How to prove if function is strictly monotonic increasing
Solution 1
Hint: induction. $a_1>a_0$, in fact $-\frac 7 8<-\frac 4 {13}$. Than given that $a_n>a_{n-1}$ let's show that $a_{n+1}>a_n$. You have $a_{n+1}=\frac {3(n+1)-7} {8+5(n+1)}=\frac {3n-4} {13+5n}, a_n={\frac {3n-7} {8+5n}}$, so you have to show that $\frac {3n-4} {13+5n}>\frac {3n-7} {8+5n} \iff(3n-4)(5n+8)>(3n-7)(13+5n)\iff 15n^2+4n-32>15n^2+4n-91$
and it's done.
Solution 2
Write $a_n$ as: $$a_n=\frac{15 n - 35}{15 n + 24} = 1 - \frac{59}{15n + 24}$$
then notice that:
$15n+24\;$ is strictly positive and increasing, thus
$\frac{59}{15n + 24}\;$ is decreasing, thus
$-\frac{59}{15n + 24}\;$ is increasing, thus
$a_n=-\frac{59}{15n + 24} + 1\;$ is increasing.
[ EDIT ] The above makes use of the known (and otherwise easy to prove) properties:
if $f(x) \gt 0$ for all $x$, then $f(x)$ is [strictly] increasing $\iff$ $\frac{1}{f(x)}$ is [strictly] decreasing;
$f(x)$ is [strictly] increasing $\iff$ $-f(x)$ is [strictly] decreasing.
Related videos on Youtube
Michael
Updated on August 16, 2022Comments
-
Michael about 1 year
I have this function and I have to prove that this function is strictly monotonic increasing
This is the equation: $$a_n = \frac{3n-7}{8+5n}$$
What would be the best way to do this?
-
Samiron about 7 yearsyou may try to use the fact that a sequence $a_n$ is increasing iff the sequence $b_n = (5/3)a_n -1 $is increasing.
-
amWhy about 7 yearsNot a good idea to ask for "the best way...", "easiest way..." Those matters are in the eye of the beholder. Besides, if you get someone willing to offer a way, you should be grateful to know one way, since you clearly haven't shown any attempt at any proof whatsoever.
-
amWhy about 7 yearsWhat is the function? I'm not going to chase links to find it. It's easy enough for you to type it into your question, if the question matters to you.
-
-
Michael about 7 yearsCould you please add more details.
-
amWhy about 7 years@Michael Beggars can't be choosers. Please ask specific questions. Have you tried to prove by induction?
-
amWhy about 7 years@Michael: the question of the hour is "Could you provide more details, in your post? If you ask a question with no details, no effort, no explanation, you have no right to insist that others give you any more than they graciously gave you, given the little you provide.
-
wythagoras about 7 yearsDid you mean $a_1>a_0$?
-
Bargabbiati about 7 yearsYou're right, I was too vague.