text: What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? Still have questions? 2LiOH (aq) + CO2 (g) --> Li2CO3 (aq) + H2O (l). We have been given a chemical equation to solve. carbon dioxide is accomplished in a chemical reaction using a sorbent known as lithium hydroxide (LiOH). ���cxy���j�������l�N�������� ($�j����i�'���]L�L�L�LjL܌����rq3\azʰS�q�A����R�>�=T�Τ q��@��4/�K}���R��ϡ�T Rr*p Should I call the police on then? 3. Balance the equation for the reaction 2. %PDF-1.5 %���� I am stuck, I cant seem to figure out the balanced equation including states for when Lithium hydroxide solution reacts with carbon dioxide gas to produce lithium carbonate solution and liquid water. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide; To enter an electron into a chemical equation use {-} or e Identify all of the phases in your answer..Can someone help my solve these and tell me how you did it? 3. Carbon Dioxide - CO 2. 6. Molar Mass of NaOH Oxidation State of NaOH. 151 0 obj <>stream Caustic Soda Lye Soda Lye Sodium Hydrate White Caustic. 136 0 obj <>/Filter/FlateDecode/ID[<46667F70CEFE13418A8C2952F5CEEFFB>]/Index[117 35]/Info 116 0 R/Length 91/Prev 116049/Root 118 0 R/Size 152/Type/XRef/W[1 2 1]>>stream In particular, lithium hydroxide was used aboard spacecraft, such as in the Apollo program, to remove carbon dioxide from the atmosphere. Write the formula for the compound that has the atoms and, or groups in the order given: 3 Fe, and two groups made up of 1 As and 4 O.? I went to a Thanksgiving dinner with over 100 guests. The Products Are Solid Lithium Carbonate And Liquid Water. The balanced chemical reaction is Ca(OH)2 + CO2 --> CaCO3 + H2O. Join Yahoo Answers and get 100 points today. Compound states [like (s) (aq) or (g)] are not required. A sample of the ore weighing 4.301 g was treated with nitric acid, which dissolved the PbCO3? The chief was seen coughing and not wearing a mask. A certain lead ore contains the compound PbCO3. %%EOF h�b�9,ܷ@(� 4. lithium hydroxide is LiOH and a solid Carbon dioxide is O2 and a gas Lithium carbonate is Li2CO3 and a solid Water is H2O is a liquid unless its frozen or a gas �r�V�X&@�$ƨq�8�� H�&����L�,@#E�ƚ� �= The balanced equation is 2LiOH + CO2 -> Li2CO3 + H2O. 2. Chem help, chemical bonds and wavelength. O). Lithium hydroxide is an attractive choice for space flight because of its high absorption capacity for carbon dioxide and the small amount of heat Start with the equation for the reaction: 2 "LiOH"+"CO"_2 \rightarrow "Li"_2"CO"_3+"H"_2"O" So one mole of lithium hydroxide reacts with one-half mole of carbon dioxide. About 921 grams of carbon dioxide will react woth 1.00 kg of lithium hydroxide. CO. 3) solid and water (H. 2. Step 1: Write the balanced equation for the reaction. Enter an equation of a chemical reaction and click 'Balance'. Some rebreathers use lithium hydroxide … You can sign in to give your opinion on the answer. By adding up atomic masses we find that one mole of lithium hydroxide weighs about 23.9 grams and one mole of carbon dioxidecweighs about … A r: H = 1, O = 16 M r: H 2 = 2, O 2 = 32, H 2 O = 18. We have been given the equation in words and we need to use the appropriate compound symbols and name and complete the balanced equation. The answer will appear below; Always use the upper case for the first character in the element name and the lower case for the second character. Solid lithium oxide reacts with liquid water to form aqueous lithium hydroxide. 117 0 obj <> endobj endstream endobj 118 0 obj <. Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment to form solid lithium carbonate and liquid water. THANK YOU! What do you think of the answers? which makes you more jittery coffee or tea? I did 155gLiOH divided by the molar … It reacts with carbon dioxide to form lithium carbonate. Other strong bases such as soda lime, sodium hydroxide, potassium hydroxide, and lithium hydroxide are able to remove carbon dioxide by chemically reacting with it. 7. SCUBA re-breathers and personal oxygen units used by firefighters and miners must also remove carbon dioxide. The International Space Station (ISS) uses lithium hydroxide canisters but it also has a newer technology that uses molecular sieves to absorb carbon dioxide. You can use parenthesis () … h�bbdbz\$W�X��+'���^ �*@�R+ Double Displacement (Acid-Base) Reactants. Get answers by asking now. hydroxide lithium - concentrated solution. To write the balanced equation for a neutralization reaction i.e. Calculate The Mass Of Carbon Dioxide Absorbed By 1.00 Kg Of Lithium Hydroxide. Lithium hydroxide canisters aren't the only CO 2 problem solver in space. Reaction Type. If it only absorbed 102 grams of carbon dioxide, what is the efficiency of that lithium hydroxide? Express your answer as a chemical equation. Sodium Hydroxide - NaOH . Lithium hydroxide react with carbon dioxide 2LiOH + CO 2 → Li 2 CO 3 + H 2 O [ Check the balance ] Lithium hydroxide react with carbon dioxide to produce lithium carbonate and water. Elle B. asked • 06/12/20 Solid lithium hydroxide is used to remove exhaled CO2 from space vehicles by forming lithium carbonate & liquid water What mass of carbon dioxide can be absorbed by 10.0 g of lithium hydroxide Gaseous carbon monoxide reacts with oxygen gas to form carbon dioxide gas. This method relies on the exothermic reaction of lithium hydroxide with carbon dioxide gas to create lithium carbonate (Li. Solid Lithium Hydroxide Is Used In Space Vehicles To Remove Exhaled Carbon Dioxide. I am stuck, I cant seem to figure out the balanced equation including states for when Lithium hydroxide solution reacts with carbon dioxide gas to produce lithium carbonate solution and liquid water. Write The Balanced Chemical Equation For This Reaction. The balanced equation for Lithium hydroxide solution + carbon dioxide gas > lithium carbonate solution + water? 2H 2 (g) + O 2 (g) → 2H 2 O (l) Step 2: Write down the relative atomic mass (A r) and the relative molecular mass (M r), for each substance in the equation. Step 4: Find the actual masses. Calculate the number of grams of carbon dioxide absorbed by 155 grams of lithium hydroxide. The balanced equation for Lithium hydroxide solution + carbon dioxide gas > lithium carbonate solution + water? endstream endobj startxref Convert known masses to moles. Procedure: 1. Molar Mass of CO2 Oxidation State of CO2. Step 3: Using A r or M r, change the moles in the equation to grams. Sodium Hydroxide + Carbon Dioxide = Sodium Carbonate + Water . Carbonic Acid Gas Dry Ice [CO2] Carbonic Anhydride R-744 E-290 E 290 E290. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. … This is what I did.. 0 Lithium hydroxide reacts with carbon dioxide to produce lithium carbonate and water. 5. The question given to us is Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water. Water is H2O is a liquid unless its frozen or a gas, For the best answers, search on this site https://shorturl.im/axBeo, (102g actually absorbed / 142 g possible ) X (100) = 71.8% efficient.


