Taylor Series Convergernce

I have a question on Taylor Series. I missed my lectures for this topic so I'm trying my best to follow the examples on my own but I just need confirmation whether I'm on the right track. I've picked this question:

i) Find the Maclaurin series for the function $f(x)=e^{\frac{1}{2}x}$, writing the series using sigma notation Maclaurin Series $x_0=0$ $$n=0:f(x)=e^{\frac{1}{2}x},f(0)=1,C_0=1$$ $$n=1:f'(x)=\frac{1}{2}e^{\frac{1}{2}x},f'(0)=\frac{1}{2},C_1=\frac{1}{2}$$ $$n=2:f''(x)=\frac{1}{4}e^{\frac{1}{2}x},f''(0)=\frac{1}{4},C_2=\frac{\frac{1}{4}}{2!}$$ $$n=3:f'''(x)=\frac{1}{8}e^{\frac{1}{2}x},f'''(0)=\frac{1}{8},C_3=\frac{\frac{1}{8}}{3!}$$ $$\sum_{n=0}^{infinity}\frac{\frac{1}{2}^nx^n}{n!}$$

ii)Apply Taylor's Theorem to find an upper bound for the 5th remainder $R_5(x)$ on the interval $-\frac{1}{2}<=x<=\frac{3}{2}$ $$R_5(x)<=\frac{M|x-x_0|^{n+1}}{(n+1)!}$$ $$[-\frac{1}{2},\frac{3}{2}]$$ Ok so I solved this by following an example from my lecture slide and when it reached this step, my lecturer substituted the right endpoint into the formula for the $x$ value because he stated that the function in the example was increasing on the right. I just copied what he did for this question, but how do you know which side the function is increasing? $$M=e^{(\frac{1}{2})(\frac{3}{2})}$$ $$R_5<=e^{\frac{3}{4}}\frac{|\frac{3}{2}|^6}{6!}$$
$$0.53586529$$

• part i) is correct. part ii) $M$ is the upper bound of $|f^{(6)}(x)|$ over the interval $\frac {1}{64} e^{\frac 34}$ – Doug M Oct 11 '17 at 21:24
• ohhh wait I think I can follow so it's the 5th derivative $\frac{1}{64}e^{\frac{1}{2}x}$ then substitute $\frac{3}{2}$ in place of $x$ But why $\frac{3}{2}$ and not $-\frac{1}{2}$? Is it correct that because it's increasing on the right? Which leads me to my next question.....you only substitute the endpoint where the function increases? – Sean Paul Oct 11 '17 at 21:31
• the nth derivative of an exponential function is an exponential function. These functions are monotonic. In this case increasing and always positive. To maximize $M$ you just need to look at the right endpoint. – Doug M Oct 11 '17 at 21:36
• So this is it? $$M=\frac{1}{64}e^{(\frac{1}{2})(\frac{3}{2})}$$ $$R_5<=\frac{1}{64}e^{\frac{3}{4}}\frac{|\frac{3}{2}|^6}{6!}$$ $5.23$X$10^{-4}$ – Sean Paul Oct 11 '17 at 21:49
• Looks right. If you have some concern, plug it all into an excel spreadsheet. – Doug M Oct 11 '17 at 22:04