natHenderson
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I need guidance on how to go about this question, I don't know where to start.
MacLaurin Series Problem
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I need guidance on how to go about this question, I don't know where to start.
Steven G
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Can you write out what g(x) equals? Do you know how to write e as an infinite series? Did you or can you try to combine the two? This is what I would look at first. Then if this doesn't help then you have to try something else. Do you know what Maclaurin series yielded f(x)? Please post back showing your work. Thanks.
natHenderson
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I'm trying to figure out the series for f(x) but I'm still stuck.
topsquark
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You don't need to know the whole function f(x), you just need to get a coefficient.
[math]g(x) = e^{2 + 3x + x^2 + \text{ ...} } = e^2 \cdot e^{3x} \cdot e^{x^2} \cdot \text{ ...}[/math]
[math]g(x) = ( e^2 ) \cdot \left ( 1 + 3x + \dfrac{9}{2} x^2 + \text{ ...} \right ) \cdot \left ( 1 + x^2 + \text{ ...} \right )[/math]
Higher terms don't give coefficients for anything less than [math]x^3[/math]. So expand this out and what's your coefficient for [math]x^2[/math]?
-Dan
[math]g(x) = e^{2 + 3x + x^2 + \text{ ...} } = e^2 \cdot e^{3x} \cdot e^{x^2} \cdot \text{ ...}[/math]
[math]g(x) = ( e^2 ) \cdot \left ( 1 + 3x + \dfrac{9}{2} x^2 + \text{ ...} \right ) \cdot \left ( 1 + x^2 + \text{ ...} \right )[/math]
Higher terms don't give coefficients for anything less than [math]x^3[/math]. So expand this out and what's your coefficient for [math]x^2[/math]?
-Dan
natHenderson
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I expanded it out and I got (11/2)e^2, is that correct?
HallsofIvy
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Yes, that is what I get. I did this in a slightly different way than Topsquark suggests. In particular, I did not use the MacLaurin polynomial for \(\displaystyle e^x\).
Instead I used the fact that the MacLaurin series for g is \(\displaystyle g(0)+ g'(0)x+ \frac{g''(0)}{2}x^2+ …\). To determine the coefficient of \(\displaystyle x^2\) we only need to calculate g''(0)/2.
Since \(\displaystyle g(x)= e^{f(x)}\), \(\displaystyle g'(x)= f'(x)e^{f(x)}\) and then \(\displaystyle g''(x)= f''(x)e^{f(x)}+ (f'(x))^2e^{f(x)}= (f''(x)+ (f'(x))^2)e^f(x)\).
In particular \(\displaystyle g''(0)= (f''(0)+ (f'(0))^2)e^{f(0)}\)
We are given that \(\displaystyle f(x)= 2+ 3x+ x^2+ \frac{1}{3}x^3+ \cdot\cdot\cdot\)
so f(0)= 2.
\(\displaystyle f'(x)= 3+ 2x+ x^2+ \cdot\cdot\cdot\) so f'(0)= 3.
\(\displaystyle f''(x)= 2+ 2x\cdot\cdot\cdot\) so f''(0)= 2.
The coefficient of \(\displaystyle x^2\) in the MacLaurin series for g(x) is \(\displaystyle \frac{g''(0)}{2}=\frac{(2+ 3^2)e^{2}}{2}= \frac{2+ 9}{2}e^2= \frac{11}{2}e^2\).
Instead I used the fact that the MacLaurin series for g is \(\displaystyle g(0)+ g'(0)x+ \frac{g''(0)}{2}x^2+ …\). To determine the coefficient of \(\displaystyle x^2\) we only need to calculate g''(0)/2.
Since \(\displaystyle g(x)= e^{f(x)}\), \(\displaystyle g'(x)= f'(x)e^{f(x)}\) and then \(\displaystyle g''(x)= f''(x)e^{f(x)}+ (f'(x))^2e^{f(x)}= (f''(x)+ (f'(x))^2)e^f(x)\).
In particular \(\displaystyle g''(0)= (f''(0)+ (f'(0))^2)e^{f(0)}\)
We are given that \(\displaystyle f(x)= 2+ 3x+ x^2+ \frac{1}{3}x^3+ \cdot\cdot\cdot\)
so f(0)= 2.
\(\displaystyle f'(x)= 3+ 2x+ x^2+ \cdot\cdot\cdot\) so f'(0)= 3.
\(\displaystyle f''(x)= 2+ 2x\cdot\cdot\cdot\) so f''(0)= 2.
The coefficient of \(\displaystyle x^2\) in the MacLaurin series for g(x) is \(\displaystyle \frac{g''(0)}{2}=\frac{(2+ 3^2)e^{2}}{2}= \frac{2+ 9}{2}e^2= \frac{11}{2}e^2\).