Infinite amount of vertical asymptotes
Is it possible that the graph of function has infinitely many vertical asymptotes?
I suppose, that it is not possible, because such function would not exist. But I need to prove it in a math-fashioned-way, and I'm clueless how to do it.
Help would be appreciated. :) Thanks.
limits graphing-functions
asked Nov 24 at 14:22
weno
817
|
show 3 more comments
Is it possible that the graph of function has infinitely many vertical asymptotes?
I suppose, that it is not possible, because such function would not exist. But I need to prove it in a math-fashioned-way, and I'm clueless how to do it.
Help would be appreciated. :) Thanks.
limits graphing-functions
asked Nov 24 at 14:22
weno
817
5
$tan x{}{}{}{}$.
– TonyK
Nov 24 at 14:24
Oh yes, that indeed has infinitely many asymptotes. Still, how do I prove that in a math-way?
– weno
Nov 24 at 14:30
1
We can even have the asymptotes accumulate in the finite. $f(x)=sum_nfrac1{(nx-1)2^n}$ has a definition gap with vertical asymptote at all $frac 1n$, while it is defined (i.e., converges) everywhere else
– Hagen von Eitzen
Nov 24 at 14:31
1
@weno I'd say that TonyK's comment is a "math-way" proof of the claim
– Hagen von Eitzen
Nov 24 at 14:31
@HagenvonEitzen Do I simply answer the question with "yes, because for example f(x) = tan x has infinitely many vertical asymptotes"?
– weno
Nov 24 at 14:36
|
show 3 more comments
Is it possible that the graph of function has infinitely many vertical asymptotes?
I suppose, that it is not possible, because such function would not exist. But I need to prove it in a math-fashioned-way, and I'm clueless how to do it.
Help would be appreciated. :) Thanks.
limits graphing-functions
asked Nov 24 at 14:22
weno
817
Is it possible that the graph of function has infinitely many vertical asymptotes?
I suppose, that it is not possible, because such function would not exist. But I need to prove it in a math-fashioned-way, and I'm clueless how to do it.
Help would be appreciated. :) Thanks.
limits graphing-functions
limits graphing-functions
asked Nov 24 at 14:22
weno
817
asked Nov 24 at 14:22
weno
817
asked Nov 24 at 14:22
weno
817
asked Nov 24 at 14:22
weno
817
asked Nov 24 at 14:22
weno
817
817
5
$tan x{}{}{}{}$.
– TonyK
Nov 24 at 14:24
Oh yes, that indeed has infinitely many asymptotes. Still, how do I prove that in a math-way?
– weno
Nov 24 at 14:30
1
We can even have the asymptotes accumulate in the finite. $f(x)=sum_nfrac1{(nx-1)2^n}$ has a definition gap with vertical asymptote at all $frac 1n$, while it is defined (i.e., converges) everywhere else
– Hagen von Eitzen
Nov 24 at 14:31
1
@weno I'd say that TonyK's comment is a "math-way" proof of the claim
– Hagen von Eitzen
Nov 24 at 14:31
@HagenvonEitzen Do I simply answer the question with "yes, because for example f(x) = tan x has infinitely many vertical asymptotes"?
– weno
Nov 24 at 14:36
|
show 3 more comments
5
$tan x{}{}{}{}$.
– TonyK
Nov 24 at 14:24
Oh yes, that indeed has infinitely many asymptotes. Still, how do I prove that in a math-way?
– weno
Nov 24 at 14:30
1
We can even have the asymptotes accumulate in the finite. $f(x)=sum_nfrac1{(nx-1)2^n}$ has a definition gap with vertical asymptote at all $frac 1n$, while it is defined (i.e., converges) everywhere else
– Hagen von Eitzen
Nov 24 at 14:31
1
@weno I'd say that TonyK's comment is a "math-way" proof of the claim
– Hagen von Eitzen
Nov 24 at 14:31
@HagenvonEitzen Do I simply answer the question with "yes, because for example f(x) = tan x has infinitely many vertical asymptotes"?
– weno
Nov 24 at 14:36
5
5
$tan x{}{}{}{}$.
– TonyK
Nov 24 at 14:24
$tan x{}{}{}{}$.
– TonyK
Nov 24 at 14:24
Oh yes, that indeed has infinitely many asymptotes. Still, how do I prove that in a math-way?
– weno
Nov 24 at 14:30
Oh yes, that indeed has infinitely many asymptotes. Still, how do I prove that in a math-way?
