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syms k x

clf;

%%%%%User Input %%%%%%%

a=@(k)-1.^k; % define coefficient sequence

f=@(x)symsum(a(k).*x.^k/k^2,k,1,inf); %define series

p=@(x,n)symsum(a(k).*x.^k/k^2,k,1,n); % define partial sum

n=5; % degree of partial sum

x1=-1;x2=1; % endpoints of plotting interval

%%%%%End User Input %%%%%%%

x=linspace(x1,x2);

subplot(1,2,1);

plot(x,f(x),'-.'); % plot limit function

hold on

plot(x,p(x,n)); % plot partial sum; leave default style

legend('series',['n = ', num2str(n)]);

% num2str converts n into a string

title(['Series \Sigma ',char(a(k)), '*x.^k/k^2 and Partial Sum' ]);

hold off

syms x

tpoly=taylor(f,x,'order',10); subplot(1,2,2);

% defines taylor polynomial

% expression

p=@(x)subs(tpoly);

% taylor polynomialfunction

approx_error= abs(p(pi/2)-f(pi/2)); % absolute difference in approximation

this is my code , but i am having trouble to generate the absolute difference

help please

Arnab Sen
on 25 Apr 2016

Hello Giannakos, This script outputs approx._output as NAN becasse f(pi/2) is evaluated to be NAN. The anonymous function 'f' is defined to be as

f(x)= -x -x^2/4-x^3/9-x^4/16-....x^i/i^2-... to infinite

Now, notice that the series is divergent whenever x>1, because the numerator grows exponentially whereas the growth of the denominator is polynomial. So the sum keeps on increasing as we proceed to infinite. Hence, MATLAB evaluates the sum to be NAN. So

f(pi/2)=NAN

Verify the appropriate definition of Taylor series. In Taylor series, ith term is something like x^i/i! and hence converges.

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