Calculate the necessary degree of expansion for a Taylor series
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I've had a couple of tasks that say something along the line of "calculate the degree of the Taylor polynomial so that the difference between it and the function is such and such". I don't know any way of getting to that point without trial and error, my intuitive guesses are usually off and it takes a lot of time to calculate the difference properly since we aren't allowed to use calculators. Here's an example:
Let $f:[0,infty)toBbbR$ be defined by $f(x)=(x-1)^2e^-2x$. Calculate $ninBbbN$ such that $T_n,2$, the Taylor polynomial of $n$-th order around the expansion point $x=2$, fulfills $$|f(x)-T_n,2(x)|le10^-3$$ for all $x$ with $|x-2|lt10^-1$.
I genuinely don't know how to approach this without just trying out different values for $n$, and I was wondering if there are any systematic ways to get a result. Any help would be appreciated.
taylor-expansion
asked Sep 3 at 10:10
pavus
495
add a comment |Â
up vote
1
down vote
favorite
I've had a couple of tasks that say something along the line of "calculate the degree of the Taylor polynomial so that the difference between it and the function is such and such". I don't know any way of getting to that point without trial and error, my intuitive guesses are usually off and it takes a lot of time to calculate the difference properly since we aren't allowed to use calculators. Here's an example:
Let $f:[0,infty)toBbbR$ be defined by $f(x)=(x-1)^2e^-2x$. Calculate $ninBbbN$ such that $T_n,2$, the Taylor polynomial of $n$-th order around the expansion point $x=2$, fulfills $$|f(x)-T_n,2(x)|le10^-3$$ for all $x$ with $|x-2|lt10^-1$.
I genuinely don't know how to approach this without just trying out different values for $n$, and I was wondering if there are any systematic ways to get a result. Any help would be appreciated.
taylor-expansion
asked Sep 3 at 10:10
pavus
495
2
You can use Taylor-Lagrange inequality for the remainder: if you can bound the derivatives at all orders, it enables you to find an estimate for this degree.
– Bernard
Sep 3 at 11:50
I didn't know you had to consider the remainder for this, thank you!
– pavus
Sep 3 at 12:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've had a couple of tasks that say something along the line of "calculate the degree of the Taylor polynomial so that the difference between it and the function is such and such". I don't know any way of getting to that point without trial and error, my intuitive guesses are usually off and it takes a lot of time to calculate the difference properly since we aren't allowed to use calculators. Here's an example:
Let $f:[0,infty)toBbbR$ be defined by $f(x)=(x-1)^2e^-2x$. Calculate $ninBbbN$ such that $T_n,2$, the Taylor polynomial of $n$-th order around the expansion point $x=2$, fulfills $$|f(x)-T_n,2(x)|le10^-3$$ for all $x$ with $|x-2|lt10^-1$.
I genuinely don't know how to approach this without just trying out different values for $n$, and I was wondering if there are any systematic ways to get a result. Any help would be appreciated.
taylor-expansion
asked Sep 3 at 10:10
pavus
495
I've had a couple of tasks that say something along the line of "calculate the degree of the Taylor polynomial so that the difference between it and the function is such and such". I don't know any way of getting to that point without trial and error, my intuitive guesses are usually off and it takes a lot of time to calculate the difference properly since we aren't allowed to use calculators. Here's an example:
Let $f:[0,infty)toBbbR$ be defined by $f(x)=(x-1)^2e^-2x$. Calculate $ninBbbN$ such that $T_n,2$, the Taylor polynomial of $n$-th order around the expansion point $x=2$, fulfills $$|f(x)-T_n,2(x)|le10^-3$$ for all $x$ with $|x-2|lt10^-1$.
I genuinely don't know how to approach this without just trying out different values for $n$, and I was wondering if there are any systematic ways to get a result. Any help would be appreciated.
taylor-expansion
taylor-expansion
asked Sep 3 at 10:10
pavus
495
asked Sep 3 at 10:10
pavus
495
asked Sep 3 at 10:10
pavus
495
asked Sep 3 at 10:10
pavus
495
asked Sep 3 at 10:10
pavus
495
495
2
You can use Taylor-Lagrange inequality for the remainder: if you can bound the derivatives at all orders, it enables you to find an estimate for this degree.
– Bernard
Sep 3 at 11:50
I didn't know you had to consider the remainder for this, thank you!
– pavus
Sep 3 at 12:02
add a comment |Â
2
You can use Taylor-Lagrange inequality for the remainder: if you can bound the derivatives at all orders, it enables you to find an estimate for this degree.
– Bernard
Sep 3 at 11:50
I didn't know you had to consider the remainder for this, thank you!
– pavus
Sep 3 at 12:02
2
2
You can use Taylor-Lagrange inequality for the remainder: if you can bound the derivatives at all orders, it enables you to find an estimate for this degree.
– Bernard
Sep 3 at 11:50
You can use Taylor-Lagrange inequality for the remainder: if you can bound the derivatives at all orders, it enables you to find an estimate for this degree.
– Bernard
Sep 3 at 11:50
I didn't know you had to consider the remainder for this, thank you!
– pavus
Sep 3 at 12:02
I didn't know you had to consider the remainder for this, thank you!
– pavus
Sep 3 at 12:02
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903728%2fcalculate-the-necessary-degree-of-expansion-for-a-taylor-series%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
You can use Taylor-Lagrange inequality for the remainder: if you can bound the derivatives at all orders, it enables you to find an estimate for this degree.
– Bernard
Sep 3 at 11:50
I didn't know you had to consider the remainder for this, thank you!
– pavus
Sep 3 at 12:02