Power Series and Estimation of Integrals

Key Questions

  • How do you use a Power Series to estimate the integral #int_0^0.01sin(x^2)dx# ?

    Assuming that you know that the power series for #sinx# is:

    #sinx=sum_(n=1)^infty ((-1)^(n-1)x^(2n-1))/(2n-1)=x-(x^3)/(3!)+(x^5)/(5!)+...#

    then we can answer this fairly quickly. If not, perhaps that can be a separate question!

    So, if:

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

    Then:

    #sinx^2=x^2-(x^2)^3/(3!)+((x^2)^5)/(5!)+((x^2)^7)/(7!)...#

    Which can be re-written as:

    #sinx^2=x^2-1/(3!)x^6+1/(5!)x^10+1/(7!)x^14...#

    So then:

    #int_0^0.01 sinx^2=int_0^0.01 x^2-1/(3!)int_0^0.01x^6+1/(5!)int_0^0.01x^10...#

    #=((x^3)/3-1/(3!)(x^7)/7+1/(5!)(x^11)/11+...)|_0^0.01#

    When we plug in zero for #x#, all the terms will disappear. So all you have to do is plug in #0.01# for x out to however many terms you want.

    Hope this helps!

    Paul Belliveau · · Oct 1 2014
  • How do you use a Power Series to estimate the integral #int_0^0.01e^(x^2)dx# ?

    Since

    #e^x=1+x+x^2/{2!}+cdots#,

    #e^{x^2}=1+x^2+x^4/{2!}+cdots#

    #int_0^{0.01}e^{x^2}dx=int_0^{0.01} (1+x^2+x^4/{2!}+cdots)dx#

    #=[x+x^3/3+x^5/{10}+cdots]_0^{0.01}#

    #=0.01+(0.01)^3/3+(0.01)^5/10+cdots#

    #approx 0.01#

    I hope that this was helpful.

    Wataru · · Oct 16 2014
  • How do you use a Power Series to estimate an integral?

    Answer:

    Integrate term by term

    Explanation:

    Replace f by its Taylor expansion. f is now a sum
    #f = sum a^n x^n#
    Integrate term by term
    #int f dx = sum n a^n x^(n-1)#

    t0hierry · 1 · Mar 31 2018

Questions

Power Series