Wikipedia:Reference desk/Archives/Mathematics/2006 July 12

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Windows 98

I'm pretty sure upgrading to a new OS won't help much. They simply have too much security holes to plug, but now that Windows 98 is no longer supported by Microsoft (do they still provide the support files they've built up so far?) my computer has become a sitting security hazard. The Windows XP article doesn't give any system requirements, but that's probably the next one to lose support. What is the latest Windows OS one can still run on a 366 MHz system without slowing it down unneccesarily? Can Microsoft legally drop support for a program millions of people use and force them to upgrade (and pay) for a new operating system when they already have one? Is some outside microsoft denouncer planning on providing support so people won't have to switch?

Yes, I probably should switch, but I don't have the cash to buy a good book let alone an overprized operating system. And no, I can't use Linux instead because the programs I use are not compatible with a Linux system. - Actin 07:18, 12 July 2006 (UTC)[reply]

Looks like you may have a problem - while support for Windows XP won't stop for a while yet (Windows ME and Windows 2000 have to go first), I severely doubt it will run well on a 366MHz machine. The official system requirements say it will, but I would be very doubtful as to how well it will run (and not do things like take two minutes from a click to the start menu opening up). Why not go for Windows 2000? The system requirements are more reasonable, it won't run well on a 366MHz, but it ought to at least run. Windows ME is also an option, however, it's a seriously lousy operating system, so I would strongly recommend against it unless you have no other choice. Another advantage of 2000 or ME is that you're likely to be able to get them at a fraction of the price of XP.

In general, however, a 366MHz is so old that it won't run any modern OS, let alone much modern software, well. I would recommend you stick with Windows 98, and accept the occasional holes in your system's security - as long as you're careful, run a firewall, don't surf on too many dodgy websites and use a good browser, you should be okay. IMO that's a much better solution than trying to upgrade your OS. — QuantumEleven 07:57, 12 July 2006 (UTC)[reply]

Switching to Windows ME won't help; it has reached end-of-life at the same time as the rest of the Windows 98 family, and in many ways it's a worse system. Software developers seem to be able do almost anything they like as far as support goes, unless they made some legally binding promise. This is not strictly a software issue. Can you still buy new parts for a Ford Model T? I don't think so!

I have sucsessfully installed XP on a machine slower than 366MHZ. It is a little sluggish, but it is functional. Using the windows powetoy TweakUI (available from the msft site) you can improve performance by diabeling fancy graphics and amnimations. It certainly is a viable option. 48v 20:00, 13 July 2006 (UTC)[reply]

Of course, not factories can't make old parts forever, but you can offer phone support on an operating system if you want. - Actin 12:45, 12 July 2006 (UTC)[reply]

But are you sure your programs cannot be run under a BSD or Linux system using Wine? Or better still, can you use free alternative software like OpenOffice.org or GIMP instead of some of your Windows apps? You might be pleasantly surprised.

Quite sure. Some will work on a Linux system, but some programs are made specifically for windows. And if I need to run a windows emulator on top of Linux I'm positive my system will crack under the pressure. Besides, I don't think I would be able to make a network connection with the other computer in my home that does use Windows XP. - Actin 12:45, 12 July 2006 (UTC)[reply]

So long as you are running Windows 98, it would be wise to avoid Microsoft apps like Word and Excel, and absolutely avoid Internet Explorer and Outlook; these tend to act as an open invitation for computer viruses. Try Opera or Firefox instead of IE, for example; they are much newer, much nicer, and much safer. Be sure to use virus protection, such as the free AVG Anti-Virus; and also use firewall software, such as the free Zone Alarm.

