Tag: maths

We have to find the coefficient of third term in Maclaurin series of $sin^2 x$.

The Maclaurin series is given by $f(x)=\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}x^k$ where $a=0$

We have $f(x)=sin^2 x$

Since we have to find the coefficient of the third term, let us take $n=8$.

$\therefore f(x)\approx\sum _{k=0}^{8}\dfrac{f^{(k)}(0)}{k!}x^k$

$f^{(0)}(x)=sin^2 x, \Rightarrow f^{(0)}(0)=0$

$f^{(1)}(x)=2:sinx:cosx, \Rightarrow f^{(1)}(0)=0$

$f^{(2)}(x)=-2sin^2x+2cos^2 x, \Rightarrow f^{(2)}(0)=2$

$f^{(3)}(x)=-8cosx:sin x, \Rightarrow f^{(3)}(0)=0$

$f^{(4)}(x)=8sin^2x-8cos^2x, \Rightarrow f^{(4)}(0)=-8$

$f^{(5)}(x)=32:sinx:cos x, \Rightarrow f^{(5)}(0)=0$

$f^{(6)}(x)=-32sin^2x+32cos^2x, \Rightarrow f^{(6)}(0)=32$

$f^{(7)}(x)=-128:sinx:cos x, \Rightarrow f^{(7)}(0)=0$

$f^{(8)}(x)=128sin^2x-128cos^2x, \Rightarrow f^{(8)}(0)=-128$

$\therefore f(x) \approx 0x^0+0x^1+\dfrac{2}{2!}x^2+\dfrac{0}{3!}x^3+\dfrac{-8}{4!}x^4+\dfrac{0}{5!}x^5+\dfrac{32}{6!}x^6+\dfrac{0}{7!}x^7+\dfrac{-128}{8!}x^8$

$\Rightarrow f(x)\approx x^2-\dfrac{1}{3}x^4+\dfrac{2}{45}x^6-\dfrac{1}{135}x^5$

Thus the coefficient of third term is $\dfrac{2}{45}$.

$\lim _{x\rightarrow 0} \dfrac{x^2e^x}{cos x-1}=?$

We need to find the limit value using Taylor series.

The Taylor series of $x^2$ is $x^2$

The Taylor series of $e^x$ is $1+x+\dfrac{1}{2}x^2+...$

The Taylor series of $cos x$ is $1-\dfrac{1}{2}x^2+\dfrac{1}{24}x^4-...$

$\lim _{x\rightarrow 0} \dfrac{x^2e^x}{cos x-1}\approx \lim _{x\rightarrow 0} \dfrac{x^2(1+x+\dfrac{1}{2}x^2+...)}{1-\dfrac{1}{2}x^2+\dfrac{1}{24}x^4-...-1}$

                              $= \lim _{x\rightarrow 0} \dfrac{x^2(1+x+\dfrac{1}{2}x^2)}{1-\dfrac{1}{2}x^2+\dfrac{1}{24}x^4-1}$

                              $= \lim _{x\rightarrow 0} \dfrac{x^2(1+x+\dfrac{1}{2}x^2)}{-\dfrac{1}{2}x^2+\dfrac{1}{24}x^4}$

                              $= \lim _{x\rightarrow 0} \dfrac{x^2(1+x+\dfrac{1}{2}x^2)}{x^2 \left(-\dfrac{1}{2}+\dfrac{1}{24}x^2 \right)}$

                              $= \lim _{x\rightarrow 0} \dfrac{1+x+\dfrac{1}{2}x^2}{-\dfrac{1}{2}+\dfrac{1}{24}x^2 }$

                              $= \lim _{x\rightarrow 0} \dfrac{1}{-\dfrac{1}{2}}$
$\lim _{x\rightarrow 0} \dfrac{x^2e^x}{cos x-1}=-2$

The Maclaurin series is given by $f(x)=\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}x^k$ where $a=0$

We have $f(x)=log (1+x)$

Since we have to find the coefficient of the third term, let us take $n=5$.

