diff --git a/_config.yml b/_config.yml
index b63745d..aafe938 100644
--- a/_config.yml
+++ b/_config.yml
@@ -109,9 +109,6 @@ theme: next
# Deployment
## Docs: https://hexo.io/docs/one-command-deployment
deploy:
- - type: git
- repo: josh@168.138.185.92:/home/josh/documents/josh-blog-deploy.git
- branch: master
- type: git
repo: josh@129.153.124.210:/home/josh/documents/josh-blog-deploy.git
branch: master
@@ -119,12 +116,6 @@ deploy:
repo: https://github.com/joshgao22/joshgao22.github.io.git
branch: master
-# MathJax [Redundant]
-# math:
-# engine: 'mathjax'
-# mathjax:
-# src: https://cdnjs.cloudflare.com/ajax/libs/mathjax/3.2.2/es5/tex-mml-chtml.min.js
-
# Local Search
search:
path: search.xml
@@ -198,12 +189,18 @@ neat_js:
related_posts:
enabled: true
+ enable_env_name: deploy
filter_threshold: 0.13
related_count: 5
weight:
title: 0.1
- description: 0
- keywords: 0
- tags: 0.01
- categories: 0.05
- text: 0
\ No newline at end of file
+ description: 0.01
+ keywords: 0.01
+ tags: 0.005
+ categories: 0.005
+ text: 0.2
+
+# config of hexo-renderer-pandoc
+pandoc:
+ args:
+ - --mathjax
diff --git a/package-lock.json b/package-lock.json
index 1a8c0a4..b6ed291 100644
--- a/package-lock.json
+++ b/package-lock.json
@@ -8,7 +8,7 @@
"name": "hexo-site",
"version": "0.0.0",
"dependencies": {
- "@next-theme/plugins": "^8.23.0",
+ "@next-theme/plugins": "^8.23.1",
"hexo": "^7.3.0",
"hexo-abbrlink": "^2.2.1",
"hexo-deployer-git": "^4.0.0",
@@ -24,13 +24,13 @@
"hexo-neat": "^1.0.9",
"hexo-related-posts": "^1.6.2",
"hexo-renderer-ejs": "^2.0.0",
- "hexo-renderer-pandoc": "^0.4.0",
+ "hexo-renderer-pandoc": "^0.5.0",
"hexo-renderer-stylus": "^3.0.1",
"hexo-server": "^3.0.0",
"hexo-tag-cloud": "2.1.*",
"hexo-theme-landscape": "^1.1.0",
- "hexo-theme-next": "^8.23.0",
- "hexo-word-counter": "^0.2.0"
+ "hexo-theme-next": "^8.23.1",
+ "hexo-word-counter": "^0.2.1"
}
},
"node_modules/@adobe/css-tools": {
@@ -39,180 +39,180 @@
"integrity": "sha512-DA5a1C0gD/pLOvhv33YMrbf2FK3oUzwNl9oOJqE4XVjuEtt6XIakRcsd7eLiOSPkp1kTRQGICTA8cKra/vFbjw=="
},
"node_modules/@algolia/client-abtesting": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-abtesting/-/client-abtesting-5.23.0.tgz",
- "integrity": "sha512-AyZ+9CUgWXwaaJ2lSwOJSy+/w0MFBPFqLrjWYs/HEpYMzBuFfGNZ7gEM9a7h4j7jY8hSBARBl8qdvInmj5vOEQ==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-abtesting/-/client-abtesting-5.20.2.tgz",
+ "integrity": "sha512-IS8JSFsDD33haaKIIFaL7qj3bEIG9GldZfb3ILW0QF3at7TcrIJYy58hrDvFee5T3p3E2aH/+wqIr0eha8jB/w==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/client-analytics": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-analytics/-/client-analytics-5.23.0.tgz",
- "integrity": "sha512-oeKCPwLBnTEPF/RWr0aaJnrfRDfFRLT5O7KV0OF1NmpEXvmzLmN7RwnwDKsNtPUHNfpJ6esP9xzkPEtJabrZ2w==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-analytics/-/client-analytics-5.20.2.tgz",
+ "integrity": "sha512-k0KxCfcX/HZySqPasKy6GkiiDuebaMh2v/nE0HHg1PbsyeyagLapDi6Ktjkxhz8NlUq6eTJR+ddGJegippKQtQ==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/client-common": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-common/-/client-common-5.23.0.tgz",
- "integrity": "sha512-9jacdC44vXLSaYKNLkFpbU1J4BbBPi/N7uoPhcGO//8ubRuVzigH6+RfK5FbudmQlqFt0J5DGUCVeTlHtgyUeg==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-common/-/client-common-5.20.2.tgz",
+ "integrity": "sha512-xoZcL/Uu49KYDb3feu2n06gALD17p5CslO8Zk3mZ7+uTurK3lgjLws7LNetZ172Ap/GpzPCRXI83d2iDoYQD6Q==",
"license": "MIT",
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/client-insights": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-insights/-/client-insights-5.23.0.tgz",
- "integrity": "sha512-/Gw5UitweRsnyb24Td4XhjXmsx8PxFzCI0oW6FZZvyr4kjzB9ECP2IjO+PdDq1A2fzDl/LXQ+u8ROudoVnXnQg==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-insights/-/client-insights-5.20.2.tgz",
+ "integrity": "sha512-fy7aCbo9y7WHt/9G03EYc471Dd5kIaM8PNP4z6AEQYr9a9X8c4inwNs6tePxAEfRHwVQi0CZ7kuVdn6/MjWx1A==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/client-personalization": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-personalization/-/client-personalization-5.23.0.tgz",
- "integrity": "sha512-ivrEZBoXfDatpqpifgHauydxHEe4udNqJ0gy7adR2KODeQ+39MQeaT10I24mu+eylIuiQKJRqORgEdLZycq2qQ==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-personalization/-/client-personalization-5.20.2.tgz",
+ "integrity": "sha512-ocL1ZXulfuXzJAwsKw2kMscKMD0rs/f4CFYu6Gjh4mK4um6rGfa1a6u1MSc4swFqRQer0wNP9Pi+kVfKhuKt5A==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/client-query-suggestions": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-query-suggestions/-/client-query-suggestions-5.23.0.tgz",
- "integrity": "sha512-DjSgJWqTcsnlXEKqDsU7Y2vB/W/VYLlr6UfkzJkMuKB554Ia7IJr4keP2AlHVjjbBG62IDpdh5OkEs/+fbWsOA==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-query-suggestions/-/client-query-suggestions-5.20.2.tgz",
+ "integrity": "sha512-Xjs4Tj1zkLCnmq1ys8RRhLQPy002I6GuT/nbHVdSQmQu4yKCI0gOFbwxHdM6yYPEuE3cJx7A4wSQjCH21mUKsg==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/client-search": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/client-search/-/client-search-5.23.0.tgz",
