Question

How do we find the Taylor series as asked in the screenshot?

Answer 1

Answer:
`\displaystyle 6e^{-\text{x}/9} = 6 - (2)/(3)\text{x} + (1)/(27)\text{x}^2- (1)/(729)\text{x}^3 + \cdots`

This is the expanded form of
`\displaystyle \sum_(n=0)^(\infty)\frac{6\left(-\text{x}/9\right)^n}{n!}`

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Step-by-step explanation:

f(x) = 6e^(-x/9) is the given function

T(x) = Taylor polynomial function

The goal is to see if we can try to model f(x) in terms of a polynomial T(x).

This means T(x) won't involve the exponential base "e", and it won't have fractional exponents either.

In your notes you should have the following about the Taylor polynomial:

`T(\text{x}) = f(a) + (f'(a))/(1!)(\text{x}-a)+ (f''(a))/(2!)(\text{x}-a)^2+ (f'''(a))/(3!)(\text{x}-a)^3+ \cdots`

So we'll need to compute the derivatives. Here's the 1st derivative

`f(x) = 6e^{-\text{x}/9}\\\\f'(x) = \frac{d}{d\text{x}}\left(6e^{-\text{x}/9}\right)\\\\f'(x) = 6e^{-\text{x}/9}*(d)/(dx)(-\text{x}/9)\\\\ f'(x) = -(2)/(3)e^{-\text{x}/9}\\\\`

Don't forget about the chain rule.

The second derivative is

`f''(x) = (d)/(dx)(f'(x))\\\\f''(x) = (d)/(dx)\left(-(2)/(3)e^{-\text{x}/9}\right)\\\\f''(x) = -(2)/(3)(d)/(dx)\left(e^{-\text{x}/9}\right)\\\\f''(x) = -(2)/(3)e^{-\text{x}/9}(d)/(dx)\left(-\text{x}/9\right)\\\\f''(x) = (2)/(27)e^{-\text{x}/9}\\\\`

I'll skip steps for the third derivative, but you should get

`f'''(x) = -(2)/(243)e^{-\text{x}/9}\\\\`

This process is repeated infinitely out; of course realistically we should only do a few derivatives or we'll be here forever.

Here's the summary derivatives we found so far

`f'(x) = -(2)/(3)e^{-\text{x}/9}\\\\f''(x) = (2)/(27)e^{-\text{x}/9}\\\\f'''(x) = -(2)/(243)e^{-\text{x}/9}\\\\`

Plug in x = 0 since the Taylor polynomial is centered around here.

`f'(0) = -(2)/(3)e^(-0/9) = -(2)/(3)\\\\f''(0) = (2)/(27)e^(-0/9) = (2)/(27)\\\\f'''(0) = -(2)/(243)e^(-0/9) = -(2)/(243)\\\\`

Then we can say:

`T(\text{x}) = f(a) + (f'(a))/(1!)(\text{x}-a)+ (f''(a))/(2!)(\text{x}-a)^2+ (f'''(a))/(3!)(\text{x}-a)^3+ \cdots`

`T(\text{x}) = f(0) + (f'(0))/(1!)(\text{x}-0)+ (f''(0))/(2!)(\text{x}-0)^2+ (f'''(0))/(3!)(\text{x}-0)^3+ \cdots`

`T(\text{x}) = 6 + (-2/3)/(1)\text{x}+ (2/27)/(2)\text{x}^2+ (-2/243)/(6)\text{x}^3+ \cdots`

`T(\text{x}) = 6 - (2)/(3)\text{x}+ (1)/(27)\text{x}^2- (1)/(729)\text{x}^3+ \cdots`

When centered around x = 0, T(x) is a pretty good replacement for f(x). Of course the further you move away from x = 0, is when the error will get worse. I recommend plotting the first few terms of T(x) and f(x) on the same xy grid. You should see the two graphs somewhat merge, or overlap, together around x = 0. The more terms you add onto T(x), the better the approximation will get.

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Another approach:

Recall that

`\displaystyle e^{\text{x}} = \sum_(n=0)^(\infty)\frac{\text{x}^n}{n!}`

This is a compact way of saying

`\displaystyle e^{\text{x}} = \frac{\text{x}^0}{0!} + \frac{\text{x}^1}{1!} + \frac{\text{x}^2}{2!} + \frac{\text{x}^3}{3!} + \cdots`

Replace every copy of x with -x/9

`\displaystyle e^{\text{x}} = \sum_(n=0)^(\infty)\frac{\text{x}^n}{n!}\\\\\\\displaystyle e^{-\text{x}/9} = \sum_(n=0)^(\infty)\frac{\left(-\text{x}/9\right)^n}{n!}\\\\`

Then multiply both sides by 6

`\displaystyle e^{-\text{x}/9} = \sum_(n=0)^(\infty)\frac{\left(-\text{x}/9\right)^n}{n!}\\\\\\\displaystyle 6e^{-\text{x}/9} = 6\sum_(n=0)^(\infty)\frac{\left(-\text{x}/9\right)^n}{n!}\\\\\\\displaystyle 6e^{-\text{x}/9} = \sum_(n=0)^(\infty)\frac{6\left(-\text{x}/9\right)^n}{n!}\\\\\\`

If you wanted, you could expand things out a bit

`\displaystyle 6e^{-\text{x}/9} = \sum_(n=0)^(\infty)\frac{6\left(-\text{x}/9\right)^n}{n!}\\\\\\\displaystyle 6e^{-\text{x}/9} = \frac{6\left(-\text{x}/9\right)^0}{0!} + \frac{6\left(-\text{x}/9\right)^1}{1!} + \frac{6\left(-\text{x}/9\right)^2}{2!} + \frac{6\left(-\text{x}/9\right)^3}{3!} + \cdots`

`\displaystyle 6e^{-\text{x}/9} = 6 - (2)/(3)\text{x} + (1)/(27)\text{x}^2- (1)/(729)\text{x}^3 + \cdots`

This matches what we found in the previous section.