Expand f around xn and demand the linear approximation vanish at xn+1:
f(xn+1)≈f(xn)+f′(xn)(xn+1−xn)=0
Why this step? We truncate Taylor after the linear term. The leftover quadratic term 21f′′(xn)(⋅)2 is the error we ignore — and that ignored term is exactly what gives quadratic convergence later.
Let the error be en=xn−r. We want to relate en+1 to en.
Taylor-expand f(r) around xn (note f(r)=0 exactly):
0=f(r)=f(xn)+f′(xn)(r−xn)+21f′′(ξ)(r−xn)2
for some ξ between r and xn (Taylor remainder).
Why this step? This is an exact equation (remainder form), not an approximation. It lets us track the true error.
Since r−xn=−en:
0=f(xn)−f′(xn)en+21f′′(ξ)en2
Divide by f′(xn) and rearrange:
f′(xn)f(xn)=en−2f′(xn)f′′(ξ)en2
Now substitute the iteration xn+1=xn−f′(xn)f(xn):
xn+1=xn−en+2f′(xn)f′′(ξ)en2
But xn−en=r, so xn+1−r=2f′(xn)f′′(ξ)en2:
Conditions for this to hold:f′(r)=0 (simple root), f′′ continuous, and starting guess "close enough" to r. If f′(r)=0 (a multiple root), convergence drops to merely linear.
State the iteration formula and what each part means geometrically.
Derive it from the tangent line.
Why is convergence quadratic? Which assumption is essential?
What happens at a multiple root? At f′(xn)=0?
Newton-Raphson iteration formula
xn+1=xn−f′(xn)f(xn)
Geometric meaning of the iteration
xn+1 is where the tangent at (xn,f(xn)) crosses the x-axis
Error recurrence for Newton-Raphson
en+1=2f′(xn)f′′(ξ)en2 (so ∣en+1∣≈C∣en∣2)
Order of convergence (simple root)
Quadratic — number of correct digits roughly doubles each step
Essential condition for quadratic convergence
f′(r)=0 (simple root), f′′ continuous, good starting guess
What kills quadratic convergence
A multiple root (f′(r)=0) drops it to linear
What happens if f′(xn)=0
Horizontal tangent → xn+1 diverges to infinity; iteration breaks
Newton iteration for a from f=x2−a
xn+1=21(xn+xna) (Babylonian average)
Newton iteration for 1/a (division-free)
xn+1=xn(2−axn)
Two derivations of the formula
Tangent-line geometry, and truncating Taylor at the linear term
Recall Feynman: explain to a 12-year-old
Imagine you're skiing down a hill blindfolded trying to reach the very bottom of the valley (where height = 0). You can feel which way the slope tilts under your skis. So you point straight down that slope and slide in a perfectly straight line until you'd hit "ground level," and stop there. You're now much closer to the bottom. Feel the new slope, slide again. Each slide gets you way, way closer — like halving then quartering then... your distance shrinks super fast. The "slope under your skis" is f′, the "how high you still are" is f, and sliding to ground level is the subtraction f/f′.
Dekho, Newton-Raphson ka core idea bahut simple hai: humein f(x)=0 solve karna hai, lekin equation tedhi-medhi hai. Toh hum cheating karte hain — kisi guess point pe curve ki jagah uski tangent line (seedhi line) le lete hain. Line ka root nikalna trivial hai, bas dekho line x-axis ko kahan kaatti hai. Wahi humara next, behtar guess ban jaata hai. Formula yaad rakho: naya = purana − height/slope, yaani xn+1=xn−f′(xn)f(xn).
Iski sabse killer baat hai quadratic convergence. Matlab error har step pe square ho ke chhota hota hai: en+1≈Cen2. Practically iska matlab — har iteration pe correct digits double ho jaate hain! Agar abhi 3 sahi digit hain, next step pe 6, phir 12. Isiliye calculators aur computers internally yahi method use karte hain (jaise 2 ya division nikalne ke liye).
Lekin do warnings: pehla, agar f′(xn)=0 ho gaya (flat tangent), toh line x-axis ko kabhi nahi kaategi aur tumhara guess infinity pe ud jaayega — toh slope check karte raho. Doosra, agar root multiple hai (yaani f′(r)=0 at the root), toh quadratic speed mar jaati hai aur sirf linear convergence milti hai. Aur haan, starting guess accha hona chahiye — galat jagah se start karoge toh method diverge ya oscillate kar sakta hai. Isliye pehle curve ka rough sketch ya bracketing kar lo, phir Newton chalao.