- "integrity": "sha512-XAYWUYUhEG4OIdo/N7H/OFFRD9fokfv3bBTky+4Y4/q07bxhnrGSUvcrU6JQ2jJTQyg6kv0ke1EIfiTO/Xxb+g==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/client-search/-/client-search-5.20.2.tgz",
+ "integrity": "sha512-2cD3RGB5byusLS0DAX1Nvl5MLiv7OoGgQrRs+94dTalqjvK8lGKzxxJhXoVojgx2qcROyIUAIDXFdTqv6NIHaA==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/ingestion": {
- "version": "1.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/ingestion/-/ingestion-1.23.0.tgz",
- "integrity": "sha512-ULbykzzhhLVofCDU1m/CqSzTyKmjaxA/z1d6o6hgUuR6X7/dll9/G0lu0e4vmWIOItklWWrhU2V8sXD0YGBIHg==",
+ "version": "1.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/ingestion/-/ingestion-1.20.2.tgz",
+ "integrity": "sha512-S593Kmhc98+5zdzGet4GrZEBEBGl4vVtqg/MPfW8dCRf9qDRNYSkhBsIzlhQe9JWiohe9oB9LW5meibwOgRmww==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/monitoring": {
- "version": "1.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/monitoring/-/monitoring-1.23.0.tgz",
- "integrity": "sha512-oB3wG7CgQJQr+uoijV7bWBphiSHkvGX43At8RGgkDyc7Aeabcp9ik5HgLC1YDgbHVOlQI+tce5HIbDCifzQCIg==",
+ "version": "1.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/monitoring/-/monitoring-1.20.2.tgz",
+ "integrity": "sha512-bW41aWLYgBv/coJUIT85mkN3kk1VBKsM8tlwB5S/s446Mgc7r8t5TX7kA8kCR2UbwDedOK51i/85/x/rM0ZXbg==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/recommend": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/recommend/-/recommend-5.23.0.tgz",
- "integrity": "sha512-4PWvCV6VGhnCMAbv2zfQUAlc3ofMs6ovqKlC/xcp7tWaucYd//piHg9CcCM4S0p9OZznEGQMRYPt2uqbk6V9vg==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/recommend/-/recommend-5.20.2.tgz",
+ "integrity": "sha512-wBMf3J1L5ogvU8p8ifHkknDXWn1zdZ2epkqpt2MkUaZynE3G77rrFU9frcO+Pu1FQJQ5xCDTKcYUUcJCDD00rg==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-common": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/requester-browser-xhr": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/requester-browser-xhr/-/requester-browser-xhr-5.23.0.tgz",
- "integrity": "sha512-bacOsX41pnsupNB0k0Ny+1JDchQxIsZIcp69GKDBT0NgTHG8OayEO141eFalNmGil+GXPY0NUPRpx+5s4RdhGA==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/requester-browser-xhr/-/requester-browser-xhr-5.20.2.tgz",
+ "integrity": "sha512-w+VMzOkIq2XDGg6Ybzr74RlBZvJQnuIdKpVusQSXCXknvxwAwbO457LmoavhZWl06Lcsk9YDx1X2k0zb+iJQmw==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0"
+ "@algolia/client-common": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/requester-fetch": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/requester-fetch/-/requester-fetch-5.23.0.tgz",
- "integrity": "sha512-tVNFREexJWDrvc23evmRgAcb2KLZuVilOIB/rVnQCl0GDbqIWJuQ1lG22HKqvCEQFthHkgVFGLYE74wQ96768g==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/requester-fetch/-/requester-fetch-5.20.2.tgz",
+ "integrity": "sha512-wpjnbvbi3A13b0DvijE45DRYDvwcP5Ttz7RTMkPWTkF1s6AHuo6O2UcwGyaogMAGa1QOOzFYfp5u4YQwMOQx5g==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0"
+ "@algolia/client-common": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
}
},
"node_modules/@algolia/requester-node-http": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/@algolia/requester-node-http/-/requester-node-http-5.23.0.tgz",
- "integrity": "sha512-XXHbq2heOZc9EFCc4z+uyHS9YRBygZbYQVsWjWZWx8hdAz+tkBX/jLHM9Xg+3zO0/v8JN6pcZzqYEVsdrLeNLg==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/@algolia/requester-node-http/-/requester-node-http-5.20.2.tgz",
+ "integrity": "sha512-YuSSdtgUt1dFBTNYUb+2TA5j0Hd0eDXE0bVISjUvTCqmoaGsGLwW+rKI7p1eLQ1r7RESwBAvUwcY1qP2Wl3Lyw==",
"license": "MIT",
"dependencies": {
- "@algolia/client-common": "5.23.0"
+ "@algolia/client-common": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
@@ -1047,16 +1047,16 @@
"integrity": "sha512-oxJlvZUHVBfnt6dNrhso0jGIkDgj7CpoGStQIif6TNlU+qELXjWYNZONcsA1vCUZdYx0qDDQYI8YFkEQCIhu/g=="
},
"node_modules/@next-theme/plugins": {
- "version": "8.23.0",
- "resolved": "https://registry.npmjs.org/@next-theme/plugins/-/plugins-8.23.0.tgz",
- "integrity": "sha512-IZylUNg+zEAZu/S8z8vYAGh0kHqTNeDlYSfnMg/nBu+4Sfkm42l7l4f2X7kWhFBuxt1MRYxIfpSBHZCnB68obQ==",
+ "version": "8.23.1",
+ "resolved": "https://registry.npmjs.org/@next-theme/plugins/-/plugins-8.23.1.tgz",
+ "integrity": "sha512-xWsAhwLaPtUiDPCMQ8ZwY4aoOEjhpFS+LHgZ08SdxjbaIbH42IG3EMC+8ohpV2zeOBv8EMBgLONw1Og0c4A/nw==",
"license": "MIT",
"dependencies": {
"@creativecommons/vocabulary": "2020.11.3",
"@fancyapps/ui": "5.0.36",
"@fortawesome/fontawesome-free": "6.7.2",
"@next-theme/pjax": "0.6.0",
- "algoliasearch": "5.23.0",
+ "algoliasearch": "5.20.2",
"animate.css": "3.1.1",
"animejs": "3.2.1",
"disqusjs": "3.0.2",
@@ -1629,24 +1629,24 @@
}
},
"node_modules/algoliasearch": {
- "version": "5.23.0",
- "resolved": "https://registry.npmjs.org/algoliasearch/-/algoliasearch-5.23.0.tgz",
- "integrity": "sha512-7TCj+hLx6fZKppLL74lYGDEltSBNSu4vqRwgqeIKZ3VQ0q3aOrdEN0f1sDWcvU1b+psn2wnl7aHt9hWtYatUUA==",
