What happens to the Taylor approximations of $\cos (x)$ when the center changes from 0 to pi?

Conditions

• Function is $\cos (x)$
• Center is a=pi

Reasoning, step by step

1. Identify the original quadratic approximation centered at 0: $1-x^2/2$.
2. Observe the animation shifting the expansion point from $x=0$ to x=pi.
3. Identify the new quadratic approximation centered at pi: -1+(x-pi)^$2/2$.
4. Note that the region of close agreement between the polynomial and the $\cos (x)$ curve moves from near 0 to near pi.
5. Conclude that changing the center point changes the Taylor polynomial itself and relocates the region where the approximation is effective.

Example

The graph updates after the speaker says to move the center to a=pi. The new quadratic label appears as -1+(x-pi)^$2/2$. The speaker says the polynomial has an (x-pi)^2 in it and is centered at pi, and that near the value of a=pi it is pretty good.

Common misconceptions

• Believing that the Taylor polynomial remains the same regardless of the chosen center.
• Assuming that the approximation is still accurate near $x=0$ after the center has been moved to x=pi.

Watch the explanation

Explore next

• Why are Taylor polynomial approximations considered local rather than global?
• What is the coefficient formula for a Maclaurin series centered at 0?