Question

In​ calculus, it can be shown that

e^x=infintesigmak=0 x^k/k!

We can approximate the value of for any x using the following sum e^x=infintesigmak=0 x^k/k!
a) Approximate e^2.2 n=3

Answer 1

Answer:
`7.39466666`

Explanation:

Setting
`x=2.2` and
`n=3`,

`e^(2.2) \approx \sum^(3)_(k=0) (2.2^(k))/(k!)=(2.2^0)/(0!)+(2.2^1)/(1!)+(2.2^2)/(2!)+(2.2^3)/(3!) \approx 7.39466666`

Answer 2

7.39466667 (8 d.p.)

Explanation:

We can approximate the value of eˣ for any x using the following sum formula:

`\boxed{e^(x) \approx \displaystyle \sum^n_(k=0)(x^k)/(k!)}`

To approximate
`e^(2.2)` with n = 3, substitute x = 2.2 and n = 3 into the given sum formula:

`e^(2.2) \approx \displaystyle \sum^3_(k=0)(2.2^k)/(k!)`

To calculate the sum, substitute k with each value from 0 to 3 and add the results together:

`\begin{aligned}e^(2.2) &\approx \displaystyle \sum^3_(k=0)(2.2^k)/(k!)\\\\&= (2.2^0)/(0!)+(2.2^1)/(1!)+(2.2^2)/(2!)+(2.2^3)/(3!)\\\\&= (1)/(1)+(2.2)/(1)+(4.84)/(2)+(10.648)/(6)\\\\&= 1+2.2+2.42+1.774666666...\\\\&=7.39466667\; \sf (8\;d.p.)\end{aligned}`

Therefore, the approximate value of
`e^(2.2)` is:

`\large \textsf{$e^(2.2)$}=\boxed{7.39466667}\; \sf (8\;d.p.)}`

Note: To obtain a more accurate approximate value, increase the value of n.