Using a Taylor's series or otherwise; derive Euler's Formula

Use the Taylor series expansion for the following three functions: f(θ) = e^(iθ), g(θ) = cos(θ) and h(θ) = sin(θ). We should find that f(θ) = e^(iθ) = 1 + iθ - (θ^2/2!) - i(θ^3/3!) + ... = Sum(θ^n/n!), g(θ) = cos(θ) = 1 - (θ^2/2!) + (θ^4/4!) - (θ^6/6!) + ... = Sum((-1)^n . (θ^2n))/2n!) and finally, g(θ) = sin(θ) = θ - (θ^3/3!) + (θ^5/5!) + ... = Sum((-1)^n . (θ^2n+1))/2n+1!). Now it is a case of manipulating a our results for our functions to match Euler's Formula. Since we know e^(iθ) = cos(θ) + isin(θ) is Euler's Formula, and that we've been asked to use a Taylor series expansion, it is just a case of algebraic manipulation, starting from either the LHS or the RHS to achieve the other part of the equation.Let's start from the LHS (for powers of θ up to 5) : e^(iθ) = 1 + iθ - (θ^2/2!) - i(θ^3/3!) + (θ^4/4!) + i(θ^5/5!) - ... = (1 - (θ^2/2!) + (θ^4/4!) - ...) + i(θ - (θ^3/3!) + (θ^5/5!) - ...) and so on. If you notice the first term corresponds to the Taylor expansion of cos(θ) and the second to the expansion of i(sin(θ)) and hence we can say that e^(iθ) = cos(θ) + isin(θ) and the derivation of Euler's Formula using a Taylor's series is complete.

MH
Answered by Mansour H. Further Mathematics tutor

3678 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

A curve has equation y=(2-x)(1+x)+3, A line passes through the point (2,3) and the curve at a point with x coordinate 2+h. Find the gradient of the line. Then use that answer to find the gradient of the curve at (2,3), stating the value of the gradient


Find the stationary points of the function z = 3x(x+y)3 - x3 + 24x


Prove e^(ix) = cos (x) + isin(x)


Prove that ∑(1/(r^2 -1)) from r=2 to r=n is equal to (3n^2-n-2)/(4n(n+1)) for all natural numbers n>=2.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences