I'm trying to add the &quot;Shader&quot; node to this drop-down menu but it's not available and I truly don't know how to figure it out or even find it otherwise lmao.  I promise I tried Googling first for about an hour, then just clicking around for an hour, but I'm stuck because I'm super-new to using Blender (I've managed to do very little in the two weeks since I started trying to learn) and tbf my ADHD is bad and it's making me feel hella stupid so I just need a kind soul to tell me exactly what to do. If it helps at all, I'm trying to make this object glossy and I'm following a Youtube tutorial to learn how to do that.  I had this problem with 2.70 and just updated to 2.80 because I figured some of you might tell me to update, but the issue persists!  And tbf (again), it seems that I might have trouble following tutorials now because 2.80 looks so different from 2.70!  Haha! Here's a screengrab of what happened when I pressed Shift + A in 2.70 (very similar to 2.80).  Here's a screengrab of what happened when I tried to search for &quot;Shader&quot; within that search function in 2.70.  And here's what happens when I try to search for &quot;shader&quot; in the search bar in 2.80 (literally the same as in 2.70; I don't know why I thought updating would fix this.  I feel so stupid right now!)  Sorry &amp; thanks in advance.  I'll answer whatever questions you have to the best of my ability, but I'm apparently an idiot so I may not know how to find what you're asking me for and anyway I'm assuming this'll be as simple as a few clicks or keystrokes.  I really appreciate it!! EDIT:  Whoever closed this question, the link provided uses v2.70 and I'm using 2.80 so everything looks completely different and I'm a complete beginner so I don't know what they're telling me to click!  So it doesn't help me at all!  I'm really frustrated and would appreciate it if you'd tell me how to do that in 2.80!  :^)
I have been following  tutorial until I started adding nodes. The problem is that he has completely different nodes than I have. Even nodes that are created at start are different (I have Material and Output and he has Render Layers and Composite). Have I missed something or should I use different nodes than he?
Would it help if you jump inside a free falling elevator? Imagine you're trapped inside a free falling elevator. Would you decrease your impact impulse by jumping during the fall? When? Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number.  The $A(p,q)$'s are known as alternating .     Can someone provide a nice proof that    $$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2?$$   I worked for a while on this today but was unsuccessful.  Summation by parts, swapping the order of summation, and approximating $H_k$ by $\log k$ were my best ideas, but I could not get any of them to work.  (Perhaps someone else can?)  I would like a nice proof in order to complete .  Bonus points for proving $A(1,2) = \frac{5}{8} \zeta(3)$ and $A(2,1) =  \zeta(3) - \frac{1}{2}\zeta(2) \log 2$, as those are the other two alternating Euler sums needed to complete my answer.   Added: I'm going to change the accepted answer to robjohn's $A(1,1)$ calculation as a proxy for the three answers he gave here.  Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn's approach is the one I was originally trying.  I am pleased to see that it can be used to do the $A(1,1)$, $A(1,2)$, and $A(2,1)$ derivations. $Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$. Question is to prove that :     $Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$.   I see that $Q_8$ is isomorphic to subgroup of $S_8$ by left multiplication action.  Hint given was to prove that stabilizer of any point contains $\{\pm 1\}$.  To prove Cayley's theorem, stating any group is isomorphic to a subgroup of $S_n$ we take action of given group on a set $A$ having same cardinality.  with that motivation I want to check if there is a Isomorphism then there is a map from $G\times A \rightarrow A$. i.e., $G$ gives a permutation group $S_A$.  I tried in same manner. Suppose $Q_8$ is isomorphic to subgroup of $S_n$ with $n\leq 7$  then it should come from a group action of $Q_8$ on a set of cardinality atmost 7.   Suppose $Q_8$ acts on a set $A$ with possible cardinality at most 7.  $stab(a)=\{g\in Q_8 : g.a=g \forall a\in A \}$  $cl(a)=\{g.a : g\in Q_8\}$  I know number of elements in class of $a$ equals to index of stabilizer.  As $cl(a)=\{g.a : g\in Q_8\}\in A$ i.e., $cl(a)\subseteq A$ and as $|A|\leq 7$ i see that $|cl(a)|\leq 7$.  But, $|cl(a)|=|Q_8:stab(a)|$ for any element $a\in A$.  So, $|Q_8:stab(a)|=|cl(a)|\leq 7$ for all $a\in A$.   So, $stab(a)$ should be non trivial subgroup of $Q_8$ if not then $|Q_8:stab(a)|=8$  non trivial subgroup (proper) of $Q_8$ contains $\{\pm1\}$.  So, In the worst case, $\{\pm 1\}\subseteq stab(a)$ for all $a\in A$.  As $Ker(\eta)=\cap_{a\in A}stab(a)$ (where $\eta$ is the action of $Q_8$ on $A$)   we see that $\{\pm 1\}\subseteq Ket(\eta)$ which means that $Ker(\eta)$ is non trivial.  thus there is no isomorphism (coming from $\eta$) between $Q_8$ and any subgroup of $S_7$.  I would be thankful if someone can check whether my approach is correct or if there is any other simple possible way.  P.S : Usually what i do to see whether two groups are isomorphic or not is to check for cardinality, abelian property, no of elements with same order and so on. But I was having no idea when i fail in all these ways. With this Group actions i could see possibility for getting a precise conclusion on Isomorphisms.I would like to Thank Mr. Jyrki Lahtonen (a user of Math.SE) who made me to get used to Group actions.  P.S $2$: If any thing is wrong in my idea, it is entirely my fault, and if anything is correct in this whole credit should go to Mr. Jyrki Lahtonen I have an idea or subject that should have a Q&amp;A, and want to propose a new site. How can I do this?  