Hi everyone! I need to weight a serie of values which x₁ range go from 280 to 4000 by the factors provided by another serie of values, which x₂ values go from 206.6 to 12400. In the graphs, the red curve is a listplot of the serie which y₁ values I want to weight by the y₂ values of the listplot of the serie in black. I want the resulting serie in the x₁ range, but with the y₃ values equal to y₂*y₁. How can I do that since the series have different intervals?

Serie in red=S and serie in black=t

S:=

...[417., .96392], [418., .92392], [419., .96354], [420., .88467], [421., 1.0067], [422., .99499], [423., .96531], [424., .96182], [425., .99312], [426., .96667], [427., .9355], [428., .94625], [429., .87766], [430., .70134], [431., .63779], [432., 1.0628], [433., .9905], [434., .91653], [435., 1.007], [436., 1.1061], [437., 1.1306], [438., .99368], [439., .95753], [440., 1.0993], [441., 1.0859], [442., 1.164], [443., 1.1823], [444., 1.1537], [445., 1.1992], [446., 1.0766], [447., 1.2257], [448., 1.2422], [449., 1.2409], [450., 1.2881], [451., 1.3376], [452., 1.2822], [453., 1.1854], [454., 1.273], [455., 1.2655], [456., 1.3088], [457., 1.3213], [458., 1.2946], [459., 1.2859], [460., 1.2791], [461., 1.3255], [462., 1.3392], [463., 1.3452], [464., 1.3055], [465., 1.2905], [466., 1.319], [467., 1.2616], [468., 1.3178], [469., 1.3247], [470., 1.2749], [471., 1.2975], [472., 1.3661], [473., 1.3144], [474., 1.3304], [475., 1.3755], [476., 1.3299], [477., 1.3392], [478., 1.3839], [479., 1.3586], [480., 1.3825], [481., 1.3836], [482., 1.3899], [483., 1.3742], [484., 1.3492], [485., 1.3457], [486., 1.0918], [487., 1.2235], [488., 1.3252], [489., 1.2492], [490., 1.3968], [491., 1.3435], [492., 1.2818], [493., 1.3719], [494., 1.3402], [495., 1.4238], [496., 1.3548], [497., 1.3788], [498., 1.3421], [499., 1.3429], [500., 1.3391], [501., 1.299], [502., 1.2991], [503., 1.3597], [504., 1.2682], [505., 1.3598], [506., 1.4153], [507., 1.3548], [508., 1.321], [509., 1.385], [510., 1.3497], [511., 1.3753], [512., 1.4125], [513., 1.3277], [514., 1.3003], [515., 1.3385], [516., 1.3514], [517., 1.1017], [518., 1.2605], [519., 1.2222], [520., 1.3349], [521., 1.3452], [522., 1.376], [523., 1.2976], [524., 1.3962], [525., 1.3859], [526., 1.3479], [527., 1.1795], [528., 1.3508], [529., 1.4142], [530., 1.3598], [531., 1.4348], [532., 1.4094], [533., 1.259], [534., 1.3491], [535., 1.3701], [536., 1.4292], [537., 1.3229], [538., 1.3896], [539., 1.3558], [540., 1.3096], [541., 1.2595], [542., 1.3714], [543., 1.3493], [544., 1.3971], [545., 1.3657], [546., 1.3536], [547., 1.3717], [548., 1.3331], [549., 1.3752], [550., 1.3648], [551., 1.3639], [552., 1.3923], [553., 1.3533], [554., 1.3802], [555., 1.3883], [556., 1.3651], [557., 1.3321], [558., 1.3613], [559., 1.2885], [560., 1.3118], [561., 1.3885], [562., 1.3225], [563., 1.3731], [564., 1.3466], [565., 1.3555], [566., 1.2823], [567., 1.3673], [568., 1.3554], [569., 1.3228], [570., 1.324], [571., 1.281], [572., 1.3534], [573., 1.3595], [574., 1.3527], [575., 1.3225], [576., 1.3118]...

