Hi,
I am a newbie at excel so I would like to know How to do the following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then the that
result in the next iteration and so on until n=28. Then it would display the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any help,
Amish

Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then the
that
result in the next iteration and so on until n=28. Then it would display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any help,
Amish

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then the
that
result in the next iteration and so on until n=28. Then it would display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any help,
Amish

No,
this does not compute what I need.

"Dana DeLouis" wrote:

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then the
that
result in the next iteration and so on until n=28. Then it would display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any help,
Amish

No,
this does not compute what I need.

Can you give a simple example, and the first few terms of the sequence.

--
Dana DeLouis


"axr0284" wrote in message
...
No,
this does not compute what I need.

"Dana DeLouis" wrote:

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then
perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the
following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the
following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then
the
that
result in the next iteration and so on until n=28. Then it would
display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any
help,
Amish

Here you go. I would use it to calculate my loan interest every day.
Daily interest percentage: 0.02 (Y)
Starting loan amount 7695 (X)
Starting day 01/11/08 (A)
ending date 01/16/08 (B)

In excel, if I do compute it manually it comes out like this
Amount Date
7695 01/11/08
7696.598684 01/12/08
7698.197701 01/13/08
7699.797049 01/14/08
7701.39673 01/15/08
7702.996743 01/16/08

I would like to automate this process so that i don't need to calculate each
date's amount to obtain the next one. I would put in the start and end data
and excel would calculate what I owe on the end date. I hope it's clearer now.

"Dana DeLouis" wrote:

No,
this does not compute what I need.

Can you give a simple example, and the first few terms of the sequence.

--
Dana DeLouis


"axr0284" wrote in message
...
No,
this does not compute what I need.

"Dana DeLouis" wrote:

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then
perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the
following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the
following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then
the
that
result in the next iteration and so on until n=28. Then it would
display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any
help,
Amish

Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

this does not compute what I need.


Hi. The sum of the first 6 terms is:

=SUM(A2:A7) -
46,193.98691

When I enter the equation:

Sub Demo()
Dim x, y, z

x = 7695
y = 0.000207756229861
z = 6 ' (ie b-a)
Debug.Print (x * ((y + 1) ^ z - 1)) / y
End Sub

It returns the same amount.

46193.98691

The value of y is probably not exact.
=7695*(1+0.02%) does not exactly match your next value.

In your table, if the next term was actually
= 7695 * (1 + .02%) = 7696.5390
and copied down, then using

y=.02/100 would give the same amount.

46,193.09116

--
HTH
Dana DeLouis



"axr0284" wrote in message
...
Here you go. I would use it to calculate my loan interest every day.
Daily interest percentage: 0.02 (Y)
Starting loan amount 7695 (X)
Starting day 01/11/08 (A)
ending date 01/16/08 (B)

In excel, if I do compute it manually it comes out like this
Amount Date
7695 01/11/08
7696.598684 01/12/08
7698.197701 01/13/08
7699.797049 01/14/08
7701.39673 01/15/08
7702.996743 01/16/08

I would like to automate this process so that i don't need to calculate
each
date's amount to obtain the next one. I would put in the start and end
data
and excel would calculate what I owe on the end date. I hope it's clearer
now.

"Dana DeLouis" wrote:

No,
this does not compute what I need.

Can you give a simple example, and the first few terms of the sequence.

--
Dana DeLouis


"axr0284" wrote in message
...
No,
this does not compute what I need.

"Dana DeLouis" wrote:

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then
perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I understand
it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the
following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the
following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and then
the
that
result in the next iteration and so on until n=28. Then it would
display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any
help,
Amish

and excel would calculate what I owe on the end date.

7702.996743 01/16/08

Sounds like you no longer want SUM

The loan at time 0 is 7695.

Is this what you are looking for?
Your 5th payment is:

=FV(0.02%,5,0,-7695)

$7,702.70

Maybe??

--
HTH
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

this does not compute what I need.


Hi. The sum of the first 6 terms is:

=SUM(A2:A7) -
46,193.98691

When I enter the equation:

Sub Demo()
Dim x, y, z

x = 7695
y = 0.000207756229861
z = 6 ' (ie b-a)
Debug.Print (x * ((y + 1) ^ z - 1)) / y
End Sub

It returns the same amount.

46193.98691

The value of y is probably not exact.
=7695*(1+0.02%) does not exactly match your next value.

In your table, if the next term was actually
= 7695 * (1 + .02%) = 7696.5390
and copied down, then using

y=.02/100 would give the same amount.

