Assignment No: 13 [ Fourier Series Transformation of Exp(2x). Submitted
By- Shashank Mishra, Date: 17.02.2014.
In[38]:= gg@x_D = Piecewise@88-Exp@-2 xD, -1 £ x £ 0<, 8Exp@2 xD, 0 £ x £ 1<<D
Out[38]=
-ã-2 x -1 £ x £ 0
ã2 x 0 £ x £ 1
0 True
In[55]:= plt@1D = Plot@gg@xD, 8x, -1, 1<,
PlotStyle ® RGBColor@1.`, 0.41000000000000003`, 0.18`D, AxesOrigin ® 80, 0<D
Out[55]=
-1.0 -0.5 0.5 1.0
-6
-4
-2
2
4
6
In[49]:= a@0D = 0.5 * NIntegrate@gg@xD, 8x, -1, 1<, AccuracyGoal ® 2, MaxRecursion ® 6D
Out[49]= 2.22045 ´ 10-16
In[200]:= hh@x_D = Sum@Evaluate@NIntegrate@gg@tD Sin@m Pi tD, 8t, -1, 1<,
AccuracyGoal ® 2, MaxRecursion ® 6DD Sin@m Pi xD, 8m, 0, 5<D;
In[201]:= plt@2D = Plot@hh@xD, 8x, -1, 1<D;
In[202]:= Show@plt@1D, plt@2DD
Out[202]=
-1.0 -0.5 0.5 1.0
-6
-4
-2
2
4
6
The error of approximating the given function using Fourier series expansion is too high. Rea-
son being since we are defining Exp[2x] as an odd function, integral value for the contant term is
zero since odd function integral over range -x to x is always zero.
Let us Solve the Half Range Fourier series expansion of Exp[2x] using cosine series and plot
them to examine the error.
The error of approximating the given function using Fourier series expansion is too high. Rea-
son being since we are defining Exp[2x] as an odd function, integral value for the contant term is
zero since odd function integral over range -x to x is always zero.
Let us Solve the Half Range Fourier series expansion of Exp[2x] using cosine series and plot
them to examine the error.
a@1D = 2 * Integrate@Exp@2 xD, 8x, 0, 1<D
Out[63]= -1 + ã2
In[100]:= an = Sum@ 2 * NIntegrate@Exp@2 xD Cos@n Pi xD, 8x, 0, 1<D Cos@ n Pi xD, 8n, 1, 5<D;
In[96]:= bn = Sum@ NIntegrate@Exp@2 xD Sin@n Pi xD, 8x, 0, 1<D Sin@ n Pi xD, 8n, 1, 5<D;
In[111]:= plt@3D = Plot@an + Ha@1D � 2L, 8x, 0, 1<, PlotStyle ® 8Red, Dashed<D
Out[111]=
0.2 0.4 0.6 0.8 1.0
2
3
4
5
6
In[112]:= plt@4D = Plot@Exp@2 xD, 8x, 0, 1<, PlotRange ® AllD;
2 Assignment 13.nb
In[113]:= Show@plt@4D, plt@3DD
Out[113]=
0.2 0.4 0.6 0.8 1.0
2
3
4
5
6
7
Therefore, we can say that cosine half series expansion of exp[2x] is aa much better approxima-
tion.
if we do the even function extension of exp[2x]
ã-2 x -1 £ x £ 0
ã2 x 0 £ x £ 1
0 True
and use only cosine terms of
Fourier series, then the coefficients of cosine series are all approaching towards to zero.
In[114]:= kk@x_D = Piecewise@88Exp@-2 xD, -1 £ x £ 0<, 8Exp@2 xD, 0 £ x £ 1<<D
Out[114]=
ã-2 x -1 £ x £ 0
ã2 x 0 £ x £ 1
0 True
In[220]:= plt@5D = PlotAPiecewiseA99ã-2 x
, -1 £ x £ 0=, 9ã2 x
, 0 £ x £ 1==, 0E, 8x, 0, 1<E
Out[220]=
0.2 0.4 0.6 0.8 1.0
2
3
4
5
6
7
Assignment 13.nb 3
In[119]:= l@0D = 0.5 * NIntegrate@kk@xD, 8x, -1, 1<, AccuracyGoal ® 2, MaxRecursion ® 6DOut[119]= 3.19453
In[213]:= hh2@x_D = Sum@2 * NIntegrate@gg@tD Cos@m Pi tD, 8t, 0, 1<D Cos@m Pi xD, 8m, 0, 5<D
The integration x->{-1,1} is reduced to 2* integration {0,1} becaue Cos() is an even fucntion and I
have extended my exp(2x) as an even function as well, therefore their product is also an Even
function.
Out[213]= 6.38906 - 2.41941 Cos@Π xD + 0.587791 Cos@2 Π xD -
0.361494 Cos@3 Π xD + 0.157839 Cos@4 Π xD - 0.133829 Cos@5 Π xD
In[210]:= plt@7D = [email protected] - 2.41941 Cos@Π xD + 0.587791 Cos@2 Π xD - 0.361494 Cos@3 Π xD +
0.157839 Cos@4 Π xD - 0.133829 Cos@5 Π xD, 8x, 0, 1<, PlotStyle ® 8Red, Dashed<D
Out[210]=
0.2 0.4 0.6 0.8 1.0
5
6
7
8
9
10
In[222]:= plt@6D = [email protected] - 2.41941 Cos@Π xD + 0.587791 Cos@2 Π xD - 0.361494 Cos@3 Π xD +
0.157839 Cos@4 Π xD - 0.133829 Cos@5 Π xD, 8x, 0, 1<, PlotStyle ® 8Red, Dashed<D
Out[222]=
0.2 0.4 0.6 0.8 1.0
2
3
4
5
6
4 Assignment 13.nb
In[223]:= Show@plt@6D, plt@5D, plt@7D , PlotRange ® AllD
Out[223]=
0.2 0.4 0.6 0.8 1.0
2
4
6
8
10
As Evident from the graph, dashed curve is of fourier series transformation using even function exten-
sion without using constant term. The graph is parallel to original graph Exp(2x). But the second graph
is an excellent approximation of the given function
In[269]:= a@100D = ReadList@"\\\\clusterfs.ceas1.uc.edu\\students\\mishrash\\Desktop\\plotdata.txt",
8Number, Number<D;
In[290]:= Maker@pts_D := Fit@pts, 81, x<, xD
In[291]:= Maker@a@100DDOut[291]= 1296.04 - 0.508211 x
In[292]:= [email protected] - 0.508211 x, 8x, -5, 5<D
Out[292]=
-4 -2 0 2 4
1294
1295
1296
1297
1298
Now we need to convert this into a piecewise function.
Assignment 13.nb 5