问:
Calculating distance between two obstacles from a Lidar vertical scan
I have a 2D Lidar scanner that is scanning vertically (see photo Graphical Representation of Data).  The center point is the scanner.  The bottom is the floor, top is the ceiling, and the left and right sides are selves.  I am trying to figure out, from the scan data, what the width of the largest box is that I can push down this isle.  The shelves can be variable length, and items on them can stick out into the isle way.

If I look at 90 and 180 degrees I can see that their is 852mm plus 1,090mm totaling 1,942mm.  But, their is no guarantee that what is scanned at 90 and 180 degrees are the limiting points.  Because it is a circle, I know tangent is involved, but I can not rap my head around the formula.  Additionally, the floor (which in this scan from about 140 degrees to 220 degrees) should not be included in the width calculation.  I'm sure a single formula will not solve this, and that a program will need to iterate through the data.  Help with the mathematics would be appreciated.

Thank you.

Download link to data below

Graphical Representation of Data
|Bearing in degrees |Distance in mm|
|------------------:|---------------:|
|0.0439|5752|
|0.3406|5752|
|0.6207|5760|
|0.9009|5752|
|1.1865|5768|
|1.4832|5792|
|1.7633|5776|
|2.0435|5776|
|2.3236|5784|
|2.6202|5800|
|2.9059|5792|
|3.186|5792|
|3.4662|5808|
|3.7628|5816|
|4.043|5816|
|4.3231|5816|
|4.6198|5816|
|4.9054|5800|
|5.1855|5808|
|5.4657|5808|
|5.7623|5808|
|6.0425|5808|
|6.3281|5816|
|6.6248|5824|
|6.9049|5824|
|7.1851|5824|
|7.4652|5808|
|7.7618|5808|
|8.042|5816|
|8.3276|5816|
|8.6243|5816|
|8.9044|5832|
|9.1846|5856|
|9.4812|5872|
|9.7614|5888|
|10.0415|5896|
|10.3271|5896|
|10.6238|5896|
|10.9039|5896|
|11.7609|5912|
|12.453|3736|
|12.7332|3696|
|13.0133|3660|
|13.5901|3676|
|15.1831|6056|
|16.1554|3740|
|17.0123|3712|
|17.309|3724|
|17.5891|3732|
|17.8748|3736|
|18.1549|3780|
|18.4515|3816|
|18.7317|3804|
|19.0118|3784|
|19.3085|3780|
|19.5886|3792|
|23.5931|3960|
|23.8733|3944|
|24.1699|3972|
|24.4501|3996|
|24.7302|3980|
|25.0104|3980|
|25.3125|4008|
|25.5927|3996|
|25.8728|3992|
|26.1694|3996|
|26.4496|4000|
|26.7297|4020|
|27.312|4052|
|30.108|4128|
|30.3882|4216|
|30.6848|4272|
|31.7175|6848|
|31.9977|6864|
|32.2778|6920|
|33.1512|6952|
|34.0741|6112|
|34.3103|7048|
|34.5905|7160|
|34.8871|7192|
|35.1672|7232|
|35.4639|7256|
|35.7495|7312|
|36.0461|7360|
|36.3263|7384|
|36.6229|7424|
|36.9031|7416|
|37.1997|7416|
|37.4799|7432|
|37.7765|7456|
|38.0621|7488|
|38.3588|7512|
|38.6389|7528|
|38.9355|7592|
|39.2322|7664|
|39.5123|7680|
|39.809|7704|
|40.0891|7736|
|40.3857|7768|
|40.6714|7808|
|40.968|7840|
|41.2482|7808|
|41.5448|7792|
|41.825|7880|
|42.1216|7960|
|42.4017|8080|
|42.6984|8144|
|42.984|8136|
|43.2806|8160|
|43.5608|8168|
|44.1376|7792|
|44.4342|7880|
|45.2966|7976|
|46.4337|1816|
|46.7303|1812|
|47.0105|1808|
|47.3071|1806|
|47.5928|1804|
|47.8894|1800|
|48.1696|1796|
|48.4662|1802|
|48.7463|1802|
|49.043|1798|
|49.3231|1798|
|49.6198|1798|
|49.9054|1814|
|50.202|1834|
|50.4822|1850|
|50.7788|1852|
|51.059|1842|
|51.9324|1840|
|52.4652|10168|
|53.042|10208|
|53.3386|9304|
|53.9209|10064|
|54.201|10264|
|54.4977|10448|
|54.7778|10536|
|55.0745|10632|
|55.3546|10736|
|55.6512|10792|
|55.9369|10848|
|56.2335|10984|
|56.5302|10920|
|58.5571|1026|
|58.9032|1022|
|59.1998|1022|
|59.48|1020|
|59.7766|1018|
|60.0623|1016|
|60.3589|1012|
|60.639|1008|
|60.9357|1006|
|61.2158|1004|
|61.5125|1002|
|61.7926|1000|
|62.0892|998|
|62.3749|994|
|62.655|990|
|62.9517|988|
|63.2318|988|
|63.5284|986|
|63.8086|984|
|64.1547|980|
|64.4348|978|
|64.7314|976|
|65.0116|974|
|65.2917|972|
|65.5884|970|
|65.874|970|
|66.4508|968|
|66.7969|922|
|67.077|914|
|67.4176|908|
|67.7142|898|
|67.9999|892|
|68.28|892|
|69.2633|1024|
|69.5599|1024|
|69.8401|1012|
|70.1202|1010|
|70.4169|1010|
|70.7025|1010|
|71.6858|1170|
|71.9165|1196|
|72.2021|1206|
|73.0591|1202|
|73.3557|1228|
|73.6359|1228|
|73.9215|1228|
|74.9652|1360|
|75.2454|1362|
|75.542|1364|
|75.7617|1368|
|76.0583|1372|
|76.3385|1372|
|76.6406|1370|
|76.9812|1366|
|77.2614|1364|
|77.5415|1366|
|77.8436|1366|
|78.1238|1360|
|78.4039|1358|
|78.7006|1356|
|78.9807|1356|
|79.2773|1354|
|79.5575|1352|
|79.8431|1352|
|80.1398|1350|
|80.4199|1350|
|80.7166|1350|
|80.9967|1346|
|81.2769|1344|
|81.5735|1344|
|81.8591|1344|
|82.1393|1344|
|82.4359|1342|
|82.7161|1342|
|83.0127|1342|
|83.2928|1342|
|83.573|1340|
|84.1553|1344|
|84.6716|1078|
|85.0122|1068|
|85.2924|1064|
|85.589|1058|
|85.8746|1056|
|86.1548|1054|
|86.4514|1050|
|86.7316|1050|
|87.0282|1050|
|87.3083|1054|
|87.5885|1056|
|87.8906|1056|
|89.0277|852|
|89.3079|852|
|89.5935|850|
|89.8901|850|
|90.1703|852|
|90.4504|852|
|90.7471|852|
|91.0272|850|
|91.3074|850|
|91.593|850|
|91.8896|850|
|92.1698|850|
|92.45|850|
|92.7466|852|
|93.0267|852|
|93.3124|852|
|93.5925|852|
|93.8892|852|
|94.1693|852|
|94.4495|854|
|94.7461|854|
|95.0262|854|
|95.3119|856|
|95.592|856|
|95.8887|856|
|96.1688|856|
|96.449|856|
|96.7456|856|
|97.0313|858|
|97.3114|858|
|97.608|860|
|97.8882|862|
|98.1683|862|
|98.4485|862|
|98.7451|862|
|99.0308|864|
|99.3109|866|
|99.6075|866|
|99.8877|862|
|100.1678|854|
|100.448|856|
|100.7446|870|
|101.0303|872|
|101.3104|862|
|101.6071|856|
|101.8872|856|
|102.1674|854|
|102.453|852|
|102.7496|854|
|103.1561|1186|
|103.4363|1174|
|103.7769|1170|
|104.0625|1164|
|104.3591|1158|
|104.6393|1160|
|104.9194|1162|
|105.2161|1166|
|105.4358|1174|
|105.7324|1178|
|106.0126|1176|
|106.2927|1178|
|106.5729|1186|
|106.875|1192|
|107.1552|1202|
|107.4353|1218|
|107.6715|1230|
|108.0121|1228|
|108.2922|1222|
|108.5889|1220|
|108.8745|1222|
|109.1547|1222|
|110.0116|1216|
|110.528|990|
|110.8081|984|
|111.1542|980|
|111.4343|972|
|111.7145|968|
|112.0111|966|
|112.2968|960|
|112.5769|962|
|112.8735|964|
|113.1537|960|
|113.4338|956|
|113.714|952|
|114.0106|956|
|114.2963|958|
|114.5764|958|
|114.873|960|
|115.1532|956|
|115.4333|948|
|115.7135|946|
|116.0156|946|
|116.2958|948|
|116.5759|952|
|116.8561|954|
|117.1527|956|
|117.4329|958|
|117.7185|962|
|117.9987|966|
|118.2788|968|
|118.5754|970|
|118.8556|974|
|119.1357|976|
|119.4214|980|
|119.6521|984|
|119.9323|988|
|120.2179|996|
|120.498|994|
|120.7782|988|
|121.0748|988|
|121.355|988|
|121.6406|988|
|122.2009|996|
|122.9041|1472|
|123.1842|1476|
|123.4644|1476|
|124.0906|1394|
|124.4366|1364|
|124.9365|1424|
|125.2332|1442|
|126.0736|1112|
|126.3593|1104|
|126.6394|1098|
|126.9196|1094|
|127.2162|1090|
|127.4963|1086|
|127.7765|1086|
|128.0621|1088|
|128.3423|1086|
|128.6389|1086|
|128.9191|1086|
|129.1992|1086|
|129.4794|1086|
|129.765|1088|
|130.0616|1090|
|130.3418|1094|
|130.6219|1094|
|130.9021|1098|
|131.1823|1104|
|131.4679|1108|
|131.7645|1116|
|131.9843|1120|
|132.2644|1122|
|132.561|1130|
|132.8412|1138|
|133.1213|1140|
|133.4015|1146|
|133.6871|1154|
|133.9838|1160|
|134.2639|1168|
|134.4836|1178|
|134.7803|1174|
|135.0604|1180|
|135.3406|1198|
|135.6207|1192|
|135.9009|1180|
|136.1865|1178|
|136.4832|1180|
|136.7633|1174|
|137.1094|1164|
|137.3895|1156|
|137.6697|1160|
|137.9498|1168|
|138.23|1168|
|138.5266|1168|
|138.8123|1170|
|139.0265|1178|
|141.2018|1366|
|141.4819|1362|
|141.7621|1348|
|142.0422|1344|
|142.3279|1342|
|142.6245|1338|
|142.9047|1330|
|143.1848|1324|
|143.465|1320|
|143.7451|1316|
|144.0417|1312|
|144.3274|1308|
|144.6075|1304|
|144.8877|1298|
|145.2338|1294|
|145.5139|1292|
|145.7941|1288|
|146.0742|1284|
|146.3708|1280|
|146.651|1274|
|146.9366|1268|
|147.2168|1268|
|147.4969|1264|
|147.7936|1262|
|148.0737|1258|
|148.3594|1252|
|148.6395|1248|
|148.9197|1246|
|149.2163|1244|
|149.4965|1240|
|149.7766|1236|
|150.0623|1232|
|150.4028|1228|
|150.683|1224|
|150.9686|1220|
|151.2488|1218|
|151.5454|1216|
|151.8256|1214|
|152.1057|1210|
|152.3859|1208|
|152.6715|1204|
|152.9681|1202|
|153.2483|1200|
|153.5284|1198|
|153.8086|1194|
|154.0887|1190|
|154.3744|1188|
|154.671|1186|
|154.9512|1184|
|155.2313|1182|
|155.5115|1178|
|155.7916|1176|
|156.0938|1174|
|156.3739|1172|
|156.7145|1168|
|156.9946|1168|
|157.2803|1166|
|157.5604|1162|
|157.8406|1160|
|158.1207|1158|
|158.4174|1156|
|158.703|1154|
|158.9832|1154|
|159.2633|1150|
|159.5435|1148|
|159.8401|1146|
|160.1202|1144|
|160.4059|1142|
|160.686|1142|
|160.9662|1140|
|161.2628|1138|
|161.543|1136|
|161.8231|1136|
|162.1088|1134|
|162.3889|1130|
|162.6691|1128|
|162.9657|1128|
|163.2458|1126|
|163.526|1124|
|163.8116|1124|
|164.0918|1122|
|164.3884|1120|
|164.6686|1118|
|164.9487|1118|
|165.2948|1116|
|165.575|1114|
|165.8551|1114|
|166.1353|1112|
|166.4374|1112|
|166.7175|1112|
|166.9977|1110|
|167.2778|1108|
|167.558|1108|
|167.8436|1106|
|168.1403|1104|
|168.4204|1104|
|168.7006|1104|
|168.9807|1104|
|169.2609|1102|
|169.5575|1100|
|169.8431|1100|
|170.1233|1100|
|170.4034|1100|
|170.6836|1098|
|170.9802|1098|
|171.2604|1098|
|171.546|1096|
|171.8262|1096|
|172.1063|1096|
|172.3865|1094|
|172.6831|1094|
|172.9688|1094|
|173.2489|1092|
|173.5291|1092|
|173.8092|1092|
|174.1058|1092|
|174.386|1090|
|174.6716|1090|
|174.9518|1092|
|175.2319|1090|
|175.5286|1090|
|175.8087|1090|
|176.0889|1090|
|176.3745|1090|
|176.6547|1090|
|176.9348|1090|
|177.2314|1090|
|177.5116|1090|
|177.7917|1090|
|178.0774|1090|
|178.3575|1090|
|178.6542|1090|
|178.9343|1090|
|179.2145|1090|
|179.4946|1090|
|179.7803|1090|
|180.0769|1090|
|180.3571|1090|
|180.6372|1090|
|180.9174|1090|
|181.203|1092|
|181.4832|1092|
