Taylor's remainder theorem — error estimation

1. Setting up the problem

WHAT we want: a usable formula for $R_n(x)$ . WHY we can't just compute it: if we knew $f(x)$ exactly we wouldn't need the approximation. So we need a bound.

2. Deriving the Lagrange remainder from scratch

We derive it from the Mean Value Theorem, generalised. Here is the cleanest route.

Derivation (Cauchy MVT trick). Fix $x$ . Treat $a$ as a variable $t$ and define $g(t) = f(x) - \sum_{k=0}^{n}\frac{f^{(k)}(t)}{k!}(x-t)^k.$

Why this step? $g(t)$ is just the remainder when we center at $t$ . Note $g(x)=0$ (Taylor polynomial centered at $x$ equals $f(x)$ ) and $g(a)=R_n(x)$ .

Differentiate $g(t)$ . Almost everything telescopes: $g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}.$

Why this step? The product-rule terms from consecutive $k$ cancel — that's the magic of the Taylor structure. Only the top derivative survives.

Now introduce $h(t) = (x-t)^{n+1}$ , so $h(x)=0$ , $h(a)=(x-a)^{n+1}$ , and $h'(t) = -(n+1)(x-t)^n$ .

Apply the Cauchy Mean Value Theorem to $g$ and $h$ on $[a,x]$ : there is $c$ between $a$ and $x$ with $\frac{g(a)-g(x)}{h(a)-h(x)} = \frac{g'(c)}{h'(c)}.$

Why this step? Cauchy MVT compares the rates of two functions; it lets the unknown $c$ appear.

Substitute everything: $\frac{R_n(x) - 0}{(x-a)^{n+1}-0} = \frac{-\frac{f^{(n+1)}(c)}{n!}(x-c)^n}{-(n+1)(x-c)^n} = \frac{f^{(n+1)}(c)}{(n+1)!}.$

The $(x-c)^n$ cancels, giving exactly the Lagrange form. $\blacksquare$

3. The error BOUND (the part you actually use)

You don't know $c$ . But you don't need it — you only need an upper bound on $|f^{(n+1)}|$ .

HOW to use it (recipe):

Pick centre $a$ , degree $n$ , target point $x$ .
Find $f^{(n+1)}$ and a bound $M = \max |f^{(n+1)}|$ on the interval $[a,x]$ .
Plug into the box.

Figure — Taylor's remainder theorem — error estimation

4. Worked examples

Worked example (A) Estimate

$e^{0.1}$ with $P_2$ , bound the error $f(x)=e^x$ , $a=0$ , $n=2$ . So $P_2(x)=1+x+\tfrac{x^2}{2}$ , giving $P_2(0.1)=1.105$ .

Bound: $f^{(3)}(t)=e^t$ . On $[0,0.1]$ the max is $e^{0.1}<e^{0.5}<2$ , so take $M=2$ . Why $M=2$ ? We need an upper bound that's easy; $e^{0.1}\approx1.105<2$ comfortably. $|R_2(0.1)|\le \frac{2}{3!}(0.1)^3 = \frac{2}{6}(0.001)\approx 3.3\times10^{-4}.$ True value $e^{0.1}=1.10517...$ , actual error $\approx1.7\times10^{-4}$ — inside our bound. ✓

5. Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine drawing a smooth hill from memory with a few pencil strokes (that's the Taylor polynomial). The remainder theorem is a rule that tells you: "your sketch is off by at most this much." To find that "this much", you check how wildly the hill bends (the next derivative) on the stretch you drew, multiply by how far you walked, and divide by a big factorial number that quickly makes the error tiny. So a few strokes can be guaranteed almost perfect for a short walk.

Flashcards

What is the definition of the Taylor remainder

R_n(x)

R_n(x)=f(x)-P_n(x)

, the exact error of the degree-

n

approximation.

State the Lagrange form of the remainder.

R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

for some

c

between

a

and

x

What is the standard error bound?

|R_n(x)|\le \dfrac{M}{(n+1)!}|x-a|^{n+1}

where

M\ge|f^{(n+1)}|

on the interval.

Which derivative and factorial appear in the bound?

The

(n+1)

-th derivative and

(n+1)!

— one step beyond the polynomial degree

n

Why can't you pick a specific value for

c

c

is unknown; you must bound

|f^{(n+1)}|

by its max

M

over the whole interval instead.

For

e^x

a=0

, why is the error easy to bound on

[0,1]

All derivatives are

e^t\le e<3

, so

M=3

works for every

n

How many terms give

e

to within

10^{-6}

Need

3/(n+1)!<10^{-6}\Rightarrow(n+1)!>3\times10^6\Rightarrow n=9

What MVT result powers the derivation?

The Cauchy (generalised) Mean Value Theorem applied to

g(t)

and

h(t)=(x-t)^{n+1}

Connections

Taylor & Maclaurin Series — the polynomial $P_n$ this bounds.
Mean Value Theorem / Cauchy Mean Value Theorem — engine of the derivation.
Radius of Convergence — series converges iff $R_n(x)\to 0$ .
Alternating Series Estimation Theorem — alternative error bound for alternating series.
Big-O and asymptotic error — $R_n=O(|x-a|^{n+1})$ near $a$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Taylor polynomial $P_n(x)$ kisi function $f(x)$ ko ek point $a$ ke aas-paas approximate karta hai — jaise thodi si pencil strokes se ek smooth curve ka sketch banana. Lekin sketch tab tak useless hai jab tak hum bata na paayein ki "galti kitni hai". Yahi kaam karta hai remainder theorem: yeh error $R_n(x)=f(x)-P_n(x)$ par ek guaranteed upper bound deta hai.

Core formula simple hai aur sab kuch $n+1$ par chalta hai: $|R_n(x)| \le \dfrac{M}{(n+1)!}|x-a|^{n+1}$ . Yahan $M$ matlab agle (next) derivative $f^{(n+1)}$ ki maximum value us interval par. Mnemonic yaad rakho — "Next derivative, Next factorial, Next power". Lagrange form mein ek unknown point $c$ aata hai jo Cauchy Mean Value Theorem se nikalta hai; hum $c$ ko exactly nahi jaante, isliye uski jagah $M$ (maximum) le lete hain — yeh sabse common confusion hai, ise mat bhoolna.

Practically yeh bahut powerful hai. Jaise example: $e$ ko $10^{-6}$ accuracy tak nikalna ho, to bound $3/(n+1)! < 10^{-6}$ solve karo, $n=9$ aata hai — bas 10 terms kaafi! Factorial neeche hone ki wajah se error super-fast chhota hota jaata hai. Isiliye chhote $|x-a|$ ke liye kuch hi terms se almost perfect answer milta hai, aur hum mathematically prove bhi kar sakte hain.

Taylor's remainder theorem — error estimation

1. Setting up the problem

2. Deriving the Lagrange remainder from scratch

3. The error BOUND (the part you actually use)

4. Worked examples

5. Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections