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CAL-04

Taylor polynomial of degree 2

T2(x) = f + f' h + ½ f'' h², h = x − a.

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SeriesTaylor

Governing equation

T2(a+h)=f+f′h+12f′′h2T_2(a+h)=f+f'h+\tfrac12 f'' h^2

h∈[−3,3]h \in \left[-3, 3\right]

where

f
f(a) (—)
f'
f'(a) (—)
f''
f''(a) (—)
h
Step h (—)
T_2
Taylor value (—)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

CAL-04 · curve
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①

Governing law — red

T2(a+h)=f+f′h+12f′′h2T_2(a+h)=f+f'h+\tfrac12 f'' h^2

h∈[−3,3]h \in \left[-3, 3\right]

h=−3h = -3T2=−9.5T_2 = -9.5

②

Domain of the variable — blue

hmin⁡=−3,hmax⁡=3h_{\min} = -3,\quad h_{\max} = 3

h∈[−3,3]h \in \left[-3, 3\right]

h=−3h = -3h=−3h = -3

③

Response curve — green

T2(a+h)=f+f′h+12f′′h2T_2(a+h)=f+f'h+\tfrac12 f'' h^2

h∈[−3,3]h \in \left[-3, 3\right]

h=−3h = -3T2=−9.5T_2 = -9.5

④

Rate of change — amber

dT2dh\frac{\mathrm{d}T_2}{\mathrm{d}h}

h∈[−3,3]h \in \left[-3, 3\right]

h=−2.962h = -2.962T2′=6.425T_2^{\prime} = 6.425

Lecture brief

Historical brief

Newton and Leibniz (1670s), Taylor, the fundamental theorem and the trapezoid rule are how change became a number. The lab differentiates, integrates and linearises in one variable. This sheet (CAL-04 — Taylor polynomial of degree 2) is the form associated with Taylor. Working symbols: ff, f′f', f′′f'', hh →\rightarrow T2T_2. The unique quadratic that matches f, f', f'' at a. Remainder ~ h³ f'''(ξ)/6.

Purpose

Purpose: compute T2T_2 from ff, f′f', f′′f'', hh in Calculus via T2(a+h)=f+f′h+12f′′h2T_2(a+h)=f+f'h+\tfrac12 f'' h^2 T2(x) = f + f' h + ½ f'' h², h = x − a. Use it when a real calculus question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given f=1.000 —f = 1.000\,\mathrm{—}, f′=0.500 —f' = 0.500\,\mathrm{—}, f′′=−2.000 —f'' = -2.000\,\mathrm{—}, h=0.200 —h = 0.200\,\mathrm{—}, the governing relation T2(a+h)=f+f′h+12f′′h2T_2(a+h)=f+f'h+\tfrac12 f'' h^2 yields T2=1.0600 —T_2 = 1.0600\,\mathrm{—}. Enter jets at a and a step h. Move a slider: the numbers are this situation, not a canned story.

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Inputs

Outputs

  • Taylor value T_21.0600 —
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CAL-04 · curve
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Narration of this film

Enter jets at a and a step h.

The unique quadratic that matches f, f', f'' at a. Remainder ~ h³ f'''(ξ)/6.

Reading speed

Watch on YouTube