| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
OK guys-
It's been a number of years since my education has 'refreshed'
my mind regarding equation solving. The following is throwing me
because of the exponents (and my text books are still boxed from
my last move).
I am trying to solve the equation for each of the variables
separately. In other words, I want to yield 4 equations that
can be solved given 3 of the variables. The LOG and EXP are
making me crazy! Can anyone help out or refresh me on the
LOG/EXP rules?
Thanks,
Mark
- Term
( 1 - ( 1 + Int Rate ) )
Principle / Payment = [ --------------------------------- ]
Int Rate
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 883.1 | three out of four | ZFC::DERAMO | I am, therefore I'll think. | Mon Jun 06 1988 19:59 | 32 |
Well, you will be able to solve it for Principle, Payment,
or Term given the other three; but not for Int Rate.
That is usually calculated iteratively -- keep closing
in on the interest rate until you get the value that
fits in the equation.
Solving for Principle or Payment in the equation is
straightforward. To solve for Term, you proceed like
this:
Prin * Intrate - Term
-------------- = 1 - (1 + IntRate)
Payment
so
- Term Prin * IntRate
(1 + IntRate) = 1 - --------------
Payment
so
Prin * IntRate
(- Term) LOG(1 + IntRate) = LOG(1 - --------------)
Payment
and finally,
Prin * IntRate
LOG(1 - --------------)
Payment
Term = - -----------------------------------
LOG(1 + IntRate)
Dan
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| 883.2 | 3-1/2 out of the way... | DELNI::MHARRIS | Mark Jay Harris, Term Srvr Mktg Mgr | Tue Jun 07 1988 00:41 | 8 |
Thanks. I knew the LOG rules would *seem* familiar! I think I
came up with the iterative process for the interest rate as well
so I guess I am well on my way.
Thanks again.
Mark
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| 883.3 | more LOG and EXP rules. | ZFC::DERAMO | I am, therefore I'll think. | Tue Jun 07 1988 10:25 | 23 |
The only LOG rule I recall using in .1 was
Y
LOG(X ) = Y LOG(X)
If you are interested in other LOG and EXP rules, here
are more:
LOG(XY) = LOG(X) + LOG(Y)
EXP( LOG(X) ) = X
LOG( EXP(X) ) = X
EXP( X + Y ) = EXP(X) EXP(Y)
Y
EXP(XY) = ( EXP(X) )
Taking LOG or EXP of both sides of one of these usually
gives another one on the list.
Dan
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| 883.4 | numbers , which larger? | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Sun Jul 26 1992 03:30 | 9 |
which is larger?
1,000,000 999,999
1,000,000 or 1,000,001
hint: no calculator needed ;-)
/Nasser
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| 883.5 | AUSSIE::GARSON | Tue Jul 28 1992 19:56 | 47 | ||
re .4
> which is larger?
>
> 1,000,000 999,999
> 1,000,000 or 1,000,001
(n+1)^(n-1)
Define f = ----------- (n >= 1)
n n^n
f 2n 2n
n+1 (n+2)^n n^n ( n(n+2) ) ( n�+2n )
---- = ----------- � ----------- = ( ---------- ) = ( ------- )
f (n+1)^(n+1) (n+1)^(n-1) ( (n+1)(n+1) ) ( n�+2n+1 )
n
0 < 1
=> n�+2n < n�+2n+1
n�+2n
=> ------- < 1
n�+2n+1
f
n+1
=> ---- < 1
f
n
=> f < f
n+1 n
But f = 3/4
2
=> f < 1, for all n >= 2
n
(n+1)^(n-1)
=> ----------- < 1, for all n >= 2
n^n
=> (n+1)^(n-1) < n^n, for all n >= 2
Putting n=1000000 gives the answer. 1000000^1000000 is the larger.
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