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Bisection method of solving an equation

It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.

I have used the equation x - cos(x) = 0 as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.

The code was written under Debian GNU/Linux in Python IDLE under Python.

$ python --version
Python 2.7.13

Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well. N

USAGE

If your maximum iterations is say 20 then you can use,

python bisection.py 20

Alternatively you can also chmod +x bisection.py and then

./bisection.py 20

LICENSE

GNU_GPL_v3.0

Totally RMS approved! While you're at it, please check out GNU and Free Software Foundation as well!

About

We use bisection method to find zeroes of an equation.

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, 'i'); if (__m === '*' || __re.test(location.href)) { injectUserscript("// Add copy buttons to all 
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}
} catch(__e) { console.warn('[Userscript:Add Copy Buttons to Code Blocks]', __e); }
})();
(function(){
try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '

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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
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, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Force GitHub README to respect dark mode\n(function() {\n var style = document.createElement('style');\n style.textContent = '\n .markdown-body {\n color-scheme: dark light;\n }\n .markdown-body pre { background: #161b22 !important; }\n .markdown-body code { background: rgba(110, 118, 129, 0.4) !important; }\n .markdown-body table th, .markdown-body table td { border-color: #30363d !important; }\n .markdown-body img { background: #0d1117; }\n .markdown-body blockquote { border-left-color: #8b949e; }\n .markdown-body hr { border-color: #30363d; }\n ';\n document.head.appendChild(style);\n})();", "GitHub Dark Mode README Fix");
}
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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
We use bisection method to find zeroes of an equation.
Topics
bisectionnumerical-analysisnumerical-methodspython
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Readme
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, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Highlight search terms from Google/DuckDuckGo/Bing referrer\n(function() {\n var ref = document.referrer;\n var terms = [];\n \n if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) {\n var url = new URL(ref);\n var q = url.searchParams.get('q') || url.searchParams.get('p');\n if (q) {\n terms = q.split(/\\s+/).filter(function(t) { return t.length > 2; });\n }\n }\n \n if (terms.length === 0) return;\n \n var style = document.createElement('style');\n style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }';\n document.head.appendChild(style);\n \n function highlight(node) {\n if (node.nodeType === 3) { // text node\n var text = node.textContent;\n var found = false;\n terms.forEach(function(term) {\n var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\\]\\\\]/g, '\\\\') + ')', 'gi');\n if (regex.test(text)) {\n found = true;\n var frag = document.createDocumentFragment();\n var parts = text.split(regex);\n parts.forEach(function(part, i) {\n if (i % 2 === 0) {\n frag.appendChild(document.createTextNode(part));\n } else {\n var span = document.createElement('span');\n span.className = 'userscript-highlight';\n span.textContent = part;\n frag.appendChild(span);\n }\n });\n node.parentNode.replaceChild(frag, node);\n }\n });\n } else if (node.nodeType === 1 && node.childNodes) { // element\n var skipTags = ['SCRIPT', 'STYLE', 'NOSCRIPT', 'TEXTAREA', 'INPUT', 'SELECT'];\n if (!skipTags.includes(node.tagName)) {\n Array.from(node.childNodes).forEach(highlight);\n }\n }\n }\n \n highlight(document.body);\n \n // Re-highlight on dynamic content\n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1 || node.nodeType === 3) highlight(node);\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Highlight Search Terms");
}
} catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); }
})();
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try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
We use bisection method to find zeroes of an equation.
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bisectionnumerical-analysisnumerical-methodspython
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, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Strip utm_, fbclid, gclid, etc. from all links on page\n(function() {\n var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content',\n 'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid',\n 'ref', 'ref_src', 'source', 'medium', 'campaign'];\n \n function cleanUrl(url) {\n try {\n var u = new URL(url, window.location.origin);\n var changed = false;\n trackingParams.forEach(function(p) {\n if (u.searchParams.has(p)) {\n u.searchParams.delete(p);\n changed = true;\n }\n });\n return changed ? u.toString() : url;\n } catch (e) {\n return url;\n }\n }\n \n function cleanLinks() {\n document.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n \n cleanLinks();\n \n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1) {\n if (node.tagName === 'A') cleanLinks();\n node.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Remove Tracking Parameters from Links");
}
} catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); }
})();
(function(){
try {
var __m = "youtube.com";
var __re = new RegExp('^' + "youtube\\.com" + '

