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 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? You assumed this question to be too elementary. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? "I expressed trigonometric functions through complex numbers and exponential as you asked" - I know how to do this, I never asked this. " Logarithms are the inverse of exponential" - yes. "which you can easily do it yourself to express them" - really? Do you know how to "easily" express inverse function through straight function in general case? Please express me logarithm through exponents. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? How it can be defined better? It is totally precise. You dio not show how to express logarithms through trigonometric functions. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? One can express trigonometric functions through hyperbolic ones, but not through logarithms. I am sorry, you just do not understand the question. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? @mathreadler it is not what I want to do. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? If someone says "it is impossible to express roots of general quintic in radicals" would you also claim the equation itself is a representation? 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? @mathreadler it is impossible to do what I am asking for on just complex numbers, independently of representation. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? equation is not an expression. Defining function by an equation is not closed form. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? what I need is the answer on my question. It is impossible to express logarithms in terms of trigonometric functions on the field of complex numbers, you cannot do it. That's why I am asking about a non-trivial extension. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? I am not asking about "derived". I am asking about "expressed in closed form". 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? So how do you express inverse trigonometric functions through non-inverse trigonometric? P.S. Definitely it is impossible on the standard complex numbers, that's why I am asking about an extension of real/complex field. 1d comment Can there be a numerical system in which logarithms can be expressed in terms of exponents in closed form? I fail to see how your examples allow to express logarithms through exponents and vice versa, -1. I wonder who had upvoted this. Apr 24 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 still I do not see the answer to this question there. Btw, why the downvotes? Apr 23 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 It would be great if there was a proof that these two series are actually equivalent on a higher level than just having coincided Ramanujan's sums. For instance, one can be shown to be able to be derived from the other using some elementary operations that are known to never affect the sum of any series. Apr 23 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 The thing is I am thinking about a non-standard numerical system for whose elements I constructed two definitions, from totally diferrent considerations. On the both definitions I defined the operation of multiplication. The first one uses a geometric approach and sutable for any two non-st numbers, the second one uses zeta regularization and suitable for powers of one distinguished element. Here the right side represent the formulas from geometric definition, the left side represents the zeta approach. It seems they coincide. Apr 23 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 I do not see examples on page 9. Apr 23 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 ?? Apr 23 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 it is defined rigorously. en.wikipedia.org/wiki/Divergent_series Apr 23 comment Why is $\sum_{k=1}^\infty -n k^{n-1}=\sum_{k=1}^\infty \left(k^n-(k-1)^n\right)$? @user1952009 what exactly? Mar 14 comment Classifying countable sets of weighted dots on a real line. @Akiva Weinberger yes! And maybe better set of rules that would be simpler but include these all.