三度和五度多项式内插法（带代码） - 二度和五度多项式内插法
设五次多项式的表达式：
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(4)
f(u)=a_0+a_1(u-u_s)+a_2(u-u_s)^2+a_3(u-u_s)^3+a_4(u-u_s)^4+a_5(u-u_s)^5 \tag 4
f(u)=a0+a1(u−us)+a2(u−us)2+a3(u−us)3+a4(u−us)4+a5(u−us)5(4)
插值的端点条件为：
{
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(5)
\begin{cases} f(u_s)=p_s\\ f'(u_s)=v_s\\ f''(u_s)=a_s\\ f(u_e)=p_e\\ f'(u_e)=v_e\\ f''(u_e)=a_e\\ \tag 5 \end{cases}
⎩
⎨
⎧f(us)=psf′(us)=vsf′′(us)=asf(ue)=pef′(ue)=vef′′(ue)=ae(5)
利用matlab符号计算功能，解得：
{
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[
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20
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8
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12
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30
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14
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[
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12
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6
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6
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5
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(6)
\begin{cases} a_0=p_s\\ a_1=v_s\\ a_2=a_s/2\\ a_3=[(20(p_e - p_s) + (8v_e + 12v_s)(u_s - u_e) + (a_e - 3a_s)(u_e^2 + u_s^2) + (6a_s - 2a_e)u_eu_s)]/[2(u_e-u_s)^3]\\ a_4=[ -(30(p_e - p_s) + (14v_e + 16v_s)(u_s - u_e) + (2a_e - 3a_s)(u_e^2 + u_s^2) + (6a_s - 4a_e)u_eu_s)]/[2(u_e-u_s)^4]\\ a_5=[(12(p_e - p_s) + (6v_e + 6v_s)(u_s - u_e) + (a_e - a_s)(u_e^2 + u_s^2) + (2a_s - 2a_e)u_eu_s)]/[2(u_e-u_s)^5]\\ \tag 6 \end{cases}
⎩
⎨
⎧a0=psa1=vsa2=as/2a3=[(20(pe−ps)+(8ve+12vs)(us−ue)+(ae−3as)(ue2+us2)+(6as−2ae)ueus)]/[2(ue−us)3]a4=[−(30(pe−ps)+(14ve+16vs)(us−ue)+(2ae−3as)(ue2+us2)+(6as−4ae)ueus)]/[2(ue−us)4]a5=[(12(pe−ps)+(6ve+6vs)(us−ue)+(ae−as)(ue2+us2)+(2as−2ae)ueus)]/[2(ue−us)5](6)
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