Linear approximation

[jwpaine]

I just had an exam, and one of the questions was: Use linear approximation to estimate the value of (0.99e^0.02)^8.


I did: (0.99e^0.02)^8 = 0.99e^0.16

I chose a linear approximation of e^x = 1 + x

0.99^8[1 + x] = 0.99^8[1 + 0.16] = 1.07038

Calculator gives 1.08285

He gave me a 0. How should I have done this differently? He said I should have used f(x,y) = x^8 * e^8y near (1,0) ??

JOhn

 

[tkhunny]

Well, here's the problem. You used a linear approximation for e^x. You did NOT use a linear approximation for (e^x)^8 and it appears you used no approximation at all for 0.99^8.

\(\displaystyle \left[0.99\cdot e^{0.02}\right]^{8}\)

Thinking...

0.02 is pretty close to zero (0.00).
0.99 is pretty close to one (1.00).

Hmm... Can we do both at the same time? Hmm... Aha!! Let's try this:

\(\displaystyle f(x)\;=\;\left[\left(1-\frac{x}{2}\right)\cdot e^{x}\right]^{8}\)

\(\displaystyle f(0.02) = \left[0.99\cdot e^{0.02}\right]^{8}\)

Start here: \(\displaystyle f(0) = \left[\left(1-\frac{0}{2}\right)\cdot e^{0}\right]^{8} = 1\)

Then here: \(\displaystyle f'(x) = 4\cdot e^{8x}\cdot \left(\frac{x}{2}-1\right)^{7}\cdot (x-1)\)

There are other ways to write this, but this one seems most convenient.

Then, \(\displaystyle f'(0) = 4\cdot e^{0}\cdot \left(0-1\right)^{7}\cdot (0-1) = 4\cdot 1 \cdot (-1) \cdot (-1) = 4\)

We're ready.

Linear approximation is f(0) + 0.02*f'(0) = 1 + 0.02*4 = 1.08

It's not all that bad. You just have to think it through and supply the missing pieces as you encounter them.

 

[jwpaine]

Thank you so much! Really appreciate it.