Infinite variables.Infinite equations

Calculus Level 5

Sequence a n a_n for non-negative integer n n is such that

∑ k = n ∞ a k k ! ( k − n ) ! = n \large \sum_{k=n}^\infty \frac{a_k k!}{(k-n)!}=n

and that lim ⁡ n → ∞ a n = 0 \displaystyle\lim_{n\to\infty}a_n=0 .

What is a 0 ? a_0?


The answer is -0.36788.

X X by X X , 17, Taiwan

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2 solutions

Relevant wiki: Taylor Series - Problem Solving

Consider a polynomial

f ( x ) = ∑ k = 0 ∞ a k x k = ∑ k = 0 ∞ k ! ( k − 0 ) ! a k x k and that f ( 1 ) = 0 f ′ ( x ) = ∑ k = 1 ∞ k a k x k − 1 = ∑ k = 1 ∞ k ! ( k − 1 ) ! a k x k f ′ ( 1 ) = 1 f ′ ′ ( x ) = ∑ k = 2 ∞ k ( k − 1 ) a k x k − 2 = ∑ k = 2 ∞ k ! ( k − 2 ) ! a k x k f ′ ′ ( 1 ) = 2 f ′ ′ ′ ′ ( x ) = ∑ k = 3 ∞ k ( k − 1 ) ( k − 2 ) a k x k − 3 = ∑ k = 3 ∞ k ! ( k − 3 ) ! a k x k f ′ ′ ′ ( 1 ) = 3 ⟹ f ( n ) ( x ) = ∑ k = n ∞ k ! ( k − n ) ! a k x k f ( n ) ( 1 ) = n \begin{aligned} f(x) & = \sum_{k=\color{#D61F06}0}^\infty a_k x^k = \sum_{k=\color{#D61F06}0}^\infty \frac {k!}{(k-{\color{#D61F06}0})!}a_k x^k & \small \color{#3D99F6} \text{and that } f(1) = \color{#D61F06} 0 \\ f'(x) & = \sum_{k=\color{#D61F06}1}^\infty k a_k x^{k-1} = \sum_{k=\color{#D61F06}1}^\infty \frac {k!}{(k-{\color{#D61F06}1})!}a_k x^k & \small \color{#3D99F6} f'(1) = \color{#D61F06} 1 \\ f''(x) & = \sum_{k=\color{#D61F06}2}^\infty k(k-1) a_k x^{k-2} = \sum_{k=\color{#D61F06}2}^\infty \frac {k!}{(k-{\color{#D61F06}2})!}a_k x^k & \small \color{#3D99F6} f''(1) = \color{#D61F06} 2 \\ f''''(x) & = \sum_{k=\color{#D61F06}3}^\infty k(k-1)(k-2)a_k x^{k-3} = \sum_{k=\color{#D61F06}3}^\infty \frac {k!}{(k-{\color{#D61F06}3})!}a_k x^k & \small \color{#3D99F6} f'''(1) = \color{#D61F06} 3 \\ \implies f^{(n)}(x) & = \sum_{k=\color{#D61F06}n}^\infty \frac {k!}{(k-{\color{#D61F06}n})!}a_k x^k & \small \color{#3D99F6} f^{(n)}(1) = \color{#D61F06} n \end{aligned}

Let T ( x ) T(x) be the Taylor series expansion of f ( x ) f(x) centered at 1. Then we have:

T ( x ) = ∑ k = 0 ∞ f k ( 1 ) k ! ( x − 1 ) k = ∑ k = 1 ∞ k k ! ( x − 1 ) k = ∑ k = 1 ∞ ( x − 1 ) k ( k − 1 ) ! = ∑ k = 0 ∞ ( x − 1 ) k + 1 k ! \begin{aligned} T(x) & = \sum_{\color{#3D99F6}k=0}^\infty \frac {f^{k}(1)}{k!} (x-1)^k = \sum_{\color{#D61F06}k=1}^\infty \frac k{k!} (x-1)^k = \sum_{\color{#D61F06}k=1}^\infty \frac {(x-1)^k}{(k-1)!} = \sum_{\color{#3D99F6}k=0}^\infty \frac {(x-1)^{k+1}}{k!} \end{aligned}

