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Hi Maple friends.

If I have a table of values, how can Maple provide the function?

For example:

x=10,15,20,25,30,35,40

y=2.01,4.87,6.50,9.875,14.00,18.90,24.40

How can Maple find out the y function with respect to x? 

Thanks in advance.

Hi,

 

I'm trying to make a piecewise function that will have period 12. That is, it repeats every 12 units across the x-axis.

 

I managed to do one cycle successfully with

plot(piecewise(0<=t and t<2, 4,2<=t and t<12,0), t=0..13);

But I'm not sure how to do it periodically. I thought about using modular arithmetic in the conditions and setting up something like

H(t)=piecewise(0<=modp1(f(t),12) and modp1(f(t),12)<2, 4,2<=modp1(f(t),12) and modp1(f(t),12)<12,0);

but this clearly doesn't work.

 

Some direction or advice would be appreciated. Thanks!

 

Hi every Body?

I will define the function f(t) in maple,how?

f(t)=1-t for 0<t<10 and t for t>10

Hi:

I will a write program in maple for plot the function f(x),while asked value n from user at first,then plot function f(x) for each value of n and finally all plots show in one figure only for different value of n,how to do it?

f(x)=1+2*(1+x)^n,x=0..5

Hi:

how i can plot the function in terms of different values of n in a figure:

f(x)=1+2*(1+x)^n

n=1,2,3,4

x-axis->x

y-axis->f(x)

 

Hi:

how i can plot the function f(x) in term of 1/x in maple?for example i have the function f(x)=x^2 and x=0..1,now i plot f(x) in term of 1/x.

Hi all

Assume that we have construct new orthogonal Hybrid function of block pulse and bernstein poly nomials as follow:

and assume that we want to approximate a function as follows:

 

how can we do this with maple????Indeed we want to optimize using this hybrid functions

note that the degree of bernstein polynomials is fix or should be fixed...and

regards

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I have a rank 1 array M of 1000 values.

I want to apply a function f on each value of M and its location giving,

[f(1,M[1]), f(2,M[2]), ... , f(1000,M[1000])]

is it possible to get this using map or map2 or map[n] or maptype (without using seq since its slowing down computation).

inotherwords can i access the member location inside a map evaluation?

Hi All

Assume that we have:

and the hybrid function with block pulses with the following form:

if we want to introduce this form to maple so that we can do:

then how can we do this????

especially if we want to approximate t or t^2 or sin(3*t) by mentioned form, how maple can help us?

 

thanks a lot for coming answers 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

We know that the set f:={(1,2),(2,a),(3,b)} can introduce a function from {1,2,3} to {2,a,b}. I want Maple to know f as a function. Is this job possible at Maple? I thought to find the Cartesian product of above latter sets and then try to select "f" as one of its subset but this did not help me to force Maple to take "f" as a function. Indeed, I want to work with this kind of function (like plotting and doing f+g, f-g, fog for two functions for example).

Thanks for your time and help.

 

Can someone please advise me how to solve the following for 'beta'. Solve function is not able to do that, or at least I dont know how.

-9999990000000000000000*cos(166*beta)*sinh(166*beta)*cosh(88*beta)^2-9999990000000000000000*cos(88*beta)^2*sin(166*beta)*cosh(166*beta)+9999990000000000000000*sinh(166*beta)*cos(88*beta)^2*cos(166*beta)+9999990000000000000000*cosh(88*beta)^2*cosh(166*beta)*sin(166*beta)+10000010000000000000000*cos(166*beta)*sinh(166*beta)+10000010000000000000000*sin(166*beta)*cosh(166*beta)+9999990000000000000000*sinh(88*beta)*cos(166*beta)*cosh(166*beta)*cosh(88*beta)-9999990000000000000000*sinh(88*beta)*sin(166*beta)*sinh(166*beta)*cosh(88*beta)+9999990000000000000000*sin(88*beta)*cos(88*beta)*sinh(166*beta)*sin(166*beta)+9999990000000000000000*cos(88*beta)*cos(166*beta)*sin(88*beta)*cosh(166*beta)-9980010000000000000000*cosh(88*beta)^2*sinh(166*beta)*cos(88*beta)^2*cos(166*beta)-9980010000000000000000*cosh(88*beta)^2*cosh(166*beta)*sin(166*beta)*cos(88*beta)^2+9980010000000000000000*sinh(88*beta)*cos(88*beta)^2*sin(166*beta)*sinh(166*beta)*cosh(88*beta)-9980010000000000000000*cos(88*beta)*cosh(88*beta)^2*sin(88*beta)*sin(166*beta)*sinh(166*beta)+9980010000000000000000*sinh(88*beta)*cosh(88*beta)*cosh(166*beta)*cos(88*beta)^2*cos(166*beta)+9980010000000000000000*cosh(88*beta)^2*cosh(166*beta)*cos(88*beta)*sin(88*beta)*cos(166*beta)-9980010000000000000000*cos(88*beta)*sinh(88*beta)*cos(166*beta)*sin(88*beta)*sinh(166*beta)*cosh(88*beta)+9980010000000000000000*cos(88*beta)*cosh(88*beta)*sin(88*beta)*sin(166*beta)*cosh(166*beta)*sinh(88*beta)=0

Hello those in Mapleprimes,

 

What I want to know is whether this is possible or not, and if possible, how should I write a code?