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# Complicating an Example by Toen (motivations for DAG)

I'm trying to read (the introduction of) a survey by Toen on Derived Algebraic Geometry, specifically the "Simplicial Presheaves and Derived Algebraic Geometry" one.

He motivates the introduction of DAG as a means to construct moduli spaces. His example is the moduli of linear representations of a group admitting a finite presentation.

Now, DAG (AFAIU) enlarges the theory of stacks in two directions: a derived bit and a stacky bit. The derived bit concerns replacing rings with more general ring-like objects. The stacky bit comes from using stacks of oo-categories.

His motivation for the derived' direction comes from the fact that in constructing a moduli space he gets a tensor product, which isn't considering higher Tor's. (I guess I'm sort of OK with that, but for no concrete reason)

His motivation for the stacky' direction comes from taking quotients of a 'rigidified' moduli problem, which admits a moduli space (the aforementioned tensor product). The problem here is the usual fact that you want to remember the isomorphism groups of objects.

Unfortunately this only motivates us to introduce stacks, not oo-stacks.

So my question is: Can we complicate this example a bit more (but hopefully not too much) so that we need to use higher stacks?

I understand that if we have already enhanced our geometry along the derived direction then we actually need oo-stacks to keep track of notions of equivalence which are weaker than isomorphisms. So a parallel question might be:

if we don't derive our geometry first, do we need to introduce higher stacks?

Comments on the derived are also very much appreciated, thanks!

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> if we don't derive our geometry first, do we need to introduce higher stacks? To get to the moduli you need to single out the solutions, what is usually picking from a larger space of candidates by requiring some conditions/equalities. This is a (projective) limit type construction hence right exact; if unsatisfactory one may repair the procedure by deriving on the left. Among the solutions, some are isomorphic, so one may need to quotient by automorphisms. This is a colimit type construction, hence left exact, so may need to right derive to get cocycles, hence sheaves, stacks, higher s. –  Zoran Skoda Feb 10 '11 at 19:05

Higher stacks arise very naturally before deriving, whenever we consider for example moduli problems of objects that have not just automorphisms but where the automorphisms have automorphisms. Obviously for this to make sense our objects need to have some categorical structure already! For example, if you want to make a moduli space of stacks of some kind, you naturally find a two-stack -- stacks are 2-categorical in nature, and your automorphisms may well have automorphisms. Or another way to think about this - suppose you have a group acting on a stack, and the action has nontrivial stabilizers when acting on automorphisms of points in your stack. Then the quotient is a 2-stack.. (Or you can take the quotient of a scheme by the action of a group-stack! eg for $G$ abelian take the quotient of a point by $BG$, aka $BBG$.)

[Edit: my favorite example of this (an algebraic 2-stack appearing as a quotient of a stack) is due to Ng\^o in his work on the fundamental lemma (Drinfeld coined the name "Ngo 2-stack"). You take a reductive group G, and consider the stack $G/G$. This stack carries a natural action of the group scheme of centralizers, but this is poorly behaved (centralizers jump in dimensions). Ngo realized there's a FLAT group scheme of dimension = rank of G that acts, the group scheme of regular centralizers (for each $g\in G$ take a regular element in $G$ with the same invariant polynomials, a totally elementary argument shows this actually acts on the centralizer of $g$..) Now take the quotient.. you get an algebraic 2-stack, which contains an open dense piece which is just the SCHEME $h/W$. This construction is behind Ngo's analysis of the Hitchin fibration.]

Thinking this way you might get the feeling that you'll probably never practically need more than 2-stacks. But once you start working in a homotopical context you run into higher stacks very quickly. There's a very natural example where you run into the full structure of $\infty$-stacks immediately, as explained by To\"en and Vaqui\'e: moduli of objects in a derived category. If you have a [nice] abelian category you can define a moduli stack of objects in it --- since objects have automorphisms, families of objects are naturally groupoids so you find a stack. Suppose now you have a derived category [technically you need to "enhance" it to a dg or $A_\infty$ category or something equivalent, but let's ignore this]. In this context the presence of negative self-exts of objects means that automorphisms may have automorphisms which may have automorphisms.. i.e. you find higher homotopy groups of the natural moduli functor (which lands now in simplicial sets, or topological spaces). In any case you can make precise sense of moduli of objects in a derived category, and it is an $\infty$-stack!

(Another example of the same nature is the moduli problem for derived categories themselves! here again you have higher Ext groups in the form of Hochschild cohomologies, which are explained homotopically e.g. in To\"en's Inventiones paper on derived Morita theory.)

In summary once you're in a derived context, your moduli functors naturally land not in sets, not in groupoids (which are the same as 1-truncated homotopy types) but in higher groupoids, aka homotopy types, aka simplicial sets. Such moduli problems may be representable by higher stacks.

