I was looking for the expansion of $e^\frac{1}{3}$ using five terms of Maclaurin's series for $e^x$ and I'm trying to estimate the accuracy of the result.
My work:
The Taylor's formula looks like this:
$$f(b) = f(a) + f'(a) (b-a) + \frac{f''(a) (b-a)^2 }{2!} + \frac{f'''(a) (b-a)^3 }{3!} + \frac{f^4(a) (b-a)^4 }{4!} + ......+ \frac{f^n(x_1) (b-a)^n }{n!} $$
But my book instructed me to use the Maclaurin's formula, which is just a special case of Taylor's formula when $ a = 0$, so the Maclaurin's formula looks like this (according to my understanding):
$$f(b) = f(0) + f'(0) (b) + \frac{f''(0) (b)^2 }{2!} + \frac{f'''(0) (b)^3 }{3!} + \frac{f^4(0) (b)^4 }{4!} + ......+ \frac{f^n(x_1) (b)^n }{n!} $$
I only need five terms of Maclaurin's formula as instructed in my book, so I got to differentiate the expression $e^x$ five times and listing them one by one. Fortunately, the 1st up to 5th derivative of $e^x$ is $e^x$.
Taking in five terms in Maclaurin's expansion of $e^x$and putting $x = \frac{1}{3}$, we have:
$$e^\frac{1}{3} = e^0 + e^0(\frac{1}{3}) + \frac{e^0(\frac{1}{3})^2}{2!} + \frac{e^0(\frac{1}{3})^3}{3!} + \frac{e^0(\frac{1}{3})^2}{4!} + R_5$$
which is equal to:
$$e^\frac{1}{3} = 1.395576132 + R_5$$
The value $1.395576132$ is from my calculator.
To determine the accuracy of the value of the expansion , given were the number of terms used, I need the so-called Lagrange's form of the reminder, which is expressed as:
$$R_n = \frac{b-a}{n!}f^n(x_1)$$
I was following the example given from the book, so I did this: $b-a = \frac{1}{3},$ $f^5 (x) = e^x.$ Plugging it into the Lagrange's form of the reminder, we get:
$$R_5 = \frac{(\frac{1}{3})^5}{5!}e^{x_1}$$
where $0 < x_1 < \frac{1}{3}$
The number $e^{x_1}$ must be less than $e^{\frac{1}{3}}$, and since $e^{\frac{1}{3}}$ < $8^{\frac{1}{3}} = 2$, it follows that:
$$R_5 < \frac{(\frac{1}{3})^5}{5!}(2) = 0.00006858$$ or $$R_5 = 0.00007$$
Hence the value of $e^{\frac{1}{3}}$, correct to four decimals, is $1.3956$.
My question is, in my book, the answer is $$R_5 < 0.00005$$ How do you get the value of $R_5$? It's so close to my answer.
$\endgroup$ 11 Answer
$\begingroup$Although completely true, your estimate that
$$e^{1/3} < 8^{1/3} = 2$$
loses a lot of information. In fact, it's easy to check (by hand, even!) that $e^{1/3} < 1.5$, since $1.5^3 = 2.25 * 1.5 > 3$. This improvement is enough to get that
$$R_5 < 0.0000515$$
Another by-hand estimate shows that $1.4^3 = 2.744 > e$, so we could improve this a touch more, to $$R_5 < 0.0000481.$$
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