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A problem I'm trying to prove by induction ...


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I'm trying to prove something but 3 units in math make it difficult for me -

I have the following function

f (x) = x (x-1) (x-2) ... (xn)

I want to prove that the function derivative at point 0 is n rally

f '(0) = n!

By my test she is n! When n is even and minus n! When it is negative.

This is what I did in the induction, something to me:

x (x-1) (x-2) ... (xk) = k!
x (x-1) (x-2) ... (xk) (xk-1) = (k + 1)! = k! * (k + 1)
Reducing ...
f '(0) = (xk-1) = (k + 1)
= (0-k-1) = (k + 1)
-k-1 = k + 1

I realize I have to somehow show that k + 1 is supposed to get minus ...


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The derivative should indeed be! N multiplied by 1 with n (just check for small values ​​of n and it will be clear).

Think of how to find a relation between the derivative of f with n = k and the derivative of f with n = k + 1 (the derivative in general, not the derivative of 0). Except that it can also be proved not by induction, but a little waving.

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Thanks for the investment,

I did do it in a similar way, but in the end I referred to the link as I had shown earlier (plus limb 1 with n possession). I reduced by the induction assumption and left with equality:

f '(0) = x (x-1) ... (xk) (xk-1) = (-1) ^ (k + 1) * (k + 1)!
f '(0) = x (x-1) ... (xk) (xk-1) = (-1) ^ (k) * k! * (-1) * (k + 1)
f '(0) = (xk-1) = (-1) * (k + 1)
-k-1 = -k-1

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