I tried this with my friends back in my youth. The "average" person can't free themselves from this position. A friend of mine that weighed little and had strong arms, however, could pull himself up the pole with his hands and free himself.
A couple of us, myself included, stayed for quite a long time trying to get free, eventually lying backward with exhaustion. There seems to be no danger involved. The position is quite uncomfortable after you stop supporting yourself with your muscles, so probably blood circulation to the feet is partly restrained.The possible causes of death from being in this position would be:I'd say dying from thirst is the most probable cause, since that will definitely eventually happen.
The position is very easy and fun to try out for yourself in controlled circumstances with at least two friends. However, this is not advice to do that
â¢ Related QuestionsFree algebra over an operad is an algebra over that operad
Let me deal with the case of a nonsymmetric operad for simplicity (so we forget about $Sigma_r$). If you have symmetric operads, it's essentially the same story, but there are more things to write down(note : I'm assuming we're dealing with a unital operad, otherwise the terminology "free algebra on $V$ is not super well-suited)Let $(C,otimes, 1)$ be a symmetric monoidal cocomplete category such that the tensor product commutes with colimits in each variable.You may think of $CmathsfVect, mathsfCh$ (chain complexes over a commutative ring, e.g. $mathbb Z$ or a field $k$), or $mathsfsSet$ for instance, or even $mathsfSet$ for more basic examples; and let $P$ be an operad in $C$.Let $Vin C$, and let $X$ be a $P$-algebra in $C$, and suppose $Vto X$ is a morphism in $C$. Then, for each $r$, you get a map $P(r)otimes V^otimes r to P(r)otimes X^otimes r$ which is naturally defined, and, since $X$ is an algebra, you can compose it with its structure maps to get a map $mu_r : P(r)otimes V^otimes rto X$.Note that for each $n_1,...,n_r$ adding up to $n$, you get that the two maps$P(r)otimes P(n_1)otimes... otimes P(n_r)otimes V^otimes nrightrightarrows X$(defined respectively by $P(r)otimes P(n_1)otimes ... otimes P(n_r)to P(n)$ followed by $mu_n : P(n)otimes V^otimes nto X$; and $P(n_1)otimes V^otimes n_1otimes ... otimes P(n_r)otimes V^n_roversetmu_n_1otimes ...otimes mu_n_rto X^otimes r$ followed by $P(r)otimes X^otimes rto X$)agree. I'll let you understand why that is (it relies on the axioms for algebras over $P$ that $X$ satisfies)All in all, we get a map $bigoplus_rgeq 0P(r)otimes V^otimes rto X$ in $C$; this map is moreover a map of $P$-algebras if we let $bigoplus_rgeq 0P(r)otimes V^otimes r$ have the "tautological structure", defined by :$$P(n)otimes (bigoplus_rgeq 0P(r)otimes V^otimes r)^otimes n cong P(n)otimes bigoplus_r_1,...,r_nP(r_1)otimes ... otimes P(r_n)otimes V^otimessum r_i
cong P(n)otimes bigoplus_kgeq 0bigoplus_r_1,...,r_n, sum r_i kP(r_1)otimes ... otimes P(r_n)otimes V^otimes k
cong bigoplus_kgeq 0bigoplus_r_1,...,r_n, sum r_i k P(n)otimes P(r_1)otimes ... otimes P(r_n)otimes V^otimes k
to bigoplus_kgeq 0P(k)otimes V^otimes k$$where all but the last line are just rearrangements of terms using the fact that $otimes$ commutes with direct sums in each variable, and the last line is given by the structure maps of $P$.Those give us maps $P(n)otimes P(V)^otimes nto P(V)$, and one checks that this gives $P(V)$ a $P$-algebra structure; for which the above map $P(V)to X$ is a $P$-algebra map.This construction establishes a natural isomorphism $hom_C(V,UX)cong hom_mathsfAlg_P(P(V),X)$, where $UX$ denotes the underlying $C$-object of $X$.(there are things to check here, for instance that's where I'd use unitality of $P$, to construct the inverse map)This is what is meant by "$P(V)$ is the free $P$-algebra on $V$" (the technical term is that $Vmapsto P(V)$ is left adjoint to the forgetful functor $mathsfAlg_Pto C$)A more down-to-earth description of this structure on $P(V)$ (in the case where we have elements, e.g. in $mathsfVect,mathsfSet$) is given by $potimes (q_1otimes x^1_1 otimes ... otimes x^1_r_1 otimes ... otimes q_notimes x^n_1otimes...otimes x^n_r_n) mapsto mu(p,q_1,...,q_n)otimes x^1_r_1otimes...otimes x_r_n^n$ where $mu : P(n)otimes P(r_1)otimes ... otimes P(r_n)mapsto P(sum r_i)$ is the structure map of $P$.One way to think about this is similar to free groups (where each element of the free group is a word on the elements of the set you started with, maybe with inverses (those correspond to the operations in $P$), and where multiplication