MHB Determine an expression using binomial theorem

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To determine the expression for f(x) = (1+x)(1+2x)(1+3x)…(1+nx) and find f'(0), focus on identifying the coefficient of x in the polynomial. The derivative at zero, f'(0), corresponds to the coefficient of x, which is a key insight for simplification. By applying the binomial theorem, the expansion can be analyzed to extract this coefficient effectively. Understanding the structure of the polynomial is essential for calculating f'(0). This approach streamlines the process of finding the desired derivative.
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Determine an expression for f(x) =(1+x)(1+2x)(1+3x)…(1 +nx),find f⸍(0) .
 
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Ernie said:
Determine an expression for f(x) =(1+x)(1+2x)(1+3x)…(1 +nx),find f⸍(0) .
Hi Ernie, and welcome to MHB!

Here's a hint that may help you. If you have a polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, then $f'(0) = a_1$. In other words, to find $f'(0)$ you only need to find the coefficient of $x$ in $f(x)$.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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