Math Pi Euler

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@Preben Alsholm 

The last part is magnificant! comparing the 3 different orders. Is it possible to very in delta as well? For example delta 0.1..0.3. In that case we get 12 different graphs? Can I add it in after delta=0.1?

 

 

@Preben Alsholm 

This one is great as well. Even up to delta^3! Now suppose I wish to evaluate Z3 at for example x=0, I can just substitute it in? 

@Rouben Rostamian  

I indeed made a major typo mistake! My humble apoligies. 

It should be:

 

restart;

de := diff(z(x),x,x) + z(x) - cos(2*x)/(1+delta*z(x)) = 0;

 

 

I changed it in your 'body' and it gives for delta the correct values now!!! But how can I check the values for delta^2?

@Rouben Rostamian 

I just got the message from my tutor that my hand solution is correct. What does Maple different in this case?

@Rouben Rostamian  I think I made a mistake in the calculation by hand I found:

z= -1/3 cos (2x) + delta (1/6 -(8 sqrt2) cos x/(45) -cos 4x /90) + delta^2( ((2sqrt2)x sin x)/45 -(sqrt2 / 90)(pi +1) cos x + 7 cos 2x /270 - ((sqrt 2)(cos 3x)) /90 - (cos 6x)/1050)

@Preben Alsholm the solution I got by hand:

z= -1/3 cos (2x) + delta (1/6 -(8 sqrt2) cos x/(45) -cos 4x /90) + delta^2( ((2sqrt2)x sin x)/45 -(sqrt2 / 90)(pi +1) cos x + 7 cos 2x /270 - ((sqrt 2)(cos 3x)) /90 - (cos 6x)/1050)

 

@tomleslie : thanks, that will do!!!

@vv I think it is working correctly. I got the same with N=1, N=2 by hand. Many thanks!

@rlopez: this is even easier then the given procedure. Is it possible to "get out" the common term: sqrt2 / (2 * exp^(x))?

 

@vv Thanks, I couldn't copy, but typed it line by line. And I get the same result. There is just one strange thing, when I did it by hand, which took me 1.5 hours, I have got some other coefficients. 

@vv 

Thanx vv. Until x=0.37 we have a reasonable good approximation. 

ytaylor:=convert(solve(series(p,x,5),y), polynom):

You used "5" for the fifth order? I experimented with different values too. Nice to see that the graph of ytaylor changes.

This helps so much to understand it more!

 

@vv 

Thanks, I see what went wrong. I tried to use the statement (order=7) in it, there was the error. (to get to the x^6), but I think that is not possible?

Could I make in the same plot the graphs of the Taylor approximation and the exactvalues of y(x)? I wish to see the differences. Do I have to use implicitplot?

The original equation was:

 ln((1+x)*y) + exp((x^2)*(y^2)) = x + cos(x), so I thought, to make it look like a function of 2 variables I define the "new" equation as p. So substituting x=0, y=1. 

But now how to find the Taylor series about x=0. 

using standard series gives an error. 

 

 

 

@acer 

Thanks for your explanation. I started wrong, by looking for the zeta function in Maple. And helpfull about the trick!

 

@tomleslie 

That is the missing link. I get the next values:2.491913069, 4.089219366, 5.384017111, 6.414587030, 7.233362705.

A conclusion should be that as Sn increases is too slow to obtain an accurate estimate of Lim (Sn), where n->00. 

@tomleslie : if we could rewrite the sum in the form:

 

Restart: Sn2:=n->add(1/k^(1/10)(sin(1/k)-1/k)+add(1/k^(11/10)),k=1..n);

And noting that the second sum can be evaluated in terms of the zeta function, can we show that the limit of Sn2 = 10.395.639 as n goes to infinity?

 

 

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