I can't get certain plugins to work in firefox. Movies and Shockwave games still go dead and a firewall gets in conflict with quite an important piece of software. - Actin 12:45, 12 July 2006 (UTC)[reply]

(Warning: POV!) Finally, start saving your pennies to purchase a Mac mini, with which you can run Mac OS X, Windows Vista, FreeBSD, or a Linux. --KSmrq^T 08:54, 12 July 2006 (UTC)[reply]

Anyway, thanks for the feedback. - Actin 12:45, 12 July 2006 (UTC)[reply]

I feel for you, Actin. I can't believe they are actually allowed to do that without providing free alternatives. - Mgm|^(talk) 08:40, 12 July 2006 (UTC)[reply]

I use to run Windows 2000 a slower processor (266, I think)--upgrading from Windows 95 (for some reason). I found it to run at a speed pretty close to Win95. I think there were enough improvements in the virtual memory handling to make up for the extra bulk. —Bradley 16:31, 12 July 2006 (UTC)[reply]

There is an advantage to an old O/S, not many people will bother writing viruses for it, either (except for Microsoft, who will probably write viruses to get you to upgrade). I am running Windows 98 on two computers, and intend to keep on doing so. Just turn your security settings up and avoid questionable downloads (like porn), and you should be fine. StuRat 20:10, 12 July 2006 (UTC)[reply]

As long as you have a decent amount of RAM (read: 192Mb+), you can run Windows 2000 rather well. Just as long as you don't bloat it with programs in the background. Our old school computers used to run Windows 2000 on 233/266Mhz processors with 256Mb of RAM and an 8Mb graphics card. x42bn6 Talk 02:09, 13 July 2006 (UTC)[reply]

I work on computers like yours. I always recomind Windows 2000 for it's superior stability and security.

Do you also recommend the spell checker ? StuRat 15:56, 13 July 2006 (UTC)[reply]

How to calculate variance from geometric and arithmetic averages?

In the textbook Corporate Finance by Brierly (sp?) and Miers (sp?) it mentions in a footnote that, for time series statistics, you can calculate the variance if you know the geometric and arithmetic averages.

I have tried to work out how to do this. I think the relationship might be:

v^2=g^2-a^2

where ^2 indicates the square v is variance g is geometric average a is arithmetic average

Am I right, or is it some other relationship please?

(I've mentioned this plus other relationships between averages in the discussion to the article Average. (I wonder if there are any other relationships bewteen averages etc that I havnt mentioned?))

Thanks

--81.104.12.70 11:00, 12 July 2006 (UTC)[reply]

Well at first I was completely taken by surprise, but then I inspected some more :it is all correct, except that g should not be the geometric average, but the quadratic mean. Evilbu 13:13, 12 July 2006 (UTC)[reply]

Thanks. Do you think quadratic mean approximates the geometric mean? Are there any transformations or any other methods I could try?

Differential Equation

I need to solve a differential equation, yet I am not sure how to go about it. Could you help? it is:

(-2*d2y/dt2) + (4*dy/dt) + y - 2sin(t) - 7 = 0.

I thought about using (y=e^λt), but it does not seem to work -- any help would be greatly appreciated. Thank you --DragonFly31 11:13, 12 July 2006 (UTC)[reply]

Problems like these are not brainteasers, there is a general procedure for this... but don't expect it to be fun.

Some questions first :

are you familiar with complex numbers? are you familiar with the concept of solving linear differential equations with a particular solution and the space of solutions of the homogenous equation?

anyway complexify your equation like this $w=u+yi$ then we have $-2{\frac {d^{2}w}{dt^{2}}}+4{\frac {dw}{dt}}+w=2e^{ti}+7$ as equation an equation like this always has a particular solution of the form $P(x)e^{ti}+C$ where $P(x)$ is a polynomial what degree? well just bet on a constant one here

that should be quite enough to solve this problem completely Evilbu 12:37, 12 July 2006 (UTC)[reply]

Although I am familiar a little with complex numbers and the concept of solving a linear diff equations for a general solution, I cannot understand how to solve them with a particular solution. Can you help me out a little more as to the procedure to completely solve this? Thanks for your help -- what you've shown me so far seems to be correct.

