KenZuuuu's blog

By KenZuuuu, history, 3 hours ago, In English

Hi everyone,

I’ve been dealing with some sequence problems where I need to check if the sequence is increasing or decreasing. Usually, I just look at u_{n+1} — u_n, but for some complex functions, it gets messy.

I’ve been trying to use the derivative f'(x) (treating n as x) to check for monotonicity. It makes things much easier, but I’m a bit worried about the domain. Since f'(x) is for x ∈ R and a sequence is only defined for n ∈ N*, is this approach always rigorous? Or are there cases where this might fail?

For instance, if I have u_n = 2^n — an, what’s the best way to find the range of a so that the sequence is strictly increasing? Should I stick to the derivative or is there a better "competitive programming" way to handle this?

Any advice would be great. Thanks!

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93 minutes ago, hide # |
 
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If $$$f(x)$$$ is monotonic then so is $$$u_n$$$. But the inverse is not necessarily true

See $$$u_n=\frac{\cos(2\pi n)}{n}$$$.

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84 minutes ago, hide # |
 
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Yes, this method can be used if computing derivatives is easier than computing differences. You're just extending a function from integers to reals, if its derivative has the same sign everywhere in a real interval (not just at integer values), it'll be monotonous on that interval whether in reals or integers.

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17 minutes ago, hide # |
 
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its not necessarily true since the derivative is the change at that point (n) for the analytic continuation, but u care about the change over an interval of length 1, so it might be decreasing at that point but u may see an overall increase, like for example: cos(2pin)exp(n) if the derivative does not change sign yea ig then u can, u can also try breaking the sequence to see if it is increasing, like for n^2/(n^2+1) u can see at 0 it is 0, it has a finite limit and since numerator and denominator grow at the same rate is most probably is monotonically inc, which it is.