Solve Logistic Function: f(x)=\frac{6}{1+(4/3)^{-x}}

[tony873004]

I did this the same way as the book example, and the class notes example, and the other problems in this section, which I got correct, but my answer for this one doesn't agree with the back of the book.

f has a limiting value 6 and passes through (0,3) and (1,4). Find the logistic function f.


f(x)=\frac{N}{1+Ab^{-x}}

Assume that for small values of x, the function is exponential.

y=Ab^x

4=Ab^1
3=Ab^0 divide these equations:

\frac{4}{3}=b^1

b=\frac{4}{3}

Solve for A

f(x)=\frac{N}{1+Ab^{-x}}

f(0)=3

3=\frac{6}{1+Ab^{-0}}

3=\frac{6}{1+A}

3+3A=6

3A=6-3

A=1

Answer:

f(x)=\frac{6}{1+(4/3)^{-x}}

Back of book answer:

f(x)=\frac{6}{1+2^{-x}}

 

[StatusX]

Assume that for small values of x, the function is exponential.

This will only give you an approximate answer, and not a very good one, since it is only really valid for x small in the sense of being very negative. And by the way, the approximation you're using isn't correct. Keep in mind that the approximation amounts to ignoring the 1 in the denominator.

If you want an exact answer, you know N, so you just need to plug in for each x and y value and solve the two resulting equations for A and b.

 

[tony873004]

Thanks, StatusX.

I'm still confused, first because the method I used was the way the book's example gave it, the lecture notes gave it, and the way I used to solve the other problems in this section. My answers and the back of the book agree there.

But trying what you said, I'd still have 2 unknowns in each of the equations. Maybe there is a way to eliminate A since it will be constant to both equations, but I'm not sure how.

 

[topsquark]

Thanks, StatusX.

I'm still confused, first because the method I used was the way the book's example gave it, the lecture notes gave it, and the way I used to solve the other problems in this section. My answers and the back of the book agree there.

But trying what you said, I'd still have 2 unknowns in each of the equations. Maybe there is a way to eliminate A since it will be constant to both equations, but I'm not sure how.

When you set up your equation for the point (0,3), note what happens to your b variable. b^0 is equal to what?

-Dan

 

[tony873004]

When you set up your equation for the point (0,3), note what happens to your b variable. b^0 is equal to what?

-Dan

b^0=1 since anything^0=1

So in dividing the equations:

b^1/1 = b^1

or

b^1/b^0 = b^{1-0}=b^1

That's what I got when I divided the equations.

 

[HallsofIvy]

I think it is likely that you misunderstood what the book was doing- perhaps it was doing a problem in which the exponential was in the numerator rather than the denominator. It is certainly not the case that "for small values of x, the function is exponential"!

You are given 3 pieces of information with which to solve for 3 unknowns, N, A, b.

The simplest is "f has a limiting value 6". As x goes to infinity, b^-x will go to 0 so the limiting value of f is N/(1+0)= 6.

f "passes through (0,3)" so f(0)= N/(1+ Ab⁰)= 6/(1+ A)= 4.
Solve that for A.

f "passes through (1,4)" so f(1)= N/(1+ Ab^-1)= 4. You've already found N and A so it is easy to solve for b.

 

[tony873004]

f "passes through (0,3)" so f(0)= N/(1+ Ab⁰)= 6/(1+ A)= 4.
Solve that for A.

4=\frac{N}{1+Ab^0}

4=\frac{6}{1+A}

4+4A=6

4A=6-4

4A=2

A=\frac{2}{4}

A=\frac{1}{2}

f "passes through (1,4)" so f(1)= N/(1+ Ab^-1)= 4. You've already found N and A so it is easy to solve for b.

y=Ab^x

3=\frac{1}{2}b^0

3=\frac{1}{2}

** Edit
Since the back of the book and I drop A from our final answers, I think we both agree it is 1.
In the part I quoted you, did you mean =3 instead of =4, or am I more lost than I think??

 

[tony873004]

I asked the teacher how to do this problem. He did it on the board, and got the same value for b that I did (1.333...). When told that the back of the book gives 2 for b, he concluded that the back of the book is wrong.

However, I'm still pondering what HallsofIvy said, that I can't assume that the function is exponential for small values of x.

It makes sense that it is simply because the book, in its examples, tells you to assume this. That's what a logistical problem is according to them. One that starts out exponential, but has a limiting value. But that's not a good reason to think that this assumption should apply to all problems. Did the book get ahead of itself in assigning a problem that it did not explain how to do in the text?

This homework is now turned in, and I got full credit on the problem, so thanks for your helpful replies, but I'm still not completely convinced that the back of the book got it wrong.

 

[HallsofIvy]

Yes, sorry about that: it should have been
"f "passes through (0,3)" so f(0)= N/(1+ Ab0)= 6/(1+ A)= 3.
Solve that for A."


Your formula is
f(x)=\frac{N}{1+Ab^{-x}}
Now that I look at it more closely, we can multiply both numerator and denominator by b^x to get
f(x)= \frac{Nb^x}{b^x+ A}
For x small (compared with N), b^x+ A will be close to A so approximately,
f(x)= \frac{N}{A}b^x
In that sense, yes, it is approximately an exponential.

 

[VietDao29]

I asked the teacher how to do this problem. He did it on the board, and got the same value for b that I did (1.333...). When told that the back of the book gives 2 for b, he concluded that the back of the book is wrong.

No, the answer at the back of the book is indeed correct.
The function:
f(x) = \frac{6}{1 + \left( \frac{4}{3} \right) ^ {-x}} does not goes through the point (1, 4), whereas the function: f(x) = \frac{6}{1 + 2 ^ {-x}} does...
After solving for N and A, you can use the fact that f(x) passes through (1, 4) (i.e f(1) = 4), and solve for b.

 

[tony873004]

No, the answer at the back of the book is indeed correct.
The function:
f(x) = \frac{6}{1 + \left( \frac{4}{3} \right) ^ {-x}} does not goes through the point (1, 4), whereas the function: f(x) = \frac{6}{1 + 2 ^ {-x}} does...
After solving for N and A, you can use the fact that f(x) passes through (1, 4) (i.e f(1) = 4), and solve for b.

You're right. I didn't even think to plot the thing and see if the points fit.

After plotting it, I can see why the assumption that it is exponential for small values of x does not work in this problem. These points are near the inflection point.

Image (book = red, mine=blue):
http://www.orbitsimulator.com/BA/function.GIF

hmmm... This textbook usually doesn't make you think. All the problems follow the template of the example probems.

It looks linear near these points, but using y=mx+b gives me 1 for m and 3 for b. So I guess I can't assume this either.

** edit... strange, the TEX in my original post still works and your TEX and Ivy's TEX worked before I posted this reply, but now I get red X's for your TEX and Ivy's TEX. I don't see a syntax error when I view the code...

 

[topsquark]

You're right. I didn't even think to plot the thing and see if the points fit.

After plotting it, I can see why the assumption that it is exponential for small values of x does not work in this problem. These points are near the inflection point.

Image (book = red, mine=blue):
http://www.orbitsimulator.com/BA/function.GIF

hmmm... This textbook usually doesn't make you think. All the problems follow the template of the example probems.

It looks linear near these points, but using y=mx+b gives me 1 for m and 3 for b. So I guess I can't assume this either.

** edit... strange, the TEX in my original post still works and your TEX and Ivy's TEX worked before I posted this reply, but now I get red X's for your TEX and Ivy's TEX. I don't see a syntax error when I view the code...

The whole system seems to be barfing. My latest post on this thread got completely lost, for example. Ah well. The programmers for the site are human too!

-Dan