1 . 牛顿迭代法是求方程近似解的一种方法.如图,方程
的根就是函数
的零点
,取初始值
的图象在点
处的切线与
轴的交点的横坐标为
的图象在点
处的切线与
轴的交点的横坐标为
,一直继续下去,得到
,它们越来越接近
.设函数
,
,用牛顿迭代法得到
,则实数
( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/49b7bff9b2431134f7683a9cc4e68acd.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8bb39857aa4d49038751a9e69d367173.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8559f5db9b978cb2bd290dbce7268629.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a02f9a42a1561e4b186eefa32be85dfb.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a24a2c53e3b0b1c08803e95419f909d3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/aec19b68e3add9d5bfcc6269a1855b87.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1fe1c31a81f198c443e71b83ca662939.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/448155ab5191a0c80a8de16d44b5aff5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6c5603d29560e66b2293cea1e3b02289.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6059ced816e65bed957376cd52d853de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5ccd4162c7d09f970cb77cadacdbe521.png)
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2 . 在概率统计中,常常用频率估计概率.已知袋中有若干个红球和白球,有放回地随机摸球
次,红球出现
次.假设每次摸出红球的概率为
,根据频率估计概率的思想,则每次摸出红球的概率
的估计值为
.
(1)若袋中这两种颜色球的个数之比为
,不知道哪种颜色的球多.有放回地随机摸取3个球,设摸出的球为红球的次数为
,则
.
(注:
表示当每次摸出红球的概率为
时,摸出红球次数为
的概率)
(ⅰ)完成下表,并写出计算过程;
(ⅱ)在统计理论中,把使得
的取值达到最大时的
,作为
的估计值,记为
,请写出
的值.
(2)把(1)中“使得
的取值达到最大时的
作为
的估计值
”的思想称为最大似然原理.基于最大似然原理的最大似然参数估计方法称为最大似然估计.具体步骤:先对参数
构建对数似然函数
,再对其关于参数
求导,得到似然方程
,最后求解参数
的估计值.已知
的参数
的对数似然函数为
,其中
.求参数
的估计值,并且说明频率估计概率的合理性.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/613f6de938db4bb3a7f98226d3a4c793.png)
(1)若袋中这两种颜色球的个数之比为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/fc5881f1ce9b4172ca346032d0fd1e3d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/54a829fdd8ec0f3b7ede883cf2c3e53b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/fadbd1d2d0294d04834dde31e0e4caaf.png)
(注:
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c74de541a96a252ca6b4bf05381a03ec.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f0a532e15e232cb4b99a8d4d07c89575.png)
(ⅰ)完成下表,并写出计算过程;
0 | 1 | 2 | 3 | |
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c74de541a96a252ca6b4bf05381a03ec.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0cf2e58249dd993ae42a7bd6d9ba0005.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0cf2e58249dd993ae42a7bd6d9ba0005.png)
(2)把(1)中“使得
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c74de541a96a252ca6b4bf05381a03ec.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0cf2e58249dd993ae42a7bd6d9ba0005.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c24095e409b025db711f14be783a406c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0807dbbfdeeaeffd987c4de037b892f5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c24095e409b025db711f14be783a406c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/acb13cf58c2aa7591391cfa8d515dc64.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c24095e409b025db711f14be783a406c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0b1aecbef5ad07da9949972dbcb9d659.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c21d19789d426d0ed871d45ac6175f66.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/889b80977780bb8caec3c90954b91a21.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b1010846eeec6c9da29640f5aa3f8738.png)
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7卷引用:浙江省杭州市2024届高三下学期4月教学质量检测数学试题
浙江省杭州市2024届高三下学期4月教学质量检测数学试题吉林省长春市实验中学2024届高三下学期对位演练考试数学试卷(一)(已下线)压轴题08计数原理、二项式定理、概率统计压轴题6题型汇总贵州省贵阳市第一中学等校2024届高三下学期三模数学试题重庆市七校联盟2024届高三下学期三诊考试数学试题(已下线)专题02 高二下期末真题精选(压轴题 )-高二期末考点大串讲(人教A版2019)山东省青岛第一中学2023-2024学年高二下学期第一次模块考试数学试题
解题方法
3 . 中国南北朝时期的著作《孙子算经》中,对同余除法有较深的研究.设
,
,![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
为整数,若
和
同时除以
所得的余数相同,则称
和
对模
同余,记为
.若
,
,则
