Math Help

[shermaine]

Hello,

I have got some doubts about sequences....
Are u able to help?

(1) Find the limits of the convergent sequnces (log10 nsqr) /n
(2) Find the limits of the convergent sequnces (1 + 3/n)power n

Where n in eaxh case is a positive integer......

(3) Find the series sum (-1) power n [ Sqr root n/(n +1)]

Use any method to determine it's convergence and divergence.....

(4) Is this differential eqn dy/dx = (3xpower2 ylny) /(2y-xpower3-xpower3 lny)
exact?

I'm also confused when it comes to having ln function......
Any one can help me on this?

 

[Styx]

mmm homework.

why are you confued with ln? can you use log? if so ln is just log to base e and it follows exactly the same rules (since it is a log function) the only thing is the inverse is not 10^x but exp(xP

 

[shermaine]

no homework.....
It's exam question.....
Any one can help me?

 

[fat-tony]

I hope it's not too late

for 1) I'm assuming you mean lim (log(n^2))/n. If this is the case, you can take the log(n^2) and change it to 2*log(n). After that, we see that n > log(n) as n->infinity. Therefore, limit = 0.

2) as n->infinity, 3/n -> 0. Then, 1^n = 1. Thus, limit as n->infinity = 1.

3) I'd use the Alternating Series test. This gives us two criteria to find: is An+1 < An, and is lim An = 0? Both of these are satisfied, so it is convergent.

4) We rearrange this a bit:

-3x^2*y*ln(y) + (2*y - x^3 - x^3*ln y)*(dy/dx) = 0.

An exact equation is in the form P(x) + Q(x)(dy/dx) = 0.

P(x) = -3x^2*y*ln(y) and Q(x) = 2*y - x^3 - x^3*ln(y)

To find out if the equation is exact, the partial derivative of P(x) wrt y must equal the partial of Q(x) wrt x.

Py = -3*x^2*ln(y) - 3*x^2
Qx = -3*x^2-3*x^2*ln(y)

Py = Qx, therefore the equation is exact.

 

[shermaine]

Sorry.......
(1) is [log base 10 n^2 /n]
How do i get the limits?
Pls advise.

 

[Optikon]

Sorry.......
(1) is [log base 10 n^2 /n]
How do i get the limits?
Pls advise.

The result posted for #1 is correct and I just want to add the intermediate step that I think Shermaine is asking about... I think...

Log is Log base 10 and not Natural Log.

We agree that Log (n^2) = 2*Log(n)
So the problem is evaluate Lim (n->InF) [2*Log(n) / n]

The numerator tends towards infinity.
The denominator tends towards infinity.

This is indeterminate and we must use L'Hospitals rule by evaluating:

Lim (n->Inf) [D(2Log(n)) / D(n)] where D is the derivative wrt n.

Doing so, D[2*Log(n)] = 1 / (LN(10)*n)
and D[(n)] = 1

now, we take Lim (n->Inf) [2/(LN(10)n)] / 1

This expression tends towards 0 and the rule stated that this limit is equivalent to the original problem so the answer is 0.

 

[Optikon]

Sorry.......
(1) is [log base 10 n^2 /n]
How do i get the limits?
Pls advise.

The result posted for #1 is correct and I just want to add the intermediate step that I think Shermaine is asking about... I think...

Log is Log base 10 and not Natural Log.

We agree that Log (n^2) = 2*Log(n)
So the problem is evaluate Lim (n->InF) [2*Log(n) / n]

The numerator tends towards infinity.
The denominator tends towards infinity.

This is indeterminate and we must use L'Hospitals rule by evaluating:

Lim (n->Inf) [D(2Log(n)) / D(n)] where D is the derivative wrt n.

Doing so, D[2*Log(n)] = 1 / (LN(10)*n)
and D[(n)] = 1

now, we take Lim (n->Inf) [2/(LN(10)n)] / 1

This expression tends towards 0 and the rule stated that this limit is equivalent to the original problem so the answer is 0.

 

[shermaine]

Hello,

Can we change to log to ln.??
Something like log base 10 n ^ 2 = ln 10/ ln 2?
How do we do it if were to change it from log to ln?
Pls advise.

 

[Styx]

ln(x) = log(x)/Log(e)

if that is of any help

 

[fat-tony]

This is indeterminate and we must use L'Hospitals rule by evaluating:

Ahh yes, I always forget that part .