# Math Challenge - May 2021

• Challenge
Problem 11
\begin{align} f(a,b)&=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\ f(a,b)&=\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)^2-2-\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\ \end{align}
replace ##\frac a b+\frac b a## with ##t##
we get,
\begin{align} f(t)&=(t^2-2)^2-2-t^2+2+t\nonumber\\ f(t)&=t^4-5t^2+t+4\nonumber \end{align}
and as we know that, ##a,b\in \mathbb{R}## and ##a,b>0##
So, using ##A.M\geq G.M## we get,
\begin{align} \frac{\dfrac{a}{b}+\dfrac{b}{a}} {2}\geq&\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}\nonumber\\ \dfrac{a}{b}+\dfrac{b}{a}\geq&2\nonumber\\ t\geq2\nonumber \end{align}
Also, ##f(t)## is always increasing for ##t\geq2## because ##f'(t)\gt0## for ##t\geq2##

Thus, the minimum value of ##f(t)=2##, when ##t=2## or ##a=b\space \forall\space a,b\gt0 \in \mathbb{R}##

Last edited:
fresh_42
Mentor
Problem 11
$$f(a,b)=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}$$
replace ##\frac a b+\frac b a## with ##t##
we get,
$$f(t)=t^4-5t^2+t+4$$
and as we know that, ##a,b\in \mathbb{R}## and ##a,b>0##
So, using ##A.M\geq G.M## we get, $$t\geq2$$
Also, ##f(t)## is always increasing for ##t\geq2## because ##f'(t)\gt0## for ##t\geq2##

Thus, the minimum value of ##f(t)=2##, when ##t=2## or ##a=b\space \forall\space a,b\gt0 \in \mathbb{R}##

For what?
How did I get ##f(t)##?

For what?
How did I get ##f(t)##?
\begin{align} f(a,b)&=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\ f(a,b)&=\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)^2-2-\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\ f(t)&=(t^2-2)^2-2-t^2+2+t\nonumber\\ f(t)&=t^4-5t^2+t+4\nonumber \end{align}

Mentor
For what?
How did I get ##f(t)##?
Yes. How did you get the polynomial, and how did you get ##t\geq 2## from AM > GM.

Yes. How did you get the polynomial, and how did you get ##t\geq 2## from AM > GM.
I edited that in my response, sorry I didn't include them initially, it takes a lot of effort for me to type this as I'm still new to this.

Yes. How did you get the polynomial, and how did you get ##t\geq 2## from AM > GM.
Also you didn't check my response for Problem 12, have I got that right?
Problem #12
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute ##x+4 \rightarrow t##
then the equation becomes,
\begin{align} y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\ y^2&=(t^2-16)\cdot(t^2-9)\nonumber \end{align}
so, for the R.H.S to be a perfect square,
the only possibilities are ##t=3,4,5##

as for product of two numbers (say ##a,b##) to be a perfect square, the only possibilities are,
if ##a=b,a=0,b=0\space or\space a=l^2,b=m^2## (where l,m are any real numbers)

if we use ##t^2-16=t^2-9##, then we won't get any solutions, so we'll have to use
##t^2-16=l^2\space \text{&} \space t^2-9=m^2##
from this we get,
##t^2=16+l^2=m^2+9##
clearly  ##t=5,-5## are  the only possibility as 3,4,5 are pythagorean triplets.

Also we can have either ##t^2-9=0## or ##t^2-16=0##
from this we get  ##t=3,4,-3,-4## 

So putting the obtained values back in ##t=x+4## we get  ##x=-9,-8,-7,-1,0,1## 

So ordered pairs ##(x,y)## are  ##(-9,-12);(-9,12);(-8,0);(-7,0);(-1,0);(0,0);(1,12);(1,-12)## 

*Edited the answer to include all values of (x,y)
Yet another edit: I forgot the ##t=-5## cases here.

Edit (iii): I wrote 144 instead of 12 for some reasons.

Last edited:
Mentor
I edited that in my response, sorry I didn't include them initially, it takes a lot of effort for me to type this as I'm still new to this.
In case you will have to write LaTeX more often, here or IRL, it is convenient to download a little script program (I use AutoHotKey), write the script, and run it. It makes LaTeX typing really easy.

Just leave out the keys you need for other purposes like Ctrl+C/V/X/A or Alt+D. But the program allows you to suspend it with a single click in case you need the regular shortcuts.

