Math Challenge - April 2020

[archaic]

$x^n−x^2=:P(x)$

This can't be as it has a positive zero.
I have shown in my second solution post that $P(x)$ is positive after its greatest zero, which is negative, given the premises, so $|P(x)|=P(x)$ for $x\geq0$.
$f(x)=c/x^n$ where $c\in(1,\,\infty)$, any number does the job, and $g(x)=1/f(x)$.
Why any $c$ in that interval does the job? Because (for $n\geq2$):
$$\lim_{x\to\infty}\left(g(x)-P(x)\right)=\lim_{x\to\infty}\left(\frac{x^n}{c}-x^n-a_{n-1}x^{n-1}+...+a_0\right)=-\infty<0$$
This tells me that, at some point, $P(x)>g(x)\Leftrightarrow \frac{1}{P(x)}<f(x)$.
EDIT: I have initially written the result of that limit as $(1-c)/c$, even in the previous post (I corrected it), sorry!

 

[fresh_42]

This is what I meant by sloppy: the limit above is infinite, not $(1-c)/c$ and one of the last occurrences of $P(x)$ should probably be $P(x)^{-1}$. A constant is not sufficient, as long as you have no control about the other coefficients. They can either exceed or remain below the borders any specific choice of $c$ gives. So you either choose a non-constant numerator, or you deal with the arbitrary terms of lower degree.

Of course, the negative zeroes are the key, but you only used the leading term of the polynomial, so all others can be arbitrary. The negative roots constrain this arbitrariness, but I cannot see how. Given any polynomial, you cannot see where the roots are, so the low order terms play a role and cannot be ignored.

The error is less in the proof, which is valid if we apply the theorem and only if, but rather in your logic.

 

[archaic]

Of course, the negative zeroes are the key, but you only used the leading term of the polynomial, so all others can be arbitrary. The negative roots constrain this arbitrariness, but I cannot see how. Given any polynomial, you cannot see where the roots are, so the low order terms play a role and cannot be ignored.

$P(x)$ can be written as $(x+|r_n|)(x+|r_{n-1}|)...(x+|r_1|)$, where $r_i$ are the negative roots, because it is a polynomial. Since $|r_i|>0$, and thus the coefficients are all positive:
$$x^n \leq x^n+a_{n-1}x^{n-1}\leq...\leq P(x)\Leftrightarrow \frac{1}{x^n}\geq\frac{1}{P(x)}\Leftrightarrow\int_1^\infty\frac{dx}{x^n}\geq\int_1^\infty\frac{dx}{P(x)}$$

 

[archaic]

This is what I meant by sloppy: the limit above is infinite, not (1−c)/c(1−c)/c(1-c)/c and one of the last occurrences of P(x)P(x)P(x) should probably be P(x)−1P(x)−1P(x)^{-1}.

Thank you, corrected it also.

 

[archaic]

A constant is not sufficient, as long as you have no control about the other coefficients. They can either exceed or remain below the borders any specific choice of $c$ gives.

But since the limit of the difference is $-\infty$, mustn't it be the case that, after some $x$, $P(x)$ is strictly bigger than $x^n/c$ forever?

 

[fresh_42]

$P(x)$ can be written as $x(x+|r_{n-1}|)...(x+|r_1|)$, where $r_i$ are the negative roots, because it is a polynomial. Since $|r_i|>0$, and thus the coefficients are all positive:
$$x^n \leq x^n+a_{n-1}x^{n-1}\leq...\leq P(x)\Leftrightarrow \frac{1}{x^n}\geq\frac{1}{P(x)}\Leftrightarrow\int_1^\infty\frac{dx}{x^n}\geq\int_1^\infty\frac{dx}{P(x)}$$

Well, you used methods which are normally beyond the scope of high school students, here the fundamental theorem of algebra, the integral comparison earlier.

It would have been much easier and straight forward with $g(x)=x^{-n+(1/2)}$.

And ... there is again a sloppiness: $P(0)\neq 0$.

