@Marcin2 in #17, solution technically doesnt mean number. I would argue the solution is +- infinity to 1/x=2/x since then both tend towards 0 (either from the left or right depending on if + or - infinity).
i got the answer as 1/1=x
@polylogarithmique said in #19:
> That is frankly misleading if not inaccurate. If you allow z to be a complex number, then the function z -> 1/z^2 does NOT have a limit at 0.
> Proof: Assume it has a limit λ. Take z=x where x is a real number and send x to 0. Then we see that λ should be +oo. Now take z=ix where x is a real number and send x to 0. This time, we see that λ should be -oo, a contradiction.
You are right, I messed up.
What I said about even and odd order poles only applies to rational functions on the real numbers f: X –> R, where X is a subset of R.
It doesn't apply to complex valued functions as you have rightly pointed out. f(x) = 1/(x^2), a function on the real numbers, has a two-sided limit at x = 0.
For complex valued functions a "two-sided limit" doesn't really make sense at all, does it?
As I understand it, the limit of a complex valued function only exists when the function approaches the same value from all angles of approach (not sure whether or not that's the correct terminology here). But again I might be mistaken.
> That is frankly misleading if not inaccurate. If you allow z to be a complex number, then the function z -> 1/z^2 does NOT have a limit at 0.
> Proof: Assume it has a limit λ. Take z=x where x is a real number and send x to 0. Then we see that λ should be +oo. Now take z=ix where x is a real number and send x to 0. This time, we see that λ should be -oo, a contradiction.
You are right, I messed up.
What I said about even and odd order poles only applies to rational functions on the real numbers f: X –> R, where X is a subset of R.
It doesn't apply to complex valued functions as you have rightly pointed out. f(x) = 1/(x^2), a function on the real numbers, has a two-sided limit at x = 0.
For complex valued functions a "two-sided limit" doesn't really make sense at all, does it?
As I understand it, the limit of a complex valued function only exists when the function approaches the same value from all angles of approach (not sure whether or not that's the correct terminology here). But again I might be mistaken.
@Thalassokrator said in #23:
> For complex valued functions a "two-sided limit" doesn't really make sense at all, does it?
> As I understand it, the limit of a complex valued function only exists when the function approaches the same value from all angles of approach (not sure whether or not that's the correct terminology here).
Yes that's right. Without mentioning angles, you can just say that the function approaches a certain fixed value when the absolute value |z| of its argument becomes arbitrarily small.
In particular a holomorphic function is a function that is differentiable with respect to its complex variable z=x+iy and since the derivative is defined as the limit ofthe difference quotient, we "see" that this is much stronger than being differentiable with respect to the real part x and imaginary part y of its variable separately. More precisely, not only f must be differentiable with respect to x and to y, but the partial derivatives df/dx and df/dy must satisfy some relations, known as the Cauchy-Riemann equations, to guarantee that the various limits of the difference quotient coincide when we approach 0 from various angles.
> For complex valued functions a "two-sided limit" doesn't really make sense at all, does it?
> As I understand it, the limit of a complex valued function only exists when the function approaches the same value from all angles of approach (not sure whether or not that's the correct terminology here).
Yes that's right. Without mentioning angles, you can just say that the function approaches a certain fixed value when the absolute value |z| of its argument becomes arbitrarily small.
In particular a holomorphic function is a function that is differentiable with respect to its complex variable z=x+iy and since the derivative is defined as the limit ofthe difference quotient, we "see" that this is much stronger than being differentiable with respect to the real part x and imaginary part y of its variable separately. More precisely, not only f must be differentiable with respect to x and to y, but the partial derivatives df/dx and df/dy must satisfy some relations, known as the Cauchy-Riemann equations, to guarantee that the various limits of the difference quotient coincide when we approach 0 from various angles.
@polylogarithmique said in #19:
> That is frankly misleading if not inaccurate. If you allow z to be a complex number, then the function z -> 1/z^2 does NOT have a limit at 0.
