Welcome to the mathematics section
of the Wikipedia reference desk.

Select a section:

WP:RD/MA

Want a faster answer?

Main page: Help searching Wikipedia

How can I get my question answered?

Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
Post your question to only one section, providing a short header that gives the topic of your question.
Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
Don't post personal contact information – it will be removed. Any answers will be provided here.
Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
Note:
- We don't answer (and may remove) questions that require medical diagnosis or legal advice.
- We don't answer requests for opinions, predictions or debate.
- We don't do your homework for you, though we'll help you past the stuck point.
- We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

March 25

Integration and physics-related question

Hello, I have a question about integration, which could not be a normal type of math type question since this is an equation related to physics. I was not sure which section I should ask, but I assumed it would be appropriate to ask here. Also, this is my first time using the reference desk, so I am not used to the style of writing an equation.

So, can you decide if my work is correct below?

To note: v(0)=0, x(0)=h, I will integrate from t=0 to t=t dv/dt=-g-bv---(1), v(t)=-(g/b)(1-e^(-bt)) is given(concluded from the former equation (1).) I want to find x(t). Multiply dt to (1) --> dv=-gdt-bdx --> Integrate --> v(t)=-gt-b(x(t)-h) --> x(t)=h-(g/b)(t-(1-e^(-bt))/b)

The answer is correct, but I'm not sure if this logic is correct. v, x is dependant on t, but is this possible? Also, if there are any Wiki pages about this topic(3 or more variables in Differential Equation?)

Thank you very very much, Luke Kern Choi 5 (talk) 15:10, 25 March 2020 (UTC)[reply]

First, the writing style here is pretty flexible, as long as you make yourself understood, follow the guidelines listed above and are reasonably polite then I, at least, won't take issue. If you plan on using Wikipedia much in the future then you might want to invest some time into learning some of the math specific markup that's available since it will make reading equations easier. See help:math as a reference for using TeX markup and select 'Math and logic' from drop down menu just below the edit box if you just need a few quick math related symbols.

On the math, I'm assuming x=v' is kind of implied in physics but I didn't see it stated anywhere. Other than that I followed everything except I got different signs for some of the terms in the final expression for x; it could be my mistake but you might want to check it. I believe the solution for v is a case of a first order linear differential equation, see Linear differential equation#First-order equation with variable coefficients for a solution to the general case. It also has constant coefficients so the section Non-homogeneous equation with constant coefficients is also applicable. To get x from v you're basically just applying the Fundamental theorem of calculus and using standard integration techniques. I guess since there are two dependent variables you might want to check Ordinary differential equation#System of ODEs, but it's not really needed in this case because you can solve for one variable at a time. --RDBury (talk) 16:03, 25 March 2020 (UTC)[reply]

I got the same signs as in the solution above. It is easier to read if you avoid

{-}\times {-}\times {-}

by pushing the first minus in, and use a shared denominator

b^{2}

:

x(t)=h+{\frac {g}{b^{2}}}(1-bt-e^{-bt}).

Using the Taylor expansion

e^{-bt}=1-bt+{\frac {1}{2}}b^{2}t^{2}-{\frac {1}{6}}b^{3}t^{3}+O(t^{4}),

it is now easy to see that

x(t)=h-{\frac {1}{2}}gt^{2}+{\frac {1}{6}}gbt^{3}+O(t^{4}).

--Lambiam 19:13, 25 March 2020 (UTC)[reply]

Thank you for answering my first question in the reference desk. For the question, it was about physics(as stated above). Although I got a little overwhelmed(because I am not used to English math words that much, I am high school level), I guess it was a good opportunity to learn! Again, thank you for your great effort, and I'll learn those writing tools. Luke Kern Choi 5 (talk) 23:49, 25 March 2020 (UTC)[reply]

FWIW, I found my mistake: reading dv/dt=g-bv, instead of dv/dt=-g-bv. --RDBury (talk) 05:54, 26 March 2020 (UTC)[reply]

By going into edit mode, you can see examples of LaTeX-style markup used for typesetting the maths formulas. --Lambiam 10:43, 26 March 2020 (UTC)[reply]

March 27

RSA encryption

I have been working on understanding RSA encryption recently, and I think I have got it. I now want to try to explain the concepts to some not particularly mathsy folk. The starting point I am using is Clock-arithmetic encryption where (say) the public key is (12,3) and the private kay is (12,9) so message 4 encrypts to 4+3 = 7 and 7 decrypts to 7+9 = 16 clock 12 = 4 the original. However despite this being "valid", it is TOO simple and folk won't decrypt by adding 9, they will decrypt by subtracting 3. So is there any other simple function out there that can work like that without going to the complexities of RSA itself? -- SGBailey (talk) 09:52, 27 March 2020 (UTC)[reply]