 

[archaic]

And ... there is again a sloppiness: $P(0)\neq 0$.

 

[Adesh]

Question 13 a) (after the error being pointed out by @archaic, he deserves the credits)

Spoiler

$$ \lim_{x\to \pi/2} \frac{x-\pi/2} {\sqrt
{1-\sin x} } $$ Let $y= x- \pi/2$ $$ \lim_{y\to 0} \frac{y}{\sqrt {
1- \sin (y+\pi/2)}} $$

$$ \lim_{y\to 0} \frac{y}{ \sqrt {
1- \cos y}} \\
\lim_{y\to 0} \frac{y}{\sqrt { 2 \sin^2 y/2}} \\

\lim_{y\to 0} \frac{1}{\sqrt 2} \frac{y} {\sqrt {\sin^2 (y/2)}} $$
Now, let’s look at a crucial fact $\sqrt{x^2} = \left ( \sqrt x \right) ^2$ but if $x$ is negative then the RHS in the above equality breaks down, therefore, if $x$ can take both +ve and -be values then we consider only +ve values that is the mod value, $|x|$.
Coming back to our limit $$ \frac{1}{\sqrt 2} \lim_{y\to 0} \frac{y}{|\sin y/2 |} $$
$$ \frac{1}{\sqrt 2} \lim_{y\to 0^-} \frac{ y/ y/2} { - \sin (y/2) / y/2} = -\sqrt 2$$ And $$ \frac {1}{\sqrt 2} \lim_{y\to 0^+} \frac{y/y/2 } { \sin (y/2) / y/2 } = \sqrt 2 $$
Hence, limit doesn’t exist.

 

[Fred Wright]

Problem #4

Spoiler

Solve$$
y'(x)=y^2(x)-2(x+\frac{1}{2})y(x) +x^2 +x+1
$$We complete the square on r.h.s.$$
y'(x)=(y-( x+\frac{1}{2}))^2 +\frac{3}{4}
$$and make the substitution,$$
v(x)=y(x)-( x+\frac{1}{2})\\
v'(x)=y'(x) - 1
$$to obtain,$$
v'(x)=v^2(x) -\frac{1}{4}$$Separate variables,$$
\int \frac{dv}{(v^2 -\frac{1}{4})}= \int dx\\

\int \frac{dv}{(v^2 -\frac{1}{4})}=\int \frac{dv}{(v+(\frac{1}{2})(v-(\frac{1}{2})}=\int[\frac{-1}{(v+\frac{1}{2})} +\frac{1}{(v-\frac{1}{2})} ]dv\\

=ln(\frac{(v-\frac{1}{2})}{(v+\frac{1}{2})})\\

\frac{(v-\frac{1}{2})}{(v+\frac{1}{2})}=Ce^x$$ where C is the constant of integration.$$

v(x)=\frac{1}{2}\frac{(1+Ce^x)}{(1-Ce^x)}\\

y(x)=\frac{1}{2}[\frac{(1+Ce^x)}{(1-Ce^x)}+2x+1]\\

y(0)=\frac{1}{1-C}$$ For the case $y(0)=0$ we take the limit of y(C) as C goes to infinity$$

lim_{C \rightarrow \infty}y(C)=x$$For the case $y(0)=1$,$$
C=1\\
y=1+x$$For the case $y(0)=2$$$
C=\frac{1}{2}\\
y(x)=\frac{1}{2}[\frac{(1+\frac{1}{2}e^x)}{(1-\frac{1}{2}e^x)}+2x+1]
$$

 

[fresh_42]

Problem #4

Solve$$
y'(x)=y^2(x)-2(x+\frac{1}{2})y(x) +x^2 +x+1
$$We complete the square on r.h.s.$$
y'(x)=(y-( x+\frac{1}{2}))^2 +\frac{3}{4}
$$and make the substitution,$$
v(x)=y(x)-( x+\frac{1}{2})\\
v'(x)=y'(x) - 1
$$to obtain,$$
v'(x)=v^2(x) -\frac{1}{4}$$Separate variables,$$
\int \frac{dv}{(v^2 -\frac{1}{4})}= \int dx\\

