Zero solvers challenge

For the ease of notation, write $g(x) = \frac{1}{e}f(x+1) = \frac{1}{e}(x+1)^4e^{x+1} = (x+1)^4e^x$ so that $g^{(100)}(0) = \frac{1}{e}f^{(100)}(1) = \frac{1}{e}(ke) = k.$

We may find this using the Maclaurin series for $g(x)$ , which in this case may be interpreted as the exponential generating function for derivatives of the form $g^{(n)}(0)$ . Toward this, $\begin{aligned} g(x) &= (x+1)^4 e^x \\ &= (x^4 + 4x^3 + 6x^2 + 4x + 1)\sum_{n=0}^\infty \frac{1}{n!}x^n \\ &= \sum_{n=0}^\infty \frac{1}{n!} x^{n+4} + \sum_{n=0}^\infty \frac{4}{n!} x^{n+3} + \sum_{n=0}^\infty \frac{6}{n!} x^{n+2} + \sum_{n=0}^\infty \frac{4}{n!} x^{n+1} + \sum_{n=0}^\infty \frac{1}{n!} x^n \\ &= P_3(x) + \sum_{n=4}^\infty \left(\frac{1}{(n-4)!} + \frac{4}{(n-3)!} + \frac{6}{(n-2)!} + \frac{4}{(n-1)!} + \frac{1}{n!}\right) x^n \qquad P_3(x) \text{ is a polynomial of degree } \leq 3 \\ &= P_3(x) + \sum_{n=4}^\infty \frac{1}{n!} \left(n(n-1)(n-2)(n-3) + 4n(n-1)(n-2) + 6n(n-1) + 4n + 1\right) x^n \end{aligned}$ Therefore, $\begin{aligned} k &= g^{(100)}(0) \\ &= 100(99)(98)(97) + 4(100)(99)(98) + 6(100)(99) + 4(100) + 1 \\ &= \boxed{98050001} \end{aligned}$

Let's write down the first few derivatives of $f(x)$

$f^{(1)} (x) = x^4 e^x + \underbrace{4x^3 e^x}_{1 \text{ time}} \\ \\ f^{(2)} (x) = x^4 e^x + \underbrace{4x^3 e^x + 4x^3 e^x}_{2 \text{ times}} + \underbrace{12x^2 e^x}_{1 \text{ time}} \\ \\ f^{(3)} (x) = x^4 e^x + \underbrace{4x^3 e^x + 4x^3 e^x + 4x^3 e^x}_{3 \text{ times}} + \underbrace{12x^2 e^x + 12x^2 e^x + 12x^2 e^x}_{3 \text{ times}} + \underbrace{24x e^x}_{1 \text{ time}} \\ \\ f^{(4)} (x) = x^4 e^x + \underbrace{4x^3 e^x + \cdots + 4x^3 e^x}_{4 \text{ times}} + \underbrace{12x^2 e^x + \cdots + 12x^2 e^x}_{6 \text{ times}} + \underbrace{24x e^x + \cdots + 24x e^x}_{4 \text{ times}} + \underbrace{24 e^x}_{1 \text{ time}} \\ \\ f^{(5)} (x) = x^4 e^x + \underbrace{4x^3 e^x + \cdots + 4x^3 e^x}_{5 \text{ times}} + \underbrace{12x^2 e^x + \cdots + 12x^2 e^x}_{10 \text{ times}} + \underbrace{24x e^x + \cdots + 24x e^x}_{10 \text{ times}} + \underbrace{24 e^x + \cdots + 24 e^x}_{5 \text{ times}}$

Here, we observe the following pattern for the number of times each term appears in the $n^{\text{th}}$ derivative of $f(x)$ :

• $x^4 e^x$ : only once $\forall \ n$ .

• $4x^3 e^x$ : $n$ times $\forall \ n$ .

• $12x^2 e^x : \begin{cases} \dbinom n2 & \text{ for } n \ge 2 \\ 0 & \text{ otherwise } \end{cases}$

• $24x e^x : \begin{cases} \dbinom n3 & \text{ for } n \ge 3 \\ 0 & \text{ otherwise } \end{cases}$

• $24 e^x : \begin{cases} \dbinom n4 & \text{ for } n \ge 4 \\ 0 & \text{ otherwise } \end{cases}$

Thus for $f^{(100)} (x)$ , we have

$f^{(100)} (x) = x^4 e^x + 100 \left( 4x^3 e^x \right) + \dbinom{100}{2} \left( 12x^2 e^x \right) + \dbinom{100}{3} \left( 24x e^x \right) + \dbinom{100}{4} \left( 24 e^x \right)$

At $x=1$ , we have

$\begin{aligned} f^{(100)} (1) &= e + 100 \left( 4e \right) + \dbinom{100}{2} \left( 12e \right) + \dbinom{100}{3} \left( 24e \right) + \dbinom{100}{4} \left( 24e \right) \\ &= e + 400 e + 59400 e + 3880800 e + 94109400 e \\ &= \boxed{98050001 e} \end{aligned}$

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Notations: $\dbinom MN = \dfrac {M!}{N! (M-N)!}$ denotes the binomial coefficient .

$LEIBNITZ ------ THEOREM$

$Nth$ differential of a product of two functions

If $y=uv$ ; where $u$ and $v$ are functions of $x$ , then

$y_n$ $= \sum_{i=1}^{n} \binom{n}{i}$ $u_n$ $v_{n-i}$

Where the subscript $n$ denotes the $nth$ derivative

$Source$ : Further Engineering Mathematics By K. A. Stroud

Programme 5 : Power Series Solution Of Differential Equations.