User:Egm6341.s11.team4.fields/HW3.6

=Problem 3.6: Prove that qn+1(n+1) =  (n+1)!=

Refer to lecture slide [[media:nm1.s11.mtg16.djvu|16-3]] for the problem statement.

Given

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(6.1)
 * $$\displaystyle q_{n+1}^{(n+1)} =  (n+1)!$$
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Objective
Prove the equality given in equation (6.1).

Solution
From lecture [[media:nm1.s11.mtg11.djvu|11-3]] we have


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(6.2)
 * $$\displaystyle q_{n+1} :=  \prod_{j=0}^n = (x-x_j) = (x-x_0)(x-x_1)\cdots(x-x_n)$$
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Since (6.2) is a polynomial we can re-write it as:


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(6.3)
 * $$\displaystyle (x-x_0)(x-x_1)\cdots(x-x_n) = x^{n+1} + c_nx^n + \cdots + c_2x^2 + c_1x + c_0$$
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Where $$ c_n $$ through $$ c_0 $$ are constants.

We now have to calculate the $$(n+1)^{th}$$ derivative.

The 1st derivative is:


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(6.4)
 * $$\displaystyle \frac{d}{dx}\big(x^{n+1} + c_nx^n + \cdots + c_2x^2 + c_1x + c_0\big)$$
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 * $$\displaystyle = (n+1)x^{n} + nc_nx^{n-1} + \cdots + 2c_2x + c_1 + 0$$


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The 2nd derivative:


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(6.5)
 * $$\displaystyle \frac{d^2}{dx^2}\big(x^{n+1} + c_nx^n + \cdots + c_2x^2 + c_1x + c_0\big)$$
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 * $$\displaystyle = n(n+1)x^{n-1} + (n-1)nc_nx^{n-2} + \cdots + 2c_2 + 0$$


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As we can see the lower orders of x are being zeroed out as we continue to take each derivative.

The nth derivative:


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(6.6)
 * $$\displaystyle \frac{d^n}{dx^n}\big(x^{n+1} + c_nx^n + \cdots + c_2x^2 + c_1x + c_0\big)$$
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 * $$\displaystyle = (2)*(3)*\cdots*(n-1)*n*(n+1)x + (1)*(2)*(3)*\cdots*(n-1)*n + 0$$


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Finally, the $$ (n+1)^{th} $$ derivative:


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(6.7)
 * $$\displaystyle \frac{d^{n+1}}{dx^{n+1}}\big(x^{n+1} + c_nx^n + \cdots + c_2x^2 + c_1x + c_0\big)$$
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 * $$\displaystyle = (1)*(2)*(3)*\cdots *(n-1)*n*(n+1) + 0 = (n+1)!$$


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Thus proving the original statement.