Friday, 27 September 2019

Difference between matrix and determinant

What is the difference between matrix and determinant?

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Matrix is one of the most important and powerful tools in mathematics which has found applications to a very large number 
of disciplines such as engineering, economics, statistics, physics, chemistry, biology, etc.
The theory of matrix is extensively used in the solution of applied business and industrial problems. Matrix is simply an ordered arrangement of elements, it is meaningless to assign a single numerical value to a matrix.
Determinants: is in the solution of the simultaneous system of linear equations. If A is a square matrix, then determinant function associates with A exactly one numerical value called the determinant. It is denoted by lAl.

A...........................lAl
square matrix                 determinant of A


Friday, 20 September 2019

Financial management and examples

What is financial management and examples?

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Financial management is concerned with the duties of the financial managers in the business form.

Examples

Finance is the art and science of managing money

Financial services are concerned with the design and delivery of advice and financial products to individual businesses and governments.
Financial managers actively manage the financial affairs of any type of business.
Example;
a. Financial and non-financial
b. Private and public
c.  Large and small
d.  Profit-seeking and not for profit.

Financial management is organized into six section

  1. Relationship of finance and related disciplines
  2. Scope of financial management
  3. The goal of financial management
  4. Agency problem
  5. Organization of the finance function
  6. The emerging role of finance managers in India

Is finance a part of economics?

The financial management can be described in the light of the two broad areas of economics (a) Macroeconomics (b) microeconomics
Macroeconomics is concerned with the institutional structure of the banking system, money, and capital markets, financial intermediaries, monetary, credit and fiscal policies and economic policies dealing with and controlling the level of activity within the economy.
Microeconomics deals with the economic decisions of individuals and organizations.

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Wednesday, 18 September 2019

How to use garch model

How to use garch model

garchmodel

> C(aapl.garch11.fit)

[1] 0.1889649
> uncmean(aapl.garch11.fit)
[1] 0.007145279
> ni.garch11 <- newsimpact(aapl.garch11.fit)
> plot(ni.garch11$zx, ni.garch11$zy, type="l", lwd=2, col="blue",main="GARCH(1,1)-News Impact", ylab=ni.garch11$yexpr, xlab=ni.garch11$xexpr)
> egarch11.spec = ugarchspec(variance.model = list(model="eGARCH",garchOrder=c(1,1)), mean.model = list(armaOrder=c(0,0)))
>
> aapl.egarch11.fit = ugarchfit(spec=egarch11.spec, data=ret)
> coef(aapl.egarch11.fit)
         mu       omega      alpha1       beta1
 0.02163174 -1.14814240  0.07408394  0.31963281
     gamma1
 0.33290745
> ni.egarch11 <- newsimpact(aapl.egarch11.fit)
> plot(ni.egarch11$zx, ni.egarch11$zy, type="l", lwd=2, col="blue",
+      main="EGARCH(1,1) - News Impact",
+      ylab=ni.egarch11$yexpr, xlab=ni.egarch11$xexpr)
> tgarch11.spec = ugarchspec(variance.model = list(model="fGARCH",submodel="TGARCH", garchOrder=c(1,1)),
+  mean.model = list(armaOrder=c(0,0)))
> aapl.tgarch11.fit = ugarchfit(spec=tgarch11.spec, data=ret)
> coef(aapl.egarch11.fit)
         mu       omega      alpha1       beta1
 0.02163174 -1.14814240  0.07408394  0.31963281
     gamma1
 0.33290745
> ni.tgarch11 <- newsimpact(aapl.tgarch11.fit)
> plot(ni.tgarch11$zx, ni.tgarch11$zy, type="l", lwd=2, col="blue",
+      main="TGARCH(1,1) - News Impact",
+      ylab=ni.tgarch11$yexpr, xlab=ni.tgarch11$xexpr)
> garch11.spec = ugarchspec(variance.model = list(garchOrder=c(1,1)),mean.model = list(armaOrder=c(0,0)),fixed.pars=list(mu = 0, omega=0.1, alpha1=0.1,
+ beta1 = 0.7))
> garch11.sim = ugarchpath(garch11.spec, n.sim=1000)

Friday, 13 September 2019

Find G.C.D of 275 and 200 in number theory

How to find G.C.D of 275 and 200 and express it in form  m.275+n.200.