$\therefore f(x)\approx\sum _{k=0}^{5}\dfrac{f^{(k)}(0)}{k!}x^k$

$f^{(0)}(x)=log (1+x), \Rightarrow f^{(0)}(0)=0$

$f^{(1)}(x)=\dfrac{1}{x+1}, \Rightarrow f^{(1)}(0)=1$

$f^{(2)}(x)=-\dfrac{1}{(x+1)^2}, \Rightarrow f^{(2)}(0)=-1$

$f^{(3)}(x)=\dfrac{2}{(x+1)^3}, \Rightarrow f^{(3)}(0)=2$

$f^{(4)}(x)=-\dfrac{6}{(x+1)^4}, \Rightarrow f^{(4)}(0)=-6$

$f^{(5)}(x)=\dfrac{24}{(x+1)^5}, \Rightarrow f^{(5)}(0)=24$

$\therefore f(x) \approx \dfrac{0}{0!}x^0+\dfrac{1}{1!}x^1+\dfrac{-1}{2!}x^2+\dfrac{2}{3!}x^3+\dfrac{-6}{4!}x^4+\dfrac{24}{5!}x^5$

$\Rightarrow f(x)\approx x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-\dfrac{1}{4}x^4+\dfrac{1}{5}x^5$

Thus the coefficient of third term is $\dfrac{1}{3}$.

$\lim _{x \rightarrow 0} \dfrac{sin x-x}{x^3}=?$. We have to find the limit value using Taylor series.

The Taylor series of $sin x$ is $sin x\approx x-\dfrac{1}{6}x^3+...$.

The Taylor series of $x$ is $x \approx x$

The Taylor series of $x^3$ is $x^3 \approx x^3$

Hence $\lim _{x \rightarrow 0} \dfrac{sin x-x}{x^3} \approx \lim _{x \rightarrow 0} \dfrac{x-\dfrac{1}{6}x^3+...-x}{x^3}$

                                            $=\lim _{x \rightarrow 0} \dfrac{-\dfrac{1}{6}x^3}{x^3}$

                                            $=\lim _{x \rightarrow 0} -\dfrac{1}{6}$

Therefore $\lim _{x \rightarrow 0} \dfrac{sin x-x}{x^3}=-\dfrac{1}{6}$

Here, $f(x)=log(cos(x))$

$f'(x)=-tan(x)$ and $f'(0)=0$

$f''(x)=-tan^2(x)-1$ and $f''(0)=-1$

$f'''(x)=-2tan(x)(tan^2(x)+1)$ and $f'''(0)=0$

$f''''(x)=-6tan^4(x)-8tan^2(x)$ and $f''''(0)=-2$

So,

$f(x)\approx \dfrac{0}{0!}x^0+\dfrac{0}{1!}x^1-\dfrac{1}{2!}x^2+\dfrac{0}{3!}x^4-\dfrac{2}{4!}x^4+........$

Therefore,
$log(cos(x))\approx-\dfrac{1}{2}x^2-\dfrac{1}{12}x^4$

Divide by $x^2$ on both the sides, we get
$\dfrac{log(cos(x))}{x^2}=-\dfrac{1}{2}-\dfrac{1}{12}x^2$

$\lim _{x\rightarrow 0}\dfrac{log(cos(x))}{x^2}=\lim _{x\rightarrow0}(-\dfrac{1}{2}-\dfrac{1}{12}x^2)=-\dfrac{1}{2}$

Using Maclaurin series, e^(5x) = 1 + 5x + (5x)^2/2 + ... so (e^(5x)-1)^5 is approximately (5x)^5. The denominator is (x^2 - (x^2 - x^6/6))^(1/3) = (x^6/6)^(1/3) = x^2 / 6^(1/3). The limit calculation leads to 25 * 6^(1/3).

$\lim _{x\rightarrow 0} \dfrac{x-tan^{-1} (x)}{x^3}=?$

Let us use Maclaurian series expansion to evaluate the limit.

The Maclaurian series of $x$ is $x$.

The Maclaurian series of $tan^{-1} (x)$ is $x-\dfrac{1}{3}x^3+\dfrac{1}{5}x^5$.

The Maclaurian series of $x^3$ is $x^3$.

$\lim _{x\rightarrow 0} \dfrac{x-tan^{-1} (x)}{x^3}=\lim _{x\rightarrow 0} \dfrac{x-(x-\dfrac{1}{3}x^3+\dfrac{1}{5}x^5)}{x^3}$

                                        $=\lim _{x\rightarrow 0} \dfrac{x-x+\dfrac{1}{3}x^3-\dfrac{1}{5}x^5}{x^3}$

                                        $=\lim _{x\rightarrow 0} \dfrac{\dfrac{1}{3}x^3-\dfrac{1}{5}x^5}{x^3}$

                                        $=\lim _{x\rightarrow 0} \dfrac{x^3(\dfrac{1}{3}-\dfrac{1}{5}x^2)}{x^3}$

                                        $=\lim _{x\rightarrow 0} (\dfrac{1}{3}-\dfrac{1}{5}x^2)$

$\lim _{x\rightarrow 0} \dfrac{x-tan^{-1} (x)}{x^3}=\dfrac{1}{3}$

  If the exponent of a negative integer is odd, then the result is a negative integer.