 #1
songoku
 2,109
 280
 Homework Statement:
 Given that ##f(x)=\frac{1}{1+x}##. Find bound of remainder ##R_3 (x)## assuming ##x1 \leq2##
 Relevant Equations:

Taylor series
Taylor Inequality
I have found the Taylor series up to 4th derivative:
$$f(x)=\frac{1}{2}\frac{1}{4}(x1)+\frac{1}{8}(x1)^2\frac{1}{16}(x1)^3+\frac{1}{32}(x1)^4$$
Using Taylor Inequality:
##a=1, d=2## and ##f^{4} (x)=\frac{24}{(1+x)^5}##
I need to find M that satisfies ##f^4 (x) \leq M##
From ##x1 \leq 2 \rightarrow 0 \leq x+1 \leq 4## , then ##0 \leq (x+1)^5 \leq 4^5##
Putting the boundary of ##(1+x)^5## to ##f^{4} (x)##, there is no upper boundary of ##f^{4} (x)## which does not make sense. And actually ##x+1## can not be zero because ##f^{4} (x)## would be undefined so my approach is certainly wrong.
How to find M? Thanks
$$f(x)=\frac{1}{2}\frac{1}{4}(x1)+\frac{1}{8}(x1)^2\frac{1}{16}(x1)^3+\frac{1}{32}(x1)^4$$
Using Taylor Inequality:
##a=1, d=2## and ##f^{4} (x)=\frac{24}{(1+x)^5}##
I need to find M that satisfies ##f^4 (x) \leq M##
From ##x1 \leq 2 \rightarrow 0 \leq x+1 \leq 4## , then ##0 \leq (x+1)^5 \leq 4^5##
Putting the boundary of ##(1+x)^5## to ##f^{4} (x)##, there is no upper boundary of ##f^{4} (x)## which does not make sense. And actually ##x+1## can not be zero because ##f^{4} (x)## would be undefined so my approach is certainly wrong.
How to find M? Thanks
Last edited: