**How to unravel Coding a recursion in MATLAB**
This output you might be getting is appropriate, and as identified in the feedback by Mad Physicist, the recursive operate you have got ought to behave this fashion.

If you have a look at the conduct of the 2 phrases, as n will get greater the preliminary subtraction may have much less of an impact on the ten*y(n) time period. So for big n, we are able to ignore 1/n.

At massive n we then count on every step will enhance our worth by roughly a issue of 10. This is what you see in your output.

As far as writing a backward recursion. By definition you want a beginning worth, so that you would want to imagine y(30) and run the recursion backward as urged in the feedback.

**Coding a recursion in MATLAB**
This output you might be getting is appropriate, and as identified in the feedback by Mad Physicist, the recursive operate you have got ought to behave this fashion.

If you have a look at the conduct of the 2 phrases, as n will get greater the preliminary subtraction may have much less of an impact on the ten*y(n) time period. So for big n, we are able to ignore 1/n.

At massive n we then count on every step will enhance our worth by roughly a issue of 10. This is what you see in your output.

As far as writing a backward recursion. By definition you want a beginning worth, so that you would want to imagine y(30) and run the recursion backward as urged in the feedback.

This output you might be getting is appropriate, and as identified in the feedback by Mad Physicist, the recursive operate you have got ought to behave this fashion.

If you have a look at the conduct of the 2 phrases, as n will get greater the preliminary subtraction may have much less of an impact on the ten*y(n) time period. So for big n, we are able to ignore 1/n.

At massive n we then count on every step will enhance our worth by roughly a issue of 10. This is what you see in your output.

As far as writing a backward recursion. By definition you want a beginning worth, so that you would want to imagine y(30) and run the recursion backward as urged in the feedback.

So, I used to be capable of reply by personal query. The code wanted would appear like this:

```
% This operate calculates the worth of y20 with a assure to have an
% absolute error lower than 10^-5
% The yn1 chosen to be excessive sufficient to ensure that is n1 = 25
% Returns the worth of y(20)
operate [x]= formulation(ok)
% RECURSION APPROXIMATION
y(ok) = 0;
n = ok:-1:20;
y(n-1) = (1./10)*(1./n - y(n));
x = y(20);
```

```
% FURTHER: I wanted to ensure y20 to have <= 10^-5 magnitude error
% I made up my mind n=25 could be my place to begin, approximating y25=0 and dealing
% backwards to n=20 as I did above.
% y(n-1)=1/10(1/n-yn) “exact solution”
% (yn-1)*=1/10(1/n-(yn)*) “approximate solution with error”
% y(n-1)-(y(n-1))*=1/10(1/n-yn)-1/10(1/n-(yn)*) calculating the error
% = 1/10((yn)*-yn)
% So,
% E(n-1)=1/10(En)
% E(n-2)=1/100(E(n-1))
% E(n-3)=1/1000(E(n-2))
% E(n-4)=1/10000(E(n-3))
% E(n-5)=1/100000(E(n-4)) ⇒ 10^(-5)
% En20=(10^-5)En25
% Therefore, if we begin with n1=25, it ensures that y20 may have 10^-5 magnitude of % the preliminary propagating error.
```