# Taylor polynomial approximation

• Sep 1st 2008, 10:07 PM
namelessguy
Taylor polynomial approximation
$e^{-5}$ $\approx$ $\sum$ $\frac{(-5)^i}{i!}$ = $\sum$ $\frac{(-1)^i 5^i}{i!}$
i runs from 0 to 9
$e^{-5}$= $\frac{1}{e^5}$ $\approx$ $\frac{1}{\sum\frac{5^i}{i!}}$
i runs from 0 to 9

Sorry, I don't know how to put the indices on the summation. The question is which formula gives the most accuracy comparing to the actual value of $e^{-5}$ correct to three digits, which is 6.74 x $10^{-3}$.
I computed using both formulas and figured out that the second formula gives the most accuracy, but I don't have a solid answer for why it is more accurate than the other.
I guess it's is because in the second formula we write down the exact formula for $e^{-5}$, then use the approximation after. On the first formula, we approximate first.
Anyone can give me some thought on this. Thank you.
• Sep 1st 2008, 11:22 PM
CaptainBlack
Quote:

Originally Posted by namelessguy
$e^{-5}$ $\approx$ $\sum$ $\frac{(-5)^i}{i!}$ = $\sum$ $\frac{(-1)^i 5^i}{i!}$
i runs from 0 to 9
$e^{-5}$= $\frac{1}{e^5}$ $\approx$ $\frac{1}{\sum\frac{5^i}{i!}}$
i runs from 0 to 9

Sorry, I don't know how to put the indices on the summation. The question is which formula gives the most accuracy comparing to the actual value of $e^{-5}$ correct to three digits, which is 6.74 x $10^{-3}$.
I computed using both formulas and figured out that the second formula gives the most accuracy, but I don't have a solid answer for why it is more accurate than the other.
I guess it's is because in the second formula we write down the exact formula for $e^{-5}$, then use the approximation after. On the first formula, we approximate first.
Anyone can give me some thought on this. Thank you.

Look at the form of the remainder for the two expansions, and what happens to the remainder in the second when you take a reciprical.

RonL
• Sep 2nd 2008, 12:04 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by namelessguy
Sorry, I don't know how to put the indices on the summation.

$$\sum_{i=0}^{9}$$ gives $\sum_{i=0}^{9}$.