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What is wrong with this code?...

Asked by: emmantop 40
convert student ode
May 06 2015
0  2

Hello,

please help check what's wrong with this code. I need the analytic solution and convert to Bessel but return error. Here is the worksheet ID_1.mw

Best regards.

this code not answer:...

Asked by: bfathi 10
May 05 2015
0  8

Hi,

this code not answer:

> whit*evalf; tx := evalf((t1+t2)*(1/2));
> t1 := 326*40; t2 := 327*40;
> y1, y2 := eval([y1, y2], Ns(t1), Ns(t2));
>while (t1 < tx) and (tx < t2) do:
yx := rhs(Ns(tx)[c2]):
if yx > 0 then
y2 := yx: t2 := tx:
else
y1 := yx: t1 := tx:
fi;
od;
>Tx = evalf(tx);
> Hx := floor((1/3600)*tx):
> Mx := floor((tx-3600*Hx)*(1/60)):
> Sx := tx-3600*Hx-60*Mx:
> Hx, Mx, Sx;
> XS := [seq(X[i], i = 0 .. 328)]: YS := [seq(Y[i], i = 0 .. 328)]:
> VxS := [seq(Vx[i], i = 0 .. 328)]: VyS := [seq(Vy[i], i = 0 .. 328)]:
> save(G,Mz,Xs,VxS,VyS, 'orbit.sav'):

 can you help me?

Integrating an equation with unknown constants....

Asked by: AmrTawfik12 5
integration
May 05 2015
1  2

This might seem easy but i am a total beginner in Maple and i really need to know how to do this. If i have an expression like this:

X^2 + 3*y + 3*z + X^7

where (y) and (z) are constants of unknown values. I need to integrate the expression over an interval for example (from 1 to 10) so that the answer would still be in terms of (y) and (z).

Would anyone please help me and give me the right expression for writing this on Maple?

Thanks in advance.

How to solve Cauchy sequences...

Asked by: laporte bernard 45
May 05 2015
0  2

(a) Show that if {an} ∞ n=1 is Cauchy then {a 2 n} ∞ n=1 is also Cauchy. (b) Give an example of a Cauchy sequence {a 2 n} ∞ n=1 such that {an} ∞ n=1 is not Cauchy

time series and global temperatures

by: eithne 2072 Maple 2015
application time-series
May 05 2015
4  1

In case anyone is interested, we recently posted a new application on the Application Center,

Time Series Analysis: Forecasting Average Global Temperatures

While interesting in itself (well, I think so, anyway), this application also provides tips and techniques for analyzing time series data in Maple, and shows how to access online data sets through the new data sets functionality in Maple 2015.

eithne

how to solve an exercise with Maple involving the ...

Asked by: laporte bernard 45
limit
May 05 2015
0  6

Show that 2^3 + x ^2 − 3x + 2 is O(x ^3 ).

How to compute direct exponentiation of the matrix...

Asked by: testht06 90
homework
May 04 2015
1  1

Hi EveryOne!

In the the answer of the question "How to find k^th root of the given matrix over finite field (at URL: http://www.mapleprimes.com/questions/203997-How-To-Find-Kth-Root-Of-The-Given-Matrix#comment215683). Carl Love helped to find k^th root of the given matrix M over GF(2m)/f(x).

Now, I need to compute direct exponentiation of the given matrix M in finite field. (If M = [mi,j], we say Direct Exponent (element-wise exponent matrix), Mdk of M is a matrix whose each element is the result of exponentiation of corresponding elements of M. If k=2, then we say Md2 is a direct square matrix of M )

Please help me!!! Thank you very much.

How to wrap code in a box for style...

Asked by: GPY 80
style block
May 04 2015
0  1

I've seen a few maple presentations where a block of code is wrapped in a box. Does anyone know how this is done?

 

Plotting a function subject to an implicit constra...

Asked by: jschulzb 150
May 04 2015
2  3

Hello, 

     I an trying to plot a function of a single variable, which is an implicit function of another variable, i.e. I want to plot F(x(t)), given that x and t are related through the implicit constraint equation f(x,t) = 0. Is there any plot stuctures in Maple that would easily let me do this? I tried implicit plot but this seems insufficient. 

     As an example, consider plotting F = x + x^2 subject to f = x + sin(x) + ln (t) = 0. I could also write this as  a function subject o a differential constraint, as is f = diff(x(t), t) + 1/t + (diff(x(t), t))*cos(x(t)) = 0 and try to use some sort of implicit DE plotting routine. 

     Any ideas?

Thanks!

Open maple worksheet...

Asked by: Annonymouse 160
worksheet incomplete-question
May 04 2015
0  0

I am running Maple in a windows virtual machine, on a mac computer.

I have a number of worksheets on its disk

Windows advised me to run its error checking utility (chkdsk)

when I try and open them it gives me a number of options:

maple text

plain text 

and maple input

 

None of these are the same as the original files. What has happened? and how can i fix it?



Error, (in DEtools/convertsys) unable to convert t...

Asked by: patient 35
dsolve numeric
May 04 2015
0  20

Hi all,

Thanks for helping me to solve the problem below using Maple.

dsys := {(1-4*(diff(ln(v(z)), z)))*(diff(u(z), z))+((3/2)*z^{-1}-2*(3* z^{-1} *(diff(ln(v(z)), z))+2*(diff(ln(v(z)),z,z )))))*u(z) = 0, -z*(diff(v(z), z))-v(z)+v(z)^(1/2)*u(z) = 0, v(0) = 1, u(0) = 1, (D(v))(0) = 1/4, (D(u))(0) = 3/8}

When trying    sol := dsolve(dsys, numeric)

I got : Error, (in DEtools/convertsys) unable to convert to an explicit first-order system.

 

Note that the analytic solution for z<=0 is:

if z>-4   then  u(z)=(1+z/4)exp(z/8) and   v(z)=exp(z/4)

else u(z)=0  and v(z)= (-4/z)exp(-1)

Regards

Length of output exceed limit 1000000...

Asked by: Sofey 10
May 03 2015
0  3

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

 (1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

 (2)

``


Download consistency_expmle_4.mw

.

how to prove with maple...

Asked by: laporte bernard 45
student
May 03 2015
0  1


> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

dynamic programming?...

Asked by: brian bovril 914
optimization proc
May 03 2015
0  2

This problem has real world applicability: Three vampires and three maidens are at the foot of a tall building and wish to get to the bar on the top floor.  The lift only holds two people (for convenience I am classing vampires as people), and needs one person to operate it.  If ever the vampires outnumber the maidens at any place, they will do something unspeakable.  How can the vampires and maidens all safety get to the top in the minimum of moves?  http://tonysmaths.blogspot.com.au/2012/05/vampires-and-maidens-problem.html

can this be solved procedurally or using optimization package?

possible manual soln:

No. m>= No. v

3m+3v, 0   [0]

2m+2v,m+v [1]

3m+2v,v  [2]

     3m, 3v [3]

3m+v , 2v  [4]

m+v  , 2m+2v [5]

#then reverse the steps

2m+2v, m+v [6]

       2v, 3m+v [7]

       3v, 3m  [8]

        v , 3m+2v [9]

    m+v, 2m+2v [10]

        0, 3m+3v [11]

 

Using angular units with trig functions...

Asked by: Dr. Ian Williams 20
convert value
May 03 2015
1  5

I'm looking at Maple as a possible alternative to Mathcad (which I've been using for years, but is now very jaded compared to other options like Maple and Mathematica).  I'm a civil engineer and for what I do, one of the better features of Mathcad is the way it handles units.  For example, if I specify an angle in degrees (say phi=30 degrees) and then ask for sin(phi), I get 0.5.  At face value, I though Maple would do the same kind of thing.  However, this doesn't appear to be the case (see attached worksheet).  The only workaround that I can see is to specify the angle in degrees (but without assigning ) and then multiply the specified value by pi/180 (to convert to radians) before passing it to the sin function.  Which is all a bit messy and not at all an attractive solution.

Am I misunderstanding the way units work in Maple and is there a clean way of specifying angles in degrees (which is what engineers work with) and using these values directy in trig functions?

Thanks in anticipation,

Ian

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