How do you do this? I have an exam in 2 hours and I know this type will be on it and I have no clue. We were taught to base it off the power series of $x^n$
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You were probably asked to write it as an Taylor or Maclaurin series. the basic principle is to find n-derivatives, it is usually just three or four.
So you find f', f", f"', and then you find the values at point x=0 (for Taylor it is x=a) and then it is just f(x)=f(0) + (1/1!) f'(0)*x + (2/2!) f"(0)*x^2 + (3/3!) f"'(0)*x^3 + R3(x),
Where Rn (in this case it is just n=3) is the remainder of the series. it can be written in Peann form "o(x^n)" or Lagrange form.
Faks
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Use the formula $\frac{1}{1 + x} = 1 - x + x^2 - x^3 + x^4 - \dots$ to conclude that $\frac{1}{3x+4} = \frac{1}{4}\frac{1}{\frac{3x}{4} + 1} = \frac{1}{4}\left(1 - \frac{3x}{4} + \frac{9x^2}{16} - \dots \right)$. Then you can square it (using cauchy product) and multiply by $x^3$.
lisyarus
- 15,517
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use the binomial theorem $$(1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + 5x^{4}+ \cdots $$ to get $$(3x+4)^{-2} = 4^{-2}(\left(1 + \frac{3x}4\right)^{-2} = \frac1{16}\left(1 - 2\frac{3x}4 + 3\left(\frac{3x}4\right)^2 - 3\left(\frac{3x}4\right)^3+ \cdots\right) \\ \frac{x^3}{(3x+4)^2} = \frac{x^3}{16}\left(1 - 2\frac{3x}4 + 3\left(\frac{3x}4\right)^2 - 3\left(\frac{3x}4\right)^3+ \cdots\right) \text{ for } -\frac4 3 < x < \frac 43 $$
good luck.
abel
- 29,170
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We have to find out the power series of $\frac{x^3}{(3x+4)^2}$.
Recall that if $|x|<1$ then $$\frac{1}{(1+x)^2}=(1+x)^{-2}=\sum_{n\geq 0}(-1)^n(n+1)x^{n}.$$
here we have $\frac{1}{(3x+4)^2}=(3x+4)^{-2}=(3x)^{-2}(1+\frac{4}{3x})^{-2}$ if $|\frac{4}{3x}|<1$ i.e. $|x|>4/3$ and
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $=4^{-2}(1+\frac{3x}{4})^{-2}$ if $|\frac{3x}{4}|<1$ i.e. $|x|<4/3$
For the first case, we have \begin{align*} \frac{1}{(3x+4)^2}=&\frac{1}{9x^2}\sum_{n\geq 0}(-1)^n(n+1)\left(\frac{4}{3x}\right)^n\\ \therefore\frac{x^3}{(3x+4)^2}=&\frac19\sum_{n\geq 0}(-1)^n\frac{(n+1)4^n}{3^n}\frac{1}{x^{n-1}}=\sum_{n\geq 0}(-1)^n\frac{(n+1)4^n}{3^{n+2}}x^{1-n}, |x|>4/3 \end{align*} For the second case we shall have \begin{align*} \frac{1}{(3x+4)^2}=&\frac{1}{16}\sum_{n\geq 0}(-1)^n(n+1)\left(\frac{3x}{4}\right)^n\\ \therefore\frac{x^3}{(3x+4)^2}=&\sum_{n\geq 0}(-1)^n\frac{(n+1)3^n}{4^{n+2}}x^{n+3}, |x|<4/3 \end{align*}
KON3
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