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Re: Math Question #1 [message #357817 is a reply to message #357810] |
Sun, 09 November 2008 08:32 |
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If Y = Sin(5X) / 2 - 2 * Cos(2X)
then as X -> 0, Y -> infinity.
If X = 0 then
Sin(5X) = Sin(0) = 0.
2 - 2Cos(2*0) = 2 - 2Cos(0) = 2 - 2(1) = 0. - Can't divide by zero!
Thus if X is almost 0, we have
Sin(5X) / 2 - 2Cos(~0) which is
Sin(5X) / 2 - 2*(~1) which is
Some number / Some other number < 1 and close to 0. This causes the whole expression to increase in value because you're dividing by a fraction.
Renguard is a wonderful initiative
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Supporters of Proposition 8 will argue California does not discriminate against gays, as the current law allows them to get married - as long as they wed a partner of the opposite sex.
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halokid wrote on Mon, 11 October 2010 08:46 |
R315r4z0r wrote on Mon, 11 October 2010 15:35 |
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the hell is that?
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[Updated on: Sun, 09 November 2008 08:33] Report message to a moderator
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Re: Math Question #1 [message #357820 is a reply to message #357817] |
Sun, 09 November 2008 09:23 |
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archerman
Messages: 348 Registered: May 2007 Location: Istanbul / Turkey
Karma: 0
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Recruit |
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CarrierII wrote on Sun, 09 November 2008 17:32 | If Y = Sin(5X) / 2 - 2 * Cos(2X)
then as X -> 0, Y -> infinity.
If X = 0 then
Sin(5X) = Sin(0) = 0.
2 - 2Cos(2*0) = 2 - 2Cos(0) = 2 - 2(1) = 0. - Can't divide by zero!
Thus if X is almost 0, we have
Sin(5X) / 2 - 2Cos(~0) which is
Sin(5X) / 2 - 2*(~1) which is
Some number / Some other number < 1 and close to 0. This causes the whole expression to increase in value because you're dividing by a fraction.
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maybe i didnt understand, but how would you know that the numerator increases more as the denominator increases less? maybe the numerator is a fraction too. sin5x is the closest to zero, and 2-2cosx is the closest to zero as well because 2-2cosx=2-2*1=~0. so its still 0/0.
sorry for my English
[Updated on: Sun, 09 November 2008 09:25] Report message to a moderator
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Re: Math Question #1 [message #357830 is a reply to message #357810] |
Sun, 09 November 2008 11:05 |
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My point about the "some number / some other number thing" was that the numerator doesn't matter, for small values of X, the denominator will always be a small fraction, and so the limit will always be +infinity even if the numerator is a fraction...
Oh: I should add (otherwise my reasoning is flawed) that for X < 33 degrees the numerator > denominator for this. I got that from trial and error informed by some graphs, but you could prove it by solving
Sin(5X) > 2 - 2CosX and find the values for which it is true.
Renguard is a wonderful initiative
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BBC news, quoting... |
Supporters of Proposition 8 will argue California does not discriminate against gays, as the current law allows them to get married - as long as they wed a partner of the opposite sex.
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halokid wrote on Mon, 11 October 2010 08:46 |
R315r4z0r wrote on Mon, 11 October 2010 15:35 |
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the hell is that?
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[Updated on: Sun, 09 November 2008 11:07] Report message to a moderator
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Re: Math Question #1 [message #357859 is a reply to message #357810] |
Sun, 09 November 2008 15:18 |
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NukeIt15
Messages: 987 Registered: February 2003 Location: Out to lunch
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Colonel |
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...because it just had to be posted.
"Arms discourage and keep the invader and plunderer in awe, and preserve order in the world as well as property. Horrid mischief would ensue were (the law-abiding) deprived of the use of them." - Thomas Paine
Remember, kids: illiteracy is cool. If you took the time to read this, you are clearly a loser who will never get laid. You've been warned.
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Re: Math Question #1 [message #358033 is a reply to message #358030] |
Tue, 11 November 2008 02:38 |
nopol10
Messages: 1043 Registered: February 2005 Location: Singapore
Karma: 0
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General (1 Star) |
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Actually, lim(5/x,x,0) (Limit of 5/x as x -> 0) is not infinity as limit of 5/x as x->0 from the negative side and the limit of 5/x as x->0 from the positive side are not equal. Therefore the limit is undefined. It is infinity only when x->0 from the positive side and negative infinity when x->0 from the negative side.
nopol10=Nopol=nopol(GSA)
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