Calculating a function value with Taylor series [on hold]
Given a function $f(x)$, how do I calculate it's value with given error range?
For example, how do I calculate $g(-0.1)$ with an error $< 0.001$ when:
$$g(x)=sin(x)+cos(x)$$
calculus taylor-expansion
edited Jan 4 at 16:43
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asked Jan 4 at 10:29
IgorIgor
315
put on hold as off-topic by RRL, Saad, mrtaurho, darij grinberg, choco_addicted Jan 5 at 15:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, mrtaurho, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Given a function $f(x)$, how do I calculate it's value with given error range?
For example, how do I calculate $g(-0.1)$ with an error $< 0.001$ when:
$$g(x)=sin(x)+cos(x)$$
calculus taylor-expansion
edited Jan 4 at 16:43
PM.
3,3982825
asked Jan 4 at 10:29
IgorIgor
315
put on hold as off-topic by RRL, Saad, mrtaurho, darij grinberg, choco_addicted Jan 5 at 15:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, mrtaurho, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
So you want to know until what order you need to take the Tayor series?
– WarreG
Jan 4 at 14:07
add a comment |
Given a function $f(x)$, how do I calculate it's value with given error range?
For example, how do I calculate $g(-0.1)$ with an error $< 0.001$ when:
$$g(x)=sin(x)+cos(x)$$
calculus taylor-expansion
edited Jan 4 at 16:43
PM.
3,3982825
asked Jan 4 at 10:29
IgorIgor
315
Given a function $f(x)$, how do I calculate it's value with given error range?
For example, how do I calculate $g(-0.1)$ with an error $< 0.001$ when:
$$g(x)=sin(x)+cos(x)$$
calculus taylor-expansion
calculus taylor-expansion
edited Jan 4 at 16:43
PM.
3,3982825
asked Jan 4 at 10:29
IgorIgor
315
edited Jan 4 at 16:43
PM.
3,3982825
asked Jan 4 at 10:29
IgorIgor
315
edited Jan 4 at 16:43
PM.
3,3982825
edited Jan 4 at 16:43
PM.
3,3982825
edited Jan 4 at 16:43
PM.
3,3982825
3,3982825
asked Jan 4 at 10:29
IgorIgor
315
asked Jan 4 at 10:29
IgorIgor
315
asked Jan 4 at 10:29
IgorIgor
315
315
put on hold as off-topic by RRL, Saad, mrtaurho, darij grinberg, choco_addicted Jan 5 at 15:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, mrtaurho, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by RRL, Saad, mrtaurho, darij grinberg, choco_addicted Jan 5 at 15:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, mrtaurho, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
So you want to know until what order you need to take the Tayor series?
– WarreG
Jan 4 at 14:07
add a comment |
So you want to know until what order you need to take the Tayor series?
– WarreG
Jan 4 at 14:07
So you want to know until what order you need to take the Tayor series?
– WarreG
Jan 4 at 14:07
So you want to know until what order you need to take the Tayor series?
– WarreG
Jan 4 at 14:07
add a comment |
2 Answers
2
active
oldest
votes
With
$$
f(x)=f(0)+frac{f^{(1)}(0)}{1!}x+cdots+frac{f^{(n)}(0)}{n!}x^n+R_n(x)
$$
Use the Lagrange form for the remainder
$$
R_n(x)=frac{f^{(n+1)}(c)}{(n+1)!}x^{(n+1)}
$$
where $c$ is between $x$ and $0$.
The derivatives of $g(x)$ are always $pmsin(x)pm cos(x)$ and so
$$
| f^{(n+1)}(c) | le 2
$$
for any $c$.
Then considering the magnitude of the error, find n such that
$$
2 frac{(0.1)^{(n+1)}}{(n+1)!} lt 0.001
$$
answered Jan 4 at 16:38
PM.PM.
3,3982825
add a comment |
The Taylor series of $g(x) = sin(x) + cos(x)$ is
$$
1+x-frac{x^2}{2!}-frac{x^3}{3!}+frac{x^4}{4!}+frac{x^5}{5!}- ...
$$
You want to evaluate the function $g(-0.1)$ such that it's error is smaller than $0.001$.
From the Taylor series you can see that the third term $frac{x^3}{3!}$ will be of that order, because $frac{-0.1^3}{6} sim 0.001$. The next term in the series will always be an order smaller so the error will be of the order of $0.0001$ or $mathcal{O}(x^4).$
What you are left with is
$$
g(x) approx 1+x-frac{x^2}{2!}-frac{x^3}{3!}.
$$
Now you just fill in $-0.1$ to retrieve the value of the function in that point.
answered Jan 4 at 14:31
WarreGWarreG
20819
2
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement thatThe next term in the series will always be an order smaller so the error will be...is false.