\int \frac{dv}{(v^2 -\frac{1}{4})}=\int \frac{dv}{(v+(\frac{1}{2})(v-(\frac{1}{2})}=\int[\frac{-1}{(v+\frac{1}{2})} +\frac{1}{(v-\frac{1}{2})} ]dv\\

=ln(\frac{(v-\frac{1}{2})}{(v+\frac{1}{2})})\\

\frac{(v-\frac{1}{2})}{(v+\frac{1}{2})}=Ce^x$$ where C is the constant of integration.$$

v(x)=\frac{1}{2}\frac{(1+Ce^x)}{(1-Ce^x)}\\

y(x)=\frac{1}{2}[\frac{(1+Ce^x)}{(1-Ce^x)}+2x+1]\\

y(0)=\frac{1}{1-C}$$ For the case $y(0)=0$ we take the limit of y(C) as C goes to infinity

I assume that means that y(x)=x is the solution.

$$lim_{C \rightarrow \infty}y(C)=x$$For the case $y(0)=1$,$$
C=1\\
y=1+x$$For the case $y(0)=2$$$
C=\frac{1}{2}\\
y(x)=\frac{1}{2}[\frac{(1+\frac{1}{2}e^x)}{(1-\frac{1}{2}e^x)}+2x+1]
$$

Have you cross checked this expression? Mine is different.

Edit: This could either mean that there is more than one solution or one of us is wrong. I checked your result and couldn't retrieve the differential equation, but I'm not sure whether I didn't make a mistake. Your shift by 1/2 makes a big mess in the calculations.

 

[QuantumQuest]

Well $2!$ is just 2.

And by the definition of derivative $$ \lim_{h\to 0} \frac{f’(a+h) - f’(a)}{h} = f”(a)$$

Yes, I (obviously) know that $2!$ is 2. I mentioned it just in case you missed it. Also, it may be useful in the context of the hint I gave.

Now, going on, the definition of derivative is what you write but you have an equality and you take the limit of the left side only, you equate it with the right hand side and then you write an equality. I don't make sense of it. In any case, if you take both limits from the start they're both $f''(a)$, at least as I see it, so, what it gives anyway?

 

[QuantumQuest]

Question 13 a)

Spoiler

$$\lim_{x\to \pi/2} \frac{x-\pi/2}{\sqrt { 1-\sin x}} $$ Let $y = x - \pi/2$ $$\lim_{y\to 0} \frac{y}{\sqrt{ 1- \sin(y+ \pi/2)}} $$. $$ \lim_{y\to 0} \frac{y}{\sqrt {1- \cos y}}$$ $$\lim_{y\to 0} \frac{y}{\sqrt{2\sin^2 y/2}} $$
$$\lim_{y \to 0} \frac{y}{\sqrt 2 \sin y/2} \\ \lim_{y\to 0} \frac{
\frac{y}{y/2} }
{\sqrt 2\frac {\sin y/2}{y/2} } $$ The denominator in the above limit goes to $\sqrt2$ and the numerator cancels out to become 2 hence the answer is $\sqrt 2$.

Not correct. It's easy to find the mistake and it has to do with some absolute value you missed.

 

[Adesh]

Now, going on, the definition of derivative is what you write but you have an equality and you take the limit only of the left side, you equate it with the right hand side and then you write an equality. I don't make sense of it. In any case, if you take both limits from the start they're both f′′(a)f″(a)f''(a), at least as I see it, so, what it gives anyway?

I can explain it this way, write everything on one side, then we can use the Theorem “limit of a sum is the sum of the limits” so, we can take the limit of that “derivative” part and can leave everything as it is with limit sign being there. Am I being legitimate here?

 

[Adesh]

Also, it may be useful in the context of the hint I gave.

