User:EGM6341.S11.team5.cavalcanti/HW3

= Problem 5 - Proof of expression of the error, $$e^{(n+1)}(x)$$ = From (meeting 16 page 3)

Given
Expression of the error such that: $$\displaystyle e^{(n+1)}(x)=f^{(n+1)}(x)-0$$.

Find
Show that: $$\displaystyle e^{(n+1)}(x)=f^{(n+1)}(x)$$.

Solution
''' We solved this problem on our own. '''

The error for the Lagrange Interpolation can be found using: $$e^{L}_{n}(x)=f(x)-f_{n}(x)$$ For simplicity, we will rewrite the equation as follows: $$\displaystyle e(x)=f(x)-f_{n}(x)$$ Taking the first derivative, $$e^{(1)}(x)=f^{(1)}(x)-f_{n}^{1}(x)$$ Taking the derivative again, $$e^{(2)}(x)=f^{(2)}(x)-f_{n}^{2}(x)$$ . . . Similarly, $$e^{(n)}(x)=f^{(n)}(x)-f_{n}^{1}(x)$$ $$e^{(n+1)}(x)=f^{(n+1)}(x)-f_{n}^{n+1}(x)$$ We know that $$f_{n}(x)$$ is a polynomial of degree n,So $$f_{n}^{(n+1)}=0$$.
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" align="center" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" align="center" |

$$\therefore e^{(n+1)}(x)=f^{(n+1)}(x)$$


 * }
 * }
 * }

Author and proof-reader
[Author] cavalcanti

[Proof-reader] Raghunathan