<a href="https://colab.research.google.com/github/G750cloud/20MA573/blob/master/Hw2(2).ipynb" target="_parent"></a>


```
Problem 1
```

$f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x)h^2+\frac{1}{6}f'''(x)h^3+\frac{1}{24}f''''(x)h^4+O(h^5)$                       （1）
  $f(x-h)=f(x)-f'(x)h+\frac{1}{2}f''(x)h^2-\frac{1}{6}f'''(x)h^3+\frac{1}{24}f''''(x)h^4-O(h^5)$                      （2）
  $f(x+2h)=f(x)+f'(x)2h+f''(x)2h^2+\frac{4}{3}f'''(x)h^3+\frac{2}{3}f''''(x)h^4+O(h^5)$                                （3）
  $f(x-2h)=f(x)-f'(x)2h+f''(x)2h^2-\frac{4}{3}f'''(x)h^3+\frac{2}{3}f''''(x)h^4-O(h^5)$                              （4）

  SO let $\frac{2}{3}((1)-(2))-\frac{1}{12}((3)-(4))$, We can get:

$O(h^4)=\left|f'(x)-\frac{\frac{2}{3}(f(x+h)-f(x-h))-\frac{1}{12}(f(x+2h)-f(x-2h))}{h}\right|$

So we get a difference approximation.


```
Problem 2
```

(1)

Since f(x) is an even function.
   
   So $f(x)=f(-x)$\
      $f'(x)=-f'(-x)$\
      $f'(0)=-f'(0)$
   
   That is $f'(0)=0$

(2)

Let f(x) be expanded at 0 by Tayler:
   $f(0+h)=f(0)+f'(0)h+\frac{1}{2}f''(0)h^2+\frac{1}{6}f'''(0)h^3+O(h^4)$

   That is: $f(h)=\frac{1}{2}f''(0)h^2+O(h^4)$
   
   So $O(h^2)=\left|\frac{2f(h)}{h^2}-f''(h)\right|$

(3)

$f(h)=\frac{1}{2}f''(0)h^2+\frac{1}{24}h^4f(0)+O(h^6)$

$f(2h)=f''(0)2h^2+\frac{2h^4}{3}f''''(0)+O(h^6)$

Since $f''(0)-\frac{C1f(h)+C2f(h)}{h^2}=O(h^2)$

So we can get the system of equations:

\begin{equation}
\left\{
             \begin{array}{lr}
            1-\frac{C1}{2}-2C2=0 &  \\
             \frac{C1}{24}+\frac{2C2}{3}=0 &  
             \end{array}
\right.
\end{equation}

So $C1=\frac{8}{3},C2=-\frac{1}{6}$

(4)

Since f(x) is an odd function.
So the odd power of the function at zero is zero.
So we are able to use the same above way to find the $f''(x)$ efficiently. 






5293e7cc631dd10c21f62d53c2490dd59190a744
ipynb
Jupyter Notebook
Hw2(2).ipynb
G750cloud/20MA573
6450c6a69542b9e1de37db2215cedfba0ba68621
MIT
Hw2(2).ipynb
G750cloud/20MA573
6450c6a69542b9e1de37db2215cedfba0ba68621
MIT
Hw2(2).ipynb
G750cloud/20MA573
6450c6a69542b9e1de37db2215cedfba0ba68621
MIT
Qwen/Qwen-72B
1. YES
2. YES