is just given by concatenation (here you simply have operations to add to the mix, instead of just concatenation)Now if you're considering a symmetric operad, everything is the same, except that in order for $P(V)$ to satisfy the $Sigma_r$-equivariance axioms, you have to mod out by the $Sigma_r$-action. The details can be tedious to write down, so I'll leave that as an exercise to you (be warned that once you understand the nonsymmetric case, dealing with those $Sigma$-details is not super-enlightening)So, for a tldr; :$P(V)$ comes with an algebra structure, which is essentially "tautological" : it is induced by the structure maps of $P$ itself; it's not the same algebra structure as one on $V$ if it already has one. In particular it is "free" on $V$ : a map of algebras $P(V)to X$ is the same data as a map in $C$: $Vto X$.An algebra structure on $V$ can be specified by a map (in $C$) $P(V)to V$ satisfying some conditions (look up "monad" if you want to see what those are)
How to play a glissando without hurting your hand?
If you play a glissando remember to relax your hand. A glissando should not cause any great pain. Even if you were playing in a large concert hall, it should not be the case that you hold your fingers so stiff that you have to make the glissando so loud for the audience. Loosen up and practice a glissando with just one finger even by just using your index finger. Downward glissandos can be played with the thumb. Make sure your instrument also is not such that it has old worn ivory key tops or something. If so, then it is the instrument. Usually the keys have a softer finish that makes playing a glissando easier. Otherwise it is the fact that you are pressing too hard. This ornament takes practice. Practice so that you are not hurting your fingers when you play the glissando. With great control, you should in fact be able to play a single note glissando scale up and down that actually sounds like someone playing a C Major scale. But, again do not play hard if it actually hurts your hand to play the glissando. Play lightly there.You can play a glissando up the piano with usually the index, middle, and ring finger positioned together. Try practicing slowly without even the piano, just in the air like you are shooing something away. Then, when you play on the piano, play the glissando lightly. The general rule is that if you play too loud it becomes more difficult to play fast. So, when you are at the pieces required tempo, try playing it softly. Hope this helps
Roth IRA: Vanguard vs Free Advisor's Mutual Fund
In your title you say free, but in the rest of the question you talk about how the adviser gains if you invest with their suggestion. Thus they are not free, they are paid based on successfully getting you to follow their suggestion and keep you following their advice. Advisers can also be paid where they charge a flat or hourly fee to make a plan, and then they are out of the picture. If you follow their advice to the letter, or only in a general sense, or throw it in the trash it doesn't impact their future earnings.The difference between the different types of fund families is where they get their customers. Given that many fund families have the same exact index funds, and similar target date funds; the big difference is fees and expenses. If a fund family requires you to go through a broker to invest in their fund, that is a layer of people that need to get paid. When picking which family to invest with the number of fund types (US Stocks, bonds, international, indexes, target date, sector funds) can be very important. The IRA (Roth or regular) can be moved from fund to fund without tax consideration, moving within the fund family makes this transaction virtually seamless. If you limit yourself to a fund family because of a suggestion made in 2015, why do you still want to be paying a fee to an outside company a decade later when you now have moved from a target date fund to some other fund within the family.As to the generic filler advice you felt you received. Nobody has any idea in the next few years which segment of the market will do well. They can only discus the past and guesses about the future. That is why many advise either an index, or a target date fund. That wasn't bad advise. The problem was the steering you to a fund family that made the adviser the most money.
So take their advice about the type of investment (Roth vs regular);target vs index; and pick the fund family that you are comfortable with regarding fees, expenses and number of investments