Does $Q(x) =1 -x + frac{x^2}{2}- frac{x^3}{3} + frac{x^4}{4} -frac{x^5}{5} +frac{x^6}{6}$ have any real...
Does $Q(x) =1 -x + frac{x^2}{2}- frac{x^3}{3} + frac{x^4}{4} -frac{x^5}{5} +frac{x^6}{6}$ have any real zeros?
begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \
&= -(1-x+x^2)+x^3(1-x+x^2)\
&=(x^3-1)(x^2-x+1)\
&=(x-1)(x^2+x+1)(x^2-x+1).
end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$?
real-analysis calculus
edited Dec 2 '18 at 20:33
egreg
178k1484201
asked Dec 2 '18 at 18:49
K.M
686312
|
show 1 more comment
Does $Q(x) =1 -x + frac{x^2}{2}- frac{x^3}{3} + frac{x^4}{4} -frac{x^5}{5} +frac{x^6}{6}$ have any real zeros?
begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \
&= -(1-x+x^2)+x^3(1-x+x^2)\
&=(x^3-1)(x^2-x+1)\
&=(x-1)(x^2+x+1)(x^2-x+1).
end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$?
real-analysis calculus
edited Dec 2 '18 at 20:33
egreg
178k1484201
asked Dec 2 '18 at 18:49
K.M
686312
2
desmos.com/calculator/ajsfiiquze. No it doesn't have any zeros. You can argue the case like this: You can argue that $x=1$ is a minimum of $Q$ and when you plug it in $Q(1)>0$ so you can't have any zeros. I think you can use the $Q''(1)=3>0$ to argue that $Q(1)$ is a min.
– Mason
Dec 2 '18 at 18:56
1
You actually get complex critical points from $(1-x^3)$ as well. No, you should not plug in those values because the complex points don't fall in your domain.
– Boshu
Dec 2 '18 at 18:58
Critical points in this case, don't include roots...I mean if you are asked to find roots, you don't consider critical points. A root for f(x) is a value for x where f(x) becomes zero. Critical points often refer to x values in the domain of f where there is a max. or min. (all these definitions are not exact).
– NoChance
Dec 2 '18 at 19:50
Are you sure your expansion is the polynomial you say it is? How did you get it the $-(1-x^3)(x^2-x+1)$?
– NoChance
Dec 2 '18 at 19:54
1
@NoChance Just expand two first terms $Q'(x)$ from OP.
– user376343
Dec 2 '18 at 22:18
|
show 1 more comment
Does $Q(x) =1 -x + frac{x^2}{2}- frac{x^3}{3} + frac{x^4}{4} -frac{x^5}{5} +frac{x^6}{6}$ have any real zeros?
begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \
&= -(1-x+x^2)+x^3(1-x+x^2)\
&=(x^3-1)(x^2-x+1)\
&=(x-1)(x^2+x+1)(x^2-x+1).
end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$?
real-analysis calculus
edited Dec 2 '18 at 20:33
egreg
178k1484201
asked Dec 2 '18 at 18:49
K.M
686312
Does $Q(x) =1 -x + frac{x^2}{2}- frac{x^3}{3} + frac{x^4}{4} -frac{x^5}{5} +frac{x^6}{6}$ have any real zeros?
begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \
&= -(1-x+x^2)+x^3(1-x+x^2)\
&=(x^3-1)(x^2-x+1)\
&=(x-1)(x^2+x+1)(x^2-x+1).
end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$?
real-analysis calculus
real-analysis calculus
edited Dec 2 '18 at 20:33
egreg
178k1484201
asked Dec 2 '18 at 18:49
K.M
686312
edited Dec 2 '18 at 20:33
egreg
178k1484201
asked Dec 2 '18 at 18:49
K.M
686312
edited Dec 2 '18 at 20:33
egreg
178k1484201
edited Dec 2 '18 at 20:33
egreg
178k1484201
edited Dec 2 '18 at 20:33
egreg
178k1484201
178k1484201
asked Dec 2 '18 at 18:49
K.M
686312
asked Dec 2 '18 at 18:49
K.M
686312
asked Dec 2 '18 at 18:49
K.M
686312
686312
2
desmos.com/calculator/ajsfiiquze. No it doesn't have any zeros. You can argue the case like this: You can argue that $x=1$ is a minimum of $Q$ and when you plug it in $Q(1)>0$ so you can't have any zeros. I think you can use the $Q''(1)=3>0$ to argue that $Q(1)$ is a min.
– Mason
Dec 2 '18 at 18:56
1
You actually get complex critical points from $(1-x^3)$ as well. No, you should not plug in those values because the complex points don't fall in your domain.
– Boshu
Dec 2 '18 at 18:58
Critical points in this case, don't include roots...I mean if you are asked to find roots, you don't consider critical points. A root for f(x) is a value for x where f(x) becomes zero. Critical points often refer to x values in the domain of f where there is a max. or min. (all these definitions are not exact).