Sir, I tried your hint. I Taylor expanded everything but you know... couldn’t get nothing

 

[QuantumQuest]

I can explain it this way, write everything on one side, then we can use the Theorem “limit of a sum is the sum of the limits” so, we can take the limit of that “derivative” part and can leave everything as it is with limit sign being there. Am I being legitimate here?

Well, to be honest, I don't quite understand. If you have an equality that holds true then either you take the limit of both sides or not. I don't really know if you wanted to write something else and you wrote it like this.

 

[QuantumQuest]

Sir, I tried your hint. I Taylor expanded everything but you know... couldn’t get nothing

So, let me help a little more. What you're interested at, is putting the third derivative into the expansion. Now, from there, you can do some math, also taking into account the given expression.

 

[Adesh]

Well, to be honest, I don't quite understand. If you have an equality that holds true then either you take the limit of both sides or not. I don't really know if you wanted to write something else and you wrote it like this.

Let’s consider this example (no relation with problem 7) $$ \lim_{h\to 0} \frac{ f(a+h) - f(a)}{h} = \frac{f”(a+h)}{h}$$ I think your argument is that equality above holds only if I take the limit both sides, I have something like this in my mind $$ \lim_{h\to 0} \frac{f(a+h) -f(a)}{h} = \frac{ f”(a+h)}{h} \\
\lim_{h\to 0} \frac{f(a+h) - f(a)}{h} - \frac{ f”(a+h)}{h} =0$$ Now, if we use the law that “limit of a sum is the sum of the limits” then we write $$ f’(a) -\lim_{h\to 0} \frac{f”(a+h)}{h} = 0 $$

 

[QuantumQuest]

Question 13 a) (after the error being pointed out by @archaic, he deserves the credits)

Spoiler

$$ \lim_{x\to \pi/2} \frac{x-\pi/2} {\sqrt
{1-\sin x} } $$ Let $y= x- \pi/2$ $$ \lim_{y\to 0} \frac{y}{\sqrt {
1- \sin (y+\pi/2)}} $$

$$ \lim_{y\to 0} \frac{y}{ \sqrt {
1- \cos y}} \\
\lim_{y\to 0} \frac{y}{\sqrt { 2 \sin^2 y/2}} \\

\lim_{y\to 0} \frac{1}{\sqrt 2} \frac{y} {\sqrt {\sin^2 (y/2)}} $$
Now, let’s look at a crucial fact $\sqrt{x^2} = \left ( \sqrt x \right) ^2$ but if $x$ is negative then the RHS in the above equality breaks down, therefore, if $x$ can take both +ve and -be values then we consider only +ve values that is the mod value, $|x|$.
Coming back to our limit $$ \frac{1}{\sqrt 2} \lim_{y\to 0} \frac{y}{|\sin y/2 |} $$
$$ \frac{1}{\sqrt 2} \lim_{y\to 0^-} \frac{ y/ y/2} { - \sin (y/2) / y/2} = -\sqrt 2$$ And $$ \frac {1}{\sqrt 2} \lim_{y\to 0^+} \frac{y/y/2 } { \sin (y/2) / y/2 } = \sqrt 2 $$
Hence, limit doesn’t exist.

Well done @Adesh.

 

[QuantumQuest]

Let’s consider this example (no relation with problem 7) $$ \lim_{h\to 0} \frac{ f(a+h) - f(a)}{h} = \frac{f”(a+h)}{h}$$ I think your argument is that equality above holds only if I take the limit both sides, I have something like this in my mind $$ \lim_{h\to 0} \frac{f(a+h) -f(a)}{h} = \frac{ f”(a+h)}{h} \\
\lim_{h\to 0} \frac{f(a+h) - f(a)}{h} - \frac{ f”(a+h)}{h} =0$$ Now, if we use the law that “limit of a sum is the sum of the limits” then we write $$ f’(a) -\lim_{h\to 0} \frac{f”(a+h)}{h} = 0 $$

No. In the example you give here there is already a limit on the LHS. What I said was about having an equality without limits - as was the case in your solution for question $7$ and you took the limit of the left side only.

