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:[tex] a_n = \frac{n^2 +1}{2n^2} [/tex]
[tex]\frac{n^2 +1}{2n^2} = \frac{n^2}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{2n^2} [/tex]
can we find "n" such that
[tex] \frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81} [/tex] ?
solve it for n
MarkFL said:I would write:
\(\displaystyle \frac{n^2+1}{2n^2}=\frac{41}{81}\)
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:
\(\displaystyle \frac{a}{b}=\frac{c}{d}\)
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:
\(\displaystyle ad=bc\)
If it a kind of shortcut for multiplying both sides by the common denominator of bd:
\(\displaystyle \frac{a}{b}bd=\frac{c}{d}bd\)
Canceling, or reducing the fractions, we get:
\(\displaystyle ad=bc\)
Can you do this with the equation I gave?
MarkFL said:You have the correct equation:
\(\displaystyle 81\left(n^2+1 \right)=41\left(2n^2 \right)\)
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:
\(\displaystyle 81n^2+81=82n^2\)
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:
\(\displaystyle a_{9}=\frac{41}{81}\)
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:
\(\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}\)
Can you see that each succeeding term is smaller than the last, and that as $n$ grows without bound $a_{n}$ asymptotically approaches \(\displaystyle \frac{1}{2}\)?
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:
\(\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}\)
And let's for now just look at this part of the $n$th term:
\(\displaystyle \frac{1}{2n^2}\)
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 \(\displaystyle \frac{1}{2n^2}\) 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:
\(\displaystyle b_{n}=\frac{1}{2n^2}\)
We find:
\(\displaystyle b_1=\frac{1}{2(1)^2}=\frac{1}{2}\)
\(\displaystyle b_2=\frac{1}{2(2)^2}=\frac{1}{8}\)
\(\displaystyle b_3=\frac{1}{2(3)^2}=\frac{1}{18}\)
\(\displaystyle b_4=\frac{1}{2(4)^2}=\frac{1}{32}\)
\(\displaystyle \vdots\)
\(\displaystyle b_{1000}=\frac{1}{2(1000)^2}=\frac{1}{2000000}\)
\(\displaystyle \vdots\)
\(\displaystyle b_{1000000}=\frac{1}{2(1000000)^2}=\frac{1}{2000000000000}\)
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:
\(\displaystyle a_n=\frac{1}{2}+b_n\)
then \(\displaystyle a_{n}\) gets closer and closer to \(\displaystyle \frac{1}{2}\).
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 \(\displaystyle \frac{n^2+1}{2n^2}\) equal to \(\displaystyle \frac{1}{2} + \frac{1}{2n^2}\) ?
MarkFL said:A property of fractions is that:
\(\displaystyle \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\)
so that:
\(\displaystyle \frac{n^2+1}{2n^2}=\frac{n^2}{2n^2}+\frac{1}{2n^2}=\frac{1}{2}+\frac{1}{2n^2}\)
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:
\(\displaystyle a_{n}=\frac{1+\frac{1}{n^2}}{2}\)
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...