max125

155 Reputation

8 Badges

6 years, 205 days

MaplePrimes Activity


These are questions asked by max125

alternatingseries.mw
I have a double about this alternating series.
According to maple this series converges:

evalf(sum((-1)^(n+1)*(ln(n)/n+1),n=1..infinity))
                          0.3401310963

However limit ln(n)/n + 1 does not equal to zero, it equals 1. Therefore the series should diverge.

Also while I am on the subject of series and limits, why is limit (-1)^n  as n goes to infinity a range between -1-I and 1 + I.

limit((-1)^(n), n=infinity)
                        -1 - I .. 1 + I

 

 

I am interested in finding the lower Riemann sum for a partition of unequal width.

The points of the partition are P = { -1, -1/4, 1/4, 3/4, 1} , and the function is f(x) = x^2.

My attempt:

restart: with(Student[Calculus1]):
RiemannSum(x^2,x={[ -1,-1/4],[-1/4, 1/4],[1/4,3/4],[3/4,1]},method=lower)

It said error range must be specified. I looked at the help page but I didnt find a specific command for entering partition points manually.

 

The problem is to find a100 for a sequence defined as follows.


https://snipboard.io/AdZKX5.jpg I tried the following command for a recursive sequence.
(I used 10 instead of 100 to test run it).

a||1:=1;
for n from 2 to 10 do a||n:=2*sum( a||i ,i=1..n)^2/(2*sum( a||i ,i=1..n)-1); od;

This approach did not work. It says "error recursive assignment"

Any ideas how to get a(100), also if its possible how to solve the recursive sequence explicitly.

I tried rsolve :

rsolve({ f(n)=(2*sum(f(i),i=1..n)^2/(2*sum(f(i),i=1..n)-1)),f(1)=1},f(n))
rsolve({ f(n)=(2*sum(f(i),i=1..n)^2/(2*sum(f(i),i=1..n)-1)),f(1)=1},f(n))

It just returned back the expression inertly.

 

integral.mw

I am trying to evaluate any which way the integral:

int(exp(-(sqrt(4*x^2+4*y^2+4*z^2)^3)), z = -sqrt(4-x^2-y^2).. sqrt(4-x^2-y^2),y=-sqrt(4-x^2)..sqrt(4-x^2),x=-2..2);

The program just hangs, so i click on 'stop current operation'.

Then I tried:

evalf(Int(exp(-(sqrt(4*x^2+4*y^2+4*z^2)^3)), z = -sqrt(4-x^2-y^2).. sqrt(4-x^2-y^2),y=-sqrt(4-x^2)..sqrt(4-x^2),x=-2..2));

It returns the integral back unevaluated.

It's true that I could use a change of variables, changing to spherical coordinates would be best here. But I would like to know if I did set up the original Cartesian integral correctly. Is there some workaround I can apply to get a numeric answer. I am satisfied with a decimal approximation. Then I can compare to the change of variable result. An exact answer would be even better of course.

Maple gives me the incorrect answer to the hundredth place. (arithmetic.mw)

>15000*(1+.06/365)^(10*365)
>                        27330.47804

I tried using an exact fraction 15000*(1+(6/100)/365)^(10*365) as well.

1 2 3 4 5 6 7 Page 2 of 8