標題:
Very hard differentiation
發問:
If y(0) = e(x2) , y(n) = differentiates y(0) n times for all positive integers n , prove that y(n+1) - 2xy(n) - 2ny(n-1) = 0
y(0) = ex2 Differentiate both sides with respect to x. y(1) = 2x ex2 = 2xy(0) Differentiate both sides with respect to x by n times, by Leibniz's Theorem, y(n+1) = 2xy(n) + 2nC1y(n-1) Therefore, y(n+1) - 2xy(n) - 2ny(n-1) = 0 for all positive integers n.
其他解答:
Very hard differentiation
發問:
If y(0) = e(x2) , y(n) = differentiates y(0) n times for all positive integers n , prove that y(n+1) - 2xy(n) - 2ny(n-1) = 0
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最佳解答:y(0) = ex2 Differentiate both sides with respect to x. y(1) = 2x ex2 = 2xy(0) Differentiate both sides with respect to x by n times, by Leibniz's Theorem, y(n+1) = 2xy(n) + 2nC1y(n-1) Therefore, y(n+1) - 2xy(n) - 2ny(n-1) = 0 for all positive integers n.
其他解答:
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