– NoChance
Dec 2 '18 at 19:50
Are you sure your expansion is the polynomial you say it is? How did you get it the $-(1-x^3)(x^2-x+1)$?
– NoChance
Dec 2 '18 at 19:54
1
@NoChance Just expand two first terms $Q'(x)$ from OP.
– user376343
Dec 2 '18 at 22:18
|
show 1 more comment
2
desmos.com/calculator/ajsfiiquze. No it doesn't have any zeros. You can argue the case like this: You can argue that $x=1$ is a minimum of $Q$ and when you plug it in $Q(1)>0$ so you can't have any zeros. I think you can use the $Q''(1)=3>0$ to argue that $Q(1)$ is a min.
– Mason
Dec 2 '18 at 18:56
1
You actually get complex critical points from $(1-x^3)$ as well. No, you should not plug in those values because the complex points don't fall in your domain.
– Boshu
Dec 2 '18 at 18:58
Critical points in this case, don't include roots...I mean if you are asked to find roots, you don't consider critical points. A root for f(x) is a value for x where f(x) becomes zero. Critical points often refer to x values in the domain of f where there is a max. or min. (all these definitions are not exact).
– NoChance
Dec 2 '18 at 19:50
Are you sure your expansion is the polynomial you say it is? How did you get it the $-(1-x^3)(x^2-x+1)$?
– NoChance
Dec 2 '18 at 19:54
1
@NoChance Just expand two first terms $Q'(x)$ from OP.
– user376343
Dec 2 '18 at 22:18
2
2
desmos.com/calculator/ajsfiiquze. No it doesn't have any zeros. You can argue the case like this: You can argue that $x=1$ is a minimum of $Q$ and when you plug it in $Q(1)>0$ so you can't have any zeros. I think you can use the $Q''(1)=3>0$ to argue that $Q(1)$ is a min.
– Mason
Dec 2 '18 at 18:56
desmos.com/calculator/ajsfiiquze. No it doesn't have any zeros. You can argue the case like this: You can argue that $x=1$ is a minimum of $Q$ and when you plug it in $Q(1)>0$ so you can't have any zeros. I think you can use the $Q''(1)=3>0$ to argue that $Q(1)$ is a min.
– Mason
Dec 2 '18 at 18:56
1
1
You actually get complex critical points from $(1-x^3)$ as well. No, you should not plug in those values because the complex points don't fall in your domain.
– Boshu
Dec 2 '18 at 18:58
You actually get complex critical points from $(1-x^3)$ as well. No, you should not plug in those values because the complex points don't fall in your domain.
– Boshu
Dec 2 '18 at 18:58
Critical points in this case, don't include roots...I mean if you are asked to find roots, you don't consider critical points. A root for f(x) is a value for x where f(x) becomes zero. Critical points often refer to x values in the domain of f where there is a max. or min. (all these definitions are not exact).
– NoChance
Dec 2 '18 at 19:50
Critical points in this case, don't include roots...I mean if you are asked to find roots, you don't consider critical points. A root for f(x) is a value for x where f(x) becomes zero. Critical points often refer to x values in the domain of f where there is a max. or min. (all these definitions are not exact).
– NoChance
Dec 2 '18 at 19:50
Are you sure your expansion is the polynomial you say it is? How did you get it the $-(1-x^3)(x^2-x+1)$?
– NoChance
Dec 2 '18 at 19:54
Are you sure your expansion is the polynomial you say it is? How did you get it the $-(1-x^3)(x^2-x+1)$?
– NoChance
Dec 2 '18 at 19:54
1
1
@NoChance Just expand two first terms $Q'(x)$ from OP.
– user376343
Dec 2 '18 at 22:18
@NoChance Just expand two first terms $Q'(x)$ from OP.