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David, could you please elaborate a bit your example of BBG? What is BBG as a stack if G is finite abelian group? What are coherent sheaves on BBG? Maybe, you know answers on some of these questions: mathoverflow.net/questions/160025/… ? –  Klim Puhov Mar 30 '14 at 1:23
BBG is the classifying (2-)stack of G-gerbes -- i.e., the functor assigning to a ring the 2-groupoid of G-gerbes (= BG-torsors) over spec R.. ie quotient stack of pt by the abelian group stack BG. I don't think it has nontrivial coherent sheaves (BG doesn't have interesting representations on vector spaces) - there are however interesting sheaves of categories on it... –  David Ben-Zvi Mar 31 '14 at 17:07

I think one place higher stacks come up is if you want to make a moduli space of objects in some derived category. Like if instead of moduli of vector bundles you want to do moduli of perfect complexes. Then your groups of automorphisms can have higher homotopy groups coming from possible negative self-Ext's of your objects.

Or again higher stacks could come up in the same classical way that higher categories come up: n-categories form an (n+1)-category. So for instance if you buy that stacks locally modelled on BG are interesting geometric objects, then if you want to consider a moduli of those it will have one more level of stackiness.

I'll put this as a tentative answer, but I still hope someone who's actually worked with this stuff will come along and have a say.

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I just thought of another example, which I think Toen did some work on: Azumaya algebras. Since Morita theory makes them naturally into a 2-category, their moduli space could be viewed as a 2-stack. –  Dustin Clausen Feb 10 '11 at 18:49

The moduli of linear categories (or abelian categories) in naturally a $2$-stack. You can take a look at the PhD thesis (in french) of Mathieu Anel (a former student of Bertrand Toën).

When you compute the tangent complex of this $2$-stack then you get the $2$-truncation of the Hochschild complex. If you consider the corresponding derived stack then you get the full Hochschild complex.

By the way, higher stacks are introduced to allow quotients, while derived schemes are introduced to allow fiber products.

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# How do I calculate a percentage?

## How do I calculate a percentage?

How To Calculate Percent

1. Determine the total or whole amount.
2. Divide the number to be expressed as a percent by the total. In most cases, you’ll divide the smaller number by the larger number.
3. Multiple the resulting value by 100.

How do I find the percentage difference between 2 numbers?

Answer: To calculate the percentage difference of any two numbers, find the absolute value of the ratio of their difference and their average and multiply that value by 100. Let us look at an example to understand this better.

How do you find the percent of change?

The first is to find the ratioof the amount of change to the original amount. If the new number is greater than the old number, then that ratio is the percent of increase, which will be a positive. If the new number is less than the old number, then that ratio is the percent of decrease, which will be a negative.

### How to calculate a percent increase or decrease?

Now multiply this number by 100: 2*100 = 200%. You’re done! You calculated difference of a number in percent, and the answer is a percentage increase of 200% percent. Enter two percent numbers to find percentage increase or percentage decrease difference: % to % ANSWER: % is percentage increase or decrease (when value is negative)

How to calculate the difference of a number in percent?

You calculated difference of a number in percent, and the answer is a percentage increase of 200% percent. Enter two percent numbers to find percentage increase or percentage decrease difference: % to % ANSWER: % is percentage increase or decrease (when value is negative)

How do you calculate percent change in enrollment?

Converting your answer from a decimal to a percentage is easy—just multiply the value by 100. Multiply 0.1019 by 100, which is 10.19. So, enrollment decreased by 10.19%. Tip: If you use this equation and end up with a negative number, it represents a percent increase.

#### What is the formula to calculate percent change?

To calculate the percent change, there are two options: either divide the new value by the original value and subtract 1 the formula = (A1-A2)-1, or subtract the new value from the original value and divide the result by the original value with the formula = (A2-A1)/A1.

How to find overall average percent change?

• Establish the Initial and Final Values Locate the starting value and final value for a given time period in your situation.
• Determine the Total Change Subtract  V  1 from  V  2.
• Determine the Percentage Change Divide the value you determined by  V  1 to get the total percentage change.

How do you find the percentage change between two numbers?

The percentage change between two numbers is found by dividing their difference by the number that is being changed from, then multiplying that result by 100%. Note: the percent change measures FROM the first value.

## How do you calculate percent change in percentage?

Calculating Percent Changes. Percent change is the change between the original value and a new value. To calculate percent change, you have to calculate the change itself (subtract one number from the other), then divide by the original number, then multiply by 100. The resulting number is the percent change.


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# Fibonacci sequence inversion

How do I get the index in the sequence from the Fibonacci number?

0 1 1 2 3 5 8 13 21 34...

For example

N(3) = 4 (starting from zero)

N(34) = 9 (starting from zero)

..

N(X) = ?

I've seen an equation in wikipedia

There are other ways to compute it?

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What's wrong with the logarithm? – Qiaochu Yuan Sep 17 '10 at 18:02
There's a sense in which the logarithm seems inelegant here, I think. – Michael Lugo Sep 17 '10 at 18:43
If a logarithm is distasteful, then you could also just iteratively compute the Fourier series until you find a number which is either greater than or equal to the number you are looking for. By doing this, you would ultimately find the index you want in time proportional to its location in the sequence. Since the Fourier series terms grow exponentially, this algorithm would be polynomial on the size of the input number (in bits), and so it isn't really that inefficient. – Mikola Sep 17 '10 at 20:11
The logarithm seems inevitable, but you can get rid of the floor function at the expense of some square roots. Just observe that $5F_n^2-2=\varphi^{2n}+\varphi^{-2n}$ and solve a quadratic equation in $\varphi^{2n}$. – Sergei Ivanov Sep 17 '10 at 20:21

As the previous answers have stated the map from $F_n \rightarrow n$ is essentially a logarithm. Since the binary representation of $F_n$ has about $c n$ bits (for the appropriate constant $c = \log_2 \phi$, where $\phi = (1+\sqrt{5})/2$), the bit complexity of calculating the logarithms is about $c' \log^2 n$, for some constant $c'$. Here's another simpler method which has the same complexity:

If you are given $F_n$ for some unknown $n$ if you knew $F_{n-1}$ with $n$ subtractions you can find out what $n$ is (run the fibonacci recursion in reverse). But $F_n/F_{n-1} \approx \phi$. So if you know $1/\phi$ to about $2n$ bits of precision you can find $F_{n-1}$ by multiplying by $1/\phi$ and rounding to the nearest integer. This calculation is certainly simpler than taking logs.

Another alternative is to take $p$-adic logarithms: the function $n \rightarrow F_n$ is $p$-adically continuous, so since $\mathbb{Z}$ is dense in $\mathbb{Z}_p$ it defines a unique $p$-adically continuous function, which by a criterion of Mahler is analytic. You can invert the function via Newton's method -- you just need to find the answer mod $p$ to get a starting point.

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Hendrik Lenstra's paper "Profinite Fibonacci Numbers" is relevant to the $p$-adic version mentioned above math.leidenuniv.nl/~hwl/papers/fibo.pdf – Victor Miller Sep 18 '10 at 2:23
By your information count, knowing the number of binary digits of $N=F_n$ still misses less than one bit to identify $n$. So there should be a practical formula based on the number of binary digits of $F_n$ and for instance the second digit... – Pietro Majer Nov 4 '13 at 15:27

Well, you can almost certainly use integer and modular arithmetic, with the Chinese remainder theorem, because the sequence is periodic for any modulus n you want. This requires some pre-computation, but probably you can predict how much in advance if you know how large a Fibonacci number ahead of time.

For example, I know the index for 34 must be a multiple of 3, just because 34 is even.

Edit: In practical terms perhaps you just sieve on a range on integers instead of using the Chinese remainder theorem (see comments)? Using the number of bits of input as the complexity measure N, you'd need a range of length proportional to N (take some easy rational upper and lower bounds to log 2/log (golden ratio)). Then looking mod 2 you can strike out at least one third of the numbers. Modulo other small primes you strike out some proportion which depends on case but is not too small. You are going to continue until only one integer from the range remains as a candidate. This really doesn't look too bad: the period mod p may be large or small, but what matters is the proportion of the time a given residue class appears.

