2 Answers | Add Yours
If you try to solve this analytically, it is almost impossible to arrive at a solution.
`tan(x) = -x/2`
`x = tan^(-1)(-x/2)`
But you can find the solution in an approximate analytical solution.
lt's write tan(x) as an taylor expansion.
Taylor expansion of f(x) around x =a is given by,
`f(x) = f(a)+(f'(a))/(1!)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+....`
let's expand tan(x) until 3rd order around x =0. So a =0
if `tan(x) = f(x)` ` f(0) = 0`
`f'(x) = sec^2(x)` ` f'(0) = 1`
`f''(x) = 2sec(x)sec(x)tan(x) = 2sec^2(x)tan(x) = 2f'(x)f(x)`
`f''(0) = 0`
`f'''(x) = 2f'(x)f'(x)+2f''(x)f(x) = 2(f'(x))^2+2f(x)f''(x)`
`f'''(0) = 2*1^2+0 = 2`
Therefore Taylor expansion of tan(x) is,
`tan(x) = 0+1/(1!)x+0+2/(3!)x^3`
`tan(x) = x + 2/(1*2*3)x^3`
`tan(x) = x + x^3/3`
now using this to sovle our problem,
`-2(x+x^3/3) = x`
`2/3x^3+3x = 0`
`x(2/3x^2+3) = 0`
this gives, either `x = 0` or `(2/3x^2+3) = 0`
But x should be a real number therefore, `2/3x^2+3 != 0`
The only answer is x = 0.
The answer is x =0
Thank you, very clear explanation
We’ve answered 319,200 questions. We can answer yours, too.Ask a question