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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
We use bisection method to find zeroes of an equation.
Topics
bisectionnumerical-analysisnumerical-methodspython
Resources
Readme
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, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Auto-enable theater mode on YouTube\n(function() {\n function tryTheater() {\n var btn = document.querySelector('button[aria-label=\"Theater mode\"], ytd-player #player button[title=\"Theater mode\"]');\n if (btn && !btn.classList.contains('activated')) {\n btn.click();\n }\n }\n \n // Try immediately\n tryTheater();\n \n // Try after navigation (SPA)\n var lastUrl = location.href;\n setInterval(function() {\n if (location.href !== lastUrl) {\n lastUrl = location.href;\n setTimeout(tryTheater, 500);\n }\n }, 1000);\n \n // Also try on player load\n var observer = new MutationObserver(tryTheater);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "YouTube Theater Mode Default");
}
} catch(__e) { console.warn('[Userscript:YouTube Theater Mode Default]', __e); }
})();
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try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
We use bisection method to find zeroes of an equation.
Topics
bisectionnumerical-analysisnumerical-methodspython
Resources
Readme
GPL-3.0 license
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, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Remove or un-stick sticky/fixed headers that block content\n(function() {\n function unstick() {\n document.querySelectorAll('header, nav, [role=\"banner\"], .header, .navbar, .sticky, .fixed-top, [style*=\"position: fixed\"], [style*=\"position:sticky\"]').forEach(function(el) {\n if (el.style.position === 'fixed' || el.style.position === 'sticky' || \n getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') {\n el.style.position = 'static';\n el.style.top = 'auto';\n el.style.zIndex = 'auto';\n }\n });\n }\n \n unstick();\n \n var observer = new MutationObserver(unstick);\n observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] });\n})();", "Kill Sticky Headers");
}
} catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); }
})();
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try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
We use bisection method to find zeroes of an equation.
Topics
bisectionnumerical-analysisnumerical-methodspython
Resources
Readme
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, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Universal Dark Mode - works on any site\n(function() {\n var enabled = true;\n \n function applyDarkMode() {\n if (!enabled) return;\n \n // Create style element if it doesn't exist\n var style = document.getElementById('universal-dark-mode-style');\n if (!style) {\n style = document.createElement('style');\n style.id = 'universal-dark-mode-style';\n document.head.appendChild(style);\n }\n \n // Dark mode CSS - inverts colors but preserves images/video\n style.textContent = '\n /* Invert everything except media */\n html {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #1a1a2e !important;\n }\n \n /* Restore images, videos, iframes, canvas */\n img, video, iframe, canvas, svg, picture, [style*=\"background-image\"] {\n filter: invert(1) hue-rotate(180deg) !important;\n }\n \n /* Preserve specific elements that should not be inverted */\n .no-dark-mode, .no-dark-mode *,\n [data-theme=\"light\"], [data-theme=\"light\"],\n .ace_editor, .ace_editor *,\n .CodeMirror, .CodeMirror *,\n .monaco-editor, .monaco-editor *,\n .markdown-body pre, .markdown-body pre *,\n .highlight, .highlight *,\n pre code, pre code * {\n filter: none !important;\n }\n \n /* Fix common UI elements */\n .modal, .popup, .dropdown-menu, .tooltip, .popover {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #2d2d44 !important;\n border-color: #444 !important;\n }\n \n /* Scrollbars */\n ::-webkit-scrollbar { background: #1a1a2e !important; }\n ::-webkit-scrollbar-thumb { background: #444 !important; }\n ::-webkit-scrollbar-thumb:hover { background: #555 !important; }\n \n /* Selection */\n ::selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ';\n }\n \n function removeDarkMode() {\n var style = document.getElementById('universal-dark-mode-style');\n if (style) style.remove();\n }\n \n // Toggle with Alt+Shift+D\n document.addEventListener('keydown', function(e) {\n if (e.altKey && e.shiftKey && e.key === 'D') {\n e.preventDefault();\n enabled = !enabled;\n if (enabled) {\n applyDarkMode();\n console.log('[Universal Dark Mode] Enabled');\n } else {\n removeDarkMode();\n console.log('[Universal Dark Mode] Disabled');\n }\n }\n });\n \n // Apply on load\n applyDarkMode();\n \n // Re-apply on dynamic content\n var observer = new MutationObserver(function(mutations) {\n if (enabled && !document.getElementById('universal-dark-mode-style')) {\n applyDarkMode();\n }\n });\n observer.observe(document.head, { childList: true });\n \n console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle');\n})();", "Universal Dark Mode");
}
} catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); }
})();
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Bisection method of solving an equation
It is a quite simple method of solving an equation numerically in cases where the exact solution is difficuilt to find. In this method we repetedly bisect an interval into halves until we reach the desired accuracy. It is a consequence of well known mean value theorem of calculus.
I have used the equation x - cos(x) = 0  as a demonstration. This problem is also on the book "Quantum Mechanics" of Schaum Outlines page 256. They have done that in FORTRAN though while my code is on Python (2.7.13). My C version of the same problem can be found here.
The code was written under Debian GNU/Linux in Python IDLE under Python.
$ python --version
Python 2.7.13
Here's the code running. The value obtained after 16 iterations can be verified with the book or other sources as well. Also note that we obtained exactly the same value using C as well.
N
USAGE
If your maximum iterations is say 20 then you can use,
python bisection.py 20
Alternatively you can also chmod +x bisection.py and then
./bisection.py 20
LICENSE
GNU_GPL_v3.0
Totally RMS approved!
While you're at it, please check out GNU and Free Software Foundation as well!
About
We use bisection method to find zeroes of an equation.
Topics
bisectionnumerical-analysisnumerical-methodspython
Resources
Readme
GPL-3.0 license
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