I am wondering how can I create a point corresponding to the maximum elevation value of a DEM. I mean, from the DEM I can easily get the maximum elevation. For the sake of a viewshed analysis, I would like to create a point feature corresponding to the spot with maximum elevation. I guess it is simple, but I am having difficulties in finding the right procedure.   I am using arcgis 10.1 (with advanced license).
I am using ArcGIS Desktop 10.1.   I am trying to find the location of the highest value pixel in my Direct Solar Radiation Raster. I can see the extent of the data in my raster but don't know where the highest value pertains to.   Is there a way to simply select pixel with highest value or create a point at highest value?
Correlations between continuous and categorical (nominal) variables Combinatorial argument for $\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$ I need to show that $$\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$$  I know that $\displaystyle \binom{n}{k}\binom{k}{i}$ is counting the number of ways to pick $k$ elements from $n$ and then $i$ elements from those $k$, so $\displaystyle \sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i}$ is counting the number of ways to do that $\forall k$.  I know that $\displaystyle \binom{n}{i} = \binom{n}{n-i}$ and that $2^{n-i}$ is the number of elements in the power set for $[n]\setminus[i]$.  How do I show that both sides are equivalent? I'm stumped on this one. Could anyone give me a pointer in the right direction? Boundary Question in $\mathbb{R}^{2}$ (Manifolds) Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result:    For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.   My thoughts so far: The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:   For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof. For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case? Determining confidence interval with one observation (for Poisson distribution) Is it possible to backup individual files with Deja Dup? In the Deja Dup preferences, it is possible to select certain folders to backup and certain folders to exclude.  I would like to be able to backup individual files in my Home folder (Such as .vimrc and .bashrc), but I don't see any way to do this.   So my question is, is it possible to either     Select individual files (instead of whole folders) to backup, or backup my Home folder, but exclude all subfolders? Is there a way to recover deleted data from a SSD / USB Flash drive?  All I could find were recovery for damaged drives, but I want to recover deleted data.  So how do I do this? Mysql Pivot rows into dynamic columns with unknown number of columns I have an output of a query as follows:  bid  | code | amount        | ----------------------------- 2915 | BF   |   -10700.00   | 2915 | YQ   |   -300.00     | 2915 | YR   |   0.00        | 2915 | YM   |   0.00        | 2915 | WO   |   -153.00     | 2915 | IN   |   -329.00     | 2915 | K3   |   0.00        | 2915 | CUTE |   -50.00      | -----------------------------   I need to convert the values in code column to individual columns with column name as corresponding code and amount shall be the value.  bid  | BF       | YQ     | ... | CUTE  | ---------------------------------------- 2915 | -10700.00| -300.00| ... | -50.00|   The issue is that the result of my initial query has dynamic codes and hence need to generate columns accordingly.  Already tried the following logic but couldn't achieve the goal:   , ,   All the above links use some expression or the other to generate the columns.   My original resulting rows are result of simple select on multiple joined tables, and has no calculation or expression scope.  Kindly suggest some pointer or hint on how to achieve this.  Many thanks in advance. In , I used quotes to demark specific terms, but after I posted the question, I wondered if I'm doing it right.  Usually, when quoting something a person says, the quotation mark goes after the comma or period, like so:     "This is my opinion," she said.   However, I sometimes use quotes not to represent something said by anyone, but instead for delineating a set term or phrase to make it clear that those words go together and none of them should be mistaken for a part of the rest of the sentence. Like this:     Should I say "stand up", "standup", or "stand-up"?   This looks weird to me:     Should I say "stand up," "standup," or "stand-up"?   ... but maybe I'm wrong about the rules.  Is it correct to use quotes in this way? And if so, are the quotes positioned correctly relative to the other punctuation marks?  