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Unity University
Department of civil and mining engineering
Course: Numerical and computational method
Assignment solution
Done by:
Name ID
1. Ephrem beyene uu57397R
2. Eden Amare uu57096R
3. Adisu getnet uu57725R
4. Nebiyu Dawit uu56428R
Submitted to: binyam(MSC)
Submission date: 25/11/2014
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1.
%this is a source code for taylor series of cos(x)
x=input('please input x in radians ');
sum=0;
% ea is relative approximation eror
ea=100;
%es is the stoping criteria
es=input('please input stoping criteria ');
k=0;
while (ea>es)
terms=((-1)^k)*(x^(2*k))/factorial(2*k);
sum=sum+terms;
%et is the relative true error
et=abs(((cos(x)-sum)/cos(x))*100);
ea=abs((terms/sum)*100);
k=k+1;
end
disp('the value of cos x obtained through taylor series is')
disp(sum)
disp('the relative true error is')
disp(et)
disp('the relative approximation error is')
disp(ea)
B. (direct copy paste from math lab)
Please input x in radians ..pi/3
Please input stopping criteria .0.5
The relative true error is
0.0071
The relative approximation error is
0.3664
The value of cos x obtained through Taylor series is
0.5000
2. f(x)=ln (x) given xi=1
Now lets come up with ln(x) nth derivative.
f(x)=ln(x)..f(1)=0
f(x)=1/xf(1)=1
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f(x)=-1/x2..f(1)=-1
f(x)=2/x3..f(1)=2
f(4)(x)=-6/x4f(4)(1)=-6
And the derivative continues like this.
(=0 ()( 0))/()! Since this is the basic Taylor series form we
can formulate ln x in
respective manner.
= (0)( 1)0 + 1( 1)1 1( 1)2
2! +
2( 1)3
3!
6( 1)4
4!+
24( 1)5
5!
120( 1)6
6! .
= 0 + ( 1) ( 1)2
2+
( 1)3
3
( 1)4
4+
( 1)5
5
( 6)6
6 ..
This can be generalized by the summation notation
= ()( )
=
Notice that k starts from one. Its so because the zeroth term is
zero and it has no effect on the final
summation.
Now the requested mathlab code looks as such;
x=input('please input number ')
Sum=0;
For k=1:1:100
terms(k+1)=(((((-1)^(k-1))*(x-1)^k))/k);%calculates terms of
taylor series for ln x.
sum=sum+terms(k+1);%calculates the value of lnx at each term
increment.
et=abs(((log(x)-sum)/log(x))*100);%calculates relative true
error
ea=abs((terms(k+1)/sum)*100);%calculates relative
approximation error
End
disp(sum)
disp(ea)
disp(et)
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Analysis of its convergence;
X values Taylor approximation Relative true error
1 0 NaN(0)
2 0.6926 0.0721
3 -8.4227e+27 7.6667e+29
4 -3.8556e+45 2.7812e+47
5 -1.2830e+58 7.9715e+59
As the table shows after x=2 the series diverges. This can be
mathematically shown by ratio test for
series;
() =(1)1 ( 1)
And
( + 1) =(1)( 1)+1
+ 1
lim
|( + 1)/() |= lim
(1)+1 (1)+1
+1*
(1)(1)
= lim
+1( 1)
=|x-1|
And for the series to converge, according to ratio test, the
term |x-1| must be less than one.by
this its interval of convergence become; 0
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et= 0.0015
The result is very accurate and it could be trusted to four
significant figures.
3. f= @ (x) (x.^3)-(2*x)-5; % this defines the function xl=0;
xu=3; ea=100;%ea is relative approximation error es=0.5;
xr=0.714285714; i=0; fprintf(' i xl xr xu f(xr)
ea(percentage) \n')
fprintf('=========================================================
\n') while(ea>es) xold=xr; %holds the old value of xr for
approximation error calculation
if f(xl)*f(xr)
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2 1.75290108 1.95638780 3.00000000 -1.42479279 10.40114432
3 1.95638780 2.04172202 3.00000000 -0.57226277 4.17952202
4 2.04172202 2.07481267 3.00000000 -0.21787301 1.59487371
5 2.07481267 2.08724175 3.00000000 -0.08125173 0.59547902
6 2.08724175 2.09185353 3.00000000 -0.03006728 0.22046378
The root is approximately
2.0919
The value of f(xr) equals
-0.0301
The relative approximation error is
0.2205
