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From: Steen Schmidt on 14 Feb 2007 10:14
reth wrote:

> Are you married, Helen? I bet you're not and never will be. :)
> reth

What in Helens post warranted such a remark? Who is the troll here...

Cheers,
Steen
From: Wes on 14 Feb 2007 13:01
In all fairness, the 89 can also numerically search for complex roots.

cZeros(ln(x)+sin(x)+x^2,x=-1-i)


-wes

From: Marcin Witek on 14 Feb 2007 16:54
Steen Schmidt wrote:
> Solve() do not provide a way to supply a starting guess, so you're
> using nSolve() for your timings.

You are not right. It's as easy as:

solve(ln(x)+sin(x)+x^2=0, x=0)

or

solve(ln(x)+sin(x)+x^2=0, x=1)

You may try it on your own.

> Since you didn't know that

I really know what am i doing.

Wit



From: Steen Schmidt on 15 Feb 2007 06:57
Marcin Witek wrote:

> > Solve() do not provide a way to supply a starting guess, so you're
> > using nSolve() for your timings.
>
> You are not right. It's as easy as:
>
> solve(ln(x)+sin(x)+x^2=0, x=0)
>
> or
>
> solve(ln(x)+sin(x)+x^2=0, x=1)
>
> You may try it on your own.
>
> > Since you didn't know that
>
> I really know what am i doing.

Sorry, it's me who isn't clear enough; when you supply a guess to
Solve(), you jump straight ahead and use the numeric solver. So when
supplying a guess you might as well just use nSolve() from the start.

Cheers,
Steen
From: Marcin Witek on 15 Feb 2007 13:55
Steen Schmidt wrote:
> Sorry, it's me who isn't clear enough; when you supply a guess to
> Solve(), you jump straight ahead and use the numeric solver. So when
> supplying a guess you might as well just use nSolve() from the start.

OK, it seems possible. But why there is difference in execution time
between solve and nsolve? I have checked this couple of times, and always
get the same results: about 9 seconds for solve and 6 seconds for
nsolve...

I am doing something like this:

Input: startTmr()->t:solve(ln(x)+sin(x)+x^2=0, x=0): checkTmr(t)
Output: 9

Input: startTmr()->t:nsolve(ln(x)+sin(x)+x^2=0, x=0): checkTmr(t)
Output: 6

3 seconds spent in parsing? ;)

Wit
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