Calculate $sin(0.2)$ to 4 decimal places with its Taylor series and the Lagrange error bound
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So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.
I understand the following:
$$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$
$$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$
By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$
Knowing that, My Lagrange error should be:
$$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
so:
$$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001
Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$
If I take n=0, then I do not know how to even start writing Rn(x):
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$
Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.
calculus taylor-expansion
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add a comment |
$begingroup$
So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.
I understand the following:
$$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$
$$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$
By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$
Knowing that, My Lagrange error should be:
$$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
so:
$$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001
Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$
If I take n=0, then I do not know how to even start writing Rn(x):
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$
Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.
calculus taylor-expansion
$endgroup$
add a comment |
$begingroup$
So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.
I understand the following:
$$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$
$$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$
By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$
Knowing that, My Lagrange error should be:
$$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
so:
$$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001
Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$
If I take n=0, then I do not know how to even start writing Rn(x):
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$
Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.
calculus taylor-expansion
$endgroup$
So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.
I understand the following:
$$sin(x)=sum_{n=0}^∞(-1)^nfrac{x^{2n+1}}{(2n+1)!}$$
$$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$
By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)leq M$$ where M (I guess?) is 1, the maximum value for $sin(x)$
Knowing that, My Lagrange error should be:
$$R_n(x)leqfrac{M(x)^{n+1}}{(n+1)!}$$
so:
$$R_n(0,2)leqfrac{(0,2)^{n+1}}{(n+1)!}leq 0,0001$$
By checking values for n, I found that for n=3, its value is 0,00006..., and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001
Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=0,2$$
If I take n=0, then I do not know how to even start writing Rn(x):
$$sin(0,2)=frac{(0,2)^{2n+1}}{(2n+1)!}=frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$
Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.
calculus taylor-expansion
calculus taylor-expansion
edited Dec 30 '18 at 13:35
AmateurMathPirate
1,336721
1,336721
asked Dec 30 '18 at 12:49
LightsongLightsong
246
246
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3 Answers
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It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.
Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$
Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.
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$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
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– Lightsong
Dec 30 '18 at 16:20
add a comment |
$begingroup$
Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
$$sin(0.2)=0.2-frac{0.2^3}{3!},$$
in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.
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add a comment |
$begingroup$
$$sin x=x-frac{x^3}{3!}+R_n$$
You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.
You'll get
$$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.
Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$
Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.
$endgroup$
$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
$endgroup$
– Lightsong
Dec 30 '18 at 16:20
add a comment |
$begingroup$
It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.
Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$
Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.
$endgroup$
$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
$endgroup$
– Lightsong
Dec 30 '18 at 16:20
add a comment |
$begingroup$
It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.
Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$
Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.
$endgroup$
It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.
Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$
Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.
answered Dec 30 '18 at 14:12
B. GoddardB. Goddard
19.8k21442
19.8k21442
$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
$endgroup$
– Lightsong
Dec 30 '18 at 16:20
add a comment |
$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
$endgroup$
– Lightsong
Dec 30 '18 at 16:20
$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
$endgroup$
– Lightsong
Dec 30 '18 at 16:20
$begingroup$
Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it.
$endgroup$
– Lightsong
Dec 30 '18 at 16:20
add a comment |
$begingroup$
Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
$$sin(0.2)=0.2-frac{0.2^3}{3!},$$
in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.
$endgroup$
add a comment |
$begingroup$
Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
$$sin(0.2)=0.2-frac{0.2^3}{3!},$$
in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.
$endgroup$
add a comment |
$begingroup$
Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
$$sin(0.2)=0.2-frac{0.2^3}{3!},$$
in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.
$endgroup$
Since you found that $n=3$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be
$$sin(0.2)=0.2-frac{0.2^3}{3!},$$
in which you use $P_3(x)$ to approximate $sin(x)$, based on your notation.
answered Dec 30 '18 at 13:18
KittyLKittyL
13.8k31534
13.8k31534
add a comment |
add a comment |
$begingroup$
$$sin x=x-frac{x^3}{3!}+R_n$$
You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.
You'll get
$$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$
$endgroup$
add a comment |
$begingroup$
$$sin x=x-frac{x^3}{3!}+R_n$$
You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.
You'll get
$$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$
$endgroup$
add a comment |
$begingroup$
$$sin x=x-frac{x^3}{3!}+R_n$$
You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.
You'll get
$$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$
$endgroup$
$$sin x=x-frac{x^3}{3!}+R_n$$
You calculated that for $n=3$ $R_n<varepsilon$ where $varepsilon$ is the error bound you have. Then $n=3$ is the degree of the Taylor polynomial that satisfies that error bound, so put $x=0.2$ into $P_n$ since that is what approximates the function, while $R_n$ tells you how good is the approximation of $P_n$.
You'll get
$$sin(0.2)=0.198669ldotssimeq P_3=0.1986666ldots$$
answered Dec 30 '18 at 13:32
Lorenzo B.Lorenzo B.
1,8602520
1,8602520
add a comment |
add a comment |
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