> Proof: Assume it has a limit λ. Take z=x where x is a real number and send x to 0. Then we see that λ should be +oo. Now take z=ix where x is a real number and send x to 0. This time, we see that λ should be -oo, a contradiction.
Let n be a fixed natural number and let f: C\{0}->C be a function given by f(z)=1/z^{n}.
Then f can not have a limit equal to -infinity or +infinity, because such situation is possible only
for functions with the set of all values contained in R sum {-infinity,+infinity}. However, for every
natural n we have lim_{z->0} f(z)=infinity (infinity from the extended set of complex numbers) .
> That is frankly misleading if not inaccurate. If you allow z to be a complex number, then the function z -> 1/z^2 does NOT have a limit at 0.
> Proof: Assume it has a limit λ. Take z=x where x is a real number and send x to 0. Then we see that λ should be +oo. Now take z=ix where x is a real number and send x to 0. This time, we see that λ should be -oo, a contradiction.
Let n be a fixed natural number and let f: C\{0}->C be a function given by f(z)=1/z^{n}.
Then f can not have a limit equal to -infinity or +infinity, because such situation is possible only
for functions with the set of all values contained in R sum {-infinity,+infinity}. However, for every
natural n we have lim_{z->0} f(z)=infinity (infinity from the extended set of complex numbers) .
Yes, here they are! The mathematical experts have been arrived finally:
Commandos like "Let n be a fixed natural number" or "Allow c to be a complex number" are the prove, that you are on mathematical drill grounds.
Assume that a small but positive number of readers knows to operate with complex numbers and know the identity i*i = - 1, what will follow for this thread?
Commandos like "Let n be a fixed natural number" or "Allow c to be a complex number" are the prove, that you are on mathematical drill grounds.
Assume that a small but positive number of readers knows to operate with complex numbers and know the identity i*i = - 1, what will follow for this thread?
#1
Clearly, there's no solution for x!=1 because x=1 cannot be solution and divison by 0 does not exist.
Clearly, there's no solution for x!=1 because x=1 cannot be solution and divison by 0 does not exist.
x! = 1 has the solution 1.
x! = 24 has the solution 4.
Which solution n has the equality 1000000! = n ?
How many solutions n has the equality
1000! Modulo n = 0 ?
x! = 24 has the solution 4.
Which solution n has the equality 1000000! = n ?
How many solutions n has the equality
1000! Modulo n = 0 ?
@Passionate_Player said in #1:
> Solve for x : 1/(x-1) + x = 1/(x-1) + 1. What is the value of x in this equation?
yes
> Solve for x : 1/(x-1) + x = 1/(x-1) + 1. What is the value of x in this equation?
yes
@e4e5f4exf4nicht said in #28:
> x! = 1 has the solution 1.
> x! = 24 has the solution 4.
>
> Which solution n has the equality 1000000! = n ?
None.
> How many solutions n has the equality
> 1000! Modulo n = 0 ?
Step 1: find out all the prime numbers up to 1000, for instance using Eratosthene's sieve. By the prime number theorem, we expect around 144 of them.
Step 2: use Legendre's formula to find out the p-adic valuation of 1000! for each prime p less than 1000.
Step 3: use the fact that the number of divisors of an integer N is the product over all primes of 1+v_p(N).
Now that I explained the method, and I leave the calculations to an interested reader.
> x! = 1 has the solution 1.
> x! = 24 has the solution 4.
>
> Which solution n has the equality 1000000! = n ?
None.
> How many solutions n has the equality
> 1000! Modulo n = 0 ?
Step 1: find out all the prime numbers up to 1000, for instance using Eratosthene's sieve. By the prime number theorem, we expect around 144 of them.
Step 2: use Legendre's formula to find out the p-adic valuation of 1000! for each prime p less than 1000.
Step 3: use the fact that the number of divisors of an integer N is the product over all primes of 1+v_p(N).
Now that I explained the method, and I leave the calculations to an interested reader.
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