E.g. if I want to write
\begin{align*}

\end{align*}

I only hit Alt+I which in the script is:
!i::
Send, \begin{{}align*{}}{Enter}{Enter}\end{{}align*{}}{Up}
Return

It may take a while to define shortcuts that you're comfortable with, but if you wrote
\left. \dfrac{\partial }{\partial }\right|_{} for the tenth time, you will appreciate a new shortcut.

jim mcnamara and kshitij
In case you will have to write LaTeX more often, here or IRL, it is convenient to download a little script program (I use AutoHotKey), write the script, and run it. It makes LaTeX typing really easy.

Just leave out the keys you need for other purposes like Ctrl+C/V/X/A or Alt+D. But the program allows you to suspend it with a single click in case you need the regular shortcuts.

E.g. if I want to write
\begin{align*}

\end{align*}

I only hit Alt+I which in the script is:
!i::
Send, \begin{{}align*{}}{Enter}{Enter}\end{{}align*{}}{Up}
Return

It may take a while to define shortcuts that you're comfortable with, but if you wrote
\left. \dfrac{\partial }{\partial }\right|_{} for the tenth time, you will appreciate a new shortcut.
Thank you so much for this, I was wondering how you all type all these so fast. I thought maybe it was just a practice thing.

Mentor
Also you didn't check my response for Problem 12, have I got that right?

Yet another edit: I forgot the ##t=-5## cases here.

Edit (iii): I wrote 144 instead of 12 for some reasons.
Sorry, that slipped through somehow.

Besides the square instead of the number, you also didn't find all solutions. The ones you have are correct though (with 12 for 144).

Last edited:
Mentor
Thank you so much for this, I was wondering how you all type all these so fast. I thought maybe it was just a practice thing.
I also have the problems and their solutions in a TeX file, so I can just copy and paste it. The time-consuming typing is so invisible. I do this in case someone solves an old problem from previous challenges and I don't remember the solution and in order to provide a solution manual if people want to practice and compare their solutions with mine.

kshitij
you also didn't find all solutions

One more case that I found is with ##t=0##, i.e., ##x=-4## & ##y=\pm12##

Mentor
One more case that I found is with ##t=0##, i.e., ##x=-4## & ##y=\pm12##
These are two. So you have 8/10 now.

These are two. So you have 8/10 now.
I'll keep thinking then

These are two. So you have 8/10 now.
I think I already have 10,
$$(x,y)=(-9,-12);(-9,12);(-8,0);(-7,0);(-4,-12);(-4,12);(-1,0);(0,0);(1,12);(1,-12)$$

Edit: I'll add an explanation for t=0 as well.

fresh_42
Problem #12 (second attempt)
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute ##x+4 \rightarrow t##
then the equation becomes,
\begin{align} y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\ y^2&=(t^2-16)\cdot(t^2-9)\nonumber \end{align}
so, for the R.H.S to be a perfect square,
the only possibilities are ##t=0,\pm3,\pm4,\pm5##

as for product of two numbers (say ##a,b##) to be a perfect square, the only possibilities are,
if ##a=b,a=0,b=0,(a=l^2\space \text{&}\space b=m^2)## or ##(a=-p^2\space \text{&}\space b=-q^2)## {where ##l,m,p,q## are any real numbers}

if we use ##t^2-16=t^2-9##, then we won't get any solutions,

From ##t^2-16=l^2\space \text{&} \space t^2-9=m^2##
we get,
##t^2=16+l^2=m^2+9##
clearly ##t=5,-5## are the only possibility as 3,4,5 are pythagorean triplets.

And from ##t^2-16=-p^2\space \text{&} \space t^2-9=-q^2##
we get,
##t^2+p^2=16## & ##t^2+q^2=9##
as mentioned earlier, ##3^2+4^2=5^2## is the only possible triplet with 3 & 4, for the above expression to be true, we have either ##(t=0,p=\pm4)## or ##(t=\pm4,p=0)## and similarly, ##(t=0,q=\pm3)## or ##(t=\pm3,q=0)##

Also we can have either ##t^2-9=0## or ##t^2-16=0##
from this we get ##t=3,4,-3,-4##

