SOLUTION: Question: Dear Sir, Please help me in giving the solution to the foll.problem as I am a chemical engineer in doing simulation softwares Please reply how to determine the value of

Algebra ->  Algebra  -> Formulas -> SOLUTION: Question: Dear Sir, Please help me in giving the solution to the foll.problem as I am a chemical engineer in doing simulation softwares Please reply how to determine the value of      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   


Question 2945: Question: Dear Sir,
Please help me in giving the solution to the foll.problem as I am a chemical engineer in doing simulation softwares
Please reply how to determine the value of THETTA in the foll.problem
((THETTA) - SIN(THETTA)) = 0.5268
Regards,
Prem.
Answer by khwang(126): Since there is no way to get the exact solution the equation x - sin x = c (constant).
We only can only use series approximations by Taylor series:
sin x ~ x -x^3/6,
To solve x - sin x = 0.5268
That is, x^3/6 = 0.5268,
so, x = (0.5268 * 6)^1/3 = 0.1536 (approximated value)
Kenny
Dear Mr.Kenny (Khwang 126),
I am sorry I couldn't able to understand the result you have given. Could u able explain it briefly along with Taylor Series? Please reply.
Prem

Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
Taylor series was one of the importance achievement of human science.

We tried to get the approximation for an arbitray function by
polynomials. Since, with polynomials, we know how todifferentiate,
integrate, find the roots,draw the graphs,etc.
The results of Taylor's series is,
for a nth differentiable function f(x)
oo
f(x) = E f^(k)(a)(x-a)^k/k! ,
k=o
(the series converges to f(x) locally at a)
when a = 0, we have
oo
f(x) = E f^(k)(0)x^k/k!
k=o
Without this series, we cannot use any calculators or computers
to find the value of sine,cosine,log or expontial, ect.
In other words, there is no modern science without Taylor series.
Now, f(x) = sin x, expand it at x = 0,
we have f'(x) = cos x, f'(x) = -sin x, f"'(x) = -cos x,
so f'(0) = 1, f"(0) = 0, f"'(0) = -1,
after simplified, we have
oo
sin x = E (-1)^(2k+1))x^(2k+1)/(2k+1)!
k=o
or sin x = x -x^3/3! + x^5 /5! -x^7/7! +....
I used the first two terms to estimate sinx before.
You also can do search to get more informations from the Internet.
Good luck.
Kenny