|181.7798|1092|
|182.0599|1092|
|182.3401|1092|
|182.6202|1092|
|182.9059|1096|
|183.2025|1096|
|183.4827|1096|
|183.7628|1096|
|184.043|1098|
|184.3231|1098|
|184.6198|1098|
|184.9054|1100|
|185.1855|1100|
|185.4657|1100|
|185.7458|1100|
|186.026|1102|
|186.3281|1102|
|186.6083|1102|
|186.8884|1102|
|187.1686|1104|
|187.4487|1106|
|187.7454|1108|
|188.031|1108|
|188.3112|1108|
|188.5913|1110|
|188.8715|1110|
|189.1681|1112|
|189.4482|1112|
|189.7339|1112|
|190.014|1114|
|190.2942|1116|
|190.5304|1118|
|190.8105|1120|
|191.0907|1120|
|191.3708|1122|
|191.6675|1124|
|191.9531|1124|
|192.2333|1126|
|192.5134|1128|
|192.7936|1128|
|193.0902|1130|
|193.3704|1132|
|193.656|1134|
|193.9362|1136|
|194.2163|1138|
|194.4965|1138|
|194.7931|1140|
|195.0732|1142|
|195.3589|1142|
|195.639|1144|
|195.9192|1146|
|196.2158|1148|
|196.496|1152|
|196.7761|1154|
|197.0618|1154|
|197.3419|1156|
|197.6385|1158|
|197.9187|1162|
|198.1989|1164|
|198.479|1164|
|198.7811|1166|
|199.0613|1170|
|199.292|1172|
|199.5776|1174|
|199.8578|1178|
|200.1379|1180|
|200.4346|1182|
|200.7147|1184|
|200.9949|1188|
|201.2915|1192|
|201.5771|1194|
|201.8573|1196|
|202.1375|1196|
|202.4341|1198|
|202.7142|1204|
|202.9999|1208|
|203.28|1210|
|203.5767|1212|
|203.8568|1214|
|204.137|1216|
|204.4336|1220|
|204.7137|1224|
|204.9994|1228|
|205.2301|1230|
|205.5103|1232|
|205.7959|1236|
|206.0925|1238|
|206.3727|1244|
|206.6528|1246|
|206.9495|1250|
|207.2296|1252|
|207.5153|1256|
|207.7954|1262|
|208.092|1264|
|208.3722|1268|
|208.6523|1272|
|208.949|1276|
|209.2291|1278|
|209.5148|1282|
|209.7949|1286|
|210.0916|1292|
|210.3113|1296|
|210.6079|1298|
|210.8881|1304|
|211.1682|1310|
|211.4484|1316|
|211.745|1318|
|212.0306|1320|
|212.3108|1324|
|212.6074|1332|
|212.8876|1338|
|213.1677|1342|
|213.4479|1348|
|213.75|1350|
|214.0302|1354|
|214.3103|1358|
|214.5465|1368|
|214.8267|1374|
|215.1233|1376|
|215.4034|1380|
|215.6836|1384|
|215.9637|1392|
|216.2604|1398|
|216.546|1404|
|216.8262|1410|
|217.1228|1414|
|217.403|1420|
|217.6831|1426|
|217.9688|1436|
|218.2654|1442|
|218.5455|1444|
|218.7762|1448|
|219.0619|1456|
|219.342|1468|
|219.6222|1490|
|219.9188|1494|
|220.4791|1474|
|220.7813|1460|
|221.0614|1454|
|221.792|1290|
|222.0776|1280|
|222.3743|1278|
|222.6544|1272|
|222.9346|1270|
|223.2312|1276|
|223.5114|1278|
|223.7915|1258|
|224.3738|1282|
|224.6539|1270|
|224.9341|1264|
|225.2307|1258|
|225.5109|1250|
|225.7965|1242|
|226.0931|1238|
|226.3733|1232|
|226.7139|1222|
|226.9995|1216|
|227.2961|1214|
|227.5763|1210|
|227.8729|1204|
|228.1531|1196|
|228.4332|1190|
|228.7299|1184|
|229.0155|1180|
|229.3121|1178|
|229.6527|1170|
|229.9329|1164|
|230.2185|1160|
|230.5151|1156|
|230.7953|1150|
|231.0754|1144|
|231.3721|1138|
|231.6522|1136|
|231.9489|1130|
|232.229|1126|
|232.5146|1122|
|232.8113|1118|
|233.1519|1112|
|233.4375|1110|
|233.7177|1106|
|234.0143|1102|
|234.2944|1098|
|234.5911|1094|
|234.8712|1090|
|235.1514|1096|
|235.448|1102|
|235.7336|1110|
|235.9644|1118|
|236.25|1126|
|236.5466|1136|
|236.8268|1144|
|237.1234|1152|
|237.4036|1142|
|237.6837|1122|
|238.0298|1112|
|238.3264|1110|
|238.6066|1108|
|238.8867|1098|
|239.1833|1098|
|239.9799|1118|
|241.2488|1052|
|241.5289|1046|
|241.8256|1042|
|242.1057|1038|
|242.4023|1036|
|242.6825|1034|
|242.9681|1030|
|243.2648|1028|
|243.5449|1028|
|243.8416|1028|
|244.1217|1026|
|244.4678|1024|
|244.6985|1026|
|244.9841|1026|
|245.2643|1028|
|245.5609|1032|
|245.8411|1034|
|246.1377|1034|
|246.4178|1034|
|246.698|1038|
|246.9946|1042|
|247.2803|1046|
|247.5769|1050|
|247.8571|1054|
|248.1372|1060|
|248.4338|1062|
|248.714|1066|
|249.527|1074|
|249.9994|1554|
|250.2795|1546|
|250.5597|1548|
|250.8563|1548|
|251.1365|1544|
|251.4331|1540|
|251.7792|1536|
|252.0593|1532|
|252.6361|1532|
|253.3722|988|
|253.6523|986|
|253.9325|986|
|254.2291|988|
|254.5587|982|
|254.8553|980|
|255.0751|996|
|255.3717|1004|
|255.6519|994|
|255.9375|990|
|256.2341|988|
|256.5143|986|
|256.7944|986|
|257.0746|986|
|257.3712|986|
|257.6514|986|
|257.937|984|
|258.2336|984|
|258.5138|984|
|258.7939|984|
|259.0741|984|
|259.3707|984|
|259.9365|984|
|261.7767|1470|
|262.0624|1468|
|262.3425|1470|
|262.6392|1468|
|262.9193|1468|
|263.1995|1468|
|263.4961|1468|
|263.7762|1466|
|264.0619|1466|
|264.342|1466|
|264.6387|1466|
|264.9188|1468|
|265.199|1468|
|265.4791|1466|
|265.7813|1474|
|266.8579|954|
|267.1381|952|
|267.4182|952|
|267.6984|954|
|267.995|950|
|268.2806|954|
|268.5608|970|
|268.8574|974|
|269.1376|966|
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|269.6979|962|
|270|962|
|270.2802|964|
|270.5603|964|
|270.8405|966|
|271.1371|968|
|271.4172|970|
|271.7029|970|
|271.9995|970|
|272.2797|970|
|272.5598|972|
|272.84|972|
|273.1366|974|
|273.4167|974|
|273.7024|976|
|273.999|976|
|274.2792|978|
|274.5593|978|
|274.8395|980|
|275.1361|982|
|275.3558|984|
|275.6525|984|
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|277.2015|1482|
|277.4817|1482|
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|278.3386|1484|
|279.0308|1316|
|279.3109|1302|
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|280.5579|1222|
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|281.4203|1228|
|281.651|1232|
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|282.5134|1240|
|282.7936|1242|
|283.0737|1244|
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|283.656|1250|
|283.9362|1250|
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|284.5129|1254|
|284.7931|1256|
|285.0732|1264|
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|285.6555|1280|
|285.9357|1286|
|286.2158|1290|
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|286.7926|1292|
|287.0728|1292|
|287.3584|1294|
|287.5891|1300|
|287.8748|1310|
|288.7317|1312|
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|290.105|1390|
|290.3906|1370|
|290.7312|1354|
|291.0114|1354|
|291.308|1346|
|291.5936|1338|
|291.8738|1340|
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|292.4506|1308|
|292.7307|1310|
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|293.6371|1280|
|293.9337|1288|
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|295.0104|1356|
|295.2466|1372|
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|296.109|1388|
|296.3892|1394|
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|298.125|10984|
|298.4052|10888|
|298.6853|10984|
|298.9655|10984|
|299.5422|9784|
|299.8279|9784|
|300.4047|10336|
|300.6848|10376|
|300.965|10376|
|301.5417|9736|
|301.8274|9792|
|302.1075|9968|
|302.4042|10040|
|302.6843|9936|
|302.9645|9808|
|303.2446|9712|
|303.5468|9632|
|303.8269|9592|
|304.1071|9584|
|304.3872|9584|
|304.6838|9552|
|304.964|9472|
|305.2496|9408|
|305.5463|9384|
|305.8264|9336|
|306.1066|9280|
|306.6833|9200|
|307.2491|8304|
|307.5458|8480|
|307.8259|8608|
|308.1061|8696|
|308.3862|8728|
|308.6829|8720|
|308.9685|8712|
|309.2487|8664|
|309.5288|8584|
|309.8254|8520|
|310.1056|8464|
|310.3857|8408|
|310.6714|8368|
|310.968|8368|
|311.2482|8296|
|311.5887|8152|
|311.8744|8112|
|312.1545|8120|
|312.4512|8104|
|312.7313|8096|
|313.0115|8064|
|313.2916|7984|
|313.5938|7920|
|313.8739|7912|
|314.1541|7864|
|314.4507|7808|
|314.7308|7776|
|315.011|7680|
|315.2966|7640|
|315.5933|7624|
|315.8734|7632|
|316.1536|7632|
|316.4337|7584|
|316.7303|7552|
|317.0105|7512|
|317.8729|7472|
|318.1531|6456|
|318.7299|6376|
|319.2957|7168|
|319.5758|7192|
|319.8724|7168|
|320.1526|7160|
|320.4327|7160|
|320.7294|7056|
|321.015|6896|
|321.2952|6904|
|321.5753|6976|
|321.8719|6976|
|322.1521|6960|
|322.4323|6912|
|322.7179|6888|
|323.0145|6904|
|323.2947|6912|
|323.5748|6912|
|323.8715|6912|
|324.1516|6872|
|324.4373|6856|
|324.7174|6816|
|325.014|6792|
|325.2942|6800|
|325.5743|6768|
|325.8545|6760|
|326.1511|6768|
|326.4368|6744|
|326.7169|6696|
|326.9971|6680|
|327.2937|6664|
|328.1561|6160|
|328.4802|6128|
|328.7769|6120|
|328.9966|6216|
|329.2932|6208|
|329.6228|6040|
|329.9194|5904|
|330.1996|5848|
|330.4797|5816|
|330.7764|5784|
|331.062|5872|
|331.5729|6304|
|331.8585|6352|
|332.1387|6352|
|332.4353|6336|
|332.7155|6328|
|332.9956|6312|
|333.2813|6232|
|333.9185|5456|
|334.1986|5376|
|334.4843|5352|
|335.061|5408|
|335.6213|6072|
|335.8575|6168|
|336.1377|6184|
|336.4343|6152|
|336.7639|6104|
|337.0605|6064|
|337.3407|6048|
|337.6208|6048|
|337.9175|6056|
|338.2031|6048|
|338.4833|6040|
|338.7634|6048|
|339.0601|6024|
|339.3402|6032|
|339.6204|6056|
|339.906|6048|
|340.2026|6032|
|340.4828|6000|
|340.7629|5984|
|341.0596|5976|
|341.3397|5952|
|341.6199|5960|
|341.9055|5944|
|342.2021|5928|
|342.4823|5928|
|342.7625|5912|
|343.0591|5904|
|343.3392|5896|
|343.6249|5888|
|343.905|5864|
|344.2017|5864|
|344.4818|5864|
|344.762|5856|
|345.0586|5856|
|345.3387|5848|
|345.6244|5840|
|345.9045|5848|
|346.2012|5848|
|346.4813|5848|
|346.7615|5856|
|347.0581|5848|
|347.3438|5856|
|347.6239|5848|
|347.9041|5840|
|348.2007|5832|
|348.4808|5840|
|348.761|5840|
|349.0466|5832|
|349.3433|5816|
|349.6234|5816|
|349.9036|5808|
|350.2002|5800|
|350.4803|5808|
|350.7605|5792|
|351.0461|5776|
|351.3428|5784|
|351.6229|5768|
|351.9031|5760|
|352.0624|5768|
|352.3425|5768|
|352.6392|5744|
|352.9193|5752|
|353.1995|5752|
|353.4796|5760|
|353.7762|5776|
|354.6222|5032|
|354.9188|5000|
|355.4791|4960|
|356.0614|5648|
|356.3416|5752|
|356.6217|5784|
|356.9183|5792|
|357.1985|5784|
|357.4841|5760|
|357.7643|5768|
|358.0609|5784|
|358.3411|5752|
|358.6212|5736|
|358.9014|5752|
|359.198|5768|
|359.4836|5760|
|359.7638|5760|

