When working with a *non-polynomial function*, we often seek a *polynomial approximation* that closely matches the original function at a specific point $x$ and in its vicinity. Polynomials are computationally easier to work with (e.g., for differentiation and integration), making them a practical choice. Using the [[Taylor Series|Taylor Series Method]], we approximate a function by a finite polynomial (truncated series), which is referred to as an $n$-th order approximation. The number of terms $n$ we include depends on the desired level of accuracy. $ g_n(x) = g(c) + g^\prime(c)(x-c) + \frac{g^{\prime\prime}(c)}{2!}(x-c)^2 + \dots+\frac{g^{(n)}(c)}{n!}(x-c)^n$ ## Derivation Let us illustrate this with an example. Suppose our original function is $f(x)$, and we want to approximate it with a cubic polynomial as good as possible around the point $x=0$. A cubic polynomial $y$ has the following general form: $ y = d + cx + bx^2 + ax^3 $ For a good approximation we want to fit the parameters $\{a,b,c,d\}$ as good as possible. We achieve this by *aligning the function value, slope, curvature, etc.* of the polynomial $y$ with the original function $f$, evaluated at the point of interest $x=0$. **Step 1:** Take derivatives of $y$ $ \begin{align} y &= ax^3 + bx^2+cx+d \\ y'&= 3ax^2 + 2bx +c \\ y''&= 6ax+2b \\ y^{(3)}&= 6a \end{align} $ **Step 2:** Evaluate derivatives at $x=0$ and equate with derivatives from original function $ \begin{align} y &= d \approx f(0)\\ y'&= c \approx f^\prime(0)\\ y''&= 2b \approx f^{\prime\prime}(0)\\ y^{(3)}&= 6a \approx f^{(3)}(0) \end{align} $ **Step 3:** Express approximations in terms of $\{a,b,c,d\}$ $ \begin{align} d= f(x), \quad c= f'(x), \quad b=\frac{1}{2}f''(x), \quad a=\frac{1}{6}f^{(3)}(x) \end{align} $ **Step 4:** Filling in representation of $\{a,b,c,d\}$ into the polynomial form from above: $ y=f(x) + f'(x)\,x + \frac{1}{2}f^{\prime\prime}(x)\,x^2 + \frac{1}{6}f^{(3)}(x)\,x^3 $ ## Example Consider $f(x)= \cos(x)$ as the original function which we want to approximate by a polynomial. The derivatives of $\cos(x)$ are: $ \begin{aligned} f(x)&=\cos(0)&=1 \\ f'(x)&=-\sin(0)&=0 \\ f''(x)&=-\cos(0)&=-1\\ f^{(3)}(x)&=\sin(0)&=0\\ \end{aligned} $ Substituting these values into the polynomial: $ \begin{align} y &=f(x) + f^\prime(x)\,x + \frac{1}{2}f^{\prime\prime}(x)\,x^2 + \frac{1}{6}f^{(3)}(x)\,x^3 \\ y&=1+0x+\frac{1}{2}*{-1}x^2 -{\frac{1}{6}* 0x^3} \\ y&=1-\frac{1}{2}x^2 \end{align} $ Thus, the cubic approximation of the cosine function at $x=0$ is $y=1-\frac{1}{2}x^2$.