Lagrange Error Value for $f(x)=frac{1}{x}$
$begingroup$
I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$
Step $1$ is to calculate the Taylor-Polynomial:
$$T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
I then used the Remainder formula $R_n(x)=frac{f^{(n+1)}(xi)}{(n+1)!} cdot (x-a)^{(n+1)}$. From this I get:
$$R_3(x)=frac{f^{(4)}(xi)}{4!} cdot (x-5)^4 =frac{-1}{xi^4} cdot (x-5)^4 = frac{-(x-5)^4}{xi^4}$$
Here's where I'm not sure how to proceed. I thought that I could make $2$ cases to try to find the Worst-Case scenario by checking which value of $xi_1=4, xi_2=6$ gives the Worst-Case scenario btu then I'm still left with either
$$R_{n_1}(x) =frac{-(x-5)^4}{4^4} text{ or } R_{n_2}(x) =frac{-(x-5)^4}{6^4}$$
but in either case I'm still left without an exact value for the error, which makes my answer incorrect.
How can I obtain an exact value for the error of the Taylor-Polynomial in the interval $x in [4,6]$?
real-analysis calculus taylor-expansion
edited Dec 17 '18 at 0:08
prt13463
1108
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
$endgroup$
add a comment |
$begingroup$
I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$
Step $1$ is to calculate the Taylor-Polynomial:
$$T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
I then used the Remainder formula $R_n(x)=frac{f^{(n+1)}(xi)}{(n+1)!} cdot (x-a)^{(n+1)}$. From this I get:
$$R_3(x)=frac{f^{(4)}(xi)}{4!} cdot (x-5)^4 =frac{-1}{xi^4} cdot (x-5)^4 = frac{-(x-5)^4}{xi^4}$$
Here's where I'm not sure how to proceed. I thought that I could make $2$ cases to try to find the Worst-Case scenario by checking which value of $xi_1=4, xi_2=6$ gives the Worst-Case scenario btu then I'm still left with either
$$R_{n_1}(x) =frac{-(x-5)^4}{4^4} text{ or } R_{n_2}(x) =frac{-(x-5)^4}{6^4}$$
but in either case I'm still left without an exact value for the error, which makes my answer incorrect.
How can I obtain an exact value for the error of the Taylor-Polynomial in the interval $x in [4,6]$?
real-analysis calculus taylor-expansion
edited Dec 17 '18 at 0:08
prt13463
1108
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
$endgroup$
$begingroup$
You can compute both the target function and the Taylor polynomial. Subtraction gives you the exact value of the error. Finding $xi$ amounts to solving a non-linear equation for which you have known the approximate location of a root.
$endgroup$
– Carl Christian
Dec 16 '18 at 22:34
add a comment |
$begingroup$
I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$
Step $1$ is to calculate the Taylor-Polynomial:
$$T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
I then used the Remainder formula $R_n(x)=frac{f^{(n+1)}(xi)}{(n+1)!} cdot (x-a)^{(n+1)}$. From this I get:
$$R_3(x)=frac{f^{(4)}(xi)}{4!} cdot (x-5)^4 =frac{-1}{xi^4} cdot (x-5)^4 = frac{-(x-5)^4}{xi^4}$$
Here's where I'm not sure how to proceed. I thought that I could make $2$ cases to try to find the Worst-Case scenario by checking which value of $xi_1=4, xi_2=6$ gives the Worst-Case scenario btu then I'm still left with either
$$R_{n_1}(x) =frac{-(x-5)^4}{4^4} text{ or } R_{n_2}(x) =frac{-(x-5)^4}{6^4}$$
but in either case I'm still left without an exact value for the error, which makes my answer incorrect.
How can I obtain an exact value for the error of the Taylor-Polynomial in the interval $x in [4,6]$?
real-analysis calculus taylor-expansion
edited Dec 17 '18 at 0:08
prt13463
1108
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
$endgroup$
I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$
Step $1$ is to calculate the Taylor-Polynomial:
$$T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
I then used the Remainder formula $R_n(x)=frac{f^{(n+1)}(xi)}{(n+1)!} cdot (x-a)^{(n+1)}$. From this I get:
$$R_3(x)=frac{f^{(4)}(xi)}{4!} cdot (x-5)^4 =frac{-1}{xi^4} cdot (x-5)^4 = frac{-(x-5)^4}{xi^4}$$
Here's where I'm not sure how to proceed. I thought that I could make $2$ cases to try to find the Worst-Case scenario by checking which value of $xi_1=4, xi_2=6$ gives the Worst-Case scenario btu then I'm still left with either
$$R_{n_1}(x) =frac{-(x-5)^4}{4^4} text{ or } R_{n_2}(x) =frac{-(x-5)^4}{6^4}$$
but in either case I'm still left without an exact value for the error, which makes my answer incorrect.