– weno
Nov 24 at 14:30
1
1
We can even have the asymptotes accumulate in the finite. $f(x)=sum_nfrac1{(nx-1)2^n}$ has a definition gap with vertical asymptote at all $frac 1n$, while it is defined (i.e., converges) everywhere else
– Hagen von Eitzen
Nov 24 at 14:31
We can even have the asymptotes accumulate in the finite. $f(x)=sum_nfrac1{(nx-1)2^n}$ has a definition gap with vertical asymptote at all $frac 1n$, while it is defined (i.e., converges) everywhere else
– Hagen von Eitzen
Nov 24 at 14:31
1
1
@weno I'd say that TonyK's comment is a "math-way" proof of the claim
– Hagen von Eitzen
Nov 24 at 14:31
@weno I'd say that TonyK's comment is a "math-way" proof of the claim
– Hagen von Eitzen
Nov 24 at 14:31
@HagenvonEitzen Do I simply answer the question with "yes, because for example f(x) = tan x has infinitely many vertical asymptotes"?
– weno
Nov 24 at 14:36
@HagenvonEitzen Do I simply answer the question with "yes, because for example f(x) = tan x has infinitely many vertical asymptotes"?
– weno
Nov 24 at 14:36
|
show 3 more comments
1 Answer
1
active
oldest
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Generally, a function has a vertical asymptote at $x$ when it can be expressed as: $f(x)=frac{a}{g(x)}mid aneq g(x)$ and $g(x)=0$, the simplest example of which is $frac{1}{x}$.
For a function to have infinitely many vertical asymptotes there must be infinitely many values of $x$ for which $g(x)=0$. There are two ways this can happen:
(1.) $g(x)$ is periodic with infinitely many zeros - i.e.
$$f(x)=ufrac{1}{per(x)}mid per:=sin,cos,tan,mod,ldots,etc.$$
(2.) $f(x)$ is a sum or product of an infinite series.
The latter case is pretty much the same as the former, as most (if not all) such infinite series have a closed form solution equivalent to (1.), for example:
$$sum_{n=0}^infty(-1)^nfrac{(2n+1)!}{x^{2n+1}}$$ is just the inverse of the Taylor series for the $sin$ function.
As TonK commented, $tan{x}$ is a straightforward example. You could also use $frac{u}{sin{x}}$, $frac{u}{cos{x}}$, $Gamma(x)$, etc.
answered Nov 24 at 15:15
R. Burton
32919
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
add a comment |
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Generally, a function has a vertical asymptote at $x$ when it can be expressed as: $f(x)=frac{a}{g(x)}mid aneq g(x)$ and $g(x)=0$, the simplest example of which is $frac{1}{x}$.
For a function to have infinitely many vertical asymptotes there must be infinitely many values of $x$ for which $g(x)=0$. There are two ways this can happen:
(1.) $g(x)$ is periodic with infinitely many zeros - i.e.
$$f(x)=ufrac{1}{per(x)}mid per:=sin,cos,tan,mod,ldots,etc.$$
(2.) $f(x)$ is a sum or product of an infinite series.
The latter case is pretty much the same as the former, as most (if not all) such infinite series have a closed form solution equivalent to (1.), for example:
$$sum_{n=0}^infty(-1)^nfrac{(2n+1)!}{x^{2n+1}}$$ is just the inverse of the Taylor series for the $sin$ function.
As TonK commented, $tan{x}$ is a straightforward example. You could also use $frac{u}{sin{x}}$, $frac{u}{cos{x}}$, $Gamma(x)$, etc.
answered Nov 24 at 15:15
R. Burton
32919
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
add a comment |
Generally, a function has a vertical asymptote at $x$ when it can be expressed as: $f(x)=frac{a}{g(x)}mid aneq g(x)$ and $g(x)=0$, the simplest example of which is $frac{1}{x}$.
For a function to have infinitely many vertical asymptotes there must be infinitely many values of $x$ for which $g(x)=0$. There are two ways this can happen:
(1.) $g(x)$ is periodic with infinitely many zeros - i.e.
$$f(x)=ufrac{1}{per(x)}mid per:=sin,cos,tan,mod,ldots,etc.$$
(2.) $f(x)$ is a sum or product of an infinite series.
The latter case is pretty much the same as the former, as most (if not all) such infinite series have a closed form solution equivalent to (1.), for example:
$$sum_{n=0}^infty(-1)^nfrac{(2n+1)!}{x^{2n+1}}$$ is just the inverse of the Taylor series for the $sin$ function.