 

[Adesh]

No. In the example you give here there is already a limit on the LHS. What I said was about having an equality without limits - as was the case in your solution for question $7$ and you took the limit of the left side only.

Are you talking about this step

limh→0f′(a+h)−f′(a)h=f”(a+θh)+θh2f”′(a+θh)f”(a)=f”(a+θh)+θh2f”′(a+θh)​

Well, it is like this $$\lim_{h\to 0} \frac{f’(a+h) - f’(a) }{h} = f”(a +\theta h) + \frac{\theta h}{2} f”’(a +\theta h) \\

f”(a) = \lim_{h\to 0} f”(a+ \theta h) +\frac{\theta h}{2} f”’(a + \theta h)$$

 

[QuantumQuest]

Are you talking about this step

Well, it is like this $$\lim_{h\to 0} \frac{f’(a+h) - f’(a) }{h} = f”(a +\theta h) + \frac{\theta h}{2} f”’(a +\theta h) \\

f”(a) = \lim_{h\to 0} f”(a+ \theta h) +\frac{\theta h}{2} f”’(a + \theta h)$$

So, in fact, you're taking both limits in the expression $\frac{f’(a+h) - f’(a)}{h} = f”(a+\theta h) + \frac{\theta h}{2} f”’(a+ \theta h)$ and not the way you wrote it in your solution to question $7$. That's what I'm talking about.

 

[Adesh]

So, in fact, you're talking both limits in the expression $f’(a+h) = f’(a) + h f”(a+ \theta h) + \frac{\theta h^2}{2} f”’(a +\theta h) \\ \frac{f’(a+h) - f’(a)}{h} = f”(a+\theta h) + \frac{\theta h}{2} f”’(a+ \theta h)$ and not the way you wrote it in your solution to $7$. That's what I'm talking about.

Yes, I thought writing the limit before the whole equation meant limit was getting applied to everything, but as I have experienced it has caused some ambiguities. I aplogizie for that unclearness, but I really meant limits being on both sides.

 

[QuantumQuest]

Yes, I thought writing the limit before the whole equation meant limit was getting applied to everything, but as I have experienced it has caused some ambiguities. I aplogizie for that unclearness, but I really meant limits being on both sides.

See again my post #56, as there was a mistake as I copied the expression which I corrected.

 

[Adesh]

See again my post #56, as there was a mistake as I copied the expression which I corrected.

See again my post #56, as there was a mistake as I copied the expression which I corrected.

Sorry but I’m unable to spot the mistake. Please help me in seeing it.

 

[QuantumQuest]

Sorry but I’m unable to spot the mistake. Please help me in seeing it.

I mean that I made a mistake in copying the expression. The one that is written now in post #56 is the one I'm talking about. And in this you took the limit of the left hand side only while you have to take the limits of both sides but as you answer in post #57 it's understood, so it's OK for this. But taking into account both limits and also, as there are further things beyond which I don't understand , I don't know where this solution leads to.

 

[Adesh]

I mean that I made a mistake in copying the expression. The one that is written now in post #56 is the one I'm talking about. And in this you took the limit of the left hand side only while you have to take the limits of both sides but as you answer in post #57 it's understood, so it's OK for this. But taking into account both limits and also, as there are further things beyond which I don't understand , I don't know where this solution leads to.

Was my explanation of why I took only the left side limit (I can do that if we take everything to one side and then we use the rule “limit of a sum is equal to the sum of limits”) unsatisfactory ?

 

[QuantumQuest]

Yes, I thought writing the limit before the whole equation meant limit was getting applied to everything, but as I have experienced it has caused some ambiguities. I aplogizie for that unclearness, but I really meant limits being on both sides.

I think that you understand my point about taking both limits by now. But, again, taking into account both limits and also, as there are further things beyond which I don't understand , I don't know where this solution leads to.

