Exponentials bearbeitet, items erweitert
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@@ -127,13 +127,18 @@ Each polynomial in the sequence is, in a sense, the best approximation possible
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\begin{itemize}
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\begin{itemize}
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\item[\textcolor{red}{\tikzcircle{3pt}}] So, how good of an approximation is a polynomial truncation of $e^{\text {? }}$. Use a calculator to compare how close $e$ is to the linear, quadratic, cubic, quartic, and quintic approximations. How many digits of accuracy do you seem to be gaining with each additional term in the series?
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\item[\textcolor{red}{\tikzcircle{3pt}}] So, how good of an approximation is a polynomial truncation of $e^{\text {? }}$. Use a calculator to compare how close $e$ is to the linear, quadratic, cubic, quartic, and quintic approximations. How many digits of accuracy do you seem to be gaining with each additional term in the series?
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\item[\textcolor{red}{\tikzcircle{3pt}}] Now, do the same thing with $1 / e$ by plugging in $x=-1$ into the series. Do you have the same results? Are you surprised?
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\item[\textcolor{red}{\tikzcircle{3pt}}] Calculate the following sum using what you know:
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$$
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\sum_{n=0}^{\infty}(-1)^n \frac{(\ln 3)^n}{n!}
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$$
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\end{itemize}
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\end{itemize}
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%-
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%-
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%- Now, do the same thing with $1 / e$ by plugging in $x=-1$ into the series. Do you have the same results? Are you surprised?
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%-
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%- Calculate the following sum using what you know:
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%$$
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%\sum_{n=0}^{\infty}(-1)^n \frac{(\ln 3)^n}{n!}
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%$$
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%- Write out the first four terms of the following series
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%- Write out the first four terms of the following series
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%$$
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%$$
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%\sum_{n=0}^{\infty}(-1)^n \frac{\pi^{2 n}}{2^n n!}
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%\sum_{n=0}^{\infty}(-1)^n \frac{\pi^{2 n}}{2^n n!}
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