Taylor series with a base point different from $0$
up vote
4
down vote
favorite
What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ 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 yesterday
Andrews
12415
asked yesterday
archaic
575
add a comment |
up vote
4
down vote
favorite
What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ 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 yesterday
Andrews
12415
asked yesterday
archaic
575
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
yesterday
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ 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 yesterday
Andrews
12415
asked yesterday
archaic
575
What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ 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 yesterday
Andrews
12415
asked yesterday
archaic
575
edited yesterday
Andrews
12415
asked yesterday
archaic
575
edited yesterday
Andrews
12415
edited yesterday
Andrews
12415
edited yesterday
Andrews
12415
12415
asked yesterday
archaic
575
asked yesterday
archaic
575
asked yesterday
archaic
575
575
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
yesterday
add a comment |
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
yesterday
4
4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
yesterday
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
yesterday
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
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
$$
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 yesterday
Arthur
109k7103186
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
yesterday
This is the point of my question by the way.
– archaic
yesterday
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
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
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
yesterday
add a comment |
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, 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 yesterday
José Carlos Santos
144k20112212
add a comment |
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
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
$$
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 yesterday
Arthur
109k7103186
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
yesterday
This is the point of my question by the way.
– archaic
yesterday
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
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
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
yesterday
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
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
$$
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 yesterday
Arthur
109k7103186
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
yesterday
This is the point of my question by the way.
– archaic
yesterday
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
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
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
yesterday
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
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
$$
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 yesterday
Arthur
109k7103186
Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
$$
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 yesterday
Arthur
109k7103186
answered yesterday
Arthur
109k7103186
answered yesterday
Arthur
109k7103186
answered yesterday
Arthur
109k7103186
109k7103186
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
yesterday
This is the point of my question by the way.
– archaic
yesterday
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
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
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
yesterday
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
yesterday
This is the point of my question by the way.
– archaic
yesterday
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
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
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
yesterday
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
yesterday
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
yesterday
This is the point of my question by the way.
– archaic
yesterday
This is the point of my question by the way.
– archaic
yesterday
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
yesterday
@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
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
– archaic
yesterday
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
yesterday
@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
yesterday
add a comment |
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, 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 yesterday
José Carlos Santos
144k20112212
add a comment |
up vote
6
down vote
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, 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 yesterday
José Carlos Santos
144k20112212
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 $sqrt{4+frac15}$, 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 yesterday
José Carlos Santos
144k20112212
If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, 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 yesterday
José Carlos Santos
144k20112212
answered yesterday
José Carlos Santos
144k20112212
answered yesterday
José Carlos Santos
144k20112212
answered yesterday
José Carlos Santos
144k20112212
144k20112212
add a comment |
add a comment |
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4
How would you find the Taylor series for $log(x)$ at $0$?
– jjagmath
yesterday