So putting the obtained values back in ##t=x+4## we get ##x=-9,-8,-7,-4,-1,0,1##

So ordered pairs ##(x,y)## are ##(-9,-12);(-9,12);(-8,0);(-7,0);(-4,-12);(-4,12);(-1,0);(0,0);(1,12);(1,-12)##

fresh_42
Problem #12 (alternate method)
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute ##x+4 \rightarrow t##
then the equation becomes,
\begin{align} y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\ y^2&=(t^2-16)\cdot(t^2-9)\nonumber\\ y^2&=t^4-25t^2+144\nonumber\\ y^2&=\left(t^2-\frac {25} 2\right)^2+144-\frac{625} 4\nonumber\\ 4y^2&=(2t^2-25)^2-49\nonumber \end{align}
from this we get,\begin{align}(7)^2+(2y)^2&=(2t^2-25)^2\nonumber\end{align}
But we know that there is only one possible pythagorean triplet with 7, i.e.,
$$(7)^2+(24)^2=(25)^2$$
So, now there are only two possible cases,

either,
##y=0## and ##2t^2-25=\pm7\Rightarrow t=\pm3,\pm4##

or,
##y=\pm12## and ##2t^2-25=\pm25\Rightarrow t=0,\pm5##

Putting the respective values of ##t## in ##x=t-4## we get,
$$(x,y)\equiv (-9,-12);(-9,12);(-8,0);(-7,0);(-4,-12);(-4,12);(-1,0);(0,0);(1,12);(1,-12)$$

fresh_42
julian
Gold Member
Do stupid things when V. sleep deprived, but have a go at #3 anyway:

\begin{align*}
I = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\end{align*}

From the standard half angle formula substitution:

\begin{align*}
\int_0^{\pi} f ( \cos x ) dx = \int_0^\infty \frac{2}{1 + t^2} f ( \frac{1 - t^2}{1 + t^2} ) dt
\end{align*}

(didn't notice this was given as a hint until after written this up!). So we can write

\begin{align*}
I & = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\nonumber \\
& = \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{1 - \frac{1 - t^2}{1 + t^2} \frac{1 - u^2}{1 + u^2} \frac{1 - v^2}{1 + v^2}} \frac{2}{1 + t^2} \frac{2}{1 + u^2} \frac{2}{1 + v^2}
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 + u^2 + v^2 + t^2 u^2 v^2} dt du dv
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 u^2 + u^2 v^2 + v^2 t^2 + 1} dt du dv
\end{align*}

where in the last step we made the substitution ##t \rightarrow 1/t##, ##u \rightarrow 1/u##, ##v \rightarrow 1/v##.

Write ##p = tu##, ##q = uv##, ##r = vt##.

\begin{align*}
J (p,q,r) & =
\left|
\frac{( \partial t , \partial u , \partial v) }{( \partial p , \partial q , \partial r) }
\right|
\nonumber \\
& = 1/
\left|
\frac{ ( \partial p , \partial q , \partial r) }{( \partial t , \partial u , \partial v)}
\right|
\nonumber \\
& = 1/
\begin{vmatrix}
u & t & 0 \\
0 & v & u \\
v & 0 & t
\end{vmatrix}
\nonumber \\
&= 1 / 2 tuv
\nonumber \\
&= \frac{1}{2 \sqrt{pqr}}
\end{align*}

So ##J (p,q,r) = \frac{1}{2 \sqrt{pqr}}## with ##0 \leq p < \infty##, ##0 \leq q < \infty##, ##0 \leq r < \infty##:

\begin{align*}
I & = 2 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{p^2 + q^2 + r^2 + 1} \frac{dp dq dr}{\sqrt{pqr}}
\nonumber \\
&= \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{x + y + z + 1} \frac{dx dy dz}{(xyz)^{3/4}} \qquad (\text{used } x = p^2 \text { etc})
\nonumber \\
& = \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{(xyz)^{3/4}} \left( \int_0^\infty e^{- \alpha (x + y + z + 1)} d \alpha \right) dx dy dz
\nonumber \\
& = \frac{1}{4} \int_0^\infty e^{- \alpha} \left( \int_0^\infty \frac{1}{x^{3/4}} e^{- \alpha x} dx \right)^3 d \alpha
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty \frac{e^{- \alpha}}{\alpha^{3/4}} d \alpha \right) \left( \int_0^\infty \frac{1}{x^{3/4}} e^{-x} dx \right)^3
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty e^{-x} x^{\frac{1}{4} - 1} dx \right)^4
\nonumber \\
& = \frac{1}{4} \left[ \Gamma \left( \frac{1}{4} \right) \right]^4
\end{align*}

romsofia and Infrared
Mentor
Do stupid things when V. sleep deprived, but have a go at #3 anyway:

\begin{align*}
I = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\end{align*}

