The following function is given; f(x)=ex.In (x+1). Use Maclaurin series approach to transform this equation into a quadratic equation (eg. determining the terms of the series for n=0,1,2 is enough). Then, calculate f(0.2) by both using the function itself and its quadratic version. Compare.

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The following function is given; f(x)=ex.In (x+1). Use Maclaurin series approach to transform this equation into a quadratic equation (eg. determining the terms of the series for n=0,1,2 is enough).  Then, calculate f(0.2) by both using the function itself and its quadratic version. Compare.

Novembruuo · Accepted Answer

Step1The given function is:f(x)=exIn(x+1)∴f(0)=0Differentiate w.r.t.x to get:f′(x)=exx+1+ln⁡(x+1)exf′(x)=(1x+1+ln(x+1))f′(0)=1f&quot;(x)=ex(1x+1+ln(x+1))+ex(−1(x+1)2+1x+1)=ex(1x+1+ln(x+1)−−1(x+1)2+1x+1)f0)=1Step2Hence the McLaurin polynomial is given by:f(x)=f(0)+f(0)x+f&quot;(0)2!x2+..Stopping at 2, it follows:f(x)=x+x22Hence f(0.2)=0.2+(0.2)22=0.22….(i)Step3Again f(x)=exIn(x+1)Substitute x=0.2 to get:f(0.2)=0.22268880523≈0.2227…(ii)From (i) and (ii) itis clear that the McLaurin expansion has decent accuracy even if only n=0,1,2 is considered and other values omitted.

The following function is given; f(x)=e^{x}.In (x+1). Use Maclaurin series

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