How can I obtain an exact value for the error of the Taylor-Polynomial in the interval $x in [4,6]$?
real-analysis calculus taylor-expansion
real-analysis calculus taylor-expansion
edited Dec 17 '18 at 0:08
prt13463
1108
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
edited Dec 17 '18 at 0:08
prt13463
1108
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
edited Dec 17 '18 at 0:08
prt13463
1108
edited Dec 17 '18 at 0:08
prt13463
1108
edited Dec 17 '18 at 0:08
prt13463
1108
1108
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
asked Dec 16 '18 at 22:09
Arthur Jr.Arthur Jr.
476
476
$begingroup$
You can compute both the target function and the Taylor polynomial. Subtraction gives you the exact value of the error. Finding $xi$ amounts to solving a non-linear equation for which you have known the approximate location of a root.
$endgroup$
– Carl Christian
Dec 16 '18 at 22:34
add a comment |
$begingroup$
You can compute both the target function and the Taylor polynomial. Subtraction gives you the exact value of the error. Finding $xi$ amounts to solving a non-linear equation for which you have known the approximate location of a root.
$endgroup$
– Carl Christian
Dec 16 '18 at 22:34
$begingroup$
You can compute both the target function and the Taylor polynomial. Subtraction gives you the exact value of the error. Finding $xi$ amounts to solving a non-linear equation for which you have known the approximate location of a root.
$endgroup$
– Carl Christian
Dec 16 '18 at 22:34
$begingroup$
You can compute both the target function and the Taylor polynomial. Subtraction gives you the exact value of the error. Finding $xi$ amounts to solving a non-linear equation for which you have known the approximate location of a root.
$endgroup$
– Carl Christian
Dec 16 '18 at 22:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the function
$$f(x) = dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$begin{array}{|c|c|}
hline text{Derivatives} & text{Values at x} =5 \hline
f(x) = dfrac{1}{x} & dfrac{1}{5} \hline
f'(x) = -dfrac{1}{x^2} & -dfrac{1}{25} \hline
f''(x)=dfrac{2}{x^3} & dfrac{2}{125} \hline
f^{(3)}(x) = -dfrac{6}{x^4} & -dfrac{6}{625} \hline
f^{(4)}(x) = dfrac{24}{x^5} & dfrac{24}{3125} \hline
end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$displaystyle left|R_n(x)right|=left|dfrac{f^{(n+1)}(xi)}{(n+1)!}right| cdot left|(x-a)^{(n+1)}right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = left|dfrac{1}{xi^5}right| cdot left|(x-5)^{4}right|$$
On the interval $4 le xi le 6$, the maximum of $left|dfrac{1}{xi^5}right|$ is at $xi = 4$, hence, the maximum error is given by
$$|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = dfrac{1}{24576} cdot max left|(x-5)^{4}right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
$endgroup$
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Consider the function
$$f(x) = dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$begin{array}{|c|c|}
hline text{Derivatives} & text{Values at x} =5 \hline
f(x) = dfrac{1}{x} & dfrac{1}{5} \hline
f'(x) = -dfrac{1}{x^2} & -dfrac{1}{25} \hline
f''(x)=dfrac{2}{x^3} & dfrac{2}{125} \hline
f^{(3)}(x) = -dfrac{6}{x^4} & -dfrac{6}{625} \hline
f^{(4)}(x) = dfrac{24}{x^5} & dfrac{24}{3125} \hline
end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$displaystyle left|R_n(x)right|=left|dfrac{f^{(n+1)}(xi)}{(n+1)!}right| cdot left|(x-a)^{(n+1)}right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = left|dfrac{1}{xi^5}right| cdot left|(x-5)^{4}right|$$
On the interval $4 le xi le 6$, the maximum of $left|dfrac{1}{xi^5}right|$ is at $xi = 4$, hence, the maximum error is given by
$$|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = dfrac{1}{24576} cdot max left|(x-5)^{4}right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
$endgroup$
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
add a comment |
$begingroup$
Consider the function
$$f(x) = dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$begin{array}{|c|c|}
hline text{Derivatives} & text{Values at x} =5 \hline
f(x) = dfrac{1}{x} & dfrac{1}{5} \hline
f'(x) = -dfrac{1}{x^2} & -dfrac{1}{25} \hline
f''(x)=dfrac{2}{x^3} & dfrac{2}{125} \hline
f^{(3)}(x) = -dfrac{6}{x^4} & -dfrac{6}{625} \hline
f^{(4)}(x) = dfrac{24}{x^5} & dfrac{24}{3125} \hline
end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$displaystyle left|R_n(x)right|=left|dfrac{f^{(n+1)}(xi)}{(n+1)!}right| cdot left|(x-a)^{(n+1)}right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = left|dfrac{1}{xi^5}right| cdot left|(x-5)^{4}right|$$
