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TheDude,
How do we know that we just have to consider the boundary points? What if there was a graph whos min was not on the boundary point?
Thanks
A circular hot plate given by the relationship x^2 + y^2 <= 4 is heated according to the spatial temperature function T(x, y) = 10-x^2 + 2x - 4y^2. Find the hottest and coldest temperatures on the plate and the points at which they occur.
Here is what I have so far:
g(x,y) = x^2 + y^2 - K where K<= 4
del f = (-2x + 2, -8y)
del g = (2x, 2y)
Applying LaGrange multiplier
1) -2x+2 = lambda(2x)
2) -8y = lambda(2y)
3) x^2 + y^2 = K
solving for eqn 2)
we get y=0 or lamba = -4
sub lambda = -4 in eqn 1
we get x = -1/3
now I am stuck, since there are infinite values of K we have to solve for. I know there is going to be points (-1/3, +/- y) and (+/-x, 0).
Another approach that gives me half the answer is solving for the global max of T(x,y), which coincidentally lies in the constraint given, but then I am at a loss at finding the local min.
Any ideas?
2.
Proof By induction
Base Case:
we need atleast 3 vertices of degree 2 to have a cycle. Example a triangle.
IH: Assume true
IS:
If graph G with n vertices contains a cycle where each vertex has a degree of 2, then a graph M with n+1 vertices contains the graph G. Therefore graph M with n+1 vertices with degree 2 also contains a cycle.
QED
This is just a rough proof to give you the idea.
hi bobbym,
I think an example might help:
Let B be the set of binary strings dened by
(a) {E,0} contained in B; and (where "E" is an empty string)
(b) for each s in the set B; s1 in set B and s10 in set B:
Show that B = C, where C is the set of all binary strings that do not have 2 consecutive
zeros.
Solution:
First we show that C is contained in B and finally B is contained in C.. Let P(n) be the statement that every binary string
of C of length at most n is in B. We use Mathematical Induction to show that P(n) is true
for all non-negative integers n.
Conditions (a) and (b) imply that P(0) and P(1) are both true. This completes the basis
step.
Induction Hyothesis: For n > 1, assume P(n - 1) is true.
Induction Step:
Let s be any binary string of C of length n. If s ends with a 1, then s = t1 where t is a binary string of C of length n-1.
Since P(n-1) is true, t is in B. By condition (b), s = t1 is in B.
If s ends with a 0, then s = t10, since s has no consecutive zeros, and t is a binary string of C of length n - 2.
Since P(n - 1) is true, t is in B.
By condition (b), s = t10 is in B.
Therefore P(n) is true.
By Mathematical Induction, P(n) is true for all positive integers n. Therefore C is contained in B.
Now we show B is contained in C. Let B(i) be the set of all strings of B that can be obtained by applying operation (b) at most i times.
For example, B(0) = {E,0} and B(1) = {E,0,1,10,01,010}.
Let Q(n) be the statement that B(n) is contained in C.
By Inspection, Q(0) is true. This completes the basis step.
Hypothesis: For n > 0, assume Q(n - 1) is true.
Induction Step:
Clearly B(n) = {E,0} union {b1|b in B(n-1)} union {b10| b in B{n-1)}.
If string b does not have any consecutive zeros, then neither does string b1.
Since B(n-1) is contained in C by our hypothesis, {b1|b in B(n-1)} is contained in C.
If string b does not have any consecutive zeros, then neither does string b10.
Since B(n-1) is contained in C by our induction hypothesis, {b10|b in B(n-1)} is contained in C. Therefore, Bn is contained in C;
that is, Q(n) is true. This completes the inductive step.
Therefore B = C.
Hi bobbym and bob bundy,
bobbym: can you share the link to proff by induction?
bob bundy: the OP asked for f(2n) = f(n+1)^2 - f(n-1)^2, your proof proves f(2n) = f(n)^2 + f(n-1)^2. I am guessing there is some algebraic manipulation to get to the identity given by OP?
Lets say we have a set A containing 600 elements = {7,9,11,7,9,11,......,11}
and another way of describing that se would be = {l,r,l,r,l,r,l,r,l.....l} where l and r are right and left shoes respectively. Then if we map them 1 to 1 we have exactly 100 pairs of size 7 shoes, size 9 shoes and size 11 shoes. This means we have a total of 300 correct pairs if arranged this way. Would this work?
Yeah it is, I cant find a good way to have a union of those sets.
hi bobbym,
I see where I went wrong in my thinking. Instad of counting the number of edges I have to count the different arrangements of the edges in a 5 vertices graph. Thanks for the link!
I also need help with this counting problem.
John ordered some boots from an online store. He was sent a crate containing 200 boots of size 7, 200 boots of size 9, 200 boots of size 11, and 300 boots are for the left boot, 300 boots are for the right foot. Prove John has at least 100 correct pairs of boots.
hi bobbym, actually what anonimystefy said is true, the way I am solving it doesnt make sense since there are 5 vertices in mine but the OP said only 4 people.
You listed out all the permutations which does make sense.
Anonimystefy how many ways would it be with 5 peolpe? !5 = 44 ways?
can you solve this using the K5 complete graph, as in a graph with 5 vertices connected to all other vertices. So wouldnt the total number of edges be 10?
sum of all degree of all vertices = 2 * edges.
so each vertex has degree 4. there are 5 vertices. 20 = 2 * edges. therefore 10 edges in total.
Can you post a pic of this?
Thanks bob!
Got it, we actually proved the pascals triangle formula for 2^n in class. I just hadnt of thought of adding in the (n-r)! on top and bottom of the summation.
3834 days = 6 * 639.
Ok I have another question I cant answer.
Prove this relationship is true.
ok so how do I divide this question into cases?
If i multiply my previous formula by (26 choose 2)+(24 choose 2)+(22 choose 2).....(2 choose2) should that take care of the parity of the cupcakes?
Its an assignment question and i need to solve it using what I have learned from class.
We haven't been taught generating functions yet, thats why I am trying to find a way to solve it without that.
@bobbym: so is my way of counting correct?
Can you explain where i went wrong?
Hmm, I have a different approach. Just wondering if this would work:
Give each person one cupcake from the 31. So you have 26 left.
Then couple those 26 cupcakes so you never give a person an even number of cupcakes.
This gives you 13 pairs of cupcakes to share between 5 people.
So then the answer would be (13+5-1) choose 5.
Hi,
I needed some help with combinations and permutations.
the question is :
John has baked 31 cupcakes for 5 different students. He wants to give them all to
his students but he wants to give an odd number of cupcakes to each one. How many
ways can he do this?
so from reading the mathisfun page on combs and perms, I get this so far:
since theres 31 cupcakes to be shared in between 5 students, the combinations for this will be (31+5-1) choose 5. But this does not take into consideration the odd number restriction. How can I do this?