Proof of Taylor's formula for polynomials

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SUMMARY

The discussion centers on the proof of Taylor's formula for polynomials, specifically for a polynomial function of order n, defined as f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n. The formula states that the coefficients c_v in the expansion f(a+h) = c_0 + c_1h + c_2h^2 + ... + c_nh^n can be expressed as c_v = \frac{1}{v!}f^v(a), linking the coefficients to the derivatives of f at the point a. The proof involves substituting h = b - a and applying the chain rule to demonstrate that differentiation with respect to h is equivalent to differentiation with respect to b = a + h.

PREREQUISITES
  • Understanding of polynomial functions and their representations
  • Familiarity with Taylor series and polynomial expansion
  • Knowledge of differentiation and the chain rule in calculus
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation of Taylor's theorem in calculus
  • Explore higher-order derivatives and their applications in polynomial approximation
  • Learn about the convergence of Taylor series for various functions
  • Investigate the implications of Taylor's formula in numerical methods
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Students of calculus, mathematicians, and anyone interested in the theoretical foundations of polynomial approximations and their applications in analysis.

JG89
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My book reads as follows:

"This is an entirely elementary algebraic formula concerning a polynomial in x or order n, say
f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n.

If we replace x by a + h = b and expand each term in powers of h, there results immediately a representation of the form f(a+h) = c_0 = c_1h + c_2h^2 + ... + c_nh^n.

Taylor's formula is the relation: c_v = \frac{1}{v!}f^v(a), for the coefficients c_v in terms of f and its derivatives at x = a. To prove this fact we consider the quantity h = b - a as the independent variable, and apply the chain rule which shows that differentiation with respect to h is the same as differentiation with respect to b = a + h."

I don't get how f'(h) = f'(a + h)?
 
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I think it's saying that if you have f'(h) and want f'(h+a), just substitute in h+a where you see h to get f'(h+a). This works because \frac{d}{dh} h+a = 1
 

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