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sol.
Applying the process of the division algorithm, 
 
275=(200).1+75.......................(1)
200=(75).2+50  ........................(2)
75 =(50).1+25  ..........................(3)
50= (25).2+0 ..............................(4)
Hence(275,200)=the last non-zero remainder in the above repeated division=25
Substituting backward, we have
25=75-(50).1   from eq.3
    =75-[200-(75).2].1   from eq.2
   =75.3-200.1
   =[275-(200).1].3-200.1   from eq.1
     =275.3-200.4
     =(3).275+(-4).200.
Hence m=3, n=-4





Friday, 6 September 2019

Four phases of the lunar cycle

What are four phases of the lunar cycle?

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There is a category name corresponding to the phases name.
print(lunar.4phases)
[1] "New"    "Waxing" "Full"   "Waning"

Eight category labels of the moon for lunar phase.

print(lunar.8phases)
[1] "New"             "Waxing crescent"
[3] "First quarter"   "Waxing gibbous" 
[5] "Full"            "Waning gibbous" 
[7] "Last quarter"    "Waning crescent"

How to get the lunar phase on a specific date?

lunar.phase(as.Date("2013-05-06"))
[1] 5.507184
 print(terrestrial.seasons)
[1] "Winter" "Spring" "Summer" "Autumn"

#Distance to the moon is returned in units of earth radii, or as a 5-level factor variable referring to the moon's perigee (at about 56 earth radii) and apogee (at about 63.8 earth radii).
> lunar.distance(as.Date("2004-03-24"))
[1] 62.87136
.>lunar.distance.mean(x, towards = -6, ..., by = c("date", "hours", "day","night"))
lunar.distance.mean(as.Date("2004-03-24"))
[1] 61.10886

lunar.illumination(x, shift = 0)


> lunar.illumination(as.Date("2004-03-24"))
[1] 0.1189133

Thursday, 5 September 2019

calculate amortization-R

 How to calculate amortization R

https://www.mathclasstutor.online

Loan= loan amount
n= the number of payments/periods
pmt= value of level payments
i= nominal interest rate convertible ic times per year
ic= interest conversion frequency per year
pf= the payment frequency- number of payments per year
t= the specified period for which the payment amount, interest paid, principal paid, and loan balance are solved for
library("FinancialMath", lib.loc="~/R/win-library/3.6")

> amort.period(Loan=100,n=5,i=.01,t=3)

           Amortization
Loan          100.00000
PMT            20.60398
Eff Rate        0.01000
Years           5.00000
At Time 3:      3.00000
Int Paid        0.60596
Princ Paid     19.99802
Balance        40.59798
> amort.period(n=5,pmt=30,i=.01,t=3,pf=12)

           Amortization

Loan         149.627429
PMT           30.000000
Eff Rate       0.010000
i^(12)         0.009954
Periods        5.000000
Years          0.416667
At Time 3:     3.000000
Int Paid       0.074535
Princ Paid    29.925465
Balance       59.925424
> amort.period(Loan=100,pmt=24,ic=1,i=.01,t=3)
           Amortization
Loan         100.000000
PMT           24.000000
Eff Rate       0.010000
Years          4.277206
At Time 3:     3.000000
Int Paid       0.537700
Princ Paid    23.462300
Balance       30.307700

