拉格朗日插值公式的證明？

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The art of doing mathematics consists in finding that special case which contains all the germs of generality

這是一個什麼樣的公式啊

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數學有好玩之處，大家一起愛數學吧!

這應該是內插法的延伸

那該如何證明

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The art of doing mathematics consists in finding that special case which contains all the germs of generality

如何用數學式詳述

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The art of doing mathematics consists in finding that special case which contains all the germs of generality

http://www.math.nsysu.edu.tw/highschool/test/95test.pdf
我在此看見的
第一題

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The art of doing mathematics consists in finding that special case which contains all the germs of generality

請問孫老師可以回答嗎

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The art of doing mathematics consists in finding that special case which contains all the germs of generality

拉格朗日插值公式，我在猜是指Lagrange's interpolation formula這個公式。
這公式主要是找一個通過n個點的n-1次多項式：
給定(x_1,y_1)、(x_2,y_2)、....、(x_n,y_n)這n個相異點，則存在唯一一個冪次至多為n-1次的函數f(x)通過這n個點，且
f(x)=y_1*[(x-x_2)*(x-x_3)*(x-x_4)*...*(x-x_n)]/[(x_1-x_2)*(x_1-x_3)*(x_1-x_4)*...*(x_1-x_n)]+y_2*[(x-x_1)*(x-x_3)*(x-x_4)*...*(x-x_n)]/[(x_2-x_1)*(x_2-x_3)*(x_2-x_4)*...*(x_2-x_n)]+y_3*[(x-x_1)*(x-x_2)*(x-x_4)*...*(x-x_n)]/[(x_3-x_1)*(x_3-x_2)*(x_1-x_4)*...*(x_1-x_n)]+...+y_n*[(x-x_1)*(x-x_2)*(x-x_3)*...*(x-x_n-1)]/[(x_n-x_1)*(x_n-x_2)*(x_n-x_3)*...*(x_n-x_n-1)]。
可以看一下 http://planetmath.org/encyclopedia/LagrangeInterpolationFormula.html 這網頁，有比較好看的數學式。

你是要問這公式的證明？或是要問這公式如何應用在你所指出的題目？

請參考
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

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孫文先 敬上

感謝你和孫老師的回應

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The art of doing mathematics consists in finding that special case which contains all the germs of generality