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Chip
Joined: 21 May 2007
Posts: 1
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 Posted: May 21, 2007 4:02 PM Post subject: Help understanding the solve function for trig stuff!! |
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Ok, when I need to solve for something like cos(x)=0, I type in the following:solve(cos(x)=0,x)
The answer I get is: x=90(2 x (backward @ sign)n6-1)
What does this mean, how can i make it so it would solve between a certain interval, like between 180 and 270 degrees. A photo of this is included for reference. Thanks for the help!
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mellamokb
Joined: 10 May 2007
Posts: 124
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| Posted: May 22, 2007 2:45 PM Post subject: |
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Hi Chip,
Welcome to TI-89 Forums!
The whole expression @n6 represents an arbitrary integer. Basically, you can plug in any whole number between -infinity to infinity into the variable @n6 and you will have a particular solution to the equation. All arbitrary integer variables are of the form @n##, where ## is just an incrementing number so that multiple arbitrary integers in a single expression or successive expressions are not confused.
You can use the with bar ("|") on your TI-89 to specify a constraint for a problem that will return infinite solutions. In your situation (assuming you are in degree mode), you would type:
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Solve(Cos(x)=0,x)|x>=180 and x<=270
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The constraint comes after the "Solve(" function. Note that the >= and <= symbols are single characters, [Diamond]+[>] and [Diamond]+[<], respectively. TI-89 Titanium users with OS 3.10+ can use compound inequalities:
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Solve(Cos(x)=0,x)|180<=x<=270
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If you need to solve a trigonometric equation or something similar that returns an infinite number of solutions, and you want to return lots of solutions (like cos(x) = 0 for -500<=x<=1000), you may find it advantageous to use this alternate form:
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Solve(Cos(x)=0 and -500<=x and x<=1000,x)
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The result will still include an arbitrary integer variable, but there will be an extra expression bounding that integer variable, so you know what range of values to plug in to see the particular solutions, something like this:
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x=90(2@n23-1) and @n23>=-2 and @n23<=6
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Then you would know you have 9 (6 - -2 + 1) solutions and you can just plug any integer from -2 to 6 in to see the individual solutions, without needing to scroll through a very long answer.
For TI-89 Titanium users, you can use:
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Solve(Cos(x)=0 and -500<=x<=1000,x)
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and you would get something like:
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x=90(2@n23-1) and -2<=@n23<=6
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Hope this helps!
~ mellamokb |
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atsign
Joined: 13 Aug 2007
Posts: 7
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| Posted: August 13, 2007 3:31 AM Post subject: |
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| Why does the arbitrary integer increment every time I solve a trig equation? It makes reading the answer more confusing. |
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mellamokb
Joined: 10 May 2007
Posts: 124
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| Posted: August 13, 2007 9:42 AM Post subject: |
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Hi atsign,
Welcome to TI-89 forums!
The arbitrary integer increments for each trig equation to represent that it's a different instance of an arbitrary integer. For instance, I believe it is possible to construct an equation whose solution would return two different arbitrary integers in the same expression, like
If @n6 was repeated in the above expression (representing a hypothetical coordinate-based trigonometric solution), and I wanted to test the solution point (4, 7), substituting 4 for @n6 would substitute in both places. I want to be able to place different numbers into the two variables.
For another example, suppose you wanted to solve three equations in list format so the solutions stay together. If you wanted to plug in different arbitrary values in each of the three equations in the list, the names of the variables would have to be unique. The basic underlying principle is the ability to substitute values in different trigonometric solutions without affecting other expressions you are working with, since the values and expressions represent different situations.
If the different numbers are confusing and you have "trained your eyes" to see @n1, for instance, you can append "|@n1=@n##" to an expression to substitute @n1 for @n##. Let's say I have a situation where the solution to the equation returned the following:
I would rather see @n1 than @n4 so I can understand what is going on, so I would write:
Which will output:
The only way to really "reset" the arbitrary integer counter back to 1 is to reset the calculator. Perhaps clearing the home screen may also reset the counter, you may want to make a trial run with this.
Hope this helps!
~ mellamokb |
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atsign
Joined: 13 Aug 2007
Posts: 7
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| Posted: October 1, 2007 2:17 AM Post subject: |
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| Well, clearing the home screen doesn't do it. But your post was very helpful, thanks. |
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mellamokb
Joined: 10 May 2007
Posts: 124
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| Posted: October 1, 2007 10:01 AM Post subject: |
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Glad to help, atsign!
~ mellamokb |
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atsign
Joined: 13 Aug 2007
Posts: 7
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| Posted: December 1, 2007 9:14 PM Post subject: |
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| I wanted to clarify that I misunderstood "clearing the home screen" to mean manually clearing it. But the manual says that using the ClrHome function would reset the arbitrary integer, so I'm sure it would work. |
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