Question: Find the Maclaurin series representation of $(1 - \frac{x}{5})^{-4}$ using the definition of the Maclaurin series $\sum_{n=0}^{\infty} \frac{F^{n}(a)}{n!}(x-a)^n$

My approach:

  1. Find $F^n(0) = \frac{(n+3)!}{5^n}$
  2. Input $F^n(0)$ into formula.
  3. Final Result: $\sum_{n=0}^{\infty} \frac{(n+3)!}{5^n * n!} * x^n$

However, answer is marked as wrong. Can someone please explain the reason for this. Also any tips to help me understand the Taylor / Maclaurin series would be much appreciated.

  • 3
    $\begingroup$ Just as a sanity check, did you remember to include the $x^n$ terms? As is written now you have calculated a series representation of $(1-\frac{1}{5})^{-4}$... $\endgroup$ – Theo Diamantakis Apr 15 at 17:41

The Maclaurin series definition isn't quite right. Instead, it should be $$\sum_{n=0}^\infty\frac{F^{(n)}(0)}{n!}x^n.$$ Consequently, we should have $$\left(1-\frac{x}{5}\right)^{-4}=\sum_{n=0}^\infty\frac{(n+3)!}{5^n\cdot n!}x^n$$ by the work that you've already done.

However, noting that $$(n+3)!=(n+3)\cdot(n+2)\cdot(n+1)\cdot n!,$$ we can write it instead as $$\left(1-\frac{x}{5}\right)^{-4}=\sum_{n=0}^\infty\frac{(n+3)(n+2)(n+1)}{5^n}x^n.$$

One way we can see that there must be something involving $x$ in the series is that $\left(1-\frac{x}{5}\right)^{-4}$ varies in value with different $x$ values, while $\sum_{n=0}^\infty\frac{(n+3)!}{5^n\cdot n!}$ is simply a constant--in particular, it turns out to be $$\left(1-\frac{1}{5}\right)^{-4}=\left(\frac45\right)^{-4}=\left(\frac54\right)^4=\frac{5^4}{4^4}=\frac{625}{256}.$$

Added: It seems that you've made an error in calculating the derivatives, unfortunately.

$$F'(x)=-4\cdot\left(1-\frac{x}{5}\right)^{-5}\cdot\frac{d\left(1-\frac{x}{5}\right)}dx=\frac15\cdot 4\cdot\left(1-\frac{x}{5}\right)^{-5}$$

$$F''(x)=-5\cdot\frac15\cdot 4\cdot\left(1-\frac{x}{5}\right)^{-6}\cdot\frac{d\left(1-\frac{x}{5}\right)}{dx}=\frac1{5^2}\cdot 5\cdot 4\cdot\left(1-\frac{x}{5}\right)^{-6}$$

$$F'''(x)=-6\cdot\frac15\cdot 5\cdot 4\cdot\left(1-\frac{x}{5}\right)^{-7}\cdot\frac{d\left(1-\frac{x}{5}\right)}{dx}=\frac1{5^3}\cdot 6\cdot 5\cdot 4\cdot\left(1-\frac{x}{5}\right)^{-7}$$

and more generally, for $n\ge 4,$ we have $$F^{(n)}(x)=\frac1{5^n}\cdot(n+3)\cdot(n+2)\cdots 5\cdot 4\cdot\left(1-\frac{x}{5}\right)^{-(n+4)}=\frac{1}{5^n}\cdot\frac{(n+3)!}{3!}\cdot \left(1-\frac{x}{5}\right)^{-(n+4)}.$$

Thus, $$F^{(n)}(0)=\frac{(n+3)!}{5^n\cdot 3!}$$ for each $n,$ and so the Maclaurin series is $$\left(1-\frac{x}{5}\right)^{-4}=\sum_{n=0}^\infty\frac{(n+3)!}{5^n\cdot 3!\cdot n!}x^n=\sum_{n=0}^\infty\frac{(n+3)(n+2)(n+1)}{5^n\cdot 3!}x^n.$$

  • $\begingroup$ Would the same concept apply to something like (1-x/3)^(-5), following your logic, I got (n+4)(n+3)(n+2)(n+1)/3^n * x^n. However, I keep getting an error. Im still new to series and sequences so my logic is incomplete, is there something wrong with what was stated? $\endgroup$ – Urmzd Apr 15 at 18:12
  • $\begingroup$ It looks like you made an error calculating the derivatives. I've adjusted my answer accordingly. $\endgroup$ – Cameron Buie Apr 15 at 18:32
  • $\begingroup$ Thank you. I'm still slightly confused as to why there is a 3!, what's the reasoning behind this? $\endgroup$ – Urmzd Apr 15 at 18:36
  • $\begingroup$ It's because $$\begin{eqnarray}(n+3)\cdot(n+2)\cdots5\cdot 4 &=&(n+3)\cdot(n+2)\cdots5\cdot 4\cdot\frac33\cdot\frac22\cdot\frac11\\ &=& \frac{(n+3)\cdot(n+2)\cdots 2\cdot 1}{3\cdot 2\cdot 1}\\ &=& \frac{(n+3)!}{3!}.\end{eqnarray}$$ $\endgroup$ – Cameron Buie Apr 15 at 18:38

$$(1 - \frac{x}{5})^{-4}=\frac{1}{(1-\frac{x}{5})^4}$$ use $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$ $$(\frac{1}{1-x})'''=\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$$ $$\frac{6}{(1-x)^4}=\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$$ or $$\frac{1}{(1-x)^4}=\frac{1}{6}\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$$ $$\frac{1}{(1-x)^4}=\frac{1}{6}\sum_{n=0}^{\infty}(n+1)(n+2)(n+3)x^{n}$$

now let $x\rightarrow \frac{x}{5}$

  • $\begingroup$ +1: I really like this approach! $\endgroup$ – Cameron Buie Apr 16 at 0:01
  • $\begingroup$ Thanks ...... . $\endgroup$ – E.H.E Apr 16 at 4:39

Let $F(x)=\left(1 - \dfrac{x}{5}\right)^{-4}.$

Using your approach,

$$(1)\quad F^{(n)}(0) = \frac{(n+3)!}{5^n3!}.$$

$(2)$ The formula for Maclaurin series is $$\sum_{n=0}^{\infty} \frac{F^{(n)}(0)}{n!}x^n.$$

$(3)$ The final result should be $$\sum_{n=0}^{\infty} \frac{(n+3)(n+2)(n+1)}{5^n 3!}x^n.$$

  • $\begingroup$ Can you explain why there is a 3! in step 1? $\endgroup$ – Urmzd Apr 15 at 18:05
  • $\begingroup$ You should be able to prove it by induction, but note that when $n=0$ it's just $1$; the first derivative gets a factor of $4$ from the exponent and the exponent becomes $-5,$ so the second derivative has a coefficient $4\times5$ and exponent $-6$, the third derivative has coefficient $4\times5\times6$...the $n^{th}$ derivative has coefficient $4\times5\times6\times...\times (n+3)$, which is $(n+3)!/3!$ $\endgroup$ – J. W. Tanner Apr 15 at 18:18

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