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January 26[edit]

Day of the week[edit]

If I was born in 2008 (a leap year) on March 1 (a Saturday), how could I find the next year when my birthday will fall on a saturday? How would I solve this in the context of a maths competition, when I cannot refer to calendars or even calculuators? Thanks. 24.92.70.160 (talk) 02:08, 26 January 2011 (UTC)[reply]

The remainder when 365 is divided by 7 is 1, so if a normal (365-day) year passes between a date and the same date the next year, the next year's date will be one day later in the week. If a leap year passes between the two dates, then the next year's date will be two days later. From those observations you should be able to figure out the answer pretty easily. —Bkell (talk) 03:03, 26 January 2011 (UTC)[reply]

We have an article on calculating the day of the week.--Shantavira|^{feed me} 07:36, 27 January 2011 (UTC)[reply]

Geometrical symbol for midpoint, slope, and length[edit]

As a part of my math work, I have to express the length, midpoint, and slopes of graphs on a plane. However, I am the typ of person who would rather use a fancy mathematical symbol or two to express what I mean, rather than words. Yes, I know about WP:HOMEWORK, but this doesn't fall under that. My question is: are there mathematical symbols for the midpoint, slope, and length of a line? Hmmwhatsthisdo (talk) 07:18, 26 January 2011 (UTC)[reply]

If A and B are points, the segment between them is usually denoted AB and the length of that segment is sometimes denoted

{\overline {AB}}

. In analytic geometry the midpoint of the segment can be denoted

{\tfrac {1}{2}}(A+B)

. I don't know a notation for the slope of a segment, but

{\tfrac {\Delta y}{\Delta x}}

is how slope is calculated and might be usable as a notation for it.

In general, both words and symbols have their purpose and you should use whatever best fits what you want to describe, rather than deliberately avoiding words. I can't guarantee you won't be marked down for being unclear. -- Meni Rosenfeld (talk) 11:06, 26 January 2011 (UTC)[reply]

Hmmwhatsthisdo - can you explain what you mean by " graphs on a plane" ? Meni has given you relevant notation for length and midpoint of the line segment between two points in the plane, but with that interpretation it is difficult to determine what you mean by "slope", as a line only has a slope relative to another line or a set of co-ordinate axes. On the other hand, if you are asking about the graph of a function, then it has a slope at each point (sometimes represented by m) and you can define an arc length along the graph (sometimes represented by s or L) - but then it is difficult to understand what you mean by the "midpoint" of a graph. Gandalf61 (talk) 13:08, 26 January 2011 (UTC)[reply]

I suppose "graphs on a plane" wasn't worded quite the way it should've. I guess the more correct term would be "linear graphs or line segments on a coordinate grid". Hmmwhatsthisdo (talk) 04:50, 27 January 2011 (UTC)[reply]

How about just using letters? -- Mesoderm (talk) 16:24, 27 January 2011 (UTC)[reply]

Well, in MS word, words in mathematical equations are just rendered... odd. They don't look wrong per se, but it doesn't really look right either IMO. Symbols, OTOH, look fine. Hmmwhatsthisdo (talk) 01:54, 28 January 2011 (UTC)[reply]

my seperable diff eq general solution does not agree with the book[edit]

I am trying to solve the differential equation y dx - ( x-2 )dy = 0 It is pretty easy to seperate this algebraically to be 1/(x-2) dx = 1/y dy and then integrate both sides to end up with y = x - 2 + C which does not agree with the back of the book which says y = C(x-2), either I am doing some algebra wrong, or I am missing something fundamental about differential equations, can someone give me some hints as to what I am missing? Thanks —Preceding unsigned comment added by 99.20.116.255 (talk) 18:10, 26 January 2011 (UTC)[reply]

You've got a log after integrating. Dmcq (talk) 18:16, 26 January 2011 (UTC)[reply]

I left out the step where I have ln|y| = ln|x+2| because I can just exponeniate the lns away -- 18:21, 26 January 2011 (UTC) —Preceding unsigned comment added by 99.20.116.255 (talk)

Classic mistake! Your D.E. solution should be ln|y|=ln|x+2| + k, where k is your constant of integration. Proceeding from there gives you the textbook answer. What you've done is to exponentiate in the same step as integrating and then added the constant of integration afterwards, which is wrong. Zunaid 18:30, 26 January 2011 (UTC)[reply]

(edit conflict) You're correct in the separation. Then you integrate to give

\int {\frac {\operatorname {d} y}{y}}=\int {\frac {\operatorname {d} x}{x-2}}\iff \ln y=\ln(x-2)+k\,,

for some constant k. Then you take the exponential of both sides. Notice that

e^{\ln y}=y\ {\text{ and }}\ e^{\ln(x-2)+k}=e^{k}\cdot e^{\ln(x-2)}=e^{k}\cdot (x-2)\,.

If you relabel c = e^k then you get y(x) = c⋅(x − 2), as the book says. But notice that e^k > 0 for all k, and so c > 0 too. So we have

y(x)=c\cdot (x-2)\ {\text{ where }}\ c>0\,.

Thanks all, I realized this after I reexamined the exponentiation step and did an edit conflict with you all. I was trying to do something like e^(ln x+2) + e^C which is wrong. I need to do e^(ln|x+2| + C) which will give me C(x+2). hooray -- 18:38, 26 January 2011 (UTC) —Preceding unsigned comment added by 99.20.116.255 (talk)

I'm not sure about that. Given what you've just written, the solution you will have is | y | = e^C⋅| x − 2 |. You will need to justify how you go from that to y = κ⋅(x − 2), where κ is a constant, for your answer to be correct. — Fly by Night (talk) 18:46, 26 January 2011 (UTC)[reply]

In particular, the constant in the solution y = c(x − 2) does not have to be positive.—Emil J. 18:52, 26 January 2011 (UTC)[reply]

It does at first, until you get rid of the absolute values. — Fly by Night (talk) 21:57, 26 January 2011 (UTC)[reply]

The solution does not have the form y = c(x − 2) in the first place until I get rid of the absolute values.—Emil J. 14:42, 28 January 2011 (UTC)[reply]

Exactly! — Fly by Night (talk) 23:02, 28 January 2011 (UTC)[reply]

The error is that the formula

\int {\frac {\operatorname {d} y}{y}}=\int {\frac {\operatorname {d} x}{x-2}}\iff \ln y=\ln(x-2)+k\,,

should be written

\int {\frac {\operatorname {d} y}{y}}=\int {\frac {\operatorname {d} x}{x-2}}\Leftarrow \ln y=\ln(x-2)+k\,,

Bo Jacoby (talk) 06:36, 29 January 2011 (UTC).[reply]

You might be interested in List of integrals#Integrals with a singularity which shows what can happen around the singularity at 0 - you can have a different constant on either side. Complex number integration is normally much simpler thankfully. Dmcq (talk) 10:50, 29 January 2011 (UTC)[reply]

Hmm... so the constant can change for positive and negative x. That's very interesting... The error didn't come from that though. It came from me forgetting to take the absolute values of the arguments. Once I realised, people had started to comment and it was too late to change. — Fly by Night (talk) 12:23, 29 January 2011 (UTC)[reply]

Well actually positive or negative x-2 but that's a quibble ;-) Dmcq (talk) 13:42, 29 January 2011 (UTC)[reply]