MHB 206.11.3.27 Tayor series 3 terms

[karush]

$\textsf{a. Find the first four nonzero terms
of the Taylor series $a=1$}$
\begin{align}
\displaystyle
f^0(x)&=6^{x} &\therefore \ \ f^0(a)&= 6 \\
f^1(x)&=6^{x}\ln(6) &\therefore \ \ f^1(a)&= 6\ln(6) \\
f^2(x)&={6^{x}\ln(6)^2} &\therefore \ \ f^2(a)&= {12\ln(6)} \\
f^3(x)&={6^{x}\ln(6)^3} &\therefore \ \ f^3(a)&= {18\ln(6)} \\
\end{align}
$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&\approx\frac{6}{0!}(x-1)^{0}
+\frac{6ln(6)}{1!}(x-1)^{1}
+\frac{12\ln(6)}{2!}(x-1)^{2}
+\frac{18\ln(6)}{3!}(x-1)^{3} \\
f\left(x\right)&\approx
6+6ln(6)(x-1)+6\ln(6)(x-1)^{2}+3\ln(6)(x-1)^{3}
\end{align}

$\textit{suggestions?}$

 

[topsquark]

$\textsf{a. Find the first four nonzero terms
of the Taylor series $a=1$}$
\begin{align}
\displaystyle
f^0(x)&=6^{x} &\therefore \ \ f^0(a)&= 6 \\
f^1(x)&=6^{x}\ln(6) &\therefore \ \ f^1(a)&= 6\ln(6) \\
f^2(x)&={6^{x}\ln(6)^2} &\therefore \ \ f^2(a)&= {12\ln(6)} \\
f^3(x)&={6^{x}\ln(6)^3} &\therefore \ \ f^3(a)&= {18\ln(6)} \\
\end{align}
$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&\approx\frac{6}{0!}(x-1)^{0}
+\frac{6ln(6)}{1!}(x-1)^{1}
+\frac{12\ln(6)}{2!}(x-1)^{2}
+\frac{18\ln(6)}{3!}(x-1)^{3} \\
f\left(x\right)&\approx
6+6ln(6)(x-1)+6\ln(6)(x-1)^{2}+3\ln(6)(x-1)^{3}
\end{align}

$\textit{suggestions?}$

Note that f^{(2)} (x) = 6^x \cdot (ln(6))^2, so f^{(2)}(1) = 6 \cdot (ln(6))^2. All of your coefficients will be of the form f^{(6)} (1) = 6 \cdot (ln(6))^n.

-Dan

Addendum: I think I see what's happened here. (ln(6))^2 \neq 2 \cdot ln(6). You are thinking of ln \left ( 6^2 \right ). They are not the same.

 

[karush]

$\textsf{a. Find the first four nonzero terms
of the Taylor series $a=1$}$
\begin{align}
\displaystyle
f^0(x)&=6^{x} &\therefore \ \ f^0(a)&= 6 \\
f^1(x)&=6^{x}\ln(6) &\therefore \ \ f^1(a)&= 6\ln(6) \\
f^2(x)&={6^{x}\ln^2(6)} &\therefore \ \ f^2(a)&= {6 \ln^2(6)} \\
f^3(x)&={6^{x}\ln^3(6)} &\therefore \ \ f^3(a)&= {6 \ln^3(6)} \\
\end{align}
$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&\approx\frac{6}{0!}(x-1)^{0}
+\frac{6ln(6)}{1!}(x-1)^{1}
+\frac{6\ln^2(6)}{2!}(x-1)^{2}
+\frac{6\ln^3 (6)}{3!}(x-1)^{3} \\
f\left(x\right)&\approx
6+6ln(6)(x-1)+3\ln^2 (6)(x-1)^{2}+\ln^3 (6)(x-1)^{3}
\end{align}
$\textit{hope this is it ... so hard to see errors}$

 

[topsquark]

$\textsf{a. Find the first four nonzero terms
of the Taylor series $a=1$}$
\begin{align}
\displaystyle
f^0(x)&=6^{x} &\therefore \ \ f^0(a)&= 6 \\
f^1(x)&=6^{x}\ln(6) &\therefore \ \ f^1(a)&= 6\ln(6) \\
f^2(x)&={6^{x}\ln^2(6)} &\therefore \ \ f^2(a)&= {6 \ln^2(6)} \\
f^3(x)&={6^{x}\ln^3(6)} &\therefore \ \ f^3(a)&= {6 \ln^3(6)} \\
\end{align}
$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&\approx\frac{6}{0!}(x-1)^{0}
+\frac{6ln(6)}{1!}(x-1)^{1}
+\frac{6\ln^2(6)}{2!}(x-1)^{2}
+\frac{6\ln^3 (6)}{3!}(x-1)^{3} \\
f\left(x\right)&\approx
6+6ln(6)(x-1)+3\ln^2 (6)(x-1)^{2}+\ln^3 (6)(x-1)^{3}
\end{align}
$\textit{hope this is it ... so hard to see errors}$

Yup! And, of course, you can verify it to a degree...pick some x's close to 1 and compare the function with the approximation. They should be reasonably close.

-Dan

 

[karush]

I'll try two more_ already getting burnt on em...(Doh)