I'm not looking for a proof of the taylor series, I want an intuition to why the quadratic term equals the orange part or if I was just wrong
My teacher explained the Taylor series saying if we create a function with all degree derivative matching the original function then our function approximate the original function. At the time it satisfied my confusion but now when I look back it feels kind of vague so I tried drawing a graph to see the details but I'm stuck trying to associate the quadratic approximation with the graph
I'm trying to build an intuition behind quadratic approximation using this graph
- red curve is the original function $f(x)$
- blue line is the slope at $x_0$
- green line is the slope at $x_1$ a point slightly to the right of $x_0$
when we do linear approximation we are estimating the original function purely on the blue line
- so when we try to approximate $f(x)$ at $x = x_1$ we end up with $y = y_0$
by adding the quadratic term we are incorporating the change in slope, which tells us that the blue line becomes the green line at $x = x_1$ therefore we need to adjust the $y$ value by the amount of change in the slope
this means adding the orange part $y_1 - y_0$, which should associate to the term $\frac{f^{\prime\prime}(x_0)}{2}(x - x_0)^2$
My problem is, I can't really see how $y_1 - y_0 = \frac{f^{\prime\prime}(x_0)}{2}(x - x_0)^2$
if we incorporate the change in slope then
- $f^{\prime}(x) = f^{\prime}(x_0) + f^{\prime\prime}(x_0)(x - x_0)$
$\begin{array}{lcl} f(x) & = & f(x_0) + (f^{\prime}(x_0) + f^{\prime\prime}(x_0)(x - x_0))(x - x_0)\\ & = & f(x_0) + f^{\prime}(x_0)(x - x_0) + f^{\prime\prime}(x_0)(x - x_0)^2 \end{array}$
which doesn't have the $\frac{1}{2}$