– zipirovich
Jan 4 at 14:57
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
With
$$
f(x)=f(0)+frac{f^{(1)}(0)}{1!}x+cdots+frac{f^{(n)}(0)}{n!}x^n+R_n(x)
$$
Use the Lagrange form for the remainder
$$
R_n(x)=frac{f^{(n+1)}(c)}{(n+1)!}x^{(n+1)}
$$
where $c$ is between $x$ and $0$.
The derivatives of $g(x)$ are always $pmsin(x)pm cos(x)$ and so
$$
| f^{(n+1)}(c) | le 2
$$
for any $c$.
Then considering the magnitude of the error, find n such that
$$
2 frac{(0.1)^{(n+1)}}{(n+1)!} lt 0.001
$$
answered Jan 4 at 16:38
PM.PM.
3,3982825
add a comment |
With
$$
f(x)=f(0)+frac{f^{(1)}(0)}{1!}x+cdots+frac{f^{(n)}(0)}{n!}x^n+R_n(x)
$$
Use the Lagrange form for the remainder
$$
R_n(x)=frac{f^{(n+1)}(c)}{(n+1)!}x^{(n+1)}
$$
where $c$ is between $x$ and $0$.
The derivatives of $g(x)$ are always $pmsin(x)pm cos(x)$ and so
$$
| f^{(n+1)}(c) | le 2
$$
for any $c$.
Then considering the magnitude of the error, find n such that
$$
2 frac{(0.1)^{(n+1)}}{(n+1)!} lt 0.001
$$
answered Jan 4 at 16:38
PM.PM.
3,3982825
add a comment |
With
$$
f(x)=f(0)+frac{f^{(1)}(0)}{1!}x+cdots+frac{f^{(n)}(0)}{n!}x^n+R_n(x)
$$
Use the Lagrange form for the remainder
$$
R_n(x)=frac{f^{(n+1)}(c)}{(n+1)!}x^{(n+1)}
$$
where $c$ is between $x$ and $0$.
The derivatives of $g(x)$ are always $pmsin(x)pm cos(x)$ and so
$$
| f^{(n+1)}(c) | le 2
$$
for any $c$.
Then considering the magnitude of the error, find n such that
$$
2 frac{(0.1)^{(n+1)}}{(n+1)!} lt 0.001
$$
answered Jan 4 at 16:38
PM.PM.
3,3982825
With
$$
f(x)=f(0)+frac{f^{(1)}(0)}{1!}x+cdots+frac{f^{(n)}(0)}{n!}x^n+R_n(x)
$$
Use the Lagrange form for the remainder
$$
R_n(x)=frac{f^{(n+1)}(c)}{(n+1)!}x^{(n+1)}
$$
where $c$ is between $x$ and $0$.
The derivatives of $g(x)$ are always $pmsin(x)pm cos(x)$ and so
$$
| f^{(n+1)}(c) | le 2
$$
for any $c$.
Then considering the magnitude of the error, find n such that
$$
2 frac{(0.1)^{(n+1)}}{(n+1)!} lt 0.001
$$
answered Jan 4 at 16:38
PM.PM.
3,3982825
edited Jan 4 at 17:03
answered Jan 4 at 16:38
PM.PM.
3,3982825
answered Jan 4 at 16:38
PM.PM.
3,3982825
answered Jan 4 at 16:38
PM.PM.
3,3982825
3,3982825
add a comment |
add a comment |
The Taylor series of $g(x) = sin(x) + cos(x)$ is
$$
1+x-frac{x^2}{2!}-frac{x^3}{3!}+frac{x^4}{4!}+frac{x^5}{5!}- ...
$$
You want to evaluate the function $g(-0.1)$ such that it's error is smaller than $0.001$.
From the Taylor series you can see that the third term $frac{x^3}{3!}$ will be of that order, because $frac{-0.1^3}{6} sim 0.001$. The next term in the series will always be an order smaller so the error will be of the order of $0.0001$ or $mathcal{O}(x^4).$
What you are left with is
$$
g(x) approx 1+x-frac{x^2}{2!}-frac{x^3}{3!}.
$$
Now you just fill in $-0.1$ to retrieve the value of the function in that point.
answered Jan 4 at 14:31
WarreGWarreG
20819
2
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement thatThe next term in the series will always be an order smaller so the error will be...is false.
