2023-05-21

Use Taylor Formula to Solve Irrational Comparing Problem in NCEE

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1, what is Taylor expansion

In mathematics, Taylor series represents a function with an infinite term addition-series, and these added terms are obtained by the derivative of the function at a certain point. Taylor series are named after Brooke Taylor, an English mathematician who published Taylor's formula in 1715. The Taylor series obtained by the derivative of the function at the zero point of the independent variable is also called the McLaughlin series, named after the Scottish mathematician Colin McLaughlin.
Wikipedia

Taylor's Mean Value Theorem 1:

If the function $f(x)$ has a $n$ order derivative at $x_0$, then there is a neighborhood of $x_0$, for the neighborhood Any $x$, have

$$f(x)=f(x_0)+f^{'}(x_0)(x-x_0)+\frac{f^{''(x_0)}}{2!}(x-x_0)^2 +\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x)\tag{1-1}$$

$$R_n(x)=o((x-x_0)^n)\tag{1-2}$$

Formula (1-1) is called the function $f(x)$ at $x_0$ (or expanded by the power of $(x-x_0)$) with Pei The $n$ order Taylor formula of the Arno residual term, the expression (1-2) of $R_n(x)$ is called the Peano residual term.

Taylor's Mean Value Theorem 2:
If the function $f(x)$ has (n+1) derivatives in a certain neighborhood $U(x_0)$ of $x_0$, then for any \ (x\in U(x_0)\), there is

$$f(x)=f(x_0)+f^{'}(x_0)(x-x_0)+\frac{f^{''(x_0)}}{2!}(x-x_0)^2 +\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x)\tag{2-1}$$

$$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}\tag{2-2} \\(\xi between x, x_0)$$

Formula (2-1) is called the function $f(x)$ at $x_0$ (or expanded by the power of $(x-x_0)$) with pull The $n$ order Taylor formula of the Grange residual term, the expression (2-2) of $R_n(x)$ is called the Lagrange residual term.

2, commonly used Taylor formula

$$e^x=1+x+\frac{x^2}{2!}+o(x^3)$$

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+o(x^3)$$

$$\frac{1}{1-x}=1+x+x^2+x^3+o(x^3)$$

$$\frac{1}{1+x}=1-x+x^2-x^3+o(x^3)$$

$$\sin x=x-\frac{x^3}{3! }+\frac{x^5}{5!}+o(x^5)$$

$$\arcsin x=x+\frac{1}{2}\times\frac{x^3}{3}+\frac{1\times3}{2\times4}\times\frac{x^5}{ 5}+\frac{1\times3\times5}{2\times4\times6}\times\frac{x^7}{7}+o(x^7)$$

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4)$$

$$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+o(x^5)$$

$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}+o(x^5) \\$$

3. Precautions

If and only when $x$ is small, the first few items of Taylor's formula can be used to estimate the original formula.

For example:
$a=e^{0.3}$, $b=\frac{\ln{1.5}}{2}+1$, $c=\sqrt{1.5}$ .

Using the first three terms of Taylor's formula, $a\approx1.35$, $b\approx1.16$, $c\approx1.25$, easy to get $a >c>b$.

In my experience, Taylor's formula works best when the value difference is around 0.1.

What if the value difference is large?

For example: $a=\sin4$, $b=\ln{4}$, $c=4^{-\frac{1}{4}}$.

Obviously,
$a<0$,$b>1$,$0<c<1$,$b>c>a$

At this time, if you use Taylor's formula, you will find that the result is not very consistent with the true value. This is because $x$ has reached 3 and 4, which is no longer applicable.

Use Taylor Formula to Solve Irrational Comparing Problem in NCEE

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