+ "version": "5.20.2",
+ "resolved": "https://registry.npmjs.org/algoliasearch/-/algoliasearch-5.20.2.tgz",
+ "integrity": "sha512-8evxG++iWyWnhng3g5RP+kwn6j+2vKLfew8pVoekn87FcfsDm92zJXKwSrU6pl+m5eAbGFhFF/gCYEQiRdbPlA==",
"license": "MIT",
"dependencies": {
- "@algolia/client-abtesting": "5.23.0",
- "@algolia/client-analytics": "5.23.0",
- "@algolia/client-common": "5.23.0",
- "@algolia/client-insights": "5.23.0",
- "@algolia/client-personalization": "5.23.0",
- "@algolia/client-query-suggestions": "5.23.0",
- "@algolia/client-search": "5.23.0",
- "@algolia/ingestion": "1.23.0",
- "@algolia/monitoring": "1.23.0",
- "@algolia/recommend": "5.23.0",
- "@algolia/requester-browser-xhr": "5.23.0",
- "@algolia/requester-fetch": "5.23.0",
- "@algolia/requester-node-http": "5.23.0"
+ "@algolia/client-abtesting": "5.20.2",
+ "@algolia/client-analytics": "5.20.2",
+ "@algolia/client-common": "5.20.2",
+ "@algolia/client-insights": "5.20.2",
+ "@algolia/client-personalization": "5.20.2",
+ "@algolia/client-query-suggestions": "5.20.2",
+ "@algolia/client-search": "5.20.2",
+ "@algolia/ingestion": "1.20.2",
+ "@algolia/monitoring": "1.20.2",
+ "@algolia/recommend": "5.20.2",
+ "@algolia/requester-browser-xhr": "5.20.2",
+ "@algolia/requester-fetch": "5.20.2",
+ "@algolia/requester-node-http": "5.20.2"
},
"engines": {
"node": ">= 14.0.0"
@@ -6446,9 +6446,9 @@
}
},
"node_modules/hexo-renderer-pandoc": {
- "version": "0.4.0",
- "resolved": "https://registry.npmjs.org/hexo-renderer-pandoc/-/hexo-renderer-pandoc-0.4.0.tgz",
- "integrity": "sha512-e913A2EZbhwbIfg+QvYUalEPHtK5qs9Rqj3ZrsQ3cYTbbiETMWhUQ7bzSlgUeh639eL0UDuBmv2vrEqj5RBchg=="
+ "version": "0.5.0",
+ "resolved": "https://registry.npmjs.org/hexo-renderer-pandoc/-/hexo-renderer-pandoc-0.5.0.tgz",
+ "integrity": "sha512-DpofabKugXwbGZdYYoVTXq8QDUd2ZSWwcqJKp4OZfLtu8sflETEb8R+Yyx1dT7S3SXlz++GV4/kwCnAhAvNVrg=="
},
"node_modules/hexo-renderer-stylus": {
"version": "3.0.1",
@@ -7126,9 +7126,9 @@
"license": "MIT"
},
"node_modules/hexo-theme-next": {
- "version": "8.23.0",
- "resolved": "https://registry.npmjs.org/hexo-theme-next/-/hexo-theme-next-8.23.0.tgz",
- "integrity": "sha512-hHRuGRCTF1HyIjvucK3Ylxr2qYB2/bFXdD/IWyPtxcVvkLgn3jcwBq4Zw/ksd0qtKNnLnjrlwWSH9Dyy9Li5mA==",
+ "version": "8.23.1",
+ "resolved": "https://registry.npmjs.org/hexo-theme-next/-/hexo-theme-next-8.23.1.tgz",
+ "integrity": "sha512-csbJPGhfflcBceJznUgWEMsAxq6Ef7ONDAD4XImPvSmczUadc+7oWEl+i1Ylg38Uu59r149p61RSZfxZ7FkosQ==",
"license": "AGPL-3.0-only"
},
"node_modules/hexo-util": {
@@ -7150,10 +7150,11 @@
}
},
"node_modules/hexo-word-counter": {
- "version": "0.2.0",
- "resolved": "https://registry.npmjs.org/hexo-word-counter/-/hexo-word-counter-0.2.0.tgz",
- "integrity": "sha512-cHFGjynprFsSrJoYEi7ar2peoavjTy7ZBDf/bQ0doXocL/ZjR42aMIcKtTi2A64KLvyTqzyb2mdKADCeclhCpQ==",
+ "version": "0.2.1",
+ "resolved": "https://registry.npmjs.org/hexo-word-counter/-/hexo-word-counter-0.2.1.tgz",
+ "integrity": "sha512-LkdXM9SykbVzZBGhCrYOonb/4FrNCQhVwnigt7pkVSa43Zejq3+VsG0kYmOZxDVlBmNCD1iokF6OZ3d6cYRa1Q==",
"hasInstallScript": true,
+ "license": "LGPL-3.0",
"dependencies": {
"cargo-cp-artifact": "^0.1"
}
diff --git a/package.json b/package.json
index 35571b3..c6cd520 100644
--- a/package.json
+++ b/package.json
@@ -12,7 +12,7 @@
"version": "7.3.0"
},
"dependencies": {
- "@next-theme/plugins": "^8.23.0",
+ "@next-theme/plugins": "^8.23.1",
"hexo": "^7.3.0",
"hexo-abbrlink": "^2.2.1",
"hexo-deployer-git": "^4.0.0",
@@ -28,12 +28,12 @@
"hexo-neat": "^1.0.9",
"hexo-related-posts": "^1.6.2",
"hexo-renderer-ejs": "^2.0.0",
- "hexo-renderer-pandoc": "^0.4.0",
+ "hexo-renderer-pandoc": "^0.5.0",
"hexo-renderer-stylus": "^3.0.1",
"hexo-server": "^3.0.0",
"hexo-tag-cloud": "2.1.*",
"hexo-theme-landscape": "^1.1.0",
- "hexo-theme-next": "^8.23.0",
- "hexo-word-counter": "^0.2.0"
+ "hexo-theme-next": "^8.23.1",
+ "hexo-word-counter": "^0.2.1"
}
}
\ No newline at end of file
diff --git a/source/_drafts/2025-06-26-josh-adv-math-1-2.md b/source/_drafts/2025-06-26-josh-adv-math-1-2.md
new file mode 100644
index 0000000..d5e48a1
--- /dev/null
+++ b/source/_drafts/2025-06-26-josh-adv-math-1-2.md
@@ -0,0 +1,35 @@
+---
+title: Josh's Summary — 工科数学分析
Part 1.2 常微分方程
+mathjax: true
+comments: true
+copyright: true
+toc:
+ enable: true
+ number: true
+ wrap: true
+ expand_all: true
+ max_depth: 4
+tags:
+ - 高等数学
+ - 数学分析
+ - 微积分
+categories:
+ - - Josh 的学习笔记
+ - 高等数学
+ - - 数学
+ - 数学分析
+abbrlink: 5a263f58
+date: 2025-06-26 10:21:02
+---
+
+# 常见等价无穷小
+
+1. $\sin x \sim x$
+2. $\tan x \sim x$
+3. $\arcsin x \sim x$
+4. $\arctan x \sim x$
+5. $1 - \cos x \sim \frac{x^2}{2}$
+6. $\ln(1+x) \sim x$
+7. $e^x - 1 \sim x$
+8. $a^x - 1 \sim x \ln a$
+9. $(1+x)^\alpha - 1 \sim \alpha x$
diff --git a/source/_posts/2023-04-14-calculas-formulas.md b/source/_posts/2023-04-14-calculas-formulas.md
deleted file mode 100644
index bd9cc11..0000000
--- a/source/_posts/2023-04-14-calculas-formulas.md
+++ /dev/null
@@ -1,194 +0,0 @@
----
-title: 高等数学常用公式
-mathjax: true
-comments: true
-copyright: true
-toc:
- enable: true