46,193.09116

--
HTH
Dana DeLouis



"axr0284" wrote in message
...
Here you go. I would use it to calculate my loan interest every day.
Daily interest percentage: 0.02 (Y)
Starting loan amount 7695 (X)
Starting day 01/11/08 (A)
ending date 01/16/08 (B)

In excel, if I do compute it manually it comes out like this
Amount Date
7695 01/11/08
7696.598684 01/12/08
7698.197701 01/13/08
7699.797049 01/14/08
7701.39673 01/15/08
7702.996743 01/16/08

I would like to automate this process so that i don't need to calculate
each
date's amount to obtain the next one. I would put in the start and end
data
and excel would calculate what I owe on the end date. I hope it's clearer
now.

"Dana DeLouis" wrote:

No,
this does not compute what I need.

Can you give a simple example, and the first few terms of the sequence.

--
Dana DeLouis


"axr0284" wrote in message
...
No,
this does not compute what I need.

"Dana DeLouis" wrote:

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then
perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I
understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the
following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the
following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and
then
the
that
result in the next iteration and so on until n=28. Then it would
display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any
help,
Amish

Sorry,
for some reason, when I ran your formula, the numbers did not come out
right but now i realize i made the mistake. Yeah eventually I would like to
calculate the accrued interest on the loan daily as well as the loan amount
and a bunch of different stuff. I'll check out the FV function also. Did not
know it existed. Thanks for all the help. I really appreciate it.
Amish

"Dana DeLouis" wrote:

and excel would calculate what I owe on the end date.

7702.996743 01/16/08

Sounds like you no longer want SUM

The loan at time 0 is 7695.

Is this what you are looking for?
Your 5th payment is:

=FV(0.02%,5,0,-7695)

$7,702.70

Maybe??

--
HTH
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

this does not compute what I need.


Hi. The sum of the first 6 terms is:

=SUM(A2:A7) -
46,193.98691

When I enter the equation:

Sub Demo()
Dim x, y, z

x = 7695
y = 0.000207756229861
z = 6 ' (ie b-a)
Debug.Print (x * ((y + 1) ^ z - 1)) / y
End Sub

It returns the same amount.

46193.98691

The value of y is probably not exact.
=7695*(1+0.02%) does not exactly match your next value.

In your table, if the next term was actually
= 7695 * (1 + .02%) = 7696.5390
and copied down, then using

y=.02/100 would give the same amount.

46,193.09116

--
HTH
Dana DeLouis



"axr0284" wrote in message
...
Here you go. I would use it to calculate my loan interest every day.
Daily interest percentage: 0.02 (Y)
Starting loan amount 7695 (X)
Starting day 01/11/08 (A)
ending date 01/16/08 (B)

In excel, if I do compute it manually it comes out like this
Amount Date
7695 01/11/08
7696.598684 01/12/08
7698.197701 01/13/08
7699.797049 01/14/08
7701.39673 01/15/08
7702.996743 01/16/08

I would like to automate this process so that i don't need to calculate
each
date's amount to obtain the next one. I would put in the start and end
data
and excel would calculate what I owe on the end date. I hope it's clearer
now.

"Dana DeLouis" wrote:

No,
this does not compute what I need.

Can you give a simple example, and the first few terms of the sequence.

--
Dana DeLouis


"axr0284" wrote in message
...
No,
this does not compute what I need.

"Dana DeLouis" wrote:

Oh wait. When you say for n=1 to (b-a), the first term is actually
f(n-1) - f(0).
See if this is better. If z is the number of terms (ie b-a) then
perhaps:
(x*((y + 1)^z - 1))/y

--
HTH :)
Dana DeLouis



"Dana DeLouis" wrote in message
...
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

Hi. You didn't quite define your F( ) function, but if I
understand it
correctly...

= (x*((y + 1)^(-A + B + 1) - y - 1))/y

--
HTH :)
Dana DeLouis
"To understand recursion, one must first understand recursion."



"axr0284" wrote in message
...
Hi,
I am a newbie at excel so I would like to know How to do the
following.
Cell A1 would contain 1 date A
Cell A2 would contain a second date B
Cell A3 would contain a number X
Cell A4 would contain a number Y
I would like cell A5 to contain a formula that will perform the
following
recursively:
Sum[F(n-1)+{F(n-1)*Y}] for n = 1 to (B - A) and F(0) = X

so if n = 1 to 20, it would iterate 20 times first using X and
then
the
that
result in the next iteration and so on until n=28. Then it would
display
the
final result in cell A5.

I have not been able to figure out how to do this. Thanks for any
help,
Amish