recherch_xi.mws

**Recherches sur
l'intégration des équations différentielles aux
différences finies, et sur leur usage dans la théorie des
hasards.**

P.S. Laplace

**PROBLEM
XI.**

*Let * a
* be a sum which Paul constitutes to an
annuity, in
a way that the interest is *
*of that which is due
to him: I suppose that, for some arbitrary reasons, one keeps each year
the
fraction*
*of this interest, so
that Paul, at the end of the first year, for example, must collect only
the
quantity*
, *this put,
if one pays him every year the sum*
, *and, consequently, more
than is due to him, and let the surplus be used to amortize the
capital,
one asks what this capital will become in the year *
x .

Let
denote this capital in
the year
*x* . At the
end of the
year Paul will be due
. Since
the sum
is paid, the capital
will be diminished by
. We
have therefore

`> `
**restart:**

`> `
**eqn11:=y(x+1)=y(x)-a/m+y(x)*(n-1)/(n*m);**

The
initial condition is
.
Therefore, the solution is given by

`> `
**ans:=rsolve({eqn11,y(1)=a},y(x));**

`> `
**normal(ans);**

We
require when the capital
will vanish. That is, for what value of
*x* is
?

`> `
**vanish:=solve(ans=0,x);**

If
we substitute, as does
Laplace,
and
then we find that this
occurs in 53.3 years.

`> `
**subs({m=20,n=10},vanish);**

`> `

`> `
**evalf(%);**

`> `

**Addendum
to the
problem:** *A
person owes the sum * a
*, and wishes to release himself at the end of
* h *
years, so
that she owes nothing in the year *
* the interest being
always *
* of the quantity due;
the question is to find what
must she give for this each year.*

With
Laplace, we let
*p* be what
she must
give each year.

`> `
**eqn11a:=y(x+1)=y(x)*(1+1/m)-p;**

`> `
**ans:=rsolve({eqn11a,y(1)=a},y(x));**

Laplace
puts
.

`> `
**subs(x=h+1,ans);**

`> `
**ans2:=solve(%,p);**

`> `
**simplify(%);**

`> `

`> `