- number: false
- wrap: true
- expand_all: true
- max_depth: 4
-abbrlink: 1823c9d1
-date: 2023-04-14 16:25:59
-tags:
- - 等价无穷小
- - 导数公式
- - Maclaurin 公式
- - 曲率
- - 积分公式
- - 三角函数
-categories:
- - [数学, 微积分]
----
-
-# 1. 常见等价无穷小
-
-||||
-| :--: | :--: | :--: |
-| $\sin x \sim x$ | $\tan x \sim x$ | $\arcsin x \sim x$ |
-| $\arctan x \sim x$ | $\displaystyle\left( 1 - \cos x \right) \sim \frac{x^2}{2}$ | $\ln (1+x) \sim x$ |
-| $\left( e^x -1 \right) \sim x$ | $\left( a^x - 1 \right) \sim x \ln a$ | $\left( \left( 1+x \right) ^\alpha -1 \right) \sim \alpha x$ |
-
-
-
-# 2. 导数基本公式
-
-| 一般函数 | 三角函数 | 反三角函数 | 双曲函数
(圆函数) |
-| :--: | :--: | :--: | :----: |
-| $\left( C \right)' = 0$ | $\left( \sin x \right)' = \cos x$ | $\displaystyle\left( \arcsin x \right)' = \frac{1}{\sqrt{1-x^2}}$ | $\displaystyle\left( \sinh x \right)' = \left(\frac{e^x-e^{-x}}{2}\right)' = \cosh x$ |
-| $\left( x^\alpha \right)'=\alpha x^{\alpha-1} \left( \alpha \in \mathbf{Z} \right)$ | $\left( \cos x \right)' = -\sin x$ | $\displaystyle\left( \arccos x \right)' = -\frac{1}{\sqrt{1-x^2}}$ | $\displaystyle\left( \cosh x \right)' = \left( \frac{e^x+e^{-x}}{2} \right)'=\sinh x$ |
-| $\displaystyle\left( \log_ax\right)'=\frac{1}{x\ln a}$ | $\displaystyle\left( \tan x \right)' = \frac{1}{\cos^2 x}=\sec^2 x$ | $\displaystyle\left( \arctan x \right)' = \frac{1}{1+x^2}$ | $\displaystyle\left( \mathrm{arsinh} \, x \right)' = \left(\ln \left( x+\sqrt{x^2 + 1} \right) \right)' = \frac{1}{x^2+1}$ |
-| $( \ln$\| $x$ \|$)'=\displaystyle\frac{1}{x}$ | $\displaystyle\left( \cot x \right)' = -\frac{1}{\sin^2 x} = -\csc^2 x$ | $\displaystyle\left( \mathrm{arccot} \, x \right)' = -\frac{1}{1+x^2}$ | $\displaystyle\left( \mathrm{arcosh} \, x \right)' = \left( \ln \left( x + \sqrt{x^2 - 1} \right) \right)' = \frac{1}{\sqrt{x^2-1}}$
-| | $\left( \sec x \right)' = \sec x \cdot \tan x$ |
-| | $\left( \csc x \right)' = \csc x \cdot \cot x$ |
-
-- Newton-Leibniz 公式:
-
- $$
- \begin{aligned}
- \left( u \cdot v \right)^{\left( n \right)} &= u^{\left( n \right)}v + nu^{\left( n-1 \right)}v' + \frac{n \left( n-1 \right)}{2!}u^{\left( n-2 \right)}v'' + \cdots + \frac{n\left( n-1 \right)\cdots\left(n-k+1\right)}{k!}u^{\left( n-k \right)}v^{(k)} + \cdots + uv^{\left( n \right)} \\
- &= \sum_{k=0}^{n}C_n^k u^{\left( u-k \right)}v^k
- \end{aligned}
- $$
-
- 可以注意到 Newton-Leibniz 公式具有类似二项式定理展开的形式。
-
-# 3. 几个初等函数的 Maclaurin 公式
-
-## 3.1 Taylor 公式的一般形式
-
-$$
-f(x)=f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + R_n
-$$
-
-其中
-
-$$
-R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} \, (\xi\,在\,x\,和\,x_0\,之间) = o\left( \left( x - x_0 \right)^n \right)(x\to x_0)
-$$
-
-## 3.2 Maclaurin 公式的一般形式
-
-$$
-f(x)=f(0)+f'(0)\,x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n
-$$
-
-其中
-
-$$
-R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} \, (\xi\,在\,0\,和\,x\,之间) = \frac{f^{(n+1)}(\theta x)}{(n+1)!} x^{n+1}(0<\theta<1) = o\left( \left( x - x_0 \right)^n \right)(x\to x_0)
-$$
-
-## 3.3 指数函数的 Maclaurin 展开式
-
-$$
-e^x=1+x+\frac{x^2}{2} + \frac{x^3}{3} + \cdots +\frac{x^n}{n!} + \frac{e^{\theta x}}{(n+1)!}(0<\theta<1)
-$$
-
-## 3.4 正弦函数的 Maclaurin 展开式
-
-$$
-\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots + (-1)^{m-1}\frac{x^{2m-1}}{(2m-1)!} + R_{2m}
-$$
-
-其中
-
-$$
-R_{2m}=\frac{\sin\left( \theta x + \frac{(2m-1)\pi}{2}\right)}{(2m+1)!}x^{2m+1} = o(x^{2m})(0<\theta <1)
-$$
-
-## 3.5 余弦函数的 Maclaurin 展开式
-
-$$
-\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots + (-1)^m\frac{x^{2m}}{2m!} + R_{2m+1}
-$$
-
-其中
-
-$$
-R_{2m+1}=\frac{\cos(\theta x + (m+1)\pi)}{(2m+2)!} x^{2m+2}=o(x^{2m+2}) (0<\theta <1)
-$$
-
-## 3.6 $\ln (1+x)$ 的 Maclaurin 展开式
-
-$$
-\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4} + \cdots + \frac{(-1)^{n-1}x^n}{n} + R_n(x)
-$$
-
-其中
-
-$$
-R_n(x) = \frac{(-1)^n}{(n+1)(1+\theta x)^{n+1}}x^{n+1} = o(x^n)(0<\theta <1)
-$$
-
-## 3.7 $(1+x)^\alpha$ 的 Maclaurin 展开式
-
-$$
-(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \cdots + \frac{\alpha(\alpha - 1)\cdots (\alpha-n+1)}{n!}x^n + R_n(x)
-$$
-
-其中
-
-$$
-R_n(x) = \frac{\alpha(\alpha -1)\cdots(\alpha-n)}{(n+1)!}(1+\theta x)^{\alpha-n-1}x^{n+1} = o(x)(0<\theta <1)
-$$
-
-## 3.8 $\frac{1}{1-x}$ 的 Maclaurin 展开式
-
-$$
-\frac{1}{1-x} = 1+x+x^2+\cdots+x^n +o(x^n)
-$$
-
-# 4. 曲率及曲率半径
-
-| 曲率 | 曲率半径 |
-| :--: | :--: |
-| $K=\left\| \dfrac{\mathrm{d}\alpha}{\mathrm{d}s}\right\| \xlongequal{\tan \alpha = y' \Rightarrow \alpha = \arctan y'} \dfrac{\left\| \frac{y''}{1+y'^2}\mathrm{d}x \right\|}{\sqrt{1+y'^2}}\mathrm{d}x = \dfrac{\left\| y'' \right\|}{\left( 1 + y' \right)^\frac{3}{2}}$ | $R = \frac{1}{K}$ |
-
-# 5. 积分常用公式
-
-| 序 | 描述 | 公式 |
-| :--: | :----: | :----------------------------------------------:|
-| 1 | 定积分的定义 | $\displaystyle\int^1_0f(x)\mathrm{d}x=I=\lim_{\lambda \to 0}\sum_{i=1}^{n}f(\xi_i)\Delta x_i$,其中 $\lambda = \max \left\{ \Delta x_1,\Delta x_2, \cdots, \Delta x_n \right\}$ |
-| 2 | 变上限积分 | $考察定积分\displaystyle\int^x_af(x)\mathrm{d}x=\int^x_af(t)\mathrm{d}t$,对 $\displaystyle\Phi(x)=\int_a^xf(t)\mathrm{d}t$,有 $\displaystyle\Phi'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\int_a^xf(t)\mathrm{d}t = f(x)$ |
