Infinite amount of vertical asymptotes












0














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







share|cite|improve this question


















  • 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
















0














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







share|cite|improve this question


















  • 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














0












0








0







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















0














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.






share|cite|improve this answer





















  • 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











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1 Answer
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1 Answer
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0














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.






share|cite|improve this answer





















  • 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
















0














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.






share|cite|improve this answer





















  • 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














0












0








0






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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