答:
Follows a MATHEMATICA script easily understood by itself.

```<code>data = Import["/home/test/Downloads/scan.csv", "Data"];
data0 = Take[data, {2, Length[data]}];

ceiling = 2000;
floor = -1000;
near = {};
For[k = 1, k <= Length[data0], k++,
 If[data0[[k, 2]] < ceiling,
   r = data0[[k, 2]];
   a = Pi data0[[k, 1]]/180 + Pi/2;
   AppendTo[near, {r Cos[a], r Sin[a]}]
 ]
]

gr0 = ListPlot[near, Axes -> False, PlotStyle -> {Black, Thick}];

near2 = {};
For[k = 1, k <= Length[near], k++,
 If[near[[k, 2]] > floor, AppendTo[near2, near[[k]]]]
]

leftmax = -20000;
rightmin = 20000;
For[k = 1, k <= Length[near2], k++,
 If[near2[[k, 1]] < 0 && near2[[k, 1]] > leftmax,
   leftmax = near2[[k, 1]]
 ]
]

For[k = 1, k <= Length[near2], k++,
 If[near2[[k, 1]] > 0 && near2[[k, 1]] < rightmin,
 rightmin = near2[[k, 1]]
 ]
]

safe = {{leftmax, ceiling}, {leftmax, floor}, {rightmin, floor},{rightmin, ceiling}, {leftmax, ceiling}};
gr2 = ListLinePlot[safe, PlotStyle -> Red, Axes -> False];
Show[gr0, gr2, PlotRange -> All, AspectRatio -> 1]
</code>```

Body:
Follows a MATHEMATICA script easily understood by itself.

data = Import["/home/test/Downloads/scan.csv", "Data"];
data0 = Take[data, {2, Length[data]}];

ceiling = 2000;
floor = -1000;
near = {};
For[k = 1, k <= Length[data0], k++,
 If[data0[[k, 2]] < ceiling,
   r = data0[[k, 2]];
   a = Pi data0[[k, 1]]/180 + Pi/2;
   AppendTo[near, {r Cos[a], r Sin[a]}]
 ]
]

gr0 = ListPlot[near, Axes -> False, PlotStyle -> {Black, Thick}];

near2 = {};
For[k = 1, k <= Length[near], k++,
 If[near[[k, 2]] > floor, AppendTo[near2, near[[k]]]]
]

leftmax = -20000;
rightmin = 20000;
For[k = 1, k <= Length[near2], k++,
 If[near2[[k, 1]] < 0 && near2[[k, 1]] > leftmax,
   leftmax = near2[[k, 1]]
 ]
]

For[k = 1, k <= Length[near2], k++,
 If[near2[[k, 1]] > 0 && near2[[k, 1]] < rightmin,
 rightmin = near2[[k, 1]]
 ]
]

safe = {{leftmax, ceiling}, {leftmax, floor}, {rightmin, floor},{rightmin, ceiling}, {leftmax, ceiling}};
gr2 = ListLinePlot[safe, PlotStyle -> Red, Axes -> False];
Show[gr0, gr2, PlotRange -> All, AspectRatio -> 1]


问:
Multivariate normal expectation $\mathbb{E}(X^2Y^2)$
Hey I have two dimensional density :
$$f_{X,Y}(x,y)=\frac{1}{2\pi\sqrt{1-\sigma^{2}}}\exp\left\{ -\frac{1}{2\left(1-\sigma^{2}\right)}\left(\begin{array}{c}
x^{2}+y^{2}-2\sigma xy\end{array}\right)\right\} $$

How to calculate $\mathbb{E}(X^2Y^2)$? I try from definition but the integrals are very strange. Is there any trick which can be useful?

答:
Your $X,Y$ are jointly standard gaussian with correlation coefficient $\rho$.  Thus it would be better to indicate your $\sigma^2$ as $\rho^2$.

Given this, your expectation is easy derived using the law of total expectation

$$\mathbb{E}[X^2Y^2]=\mathbb{E}[\mathbb{E}(X^2Y^2|X)]=\mathbb{E}[X^2\mathbb{E}(Y^2|X)]$$

The quantity

$$\mathbb{E}[Y^2|X]$$

is the conditional second moment...and it is well known how to factorize the joint density function

$$f_{XY}(x,y)=f_X(x)\cdot f_{Y|X}(y|x)$$

into 2 gaussian densities in order to get it.

It results to me

$$\mathbb{E}[X^2Y^2]=1+2\rho^2$$

... no integrals or weird calculations are needed.

ContentLicense: CC BY-SA 4.0


ContentLicense: CC BY-SA 3.0

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.

Body: Direct proof: $10^n=(11-1)^n=11(\dots)+(-1)^n$


问:
Linear Algebra - Find matrice of linear transformation in two different bases
We define the linear trasformation: $T:\mathbb{R}_2[x] \to \mathbb{R}_2[x]$, $T(p(x)) = p(x) + (x+1)p'(x)$. We have two basis:$B=\{1,x,x^2\}$ and $B'=\{1,1+x,(1+x)^2\}$. Find $[T]_{B'}$.

I tried to use two methods to solve this question just to verify if my answer is correct.

We know that:
$$[T]_B = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 2 & 3 \end{pmatrix}$$

The first method is to find $[T]_{B'}$ directly:
$$[T]_B' = \begin{pmatrix} T(1)_{B'} & T(1+x)_{B'} & T(1+2x+x^2)_{B'}  \end{pmatrix} = \begin{pmatrix} 1 & 0 &0 \\0 & 2 & 0 \\ 0& 0& 3  \end{pmatrix}$$

The second method is to use the formula: $[T]_{B'} = P_{B' \leftarrow B}[T]_BP_{B \leftarrow B'} $ where P is change of basis matrix.

$$P_{B \leftarrow B'} = \begin{pmatrix} 1 & 0 &0 \\1 & 1 & 0 \\ 1 & 2 & 1  \end{pmatrix}$$
$$[T]_{B'} = P_{B \leftarrow B'}^{-1} [T]_B P_{B \leftarrow B'} = \begin{pmatrix} 1 & 0 & 0 \\2 & 2 & 0 \\ 0 & 4 &3  \end{pmatrix}$$

However, I should obtain the same result. So I am not really sure where I did wrong? Because I have always used both methods to solve similar question. Not sure why it is not working this time

Body: Note that, since $T(1)=1$, $T(x)=2x+1$, and $T(x^2)=3x^2+2x$, you actually have$$[T]_B=\begin{bmatrix}1&1&0\\0&2&2\\0&0&3\end{bmatrix}.$$


问:
Prove that the integral of a monotone function is between a given area interval
So I have to prove the following:

If $f$ is monotone in an interval $I:\{a\leq x\leq b\}$, $\int_a^bf(x)dx$ is between $f(a)(b-a)$ and $f(b)(b-a)$.

I clearly see the geometric interpretation of this. The previous statement is saying that the area represented by the integral is "between" the area of a rectangle with height $f(a)$ and base $(b-a)$, and a rectangle with height $f(b)$ and base $(b-a)$. Obviously the function could be strictly increasing, decreasing or just being a constant (as far as I know that's the definition of a monotone function), as such I can visualize that the statement holds, however I don't seem to be able to put this into math notation and actually prove it.

I have tried to use a Riemann sum to solve it, I've also tried to use upper and lower integrals but I didn't really know what to do with those. Any help or insight on how to tackle this would be truly appreciated.

EDIT 1:

So, from the comment I came up with the following (for the case in which we have a strictly increasing function):

If $f(b)>f(a)$, then $\int_a^b[f(b)-f(a)dx]\geq 0$, since the integrand is positive, while $\int_a^b[f(a)-f(x)]dx\leq 0$, because the integrand is negative. From this we get the following inequality:

$\int_a^b[f(a)-f(x)]dx\leq 0\leq\int_a^b[f(b)-f(a)]dx$

By using properties of the integrals, we simplify as:
$\int_a^bf(a)-\int_a^bf(x)dx\leq 0\leq\int_a^bf(b)dx-\int_a^bf(x)$

It is then obvious,

$\int_a^bf(a)\leq\int_a^bf(x)\leq\int_a^bf(b)$

For this last part I am not really sure how to convert the integrals of the function evaluated at the lower and upper bound of the integral into $f(a)(b-a)$ and $f(b)(b-a)$. I can see that the height will always be the same and the base is just going to be $(b-a)$. Maybe a Riemann sum of the previous integral would help? I have tried doing this:

$\int_a^bf(a)=\lim_{x\to \infty}\sum_{k=1}^nf(\xi)\Delta_kx$. Since we have that for every $\xi$, $f(\xi)$ will be the same as $f(a)$, then by properties of limits and sums (because f(a) is a constant), $\lim_{n\to \infty}\sum_{k=1}^nf(a)\Delta_kx=f(a)\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$, and since the sum of all the little sub-intervals that compose $\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$ is $(b-a)$, we finally have $f(a)(b-a)$. We then do the same for $f(b)$. This leads us to,

$f(a)(b-a)\leq\int_a^bf(x)\leq f(b)(b-a)$

答:
Here is a proof using Riemann sums.

Since $f$ is Riemann integrable, for any $\epsilon > 0$ there is a partition $P_\epsilon: a = x_0 < x_1 < \ldots < x_n =b$ such that for any choice of tags $\xi_j \in [x_{j-1}, x_j]$, the Riemann sum $S(P_\epsilon,f)$ satisfies

$$\tag{1}\int _a^bf(x) \, dx - \epsilon \leqslant S(P_\epsilon,f) = \sum_{j=1}^n f(\xi_j) (x_j-x_{j-1}) \leqslant \int_a^bf(x) + \epsilon,$$

Assume WLOG that $f$ is monotone increasing (otherwise we can consider $-f$). We have $f(a) \leqslant f(x) \leqslant f(b)$ for all $x \in [a,b]$, and it follows that for any Riemann sum

$$\tag{2}f(a)(b-a) = \sum_{j=1}^n f(a)(x_j-x_{j-1})\leqslant \underbrace{S(P,f)}_{\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1})}  \leqslant \sum_{j=1}^n f(b)(x_j-x_{j-1})\leqslant f(b)(b-a) $$

Supoose $\int_a^b f(x) \,dx > f(b)(b-a)$. Taking $0 < \epsilon < \int_a^b f(x) \, dx - f(b)(b-a)$, it follows from (1) that

$$f(b)(b-a) < \int_a^b f(x) \, dx - \epsilon \leqslant S(P_\epsilon,f)$$

However, as (2) holds for any partition and corresponding Riemann sum we get the contradiction

$$S(P_\epsilon,f) \leqslant f(b)(b-a) < S(P_\epsilon,f)$$

By a similar argument we get a contradiction if $\int_a^b f(x) \,dx < f(a)(b-a)$ and, therefore, we conclude

$$f(a)(b-a) \leqslant \int_a^b f(x) \, dx \leqslant f(b)(b-a)$$


问:
Substitution in $\int \sqrt{1-x^2}\, dx$
Im not asking ‘How to evaluate this integral’, I just want to know why putting while making the u-substitution $$x=\sin(u), dx=\cos(u)$$

This is weird to me because all what i know is that we put $u$ equal to $x$  times something (or whatever you want) in the integral, but here we’ve putted
$x$ equal to $\sin(u)$ .

So why x and not $u$?
and why specifically sine why not cosine or tangent or anything else?

答: Since you want to get the $x$ out of the square root, and $1-x^2$ must be a square of something.Also, $0 \le 1-x^2 \le 1$. So the something should be $\sin u$ or $\cos u$. This is a natural choice. There could be other choices, but this seems the best one so far.


问:
A countable family in $\mathcal P (\mathbb N)$ such that every set in $\mathbb N$ contains some element of the family.
Does there exists a countable family of infinite sets $\{A_n:n\in\mathbb N\}\subset\mathcal P(\mathbb N)$ satisfying the following property:
$$\text{For every infinite set }I\in\mathcal P(\mathbb N),\text{ there is }n\in\mathbb N\text{ such that }A_n\subset I\text{ ?}$$

If we require the family to have cardinality $\mathfrak c$, then the question is trivial, but for the countable case I'm stuck.

答:
Here is a slightly different way to construct a counterexample, one that I find a little simpler. Let $\mathscr{A}=\{A_n:n\in\Bbb N\}$ be a family of infinite subsets of $\Bbb N$. If $n\in\Bbb N$, and distinct $x_k,y_k\in A_k$ have been chosen for each $k<n$, let $X_n=\{x_k:k<n\}$ and $Y_n=\{y_k:k<n\}$. Then let

$$x_n=\min\big(A_n\setminus(X_n\cup Y_n)\big)\,,$$

set $X_{n+1}=X_n\cup\{x_n\}$, let

$$y_n=\min\big(A_n\setminus(X_{n+1}\cup Y_n)\big)\,,$$

and continue. Let $X=\{x_n:n\in\Bbb N\}$ and $Y=\{y_n:n\in\Bbb N\}$; clearly $X$ and $Y$ are disjoint, infinite subsets of $\Bbb N$ such that $x_n\in X\cap A_n$ and $y_n\in Y\cap A_n$ for each $n\in\Bbb N$. Thus, $x_n\in A_n\setminus Y$ and $y_n\in A_n\setminus X$ for each $n\in\Bbb N$, and therefore neither $X$ nor $Y$ contains any member of $\mathscr{A}$.


问:
Volume of solid of revolution by revolving the region $y=x^2$,$x=0$,$y=9$
Find the volume generated by revolving the region shown below:

I have done as follows:
Revolve the rectangle formed by the vertices $(0,0),(3,0),(3,9),(0,9)$ about $X$ axis.
The required volume generated is the volume of cylinder of radius $9$ and height $3$ which is $$V_1=\pi (9)^2(3)=243\pi$$

Now revolve the region formed by $y=x^2$,$x=3$ and $X$ axis about the $X$ axis.
By Disk method the volume is:
$$V_2=\pi \int_{0}^{3}(x^2)^2dx=\frac{243\pi}{5}$$
So the required volume is
$$V_1-V_2=\frac{972\pi}{5}$$
Is there any other approach?

答:
Calculating surface or volumes of revolution can often be done using Guldin's theorems (or Pappus centroid I think in english).

$V=2\pi\times S\times \operatorname{dist}(G,axis)$

Where $S$ is the area of the surface to revolve and $G$ its centroid.

Though using the moments $\begin{cases}M_x=\iint y\mathop{dx}\mathop{dy}\\M_y=\iint x\mathop{dx}\mathop{dy}\end{cases}\quad$ we can get rid of $S$ since $\begin{cases}x_G=\frac 1SM_y\\y_G=\frac 1SM_x\end{cases}$

In the present case, the axis of revolution is $(Ox)$ thus $\operatorname{dist}(G,axis)=y_G$ and the volume is given by $$V=2\pi\times S\times y_G=2\pi\times M_x=2\pi\int_0^3 \int_{x^2}^9 y\mathop{dy}\mathop{dx}=\frac{972\pi}{5}$$

Notice this is the "disk method" referenced in Math Lover's answer.


问:
$\frac{\partial}{\partial y_i}(f(x-y))=-\frac{\partial f}{\partial y_i}((x-y))$
If I have $f:\mathbb{R^n} \rightarrow \mathbb{R}$, a sufficiently regular function. Then how to prove that:

$\frac{\partial}{\partial y_i}(f(x-y))=-\frac{\partial f}{\partial y_i}((x-y))$,

where $y_i$ is the $i^{th}$ component of vector $y \in \mathbb{R^n}$. And the second term means:the partial derivative of $f$ with respect to $y_i$ (applied) to $x-y$

答:
Differentiating $f$ with respect to $y_i$ at the point $(x-y)$ can be done using the chain rule for higher dimension. It states that
$$\frac{\partial}{\partial y_i} (f(x-y)) = \sum_{k=1}^n \frac{\partial f}{\partial y_k}(x-y) \cdot \frac{\partial (x_k-y_k)}{\partial y_i}(y) = -\frac{\partial f}{\partial y_i}(x-y)$$
where the $y_k$ are the variables you can put into $f$. Therefore the $i$-the partial derivative of the function $f \circ (x-y)$ equals the negative of the $i$-th partial derivative of $f$ at the point $(x-y)$ which was to be shown


问:
Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$
Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test?

If I remove the first term I can see $\dfrac{u_n}{u_{n+1}}=\dfrac{(2n+2)(2n+3)}{(2n+1)^2}
\\=\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^2}
\\={\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^{-2}}\\=\left(1+\dfrac{5}{2n}+\dfrac{3}{2n^2}\right)\left(1-\dfrac{1}{n}\ldots\right)\\=1+\dfrac{3}{2n}+O\left(\dfrac{1}{n^2}\right)$

So the series is convergent.

Is it a correcct attempt?

答:
Ratio test is not enough.

Example 1:

$$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$

$$\dfrac{u_n}{u_{n+1}}=\dfrac{\left(\dfrac{1}{n}\right)}{\left(\dfrac{1}{n+1}\right)}=\frac{n+1}{n}=1+\frac{1}{n}$$

Ratio will be $1$ if $n--->\infty$

But we know that S series is divergent.

So the ratio test says us If the ratio is 1.  maybe the series will be convergent. We need more test to determine the result.

I will give you a method how you can approach such series.

$$f(x)=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{x^5}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{x^7}{7}+\cdots$$

$$f'(x)=1+\dfrac{1}{2}x^2+\dfrac{1\cdot 3}{2\cdot 4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\cdots$$

$$f''(x)=x+\dfrac{1\cdot 3}{2}x^3+\dfrac{1\cdot 3}{2\cdot 4}x^5+\cdots$$

$$\int_0^x \frac{f''(x)}{x} dx=x+\dfrac{1}{2}x^3+\dfrac{1\cdot 3}{2\cdot 4}x^5+\cdots$$

$$\int_0^x \frac{1}{x}\int_0^x \frac{f''(t)}{t} dt dx=f(x)$$

$$\int_0^x \frac{f''(x)}{x}dx=xf'(x)$$

$$ \frac{f''(x)}{x}=xf''(x)+f'(x)$$

$$ \int_0^x \frac{f''(x)}{f'(x)} dx=\int_0^x\frac{x}{1-x^2} dx$$

$$ \ln f'(x)=-\frac{1}{2}\ln{(1-x^2)} $$

$$  f'(x)=\frac{1}{\sqrt{1-x^2}} $$

$$  f(x)=\arcsin(x) $$

$$\arcsin(x)=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{x^5}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{x^7}{7}+\cdots$$

$x=1$

$$\arcsin(1)=1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$$

$$\dfrac{\pi}{2}=1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$$

You can use the same method and to show the series $$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$ is divergent.

$$f(x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$

$$f'(x)=1+x+x^2+x^3+x^4\cdots=\frac{1}{1-x}$$
$$\int_0^x f'(x) dx=f(x)=-\ln{(1-x)}$$

$x=1$

$$f(1)=-\ln{0}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$


答:
We have a function
$$f:\quad{\mathbb R}^n\to{\mathbb R}, \qquad u\mapsto f(u)$$
of one vector variable, and form the new function
$$g(x,y):=f(x-y)$$
of two vector variables. Expanded this means
$$g(x_1,\ldots, x_n,y_1,\ldots, y_n):=f(x_1-y_1,\ldots, x_n-y_n)\ .$$

By the chain rule we then have
$${\partial g\over\partial y_i}={\partial f\over\partial u_i}\biggr|_{(x-y)}\cdot(-1)=-f_{.i}(x-y)\ ,$$
where $f_{.i}$ denotes the $i^{\rm th}$ component of the gradient.

Note that writing $((\ldots))$ is the same as $(\ldots)$, so that your equation says $\Phi=-\Phi$ for a certain $\Phi$. This cannot generally be true in your situation. The actual problem you have is one of notation: using the same letter $f$ for various purposes.

Body:
If I have $f:\mathbb{R^n} \rightarrow \mathbb{R}$, a sufficiently regular function. Then how to prove that:

$\frac{\partial}{\partial y_i}(f(x-y))=-\frac{\partial f}{\partial y_i}((x-y))$,

where $y_i$ is the $i^{th}$ component of vector $y \in \mathbb{R^n}$. And the second term means:the partial derivative of $f$ with respect to $y_i$ (applied) to $x-y$


答:
The 'textbook answer' to your question is that making the substitution $x=\sin\theta$ allows us to apply the identity $\cos^2+\sin^2=1$. Any integral of the form
$$
\int f\left(\sqrt{1-x^2}\right) \, dx \, ,
$$
can be attacked by making this substitution. If $x=\sin\theta$, then $\sqrt{1-x^2}=\cos\theta$ and $dx=\cos\theta d\theta$, meaning that the integral is transformed to
$$
\int f(\cos\theta)\cos\theta \, d\theta \, ,
$$
which often makes things simpler. When I was first learning about integration by substitution, I found this explanation unconvincing for three reasons:

If $x=\sin\theta$, then $\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$. And $\sqrt{\cos^2\theta}=|{\cos\theta}|$, not $\cos\theta$.