t:=

[[206.6000, 0.9110470095e-3], [210.1000, 0.8504840202e-3], [213.8000, 0.8153191021e-3], [217.5000, 0.7517070516e-3], [221.4000, 0.7035057451e-3], [225.4000, 0.6694785180e-3], [229.6000, 0.6318423899e-3], [233.9000, 0.5994622965e-3], [238.4000, 0.5764355678e-3], [243.1000, 0.5331430203e-3], [248.0000, 0.4950946716e-3], [253.0000, 0.4422658739e-3], [258.3000, 0.4023846661e-3], [263.8000, 0.3732428357e-3], [269.5000, 0.3444825682e-3], [275.5000, 0.3307336604e-3], [281.8000, 0.3570645679e-3], [288.3000, 0.4052774535e-3], [295.2000, 0.4741877081e-3], [302.4000, 0.5345931300e-3], [310.0000, 0.6002211844e-3], [317.9000, 0.7217256785e-3], [326.3000, 0.9176207886e-3], [335.1000, 0.1271492339e-2], [344.4000, 0.1679935813e-2], [354.2000, 0.1976514731e-2], [364.7000, 0.2169413297e-2], [375.7000, 0.2420056977e-2], [387.5000, 0.2698356864e-2], [399.9000, 0.3092320191e-2], [413.3000, 0.3891714441e-2], [427.5000, 0.4875961623e-2], [442.8000, 0.6098167657e-2], [459.2000, 0.8645631738e-2], [476.9000, 0.1639763449e-1], [495.9000, 0.1611551307e-1], [516.6000, 0.1102654621e-1], [539.1000, 0.5287562153e-2], [563.6000, 0.3253578980e-2], [590.4000, 0.2176718340e-2], [619.9000, 0.1510939515e-2], [652.5000, 0.1117755662e-2], [688.8000, 0.8568924681e-3], [729.3000, 0.6693019750e-3], [774.9000, 0.5484788449e-3], [826.6000, 0.4632457834e-3], [885.6000, 0.3982461677e-3], [953.7000, 0.3505755460e-3], [1033.000, 0.3130972421e-3], [1127.000, 0.2814970511e-3], [1240.000, 0.2609345123e-3], [1305.000, 0.2500612658e-3], [1378.000, 0.2397447628e-3], [1459.000, 0.2312899569e-3], [1550.000, 0.2268792462e-3], [1653.000, 0.2197119097e-3], [1771.000, 0.2106962051e-3], [1907.000, 0.2046935228e-3], [2066.000, 0.2003490235e-3], [2254.000, 0.1974613611e-3], [2480.000, 0.1948841680e-3], [2755.000, 0.1957355188e-3], [3100.000, 0.1937027704e-3], [3542.000, 0.1867538618e-3], [4133.000, 0.1849393481e-3], [4959.000, 0.1835339010e-3], [6199.000, 0.1838769393e-3], [8266.000, 0.1869179329e-3], [12400.00, 0.1990760933e-3]]

Can anyone coax Maple to solve this reccurence relation? It seems harmless enough but Maple is struggng a bit with "hypergeomsols."

f := c -> (2*n-c)*f(c-1) - (c-1)*n*f(c-2);

f(0) := 1;

f(1) := 2*n-1;

Determine using determinants the range of values of a (if any) such that
f(x,y,z)=4x^2+y^2+2z^2+2axy-4xz+2yz
has a minimum at (0,0,0).

From the theory, I understand that if the matrix corresponding to the coefficients of the function is positive definite, the function has a local min at the point. But, how do I get the range of values of a such that f is a min? Is this equivalent to finding a such that det(A) > 0?

2.

Now modify the function to also involve a parameter b: g(x,y,z)=bx^2+2axy+by^2+4xz-2a^2yz+2bz^2. We determine conditions on a and b such that g has a minimum at (0,0,0).
By plotting each determinant (using implicitplot perhaps, we can identify the region in the (a,b) plane where g has a local minimum.

Which region corresponds to a local minimum?

Now determine region(s) in the (a,b) plane where g has a local maximum.

I don't understand this part at all..

I've got the following matrix :

A:=[<a,a-1,-b>|<a-1,a,-b>,<b,b,2a-1>] where <> are the column elements of A, a is  a real number defined on [0,1] and b^2=2a(1-a) 

a) to show A is an orthogonal matrix, I understand that I need A.Transpose(A)=Identity(3*3) but is there a way in which I can let a take a random real numbered value between 0 and 1? The rand() only returns an integer within a range. Directly multiplying A and Transpose(A) will return an expression in a, so what's the right approach?

b) from a) we can infer that A is a matrix that describes a rotation in e1,e2,e3 where these are the standard bases vectors in R3. How can I determine the rotation axis? The hint I've been given says I need to consider the Eigenvalues and eigen vectors but I don't quite understand how.

