# Math Challenge - April 2020

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QuantumQuest
Gold Member
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.

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

$$\lim_{x\to \pi/2} \frac{x-\pi/2} {\sqrt {1-\sin x} }$$ 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.

• QuantumQuest
Gold Member
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.

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.
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
Gold Member

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.

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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
Gold Member
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.

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.

QuantumQuest
Gold Member

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.

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
Gold Member
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.

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
Gold Member
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.

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

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 $$3 f”’(a) = \frac{f”’(a)}{\theta} \\ 3 \theta = 1 \\ \theta = 1/3$$

• QuantumQuest
julian
Gold Member
Problem 6:

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##.

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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.

• fresh_42
QuantumQuest
Gold Member
Qustiom7:
This solution is achieved only by the hint given by @QuantumQuest .

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
Gold Member
Problem 6:

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)##.

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.
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.

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.
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.
I am not exactly sure, but maybe I can use a recursive algorithm. I'll think about it later.

Infrared
Gold Member
@archaic If ##n## is even, then ##p_n## achieves a minimum at some point ##x_0##. Think about how to use the fact that ##p_n'(x_0)=0##.

@archaic If ##n## is even, then ##p_n## achieves a minimum at some point ##x_0##. Think about how to use the fact that ##p_n'(x_0)=0##.
I think that I have thought of that but didn't know how to prove its uniqueness (the minimum).

Infrared