Question #e8dd9

3 Answers
Dec 8, 2016

Answer:

See below.

Explanation:

Given a model

#y(a,t)=a_1t+a_2+a_3sin(omega_0t)#

and a set of points

#P=((t_1,y_1),(t_2,y_2),cdots,(t_n,y_n))#

We will try to adjust the model to the #P# table, by choosing #a_i# such that is minimum the error defined as

#E(a)=sum_(i=1)^n(y(a,t_k)-y_k)^2#

The determination of #a=(a_1,a_2,a_3)# is done by solving the linear system obtained through the stationary conditions

#(partial E(a))/(partial a_i) = 0# for #i=1,2,3#

so we have to solve the linear system

#M a = b#

where

#M = ( (sum t_k^2, sum t_k,sum sin(omega_0t_k)t_k), ( sum t_k,n,sum sin(omega_0t_k)), (sum sin(omega_0t_k)t_k,sum sin(omega_0t_k),sum sin(omega_0t_k)^2))#

and

#b=((sum t_k y_k),(sum y_k),(sum sin(omega_0 t_k)y_k))#

Putting numeric values into the formulas we get

#M=((540, 60, -12),(60, 10, 0),(-12, 0, 4))#
#b=((2430.), (350.), (10.))#

obtaining #{a_1 = 2.5, a_2 = 20, a_3 = 10#

Attached a plot showing the results.

enter image source here

Answering the requested items:

a) Using an square minimum error criteria, the parameters are:

#m = a_1 = 2.5#
#b = a_2 = 20#
#A = a_3 = 10#

b) The stock appreciation is linked to a positive rate so

#(df)/(dt) = m + A omega_0 cos(omega_0 t) > 0#

This is attained for

#0 le t le 3.95# and #8.049 le t le 12#

c) The stock loses value for #3.95 le t le 8.049#

Dec 9, 2016

Given that #f(t)# the value of a stock is modeled as

#f(t)=mt+b+Asin((pit)/6)# where t is in months since Jan1

Let us consider

#"At Jan1 "t= 0 and f(0)=$20.00#

#"At Apr1 "t= 3and f(3)=$37.50#

#"At Jul1 "t= 6 and f(6)=$32.50#

#"At Oct1 "t= 9 and f(6)=$35.00#

#"At Jan1 "t= 12 and f(6)=$50.00#

so
#"At Jan1 "t= 0 and f(0)=$20.00#

#=>mxx0+b+Asin(0)=20=>b=20.......(1)#

#"At Apr1 "t= 3and f(3)=$37.50#

#=>mxx3+20+Asin((pi*3)/6)=37.50#

#=>3m+20+A=37.50#

#=>3m+A=37.50-20.00#

#=>3m+A=17.50............................(2)#

#"At Jul1 "t= 6 and f(6)=$32.50#

#=>6xxm+20+Asin((pi*6)/6)=35.00#

#=>6xxm+20+0=35#

#=>m=15/6=2.5..........................(3)#

From (2) and (3) we get

#=>3xx2.5+A=17.50#

#=>A=10.00#

Hence
(a)
#color(red)(m=2.50" "b=20 and A=10)#
(b)

Now the given equation becomes

#f(t)=2.5t+20+10sin((pit)/6)#

Differentiating this equation w,r to t we get the rate of change of the price of stock as

#f'(t)=d/(dt)(2.5t)+d/(dt)(20)+d/(dt)(10sin((pit)/6))#

#=>f'(t)=2.5+(10pi)/6cos((pit)/6)#
In this equation the value of #cos((pit)/6)# varies with time.
#cos((pit)/6)# has got maximum positive value at t=0 and t =12 as #cos 0 and cos (2pi) = 1#
So rate of positive change of price of stock will occur in this period i.e. Jan every year this means the stock appreciate s most in this period.

(c)

Taking #f'(t)=0# we have

#2.5+(10pi)/6cos((pit)/6)=0#

#=>cos((pit)/6)=-1.5/pi#

#=>t=6/picos^-1(-1.5/pi)~~3.95~~4->color(red)"month of May"#

when t =4

#f''(4)=-(10pi^2)/36sin((pi*4)/6)<0->"indicates maximum"#

Again #t = 6/pi(2pi-cos^-1(-1.5/pi))=12-6/picos^-1(-1.5/pi)#

#=12-3.95=8.05->" indicates month of September" #

when t =8

#f''(8)=-(10pi^2)/36sin((pi*8)/6)>0->color(red)"indicates Minimum"#

So during the period May -September growth of price gets diminished i.e the price of the stock is actually losing in this period.
enter image source here

Dec 17, 2016

Answer:

Alternate solution for part (c)

Explanation:

For Solution posted by @dk_ch
The modelling equation becomes

#f(t)=2.5t+20+10sin((pit)/6)#

Using inbuilt graphics tool the plotted equation looks like

enter image source here
The maximum and minimum of the curve are located for #t=3.951# and #8.049# respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.

--.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-

For Solution posted by @Cesareo R

#f(t)=2.25t+18.49+7.29sin((pit)/6)#, values rounded to two decimal places

#m = a_1 = 2.25439#, #b = a_2 = 18.49#, #A = a_3 = 7.29285#
enter image source here
Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for #t=4.204# and #7.796# respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.