Calculate Level Annuity

> annuity.level(pv=NA,fv=101.85,n=10,pmt=8,i=NA,ic=1,pf=1,imm=TRUE)
         Level Annuity
PV          61.0029569
FV         101.8500000
PMT          8.0000000
Eff Rate     0.0525943
Years       10.0000000
> annuity.level(pv=80,fv=NA,n=15,pf=2,pmt=NA,i=.01,imm=FALSE)
         Level Annuity
PV        80.000000000
FV        86.198615501
PMT        5.521069443
Eff Rate   0.010000000
i^(2)      0.009975124
Periods   15.000000000
Years      7.500000000

Calculate Arithmatic annuity

> annuity.arith(pv=NA,fv=NA,n=20,p=100,q=4,i=.03,ic=1,pf=2,imm=TRUE)
         Arithmetic Annuity
PV             2.338128e+03
FV             3.142248e+03
P              1.000000e+02
Q              4.000000e+00
Eff Rate       3.000000e-02
i^(2)          2.977831e-02
Periods        2.000000e+01
Years          1.000000e+01
annuity.arith(pv=NA,fv=3000,n=20,p=100,q=NA,i=.05,ic=3,pf=2,imm=FALSE)
         Arithmetic Annuity
PV             1.827106e+03
FV             3.000000e+03
P              1.000000e+02
Q              1.664438e+00
Eff Rate       5.083796e-02
i^(3)          5.000000e-02
i^(2)          5.020776e-02
Periods        2.000000e+01
Years          1.000000e+01
annuity.geo(pv=NA,fv=100,n=10,p=9,k=.02,i=NA,ic=2,pf=.5,plot=TRUE)

         Geometric Annuity

PV            9.669279e+01
FV            1.000000e+02
P             9.000000e+00
K             2.000000e-02
Eff Rate      1.682984e-03
i^(2)         1.682276e-03
i^(0.5)       1.684400e-03
Periods       1.000000e+01
Years         2.000000e+01
Geometric Annuity

 perpetuity.geo(pv=1000,p=5,k=NA,i=.04,ic=1,pf=1,imm=FALSE)

         Geometric Perpetuity

PV                   1.00e+03
P                    5.00e+00
K                    3.48e-02
Eff Rate             4.00e-02

Tuesday, 3 September 2019

Abelian group in module

How to say every abelian group G is a module over the ring of integers I?

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Let G be an abelian group, the operation in G being denoted by+ and the identity element of G by 0. For any integer n and for any element an of G  we define na in a several ways:
If n is a positive integer, we define na=a+a+.....nterms.if n=0, we define 0a=0 where 0 on the right-hand side is the identity of G.
If n is a negative number, we can say n=-m where m is +ve integer, we define  (-m)a=-(ma) where -ma denoted  the inverse of ma in G.It can be easily seen that -(ma)=m(-a).

Monday, 2 September 2019

Cross product of two matrix

How to find the cross product of two matrix

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> C_mat = matrix(c(1, 2, 3, 4, 5, 7), 2, 3)
>
> D_mat = matrix(c(2, 2, 2, 2, 2, 2), 2, 3)
>
> crossprod(C_mat, D_mat)
     [,1] [,2] [,3]
[1,]    6    6    6
[2,]   14   14   14
[3,]   24   24   24
>
> t(C_mat) %*% D_mat
     [,1] [,2] [,3]
[1,]    6    6    6
[2,]   14   14   14
[3,]   24   24   24
>
> all.equal(crossprod(C_mat, D_mat), t(C_mat) %*% D_mat)
[1] TRUE
> crossprod(C_mat, C_mat)
     [,1] [,2] [,3]
[1,]    5   11   19
[2,]   11   25   43
[3,]   19   43   74
>
> t(C_mat) %*% C_mat
     [,1] [,2] [,3]
[1,]    5   11   19
[2,]   11   25   43
[3,]   19   43   74
>
> all.equal(crossprod(C_mat, C_mat), t(C_mat) %*% C_mat)
[1] TRUE

Latin Cubes

What is Latin Cubes? The practical applications of Latin crops and related designs are factorial experiment.factorial experiments for me...