When defining a variable, is it really correct to write 'let $x=\ldots$' as opposed to 'let $\theta=\ldots$'?

Integration by substitution is meant to be the chain rule in reverse. But here, it doesn't look like we are applying the reverse chain rule.

It is certainly possible to answer these objections, but the full answer is quite lengthy. Let me try to address them as concisely as I can—and I'll provide links if you're interested in the details:

As user170231 has mentioned in the comments, when we make the substitution $x=\sin\theta$, what we are really doing is making the substitution $\theta=\arcsin x$. This means that, by definition, $\theta$ is between $-\pi/2$ and $\pi/2$ radians. And so $\cos\theta$ is positive. In general, to integrate
$$
\int f(g(x)) \, dx \, ,
$$
we can make the substitution $u=g(x)$, provided that $g$ is invertible. Then, we always rewrite this as $x=g^{-1}(u)$. As a shortcut, we say things like 'let $x=\sin u$', where it is understood that we are actually making the substitution $u=\arcsin x$.

I can't offer you a definitive answer to this question, but it does seem to me like it is a mild abuse of notation to define a variable by using it in an expression. However, we all know what was intended, and so this small point can be disregarded.

We are still doing the chain rule in reverse, even though it doesn't look like we are. First, if you are unfamiliar with the links between substitution and the chain rule, then see here. For more details, see here.

Title: Substitution in $\int \sqrt{1-x^2}\, dx$

Body:
Im not asking ‘How to evaluate this integral’, I just want to know why putting while making the u-substitution $$x=\sin(u), dx=\cos(u)$$

This is weird to me because all what i know is that we put $u$ equal to $x$  times something (or whatever you want) in the integral, but here we’ve putted
$x$ equal to $\sin(u)$ .

So why x and not $u$?
and why specifically sine why not cosine or tangent or anything else?


问:
Efficiently finding $k$th derivative of power of rational function
Suppose I have the rational function
$$g(x) = \frac{x+c+2cx}{1-cx}$$
where $c$ is some constant and I want to find the $k$th derivative of $f(x)=(g(x))^n$ at $x=0$. I know that this is feasible for very small $k,n$ using product rule repeatedly since the $k$th derivative of $g$ itself is relatively simple; $g^{(i)}(x)$ can be easily found from writing $g(x)$ as
$$g(x) = 2-\frac1c-\frac{1/c+c+2}{x-\frac1c}$$
which is easy to work with. But when $n,k$ are larger, the product becomes an annoying sum:
$$\left(g(x)^n\right)^{(k)}=\left(ng(x)^{n-1}\cdot g'(x)\right)^{(k-1)}=\left(n(n-1)g(x)^{n-2}\cdot g'(x)+ng(x)^{n-1}\cdot g''(x)\right)^{(k-2)}$$
and it seems to only get worse from there. Obviously we're able to extract a factor of $n$ or something but that doesn't really seem to help. Once it's properly expanded, it should be easy just to substitute in $c$ for $g(0)$ and $-\frac{(-1)(-2)\cdots (-i+1)\left(1/c+c+2\right)}{\left(-\frac1c\right)^i}$ in for $g^{(i)}(0)$, so the finish should not be too bad. I was thinking maybe it's possible to force some "combinatorial" identity. Any tips? Thanks.

答:
You might try using Faa di Bruno's formula.

$$ \dfrac{d^k}{dx^k} g(x)^n = \sum_{\pi \in \Pi} \frac{n!}{(n-|\pi|)!} g(x)^{n - |\pi|}
\prod_{B \in \pi} g^{(|B|)}(x) $$
where $\Pi$ is the set of partitions of $\{1,\ldots,k\}$ of cardinality at most $n$, $|\pi|$ is the cardinality of the partition $\pi$, and $|B|$ is the cardinality of the block $B$.


问:
How to find angle at centroid of triangle by its edge lengths?
I need to write a program that takes length of a triangle's edges and calculates the angle $\angle APB$ ($P$ is the centroid of the triangle). Thanks for any help or clue.

Body:
$$
PA^2=\frac {2b^2+2c^2-a^2}9,\quad PB^2=\frac {2a^2+2c^2-b^2}9\\
\implies \cos(\angle APB)=\frac {PA^2+PB^2-c^2}{2\cdot PA \cdot PB}
=\frac {a^2+b^2-5c^2}{2\sqrt{(2b^2+2c^2-a^2)(2a^2+2c^2-b^2)}}.
$$


答:
Note that the objective of the substitution is to simplify the integrand, i.e. removing the square-root operation. So, any substitution that achieves that is applicable, including

$$x= \sin{t},\>\>\>x=\cos{t},\>\>\>x=\tanh{t},\>\>\>x=\operatorname{sech}{t} \>\>\>etc.$$


问:
Factorial induction proof with recursion, no idea where to go
So I've been stuck on this one problem from my Foundations of Computing class where I'm given a function $T(n)$ and the base cases are defined already and a function is given to find the other $n$ values greater than the base case. This is how it looks.

Consider the following recursively defined function:

$$T(n)=\begin{cases}
3,&\text{if }n=1\\
6,&\text{if }n=2\\
18,&\text{if }n=3\\
T(n-3)(n^3-3n^2+2n),&\text{otherwise.}
\end{cases}$$

Use induction to prove that $T(n)=3(n!)$ for all $n\ge 1$.

I have no clue how to even get started on this. We never covered factorial induction proofs, let alone with recursion in my class so I would really appreciate some help with this.

Thanks :)

答:
There’s nothing special about the fact that a factorial is involved. It’s immediate from the definition of $T$ that $T(n)=3n!$ for $n=1,2,3$; those are your base cases for this strong induction. For the induction step you simply have to use the definition of $T$ to show that if $T(k)=3k!$ for $k=1,\ldots,n-1$, where $n>3$, then $T(n)=3n!$. That definition says that

$$T(n)=T(n-3)(n^3-3n^2+2n)\,,$$

so your task is to use the induction hypothesis and a little algebra to show that the righthand side simplifies to $3n!$. HINT: It will be helpful to factor $n^3-3n^2+2n$.


问:
Composition of two axis-angle rotations
Please note that I am not referring to Euler angles of the form (α,β,γ). I am referring to the axis-angle representation, in which a unit vector indicates the direction axis of a rotation and a scalar the magnitude of the rotation.

Let $(\hat{n_1},\theta_1)$ refer to the first rotation and $(\hat{n_2},\theta_2)$ refer to the second rotation. What is the value of the first rotation followed by the second rotation, in axis-angle representation?

I understand that the composition of two rotations represented by quaternions $q_1$ and $q_2$ is equal to their product $q_2q_1$. Is there a way to find the composition of axis-angle rotations (without having to convert them to quaternions, multiply them, and convert them back to axis-angle) in a similar manner? Is there a simplified formula for this operation?

答:
I'm not sure if this is simpler or more complex.  This reworks the formulae to combine the angles before taking the sin/cos.  I had started with this... https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation#The_composition_of_spatial_rotations and ended up feeding the partials to wolfram alpha, and getting equivalent functions.

https://github.com/d3x0r/stfrphysics#live-demos implement this method; it is equivalent.

given

${Q} = [n,\theta]$ n is a normal vector/axis of rotation

${P} = [n,\theta]$ $\theta$ is the angle of rotation - often positive.

composite result, of Q rotated around P.

$ R_{\theta} = 2 cos^{-1}(\frac { {\cos (\frac {{ Q_\theta} - P_{\theta}} 2)}( 1 - (Q_n \cdot P_n) )  + {\cos (\frac {{ Q_\theta} + P_{\theta}} 2) }(1+(Q_n \cdot P_n)) } 2 ) $

$ R_n = ( Q_n \times P_n ) ({\cos (\frac {{ Q_\theta} - P_{\theta}} 2)}-{ \cos (\frac {{ Q_\theta} + P_{\theta}} 2)   }) + P_n ({\sin (\frac {{ Q_\theta} + P_{\theta}} 2)}+{\sin (\frac {{ Q_\theta} - P_{\theta}} 2)}) + Q_n ({\sin (\frac {{ Q_\theta} + P_{\theta}} 2)}-{\sin (\frac {{ Q_\theta} - P_{\theta}} 2)}) $

-or-

$ A = Q_n \cdot P_n $

$ B = \cos \frac {{ Q_\theta} + P_{\theta}} 2 $

$ C = \cos \frac {{ Q_\theta} - P_{\theta}} 2 $

$ D = \frac { C( 1 - A )  + B(1+A) } 2 $

$ {Result}_{\theta} = 2 \arccos( D ) $

if $ {Result}_{\theta} = 0 $

then the two rotations are co-incidental the axis is left unmodified.

else

$ E = \sin \frac {{ Q_\theta} + P_{\theta}} 2 $

$ F = \sin \frac {{ Q_\theta} - P_{\theta}} 2 $

$ G = ( Q_n \times P_n ) (C-B) + P_n (E+F) + Q_n (E-F) $

$ {Result}_n = \frac G  {||G||} $

$ {Result} = Result_{\theta} {Result}_n  $

Title: Composition of two axis-angle rotations


问:
Prove that Newtons Method works for a concrete polynomial
Can somebody help me with following:
Let $f(x)=x^5-x+1$ and let $x$ be the limit of the Newton series $x_n=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}$.
How do I prove that $x$ is a root of $f$, i.e. $f(x)=0$.

I have tried to proove that $|f(x_N)|<\frac{1}{n}$ for some $N$ big enough, by using the triangle inequality, but without succes.

答:
hint

Assume that $$\lim_{n\to+\infty}x_n=x \in \Bbb R$$

$ f $ and its derivative  $ f' $ are continuous at $ \Bbb R $.

So

$$\lim_{n\to+\infty}x_{n-1}=x$$
$$\lim_{n\to+\infty}f(x_{n-1})=f(x)$$
$$\lim_{n\to+\infty}f'(x_{n-1})=f'(x)$$

and
$$\lim_{n\to+\infty}x_n=x=x-\frac{f(x)}{f'(x)}$$

By MVT, it is easy to check that the root of $ f(x)=0 $ is in $ [-2,-1]$, so $ f'(x)\ne 0$.


问:
Prove this vector identity
Let $f$ and $g$: $\mathbb R^3 \rightarrow \mathbb R$ be $C^{1}$ scalar functions. Prove that</p>

<p>$$ \nabla \left( \frac{f}{g} \right) = \frac{1}{g^2}\left( g\nabla f - f\nabla g \right)$$ $$g \neq 0$$  </p>

答:
The $i$-th component of the gradient is:

$$
\frac{\partial }{\partial x_i} \left(\frac{f}{g}\right) = \left(\frac{\partial f}{\partial x_i} g - \frac{\partial g}{\partial x_i} f\right)\frac{1}{g^2},
$$
thus
\begin{align}\nabla \left(\frac{f}{g}\right) &= \left(\frac{\partial }{\partial x_1} \left(\frac{f}{g}\right), \frac{\partial }{\partial x_2} \left(\frac{f}{g}\right), \frac{\partial }{\partial x_3} \left(\frac{f}{g}\right)\right)
\\
&= \frac{1}{g^2} \left(\frac{\partial f}{\partial x_1} g - \frac{\partial g}{\partial x_1} f , \frac{\partial f}{\partial x_2} g - \frac{\partial g}{\partial x_2} f, \frac{\partial f}{\partial x_3} g - \frac{\partial g}{\partial x_3} f\right)
\\
&=\frac{1}{g^2} \left( \left(\frac{\partial f}{\partial x_1} g,\frac{\partial f}{\partial x_2} g, \frac{\partial f}{\partial x_3} g\right) - \left(\frac{\partial g}{\partial x_1} f,\frac{\partial g}{\partial x_2} f, \frac{\partial g}{\partial x_3} f\right)\right)
\\
& =  \frac{1}{g^2}\left(g\nabla f - f\nabla g\right),
\end{align}
and I believe there should be a parenthesis in your expression.

Title: Prove this vector identity


问:
Why does I^I = 1/5?
I am stuck on the idea $i^i$ equals $1 \over 5$ I have seen proof of it, but I think the explanations don't make sense, I understand the maths itself but what I don't understand is how you assume it is a number but it is the square root of $-1$ and thus is I but also undefined because no numbers squared achieve, therefore you can substitute undefined^undefined, and a number divided by zero is undefined so what we now get is that:
${({5 \over 0})^{5 \over 0}} = 0.2$ which isn't logical so does anyone knows proof of $i^i=0.2$ without assuming it is a number I would imagine you can't but please tell me why you can assume $i$ to be a number?

答:
It is a pretty long story.

It is true that there is no real number $r$ such that $r^2=-1$ hence $-1$ has no square root. We will not contradict that.

But it turns out that if you accept a special symbol denoted $i$ and form expressions like

$$r+is$$ where $r,s$ are two real numbers, you define a whole new arithmetic that has extraordinary and very useful properties.

We define the addition and multiplication as follows:

$(r+is)+(r'+is')=(r+r')+i(s+s')$,

$(r+is)(r'+is')=(rr'-ss')+i(rs'+r's)$.

(as arithmetic on the reals is known, these formulas extend the arithmetic to those special expressions).

In particular, $i^2=i\cdot i=(0+i\cdot1)(0+i\cdot1)=-1+i\cdot0=-1$. There is no contradiction with what is written above, as $i$ is not a real number, just a conventional symbol.

From these definitions, it is possible to define exponentiation between these special expressions. I have to skip the details, as they are very technical. In particular, that leads to the famous "Euler's identity",

$$\color{green}{e^{i\pi}=-1}.$$

From the latter, we can draw

$$e^{i\pi/2}=i,$$

then

$$\log(i)=i\frac\pi2$$

and finally
$$i^i=e^{i\log(i)}=e^{i\cdot i\pi/2}=e^{-\pi/2}.$$

Final remark: those special expressions are known as complex numbers.

The numerous ones who are questioning the "existence" of the complex numbers are on a wrong track. A complex number is just a handy notation for a pair of reals, using the above computation rules. So if you don't question the existence of reals, you have no reason to question that of the complex.

Equivalently,
$$(0,1)^{(0,1)}=\left(e^{-\pi/2},0\right).$$

Title: Why does I^I = 1/5?


问:
What is the image of $f:A\to B\times C\cup\{\emptyset\}$?
Let $f:A\to B\times C\cup\{\emptyset\}$ be any function. My question is simple. Given some $a\in A$, is this function $f:A\to B\times C\cup\{\emptyset\}$ returning either a pair $(a,b)\in A\times B$ or the empty set; or is this function returning a pair $(a,b)\in A\times B\cup\{\emptyset\}$, where $b$ could be the empty set?

答: So the function is written as $f: A \to B \times C \cup \{\emptyset\}$ which means that the range of values is $B \times C \cup \{\emptyset\}$ (because that is the set "right of the arrow"). Now this set of possible values is a union of two sets meaning that the output $f(a)$ for some $a$ is in either one of those two sets (normally it could also be an element of both of the sets in the union, but here the two sets $B \times C$ and $\{\emptyset\}$ are disjoint). Therefore $f(a)$ is going to give you either a pair $(b,c) \in B \times C$ or a value in the set $\{\emptyset\}$ which can only be $f(a) = \emptyset$.


问:
An example of a generalized Cantor set with positive Lebesgue measure
I want to know if there exist a set $ X\subset \mathbb R$ such that $X$ is

$i)$ Perfect

$ii)$ Compact

$iii)$ Has empty interior

$iv)$ Totally disconnected

$v)$ Is not countable

But $X$ has positive Lebesgue measure.

The sets that are defined with the above properties are called generalized Cantor sets.

Please could you tell me how to construct an explicit example?

答:
Yes there is.  Simply modify the construction of the standard ternary set in such a way that you remove an increasing proportion of the remaining intervals with each step.

Note that it's not enough to simply remove a greater proportion than 1/3.  You've got to remove an increasing proportion so that the total length of the removed intervals is less than one.

There's a nice discussion of this in Stromberg's Classical Real Analysis.


问:
Is that $[\mathbb{C} : \mathbb{R}] = 2$ and is that $[\mathbb{C} : \mathbb{Q}]$ infinite?
Since, $\mathbb{C} = \mathbb{R}(i)$ where $i = \sqrt{-1}$.

So, $[\mathbb{C} : \mathbb{R}] = [\mathbb{R}(i) : \mathbb{R}] = 2$.

Which $\mathbb{C}$ is a finite field extention of $\mathbb{R}$.

Since, there are infinite number of algebric numbers in $\mathbb{C}$.

So, $[\mathbb{C} : \mathbb{Q}]$ is infinite.

Which means $\mathbb{C}$ is an infinite field extention of $\mathbb{Q}$.

Are they correct ?

答:
Remember that if $[F : K]=n\in\mathbb{N}$, $F$ is a vector space over $K$, with dimension $n$. So if $[\mathbb{R} : \mathbb{Q}]=n$, then $\mathbb{R}=\mathbb{Q}\times ... \times\mathbb{Q}$ $n$ times.
That would imply that $\mathbb{R}$ is countable.

Thus,$[\mathbb{R} : \mathbb{Q}]=\infty$ and $[\mathbb{C} : \mathbb{Q}]=\infty$.


问:
If $m$ and $n$ are integers with $\gcd(m,n) = 1$, prove that $\sigma(mn)= \sigma(m)\sigma(n)$.
If $m$ and $n$ are integers with $\gcd(m,n) = 1$, prove that $\sigma(mn)= \sigma(m)\sigma(n)$. I am thinking about using the formula for $\sigma(p^k)$ where $p$ is prime.

It follows from the geometric series that

$$\sigma(p^k) = 1 + p + p^2 +\cdots+p^k = \frac{p^{k+1}-1}{p-1}$$

Any Help?

答:
Let $m = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ and $n = q_1^{d_1}q_2^{d_2}...q_i^{d_i}$ for $p, q$ prime and $e, d \in \mathbb{Z}$.

Then we have
$$\sigma(mn) = \sigma(p_1^{e_1}p_2^{e_2}...p_k^{e_k}*q_1^{d_1}q_2^{d_2}...q_i^{d_i})$$

We know the divisors of $p_1^{e_1}$ are $1, p_1, p_1^2, p_1^3,..., p_1^{e_1}.$ This holds for any prime to any power.

Since the gcd$(m,n) = 1$, we also know $m$ and $n$ share no common divisors

So we have $$=[1+p_1+p_1^2+...+p_1^{e_1}][1+p_2+p_2^2+...+p_2^{e_2}]...[1+p_k+p_k^2+...+p_k^{e_k}] * [1+q_1+q_1^2+...+q_1^{d_1}][1+q_2+q_2^2+...+q_2^{d_2}]...[1+q_i+q_i^2+...+q_i^{d_i}]$$
The first half of the statement above is just the divisors of $m$ and the second half is the divisors of $n$

Then
$$= \sigma(m)\sigma(n) $$

The proof is bidirectional


问:
Why is the distribution of values of this random matrix product (seemingly) independent of dimension?
I'm investigating the behavior of the value $|x A y|$ where $x, y \in \mathbb{C}^N$ have unit 2-norm and are uniformly sampled from the unit ball, and $A \in \mathbb{C}^{N \times N}$ has elements sampled from the complex standard normal distribution.

When I plot an empirical histogram of the magnitude of this product for varying dimension $N$, it seems that the distribution of values is independent of $N$. For example, in Julia:

```<code>julia> using Random, Plots, LinearAlgebra
julia> CF = Complex{Float32}
julia> histogram(norm.([normalize(randn(CF, n))' * randn(CF, (n, n)) * normalize(randn(CF, n)) for _ in 1:2000, n in [1, 8, 64, 512]]), layout = (4, 1))
</code>```

gives this result:

Is there a proof or at least an intuition on why this would be the case? I tried reading a bit on random matrix theory but it was a bit too advanced for me and I'm guessing what's going on here isn't that complicated.

答:
Double-check my arithmetic, but here's the basic reason for what you observed.

Write$$x=a+ib,\,y=c+id,\,A=U+iV$$with $a,\,b,\,c,\,d\in\Bbb R^N,\,U,\,V\in\Bbb R^{N\times N}$ so $a^2+b^2=c^2+d^2=1$, with vectors in $\Bbb R^N$ squared in the sense $z^2:=z\cdot z$. It suffices to show that if we fix $a,\,b,\,c,\,d$ the conditional distribution of$$|x^\dagger Ay|^2=[\color{red}{a^TUc+b^TUd-a^TVd+b^TVc}]^2+[\color{blue}{a^TUd-b^TUc+a^TVc+b^TVd}]^2$$is $N$-independent. With a value of $\sigma^2$ that depends on how you normalize complex Gaussians (I'll leave you to work it out), $U,\,V\sim\mathcal{N}(0,\,\sigma^2I_N)$ are independent. The red vector is $N(0,\,R\sigma^2)$-distributed, with$$R:=\operatorname{tr}[(ac^T+bd^T)(ca^T+db^T)+(bc^T-ad^T)(cb^T-da^T)].$$You can treat the blue vector similarly. So $|x^\dagger Ay|$ is an $N$-independent multiple of a $\chi_2$ distribution, which peaks at $1$, roughly in line with your empirical results (although the fact that $a,\,b,\,c,\,d$ are stochastic may cause the peak to drift somewhat). (It's also a special case of the Rademacher distribution.)


问:
Showing sequence convergence
Let $\{a_n\}$ for $n \geq 1$ be a sequence of real numbers defined by the recurrence relation $a_1 > 3$ and $\frac{a_{n+1} \cdot (a_n + 1)}{a_n + 9} = 1$. Prove that

(i) $\{a_{2n}\}$ for $n \geq 1$ is increasing and bounded above.

(ii) $\{a_{2n - 1}\}$ for $n \geq 1$ is decreasing and bounded below.

(iii) If $\lim_{n \rightarrow \infty} a_{2n} = s$ and $\lim_{n \rightarrow \infty} a_{2n-1} = t$ then $s = t$ (in other words, both sequences converge to the same number), and then deduce that $\{a_n\}$ converges to $3$.

My attempt:

(i) You can see that $a_{2n} = \frac{a_{2n-1}+9}{a_{2n-1}+1}$.

Also, a careful look would imply that it is bounded above by $3$, which can be shown by using induction:

Goal. Show that $a_{2n} \leq 3$ $\forall n\geq 1$.

Base. Suppose $n = 1$. You will see that $a_{2} - 3 = \frac{a_{1}+9}{a_{1}+1} - 3 = \frac{-2a_1 + 6}{a_{1}+1}$. But the denominator is bigger than $0$ and the numerator is less than $0$ because $a_1 > 3$. This shows that $a_{2} - 3 \leq 0$ and $a_{2} \leq 3$.

Inductive. Assume that $a_{2n} \leq 3$. We show that $a_{2(n+1)} \leq 3$.

$$a_{2n + 2} - 3 = \frac{a_{2n+1}+9}{a_{2n+1}+1} - 3 = \frac{\frac{a_{2n} + 9}{a_{2n} + 1} + 9}{\frac{a_{2n} + 9}{a_{2n} + 1} + 1} - 3 = \frac{5a_{2n} + 9}{a_{2n} + 5} - 3 = \frac{2a_{2n} - 6}{a_{2n} + 5} \leq 0$$

Therefore $a_{2n}$ is bounded above by $3$.

Goal. Show that $a_{2n} \leq a_{2(n+1)}$ $\forall n\geq 1$.

Base. Suppose $n = 1$. Then $a_{2} \leq 3$ so

$$a_{2} - a_{4} = a_{2} - \frac{5a_{2} + 9}{a_{2} + 5} = \frac{a_{2}^2 - 9}{a_{2} + 5} \leq 0$$

Inductive. Assume that $a_{2n} \leq a_{2n+2}$. We show that $a_{2n+2} \leq a_{2n + 4}$.

$$a_{2n + 2} - a_{2n + 4} =a_{2n + 2} - \frac{a_{2n+3}+9}{a_{2n+3}+1} = a_{2n + 2} - \frac{\frac{a_{2n+2} + 9}{a_{2n+2} + 1} + 9}{\frac{a_{2n+2} + 9}{a_{2n+2} + 1} + 1}  = a_{2n + 2} - \frac{5a_{2n+2} + 9}{a_{2n+2} + 5} = \frac{a_{2n+2}^2 - 9}{a_{2n+2} + 5} \leq 0$$

This follows from the fact that $a_{2n + 2} \leq 3$ as shown in the bounded above induction proof.

(ii) You can see that $a_{2n-1} = \frac{a_{2n-2}+9}{a_{2n-2}+1}$.

As before it looks to me that the $a_{2n-1}$ is bounded below by $3$. Once again this can be shown by using induction:

Goal. Show that $a_{2n-1} \geq 3$ $\forall n\geq 1$.

Base. Suppose $n = 1$ then $a_{1} > 3$ (this has been directly given in the problem).

Inductive. Assume that $a_{2n-1} \geq 3$. We show that $a_{2(n+1)-1} \geq 3$.

$$a_{2n + 1} - 3 = \frac{a_{2n}+9}{a_{2n}+1} - 3 = \frac{\frac{a_{2n-1} + 9}{a_{2n-1} + 1} + 9}{\frac{a_{2n-1} + 9}{a_{2n-1} + 1} + 1} - 3 = \frac{5a_{2n-1} + 9}{a_{2n-1} + 5} - 3 = \frac{2a_{2n-1} - 6}{a_{2n-1} + 5} \geq 0$$

Goal. Show that $a_{2n-1} \geq a_{2n+1}$ $\forall n\geq 1$.

Base. Suppose $n = 1$. It is clear that $a_{1} > 3$ and $a_3 = \frac{5a_{1} + 9}{a_{1} + 5}$. So

$$a_{1} - a_{3} = a_{1} - \frac{5a_{1} + 9}{a_{1} + 5} = \frac{a_{1}^2 - 9}{a_{1} + 5} > 0$$

Inductive. Assume that $a_{2n-1} \geq a_{2n+1}$. We show that $a_{2n+1} \geq a_{2n + 3}$.

$$a_{2n + 1} - a_{2n + 3} =a_{2n + 1} - \frac{a_{2n+2}+9}{a_{2n+2}+1} = a_{2n + 1} - \frac{\frac{a_{2n+1} + 9}{a_{2n+1} + 1} + 9}{\frac{a_{2n+1} + 9}{a_{2n+1} + 1} + 1}  = a_{2n + 1} - \frac{5a_{2n+1} + 9}{a_{2n+1} + 5} = \frac{a_{2n+1}^2 - 9}{a_{2n+1} + 5} \geq 0$$

This follows from the fact that $a_{2n + 1} \geq 3$ as shown in the bounded below induction proof.

(iii) From the previous parts, it is clear that the two sequences $a_{2n}$ and $a_{2n-1}$ converge to some limit. Let $\lim_{n \rightarrow \infty} a_{2n} = s$ and $\lim_{n \rightarrow \infty} a_{2n-1} = t$.

$$s = \lim_{n \rightarrow \infty} a_{2n} = \lim_{n \rightarrow \infty} a_{2(n+1)} = \lim_{n \rightarrow \infty} \frac{5a_{2n} + 9}{a_{2n} + 5} = \frac{5s + 9}{s + 5}$$

$s^2 = 9 \implies s = 3$ or $s = -3$.

$$t = \lim_{n \rightarrow \infty} a_{2n-1} = \lim_{n \rightarrow \infty} a_{2n+1} = \lim_{n \rightarrow \infty} \frac{5a_{2n-1} + 9}{a_{2n-1} + 5} = \frac{5t + 9}{t + 5}$$

$t^2 = 9 \implies t = 3$ or $t = -3$.

But $s, t > 0$? (not sure about this).

So $s = t = 3$.

Since two subsequences of $\{a_{n}\}$ converge to the same number $3$, $\{a_{n}\}$ must also converge to $3$ (not sure about this either).

QED.

Is all this correct? I am not sure if my reasoning that $\{a_{n}\}$ converges to $3$ is correct. I was told by somebody that I should "consider expressing $a_{n+2}$ as a function of $a_{n}$ and show that $a_{2} < 3$." Not sure how that would work though.

Any assistance is much appreciated.

Title: Showing sequence convergence

Body:
hint

Let $$f(x)=\frac{x+9}{x+1}=1+\frac{8}{x+1}$$

$$g(x)=f(f(x))=\frac{5x+9}{x+5}$$

If $ a_{2n}=v_n$, we have
$$v_{n+1}=g(v_n)$$

So, if $ (v_n) $ goes to $ L $, then by continuity
$$g(L)=L=\frac{5L+9}{L+5}$$
and
$$L=\pm 3$$

the same for $ (a_{2n+1})$.


