exponential equation for?

Question

how would you get an exponential equation for a line that goes through 5 points given those points?

for arguments sake the points can be

1,1
2,3
3,6
4,10
5,15
*yes this will be a simple looking equation, but i need to know how to get there*

Answer 1

if, by "exponential equation", you mean:

y = a * b^(cx) + d >... (iii

start by substituting (x,y) pairs into (iii
and determing what a,b,c,d are ....

just wild-*** guessing,, I don't think that
ANY a,b,c,d will fit the points you give as
"argument's sake" ... unless you go to "complex numbers"

1,1

1 = a*b^c +d
1-d = a*(b^c)
(1-d) / a = b^c

- - - -
2,3
2 = a*(b^2c) + d
(2-d) / a = (b^2c) = (b^c)^2
(2-d) / a = (b^2c) = (b^c)^2 = ((1-d) / a)^2 {from 1,1 calc}

(multiplying eqn above by a^2 on both side)
a * (2-d) = (1-d)^2
2a -da = 1 -2d + d^2
d^2 -2d+da +1-2a = 0

d^2 + (a-2)d +(1-2a) = 0

quadratic eqn formula usage here to solve for "d"
in terms of "a" ...

My intuitiion tell me that a "exponential equation"
will NOT pass thru these point ...

possibly a "linear combination of exponential eqn's"
might do this, but it will NOT be "simple"

Answer 2

A general form for the exponential equation is
y = a b^(cx + d) + e

You have 5 pairs for (x,y)
(1,1)
(2,3)
(3,6)
(4,10)
(5,15)

If you subs. them,
1 = a b^(c + d) + e
3 = a b^(2c + d) + e
6 = a b^(3c + d) + e
10 = a b^(4c + d) + e
15 = a b^(5c + d) + e

Maybe if you do this system, you'll probably get some undefined or imaginary values for a, b, c, d, and/or e.

My experience tells me that these set of points will pass through a parabolic equation which is in the form
y = a(x - h)² + k

My experience also tells me that if this is a parabola, then you will only need 3 pairs. The extra 2 are just for checking.
1 = a(1 - h)² + k
3 = a(2 - h)² + k
6 = a(3 - h)² + k

This is easier to solve, so
1 = a(h² - 2h + 1) + k
3 = a(h² - 4h + 4) + k
6 = a(h² - 6h + 9) + k

thus,
1 = ah² - 2ah + a + k
3 = ah² - 4ah + 4a + k
6 = ah² - 6ah + 9a + k

Subtracting the first equation from the last 2 equations, we get
2 = -2ah + 3a
5 = -4ah + 8a

Multiply the first equation by -2:
-4 = 4ah - 6a
5 = 4ah + 8a

Add them:
1 = 2a
and
a = 1/2

Therefore,
5 = -4ah + 8a
5 = -4(1/2)h + 8(1/2)
5 = -2h + 4
-2h = 1
h = -1/2

Therefore,
1 = a(1 - h)² + k
1 = (1/2)(1 + 1/2)² + k
1 = (1/2)(3/2)² + k
1 = (1/2)(9/4) + k
1 = 9/8 + k
k = -1/8

Therefore,
y = a(x - h)² - 1/8
y = (x + 1/2)²/2 - 1/8
y = 4(x + 1/2)²/8 - 1/8
y = [(2x + 1)² - 1]/8
y = (4x² + 4x + 1 - 1)/8
y = (4x² + 4x)/8
y = (x² + x)/2
y = x(x + 1)/2

I have made a parabola for you...
If you really insist for the exponential, then it is up to you to solve the 5-equation system I gave you.

^_^

Answer 3

I have two thoughts on this ...

First, the example you gave increases rather slowly for an exponential. The numbers you use are a Fibonacci series. You can use this equation:

y(1) = 1
y(i) = y(i-1) + i, for i = 2, 3 ,4, ...

This will generate that entire sequence of points.

Next idea. If you have some real data that you think is exponential, and you want to generate the function. I suggest you take the natural log of the data values (the y-values), and then do a linear least squares fit. When you finish, convert it back, and you'll have an exponential curve.

If you graph the data on "semi-log paper", you ought to see a straight line, or a scatter diagram that can support a least-squares fit.

This is based on the relation e^(ln x) = x, where ln is the natural (Napierian) log (base e).

If this doesn't make sense at first, study what I wrote, and think about it. Then hopefully you'll figure it out.

Answer 4

Whats the answer