N-th Taylor Polynomial

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Let $f$ and $f'$, $f''$, . . . , $f^(n)$ be continuous in a closed interval [$a, b$] containing a point $c$.
Write down the $n$-th Taylor polynomial of $f(x)$ around $x = c$.



So I know that the Taylor series is
$$sum_k=0^infty fracf^(k)(a)k! (x-a)^k = f(a) + f'(a)(x-a)+fracf''(a)2!(x-a)^2+...+fracf^(n)(a)n!(x-a)^n$$



Since this is worth only one mark, I was wondering do I just sub in $x=c$ or am I just not thinking right?




sequences-and-series taylor-expansion




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  • 2




    You sup $a=c$, also your summation has to go up to $n$ and not $infty$.
    – Cornman
    Aug 14 at 7:10















up vote
0
down vote

favorite












Let $f$ and $f'$, $f''$, . . . , $f^(n)$ be continuous in a closed interval [$a, b$] containing a point $c$.
Write down the $n$-th Taylor polynomial of $f(x)$ around $x = c$.



So I know that the Taylor series is
$$sum_k=0^infty fracf^(k)(a)k! (x-a)^k = f(a) + f'(a)(x-a)+fracf''(a)2!(x-a)^2+...+fracf^(n)(a)n!(x-a)^n$$



Since this is worth only one mark, I was wondering do I just sub in $x=c$ or am I just not thinking right?




sequences-and-series taylor-expansion




share|cite|improve this question
















  • 2




    You sup $a=c$, also your summation has to go up to $n$ and not $infty$.
    – Cornman
    Aug 14 at 7:10













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f$ and $f'$, $f''$, . . . , $f^(n)$ be continuous in a closed interval [$a, b$] containing a point $c$.
Write down the $n$-th Taylor polynomial of $f(x)$ around $x = c$.



So I know that the Taylor series is
$$sum_k=0^infty fracf^(k)(a)k! (x-a)^k = f(a) + f'(a)(x-a)+fracf''(a)2!(x-a)^2+...+fracf^(n)(a)n!(x-a)^n$$



Since this is worth only one mark, I was wondering do I just sub in $x=c$ or am I just not thinking right?




sequences-and-series taylor-expansion




share|cite|improve this question












Let $f$ and $f'$, $f''$, . . . , $f^(n)$ be continuous in a closed interval [$a, b$] containing a point $c$.
Write down the $n$-th Taylor polynomial of $f(x)$ around $x = c$.



So I know that the Taylor series is
$$sum_k=0^infty fracf^(k)(a)k! (x-a)^k = f(a) + f'(a)(x-a)+fracf''(a)2!(x-a)^2+...+fracf^(n)(a)n!(x-a)^n$$



Since this is worth only one mark, I was wondering do I just sub in $x=c$ or am I just not thinking right?





sequences-and-series taylor-expansion





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 14 at 7:08









oka

294




294







  • 2




    You sup $a=c$, also your summation has to go up to $n$ and not $infty$.
    – Cornman
    Aug 14 at 7:10













  • 2




    You sup $a=c$, also your summation has to go up to $n$ and not $infty$.
    – Cornman
    Aug 14 at 7:10








2




2




You sup $a=c$, also your summation has to go up to $n$ and not $infty$.
– Cornman
Aug 14 at 7:10





You sup $a=c$, also your summation has to go up to $n$ and not $infty$.
– Cornman
Aug 14 at 7:10











1 Answer
1






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up vote
2
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The polynomial you are looking for reads:



$$sum_k=0^n fracf^(k)(c)k! (x-c)^k. $$






share|cite|improve this answer




















  • Thanks a lot Fred!
    – oka
    Aug 14 at 7:13










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













The polynomial you are looking for reads:



$$sum_k=0^n fracf^(k)(c)k! (x-c)^k. $$






share|cite|improve this answer




















  • Thanks a lot Fred!
    – oka
    Aug 14 at 7:13














up vote
2
down vote













The polynomial you are looking for reads:



$$sum_k=0^n fracf^(k)(c)k! (x-c)^k. $$






share|cite|improve this answer




















  • Thanks a lot Fred!
    – oka
    Aug 14 at 7:13












up vote
2
down vote










up vote
2
down vote









The polynomial you are looking for reads:



$$sum_k=0^n fracf^(k)(c)k! (x-c)^k. $$






share|cite|improve this answer












The polynomial you are looking for reads:



$$sum_k=0^n fracf^(k)(c)k! (x-c)^k. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 14 at 7:12









Fred

37.9k1238




37.9k1238











  • Thanks a lot Fred!
    – oka
    Aug 14 at 7:13
















  • Thanks a lot Fred!
    – oka
    Aug 14 at 7:13















Thanks a lot Fred!
– oka
Aug 14 at 7:13




Thanks a lot Fred!
– oka
Aug 14 at 7:13












 

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