Given the series of numbers: {1.,0.612547,0.466856,0.375214,0.311459,0.264691,0.229088,0.201197,0.17884,0.160573,0.145406,0.132639,0.121764,0.112404,0.104273,0.0971538,0.0908741,0.0852989,0.08032,0.0758497,0.0718166,0.0681615,0.0648355,0.0617975,0.0590128

Given the series of numbers:
{1.,0.612547,0.466856,0.375214,0.311459,0.264691,0.229088,0.201197,0.17884,0.160573,0.145406,0.132639,0.121764,0.112404,0.104273,0.0971538,0.0908741,0.0852989,0.08032,0.0758497,0.0718166,0.0681615,0.0648355,0.0617975,0.0590128,0.056452,0.05409,0.0519052,0.0498791,0.0479955,0.0462403,0.0446012,0.0430675,0.0416295,0.0402789,0.0390081,0.0378105,0.0366801,0.0356115,0.0346001,0.0336414,0.0327315}

What is the 100th term to 12-decimal places? How is this sequence made?

Expert Solution

Solution:

%plot of the given data and fitted curve in matlab

close all
clear all

y=[1,0.612547,0.466856,0.375214,0.311459,0.264691,0.229088,...
0.201197,0.17884,0.160573,0.145406,0.132639,0.121764,0.112404,...
0.104273,0.0971538,0.0908741,0.0852989,0.08032,0.0758497,0.0718166,0.0681615,0.0648355,0.0617975,0.0590128,0.056452,0.05409,...
0.0519052,0.0498791,0.0479955,0.0462403,0.0446012,0.0430675,...
0.0416295,0.0402789,0.0390081,0.0378105,0.0366801,0.0356115,...
0.0346001,0.0336414,0.0327315];
x=1:length(y);
p = polyfit(x,y,5) %polynomial of degree 5 fit of the data
y1 = polyval(p,x);
figure
plot(x,y,'o')
hold on
plot(x,y1)
hold off

%output

p =

 Columns 1 through 5:

  -0.00000019943   0.00002447160  -0.00113191903   0.02452586379  -0.25205830228

 Column 6:

   1.10494242333

hence the general term of the given sequence

substituing n =100 to get 100th term of the sequence is given by

 -457.9455002640371

Colby Messinger answered 2 months ago

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