-| 3. | — | $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\int^{\beta(x)}_{\alpha(x)}f'(t)\mathrm{d}t = f\left[ \beta\left(x\right)\right]\beta'(x)-f[\alpha(x)]\alpha'(x)$ |
-| 4 | 上下限对称的积分 | 若 $f(x)$ 为偶函数,则 $\displaystyle\int_{-a}^af(x)\mathrm{d}x=2\int_0^af(x)\mathrm{d}x$;若 $f(x)$ 为奇函数,则 $\displaystyle\int_{-a}^af(x)\mathrm{d}x=0$ |
-| 5 | 周期函数的积分 | $\displaystyle f(x+T)=f(x)\Rightarrow\int_0^{nT}f(x)\mathrm{d}x=n\int^T_0f(x)\mathrm{d}x$ |
-| 6 | 分部积分法 | $\displaystyle\int uv'\mathrm{d}x=uv - \int u'v\mathrm{d}x \Leftrightarrow \int u\mathrm{d}v = uv - \int v\mathrm{d}u$ |
-
-# 6. 基本积分表
-
-| 序 | 积分式 |
-| :--: | :-------------------------: |
-| 1 | $\displaystyle\int k\mathrm{d}x=kx+C$,特别地 $\displaystyle\int\mathrm{d}x=x+C$ |
-| 2 | $\displaystyle\int x^\mu\mathrm{d}x=\frac{x^{\mu+1}}{\mu+1}+C(\mu\ne-1)$,特别地 $\displaystyle\int \frac{\mathrm{d}x}{x^2}=-\frac{1}{x}+C,\int\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{x}+C$ |
-| 3 | $\displaystyle\int\frac{\mathrm{d}x}{x}=\ln \|x\|+C$ |
-| 4 | $\displaystyle\int\frac{\mathrm{d}x}{1+x^2} = \arctan x +C$ |
-| 5 | $\displaystyle\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin x+C$ |
-| 6 | $\displaystyle\int\sin x\mathrm{d}x = -\cos x+C, \int\cos x\mathrm{d}x = \sin x +C$ |
-| 7 | $\displaystyle\int\frac{\mathrm{d}x}{\cos^2x} = \int\sec^2x\mathrm{d}x = \tan x +C, \int\frac{\mathrm{d}x}{\sin^2x}=\int\csc^2x\mathrm{d}x = -\cot x +C$ |
-| 8 | $\displaystyle\int a^x\mathrm{d}x = \frac{a^x}{\ln a} + C$,特别地 $\displaystyle\int e^x\mathrm{d}x = e^x +C$
-| 9 | $\displaystyle\int \sinh x\mathrm{d}x = \cosh x + C,\int \cosh x \mathrm{d}x=\sinh x +C$ |
-
-# 7. 三角函数相关公式
-
-| 序 | 公式 |
-| :--: | :-------------------------: |
-| 1 | $\sin^2x+\cos^2x=1, \tan^2x+1=\sec^2x,\cot^2x+1=\csc^2x$ |
-| 2 | $\begin{aligned}&\cosh^2x-\sinh^2x=1,\\ &\sinh(x\pm y)=\sinh x\cdot\cosh y \pm \sinh y\cdot\cosh x, \cosh(x\pm y) = \cosh x\cdot\cosh y,\\ &\sinh2x = 2\sinh x\cdot \cosh x, \cosh 2x = \cosh^2x+\sinh^2x = 2\cosh^2x - 1\end{aligned}$ |
-| 3
积化和差公式 | $\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)],\cos\alpha\sin\beta = \frac12[\sin(\alpha+\beta) - \sin(\alpha-\beta)]$
$\cos\alpha\cos\beta = \frac12[cos(\alpha+\beta) + cos(\alpha - \beta)], \sin\alpha\sin\beta=-\frac12[\cos(\alpha+\beta) - \cos(\alpha-\beta)]$ |
-| 4
和差化积公式 | $\displaystyle\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right), \sin\alpha - \sin\beta = 2\sin\left(\frac{\alpha-\beta}{2}\right) \cos\left(\frac{\alpha+\beta}{2}\right)$
$\displaystyle\cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right), \cos\alpha - \cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)$ |
-| 5
三角代换 | $\displaystyle\sqrt{a^2-x^2}\xrightarrow{x=a\sin t}a\cos t, \sqrt{a^2+x^2} \xrightarrow{x=a\tan t}a\sec t, \sqrt{x^2-a^2}\xrightarrow{x=a\sec t}a\tan t \left( a>0,t\in\left( -\frac{\pi}{2} , \frac{\pi}{2} \right) \right)$ |
-| 6 | $\displaystyle\int\sqrt{a^2-x^2}\mathrm{d}x = \frac{a^2}{2}\arcsin\frac xa + \frac12x\sqrt{a^2-x^2} + C(a>0)$ |
-| 7 | $\displaystyle\int\sqrt{x^2+a^2}\mathrm{d}x = \frac x2\sqrt{x^2+a^2} + \frac{a^2}2\ln \left\| x+\sqrt{x^2+a^2} \right\|+C(a>0)$ |
-| 8 | $\displaystyle\int\frac{1}{\sqrt{x^2+a^2}}\mathrm{d}x = \ln\left\| x+ \sqrt{x^2+a^2} \right\| +C$, $\displaystyle\int\frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x = \ln\left\| x + \sqrt{x^2+a^2}\right\|(a>0)$,特别地,$\displaystyle\int\frac{1}{\sqrt{x^2+1}}\mathrm{d}x=\left(x+\sqrt{x^2+1}\right) = \mathrm{arsinh}\, x$, $\displaystyle\int\frac{1}{\sqrt{x^2-1}}\mathrm{d}x = \ln\left( x+\sqrt{x^2-1}\right) = \mathrm{arcosh}\, x$ |
-| 9 | $\displaystyle\int\sec x\mathrm{d}x = \int\frac{1}{\cos x}\mathrm{d}x = \int\frac{\sec^2x+\sec x \tan x}{\sec x + \tan x}\mathrm{d}x = \int\frac{1}{\sec x + \tan x}\mathrm{d}(\sec x + \tan x) = \ln \left\| \sec x + \tan x \right\| + C$ |
-| 10 | $\displaystyle\int\csc x\mathrm{d}x = \int\frac{1}{\sin x}\mathrm{d}x = \ln\left\| \tan\frac x2 \right\| + C = \frac 12 \ln \left\| \frac{1-\cos x}{1 + \cos x}\right\| + C$ |
-| 11 | $\displaystyle\int\tan x\mathrm{d}x = \int\frac{\sin x}{\cos x}\mathrm{d}x = -\int\frac{1}{\cos x}\mathrm{d}(\cos x) = -\ln \left\| \cos x\right\| +C$ |
-| 12 | $\displaystyle\int\frac{1}{\tan x}\mathrm{d}x = \int\frac{\cos x}{\sin x}\mathrm{d}x = \int\frac{1}{\sin x}\mathrm{d}(\sin x) = \ln\left\| \sin x \right\| + C$ |
-| 13 | $\displaystyle\int\arctan x\mathrm{d}x = x\arctan x - \ln \left\| 1+x^2 \right\| +C$ |
-| 14
Wallis 公式 | $\displaystyle\int^{\frac{\pi}{2}}_{0}\sin^nx\mathrm{d}x = \int^{\frac{\pi}{2}}_{0}\cos^nx\mathrm{d}x = \begin{cases} \displaystyle\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12\cdot\frac\pi2,&n为偶数 \\ \displaystyle\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23\cdot1,&n为奇数 \end{cases} \\ = \begin{cases} \displaystyle\frac{(n-1)!!}{n!!}\cdot\frac\pi2,&n为偶数 \\ \displaystyle\frac{(n-1)!!}{n!!},&n为奇数 \end{cases}$ |
