User:Lovepeacejoy404/sandbox

Mathematical treatment

At time t = 0 a loan of an amount $P_{0}$ is disbursed . This sum is repaid with constant installments of amount R paid starting from time t = 1, therefore if we consider the year as a unit of measurement then R is the annual installment . Indicating with r the constant interest rate with which the interest on the residual debt is calculated $P_{t}$ we have that:

$\ P_{t+1}=P_{t}+rP_{t}-R=(1+r)P_{t}-R$

Calculating $P_{1}$ i.e. the loan after 1 year is:

$\ P_{1}=(1+r)P_{0}-R$

Calculating $P_{2}$ i.e. the loan after 2 year is:

$\ P_{2}=(1+r)P_{1}-R=(1+r)^{2}P_{0}-R(1+r)-R$

Calculating $P_{3}$ i.e. the loan after 3 year is:

$\ P_{3}=(1+r)P_{2}-R=(1+r)^{3}P_{0}-R(1+r)^{2}-R(1+r)-R$

Therefore the loan at time t will be:

$\ P_{t}=(1+r)P_{t-1}-R=(1+r)^{t}P_{0}-R(1+r)^{t-1}-....-R(1+r)^{2}-R(1+r)-R$

Place:

$\ S_{t}=:-R(1+r)^{t-1}-....-R(1+r)^{2}-R(1+r)-R$

We rewrite $P_{t}$ like:

$\ P_{t}=(1+r)^{t}P_{0}+S_{t}$

Multiplying both sides of the equation $S_{t}$ for $-(1+r)$ we have:

$\ -(1+r)S_{t}=R(1+r)^{t}+....+R(1+r)^{3}+R(1+r)^{2}+R(1+r)$

By adding the 2 equations member by member, we obtain:

$\ S_{t}-(1+r)S_{t}=R(1+r)^{t}-R$

From which we obtain:

$\ S_{t}=-{\dfrac {R}{r}}\left((1+r)^{t}-1\right)$

Then the loan at time t is equal to:

$\ P_{t}=(1+r)^{t}P_{0}+S_{t}=(1+r)^{t}\left(P_{0}-{\dfrac {R}{r}}\right)+{\dfrac {R}{r}}$

Considering the succession $P_{t}$ continuous if you can calculate the derivative to see when it is increasing or decreasing. Therefore it turns out:

$\ {\dfrac {d(P(t))}{dt}}=(1+r)^{t}log(1+r)\left(P_{0}-{\dfrac {R}{r}}\right)$

whereby the derivative is positive and the function is increasing by $R<=P_{0}r$ and therefore in this case the loan would never be extinguished, while for $R>P_{0}r$ the derivative is negative, the function is decreasing so that after a certain time the loan is extinguished. Wanting to calculate after how long the loan expires, the condition must be imposed:

$P_{t}=(1+r)^{t}\left(P_{0}-{\dfrac {R}{r}}\right)+{\dfrac {R}{r}}=0$

from which the exponential equation is obtained:

$(1+r)^{t_{*}}=\left({\dfrac {R}{R-P_{0}r}}\right)$

therefore passing to the logarithms we obtain that the time $t_{*}$ i at which the loan is extinguished is:

$t_{*}={\dfrac {\log \left({\dfrac {R}{R-P_{0}r}}\right)}{\log(1+r)}}$

while the annual installment to pay off the loan in an annual time $t_{*}$ at interest rate r is:

$R={\frac {{P_{0}}r\,{{\left(r+1\right)}^{t_{*}}}}{{{\left(r+1\right)}^{t_{*}}}-1}}$

Using the wxMaxima program to calculate the monthly payment of a loan of € 100,000 at the rate of 2% over 20 years, a monthly payment of € 505.88 is obtained:

M_0 : 100000 ;
t : 20 * 12 ;
r : 0.02 / 12 ;
R : ( M_0 * r * ( 1 + r ) ^ t ) / (( 1 + r ) ^ t -1 ) ;   

( M_0 ) 100000 
( t ) 240       
( r ) 0.001666666666666666      
( R ) 505.8833350451002

Using the wxMaxima program to calculate the years required for a loan of € 136,000 at the rate of 3.5% with a monthly payment of € 616, we obtain a time of 30 years:

P_0 : 136000 ;
R : 616 * 12 ;
r : 0.035 ;
t : log ( R / ( R - P_0 * r )) / log ( 1 + r );

( P_0 ) 136000
( R ) 7392      
( r ) 0.035     
( t ) 30.01777598822768

Finally, wanting to calculate the annual interest rate of a loan of € 136,000 with a monthly payment of € 616 for 30 years, using wxMaxima both real and complex solutions are obtained but only the real positive one must be considered, which is equal to r = 0.035 i.e. r = 3.5%:

P_0 : 136000 ;
R : 616 * 12 ;
t : 30 ;
to_poly_solve ([ R = ( P_0 * r * ( 1 + r ) ^ t ) / (( 1 + r ) ^ t -1 )], [ r ]);   

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