Taylor Polynomial Approximations.

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Taylor polynomial approximations involve using polynomials of increasing degree to approximate functions at a specific point. The process begins with a linear approximation and incorporates derivatives to refine the estimate, resulting in a series that sums all previous approximations. This summation is crucial because higher-degree terms provide more accurate representations of the function, capturing its behavior more effectively. Simply using the highest degree term would not yield a reliable approximation of the original function. Understanding this cumulative approach is essential for grasping the utility of Taylor polynomials in mathematical analysis.
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Hello,

I'm new here, nice to meet you guys i was in class today and just didn't understand the taylor polynomial approximation, the professor started out approximating a function by polynomials of degree N, he first showed us how a linear polynomial was a crude approximation of the function at a given point but thing i don't get is when he moved on to the next approximation by using the first derivative, than the second...i just didnt get why the final formula added up all the approximations together?...

I just didnt get why like we had to sum up all the approximations rather than just making a formula for the last approximation of the function?...if I am not making any sense I am sorry but if you guys can explain the taylor polynomial a bit better i would greatly appreciate it.



Thanks,

Moe.
 
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I wrote a little bit about this some time ago... does this help you get started? https://www.physicsforums.com/blog.php?b=1758
 
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Not sure what you mean but if you take the Taylor polynomials for f(x)=e^x about x=0 (also called the Maclaurin polynomials for e^x) they are:

P_0(x) = 1
P_1(x) = 1 + x
P_2(x) = 1 + x + (1/2)x^2
P_3(x) = 1 + x + (1/2)x^2 + (1/6)x^3
etc.

So the first k+1 terms of a Taylor polynomial of degree n is the Taylor polynomial for the same function about the same point of degree k.

If you mean that we should just take the highest degree terms of the Taylor polynomials, in this example 1, x, (1/2)x^2, (1/6)x^3 etc. these terms alone would NOT be good approximations of the original function.
 
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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