From a4f1c539c345210d663573c5c502e85d760fcaad Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E1=B5=87=E1=B5=92?= <40080007@qq.com>
Date: Wed, 24 Apr 2024 10:07:43 +0800
Subject: [PATCH] feat: test

---
 _posts/2024-04-23-Fast-inverse-square-root.md | 66 +++++--------------
 1 file changed, 18 insertions(+), 48 deletions(-)

diff --git a/_posts/2024-04-23-Fast-inverse-square-root.md b/_posts/2024-04-23-Fast-inverse-square-root.md
index 42b4fc8..69db25e 100644
--- a/_posts/2024-04-23-Fast-inverse-square-root.md
+++ b/_posts/2024-04-23-Fast-inverse-square-root.md
@@ -31,6 +31,18 @@ float Q_rsqrt( float number )
 }
 ```
 
+## IEEE 754
+
+科学计数法
+
+normalised numbers
+denormalised numbers
+NaN
+0 & -0
+
+由 IEEE 754 可知 float 的 long representation 为 $$(E*2^{23} + M)$$
+
+
 ## 牛顿迭代
 
 牛顿迭代公式：
@@ -58,17 +70,14 @@ $$
 \end{aligned}
 $$
 
+泰勒级数展开式：
 
-## IEEE 754
-
-科学计数法
-
-normalised numbers
-denormalised numbers
-NaN
-0 & -0
+$$ f(x) = f(x_0) + f^{\prime}(x_0)(x-x_0) + \tfrac{f^{\prime\prime}(x_0)(x-x_0)^2}{2!} + ··· +  \tfrac{f^{(n)}(x_0)(x-x_0)^n}{n!} + R_n(x) $$ 
 
-由 IEEE 754 可知 float 的 long representation 为 $$(E*2^{23} + M)$$
+牛顿迭代提高精度的几种方式：
+1. 选取一个合适的初始值 $$x_0$$
+2. 增加迭代次数
+3. 多阶迭代
 
 ## 推导
 
@@ -109,45 +118,6 @@ $$
 $$
 
 
-$\log_2 10$
-
-$$\lg_2 10$$
-
-$$\lg 10$$
-
-$$lg 10$$
-
-$$\ln 10$$
-
-$$ln 10$$
-
-$$\log_2 10$$
-
-$$log_2 10$$
-
-$\sqrt{3x-1}+(1+x)^2$
-
-$$\sqrt3$$
-
-$$\sqrt{3}$$
-
-$$\sqrt{3x-1}$$
-
-$$\sqrt{3x-1}+(1+x)^2$$
-
-$10^{10}$
-
-$\frac{4n^3-6n^2-10n-2k^3-3k^2+8k+12kn+c}{12}$
-
-$$\frac{4n^3-6n^2-10n-2k^3-3k^2+8k+12kn+c}{12}$$
-
-$$\frac{4}{12}$$
-
-$$\tfrac{4}{12}$$
-
-$$\dfrac{4}{12}$$
-
-$$\cfrac{4}{12}$$
 
 ## 总结