We note that

a 0 = T ( 0 ) = ∑ k = 0 ∞ ( − 1 ) k + 1 k ! = − ∑ k = 0 ∞ ( − 1 ) k k ! = − 1 e ≈ − 0.368 \begin{aligned} a_0 = T(0) & = \sum_{k=0}^\infty \frac {(-1)^{k+1}}{k!} = - \sum_{k=0}^\infty \frac {(-1)^k}{k!} = - \frac 1e \approx \boxed {-0.368} \end{aligned}

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We still need to show that ∑ k = 0 ∞ a k ( 1 + ε ) k \sum_{k=0}^\infty a_k(1+\varepsilon)^k converges for some ε > 0 \varepsilon>0

All we know immediately is that the power series for f ( x ) f(x) converges on ( − 1 , 1 ] (-1,1] . Therefore it is possible that T ( x ) T(x) , as the Taylor series for f ( x ) f(x) about x = 1 x=1 , only converges to f ( x ) f(x) at x = 1 x=1 and nowhere else. The series result I mention above is precisely what we need to show to imply the Taylor series has a positive radius of convergence, from which the rest of your proof follows.

At the moment, I'm still trying to see if this can be recovered from the given assumptions, but I don't see how. If someone can show it from the given assumptions, it would actually fix both of our proofs--yours immediately and mine after a slight comment.

Brian Moehring - 2 years, 10 months ago
Brian Moehring
Aug 4, 2018

Incomplete Solution.... Please See My Report For Details

Note that ∑ n = 0 ∞ ∑ k = n ∞ ( k ! ( k − n ) ! a k ⋅ ( − 1 ) n n ! ) = ∑ n = 0 ∞ ( ∑ k = n ∞ k ! ( k − n ) ! a k ) ( − 1 ) n n ! = ∑ n = 0 ∞ n ⋅ ( − 1 ) n n ! = ( − 1 ) ⋅ ∑ n = 0 ∞ ( − 1 ) n n ! = − e − 1 \begin{aligned} \sum_{n=0}^\infty \sum_{k=n}^\infty \left(\frac{k!}{(k-n)!} a_k \cdot \frac{(-1)^n}{n!}\right) &= \sum_{n=0}^\infty \left(\sum_{k=n}^\infty \frac{k!}{(k-n)!} a_k\right) \frac{(-1)^n}{n!} \\ &= \sum_{n=0}^\infty n \cdot \frac{(-1)^n}{n!} \\ &= (-1) \cdot \sum_{n=0}^\infty \frac{(-1)^n}{n!} \\ &= -e^{-1} \end{aligned} and ∑ k = 0 ∞ ∑ n = 0 k ( k ! ( k − n ) ! a k ⋅ ( − 1 ) n n ! ) = ∑ k = 0 ∞ a k ( ∑ n = 0 k ( k n ) ( − 1 ) n ) = ∑ k = 0 ∞ a k ⋅ { 1 if k = 0 0 otherwise = a 0 \begin{aligned} \sum_{k=0}^\infty \sum_{n=0}^k \left(\frac{k!}{(k-n)!} a_k \cdot \frac{(-1)^n}{n!}\right) &= \sum_{k=0}^\infty a_k \left(\sum_{n=0}^k \binom{k}{n} (-1)^n\right) \\ &= \sum_{k=0}^\infty a_k \cdot \begin{cases} 1 & \text{ if } k=0 \\ 0 & \text{ otherwise} \end{cases} \\ &= a_0 \end{aligned}

Therefore, if we can interchange the sums then we have a 0 = − e − 1 ≈ − 0.367879441 a_0 = -e^{-1} \approx \boxed{-0.367879441}

Unfortunately, it looks like I need to know something about as strong as ∑ k = 0 ∞ 2 k ∣ a k ∣ < ∞ \sum_{k=0}^\infty 2^k|a_k| < \infty to make this conclusion, so I don't know if it's always true...

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