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You shouldn't need much precomputation. Just knowing the number of digits of the Fibonacci number in question, you can compute n to within a constant factor, and after that it shouldn't be hard to use modular considerations to pin the index down precisely (if you really, really want to avoid using logarithms for some reason). – Qiaochu Yuan Sep 17 '10 at 18:49
If you wanted to compute this quickly, logs are hard to compute exactly. – Michael Lugo Sep 17 '10 at 18:58
In particular the sequence is periodic with period at most $n^2-1$ modulo any integer $n \ge 1$ (the proof is elementary and nearly obvious). So to do CRT you just need a bunch of primes (or maybe prime powers?). Shouldn't be much bigger than $n^2$? – dvitek Sep 18 '10 at 0:14

You can find the index using only integer arithmetic.

Since $F_n$ is monotone one can use Binary search using the fact that $F_i$ can be computed in $O(\log(i))$.

This method works for all monotone sequences and can be used to check if a number is in the sequence.

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For what it's worth, this is essentially A072649 in Sloane (http://oeis.org/A072649). All the formulas given there for that formula either are in terms of the logarithm or assume the Fibonacci numbers are already known and just search that sequence.

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Any such formula has a problem in that two consecutive elements of the sequence are equal to 1. So any formula applies for sufficiently large members of the sequence.

Another question is whether you know that the number is a Fibonacci number and want to find the index, or whether the question involves detecting whether the number is a Fibonacci number and also determining its position in the sequence.

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Invert the formula $F(n)= (r^n - (1-r)^n)/\sqrt{5}$ where $r=(1+\sqrt{5})/2$ by the Lagrange Inversion Formula (LIF). Let $X=\sqrt{5}*F(n)$ and $s=1-r$ so $X=r^n - s^n$. So

$$X=\sum_{j=0}^\infty \frac{(\ln(r))^j - (\ln(s))^j}{j! n^j}$$

Then n = X + sum of X^j * sum over all sequences (b(2),b(3),....,) of nonnegative integers of (-1)^(sum of b(i) from i=2 to j) * ( (sum of b(i) from i=2 to j) + j-1)!/(j!) * product from i=2 to j of (((ln(r))^i - (ln(s))^i)/i!)^b(i) / (b(i)!) ) such that sum of (i-1)*b(i) from i=2 to j equals j-1 and pray for convergence everywhere.

I got this form of the LIF from page 264 of G.P. Egorychev's book, "Integral Representations of Combinatorial Sums"

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@resolvent: the LaTeX support exists for a reason. Use it and your formulae will appear much nicer (I fixed up the first paragraph for ya!). Also, proper handling of equation makes it clearer what you meant. For the displayed equation for $X = \ldots$ above, I had to guess whether you meant that $n^j$ factor to go in the denominator or the numerator. I probably guessed wrong, but it wasn't clear from how you written it in in-line text form. Please try to see if you can fix up the second paragraph? – Willie Wong Oct 4 '10 at 16:58
(I mean, it is okay if it is some simple things, but with all the sums and square roots floating around, you really should take advantage of the LaTeX display.) – Willie Wong Oct 4 '10 at 17:00


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### Home > CALC > Chapter 12 > Lesson 12.2.2 > Problem12-83

12-83.

$\vec{v}(0.5)=\langle -\pi\sin(\pi/2),\frac{0.5^2+1}{(1.5)^2}\rangle=\langle -\pi,\frac{5}{9}\rangle$

Sketch this vector, then use the Pythagorean Theorem to determine its magnitude.


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# How do you evaluate 9^2+5times4?

Nov 14, 2016

${9}^{2} + 5 \times 4 = \textcolor{b l u e}{101}$

#### Explanation:

The order in which to evaluate a math problem can be remembered using the acronym PEMDAS.