From the standard half angle formula substitution:

\begin{align*}
\int_0^{\pi} f ( \cos x ) dx = \int_0^\infty \frac{2}{1 + t^2} f ( \frac{1 - t^2}{1 + t^2} ) dt
\end{align*}

(didn't notice this was given as a hint until after written this up!). So we can write

\begin{align*}
I & = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\nonumber \\
& = \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{1 - \frac{1 - t^2}{1 + t^2} \frac{1 - u^2}{1 + u^2} \frac{1 - v^2}{1 + v^2}} \frac{2}{1 + t^2} \frac{2}{1 + u^2} \frac{2}{1 + v^2}
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 + u^2 + v^2 + t^2 u^2 v^2} dt du dv
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 u^2 + u^2 v^2 + v^2 t^2 + 1} dt du dv
\end{align*}

where in the last step we made the substitution ##t \rightarrow 1/t##, ##u \rightarrow 1/u##, ##v \rightarrow 1/v##.

Write ##p = tu##, ##q = uv##, ##r = vt##.

\begin{align*}
J (p,q,r) & =
\left|
\frac{( \partial t , \partial u , \partial v) }{( \partial p , \partial q , \partial r) }
\right|
\nonumber \\
& = 1/
\left|
\frac{ ( \partial p , \partial q , \partial r) }{( \partial t , \partial u , \partial v)}
\right|
\nonumber \\
& = 1/
\begin{vmatrix}
u & t & 0 \\
0 & v & u \\
v & 0 & t
\end{vmatrix}
\nonumber \\
&= 1 / 2 tuv
\nonumber \\
&= \frac{1}{2 \sqrt{pqr}}
\end{align*}

So ##J (p,q,r) = \frac{1}{2 \sqrt{pqr}}## with ##0 \leq p < \infty##, ##0 \leq q < \infty##, ##0 \leq r < \infty##:

\begin{align*}
I & = 2 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{p^2 + q^2 + r^2 + 1} \frac{dp dq dr}{\sqrt{pqr}}
\nonumber \\
&= \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{x + y + z + 1} \frac{dx dy dz}{(xyz)^{3/4}} \qquad (\text{used } x = p^2 \text { etc})
\nonumber \\
& = \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{(xyz)^{3/4}} \left( \int_0^\infty e^{- \alpha (x + y + z + 1)} d \alpha \right) dx dy dz
\nonumber \\
& = \frac{1}{4} \int_0^\infty e^{- \alpha} \left( \int_0^\infty \frac{1}{x^{3/4}} e^{- \alpha x} dx \right)^3 d \alpha
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty \frac{e^{- \alpha}}{\alpha^{3/4}} d \alpha \right) \left( \int_0^\infty \frac{1}{x^{3/4}} e^{-x} dx \right)^3
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty e^{-x} x^{\frac{1}{4} - 1} dx \right)^4
\nonumber \\
& = \frac{1}{4} \left[ \Gamma \left( \frac{1}{4} \right) \right]^4
\end{align*}
The integral is known as Watson integral. Its value is
$$\int_0^\pi \int_0^\pi \int_0^\pi \dfrac{1}{1-\cos x\,\cos y\,\cos z}\,dx\,dy\,dz=\dfrac{1}{4}\,\Gamma\left(\dfrac{1}{4}\right)^4=2\pi {\overline\omega}^2=2G^2\pi^3\approx 43.198$$
with the Gauß constant ##G=\displaystyle{\dfrac{2}{\pi}}\int_0^1\dfrac{ds}{\sqrt{1-s^4}}\,.##

romsofia
I have been trying problem #14 for a long long time, I hope that I finally got it right,

$$f(x)=a_nx^n+\ldots+a_1x+a_0$$

(case I)

Let ##a_n>0##, then ##f(x)>0 \space \forall \space x \in\mathbb{R}##
And ##f(x) \to +\infty## when ##x \to \pm \infty## {as ##f(x)## has no real roots, thus n is even}

Also, the coefficient of ##x^n## in ##F(x)## is also ##a_n##
##\therefore F(x) \to +\infty## when ##x \to \pm \infty## {as n is even}
##\therefore## maximum value of F(x) is ##+\infty##

