Number of zeroes of a function
$begingroup$
$a_i$, $i=1,2018$ are distinct real numbers.
Find number of zeroes of a function:
$$f(x) = frac{1}{x-a_1} + frac{1}{x-a_2} + ... + frac{1}{x-a_{2018}}$$
Problem is in the limits and continuity section ...
real-analysis
$endgroup$
add a comment |
$begingroup$
$a_i$, $i=1,2018$ are distinct real numbers.
Find number of zeroes of a function:
$$f(x) = frac{1}{x-a_1} + frac{1}{x-a_2} + ... + frac{1}{x-a_{2018}}$$
Problem is in the limits and continuity section ...
real-analysis
$endgroup$
$begingroup$
No idea. What happens for two distinct real numbers? How about three?
$endgroup$
– Will Jagy
Jan 16 at 20:04
add a comment |
$begingroup$
$a_i$, $i=1,2018$ are distinct real numbers.
Find number of zeroes of a function:
$$f(x) = frac{1}{x-a_1} + frac{1}{x-a_2} + ... + frac{1}{x-a_{2018}}$$
Problem is in the limits and continuity section ...
real-analysis
$endgroup$
$a_i$, $i=1,2018$ are distinct real numbers.
Find number of zeroes of a function:
$$f(x) = frac{1}{x-a_1} + frac{1}{x-a_2} + ... + frac{1}{x-a_{2018}}$$
Problem is in the limits and continuity section ...
real-analysis
real-analysis
asked Jan 16 at 20:01
user626177
$begingroup$
No idea. What happens for two distinct real numbers? How about three?
$endgroup$
– Will Jagy
Jan 16 at 20:04
add a comment |
$begingroup$
No idea. What happens for two distinct real numbers? How about three?
$endgroup$
– Will Jagy
Jan 16 at 20:04
$begingroup$
No idea. What happens for two distinct real numbers? How about three?
$endgroup$
– Will Jagy
Jan 16 at 20:04
$begingroup$
No idea. What happens for two distinct real numbers? How about three?
$endgroup$
– Will Jagy
Jan 16 at 20:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint From
$$P(x)=prodlimits_{i=1}^{2018}(x-a_i) text{ and }
P'(x)=sumlimits_{i=1}^{2018}prodlimits_{k=1,kne i}^{2018}left(x-a_kright)$$
then
$$f(x)=frac{P'(x)}{P(x)}=sumlimits_{i=1}^{n}frac{1}{x-a_i}$$
and the answer is in $P'(x)$ and Rolle's theorem.
Another, not as easy as the previous one, hint is to use intermediate value theorem (aka IVT). E.g. let's assume $a_1<a_2<...<a_{2018}$ and let's consider 2 adjacent $a_{i}<a_{i+1}$. It's not too difficult to see that
$$limlimits_{xrightarrow a_i^{+}}f(x)rightarrow +infty$$
$$limlimits_{xrightarrow a_{i+1}^{-}}f(x)rightarrow -infty$$
or, there $exists varepsilon>0$ s.t. $f(a_i +varepsilon)>0$ and $f(a_{i+1}-varepsilon)<0$. Also, $f(x)$ is continuous in $[a_i +varepsilon, a_{i+1}-varepsilon]$. By IVT, $f(x)$ has a zero in $[a_i +varepsilon, a_{i+1}-varepsilon]$. It remains to show this zero is unique, between $a_i$ and $a_{i+1}$.
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
1
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
1
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
1
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
1
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
|
show 6 more comments
$begingroup$
============================================
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint From
$$P(x)=prodlimits_{i=1}^{2018}(x-a_i) text{ and }
P'(x)=sumlimits_{i=1}^{2018}prodlimits_{k=1,kne i}^{2018}left(x-a_kright)$$
then
$$f(x)=frac{P'(x)}{P(x)}=sumlimits_{i=1}^{n}frac{1}{x-a_i}$$
and the answer is in $P'(x)$ and Rolle's theorem.
Another, not as easy as the previous one, hint is to use intermediate value theorem (aka IVT). E.g. let's assume $a_1<a_2<...<a_{2018}$ and let's consider 2 adjacent $a_{i}<a_{i+1}$. It's not too difficult to see that
$$limlimits_{xrightarrow a_i^{+}}f(x)rightarrow +infty$$
$$limlimits_{xrightarrow a_{i+1}^{-}}f(x)rightarrow -infty$$
or, there $exists varepsilon>0$ s.t. $f(a_i +varepsilon)>0$ and $f(a_{i+1}-varepsilon)<0$. Also, $f(x)$ is continuous in $[a_i +varepsilon, a_{i+1}-varepsilon]$. By IVT, $f(x)$ has a zero in $[a_i +varepsilon, a_{i+1}-varepsilon]$. It remains to show this zero is unique, between $a_i$ and $a_{i+1}$.
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
1
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
1
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
1
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
1
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
|
show 6 more comments
$begingroup$
Hint From
$$P(x)=prodlimits_{i=1}^{2018}(x-a_i) text{ and }
P'(x)=sumlimits_{i=1}^{2018}prodlimits_{k=1,kne i}^{2018}left(x-a_kright)$$
then
$$f(x)=frac{P'(x)}{P(x)}=sumlimits_{i=1}^{n}frac{1}{x-a_i}$$
and the answer is in $P'(x)$ and Rolle's theorem.