${9}^{2} + 5 \times 4$

First apply the exponent.

$81 + 5 \times 4$

Now multiply $5$ times $4$.

$81 + 20$

Add $81 \mathmr{and} 20$

$101$


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# Question 5fd98

Jan 10, 2017

["OH"^(-)] = 2.24 * 10^(-5)"M"

#### Explanation:

Your starting point here will be the definition of a solution's $\text{pOH}$, which as you know is defined as the negative log base $10$ of the concentration of hydroxide anions, ${\text{OH}}^{-}$

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))

Now, in order to solve for the concentration of hydroxide anions, you must rearrange this equation as

log(["OH"^(-)]) = - "pOH"

and rewrite it using exponents of base $10$

10^log(["OH"^(-)]) = 10^(-"pOH")

This is equivalent to

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[\text{OH"^(-)] = 10^(-"pOH}\right)}}}$

Now all you have to do is to plug in the value given to you for the $\text{pOH}$ of the solution

["OH"^(-)] = 10^(-4.65) = color(darkgreen)(ul(color(black)(2.24 * 10^(-5)"M")))#

I'll leave the answer rounded to three sig figs.


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# Network meta analysis without standard deviations

I am trying to compute Cohen's in some studies that I gathered in a Systematic Literature Review.

Unfortunately, almost none of them report the standard deviation (nor df/t, or something for back-computing Cohen's d). Thus, I cannot neither compute Cohen's nor calculate the standard error of each study.

However, the studies commonly report the means and the sample sizes.

I was thinking about using the response ratio (mean1/mean2) as the effect size, and weighting each study according to its sample size (instead of its standard error as usual).

Question 1: do you know any reference where they have done exactly that, and where they acknowledge to have obtained similar results to those with standard error weighting?

Question 2: do you know whether it is posible doing that with network meta analysis in R? I have studies comparing A vs B, and studies comparing A vs C.

• Sample size can be a reliable weighting factor only for categorical variables, but you might proceed with using point effect estimates and sample sizes nonetheless, with a hypothesis-generating scope. You can surely use them as you suggest in R with netmeta or mvmeta packages. – Joe_74 Sep 27 '18 at 10:53

In principle weighting by the sample size makes sense for a difference between groups (or for Cohen's D), if you assume the standard deviation to be the same across studies. This is because the SD of the sampling distribution of a difference between two means in both groups is $$\sqrt{\text{SD}_1^2/n_1+\text{SD}_2^2/n_2}$$ and if $$n_1=n_2=N/2$$ and $$\text{SD}_1=\text{SD}_2=\text{SD}$$, then that simplifies $$\text{SD} \sqrt{4/N}$$. So, to get inverse variance weights, you would want something close to $$N/(4\times\text{SD}^2)$$. If the SD is the same in all trials and the trials have 1:1 allocation, then weighting by N gets you very close (there's of course issues like the SD being estimated in practice etc.) to inverse variance weights.


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# chainer.functions.normalize¶

chainer.functions.normalize(x, eps=1e-05, axis=1)[source]

Normalize input by L2 norm.

This function implements L2 normalization on a sample along the given axis/axes. No reduction is done along the normalization axis.

In the case when axis=1 and $$\mathbf{x}$$ is a matrix of dimension $$(N, K)$$, where $$N$$ and $$K$$ denote mini-batch size and the dimension of the input vectors, this function computes an output matrix $$\mathbf{y}$$ of dimension $$(N, K)$$ by the following equation:

$\mathbf{y}_i = {\mathbf{x}_i \over \| \mathbf{x}_i \|_2 + \epsilon}$

eps is used to avoid division by zero when norm of $$\mathbf{x}$$ along the given axis is zero.

The default value of axis is determined for backward compatibility.

Parameters
• x (Variable or N-dimensional array) – multi-dimensional output variable. The first dimension is assumed to be the mini-batch dimension.

• eps (float) – Epsilon value for numerical stability.

• axis (int or tuple of ints) – Axis along which to normalize.

Returns

The output variable which has the same shape as $$x$$.

Return type

Variable