 

[Adesh]

I think that you understand my point about taking both limits by now. But, again, taking into account both limits and also, as there are further things beyond which I don't understand , I don't know where this solution leads to.

Am I allowed to explain my solution or do I need to come up with a new one?

 

[QuantumQuest]

Am I allowed to explain my solution or do I need to come up with a new one?

It's up to you but if you want to keep the current solution you gave, you must do the appropriate corrections and present it in a more understandable way and, of course, reach the correct result. Alternatively, I'd recommend to take into account my hint and work that way which leads to a nice solution.

 

[Adesh]

Qustiom7:
This solution is achieved only by the hint given by @QuantumQuest .

Spoiler

By Taylor Expansion:
$$f(a+h) = f(a) + hf ‘(a) + \frac{h^2}{2!} f’’(a) + \frac{h^3}{3!} f’’’(a) +... $$
From the expression which is given we can write $$ f(a+h) = f(a+h) - \frac{h^2}{2!} f”(a +\theta h) + \frac{h^2}{2!} f”(a) \frac{h^3}{3!} + f’’’(a) +... $$ As $h$ goes to zero we can ignore higher terms, hence we can write
$$\frac{h^2}{2!} \left ( f”(a+\theta h) - f”(a)\right) = \frac{h^3}{3!} f”’(a) \\
\frac{6}{2} \left( \frac {
f”(a+\theta h) - f”(a) }
{ \theta h} \right) = \frac{f”’(a)}{\theta} $$
In the limit as $h$ goes to zero we have $$ 3 f”’(a) = \frac{f”’(a)}{\theta} \\
3 \theta = 1 \\
\theta = 1/3 $$

 

[julian]

Problem 6:

Spoiler

As $\ln (x)$ is a convex up function we have:

$\ln (m_1 a_1 + m_2 a_2) \geq m_1 \ln (a_1) + m_2 \ln (a_2)$

Taking the exponential, we obtain:

$m_1 a_1 + m_2 a_2 \geq a_1^{m_1} a_2^{m_2} .$

Equality holds if $m_1 = 0$ or $m_2 = 0$. Also, equality holds if $a_1 = a_2$.

 

[Fred Wright]

I assume that means that y(x)=x is the solution.

Have you cross checked this expression? Mine is different.

Edit: This could either mean that there is more than one solution or one of us is wrong. I checked your result and couldn't retrieve the differential equation, but I'm not sure whether I didn't make a mistake. Your shift by 1/2 makes a big mess in the calculations.

Hi fresh_42,
I simplified my expression for $y(0)=2$ to $y=x + \frac{1}{(1-\frac{1}{2} e^x)}$. My solution gives,$$
y(0)=0\\
y=x\\
y(0)=1\\
y=x+1\\
y(0)=2\\
y=x + \frac{1}{(1-\frac{1}{2} e^x)}
$$I cross checked these solutions and they check out.

 

[QuantumQuest]

Qustiom7:
This solution is achieved only by the hint given by @QuantumQuest .

Spoiler

By Taylor Expansion:
$$f(a+h) = f(a) + hf ‘(a) + \frac{h^2}{2!} f’’(a) + \frac{h^3}{3!} f’’’(a) +... $$
From the expression which is given we can write $$ f(a+h) = f(a+h) - \frac{h^2}{2!} f”(a +\theta h) + \frac{h^2}{2!} f”(a) \frac{h^3}{3!} + f’’’(a) +... $$ As $h$ goes to zero we can ignore higher terms, hence we can write
$$\frac{h^2}{2!} \left ( f”(a+\theta h) - f”(a)\right) = \frac{h^3}{3!} f”’(a) \\
\frac{6}{2} \left( \frac {
f”(a+\theta h) - f”(a) }
{ \theta h} \right) = \frac{f”’(a)}{\theta} $$
In the limit as $h$ goes to zero we have $$ 3 f”’(a) = \frac{f”’(a)}{\theta} \\
3 \theta = 1 \\
\theta = 1/3 $$

The solution is essentially correct. Just two comments. First, you must correct the typo in the addition and the multiplication in the end of the second expression in order to be correctly written for the people that will read your solution. As a second comment, it's best to write the expansion like this

$f(a+h) = f(a) + hf ‘(a) + \frac{h^2}{2!} f’’(a) + \frac{h^3}{3!} f’’’(a + \theta_1h)$

as what is given is that $f$ is a three times differentiable function. At the limit where $h \rightarrow 0$ $\hspace{1cm}$ $\theta_1h$ vanishes.

