Question

Approximate the value of e^3x when x=1 using the first six terms of the Maclaurin expansion

Answer 1

Given: A function

`f(x)=e^(3x)`

Required: To approximate the value of the function when x=1 using the first six terms of the Maclaurin expansion.

Explanation: Maclaurin's expansion is a special case of Taylor's theorem. The terms of the expansion are as follows:

`f(x)+(f^(\prime)(x))/(1!)+(f^(\prime\prime)(x))/(2!)+...`

So we need to find the following terms at x=1

`\begin{gathered} f(1)=e^3 \\ f^(\prime)(1)=3e^3 \\ f^(\prime)^(\prime)(1)=9e^3 \\ f^(\prime)^(\prime)^(\prime)(1)=27e^3 \end{gathered}`

Further

`e^x=1+x+(x^2)/(2!)+...`

We can write the given function as

`e^(3x)=e^(-3)e^(3(x+1))`

The series would be

`e^(3(x+1))=1+3(x+1)+(3^2(x+1)^2)/(2)+(3^3(x+1)^3)/(6)+(3^4(x+1)^4)/(24)+(3^5(x+1)^5)/(120)`

Solving further at x=1 gives

`e^(3(x+1))=1+3(x+1)+(9)/(2)(x+1)^2+(27(x+1)^3)/(6)+(81(x+1)^4)/(24)+(243(x+1)^5)/(120)`

Hence the required function is-

`e^(-3)e^(3(x+1))=e^(-3)(1+3(x+1)+(9)/(2)(x+1)^2+(27(x+1)^3)/(6)+(81(x+1)^4)/(24)+(243(x+1)^5)/(120))`

Now for x=1 we have

`=e^(-3)(1+6+18+36+54+64.8)`

Which gives

`\begin{gathered} =0.0497*179.8 \\ =8.93606 \end{gathered}`

Final Answer: 8.93606