Related: Characterize the groups $G$ for which the map $\iota: G \to G$, sending $x \mapsto x^{-1}$ for all $x \in G$, is an automorphism of $G$ I have trouble on this question: "Characterize the groups G for which the map $$\iota: G \to G$$ sending $$x \mapsto x^{-1}$$ for all $x \in G$, is an automorphism of $G$." From  post I have seen that $G$ must be abelian. Can anyone propose another characteristic of $G$ that I can derive from the given conditions besides abelian? Thank you so much!! Code formatting - tag/block rather than indentation - code block more obvious This is a good news.    This is good news.    Which one is correct? Clarify the difference Repeatedly rolling a six sided die four times and summing the highest three results gives you a distribution with what mean and standard deviation? Repeatedly rolling a six sided die four times and summing the highest three results gives you a distribution with what mean and standard deviation?  I've only taken AP statistics, but I would like to learn how to do this. Extended Euclidean Algorithm: backward vs. forward How to mount a NTFS partition in /etc/fstab? I have two partitions that I want to mount on startup: /dev/sda3   /mnt/devel  ext4    defaults    0   2 /dev/sda2   /mnt/excess ntfs    defaults    0   2  The ext4 partition mounts fine (owned by me, writable only by me), but the NTFS mounts owned by root with R\W permission for all. How to fix this? drwxr-xr-x  7 amanda amanda 4096 2012-03-14 19:07 devel drwxrwxrwx  1 root   root   4096 2012-03-14 22:38 excess Fastest Square Root Algorithm What is the fastest algorithm for finding the square root of a number?  I created one that can find the square root of "$987654321$" to $16$ decimal places in just $20$ iterations   I've now tried Newton's method as well as my own method (Newtons code as seen below)  What is the fastest known algorithm for taking the second root of a number?  My code for Newton's Method (*Edit: there was an error in my code, it is fixed in the comments below):        a=2   //2nd root     b=97654321   //base     n=1   //initial guess     c=0   //current iteration (this is a changing variable)     r=500000   //total number of iterations to run     while (c&lt;r)      {         m = n-(((n^a)-b)/(a*b))  //Newton's algorithm         n=m         c++;         trace(m + "  &lt;--guess   ...   iteration--&gt;  " + c)     } In the biblatex doc, it says ( Version 1.7, p.11):     thesis      A thesis written for an educational institution to satisfy the   requirements for a degree. Use the type field to specify the type of   thesis.      Required fields: author, title, type, institution, year/date      Optional fields: subtitle, titleaddon, language, note, location,   month, isbn, chapter, pages, pagetotal, addendum, pubstate, doi,   eprint, eprintclass, eprinttype, url, urldate   I would like to know if there is any "canonical" way to add the name of the supervisor(s) in a field of the .bib file (no .bbl manipulation please, I want my bibliography to be usable "as it" in several documents).  The expect result could be something like  :     Meand Myself. “A brilliant result, explained”. PhD thesis**, under the   joint supervision of Pr. Toto and Pr. Tata**. Université Paris 24,   Nov. 2013.   The "under the (joint) supervision" could be a string that would be declined in four ways:   under the supervision of  under the joint supervision of and the abridged forms Given an alphabet of size $n$, how many strings of length $c$ contain every single letter of the alphabet at least once?  I first attempted to use a recurrence relation to work it out:  $$ T(c) = \left\{ \begin{array}{cr} 0 &amp;\mbox{ if $c&lt;n$} \\ n! &amp;\mbox{ if $c = n$} \\ T(c-1) \cdot n \cdot c &amp;\mbox{ if $c &gt; n$} \end{array} \right. $$  As there's no strings that contain every letter if c &lt; n, and if c = n then it's just all permutations. When c > n you can take any string of size (c-1) that contains all letters (of which there are $T(c-1)$ to choose from), you choose which letter to add (of which there are $n$ choices) and there are $c$ different positions to put it. However, this gives out results that are larger than $n^c$ (the total number of strings), so it can't be right, and I realised it was because you could count some strings multiple times, as you can make them taking different inserting steps.   Then I thought about being simpler: you choose n positions in the string, put each letter of the alphabet in one of those positions, then let the rest of the string be anything:  $$ {c\choose{n}} \cdot n! \cdot n^{c-n} $$  But again this counts strings multiple times.  I've also considered using multinomial coefficients, but as we don't know how many times each letter appears in the string it seems unlikely they would be much help. I've also tried several other methods, some complicated and some simple, but none of them seem to work.  How would you go about working out a formula for this? I'm sure there's something simple that I'm missing. Since the Earth has a magnetic field, can it, in theory, be run through a conductive metal coil to create electricity? Formatting a bootable USB that thinks it's a CD-ROM After upgrading to Windows 10, I remember that I had an old bootable recovery usb drive from the manufacturer that I no longer needed. I want to format it but it's showing up as a CD-ROM and format options are not available.  I also tried formatting it in a live linux mint environment, but linux cannot even mount the cd-rom.  How can I format this recovery usb that is showing up as a cd-rom? I was making a scene in blender and I wanted to change the texture from "Material_1" to "Material_2". It lets me change it but when I go to rendered view the part looks the same. can someone please help me I would like to ask a question about Java that doesn't have a specific answer. It's a question to see what people think and to share their opinion.    Will Stack Overflow tolerate this, or is there another Stack Exchange resource where I may ask a question like this? Cannot set LC_CTYPE to default locale: No such file or directory Computing convex hull of points using shapely How to build a dmg Mac OS X file (on a non-Mac platform)? Is it possible to build a .dmg file (for distributing apps) from a non-Mac platform? And if yes, how?