Maximizing an evolutionary biology equation (vector calculus)

In summary, for a Gaussian landscape, the log-fitness change caused by a mutation of size r in chemotype element i is given by Q_i(r) = -\vec{k} \cdot S \cdot \hat{r_i}r - \dfrac{1}{2} \hat{r_i} \cdot S \cdot \hat{r_i}r^2. To find the largest possible gain in log-fitness achievable by mutating chemotype element i, the paper suggests maximizing Q_i(r) with respect to r. After finding the derivative of Q_i(r) and setting it equal to 0, the paper arrives at the solution r = \dfrac{-\vec{k} \cdot S \cdot \
  • #1
mliuzzolino
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Homework Statement



For a Gaussian landscape, the log-fitness change caused by a mutation of size r in chemotype element i is

[itex] Q_i(r) = -\vec{k} \cdot S \cdot \hat{r_i}r - \dfrac{1}{2} \hat{r_i} \cdot S \cdot \hat{r_i}r^2 [/itex].

To find the largest possible gain in log-fitness achievable by mutating chemotype element i, maximize [itex] Q_i(r) [/itex] with respect to r.


Homework Equations



The solution is:

[itex] \Theta _i = \dfrac{|\vec{k} \cdot S \cdot \hat{r_i}|^2}{2\hat{r_i} \cdot S \cdot \hat{r_i}} [/itex]

The Attempt at a Solution



[itex] Q_i(r)' = -\vec{k} \cdot S \cdot \hat{r_i} - \hat{r_i} \cdot S \cdot \hat{r_i}r = 0 [/itex]

[itex] \hat{r_i} \cdot S \cdot \hat{r_i} r = -\vec{k} \cdot S \cdot \hat{r_i} [/itex]

[itex] r = \dfrac{-\vec{k} \cdot S \cdot \hat{r_i}}{\hat{r_i} \cdot S \cdot \hat{r_i}} [/itex]

It's been forever since I've dealt with vector calculus so I know that I'm approaching this entirely the wrong way. Any points in the right direction will be greatly appreciated!
 
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  • #2
mliuzzolino said:

Homework Statement



For a Gaussian landscape, the log-fitness change caused by a mutation of size r in chemotype element i is

[itex] Q_i(r) = -\vec{k} \cdot S \cdot \hat{r_i}r - \dfrac{1}{2} \hat{r_i} \cdot S \cdot \hat{r_i}r^2 [/itex].
It looks like you put a great deal of effort into formatting the equation above, but I'm having a hard time understanding what it says. If you "dot" two vectors, you get a scalar, but you can't dot that scalar with another vector. In other words, an expression such as ##\vec{u} \cdot \vec{v} \cdot \vec{w}## doesn't make sense.

Also, is S a scalar? How you wrote it suggests that it is.
mliuzzolino said:
To find the largest possible gain in log-fitness achievable by mutating chemotype element i, maximize [itex] Q_i(r) [/itex] with respect to r.

Homework Equations



The solution is:

[itex] \Theta _i = \dfrac{|\vec{k} \cdot S \cdot \hat{r_i}|^2}{2\hat{r_i} \cdot S \cdot \hat{r_i}} [/itex]

The Attempt at a Solution



[itex] Q_i(r)' = -\vec{k} \cdot S \cdot \hat{r_i} - \hat{r_i} \cdot S \cdot \hat{r_i}r = 0 [/itex]

[itex] \hat{r_i} \cdot S \cdot \hat{r_i} r = -\vec{k} \cdot S \cdot \hat{r_i} [/itex]

[itex] r = \dfrac{-\vec{k} \cdot S \cdot \hat{r_i}}{\hat{r_i} \cdot S \cdot \hat{r_i}} [/itex]

It's been forever since I've dealt with vector calculus so I know that I'm approaching this entirely the wrong way. Any points in the right direction will be greatly appreciated!
 
  • #3
Sorry! I forgot to state that S is a symmetric positive definite matrix. I believe that the operation will just be taking the dot product of [itex] \vec{k} [/itex] and S, and then using that as the scalar weight on [itex] \vec{k} [/itex].

This is for a research project and I'm just going through old literature trying to rederive the equations so that I can better understand what's going on, and I kind of mindlessly transcribed it exactly as it was in the paper (with two dots). I'm not sure of the rationale behind putting the two dots in the paper, but it's there nonetheless.

Hope this helps explain it better...
 
  • #4
mliuzzolino said:

Homework Statement



For a Gaussian landscape, the log-fitness change caused by a mutation of size r in chemotype element i is

[itex] Q_i(r) = -\vec{k} \cdot S \cdot \hat{r_i}r - \dfrac{1}{2} \hat{r_i} \cdot S \cdot \hat{r_i}r^2 [/itex].

To find the largest possible gain in log-fitness achievable by mutating chemotype element i, maximize [itex] Q_i(r) [/itex] with respect to r.


Homework Equations



The solution is:

[itex] \Theta _i = \dfrac{|\vec{k} \cdot S \cdot \hat{r_i}|^2}{2\hat{r_i} \cdot S \cdot \hat{r_i}} [/itex]

The Attempt at a Solution



[itex] Q_i(r)' = -\vec{k} \cdot S \cdot \hat{r_i} - \hat{r_i} \cdot S \cdot \hat{r_i}r = 0 [/itex]

[itex] \hat{r_i} \cdot S \cdot \hat{r_i} r = -\vec{k} \cdot S \cdot \hat{r_i} [/itex]

[itex] r = \dfrac{-\vec{k} \cdot S \cdot \hat{r_i}}{\hat{r_i} \cdot S \cdot \hat{r_i}} [/itex]

It's been forever since I've dealt with vector calculus so I know that I'm approaching this entirely the wrong way. Any points in the right direction will be greatly appreciated!

If I understand correctly, you have an expression of the form
[tex]Q(r) = -a r - \frac{1}{2} b r^2 \\
\text{where } a = \vec{k} \cdot S \hat{r}_i,\text{ and } b = \hat{r}_i \cdot S \hat{r}_i[/tex]
with ##a, b## being constants, independent of ##r##. Maximizing Q(r) is a simple excercise in univariate calculus, and you did it correctly. Why do you think you have made an error?
 
  • #5
Unfortunately, my result does not seem to match with the solution arrived at in the paper which is provided in 2. Homework Equations .
 
  • #6
Mark44 said:
It looks like you put a great deal of effort into formatting the equation above, but I'm having a hard time understanding what it says. If you "dot" two vectors, you get a scalar, but you can't dot that scalar with another vector. In other words, an expression such as ##\vec{u} \cdot \vec{v} \cdot \vec{w}## doesn't make sense.

Also, is S a scalar? How you wrote it suggests that it is.

My mistake in the previous reply to you. The expression should be [itex] \vec{k} \cdot S \cdot \vec{k}^T [/itex]. Let [itex] \vec{k} [/itex] be a 1 x N matrix and S an N x N matrix. The dot product of S and [itex] \vec{k}^T [/itex] will result in an N x 1 matrix which is then dotted with the 1 x N [itex] \vec{k} [/itex] matrix, resulting in a scalar.
 
  • #7
mliuzzolino said:
Unfortunately, my result does not seem to match with the solution arrived at in the paper which is provided in 2. Homework Equations .

I think that reason for the mis-match is that you are not computing the same thing the paper is computing. You calculated the best value of ##r##; the paper calculated the maximum value of ##Q##. Can you see now what you need to do?

BTW: when replying, always use the "quote" button; otherwise nobody can figure out which message you are responding to.
 
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  • #8
Ray Vickson said:
I think that reason for the mis-match is that you are not computing the same thing the paper is computing. You calculated the best value of ##r##; the paper calculated the maximum value of ##Q##. Can you see now what you need to do?

BTW: when replying, always use the "quote" button; otherwise nobody can figure out which message you are responding to.

Ah. I have it figured out now. I can't believe I overlooked such an elementary concept...

Thank you Ray!
 

1. What is an evolutionary biology equation?

An evolutionary biology equation is a mathematical representation of the evolutionary process, taking into account factors such as genetic variation, natural selection, and population dynamics.

2. What is vector calculus?

Vector calculus is a branch of mathematics that deals with vectors and their operations, such as differentiation and integration. It is often used to model physical phenomena, including evolutionary biology.

3. Why is maximizing an evolutionary biology equation important?

Maximizing an evolutionary biology equation allows us to better understand how different factors influence the evolution of species, and can help us make predictions about future changes in populations.

4. How is vector calculus used in evolutionary biology?

Vector calculus is used to model and analyze the changes in a population's genetic makeup over time. By using vector calculus, we can quantify the rate of change in a population's traits and determine how natural selection and other factors are impacting its evolution.

5. What are some practical applications of maximizing an evolutionary biology equation?

Maximizing an evolutionary biology equation can be used to study the effects of environmental changes on a population's evolution, to predict the spread of diseases, and to inform conservation efforts for endangered species.

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