On the interval $4 le xi le 6$, the maximum of $left|dfrac{1}{xi^5}right|$ is at $xi = 4$, hence, the maximum error is given by
$$|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = dfrac{1}{24576} cdot max left|(x-5)^{4}right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
$endgroup$
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
add a comment |
$begingroup$
Consider the function
$$f(x) = dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$begin{array}{|c|c|}
hline text{Derivatives} & text{Values at x} =5 \hline
f(x) = dfrac{1}{x} & dfrac{1}{5} \hline
f'(x) = -dfrac{1}{x^2} & -dfrac{1}{25} \hline
f''(x)=dfrac{2}{x^3} & dfrac{2}{125} \hline
f^{(3)}(x) = -dfrac{6}{x^4} & -dfrac{6}{625} \hline
f^{(4)}(x) = dfrac{24}{x^5} & dfrac{24}{3125} \hline
end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$displaystyle left|R_n(x)right|=left|dfrac{f^{(n+1)}(xi)}{(n+1)!}right| cdot left|(x-a)^{(n+1)}right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = left|dfrac{1}{xi^5}right| cdot left|(x-5)^{4}right|$$
On the interval $4 le xi le 6$, the maximum of $left|dfrac{1}{xi^5}right|$ is at $xi = 4$, hence, the maximum error is given by
$$|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = dfrac{1}{24576} cdot max left|(x-5)^{4}right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
$endgroup$
Consider the function
$$f(x) = dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$begin{array}{|c|c|}
hline text{Derivatives} & text{Values at x} =5 \hline
f(x) = dfrac{1}{x} & dfrac{1}{5} \hline
f'(x) = -dfrac{1}{x^2} & -dfrac{1}{25} \hline
f''(x)=dfrac{2}{x^3} & dfrac{2}{125} \hline
f^{(3)}(x) = -dfrac{6}{x^4} & -dfrac{6}{625} \hline
f^{(4)}(x) = dfrac{24}{x^5} & dfrac{24}{3125} \hline
end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= frac{1}{5}-frac{(x-5)}{25}-frac{(x-5)^2}{125}-frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$displaystyle left|R_n(x)right|=left|dfrac{f^{(n+1)}(xi)}{(n+1)!}right| cdot left|(x-a)^{(n+1)}right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = left|dfrac{1}{xi^5}right| cdot left|(x-5)^{4}right|$$
On the interval $4 le xi le 6$, the maximum of $left|dfrac{1}{xi^5}right|$ is at $xi = 4$, hence, the maximum error is given by
$$|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right| = dfrac{1}{24576} cdot max left|(x-5)^{4}right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
edited Dec 17 '18 at 0:29
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
answered Dec 17 '18 at 0:02
MooMoo
5,63131020
5,63131020
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
add a comment |
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
I'm not sure how you got the very last step. I haven't learned the max notation.
$endgroup$
– Arthur Jr.
Dec 17 '18 at 11:02
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
All it is asking is what is the maximum result that I can get from the absolute value of that quantity. Since the power is even, I can drop the absolute value sign and then just ask, what is the max of $(x-5)^4$ over the range. we can see that that occurs at the endpoints, so we can choose $x = 4$ for a max of one. This tells us the absolute worst error, not the exact error and at times can be way worse that the actual error. Sometimes, this process can also fail for certain types of problems. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:32
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
Yes. It's clear now. But if you say that the max occurs when $xi = 4$ then we would have to substitute that into $|R_3(x)| = left|dfrac{f^{(4)}(xi)}{(4)!}right| cdot left|(x-5)^{4}right|$, correct?
$endgroup$
– Arthur Jr.
Dec 17 '18 at 12:34
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
We are trying to find the worst case maximum, so that we have the best estimate. Because of this, we have to do for each absolute value. Sometimes, it is the same value and sometimes not. You will sometimes see the first value as $M$. Lastly, the form given allows us to find the error for any $x$ that they might give. Clear?
$endgroup$
– Moo
Dec 17 '18 at 12:51
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
$begingroup$
@ArthurJr.: This is a nice write up that is worth reading/reviewing people.csail.mit.edu/ddeford/Taylor_ERROR.pdf
$endgroup$
– Moo
Dec 17 '18 at 13:04
add a comment |
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You can compute both the target function and the Taylor polynomial. Subtraction gives you the exact value of the error. Finding $xi$ amounts to solving a non-linear equation for which you have known the approximate location of a root.
$endgroup$
– Carl Christian
Dec 16 '18 at 22:34