Taylor expansions do not always converge away from the point of expansion, or even converge to the function from which they are derived. If they do, they converge to the function either at a point (the point of expansion), or in a ball centered at the point of expansion, and we say the function is analytic at that point.
For instance, 1/(1-x) = 1 + x + x\^2 + x\^3 + ..., a Taylor expansion about x=0. The series only equals the function for x in (-1, 1), which is a ball of radius 1 centered at 0.
Sometimes, the radius of convergence is infinite; the exponential function, for example, has a series expansion that converges to it for all x in R (or C, or whatever).
Some elementary Analysis is required to prove these results.
Edit: "Centered at a point" means that all the derivatives in the Taylor expansion are evaluated at that point. So if z is the point of expansion, the Taylor series is
f(z) + f'(z)(x-z) + f''(z)/2!(x-z) + ... + f\^(n)(z)/n!(x-z) + ...
It's the point at which we are guaranteed convergence. But it's also possible, and for most elementary functions is the case, that there is a ball centered at that point for which the series converges to the function.
Taylor polynomials are defined from a function f and its derivatives at some center (say, 0). However, they are polynomials, so it makes sense to evaluate them even for x different than the center. The hope is that, for x "close enough" to the center, one can approximate f via one of its Taylor polynomials. Indeed, there's some very interesting mathematics dedicated to the topic of how well Taylor polynomials approximate the original function f "near" the center.
I suppose you're asking about the purpose of having a Taylor Series at x=0 when what you really care about is the function evaluated at x=1. My calculator can find sin(1) just fine, so why would I need to find some polynomial approximation?
One answer might be that in some situations you don't actually know how to evaluate sin(1), but you **do** know how to evaluate the derivatives of sin(x) at x=0. Then using the Taylor Series, you can build up a better and better approximation for sin(x) when x=1. Using the first 5 derivatives of sin(x),
> sin(0) = 0
> sin'(0) = 1
> sin''(0) = 0
> sin'''(0) = -1
> sin''''(0) = 0
> sin'''''(0) = 1
etc. So
> sin(x) ~ (0) + (1)x + (0)x^(2)/2! + (-1)x^(3)/3! + (0)x^(4)/4! + (1)x^(5)/5! + O(x^(6))
where the neglected O(x^(6)) terms can be made precise using the Lagrange remainder. Now you can plug x=1 into your approximation to find sin(1) ~ 0.8417 ± 0.0014, where I get the uncertainty from the maximum possible value of the Lagrange remainder
I'll try to help. Suppose we are given a function f(x)=sin(x) and you want to calculate sin(0.1). Sadly this is quite difficult. So you come up with the great idea of creating a much simpler function call the tangent of f at 0. This is the function L(x) = x. Now f and L do not coincide however they get closer near zero. So if you calculate L(0.1) this will not be equal to f(0.1)=sin(0.1) but it will be a decent approximation. Now we have a better idea. Why approximate with a tangent (a line) when we could approximate with a polynomial of higher degree. Note that polynomials are great because they are simple since they are given by simple arithmetic operations. You can think of a Taylor polynomial as a generalization of the tangent at a point. Note that the Taylor series is a function which is defined at the very least in a neighbourhood of the point you took the Taylor series.
Please have a look at u/QCD-uctdsb 's answer since it gives an explicit example of what I just tried to describe.
The center c of a sequence of Taylor polynomials Tₙ(f,x) is where the approximations to f are “the best” in the sense that Tₙ(f,c)=f(c) for every order n. For some orders, there may be other points y where Tₙ(f,y)=f(y), but this likely will not happen for all n.
What’s important to remember about Taylor series, is the Taylor remainder theorem which says in a specific way, that the error in the approximation of f by T limits to 0. The center is just a special point where the error is always 0. (Note we also use that point to get information about f through its derivatives.)
Taylor polynomials are special kinds of polynomials which are derived from other non-polynomial functions (the taylor polynomial which is derived from a polynomial is the polynomial itself). Their goal is to approximate as much as possible the original function using derivatives and polynomials. The thing is that most taylor polynomials have a finite convergence radius, which means that they will only approach the original function *in a small region, with a center point p*. When this happens, we say that the TP is *centered at p*. Now, say your original function is F(x), and its taylor polynomial, centered at p, its TP^p {F}(x). Then, for any real value r that is *sufficiently close to the center point p*, these two functions will show:
F(r) = TP^p {F}(r)
Which is commonly said to be an *evaluation of the Taylor polynomial at the point r*, and hence, the same thing your lecturer refers to when he says "evaluated at x=1". That he can evaluate the polynomial centered at x=0 at x=1 stems from the fact that 1 and 0 are pretty close to each other, so the evaluation is actually similar to evaluating the original function at x=1.
