Question

Find the first six terms of the expansion of the function e'log(1 + y) in a taulors series in the neighbourhood poimt (0.0)​

Answer 1

Answer:

The first six terms of the expansion is $e^{x} \log (1+y)=y+\frac{1}{2}\left(2 x y-y^{2}\right)+\frac{1}{6}\left(3 x^{2} y-3 x y^{2}+2 y^{3}\right)$

Step-by-step explanation:

Let $f(x, y)=e^{x} \log (1+y)$

The function  evaluated at (0,0) and its partial derivatives are:

$f(x, y) &=e^{x} \log (1+y) & & \rightarrow 0$

$f_{x} &=e^{x} \log (1+y) & & \rightarrow 0$

$f_{y} &=e^{x} \cdot 1 /(1+y) & & \rightarrow 1$

$f_{x x} &=e^{x} \log (1+y) & & \rightarrow 0$terms

$f_{x y} &=e^{x} / 1+y & & \rightarrow 1$

$f_{y y} &=-e^{x} /(1+y)^{2} & & \rightarrow-1$

$f_{x x x} &=e^{x} \log (1+y) & & \rightarrow 0$

$f_{x x y} &=e^{x} /(1+y) & & \rightarrow 1$

$f_{x y y} &=-e^{x} /(1+y)^{2} & & \rightarrow-1$

$f_{y y y} &=2 e^{x} /(1+y)^{3} & & \rightarrow 2$

By MacLaurin's theorem

$\begin{aligned}&f(x, y)=f(0,0)+\left[x f_{x}(0,0)+y f_{y}(0,0)\right] \\&+\frac{1}{2 !}\left[x^{2} f_{x x}(0,0)+2 x y f_{x y}(0,0)+y^{2} f_{y y}(0,0)\right] \\&+\frac{1}{3 !}\left[x^{3} f_{x x x}(0,0)+3 x^{2} y f_{x x y}(0,0)\right. \\&\left.\quad+3 x y^{2} f_{x y y}(0,0)+y^{3} f_{y y y}(0,0)\right]+\cdots \cdots \cdots\end{aligned}$

Substituting all the values in the expansion of f(x, y),

$e^{x} \log (1+y)=y+\frac{1}{2}\left(2 x y-y^{2}\right)+\frac{1}{6}\left(3 x^{2} y-3 x y^{2}+2 y^{3}\right)$

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