The Taylor series expansion of \[\frac{{\sin {\text{x}}}}{{{\text{x}} - \pi }}\] at \[{\text{x}} = \pi \] is given by
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
Answer: Option B
This Question Belongs to Engineering Maths >> Calculus
Join The Discussion
Comments (1)
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
The correct option is. D. −1+(x−π)23!-..
We want the **Taylor (Maclaurin-type) series expansion** of
$$
f(x) = frac{sin x}{x - pi}, quad text{about } x = pi.
$$
Let
$$
h = x - pi quad Rightarrow quad x = pi + h.
$$
Then
$$
f(x) = frac{sin(pi + h)}{h}.
$$
$$
sin(pi + h) = -sin h.
$$
So
$$
f(x) = frac{-sin h}{h}.
$$
Use series for $sin h$
$$
sin h = h - frac{h^3}{3!} + frac{h^5}{5!} - cdots
$$
So
$$
frac{sin h}{h} = 1 - frac{h^2}{3!} + frac{h^4}{5!} - cdots
$$
Thus
$$
f(x) = -left(1 - frac{h^2}{6} + frac{h^4}{120} - cdots right).
$$
Substitute back $h = x - pi$
$$
f(x) = -1 + frac{(x - pi)^2}{6} - frac{(x - pi)^4}{120} + cdots
$$
(Taylor series of $tfrac{sin x}{x-pi}$ at $x=pi$):**
$$
frac{sin x}{x-pi} = -1 + frac{(x - pi)^2}{6} - frac{(x - pi)^4}{120} + frac{(x - pi)^6}{5040} - cdots
$