I have the function:   f(x,y) = 1/(sqrt(2*Pi)) * e^{-1/2(x^2+y^2)}

I need to take the total derivative of this function in maple, but I don't know the syntax.

When plotting a continous function that looks something like a square wave, for example like this:

f:=x->piecewise(0<x<Pi,0,Pi<x<2*Pi,Pi);
a:=0: b:=2*Pi: p:=b-a:
fp:=f(x-floor( (x-a)/p)*p);
plot(fp,x=-6*Pi..6*Pi,discont=true);

Is there some way to show dashed vertical lines at the points of discontinuity?

So i got a procedure test, she is kind of numeric, i whant to optimaze test([.5, .5, .5], 1, 3, 100, 100, true, [x, 0, 0, 0, 0])=0 by x. But optinization substitutes x like a symbol, i tryed all methods but they all do the same.

f := proc (x) options operator, arrow; abs(test([.5, .5, .5], 1, 3, 100, 100, true, [x, 0, 0, 0, 0])-.4) end proc; Minimize(f(x));
%;
Error, (in test) cannot determine if this expression is true or false: 0 < -43.0+100*x

Can i some how use optinization on such procedure?

file link  - >    primset.mw

A created an overly complicated worksheet thaI can no longer open.  It's about 17 Mb, has several interactive plots and equations in it.  Is there a way to force Maple to open a worksheet without executing it?

(Lesson learned:  split this up into separate worksheets next time.)

I wish to use closed Newton-Cotes with n=2, also known as Simpson's Rule to numerically integrate an improper integral.

If it matters the integrand is (cos(2x))/(x^1/3), integrating between x=0..1

I've tried a few different (but similar) code but to no avail. Here is some stuff I've tried:

1.

with(Student[NumericalAnalysis]):

with(Student[Calculus 1]):

Simp1 := ApproximateInt(cos(2*x)/x^(1/3), x = 0 .. 1, method = newtoncotes[2]);

This gives me an output message that says "Float(infinity)".

2.

with(Student[NumericalAnalysis]):

with(Student[Calculus 1]):

Simp2 := int(exp(-x)/sqrt(1-x), x = 0 .. 1);

This doesn't have Simpson's rule as an option.

I think I'm on the right track with my first try, since I guess it wasn't tecnically an error message, but I'm not sure how to alter the code accordingly to get a numerical value instead. Thanks for any help.

I am trying to numerically double integrate x^2+sqrt(y), with the bounds y=0..x and x=1..1.5.

Then I tried the following code:

int(int(x^2+sqrt(y),method=trapezoid,y=0..x),method=trapezoid,x=1..1.5);

I know how to write the code if instead of a 'x' in my upper limit for my integral, I had a real number, but I'm not sure how to remedy to code in order make it work. Any help would be appreciated. Thanks!

Thanks.

Download dsolve_question.mw

Hi,

I got the Real and Imaginary of an expression J1 

assume(d,real):

Gamma:=0.04:tau:=10*Pi:j:=0:

J1:=(exp((1-I*d)*Gamma*tau)-1)/((1-I*d));

J1mod:=simplify((Re(J1))^2+(Im(J1))^2): (I works here this amont is real)

################

but when I change the expression  for J1 to be

J1:=((2*e^(-2^(-j-1)*(1-I*d))-e^(-2^(-j)*(1-I*d))-1)*exp((1-I*d)*Gamma*tau)-1)/((1-I*d)):

J1mod:=simplify((Re(J1))^2+(Im(J1))^2): 

J1mod here is complex(I dont know why? it doesnt separate the real and the im )

Any comments will help

Thanks

Hello everybody,

can somebody tell me how to use subscripts in textmode? 

Thanks for your help!

Picard

OK, what am I doing wrong here?

This polynomial can be factored by almost anyone who knows algebra:

factor(x^2+y^2+2*xy);

but Maple refuses:

 x²  + y²  + 2 xy

but if I pose the question this way:

ans := expand((x+y)^2);

factor(ans);

(x + y)²

I get the expected result.  What is wrong with how I am using the factor?

If we have a piecewise continous 2*Pi-periodic function

h(t)=e^(2*t) when 0 < t < 2*Pi

How can we plot it? It should look something like this:

I.e. I want to be able to just set an interval and then it automatically plots the function with the specified periodicity.