Solve for X

hello all!

How do I solve the below equation in excel?

[5.2/(1+x)^2] + [12.1/(1+x)^3] + [16.7/(1+x)^4] + [16.8/
(1+x)^5] + [17.1/(1+x)^6] = 0


Thanks!

Kunal

Kunal wrote:
hello all!

How do I solve the below equation in excel?

[5.2/(1+x)^2] + [12.1/(1+x)^3] + [16.7/(1+x)^4] + [16.8/
(1+x)^5] + [17.1/(1+x)^6] = 0


Thanks!

Kunal

One way might be to use the solver or goal seek.

If you have it try:

Tools - Goal Seek

Make 'x' be any given cell, and set your formula up so that it tries to make
the formula = 0 by changing 'x'.

HTH,

Alan.

"Kunal" wrote in message
...
hello all!

How do I solve the below equation in excel?

[5.2/(1+x)^2] + [12.1/(1+x)^3] + [16.7/(1+x)^4] + [16.8/
(1+x)^5] + [17.1/(1+x)^6] = 0


Thanks!

Kunal

I do not think your equation cannot be simply solved by excel because the
roots
appear to be imaginary, specifically
(-2.2629 + 0.813758 I), (-2.2629 - 0.813758 I), (-0.9005 + 1.2029 I) and
(-0.9005 - 1.2029 I).

a/(1+x)^2 + b/(1+x)^3 + c/(1+x)^4 + d/(1+x)^5 + e/(1+x)^6 = 0 may be
considered the same as solving for
a x^4 + (4 a + b) x^3 + (6 a + 3 b + c) x^2 + (4 a + 3 b + 2 c + d) x + a +
b + c + d + e = 0. As this is now a quartic, you could perhaps use an add in
such as
polynomials.zip from http://www.tushar-mehta.com/ ,which gives the results
as above.

mows
Excel XP SP2 / Win XP SP1

I get 3277.8
File on its way to you.
Bernard
www.stfx.ca/people/bliengme

"Kunal" wrote in message
...
hello all!

How do I solve the below equation in excel?

[5.2/(1+x)^2] + [12.1/(1+x)^3] + [16.7/(1+x)^4] + [16.8/
(1+x)^5] + [17.1/(1+x)^6] = 0


Thanks!

Kunal

I might be wrong, but I think the equation goes to zero as the limit of x
approaches infinity.
With x's in the denominator, as x gets larger, each term tends towards zero.

equ =
17.1/(1 + x)^6 + 16.8/(1 + x)^5 + 16.7/(1 + x)^4 + 12.1/(1 + x)^3 + 5.2/(1 +
x)^2

Limit[equ, x - Infinity]
0

vs..

Limit[equ, x - 3277.8]
4.840405263255477*^-7

You could make the numbers rational, and use..
v = (679 + 1073*x + 842*x^2 + 329*x^3 + 52*x^4)/(10*(1 + x)^6)


x /. NSolve[v == 0]

-2.262940780472177 + 0.8137579106562657*I,
-2.262940780472177 - 0.8137579106562657*I,
-0.9005207579893606 + 1.2028992293600107*I,
-0.9005207579893606 - 1.2028992293600107*I

(same solution as Mows)

In Excel XP, there is a workaround for using Solver to find Imaginary
solutions. It works ok sometimes, but it is not the best.
Have two adjustable cells called "real" and "imag"
Combine these two numeric cells into a string =COMPLEX(A1,A2)
Then use the engineering function like =IMPOWER() and =IMSUM(), ...etc
Try to set the final target cell to zero." IMABS() "
Not the best, but it can work in a pinch.

--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


"Bernard Liengme" wrote in message
...
I get 3277.8
File on its way to you.
Bernard
www.stfx.ca/people/bliengme

"Kunal" wrote in message
...
hello all!

How do I solve the below equation in excel?

[5.2/(1+x)^2] + [12.1/(1+x)^3] + [16.7/(1+x)^4] + [16.8/
(1+x)^5] + [17.1/(1+x)^6] = 0


Thanks!

Kunal

Hello Bernard,

I am interested in how you get your answer, as I cannot repeat it. As I see
it, the function

[5.2/(1+x)^2] + [12.1/(1+x)^3] + [16.7/(1+x)^4] + [16.8/(1+x)^5] +
[17.1/(1+x)^6] = 0

has no negative values and does not cross the axis, therefore has no real
roots, it approaches 0 as x- + / - infinity. As x tends to -1 the functions
value becomes infinity.
As Dana says, setting x = 3277.8 (or indeed -3277.47308) the functions
value is 0.000000484. Choosing a higher x will return a value closer but not
equal to 0

Indeed, for large x the function approximates to 5.2 / x^2
For x approaching -1, the function is dominated by 17.1 / (1 + x)^6

I am not a solver expert, so possibly the easiest other way to tackle it is
to solve for the quartic

a x^4 + (4 a + b) x^3 + (6 a + 3 b + c) x^2 + (4 a + 3 b + 2 c + d) x +
(a + b + c + d + e) = 0

where a, b, c, d and e are the coeffs in
a/(1+x)^2 + b/(1+x)^3 + c/(1+x)^4 + d/(1+x)^5 + e/(1+x)^6 = 0

I suspect that it is now possible, with quite a bit of tedium, to find the
reducing cubic and so on using the likes of IMPRODUCT() & etc
from the Analysis ToolPak to eventually solve the quartic. The other
even simpler way is to cheat.
Use the addin from http://www.tushar-mehta.com/ I mentioned before!


Best regards
Peter -- (polygon moments / Greens theorem)

mows
Excel XP SP2 / Win XP SP1