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- Homework Statement
- Given that ##f(x)=\frac{1}{1+x}##. Find bound of remainder ##R_3 (x)## assuming ##|x-1| \leq2##

- Relevant Equations
- Taylor series

Taylor Inequality

I have found the Taylor series up to 4th derivative:

$$f(x)=\frac{1}{2}-\frac{1}{4}(x-1)+\frac{1}{8}(x-1)^2-\frac{1}{16}(x-1)^3+\frac{1}{32}(x-1)^4$$

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 ##|x-1| \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

$$f(x)=\frac{1}{2}-\frac{1}{4}(x-1)+\frac{1}{8}(x-1)^2-\frac{1}{16}(x-1)^3+\frac{1}{32}(x-1)^4$$

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 ##|x-1| \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

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