MS-C1650 - Numerical Analysis, Lecture, 23.4.2024-31.5.2024

Transcript 02

• Page 7: ja -> and

• Page 3: In the 2nd row below the Definition, the product should range from j=0,...,n such that j is not equal to i, that is:
  

\varphi_i (x) = \prod_{j=0, j\not=i}^{n} \frac{x - x_j}{x_i - x_j}

• Page 4: 3rd row and 3rd column of the lower triangular matrix should be (x_2 - x_0)(x_2 - x_1).
• On page 4, the lower triangular matrix should be multiplied with (a_0, a_1, ..., a_n)^T and equated to (y_0, y_1, ..., y_n)^T.

• Page 1: 3rd row and 3rd column of the lower triangular matrix should be (x_2 - x_0)(x_2 - x_1).
• Page 4: The last statement is not correct. The highest order term is actually (r_{k-1} - q_{k-1}) / (x_k-x_0), where r_{k-1}=f[x_1,x_2,...,x_k]= r(x)^{(k-1)}/((k-1)! and q_{k-1}=f[x_0,x_1,...,x_{k-1}])=q(x)^{(k-1)}/((k-1)!), which can be seen by taking k derivatives of the equation p(x)=q(x)+((x-x_0)/(x_k-x_0))(r(x)-q(x)) and noting that p_k=f[x_0,x_1,...,x_k]=(p(x))^{(k)}/(k!). This argument would complete the proof and is left as an exercise (hint: Leibniz formula).

Transcript 05

• Page 4: vakio -> constant
• Page 4: Should be s''(x) instead of s''(x_0).
• Page 5: There are parentheses missing on both sides of the last equation.
• Page 6: In the 1st equation, the last summand should be (h/6) z_{i+1}.
• Page 6: In the 2nd equation, the last summand should be f_{i-1}.
• Page 6: In the 4th equation, it should be b_{n-1}.

Transcript 08

• Page 7: In the first formula it should be (2n)! and not (2n!).
  

Transcript 09

• Page 1: In Euler's method, there should also be written t_{k+1} = t_k + h.
  
• Page 4: The formula derived by Taylor and Euler should be
  
• d_{k+1} = d_k +h [f(t_k, y(t_k)) - f(t_k, y_k)] + (h^2)/2 y''(\xi_k)
  

Transcript 10

• Page 4: There is an 'm' missing in the product formula under the integral. It should read \prod_{j=0, j\not=l}^m.
• Page 5: There is a square bracket ']' missing in the end of the first formula.
• The second line of the iteration below should read:
  

y_2 = 3 y_1 - 2 y_0 = 1 + 3 \delta

• The third line of the iteration below should read:
  

y_{k} = 3 y_{k-1} - 2 y_{k-2} = 1 + (2^k - 1) \delta