问:
How can I solve this maximization problem with inequality constraint on the Frobenius norm?
$$
\max_{C\in \mathbb{S}^n} x^TCx\\
s.t. \|C\|_{tr}^2\leq b
$$
where $\|C\|_{tr}^2=tr(C^TC)=tr(CC)$, $C$ is a symmetric matrix.

答:
Through some calculations, I have an idea as follows:
$$
L=x^TCx-\lambda(\|C\|_{tr}^2-b)
$$
Calculating $\frac{\partial L}{\partial C}$ and $\frac{\partial L}{\partial \lambda}$, we have
$$
\frac{\partial L}{\partial C}=xx^T=\lambda C\\
\frac{\partial L}{\partial \lambda}=tr(CC^T)=b
$$
that is
$$
C=\frac{xx^T}{\lambda}
$$
plugging the term into the second equation, we have
$$
tr(xx^T/\lambda)=\sqrt{b}\\
tr(xx^T/\lambda)=1=\sum{x_i^2}/\lambda\\
\lambda = \|x\|^2=\sum{x_i^2}
$$
Hence, the solution is
$$
C^*=\sqrt{b}*\frac{xx^T}{\|x\|^2}
$$

OwnerDisplayName: Liangyu Min


答:
Rephrasing slightly, given vector ${\rm a} \in \Bbb R^n \setminus \{ \Bbb 0_n \}$ and scalar $b >0$,

$$\begin{array}{ll} \text{maximize} & {\rm a}^\top {\rm X} \, {\rm a}\\ \text{subject to} & \| {\rm X} \|_{\text{F}} \leq b\end{array}$$

Using the cyclic property of the trace, the Cauchy-Schwarz inequality and $\| {\rm a} {\rm a}^\top \|_{\text{F}} = \| {\rm a} \|_2^2$,

$${\rm a}^\top {\rm X} \, {\rm a} = \mbox{tr} \left( {\rm a}^\top {\rm X} \, {\rm a} \right) = \mbox{tr} \left( {\rm a} {\rm a}^\top {\rm X} \right) = \langle {\rm a} {\rm a}^\top, {\rm X} \rangle \leq \| {\rm a} {\rm a}^\top \|_{\text{F}} \underbrace{\| {\rm X} \|_{\text{F}}}_{\leq b} \leq b \| {\rm a} \|_2^2$$

where the maximum is attained at

$$\bar{{\rm X}} := b \underbrace{\frac{{\rm a} {\rm a}^\top}{\,\,\,\| {\rm a} {\rm a}^\top \|_{\text{F}}}}_{=: {\rm P}_{\rm a}} = b \left( \frac{{\rm a}}{\,\,\,\| {\rm a} \|_2} \right) \left( \frac{{\rm a}}{\,\,\,\| {\rm a} \|_2} \right)^\top = b \, {\rm P}_{\rm a}$$

where ${\rm P}_{\rm a}$ is a rank-$1$ projection matrix.

Title: How can I solve this maximization problem with inequality constraint on the Frobenius norm?

Body:
$$
\max_{C\in \mathbb{S}^n} x^TCx\\
s.t. \|C\|_{tr}^2\leq b
$$
where $\|C\|_{tr}^2=tr(C^TC)=tr(CC)$, $C$ is a symmetric matrix.

OwnerDisplayName: Rodrigo de Azevedo


问:
Substitution in $\int \sqrt{1-x^2}\, dx$
Im not asking ‘How to evaluate this integral’, I just want to know why putting while making the u-substitution $$x=\sin(u), dx=\cos(u)$$

This is weird to me because all what i know is that we put $u$ equal to $x$  times something (or whatever you want) in the integral, but here we’ve putted
$x$ equal to $\sin(u)$ .

So why x and not $u$?
and why specifically sine why not cosine or tangent or anything else?

答:
The substitution is often driven by the domain of definition of the function under the integration symbol, this is because to be able to conserve the value of the integral, the change of variable must be bijective.

In our case for $\sqrt{1-x^2}$ to be well defined we must have $x\in[-1,1]$

So the choice of a function with a range in $[-1,1]$ is compulsory, there are several choices, for instance $\sin(x),\cos(x),\tanh(x)$.

Theoretically all would give the same end result, but some may lead to a nice result after substitution, for instance here we have the relations $\begin{cases}1-\tanh(x)^2=\frac 1{\cosh(x)^2}\\\cos(x)^2+\sin(x)^2=1\end{cases}$

In both cases $\sqrt{1-x^2}$ will transform to $\sqrt{(something)^2}$ and will simplify, but $\sin,\cos$ will result in rather nice $\int \cos^2,-\int \sin^2$ integrals while the last change will result in a less nice $\int \frac 1{\cosh^3}$.

Although notice that in the end, if you lead calculations to the end:

the $\sin$ change will result in $\, \frac 12x\sqrt{1-x^2}+\frac 12\arcsin(x)$

the $\tanh$ change will result in $\, \frac 12x\sqrt{1-x^2}+\arctan(\frac{x+1}{\sqrt{1-x^2}})$

It may appear different, but they just differ by a constant value $\frac{\pi}4$, so the choice of the variable substitution doesn't change the result, just take the one that leads to the simpler calculation.

Here are two other examples
$\displaystyle \int x\sqrt{\frac{x-1}{x+1}}\mathop{dx}\, $ and $\, \displaystyle \int \frac{x^4}{\sqrt{x^2-9}}\mathop{dx}$

Notice that for the first one, we need $x-1\ge 0\implies x\ge 1$ and for the second one $x\ge 3$ or $\frac x3\ge 1$.

Both suggest a function whose range is $[1,+\infty)$, and we know that $\cosh$ is such a function, also since it is a trig function the relation $\cosh^2-\sinh^2=1$ also leads to nice simplifications. And indeed, such a substitution allows to solve both integrals effectively.


问:
For a finite structure $\mathfrak A$ in an *infinite* language, is $Th \mathfrak A$ decidable?
In Enderton's logic, Theorems 26C and 26D, the assumption is that the first order language is finite.

Theorem 26C: For a finite structure $\mathfrak A$ in a finite language, $Th \mathfrak A$ is decidable.

Theorem 26D: For a finite language, $\{\sigma | \sigma \text{ has a finite model} \}$ is effectively enumerable.

Is a finite language a necessary assumption in these two theorems? So long as the universe of some structure is finite, one could build a tree from the finite sentence $\sigma$ and evaluate all the leaves with the values from the universe.

If a sentence of the form $\forall c \phi(c)$, where c referred to constants is allowed, then a finite language would be a necessary assumption. But I believe that $\forall$ can only range over variables.

答:
Note that in order to make sense of computability-flavored questions around infinite languages, we need to think of symbols as being represented by natural numbers in some reasonable way. This is something we can ignore when the language is finite, but with infinite languages it's important.

Theorem $26D$ makes no real use of the finiteness of the language. This is because every sentence can only use finitely many symbols. Really the only way complexity can enter the picture is if the language we're looking at is itself of high complexity, in which case the set of sentences in that language itself (not even thinking about the set of sentences in that language with finite models, yet) will be of high complexity. But this has nothing to do with the finite model property. A bit more precisely, as long as $\Sigma$ is a computable language, the set of $\Sigma$-sentences which have finite models is c.e. - given a computable language $\Sigma$ and a $\Sigma$-sentence $\sigma$, let $\Sigma_0$ be the finite fragment of $\Sigma$ which actually appears in $\sigma$ and search through the finite $\Sigma_0$-structures for one satisfying $\sigma$.

Theorem $26C$, by contrast, absolutely does rely on a finite language. Take for example a language with a unary relation symbol $U_i$ for each $i\in\mathbb{N}$ (note that this language is computable!). For each set $X\subseteq\mathbb{N}$, let $\mathfrak{M}_X$ be the structure whose universe consists of a single element, such that $U_i$ holds on that element iff $i\in X$. Then $Th(\mathfrak{M}_X)$ computes $X$ since we have $$i\in X\iff\mathfrak{M}_X\models\exists xU_i(x).$$ So now just pick some really complicated $X$.

Informally speaking, the infinite language lets us "access" each of the bits of $X$ (using one symbol per bit), and so code the entire complexity of $X$ into the theory of the resulting structure. If we tried to do this with only finitely many $U_i$s, we'd only get finitely much information about $X$, which - regardless of how complicated $X$ is - would be computable. And the fact that every finite structure in a finite language has a computable theory demonstrates that there's no clever way around this.


问:
What is $\lim \int_{\mathbb{R}} \frac{1}{n} \mathrm{d}m$?
I have this probably silly doubt about what is $\displaystyle\lim_{n \to \infty} \int_{\mathbb{R}} \dfrac{1}{n} \mathrm{d}m$.

Here I have the Lebesgue integral.

If I'm calculating correctly, each integral in the sequence is $\infty$, so I have a constant sequence of $\infty$, and a constant sequence converges to any of its elements. (?) (This may fail for infinity, but I don't know how to verify.)

Looking at other way, I can take $\dfrac{1}{n}$ out of the integral, and $\int_{\mathbb{R}^n}1\mathrm{d}m=\infty$, but $\lim \frac{1}{n} = 0$, so I have $0.\infty$, which in measure theory is $0$.

I'm really confused, any help would be appreciated.

答:
Even if you take the $\frac1n$ out of the integral, you have
$$
\lim_{n\to\infty}\int_{\Bbb R^n}\frac1ndm
=\lim_{n\to\infty}\left(\frac1n\int_{\Bbb R^n}dm\right)
=\lim_{n\to\infty}\frac\infty n
=\lim_{n\to\infty}\infty
=\infty.
$$

Rewriting $\lim_{n\to\infty}\left(\frac1n\int_{\Bbb R^n}dm\right)$ as $(\lim_{n\to\infty}\frac1n)(\lim_{n\to\infty}\int_{\Bbb R^n}dm)$ is only valid if the resulting product (in this case $0\times\infty$) is not indeterminate. See: when is the limit of a product the product of the limits?


问:
Doubt over a term in the definition
I was studying a book Handbook of Product Graphs by Richard Hammack, Wilfried  Imrich, Sandi Klavžar. There is a definition of Antipodal Graph.

A graph $G$ is called antipodal if there exists a vertex $v$ to any vertex $u$ $\in$ $V(G)$ such that  $V(G) = I(u,v)$.

What does $I(u,v)$ here stands for.

What I know about the antipodal graph of a graph $G$ is that its a graph containing vertices of $G$ and those vertices are adjacent in antipodal graph whose distance equals diameter of $G$.
Thanks a lot for help.

答: It’s defined near the top of page $9$, according to the Look Inside feature at Amazon: $I(u,v)$, the interval between $u$ and $v$, is the set of vertices lying on shortest paths between $u$ and $v$, including the vertices $u$ and $v$ themselves. For example, an even cycle $C_{2n}$ is antipodal, because if $u$ is any vertex, we may take $v$ to be the diametrically opposite vertex at distance $n$ from $u$, and $I(u,v)$ will then be the whole graph: the two paths from $u$ to $v$ are both of minimum length, and every vertex of $C_{2n}$ lies on one of them.

Title: Doubt over a term in the definition


问:
Differentiating Lebesgue Decomposition of a signed measure
Let $\{\mu_n\}$ be measures on $(X,M)$ with $\sup\{\mu_n\}(X) \leq M < \infty$.

Let $\mu = \sum_{n \geq 1 } 2^{-n} \mu_n$

Let $\nu$ be a $\sigma-$finite measure.

Suppose $\mu_n= \mu_{n,0} + \mu_{n,1}$ is the Lebesgue decomposition of $\mu_n$ with respect to $\nu$.

Find the Lebesgue decomposition of $\mu = \mu_0 +\mu_1$ and $\frac{d\mu_0}{d\nu}$.

My attempt:

By Lebesgue's decomposition theorem, we know that $\mu_{n,0} \ll \nu$ and $\mu_{n,1} \perp \nu$.

By breaking down the definitions, this means that

if $\nu(E) = 0$ ,then $\mu_{n,0}(E)=0$. So $\mu_0(E) := \sum_{n \geq 1 } 2^{-n} \mu_{n,0}(E)=0.$

There exists disjoint $A,B \in M$ such that $A \cup B=X$ and $\mu_{n,1}(A)=0$ and $\nu(B)=0$. So $\mu_1(A):=  \sum_{n \geq 1 } 2^{-n} \mu_{n,1}(A)=0.$ And $\mu_1 \perp \nu$

Are we done finding the Lebesgue Decomposition here? I feel like just wrote down the definitions, not sure if I missed anything.

Now to find $\frac{d\mu_0}{d\nu}$ (the part I'm stuck on), I showed that $\mu$ is a finite measure. I don't know if Radon Nikodym Theorem is relevant here. By definition, we have $d \mu_0 = \frac{d\mu_0}{d\nu} d\nu$. Not sure how to proceed from here.

Thank you in advance.

答:
You have an issue in 2. The $A,B$ actually depend on $n$, so your proof does not work. Fix this by taking $A_n$ and $B_n$ and noting that $\nu(\cup_n B_n) = 0$ and $\mu_1(\cap_n A_n) = 0$.

As for $\frac{d\mu_0}{d\nu}$, for any positive measurable $f$,
$$\int f d\mu_0 = \sum_n2^{-n}\int f d\mu_{n,0} =\sum_n2^{-n}\int f \frac{d\mu_{n,0}}{d\nu} d\nu $$
$$
= \int f \left(\sum_n 2^{-n} \frac{d\mu_{n,0}}{d\nu}\right)d\nu.
$$
Thus $\frac{d\mu_0}{d\nu} = \sum_n 2^{-n} \frac{d\mu_{n,0}}{d\nu}$. In general Radon Nikodym derivatives add and multiply by constants the same way that measures do.


答:
Another hint.  If we know
$$
\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C
\tag1$$
then it is not too hard to write your integral in terms of $(1)$.
Instead of fiddling around with your integral to get $(1)$ somehow, we can just substitute $x = \sin\theta$ and let the algebra do it for us.


问:
Define a relation $R$ on $\mathbb{Z}$ by $aRb$ iff $3a - 5b$ is even. Prove $R$ is an equivalence relation and describe equivalence classes.
I think I got most of the proof, but feel free to critique anything you would like :).

First, notice that $aRa$ means $3a - 5a = a(3 - 5) = 2(-a)$. This is even, thus, $R$ is reflexive.

Second we show that $aRb \implies bRa$. Well, $aRb$ means that $3a - 5b$ is even. Then we have $3a - 5b = 2k$ for some integer k. Rearrange like so
$$
\begin{equation} \begin{split}
3a - 5b &= 2k \\
-5b &= 2k - 3a \\
-5b &= 2(k - a) + (-a)
\end{split} \end{equation}
$$
Since, we stated that the LHS and RHS were both even in the beginning and still are (i think?), this forces $a$ and $b$ to be even. Thus, $3b - 5a$ is even because we are dealing with two even numbers. Therefore $bRa$ which means $R$ is symmetric.