– zipirovich
Jan 4 at 14:57
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
add a comment |
The Taylor series of $g(x) = sin(x) + cos(x)$ is
$$
1+x-frac{x^2}{2!}-frac{x^3}{3!}+frac{x^4}{4!}+frac{x^5}{5!}- ...
$$
You want to evaluate the function $g(-0.1)$ such that it's error is smaller than $0.001$.
From the Taylor series you can see that the third term $frac{x^3}{3!}$ will be of that order, because $frac{-0.1^3}{6} sim 0.001$. The next term in the series will always be an order smaller so the error will be of the order of $0.0001$ or $mathcal{O}(x^4).$
What you are left with is
$$
g(x) approx 1+x-frac{x^2}{2!}-frac{x^3}{3!}.
$$
Now you just fill in $-0.1$ to retrieve the value of the function in that point.
answered Jan 4 at 14:31
WarreGWarreG
20819
2
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement thatThe next term in the series will always be an order smaller so the error will be...is false.
– zipirovich
Jan 4 at 14:57
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
add a comment |
The Taylor series of $g(x) = sin(x) + cos(x)$ is
$$
1+x-frac{x^2}{2!}-frac{x^3}{3!}+frac{x^4}{4!}+frac{x^5}{5!}- ...
$$
You want to evaluate the function $g(-0.1)$ such that it's error is smaller than $0.001$.
From the Taylor series you can see that the third term $frac{x^3}{3!}$ will be of that order, because $frac{-0.1^3}{6} sim 0.001$. The next term in the series will always be an order smaller so the error will be of the order of $0.0001$ or $mathcal{O}(x^4).$
What you are left with is
$$
g(x) approx 1+x-frac{x^2}{2!}-frac{x^3}{3!}.
$$
Now you just fill in $-0.1$ to retrieve the value of the function in that point.
answered Jan 4 at 14:31
WarreGWarreG
20819
The Taylor series of $g(x) = sin(x) + cos(x)$ is
$$
1+x-frac{x^2}{2!}-frac{x^3}{3!}+frac{x^4}{4!}+frac{x^5}{5!}- ...
$$
You want to evaluate the function $g(-0.1)$ such that it's error is smaller than $0.001$.
From the Taylor series you can see that the third term $frac{x^3}{3!}$ will be of that order, because $frac{-0.1^3}{6} sim 0.001$. The next term in the series will always be an order smaller so the error will be of the order of $0.0001$ or $mathcal{O}(x^4).$
What you are left with is
$$
g(x) approx 1+x-frac{x^2}{2!}-frac{x^3}{3!}.
$$
Now you just fill in $-0.1$ to retrieve the value of the function in that point.
answered Jan 4 at 14:31
WarreGWarreG
20819
answered Jan 4 at 14:31
WarreGWarreG
20819
answered Jan 4 at 14:31
WarreGWarreG
20819
answered Jan 4 at 14:31
WarreGWarreG
20819
20819
2
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement thatThe next term in the series will always be an order smaller so the error will be...is false.
– zipirovich
Jan 4 at 14:57
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
add a comment |
2
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement thatThe next term in the series will always be an order smaller so the error will be...is false.
– zipirovich
Jan 4 at 14:57
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
2
2
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement that
The next term in the series will always be an order smaller so the error will be... is false.– zipirovich
Jan 4 at 14:57
Sorry, but your answer is at least misleading. "The next term" is NOT the same as "the remainder (the sum of all remaining terms)". That's true for series satisfying the Alternating Series Test, and often true for rapidly convergent series, but it's not true in general. So your statement that
The next term in the series will always be an order smaller so the error will be... is false.– zipirovich
Jan 4 at 14:57
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
I'm saying for this example the next term will be an order smaller. I'm not stating anything general here.
– WarreG
Jan 4 at 15:22
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
Yes, the next term will be smaller. But technically speaking, in general that's irrelevant to the question: the error is given by the remainder, not by the next term. If you claim that in this series the error $=$ the remainder can be bounded by the next term, you have to justify this claim. Otherwise, it's wrong.
– zipirovich
Jan 4 at 15:24
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
You are right. But in this case, because $-1 < x <1$ the value will converge very rapidly to the exact value. Because each following term is guaranteed to be smaller you can approximately guess what order you have to use for the Taylor series, where the question actually is about. I leave it to you then to write an answer with the total remainder.
– WarreG
Jan 4 at 15:35
add a comment |
So you want to know until what order you need to take the Tayor series?
– WarreG
Jan 4 at 14:07