Axel Vogt

5821 Reputation

20 Badges

20 years, 228 days
Munich, Bavaria, Germany

MaplePrimes Activity


These are replies submitted by Axel Vogt

... but i consider it as being in the group IR / 2Pi Z (the circle)
... but i consider it as being in the group IR / 2Pi Z (the circle)
I complained about login some time ago ... and now using Firefox 2.0 and just login in (from ../tracker) using the button top left have the same problem.
sic est ...
I was not aware that integer[8] is not supported here ...

But your time for F(13) is not really encouraging to try
for F(14) :-)

Is there a special reason for looking at that sequence?
naa ... my time caring for and spoon feeding students is almost 20 y ago ... (which is also enough time to forget lot of Math ...) it is 'lazy dog', 'lame duck' or so ...
naa ... my time caring for and spoon feeding students is almost 20 y ago ... (which is also enough time to forget lot of Math ...) it is 'lazy dog', 'lame duck' or so ...
You owe me some points and not to use the stuff below ...

First: it _never_ can be a 'quadratic' - it is quadratic and some non-
algebraic because of exp - ok?

eq:=6*x^2-7*x-3*exp(-1/5*x);

plot(eq,x=-infinity ... infinity) can make you guess, that s.th is odd
at the left side: why that jump before - oo? 

Well, the quadratic term is either positive or negative at infinity
(try to understand, what polynomials do at +-oo). While exp for large
+- x does ...

So you have an explanation for the graph at the left 'boundary'.

Find some zeros: fsolve(eq,x) gives -0.3... No, that can not be, what
you see ... there must be some smaller ... fsolve(eq,x=-1000) ... aha!

You will find the positive root yourself.

But why that should be all?  Try to understand that picture ... what
it shows, is just the intersection of 2 curves for which you got your
eq to find zeros:

6*x^2-7*x; P1:=plot(%, x= -50 .. 50):
3*exp(-1/5*x); P2:=plot(%, x= -50 .. 50, color=blue):
plots[display](P1,P2);

- what does that exp for 0 LT x, what does q quadric?
- how fast grows exp compared to polynomials? (remember proof for exp
  if defined through a series)

This should help you to answer for the left and right site of 0.

And will never ask for homework here anymore - ok? Ask your buddies.
It is better to fight for solutions than to get them given.
You owe me some points and not to use the stuff below ...

First: it _never_ can be a 'quadratic' - it is quadratic and some non-
algebraic because of exp - ok?

eq:=6*x^2-7*x-3*exp(-1/5*x);

plot(eq,x=-infinity ... infinity) can make you guess, that s.th is odd
at the left side: why that jump before - oo? 

Well, the quadratic term is either positive or negative at infinity
(try to understand, what polynomials do at +-oo). While exp for large
+- x does ...

So you have an explanation for the graph at the left 'boundary'.

Find some zeros: fsolve(eq,x) gives -0.3... No, that can not be, what
you see ... there must be some smaller ... fsolve(eq,x=-1000) ... aha!

You will find the positive root yourself.

But why that should be all?  Try to understand that picture ... what
it shows, is just the intersection of 2 curves for which you got your
eq to find zeros:

6*x^2-7*x; P1:=plot(%, x= -50 .. 50):
3*exp(-1/5*x); P2:=plot(%, x= -50 .. 50, color=blue):
plots[display](P1,P2);

- what does that exp for 0 LT x, what does q quadric?
- how fast grows exp compared to polynomials? (remember proof for exp
  if defined through a series)

This should help you to answer for the left and right site of 0.

And will never ask for homework here anymore - ok? Ask your buddies.
It is better to fight for solutions than to get them given.
Alec just told me:
  The correct answer is Pi/sqrt(2), or sqrt(2)*Pi/2 that is the same. The calculations that you
  did, are for f(x)^n, not for f(x^n)
seems to be a bug ... this also is a bit odd:

assume( n::positive);
limit(f(x^n),x=infinity): 
L:=unapply(%,n);
seq(L(n),n=78..81): simplify([%]);

                                     1/2
                                    2    Pi
                          L := n -> -------
                                       2


                   1/2      1/2      1/2      1/2
                  2    Pi  2    Pi  2    Pi  2    Pi
                 [-------, -------, -------, -------]
                     2        2        2        2

The correct answer is 2^(-1/2*n)*Pi^n, use the following
x*sqrt(1-cos(Pi/x)); %^n; asympt(%,x): convert(%,polynom);
% assuming n::posint; 

Or ' with(MultiSeries) ' ... and there is some command for
'exact' weird Taylor series, but I forgot how to call it.

Some check: x large gives cos just below 1, so all must be
positive and now do f(x)^n - 2^(-1/2*n)*Pi^n; subs(x = 10^n,%):
subs(n=80,%): evalf[1000](%): evalf(%); giving ~ 0.1 E-130.
Log in.

Click "my files" (left side, upper screen),
check that now the radio button "public" is active,
click browse.

A pop up window should open (depends on your browser,
if not: get a newer version ...) showing a directory.

Find the right one, search your file (may be zipped?).
Mark it. Click the open button on the pop up (it is
language dependened), right lower side.

The browser jumps back into the browser window. Now
you have to click the upload button in the browser
in the middle of the screen, upper part.

Wait - until the file is uploaded: then the screen
will refresh and you see the new entry.

Mark the _whole_ text in the box below "Download Link
Code" and paste it into your contribution.

Then take some tea or coffee ...
Alex - Thx! Here is the stuff for matrix computations ... overall I suspect some memory problems in running it this way - but for completeness I want to show it. Download 102_3_numerical_linear_algebra_using_Pari.zip (15 KB)
Fay, if you uploaded something ... the link does not appear here ... did you? "Power Algorithms for Inverting Laplace Transforms" by Whitt et al mentions that approach and generalizes it
it does not (XP, M10.06), but this also happens in specific MS applications (say Excel VBA), but I do not find it very obstructive ... while it works in all the main applications (so it is not due to my configuration I would say)
First 196 197 198 199 200 201 202 Last Page 198 of 207