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I have the exponents $k^4+k^5+k^6$. Is there any way to simplify this into one exponent? I'm trying to find a way to simplify a sequence of increasing exponents into one, but am not sure how. With some guess and checking, I got $k^4+k^5+k^6 \approx k^{6.13368}$, but why does this sequence simplify to approximately $k^{6.13368}$, and how can I simplify other sequences, like $k^6+k^7+k^8+k^9$?

Blackhole927
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  • No. It doesn’t simplify. Why do you think it does? At best you have Big O approximations. – Eric Aug 16 '23 at 03:29
  • For which $k$ did you find this approximation to work? If I plot the two functions, they look nothing alike. – CW279 Aug 16 '23 at 03:30
  • K = 5 got within 0.7 of the answer, and I asked here because I didn't know if it simplified. I'm not good with exponents, so I thought I would ask here in hopes that there was something I was missing. – Blackhole927 Aug 16 '23 at 03:34
  • The bigger $k$ gets, the more it's going to look like just the term with the highest exponent. Depending on how good an approximation your need, once$k$ gets bigger than, say, $100$, you can drop all but the highest exponent term. – JonathanZ Aug 16 '23 at 04:45
  • You can write it as $(k^7-k^4)/(k-1)$, if that's any help. – Gerry Myerson Aug 16 '23 at 07:00

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There's no way to simplify the expression $P(k)=k^2+k^4+k^6$. However, you can ask a related question: given some particular $k=k_0$, find the exponent $a$ such that

$$ (k_0)^a=P(k_0) $$

Taking the logarithm of both sides and solving for $a$ gives

$$ a=\frac{\ln{P(k_0)}}{\ln{k_0}} $$

Substituting $k_0=5$ into the above expression, I find $a=6.133656\dots$, which is in agreement with the exponent you found from trial-end-error.

CW279
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