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Taylor expansion centering question

  1. Sep 13, 2011 #1
    what does it mean to say taylor expansion of ex centered at 0?

    does it mean that the sum of the expansion will give me the value that the function ex will take when x = 0 ?

    so its e0 = 1?



    also, how do we know what value to center on? because as i encounter taylor series in my undergrad years, it seems that the professors all just say

    "taylor expanding this particular function, we have.. blah blah blah.."

    they don't seem to say centered about what? is it always assume to be centered upon 0?

    thanks!
     
  2. jcsd
  3. Sep 13, 2011 #2

    chiro

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  4. Sep 13, 2011 #3
    haha..

    as a matter of fact, i actually read that page before coming here, in part to double check whether i understood correctly

    so if i understood correctly, taylor series is something like this

    i have function say 10

    taylor expansion does something like 1+3+5 which is approximately 10





    lets take the example in the wiki page, where the first graph showed sin x
    is it saying that the first term, which is the first degree polynomial, is the red straight line in the first graph of the wiki page?

    so that the green, which is degree 3, is solely due to just the term in taylor expansion with power 3? i.e, it doesn't include the first term, 2nd term of taylor expansion. or does it include every smaller degree terms before it? since they say i have better approximation as i increase the degrees, i assume it includes every smaller degree terms before it?

    then what about 2,4,6 etc? why aren't they showed?


    but with regards to the centered at what point, is my point in post 1 correct? wiki doesn't seem to explain much about centering.

    what does it mean?
     
    Last edited: Sep 13, 2011
  5. Sep 13, 2011 #4
    Degrees 2,4,6 etc are not shown on the graph because the Taylor expansion for sinx doesn't include those terms.

    sinx=x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

    Yes you are correct that when they show degree 3 on the graph it includes the lower degree terms as well.

    Now about the power series being centered at a point, consider the function:

    1/(1-x)=1 + x + x^2 + x^3 + ...

    You've probably seen this in your calculus course, so you know that equality only holds for |x|<1. This means that the power series for 1/(1-x) has a radius of convergence of 1. The centre, in this case 0, is the centre of the interval from which that radius extends.

    If we want to change the centre we can, for example lets change the centre to 2. Then we'll have:

    1/[1-(x-2)]= 1 + (x-2) + (x-2)^2 + (x-2)^3 + ...

    This series has the same radius of convergence, namely 1, but it will be valid for different values of x than the original series. This series is valid for:

    |x-2|< 1 or 1 < x < 3.

    In the case of sinx or e^x, their Taylor series are convergent for the whole real line, so in most cases we can just choose 0 as the centre since it is the simplest.

    I hope that helped a little bit, let me know if I can clarify anything for you.
     
  6. Sep 13, 2011 #5

    HallsofIvy

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    ??? "10" is not a function, it is a number. It makes no sense to talk about a Taylor series for a number.



    Yes, every higher degree includes the lower degree terms- this a a polynomial, not a mononial. Taylor series, by the way, have the nice property that if you increase the degree, the coefficients of all lower powers stay the same. Some other ways of approximating functions by polynomials don't have that property- to increase the degree of the polynomial, you have to recalculate all coefficients.

    sin(x) is an odd function- sin(-x)= -sin(x). Since the only "odd" polynomials are those with only odd power (that's where the term came from) all the coefficients of the even powers are 0. "Adding" an even power term doesn't change the polynomial.

    Another way to see that is to observe that all even derivatives of sin(x) are either sin(x) or -sin(x), and sin(0)= 0.


    The Taylor series of a function f(x), centered at x= a, is, as that Wikipedia article says,
    [tex]\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x- a)^n[/tex]
    It isn't really necessary to state the "center" since it is clearly shown in that "(x- a)".

    The Taylor series for [itex]e^x[/itex], centered at x= 0, is NOT just the single number [itex]e^0= 1[/itex], it is, rather, the infinite power series
    [tex]e^0+ \frac{e^0}{1!}(x- 0)+ \frac{e^0}{2!}(x-0)^2+ \cdot\cdot\cdot+ \frac{e^0}{n!}(x-0)^n+ \cdot\cdot\cdot[/tex]
    [tex]= 1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot[/tex]
    since all derivatives of [itex]e^x[/itex] are [itex]e^x[/itex] and, of course, [itex]e^0= 1[/itex].
    (That includes both even and odd powers because [itex]e^{x}[/itex] is neither "even" nor "odd" function: [itex]e^{-x}\ne e^x[/itex], [itex]e^{-x}\ne -e^x[/itex].)

    The Taylor series for [itex]e^x[/itex], centered at x= 1, instead, would be
    [tex]= e+ e(x-1)+ \frac{e}{2}(x- 1)^2+ \cdot\cdot\cdot+ \frac{e}{n!}(x-1)^n+ \cdot\cdot\cdot[/tex]
    since now we evaluate the evaluate the derivative at x= 1 and [itex]e^1= e[/itex].

    By the way, it is interesting to look at Taylor series for polynomials. Suppose [itex]f(x)= x^4- 3x^2+ 2x+ 1[/itex]. f(0)= 1, f'(0)= 2, f''(0)= -6, [itex]f^(3)(x)= 0[/itex], and [itex]f^(4)(0)= 24[/itex] and all higher derivatives are identically 0 so the Taylor polynomials of degrees 1 through 4, centered at x= 0, are degree 1: [itex]f(0)+ f'(0)x= 1+ 2x[/itex], degree 2: [itex]f(0)+ f'(0)x+ (f''(0)/2)x^2= 1+ 2x- 3x^2[/itex], degree 3: [itex]f(0)+ f'(0)x+ (f''(0)/2)x^2+ (f'''(0)/3!)x^3= 1+ 2x- 3x^2[/itex], degree 4: [itex]f(0)+ f'(0)x+ (f''(0)/2)x^2+ (f'''(0)/3!)x^3+ (f^(4)(0)/4!)x^4= 1+ 2x- 3x^2+ x^4[/itex]. That is, the Taylor polynomials of degree less than 4 are just the terms up to that degree while the Taylor series of degree 4 or higher (including the "infinite" series) is just the polynomial itself.

    But suppose we wanted the Taylor series for that polynomial centered at 1 rather than 0:
    f(1)= 1- 3+ 2+ 1= 1, f'(1)= 4- 6+ 2= 0, f''(1)= 12- 6= 6, f'''(1)= 24, and [itex]f^(4)(1)= 24[/itex] so the Taylor series is
    [tex]1+ (0/1)(x- 1)+ (6/2)(x- 1)^2+ (24/6)(x- 1)^3+ (24/24)(x- 1)^4[/tex][tex]= 1+ 3(x-1)^2+ 4(x-1)^3+ (x-1)^4[/tex]

    And, indeed, if you multiply it out, you will see that
    [tex]1+ 3(x-1)^2+ 4(x-1)^3+ (x-1)^4[/tex][tex]= 1+ 2x- 3x^2+ x^4[/tex]

    That's a convenient way of "shifting" a polynomial.
     
    Last edited: Sep 13, 2011
  7. Sep 14, 2011 #6
    wwow...


    thanks a lot everyone. now i need to digest
     
  8. Sep 15, 2011 #7

    morrobay

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    Is there a limit on how far the polynomial can be shifted ?
    For example can the Taylor series expansion still generate the original function
    if the polynomial is centered at 2,3,4...8,9 ?
     
    Last edited: Sep 15, 2011
  9. Sep 16, 2011 #8
    You can take any a in the domain of the function that you like, however the Taylor expansion won't ever 'generate' the original function, it simply approximates it in polynomial form
     
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