 

[QuantumQuest]

Problem 6:

Spoiler

As $\ln (x)$ is a convex up function we have:

$\ln (m_1 a_1 + m_2 a_2) \geq m_1 \ln (a_1) + m_2 \ln (a_2)$

Taking the exponential, we obtain:

$m_1 a_1 + m_2 a_2 \geq a_1^{m_1} a_2^{m_2} .$

Equality holds if $m_1 = 0$ or $m_2 = 0$. Also, equality holds if $a_1 = a_2$.

Yes, I see what you mean but in order for other people to understand it, I think that it is best to show how to utilize the derivatives and the appropriate theorem in order to reach the point where $\ln (m_1 a_1 + m_2 a_2) \geq m_1 \ln (a_1) + m_2 \ln (a_2)$.

 

[archaic]

Not yet a full solution; missing a detail. I'd be grateful if someone could give me a hint for 3) of the first part.

Spoiler: Question 8

We know that $p_n(0)=1$ for all $n$, $p_0(x)=1$ has no zeroes, and $p_1(x)$ has only one zero (linear function). For the rest, I suppose that $n\geq2$.
Let $n$ be an even natural number.
1) $\forall x\geq0,\,p_n(x)>0$ since we have a sum of positive numbers.
2) $\forall x\in[-1,\,0),\,p_n(x)>0$.
I can write $p_n(x)$ like this:
$$p_n(x)=1+\sum_\underset{\text{step}=2}{k=1}^{n-1}\left(\frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}\right)$$
I am going to prove that the parenthesis is always positive:
$$\begin{align*}
x^k\geq |x|^{k+1}&\Leftrightarrow\frac{x^k}{k!}>\frac{x^k}{(k+1)!}\geq\frac{|x|^{k+1}}{(k+1)!} \\
&\Leftrightarrow\frac{x^k}{k!}>\frac{|x|^{k+1}}{(k+1)!}\\
&\Leftrightarrow\frac{x^k}{k!}>\frac{x^{k+1}}{(k+1)!}>-\frac{x^k}{k!}\\
&\Leftrightarrow-\frac{x^k}{k!}<-\frac{x^{k+1}}{(k+1)!}<\frac{x^k}{k!}\\
&\Leftrightarrow0<\frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}
\end{align*}$$
3) $\forall x<-1,\,p_n(x)>0$. (WIP)
Since $p_n(x)>0$ for all real values of $x$, it has no real zeroes when $n$ is even.

Let $n$ be an odd natural number.
I notice that $p'_n(x)=p_{n-1}(x)$. We have that $n-1$ is even, so $p_{n-1}>0$ for $x\in\mathbb{R}$. From this I can say that $p_n(x)$ is a bijection, since it is constantly growing, with no absolute/local maximum or minimum. $(*)$
We also have that:
$$\lim_{x\to\pm\infty}p_n(x)=\lim_{x\to\pm\infty}\frac{x^n}{n!}=\pm\infty$$
1) This tells me that there exists a real number $N$, such that $p_n(N)p_n(-N)<0$.
2) Since $p_n(x)$ is a polynomial, it is continuous over all $\mathbb{R}$.
Using 1) and 2), I conclude from the intermediate value theorem that there exists a real number $c$ such that $p_n(c)=0$. Moreover, using $(*)$, I also conclude that $c$ is unique.

Conclusion:
If $n$ is even, then $p_n(x)$ has no real zeroes. Else, it has only one.