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. . . . . . Section 2.8 Linear Approximation and Differentials V63.0121.034, Calculus I October 12, 2009 Announcements I Midterm Wednesday on Sections 1.1–2.4
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. . . . . .
Section2.8LinearApproximationand
Differentials
V63.0121.034, CalculusI
October12, 2009
Announcements
I MidtermWednesdayonSections1.1–2.4
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32
andf′
(π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32
andf′
(π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32
andf′
(π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈ 0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈ 0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈ 0.87475
Calculatorcheck: sin(61◦) ≈ 0.87462.
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!.y = L2(x) =
√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
. .verylittledifference!
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=
36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
Dividingwithoutdividing?ExampleSupposeI haveanirrationalfearofdivisionandneedtoestimate577÷ 408. I write
577408
= 1 + 1691
408= 1 + 169× 1
4× 1
102.
ButstillI havetofind1
102.
SolutionLet f(x) =
1x. Weknow f(100) andwewanttoestimate f(102).
f(102) ≈ f(100) + f′(100)(2) =1
100− 1
1002(2) = 0.0098
=⇒ 577408
≈ 1.41405
Calculatorcheck:577408
≈ 1.41422.
. . . . . .
Dividingwithoutdividing?ExampleSupposeI haveanirrationalfearofdivisionandneedtoestimate577÷ 408. I write
577408
= 1 + 1691
408= 1 + 169× 1
4× 1
102.
ButstillI havetofind1
102.
SolutionLet f(x) =
1x. Weknow f(100) andwewanttoestimate f(102).
f(102) ≈ f(100) + f′(100)(2) =1
100− 1
1002(2) = 0.0098
=⇒ 577408
≈ 1.41405
Calculatorcheck:577408
≈ 1.41422.
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
Questions
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
. . . . . .
Answers
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
Answer
I 100miI 150miI 600mi(?) (Isitreasonabletoassume12hoursatthesamespeed?)
. . . . . .
Answers
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
Answer
I 100miI 150miI 600mi(?) (Isitreasonabletoassume12hoursatthesamespeed?)
. . . . . .
Questions
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
. . . . . .
Answers
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
Answer
I $100I $150I $600(?)
. . . . . .
Questions
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
. . . . . .
Answers
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
AnswerTheslopeofthelineis
m =riserun
Wearegivena“run”of dx, sothecorresponding“rise”is mdx.
. . . . . .
Answers
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
AnswerTheslopeofthelineis
m =riserun
Wearegivena“run”of dx, sothecorresponding“rise”is mdx.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
f(x + ∆x) − f(x)︸ ︷︷ ︸∆y
≈ f′(x)∆x︸ ︷︷ ︸dy
Rename ∆x = dx, sowecanwritethisas
∆y ≈ dy = f′(x)dx.
AndthislooksalotliketheLeibniz-Newtonidentity
dydx
= f′(x) . .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
Linearapproximationmeans ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
f(x + ∆x) − f(x)︸ ︷︷ ︸∆y
≈ f′(x)∆x︸ ︷︷ ︸dy
Rename ∆x = dx, sowecanwritethisas
∆y ≈ dy = f′(x)dx.
AndthislooksalotliketheLeibniz-Newtonidentity
dydx
= f′(x) . .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
Linearapproximationmeans ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Usingdifferentialstoestimateerror
If y = f(x), x0 and ∆x isknown, andanestimateof∆y isdesired:
I Approximate: ∆y ≈ dyI Differentiate:
dy = f′(x)dxI Evaluateat x = x0 and
dx = ∆x.
. .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
Why?
Whyuselinearapproximations dy whentheactualdifference ∆yisknown?
I Linearapproximationisquickandreliable. Finding ∆yexactlydependsonthefunction.
I Theseexamplesareoverlysimple. Seethe“AdvancedExamples”later.
I Inreallife, sometimesonly f(a) and f′(a) areknown, andnotthegeneral f(x).
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
MidtermFacts
I Coverssections1.1–2.4(Limits, Derivatives,DifferentiationuptoQuotientRule)
I CalculatorfreeI Hasabout7problemseachcouldhavemultipleparts
I Somefixed-response,somefree-response
I Tostudy:I outlineI doproblemsI metacognitionI askquestions!
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
GravitationPencilsdown!
Example
I Dropa1 kgballofftheroofoftheSilverCenter(50mhigh).Weusuallysaythatafallingobjectfeelsaforce F = −mgfromgravity.
I Infact, theforcefeltis
F(r) = −GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
I At r = re theforcereallyis F(re) =GMmr2e
= −mg.
I Whatisthemaximumerrorinreplacingtheactualforcefeltatthetopofthebuilding F(re + ∆r) bytheforcefeltatgroundlevel F(re)? Therelativeerror? Thepercentageerror?
. . . . . .
GravitationPencilsdown!
Example
I Dropa1 kgballofftheroofoftheSilverCenter(50mhigh).Weusuallysaythatafallingobjectfeelsaforce F = −mgfromgravity.
I Infact, theforcefeltis
F(r) = −GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
I At r = re theforcereallyis F(re) =GMmr2e
= −mg.
I Whatisthemaximumerrorinreplacingtheactualforcefeltatthetopofthebuilding F(re + ∆r) bytheforcefeltatgroundlevel F(re)? Therelativeerror? Thepercentageerror?
. . . . . .
SolutionWewonderif ∆F = F(re + ∆r) − F(re) issmall.
I Usingalinearapproximation,
∆F ≈ dF =dFdr
∣∣∣∣re
dr = 2GMmr3e
dr
=
(GMmr2e
)drre
= 2mg∆rre
I Therelativeerroris∆FF
≈ −2∆rre
I re = 6378.1 km. If ∆r = 50m,
∆FF
≈ −2∆rre
= −2 506378100
= −1.56×10−5 = −0.00156%
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2.
So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Illustrationofthepreviousexample
.
.2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
.
.2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
..(94 ,
32).
.(2, 17/12)
..(289144 ,
1712
)..(2, 577408
)
. . . . . .
Illustrationofthepreviousexample
..2
..(94 ,
32).
.(2, 17/12)..(289144 ,
1712
)
..(2, 577408
)
. . . . . .
Illustrationofthepreviousexample
..2
..(94 ,
32).
.(2, 17/12)..(289144 ,
1712
)..(2, 577408
)
