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Hi,

I have a first order differential eq. for some variable say $r(x)$, where $x$ is the independent variable.

After solving this differential equation numerically, I want to use its solution in other expression for $r(x)$ and plot the expession with $x$.

Please let me know how to do it.

Thanks in advance.

 

 

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Probability_density_normalization.mw

In this code I'm trying to separately normalize two independent probability densities and then combine them to get the joint probability density that's normalized and then use it to calculate the probability that the two variables are equal. fD(x) is a Gaussian divided by x^2 and fA(x) is a Gaussian. The first problem occurs when I'm checking the normalization of the joint probability density by doing the double integral over all space for fD(x)*fA(y)dxdy, I get weird vanishing number when the parameter "hartree" takes a certain value, namely 27.211. If I change hartree to 27 or 1 or 2 it all worked, but 27.211 is not good. Also later when I do a single integral over all space for fD(x)*fA(x)dx to get the probability that these two are equal, I find the result is dependent on hartree. This hartree thing is a unit conversion in my physical problem and in principle should not interfere with either the normalization or the probability result at all. I suspect this is a coding bug but I can't find what it is. I'd appreciate any input.

Thank you very much!

 

Edit: I found out that the problem with the double integral normalization may have something to do with the discretization for numerical evaluation of the integral, since if I change the lower bound to 1/hartree and upper to 10/hartree then it's fine, however if I use lower bound at 1/hartree and upper at 5/hartree it doesn't work, although the distribution has no value between 5/hartree and 10/hartree. However after this is fixed I still have the problem with the single integral over all space for fD(x)*fA(x)dx changing with hartree. Well as a probability I would expect the integral to be bound between 0 and 1, but since it almost linearly depends on hartree, at hartree around 27 I would get the integral value to be about 25, which doesn't make sense. In fact, I now suspect it is not Maple, but my calculation of the probability of the two random variables taking the same value is wrong, I'd appreciate it very much if someone can confirm this.

I would have love to attach a document because I try pasting it but it is not allowed I want to integrate something of this nature;... I don't even get how to write anything here maybe because am using a phone.53e77f9f0cf21cc29fd9d4e8.pdf 

This is the paper i'm working on,

1) I couldn't get 11a and 11b on page 1918.

2) I don't know how to integrate 13b to 13e. Please somebody help my career I will never forget it.

My e-mail is foyt22@gmail.com

Hi fellow Maple users,

I'm trying to solve an eigenvalue problem of Ax=wx, where A is a 6 by 6 Hermitian matrix with two parameters x and y. I want to solve it for w and then plot3d it with x and y as unknowns. The way I have been doing is first find the characteristic equation Determinant(A-wI)=0 and then solve it for w, and then plot3d the solutions within a range for x and y. My problem is sometimes solve(Determinant(A-wI)=0,w) would give me the 6 solutions expressed in x and y, but sometimes when the numbers in A are changed it will only give me a Rootof solution with which I cannot plot. I'm wondering if there is a better way to do this. I'm actually not very interested in the symbolic solution of w expressed in x and y, just the plot, so if there is a numerical alternative it's good too.

Thank you in advance!

how to show chain rules result when diff this

Eq1 := f(x,g(x,t)) + f(x,y);
diff(Eq1, x);

 http://math.stackexchange.com/questions/372093/chain-rule-definition-f-fx-gx-y

https://drive.google.com/file/d/0Bxs_ao6uuBDUanVWYm1SMWc4R3M/view?usp=sharing

 

Hoe do you plot two vectors of data as a bar plot or historygram. I tried the statistics package but could not plot a bar plot that shows the proper relation of numbers in vector x to numbers in relation to vector y. Vector x contained years as data, and vector y contained for example crime data per year.

with(Groebner):
T := lexdeg([x,y,z],[e1,e2]);
intermsof1 := y;
intermsof2 := -z;
GB := Basis([e1-intermsof1, e2-intermsof2], 'tord',T);
result := NormalForm(y^2-x*z, GB,'tord', T);
result := NormalForm(y^2-x*z, GB, T);

originally Basis do not have error when without parameter 'tord'

after it has argument error, it has to be added extra parameter tord

NormalForm has the same error too.

i do not understand why it has error, how to solve?

i just want to express y^2-x*z in terms of y and -z

 

 

hi everyone,
I wanna to solve this equation using maple 12 but I receive this warning message

cos(x)*cosh(x)=1

cos(x)*cosh(x)=-1

tan(x)=tanh(x)

thanks in advance

(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

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


"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

If a dosage Q units of a certain drug is administrated to an individual, then the amount remaining in the bloodstream at the end of t minutes is given by Q*exp^-ct, where c>0. Suppose this same dosage is given at successive T-minute intervals.

 

a) Show that the amount A(k) of the drug is given by A(k) = ∑n=0k-1 Q*exp(^-ncT).

b) Find an upper bound for the amount of the drug in the bloodsteam after any number of doses.

c) Find the smallest time between doses that will ensure that A(k) does not exceed a certain level M for M>Q.

worksheet/expressions/copypasteMaple

Gerschgorin := proc (A::Matrix) local Delta, m, n, AA, R, C, i, c, eig, P, Plt; Delta := proc (i, j) if i = j then 0 else 1 end if end proc; m, n := LinearAlgebra[Dimension](A); AA := Matrix(m, n, proc (i, j) options operator, arrow; Delta(i, j)*abs(A[i, j]) end proc); R := evalm(`&*`(AA, Vector(m, 1))); C := {seq(('plottools[circle]')([Re(A[i, i]), Im(A[i, i])], R[i], color = violet), i = 1 .. m)}; c := {seq(('plottools[point]')([Re(A[i, i]), Im(A[i, i])], color = blue, symbol = diamond), i = 1 .. m)}; eig := evalf(LinearAlgebra[Eigenvalues](A)); P := {seq(('plottools[point]')([Re(eig[i]), Im(eig[i])], color = red, symbol = box), i = 1 .. m)}; Plt := `union`(`union`(C, c), P); plots[display](eval(Plt), scaling = constrained) end proc

 

A := Matrix([[5, 8, 4, -3], [8, -9, 7, 5], [0, 4, 4, 2], [5, -5, 9, -9]]); evalf(LinearAlgebra[Eigenvalues](A), 3); Gerschgorin(A)

worksheet/expressions/pasteMathML

 

F := Matrix([[2, -1/2, -1/3, 0], [0, 6, 1, 0], [1/3, -1/3, 5, 1/3], [-1/2, 1/4, -1/4, 4]]); evalf(LinearAlgebra[Eigenvalues](F)); Gerschgorin(F)

Could you print A & F ?

 

regards

 

 

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