How to form a function using 'for loop' without 'sym' command

1 次查看(过去 30 天)
i tried like this
m = 150;
Aos = 40;
nAi = 0;
for i=1:2:m
nAi = @(phi)nAi+cos(i*phi);
end
nA = vpa((nAi+Aos), 4)
Iam getting this error
Operator '+' is not supported for operands of type 'function_handle'.
Error in ll (line 8)
nA = vpa(nAi+Aos, 4)
  2 个评论
Jan
Jan 2021-9-6
You did not mention, what you try to achieve. All we see is the failing code. This is not enough information to guess, what you want to get instead.
Bathala Teja
Bathala Teja 2021-9-7
Its clearly there. i want to form nA interms of phi.
But i want to evalute without sym. Any idea.....

请先登录,再进行评论。

回答(2 个)

Abolfazl Chaman Motlagh
Assuming you want nA be function of phi, you can use symsum for summuation :
m = 150;
Aos = 40;
syms phi k
nAi =@(phi) symsum(cos(2*k*phi),k,1,m/2);
nA =@(phi) vpa((nAi(phi)+Aos), 4);
% example
nA(pi)
ans = 
115.0

Walter Roberson
Walter Roberson 2021-9-6
m = 150;
Aos = 40;
nAi = @(phi) zeros(size(phi));
for i=1:2:m
nAi = @(phi)nAi(phi)+cos(i*phi);
end
nA = @(phi) nAi(phi)+Aos
nA = function_handle with value:
@(phi)nAi(phi)+Aos
You asked to avoid sym, so it seems likely to me you want to avoid using vpa() as well. But if using a final vpa is okay, then
vpa(nA, 4)
ans = 
  2 个评论
Bathala Teja
Bathala Teja 2021-9-7
编辑:Bathala Teja 2021-9-7
How to show result like this(ans which you shared) without vpa??
And you got result in the form of phi(means symbol), how to get symbol in place of phi??
Walter Roberson
Walter Roberson 2021-9-7
With vpa(), the symbol for phi appears if you are using LiveScript, but not for traditional .m files.
To get the result without using vpa:
m = 150;
Aos = 40;
nA = str2func("@(phi) " + strjoin(compose("cos(%d*theta)", 1:2:m), " + ") + " + " + string(Aos))
nA = function_handle with value:
@(phi)cos(1*theta)+cos(3*theta)+cos(5*theta)+cos(7*theta)+cos(9*theta)+cos(11*theta)+cos(13*theta)+cos(15*theta)+cos(17*theta)+cos(19*theta)+cos(21*theta)+cos(23*theta)+cos(25*theta)+cos(27*theta)+cos(29*theta)+cos(31*theta)+cos(33*theta)+cos(35*theta)+cos(37*theta)+cos(39*theta)+cos(41*theta)+cos(43*theta)+cos(45*theta)+cos(47*theta)+cos(49*theta)+cos(51*theta)+cos(53*theta)+cos(55*theta)+cos(57*theta)+cos(59*theta)+cos(61*theta)+cos(63*theta)+cos(65*theta)+cos(67*theta)+cos(69*theta)+cos(71*theta)+cos(73*theta)+cos(75*theta)+cos(77*theta)+cos(79*theta)+cos(81*theta)+cos(83*theta)+cos(85*theta)+cos(87*theta)+cos(89*theta)+cos(91*theta)+cos(93*theta)+cos(95*theta)+cos(97*theta)+cos(99*theta)+cos(101*theta)+cos(103*theta)+cos(105*theta)+cos(107*theta)+cos(109*theta)+cos(111*theta)+cos(113*theta)+cos(115*theta)+cos(117*theta)+cos(119*theta)+cos(121*theta)+cos(123*theta)+cos(125*theta)+cos(127*theta)+cos(129*theta)+cos(131*theta)+cos(133*theta)+cos(135*theta)+cos(137*theta)+cos(139*theta)+cos(141*theta)+cos(143*theta)+cos(145*theta)+cos(147*theta)+cos(149*theta)+40

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Calculus 的更多信息

产品


版本

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by