The derivation presented in the link below states that the $P_{n+1}(x)$ Taylor polynomial must be more accurate than $P_n(x)$. Since the difference between these polynomials is just the $n+1$ term, then the remaining $R_n(x)$ must not exceed that term.


I'm having trouble to understand this. I plotted a few elementary Taylor polynomials in order to visualize that. These can be seen in the link below.

Taylor polynomials for $e^x$, centered in zero, with $n=1,2,3$

Now, let's make $T_1(x) = P_n(x)$ and $T_2(x) = P_{n+1}(x)$.

Referring to the plot, the error for $P_n(1.5) = T_1(1.5)$ is roughly $f(1.5) - T_1(1.5) \approx 4.5 - 2.5 \approx 2$.

The $n+1$ term for $x=1.5$ can be computed as $T_{n+1}(1.5) -T_n(1.5) \approx 3.625 - 2.5 \approx 1.125$.

Or, directly, $(1/2)*x^2 = (1/2)*1.5^2 = 1.125$

So, the error for the $P_n(1.5)$ is larger than the last term $n+1$, which contradicts the statement presented in the derivation of the error bound.

What am I getting wrong? Also, why exactly the error for $P_n(x)$ is bounded by the last term of $P_{n+1}(x)$? That's not clear to me at all. Is there any way to visualize that clearer?


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