Taylor series with a base point different from $0$
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up vote
4
down vote
favorite
What's the need for $f(x) = sum_k=0^inftyfracf^(k)(a)k!(x-a)^k$ if we already have the formula at $0$?
Isn't the $(x-a)$ just making the $a$ as the new origin?
When is this formula more useful than that at $0$?
functions taylor-expansion
edited Dec 5 at 15:56
Andrews
318117
asked Dec 5 at 14:58
archaic
575
add a comment |
up vote
4
down vote
favorite
What's the need for $f(x) = sum_k=0^inftyfracf^(k)(a)k!(x-a)^k$ if we already have the formula at $0$?
Isn't the $(x-a)$ just making the $a$ as the new origin?
When is this formula more useful than that at $0$?
functions taylor-expansion
edited Dec 5 at 15:56
Andrews
318117
asked Dec 5 at 14:58
archaic
575
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
Dec 5 at 15:03
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
What's the need for $f(x) = sum_k=0^inftyfracf^(k)(a)k!(x-a)^k$ if we already have the formula at $0$?
Isn't the $(x-a)$ just making the $a$ as the new origin?
When is this formula more useful than that at $0$?
functions taylor-expansion
edited Dec 5 at 15:56
Andrews
318117
asked Dec 5 at 14:58
archaic
575
What's the need for $f(x) = sum_k=0^inftyfracf^(k)(a)k!(x-a)^k$ if we already have the formula at $0$?
Isn't the $(x-a)$ just making the $a$ as the new origin?
When is this formula more useful than that at $0$?
functions taylor-expansion
functions taylor-expansion
edited Dec 5 at 15:56
Andrews
318117
asked Dec 5 at 14:58
archaic
575
edited Dec 5 at 15:56
Andrews
318117
asked Dec 5 at 14:58
archaic
575
edited Dec 5 at 15:56
Andrews
318117
edited Dec 5 at 15:56
Andrews
318117
edited Dec 5 at 15:56
Andrews
318117
318117
asked Dec 5 at 14:58
archaic
575
asked Dec 5 at 14:58
archaic
575
asked Dec 5 at 14:58
archaic
575
575
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
Dec 5 at 15:03
add a comment |
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
Dec 5 at 15:03
4
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
Dec 5 at 15:03
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
Dec 5 at 15:03
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac(x-1)^22 + frac(x-1)^33 - cdots
$$
Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.
answered Dec 5 at 15:03
Arthur
110k7104186
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
2
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
3
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
add a comment |
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt4+frac15$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac(x-1)^22 + frac(x-1)^33 - cdots
$$
Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.
answered Dec 5 at 15:03
Arthur
110k7104186
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
2
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
3
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
add a comment |
up vote
7
down vote
accepted
Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac(x-1)^22 + frac(x-1)^33 - cdots
$$
Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.
answered Dec 5 at 15:03
Arthur
110k7104186
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
2
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
3
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac(x-1)^22 + frac(x-1)^33 - cdots
$$
Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.
answered Dec 5 at 15:03
Arthur
110k7104186
Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac(x-1)^22 + frac(x-1)^33 - cdots
$$
Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.
answered Dec 5 at 15:03
Arthur
110k7104186
answered Dec 5 at 15:03
Arthur
110k7104186
answered Dec 5 at 15:03
Arthur
110k7104186
answered Dec 5 at 15:03
Arthur
110k7104186
110k7104186
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
2
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
3
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
add a comment |
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
2
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
3
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
This is the point of my question by the way.
– archaic
Dec 5 at 15:08
2
2
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
@archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
– Arthur
Dec 5 at 15:12
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
Dec 5 at 15:14
3
3
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
@archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
– Shufflepants
Dec 5 at 15:32
add a comment |
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt4+frac15$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
add a comment |
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt4+frac15$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
add a comment |
up vote
6
down vote
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt4+frac15$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt4+frac15$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
answered Dec 5 at 15:02
José Carlos Santos
146k22116216
146k22116216
add a comment |
add a comment |
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4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
Dec 5 at 15:03