## 30 Reputation

6 years, 275 days

## Magnificant!...

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?

## Variations of constants 2...

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?

## indeed typo mistake...

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?

## by hand solution 2...

@Rouben Rostamian

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

## by hand solution...

@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)

## solution by hand...

@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)

## syntax...

@tomleslie : thanks, that will do!!!

## @vv I think it is working correctly...

@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 gi...

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

## @vv When I try to run your solution...

@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 hav...

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...

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?

## Implicit diff & Taylor series...

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.

## Slow and fast...

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

## evaluation of a sum and the zeta functio...

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: