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Hi to everyone Here i am studying Taylor series.

$$f(x)=c_0 + c_1 (x-a) + c_2 (x-a)^2+ ...$$

$$ f(x)= f(a) + \frac{df(a)/dx}{1!}(x-a)^1 + \frac{d^2f(a)/dx^2}{2!}(x-a)^2 ...$$

Well my problem is that this is derived from a polynomial function but can be applied to almost any function. How come?

user96369
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1 Answers1

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The fact that it's based on an infinite polynomial means that it can adjust an infinite number of times in approaching the original function. Compare the graphs (I personally like https://www.desmos.com/calculator) of

$sinx ≈ x$

$sinx ≈ x - \frac{x^3}{3!}$

$sinx ≈ x - \frac{x^3}{3!} + \frac{x^5}{5!}$

$sinx ≈ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}$

EDIT: How do Taylor polynomials work to approximate functions? does a good job of explaining why Taylor series uses a certain series to approach the function rather than a different series.

If $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5x^5 ...$ all the way to infinity (so that the polynomial can bend an infinite number of times and approach every single point on the original function), then $f(0) = a_0$

$f'(x) = a_1 + 2a_2x + 3a_3x^2+4a_4x^3+5a_5x^4 + ...$ in which case $f'(0) = a_1$

Further, $f''(x) = 2a_2 + 6a_3x+12a_4x^2 + 20a_5x^3 + ...$ in which case $f''(0) = 2a_2$

$f'''(x) = 6a_3 + 24a_4x+ 60 a_5x^2+...$ in which case $f'''(0) = 6a_3$

$f''''(x) = 24a_4 + 120a_5x + ...$ in which case $f''''(0) = 24a_4$

Now we have a pattern for all of the coefficients $a_n = \frac{f^n(0)}{n!}$ that we can plug back into the original $f(x)$

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Simpson17866
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  • I just tried linear aproximation first two expression are same however we can continue with taylor expansion. Again I do not see any aproximation while deriving the expansion. Can u give me not the imagination but the theory of it. – user96369 May 13 '16 at 19:17