In the Taylor Expansion, it looks like they've made x_0 = x_1, x_2 and made x = x_1 - a, x_2 - a. I don't understand how they can use that substitution for x_0 since it is a constant, and x_1 and x_2 are variables that can change
In the Taylor Expansion, it looks like they've made x_0 = x_1, x_2 and made x = x_1 - a, x_2 - a. I don't understand how they can use that substitution for x_0 since it is a constant, and x_1 and x_2 are variables that can change
Related questions
Question
In the Taylor Expansion, it looks like they've made x_0 = x_1, x_2 and made x = x_1 - a, x_2 - a.
I don't understand how they can use that substitution for x_0 since it is a constant, and x_1 and x_2 are variables that can change.
Transcribed Image Text:f(x) = f(x) + f'(x)(x − x₁) + ƒ" (x¹) ( x − x₁)²
-
2!
=
1¹"'(x0)(x − xo)² + 1"'"(x0)(x − x3)² + ...
+
3!
4!
Σ
n=0
f(n)(x₁) (x − xo)”.
n!
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 1 images
