#1
Ok I posted this in the maths thread, but it's for homework for tomorrow and I still have other stuff to do, so I kind of need an answer ASAP.

The question is this.

If tn=100

tn=.5(tn-1)+ 10

So the sequence is,

100,60,40,30,25,22.5 ..etc

And T(inifinity) = 20

How can I determine the limit (20) just by the number in the formula..

I tried square root of 100 + 10 which = 20 which would be right but it just seemed like a coinicidence.

I promise I'll delete this thread, I just need an answer or a bit of help.


Thanks in advance,
Shaun
#2
Is this infinite series or geometric series?

EDIT: The formula you have there is wrong, bud....

IF tn=.5(tn-1)+ 10
and tn=100
then...
100=.5(100-1)+10
which equals:
90=49.5
Last edited by mamosa at Sep 14, 2009,
#3
Quote by mamosa
Is this infinite series or geometric series?

Geometric I guess?

I don't have that great of a handle on it, as we just started it today.
Quote by mamosa
Is this infinite series or geometric series?

EDIT: The formula you have there is wrong, bud....

I'm just copying what's in my binder...

tn=tn-1 + x

?
#4
And you need to be more specific or reword the problem. What limit are you talking about? Limits as in calculus? And what do you mean "the number in the formula"?
#5
mamosa's edit is right. You make no sense.

SHEdit: OH!

T(n) == T(n-1) + x, as in subscript n. Okay. Gotcha.

[IN PHIL WE TRUST]


Quote by Trowzaa
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#6
Quote by SteveHouse
mamosa's edit is right. You make no sense.

SHEdit: OH!

T(n) == T(n-1) + x, as in subscript n. Okay. Gotcha.

yesss. subscript n.
Quote by mamosa
And you need to be more specific or reword the problem. What limit are you talking about? Limits as in calculus? And what do you mean "the number in the formula"?

Limit as in, as long as the sequence continues, it would remain at 20. 20 = the limit.

No matter how many terms it will end up at 20...
Last edited by ShaunDiel at Sep 14, 2009,
#7
Limit of T (as x->infinity) == 20.

I still don't get it though Too far removed from AP Calc.

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Quote by Trowzaa
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#8
Quote by SteveHouse
Limit of T (as x->infinity) == 20.

I still don't get it though Too far removed from AP Calc.

Yeah you wrote what I should have typed originally. I'm in a bit of a rush but the formula you just typed is what's in my scribbler, only with the little infinity sign( oo ) ?
#9
Quote by ShaunDiel
Yeah you wrote what I should have typed originally. I'm in a bit of a rush but the formula you just typed is what's in my scribbler, only with the little infinity sign( oo ) ?

I'm sorry, I'm enthralled by your boob with two nipples O_o

[IN PHIL WE TRUST]


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#11
Good call SteveHouse
I'm taking calc now and I could figure out what the **** was going on.

So:

100=t(n-1) +10
90=t(n-1)
90=tn-t
And since tn=100...
90=100-10

Therefore, n=10.
#12
Quote by mamosa
Good call SteveHouse
I'm taking calc now and I could figure out what the **** was going on.

So:

100=t(n-1) +10
90=t(n-1)
90=tn-t
And since tn=100...
90=100-10

Therefore, n=10.

I'm not sure I get this...

The sequence I gave was checked by my teacher and is right.

t= term
n= number

so tn would mean term variable?

ugh idk.
#13
t(n) means the term before anything happened. It is the original number.

EDIT: I made a huge ****ing mistake! Damn non-subscript text

I'll get on this in a sec.
Last edited by mamosa at Sep 14, 2009,
#14
Quote by mamosa
t(n) means the term before anything happened. It is the original number.

EDIT: I made a huge ****ing mistake!

Yea i thought it might have been me. I was sitting here reading your post because I thought you were right shitting my pants.
#15
I know what it is!

If T(n)=100, then T(n-1) is equal to 2 times 10 less than T(n). I misread the problem a n-1 when it really refers to (n-1). So therefore, T(n-1) is 180.

Why? Because you have T(n). T(n-1) is the number before T(n). You'll notice the number keep getting lower, so t(n-1) is higher than T(n).

So...

100=.5(180)+10! There!

Now, what you need to do is find out exactly what "n" is. (n is the number of stages. So the very first number, let's say (hypothetically) could be 5,000. So then you apply the rule and you multiply it by 1/2 and add 10. All the way until you get to 180. So you need to find out what n is. Obviously n=20, so that means T(n) was the 20th stage of the function).
#16
Quote by mamosa
I know what it is!

If T(n)=100, then T(n-1) is equal to 2 times 10 less than T(n). I misread the problem a n-1 when it really refers to (n-1). So therefore, T(n-1) is 180.

Why? Because you have T(n). T(n-1) is the number before T(n). You'll notice the number keep getting lower, so t(n-1) is higher than T(n).

So...

100=.5(180)+10! There!

Now, what you need to do is find out exactly what "n" is. (n is the number of stages. So the very first number, let's say (hypothetically) could be 5,000. So then you apply the rule and you multiply it by 1/2 and add 10. All the way until you get to 180. So you need to find out what n is. Obviously n=20, so that means T(n) was the 20th stage of the function).

Um... the first number is 100 t1=100?

Term 1 = 100..

100 is the first number of the sequence. Am i wrong or did you misunderstand me..

Also, I appreciate the help so much.
#17
But if T(n) is the first number in the sequence, there can be no T(n-1). T(n-1) comes before T(n), so if T(n) is the first number, then T(n-1) DNE.