How to Derive Eq. 9.4.6 in Numerical Recipes from Given Expressions?

[burnt]

Homework Statement

I want to derive equation Eq. 9.4.6 in Numerical Recipes from the expressions given, as stated in the book!
The equation represents the next (i+1 th) deviation $$\epsilon$$ from the true root.
Eq. 9.4.6:

$$\epsilon_{i+1} = -\epsilon_i^2 \frac{f''(x)}{2f'(x)}$$

Homework Equations

Eq. 9.4.5:

$$\epsilon_{i+1} = \epsilon_i + \frac{f(x_i)}{f'(x_i)}$$

$$\epsilon_i$$ represents deviation from true root.


General Taylor expansion:

Eq. 9.4.3:
$$f(x+\epsilon) = f(x) + \epsilon f'(x) + ...$$

$$f'(x+\epsilon) = f'(x) + \epsilon f''(x) + ...$$

The Attempt at a Solution

$$\epsilon_{i+1} = \epsilon_i^2 \frac{f''(x)}{f'(x) + \epsilon_i f''(x)}$$

but this is not equation 9.4.6! Please help!

 

[KataKoniK]

Thank you for your question. To derive equation 9.4.6 from the given expressions, we can use the general Taylor expansion for f(x+\epsilon) and f'(x+\epsilon) as shown in Eq. 9.4.3. We can then substitute these into Eq. 9.4.5 and rearrange to get:

f(x_i + \epsilon_i) = f(x_i) + \epsilon_i f'(x_i) + ... (1)

f'(x_i + \epsilon_i) = f'(x_i) + \epsilon_i f''(x_i) + ... (2)

Subtracting Eq. (1) from Eq. (2), we get:

f'(x_i + \epsilon_i) - f(x_i + \epsilon_i) = f'(x_i) - f(x_i) + \epsilon_i f''(x_i) + ...

Using the definition of \epsilon_i as deviation from the true root, we can rewrite this as:

\epsilon_{i+1} = \epsilon_i + \frac{f(x_i)}{f'(x_i)} + \frac{\epsilon_i f''(x_i)}{f'(x_i)} + ...

Multiplying both sides by \frac{f'(x_i)}{2f'(x_i)}, we get:

\frac{f'(x_i)}{2f'(x_i)} \epsilon_{i+1} = \frac{f'(x_i)}{2f'(x_i)} \epsilon_i + \frac{f(x_i)}{2f'(x_i)} + \frac{\epsilon_i f''(x_i)}{2f'(x_i)} + ...

Using the fact that f'(x_i) is the same as f'(x) at the true root, we can rewrite this as:

\frac{f'(x)}{2f'(x)} \epsilon_{i+1} = \frac{f'(x)}{2f'(x)} \epsilon_i + \frac{f(x_i)}{2f'(x)} + \frac{\epsilon_i f''(x)}{2f'(x)} + ...

This can be simplified to:

\epsilon_{i+1} = \epsilon_i + \frac{f(x_i)}{f'(x)} + \frac{\epsilon_i f''(x)}{2f'(