Let f(x) = x²-6. Use Newton's method Pn=Pn-1-Pn-1/pn-1 with po=3 to find p3

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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This is a two part question please help. Part a is one image, b is the second image
Let \( f(x) = x^2 - 6 \). Use Newton's method 

\[ p_n = p_{n-1} - \frac{p_{n-1}}{p'_{n-1}} \]

with \( p_0 = 3 \) to find \( p_3 \).
Transcribed Image Text:Let \( f(x) = x^2 - 6 \). Use Newton's method \[ p_n = p_{n-1} - \frac{p_{n-1}}{p'_{n-1}} \] with \( p_0 = 3 \) to find \( p_3 \).
The question presented is: "Which of these is a fixed point equation for \( f(x) = x^4 + 2x^2 - x - 3 \)." 

A fixed point equation is derived from setting \( f(x) = x \).

Mathematically, this means finding \( x \) such that:

\[ x = x^4 + 2x^2 - x - 3 \]

Rearranging the equation, we have:

\[ 0 = x^4 + 2x^2 - 2x - 3 \]

To explore fixed points further, you can rewrite the function as:

\[ x = g(x) \]

where \( g(x) \) is any transformation of \( f(x) \) that equals \( x \) when solved. Here, \( f(x) \) and the related transformations could be analyzed to identify \( g(x) \).
Transcribed Image Text:The question presented is: "Which of these is a fixed point equation for \( f(x) = x^4 + 2x^2 - x - 3 \)." A fixed point equation is derived from setting \( f(x) = x \). Mathematically, this means finding \( x \) such that: \[ x = x^4 + 2x^2 - x - 3 \] Rearranging the equation, we have: \[ 0 = x^4 + 2x^2 - 2x - 3 \] To explore fixed points further, you can rewrite the function as: \[ x = g(x) \] where \( g(x) \) is any transformation of \( f(x) \) that equals \( x \) when solved. Here, \( f(x) \) and the related transformations could be analyzed to identify \( g(x) \).
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