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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): \(\displaystyle f'(1)=-\frac{1}{4}\) \(\displaystyle f'(2)=\frac{1}{8}\) \(\displaystyle f'(3)=-\frac{1}{16}\) \(\displaystyle f'(4)=\frac{1}{32}\) \(\displaystyle f'(5)=-\frac{1}{64}\)
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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 $x=1$?
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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.
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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: $$ R_{5,0}(1)=f(1)-\sum_{k=0}^{5}\frac{f^{(k)}(0)}{k!} $$
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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: \(\displaystyle f^{(1)}(1)\) \(\displaystyle f^{(2)}(1)\) \(\displaystyle f^{(3)}(1)\) \(\displaystyle f^{(4)}(1)\) \(\displaystyle f^{(5)}(1)\)
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