How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate #f(1/3)# given #f(x)=int_0^x sin(t^2) dt# ?

Question

Christopher Allers · Accepted Answer

# f(1/3) ~~0.01233 \ \ \ # (5dp)

The Maclaurin series for #sinx# is given by: # sinx= sum_(n=0)^oo (-1)^n/((2n+1)!)x^(2n+1) # # \ \ \ \ \ \ \ = x - x^3/(3!) + x^5/(5!) - x^7/(7!) + x^9/(9!) - ... # And so we can write the integrand #sin(t^2)# as : #sint^2 =(t^2) - (t^2)^3/(3!) + (t^2)^5/(5!) - (t^2)^7/(7!) + (t^2)^9/(9!) - ...# #\ \ \ \ \ \ \ \ =t^2 - t^6/(3!) + t^10/(5!) - t^14/(7!) + t^18/(9!) - ...# And so expressing the integrand as an infinite power series, our function #f(x)# becomes # f(x) = int_0^x sin t^2 \ dt # # \ \ \ \ \ \ \ = int_0^x t^2 - t^6/(3!) + t^10/(5!) - t^14/(7!) + t^18/(9!) - ... \ dt # Note: mathematical vigor will show that in some cases we can do this, and in this specific case the method is valid. However, this assumes that we can actually replace the integrand with an infinite power series and the integral remain valid. Term by term, we now integrate to obtain: # f(x) = [t^3/3 - t^7/(7(3!)) + t^11/(11(5!)) - t^15/(15(7!)) + t^19/(19(9!)) - ... ]_0^x # # \ \ \ \ \ \ = x^3/3 - x^7/(7(3!)) + x^11/(11(5!)) - x^15/(15(7!)) + x^19/(19(9!)) - ... # Thus, the initial pair of non-zero terms is: # f(x) = x^3/3 - x^7/42 + ... # And so using just these terms, we can approximate #f(1/3)# as # f(1/3) ~~ (1/3)^3/3 - (1/3)^7/42# # \ \ \ \ \ \ \ \ \ \ \ =1133/91854 # # \ \ \ \ \ \ \ \ \ \ \ =0.01233479# # \ \ \ \ \ \ \ \ \ \ \ =0.01233 \ \ \ # (5dp)