Complex Analysis: Power Functions

  Рет қаралды 3,655

Maths 505

Maths 505

Күн бұрын

Пікірлер: 12
@TanmaY_TalK
@TanmaY_TalK Жыл бұрын
5:19 in this observation where a is constant and z is variable =! 0; it is no more power function it becomes polynomial function
@cameronspalding9792
@cameronspalding9792 Жыл бұрын
@21:00 a+b must be an integer
@michaelbaum6796
@michaelbaum6796 Жыл бұрын
Brilliant explanation 👍
@cameronspalding9792
@cameronspalding9792 Жыл бұрын
Technically this function is only multivalved when considered as a function of a: once you’ve fixed a value of log(a), f is single valued
@Grecks75
@Grecks75 4 ай бұрын
Even with a fixed value of a (and that is what we're talking about here, f(z) is only a function of z), you still need to specify which branch of the complex log function you're using in _your_ definition of the f(z)=a^z exponential function. After making (and documenting) that decision, your exponential function will be single-valued. But depending on that choice, the f(z)=a^z function will take on different values for a given z. On the other hand, if you don't specify which branch of the log function you're using in your definition, the f(z)=a^z function can be considered a multi-valued function of the single variable z in the same way as the complex log function is considered a multi-valued function of its single argument.
@cameronspalding9792
@cameronspalding9792 4 ай бұрын
@@Grecks75 I don’t think that’s right. f(z) = a^z = exp(log(a)*z). The only ambiguity is which value of log(a) you choose, f should be single valued because exp is single valued
@Grecks75
@Grecks75 4 ай бұрын
@@cameronspalding9792 I agree with everything you said. Once you fix a specific value for log(a) (this includes chosing a branch for log), the function becomes single-valued. As long as you only fix the value of a but do not fix a specific branch of log, the resulting function will still be multi-valued because of the ambiguity in the value of log(a). I think we both mean the same thing, but have focused on different aspects of the resulting concatenated function.
@041-tutulmalakar7
@041-tutulmalakar7 Жыл бұрын
Integrate x^3exp(-2x)sinx from 0 to infinity using differentiation under integral sign . I find this problem in Advance calculus explore book's gamma function chapter
@arkadelik
@arkadelik Жыл бұрын
emperor 👑
@javierpacheco2895
@javierpacheco2895 Жыл бұрын
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