Calculus Option Notes

Calculus Option Notes [RR] IB Math HL Watch this if you need motivation : https:!!!"youtu#e"com!atch$v%&'(N)*y+,-.

Sources used: HAESE Mathematics HL Calculus Option workbook Convergence thread on I Survival !orum http:""www#ibsurvival#com"topic"$$$%&'tests'!or'convergence" A( Calculus ) *inal +eview Sheet https:""cha!!eemath#wikispaces#com"!ile"view"*inal,$-+eview,$-Sheet,$-',$-+ahn ,$-.E/#pd!"0$123-4&0"*inal,$-+eview,$-Sheet,$-',$,$-.E/ #pd!"0$123-4&0"*inal,$-+eview,$-Sheet,$-',$- +ahn,$-.E/ +ahn,$-.E/#pd! #pd! "u"Sushi5oob6s teacher6s mindmap https:""www#reddit#com"r"IO"comments"074dwc"m89teachers9mathematics9hl9calculus9o ption9p%"

"u"re9; 6s pla8list o! Calculus videos made b8 (ro!essor Leonard https:""www#reddit#com"r"IO"comments"07<<8p"lectures9on9the9calculus9option9b89pro!e ssor"

=hat the >uestion asks:

=hat 8ou have to do ? 5otes:

/"0 1 Basics Show that

( x) l i mf

e@ists

( x) l i m f

*ind

( x) l i m f

and

 x → a+¿

 x →a

the limit

 x → a− ¿

as it reaches the point !rom the le!t side and !rom the right sideB and show that the8 are e>ual# Show that !@B is continuous

Show that

( x) l i mf

e@ists and that it is

→a  x

e>ual to !aB# Show that !@B is di!!erentiable at a

Show that !@B is continuous at a and: (a+ h)−f(a) f l i m and −¿ h→ a

l i m h→ a +¿

h

( a+ h)−f( a) f h

e@ist and are e>ual#

/"2 3ho! that se4ies 5a n6 is conve47ent  dive47ent [Infinite 3e4ies] : 8 comparison test

*ind a series bnD where bnD F anD and anDbnD F - B# I! bnD converges anD is !orced to converge as well# Ghis can also be used to test !or divergence# I! bnD is divergent and anD F bnD anD is divergent as well# =hen to use this:

1#4 se the s>ueee theorem to evaluate a limit lim  x → -

g ( x )

4# *ind a !@B and a h@B such that: !@B J g@B J h@B # and

( x)=l =l ( x) l i mf i mh  x→ a

 x →a

$# I! both o! these conditions appl8

( x)=l l i mg  x→ a

8 limit comparison test

*ind a series bnD !or which 8ou know whether it converges or diverges and anDbnD F - B I!

l i m n→∞

an

= c  where c is a positive !inite

bn

value both series will either converge or diverge#

8 integral test:

Let ! be a positive decreasing on 4  and let an N !nB# Ghen the series

∑a

∞

∞

K5ote anD is the same as

∑a

n

is convergent i!

n= 4 n



∫ f( x) dx is divergent 1

∞

∑a

1

∞

convergent# I!

∫ f( x) dx is also

n

is also divergent#

n= 4

8 ratio test: Ksed to test i! a series is absolutel8 convergent divergent or inconclusive

*ind an?4# istribute a ? 4 !or all the n6s in anB Calculate  L=lim n→∞

A series is a#solutely conve47ent i! ∞ ∑ |a n| is also convergent# Otherwise n= 4

| |  an+4 an

I! L4 the series is absolutel8 convergent thus convergent

conditionally conve47ent#  I! LP4 the series is divergent I! LN4 the series ma8 be divergent conditionall8 convergent or absolutel8 convergent

5th term test

'

Qemoetric series

'

4

p'series

∑n

 p

is convergent i! p P 4 otherwise it is

divergent# I! pN4 it is the harmonic series

5OGE: ('series is use!ul when testing !or convergence using the other tests# E#g# it can be used !or comparison test as it is known that the p series converges " diverges# Absolute value test

'

/"8: 9iffe4ential ;uations <0 Show that a general solution !or di!!erential e>uation

4# *ind derivative o! general solution provided $# (lug in derivative !ound !or di!!erential e>uation given

raw a slope !ield

'

KInitial value problem IRG

Go appro@imate a value !or 8cB: 4# Start with writing down intial condition @- 8$# Obtain @4 b8 adding step sie @ to intial condition @- # @4N @- ? h %# Obtain 84 b8:

Qiven an initial condition use Euler6s method o! numerical integration with a step sie o! h to !ind an appro@imate value !or 8cB

 y 4= y - + h (

dy  )  where d8 over d@ is !ound dx

using the values @- and 8-  and h is step sie# 0# +epeat this process until 8cB is obtained# i#e# 8n?4 N 8n ? hf(xn, yn ) , @n?4 N @n?h  o all o! that on calculator KSeperable di!!erential e>uations Integrate a !unction dy  f  ( x ) = or something such dx g ( y )

4# Qet e>uation in the !orm o!: dy g ( x )⋅ = f  ( x ) dx

$# Integrate both sides with respect to @ dy

∫ g ( x )⋅dx dx =∫ f ( x )dx

%# Ghis gives

∫ g ( x )⋅dy ∫ f ( x ) dx 0# Integrate both sides# Sometimes it6s nice to simpli!8 the constant o! integration# E#g# 4

dy

4

4

∫  y dx dx =∫ ( 4 +  x−4 − x + 4 )dx 4

∫  y  dy = x + ln | x + 4|−ln | x −4| + c let lncB N A ln  y = x + ln ( A

| x x−+ |) 4 4

Ghere!ore  x

 y = Ae (

 x −4 )  x + 4

Solve a homogeneous e>uation

4# se substitution 8Nv@

It can be deduced that an e>uation is homogeneous when 8ou cannot simpl8 get the e>uation in the !orm

$# 8 product rule

dy g ( x )⋅ = f  ( x ) dx

A homogeneous e>uation is o! the !orm dy  y =f  ( ) where 8N8@B dx  x

where v is a !unction o! @#

dy dv =  dx + v dx dx

%# Compare this to di!!!erential e>uation dy  y =f ( ) given# dx  x

0# Qet this in the !orm o! a seperable di!!erntial e>uation where one side is integrated with respect to v and one side is integrated with respect to @

Solve a di!!erential e>uation in the !orm: dy + P ( x ) y =Q ( x ) dx

4# *ind (@B $# Integrating !actor I@B is !ound b8: ∫ P( x ) dx

 I ( x )=e

=hich is not seperable

 in !ormula bookletB

%# Calculate this integrating !actor 0# Multipl8 the !ull di!!erential e>uation b8 I@B 3# Simpli!8 the LHS to obtain  I ( x ) y =∫ I ( x ) Q ( x ) dx + c

2# Integrate to obtain general solution

/"= Riemann sums> ?*undamental theo4em of calculus?> Imp4ope4 inte74als *ind the area under a curve using riemann sums

4# *ind delta @ given b8:

Δ x =

Identities:

b

integral

n

∑ c =cn ∑ i= n

∑ i= i = 4$

# Lower and upper

values o! interval Kab  where b P a#

n (n +4 )

i= 4

∫ f ( x ) dx a

i= 4 n

b− a a and b are given b8 de!inite n

$# *ind @iT  given b8:

$

 x i∗= a +Δ x ( i ) n ( n + 4 )( $ n+ 4) 2

%# +iemann sum is given b8: b

n

∫ f ( x ) dx =lim ∑ f ( x ∗)Δ x →∞ a

n

i

i= 4

0# Calculate the area b8 using the identities# /ou will want to move an8thing 6n'  out o! the sum and keep 6i'  in the sum#

Lower and upper values !or +iemann Sum

'

*undamental theorem o! calculus

'

Evaluate an improper integral o! the !orm ∞

4# =rite integral in the !orm o!: b

∫ f  ( x ) dx

lim

a

b→∞

O!ten includes techni>ues such as integration b8 parts limit laws l6Hopital6s +ule and integration b8 substitution#

∫ f  ( x ) dx a b

$# Integrate

∫ f  ( x ) dx a

%# Move all that does not have b in it to the le!t hand side o! the limit as it won6t be a!!ected b8 b tending towards in!init8B 0# Evaluate the limit 3# Solve improper integral etermine whether an imp4ope4 inte74al o! If inte74al can #e found easily: ∞

the !orm

∫ f ( x ) dx

is convergent or

4# Integrate what is given

a

divergent#

b

$# Convergent i!