Now,
\begin{align} F(x)&=f(x)+h\cdot f'(x)+h^2\cdot f''(x)+\ldots+h^n\cdot f^{(n)}(x)\nonumber\\ F(x)&=f(x)+h\left(f'(x)+h\cdot f''(x)+\ldots+h^{(n-1)}\cdot f^{(n)}(x)\right)\nonumber\\ F(x)&=f(x)+h\cdot F'(x) \end{align}

Let the minimum value of ##F(x)## is at ##x=a##, thus ##F'(a)=0##, so using equation (1),
$$F(a)=f(a)>0\space \left(\text{as}\space f(x)>0 \space \forall \space x \in\mathbb{R}\right)$$
##\therefore## minimum value of ##F(x)## is ##F(a)## which is greater than zero.
##\therefore F(x)>0 \space \forall \space x \in\mathbb{R}##

(case II)

Similarly, if ##a_n<0##, then ##f(x)<0 \space \forall \space x \in\mathbb{R}##
And ##f(x) \to -\infty## when ##x \to \pm \infty##

##\therefore F(x) \to -\infty## when ##x \to \pm \infty##
##\therefore## minimum value of F(x) is ##-\infty##

Let the maximum value of ##F(x)## is at ##x=b##, thus ##F'(b)=0##, and using equation (1),
$$F(b)=f(b)<0\space \left(\text{as}\space f(x)<0 \space \forall \space x \in\mathbb{R}\right)$$
##\therefore## maximum value of ##F(x)## is ##F(b)## which is less than zero.
##\therefore F(x)<0 \space \forall \space x \in\mathbb{R}##

Thus, we can see that ##F(x)## is never equal to zero for any real value of ##x##
##\therefore## it doesn't have any real zeros.

Last edited:
fresh_42
Problem 13
$$\underbrace{\left|x-\dfrac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\right|}_{=:f(x)}\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$$
let,
\begin{align} f(x)&=x-\frac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\nonumber\\ f'(x)&=1-\frac{\cos(x)(14+\cos(x))(9+6\cos(x))-\sin^2(x)(9+6\cos(x))+6\sin^2(x)(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{(9+6\cos(x))^2-(14\cos(x)+\cos^2(x))(9+6\cos(x))+(1-\cos^2(x))(9+6\cos(x))-6(1-\cos^2(x))(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{81+36\cos^2(x)+108\cos(x)-126\cos(x)-84\cos^2(x)-9\cos^2(x)-6\cos^3(x)+9+6\cos(x)-9\cos^2(x)-6\cos^3(x)-84-6\cos(x)+84\cos^2(x)+6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{6-18\cos(x)+18\cos^2(x)-6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{6(1-\cos(x))^3}{(9+6\cos(x))^2}\nonumber \end{align}
We can see that ##f'(x)>0 \space \forall \space x \in \mathbb{R}##
##\therefore f(x)## is always increasing

Thus, the minimum value of ##f(x)\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]## is 0 at ##x=0##
And, maximum value is,
\begin{align} f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{\frac{1}{\sqrt2}\left(14+\frac{1}{\sqrt2}\right)}{(9+3\sqrt2)}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)}{(6(3+\sqrt2))}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)(3-\sqrt2)}{42}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{42\sqrt2-28+3-\sqrt2}{42}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{41\sqrt2}{42}+\frac{25}{42}\nonumber \end{align}
On putting the values of ##\pi## and ##\sqrt2##, we get
$$f(x)_{max}=9.7261908 \times 10^{-5}$$
##\therefore f(x)\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]##

Last edited:
Mentor
I have been trying problem #14 for a long long time, I hope that I finally got it right,

$$f(x)=a_nx^n+\ldots+a_1x+a_0$$

(case I)

Let ##a_n>0##, then ##f(x)>0 \space \forall \space x \in\mathbb{R}##
And ##f(x) \to +\infty## when ##x \to \pm \infty## {as ##f(x)## has no real roots, thus n is even}

Also, the coefficient of ##x^n## in ##F(x)## is also ##a_n##
##\therefore F(x) \to +\infty## when ##x \to \pm \infty## {as n is even}
##\therefore## maximum value of F(x) is ##+\infty##

Now,
\begin{align} F(x)&=f(x)+h\cdot f'(x)+h^2\cdot f''(x)+\ldots+h^n\cdot f^{(n)}(x)\nonumber\\ F(x)&=f(x)+h\left(f'(x)+h\cdot f''(x)+\ldots+h^{(n-1)}\cdot f^{(n)}(x)\right)\nonumber\\ F(x)&=f(x)+h\cdot F'(x) \end{align}

