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EI

Graphs of logistic function (Figures 2 and 3) look suspiciously similar to the graph of the hyperbolic tangent function (Figure 3.11.3). Explain the similarity by showing that the logistic function given by Equation 7 can be written as $P(t) = \frac {1}{2} M[ 1 + \tanh(\frac {1}{2} k(t -c))]$where $c = (\ln A)/k.$ Thus the logistic function is really just a shifted hyperbolic tangent.

$$\frac{K}{2} \cdot \frac{2}{1+e^{-2 u}}=\frac{K}{1+e^{-2 u}}=\frac{K}{1+A e^{-k t}}=P(t)$$

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Kristen K.

University of Michigan - Ann Arbor

Michael J.

Idaho State University

Boston College

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Video Transcript

Okay. And this question rest to relate the hyperbolic tangent function to the logistic function. So Pft equals one half mm. Times one plus 10 H. One half. Okay T minus C. And I'm going to evaluate this to show that's the same as the equation for the logistic function. So we know that hyperbolic tan equals hyperbolic sine over hyperbolic cosign. Which can be evaluated as E. To the X minus E. To the negative X. Divided by E. To the X. Plus E. To the negative X. So we're gonna plug in that one half K. T minus C. In there to get one half. M. One yes. Eat the one half. Okay T minus E. My C. E. To the negative one half. Okay. Team on the sea divided by the E. Two and one half K. Timur. C plus E. To the negative one half. Okay. Team. RC Going to rewrite this one as a fraction so that we can combine like terms. So this becomes 1/2. Um We have Eat the 1/2. Okay do you wanna see plus E. To the negative one half? Okay. Team I see Plus E. To the 1/2. Okay T minus C minus E. To the negative one half K. T. You wanna see all over E. to the 1/2 k. She want a C minus E. To the negative one half. K. T. M. I. C. And then we can see that these cancel out and there's gonna be two of these ease. They'll cancel it with the one half. So we got em times E. To the 1/2. Okay. Team on the sea Divided by E. to the 1/2. Okay. T. M. I C minus E. To the negative. Actually this plus -1 F. K. Timor Sea. And now to simplify this bit of multiply bye. E. to the negative 1/2 K. Team on a. C. Over. Eat a negative 1/2 k. Team on the sea. two can't allow some of these exponents. So this gives us em we have a one in the numerator because those exponents cancel out. We have a one in the denominator and then plus E. To the negative K. Team on the sea equals mm over one plus eat a negative K. T. E. Two to the K. C. Now C. Equals Ellen of a over K. So this is gonna simplify to be I am over one plus a eat a negative K. T. Which equals pft. Uh huh.

EI

Topics

Differential Equations

Kristen K.

University of Michigan - Ann Arbor

Michael J.

Idaho State University

Boston College

Lectures

Join Bootcamp