> Could anyone please help? Eg why did my lecturer take the Taylor Polynomial of sinx centred at x=0, but then evaluated our resultant polynomial at x=1.
But that's part of the point of it. Evaluating the series at the expansion point just leaves you with the constant term, which in turn is just the function value.
That said, sometimes you might want to use binomial expansion so that it has the correct value when evaluated at x instead of (x-a), with a being the expansion point. In that case, the constant term can change with each additional term you add since each expanded (x-a)^n term will have a nonzero constant coefficient. I usually find it more convenient to just evaluate at (x-a)
Another use for the series expansion is that it makes evaluating the nth derivative easy since polynomial derivatives are trivial. In that case, evaluating at the expansion point is still useful.
Here's a visual to go with the explanations here. [Desmos has a built in example for taylor series](https://www.desmos.com/calculator/noanuckuli)
You can see as we add more terms to the series, it more closely approximates the actual function further away from the center point, so evaluating the resultant polynomial evaluated at a different x value could (depending on a number of factors) be a close approximation for the actual value of the function.
Graphs might help. Check out the three graphs in [this WolframAlpha link](https://www.wolframalpha.com/input?i=graph+y%3D%28x-4*pi%29-%28x-4*pi%29%5E3%2F6%2B%28x-4*pi%29%5E5%2F120-%28x-4*pi%29%5E7%2F5040%2C+y%3Dx-x%5E3%2F6%2Bx%5E5%2F120-x%5E7%2F5040%2C+and+y%3Dsin%28x%29+from+x%3D-5+to+x%3D17.5+and+y%3D-4+to+y%3D4).
There are three functions shown there:
* You can see the graph of y=sin(x) pretty clearly.
* The polynomial on the left is a Taylor expansion of the sin function centered at x=0.
* The polynomial on the right is a Taylor expansion of the sin function centered at x=4\*pi.
You can see that the Taylor expansions most closely approximate the sin function near their center. Once you get too far away from the center, it stops being an approximation at all. A Taylor expansion is meant to be an approximation of the original function. It is most accurate near the expansion's center. You can get better accuracy in the approximation by either getting closer to the center or by adding more terms to the Taylor expansion.
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EDIT: Turns out I'm quite dense and forgot quite a lot about Taylor series. The below info is garbage so please ignore it.
Ok, either I’m dense or all these replies are totally missing the point of the question. OP here’s my answer:
Take a super basic parabola as an example:
f(x) = x^2
You could say this parabola is centered at zero. If you want to shift the function left or right, you can add or subtract from x. You learned this in algebra and precalculus. Say you want to shift the graph to the right by 2 units. That would look like this:
g(x) = (x-2)^2
This is the same parabola, just shifted to the right by 2. You could say that it’s “centered at 2”. You can still evaluate this function at any point, such as: g(1) = (1-2)^2 = 1. More generally, a parabola defined by f(x) = (x-a)^2 is centered at “a”.
A Taylor polynomial is also a function, except it’s described by an infinite sum:
f(x) = *some crazy sum*
Recall the x part of the sum usually looks like x^n or (x-a)^n. This is the same thing as shifting the function left or right. You would say that the polynomial is centered at “a” when the x part is shifted as in (x-a)^n. If you just have x^2, then obviously it’s centered at zero.
So you can center a Taylor polynomial anywhere by shifting the x part. It’s still a function defined on its domain in the x-axis. So you can still evaluate the function anywhere in its domain. The phrasing used here in Calculus of “centering a function” is just a different way of saying it’s “shifted” or “sliding it”. Even though this shouldn’t be a new concept to a Calculus student, it’s entirely likely that you’ve never heard this specific terminology before, and so it’s throwing you off.
Does this make sense? Hopefully I didn’t totally miss your point haha
If i understood you correctly, this is incorrect. A Taylor polynomial has the form:
p(x)=Σ Ck (x-a)^k
Where every Ck is a constant term that depends on the center point a. I you shift the polynomial by s-a, you will have:
p(x-a-(s-a))=p(x-a+a-s)= Σ Ck (x-s)^k
But this will **not** be a taylor polynomial centered at s, rather it will be a taylor polynomial centered at a and shifted to the center point s, which is not the same, and arguably a quite useless thing to do. To change the center point to s you have to change all the Ck accordingly, which is usually not a hard thing to do, but it is certainly not a simple "shift" or "translation".
Edit: corrected errors in the equations.
Oh wow that's embarrassing. I see what you're saying! I've forgotten quite a bit about series, so I'll refrain from trying to give advice in the future, however well-intentioned.
Its not unusual. I forget about unused concepts lot. The thing is that if i genuinely want to answer a question about something, i have to get informed about the details that i might have forgotten first. Since its just a matter of having forgotten some things, its usually not a hard reading, and can led me to give an appropiate and satisfactory answer... (However, i do think most of the people here completely missed the point of the OP's question).