Now for transitive. We have show that $(aRb \wedge bRc) \implies aRc$. Since $aRb \wedge bRc$, we have $3a - 5b$ even and $3b - 5c$ even. Note when we add, the sum is even. Observe
$$
(3a - 5b) + (3b - 5c) \\
3a - 2b - 5c \\
2(-b) + (3a - 5c)
$$
$2(-b)$ is even and this forces $3a - 5c$ to be even. Therefore $R$ is transitive. QED.

For the equivalences classes we know that if $a$ is even we have that $b$ must be even. Similarly, if $a$ is odd then $b$ is odd. Therefore, the equivalence classes are the set of odd integers and the set of even integers.

My question is on proving that $R$ is symmetric. Is it mathematically sound?

答:
Suppose that $a\sim b$.  That is to say, $3a-5b$ is even.  That is to say, $3a-5b$ is equal to $2$ times some integer, we'll call it $k$ so $3a-5b=2k$

We ask whether or not this implies that $b\sim a$, that is if $3b-5a$ can be written as $2$ times an integer as well (note, not necessarily the same integer as before)

$$\begin{array}{l|l}~~~~3b-5a&\text{original}\\=3b+0-5a+0&\text{add zero twice}\\=3b+(-8b+8b)-5a+(8a-8a)&\text{replace zeroes}\\=(3b-8b)+8b+(-5a+8a)-8a&\text{adjust parentheses}\\=(3a-5b)-8a+8b&\text{simplify and rearrange}\\=2k-8a+8b&\text{use hypothesis}\\=2(k-4a+4b)&\text{factor out two}\end{array}$$

This shows that $3b-5a$ can also be written as $2$ times an integer and so is also even.

Similarly, for transitivity, we suppose $3a-5b$ is even and $3b-5c$ is even and we ask about $3a-5c$.

$3a-5c = 3a+0-5c=3a-5b+3b+2b-5c = (3a-5b)+(3b-5c)+2b$ is the sum of three even numbers and thus even as well.


答: In measure theory, there are some important theorems that justify interchange $\lim \int = \int \lim$.  But the hypotheses of those theorems are important, too.  Your example does not satisfy the needed hypotheses.

Title: What is $\lim \int_{\mathbb{R}} \frac{1}{n} \mathrm{d}m$?

Body:
I have this probably silly doubt about what is $\displaystyle\lim_{n \to \infty} \int_{\mathbb{R}} \dfrac{1}{n} \mathrm{d}m$.

Here I have the Lebesgue integral.

If I'm calculating correctly, each integral in the sequence is $\infty$, so I have a constant sequence of $\infty$, and a constant sequence converges to any of its elements. (?) (This may fail for infinity, but I don't know how to verify.)

Looking at other way, I can take $\dfrac{1}{n}$ out of the integral, and $\int_{\mathbb{R}^n}1\mathrm{d}m=\infty$, but $\lim \frac{1}{n} = 0$, so I have $0.\infty$, which in measure theory is $0$.

I'm really confused, any help would be appreciated.


问:
Subspace consisting of normal operators
$V$ is an $n$-dimensional inner product space. Let $L=L(V,V)$ be a vector space of all linear operators on $V$, and let $T\in V$ be a normal operator. If the char. poly of $T$ splits, show that there is an $n$-dimensional subspace of $L$ containing $T$ and consisting of normal operators. Unsure about how the characteristic polynomial of a normal operator here is relevant.

答: Hint: Consider the subspace spanned by the powers of $T$ (including the identity matrix, which is treated as the "$0$th power").

Title: Subspace consisting of normal operators


问:
$\sum_{n=1}^{\infty}\frac{n+1}{n^2+\sqrt{n}+n^p}$ diverges for what p?
So, here's question: For what values of $p$ does $$\sum_{n=1}^{\infty}\frac{n+1}{n^2+\sqrt n+n^p}$$ diverge? I think it's using a test to compare to $\sum \frac{1}{n}$ and it diverges for $p<3$?

答:
You have the right idea, but you are incorrect to assume the $p$ test only works for $p\in\mathbb{N}$. In fact, the series will diverge for $p\leq 2$ and converge for $p>2$.

Proof: Let $p\leq 2$. Then

$$\frac{n+1}{n^2+\sqrt n+n^p}>\frac{n+0}{n^2+n^2+n^2}=\frac{1}{3n}$$

Thus

$$\sum_{n=1}^\infty \frac{n+1}{n^2+\sqrt n+n^p}\geq \sum_{n=1}^\infty \frac{1}{3n}=\infty$$

For $p>2$ We have that

$$\frac{n+1}{n^2+\sqrt n+n^p}<\frac{n+n}{0+0+n^p}=\frac{2}{n^{p-1}}$$

Since $p-1>1$ we have that

$$\sum_{n=1}^\infty \frac{n+1}{n^2+\sqrt n+n^p}\leq \sum_{n=1}^\infty\frac{2}{n^{p-1}}<\infty$$


问:
Solving for $U$,$\Sigma$ and $V$ in SVD of a matrix
Knowing from theory that fro a matrix $A$ we $A = U\Sigma V^{T}$

I want to solve for $U$,$V$ and $\Sigma$.
My effort is the following but I don't if I am correct.
\begin{align*}
A &= U\Sigma V^{T} \\
U^{T}A&=  \Sigma V^{T}\\
U^{T}&= \Sigma V^{T}A^{-1}\\
U&= \Sigma^{-1}V A
\end{align*}
and

\begin{align*}
A &= U\Sigma V^{T} \\
A\Sigma^{-1}&=  UV^{T}\\
\Sigma^{-1}&= A^{-1} U V^{T}\\
\Sigma&= U^{T}A V
\end{align*}

\begin{align*}
A &= U\Sigma V^{T} \\
A V  &= \Sigma  U \\
V &=\Sigma U^{T}A ^{-1}
\end{align*}

Along with $A^{T}=V\Sigma U^{T} $
\begin{align*}
A&=USV^{T}  \\
A^{T}A &= V\Sigma U^{T}U\Sigma V^{T} \\
A^{T}A &= V\Sigma^{2}V^{T} \\
A^{T}AV &= V\Sigma^{2}
\end{align*}

答: I'd be really careful with $A^{-1}$, since the SVD decomposition exists for matrices with different dimensions. The third section is what tells em the morst information, mainly that $V$ contains the eigenvectors of $A{*} A$. If you try to simplify $A A{*}$, you'll find similar information about $U$.


问:
Specific Case For Jordan Block
Let $F$ be a field, and take $\lambda\in F$ with $n\geq 1$. Let $A = (a_{ij})\in M_n(F)$ be an upper triangular matrix, where $a_{ii}=\lambda$ for all $i$. Show that the Jordan Canonical Form of $A$ has a single block if and only if $a_{12}a_{23}....a_{(n-1)(n)}$ is nonzero. I'm unsure of how to use the fact that $A$ is upper triangular; would Schur's lemma be helpful here?

答: Hint: If $J$ is the Jordan form of $A$, then $A - \lambda I$ and $J - \lambda I$ will have the same rank.

Title: Specific Case For Jordan Block


问:
Prove this vector identity
Let $f$ and $g$: $\mathbb R^3 \rightarrow \mathbb R$ be $C^{1}$ scalar functions. Prove that</p>

<p>$$ \nabla \left( \frac{f}{g} \right) = \frac{1}{g^2}\left( g\nabla f - f\nabla g \right)$$ $$g \neq 0$$  </p>

答:
I will assume $f, g : \mathbb{R}^3 \to \mathbb{R}$.

We have

$$\nabla\bigg(\frac{f}{g}\bigg) = \bigg\langle \frac{\partial f/g}{\partial x},\frac{\partial f/g}{\partial y}, \frac{\partial f/g}{\partial z} \bigg \rangle $$

By the quotient rule, the above becomes

$$\bigg \langle \bigg(\frac{\partial f}{\partial x} g - \frac{\partial g}{\partial x}f\bigg)\frac{1}{g^2},\bigg(\frac{\partial f}{\partial y} g - \frac{\partial g}{\partial y}f\bigg)\frac{1}{g^2},\bigg(\frac{\partial f}{\partial z} g - \frac{\partial g}{\partial z}f\bigg)\frac{1}{g^2} \bigg \rangle $$

which is exactly

$$\frac{1}{g^2}(g\nabla f - f\nabla g)$$

Title: Prove this vector identity


问:
Prove that if $f$ and $g$ are both injective, then so is $f \times g$
Let $f: A \rightarrow C$ and $g: B \rightarrow D$. Defined $f \times g: A \times B \rightarrow C \times D$ to be $(f \times g)(a,b) = (f(a),g(b))$. Prove that if $f$ and $g$ are both injective, then so is $f \times g$.

This is what I need to prove. I have a proof here, but I'm not sure how to interpret the definition of injectivity with the product of two functions. Does it make sense to say that $(f(a), g(c)) \neq (f(b), g(d)) \Rightarrow (a,c) \neq (b,d)$? or should I approach this proof separately such as $(f(a), g(c)) \Rightarrow a \neq c$?

Here's my proof:

Assume $f$ and $g$ are both injective. we need to prove for any $x,y \in A$ and $x,y \in B$ such that $x \neq y$. Since $f$ and $g$ are injective, we have $f(x) \neq f(y) \Rightarrow x \neq y$ and $g(x) \neq g(y) \Rightarrow x \neq y$. Let us take $(f \times g)(a,c)$ and $(f \times g)(b,d)$. Then, we have $(f \times g)(a,c) = (f(a),g(c)) = (a, c)$ and $(f \times g)(b,d) = (f(b), g(d)) = (b, d)$. By the definition of injectivity, we have $(f(a),g(c)) \neq (f(b),g(d)) \Rightarrow (a,c) \neq (b,d)$. Thus, $f \times g$ is injective.

答:
Taking a step back, injectivity of a mapping $\phi$ means that
$$
(\phi(x_1) = \phi(x_2)) \mbox{ implies } x_1 = x_2. \quad (*)
$$
So, yes, if $x_1 = (a_1, b_1)$ and $x_2 = (a_2, b_2)$, and if we define
$$
\phi( (a, b) ) = (f(a), g(b)),
$$
then--as you correctly think--we interpret the equalities in $(*)$ as equalities of ordered pairs.


问:
Ring Isomorphism between integer ring and polynomial ring
I want to check if $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})[x]/(x^2)$. To begin with, I know that both rings have the same cardinality, and I tried to explicitly map each element in $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ to an element in $(\mathbb{Z}/2\mathbb{Z})[x]/(x^2)$. However, this didn't seem to work as the function $f$ I defined didn't satisfy $f(ab) = f(a)f(b)$ (due to $f[(0,1)(1,0)]$ not mapping to the right element). Is this enough to show that these rings aren't isomorphic?

答:
As the comment on your post states: it is not enough to show that a particular map $f$ fails to be an isomorphism.

Typically, one shows that two rings (or more generally, two mathematical objects) fail to be isomorphic by showing that one ring has an "essential" property that the other ring does not have. In this case, $\Bbb Z_2[x]/(x^2)$ has a nilpotent element, but $\Bbb Z_2 \times \Bbb Z_2$ does not.

For a detailed proof, we might state the following:

Suppose (for the purpose of contradiction) that $f:\Bbb Z_2^2 \to \Bbb Z_2[x]/(x^2)$ is a ring isomorphism. There exists an element $a \in \Bbb Z_2^2$ such that $f(a) = x$. It follows that
$$
f(a^2) = f(a)^2 = x^2 = 0.
$$
Because $f$ is a ring isomorphism, we have $f(a^2) = 0 \implies a^2 = f^{-1}(0) = 0$. However, $a \in \Bbb Z_2^2$ can only satisfy $a^2 = 0$ if $a = 0$. Thus, we have $x = f(a) = f(0) = 0$, which is a contradiction.


问:
Itô Isometry proof
Let $\{\phi_n\}$ a sequencie of functions in the space $\mathcal{V}(S,T)$ of those functions that verify:

$f(t,\omega)$ is $\mathcal{B} \times\mathcal{F}$ mesurable.

$\mathbb{E}[\int_S^T|f(t,\omega)|^2 dt] < \infty$.

$ f$ is $\mathcal{F}_t$-adapted.
Moreover, for each $n$, $E[(\int_{S}^{T} \phi_n(t,\omega) dB_t)^2] = E[\int_{S}^{T}\phi_n^2(t,\omega) dt]$.
Finally,let $f \in \mathcal{V}$ and supose $\lim_n E[\int_S^T(f(t,\omega)-\phi_n(t,\omega))^2 ds] = 0.$
I want to prove:
$E[(\int_{S}^{T} f(t,\omega) dB_t)^2] = E[\int_{S}^{T}f^2(t,\omega) dt]$.

答:
We can use almost surely $\phi $ instead of $f(t,w)$ ,and$\phi $ is bounded and elementary function. let's define $\phi(t,w)=\Sigma \phi_i$

for $E[(\int_S^T\phi(t,w)dB_t)^2]$ first discreitize , define $\Delta B_i=B_{t_{i+1}}-B_{t_i},\Delta B_j=B_{t_{j+1}}-B_{t_j}$ over intervals $[t_{j},t_{j+1}],[t_{i},t_{i+1}]$
so $$E[\phi_i\phi_j\Delta B_i\Delta B_j]=\\\begin{cases}0 & i<j\\E[\phi_i^2(\Delta B_i)^2] & i=j\\0  &i>j\end{cases}\\=\begin{cases}0 & i<j\\E[\phi_i^2(\Delta t_i)] & i=j\\0  &i>j\end{cases}\\
=\begin{cases}0 & i<j\\E[\phi_i^2]\Delta t_i & i=j\\0  &i>j\end{cases}$$ with respect that $$[t_{j},t_{j+1}],[t_{i},t_{i+1}]$$ have not cross secton when $ i\neq j$  now
$$E[(\int_{S}^{T}\phi^2(t,w) dB_t)^2]=\\\Sigma_{i}\Sigma_{j}E[\phi_i\phi_j\Delta B_i\Delta B_j]=\\\Sigma_{i=j}E[\phi_i\phi_j\Delta B_i\Delta B_j]+\underbrace{\Sigma_{i\neq j}E[\phi_i\phi_j\Delta B_i\Delta B_j]}_{0}\\
= \Sigma_{i=j}E[\phi_i\phi_i\Delta B_i\Delta B_i]\\=\Sigma_{i=j}E[(\phi_i)^2(\Delta B_i)^2]\\=
\Sigma_{i=j}E[(\phi_i)^2(\Delta t_i)]\\=
\Sigma_{i=j}E[(\phi_i)^2]\Delta t_i\\=E[(\int_S^T(\phi(t,w))^2dt]$$
Remark $(\Delta B_t)^2\to \Delta t$  and $\Delta B_i,\Delta B_j$ are independent when $i\neq j$

Title: Itô Isometry proof


问:
Mean curvature of compact surface not homeomorphic to a sphere
I'm using Do Carmo's book to self-study differential geometry, where I encounter Exercise 1 on page 282 (Section 4-5), which talks about the Gaussian curvature of points on the surface.

I'm wondering whether the conclusion can be generalized to mean curvature. For instance, for any compact surface in $\mathbb{R}^{3}$ that's not homeomorphic to a sphere, does there always exist a point with zero mean curvature, i.e $H=0$?

答: Yes, your hunch is right. Take a very long, thin torus. What do you think about $g\ge 2$?


