andreask said:Can somebody help me with this problem? I have to proof if 41/81 is a part of attached sequence. Step by step guide would be useful. Thank you!
Amer said:a_n = \frac{n^2 +1}{2n^2}
\frac{n^2 +1}{2n^2} = \frac{n^2}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{2n^2}
can we find "n" such that
\frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81} ?
solve it for n
MarkFL said:I would write:
$$\frac{n^2+1}{2n^2}=\frac{41}{81}$$
Now, cross-multiply and then see if you can find a natural number solution.
andreask said:Can you continiue?
And many thanks
Yes,i understand this!MarkFL said:I would rather help you continue, so that you learn more. :D
What do you get when you cross-multiply? If you are unfamiliar with this method, if you have:
$$\frac{a}{b}=\frac{c}{d}$$
then cross-multiplying, which means to multiply the numerators on each side by the denominators on the other, and then equating both products, will give you:
$$ad=bc$$
If it a kind of shortcut for multiplying both sides by the common denominator of bd:
$$\frac{a}{b}bd=\frac{c}{d}bd$$
Canceling, or reducing the fractions, we get:
$$ad=bc$$
Can you do this with the equation I gave?
MarkFL said:You have the correct equation:
$$81\left(n^2+1 \right)=41\left(2n^2 \right)$$
Now, distribute on both sides, then see if you can solve for $n$...
And no, I do not know of any online resources for studying sequences specifically.
andreask said:81n2+1 = 82n2
1 = 82n2-81nn
1= n2
I don't know i I am right...or what to do...
81 = n2 ??MarkFL said:When you distribute the $81$ on the left side, you need to do so to both terms within the parentheses, as follows:
$$81n^2+81=82n^2$$
Next, subtract $81n^2$ from both sides, and what are we left with?
andreask said:81 = n2 ??
MarkFL said:Yes, good! :D
So, what is the positive root? What is the positive value of $n$?
andreask said:9 of course.
But what have i proved?That 41/81 is a 9th element of sequence?
MarkFL said:Yes, you have shown that:
$$a_{9}=\frac{41}{81}$$
Seems I am to stupid for sequences...MarkFL said:Without using calculus, to explore the monotonicity of the sequence, I would write the $n$th term as suggested above by Amer:
$$a_{n}=\frac{1}{2}+\frac{1}{2n^2}$$
Can you see that each succeeding term is smaller than the last, and that as $n$ grows without bound $a_{n}$ asymptotically approaches $$\frac{1}{2}$$?
You could plot the first few terms, along with the asymptote.
You could make up another such problem just as easily as I can. :D
andreask said:Seems I am to stupid for sequences...
MarkFL said:No, that's not the case at all...let's look at the $n$th term:
$$a_{n}=\frac{1}{2}+\frac{1}{2n^2}$$
And let's for now just look at this part of the $n$th term:
$$\frac{1}{2n^2}$$
Would you agree that when the numerator of a fraction is fixed, that is remains constant, and the denominator gets bigger and bigger, then the value of the fraction is getting smaller and smaller?
andreask said:Yes,i agree!
I think that this sequence is limited too 0,5, but i don't know how to prove it...i have started a master degree with 38,my math was very good at high school,but now...now is all at higher level,i really don't know how would i master this all things...
MarkFL said:What does $$\frac{1}{2n^2}$$ approach as $n$ grows without bound?
andreask said:1/2 ?
I really don't know how to get it...
And i really apreciate that you have time for me,thank you
MarkFL said:Let's look at the first few term of this fraction...let:
$$b_{n}=\frac{1}{2n^2}$$
We find:
$$b_1=\frac{1}{2(1)^2}=\frac{1}{2}$$
$$b_2=\frac{1}{2(2)^2}=\frac{1}{8}$$
$$b_3=\frac{1}{2(3)^2}=\frac{1}{18}$$
$$b_4=\frac{1}{2(4)^2}=\frac{1}{32}$$
$$\vdots$$
$$b_{1000}=\frac{1}{2(1000)^2}=\frac{1}{2000000}$$
$$\vdots$$
$$b_{1000000}=\frac{1}{2(1000000)^2}=\frac{1}{2000000000000}$$
Do you see that as $n$ gets bigger and bigger, $b_{n}$ gets smaller and smaller? If we could let $n$ go to infinity, then what would $b_{n}$ be?
MarkFL said:Yes, $b_{n}$ gets closer and closer to zero, and since:
$$a_n=\frac{1}{2}+b_n$$
then $$a_{n}$$ gets closer and closer to $$\frac{1}{2}$$.
Does this make sense?
andreask said:Yes,it makes sense ,of course. Only problem si,how to write this in a proper way for my profesor? :pAnd why is $$\frac{n^2+1}{2n^2}$$ equal to $$\frac{1}{2} + \frac{1}{2n^2}$$ ?
MarkFL said:A property of fractions is that:
$$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$
so that:
$$\frac{n^2+1}{2n^2}=\frac{n^2}{2n^2}+\frac{1}{2n^2}=\frac{1}{2}+\frac{1}{2n^2}$$
MarkFL said:From what course is this problem? If it is for a course prior to calculus, then you will simply need to explain it as best you can in your own words. You could also choose to write the $n$th term as:
$$a_{n}=\frac{1+\frac{1}{n^2}}{2}$$
Now it is easy to see that the denominator is fixed, and the numerator is getting smaller and smaller as $n$ gets larger and larger, that is, the sequence is strictly decreasing, i.e., it is monotonic.
MarkFL said:I would think the argument I gave in my post previous to this would suffice then.
andreask said:I agree...
But,now i have new problem,about Mathematical induction,should i open new Thread?
MarkFL said:Yes, and I would post it in our Discrete Mathematics sub-forum.
andreask said:I still have one question...What kind of type this sequence is? Arithmetic or geometric? I can't figure it out...
MarkFL said:An arithmetic sequence has the property:
$$a_{n+1}-a_{n}=d$$
While a geometric sequence would have the property:
$$\frac{a_{n+1}}{a_{n}}=r$$
Note: Both $d$ and $r$ are constants that do not depend on $n$. Does this sequence satisfy either property?
andreask said:It seems its arthmetic
MarkFL said:I find:
$$a_{n+1}-a_{n}=\frac{(n+1)^2+1}{2(n+1)^2}-\frac{n^2+1}{2n^2}=-\frac{2n+1}{2n^2(n+1)^2}$$
Thus, the difference between two succeeding terms is a function of $n$, and so the sequence is not arithmetic.
andreask said:And its not geometric?
MarkFL said:What do you find when you compute the ratio I gave above?
