微積分
在微積分中, 拉格朗日中值定理是羅爾中值定理的推廣,同時也是柯西中值定理的特殊情形。
1.文字敘述
![拉格朗日定理](/img/b/cd4/wZwpmLxcTNyUjNwMDO4EDN0UTMyITNykTO0EDMwAjMwUzLzgzLzIzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
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如果函式 滿足:1) 在閉區間 上連續;2) 在開區間 內可導;那么在 內至少有一點 ,使等式
![拉格朗日定理](/img/9/3da/wZwpmLzgTM5QDOyYjMzEzM1UTM1QDN5MjM5ADMwAjMwUzL2IzLxYzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
成立。
2.邏輯語言的敘述
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若函式 滿足:
![拉格朗日定理](/img/d/3fb/wZwpmLzETM0kDM3EzMzEzM1UTM1QDN5MjM5ADMwAjMwUzLxMzL3EzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
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則
![拉格朗日定理](/img/0/b76/wZwpmL0ETMyIDNzQTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL0EzL4MzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![圖1.拉格朗日中值定理的幾何意義](/img/0/0b8/wZwpmL3EzM4ITMzkDNzEzM1UTM1QDN5MjM5ADMwAjMwUzL5QzL1YzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
3.證明
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令 ,那么
![拉格朗日定理](/img/9/ce6/wZwpmLyATN5UjM4QTOwADN0UTMyITNykTO0EDMwAjMwUzL0kzL1AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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1) 在 上連續,
![拉格朗日定理](/img/9/ce6/wZwpmLyATN5UjM4QTOwADN0UTMyITNykTO0EDMwAjMwUzL0kzL1AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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2) 在 上可微(導),
![拉格朗日定理](/img/2/fef/wZwpmLzYzN3ATO1EzMzEzM1UTM1QDN5MjM5ADMwAjMwUzLxMzL1QzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
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![拉格朗日定理](/img/d/085/wZwpmL1ITN1YjN1UTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL1EzL4IzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
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3 ,由羅爾定理,存在一點 ,使得 。即 。
數論
1.內容
四平方和定理(Lagrange's four-square theorem) 說明每個正整數均可表示為4個整數的平方和。它是費馬多邊形數定理和華林問題的特例。注意有些整數不可表示為3個整數的平方和,例如7。
2.歷史
![拉格朗日定理](/img/9/bfa/wZwpmL3QTN1YDM4QTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL0EzLzUzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
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1. 1743年,瑞士數學家歐拉發現了一個著名的恆等式:。根據上述歐拉恆等式或四元數的概念可知如果正整數 和 能表示為4個整數的平方和,則其乘積 也能表示為4個整數的平方和。於是為證明原命題只需證明每個素數可以表示成4個整數的平方和即可。
![拉格朗日定理](/img/5/fa3/wZwpmL2ADOzEDMwUTMwEDN0UTMyITNykTO0EDMwAjMwUzL1EzL1gzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
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2. 1751年,歐拉又得到了另一個一般的結果。即對任意奇素數 ,同餘方程 必有一組整數解 滿足 , (引理一)。
至此,證明四平方和定理所需的全部引理已經全部證明完畢。此後,拉格朗日和歐拉分別在1770年和1773年作出最後的證明。
群論
拉格朗日定理是群論的定理,利用陪集證明了子群的階一定是有限群的階的約數值。
1.定理內容
![拉格朗日定理](/img/5/ea1/wZwpmLxAzN2MjM5cTO4kzM0UTMyITNykTO0EDMwAjMwUzL3kzL2AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![拉格朗日定理](/img/0/baf/wZwpmLwIDO4ETM5kDM5czN0UTMyITNykTO0EDMwAjMwUzL5AzLwEzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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敘述:設H是有限群 的子群,則 的階整除 的階。
![拉格朗日定理](/img/0/baf/wZwpmLwIDO4ETM5kDM5czN0UTMyITNykTO0EDMwAjMwUzL5AzLwEzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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定理的證明是運用 在 中的左陪集。 在 中的每個左陪集都是一個等價類。將 作左陪集分解,由於每個等價類的元素個數都相等,都等於 的元素個數( 是 關於 的左陪集),因此 的階(元素個數)整除 的階,商是 在 中的左陪集個數,叫做 對 的指數,記作 。
陪集的等價關係
![拉格朗日定理](/img/8/37f/wZwpmLwMDOyUzN0cTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL3EzLyUzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
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定義二元關係 : 。下面證明它是一個等價關係。
![拉格朗日定理](/img/5/84d/wZwpmLwcDO2MzN5QDNzEzM1UTM1QDN5MjM5ADMwAjMwUzL0QzL1AzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
1) 自反性: ;
![拉格朗日定理](/img/f/a2d/wZwpmL1UjMyAjN3YzMzEzM1UTM1QDN5MjM5ADMwAjMwUzL2MzL0UzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![拉格朗日定理](/img/0/d66/wZwpmLzUTOzMjMyYjMzEzM1UTM1QDN5MjM5ADMwAjMwUzL2IzL0czLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
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2) 對稱性: ,因此 ,因此 ;
![拉格朗日定理](/img/7/1e5/wZwpmL4MzN5UDMwkDNzEzM1UTM1QDN5MjM5ADMwAjMwUzL5QzL3AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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3) 傳遞性: ,因此 ,因此 。
![拉格朗日定理](/img/1/c7f/wZwpmLwADO5UzMwAzMzEzM1UTM1QDN5MjM5ADMwAjMwUzLwMzL0MzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
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可以證明, 。因此左陪集是由等價關係 確定的等價類。
![拉格朗日定理](/img/e/89b/wZwpmL1gzN1YzN1kTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL5EzLwgzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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拉格朗日定理說明,如果商群 存在,那么它的階等於 對 的指數 。
![拉格朗日定理](/img/c/ed5/wZwpmL0UDM5AzN2cTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL3EzL3YzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
2.推論
![拉格朗日定理](/img/5/ea1/wZwpmLxAzN2MjM5cTO4kzM0UTMyITNykTO0EDMwAjMwUzL3kzL2AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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![拉格朗日定理](/img/5/ea1/wZwpmLxAzN2MjM5cTO4kzM0UTMyITNykTO0EDMwAjMwUzL3kzL2AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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由拉格朗日定理可立即得到:由有限群 中一個元素 的階數整除群 的階(考慮由 生成的循環群)。
3.逆命題
![拉格朗日定理](/img/5/ea1/wZwpmLxAzN2MjM5cTO4kzM0UTMyITNykTO0EDMwAjMwUzL3kzL2AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
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![拉格朗日定理](/img/8/d7e/wZwpmL1QTM5gjM4QTOwADN0UTMyITNykTO0EDMwAjMwUzL0kzLwczLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![拉格朗日定理](/img/1/884/wZwpmLxYTO3UjN0cTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL3EzL1QzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![拉格朗日定理](/img/1/884/wZwpmLxYTO3UjN0cTMzEzM1UTM1QDN5MjM5ADMwAjMwUzL3EzL1QzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
拉格朗日定理的逆命題並不成立。給定一個有限群 和一個整除 的階的整數 , 並不一定有階數為 的子群。最簡單的例子是4次交替群 ,它的階是12,但對於12的因數6, 沒有6階的子群。對於這樣的子群的存在性,柯西定理和西洛定理給出了一個部分的回答。