问:
Is it possible to find isometric mapping of discrete points on two hyperbolic spaces with different curvature?
I understand that hyperbolic spaces with different curvatures are not isometric. However, I wonder if this is also true for finite, discrete points on two hyperbolic spaces with different curvatures. Specifically when the number of points is n = 3 or n = 4, what's the conclusion?

Formally, consider two two-dimensional hyperbolic spaces $\mathbb{H}_2^{-K_1}, \mathbb{H}_2^{-K_2}$ with curvatures $-K_1 \neq -K_2$. Let ABC be a triangle on $\mathbb{H}_2^{-K_1}$, then is it possible to find A'B'C' on $\mathbb{H}_2^{-K_2}$, such that $\overline{AB} = \overline{A'B'}$, $\overline{AC} = \overline{A'C'}$ and $\overline{BC} = \overline{B'C'}$? Intuitively, mimicing what we would do in Euclidean space, one can first find two points $A',B'$ such that $\overline{AB} = \overline{A'B'}$; then draw two circles centered at $A', B'$ with radius $\overline{AC}, \overline{BC}$ respectively; and find the intersection point as $C'$. However, I'm not sure whether this procedure makes sense in hyperbolic space and seek for a theoretical answer here.

Also, if it's indeed possible for n = 3, then what about n = 4? n = 5?

Just in case someone would like to do analytical calculations, here are some basics for hyperbolic spaces with curvature $\neq -1$. Using Poincare disk model, the two hyperbolic spaces $\mathbb{H}_2^{-K_1}, \mathbb{H}_2^{-K_2}$ aforementioned can be characterized as same set of points: $\{z \in \mathbb{C}_2||z|\leq 1\}$; but with different distance metric: $d_1(z_1, z_2) = \frac{1}{\sqrt{K_1}}atanh(|\frac{z_1-z_2}{1-\bar{z_1}z_2}|), d_2(z_1, z_2) = \frac{1}{\sqrt{K_2}}atanh(|\frac{z_1-z_2}{1-\bar{z_1}z_2}|)$. (As you can see, the distance metric is only different up to scaling.)

答:
All of the standard axioms of Euclid for planar geometry hold as well in $\mathbb H_2^{-K}$, except for the parallel axiom (or any of its equivalents). So your construction of triangles holds as well.

And that construction in fact shows that for a given numeric triple $(s_1,s_2,s_3)$ of positive numbers, that triple forms the side lengths of some triangle in $\mathbb H_2^{-K}$ if and only if it forms the side lengths of some Euclidean triangle, which holds if and only if the three "strict triangle inequalities" are satisfied:
$$s_1 < s_2 + s_3
$$
$$s_2 < s_3 + s_1
$$
$$s_3 < s_1 + s_2
$$
If you look in detail at your geometric argument drawing those circles, those three inequalities are exactly what ensures that the two circles centered at $A',B'$ with radii $\overline{AC}$, $\overline{BC}$ actually do intersect at two points.


问:
The graph of $x^{n}+y^{n}=r^{n}$ for sufficiently large $n$
The graph of the function $x^{n}+y^{n}=r^{n}$ for certain large values of $n$ looks suspiciously like a square.

See this page from wolframalpha. Have any results been proven regarding this observation? What do we call this figure anyway?

答: Sometimes it's called a superellipse - see, e.g., http://en.wikipedia.org/wiki/Superellipse</p>


答:
Any function $f$ on some domain $D$ called injective iff

$f(x)=f(y)$ implies $x=y$ for all $x,y\in D$.

Now, back to your problem:

$(f\times g)(a,b)=(f\times g)(c,d)\implies (f(a),g(b))=(f(c),g(d))\implies f(a)=f(c),$ and $g(b)=g(d)$. Now, by injectivity of $f$ and $g$,
$a=c, b=d$. Consequently, $(a,b)=(c,d)$.

Body:
Let $f: A \rightarrow C$ and $g: B \rightarrow D$. Defined $f \times g: A \times B \rightarrow C \times D$ to be $(f \times g)(a,b) = (f(a),g(b))$. Prove that if $f$ and $g$ are both injective, then so is $f \times g$.

This is what I need to prove. I have a proof here, but I'm not sure how to interpret the definition of injectivity with the product of two functions. Does it make sense to say that $(f(a), g(c)) \neq (f(b), g(d)) \Rightarrow (a,c) \neq (b,d)$? or should I approach this proof separately such as $(f(a), g(c)) \Rightarrow a \neq c$?

Here's my proof:

Assume $f$ and $g$ are both injective. we need to prove for any $x,y \in A$ and $x,y \in B$ such that $x \neq y$. Since $f$ and $g$ are injective, we have $f(x) \neq f(y) \Rightarrow x \neq y$ and $g(x) \neq g(y) \Rightarrow x \neq y$. Let us take $(f \times g)(a,c)$ and $(f \times g)(b,d)$. Then, we have $(f \times g)(a,c) = (f(a),g(c)) = (a, c)$ and $(f \times g)(b,d) = (f(b), g(d)) = (b, d)$. By the definition of injectivity, we have $(f(a),g(c)) \neq (f(b),g(d)) \Rightarrow (a,c) \neq (b,d)$. Thus, $f \times g$ is injective.


问:
Question involving eigenspaces and generalized eigenspaces
Question image

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be an eigenvalue of $T$ with corresponding eigenspace and generalized eigenspace $E_{\lambda}$ and $K_{\lambda}$, respectively. Let $U$ be an invertible operator on $V$ that commutes with $T$ (i.e. $TU = UT$). Prove that $U(E_{\lambda}) = E_{\lambda}$ and $U(K_\lambda) = K_{\lambda}$.

I need to solve the question given above; however, I am unsure of how to proceed exactly; am I required to use the invertibility of $U$ to somehow show the two equalities? Any help would be appreciated!

答:
Hint: Note that $x \in E_\lambda$ if and only if $(T - \lambda I) x = 0$ (where $I$ denotes the identity operator). Show that if $x \in E_\lambda$, we also have $(T - \lambda I) Ux = 0$.

In the same vein, note that $x \in K_\lambda$ if and only if there is a positive integer $k$ for which $(T - \lambda I)^k x = 0$


问:
can we prove that is an increasing function
I have a function $f: \mathbb{R}\to [0,+\infty)$
wich is continuous, convex, even and
f(t)=0 if and only if t=0

Can we prove that f is increasing

答:
Assuming you mean increasing on $[0, \infty)$ (it's decreasing elsewhere)

The definition of convex is: of any point $s< t$ then line segment connecting $(s,f(s))$ to $(t, f(t))$ will be above the graph.

So as $f(0)=0$ for any $w,t$ so that $0 < w < t$ we must have the $f(w)$ is below the line segment from $(0,0)$ to $(t, f(t))$.

So $f(t) < $ the $y$ value of the line segment at $x=w$ $< $ then endpoint of the segment as $x = t$ (as the line segment has positive slope) $= f(t)$.

So $0< w< t\implies f(w) < f(t)$.  So $f$ is increasing on $[0, \infty)$.

.....

If you wan to be formal:

The equation for the line segment from $(0,0)$ to $(t,f(t))$ is

$y_x - 0 = \frac {f(t) -0}{t-0}(x- 0) = \frac {f(t)}t x$.

And because $f$ is concave and $w < t$ we have  $\underbrace{f(w) < y_w}_{\text{because }f\text{ is convex}}=\frac {f(t)}t w=\underbrace{f(t)\cdot \frac tw< f(t)}_{\text{because } w< t\text{ so } \frac wt < 1}$

====

Oh, for pete's sake.....

Look.  Just draw a picture.

Mark off $f(0) = 0$ on the graph.  Then mark off $f(c) = K > 0$ on the graph for some $c > 0$.  Now draw a line for $(0,0)$ to $(c,K)$.  Not it is a line going upward.

That means that all the points of the line segment in that section are less than or equal to $K$ (and only equal to $K$ at the point $(c,K)$.

Now draw the graph of the function between these endpoints, $(0,0)$ and $(c,K)$.  Because we are told the graph is convex, the graph must appear below the line.

Now pick a point $x$ on the $x$ axis so that $0\le x < c$.  $f(x) \ge 0=f(0)$ (because the function only maps to $[0,\infty)$.) and $f(x) > 0$ if $x\ne 0$. Now look at the point $(x,y_x)$ on the line segment. Because the function is convex and the graph lies below the line we have:

$0 \le f(x) \le y_x$ with $0 = f(x) = y_0$ only when $x = 0$. Otherwise we have $0 < f(x) < y_x$.

But the line is pointing upward.  So if $x < c$ we have $y_x < K=f(c)$.

ANd that's that.

If $0 \le x < c$ then $0\le f(x) \le y_x < K =f(c)$ or with the extaneous bits trimmed off:

If $x \in [0,\infty)$ and $x < c$ then $f(x) < f(c)$.

And that is the definition of $f$ is increasing on $[0,\infty)$.


问:
Truth of a sentence in predicate logic
I'm working through the exercises in this Logic problem set and I'm struggling to understand ex. 7.5:

Determine whether the following sentence is logically true in predicate logic:

There is someone such that, if he or she is asleep, everyone is asleep.

The solutions section suggests this is True and gives a formal 'Natural Deduction' proof that I don't really understand (it probably doesn't help that I don't have a copy of the book the solutions keep referring to).

Is there another way to approach this?

答:
Case 1: There is someone who is not asleep. In that case, let $p$ be the person who is not asleep.

Then we see the following implication holds: if $p$ is asleep, then everyone is asleep. This is vacuously true, since $p$ is not asleep.

Then there indeed exists some person $p$, such that if $p$ is asleep, everyone is asleep.

Case 2: There is not anyone who is not asleep. In that case, everyone is asleep. Let $p$ be some person.

Then we see that the following implication holds: if $p$ is asleep, then everyone is asleep. This is true because everyone is, in fact, asleep.

Then there indeed exists some person $p$, such that if $p$ is asleep, everyone is asleep.

Title: Truth of a sentence in predicate logic


答:
This is a sometimes odd-sounding feature of existentially quantified implications.

If we express the statement formally it's: $$\exists x (A(x) \to \forall y\, A(y)).$$
Consider two cases:

everyone is asleep, or

not everyone is asleep.

In case 1, $\forall y\, A(y)$ holds, and hence the implication $A(x) \to \forall y\, A(y)$ would hold (without worrying about whether $x$ is asleep or not),

In case 2, we can find someone so that $A(x)$ is false, hence again the implication must be true.

Title: Truth of a sentence in predicate logic

Body:
I'm working through the exercises in this Logic problem set and I'm struggling to understand ex. 7.5:

Determine whether the following sentence is logically true in predicate logic:

There is someone such that, if he or she is asleep, everyone is asleep.

The solutions section suggests this is True and gives a formal 'Natural Deduction' proof that I don't really understand (it probably doesn't help that I don't have a copy of the book the solutions keep referring to).

Is there another way to approach this?


问:
trace inequality of positive definite matrices and diagonal matrices
Assume that $A,B$ are $\mathbb{R}^{n\times n}$ positive semi-definite matrices and $C\in\mathbb{R}^{n\times n}$ is a diagonal matrix with positive diagonal entries taking values from $[0,1]$. Is it true that

$$tr(ACBC)\leq tr(AB)$$

答:
No, here's a quick counterexample.
$$
A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}
$$
$$
B = \begin{pmatrix} 1 & 1 \\ 1& 1 \end{pmatrix}
$$
$$
C = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
$$
Then $\mathrm{Tr}[AB] = 0$ but $\mathrm{Tr}[ACBC]=1$.


问:
Expected value/variance with respect to two randomly selected 2-digit numbers
I have two two-digit numbers. Each digit (4 in total) is chosen independently and can be either 2, 3, 4, 5, or 6.

For example, the two numbers can be 35 and 22.

What's the expected value of the product of these two numbers? What about the variance of their sums?

I first found the number of possible combinations: there are 5 random digits to choose from, so $5! \cdot 2$ to account for the two numbers. What do I do from here?

答:
Let $X_1, X_2, X_3, X_4$ be the four digits, so your numbers are $Y_1 = 10 X_1 + X_2$ and $Y_2 = 10 X_3 + X_4$.  The product is $$Y_1 Y_2 = 100 X_1 X_3 + 10 X_1 X_4 + 10 X_2 X_3 + X_2 X_4$$  Then $$ \eqalign{\mathbb E[Y_1 Y_2] &= 100\; \mathbb E[X_1 X_3] + 10 \;\mathbb E[X_1 X_4] + 10\; \mathbb E[X_2 X_3] + \mathbb E[X_2 X_4]\cr
&= 100\; \mathbb E[X_1] \mathbb E[ X_3] + 10\; \mathbb E[X_1] \mathbb E[X_4] + 10 \;\mathbb E[X_2]\mathbb E[ X_3] + \mathbb E[X_2 ] \mathbb E[X_4]}$$
using the linearity of expected value and the fact that independent random variables are uncorrelated.

Similarly, use the formulas for variance of a sum of independent random variables and a multiple of a random variable.


答:
This is a classic example of the mismatch between the meaning of if in classical logic and its ordinary intuitive meaning. It is called the drinker's paradox.

Let's use the fact that $\varphi \to \psi$ is equivalent to $\lnot \varphi \lor \psi$ in classical logic to paraphrase the sentence in natural language and avoid the troublesome if.

```<code>There exists a person w such that w is not asleep or everyone is asleep.
</code>```

Let $c$ be an arbitrary person in our group of people.

Suppose everyone is asleep, then we pick $w$ to be $c$. Since everyone is asleep, this value of $w$ satisfies the condition.

Suppose at least one person is awake, call them $a$. Then we pick $a$ as $w$.

I think the reason why this is counterintuitive is that the choice of $w$ depends on who happens to be asleep right now or not, but the phrasing of the question in natural language implies that a fixed $w$ has some special property that someone causally determines which configurations of awake and asleep people are possible.

The paradox goes away completely if we give the conjunction or wide scope over exists. We can do this unambiguously in natural language by reordering the dijsuncts. Reordering the dijsuncts, and changing the order of the disjunction and existential quantifier in this specific case are both allowed in classical logic.

```<code>Everyone is asleep or there's a person who is not asleep.
</code>```

Title: Truth of a sentence in predicate logic

Body:
This is a classic example of the mismatch between the meaning of if in classical logic and its ordinary intuitive meaning. It is called the drinker's paradox.

Let's use the fact that $\varphi \to \psi$ is equivalent to $\lnot \varphi \lor \psi$ in classical logic to paraphrase the sentence in natural language and avoid the troublesome if.

There exists a person w such that w is not asleep or everyone is asleep.

Let $c$ be an arbitrary person in our group of people.

Suppose everyone is asleep, then we pick $w$ to be $c$. Since everyone is asleep, this value of $w$ satisfies the condition.

Suppose at least one person is awake, call them $a$. Then we pick $a$ as $w$.

I think the reason why this is counterintuitive is that the choice of $w$ depends on who happens to be asleep right now or not, but the phrasing of the question in natural language implies that a fixed $w$ has some special property that someone causally determines which configurations of awake and asleep people are possible.

The paradox goes away completely if we give the conjunction or wide scope over exists. We can do this unambiguously in natural language by reordering the dijsuncts. Reordering the dijsuncts, and changing the order of the disjunction and existential quantifier in this specific case are both allowed in classical logic.

Everyone is asleep or there's a person who is not asleep.