Question

What is the local linearization of `e^sin(x)` near x=1?

Answer 1

`f(x) ~~ 1.2534x+1.0664`

Explanation:

Let:

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

The linear approximation of a function `f(x)` about `x=1` is given by the linear terms of the Taylor Series about the point `x=a`, ie the terms given by:

`f(x) ~~ f(a) + f'(a)(x-a)`

Differentiating `f(x)` using the chain rule, we have:

`f'(x) = e^(sinx) (cosx)`

So with `x=1` we have:

`f(1) = e^(sin1)`
`" " = 2.3197768 ...`

`f'(1) = e^(sin1) (cos1)`
`" " = 1.2533897 ...`

Hence, the linear approximation near `x=1` is given by:

`f(x) ~~ 2.3198 + 1.2534(x-1)`
`" " = 2.3198 + 1.2534x-1.2534`
`" " = 1.2534x+1.0664`

Where we have rounded to 4dp.

Example:

Consider the case `x=1.01` which is near `x=1`, Then:

`f(1.01) = e^(sin(1.01))`
`" " = 2.332246 ...`

And the linear approximation gives us:

`f(1.01) ~~ 1.2534(1.01)+1.0664`
`" " = 1.265934+1.0664`
`" " = 2.332334`

Which is correct within `3dp`