Respuesta :
Answer: [tex]\displaystyle 6e^{-\text{x}/9} = 6 - \frac{2}{3}\text{x} + \frac{1}{27}\text{x}^2- \frac{1}{729}\text{x}^3 + \cdots[/tex]
This is the expanded form of [tex]\displaystyle \sum_{n=0}^{\infty}\frac{6\left(-\text{x}/9\right)^n}{n!}[/tex]
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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:
[tex]T(\text{x}) = f(a) + \frac{f'(a)}{1!}(\text{x}-a)+ \frac{f''(a)}{2!}(\text{x}-a)^2+ \frac{f'''(a)}{3!}(\text{x}-a)^3+ \cdots[/tex]
So we'll need to compute the derivatives. Here's the 1st derivative
[tex]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}*\frac{d}{dx}(-\text{x}/9)\\\\ f'(x) = -\frac{2}{3}e^{-\text{x}/9}\\\\[/tex]
Don't forget about the chain rule.
The second derivative is
[tex]f''(x) = \frac{d}{dx}(f'(x))\\\\f''(x) = \frac{d}{dx}\left(-\frac{2}{3}e^{-\text{x}/9}\right)\\\\f''(x) = -\frac{2}{3}\frac{d}{dx}\left(e^{-\text{x}/9}\right)\\\\f''(x) = -\frac{2}{3}e^{-\text{x}/9}\frac{d}{dx}\left(-\text{x}/9\right)\\\\f''(x) = \frac{2}{27}e^{-\text{x}/9}\\\\[/tex]
I'll skip steps for the third derivative, but you should get
[tex]f'''(x) = -\frac{2}{243}e^{-\text{x}/9}\\\\[/tex]
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
[tex]f'(x) = -\frac{2}{3}e^{-\text{x}/9}\\\\f''(x) = \frac{2}{27}e^{-\text{x}/9}\\\\f'''(x) = -\frac{2}{243}e^{-\text{x}/9}\\\\[/tex]
Plug in x = 0 since the Taylor polynomial is centered around here.
[tex]f'(0) = -\frac{2}{3}e^{-0/9} = -\frac{2}{3}\\\\f''(0) = \frac{2}{27}e^{-0/9} = \frac{2}{27}\\\\f'''(0) = -\frac{2}{243}e^{-0/9} = -\frac{2}{243}\\\\[/tex]
Then we can say:
[tex]T(\text{x}) = f(a) + \frac{f'(a)}{1!}(\text{x}-a)+ \frac{f''(a)}{2!}(\text{x}-a)^2+ \frac{f'''(a)}{3!}(\text{x}-a)^3+ \cdots[/tex]
[tex]T(\text{x}) = f(0) + \frac{f'(0)}{1!}(\text{x}-0)+ \frac{f''(0)}{2!}(\text{x}-0)^2+ \frac{f'''(0)}{3!}(\text{x}-0)^3+ \cdots[/tex]
[tex]T(\text{x}) = 6 + \frac{-2/3}{1}\text{x}+ \frac{2/27}{2}\text{x}^2+ \frac{-2/243}{6}\text{x}^3+ \cdots[/tex]
[tex]T(\text{x}) = 6 - \frac{2}{3}\text{x}+ \frac{1}{27}\text{x}^2- \frac{1}{729}\text{x}^3+ \cdots[/tex]
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
[tex]\displaystyle e^{\text{x}} = \sum_{n=0}^{\infty}\frac{\text{x}^n}{n!}[/tex]
This is a compact way of saying
[tex]\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[/tex]
Replace every copy of x with -x/9
[tex]\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!}\\\\[/tex]
Then multiply both sides by 6
[tex]\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!}\\\\\\[/tex]
If you wanted, you could expand things out a bit
[tex]\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[/tex]
[tex]\displaystyle 6e^{-\text{x}/9} = 6 - \frac{2}{3}\text{x} + \frac{1}{27}\text{x}^2- \frac{1}{729}\text{x}^3 + \cdots[/tex]
This matches what we found in the previous section.
