Finding the first-order taylor polynomial

[the7joker7]

Homework Statement

Basically, I have a differential equation. One of the elements of it is...

F(P) = 0.2P(1 - (P/10))

And I need to replace it with it's first-order Taylor polynomial centered at P=10.

The Attempt at a Solution

I haven't done Taylor polynomial stuff in over a year so I went and looked it up...as near as I could tell, an FOTP of an equation was the equation plus the derivative of that equation times (x - a). Is this accurate?

If so, this is fairly simple, as I can find the derivative of F(P) as 0.2 - 0.04P. but what do I do with the (x - a) part? I think for my purposes it would be (x - P), but still, how do I treat this?

 

[quasar987]

Homework Statement

Basically, I have a differential equation. One of the elements of it is...

F(P) = 0.2P(1 - (P/10))

And I need to replace it with it's first-order Taylor polynomial centered at P=10.

The Attempt at a Solution

I haven't done Taylor polynomial stuff in over a year so I went and looked it up...as near as I could tell, an FOTP of an equation was the equation plus the derivative of that equation times (x - a). Is this accurate?

You do not take the FOTP of an equation, you take the FOTP of a function. Here the function is F and the variable is P.

In general, if f is a function and we write x the variable on which it depends, its FOTP at the point a is the function

x\mapsto f(a)+f'(a)(x-a)

In your case, f=F, x=P and a=10.

I leave to you the pleasure of finding the FOTP of F at 10

 

[chrisk]

The value a is the point you are expanding around. In your case a = 10. You expand to first order so the differential equation can be solved by analytical methods. Let x = P. Then the equation is

F(x) = 0.2x(1 - (x/10))

The first order expansion is then

F(x)\approx F(10) + F'(10)(x-10)