Answer to Question #156891 in Calculus for Ayush

Question

Expand e

ax cos bx by Maclaurins theorem. Hence prove that

e

x cos α

cos(x sin α) = 1 + x cos α +

x

2

2!

cos 2α +

x

3

3!

cos 3α + ⋯ ...

Answer 1

$Solution: By~ Maclurin's~ theorem, We~ have \\y=(y)_0+\frac{x}{1!}(y_1)_0+\frac{x^2}{2!}(y_2)_0+\frac{x^3}{3!}(y_3)_0+ \frac{x^4}{4!}(y_4)_0+..........(1) \\or \\f(x)=f(0)+\frac{x}{1!}f'(0)+ \frac{x^2}{2!}f''(0)+ \frac{x^3}{3!}f'''(0)+\frac{x^4}{4!!}f'''''(0)+...........\frac{x^n}{n!}f^n(0)$

$\\Now~we~have ~to~find ~(y)_0,(y_1)_0 ~and~so~on... \\ y=e^{ax}cos(bx) ~\Rightarrow (y)_0=e^{a(0)}cos ~(b(0))=1 \\(y_1)=ae^{ax}cos(bx)+e^{ax}(-b ~sin(bx))=ae^{ax}cos(bx)-e^{ax}(b ~sin(bx)) \\ \therefore y_1=ay-be^{ax}sin(bx) \\ \Rightarrow (y_1)_0=a(1)-be^{a(0)}sin(0)=a$

$y_2=ay_1-(abe^{ax}sin(bx)+e^{ax}b^2cos(bx)) \\~~~~~=ay_1-abe^{ax}sin(bx)-e^{ax}b^2cos(bx) \\~~~~~=ay_1-abe^{ax}sin(bx)-b^2e^{ax}cos(bx) \\~=ay_1-a(ay-y_1)-b^2y[Since~y_1=ay-be^{ax}sin(bx) ~\Rightarrow be^{ax}sin(bx)=ay-y_1 ] \\~=ay_1-a^2y-ay_1-b^2y \\~=2ay_1-(a^2+b^2)y \\ \Rightarrow ~(y_2)_0=2a(a)-(a^2+b^2)(1)=2a^2-a^2-b^2=a^2-b^2$

$y_3=2ay_2-(a^2+b^2)y_1 \\ \Rightarrow (y_3)_0=2a(a^2-b^2)-(a^2+b^2)a=2a^3-2ab^2-a^3-ab^2=a^3-3ab^2 \\ \Rightarrow (y_3)_0=a(a^2-3b^2) \\Smilarly ~we~ can~ find~ next~ derivatives ~at ~point~ 0 \\Putting ~all~ these~ values~ in~ Maclurin's ~Theorem~i.e. ~in~equation~(1),$

$y=(y)_0+\frac{x}{1!}(y_1)_0+\frac{x^2}{2!}(y_2)_0+\frac{x^3}{3!}(y_3)_0+ \frac{x^4}{4!}(y_4)_0+....... \\\therefore e^{ax} cos(bx)=1+\frac{x}{1!}(a)+\frac{x^2}{2!}(a^2-b^2)+\frac{x^3}{3!}a(a^2-3b^2)+.... \\~~~~e^{ax} cos(bx)=1+\frac{a}{1!}(x)+\frac{a^2-b^2}{2!}(x^2)+\frac{a(a^2-3b^2)}{3!}x^3+.............(2)$

$Hence~we~proved~the~ first~part~of~the~example. \\Now~to~prove~the~ second ~part~of~the~example~we~put~ a=cos~\alpha~and ~b=sin~\alpha~in~equation~(2) \\e^{x~cos~\alpha } cos(x~sin~\alpha )=1+\frac{cos~\alpha}{1!}(x)+\frac{cos^2~\alpha-sin^2~\alpha}{2!}(x^2)+\frac{cos~\alpha(cos^2~\alpha-3sin^2~\alpha)}{3!}x^3+............. \\Simplifying , we~ get \\\\e^{x~cos~\alpha } cos(x~sin~\alpha )=1+x~cos~\alpha+\frac{x^2}{2!}(cos~2\alpha)+\frac{x^3}{3!}(cos~3\alpha)+............. \\Hence~ proved.$