Question

How do you use linear Approximation to find the value of `(1.01)^10`?

Answer 1

`(1.01)^10 ~= 1.10`

with two significant digits.

Explanation:

Consider the function:

`f(x) = (1+x)^a` with `a > 0`

We have:

`f(0) = 1`

`f'(x) = a(1+x)^(a-1)`

`f'(0) = a`

Develop in MacLaurin series and stop at the linear term:

`(1+x)^a = f(0)+ (f'(0))/(1!) x +R_2(x) = 1+ax+R_1(x)`

Then we have for `x=1/100` and `a=10`:

`(1.01)^10 = 1+10*1/100 +R =1.1+R`

Consider now the second derivative:

`f''(x) = a(a-1)(1+x)^(a-2) = 90(1+x)^8`

For `x > 0` this is an increasing function, so we have:

`f''(x) < f''(1/100)` for `0 < x < 1/100`

or:

`abs(f''(x)) <= 90(1.01)^8`

So, using Lagrange's formula:

`abs(R_n(x)) <= absx^(n+1)/((n+1)!)max_(xi in (0,x)) abs (f^((n))(xi)`

we have:

`absR < 45/100^2(1.01)^8 ~=0.005`

Answer 2

`(1.01)^10 ~~1.1`
Using a calculator you should get`1.104622125...` , so the approximation is quite good.

Explanation:

Here's a slightly simpler, less comprehensive answer.

We let `f(x) = x^10`. Then `f'(x) = 10x^9`.

Now use the linear approximation formula, which is `y = f(a) + f'(a)(x - a` for a function `f(x)` approximated at a point close to `x = a`.

Our point will obviously be `a= 1`.

Also, `f(1) = 1`, `f'(1) = 10` and `x - a = 1.01 - 1 = 0.01`

Plugging this into our formula, we have:

`y = 1+ 10(0.01) = 1.1`

Note that this is a rough approximation. The further away you get from the point `x = a`, the less precise the approximation becomes.

Hopefully this helps!