stem-plot difference of equations

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Tony
Tony 2013 年 12 月 16 日
コメント済み: Walter Roberson 2013 年 12 月 31 日
Here is the question
Find numerically and stem-plot the solutions of the difference equations [1<=k<=10]
a. y(k+1) - 3*k*y(k) = 0
b. y(y+1) - k*y(k) = 1
assume that y(1)=1
not sure if i am looking at this properly but here is what i was thinking. a and b do not have the same y values after 1 since in order to get 0 for the first one y(2) needs to be 3 while for equation b y(2) needs to be 2. So solve them differently and then plot? i am not sure which would be my x and y for stem plot as well. This looks simple but its confusing me not sure if am over thinking.
Thank you for your time
  3 件のコメント
Tony
Tony 2013 年 12 月 16 日
that is the main question sorry if it was not but i am not sure if its suppose to be answered separate or together
Walter Roberson
Walter Roberson 2013 年 12 月 24 日
Are you sure that (b) should have y(y+1) on the left side, and not y(k+1) ??

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採用された回答

Walter Roberson
Walter Roberson 2013 年 12 月 16 日
Once you have the set of values x1 with y1, and x2 with y2, you can use interp1() to interpolate each set of data over a common set of times; then you can stem().
For example,
y1_interp = interp1(x1, y1, 1:10)
y2_interp = interp1(x2, y2, 1:10)
plot(1:10, y1_interp, 1:10, y2_interp)
  4 件のコメント
Tony
Tony 2013 年 12 月 31 日
編集済み: Tony 2013 年 12 月 31 日
edit actually doesn't seem to display properly with stem
second edit - its good thank you happy holiday
Walter Roberson
Walter Roberson 2013 年 12 月 31 日
At the command line give the command
format long g

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その他の回答 (1 件)

Youssef  Khmou
Youssef Khmou 2013 年 12 月 24 日
Tony, try this iterative method :
y1=zeros(10,1);
y2=zeros(10,1);
y1(1)=1;y2(1)=1; % initial conditions
for t=1:9
y1(t+1)=3*t*y1(t);
y2(t+1)=1+t*y2(t);
end
k=1:10;
figure, stem(k,y1), hold on,stem(k,y2,'r')

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