As TonK commented, $tan{x}$ is a straightforward example. You could also use $frac{u}{sin{x}}$, $frac{u}{cos{x}}$, $Gamma(x)$, etc.
answered Nov 24 at 15:15
R. Burton
32919
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
add a comment |
Generally, a function has a vertical asymptote at $x$ when it can be expressed as: $f(x)=frac{a}{g(x)}mid aneq g(x)$ and $g(x)=0$, the simplest example of which is $frac{1}{x}$.
For a function to have infinitely many vertical asymptotes there must be infinitely many values of $x$ for which $g(x)=0$. There are two ways this can happen:
(1.) $g(x)$ is periodic with infinitely many zeros - i.e.
$$f(x)=ufrac{1}{per(x)}mid per:=sin,cos,tan,mod,ldots,etc.$$
(2.) $f(x)$ is a sum or product of an infinite series.
The latter case is pretty much the same as the former, as most (if not all) such infinite series have a closed form solution equivalent to (1.), for example:
$$sum_{n=0}^infty(-1)^nfrac{(2n+1)!}{x^{2n+1}}$$ is just the inverse of the Taylor series for the $sin$ function.
As TonK commented, $tan{x}$ is a straightforward example. You could also use $frac{u}{sin{x}}$, $frac{u}{cos{x}}$, $Gamma(x)$, etc.
answered Nov 24 at 15:15
R. Burton
32919
Generally, a function has a vertical asymptote at $x$ when it can be expressed as: $f(x)=frac{a}{g(x)}mid aneq g(x)$ and $g(x)=0$, the simplest example of which is $frac{1}{x}$.
For a function to have infinitely many vertical asymptotes there must be infinitely many values of $x$ for which $g(x)=0$. There are two ways this can happen:
(1.) $g(x)$ is periodic with infinitely many zeros - i.e.
$$f(x)=ufrac{1}{per(x)}mid per:=sin,cos,tan,mod,ldots,etc.$$
(2.) $f(x)$ is a sum or product of an infinite series.
The latter case is pretty much the same as the former, as most (if not all) such infinite series have a closed form solution equivalent to (1.), for example:
$$sum_{n=0}^infty(-1)^nfrac{(2n+1)!}{x^{2n+1}}$$ is just the inverse of the Taylor series for the $sin$ function.
As TonK commented, $tan{x}$ is a straightforward example. You could also use $frac{u}{sin{x}}$, $frac{u}{cos{x}}$, $Gamma(x)$, etc.
answered Nov 24 at 15:15
R. Burton
32919
answered Nov 24 at 15:15
R. Burton
32919
answered Nov 24 at 15:15
R. Burton
32919
answered Nov 24 at 15:15
R. Burton
32919
32919
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
add a comment |
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
Hello, thanks for sending this! I have one question though. :) In the first sentence, did you mean that g(x) = 0 or g(x) =/= 0? :) Confused here.
– weno
Nov 24 at 19:51
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I have one more question: what is the $ u $ in the first example?
– weno
Nov 24 at 19:54
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
I did mean $g(x)=0$, basically, this is why you're not supposed to divide by 0, and why some people claim that $1/0=infty$. As for $u$, it can be whatever you want it to be - a constant, another function of $x$ - as long as it isn't equal to whatever's in the denominator (because then $f(x)$ would be 1).
– R. Burton
Nov 24 at 21:03
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
This answer is terribly muddled! "There are two ways this can happen"? Where did that come from?
– TonyK
Nov 26 at 0:23
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
Generally, any problem involving an infinite amount of something can be solved using an infinite series. There are two ways because it is intuitively obvious that $tan{x}$ has infinitely many vertical asymptotes, but arbitrarily complicated functions can be built using infinite sums.
– R. Burton
Nov 26 at 4:13
add a comment |
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5
$tan x{}{}{}{}$.
– TonyK
Nov 24 at 14:24
Oh yes, that indeed has infinitely many asymptotes. Still, how do I prove that in a math-way?
– weno
Nov 24 at 14:30
1
We can even have the asymptotes accumulate in the finite. $f(x)=sum_nfrac1{(nx-1)2^n}$ has a definition gap with vertical asymptote at all $frac 1n$, while it is defined (i.e., converges) everywhere else
– Hagen von Eitzen
Nov 24 at 14:31
1
@weno I'd say that TonyK's comment is a "math-way" proof of the claim
– Hagen von Eitzen
Nov 24 at 14:31
@HagenvonEitzen Do I simply answer the question with "yes, because for example f(x) = tan x has infinitely many vertical asymptotes"?
– weno
Nov 24 at 14:36