Question

T_{n}left(frac{1}{1+x} ight)=f(0)+f^{prime}(0) x+frac{f^{prime prime}(0)}{2!} x^{2}+cdots Step 6: Substituting the values we found earlier: T_{n}left(frac{1}{1+x} ight)=1-x+frac{2}{2!} x^{2}-frac{6}{3!} x^{3}+cdots Step 7: Simplify the terms: T_{n}left(frac{1}{1+x} ight)=1-x+x^{2}-frac{3}{2} x^{3}+cdots Final Answer Step 8: Finally, we can rewrite the Taylor polynomial in the given form: T_{n}left(frac{1}{1+x} ight)=sum_{k=0}^{n}(-1)^{k} x^{k}

          T_{n}left(frac{1}{1+x}
ight)=f(0)+f^{prime}(0) x+frac{f^{prime prime}(0)}{2!} x^{2}+cdots
Step 6: Substituting the values we found earlier:
T_{n}left(frac{1}{1+x}
ight)=1-x+frac{2}{2!} x^{2}-frac{6}{3!} x^{3}+cdots
Step 7: Simplify the terms:
T_{n}left(frac{1}{1+x}
ight)=1-x+x^{2}-frac{3}{2} x^{3}+cdots
Final Answer
Step 8: Finally, we can rewrite the Taylor polynomial in the given form:
T_{n}left(frac{1}{1+x}
ight)=sum_{k=0}^{n}(-1)^{k} x^{k}

$Tnleft(frac11+x ight)=f(0)+f^prime(0) x+fracf^prime prime(0)2! x^2+cdots Step 6: Substituting the values we found earlier: Tnleft(frac11+x ight)=1-x+frac22! x^2-frac63! x^3+cdots Step 7: Simplify the terms: Tnleft(frac11+x ight)=1-x+x^2-frac32 x^3+cdots Final Answer Step 8: Finally, we can rewrite the Taylor polynomial in the given form: Tnleft(frac11+x ight)=sumk=0^n(-1)^k x^k$

Added by Mubdikhan

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