Let the minimum value of ##F(x)## is at ##x=a##, thus ##F'(a)=0##, so using equation (1),
$$F(a)=f(a)>0\space \left(\text{as}\space f(x)>0 \space \forall \space x \in\mathbb{R}\right)$$
##\therefore## minimum value of ##F(x)## is ##F(a)## which is greater than zero.
##\therefore F(x)>0 \space \forall \space x \in\mathbb{R}##

(case II)

Similarly, if ##a_n<0##, then ##f(x)<0 \space \forall \space x \in\mathbb{R}##
And ##f(x) \to -\infty## when ##x \to \pm \infty##

##\therefore F(x) \to -\infty## when ##x \to \pm \infty##
##\therefore## minimum value of F(x) is ##-\infty##

Let the maximum value of ##F(x)## is at ##x=b##, thus ##F'(b)=0##, and using equation (1),
$$F(b)=f(b)<0\space \left(\text{as}\space f(x)<0 \space \forall \space x \in\mathbb{R}\right)$$
##\therefore## maximum value of ##F(x)## is ##F(b)## which is less than zero.
##\therefore F(x)<0 \space \forall \space x \in\mathbb{R}##

Thus, we can see that ##F(x)## is never equal to zero for any real value of ##x##
##\therefore## it doesn't have any real zeros.
You can abbreviate the second case by simply mention that we can use ##-f(x)## instead.

kshitij
You can abbreviate the second case by simply mention that we can use ##-f(x)## instead.
As I said that I was trying this problem for a long time, I didn't think much after I got an idea about how to prove it, I was just too excited to post it here

fresh_42
Mentor
Problem 13
$$\underbrace{\left|x-\dfrac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\right|}_{=:f(x)}\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$$
let,
\begin{align} f(x)&=x-\frac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\nonumber\\ f'(x)&=1-\frac{\cos(x)(14+\cos(x))(9+6\cos(x))-\sin^2(x)(9+6\cos(x))+6\sin^2(x)(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{(9+6\cos(x))^2-(14\cos(x)+\cos^2(x))(9+6\cos(x))+(1-\cos^2(x))(9+6\cos(x))-6(1-\cos^2(x))(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{81+36\cos^2(x)+108\cos(x)-126\cos(x)-84\cos^2(x)-9\cos^2(x)-6\cos^3(x)+9+6\cos(x)-9\cos^2(x)-6\cos^3(x)-84-6\cos(x)+84\cos^2(x)+6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{6-18\cos(x)+18\cos^2(x)-6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\ f'(x)&=\frac{6(1-\cos(x))^3}{(9+6\cos(x))^2}\nonumber \end{align}
We can see that ##f'(x)>0 \space \forall \space x \in \mathbb{R}##
##\therefore f(x)## is always increasing

Thus, the minimum value of ##f(x)\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]## is 0 at ##x=0##
And, maximum value is,
\begin{align} f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{\frac{1}{\sqrt2}\left(14+\frac{1}{\sqrt2}\right)}{(9+3\sqrt2)}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)}{(6(3+\sqrt2))}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)(3-\sqrt2)}{42}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{42\sqrt2-28+3-\sqrt2}{42}\nonumber\\ f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{41\sqrt2}{42}+\frac{25}{42}\nonumber \end{align}
On putting the values of ##\pi## and ##\sqrt2##, we get
$$f(x)_{max}=9.7261908 \times 10^{-5}$$
##\therefore f(x)\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]##
You should have used the approximations that I gave in the problem statement, not a calculator so that the final conclusion would be
Now ##\pi/4= 0.7853975+\dfrac{\delta}{4} < 0.7854## and ##\dfrac{41\sqrt{2}-25}{42}=\dfrac{41\cdot 1.41421-25+41 \varepsilon }{42}>\dfrac{32.98261}{42}>0.7853\,,## i.e. ##0\leq f(x)<0.7854-0.7853=10^{-4}.##
but, yes, this is correct.

The reason is: If you use a calculator, then you make implicitly the assumption, that it is more precise than the values you have been given. Well, this is probably correct, as long as you didn't use a slide rule. Nevertheless, it is an assumption about a device you have no control of and you should be aware of it, e.g. if you write a protocol of an experiment.

kshitij