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: $$ R_{5,0}(1)=f(1)-\sum_{k=0}^{5}\frac{f^{(k)}(0)}{k!} $$