a function is not defined at a point, the question of its possible continuity or differentiability at that point cannot come up.

Notice also that the help page for isdifferentiable specifically refers to piecewise defined functions.
?isdifferentiable
f:=piecewise(x=3,7,1/(x-3)):

isdifferentiable(f,x,1);
false
isdifferentiable(1/(x-3),x,1);
true

Finding C...

Integrate eqn1 from s = 0 to s = 1.5:

int(15*(diff(theta(s), s, s)), s = 0 .. 1.5) = int(N3*cos(theta(s))-N1*sin(theta(s)), s = 0 .. 1.5);

Now the requirements eqn4, eqn5 and bcns imply that the right hand side is just N3.

Thus 15*(D(theta)(1.5)-D(theta)(0)) = N3.

But from the equation for diff(theta(s),s) we find (taking signs into account!):

D(theta)(0) = - sqrt( (2/15)*N1 + C)

D(theta)(1.5) = sqrt( (2/15)*N1 + C)

By combining these results, C is easily found.

Finding C...

Integrate eqn1 from s = 0 to s = 1.5:

int(15*(diff(theta(s), s, s)), s = 0 .. 1.5) = int(N3*cos(theta(s))-N1*sin(theta(s)), s = 0 .. 1.5);

Now the requirements eqn4, eqn5 and bcns imply that the right hand side is just N3.

Thus 15*(D(theta)(1.5)-D(theta)(0)) = N3.

But from the equation for diff(theta(s),s) we find (taking signs into account!):

D(theta)(0) = - sqrt( (2/15)*N1 + C)

D(theta)(1.5) = sqrt( (2/15)*N1 + C)

By combining these results, C is easily found.

Combining with min or max...

A minor modification (admittedly making the technique more complicated)  makes it possible to avoid determining the points of intersection:

plottools[transform]((x,y) -> [x,y + x + 5])(plot(min(x^2-x-5,0),x=-3..4,filled=true));

Combining with min or max...

A minor modification (admittedly making the technique more complicated)  makes it possible to avoid determining the points of intersection:

plottools[transform]((x,y) -> [x,y + x + 5])(plot(min(x^2-x-5,0),x=-3..4,filled=true));

Wildly oscillating...

Your integrand is not only oscillating, it is oscillating wildly.

Even on a much smaller interval you will have problems.

This works though:

evalf(Int(sin(x^19),x=0..2,method = _d01akc,maxintervals=10000));
0.08028631940

See ?evalf,Int

Preben Alsholm

Define C to be a Vector in the beginning...

I don't know what you are trying to do, but I would start like this:

restart;
C:=Vector(3):
A := evalf(Matrix([[2, -1, sqrt(2)], [3, 2, -3], [3, sqrt(2), -15/7]]));
B := evalf(<5+7*sqrt(2), -24, -12-3*sqrt(2)>);
N := 3: t := 1e-6:
P := evalf((1/2)*sqrt(t*N)):

You don't need the packages linalg and student, and they are superseded anyway by the LinearAlgebra and the Student packages.

Preben Alsholm

Define C to be a Vector in the beginning...

I don't know what you are trying to do, but I would start like this:

restart;
C:=Vector(3):
A := evalf(Matrix([[2, -1, sqrt(2)], [3, 2, -3], [3, sqrt(2), -15/7]]));
B := evalf(<5+7*sqrt(2), -24, -12-3*sqrt(2)>);
N := 3: t := 1e-6:
P := evalf((1/2)*sqrt(t*N)):

You don't need the packages linalg and student, and they are superseded anyway by the LinearAlgebra and the Student packages.

Preben Alsholm

Removal of evalf and replacement of subs...

The removal of evalf in my previous comment could be replaced by the version below, which seems to take care of the title problem.

The first change to `plots/animate` is the same as before.

`plots/animate`:=subs(subs=((x,y)->eval(y,x)),eval(`plots/animate`)):
`plots/animate`:=subs(evalf=(proc(x::uneval) local a;a:=[eval(x)];if _npassed>1 then evalf(a[1],_rest) else op(a) end if;end proc), eval(`plots/animate`)):
`plots/animate`:=subs(numeric=realcons,eval(`plots/animate`)):
﻿

Preben Alsholm

Removal of evalf and replacement of subs...

The removal of evalf in my previous comment could be replaced by the version below, which seems to take care of the title problem.

The first change to `plots/animate` is the same as before.

`plots/animate`:=subs(subs=((x,y)->eval(y,x)),eval(`plots/animate`)):
`plots/animate`:=subs(evalf=(proc(x::uneval) local a;a:=[eval(x)];if _npassed>1 then evalf(a[1],_rest) else op(a) end if;end proc), eval(`plots/animate`)):
`plots/animate`:=subs(numeric=realcons,eval(`plots/animate`)):
﻿

Preben Alsholm

A workaround...

Besides finding 'a' in Alec Mihailov's original post you can find the value of the sum b by using the exact values of sums that are related:

b:=Sum(ln(n)/n/(n-1),n=2..infinity);
infinity
-----
\
)      ln(n)
/     ---------
-----  n (n - 1)
n = 2
c2:=Sum(ln(n)/n^2,n=2..infinity);
infinity
-----
\
)    ln(n)
/     -----
-----    2
n = 2   n
c3:=Sum(ln(n)/n^3,n=2..infinity);
infinity
-----
\
)    ln(n)
/     -----
-----    3
n = 2   n
b0:=factor(combine(b-c2-c3));
infinity
-----
\
)      ln(n)
/     ----------
-----   3
n = 2  n  (n - 1)
evalf(b0+value(c2+c3));
1.257746887

Preben Alsholm

Old bug in animate...

Yes, that bug is old. I told Maple about it a long time ago, on October 10 2007.

"We have entered this issue into our internal bug-tracker. Thank you for bringing this to our attention."

The Maple version then must have been Maple 11. But nothing changed in Maple 12 or 13.

I still (!) don't have Maple 14.

The remedy I use, and which I also suggested at the time, is to replace the occurrences of subs with eval.

I have used the following redefinition with success and seemingly with no bad side effects:

`plots/animate`:=subs(subs=((x,y)->eval(y,x)),eval(`plots/animate`)):

Preben Alsholm

Old bug in animate...

Yes, that bug is old. I told Maple about it a long time ago, on October 10 2007.

"We have entered this issue into our internal bug-tracker. Thank you for bringing this to our attention."

The Maple version then must have been Maple 11. But nothing changed in Maple 12 or 13.

I still (!) don't have Maple 14.

The remedy I use, and which I also suggested at the time, is to replace the occurrences of subs with eval.

I have used the following redefinition with success and seemingly with no bad side effects:

`plots/animate`:=subs(subs=((x,y)->eval(y,x)),eval(`plots/animate`)):

Preben Alsholm

Equality...

Sum(ln(1+1/n)/n,n=1..N) = Sum(ln(n)/n/(n-1),n=2..N+1)+ln(N+1)/(N+1);

so the two infinite series have the same sum. Convergence follows by comparison to Sum(ln(n)/n^2,n=2..infinity), which is seen to be convergent by using e.g. the integral test.

Preben Alsholm

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