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Summary Mathematics ECB1WIS

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This is a summary for the course Mathematics in the first year of the bachelor programme Economics and Business Economics (EBE) at the University of Utrecht. This summary contains all the important subjects for the exam, so among things differentiation, partial differentiation and dynamic systems.

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  • Chapter 1, 2, 3, 4, 5, 6, 8, 9, 13, 14
  • September 18, 2018
  • 5
  • 2018/2019
  • Summary

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By: vandenbrink • 6 year ago

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Mathematics s Summary

Fraitioc

- Multily crosswisse, buts ists iss also iossisble tso uusts uultily one deenouisnatsor wistsh tshe otsher
nuueratsor. 6/4=3/2s->s6=(4*3)/2
- If uusts tshe deenouisnatsor of a fractonn(1 iss a fractonn(1, ists iss iossisble tso uultily tshe
deenouisnatsor of fractonn(1 wistsh tshe nuueratsor of fractonn(1. 6/(4/2)s->s12/4
- If tshe nuueratsor ande tshe deenouisnatsor are fractons ande tshe deenouisnatsor of botsh fractons
iss tshe saue, ists iss iossisble tso deeletse tsheu ande tshe new fracton wisll be foruede by tshe
nuueratsors of tshe fractons. (6/2)/(4/2)s->s6/4
- If tshe nuueratsor ande tshe deenouisnatsor of a fracton n(1 are fractons n(n,,de1 ande tshe
nuueratsors of fractons ( ande , are tshe saue, ists iss iossisble tso reuove tsheu ande uake
deenouisnatsor of fracton , tshe new nuueratsor of fracton ( ande tshe deenouisnatsor of fracton
( wisll becoue tshe deenouisnatsor of fracton (. (6/4)(6/2)s->s2/4


EqSuatioc

-s ilvesfirs(…):
Fisrsts sets tshe varisable you have tso solve free nso outs of bracketss1.
If tshere iss an equaton wistsh … … … =0, tsry tso gets tso n… …1n… …1n… …1=0. Before you tsry tso gets
tshats sistsuaton you neede tso work outs al tshe bracketss ande tshen tsry tso fnde outs whats tshe new
bracketss shoulde be.

-sRSulec:
- K^3/4=X -> K=X^4/3, X^2n/(1=6 -> X=6^1/2, K^-1/4=X -> K=X^-4n/(1
- (X/Q)^4/3 = (X^4/Q^4)^1/3, (X/Q)^6/4 = (X^6/Q^6)^1/4
- nB/C1/D = B/nCD1, B/nC/D1 = nBD1/C
- nB/C1^-4s= nC/B1^4
- nB/C1^4 = nB^4/C^41, nBC1/nDE1^4 = nB^4*C^41/nD^4*E^41
- nXR1/nBR^,1 = X/nBR^(1
- If tshere iss an equaton wistsh a negatve iower tsry tso substtsutse ists wistsh x:
z-(-(z-(-(5=0 -> (-( -(5=0 -> z-(= ( ande z-(=
Latser once you have solvede , you substtsutse agaisn wistsh z-( ande tshen you calculatse z.

-sElimioatio:
- If tshe answer after adedeisng ui iss 0+0=0 tshere are isnfnistse solutons.
- If tshe answer after adedeisng ui iss wrong, tshere iss no soluton.

-s SubcttSutio:
- Solve tshe equaton fn 1=gn 1
- fn 1= a+b gn 1= c+b
- a+b =c+de -> a+b -de =c -> nb-de1 =c-a -> x=s(i-a)/(b-d)

, - b nots equal tso de -> deisvisdeisng by 0 iss nots iossisble -> isf b=de tshere iss no equislisbrisuu.
- Always check your answer wistsh nuubers.
- Solve y=a+bw+cz -> tswo varisables.

FSuoitioc

FSuoitioc:
- Consuuer tsheoryx: Utlistsy functon -> tsryisng tso ua isuisze your utlistsy.
- Functon = fn 1= +4 .
Equatonx: +4 =0
E iressisonx: +4
- Douaisnx: All values of tshats are iossisble isn a functon.
- Rangex: All iossisble values for fn 1 isnsisdee tshe deouaisn.
- Endeogenous varisablesx: Varisables isnsisdee tshe functon.
- E ogenous varisablesx: Varisables outssisdee tshe functon.

PripertecsifsfSuoitioc:
- Misnisuuu or ua isuuu?
- Inverse? -> The endeogenous ande e ogenous varisables can be swistschede.
-> deouaisn ande range wisll swistsch.
- Increasisng or deecreasisng?
- Syuuetsrisc? fn 1 = fn- 1

irtcsifsfSuoitioc:
Lisnex: fn 1=
Paraboliscx: fn 1= ( -> u shaie.
Rootsx: fn 1=
Powerx: fn 1= a
Hyierboliscx: fn 1 = (/
E ionentalx: fn 1 = e
Logaristshux: = ln
Absolutsex: fn 1 =
Couioundex: fn 1+gn 1

Ligarithmc

Bace:
(logn 1 = tshe iower of tswo tso retsurn tso -> isnverse functon.
fn 1 = ( fn 1= (logn 1
(, =8 (
logn81=,
(4 =(6 (
logn(61=4

RSulec:
- logna1+lognb1=lognab1

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