Taylor polynomial remainder theorem, why is $R_n^n+1(x) = f^n+1(x)$?

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I am trying to grasp the concept of Taylor remainder.



Online lecture taught me that in equation



$R_n^(n+1)(x) = f^(n+1)(x) + T_n^(n+1)(x)$,



$T_n^(n+1)(x)$ becomes zero because in general nth polynomial when taken derivative n+1 times becomes zero. For example, $x^2$ when taken derivatives 3 times it becomes zero, 2x, 2, then 0.



So the lecture goes $R_n^(n+1)(x) = f^(n+1)(x)$.



But I am not getting it because why not $R_n^(n+1)$ and $f^(n+1)$ also do not become zero?



I don't quite get the idea of nth derivative of f function becoming error itself, though I am not sure if my interpretation is even right.




power-series




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  • What do $R_n^(n+1)(x)$ etc. denote?
    – Lord Shark the Unknown
    Aug 13 at 6:23










  • $R_n^(n+1)(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:25










  • $T_n^(n+1)(x)$ denotes taylor polynomial when taken (n+1)th derivative.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:26














up vote
0
down vote

favorite












I am trying to grasp the concept of Taylor remainder.



Online lecture taught me that in equation



$R_n^(n+1)(x) = f^(n+1)(x) + T_n^(n+1)(x)$,



$T_n^(n+1)(x)$ becomes zero because in general nth polynomial when taken derivative n+1 times becomes zero. For example, $x^2$ when taken derivatives 3 times it becomes zero, 2x, 2, then 0.



So the lecture goes $R_n^(n+1)(x) = f^(n+1)(x)$.



But I am not getting it because why not $R_n^(n+1)$ and $f^(n+1)$ also do not become zero?



I don't quite get the idea of nth derivative of f function becoming error itself, though I am not sure if my interpretation is even right.




power-series




share|cite|improve this question




















  • What do $R_n^(n+1)(x)$ etc. denote?
    – Lord Shark the Unknown
    Aug 13 at 6:23










  • $R_n^(n+1)(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:25










  • $T_n^(n+1)(x)$ denotes taylor polynomial when taken (n+1)th derivative.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:26












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to grasp the concept of Taylor remainder.



Online lecture taught me that in equation



$R_n^(n+1)(x) = f^(n+1)(x) + T_n^(n+1)(x)$,



$T_n^(n+1)(x)$ becomes zero because in general nth polynomial when taken derivative n+1 times becomes zero. For example, $x^2$ when taken derivatives 3 times it becomes zero, 2x, 2, then 0.



So the lecture goes $R_n^(n+1)(x) = f^(n+1)(x)$.



But I am not getting it because why not $R_n^(n+1)$ and $f^(n+1)$ also do not become zero?



I don't quite get the idea of nth derivative of f function becoming error itself, though I am not sure if my interpretation is even right.




power-series




share|cite|improve this question












I am trying to grasp the concept of Taylor remainder.



Online lecture taught me that in equation



$R_n^(n+1)(x) = f^(n+1)(x) + T_n^(n+1)(x)$,



$T_n^(n+1)(x)$ becomes zero because in general nth polynomial when taken derivative n+1 times becomes zero. For example, $x^2$ when taken derivatives 3 times it becomes zero, 2x, 2, then 0.



So the lecture goes $R_n^(n+1)(x) = f^(n+1)(x)$.



But I am not getting it because why not $R_n^(n+1)$ and $f^(n+1)$ also do not become zero?



I don't quite get the idea of nth derivative of f function becoming error itself, though I am not sure if my interpretation is even right.





power-series





share|cite|improve this question











share|cite|improve this question




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asked Aug 13 at 6:13









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  • What do $R_n^(n+1)(x)$ etc. denote?
    – Lord Shark the Unknown
    Aug 13 at 6:23










  • $R_n^(n+1)(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:25










  • $T_n^(n+1)(x)$ denotes taylor polynomial when taken (n+1)th derivative.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:26
















  • What do $R_n^(n+1)(x)$ etc. denote?
    – Lord Shark the Unknown
    Aug 13 at 6:23










  • $R_n^(n+1)(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:25










  • $T_n^(n+1)(x)$ denotes taylor polynomial when taken (n+1)th derivative.
    – ê°•ìŠ¹íƒœ
    Aug 13 at 6:26















What do $R_n^(n+1)(x)$ etc. denote?
– Lord Shark the Unknown
Aug 13 at 6:23




What do $R_n^(n+1)(x)$ etc. denote?
– Lord Shark the Unknown
Aug 13 at 6:23












$R_n^(n+1)(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error.
– ê°•ìŠ¹íƒœ
Aug 13 at 6:25




$R_n^(n+1)(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error.
– ê°•ìŠ¹íƒœ
Aug 13 at 6:25












$T_n^(n+1)(x)$ denotes taylor polynomial when taken (n+1)th derivative.
– ê°•ìŠ¹íƒœ
Aug 13 at 6:26




$T_n^(n+1)(x)$ denotes taylor polynomial when taken (n+1)th derivative.
– ê°•ìŠ¹íƒœ
Aug 13 at 6:26















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