Are there other alternatives for delineating terms without the use of quotes? Boats from Manaus, Brazil to Leticia, Colombia? I've heard of and met a few people who have done this route, and I hope to do it at the end of this year or early next year.  However, the descriptions were mostly sketchy, and I'm wondering if it's a regularly scheduled service, or more ad-hoc.  So I guess the question is, how can I find out which boats go from Manaus to Leticia, and when they go?  Ideally this wouldn't be a tour, but a 'locals' boat. $$\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}= 5$$ How to manage users and groups using GUI? Just switched from 10.04 to 11.10. I'm wondering about the limited options in the GUI for managing users and groups. Where is the old GUI to change groups, id's, etc. I know that I can do this from cli, that's no problem. But I want this comfortable GUI back. How can I have colored and underlined links with hyperref? I have got the Client Cross Frame Scripting Attack alert in the recent CheckMarx scanner. However, it points to an auto generated ID.   I've seen the following links however, the I'm not sure about relavancy of the soluions given i.e.,          Regarding my visualforce component, I'm doing just one DML operation on a custom setting from a callout response. Also, I'm not passing any attribute to the component.   To give a background of the component, its just a list view of records from a different system retrieved through a callout on load via JavaScript.  Let me know if in case anything else is needed.   Thanks in Advance! Peace! My furnace pilot light keeps going out so I changed the thermocouple. Now the furnace will come on and heat up to the degree set on my thermostat, then the pilot light will go out and the furnace will not come back on.    Could the thermostat be bad and cause the pilot light to go out? How to pass multiple operator properties via UI layout? Basically, I want to have a single operator which can do multiple things, based on some attributes values.  class MyPanel(bpy.types.Panel):     # (...)     layout = self.layout     row = layout.row()     row.operator("my.button", text="Button text")  class MY_BUTTON_OT_Button(bpy.types.Operator):     bl_idname = "my.button"     bl_description = "Button description"     bl_label = "Button"     foo = bpy.props.IntProperty()     bar = bpy.props.BoolProperty()      if bar:         # Do something with foo     else:         # Do something else :P   I know how to set one attribute, like so:      row.operator("my.button", text="Button text").foo=5   But what about multiple arguments? I tried several things that didn't worked, like (e.g.):   With setattr(): row.operator("my.button", text="Button text").setattr(foo=5,bar=True)   With a custom method setVal() in the operator class: row.operator("my.button", text="Button text").setVal(foo=5,bar=True)   Maybe I must use a single EnumProperty? I hope to avoid this, so my question is: Is it possible to pass multiple custom arguments to set attributes in an operator class? And if yes, how ?  Thanks for your help. Multiplication operator has no no eigenvalues Statement: Let $M_x$ denote the multiplicative operator acting on $L^2([0,1], \, dx)$ by $M_x(f) = xf$. Show that $M_x$ has no eigenvalues  Attempt: Let $M_x(f) = xf$ then we should have $M_x(f) = \lambda f$  then we get $M_x(f) = \lambda f \Rightarrow xf = \lambda f \Rightarrow xf-\lambda f \Rightarrow (x-\lambda)f = 0$  Now $f \neq 0$ so that implies that $(x-\lambda) = 0$ but this is achieved when $x = \lambda$, so I don't see how it doesn't have eigenvalue, when we achieve the desired result. Find Orthogonal complement What spell targets count for the Graviturgy Magic wizard's Gravity Well feature? The  feature from the  (EGtW, p. 185) triggers &quot;whenever you cast a spell on a creature&quot;, and has an effect &quot;if the target is willing to move, the spell hits it with an attack, or it fails a saving throw against the spell.&quot; It's not clear to me what kinds of spells can actually trigger this feature. Based on the singular phrasing of &quot;on a creature&quot; and the later use of the word &quot;target&quot;, I would normally read this as only applying to spells that explicitly target creatures (and perhaps only those with a single target, in contrast to the phrasing of the  feature of the  wizard, for example). However, spells that target a location can use the word target in a different sense (see ). What's actually going on here? Does this feature only work on targeted spells? Or does it function for area of effect spells as well? Or is this just an ambiguous phrasing that will rely on the DM's ruling? Prove or disprove: if $\sum\limits_{n=0}^∞a_n$ converges and $\lim\limits_{n\to\infty}b_n =0$, then $\sum\limits_{n=0}^{\infty}a_n b_n$ converges.