 

The following code works properly:

 

U:=(x,y)->(x^theta+y^theta+X)^(1/theta);

diff(U(x,y),x)/diff(U(x,y),y)=p/q;simplify(%);

But, what I want to ask is this. As for the part of simplify(%), I want to do it with a way which has me 

feel more being from the former to latter.

That is, if ,for example, "diff(U(x,y),x)/diff(U(x,y),y)=p/q;@simplify;" works, it is better to me, though this does not work.

As the second part, @simplify, receives the result of the first part"diff(U(x,y),x)/diff(U(x,y),y)", it seems more 

natural to me than to write simplify(%).

 

Can't I do this, in a meaning, reversal of operator to argument?

 

taro

Hi every body:

i will plot this function in maple in domain x=0..1,but when use the plot commond i see this warning:

Warning, unable to evaluate the function to numeric values in the region; complex values were detected

how can I plot it?

eq := 4.728139762*10^(-14)*sqrt((-5.905404581*10^83*x^4-1.382542004*10^88*x^2+2.271592177*10^92+1.631888578*10^79*x^6+2.795756904*10^23*sqrt(-1.317535223*10^121*x^8+6.370084310*10^125*x^6-7.852926774*10^129*x^4+2.223707894*10^132*x^2-1.592721566*10^134))^(1/3)*(-(1.114348319*10^53*I)*x^4-6.433693022*10^52*x^4+(2.688368867*10^57*I)*x^2+1.552130489*10^57*x^2+5.072945110*10^26*x^2*(-5.905404581*10^83*x^4-1.382542004*10^88*x^2+2.271592177*10^92+1.631888578*10^79*x^6+2.795756904*10^23*sqrt(-1.317535223*10^121*x^8+6.370084310*10^125*x^6-7.852926774*10^129*x^4+2.223707894*10^132*x^2-1.592721566*10^134))^(1/3)-3.723246850*10^61-6.448852713*10^61*I+(1.732050808*I)*(-5.905404581*10^83*x^4-1.382542004*10^88*x^2+2.271592177*10^92+1.631888578*10^79*x^6+2.795756904*10^23*sqrt(-1.317535223*10^121*x^8+6.370084310*10^125*x^6-7.852926774*10^129*x^4+2.223707894*10^132*x^2-1.592721566*10^134))^(2/3)+1.231339558*10^31*(-5.905404581*10^83*x^4-1.382542004*10^88*x^2+2.271592177*10^92+1.631888578*10^79*x^6+2.795756904*10^23*sqrt(-1.317535223*10^121*x^8+6.370084310*10^125*x^6-7.852926774*10^129*x^4+2.223707894*10^132*x^2-1.592721566*10^134))^(1/3)-1.*(-5.905404581*10^83*x^4-1.382542004*10^88*x^2+2.271592177*10^92+1.631888578*10^79*x^6+2.795756904*10^23*sqrt(-1.317535223*10^121*x^8+6.370084310*10^125*x^6-7.852926774*10^129*x^4+2.223707894*10^132*x^2-1.592721566*10^134))^(2/3)))/(-5.905404581*10^83*x^4-1.382542004*10^88*x^2+2.271592177*10^92+1.631888578*10^79*x^6+2.795756904*10^23*sqrt(-1.317535223*10^121*x^8+6.370084310*10^125*x^6-7.852926774*10^129*x^4+2.223707894*10^132*x^2-1.592721566*10^134))^(1/3)

with regards...

Hi All. Hope all is well.

Assume that we have partitioned [0,a], into N equidistant subintervals and in each subinterval we have M sets of polynomials of arbitrary form[say bij(t)](a.e Taylor series, or Bernstein series,…)

for Example with N=4, M=3 and by Taylor series we have:

 

now we want to approximate a function, asy f(t), in this interval with following form:

 

If we have:

(Tau is a constant number)
then: How can  we find L and Z matrices using maple? Is it any way? (or other softwares?)

Regards

 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

 

The help documents read,

 The function unames returns an expression sequence consisting of all the active names in the current Maple session which are ``unassigned names''.

 

But what unames() returns is obviously not the contents one expects:

 

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