Higher-Order Derivative and Preliminary of Taylor Series

Approximation by Using Polynomials

Using functions such as trig functions and logarithm functions to depict problems is sometimes not intuitive. When handling with some limits and derivatives, it is also troublesome. A good approximation can make the problem easier since we use some else functions to replace the original disharmony item.

When a function is infinite differentiable at somewhere, we can use a series of exponents of ${\displaystyle x}$ to make the approximation.

For example, use ${\displaystyle c_{1}x}$ to approximate the property of increasing or decreasing, use ${\displaystyle c_2{x}^2}$ to approximate the property of concave up or down, and etc. Finally add them together and we will get a great approximation at a given position. So we have the form:

${\displaystyle P(x)=c_0+c_{1}x+c_2{x}^2+\cdots +c_n{x}^n}$

So that we can use the parameter to control the final function to approach the origin function.

To find the value of parameters, we let ${\displaystyle c_0}$ equals to the original function's value ${\displaystyle f(x)}$, then let ${\displaystyle \frac{dP}{dx}=\frac{df}{dx}}$, ${\displaystyle \frac{d^{2}P}{d{P}^2}=\frac{d^{2}f}{d{x}^2}}$, and ${\displaystyle \frac{d^{2}P}{d{P}^n}=\frac{d^{n}f}{d{x}^n}}$. To satisfy these, we have ${\displaystyle c_n=\frac{d^{n}f}{d{x}^n}\cdot \frac{1}{n!}}$. The larger the ${\displaystyle n}$ is, the better the approximation is. When the given ${\displaystyle x}$ is not ${\displaystyle 0}$, we let ${\displaystyle x=x-a}$, so we are finding the approximation at ${\displaystyle x=a}$.

At last, we have the common form of the Taylor Series:

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{d^{n}f}{d{x}^n}(a)\cdot \frac{1}{n!}\cdot (x-a{)}^n}$

That is the approximation to ${\displaystyle f(x)}$ at ${\displaystyle x=a}$.

Taylor Series And Equivalent Infinitesimals

In the section of infinitesimals, some common equivalent infinitesimals are given. From the Taylor series, we can know how they have been given.

Take ${\displaystyle x-\sin x\sim -\frac{1}{6}{x}^3}$ for example, it is actually the transform of the quadric Taylor expansion for ${\displaystyle \sin x}$ when ${\displaystyle x=0}$.

We can use Taylor expansion to find some other equivalent infinitesimals to solve many troublesome limits.