的值可以是( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c94bb12cee76221e13f9ef955b0aab1.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/344cc24b575f4fd1ea7fe8ce5612fa9a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c94bb12cee76221e13f9ef955b0aab1.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c94bb12cee76221e13f9ef955b0aab1.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/69c8febfc211a5b23fc2450fee31aef8.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/57e718af12979e6ada207941bbb011fc.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1eef4945466d2995eefccf8d01164485.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c94bb12cee76221e13f9ef955b0aab1.png)
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解题方法
4 . 如图,将两个相同大小的圆柱垂直放置,两圆柱的底面直径与高相等,且中心重合,它们所围成的几何体称为“牟合方盖”,已知两圆柱的高为2,则该“牟合方盖”内切球的体积为( )
A.![]() | B.![]() | C.![]() | D.![]() |
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5卷引用:安徽省六安第一中学2023-2024学年高三下学期期末质量检测卷(二)数学试题
安徽省六安第一中学2023-2024学年高三下学期期末质量检测卷(二)数学试题(已下线)模块三 易错点3 不会从情境题中抽象出数学图形辽宁省沈阳铁路实验中学2024届高三第八次模拟考试数学试题云南省2024届高三学期”3_3_3“高考备考诊断性联考卷(二)数学试题(已下线)专题07 球与几何体的切、接及立体几何最值问题-期末考点大串讲(苏教版(2019))
5 . 某艺术吊灯如图1所示,图2是其几何结构图.底座
是边长为
的正方形,垂直于底座且长度为6的四根吊挂线
,
,
,
一头连着底座端点,另一头都连在球
的表面上(底座厚度忽略不计),若该艺术吊灯总高度为14,则球
的体积为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/411b38a18046fea8e9fab1f9f9b80a5f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4e2031d209711b058f3d278ede3c1d33.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2777840758e70e7dbbc18cef8f3d6d2b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e0a851907ada2ac2c3c4880a6736d28a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9d88bf46ad08f9677c37eed1d0369329.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/22adbc0da438220f9cace11b629d799b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1dde8112e8eb968fd042418dd632759e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1dde8112e8eb968fd042418dd632759e.png)
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3卷引用:云南省昆明市2023-2024学年高三三模数学试题
2024高三·全国·专题练习
名校
6 . 三星堆古遗址作为“长江文明之源“,被誉为人类最伟大的考古发现之一.3号坑发现的神树纹玉琮,为今人研究古蜀社会中神树的意义提供了重要依据.玉琮是古人用于祭祀的礼器,有学者认为其外方内圆的构造,契合了古代“天圆地方”观念,是天地合一的体现,如图,假定某玉琮形状对称,由一个空心圆柱及正方体构成,且圆柱的外侧面内切于正方体的侧面,圆柱的高为12cm,圆柱底面外圆周和正方体的各个顶点均在球O上,则球O的表面积为 _________
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7 . 古希腊数学家阿波罗尼斯的著作《圆锥曲线论》中有这样一个结论:平面内与两点距离的比为常数
(
)的点的轨迹是圆,后人称这个圆为阿波罗尼斯圆.已知点
,
,动点
满足
,若点
的轨迹与圆
:
(
)有且仅有三条公切线,则
( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/df64046e91b047037f19e4032e3b6de3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/472393b18c7880e73b40e31fbe2d951c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/19b62194097ac66a5093c57fca2f5b4f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/12e2db2888cf296b8f56c3702dcd8435.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/aee82283f06cedef32eb15b87964f5d2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9cdb7b37d9b96793ec84d60bef24586f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c5db41a1f31d6baee7c69990811edb9f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a8a2b1f5228d0dd50ae4c229b8c64da4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/45ba6b6aa6c3f9faba6b03bc193a6e61.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/258e40e3883271c3580c1d3c805dcac6.png)
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8 . 设
是函数
的有限实数集,
是定义在
上的函数,若
的图象绕坐标原点逆时针旋转
后与原图象重合,则在以下各项中,
的取值不可能是( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e105760638b22b26ff8bec4354255e4c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ed2d1ecae9c649cc3c89f9ce0c063208.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e105760638b22b26ff8bec4354255e4c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ff7a3159579864a8ea0ab42005144864.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e4c4a8dd0b01e7ebd32f3080f93f453e.png)
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解题方法
9 . 拉格朗日中值定理是微分学的基本定理之一,其内容为:如果函数
在闭区间
上的图象连续不断,在开区间
内的导数为
,那么在区间
内存在点
,使得
成立.设
,其中
为自然对数的底数,
.易知,
在实数集
上有唯一零点
,且
.