-| 15 | $\displaystyle\int_0^\pi\sin^nx\mathrm{d}x = 2\int^\frac\pi2_0\sin^nx\mathrm{d}x,\int_0^\pi\cos^nx\mathrm{d}x = \begin{cases} 2\displaystyle\int^\frac\pi2_0\cos^nx\mathrm{d}x,&n为偶数 \\ 0,&n为奇数 \end{cases}$
$\displaystyle\int_0^{2\pi}\sin^nx = \int_0^{2\pi}\cos^nx\mathrm{d}x = \begin{cases} 4\displaystyle\int^\frac\pi2_0\sin^nx\mathrm{d}x,&n为偶数 \\ 0,&n为奇数 \end{cases}$
$\displaystyle\int_0^\frac\pi2f(\sin x)\mathrm{d}x = \int_0^\frac\pi2f(\cos x)\mathrm{d}x$, $\displaystyle\int_0^\pi f(\sin x)\mathrm{d}x \ne \int_0^\pi f(\cos x)\mathrm{d}x$
$\displaystyle\int_0^\pi\pi f(\sin x)\mathrm{d}x = \frac\pi2\int_0^\pi f(\sin x)\mathrm{d}x = \pi\int_0^\frac\pi2 f(\sin x)\mathrm{d}x$ |
-| 16
万能公式 | 令 $u=\displaystyle\tan\frac x2,x=2\arctan u$,则 $\displaystyle\sin x = \frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}, \mathrm{d}x = \frac{2}{1+u^2}\mathrm{d}u$ |
-| 17 | $\displaystyle\int \frac{a\cos x+ b\sin x}{c\cos x + d \sin x} = A\int\mathrm{d}x+B\int\frac{\mathrm{d}(c\cos x+d \sin x)}{c \cos x + d \sin x}$
(设 $a\cos x + b\sin x = A(c \cos x + d \sin x) + B(c \cos x + d \sin x)'$,由待定系数法求出A、B)$ |
diff --git a/source/_posts/2023-04-14-josh-adv-math-1-1.md b/source/_posts/2023-04-14-josh-adv-math-1-1.md
new file mode 100644
index 0000000..a01f91f
--- /dev/null
+++ b/source/_posts/2023-04-14-josh-adv-math-1-1.md
@@ -0,0 +1,285 @@
+---
+title: Josh's Summary — 工科数学分析
Part 1.1 基本微分与积分
+mathjax: true
+comments: true
+copyright: true
+toc:
+ enable: true
+ number: false
+ wrap: true
+ expand_all: true
+ max_depth: 4
+abbrlink: 1823c9d1
+date: 2023-04-14 16:25:59
+tags:
+ - 等价无穷小
+ - 导数公式
+ - Maclaurin 公式
+ - 曲率
+ - 积分公式
+ - 三角函数
+categories:
+ - - Josh 的学习笔记
+ - 微积分
+ - - 数学
+ - 微积分
+---
+
+$\newcommand\diff{\mathop{}\!\mathrm{d}}$
+
+# 1. 常见等价无穷小
+
+||||
+| :--: | :--: | :--: |
+| $\sin x \sim x$ | $\tan x \sim x$ | $\arcsin x \sim x$ |
+| $\arctan x \sim x$ | $\displaystyle\left( 1 - \cos x \right) \sim \dfrac{x^2}{2}$ | $\ln (1+x) \sim x$ |
+| $\left( e^x -1 \right) \sim x$ | $\left( a^x - 1 \right) \sim x \ln a$ | $\left( \left( 1+x \right) ^\alpha -1 \right) \sim \alpha x$ |
+
+
+
+# 2. 导数基本公式
+
+| 一般函数 | 三角函数 | 反三角函数 | 双曲函数
(圆函数) |
+| :--: | :--: | :--: | :----: |
+| $\left( C \right)' = 0$ | $\left( \sin x \right)' = \cos x$ | $\displaystyle\left( \arcsin x \right)' = \dfrac{1}{\sqrt{1-x^2}}$ | $\displaystyle\left( \sinh x \right)' = \left(\dfrac{e^x-e^{-x}}{2}\right)' = \cosh x$ |
+| $\left( x^\alpha \right)'=\alpha x^{\alpha-1} \left( \alpha \in \mathbb{Z} \right)$ | $\left( \cos x \right)' = -\sin x$ | $\displaystyle\left( \arccos x \right)' = -\dfrac{1}{\sqrt{1-x^2}}$ | $\displaystyle\left( \cosh x \right)' = \left( \dfrac{e^x+e^{-x}}{2} \right)'=\sinh x$ |
+| $\displaystyle\left( \log_ax\right)'=\dfrac{1}{x\ln a}$ | $\displaystyle\left( \tan x \right)' = \dfrac{1}{\cos^2 x}=\sec^2 x$ | $\displaystyle\left( \arctan x \right)' = \dfrac{1}{1+x^2}$ | $\displaystyle\left( \mathrm{arsinh} \, x \right)' = \left(\ln \left( x+\sqrt{x^2 + 1} \right) \right)' = \dfrac{1}{x^2+1}$ |
+| $( \ln$\| $x$ \|$)'=\displaystyle\dfrac{1}{x}$ | $\displaystyle\left( \cot x \right)' = -\dfrac{1}{\sin^2 x} = -\csc^2 x$ | $\displaystyle\left( \mathrm{arccot} \, x \right)' = -\dfrac{1}{1+x^2}$ | $\displaystyle\left( \mathrm{arcosh} \, x \right)' = \left( \ln \left( x + \sqrt{x^2 - 1} \right) \right)' = \dfrac{1}{\sqrt{x^2-1}}$
+| | $\left( \sec x \right)' = \sec x \cdot \tan x$ |
+| | $\left( \csc x \right)' = \csc x \cdot \cot x$ |
+
+- Newton-Leibniz 公式:
+
+ $$
+ \begin{aligned}
+ \left( u \cdot v \right)^{\left( n \right)} &= u^{\left( n \right)}v + nu^{\left( n-1 \right)}v' + \dfrac{n \left( n-1 \right)}{2!}u^{\left( n-2 \right)}v'' + \cdots + \dfrac{n\left( n-1 \right)\cdots\left(n-k+1\right)}{k!}u^{\left( n-k \right)}v^{(k)} + \cdots + uv^{\left( n \right)} \\
+ &= \sum_{k=0}^{n}C_n^k u^{\left( u-k \right)}v^k
+ \end{aligned}
+ $$
+
+ 可以注意到 Newton-Leibniz 公式具有类似二项式定理展开的形式。
+
+# 3. 几个初等函数的 Maclaurin 公式
+
+## 3.1 Taylor 公式的一般形式
+
+$$\begin{aligned}
+f(x)&=f(x_0) + f'(x_0)(x-x_0) + \dfrac{f''(x_0)}{2!}(x-x_0)^2 + \cdots + \dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n + R_n
+\\
+&= \sum_{k=0}^n \dfrac{f^{(n)}(x_0)}{n!}(x - x_0)^n + R_n
+\end{aligned}$$
+
+其中
+
+$$
+R_n = \dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} \xlongequal{x\to x_0} o\left( \left( x - x_0 \right)^n \right), \quad \xi\,在\,x\,和\,x_0\,之间
+$$
+
+## 3.2 Maclaurin 公式的一般形式
+
+$$\begin{aligned}
+f(x) &= f(0)+f'(0)\,x + \dfrac{f''(0)}{2!}x^2 + \cdots + \dfrac{f^{(n)}(0)}{n!}x^n + R_n
+\\
+&= \sum_{k=0}^n \dfrac{f^{(n)}(0)}{n!}x^n + R_n
+\end{aligned}$$
+
+其中
+
+$$
+R_n = \dfrac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} = \dfrac{f^{(n+1)}(\theta x)}{(n+1)!} x^{n+1} \xlongequal{x\to x_0} o\left( \left( x - x_0 \right)^n \right), \quad \xi\,在\,0\,和\,x\,之间, 0<\theta<1
+$$
+
+## 3.3 指数函数的 Maclaurin 展开式
+
+$$
+e^x=1+x+\dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots +\dfrac{x^n}{n!} + \dfrac{e^{\theta x}}{(n+1)!}, \quad 0<\theta<1
+$$
+
+## 3.4 正弦函数的 Maclaurin 展开式
+
+$$
+\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} + \cdots + (-1)^{m-1}\dfrac{x^{2m-1}}{(2m-1)!} + R_{2m}