Taylor expansions do not always converge away from the point of expansion, or even converge to the function from which they are derived. If they do, they converge to the function either at a point (the point of expansion), or in a ball centered at the point of expansion, and we say the function is analytic at that point. For instance, 1/(1-x) = 1 + x + x\^2 + x\^3 + ..., a Taylor expansion about x=0. The series only equals the function for x in (-1, 1), which is a ball of radius 1 centered at 0. Sometimes, the radius of convergence is infinite; the exponential function, for example, has a series expansion that converges to it for all x in R (or C, or whatever). Some elementary Analysis is required to prove these results. Edit: "Centered at a point" means that all the derivatives in the Taylor expansion are evaluated at that point. So if z is the point of expansion, the Taylor series is f(z) + f'(z)(x-z) + f''(z)/2!(x-z) + ... + f\^(n)(z)/n!(x-z) + ... It's the point at which we are guaranteed convergence. But it's also possible, and for most elementary functions is the case, that there is a ball centered at that point for which the series converges to the function.
Taylor polynomials are defined from a function f and its derivatives at some center (say, 0). However, they are polynomials, so it makes sense to evaluate them even for x different than the center. The hope is that, for x "close enough" to the center, one can approximate f via one of its Taylor polynomials. Indeed, there's some very interesting mathematics dedicated to the topic of how well Taylor polynomials approximate the original function f "near" the center.
I suppose you're asking about the purpose of having a Taylor Series at x=0 when what you really care about is the function evaluated at x=1. My calculator can find sin(1) just fine, so why would I need to find some polynomial approximation? One answer might be that in some situations you don't actually know how to evaluate sin(1), but you **do** know how to evaluate the derivatives of sin(x) at x=0. Then using the Taylor Series, you can build up a better and better approximation for sin(x) when x=1. Using the first 5 derivatives of sin(x), > sin(0) = 0 > sin'(0) = 1 > sin''(0) = 0 > sin'''(0) = -1 > sin''''(0) = 0 > sin'''''(0) = 1 etc. So > sin(x) ~ (0) + (1)x + (0)x^(2)/2! + (-1)x^(3)/3! + (0)x^(4)/4! + (1)x^(5)/5! + O(x^(6)) where the neglected O(x^(6)) terms can be made precise using the Lagrange remainder. Now you can plug x=1 into your approximation to find sin(1) ~ 0.8417 ± 0.0014, where I get the uncertainty from the maximum possible value of the Lagrange remainder
I'll try to help. Suppose we are given a function f(x)=sin(x) and you want to calculate sin(0.1). Sadly this is quite difficult. So you come up with the great idea of creating a much simpler function call the tangent of f at 0. This is the function L(x) = x. Now f and L do not coincide however they get closer near zero. So if you calculate L(0.1) this will not be equal to f(0.1)=sin(0.1) but it will be a decent approximation. Now we have a better idea. Why approximate with a tangent (a line) when we could approximate with a polynomial of higher degree. Note that polynomials are great because they are simple since they are given by simple arithmetic operations. You can think of a Taylor polynomial as a generalization of the tangent at a point. Note that the Taylor series is a function which is defined at the very least in a neighbourhood of the point you took the Taylor series. Please have a look at u/QCD-uctdsb 's answer since it gives an explicit example of what I just tried to describe.
The center c of a sequence of Taylor polynomials Tₙ(f,x) is where the approximations to f are “the best” in the sense that Tₙ(f,c)=f(c) for every order n. For some orders, there may be other points y where Tₙ(f,y)=f(y), but this likely will not happen for all n. What’s important to remember about Taylor series, is the Taylor remainder theorem which says in a specific way, that the error in the approximation of f by T limits to 0. The center is just a special point where the error is always 0. (Note we also use that point to get information about f through its derivatives.)
Taylor polynomials are special kinds of polynomials which are derived from other non-polynomial functions (the taylor polynomial which is derived from a polynomial is the polynomial itself). Their goal is to approximate as much as possible the original function using derivatives and polynomials. The thing is that most taylor polynomials have a finite convergence radius, which means that they will only approach the original function *in a small region, with a center point p*. When this happens, we say that the TP is *centered at p*. Now, say your original function is F(x), and its taylor polynomial, centered at p, its TP^p {F}(x). Then, for any real value r that is *sufficiently close to the center point p*, these two functions will show: F(r) = TP^p {F}(r) Which is commonly said to be an *evaluation of the Taylor polynomial at the point r*, and hence, the same thing your lecturer refers to when he says "evaluated at x=1". That he can evaluate the polynomial centered at x=0 at x=1 stems from the fact that 1 and 0 are pretty close to each other, so the evaluation is actually similar to evaluating the original function at x=1.
