62 lines
2.1 KiB
TeX
62 lines
2.1 KiB
TeX
\beginsection{28.15}
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Prove Leibniz' rule
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$$(fg)^{(n)}=\sum_{k=0}^n\left({n\atop k}\right)f^{(k)}(a)g^{(n-k)}(a)$$
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{\it Hint:} Use mathematical induction.
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For $n=1$, apply Theorem 28.3(iii).
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\medskip
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The following equalities will be used.
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$$\left({n\atop k}\right)+\left({n\atop k-1}\right)=\left({n+1\atop k}\right)
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\eqno\hbox{(A)}$$
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$$\left({n\atop0}\right)=1\eqno\hbox{(B)}$$
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$$\left({n\atop n}\right)=1\eqno\hbox{(C)}$$
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Theorem 28.3(iii) is not really needed.
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Just using the case $n=0$ for induction step 1.
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For induction step 2, show that Leibniz' rule is true for $n+1$ whenever it is
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true for $n$.
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$$\eqalign{
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(fg)^{(n+1)}&=((fg)^{(n)})^\prime\cr
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&=\left(\sum_{k=0}^n\left({n\atop k}\right)
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f^{(k)}g^{(n-k)}\right)^\prime\cr
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&=\sum_{k=0}^n\left({n\atop k}\right)
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\left[f^{(k)}\left(g^{(n-k)}\right)^\prime
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+\left(f^{(k)}\right)^\prime g^{(n-k)}\right]\cr
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&=\sum_{k=0}^n\left({n\atop k}\right)
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\left[f^{(k)}g^{(n+1-k)}+f^{(k+1)}g^{(n-k)}\right]\cr
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&=\sum_{k=0}^n\left({n\atop k}\right)f^{(k)}g^{(n+1-k)}
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+\sum_{k=0}^n\left({n\atop k}\right)f^{(k+1)}g^{(n-k)}
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}$$
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Use (B) to extract the $k=0$ term of the sum on the left and use (C)
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to extract the $k=n$ term of the sum on the right.
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$$\eqalign{
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(fg)^{(n+1)}&=
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f^{(0)}g^{(n+1)}
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+\sum_{k=1}^n\left({n\atop k}\right)f^{(k)}g^{(n+1-k)}
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+\sum_{k=0}^{n-1}\left({n\atop k}\right)f^{(k+1)}g^{(n-k)}
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+f^{(n+1)}g^{(0)}
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}$$
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Here is the main trick.
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The index $k$ is just a dummy variable that takes on a range of values.
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We can shift the range of $k$ if we balance the math where $k$ is used.
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For the sum on the right, change $k$ so that it runs from
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1 to $n$.
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$$\eqalign{
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(fg)^{(n+1)}&=
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f^{(0)}g^{(n+1)}
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+\sum_{k=1}^n\left({n\atop k}\right)f^{(k)}g^{(n+1-k)}
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+\sum_{k=1}^n\left({n\atop k-1}\right)f^{(k)}g^{(n+1-k)}
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+f^{(n+1)}g^{(0)}
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}$$
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Note that the $n$ in the binomial factor does not change because $n$ does
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not depend on $k$.
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Next use (A) to combine the two sums.
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$$\eqalign{
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(fg)^{(n+1)}&=
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f^{(0)}g^{(n+1)}
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+\sum_{k=1}^n\left(n+1\atop k\right)f^{(k)}g^{(n+1-k)}
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+f^{(n+1)}g^{(0)}
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}$$
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Finally, use (B) and (C) to combine all the terms.
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$$(fg)^{(n+1)}=\sum_{k=0}^{n+1}
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\left({n+1\atop k}\right)f^{(k)}g^{(n+1-k)}$$ |