Mixing
error term of taylor series
$begingroup$
I've been posed the following questions and i'm struggling to solve (b) and (c):
I'm not exactly sure how to use taylor's theorem in this case, any suggestions?
calculus
asked Jan 16 at 4:49
lohboyslohboys
958
$endgroup$
add a comment |
$begingroup$
I've been posed the following questions and i'm struggling to solve (b) and (c):
I'm not exactly sure how to use taylor's theorem in this case, any suggestions?
calculus
asked Jan 16 at 4:49
lohboyslohboys
958
$endgroup$
$begingroup$
Last question: which form does the remainder have in the version of the Taylor theorem you learned?
$endgroup$
– Raskolnikov
Jan 16 at 22:51
$begingroup$
$(x-c)^n$ for a taylor series centered around $c$
$endgroup$
– lohboys
Jan 16 at 23:02
add a comment |
$begingroup$
I've been posed the following questions and i'm struggling to solve (b) and (c):
I'm not exactly sure how to use taylor's theorem in this case, any suggestions?
calculus
asked Jan 16 at 4:49
lohboyslohboys
958
$endgroup$
I've been posed the following questions and i'm struggling to solve (b) and (c):
I'm not exactly sure how to use taylor's theorem in this case, any suggestions?
calculus
calculus
asked Jan 16 at 4:49
lohboyslohboys
958
asked Jan 16 at 4:49
lohboyslohboys
958
edited Jan 16 at 21:33
lohboys
asked Jan 16 at 4:49
lohboyslohboys
958
asked Jan 16 at 4:49
lohboyslohboys
958
asked Jan 16 at 4:49
lohboyslohboys
958
958
$begingroup$
Last question: which form does the remainder have in the version of the Taylor theorem you learned?
$endgroup$
– Raskolnikov
Jan 16 at 22:51
$begingroup$
$(x-c)^n$ for a taylor series centered around $c$
$endgroup$
– lohboys
Jan 16 at 23:02
add a comment |
$begingroup$
Last question: which form does the remainder have in the version of the Taylor theorem you learned?
$endgroup$
– Raskolnikov
Jan 16 at 22:51
$begingroup$
$(x-c)^n$ for a taylor series centered around $c$
$endgroup$
– lohboys
Jan 16 at 23:02
$begingroup$
Last question: which form does the remainder have in the version of the Taylor theorem you learned?
$endgroup$
– Raskolnikov
Jan 16 at 22:51
$begingroup$
Last question: which form does the remainder have in the version of the Taylor theorem you learned?
$endgroup$
– Raskolnikov
Jan 16 at 22:51
$begingroup$
$(x-c)^n$ for a taylor series centered around $c$
$endgroup$
– lohboys
Jan 16 at 23:02
$begingroup$
$(x-c)^n$ for a taylor series centered around $c$
$endgroup$
– lohboys
Jan 16 at 23:02
add a comment |
1 Answer
1
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votes
$begingroup$
So, the third order term $R_2(x)$ should be of the form $$frac{f'''(z)}{3!}(x-16)^3$$ for some $z$ in between $x$ and $16$ by Taylor's theorem.
The third derivative of $x^{1/2}$ can be evaluated to $$f'''(z) = frac{3}{8}z^{-5/2} ; .$$
So you need to look for which values of $x$, $|R_2(x)|<0.01$. Can you take it from here with the hint in (c)?
EDIT: To complete the answer, since OP already solved the problem,
$$left|frac{f'''(z)}{3!}(x-16)^3right|=left|frac{3}{3! 8}z^{-5/2}(x-16)^3right|leqleft|frac{1}{2^{14}}p^3right|$$
where in the last step I've made use of the fact that for $z>16$,
$$z^{-5/2}<16^{-5/2} = 2^{-10}$$
and where I also put $x=16+p$. Then requesting $|R_2(x)|<0.01$ amounts to
$$p<sqrt[3]{0.01 cdot 2^{14}} approx 5.4719 ; .$$
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
1
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
add a comment |
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
So, the third order term $R_2(x)$ should be of the form $$frac{f'''(z)}{3!}(x-16)^3$$ for some $z$ in between $x$ and $16$ by Taylor's theorem.
The third derivative of $x^{1/2}$ can be evaluated to $$f'''(z) = frac{3}{8}z^{-5/2} ; .$$
So you need to look for which values of $x$, $|R_2(x)|<0.01$. Can you take it from here with the hint in (c)?
EDIT: To complete the answer, since OP already solved the problem,
$$left|frac{f'''(z)}{3!}(x-16)^3right|=left|frac{3}{3! 8}z^{-5/2}(x-16)^3right|leqleft|frac{1}{2^{14}}p^3right|$$
where in the last step I've made use of the fact that for $z>16$,
$$z^{-5/2}<16^{-5/2} = 2^{-10}$$
and where I also put $x=16+p$. Then requesting $|R_2(x)|<0.01$ amounts to
$$p<sqrt[3]{0.01 cdot 2^{14}} approx 5.4719 ; .$$
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
1
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
add a comment |
$begingroup$
So, the third order term $R_2(x)$ should be of the form $$frac{f'''(z)}{3!}(x-16)^3$$ for some $z$ in between $x$ and $16$ by Taylor's theorem.
The third derivative of $x^{1/2}$ can be evaluated to $$f'''(z) = frac{3}{8}z^{-5/2} ; .$$
So you need to look for which values of $x$, $|R_2(x)|<0.01$. Can you take it from here with the hint in (c)?
EDIT: To complete the answer, since OP already solved the problem,
$$left|frac{f'''(z)}{3!}(x-16)^3right|=left|frac{3}{3! 8}z^{-5/2}(x-16)^3right|leqleft|frac{1}{2^{14}}p^3right|$$
where in the last step I've made use of the fact that for $z>16$,
$$z^{-5/2}<16^{-5/2} = 2^{-10}$$
and where I also put $x=16+p$. Then requesting $|R_2(x)|<0.01$ amounts to
$$p<sqrt[3]{0.01 cdot 2^{14}} approx 5.4719 ; .$$
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
1
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
add a comment |
$begingroup$
So, the third order term $R_2(x)$ should be of the form $$frac{f'''(z)}{3!}(x-16)^3$$ for some $z$ in between $x$ and $16$ by Taylor's theorem.
The third derivative of $x^{1/2}$ can be evaluated to $$f'''(z) = frac{3}{8}z^{-5/2} ; .$$
So you need to look for which values of $x$, $|R_2(x)|<0.01$. Can you take it from here with the hint in (c)?
EDIT: To complete the answer, since OP already solved the problem,
$$left|frac{f'''(z)}{3!}(x-16)^3right|=left|frac{3}{3! 8}z^{-5/2}(x-16)^3right|leqleft|frac{1}{2^{14}}p^3right|$$
where in the last step I've made use of the fact that for $z>16$,
$$z^{-5/2}<16^{-5/2} = 2^{-10}$$
and where I also put $x=16+p$. Then requesting $|R_2(x)|<0.01$ amounts to
$$p<sqrt[3]{0.01 cdot 2^{14}} approx 5.4719 ; .$$
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
So, the third order term $R_2(x)$ should be of the form $$frac{f'''(z)}{3!}(x-16)^3$$ for some $z$ in between $x$ and $16$ by Taylor's theorem.
The third derivative of $x^{1/2}$ can be evaluated to $$f'''(z) = frac{3}{8}z^{-5/2} ; .$$
So you need to look for which values of $x$, $|R_2(x)|<0.01$. Can you take it from here with the hint in (c)?
EDIT: To complete the answer, since OP already solved the problem,
$$left|frac{f'''(z)}{3!}(x-16)^3right|=left|frac{3}{3! 8}z^{-5/2}(x-16)^3right|leqleft|frac{1}{2^{14}}p^3right|$$
where in the last step I've made use of the fact that for $z>16$,
$$z^{-5/2}<16^{-5/2} = 2^{-10}$$
and where I also put $x=16+p$. Then requesting $|R_2(x)|<0.01$ amounts to
$$p<sqrt[3]{0.01 cdot 2^{14}} approx 5.4719 ; .$$
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
edited Jan 17 at 16:38
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
answered Jan 16 at 23:21
RaskolnikovRaskolnikov
12.6k23571
12.6k23571
1
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
add a comment |
1
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
1
1
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
$begingroup$
for part (c), is the domain $16<x<21.47$?
$endgroup$
– lohboys
Jan 17 at 0:07
add a comment |
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$begingroup$
Last question: which form does the remainder have in the version of the Taylor theorem you learned?
$endgroup$
– Raskolnikov
Jan 16 at 22:51
$begingroup$
$(x-c)^n$ for a taylor series centered around $c$
$endgroup$
– lohboys
Jan 16 at 23:02