时,
;
(2)从图形上看,函数
的零点就是函数
的图象与
轴交点的横坐标.直接求解
的零点
是困难的,运用牛顿法,我们可以得到
零点的近似解:先用二分法,可在
中选定一个
作为
的初始近似值,使得
,然后在点
处作曲线
的切线,切线与
轴的交点的横坐标为
,称
是
的一次近似值;在点
处作曲线
的切线,切线与
轴的交点的横坐标为
,称
是
的二次近似值;重复以上过程,得
的近似值序列
.
①当
时,证明:
;
②根据①的结论,运用数学归纳法可以证得:
为递减数列,且
.请以此为前提条件,证明:
.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8ca6d68f1de3e70696f1d5d60affe6ef.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/63313f7ac7402fcb5a9a840db64c6f08.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/090a91e4f3c8930674f98a9fa527709b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/63313f7ac7402fcb5a9a840db64c6f08.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/071a7e733d466949ac935b4b8ee8d183.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d59685311c7aa9ca98b1fdbabde40171.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15432e3c4e6c1d9cde98ec9187d162c2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6dcd143a57a268a5a8ef486e2a4d5c0a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/00108fe668a98c905f3f92b720e35a0b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cf3ed15aa3dcc4211fb520b5b942c989.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c8e356055d318b6d336e9e33a1e78aad.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/70142f9c28dc50c8ab41e71b19d18fb9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9a8488679e2fa13e44ffa5b4d802848d.png)
(2)从图形上看,函数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15432e3c4e6c1d9cde98ec9187d162c2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15432e3c4e6c1d9cde98ec9187d162c2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/de261e9b4defbc0be6440397031a87b8.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/79b752f0f189e5d8666daea73e145dff.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/168e68d052280fe48e1a3a6de67c6f2d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8559f5db9b978cb2bd290dbce7268629.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0afb80007983e5b99dcdeebf87d18ff4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c814128ea2139e33db94ea590e7c2223.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c814128ea2139e33db94ea590e7c2223.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a24a2c53e3b0b1c08803e95419f909d3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0afb80007983e5b99dcdeebf87d18ff4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/aec19b68e3add9d5bfcc6269a1855b87.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/aec19b68e3add9d5bfcc6269a1855b87.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/87529d4cadc1e84f72d462cb8e3afac0.png)
①当
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7c1a778faac194e8de4d5178454bd04c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6f274881a6ad83e68c9b6652ebf4dc09.png)
②根据①的结论,运用数学归纳法可以证得:
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c1fd18a909cecbaee7115d6b15631d83.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2adb4f1a98a9db3b5d4e4cfc7560fdb1.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/fee28be9d207a3d3eed938484f980195.png)
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名校
10 . 阅读材料一:“装错信封问题”是由数学家约翰·伯努利(Johann Bernoulli,1667~1748)的儿子丹尼尔·伯努利提出来的,大意如下:一个人写了
封不同的信及相应的
个不同的信封,他把这
封信都装错了信封,问都装错信封的这一情况有多少种?后来瑞士数学家欧拉(Leonhard Euler,1707~1783)给出了解答:记都装错
封信的情况为
种,可以用全排列
减去有装正确的情况种数,结合容斥原理可得公式:
,其中
.
阅读材料二:英国数学家泰勒发现的泰勒公式有如下特殊形式:当
在
处
阶可导,则有:
,注
表示
的
阶导数,该公式也称麦克劳林公式.阅读以上材料后请完成以下问题:
(1)求出
的值;
(2)估算
的大小(保留小数点后2位),并给出用
和
表示
的估计公式;
(3)求证:
,其中
.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/66d4e8502106802f1485c3b0f28f2664.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a8412f5256b2b370e421c07f18cc732.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e4403d632f9a81e52c6cd135c6834bc2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a37a59558292ad6b3d0978bfd7484990.png)
阅读材料二:英国数学家泰勒发现的泰勒公式有如下特殊形式:当
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/09f86f37ec8e15846bd731ab4fcdbacd.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bb45f673c56a289ea78831c9237e8d20.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6ce152ca98ac7e21237e00667f005b62.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/35993bd1db970330494665d925c0be7a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/09f86f37ec8e15846bd731ab4fcdbacd.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
(1)求出
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/395c6efaa63dcd4ee513323d51c6a7eb.png)
(2)估算
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2598975ac1edb754817eada15b9a473e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/041a7c8fc017f596542c5e6ec7d1c40b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/66d4e8502106802f1485c3b0f28f2664.png)
(3)求证:
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ca08ded0d1136421f0a81517f5c2fc9d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a37a59558292ad6b3d0978bfd7484990.png)
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