+$$
+
+其中
+
+$$
+R_{2m}=\dfrac{\sin\left( \theta x + \dfrac{(2m-1)\pi}{2}\right)}{(2m+1)!}x^{2m+1} = o(x^{2m}), \quad 0<\theta<1
+$$
+
+## 3.5 余弦函数的 Maclaurin 展开式
+
+$$
+\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + \cdots + (-1)^m\dfrac{x^{2m}}{2m!} + R_{2m+1}
+$$
+
+其中
+
+$$
+R_{2m+1}=\dfrac{\cos(\theta x + (m+1)\pi)}{(2m+2)!} x^{2m+2}=o(x^{2m+2}), \quad 0<\theta<1
+$$
+
+## 3.6 $\ln (1+x)$ 的 Maclaurin 展开式
+
+$$
+\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} -\dfrac{x^4}{4} + \cdots + \dfrac{(-1)^{n-1}x^n}{n} + R_n(x)
+$$
+
+其中
+
+$$
+R_n(x) = \dfrac{(-1)^n}{(n+1)(1+\theta x)^{n+1}}x^{n+1} = o(x^n), \quad 0<\theta<1
+$$
+
+## 3.7 $(1+x)^\alpha$ 的 Maclaurin 展开式
+
+$$
+(1+x)^\alpha = 1 + \alpha x + \dfrac{\alpha(\alpha-1)}{2!}x^2 + \cdots + \dfrac{\alpha(\alpha - 1)\cdots (\alpha-n+1)}{n!}x^n + R_n(x)
+$$
+
+其中
+
+$$
+R_n(x) = \dfrac{\alpha(\alpha -1)\cdots(\alpha-n)}{(n+1)!}(1+\theta x)^{\alpha-n-1}x^{n+1} = o(x), \quad 0<\theta<1
+$$
+
+## 3.8 $\dfrac{1}{1-x}$ 的 Maclaurin 展开式
+
+$$
+\dfrac{1}{1-x} = 1+x+x^2+\cdots+x^n +o(x^n)
+$$
+
+# 4. 曲率及曲率半径
+
+| 曲率 | 曲率半径 |
+| :--: | :--: |
+| $K=\left\| \dfrac{\diff \alpha}{\diff s}\right\| \xlongequal{令\ \tan \alpha = y', 则\ \alpha = \arctan y'} \dfrac{\left\| \dfrac{y''}{1+y'^2}\diff x \right\|}{\sqrt{1+y'^2}}\diff x = \dfrac{\left\| y'' \right\|}{\left( 1 + y' \right)^{3/2}}$ | $R = \dfrac{1}{K}$ |
+
+# 5. 积分常用公式
+
+| 序 | 描述 | 公式 |
+| :--: | :----: | :----------------------------------------------:|
+| 1 | 定积分的定义 | $\displaystyle\int^1_0f(x)\diff x=I=\lim_{\lambda \to 0}\sum_{i=1}^{n}f(\xi_i)\Delta x_i$,其中 $\lambda = \max \left\{ \Delta x_1,\Delta x_2, \cdots, \Delta x_n \right\}$ |
+| 2 | 变上限积分 | $考察定积分\displaystyle\int^x_af(x)\diff x=\int^x_af(t)\diff t$,对 $\displaystyle\Phi(x)=\int_a^xf(t)\diff t$,有 $\displaystyle\Phi'(x)=\dfrac{\diff }{\diff x}\int_a^xf(t)\diff t = f(x)$ |
+| 3 | 变上限积分的扩展 | $\displaystyle\dfrac{\diff }{\diff x}\int^{\beta(x)}_{\alpha(x)}f'(t)\diff t = f\left[ \beta\left(x\right)\right]\beta'(x)-f[\alpha(x)]\alpha'(x)$ |
+| 4 | 上下限对称的积分 | 若 $f(x)$ 为偶函数,则 $\displaystyle\int_{-a}^af(x)\diff x=2\int_0^af(x)\diff x$;若 $f(x)$ 为奇函数,则 $\displaystyle\int_{-a}^af(x)\diff x=0$ |
+| 5 | 周期函数的积分 | $\displaystyle f(x+T)=f(x)\Rightarrow\int_0^{nT}f(x)\diff x=n\int^T_0f(x)\diff x$ |
+| 6 | 分部积分法 | $\displaystyle\int uv'\diff x=uv - \int u'v\diff x \Leftrightarrow \int u\diff v = uv - \int v\diff u$ |
+
+# 6. 基本积分表
+
+| 序 | 积分式 |
+| :--: | :-------------------------: |
+| 1 | $\displaystyle\int k\diff x=kx+C$,特别地 $\displaystyle\int\diff x=x+C$ |
+| 2 | $\displaystyle\int x^\mu\diff x=\dfrac{x^{\mu+1}}{\mu+1}+C(\mu\ne-1)$,特别地 $\displaystyle\int \dfrac{1}{x^2}\diff x=-\dfrac{1}{x}+C,\int\dfrac{1}{\sqrt{x}}\diff x=2\sqrt{x}+C$ |
+| 3 | $\displaystyle\int\dfrac{1}{x}\diff x=\ln \|x\|+C$ |
+| 4 | $\displaystyle\int\dfrac{1}{1+x^2}\diff x = \arctan x +C$ |
+| 5 | $\displaystyle\int\dfrac{1}{\sqrt{1-x^2}}\diff x=\arcsin x+C$ |
+| 6 | $\displaystyle\int\sin x\diff x = -\cos x+C, \int\cos x\diff x = \sin x +C$ |
+| 7 | $\displaystyle\int\dfrac{\diff x}{\cos^2x} = \int\sec^2x\diff x = \tan x +C, \int\dfrac{\diff x}{\sin^2x}=\int\csc^2x\diff x = -\cot x +C$ |
+| 8 | $\displaystyle\int a^x\diff x = \dfrac{a^x}{\ln a} + C$,特别地 $\displaystyle\int e^x\diff x = e^x +C$
+| 9 | $\displaystyle\int \sinh x\diff x = \cosh x + C,\int \cosh x \diff x=\sinh x +C$ |
+
+# 7. 三角函数相关公式
+
+| 序 | 公式 |
+| :--: | :-------------------------: |
+| 1 | $\sin^2x+\cos^2x=1, \tan^2x+1=\sec^2x,\cot^2x+1=\csc^2x$ |
+| 2 | $\begin{aligned}&\cosh^2x-\sinh^2x=1,\\ &\sinh(x\pm y)=\sinh x\cdot\cosh y \pm \sinh y\cdot\cosh x, \cosh(x\pm y) = \cosh x\cdot\cosh y,\\ &\sinh2x = 2\sinh x\cdot \cosh x, \cosh 2x = \cosh^2x+\sinh^2x = 2\cosh^2x - 1\end{aligned}$ |
+| 3
积化和差公式 | $\sin\alpha\cos\beta=\dfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)],\cos\alpha\sin\beta = \dfrac12[\sin(\alpha+\beta) - \sin(\alpha-\beta)]$
$\cos\alpha\cos\beta = \dfrac12[cos(\alpha+\beta) + cos(\alpha - \beta)], \sin\alpha\sin\beta=-\dfrac12[\cos(\alpha+\beta) - \cos(\alpha-\beta)]$ |
+| 4
和差化积公式 | $\displaystyle\sin\alpha + \sin\beta = 2\sin\left(\dfrac{\alpha+\beta}{2}\right) \cos\left(\dfrac{\alpha-\beta}{2}\right), \sin\alpha - \sin\beta = 2\sin\left(\dfrac{\alpha-\beta}{2}\right) \cos\left(\dfrac{\alpha+\beta}{2}\right)$
$\displaystyle\cos\alpha + \cos\beta = 2\cos\left(\dfrac{\alpha+\beta}{2}\right) \cos\left(\dfrac{\alpha-\beta}{2}\right), \cos\alpha - \cos\beta = -2\sin\left(\dfrac{\alpha+\beta}{2}\right) \sin\left(\dfrac{\alpha-\beta}{2}\right)$ |
+| 5
三角代换 | $\displaystyle\sqrt{a^2-x^2}\xrightarrow{x=a\sin t}a\cos t, \sqrt{a^2+x^2} \xrightarrow{x=a\tan t}a\sec t, \sqrt{x^2-a^2}\xrightarrow{x=a\sec t}a\tan t \left( a>0,t\in\left( -\dfrac{\pi}{2} , \dfrac{\pi}{2} \right) \right)$ |
+| 6 | $\displaystyle\int\sqrt{a^2-x^2}\diff x = \dfrac{a^2}{2}\arcsin\dfrac xa + \dfrac12x\sqrt{a^2-x^2} + C(a>0)$ |
+| 7 | $\displaystyle\int\sqrt{x^2+a^2}\diff x = \dfrac x2\sqrt{x^2+a^2} + \dfrac{a^2}2\ln \left\| x+\sqrt{x^2+a^2} \right\|+C(a>0)$ |