∫ f  ( x ) dx

e@ists !or all b

a b

where a J b   and i!

lim b→∞

∫ f  ( x ) dx

e@ists#

a

Otherwise divergent#

By compa4ison test: *or convergence: *ind g@B that is greater than or e>ual to !@B !or all @ F a# ∞

∫ g ( x ) dx

I! ∞

is convergent so is

a

∫ f ( x ) dx a

*or divergence: *ind g@B that is smaller or e>ual to !@B !or all @ F a# ∞

I!

∫ g ( x ) dx a

∞

is divergent so is

∫ f  ( x ) dx a

O!ten uses p series as convergence and divergence is known

1#2 etermine whether or not Rolle?s theo4em 4# (lug in values a and b in !unction !@B applies to the !unction ! on the given interval Kab $# I! both result in - +olle6s theorem guarantees that there is at least one point c at which !6cBN-# Show that the given !unction ! satis!ies the Mean @alue Aheo4em M@A on the given interval Kab# *ind all values o! c such that !bB'!aB N !6cB @ b'aB

4# Show that !@B is continuous on Kab and di!!erentiable on abK then MRG guarantees !bB'!aB N !6cB @ b'aB !or some value c U abK#

K !bB'!aB N !6cB @ b'aB basicall8 means that $# Ralues !or !6cB can be !ound b8 plugging there must be a value c U abK where the in values in !bB'!aB N !6cB @ b'aB# gradient !6cB is e>ual to the gradient f ( b )− f ( a ) 7oining two outer values o! b− a interval together  *ind the nth de74ee Aaylo4 polynomial appro@imation to !@B about @Na

=atch (ro!essor Leonard 6s video about Ga8lor Series and Maclaurin Series Ki!!icult to summarie

se o! error term Ga8lor6s !ormula with remainder

'

Maclaurin series !or e>uations given in !ormula booklet

'

Ga8lor series developed !rom di!!erentials

'

se o! substitution products integration and ' di!!erentiation to obtain other series#

/"D l?Hopital?s 4ule> evaluation of limits of fo4m fE  7E Evaluate limit o! the !orm !@B"g@B using l6Hopital6s

I! given a limit o! the !orm !@B"g@B which results indeterminate !orm either -"- in!init8 over in!init8 or - times in!init8B *ind !6@B and g6@B# Ghe limit o! !6@B " g6@B e>uals the limit o! !@B"g@B provided limit o! !6@B"g6@B e@ists#

3umma4y a#out se4ies and stuff liFe that Concept Alternating series

E@plained Germs in series are alternatel8 positive and negative# ∞

∑ (−4 ) −

I!

n

4

bn=b 4−b$ + b%−## . satis!ies

n= 4

- J bn?4 J bn !or all n U VZ ?  and i! lim b n=-  then the series is convergent# n→∞

? others (ower series

A series o! the !orm: ∞

n

$

lim cn x = c - + c 4 x + c $ x + ## . n=-

Ghree possibilities !or convergence o! a ∞ $ n lim cn ( x − a) =c -+ c 4 ( x− a )+ c $( x − a) + ## . n=-

' Series converges onl8 when @ N a ' Ghe series converges !or all n U |R+ ' Ghere e@ists a radius o! convergence + U |R+, such that series converges i! V@'aV  + and diverges i! V@'aV P + +adius o! convergence

A value + which is the greatest number such that the series converges !or all @ U |R such that V@'aV  + and diverges !or all @ U |R such that V@'aV P +

Interval o! convergence

Interval o! convergence I  is the set o! all points @ !or which the power series converges#

Gruncating

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Conditional vs absolute convergence

A series

∞

∑a

n

is absolutel8 convergent i!

n= 4

∞

the series o! absolute values

∑ |a | n

is

n= 4

convergent# I! an F - !or all n absolute convergence is the same as convergence# ∞

A series such as

∑ n= 4

(−4)n− n

4

which is

convergent but not absolutel8 convergent is called conditionall8 convergent# I! a series is absolutel8 convergent then it6s convergent#

inomial series

'

.inds o! continuit8

Continuous i! !aB is de!ined lim o! !@B !rom positive side and lim o! !@B !rom negative side e@ist and are e>ual and i! lim o! @ tending towards a o! !@B e>uals to !aB# I! conditions !ail !@B is discontinuous at @Na# I! lim o! @ tending to a o! !@B e@ists then ! has a 4emova#le discontinuity  at @Na# Otherwise essential discontinuity#

*undamental theorem o! calculus

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Monotone se>uence

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Monotone convergence

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(roduct o! series

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Note on tests: Compa4ison tests can #e done fo4 inte74als> limits> se;uences and se4ies