Recent content by tarheelborn

But no node is isolated by the hypothesis. I'm not sure what you mean about k+1 edges not changing since k is variable. I am simply not seeing this at all...

Yes, you are absolutely right. It has been a VERY long day. Following what you said re the base case, we can say that G has a match of size at least 2/(1+1) = 1, which is clear. Now we could suppose that the theorem holds for all graphs with k or fewer edges and consider a graph with k+1 edges...

Homework Statement Prove that every graph G without isolated vertices has a matching of size at least n(G)/(1+∆(G)). (Hint: Apply induction on e(G)). Homework Equations n(G) = size of the vertex set of G and ∆(G)= maximum degree of v in G The Attempt at a Solution For the base...

We could make a closed trail shorter by deleting a vertex. If we delete a vertex, the edges incident to that vertex are also deleted, so the trail would be shorter. But this action wouldn't necessarily affect the degree of vertices, would it?

Homework Statement All vertices in a closed trail have even degree. Homework Equations The Attempt at a Solution Intuitively, I know this statement is true, but I can't seem to see a clear way to show it. I know that a closed trail is a path that connects vertices, so one would follow an...

The ith component would need odd vertices. Without that condition the graph could not be composed into k trails. Is that what you mean?

I think the ith component must have odd vertices; otherwise we have a circuit, not a trail.

You are losing me here somewhat... I see that each trail has odd-valent vertices at its endpoints, so there would be two odd-valent vertices per trail. Then a graph with 2k odd-valent vertices would decompose into k trails provided the graph was Eulerian?

A trail contributes even degree to every vertex, except that non-closed trails contribute odd degree to their endpoints. Thus a single trail can have, at most, two vertices of odd degree, so the composition of G contains 2k odd vertices, divided by 2 vertices per trail = k trails. That should do...

No, a connected component cannot have an odd number of odd-valent vertices. Then could we say that since G cannot have an odd number of odd-valent vertices, there must be some (at least one) even-valent vertices which would indicate some (at least one) trails?

Homework Statement Use ordinary induction on k or on the number of edges (one by one) to prove that a connected graph with 2k odd vertices composes into k trails if k > 0. Does this remain true without the connectedness hypothesis? Homework Equations The Attempt at a Solution If k...

Oops, that is what I mean. Thank you very much.

Homework Statement I need to prove that \sum_{n=1}^{∞}[1−Cos(n−1z)] is entire. Homework Equations The Attempt at a Solution I know that I need to show that the series is differentiable for its whole domain, but I am not sure how to do that. Should I try to use the ratio test?

I finally managed to get this using ParametricPlot. It worked beautifully; just took some tweaking! Thank you.

I cannot seem to make mathematica plot the following equation correctly: r[t_] = 3 Cos[t] + iSin[t]; plotbeta = PolarPlot[r[t], {t, -Pi/2, Pi/2}] I have used the capital I to no avail; I have used * to indicate multiplication. It plots the curve without a problem when I leave out the i...