Another, not as easy as the previous one, hint is to use intermediate value theorem (aka IVT). E.g. let's assume $a_1<a_2<...<a_{2018}$ and let's consider 2 adjacent $a_{i}<a_{i+1}$. It's not too difficult to see that
$$limlimits_{xrightarrow a_i^{+}}f(x)rightarrow +infty$$
$$limlimits_{xrightarrow a_{i+1}^{-}}f(x)rightarrow -infty$$
or, there $exists varepsilon>0$ s.t. $f(a_i +varepsilon)>0$ and $f(a_{i+1}-varepsilon)<0$. Also, $f(x)$ is continuous in $[a_i +varepsilon, a_{i+1}-varepsilon]$. By IVT, $f(x)$ has a zero in $[a_i +varepsilon, a_{i+1}-varepsilon]$. It remains to show this zero is unique, between $a_i$ and $a_{i+1}$.
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
1
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
1
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
1
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
1
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
|
show 6 more comments
$begingroup$
Hint From
$$P(x)=prodlimits_{i=1}^{2018}(x-a_i) text{ and }
P'(x)=sumlimits_{i=1}^{2018}prodlimits_{k=1,kne i}^{2018}left(x-a_kright)$$
then
$$f(x)=frac{P'(x)}{P(x)}=sumlimits_{i=1}^{n}frac{1}{x-a_i}$$
and the answer is in $P'(x)$ and Rolle's theorem.
Another, not as easy as the previous one, hint is to use intermediate value theorem (aka IVT). E.g. let's assume $a_1<a_2<...<a_{2018}$ and let's consider 2 adjacent $a_{i}<a_{i+1}$. It's not too difficult to see that
$$limlimits_{xrightarrow a_i^{+}}f(x)rightarrow +infty$$
$$limlimits_{xrightarrow a_{i+1}^{-}}f(x)rightarrow -infty$$
or, there $exists varepsilon>0$ s.t. $f(a_i +varepsilon)>0$ and $f(a_{i+1}-varepsilon)<0$. Also, $f(x)$ is continuous in $[a_i +varepsilon, a_{i+1}-varepsilon]$. By IVT, $f(x)$ has a zero in $[a_i +varepsilon, a_{i+1}-varepsilon]$. It remains to show this zero is unique, between $a_i$ and $a_{i+1}$.
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
$endgroup$
Hint From
$$P(x)=prodlimits_{i=1}^{2018}(x-a_i) text{ and }
P'(x)=sumlimits_{i=1}^{2018}prodlimits_{k=1,kne i}^{2018}left(x-a_kright)$$
then
$$f(x)=frac{P'(x)}{P(x)}=sumlimits_{i=1}^{n}frac{1}{x-a_i}$$
and the answer is in $P'(x)$ and Rolle's theorem.
Another, not as easy as the previous one, hint is to use intermediate value theorem (aka IVT). E.g. let's assume $a_1<a_2<...<a_{2018}$ and let's consider 2 adjacent $a_{i}<a_{i+1}$. It's not too difficult to see that
$$limlimits_{xrightarrow a_i^{+}}f(x)rightarrow +infty$$
$$limlimits_{xrightarrow a_{i+1}^{-}}f(x)rightarrow -infty$$
or, there $exists varepsilon>0$ s.t. $f(a_i +varepsilon)>0$ and $f(a_{i+1}-varepsilon)<0$. Also, $f(x)$ is continuous in $[a_i +varepsilon, a_{i+1}-varepsilon]$. By IVT, $f(x)$ has a zero in $[a_i +varepsilon, a_{i+1}-varepsilon]$. It remains to show this zero is unique, between $a_i$ and $a_{i+1}$.
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
edited Jan 16 at 21:37
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
answered Jan 16 at 20:10
rtybasertybase
11.5k31534
11.5k31534
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
1
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
1
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
1
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
1
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
|
show 6 more comments
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
1
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
1
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
1
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
1
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
$begingroup$
I think it's supposed to be done without derivatives or using more advanced tools
$endgroup$
– user626177
Jan 16 at 20:11
1
1
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
$begingroup$
It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ...
$endgroup$
– rtybase
Jan 16 at 20:24
1
1
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
$begingroup$
Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want.
$endgroup$
– Andreas
Jan 16 at 20:29
1
1
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
$begingroup$
@someone the way to figure this out with minimal theory is to draw a careful graph for, say, $frac{1}{x-1} + frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $frac{1}{x-1} + frac{1}{x-2}+ frac{1}{x-3}.$
$endgroup$
– Will Jagy
Jan 16 at 20:34
1
1
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
$begingroup$
@someone I may have another approach, but I need to know if you are familiar with intermediate value theorem?
$endgroup$
– rtybase
Jan 16 at 20:37
|
show 6 more comments
$begingroup$
============================================
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
add a comment |
$begingroup$
============================================
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
add a comment |
$begingroup$
============================================
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
$endgroup$
============================================
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
answered Jan 16 at 21:06
Will JagyWill Jagy
104k5103202
104k5103202
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
add a comment |
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
$begingroup$
yeah, got it now, from your and other answer
$endgroup$
– user626177
Jan 16 at 21:14
add a comment |
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$begingroup$
No idea. What happens for two distinct real numbers? How about three?
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– Will Jagy
Jan 16 at 20:04