+| 8 | $\displaystyle\int\dfrac{1}{\sqrt{x^2+a^2}}\diff x = \ln\left\| x+ \sqrt{x^2+a^2} \right\| +C$, $\displaystyle\int\dfrac{1}{\sqrt{x^2-a^2}}\diff x = \ln\left\| x + \sqrt{x^2-a^2}\right\|(a>0)$
特别地,$\displaystyle\int\dfrac{1}{\sqrt{x^2+1}}\diff x=\ln\left(x+\sqrt{x^2+1}\right) = \mathrm{arsinh}\, x$, $\displaystyle\int\dfrac{1}{\sqrt{x^2-1}}\diff x = \ln\left( x+\sqrt{x^2-1}\right) = \mathrm{arcosh}\, x$ |
+| 9 | $\displaystyle\int\sec x\diff x = \int\dfrac{1}{\cos x}\diff x = \int\dfrac{\sec^2x+\sec x \tan x}{\sec x + \tan x}\diff x = \int\dfrac{1}{\sec x + \tan x}\diff (\sec x + \tan x) = \ln \left\| \sec x + \tan x \right\| + C$ |
+| 10 | $\displaystyle\int\csc x\diff x = \int\dfrac{1}{\sin x}\diff x = \ln\left\| \tan\dfrac x2 \right\| + C = \dfrac 12 \ln \left\| \dfrac{1-\cos x}{1 + \cos x}\right\| + C$ |
+| 11 | $\displaystyle\int\tan x\diff x = \int\dfrac{\sin x}{\cos x}\diff x = -\int\dfrac{1}{\cos x}\diff (\cos x) = -\ln \left\| \cos x\right\| +C$ |
+| 12 | $\displaystyle\int\dfrac{1}{\tan x}\diff x = \int\dfrac{\cos x}{\sin x}\diff x = \int\dfrac{1}{\sin x}\diff (\sin x) = \ln\left\| \sin x \right\| + C$ |
+| 13 | $\displaystyle\int\arctan x\diff x = x\arctan x - \ln \left\| 1+x^2 \right\| +C$ |
+| 14
Wallis 公式 | $\displaystyle\int^{\pi/2}_{0}\sin^nx\diff x = \int^{\pi/2}_{0}\cos^nx\diff x = \begin{cases} \displaystyle\dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdot\cdots\cdot\dfrac12\cdot\dfrac\pi2,&n为偶数 \\ \displaystyle\dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdot\cdots\cdot\dfrac23\cdot1,&n为奇数 \end{cases} \\ = \begin{cases} \displaystyle\dfrac{(n-1)!!}{n!!}\cdot\dfrac\pi2,&n为偶数 \\ \displaystyle\dfrac{(n-1)!!}{n!!},&n为奇数 \end{cases}$ |
+| 15 | $\displaystyle\int_0^\pi\sin^nx\diff x = 2\int^{\pi/2}_0\sin^nx\diff x,\int_0^\pi\cos^nx\diff x = \begin{cases} 2\displaystyle\int^{\pi/2}_0\cos^nx\diff x,&n为偶数 \\ 0,&n为奇数 \end{cases}$
$\displaystyle\int_0^{2\pi}\sin^nx = \int_0^{2\pi}\cos^nx\diff x = \begin{cases} 4\displaystyle\int^{\pi/2}_0\sin^nx\diff x,&n为偶数 \\ 0,&n为奇数 \end{cases}$
$\displaystyle\int_0^{\pi/2}f(\sin x)\diff x = \int_0^{\pi/2}f(\cos x)\diff x$, $\displaystyle\int_0^\pi f(\sin x)\diff x \ne \int_0^\pi f(\cos x)\diff x$
$\displaystyle\int_0^\pi\pi f(\sin x)\diff x = \dfrac\pi2\int_0^\pi f(\sin x)\diff x = \pi\int_0^{\pi/2} f(\sin x)\diff x$ |
+| 16
万能公式 | 令\ $u=\displaystyle\tan\dfrac x2,x=2\arctan u$,则\ $\displaystyle\sin x = \dfrac{2u}{1+u^2},\cos x=\dfrac{1-u^2}{1+u^2}, \diff x = \dfrac{2}{1+u^2}\diff u$ |
+| 17 | $\displaystyle\int \dfrac{a\cos x+ b\sin x}{c\cos x + d \sin x} = A\int\diff x+B\int\dfrac{\diff (c\cos x+d \sin x)}{c \cos x + d \sin x}$
(设 $a\cos x + b\sin x = A(c \cos x + d \sin x) + B(c \cos x + d \sin x)'$,由待定系数法求出 $A, B$) |
+
+# 8. 定积分的几何应用
+
+## 8.1 平面
+
+### 8.1.1 直角坐标
+
+1. 直角梯形的面积
+
+$$\begin{equation*}
+A = \int_a^b [f(x_1) - f(x_2)] \diff x
+\end{equation*}$$
+
+2. 曲边梯形的面积:对于参数方程 $\begin{cases}
+ x = \varphi(x) \\ y = {\psi(x)}
+ \end{cases}$,曲边梯形的面积(变换上下限)
+
+ $$\begin{equation*}
+ A = \int_a^b y\diff x = \int_{t_1}^{t_2} \psi(t) \diff \varphi(t) = \int_{t_1}^{t_2} \psi(t)\varphi'(t) \diff t
+ \end{equation*}$$
+
+3. 椭圆的面积:椭圆 $\begin{cases}
+ x = a \cos t \\ y = b \sin t
+ \end{cases}$,即 $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ 的面积
+
+ $$\begin{equation*}
+ A = 4 \int_0^a y \diff x = 4ab \int_0^{\pi/2} (1-\cos t)^2 \diff t = 3\pi a^2
+ \end{equation*}$$
+
+4. 摆线的面积:摆线 $\begin{cases}
+ x = a (t - \sin t) \\ y = a (1 - \cos t)
+ \end{cases}$ 的一拱与 $x$ 轴所围图形的面积
+
+ $$\begin{equation*}
+ A = \int_0^2\pi a y \diff x = a^2 \int_0^{2\pi} (1 - \cos t)^2 \diff t = 3\pi a^2
+ \end{equation*}$$
+
+### 8.1.2 极坐标
+
+1. 扇形的面积:由 $\rho = \rho(\theta)$ 及射线 $\theta = \alpha$ 和 $\theta = \beta$ 所围曲边扇形的面积
+
+ $$\begin{equation*}
+ A = \int_\alpha^\beta \frac{1}{2} \rho^2(\theta) \diff \theta = \frac{1}{2} \int_\alpha^\beta \rho^2 \diff \theta
+ \end{equation*}$$
+
+2. 阿基米德螺线:阿基米德螺线 $\rho = a \theta (a > 0)$ 对应 $\theta$ 从 $0$ 变换到 $2\pi$ 所围图形的面积
+
+ $$\begin{equation*}
+ A = \int_0^{2\pi} \frac{1}{2} (a\theta)^2\diff \theta = \frac{4}{3} \pi^3 a^2
+ \end{equation*}$$
+
+3. 双纽线:双纽线 $\rho^2 = a^2 \cos 2\theta$ 所围平面图形的面积
+
+ $$\begin{equation*}
+ A = 4 \int_0^{\pi/4} a^2 \cos 2\theta \diff \theta = a^2
+ \end{equation*}$$
+
+4. 心形线:心形线 $\rho = a(a+\cos\theta), a>0$ 所围平面图形的面积
+
+ $$\begin{equation*}
+ A = 2 \cdot \frac{1}{2} \int_0^\pi a^2 (1+ \cos \theta)^2 \diff \theta = \frac{3}{2}\pi a^2
+ \end{equation*}$$
+
+5. 曲线的弧长
+
+ | 弧微分 | 直角坐标 | 参数方程 | 极坐标 |
+ | :--: | :--: | :--: | :--: |
+ | $\diff s = \sqrt{(\diff x)^2 + (\diff y)^2}$ | $s = \displaystyle\int_a^b \sqrt{1 + y'^2}\diff x$ | $s = \displaystyle\int_a^b \sqrt{\varphi'^2(t) + \psi'^2(t)}\diff t$ | $s = \displaystyle\int_\alpha^\beta \sqrt{\rho^2{\theta} + \rho'^2(\theta)} \diff\theta$
+
+## 8.2 空间
+
+1. 已知平行截面面积求立体体积:令 $A(x)$ 为截面面积,则立体体积
+
+ $$\begin{equation*}
+ V = \int_a^b A(x) \diff x
+ \end{equation*}$$
+
+2. 旋转体体积
+
+ | 薄片法 | 柱壳法 |
+ | :--: | :--: |
+ | $V = \displaystyle\int_a^b \pi f^2(x) \diff x$ | $V = \displaystyle\int_a^b 2\pi x \left| f(x) \right|\diff x$ |