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Say we want to find the Taylor polynomial of order $2$ for $\sqrt{2+x}$ around $x = 10$.

Simply plugging in the numbers for a Taylor sum $\sum \dfrac{f^k(10)(x - 10)^k}{k!} $ with $n = 2$ then gives us a polynomial approximation of $\sqrt{2+x}$. Is it possible to give a different polynomial expansion than the one from the formula of $\sqrt{2+x}$ of order $2$ around $x = 10$? I personally don't understand how a different polynomial expression with the same order around the same $x$ is to be found. However, the problem I've found asks this precise question.

(Sorry for poor formatting)

an4s
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Blink
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  • The point, I think, is that the Taylor series of a function at a point is unique. – Robert Israel Mar 13 '18 at 19:07
  • Look here for the proof of uniqueness https://math.stackexchange.com/questions/1923624/taylors-polynomial-uniqueness-proof-why-are-these-limits-inferable – user Mar 13 '18 at 19:12
  • @KafkaBoi Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 17 '18 at 22:57

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The Taylor polynomial of order $2$ of the function $f(x):=\sqrt{2+x}$ is the unique polynomial $j_2$ of degree $\leq 2$ satisfying $$f(10+t)-j_2(t)= o(t^2)\qquad (t\to0)\ .$$ It gives a good approximation to the function value $f(x)$ when $x=10+t$ is near $10$, and the quality of this approximation increases tremendously as $t\to0$.

But, depending on your purpose, there are other polynomials of degree $\leq2$ approximating $f$ very well in the neighborhood of $x=10$, e.g. the polynomial $$h(x):=1.71446 + 0.205225 x - 0.00302609 x^2\ .$$ The following figure shows that that the maximal errror of $h$ in the $x$-interval $[8,12]$ is much smaller than the error of $j_2$.

enter image description here

Christian Blatter
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