Sum of series with differentaition

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SHARAD KUMAR UPADHYAY
SHARAD KUMAR UPADHYAY 2019 年 2 月 5 日
コメント済み: Walter Roberson 2019 年 2 月 6 日
%% I am try to sum a series with respect to k, where function f has 2*k-1 times derivative too. As below program i am using and then error occured.
sysms k t
f = zeta(2*k)*(4^k-2)/4^k*diff(t^(-1/2),t,2*k-1)
pretty(f)
F = symsum(f,k,1,10)
%% I am facing error as
%%Error using sym/diff (line 71)
%%The second and third arguments must either be variables or a variable and a nonnegative integer specifying the number of differentiations.
%%Error in main (line 13)
%%f = zeta(2*k)*(4^k-2)/4^k*diff(t^(-1/2),t,2*k-1)

回答 (1 件)

John D'Errico
John D'Errico 2019 年 2 月 5 日
There is no reason to use symsum on this.
syms t
f = sym(0);
for k = 1:10
f = f + zeta(2*k).*(4.^k-2)/4.^k*diff(t^(-1/2),t,2*k-1);
end
vpa(f,5)
ans =
- 0.41123/t^(3/2) - 1.7757/t^(7/2) - 29.105/t^(11/2) - 1051.8/t^(15/2) - 67239.0/t^(19/2) - 6.7119e6/t^(23/2) - 9.6501e8/t^(27/2) - 1.8891e11/t^(31/2) - 4.8314e13/t^(35/2) - 1.5642e16/t^(39/2)
diff cannot handle a sumbolic order of dfferentiation anyway, so a simple loop is entirely adequate.
  3 件のコメント
SHARAD KUMAR UPADHYAY
SHARAD KUMAR UPADHYAY 2019 年 2 月 6 日
%% when i am trying this
syms y a t Q
FT = sym(0 );
y= a/t
for k = 1:5
FT = FT + zeta(2*k).*(4.^k-2)/4.^k*diff(y^(-1/2),y,2*k-1 );
end
vpa(FT,5 )
chi=(y^(1/2)+FT)*t^(1/2)/Q;
vpa(chi,3)
%%Then I found the error as
%%Error using sym/diff (line 71)
%%The second and third arguments must either be variables or a variable and a nonnegative integer specifying the number of differentiations.
%%Error in Im_Xi (line 6)
%%FT = FT + zeta(2*k).*(4.^k-2)/4.^k*diff(y^(-1/2),y,2*k-1 );
Walter Roberson
Walter Roberson 2019 年 2 月 6 日
your y is an expression not a variable . you cannot differentiate with respect to an expression

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