functional equation (1)

2 posters

jacks mathematics forum :: MATHEMATICS :: CALCULUS

Page 1 of 1

functional equation (1)

Post Admin Sun Jul 24, 2011 8:39 am

$functional equation (1) Gif.latex?\hspace{-15}$\textbf{\underline{\underline{Questions%20on%20functional%20equation}}}.\\\\%20(1):%20If%20$f:\mathbb{R}\rightarrow%20\mathbb{R}.$%20and%20$f(x)-2f(-x)=x^2$.%20then%20$f(x)=$\\\\%20(2):%20If%20$f:\mathbb{R}\rightarrow%20\mathbb{R}.$%20and%20$x.f(x)+f(1-x)=x^2$.%20then%20$f(x)=$\\\\%20(3):%20If%20$f:\mathbb{R}\rightarrow%20\mathbb{R}.$%20and%20$f(x)+2f\left(\frac{1}{x}\right)=x^2$$

Admin: Admin; Posts : 103
Join date : 2011-02-01
Location : India

funct. eqns.

Post mohit Sun Jul 24, 2011 1:58 pm

$functional equation (1) Gif.latex?\hspace{-15}(1)::;f(x)=-x^2\\\\%20(3):::f(x)=\frac{1}{3}$

(by Mohit)

mohit: Posts : 8
Join date : 2011-07-23

Re: functional equation (1)

Post Admin Sun Jul 24, 2011 9:11 pm

$functional equation (1) Gif.latex?\hspace{-15}\mathbf{(1)}:::f(x)-2f(-x)=x^2....................................(1)$\\\\%20Now%20Replacing%20$x\rightarrow%20-x,\;$%20we%20get\\\\%20$f(-x)-2f(x)=(-x)^2=x^2.......................(2)\times%202$\\\\%20$2f(-x)-4f(x)=2x^2..................................(2)$\\\\%20Now%20Add%20Equ..$

Admin: Admin; Posts : 103
Join date : 2011-02-01
Location : India

Re: functional equation (1)

Post Admin Mon Jul 25, 2011 12:50 am

$functional equation (1) Gif.latex?\hspace{-15}(3):::%20f(x)+2f\left(\frac{1}{x}\right)=x^2.......................(1)$\\\\%20Now%20Replace%20$x\rightarrow%20\frac{1}{x},$%20we%20Get\\\\%20$f\left(\frac{1}{x}\right)+2f(x)=\frac{1}{x^2}................................(2)\times%202$\\\\%20Now%20Sub%20$(2)-(1),$%20We%20Get\\\\%20$3f(x)=\frac{2}{x^2}-x^2$\\\\%20$\boxed{\boxed{f(x)=\frac{1}{3}$

Admin: Admin; Posts : 103
Join date : 2011-02-01
Location : India

Re: functional equation (1)

Post Admin Mon Jul 25, 2011 1:02 am

$functional equation (1) Gif.latex?\hspace{-15}$\boldsymbol{(2)::}$%20$If%20$f(x)$%20is%20a%20polynomial%20function.%20then%20$f(x)$%20must%20be%20in%20the%20linear%20form.\\\\%20So%20$f(x)=ax+b$%20and%20$f(1-x)=a(1-x)+b$,%20Now%20Put%20values%20in\\\\%20equation.,%20We%20Get\\\\%20$x.f(x)+f(1-x)=x^2$\\\\%20$x.(ax+b)+a(1-x)+b=x^2$\\\\%20$ax^2+(b-a).x+(a+b)=x^2+0.x+0$\\\\%20So%20$a=1,b=1\Leftrightarrow%20a+b=2$\\\\%20But%20Here%20$a+b=0$(after%20camparing%20Const.)\\\\%20So%20no%20such%20polynomial%20$f(x)$%20exists$

Admin: Admin; Posts : 103
Join date : 2011-02-01
Location : India

Re: functional equation (1)

Post Sponsored content

Sponsored content