In order to keep notation a bit lighter, let us take an easier example, consider this linear(!) differential equation : ${\frac {d^{2}y}{dt^{2}}}+4y=1$

One easily sees that the simple constant 1 is a solution. This is called a particular solution. Let us call it f. But is this the only solution? Well suppose there is another solution g. Let us, to keep everything as simple as possible say $w=f-g$

well one easily checks that ${\frac {d^{2}w}{dt^{2}}}+4w=0$ this equation is rather nice, it is homogenous and linear , one can multiply any function with a real constant or add two up, and one still obtains a (new) solution. The solutions form a vector space in other words. Well what is this space. As this equation is of second degree, it is a theorem that they are two dimensional. Thus remains the ask of finding a basis : two independent solutions. If $u_{1}$ and $u_{2}$ happen to be such independent solutions, the general solution will look like $f+C_{1}u_{1}+C_{2}u_{2}$

Evilbu 11:35, 13 July 2006 (UTC)[reply]

so... what would be the answer?... --DragonFly31 14:53, 12 July 2006 (UTC)[reply]

Well look : solving differential equations is in general not easy, there is no method, most of times luck, ingenuity is the only way to get a real 'formula', usually they just use computers to approximate. However there are a couple of cases (and they do occur quite frequently, for instance in physics) where it is simple routine. But that does mean you have to study to be able to do thel. What I am hinting at, is that you also know how to solve homogenous differential linear equations like this :

$a{\frac {d^{2}y}{dt^{2}}}+b{\frac {dy}{dt}}+c\ y=0$

with a , b and c constant. Have you ever seen those, and can you do them?

Evilbu 11:35, 13 July 2006 (UTC)[reply]

Please, everyone, sign your posts, as stated at the top of this page. Thanks.

DragonFly31, what you are seeing here is a reluctance to do your homework for you. If this is not for homework, you can persuade us by explaining where you encountered this equation and why you need to solve it. In either case we would help more if we could see any evidence that you were trying to do anything for yourself. I have elsewhere told you enough for you to solve this on your own. Saying "so... what would be the answer?" tells us you are not serious about understanding and working on this, you just want a solution served on a silver platter. --KSmrq^T 23:26, 12 July 2006 (UTC)[reply]

I like to think about these logicaly. Your first instinct to use e^x or something like that often makes sense. But think about why. y(x)=e^x is an amazing function because the value of the function equals the value of it's nth derivative for all n and all x. So often, you can get all of those complex derivatives to cancel out, because they are all the same. This time, you have to look at what your equation says. It tells you that you need some function of y or t that will somehow "cancel out" the sin(t) term and the y term. As a jumping off point, you know this can't be e^(g(x)) because you can't take derivatives of that and get a sin term. Think about this logically, really analyze what the equation is telling you. Yes, there are protocals to follow; there must be, because your TI-89 can crunch through it and give you an implicit result, but you're doing yourself a great disservice if you don't first think about what the equation is saying. I don't have a user name, I just like math

Signing with ~~~~ also applies to anonymous editors (IP addresses). No one asked you to create a user account (despite all of the benefits and the lack of clear drawbacks), but that doesn't mean you shouldn't sign your posts. -- Meni Rosenfeld (talk) 12:46, 13 July 2006 (UTC)[reply]

Like the others, I won't volunteer your homework solution to you, but I can give some pointers, which I believe is of more use. First, start by solving the homogenous equation (with the terms that don't contain y equal to zero). Just as you wrote, you should try $y=e^{\lambda t}$ . Don't worry if your $\lambda$ -values turn out strange (they contain a root-expression). Now you have the homogenous solution $y=Ae^{\lambda _{1}t}+Be^{\lambda _{2}t},$ where A and B are arbitrary constants.

Next, you need to find the particular solution, which is any other solution to the entire equation. This is typically on a similar form to the terms you zeroed away earlier:

y_{p}=C\sin(t)+D\cos(t)+E

, where C, D and E are once again constants. The clue now is to differentiate this equation twice and put these expressions into the original differential equation. You will then be able to find constants C, D and E by realizing that all terms containing sin(t) must equal zero, as well as the cos(t)-terms and the terms that are constant with regard to t. That procedure produce three equations with three parameters, which you should be able to solve. If you can reach this answer, your quest will be quite well done (but remember that it is the path to the answer, and not your answer in its own right that proves mathemathical skill).

Cheers, Eivind

For future reference, the OP may want to bookmark Physics Forums, which has a section devoted to answering questions about differential equations. ---CH 02:02, 14 July 2006 (UTC) (Not a poster at that forum, BTW, but I've been impressed with some the more helpful contributions in this particular section)[reply]