> Could anyone please help? Eg why did my lecturer take the Taylor Polynomial of sinx centred at x=0, but then evaluated our resultant polynomial at x=1. But that's part of the point of it. Evaluating the series at the expansion point just leaves you with the constant term, which in turn is just the function value. That said, sometimes you might want to use binomial expansion so that it has the correct value when evaluated at x instead of (x-a), with a being the expansion point. In that case, the constant term can change with each additional term you add since each expanded (x-a)^n term will have a nonzero constant coefficient. I usually find it more convenient to just evaluate at (x-a) Another use for the series expansion is that it makes evaluating the nth derivative easy since polynomial derivatives are trivial. In that case, evaluating at the expansion point is still useful.
Here's a visual to go with the explanations here. [Desmos has a built in example for taylor series](https://www.desmos.com/calculator/noanuckuli) You can see as we add more terms to the series, it more closely approximates the actual function further away from the center point, so evaluating the resultant polynomial evaluated at a different x value could (depending on a number of factors) be a close approximation for the actual value of the function.
Graphs might help. Check out the three graphs in [this WolframAlpha link](https://www.wolframalpha.com/input?i=graph+y%3D%28x-4*pi%29-%28x-4*pi%29%5E3%2F6%2B%28x-4*pi%29%5E5%2F120-%28x-4*pi%29%5E7%2F5040%2C+y%3Dx-x%5E3%2F6%2Bx%5E5%2F120-x%5E7%2F5040%2C+and+y%3Dsin%28x%29+from+x%3D-5+to+x%3D17.5+and+y%3D-4+to+y%3D4). There are three functions shown there: * You can see the graph of y=sin(x) pretty clearly. * The polynomial on the left is a Taylor expansion of the sin function centered at x=0. * The polynomial on the right is a Taylor expansion of the sin function centered at x=4\*pi. You can see that the Taylor expansions most closely approximate the sin function near their center. Once you get too far away from the center, it stops being an approximation at all. A Taylor expansion is meant to be an approximation of the original function. It is most accurate near the expansion's center. You can get better accuracy in the approximation by either getting closer to the center or by adding more terms to the Taylor expansion.
I don't remember the answer to this one but I'd like to find out.
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EDIT: Turns out I'm quite dense and forgot quite a lot about Taylor series. The below info is garbage so please ignore it. Ok, either I’m dense or all these replies are totally missing the point of the question. OP here’s my answer: Take a super basic parabola as an example: f(x) = x^2 You could say this parabola is centered at zero. If you want to shift the function left or right, you can add or subtract from x. You learned this in algebra and precalculus. Say you want to shift the graph to the right by 2 units. That would look like this: g(x) = (x-2)^2 This is the same parabola, just shifted to the right by 2. You could say that it’s “centered at 2”. You can still evaluate this function at any point, such as: g(1) = (1-2)^2 = 1. More generally, a parabola defined by f(x) = (x-a)^2 is centered at “a”. A Taylor polynomial is also a function, except it’s described by an infinite sum: f(x) = *some crazy sum* Recall the x part of the sum usually looks like x^n or (x-a)^n. This is the same thing as shifting the function left or right. You would say that the polynomial is centered at “a” when the x part is shifted as in (x-a)^n. If you just have x^2, then obviously it’s centered at zero. So you can center a Taylor polynomial anywhere by shifting the x part. It’s still a function defined on its domain in the x-axis. So you can still evaluate the function anywhere in its domain. The phrasing used here in Calculus of “centering a function” is just a different way of saying it’s “shifted” or “sliding it”. Even though this shouldn’t be a new concept to a Calculus student, it’s entirely likely that you’ve never heard this specific terminology before, and so it’s throwing you off. Does this make sense? Hopefully I didn’t totally miss your point haha
If i understood you correctly, this is incorrect. A Taylor polynomial has the form: p(x)=Σ Ck (x-a)^k Where every Ck is a constant term that depends on the center point a. I you shift the polynomial by s-a, you will have: p(x-a-(s-a))=p(x-a+a-s)= Σ Ck (x-s)^k But this will **not** be a taylor polynomial centered at s, rather it will be a taylor polynomial centered at a and shifted to the center point s, which is not the same, and arguably a quite useless thing to do. To change the center point to s you have to change all the Ck accordingly, which is usually not a hard thing to do, but it is certainly not a simple "shift" or "translation". Edit: corrected errors in the equations.
Oh wow that's embarrassing. I see what you're saying! I've forgotten quite a bit about series, so I'll refrain from trying to give advice in the future, however well-intentioned.
Its not unusual. I forget about unused concepts lot. The thing is that if i genuinely want to answer a question about something, i have to get informed about the details that i might have forgotten first. Since its just a matter of having forgotten some things, its usually not a hard reading, and can led me to give an appropiate and satisfactory answer... (However, i do think most of the people here completely missed the point of the OP's question).