MHB Taylor Series Expansion and Radius of Convergence for $f(x)=x^4-3x^2+1$

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The Taylor series for the function f(x) = x^4 - 3x^2 + 1 centered at a = 1 is given as -1 - 2(x-1) + 3(x-1)^2 + 4(x-1)^3 + (x-1)^4. Since this series is a polynomial, it converges for all values of x. Therefore, the radius of convergence is infinite, expressed as (-∞, ∞). The discussion emphasizes that polynomial functions have no restrictions on convergence. Understanding the nature of the function confirms the findings on convergence and radius.
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find the taylor series for $f(x)=x^4-3x^2+1$ centered at $a=1$. assume that f has a power series expansion. also find the associated radius of convergence.

i found the taylor series. its $-1-2(x-1)+3(x-1)^2+4(x-1)3+(x-1)^4$ but how do i find the radius of convergence?
 
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Since your Taylor series is a polynomial in $x$ (or $x-1$), it converges for all $x$. So the radius of convergence is $(-\infty, \infty)$.
 
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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