laporte bernard

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These are questions asked by laporte bernard

(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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