MHB Polynomial approximation of e to the x

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The discussion focuses on the polynomial approximation of the function e^x near x = 2 using Taylor's theorem. The approximation is expressed as a series involving derivatives evaluated at x = 2. There is a need to control the term |(x - 2)^(N + 1)| to ensure accuracy, leading to the condition |(x - 2)| ≤ 1, which restricts x to the interval [1, 3]. The original poster finds the next steps unclear and requests clarification on how to proceed with this approximation. Understanding the bounds on x is crucial for ensuring the validity of the polynomial approximation.
tmt1
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I am examining the polynomial approximation for $e^x$ near $x = 2$.

From Taylor's theorem:

$$e^x = \sum_{n = 0}^{\infty} \frac{e^2}{n!} (x - 2)^n + \frac{e^z}{(N + 1)! } (x - 2)^{N - 1}$$

Now, I don't get the next part:

We need to keep $\left| (x - 2)^{N + 1} \right|$ in check so we can specify $\left| (x - 2) \right| \le 1$ so $x \in [1,3]$.
 
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The question is unclear, could you please rephrase it ?
 

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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