taylor series method expansion

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PJS KUMAR 2018-9-22

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/420289-taylor-series-method-expansion

评论： Torsten 2018-9-27

采纳的回答： PJS KUMAR

在 MATLAB Online 中打开

I wrote the following code

function yb=taylor(f,a,b,ya,n)
syms x y;
h=(b-a)/n;
y(1)=ya;
y(2)=f;
ht=h.^(0:5)./factorial(0:5)
for i=2:3
      y(i+1)=diff(y(i),x);
end
y

I got the output as follows

>> f=@(x)x^2+y^2;
>> taylor(f,0,0.8,1.5,4)
ht =
    1.0000    0.2000    0.0200    0.0013    0.0001    0.0000
y =
[ 3/2, x^2 + y(x)^2, 2*x + 2*y(x)*diff(y(x), x), 2*diff(y(x), x)^2 + 2*y(x)*diff(y(x), x, x) + 2]

Suggest me how to perform the following actions

1) I want to evaluate y at different values of x

2) Multiply ht and y elements and sum

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PJS KUMAR 2018-9-25

编辑：PJS KUMAR 2018-9-25

在 MATLAB Online 中打开

I wrote the following code for Taylor series expansion

function yb=taylor(f,a,b,ya,n)
syms x y;
h=(b-a)/n;
y(1)=ya;
y(2)=f;
ht=h.^(0:3)./factorial(0:3)
for i=2:3
      y(i+1)=diff(y(i),x);
end
s=sum(ht.*y)

I run code and got the following output

>> syms x y(x);
>> f=@(x)x^2+y^2;
>> taylor(f,1,3,0.8,4)
s =
x/4 + (y(x)*diff(y(x), x))/4 + diff(y(x), x)^2/24 + y(x)^2/2 + (y(x)*diff(y(x), x, x))/24 + x^2/2 + 101/120

Now i want to evaluate the output 's' at different values of x and y, where x is from a to b and y(a)=0.8

Walter Roberson 2018-9-25

在 MATLAB Online 中打开

y' = x^2+y^2, for the values of x = 1 (0.5) 3 with y(1)=0.8

I am having difficulty understanding the initial conditions. Could you confirm that you are referring to

syms y(x)
>> simplify(dsolve(diff(y,x) == x^2+y^2, y(1)==0.8))
ans =
piecewise(C5 ~= 0, (x*(4*besselj(-1/4, 1/2)*besselj(-3/4, x^2/2) + 4*besselj(1/4, 1/2)*besselj(3/4, x^2/2) + 5*besselj(-3/4, 1/2)*besselj(3/4, x^2/2) - 5*besselj(3/4, 1/2)*besselj(-3/4, x^2/2)))/(4*besselj(1/4, 1/2)*besselj(-1/4, x^2/2) - 4*besselj(-1/4, 1/2)*besselj(1/4, x^2/2) + 5*besselj(-3/4, 1/2)*besselj(-1/4, x^2/2) + 5*besselj(3/4, 1/2)*besselj(1/4, x^2/2)))

and you want taylor expansion of that without having solved the differential equation ?

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采纳的回答

PJS KUMAR 2018-9-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/420289-taylor-series-method-expansion#answer_338305

在 MATLAB Online 中打开

function D = dy(x,y)
  f = x^2+y^2;
  df = 2*x+2*y*f;
  d2f = 2+2*(f^2+y*df);
  d3f = 2*(2*f*df+f*df+y*d2f);
  d4f = 2*(2*(df^2+f*d2f)+df^2+f*d2f+f*d2f+y*d3f);
  D = [f df d2f d3f d4f];
end

suggest me to change the above code as given below, so that the differentiation is calculated by using diff function, without giving the differentiation directly in the program.

function D = dy(x,y)
  f = x^2+y^2;
  df = diff(f);
  d2f = diff(df);
  ---
  ---

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PJS KUMAR 2018-9-25

在 MATLAB Online 中打开

I wrote the program as below

function main
 x = linspace(1,3,5);
 % y = zeros(numel(x));
 h = x(2)-x(1);
 y(1) = 0.8;
 for i = 1:numel(x)-1
   D = dy(x(i),y(i));
    y(i+1) = y(i)+h*D(1)+h^2/2*D(2)+h^3/6*D(3)+h^4/24*D(4)+h^5/120*D(5);
 end
[x' y'] 
end
function D = dy(xnum,ynum)
  syms x y(x)
  f = x^2+y^2;
  df = diff(f,x);
  d2f = diff(df,x);
  d3f = diff(d2f,x);
  d4f = diff(d3f,x);
  fnum = double(subs(f,[x,y],[xnum,ynum]));
  dfnum = double(subs(df,[x,y],[xnum,ynum]));
  d2fnum = double(subs(d2f,[x,y],[xnum,ynum]));
  d3fnum = double(subs(d3f,[x,y],[xnum,ynum]));
  d4fnum = double(subs(d4f,[x,y],[xnum,ynum]));
  D = [fnum,dfnum,d2fnum,d3fnum,d4fnum];
end

It is giving error as

>> main
Index exceeds matrix dimensions.
Error in sym/subs>normalize (line 193)
        Y{i} = fun2sym(Y{i},argnames(X{i}));
Error in sym/subs>mupadsubs (line 147)
[X2,Y2,symX,symY] = normalize(X,Y); %#ok
Error in sym/subs (line 135)
    G = mupadsubs(F,X,Y);
Error in main>dy (line 20)
  fnum = double(subs(f,[x,y],[xnum,ynum]));
Error in main (line 7)
   D = dy(x(i),y(i));

kindly correct the program

Walter Roberson 2018-9-26

As I tried to get you to understand before:

If you are using a symbolic variable with a particular name, then you should strongly avoid using the same variable name for a different purpose, such as storing a list of particular locations to execute at, or such as storing the results of calculating the taylor series at particular locations.

Torsten 2018-9-27

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n is undefined.
You use i as a loop variable in both nested loops.
Walter's remark.

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Walter Roberson 2018-9-22

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/420289-taylor-series-method-expansion#answer_337914

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As I already explained to you, because you have

y = [ 3/2, x^2 + y(x)^2, 2*x + 2*y(x)*diff(y(x), x), 2*diff(y(x), x)^2 + 2*y(x)*diff(y(x), x, x) + 2]

then the y on the left side refers to the same thing as the y on the right side, and so y(x) on the right side signifies array indexing. Your y vector is 5 elements long, so the only valid values of x for this would be 1, 2, 3, 4, or 5. Each y(x) would resolve to a numeric scalar value, and diff() of a numeric scalar value is empty. Any operation involving an empty array returns an empty array, so all of the entries except the first two are going to disappear, leaving you only [3/2, x^2 + y(x)^2] to work with for preliminary consistency.

Now, with that subset, can x = 1 be made self-consistent? That would require that y(1) be 3/2, which seems plausible. With the subset, can x = 2 be made self-consistent? That would require that y(2) = x^2 + y(x)^2 = 2^2 + y(2)^2 . Rewriting as z = 4 + z^2 we can see that has only complex roots 1/2-1i*sqrt(15)*(1/2), 1/2+1i*sqrt(15)*(1/2) . Is that acceptable, to force y(2) to be complex valued?