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Hello, I was trying to solve the following exercises but I encountered some difficulties. Given the function $$\displaystyle f(x)=\frac{1}{1+x}$$ compute the reminder when x = 1 and n = 5 using the following formula: $$\displaystyle Rn(a)=f(x)-\sum_{k=0}^{n}\frac{f^{(n)}(a)}{k!}$$ I have already computed the derivates at point 1 but cannot get the correct result (which should be 1/2): $$\displaystyle f'(1)=-\frac{1}{4}$$ $$\displaystyle f'(2)=\frac{1}{8}$$ $$\displaystyle f'(3)=-\frac{1}{16}$$ $$\displaystyle f'(4)=\frac{1}{32}$$ $$\displaystyle f'(5)=-\frac{1}{64}$$

Havel29 said: You both are right, the question was more of a theoretical one. Since the formula I posted was given, one had to understand a=0, and therefore calculate the sum as @CaptainBlack stated, which correctly results in 1/2. Click to expand... Well, what have you done? What are the first five derivatives of $\frac{1}{1+x}$ evaluated at $x=0$? ...and would the difficulty you are experiencing be related to the Taylor series of $\frac{1}{1+x}$ about $0$ not converging at $x=1$?
Havel29 said: Hello, I was trying to solve the following exercises but I encountered some difficulties. Given the function $$\displaystyle f(x)=\frac{1}{1+x}$$ compute the reminder when x = 1 and n = 5 using the following formula: $$\displaystyle Rn(a)=f(x)-\sum_{k=0}^{n}\frac{f^{(n)}(a)}{k!}$$ I have already computed the derivates at point 1 but cannot get the correct result (which should be 1/2): $$\displaystyle f'(1)=-\frac{1}{4}$$ $$\displaystyle f'(2)=\frac{1}{8}$$ $$\displaystyle f'(3)=-\frac{1}{16}$$ $$\displaystyle f'(4)=\frac{1}{32}$$ $$\displaystyle f'(5)=-\frac{1}{64}$$ Click to expand... Please check the question you are trying to ask. From what you post I am guessing you are dealing with something like: This is the Taylor polynomial of degree $n$ developed about $a$: $$P_{n,a}(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k$$ Those exponents $^{(k)}$ denote derivatives of order $k$, while the exponent $^k$ denotes a normal power. The remainder is then: $$R_{n,a}(x)=f(x)-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k$$ You want the remainder when $n=1$ and $n=5$: $$R_{5,a}(1)=f(1)-\sum_{k=0}^{5}\frac{f^{(k)}(a)}{k!}(1-a)^k$$ Which depends on the point $a$ about which the Taylor polynomial is generated, if $a=0$ then the remainder is: $$R_{5,0}(1)=f(1)-\sum_{k=0}^{5}\frac{f^{(k)}(0)}{k!}$$
Havel29 said: Hello, I was trying to solve the following exercises but I encountered some difficulties. Given the function $$\displaystyle f(x)=\frac{1}{1+x}$$ compute the reminder when x = 1 and n = 5 using the following formula: $$\displaystyle Rn(a)=f(x)-\sum_{k=0}^{n}\frac{f^{(n)}(a)}{k!}$$ I have already computed the derivates at point 1 but cannot get the correct result (which should be 1/2): $$\displaystyle f'(1)=-\frac{1}{4}$$ $$\displaystyle f'(2)=\frac{1}{8}$$ $$\displaystyle f'(3)=-\frac{1}{16}$$ $$\displaystyle f'(4)=\frac{1}{32}$$ $$\displaystyle f'(5)=-\frac{1}{64}$$ Click to expand... You misinterpreted the meaning of n=5. It means the 5th derivative and evaluated at x=1. $$\displaystyle f'(1)$$ $$\displaystyle f''(1)$$ $$\displaystyle f'''(1)$$ $$\displaystyle f'''''(1)$$ $$\displaystyle f''''''(1)$$ or the short hand notation: $$\displaystyle f^{(1)}(1)$$ $$\displaystyle f^{(2)}(1)$$ $$\displaystyle f^{(3)}(1)$$ $$\displaystyle f^{(4)}(1)$$ $$\displaystyle f^{(5)}(1)$$