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February 2

What's the technical term for this ratios?

Lambiam Question: In two vessels A and B, there is mixture of milk and water. The ratio of milk and water in these vessels is 5:2 and 8:5, respectively. In what ratio, these mixtures be mixed together, so that the ratio of milk and water in the new mixture becomes 9:4?

I got answer: 7:2

In total problem, there are two mixture ratios i.e 9:4 and 7:2. I want technical term for this ratios. Rizosome (talk) 06:57, 2 February 2021 (UTC)[reply]

@Rizosome: Your answer is wrong; it yields (35 + 16):(14 + 10) = 51:24 = 17:8. Also, it is considered bad etiquette to be pinging a user about something that does not specifically concern them. The correct answer is 13:2. You'd also assist others answering by using correct grammar; "this ratios" should be "these ratios", for example. There's no mathematical term for these ratios; scientifically you could be talking about volume fraction, mole fraction, mass fraction (chemistry), or many others. However, the formulation of this problem as a system of linear equations in two variables is all you need to solve this problem in all of those cases.--Jasper Deng (talk) 07:20, 2 February 2021 (UTC)[reply]

I also get the answer 7:2. I get easily confused when working with ratios and therefore prefer to work with fractions instead of ratios. If a and b are the milk fractions in the original mixtures, and a new mixture is created by mixing these, using a fraction x from A and a fraction (1 − x) from B, the milk fraction in the new mixture will be x·a + (1 − x)·b. If a target is set, requiring the milk fraction in the new mixture to be t, we need to solve the equation x·a + (1 − x)·b = t. Assuming that a ≠ b, the solution is given by x = (t−b)/(a-b). This solution is only valid if 0 ≤ x ≤ 1; otherwise the target cannot be met.

To apply this algebraic solution we need to convert between ratios and fractions. A milk–water ratio m:w corresponds to a milk fraction m⁄m+w, and a mixing fraction p⁄q corresponds to a mixing ratio p:(q−p). So for the exercise, we have:

a = 5⁄7; b = 8⁄13; t = 9⁄13.

x = (9⁄13 − 8⁄13)/(5⁄7 - 8⁄13) = (1⁄13)/(9⁄7·13) = 7⁄9.

This results in a mixing ratio of 7:(9−7) = 7:2.

If there are names for keeping these ratios apart in this confusing mix of ratios, I am not aware of it. The required ratio 9:4 is a target to be achieved in the mixing process, so you might call it the target ratio; then 5:2 and 8:5 are, naturally, the two source ratios. You might call 7:2 the mixing ratio, being the ratio employed in the mixing process. As JD stated, one should not call spirits by name from the vasty deep unless the nature of the question requires specific expertise. Moreover, the notification works only if you sign your post in the same edit in which you use the template. Adding it later has no effect. --Lambiam 09:45, 2 February 2021 (UTC)[reply]

I did it thus: 5a + 8b = 9; 2a + 5b = 4 → a = 13/9, b = 2/9; hence my answer is 13:2. Maybe in the morning I'll try following Lambiam's reasoning. —Tamfang (talk) 08:20, 5 February 2021 (UTC)[reply]

And now I see my (and presumably Jasper Deng's) error: failure to normalize – my 'unit' portions from each vessel are not units. —Tamfang (talk) 08:42, 5 February 2021 (UTC)[reply]

A very common pitfall, which is why I always make a beeline to fractions. --Lambiam 09:13, 5 February 2021 (UTC)[reply]

limit at infinity of constant

What happens if I take the limit at infinity of a function which is constant for all finite values but 0 at infinity? For example, the area of an annulus with constant chord length.

Duomillia (talk) 22:00, 2 February 2021 (UTC)[reply]

You're considering that as a function of, say, the inner radius? I'm not sure how you arrived at the conclusion that that's 0 when the inner radius is ∞. Not really sure what that means, actually.

In any case, there is no reason in principle that functions defined on the extended real numbers have to be continuous, so it is allowed for

f(\infty )

to be different from

\lim _{x\rightarrow \infty }f(x)

. --Trovatore (talk) 22:50, 2 February 2021 (UTC)[reply]

(ec) What is the domain of the function? In the usual understanding of infinity, it is not an element of the domain of a function of which we want to determine its limit at infinity. Usual function domains are

R

and

N

, or subsets of these sets. In all these cases, all elements of the domain are finite. Assuming the domain is

R \cup {\infty}

, I don't think anyone has cared to define the meaning of "limit at infinity" for this case. Let the function value for finite arguments be

a

. Using the usual definition of "limit at infinity" for the domain

R

(the limit at infinity is

L

if for all

ε > 0

there exists

c \in R

such that

| f (x) - L | < ε

whenever

x > c

, and applying it blindly, without any adjustment, we see that the limit is 0 – assuming that

\infty > c

for all

c \in R

. However, that does not jibe with the usual terminology of "the limit of

f

as

x

approaches infinity", which implies

x

does not reach it. In the usual treatment of limits, the point "at which" the limit is taken is excluded from the range of

x

. Normally it is not necessary to specify that for limits at infinity, because infinity is not an element of the domain anyway. Therefore it seems much more reasonable, if infinity is included in the function domain, to replace "whenever

x > c

" by "whenever

c < x < \infty

", analogous to a one-sided limit for

x ↑ 0

, where we have "whenever

-δ < x < 0

". Using this more reasonable definition, the limit is

a

. --Lambiam 23:36, 2 February 2021 (UTC)[reply]

I think it's clear that this ∞ is in the sense of the extended real numbers, which have a completely standard topology and notion of limit. There is no reason ∞ can't be in the domain of the function. --Trovatore (talk) 05:12, 3 February 2021 (UTC)[reply]

Clarification: So, when I talk about a circle (or annulus) with infinite radius... I am imagining looking at the side of the curve getting less and less curvy, getting closer to a straight line. (Supposing, then, that the centre of the circle is at infinity and the edge is somewhere finite)

Duomillia (talk) 04:17, 3 February 2021 (UTC)[reply]

Oh, I see. But then what does it mean to say it has the same chord length as your finite examples? --Trovatore (talk) 05:11, 3 February 2021 (UTC)[reply]