– user376343
Dec 2 '18 at 22:18
|
show 1 more comment
2 Answers
2
active
oldest
votes
The derivative has a single real root, namely $1$. The other two factors never vanish on $mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-infty$ and $infty$ are both $infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}+frac{1}{6}=frac{30-20+15-12+10}{60}=frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
answered Dec 2 '18 at 20:42
egreg
178k1484201
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
2
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
add a comment |
egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $ln(x+1)$:
$$ln(x+1)=x-frac12x^2+frac13x^3-frac14x^4+cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-ln(x+1)+O(x^7)+1.$$
As $xto-1$ (here, the series expansion works), $ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $xtoinfty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
answered Dec 2 '18 at 21:29
YiFan
2,6051421
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
add a comment |
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2 Answers
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2 Answers
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oldest
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The derivative has a single real root, namely $1$. The other two factors never vanish on $mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-infty$ and $infty$ are both $infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}+frac{1}{6}=frac{30-20+15-12+10}{60}=frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
answered Dec 2 '18 at 20:42
egreg
178k1484201
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
2
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
add a comment |
The derivative has a single real root, namely $1$. The other two factors never vanish on $mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-infty$ and $infty$ are both $infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}+frac{1}{6}=frac{30-20+15-12+10}{60}=frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
answered Dec 2 '18 at 20:42
egreg
178k1484201
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
2
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
add a comment |
The derivative has a single real root, namely $1$. The other two factors never vanish on $mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-infty$ and $infty$ are both $infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}+frac{1}{6}=frac{30-20+15-12+10}{60}=frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
answered Dec 2 '18 at 20:42
egreg
178k1484201
The derivative has a single real root, namely $1$. The other two factors never vanish on $mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-infty$ and $infty$ are both $infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}+frac{1}{6}=frac{30-20+15-12+10}{60}=frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
answered Dec 2 '18 at 20:42
egreg
178k1484201
answered Dec 2 '18 at 20:42
egreg
178k1484201
answered Dec 2 '18 at 20:42
egreg
178k1484201
answered Dec 2 '18 at 20:42
egreg
178k1484201
178k1484201
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
2
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
add a comment |
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
2
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
I was wondering what exactly is meant by "the other two factors (complex roots) never vanish on $mathbb{R}$?
– K.M
Dec 2 '18 at 20:47
2
2
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
@K.M The factors are $x-1$, $x^2+x+1$ and $x^2-x+1$; one of them has a real root, the other two don't. I fixed a bit your algebra.
– egreg
Dec 2 '18 at 20:49
add a comment |
egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $ln(x+1)$:
$$ln(x+1)=x-frac12x^2+frac13x^3-frac14x^4+cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-ln(x+1)+O(x^7)+1.$$
As $xto-1$ (here, the series expansion works), $ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $xtoinfty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
answered Dec 2 '18 at 21:29
YiFan
2,6051421
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
add a comment |
egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $ln(x+1)$:
$$ln(x+1)=x-frac12x^2+frac13x^3-frac14x^4+cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-ln(x+1)+O(x^7)+1.$$
As $xto-1$ (here, the series expansion works), $ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $xtoinfty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
answered Dec 2 '18 at 21:29
YiFan
2,6051421
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
add a comment |
egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $ln(x+1)$:
$$ln(x+1)=x-frac12x^2+frac13x^3-frac14x^4+cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-ln(x+1)+O(x^7)+1.$$
As $xto-1$ (here, the series expansion works), $ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $xtoinfty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
answered Dec 2 '18 at 21:29
YiFan
2,6051421
egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $ln(x+1)$:
$$ln(x+1)=x-frac12x^2+frac13x^3-frac14x^4+cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-ln(x+1)+O(x^7)+1.$$
As $xto-1$ (here, the series expansion works), $ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $xtoinfty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
answered Dec 2 '18 at 21:29
YiFan
2,6051421
edited Dec 3 '18 at 1:41
answered Dec 2 '18 at 21:29
YiFan
2,6051421
answered Dec 2 '18 at 21:29
YiFan
2,6051421
answered Dec 2 '18 at 21:29
YiFan
2,6051421
2,6051421
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
add a comment |
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
Is $O(x^7)$ supposed to be the error?
– K.M
Dec 2 '18 at 21:56
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
@K.M what do you mean? It's the "extra trailing terms" that the series expansion for $ln(1+x)$ has compared to $Q(x)$.
– YiFan
Dec 2 '18 at 21:58
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
This expansion of $ln (x+1)$ is correct near $1$, specifically in [-1,1) only, it is not for all values of $x$.
– NoChance
Dec 3 '18 at 0:49
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
@NoChance Yes of course you're right, that's the radius of convergence for the series. Edited to reflect that.
– YiFan
Dec 3 '18 at 1:41
add a comment |
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2
desmos.com/calculator/ajsfiiquze. No it doesn't have any zeros. You can argue the case like this: You can argue that $x=1$ is a minimum of $Q$ and when you plug it in $Q(1)>0$ so you can't have any zeros. I think you can use the $Q''(1)=3>0$ to argue that $Q(1)$ is a min.
– Mason
Dec 2 '18 at 18:56
1
You actually get complex critical points from $(1-x^3)$ as well. No, you should not plug in those values because the complex points don't fall in your domain.
– Boshu
Dec 2 '18 at 18:58
Critical points in this case, don't include roots...I mean if you are asked to find roots, you don't consider critical points. A root for f(x) is a value for x where f(x) becomes zero. Critical points often refer to x values in the domain of f where there is a max. or min. (all these definitions are not exact).
– NoChance
Dec 2 '18 at 19:50
Are you sure your expansion is the polynomial you say it is? How did you get it the $-(1-x^3)(x^2-x+1)$?
– NoChance
Dec 2 '18 at 19:54
1
@NoChance Just expand two first terms $Q'(x)$ from OP.
– user376343
Dec 2 '18 at 22:18