Project One: C Bump functions

Transcription

1 Project One: C Bump functions James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University November 2, 2018

2 Outline 1 2 The Project

3 Let s recall what the bump functions are: Define f (x) = { 0, x 0 e 1/x, x > 0 Let (x n ) be any sequence with x n 0 which converges to 0. Then f (x n ) 0 also. So 0 is in S(0). Let (x n ) with x n > 0 be any sequence with x n 0. Then f (x n ) = e 1/xn. It is easy to see 1/x n with x n > 0 always has an unbounded limit which we usually call. If you recall from calculus, e y 0 as y. Thus, f (x n ) 0 here for these sequences. If (x n ) was a sequence with infinitely many terms both positive and negative which still converged to 0, then the subsequence of positive terms, (x 1 n ), converges to 0 and the subsequence of negative terms, (x 2 n ), also converges to 0. Further, f (x 1 n ) 0 and f (x 2 n ) 0 as well. With a little work we can see f (x n ) 0 too. We see the only cluster point at 0 is 0. Thus, lim xn 0 f = lim x n 0 f = 0.

4 This function has many properties which you are going to prove! Note This function is continuous at 0 This function is infinitely differentiable and f (n) (0) = 0 for all n. Note this tells us all the derivatives of f are actually continuous. f is so flat at 0 its Taylor Series expansion at 0 does not match f at all! Even though 0 is a global minimum of f occurring at 0 and other points, the second derivative test will fail there.

5 So now we can build some more things. { 0, x a f a (x) = e 1/(x a), x > a A similar analysis shows f a has a zero n th order derivative at a for all n and has a continuous n th order derivative at a for all n. g b (x) = { 0, x b e 1/(x b), x < b A similar analysis shows g b has a zero n th order derivative at b for all n and has a continuous n th order derivative at b for all n.

6 We can then show for any a < b, h a b (x) = f a (x) g b (x) = = 0, x a e 1 x b e 1 x a, a < x < b 0, x b 0, x a b a e (x a)(x b), a < x < b 0, x b A similar analysis shows h ab has a zero n th order derivative at a and b for all n which is continuous a and b for all n. Now multiply h ab by e 2 a+b to get Hab (x) = e 2 a+b hab (x). H ab (x) is an infinitely differentiable function whose value is 0 off of the interval (a, b) and whose maximum value is +1. It is called a C bump function. The closure of the set of x values where H ab (x) > 0 is then [a, b] which is a compact set. This closure is called the support of H ab (x) and so H ab (x) is a C function with compact support. Very important!

7 To plot h use the code below in the function Plottingh(a,b,eps,P). The arguments for the function are a and b which are used to define f a and g b, eps which is used to stay away for the points a and b in the plots and P which is used to set up the linspace command. In the MatLab window type Plottingh(-1,1,.01,72);. This generates the plot which needs labels and a title, of course, which you will have to add.

8 Just type in this function and save as the file Plottingh.m in your current directory. f u n c t i o n P l o t t i n g h ( a, b, eps, P) f a x, a ) exp ( 1/(x a ) ) ; gb x, b ) exp ( 1 / ( x b ) ) ; XA = l i n s p a c e ( a 2,a eps, P/3) ; 5 XM = l i n s p a c e ( a+eps, b eps, P/3) ; XB = l i n s p a c e ( b+eps, b+2,p/3) ; X = [XA,XM,XB ] ; [m, n ] = s i z e (X) ; f o r i =1:n 10 i f (X( i ) < a ) h ( i ) = 0 ; end i f (X( i ) > b ) h ( i ) = 0 ; 15 end i f (X( i ) >a & X( i ) < b ) h ( i ) = f a (X( i ), a ) gb (X( i ), b ) ; end end 20 p l o t (X, h ) ; end

9 Let s look at derivatives of f a. (1): For x a, the derivative f a is easy to calculate f a (x) = { 0, x < a (e 1/(x a) ) = 1 (x a) e 1/(x a), 2 x > a We can see lim x a f a (x) = 0 and from the right, using y = 1/(x a), we have ( ) lim f 1 x a + a (x) = lim x a + (x a) 2 e 1/(x a) y 2 = lim y e y = 0 using our recent homework. Since the right hand and left hand limits match, we see f a (a) = 0.

10 (2): f (2) a (x) = { ( (e 1/(x a) ) (2) = 0, x < a 2 (x a) (x a) 4 ) e 1/(x a), x > a We can see lim x a f a (x) (2) = 0 and from the right, using y = 1/(x a), we have lim f (2) 2y 3 + y 4 x a + a (x) = lim y e y = 0 using our recent homework. Since the right hand and left hand limits match, we see f (2) a (a) = 0. You should be able to see that the basis step for your induction proof is Step (1) and for the induction step, you recognize that at the k th step, the lim x a+ calculation will have the form lim f (k) p k (y) x a + a (x) = lim y e y = 0 where p k (y) is a polynomial in the variable y = 1/(x a).

11 It doesn t matter what this polynomial looks like so you don t need to fuss about finding its actual form. What matters is that it is a polynomial. So convince yourself that this is true by taking a few more derivatives. Then to complete the induction step, you assume f a (k) (x) has the form p k (y) e for y = 1/(x a) and then the rest follows from using the product y rule to calculate f a (k+1) (x).

12 The Project Due Date: Friday November 30, 2018 You are going to do the work to prove the stuff we talk about in the background slides. Do a great job on this. I want to see you put all the stuff we have been doing together. Write well on one side of the paper and treat this like a full report. 1 f a Prove f a is differentiable for all x and in particular f a (a) = 0 Prove f a is twice differentiable for all x and in particular f a (a) = 0 Now look at what you have done and see if you can figure out how to scale your argument up to show f a has an n th derivative with f a (n) (a) = 0. The comments earlier will help with this. 2 Do the same thing for g b. 3 Now handle h ab but note it is easier because we can use the product rule. 4 Plot a nice selection of h ab s to get the feel of how they look. You will have to be careful with the plots as you can t ask MatLab etc to plot near the place where h become 0.