How to find the upper bound of the error in given Taylor polynomial?
$begingroup$
We used this polynomial $P_2(x)=1-frac{1}{2}*x^2$ to approximate the function
$f(x)=cos x$ in the interval $[frac{-1}{2},frac{1}{2}]$
How do I find the upper bound of the error?
calculus polynomials numerical-methods taylor-expansion
edited Dec 12 '18 at 9:37
Bernard
124k741118
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
$endgroup$
add a comment |
$begingroup$
We used this polynomial $P_2(x)=1-frac{1}{2}*x^2$ to approximate the function
$f(x)=cos x$ in the interval $[frac{-1}{2},frac{1}{2}]$
How do I find the upper bound of the error?
calculus polynomials numerical-methods taylor-expansion
edited Dec 12 '18 at 9:37
Bernard
124k741118
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
$endgroup$
add a comment |
$begingroup$
We used this polynomial $P_2(x)=1-frac{1}{2}*x^2$ to approximate the function
$f(x)=cos x$ in the interval $[frac{-1}{2},frac{1}{2}]$
How do I find the upper bound of the error?
calculus polynomials numerical-methods taylor-expansion
edited Dec 12 '18 at 9:37
Bernard
124k741118
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
$endgroup$
We used this polynomial $P_2(x)=1-frac{1}{2}*x^2$ to approximate the function
$f(x)=cos x$ in the interval $[frac{-1}{2},frac{1}{2}]$
How do I find the upper bound of the error?
calculus polynomials numerical-methods taylor-expansion
calculus polynomials numerical-methods taylor-expansion
edited Dec 12 '18 at 9:37
Bernard
124k741118
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
edited Dec 12 '18 at 9:37
Bernard
124k741118
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
edited Dec 12 '18 at 9:37
Bernard
124k741118
edited Dec 12 '18 at 9:37
Bernard
124k741118
edited Dec 12 '18 at 9:37
Bernard
124k741118
124k741118
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
asked Dec 12 '18 at 9:16
Betim ShalaBetim Shala
1314
1314
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The cosine series is alternating. This means that the approximation error is bounded by the next term in the series, $dfrac{x^4}{24}$.
This is a consequence of the Leibniz test for the convergence of alternating series. You need to show that the sequence $dfrac{x^{2k}}{(2k)!}$, $kge 1$, is falling, the claim is then a consequence.
The same argument about the partial sums of an alternating series tells you that
$$
frac{x^4}{24}-frac{x^6}{720}le cos(x)-1+frac{x^2}2le frac{x^4}{24}
$$
so that this error bound can not be much improved over the given interval.
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The cosine series is alternating. This means that the approximation error is bounded by the next term in the series, $dfrac{x^4}{24}$.
This is a consequence of the Leibniz test for the convergence of alternating series. You need to show that the sequence $dfrac{x^{2k}}{(2k)!}$, $kge 1$, is falling, the claim is then a consequence.
The same argument about the partial sums of an alternating series tells you that
$$
frac{x^4}{24}-frac{x^6}{720}le cos(x)-1+frac{x^2}2le frac{x^4}{24}
$$
so that this error bound can not be much improved over the given interval.
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
add a comment |
$begingroup$
The cosine series is alternating. This means that the approximation error is bounded by the next term in the series, $dfrac{x^4}{24}$.
This is a consequence of the Leibniz test for the convergence of alternating series. You need to show that the sequence $dfrac{x^{2k}}{(2k)!}$, $kge 1$, is falling, the claim is then a consequence.
The same argument about the partial sums of an alternating series tells you that
$$
frac{x^4}{24}-frac{x^6}{720}le cos(x)-1+frac{x^2}2le frac{x^4}{24}
$$
so that this error bound can not be much improved over the given interval.
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
add a comment |
$begingroup$
The cosine series is alternating. This means that the approximation error is bounded by the next term in the series, $dfrac{x^4}{24}$.
This is a consequence of the Leibniz test for the convergence of alternating series. You need to show that the sequence $dfrac{x^{2k}}{(2k)!}$, $kge 1$, is falling, the claim is then a consequence.
The same argument about the partial sums of an alternating series tells you that
$$
frac{x^4}{24}-frac{x^6}{720}le cos(x)-1+frac{x^2}2le frac{x^4}{24}
$$
so that this error bound can not be much improved over the given interval.
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
$endgroup$
The cosine series is alternating. This means that the approximation error is bounded by the next term in the series, $dfrac{x^4}{24}$.
This is a consequence of the Leibniz test for the convergence of alternating series. You need to show that the sequence $dfrac{x^{2k}}{(2k)!}$, $kge 1$, is falling, the claim is then a consequence.
The same argument about the partial sums of an alternating series tells you that
$$
frac{x^4}{24}-frac{x^6}{720}le cos(x)-1+frac{x^2}2le frac{x^4}{24}
$$
so that this error bound can not be much improved over the given interval.
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
edited Dec 12 '18 at 12:06
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
answered Dec 12 '18 at 9:28
LutzLLutzL
60.1k42057
60.1k42057
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
add a comment |
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
thanks for the solution, but the problem is that I'm having problem how to come to that solution.
$endgroup$
– Betim Shala
Dec 12 '18 at 9:36
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
$begingroup$
I hope that you know about the convergence tests for series?
$endgroup$
– LutzL
Dec 12 '18 at 9:45
add a comment |
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