Oh, Integrals ? How many ?

Calculus Level 2

Find the coefficient of t n t^n in the expansion of . . . . . . n integrals d t d t n d t s \displaystyle \underbrace{\int\int\int... ... \int \int}_\text{ n integrals } \underbrace{\text{d}t\text{ }\ldots\text{ }\ldots\text dt}_{\text{n}\text{ } dt's}

Assume all constants are equal to 1. 1.


You might like Wow , really many Integrals !

1 n \dfrac{1}{n} n n 1 \dfrac{n}{n-1} 1 n ! \dfrac{1}{n!} 1 ( n 1 ) ! \dfrac{1}{(n-1)!}

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1 solution

Trevor B.
Oct 9, 2014

Taking a look at the first few terms in the sequence, we find this. d t d t = t + C d t = 1 2 t 2 + C t + D \int\int\text{ }dt\text{ }dt=\int t+C\text{ }dt=\dfrac{1}{2}t^2+Ct+D d t d t d t = t + C d t d t = 1 2 t 2 + C t + D d t = 1 6 t 3 + C 2 t 2 + D t + F . \int\int\int\text{ }dt\text{ }dt\text{ }dt=\int\int t+C\text{ }dt\text{ }dt=\int\dfrac{1}{2}t^2+Ct+D\text{ }dt=\dfrac{1}{6}t^3+\dfrac{C}{2}t^2+Dt+F. It appears that the coefficient of the n n th term in the sequence is equal to a constant times the reciprocal of n ! n! In fact, the n n th term in the sequence is equal to this. t n n ! + i = 1 n 1 C i t i i ! \dfrac{t^{n}}{n!}+\sum_{i=1}^{n-1}\dfrac{C_it^i}{i!} Because all constants are equal to 1 , 1, the t n t^n term is equal to t n n ! \dfrac{t^n}{n!} with a coefficient of 1 n ! \dfrac{1}{n!} .

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