The scheme you are rejecting, for good reasons, is known as a Caesar cipher, the oldest and simplest of the class of substitution ciphers, and the first any amateur cryptologist will try when attempting to decode a ciphertext. Not only is it simple, it is also blindingly obvious to anyone who knows the public encryption method how to compute its inverse, the supposedly secret decryption method. This applies not only to the Caesar ciphers, but to all substitution ciphers. The whole point of public key cryptography is in the difficulty of inverting the public encryption method. Coming up with a method that is not trivially invertible is not a simple matter. The basic RSA encryption–decryption algorithms are actually very simple. Can we assume that your not very mathsy folk might understand "raising to a power" (repeated multiplication) in clock arithmetic, as well as generalize the latter from the modulus 12 to the modulus 22? Then the public key can be 3 and the private (secret) key 7. For example, encrypting the message 13 gives 13³ modulo 22 = 2197 modulo 22 = 19. To decrypt the message, compute 19⁷ modulo 22 = 893871739 modulo 22 = 13. (In practice one would use a fast exponentiation method.) The only suspension of disbelief required is that it is hard to find the secret key given the public one. In this toy version of RSA it is very easy, but only because the key length is ridiculously small: less than 4.5 bits, so it is almost effortless to bruteforce this. If the key is long enough, bruteforcing becomes infeasible. --Lambiam 16:28, 27 March 2020 (UTC)[reply]

I think the first thing is to explain the difference between symmetric and asymmetric encryption, which can be done without any math. (Based on the axiom that if something can be explained to the general public, then it will have already been done on YouTube with cartoons, I found "Asymmetric encryption - Simply explained" which does so by comparing it to a mailbox.) Regardless of how simple it is, the Caesar cipher is an example of symmetric encryption while RSA is asymmetric, and so yes, it does kind of miss the point. The main feature of RSA is not that it's difficult to crack, that's a given for any encryption method, but that it remains difficult to crack even if you know the encryption key. I think there is a reason that, despite encryption being in use for millennia, the RSA scheme didn't appear until the 1970's. (Fun fact, the R(ivest), S(hamir) and A(dleman) of RSA are all still living.) Partly it's computers, but the main takeaway for me is that it depends on the existence of mathematical operations that are easy to perform but difficult to undo; for RSA it's the fact that multiplication of large integers is relatively easy while undoing that operation (i.e. factoring) is difficult. Such operations are not very common, and even if you have found one it take quite a bit of cleverness to turn it into an encyption method. --RDBury (talk) 09:34, 28 March 2020 (UTC)[reply]

Thanks L & RDB. I think the clock powers will do nicely. -- SGBailey (talk) 10:07, 28 March 2020 (UTC)[reply]

I don't see how to explain RSA without actually explaining RSA. But the old puzzle with lockboxes and keys that inspired the Shamir three-pass protocol is a good way to explain what RSA tries to do. See the "double lock" scheme at b:Cryptography/Public_Key_Overview for explanation. 2601:648:8202:96B0:E0CB:579B:1F5:84ED (talk)` —Preceding undated comment added 01:40, 29 March 2020 (UTC)[reply]

March 30

ODE

I have an equation of the form dP/dt = bP² where b is a constant.

How do I find the time taken for P to change from P0 to P1 (both are positive)?

Thank you

2A01:E34:EF5E:4640:35D3:546B:6E11:13A9 (talk) 17:07, 30 March 2020 (UTC)[reply]

Step 1:

Solve the differential equation. When you do this, you will first get an equation defining an algebraic relation between P and t. The equation should also have an constant of integration. You can solve this equation for P, giving you a formula for P as a function of t. However, for performing the next steps, this is not actually necessary.

Step 2:

Create two new equations from the one you have. In one, substitute P₀ for P and t₀ for t. In the other, do the same but now with P₁ fand t₁.

Step 3:

Manipulate these equations to get an equation of the form t₁ − t₀ = ..., in which there is no t in the right-hand side. The constant of integration should also disappear in the process, otherwise you did something wrong. Then the right-hand side is a formula for the time taken for the change.

If you get stuck, let us know. --Lambiam 18:36, 30 March 2020 (UTC)[reply]

To be honest I was stuck as soon as I realised it was a differential equation. Numerical Recipes, my usual "go to" for this kind of thing wasn't much help. 2A01:E34:EF5E:4640:35D3:546B:6E11:13A9 (talk) 18:56, 30 March 2020 (UTC)[reply]

Relevant links for step 1:

Non-linear differential equations#Ordinary differential equations

Separation of variables

-- ToE 18:53, 30 March 2020 (UTC)[reply]

Yeah, basically separate the variables so you have